Arihant Biology 34 Years NEET PYQ.pdf

arihant biology 34 years neet

1. t.me/Exam_Sakha_ O f f i c i a l

3. Arihant Prakashan (Series), Meerut
B
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4
( )YEARS’ 1988-2021
Chapterwise Topicwise
Complete Collection of all Questions
asked in last 34 years’ in NEET &CBSE AIPMT
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4. No part of this publication may be re-produced, stored in a retrieval system or
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5. PREFACE
Whenever a student decides to prepare for any examination his/her first
and foremost curiosity is to know about the type of questions that are
expected in the exam. This becomes more important in the context of
competitive entrance examinations where there is neck-to-neck
competition.
We feel great pleasure in presenting before you this book containing
Error Free Chapterwise Topicwise Solutions of CBSE AIPMT/NEET
Biology Questions from the years 1988 to 2021.
It has been our efforts to provide correct solutions to the best of our
knowledge and opinion. Detailed explanatory discussions follow the
answers. Discussions are not just sketchy–rather, have been drafted in a
manner that the students will surely be able to answer some other
related questions too ! Going through this book, the students would be
able to have the complete idea of the questions being asked in the test.
We hope this chapterwise solved papers would be highly beneficial to
the students. We would be grateful if any discrepancies or mistakes in
the questions or answers are brought to our notice so that these could
be rectified in subsequent editions.
Publisher

6. 1. The Living World 1-3
2. Biological Classification 4-17
3. Plant Kingdom 18-30
4. Animal Kingdom 31-48
5. Morphology of Flowering Plants 49-59
6. Anatomy of Flowering Plants 60-70
7. Structural Organisation in Animals 71-77
8. Cell : The Unit of Life 78-93
9. Biomolecules 94-106
10. Cell Cycle and Cell Division 107-115
11. Transport in Plants 116-121
12. Mineral Nutrition 122-126
13. Photosynthesis 127-135
14. Respiration in Plants 136-141
15. Plant Growth and Development 142-150
16. Digestion and Absorption 151-159
17. Breathing and Exchange of Gases 160-166
18. Body Fluids and Circulation 167-175
19. Excretory Products and Their Elimination 176-183
20. Locomotion and Movement 184-189
21. Neural Control and Coordination 190-198
22. Chemical Coordination and Integration 199-207
23. Reproduction in Organisms 208-211
CONTENTS

7. 24. Sexual Reproduction in Flowering Plants 212-222
25. Human Reproduction 223-233
26. Reproductive Health 234-239
27. Principles of Inheritance and Variation 240-262
28. Molecular Basis of Inheritance 263-279
29. Evolution 280-295
30. Human Health and Diseases 296-308
31. Strategies for Enhancement in Food Production 309-316
32. Microbes in Human Welfare 317-323
33. Biotechnology: Principles and Processes 324-332
34. Biotechnology and its Applications 333-337
35. Organisms and Population 338-348
36. Ecosystem 349-357
37. Biodiversity and Its Conservation 358-365
38. Environmental Issues 366-376

8. CLASS 11th
UNIT I Diversity in Living World
What is living? ; Biodiversity, Need for classification, Three domains of life, Taxonomy and Systematics, Concept of species and
taxonomical hierarchy, Binomial nomenclature, Tools for study of Taxonomy – Museums, Zoos, Herbaria, Botanical gardens. Five
kingdom classification, salient features and classification of Monera, Protista and Fungi into major groups, Lichens, Viruses and
Viroids. Salient features and classification of plants into major groups-Algae, Bryophytes, Pteridophytes, Gymnosperms and
Angiosperms (three to five salient and distinguishing features and at least two examples of each category), Angiosperms-
classification up to class, characteristic features and examples). Salient features and classification of animals-nonchordate upto
phyla level and chordate up to classes level (three to five salient features and at least two examples).
UNIT II Structural Organisation in Animals and Plants
Morphology and modifications, Tissues, Anatomy and functions of different parts of flowering plants, Root, stem, leaf,
inflorescence- cymose and recemose, flower, fruit and seed (To be dealt along with the relevant practical of the Practical
Syllabus). Animal tissues, Morphology, anatomy and functions of different systems (digestive, circulatory, respiratory, nervous
and reproductive) of an insect (cockroach). (Brief account only)
UNIT III Cell Structure and Function
Cell theory and cell as the basic unit of life Structure of prokaryotic and eukaryotic cell, Plant cell and animal cell, Cell envelope,
cell membrane, cell wall, Cell organelles-structure and function, Endomembrane system-endoplasmic reticulum, Golgi bodies,
lysosomes, vacuoles, mitochondria, ribosomes, plastids, micro bodies, Cytoskeleton, cilia, flagella, centrioles (ultra structure and
function), Nucleus-nuclear membrane, chromatin, nucleolus. Chemical constituents of living cells Biomolecules-structure and
function of proteins, carbodydrates, lipids, nucleic acids, Enzymes-types, properties, enzyme action. B Cell division Cell cycle,
mitosis, meiosis and their significance.
UNIT IV Plant Physiology
Transport in plants Movement of water, gases and nutrients, Cell to cell transport-Diffusion, facilitated diffusion, active transport,
Plant– water relations – Imbibition, water potential, osmosis, plasmolysis, Long distance transport of water – Absorption,
apoplast, symplast, transpiration pull, root pressure and guttation, Transpiration-Opening and closing of stomata, Uptake and
translocation of mineral nutrients-Transport of food, phloem transport, Mass flow hypothesis, Diffusion of gases (brief mention).
Mineral nutrition Essential minerals, macro and micronutrients and their role, Deficiency symptoms, Mineral toxicity, Elementary
idea of Hydroponics as a method to study mineral nutrition, Nitrogen metabolism Nitrogen cycle, biological nitrogen fixation.
Photosynthesis Photosynthes is as a means of Autotrophic nutrition, Site of photosynthesis take place, pigments involved in
Photosynthesis (Elementary idea), Photochemical and biosynthetic phases of photosynthesis, Cyclic and non-cyclic and
photophosphorylation, Chemiosmotic hypothesis, Photorespiration C3 and C4 pathways, Factors affecting photosynthesis.
Respiration Exchange gases, Cellular respiration-glycolysis, fermentation (anaerobic), TCA cycle and electron transport system
(aerobic), Energy relations Number of ATP molecules generated, Amphibolic pathways, Respiratory quotient.
Plant growth and development Seed germination, Phases of Plant growth and plant growth rate, Conditions of growth,
Differentiation, dedifferentiation and redifferentiation, Sequence of developmental process in a plant cell, Growth regulators-
auxin,gibberellin, cytokinin, ethylene, ABA Seed dormancy, Vernalisation, Photoperiodism.
SYLLABUS

9. UNIT V Human Physiology
Digestion and absorption, Alimentary canal and digestive glands, Role of digestive enzymes and gastrointestinal hormones,
Peristalsis, digestion, absorption and assimilation of proteins, carbohydrates and fats, Caloric value of proteins, carbohydrates
and fats, Egestion, Nutritional and digestive disorders – PEM, indigestion, constipation, vomiting, jaundice, diarrhea.
Breathing and Respiration Respiratory organs in animals (recall only), Respiratory system in humans, Mechanism of breathing
and its regulation in humans-Exchange of gases, transport of gases and regulation of respiration Respiratory volumes, Disorders
related to respiration-Asthma, Emphysema, Occupational respiratory disorders.
Body fluids and circulation Composition of blood, blood groups, coagulation of blood, Composition of lymph and its function,
Human circulatory system-Structure of human heart
and blood vessels, Cardiac cycle, cardiac output, ECG, Double circulation, Regulation of cardiac activity, Disorders of circulatory
system-Hypertension, Coronary artery disease, Angina pectoris, Heart failure.
Excretory products and their elimination Modes of excretion- Ammonotelism, ureotelism, uricotelism, Human excretory system-
structure and fuction, Urine formation, Osmoregulation, Regulation of kidney function-Renin-angiotensin, Atrial Natriuretic
Factor, ADH and Diabetes insipidus, Role of other organs in excretion, Disorders, Uraemia, Renal failure, Renal calculi, Nephritis,
Dialysis and artificial kidney.
Locomotion and Movement Types of movement- ciliary, fiagellar, muscular, Skeletal muscle- contractile proteins and muscle
contraction, Skeletal system and its functions (To be dealt with the relevant practical of Practical syllabus), Joints, Disorders of
muscular and skeletal system-Myasthenia gravis, Tetany, Muscular dystrophy, Arthritis, Osteoporosis, Gout.
Neural control and coordination Neuron and nerves, Nervous system in humans- central nervous system, peripheral nervous
system and visceral nervous system, Generation and conduction of nerve impulse, Reflex action, Sense organs, Elementary
structure and function of eye and ear.
Chemical coordination and regulation Endocrine glands and hormones, Human endocrine system-Hypothalamus, Pituitary,
Pineal, Thyroid, Parathyroid, Adrenal, Pancreas, Gonads, Mechanism of hormone action (Elementary Idea), Role of hormones as
messengers and regulators, Hypo-and hyperactivity and related disorders (Common disorders e.g., Dwarfism, Acromegaly,
Cretinism, goiter, exopthalmic goiter, diabetes, Addison's disease). (Important, Diseases and disorders mentioned above to be
dealt in brief.)
CLASS 12th
UNIT I Reproduction
Reproduction in organisms Reproduction, a characteristic feature of all organisms for continuation of species, Modes of
reproduction– Asexual and sexual, Asexual reproduction, Modes-Binary fission, sporulation, budding, gemmule, fragmentation,
vegetative propagation in plants.
Sexual reproduction in flowering plants Flower structure, Development of male and female gametophytes, Pollination-types,
agencies and examples, Outbreeding devices, Pollen-Pistil interaction, Double fertilization, Post fertilization events-
Development of endosperm and embryo, Development of seed and formation of fruit, Special modes-apomixis, parthenocarpy,
polyembryony, Significance of seed and fruit formation.
Human Reproduction Male and female reproductive systems, Microscopic anatomy of testis and ovary, Gametogenesis-
spermatogenesis and oogenesis, Menstrual cycle, Fertilisation, embryo development upto blastocyst formation, implantation,
Pregnancy and placenta formation (Elementary idea), Parturition (Elementary idea), Lactation (Elementary idea).
Reproductive health Need for reproductive health and prevention of sexually transmitted diseases (STD), Birth control-Need and
Methods, Contraception and Medical Termination of Pregnancy (MTP), Amniocentesis, Infertility and assisted reproductive
technologies – IVF, ZIFT, GIFT (Elementary idea for general awareness).

10. UNIT II Genetics and Evolution
Heredity and variation Mendelian Inheritance, Deviations from Mendelism-Incomplete dominance, Co-dominance, Multiple
alleles and Inheritance of blood groups, Pleiotropy, Elementary idea of polygenic inheritance, Chromosome theory of
inheritance, Chromosomes and genes, Sex determination-In humans, birds, honey bee, Linkage and crossing over, Sex linked
inheritance-Haemophilia, Colour blindness, Mendelian disorders in humans-Thalassemia, Chromosomal disorders in humans,
Down's syndrome, Turner's and Klinefelter's syndromes.
Molecular basis of Inheritance Search for genetic material and DNA as genetic material, Structure of DNA and RNA, DNA
packaging, DNA replication, Central dogma, Transcription, genetic code, translation, Gene expression and regulation-Lac
Operon, Genome and human genome project, DNA finger printing.
Evolution Origin of life, Biological evolution and evidences for biological evolution from Paleontology, comparative anatomy,
embryology and molecular evidence), Darwin's contribution, Modern Synthetic theory of Evolution, Mechanism of evolution-
Variation (Mutation and Recombination) and Natural Selection with examples, types of natural selection, Gene flow and genetic
drift, Hardy-Weinberg's principle, Adaptive Radiation, Human evolution.
UNIT III Biology and Human Welfare
Health and Disease, Pathogens, parasites causing human diseases (Malaria, Filariasis, Ascariasis, Typhoid, Pneumonia, common
cold, amoebiasis, ring worm), Basic concepts of immunology-vaccines, Cancer, HIV and AIDS, Adolescence, drug and alcohol
abuse.
Improvement in food production Plant breeding, tissue culture, single cell protein, Biofortification, Apiculture and Animal
husbandry. Microbes in human welfare In household food processing, industrial production, sewage treatment, energy
generation and as biocontrol agents and biofertilizers.
UNIT IV Biotechnology and Its Applications
Principles and process of Biotechnology Genetic engineering (Recombinant DNA technology). Applications of Biotechnology in
health and agriculture Human insulin and vaccine production, gene therapy, Genetically modified organisms-Bt crops,
Transgenic Animals, Biosafety issues-Biopiracy and patents.
UNIT V Ecology and Environment
Organisms and environment Habitat and niche, Population and ecological adaptations, Population interactions-mutualism,
competition, predation, parasitism, Population attributes-growth, birth rate and death rate, age distribution.
Ecosystem Patterns, components, productivity and decomposition, Energy flow, Pyramids of number, biomass, energy, Nutrient
cycling (carbon and phosphorous), Ecological succession, Ecological Services-Carbon fixation, pollination, oxygen release.
Biodiversity and its conservation Concept of Biodiversity, Patterns of Biodiversity, Importance of Biodiversity, Loss of Biodiversity,
Biodiversity conservation, Hotspots, endangered organisms, extinction, Red Data Book, biosphere reserves, National parks and
sanctuaries.
Environmental issues Air pollution and its control, Water pollution and its control, Agrochemicals and their effects, Solid waste
management, Radioactive waste management, Greenhouse effect and global warning, Ozone depletion, Deforestation, Any
three case studies as success stories addressing environmental issues.

11. 01The contrasting characteristics
generally in a pair used for
identification of animals in
taxonomic key are referred to as
[NEET (Odisha) 2019]
(a) lead (b) couplet
(c) doublet (d) alternate
Ans.(b)
Couplet is the contrasting
characteristic generally in a pair used
for identification of animals in
taxonomic key. It represents the choice
made between two opposite options,
each half of a couplet is called lead.
Separate taxonomic keys are required
for each taxonomic category like family,
genus species.
02Match the items given in Column I
with those in Column II and select
thecorrectoption given below
[NEET 2018]
Column I Column II
1. Herbarium (i) It is a place having a
collection of
preserved plants and
animals.
2. Key (ii) A list that
enumerates
methodically all the
species found in an
area with brief
description aiding
identification.
Column I Column II
3. Museum (iii) It is a place where
dried and pressed
plant specimens
mounted on sheets
are kept.
4. Catalogue (iv) A booklet containing
a list of characters
and their alternates
which are helpful in
identification of
various taxa.
1 2 3 4
(a) (ii) (iv) (iii) (i)
(b) (iii) (ii) (i) (iv)
(c) (i) (iv) (iii) (ii)
(d) (iii) (iv) (i) (ii)
Ans.(d)
Herbariumis a place where dried and
pressed plant specimens, mounted on
sheets are kept systematically. It is a
repository or store house for future
use.Keyis a booklet containing list of
characters and their alternates which
are helpful in identification of various
taxa-class, order, family, genus and
species.
Museumis an institution where artistic
and educational materials are
exhibited to the public. The materials
available for observation and study are
called a collection.
Catalogueis a list or register that
enumerates methodically all the
species found in a particular place. It
often possesses brief description of
species that aids in identification.
Therefore, option (d) is correct.
03The label of a herbarium sheet does
not carry information on
[NEET 2016, Phase II]
(a) date of collection
(b) name of collector
(c) local names
(d) height of the plant
Ans.(d)
Herbarium is a collection of plants that
usually have been dried, pressed,
preserved plant on sheets and are
arranged in accordance with any
accepted system of classification for
future reference and study. It does not
have information on height of the plant.
04Study the four statements (I-IV)
given below and select the two
correct ones out of them :
[NEET 2016, Phase II]
I. Definition of biological species
was given by Ernst Mayr.
II. Photoperiod does not affect
reproduction in plants.
III. Binomial nomenclature system
was given by RH Whittaker.
IV. In unicellular organisms, reproduc-
tion is synonymous with growth.
The two correct statements are
(a) II and III (b) III and IV
(c) I and IV (d) I and II
Ans.(c)
Statements I and IV are correct. The
correct form of II and III are as follows:
(II) Photoperiod does affect the
reproduction in plants.
(III) Binomial nomenclature was given by
Carolus Linnaeus.
TheLivingWorld
01
Diversity and Taxonomy
TOPIC 1

12. 05Biological organisation starts with
[CBSE AIPMT 2007]
(a) sub-microscopic molecular level
(b) cellular level
(c) organismic level
(d) atomic level
Ans.(a)
Biological organisation starts with
sub-microscopic molecular level, where
four types of molecules, i.e.
carbohydrates, lipids, proteins and
nucleic acids are organised into
organelles of cell.
06The living organisms can be
un-exceptionally distinguished
from the non-living things on the
basis of their ability for
[CBSE AIPMT 2007]
(a) responsiveness to touch
(b) interaction with the environment
and progressive evolution
(c) reproduction
(d) growth and movement
Ans.(b)
All living organisms interact with their
environment and shows progressive
evolution. They can sense and respond
to environmental uses. On the other
hand reproduction, growth and
movement cannot be all inclusive
defining properties of living organisms.
07Which one of the following is an
example of negative feedback
loop in humans?
[CBSE AIPMT 2007]
(a) Constriction of skin blood vessels
and contraction of skeletal
muscles when it is too cold
(b) Secretion of tears after falling of
sand particles into the eye
(c) Salivation of mouth at the sight of
delicious food
(d) Secretion of sweat glands and
constriction of skin blood vessels
when it is too hot
Ans.(a)
Skin blood vessels constrict and
skeletal muscles contract due to the
cold is an example of negative feedback
mechanism of homeostasis.
08Carbohydrates the most abundant
biomolecules on earth, are
produced by[CBSE AIPMT 2005]
(a) all bacteria, fungi and algae
(b) fungi, algae and green plant cells
(c) some bacteria, algae and green
plant cells
(d) viruses, fungi and bacteria
Ans.(c)
Some photosynthetic bacteria such as
Rhodopseudomonas can prepare
carbohydrates. But during this type of
food synthesisO
2
is not evolved
because, in this case hydrogen donor is
other thanH O
2
.
Algae (green and blue-green) and all
green plant cells prepare their food
(carbohydrate) through photosynthesis.
Here, hydrogen ions are donated by
water molecules by the process of
photolysis of water, i.e.O
2
is released
during this type of food synthesis.
09More than 70% of world’s fresh
water is contained in
(a) antarctica[CBSE AIPMT 2005]
(b) greenland
(c) glaciers and mountains
(d) polar ice
Ans.(d)
Three fourth surface of earth (about
71% of total) is occupied by oceans
which contain 97.5% of total water. This
is marine water with about 3.5% salt
contents.
Rest water, i.e. 2.5% is freshwater
which occurs on land. Most amount of
this water (about 1.97%) occurs as
Frozen ice caps and glaciers and 0.5%
fresh water occurs as ground water.
10There is no life on moon due to
the absence of[CBSE AIPMT 2002]
(a)O
2
(b) water
(c) light (d) temperature
Ans.(b)
Water is an essential constituent of
cytoplasm of all living organisms. It helps
in distribution of substances within the
organism, elimination of waste products,
maintenance of body temperature, etc.
It is absent on the moon.
Anaerobic organisms that can live in
the absence ofO
2
. Light and
temperature are already known to exist
on the moon.
11The most important feature of all
living systems is to
[CBSE AIPMT 2000]
(a) utilise oxygen to generate energy
(b) replicate the genetic information
(c) produce gametes
(d) utilise solar energy for metabolic
activities
Ans.(b)
Reproduction is necessary for
continuity of life. However, production
of gametes is not only method for this.
A number of organisms reproduce
asexually. In any case, cell division is
necessary which involves replication of
DNA.
12Homeostasis is[CBSE AIPMT 1991]
(a) tendency to change with change in
environment
(b) tendency to resist change
(c) disturbance in regulatory
control
(d) plants and animals extracts used
in homeopathy
Ans.(b)
Homeostasis (Gr. homeos = similar;
stasis = standing) is the tendency of
maintaining a relatively stable internal
physiological environment in an
organism, or steady-state equilibrium in
a population or ecosystem. It is carried
out by regulatory mechanisms which
coordinate internal functions such as
providing nutrients to cells and
transporting substances.
13Employment of hereditary
principles in the improvement of
human race is[CBSE AIPMT 1990]
(a) Euthenics (b) Eugenics
(c) Euphenics (d) Ethnology
Ans.(b)
Eugenics refers to improvement of
human race by modifying fertility or
employing the hereditary principles.
2 NEETChapterwise Topicwise Biology

13. 14Select the correctly written
scientific name of Mango which
was first described by Carolus
Linnaeus.[NEET (Naional) 2019)]
(a)Mangifera indicaLinn
(b)Mangifera indica
(c)Mangifera Indica
(d)Mangifera indicaCar. Linn.
Ans.(a)
The correct form of writing the
scientific name of mango as described
by Carolus Linnaeus isMangifera indica
Linn.
As per binomial nomenclature rules, the
name of an organism contains a generic
and specific name. The former begins
with capital letter while the later begins
with small letter. The name of
taxonomist is written in Roman script
and it is written in abbreviated form.
15Which of the following is against
the rules of ICBN?
[NEET (Odisha) 2019]
(a) Handwritten scientific names
should be underlined
(b) Every species should have a
generic name and a specific
epithet
(c) Scientific names are in Latin and
should be italicised
(d) Generic and specific names
should be written starting with
small letters
Ans.(d)
Statement (d) is against the rules of
ICBN because the universal rule of
nomenclature is that the first word
denoting the genus starts with a
capital letter while the specific epithet
starts with a small letter. It can be
illustrated with the example of
Mangifera indica.
16Nomenclature is governed by
certain universal rules. Which one
of the following is contrary to the
rules of nomenclature?
[NEET 2016, Phase I]
(a) The first word in a biological
name represents the genus name
and the second is a specific
epithet
(b) The names are written in Latin and
are Italicised
(c) When written by hand, the names
are to be underlined
(d) Biological names can be written in
any language
Ans.(d)
Biological names originate from latin
language and are printed in italics.
The Living World 3
Binomial Nomenclature
TOPIC 2

14. 01Which of the following statement
is correct? [NEET 2021]
(a) Fusion of two cells is called
karyogamy
(b) Fusion of protoplasm between
two motile or non-motile gametes
is called plasmogamy
(c) Organisms that depend on living
plants are called saprophytes
(d) Some of the organisms can fix
atmospheric nitrogen in
specialised cells called sheath
cells
Ans.(b)
Statement in option (b) is correct.
Plasmogamy, the fusion of two
protoplasts (the contents of the two
cells), brings together two compatible
haploid nuclei. At this point, two parent
cells are present in the same cell, but
the nuclei have not yet fused.
Incorrect statements can be corrected
as Organisms that can fix atmospheric
nitrogen in specialised cells are called
heterocyst.
Karyogamy is nothing but the fusion of
two nuclei means production of diploid
cell (2ncondition).
Organisms that depends on living plants
are called heterotrophs.
02The size of Pleuropneumonia Like
Organism (PPLO) is
[NEET (Oct.) 2020]
(a) 0.02μm (b) 1-2 μm
(c) 10–20μm (d) 0.1μm
Ans.(d)
The size of various organisms/cells are
Pleuropneumonia Like Organism
(PPLO)–0.1μmViruses–0 02.–0.2μm
Bacterial cell–1–2μm
Eukaryotic cell–10–20μm
03Which of the following is incorrect
about cyanobacteria?
[NEET (Oct.) 2020]
(a) They are photoautotrophs
(b) They lack heterocysts
(c) They often form blooms in
polluted water bodies
(d) They have chlorophyll-asimilar to
green plants
Ans.(b)
Cyanobacteria or blue-green algae are
photosynthetic organisms which
perform oxygenic photosynthesis. They
have the ability of nitrogen fixation due
to the presence of large pale cells called
heterocyst in their filaments. Due to the
presence of thick walls, heterocysts are
impermeable to oxygen.
04Oxygen isnotproduced during
photosynthesis by[NEET 2018]
(a)Cycas
(b)Nostoc
(c) Green sulphur bacteria
(d)Chara
Ans.(c)
Green sulphur bacteriaare anaerobic
bacteria. They do not evolve oxygen
during photosynthesis. Such type of
photosynthesis is known as anoxygenic
photosynthesis.
They do not use water as a source of
reducing power. Instead, hydrogen is
obtained from hydrogen sulphide.
H S 2[H] S
2
Hydrogen
sulphide
→ +
6CO + 12H C H O + 6H O
2 2 6 12 6 2
Enzymes
Light
→
Concept EnhancerGreen sulphur
bacteria, e.g.Chlorobium limicola,
possesses bacteriophaeophytin as
photosynthetic pigment.
Cycasis a gymnosperm,Nostocis a
blue-green algae andCharais a green
algae. All of these produce oxygen
during photosynthesis.
05Which of the following organisms
are known as chief producers in
the oceans? [NEET 2018]
(a) Cyanobacteria (b) Diatoms
(c) Dinoflagellates (d) Euglenoids
Ans.(b)
Diatomsare chief producers in the
oceans and they contribute 40% of
marine primary productivity. They
constitute a major group of unicellular
eukaryotic microalgae and are among
the most common types of
phytoplanktons.
The other given organisms also exhibit
autotrophic mode of nutrition.
06Which of the following are found
in extreme saline conditions?
(a) Archaebacteria [NEET 2017]
(b) Eubacteria
(c) Cyanobacteria
(d) Mycobacteria
Biological
Classification
02
Kingdom-Monera
TOPIC 1

15. Ans.(a)
Archaebacteria are the most primitive
form of bacteria. These live in diverse
habitat, e.g. extreme hot temperature,
saline condition, variable pH, etc. Saline
bacteria are called Halophiles (e.g.
Halobacterium, Halococcus).
Concept EnhancerThe ability of
archaebacteria to survive in such
conditions is due to the presence of
branched lipid chain in their membrane,
which reduces the fluidity of their
membrane.
07Which among the following are the
smallest living cells, known
without a definite cell wall,
pathogenic to plants as well as
animals and can survive without
oxygen? [NEET 2017]
(a)Bacillus
(b)Pseudomonas
(c)Mycoplasma
(d)Nostoc
Ans.(c)
Mycoplasma is triple layered smallest
living cells. It does not have definite cell
wall. It is an anaerobic organism. It
cause diseases in plants (little leaf of
brinjal) as well as in animals
(pleuromorphic pneumonia in man).
08The primitive prokaryotes
responsible for the production of
biogas from the dung of ruminant
animals, include the
[NEET 2016 Phase I]
(a) thermoacidophiles
(b) methanogens
(c) eubacteria
(d) halophiles
Ans.(b)
Methanogens are group of obligate
anaerobic ancient and primitive
bacteria. They are involved in
methanogenesis and produce methane
gas in ruminant of cattles.
09Which one of the following
statements is wrong?
[NEET 2016 Phase I]
(a) Golden algae are also called
desmids
(b) Eubacteria are also called false
bacteria
(c)Phycomycetesare also called algal
fungi
(d) Cyanobacteria are also called
blue-green algae
Ans.(b)
Eubacteria are true bacteria which
exhibit all true characteristic features
of group Eubacteria.
10Methanogens belong to
[NEET 2016 Phase II]
(a) eubacteria (b) archaebacteria
(c) dinoflagellates (d) slime moulds
Ans.(b)
Methanogens belong to archaebacteria.
It contains three major classes of
primitive bacteria, i.e. methanogens,
halophilic and thermoacidophilic.
Methanogensare strict anaerobes,
present in the gut of several ruminant
animals (e.g. cows, etc.) and are
responsible for production of methane
gas from the dung of these animals.
Concept EnhancerHalophilic bacteria
usually occur in salt rich substrate like
salt marshes, etc. and are aerobic
chemoheterotrophs. Thermophilic
bacteria have the dual ability to tolerate
high temperature as well as high
acidity. These are basically
chemosynthetic.
11Chromatophores take part in
[CBSE AIPMT 2015]
(a) photosynthesis (b) growth
(c) movement (d) respiration
Ans.(a)
Chromatophores are found in members
of phototrophic bacteria. They contain
bacteriochlorophyll pigments and
carotenoids and take part in
photosynthesis. In purple bacteria,
such asRhodospirillum rubrum, the
light-harvesting proteins are intrinsic to
the chromatophore membranes.
However, in green sulphur bacteria,
they are arranged in specialised
antenna complexes called
chlorosomes.
12Archaebacteria differ from
eubacteria in[CBSE AIPMT 2014]
(a) cell membrane structure
(b) mode of nutrition
(c) cell shape
(d) mode of reproduction
Ans.(a)
Archaebacteria different from
eubacteria in that eubacteria have cell
membrane composed mainly of
glycerol-ester lipids, while
archaebacteria have membrane made
up of glycerol-ether lipid.
Ether lipids are chemically more
resistant then ester lipids. This stability
helps archaebacteria to survive at high
temperture and in very acidic or alkaline
environment.
13Besides paddy fields, cyanobacteria
are also found inside vegetative
part of [NEET 2013]
(a)Pinus (b)Cycas
(c)Equisetum (d)Psilotum
Ans.(b)
Cyanobacteria within the coralloid roots
ofCycasare chemoheterotrophic and
specifically adapted to life in symbiosis.
Only a few species of cyanobacteria
form associations withCycas.Pinusis a
gymnosperm.Equisetumbelongs to
vascular plants and to horse tail family.
Psilotumbelongs to
division–Pteridophyta and is a fern-like
plant.
14Pigment-containing membranous
extensions in some cyanobacteria
are [CBSE AIPMT 2012]
(a) heterocysts
(b) basal bodies
(c) pneumatophores
(d) chromatophores
Ans.(d)
Cyanobacteria contain chlorophyll but
the chlorophyll is not located in
chloroplasts, rather it is found in
chromatophores which are infolding of
the plasma membrane, where
photosynthesis is carried out.
Heterocystsare specialised nitrogen
fixing cells formed by some filamentous
cyanobacteria such asNostoc.Abasal
bodyis an organelle formed from a
centriole and a short cylindrical array of
microtubules.
Biological Classification 5
Lipoprotein membrane
Soluble protein
dsDNA
Ribosomes
Soluble
RNA
(Three layers)

16. Pneumatophores are lateral roots that
grow upward for varying distance and
function as the site of oxygen intake for
the submerged primary roots of
mangrove plants.
15Which of the following are likely to
be present in deep sea water?
[CBSE AIPMT 2012]
(a) Archaebacteria
(b) Eubacteria
(c) Blue-green algae
(d) Saprophytic fungi
Ans.(a)
Archaebacteria can flourish in hot
springs and deep sea hypothermal
vents. Eubacteria are true bacteria,
characterised by the presence of rigid
cell wall and if motile a flagellum. Most
fungi are heterotrophic and absorb
soluble organic matter from dead
substrates and hence, are called
saprophyte. The cyanobacteria have
chlorophyll-asimilar to green plants
and are photosynthetic autotrophs.
16The cyanobacteria are also
referred to as[CBSE AIPMT 2012]
(a) protists
(b) golden algae
(c) slime moulds
(d) blue-green algae
Ans.(d)
Cyanobacteria, also known as
blue-green algae (BGA) are most
primitive prokaryotic organisms. These
are considered to be the most ancient
of all the chlorophyll bearing organisms
on earth.
17Nuclear membrane is absent in
[CBSE AIPMT 2012]
(a)Penicillium(b)Agaricus
(c)Volvox (d)Nostoc
Ans.(d)
Nostocis a prokaryote. Prokaryotic
cells lack membrane bound organelles
and well organised nucleus, i.e. nuclear
envelope is absent.
Penicillium, AgaricusandVolvoxare
eukaryotic.
18In eubacteria, a cellular component
that resembles eukaryotic cells is
[CBSE AIPMT 2011]
(a) nucleus
(b) ribosomes
(c) cell wall
(d) plasma membrane
Ans.(d)
The plasma membrane of eubacteria
resembles to that of eukaryotic cell. It
is made of phospholipid, protein and
some amount of polysaccharides.
However, it lacks sterol, the
characteristic of eukaryotic cell
membrane. Instead, there is sterol like
compounds called hopanoid.
19Organisms called methanogens
are most abundant in a
[CBSE AIPMT 2011]
(a) cattle yard (b) polluted stream
(c) hot spring (d) sulphur rock
Ans.(a)
Methanogens are present in the gut of
several ruminants animals such as
cows and buffaloes and they are
responsible for the production of
methane (biogas) from the dung of
these animals. Thus, they are most
abundant in a cattle yard.
20A prokaryotic autotrophic nitrogen
fixing symbiont is found in
[CBSE AIPMT 2011]
(a)Cycas(b)Cicer(c)Pisum(d)Alnus
Ans.(a)
Thecoralloid rootofCycasis
symbiotically associated with nitrogen
fixing blue-green algae,Anabaena
cycadaeandNostoc punctiforme. These
blue green-algae (cyanobacteria) are
prokaryotic, photosynthetic and
autotrophic.
21Maximum nutritional diversity is
found in the group
[CBSE AIPMT 2010]
(a) Fungi (b) Animalia
(c) Monera (d) Plantae
Ans.(c)
Maximum nutritional diversity is shown
by the members of kingdom–Monera.
Some of them are autotrophic (e.g.
photosynthetic autotrophic or
chemosynthetic autotrophic) while the
vast majority are heterotrophs (e.g.
saprotrophic or parasitic). Ecologically,
these may be producers or
decomposers.
22Some hyperthermophilic
organisms that grow in highly
acidic habitats belong to the two
groups called[CBSE AIPMT 2010]
(a) eubacteria and archaea
(b) cyanobacteria and diatoms
(c) protists and mosses
(d) liverworts and yeasts
Ans.(a)
Thermophiles live in very hot places,
with temperature ranging from
60°-80°C. Many thermophiles (some
eubacteria and archaebacteria) are
autotrophs and have metabolisms
based on sulphur.
Some thermophilic archaebacteria
form the basis of food webs around
deep-sea thermal vents, where they
must withstand extreme temperature
and pressures. Archaebacteria can
grow in highly acidic (pH = 0.7) and very
basic (pH = 11) environments.
23Thermococcus, Methanococcus
andMethanobacteriumexemplify
[CBSE AIPMT 2008]
(a) archaebacteria that contain
protein homologous to eukaryotic
core histones
(b) archaebacteria that lack any
histones resembling those found
in eukaryotes but whose DNA is
negatively supercoiled
(c) bacteria whose DNA is relaxed or
positively supercoiled but which
have a cytoskeleton as well as
mitochondria
(d) bacteria that contain a
cytoskeleton and ribosomes
Ans.(a)
Inspection of domain Archaea shows
that two sub-divisions exist, the
Euryarchaeota and the Crenarchaeota.
The Euryarchaeota includes
Methanobacterium, Methanococcus,
Thermococcus.
24Bacterial leaf blight of rice is
caused by a species of
[CBSE AIPMT 2008]
(a)Xanthomonas
(b)Pseudomonas
(c)Alternaria
(d)Erwinia
Ans.(a)
Disease Causing Organism
Leaf blight of rice —Xanthomonas
oryzae
Red strip of suga —Pseudomonas
canerubrilineans
Fire blight of apple —Erwinia amylovora
Early blight of potato —Alternaria solani
6 NEETChapterwise Topicwise Biology

17. 25Which one of the following
statements aboutMycoplasmais
wrong? [CBSE AIPMT 2007]
(a) They are also called PPLO
(b) They are pleomorphic
(c) They are sensitive to penicillin
(d) They cause disease in plants
Ans.(c)
Penicillin acts on cell wall and
Mycoplasmalacks cell wall. Thus
Mycoplasmais not sensitive to
penicillin.Mycoplasmaare inhibited by
metabolic inhibitors like
chloramphenicol and tetracyclin.
26Barophillic prokaryotes
[CBSE AIPMT 2005]
(a) grow slowly in highly alkaline
frozen takes at high altitudes
(b) occur in water containing high
concentrations of barium
hydroxide
(c) grow and multiply in very deep
marine sediments
(d) readily grown and divides in sea
water enriched in any soluble salt
of barium
Ans.(c)
Barophillic prokaryotes grow and
multiply in very deep marine sediments.
27A free living nitrogen fixing
cyanobacterium which can also
form symbiotic association with the
water fernAzollais
[CBSE AIPMT 2004]
(a)Tolypothrix(b)Chlorella
(c)Nostoc (d)Anabaena
Ans.(d)
Anabaenais a free living nitrogen fixing
cyanobacterium which can form
symbiotic association with water fern
Azolla.
28Chromosomes in a bacterial cell
can be 1-3 in number and
[CBSE AIPMT 2003]
(a) can be circular as well as linear
within the same cell
(b) are always circular
(c) are always linear
(d) can be either circular or linear, but
never both within the same cell
Ans.(b)
Bacterial chromosomes are circular
DNA molecules.
29Organisms which obtain energy by
the oxidation of reduced inorganic
compounds are called
[CBSE AIPMT 2002]
(a) homoautotrophs
(b) chemoautotrophs
(c) saprozoic
(d) coproheterotrophs
Ans.(b)
The organisms obtaining energy by
chemical reactions independent of light
are called chemotrophs. The reductants
obtained from the environment may be
inorganic (in case of chemoautotrophs)
or organic (in case of
chemoheterotrophs).
Photoautotrophs organisms that make
their own food by photosynthesis, using
the light energy.
Saprozoic organisms obtain food from
dead and decaying matter.
30In bacteria, plasmid is
[CBSE AIPMT 2002]
(a) extra-chromosomal material
(b) main DNA
(c) non-functional DNA
(d) repetitive gene
Ans.(a)
Plasmid is an extrachromosomal
material capable of replicating
independently from the main
chromosome. Plasmids usually possess
antibiotic resistance genes.
31What is true for archaebacteria?
[CBSE AIPMT 2001]
(a) All halophiles
(b) All photosynthetic
(c) All fossils
(d) Oldest living beings
Ans.(d)
The archaebacteria are able to flourish in
extreme conditions of environment that
are believed to have existed on the
primitive earth. It is believed that these
represent the early forms of life. Hence,
archaebacteria are called ‘‘oldest living
beings’’.
32What is true for cyanobacteria?
[CBSE AIPMT 2001]
(a) Oxygenic with nitrogenase
(b) Oxygenic without nitrogenase
(c) Non-oxygenic with nitrognase
(d) Non-oxygenic without nitrogenase
Ans.(a)
Cyanobacteria (Gk.Kyanos= dark blue;
bakterion= a staff) also known as
blue-green algae. It is a very important
group of photosynthetic bacteria in the
history of life on earth.
Thecyanobacteria fix atmospheric
nitrogen through the help of enzyme
nitrogenase and also show oxygenic
photosynthetic.
33What is true for photolithotrophs?
[CBSE AIPMT 2001]
(a) Obtain energy from radiations
and hydrogen from organic
compounds
(b) Obtain energy from radiations
and hydrogen from inorganic
compounds
(c) Obtain energy from organic
compounds
(d) Obtain energy from inorganic
compounds
Ans.(b)
Photolithotrophs used light as energy
and inorganic electron donor (like
H , H S)
2 2
as hydrogen source. Purple
and green sulphur bacteria are
examples of photolithotrophs.
34Photosynthetic bacteria have
pigments in[CBSE AIPMT 1999]
(a) leucoplasts
(b) chloroplasts
(c) chromoplasts
(d) chromatophores
Ans.(d)
In photosynthetic bacteria, small
particles of60μmdiameter, called
chromatophores, are present. These
are attached to the inner surface of
the cell membrane, have no limiting
membrane and possess
bacteriochlorophyll.
Chloroplast, leucoplast and
chromoplasts are plastids present in
eukaryotic cells.
(a)ChromoplastsColoured plastids
except green, give different type
of colour appearance to different
parts of the plant.
(b)ChloroplastsGreen plastids take
part in the process of
photosynthesis.
(c)LeucoplastsColourless plastids,
mainly function as store house of
various types of food.
Biological Classification 7

18. 35A few organisms are known to
grow and multiply at temperatures
of 100-105°C. They belong to
[CBSE AIPMT 1998]
(a) marine archaebacteria
(b) thermophilic sulphur bacteria
(c) hot-spring blue-green algae
(cyanobacteria)
(d) thermophilic, subaerial fungi
Ans.(a)
The cell membrane of archaebacteria is
consists of branched chain lipids, long
chain branched alcohols, phytanals,
ether linked to glycerol. This helps them
to withstand extreme conditions and
temperature.
36The hereditary material present in
the bacteriumEscherichia coliis
[CBSE AIPMT 1997, 98]
(a) single stranded DNA
(b) deoxyribose sugar
(c) double stranded DNA
(d) single stranded RNA
Ans.(c)
Bacterial chromosome is single
circular, double-stranded DNA
molecule.
37Bacterium divides every 35
minutes. If a culture containing10
5
cells per mL is grown for 175
minutes, what will be the cell
concentration per mL after 175
minutes? [CBSE AIPMT 1998]
(a)5 10
5
×cells (b)35 10
5
×cells
(c)32 10
5
×cells (d)175 10
5
×cells
Ans.(c)
1 10 2 10 4 10
5 5 535 70
× → × → ×
min min
→ × → ×
105 140
8 10 16 10
5 5min min
→ ×
175
32 10
5min
38The site of respiration in bacteria
is [CBSE AIPMT 1997]
(a) episome
(b) mesosome
(c) ribosome
(d) microsome
Ans.(b)
The cytoplasmic membrane of bacteria
is invaginated at certain places into the
cytoplasm in the form of tubules, which
are called mesosomes; on their surface
are found enzymes associated with
respiration.
Mesosome works as mitochondia in
bacterial cell.
39In bacterial chromosomes, the
nucleic acid polymers are
[CBSE AIPMT 1996]
(a) linear DNA molecule
(b) circular DNA molecule
(c) of two types — DNA and RNA
(d) linear RNA molecule
Ans.(b)
Nucleoid or chromatin body or
genophore in bacteria occupies 10-20%
of cell, is present near the centre of
cell. It consists of a single, circular DNA
molecule in which all the genes are
linked. It is over a thousand times
longer than the cell itself and is,
therefore, highly folded. It lacks the
histone proteins.
40Sex factor in bacteria is
[CBSE AIPMT 1996]
(a) chromosomal replicon
(b) F-replicon
(c) RNA
(d) sex-pilus
Ans.(b)
Bacterial strains such as ofE. colishow
sexual differences. Each male cell
possesses a sex factor or fertility factor
calledF-factor. Infact, it is a small
circular piece of DNA, self-replicating
like bacterial chromosome but only 1/100
in size. TheF-factorcodes for the
protein of a special type of pilus, the sex
pilus which enables cell to cell contact
and transfer of genetic material through
a conjugation tube.
41The plasmid[CBSE AIPMT 1995]
(a) helps in respiration
(b) genes found inside nucleus
(c) is a component of cell wall of
bacteria
(d) is the genetic part in addition to
DNA in microorganisms
Ans.(d)
Plasmids are small, self-replicating,
extrachromosomal, non-essential
genetic elements in bacteria. Each
plasmid has a ring of circular,
supercoiled, double stranded DNA.
They carry genes for replication and for
one or more cellular non-essential
functions. These are also called
minichromosomes or dispensable
autonomous elements.
42Temperature tolerance of thermal
blue-green algae is due to
[CBSE AIPMT 1994]
(a) cell wall structure
(b) cell organisation
(c) mitochondrial structure
(d) homopolar bonds in their proteins
Ans.(a)
Cyanobacteria provide a good example
of the adaptability of life to extremes of
environment (high temperature of hot
springs and low temperature of polar
regions). It is due to their gelatinous
sheath, that can withstand long periods
of desiccation. The compactness of
protein molecules and their bonds in
the protoplasm also help the cells to
face the extremes.
43Escherichia coliis used extensively
in biological research as it is
[CBSE AIPMT 1993]
(a) easily cultured (b) easily available
(c) easy to handle
(d) easily multiplied in host
Ans.(a)
Escherichia coli, a commensal
bacterium is most studied bacterium
and widely used in research because it
is easily cultured on minimal medium
and has fast rate of multiplication and
short generation time.
44Genophore bacterial genome or
nucleoid is made of
[CBSE AIPMT 1993]
(a) histones and non-histones
(b) RNA and histones
(c) a single double stranded DNA
(d) a single stranded DNA
Ans.(c)
In bacteria nucleoid or genophore is
haploid and consists of single, naked,
double stranded, circular ring like highly
folded supercoiled DNA with no free
ends, no histone proteins.
The nucleoid ofE. colihas a central
core of RNA surrounded by about 50
super coilings of DNA which is then
associated with some basic proteins
but never histones. Some proteins
like polyamines rich in alanine are also
found associated with DNA.
45Bacteria lack alternation of
generation because there is
[CBSE AIPMT 1992, 91]
(a) neither syngamy nor reduction
division
(b) distinct chromosomes are absent
8 NEETChapterwise Topicwise Biology

19. (c) no conjugation
(d) no exchange of genetic material
Ans.(a)
Bacteria reproduces asexually by
transverse binary fission, conidia,
budding, cyst and sporulation. No true
sexual reproduction (involving formation
of gametes, their fusion and meiosis) is
known to occur in bacteria.
However, in bacteria the transfer of
genetic material from donor to recipient
cell to bring genetic
recombinations/variations is reported
that occurs not through gametes/sex
cells, but by other methods like
conjugation, transduction and
transformation. It does not result in any
multiplication of cells.
46Which one belongs to the Monera?
[CBSE AIPMT 1990]
(a)Amoeba (b)Escherichia
(c)Gelidium (d)Spirogyra
Ans.(b)
The kingdom–Monera (Gr.monera=
simple) includes simple, prokaryotic
primitive organisms. It includes bacteria,
archaebacteria, Actinomycetes,
Mycoplasma, spirochaetes, rickettsiae,
chlamydiae and cyanobacteria.
Escherichia coliis the most studied
bacterium.E. coliis an enteric bacteria,
found in entire colon, secretes
vitamin-K,B , B , B
36 12
and folic acid.
47Ciliates differ from all other
protozoans in [NEET 2018]
(a) using pseudopodia for capturing
prey
(b) having a contractile vacuole for
removing excess water
(c) using flagella for locomotion
(d) having two types of nuclei
Ans.(d)
Ciliates differ from all other protozoans
inhaving two types of nuclei.
These two nuclei are usually of different
size,i.e.one is meganucleus and the
other is micronucleus. The former
controls metabolism whereas the latter
is concerned with reproductions, e.g.
paramecium.
In other protozoans, likeAmoeba,
single nucleus is present which is
involved in metabolism and
reproduction.
Other options are incorrect because
Ciliates use filtre feeding mechanism
for obtaining food.
Like other protozoans, they also
possess contractile vacuoles.
Ciliates use cilia for locomotion.
48Chrysophytes, euglenoids,
dinoflagellates and slime moulds
are included in the kingdom
[NEET 2016, Phase I]
(a) Protista
(b) Fungi
(c) Animalia
(d) Monera
Ans.(a)
All single celled eukaryotic organisms
like chrysophytes [diatoms and
desmids], euglenoids[Euglena],
dinoflagellates and slime moulds are
included in kingdom–Protista.
49Select the wrong statement.
[NEET 2016, Phase II]
(a) The walls of diatoms are easily
destructible
(b) ‘Diatomaceous earth’ is formed
by the cell walls of diatoms
(c) Diatoms are chief producers in
the oceans
(d) Diatoms are microscopic and
float passively in water
Ans.(a)
Diatoms are single celled plant like
protists that produce intricately
structured cell walls made of nano(–)
silica(SiO )
2
. Thus, the walls are
indestructible. Hence, only option (a) is
wrong and rest of the options are
correct.
50Pick up the wrong statement.
[CBSE AIPMT 2015]
(a) Cell wall is absent in Animalia
(b) Protista have photosynthetic and
heterotrophic modes of nutrition
(c) Some fungi are edible
(d) Nuclear membrane is present in
Monera
Ans.(d)
In Protista kingdom members exhibit
both autotrophic as well as
heterotrophic nutrition. Animal cells
lack cell wall and there are a few fungi
that are edible. Monera is the kingdom
that contains unicellular organisms
with a prokaryotic cell organisation,
i.e. which lacks nuclear membrane and
other membrane bound organelles.
51In which group of organisms the
cell walls form two thin
overlapping shells which fit
together? [CBSE AIPMT 2015]
(a) Chrysophytes (b) Euglenoids
(c) Dinoflagellates (d) Slime moulds
Ans.(a)
Chrysophytes are placed under the
kingdom– Protista. This group includes
diatoms and golden algae (desmids).
Most of them are photosynthetic. In
diatoms, the cell walls form two thin
overlapping cells, which fit together as
in a soap box.
52What is common about
Trypanosoma, Noctiluca,
MonocystisandGiardia?
[CBSE AIPMT 2006]
(a) These are all unicellular protists
(b) They have flagella
(c) They produce spores
(d) These are all parasites
Ans.(a)
Trypanosoma, Noctiluca, Monocystis
andGiardiaare all unicellular protists.
Trypanosoma gambienseis the single
celled, parasitic zooflagellate causing
trypanosomiasisorsleeping sickness.
Giardiaor the ‘Grand old man of the
intestine’ is a parasitic flagellate
occurring in the intestine of man and
other animals and causes giardiasis or
diarrhoea (i.e. very loose and frequent
stool containing large quantity of fat).
Noctilucais a marine, colourless
dinoflagellate. It is a voracious predator
and has a long, motile tentacle, near the
base of which, its single short flagellum
emerges.
Monocystisis a microscopic, unicellular
endoparasitic protozoan found in the
coelom and seminal vesicles of
earthworm. As it is an endoparasite, it
does not possess any special structure
for locomotion.
53Auxospores and hormocysts are
formed respectively by
[CBSE AIPMT 2005]
(a) several diatoms and a few
cyanobacteria
(b) several cyanobacteria and several
diatoms
(c) some diatoms and several
cyanobacteria
(d) some cyanobacteria and many
diatoms
Biological Classification 9
Kingdom-Protista
TOPIC 2

20. Ans.(a)
Auxospores and hormocysts are formed
by several diatoms and a few
cyanobacteria respectively.
Bacillariophyceae members (diatoms)
are microscopic, eukaryotic, unicellular
or colonial coccoid algae. These algae
are sexually reproduced by the
formation of auxospores in most cases.
Bozi(1914) andFermi(1930) reported
that short sections of living cells at the
tips of the trichomes ofWertiella lanosa
become invested by a thick, lamellated,
pigmented sheath. Such mullticellular
spore like structures function as
perennating bodies. They are specially
modified hormogones and are called
hormospores or hormocysts.
54Which of the following unicellular
organism has a macro-nucleus for
trophic function and one or more
micro-nuclei for reproduction?
[CBSE AIPMT 2005]
(a)Euglena (b)Amoeba
(c)Paramecium(d)Trypanosoma
Ans.(c)
Parameciumis a heterokaryotic
organism, i.e. it has two nuclei near the
cytostome (oral-shaped opening called
mouth). The macronucleus, which is a
conspicuous larger ellipsoidal
vegetative nucleus, divides amitotically
and controls the vegetative characters
and micronucleus is a small compact
reproductive nucleus which divides
mitotically and controls the
reproduction.
55When a freshwater protozoan
possessing a contractile vacuole is
placed in a glass containing
marine water, the vacuole will
[CBSE AIPMT 2004]
(a) increase in number
(b) disappear
(c) increase in size
(d) decrease in size
Ans.(b)
Fresh water protozoans live in
hypotonic environment so, for
regulation of excess of water which
comes in the protoplasm through the
process of endosmosis, contractile
vacuoles have developed. When these
protozoans are placed in marine water,
i.e. hypertonic water, the contractile
vacuoles become disappear because
the process of endosmosis does not
occur and thus, water does not come in
the protoplasm.
56The chief advantage of
encystment to anAmoebais
[CBSE AIPMT 2003]
(a) the chance to get rid of
accumulated waste products
(b) the ability to survive during
adverse physical conditions
(c) the ability to live for some time
without ingesting food
(d) protection from parasites and
predators
Ans.(b)
Encystment ofAmoebais occurred
regularly to tide over unfavourable
conditions like drought and extreme
temperature, etc. During these
conditions, theAmoebaforms a
covering or cyst wall around itself.
Thus, it is an adaptation to sunrise
during adverse conditions
(extranuclear inheritance adverse
condition).
57In which animal, dimorphic
nucleus is found?
[CBSE AIPMT 2002]
(a)Amoeba
(b)Trypanosoma gambiense
(c)Plasmodium vivax
(d)Paramecium caudatum
Ans.(d)
Parameciumis heterokaryotic, it
possesses a dimorphic nuclear
apparatus (a single large macronucleus
which controls metabolism and one or
more small micronuclei concerned with
reproduction).
58Extranuclear inheritance occurs in
[CBSE AIPMT 2001]
(a) killer strain inParamecium
(b) colour blindness
(c) phenylketonuria
(d) Tay Sachs disease
Ans.(a)
Parameciumexhibits cytoplasmic
inheritance due to the presence of
Kappa particles (self replicating bodies
that produce toxin called paramecin).
Besides binary fission and conjugation
other reproductive processes that
occurs inParameciumare autogamy,
endomixis and cytogamy.
Tay Sachs diseaseIt is a rare
autosomal recessive genetic disorder.
Colour blindnessIt is X-chromosome
related disease. So, it is a sex-linked
disease.
PKUautosomal recessive genetic
disorder characterised by homozygous
or compound heterozygous mutation in
the gene.
59Which of the following organisms
possesses characteristics of both
a plant and an animal?
[CBSE AIPMT 1995]
(a) Bacteria (b)Mycoplasma
(c)Euglena (d)Paramecium
Ans.(c)
Euglenais a connecting link between
animals and plants.Euglenacontains
chlorophyll, yet it resembles animals,
because it feeds like animals in the
absence of sunlight. It resembles the
ancestral form from which the plants
and animals evolved.
60Macro and micronucleus are the
characteristic feature of
[CBSE AIPMT 1995, 2002, 05]
(a)ParameciumandVorticella
(b)OpelinaandNictothirus
(c)HydraandBallantidium
(d)VorticellaandNictothirus
Ans.(a)
ParameciumandVorticellahave
dimorphic nuclei (heterokaryotic).
Micronucleusthat is inactive except
during cell division and houses the
master copy of the genome.
Macronucleuscontrols daily synthetic
activities or on going metabolic
functions of the cell and asexual
reproduction. Macronucleus contains
multiple copies of DNA.
61When a freshwater protozoan
possessing a contractile vacuole,
is placed in a glass containing
marine water, the vacuole will
[CBSE AIPMT 2004]
(a) increase in number
(b) disappear
(c) increase in size
(d) decrease in size
Ans.(b)
Freshwater protozoans live in hypotonic
solution so, for regulation of excess of
water which comes in the protoplasm
through the process of endosmosis,
contractile vacuoles have developed.
When these protozoans are placed in
marine water, i.e. hypertonic water, the
contractile vacuoles disappear because
the process of endosmosis does not
happen and thus, water does not come
in the protoplasm.
10 NEETChapterwise Topicwise Biology

21. 62In Protozoa likeAmoebaand
Paramecium,an organelle is
found for osmoregulation which is
[CBSE AIPMT 2002]
(a) contractile vacuole
(b) mitochondria
(c) nucleus
(d) food vacuole
Ans.(a)
Unicellular organisms such asAmoeba,
Parameciumhave some organelles
called contractile vacuole for excretion.
These are freshwater animals, i.e. they
live in hypotonic solution. Therefore,
water flows from outside to inside of
the body of the organism.
The contractile vacuoles in these
organisms collect this excess water and
gradually increase in size. When the
vacuoles reach a critical size they
contract, squeezing out their contents
through the process of simple
diffusion.
63Excretion inAmoebaoccurs
through [CBSE AIPMT 1995]
(a) lobopodia
(b) uroid portion
(c) plasma membrane
(d) contractile vacuole
Ans.(d)
Endoplasm ofAmoebain the posterior
part contains a single clear rounded and
pulsating contractile vacuole.
Contractile vacuole is analogous to
uriniferous tubules of frog, it functions
in excretion and osmoregulation.
64Protistan genome has
[CBSE AIPMT 1994]
(a) membrane bound nucleoproteins
embedded in cytoplasm
(b) free nucleic acid aggregates
(c) gene containing nucleoproteins
condensed together in loose mass
(d) nucleoprotein in direct contact
with cell substance
Ans.(a)
Eukaryotic (e.g. protistan) genome is
organised in the form of nucleus. It is
differentiated into nuclear envelope,
chromatin, one or more nucleoli and
nucleoplasm. Nuclear DNA is linear,
associated with histone proteins. A
small quantity of DNA is also found in
the plastids and mitochondria. In
contrast, prokaryotic (bacterial) DNA is
circular and lies freely in the cytoplasm.
65Entamoeba colicauses
[CBSE AIPMT 1994]
(a) pyorrhoea
(b) diarrhoea
(c) dysentery
(d) None of these
Ans.(d)
Entamoeba coliis the common parasitic
genera of phylum—Protozoa. It
harbours the upper part of large
intestine (colon) and very often in the
liver, brain and testes and causes
constipation.
66Protists obtain food as
[CBSE AIPMT 1994]
(a) photosynthesisers, symbionts and
holotrophs
(b) photosynthesisers
(c) chemosynthesisers
(d) holotrophs
Ans.(a)
Protistans have variable modes of
nutrition. They are photosynthetic
heterotrophic, i.e. saprophytic,
parasitic and ingestive.
67African sleeping sickness is due to
[CBSE AIPMT 1991]
(a)Plasmodium vivaxtransmitted by
tse-tse fly
(b)Trypanosoma lewsiitransmitted by
bed bug
(c)Trypanosoma gambiense
transmitted byGlossina palpalis
(d)Entamoeba gingivalisspread by
house fly
Ans.(c)
The disease African sleeping sickness
is caused byTrypanosoma gambiense
and this is transmitted by tse-tse fly
(Glossina palpalis).
68InAmoebaandParamecium
osmoregulation occurs through
[CBSE AIPMT 1991, 95, 2002]
(a) pseudopodia
(b) nucleus
(c) contractile vacuole
(d) general surface
Ans.(c)
AmoebaandParameciumcontains two
contractile vacuoles (anterior and
posterior, the latter being faster) for
osmoregulation, i.e. maintaining water
balance in the body.
69Genetic information in
Parameciumis contained in
[CBSE AIPMT 1990]
(a) micronucleus
(b) macronucleus
(c) Both (a) and (b)
(d) mitochondria
Ans.(a)
Ciliates (Paramecium) show nuclear
dimorphism, large macronucleus
controls metabolic activities and
growth. It is called vegetative nucleus.
Micronucleus contains genetic
informatioin and thus, takes part in
reproduction.
70Plasmodium, the malarial parasite,
belongs to class
[CBSE AIPMT 1990]
(a) Sarcodina
(b) Ciliata
(c) Sporozoa
(d) Dinophyceae
Ans.(c)
Plasmodium,the malarial parasite
belongs to class—Sporozoa.
Sporozoans are intracellular parasites,
reproduce by multiple fission and life
cycle may include the two different
hosts.
71What is true aboutTrypanosoma?
[CBSE AIPMT 1990]
(a) Polymorphic
(b) Monogenetic
(c) Facultative parasite
(d) Non-pathogenic
Ans.(a)
Trypanosomais an obligate parasite, it
is digenetic polymorphic (Trypanosoma
is adult form in human, whereas,
crithidal and leptomonal are
developmental forms in tse-tse fly).
72Trypanosomabelongs to class
[CBSE AIPMT 1989]
(a) Sarcodina (b) Zooflagellata
(c) Ciliata (d) Sporozoa
Ans.(b)
Zooflagellates are protozoan parasites
which possess one to several flagella
for locomotion. They are generally
uninucleate (occasionally
multinucleate), body is covered by a
firm pellicle, e.g.Trypanosoma,
Leishmania, Trichomonas,etc.
Biological Classification 11

22. 73Select thewrongstatement.
[NEET 2018]
(a) Pseudopodia are locomotory and
feeding structures in sporozoans
(b) Mushrooms belong to
Basidiomycetes
(c) Cell wall is present in members of
Fungi and Plantae
(d) Mitochondria are the powerhouse
of the cell in all kingdoms except
Monera
Ans.(a)
Sporozoansare endoparasites. They
lack locomotory organelles like cilia,
flagella, pseudopodia, etc.,e.g.
Plasmodium. Pseudopodia are found in
amoeboid protozoans,e.g., Amoeba,
Entamoeba,etc. Therefore, statement
(a) is wrong while rest of the
statements are correct.
74Which of the following would
appear as the pioneer organisms
on bare rocks?[NEET 2016, Phase I]
(a) Liverworts (b) Mosses
(c) Green algae (d) Lichens
Ans.(d)
In primary succession on rocks, lichens
secrete acids to dissolve rock, helps in
weathering and soil formation. So,
lichens are pioneer species to colonise
the bare rock.
75One of the major components of
cell wall of most fungi is
[NEET 2016, Phase I]
(a) peptidoglycan (b) cellulose
(c) hemicellulose (d) chitin
Ans.(d)
Cell wall of most fungi is made up of
chitin. Chemically it is N-acetyl
glucosamine. It is also found in the
exoskeleton of insects.
76Which one of the following is
wrong for fungi?
[NEET 2016, Phase II]
(a) They are eukaryotic
(b) All fungi possess a purely
cellulosic cell wall
(c) They are heterotrophic
(d) They are both unicellular and
multicellular
Ans.(b)
In fungi, cell wall contains chitin or
cellulose along with other
polysaccharides, proteins and lipids.
Only in some fungi, e.g.Phytophthoraor
other oomycetes a purely cellulosic cell
wall is present. Hence, only option (b) is
wrong, rest of the options are correct.
Concept EnhancerChitin is chemically
N-acetyl glucosamine. The exoskeleton
of insects also contains this chemical.
77Choose the wrong statement.
[CBSE AIPMT 2015]
(a)Penicilliumis multicellular and
produces antibiotics
(b)Neurosporais used in the study of
biochemical genetics
(c) Morels and truffles are poisonous
mushrooms
(d) Yeast is unicellular and useful in
fermentation
Ans.(c)
All statements are correct, except
statement (c), which can be corrected as
Morels and truffles are edible and
members of Ascomycetes in fungi.
78The imperfect fungi which are
decomposers of litter and help in
mineral cycling belong to
[CBSE AIPMT 2015]
(a) Deuteromycetes (b) Basidiomycetes
(c) Phycomycetes (d) Ascomycetes
Ans.(a)
The imperfecti fungi which are
decomposers of litter and help in
mineral cycling belong to
Deuteromycetes. They are fungi which
do not fit into the commonly established
taxonomic classification of fungi. They
include all those fungi in which the
perfect stage (sexual stage) is not
reported.
79The highest number of species in
the world is represented by
[CBSE AIPMT 2012]
(a) fungi (b) mosses
(c) algae (d) lichens
Ans.(a)
Fungi represent the highest number of
species in the world. Around 100000
species of fungi have been formally
described by taxonomists but the global
biodiversity of kingdom—Fungi is not
fully understood.
80Which one of the following has
haplontic life cycle?
[CBSE AIPMT 2009]
(a)Funaria
(b)Polytrichum
(c)Ustilago
(d) Wheat
Ans.(c)
Ustilagohas haplontic life cycle. In
their sexual phase, only zygospore is
diploid structure. All others are
haploid, such a sexual cycle is termed
as haploid or haplontic.
81Which one is the wrong pairing
for the disease and its causal
organism? [CBSE AIPMT 2009]
(a) Late blight of potato
—Alternaria solani
(b) Black rust of wheat
—Puccinia graminis
(c) Loose smut of wheat
—Ustilago nuda
(d) Root-knot of vegetables
—Meloidogynesp.
Ans.(a)
The causative agent of late blight of
potato is fungusPhytophthora infestans,
class–Oomycetes,
order–Peronosporales,
family–Pythiaceae. In India, the late
blight of potato is a seed borne
disease.
82Trichoderma harzianumhas
proved a useful microorganism
for [CBSE AIPMT 2008]
(a) bioremediation of contaminated
soils
(b) reclamation of wastelands
(c) gene transfer in higher plants
(d) biological control of soil-borne
plant pathogens
Ans.(d)
Some common fungal inhabitants of
soil help to combat diseases caused by
soil borne plant pathogens. These
includeTrichoderma harzianumwhich
are found in damp soils. They have an
inhibitory effect on the growth of the
mycelium ofPythium.They serve to
supress fungi causing damping off
disease of the seedlings and thereby
influence favourably the growth of
crops.
12 NEETChapterwise Topicwise Biology
Kingdom-Fungi
TOPIC 3

23. 83Cellulose is the major component
of cell walls of[CBSE AIPMT 2008]
(a)Pythium (b)Xanthomonas
(c)Pseudomonas(d)Saccharomyces
Ans.(a)
Cellulose does occur in cell walls of
Oomycetes (e.g.Pythium) and
Hyphochytridiomycetes. Fungal cell
wall contains 80–90% carbohydrates,
the remainder being proteins and lipids.
The typical feature of fungal cell wall is
presence of chitin.
84Which of the following is a slime
mold? [CBSE AIPMT 2007]
(a)Rhizopus (b)Physarum
(c)Thiobacillus(d)Anabaena
Ans.(b)
The genusPhysarumwith about 100
species is the largest and
best-studied slime mold in the
class–Myxomycetes.
85Ergot of rye is caused by a species
of [CBSE AIPMT 2007]
(a)Phytophthora(b)Uncinula
(c)Ustilago (d)Claviceps
Ans.(d)
The fungusClaviceps purpureais
responsible forergot disease of ryewhich
lowers the yield of rye plant.
86Which pair of the following
belongs to Basidiomycetes?
[CBSE AIPMT 2007]
(a) Birds nest fungi and puff balls
(b) Puff balls andClaviceps
(c)Pezizaand stink horns
(d)Morchellaand mushrooms
Ans.(a)
Birds nest fungi (Nidulariales) and puff
ball fungi (Lycoperdales) belongs to
Basidiomycetes. The common example
of class–Basidiomycetes are smut, rusts,
the mushrooms, the toad stools, the
puff balls and the pore fungi.
87The thalloid body of a slime mold
(Myxomycetes) is known as
[CBSE AIPMT 2006]
(a)Plasmodium(b) fruiting body
(c) mycelium (d) protonema
Ans.(a)
The thalloid body of a slime mould is
known asPlasmodium. The members of
Myxomycetes are called slime molds
because they contain and secrete
slime. They are included in lower fungi.
Their somatic phase is a multinucleate,
diploid holocarpicPlasmodium(a
product of syngamy).
InPlasmodium, propagation occurs
through fission or thick walled cysts or
sclerotium like structures.
Reproduction takes place by the
formation of uninucleate, thick walled
resting spores which are produced
within minute fruiting bodies like
structures, i.e. the sporangia.
Fruiting bodies and mycelium are
absent in lower fungi.Protonema is not
formed in fungi.
88Which of the following
environmental conditions are
essential for optimum growth of
Mucoron a piece of bread?
[CBSE AIPMT 2006]
(i) Temperature of about 25°C
(ii) Temperature of about 5°C
(iii) Relative humidity of about 5%
(iv) Relative humidity of about 95%
(v) A shady place
(vi) A brightly illuminated place
Choose the answer from the
following options
(a) (i), (iv) and (v) only
(b) (ii), (iv) and (v) only
(c) (ii), (iii) and (vi) only
(d) (i), (iii) and (v) only
Ans.(a)
Mucorshows the best growth on a
piece of bread at a temperature of
about 25°C, relative humidity of about
95% in a moist and shady place.Mucor
is a saprophytic fungus belonging to
the order–Mucorales and
family–Mucoraceae and grows on
decaying dung and on some food
stuffs.
89There exists a close association
between the alga and the fungus
within a lichen. The fungus
[CBSE AIPMT 2005]
(a) provides protection, anchorage
and absorption for the alga
(b) provides food for the alga
(c) fixes the atmospheric nitrogen for
the alga
(d) release oxygen for the alga
Ans.(a)
Lichen is a symbiotic association
between a fungus and an algae. The
fungal partner of lichen helps in the
absorption of water and mineral to algal
partner. It also provides protection and
anchorage to algal partner of lichen. In
exchange of this, the fungal partner
absorbs prepared food material from
algal partner. This food material
is prepared by the algal partner of
lichen through the process of
photosynthesis.
90Lichens are well known
combination of an alga and a
fungus where fungus has
[CBSE AIPMT 2004]
(a) a saprophytic relationship with the
alga
(b) an epiphytic relationship with the
alga
(c) a parasitic relationship with the
alga
(d) a symbiotic relationship with the
alga
Ans.(d)
Lichen is a symbiotic association
between a fungus and an alga. The
fungal part is called mycobiont while
the algal part is calledphycobiont. The
fungi absorb mineral and water to algae
and the algae synthesise food by
photosynthesis.
91During the formation of bread it
becomes porous due to the
release ofCO
2
by the action of
[CBSE AIPMT 2002]
(a) yeast (b) bacteria
(c) virus (d) protozoans
Ans.(a)
Strains ofSaccharomyces cerevisiae
are extensively used for leavening of
bread. During fermentation, the
yeasts produce alcohol andCO
2
which
leaves the bread porous.
92Which fungal disease spreads by
seed and flowers?
[CBSE AIPMT 2002]
(a) Loose smut of wheat
(b) Corn stunt
(c) Covered smut of barley
(d) Soft rot of potato
Biological Classification 13

24. Ans.(a)
Fungal disease, loose smut of wheat
spreads by seed and flowers. The
causal organism of this disease is
Ustilagofungus. It is an internal
parasite. It has a dikaryotic mycelium
which remains within the intercellular
spaces of the host tissue.
This fungus infects the ovary of the
host flower as a result of which the
masses of teliospores or brand spores
are formed in place of grains.
Teliospores are not surrounded by any
wall hence, called loosesmut.
93Plant decomposers are
[CBSE AIPMT 2001]
(a) Monera and Fungi
(b) Fungi and Plants
(c) Protista and Animalia
(d) Animalia and Monera
Ans.(a)
Decomposers are living components
chiefly the bacteria and fungi that
breakdown the complex compounds of
dead protoplasm of producers and
consumers absorb some products and
release others.
94Adhesive pad of fungi penetrates
the host with the help of
[CBSE AIPMT 2001]
(a) mechanical pressure and enzymes
(b) hooks and suckers
(c) softening by enzymes
(d) only by mechanical pressure
Ans.(a)
Cell wall degrading enzymes
(cellulolytic, pectolytic) as well as
mechanical pressure of adhesive pad
(appressorium) help the fungus in
penetrating the host.
95In fungi stored food material is
[CBSE AIPMT 2000]
(a) glycogen (b) starch
(c) sucrose (d) glucose
Ans.(a)
Glycogen, also known as ‘animal starch’,
is the chief storage polysaccharide of
animal cells and most of the fungi
(though food is also stored as oil
globules in some fungi).
Starch is a complex water insoluble
polysaccharide carbohydrate chiefly
found in green plants as their principal
energy (food) source.
Glucose is the most widely distributed
hexose sugar. It is an aldohexose reducing
sugar. It is found in blood muscles and
brain and works as energy fuel.
Sucrose is a non-reducing disaccharide
consists of one glucose and one
fructose molecules. It is one of the
abundant transport sugar in plants.
96Black rust of wheat is caused by
[CBSE AIPMT 2000]
(a)Puccinia (b)Mucor
(c)Aspergillus(d)Rhizopus
Ans.(a)
Black rust of wheat is caused by
Puccinia graminis tritici.This is the
potential cause of enormous economic
loss in all wheat growing regions of the
world.Puccinia graminis triticiusually
passes its life cycle on two different
hosts, wheat and barbery.
97Which of the following is the use of
lichens in case of pollution?
[CBSE AIPMT 1999]
(a) Lichens are not related with
pollution
(b) They act as bioindicators of
pollution
(c) They treat the polluted water
(d) They promote pollution
Ans.(b)
Growth of lichens on trees is inhibited by
air pollution. Hence, atmospheric
pollution causes decrease in their
populations. So, lichens are biological
indicators of pollution.
98Columella is a specialised
structure found in the sporangium
of [CBSE AIPMT 1999]
(a)Ulothrix (b)Rhizopus
(c)Spirogyra (d) None of these
Ans.(b)
InRhizopus, the central non-sporiferous
region of sporangium is called
columella.
99Pucciniaforms[CBSE AIPMT 1998]
(a) uredia and aecia on wheat leaves
(b) uredia and telia on wheat leaves
(c) uredia and aecia on barbery leaves
(d) uredia and pycnia on barbery leaves
Ans.(b)
Puccinia graminis tritici(fungus) causes
black rust of wheat. It forms
Urediospores (uredia) and teleutospores
(telia) on wheat leaves.
100Most of the lichens consist of
[CBSE AIPMT 1997]
(a) blue-green algae and
Basidomycetes
(b) blue-green algae and
Ascomycetes
(c) red algae and Ascomycetes
(d) brown algae and Phycomycetes
Ans.(b)
Lichens consist of the fungal
component mycobiont, mainly
Ascomycotina (only a few
Basidiomycotina and Deuteromycotina)
and the algal component, phycobiont
which are mostly blue-green algae
(Nostoc, Scytonema) or green algae
(Trebouxia, Trentophila, etc.)
101Which one of the following is not
true about lichens?
[CBSE AIPMT 1996]
(a) Their body is composed of both
algal and fungal cells
(b) Some form food for reindeers in
Arctic regions
(c) Some species can be used as
pollution indicators
(d) These grow very fast at the rate
of about 2 cm per year
Ans.(d)
Statement (d) is incorrect because
lichens show very slow growth. Their
size and slow rate of growth suggest
that some lichens in the Arctic are
4000 years ago.
102Which of the following is not
correctly matched?
[CBSE AIPMT 1995]
(a) Root knot disease
—Meloidogyne javanica
(b) Smut of bajra
—Tolysporium penicillariae
(c) Covered smut of barley
—Ustilago nuda
(d) Late blight of potato
—Phytophthora infestans
Ans.(c)
Option ‘c’ is mismatched because the
smuts in which sori are covered by the
membranous covering or peridium are
called covered smuts. These are
caused byUstilago hordei, in which the
sorus or smut ball is covered by a
peridium of host cells.Ustilago nudais
responsible for causing loose smut of
barley.
14 NEETChapterwise Topicwise Biology

25. Biological Classification 15
103The chemical compounds
produced by the host plants to
protect themselves against fungal
infection is[CBSE AIPMT 1995]
(a) phytotoxin (b) pathogen
(c) phytoalexins (d) hormone
Ans.(c)
Phytoalexins are chemical substances
produced by plants in response to
fungal infection and are toxic to fungi.
104White rust disease is caused by
[CBSE AIPMT 1995]
(a)Claviceps
(b)Alternaria
(c)Phytophthora
(d)Albugo candida
Ans.(d)
Albugo candida(Oomycetes) is an
obligate parasite commonly found
parastising a wide range of crucifers. It
causes a disease called white rust or
blister rustof crucifers resulting in
economically significant losses in the
yield of turnip, rape and mustard.
105Ustilagocaused plant diseases are
called smuts because
[CBSE AIPMT 1994]
(a) they parasitise cereals
(b)Myceliumis black
(c) they develop sooty masses of
spores
(d) affected parts become completely
black
Ans.(c)
The genusUstilago(L.ustus= burnt)
includes the group of fungi producing
black, sooty powder mass of spores on
the host plant parts imparting them a
'burnt' appearance, hence, the name,
the black dusty masses of spores
produced by these fungi resemble soot
or smut, so these are also known as
smut fungi.
106Mycorrhiza represents
[CBSE AIPMT 1994, 96, 2003]
(a) antagonism (b) endemism
(c) symbiosis (d) parasitism
Ans.(c)
Mycorrhiza (mykes= mushroom +rhiza=
root) represents a symbiotic
association of fungi with the roots of
higher plant. Mycorrhiza meaning
fungus root is an infected root system
arising from the rootlets of a seed
plant.
In ectomycorrhiza, the ultimate
absorbing rootlets of the root system
are completely surrounded by a distinct
mantle or sheath of fungal tissue. In
endomycorrhiza, there is no such
sheath. Most of the fungus is within the
root and may be intracellular as well as
intercellular.
107Absorptive heterotrophic nutrition
is exhibited by[CBSE AIPMT 1990]
(a) algae (b) fungi
(c) bryophytes (d) pteridophytes
Ans.(b)
Fungi are heterotrophic, e.g. these
require an organic source of carbon,
also require some source of nitrogen,
inorganic ions(K , Mg )
+ +
, trace
elements (Fe, Zn, Cu) and growth
factors like vitamins. Fungi may act as
saprobes and parasites. They obtain
nutrition from host by means of special
structures called haustoria and exhibit
absorptive or holophytic type of
nutrition.
108Lichens indicateSO
2
pollution
because they[CBSE AIPMT 1989, 92]
(a) show association between algae
and fungi
(b) grow faster than others
(c) are sensitive toSO
2
(d) flourish inSO
2
rich environment
Ans.(c)
Lichens, the composite organisms
consist of a specific fungus living in
symbiotic association with one or
sometimes, two species of algae.
Lichens are world-wide in distribution.
These are pioneer colonisers of barren
rocks and mountains. Being extremely
sensitive toSO
2
, the lichens especially
epiphytic lichens serve as bioindicators
air pollution.
109Which of the following statements
about inclusion bodies is
incorrect? [NEET (Sep.) 2020]
(a) These are involved in ingestion of
food particles
(b) They lie free in the cytoplasm
(c) These represent reserve material
in cytoplasm
(d) They are not bound by any
membrane
Ans.(a)
Option (b),(c) and (d) are correct
whereas option (a) is incorrect because
Inclusion bodies are nuclear or
cytoplasmic aggregates of proteins.
They represent sites of viral
multiplication in a bacterium or a
eukaryotic cell and usually consist of
viral capsid proteins.These are not
involved in ingestion of food particles.
110Which of the following statements
is incorrect?[NEET (National) 2019]
(a) Viruses are obligate parasites
(b) Infective constituent in viruses is
the protein coat
(c) Prions consist of abnormally
folded proteins
(d) Viroids lack a protein coat
Ans.(b)
The statement “infective constituent in
viruses is protein coat” is incorrect. The
correct information about the
statement is as follows. Viruses infect
their host organisms through their
genetic material, i.e either DNA or RNA
and not protein. They take over the
biosynthetic machinery of the host cell
and produce chemicals required for
their own multiplication. Rest
statements are correct.
111Viroids differ from viruses in
having [NEET 2017]
(a) DNA molecules with protein coat
(b) DNA molecules without protein
coat
(c) RNA molecules with protein coat
(d) RNA molecules without protein
coat
Ans.(d)
Viroids differ from viruses in having
RNA molecules without protein coat.
Viruses on the other hand posses DNA
or RNA with a protein coat as their
genetic material.
Viruses can infect a wide range of
organisms including plants, animals or
bacteria, while viroids infect only
plants.
112Which of the following statements
is wrong for viroids?
[NEET 2016 Phase I]
(a) They are smaller than viruses
(b) They cause infections
(c) Their RNA is of high molecular
weight
(d) They lack a protein coat
Virus and Viroids
TOPIC 4

26. 16 NEETChapterwise Topicwise Biology
Ans.(c)
Viroids are infectious, non-protein-
coding, highly structured with small
circular RNA’s, which have the ability to
replicate autonomously. These contain
RNA of low molecular weight and
induce diseases in higher plants.
113Select wrong statement.
[CBSE AIPMT 2015]
(a) The viroids were discovered by DJ
Ivanowski
(b) WM Stanley showed that viruses
could be crystallised
(c) The term ‘Contagium vivum
fluidum’ was coined by MW
Beijerinek
(d) Mosaic disease in tobacco and
AIDS in human being are caused
by viruses
Ans.(a)
All statements are correct except the
statement (a), which can be corrected as
Viroids were discovered by TO Diener in
1971 as a new infectious agent that was
smaller than virus.
114Which of the following shows
coiled RNA strand and
capsomeres?[CBSE AIPMT 2014]
(a) Polio virus
(b) Tobacco mosaic virus
(c) Measles virus
(d) Retrovirus
Ans.(b)
In TMV RNA is single stranded (ss)
helically coiled structure containing
about 2130 capsomeres, which is a
basic subunit of capsid (an outer
covering of protein that protects the
genetic material of a virus).
There are about 16 capsomeres present
in each helical turn.
115Which statement is wrong for
viruses? [CBSE AIPMT 2012]
(a) All are parasites
(b) All of them have helical symmetry
(c) They have ability to synthesise
nucleic acids and proteins
(d) Antibiotics have no effect on them
Ans.(b)
The nucleocapsids of viruses are
constructed in highly symmetric ways.
Two types of symmetry are recognised
in viruses, which correspond to the two
primary shapes, rod and spherical.
Rod-shaped viruses have helical
symmetry and spherical viruses have
icosahedral symmetry.
116Virus envelope is known as
[CBSE AIPMT 2010]
(a) capsid
(b) virion
(c) nucleoprotein
(d) core
Ans.(a)
Structurally viruses are very diverse
varying widely in size, shape and
chemical composition. The nucleic acid
of virus is always located within the
virion particle and surrounded by a
protein shell called capsid.
The protein coat is composed of a
number of individual protein molecules
called structural subunits. The
complete complex of nucleic acid and
proteins, packaged in the virion is called
the virus nucleocapsid.
117TO Diener discovered a
[CBSE AIPMT 2009]
(a) free infectious RNA
(b) free infectious DNA
(c) infectious protein
(d) bacteriophage
Ans.(a)
Viroids are small, circular, single
stranded free infectious RNA molecules
that are the smallest known pathogens.
The extracellular form of the viroid is
naked RNA, i.e. there is no protein
capsid of any kind.
These RNA molecule contains no
protein encoding genes and therefore,
the viroid is totally dependent on host
for its replication. No viroid diseases of
animals are known and the precise
mechanisms by which viroids cause
plant diseases remain unclear.
118Viruses that infect bacteria,
multiply and cause their lysis are
called [CBSE AIPMT 2004]
(a) lysozymes (b) lytic
(c) lipolytic (d) lysogenic
Ans.(b)
When bacteriophage infects a
bacterium, it entirely depends on the
host for its multiplication. It utilises the
host machinery for replication and
produce a large number of progeny
(phage particles). The bacterium cell
undergoes lysis and dies to liberate a
large number of these phage particles
which are ready to start another cycle
by infecting new bacterial cell. This
cycle is known as lytic cycle.
119Which of the following statements
is not true for retroviruses?
[CBSE AIPMT 2004]
(a) DNA is not present at any stage in
the life cycle of retroviruses
(b) Retroviruses carry gene for RNA
dependent DNA polymerase
(c) The genetic material in mature
retroviruses is RNA
(d) Retroviruses are causative agents
for certain kinds of cancer in man
Ans.(a)
Retroviruses are so, named because
they contain enzyme reverse
transcriptase orRNA dependent DNA
polymerase.The genetic material of
these viruses is RNA, e.g. Rous
Sarcoma Virus.
120Viruses are no more ‘alive’ than
isolated chromosomes because
[CBSE AIPMT 2003]
(a) both require the environment of a
cell to replicate
(b) they require both RNA and DNA
(c) they both need food molecules
(d) they both require oxygen for
respiration
Ans.(a)
Viruses are non-cellular obligate
parasite. In the free state they are just
like the particles. They do not have their
own metabolic machinery and use
host’s machinery for multiplication.
121Which one of the following
statements about viruses is
correct? [CBSE AIPMT 2003]
(a) Nucleic acid of viruses is known
as capsid
(b) Viruses possess their own
metabolic system
(c) All viruses contain both RNA and
DNA
(d) Viruses are obligate parasites
18 mm
2.3 cm
Capsomeres
RNA
strand

27. Biological Classification 17
Ans.(d)
Viruses arenon-cellular, obligate
parasites.They have DNA or RNA as
genetic material (never both). Genetic
material of virus is covered in protein
coat, known ascapsid.Viruses do not
contain their own metabolic system
instead they occupy host’s metabolic
system after entrance in them.
122Tobacco mosaic virus is a tubular
filament of size[CBSE AIPMT 2003]
(a)700 30×nm
(b)300 10×nm
(c)300 5×nm
(d)300 18×nm
Ans.(d)
TMV is elongated rod-like, 3000Å
(300 nm) long and 180Å (18 nm) in
diameter.
123Interferons are synthesised in
response to[CBSE AIPMT 2001]
(a) Mycoplasma (b) bacteria
(c) viruses (d) fungi
Ans.(c)
Cells infected by virus produce
interferon which is an antiviral protein. It
spreads to neighbouring cells and
makes them resistant to virus infections
by inhibiting viral growth.
124Cauliflower mosaic virus contains
[CBSE AIPMT 2001]
(a)ssRNA
(b)dsRNA
(c)dsDNA
(d)ssDNA
Ans.(c)
Caulimovirus(Cauliflower Mosaic Virus)
contains double stranded (ds) DNA.
Influenza viruscontains single
stranded RNA (ssRNA).
Parvoviruscontains single stranded
DNA (ssDNA).
125Which one of the following
statements about viruses is
correct? [CBSE AIPMT 1997]
(a) Viruses possess their own
metabolic system
(b) Viruses contain either DNA or
RNA
(c) Viruses are facultative parasites
(d) Viruses are readily killed by
antibiotics
Ans.(b)
Viruses contain only one type of
nucleic acid DNA or RNA. These are
obligate parasites; do not possess
metabolic machinery and are not
readily killed by antibiotics.
126Influenza virus has
[CBSE AIPMT 1996]
(a) DNA
(b) RNA
(c) Both (a) and (b)
(d) Only proteins and no nucleic
acids
Ans.(b)
Influenza virus is single (–) stranded
RNA virus, which cannot serve directly
asmRNA but rather as templates for
mRNA synthesisviaa viral
transcriptase. Influenza virus
(orthomyxo virus) infects the
respiratory tract and cause influenza.
127Tobacco Mosaic Virus (TMV) genes
are [CBSE AIPMT 1994]
(a) double stranded RNA
(b) single stranded RNA
(c) polyribonucleotides
(d) proteinaceous
Ans.(b)
Tobacco Mosaic Virus (TMV) is
elongated, rod shaped, most thoroughly
studied plant virus, with 95% protein
and 5% RNA by weight. RNA is genomic,
i.e. genetic material which is single
stranded, linear, helically coiled, 5mμin
length with 6500 nucleotides long.

28. 01Which of the following algae
contains mannitol as reserve food
material ? [NEET 2021]
(a)Ectocarpus(b)Gracilaria
(c)Volvox (d)Ulothrix
Ans.(a)
Ectocarpusis a cosmopolitan marine
brown seaweed. Mannitol is stored as a
food reserve inEctocarpus.
Other options can be explained as :
lGracilariais also a type of red algae
that is notable for its economic
importance. Reserved food found in
Gracilariais in the form of floridean
starch, which is similar to amylopectin
and glycogen in structure.
lInVolvox(green algae), stored food
material is starch and the major
pigments are chlorophyllaandd.
Some may store food as oil droplets.
lUlothrixis a genus of non-branching
filamentous green algae, starch
molecule is the reserved food.
02Which of the following algae
produces carrageen?[NEET 2021]
(a) Green algae
(b) Brown algae
(c) Red algae
(d) Blue-green algae
Ans.(c)
Carrageen is a common name give to
polysaccharides (carbohydrates) that
are extracted from seaweeds like red
algae. Carrageen is known for its gelling
properties and it is one of the industrial
source of carrageenan that is utilised
as a stabilizer and thickner of milk
products. It can be harmful to immune
system, in severe cases it leads to
internal bleeding.
03Phycoerythrin is the major
pigment in [NEET (Oct.) 2020]
(a) red algae
(b) blue-green algae
(c) green algae
(d) brown algae
Ans.(a)
Phycoerythrin is the major pigment in
red algae or rhodophytes. The
photosynthetic pigments in red algae
include chlorophyll-a, carotenoids and
phycobilins. Phycoerythrin belongs to
the phycobilins. These pigments are
soluble in water.
04Which of the following pairs is of
unicellular algae?[NEET (Sep.) 2020]
(a)GelidiumandGracilaria
(b)AnabaenaandVolvox
(c)ChlorellaandSpirulina
(d)LaminariaandSargassum
Ans.(c)
ChlorellaandSpirulinaare unicellular
algae as they are rich in proteins and
hence used as food supplements by
space travellers.Gelidium,Gracilaria,
LaminariaandSargassumare
multicellular.Volvoxis colonial.
05Which one iswronglymatched?
[NEET 2018]
(a) Gemma cups –Marchantia
(b) Biflagellate zoospores – Brown
algae
(c) Uniflagellate gametes –
Polysiphonia
(d) Unicellular organism –Chlorella
Ans.(c)
Polysiphoniais a red algae. In it sexual
reproduction is of oogamous type. The
male sex organ, spermatangium
produces non-flagellate male gametes.
InBrown algae,sexual reproduction
varies from isogamy, anisogamy to
oogamy. In isogamy and anisogamy
both the gametes are motile while in
oogamy only male gametes are motile.
These motile gametes have two
unequal laterally attached flagella.
Chlorellais a unicellular organism. It
is green algae belonging to class
Chlorophyta. InMarchantia, gemma
cups are found on its dorsal surface.
It contains gammae which help in
vegetative propagation.
06Zygotic meiosis is characteristic
of [NEET 2017]
(a)Marchantia(b)Fucus
(c)Funaria (d)Chlamydomonas
Ans.(d)
Zygotic meiosis is represented in the
haplontic life cycle of many algae
including
Chlamydomonas. In such a life cycle, all
cells are haploid except zygote. This is
because meiosis occurs in the zygote
itself resulting into four haploid cells
that give rise to hapolid plants.
07An example of colonial alga is
(a)Chlorella [NEET 2017]
(b)Volvox
(c)Ulothrix
(d)Spirogyra
PlantKingdom
03
Algae
TOPIC 1
Gametogenesis
Syngamy
Zygote
(2 )n Meiosis
Spores
( )n
A
B
Haplontic
Gametophyte
( )n
Life cycle pattern : Haplontic

29. Ans.(b)
Volvoxis a fresh water green hollow
ball-like colonial alga. Its colony has a
fixed number of cells (500-60000). It is
called coenobium.
08Which one of the following
statements is wrong ?
[NEET 2016, Phase II]
(a) Algae increase the level of
dissolved oxygen in the immediate
environment
(b) Algin is obtained from red algae and
carrageenan from brown algae
(c) Agar-agar is obtained from
GelidiumandGracilaria
(d)LaminariaandSargassumare
used as food
Ans.(b)
Algin extracted from brown algae, e.g.
Laminaria, etc. is a hydrocolloid used
in shaving creams, jellies, flameproof
plastic, etc. Carrageenan is extracted
from red algae likeChondrusand used
as emulsifier and clearing agent.
Thus, only option (b) is incorrect and
all other options are correct.
09Which one is a wrong statement?
[CBSE AIPMT 2015]
(a) Archegonia are found in
Bryophyta, Pteridophyta and
Gymnosperms
(b)Mucorhas biflagellate zoospores
(c) Haploid endosperm is typical
feature of gymnosperms
(d) Brown algae have chlorophyll-a
andc, and fucoxanthin
Ans.(b)
All the statements are correct except
the statement (b).Mucor(fungus)
belongs to the class—Zygomycetes. The
members of Zygomycetes bear
non-motile non-flagellated gametes.
10Which one of the following shows
isogamy with non-flagellated
gametes? [CBSE AIPMT 2014]
(a)Sargassum (b)Ectocarpus
(c)Ulothrix (d)Spirogyra
Ans.(d)
Isogamy with non-flagellated gametes
is seen inSpirogyra. It can reproduce
both by sexual and asexual (vegetative)
means.
They reproduce sexually by conjugation
in which two non-flagellated
morphologically similar but
physiologically different gemetes
(isogamous) fuse together. One
filament acts as male gamete and
passes through the conjugation tube of
another filament which acts as female
gamete.
11Which one of the following is
wrong aboutChara?
[CBSE AIPMT 2014]
(a) Upper oogonium and lower round
antheridium
(b) Globule and nucule present on the
same plant
(c) Upper antheridium and lower
oogonium
(d) Globule is male reproductive
structure
Ans.(c)
Both antheridium and oogonium are the
male and the female reproductive
structures respectively. They have
sterile jackets on their surface. InChara
globule(antheridium) is present on
lower side, while thenucule(oogonium)
is present on upper side of sterile
vegetative (leaf-like) structure.
12An alga which can be employed as
food for human being is
[CBSE AIPMT 2014]
(a)Ulothrix
(b)Chlorella
(c)Spirogyra
(d)Polysiphonia
Ans.(b)
Chlorellais a potential food source
because it is high in protein and other
essential nutrients when dried, it
contains about 45% protein, 20% fat,
20% carbohydrate, 5% fibre and 10%
minerals and vitamins.
13Select the wrong statement.
[NEET 2013]
(a) Isogametes are similar in
structure, function and behaviour
(b) Anisogametes differ either in
structure, function and behaviour
(c) In oomycetes female gamete is
smaller and motile, while male
gamete is larger and non-motile
(d)Chlamydomonasexhibits both
isogamy and anisogamy andFucus
shows oogamy
Ans.(c)
Statement (c) is wrong as oomycetes
include water moulds, white rusts
and downy mildews. In these, female
gamete is large and non-motile,
whereas, male gamete is small and
non-motile. Isogametes are found in
algae likeUlothrix, Chlamydomonas,
Spirogyra, etc. which are similar in
structure, function and behaviour.
Anisogametes are found in
Chlamydomonasin which one gamete
is larger and non-motile and the
other one is motile and smaller.
Oogamyis the fusion of non-motile egg
with motile sperm. The gametes, differ
both morphologically as well as
physiologically. It occurs in
Chlamydomonas,Fucus, Chara,
Volvox,etc.
14Algae have cell wall made up of
[CBSE AIPMT 2010]
(a) cellulose, galactans and mannans
(b) hemicellulose, pectins and
proteins
(c) pectins, cellulose and proteins
(d)cellulose, hemicellulose and
pectins
Plant Kingdom 19
Conjugation tube
Female like gamete
Male like gamete
(b)
(a)
Zygospore
Zygospore wall
(c)
Protoplasmic strands
Mucilage
Daughter
colony
Cells
Flagella
Volvox

30. 20 NEETChapterwise Topicwise Biology
Ans.(a)
Algae have cell wall made up of
cellulose, galactans and mannans.
Like plants, algae have cell walls
containing either polysaccharides
such as cellulose (a glucan) or a
variety of glycoproteins or both.
The inclusion of additional
polysaccharide in algal cell walls is used
as a feature for algal taxonomy.
Mananas form microfibrils in the cell
walls of a number of marine green algae
including those from the genera
Codium,Acetabulariaas well as in the
walls of some red algae likePorphyra.
15Mannitol is the stored food in
[CBSE AIPMT 2009]
(a)Chara (b)Porphyra
(c)Fucus (d)Gracilaria
Ans.(c)
Fucusbelongs to class–Phaeophyceae,
in which reserve food is found in form
of laminarin, mannitol and oil.
Charabelongs to class–Chlorophyceae,
in which reserve food is found in form of
starch and oil.
PorphyraandGracillariabelongs to
class—Rhodophyceae, in which reserve
food is found in form of floridean starch
and Galactan-SO
4
polymers.
16If you are asked to classify the
various algae into distinct groups,
which of the following characters
you should choose?
[CBSE AIPMT 2007]
(a) Types of pigments present in the
cell
(b) Nature of stored food materials in
the cell
(c) Structural organisation of thallus
(d) Chemical composition of the cell
wall
Ans.(a)
Types of pigments present in the cell of
algae is the most important character
for classification.
17Sexual reproduction inSpirogyrais
an advanced feature because it
shows [CBSE AIPMT 2003]
(a) physiologically differentiated sex
organs
(b) different size of motile sex organs
(c) same size of motile sex organs
(d) morphologically different sex
organs
Ans.(a)
InSpirogyra,the sexual reproduction
involves the fusion of two
morphologically identical isogametes and
physiologically dissimilar anisogametes.
This is a case of primitive anisogamy. In
this the active gamete is known as the
male and the passive as the female.
18A research student collected certain
alga and found that its cells
contained both chlorophyll-a,b,cand
chlorophyll-das well as
phycoerythrin. The alga belongs to
[CBSE AIPMT 2000]
(a) Rhodophyceae (b) Bacillariophyceae
(c) Chlorophyceae (d) Phaeophyceae
Ans.(a)
Members of Rhodophyceae (red algae)
contains Chlorophyll-a,d,
r-phycoerythrin,r-phycocyanin,αand
β-carotene pigments.
Members of Chlorophyceae (green algae)
contain chlorophyll-a, bandβ-carotene
pigments.
Members of Bacillariophyceae (diatoms)
contain chlorophyll-a, c,β-carotene,
α-carotene pigments.
Members of Cyanophyceae
(cyanobacteria, blue-green algae) contain
chlorophyll-a,c-phycocyanin,
c-phycoerythrin andβ-carotene pigments.
19Ulothrix canbe described as a
[CBSE AIPMT 1998]
(a) non-motile colonial alga lacking
zoospores
(b) filamentous alga lacking flagellated
reproductive stages
(c) membranous alga producing
zoospores
(d) filamentous alga with flagellated
reproductive stages
Ans.(d)
Ulothrixis a freshwater, filamentous
green algae, found in rather cold flowing
water. Sexual reproduction inUlothrixis
isogamous type, i.e. it takes place
between two morphological similar
motile, flagellated male and female
gametes which come from different
filaments.
20Ulothrixfilaments produce
[CBSE AIPMT 1997]
(a) isogametes
(b) anisogametes
(c) heterogametes
(d) basidiospores
Ans.(a)
Ulothrixbelongs to green algae.
Sexual reproduction inUlothrixtakes
place by the union of isogametes
which are motile, biflagellate,
morphologically similar gametes.
Approximately 8-32 isogametes are
produced from a mother cell. Two
gametes come from two different
filament, fuse and form a diploid
zygote.
21Brown algae is characterised by
the presence of
[CBSE AIPMT 1997]
(a) phycocyanin
(b) phycoerythrin
(c) fucoxanthin
(d) haematochrome
Ans.(c)
In addition to chlorophyll-a, brown
algae posses special carotenoids and
fucoxanthin. It is due to the
fucoxanthin (brown pigment) that
these algae appear brown.
Phycocyanin and phycoerythrin are
phycobilins which are found in red
algae (phycocyanin-r, phycoerythrin-r)
and blue-green algae (phycocyanin-c,
phycoerythrin-c).
22An alga very rich in protein is
[CBSE AIPMT 1997]
(a)Spirogyra
(b)Ulothrix
(c)Oscillatoria
(d)Chlorella
Ans.(d)
DriedChlorella pyrenoidosacontains
approxi- mately 50-55% crude protein
(more than that in dried beef,
soyabean meal and dried yeast).
23Blue-green algae belong to
[CBSE AIPMT 1996]
(a) eukaryotes
(b) prokaryotes
(c) Rhodophyceae
(d) Chlorophyceae
Ans.(b)
Blue-green algae or cyanobacteria are
the largest Gram negative, aerobic,
photoautotrophic, nitrogen-fixing,
simplest chlorophyll containing
thallophytes/ prokaryotes. They
neither have a definite nucleus nor
definite plastid with grana. They also
lack flagella, chlorophyll-b,
mesosome, meiosis and membrane
bound organelles (except ribosome of
70S type).

31. 24Agar is commercially obtained
from [CBSE AIPMT 1995]
(a) red algae (b) green algae
(c) brown algae (d) blue-green algae
Ans.(a)
Agaris a gelatinous, sulphated
non-nitrogenous, tasteless, odourless
mucopolysaccharide obtained from
middle lamella of cell wall of marine red
algae likeGracillaria, Gelidium, Gigartina,
etc. commonly known as agarophytes. It
is used as solidifying agent in the culture
medium, as luxative stabiliser or
thickener in preparing jams, jellies,
creams, ice creams, bakery products
and as luxative in drug industry.
25The absence of chlorophyll, in the
lowermost cell ofUlothrix,shows
[CBSE AIPMT 1995]
(a) functional fission
(b) tissue formation
(c) cell characteristic
(d) beginning of labour division
Ans.(d)
Ulothrixis an advanced alga, with three
types of cells—green dome-shaped
apical cell, green intercalary cell and
basal non-green cell called holdfast.
Holdfast or basal cell is for attachment,
it has nucleus and cytoplasm, its
presence shows the beginning of
division of labour.
26In Chlorophyceae, sexual
reproduction occurs by
[CBSE AIPMT 1994]
(a) isogamy and anisogamy
(b) isogamy, anisogamy and oogamy
(c) oogamy only
(d) anisogamy and oogamy
Ans.(b)
In Chlorophyceae, three types of sexual
reproduction occurs, i.e. isogamy,
anisogamy and oogamy.
Isogamy involves the fusion of those
gametes which are similar in size,
shape and structure, e.g.
Chlamydomonas debaryana.
In anisogamy gametes differ
morphologically and also behave
differently, e.g.Chlamydomonas braunii.
In oogamy, fusion between motile and
non-motile gametes takes place, e.g.
Chlamydomonas coccifera.
27Which of the following cannot fix
nitrogen? [CBSE AIPMT 1994]
(a)Nostoc (b)Azotobacter
(c)Spirogyra (d)Anabaena
Ans.(c)
Spirogyrais a free floating, filamentous,
green, freshwater alga, popularly called
pond silk. It has no role in nitrogen
fixation. It forms a green slimy mass on
the surface of standing and stagnant
water of ponds during spring season,
hence also called pond scum.
Whereas, blue-green algae likeNostoc,
Anabaenaare the important nitrogen
fixing cyanobacteria, so these are used
as biofertiliser.Azotobacteris the
aerobic free living/non-symbiotic
nitrogen fixer.
28Chloroplast ofChlamydomonasis
[CBSE AIPMT 1993]
(a) stellate (b) cup-shaped
(c) collar-shaped (d) spiral
Ans.(b)
InChlamydomonas,chloroplast is single
and cup-shaped. Chloroplasts are the
pigment (chl-aand chl-b) containing
bodies present in green algae. The
green colouration of the members of
chlorophyta is due to the presence of
excess of chlorophyll in the
chloroplasts. The chloroplasts are well
defined bodies met within every cell of
the members of this class, though
number and shape of the chloroplasts
varies in different orders of the class.
29InUlothrix/Spirogyra,reduction
division (meiosis) occurs at the
time of [CBSE AIPMT 1993]
(a) gamete formation
(b) zoospore formation
(c) zygospore germination
(d) vegetative reproduction
Ans.(c)
InUlothrix/Spirogyra,meiosis takes place
at the time of zygospore germination. It
takes place when (+) and (–)
plants/filaments results in the formation
of diploid zygote (2x). Zygote is
tetraflagellated, it secretes a thick wall
and becomes non-motile to form diploid
zygospore.
Under favourable conditions, zygospore
undergoes zygotic meiosis to form
motile tetraflagellated zoomeiospores
or non-motile aplanomeiospores. Each
meiospore (haploid) germinates to new
filament of (+) (–) strain.
30Pyrenoids are the centres for
formation of[CBSE AIPMT 1993]
(a) porphyra (b) enzymes
(c) fat (d) starch
Ans.(d)
Pyrenoid is a seat of synthesis and
storage of starch present in the
chloroplast of algae. A pyrenoid has a
core of protein around which starch is
deposited as layers.
31A plant in which sporophytic
generation is represented by
zygote is [CBSE AIPMT 1992]
(a)Pinus (b)Selaginella
(c)Chlamydomonas(d)Dryopteris
Ans.(c)
InChlamydomonas, the plant body is
haploid, and represents gametophyte.
It reproduces asexually through the
formation of zoospores and sexually
through gametes. Gametes (haploid)
fuse to produce diploid zygote,
representing the sporophytic
generation. The zygote secretes a wall
around it to become a resting
zygospore (diploid). The zygote and
zygospore are the only diploid structure
which represents the diplophase.
32The product of conjugation in
Spirogyraor fertilisation of
Chlamydomonasis
[CBSE AIPMT 1991]
(a) zygospore (b) zoospore
(c) oospore (d) carpospore
Ans.(a)
Zygospore (zygote) is the fusion
product of two gametes. It infact,
represents the resting stage formed
after withdrawl of flagella and
formation of a thick wall around the
freshly formed zygote. Zygospore is
spherical with thick, smooth or stellate
wall and contains fats and reserve food
materials other than starch. It can
resist unfavourable conditions.
InChlamydomonas,zygospore is the
resultant of isogamy, anisogamy or
oogamy. InSpirogyrasexual
reproduction occurs through
conjugation, which may be scalariform
or lateral. The resulting zygote secretes
a thick wall called zygospore (having 3
layers thick wall, diploid nucleus and
abundant food reserves in the form of
oil and starch).
33Sexual reproduction involving
fusion of two cells in
Chlamydomonasis
[CBSE AIPMT 1988]
(a) isogamy (b) homogamy
(c) somatogamy (d) hologamy
Plant Kingdom 21

32. Ans.(d)
InChlamydomonas,hologamy involves
the fusion of two young individuals
directly, e.g.C. snowiaeand isogamy
involves fusion of gametes which are
similar in size, structure and physiology,
e.g.C. euganetos.
34Acetabulariaused in Hammerling’s
nucleocytoplasmic experiments is
[CBSE AIPMT 1988]
(a) unicellular fungus
(b) multicellular fungus
(c) unicellular uninucleate green alga
(d) unicellular multinucleate green
alga
Ans.(c)
Acetabulariais the largest uninucleated
green marine alga popularly called
mermaids wine glass umbrella plant.
It has a cap, stalk and rhizoidal base and
nucleus lies in the base. Danish
biologistJ Hammerling(1953) by his
grafting experiments involving
exchange of nucleus inAcetabularia
proved the role of nucleus in heredity,
growth, morphology, differentiation and
morphogenesis.
35Gemmae are present in
[NEET 2021]
(a) mosses
(b) pteridophytes
(c) some gymnosperms
(d) some liverworts
Ans.(d)
Some liverworts reproduce asexually by
fragmentation of thalli or by the
formation of gemmae. (green,
multicelluar asexual buds). The
gemmae are held in special organs
known as gemma cups and are
dispersed by rainfall.Gemmae becomes
detached from parent body and
germinate to form new individuals.
36Which of the following is
responsible for peat formation?
[CBSE AIPMT 2014]
(a)Marchantia(b)Riccia
(c)Funaria (d)Sphagnum
Ans. (d)
Peat is mainly an accumulation of
partially decayed vegetation or organic
matter andSphagnumaccumulations
can store water, since both living and
dead plants can hold large quantities
of water and living matter (like meat)
for long distance transport inside their
cells hence, it is responsible for peat
formation.
37Which one of the following is
common to multicellular fungi,
filamentous algae and protonema
of mosses? [NEET 2013]
(a) Diplontic life cycle
(b) Members of kingdom–Plantae
(c) Mode of nutrition
(d) Multiplication by fragmentation
Ans.(d)
Multicellular fungi, filamentous algae
and protonema of mosses all show
multiplication by fragmentation.
38Archegoniophore is present in
[CBSE AIPMT 2011]
(a)Chara (b)Adiantum
(c)Funaria (d)Marchantia
Ans.(d)
InMarchantia, a bryophyte, the
archegonia (female sex organs) are
borne on special branches called
archegoniophoreor female
receptacles. Each archegoniophore has
rows of archegonia protected by
involucre or perichaetium.
39Male and female gametophytes
are independent and free-living in
[CBSE AIPMT 2010]
(a) mustard (b) castor
(c)Pinus (d)Sphagnum
Ans.(d)
InSphagnum, male and female
gametophytes are independent and
free living. In bryophytes, the most
conspicuous phase in life cycle is the
gametophyte. It is independent and
concerned with reproduction.
40In the prothallus of a vascular
cryptogam, the antherozoids and
eggs mature at different times, as
a result [CBSE AIPMT 2007]
(a) there is no change in success rate
of fertilisation
(b) there is high degree of sterility
(c) one can conclude that the plant is
apomictic
(d) self fertilisation is prevented
Ans.(d)
In the prothallus of a vascular
cryptogam, the antherozoids and eggs
mature at different times. As a result
self fertilisation is prevented.
41Spore dissemination in some
liverworts is aided by
[CBSE AIPMT 2007]
(a) elaters
(b) indusium
(c) calyptra
(d) peristome teeth
Ans.(a)
Elaters are hygroscopic in nature and
help in dispersal of spores.
42Peat moss is used as a packing
material for sending flowers and
live plants to distant places
because [CBSE AIPMT 2006]
(a) it is hygroscopic
(b) it reduces transpiration
(c) it serves as a disinfectant
(d) it is easily available
Ans.(a)
Sphagnumis a bryophyte, commonly
called asbog mossorpeat moss. It is
hygroscopic and possesses a
remarkable water holding capacity.
Hence, it is used as a packing material
in the transportation of flowers, live
plants, tubers, bulbs, seedlings, etc. It
is also used in seed-beds and in
moss-sticks.
43The antherozoids ofFunariaare
[CBSE AIPMT 1999]
(a) aciliated
(b) biflagellated
(c) multiciliated
(d) monociliated
Ans.(b)
The antherozoids ofFunariaare spirally
coiled and bear two equal flagella at
anterior end.
22 NEETChapterwise Topicwise Biology
Bryophytes
TOPIC 2
Flagella
Nucleus
Antherozoids (Funaria)

33. 44Dichotomous branching is found in
[CBSE AIPMT 1999]
(a) fern
(b)Funaria
(c) liverworts
(d)Marchantia
Ans.(d)
Though many liverworts are
dichotomously branched but some of
the leafy liverworts are not. However,
Marchantiais a liverwort which is
dichotomously branched.
45Bryophytes comprise
[CBSE AIPMT 1999]
(a) sporophyte of longer duration
(b) dominant phase of sporophyte
which is parasitic
(c) dominant phase of gametophyte
which produces spores
(d) small sporophyte phase generally
parasitic on gametophyte
Ans.(d)
Bryophyta is a group of thalloid,
non-vascular, cryptogams which have
gametophytic (haploid phase) as
dominant phase. It bears diploid
sporophytic phase which takes food
from gametophytic phase, thus
behaves as parasite on gametophyte.
46Which of the following is true
about bryophytes?
[CBSE AIPMT 1999]
(a) They possess archegonia
(b) They contain chloroplast
(c) They are thalloid
(d) All of the above
Ans.(d)
Bryophytes are non-vascular
cryptogams, their main plant body is
gametophytic (haploid) which is a
thalloid structure. It contains
chlorophyll for the process of
photosynthesis. Thalloid plant body
bear archegonia as female sex organs.
47Bryophytes are dependent on
water because[CBSE AIPMT 1998]
(a) water is essential for fertilisation
for their homosporous nature
(b) water is essential for their
vegetative propagation
(c) the sperms can easily reach up to
egg in the archegonium
(d) archegonium has to remain filled
with water for fertilisation
Ans.(c)
The antherozoids (sperms) of
bryophytes are flagellated (motile) and
need a film of water to swim through for
reaching the archegonium. Bryophyta
is a group of cryptogams. Main plant
body of bryophytes is gametophytic
(haploid). It bearing male and female sex
organs.
48Bryophytes can be separated from
algae because they
[CBSE AIPMT 1997]
(a) are thalloid forms
(b) have no conducting tissue
(c) possess archegonia with outer layer
of sterile cells
(d) contain chloroplasts in their cells
Ans.(c)
Archegonia is female sex organ in
bryophytes. It is a flask-shaped
multicellular organ. It is composed of a
cylindrical upper portion called neck
with a single layer of sterile cells called
Neck Canal Cells (NCC) and a lower
swollen sac-like portion called venter. It
also has layer or layers of sterile cells.
Venter encloses a larger egg cell and a
smaller (just above to egg) Venter Canal
Cell (VCC).
49Multicellular branched rhizoids and
leafy gametophytes are
characteristics of
[CBSE AIPMT 1997]
(a) all bryophytes
(b) some bryophytes
(c) all pteridophytes
(d) some pteridophytes
Ans.(b)
In pteridophytes and gymnosperms,
gameto- phyte generation is reduced,
sporophyte is well-developed. In
bryophytes, gametophyte constitutes
the main well developed generation but
in mosses, (e.g.Funaria), it is foliose.
50In which one of these the elaters
are present along with mature
spores in the capsule (to help in
spore dispersal)?
[CBSE AIPMT 1996]
(a)Riccia (b) Marchantia
(c)Funaria (d)Sphagnum
Ans.(b)
InMarchantia, capsule (the part of
sporophyte) contains elaters (2n) and
spores (n) in tetrads.
Elaters are diploid, spindle shaped
hygroscopic elongated structures with
2 spiral bands. They show twisting
movement and assist in spore dispersal
on maturity.
51The plant body of moss (Funaria) is
[CBSE AIPMT 1995, 2006]
(a) completely sporophyte
(b) completely gametophyte
(c) predominantly sporophyte with
gametophyte
(d) predominantly gametophyte with
sporophyte
Ans.(d)
Funaria(green moss) and other
bryophytes show alternation of
generation with haploid gametophytic
(n) and diploid sporophytic phases( )2n.
Plant body represents the
gametophytic phase, which reproduces
by producing gametes and on fusion
form zygote.
Zygote develops into sporophyte and
produces haploid meiospores which on
germination form gametophyte.
Sporophyte in mosses is differentiated
into foot, seta and capsule.
52Unique features of bryophytes is
that they [CBSE AIPMT 1994]
(a) produce spores
(b) have sporophyte attached to
gametophyte
(c) lack roots
(d) lack vascular tissues
Ans.(b)
The main plant body of bryophytes is
gametophytic which is independent and
may be thallose or foliose. The
sporophyte is differentiated into foot,
seta and capsule and is partially or fully
dependent upon the gametophyte.
53The plant group that produces
spores and embryo but lacks
vascular tissues and seeds is
[CBSE AIPMT 1992]
(a) Pteridophyta (b) Rhodophyta
(c) Bryophyta (d) Phaeophyta
Ans.(c)
Bryophytes consist of thalloid body,
attached to the substratum by hair-like
structures called rhizoids (true roots
are absent), these lack vascular tissue
(xylem and phloem) and require water at
the time of fertilisation. Bryophytes
exhibit alternation of generation.
Plant Kingdom 23

34. The haploid gametophyte (producing
gametes for sexual reproduction)
alternates with diploid sporophyte
(producing spores for asexual
reproduction). Production of large
number of spores is for increasing the
chances of survival and is an adaptation
to land conditions.
54Which one has the largest
gametophyte?[CBSE AIPMT 1991]
(a)Cycas (b) Angiosperm
(c)Selaginella(d) Moss
Ans.(d)
Of the given options, moss has the
largest gametophyte. It get reduced in
the order asSelaginella, Cycas,
angiosperms. Gametophyte begins with
the haploid spore and ends with the
formation of haploid gametes.
Gametophyte gives rise to the
sporophyte (through sexual
reproduction) and sporophyte gives rise
to gametophyte. As one moves from
thallophyte→bryophyte→pteridophyte
→gymnosperms→angiosperms, there
is development in the sporophyte and
reduction in the gametophyte.
55Bryophytes are amphibians
because [CBSE AIPMT 1991, 96]
(a) they require a layer of water for
carrying out sexual reproduction
(b) they occur in damp places
(c) they are mostly aquatic
(d) All of the above
Ans.(a)
Bryophytes are called amphibians
(amphibians of plant kingdom). They are
first amongst land plants which occur in
damp and shady habitats. As vascular
tissues are absent, male gametes
require a layer of water for swimming
and fertilisation.
56Apophysis in the capsule of
Funariais [CBSE AIPMT 1990]
(a) lower part (b) upper part
(c) middle part (d) fertile part
Ans.(a)
Capsule (the body containing spores) of
Funariais differentiated
into–operculum (cap shaped 2-3 layered
thick lid on the top); theca (middle fertile
part) and apophysis (sterile, solid basal
portion of the capsule having
chloroplasts).
57Moss peristome takes part in
[CBSE AIPMT 1990]
(a) spore dispersal
(b) photosynthesis
(c) protection
(d) absorption
Ans.(a)
Peristome functions in the dispersal of
the spores. Peristome constitutes
rings of teeth like projections at the rim
of the capsule of the mosses. In
Funaria, peristome are 32 in number,
arragned in two rings of 16 each
(a) outer exostome and (b) inner
endostome.
58Protonema occurs in the life cycle
of [CBSE AIPMT 1990, 93]
(a)Riccia
(b)Funaria
(c)Chlamydomonas
(d)Spirogyra
Ans.(b)
Protonema represents the juvenile
stage of moss (Funaria). It is
much-branched, green filamentous
structure formed by the germination of
spores, under favourable conditions.
The protonema consists of (a) some
slender rhizoids (b) a number of aerial
green prostrate branches bearing small
lateral buds which grow up into new
moss gametophores.
59Sperms of bothFunariaandPteris
were released together near the
archegonia ofPteris. OnlyPteris
sperms enter the archegonia as
[CBSE AIPMT 1989]
(a)Pterisarchegonia repelFunaria
sperms
(b)Funariasperms get killed byPteris
sperms
(c)Funariasperms are less mobile
(d)Pterisarchegonia release
chemical to attract its sperms
Ans.(d)
InDryopterisandPteris,when
fertilisation occurs, sperms are
attracted by the chemical diffusing into
the water from the mucilage exuded by
the open necks of archegonia of the
older prothalli, some of them make
their way down the canal to the egg in
the venter and only one of these enters
the egg to accomplish fertilisation.
60In bryophytes and pteridophytes,
transport of male gametes
requires [NEET 2016, Phase I]
(a) insects (b) birds
(c) water (d) wind
Ans.(c)
In several primitive simple plants–like
algae, bryophytes and pteridophytes,
water is the medium through which
male gametes are transferred to the
female reproductive organ or gamete to
bring about fertilisation.
61Which one of the following is a
correct statement?[NEET 2013]
(a) Pteridophyte gametophyte has a
protonemal and leafy stage
(b) In gymnosperms female
gametophyte is free-living
(c) Antheridiophores and
archegoniophores are present in
pteridophytes
(d) Origin of seed habit can be traced
in pteridophytes
Ans.(d)
Origin of seed habitat can be traced in
pteridophytes. Some pteridophytes like
SelaginellaandSalviniaare
heterosporous as they produce two
kinds of spores micro (small) spores and
macro (large) spores, which germinate
and give rise to male and female
gametophyte respectively.
The female gametophyte in these plants
are retained on the parent sporophytes
for variable periods. The development of
the zygote into young embryos takes
place within the female gametophyte.
This event is the precursor to the seed
habit and considered to an important
step in evolution.
62Which one of the following is a
vascular cryptogam?
[CBSE AIPMT 2009]
(a)Equisetum (b)Ginkgo
(c)Marchantia(d)Cedrus
Ans.(a)
Pteridophytes are also called vascular
cryptogams as these have a well
developed vascular system but are
non-flowering plants. e.g.Equisetum.
24 NEETChapterwise Topicwise Biology
Pteridophytes
TOPIC 3

35. 63In which one of the following, male
and female gametophytes don’t
have free living independent
existence?[CBSE AIPMT 2008]
(a)Pteris (b)Funaria
(c)Polytrichum(d)Cedrus
Ans.(a)
InPteris(alsoDryopteris) the spore
germinates to produce the prothallus.
The prothallus is a small, green flat,
surface loving, thallus-like object. It is
monoecious and bears sex organs on
the ventral side. The antheridia (male
sex organs) arise among the rhizoids
towards the posterior side of the
prothallus and are emergent. The
archegonia develop in central cushion
behind the apical notch. In these plants
male and female, gametophytes do
have free living independent existence.
All species ofPolytrichumare
dioecious. The antheridia and
archegonia are borne on different
gametophore. The plant body is an
erect leafy shoot but is not the entire
gametophyte. The leafy shoot arise
from protonema (the juvenile stage).
The leafy gametophore ofFunaria
reproduces sexually by formation of
antheridia and archegonia. The
antheridia are formed at the summit of
a relatively small, thin, leafy shoot,
which develops first. The female branch
arises later as a lateral outgrowth from
the base of parent male shoot.
Cedrusis a gymnosperm in which main
plant body is a sporophyte on which
reduced type of gametophytes are
formed.
64Which one of the following is
heterosporous?
[CBSE AIPMT 2008]
(a)Dryopteris(b)Salvinia
(c)Adiantum (d)Equisetum
Ans.(b)
From the followingSalviniais
heterosporous.Heterospory is the
production of spores of two different
sizes and two different developmental
patterns. Small spores are called
microspores and larger as megaspore.
Microspores germinate to produce the
male gametophyte or
microgametophyte that bear male sex
organs while, megaspore germinates to
form female gametophyte or
megagametophyte that bears
archegonia or female sex organs. It is
most important evolutionary
development in the vascular plants
because it has ultimately lead to seed
development, e.g. Selaginella, Marselia,
Salvinia, Azolla, Isoetes.
Dryopterisis homosporous and 32–64
haploid spores are produced in each
sporangium.
Adiantumis also homosporous. The
spores are the pioneer structures of
the gametophytic generation.
65Which of the following
propagates through leaf-tip?
[CBSE AIPMT 2004]
(a) Walking fern
(b) Sproux-leaf plant
(c)Marchantia
(d) Moss
Ans.(a)
Adiantumis also called walking fern. In
Adiantumthe tips of the leaves, on
coming in contact with the soil, give
out adventitious roots which, in turn,
produce new leaves and develop into
new plants.
66Which one the following pairs of
plants are not seed producers?
[CBSE AIPMT 2003]
(a)FicusandChlamydomonas
(b)PunicaandPinus
(c) Fern andFunaria
(d)FunariaandFicus
Ans.(c)
Fernis a pteridophyte andFunariais a
bryophyte, both these produce spores
but not seeds. Phanerogams is a group
of plants which produce seeds and
flowers. It includes gymnospermic and
angiospermic plants. While
cryptogams is another group of plants
which do not produce seeds and
flowers, e.g. algae, fungi, bryophytes
and pteridophytes.
67In ferns meiosis occurs when
[CBSE AIPMT 2000]
(a) spore germinates
(b) gametes are formed
(c) spores are formed
(d) antheridia and archegonia are
formed
Ans.(c)
In the ferns sporangium is a diploid
structure. It bears diploid spore
mother cells which undergo meiosis and
produce haploid spores. Each spore
contains an outer thick brown wall called
exine and a thin inner wall called intine.
Spore is the first cell of gametophyte.
On germination, it gives rise to a haploid
gametophytic stage.
68The ‘walking fern’ is so named
because [CBSE AIPMT 1998]
(a) it is dispersed through the agency
of walking animals
(b) it propagates vegetatively by its
leaf tips
(c) it knows how to walk by itself
(d) its spores are able to walk
Ans.(b)
Adiantumis called walking fern. The tips
of its leaves, on coming in contact with
the soil, gives out adventitious roots
which in turn produce new leaves and
develop into new plants.
69A well developed archegonium with
neck consisting of 4-6 rows of
neck canal cells, characterises
[CBSE AIPMT 1995]
(a) gymnosperms only
(b) bryophytes and pteridophytes
(c) pteridophytes and gymnosperms
(d) gymnosperms and flowering plants
Ans.(b)
Archegonium is the flask-shaped female
reproductive body of bryophytes and
pteridophytes. Archegonium usually
consists of a tubular neck and a swollen
venter. Neck is made up of 4-6 vertical
row of cells and encloses 6-10 neck
canal cells in bryophytes and 4 vertical
rows in pteridophytes enclosing 1-4 neck
canal cells. Venter has 1-2 layer but it is
wall-less in pteridophytes.
70Which one of the following is not
common betweenFunariaand
Selaginella? [CBSE AIPMT 1992]
(a) Archegonium
(b) Embryo
(c) Flagellate sperms
(d) Roots
Ans.(d)
Roots are not the common structure in
FunariaandSelaginella. Funaria(moss)
andSelaginella(pteridophytes) can be
related with the presence of
archegonium, embryo, flagellated
sperms.
Plant Kingdom 25

36. InFunaria,the plant is attached to the
substratum by means of root-like
structures called rhizoids, which are
multicellular, branched, have oblique
cross wall. InSelaginella,special
leafless positively geotropic structures
called rhizophores arise from the stem
at the point of branching. Rhizophore
resembles the stem in some characters
and roots in other characters and was
regarded asorgan-sui-genesis(organ of
independent origin).
71Pteridophytes differ from mosses/
bryophytes in possessing
[CBSE AIPMT 1993]
(a) independent gametophyte
(b) well developed vascular system
(c) archegonia
(d) flagellate spermatozoids
Ans.(b)
Pteridophytes are most primitive
vascular flowerless, spore producing
cryptogamic land plants, commonly
called vascular amphibians/botanical
snakes/spore producing seedless
trachaeophytes. They are first vascular
land plants to have independent
sporophyte diploid plant body with true
root, stem and leaves. In contrast
bryophytes, the amphibians of plant
kingdom are devoid of true roots, stem
and leaves, with no vascular supply but
root-like, non-vascular rhizoids,
leaf-like and stem-like structures are
present.
72Evolutionary important character
ofSelaginellais[CBSE AIPMT 1989]
(a) heterosporous nature
(b) rhizophore
(c) strobili
(d) ligule
Ans.(a)
Heterospory; i.e. the production of two
different types of spores : larger
macrospores and smaller microspores
is a character of evolutionary
significance in pteridophyte
(Selaginella) because seed habit
(characteristic feature of gymnosperms
and angiosperms) and differentiation of
spores on the basis of sex is believed to
have originated from heterosporous
condition.
73Which of the following statements
is incorrect about gymnosperms?
[NEET (Oct.) 2020]
(a) They are heterosporous
(b) Male and female gametophytes
are free living
(c) Most of them have narrow leaves
with thick cuticle
(d) Their seeds are not covered
Ans.(b)
Statement (b) is incorrect. It can be
corrected as In gymnosperms, the male
and the female gametophyte do not
have an independent free-living
existence. The male gametophyte
remains within the sporangia, retained
on the sporophytes and is dependent on
sporophyte.
74Which one of the following
statements iscorrect?[NEET 2018]
(a) Horsetails are gymnosperms
(b)Selaginellais heterosporous,
whileSalviniais homosporous
(c) Ovules are not enclosed by ovary
wall in gymnosperms
(d) Stems are usually unbrancned in
bothCycasandCedrus.
Ans.(c)
Ingymnosperms, ovules are not
enclosed by ovary wall. Seeds do not
occur inside a fruit. They are naked.
Horsetailis the common name of
Equisetum.
Pteridophytes likeSelaginellaand
Salviniaare heterosporous and possess
two types of spores,i.e.microspores
and megaspores.Cycashas an
unbranched columnar stem while
Cedruspossess branched stem.
Therefore, only statement (c) is correct.
75Select the correct statement.
[NEET 2016, Phase I]
(a)Salvinia,GinkgoandPinusall are
gymnosperms
(b)Sequoiais one of the tallest trees
(c) The leaves of gymnosperms are
not well adapted to extremes of
climate
(d) Gymnosperms are both
homosporous and heterosporous
Ans.(b)
Sequoiais one of the tallest tree
species, known as red wood tree. It is a
gymnospermic plant.
Salviniais an angiosperm, butGinkgo
andPinusare gymnosperms.
Gymnosperms are well adapted to
extremes of climate and are
heterosporous.
76Conifers are adapted to tolerate
extreme environmental conditions
because of[NEET 2016, Phase II]
(a) broad hardy leaves
(b) superficial stomata
(c) thick cuticle
(d) the presence of vessels
Ans.(c)
Conifers are gymnosperms. Their
leaves show xerophytic adaptations.
The leaves are like needle with thick
walled single layered epidermal cells
covered with thick cuticle. This enables
them to tolerate extreme climatic
conditions.
77Read the following statements
and answer the question which
follows them
I. In liverworts, mosses and ferns
gametophytes are free living.
II. Gymnosperms and some ferns
are heterosporous.
III. Sexual reproduction inFucus,
VolvoxandAlbugois oogamous.
IV. The sporophyte in liverworts is
more elaborate than that in
mosses.
How many of the above statements
are correct? [NEET 2013]
(a) One (b) Two
(c) Three (d) Four
Ans.(c)
Statement I, II and III are correct.
Statement IV is incorrect and can be
corrected as the sporophyte in mosses
is more elaborate than in liverworts.
26 NEETChapterwise Topicwise Biology
Gymnosperms
TOPIC 4

37. 78CycasandAdiantumresemble
each other in having[NEET 2013]
(a) seeds (b) motile sperms
(c) cambium (d) vessels
Ans.(b)
Multiciliated motile sperms are found in
bothCycas(gymnosperm) andAdiantum
or walking fern (pteridophyte).
Seeds and cambium are present in
Cycas(quite common in gymnosperms)
but absent in pteridophytes. Vessels
are absent in both.
79Gymnosperms are also called soft
wood spermatophytes because they
lack [CBSE AIPMT 2012]
(a) cambium
(b) phloem fibres
(c) thick-walled tracheids
(d) xylem fibres
Ans.(d)
Gymnosperms lack xylem fibres. Large
amount of parenchymatous cells are
present with secondary xylem
tracheids. So, these are also known as
softwood spermatophytes.
80The gametophyte is not an
independent, free living
generation in[CBSE AIPMT 2011]
(a)Adiantum (b)Marchantia
(c)Pinus (d)Polytrichum
Ans.(c)
In gymnosperms, (e.g.Pinus, Cycas,
etc.) the male and female
gametophytes do not have an
independent free-living existence. They
remain within the sporangia retained on
the sporophytes.
81Select one of the following pairs of
important features distinguishing
GnetumfromCycasandPinusand
showing affinities with
angiosperms[CBSE AIPMT 2008]
(a) absence of resin duct and leaf
venation
(b) presence of vessel elements and
absence of archegonia
(c) perianth and two integuments
(d) embryo development and apical
meristem
Ans.(b)
The presence of vessels in the xylem is
an angiospermic character found in
Gnetumwhich distinguish it fromCycas
andPinus.
Gnetumresembles angiosperms in
many other aspects also like
(i) The leaves inGnetumhave
reticulate venation that is an
angiospermic character.
(ii) InGnetumfemale gametophyte is
only partly cellular before
fertilisation and becomes
completely cellular only after
fertilisation. Some of the free
nuclei act as eggs as there are no
archegonia.
The short apices inGnetumand
angiosperms show a distinct tunica and
corpus configuration.
The cortex of stem ofPinusis traverse
by large resin ducts or canals. Each duct
or canal is lined by a layer of thin walled
parenchymatous glandular secretory
cells constituting epithelium. The
epithelial cells secrete resin into canal.
Resin is the chief source of terpentine.
82Flagellated male gametes are
present in all the three of which
one of the following sets?
[CBSE AIPMT 2007]
(a)Anthoceros,FunariaandSpirogyra
(b)Zygnema,SaprolegniaandHydrilla
(c)Fucus,MarseliaandCalotropis
(d)Riccia,DryopterisandCycas
Ans.(d)
Flagellated male gametes are present
inRiccia, DryopterisandCycas.
83In gymnosperms, the pollen
chamber represents
[CBSE AIPMT 2007]
(a) a cell in the pollen grain in which
the sperms are formed
(b) a cavity in the ovule in which
pollen grains are stored after
pollination
(c) an opening in the
megagametophyte through which
the pollen tube approaches the
egg
(d) the microsporangium in which
pollen grains develop
Ans.(b)
In gymnosperms, the pollen chamber is
a cavity in the ovule in which pollen
grains are stored after pollination.
84Which one of the following is a
living fossil?
[CBSE AIPMT 2004, 03]
(a)Cycas (b) Moss
(c)Saccharomyces(d)Spirogyra
Ans.(a)
The Cycadales is an ancient order of
gymnosperms exhibiting several
primitive features—now having only a
few living representative of once a large
group of plants that glorified during the
Mesozoic era. Therefore,Cycasis called
as living fossil.
85Which one pair of examples will
correctly represent the grouping
spermatophyta according to one
of the schemes of classifying
plants? [CBSE AIPMT 2003]
(a)Rhizopus, Triticum
(b)Ginkgo, Pisum
(c)Acacia,sugarcane
(d)Pinus, Cycas
Ans.(b)
Ginkgois a gymnospermic plant so it
comes before angiospermic plant,
Pisum(pea).Rhizopusis a fungus and
Triticum(wheat) is an angiospermic
plant.
Option (c) has both angiospermic plants
and option (d) has both gymnospermic
plants.
86Which of the following plants
produces seeds but not flowers?
[CBSE AIPMT 2002]
(a) Maize (b) Mint
(c) Peepal (d) Pinus
Ans.(d)
Pinusis a gymnospermic plant. Ovules
ofPinusare uncovered which lie on the
megasporophyll, hence, these plants do
not have flowers. However, it produces
seeds (from ovule after fertilisation) like
other three plants mentioned, all of
which are angiosperms.
87Cycashas two cotyledons but not
included in angiosperms because
of [CBSE AIPMT 2001]
(a) naked ovules
(b) seems like monocot
(c) circinate ptyxis
(d) compound leaves
Ans.(a)
Gymnosperms (Gk.gymnos-naked;
sperma-seed e.g. naked seed plants) is
a group of phanerogams which have
naked ovules, i.e. structure (ovules)
that eventually become the seeds after
fertilisation and not completely
enclosed by the tissues of the parent
individual.
Plant Kingdom 27

38. 28 NEETChapterwise Topicwise Biology
88The largest ovules, largest male
and female gametes and largest
plants are found among
(a) angiosperms[CBSE AIPMT 2000]
(b) tree ferns and some monocots
(c) gymnosperms
(d) dicotyledonous plants
Ans.(c)
Cycaswith largest antherozoids and
ovule and the largest plantSequoia
belong to gymnosperms.
Gymnosperm is a group of naked
seeded plants, i.e. their ovules are not
enclosed by ovary walls. Ovules of
gymnosperms are directly borne on the
megasporophyll.
89In which of the following would
you place the plants having
vascular tissue, lacking seeds?
[CBSE AIPMT 1999]
(a) Algae (b) Bryophytes
(c) Pteridophytes (d) Gymnosperms
Ans.(c)
Pteridophytes and gymnosperms do
have vascular tissues. However,
gymnosperms bear seeds while
pteridophytes not bear seeds. Algae
and bryophytes do not possess vascular
tissues.
90Largest sperms in the plant world
are found in[CBSE AIPMT 1998]
(a)Pinus (b) Banyan
(c)Cycas (d)Tsuja
Ans.(c)
Cycasis a gymnospermic plant. It has
the biggest sperms (antherozoids) and
ovules in the plant world.
91Which one of the following
statements aboutCycasis
incorrect? [CBSE AIPMT 1998]
(a) It does not have a well organised
female flower
(b) It has circinate vernation
(c) Its xylem is mainly composed of
xylem vessels
(d) Its roots contain some blue-green
algae
Ans.(c)
Statements (c) is incorrect regarding
Cycasas Vessels are absent from the
xylem of all gymnosperms (except
Gnetales).Cycasbelongs to Cycadales
(not Gnetales).
92Seed habit first originated in
(a) certain ferns[CBSE AIPMT 1996]
(b) certain pines
(c) certain monocots
(d) primitive dicots
Ans.(a)
The tendency towards seed formation
is called seed habit. It was developed in
fossil gymnosperm of group
Cycadofilicales (pteridosperms), i.e.
seed ferns, e.g.Lyginopteriswhich
bears characters of cycads and ferns
both. Seed habit is shown by few
pteridophytes likeSelaginella, Marselia,
Isoetes,etc. which exhibit heterospory.
93The ‘wing’ ofPinusseed is derived
from [CBSE AIPMT 1994]
(a) testa
(b) testa and tegmen
(c) surface of ovuliferous scale
(d) All of the above
Ans.(c)
The winges of seed ofPinusis thin,
membranous diploid and develops
jointly from the basal upper surface
(adaxial) of ovuliferous scale and outer
layer of integument of the ovule.
94Pinusdiffers from mango in having
(a) tree habit[CBSE AIPMT 1993]
(b) green leaves
(c) ovules not enclosed in ovary
(d) wood
Ans.(c)
Gymnosperms, (e.g.Pinus) are
commonly called as naked seeded
plants since their ovules (which later
become seed) are not covered and lie
naked on the surfaces of specialised
leaves (megasporophylls or ovuliferous
scales) arranged into cones.
Thus, gymnosperms are also known as
seeded plants without flowers or
phanerogams without ovary. In
contrast, angiosperms are seed
bearing, flowering vascular plants in
which seeds are enclosed in fruits, and
are called as phanerogams with ovary
or seeded flowering plants.
95Which one is the most advanced
from evolutionary view point?
[CBSE AIPMT 1993]
(a)Selaginella
(b)Funaria
(c)Chlamydomonas
(d)Pinus
Ans.(d)
From the evolutionary point of view, the
given options can be arranged as
Chlamydomonas, Funaria, Selaginella,
and Pinus.
Pinus,i.e. gymnosperms are the most
evolved seed bearing phanerogamic
vascular sporophytic plants, after
angiosperms (most advanced group of
plants).
Pteridophytes (e.g.Selaginella) are
spore bearing non-seeded vascular
cryptogams. Algae, bryophytes and
pteridophytes resemble each other in
dependence on water for fertilisation.
96A plant having seeds but lacking
flowers and fruits belongs to
[CBSE AIPMT 1992]
(a) pteridophytes
(b) mosses
(c) ferns
(d) gymnosperms
Ans.(d)
Gymnosperms (GkGymno=naked ;
sperma-seed) are commonly known as
naked seed plants because their ovules
(which later become seeds) are not
covered and lie naked on the surfaces of
specialised leaves called
megasporophylls or ovuliferous scales,
arranged into cones, flowers are absent;
seed may have two, (e.g.Cycas) or more
(e.g.Pinus) cotyledons.
97InPinus,the pollen grain has 6
chromosomes then its endosperm
will have the chromosome
[CBSE AIPMT 1992]
(a) 12 (b) 18
(c) 6 (d) 24
Ans.(c)
In gymnosperms (Pinus) both the pollen
grains and endosperm are haploid
structure, formed before fertilisation. If
the pollen grain has haploid number of
chromosome equal to 6, then its
endosperm will also have the same
number of chromosome, i.e. 6.
98Resin and terpentine are obtained
from [CBSE AIPMT 1992]
(a)Cycas (b)Pinus
(c)Cedrus (d)Abies
Ans.(b)
Pinus roxburghiis a source of resin and
terpentine, obtained after distillation.
Terpentine is utilised in varnishes and
paints.

39. 99InPinus/gymnosperms, the
haploid structure are
[CBSE AIPMT 1989]
(a) megaspore, endosperm and
embryo
(b) megaspore, pollen grain and
endosperm
(c) megaspore, integument and root
(d) pollen grain, leaf and root
Ans.(b)
In gymnosperms, the megaspore
(haploid) is first cell of female
gametophyte and undergoes repeated
divisions to form a multicellular female
gametophyte, which because of
abundant food reserves, serves as
endosperm. So, being produced before
fertilisation, endosperm is haploid in
gymnosperms.
Microspores or pollen grains are the
first cell of the male gametophyte and
are haploid in nature.
100InPinus/Cycas/gymnosperms, the
endosperm is[CBSE AIPMT 1988]
(a) triploid (b) haploid
(c) diploid (d) tetraploid
Ans.(b)
In gymnosperms, e.g.Pinus, Cycas,
endosperm develops before
fertilisation and is haploid in nature. In
angiosperms, endosperm is triploid (3n)
and formed after double fertilisation.
101Male and female gametophytes do
not have an independent
free-living existence in
[NEET (Oct.) 2020]
(a) pteridophytes (b) algae
(c) angiosperms (d) bryophytes
Ans.(c)
Male and female gametophyte do not
have an independent free-living
existence in gymnosperms and
angiosperms. In them they remains
within the sporangia retained on the
sporophytes. The pollen grain is
released from the microsporangium
and are carried in air currents and come
in contact with the opening of the
ovules borne on megasporophylls. The
pollen tube carrying the male gametes
grows towards archegonia in the ovules
and discharge their contents near the
mouth of the archegonia.
102Which is the most common type
of embryo sac in angiosperms?
[NEET (Odisha) 2019]
(a) Tetrasporic with one mitotic
stage of divisions
(b) Monosporic with three sequential
mitotic divisions
(c) Monosporic with two sequential
mitotic divisions
(d) Bisporic with two sequential
mitotic divisions
Ans.(b)
The most common type of female
gametophyte (embryo sac) in
angiosperms is the monosporic embryo
sac in which the embryo sac develops
from a single functional megaspore (n)
while the other three megasproes
degenerate. The functional megaspore
undergoes three sequential mitotic
divisions and gives rise to 8-nucleate
and 7-celled mature embryo sac.
103Male gametophyte with least
number of cells is present in
[CBSE AIPMT 2014]
(a)Pteris
(b)Funaria
(c)Lilium
(d)Pinus
Ans.(c)
Lilium(angiosperm) possesses the
male gametophyte with least number of
cells.
The number of cells in male
gametophyte shows the pattern of
reduction from bryophytes to
angiosperms. In angiosperms, it is
reduced to about 2-3 celled and called
as pollen grains.
The number of cells in male
gametophyte decreases in the
following order
Funaria > Pteris > Pinus > Lilium
104Compared with the gametophytes
of the bryophytes, the
gametophytes of vascular plants
tends to be[CBSE AIPMT 2011]
(a) larger but to have smaller sex
organs
(b) larger and to have large sex
organs
(c) smaller and to have smaller sex
organs
(d) smaller but to have larger sex organs
Ans.(c)
As we proceed from thallophyta to
angiosperms, there is gradual reduction
in gametophyte with reduced size of sex
organs. From thallophyta to
angiosperms, there is progressive
elaboration of sporophytes.
Phase
Thallo
-phyta
Bryop
-hyta
Pterido-
phyta
Gymno
sperms
Angio-
sperms
Gameto-
phyte (%)
90 75 50 25 10
Sporop
-hyte (%)
10 25 50 75 90
105Conifers differ from grasses in the
[CBSE AIPMT 2006]
(a) lack of xylem tracheids
(b) absence of pollen tubes
(c) formation of endosperm before
fertilisation
(d) production of seeds from ovules
Ans.(c)
The conifers (gymnosperm) differ from
the grasses (angiosperm) in the
formation ofendosperm before
fertilisation. Infact, in gymnosperms,
the endosperm is a haploid tissue as it
formed before fertilisation, while in
angiosperms, endosperm is formed
after fertilisation as a result of triple
fusion or double fertilisation, thus, it is a
triploid tissue.
In both conifers and grasses seeds are
produced from ovules.
Xylem tracheids are present in both
conifers and grasses.
Pollen tubes are also formed in both
conifers and grasses.
106Angiosperms have dominated the
land flora primarily because of
their [CBSE AIPMT 2004]
(a) power of adaptability in diverse
habitat
(b) property of producing large
number of seeds
(c) nature of some pollination
(d) domestication by man
Ans.(a)
Angiosperms are well adapted to
terrestrial life and occur in diverse
habitats like cold tundra to hot tropical
and even desert areas. They also thrive
well in aquatic habitat. Hence, they
being the most successful, to have
dominated the land flora.
Plant Kingdom 29
Angiosperms
TOPIC 5

40. 30 NEETChapterwise Topicwise Biology
107Which of the following is without exception in
angiosperms? [CBSE AIPMT 2002]
(a) Presence of vessels
(b) Double fertilisation
(c) Secondary growth
(d) Autotrophic nutrition
Ans.(b)
A few plants, (e.g.Rafflesia) are parasitic. Some angiosperm
genera are vesselless. Secondary growth does not take place in
a large variety of angiosperms. However, double fertilisation is
met with amongst all angiosperms.
In this one male gamete fuses with egg nucleus( )n. This is
known as syngamy and the second male gamete fuses with the
secondary nucleus or polar nuclei( )2n. This is known as triple
fusion. Together these syngamy and triple fusion is known as
double fertilisation.
108Match items in column I with those in column II.
Column I Column II
A. Peritrichous flagellation 1.Ginkgo
B. Living fossil 2. Macrocystis
C. Rhizophore 3. Escherichia coli
D. Smallest flowering plant 4.Selaginella
E. Largest perennial alga 5.Wolffia
Select the correct answer from the following.
[CBSE AIPMT 2005]
A B C D E A B C D E
(a) 3 1 4 5 2 (b) 2 1 3 4 5
(c) 5 3 2 5 1 (d) 1 2 5 3 2
Ans.(a)
Column I Column II
A. Peritrichous flagella
(flagella all over the body)
1.Escherichia coli
(a bacterium)
B. Living fossil 2. Ginkgo biloba
(maiden hair tree)
C. Rhizophore (a form of aerial
adventitious roots)
3.Selaginella
(a pteridophyte)
D. Smallest flowering plant 4.Wolffia
E. Largest perennial algae 5.Macrocystis
109Life cycle ofEctocarpusandFucusrespectively are
[NEET 2017]
(a) Haplontic, Diplontic (b) Diplontic, Haplodiplontic
(c) Haplodiplontic, Diplontic (d) Haplodiplontic, Haplontic
Ans.(c)
EctocarpusandFucusrespectively show haplodiplontic and
diplontic life cycle.
Concept EnhancerInEctocarpussporic meiosis occurs and
haploid biflagellate meiozoospores are formed. They germinate
to produce gametophytic thalli. The gametophytes liberate
gametes, which fuse to form diploid zygote, which gives rise to a
diploid plant.
110Metagenesis refers to [CBSE AIPMT 2015]
(a) the presence of different morphic forms
(b) alternation of generation between asexual and sexual
phases of an organism
(c) occurrence of a drastic change in form during
post-embryonic development
(d) the presence of a segmented body and parthenogenetic
mode of reproduction
Ans.(b)
Metagenesis in an organism refers to the reproduction
characterised by the alteration of a sexual generation and a
generation that reproduces asexually, i.e. alteration of
generations.
111Which one of the following is considered important in
the development of seed habit?[CBSE AIPMT 2009]
(a) Dependent sporophyte
(b) Heterospory
(c) Haplontic life cycle
(d) Free-living gametophyte
Ans.(b)
Heterosporyis the production of spores of two different sizes
and of two different developmental patterns. It is the most
important evolutionary development in the vascular plants
because it has ultimately lead to seed development, which is
seen in,Selaginella,Salvinia,Azolla, etc.
Plant Life Cycle and Alternation
of Generations
TOPIC 6
u
v
uv u uv
Diploid
sporophyte
Meiosis
Sexual
cycle
Syngamy
Mature
gametophyte
Parthenogenic
cycle
Immature
gametophyte
Haploid
partheno
sporophyte
Haplodiplontic life cycle ofEctocarpus

41. 01In case of poriferans, the
spongocoel is lined with flagellated
cells called [NEET 2017]
(a) ostia
(b) oscula
(c) choanocytes
(d) mesenchymal cells
Ans.(c)
The body wall of a common sponge
consists of three layer, i.e.pinacoderm,
choanodermandmesophyll layer.
Choanoderm is inner cellular layer
which consists of highly specialised
flagellated cells called choanocytes.
The beating of their flagella creates
water current.
02Body having meshwork of cells,
internal cavities lined with food
filtering flagellated cells and
indirect development are the
characteristics of phylum
[CBSE AIPMT 2015]
(a) Coelenterata
(b) Porifera
(c) Mollusca
(d) Protozoa
Ans.(b)
In Porifera (sponges), bodies are
asymetrical. Body lacks tissue or
organs, but from a meshwork of cells
surrounding channels that open to the
outside through pores, and that
expand into internal cavities lined with
food filtering flagellated cells
(choanocytes).
03Which one of the following
statements about all the four
Spongilla,leech, dolphin and
penguin is correct?
[CBSE AIPMT 2010]
(a) Penguin is homiothermic, while
the remaining three are
poikilothermic
(b) Leech is a freshwater form, while
all others are marine
(c)Spongillahas special collared cells
called choanocytes, not found in
the remaining three
(d) All are bilaterally symmetrical
Ans.(c)
Spongillabelongs to phylum–Porifera, in
which, choanocytes are the
characteristic cells, these are absent in
leech, dolphin and pengiun. These
distinctive cells line the interior body
walls of sponges.
These cells have a central flagellum that
is surrounded by a collar of microvilli.
Choanocytes are versatile cells.
Their flagella beat to create the active
pumping of water through the sponge,
while the collars of the choanocytes are
the primary areas where nutrients are
absorbed into the sponge.
04Syconbelongs to a group of
animals which are best described
as [CBSE AIPMT 2003]
(a) multicellular with a gastrovascular
system
(b) multicellular having tissue
organisation, but no body cavity
(c) unicellular or acellular
(d) multicellular without any tissue
organisation
Ans.(d)
Syconbelongs to phylum–Porifera. The
porifers are most primitive group of
multicellular animals. They have no
tissue grade of organisation and
represent cell aggregated body plan,
hence, included in the
sub-kingdom—Parazoa.
05The canal system is a
characteristic feature of
[CBSE AIPMT 1999]
(a) echinoderms
(b) helminthes
(c) coelenterates
(d) sponges
Ans.(d)
Sponges possess an extensive system
of interconnected cavities called canal
system, which typically consists of
incurrent canals, radial canals,
excurrent canals and spongocoel. The
system is useful for nutrition,
respiration and excretion.
06What is true about all sponges
without exception?
[CBSE AIPMT 1996]
(a) They are all marine
(b) They have flagellated collar cells
(c) They have a mixed skeleton
consisting of spicules and spongin
fibres
(d) They reproduce only asexually by
budding
Ans.(b)
Sponges are all aquatic, mostly marine
but few fresh water forms also exist.
Choanocytes or collar cells are only
present in sponges.
AnimalKingdom
04
Phylum–Porifera
TOPIC 1

42. Sponges usually have a skeleton
consisting of spicules or spongin fibres.
Sponges reproduce asexually by
budding, gemmules and reproduce
sexually too.
07The simplest type of canal system
in Porifera is[CBSE AIPMT 1992]
(a) ascon type (b) leucon type
(c) sycon type (d) radial type
Ans.(a)
The simplest type of canal system in
Porifera is ascon type.
08Classification of Porifera is based
on [CBSE AIPMT 1991]
(a) branching (b) spicules
(c) reproduction (d) symmetry
Ans.(b)
Classification of sponges is primarily
based on skeleton which are spicules.
These belongs to phylum– Porifera.
09Which one of the following living
organisms completely lacks a cell
wall? [CBSE AIPMT 2014]
(a) Cyanobacteria
(b) Sea-fan (Gorgonia)
(c)Saccharomyces
(d) Blue-green algae
Ans.(b)
Sea-fan (Gorgonia) belongs to: kingdom
–Animalia, phylum–Cnidaria and order–
Gorgonacea. As, it is on animal, thus it
lacks cell wall unlike cyanobacteria or
blue–green algae (kingdom–Monera)
have cell wall which is composed of
peptidoglycan.
Saccharomycesis a genus of kingdom
fungi which includes many species of
yeast. Their cell wall is made up of
chitin.
10Select the taxon mentioned that
represents both marine and
freshwater species
(a) Echinoderms[CBSE AIPMT 2014]
(b) Ctenophora
(c) Cephalochordata
(d) Cnidaria
Ans.(d)
Cnidarian members are found both in
freshwater and marine environments,
while members of Ctenophora,
Cephalochordata and Echinodermata
are found exclusively in marine
environment.
11In which animal nerve cell is
present but brain is absent?
[CBSE AIPMT 2002]
(a) Sponge (b) Earthworm
(c) Cockroach (d)Hydra
Ans.(d)
The neurons ofHydrasynapse with
each other and other body cells. As a
result, it responds to external stimuli.
However, there is no brain inHydrato
coordinate the responses.
12InHydra,waste material of food
digestion and nitrogenous waste
material are removed from
[CBSE AIPMT 2001]
(a) mouth and mouth
(b) body wall and body wall
(c) mouth and body wall
(d) mouth and tentacles
Ans.(c)
InHydra, undigested residues are
egested from coelentron through
mouth, while nitrogenous excretory
product (ammonia) is removed through
general body surface.
Hydraare solitary, sessile, freshwater
coelenterates. They are found in lakes,
ponds stream.Hydrais an ammonotelic
animal i.e., it excretes waste material in
the form of ammonia.
13Most appropriate term to describe
the life cycle ofObeliais
[CBSE AIPMT 1998]
(a) neoteny (b) metagenesis
(c) metamorphosis
(d) All of these
Ans.(b)
True alternation of generations is
alternation between haploid
gametophyte and diploid sporophyte
phase. InObelia,the asexual polypoid
phase is alternate with sexual medusoid
phase and this type of life cycle is called
metagenesis.
NeotenyIt is also called Juvenilisation,
is one of the two ways by which
paedomorphism can arise.
PaedomorphismIt is the retention by
adults of traits previously seen only in
young, and is a subject studied in the
development biology.
MetamorphosisIt is a biological
process by which an animal physically
develops after birth or notching,
involving a conspicous and relatively
abrupt change in the animal body
structure through cell growth and
differentiation. e.g., in some insects,
molluses etc.
14Point out a non-parasite
[CBSE AIPMT 1994]
(a) tapeworm (b) mosquito
(c) leech (d) sea anemone
Ans.(d)
Sea anemone is a coelenterate found in
marine water, attached to the empty
shell of gastropods already occupied by
hermit crab, this association is
symbiosis not parasitism.
Tapeworm, mosquito and leech are
parasites.
15Special character of coelenterates
is [CBSE AIPMT 1994]
(a) polymorphism
(b) nematocysts
(c) flame cells
(d) hermaphroditism
Ans.(b)
The most characteristic feature of
coelenterates is the presence of
nematocysts or stinging cells.
Nematocysts are mainly concerned
with food capture, defence and
attachment.
16Which one of the following
animals possesses nerve cells but
no nerves? [CBSE AIPMT 1993]
(a)Hydra
(b) Tapeworm
(c) Earthworm
(d) Frog’s tadpole
32 NEETChapterwise Topicwise Biology
Phylum–Cnidaria
(Coelenterata) and
TOPIC 2
Cylinder bore
Osculum
Ostium
Central cavity
(spongocoel)
Flagellated
collar cells
Water
current
Ascon type canal system (Porifera)

43. Ans.(a)
Hydrahas nerve cells but no nerves.
Hydrapossesses a very primitive
nervous system consisting of a synaptic
network of bipolar and multipolar nerve
cells.
17Budding is a normal mode of
asexual reproduction in
[CBSE AIPMT 1993]
(a) starfish andHydra
(b)Hydraand sponges
(c) tapeworm andHydra
(d) sponge and starfish
Ans.(b)
Hydraand sponges reproduce asexually
by exogenous budding, a type of
vegetative propagation.
18Jelly fish belongs to class
[CBSE AIPMT 1989]
(a) Hydrozoa
(b) Scyphozoa
(c) Anthozoa
(d) None of these
Ans.(b)
Aurelia(jellyfish) belongs to
class—Scyphozoa of phylum—Cnidaria.
It is without skeleton.
19Organ pipe coral is
[CBSE AIPMT 1988]
(a)Tubipora (b) Astraea
(c)Helipora (d)Fungia
Ans.(a)
Organ pipe coral is the common name
ofTubipora.This is coelenterate and
comes under class—Anthrozoa.
20Biradial symmetry and lack of
cnidoblasts are the characteristics
of [CBSE AIPMT 2006]
(a) Starfish and sea anemone
(b)CtenoplanaandBeroe
(c)AureliaandParamecium
(d)Hydraand starfish
Ans.(b)
CtenoplanaandBeroelack cnidoblasts
and have biradial symmetry. These
belong to phylum–Ctenophora.
Hydra, sea anemone andAureliaare
coelenterates which have cnidoblasts.
Although sea anemone has biradial
symmetry.
21Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. Gregarious,
polyphagous pest
1.Asterias
B. Adult with radi
symmetry and larva
with bilateral
symmetry
2. Scorpion
C. Book lungs 3. Ctenoplana
D. Bioluminescence 4. Locusta
A B C D
(a) 4 1 2 3
(b) 3 2 1 4
(c) 2 1 3 4
(d) 1 3 2 4
Ans.(a)
The correct option is (a). It can be
explained as follows
Locustais a gregareous pest.
In echinoderms, adults are radially
symmetrical but larvae are bilaterally
symmetrical.
Scorpions respire through book lungs.
Bioluminescence is well-marked in
ctenophores.
22Bilaterally symmetrical and
acoelomate animals are
exemplified by[NEET (Sep.) 2020]
(a) Platyhelminthes (b) Aschelminthes
(c) Annelida (d) Ctenophora
Ans.(a)
Platyhelminthes are bilaterally
symmetrical, triploblastic and
acoelomate animals with organ level of
organisation.
Aschelminthes are bilaterally
symmetrical, triploblastic and
pseudocoelomate with organ system
grade of body organisation. Annelida
are bilaterally symmetrical, triploblastic
and truly schizocoelomate. Ctenophora
are biradially symmetrical,triploblastic
and acoelomates.
23Planariapossess high capacity of
[CBSE AIPMT 2014]
(a) metamorphosis
(b) regeneration
(c) alternation of generation
(d) bioluminescence
Ans.(b)
Planariais a flatworm belonging to
phylum–Platyhelminthes. They are the
simplest form of multicellular animal.
They have high capacity of regeneration
of new tissue at the wound siteviacell
proliferation (blastema formation) and
the remodelling of pre-existing tissue
to restore symmetry and proportion.
This is due the neoblast cells.
These cells are usually scattered
through out the body and are able to
participate in any type of development.
The regenerative capacity of different
body sections is an indicator of the
presence of different numbers of
neoblast cells.
24One example of animal having a
single opening to the outside that
serves both as mouth as well as
anus is [CBSE AIPMT 2010]
(a)Octopus (b)Asterias
(c)Ascidia (d)Fasciola
Ans.(d)
Fasciola hepatica(sheep liver fluke)
belongs to phylum–Platyhelminthes.
These worms have incomplete
alimentary canal, there is a single
opening for both ingestion and
egestion. This is also called as blind sac
body plan.
25Which one of the following kinds
of animals are triploblastic?
[CBSE AIPMT 2010]
(a) Flatworms
(b) Sponges
(c) Ctenophores
(d) Corals
Ans.(a)
Flatworms(phylum–Platyhelminthes)
are triploblastic animals. The cells of
the body wall are arranged in three
layers. i.e. ectoderm, mesoderm and
endoderm.
Sponges, ctenophores and corals are
diploblastic animals.
26During its life cycle,Fasciola
hepatica(liver fluke) infects its
intermediate host and primary
host at the following larval stages
respectively[CBSE AIPMT 2003]
(a) metacercaria and cercaria
(b) miracidium and metacercaria
(c) redia and miracidium
(d) cercaria and redia
Animal Kingdom 33
Phylum-Platyhelminthes
TOPIC 4
Ctenophora
TOPIC 3

44. Ans.(b)
Liver fluke (Fasciola hepatica) is a
‘digenetic’ endoparasite, i.e. its life
cycle completes within two hosts. The
primary host is sheep and the
secondary or intermediate host is fresh
water gastropod, snail. Fasciola
hepatica infects its intermediate host
at miracidium stage and its primary
host at metacercaria stage.
27Which one belongs to
Platyhelminthes?[CBSE AIPMT 1994]
(a)Schistosoma(b)Trypansoma
(c)Plasmodium(d)Wuchereria
Ans.(a)
Schistosomais the commonhuman
blood flukeand belongs to
Platyhelminthes. It is a genus of
trematodes.
It is a parasitic flatworm responsible for
a highly significant group of infections
in humans termed as Schistosomiasis.
28What is true aboutTaenia
saginata? [CBSE AIPMT 1993]
(a) Life history has pig as
intermediate host
(b) There are two large suckers on
scolex
(c) Rostellar hooks are absent
(d) Rostellum has double circle of
hooks
Ans.(c)
Taenia saginatado not possess
rostellum and rostellar hooks.
29What is correct aboutTaenia?
[CBSE AIPMT 1992]
(a) Male organs occur in posterior
proglottids
(b) Male organs occur in anterior
proglottids
(c) Female organs occur in anterior
proglottids
(d) Mature proglottids contain both
male and female organs
Ans.(d)
Body ofTaeniais divided into scolex,
neck and strobilla. Strobilla is the main
body and made of proglottids. A
proglottid is a unit of body enclosing a
complete set of genitalia.
Mature proglottids are in the middle
having both male and female
reproductive organs.
30The excretory structures of flat
worms/Taeniaare
[CBSE AIPMT 1991]
(a) flame cells
(b) protonephridia
(c) Malpighian tubules
(d) green glands
Ans.(a)
Excretory structures of
flatworms/Taeniaare flame cells.
Longitudinal and cross-connecting
excretory canals drain fluid from flame
cells in each proglottid. Main excretory
products are ammonia and fatty acids.
31Bladderworm/cysticercus is the
larval stage of[CBSE AIPMT 1991]
(a) tapeworm (b) roundworm
(c) pinworm (d) liver fluke
Ans.(a)
Bladderworm/Cysticercus is the larval
stage of tapeworm. It is found in the
muscles of pig and this is the stage
through which man gets infected by
eating raw or poorly cooked ‘measly
pork’.
32Transfer ofTaeniato secondary
host occurs as
[CBSE AIPMT 1989, 90]
(a) oncosphere
(b) cysticercus
(c) morula
(d) egg
Ans.(a)
Oncospheres pass through faecal
matter of man. Secondary host pig
acquires infection by ingesting the
oncospheres.
33Which one of the following
statements about certain given
animals is correct?
[CBSE AIPMT 2010]
(a) Round worms (Aschelminthes) are
pseudocoelomates
(b) Molluscs are acoelomates
(c) Insects are pseudocoelomates
(d) Flatworms (Platyhelminthes) are
coelomates
Ans.(a)
Round worms(phylum–Aschelminthes)
are pseudocoelomates, false coelom is
derived from embryonic blastocoel.
Flat worms(phylum–Platyhelminthes)
are acoelomate animals.
Molluscs and insects
(phylum–Arthropoda) are coelomate
animals.
34Which one of the following groups
of animals is bilaterally symmetrical
and triploblastic?
[CBSE AIPMT 2009]
(a) Coelenterates (cnidarians)
(b) Aschelminthes (roundworms)
(c) Ctenophores
(d) Sponges
Ans.(b)
Aschelminthesare bilateral
symmetrical and triploblastic animals,
e.g.Ascaris.
Coelenteratesare radially symmetrical
and diploblastic animals, e.g.Obelia.
Ctenophoresare radially symmetrical
and diploblastic animals, e.g.
Ctenoplana.
Spongesare asymmetrical or radially
symmetrical and diploblastic animals,
e.g.Sycon.
35Ascarisis characterised by
[CBSE AIPMT 2008]
(a) absence of true coelom but
presence of metamerism
(b) presence of neither true coelom
nor metamerism
(c) presence of true coelom but
absence of metamerism
(d) presence of true coelom and
metamerism (metamerisation)
Ans.(b)
Ascarisis endoparasite of man. It
inhabits the small intestine more
frequently of children than of adults.
The body is elongate, cylindrical and
gradually tapering at both ends. There
is no metameric segmentation. In
Ascarisbetween body wall and visceral
organs is a spacious fluid filled cavity.
This cavity is not true coelom as it is not
lined by coelomic epithelium, has no
relation with reproductive and
excretory organs and develops from
blastocoel. This body cavity is referred
as pseudocoel.
34 NEETChapterwise Topicwise Biology
Phylum-Aschelminthes
TOPIC 5

45. 36What iscommon betweenAscaris
lumbricoidesandAnopheles
stephensi? [CBSE AIPMT 2000]
(a) Hibernation
(b) Metamerism
(c) Anaerobic respiration
(d) Sexual dimorphism
Ans.(d)
BothA. lumbricoidesandA. stephensi;
have different males and females and it
is possible to distinguish between them
morphologically.
37Ascarislarva is called
[CBSE AIPMT 1992]
(a) cysticercus (b) rhabditiform
(c) hexacanth (d) onchosphere
Ans.(b)
Ascarislarva is calledrhabditoidor
rhabditiformdue to its close
resemblance withRhabditis.
38Ascaris lumbricoidesinfection
occurs through[CBSE AIPMT 1991]
(a) sole of uncovered feet
(b) contaminated cooked measly pork
(c) improperly cooked measly pork
(d) from air through inhalation
Ans.(b)
The transmission of infective stage
through embryonated egg ofAscaris
lumbricoidestakes place by
contaminated cooked measly pork and
contaminated water.
39Match the List-I with List-II.
[NEET 2021]
List-I List-II
A. Metamerism 1. Coelenterata
B. Canal system 2. Ctenophora
C. Comb plates 3. Annelida
D. Cnidoblasts 4. Porifera
Choose the correct answer from
the options given below.
A B C D
(a) 4 3 1 2
(b) 3 4 1 2
(c) 3 4 2 1
(d) 4 1 2 3
Ans.(c)
(c) (A)- (3), (B)-(4),(C)-(2),(D)-(1)
The annelid worms were thought to
have evolved from a coelomate
worm-like ancestor which developed
metameric segmentation or
metamerism and the segments were
termed as somites or metameres.
Sponges or porifera have a water
transport or canal system. Water enters
viaminute pore (ostia) in the body wall
into the central cavity spongocoel from
where it goes outviaosculum.
The body of ctenophores bear eight
external rows of ciliated comb plates
which helps in locomotion.
The name Cnidaria is derived from
cnidocytes or cnidoblast that are found
on the tentacles and body of the
organism.
40Which of the following options
does correctly represent the
characteristic features of
phylum–Annelida?
[NEET (Oct.) 2020]
(a) Triploblastic, unsegmented body
and bilaterally symmetrical
(b) Triploblastic, segmented body and
bilaterally symmetrical
(c) Triploblastic, flattened body and
acoelomate condition
(d) Diploblastic, mostly marine and
radially symmetrical
Ans.(b)
Animals belonging to phylum-Annelida
are triploblastic, bilaterally symmetrical
and metamerically segmented.
They exhibit organ system level of body
organisation with presence of coelom.
They may be aquatic (marine and
freshwater) or terrestrial, free-living
and sometimes parasitic.
41Which of the following animals are
true coelomates with bilateral
symmetry? [NEET (Odisha) 2019]
(a) Adult echinoderms
(b) Aschelminthes
(c) Platyhelminthes
(d) Annelids
Ans.(d)
Annelids are true coelomates with
bilateral symmetry. These exhibit
organ-system level of body
organisation with true coelom. They are
triploblastic, metamerically segmented
and coelomate animals, e.g.
earthworm.
42Pheretimaand its close relatives
derive nourishment from
[CBSE AIPMT 2012]
(a) sugarcane roots
(b) decaying fallen leaves and soil
organic matter
(c) soil insects
(d) small pieces of fresh fallen leaves
of maize
Ans.(b)
Food of earthworm (Pheretimasp.)
consists of organic matter, fallen
decaying leaves, algae, etc. present in
the soil. Food is swallowed along with
soil by sucking action.
43One very special feature in the
earthwormPheretimais that
[CBSE AIPMT 2011]
(a) the typhlosole greatly increases
the effective absorption area of
the digested food in the intestine
(b) the S-shaped setae embedded in
the integument are the defensive
weapons used against the
enemies
(c) it has a long dorsal tubular heart
(d) fertilisation of eggs occurs inside
the body
Ans.(a)
A pair of short and conical intestinal
caecae project from the intestine on
the 26th segment. The characteristic
feature of the intestine between 26-35
segments is the presence of internal
median fold of dorsal wall called
typhlosole.
This increases the effective area of
absorption in the intestine.
44If a live earthworm is pricked with
a needle on its outer surface
without damaging its gut, the fluid
that comes out is
[CBSE AIPMT 2009]
(a) excretory fluid
(b) coelomic fluid
(c) haemolymph
(d) slimy mucus
Ans.(b)
The body cavity of earthworm is true
coelom (schizocoel) as it is formed by
the division of mesoderm. The coelom
is filled with milky, alkaline coelomic
fluid, which contains different types of
corpuscles. Thus, if alive earthworm is
prickled with a needle on its outer
surface, the coelomic fluid will come out.
Animal Kingdom 35
Phylum-Annelida
TOPIC 6

46. Excretory fluidsExcretory system
regulate the chemical composition of
body fluids by removing metabolic
waste substances.
HaemolymphIt is a fluid in the open
circulatory system of arthropods, e.g.
spiders, crustaceans etc.
It is analogous to the fluids and cells
making both blood and interstitial fluid.
Slimy mucusMucus trap bacteria cell
debris and prevents it from entering
into lungs and in their body parts.
45Which one of the following is not a
characteristic of
phylum–Annelida?
[CBSE AIPMT 2008]
(a) Closed circulatory system
(b) Segmentation
(c) Pseudocoelom
(d) Ventral nerve cord
Ans.(c)
Name of the phylum–Annelida was first
coined byLamarck. The body of
annelids is elongated, bilaterally
symmetrical, triploblastic, truely
coelomate and metamerically
segmented into similar metameres. The
coelom is true, schizocoelous. Blood
vascular system is closed. The nervous
system is with a pair of cerebral ganglia
and a double ventral nerve cord bearing
ganglia and lateral nerves in each
segment.
The blood vascular system consists of
blood vessels and capillaries. Blood is
composed of fluid plasma and
colourless corpuscles, physiologically
comparable to the leucocytes of
vertebrates.
Pseudocoelom is the body cavity of
Aschelminthes (roundworm).
46Earthworms have no skeleton but
during burrowing, the anterior end
becomes turgid and acts as a
hydraulic skeleton. It is due to
[CBSE AIPMT 2008]
(a) coelomic fluid (b) blood
(c) gut peristalsis (d) setae
Ans.(a)
The body cavity (coelom) of earthworm is
filled with an alkaline, colourless or milky
coelomic fluid containing water, salts
some proteins and four types of
coelomic corpuscles i.e. phagocytes,
mucocytes, circular nucleated cells and
chloragogen cells. During burrowing the
coelomic fluid becomes turgid and acts
as hydraulic skeleton.
Earthworm (Pheretima posthuma) living
in burrows made in moist earth. The
body shows metameric segmentation.
About the middle of each segment
there is a ring of tiny curved bristles
called setae or chaetae, formed of a
horny nitrogenous organic substance
known as chitin. The setae and
musculature serve for locomotion as
well as for anchoring body firmly in
burrow.
The blood of earthworm is composed of
a fluid plasma and colourless coruscles,
physiologically comparable to the
leucocytes of vertebrates.
47Which one of the following is
correct matching pair of a body
feature and the animal possessing
it? [CBSE AIPMT 2007]
(a) Post-anal tail —Octopus
(b) Ventral central — Leech
nervous system
(c) Pharyngeal gills slits
—Chameleonabsent in embryo
(d) Ventral heart — Scorpion
Ans.(b)
The nervous system of leech consists of
ventral central nervous system,
peripheral nervous system and
sympathetic nervous system.
48In which of the following
chlorocruorin pigment is found?
(a) Annelida[CBSE AIPMT 2001]
(b) Echinodermata
(c) Insecta
(d) Lower Chordata
Ans.(a)
Chlorocruorin is a respiratory pigment
(green, fluorescing red) dissolved in the
plasma of some polychaete worms
(annelid).
49Functionwise, just as there are
nephridia in an earthworm, so are
[CBSE AIPMT 1996]
(a) parotid glands in toad
(b) statocysts in prawn
(c) flame cells in liver fluke
(d) myotomes in fish
Ans.(c)
Flame cells in liver fluke are excretory
organs as nephridia in an earthworm.
50True coelom is the space lying
between the alimentary canal and
body wall enclosed by the layers of
[CBSE AIPMT 1996]
(a) ectoderm on both sides
(b) endoderm on one side and
ectoderm on the other
(c) mesoderm on one side and
ectoderm on the other
(d) mesoderm on both sides
Ans.(d)
Coeloms are secondary body cavities
bounded on all sides by mesodermal
peritoneum. The true coelom arises
within the mesoderm itself.
51Coelom derived from blastocoel is
known as [CBSE AIPMT 1994]
(a) enterocoelom
(b) schizocoelom
(c) pseudocoelom
(d) haemocoelom
Ans.(c)
Coelom derived from blastocoel is
pseudocoelom. These are cavities not
entirely lined by peritoneum (thin
cellular membrane derived from
mesoderm).
Embryologically pseudocoel may be a
persistent blastocoel. Such type of
coelom is found inNematohelminthes.
52Which one assists in locomotion?
[CBSE AIPMT 1993]
(a) Trichocysts inParamecium
(b) Pedicellariae of starfish
(c) Clitellum inPheretima
(d) Posterior sucker inHirudinaria
Ans.(d)
InHirudinarialocomotion takes place by
looping and swimming in which its
posterior suckers provide help as setae
are not present for locomotion.
53Blood ofPheretimais
[CBSE AIPMT 1990]
(a) blue with haemocyanin in
corpuscles
(b) blue with haemocyanin in plasma
(c) red with haemoglobin in
corpuscles
(d) red with haemoglobin in plasma
Ans.(d)
Blood ofPheretimais red in colour and
respiratory pigment haemoglobin is
dissolved in blood plasma.
54Pheretima posthumais highly
useful as [CBSE AIPMT 1990]
(a) their burrows make the soil loose
(b) they make the soil porous, leave
their castings and take organic
debris in the soil
36 NEETChapterwise Topicwise Biology

47. (c) they are used as fish meal
(d) they kill the birds due to
biomagnification of chlorinated
hydrocarbons
Ans.(b)
Pheretima posthumais useful for
farmers as they enrich the soil by their
excretory wastes and make the soil
porous.
55Earthworms are[CBSE AIPMT 1989]
(a) useful
(b) harmful
(c) more useful than harmful
(d) more harmful
Ans.(a)
Earthworms arefriends of farmers
because they enrich the soil by
nephridial excretion, it increases the
fertility of soil. They also help in
ploughing of fields, make the soil
porous. Earthworms are also used for
dissection in laboratories.
56Photoreceptors of earthworm occur
on [CBSE AIPMT 1989]
(a) clitellum (b) many eyes
(c) dorsal surface (d) lateral sides
Ans.(c)
Photoreceptors (with L-shaped lens or
optic organelles) are present on the
surface of skin on dorsal side.
Earthworm hasno eyes,
photoreceptors are used to judge
intensity and duration of light, but do
not have the capacity of vision.
57Which one of the following
belongs to the family-Muscidae ?
[NEET 2021]
(a) Firefly (b) Grasshopper
(c) Cockroach (d) Housefly
Ans.(d)
Housefly belong to Muscidae family.
Muscidae are a family of flies found in
superfamily-Muscoidea. The
family-Muscidae is a large dipteran
family comprised of more than 5000
species.
Other options can be explained as:
Firefly belong to Lampyridae family
Grasshopper belong to Acrididae family
Cockroach belong to Blattidae family.
58Match the following organisms
with their respective
characteristics.
[NEET (National) 2019]
Column I Column II
A.Pila (i) Flame cells
B.Bombyx (ii) Comb plates
C.Pleurobrachia(iii) Radula
D.Taenia (iv) Malpighian
tubules
Select the correct option from the
following :
A B C D
(a) (iii) (iv) (ii) (i)
(b) (ii) (iv) (iii) (i)
(c) (iii) (ii) (iv) (i)
(d) (iii) (ii) (i) (iv)
Ans.(a)
(A)–(iii), (B)–(iv), (C)–(ii), (D)–(i)
Pilaor apple snail contains a file-like
rasping organ called radula for feeding.
Bombyxor silkworm is an arthropod in
which excretion occurs through
Malpighian tubules. The body of
ctenophorePleurobranchiabears eight
rows of ciliated comb plates, which help
in locomotion.
InTaenia, excretion occurs through
specialised cells called flame cells
which contain a protonephridia.
59Which of the following features is
not present in the
phylum–Arthropoda?
[NEET 2016, Phase I]
(a) Metameric segmentation
(b) Parapodia
(c) Jointed appendages
(d) Chitinous exoskeleton
Ans.(b)
Parapodia are present in aquatic
annelids likeNereis,which help them in
swimming. Other three features, i.e.
metameric segmentation, jointed
appendages and chitinous exoskeleton
are present in phylum—Arthropoda. Out
of these metameric segmentation is
visible as tagmetisation.
60Match column I with column II for
housefly classification and select
the correct option using the codes
given below:[NEET 2016, Phase II]
Column I Column II
A. Family 1. Diptera
B. Order 2. Arthropoda
C. Class 3. Muscidae
D. Phylum 4. Insecta
Codes
A B C D
(a) 3 1 4 2
(b) 3 2 4 1
(c) 4 3 2 1
(d) 4 2 1 3
Ans.(a)
Classification of housefly
A. Family – Muscidae
B. Order – Diptera
C. Class – Insecta
D. Phylum – Arthropoda
Short TrickThe question can be easily
solvedviaelimination technique as (D)
phylum is given with (2) Arthropoda
combination in only option (a). This
easily eliminates other options as
correct answer. Thus, saves your time
too.
61Which group of animals belong to
the same phylum? [NEET 2013]
(a) Malarial parasite,Amoeba,
mosquito
(b) Earthworm, pinworm, tapeworm
(c) Prawn, scorpion,Locusta
(d) Sponge, sea anemone, starfish
Ans.(c)
Prawn,Scorpion,Locustabelong to
phylum– Arthropoda therefore, it is
correct. Others can be corrected as
Malarial parasite (Plasmodium vivax) and
Amoebabelong to phylum–Protozoa.
Mosquito—Phylum –Arthropoda
Earthworm — Phylum –Annelida
Pinworm and Tapeworm –Phylum–
Aschelminthes
Sponge—Phylum—Porifera, Sea
anemone –
Phylum — Coelenterata
Starfish — Phylum-Echinodermata
62One of the representative of
Phylum– Arthropoda is[NEET 2013]
(a) cuttle fish (b) silver fish
(c) puffer fish (d) flying fish
Animal Kingdom 37
Phylum-Arthropoda
TOPIC 7

48. Ans.(b)
Silver fish — Arthropoda (phylum)
Silver fish is a small, wingless insect in
the order Thysanura.
Cuttle fish — Mollusca
Putter fish — Chordata (phylum),
class — Pisces
Flying fish — Pisces.
63Which of the following are
correctly matched with respect to
their taxonomic classification?
[NEET 2013]
(a) Flying fish, cuttle fish, silver
fish—Pisces
(b) Centipede, millipede, spider,
scorpion—Insecta
(c) House fly, butterfly, tse-tse fly,
silver fish—Insecta
(d) Spiny anteater, sea urchin, sea
cucumber—Echinodermata
Ans.(c)
Option (c) is correctly matched
housefly, butterfly, tse-tse fly, silver
fish–Insecta of phylum–Arthropoda.
Others can be corrected as
Flying fish, class – Osteichthyes of
phylum – Pisces, cuttle fish (Sepia) of
phylum – Mollusca
Silver fish, class–Insecta of phylum
– Arthropoda
Sea urchin and sea cucumber belong to
Echinodermata.
Spider and scorpion belong to class
– Arachnida of phylum−Arthropoda.
Centipede belongs to class – Chilopoda
of phylum – Arthropoda.
Millipede belongs to class – Diplopeda
of phylum – Arthropoda.
Spring anteater belongs to phylum –
Mammalia.
64Which one of the following have
the highest number of species in
nature? [CBSE AIPMT 2011]
(a) Insects
(b) Birds
(c) Angiosperms
(d) Fungi
Ans.(a)
More than 70% of all the species
recorded are animals. Among animals,
insects are the most species rich
taxonomic group, making more than
70% of the total. It means out of every
10 animals on this planet, 7 are insects.
65Which of the following is correctly
states as it happens in the
common cockroach?
[CBSE AIPMT 2011]
(a) Oxygen is transported by
haemoglobin in blood
(b) Nitrogenous excretory product is
urea
(c) The food is ground by mandibles
and gizzard
(d) Malpighian tubules are excretory
organs projecting out from the
colon
Ans.(c)
In cockroach, mandibles are a pair of
hard, strong, large, dark coloured
triangular, structures which move in
horizontal motion and crush food
between them.
Gizzard or proventriculus has an outer
layer of thick circular muscles and thick
inner cuticle forming six highly
chitinous plate called teeth. The gizzard
acts as the grinding chamber and helps
in grinding the food particles.
66Which one of the following groups
of three animals each is correctly
matched with their one
characteristic morphological
feature? [CBSE AIPMT 2008]
Animals
Morphological
Feature
(a) Liver fluke, sea
anemone, sea
cucumber
— Bilateral
symmetry
(b) Centipede,
prawn, sea
urchin
— Jointed
appendages
(c) Scorpion, spider,
cockroach
— Ventral solid
central nervous
system
(d) Cockroach,
locust, Taenia
— Metameric
segmentation
Ans.(c)
Characteristic Animal
Bilateral
symmetry
—Liver fluke,Taenia
Jointed
appendages
—Prawn, cockroach,
scorpion
Ventral solid
central
nervous system
—Scorpion, spider,
Cockroach
Metameric
segmentation
—Annelids
Radial
symmetry
—Sea anemone
67Which one of the following is the
true description about an animal
concerned?[CBSE AIPMT 2008]
(a) Earthworm— The alimentary canal
consists of a sequence
of pharynx,
oesophagus, stomach,
gizzard and intestine
(b) Frog — Body divisible into three
regions— head, neck
and trunk
(c) Rat — Left kidney is slightly
higher in position than
the right one
(d) Cockroach — 10 pairs of spiracles (2
pairs on thorax and 8
pairs on abdomen)
Ans.(d)
The respiratory system of cockroach
consists of tracheae, tracheoles and
spiracles. The main tracheal trunks
open to exterior on body surface
through 10 pairs of segmentally
arranged spiracles. Two pair of
spiracles are thoracic (one between pro
and mesothorax and other between
meso and metathorax). Eight pairs of
spiracle are abdominal (one pair in each
of the first eight abdominal segments).
Alimentary canal of earthworm is
complete and functionally regioned into
buccal chamber, pharynx, oesophagus,
gizzard, stomach and intestine.
Frog’s body has two main parts, i.e.,
head and trunk. Absence of neck and
tail provide convenience in both
hopping and swimming.
68Which one of the following phyla is
correctly matched with its two
general characteristics?
[CBSE AIPMT 2008]
(a) Arthropoda —Body divided into
head, thorax and
abdomen and
respiration by
tracheae
(b) Chordata — Notochord at some
stage and separate
anal and urinary
openings to the
outside
(c) Echinodermata—Pentamerous radial
symmetry and mostly
internal fertilisation
(d) Mollusca — Normally oviparous
and development
through a
trochophore or
veliger larva
38 NEETChapterwise Topicwise Biology

49. Ans.(a)
Arthropoda is the largest phylum of
animal kingdom. Body of an arthropod
is divisible into head, thorax and
abdomen. Head and thorax often fused
to form cephalothorax. The respiration
takes place by general body surface,
gills, tracheae or book lungs.
Molluscs are mostly dioecious or
monoecious, one or more gonads with
gonoducts, opening into renal ducts or
to exterior. The fertilisation is external
or internal, development direct or
indirect through free larval forms.
Echinoderms have a pentamerous
radial symmetry derived from an
original bilateral symmetry. The
fertilisation is external, development
indirect through free-swimming larval
forms.
Chordates are sharply distinguished
from non-chordates by presence of
notochord, dorsal tubular central
nervous system and pharyngeal gills
slits.
69What is true aboutNereis,
scorpion, cockroach and silver
fish? [CBSE AIPMT 2007]
(a) They all have jointed paired
appendages
(b) They all possess dorsal heart
(c) None of them is aquatic
(d) They all belong to the same
phylum
Ans.(c)
Nereisliving in burrows in sand or mud
often with clams. Scorpion are
abundant in deserts. Cockroach are
found in warmth, dampness and plenty
of organic food to devour.Lepisma
(silver fish) residing in damp cool places
and feeding on starch of starchy
matter.
70Two common characters found in
centipede, cockroach and crab are
[CBSE AIPMT 2006]
(a) compound eyes and anal cerci
(b) jointed legs and chitinous
exoskeleton
(c) green gland and tracheae
(d) book lungs and antennae
Ans.(b)
Crab, centipede and cockroach belong
to phylum–Arthropoda. These have
jointed appendages and chitinous
exoskeleton.
71From the following statements
select the wrong one.
[CBSE AIPMT 2005]
(a) Millipedes have two pairs of
appendages in each segment of
the body
(b) Prawn has two pairs of antennae
(c) Animals belonging to
phylum–Porifera are exclusively
marine
(d) Nematocysts are characteristic of
the phylum–Cnidaria
Ans.(b)
Prawn does not have two pairs of
antennae instead it has one pair of
antennae and one pair of antennules.
72In Arthropoda, head and thorax are
often fused to form
cephalothorax, but in which one of
the following classes, is the body
divided into head, thorax and
abdomen? [CBSE AIPMT 2004]
(a) Insecta
(b) Myriapoda
(c) Crustacea
(d) Arachnida and Crustacea
Ans.(a)
An arthropod body consists of head,
thorax and abdomen, in some cases
head and thorax may be fused to form
cephalothorax. Class—Insecta have
body divided into head, thorax and
abdomen.
73Which one of the following is
correct matching pair of an animal
and a certain phenomenon it
exhibits? [CBSE AIPMT 2003]
(a)Chameleon— Mimicry
(b)Taenia— Polymorphism
(c)Pheretima— Sexual dimorphism
(d)Musca — Complete
metamorphosis
Ans.(d)
The young forms of housefly (maggot,
pupas) are entirely different from the
adult, the metamorphosis being
complete (holometabolic
metamorphosis). Metamorphosis is a
process during which an animal
undergoes a comparatively rapid
changes from their larval stages to
adult form.
74What is common among silver
fish, scorpion, crab and honeybee?
[CBSE AIPMT 1997]
(a) Compound eyes
(b) Poison glands
(c) Jointed appendages
(d) Metamorphosis
Ans.(c)
The main characteristics of phylum—
Arthropoda are as follows
(i) Jointed appendages, present in
some or all somite or segments, but
often modified for specialised
functions like walking, clinging,
jumping, etc.
(ii) Bilateral symmetry.
(iii) Exoskeleton of cuticle.
(iv) Complex muscular system.
(v) Reduced coelom.
(vi) Complete digestive system.
(vii) Open circulatory system.
(viii) Respiration by body surface, gills,
trachea (air tubes), or book lungs.
(ix) Paired excretory glands called
coxal, antennal or maxillary glands
present in some, homologous to
metameric nephridial system of
annelids, some with other excretory
organs called Malpighian tubules.
(x) Nervous system with dorsal brain
connected by a ring around the
gullet to a double nerve chain of
ventral ganglia.
(xi) Sexes usually separate.
75A larval stage occurs in the life
history of all members of the
group [CBSE AIPMT 1993]
(a) frog, lizard and cockroach
(b)Ascaris, housefly and frog
(c) housefly, earthworm and mosquito
(d) butterfly, frog and mosquito
Ans.(d)
Lizard, cockroach,Ascaris,earthworm
shows direct development in their life
cycle. Whereas, butterfly have catterpillar,
frog have tadpole and mosquito have
wriggler larvae in their life cycle.
76AdultCulexandAnophelescan be
distinguished with the help of
[CBSE AIPMT 1992]
(a) mouth parts/colour
(b) sitting posture
(c) antennae/wings
(d) feeding habits
Animal Kingdom 39

50. Ans.(b)
The body ofAnophelesmosquito makes
an angle of 45° while sitting whereas
the body ofCulexmosquito lies parallel
to the surface.
77Male and female cockroaches can
be distinguished externally
through [CBSE AIPMT 1991]
(a) anal styles in male
(b) anal cerci in female
(c) anal style and antennae in females
(d) Both (b) and (c)
Ans.(a)
Anal style, a pair of small, spine–like
unjointed structures are present on
sternite of 9th segment in males only.
78Metamorphosis of insects is
regulated through hormone
[CBSE AIPMT 1991]
(a) pheromone (b) thyroxine
(c) ecdysone (d) All of these
Ans.(c)
Metamorphosis is regulated by
ecdysone hormone secreted by
prothoracic glands. This hormone was
isolated in a crystalline form in 1954 by
ButenandtandKarlson.
79An insect regarded as greatest
mechanical carrier of diseases is
[CBSE AIPMT 1991]
(a)Pediculus (b)Cimex
(c)Musca (d)Xenopsylla
Ans.(c)
Musca(house fly) is the carrier of many
disease as anthrax, trachoma,
diarrhoea, tuberculosis, leprosy,
gaugrene, plague, gonorrhoea, typhoid,
cholera and dysentery.
80Ecdysis is shedding of
[CBSE AIPMT 1990]
(a) stratum corneum
(b) epidermis
(c) dermis
(d) stratum malpighi
Ans.(a)
Ecdysis is the removal of outermost
partially cornified layer of stratum
corneum, which is then replaced by the
cells formed by stratum germinativum.
81Malpighian tubules are
[CBSE AIPMT 1990]
(a) excretory organs of insects
(b) excretory organs of annelids
(c) respiratory organs of insects
(d) respiratory organs of annelids
Ans.(a)
Malpighian tubules are the excretory
organs of insects. These are
unbranched tubules lying almost freely
in haemocoel and open into alimentary
canal. The main function of these
tubules is to absorb nitrogenous waste
product.
82Kala-azar and oriental sore are
spread by [CBSE AIPMT 1990]
(a) housefly (b) bed bug
(c) sand fly (d) fruit fly
Ans.(c)
Kala-azar and oriental sore both the
disease are caused byLeishmania.It is
digenetic and intermediate host is sand
fly belonging to the genusPhlebotomus.
Leishmania donovanicauseskala-azar
orvisceral leishmaniasiswhich is also
calleddum-dum fever, infection occurs
chiefly in spleen and liver, secondarily in
bone marrow and intestinal villi.
L. tropicacausesoriental soreor
cutaneous leishmaniasisin man.
83Silk thread is obtained from silk
moth during[CBSE AIPMT 1988]
(a) pupal stage (b) larval stage
(c) nymph stage (d) adult stage
Ans.(a)
Caterpillar feeds on mulberry leaves. Its
salivary gland secretes liquid silk. Silk is
obtained from (pupa, chrysalis). Ripe
cocoons are treated with boiling water
to kill the moth before hatching.
84Match the following.[NEET 2021]
List-I List-II
A.Physalia 1. Pearl oyster
B.Limulus 2. Portuguese Man
of War
C.Ancylostoma3. Living fossil
D.Pinctada 4. Hookworm
Choose the correct answer from
the options given below.
A B C D
(a) 2 3 1 4
(b) 4 1 3 2
(c) 2 3 4 1
(d) 1 4 3 2
Ans.(c)
(A)-(2), (B)-(3, (C)-(4), (D)-(1)
lPhysaliabelongs to
phylum-Coelenterata (Cnidaria) and is
commonly known as Portuguese
man-of-war.
lLimulusbelongs to
phylum-Arthropoda and is a living fossil
and commonly termed as king crab.
lAncylostomabelongs to
phylum-Aschelminthes and is
commonly referred to as hookworm.
lPinctadabelongs to phylum-Mollusca
and is commonly called pearl oyster.
85In which one of the following, the
genus name, its two characters
and its phylum are not correctly
matched, whereas the remaining
three are correct?
[CBSE AIPMT 2012]
Genus Name Two characters Phylum
(a)Pila (i) Body
segmented
Mollusca
(ii) Mouth with
Radula
(b)Asterias(i) Spiny skinnedEchinoder-
mata
(ii) Water
vascular
system
(c)Sycon (i) Pore bearing Porifera
(ii) Canal system
(d)Periplaneta(i) Jointed
appendages
Arthropo-da
(ii) Chitinous
exoskeleton
Ans.(a)
Option (a) is incorrect because molluscs
are bilaterally symmetrical,
triploblastic, coelomate, soft bodied
animals. Their soft body is covered by a
calcareous shell and is unsegmented
with a distinct head, muscular foot and
visceral hump.
e.g.Pila(apple snail),Sepia(cuttle fish),
Pinctada(pearl oyster), etc.
86Which one of the following is a
matching set of phylum and its
three examples?
[CBSE AIPMT 2006]
(a) Cnidaria —Bonellia, Physalia,
Aurelia
(b) Platyhelminthes —Planaria,
Schistosoma, Enterobius
(c) Mollusca —Loligo, Teredo, Octopus
(d) Porifera—Spongilla, Euplectella,
Pennatula
40 NEETChapterwise Topicwise Biology
Phylum-Mullusca
TOPIC 8

51. Ans.(c)
Loligo, TeredoandOctopusare the
members of phylum—Mollusca.
Loligois commonly called squid or sea
arrow and is gregarious, fast swimmer
in the open water of the sea and is
carnivorous, feeding on crabs and
fishes.
Octopus(devil fish) is found at the
bottom of the sea. It is nocturnal and
feeds on crabs, fishes and other
molluscs.
Teredoor shipworm is a marine bivalve
which has small anterior shell and long
slender body with a small foot
functioning as adhesive structure.
87Closed circulatory system occurs
in [CBSE AIPMT 1994]
(a) snail
(b) cockroach
(c) cuttle fish
(d) All of these
Ans.(c)
Closed circulatory system occurs in
cuttle fish, i.e.Sepia, cephalopods are
the only molluscs with a closed
circulatory system. They have two gill
hearts (also known as bronchial hearts)
that move blood through capillaries of
the gill. A single systemic heart pumps
the oxygenated blood through rest of
the body.
88Eye of the molluscan group that
resembles vertebrate eye is
[CBSE AIPMT 1992]
(a) Bivalvia (b) Gastropoda
(c) Pelecypoda (d) Cephalopoda
Ans.(d)
Cephalopoda contains the most
specialised molluscs including squids,
Octopus,cuttle fish. In these animals
well developed eyes are present which
resembles vertebrate eyes.
89A wood boring mollusc/shipworm
is [CBSE AIPMT 1988]
(a)Chiton
(b)Teredo
(c)Limax
(d)Patella
Ans.(b)
Teredocomes under class—Pelecypoda
(bivalvia). It is commonly known as
‘shipworm’. It is destructive to wood in
sea water.
90Read the following statements.
[NEET 2021]
I. Metagenesis is observed in
helminths.
II. Echinoderms are triploblastic
and coelomate animals.
III. Round worms have
organ-system level of body
organisation.
IV. Comb plates present in
ctenophores help in digestion.
V. Water vascular system is
characteristic of echinoderms.
Choose the correct answer from
the options given below.
(a) III, IV and V are correct
(b) I, II and III are correct
(c) I, IV and V are correct
(d) II, III and V are correct
Ans.(d)
Statements II, III and V are correct,
while statements I and IV are incorrect.
Incorrect statement can be corrected
as-
Comb plates are present in
ctenophores which help in locomotion
or swimming and not in digestion.
Metagenesis is the alternation of
generations between sexual and
asexual reproduction. In helminths
metagenesis is not observed.
91Match the following genera with
their respective phylum
[NEET (Odisha) 2019]
Column I Column II
1.Ophiura(i) Mollusca
2.Physalia(ii) Platyhelminthes
3.Pinctada(iii) Echinodermata
4.Planaria(iv) Coelenterata
Select the correct option from the
following
1 2 3 4
(a) (iv) (i) (iii) (ii)
(b) (iii) (iv) (i) (ii)
(c) (i) (iii) (iv) (ii)
(d) (iii) (iv) (ii) (i)
Ans.(b)
The correct matches are
1.Ophiura(Brittle star) (iii) Echinodermata
2.Physalia(Portuguese
man of war)
(iv) Coelenterata
3.Pinctada
(Pearl oyester)
(ii) Mollusca
4.Planaria(Flatworm) (ii) Platyhelminthes
92The animal with bilateral
symmetry in young stage and
radial pentamerous symmetry in
the adult stage belong to the
phylum [CBSE AIPMT 2004]
(a) Annelida (b) Mollusca
(c) Cnidaria (d) Echinodermata
Ans.(d)
Echinoderms are triploblastic animals
with organ system level of organisation.
Larval forms possess bilateral symmetry
while adults have radial symmetry.
93Radial symmetry is usually
associated with[CBSE AIPMT 1996]
(a) aquatic mode of life
(b) lower grade of organisation
(c) creeping mode of locomotion
(d) sedentary mode of life
Ans.(b)
In radial symmetry body is in the form of
flat or tall cylinder and can be divided
into similar halves by more than two
planes passing through one main axis. It
is found in some sponges, hydras,
jellyfish, sea urchin, etc.
94The organisms attached to the
substratum generally, possess
[CBSE AIPMT 1995]
(a) radial symmetry
(b) one single opening of digestive
canal
(c) asymmetrical body
(d) cilia on surface to create water
current
Ans.(a)
The organisms attached to the
substratum, generally possess radial
symmetry.
95Radial symmetry is often exhibited
by animals having
[CBSE AIPMT 1994]
(a) one opening of alimentary canal
(b) aquatic mode of living
(c) benthos/sedentary
(d) ciliary mode of feeding
Animal Kingdom 41
Phylum-Echinodermata
TOPIC 9

52. Ans.(b)
Radial animals are usually sessile, freely
floating or weakly swimming.
96Tube feet occur in
[CBSE AIPMT 1994]
(a) cockroach (b) starfish
(c) cuttle fish (d) cat fish
Ans.(b)
In starfish locomotion takes place by
externaltube feet, connected with
water vascular system.
97Aristotle’s lantern occurs in class
[CBSE AIPMT 1992]
(a) Echinoidea (b) Asteroidea
(c) Holothuroidea (d) Ophiuroidea
Ans.(a)
A biting or masticatory apparatus or
Aristotle’s lantern’is present in the
members of class—Echinoidea.
98Star fish belongs to
[CBSE AIPMT 1992]
(a) Asteriodea (b) Ophiuroidea
(c) Holothuroidea (d) Crinoidea
Ans.(a)
Starfish belongs to class—Asteroidea.
99Which one occurs in
Echinodermata?[CBSE AIPMT 1991]
(a) Bilateral symmetry
(b) Radial symmetry
(c) Porous body
(d) Soft skin
Ans.(b)
Echinoderms are triploblastic and
radially symmetrical but their larvae are
bilaterally symmetrical.
100All vertebrates are chordates, but
all chordates are not vertebrates,
why? [NEET (Oct.) 2020]
(a) Notochord is replaced by vertebral
column in adult of some
chordates
(b) Ventral hollow nerve cord remains
throughout life in some chordates
(c) All chordates possess vertebral
column
(d) All chordates possess notochord
throughout their life
Ans.(a)
The members of subphylum–Vertebrata
possess notochord during the
embryonic period. The notochord is
replaced by a cartilaginous or bony
vertebral column in the adult. Thus, all
vertebrates are chordates but, all
chordates are not vertebrates.
101Which of the following statements
are true for the phylum–Chordata?
[NEET (Sep.) 2020]
I. In Urochordata notochord
extends from head to tail and it
is present throughout their life.
II. In Vertebrata, notochord is
present during the embryonic
period only.
III. Central nervous system is dorsal
and hollow.
IV. Chordata is divided into 3
subphyla, Hemichordata,
Tunicata and Cephalochordata.
(a) III and I
(b) I and II
(c) II and III
(d) IV and III
Ans.(c)
Statement II and III are correct.
Statement I and IV are incorrect and
can be corrected as
In Urochordata, notochord is present
only in larval tail, while in
Cephalochordata, it extend from head
to tail region and is persistent
throughout their life.
Phylum–Chordata is divided into three
subphyla i.e. Urochordata or Tunicata,
Cephalochordata and Vertebrata.
102Consider the following features.
[NEET (National) 2019]
A. Organ system level of
organisation
B. Bilateral symmetry
C. True coelomates with
segmentation of body
Select the correct option of
animal groups which possess all
the above characteristics.
(a) Annelida, Arthropoda and Mollusca
(b) Arthropoda, Mollusca and
Chordata
(c) Annelida, Mollusca and Chordata
(d) Annelida, Arthropoda and
Chordata
Ans.(d)
All the three animal groups namely
Annelida, Arthropoda and Chordata
possess organ system level of
organisation, bilateral symmetry and
true coelom with segmented body.
Molluscans are also bilaterally
symmetrical and show organ system
grade of organisation but they do not
possess segmented body.
103An important characteristic that
hemichordates share with
chordates is [NEET 2017]
(a) absence of notochord
(b) ventral tubular nerve cord
(c) pharynx with gill slits
(d) pharynx without gill slits
Ans.(c)
The important characteristic that
hemichordates share with chordates is
pharynx with gillslits. These gillslits are
narrow openings in the pharynx. The
position of these pharyngeal gillslits is
lateral in chordates, while dorsal in
hemichordates.
104Choose the correct statement.
[NEET 2016, Phase II]
(a) All mammals are viviparous
(b) All cyclostomes do not possess
jaws and paired fins
(c) All reptiles have a
three-chambered heart
(d) All Pisces have gills covered by an
operculum
Ans.(b)
Cyclostomata is a class belonging to
section 1-Agnatha of subphylum–
Vertebrata.
It is a paraphytelic superclass of jawless
fishes. They lack paired fins too. Thus,
option (b) is correct.
105In which of the following notochord
is present in embryonic stage?
[CBSE AIPMT 2002]
(a) All chordates (b) Some chordates
(c) Vertebrates (d) Non-chordates
Ans.(a)
Notochord is the primary axial
supportive structure present in all
chordate embryos as well as in many
adults. In a vast majority, it is replaced
by vertical column in adults.
Vertebrates come under the category
of chordate. Non-chordates do not
contain notochord.
42 NEETChapterwise Topicwise Biology
Phylum-Chordata
TOPIC 10

53. 106Besides Annelida and Arthropoda,
the metamerism is exhibited by
[CBSE AIPMT 1995]
(a) Cestoda
(b) Chordata
(c) Mollusca
(d) Acanthocephala
Ans.(b)
Chordates also exhibit metameric
segmentation. This is because of
repetition of homologous structures
along the length of an animal.
107A common characteristic of all
vertebrates is[CBSE AIPMT 1994]
(a) presence of skull
(b) division of body into head, neck,
trunk and tail
(c) presence of two pairs of
functional appendages
(d) body is covered with an
exoskeleton
Ans.(a)
Vertebrates have well developed
cranium (skull) hence, they are also
called as Craniata.
108All chordates possess
[CBSE AIPMT 1994]
(a) exoskeleton
(b) limbs
(c) skull
(d) axial skeletal rod of notochord
Ans.(d)
Phylum—Chordata was created by
Balfourin 1880. This refers to the
presence of a shoft, supporting rod-like
structure along the back called
‘notochord’.
109All vertebrates possess
[CBSE AIPMT 1993]
(a) renal portal system
(b) dorsal, hollow, central nervous
system
(c) four chambered ventral heart
(d) pharyngeal gill slits
Ans.(b)
Dorsal, hollow, central nervous system
is present in all vertebrates.
110Penguin occurs in
[CBSE AIPMT 1990]
(a) Australia
(b) Antarctica
(c) Africa
(d) America
Ans.(b)
Penguins are found in Antarctica (South
pole). They have paddle like wings and
cannot fly. Penguins are marine and lay
eggs in ice.
111A chordate character is
[CBSE AIPMT 1989]
(a) gills
(b) spiracles
(c) post-anal tail
(d) chitinous exoskeleton
Ans.(c)
Presence of post-anal tail is one of the
characters of chordate.
112Necturusis [CBSE AIPMT 1988]
(a) hell bender
(b) congo eel
(c) mud puppy
(d) blind worm
Ans.(c)
Common name ofNecturusis mud
puppy. This comes under
order–Urodela.
113Typhlosis [CBSE AIPMT 1988]
(a) sea snake (b) glass snake
(c) blind snake (d) grass snake
Ans.(c)
Typhlosis blind snake and it is
non-poisonous.
114Match the following columns and
select the correct option from the
codes given belows.
[NEET (Oct.) 2020]
Column I Column II
A.Aptenodytes1. Flying fox
B.Pteropus 2. Angel fish
C.Pterophyllum3. Lamprey
D.Petromyzon 4. Penguin
Codes
A B C D
(a) 3 4 2 1
(b) 3 4 1 2
(c) 4 1 2 3
(d) 2 1 4 3
Ans.(c)
The option (c) is the correct match
which is as follows
Aptenodytesis penguin
Pteropusis flying fox
Pterophyllumis angel fish
Petromyzonis lamprey
115Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. 6-15 pairs of gill
slits
1.Trygon
B. Heterocercal
caudal fin
2. Cyclostomes
C. Air bladder 3. Chondrichthyes
D. Poison sting 4. Osteichthyes
A B C D
(a) 3 4 1 2
(b) 4 2 3 1
(c) 1 4 3 2
(d) 2 3 4 1
Ans.(d)
Option (d) is correct. It can be explained
as follows
Cyclostomes have an elongated body
bearing 6-15 pairs of gill slits for
respiration. Air bladder is present in
bony fishes belonging to
class-Osteichthyes which regulates
buoyancy.
Trygon, a cartilaginous fish, possesses
poison sting.
Heterocercal caudal fin is present in
members of class-Chondrichthyes.
116Which of the following
characteristic features always
holds true for the corresponding
group of animals?
[NEET 2016, Phase I]
(a) Viviparous Mammalia
(b) Possess a mouth
with an upper and a
lower jaw
Chordata
(c) 3-chambered heart
with one
incompletely
divided ventricle
Reptilia
(d) Cartilaginous – Chondrichthyes
endoskeleton
Ans.(d)
Reptiles have 3-chambered heart
except crocodiles. Mammals are
viviparous except prototherian
mammals; chordates have jaws except
protochordates and cyclostomes.
Animal Kingdom 43
Super-Class-Pisces
TOPIC 11

54. 117A jawless fish, which lays eggs in
fresh water and whose
ammocoetes larvae after
metamorphosis return to the
ocean is [CBSE AIPMT 2015]
(a)Eptatretus(b)Myxine
(c)Neomyxine (d)Petromyzon
Ans.(d)
Petromyzon(the lamprey) belongs to the
section Agnatha of the
subphylum—Vertebrata. They have long,
greenish brown, cylindrical body with
smooth scaleless, slimy skin, jawless
mouth, etc. They lay eggs in freshwater,
but their ammocoete larvae (lower)
after metamorphosis return to the
ocean.
118A marine cartilaginous fish that can
produce electric current is
[CBSE AIPMT 2014]
(a)Pristis (b)Torpedo
(c)Trygon (d)Scoliodon
Ans.(b)
Torpedois a marine cartilaginous fish
which produces 8-220 volt electric
charge (current) depending on species.
Their electric organs are modified
lateral muscle plates innervated by
cranial nerves.
Trygon(sting ray) resembles electric ray
in many aspects but is devoid of
electricity discharging (or producing)
organs.
Scoliodon(dog fish) is known for its
great sense of smell.
Pristisor common saw fish (also known
as carpenters shark) is characterised by
a long, narrow, flattened rostrum lined
with sharp transverse teeth to
resembles a saw.
119Match the name of the animal
(column I) with one characteristics
(column II) and the phylum/class
(column III) to which it belongs
[NEET 2013]
Column IColumn II Column III
(a)PetromyzonEctoparasite Cyclostomata
(b)IchthyophisTerrestrial Reptilia
(c)LimulusBody
covered by
chitinous
exoskeleton
Pisces
(d)AdamsiaRadially
symmetrical
Porifera
Ans.(a)
Option (a) is correctly matched as
Petromyzon(lamprey) is an ectoparasite
on fishes, which belongs to
Cyclostomata. Other can be corrected
asIchthyophisis a limbless amphibian.
Limulus(king crab) is a living fossil,
which belongs to Arthropoda.Adamasia
having polyp body form is a
coelenterate.
120Which one of the following groups
of animals is correctly matched
with its one characteristic feature
without even a single exception?
[CBSE AIPMT 2011]
(a) Chordata – Possess a mouth
provided with an
upper and a lower
jaw
(b) Chondrichthyes –Possess
cartilaginous
endoskeleton
(c) Mammalia – Give birth to young
ones
(d) Reptilia – Possess
3-chambered heart
with one
incompletely
divided ventricle
Ans.(b)
The members of class–Chondrichthyes
are marine animals with streamlined
body and have cartilaginous
endoskeleton.
ChordataThese possessing a
notochord, a hollow dorsal nerve cord,
pharyngeal slits, an endostyle and a
post and tail for atleast some period of
their life cycle.
MammaliaThese are a clade of
endothermic amniotes distinguished
from reptiles and birds by the
possession of hair, three middle ear
bones, mammary glands, and a
neocortex.
ReptileThese are an evolutionary
clade of animals, comprising today’s
turtles, crocodilians, snakes, lizards and
tuatara, their extinct relatives and some
of the extinct ancestors of mammals.
121What will you look for to identify
the sex of the following?
[CBSE AIPMT 2011]
(a) Male frog – A copulatory pad
on the first digit
of the hindlimb
(b) Female
cockroach
–Anal cerci
(c) Male shark – Claspers borne
on pelvic fins
(d) Female
Ascaris
– Sharply curved
posterior end
Ans.(c)
A clasper is a male anatomical structure
found in some groups of animals and
used in mating. Male cartilaginous fish
like shark have claspers formed from
the posterior portion of their pelvic fin
which serves as intermittent organs
used to channel semen into the
female’s cloaca during mating.
122Which one of the following pairs of
animal comprises ‘jawless fishes’?
[CBSE AIPMT 2009]
(a) Lampreys and eels
(b) Mackerals and rohu
(c) Lampreys and hag fishes
(d) Guppies and hag fishes
Ans.(c)
Lampreys andMyxine(hag fish) belong
to the class—Cyclostomata,
group–Agnatha of Vertebrata. Agnatha
have mouth without jaws. In these,
mouth is ventral, suctorial and circular.
123Which one of the following is an
exotic Indian fish?
[CBSE AIPMT 1996]
(a)Catla catla
(b)Heteropneustes fossilis
(c)Cyprinus caprio
(d)Labeo rohita
Ans.(c)
Cyprinus capriois the exotic breed. It is
also known as common carp. It is a
widespread fresh water fish of
eutrophic water lakes and rivers in
Europe and Asia.
124Fish which can be used in biological
control of mosquitoes/larvicidal
fish is[CBSE AIPMT 1989, 99, 2001]
(a) eel
(b) carp
(c) cat fish
d)Gambusia
Ans.(d)
Gambusia(mosquito fish) eats the
larvae of mosquito, so it is used in
biological controlof mosquito.
44 NEETChapterwise Topicwise Biology

55. 125In which one of the following the
genus name, its two characters and
its class/phylum are correctly
matched? [CBSE AIPMT 2011]
Genus Two characters
Class/
phylum
(a)Salamandra(i) A tympanum
represents
ear
Amphibia
(ii) Fertilisation
is external
(b)Pteropus(i) Skin
possesses
hair
Mammalia
(ii) Oviparous
(c)Aurelia(i) Cnidoblast Coelenter
ata
(ii) Organ level
of
organisation
(d)Ascaris(i) Body
segmented
Annelida
(ii) Males and
females
distinct
Ans.(a)
Salamandra(salamander) is a member
of class–Amphibia. It has a tympanum
representing the ear and fertilisation is
external.
126The presence of gills in the
tadpole of frog indicates that
[CBSE AIPMT 2004]
(a) fishes were amphibious in the past
(b) fishes evolved from frog like
ancestors
(c) frogs will have gills in future
(d) frogs evolved from gilled
ancestors
Ans.(d)
According tobiogenetic law of Ernst
Haeckel(1860)Ontogeny repeats
Phylogeny.Ontogeny is the life history
of an organism, while phylogeny is the
evolutionary history of the race of that
organism. In other words we can say ‘an
organism repeats its ancestral history
during its development’. Hence,
resemblance of Amphibia to fish is seen
in most systems of the body, both are
cold-blooded, both respire by gills (as
tadpole of frog), both usually lay eggs in
water leading to the conclusion that
amphibians have originated from fishes.
127Mucus helps frog in forming
[CBSE AIPMT 1993]
(a) thick skin (b) dry skin
(c) smooth skin (d) moist skin
Ans.(d)
Mucus helps frog in keeping the skin
moist that helps in cutaneous
respiration when the frog is in
hibernation or estivation.
128Bull frog of India is
[CBSE AIPMT 1992]
(a)Rana tigrina
(b)R. sylvatica
(c)R. ecutesbeiana
(d)R. esculenta
Ans.(a)
Indian bull frog isRana tigrina.
129Skin is a respiratory organ in
[CBSE AIPMT 1990]
(a) lizards
(b) birds
(c) primitive mammals
(d) frog
Ans.(d)
Frog is an amphibian and its skin is well
adapted or in other words acts as a
secondary respiratory organ when it is
in water.
130Which is not a true amphibian
animal? [CBSE AIPMT 1988]
(a) Salamander
(b) Toad
(c) Tortoise
(d) Frog
Ans.(c)
Tortoise is a reptile of
sub–class—Anapsidaand
order—Chetomia or Testudinata.
131Fire bellied toad is
[CBSE AIPMT 1988]
(a)Amphiuma
(b)Bombinator
(c)Necturus
(d)Salamandra
Ans.(b)
Fire bellied toad isBombinatorand this
is an Anuran.
132Which of the following pairs are
correctly matched?
[CBSE AIPMT 2007]
Animals Morphological features
(i) Crocodile — Four-chambered
heart
(ii) Sea urchin — Parapodia
(iii) Obelia — Metagenesis
(iv) Lemur — Thecodont
(a) (i), (iii) and (iv)
(b) (ii), (iii) and (iv)
(c) Only (i) and (iv)
(d) Only (i) and (ii)
Ans.(a)
Exceptionally, crocodile has
four-chambered heart. InObeliathe
alternation of generation is called
metagenesis in which an asexual
polypoid generation appears to
alternate regularly with a sexual
medusoid generation.
Thecodont dentition is found in
mammals.
133In which of the following animal,
post-anal tail is found?
[CBSE AIPMT 2001]
(a) Earthworm
(b) Lower invertebrate
(c) Scorpion
(d) Snake
Ans.(d)
In snakes, post-anal tail is found.
Snakes belong to class—Reptilia.
Jurassic period (Mesozoic era) is known
as Golden age of reptiles. The study of
snake is known as serpentology.
134Which one of the following
organisms bears hollow and
pneumatic long bones?
[NEET 2021]
(a)Neophron
(b)Hemidactylus
(c)Macropus
(d)Ornithorhynchus
Animal Kingdom 45
Class-Reptilia
TOPIC 13
Class-Aves
TOPIC 14
Class-Amphibia
TOPIC 12

56. Ans.(a)
Pneumatic bones are hollow bones
found in birds, which enables them to
fly.Neophronis a bird.
Other options are incorrect because
lHemidactylusis a reptile.
lMacropusis a mammal.
lOrnithorhynchusis a mammal.
135Match the following (Columns)
group of organisms with their
respective distinctive
characteristics and select the
correct option from the codes
given belows[NEET (Oct.) 2020]
Column-I
(Organisms)
Column-II
(Characteristics)
A. Platyhelminthes 1.Cylindrical body with
no segmentation
B. Echinoderms 2.Warm blooded
animals with direct
development
C. Hemichordates 3.Bilateral symmetry
with incomplete
digestive system
D. Aves 4. Radial symmetry
with indirect
development
Codes
A B C D
(a) 3 4 1 2
(b) 2 3 4 1
(c) 4 1 2 3
(d) 1 2 3 4
Ans.(a)
The option (a) is the correct match
which is as follows
Platyhelminthes are bilaterally
symmetrical with incomplete digestive
system, e.g.Taenia.
Echinoderms are radially symmetrical
with indirect development, e.g. star
fish, sea urchin etc.
lHemichordates are cylindrical bodied
animal with no segmentation, e.g.
Balanoglossus.Aves are
warm-blooded animals with direct
development like pigeon.
136Which one of the following in
birds, indicates their reptilian
ancestry? [CBSE AIPMT 2008]
(a) Scales on their hindlimbs
(b) Four chambered heart
(c) Two special chambers crop and
gizzard in their digestive tract
(d) Eggs with a calcareous shell
Ans.(d)
Nearly a century agoTH Huxleycalled
birds ‘glorified reptiles’ thereby meaning
that birds have evolved from some
reptilian ancestor. Both birds and
reptiles lay the same type of eggs,
which are deposited outside water.
Eggs are large and telolecithal. The
ovum is surrounded by albumen, an egg
membrane and a thick hard calcareous
shell, which are all secreted by special
glands located in the walls of oviduct.
Like mammals birds also have complete
four chambered heart with double
circulation, in which there is no mixing
of pure and impure bloods. Whereas,
the ventricle is imperfectly divided in
reptiles, resulting in partial mixing of
blood.
137Which of the following is not found
in birds? [CBSE AIPMT 1999]
(a) Hind limb (b) Pectoral girdle
(c) Pelvic girdle (d) Fore limb
Ans.(d)
Fore limbs of birds are modified into
wings.
138The long bones are hollow and
connected by air passage. They are
the characteristics of
[CBSE AIPMT 1998]
(a) Aves
(b) mammals
(c) Reptilia
(d) land vertebrates
Ans.(a)
The bones of birds are pneumatic, (i.e.
they have air cavities) to reduce weight
which help them in flying.
139Feet of kingfisher are modified for
[CBSE AIPMT 1988]
(a) wading (b) perching
(c) running (d) catching
Ans.(a)
Kingfisher is fish-eating bird and its
feet are modified for wading. It is
grouped into order–Coraciifermers,
which comes in family– Alcedinidae.
140Bird vertebrae are
[CBSE AIPMT 1988]
(a) acoelous (b) heterocoelous
(c) amphicoelous (d) procoelous
Ans.(b)
Bird vertebrae are heterocoelous or
saddle shaped.
141Pneumatic bones are expected to
be found in[CBSE AIPMT 1996]
(a) pigeon (b) house lizard
(c) frog’s tadpole (d) flying fish
Ans.(a)
In birds like pigeon bones are
pneumatic or hollow and have no
marrow, thus helps in reducing the body
weight.
142The flightless bird cassowary is
found in [CBSE AIPMT 1996]
(a) Mauritius (b) Australia
(c) New Zealand (d) Indonesia
Ans.(b)
Cassowary (Casuarius) is found in
Australia and New Guinea.
143What is common between ostrich,
penguin and kiwi?
[CBSE AIPMT 1993]
(a) Running birds (b) Migratory birds
(c) Flightless birds (d) Four toed birds
Ans.(c)
Ostrich (Struthio), penguin (Aptenodytes)
and kiwi (Apteryx) all are flightless birds.
144Sound box of birds is called
[CBSE AIPMT 1992]
(a) pygostyle (b) larynx
(c) syrinx (d) synsacrum
Ans.(c)
Sound box of birds is called syrinx, this
produces voice. It lies at or near
junction of trachea and bronchi.
145Flight muscles of bird are attached
to [CBSE AIPMT 1989]
(a) clavicle (b) keel of sternum
(c) scapula (d) coracoid
Ans.(b)
Sternum of birds is large, usually with a
vertical, midneutralkeelwhich provides
attachment of flight muscles.
146Wish bone of birds is formed from
[CBSE AIPMT 1989]
(a) pelvic girdle
(b) skull
(c) hindlimbs
(d) pectoral girdle/clavicles
Ans.(d)
Both clavicles and a single interclavicle
fused to form a V-shaped bone called
furculaorwish boneormerry-thought
bone.
46 NEETChapterwise Topicwise Biology

57. 147Both male and female pigeons
secrete milk, through
[CBSE AIPMT 1988]
(a) salivary glands
(b) modified sweat glands
(c) crop
(d) gizzard
Ans.(c)
Both male and female pigeon secrete
milk through crop during breeding
season.
148Which among these is the correct
combination of aquatic mammals?
[NEET 2017]
(a) Seals, Dolphins, Sharks
(b) Dolphins, Seals,Trygon
(c) Whales, Dolphins, Seals
(d)Trygon, Whales, Seals
Ans.(c)
Among the given options, option (c)
contains all aquatic mammals. Whales
are inhabitants of the open sea, while
seal (Phoca) is a marine carnivore.
Dolphins are found in rivers.Trygonand
sharks are fishes, which belong to
chondrichthyes class of
superclass–Pisces.
149Which one of the following
characteristics is not shared by
birds and mammals?
[NEET 2016, Phase I]
(a) Breathing using lungs
(b) Viviparity
(c) Warm blooded nature
(d) Ossified endoskeleton
Ans.(b)
Mammals are viviparous while birds are
oviparous. Viviparous means giving
birth to offspring that develops within
the mother’s body. Oviparous means
producing eggs that hatch outside the
body of mother.
150Compared to those of humans, the
erythrocytes in frog are
[CBSE AIPMT 2012]
(a) without nucleus but with
haemoglobin
(b) nucleated and with haemoglobin
(c) very much smaller and fewer
(d) nucleated and without
haemoglobin
Ans.(b)
RBCs or erythrocytes of frog are oval,
disc–like biconvex, have centrally
placed nucleus and with haemoglobin.
Usually in mammals (including human),
RBCs are circular and non-nucleated
except those of family–Camilladae
(camels, llamas, etc), which have
nucleated RBCs.
151What is common between parrot,
platypus and kangaroo?
[CBSE AIPMT 2007]
(a) Homeothermy
(b) Toothless jaws
(c) Functional post-anal tail
(d) Oviparity
Ans.(a)
Parrot (birds), platypus and kangaroo
(both mammal) are homeothermic
animals.
152In which one of the following sets of
animals do all the four give birth to
young ones?[CBSE AIPMT 2006]
(a) Lion, bat, whale, ostrich
(b) Platypus, penguin, bat,
hippopotamus
(c) Shrew, bat, cat, kiwi
(d) Kangaroo, hedgehog, dolphin, loris
Ans.(d)
Kangaroo, hedgehog, dolphin and loris
are all mammals. These give birth to
young ones.
153Which one of the following
characters is not typical of the
class–Mammalia?
[CBSE AIPMT 2005]
(a) Seven cervical vertebrae
(b) Thecodont dentition
(c) Ten pairs of cranial nerves
(d) Alveolar lungs
Ans.(c)
Ten pairs of cranial nerves are not
common in mammals but common in
frog. In mammals 12 pairs of cranial
nerves are found.
154A terrestrial animal must be able
to [CBSE AIPMT 2004]
(a) excrete large amounts of water in
urine
(b) conserve water
(c) actively pump salts out through
the skin
(d) excrete large amounts of salts in
urine
Ans.(b)
Since, terrestrial animals do not have
automatic access to either fresh or salt
water, they must regulate water
content in other ways, balancing off
gains and losses.
155Given below are four matchings of
an animal and its kind of respiratory
organ [CBSE AIPMT 2003]
(i) silver fish — trachea
(ii) scorpion — book lung
(iii) sea squirt — pharyngeal gills
(iv) dolphin — skin
The correct matchings are
(a) (ii) and (iv) (b) (iii) and (iv)
(c) (i) and (iv) (d) (i), (ii) and (iii)
Ans.(d)
Dolphins are mammals which respire by
lungs, which are situated next to the
heart in the thorax cavity.
156Trachea of cockroach and
mammal are similar in having
[CBSE AIPMT 1993]
(a) paired nature
(b) non-collapsible walls
(c) ciliated inner lining
(d) origin from head
Ans.(b)
The trachea of cockroach and
mammals is lined with spiral thickening
of cuticle called intina which prevents
the tracheal tubes from collapsing.
157What is common in whale, bat and
rat?[CBSE AIPMT 1993, 2000, 04]
(a) Absence of neck
(b) Muscular diaphragm between
thorax and abdomen
(c) Extra-abdominal testes to avoid
high temperature of body
(d) Presence of external ears
Ans.(b)
The presence of diaphragm is a
characteristic feature of mammals,
diaphragm acts as a separater between
thorax and abdomen. Whale, bat and rat
all are mammals and have diaphragm.
158Gorilla, chimpanzee, monkeys and
human belong to the same
[CBSE AIPMT 1993]
(a) species
(b) genus
(c) family
(d) order
Animal Kingdom 47
Class-Mammalia
TOPIC 15

58. Ans.(d)
Gorilla, chimpanzee, monkeys and
humans these all belong to same
orderPrimata of class—Mammalia.
159The cervical vertebrae in humans is
[CBSE AIPMT 1993]
(a) same as in whale
(b) more than that in rabbit
(c) double than that of horse
(d) less than that in giraffe
Ans.(a)
The number of cervical vertebrae is
constant, i.e. 7 in number in all
mammals except sea cow and sloth. So,
the cervical vertebrae in humans is
same as in whale.
160An egg laying mammal is
[CBSE AIPMT 1992, 2000]
(a) kangaroo (b) platypus
(c) koala (d) whale
Ans.(b)
Egg laying mammals are primitive
mammals classified under
sub-class–Prototheria. These are
oviparous, reptile-like mammals,
confined to Australian region.
Ornithorhynchus,(duck billed platypus)
is found is Australia and New Zealand is
an egg laying mammal.
161Kidney of adult rabbit is
[CBSE AIPMT 1991]
(a) pronephros
(b) metanephros
(c) mesonephros
(d) opisthonephros
Ans.(b)
Kidney of adult rabbit is metanephros
which is divided into cortex and medulla
and its duct is useless. This is most
advanced type of kidney found in man
and rabbit.
162Eutherians are characterised by
[CBSE AIPMT 1989]
(a) hairy skin
(b) true placentation
(c) ovoviviparity (d) glandular skin
Ans.(b)
Infraclass—Eutheria includes viviparous
placental mammals. They give birth to
young ones, which gets nourishment
through placenta inside the body of its
mother.
163Hair occur in all mammals except
those of [CBSE AIPMT 1988]
(a) Rodentia (b) Chiroptera
(c) Primata (d) Cetacea
Ans.(d)
Order—Cetacea of mammals consists of
aquatic mammals in which hairs are
present only on snout, e.g. porpoise,
killer whale, dolphin, blue whale, sperm
whale, etc.
48 NEETChapterwise Topicwise Biology

59. 01Identify the incorrect statement.
[NEET (Sep.) 2020]
(a) Sapwood is involved in the
conduction of water and minerals
from root to leaf
(b) Sapwood is the innermost
secondary xylem and is lighter in
colour
(c) Due to deposition of tannins,
resins. oils, etc., heartwood is
dark in colour
(d) Heartwood does not conduct
water but gives mechanical
support
Ans.(b)
Statement in option (b) is incorrect and
can be corrected as :
Sapwood is outermost secondary xylem
and is lighter in colour. There cell walls
are not lignified and there is no
deposition of organic compounds.
Sapwood is involved in the conduction
of water and minerals from root to leaf.
02The roots that originate from the
base of the stem are
[NEET (Sep.) 2020]
(a) primary roots
(b) prop roots
(c) lateral roots
(d) fibrous roots
Ans.(d)
The roots that originates from the base
of the stem are fibrous roots. In fibrous
root system, the primary roots stop
growing and becomes rudimentary,
while new roots for anchorage and
absorption develop from the base of
the stem. This type of root formation is
observed in monocot like wheat plant.
03Pneumatophores occur in
[NEET 2018]
(a) carnivorous plants
(b) free-floating hydrophytes
(c) halophytes
(d) submerged hydrophytes
Ans.(c)
Pneumatophores are breathing or
respiratory roots which are found in
halophyteslike mangroves. Halophytes
grow in saline swamps, therefore
respiratory roots come out of water and
pick up oxygen for respiration. Excess
CO
2
is also given out. It occurs through
small pores, called lenticles.
Carnivorous plants, free-floating
hydrophytes and submerged
hydrophytes do not possess
pneumatophores.
04Plants, which produce
characteristic pneumatophores
and show vivipary belong to
[NEET 2017]
(a) mesophytes
(b) halophytes
(c) psammophytes
(d) hydrophytes
Ans.(b)
Plants that produce pneumatophores,
i.e. negatively geotropic roots and show
vivipary, i.e. germination of seeds inside
the fruits are halophytes. These plants
are adapted to grow in highly saline
areas such as mangroves.
Pneumatophores help these plants in
respiration as they do not get sufficient
oxygen from the soil. On the other hand
vivipary aids in perennation.
05Roots play insignificant role in
absorption of water in
[CBSE AIPMT 2015]
(a) sunflower (b)Pistia
(c) pea (d) wheat
Ans.(b)
Pistiais a hydrophyte where absorption
of water by root is insignificant.
06Roots of which plant contains an
oxidising agent?[CBSE AIPMT 2001]
(a) Carrot (b) Soyabean
(c) Mustard (d) Radish
Ans.(b)
Leghaemoglobinis present in roots of
soyabean plants. It is an oxygen-binding
haem protein, which is present in the
cytoplasm of infected nodule cells at
high concentration and gives the
nodules a pink colour. It’s main function
is to help in the transport ofO
2
to the
respiring symbiotic bacterial cells in a
manner analogous to haemoglobin,
which transportsO
2
to respiring tissues
in animals.
07The plant, which bears clinging
roots, is [CBSE AIPMT 1999]
(a)Trapa
(b) orchid
(c) screw pine
(d)Podostemon
Morphologyof
FloweringPlants
05
Root
TOPIC 1

60. Ans.(b)
Orchids have clinging roots that are
modified adventitious roots meant for
providing mechanical support. These
arise from the axils of leaves or nodes
of the stem and pierce the substratum
plant to facilitate fixation.
08The plant which bears clinging
roots is
[CBSE AIPMT 1999]
(a)Podostemon(b) orchid
(c)Trapa (d) Screwpine
Ans.(b)
Clinging roots are modified adventitious
roots meant for providing mechanical
support. These arise from the axils of
leaves or nodes of the stem and pierce
the substratum plant to facilitate
fixation e.g. orchids, ivy.
09Buttress roots are found in
[CBSE AIPMT 1995]
(a)Sorghum (b) Banyan
(c)Terminalia(d)Pandanus
Ans.(c)
Buttress rootsrefer to irregular,
broad–like the wood plants that arise
from basal parts of main stem of older
plants, and spread in different directions
in the soil, e.g.Ficus religiosa(peepal),
Terminalia, Bombax, etc. These roots
aids in providing mechanical support to
the trees.
10Velamen is found in
[CBSE AIPMT 1991]
(a) roots of screwpine
(b) aerial and terrestrial roots of
orchids
(c) leaves ofFicus elastica
(d) aerial roots of orchids
Ans.(d)
The aerial roots of orchids, (e.g.Vanda)
are surrounded by a spongy tissue,
(velamen) which is hygroscopic and
absorbs moisture from the surrounding
air.
11InBougainvilleathorns are the
modifications of [NEET 2017]
(a) stipules
(b) adventitious root
(c) stem
(d) leaf
Ans.(c)
InBougainvillea, thorns are the
modifications of stem. They are stiff,
sharp structures, which have lost their
growing point and become hard. They
reduce transpiration as well as
browsing by animals.
12Stems modified into flat green
organs performing the functions
of leaves are known as
[NEET 2016, Phase I]
(a) phyllodes (b) phylloclades
(c) scales (d) cladodes
Ans.(b)
Phylloclades are aerial modified stem,
in which stem becomes thick, fleshy
succulent, green and perform the
function of photosynthesis. The leaves
are reduced to spines in this.
13Which of the following is not a
stem modification?
[NEET 2016, Phase I]
(a) Thorns ofCitrus
(b) Tendrils of cucumber
(c) Flattened structures ofOpuntia
(d) Pitcher ofNepenthes
Ans.(d)
Pitcher ofNepenthesis modified leaf. It
helps to trap insects, in insectivorous
plants.
14An example of edible underground
stem is [CBSE AIPMT 2014]
(a) carrot (b) groundnut
(c) sweet potato (d) potato
Ans.(d)
Potato (Solanum tuberosum) is an edible
underground stem which become
fleshy and tuberous as a result of food
storage.
15The ‘eyes’ of the potato tuber are
[CBSE AIPMT 2011, 01]
(a) flower buds (b) shoot buds
(c) axillary buds (d) root buds
Ans.(c)
Eyes of potato tubers are axillary buds.
As tuber is oval or spherical swollen
underground modified stem lacking
adventitious roots.
It possesses a number of spirally
arranged depressions called eyes. Each
eye represents node and consists of 1-3
axillary buds in the axil of small scally
leaves.
16Vegetative propagation in mint
occurs by [CBSE AIPMT 2009]
(a) runner (b) offset
(c) rhizome (d) sucker
Ans.(d)
The vegetative propagation in mint
(Mentha arvensis) occurs by sucker.
Sucker is a non-green underground
specialised stem developing from the
underground base of an erect shoot or
crown.
17Vegetative reproduction ofAgave
occurs through[CBSE AIPMT 1991]
(a) rhizome (b) stolon
(c) bulbils (d) sucker
Ans.(c)
Bulbil is a condensed axillary bud meant
for vegetative propagation, e.g.
DioscoreaandAgave. In ananas, bulbils
are associated with flowers and fruits.
Tiny secondary bulb that forms in the
angle between a leaf and stem or in
place of flowers. in certain plants.
Bulbills are called offsets when
full-sized, fall or are removed and
planted to produce new plants.
18New banana plants develop from
[CBSE AIPMT 1990]
(a) rhizome (b) sucker
(c) stolon (d) seed
Ans.(b)
Sucker is the subaerial modification of
stem, which originates from the basal
and underground portion of the main
stem but it grows obliquely upwards
and give rise to leafy shoot or a new
plant, e.g.Musa(banana),
Chrysanthemum, mint.
19Large, empty colourless cells of
the adaxial epidermis along the
veins of grass leaves are
[NEET (Oct.) 2020]
(a) lenticels
(b) guard cells
(c) bundle sheath cells
(d) bulliform cells
Ans.(d)
Grass leaves are isobilateral or typical
monocot leaves. In such leaves, at
places, the upper or adaxial epidermis
contain groups of larger, thin-walled,
50 NEETChapterwise Topicwise Biology
Stem
TOPIC 2
Leaf
TOPIC 3

61. colourless cells called bulliform cells.
These cells are highly vacuolated and
can store water, if available. However,
these cells lose water and become
flaccid in case of water deficiency.
20Which of the following shows
whorled phyllotaxy?
[NEET (Odisha) 2019]
(a) Mustard (b) China rose
(c)Alstonia (d)Calotropis
Ans.(c)
Alstoniashows whorled phyllotaxy. In
whorled phyllotaxy, more than two
leaves arise at a node and form a whorl.
Mustard and China rose show alternate
phyllotaxy.Calotropisshows opposite
phyllotaxy.
21Phyllode is present in
(a)Asparagus [CBSE AIPMT 2012]
(b)Euphorbia
(c) AustralianAcacia
(d)Opuntia
Ans.(c)
Modified petiole is calledphyllode
present inAustralian Acacia. Phyllodes
develop usually vertically and possess
fewer stomata hence, reduce
transpiration.
22Choose the correct match.
Bladderwort, Sundew, Venus fly
trap [CBSE AIPMT 2002]
(a)Nepenthes,Dionea,Drosera
(b)Nepenthes,Utricularia,Vanda
(c)Utricularia,Drosera,Dionea
(d)Dionea,Trapa,Vanda
Ans.(c)
Utricularia, DroseraandDionaeaare
also known as bladderwort, sundew and
venus fly trap, respectively.
23Hypanthodium is
[CBSE AIPMT 1994]
(a) thalamus (b) fruit
(c) inflorescence (d) ovary
Ans.(c)
Hypanthodium is special type of
inflorescence in which the axis
becomes fleshy and pear-shaped with a
hollow cavity inside. At the base lie
female flowers and towards the apical
opening, lie the male flowers, e.g.Ficus.
24A family delimited by type of
inflorescence is[CBSE AIPMT 1990]
(a) Fabaceae (b) Asteraceae
(c) Solanaceae (d) Liliaceae
Ans.(b)
Asteraceae (Compositae) is
characterised by the inflorescence
head or capitulum, consisting of a few
or large number of flowers closely
arranged on an axis surrounded by
involucral bracts.
The whole head or capitulum with
racemose arrangement is apparently
similar to a single flower.
25Correct position of floral parts
over thalamus in mustard plant is
[NEET (Oct.) 2020]
(a) gynoecium occupies the highest
position, while the other parts are
situated below it
(b) margin of the thalamus grows
upward, enclosing the ovary
completely and other parts arise
below the ovary
(c) gynoecium is present in the
centre and other parts cover it
partially
(d) gynoecium is situated in the
centre and other parts of the
flower are located at the rim of
the thalamus, at the same level
Ans.(a)
Option (a) is correct. It can be explained
as
A flower, has four different whorls
arranged successively on the swollen
end of the stalk or pedicel called
thalamus or receptacle. In mustard,
flower is hypogymous,i.e.,the
gynoecium occupies the highest
position while the other parts are
situated below it. The ovary in such
flowers is said to be superior.
26Ray florets have[NEET (Sep.) 2020]
(a) superior ovary
(b) hypogynous ovary
(c) half inferior ovary
(d) inferior ovary
Ans.(d)
Ray florets have inferior ovary as
epigynous flowers have an inferior
ovary with the flattened thalamus as
seen in the ray florets of sunflowers. In
epigynous flowers,all the floral parts
are present above the gynoecium.Ovary
has been covered completely by
thalamus, e.g. guava, apple, sunflower.
27The ovary is half inferior in
[NEET (Sep.) 2020]
(a) mustard (b) sunflower
(c) plum (d) brinjal
Ans.(c)
The ovary is half inferior in plum. It is a
perigynous condition in which ovary is
at the top (origin wise) and other whorls
are at the same level (position wise).
There are three possible arrangements
of attachment of floral whorls with
thalamus where thalamus is disc-like,
cup-shaped and flask-shaped.
28Placentation in which ovules
develop on the inner wall of the
ovary or in peripheral part, is
[NEET (National) 2019]
(a) axile (b) parietal
(c) free central (d) basal
Ans.(b)
In parietal placentation, the ovules
develop on the inner wall of the ovary or
in peripheral part. In this type, the ovary
is compound or syncarpous type which
is either unilocular or falsely two or
more locular, e.g. mustard,Argemone,
Fumaris, etc. Axial placentation is when
ovules are placed along the central axis
of the ovary.
Basal placentation is when ovules are
placed at the base of the ovary.
Free-central placentation is when
placenta develops at the centre of
ovary as prolongation of floral axis. On
this axis, the ovules are attached.
29Bicarpellary ovary with obliquely
placed septum is seen in
[NEET (Odisha) 2019]
(a)Brassica (b)Aloe
(c)Solanum (d)Sesbania
Morphology of Flowering Plants 51
Inflorescence
TOPIC 4
Flower
TOPIC 5
Inflorescence
Seeds
Ray floret
Disk
floret
Leaves
Stem
Root
Disk floret
Ray floret
Inflorescence
Involucral bractReceptacle
Stigma
Style
Anther
Corolla
Sepal
ovaryBract
Disk floret
Sepal
Ovary
Bract
Corolla
Ray floret

62. Ans.(c)
Solanum(family Solanaceae) has
bicarpellary ovary with obliquely placed
septum.Brassica(family Brassicaceae)
has bicarpellary ovary with false
septum. InSesbania(subfamily
Fabaceae), ovary is monocarpellary.
Aloe(family Liliaceae) shows
tricarpellary ovary.
30Match the placental types (Column
I) with their examples (Column II)
[NEET (Odisha) 2019]
Column I Column II
1. Basal (i) Mustard
2. Axile (ii) China rose
3. Parietal (iii) Dianthus
4. Free-central (iv) Sunflower
Select the correct option from the
following
1 2 3 4
(a) (ii) (iii) (iv) (i)
(b) (i) (ii) (iii) (iv)
(c) (iv) (ii) (i) (iii)
(d) (iii) (iv) (i) (ii)
Ans.(c)
The correct matches are
The placental typesExamples
Basal Sunflower
Axile China rose
Parietal Mustard
Free-central Dianthus
31The standard petal of a
papilionaceous corolla is also
called [NEET 2016, Phase I]
(a) pappus (b) vexillum
(c) corona (d) carina
Ans.(b)
The standard or large upper petal of a
papilionaceous corolla is also called
vexillum.
32Tricarpellary, syncarpous
gynoecium is found in flowers of
[NEET 2016, Phase I]
(a) Solanaceae (b) Fabaceae
(c) Poaceae (d) Liliaceae
Ans.(d)
Liliaceae represents G
( )3
. This family
includes plants like garlic, onion, tulip,
indigo, etc.
33Free-central placentation is found
in [NEET 2016, Phase II]
(a)Dianthus (b) Argemone
(c) Brassica (d) Citrus
Ans.(a)
The plantDianthushas free-central
placentation in its ovary. Ovules are
attached on the main axis of the
placenta.
34Radial symmetry is found in the
flowers of[NEET 2016, Phase II]
(a)Brassica (b)Trifolium
(c)Pisum (d)Cassia
Ans.(a)
TheBrassicaflower can be cut into two
equal halves from any plane. So, it
shows the radial symmetry. These
flowers are referred to as
actinomorphic flowers.
Other flowers given in the option
(Trifolium, PisumandCassia) can be cut
into two equal halves only at one plane
so they are calledzygomorphic flowers.
These three flowers belong to family–
Fabaceae.
35How many plants among
Indigofera, Sesbania, Salvia, Allium,
Aloe, mustard, groundnut, radish,
gram and turnip have stamens
with different lengths in their
flowers? [NEET 2016, Phase II]
(a) Three (b) Four
(c) Five (d) Six
Ans.(b)
OnlySalvia, mustard, radish and turnip
have stamens of different length in
their flowers.
Hence, correct answer is (b).
36The term ‘polyadelphous’ is related
to [NEET 2016, Phase II]
(a) gynoecium (b) androecium
(c) corolla (d) calyx
Ans.(b)
The term polyadelphous is related to
androecium. In this condition, many
stamens are present in more than 2
groups which are separated from each
other.
37Flowers are unisexual in
[CBSE AIPMT 2015]
(a) pea (b) cucumber
(c) China rose (d) onion
Ans.(b)
Cucumber is a member of the
family—Cucurbitaceae. In the members
of this family, the flowers are unisexual,
with male and female flowers on
different plants (dioecious) or on the
same plant (monoecious).
38Axile placentation is present in
[CBSE AIPMT 2015]
(a) dianthus (b) lemon
(c) pea (d) Argemone
Ans.(b)
Placentation refers to the arrangement
of ovules inside the ovary.
In axile placentation, the ovules are
axial as shown in the figure below.
Examples include, lemon, China rose
and tomato.
39Among China rose, mustard,
brinjal, potato, guava, cucumber,
onion and tulip, how many plants
have superior ovary?
[CBSE AIPMT 2015]
(a) Five (b) Six
(c) Three (d) Four
Ans.(b)
A superior ovary is an ovary attached to
the receptacle above the attachment of
other floral parts. In this case the flower
is said to be hypogynous, e.g. China
rose, mustard, brinjal, potato, onion and
tulip. Guava and cucumber have inferior
ovaries or epigynous flowers.
So, out of 8 given names, 6 have
superior ovaries.
40Placenta and pericarp are both
edible portions in
[CBSE AIPMT 2014]
(a) apple (b) banana
(c) tomato (d) potato
Ans.(c)
Tomato is botanically a fruit but it is
considered as a vegetable for various
purposes. Its edible parts are both
placenta and pericarp.
52 NEETChapterwise Topicwise Biology
Locule
Placenta
Ovules
Ovary wall
Free-central placentation

63. Placenta is the connecting tissue which
supply nutrition for embryo and
pericarp is the remains of ovular wall. It
is rich in lycopene.
41When the margins of sepals or
petals overlap one another
without any particular direction,
the condition is termed as
[CBSE AIPMT 2014]
(a) vexillary (b) imbricate
(c) twisted (d) valvate
Ans.(b)
Imbricate aestivation is the
arrangements of five petals being
arranged in such a way that one petal is
completely external and another petal
is completely internal, while three
petals are partially external and partially
internal, e.g.Cassia, Callistemonand
Caesalpinia.
42Among bitter gourd, mustard,
brinjal, pumpkin, China rose, lupin,
cucumber, sunnhemp, gram,
guava, bean, chilli, plum,Petunia,
tomato, rose,Withania, potato,
onion, aloe and tulip, how many
plants have hypogynous flower?
[NEET 2013]
(a) Six (b) Ten
(c) Fifteen (d) Eighteen
Ans.(c)
All the given plants except bitter gourd,
pumpkin, cucumber, guava, plum and
rose are hypogynous flower, i.e. 15.
Hypogynous flower have gynoecium
present above all other floral parts and
with superior ovary.
43In China rose the flowers are
[NEET 2013]
(a) actinomorphic, hypogynous with
twisted aestivation
(b) actinomorphic, epigynous with
valvate aestivation
(c) zygomorphic, hypogynous with
imbricate aestivation
(d) zygomorphic, epigynous with
twisted aestivation
Ans.(a)
Actinomorphic (star-shaped) can be
divided into 3 or more identical
sections, which are related to each
other by rotation about the centre of
the flower, e.g. China rose.
Zygomorphic flowers can be divided by
only a single plane into two mirror
image halves, e.g. orchids.
Valvate aestivationThe sepals or
petals in a whorl just touch one another
at the margin, e.g.Calotropis.
Twisted aestivationOne margin of the
appendage overlaps that of the next
one, e.g. China rose. In Imbricate
aestivation the margins of sepals or
petals overlap but not necessarily in
specific direction, e.g.Cassia.
44Placentation in tomato and lemon
is [CBSE AIPMT 2012]
(a) parietal (b) free central
(c) marginal (d) axile
Ans.(d)
Lemon orCitrus(family–Rutaceae),
tqmato orLycopersiconsp.
(family–Solanaceae), China rose or
Hibiscus(family–Malvaceae) etc. have
axile placentation. It occurs in
multicarpellary, syncarpous ovary.
Inward growth of margins of carpels
forms a multicarpellary condition which
contains an axis in the centre.
Placentae arise from this central axis
which bear ovules.
45Cymose inflorescence is present
in [CBSE AIPMT 2012]
(a)Solanum (b)Sesbania
(c)Trifolium (d)Brassica
Ans.(d)
Cymose inflorescence seen inBrassica
is also called definite or determinate
inflorescence because the growing
point of the peduncle is used up in the
formation of a flower. Further growth of
the flowering axis is continued by one or
more lateral branches (peduncles)
which also end in flowers. This type of
inflorescence is found in
family–Solanaceae.
46The gynoecium consists of many
free pistils in flowers of
[CBSE AIPMT 2012]
(a)Aloe (b) tomato
(c)Papaver (d)Michelia
Ans.(d)
Apocarpous condition arises when the
number of carpels is two or more and
they are free from each other. e.g.
Clematis, Michelia(Magnoliaceae),
Aconitum, Ranunculus(buttercup), etc.
47Which one of the following
statements is correct?
[CBSE AIPMT 2011]
(a) Seeds of orchids have oil-rich
endosperm
(b) Placentation in primrose is basal
(c) Flower of tulip is a modified shoot
(d) In tomato, fruit is a capsule
Ans.(c)
The correct statement is ‘c’ because
flower is highly condensed and modified
shoot meant for sexual reproduction
(Dr. Goethe; 1790). During the course of
evolution, the nodes of the axis of shoot
came in contact so, that internodes got
reduced, and leaves get modified and
specialised to form floral leaves.
48Flowers are zygomorphic in
[CBSE AIPMT 2011]
(a) gulmohur (b) tomato
(c) datura (d) mustard
Ans.(a)
When a flower can be divided into two
similar halves only in one particular
vertical plane, it is called zygomorphic,
e.g. bean, pea, gulmohur,Cassia,etc.
49The ovary is half inferior in flowers
of [CBSE AIPMT 2011]
(a) cucumber (b) cotton
(c) guava (d) peach
Ans.(d)
If gynoecium is situated in the center
and other parts of the flowers are
located on the rim of the thalamus
almost at the same level, it is called
perigynous. The ovary here is said to be
half inferior, i.e. plum rose, peach, etc.
50Keel is characteristic of the
flowers of [CBSE AIPMT 2010]
(a) gulmohur (b)Cassia
(c)Calotropis(d) bean
Ans.(d)
In bean-family the two-in-one petal is
called the keel, like the keel of a boat.
Bean blossoms with this configuration
are said to be papilionaceous.
51In unilocular ovary with a single
ovule, the placentation is
[CBSE AIPMT 2010]
(a) marginal
(b) basal
(c) free central
(d) axile
Morphology of Flowering Plants 53

64. Ans.(b)
Inbasal placentation, ovary is
bicarpellary, syncarpous and unilocular
and a single ovule is borne at the base
of ovary, e.g. marigold.
Inmarginal placentation, the ovary is
simple, unilocular and the ovules are
arranged along the margin of the
unilocular ovary, e.g. pea.
Infree central placentation, ovary is
unilocular and the placenta bearing
ovules arise from the central axis, e.g.
Stellaria.
Inaxile placentation, ovary is two or
more chambered, usually as many as
the number of carpels, e.g.Petunia.
52The technical term used for the
androecium in a flower of China
rose (Hibiscus rosa sinensis) is
[CBSE AIPMT 2010]
(a) monadelphous (b) diadelphous
(c) polyandrous (d) polyadelphous
Ans.(a)
Inmonadelphouscondition, all
filaments become fused and form a
group, while anther remain free, e.g.
China rose,Achyranthes, etc. In
diadelphous, two separate bundles of
united filaments are formed, e.g. pea. In
polyadelphous, more than two separate
bundles of filaments are formed, e.g.
Ricinus.
53An example of axile placentation is
[CBSE AIPMT 2009]
(a)Argemone (b)Dianthus
(c) lemon (d) marigold
Ans.(c)
Lemon (Citrussp.) belongs to
family–Rutaceae, characterised by axile
placentation.
Argemone belongs to
family–Papaveraceae, contains parietal
placentation.Dianthusbelongs to
family–Caryophyllaceae, contains
free-central placentation. Marigold
belongs to family–Asteraceae, contains
basal placentation.
54What type of placentation is seen
in sweet pea?[CBSE AIPMT 2006]
(a) Axile (b) Free central
(c) Marginal (d) Basal
Ans.(c)
In sweet pea(Pisum sativum),the
placentation ismarginal,in which, the
placenta develops along the junction of
two carpels, in a unilocular ovary.
Inbasal placentation,the ovules are
few or reduced to one and are borne at
the base of ovary e.g. Compositae.
Inaxile placentation,margins of
carpels fold inwards, fusing together in
centre of ovary to form a single central
placenta. Ovary is divided into as many
locules, as there are carpels, e.g.
Hibiscus, Asphodelus.Free-central
placentationpossesses a placenta
arises as a central upgrowth from ovary
base, e.g.Stellaria.
55Long filamentous threads
protruding at the end of the young
cob of maize are
[CBSE AIPMT 2006]
(a) styles (b) ovaries
(c) hairs (d) anthers
Ans.(a)
In a cob of maize, each ovary has a long
silky (hairy)style,called ascorn silk.
Collectively these styles protrude at the
end of a young cob. The grains are
formed on the cob which remain
covered by the leafy bracts.
56Bicarpellary gynoecium and
oblique ovary occurs in
[CBSE AIPMT 2001]
(a) mustard (b) banana
(c)Pisum (d) brinjal
Ans.(d)
Brinjal belongs to family–Solanaceae.
Obliquely placed bicarpellary ovary is
characteristic of Solanaceae.
57The type of placentation in which
ovary is syncarpous, unilocular and
ovules on sutures is called
[CBSE AIPMT 1999]
(a) apical placentation
(b) parietal placentation
(c) marginal placentation
(d) superficial placentation
Ans.(b)
Parietal placentation occurs in
bicarpellary to polycarpellary
syncarpous pistils in which the ovary is
unilocular. The placentae are formed
along the fused margins of the carpels
from where the ovules arise.
58Angiosperm to which the largest
flowers belong is[CBSE AIPMT 1999]
(a) total stem parasite
(b) partial stem parasite
(c) total root parasite
(d) partial root parasite
Ans.(c)
Rafflesia,atotal root parasite,
produces largest flower in the world.
59Floral features are chiefly used in
angiosperms identification
because [CBSE AIPMT 1998]
(a) flowers are of various colours
(b) flowers can be safely pressed
(c) reproductive parts are more
stable and conservative than
vegetative parts
(d) flowers are nice to work with
Ans.(c)
Vegetative parts are relatively less
stable and exhibit changes due to the
environmental factors quite readily. On
the other hand, floral features are more
conservative and can be relied upon. On
the basis of reproductive parts of
different flowers Linnaeus classified
plants into different groups.
60Which one yields fibre?
[CBSE AIPMT 1988]
(a) Coconut (b) Oak
(c) Teak (d) Sissoo
Ans.(a)
On the basis of their origin, commercial
fibres are of three types : surface fibres
(fibres obtained from seeds), e.g.
cotton, coconut (coir), Calotropis,bast
fibres (fibres present in phloem), e.g.
Cannabis(hemp),Linum(flax), jute,
sunnhemp, and leaf fibres (fibres
extracted from leaves), e.g.Agave(sisal
hemp),Musa(masnila hemp).
54 NEETChapterwise Topicwise Biology
Ovary wall
Ovules
Placenta
Locule
Obliquely placed bicarpellary ovary
Ovary wall
Locule
Placenta
Ovule
Parietal placentation

65. 61Identify the correct features of
mango and coconut fruits.
[NEET (Oct.) 2020]
(i) In both fruit is a drupe
(ii) Endocarp is edible in both
(iii) Mesocarp in coconut is fibrous
and in mango it is fleshy
(iv) In both, fruit develops from
monocarpellary ovary
Select the correct option.
(a) (i), (iii) and (iv) (b) (i), (ii) and (iii)
(c) (i) and (iv) (d) (i) and (ii)
Ans.(a)
Both mango and coconut are drupe,
belonging to the class of simple
succulent fruits. These are also called
stone fruit as the endocarp is stony and
non-edible. These fruits develop from
monocarpellary ovary. In coconut,
epicarb is membranous and mesocarp
is fibrous. In mango, mesocarp is fleshy
and pulpy.
Thus, statements (i), (iii), (iv) are correct
while (ii) is incorrect.
62An aggregate fruit is one which
develops from[CBSE AIPMT 2014]
(a) multicarpellary syncarpous
gynoecium
(b) multicarpellary apocarpus
gynoecium
(c) complete inflorescence
(d) multicarpellary superior ovary
Ans.(b)
Aggregate fruits or etario are those
fruits that develops from the
multicarpellary apocarpus gynoecium
(ovary). In contrast, a simple fruit
develops from one ovary.
Aggregate fruit may also be called
accessory fruitsin which part of the
flower other than the ovary become
fleshy and form part of the fruit, e.g.
raspberry, etc.
63How many plants in the list given
below have composite fruits that
develop from an inflorescence?
[CBSE AIPMT 2012]
Walnut, poppy, radish, fig,
pineapple, apple, tomato, mulberry
(a) Four (b) Five
(c) Two (d) Three
Ans.(d)
Fig, pineapple and mulberry are
composite fruits.
SNPlant
Botanical
Name
Fruit
Inflores-
cence
1. Fig Ficus
carica
Sycous Hypant-h
odium
2. PineappleAnnanas
sativus
Sorosis Spike
3. MulberryMorussp. Sorosis Catkin
64The coconut water and the edible
part of coconut are equivalent to
[CBSE AIPMT 2012]
(a) endosperm
(b) endocarp
(c) mesocarp
(d) embryo
Ans.(a)
The coconut water obtained from the
coconut is the free nuclear endosperm
(made up of thousands of nuclei) and
the surrounding white kernel is the
cellular endosperm.
65A drupe develops in
[CBSE AIPMT 2011, 1994]
(a) wheat (b) pea
(c) tomato (d) mango
Ans.(d)
In mango, coconut, plum, etc. the fruit
is known as drupe (stony fruit). They
develop from monocarpellary, superior
ovaries and are one seeded. In mango,
the pericarp is well differentiated into
an outer thin epicarp, a middle fleshy
edible mesocarp and an inner stony
hard endocarp.
66A fruit developed from
hypanthodium inflorescence is
called [CBSE AIPMT 2009]
(a) hesperidium
(b) sorosis
(c) syconus
(d) caryopsis
Ans.(c)
Syconusfruit develops from
hypanthodium inflorescence, e.g.Ficus
carica,F. religiosa,F. benghalensis. The
flask shaped receptacle encloses
female flowers that gives rise to
achene-like fruitlets. This fruit
possesses a small pore protected by
scaly leaves. The receptacle that
becomes fleshy is edible.
67The fruit is chambered, developed
from inferior ovary and has seeds
with succulent testa in
[CBSE AIPMT 2008]
(a) pomegranate (b) orange
(c) guava (d) cucumber
Ans.(a)
Balaustais special type of false or
pseudocarpic berry, that develops from
multilocular, syncarpous inferior ovary.
The whole fruit is enclosed by a hard
rind made up of exocarp (epicarp fused
with thalamus) and part of mesocarp.
Plate-like infoldings are developed by
mesocarp. The papery endocarp covers
the individual group of seeds. The
seeds possess bright red juicy testa
that form edible part of fruit, e.g.
pomegranate.
The fruit of cucumber ispepo. In this
the exocarp is not separable from
mesocarp and the seeds from
placentae.
The fruit of guava isberry. A berry is
pulpy, indehiscent, few to multiseeded
fruit derived from multicarpellary
syncarpous gynoecium.
The fruit of orange is hesperidium. It
develops from multicarpellary,
syncarpous, multilocular, superior
ovary with axile placentation.
68The fleshy receptacle of syconous
of fig encloses a number of
[CBSE AIPMT 2008]
(a) achenes (b) samaras
(c) berries (d) mericarps
Ans.(a)
Syconusis a composite fruit that
develops from hypanthodium
inflorescence, e.g.Ficus carica, Ficus
benghalensis.The flask-shaped
receptacle encloses female flowers
that give rise to achene-like fruitlets.
This fruit possesses a small pore
protected by scaly leaves. The
receptacle that becomes fleshy is
edible.
Samarais a single seeded, dry
indehiscent fruit. Its pericarp becomes
membranous and flat-like wings that
help in dispersal, e.g.Ulmus, Holoptelia
indica.
Aberryis a pulpy-indehiscent, few to
multiseeded fruit derived from
multicarpellary syncarpous gynoecium.
The fleshy pericarp of berry consists of
three parts i.e. epicarp that make the
rind of fruit, mesocarp and endocarp.
Morphology of Flowering Plants 55
Fruit
TOPIC 6

66. Schizocarpic fruitsare simple, dry,
multiseeded fruits, which breakup into
single seeded parts. The single seeded
parts, which further do not dehisce are
called mericarps.
69Dry indehiscent single-seeded
fruit formed from bicarpellary
syncarpous inferior ovary is
[CBSE AIPMT 2008]
(a) caryopsis (b) cypsela
(c) berry (d) cremocarp
Ans.(b)
Cypsela is dry indehiscent single
seeded fruit that develops from
unilocular, single ovulate inferior ovary
of bicarpellary syncarpous gynoecium
possessing basal placentation. The fruit
wall develops from pericarp and
thalamus and is thin and remains
attached to the seed at one point,
e.g.Helianthus.
Caryopsis is dry indehiscent, small,
single seeded fruit develop from
unilocular, single ovuled, superior ovary
of multicarpellary gynoecium. It differs
from typical achenes as their pericarp is
completely fused with the seed coat
testa, e.g. Poaceae.
Cremocarp are bilocular and two
seeded schizocarpic fruits (small, dry)
developed from inferior ovary of
bicarpellary, syncarpous gynoecium
possessing persistant stylopodium, e.g.
Apiaceae.
Berry is a fleshy, indehiscent few to
multiseeded fruit derived from
multicarpellary syncarpous gynoecium.
The fleshy pericarp of berry consists of
epicarp, mesocarp and endocarp.
70Pineapple (ananas) fruit develops
from [CBSE AIPMT 2006]
(a) a multipistillate syncarpous
flower
(b) a cluster of compactly borne
flowers on a common axis
(c) a multilocular monocarpellary
flower
(d) a unilocular polycarpellary flower
Ans.(b)
The fruit ofAnanas comosus(pineapple
or ananas) is sorosis, (a type of multiple
fruits), developing from spike, spadix or
catkin. In this type, the flowers
associate by their succulent tepals, the
axis bearing them grows and becomes
fleshy or woody, thus, the whole
inflorescence turns into a compact
fruit.
71Juicy hair-like structures
observed in the lemon fruit
develop from[CBSE AIPMT 2003]
(a) mesocarp and endocarp
(b) exocarp
(c) mesocarp
(d) endocarp
Ans.(d)
Lemon is a hesperidium type of fruit.
Epicarp of this fruit contains many oil
glands. Below epicarp is present a
fibrous part which fuses with epicarp,
this is known as mesocarp. While
endocarp projects inwards and forms
distinct chambers. Many unicellular
juicy hairs are present on the inner
side of endocarp which are edible
parts of this fruit.
72Edible part in mango is
[CBSE AIPMT 2002, 04]
(a) mesocarp (b) epicarp
(c) endocarp (d) epidermis
Ans.(a)
Mango is a drupe fruit which develops
from a monocarpellary, syncarpous,
unilocular and superior ovary. Epicarp
of mango fruit forms skin while
mesocarp is fleshy and fibrous which is
edible part of this fruit. Endocarp is
hard and stony.
73.Geocarpic fruits are produced by
[CBSE AIPMT 2000, 02]
(a) onion (b) watermelon
(c) ground nut (d) carrot
Ans.(c)
Geocarpy refers to the ripening of fruits
underground. In the case of ground nut,
the young fruits are pushed into the soil
as a result of post-fertilisation
curvature of the stalk.
74Geocarpic fruit is
[CBSE AIPMT 2002]
(a) potato (b) groundnut
(c) onion (d) garlic
Ans.(b)
The groundnut fruits ripen
underground, the young fruits being
pushed into the soil by a
post-fertilisation curvature of the stalk.
75Which is correct pair for edible
part? [CBSE AIPMT 2001]
(a) Tomato — Thalamus
(b) Maize — Cotyledons
(c) Guava — Mesocarp
(d) Date palm — Pericarp
Ans.(d)
Each fruit of date palm(Phoenix
dactylifera)is one seeded oblong berry.
Fleshy pericarp of date palm is edible.
We eat mesocarp of tomato,
endosperm of maize and thalamus and
pericarp of guava.
76Edible part of banana is
[CBSE AIPMT 2001]
(a) epicarp
(b) mesocarp and less developed
endocarp
(c) endocarp and less developed
mesocarp
(d) epicarp and mesocarp
Ans.(b)
Fleshy mesocarp and rudimentary
endocarp of banana are edible.
Banana (Musa sapientum) is a perennial
herb and belongs to family–Musaceae.
A fully ripened fruit of banana contains
near about 75.6% moisture, 20.4%
sugars (mainly glucose and fructose),
1.2% starch, 2% fats, 1.22% protein, 6%
crude fibres and 0.8% ash
characteristic of Solanaceae.
77Edible part in litchi is
[CBSE AIPMT 1999, 2005, 06]
(a) mesocarp (b) fleshy aril
(c) endosperm (d) pericarp
Ans.(b)
Arilis a fleshy covering on the seed,
arising as an upgrowth of the funicle or
base of the ovule. It is the edible part of
litchi.
78Which plant will lose its economic
value if its fruits are produced by
induced parthenocarpy?
[CBSE AIPMT 1997]
(a) Grape (b) Pomegranate
(c) Banana (d) Orange
Ans.(b)
Testa is the edible part in pomegranate.
It is not formed if fruits are produced by
parthenocarpy (no seeds will be
formed). Fruits of banana, grape and
orange have seeds, so induced
parthenocarpy in these fruits is
beneficial.
79Which one of the following is a
true fruit?[CBSE AIPMT 1996]
(a) Apple (b) Pear
(c) Cashewnut (d) Coconut
56 NEETChapterwise Topicwise Biology

67. Ans.(d)
Coconut is a true fruit, when a fruit
develops from ripened ovary, it is called
true fruitas in majority of fruits. Afalse
fruitdevelops from any part of the
flower except the ovary, e.g. apple,
pear, cashewnut and all composite
fruits (mulberry, pine apple).
80Which part of the coconut
produces coir?[CBSE AIPMT 1996]
(a) Seed coat (b) Mesocarp
(c) Epicarp (d) Pericarp
Ans.(b)
In coconut (drupe or stone fruit),
epicarp is thin, mesocarp is fibrous,
produces coir, endocarp bears three
eye spots and encloses a single seed
with brown testa, oily endosperm,
embryo and watery fluid.
81Mango juice is got from
[CBSE AIPMT 1989]
(a) epicarp
(b) mesocarp
(c) endocarp
(d) pericarp and thalamus
Ans.(b)
In mango (a drupe) edible part is
mesocarp, which is fibrous, pulpy and
juicy.
82Fruit of groundnut is
[CBSE AIPMT 1988]
(a) legume (b) caryopsis
(c) berry (d) nut
Ans.(a)
Legume or pod develops from
monocarpellary gynoecium. It is a dry
dehiscent fruit occurs in Leguminosae
(pea, gram, bean and groundnut).
83The body of the ovule is fused
within the funicle at
[NEET (Sep.) 2020]
(a) micropyle (b) nucellus
(c) chalaza (d) hilum
Ans.(d)
The attachment point of funicle and
body of ovule is known as hilum. It is the
point where ovule attaches to the base.
It is generally present as an eye of the
seed as it is present as a scar.
84Persistent nucellus in the seed is
known as [NEET (National) 2019)
(a) perisperm (b) hilum
(c) tegmen (d) chalaza
Ans.(a)
Persistent nucellus in the seed is known
as perisperm.
The albuminous seeds usually retain a
part of endosperm as it is not
completely used up during embryo
development.
But in some seeds, remnants of
nucellus are also persistent, e.g. black
pepper and beet.
85The morphological nature of the
edible part of coconut is
[NEET 2017]
(a) perisperm (b) cotyledon
(c) endosperm (d) pericarp
Ans.(c)
The edible part of coconut is
endosperm. Coconut water is free
nuclear endosperm and white kernel is
the cellular endosperm.
86Which one of the following
statements is correct?
[CBSE AIPMT 2014]
(a) The seed in grasses is not
endospermic
(b) Mango is a parthenocarpic fruit
(c) A proteinaceous aleurone layer is
present in maize grain
(d) A sterile pistil is called a
staminode
Ans.(c)
A proteinaceous aleurone protein layer
of maize start developing
approximately 10-15 days after
pollination in stack that take 40 day for
the aleurone to mature completely.
Thus is the correct statement. Correct
sentences for other options are
(a) Seed in grasses are endospermic
(b) Banana is a parthanocarpic fruit
but mango is not a parthenocarpic
fruit.
(c) Sterile pistil is called pistillode.
87Seed coat is not thin,
membranous in [NEET 2013]
(a) maize (b) coconut
(c) groundnut (d) gram
Ans.(b)
Seed coat is thick in coconut seed and
thin, membranous in groundnut, gram
and maize seeds.
88Perisperm differs from endosperm
in [NEET 2013]
(a) being a haploid tissue
(b) having no reserve food
(c) being a diploid tissue
(d) its formation by fusion of
secondary nucleus with several
sperms
Ans.(b)
The main difference between
perisperm and endosperm is that
perisperm is present in seeds and
endosperm is present in developing
embryo as its reserved food which is
completely consumed by it during
development.
89Cotyledons and testa are edible
parts of [CBSE AIPMT 2009]
(a) groundnut and pomegranate
(b) walnut and tamarind
(c) french bean and coconut
(d) cashew nut and litchi
Ans.(a)
Cotyledons and testa are edible parts of
groundnut and pomegranate
respectively.
The edible part of walnut is cotyledon,
tamarind-mesocarp, french
bean-seeds, coconut-endosperm,
testa, cotyledons and embryo,
cashewnut-cotyledons and fleshy
pedicels and of litchi is fleshy aril.
90An example of a seed with
endosperm, perisperm and
caruncle is[CBSE AIPMT 2009]
(a) cotton (b) coffee
(c) lily (d) castor
Ans.(d)
The seeds of castor (Ricinus communis,
family–Euphorbiaceae) are
endospermic dicot seeds. They
possess, endosperm which acts as the
food storage tissue of seed. They also
possess perisperm and caruncle.
91In a cereal grain the single
cotyledon of embryo is
represented by[CBSE AIPMT 2006]
(a) scutellum (b) prophyll
(c) coleoptile (d) coleorhiza
Ans.(a)
In a cereal grain (e.g. wheat), the single
cotyledon of embryo is represented by
thescutellum.Scutellum is specialised
for nutrient absorption from the
endosperm.
Morphology of Flowering Plants 57
Seed
TOPIC 7

68. Coleoptileis a modified ensheathing
leaf that covers and protects the young
primary leaves of a grass seedling.
Coleorhizais a sheath like structure
found on the radicle which covers and
protects it during the growth into the
soil.
92The embryo in sunflower has
[CBSE AIPMT 1998]
(a) one cotyledon
(b) two cotyledons
(c) many cotyledons
(d) no cotyledon
Ans.(b)
Sunflower is a dicotyledonous plant, so
the number of cotyledons in sunflower
is two. Monocotyledons contain only
one cotyledon in their embryo. It belong
to order–Asterales and
family–Asteraceae.
93Heterospory and seed habit are
often discussed in relation to a
structure called[CBSE AIPMT 1997]
(a) spathe
(b) bract
(c) petiole
(d) ligule
Ans.(d)
Heterospory and seed habit are
discussed mostly with respect to ligule.
Development of two types of spores
(microspores and megaspores) is called
heterospory. Though all seed-bearing
plants are heterosporous, it evolved in
some pteridophytes. In early
heterosporous plants, megaspores
were released from the parent.
But in seed plants, these are retained
and fertilised to become seed. This habit
is seen inSelaginellawhich bears a small
multicellular scale-like structure called
ligule at the base of leaf on adaxial side.
94Plant having column of vascular
tissues, bearing fruits and having
a tap root system is
[CBSE AIPMT 1994]
(a) monocot
(b) dicot
(c) gymnosperm or dicot
(d) gymnosperm or monocot
Ans.(b)
Dicotyledoneae is the group of
angiosperm with two cotyledons, flower
bi or pentamerous, leaves net-viened,
stem with open collateral vascular
bundle arranged in a ring and roots form
tap root system.
95Vivipary is[CBSE AIPMT 1992]
(a) seed germination with
subterranean cotyledons
(b) seed germination with
epiterranean cotyledons
(c) fruit development without
pollination
(d) seed germination inside the fruit
while attached to the plant
Ans.(d)
Germination of seeds inside the fruit
which is still attached to the parent tree
is called vivipary. It is a special type of
seed germination occurring in plants
growing in sea coasts and salt lakes,
(e.g. man-groves). These seeds lack any
dormant period.
96Oil reserve of groundnut is present
in [CBSE AIPMT 1990]
(a) embryo
(b) cotyledons
(c) endosperm
(d) underground tubers
Ans.(b)
Cotyledons are the leaves of embryo,
usually store food for the use of embryo
during development. Groundnut, a dicot
stores food in the form of oil reserve in
the cotyledons.
97Vivipary is characteristics of
[CBSE AIPMT 1990]
(a) mesophytes (b) xerophytes
(c) hygrophytes (d) halophytes
Ans.(d)
Viviparyis a special type of seed
germination, known to occur in
halophytes, e.g.Rhizophorawhere
seeds germinate inside the fruit, while
it is still attached to the parent plant.
The embryo get nourishment from the
parent plant and grows out of seed as
well as fruit. Projecting out in the form
of a green seedling.
98Match the Column -I with
Column -II. [NEET 2021]
Column I Column II
A. 1. Brassicac
eae
B. 2. Liliaceae
C. 3. Fabaceae
D. 4. Solanacea
e
Select the correct answer from
the options given below.
A B C D
(a) 3 4 2 1
(b) 1 2 3 4
(c) 2 3 4 1
(d) 4 2 1 3
Ans.(a)
(A)-(3), (B)-(4), (C)-(2), (D)-(1)
Fabaceae was earlier called
Papilionoideae. Flowers of Fabaceae
are zygomorphic (%),
hermaphrodite (O
+
_
), have 5 fused
sepals(K )
(5)
,
have 5 petals out of which 2 are fused
( ( ))1 2 2+ + have 9 united stamens and 1
free stamen (C9)+1) and have 1 superior
ovary (G
1
).
This way the floral formula is made and
thus, similarly for Brassicaceae,
Liliaceae and Solanaceae.
99Which of the following is the
correct floral formula of Liliaceae?
[NEET (Oct.) 2020]
(a)
(b)
(c)
(d)
58 NEETChapterwise Topicwise Biology
Important
Angiospermic Families
TOPIC 8
%O K C A
(5)1+2+(2)(9)+11
G
+
⊕O K C A
(5)(5)52
G
+
⊕O P A
(3+3)3+33
G
+
⊕O K C A
2+24 2–4(2)
G
+
+
% O C A G
1+2+(2) (9)+1 1
+
rO K C A
(5) (5) 1 1G
+
Br O P Ar
(3+3) 3+3 (3)G
+
rO K C A
(5) (5) 5 (2)G

69. Ans.(c)
Floral formula of Liliaceae is
It represents bracteate, actinomorphic,
bisexual flowers having six tepals and
six epitepalous stamens. Ovary is
superior with three lobed stigma.
100Vexillary aestivation is
characteristic of the family
[CBSE AIPMT 2012]
(a) Fabaceae (b) Asteraceae
(c) Solanaceae (d) Brassicaceae
Ans.(a)
Vexillary aestivation of corolla is a
characteristic of family–Fabaceae. In
corolla, the posterior petal called
vexillumis largest, two lateral, curved
petals are calledwingsand two
anterior, boat-shaped petals are called
keels.
101The correct floral formula of chilli
is [CBSE AIPMT 2011]
(a)⊕
+
O
_
K C A G
( ) ( ) ( )5 5 5 2
(b)⊕
+
O
_
K C A G
( ) ( ) ( )5 5 5 2
(c)⊕
+
O
_
K C A G
5 5 5 2( )
(d)⊕
+
O
_
K C A G
( ) ( )5 5 5 2
Ans.(a)
Floral formula of chilli (Capsicum annum)
is⊕
+
O_
K C A G
(5) (5) 5 (2)
. It belongs to
family–Solanaceae. In this family,
flower is actinomorphic, complete and
bisexual, calyx has five sepals which are
gamosepalous showing valvate
aestivation, corolla has five petals
which are polypetalous showing valvate
aestivation, androecium has five free
stamens (polyandrous) but epipetalous,
gynoecium is bicarpellary, syncarpous,
bilocular with superior ovary having
axile placentation.
102Which of the following is a correct
pair? [CBSE AIPMT 2002]
(a)Cuscuta— Parasite
(b)Dischidia— Insectivorous
(c)Opuntia— Predator
(d)Capsella— Hydrophyte
Ans.(a)
Cuscuta,commonly known as dodder or
amarbel, is a parasitic plant. It belongs
to order Solanales and
family–Convolvulaceae.
103Match the following and indicate
which is correct?
[CBSE AIPMT 2000]
(a) Cucurbitaceae — Orange
(b) Malvaceae — Cotton
(c) Brassicaceae — Wheat
(d) Leguminosae — Sunflower
Ans.(b)
Sunflower belongs to Asteraceae
(Compositae), orange to Rutaceae wheat
to Poaceae (Gramineae) while cotton
(Gossypium) belongs to Malvaceae.
104Tetradynamous stamens are
found in family
[CBSE AIPMT 1995, 2001]
(a) Malvaceae
(b) Solanaceae
(c) Cruciferae
(d) Liliaceae
Ans.(c)
Tetradynamous stamens refer to four
long and two short stamens arranged in
a flower, e.g. members of Cruciferae
(Brassica).
105Pulses are obtained from
[CBSE AIPMT 1993]
(a) Fabaceae
(b) Asteraceae
(c) Poaceae
(d) Solanaceae
Ans.(a)
Fabaceae (Leguminaceae) is
economically important family
containing a number of legumes or
pulses, e.g. gram (Cicer arietinum),
green gram (Vigna radiata), black gram
(Phaseolus mungo), pigeon pea (Cajanus
cajan), soyabean, etc.
106Epipetalous and syngenesious
stamens occur in[CBSE AIPMT 1991]
(a) Solanaceae (b) Brassicaceae
(c) Fabaceae (d) Asteraceae
Ans.(d)
In disc florets of family–Asteraceae
(Compositae), androecium consists of
five stamens, which are epipetalous,
with free filaments and fused anthers
(e.g. syngenesious), dithecous, introse
and dehiscing longitudinally.
107Botanical name of cauliflower is
[CBSE AIPMT 1991]
(a)Brassica oleraceavar. capitata
(b)Brassica campestris
(c)Brassica oleraceavar.botrytis
(d)Brassica oleraceavar.gemmifera
Ans.(c)
Brassica oleraceavar.botrytis
(cauliflower = phulgobhi),B. oleracea
var.capitata(cabbage = bandgobhi),B.
oleraceavar.caulorapa(knol khol =
ganth gobhi),B. campestrisvar. sarson
(yellow mustard),B. rapa(turnip).
108Floral formula of tomato/tobacco
is [CBSE AIPMT 1989, 92]
(a)⊕o
+ 4 – 5 10 (2)
ä
K A G
(b)⊕o
+ 2+ 2 4 2+ 4 1
ä
K C A G
(c)⊕o
+2 3 1
ä
P A G
(d) Br⊕o
+
(5) (5) 5 (2)
ä
K C A G
Ans.(d)
Tomato and tobacco are the members
of family–Solanaceae, with floral
formula
Morphology of Flowering Plants 59
Br O P A(3+3) (3+3)(3)G
+
+

70. 01Match the List-I with List - II.
[NEET 2021]
List-I List-II
A.Cells with active cell
division capacity
1. Vascular
tissues
B.Tissue having all cells
similar in structure
and function
2. Meristema
tic tissue
C.Tissue having
different types of
cells
3. Sclereids
D.Dead cells with highly
thickened walls and
narrow lumen
4. Simple
tissue
Select the correct answer from the
options given below.
A B C D
(a) 2 4 1 3
(b) 4 3 2 1
(c) 1 2 3 4
(d) 3 2 4 1
Ans.(a)
(A)-(2), (B)-(4), (C)-(1), (D)-(3)
Meristematic cellsare totipotent and
are capable of continued cell division.
Division of meristematic cells provides
new cells for expansion and
differentiation of tissues and the
initiation of new organs, providing the
basic structure of the plant body.
Hence, they have cells with active cell
division capacity.
Permanent tissues having all cells
similar in structure and function are
calledsimple tissue.
Vascular tissuesare formed of more
than one cell type, found in vascular
plants.
The primary components of vascular
tissues are the xylem and phloem.
These two tissues transport fluid and
nutrients internally.
Sclereidsare a kind of sclerenchyma
cells that are irregular or short. These
are dead cells. Their walls are irregular,
very thick and their lumen is very
narrow. They do not conduct any
metabolic activities. They exhibit
different types of lignin depositions and
also have pits. Sclereids are found in
hard parts such as hard seed coats,
endocarp of coconut. They are also
referred to as stone cells.
02Regeneration of damaged growing
grass following grazing is largely
due to [NEET (Odisha) 2019]
(a) lateral meristem
(b) apical meristem
(c) intercalary meristem
(d) secondary meristem
Ans.(c)
Regeneration of damaged growing
grass following grazing is largely due to
intercalary meristem. It is the meristem
which occurs between mature tissues. It
is found in grasses and regenerates parts
damaged by the grazing herbivores.
03Root hairs develop from the region
of [NEET 2017]
(a) maturation
(b) elongation
(c) root cap
(d) meristematic activity
Ans.(d)
The root is covered at the apex by a
thimble-like structure called the root
cap. It protects the tender apex of the
root as it makes its way through the
soil. A few millimetres above the root
cap is the region of meristematic
activity. The cells of this region are very
small, thin-walled and with dense
protoplasm. They divide repeatedly.
The cell proximal to this region undergo
rapid elongation and enlargement and
are responsible for the growth of the
root in length. This region is called the
region of elongation. The cells of the
elongation zone gradually differentiate
and mature. Hence, this zone, proximal.
From this region, some of the
epidermal cells form very fine and
delicate, thread-like structure called
root hairs. These root hairs absorb
water and minerals from the soil.
04Which of the following is made up
of dead cells? [NEET 2017]
(a) Xylem parenchyma
(b) Collenchyma
(c) Phellem
(d) Phloem
Anatomyof
FloweringPlants
06
Tissues and Tissue System
TOPIC 1

71. Ans.(c)
Phellem or cork is a tissue formed on
the outer side of cork cambium. It is
composed of dead cells. The cell wall
become impermeable due to
suberisation.
05Specialised epidermal cells
surrounding the guard cells are
called [NEET 2016, Phase I]
(a) subsidiary cells
(b) bulliform cells
(c) lenticels
(d) complementary cells
Ans.(a)
Few epidermal cells, in the vicinity of
the guard cells become specialised in
their shape and size and are known as
subsidiary cells. These cells are devoid
of chloroplasts.
The stomatal aperture, guard cells and
the surrounding subsidiary cells are
together calledstomatal apparatus.
06Tracheids differ from other trachery
elements in[CBSE AIPMT 2014]
(a) having casparian strips
(b) being imperforate
(c) lacking nucleus
(d) being lignified
Ans.(b)
Tracheids and vessels both are called
tracheary elements because their main
function is conduction of sap.
Tracheids differ from other trachaery
elements in being imperforate.
Tracheids are the specific cells which
have pits to support upwards and
lateral conduction ofwater sap.
Tracheid are comparatively short and
single cell, while vessels have more
than one cell and up to 10 cm long.
07The common bottle cork is a
product of [CBSE AIPMT 2012]
(a) dermatogen
(b) phellogen
(c) xylem
(d) vascular cambium
Ans.(b)
The cork cambium or phellogen cells
divide periclinally cutting off cells
towards the outside and inside. The
cells cut off towards the out side
becomes suberised and dead. These
are compactly packed in radial rows
without intercellular spaces and form
corkorphellem.
Cork is impervious to water due to the
suberin and provides protection to
underlying tissues. The cells cut off
from cork cambium towards inside add
to the cortex and are called secondary
cortex cells orphelloderm cells.
08Companion cells are closely
associated with[CBSE AIPMT 2012]
(a) sieve elements (b) vessel elements
(c) trichomes (d) guard cells
Ans.(a)
Companion cellsare characteristic
elements of phloem tissue associated
with the sieve tubes (sieve elements) in
the angiosperms. The sieve tubes and
companion cells are related
ontogenically as they develop from the
same mother cell. The companion cells
and sieve tubes maintain close
cytoplasmic connections through
plasmodesmata.
09Closed vascular bundles lack
[CBSE AIPMT 2012]
(a) ground tissue
(b) conjunctive tissue
(c) cambium
(d) pith
Ans.(c)
Closed vascular bundles lack cambium.
In dicot stems, cambium is present
between phloem and xylem. Such
vascular bundles because of the
presence of cambium, possess the
ability to form secondary xylem and
phloem tissues and hence, are called
open vascular bundles. On the contrary,
vascular bundles in monocots have no
cambium. Hence, they do not form
secondary tissues, and are referred to
as closed.
10Ground tissue includes
[CBSE AIPMT 2011]
(a) all tissues except epidermis and
vascular bundles
(b) epidermis and cortex
(c) all tissues internal to endodermis
(d) all tissues external to endodermis
Ans.(a)
All tissues except epidermis and
vascular bundles constitute the ground
tissue or fundamental tissue. It
consists of simple tissues such as
parenchyma, collenchyma and
sclerenchyma. It includes cortex,
pericycle, medullary rays, in leaves the
ground tissue consists of mesophyll.
11The chief water conducting
elements of xylem in
gymnosperms are
[CBSE AIPMT 2010]
(a) vessels
(b) fibres
(c) transfusion tissue
(d) tracheids
Ans.(d)
The tracheids are elongated, angular
dead cells with hard lignified wide
lumen and narrow end walls. The walls
of tracheids possess different types of
thickenings and the unthickened areas
of its wall allow the rapid movement of
water from one tracheid to another.
Tracheids are the characteristic cell
types of xylem tissues in gymnosperms
and pteridophytes, where they are the
chief elements of water conduction.
12Which one of the following is not a
lateral meristem?
[CBSE AIPMT 2010]
(a) Intrafascicular cambium
(b) Interfascicular cambium
(c) Phellogen
(d) Intercalary meristem
Ans.(d)
Intercalary meristems are the portions
of apical meristems which are
separated from the apex during the
growth of axis and formation of
permanent tissues . It is present mostly
at the base of node, (e.g.Mentha viridis),
base of internode, (e.g. stem of many
monocotsviz,wheat, paddy, grasses;
pteridophytes likeEquisetum) or at the
base of leaf, (e.g.Pinus).
13Palisade parenchyma is absent in
leaves of [CBSE AIPMT 2009]
(a)Sorghum (b) mustard
(c) soyabean (d) gram
Ans.(a)
Sorghum(family–Poaceae) is a monocot
plant. The leaves of monocot do not
contain palisade parenchyma because
the mesophyll of monocot leaf is not
differentiated into palisade and spongy
parenchyma, all being thin walled,
chlorophyllous and irregularly
compactly arranged with fewer
intercellular spaces.
Palisade cells occur in dicotyledonous
plants and also in the net-veined
monocots, the Araceae and
Dioscoreaceae.
Anatomy of Flowering Plants 61

72. 14The annular and spirally thickened
conducting elements generally
develop in the protoxylem when
the root or stem is
[CBSE AIPMT 2009]
(a) maturing (b) elongating
(c) widening (d) differentiating
Ans.(c)
Vessels or tracheae are made up of a
row of cells, placed one above the
other, with their intervening walls
absent or variously pored. The walls of
vessels are lignified and hard but not
very thick. The cell cavity or the lumen
is wide. The thickening may be annular,
spiral, scalariform, reticulate and
pitted.
15In barley stem, vascular bundles
are [CBSE AIPMT 2009]
(a) open and scattered
(b) closed and scattered
(c) open and in a ring
(d) closed and radial
Ans.(b)
The vascular bundles inHordeum
vulgare(barley) plant are scattered in
ground tissues, many in number and
vary in size-smaller towards periphery
and bigger towards centre of the
ground tissue, oval or rounded in
outline, conjoint, collateral and closed.
16Reduction in vascular tissue,
mechanical tissue and cuticle is
characteristic of
[CBSE AIPMT 2009]
(a) xerophytes (b) mesophytes
(c) epiphytes (d) hydrophytes
Ans.(d)
In hydrophytes, vascular tissue and
mechanical tissue are reduced. Cuticle
is either completely absent or if
present, it is thin and poorly developed.
In xerophytes, cuticle is heavy and well
developed. Vascular tissue and
mechanical tissue are well developed
and differentiated.
In mesophytes, cuticle in aerial part is
moderately developed. Vascular and
mechanical tissues are fairly developed
and well differentiated.
17The length of different internodes
in a culm of sugarcane is variable
because of[CBSE AIPMT 2008]
(a) shoot apical meristem
(b) position of axillary buds
(c) size of leaf lamina at the node
below each internode
(d) intercalary meristem
Ans.(d)
Intercalary meristem is present away
from apical meristem in primary
permanent tissue. It is present at the
base of internodes, e.g. in
family–Gramineae or at the base of
leaves, e.g.Pinusor at the base of node,
e.g.Mentha.Intercalary meristem is
responsible for increase in length.
The shoot apical meristem present at
the apices of shoot, is self determining
and autonomous organising centre. The
primary growth and differentiation of
primary tissues is entirely due to this
meristem as it continuously divides
giving rise to new cells. These are the
apical meristems that increase the
length of plant but not internodes
variability.
18Vascular tissues in flowering
plants develop from
[CBSE AIPMT 2008]
(a) phellogen
(b) plerome
(c) periblem
(d) dermatogen
Ans.(b)
Histogen theory for shoot apical
meristem has been proposed by
Hanstein (1870). It advocates that there
are three distinct meristematic zones
(layers) called dermatogen, periblem
and plerome. The dermatogen is the
outermost histogen giving rise to
epidermis, periblem is the middle one
producing the cortex and plerome is the
innermost resulting in central cylinder,
(i.e. vascular tissue).
Cork cambium (phellogen) is the
secondary lateral meristem found in
outer cortical region. Its cells divide
periclinally cutting off cells towards the
outside (forming cork or phellem) and
inside (forming secondary cortex or
phelloderm).
19Passage cells are thin walled cells
found in [CBSE AIPMT 2007]
(a) endodermis of roots facilitating
rapid transport of water from
cortex to pericycle
(b) phloem elements that serve as
entry points for substances for
transport to other plant parts
(c) testa of seeds to enable
emergence of growing embryonic
axis during seed germination
(d) central region of style through
which the pollen tube grows
towards the ovary
Ans.(a)
In roots, endodermis is the innermost
layer of cortex. Some of the
endodermal cells present opposite to
the xylem patches are thin walled and
are calledpassage cellsortransfusion
cells.
Passage cells help in transfer of water
and dissolved salts from cortex directly
into the xylem and ultimately to the
pericycle.
20A common structural feature of
vessel elements and sieve tube
elements are[CBSE AIPMT 2006]
(a) pores on lateral walls
(b) presence of p-protein
(c) enucleate condition
(d) thick secondary walls
Ans.(a)
The vessels are nucleated and the sieve
tube elements are enucleated.
The wall of both the vessel and sieve
tube elements are perforated by large
openings. Due to these adaptation the
cell to cell contact is possible.
21Chlorenchyma is known to develop
in the [CBSE AIPMT 2003]
(a) pollen tube ofPinus
(b) cytoplasm ofChlorella
(c) mycelium of a green mould such
asAspergillus
(d) spore capsule of a moss
Ans.(d)
The apophysis of moss capsule
contains chloroplast bearing
parenchymatous cells, called as
chlorenchyma. Due to the presence of
chloroplasts, chlorenchyma cells have
the ability to prepare food by the
process of photosynthesis.
22The apical meristem of the root is
present [CBSE AIPMT 2003]
(a) in all the roots
(b) only in radicals
(c) only in tap roots
(d) only in adventitious roots
Ans.(a)
Apical meristems are primary
meristems which are located in the
growing points (roots and shoot
apices), as well as buds in the axils of
leaves. The various organs increase in
length due to the activity of apical
meristem.
62 NEETChapterwise Topicwise Biology

73. 23The cells of the quiescent centre
are characterised by
[CBSE AIPMT 2003]
(a) dividing regularly to add to tunica
(b) having dense cytoplasm and
prominent nuclei
(c) having light cytoplasm and small
nuclei
(d) dividing regularly to add to the
corpus
Ans.(c)
The region of quiescent centre was
discovered byClowes(1958). The cells
of quiescent centre have lower
concentration of DNA, RNA and protein
as compared to other cells in the root
apex. These cells also have fewer
mitochondria, little endoplasmic
reticulum and very small nuclei and
nucleoli.
24Vessels are found in
[CBSE AIPMT 2002]
(a) all angiosperms and some
gymnosperms
(b) most of angiosperms and few
gymnosperms
(c) all angiosperms and few
gymnosperms and some
pteridophytes
(d) all pteridophytes
Ans.(b)
Most angiosperms have vessels except
a few, (e.g.Drimys, Tetracentron,
Trochodendron). The gymnosperms, as
a rule, lack vessels but these are found
in the order–Gnetales. Vessels are the
constituent of xylem. They are
composed of row of cells placed one
above the other. Transverse wall of
these cells is absent due to the
dissolution.
25Four radial vascular bundles are
found in [CBSE AIPMT 2002]
(a) dicot root (b) monocot root
(c) dicot stem (d) monocot stem
Ans.(a)
Roots have radial vascular bundles
while stems have conjoint vascular
bundles. Dicot roots have 3-6 vascular
bundles while monocot roots have more
than 6 vascular bundles.
26Axillary bud and terminal bud are
derived from the activity of
[CBSE AIPMT 2002]
(a) lateral meristem
(b) intercalary meristem
(c) apical meristem
(d) parenchyma
Ans.(c)
It is the shoot apical meristem which
gives rise to lateral buds. The lateral
buds however remain suppressed due
to the apical dominance.
27Which of the following statements
is true? [CBSE AIPMT 2002]
(a) Vessels are multicellular with
narrow lumen
(b) Tracheids are multicellular with
narrow lumen
(c) Vessels are unicellular with wide
lumen
(d) Tracheids are unicellular with
wide lumen
Ans.(a)
Each vessel is made up of a number of
components called ‘vessel members’
arranged end-to-end running parallel to
the long axis of the organ in which it lies.
28What happens during
vascularisation in plants?
[CBSE AIPMT 2000]
(a) Differentiation of procambium is
immediately followed by the
development of secondary xylem
and phloem
(b) Differentiation of procambium
followed by the development of
xylem and phloem
(c) Differentiation of procambium,
xylem and phloem is simultaneous
(d) Differentiation of procambium
followed by the development of
primary phloem and then by
primary xylem
Ans.(c)
From the procambium, primary xylem
and phloem arise simultaneously.
29Which of the following meristems
is responsible for extrastelar
secondary growth in
dicotyledonous stem?
[CBSE AIPMT 1998]
(a) Intrafascicular cambium
(b) Interfascicular cambium
(c) Intercalary meristem
(d) Phellogen
Ans.(d)
Phellogen or cork cambium is a part of
periderm. It presents between phellem or
cork towards outerside and phelloderm or
secondary cortex towards inner side.
Phellogen appearing in the cortical
regions cuts off new cells for extrastelar
secondary growth—cork on the outer side
and secondary cortex on the inner side.
30A leaf primordium grows into the
adult leaf lamina by means of
[CBSE AIPMT 1998]
(a) apical meristem
(b) lateral meristem
(c) marginal meristem
(d) at first by apical meristem and
later largely by marginal meristem
Ans.(d)
Apical meristem is present at the tips of
stems, roots and leaves, it takes part in
initial growth or elongation of roots,
stems and leaves. Marginal meristem or
plate meristem has parallel layers of
cells which divide anticlinally in two
planes as in growing flat organs
like-leaves.
31At maturity which of the following
is enucleate?[CBSE AIPMT 1997]
(a) Sieve cell (b) Companion cell
(c) Palisade cell (d) Cortical cell
Ans.(a)
Phloem (complex tissue) is composed
of companion cells, phloem
parenchyma, phloem fibre and sieve
tube cells. Sieve tube cells are
cylindrical and tube-like structure
which are involved in transport of
organic solute. In sieve cells nucleus is
evident in the younger stage but
disappears in mature stage.
32What is not true about sclereids?
[CBSE AIPMT 1996]
(a) These are parenchyma cells with
thickened lignified walls
(b) These are elongated and flexible
with tapered ends
(c) These are commonly found in the
shells of nuts and in the pulp of
guava, pear, etc
(d) These are also called the stone
cells
Ans.(a)
Sclereidsare broad sclerenchyma cells
which may be oval, spherical, cylindrical,
or stellate in structure. Sclereids
develop from sclerenchyma cells, occur
Anatomy of Flowering Plants 63
Phloem
Xylem
Radial vascular bandies in dicot root

74. singly or in groups to provide stiffness.
These may be of different types, such as
brachysclereids (stone cells) found in grit
of pear, apple, macrosclereids
(columnar), e.g. legume seeds,
astrosclereids (star shaped), e.g. tea
leaves, etc.
33Bordered pits are found in
[CBSE AIPMT 1993]
(a) sieve cells
(b) vessel wall
(c) companion cells
(d) sieve tube wall
Ans.(b)
Pits are the depressions or cavity
formed in the cell wall that are found in
the sclerenchyma, thick walled
parenchyma cells and the tracheary
elements, (i.e. vessels and tracheids).
Simple pits are uniform while bordered
pits are the flask shaped depressions.
34Which is correct about transport
or conduction of substances?
[CBSE AIPMT 1991, 97]
(a) Organic food moves up through
phloem
(b) Organic food moves up through
xylem
(c) Inorganic food moves upwardly
and downwardly through xylem
(d) Organic food moves upwardly and
downwardly through phloem
Ans.(d)
Sieve tubes are the conducting
elements of phloem that carry the
organic nutrients upwardly and
downwardly. Organic nutrients
absorbed through roots reaches the
shoot apex and other parts through
phloem cells. Moreover, nutrients are
also translocated downwardly through
phloem as demonstrated by girdling
experiment (Malpighi; 1671) in which
removal of ring of bark (including
phloem) prevents downward
translocation of food and causing
starvation and death of roots.
35Angular collenchyma occurs in
[CBSE AIPMT 1991]
(a)Cucrbita (b)Tagetes
(c)Althaea (d)Salvia
Ans.(b)
Depending upon the thickening,
collenchyma is of three types—(a)
Angular—thickening at the angles, e.g.
stem of tomato,Datura, Tagetes
(marigold), (b) Lamellar—thickening on
tangential walls, e.g. stem of sunflower,
(c) Lacunate—thickening on the walls
bordering intercellular spaces, e.g.
stem ofCucurbita.
36An organised and differentiated
cellular structure having
cytoplasm but no nucleus is
[CBSE AIPMT 1991]
(a) vessels (b) xylem parenchyma
(c) sieve tubes (d) tracheids
Ans.(c)
Sieve tubes are food conducting
elements of phloem. They possess
nucleus in the young stage but
disappears in mature ones. The central
part of sieve tube has an organised and
differentiated cellular structure with a
network of cytoplasmic strands, though
the peripheral cytoplasm is thin and
tonoplast is absent.
37Collenchyma occurs in the stem
and petioles of[CBSE AIPMT 1990]
(a) xerophytes (b) monocots
(c) dicot herbs (d) hydrophytes
Ans.(c)
Collenchymais a simple permanent
tissue, made of elongated living cells
having thickened pectocellulosic walls.
It is a living mechanical tissue which
provides both mechanical strength and
elasticity and allow them to grow in
size. Collenchyma is found in epidermis
of dicot stem and petioles and
abundant in climbing stems.
38Monocot leaves possess
[CBSE AIPMT 1990]
(a) intercalary meristem
(b) lateral meristem
(c) apical meristem
(d) mass meristem
Ans.(a)
Intercalary meristems are derived from
the apical meristems and separated
from the same by permanent cells.
These meristems occur at leaf bases
and above or below the nodes (e.g.
grass, mint). They help in elongation of
leaves and internodes besides allowing
the prostrate stems to become erect.
39Organisation of stem apex into
corpus and tunica is determined
mainly by [CBSE AIPMT 1989]
(a) planes of cell division
(b) regions of meristematic activity
(c) rate of cell growth
(d) rate of shoot tip growth
Ans.(a)
Tunica(outer zone of shoot apex) forms
protoderm which through anticlinal
divisions gives rise to epidermis of stem
and leaves.Corpusis inner mass of
meristem where cells undergo divisions
in different planes to form procambium
and ground meristem.
40Sieve tubes are suited for
translocation of food because
they possess[CBSE AIPMT 1989]
(a) bordered pits
(b) no ends walls
(c) broader lumen and perforated
cross walls
(d) no protoplasm
Ans.(c)
Sieve tubes function as the food
conducting elements of phloem which
are elongated tubular channels formed
by end to end union of numerous cells.
Sieve tubes have broader lumen, thin
walls. Septa present between sieve tube
cells are called sieve plates, they
possess a number of perforations called
sieve pores or sieve pits.
41Death of protoplasm is a
prerequisite for a vital function
like [CBSE AIPMT 1989]
(a) transport of sap
(b) transport of food
(c) absorption of water
(d) gaseous exchange
Ans.(a)
Certain cells get lignified, leading to
death of protoplasm such as xylem
cells. Xylem cells are dead, i.e. devoid of
protoplasm, and performs the function
of conducting water or sap inside the
plant.
42Out of diffuse porous and ring
porous woods, which is correct?
[CBSE AIPMT 1989]
(a) Ring porous wood, carries more
water for short period
(b) Diffuse porous wood carries more
water
(c) Ring porous wood carries more
water when need is higher
(d) Diffuse porous wood is less
specialised but conducts water
rapidly through out
64 NEETChapterwise Topicwise Biology

75. Ans.(c)
In dicots, large sized vessels are
arranged in two ways—ring porous
(large sized vessels in early wood, e.g.
Quercus) and diffuse porous (large sized
vessels distributed throughout, e.g.
Azadirachta). Ring porous vessels are
more efficient and advanced as they
provide quicker translocation when
water requirement is maximum.
43Which meristem helps in
increasing girth?
[CBSE AIPMT 1988]
(a) Lateral meristem
(b) Intercalary meristem
(c) Primary meristem
(d) Apical meristem
Ans.(a)
Lateral meristem occurs on the sides
and helpful in increasing girth of stem
and root, e.g. vascular cambium,
phellogen (cork cambium).
44Tunica corpus theory is connected
with [CBSE AIPMT 1988]
(a) root apex
(b) root cap
(c) shoot apex
(d) secondary growth
Ans.(c)
Tunica corpus theory, given bySchmidt
(1927) is related with shoot apex or stem
apical meristem. According to it, tunica
is the outer zone of shoot apex while
corpus is inner zone. Tunica forms
protoderm that give rise to epidermis of
stem and leaves. Corpus is the inner
mass which undergoes divisions to form
procambium and ground meristem.
45In dicot root, the vascular
cambium originates from
[NEET (Odisha) 2019]
(a) tissue located below the phloem
bundles and a portion of pericycle
tissue above protoxylem
(b) cortical region
(c) parenchyma between endodermis
and pericycle
(d) intrafascicular and interfascicular
tissue in a ring
Ans.(a)
In dicot root, the vascular cambium
originates from tissues located below
the phloem bundles and a portion of
pericycle tissue above protoxylem.
Vascular cambium is the meristematic
layer that is responsible for cutting off
vascular tissues (xylem and phloem). In
young stem, it is present in patches as a
single layer between the xylem and
phloem.
46Casparian strips occur in
[NEET 2018]
(a) cortex (b) pericycle
(c) epidermis (d) endodermis
Ans.(d)
Casparian strips are found in
endodermisof roots. It is a band of
thickening which runs along the radial
and tangential walls of endodermal
cells. It is made up of suberin and lignin.
Casparian strips prevent plasmolysis of
endodermal cells.
Cortexis found below epiblema. It is
made up of thin-walled parenchymal
cells.Epidermisis the outermost layer
made up of thin-walled flattened and
slightly elongated parenchymal cells.
Pericycleis found below endodermis
and it is made of parenchymatous cells.
47The balloon-shaped structures
called tyloses[NEET 2016, Phase II]
(a) originate in the lumen of vessels
(b) characterise the sapwood
(c) are extensions of xylem
parenchyma cells into vessels
(d) are linked to the ascent of sap
through xylem vessels
Ans.(c)
The tyloses are the structures found in
the woody tissues of dicot stems.
These are the extensions of xylem
parenchyma cells into the vessel
elements.
48Cortex is the region found
between [NEET 2016, Phase II]
(a) epidermis and stele
(b) pericycle and endodermis
(c) endodermis and pith
(d) endodermis and vascular bundle
Ans.(a)
Cortex tissue is found in between the
epidermis and stele. It is multilayered
and is made up of parenchymatous cells
with big intercellular spaces.
49Anatomically fairly old
dicotyledonous root is
distinguished from the
dicotyledonous stem by
[CBSE AIPMT 2009]
(a) absence of secondary xylem
(b) absence of secondary phloem
(c) presence of cortex
(d) position of protoxylem
Ans.(d)
In dicotyledonous root, the condition of
xylem is exarch as the protoxylem is
away from the centre and metaxylem is
towards the centre. In dicotyledonous
stem, (e.g.Cucurbita), the condition of
xylem is endarch as the metaxylem is
away from the centre and protoxylem is
towards the centre.
50In a woody dicotyledonous tree
which of the following parts will
mainly consist of primary tissues?
[CBSE AIPMT 2005]
(a) All parts
(b) Stem and root
(c) Flowers, fruits and leaves
(d) Shoot tips and root tips
Anatomy of Flowering Plants 65
Internal Structure
of Dicot Plants
TOPIC 2
Parenchyma
Structure of tyloses
in woody tissue
(a) in longitudinal section (b) in cross section
Epiblema
Cortex
Endodermis
Pericycle
Phloem
Protoxylem
Metaxylem
Pith
Root hair
Stele
Structure of a portion of
TS of Gram root

76. Ans.(d)
Primary tissues are those meristematic
tissues which are derived directly from
embryonal tissues.
e.g. shoot apex and root apex.
51Ectophloic siphonostele is found
in [CBSE AIPMT 2005]
(a)Adiantumand Cucurbitaceae
(b)OsmundaandEquisetum
(c)MarsileaandBotrychium
(d)Dicksoniaand maiden hair fern
Ans.(b)
In the ectophloic siphonostele, the
xylem surrounds pith and this xylem is
surrounded by phloem, pericycle and
endodermis respectively. e.g.Osmunda
andEquisetum.
52In a longitudinal section of root,
starting from the tip upward, the
four zones occur in the following
order [CBSE AIPMT 2004]
(a) root cap, cell division, cell
enlargement, cell maturation
(b) root cap, cell division, cell
maturation, cell enlargement
(c) cell division, cell enlargement, cell
maturation, root cap
(d) cell division, cell maturation, cell
enlargement, root cap
Ans.(a)
In a longitudinal section of root, starting
from the tip upward the four zones
occur in the following order
Root cap→Zone of cell division→Zone
of cell enlargement→Zone of cell
maturation
53Pericycle of roots produces
[CBSE AIPMT 1990]
(a) mechanical support
(b) lateral roots
(c) vascular bundles
(d) adventitious buds
Ans.(b)
In roots, pericycle lies below
endodermis and is made of one or more
layers of parenchymatous cells.
Pericycle gives rise to lateral roots, root
branches of vascular cambium and
whole of cork cambium.
54Where do the casparian bands
occur?[CBSE AIPMT 1990, 94, 99]
(a) Epidermis (b) Endodermis
(c) Pericycle (d) Phloem
Ans.(b)
In roots, the inner most layer of cortex,
i.e. endodermis consists of compactly
arranged barrel shaped cells that
possess ligno-suberin thickenings
calledcasparian strips.In dicot stems,
endodermis is called starch sheath, it
does not contain casparian strips.
55A bicollateral vascular bundle is
characterised by
[CBSE AIPMT 1992]
(a) phloem being sandwitched
between xylem
(b) transverse splitting of vascular
bundle
(c) longitudinal splitting of vascular
bundle
(d) xylem being sandwitched between
phloem
Ans.(d)
Bicollateral vascular bundles are
conjoint bundles having phloem both on
the outer and inner side of xylem, e.g.
Cucurbita. These vascular bundle
generally seen in Solanaceae (the
potato family) and cucurbitaceae (the
cucumber family). In this situation
phloem is present on both outside and
inside the xylem.
56Select the correct pair.[NEET 2021]
(a) Large colourless empty cells in
the epidermis of grass leaves−
Subsidiary cells
(b) In dicot leaves, vascular bundles
are surrounded by large
thick-walled cells−Conjunctive
tissue
(c) Cells of medullary rays that form
part of cambial ring−
Interfascicular cambium
(d) Loose parenchyma cells rupturing
the epidermis and forming a
lens-shaped opening in bark−
Spongy parenchyma
Ans.(c)
Medullary rays (pith rays or wood rays)
are sheets or ribbons of cells running
from the inside of the plant to the
outside. That is, they run at right angles
to the xylem and phloem, which run
vertically. While the plant is alive, these
medullary cells are alive. In dicot stems,
the cambium which is present between
primary xylem and primary phloem is
called intrafascicular cambium. The
cells of medullary rays near these
intrafascicular cambium become
meristematic and form interfascicular
cambium. This leads to the formation of
a continuous ring of cambium.
Other options can be corrected as :
Few epidermal cells in the vicinity of
guard cells become specialised in their
shape and size and are called subsidiary
cells.
The parenchymatous cells which lie
between xylem and the phloem are
called conjunctive tissue.
A spongy layer of irregular chlorophyll
bearing cells interspersed with air
spaces that fills the interior part of leaf
below the palisade layer is called
spongy parenchyma.
57Match the List -I with List - II.
List-I List-II
A. Lenticels 1. Phellogen
B. Cork cambium 2. Suberin
deposition
C. Secondary
cortex
3. Exchange of
gases
D. Cork 4. Phelloderm
Choose the correct answer from the
options given below.[NEET 2021]
A B C D
(a) 4 1 3 2
(b) 3 1 4 2
(c) 2 3 4 1
(d) 4 2 1 3
Ans.(b)
(A)-(3), (B)-(1), (C)-(4), (D)-(2)
Lenticelspermit the exchange of gases
between the environment and the
internal tissue spaces of the organs
(stems and some fruits). They permit
the entrance of oxygen and
simultaneously the output of carbon
dioxide and water vapour. Thus, they
are responsible for gaseous exchange.
Cambium,called the phellogen or cork
cambium, is the source of the periderm,
a protective tissue that replaces the
epidermis when the secondary growth
66 NEETChapterwise Topicwise Biology
Pith
Xylem
Phloem
Ectophloic siphonostele
Secondary Growth
TOPIC 3

77. displaces, and ultimately destroys, the
epidermis of the primary plant body.
Phellodermis the parenchymatous
tissue which originates from the
phellogen towards its inner side known
as the secondary cortex. It is a living
tissue having a cellulosic cell wall.
Cork tissue,consisting of dead cells
surrounded by alternating layers of
suberin and wax, has a particularly high
suberin content. Cork cells are found in
a secondary protective layer (periderm)
in the bark of trees.
58Which of the following statements
about cork cambium is incorrect?
[NEET (Oct.) 2020]
(a) It forms secondary cortex on its
outerside
(b) It forms a part of periderm
(c) It is responsible for the formation
of lenticels
(d) It is a couple of layers thick
Ans.(a)
Cork cambium is a meristematic tissue
involved in secondary growth. It is also
called phellogen. It is few layer thick
and it cut off cells into an outer layer
and an inner layer. The former
differentiates into cork or phellem while
the latter differentiate into secondary
cortex or phelloderm. The phellogen,
phellem and phelloderm are collectively
known as periderm. In woody trees,
phellogen cuts off closely arranged
parenchymatous cells which ruptures
the epidermis and form lenticels. Thus,
statement 1 is incorrect while other are
correct.
59Which of the statements given
below is not true about formation
of annual rings in trees?
[NEET (National) 2019]
(a) Differential activity of cambium
causes light and dark bands of
tissue early and late wood,
respectively
(b) Activity of cambium depends
upon variation in climate
(c) Annual rings are not prominent in
trees of temperate region
(d) Annual ring is a combination of
springwood and autumnwood
produced in a year
Ans.(c)
The statement “annual rings are not
prominent in trees of temperate region’’
is incorrect. Correct information about
the statement is as follows:
Annual rings are formed due to the
seasonal activity of cambium. In the
plants of temperate region, cambium is
highly active in spring and less active in
autumn season. Hence, prominent
rings are formed in these plants having
light and dark bands of tissue. Rest
statements are correct about the
formation of annual rings in trees.
60Plants having little or no
secondary growth are[NEET 2018]
(a) conifers
(b) deciduous angiosperms
(c) grasses
(d) cycads
Ans.(c)
Secondary growth occurs due to the
presence of vascular cambium.Grasses
are monocot and lacks vascular
cambium. Therefore, they do not show
secondary growth.
Deciduous angiospermsare usually
woody dicot plants and show secondary
growth.Conifersandcycadsare
gymnosperms and usually show
anomalous secondary growth.
61Secondary xylem and phloem in
dicot stem are produced by
[NEET 2018]
(a) phellogen
(b) vascular cambium
(c) apical meristems
(d) axillary meristems
Ans.(b)
Secondary vascular tissues,i.e.,
secondary xylem and pholem are
formed by thevascular cambium. It is
produced by two types of meristems;
fascicular or intrafascicular and
interfascicular cambium.
Intrafascicular cambium is a primary
meristem which occurs as strips in
vascular bundles.
It divides to form secondary phloem on
outer side and secondary xylem on the
inner side. Interfascicular cambium
arises secondarily from the cells of
medullary rays.
Phellogenorcork cambiumis produced
in the outer cortical cells of dicot
stems. It is helpful in increasing the
girth.Apical meristemsare present at
the tips of stem, root and their
branches. They are responsible for
increase in length of the plant.Axillary
meristemis found in axillary buds.
These cells are left behind from shoot
apical meristem during the formation of
leaves and elongation of stems.
62Plants having little or no
secondary growth are[NEET 2018]
(a) conifers
(b) deciduous angiosperms
(c) grasses
(d) cycads
Ans.(c)
Secondary growth occurs due to the
presence of vascular cambium.Grasses
are monocot and lacks vascular
cambium. Therefore, they do not show
secondary growth.
Deciduous angiospermsare usually
woody dicot plants and show secondary
growth.Conifersandcycadsare
gymnosperms and usually show
anomalous secondary growth.
63Identify the wrong statement in
context of heartwood.[NEET 2017]
(a) Organic compounds are deposited
in it
(b) It is highly durable
(c) It conducts water and minerals
efficiently
(d) It comprises dead elements with
highly lignified walls
Ans.(c)
Heartwood also called duramen is the
central dead wood of trees. It
comprises of dead, lignified cells
containing organic compounds, e.g.
tannins or other substances. These
substances make it darker in colour and
aromatic. Heartwood is strong durable
and resistant to decay. It does not
conduct water and minerals because of
the presence of dead elements.
Thinking Process The conduction of
water and minerals is carried out by
sapwood, because it contains living
cells.
64The vascular cambium normal
gives rise to [NEET 2017]
(a) phelloderm
(b) primary phloem
(c) secondary xylem
(d) periderm
Ans.(c)
Vascular cambium located between
xylem and phloem in the stems and
roots of vascular plants. It produces
secondary xylem towards the pith and
secondary phloem towards the bark.
Phellogen is made of narrow thin-walled
and nearly rectangular cells. phellogen
cuts off cells on both sides. The outer
cells differentiate into cork or phellem
while inner cells differentiate into
Anatomy of Flowering Plants 67

78. 68 NEETChapterwise Topicwise Biology
secondary cortex or phelloderm. The
phellogen, phellem and phelloderm are
collectively known as periderm.
65Read the different components
from I-IV in the list given below and
tell the correct order of the
components with reference to their
arrangement from outer side to
inner side in a woody dicot stem.
[CBSE AIPMT 2015]
I. Secondary cortex
II. Wood
III. Secondary phloem
IV. Phellem
The correct order is
(a) III, IV, II, I (b) I, II, IV, III
(c) IV, I, III, II (d) IV, III, I, II
Ans.(c)
The correct order of arrangement of
the given components from outside to
inside in a woody dicot stem is as
follows
Phellem→Secondary cortex→
Secondary phloem→Wood
66You are given a fairly old piece of
dicot stem and a dicot root. Which
of the following anatomical
structures will you use to
distinguish between the two?
[CBSE AIPMT 2014]
(a) Secondary xylem
(b) Secondary phloem
(c) Protoxylem
(d) Cortical cells
Ans.(c)
We will observe the protoxylem of the
dicot stem and dicot root to distinguish
between them. In dicot stem the
protoxylem is present towards center
(pith) and metaxylem is present towards
the periphery of the organ. This type of
xylem is known as endarch.
In rootprotoxylem is present towards
periphery and metaxylem is present
towards center that is called exarch.
67Interfascicular cambium develops
from the cells of[NEET 2013]
(a) medullary rays
(b) xylem parenchyma
(c) endodermis
(d) pericycle
Ans.(a)
In dicot stems, the cells of cambium
present between primary xylem and
primary phloem is the intrafascicular
cambium. The cells ofmedullary rays,
adjoining these intrafascicular cambium
become meristematic and form the
interfascicular cambium.
Xylem parenchyma are living and
thin-walled and their cell walls are made
up of cellulose. In dicot roots the
innermost layer of the cortex is called
endodermis. Next to endodermis lies a
few layers of thick walled
parenchymatous cells called as
pericycle.
68Age of a tree can be estimated by
[NEET 2013]
(a) its height and girth
(b) biomass
(c) number of annual rings
(d) diameter of its heartwood
Ans.(c)
Age of a tree can be estimated by
counting the number ofannual rings.
This study is known as
dendrocronology. The two kinds of
woods that appear as alternate
concentric rings, constitute an annual
ring heartwood comprises dead
elements with highly lignified walls that
give mechanical support to the stem.
The height and girth of a tree increases
due to the activity of vascular cambium.
69The cork cambium, cork and
secondary cortex are collectively
called [CBSE AIPMT 2011]
(a) phellogen (b) periderm
(c) phellem (d) phelloderm
Ans.(b)
The periderm is a secondary protective
structure and is made up of cork
cambium (phellogen), cork (phellem)
and secondary cortex (phelloderm).
70Heartwood differs from sapwood
in [CBSE AIPMT 2010]
(a) presence of rays and fibres
(b) absence of vessels and
parenchyma
(c) having dead and non-conducting
elements
(d) being susceptible to pests and
pathogens
Ans.(c)
As a result of continued secondary
growth in subsequent years, the older
part of secondary xylem or wood
becomes non-functional (dead) as it
loses the power of conduction. The
cells of this wood are filled with resins
or tannins produced by adjacent
functional cells. The activities of
vessels become blocked by tyloses. Due
to these activities, non-functional,
secondary xylem becomes hard,
durable and blackish in colour, called
heartwood. Now, the function of
secondary xylem (water and mineral
conduction from roots) is performed by
outer younger-rings of xylem which is
called sapwood.
71For a critical study of secondary
growth in plants which one of the
following pairs is suitable?
[CBSE AIPMT 2007]
(a) Sugarcane and sunflower
(b) Teak and pine
(c) Deodar and fern
(d) Wheat and maiden hair fern
Ans.(b)
The increase in diameter or thickness is
due to the formation of secondary
tissues as a result of the activities of
vascular cambium and cork cambium.
This secondary growth is characteristic
of dicot stem and root. Thus,
sugarcane, pine, ferns, wheat, etc.
cannot be used to study secondary
growth.
Metaxylem
(outer side)
Protoxylem
(inside)
Protoxylem
(outer sides)
Metaxylem
(inside)
TS of dicot stem TS of dicot root
Resin duct
Cork
(phellem)
Cork cambium
(phellogen)
Cortex
endodermis
Sclerenchy
matous pericycle
Primary phloem
Secondary
phloem
Cambium
Secondary xylem
Medullary rays
Autumn wood
Spring wood
Metaxylem
Protoxylem
Pith
Primary
Xylem
Annual
rings
TS of a typical dicot stem
showing secondary growth

79. 72Main function of lenticel is
[CBSE AIPMT 2002]
(a) transpiration
(b) guttation
(c) gaseous exchange
(d) bleeding
Ans.(c)
The primary function of lenticels is
gaseous exchange. Lenticel respiration
generally seen in stem of dicotyledons
plants. Transpiration takes place mostly
through stomata. Guttation and
bleeding takes place through
hydathodes.
73As the secondary growth takes
place (proceeds) in a tree,
thickness of[CBSE AIPMT 1994]
(a) heartwood increases
(b) sapwood increases
(c) both increase
(d) both remain the same
Ans.(a)
As a result of continous secondary
growth in subsequent year, the older
part of secondary xylem becomes
non-functional. Due to this activities of
vessels become blocked by bladder like
ingrowths which are called tyloses. Due
to this non-functional xylem becomes
hard and blackish in colour called
duramen or heartwood.
Now, the function of secondary xylem is
continoued by younger rings called
sapwoodoralburnumwith the passage
of time and addition of new outer rings
of secondary xylem more rings of
sapwood changes into heartwood. This
is why the heartwood increases in
diameter year after year but the
sapwood remains almost in the same
thickness.
74Procambium forms
[CBSE AIPMT 1994]
(a) only primary vascular bundles
(b) only vascular cambium
(c) only cork cambium
(d) primary vascular bundles and
vascular cambium
Ans.(d)
The meristematic tissue which forms
the primary xylem and phloem is known
as procambium. The term procambium
is used to indicate the meristematic
tissue that give rise to the morphological
vascular units.
75Abnormal/anomalous secondary
growth occurs in[CBSE AIPMT 1993]
(a)Dracaena (b) ginger
(c) wheat (d) sunflower
Ans.(a)
Secondary growth in monocotyledons is
rather rare, it is commonly seen in
woody monocotyledons such as
Dracaena, Aloe, Agave,etc.InDracaena
sp exceptionally large secondary
growth in thickness occurs that begins
with the formation of secondary
meristematic tissue—the cambium, in
the parenchyma outside the primary
bundles. Moreover, cork inDracaena
appears in seriated bands without the
formation of cork cambium (phellogen)
and is known as storied cork.
76A narrow layer of thin walled cells
found between phloem/bark and
wood of a dicot is[CBSE AIPMT 1993]
(a) cork cambium
(b) vascular cambium
(c) endodermis
(d) pericycle
Ans.(b)
Vascular cambium is formed by strips
of fascicular cambium and
interfascicular cambium. It consists of
narrow layer of thin walled cells found
between phloem and xylem tissues.
Vascular cambium helps in secondary
growth in dicot root and stem.
77Periderm is produced by
[CBSE AIPMT 1993]
(a) vascular cambium
(b) fascicular cambium
(c) phellogen
(d) intrafascicular cambium
Ans.(c)
Phellogen or cork cambium which
develops secondarily from some outer
layer of cortex (or pericycle) divides on
the outside as well as inside to form
respectively cork or phellem and
secondary cortex or phelloderm. Cork,
cork cambium and secondary cortex are
together called periderm.
78Which exposed wood will decay
faster? [CBSE AIPMT 1993]
(a) Sapwood
(b) Softwood
(c) Wood with lot of fibres
(d) Heartwood
Ans.(a)
In old trees, secondary xylem or wood
gets differentiated into outer light
coloured functional sapwood or
alburnum and inner dark coloured
non-functionalheartwoodorduramen.
Heartwood is stronger and highly
durable because of presence of oils,
resins, gums, tannins and tyloses which
are plugged into the tracheids and
vessel elements.
As the secondary growth takes place
size of heart wood increases because
of conversion of inner alburnum
(sapwood) into it.
79Commercial cork is obtained from
[CBSE AIPMT 1991]
(a)Berberis/Barberry
(b)Salix/Willow
(c)Quercus/Oak
(d)Betula/Birch
Ans.(c)
Cork or phellem develops from cork
cambium and is made up of dead
suberised and rectangular cells which
are filled with air and tannins. Cork is
protective and is obtained
commercially fromQuercus suber(cork
oak, bottle cork).
80Cork cambium and vascular
cambium are[CBSE AIPMT 1990]
(a) parts of secondary xylem and
phloem
(b) parts of pericycle
(c) lateral meristems
(d) apical meristems
Ans.(c)
Lateral meristems are present along
the lateral sides of stem and root. They
divide only in radial direction. The
cambium of vascular bundles
(fascicular, intrafascicular) and the cork
cambium or phellogen belongs to this
category and are found in dicotyledons
and gymnosperms.
81For union between stock and
scion in grafting which one is the
first to occur?[CBSE AIPMT 1990]
(a) Formation of callus
(b) Production of plasmodesmata
(c) Differentiation of new vascular
tissues
(d) Regeneration of cortex and
epidermis
Anatomy of Flowering Plants 69

80. 70 NEETChapterwise Topicwise Biology
Ans.(a)
In grafting, cambium bearing shoot
(scion) of one plant is joined to cambium
bearing stump (root system = stock) of
related plant through different union
like tongue grafting, wedge grafting,
etc.
The union of scion and stock leads to
irregular, unorganised and
undifferentiated mass of actively
dividing cells known as callus.
82Vascular cambium produces
[CBSE AIPMT 1990, 92]
(a) primary xylem and primary phloem
(b) secondary xylem and secondary
phloem
(c) primary xylem and secondary
phloem
(d) secondary xylem and primary
phloem
Ans.(b)
Vascular cambium is formed by
fascicular and interfascicular cambium.
It leads to secondary growth in dicot
roots and dicot stems. Cells of vascular
cambium, known asfusiform initially
produce secondary phloem on the
outside and secondary xylem on the
inner side.
Youngest xylem layer occur inner to
vascular cambium while oldest layer of
secondary xylem is found outside the
primary xylem or towards pith. In case
of phloem, youngest layer of secondary
phloem lies just outside the vascular
cambium while oldest layer is towards
outside, inner to primary phloem.
83Cork is formed from
[CBSE AIPMT 1988]
(a) cork cambium (phellogen)
(b) vascular cambium
(c) phloem
(d) xylem
Ans.(a)
Cells of cork cambium (phellogen) divide
on the outside as well as inside to form
respectively cork or phellem and
secondary cortex or phelloderm. Cork or
phellem is made up of dead suberised
and rectangular cells, filled with air and
tannins, it is protective in nature.
84The transverse section of a plant
shows following anatomical
features. [NEET (Sep.) 2020]
I. Large number of scattered
vascular bundles surrounded
by bundle sheath.
II. Large conspicuous
parenchymatous ground
tissue.
III. Vascular bundles conjoint and
closed.
IV. Phloem parenchyma absent.
Identify the category of plant and
its part.
(a) Monocotyledonous root
(b) Dicotyledonous stem
(c) Dicotyledonous root
(d) Monocotyledonous stem
Ans.(d)
The transverse section of
monocotyledonous stem shows
following anatomical features :
The monocot stem have conjoint and
closed vascular bundles, scattered in
the ground tissue containing the
parenchyma. Each vascular bundle is
surrounded by sclerenchymatous
bundle-sheath cells. Phloem
parenchyma and medullary rays are
absent in monocot stems.
85Water containing cavities in
vascular bundles are found in
[CBSE AIPMT 2012]
(a) sunflower (b) maize
(c)Cycas (d)Pinus
Ans.(b)
In monocot stem likeZea mays, vascular
bundles are conjoint, collateral and
closed. In vascular bundles, the
lowermost protoxylem vessels and
xylem parenchyma cells dissolve forming
a water containing schisolysigenous
cavity calledprotoxylem cavityorlacuna
orlysigenous cavity. Protoxylem cavity
and protophloem may be absent in the
smaller vascular bundles in maize.
86What is true about a monocot
leaf? [CBSE AIPMT 1990]
(a) Reticulate venation
(b) Absence of bulliform cells from
epidermis
(c) Mesophyll not differentiated into
palisade and spongy tissues
(d) Well diffferentiated mesophyll
Ans.(c)
Monocot leaves are characterised by
parallel venation, leaves are isobilateral
(both the surfaces equally green),
amphistomatic with dumb bell-shaped
guard cells. The upper epidermis
possesses groups of larger sized thin
walled vacuolate cells called bulliform
or motor cells. Mesophyll is
undifferentiated and consists of
isodiametric cells enclosing
intercellular spaces. Midrib region does
not contain mesophyll and possesses
number of parallel vascular bundles.
87Pith and cortex do not
differentiate in[CBSE AIPMT 1988]
(a) monocot stem (b) dicot stem
(c) monocot root (d) dicot root
Ans.(a)
In monocot stem, ground tissue is
undifferentiated, thus endodermis,
pericycle cortex and pith are not
recognisable. Ground tissue consists of
only parenchyma cells that store food.
Internal Structure or
Monocot Plants
TOPIC 4

81. 01Identify the types of cell junctions
that help to stop the leakage of
the substances across a tissue
and facilitation of communication
with neighbouring cellsviarapid
transfer of ions and molecules.
[NEET 2021]
(a) Gap junctions and adhering
junctions, respectively
(b) Tight junctions and gap junctions,
respectively.
(c) Adhering junction and tight
junctions, respectively
(d) Adhering junctions and gap
junctions, respectively
Ans.(b)
Tight junction helps to stop the leakage
of the substances across a tissue and
Gap junctionare channels that
physically connect neighbouring cells,
mediating the rapid exchange of small
molecules or ions.
Adhering junctionThey are cell-cell
junction complexes that make
important contribution to
embryogenesis and tissue
homeostasis.
02Goblet cells of alimentary canal
are modified from
[NEET (Sep.) 2020]
(a) columnar epithelial cells
(b) chondrocytes
(c) compound epithelial cells
(d) squamous epithelial cells
Ans.(a)
Goblet cells of alimentary canal are
modified from columnar epithelial cells,
which secrete mucus.
These cells are found in the lining of
organs like intestine and respiratory
tract.
These secretes mucin (glycoprotein)
and maintains the layer of mucus. As
each cell secretes mucin for mucous
production it is called as a unicellular
mucous gland.
03Cuboidal epithelium with brush
border of microvilli is found in
[NEET (Sep.) 2020]
(a) ducts of salivary glands
(b) proximal convoluted tubule of
nephron
(c) Eustachian tube
(d) lining of intestine
Ans.(b)
Cuboidal epithelium with brush border
of microvilli is found in proximal
convoluted tubule of nephron. The
epithelium fills the lumen, and the
microvilli increases the surface area by
30-40 fold.
A brush border is a name for the
microvilli-covered surface of simple
cuboidal epithelium and simple
columnar epithelium cells found in
certain locations of the body.
The proximal convoluted tubule of the
vertebrate nephron lies between
Bowman’s capsule and the loop of
Henle and functions especially in the
absorption of sugar, sodium and
chloride ions and water from the
glomerular filtrate.
04Mach the following cell structures
with their characteristic features.
[NEET (Odisha) 2019]
Column I Column II
1. Tight
junctions
(i) Cement
neighbouring cells
together to form
sheet
2. Adhering
junctions
(ii) Transmit information
through chemical to
another cells
3. Gap
junctions
(iii) Establish a barrier to
prevent leakage of
fluid across epithelial
cells
4. Synaptic
junctions
(iv) Cytoplasmic
channels to facilitate
communication
between adjacent
cells
Select the correct option from the
following
1 2 3 4
(a) (ii) (iv) (i) (iii)
(b) (iv) (ii) (i) (iii)
(c) (iii) (i) (iv) (ii)
(d) (iv) (iii) (i) (ii)
Ans.(c)
The correct matches are
1. Tight
junctions
(iii) Establish a barrier to
prevent leakage of
fluid across epithelial
cells.
2. Adhering
junctions
(i) Cement neighbouring
cells together to form
sheet.
StructuralOrganisation
inAnimals
07
Epithelial Tissue
TOPIC 1

82. 3. Gap
junctions
(iv) Cytoplasmic
channels to facilitate
communication
between adjacent
cells
4. Synaptic
junctions
(ii) Transmit information
through chemical to
another cells.
05The ciliated epithelial cells are
required to move particles or
mucus in a specific direction.
In humans, these cells are mainly
present in[NEET (National) 2019]
(a) Fallopian tubes and pancreatic
duct
(b) eustachian tube and salivary duct
(c) bronchioles and Fallopian tubes
(d) bile duct and bronchioles
Ans.(c)
In humans, ciliated epithelial cells are
present in the bronchioles and Fallopian
tube. In bronchioles, these cells help in
the movement of mucus and in
Fallopian tube, these are required to
move the egg towards uterus.
On the other hand, salivary and
pancreatic ducts are lined by simple
cuboidal epithelium. Bile duct is lined by
simple columnar epithelium and
Eustachian tube is lined by columnar
epithelium having ciliated cells.
06Which type of tissue correctly
matches with its location?
[NEET 2016, Phase I]
Tissue Location
(a) Areolar tissue Tendons
(b) Transitional
epithelium
Tip of nose
(c) Cuboidal
epithelium
Lining of
stomach
(d) Smooth muscle Wall of intestine
Ans.(d)
Columnar epithelium is present in the
lining of stomach.
Tendon is dense connective tissue and
connects muscle to bone. Tip of nose
consists of elastic cartilage.
07The function of the gap junction is
to [CBSE AIPMT 2015]
(a) performing cementing to keep
neighbouring
cells together
(b) facilitate communication between
adjoining cells by connecting the
cytoplasm for rapid transfer of
ions, small molecules and some
large molecules
(c) separate two cells from each
other
(d) stop substance from leaking
across a tissue
Ans.(b)
A gap junction may nexus or macula
communicans. These are specialised
intercellular connection between a
multitude of animal cell types. They
directly connect the cytoplasm of two
cells, which allows various molecules,
ions and electrical impulses to directly
pass through a regulated gate between
cells.
08Choose the correctly matched
pair. [CBSE AIPMT 2014]
(a) Inner lining of salivary
ducts– Ciliated epithelium
(b) Moist surface of buccal
cavity– Glandular epithelium
(c) Tubular parts of
nephrons – Cuboidal epithelium
(d) Inner surface of
bronchioles – Squamous
epithelium
Ans.(c)
Cuboidal epithelium is present in the
tubular parts of nephron (PCT and DCT).
It consists of short, cube-shaped cells
with round nuclei located in the centre
of the cell. These cells often forms
microvilli to increase the absorptive
surface area of the cell.
Other correctly matched points are
Inner lining of salivary ducts is lined by
compound epithelium.
Moist surface of buccal cavity is lined by
compound epithelium.
nner surface of bronchioles is lined by
ciliated epithelium.
09In which one of the following
preparations are you likely to
come across cell junctions most
frequently?[CBSE AIPMT 2007]
(a) Ciliated epithelium
(b) Thrombocytes
(c) Tendon
(d) Hyaline cartilage
Ans.(a)
Specialised cell junctions occur at many
points of cell-cell and cell matrix
contact in all tissues, but they are
particularly important and plentiful in
epithelium.
10The cell junctions called tight,
adhering and gap junctions are
found in [CBSE AIPMT 2009]
(a) muscular tissue
(b) connective tissue
(c) epithelial tissue
(d) neural tissue
Ans.(c)
In epithelial tissue, the adjacent cells
form ion-rich gap or cell junctions for
intercellular communication and
chemical exchange. These junctions
probably do not provide physical
support.
11The ciliated columnar epithelial
cells in humans are known to
occur in[CBSE AIPMT 2011, 09]
(a) bronchioles and Fallopian tubes
(b) bile duct and oesophagus
(c) Fallopian tubes and urethra
(d) eustachian tube and stomach
lining
Ans.(a)
The columnar epithelium is composed
of a single layer of tall and slender cells.
If the columnar cells bear cilia on their
free surface, they are called ciliated
columnar epithelium. They are mainly
present in the inner surface of hollow
organs like bronchioles and Fallopian
tubes. Their function is to move
particles or mucus in a specific
direction over the epithelium.
12The kind of epithelium which
forms the inner walls of blood
vessels is [CBSE AIPMT 2010]
(a) cuboidal epithelium
(b) columnar epithelium
(c) ciliated columnar epithelium
(d) squamous epithelium
Ans.(d)
Simple squamous epithelium consists
of only one layer of flat, scale like cells
usually polygonal cells, which are
closely fitted together like the tiles of a
mosaic. It is also known as pavement
epithelium. It forms lining of blood
vessels, lymph vessels, heart,
peritoneum, pleura, Bowman’s capsule,
etc.
12Simple epithelium is a tissue in
which the cells are
[CBSE AIPMT 2000]
(a) hardened and provide support to
the organ
(b) cemented directly to one another
to form a single layer
72 NEETChapterwise Topicwise Biology

83. (c) continuously dividing to provide
form to an organ
(d) loosely connected to one another
to form an irregular organ
Ans.(b)
Simple epithelium is a tissue in which
the cells are cemented directly to one
another to form a single layer. An
epithelium is a sheet or tube of firmly
adherent cells with minimum
(practically negligible) material and
space between them.
13Stratum germinativum is an
example of which kind of
epithelium?[CBSE AIPMT 1997]
(a) Cuboidal (b) Ciliated
(c) Columnar (d) Squamous
Ans.(c)
Stratum germinativum (Stratum
Malpighi/ Stratum cylindricum) consists
of columnar cells resting upon a
common basement membrane.
14Basement membrane is made up
of [CBSE AIPMT 1997]
(a) only epidermal cells
(b) only endodermal cells
(c) Both (a) and (b)
(d) no cell at all, but is a product of
epithelial cells
Ans.(d)
The cells of epithelial tissues rest upon
a thin layer composed of protein bound
mucopoly- saccharides and
glycoproteins, both secreted by
epithelial cells (hence, product of
epithelial cell) along with a layer of
collagen fibres of the underlying
connective tissue.
15Epithelial tissue with thin flat cells
appearing like packed tiles occurs
on [CBSE AIPMT 1994]
(a) inner lining of cheek
(b) inner lining of stomach
(c) inner lining of Fallopian tubes
(d) inner lining of ovary
Ans.(d)
Epithelial tissue with thin flat cells
appearing like packed tiles occurs on
inner lining of ovary. It is called germinal
epithelium.
16The alveolar epithelium in the lung
is [CBSE AIPMT 1990]
(a) non-ciliated columnar
(b) non-ciliated squamous
(c) ciliated columnar
(d) ciliated squamous
Ans.(b)
The wall of alveoli is highly vascularised.
It is surrounded by flattened,
non-ciliated squamous cells to increase
the surface area. Simple columnar
ciliated cells line the few portions of
upper respiratory tract, oviduct,
Fallopian tube and neurocoel of CNS,
whereas non-ciliated columnar
epithelium lines stomach, intestine,
digestive glands and gall bladder.
17Choose the correctly matched
pair. [CBSE AIPMT 2014]
(a) Tendon – Specialised connective
tissue
(b) Adipose tissue – Dense connective
tissue
(c) Areolar tissue – Loose connective
tissue
(d) Cartilage– Loose connective
tissue
Ans.(c)
Areolar tissue is the body’s loose
connective tissue, and provide
flexibility and cushioning. Adipose
tissue is also loose connective tissue
while, the tendon is a dense connective
tissue, which connect the muscles with
the bone. Cartilage is composed of
specialised connective tissue called
chondrocytes that produce a large
amount of extracellular matrix
composed of collagen fibre.
18The kind of tissue that forms the
supportive structure in our pinna
(external ears) is also found in
[CBSE AIPMT 2009]
(a) vertebrae
(b) nails
(c) ear ossicles
(d) tip of the nose
Ans.(d)
Yellow fibrous cartilage tissue is found
in pinna (external ear). It is also found at
the tip of the nose.
19Which one of the following
contains the largest quantity of
extracellular material?
[CBSE AIPMT 2003]
(a) Myelinated nerve fibres
(b) Striated muscle
(c) Areolar tissue
(d) Stratified epithelium
Ans.(c)
Loose connective tissue, also called
areolar connective tissue, is the ‘packing
material’ of the body that anchors blood
vessels, nerves and body organs. It
contains fibroblasts that synthesise the
fibres and ground substance of
connective tissue and wandering
macrophages that phagocytise pathogens
or damaged cells. The different fibre
types include strong collagen fibres and
thin elastic fibres formed of the protein
elastin. Adipose (fat) tissue is considered
a type of loose connective tissue.
20Compared to blood our lymph has
[CBSE AIPMT 2009, 1989]
(a) no plasma
(b) plasma without proteins
(c) more WBCs and no RBCs
(d) more RBCs and less WBCs
Ans.(c)
Lymph can be defined as blood minus
RBCs, also it contains more WBCs.
Lymph is a clear, colourless fluid,
similar to plasma but with less protein.
It is a mobile connective tissue like,
blood and is formed by the filtration of
blood. Microscopic examination of
lymph depicts that it contains a large
number of leucocytes (mostly
lymphocytes), ranging from 500-75,000
per cubic mm. But platelates are not
present in lymph.
21The most active phagocytic white
blood cells are[CBSE AIPMT 2008]
(a) neutrophils and eosinophils
(b) lymphocytes and macrophages
(c) eosinophils and lymphocytes
(d) neutrophils and monocytes
Ans.(d)
Neutrophils and monocytes are
phagocytic white blood cells.
Monocytes are largest of all leucocytes
and generally change into macrophages
after entering tissue spaces.
Neutrophils are most numerous of all
leucocytes, and have many lobed
nucleus.
Eosinophils are granular white blood
cells. Their number increases in people
with allergic conditions such as asthma
or hay fever. They are non-phagocytic
and seem to play part in immune system.
Lymphocytes are non-motile and
non-phagocytic. These are found as B
and T-lymphocytes. B-lymphocytes
secrete antibodies to destroy microbes.
T-lymphocytes either directly attack
the antigens or stimulate
B-lymphocytes to produce antibodies.
Structural Organisation in Animals 73
Connective Tissue
TOPIC 2

84. 22Which type of white blood cells
are concerned with the release of
histamine and the natural
anticoagulant heparin?
[CBSE AIPMT 2008]
(a) Neutrophils (b) Basophils
(c) Eosinophils (d) Monocytes
Ans.(b)
Thebasophilsare probably like mast
cells of connective tissue. They release
heparin, (a natural anticoagulant),
histamine and serotonin. Their nucleus
is usually three lobed and their granules
take basic stain strongly.
Monocytesare largest of all types of
leucocytes. Their nucleus is
bean-shaped. They are motile and
phagocytic in nature and engulf
bacteria and cellular debris. Generally,
they change into macrophages after
entering tissue spaces.
Eosinophilshave two-lobed nucleus.
They are non-phagocytic and help in
dissolving blood clot. Their number
increases in people with allergic
conditions such as asthma or hay fever.
Neutrophilsare most numerous of all
leucocytes. They eat harmful germs and
are therefore, phagocytic in nature.
Their nucleus is many lobed and stain
weakly with both acid and basic stains.
23Which one of the following
mammalian cells is not capable of
metabolising glucose to
carbon-dioxide aerobically?
[CBSE AIPMT 2007]
(a) White blood cells
(b) Unstriated muscle cells
(c) Liver cells
(d) Red blood cells
Ans.(d)
Cell organelles and nucleus are absent
in mature red blood cells. Therefore,
aerobic respiration do not take place
red blood cells.
24A drop of each of the following, is
placed separately on four slides.
Which of them will not coagulate?
[CBSE AIPMT 2007]
(a) Blood plasma
(b) Blood serum
(c) Sample from the thoracic duct of
lymphatic system
(d) Whole blood from pulmonary vein
Ans.(b)
Serum will not coagulate if placed
separately on a slides. This is because
in blood, the serum is the component
that does not contain blood cell (WBCs
or RBCs) nor any clotting factor. It is the
blood plasma without the fibrinogens.
25Areolar connective tissue joins
[CBSE AIPMT 2006]
(a) integument with muscles
(b) bones with muscles
(c) bones with bones
(d) fat body with muscles
Ans.(a)
Loose or areolar connective tissue is the
most generalised connective tissue,
which is spread extensively throughout
the body under the skin and epithelium,
around and in between the muscles,
nerves and blood vessels, between lobes
and lobules of compound glands, in the
sub-mucosa of respiratory and
gastrointestinal tracts.
It functions mainly for binding the parts
together but also helps in sliding
movement of epithelia, muscles and
other parts and also forms the internal
histological framework or stroma of
many solid organs.
26Mast cells of connective tissue
contain [CBSE AIPMT 2004]
(a) vasopressin and relaxin
(b) heparin and histamine
(c) heparin and calcitonin
(d) serotonin and melanin
Ans.(b)
Mast cells are granulated wandering
leucocyte cells. Their granules contain
histamine which is vasodilator and
heparin (anticoagulant). These take part
in body defence and allergic reaction.
27Which cartilage is present at the
end of long bones?
[CBSE AIPMT 2002]
(a) Calcified cartilage
(b) Hyaline cartilage
(c) Elastic cartilage
(d) Fibrous cartilage
Ans.(b)
Hyaline cartilageis present at the end
of long articular bones. It provides a
smooth articular surface to permit
movement at joints. Elastic cartilage is
found where support with flexibility is
needed such as in external ears.
Fibrous cartilageis a very tough
substance and is used in places of the
body where shock absorbers are
needed, e.g. discs between the
vertebrae and in the knee joint.
28During an injury nasal septum gets
damaged and for its recovery
which cartilage is preferred?
[CBSE AIPMT 2001]
(a) Hyaline cartilage
(b) Elastic cartilage
(c) Calcified cartilage
(d) Fibrous cartilage
Ans.(a)
Hyaline cartilage forms nasal septum,
larynx, trachea and hyoid apparatus.
29The polysaccharide present in the
matrix of cartilage is known as
[CBSE AIPMT 2000]
(a) cartilagin (b) ossein
(c) chondriotin (d) casein
Ans.(c)
The chief component of ground
substance of cartilage is
chondromucoprotein which is formed
of chondriotin sulphate, keratin
sulphate and hyaluronic acid.
30Formation of cartilage bones
involves [CBSE AIPMT 1993]
(a) deposition of bony matter by
osteoblasts and resorption by
chondroclasts
(b) deposition of bony matter by
osteoclasts and resorption by
chondroblasts
(c) only deposition of bony matter by
osteolasts only
(d) deposition of bony matter by
osteoblasts only
Ans.(d)
Osteoblasts are bone forming cells
while osteoclasts are bone-destroying
cells. Ossification or osteogenesis is
the process of bone formation.
Chondroblastsare cartilage forming
cells.
31Component of blood responsible
for producing antibodies is
[CBSE AIPMT 1992]
(a) thrombocytes (b) monocytes
(c) erythrocytes (d) lymphocytes
Ans.(d)
Lymphocytes (20-25%) are the second
most abundant leucocytes. These
protect us from pathogens and are
involved in the production of
antibodies. These lymphocytes are of
two types, which are known as
B-lymphocytes and T-lymphocytes.
74 NEETChapterwise Topicwise Biology

85. 32Histamine secreting cells are
found in [CBSE AIPMT 1989]
(a) connective tissue
(b) lungs
(c) muscular tissue
(d) nervous tissue
Ans.(a)
Histamineis a vasodilator, involved in
allergic and inflammatory reactions. It is
secreted by mast cells, these are
modified basophils of blood that occur in
areolar tissue i.e. the loose connective
tissue.
33Haversian canal occurs in
[CBSE AIPMT 1989]
(a) humerus (b) pubis
(c) scapula (d) clavicle
Ans.(a)
Haversiancanal, named after ‘Clopton
Havers’ are fine channels parallel to the
long axis of mammalian bone
containing blood vessels, nerve fibres,
connective tissue and occasionally
lymphatic vessels. Haversian system is
found in long bones of mammals
(humerus among given options) and
absent in spongy bones of mammals.
34Mineral found in red pigment of
vertebrate blood is
[CBSE AIPMT 1989]
(a) magnesium (b) iron
(c) calcium (d) copper
Ans.(b)
Haemoglobin constitutes about 33% of
red blood cells. It is a conjugated
protein, composed of a protein called
globin and anFe
2+
porphyrin complex
called heme. The mineral present in the
red pigment (haemoglobin) of
vertebrate blood is iron.
35Which of the following statements
wrongly represents the nature of
smooth muscle? [NEET 2021]
(a) These muscles have no striations
(b) They are involuntary muscles
(c) Communication among the cells is
performed by intercalated discs
(d) These muscles are present in the
wall of blood vessels
Ans.(c)
Statement in option (c) is incorrect and
can be corrected as
Intercalated discs are not found in
smooth muscles these are found in
cardiac muscles.
Smooth muscle cells are
spindle-shaped, have a single nucleus
and do not show striations. These
involuntary muscles are found on the
walls of internal organs such as blood
vessels.
36Which of the following is not
exclusively supplied with
involuntary muscles?
[CBSE AIPMT 1998]
(a) Muscular coats of blood vessels
(b) Muscles of the ducts of glands
(c) Muscles of iris
(d) Muscles of urethra
Ans.(b)
Muscles of the ducts of glands are not
exclusively supplied with involuntary
muscles. Smooth muscles are
involuntary muscles and are found in
the posterior part of oesophagus,
stomach, urinogenital tract, blood
vessels, iris of eye and dermis of skin.
37Characteristics of smooth muscle
fibres are [CBSE AIPMT 1990]
(a) spindle-shaped, unbranched,
unstriated, uninucleate and
involuntary
(b) spindle-shaped, unbranched,
unstriped, multinucleate and
involuntary
(c) cylindrical, unbranched, unstriped,
multinucleate and involuntary
(d) cylindrical, unbranched,
unstriated, multinucleate and
voluntary
Ans.(a)
Smooth muscle fibres are
spindle-shaped, thick in the middle and
thin at either ends, uninucleated, no
sarcolemma, contraction is slow,
involuntary under the control of
autonomous nervous system. These
muscles are also known as visceral
muscles, non-striated, non-skeletal or
involuntary muscles.
38Which of the following
characteristic is incorrect with
respect to cockroach?[NEET 2021]
(a) A ring of gastric caeca is present
at the junction of midgut and
hindgut
(b) Hypopharynx lies within the cavity
enclosed by the mouth parts
(c) In females, 7
th
- 9
th
sterna
together form a genital pouch
(d) 10
th
abdominal segment in both
sexes, bears a pair of anal cerci
Ans.(a)
In cockroach ,a ring of 6-8 blind tubules
called hepatic or gastric caeca is
present at the junction of foregut and
midgut which secrete digestive juice to
facilitate digestion.
Rest statements are correct.
39Following are the statements
about prostomium of earthworm.
[NEET 2021]
I. It serves as a covering for
mouth.
II. It helps to open cracks in the
soil into which it can crawl.
III. lt is one of the sensory
structures.
IV. It is the first body segment.
Choose the correct answer from
the options given below
(a) I, II and III are correct
(b) I, II and IV are correct
(c) I, II, III and IV are correct
(d) II and III are correct
Ans.(a)
Statements I, II and III are correct.
Earthworms have cylindrical body.
Anterior end consists of the mouth and
the prostomium, a lobe which serves as
a covering for the mouth and as a
wedge to force open cracks in the soil.
The prostomium is sensory in function.
Statement IV is incorrect and can be
corrected as :
The first body segment is called the
peristomium.
Structural Organisation in Animals 75
Muscular Tissue
and Nervous Tissue
TOPIC 3
Structural Organisation
in Some Animals
TOPIC 4

86. 40In cockroach, identify the parts of
the foregut in correct sequence.
[NEET (Oct.) 2020]
(a) Mouth→Oesophagus→Pharynx
→Crop→Gizzard
(b) Mouth→Crop→Pharynx→
Oesophagus→Gizzard
(c) Mouth→Gizzard→Crop→
Pharynx→Oesophagus
(d) Mouth→Pharynx→Oesophagus
→Crop→Gizzard
Ans.(d)
In cockroach, the parts of the foregut in
correct sequence are
Mouth→Pharynx→Oesophagus→
Crop→Gizzard
The alimentary canal in cockroach has
three regions, i.e. foregut, midgut and
hindgut. The foregut comprises of the
mouth, which opens into a short tubular
pharynx, leading to a narrow tubular
passage called oesophagus. This in turn
opens into a sac-like structure called
crop used for storing food. The crop is
followed by gizzard or proventriculus
which helps in grinding the food
particles.
41Match the following columns with
reference to cockroach and select
the correct option from the codes
given belows.[NEET (Oct.) 2020]
Column I Column II
A. Grinding of the
food particles
1. Hepatic caecal
B. Secrete gastric
juice
2. 10th segment
C. 10 pairs 3. Proventriculus
D. Anal cerci 4. Spiracles
5. Alary muscles
Codes
A B C D
(a) 3 1 4 2
(b) 4 3 5 2
(c) 1 4 3 2
(d) 2 3 1 4
Ans.(a)
The option (a) is the correct match with
reference to cockroach which is as
follows
Grinding of the food particles is done by
proventriculus or gizzard.
Hepatic caeca is a ring of 6-8 blind
tubules present at the junction of
foregut and midgut. It secretes
digestive juice.
There are 10 pairs of small holes called
spiracles present on the lateral side of
the body which are part of respiratory
system.
Anal cerci is a pair of jointed
filamentous structure in both sexes in
their 10th segment.
42If the head of cockroach is
removed, it may live for few days
because [NEET (Sep.) 2020]
(a) the cockroach does not have
nervous system
(b) the head holds a small proportion
of a nervous system, while the
rest is situated along the ventral
part of its body
(c) the head holds a 1/3rd of a nervous
system, while the rest is situated
along the dorsal part of its body.
(d) the supra-oesophageal ganglia of
the cockroach are situated in
ventral part of abdomen
Ans.(b)
Option (b) is correct because the
nervous system of cockroach consists
of a series of fused segmentally
arranged ganglia joined by paired
longitudinal connectives on the ventral
side. Three ganglia, i.e. in the throax
and six in the abdomen. In this way
cockroach nervous system is spread
throughout the body. The head holds a
bit of a nervous system, while the rest is
situated along the ventral (belly side)
part of its body. Therefore if the head
region of a cockroach is removed it may
live for few days.
43Which of the following statements
is incorrect?[NEET (Odisha) 2019]
(a) Cockroaches exhibit mosaic
vision with less sensitivity and
more resolution
(b) A mushroom-shaped gland is
present in the 6-7th abdominal
segments of male cockroach
(c) A pair of spermatheca is present
in the 6th segment of female
cockroach
(d) Female cockroach possesses
sixteen ovarioles in the ovaries
Ans.(a)
Statement (a) is incorrect because
cockroach vision is very sensitive but
provides less resolution. Such vision is
called mosaic vision. It has compound
eye and each eye contains about 2000
ommatidia.
44Select the correct sequence of
organs in the alimentary canal of
cockroach starting from mouth
[NEET (National) 2019]
(a) Pharynx→Oesophagus→Gizzard
→Crop→Ileum→Colon→
Rectum
(b) Pharynx→Oesophagus→Gizzard
→Ileum→Crop→Colon→
Rectum
(c) Pharynx→Oesophagus→Ileum
→Crop→Gizzard→Colon→
Rectum
(d) Pharynx→Oesophagus→Crop→
Gizzard→Ileum→Colon→
Rectum
Ans.(d)
The correct sequence of organs in the
alimentary canal of cockroach starting
from mouth is
Pharynx→Oesophagus→Crop→
Gizzard→Ileum→Colon→Rectum.
Both crop and gizzard are the
structures of foregut. The former
serves as the food reservoir while the
latter helps to masticate the food due
to the presence of six chitinous teeth in
it.
45Select the correct route for the
passage of sperms in male frogs.
[NEET 2017]
(a) Testes→Bidder’s canal→Kidney
→Vasa efferentia→Urinogenital
duct→Cloaca
(b) Testes→Vasa efferentia→
Kidney→Seminal vesicle→
Urinogential duct→Cloaca
(c) Testes→Vasa efferentia→
Bidder’s canal→Ureter→Cloaca
(d) Testes→Vasa efferentia→
Kidney→Bidder’s canal→
Urinogenital duct→Cloaca
Ans.(d)
In male frogs, germinal epithelium of
seminiferous tubules produce sperms,
which are transferred to kidneyviavasa
efferentia, from the kidney, these enter
into Bidder’s canal from where, the
sperms are carried to the transverse
collecting tubules, longitudinal
collecting tubule and then to
urinogenital duct. The later carries the
sperms to seminal vesical. where, they
are stored temporarily. From here,
sperms are carried to cloaca and then
these shed into water.
76 NEETChapterwise Topicwise Biology

87. 46Frog’s heart when taken out of the
body continues to beat for
sometime.
Select the best option from the
following statements.[NEET 2017]
I. Frog is a poikilotherm.
II. Frog does not have any
coronary circulation.
III. Heart is myogenic in nature.
IV. Heart is autoexcitable.
(a) Only III
(b) Only IV
(c) I and II
(d) III and IV
Ans.(d)
Frog heart is myogenic and
autoexcitable. In this conditions,
contraction of the heart originates
within the muscles itself. When
muscles are contracting. They are
releasing heat, which keeps the
electrochemical reactions in muscles
going so the muscles of heart keep
contracting after the removal of heart
from the body.
47Which of the following features is
not present inPeriplaneta
americana?[NEET 2016, Phase I]
(a) Indeterminate and radial cleavage
during embryonic development
(b) Exoskeleton composed of
N-acetylglucosamine
(c) Metamerically segmented body
(d) Schizocoelom as body cavity
Ans.(a)
Cockroach has determinate cleavage
during embryonic development and it
develops into nymph, which is a fully
developed cockroach except its size as
it is much smaller than the adult one.
48In male cockroaches, sperms are
stored in which part of the
reproductive system?
[NEET 2016, Phase II]
(a) Seminal vesicles
(b) Mushroom glands
(c) Testes
(d) Vas deferens
Ans.(a)
In male cockroaches, sperms are
stored in the seminal vesicles of the
reproductive system. The seminal
vesicles are small sacs present on the
ventral surface of the anterior of the
ejaculatory duct.
49The body cells in cockroach
discharge their nitrogenous waste
in the haemolymph mainly in the
form of [CBSE AIPMT 2015]
(a) ammonia
(b) potassium urate
(c) urea
(d) calcium carbonate
Ans.(b)
Insects including cockroach excrete
nitrogenous waste in the form of
soluble potassium urate which is
liberated into the haemolymph and
taken up by the cells lining the
Malpighian tubules. The tubules
facilitate the absorption of urate by
stirring up the blood. In the cells of the
tubule the potassium urate reacts with
water and carbon dioxide (from
respiration) to form potassium
hydrogen carbonate and uric acid. The
former is absorbed back into the blood,
but the later is excreted.
50What external changes are visible
after the last moult of a cockroach
nymph? [NEET 2013]
(a) Mandibles become harder
(b) Anal cerci develop
(c) Both fore wings and hind wings
develop
(d) Labium develops
Ans.(c)
In cockroach, development is pauro
metabolous. The nymph grows by
moulting about 13 times to reach the
adult form. The next to last nymphal
stage has wing pads but only adult
cockroaches have wings.
Analcerciare a pair of jointed
filamentous structures in 10
th
segment
of both sexeslabiumis a lower lip and a
pair of mandibles are present in
mouthparts of cockroach.
51Select the correct statement from
the ones given below with respect
toPeriplaneta americana.
[CBSE AIPMT 2012]
(a) Nervous system located dorsally,
consists of segmentally arranged
ganglia joined by a pair of
longitudinal connectives
(b) Males bear a pair of short
thread-like anal styles
(c) There are 16 very long Malpighian
tubules present at the junctions of
midgut and hindgut
(d) Grinding of food is carried out only
by the mouth parts
Ans.(b)
In cockroach, the male and females can
be identified easily with the presence or
absence of one pair of anal styles in the
posterior region. Theanal stylesare
unjointed, thread-like structures
present on the 9
th
sternite of male.
52Which one of the following
correctly describes the location of
some body parts in the earthworm
Pheretima? [CBSE AIPMT 2009]
(a) Two pairs of accessory glands in
16-18 segments
(b) Four pairs of spermathecae in 4-7
segments
(c) One pair of ovaries attached at
intersegmental septum of 14th
and 15th segments
(d) Two pairs of testes in 10th and
11th segments
Ans.(d)
In earthworm, two pairs of testes are
found in 10th and 11th segments,
accessory glands in 17th and 19th
segments, four pairs of spermatheca
from 6th-9th segment and one pair of
ovaries in 13th segment.
53If a piece of bone such as femur of
frog is kept in dilute HCl for about
a week. It will[CBSE AIPMT 2000]
(a) assume black colour
(b) shrink in size
(c) turn flexible
(d) crack into pieces
Ans.(c)
When a bone is kept in dilute acid for a
few days, the salts get dissolved,
leaving only soft and flexible organic
matrix (decalcification). If kept in KOH,
it remains unaffected, only the
surrounding muscles and connective
tissue get dissolved so that the bone
now appears clean. If burned, the
organic matter burns off producing
smoke and the ash (mineral matter) is
left behind.
54In frog, the surface of attachment
of tongue is[CBSE AIPMT 1997]
(a) sphenoid (b) palatine
(c) pterygoid (d) hyoid apparatus
Ans.(d)
Main part of hyoid apparatus is a broad,
flat and squarish plate, also called
basilingual plate, because it provides
attachment and support to the tongue.
Structural Organisation in Animals 77

88. 01What will be the direction of flow
of water when a plant cell is
placed in a hypotonic solution?
[NEET (Odisha) 2019]
(a) Water will flow in both directions
(b) Water will flow out of the cell
(c) Water will flow into the cell
(d) No flow of water in any direction
Ans.(c)
The behaviour of the plant cells with
regard to water movement depends on
the surrounding solution. When a plant
cell is placed in hypotonic solution then
the water will flow into the cell and the
cell will swell.
02Which of the following organic
compounds is the main
constituent of lecithin?
[NEET (Odisha) 2019]
(a) Arachidonic acid
(b) Phospholipid
(c) Cholesterol
(d) Phosphoprotein
Ans.(b)
Phospholipids are main constituents of
lecithin. These molecules are
composed of choline and inositole. It is
found in all living cells as a major
component of cell membrane.
03The concept of ‘Omnis
cellula-e-cellula’ regarding cell
division was first proposed by
[NEET (National) 2019]
(a) Theodor Schwann
(b) Schleiden
(c) Aristotle
(d) Rudolf Virchow
Ans.(d)
Rudolf Virchow proposed the concept
of“omnis cellula-e-cellula’’, i.e. all cells
are derived from the pre-existing cells.
Schleiden and Theodor Schwann jointly
put forward the cell theory in 1839.
04Which one of the following
elements is responsible for
maintaining turgor in cells?
[NEET 2018]
(a) Potassium (b) Sodium
(c) Magnesium (d) Calcium
Ans.(a)
Among the given elements, potassium
(K )
+
is responsible for maintaining
turgor pressure in cell because it
regulates the proton pumps involved in
opening and closing of stomata.
Magnesium(Mg )
2 +
is a constituent of
chlorophyll pigment which helps in
photosynthesis in green plants. Calcium
(Ca )
2 +
provides selective permeability
to the cell membrane. All of these, i.e.
K
+
, Ca
2 +
andMg
2 +
are essential
elements.
Sodium(Na )
+
is involved in membrane
permeability. It is a non-essential
element.
05Which among the following is not
a prokaryote? [NEET 2018]
(a)Nostoc (b)Mycobacterium
(c)Saccharomyces(d)Oscillatoria
Ans.(c)
Among the given options,
Saccharomycesis a fungus,i.e.it is a
eukaryote. They possess a well defined
nucleus and other cell organelles.
NostocandOscillatioriaare
cyanobacteria whileMycobacteriumis
a true bacterium. Cyanobacteria and
bacteria both are prokaryotes as they
lack a well-defined nucleus and other
cell organelles.
06Which of the following
components provides sticky
character to the bacterial cell ?
[NEET 2017]
(a) Cell wall
(b) Nuclear membrane
(c) Plasma membrane
(d) Glycocalyx
Ans.(d)
Glycocalyx is the outer most mucilage
layer of the cell envelope. It gives sticky
character to the bacterial cell .
07Select the wrong statement.
[NEET 2016, Phase II]
(a) Bacterial cell wall is made up of
peptidoglycan
(b) Pili and fimbriae are mainly
involved in motility of bacterial
cells
(c) Cyanobacteria lack flagellated
cells
(d)Mycoplasmais a wall-less
microorganism
Ans.(b)
Fimbriae or piliare fine hair like
appendages used by bacteria for
attachment rather than motility. These
are formed of protein called pilin.
Piliare longer than fimbriae and are one
or two per cells.
Some special type of pili called sex pili
are present in certain strains of
bacteria which help the bacterium for
forming conjugation canal during sexual
reproduction by conjugation method.
Fimbriae also help bacterium for cell to
cell adhering and colonisation.
08The structures that help some
bacteria to attach to rocks and/or
host tissues are[CBSE AIPMT 2015]
(a) rhizoids (b) fimbriae
(c) mesosomes (d) holdfast
Cell:TheUnitofLife
08
The Cell
TOPIC 1

89. Cell: The Unit of Life 79
Ans.(b)
Fimbriae are small bristle like fibres
sprouting out of the cell. In some
bacteria, they are known to help is
attachment to rocks in streams and
also to the host tissues.
09Which of the following structure is
not found in a prokaryotic cell?
[CBSE AIPMT 2015]
(a) Nuclear envelope
(b) Ribosome
(c) Mesosome
(d) Plasma membrane
Ans.(a)
In a prokaryotic cell, nuclear envelope is
not found. It means genetic material
(DNA) is not enclosed by any envelope
and lies in direct contact with the
cytoplasm.
10A protoplast is a cell
[CBSE AIPMT 2015]
(a) without plasma membrane
(b) without nucleus
(c) undergoing division
(d) without cell wall
Ans.(d)
A protoplast is a cell without cell wall. It
is a plant, bacterial or fungal cell that
had its cell wall completely or partially
removed using either mechanical or
enzymatic means.
11Select the correct statement from
the following regarding cell
membrane. [CBSE AIPMT 2012]
(a)Na
+
andK
+
ions move across cell
membrane by passive transport
(b) Proteins make up 60 to 70% of the
cell membrane
(c) Lipids are arranged in a bilayer
with polar heads towards the
inner part
(d) Fluid mosaic model of cell
membrane was proposed by
Singer and Nicolson
Ans.(d)
In 1972, Singer and Nicolson proposed
fluid mosaic model for internal
structure of plasma membrane. This is
most widely accepted model for plasma
membrane.
According to this model, a membrane
consists of a continuous bilayer of
phospholipids with their polar
hydrophilic ends on the outer surfaces
and two non-polar hydrohobic tails of
each phospholipid molecule point
inwards. The globular alpha proteins do
not form continuous layer but are
embedded randomly in the lipid bilayer or
superficially attached.
12Which one of the following
organisms is not an example of
eukaryotic cells?
[CBSE AIPMT 2011]
(a)Escherichia coli
(b)Euglena viridis
(c)Amoeba proteus
(d)Paramecium caudatum
Ans.(a)
The bacteriumE. coliis a prokaryote. It
is a Gram-negative, facultatively
anaerobic, rod-shaped bacterium. It is
the most widely studied prokaryotic
model organism.
13Which one of the following also
acts as a catalyst in a bacterial
cell? [CBSE AIPMT 2011]
(a)snRNA (b) hnRNA
(c) 23SrRNA (d) 5S rRNA
Ans.(c)
23SrRNA in bacteria is the enzyme
ribozyme for the formation of peptide
bond. 23SrRNA is found in large
sub-unit (70S) of ribosome of bacteria.
14Which one of the following
structures between two adjacent
cells is an effective transport
pathway? [CBSE AIPMT 2010]
(a) Plasmodesmata
(b) Plastoquinones
(c) Endoplasmic reticulum
(d) Plasmalemma
Ans.(a)
The primary cell wall contains many
small openings or pores situated in
the primary pit fields. The cytoplasm
of adjacent cells communicates by
means of cytoplasmic bridges called
plasmodesmata. The plasmodesmata
permit translocation of fluid and
passage of solutes between cells.
15The plasma membrane consists
mainly of [CBSE AIPMT 2010]
(a) phospholipids embedded in a
protein bilayer
(b) proteins embedded in a
phospholipid bilayer
(c) proteins embedded in a polymer
of glucose molecules
(d) proteins embedded in a
carbohydrate bilayer
Ans.(b)
According to fluid mosaic model given
bySingerandNicolson(1972), plasma
membrane consists of a continuous
bilayer of phospholipid molecules, in
which globular proteins are embedded.
This arrangement corresponds to
icebergs (proteins) floating in a sea of
phospholipids.
Proteins stay in the membrane because
they have regions of hydrophobic amino
acids which interact with fatty acid tails
to exclude water. Rest of the molecule
is hydrophilic, which faces into or
outward, both of which are aqueous
environment.
16Middle lamella is mainly composed
of [CBSE AIPMT 2009]
(a) hemicellulose
(b) muramic acid
(c) calcium pectate
(d) phosphoglycerides
Ans.(c)
The middle lamella is cementing layer
between the cells. It is made up of Ca
and Mg pectates. The basic chemical
unit of pectin is galacturonic acid
which have the capability of salt
formation with calcium and
magnesium (an acid base reaction).
17Plasmodesmata are
[CBSE AIPMT 2009]
(a) lignified cemented layers between
cells
(b) locomotory structures
(c) membranes connecting the
nucleus with plasmalemma
(d) connections between adjacent
cells
Ans.(d)
The primary cell wall contains many
small openings or pores situated in
primary pit fields. The cytoplasm of
adjacent cells communicates through
the pores by means of cytoplasmic
bridges calledplasmodesmata. The
plasmodesmata permit circulation of
fluid and passage of solutes between
cells.
18Keeping in view the ‘fluid mosaic
model’ for the structure of cell
membrane, which one of the
following statements is correct
with respect to the movement of

90. lipids and proteins from one lipid
monolayer to the other (described
as flip-flop movement)?
[CBSE AIPMT 2008]
(a) Both lipids and proteins can
flip-flop
(b) While lipids can rarely flip-flop,
proteins cannot
(c) While proteins can flip-flop, lipids
cannot
(d) Neither lipids, nor proteins can
flip-flop
Ans.(b)
Mobility of membrane proteins due to
the fluid property of lipid bilayer was
demonstrated by classical experiment
of D Frye and M Edidin (1970). Lipid
molecules very rarely migrate from one
lipid monolayer to other monolayer of
lipid bimolecular layer.
Such a type of movement is called
flip-flop or transbilayer movement and
occurs once a month for any individual
lipid molecule. But protein can never
perform flip-flop movement.
19A major breakthrough in the
studies of cells came with the
development of electron
microscope. This is because
[CBSE AIPMT 2006]
(a) the resolving power of the electron
microscope is 200-350 nm as
compared to 0.1-0.2 for the light
microscope
(b) electron beam can pass through
thick materials, whereas light
microscopy required thin sections
(c) the electron microscope is more
powerful than the light microscope
as it uses a beam of electrons
which has wavelength much longer
than that of photons
(d) the resolution power of the
electron microscope is much
higher than that of the light
microscope
Ans.(d)
The resolution power of the electron
microscope is much higher than that of
the light microscope.
As an average the resolving power of a
light microscope is 0.25μm–0.3μmwhile
that of electron microscope is 2–10Å
though theoritically, it is 0.25Å. The
magnification range of light microscope
is 2000–4000 while of electron
microscope is 100000–300000.
20According to widely accepted ‘fluid
mosaic model’, cell membranes are
semi-fluid, where lipids and
integral proteins can diffuse
randomly. In recent years, this
model has been modified in several
respects. In this regard, which of
the following statements is
incorrect? [CBSE AIPMT 2005]
(a) Proteins in cell membranes can
travel within the lipid bilayer
(b) Proteins can also undergo flip-flop
movements in the lipid bilayer
(c) Proteins can remain confined
within certain domains of the
membrane
(d) Many proteins remain completely
embedded within the lipid bilayer
Ans.(b)
Statement (b) is incorrect because
flip-flop or transmembrane movement
is due to the migration of lipid molecules
from one lipid monolayer to other
monolayer of lipid bilayer.
21A student wishes to study the cell
structure under a light microscope
having 10X eyepiece and 45X
objective. He should illuminate the
object by which one of the
following colours of light so as to
get the best possible resolution?
[CBSE AIPMT 2005]
(a) Blue (b) Green
(c) Yellow (d) Red
Ans.(a)
Resolving power or resolution is the
ability of the lens to distinguish fine
details and structure. Specifically, it
refers to the ability of the lenses to
distinguish between two points a
specified distance apart. Resolving
power depends on two factors :
(a) Wavelength of light used for
illumination.
(b) Power of objective lenses.
Resolving power
=
Wavelength of light
NA2×
Since, the limit of resolving power of a
microscope is fixed by the structure of
light, the shortest wavelength of visible
light will give the maximum resolution.
Among yellow, green, red and blue light
colour. Blue (500 nm) have shortest
wavelength so, it will give best
resolution.
22The main difference in Gram (+)ve
and Gram (–)ve bacteria resides in
their[CBSE AIPMT 1990, 2001]
(a) cell wall (b) cell membrane
(c) cytoplasm (d) flagella
Ans.(a)
Gram stain is a differential stain that
differentiates bacteria into two
groups —Gram +ve and Gram –ve. The
basis of this differentiation lies in the
composition (lipid contents) and
thickness of cell wall of these bacteria.
Bacteria are called Gram +ve, if they
retain the crystal violet colour even
after alcohol washing. Whereas, cell
wall of Gram –ve bacteria is thin, rich in
lipids and decolourise the crystal violet
colour of Gram stain.
23In ‘fluid mosaic model’ of plasma
membrane, [CBSE AIPMT 2002]
(a) upper layer is non-polar and
hydrophilic
(b) upper layer is polar and
hydrophobic
(c) phospholipids form a bimolecular
layer in middle part
(d) proteins form a middle layer
Ans.(c)
According to the fluid mosaic model,
the cell membrane consists of a highly
viscous fluid matrix of two layers of
phospholipid molecules. Protein
molecules (or their complexes) occur
in membrane but not in continuous
layer.
24DNA is mainly found in
[CBSE AIPMT 1999]
(a) nucleus (b) cytoplasm
(c) Both (a) and (b) (d) nucleolus
Ans.(a)
Most of the amount of DNA is found
in nucleus. Though some amount of
DNA is found in chloroplast and
mitochondria also .
25The eukaryotic genome differs
from the prokaryotic genome
because [CBSE AIPMT 1999]
(a) DNA is complexed with histones
in prokaryotes
(b) repetitive sequences are present
in eukaryotes
(c) genes in the former cases are
organised into operons
(d) DNA is circular and single
stranded in prokaryotes
80 NEETChapterwise Topicwise Biology

91. Ans.(b)
A major component (20–50%) of the
eukaryotic genome consists of DNA
which does not code for any protein.
This portion consists of certain base
sequences which are repeated many
times (hence, called repetitive DNA).
DNA of prokaryotes does not contain
histones nor it is single stranded.
26Which is correct about cell theory
in view of current status of our
knowledge about cell structure?
[CBSE AIPMT 1993]
(a) It needs modification due to
discovery of subcellular
structures like chloroplasts and
mitochondria
(b) Modified cell theory means that all
living beings are composed of
cells capable of reproducing
(c) Cell theory does not hold good
because all living beings do not
have cellular organisation
(e.g. viruses)
(d) Cell theory means that all living
objects consist of cells whether or
not capable of reproducing
Ans.(c)
Cell theory proposed bySchleidenand
Schwannstates that all living
organisms whether animal or plants are
made up of cells and have similar
organisation. It is the basic unit of
structure and function. Exception to
cell theory are the viruses,
mycoplasma, viroids, bacteria which
are acellular organisms, i.e. lacks
cellular organisations. Moreover,
coenocytic forms likeParamecium,
Rhizopus,have more than one nuclei
and are exception to cell theory.
27Name of Schleiden and Schwann
are associated with
[CBSE AIPMT 1993]
(a) protoplasm as the physical basis
of life
(b) cell theory
(c) theory of cell lineage
(d) nucleus functions as control
center of cell
Ans.(b)
Matthias J Schleiden (a German
botanist, 1938). Theodor Schwann
(1939), a German zoologist in 1939,
jointly proposed the cell theory.
28Cell recognition and adhesion
occur due to biochemicals of cell
membranes named
[CBSE AIPMT 1993]
(a) proteins
(b) lipids
(c) Both (a) and (b)
(d) glycoproteins and glycolipids
Ans.(d)
In the cell membrane, oligosaccharides
do not occur freely but are attached to
the external surface of phospholipids
and proteins forming glycolipids and
glycoproteins respectively. They form
cell coat (glycocalyx) which acts as
recognition centre, site for attachment
and provides antigen specificity to cell
membranes, blood grouping and
matching of tissues in transplantation
of organs.
29Genophore/bacterial genome or
nucleoid is made of
[CBSE AIPMT 1993]
(a) histones and non-histones
(b) RNA and histones
(c) a single double stranded DNA
(d) a single stranded DNA
Ans.(c)
Genophore refers to nucleoid or
bacterial genome, made of single,
double stranded DNA. It is
supercoiled with the help of RNA and
polyamines forming a circular genetic
material complex.
30Balbiani rings (puffs) are sites of
[CBSE AIPMT 1993]
(a) DNA replication
(b) RNA and protein synthesis
(c) synthesis of polysaccharides
(d) synthesis of lipids
Ans.(b)
In polytene chromosomes (salivary
gland chromosomes). Large swellings
are called puffs or Balbiani rings, named
after their discoverer. In such rings,
DNA is active, uncoiled for rapid
transcription of RNA or protein
synthesis.
31Angstrom (Å) is equal to
[CBSE AIPMT 1992]
(a) 0.01μm
(b) 0.001μm
(c) 0.0001μm
(d) 0.00001μm
Ans.(c)
Angstrom (Å) = 0.0001μm
1 Å =10
10−
M =10
8−
cm =10
7−
mm =10
4−
μ.
32Addition of new cell wall particles
amongst the existing ones is
[CBSE AIPMT 1991]
(a) deposition
(b) apposition
(c) intussusception
(d) aggregation
Ans.(b)
Appositionor accretion is defined as
the addition of new cell wall particles
amongst the existing one, such as
deposition of secondary walls in layers
from outside over the existing primary
wall.
Intussusception can be demonstrated
as the internal growth of the primary
wall which occurs during the growth
period of the cell resulting in the
increase in volume of cell wall.
DepositionMolecules settling out of a
solution.
AggregationDirect mutual attraction
between particles or Aggregation of
soil granules to form soil structure.
33Resolution power is the ability to
[CBSE AIPMT 1991]
(a) distinguish two trees
(b) distinguish two close objects
(c) distinguish amongst organelles
(d) magnify image
Ans.(b)
Resolution power is the ability of a
microscope to distinguish between two
points that are closely situated, i.e. the
smallest distance by which two objects
lying closely can be separated. Higher
resolution makes image clear.
34Cell wall shows
[CBSE AIPMT 1991]
(a) complete permeability
(b) semi-permeability
(c) differential permeability
(d) impermeability
Ans.(a)
Cell wall is the structural, functional and
heritable unit of living organisms. It is
non-living, porous, permeable, inert,
hydrophilic, inelastic, rigid,
semi-transparent protective covering
around the plasmalemma.
Cell: The Unit of Life 81

92. 35Fluid mosaic model of cell
membrane was put forward by
[CBSE AIPMT 1991]
(a) Danielli and Davson
(b) Singer and Nicolson
(c) Garner and Allard
(d) Watson and Crick
Ans.(b)
Fluid mosaic model of plasma
membrane was proposed bySJ Singer
andGL Nicolson(1972).
36Magnification of compound
microscope is not connected with
[CBSE AIPMT 1990]
(a) numerical aperture
(b) focal length of objective
(c) focal length of eye piece
(d) tube length
Ans.(a)
Magnificationis the power of
enlargement. It is the ratio of the size of
an object seen under microscope to the
actual size observed without
microscope. Magnification depends on
focal length of lenses and length of body
tube.
It does not depend on numerical
aperture of objective lens and the
nature of light being used for
illumination. The total magnification
of a microscope is determined by
multiplying the magnifying power of
the objective lens by that of the eye
piece.
37Electron microscope has a high
resolution power. This is due to
[CBSE AIPMT 1990, 92]
(a) electromagnetic lenses
(b) very low wavelength of electron
beam
(c) low wavelength of light source
used
(d) high numerical aperture of glass
lenses used
Ans.(b)
Resolving poweris defined as the
ability of an optical system or objective
lens to distinguish two closely placed
points as two distinct separate points. It
depends on wavelength of light and
numerical aperture, as limit of resolution
(L )
0.61
NA
m=
λ
. In electron microscope,
higher resolution is provided by the low
wavelength of electrons.
38Plasma membrane is made up of
[CBSE AIPMT 1989]
(a) proteins and carbohydrates
(b) proteins and lipids
(c) proteins, lipids and carbohydrates
(d) proteins, some nucleic acid and
lipids
Ans.(c)
Plasma membrane is living, quasifluid,
trilaminar membrane, usually consists
of proteins (44-76%), lipids (20-53%),
water (20%) and carbohydrates (1-8%).
39Nucleoproteins are synthesised in
[CBSE AIPMT 1989]
(a) nucleoplasm
(b) nuclear envelope
(c) nucleolus
(d) cytoplasm
Ans.(d)
Nucleoproteins are the conjugated
proteins. These include
ribonucleoproteins and occur in
ribosomes. Deoxyribonucleoproteins
occur in chromosomes.
40According to fluid mosaic model,
plasma membrane is composed of
[CBSE AIPMT 1988]
(a) phospholipids and
oligosaccharides
(b) phospholipids and hemicellulose
(c) phospholipids and integral
proteins
(d) phospholipids, extrinsic proteins
and intrinsic proteins
Ans.(d)
In fluid mosaic model lipid bilayer is
composed of phospholipids with their
polar hydrophilic ends on the outer
surfaces and two non-polar
hydrophobic tails of each phospholipid
molecule point inwards. The globular
alpha proteins do not form continuous
layer but are embedded irregularly in
the lipid bilayer (called integral or
intrinsic proteins) or superficially
attached (extrinsic or peripheral
proteins).
41Element necessary for the middle
lamella [CBSE AIPMT 2001]
(a) Ca
(b) Zn
(c) K
(d) Cu
Ans.(a)
Presence of cell wall is the
characteristic feature of all plant cells.
Cell wall consisted of three layers,
middle lamella or middle layer, primary
layer and secondary layer. The position
of middle lamella is between two
primary walls of different cells and thus
functions as cementing layer between
these two cells. Main constituents of
middle lamella are calcium and
magnesium pectate.
42In plants, inulin and pectin are
[CBSE AIPMT 2001]
(a) reserve materials
(b) wastes
(c) excretory material
(d)insect-attracting material
Ans.(a)
Inulin a polymer of fructose, is used as a
stored food, particularly in roots and
tubers of family–Compositae. Pectin is a
mucopolysaccharide which is found in
cell wall of plants. During the time of fruit
ripening, the pectin hydrolyses and gives
rise to the constituents of sugar.
43Match the List-I with List-II.
List-I List-II
A. Cristae 1. Primary constriction
in chromosome
B. Thylakoids 2. Disc-shaped sacs in
Golgi apparatus
C. Centromere 3. Infoldings in
mitochondria
D. Cisternae 4. Flattened
membranous sacs in
stoma of plastids
Choose the correct answer from
the options given below.
[NEET 2021]
A B C D
(a) 4 3 2 1
(b) 1 4 3 2
(c) 3 4 1 2
(d) 2 3 4 1
Ans.(c)
(A)-(3), (B)-(4), (C)-(1), (D)-(2)
The inner membrane of mitochondria
forms a number of infolding of the
cristae.
82 NEETChapterwise Topicwise Biology
The Cell Organelles
TOPIC 2

93. These dramatically increases the
surface area available for hosting the
enzymes responsible for cellular
respiration.
The lamellae, in chloroplast after
separation from the inner membrane,
usually take the form of closed,
flattened, ovoid sacs, thethylakoids,
which lie closely packed in piles, the
grana.
Primary constriction in the
chromosome forms thecentromere.
Acisternaeare series of flattened,
curved membrane saccules of the
endoplasmic reticulum and Golgi
apparatus.
44Which of the following is an
incorrect statement?[NEET 2021]
(a) Mature sieve tube elements
possess a conspicuous nucleus
and usual cytoplasmic organelles
(b) Microbodies are present both in
plant and animal cells
(c) The perinuclear space forms a
barrier between the materials
present inside the nucleus and
that of the cytoplasm
(d) Nuclear pores act as passages for
proteins and RNA molecules in
both directions between nucleus
and cytoplasm
Ans.(a)
Mature sieve tube elements contain
structural phloem specific proteins
(P-proteins), mitochondria, ER, and
sieve elements plastids but not
conspicuous nucleus.
45The organelles that are included in
the endomembrane system are
[NEET 2021]
(a) endoplasmic reticulum,
mitochondria, ribosomes and
lysosomes
(b) endoplasmic reticulum, Golgi
complex, lysosomes and vacuoles
(c) Golgi complex, mitochondria,
ribosomes and lysosomes
(d) Golgi complex, endoplasmic
reticulum, mitochondria and
lysosomes
Ans.(b)
The endomembrane system is a group
of membranes and organelles in
eukaryotic cells that works together to
modify, package and transport protein
and lipids.
The endomembrane system include-
Nuclear envelop
Endoplasmic reticulum
Golgi apparatus
Lysosomes
Vacuoles
Plasma membrane
46When the centromere is situated
in the middle of two equal arms of
chromosomes, the chromosome
is referred as [NEET 2021]
(a) metacentric (b) telocentric
(c) sub-metacentric (d) acrocentric
Ans.(a)
Metacentricchromosomes have the
centromere in the center, such that both
sections are of equal length, e.g. human
chromosome 1 and 3.
Other options can be explained as :
Telocentricchromosomes have the
centromere at the very end of the
chromosome.
Sub-metacentricchromosomes have
the centromere slightly offset from the
center leading to a slight asymmetry in
the length of the two sections.
Acrocentricchromosomes have
centromere which is severely offset
from the center leading to one very long
and one very short section.
47Inclusion bodies of blue-green,
purple and green photosynthetic
bacteria are[NEET (Oct.) 2020]
(a) contractile vacuoles
(b) gas vacuoles
(c) centrioles
(d) microtubules
Ans.(b)
Gas vacuoles are the inclusion bodies in
many aquatic prokaryotes like
blue-green, purple and green
photosynthetic bacteria. These are
generally small, hollow cylindrical
structure which facilitates air
permeability. Gas vacuoles are
membrane bound inclusion bodies that
contain an array of substructures
referred to as gas vesicles. The
membrane of gas vacuoles is rigid,
impermeable to water and freely
permeable to all gases.
48The biosynthesis of ribosomal
RNA occurs in[NEET (Oct.) 2020]
(a) ribosomes (b) Golgi apparatus
(c) microbodies (d) nucleolus
Ans.(d)
The biosynthesis of ribosomal RNA
occurs in nucleolus of nucleus. It helps
the nucleus of the cell to control cell
metabolism and other activities. The
other two types of RNA, i.e.mRNA and
tRNA are also synthesised here.
49Match the following columns and
select the correct option from the
codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. Smooth
Endoplasmic
Reticulum
1. Protein
synthesis
B. Rough
endoplasmic
reticulum
2. Lipid
synthesis
C. Golgi complex 3. Glycosylation
D. Centriole 4. Spindle
formation
Codes
A B C D
(a) 2 1 3 4
(b) 3 1 2 4
(c) 4 2 1 3
(d) 1 2 3 4
Ans.(a)
The option (a) is the correct match
which is as follows
Smooth Endoplasmic reticulum is the
major site for synthesis of lipid.
Rough Endoplasmic reticulum is
actively involved in protein synthesis
and secretion.
Golgi complex is an important site of
formation of glycoproteins and
glycolipids, i.e. glycosylation.
Centrioles help in spindle formation in
the cell.
50Which of the following elements
helps in maintaining the structure
of ribosomes?[NEET (Oct.) 2020]
(a) Magnesium (b) Zinc
(c) Copper (d) Molybdenum
Cell: The Unit of Life 83
Centromeres
Long arms
Short arms
Short arms
Centromeres
Secondary constriction
Satellite
Types of chromosomes

94. Ans.(a)
Each ribosome consist of two unequal
subunits, a larger and a smaller one.
Mg
2+
ions are required for binding the
two subunits. Below0 0001 M Mg
2
⋅
+
, the
two subunits dissociate while above
this strength, the two subunits form
the dimer.
51Which is the important site of
formation of glycoproteins and
glycolipids in eukaryotic cells?
(a) Peroxisomes[NEET (Sep.) 2020]
(b) Golgi bodies
(c) Polysomes
(d) Endoplasmic reticulum
Ans.(b)
Golgi bodies are site of formation of
glycoproteins and glycolipids in
eukaryotic cells. Glycoproteins are simply
proteins with a sugar attached to them.
The sugars can be attached to a protein in
two locations in the cell, the endoplasmic
reticulum, which produces N-linked
sugars, and the Golgi apparatus, which
produces O-linked sugars. Glycolipids are
components of cellular membranes
comprised of a hydrophobic lipid tail and
one or more hydrophilic sugar groups
linked by a glycosidic bond. Their role is to
maintain the stability of the cell
membrane .
52Match the Column I with Column II.
[NEET (Odisha) 2019]
Column I Column II
1. Golgi
apparatus
(i) Synthesis of
protein
2. Lysosomes (ii) Trap waste and
excretory products
3. Vacuoles (iii) Formation of
glycoproteins and
glycolipids
4. Ribosomes (iv) Digesting
biomolecules
Select the correct option from the
following
1 2 3 4
(a) (iii) (iv) (ii) (i)
(b) (iv) (iii) (i) (ii)
(c) (iii) (ii) (iv) (i)
(d) (i) (ii) (iv) (iii)
Ans.(a)
The correct matches are
Golgi apparatus – Formation of
glycoproteins and glycolipids
Lysosomes – Digesting biomolecules
Vacuoles – Trap waste and excretory
products
Ribosomes – Synthesis of protein
53Which of the following cell
organelles is present in the
highest number in secretory
cells? [NEET Odisha) 2019]
(a) Mitochondria
(b) Golgi complex
(c) Endoplasmic reticulum
(d) Lysosomes
Ans.(b)
Golgi complex (Golgi apparatus) is a
cell organelle present in highest
number in secretory cells. These are
the site of modification, packaging
and secretions of secretory proteins
and glycoproteins outside the cell.
54Non-membranous nucleoplasmic
structures in nucleus are the site
for acitive synthesis of
[NEET (Odisha) 2019]
(a) protein synthesis (b)mRNA
(c)rRNA (d) tRNA
Ans.(c)
Non-membranous nucleoplasmic
structure in the nucleus of the cell are
the site for active synthesis ofrRNA.
These structures are called nucleolus.
Larger and more numerous nucleoli
are present in the cell actively
carrying out protein synthesis.
55Which of the following pairs of
organelles does not contain
DNA? [NEET (National) 2019]
(a) Chloroplast and Vacuoles
(b) Lysosomes and Vacuoles
(c) Nuclear envelope and
Mitochondria
(d) Mitochondria and Lysosomes
Ans.(b)
Lysosomes and vacuoles do not
contain DNA. Lysosomes are single
membrane bound small vesicles
which contain hydrolytic enzymes.
Vacuoles are a large membranous sac
found in the cytoplasm. These contain
substances that are not essentially
useful for the cell like water, sap,
excretory products and other
materials. Chloroplast and
mitochondria are semi-autonomous
organelles because they contain their
own DNA and are believed to be
prokaryotic symbionts.
56Which of the following
statements is not correct?
[NEET (National) 2019]
(a) The hydrolytic enzymes of
lysosomes are active under acidic
pH
(b) Lysosomes are membrane bound
structures
(c) Lysosomes are formed by the
process of packaging in the
endoplasmic reticulum
(d) Lysosomes have numerous
hydrolytic enzymes
Ans.(c)
The statement “lysosomes are formed
by the process of packaging in the
endoplasmic reticulum” is incorrect.
The correct form of the statement is
‘lysosomes are actually formed by the
budding off from thetrans-face of
Golgi bodies.
These membrane bound structures
contain hydrolytic enzymes whose
precursors are synthesised by rough
endoplasmic reticulum. Rest
statements are correct.
57The shorter and longer arms of a
submetacentric chromosome are
referred to as
[NEET (National) 2019]
(a) p-arm and q-arm, respectively
(b) q-arm and p-arm, respectively
(c) m-arm and n-arm, respectively
(d) s-arm and l-arm, respectively
Ans.(a)
The shorter and longer arms of
submetacentric chromosome are
designated as p and q arm,
respectively. Here, ‘p’ signifies petite
or short. In a submetacentric
chromosome, centromere is located
near the centre due to which the two
arms appear unequal in length.
58Which of the following
statements regarding
mitochondria is incorrect?
[NEET (National) 2019]
(a) Enzymes of electron transport
are embedded in outer membrane
(b) Inner membrane is convoluted
with infoldings
(c) Mitochondrial matrix contains
single circular DNA molecule and
ribosomes
(d) Outer membrane is permeable to
monomers of carbohydrates, fats
and proteins
84 NEETChapterwise Topicwise Biology

95. Ans.(a)
The statement ‘‘enzymes of electron
transport are embedded in outer
membrane’’ is incorrect. The correct
form of statement is
Enzymes of electron transport are
embedded in the inner membrane of
mitochondria. An electron transport
chain is a series of coenzymes and
cytochromes that take part in the
passage of electrons from a chemical
to its ultimate acceptor. Rest
statements are correct.
59The Golgi complex participates in
[NEET 2018]
(a) respiration in bacteria
(b) formation of secretory vesicles
(c) fatty acid breakdown
(d) activation of amino acid
Ans.(b)
Golgi complexparticipates in the
formation of secretory vesicles. It is a
cytoplasmic structure found in
eukaryotic cells. It is made up of four
parts; cisternae, tubules, vesicles and
vacuoles.
The forming face or cisternae receives
vesicles from endoplasmic reticulum.
Their contents pass through various
cisternae with the help of coated
vesicles and intercisternal connectives.
They ultimately reach the maturing face
where they are budded off as, coated
secretory or Golgian vesicles or
vacuoles.
Inbacteria, respiration occurs with the
help of mesosomes. Thebreakdown of
fatty acidoccurs in peroxisomes and
mitochondria.Activation of amino
acidis an important step of protein
synthesis and it occurs in cytoplasm. In
this process, amino acids get attached
totRNA molecules.
60Which of the following is true for
nucleolus? [NEET 2018]
(a) It takes part in spindle formation
(b) It is a membrane-bound structure
(c) Larger nucleoli are present in
dividing cells
(d) It is a site for active ribosomal
RNA synthesis
Ans.(d)
Nucleolus is a naked, round or slightly
irregular structure in nucleus. It lacks a
membrane and its contents are in direct
contact with the nucleoplasm. It is a
site for activeribosomal RNA(rRNA)
synthesis.
Microtubules take part in thespindle
formation.Mitochondria, vacuoles and
plastids, etc. aremembrane-bound
structures. Thedividing cellspossess a
large number of mitochondria.
61Nissl bodies are mainly composed
of [NEET 2018]
(a) nucleic acids and SER
(b) DNA and RNA
(c) proteins and lipids
(d) free ribosomes and RER
Ans.(d)
Nissl granules are found in the cell-body
of neurons. These granules are
composed ofRough Endoplasmic
Reticulum(RER) that bears free
ribosomes. The latter acts as the site of
protein synthesis. These granules were
named after its discovererFranz Nissl.
62Select theincorrectmatch.
[NEET 2018]
(a) Submetacentric
chromosomes
– L-shaped
chromosomes
(b) Allosomes – Sex
chromosomes
(c) Lampbrush
chromosomes
– Diplotene
bivalents
(d) Polytene
chromosomes
– Oocytes of
amphibians
Ans.(d)
Polytene chromosomes are giant
chromosomes that are quite common in
the salivary glands of insects therefore
they are popularly called as salivary
chromosomes.
TheLampbrush chromosomesare
highly elongated special kind of synapsed
mid-prophase or diplotene chromosome
that are bivalents.Sex chromosomes
are also called as allosomes. They
determine the sex of an organism.
Submetacentric chromosomeshave a
submedian centromere. They appear
L-shaped during metaphase.
Therefore, except option (d), all are
correctly matched.
63Which one of the following
events doesnotoccur in rough
endoplasmic reticulum?
[NEET 2018]
(a) Cleavage of signal peptide
(b) Protein glycosylation
(c) Protein folding
(d) Phospholipid synthesis
Ans.(d)
Phospholipid synthesisdoes not
occur in RER. It occurs inside
Smooth Endoplasmic Reticulum
(SER). Asignal peptideis a short
peptide present at the N-terminus
of the newly synthesised proteins. It
targets them to the ER and is then
cleaved off. RER synthesises
proteins.It bears enzymes for
modifying polypeptides synthesised
by attached ribosomes,e.g.
glycosylation.
64Many ribosomes may associate
with a singlemRNA to form
multiple copies of a polypeptide
simultaneously. Such strings of
ribosomes are termed as
[NEET 2018]
(a) plastidome (b) polyhedral bodies
(c) polysome (d) nucleosome
Ans.(c)
Polysomeis a string of ribosomes
associated with a singlemRNA.
Polysome helps to produce a number
of copies of the same polypeptide.
Nucleosomeis the unit of eukaryotic
DNA that consists of a DNA segment
wrapped around a core of eight
histone proteins. Nucleosome chain
gives a ‘beads on string’ appearance
under electron microscope.
Plastidomerefer to all the plastids of
a cell which work as a functional unit.
65Which of the following cell
organelles is responsible for
extracting energy from
carbohydrates to form ATP?
[NEET 2017]
(a) Lysosome (b) Ribosome
(c) Chloroplast (d) Mitochondrion
Ans.(d)
Mitochondria is referred as ‘power
house of the cell’. It contains the
enzymes for cellular respiration. It
oxidises carbohydrate to produce
ATP molecules in the process of
aerobic respiration.
Cell: The Unit of Life 85
Perinucleolar
chromatin
Intranucleolar
chromatin
Matrix
(Pars amorpha)
Granular portion
(Ribosomal
precursor)
Fibrillar portion
(RNA fibrils)
Structure of nucleolus

96. Thinking process Mictochondria is a
double membrane bound
semi-autonomous cell organelles. The
number of mitochondria per cell is more
in metabolically active cells.
66Water soluble pigments found in
plant cell vacuoles are
[NEET 2016, Phase I]
(a) chlorophylls (b) carotenoids
(c) anthocyanins (d) xanthophylls
Ans.(c)
Anthocyanins are water soluble
vacuolar pigments that may appear red,
purple or blue depending on pH. It is
impermeable to cell membranes of
plants and can leak out only when
membrane is damaged or dead.
67Which one of the following cell
organelles is enclosed by a single
membrane? [NEET 2016, Phase I]
(a) Chloroplasts (b) Lysosomes
(c) Nuclei (d) Mitochondria
Ans.(b)
Nuclei, mitochondria and chloroplasts
are all double membrane bound
organelles. Lysosomes are single
membrane bound organelle.
68Mitochondria and chloroplast are
I. semi-autonomous organelles.
II. formed by division of
pre-existing organelles and
they contain DNA but lack
protein synthesising machinery.
Which one of the following
options is correct?
[NEET 2016, Phase I]
(a) II is true, but I is false
(b) I is true, but II is false
(c) Both I and II are false
(d) Both I and II are true
Ans.(b)
Mitochondria and chloroplast are
semi-autonomous organelles which
contains DNA, ribosomes (70S), etc.
They are capable of self-replication so
called semi-autonomous.
69Microtubules are the constituents
of [NEET 2016, Phase I]
(a) spindle fibres, centrioles and cilia
(b) centrioles, spindle fibres and
chromatin
(c) centrosome, nucleosome and
centrioles
(d) cilia, flagella and peroxisomes
Ans.(a)
Microtubules are structures present in
cilia, flagella, centrioles and spindle
fibres. They are also the part of fibres
found in cytoskeleton.
70A cell organelle containing
hydrolytic enzyme is
[NEET 2016, Phase II]
(a) lysosome
(b) microsome
(c) ribosome
(d) mesosome
Ans.(a)
Lysosomes are membrane bound cell
organelles. These contain many
hydrolytic enzymes which work at high
pH. They bring about the intracellular
digestion of cell debris and worn and
torned cell organelles. These loose
their existence while doing so, that is
why they are also called as suicidal
bags.
71Select the mismatch.
[NEET 2016, Phase II]
(a) Gas vacuoles — Green bacteria
cells
(b) Large central vacuoles — Animal
cells
(c) Protists — Eukaryotes
(d) Methanogens — Prokaryotes
Ans.(b)
Animal cells do not have large central
vacuole. Instead, these have 2-3 small
vacuoles. The presence of such large
central vacuoles is the characteristics
feature of plant cells.
Concept EnhancerThe presence of
large vacuole is an indication of
irregular growth, i.e. growth in cell
membrane is synchronised with growth
in protoplasmic content.
72Match the columns and identify
the correct option.
[CBSE AIPMT 2015]
Column I Column II
A. Thylakoids 1. Disc-shaped
sacs in Golgi
apparatus
B. Cristae 2. Condensed
structure of DNA
C. Cisternae 3. Flat
membranous
sacs in stroma
D. Chromatin 4. Infoldings in
mitochondria
Codes
A B C D
(a) 4 3 1 2
(b) 3 4 1 2
(c) 3 1 4 2
(d) 3 4 2 1
Ans.(b)
The columns can be matched correctly
as follows
Column I Column II
A. Thylakoids 3. Flat membranous
sacs in stroma
B. Cristae 4. Infoldings in
mitochondria
C. Cisternae 1. Disc-shaped sacs
in Golgi apparatus
D. Chromatin 2. Condensed
structure of DNA
73Balbiani rings are sites of
[CBSE AIPMT 2015]
(a) lipid synthesis
(b) nucleotide synthesis
(c) polysaccharide synthesis
(d) RNA and protein synthesis
Ans.(d)
A Balbiani ring is a large chromosome
puff. Balbiani rings are diffused
uncoiled regions of the polytene
chromosome that are sites of RNA
transcription and protein synthesis.
74Which of the following are not
membrane bound?
[CBSE AIPMT 2015]
(a) Vacuoles
(b) Ribosomes
(c) Lysosomes
(d) Mesosomes
Ans.(b)
Ribosomes are non-membranous
particles these are simple aggregations
of RNA (rRNA) and proteins.
75Cellular organelles with
membranes are[CBSE AIPMT 2015]
(a) nuclei, ribosomes and
mitochondria
(b) chromosomes, ribosomes and
endoplasmic
reticulum
(c) endoplasmic reticulum,
ribosomes and nuclei
(d) lysosomes, Golgi apparatus and
mitochondria
86 NEETChapterwise Topicwise Biology

97. Ans.(d)
Membrane bound organelles include
lysosomes, endoplasmic reticulum,
Golgi apparatus, mitochondria,
chloroplasts, vacuoles, nucleus.
Non-membrane bound organelles
include ribosomes, centrioles,
microtubules.
76The solid linear cytoskeletal
elements having a diameter of 6
nm and made up of a single type
of monomer are known as
[CBSE AIPMT 2014]
(a) microtubules
(b) microfilaments
(c) intermediate filaments
(d) lamins
Ans.(b)
Microfilaments (actin filament) are the
thinnest filaments of the cytoskeleton.
They are found in the cytoplasm of the
eukaryotic cells.
They constitute the cytoskeleton
through which the cells acquire shape
their diameter is approximately 6 nm
(avg.) These are the polymer of actin
sub-units.
77The osmotic expansion of a cell
kept in water is chiefly regulated
by [CBSE AIPMT 2014]
(a) mitochondria (b) vacuoles
(c) plastids (d) ribosomes
Ans.(b)
The osmotic expansion of a cell kept in
water is chiefly regulated by vacuole. It
is the single and large organelle which
constitutes about 20% of plant cells
and is small and multiple in animal cells.
Vacuole store water and
macromolecules including ions, sugar,
amino acid, protein and carbohydrates.
The membrane that surrounds the
vacuole is called tonoplast. The vacuole
contains cell sap in it.
The cell sap has high osmotic pressure
which regulate turgor pressure in plant
cells.
78Match the following and select the
correct answer.[CBSE AIPMT 2014]
Column I Column II
A. Centriole 1. Infoldings in
mitochondria
B. Chlorophyll 2. Thylakoids
C. Cristae 3. Nucleic acids
D. Ribozymes 4. Basal body of cilia
or flagella
Codes
A B C D
(a) 4 2 1 3
(b) 1 2 4 3
(c) 1 3 2 4
(d) 4 3 1 2
Ans.(a)
(a)CentrioleIn organism with flagella
and cilia, the position of these
organelles is determined by the
mother centriole which become the
basal body.
(b)ChlorophyllChlorophyll molecules
are specially arranged in and around
photosystem that are embedded in
the thylakoid membrane of
chloroplast.
(c)CristaeThese are folds in the inner
membraneof mitochondria which
provides a large amount of surface
area for chemical reaction.
(d)Ribozymes(Ribonucleic acid enzyme)
is an RNA molecule that is capable of
catalysing specific biochemical
reactions of nucleic acids
79Which one of the following
organelles in the figure correctly
matches with its function?
[NEET 2013]
(a) Rough endoplasmic reticulum,
formation of glycoproteins
(b) Golgi apparatus, protein synthesis
(c) Golgi apparatus, formation of
glycolipids
(d) Rough endoplasmic reticulum,
protein synthesis
Ans.(d)
Rough Endoplasmic Reticulum (RER) –
Protein synthesis
Smooth Endoplasmic Reticulum (SER) –
Lipid synthesis.
Golgi apparatus – Important site of
formation glycoproteins and glycolipids.
80A major site for synthesis of lipids
is [NEET 2013]
(a) RER (b) SER
(c) symplast (d) nucleoplasm
Ans.(b)
The Smooth Endoplasmic Reticulum
(SER) is the major site for synthesis of
lipids. RER is actively involved in
protein synthesis and secretion.
Nucleoplasm is the site for active
ribosomal RNA synthesis. Symplast is
the system of interconnected
protoplast through which water
movement occurs.
81The Golgi complex plays a major
role [NEET 2013]
(a) in trapping the light and
transforming it into chemical
energy
(b) in digesting proteins and
carbohydrates
(c) as energy transferring organelles
(d) in post translational modification
of proteins and glycosidation of
lipids
Ans.(d)
Golgi complex plays a major role in post
translational modification of proteins
and glycosidation of lipids.
Chloroplasts contain chlorophyll which
traps light and transform into chemical
energy. Lysosomes are involved in
digesting proteins, fats and
carbohydrates. Mitochondria are
energy transferring organelles.
82Ribosomal RNA is actively
synthesised in[CBSE AIPMT 2012]
(a) lysosomes (b) nucleolus
(c) nucleoplasm (d) ribosomes
Ans.(b)
In eukaryotes, the site of synthesis of
most of the ribosomal RNA (rRNA) is
nucleolus. The nucleolar organiser
contains many copies of ribosomal
DNA (repetitive DNA). The RNA cistron
of nucleolar DNA forms 45S precursor
with the help of RNA polymerase. This
45S RNA undergoes cleavage with the
help of nucleases to give 18S, 28S and
5.8SrRNA units. Out of different
rRNAs, the 5SrRNA is not synthesised
in nucleolus. It is synthesised outside
it.
83What is true about ribosomes?
[CBSE AIPMT 2012]
(a) The prokaryotic ribosomes are
80S, where S stands for
sedimentation coefficient
(b) These are composed of
ribonucleic acid and proteins
Cell: The Unit of Life 87
Nucleus
Rough endoplasmic
reticulum
Golgi apparatus

98. (c) These are found only in eukaryotic
cells
(d) These are self-splicing introns of
some RNAs
Ans.(b)
Ribosomes are large,
non-membranous, RNA protein
complexes which are necessary for
protein synthesis. In prokaryotes, 70S
type of ribosomes are found while 80S
type of ribosomes are found in
eukaryotes.
84Important site for formation of
glycoproteins and glycolipids is
[CBSE AIPMT 2011]
(a) Golgi apparatus
(b) plastid
(c) lysosome
(d) vacuole
Ans.(a)
The Golgi apparatus principally
performs the function of packaging
materials. Golgi apparatus is the main
site of formation of glycoproteins and
glycolipids.
85Peptide synthesis inside a cell
takes place in[CBSE AIPMT 2011]
(a) mitochondria (b) chromoplast
(c) ribosomes (d) chloroplast
Ans.(c)
The cellular factory responsible for
synthesising proteins (peptide
synthesis) is the ribosome.
86Which one of the following has its
own DNA? [CBSE AIPMT 2010]
(a) Mitochondria
(b) Dictyosome
(c) Lysosome
(d) Peroxisome
Ans.(a)
In mitochondria, the inner membrane
space is filled with a matrix which
contains dense granules along with
ribosomes and mitochondrial DNA. The
mitochondrial DNA is circular in nature.
Their number varies from 2-6. Besides
DNA, a mitochondrion has RNA and its
ribosomes (70S) also.
Thus, a complete protein synthesising
machinery is present in mitochondria,
so mitochondria is semi autonomous
organelle in nature. Dictyosome,
lysosome and peroxisome do not have
their own DNA.
87Cytoskeleton is made up of
[CBSE AIPMT 2009]
(a) calcium carbonate granules
(b) callose deposits
(c) cellulosic microfibrils
(d) proteinaceous filaments
Ans.(d)
The cytoplasm of all eukaryotic cells is
criss-crossed by a network of protein
fibres that support the shape of the cell
and here organelles are anchored at
fixed locations. It is a dynamic system
which includes three types
proteinaceous filaments—actin
filaments, microtubule and intermediate
filament.
88The two sub units of ribosome
remain united at a critical ion level
of [CBSE AIPMT 2008]
(a) copper (b) manganese
(c) magnesium (d) calcium
Ans.(c)
Magnesium is constituent of chlorophyll,
middle lamella, and connected with
phosphate transfer in respiration. It is
concerned with binding of ribosomes,
DNA and RNA synthesis.
Manganese activates enzymes of
respiration, photosynthesis and
nitrogen metabolism performing
oxidation, reduction, decarboxylation,
photolysis of water, etc.
Copper is activator of plastocyanin,
cytochrome oxidase, RuBP carboxylase
and many other enzymes. It functions in
electron transfer, maintenance of
carbohydrate/nitrogen balance,
chlorophyll synthesis, etc.
Calcium is constituent of middle lamella,
activator of enzymes connected with
chromosome formation and many
aspects of metabolism.
89Vacuole in a plant cell
[CBSE AIPMT 2008]
(a) is membrane-bound and contains
storage proteins and lipids
(b) is membrane-bound and contains
water and excretory substances
(c) lacks membrane and contains air
(d) lacks membrane and contains
water and excretory substances
Ans.(b)
The vacuoles of plant cells are bounded
by a single semipermeable membrane
known as tonoplast.
These vacuoles contain water, phenol,
flavonols, anthocyanins, alkaloids and
storage products such as sugars and
proteins.
The vacuoles of animal cells are
bounded by a lipoproteinaceous
membrane and their function is
storage, transmission of materials and
maintenance of internal pressure of
cell.
90In germinating seeds, fatty acids
are degraded exclusively in the
[CBSE AIPMT 2008]
(a) proplastids
(b) glyoxysomes
(c) peroxisomes
(d) mitochondria
Ans.(b)
Glyoxysomesare found to occur in the
cells of yeast,Neurosporaand oil rich
seeds of many higher plants. During
germination of oily seeds, the stored
lipid molecules of glyoxysomes are
hydrolysed by the enzyme lipase to
glycerol and fatty acids. The long chain
fatty acids are then broken down by
successive removal of two carbon
fragments in the process of
β-oxidation.
Peroxisomesare present in all
photosynthetic cells of higher plants in
etiolated leaf tissue. It is the site of
hydrogen peroxide metabolism and
glycolate cycle.
Mitochondriais the site of aerobic
respiration in eukaryotic cell. It is
called power house of the cell.
91Select the wrong statement from
the following[CBSE AIPMT 2007]
(a) Both chloroplasts and
mitochondria contain an inner
and an outer membrane
(b) Both chloroplasts and
mitochondria have an internal
compartment, the thylakoid
space bounded by the thylakoid
membrane
(c) Both chloroplasts and
mitochondria contain DNA
(d) The chloroplasts are generally
much larger than mitochondria
Ans.(b)
Thylakoid space which is known as
lumen is present only in chloroplasts.
The inner membrane of mitochondria
is folded to form cristae.
88 NEETChapterwise Topicwise Biology

99. Cell: The Unit of Life 89
92Which of the following statements
regarding mitochondrial
membrane is not correct?
[CBSE AIPMT 2006]
(a) The enzymes of the electron
transfer chain are embedded in
the outer membrane
(b) The inner membrane is highly
convoluted forming a series of
infoldings
(c) The outer membrane resembles a
sieve
(d) The outer membrane is
permeable to all kinds of
molecules
Ans.(a)
In mitochondria, the enzymes of
electron transport chain are found in
the inner membrane while outer
membrane contains enzymes involved
in mitochondrial lipid synthesis and
those enzymes that convert lipid
substrates into other forms that are
subsequently metabolised in the
matrix.
The outer membrane resembles a sieve
that is permeable to all molecules of
10,000 daltons mole weight or less,
including small proteins.
The inner membrane is impermeable
and highly convoluted, forming a
series of infoldings, known as cristae,
in the matrix space.
93The main organelle involved in
modification and routing of newly
synthesised proteins to their
destinations is[CBSE AIPMT 2005]
(a) chloroplast
(b) mitochondria
(c) lysosome
(d) endoplasmic reticulum
Ans.(d)
Porter coined the name Endoplasmic
Reticulum (ER). It is a network of
tubules, vesicles and cisternae within
an eukaryotic cell (absent in prokaryotic
cells). Two types of ER are recognised
on the basis of presence/absence of
ribosomes on the wall of the ER.
(i)Smooth ERIt does not have
ribosomes. Smooth ER helps in the
synthesis of lipid and glycogen.
(ii)Rough ERWall of this ER contains
ribosomes. Rough ER is involved in
protein synthesis and transfer.
Protein synthesis takes place in
ribosomes attached on wall of ER.
Newly formed protein enters within the
cavity of rough ER and follows following
path :
Protein→Cavity of rough
ER→Cavity of smooth ER→Golgi
membrane→Lysosomes or
transport vesicles or secretory
granules.
94Protein synthesis in an animal cell
occurs[CBSE AIPMT 2005, 2000]
(a) only on the ribosomes present in
cytosol
(b) only on ribosomes attached to the
nuclear envelope and
endoplasmic reticulum
(c) on ribosomes present in the
nucleolus as well as in cytoplasm
(d) on ribosomes present in
cytoplasm as well as in
mitochondria
Ans.(d)
Protein synthesis is taken place on
ribosomes. In an eukaryotic cell
ribosomes are present in cytoplasm,
mitochondria and chloroplasts. So,
in these places protein synthesis also
takes place.
95Chlorophyll in chloroplast is
located in[CBSE AIPMT 2005]
(a) grana
(b) pyrenoid
(c) stroma
(d) Both (a) and (c)
Ans.(a)
Chlorophyll is a specialised light
absorbing pigment which is found in the
inner wall of granum. Each granum is a
flat, sac-like structure in which light
reaction of photosynthesis takes place.
96Extra nuclear inheritance is a
consequence of presence of
genes in [CBSE AIPMT 2004]
(a) mitochondria and chloroplasts
(b) endoplasmic reticulum and
mitochondria
(c) ribosomes and chloroplast
(d) lysosomes and ribosomes
Ans.(a)
Extra nuclear or extra chromosomal or
cytoplasmic or organellar inheritance is
a consequence of presence of genes in
mitochondria and chloroplast. Extra
chromosomal units function either
independently or in collaboration with
nuclear genetic system.
97In chloroplasts, chlorophyll is
present in the[CBSE AIPMT 2004]
(a) outer membrane
(b) inner membrane
(c) thylakoids
(d) stroma
Ans.(c)
The thylakoids of chloroplast are
flattened vesicles arranged as a
membranous network within the
stroma. 50% of chloroplast proteins
and various components involved
(namely chlorophyll, carotenoids and
plastoquinone) are present in thylakoid
membranes that are involved in
photosynthesis.
98Flagella of prokaryotic and
eukaryotic cells differ in
[CBSE AIPMT 2004]
(a) type of movement and placement
in cell
(b) location in cell and mode of
functioning
(c) microtubular organisation and
type of movement
(d) microtubular organisation and
function
Ans.(c)
Flagella of prokaryotic and eukaryotic
species differ in microtubular
organisation and type of movement.
In eukaryotes the arrangement is
( )9 2+and specialised while in
prokaryotes arrangement is( )9 0+
and is simple.
99Ribosomes are produced in
[CBSE AIPMT 2002]
(a) nucleolus
(b) cytoplasm
(c) mitochondria
(d) Golgi body
Ans.(a)
The proteins required for the formation
of ribosome are synthesised within the
cytoplasm through the process of
translation.
These proteins are later shifted to
nucleus and then to nucleolus where
the RNA and proteins are assembled
into ribosomal sub-units. In prokaryotes
(bacteria) ribosomes are synthesised in
cytoplasm.
In eukaryotes, the process takes place
both in the cell cytoplasm and in the
nucleoluswhich is a region within the
cell nucleus.

100. 100Microtubules are absent in
[CBSE AIPMT 2001]
(a) mitochondria (b) centriole
(c) flagella (d) spindle fibres
Ans.(a)
Microtubules are present only in
eukaryotes; and are component of cilia
and flagella as well as spindle (during
cell division). They are straight, hollow
rods measuring about 25 nm in
diameter and 200 nm to 25μm in
length. Microtubules give shape and
support to the cell.
101Lysosomes are reservoirs of
[CBSE AIPMT 2000]
(a) RNA and protein
(b) fats
(c) secretory glycoproteins
(d) hydrolytic enzymes
Ans.(d)
Lysosomes were discovered by
Christian de Duve(1955) from rat liver.
Matile(1964) discovered lysosomes in
plants. Generally, lysosomes are 0.2–0.8
μin size, irregular membranous vesicles
filled with hydrolytic enzymes. They are
polymorphic. About 40 enzymes (all
hydrolytic) are present in lysosomes.
These include proteases, nucleases,
glycosidases, lipases, phospholipases,
phosphatases and sulphatases.
102The cell organelle involved in
glycosylation of protein is
[CBSE AIPMT 2000]
(a) ribosome
(b) peroxisome
(c) endoplasmic reticulum
(d) mitochondria
Ans.(c)
The proteins synthesised by the
ribosomes bound to ER are passed into
the lumen of ER where an
oligosaccharide is added to them,
(i.e. these are glycosylated).
103Some of the enzymes which are
associated in converting fats into
carbohydrates, are present in
[CBSE AIPMT 1999]
(a) liposomes (b) Golgi bodies
(c) microsomes (d) glyoxysomes
Ans.(d)
Besides catalase, the glyoxysomes
contain enzymes for the glyoxylate
cycle through which fats are converted
into carbohydrates.
Microsomes are product of
homogenisation of ER.
Liposomesare artificially produced
lipid bilayers, 25 nm or more in
diameter. Golgi body is a dynamic
eukaryotic organalle, consisted of
cisternae, vesicles and tubules.
104The proteins are synthesised at
[CBSE AIPMT 1999]
(a) ribosomes (b) mitochondria
(c) centrosomes (d) Golgi bodies
Ans.(a)
During protein synthesis, smaller
sub-units of ribosomes attach to
mRNA. The ribosomes provide space as
well as enzyme for the synthesis of
proteins. Therefore, these are known as
protein factories or workbenches of
protein.
105Which of the following organ has
single membrane?
[CBSE AIPMT 1999]
(a) Nucleus
(b) Cell wall
(c) Mitochondria
(d) Spherosomes
Ans.(d)
Cell wall does not have a membrane.
The mitochondria and nucleus are
surrounded by double membraned
envelope.
Spherosomesare single membrane
bound, spherical structures in plant cell
cytoplasm. These are apparently
centres of lipid synthesis and
accumulation.
106Microtubule is involved in the
[CBSE AIPMT 1998]
(a) cell division
(b) membrane architecture
(c) muscle contraction
(d) DNA recognition
Ans.(a)
Microtubules are one of the essential
protein filaments of the cytoskeletons
of probably all eukaryotic cells and their
cilia, flagella, basal bodies, centrioles
and mitosis and meiosis spindles. Each
microtubule is made up of a hollow
cylinder of 13 protofilaments of the
tubulin protein. The diameter of each
microfibril is 25 nm. The function of
microtubule is to guide organelle and
chromosome movement in the cell,
cause cell elongation and help in
movements of cilia/flagella.
107Centromere is a part of
[CBSE AIPMT 1997]
(a) ribosomes
(b) chromosome
(c) mitochondria
(d) endoplasmic reticulum
Ans.(b)
In every chromosome, there is a small
region called primary constriction in
which there is a centromere where two
sister chromatids are held together and
spindle fibres get attached during cell
dizvision.
108The mechanism of ATP formation
both in chloroplast and
mitochondria is explained by
[CBSE AIPMT 1997]
(a) Relay Pump Theory of Godlewski
(b) Cholodny-Went’s Model
(c) Chemiosmotic Theory
(d) Munch’s Mass Flow Hypothesis
Ans.(c)
As per Peter Mitchell’s chemiosmotic-
coupling hypothesis, outward pumping
of protons across the inner chloroplast
or mitochondrial membrane results in
accumulation of protons between outer
membrane and inner membrane. A
proton gradient is thus established. As
protons now flow back, passively down
the gradient, the proton motive force is
utilised to synthesise ATP.
109Protein synthesis in an animal cell
takes place[CBSE AIPMT 1997]
(a) only in cytoplasm
(b) in the nucleolus as well as in the
cytoplasm
(c) in the cytoplasm as well as in
mitochondria
(d) only on ribosomes attached to
nucleus
90 NEETChapterwise Topicwise Biology
Telomeres
Chromonemata
Centromere
Chromatid
Chromosome

101. Ans.(c)
Protein synthesis is a complex
process, it essentially involves DNA for
the synthesis ofmRNA (transcription)
which contains information for the
synthesis of proteins (translation).
The process of translation takes place
on ribosomes which are found in
cytoplasm (in attached form on ER)
and in mitochondria (in the free form).
110Genes located on mitochondrial
DNA [CBSE AIPMT 1997]
(a) generally show maternal
inheritance
(b) are always inherited from the
male parent
(c) show biparental inheritance like the
nuclear genes
(d) are not inherited
Ans.(a)
Mitochondria are found only in
eukaryotic cells, they contain a single
circular double stranded DNA
molecule (mtDNA). Available evidences
show that mitochondria of female
parent are transferred to progeny
during fertilisation. Recent studies
have shown that factors responsible
for cytoplasmic male sterility are
located in mitochondrial DNA.
111Lysosomes have a high content
of [CBSE AIPMT 1996]
(a) hydrolytic enzymes
(b) lipoproteins
(c) polyribosomes
(d) DNA ligases
Ans.(a)
Lysosomes or suicidal bags are filled
with about 40 types of acid hydrolases
(digestive enzymes) like acid
proteases, acid nucleases, acid
phosphatases, acid sulphatases, acid
lipases, acid glycosidases working at
an optimum pH≤5 for controlling
intracellular digestion of
macromolecules.
112The function of rough
endoplasmic reticulum is
[CBSE AIPMT 1995]
(a) fat synthesis (b) lipid synthesis
(c) protein synthesis
(d) steroid synthesis
Ans.(c)
Rough Endoplasmic Reticulum (RER)
provides surface for ribosomes for
synthesis of secretory (serum
proteins), lysosomal and membrane
proteins, glycoproteins and helps in
packaging of polypeptide chains into
enzymes/proteins. It also provides
membrane to Golgi bodies for forming
vesicles and lysosomes.
113The prokaryotic flagella possess
[CBSE AIPMT 1995]
(a) unit membrane enclosed fibre
(b) protein membrane enclosed fibre
(c) ‘9+2’ membrane enclosed structure
(d) helically arranged protein molecule
Ans.(d)
Prokaryotic (bacterial) flagellum is made
up of flagellin protein arranged helically.
It do not show 9+2 organisation and
ATPase activity. These flagella do not
beat but rotate like a propellar that
brings about backward pushing of water.
Gram+ve bacteria having two rings in the
basal body. Gram−ve bacteria
have four rings. The L, P, M and S rings.
114The desmosomes are concerned
with [CBSE AIPMT 1995]
(a) cytolysis (b) cell division
(c) cell adherence (d) cellular excretion
Ans.(c)
Desmosomes(macula adherens)
consists of intercellular thickening
materials, disc-shaped intracellular
thickening adjacent to each membrane
with tonofibrils. These act as
intercellular cementing material, adhere
cells together at places like spot welding.
115Inner membrane convolutions of a
mitochondrion are known as
[CBSE AIPMT 1994]
(a) lamellae (b) thylakoids
(c) grana (d) cristae
Ans.(d)
Mitochondrial inner membrane is
convoluted several times to form cristae.
116Mitochondrial cristae are sites of
[CBSE AIPMT 1994]
(a) breakdown of macromolecules
(b) protein synthesis
(c) phosphorylation of flavoproteins
(d) oxidation-reduction reactions
Ans.(d)
Mitochondrial cristae bear the functional
unit, i.e. oxysomes, Fernandes and Moran
particles(F F
0 1
−) particles or electron
transport particle. Since, inner
membrane is impermeable to ATP, thus
ATP is synthesised on oxysomes having
ATPase inF
1
(stalk) and proton channels
onF
0
(base). ATPase helps in oxidative
phosphorylation, synthesise ATP
through electron transport system by
undergoing oxidation reduction
reactions.
117Organelle having flattened
membrane bound cisternae and
lying near the nucleus is
[CBSE AIPMT 1994]
(a) Golgi apparatus
(b) mitochondrion
(c) centriole
(d) nucleolus
Ans.(a)
Golgi body or dictyosome has a stack of
single membrane bound cisternae with
swollen ends, network of tubules and
vesicles. Cisternae are parallel
membrane lined narrow sacs which are
interconnected.
Golgi body has two faces- concave or
distal or maturing (M) face ortransface
towards cell membrane andcisor
convex or proximal or forming (F) face
towards rough ER and nuclear
membrane. New cisternae are formed
from SER and added from F-face.
118Cell organelles having
hydrolases/digestive enzymes are
[CBSE AIPMT 1994]
(a) peroxisomes
(b) lysosomes
(c) ribosomes
(d) mesosomes
Ans.(b)
Lysosomes (or suicidal bags or cellular
house keepers or scavenger of cell) are
single membrane bound, spherical
microbodies, filled with different types
of acid hydrolases (digestive enzymes)
working at pH≤5 and can digest almost
every type of organic matter except
cellulose.
Primary lysosome (storage granules)
unites with food vacuole (phagosome)
forming secondary lysosomes
(heterophagosomes or digestive
vacuoles) which cause intracellular
digestion (heterophagy).
119Organelle/organoid involved in
genetic engineering is
[CBSE AIPMT 1994]
(a) plasmid
(b) mitochondrion
(c) Golgi apparatus
(d) lomasome
Cell: The Unit of Life 91

102. Ans.(a)
Plasmids are small, self-replicating
extra chromosomal, non-essential
genetic/DNA elements. Plasmid
consists of a ring of circular,
supercoiled double stranded naked DNA
carrying genes for replication and for
one or more cellular non-essential
functions.
They are ideal vectors for genetic
engineering, gene cloning since, they
are self-replicating, carry non-essential
genes and has a restriction site for one
or more restriction endonucleases.
120In plant cells, peroxisomes are
associated with[CBSE AIPMT 1993]
(a) photorespiration
(b) phototropism
(c) photoperiodism
(d) photosynthesis
Ans.(a)
Peroxisomesare microbodies arising
from ER and containing enzymes for
peroxide formation. In plants, they
occur in green mesophyll cells of
leaves ofC
3
plants and are involved in
photorespiration through interacting
with chloroplast and mitochondria. In
animals, they are involved in lipid
synthesis, purine catabolism,
gluconeogenesis, etc.
121Membranous bag with hydrolytic
enzymes which is used for
controlling intracellular digestion
of macromolecules is
[CBSE AIPMT 1993]
(a) endoplasmic reticulum
(b) nucleosome
(c) lysosome
(d) phagosome
Ans.(c)
The intracellular digestion, i.e. the
breakdown of substances within the
cytoplasm of a cell is controlled by
lysosomes.
Intracellular digestion occurs in
animals that lack a digestive track.
e.g. in Pycnogonida, Mollusca,
Cnidaria and Porifera.
122Golgi apparatus is absent in
[CBSE AIPMT 1993]
(a) higher plants
(b) yeast
(c) bacteria and blue-green algae
(d) None of the above
Ans.(c)
Golgi bodies are absent in prokaryotic
cells, (i.e. bacteria, cyanobacteria,
mycoplasma), in mature RBC, mature
sperms, mature eggs, sperms of
bryophytes, pteridophytes and mature
sieve tubes. In contrast, active
eukaryotic cells are rich in Golgi bodies.
Maximum number (25000) of Golgi bodies
are found in rhizoidal cells ofChara.
123All plastids have similar structure
because they can
[CBSE AIPMT 1992]
(a) store starch, lipids and proteins
(b) get transformed from one type to
another
(c) perform same function
(d) be present together
Ans.(b)
All plastids are similar in structure
because these are interconvertible and
get transformed from one type to
another. Leucoplasts are formed from
proplastids and leucoplasts; chloroplasts
can arise from pre-existing chloroplasts,
proplastids and leucoplasts and
chromoplasts can develop from
proplastids, leucoplasts and
chloroplasts.
124Oxysomes orF F
0 1
−particles occur
on [CBSE AIPMT 1992]
(a) thylakoids
(b) mitochondrial surface
(c) inner mitochondrial membrane
(d) chloroplast surface
Ans.(c)
Oxysomes are the elementary particles
orF F
0 1
−or Fernandes-Moran particle
present on the inner membrane of
mitochondria. They are about10 10
4 5
−in
number and has a base ofF
0
sub-unit
toward C-face(11×1.5 nm), a stalk
(0.5 3.5 nm)− and a head orF
1
sub-unit
(8.5–10 nm diameter) towards matrix or
M-face.
125An outer covering membrane is
absent over [CBSE AIPMT 1992]
(a) nucleolus (b) lysosome
(c) mitochondrion (d) plastids
Ans.(a)
Nucleolus was discovered by Fontana
(1781) and named by Bowman (1840) is a
naked roughly rounded darkly stained
structure. It is attached to chromatin
at specific spot called nucleolar
organiser region or NOR. Nucleolus
constitute 35% mass of nucleus and is
the largest part of nucleus.
126Ribosomes are the centre for
[CBSE AIPMT 1992]
(a) respiration
(b) photosynthesis
(c) protein synthesis
(d) fat synthesis
Ans.(c)
Ribosomes are smallest,
membraneless sub- microscopic
organelles, called as protein factories.
They act as a template, bringing
together different components
involved in the protein synthesis.
127Which one isapparato reticolare
interno?
[CBSE AIPMT 1992]
(a) Golgi apparatus
(b) Endoplasmic reticulum
(c) Microfilaments
(d) Microtubules
Ans.(a)
Golgi body (dictyosomes, lipochondria)
is a stack of flattened membrane
bound sac-like body. They form
internal reticulare apparatus (apparato
reticolare interno).
128Experiments onAcetabulariaby
Hammerling proved the role of
[CBSE AIPMT 1992]
(a) cytoplasm in controlling
differentiation
(b) nucleus in heredity
(c) chromosomes in heredity
(d) nucleocytoplasmic ratio
Ans.(b)
J Hammerling(1953) carried the
grafting experiments involving
exchange of nucleus (located at the
base) inAcetabularia. He proved the
role of nucleus in heredity, growth,
etc.
92 NEETChapterwise Topicwise Biology

103. Cell: The Unit of Life 93
129Ribosomes were discovered by [CBSE AIPMT 1991]
(a) Golgi
(b) Porter
(c) de Robertis
(d) Palade
Ans.(d)
Ribosomes were first observed byClaude(1941), he called them
as microsomes.RobinsonandBrown(1950) noticed them in
plant cells of bean roots andPalade(1955) detected them in
animal cells and called these structures as ribosomes.
130Polyribosomes are aggregates of[CBSE AIPMT 1989]
(a) ribosomes andrRNA
(b) onlyrRNA
(c) peroxisomes
(d) several ribosomes held together by string ofmRNA
Ans.(d)
Polyribosomes or ergasomes are formed by the combination of
6-8 ribosomes attached on a single-strand ofmRNA.mRNA
brings about polymerisation of a specific protein molecule, with
the help of ribosomes, from amino acid molecules found in the
cytosol.
131Organelles can be separated from cell homogenate
through [CBSE AIPMT 1989]
(a) chromatography
(b) X-rays diffraction
(c) differential centrifugation
(d) auto-radiography
Ans.(c)
Differential centrifugationis the mechanical separation of
individual sub-cellular components from homogenate in
centrifuge at low speed.
Depending upon the size, specific gravity, mass, density,
different organelles are separated and settled at the bottom of
the centrifuge tube at different centrifugal speeds.
Ultracentrifuges have 50,000-1,00,000 rpm and are used for the
separation of minute cell organelles and constitutents on the
basis of different densities.

104. 01Which of the following are not
secondary metabolites in plants?
[NEET 2021]
(a) Morphine, codeine
(b) Amino acids, glucose
(c) Vinblastine, curcumin
(d) Rubber, gums
Ans.(b)
Primary metabolites are compounds
that are directly involved in the growth
and development of a plant like amino
acids and sugars (glucose). Primary
metabolites prominently function
growth, development and reproduction
of cell. Secondary metabolites are
compounds produced in other
metabolic pathways that, although
important, are not essential for the
functioning of the plant.
02Floridean starch has structure
similar to [NEET (Sep.) 2020]
(a) amylopectin and glycogen
(b) mannitol and algin
(c) laminarin and cellulose
(d) starch and cellulose
Ans.(a)
Floridean starch has structure similar
to amylopectin and glycogen as both
are made fromα-D glucose monomers.
The key difference between
amylopectin and glycogen is,
amylopectin is a soluble form of starch,
while glycogen is an insoluble form of
starch.
03Which two functional groups are
characteristic of sugars?
[NEET 2018]
(a) Carbonyl and phosphate
(b) Carbonyl and methyl
(c) Hydroxyl and methyl
(d) Carbonyl and hydroxyl
Ans.(d)
Sugarsare chemically carbohydrates.
They are polyhydroxy aldoses, ketoses
and their condensation products.
Aldoses bear a terminal aldehyde or
CHOgroup while ketoses have an
internal ketone orCOgroup.
Thus, they possess two functional
groups, i.e. carbonyl and hydroxyl.
04The chitinous exoskeleton of
arthropods is formed by the
polymerisation of
[CBSE AIPMT 2015]
(a) keratin sulphate and chondroitin
sulphate
(b) D-glucosamine
(c) N-acetyl glucosamine
(d) lipoglycans
Ans.(b)
The chitinous exoskeleton of
arthropods is formed by the
polymerisation of N-acetyl
glucosamine, which is a derivative of
glucose. It is also a characteristic
component of the cell wall of fungi, the
radulae of molluscs and the beaks and
internal shells of cephalopods,
including squid and octopuses.
05Which one of the following is a
non-reducing carbohydrate?
[CBSE AIPMT 2014]
(a) Maltose
(b) Sucrose
(c) Lactose
(d) Ribose 5-phosphate
Ans.(b)
Sucrose is a disaccharide of glucose
and fructose. It is a non-reducing sugar
as it do not contain any free anomeric
carbon atom. Maltose is a disaccharide
of 2 glucose units. Its first glucose
residue cannot undergo oxidation,
whereas, second residue can undergo
oxidation because it has a reactive free
anomeric carbon atom. Hence, it is a
reducing sugar.
Lactose and ribose-5-phosphate are
also reducing in nature due to the
presence of a free ketonic or aldehyde
group.
06Macromolecule chitin is
[NEET 2013]
(a) nitrogen containing
polysaccharide
(b) phosphorus containing
polysaccharide
(c) sulphur containing polysaccharide
(d) simple polysaccharide
Ans.(a)
Macromolecule chitin is a complex
polysaccharide containing amino
sugars and chemically modified sugars,
(e.g. glucosamine, N-acetyl
galactosamine, etc). Polysaccharides
are long carbohydrate molecules of
monosaccharide units joined together
by glycosidic bonds. They have a
general formulaC (H O)
x2y
. Chitin is the
main component of the cell wall of
fungi, the exoskeletons of arthropods,
insects and radulae of molluscs, etc.
07Which one of the following pairs is
wrongly matched?
[CBSE AIPMT 2009]
(a) Detergents — Lipase
(b) Alcohol — Nitrogenase
(c) Fruit juice — Pectinase
(d) Textile—Amylase
Ans.(b)
The wrongly matched pair is ‘b’ because
alcohol (ethyl alcohol) can be produced
by fermentation of any carbohydrate,
containing a fermentable sugar.
Biomolecules
09
Carbohydrates
TOPIC 1

105. Biomolecules 95
The development of synthetic alcohol
follows following reaction
08About 98 per cent of the mass of
every living organism is composed of
just six elements including carbon,
hydrogen, nitrogen, oxygen and
[CBSE AIPMT 2007]
(a) phosphorus and sulphur
(b) sulphur and magnesium
(c) magnesium and sodium
(d) calcium and phosphorus
Ans.(a)
About 98% of the mass of every living
organism including bacterium and human
beings is composed of just six elements,
i.e. Carbon (C), Hydrogen (H), Nitrogen (N),
Oxygen (O), Phosphorus (P) and Sulphur (S).
Charagaff(1950) suggested that despite
wide compositional variations exhibited
by different types of DNA the total
amount of purines equaled the total
amount of pyrimidines(A G T C)+ = + .
09Which of the following is a reducing
sugar?
[CBSE AIPMT 2002]
(a) Galactose
(b) Gluconic acid
(c)β-methyl galactoside
(d) Sucrose
Ans.(a)
Glucose, fructose, mannose, galactose
are hexose monosaccharides. The
monosaccharides have free aldehyde or
ketone group which can reduceCu
2 +
to
Cu. Therefore, these are called reducing
sugars.
10Most abundant organic compound on
earth is [CBSE AIPMT 2001, 04]
(a) protein (b) cellulose
(c) lipids (d) steroids
Ans.(b)
Cellulose is fibrous polysaccharide that
forms the structural component of plant
cell wall, some primitive fungi and tunic of
ascidians. Cellulose is the most abundant
organic substance on earth. It can be
digested by only a few microbes present in
the gut of ruminants and white ants.
11Lactose is composed of
[CBSE AIPMT 1998]
(a) glucose+glucose
(b) glucose+fructose
(c) fructose+galactose
(d) glucose+galactose
Ans.(d)
Lactose(C H O )
12 22 11
is a disaccharide
found in mammalian milk. It comprises
of galactose and glucose units which
are linked together byβ, 1-4 glycosidic
bonds. It is a reducing sugar.
12Cellulose, the most important
constituent of plant cell wall is
made of [CBSE AIPMT 1998]
(a) unbranched chain of glucose
molecules linked byα1,
4-glycosidic bond
(b) branched chain of glucose
molecules linked byβ1,
4-glycosidic bond in straight
chain andα, ,1 6-glycosidic bond
at the site of branching
(c) unbranched chain of glucose
molecules linked byβ1,
4-glycosidic bond
(d) branched chain of glucose
molecules linked byα1,
6-glycosidic bond at the site of
branching
Ans.(c)
Cellulose(C H O )
6 105n
is the most
abundant organic polymer. It is a
polysaccharide and consists of long
unbranched chains of glucose residues
linked byβ, 1-4 glycosidic bonds. In
plants, cellulose is formed from sugar.
It serves as building material in the
formation of cell wall.
13In which one of the following
groups, all the three are examples
of polysaccharides?
[CBSE AIPMT 1996]
(a) Starch, glycogen, cellulose
(b) Sucrose, maltose, glucose
(c) Glucose, fructose, lactose
(d) Galactose, starch, sucrose
Ans.(a)
Starch and glycogen are storage
polysaccharide and cellulose is a
structural polysaccharide. Starch has
straight chain of 1,4α-D pyranose
glucose units and side chains or
amylopectin part of 2-200 thousand
glucose units that are attached to
straight chains by 1,6α-D glycosidic
bonds.
Glycogen is long chain branched
polymer of 5-300 thousand glucose
units, straight chain linked byα-D 1-4
glycosidic bonds and side chains
throughα-D1 6→glycosidic bonds.
Cellulose, a fibrous polysaccharide
has a linear chain of 6-10 thousand
1 4→linkedβ-pyranose glucose
chain.
14Glycogen is a polymer of
[CBSE AIPMT 1992]
(a) galactose (b) glucose
(c) fructose (d) sucrose
Ans.(b)
A glycogen molecule is a long highly
branched chain of about 30000 to
100000α-D glucose units joined by
glycosidic bonds. It is storage form of
glucose, popularly called animal
starch. It provides energy to animals,
fungi and bacteria.
15Living cell contains 60-75%
water. Water present in human
body is [CBSE AIPMT 1992]
(a) 60-65%
(b) 50-55%
(c) 75-80%
(d) 65-70%
Ans.(d)
In humans, about two-third of body is
formed of water. About 65-70% of
human body is water, of these about
55% (20-22 L) is confined to cells as
intracellular water.
16Which one of the following is the
most abundant protein in the
animals? [NEET (Sep.) 2020]
(a) Collagen
(b) Lectin
(c) Insulin
(d) Haemoglobin
Ans.(a)
Collagen is the most abundant
protein in animal world and RuBisCO
is the most abundant protein in the
whole of the biosphere because it is
present in every plant that undergoes
photosynthesis and molecular
synthesis through the Calvin cycle.
Proteins
TOPIC 2
Starch Glucose
Enzyme Yeast
Hydrolysis
(50-60°C)
Glycolysis
(30-35°C)
Pyruvate
Yeast (30-35°C)
(Pyruvate
decarboxylase)
Ethanol
(Recovered by distillation)
Yeast
(30-35°C)
Alcohol
dehydrogenase
Aldehyde

106. 17Identify the substances having
glycosidic bond and peptide bond,
respectively in their structure.
[NEET (Sep.) 2020)
(a) Glycerol, trypsin
(b) Cellulose, lecithin
(c) Inulin, insulin
(d) Chitin, cholesterol
Ans.(c)
Inulin is a fructan (polysaccharide of
fructose). Adjacent fructose units are
linked through glycosidic bond. Insulin
is a protein composed of 51 amino
acids. Adjacent amino acids are
attached through peptide bond.
18Identify the basic amino acid from
the following.[NEET (Sep.) 2020]
(a) Glutamic acid (b) Lysine
(c) Valine (d) Tyrosine
Ans.(b)
Option (b) is correct as lysine is a basic
amino acid because it’s side chain
contain nitrogen and resemble
ammonia, which is a base. Valine is a
neutral amino acid, glutamic acid is an
acidic amino acid, while tyrosine is an
aromatic amino acid.
19“Ramachandran plot” is used to
confirm the structure of
[NEET (Odisha) 2019]
(a) RNA (b) proteins
(c) triacylglycerides (d) DNA
Ans.(b)
‘Ramachandran plot’ is used to confirm
the structure of proteins.
Ramachandran plot is a plot of the
angles-phi (ϕ) and psi(ψ)of amino acids
found in a peptide chain. This plot was
developed by GN Ramachandran, an
Indian Scientist in 1963.
20Which of the following is an amino
acid derived hormone?[NEET 2018]
(a) Estradiol (b) Ecdysone
(c) Epinephrine (d) Estriol
Ans.(c)
Among the following,epinephrineis an
amino acid derived hormone. It is a
catecholamine which is produced in the
chromaffin cells of adrenal medulla
from amino acids tyrosine. On the other
hand, estradiol and estriol are steroid
hormone that are involved in the
regulation of estrous and menstrual
cycles.
Ecdysone is also a steroid hormone that
controls moulting in insects.
21Which of the following is the least
likely to be involved in stabilising
the three-dimensional folding of
most proteins?
[NEET 2016, Phase II]
(a) Hydrogen bonds
(b) Electrostatic interaction
(c) Hydrophobic interaction
(d) Ester bonds
Ans.(d)
Ester bonds are the least likely to be
involved in stabilising the 3D folding of
most proteins. A long protein chain gets
folded upon itself like a hollow woolen
ball, giving rise to a tertiary structure
(3D) structure.
This structure is stabilised by several
type of bonds, i.e. hydrogen bonds,
ionic bonds, Vander waal’s interactions,
covalent bonds and hydrophobic bonds.
Ester bond is formed between sugar
and phosphate in a nucleotide and is
not involved in stability of a polypeptide
chain. Thus, option (d) is correct.
22Which one is the most abundant
protein in the animal world?
[CBSE AIPMT 2012]
(a) Trypsin
(b) Haemoglobin
(c) Collagen
(d) Insulin
Ans.(c)
Collagen is the most abundant protein
(structural protein) in the animal world
while Ribulose Bisphosphate
Carboxylase Oxygenase (RuBisCO) is the
most abundant protein in the whole of
the plant world.
23Which one out of A-D given below
correctly represents the structural
formula of the basic amino acid?
[CBSE AIPMT 2012]
A B C D
(a) C (b) D
(c) A (d) B
Ans.(c)
Structure D represents basic amino
acid lysine whereas structure A
represents glutamic acid (acidic amino
acid) and structure B represents
alcoholic amino acid serine.
24Which of the following is the
simplest amino acid?
[CBSE AIPMT 2005]
(a) Alanine (b) Asparagine
(c) Glycine (d) Tyrosine
Ans.(c)
Proteins are polymers of amino acids in
which amino acids are joined by peptide
bonds. Glycine is the simplest amino
acid.
25Collagen is[CBSE AIPMT 2002]
(a) fibrous protein
(b) globular protein
(c) lipid
(d) carbohydrate
Ans.(a)
Collagen is a major fibrous protein of
connective tissue, occurring as white
fibres produced by fibroblast.
26Which is an essential amino acid?
[CBSE AIPMT 2000]
(a) Serine (b) Aspartic acid
(c) Glycine (d) Phenylalanine
Ans.(d)
For human beings, eight amino acids
are essential : Leucine, isoleucine,
lysine, methionine, phenylalanine,
threonine, tryptophan and valine. So,
these are known as essential amino
acids.Cannot be synthesised in animal
body.
27Conjugated proteins containing
carbohydrates as prosthetic group
are known as[CBSE AIPMT 2000]
(a) chromoproteins
(b) glycoproteins
(c) lipoproteins
(d) nucleoproteins
Ans.(b)
Glycoproteins(mucoproteins) are
conjugated protein having a simple or
complex sugar (galactose) residue at
their N-terminal end. Glycoprotein is
found in egg white, mucin, antibody IgG,
cell membrane, saliva, synovial fluid,
heparin. Lipoproteins contain lipids and
nucleoproteins contain nucleic acid as
the prosthetic group.
96 NEETChapterwise Topicwise Biology
H C COOH 
CH
2
NH
2
CH
2
C
OHO
H C COOH 
CH
2
NH
2
OH
CH
2
CH OH
2
NH
2
CH
2H C COOH 
CH
2
NH
2
CH
2
CH
2
CH
2
NH
2

107. 28What is common among amylase,
rennin and trypsin?
[CBSE AIPMT 1997]
(a) These are all proteins
(b) These are proteolytic enzymes
(c) These are produced in stomach
(d) These act at a pH lower than 7
Ans.(a)
All three are proteins enzymes amylase,
rennin and trypsin acts respectively on
the substrates starch, lactose and
protein. Amylase acts at an optimum
pH—6.8-7, rennin at pH-2 and trypsin at
an optimum pH—8.5.
29Most diverse macromolecules,
found in the cell both physically and
chemically are[CBSE AIPMT 1996]
(a) proteins (b) carbohydrates
(c) nucleic acids (d) lipids
Ans.(a)
Proteins are the most diverse among
organic compounds. Among the
biomolecules, proteins constitute 9-12%,
fat 1-3%, carbohydrates 1-2%, minerals
1-3%, nucleic acids 2% and water 60-75%.
30The pyrenoids are made up of
[CBSE AIPMT 1995]
(a) proteinaceous centre and starchy
sheath
(b) core of protein surrounded by fatty
sheath
(c) core of starch surrounded by
sheath of protein
(d) core of nucleic acid surrounded by
protein sheath
Ans.(a)
Pyrenoids consist of dense proteinaceous
area surrounded by starchy sheath. These
are unique to chloroplast of algae and are
associated in polymerising sugars into
reserve polymers (starch).
31Amino acids are produced from
[CBSE AIPMT 1992]
(a) proteins (b) fatty acids
(c) essential oils (d)α-keto acids
Ans.(d)
Amino acids are smallest structural units
of proteins. Plants and some
microorganisms can synthesise these
amino acids from inorganic
nitrogen/α-keto acids by reductive
amination and transamination. Amino
acids besides bearing a carboxyl group,
possess an amino group attached to
α-carbon, hence, called asα-amino
acids.
32Following are the statements
with reference to lipids.
[NEET 2021]
I. Lipids having only single
bonds are called unsaturated
fatty acids.
II. Lecithin is a phospholipid.
III. Trihydroxy propane is
glycerol.
IV. Palmitic acid has 20 carbon
atoms including carboxyl
carbon.
V. Arachidonic acid has 16
carbon atoms.
Choose the correct answer from
the options given below
(a) I and III (b) III and IV
(c) II and III (d) II and V
Ans.(c)
Statements I, IV and V are incorrect,
while II and III are correct.
The incorrect statements can be
corrected as
Lipid having only single bonds are
called saturated fatty acids.
Palmitic acid has 16 carbon atoms
including carboxyl carbon.
Arachidonic acid has 20 carbon atoms.
33Which of the following are not
polymeric ? [NEET 2017]
(a) Nucleic acid
(b) Proteins
(c) Polysaccharides
(d) Lipids
Ans.(d)
Among the given options except lipids
all are polymers. These are formed by
the polymerisation of monomers. The
basic unit of lipid are fatty acids and
glycerol molecules that do not form
repetitive chains. Instead they form
triglycerides from three fatty acids and
one glycerol molecules.
34A typical fat molecule is made up
of
[NEET 2016, Phase I]
(a) One glycerol and three fatty acid
molecules
(b) One glycerol and one fatty acid
molecule
(c) Three glycerol and three fatty acid
molecules
(d) Three glycerol molecules and one
fatty acid molecule
Ans.(a)
A typical fat molecule is triglyceride
formed by esterification of one glycerol
and three fatty acid molecules. The
three fatty acids can be of same type or
different depending on the type of the
fat molecules.
35A phosphoglyceride is always made
up of [NEET 2013]
(a) only a saturated fatty acid
esterified to a glycerol molecule
to which a phosphate group is also
attached
(b) only an unsaturated fatty acid
esterified to a glycerol molecule
to which a phosphate group is also
attached
(c) a saturated or unsaturated fatty
acid esterified to a glycerol
molecule to which a phosphate
group is also attached
(d) a saturated or unsaturated fatty
acid esterified to a phosphate
group, which is also attached to a
glycerol molecule
Ans.(c)
A fat is formed of two kinds of smaller
molecules, i.e. glycerol and fatty acids.
In making a fat free fatty acid molecules
join to glycerol by an ester linkage. A
fatty acid has a long carbon skeleton,
usually 16 or 18 carbon atoms in length.
If there are no double bonds between
carbon atoms composing the chain,
then as many hydrogen atoms as
possible are bonded to the carbon
skeleton.This is called saturated fatty
acid. Unsaturated fatty acid has one or
more double bonds. Phosphoglycerides
are esters of two fatty acids,
phosphoric acid and a trifunctional
alcohol glycerol. The fatty acids are
attached to the glycerol at the 1 and 2
position on glycerol through ester
bonds.
36Which one of the following is not a
constituent of cell membrane?
[CBSE AIPMT 2007]
(a) Cholesterol
(b) Glycolipids
(c) Proline
(d) Phospholipids
Biomolecules 97
Lipids
TOPIC 3

108. Ans.(c)
The lipids of cell membrane are of three
types namely phospholipids, glycolipids
and sterols. The sterol found in the
membrane may be cholesterol
(animals), phytosterol (plants) or
ergosterol (microorganisms). Cell
membrane is made up of lipid+protein
(60%) and carbohydrates (2-10%).
Proline is an amino acid.
37Lipids are insoluble in water
because lipid molecules are
[CBSE AIPMT 2002]
(a) hydrophilic (b) hydrophobic
(c) neutral (d) Zwitter ions
Ans.(b)
Bloor(1943) first time used the term
‘lipid’. These are the compounds of C,
H, O but the ratio of H and O is more
than 2 : 1 that is the ratio of oxygen is
lesser as compared to carbohydrates.
Lipids are insoluble in water but soluble
in non-polar solvents such as benzene,
chloroform, etc. Common lipid
occurring in a cell is phospholipid. It
contains a hydrophilic (polar) head and a
hydrophobic (non-polar) tail.
38Which steroid is used for
transformation?
[CBSE AIPMT 2002]
(a) Cortisol
(b) Cholesterol
(c) Testosterone
(d) Progesterone
Ans.(b)
Cholesterol forms a major component
of animal cell membranes liposomes
(artificially created spheres surrounded
by a phospholipid bilayer like a
membrane) which are used for
transformation (transgenics).
39Spoilage of oil can be detected by
which fatty acid?
[CBSE AIPMT 2001]
(a) Oleic acid (b) Linolenic acid
(c) Linoleic acid (d) Erucic acid
Ans.(d)
Erucic acid occurs as glycerides in
vegetable oils. It is a monounsaturated
omega-9-fatty acid. It is found in
rapseed, mustared seed and wallflower
seed.
40Match the List-I with List-II.
[NEET 2021]
List-I List-II
A. Protein 1. C ==C double
bonds
B. Unsaturated fatty
acid
2. Phosphodiester
bonds
C. Nucleic acid 3. Glycosidic bonds
D. Polysaccharide 4. Peptide bonds
Choose the correct answer from
the options given below.
A B C D
(a) 4 1 2 3
(b) 1 4 3 2
(c) 2 1 4 3
(d) 4 3 1 2
Ans.(a)
(A)-(4), (B)-(1), (C)-(2), (D)-(3)
Proteinsare polypetides, they are
linear chain of amino acids linked by
peptide bond.
Unsaturated fatty acidsare carbon
chains containing one or more double
bonds with terminal carboxylic acid.
The two sugar molecules ofnucleic
acidsare linked togethervia
phosphodiester bond.
Polysaccharidesare long chain of sugar
molecules joined with a covalent bond,
i.e. glycosidic linkage.
41Match the items in Column I with
those in Column II.
[NEET (Oct.) 2020]
Column I Column II
A. Aquaporin (i) Amide
B. Asparagine (ii) Polysaccharide
C. Abscisic acid (iii) Polypeptide
D. Chitin (iv) Carotenoids
Select the correct option.
A B C D
(a) (iii) (i) (iv) (ii)
(b) (ii) (iii) (iv) (i)
(c) (ii) (i) (iv) (iii)
(d) (iii) (i) (ii) (iv)
Ans.(a)
Option (a) is correct match which is as
follows
Chemically, aquaporins are major
intrinsic proteins that form pores in the
membranes of cell. Asparagine is
beta-amide derivative of aspartic acid.
Abscisic acid is an apo-carotenoid.
Chitin is a heteropolysaccharide
consisting of two types of
monosaccharide monomers.
42DNA precipitation out of a mixture
of biomolecules can be achieved
by treatment with
[NEET (National) 2019]
(a) chilled ethanol
(b) methanol at room temperature
(c) chilled chloroform
(d) isopropanol
Ans.(a)
Chilled ethanol is used to precipitate
DNA out of a mixture of biomolecules.
Low temperature protects the DNA by
slowing down the activity of enzymes
that could break it apart and ethanol
helps in the quick precipitation of DNA.
43Which one of the following
statements is wrong?
[NEET 2016, Phase I]
(a) Cellulose is a polysaccharide
(b) Uracil is a pyrimidine
(c) Glycine is a sulphur containing
amino acid
(d) Sucrose is a disaccharide
Ans.(c)
Glycine is the simplest amino acid in
which functional group ‘R’ is replaced by
hydrogen atom (H).
44Which of the following
biomolecules does have a
phosphodiester bond?
[CBSE AIPMT 2015]
(a) Fatty acids in a diglyceride
(b) Monosaccharides in a
polysaccharide
(c) Amino acids in a polypeptide
(d) Nucleic acids in a nucleotide
Ans.(d)
Phosphodiester bond is in responsible
for linking nucleotides in nucleic acid
(DNA and RNA).
98 NEETChapterwise Topicwise Biology
Nucleic Acids
TOPIC 4

109. 45Given below is the diagrammatic
representation of one of the
categories of small molecular
weight organic compounds in the
living tissues. Identify the category
shown and the one blank
componentXin it
[CBSE AIPMT 2012]
Category Component
(a) Cholesterol — Guanine
(b) Amino acid —NH
2
(c) Nucleotide — Adenine
(d) Nucleoside — Uracil
Ans.(d)
Nucleosideis made up of ribose sugar
and nitrogenous base only. Uracil forms
nucleoside with only ribose sugar. So,
the option with category nucleoside
component uracil is correct.
46Which one of the following
structural formula of two organic
compounds is correctly identified
along with its related function?
[CBSE AIPMT 2011]
(a)A — Triglyceride-
major
— Source of
energy
(b)B — Uracil — A component
of DNA
(c)A — Lecithin — A component
of cell
membrane
(d)B — Adenine — A nucleotide
that makes
up nucleic
acids
Ans.(c)
Lecithin is a phospholipid composed of
choline and inositol. It is found in all
living cells as a major component of cell
membrane.
47Which one of the following is the
correct matching of three items
and their grouping category?
[CBSE AIPMT 2009]
Items Group
(a) Malleus, incus,
cochlea
— Ear ossicles
(b) Ilium, ischium,
pubis
— Coxal bones of
pelvic girdle
(c) Actin, myosin,
rhodopsin
— Muscle
proteins
(d) Cytosine,
uracil,
thymine
— Pyrimidines
Ans.(d)
There are total five nitrogenous bases
found in nucleic acids. Out of these
adenine, guanine(purines) and
cytosine, thymine(pyrimidines) are
present in DNA, while RNA contains
uracilin place ofthymine(both
pyrimidines) along with rest 3 similar to
DNA.
48Which one of the following pairs of
nitrogenous bases of nucleic acids,
is wrongly matched with the
category mentioned against it?
[CBSE AIPMT 2008]
(a) Thymine, Uracil — Pyrimidines
(b) Uracil, Cytosine — Pyrimidines
(c) Guanine, Adenine — Purines
(d) Adenine, Thymine — Purines
Ans.(d)
DNA and RNA the principal genetic
materials of living organisms are
chemically called nucleic acids.
These are polymers of nucleotides.
Each nucleotide is composed of
phosphoric acid, pentose sugar and
nitrogenous base. The nitrogenous
bases are of two types, i.e. purine and
pyrimidines.
Purines are heterocyclic and two
rings, e.g. adenine, guanine.
Pyrimidines are single ring compound,
e.g. thymine, cytosine, uracil.
49In the DNA molecule
[CBSE AIPMT 2008]
(a) the total amount of purine
nucleotides and pyrimidine
nucleotides is not always equal
(b) there are two strands which run
parallel in the5 3′ → ′direction
(c) the proportion of adenine in
relation to thymine varies with the
organism
(d) there are two strands which run
antiparallel one in5 3′ → ′direction
and other in3 5′ → ′
Ans.(d)
In DNA molecule the adjacent
deoxyribonucleotides are joined in a
chain by phosphodiester bridges or
bonds, which link the5′carbon of
deoxyribose of one mononucleotide
unit with3′carbon of deoxyribose of
next mononucleotide unit.
According toWatsonandCrickDNA
molecule consists of two such
polynucleotide chains wrapped helically
around each other, with the sugar
phosphate chain on the outside and
purine and pyrimidine on the inside of
helix.
The two strands run antiparallel, i.e. one
strand has phosphodiester linkage in3′
→ ′5direction while other strand has
phosphodiester linkage in5 3′→ ′
direction.
Biomolecules 99
R C O CH
2  
O CH O C R
2  
O
CH O P O CH CH
2 2 2    
O
OH N
CH
3
CH
3
CH
2
(A)



 






N
NH
NH
2
N
N
(B)
3
4
5
6
1
2
H
N
9
N
8
7
N
N
Purine
OHOH
HOCH
2
Uracil
O
Nucleoside Uridine
3
4
5
6
1
2
N
N
Pyrimidine
HOCH
2
OH OH
O
X

110. 100 NEETChapterwise Topicwise Biology
Chargaff(1950) suggested that despite wide compositional variations exhibited by different types of DNA the total amount of purines
equaled the total amount of pyrimidines (A + G = T + C).
50The two polynucleotide chains in DNA are [CBSE AIPMT 2007]
(a) parallel (b) discontinuous (c) antiparallel (d) semiconservative
Ans.(c)
In 1953James WatsonandFrancis Cricksuggested that in a DNA molecule there are two polynucleotide chains arranged antiparallel or
in opposite directions.
51One turn of the helix in a B-form DNA is approximately
[CBSE AIPMT 2006]
(a) 0.34 nm (b) 3.4 nm (c) 2 nm (d) 20 nm
Ans.(b)
B-DNA is helical structure with 20Å diameter and the distance between the two base pairs is 3.4Å and there are 10 base pairs in each turn
or pitch (one round). Hence, one turn of the helix is approximately 34Å or 3.4 nm (10Å = 1.0 nm). Z-DNA (in comparision to B-DNA) is left
handed double helical structure in which double helix winds to left inzig-zagpattern (instead of right, like B-DNA).
52Nucleotides are building blocks of nucleic acids. Each nucleotide is a composite molecule formed by
[CBSE AIPMT 2005, 1991]
(a) base-sugar-phosphate (b) base-sugar-OH (c) (base-sugar-phosphate )
n
(d) sugar-phosphate
Ans.(a)
Nucleotides are the building blocks of nucleic acids (DNA and RNA). A single nucleotide is composed of a phosphate molecule, a five
carbon sugar (either ribose or deoxyribose) and a purine (adenine or guanine) or a pyrimidine (thymine or cytosine or uracil) nitrogenous
base.
53ATP is a [CBSE AIPMT 2000]
(a) nucleotide (b) nucleosome
(c) nucleoside (d) purine
Ans.(a)
A nucleotide contains (a) a 5-C sugar (b) a phosphate molecule (c) a nitrogenous base. ATP is also a nucleotide. It also has a 5-C
sugar (ribose), 3 phosphate molecules and a nitrogenous base (adenine).
P
O
N
H
2´
3´
O
4´
1´
OH C
2
5´
O
O
2´
3´
1´
H
HN
O
O CH
3
NH
HN
H
Hydrogen
bonds
H
O
4´
3´
2´
O
CH
2
O
P
HO
O
–
O
–
4´
3´
2´
1´
CH
2
5´
O
P
HO
O
–
O
–
Guanine
Cytosine
Pentose
sugar
Phosphate
Adenine
Nucleotides
H
H
N
HO
O
Thymine
O
O
N
N
N
N
N
N
N
O
Chemical structures of base pairs and sugar phosphate chains

111. 54Which one of the following amino acids is an essential part ofhuman diet ? [CBSE AIPMT 2000]
(a) Glycine (b) Phenylalanine (c) Serine (d) Aspartic acid
Ans.(b)
For human beings, eight amino acids are essential. These are leucine, isoleucine, lysine, methionine, phenylalanine, threonine,
tryptophan and valine.
55Radioactive thymidine when added to the medium surroundingliving mammalian cells gets incorporated into the newly
synthesised DNA. Which of the following types of chromatin is expected to become radioactive if cells are exposed
radioactive thymidine as soon as they enter the S-phase? [CBSE AIPMT 1998]
(a) Heterochromatin (b) Euchromatin (c) Both (a) and (b)
(d) Neither (a) nor (b) but only the nucleolus
Ans.(b)
In the beginning of S-phase, DNA replication occurs. DNA replication can occur in diffuse/less tightly coiledeuchromatin. So active DNA
stains light in colour when stained with acetocarmin and feulgen reagent in comparison to heterochromatin.
56DNA synthesis can be specifically measured by estimating the incorporation of radio labelled[CBSE AIPMT 1997]
(a) uracil (b) adenine (c) thymidine (d) deoxyribose sugar
Ans.(c)
DNA consists of deoxyribose sugar, phosphate molecules and nitrogenous bases-adenine, guanine, cytosine and thymine, whereas RNA
consists of ribose sugar, phosphate molecules and nitrogenous bases-adenine, guanine, cytosine and uracil.
Thus, estimating the incorporation of radiolabelled thymine can measure DNA synthesis and radiolabelled uracil can measure RNA
synthesis, as all other nitrogenous bases are similar in both DNA and RNA.
57The nitrogenous organic base purine occurring in RNA is [CBSE AIPMT 1996]
(a) cytosine (b) thymine (c) guanine (d) uracil
Ans.(c)
Purines are 9-membered double ring nitrogen bases which possess nitrogen at 1, 3, 7 and 9 positions, e.g. adenine (A), guanine (G). These
purines are present in both DNA and RNA.
58Two free ribonucleotide units are interlinked with [CBSE AIPMT 1995]
(a) peptide bond (b) covalent bond (c) hydrogen bond (d) phosphodiester bond
Ans.(d)
The bonds that exist between the phosphate group of one nucleotide and hydroxyl group of sugar (ribose or deoxyribose) of the adjacent
nucleotide is known as phosphodiester bond.
Biomolecules 101
Adenine + sugar (ribose)→Adenosine
Adenosine + 1 phosphate→Adenosine Monophosphate (AMP)
AMP + 1 phosphate→Adenosine Diphosphate (ADP)
O
||
C
C
N
C
N
C
NH
|
2
3
4
5
6
1
2
H
N
9
C H
N
8
7
H
H H
H
OHOH
Ribose
Adenine
O
2'
1'
3'
4'
5'
CH O—P—O P—O P—OH
2
~~
|
OH
O
||
|
OH
O
||
|
OH
AMP
ADP
ATP
Adenine–Ribose–O–P=AMP
Adenine–Ribose–O—P—O P=ADP
Adenine–Ribose–O—P—O P—O P=ATP
~
~ ~
The chemical structure of AMP, ADP and ATP
Pyrophosphate

112. 59Which is wrong about nucleic
acids? [CBSE AIPMT 1993]
(a) DNA is single stranded in some
viruses
(b) RNA is double stranded
occasionally
(c) Length of one helix is 45 Å in
B-DNA
(d) One turn of Z-DNA has 12 bases
Ans.(c)
Length of one helix or the pitch per turn
of helix is 34 Å in B-DNA, 25 Å in A-DNA
and 46 Å in Z-model of DNA. In Z-DNA
sugar moieties are seen in opposite
direction. So the 3-5 diester bond forms
zig-zag structure in Z-DNA.
60Adenine is [CBSE AIPMT 1992]
(a) purine (b) pyrimidine
(c) nucleoside (d) nucleotide
Ans.(a)
Nitrogen bases are of two
types-purines and pyrimidines.Purines
are 9 membered double ring nitrogen
bases which possess nitrogen at 1, 3, 7
and 9 positions, e.g. adenine (A),
guanine (G). Pyrimidines are
6-membered nitrogen bases that
contain nitrogen at 1 and 3 positions,
e.g. cytosine (C), thymine (T), uracil (U).
61Which is distributed more widely
in a cell? [CBSE AIPMT 1992]
(a) DNA (b) RNA
(c) Chloroplasts (d) Spherosomes
Ans.(b)
Ribonucleic Acid (RNA) is a single chain
polyribonucleotide which functions as
carrier of coded genetic information
from DNA to cytoplasm, and takes part
in protein and enzyme synthesis. RNA is
more common and abundant than DNA.
There are six types of RNAs-ribosomal
(most abundant), transfer RNA (15% to
total RNA), messenger RNA (2-5%),
small sized nuclear RNA, small
cytoplasmic RNA, and genetic RNA (in
viruses called riboviruses).
62The basic unit of nucleic acid is
[CBSE AIPMT 1991]
(a) pentose sugar (b) nucleoid
(c) nucleoside (d) nucleotide
Ans.(d)
Nucleic acids (DNA and RNA) are
linear mixed polymers ofnucleotides,
so also called polynucleotides.
Nucleic acids were first discovered
byMiescher(1868-69) as nuclein and
named nucleic acid byAltmann
(1889). They are formed of C, H, O, N
and P comprising nitrogenous
heterocyclic basesvizpurines or
pyrimidines, pentose sugar and
phosphoric acid.
63DNA is composed of repeating
units of
[CBSE AIPMT 1991]
(a) ribonucleosides
(b) deoxyribonucleosides
(c) ribonucleotides
(d) deoxyribonucleotides
Ans.(d)
Nucleic acids are repeating units of
nucleotides, DNA is formed of the
repeating units of deoxyribonucleotides
and RNA is a mixed polymer chain of
ribonucleotides.
64In RNA, thymine is replaced by
[CBSE AIPMT 1991]
(a) adenine
(b) guanine
(c) cytosine
(d) uracil
Ans.(d)
Uracil forms nucleoside with only ribose
sugar while thymine forms the same
with only deoxyribose sugar. Other
nitrogen bases (i.e. adenine, guanine,
cytosine) produce nucleosides with
both sugars.
65RNA does not possess
[CBSE AIPMT 1988]
(a) uracil (b) thymine
(c) adenine (d) cytosine
Ans.(b)
Ribonucleic Acid (RNA) contains
pyrimidine bases-cytosine and uracil
and purine bases-adenine and guanine.
Thymine is the pyrimidine base
present only in DNA and uracil is
present only in RNA, though other
nitrogenous bases remain the same
in both DNA and RNA.
66In double helix of DNA, the two
DNA strands are [CBSE AIPMT
1988]
(a) coiled around a common axis
(b) coiled around each other
(c) coiled differently
(d) coiled over protein sheath
Ans.(a)
Deoxyribo Nucleic Acid (DNA) is helically
coiled macromolecule, made up of two
antiparallel polydeoxyribonucleotide
chains held together by hydrogen
bonds.
The chains are interlocked and are
coiled around a common axis. DNA has
a diameter of 20 Å. One turn of spiral
has a distance of 34 Å containing 10
nucleotides in each turn.
67Match the following columns.
[NEET (Sep.) 2020]
Column I Column II
A. Inhibitor of catalytic
activity
1. Ricin
B. Possess peptide
bonds
2. Malonate
C. Cell wall material in
fungi
3. Chitin
D. Secondary
metabolite
4. Collagen
Choose the correct option.
A B C D
(a) 3 1 4 2
(b) 3 4 1 2
(c) 2 3 1 4
(d) 2 4 3 1
Ans.(d)
Option (d) is the correct. It can be
explained as follows.
Malonate is the competitive inhibitor of
catalytic activity of succinic
dehydrogenase. Collagen is
proteinaceous in nature and possesses
peptide bonds.
Chitin is a homopolymer present in the
cell wall of fungi and exoskeleton of
arthropods.
Abrin and ricin are toxins, secondary
metabolites.
102 NEETChapterwise Topicwise Biology
Enzymes
TOPIC 5
NH
O
H
N
Uracil
O

113. 68Prosthetic groups differ from
coenzymes in that
[NEET (Odisha) 2019]
(a) they require Gmetal ions for their
activity
(b) they (prosthetic groups) are
tightly bound to apoenzymes
(c) their association with
apoenzymes is transient
(d) they can serve as cofactors in a
number of enzyme catalysed
reactions
Ans.(b)
Prosthetic groups are organic
compounds and are distinguished from
other cofactors in that they are tightly
bound to the apoenzyme. For example,
in peroxidase and catalase, which
catalyse the breakdown of hydrogen
peroxide to water and oxygen, haem is
the prosthetic group and it is a part of
the active site of the enzyme.
69Which of the following is a
commercial blood cholesterol
lowering agent?
[NEET (National) 2019]
(a) Statin (b) Streptokinase
(c) Lipases (d) Cyclosporin A
Ans.(a)
Statins are commercial blood
cholesterol lowering agent as they
competitively inhibit the enzymes
involved in cholesterol synthesis. They
are obtained from a yeast,Monascus
purpureus.
Streptokinase is a thrombolytic agent
which is used to treat pulmonary
embolism and myocardial infarction.
Lipases help to digest fat molecules
while cyclosporin A is an
immunosuppressant.
70Consider the following statement
[NEET (National) 2019]
(A) Coenzyme or metal ion that is
tightly bound to enzyme
protein is called prosthetic
group.
(B) A complete catalytic active
enzyme with its bound
prosthetic group is called
apoenzyme. Select the
correct option.
(a) (A) is true but (B) is false
(b) Both (A) and (B) are false
(c) (A) is false but (B) is true
(d) Both (A) and (B) are true
Ans.(a)
Statement A is true but B is false.
Correct information about statement B
is as follows.
A complete catalytic, active enzyme
with its bound prosthetic group is called
holoenzyme. An apoenzyme is an
inactive enzyme which gets activated
by the binding of an organic or inorganic
cofactor.
71Select thecorrectmatch
[NEET 2018]
(a) TH Morgan – Transduction
(b)F
2
×Recessive parent – Dihybrid
cross
(c) Ribozyme – Nucleic acid
(d) G Mendel – Transformation
Ans.(c)
Ribozymesare RNA molecules having
enzymatic activity, i.e. they are capable
of catalysing specific biochemical
reactions. Hence, they are nucleic acids
with enzymatic function.
TH Morganis known as the ‘Father of
Experimental Genetics’. He worked on
linkage, crossing over, linkage maps,
etc.
Indihybrid cross,two allelic pairs are
used for crossing.
Mendelis considered as the ‘Father of
Genetics’. He proposed the laws of
inheritance.
72Which one of the following
statements is correct, with
reference to enzymes?[NEET 2017]
(a) Apoenzyme=Holoenzyme+
Coenzyme
(b) Holoenzyme=Apoenzyme+
Coenzyme
(c) Coenzyme=Apoenzyme+
Holoenzyme
(d) Holoenzyme=Coenzyme+
Cofactor
Ans.(b)
Holoenzyme It is a conjugate
catalytically active enzyme together
with its coenzyme.
ApoenzymeThe protein part of
catabolically active enzyme is called
apoenzyme.
CoenzymeSome enzymes require
additional organic or metallo-organic
molecules for their activity. These
molecules are called coenzyme.
So, holoenzyme is apoenzyme together
with coenzyme hence option (b) is
correct.
73Which of the following describes
the given graph correctly?
[NEET 2016, Phase II]
(a) Endothermic reaction with energy
Ain the presence of enzyme andB
in the absence of enzyme
(b) Exothermic reaction with energyA
in the presence of enzyme andB
in the absence of enzyme
(c) Endothermic reaction with energy
Ain the absence of enzyme andB
in the presence of enzyme
(d) Exothermic reaction with energyA
in the absence of enzyme andBin
the presence of enzyme
Ans.(b)
The graph shows the exothermic
reaction A in the presence of enzyme as
it lowers down the activation energy
substantially.
The B graph shows this reaction in the
absence of enzyme when activation
energy is quite high. Thus, option (b) is
correct.
74A non-proteinaceous enzyme is
[NEET 2016, Phase II]
(a) Iysozyme (b) ribozyme
(c) ligase
(d) deoxyribonuclease
Ans.(b)
Ribozyme is a form of ribosomal RNA
(23 SrRNA) which acts as a catalyst in
splicing of RNA during protein
synthesis. It is the only non-protein
enzyme known so far, rest all the
enzymes are proteinaceous. Hence,
option (b) is correct.
75Select the option which is not
correct with respect to enzyme
action. [CBSE AIPMT 2014]
(a) Substrate binds with enzyme as its
active site
(b) Addition of lot of succinate does
not reverse the inhibition of
succinic dehydrogenase by
malonate
Biomolecules 103
Substrate
A
B
Reaction
Product
Potential energy

114. (c) A non-competitive inhibitor binds
the enzyme at a site distinct from
that which binds the substrate
(d) Malonate is a competitive inhibitor
of succinic dehydrogenase
Ans.(b)
Option (b) is incorrect with respect to
enzyme action because addition of a lot
of succinate reverse the inhibition of
succinic dehydrogenase by malonate.
Inhibition of succinic dehydrogenase by
malonate is an example of competitive
inhibition.
Competitive inhibition occurs when
enzyme and inhibitor both have more or
less similar structure and are present in
higher concentration.
Thus, both enzyme and inhibitor
compete for the active site of enzyme
resulting in the decrease of the
enzymatic activity.
76Transition state structure of the
substrate formed during an
enzymatic reaction is
(a) transient but stable[NEET 2013]
(b) permanent but unstable
(c) transient and unstable
(d) permanent and stable
Ans.(c)
The substrate binds to the enzyme at
its active site forming an enzyme
substrate complex. This complex
formation is a transient and unstable
phenomenon. Very soon, the product
is released from the active site. It is
the fact that all other intermediate
structural states are unstable.
Stability is related to energy status of
the molecule or the structure.
77The essential chemical
components of many coenzymes
are [NEET 2013]
(a) proteins
(b) nucleic acids
(c) carbohydrates
(d) vitamins
Ans.(d)
Essential chemical components of
many coenzymes are vitamins, e.g.
coenzyme Nicotinamide Adenine
Dinucleotide (NAD) and NADP contain
the vitamin niacin. Proteins, nucleic
acids and carbohydrates are not
enzymatic biomolecules.
78The curve given below shows
enzymatic activity with relation to
three conditions (pH, temperature
and substrate concentration).
[CBSE AIPMT 2011]
What do the two axis (XandY)
represent?
X-axis Y-axis
(a) Temperature — Enzyme activity
(b) Substrate
concentration
— Enzymatic
activity
(c)Enzymatic
activity
— Temperature
(d)Enzymatic
activity
— pH
Ans.(a)
X-axis represents temperature and
Y-axis represent enzyme activity. All
enzyme act at an optimum
temperature, above and below this
temperature, the enzyme activity
declines.
79Modern detergents contain
enzyme preparation of
[CBSE AIPMT 2008]
(a) acidophiles
(b) alkaliphiles
(c) thermoacidophiles
(d) thermophiles
Ans.(b)
Modern detergents contains enzyme
preparation of alkaliphiles. Detergents
represent the largest industrial
application of enzymes amounting to
25–30% of total sales of enzyme.
The enzymes used in detergents must
be cost effective, safe to use and be
able to perform the task in the
presence of anionic and non-ionic
detergents, soaps, oxidants, etc at pH
between 8–10.5%.
The chief enzymes used are
proteases,α-amylase and
sometimes cellulase.
80A competitive inhibitor of succinic
dehydrogenase is
[CBSE AIPMT 2008]
(a) malonate
(b) oxaloacetate
(c)α-ketoglutarate
(d) malate
Ans.(a)
Succinic dehydrogenase oxidised the
succinate to fumarate. Malonate, an
analogue of succinate, which is a strong
competitive inhibitor of succinate
dehydrogenase and therefore, blocks
the activity of citric acid cycle in
eukaryotes.
In Krebs’ cycle the reversible hydration
of fumarate to malate is catalysed by
fumarase enzyme.
81An organic substance bound to an
enzyme and essential for its
activity is called
[CBSE AIPMT 2006]
(a) holoenzyme (b) apoenzyme
(c) isoenzyme (d) coenzyme
Ans.(d)
Coenzymeis an organic non-protein
molecule that associates with an
enzyme molecule in catalysing
biochemical reactions. It usually
participates in the substrate enzyme
interaction by donating or accepting
certain chemical groups.
Holoenzymeis a complex comprising of
enzyme molecule and its cofactor. The
enzyme is catalytically active in this
state.
Apoenzymeis an inactive enzyme that
must associate with a specific cofactor
molecule in order to function.
Isoenzymeorisozymeis one of the
several forms of an enzyme that catalyse
the same reaction but differ from each
other in such properties as substrate
affinity and maximum rates of enzyme
substrate reaction.
82An enzyme that can stimulate
germination of barley seeds is
[CBSE AIPMT 2006]
(a)α-amylase (b) lipase
(c) protease (d) invertase
Ans.(a)
Barley seeds are rich in carbohydrate
(starch). The starch is hydrolysed by
α-amylase to monosaccharides unit at
the time of germination of seeds.
104 NEETChapterwise Topicwise Biology
Y-
axis
X-axis

115. 83Which one of the following
hydrolyses internal
phosphodiester bonds in a
polynucleotide chain?
[CBSE AIPMT 2005]
(a) Lipase (b) Protease
(c) Endonuclease (d) Exonuclease
Ans.(c)
DNase (deoxyribonuclease) or simply
nuclease is an enzyme which breaks
down DNA by hydrolysis of the
phosphodiester bonds of its
sugar-phosphate back bone.
Depending on the position of
hydrolysing phosphodiester bonds,
nucleases are of two types :
(i) Endonucleases
(ii) Exonucleases
Endonucleases hydrolyse internal
phosphodiester bonds in a
polynucleotide chain (i.e. DNA). While
exonucleases hydrolyse terminal
phosphodiester bonds in a
polynucleotide chain (i.e. DNA).
84Which of the following statements
regarding enzyme inhibition is
correct? [CBSE AIPMT 2005]
(a) Competitive inhibition is seen
when a substrate competes with
an enzyme for binding to an
inhibitor protein
(b) Competitive inhibition is seen
when the substrate and the
inhibitor compete for the active
site on the enzyme
(c) Non-competitive inhibition of an
enzyme can be overcome by
adding large amount of substrate
(d) Non-competitive inhibitors often
bind to the enzyme irreversibly
Ans.(b)
A competitive inhibitor competes with
substrate molecule for occupying the
active site of an enzyme. These
inhibitors have structural resemblance
with substrate molecules due to which
they easily bind with active site of an
enzyme and form an enzyme-inhibitor
complex.
E + I EI complex
(enzyme)(inhibitor)
→
85Enzymes, vitamins and hormones
can be classified into a single
category of biological chemicals,
because all of these
[CBSE AIPMT 2005]
(a) help in regulating metabolism
(b) are exclusively synthesised in the
body of a living organism as at
present
(c) are conjugated proteins
(d) enhance oxidative metabolism
Ans.(a)
Enzymes, vitamins and hormones are
classified into a single category of
biological chemical because all these
help in regulation of metabolism.
Enzyme is a proteinaceous catalyst
produced by a cell and responsible for
the high rate and specificity of one or
more intercellular or intracellular
biochemical reactions. Vitamin is an
organic substance which generally
synthesised by plants (exception
vitamin-D). Absence of a vitamin from
the diet for sufficient time gives
symptoms of a resulting deficiency
disease. Hormones are chemical
messengers which on secretion bring
about a specific and adaptive
physiological response.
86The catalytic efficiency of two
different enzymes can be
compared by the
[CBSE AIPMT 2005]
(a) formation of the product
(b) pH optimum value
(c)K
m
value
(d) molecular size of the enzyme
Ans.(c)
K
m
,Michaelis-Menten constantis
defined as substrate concentration at
which reaction velocity of enzyme
catalysed reaction( )V
0
is half of the
maximum velocity of this reaction( )
max
V
, (i.e.K
m
= ½V
max
).K
m
can vary greatly
from enzyme to enzyme and even for
the different substrates of the same
enzyme.
87The major role of minor elements
inside living organisms is to act as
[CBSE AIPMT 2003]
(a) binder of cell structure
(b) cofactors of enzymes
(c) building blocks of important amino
acids
(d) constituent of hormones
Ans.(b)
Though trace elements are required for
various uses, most of these have a
significant role in enzyme activities (e.g.
zinc activates carboxylases, carbonic
anhydrase and various dehydrogenases).
88Hydrolytic enzymes which act at
low pH are called as
[CBSE AIPMT 2002]
(a) proteases (b)α-amylases
(c) hydrolases (d) peroxidases
Ans.(c)
Lysosomes are the reservoirs of acid
hydrolases showing optimum activity at
pH 5.0 maintained within the lysosome.
These include proteases, nucleases,
glycosidases, lipase, etc. Among these
protease act a very low pH, i.e. 2.
89Cytochrome is
[CBSE AIPMT 2001]
(a) metallo flavoprotein
(b) Fe containing porphyrin pigment
(c) glycoprotein
(d) lipid
Ans.(b)
Cytochrome is the respiratory pigment.
It is composed of protein, iron and
porphyrin ring. It functions as an
enzyme in the respiratory chain. Unlike
haemoglobin the metal atom in the
porphyrin ring must change it’s valency
for the molecule to function.
Cytochromes are basically located in
inner mitochondrial membranes and
thylakoids of chloroplasts.
90Enzymes enhance the rate of
reaction by[CBSE AIPMT 2000]
(a) forming a reactant-product
complex
(b) changing the equilibrium point of
the reaction
(c) combining with the product as soon
as it is formed
(d) lowering the activation energy of
the reaction
Ans.(d)
Enzymes enhance the rate of a reaction
by just lowering the activation energy
(the energy required for substances to
react and get converted into product) of
a reaction.
Biomolecules 105
Activation energy
without
catalyst
Activation
energy
with catalyst
Reactants
Products
Low
Energy
content
of
molecules
Progress of reaction
High

116. 106 NEETChapterwise Topicwise Biology
91Feedback inhibition of an
enzymatic reaction is caused by
(a) end product[CBSE AIPMT 2000]
(b) substrate
(c) enzyme
(d) rise in temperature
Ans.(a)
In feedback inhibition, the product of
an enzyme-catalysed reaction
accumulates and acts as inhibitor of
the reaction.
92Cofactor (coenzyme) is a part of
holoenzyme it is[CBSE AIPMT 1997]
(a) loosely attached inorganic part
(b) accessory non-protein substance
attached firmly
(c) loosely attached organic part
(d) None of the above
Ans.(c)
Coenzyme/cofactors are organic
substances (often vitamins) which are
loosely attached with apoenzymes. A
holoenzyme is a conjugated enzyme.
It has a proteinaceous part called
apoenzymeand a non-proteinaceous
part calledcofactor.
Holoenzyme Apoenzyme
(conjugated enzyme) (prote
a
in part)
+ Cofactor
(non-protein part)
Cofactor is very necessary for the
activity of holoenzyme.
Cofactors can be separated from
enzyme by dialysis. Cofactors may be
inorganic (i.e. metal activators) or
organic (i.e. coenzymes and prosthetic
group).
93An enzyme brings about
[CBSE AIPMT 1993]
(a) decrease in reaction time
(b) increase in reaction time
(c) increase in activation energy
(d) reduction in activation energy
Ans.(d)
Enzymes act by reducing the amount of
activation energy. The binding energy is
the source of energy used by enzyme to
lower the activation energy of reaction.
Activation energy is the minimum
energy required from outside to
overcome the energy barrier of
reactants. Enzymes lower energy of
activation by two ways-bringing
reactants molecules together, and
developing strain in bonds of reactants.
94Enzymes having slightly different
molecular structure but
performing identical activity are
[CBSE AIPMT 1991]
(a) homoenzymes (b) isoenzymes
(c) apoenzymes (d) coenzymes
Ans.(b)
Isoenzymes (analogous enzymes) are
multiple molecular forms of same
enzyme with same substrate activity
and found in the same organism. These
enzymes are formed by different genes,
have differentK
m
constants for their
substrates and operate at different pH.
About 100 enzymes are known to have
isoenzymes, e.g. LDH
(Lacto-Dehydrogenase acting on
pyruvate to form lactate) has 5
isoenzymes in man,α-amylase in wheat
has 16 isoenzymes.
95Which of the following is not a part
of enzyme but it activates the
enzyme? [CBSE AIPMT 1989]
(a) K (b) C
(c) N (d) Si
Ans.(a)
Potassium(K )
+
is an essential
element, loosely held to the
apoenzyme part of the enzyme.K
+
is
an inorganic cofactor (metal
activator) of enzyme pyruvate kinase.
Glucose + ATP
Hexo-
kinase
(enzyme)
Glucose-6-phosphate
(end product)
+
ADP
Inhibits

117. 01The centriole undergoes
duplication during[NEET 2021]
(a) S-phase (b) prophase
(c) metaphase (d) G
2
-phase
Ans.(a)
During S phase or synthesis phase of
interphase replication of DNA and
synthesis of histone protein,
centromere and centrioles occur.
During the S phase, DNA replication
begins in the nucleus, and the
centriole duplicates in the cytoplasm
of the cell.
02Match the List-I with List-II.
[NEET 2021]
List-I List-II
A. S-phase 1. Proteins are
synthesised
B. G
2
-phase 2. Inactive phase
C. Quiescent
stage
3. Interval between
mitosis and initiation
of DNA replication
D. G
1
-phase 4. DNA replication
Choose the correct answer from
the options given below.
A B C D
(a) 3 2 1 4
(b) 4 2 3 1
(c) 4 1 2 3
(d) 2 4 3 1
Ans.(c)
(A)-(4), (B)-(1), (C)-(2), (D)-(3)
During DNA replication, the unwinding
of strands leaves a single strand
vulnerable. In the eukaryotic cell cycle,
chromosome duplication occurs during
‘S phase’ (the phase of DNA synthesis)
and chromosome segregation occurs
during ‘M phase’ (the mitosis phase).
During the G
2
phase, extra protein is
often synthesised, and the organelles
multiply until there are enough for two
cells. Other cell materials such as lipids
for the membrane may also be
produced.
The cell is in aquiescent(inactive)stage
that occurs when cells exit the cell
cycle. Some cells enter G
0
temporarily
until an external signal triggers the
onset of G
1
. Other cells that never or
rarely divide, such as mature cardiac
muscle and nerve cells, remain in G
0
permanently.
G
1
phase corresponds to the interval
between mitosis and initiation of DNA
replication. During G
1
phase the cell is
metabolically active and continuously
grows but does not replicate its DNA.
03Attachment of spindle fibres to
kinetochores of chromosomes
becomes evident in
[NEET (Oct.) 2020]
(a) anaphase (b) telophase
(c) prophase (d) metaphase
Ans.(d)
During the metaphase stage of cell
cycle, spindle fibres originating from
the centrosomes attaches to the
kinetochore of chromosomes.
Kinetochore is a disc-shaped structure
at the surface of centromere through
which the sister chromatids are held
together. During metaphase, the
chromosomes arrange themselves at
the equator on metaphasic plate. Due
to this arrangement, the attachment of
spindle fibres to kinetochore is clearly
visible.
04Identify the correct statement
with regard toG
1
-phase (Gap 1) of
interphase.[NEET (Sep.) 2020]
(a) Reorganisation of all cell
components, takes place.
(b) Cell is metabolically active, grows
but does not replicate its DNA
(c) Nuclear division takes place
(d) DNA synthesis or replication takes
place
Ans.(b)
The statement in option (b) is correct
with regard to G
1
- phase of interphase
because during G
1
-phase the cell is
metabolically active and continuously
grows but does not replicate its DNA.
DNA synthesis takes place in S-phase.
Nuclear division occurs during
karyokinesis.
Reorganisation of all cell components
takes place in M-phase.
05Some dividing cells exist the cell
cycle and enter vegetative inactive
stage. This is called quiescent
stage(G )
0
. This process occurs at
the end of [NEET (Sep.) 2020]
(a)G
1
-phase (b) S-phase
(c)G
2
-phase (d) M-phase
Ans.(d)
Some dividing cells exit the cell cycle
and enter vegetative inactive stage,
called quiescent stage (G
0
). This
process occurs at the end of M-phase
CellCycleand
CellDivision
10
The Cell Cycle
TOPIC 1

118. and beginning of G
1
-phase. Cells enter
G
0
for varying amounts of time, and
some cells enter the G
0
-phase and stay
there forever. This is because once
they reach maturity, like nerve and
heart cells they do not divide again, so
they stay in the G
0
-phase.
06Cells inG
0
phase
[NEET (National) 2019]
(a) enter the cell cycle
(b) suspend the cell cycle
(c) terminate the cell cycle
(d) exit the cell cycle
Ans.(d)
G
0
phase is the stage in which the cells
exit the cell cycle. It is the resting or
quiescent phase in which the cells do
not divide. It is the permanent state for
some cells, e.g., neurons.
07The correct sequence of phases
of cell cycle is
[NEET (National) 2019]
(a)G G S M
1 2
→ → →
(b)S G G M
1 2
→ → →
(c)G S G M
1 2
→ → →
(d)M G G S
1 2
→ → →
Ans.(c)
The correct sequence of phases of cell
cycle is
G S G M
1 2
→ → →
HereG
1
andG
2
represent first and
second growth phase, respectively.
S-phase represents synthesis phase
during which DNA replicates. M-phase
is mitotic phase during which cell
begins to divide.
08When cell has stalled DNA
replication fork, which checkpoint
should be predominantly
activated?[NEET 2016, Phase II]
(a)G S
1
/
(b)G M
2
/
(c) M
(d) BothG M
2
/and M
Ans.(a)
Stalled forks activate checkpoint
signaling and pause replication. Since,
G
1
/S checkpoint checks DNA damage,
cells size prior to S-phase (i.e. DNA
replication phase), this checkpoint
would be activated by stalled DNA
replication fork.
09During cell growth, DNA synthesis
takes place in[NEET 2016, Phase II]
(a) S-phase (b)G
1
-phase
(c)G
2
-phase (d) M-phase
Ans.(a)
In the cycle of cell division,interphase is
the longest phase consisting ofG , S,
1
G
2
-phases. In this phase cell prepares
itself for cell division. In S or synthetic
phase DNA duplication (synthesis) takes
place.
10During which phase(s) of cell cycle,
amount of DNA in a cell remains at
4C level if the initial amount is
denoted as 2C?[CBSE AIPMT 2014]
(a)G and G
0 1
(b)G
1
and S
(c) OnlyG
2
(d)G
2
and M
Ans.(d)
During the S or synthetic phase, the
DNA content doubles, i.e., from 2C to
4C for all diploid cells. TheG
2
phase
follows the S-phase and is called
second growth phase or pre mitotic gap
phase. InG
2
phase the synthesis of DNA
stops therefore, the DNA level remains
4C if initial was 2C.
However, the formation of RNA and
protein continue as they are required
for the multiplication of cell
organelles, spindle formation and cell
growth. This amount becomes half
(i.e.) 2C only during anaphase
(in mitosis) when chromosomes
separate.
11In S-phase of the cell cycle
[CBSE AIPMT 2014, 2000, 1996]
(a) amount of DNA doubles in each
cell
(b) amount of DNA remains same in
each cell
(c) chromosome number is increased
(d) amount of DNA is reduced to half
in each cell
Ans.(a)
S-phase is the synthesis phase in which
the cell synthesise a replica of its
genome, i.e. DNA replication occurs by
DNA polymerase.
DNA replication along with the
synthesis of histone proteins results in
the duplication of chromosomal
material, i.e., amount of DNA doubles
in each cell.
Amount of DNA remains unchanged
duringG
1
-phase or post mitotic gap
and/orG
2
-phase or pre mitotic phase.
12Given below is a schematic
break-up of the phases/stages of
cell cycle
Which one of the following is the
correct indication of the
stage/phase in the cell cycle?
[CBSE AIPMT 2009]
(a) B-metaphase
(b) C-karyokinesis
(c) D-synthetic phase
(d) A-cytokinesis
Ans.(c)
Cell cycle completes in two steps, i.e.
interphase and M-phase. Interphase is
completed in three successive stages
G
1
-phase (post mitotic phase), S-phase
(synthetic phase) andG
2
-phase
(premitotic or post synthetic phase). In
the given figure, D is representing the
S-phase (synthetic phase) of cell cycle.
13At what stage of the cell cycle are
histone proteins synthesised in a
eukaryotic cell?
[CBSE AIPMT 2005]
(a) DuringG
2
-stage of prophase
(b) During S-phase
(c) During entire prophase
(d) During telophase
Ans.(b)
During S-phase of cell cycle synthesis
of histone proteins takes place because
at this stage the amount of DNA per cell
get double to that of somatic number.
Histone proteins are basic proteins and
are used in packing of eukaryotic
(absent in prokaryotes) DNA. DNA and
histones together comprise chromatin,
forming bulk of the eukaryotic
chromosomes. Histones are of five
major kindsH1, H2A, H2B, H3andH4. H1
histones link neighbouring
nucleosomes.
14In the somatic cell cycle
[CBSE AIPMT 2004]
(a) inG
1
-phase DNA content is double
the amount of DNA present in the
original cell
108 NEETChapterwise Topicwise Biology
Mitosis
Interphase
A
BC
D

119. (b) DNA replication takes place in
S-phase
(c) a short interphase is followed by a
long mitotic phase
(d)G
2
-phase follows mitotic phase
Ans.(b)
DNA replication occurs during S-phase
of the mitotic cycle where it gets
doubled as compared to that in the
original cell.
15Which of the following occurs
more than one and less than five
in a chromosome?
[CBSE AIPMT 2002]
(a) Chromatid
(b) Chromosome
(c) Centromere
(d) Telomere
Ans.(d)
A chromosome has one centromere,
may have many chromomeres, two
chromatids; but four telomeres (two
each at the opposite ends of each
chromatid).
16During cell division in apical
meristem the nuclear membrane
appears in [CBSE AIPMT 1997]
(a) metaphase (b) anaphase
(c) telophase (d) cytokinesis
Ans.(c)
During telophase, nuclear envelope
initially reforms around each
chromosome individually which later on
fuse to form complete nuclear
envelope.
MetaphaseChromosomes are
arranged on equatorial plate.
AnaphaseChromosomes split
longitudinally. Chromatids migrate
towards opposite poles.
CytokinesisDivision of cytoplasm.
17The fruit fly has 8 chromosomes
(2n) in each cell. During interphase
of mitosis, if the number of
chromosomes at G
1
-phase is 8,
what would be the number of
chromosomes after S-phase ?
(a) 8 (b) 16 [NEET 2021]
(c) 4 (d) 32
Ans.(a)
During S phase or synthetic phase of
interphase, replication of DNA and
synthesis of histone protein,
centromere and centrioles occur, but
the number of chromosomes remains
same from beginning till the end of S
phase.
Hence, number of chromosome will
remain 8 after the S phase in fruitfly.
18In a mitotic cycle, the correct
sequence of phases is
[NEET (Oct.) 2020]
(a)S,G ,G ,M
1 2
(b)G ,S,G ,M
1 2
(c)M, G , G , S
1 2
(d)G , G , S, M
1 2
Ans.(b)
In a mitotic cycle, the correct sequence
of phases isG
1
, S,G
2
, M. The first three
phases, i.e.G
1
, S,G
2
occurring during
interphase whereas the M-phase is the
period of actual cell division. The major
event occurring in each phase is
tabulated below
Phases Activities
G
1
-phase Cell becomes metabolically
active, enzymes and proteins
required for replication are
synthesised.
S-phase Synthesis or replication of DNA
occurs so that amount of DNA
per cell gets doubled.
G
2
-phase Proteins required for mitosis are
synthesised while the growth
of cell continues.
M-phase Cell divides to form daughter
cells.
19Match the following (Columns)
events that occur in their
respective phases of cell cycle
and select the correct option from
the codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. G
1
-phase 1. Cell grows and
organelle
duplication
B. S phase 2. DNA replication
and chromosome
duplication
C. G
2
-phase 3. Cytoplasmic
growth
D. Metaphase
in M-phase
4. Alignment of
chromosomes
Codes
A B C D
(a) 2 3 4 1
(b) 3 4 1 2
(c) 4 1 2 3
(d) 1 2 3 4
Ans.(d)
20After karyogamy followed by
meiosis, spores are produced
exogenously in [NEET 2018]
(a)Agaricus (b)Alternaria
(c)Neurospora(d)Saccharomyces
Ans.(a)
AgaricusMeiospores are produced
exogenously after karyogamy and
meiosis. It belongs to Basidiomycetes.
Alternariabelongs to the
Deuteromycetes class of fungi. The
fungi of this class lack sexual
reproduction. Therefore, sexual spores
are not formed.
NeurosporaandSaccharomycesbelong
to Ascomycetes class of fungi. They
produce ascopores as meiospores.
Their ascospores are produced
endogenously.
21Anaphase Promoting Complex
(APC) is a protein degradation
machinery necessary for proper
mitosis of animals cells. If APC is
defective in a human cells, which
of the following is expected to
occur? [NEET 2017]
(a) Chromosomes will not condense
(b) Chromosomes will be fragmented
(c) Chromosomes will not segregate
(d) Recombination of chromosome
arms will occur
Ans.(c)
If anaphase promoting complex is
defective in a human cell, the
chromosomes will not segregate during
anaphase of mitosis. APC triggers the
Cell Cycle and Cell Division 109
Mitosis
TOPIC 2
Ascospores
Ascus
Basidiospores
Sterigma
Basidium
Ascospores
Basidiospores

120. transition from metaphase to anaphase
by tagging specific proteins for
degradation.
Concept EnhancerAnaphase stage of
mitosis is characterised by two events
(a) Splitting of centromeres and
segregation of chromosomes.
(b) Movement of chromatids towards
the opposite poles.
22Which of the following options
gives the correct sequences of
events during mitosis ?[NEET 2017]
(a) Condensation→nuclear
membrane disassembly→
crossing over→segregation→
telophase
(b) Condensation→nuclear
membrane disassembly→
arrangement at equator→
centromere division→
segregation→telophase
(c) Condensation→crossing over→
nuclear membrane disassembly→
segregation→telophase
(d) Condensation→arrangement at
equator→centromere division→
segregation→telophase
Ans.(b)
During mitosis following events occurs
as follows
Condensation of chromosomal
material, which takes place at an early
prophase stage. During late prophase
nuclear membrane disintegrates.
Then chromosomes get arranged at
equator in the metaphase stage. After
that splitting ofcentromereand
segregation of chromosomesoccur
in the anaphase stage. In telophase
stage chromosomes move to opposite
poles of the cell. It is last stage of
mitosis.
23Which of the following is not a
characteristic feature during
mitosis in somatic cells?
[NEET 2016, Phase I]
(a) Disappearance of nucleolus
(b) Chromosome movement
(c) Synapsis
(d) Spindle fibres
Ans.(c)
Synapsis is pairing of homologous
chromosomes. It occurs during
zygotene stage of meiosis. The
homologous chromosomes come
closer leading to cross over in the
next stage called pachytene. These
are not observed during mitosis.
24Spindle fibres attach on to
[NEET 2016, Phase I]
(a) kinetochore of the chromosome
(b) centromere of the chromosome
(c) kinetosome of the chromosome
(d) telomere of the chromosome
Ans.(a)
Spindle fibres attach to kinetochores of
chromosomes during cell division. They
help the chromosomes/chromatids to
get separated to the two daughter cells,
towards opposite poles.
25The complex formed by a pair of
synapsed homologous
chromosomes is called[NEET 2013]
(a) equatorial plate (b) kinetochore
(c) bivalent (d) axoneme
Ans.(c)
The process of pairing of homologous
chromosomes is called synapsis. Each
pair of synapsed homologous
chromosome calledbivalent.
26A stage in cell division is shown in
the figure. Select the answer
which gives correct identification
of the stage with its characteristic
mentioned. [NEET 2013]
(a) Telophase — Nuclear envelope
reforms, Golgi
complex reforms
(b) Late anaphase— Chromosomes move
away from equatorial
plate, Golgi complex
not present
(c) Cytokinesis — Cell plate formed,
mitochondria
distributed between
two daughter cells
(d) Telophase — Endoplasmic
reticulum and
nucleolus not
reformed yet
Ans.(a)
Telophaseis reverse ofprophase. The
chromosome that have reached their
respective poles decondense, i.e.,
nuclear envelop reforms, Golgi complex
reforms, etc. In late anaphase
centromeres split and chromatid
separate and chromatid move to
opposite poles. Cytokinesis is process
in which cell itself is divided two
daughter cells.
27Select the correct option with
respect to mitosis.
[CBSE AIPMT 2011]
(a) Chromatids start moving towards
opposite poles in telophase
(b) Golgi complex and endoplasmic
reticulum are still visible at the
end of prophase
(c) Chromosomes move to the
spindle equator and get aligned
along equatorial plate in
metaphase
(d) Chromatids separate but remains
in the centre of the cell in
anaphase
Ans.(c)
Inmetaphaseof mitosis, spindle fibres
attach to kinetochore of chromosomes.
Chromosomes are moved to spindle
equator and get aligned along
metaphasic plate through spindle fibres
to both pole.
28During mitosis ER and nucleolus
begin to disappear at
[CBSE AIPMT 2010]
(a) late prophase
(b) early metaphase
(c) late metaphase
(d) early prophase
Ans.(d)
In mitosis,prophaseis the longest
phase of karyokinesis. In early
prophase, nuclear membranes,
nucleolus start disintegrating. Cells
cytoskeleton, Golgi complex, ER, etc.
disappear.
29Which stages of cell division do
the following figures A and B
represent respectively?
[CBSE AIPMT 2010]
110 NEETChapterwise Topicwise Biology
A
B

121. (a) Metaphase – Telophase
(b) Telophase – Metaphase
(c) Late anaphase – Prophase
(d) Prophase – Anaphase
Ans.(c)
In the given figures
A. Late anaphase is characterised by
following events
(i) Centromeres split and chromatids
separate.
(ii) Chromatids move to opposite poles.
B. Prophase is characterised by
centriole separation.
30Centromere is required for
[CBSE AIPMT 2005]
(a) movement of chromosomes
towards poles
(b) cytoplasmic cleavage
(c) crossing over
(d) transcription
Ans.(a)
The arms of chromosome are known as
chromatids. These arms are held
together at a point called the
centromere (or primary constriction).
Centromere occurs any where along the
length of chromosome. During cell
division spindle fibres are attached to
centromere and help in the movement
of chromosomes towards the poles.
31If you are provided with root tips
of onion in your class and are
asked to count the chromosomes
which of the following stages can
you most conveniently look into?
[CBSE AIPMT 2004]
(a) Metaphase
(b) Telophase
(c) Anaphase
(d) Prophase
Ans.(a)
At metaphase, the chromosomes are
clearly visible as composed of two
closely associated halves
(chromatids) and the chromosomes
have undergone maximum
contraction so, these can be counted
conveniently. In metaphase
chromosome align in the equator of
the cell before being separated into
each of the two daughter cells.
32Which one of the following
precedes reformation of the
nuclear envelope during M-phase
of the cell cycle?
[CBSE AIPMT 2004]
(a) Decondensation from
chromosomes and reassembly of
the nuclear lamina
(b) Transcription from chromosomes
and reassembly of the nuclear
lamina
(c) Formation of the contractile ring
and formation of the
phragmoplast
(d) Formation of the contractile ring
and transcription from
chromosomes
Ans.(a)
Attelophasestage, nuclear membrane
vesicles associate with the surface of
individual chromosomes and fuse to
reform the nuclear membranes, which
partially enclose clusters of
chromosomes before coalescing to
reform the complete nuclear envelope.
During this process the nuclear pores
reassemble and the dephosphorylated
reassociate to form the nuclear lamina.
One of the lamina proteins (lamina-B)
remains with the nuclear membrane
fragments throughout mitosis and may
help nuclear reassembly. After the
nucleus reforms, the chromosome
decondense and RNA synthesis
resumes, causing the nucleolus to
reappear.
33Mitotic spindle is mainly
composed of which protein?
[CBSE AIPMT 2002]
(a) Actin (b) Myosin
(c) Actomyosin (d) Myoglobin
Ans.(a)
Spindles formed during mitosis and
meiosis are nothing but microtubule
complex. Microtubules are made up
of small units of tubulin which has
amino acid composition similar to
actin.
34Best material for the study of
mitosis in laboratory is
[CBSE AIPMT 2002]
(a) anther (b) root tip
(c) leaf tip (d) ovary
Ans.(b)
Root tips have active meristematic
zone where cells divide mitotically
leading to increase in the length of the
roots. This is the best site for the study
of mitosis, e.g. onion root tips.
35During cell division, the spindle
fibres attach to the chromosome
at a region called
[CBSE AIPMT 2000]
(a) Chromocentre (b) kinetochore
(c) centriole (d) chromomere
Ans.(b)
Duringlate prophase, specialised
structures called kinetochores develop
on either surface of the centromere.
Chromosomal fibres get attached to
kinetochore.
Chromomeresare beaded structures
on the chromosomes which are found
particularly at the prophase-I
(particularly at leptotene) of meiosis-I.
Chromocentreis developed due to the
fusion of centromeric regions of all the
chromosomes of a cell.
Centrosomeis found in animal cells
(absent in higher plant cells). It is found
near the nucleus. Each centrosome is
made up of two centrioles and each
centriole is a cylindrical structure with a
diameter of 1500-1800 Å and is made up
of nine fibrils.
36How many mitotic divisions are
needed for a single cell to make
128 cells? [CBSE AIPMT 1997]
(a) 7 (b) 14
(c) 28 (d) 64
Ans.(a)
A single mitotic division results in the
production of two cells from single
cell.
37Which one of the following
structures will not be common to
mitotic cells of higher plants?
[CBSE AIPMT 1997]
(a) Cell plate (b) Centriole
(c) Centromere (d) Spindle fibres
Ans.(b)
Centrosome is found in animals,
Euglena, Nitella,some fungi and
members of dinoflagellate. It is found
near the nucleus.
Centriole is not common to mitotic cell
of higher plants. Main function of
centrosome is at the time of cell
division when the two centrioles
separate and move on two poles. Aster
and spindle are formed from it which
Cell Cycle and Cell Division 111
O
D
1
O
D
2D
3 D
4 D
5 D
6 D
7 D
8
O O O O O O O
1 2 4 8 16 32 64 128

122. help in the movement of chromatids.
They form basal body, cilia, flagella, etc.
Centriole is rich in tubulin and ATPase.
Centrioles replicate inG
2
-phase of
interphase of cell cycle but do not
initiate cell division.
38The point, at which polytene
chromosomes appear to be
attached together, is called
[CBSE AIPMT 1995]
(a) centriole (b) centromere
(c) chromomere (d) chromocentre
Ans.(d)
Polytene chromosomes are infact
formed by pairing of two somatic
homologous chromosomes which
undergo repeated endomitosis, forming
a number of strands. These strands
remain attached to a common large
chromocentre of all polytene
chromosomes and are rich in
heterochromatin.
39Best stage to observe shape, size
and number of chromosomes is
[CBSE AIPMT 1994]
(a) interphase (b) metaphase
(c) prophase (d) telophase
Ans.(b)
Metaphase can be characterised by
the chromosomes that are least
coiled which show maximum
condensation and are shortest in
length. It is the best stage to study the
structure, size and number of
chromosome in a cell.
Idiogram/karyotype of chromosomes
is prepared at metaphase.
40In salivary gland
chromosomes/polytene
chromosomes pairing is
[CBSE AIPMT 1993]
(a) absent
(b) occasional
(c) formed between non-homologous
chromosomes
(d) formed between homologous
chromosomes
Ans.(d)
Polytene chromosomes/salivary gland
chromosomes was reported by Balbiani
(1881) from cells of salivary glands of
Chironomuslarva (insect of Diptera
group). The polytene chromosomes
become giant due to the
endoduplication, i.e., repeated
replication of chromatids without their
separation and cytokinesis. In fact, each
polytene chromosome is formed by
pairing of two somatic homologous
chromosomes which undergo repeated
endomitosis to form numerous strands
attached to a common large
chromocentre.
41Number of chromatids at
metaphase is[CBSE AIPMT 1992]
(a) two each in mitosis and meiosis
(b) two in mitosis and one in meiosis
(c) two in mitosis and four in meiosis
(d) one in mitosis and two in meiosis
Ans.(a)
In metaphase, chromosomes are thick,
shortest least coiled and minimum in
size. Each chromosome has its both
chromatids attached at centromere,
oriented at the equator of spindle
apparatus.
In meiotic metaphase, each
chromosome with two chromatids in a
bivalent is connected to the spindle
pole of its side by a kinetochore
microtubule instead of two as in
metaphase of mitosis.
42Mitotic anaphase differs from
metaphase in possessing
[CBSE AIPMT 1991]
(a) same number of chromosomes
and same number of chromatids
(b) half number of chromosomes and
half number of chromatids
(c) half number of chromosomes and
same number of chromatids
(d) same number of chromosomes
and half number of chromatids
Ans.(d)
Mitotic metaphase is the best stage to
observe the structure, size and number
of chromosomes in a cell. Centromeres
of all chromosomes lie closely at
equator and their arms in different
directions towards poles.
Chromosomes are shortest in
metaphase but thickest in anaphase.
In anaphase, centromere of each
chromosome divides so that each sister
chromatid now has its own centromere.
Thus, mitotic anaphase differs from
metaphase in possessing same number
of chromosomes and half number of
chromatids.
43A bivalent consists of
[CBSE AIPMT 1989]
(a) two chromatids and one
centromere
(b) two chromatids and two
centromeres
(c) four chromatids and two
centromeres
(d) four chromatids and four
centromeres
Ans.(c)
Each pair of homologous chromosome
carrying one maternal and one paternal
chromosome of similar type is called
bivalent. Each chromosome has two
sister chromatids and a centromere.
Thus, bivalents possesses four
chromatids, two centromeres. This
bivalent with four chromatids is called
pachytene tetrad (quadrivalent).
44Which stage of meiotic prophase
shows terminalisation of
chiasmata as its distinctive
feature?
[NEET 2021]
(a) Leptotene (b) Zygotene
(c) Diakinesis (d) Pachytene
Ans.(c)
Diakinesis is the final stage of meiotic
prophase 1. In this stage the two
homologous chromosomes do not
separate completely but remain
attached together at one or more
points as indicated by ‘X’ arrangement
known as chiasmata. The
displacement of chiasmata is termed
as terminalisation of chiasmata
which is completed in diakinesis
phase.
45Which of the following stages of
meiosis involves division of
centromere? [NEET 2021]
(a) Metaphase-I (b) Metaphase-lI
(c) Anaphase-II (d) Telophase-II
Ans.(c)
DuringanaphaseII, each pair of
chromosomes is separated into two
identical, independent chromosomes.
The chromosomes are separated by a
structure called the mitotic spindle
made up of many long proteins called
microtubules, which are attached to a
chromosome at one end and to the pole
of a cell at the other end. The sister
chromatids are separated
simultaneously at their centromeres. The
separated chromosomes are then
112 NEETChapterwise Topicwise Biology
Meiosis
TOPIC 3

123. pulled by the spindle to opposite poles
of the cell. Thus, the centromere splits,
freeing the sister chromatids from each
other. Other options can be explained
as:
Inmetaphase I, the homologous pair of
chromosomes align on either side of
the equatorial plate.
Duringmetaphase II, the centromeres
of the paired chromatids align along the
equatorial plate in both cells.
Duringtelophase II, the two groups of
chromosome once again get
enclosed by nuclear envelope.
46During meiosis 1, in which stage
synapsis takes place?
[NEET (Oct.) 2020]
(a) Pachytene (b) Zygotene
(c) Diplotene (d) Leptotene
Ans.(b)
During zygotene stage of meiosis-I,
chromosomes start pairing together
and this process of association is called
synapsis. Such paired chromosomes
are called homologous chromosomes.
47Dissolution of the synaptonemal
complex occurs during
[NEET (Sep.) 2020]
(a) zygotene (b) diplotene
(c) leptotene (d) pachytene
Ans.(b)
Dissolution of the synaptonemal
complex occurs during diplotene
stage of prophase-I of meiosis-I.
Prophase of meiosis-I is long and
complex. It is comprised of
leptotene, zygotene, pachytene,
diplotene and diakinesis. During
diplotene, at most places
synaptonemal complex dissolves.
48Match the following columns with
respect to meiosis.
[NEET (Sep.) 2020]
Column I Column II
A. Zygotene 1. Terminalisation
B. Pachytene 2. Chiasmata
C. Diplotene 3. Crossing over
D. Diakinesis 4. Synapsis
Select the correct option.
A B C D
(a) 4 3 2 1
(b) 1 2 4 3
(c) 2 4 3 1
(d) 3 4 1 2
Ans.(a)
The correct option is (a). It can be
explained as follows
During zygotene phase the homologous
chromosomes pair or come together
and forms synapsis.
Crossing over takes place during
pachytene stage and at each point of
crossing over a chiasma is formed
between non-sister chromatids of
homologous chromosomes.
Chiasmata is the point of contact
between the two non sister chromatids
of homologous chromosomes,
chiasmata becomes visible during
diplotene stage.
Terminalisation of chiasmata gets
completed during diakinesis phase
where chromosomes gets freely
distributed in the cytoplasm.
49After meiosis-I, the resultant
daughter cells have
[NEET (Odisha) 2019]
(a) same amount of DNA as in the
parent cell in
S-phase
(b) twice the amount of DNA in
comparison to haploid gamete
(c) same amount of DNA in
comparison to haploid gamete
(d) four times the amount of DNA in
comparison to haploid gamete
Ans.(b)
After meiosis-I, the resultant daughter
cells have twice the amount of DNA in
comparison to haploid gamete.
Meiosis-I causes segregation of
homologous pairs of chromosomes.
However, each chromosome is
double-stranded, having two sister
chromatids due to DNA replication
before meiosis began.
50In meiosis crossing over is
initiated at[NEET 2016, Phase I]
(a) leptotene (b) zygotene
(c) diplotene (d) pachytene
Ans.(d)
Leptotene — Condensation of
chromatin
Zygotene — Synapsis of homologous
chromosomes
Pachytene — Crossing over
Diplotene — Dissolution of
synaptonemal complex and appearance
of chiasmata
Diakinesis — Terminalisation of
chiasmata
51Match the stages of meiosis in
column I to their characteristic
features in column II and select
the correct option using the codes
given below[NEET 2016, Phase II]
Column I Column II
A. Pachytene 1. Pairing of
homologous
chromosomes
B. Metaphase-I 2. Terminalisation
of chiasmata
C. Diakinesis 3. Crossing-over
takes place
D. Zygotene 4. Chromosomes
align at
equatorial plate
Ans.(a)
Various phases of meiosis and their
characteristic features are
Pachytene — Crossing-over takes place
Metaphase-I — Chromosomes align at
equatorial plate
Diakinesis — Terminalisation of
chiasmata
Zygotene — Pairing of homologous
chromosomes
52Arrange the following events of
meiosis in correct sequences.
[CBSE AIPMT 2015]
I. Crossing over
II. Synapsis
III. Terminalisation of chiasmata
IV. Disapperance of nucleolus
(a) II, I, IV, III (b) II, I, III, IV
(c) I, II, III, IV (d) II, III, IV, I
Ans.(b)
The correct sequence of events of
meiosis are
Synapsis in zygotene→Crossing
over in pachytene→Terminalisation
of chaismata in diplotene→
Disapperance of nucleolus in diakinesis.
53The enzyme recombinase is
required at which stage of
meiosis? [CBSE AIPMT 2014]
(a) Pachytene (b) Zygotene
(c) Diplotene (d) Diakinesis
Ans.(a)
Crossing overis an enzymatic process
occurring during the pachytene stage
of prophase-I. The enzyme involved in
this process is called recombinase
which aids in the recombination of
genes between homologous
chromosomes.
Cell Cycle and Cell Division 113

124. During zygotene stage, homologous
chromosomes pair up by a process
called synapsis and form a complex
bivalent structure. Diplotene is marked
by the dissolution of synaptonemal
complex and chaisma formation. While
diakinesis is marked by terminalisation
of chiasmata (i.e. chiasmata shifts
towards periphery of chromosome).
54Meiosis takes place in[NEET 2013]
(a) meiocyte (b) conidia
(c) gemmule (d) megaspore
Ans.(a)
In diploid organisms, specialised cells
called meiocytes (gamete mother cell)
undergo meiosis. Conidia and
gemmules are asexual reproductive
structures found inPenicilliumand
sponge respectively. Megaspores are
female gametes in plants which
undergo sexual reproduction.
55During gamete formation, the
enzyme recombinase participates
during [CBSE AIPMT 2012]
(a) metaphase-I (b) anaphase-II
(c) prophase-I (d) prophase-II
Ans.(c)
The pachytene stage of prophase-I of
meiosis-I is characterised by the
appearance of recombination nodules,
the sites at whichcrossing overoccurs
between non-sister chromatids of the
homologous chromosomes. Crossing
over is the exchange of genetic material
between two homologous
chromosomes. It is also an enzyme
mediated process and the enzyme
involved is called recombinase.
56Given below is the representation
of a certain event at a particular
stage of a type of cell division.
Which is this stage?
[CBSE AIPMT 2012]
(a) Prophase-I during meiosis
(b) Prophase-II during meiosis
(c) Prophase of mitosis
(d) Both prophase and metaphase of
mitosis
Ans.(a)
During zygotene stage of prophase-I of
meiosis-I, bivalent chromosomes
clearly appear as tetrads. Pachytene
stage is characterised by the
appearance of recombination nodules,
the sites at which crossing over
(exchange of genetic material) occurs
between non-sister chromatids of the
homologous chromosomes.
57Synapsis occurs between
[CBSE AIPMT 2009]
(a) a male and a female gamete
(b)mRNA and ribosomes
(c) spindle fibres and centromere
(d) two homologous chromosomes
Ans.(d)
In zygotene of prophase-I, homologous
chromosomes pair up. This process is
called synapsis. One chromosome of
the pair is from the male parent and one
from the female parent.
58Crossing over that results in
genetic recombination in higher
organisms occur between
[CBSE AIPMT 2004]
(a) sister chromatids of bivalent
(b) non-sister chromatids of a
bivalent
(c) two daughter nuclei
(d) two different bivalents
Ans.(b)
The process of crossing over takes
place in pachytene stage of prophase-I
of meiosis-I. In this process some
genes of two non-sister chromatids of a
bivalent are exchanged.
The process of crossing over is
depicted
59The exchange of genetic material
between chromatids of paired
homologous chromosomes during
first meiotic division is called
[CBSE AIPMT 1996]
(a) transformation (b) chiasmata
(c) crossing over (d) synapsis
Ans.(c)
In pachytene stage of prophase-I of
meiosis, there is breakage and
reunion of chromatids, it results in
exchange of segments between non-
sister chromatids of a bivalent,
known as crossing over. It leads to
recombination of linked genes/alleles
and is a major source of continuous
type of genetic variations in sexually
reproducing organisms.
60Lampbrush chromosomes occur
during [CBSE AIPMT 1995]
(a) prophase of mitosis
(b) diplotene of meiosis
(c) metaphase of meiosis
(d) interphase
Ans.(b)
Lampbrush chromosomes were
reported by W Flemming (1882) and
described by Ruckert (1892) from nuclei
of yolk rich primary oocytes of newts
and frog (amphibians). These are also
found in spermatocytes of many
animals. These are found in permanent
diplotene stage of meiosis and do not
undergo cell cycle.
Each such chromosome has a double
main axis made up of DNA and histones.
The chromosomes are coiled and held
at many places forming cross like
structure called chiasmata. Loops
arising laterally has uncoiled DNA which
helps in rapid transcription and yolk
synthesis.
61Meiosis has evolutionary
significance because it results in
[CBSE AIPMT 1994]
(a) genetically similar daughters
(b) four daughter cells
(c) eggs and sperms
(d) recombinations
Ans.(d)
Recombination takes place in meiosis
but stillMeiosismaintains the
chromosome number constant. It
produces haploid gametes by reducing
the chromosome number to half.
Crossing overproduces new
combination of linked genes and is
major source of genetic variation.
114 NEETChapterwise Topicwise Biology
123
Sister chromatids of one
duplicate chromosome
Synapsis
Sister chromatids of chromosome’s homologue
Adjacent
non-sister
chromatids
Centromere
Chiasmata
Crossing
over
Recombinant
chromosomes
Crossing Over Process

125. Also, distribution of bivalents which is
at random in metaphase-I provides the
secondary source of genetic variation
in the organisms and is essential for
speciation and evolution.
62Meiosis-II performs
[CBSE AIPMT 1993]
(a) separation of sex chromosomes
(b) synthesis of DNA and centromeres
(c) separation of homologous
chromosomes
(d) separation of chromatids
Ans.(d)
Meiosis-II is homotypic or equational
division similar to mitosis but occurs in
haploid nuclei. Meiosis-II is essential to
separate out the chromatids of diad
chromosomes to bring real haploidy in
amount of DNA. It also increases the
number of daughter cells though the
chromosome number remains the same
in daughter cells as produced after
meiosis-I.
63In meiosis, the daughter cells
differ from parent cell as well as
amongst themselves due to
[CBSE AIPMT 1991]
(a) segregation, independent
assortment and crossing over
(b) segregation and crossing over
(c) independent assortment and
crossing over
(d) segregation and independent
assortment
Ans.(b)
The daughter cells differ from parent
cell as well as amongst themselves due
to the segregation and crossing over
taking place in them. Meiosis-I brings
gene recombinations and haploidy of
number of chromosomes. Crossing
over during pachytene produces new
combination of genes and is the major
source of new genetic variations in the
sexually reproducing organisms.
64Meiosis-I is reductional division.
Meiosis-II is equational division due
to [CBSE AIPMT 1988]
(a) pairing of homologous
chromosomes
(b) crossing over
(c) separation of chromatids
(d) disjunction of homologous
chromosomes
Ans.(c)
Meiosis-I is called heterotypic division
as the two chromatids of a
chromosome became genetically
different due to the crossing over.
Number of chromosomes is reduced to
half, hence, called reduction division.
Meiosis-II is just like mitosis but occurs
in haploid nuclei, it is called homotypic
or equational division as the
chromosomes are distributed equally
into daughter cells and chromosome
number remains the same in daughter
cells.
Cell Cycle and Cell Division 115

126. 01The main difference between
active and passive transport
across cell membrane is
[NEET (Odisha) 2019]
(a) passive transport is non-selective
whereas active transport is
selective
(b) passive transport requires a
concentration gradient across a
biological membrane whereas
active transport requires energy
to move solutes
(c) passive transport is confined to
anionic carrier proteins whereas
active transport is confined to
cationic channel proteins
(d) active transport occurs more
rapidly than passive transport
Ans.(b)
The main difference between active
and passive transport across the cell is
that passive transport requires
concentration gradient across
biological membrane whereas during
active transport, the movement of
molecules is from high concentration to
low concentration that means they
move against the concentration
gradient by using ATP.
02Which of the following is not a
feature of active transport of
solutes in plants?
[NEET (Odisha) 2019]
(a) Occurs against concentration
gradient
(b) Non-selective
(c) Occurs through membranes
(d) Requires ATP
Ans.(b)
Option (b) is not a feature of active
transport of solutes in plants. Active
transport of solutes in plants is carried
out by membrane proteins. Like
enzymes, the carrier proteins are very
specific (i.e. selective) in what they
carry across the membranes. Active
transport uses energy (ATP) to pump
molecules against a concentration
gradient.
03Water entering root due to diffusion
is part of [CBSE AIPMT 1996]
(a) endosmosis
(b) osmosis
(c) passive absorption
(d) active absorption
Ans.(c)
Most of the water is absorbed by the
plants passively. The force for this type
of water absorption originates in the
aerial parts of the plant due to the loss
of water in transpiration. The root
serves only as pathway and is not
actively involved in this process.
04An innovative professor who
wanted to give a live demonstration
of a physiological process, filled a
glass bottle with previously
moistened mustard seeds and
water. He screwcapped the bottle
and kept it away in a corner and
resumed his lecture. Towards the
end of his lecture there was a
sudden explosion with glass pieces
of bottle thrown around.
Which of the following phenomena
did the professor want to
demonstrate?[CBSE AIPMT 1990]
(a) Diffusion (b) Osmosis
(c) Anaerobic respiration
(d) Imbibition
Ans.(d)
If certain substances for example seeds
are dropped in water, within a few hour
they absorbs the water and swell up
considerably. These substances are
called imbibants and the phenomenon
is known as imbibition. The blast
occurred due to increasing imbibitional
pressure whose magnitude is extremely
high.
05The process responsible for
facilitating loss of water in liquid
form from the tip of grass blades
at night and in early morning is
[NEET (Sep.) 2020]
(a) root pressure
(b) imbibition
(c) plasmolysis
(d) transpiration
Ans.(a)
Root pressure is positive hydrostatic
pressure. It develops in tracheary
element at night and in early morning.
Which results loss of water in liquid
form from tip of grass blades.
Imbibition is the process by which
colloidal soild particles absorb water
and swell without being dissolved.
Which results loss ofwater in liquid
form from tip of grass blades.
Plasmolysis is the process in which
cells lose water in a hypertonic solution.
Transpiration is the loss of water
from a plant in the form of water
vapour.
TransportinPlants
11
Means of Transport
TOPIC 1
Plant-Water Relation
TOPIC 2

127. Transport in Plants 117
06Xylem translocates
[NEET (National) 2019]
(a) water and mineral salts only
(b) water, mineral salts and some
organic nitrogen only
(c) water, mineral salts, some organic
nitrogen and hormones
(d) water only
Ans.(c)
Xylem in plants helps in the
translocation of water, mineral salts,
some organic nitrogen and hormones
from the soil to the aerial parts of the
plant. This process is known as the
ascent of sap and it involves four major
forces namely root pressure, capillarity,
transpirational pull and cohesion and
adhesion of water molecule and
cell wall.
07The water potential of pure water
is [NEET 2017]
(a) zero
(b) less than zero
(c) more than zero, but less than one
(d) more than one
Ans.(a)
The value of water potential of pure
water is highest and it is zero. Water
molecules possess kinetic energy.
Greater the concentration of water in a
system, greater is its kinetic energy or
its water potential.
If we add solute in water, the solution
has fewer free water molecule and
concentration of water decreases thus
reducing its water potential.
08Two cellsAandBare contiguous.
CellAhas osmotic pressure 10
atm, turgor pressure 7 atm and
diffusion pressure deficit 3 atm.
CellBhas osmotic pressure 8 atm,
turgor pressure 3 atm and
diffusion pressure deficit 5 atm.
The result will be
[CBSE AIPMT 2007]
(a) movement of water from cellB-A
(b) no movement of water
(c) equilibrium between the two
(d) movement of water from cellA-B
Ans.(d)
The water moves from lower DPD to
higher DPD, i.e. from cell A to cell B.
This is because when a cell is placed in
pure water, the water enters into the
cell as a result of the diffusion pressure
deficient (DPD) of cell sap.
DPD OP TP= − .
09Water enters a cell due to
[CBSE AIPMT 2001]
(a) OP (b) SP (c) TP (d) WP
Ans.(b)
Suction pressure (also called DPD) is
believed to suck water (cause
movement of water molecule) from
pure solvent/hypotonic solution. It is a
measure of the ability of the cells to
absorb water. BS Meyer (1938) coined
the term suction pressure or DPD.
10In soil, water available for roots
(to plants) is[CBSE AIPMT 1999, 91]
(a) capillary water
(b) hygroscopic water
(c) gravitational water
(d) chemically bound water
Ans.(a)
Water is present in the space between
the soil particles. A large proportion of
water is retained between the soil
particles against the gravitational
force. This is called capillary water. It is
readily available to plants for absorption
by roots.
11The water potential and osmotic
potential of pure water are
[CBSE AIPMT 1998]
(a) 100 and zero (b) zero and zero
(c) 100 and 200 (d) zero and 100
Ans.(b)
The term water potential indicates the
net tendency of any system to donate
water to its surroundings. The water
potential of pure water at atmospheric
pressure is zero. Any addition of solute
to this water reduces its water potential
and makes its value negative. The
osmotic potential of pure water also
would be zero.
12If turgidity of a cell surrounded by
water increases, the wall pressure
will [CBSE AIPMT 1997]
(a) increase
(b) decrease
(c) fluctuate
(d) remain unchanged
Ans.(a)
If a plant cell is placed in a hypotonic
solution/pure water, water starts
moving in by endosmosis. As the
volume of the protoplast increases, it
begins to exert pressure against the
cell wall (turgor pressure). The cell wall
exerts equal and opposite pressure
(wall pressure) on the protoplast.
13Osmotic pressure in the leaf cells is
positive during[CBSE AIPMT 1997]
(a) excessive transpiration
(b) low transpiration
(c) excessive absorption
(d) guttation
Ans.(a)
Osmotic pressure in the leaf cells is
positive during excessive transpiration.
Transpiration is a physical and
physiological process in which aerial
parts of a living plant remove water in
the form of vapours. As the water is
lost from the leaf surface by
transpiration, osmotic pressure of the
leaf cells increases.
14The movement of water from one
cell of the cortex to the adjacent
one in roots is due to
(a) accumulation of inorganic salts in
the cells
(b) accumulation of organic
compounds in the cells
(c) chemical potential gradient
(d) water potential gradient
Ans.(d)
Water potential is the difference in the
chemical potential per unit molal
volume of water in a system and that of
pure water at the same temperature
and pressure. Water always moves
from the area of high water potential to
the area of low water potential.
15The direction and rate of water
movement from cell to cell is
based on [CBSE AIPMT 1992]
(a) WP (b) TP
(c) DPD (d) incipient plasmolysis
Ans.(c)
The difference between the diffusion
pressure of the solution and its solvents
at a particular temperature and
atmospheric condition is called DPD.
The direction and rate of water
movement from cell to cell is based on
DPD (Diffusion Pressure Deficit).
16Water potential can be obtained
by [CBSE AIPMT 1991]
(a) OP + TP (b) OP = WP
(c)Ψ Ψ
S P
+ (d) OP – DPD
Ans.(c)
Water potential is the difference
between the free energy of water in a
system and free energy of pure water. It
can be obtained byΨ Ψ Ψ
W S P
= + . It is a
function of solute potential and
pressure potential.

128. 17Mainly conduction of water in an
angiosperm occurs through
[CBSE AIPMT 1990]
(a) tracheids (b) xylem vessels
(c) sieve tubes (d) All of these
Ans.(b)
The water absorbed by roots has to be
conducted upward so as to meet the
needs of tissues there. Water moves up
through the lumen of xylem vessels.
18Root system in a plant is well
developed [CBSE AIPMT 1990]
(a) due to deficiency of auxins
(b) due to deficiency of cytokinins
(c) due to deficiency of minerals
(d) for increased absorption of water
Ans.(d)
Water in land plants is mainly absorbed
through the roots, especially at the tips
in the region of root hairs. Therefore,
root system in a plant is well developed,
for increasing absorption of water.
19Match the List-I with List-II.
[NEET 2021]
List-I List-II
A. Cohesion 1. More attraction in
liquid phase
B. Adhesion 2. Mutual attraction
among water
molecules
C. Surface
tension
3. Water loss in liquid
phase
D. Guttation 4. Attraction towards
polar surfaces
Choose the correct answer from
the options given below.
A B C D
(a) 2 4 1 3
(b) 4 3 2 1
(c) 3 1 4 2
(d) 2 1 4 3
Ans.(a)
(A)-(2), (B)-(4), (C)-(1), (D)-(3)
Cohesionis the property of like molecules
(of the same substance) to stick to each
other due to mutual attraction, e.g.
attraction of water molecules.
Adhesion is the property of different
molecules or surfaces to cling to each
other. Adhesion is the attraction of
molecules of one kind for molecules of
a different kind, and it can be quite
strong for water, especially with other
molecules bearing positive or negative
charges.
At liquid-air interfaces,surface tension
results from the greater attraction of
liquid molecules to each other
(cohesion) than to the molecules in the
air (adhesion). The process by which
plants loose water in liquid form usually
from tips and margins of leaves is called
guttation.
20A few drops of sap were collected
by cutting across a plant stem by a
suitable method. The sap was
tested chemically. Which one of
the following test results indicates
that it is phloem sap?
[NEET 2016, Phase II]
(a) Acidic
(b) Alkaline
(c) Low refractive index
(d) The absence of sugar
Ans.(b)
The phloem sap is alkaline in nature. It
continuously pumps protons from its
companion cells to the other neighbouring
cells during transport of sugars.
21Root pressure develops due to
[CBSE AIPMT 2015]
(a) active absorption
(b) low osmotic potential in soil
(c) passive absorption
(d) increase in transpiration
Ans.(a)
Root pressure is the positive pressure
that develops in the roots of plants by
the active absorption of nutrients from
the soil. When the nutrients are actively
absorbed by root hairs, water (along
with minerals) increases the pressure in
the xylem. This pressure pushes water
up to small heights.
22A column of water within xylem
vessels of tall trees does not
break under its weight because of
[CBSE AIPMT 2015]
(a) dissolved sugars in water
(b) tensile strength of water
(c) lignification of xylem vessels
(d) positive root pressure
Ans.(b)
A column of water within xylem vessels
of tall trees does not break under its
weight because of high tensile strength
of water, i.e. an ability to resist a pulling
force. This high tensile property
depends on cohesion, adhesion and
surface tension property of water. Due
to these forces only transpiration
driven ascent of xylem sap occurs.
23Which of the following criteria
does not pertain to facilitated
transport? [NEET 2013]
(a) Requirement of special membrane
proteins
(b) High selectivity
(c) Transport saturation
(d) Uphill transport
Ans.(d)
Uphill transport is a process in which
diffusion of a component occurs from a
less concentrated stream to a more
concentrated permeable stream.
Facilitated transport is a form of passive
transport in which materials are moved
across the plasma membrane by a
transport protein down their
concentration gradient.
It requires integral membrane proteins
and highly selective biological
membranes to cross. Saturation occurs
in facilitated, diffusion because not
enough carriers may be available to
handle all the free solute molecules.
The rate of movement may reach a
maximum. Thus, uphill transport does
not pertain to faciliated transport.
24The rupture and fractionation do
not usually occur in the water
column in vessel/tracheids during
the ascent of sap because of
[CBSE AIPMT 2008]
(a) lignified thick walls
(b) cohesion and adhesion
(c) weak gravitational pull
(d) transpiration pull
Ans.(b)
The vertical conduction of water from
root to aerial parts of plant is called
ascent of sap. The water molecules
remain joined to each other due to a
force of attraction called cohesion force.
This attraction is due to the presence of
hydrogen bonds between them.
The magnitude of this force is very high
therefore, continuous water column in
the xylem cannot be broken easily due
to the force of gravity or other
obstructions offered by internal tissues
in the upward movement of water. This
adhesive property of water i.e.,
attraction between the water
molecules and the walls of xylem also
ensures the continuity of water column
in xylem.
118 NEETChapterwise Topicwise Biology
Ascent of Sap
TOPIC 3

129. 25Meaningful girdling (ringing)
experiment cannot be performed
within sugarcane because
[CBSE AIPMT 1992]
(a) its phloem is situated interior to
xylem
(b) its stem surface is covered with
waxy coating
(c) its vascular bundles are not
present in a ring
(d) its stem is thin
Ans.(c)
In ringing experiment removal of bark
breaks the continuity of epidermis,
cortex and phloem. It shows clearly that
epidermis, cortex and phloem do not
take part in transport of sap or water.
In sugarcane (monocot) the vascular
bundles are scattered, therefore,
ringing experiment is not applicable.
26Guttation is caused by
[CBSE AIPMT 1992]
(a) transpiration
(b) osmosis/DPD
(c) root pressure
(d) osmotic pressure
Ans.(c)
Guttation refers to the exudation of
droplets of liquid water from the
margins and tip of leaves through a
group of cells called hydathodes.
Guttation depends on root pressure.
27Soil water available to roots is
[CBSE AIPMT 1991]
(a) surface water
(b) hygroscopic water
(c) gravitational water
(d) capillary water
Ans.(d)
Water, which is mainly held between
the spaces of soil particles forming a
septum of capillaries, is called capillary
water. It is most important form of
water which is mainly taken by the
plants.
28Grass leaves curl inwards during
very dry weather.Select the most
appropriate reason from the
following[NEET (National) 2019]
(a) Flaccidity of bulliform cells
(b) Shrinkage of air spaces in spongy
mesophyll
(c) Tyloses in vessels
(d) Closure of stomata
Ans.(a)
Flaccidity of bulliform cells is the most
appropriate reason for the curling of
grass leaves during dry weather.
Bulliform cells are present between the
epidermal cells of the leaf and they help
to minimise the water loss due to
transpiration during water stress
period.
29Stomatal movement isnot
affected by
(a)O
2
concentration[NEET 2018]
(b) Light
(c) Temperature
(d)CO
2
concentration
Ans.(a)
Stomatal movement is not affected by
CO
2
concentration.Stomata are tiny
pore complexes found in the epidermis
of leaves and other soft aerial parts.
They are meant for the gaseous
exchange but are also the main source
of transpiration. Stomatal movements
are affected by many factors likelight,
temperatureandCO
2
concentration.In
the majority of plants, the stomata are
open in light and close in darkness.
Normally high temperature above30°C
reduces stomatal opening in many
species. LowCO
2
concentration usually
induces opening of stomata while high
CO
2
concentration closes the same.
30Stomatal movement isnot
affected by [NEET 2018]
(a)O
2
concentration
(b) Light
(c) Temperature
(d)CO
2
concentration
Ans.(a)
Stomatal movement is not affected by
O
2
concentration. Stomata are tiny
pore complexes found in the epidermis
of leaves and other soft aerial parts.
They are meant for the gaseous
exchange but are also the main source
of transpiration. Stomatal movements
are affected by many factors likelight,
temperatureandCO
2
concentration.In
the majority of plants, the stomata are
open in light and close in darkness.
Normally high temperature above30°C
reduces stomatal opening in many
species. LowCO
2
concentration usually
induces opening of stomata while high
CO
2
concentration closes the same.
31Stomata in grass leaf are
(a) rectangular [NEET 2018]
(b) kidney-shaped
(c) dumb-bell-shaped
(d) barrel-shaped
Ans.(c)
Epidermis of all green aerial parts of
plants contain minute opening called
stomata. It is surrounded by guard cells
and neighbouring subsidiary cells
collectivity termed as stomatal
apparatus. Kidney-shaped or
bean-shaped guard cells are present in
dicots only, while in monocots like
grasses, these cells are dumb-bell
shaped. Guard cells differ from rest of
the cells in shape, size and thickenings.
32Which of the following facilitates
opening of stomatal aperture ?
(a) Contraction of outer wall of guard
cells [NEET 2017]
(b) Decrease in turgidity of guard cells
(c) Radial orientation of cellulose
microfibrils in the cell wall of
guard cells
(d) Longitudinal orientation of
cellulose microfibrils in the cell
wall of guard cells
Ans.(c)
Stomata are tiny pore complexes. Each
stomata is surrounded by two
specialised green epidermal cells called
guard cells. The opening of the stoma is
facilitated by the orientation of the
microfibril in the cell walls of the guard
cells. Cellulose microfibrils are oriented
radially rather than longitudinally
making it easier for the stoma to open.
Transport in Plants 119
Transpiration
TOPIC 4
ring of bank and
phloem removed
A swelling
above the ring
Leave for
a week
Epidermal
cells
Subsidiary
cells
Chloroplast
Guard cells
Stomatal
aperture
(a) (b)
Bean-shaped guard cells Dumb-bell guard cells
Nucleus
Stomatal apparatus

130. 33Water vapour comes out from the
plant leaf through the stomatal
opening. Through the same
stomatal opening carbon dioxide
diffuses into the plant during
photosynthesis. Reason out the
above statements using the
following options.
[NEET 2016, Phase I]
(a) Both processes can happen
together because the diffusion
coefficient of water andCO
2
is
different
(b) The above processes happen only
during night time
(c) One process occurs during day
time and the other at night
(d) Both processes cannot happen
simultaneously
Ans.(a)
Diffusion of water vapour andCO
2
are
independent process. Their diffusion
depends on the difference in their
partial pressure in the atmosphere as
well as inside the leaves.
34In land plants, the guard cells
differ from other epidermal cells in
having [CBSE AIPMT 2011]
(a) mitochondria
(b) endoplasmic reticulum
(c) chloroplasts
(d) cytoskeleton
Ans.(c)
The guard cells of stomata in land
plants are specialised epidermal cells
which contain chloroplasts. In rest of
epidermal cells, chloroplasts are
absent. The chloroplasts of guard
cells are capable of poor
photosynthesis as there is absence
of NADP reductase enzyme.
35Guard cells help in
[CBSE AIPMT 2009]
(a) protection against grazing
(b) transpiration
(c) guttation
(d) fighting against infection
Ans.(b)
Guard cells help in transpiration.
Stomata are surrounded by two
specialised epidermal cells, called
guard cells. Because of their small size
guard cells are rapidly influenced by
turgor changes and thus regulate the
opening and closing of stomata.
Stomata are involved in releasing water
vapour into the atmosphere. This
process is known as transpiration.
36Stomata of a plant open due to
[CBSE AIPMT 2003]
(a) influx of hydrogen ions
(b) influx of calcium ions
(c) influx of potassium ions
(d) efflux of potassium ions
Ans.(c)
Accumulation ofK
+
ions in the guard
cells during the day time is responsible
for migration of water molecules from
subsidiary cells to guard cells. This
increases turgidity of guard cells and
thus stomata open.
37Opening and closing of stomata is
due to [CBSE AIPMT 2002]
(a) hormonal change in guard cells
(b) change in turgor pressure of guard
cells
(c) gaseous exchange
(d) respiration
Ans.(b)
When the guard cells become turgid,
the stomata open. On the other hand, if
the guard cells loose water, these
become flaccid, the inner walls sag and
the pore closes.
Thus, stomatal movement occurs due
to changes in turgor pressure of guard
cells.
38Glycolate induces opening of
stomata in [CBSE AIPMT 2001]
(a) presence of oxygen
(b) lowCO
2
concentration
(c) highCO
2
concentration
(d) absence ofCO
2
Ans.(b)
Light causes photosynthesis which
leads to reduction inCO
2
concentration
→synthesis of glycolate→oxidation of
glycolate→ATP synthesis→activation
ofK
+
pump→movement ofK
+
in guard
cell→movement of water into guard
cells→swelling of guard cell→opening
of stomata.
39Which of the following is used to
determine the rate of transpiration
in plants? [CBSE AIPMT 1994]
(a) Porometer
(b) Potometer
(c) Auxanometer
(d) Tensiometer
Ans.(b)
Potometer is an apparatus for
measuring the rate of transpiration. It is
also known as Transpirometer.
40Which of the following is an
effective adaptation for better gas
exchange in plants?
[CBSE AIPMT 1993]
(a) Presence of multiple epidermis
(b) Presence of hair on the lower
epidermis
(c) Presence of waxy cuticle covering
the epidermis of the leaves
(d) The location of the stomata
primarily on the lower surface of
the leaf, the side turned away
from the direct sun rays
Ans.(d)
Stomata are meant for the gaseous
exchange but are also the main source
of transpiration. The presence of
stomata on the lower surface of leaf,
the side turned away from the direct
sun rays is an effective adaptation for
better gaseous exchange in plants.
41Conversion of starch to organic
acid is essential for
[CBSE AIPMT 1992]
(a) stomatal closure
(b) stomatal opening
(c) stomatal initiation
(d) stomatal growth
Ans.(b)
The starch hydrolyse into glucose
1-phosphate so that, osmotic potential
becomes lower in guard cells. As a
result of this water enters into the
guard cell by osmotic diffusion from
surrounding epidermal and mesophyll
cells. Guard cells become turgid and
stomata will open. The starch-sugar
interconversion theory about the
mechanism of stomatal movement was
given by Seyere (1923) and modified by
Steward (1964).
42In guard cells when sugar is
converted into starch the stomatal
pore [CBSE AIPMT 1992]
(a) opens fully
(b) opens partially
(c) closes completely
(d) remains unchanged
Ans.(c)
According to starch-sugar
interconversion theory, during night the
120 NEETChapterwise Topicwise Biology

131. Transport in Plants 121
glucose 1-phosphate (sugar) converts
into starch in guard cells and thus
increasing the osmotic potential. The
guard cells release water, become
flaccid and stomata closes.
43In a terrestrial habitat which of the
following is affected by
temperature and rainfall
condition? [CBSE AIPMT 1989]
(a) Translocation
(b) Transpiration
(c) Transformation
(d) Thermodenaturation
Ans.(b)
Transpiration increases with increase in
temperature and decreases with rainfall
conditions. Transpiration will also
depends on number of stomata as more
stomata may provide more pores for
transpiration.
44Basis of stomatal opening is
[CBSE AIPMT 1988]
(a) exosmosis
(b) endosmosis
(c) decrease in cell sap concentration
(d) plasmolysis of guard cells
Ans.(b)
The entry of water into a cell when
placed in less concentrated solution is
called endosmosis. Due to increase in
osmotic pressure and diffusion
pressure deficit of guard cells
endosmosis of water from the
surrounding epidermal and mesophyll
cells takes place into the guard cells.
The guard cells swell and the stomata
open.
45Stomata in angiosperms open and
close due to[CBSE AIPMT 1988]
(a) their genetic constitution
(b) effect of hormones
(c) change of turgor pressure in
guard cells
(d) pressure of gases inside the
leaves
Ans.(c)
The turgor pressure of guard cell
increases due to the osmotic diffusion
of water from the surrounding
epidermal and mesophyll cells. Thus,
the guard cells swell and stomata open.
46The spraying of phenyl mercuric
acetate in leaves
[CBSE AIPMT 1988]
(a) increases transpiration
(b) reduces transpiration
(c) increases rate of photosynthesis
(d) causes guttation
Ans.(b)
Phenyl mercuric acetate is an
antitranspirant. It covers the stomata
as a film and resist the diffusion of
water therefore, reduces the rate of
transpiration. Other antitranspirants
include Abscisic Acid (ABA) and aspirin.
47Select the incorrect statement.
[NEET (Oct.) 2020]
(a) Transport of molecules in phloem
can be bidirectional
(b) Movement of minerals in xylem is
unidirectional
(c) Unloading of sucrose at sink does
not involve the utilisation of ATP
(d) Elements most easily mobilised in
plants from one region to another
are phosphorus, sulphur, nitrogen
and potassium
Ans.(c)
Statement (c) is incorrect. It can be
corrected as Unloading of sucrose at
sink does require utilisation of ATP.
Long distance transport of sucrose
from the source to sink is carried out by
phloem. It is a bidirectional movement.
Both loading of food at the source and
unloading of the same at sink region are
active processes, i.e. require energy in
the form of ATP.
48What is the direction of movement
of sugars in phloem?
[NEET (National) 2019]
(a) Upward
(b) Downward
(c) Bidirectional
(d) Non-multidirectional
Ans.(c)
Sugars show bidirectional movements
in phloem unlike unidirectional
movement of water in xylem. In phloem,
sugars move in both directions
depending upon the source-sink
relationship.
Initially leaves act as source of ‘food
from where it moves to the sink (parts
requiring food). Later, the food in sink is
mobilised towards the growing buds of
the plant.
49The translocation of organic
solutes in sieve tube members is
supported by[CBSE AIPMT 2006]
(a) P-proteins
(b) mass flow involving a carrier and
ATP
(c) cytoplasmic streaming
(d) root pressure and transpiration
pull
Ans.(b)
According to mass flow hypothesis the
transport of organic solutes takes place
from source to sink, this transport also
depends on metabolic energy.
According to cytoplasmic streaming
hypothesis (which was given by de
Vries, 1885) the transport of organic
solutes takes place by combination of
diffusion and cytoplasmic streaming.
Cytoplasmic streaming carry organic
solutes from one and to the other end
of sieve tube.
P-proteins has a role as defence
against phloem feeding insects and
sealing of damaged sieve tubes.
50Loading of phloem is related to
[CBSE AIPMT 2001]
(a) increases of sugar in phloem
(b) elongation of phloem cell
(c) separation of phloem parenchyma
(d) strengthening of phloem fibre
Ans.(a)
Sugar synthesied in leaves (source) is
‘loaded’ in phloem through which it is
transported to all other parts of plants
where it is required (sinks).
51Translocation of carbohydrate
nutrients usually occurs in the
form of [CBSE AIPMT 1992]
(a) glucose (b) maltose
(c) starch (d) sucrose
Ans.(d)
In plants, translocation, i.e. the
movement of organic nutrients from
the region of supply to the region of
sink or utilisation occurs through
phloem (sieve tube/sieve cells) tissue.
Translocated organic nutrients
constitute 10-26% carbohydrates
(usually sucrose) and 1% nitrogenous
components (mostly amino acids).
Phloem Transport
TOPIC 5

132. 01Match the following concerning
essential elements and their
functions in plants.
[NEET (Sep.) 2020]
Column I Column II
A. Iron 1. Photolysis of water
B. Zinc 2. Pollen germination
C. Boron 3. Required for
chlorophyll
biosynthesis
D. Manganese 4. IAA biosynthesis
Select the correct option.
A B C D
(a) 4 3 2 1
(b) 3 4 2 1
(c) 4 1 2 3
(d) 2 1 4 3
Ans.(b)
The correct option is (b). It can be
explained as follows
Iron – essential for the formation of
chlorophyll.
Zinc – needed for synthesis of auxin
Boron – have a role in pollen grain
germination
Manganese – is involved in the splitting
of water to liberate O
2
during
photosynthesis.
02In which of the following forms is
iron absorbed by plants?
[NEET 2018]
(a) Free element
(b) Ferrous
(c) Ferric
(d) Both ferric and ferrous
Ans.(c)
According to NCERT, plants absorb iron
mostly in the form offerric(Fe )
3 +
ions.
However, plants in acidic soil can
absorb iron inferrous(Fe )
2 +
as well as
ferric(Fe )
3 +
form. It is an important
constituent of proteins involved in the
transfer of electrons like ferredoxin and
cytochromes. It is reversibly oxidised
fromFe
2 +
toFe
3 +
during electron
transfer. It activates catalase enzyme.
It is essential for the formation of
chlorophyll.
03In which of the following, all three
are macronutrients?
[NEET 2016, Phase I]
(a) Iron, copper, molybdenum
(b) Molybdenum, magnesium,
manganese
(c) Nitrogen, nickel, phosphorus
(d) Boron, zinc, manganese
Ans.(c)
None of the option is correct w.r.t.
question. The option (a) seems to be
more appropriate.
None of the option consist of all three
macronutrients, But option (c) have
nitrogen and phosphorus which are
macronutrients, but nickel is a
micronutrients.
04Which is essential for the growth
of root tip?[NEET 2016, Phase II]
(a) Zn
(b) Fe
(c) Ca
(d) Mn
Ans.(c)
Calcium is needed to the growing root
tip. It is required in the formation of
middle lamella of the cell wall present in
the form of calcium pectate. Thus,
correct answer is (c).
05Deficiency symptoms of nitrogen
and potassium are visible first in
[CBSE AIPMT 2014]
(a) senescent leaves
(b) young leaves
(c) roots
(d) buds
Ans.(a)
Deficiency of both nitrogen and (N)
potassium (K) is first visible in senescent
(older) leaves, due to the deficiency
symptoms of N chlorosis occurs while,
the deficiency of K causes the inhibition
of protein synthesis and scorching of
older leaves.
06Which one of the following is
correctly matched?
[CBSE AIPMT 2012]
(a) Passive transport of nutrients –
ATP
(b) Apoplast – Plasmodesmata
(c) Potassium – Readily
immobilisation
(d) Bakane of rice seedlings – F Skoog
Ans.(c)
Active transport of nutrients require
ATP. Symplast includes translocation
through plasmodesmata. Bakane
disease of rice was found by Hori (1918)
to be caused due toGibberella fujikuroi.
None of the options is correct. Option
(c) may be correct if statement be read
as “Potassium readily mobilisation”
instead of “potassium readily
immobilisation.
07Which one of the following
elements in plants is not
remobilised?[CBSE AIPMT 2011]
(a) Calcium
(b) Potassium
(c) Sulphur
(d) Phosphorus
MineralNutrition
12
Inorganic Nutrients
TOPIC 1

133. Ans.(a)
Element like calcium are a part of
structural component of the cell and
hence, are not released. The
deficiency symptoms tend to appear
first in the young tissues whenever,
the elements are not demobilised.
08Which one of the following is not a
micronutrient?
[CBSE AIPMT 2010, 03]
(a) Molybdenum (b) Magnesium
(c) Zinc (d) Boron
Ans.(b)
Magnisium is an essential
macrountrient found from 0.2-0.4% dry
matter and is necessary for normal
plant growth.
Magnesium has an important role in
photosynthesis because it forms the
central atom of chlorophyll.
09Manganese is required in
[CBSE AIPMT 2009]
(a) nucleic acid synthesis
(b) plant cell wall formation
(c) photolysis of water during
photosynthesis
(d) chlorophyll synthesis
Ans.(c)
In plants, manganese is absorbed in the
form of manganous ions (Mn
2+
). It
activates many enzymes involved in
photosynthesis, respiration and
nitrogen metabolism. The best defined
function of manganese is in the
splitting of water or photolysis of water
to liberate oxygen during
photosynthesis.
10Which one of the following
elements is not an essential
micronutrient for plant growth?
[CBSE AIPMT 2007]
(a) Mn (b) Zn (c) Cu (d) Ca
Ans.(d)
Ca is essential macronutrient for plant
growth. It is constituent of middle
lamella, activator of enzymes,
connected with chromosome formation
and many aspects of metabolism.
11A plant requires magnesium for
[CBSE AIPMT 2007]
(a) holding cells together
(b) protein synthesis
(c) chlorophyll synthesis
(d) cell wall development
Ans.(c)
Magnesium is an important constituent
of chlorophyll molecule. Thus, a plant
requires magnesium for chlorophyll
synthesis.
12The deficiencies of micronutrients,
not only affects growth of plants
but also vital functions such as
photosynthetic and mitochondrial
electron flow. Among the list given
below, which group of three
elements shall affect most, both
photosynthetic and mitochondrial
electron transport?
[CBSE AIPMT 2005]
(a) Co, Ni, Mo (b) Ca, K, Na
(c) Mn, Co, Ca (d) Cu, Mn, Fe
Ans.(d)
Micronutrients are minerals obtained
from the soil and present in plant
tissues at concentrations usually less
than 3μmolg
–1
dry matter. Cu (copper),
Mn (manganese) and Fe (iron) are those
micronutrients which affect both
photosynthesis and mitochondrial
electron transport because they are the
main constituents of various electron
carriers.
13Boron in green plants assists in
[CBSE AIPMT 2003]
(a) sugar transport
(b) activation of enzymes
(c) acting as enzyme cofactor
(d) photosynthesis
Ans.(a)
Boron is required by plants for (i) uptake
and utilisation ofCa
2 +
, (ii) pollen
germination and cell differentiation
(iii) carbohydrate translocation.
14Grey spots of oat are caused by
deficiency of[CBSE AIPMT 2003]
(a) Fe (b) Cu
(c) Zn (d) Mn
Ans.(d)
Grey spots of oat are caused by the
deficiency of manganese (Mn). It is a
trace element, required in very small
amount. Manganese exists in the soil in
the form of bivalents, trivalents.
15The major portion of the dry
weight of plants comprises of
[CBSE AIPMT 2003]
(a) carbon, hydrogen and oxygen
(b) nitrogen, phosphorus and
potassium
(c) calcium, magnesium and sulphur
(d) carbon, nitrogen and hydrogen
Ans.(a)
The four most common elements in
living organisms are H, C, O, N. These
are also called as framework element.
16Passive absorption of minerals
depend on [CBSE AIPMT 2001]
(a) temperature
(b) temperature and metabolic
inhibitor
(c) metabolic inhibitor
(d) humidity
Ans.(a)
Metabolic inhibitors affect active
absorption. Humidity does not affect
absorption of minerals as much as
temperature.
The movement of mineral ions into the
root cells by the process of simple
diffusion is called aspassive
absorption.This is spontaneous
process and does not require energy.
17Zinc as a nutrient is used by the
plants in the form of
[CBSE AIPMT 2000]
(a) Zn (b) Zn
2+
(c) ZnO (d) ZnSO
4
Ans.(b)
Zinc is taken up by the plants in the
form ofZn
2+
. It is required for
biosynthesis of chlorophyll in some
plants. Deficiency of Zn is shown by a
reduction in internodal growth as a
result plants display a rosette habit of
growth in which the leaves form a
circular cluster radiating at or close to
the ground.
18The plants grown in magnesium
deficient but urea sprayed soil
would show[CBSE AIPMT 2000]
(a) deep green foliage
(b) early flowering
(c) yellowing of leaves
(d) loss of pigments in petals
Ans.(c)
Nitrogen is the constituent of amino
acids, proteins, nucleic acids,
nucleotides, coenzymes, hexosamines
etc. Deficiency of nitrogen rapidly
inhibits the plant growth and yellowing
of the leaves (chlorosis) magnesium has
the specific role in the activation of
enzymes, taking part in photosynthesis
and respiration. It also forms a part of
Mineral Nutrition 123

134. the ring structure of the chlorophyll
molecule. Deficiency of Mg causes
chlorosis, i.e. yellowing of leaves.
Thus a plant growing in
magnesium-deficient soil would show
chlorosis inspite of being sprayed with
urea (nitrogen).
19Which of the following is not
caused by deficiency of mineral
nutrition? [CBSE AIPMT 1997]
(a) Necrosis
(b) Chlorosis
(c) Etiolation
(d) Shortening of internodes
Ans.(c)
The plants grown in dark are said to be
etiolated. They lack chlorophyll and,
therefore, appear yellow. Thus, it is not
caused by deficiency of mineral
nutrient.
Chlorosisis resulted due to the partial
failure of development of chlorophyll
which causes an abnormal colour to
plant tissues.
Uncontrolled death of tissues/cells is
called as necrosis.
Shortening of internodes is due to the
deficiency ofZn
2 +
ions.
20The core metal of chlorophyll is
[CBSE AIPMT 1997]
(a) iron (b) magnesium
(c) nickel (d) copper
Ans.(b)
Chlorophyll molecule, is made up of
porphyrin ring which is a structure with
alternating single and double bonds
containing four small pyrrole rings. It
has magnesium atom at the centre.
21Which one of the following is not
an essential element for plants?
[CBSE AIPMT 1996]
(a) Potassium (b) Iron
(c) Iodine (d) Zinc
Ans.(c)
Essential elements are those element
which are absolutely necessary for
supporting normal growth and
reproduction of plants, the elements
are specific in their action and are
directly involved in the nutrition of the
plant. Potassium is macroelement, zinc
and iron are microelement. Iodine and
sodium are essential for animals and
are not required by most of the plants
however, iodine is found in certain
marine algae.
22Which one of the following is a
micronutrient for plants?
[CBSE AIPMT 1996]
(a) Calcium
(b) Magnesium
(c) Manganese
(d) Nitrogen
Ans.(c)
Micronutrients are needed in very small
amounts by plants, e.g. Manganese,
copper, molybdenum, zinc, boron and
chlorine.
23Phosphorus and nitrogen ions
generally get depleted in soil
because they usually occur as
[CBSE AIPMT 1989]
(a) neutral ions
(b) negatively charged ions
(c) positively charged ions
(d) both positively and negatively
charged but disproportionate
mixture
Ans.(b)
In the soil, phosphorus and nitrogen are
present as negatively charged ions, e.g.
H PO
2 4
−
,NO
2
−
,NO
3
−
ions. These are usually
supplied by fertilisers as urea.
24The four elements making 99% of
living system are
[CBSE AIPMT 1994]
(a) CHOS (b) CHOP
(c) CHON (d) CNOP
Ans.(c)
Carbon, hydrogen, oxygen and nitrogen
are called as big four of the cell. C is
18%, O is 65%, H is 10% and N is 2.5%.
These are principal non-metal elements
and form 95% of total cellular materials.
25Mineral associated with
cytochrome is[CBSE AIPMT 1991]
(a) Cu (b) Mg
(c) Fe and Mg (d) Fe and Cu
Ans.(d)
Cytochromes are iron-porphyrin (haem)
proteins discovered byMac Cunn.
Cytochromes are infact, the conjugated
proteins, composed of a protein
molecule and a non-protein group, i.e.
inorganic factor iron. It is to be noted,
that cell cytochromes have iron only,
though cyt-a
3
possesses both Fe and
Cu. Fe has a role in picking up of
electrons and Cu hands over the
electrons to oxygen.
26Match the Column -I with
Column-II. [NEET 2021]
Column I Column II
A.Nitrococcus1. Denitrification
B.Rhizobium2. Conversion of
ammonia to
nitrite
C.Thiobacillus3. Conversion of
nitrite to nitrate
D.Nitrobacter4. Conversion of
atmospheric
nitrogen to
ammonia
Choose the correct answer from
options given below.
A B C D
(a) 2 4 1 3
(b) 1 2 3 4
(c) 3 1 4 2
(d) 4 3 2 1
Ans.(a)
(A)-(2), (B)-(4), (C)-(1), (D)-(3)
Nitrification is the process of
conversion of ammonia into nitrites and
nitrites into nitrates. It is facilitated by
microorganism like
NitrococcusandNitrobacter.
Rhizobiumis involved in biological
nitrogen fixation, i.e. it converts
atmospheric nitrogen to ammonia.
These are symbiotically associated with
the roots of leguminous plants.
Thiobacillusare involved in conversion of
nitrate to nitrogen gas by the process
called denitrification
27InGlycine max, the product of
biological nitrogen fixation is
transported from the root nodules
to other parts as
[NEET (Oct.) 2020]
(a) ammonia
(b) glutamate
(c) nitrates
(d) ureides
Ans.(c)
InGlycine max(Soyabean), the product
of biological nitrogen fixation is
transported from the root nodules to
other parts as nitrate.
124 NEETChapterwise Topicwise Biology
Nitrogen Metabolism
TOPIC 2

135. 28The product(s) of reaction
catalysed by nitrogenase in root
nodules of leguminous plants
is/are [NEET (Sep.) 2020]
(a) nitrate only
(b) ammonia and oxygen
(c) ammonia and hydrogen
(d) ammonia only
Ans.(c)
The products of reaction catalysed by
nitrogenase in root nodules of
leguminous plants are ammonia and
hydrogen. The reaction is as follows
N 8e 8H 16ATP 2NH + H
2 3 2
+ + + →
− + +
+ 16ADP + 16P
i
SymbioticRhizobiumbacteria invade
the roots of leguminous plants and
form root nodules in which they fix
nitrogen, supplying both to the bacteria
and the plants.
29Which of the following bacteria
reduces nitrate in soil into
nitrogen? [NEET (Odisha) 2019]
(a)Nitrobacter(b)Nitrococcus
(c)Thiobacillus(d)Nitrosomonas
Ans.(c)
Thiobacillus reduces nitrate in soil into
nitrogen. The process is called
denitrification.
On the other hand,Nitrosomonasand
Nitrococcusoxidise ammonia into
nitrite. The bacterium,Nitrobacter
oxidises nitrite to nitrate. These
processes together are known as
nitrification.
30Thiobacillusis a group of bacteria
helpful in carrying out
[NEET (National) 2019]
(a) chemoautotrophic-fixation
(b) nitrification
(c) denitrification
(d) nitrogen-fixation
Ans.(c)
Thiobacillusbacteria help to carry out
denitrification during nitrogen cycle.
This bacteria brings about the
reduction of nitrate to free nitrogen
(N
2
).NitrosomonasandNitrobacterare
chemoautotrophs that cause
nitrification.
31Select the mismatch.[NEET 2017]
(a)Frankia —Alnus
(b)Rhodospirillum— Mycorrhiza
(c)Anabaena — Nitrogen fixer
(d)Rhizobium — Alfa-alfa
Ans.(b)
Among the given options, only option (b)
is mismatched.Rhodospirillumis a free
living nitrogen fixing bacteria.
Mycorrhiza is the symbiotic association
of a fungus with the root of a higher
plant.
32During biological nitrogen fixation,
inactivation of nitrogenase by
oxygen poisoning is prevented by
[CBSE AIPMT 2015]
(a) leghaemoglobin
(b) xanthophyll
(c) carotene
(d) cytochrome
Ans.(a)
During biological nitrogen fixation,
inactivation of nitrogenase by oxygen
poisioning is prevented by
leghaemoglobin.
It is a red-pigment that is filled outside
the peribacteroid space in the cytosol
of nodule cells. It is similar to
haemoglobin of red blood cells. It has
the ability to combine very rapidly with
oxygen and thus acts as a very efficient
O
2
scavenger.
33The first stable product of fixation
of atmospheric nitrogen in
leguminous plants is[NEET 2013]
(a)NO
2
−
(b) ammonia
(c)NO
3
−
(d) glutamate
Ans.(b)
The process of conversion of nitrogen
(N
2
) to ammonia is termed as nitrogen
fixation.
N N H N H 2NH
2 2 2 2 4 3
→ → →
(Nitrogen) (Dimide) (Hydrazine)
(Ammonia)
2NH 3O 2NO 2H 2H O
3 2 2 2
+ → + +
− +
2NO O 2NO
2 2 3
+ →
−
In reductive animation ammonia reacts
withα-ketoglutaric acid and forms
glutamic acid
α-ketoglutaric acidNH NADPH
4
+ +
+
→ + +
Dehygrogenae
Glutamate
Glutamate H O NADP
2
34Which one of the following helps
in absorption of phosphorus from
soil by plants?[CBSE AIPMT 2011]
(a)Rhizobium (b)Frankia
(c)Anabaena (d)Glomus
Ans.(d)
Glomusis a genus of arbuscular
mycorrhizal fungi, which form symbiotic
relationships with plant roots. It is a
longest Genus of AM fungi but it is
currently defined as non-monphylactic.
Mo, Zn and B are micronutrients.
35The function of leghaemoglobin in
the root nodules of legumes is
[CBSE AIPMT 2011]
(a) oxygen removal
(b) nodule differentiation
(c) expression ofnifgene
(d) inhibition of nitrogenase activity
Ans.(a)
Leghaemoglobin is an oxygen
scavenger. It protects the nitrogen
fixing enzyme nitrogenase from
oxygen.
36Nitrifying bacteria
[CBSE AIPMT 2011]
(a) convert free nitrogen to nitrogen
compounds
(b) convert proteins into ammonia
(c) reduce nitrates to free nitrogen
(d) oxidise ammonia to nitrates
Ans.(d)
Nitrifying bacteria (one of the
chemosynthetic bacteria) which oxidise
ammonia to nitrites and obtain energy
for the preparation of food. This
oxidation occurs in two steps. In the
first step, ammonia is oxidised to nitrite
by nitrite bacteria (e.g.Nitrosomonas
andNitrococcus). In the second step,
nitrite is oxidised to nitrate by nitrate
bacteria (e.g.Nitrocystisand
Nitrobacter).
37An element playing important role
in nitrogen fixation is
[CBSE AIPMT 2010]
(a) molybdenum (b) copper
(c) manganese (d) zinc
Ans.(a)
Molybdenum is absorbed as molybdate
by plants. It is involved in nitrogen
metabolism including nitrogen fixation.
It is a component of enzyme
nitrogenase and acts as enzyme
activator. Its deficiency causes
chlorosis and necrosis, whiptail of
cauliflower and premature leaf fall.
Copperis absorbed by the plant in ionic
form. Its deficiency causes necrosis,
die back inCitrus, reclamation in
legumes.
Manganeseis absorbed by the plants as
bivalent ion. It acts as enzyme
activator. Its deficiency causes
interveinal chlorosis as well as
Mineral Nutrition 125

136. 126 NEETChapterwise Topicwise Biology
yellowing of starch and their
subsequent degenerate.
Zincis needed for biosynthsis of IAA
and also acts as enzyme activator. Its
deficiency causes chlorosis, little
leaf,rosette, white bud of maize and
mottling of leaves.
38The common nitrogen-fixer in
paddy fields is[CBSE AIPMT 2010]
(a)Rhizobium
(b)Azospirillum
(c)Oscillatoria
(d)Frankia
Ans.(b)
Azospirillumis a nitrogen fixing
bacterium in paddy fields. It is very
useful soil and root bacterium. It is
associative symbioticN
2
-fixing
bacteria. When it is added to the soil, it
multiplies in millions and can supply
20-40 Kg of nitrogen per hectare per
season. It also producess growth
promoting substances like Indole
Acetic Acid (IAA), gibberellins(GA )
3
and
promotes root proliferation. These
substances improve the plant growth
and yield.
39Nitrogen fixation in root nodules
ofAlnusis brought about by
[CBSE AIPMT 2009, 08]
(a)Bradyrhizobium(b)Clostridium
(c)Frankia (d)Azorhizobium
Ans.(c)
Nitrogen is the most critical element.
Certain non-leguminous plants also
form nodules to fix nitrogen. The best
known example in temperate region is
alder (Alnussp). The bacteria involved in
nodule formation is an Actinomycetes
theFrankia.
Clostridiumis anaerobic, saprotrophic
free-living nitrogen fixing bacteria.
Bradyrhizobiumsp are symbiont in
plants ofParaspaniaand soyabean.
TheAzorhizobiumforms both stem and
root nodules inSesbania(aquatic plant).
40Which of the following is a
flowering plant with nodules
containing filamentous nitrogen
fixing microorganism?
[CBSE AIPMT 2007]
(a)Casuarina equisetifolia
(b)Crotalaria juncea
(c)Cycas revoluta
(d)Cicer arietinum
Ans.(a)
TheCasuarinatree has nitrogen fixing
root nodules that harbor a filamentous
actinomycete nitrogen fixing organism
calledFrankia.
41A free living, nitrogen fixing
cyanobacterium which can also
form symbiotic association with
the water fernAzollais
[CBSE AIPMT 2004]
(a)Tolypothrix
(b)Chlorella
(c)Nostoc
(d)Anabaena
Ans.(d)
Anabaenais a free living, nitrogen fixing
cyanobacterium which can form
symbiotic association with water fern
Azolla.
42Enzyme involved in nitrogen
assimilation[CBSE AIPMT 2001]
(a) nitrogenase
(b) nitrate reductase
(c) transferase
(d) transaminase
Ans.(a)
In the process of biological nitrogen
fixation, the dinitrogen molecule is
reduced by the addition of pairs of
hydrogen in the presence of enzyme
nitrogenase. Enzyme nitrate reductase
and nitrite reductase come into picture
at later stage
(for nitrate assimilation).
43In plants inulin and pectin are
[CBSE AIPMT 2001]
(a) reserved food material
(b) wastes
(c) secretory material
(d) insect attracting material
Ans.(a)
Inulin a polymer of fructose, is used as
a food store, particularly in roots and
tubers of family–Compositae. Pectin is
a mucopolysaccharide which is found in
cell wall of plants. During the time of
food ripening, the pectin becomes
hydrolyse and gives rise the
constituents of sugar.
44Which aquatic fern performs
nitrogen fixation?
[CBSE AIPMT 2001]
(a)Azolla (b)Nostoc
(c)Salvia (d)Salvinia
Ans.(a)
The leaves ofAzollacontain colonies of
Anabaena azollaewhich have the
capacity to fix atmospheric nitrogen.
45Which of the following is a free
living aerobic non-photosynthetic
nitrogen fixer?[CBSE AIPMT 1997]
(a)Rhizobium
(b)Azotobacter
(c)Azospirillum
(d)Nostoc
Ans.(b)
Azotobacteris a free living, aerobic
non-photosynthetic, i.e. saprophytic
bacteria. It retains the capability of
fixing atmospheric nitrogen, i.e. fixation
of atmospheric dinitrogen into
ammonia.
46Which of the following can fix
atmospheric nitrogen?
[CBSE AIPMT 1995]
(a)Albugo
(b)Cystopus
(c)Saprolegnia
(d)Anabaena
Ans.(d)
Blue-green algae (BGA) are the only
organisms, capable of performing
oxygenic photosynthesis and fixation of
nitrogen, e.g.Anabaena,Nostocwhich
produce a specialized type of cell,
calledheterocystwithin whichN
2
fixation occurs.
47Minerals absorbed by roots move
to the leaf through
[CBSE AIPMT 1988]
(a) xylem
(b) phloem
(c) sieve tubes
(d) None of these
Ans.(a)
Mineral ions accumulated in the root
hairs passes into the cortex and finally
reach the xylem from where these are
carried along with water to other parts
of the plant along the transpiration
stream. Like organic solutes, minerals
can move upwards, downwards
(bidirectional movement) as well as
laterally.

137. 01One scientist culturedCladophora
in a suspension ofAzotobacterand
illuminated the culture by splitting
light through a prism. He observed
that bacteria accumulated mainly
in the region of
[NEET (Odisha) 2019]
(a) violet and green light
(b) indigo and green light
(c) orange and yellow light
(d) blue and red light
Ans.(d)
Engelmann used a prism to split light
into its spectral components and then
illuminated a green alga,Cladophora
placed in a suspension of aerobic
bacteria (Azotobacter). The bacteria
were used to detect the sites of oxygen
evolution.
He observed that bacteria mainly
accumulated in the region of blue and
red light of the split spectrum, thus
giving the first action spectrum of
photosynthesis.
02In a chloroplast the highest
number of protons are found in
[NEET 2016, Phase I]
(a) lumen of thylakoids
(b) inter membrane space
(c) antennae complex
(d) stroma
Ans.(a)
Proton concentration is higher in the
lumen of thylakoid due to photolysis of
water,H
+
pumping and NADP reductase
activity which occurs in stroma of the
chloroplast.
03Of the total incident solar radiation
the proportion of PAR is
[CBSE AIPMT 2011]
(a) about 60% (b) less than 50%
(c) more than 80% (d) about 70%
Ans.(b)
PAR(Photosynthetically Active
Radiation) designates the spectral
range of solar radiation from 400-700
nm that photosynthetic organisms are
able to use in the process of
photosynthesis. Of the total incident
solar radiation the proportion of PAR is
less than 50%.
04.Stroma in the chloroplasts of
higher plants contains
[CBSE AIPMT 2009]
(a) light-independent reaction
enzymes
(b) light-dependent reaction enzymes
(c) ribosomes
(d) chlorophyll
Ans.(a)
In higher plants, enzymes for light
independent reactions (dark reactions)
are present in the stroma of
chloroplasts.
Light dependent reaction occurs in
grana of chloroplast.
Ribosomesare necessary for protein
synthesis.
Chlorophyllis green photosynthetic
pigment found in chloroplasts.
05Carbohydrates are commonly
found as starch in plant storage
organs. Which of the following five
properties of starch (A–E) make it
useful as a storage material?
A. easily translocated
B. chemically non-reactive
C. easily digested by animals
D. osmotically inactive
E. synthesised during
photosynthesis
The useful properties are
[CBSE AIPMT 2008]
(a) B and C (b) B and D
(c) A, C and E (d) A and E
Ans.(c)
Option (c) is correct. As starch is a high
molecular weight polymer of D-glucose
inα1→4 linkage. It is synthesised in
chloroplasts as one of the stable end
products of photosynthesis. It is most
abundant and common storage
polysaccharide in plants hence, most
staple food for man and herbivores.
It is a mixture of two types of glucose
homopolysaccharideviz, amylose and
amylopectin. During day time the starch
synthesis in chloroplast is coordinated
with sucrose synthesis in cytosol.
Typically about 90% of total solute
carried in phloem is the carbohydrate
sucrose, a disaccharide.
This is relatively inactive and highly
soluble sugar playing little direct role in
metabolism and so, making an ideal
transport sugar.
06Chlorophyll in chloroplasts is
located in[CBSE AIPMT 2004]
(a) outer membrane
(b) inner membrane
(c) thylakoids (d) stroma
Ans.(c)
The thylakoids of chloroplast are
flattened vesicles arranged as a
membranous network within the
stroma. 50% of chloroplast proteins
and various components involved
(namely chlorophyll, carotenoids and
plastoquinone) in photosynthesis are
present in thylakoid membranes.
Photosynthesis
13
Site and Pigments
Involved in Photosynthesis
TOPIC 1

138. 07Which fractions of the visible
spectrum of solar radiations are
primarily absorbed by carotenoids
of the higher plants?
[CBSE AIPMT 2003]
(a) Violet and blue (b) Blue and green
(c) Green and red (d) Red and violet
Ans.(a)
Carotenoidsare a group of yellow, red
and orange pigments which function as
accessory pigments and protect
chlorophyll molecules from destruction
by intensive light rays. Carotenoids
have three absorption peaks in the
blue-violet range of the spectrum.
08Which element is located at the
centre of the porphyrin ring in
chlorophyll?[CBSE AIPMT 2003]
(a) Manganese (b) Calcium
(c) Magnesium (d) Potassium
Ans.(c)
Magnesium is at the centre of the
porphyrin ring in chlorophyll. The
general structure of chlorophyll was
elucidated by Hand Fischer in 1940.
09Stomata of CAM plants
[CBSE AIPMT 2003]
(a) open during the night and close
during the day
(b) never open
(c) are always open
(d) open during the day and close at
night
Ans.(a)
CAM (Crassulacean Acid Metabolism)
plants open stomata only at night (when
temperature is low and humidity is high)
to cause lesser loss of water (e.g.
Agave, Opuntia,etc.).So, CAM
photosynthesis is a carbon fixation
pathway that evolved in some plants as
an adaptation to arid condition.
10The first step of photosynthesis is
[CBSE AIPMT 2000]
(a) excitation of electron of
chlorophyll by a photon of light
(b) formation of ATP
(c) attachment ofCO
2
to 5 carbon
sugar
(d) ionisation of water
Ans.(a)
The entire process of photosynthesis is
driven by light energy coming from the
sun. This energy is first captured by
chlorophyll molecules and later on
utilised for the synthesis of ATP
(chemical energy) molecules which are
later utilised in the dark reaction, i.e.,
Calvin cycle.
11Chlorophyll-amolecule at its
carbon atom 3 of the pyrrole
ring-II has one of the following
[CBSE AIPMT 1996]
(a) aldehyde group
(b) methyl group
(c) carboxyl group
(d) magnesium
Ans.(a)
Chlorophyll has a tetrapyrrole porphyrin
head and a long chain alcohol called
phytol tail. Each pyrrole is a 5 member
ring with one nitrogen and four carbon.
A non-ionic Mg atom lies in the centre
of porphyrin, attached to nitrogen
atoms of pyrrole rings. Chlorophyll-a
has methyl group at carbon 3 of pyrrole
ring and chlorophyll-bhas formyl
(aldehyde) group attached to this atom.
12Pigment acting as a reaction
centre during photosynthesis is
[CBSE AIPMT 1994]
(a) carotene
(b) phytochrome
(c)P
700
(d) cytochrome
Ans.(c)
Photosynthetic pigment molecules (e.g.
P , P
700 680
) are able to convert light energy
into chemical energy. These pigment
molecules which together forms the
photosynthetic units, possess
photocentres (reaction centre = trap
centre) surrounded by harvesting
molecules differentiated into core
molecules and antenna molecules.
13Nine-tenth of all photosynthesis of
world (85-90%) is carried out by
[CBSE AIPMT 1994]
(a) large trees with millions of
branches and leaves
(b) algae of the ocean
(c) chlorophyll containing ferns of the
forest
(d) scientists in the laboratories
Ans.(b)
90% of total photosynthesis is carried
out by aquatic plants, chiefly algae (80%
in oceans and 10% in freshwater). 10%
of total photosynthesis is performed by
land plants.
14Maximum solar energy is trapped
by
[CBSE AIPMT 1993]
(a) planting trees
(b) cultivating crops
(c) growing algae in tanks
(d) growing grasses
Ans.(d)
Maximum solar energy is trapped by
growing grasses, as they have the
largest surface area for absorption.
Limited number of algal individual are
growing in tank so, they absorb limited
amount of light.
15Chlorophyll-aoccurs in
[CBSE AIPMT 1992]
(a) all photosynthetic autotrophs
(b) in all higher plants
(c) all oxygen liberating autotrophs
(d) all plants except fungi
Ans.(b)
Chlorophyll-a(C H O N Mg
557254
) is a bluish
green pigment, it is the primary
photosynthetic pigment or universal
photosynthetic pigment that occurs in
all plants except photoautotrophic
bacteria, i.e. found in all oxygenic
photoautotrophs.
16Photosynthetic pigments found in
the chloroplasts occur in
[CBSE AIPMT 1991]
(a) thylakoid membranes
(b) plastoglobules
(c) matrix
(d) chloroplast envelope
Ans.(a)
Photosynthetic pigments are those
pigments which occur on
photosynthetic thylakoids of
chloroplasts and absorb light energy for
the purpose of photosynthesis. These
are mainly of two types—chlorophylls
and carotenoids.
17The size of chlorophyll molecule is
[CBSE AIPMT 1988]
(a) head15 15×Å, tail 25 Å
(b) head20 20×Å, tail 25 Å
(c) head15 15×Å, tail 20 Å
(d) head10 12×Å, tail 25 Å
Ans.(c)
A chlorophyll molecule consists of two
parts, the porphyrin ring (head)15 15×Å
and a phytol tail (20 Å).
128 NEETChapterwise Topicwise Biology

139. 18Which of the following statement
is incorrect? [NEET 2021]
(a) Both ATP and NADPH + H
+
are
synthesised during non-cyclic
photophosphorylation
(b) Stroma lamellae have PS-I only
and lack NADP reductase
(c) Grana lamellae have both PS-I and
PS-II
(d) Cyclic photophosphorylation
involves both PS-I and PS-II
Ans.(d)
Statement in option (d) is incorrect and
can be corrected as
Only photosystem I is involved in cyclic
photophosphorylation process. Cyclic
photophosphorylation is a process in
which an electron expelled by the
excited photocentre is returned to it
after passing through a series of
electron carriers. The excited electron
does not pass on to NADP
+
but is
cycled back to the PS I complex through
the electron transport chain.
Non-cyclic photophosphorylation
involves both photosystems I and II. The
electron follows a non-cyclic pathway in
it. The representation of it is also called
Z scheme.
19In light reaction, plastoquinone
facilitates the transfer of
electrons from[NEET (Sep.) 2020]
(a) Cyt-b
6
fcomplex to PS-I
(b) PS-I to NADP
+
(c) PS-I to ATP synthase
(d) PS-II to Cyt-b
6
fcomplex
Ans.(d)
In light reaction, plastoquinone
facilitates the transfer of electrons
from PS II to cytochromeb f
6
complex
(non-cyclic photophosphorylation)
process of light reaction starts with PS
II (680 nm). When sunlight falls on the
reaction center (chlorophyll,a) it
abosrbs 680 nm wavelength of red light
causing electrons to become excited
and jump into an orbit farther from the
atomic nucleus. These electrons are
picked by the electron acceptor which
passes them to an electron transport
system consisting of cytochromeb
6
complex.
20Which of the following is not a
product of light reaction of
photosynthesis? [NEET 2018]
(a) NADPH (b) NADH
(c) ATP (d) Oxygen
Ans.(b)
During light reaction of photosynthesis
NADPH, ATP and oxygen are formed.
Oxygenis liberated by the photolysis of
water.
4H O 4H + 4OH
2
+
r
−
4OH
Mn , Ca , Cl
Oxygen -Evolving complex
2 + 2 +
−
→
−
2H O + O 4
2 2
↑ +
−
e
The electrons released during
photolysis of water are picked up byP
680
photocentre of PS-II. On receiving light
energy photocentre expels an electron
which passes over a series of carriers.
As a result assimilatory power, i.e.ATP
andNADPHis produced.NADHis
formed during respiration.
21In photosynthesis, the
light-independent reactions take
place at [CBSE AIPMT 2015]
(a) thylakoid lumen
(b) photosystem-I
(c) photosystem-II
(d) stromal matrix
Ans.(d)
The light-independent reactions (or
dark reactions) take place in the
stromal matrix of the chloroplasts.
In light independent reactions, carbon
dioxide is reduced to glucose
(carbohydrate) by the hydrogen in
NADPH by using the chemical energy
stored in ATP. This reaction takes place
in the presence of a substance called
RuDP.
22Anoxygenic photosynthesis is
characteristic of
[CBSE AIPMT 2014]
(a)Rhodospirillum(b)Spirogyra
(c)Chlamydomonas(d)Ulva
Ans.(a)
Anoxygenic photosynthesis (in whichO
2
is not released) is seen inRhodospirillum
which is a purple non-sulphur bacteria. It
helps an organism to trap light energy and
store it as chemical energy.
Other than this anoxygenic
photosynthesis commonly occurs in
purple non-sulphur bacteria, green
sulphur/non-sulphur bacteria, and
heliobacteria, etc.
23A process that makes important
difference betweenC
3
and
C
4
-plants is[CBSE AIPMT 2012]
(a) transpiration (b) glycolysis
(c) photosynthesis
(d) photorespiration
Ans.(d)
Photorespirationis a light dependent
process which occurs inC
3
-plants. It is
opposite to photosynthesis because
during this process, uptake ofO
2
and
release ofCO
2
take place. Due to the
presence of Kranz anatomy,C
4
-plants
do not show photorespiration.
24Oxygenic photosynthesis occurs in
[CBSE AIPMT 2009]
(a)Chromatium(b)Oscillatoria
(c)Rhodospirillum(d)Chlorobium
Ans.(b)
Oscillatoria is a photosynthetic
cyanobacterium. In this cyanobacteria
during photosynthesis water is electron
donor and oxygen is a byproduct,
i.e., oxygenic photosynthesis occurs.
RhodospirillumandChlorobiumare
non-oxygenic photosynthetic, purple
non-sulphur and green- sulphur
bacteria.
Chromatiumis purple sulphur
bacterium, also a non-oxygenic
photosynthetic.
25Cyclic-photophosphorylation
results in the formation of
[CBSE AIPMT 2009]
(a) NADPH
(b) ATP and NADPH
(c) ATP, NADPH andO
2
(d) ATP
Ans.(d)
Cyclic-photophosphorylation involves
only pigment system-I and results in the
formation of ATP only. When the
photons activate PS-I, a pair of
electrons are raised to a higher energy
level. They are captured by primary
acceptor which passes them on to
ferredoxin, plastoquinone, cytochrome
complex, plastocyanin and finally back
to reaction centre of PS-I, i.e.,P
700
.
At each step of electron transfer, the
electrons lose potential energy. Their
trip down hill is caused by the transport
chain to pumpH
+
across the thylakoid
membrane. The proton gradient thus
established is responsible for forming
ATP (2 molecules). No reduction of
NADP to NADPH +H
+
.
Photosynthesis 129
Light Reaction
TOPIC 2

140. 26The first acceptor of electrons
from an excited chlorophyll
molecule of photosystem-II is
[CBSE AIPMT 2007, 08]
(a) cytochrome
(b) iron-sulphur protein
(c) ferredoxin
(d) quinone
Ans.(d)
Plastoquinone is the first acceptor of
electrons from an excited chlorophyll
molecule of photosystem-II.
27In photosystem-I the first electron
acceptor is[CBSE AIPMT 2006]
(a) cytochrome
(b) plastocyanin
(c) an iron-sulphur protein
(d) ferredoxin
Ans.(c)
In photosystem-I, the primary electron
acceptor is probably a Fe-S protein.
The reduced primary acceptor
transfers the electrons to secondary
electron acceptor (most probablyP
430
).
The sequence of electron transfer is as
follows :
P — A — A
700
(Chl -a )
e
1
(Phyloquinone)
e
2
(Fe -S
p
+
− −
→ →
rotein)
e
3
(P )
— A
430
→
−
The reducedP
430
passes its electrons to
ferredoxin (Fd) present at outer surface
of thylakoid membrane.
28Which of the following absorb light
energy for photosynthesis?
[CBSE AIPMT 2002]
(a) Chlorophyll (b) Water molecule
(c)O
2
(d) RuBP
Ans.(a)
Chlorophyll molecule absorbs light for
photosynthesis.H O
2
molecules provide
H
+
ions and electrons during
photosynthesis.O
2
is liberated during
photosynthesis. RuBP (Ribulose 1,
5-bisphosphate) reacts withCO
2
during
dark reaction of photosynthesis. This
process takes place in the presence of
enzyme RuBisCO.
29Which pigment system is
inactivated in red drop?
[CBSE AIPMT 2001]
(a) PS-I and PS-II (b) PS-I
(c) PS-II (d) None of these
Ans.(c)
The fall in photosynthetic yield beyond
red region of spectrum (680 nm) is
called red drop. Reaction centre of PS-II
isP
680
while that of PS-I isP
700
. So in the
red drop reaction PS-II is inactivated.
30Photochemical reactions in the
chloroplast are directly involved in
[CBSE AIPMT 2000]
(a) formation of phosphoglyceric acid
(b) fixation of carbon dioxide
(c) synthesis of glucose and starch
(d) photolysis of water and
phosphorylation of ADP to ATP
Ans.(d)
CO
2
is fixed in the stroma of the
chloroplast leading to the synthesis of
PGA from which glyceraldehyde
phosphate is formed. From
glyceraldehyde phosphate, sugar and
starch are formed.
All these do not require light. However,
photolysis of water and
phosphorylation of ADP to ATP requires
light energy.
31Protochlorophyll differs from
chlorophyll in lacking
[CBSE AIPMT 1998]
(a) 2 hydrogen atoms in one of its
pyrrole rings
(b) 2 hydrogen atoms in two of its
pyrrole rings
(c) 4 hydrogen atoms in one of its
pyrrole rings
(d) 4 hydrogen atoms in two of its
pyrrole rings
Ans.(a)
Protochlorophyll differs from
chlorophyll in lacking 2 hydrogen atoms
in one of its pyrrole rings.
32NADPH is generated through
[CBSE AIPMT 1997]
(a) photosystem-I
(b) photosystem-II
(c) anaerobic respiration
(d) glycolysis
Ans.(b)
NADPH is generated through
photosystem-II. In non-cyclic
photophosphorylation (which involves
both PS-I and II) protons released from
photolysis and electrons emitted from
P
700
are ultimately passed on to NADP
+
resulting in the formation of NADPH.
In cyclic photophosphorylation (which
involves only PS-I) electrons flow in a
cyclic manner but there is no net
formation of NADPH andO
2
.
33Which one occurs both during
cyclic and non-cyclic modes of
photophosphorylation?
[CBSE AIPMT 1994]
(a) Involvement of both PS-I and PS-II
(b) Formation of ATP
(c) Release ofO
2
(d) Formation of NADPH
Ans.(b)
Cyclic photophosphorylation is that
type of light energised ATP synthesis in
which electron expelled by excited
photocentre does not return to them. It
involves two Photochemical Systems
(PS-I and PS-II) and produces
assimilatory power (ATP and NADPH).
In both, cyclic and non-cyclic
photophosphorylation, formation of ATP
takes place.
34Formation of ATP in
photosynthesis and respiration is
an oxidation process which
utilises the energy from
[CBSE AIPMT 1992]
(a) cytochromes (b) ferredoxin
(c) electrons (d) carbon dioxide
Ans.(c)
Cytochromes (Keilin; 1925) are the
electron transport intermediates
containing heme (or related prosthetic
groups) in which the iron undergoes
valency changes during electron
transfer and produces energy (ATP) in
both photosynthesis and respiration.
35Photosystem-II occurs in
[CBSE AIPMT 1992]
(a) stroma (b) cytochrome
(c) grana
(d) mitochondrial surface
Ans.(c)
PS-II is present in appressed part of
granal thylakoids. PS-I is present in
stroma thylakoids and non-appressed
parts of granal thylakoids.
36Ferredoxin is a constituent of
[CBSE AIPMT 1991]
(a) PS-I (b) PS-II
(c) Hill reaction (d)P
680
Ans.(a)
Ferredoxin(Fd) is a soluble protein
which acts as electron carrier and
forms a constituent of PS-I. Ferredoxin
passes electrons to reductase complex
which helps in reducing NADP
+
to
NADPH (a strong reducing agent).
130 NEETChapterwise Topicwise Biology

141. 37NADP
+
is reduced to NADPH in
[CBSE AIPMT 1988]
(a) PS-I (b) PS-II
(c) Calvin cycle
(d) Non-cyclic photophosphorylation
Ans.(d)
In photosynthesis during non-cyclic
photophosphorylation involving both
PS-I and PS-II, electrons released
during photolysis of water are
transfered to PS-II and then PS-I via a
series of electron carriers.P
700
of PS-I
releases electron after absorbing light
energy. This electron passes through
chlorophyll X, Fe-S, ferredoxin and
finally to NADP
+
.NADP
+
combines with
H
+
(released during photolysis) with the
help of NADP reductase to form
NADPH.
NADP + 2e H NADPH
+
NADP
reductase− +
+ →
38The first stable product of CO
2
fixation inSorghumis[NEET 2021]
(a) pyruvic acid
(b) oxaloacetic acid
(c) succinic acid
(d) phosphoglyceric acid
Ans.(b)
Carbon fixation or carbon assimilation
is the process by which inorganic
carbon (particularly in the form of
carbon dioxide) is converted to organic
compounds by living organisms. The
compounds are then used to store
energy and as structure for other
biomolecules.
Most of the plants that are adapted to
dry tropical regions form C-4 acid i.e.
oxalic acid as their first stable product.
These plants are called C
4
plants.
Sugarcane, maize,Sorghum, etc. are
the examples of these plants.
39Which of the following statements
is incorrect?[NEET (Oct.) 2020]
(a) RuBisCO is a bifunctional enzyme
(b) InC
4
plants the site of RuBisCO
activity is mesophyll cell
(c) The substrate molecule for
RuBisCO activity is a 5-carbon
compound
(d) RuBisCO action requires ATP and
NADPH
Ans.(b)
Statement (b) is incorrect and can be
corrected as InC
4
plants, Kranz anatomy
in leaf is found due to the presence of
two type of cellsviz., mesophyll cells
and bundle sheath cells.
The mesophyll cells are specialised to
perform light reaction, evolveO
2
and
produce assimilatory power. The bundle
sheath cells possess RuBisCO and thus,
perform RuBisCO activity at this site.
40The oxygenation activity of
RuBisCO enzyme in
photorespiration leads to the
formation of[NEET (Sep.) 2020]
(a) 1 molecule of 3-C compound
(b) 1 molecule of 6-C compound
(c) 1 molecule of 4-C compound and 1
molecule of 2-C compound
(d) 2 molecules of 3-C compound
Ans.(a)
The oxygenation activity of RuBisCO
enzyme in photorespiration leads to the
formation of 1 molecule of 3C
compound (phosphoglycerate). InC
3
plants during oxygen fixation, one
molecule of PGA(3C) and one molecule
of 2-phosphoglycolate(2C) are formed.
The latter is then converted back to
PGA in the photorespiratory cycle.
Photorespiration occurs at high
concentration of oxygen and
temperature in the environment.
41In Hatch and Slack pathway, the
primaryCO
2
acceptor is
[NEET (Odisha) 2019]
(a) oxaloacetic acid
(b) phosphoglyceric acid
(c) phosphoenol pyruvate
(d) RuBisCO
Ans.(c)
In Hatch and Slack pathway, the primary
CO2acceptor is phosphoenol pyruvate.
This occurs in C4-plants. Phosphoenol
pyruvate, a 3-carbon compound,
accepts CO2and forms oxaloacetic acid
which is a 4-carbon compound.
42Phosphoenol Pyruvate (PEP) is the
primaryCO
2
acceptor in
[NEET 2017]
(a)C
3
-plants (b)C
4
-plants
(c)C
2
-plants (d)C
3
andC
4
-plants
Ans.(b)
Phosphoenol Pyruvate (PEP) is found in
the mesophyll cell, which accepts the
atmosphericCO
2
inC
4
-plants and
converts it to oxalo acetate ––aC
4
compound. It is the first stable
compound ofC
4
-plants.
Concept EnhancerC
4
-plants possess
special adaptation anatomy in their
leaves to cope up the photorespiratory
losses. There are dimorphic chloroplast
present in them—agranal in bundle
sheath cells and granal in mesophyll
cells.
43A plant in your garden avoids
photorespiratory losses, has
improved water use efficiency,
shows high rates of
photosynthesis at high
temperatures and has improved
efficiency of nitrogen utilisation. In
which of the following
physiological groups would you
assign this plant?
[NEET 2016, Phase I]
(a)C
4
(b) CAM
(c) Nitrogen fixer (d)C
3
Ans.(a)
This plant is aC
4
-plant as these group of
plants shows little photorespiration,
efficient in binding toCO
2
even at low
concentrations, better utilisation of
water as well as high rates of
photosynthesis even at high
temperatures, i.e. tropical region.
Besides, they can also tolerate excess of
salts due to presence of organic acids.
44PGA as the firstCO
2
-fixation
product was discovered in
photosynthesis of
[CBSE AIPMT 2010]
(a) bryophyte (b) gymnosperm
(c) angiosperm (d) alga
Ans.(d)
The use of radioactive
14
CbyMelvin
Calvinin algal (Chlorella) photosynthesis
studies led to the discovery that the
firstCO
2
fixation product was a
3-carbon organic acid. The first product
identified was 3-phosphoglyceric acid
(PGA).
45C
4
-plants are more efficient in
photosynthesis thanC
3
-plants due
to [CBSE AIPMT 2010, 08]
(a) higher leaf area
(b) presence of larger number of
chloroplasts in the leaf cells
(c) presence of thin cuticle
(d) lower rate of photorespiration
Photosynthesis 131
Dark Reaction
TOPIC 3

142. Ans.(b)
C
4
-plants are more efficient in
photosynthesis thanC
3
-plants but use
more energy. They possess the larger
number of chloroplasts in the leaf cells.
In the leaves of C
4
-plants, the vascular
bundles are surrounded by bundle
sheath cells which in turn are
surrounded by mesophyll cells.
Chloroplast in bundle sheath cells are
larger and always contain grana,
whereas chloroplasts in mesophyll cells
are smaller.
46In the leaves ofC
4
-plants, malic
acid formation duringCO
2
-fixation
occurs in the cells of
[CBSE AIPMT 2007, 08]
(a) mesophyll (b) bundle sheath
(c) phloem (d) epidermis
Ans.(a)
The oxalic acid is reduced to malic acid
in mesophyll cells, from chloroplast of
mesophyll cells the malic acid is
transferred to the chloroplast of bundle
sheath cells where, it is decarboxylated to
formCO
2
and pyruvic acid.
47As compared to aC
3
-plant, how
many additional molecules of ATP
are needed for net production of
one molecule of hexose sugar by
C
4
-plants [CBSE AIPMT 2005]
(a) 2 (b) 6 (c) 12 (d) zero
Ans.(c)
InC
4
-plants everyCO
2
molecule has
to be fixed twice, so these plants are
needed more energy for the
synthesis of hexose sugar molecules
thanC
3
-plants in whichCO
2
has to be
fixed only once. 18 ATP molecules are
required byC
3
-plants for the
synthesis of one molecule of hexose
sugar while 30 ATP molecules are
needed by theC
4
-plants for the
same. Thus,C
4
-plants have a need of
12 ATP molecules extra than
C
3
-plants for the synthesis of one
molecule of hexose sugar.
48Photosynthesis inC
4
-plants is
relatively less limited by
atmosphericCO
2
levels because
[CBSE AIPMT 2005]
(a) effective pumping ofCO
2
into
bundle sheath cells
(b) RuBisCO inC
4
-plants has higher
affinity forCO
2
(c) four carbon acids are the primary
initialCO
2
-fixation products
(d) the primary fixation ofCO
2
is
mediatedviaPEP carboxylase
Ans.(d)
The fixation ofCO
2
inC
4
-plants takes
place in two places and by two different
organic compounds. Phosphoenol
Pyruvate (PEP) is found in mesophyll
cells which primarily fixes atmospheric
CO
2
into oxalo acetic acid (4C).
RuBisCO is present in bundle sheath
cells where final fixation ofCO
2
in
hexose sugars takes place.CO
2
is
primarily fixed by PEP carboxylase
because this enzyme has greater
affinity toCO
2
than RuBisCO.
49InC
3
-plants, the first stable
product of photosynthesis during
the dark reaction is
[CBSE AIPMT 2004]
(a) malic acid
(b) oxaloacetic acid
(c) 3-phosphoglyceric acid
(d) phosphoglyceraldehyde
Ans.(c)
InC
3
-plants the first stable product
formed during dark reaction is
3-phosphoglyceric acid. Since, it is a 3
carbon compound hence, the pathway
is referred asC
3
-pathway. Oxalo Acetic
Acid (OAA) is the first stable compound
inC
4
-plants. It is a 4C compound.
50In sugarcane plant
14
2
COis fixed
in malic acid, in which the enzyme
that fixesCO
2
is
[CBSE AIPMT 2003]
(a) fructose phosphatase
(b) ribulose bisphosphate carboxylase
(c) phosphoenol pyruvic acid
carboxylase
(d) ribulose phosphate kinase
Ans.(c)
InC
4
-plants,CO
2
is taken up by
Phosphoenol- Pyruvate (PEP) and the
reaction being catalysed by PEP
carboxylase.
51In photosynthesis energy from
light reaction to dark reaction is
transferred in the form of
[CBSE AIPMT 2002]
(a) ADP (b) ATP
(c) RuDP (d) chlorophyll
Ans.(b)
As a result of light reaction, oxygen,
NADPH and ATP are formed. Oxygen is
released into the atmosphere while
NADPH and ATP are utilised for
reduction ofCO
2
to carbohydrate in
dark reaction.
52Which pair is wrong?
[CBSE AIPMT 2001]
(a)C
3
—Maize
(b)C
4
—Kranz anatomy
(c) Calvin cycle—PGA
(d) Hatch and Slack Pathway—Oxalo
acetic acid
Ans.(a)
Maize is aC
4
-plant.C
4
-plants have
Kranz type anatomy of leaves.
PGA (3-Phosphoglyceric Acid) is formed
during Calvin cycle.
OAA (Oxalo Acetic Acid) a 4C compound
is formed during Hatch and Slack cycle
(C
4
cycle).
53How many turns of Calvin cycle
yield one molecule of glucose?
[CBSE AIPMT 2000]
(a) 8 (b) 2 (c) 6 (d) 4
Ans.(c)
Conversion ofCO
2
to simple (reduced)
organic compounds is calledCO
2
assimilation orCO
2
fixation or carbon
fixation. This fixation pathway was
elucidated in the early 1950s by Melvin
Calvin and Coworkers and is often
called as Calvin cycle.
Since, one molecule of carbon is fixed
in one turn of the Calvin cycle. So, six
turns of the cycle are required to fix the
glucose molecule containing 6 carbon
atoms.
54Fixation of oneCO
2
molecule
through Calvin cycle requires
[CBSE AIPMT 2000]
(a) 1 ATP and 2NADPH
2
(b) 2 ATP and 2NADPH
2
(c) 3 ATP and 2NADPH
2
(d) 2ATP and 1NADPH
2
Ans.(c)
2 ATP are required during conversion of
PGA to 1, 3 diphosphoglyceric acid and 1
132 NEETChapterwise Topicwise Biology
CO + PEP
2
PEP carboxylase
(Mesophyll cells of
C -plant)
4
Oxalo acetic acid
Malic acid
+
NADP
+
NADPH
Dehydro
genase

143. ATP during conversion of glyceraldehyde
phosphate to ribulose biphosphate. 2
NADPH
2
molecules are utilised for
converting 1, 3 diphosphoglyceric acid
to glyceraldehyde phosphate.
55Which one of the following is
represented by Calvin cycle?
[CBSE AIPMT 1996]
(a) Reductive carboxylation
(b) Oxidative carboxylation
(c) Photophosphorylation
(d) Oxidative phosphorylation
Ans.(a)
In dark phase or Calvin cycle, carbon
dioxide is assimilated with the help of
assimilatory power (ATP and NADPH
2
)
to produce organic acid. The cycle
involves reduction of carbon involving
carboxylation, glycolytic reversal and
regeneration of RuBP.C
3
cycle is also
known as reductive pentose pathway or
Photosynthetic Carbon Reduction
(PCR).
56C
4
-cycle was discovered by
[CBSE AIPMT 1994]
(a) Hatch and Slack (b) Calvin
(c) Hill (d) Arnon
Ans.(a)
C
4
pathway or dicarboxylic acid pathway
is an alternative path ofCO
2
-fixation in
photosynthesis. It was discovered by
MD Hatch and CR Slack in 1967, so also
known as Hatch- Slack cycle.
57The carbon dioxide acceptor
in Calvin cycle/C
3
-plants is
[CBSE AIPMT 1993, 95, 96, 99]
(a) Phosphoenol Pyruvate (PEP)
(b) Ribulose 1,5-Diphosphate (RuDP)
(c) Phosphoglyceric Acid (PGA)
(d) Ribulose Monophosphate (RMP)
Ans.(b)
InC
3
-plants,CO
2
combines with
ribulose biphosphate (acceptor
molecule) in the presence of RuBisCO
(RuBP carboxylase) and form two
molecules of 3-Phosphoglyceric acid or
PGA (first stable product).
58Which one is aC
4
-plant?
[CBSE AIPMT 1992]
(a) Papaya (b) Pea
(c) Potato (d) Maize/Corn
Ans.(d)
The plants in which the first stable
product of dark reaction of
photosynthesis is a 4-carbon
compound are calledC
4
-plants, e.g.
sugarcane, maize, sorghum, etc. These
plants show characteristic Kranz
anatomy. The firstCO
2
acceptor in
these plants is Phosphoenol Pyruvate
(PEP).
59The enzyme that catalyses initial
carbon dioxide fixation in
C
4
-plants is
[CBSE AIPMT 1992, 2002]
(a) RuBP carboxylase
(b) PEP carboxylase
(c) carbonic anhydrase
(d) carboxydismutase
Ans.(b)
InC
4
-plants, mesophyll cells fix carbon
dioxide with the help of
phosphoenol-pyruvate (the first
acceptor) in the presence of PEP
carboxylase to a compound oxaloacetic
acid (first product).
60Dark reactions of photosynthesis
occur in [CBSE AIPMT 1991]
(a) granal thylakoid membranes
(b) stromal lamella membranes
(c) stroma outside photosynthetic
lamellae
(d) periplastidial space
Ans.(c)
Light reaction of photosynthesis
occurs in granal thylakoid membranes
of chloroplast while dark reaction
occurs in the stroma or matrix, i.e.
outside the photosynthetic lamellae of
chloroplast.
61Which technique has helped in
investigation of Calvin cycle?
[CBSE AIPMT 1991]
(a) X-ray crystallography
(b) X-ray technique
(c) Radioactive isotope technique
(d) Intermittent light
Ans.(c)
Calvin, Benson and Basshan utilisedC
14
(with long life) to trace the path of
carbon in photosynthesis. Calvin was
awarded Nobel Prize in 1961 in
recognition to his work withC
14
isotope.
He discovered the cycle involved in
carbon assimilation, known as Calvin
cycle orC
3
-cycle.
62Kranz anatomy is typical of
[CBSE AIPMT 1990, 95]
(a)C
4
-plants (b)C
3
-plants
(c)C
2
-plants (d) CAM plants
Ans.(a)
Leaves ofC
4
-plants (e.g. sugarcane,
maize) are characterised by Kranz
anatomy in which the mesophyll is
undifferentiated and its cells occur in
concentric layers around vascular
bundles.
Vascular bundles are surrounded by
large sized bundle sheath cells which
are arranged in a wreath-like manner
(Kranz- wreath).
63The first carbon dioxide acceptor
inC
4
-plants is
[CBSE AIPMT 1990, 92]
(a) phosphoenol-pyruvate
(b) ribulose 1,5-diphosphate
(c) oxalo acetic acid
(d) phosphoglyceric acid
Ans.(a)
InC
4
-plants, phosphoenol-pyruvate is
the first acceptor ofCO
2
while ribulose
bi-phosphate is the second acceptor.
Oxalo Acetic Acid (OAA) is the first
product ofC
4
-cycle.
64InC
4
-plants, Calvin cycle operates
in [CBSE AIPMT 1989]
(a) stroma of bundle sheath
chloroplasts
(b) grana of bundle sheath
chloroplasts
(c) grana of mesophyll chloroplasts
(d) stroma of mesophyll chloroplasts
Ans.(a)
C
4
-plants possess two types of
chloroplasts granal in mesophyll cells
and agranal in bundle sheath cells.
Mesophyll cells are specialised to
perform light reaction and bundle
sheath cells possess RuBisCO, hereCO
2
is fixed through Calvin cycle.
65A very efficient converter of solar
energy with net productivity of 2-4
kg/m
2
or more is the crop of
[CBSE AIPMT 1989]
(a) wheat (b) sugarcane
(c) rice (d) bajra
Ans.(b)
InC
4
-plants, (e.g. maize, sugarcane,
sorghum) optimum temperature of
photosynthesis is 30–45°C. InC
4
-plants,
rate of net photosynthesis in full
sunlight is (40–80 mgCO
2
/dm
2
/hr) which
is more than the rate of net
photosynthesis (15–35 mgCO
2
/dm
2
/hr)
at optimum sunlight inC
3
-plants.
Photosynthesis 133

144. 66Carbon dioxide joins the
photosynthetic pathway in
[CBSE AIPMT 1988]
(a) PS-I (b) PS-II
(c) light reaction (d) dark reaction
Ans.(d)
In dark reaction of photosynthesis,
reducing agent (NADPH) and source of
energy (ATP) formed during light
reaction, are utilised in the conversion
ofCO
2
to carbohydrates.
67During non-cyclic
photophosphorylation, when
electrons are lost from the
reaction centre at PS-II, what is
the source which replaces these
electrons? [NEET (Oct.) 2020]
(a) Oxygen (b) Water
(c) Carbon dioxide (d) Light
Ans.(b)
During non-cyclic
photophosphorylation, electrons
expelled by the excited PS-II
photocentre does not return to it.
Therefore, it requires an external
electron donor and that purpose is
served by water.
H O
2
undergo photolysis and the
electrons thus released are picked up
by PS-II(P )
680
and handed over to PS-I or
P
700
.
68The process which makes major
difference betweenC
3
and
C
4
-plants is[NEET 2016, Phase II]
(a) glycolysis
(b) Calvin cycle
(c) photorespiration
(d) respiration
Ans.(c)
Photorespiration is the process which
makes a difference between theC
3
and
C
4
-plants. In this process, there is a
continuous loss of carbon fixed in the
form ofCO
2
.
It occurs due to the highO
2
content,
high temperature conditions in which
RuBP carboxylase starts working as
RuBP oxygenate and normal
photosynthesis does not occur.
69The correct sequence of cell
organelles during photorespiration
is [CBSE AIPMT 2012]
(a) chloroplast–Golgi
bodies–mitochondria
(b) chloroplast-rough endoplasmic
reticulum- dictyosomes
(c) chloroplast–mitochondria
–peroxisome
(d) chloroplast–vacuole–peroxisome
Ans.(c)
None of the option is correct.
Photorespiration required three cell
organelles in sequence of chloroplast,
peroxisome and mitochondria.
Option (c) may be correct if be read as
said sequence.
70During photorespiration, the
oxygen consuming reaction(s)
occur in [CBSE AIPMT 2006]
(a) stroma of chloroplasts and
peroxisomes
(b) grana of chloroplasts and
peroxisomes
(c) stroma of chloroplasts
(d) stroma of chloroplasts and
mitochondria
Ans.(a)
The first reaction of photorespiration
occurs in stroma of chloroplast. In this
reaction the RuBP (Ribulose 1-5
biphosphate) consumes one oxygen
molecule in presence of enzyme
RuBisCO.
In peroxisome the glycolate transferred
from chloroplast takes upO
2
and
formed the glyoxylate whereas, the
H O
2 2
released as byproduct.
71Which one of the following is
wrong in relation to
photorespiration?
[CBSE AIPMT 2003]
(a) It is a characteristic ofC
3
-plants
(b) It occurs in chloroplasts
(c) It occurs in day time only
(d) It is a characteristic ofC
4
-plants
Ans.(d)
Dicker andTio(1959) discovered
photorespiration in tobacco plant. It is a
light dependent process of oxygenation
of Ribulose Bisphosphate (RuBP).
During this processCO
2
is liberated and
O
2
is consumed.C
4
-plants avoid
photorespiration by following Hatch
Slack pathway.
72Which enzyme is most abundantly
found on earth?[CBSE AIPMT 1999]
(a) Catalase (b) RuBisCO
(c) Nitrogenase (d) Invertase
Ans.(b)
RuBisCO (RuBP carboxylase) is the most
abundant protein on this planet.
RuBisCO constitutes 16% of chloroplast
protein. It is required forCO
2
fixation
with RuBP (Ribulose Biphosphate) in
Calvin cycle.
73Photorespiration is favoured by
[CBSE AIPMT 1996]
(a) highO
2
and lowCO
2
(b) low light and highO
2
(c) low temperature and highO
2
(d) lowO
2
and highCO
2
Ans.(a)
Photorespiration is light induced
oxidation of photosynthetic
intermediates with the help of oxygen.
It is stimulated by highO
2
concentration
or lowCO
2
, high light intensity, high
temperature and ageing of leaf.
74The substrate for photorespiration
is [CBSE AIPMT 1989]
(a) ribulose bis-phosphate
(b) glycolate
(c) serine
(d) glycine
Ans.(b)
Photorespiration is the oxidation of
photosynthetic intermediate without
production ofCO
2
, ATP andNADH
2
. The
substrate for photorespiration is a
2-carbon compound glycolic acid
(glycolate).
75With reference to factors
affecting the rate of
photosynthesis, which of the
following statements is not
correct? [NEET 2017]
(a) Light saturation forCO
2
-fixation
occurs at10%of full sunlight
(b) Increasing atmosphericCO
2
concentration upto 0.05% can
enhanceCO
2
-fixation rate
134 NEETChapterwise Topicwise Biology
Photorespiration
TOPIC 4
Factors Affecting
Photosynthesis
TOPIC 5

145. (c)C
3
-plants respond to higher
temperature with enhanced
photosynthesis, whileC
4
-plants
have much lower temperature
optimum
(d) Tomato is a greenhouse crop,
which can be grown inCO
2
enriched atmosphere for higher
yield
Ans.(c)
InC
4
-plants, the initial fixation ofCO
2
occurs in mesophyll cells. The primary
acceptor ofCO
2
is Phosphoenol
Pyruvate (PEP). It combines with CO2in
the presence of enzyme PEP
carboxylase to form the first stable
product, i.e. Oxalo Acetic Acid (OAA).
Where asC
3
-plants lackPEPcarboxylase
enzyme. They possess RuBisCO
enzyme. This enzyme can work as both
carboxylase (fixation ofCO
2
) and
oxygenase (fixation ofO
2
). RuBisCO has
a much greater affinity forCO
2
than for
O
2
and the binding is competitive. At
higher temperature, its affinity forCO
2
decrease and it works as oxygenase.
Therefore, at higher temperature
photosynthesis decrease inC
3
-plants,
while inC
4
-plants it increases.
76Emerson’s enhancement effect
and red drop have been
instrumental in the discovery of
[NEET 2016, Phase I]
(a) two photosystems operating
simultaneously
(b) photophosphorylation and cyclic
electron transport
(c) oxidative phosphorylation
(d) photophosphorylation and
non-cyclic electron transport
Ans.(a)
Emerson performed photosynthetic
experiment onChlorella. He provided
monochromatic light of more than 680
nm and observed decrease in rate of
photosynthesis known as red drop.
Later, he provided synchronised light of
680 nm and 700 nm and observed
increase in rate of photosynthesis,
known as enhancement effect. This
experiment led to discovery of two
photosystems –PS-I and PS-II operating
in photosynthesis.
77The oxygen evolved during
photosynthesis comes from water
molecules. Which one of the
following pairs of elements
involved in this reaction?
[NEET 2016, Phase I]
(a) Manganese and chlorine
(b) Manganese and potassium
(c) Magnesium and molybdenum
(d) Magnesium and chlorine
Ans.(a)
Photolysis of water during
photosynthesis evolve nascent oxygen
in the presence of manganese, calcium
and chloride ions.
78Plants adapted to low light
intensity have[CBSE AIPMT 2004]
(a) larger photosynthetic unit size
than the sun plants
(b) higher rate ofCO
2
fixation than the
sun plants
(c) more extended root system
(d) leaves modified to spines
Ans.(a)
Shade tolerant plants have lower
photosynthetic rates and hence, lower
growth rates. On the other hand, these
plants have larger photosynthetic unit
size than the sun plants.
79The principle of limiting factors
was proposed by
[CBSE AIPMT 1996]
(a) Blackmann (b) Hill
(c) Arnon (d) Liebig
Ans.(a)
The principle of limiting factors was
given by Blackmann, a British plant
physiologist in 1905, according to him,
light intensity, carbon dioxide
concentration and temperature are the
limiting factors in photosynthesis.
When a process is conditioned as to its
rapidity by a number of separate
factors, the rate of the process is
limited by the pace of the slowest
factor.
80Photosynthetically active radiation
is represented by the range of
wavelength
[CBSE AIPMT 1996, 2004, 05]
(a) 340-450 nm (b) 400-700 nm
(c) 500-600 nm (d) 400-950 nm
Ans.(b)
Photosynthetically Active Region (PAR)
of solar radiation is visible region. It
consists of radiations having
wavelength betwen 400 to 700 nm.
Green plants use this wavelength in the
process of manufacture of food, i.e.
photosynthesis.
81A photosynthesising plant is
releasing
18
Omore than the
normal. The plant must have been
supplied with[CBSE AIPMT 1993]
(a)O
3
(b)H O
2
with
18
O
(c)CO
2
with
18
O
(d)C H O
6 12 6
with
18
O
Ans.(b)
Ruben, Hassid and Kamen (1941) using
heavy isotope of oxygen,O
18
in water,
found that oxygen evolved in
photosynthesis comes from water.
Evolution of oxygen does not require
carbon dioxide.
82At a temperature above 35°C
[CBSE AIPMT 1992]
(a) rate of photosynthesis will decline
earlier than that of respiration
(b) rate of respiration will decline
earlier than that of
photosynthesis
(c) there is no fixed pattern
(d) both decline simultaneously
Ans.(a)
Optimum temperature for
photosynthesis is 10°-25°C for
C
3
-plants and 30°-45°C forC
4
-plants.
Optimum temperature for respiration is
20°C–30°C, i.e. respiration has a higher
temperature optimum than
photosynthesis and thus declines later.
83During monsoon, the rice crop of
Eastern states of India shows
lesser yield due to limiting factor
of [CBSE AIPMT 1991]
(a)CO
2
(b) light
(c) temperature (d) water
Ans.(b)
According to the principle of limiting
factor, the rate of the process is limited
by the pace of the slowest factor. Light
intensity varies with latitude, altitude,
season, topography, presence or
absence of interceptors like cloud,
dust, fog, humidity, etc. In Eastern
states, low light intensity during
monsoon results in low photosynthesis
and hence, lesser yield.
Photosynthesis 135

146. 01Conversion of glucose to
glucose-6-phosphate, the first
irreversible reaction of glycolysis, is
catalysed by[NEET (National) 2019]
(a) hexokinase
(b) enolase
(c) phosphofructokinase
(d) aldolase
Ans.(a)
Conversion of glucose to
glucose-6-phosphate during glycolysis
is catalysed by the enzyme hexokinase.
During this step, glucose is
phosphorylated to glucose-6-phosphate
by ATP. It is the first step of activation
phase of glycolysis.
02Which of the following
biomolecules is common to
respiration-mediated breakdown
of fats, carbohydrates and
proteins? [NEET 2016, Phase II]
(a) Glucose-6-phosphate
(b) Fructose 1,6-bisphosphate
(c) Pyruvic acid
(d) Acetyl Co-A
Ans.(d)
Carbohydrates, fats and proteins, all
can be used as a substrate in cellular
respiration. All of them first get
converted to acetyl Co-A to enter into
Krebs’ cycle of aerobic cellular
respiration. Thus, it is the common
factor of respiration entering Krebs’
cycle after breakdown of
carbohydrates, fats and proteins.
03In glycolysis, during oxidation
electrons are removed by
[CBSE AIPMT 2004]
(a) ATP
(b) glyceraldehyde-3-phosphate
(c)NAD
+
(d) molecular oxygen
Ans.(c)
When 3-phosphoglyceraldehyde is
converted into 1,3 diphosphoglyceric
acid, two electrons and two protons are
released which are utilised to convert
NAD
+
to NADH and one H
+
.
NAD + 2H + 2e NADH + H
+ + + −
→ .
04In animal cells, the first stage of
glucose breakdown is
[CBSE AIPMT 1994]
(a) Krebs’ cycle
(b) glycolysis
(c) oxidative phosphorylation
(d) ETC
Ans.(b)
Glycolysis refers to the sequence of
reactions by which glucose is
degraded anaerobically into pyruvic
acid inside cytoplasm. The net gain of
molecules of ATP during glycolysis is 2.
05End product of glycolysis is
[CBSE AIPMT 1990]
(a) acetyl Co-A
(b) pyruvic acid
(c) glucose 1-phosphate
(d) fructose 1-phosphate
Ans.(b)
During glycolysis one molecule of
glucose is broken down into two
molecules of pyruvic acid in the
cytoplasm. Glycolysis is the common
path of aerobic and anaerobic
respiration.
06EMP can produce a total of
[CBSE AIPMT 1990]
(a) 6 ATP (b) 8 ATP
(c) 24 ATP (d) 38 ATP
Ans.(b)
EMP (Embden-Meyerhof Paranas
Pathway) refers to glycolysis, in which
one molecule of glucose is broken down
to two molecules of pyruvic acid. In this
process there is a gain of 2NADH
(2×3 ATP) and 2 ATP, i.e. total 8 ATP.
07Incomplete oxidation of glucose
into pyruvic acid with several
intermediate steps is known as
[CBSE AIPMT 1988]
(a) TCA-pathway (b) glycolysis
(c) HMS-pathway (d) Krebs’ cycle
Ans.(b)
Glycolysis is the sequence of enzyme
mediated reactions by which glucose is
degraded anaerobically into pyruvic acid
in cell cytoplasm. The net gain of
molecules of ATP during glycolysis is two.
08Which of the following statement
is incorrect? [NEET 2021]
(a) During aerobic respiration, role of
oxygen is limited to the terminal
stage
(b) In ETC (Electron Transport Chain),
one molecule of NADH + H
+
gives
rise to 2 ATP molecules and one
FADH
2
gives rise to 3 ATP molecules
(c) ATP is synthesised through
complex V
(d) Oxidation-reduction reactions
produce proton gradient in
respiration
Ans.(b)
Oxidation of one molecule of NADH
gives rise to 3 molecules of ATP and
one molecule of FADH
2
produces 2
molecules of ATP.
RespirationinPlants
14
Glycolysis
TOPIC 1
Krebs Cycle and Electron
Transport Chain
TOPIC 2

147. Respiration in Plants 137
NADH and FADH
2
are two different
types of electron donors. They differ in
the ways they feed electron during
electron transport chain. NADH feeds
its electrons into the electron transport
chain at the beginning (Complex I).
FADH
2
feeds into the electron
transport chain at Complex II (at a lower
energy level down the chain). The high
energy electrons from NADH have
sufficient energy to result in 3 ATP
whereas the lower energy electrons in
FADH
2
have energy for 2 ATP production.
09Pyruvate dehydrogenase activity
during aerobic respiration requires
[NEET (Oct.) 2020]
(a) calcium (b) iron
(c) cobalt (d) magnesium
Ans.(d)
Pyruvate dehydrogenase enzyme is
involved in the conversion of pyruvate
to acetyl Co-A, after the completion of
glycolysis and before the start of Krebs’
cycle. This enzyme is made up of
decarboxylase, lipoic acid,
transacetylase andMg
2+
ion. The
reaction occur in following way
Pyruvate+ +  → 
+
NAD Co- A
+ Pyruvate
dehydrogenase, Mg
2
Acetyl Co- A NADH H CO
2
+ + +
+
In this reactionMg
2+
acts as a
cofactor.
10The number of substrate level
phosphorylation in one turn of
citric acid cycle is
[NEET (Sep.) 2020]
(a) one (b) two (c) three (d) zero
Ans.(a)
The number of substrate level
phosphorylation in one turn of citric
acid cycle is 1. During Krebs’ or citric
acid cycle, succinyl-Co-A is acted upon
by enzyme succinyl-Co-A synthetase to
form succinate (a 4C compound). The
reaction releases sufficient energy to
form ATP (in plants) or GTP (in animals)
by substrate-level phosphorylation.
GTP can form ATP through a coupled
reaction. Succinyl Co-A+GDP/ADP+H
3
PO
4
synthetase. Succinyl Co-A
Succinate +Co-A+GTP/ATP.
11Where is respiratory Electron
Transport System (ETS) located in
plants? [NEET (Odisha) 2019]
(a) Mitochondrial matrix
(b) Outer mitochondrial membrane
(c) Inner mitochondrial membrane
(d) Intermembrane space
Ans.(c)
Respiratory Electron Transport System
(ETS) in plants is located in inner
mitochondrial membrane. It serves as
the site of oxidative phosphorylation
through the action of ATP synthase.
12Which one of these statements is
incorrect? [NEET 2018]
(a) Glycolysis operates as long as it is
supplied with NAD that can pick
up hydrogen atoms.
(b) Glycolysis occurs in cytosol
(c) Enzymes of TCA cycle are present
in mitochondrial matrix
(d) Oxidative phosphorylation takes
place in outer mitochondrial
membrane
Ans.(d)
Oxidative phosphorylation is the
process of ATP formation due to the
transfer of electrons from NADH or
FADH
2
to oxygen molecule(O )
2
by a
series of electron carriers. This process
occurs in theinner mitochondrial
membranebecause of its less
permeability, presence of ETC proteins
and ATP synthase.
The rest three statements are correct.
13What is the role ofNAD
+
in
cellular respiration?[NEET 2018]
(a) It is a nucleotide source of ATP
synthesis
(b) It functions as an electron carrier
(c) It functions as an enzyme
(d) It is the final electron acceptor for
anaerobic respiration
Ans.(b)
NAD
+
functions as anelectron carrier
in cellular respiration. NAD is an
oxidising agent which accept electrons
and then transfer them to the Electron
Transport System (ETS). As a result,
3ATP molecules are formed.
14Which statement is wrong for
Krebs’ cycle? [NEET 2017]
(a) There are three points in the cycle
whereNAD
+
is reduced to NADH
+
+
H
(b) There is one point in the cycle
where FAD
+
is reduced toFADH
2
(c) During conversion of succinyl
Co-A to succinic acid, a molecule
of GTP is synthesised
(d) The cycle starts with
condensation of acetyl group
(acetyl Co-A) with pyruvic acid to
yield citric acid
Ans.(d)
Option (d) is incorrect, which can be
corrected as
Krebs’ cycle starts with the
condensation of acetyl group with
oxaloacetic acid and water to yield
citric acid.
During conversion of succinic acid to
fumaric acid FAD
+
is reduced toFADH
2
.
During conversion of pyruvic acid to
acetylCo- A, isocitrate to oxalosuccinic
acid andα-ketoglutaric acid to succinyl
Co-ANAD
+
is reduced toNADPH+H
+
.
15Oxidative phosphorylation is
[NEET 2016, Phase II]
(a) formation of ATP by transfer of
phosphate group from a substrate
to ADP
(b) oxidation of phosphate group in ATP
(c) addition of phosphate group to ATP
(d) formation of ATP by energy
released from electrons removed
during substrate oxidation
Ans.(a)
Oxidative phosphorylation is the
process of formation of ATP from ADP
and inorganic phosphate(P)
i
in the
presence of oxygen. It occurs mainly in
the Electron Transport Chain (ETC) of
cellular respiration.
16Which of the metabolites is
common to respiration mediated
breakdown of fats, carbohydrates
and proteins? [NEET 2013]
(a) Glucose-6-phosphate
(b) Fructose 1, 6-bisphosphate
(c) Pyruvic acid
(d) Acetyl Co-A
Ans.(d)
Acetyl Co-A is common to respiration
mediated breakdown of fats,
carbohydrates and proteins.
Glucose and fructose are
phosphorylated to give rise to
glucose-6-phosphate by the activity of
the enzyme hexokinase.
Glucose-6-phosphate is then converted
into fructose-6-phosphate and further
to fructose 1-6-bisphosphate. Pyruvic
acid is the end product of glycolysis.
17The three boxes in this diagram
represent the three major
biosynthetic pathways in aerobic
respiration. Arrows represent net
reactants or products[NEET 2013]

148. Arrows numbered 4, 8 and 12 can
all be
(a) NADH (b) ATP
(c)H O
2
(d)FAD
+
orFADH
2
Ans.(b)
PathwayAis glycolysis, pathwayBis
the Krebs’ cycle and pathwayCis
oxidative phosphorylation
Arrow 1 — ADP or NAD
+
Arrow 2 — Pyruvate
Arrow 3 — NADH
Arrow 4 — ATP
Arrow 5 — ADP, NAD
+
or FAD
Arrow 6 and 7 — FADH
2
and NADH
(either one can be 6 or 7)
Arrow 8 — ATP or CO
2
Arrow 9 and 10 — O
2
and ADP (either
one can be 9 or 10)
Arrow 11 and 12 — H
2
O and ATP
(either onecan be 11 or 12)
18Aerobic respiratory pathway is
appropriately termed
[CBSE AIPMT 2009]
(a) catabolic (b) parabolic
(c) amphibolic (d) anabolic
Ans.(c)
Anamphibolic pathwayis a biochemical
pathway that serves both anabolic and
catabolic processes. An important
example of an amphibolic pathway is
the Krebs’ cycle, which involves both
the catabolism of carbohydrates and
fatty acid and the synthesis of anabolic
precursors for amino acid synthesis,eg,
α-ketogluturate and oxalo acetate.
19The chemiosmotic coupling
hypothesis of oxidative
phosphorylation proposes that
Adenosine Triphosphate (ATP) is
formed because
[CBSE AIPMT 2008]
(a) high energy bonds are formed in
mitochondrial proteins
(b) ADP is pumped out of the matrix
into the intermembrane space
(c) a proton gradient forms across
the inner membrane
(d) there is a change in the
permeability of the inner
mitochondrial membrane toward
Adenosine Diphosphate (ADP)
Ans.(c)
The production of ATP with the help of
energy liberated during oxidation of
reduced coenzymes and terminal
oxidation is called oxidative
phosphorylation.Peter Mitchell(1961)
gave a hypothesis known as chemiosmotic
hypothesis for ATP synthesis.
According to this when electrons flow
from dual proton, electron carrier to a
non-hydrogen carrier theH
+
are
released and expelled into the
intermembrane space and thus creates
a proton gradient with higher
concentration ofH
+
in the inter
membranous space than the matrix.
Due to the proton motive force the
protons flow back and energy liberated
during this back flow of protons
activate ATPase present inF
1
head to
synthesize ATP.
20The overall goal of glycolysis,
Krebs’ cycle and the electron
transport system is the formation
of [CBSE AIPMT 2007]
(a) ATP in small stepwise units
(b) ATP in one large oxidation reaction
(c) sugars
(d) nucleic acids
Ans.(a)
Glycolysis, Krebs’ cycle and electron
transport system are meant for ATP
synthesis in different steps. ATP is the
energy currency of cell.
21All enzymes of TCA cycle are
located in the mitochondrial
matrix except one which is located
in inner mitochondrial membranes
in eukaryotes and in cytosol in
prokaryotes. This enzyme is
[CBSE AIPMT 2007]
(a) lactate dehydrogenase
(b) isocitrate dehydrogenase
(c) malate dehydrogenase
(d) succinate dehydrogenase
Ans.(d)
Succinate dehydrogenase enzyme is
present on inner membrane of
mitochondria and catalyses the
oxidation of succinate to fumarate.
22During which stage, in the
complete oxidation of glucose are
the greatest number of ATP
molecules formed from ADP
[CBSE AIPMT 2005]
(a) glycolysis (b) Krebs’ cycle
(c) conversion of pyruvic acid to
acetyl Co-A
(d) electron transport chain
Ans.(d)
The last step of aerobic respiration is
the oxidation2 of reduced coenzymes,
i.e.,NADH
2
andFADH
2
by molecular
oxygen through FAD, ubiquinone, cyt.-b,
cyt.-c, cyt-c
1
, cyt.-aand cyt.-a
3
. By
oxidation of 1 molecule ofNADH
2
, 3 ATP
molecules are produced and by
oxidation of 1 molecule of FADH
2
2ATP
molecules are Produced.
In glycolysis 2 ATP molecules are
produced from ADP.
Further 2NADH
2
produced, give
2 3 6× =ATP, on oxidative
phosphorylation. Similarly in Krebs’
cycle 2 ATP molecules are produced.
So the greatest number of ATP
molecules are produced in the
electron transport chain.
23Chemiosmotic theory of ATP
synthesis in the chloroplast and
mitochondria is based on
[CBSE AIPMT 2005]
(a) membrane potential
(b) accumulation ofNa
+
ions
(c) accumulation ofK
+
ions
(d) proton gradient
Ans.(d)
Chemiosmotic hypothesis for oxidative
phosphorylation (ATP synthesis) was
proposed byPeter Mitchellin 1961, for
this he was awarded Nobel Prize in
1978. This theory is based on proton
gradient.
24Which one of the following
concerns photophosphorylation?
[CBSE AIPMT 2003]
(a) AMP + inorganicPO
4
→
Light energy
ATP
(b) ADP + AMP→
Light energy
ATP
(c) ADP + inorganicPO
4
→
Light energy
ATP
(d) ADP + inorganicPO ATP
4
→
Ans.(c)
Phosphorylation refers to the process
in which ATP is made when energy is
used to bind another phosphate to ADP.
Photophosphorylation reactions are
part of both respiration and
photosynthesis.
138 NEETChapterwise Topicwise Biology
Glucose
Pathway A
1
Pathway
B
Pathway C11
5
2
7
6
10
9
4
8
12
3

149. 25In which one of the following do
the two names refer to one and the
same thing?[CBSE AIPMT 2003]
(a) Tricarboxylic acid cycle and urea
cycle
(b) Krebs’ cycle and Calvin cycle
(c) Tricarboxylic acid cycle and citric
acid cycle
(d) Citric acid cycle and Calvin cycle
Ans.(c)
Tricarboxylic acid cycle is also known as
citric acid cycle. This is an aerobic
process, that takes place in the matrix
of mitochondria. Kreb that discovered
this cycle in 1937. So, this is also known
as Krebs’ cycle.
26The mechanism of ATP formation
both in chloroplast and
mitochondria is explained by
[CBSE AIPMT 1997]
(a) relay pump theory of Godlewski
(b) Munch’s pressure/mass flow model
(c) chemiosmotic theory of Mitchell
(d) Cholondy-Went’s model
Ans.(c)
In chemiosmotic- coupling hypothesis,
outward pumping of protons across the
inner chloroplast or mitochondrial
membrane results in accumulation of
protons between outer membrane and
inner membrane. A proton gradient is
thus established. As protons now flow
back passively down the gradient, the
proton motive force is utilised to
synthesise ATP.
27In Krebs’ cycle FAD participates as
electron acceptor during the
conversion of[CBSE AIPMT 1997]
(a) succinyl Co-A to succinic acid
(b)α-ketoglutarate to succinyl Co-A
(c) succinic acid to fumaric acid
(d) fumaric acid to malic acid
Ans.(c)
Electrons and H-ions during oxidation
of succinic acid to fumaric acid, are
taken up by FAD which gets reduced to
FADH
2
.
28Oxidative phosphorylation involves
simultaneous oxidation and
phosphorylation to finally form
[CBSE AIPMT 1996]
(a) pyruvate (b) NADP
(c) DPN (d) ATP
Ans.(d)
In oxidative phosphorylation ATP is
formed as the electrons are transferred
from NADH or FADH
2
toO
2
by a series
of electron carriers, located in the inner
membrane of mitochondria.
29Krebs’ cycle occurs in
[CBSE AIPMT 1996]
(a) mitochondria
(b) cytoplasm
(c) chloroplast
(d) ribosomes
Ans.(a)
Krebs’ cycle occurs inside the matrix of
mitochondria. The cycle is also named
ascitric acid cylceortricarboxylic
acid cycle.It includes stepwise
oxidative and cyclic degradation of
activated acetate derived from pyruvic
acid.
30Which of the following is essential
for conversion of pyruvic acid into
acetyl Co-A?[CBSE AIPMT 1995]
(a) LAA (b) NAD
+
(c) TPP (d) All of these
Ans.(d)
The oxidative decarboxylation of
pyruvate into acetyl Co-A involves the
presence of atleast five essential
co-factors and an enzyme complex. The
co-factors involved are Mg ions,
Thiamine Pyrophosphate (TPP), NAD
+
,
Coenzyme-A (Co-A) and lipoic acid.
31ATP is injected in cyanide
poisoning because it is
[CBSE AIPMT 1994]
(a) necessary for cellular functions
(b) necessary forNa – K
+ +
pump
(c)Na – K
+ +
pump operates at the
cell membranes
(d) ATP breaks down cyanide
Ans.(a)
Cyanide is a deadly poison. It stops
respiration by inhibiting electron flow
from cyt.-btocyt.-c
1
. ATP is the energy
currency of cell is injected in cyanide
poisoning because, it is necessary for
cellular functions.
32Out of 38 ATP molecules produced
per glucose, 32 ATP molecules are
formed fromNADH / FADH
2
in
[CBSE AIPMT 1993]
(a) respiratory chain
(b) Krebs’ cycle
(c) oxidative decarboxylation
(d) EMP
Ans.(a)
Respiratory chain helps in forming 32
ATP molecules from NADH/FADH
2
molecules. In which Oxidative
phosphorylation is the synthesis of
energy rich ATP molecules with the help
of energy liberated during oxidation of
reduced coenzymes (NADH
2
,FADH
2
).
produced in glycolysis and Krebs’ cycle.
A total of10NADH
2
and2FADH
2
molecules are formed in aerobic
respiration. They help in formation of 32
or 34ATP molecules.
33End product of citric acid/Krebs’
cycle is [CBSE AIPMT 1993]
(a) citric acid (b) lactic acid
(c) pyruvic acid (d)CO H O
2 2
+
Ans.(d)
Krebs’ cycle or citric acid cycle that
takes place in the matrix of
mitochondrion begins by linking acetyl
Co-A to oxaloacetic acid forming citric
acid. In the presence of various
enzymes, cycle continues through the
formation of various intermediates and
release ofCO
2
andH O
2
as end-products.
34Link between glycolysis, Krebs’
cycle andβ-oxidation of fatty acid
or carbohydrate and fat
metabolism is
[CBSE AIPMT 1992, 90]
(a) oxaloacetic acid (b) succinic acid
(c) citric acid (d) acetyl Co-A
Ans.(d)
The pyruvic acid synthesised from
glycolysis enters into mitochondria and
undergoes oxidative decarboxylation to
produceCO
2
andNADH
2
.
The product combines with
coenzyme-A to form acetyl Co-A. It is
the connecting link between glycolysis,
Krebs’ cycle and fat oxidation.
35Oxidative phosphorylation is
production of[CBSE AIPMT 1992]
(a) ATP in photosynthesis
(b) NADPH in photosynthesis
(c) ATP in respiration
(d) NADH in respiration
Respiration in Plants 139
H
|
C—C
|
H
Succinate
dehydrogenase
+ FAD
2+
C—
O
–
O
H
|
C—
|
H
O
–
O
Succinic acid
H
|
C—CC—
O
–
O
H
|
C
O
–
O
Fumaric acid
+ FADH
2

150. Ans.(c)
Oxidative phosphorylation is the
synthesis of ATP from ADP and
inorganic phosphate which occurs with
the help of energy obtained from
oxidation of reduced coenzymes
formed in cellular respiration.
36Terminal cytochrome of
respiratory chain which donates
electrons to oxygen is
[CBSE AIPMT 1992]
(a) cyt-b (b) cyt-c
(c) cyt-a
1
(d) cyt-a
3
Ans.(d)
The ETS system contains various
electron carriers such as
cytochromes. The correct sequence of
electron carrier/acceptor in ATP
synthesis is cyt-b, cyt-c
1
, cyt-c, cyt
( and cyt -
3
a a ). Cyt-a
3
is the terminal
cytochrome, it possess two copper
centers, which help in transfer of
electron to oxygen.
37NADP
+
is reduced to NADPH in
[CBSE AIPMT 1988]
(a) HMP (b) Calvin cycle
(c) glycolysis (d) EMP
Ans.(a)
Pentose Phosphate Pathway (PPP) or
Hexose Monophosphate Shunt (HMP) or
phosphogluconate pathway occurs in
the cytosol of mammalian cells. It
involves oxidation of glucose toCO
2
and water through a series of reactions
in which NADP is reduced to NADPH.
Complete breakdown of one molecule
of glucose forms 12 NADPH equal to
36 ATP molecules.
38Respiratory Quotient (RQ) value of
tripalmitin is[NEET (National) 2019]
(a) 0.7 (b) 0.07
(c) 0.09 (d) 0.9
Ans.(a)
The RQ value of tripalmitin is 0.7. It can
be calculated as follows
Respiratory Quotient (RQ)
=
Amount of CO released
Amount of O Consumed
2
2
2 (C H O ) 145 O 102CO 98 H O
5198 6 2 2 2
+ → +
Tripalmitin
RQ= =
102 CO
1450 O
0.7
2
2
It is to note that RQ of common fats is
usually less than 1 under aerobic
conditions.
39How many ATP molecules could
maximally be generated from one
molecule of glucose, if the
complete oxidation of one mole of
glucose toCO
2
andH O
2
yields 686
kcal and the useful chemical
energy available in the high energy
phosphate bond of one mole of
ATP is 12 kcal?[CBSE AIPMT 2006]
(a) 30 (b) 57 (c) 1 (d) 2
Ans.(b)
One mole of ATP liberates 12 kcal of
energy. so 686 kcal will be liberated by
686 12 57 1/ .=ATP molecules.
40How many ATP molecules are
produced by aerobic oxidation of
one molecule of glucose?
[CBSE AIPMT 2002]
(a) 2 (b) 4 (c) 38 (d) 34
Ans.(c)
A total of 38 molecules of ATP are
produced during aerobic respiration of
one molecule of glucose
Summary of ATP synthesis
8 ATP from glycolysis.
6 ATP from acetyl Co-A.
24 ATP from Krebs’ cycle.
Total=38ATP from aerobic oxidation of
one molecule of glucose.
41Net gain of ATP molecules during
aerobic respiration is
[CBSE AIPMT 1999]
(a) 36 molecules (b) 38 molecules
(c) 40 molecules (d) 48 molecules
Ans.(b)
During aerobic respiration, 38 ATP
molecules are gained. If specifically
aerobic respiration in eukaryote is
asked, then the answer would be 36 ATP
because 2 ATP molecules are produced
byFADH
2
which accepts theH
+
from 2
NADH molecules produced in glycolysis.
42Respiratory quotient (RQ) for fatty
acid is [CBSE AIPMT 1995]
(a) > 1 (b) < 1
(c) 1 (d) 0
Ans.(b)
Respiratory Quotient (RQ)
=
Volume of CO formed
Volume of O utilised
2
2
In fats, large amount ofO
2
is used to
combine withH ,
2
so output ofCO
2
is
less and RQ is only 0.70, i.e., less than
unity.
43Respiratory substrate yielding
maximum number of ATP
molecule is[CBSE AIPMT 1994]
(a) ketogenic amino acids (b) glucose
(c) amylose (d) glycogen
Ans.(b)
Respiratory substrate yielding maximum
number of ATP molecules is glucose.
One glucose molecule on aerobic
respiration yields 36 ATP molecules.
44Maximum amount of energy/ATP
is liberated on oxidation of
[CBSE AIPMT 1994]
(a) fats (b) proteins
(c) starch (d) vitamins
Ans.(a)
Fats or lipids are second to
carbohydrates as a source of energy.
By weight, each gram mol of fat yields
about 9.3 kcal of energy, i.e. more than
double of that yielded by glucose.
45Apparatus to measure rate of
respiration and RQ is
[CBSE AIPMT 1992]
(a) auxanometer (b) potometer
(c) respirometer (d) manometer
Ans.(c)
Respirometeris an instrument used to
measure the rate of respiration and also
Respiratory Quotient (RQ). The most
common respirometer is Ganong’s
respirometer.
46When one glucose molecule is
completely oxidised, it changes
[CBSE AIPMT 1992]
(a) 36 ADP molecules into 36 ATP
molecules
(b) 38 ADP molecules into 38 ATP
molecules
(c) 30 ADP molecules into 30 ATP
molecules
(d) 32 ADP molecules into 32 ATP
molecules
Ans.(b)
In aerobic respiration or biological
oxidation of one glucose molecule, 38
140 NEETChapterwise Topicwise Biology
Respiratory Quotient and
Respiratory Balance Sheet
TOPIC 3

151. ADP molecules change into 38 ATP
molecules, where donor phosphate is
inorganic phosphate. ATP molecules
are the energy currency of the cell, i.e.
the common immediate source of
energy in cellular activity.
47Which one of the following
statements about cytochrome 450
is wrong? [CBSE AIPMT 1999]
(a) It contains iron
(b) It is a coloured cell
(c) It has an important role in
metabolism
(d) It is an enzyme involved in
oxidation reactions
Ans.(b)
Cytochrome is not a coloured cell,
instead this is a respiratory
pigment-mixture of iron and protein
which are electron acceptors.
Cytochrome are membrane bound
hemeproteins contains heme groups
and are primarily responsible for the
generation of ATPviaelectron
transport.
48RQ is [CBSE AIPMT 1988]
(a) C/N (b) N/C (c)CO
2
/O
2
(d)O
2
/CO
2
Ans.(c)
Respiratory Quotient (RQ) is the ratio of
volume ofCO
2
evolved to the volume of
oxygen consumed per unit time per
unit weight. Therefore, RQ =CO
2
/O
2
.
It is useful in knowing the type of
respiration, major transformations and
respiratory substrate.
49In which one of the following
processesCO
2
is not released?
[CBSE AIPMT 2014]
(a) Aerobic respiration in plants
(b) Aerobic respiration in animals
(c) Alcoholic fermentation
(d) Lactate fermentation
Ans.(d)
Lactic acid fermentation is process by
which glucose, fructose and sucrose
are converted into energy and the
metabolite lactate. It is an anaerobic
fermentation reaction that occurs in
some bacteria and animal cells and
allows glycolysis to continue by
ensuring that NADH is returned to its
oxidised state(NAD )
+
.
50The energy-releasing metabolic
process in which substrate is
oxidised without an external
electron acceptor is called
[CBSE AIPMT 2010, 08]
(a) glycolysis
(b) fermentation
(c) aerobic respiration
(d) photorespiration
Ans.(b)
Fermentation takes place in the lack of
oxygen (when the electron transport
chain is unusable) and becomes the
cell’s primary means of ATP (energy)
production. It turns NADH and pyruvate
in the glycolysis intoNAD
+
and various
small molecules depending on the type
of fermentation. In the presence ofO
2
,
NADH and pyruvate are used to
generate ATP in respiration. It is called
oxidative respiration.
51In alcoholic fermentation
[CBSE AIPMT 2003]
(a) oxygen is the electron acceptor
(b) triose phosphate is the electron
donor while acetaldehyde is the
electron acceptor
(c) triose phosphate is the electron
donor while pyruvic acid is the
electron acceptor
(d) there is no electron donor
Ans.(b)
Inalcoholic fermentation,
(a) NADH (formed during conversion
of triose-3 phosphate to 3
phosphoglycerate) is oxidised to
NAD
+
(b) electrons are accepted by
acetaldehyde formed by
decarboxylation of pyruvate.
52Fermentation is anaerobic
production of[CBSE AIPMT 1996]
(a) Protein and acetic acid
(b) alcohol, lactic acid or similar
compounds
(c) ethers and acetones
(d) alcohol and lipoproteins
Ans.(b)
Fermentation is defined as anaerobic
break down of carbohydrates and other
organic compounds to form aldehyde,
alcohol and organic acids (lactic acid)
with the help of microorganisms or
their enzymes.
53Fermentation products of yeast
are
[CBSE AIPMT 1994]
(a)H O + CO
2 2
(b) methyl alcohol+CO
2
(c) methyl alcohol+H
2
O
(d) ethyl alcohol+CO
2
Ans.(d)
Yeast cells undergo alcoholic
fermentation in which glucose is first
converted into pyruvic acid. In the
presence of pyruvic decarboxylase, it is
changed into acetaldehyde. Alcohol
dehydrogenase changes it to ethyl
alcohol andCO
2
.
54Life without air would be
[CBSE AIPMT 1993]
(a) reductional
(b) free from oxidative damage
(c) impossible
(d) anaerobic
Ans.(d)
Anaerobic means ‘in the absence of
molecular oxygen’, so life without air
would be anaerobic. The atmosphere of
earth at the time of origin of life was
without free oxygen atoms, so the
primitive atmosphere was reducing.
55Out of 36 ATP molecules produced
per glucose molecule during
respiration[CBSE AIPMT 1991]
(a) 2 are produced outside glycolysis
and 34 during respiratory chain
(b) 2 are produced outside
mitochondria and 34 inside
mitochondria
(c) 2 during glycolysis and 34 during
Krebs’ cycle
(d) all are formed inside mitochondria
Ans.(b)
A total of 38 ATP molecules are
produced per glucose molecule during
respiration. Out of which, 2 ATP are
produced outside mitochondria
(i.e. glycolysis in cytoplasm) and 36 ATP
inside mitochondria (i.e. 2 ATP through
Krebs’ cycle and 34 ATP from
NADH/FADH
2
through respiratory
chain). In contrast, in some cells the
number of ATP produced inside
mitochondria equals to 34 and thus,
there is a net synthesis of 36 ATP
molecules.
Respiration in Plants 141
Fermentation
TOPIC 4
Glucose
Glycolysis
2 pyruvate
P +ADP+2P
+
ATP + P P P
NAD
2NADH
+
2 lactate
+2H
+

152. 01The plant hormone used to
destroy weeds in a field is
[NEET 2021]
(a) IAA (b) NAA
(c) 2,4-D (d) IBA
Ans.(c)
2, 4 D hormone is used as a herbicide to
destroy weeds.
Auxins like IAA and IBA are used to
induce parthenocarpy. IAA also
stimulate nodule formation. Auxin like
NAA is used to increase dwarf shoots.
02Plants follow different pathways in
response to environment or
phases of life to form different
kinds of structures. This ability is
called [NEET 2021]
(a) elasticity
(b) flexibility
(c) plasticity
(d) maturity
Ans.(c)
Plant plasticity refers to the ability to
modify itself by forming different kind
of structures to adapt and cope with
changes in its environment. It can be
intrinsic plasticity or extrinsic plasticity.
In both the cases plants shows
heterophylly along with other
morphological features, e.g. in the
leaves Larkspur and buttercup.
03Inhibitory substances in dormant
seeds cannot be removed by
subjecting seeds to
[NEET (Oct.) 2020]
(a) gibberellic acid
(b) nitrate
(c) ascorbic acid
(d) chilling conditions
Ans.(c)
Presence of inhibitory substances in
dormant seeds can be removed by
subjecting seeds to
(i) growth hormones like gibberellic
acid, cytokinins,
(ii) by stratification in which seeds
requiring low temperature are first
allowed to imbibe water and then
exposed to low temperature.
(iii) low concentration application of
nitrates because it promote seed
germination.
Ascorbic acid is known to cause seed
dormancy as it is a potent chemical
inhibitor. Thus, option (c) is incorrect.
04Match the following concerning
the activity/ function and the
phytohormone involved.
[NEET (Oct.) 2020]
Column I Column II
A. Fruit ripener (i) Abscisic acid
B. Herbicide (ii) GA
3
C. Bolting agent (iii) 2, 4-D
D. Stress hormone (iv) Ethephone
Select the correct option.
A B C D
(a) (ii), (iii), (iv), (i)
(b) (iii), (iv), (ii), (i)
(c) (iv), (iii), (ii), (i)
(d) (iv), (ii), (i), (iii)
Ans.(c)
Options (c) is the correct match which is
as follows
The function of different
phytohormones include ethephon is a
commercial derivative of ethylene and
it is used to ripen fruits. 2, 4-D is a
synthetic auxin and it is used as
herbicide against dicotyledonous
weeds. Gibberellic acid induces bolting
in plants, i.e. the promotion of
internodal elongation just prior to their
reproductive phase.
Abscisic acid is a stress hormone as it
the plants to overcome unfavourable
conditions by inhibiting growth.
05A species which was introduced
for ornamentation but has
become a trouble some weed in
India [NEET (Oct.) 2020]
(a)Parthenium hysterophorus
(b)Eichhornia crassipes
(c)Prosopis juliflora
(d)Trapa spinosa
Ans.(b)
Eichhornia crassipes, (water hyacinth)
was introduced in several tropical
countries including India for
ornamentation but later this exotic
species became a trouble-some
aquatic weed.
PlantGrowthand
Development
15
Plant Growth and
Plant Hormones
TOPIC 1

153. Plant Growth and Development 143
This free floating weed clogged rivers
and lakes and threatened the survival of
many native species to the point of
extinction. This species was called
‘Terror of Bengal’ in India.
06Which of the following is not an
inhibitory substance governing
seed dormancy?[NEET (Sep.) 2020]
(a) Abscisic acid
(b) Phenolic acid
(c)Para-ascorbic acid
(d) Gibberellic acid
Ans.(d)
Gibberellic acid is not an inhibitory
substance governing seed dormancy
because gibberellic acid promotes
growth and elongation of cells. It
affects decomposition of plants and
helps plants grow if used in small
amounts, but eventually plants develop
tolerance to it. Action of ABA is
counteracted by GA, which promotes
seed germination at appropriate time.
Abscisic acid, phenolic acid and
para-ascorbic acid are inhibitory
substances that causes seed dormancy
as they occur in the seed coats and
cotyledons of the embryos, e.g., apple,
peach, ash,Cucurbita, iris,Xanthium.
07The process of growth is
maximum during [NEET (Sep.)
2020]
(a) lag phase (b) senescence
(c) dormancy (d) log phase
Ans.(d)
The process of growth is maximum
during log phase (exponential phase )
because during log phase the growth
rate of the cells gradually increases, at
a maximum rate. In exponential growth,
the initial growth is slow (lag phase) and
thereafter it increases rapidly.
08Name the plant growth regulator
which upon spraying on sugarcane
crop, increases the length of
stem, thus increasing the yield of
sugarcane crop.[NEET (Sep.) 2020]
(a) Gibberellin (b) Ethylene
(c) Abscisic acid (d) Cytokinin
Ans.(a)
The correct option is (a) because
spraying on sugarcane crop with
gibberellins increases the length of the
stem, thus increasing the yield by as
much as 20 tonnes per acre.
09Removal of shoot tips is very
useful technique to boost the
production of tea leaves. This is
because [NEET (Odisha) 2019]
(a) gibberellins prevent bolting and
are inactivated
(b) auxins prevent leaf drop at early
stages
(c) effect of auxins is removed and
growth of lateral buds is enhanced
(d) gibberellins delay senescence of
leaves
Ans.(c)
Removal of shoot tips is a very useful
technique to boost the production of
tea leaves. This is because effect of
auxin is removed and growth of lateral
bud is enhanced. This phenomenon in
most higher plants in which growing
apical buds (shoot tips) inhibit growth of
lateral buds due to effects of auxin is
called apical dominance.
10In order to increase the yield of
sugarcane crop, which of the
following plant growth regulators
should be sprayed?
[NEET (Odisha) 2019]
(a) Ethylene (b) Auxins
(c) Gibberellins (d) Cytokinins
Ans.(c)
In order to increase the yield of
sugarcane crop, gibberellins should be
sprayed.Sugarcane stores
carbohydrates as sugar in their stems.
Spraying sugarcane crop with
gibberellins increases the length of the
stem, thus increasing the yield by as
much as 20 tonnes per acre.
11It takes very long time for
pineapple plants to produce
flowers. Which combination of
hormones can be applied to
artificially induce flowering in
pineapple plants throughout the
year to increase yield?
[NEET (National) 2019]
(a) Gibberellin and Cytokinin
(b) Gibberellin and Abscisic acid
(c) Cytokinin and Abscisic acid
(d) Auxin and Ethylene
Ans.(d)
Auxin and ethylene can be applied to
artificially induce flowering in pineapple
plants throughout the year to increase
yield.
Auxin induces flowering in pineapple
and ethylene helps to synchronise
flower and fruit growth in this plant.
Though in other cases, ethylene causes
fading of flowers.
12Fruit and leaf drop at early stages
can be prevented by the
application of [NEET 2017]
(a) cytokinins (b) ethylene
(c) auxins (d) gibberellic acid
Ans.(c)
Auxin delay abscission of leaves and
fruits at early stages. Whenever leaf or
fruit fall occurs, the organ concerned
stops producing auxin. However, it
promotes abscission of older, mature
leaves and fruits.
13TheAvenacurvature is used for
bioassay of[NEET 2016, Phase I]
(a) GA
3
(b) IAA
(c) Ethylene (d) ABA
Ans.(b)
Bioassay is a quantitative and
qualitative test used to determine the
nature and function of a biochemical by
using living material, e.g.Avena
curvature test is used as bioassay
usually for auxins (Indole Acetic Acid).
14You are given a tissue with its
potential for differentiation in an
artificial culture. Which of the
following pairs of hormones would
you add to the medium to secure
shoots as well as roots?
[NEET 2016, Phase II]
(a) IAA and gibberellin
(b) Auxin and cytokinin
(c) Auxin and abscisic acid
(d) Gibberellin and abscisic acid
Ans.(b)
When a tissue with a potential of
differentiation is grown in an artificial
medium containing auxin and cytokinin
in a specific ratio, it starts
differentiating.
Thus, root and shoot differentiation
occurs. Auxin initiate root formation
while cytokinin starts shoot formation.
15Auxin can be bioassayed by
[CBSE AIPMT 2015]
(a) Avena coleoptile curvature
(b) hydroponics
(c) potometer
(d) lettuce hypocotyl elongation

154. Ans.(a)
Auxin is a phytohormone that is often
bioassayed byAvenacoleoptile
curvature test. The angle of curvature
of a decapitated oat coleoptile is
measured after placing an agar block
containing auxin on one side. The ability
of auxin to stimulate shoot growth is
then measured.
16Dr. F Went noted that if coleoptile
tips were removed and placed on
agar for one hour, the agar would
produce a bending when placed on
one side of freshly cut coleoptile
stumps. Of what significance is this
experiment?[CBSE AIPMT 2014]
(a) It made possible the isolation and
exact identification of auxin
(b) It is the basis for quantitative
determination of small amounts
of growth-promoting substances
(c) It supports the hypothesis that
IAA is auxin
(d) It demonstrated polar movement
of auxins
Ans.(b)
Dr. F Went isolated auxin fromAvena
coleoptile tip. His experiment
demonstrated the polar movement of
auxins, i.e. it showed that the plants
grow towards light in response to a
signal generated in the tip of coleoptile
by a plant hormone auxin.
17Senescence as an active
developmental cellular process in
the growth and functioning of a
flowering plant, is indicated in
[CBSE AIPMT 2008]
(a) vessels and tracheid
differentiation
(b) leaf abscission
(c) annual plants
(d) floral parts
Ans.(b)
Abscission is natural shedding of
leaves, foliage branches, fruits, floral
parts, etc. According toLeopold(1967)
abscission is a senescence
phenomenon.Senescenceis known as
‘the sum of deteriorative processes
which naturally terminate the functional
life of an organism.
Senescence is not confined only to
whole plant, it may be limited to a
particular plant organ such as leaf and
flower or cells such as phloem and
xylem.
Senescence as an active
developmental cellular process in the
growth and functioning of a flowering
plant, is indicated in leaf abscission.
Whole plant senescence also known as
all senescence occurs in annuals, (e.g.,
rice, wheat, gram, mustard), biennials
(e.g. henbane) or perennials.
18Some of the growth regulators
affect stomatal opening. Closure
of stomata is brought about by
[CBSE AIPMT 1994]
(a) indole butyric acid
(b) abscisic acid
(c) kinetin
(d) gibberellic acid
Ans.(b)
Abscisic acid promotes reversal of
H K
+ +
º
pump and increasing
availability ofH
+
inside the guard cell
cytoplasm. Loss ofK
+
decreases
osmotic concentration of guard cells as
compared to adjacent epidermal cells.
Due to the exosmosis the turgidity of
guard cells decreases and it closes the
pore of stoma.
19Which one of the following growth
regulators is known as ‘stress
hormone’?[CBSE AIPMT 2014, 1993]
(a) abscsic acid
(b) Ethylene
(c)GA
3
(d) Indole acetic acid
Ans.(a)
Abscisic Acid (ABA) is also known as
‘stress hormone’ or dormin because it
is produced in much higher amounts,
when plants are subjected to various
kinds of stresses.
It often gives plant organs a signal that
they are undergoing physiological
stresses such as lack of water, saline
soil, cold temperature and frost. ABA
often cause responses that help plants
and protect against these stresses.
20During seed germination its stored
food is mobilised by[NEET 2013]
(a) ethylene
(b) cytokinin
(c) ABA
(d) gibberellin
Ans.(d)
Gibberellin induces aleurone cells to
secrete enzyme that break stored food
in seed.
Cytokinines promote nutrient
mobilisation which helps in the delay of
leaf senescence. ABA plays an
important role in seed development,
maturation and dormancy. Ethylene
induces fruit ripening, breaks seed
dormancy.
21Phototropic curvature is the result
of uneven distribution of
[CBSE AIPMT 2010]
(a) gibberellin (b) phytochrome
(c) cytokinins (d) auxin
Ans.(d)
Darwin and his sonFrancisused
germinating oat (Avena sativa) and
canary grass (Phalaris canariensis)
seedling in their experiments and
hypothesised that when shoots were
illuminated from one side, they bent
forward the light in response to an
‘influence’, that was transmitted
downward from its source at the tip of
the shoot. Paal concluded that the tip
secretes a substance which promotes
the growth of part below it.
In 1926, F Went discovered that some
unidentified compound probably
caused curvature of oat coleoptile
towards light, i.e., phototropism. The
compound (auxin) found by Went is
relatively abundant in coleoptile tips.
22Coiling of graden pea tendrils
around any support is an example
of [CBSE AIPMT 2010, 1995, 91]
(a) thigmotaxis (b) thigmonasty
(c) thigmotropism (d) thermotaxis
Ans.(c)
Thigmotropism movements are due to
the contact with a foreign body. In
twiners and lianas, there is less growth
on the idea of contact and more growth
on the side of branch away from the
contact. Coiling of garden pea tendrils
arround any support is an example of
thigmotropism.
23One of the synthetic auxin is
[CBSE AIPMT 2009]
(a) NAA (b) IAA (c) GA (d) IBA
Ans.(a)
NAA (Naphthalene Acetic Acid) and
2,4-D (2,4-dichlorophenoxy acetic acid)
are synthetic auxins.
The term auxin is applied to the
indole-3-acetic acid (IAA) and to other
natural and synthetic compounds
having certain growth regulating
properties.
144 NEETChapterwise Topicwise Biology

155. IAA and IBA (Indole Butyric Acid) have
been isolated from plants. All these
auxins have been used extensively in
agricultural and horticultural
practices.
24Which one of the following acids is
a derivative of carotenoids?
[CBSE AIPMT 2009]
(a) Indole-butyric acid
(b) Indole-3-acetic acid
(c) Gibberellic acid
(d) Abscisic acid
Ans.(d)
Abscisic acid is a terpenoid, which is a
derivative of steroid (carotenoid).
Indole butyric acid and indole-3-acetic
acid are auxins which are weak organic
acids. Gibberellic acid (gibberellin) is a
terpene.
25Opening of floral buds into flowers,
is a type of[CBSE AIPMT 2007]
(a) autonomic movement of
locomotion
(b) autonomic movement of variation
(c) paratonic movement of growth
(d) autonomic movement of growth
Ans.(d)
Opening of floral buds into flower is a
type ofautonomic movementof growth
(nastic movement). This is
non-directional movement in which the
response is determined by the
structure of the responsive organ and
not to the direction of stimulus. Greater
growth on one side causes the organ to
bend to the opposite side.
26‘Foolish seedling’ disease of rice
led to the discovery of
[CBSE AIPMT 2007]
(a) GA (b) ABA
(c) 2, 4 D (d) IAA
Ans.(a)
Gibberellins (GA) were first observed
from the fungusGibberella fujikuroi, the
causal organism of foolish seedling
disease of rice plants in Japan by
Kurasawa in 1926.
27Which one of the following pairs,
is not correctly matched ?
[CBSE AIPMT 2007]
(a) Abscisic acid —Stomatal closure
(b) Gibberellic acid —Leaf fall
(c) Cytokinin —Cell division
(d) IAA — Cell wall elongation
Ans.(b)
Gibberellins help in cell growth of stem,
leaves and other aerial parts.
28How does pruning help in making
the hedge dense?
[CBSE AIPMT 2006]
(a) It frees axillary buds from apical
dominance
(b) The apical shoot grows faster
after pruning
(c) It releases wound hormones
(d) It induces the differentiation of
new shoots from the rootstock
Ans.(a)
Pruning helps in making the hedge
dense as it frees the axillary buds from
apical dominance. In fact, the apices of
the plant axis, (e.g. shoot apex) has the
highest concentration of auxin which
suppresses the axillary buds while
promotes the growth of apical bud.
When the shoot apex is cut down
through pruning, the axillary buds grow
and the hedge becomes dense.
29An enzyme that can stimulate
germination of barley seeds is
[CBSE AIPMT 2006]
(a) lipase (b) protease
(c) invertase (d)α-amylase
Ans.(d)
Barley seeds are rich in carbohydrate
(starch). The starch is hydrolysed by
α-amylase to monosaccharides unit
at the time of germination of seeds.
30Treatment of seed at low
temperature under moist
conditions to break its dormancy
is called [CBSE AIPMT 2006]
(a) vernalisation (b) chelation
(c) stratification (d) scarification
Ans.(c)
Stratification involves the treatment of
seed at low temperature (5–10°C) under
sufficiently moist conditions to break
its dormancy and to induce
germination.
Scarification involves any damage or
breakage of seed coat by physical
methods, (e.g. use of scalpel, wooden
hammer, etc.) or chemical methods
(use of mild acids) to break seed dormancy.
Vernalisation and Chelation are the
chill treatment of plant in its early
stages of life history to stimulate or
induce early flowering.
31Cell elongation in internodal
regions of the green plants takes
place due to[CBSE AIPMT 2004]
(a) indole acetic acid
(b) cytokinins
(c) gibberellins
(d) ethylene
Ans.(c)
Gibberellin(GA )
3
promotes internodal
elongation in a wide range of species.
This internodal elongation phenomenon
is known as bolting.
Gibberellin is a plant growth hormone
which was first time obtained from a
fungusGibberella fujikuroi (Fusarium
moniliformi).
32Differentiation of shoot is
controlled by[CBSE AIPMT 2003]
(a) high gibberellin—cytokinin ratio
(b) high auxin—cytokinin ratio
(c) high cytokinin—auxin ratio
(d) high gibberellin—auxin ratio
Ans.(c)
Ratio of cytokinins to auxins controls
differentiation. If both of these are
present in equal quantities, the cells
divide but fail to differentiate. If there is
more cytokinin than auxin, shoot buds
develop. If there is more auxin than
cytokinin, roots develop.
33Plants deficient of element zinc,
show its effect on the
biosynthesis of plant growth
hormone
[CBSE AIPMT 2003]
(a) abscisic acid
(b) auxin
(c) cytokinin
(d) ethylene
Ans.(b)
Deficiency of zinc effects biosynthesis
of auxin. It is characterised by a
reduction in internodal growth due to
which plant develops in rosette habit.
The leaves may also be small and
distorted. These results are due to loss
of capacity to produce Indole Acetic
Acid (IAA).
34Hormone responsible for
senescence is[CBSE AIPMT 2001]
(a) ABA
(b) auxin
(c) GA
(d) cytokinin
Plant Growth and Development 145

156. Ans.(a)
Senescence is an active genetically
controlled developmental process in
which cellular structure and
macromolecules are broken down and
translocated away from the senescing
organ (typical leaves) to actively
growing region that serve as nutrient
sinks. Senescence is initiated by
environmental cues and is regulated by
the hormones, e.g. ABA (Abscisic Acid).
Higher amount of ABA stops protein
and RNA synthesis thus accelerating
the senescence.
35Which of the following prevents
fall of fruits?[CBSE AIPMT 2001]
(a)GA
3
(b) NAA
(c) Ethylene (d) Zeatin
Ans.(b)
NAA (Naphthalene Acetic Acid) is a
synthetic auxin hormone which is useful
for preventing pre-harvest fruit drop of
tomatoes.
36Which breaks bud dormancy of
potato tuber?[CBSE AIPMT 2001]
(a) Gibberellin (b) IAA
(c) ABA (d) Zeatin
Ans.(a)
Gibberellinsovercome the natural
dormancy of buds, seeds, tubers, etc. In
this way, these are antagonistic to ABA.
IAA(Indole Acetic Acid) is the principal
naturally occurring auxin, found in all
plants including fungi. It helps in
eradication of weeds, root initiation and
production of parthenocarpic fruits.
ABA (Abscisic Acid) is the natural
growth inhibitor.Zeatinis a naturally
occurring cytokinin that stimulates
mature plant cells to divide when added
to a culture medium along with an
auxin.
37What reason will you assign for
coconut milk used in tissue
culture?[CBSE AIPMT 2000, 03]
(a) Gibberellins (b) Cytokinins
(c) Auxins (d) Ethylene
Ans.(b)
Skoog (1954-1956) observed that
coconut milk contained a substance
which stimulated cell division. The
substance was later on called cytokinin.
The most widely occurring cytokinin in
plants is isopentanyladenine (IPA). IPA
has been isolated fromPseudomonas
tumefaciens.
38The closing and opening of the
leaves ofMimosa pudicais due to
[CBSE AIPMT 1999]
(a) thermonastic movement
(b) hydrotropic movement
(c) seismonastic movement
(d) chemonastic movement
Ans.(c)
Seismonastic movementsare nastic
movements of turgor in response to
stimulus of shock (like
touch/mechanical/electrical/thermal/c
hemical shock). On touchingMimosa
pudica,its leaves droop down and the
stimulus travels at the speed of 1
cm/sec.
39Which combination of gases is
suitable for fruit ripening?
[CBSE AIPMT 1998]
(a) 80%CO
2
and 20%CH
2
(b) 80%CH
4
and 20%CO
2
(c) 80%CO
2
and 20%O
2
(d) 80%C H
2 4
and 20%CO
2
Ans.(d)
Ethylene is a gaseous hormone which
promotes ripening of fruits. Methionine
amino acid is precursor molecule for
ethylene synthesis. Ethylene synthesis
takes place in all parts of a plant such
as roots, stems, leaves, fruits, seeds,
etc.
40A plant hormone used for inducing
morphogenesis in plant tissue
culture is [CBSE AIPMT 1998]
(a) gibberellins (b) cytokinins
(c) ethylene (d) abscisic acid
Ans.(b)
Ratio of cytokinin to auxin controls cell
differentiation. If there is more
cytokinin than auxin, shoot buds
develop. Relatively more auxin than
cytokinins leads to the development of
roots. Abscisic Acid (ABA) is known as
natural plant growth inhibitor.
Gibberellin stimulates stem elongation,
leaf expansion, bolting, flowering, etc.
Ethylene is a fruit ripening hormone.
41Gibberellins induce
[CBSE AIPMT 1997]
(a) flowering
(b) production of hydrolysing
enzymes in germinating seeds
(c) cell division
(d) hasten leaf senescence
Ans.(b)
During seed germination especially of
cereals gibberellins stimulate the
production of hydrolytic enzymes like
amylases, lipases, ribonucleases. These
enzyme solubilise the reserve food of
the seed.
42Ethylene gas is used for
[CBSE AIPMT 1995]
(a) growth of plants
(b) delaying fruit’s abscission
(c) ripening of fruits
(d) stopping the leaf abscission
Ans.(c)
Climacteric fruits are fleshy fruits which
show a sudden sharp rise of respiration
rate at the time of ripening. Ethylene is
used to induce artificial ripening of
these fruits, e.g. apple, mango, banana.
43Movement of auxin is
[CBSE AIPMT 1994]
(a) centripetal (b) basipetal
(c) acropetal (d) Both (b) and (c)
Ans.(d)
Went (1928) reported that auxin is
transported basipetally, i.e., it moves
from apical to basal end. However,
McCready and Jacobs (1963) working on
petiole segments ofPhaseolus vulgaris
observed acropetal movement of auxin
but such type of movement occurs very
little and directly dependent with the
presence of oxygen.
Thus, recent studies have indicated
that the polar movement of auxin is an
active transport.
44Removal of apical bud results in
[CBSE AIPMT 1993, 2000]
(a) formation of new apical bud
(b) elongation of main stem
(c) death of plant
(d) formation of lateral branching
Ans.(d)
Apical dominanceis the phenomenon in
which the presence of apical bud does
not allow the nearby lateral buds to
grow. When the apical bud is removed
the lateral buds sprout.
45Klinostat is employed in the study
of [CBSE AIPMT 1993]
(a) osmosis
(b) growth movements
(c) photosynthesis
(d) respiration
146 NEETChapterwise Topicwise Biology

157. Ans.(b)
Clinostat/klinostat is an instrument
which can nullify the effect of gravity
and allow a plant to grow horizontally by
slowly rotating it.
Rotating clinostat do not show any
bending because the gravitation
stimulus in this case is not unilateral as
it affects all the sides of the rotating
organs equally, whereas plant kept in
unrotated/fixed clinostat bends
downwards showing positive
geotropism.
46Cytokinins [CBSE AIPMT 1992]
(a) promote abscission
(b) influence water movement
(c) help retain chlorophyll
(d) inhibit protoplasmic streaming
Ans.(c)
Cytokinin retards the process of
chlorophyll degradation. Leaf discs are
taken in two lots. In one lot cytokinin is
provided. After 48–72 hrs the leaf discs
are compared for chlorophyll content.
The leaf disc of cytokinin containing lot
has high chlorophyll content.
47Which is employed for artificial
ripening of banana fruits?
[CBSE AIPMT 1992]
(a) Auxin
(b) Cumarin
(c) Ethylene
(d) Cytokinin
Ans.(c)
Ethylene is a gaseous hormone that
induces ripening and maturity of fruits.
When applied as foliar spray ethylene
accelerates maturity and induces
uniform ripening in banana, pineapple,
fig, etc. It also induces fruiting in
ornamental plants and causes
preharvest defoliation in nursery stock.
48Bananas can be prevented from
over-ripening by
[CBSE AIPMT 1992]
(a) maintaining them at room
temperature
(b) refrigeration
(c) dipping in ascorbic acid solution
(d) storing in a freezer
Ans.(c)
Ascorbic acid (vitamin-C) prevents over
ripening of banana and other fruits
because it is an antioxidant.
49Dwarfness can be controlled by
treating the plant with
[CBSE AIPMT 1992, 2002]
(a) cytokinin (b) gibberellic acid
(c) auxin (d) antigibberellin
Ans.(b)
The most important effect of GA is the
stem elongation, in which GA induces
internodal elongation or sub-apical
elongation.
It has been confirmed on several
plants such as pea, bean, tomato,
cabbage, etc. where a significant
elongation of internodes is reported.
Genetically dwarf plants like pea and
maize show normal size in the
presence of gibberellins.
50A chemical believed to be involved
in flowering is
[CBSE AIPMT 1991, 95]
(a) gibberellin (b) kinetin
(c) florigen (d) IBA
Ans.(c)
Chailakhyan (1936) proposed that
photoperiodic induction produces a
chemical complex ‘florigen’ for flowering.
It is synthesised in the older leaves and
then transferred to the growing region
where it initiates the floral bud initiation.
However, florigen has not been
extracted, nor identified till now.
51Abscisic acid causes
[CBSE AIPMT 1991]
(a) stomatal closure
(b) stem elongation
(c) leaf expansion
(d) root elongation
Ans.(a)
Application of minute quantity of
abscisic acid to leaves shall reduce
transpiration to a great extent through
partial closure of stomata.
52The hormone responsible for
apical dominance is
[CBSE AIPMT 1991, 92]
(a) IAA (b) GA
(c) ABA (d) florigen
Ans.(a)
Apical dominanceis the phenomenon in
which the presence of apical bud does
not allow the nearby lateral buds to
grow. This is characteristically caused
by high auxin concentration.
53Which of the following movement
is not related to auxin level?
[CBSE AIPMT 1990]
(a) Bending of shoot towards light
(b) Movement of root towards soil
(c) Nyctinastic leaf movements
(d) Movement of sunflower head
tracking the sun
Ans.(c)
Phototropic movements, e.g. bending
of shoot toward light, movement of
sunflower head tracking the sun, etc.
and geotropic movement,
e.g. movement of root towards soil are
mediated through differential
distribution of IAA.
Nyctinastic leaf movements are
affected by diurnal variation of light
intensity and temperature, such as
elliptical up and down movement of the
two lateral leaflet.
54Phototropic and geotropic
movements are linked to
[CBSE AIPMT 1990]
(a) gibberellins (b) enzymes
(c) auxins (d) cytokinins
Ans.(c)
Differential distribution of indole
3-acetic acid produces tropical plant
responses like phototropism and
geotropism. Phototropism is directional
growth movement of curvature induced
by direction of light while geotropism is
directional movement of curvature
caused by the unilateral application of
force of gravity.
55Phytohormones are
[CBSE AIPMT 1990]
(a) chemicals regulating flowering
(b) chemicals regulating secondary
growth
(c) hormones regulating growth from
seed to adulthood
(d) regulators synthesised by plants
and influencing physiological
processes
Ans.(d)
Phytohormones (Thimann; 1948) are the
plant hormones, i.e. the organic
substances which are naturally
produced in plants, control the growth
or other physiological functions at a
site away from their place of synthesis
and active in extremely minute
quantities.
Plant Growth and Development 147

158. 148 NEETChapterwise Topicwise Biology
56Highest auxin concentration
occurs
[CBSE AIPMT 1990]
(a) in growing tips
(b) in leaves
(c) at base of plant organs
(d) in xylem and phloem
Ans.(a)
Boysen-Jensen (1913), Paal (1919) and
Went (1928) reported that stem tip is
the seat of growth regulating centre.
Auxin shows polar transport from stem
apex to base and from these to root
apex.
57Abscisic acid controls
[CBSE AIPMT 1990, 93, 99, 2000]
(a) cell division
(b) leaf fall and dormancy
(c) shoot elongation
(d) cell elongation and wall formation
Ans.(b)
Abscisic Acid (ABA) is called stress
hormone or dormin. It is a growth
retarding hormone which induces
dormancy, promotes ageing and
abscission of fruits, leaves and flowers.
It also causes closure of stomata and
overcome the conditions of stress.
58Mowing grass lawn facilitates
better maintenance because
[CBSE AIPMT 1989]
(a) wounding stimulates regeneration
(b) removal of apical dominance and
stimulation of intercalary
meristem
(c) removal of apical dominance
(d) removal of apical dominance and
promotion of lateral meristem
Ans.(d)
Apical dominance of terminal bud is due
to the secretion of auxin (IAA) by it.
According to Thimann and Skoog (1933)
removal of apical bud causes sprouting
of lateral buds with stimulation of
intercalary meristem and this is the
reason that mowing grass lawn
facilitates better maintenance.
59Leaf fall can be prevented with the
help of [CBSE AIPMT 1989]
(a) abscisic acid
(b) auxins
(c) florigen
(d) cytokinins
Ans.(d)
Cytokinin retards senescence and
ageing of leaves by preventing
disappearance of chlorophyll and
degradation of proteins that occur with
the ageing process of leaves.
60Which of the following hormones
can replace vernalisation?
[CBSE AIPMT 1989]
(a) Auxin (b) Cytokinin
(c) Gibberellins (d) Ethylene
Ans.(c)
Vernalisation refers to the application
of low temperature to moistened seeds
and young plants, causing shortening of
vegetative phase and initiation of
reproductive phase.
Chailakhyan (1968) reported that under
long-day conditions vernalin hormone
turn into gibberellin and thus, in some
plants, the requirement for
vernalisation is overcome by
gibberellins.
61Leaves of many grasses are
capable of folding and unfolding
because they[CBSE AIPMT 1989]
(a) are very thin
(b) are isobilateral
(c) have specialised bulliform cells
(d) have parallel vascular bundles
Ans.(c)
Leaves of monocots are characterised
as isobilateral (equally green on both
the surfaces), amphistomatic (stomata
on both surface), dumb bell-shaped
guard cells.
The upper epidermis possesses groups
of larger sized thin walled vacuolate
cells called bulliform or motor cells.
Bulliform cells help in rolling of leaves
during water stress or drought.
62Movement of leaves of sensitive
plant,Mimosa pudicais due to
[CBSE AIPMT 1988]
(a) thermonasty (b) seismonasty
(c) hydrotropism (d) chemonasty
Ans.(b)
In seismonastic movement, response is
made to mechanical shocks such as
blows, shaking or pressure. InMimosa,
turgor changes occur in thin walled cells
of pulvinus (lower side) and pulvinnules
(upper side), causing folding of
pinnules, drooping of compound leaves.
63Out or excised leaves remain
green for long if induced to root or
dipped in [CBSE AIPMT 1988]
(a) gibberellins (b) cytokinins
(c) auxins (d) ethylene
Ans.(b)
In Richmond-Lang effect, cytokinin
delays senescence of leaves. As
cytokinin treated detached leaves
remain green after a period of twenty
days whereas controlled leaves were
completely yellow and drying at tips and
margins.
It was concluded that cytokinin was
able to postpone for a number of days
the disappearance of chlorophyll and
degradation of proteins that normally
occur with the ageing process of
leaves.
64Hormone primarily connected with
cell division is
[CBSE AIPMT 1988, 91]
(a) IAA
(b) NAA
(c) cytokinin/zeatin
(d) gibberellic acid
Ans.(c)
Cell division is by far the most
characteristic property associated with
cytokinins, though cytokinin never acts
alone, as in combination with auxins,
cytokinins stimulate cell division even in
non-meristematic tissues.
65The site of perception of light in
plants during photoperiodism is
[NEET 2021]
(a) shoot apex (b) stem
(c) axillary bud (d) leaf
Ans.(d)
The response of plants to periods of
day/night is termed as photoperiodism.
The site of perception of photoperiod is
leaf. The hormone florigen is
responsible for inducing flowering as it
migrates from leaves to shoot apices
on induction of required photoperiods.
Photoperiodism,
Vernalisation
and Senescence
TOPIC 2

159. 66What is the site of perception of
photoperiod necessary for
induction of flowering in plants?
[NEET (National) 2019]
(a) Pulvinus (b) Shoot apex
(c) Leaves (d) Lateral buds
Ans.(c)
For the induction of flowering in
plants, photoperiod stimulus is
percieved by the leaves of plants. As
a result, floral hormones are
produced in the leaves which are
then translocated to the apical part
and subsequently cause the initiation
of floral primordial growth.
67Phytochrome is a
[NEET 2016, Phase II]
(a) flavoprotein
(b) glycoprotein
(c) lipoprotein
(d) chromoprotein
Ans.(d)
Phytochrome is a chromoprotein which
exist in two forms,P
r
andP
fr
. These are
inter-convertible. When plants get red
right this protein gets converted intoP
fr
formP
r
andvice versa. It controls the
photoperiodism in the plants.
68A few normal seedlings of tomato
were kept in a dark room. After a
few days they were found to have
become white-coloured like
albinos. Which of the following
terms will you use to describe
them? [CBSE AIPMT 2014]
(a) Mutated
(b) Embolised
(c) Etiolated
(d) Defoliated
Ans.(c)
Etiolation is a process in which
flowering plants are grown in partial or
complete absence of light. Etiolation is
mainly characterised by long and weak
stem and smaller, sparse pale yellow
colour of leaves due to the longer
internodes. Thus due to this tomato
seeding became white coloured.
69Importance of day length in
flowering of plants was first
shown in [CBSE AIPMT 2008]
(a)Lemna (b) tobacco
(c) cotton (d) Petunia
Ans.(b)
Photoperiodismwas first discovered by
Garner and Allard (1920, 1922). They
observed that maryland mammoth
variety of tobacco could be made to
flower only by reducing the light hours
with artificial darkning.
On the basis of photoperiodic response
to flowering plants have been divided
into short day plants (tobacco), long day
plant (e.g. wheat, hanbane), short long
day plants. (e.g.Campanula), long short
day plants (e.g.Bryophyllum)
intermediate plants (e.g. wild kidney
bean) and day neutral plants
(e.g. cotton).
70The wavelength of light absorbed
byP
r
form of phytochrome is
[CBSE AIPMT 2007]
(a) 640 nm (b) 680 nm
(c) 720 nm (d) 620 nm
Ans.(b)
When P
r
absorbs red light (650-670 nm)
it is converted into P
fr
form and when
P
fr
absorbs far red light (730-735 nm) it
is converted back into P
r
form.
71One set of a plant was grown at 12
hr day and 12 hr night period
cycles and it flowered while in the
other set night phase was
interrupted by flash of light and it
did not produce flower. Under
which one of the following
categories will you place this
plant? [CBSE AIPMT 2004]
(a) Long-day
(b) Darkness neutral
(c) Day neutral (d) Short-day
Ans.(d)
The condition shows that the plant
require photo-period shorter than the
critical day length. This plant needs
uninterrupted dark period for flowering.
Therefore, it is a short-day plant and
these do not flower if the dark period is
interrupted with flashes of light.
72Which one is a long-day plant?
[CBSE AIPMT 2001]
(a) Tobacco (b)Glycine max
(c)Mirabilis jalapa(d) Spinach
Ans.(d)
Plants which require long-day
photoperiod for flowering and a small
dark period for vegetation are known as
long-day plants, e.g. spinach.
73Proteinaceous pigment which
control activities concerned with
light [CBSE AIPMT 2001]
(a) phytochrome (b) chlorophyll
(c) anthocyanin (d) carotenoids
Ans.(a)
Phytochromesare the plant
chromoproteins, containing protein
pigment existing in two
inter-convertible forms—P
r
(absorbs red
light- 660 nm) andP
fr
(absorbs far red
light-730 nm). It controls flowering,
seed dormancy, etc.
74The method that renders the seed
coat permeable to water so that
embryo expansion is not physically
retarded, is[CBSE AIPMT 2000]
(a) vernalisation (b) stratification
(c) denudation (d) scarification
Ans.(d)
In many plants, the seed coats are quite
tough and provide mechanical
resistance to the growth of the
embryos. Scarification done by
abrasion through machine, threshing,
filing, etc this process is done to
rupture or weaken the seed coat and
promote germination.
75The response of different
organisms to environmental
rhythms of light and darkness is
called [CBSE AIPMT 1998]
(a) phototaxis (b) photoperiodism
(c) phototropism (d) vernalisation
Ans.(b)
Photoperiodism is the term to denote
a biological response to changes in
the ratio of light and darkness in a 24
hrs cycle.
76The pigment, that absorbs red and
far-red light in plants, is
[CBSE AIPMT 1995, 2002]
(a) xanthophyll (b) cytochrome
(c) phytochrome (d) carotene
Ans.(c)
Phytochrome is a type of pigment
which absorbs red or far-red light and
its absorbing region is closely
associated with protein. The
phytochrome pigment is found to be
present in two photoreversible forms
P
r
(P
660
) andP (P )
fr730
⋅
P P
r fr
(inactive)
Far red
Red light
(active)
Plant Growth and Development 149

160. 77What will be the effect on
phytochrome in a plant subjected
to continuous red light?
[CBSE AIPMT 1997]
(a) Level of phytochrome decreases
(b) Phytochrome is destroyed
(c) Phytochrome synthesis increases
(d) Destruction and synthesis of
phytochrome remain in
equilibrium
Ans.(b)
Continuous exposure to red light
causes (a) conversion ofP
r
-P
fr
which is
rapidly destroyed, (b) inhibition of
synthesis ofP
r
. Thus, total amount of
phytochrome is decreased.
78If a tree, flowers thrice in a year
(Oct., Jan. and July) in Northern
India, it is said to be
[CBSE AIPMT 1997]
(a) photosensitive but
thermoinsensitive
(b) thermosensitive but
photoinsensitive
(c) hoto and thermosensitive
(d) photo and thermoinsensitive
Ans.(d)
Since, flowering can take place during
any part of the year, therefore, the plant
is not sensitive to photoperiod and
temperature.
79In short-day plants, flowering is
induced by [CBSE AIPMT 1992]
(a) photoperiod less than 12 hrs
(b) photoperiod below a critical
length and uninterrupted long
night
(c) long night
(d) short photoperiod and interrupted
long night
Ans.(b)
In short-day plants, flowering is induced
when the day length do not exceed a
certain critical value, the day length
required is less than a certain critical
length. Short-day plants may be more
correctly called long night plants as a
certain minimum of uninterrupted dark
period in 24 hrs is necessary for their
flowering. Short-day plants will not
flower if the dark period is less than a
critical length.
80Flowering dependent on cold
treatment is[CBSE AIPMT 1992]
(a) cryotherapy
(b) cryogenics
(c) cryoscopy
(d) vernalisation
Ans.(d)
Vernalisationis a process of shortening
of juvenile or vegetative phase and
hastening flowering by a previous cold
treatment.
81Which one increases in the
absence of light?
[CBSE AIPMT 1989]
(a) Uptake of minerals
(b) Uptake of water
(c) Elongation of internodes
(d) Ascent of sap
Ans.(c)
In general intense light retards growth
in plants. High light intensities induce
dwarfening of the plant. Absence of
light reduces the overall growth,
photosynthesis, uptake of minerals and
ascent of sap. However, the elongation
of internodes is seen to occur in the
absence of light.
82Phytochrome is involved in
[CBSE AIPMT 1988]
(a) phototropism
(b) photorespiration
(c) photoperiodism
(d) geotropism
Ans.(c)
Phytochrome is a chromoprotein
(photosensitive pigment) that exists in
two states,P
r
(red) orP
660
andP
fr
(far red)
orP
730
.
Phytochrome is involved in
photomorphogenetic responses, seed
germination, bud dormancy, synthesis
of gibberellin and ethylene and
photoperiodism.
150 NEETChapterwise Topicwise Biology

161. 01Sphincter of Oddi is present at
[NEET 2021]
(a) ileo-caecal junction
(b) junction of hepato-pancreatic duct
and duodenum
(c) gastro-oesophageal junction
(d) junction of jejunum and duodenum
Ans.(b)
Sphincter of Oddi is the smooth muscle
or a muscular valve that surrounds the
end portion of the common bile duct and
pancreatic duct (hepato-pancreatic
duct). It controls the flow of digestive
juices into the intestine.
02Identify the correct statement with
reference to human digestive
system. [NEET (Sep.) 2020]
(a) Serosa is the innermost layer of the
alimentary canal
(b) Ileum is a highly coiled part
(c) Vermiform appendix arises from
duodenum
(d) Ileum opens into small intestine
Ans.(b)
The option (b) is correct as ileum is a
highly coiled tube with reference to
human digestive system. Other option
can be corrected as
Serosa is the outermost layer of the
alimentary canal. A narrow finger-like
tubular projection, the vermiform
appendix arises from caecum part of
large intestine. Ileum opens into the
large intestine.
03Match the following structures with
their respective location in organs.
[NEET (National) 2019]
Column I Column II
A. Crypts of Lieberkuhn (i) Pancreas
B. Glisson's capsule (ii) Duodenum
C. Islets of Langerhans (iii) Small
intestine
D. Brunner's glands (iv) Liver
Select the correct option from the
following
A B C D
(a) (ii) (iv) (i) (iii)
(b) (iii) (iv) (i) (ii)
(c) (iii) (ii) (i) (iv)
(d) (iii) (i) (ii) (iv)
Ans.(b)
(A)–(iii), (B)–(iv), (C)–(i), (D)–(ii)
Crypts of Lieberkuhn are simple, tubular
intestinal glands which occur throughout
the small intestine between the villi.
They secrete digestive enzymes and
mucus. Glission’s capsule is the inner
thin layer of connective tissue in liver.
Islets of Langerhans constitute the
endocrine part of pancreas which
secrete hormones.
Brunner’s glands are located in the
submucosa of duodenum and they open
into the crypts of Lieberkuhn.
04Which one of the following terms
describe human dentition?
[NEET 2018]
(a) Pleurodont, Monophyodont, Homodont
(b) Thecodont, Diphyodont, Heterodont
(c) Thecodont, Diphyodont, Homodont
(d) Pleurodont, Diphyodont, Heterodont
Ans.(b)
The terms,thecodont,diphyodontand
heterodontdescribe human dentition. In
men, two types of teeth are found, milk
or deciduous teeth and permanent
teeth. Thus, they havediphyodont
teeth.The teeth arethecodont, i.e. they
remain embedded in the sockets of the
jaw bones. Men have four types of teeth;
incisors, canine, premolars and molars,
i.e.,heterodont teeth.
05Conversion of milk to curd
improves its nutritional value by
increasing the amount of
[NEET 2018]
(a) vitamin-B
12
(b) vitamin-A
(c) vitamin-D
(d) vitamin-E
Ans.(a)
Conversion of milk to curd improves its
nutritional value by increasing the
amount ofvitamin-B
12
.
Vitamin-Ais found in milk, carrot,
tomato, etc. Skin can synthesise
vitamin-Din the presence of sunlight.
Vitamin-Eis found in wheat, green
leafy vegetables, etc.
06A baby boy aged two years is
admitted to play school and passes
through a dental check-up. The
dentist observed that the boy had
twenty teeth. Which teeth were
absent? [NEET 2017]
(a) Incisors
(b) Canines
(c) Premolars
(d) Molars
Digestionand
Absorption
16
Components of Food
and Digestive System
TOPIC 1

162. Ans.(c)
In human beings, after birth the first set
of teeth that develops are deciduous
teeth or temporary teeth. These are 20
in number. The dental formula of child is
2102/2102.
Thus, they have 2 incisors, 1 canine, 0
premolars and 2 molars. Therefore,
the baby boy would not have
premolars.
07Which cells of ‘Crypts of
Lieberkuhn’ secrete antibacterial
lysozyme ? [NEET 2017]
(a) Argentaffin cells
(b) Paneth cells
(c) Zymogen cells
(d) Kupffer cells
Ans.(b)
The mucosa present in between the
bases of villi of small intestine (Crypts
of Lieberkuhn) contain paneth, which
secrete antibacterial lysozyme.
Concept EnhancerKupffer cells are
phagocytic cells of liver.
Zymogen cells produce enzyme.
Argentaffin cells produce hormones.
08The hepatic portal vein drains
blood to liver from[NEET 2017]
(a) heart (b) stomach
(c) kidneys (d) intestine
Ans.(d)
In the hepatic portal system, the hepatic
veins takes blood from intestine to the
liver. This way, it takes all the nutrients
absorbed from intestine to the liver first,
where screening and storing of nutrition
takes place.
Concept Enhancer
The portal systemis a system of veins in
which vein takes blood to some
organ/tissue of the body other than
heart. In this, the vein has capillary
network at it’s both ends.
There is one more portal system in
human body named hypophyseal portal
system present in the hypothalamus,
which brings neuro secretions of
hypothalamus to pituitary gland.
The renal portal systemis found in
fishes and amphibians. It supplies blood
from posterior region of the body to the
kidneys by renal portal veins to remove
waste products before sending it to
heartviarenal veins and post canal
veins.
09Which of the following guards the
opening of hepatopancreatic duct
into the duodenum?
[NEET 2016, Phase I]
(a) Ileocaecal valve
(b) Pyloric sphincter
(c) Sphincter of Oddi
(d) Semilunar valve
Ans.(c)
Sphincter of Oddi guards the opening
of hepatopancreatic duct into the
duodenum. Hepatopancreatic duct
brings secretion of liver as well as
pancreas to the duodenum.
10The primary dentition in human
differs from permanent dentition in
not having one of the following
type of teeth[CBSE AIPMT 2015]
(a) Canine (b) Premolars
(c) Molars (d) Incisors
Ans.(b)
There are four classes of teeth, i.e.
incisors, canines, premolars and molars.
There are no premolars in primary
dentition (deciduous or boby teeth).
These are found only in permanent
dentition (adult teeth).
11Two friends are eating together on
a dining table. One of them
suddenly starts coughing while
swallowing some food. This
coughing would have been due to
improper movement of
[CBSE AIPMT 2011]
(a) diaphragm (b) neck
(c) tongue (d) epiglottis
Ans.(d)
The epiglottis is a flap that is made of
elastic cartilage tissue covered with a
mucous membrane, attached to the
entrance of the larynx. It prevents the
entry of food into the larynx and directs
it to the oesophagus.
Due to the improper movement of
epiglottis, one may suddenly start
coughing while swallowing some food.
12Epithelial cells of the intestine
involved in food absorption have on
their surface[CBSE AIPMT 2005]
(a) pinocytic vesicles
(b) phagocytic vesicles
(c) zymogen granules
(d) microvilli
Ans.(d)
The mucosa and sub-mucosa of small
intestine are thrown into folds. Surfaces
of these folds are covered by fine,
finger-like projections of the epithelium.
These projections are called villi. In
addition, the epithelial cells of the villi
are covered on their exposed surface by
cytoplasmic projections called microvilli.
13The richest sources of vitamin–B
12
are [CBSE AIPMT 2004]
(a) goat’s liver andSpirulina
(b) chocolate and green gram
(c) rice and hen’s egg
(d) carrot and chicken’s breast
Ans.(a)
Vitamin-B
12
(cyanocobalamin) is the only
vitamin which is not found in vegetables.
It is present in animal protein such as
meat, liver, fish andSpirulina(single cell
protein). It promotes DNA synthesis,
maturation of RBCs and myelin formation.
14Which one is correctly matched ?
[CBSE AIPMT 2001]
(a) Vit-E–Tocopherol
(b) Vit-D–Riboflavin
(c) Vit-B–Calciferol
(d) Vit-A–Thiamine
Ans.(a)
Option (a) is correctly matched.
Thiamine, riboflavin, calciferol,
tocopherol are also known as vitamin-B
1
,
vitamin-B
2
, vitamin-D
2
and vitamin-E
respectively.
15The layer of cells that secrete
enamel of tooth is
[CBSE AIPMT 1998]
(a) dentoblast (b) amiloblast
(c) osteoblast (d) odontoblast
Ans.(d)
The pulp cavity contains a mass of dense
but soft connective tissue which is
called pulp. A single layer of odontoblast
cells is lined by the pulp cavity.
152 NEETChapterwise Topicwise Biology
Villi
Lacteal
Capillaries
Artery
Crypts of
Lieberkuhn
Vein
A section of small intestinal mucosa
showing villi and the Crypts of

163. These cells secrete enamel which is a
bluish white, shiny translucent and the
hardest substance of the body.
16A dental disease characterised by
moltting of teeth is due to the
presence of a certain chemical
element in drinking water. Which of
the following is that element?
[CBSE AIPMT 1995]
(a) Mercury (b) Chlorine
(c) Fluorine (d) Boron
Ans.(c)
Increased amount of fluorine in drinking
water causes fluorosis, responsible for
moltting of teeth.
17Brunner’s glands occur in
[CBSE AIPMT 1992]
(a) sub-mucosa of duodenum
(b) sub-mucosa of stomach
(c) mucosa of oesophagus
(d) mucosa of ileum
Ans.(a)
Brunner’s glandsare convoluted and
branched glands found only in duodenum
and located in sub-mucosa.
18In man the zymogen or chief cells
are mainly found in
[CBSE AIPMT 1990]
(a) cardiac part of stomach
(b) pyloric part of stomach
(c) duodenum
(d) fundic part of stomach
Ans.(d)
Chief cells or zymogen are mainly found
in fundic part of stomach which secretes
two proenzymes, pepsinogen and
prorennin and an enzyme gastric lipase.
19Whartson’s duct is associated with
[CBSE AIPMT 1988]
(a) sub-lingual salivary duct
(b) parotid salivary gland
(c) sub-maxillary salivary gland
(d) Brunner’s glands
Ans.(c)
Whartson’s duct is associated with
sub-maxillary salivary gland. These lie
beneath the jaw angles, their secretion is
carried by Whartson’s duct which open
below the tongue. These are compound
acinar gland.
20Duct leading from parotid gland
and opening into vestibule is
[CBSE AIPMT 1988]
(a) Haversian duct
(b) Stenson’s duct
(c) Wolffian duct
(d) Infra-orbital duct
Ans.(b)
Parotid glandsare largest salivary
glands, present just below the external
ear. These are compound tubulo-acinar
glands. Saliva is secreted by Stenson’s
duct which open opposite to the second
upper molar tooth.
21Lamina propria is connected with
[CBSE AIPMT 1988]
(a) acini
(b) liver
(c) Graafian follicle
(d) intestine
Ans.(d)
Lamina propria of ileum shows yellow
coloured oval, granular masses of lymph
nodules called ‘Peyer’s patches.’
22Succus entericus is referred to as
[NEET 2021]
(a) pancreatic juice
(b) intestinal juice
(c) gastric juice
(d) chyme
Ans.(b)
Succus entericus also known as
intestinal juice. It is a fluid secreted in
small intestine in small quantity. The
secretion of the brush border cells of the
mucosa along with the secretions of
goblet cells constitute succus entericus.
It consist of various enzymes like
lipases, disaccharides, nucleosidases
etc. and mucus.
23Intrinsic factor that helps in the
absorption of vitamin-B
12
is
secreted by [NEET (Oct.) 2020]
(a) goblet cells (b) hepatic cells
(c) oxyntic cells (d) chief cells
Ans.(c)
Parietal cells or oxyntic cells secrete HCl
and intrinsic factor. These intrinsic
factors are essential for absorption of
vitamin-B
12
. Goblet cells secrete mucus.
Peptic or chief-cells secrete the
proenzyme pepsinogen. Hepatic cells
secrete bile.
24The proteolytic enzyme renin is
found in [NEET (Oct.) 2020]
(a) intestinal juice (b) bile juice
(c) gastric juice (d) pancreatic juice
Ans.(c)
The proteolytic enzyme rennin is found
in gastric juice of infants which helps in
the digestion of milk proteins, casein
into paracasein.
25The enzyme enterokinase helps in
conversion of[NEET (Sep.) 2020]
(a) trypsinogen into trypsin
(b) caseinogen into casein
(c) pepsinogen into pepsin
(d) protein into polypeptides
Ans.(a)
The correct option is (a) because
the enzyme enterokinase helps in
conversion of trypsinogen into trypsin.
Trypsinogen is activated by an
enzyme, enterokinase, secreted by
the intestinal mucosa into active
trypsin. Trypsinogen is a zymogen
released from pancreas.
26Match the items given in Column I
with those in Column II and choose
the correct option.
[NEET (Odisha) 2019]
Column I Column II
1. Rennin (i) Vitamin-B
12
2. Enterokinase (ii) Facilitated
transport
3. Oxyntic cells (iii) Milk proteins
4. Fructose (iv) Trypsinogen
1 2 3 4
(a) (iii) (iv) (ii) (i)
(b) (iv) (iii) (i) (ii)
(c) (iv) (iii) (ii) (i)
(d) (iii) (iv) (i) (ii)
Ans.(d)
The correct matches are
1. Rennin is a proteolytic enzyme that
causes
coagulation of milk.
2. Enterokinase converts trypsinogen
into its
active form trypsin.
3. Oxyntic cells (also called parietal
cells) during digestion release
stomach acid to allow release of
vitamin-B
12
from food.
4. Fructose is absorbed by facilitated
transport into the blood capillaries.
Digestion and Absorption 153
Functioning of
Digestive System
TOPIC 2

164. 27Identify the cells whose secretion
protects the lining of
gastrointestinal tract from various
enzymes. [NEET (National) 2019]
(a) Goblet cells (b) Oxyntic cells
(c) Duodenal cells (d) Chief cells
Ans.(a)
Secretions of goblet cells protect the
lining of gastrointestinal tract from
various enzymes. These cells secrete
mucus which along with bicarbonate ions
helps in the lubrication and protection of
the mucosal epithelium from the
excoriation by the highly concentrated
HCl. On the other hand, oxyntic or
parietal cells secrete hydrochloric acid.
Chief cells or peptic cells secrete
proenzymes-pepsinogen and prorenin.
28Which of the following options best
represents enzyme composition of
pancreatic juice? [NEET 2017]
(a) Amylase, peptidase, trypsinogen,
rennin
(b) Amylase, pepsin, trypsinogen, maltase
(c) Peptidase, amylase, pepsin, rennin
(d) Lipase, amylase, trypsinogen,
procarboxypeptidase
Ans.(d)
Pancreas consist of exocrine and
endocrine part. Exocrine part secrets
alkaline pancreatic juice. This juice
contains trypsinogen,
chymotrypsinogen, procarboxypeptidase,
lipase, amylase, elastase.
Concept EnhancerRenin and pepsin
enzymes are present in gastric juice.
Maltase is present in the intestinal juice.
29In the stomach, gastric acid is
secreted by the[NEET 2016, Phase I]
(a) parietal cells
(b) peptic cells
(c) acidic cells
(d) gastrin secreting cells
Ans.(a)
In stomach, gastric acid (HCl) is secreted
by parietal cells of gastric gland. It
makes the medium of food in stomach
acidic for stimulation of proteolytic
enzymes of stomach.
30Which hormones do stimulate the
production of pancreatic juice and
bicarbonate?[NEET 2016, Phase II]
(a) Angiotensin and epinephrine
(b) Gastrin and insulin
(c) Cholecystokinin and secretin
(d) Insulin and glucagon
Ans.(c)
Cholecystokinin (CCK) and secretin are
the peptide hormones that stimulate the
production of pancreatic juice and
bicarbonates within the alimentary canal.
Secretin acts on the exocrine pancreas
and stimulates the secretion of water
and bicarbonate ions.
CCK acts on both pancreas and gall
bladder and stimulates the secretion of
pancreatic enzymes and bile juice
respectively. Hence, option (c) is
correct.
31The enzyme that is not present in
succus entericus is
[CBSE AIPMT 2015]
(a) maltase (b) nucleases
(c) nucleosidase (d) lipase
Ans.(b)
Succus entericus or intestinal digestive
juice contains a variety of enzymes like
disaccharidases (e.g. maltase),
dipeptidases, lipases, nucleosidases.
Nucleases are enzymes present in
pancreatic juice that break nucleic acids
into nucleotides.
32The initial step in the digestion of
milk in humans is carried out by?
[CBSE AIPMT 2014, 11]
(a) Lipase (b) Trypsin
(c) Rennin (d) Pepsin
Ans.(d)
In humans, the milk protein digesting
enzyme in stomach is pepsin. In calves it
is rennin. It is also present in small
amounts in human infants but not
adults. Pepsin acts on water soluble
‘caseinogen (milk protein) to form
soluble ‘casein’. This combines with
calcium salts to form insoluble calcium
paracaseinate, which gets readily
digested enzymatically.
33Fructose is absorbed into the blood
through mucosa cells of intestine
by the process called
[CBSE AIPMT 2014]
(a) active transport
(b) facilitated transport
(c) simple diffusion
(d) co-transport mechanism
Ans.(b)
Fructose is absorbed into the blood
through mucosa cells of intestine by
the process called facilitated
transport thus, facilitated transport is
the process of spontaneous passive
transport of the molecules or ions
across a biological membranevia
specific transmembrane integral
protein.
34Select the correct match of the
digested products in humans given
in column I with their absorption
site and mechanism in column II.
[NEET 2013]
Column I Column II
(a)Glycine and
glucose
Small intestine and
active absorption
(b)Fructose and
Na
+
Small intestine
passive absorption
(c)Glycerol and
fatty acids
Duodenum and move
as chilomicrons
(d)Cholesterol
and maltose
Large intestine and
active absorption
Ans.(a)
Amino acids, monosaccharides like
glucose, electrolytes like Na
+
are
absorbed into the blood by active
transport. Fructose and some amino
acids are absorbed with the help of the
carrier ions like Na
+
by facilitated
transport. Fatty acid and glycerol cannot
be absorbed into the blood. They are
first incorporated into small droplets
called micelles, which move into the
intestinal mucosa.
35If for some reason our goblet cells
are non-functional, this will
adversely affect[CBSE AIPMT 2010]
(a) production of somatostatin
(b) secretion of sebum from the
sebaceous glands
(c) maturation of sperms
(d) smooth movement of food down the
intestine
Ans.(d)
Goblet cells are something like a
wineglass that present in the columnar
epithelium of the mammalian intestine
and secrete mucin, a mucoprotein that
forms mucus when in solution. If Goblet
cells become non-functional, this will
adversely affect smooth movement of
food down the intestine due to the
absence of mucin.
36Carrier ions likeNa
+
facilitate the
absorption of substance like
[CBSE AIPMT 2010]
(a) amino acids and glucose
(b) glucose and fatty acids
(c) fatty acids and glycerol
(d) fructose and some amino acids
154 NEETChapterwise Topicwise Biology

165. Ans.(d)
Active transport occurs with the help of
energy, usually against concentration
gradient. For this, cell membrane
possesses carriers and gated channels.
Active transport of one substance is
often accompanied by permeation of
other substances.
The phenomenon is calledsecondary
active transport. It is of two main types,
i.e. Co-transport, (e.g. glucose and some
amino acids along with inward pushing of
excess Na
+
) and counter transport
(Ca and H
2+ +
movement outwardly as
excess Na
+
passes inwardly).
37A young infant may be feeding
entirely on mother’s milk, which is
white in colour but the stools, which
the infant passes out is quite
yellowish. What is this yellow colour
due to? [CBSE AIPMT 2009]
(a) Intestinal juice
(b) Bile pigments passed through bile
juice
(c) Undigested milk protein casein
(d) Pancreatic juice poured into
duodenum
Ans.(b)
The stools, which the infant passes out
is quite yellowish due to the bile
pigments. These bile pigments are
released in the bile juice.
38Which one of the following pairs of
food components in humans
reaches the stomach totally
undigested?[CBSE AIPMT 2009]
(a) Protein and starch
(b) Starch and fat
(c) Fat and cellulose
(d) Starch and cellulose
Ans.(c)
In humans, starch is digested in
buccopharyngeal cavity. Cellulose is not
digested in the humans because
cellulose containsβ-1, 4-linkage and
vertebrates themselves do not possess
any enzyme capable of hydrolysingβ-1,
4-linkages. Protein is digested in
stomach and fat in small intestine. Thus,
in the given options, fat and cellulose
reach totally undigested in the stomach
of humans.
39Which one of the following
statement is true regarding
digestion and absorption of food in
humans? [CBSE AIPMT 2009]
(a) Oxyntic cells in our stomach secrete
the proenzyme pepsinogen
(b) Fructose and amino acids are
absorbed through intestinal mucosa
with the help of carrier ions likeNa
+
(c) Chylomicrons are small lipoprotein
particles that are transported from
intestine into blood capillaries
(d) About 60% of starch is hydrolysed by
salivary amylase in our mouth
Ans.(c)
Chylomicrons are lipoprotein particles
synthesised by intestinal epithelial cells
and consisting mainly of triglycerides.
Chylomicrons are the form, in which
dietary fat is transported in the
circulatory system.
40Which one of the following is the
correct matching of the site of
action on the given substrate, the
enzyme acting upon it and the end
product? [CBSE AIPMT 2008]
(a) Duodenum: Triglycerides trypsin
monoglycerides
(b) Small intestine: Starchα-amylase
disaccharide (maltose)
(c) Small intestine: Proteins pepsin
amino acids
(d) Stomach: Fats, Lipase micelles
Ans.(b)
In small intestine food meats with the
pancreatic juice containingα-amylase,
which converts starch into maltose,
isomaltose andα-dextrins in small
intestine.
Starch Maltose
Pancreatic-amylase
→
α
(disaccharide)
The pancreatic juice also contains
proenzymes trypsinogen,
chymotrypsinogen and
procarboxypeptidase. The trypsinogen is
converted to active trypsin in intestine
by enterokinase of intestinal juice. The
trypsin converts proteins into large
peptides and the large peptides are
converted to dipeptide and amino acids
by carboxypeptidase.
41What will happen if the secretion of
parietal cells of gastric glands is
blocked with an inhibitor?
[CBSE AIPMT 2008]
(a) Gastric juice will be deficient in
chymosin
(b) Gastric juice will be deficient in
pepsinogen
(c) In the absence of HCl secretion,
inactive pepsinogen is not converted
into the active enzyme pepsin
(d) Enterokinase will not be released
from the duodenal mucosa and so
trypsinogen is not converted to
trypsin
Ans.(c)
The parietal cells (oxyntic cells) are large
and most numerous on the side walls of
gastric glands. These secrete hydrochloric
acid and castle intrinsic factor. The peptic
cells (zymogen) of gastric glands secrete
gastric digestive enzymes as
proenzymes-pepsinogen and prorennin
and small amount of gastric amylase and
gastric lipase. The hydrochloric acid
maintains a strongly acidic pH of about
1.5-2.5 in the stomach. HCl converts
pepsinogen and prorennin to pepsin and
rennin respectively.
42Secretin and cholecystokinin are
digestive hormones. They are
secreted in[CBSE AIPMT 2005]
(a) oesophagus
(b) ileum
(c) duodenum
(d) pyloric stomach
Ans.(c)
Secretin and cholecystokinin (CCK) are
two main gastrointestinal (GI) hormones
secreted in duodenum of alimentary
canal. CCK stimulates gall bladder
contraction and thus increases the flow
of bile salts into the intestine.
Secretin stimulates the release of an
alkaline pancreatic fluid that neutralises
stomach acid as it enters the intestine.
43Duodenum has characteristic
Brunner’s glands which secrete two
hormones called
[CBSE AIPMT 2004]
(a) kinase, oestrogen
(b) secretin, cholecystokinin
(c) prolactin, parathormone
(d) estradion, progesterone
Ans.(b)
Brunner’s glandsecrete large amount of
mucus and bicarbonates to protect
duodenal mucosa and to neutralise the
acidic chyme. It also secretes two
hormones :
(a) Secretin
(b) Cholecystokinin (CCK)
These stimulate:
(i) Secretion of pancreatic juice by
pancreas.
(ii) Release of bile from gall bladder.
(iii) Formation of bile by liver and
pancreatic juice.
Digestion and Absorption 155

166. 44During prolonged fasting, in what
sequence are the following organic
compounds used up by the body?
[CBSE AIPMT 2003]
(a) First carbohydrates, next proteins
and lastly lipids
(b) First proteins, next lipids and lastly
lcarbohydrates
(c) First carbohydrates, next fats and
lastly proteins
(d) First fats, next carbohydrates and
lastly proteins
Ans.(c)
During prolonged fasting, first of all
carbohydrates are utilised which include
glycogen stored in liver. This is followed
by the breakdown of adipose tissue, thus
providing lipids and lastly the body
utilises proteins.
45A certain person eats boiled
potato; one of the food component
in it is [CBSE AIPMT 2000]
(a) lactose which is indigestible
(b) starch which does not get digested
(c) cellulose which is digested by
intestinal cellulase
(d) DNA which gets digested by
pancreatic DNAase
Ans.(d)
Anything which cannot be digested
cannot serve as ‘food’. Therefore, starch
and lactose in the present case have
been automatically deleted.
Cellulose cannot be digested by human
beings, thus option (c) also stands
rejected. Pancreatic juice can digest
DNA which is the component of every
cell.
46Cholecystokinin and duocrinin are
secreted by [CBSE AIPMT 1999]
(a) adrenal cortex (b) thyroid gland
(c) pancreas (d) intestine
Ans.(d)
Both cholecystokinin and duocrinin
are hormones secreted by the
intestine, while the former stimulates
the gall bladder to release bile and
pancreas to release enzyme mixture,
the latter regulates the release of
mucus from Brunner’s glands.
47Which part of body secretes the
hormone secretin?
[CBSE AIPMT 1999]
(a) Oesophagus (b) Duodenum
(c) Stomach (d) Ileum
Ans.(b)
Secretin is a polypeptide hormone
secreted by the mucosa of duodenum
and jejunum.
It perform two functions : (a) It
stimulates sodium bicarbonate from
the pancreas which neutralises the
acid in the chyme so that it will not
damage the wall of the small intestine.
(b) It increases the rate of bile
secretion in the liver.
48The hormone that stimulates the
stomach to secrete gastric juice is
[CBSE AIPMT 1998]
(a) gastrin
(b) renin
(c) enterokinase
(d) enterogasterone
Ans.(a)
The stomach controls the production of
gastric juice by means of a digestive
hormone called gastrin. It is produced by
endocrine (hormone secreting) cells that
are scattered throughout the epithelium
of the stomach.
49Lactose is composed of
[CBSE AIPMT 1998]
(a)glucose+ fructose
(b)glucose + glucose
(c)glucose + galactose
(d)fructose + galactose
Ans.(c)
Lactose(C H O )
12 22 11
is a disaccharide
found in mammalian milk. It comprises
galactose and glucose units which are
linked together byβ, 1-4 glycosidic
bonds. It is a reducing sugar.
50In vertebrates lacteals are found in
(a) ileum [CBSE AIPMT 1998]
(b) ischium
(c) oesophagous
(d) ear
Ans.(a)
Lactealsare found in ileum they are
lymph vessels draining villi of vertebrate
small intestine. After digestion,
reconstituted fats are released into
lacteals as chylomicrons.
51The contraction of gall bladder is
due to [CBSE AIPMT 1998]
(a) gastrin
(b) secretin
(c) cholecystokinin
(d) enterogasterone
Ans.(c)
Cholecystokinin (also called
pancreozymin) is a hormone of mucosa
of small intestine. It is released in
response to chyme. It causes pancreas
to release pancreatic enzymes and gall
bladder to eject bile.
52If pancreas is removed, the
compound which remain
undigested is[CBSE AIPMT 1997]
(a) carbohydrates
(b) fats
(c) proteins
(d) All of these
Ans.(d)
Pancreas secretes pancreatic juice
which contain enzymes that acts on
proteins, carbohydrates and fats.
Enzymes of pancreatic juice are
(a) Pancreatic amylase which acts on
starch and glycogen
(polysaccharides).
(b) Trypsin, chymotrypsin and
carboxypeptidases, which act on
proteins and lipase which acts on
triglycerides and converts it into
fatty acids and glycerol.
If pancreas is removed from the body,
the digestion of all these would not occur.
53Which one of the following vitamin
can be synthesised by bacteria
inside the gut?[CBSE AIPMT 1997]
(a)B
1
(b) C
(c) D (d) K
Ans.(d)
Vitamin-K
2
(menaquinone) It is formed
by bacteria in the gut, while vitamin-K
1
(phylloquinone) is found in green plant
leaves.
Vitamin-B
1
(thiamine) It acts as
TPP-coenzyme for decarboxylases.
Vitamin-C(ascorbic acid). It helps in
development of teeth gums and
maintenance of capillary wall.
Vitamin-D(calciferol). It helps in
maintenance of calcium and
phosphorus balance within the body.
54Which one of the following is a
matching pair of a substrate and its
particular digestive enzyme?
[CBSE AIPMT 1996]
(a) Maltose — Maltase
(b) Lactose — Rennin
(c) Starch — Steapsin
(d) Casein — Chymotrypsin
156 NEETChapterwise Topicwise Biology

167. Ans.(a)
Intestinal juices contain a number of
oligosaccharidase which hydrolyse the
specific oligosaccharides into their
monosaccharides. Maltase is one of
them, which hydrolyses maltose into two
glucose molecules.
55The enzyme enterokinase helps in
the conversion of
[CBSE AIPMT 1995]
(a) pepsinogen into pepsin
(b) trypsinogen into trypsin
(c) caseinogen into casein
(d) proteins into polypeptides
Ans.(b)
Enterokinase helps in conversion of
trypsinogen into trypsin in small
intestine, which is an endoproteolytic
enzyme and hydrolyses the peptones
and proteoses into peptides.
56Rennin acts on
[CBSE AIPMT 1994, 2000]
(a) milk changing casein into calcium
paracaseinate at 7.2–8.2 pH
(b) protein in stomach
(c) fat in intestine
(d) milk changing casein into calcium
paracaseinate at 1-3 pH
Ans.(d)
Stomach secretes gastric juice pH
[1–3.5] which contains prorennin
secreted by the zymogen cells. Inactive
prorennin is converted into rennin by
HCl. Rennin acts on casein, a protein
milk changing it into calcium
paracaseinate, it is known as curdling of
milk.
57Inhibition of gastric and stimulation
of gastric, pancreatic and bile
secretions are controlled by
hormones [CBSE AIPMT 1994]
(a) gastrin, secretin, enterokinin and
cholecystokinin
(b) enterogasterone, gastrin,
pancreozymin and cholecystokinin
(c) gastrin, enterogasterone,
cholecystokinin and pancreozymin
(d) secretin, enterogasterone, gastrin
and enterokinin
Ans.(b)
Enterogasteronehormone secreted by
mucosa of duodenum inhibits secretion
of gastric juices and slows down the
gastric movements.
Gastrinis the hormone secreted by
G-cells/argentaffin cells of pyloric region
of stomach and stimulates the gastric
glands to secrete gastric juices.
Pancreozyminis secreted by mucosa of
duodenum and stimulates the acinal
cells of pancreas to secrete pancreatic
enzymes.
Cholecystokinin is secreted by cells of
mucosa of duodenum and stimulates
contraction of gall bladder to release
bile.
58Most of the fat digestion occurs in
[CBSE AIPMT 1993]
(a) rectum (b) stomach
(c) duodenum (d) small intestine
Ans.(d)
Fats are emulsified in small intestine by
the detergent action of bile salts.
Emulsification of fat converts large fat
droplets into large number of small
droplets, which provide larger surface
area to lipases.
Then pancreatic lipase (steapsin) which
is principal fat digesting enzyme, digest
about 2/3
rd
of fats in these stages.
Then intestinal lipase hydrolyses
some tri, di and monoglycerides to
fatty acids and glycerol molecules. So,
the most of fat digestion occurs in
small intestine.
59Secretion of gastric juice is
stopped by [CBSE AIPMT 1993]
(a) gastrin
(b) pancreozymin
(c) cholecystokinin
(d) enterogasterone
Ans.(d)
Enterogasterone is produced by small
intestine and slows down the secretion
of gastric juice and decreases the
gastric movements.
60Where is protein digestion
accomplished?[CBSE AIPMT 1991]
(a) Stomach (b) Ileum
(c) Rectum (d) Duodenum
Ans.(b)
The cells that line the ileum contain the
protease and carbohydrase enzymes
responsible for the final stages of
protein and carbohydrate digestion.
These enzymes are present in the
cytoplasm of the epithelial cells.
61Release of pancreatic juice is
stimulated by[CBSE AIPMT 1990, 89]
(a) enterokinase (b) cholecystokinin
(c) trypsinogen (d) secretin
Ans.(d)
Secretinis secreted byδ-cells of mucosa
of duodenum which stimulates pancreas
and controls the volume of pancreatic
juice including water and electrolytes.
62Pancreas produces
[CBSE AIPMT 1991]
(a) three digestive enzymes and one
hormone
(b) three digestive enzymes and two
hormones
(c) two digestive enzymes and one
hormone
(d) three digestive enzymes and no
hormone
Ans.(b)
Pancreas produces pancreatic juice
which contains trypsinogen,
chymotrypsin, carboxypeptidases,
lipase, pancreaticα-amylase, elastage,
nucleases. Out of these enzymes, the
first three are concerned with protein
digestion which finally converts protein
into small peptides. Pancreas also
secretes insulin and glucagon hormones
which acts antagonastically in
controlling the blood sugar level.
63Emulsification of fat will not occur
in the absence of
[CBSE AIPMT 1990]
(a) lipase
(b) bile pigments
(c) bile salts
(d) pancreatic juice
Ans.(c)
Bile is a watery greenish fluid containing
bile salts, bile pigments, cholesterol and
phospholipid. Bile salts play an
important role in digestion of fats.
Therefore in their absence
emulsification of fat cannot take place.
64Kwashiorkor disease is due to
[NEET (Odisha) 2019]
(a) simultaneous deficiency of proteins
and fats
(b) simultaneous deficiency of protein
and calories
(c) deficiency of carbohydrates
(d) protein deficiency not accompanied
by calorie deficiency
Digestion and Absorption 157
Nutritional and
Digestive Disorders
TOPIC 3

168. Ans.(d)
Kwashiorkor disease is due to protein
deficiency not accompanied by calorie
deficiency in the children of age 1-5
years. It’s symptoms are weak muscle,
thin limbs, retarded growth of the body
and brain, swelling of legs due to
retention of water (oedema), reddish
hair, pot belly, etc.
65Good vision depends on adequate
intake of carotene rich food.
Select the best option from the
following statements.[NEET 2017]
I. Vitamin-A derivatives are formed
from carotene.
II. The photopigments are
embedded in the membrane
discs of the inner segment.
III. Retinal is a derivative of
vitamin-A.
IV. Retinal is a light absorbing part
of all the visual photopigments.
(a) (I) and (II) (b) (I), (III) and (IV)
(c) (I) and (III) (d) (II), (III) and (IV)
Ans.(b)
Vitamin-A is a group of unsaturated
nutritional organic compounds that
includes retinol, retinal, retinoic acid and
β-carotene.
Vitamin-A is needed by the retina of eye
in the form of retinal, which combines
with protein opsin to form rhodopsin, the
light absorbing molecule.
66Anxiety and eating spicy food
together in an otherwise normal
human, may lead to
[CBSE AIPMT 2012]
(a) indigestion (b) jaundice
(c) diarrhoea (d) vomiting
Ans.(a)
Unhealthy eating habits together with
anxiety, stress or panic attacks may
cause indigestion, stomach ache,
stomach palpitations, nausea, etc.
67When breast feeding is replaced by
less nutritive food low in proteins
and calories; the infants below the
age of one year are likely to suffer
from [CBSE AIPMT 2009]
(a) marasmus (b) rickets
(c) kwashiorkor (d) pellagra
Ans.(c)
Prolonged starvation causes marasmus
disease due to a generalised wasting of
body because of both energy and protein
deficiency. The body becomes lean and
weak, eyes depressed and skin wrinkled.
Kwashiorkor is a disease caused by
continued deficiency of proteins in diet
although energy intake may be adequate.
Rickets (in children) occurs due to the
deficiency of vitamin-D.
Pellagra occurs due to the deficiency
of nicotinamide (vitamin-B
5
).
68Which one of the following is a
fat-soluble vitamin and it’s related
deficiency disease ?
[CBSE AIPMT 2007]
(a) Ascorbic acid — Scurvy
(b) Retinol — Xerophthalmia
(c) Cobalamine — Beri-beri
(d) Calciferol — Pellagra
Ans.(b)
Xerophthalmiais caused due to the
deficiency of vitamin-A (retinol). Retinol
and calciferol are fat soluble vitamins
but pellagra is not the deficiency disease
of calciferol.
69A patient is generally advised to
specially, consume more meat,
lentils, milk and eggs in diet only
when he suffers from
[CBSE AIPMT 2005]
(a) kwashiorkor (b) rickets
(c) anaemia (d) scurvy
Ans.(a)
The deficiency of proteins within the
body is responsible for a disease, known
askwashiorkor. So, a kwashiorkor
diseased patient is generally advised to
specially, consume more meat, lentils,
milk and eggs because these are rich
sources of protein.
70Which group of three of the
following five statements (A–E)
contains all three correct
statements regarding beri-beri ?
[CBSE AIPMT 2005]
A. A crippling disease prevalent
among the native population of
sub-Sahara Africa.
B. A deficiency disease caused by
lack of thiamine (vitamin-B
1
).
C. A nutritional disorder in infants
and young children when the diet
is persistently deficient in
essential protein.
D. Occurs in those countries where
the staple diet is polished rice.
E. The symptoms are pain from
neuritis, paralysis, muscle
wasting, progressive oedema,
mental deterioration and finally
heart failure.
(a) A, B and D (b) B, C and E
(c) A, C and E (d) B, D and E
Ans.(d)
The deficiency of vitamin-B
1
or thiamine
causes the disease beri-beri. This
disease occurs in those countries where
the staple diet is polished rice.
The symptoms of this disease are pain
from neuritis, paralysis, muscle wasting,
progressive oedema, mental
deterioration and finally heart failure.
71Which one of the following is the
correct matching of a vitamin, its
nature and its deficiency disease ?
[CBSE AIPMT 2004]
(a) Vitamin-A—Fat soluble—Night
blindness
(b) Vitamin-K—Fat soluble—Beri-beri
(c) Vitamin-A—Fat soluble—Beri-beri
(d) Vitamin-K—Water soluble—Pellagra
Ans.(a)
Fat soluble vitamins are A, D, E and K.
Deficiency of vitamin-A leads to night
blindness or nyctalopia because
vitamin-A is essential for synthesis of
visual pigments (rhodopsin).
72Which one of the following pairs is
not correctly matched?
[CBSE AIPMT 2003, 04]
(a) Vitamin-B
12
— Pernicious anaemia
(b) Vitamin-B
1
— Beri-beri
(c) Vitamin-C — Scurvy
(d) Vitamin-B
2
— Pellagra
Ans.(d)
Pellagra (ItalianPelle=skin,agra=
rough) is a skin disease which is caused
by the deficiency of vitamin-B
3
or niacin.
Pellagra is especially frequent among
the people eating food with low
tryptophan (an essential amino acid).
158 NEETChapterwise Topicwise Biology

169. 73Stool of a person contains whitish
grey colour due to malfunction of
which type of organ ?
[CBSE AIPMT 2002]
(a) Pancreas (b) Spleen
(c) Kidney (d) Liver
Ans.(d)
Bilirubin is broken down to urobilinogen
and stereobilinogen. Yellowish brown
colour of stool is due to the
stereobilinogen.
Due to the malfunctioning of liver,
insufficient production of
stereobilinogen leads to white stool.
74In a person of advanced age, the
hair become thinner gradually. It
happens because of decrease in
[CBSE AIPMT 2000]
(a) synthesis of glucose
(b) synthesis of proteins
(c) energy availability
(d) blood supply
Ans.(b)
In vertebrates,α-keratin (a protein)
constitutes almost the entire dry weight
of hair. Therefore, as the person ages,
metabolism decreases, synthesis of
protein decreases leading to thinning of
hair.
75Which of the following is
mismatched? [CBSE AIPMT 1999]
(a) Vitamin-K — Beri-beri
(b) Vitamin-D — Rickets
(c) Vitamin-C — Scurvy
(d) Vitamin-A — Xerophthalmia
Ans.(a)
Beri-beri is caused by the deficiency of
vitamin-B
1
(thiamine).
Vitamin-K is also known as
antihaemorrhagic factor.
The main sources of vitamin-K are green
leafy vegetables such as cauliflower,
cabbage, spinach, etc. It is also found in
animal sources like egg yolk, liver, etc.
Vitamin-K is essential for blood clotting
and deficiency of it causes haemorrhage.
76Which one of the following is a
protein deficiency disease?
[CBSE AIPMT 1998]
(a) Eczema (b) Cirrhosis
(c) Kwashiorkor (d) Night blindness
Ans.(c)
Kwashiorkor is a protein deficiency
disease (no calorie deficiency but
structural). Its common symptoms are
wasting of muscles, thinning of limbs,
failure of growth and brain development
and diarrhoea.
77For person suffering from high
blood cholesterol, the physicians
recommend [CBSE AIPMT 1996]
(a) pure ‘deshi ghee’ or butter
(b) vegetable oil such as groundnut oil
(c) red meat with layers of fats
(d) vanaspati margarine
Ans.(b)
A patient of high blood cholesterol is
suggested to take unsaturated fats as
vegetable oils. Such as ground nut oil
because high intake of saturated fat
causes high blood cholesterol which
ultimately gets deposited in the walls of
arteries causing their blockage resulting
in various cardiac-disease. Ghee, butter,
red meat vanaspati, they all are rich
sources of saturated fats.
78The vitamin-C or ascorbic acid
prevents [CBSE AIPMT 1995]
(a) rickets
(b) pellagra
(c) scurvy
(d) antibody synthesis
Ans.(c)
Vitamin-C or ascorbic acid prevents
scurvy (failure to form connective
tissue). Characterised by bleeding gums,
anaemia, loose teeth, painful and
swollen joints, delayed healing of
wounds and emaciation.
79Calcium deficiency occurs in the
absence of vitamin
[CBSE AIPMT 1994]
(a) D (b) C
(c) E (d) B
Ans.(a)
Vitamin-D (calciferol/antirachitic factor)
mainly helps in Ca/P balance in the body
fluids. It increases absorption of calcium
from intestine so, it is necessary for
formation of healthy bones and teeth.
Deficiency of vitamin-D causes
increased loss ofCa
2+
in urine, so, no
Ca
2+
gets deposited in the bones. This
cause rickets in children and in pregnant
woman it causes osteomalacia.
80Which of the following pair is
characterised by swollen lips, thick
pigmented skin of hands and legs
and irritability?
[CBSE AIPMT 1993, 94, 96]
(a) Thiamine — Beri-beri
(b) Protein — Kwashiorkor
(c) Nicotinamide — Pellagra
(d) Iodine — Goitre
Ans.(c)
Pellagrais characterised by swollen lips,
pigmented skin of hands, legs and
irritability. This disease is caused by the
deficiency of vitamin-B
3
or niacin. This is
pellagra protective vitamin and can be
synthesised in the body from amino acid
tryptophan.
Digestion and Absorption 159

170. 01The figure shows a diagrammatic
view of human respiratory system
with labelsA, B, CandD. Select the
option, which gives correct
identification and main function
and/or characteristic.[NEET 2013]
(a)A–trachea-long tube supported by
complete cartilaginous rings for
conducting inspired air
(b) B–pleural membrane-surround ribs
on both sides to provide cushion
against rubbing
(c) C–alveoli-thin walled vascular
bag-like structures for exchange of
gases
(d)D–lower end of lungs-diaphragm pulls
it down during inspiration
Ans.(c)
C–Alveoli are thin-walled vascular bag-like
structures for exchange of gases.
A–trachea or wind pipe is an air
conducting tube through which transport
of gases takes place.B–pleural
membrane is double layered which
reduces friction on the lung surface.
D–diaphragm is involved in the inspiration
and expiration process of breathing.
02The figure given below shows a
small part of human lung where
exchange of gas takes place. In
which one of the options given
below, the one part A, B, C or D is
correctly identified along with its
function. [CBSE AIPMT 2011]
(a)A — Alveolar
cavity
— main site of
exchange of
respiratory
gases
(b)D — Capillary
wall
— exchange of
gases takes
place here
(c)B — Red blood
cell
— transport of
mainly
haemoglobin
(d)C — Arterial
capillary
— passes oxygen
to tissues
Ans.(a)
Option (a) is correctly mentioned as
alveoli which are the primary sites of
exchange of gases. The exchange of
gases (O
2
and CO
2
) between the alveoli
and the blood occurs by simple diffusion.
03Which one of the following organs
in the human body is most affected
due to shortage of oxygen?
[CBSE AIPMT 1999]
(a) Intestine (b) Skin
(c) Kidney (d) Brain
Ans.(d)
The brain cells are highly specialised.
They cannot regenerate and respire
withoutO
2
. Therefore, the shortage ofO
2
leads to death of brain cells.
04In alveoli of the lungs, the air at the
site of gas exchange, is separated
from the blood by
[CBSE AIPMT 1997]
(a) alveolar epithelium only
(b) alveolar epithelium and capillary
endothelium
(c) alveolar epithelium, capillary
endothelium and tunica adventitia
(d) alveolar epithelium, capillary
endothelium, a thin layer of tunica
media and tunica adventitia
Ans.(b)
The wall of the capillaries consists of
only tunica internae which is made up of
simple squamous endothelium. The wall
of alveoli is also very thin, consisting of
squamous epithelium.
Breathingand
ExchangeofGases
17
Human Respiratory
System: Structure
TOPIC 1
A
B
C
D
Bronchus
Cut end of rib
Lung
Heart
D
B
A
C

171. Breathing and Exchange of Gases 161
05Select the correct events that
occur during inspiration.
[NEET (Sep.) 2020]
I. Contraction of diaphragm.
II. Contraction of external
inter-costal muscles.
III. Pulmonary volume decreases.
IV. Intra pulmonary pressure
increases.
(a) III and IV (b) I, II and IV
(c) Only IV (d) I and II
Ans.(d)
Statement I and II are correct as during
inspiration, the contraction of
diaphragm occurs which pulls it
downward, while the external intercostal
muscles contract and lifts up the ribs
and sternum. This increases the size of
the thoracic cavity and decreases the
pressure inside. As a result, air rushes in
and fills the lungs.
Statement III and IV are incorrect
because during inspiration, the volume
of the thoracic cavity increases. This
causes a similar increase in pulmonary
volume. An increase in pulmonary
volume decreases the intrapulmonary
pressure to less than the atmospheric
pressure which forces the air from
outside to move into the lungs.
06Select the correct statement.
[NEET (Odisha) 2019]
(a) Expiration occurs due to external
intercostal muscles
(b) Intrapulmonary pressure is lower
than the atmospheric pressure during
inspiration
(c) Inspiration occurs when atmospheric
pressure is less than intrapulmonary
pressure
(d) Expiration is initiated due to
contraction of diaphragm
Ans.(b)
Statement (b) is correct as
intrapulmonary pressure is lower than
the atmospheric pressure during
inspiration. Other statements can be
corrected as
(a) Inspiration occurs due to external
intercoastal muscles.
(c) Inspiration occurs when
atmospheric pressure is more than
intrapulmonary pressure.
(d) Inspiration is initiated due to
contraction of diaphragm.
07Air is breathed through
[CBSE AIPMT 1994]
(a) trachea—lungs—larynx—pharynx—
alveoli
(b) nose—larynx—pharynx—bronchus—
alveoli—bronchioles
(c) nostrils—pharynx—larynx—trachea—
bronchi—bronchioles—alveoli
(d) nose—mouth—lungs
Ans.(c)
In mammalian respiratory system, air is
breathed through nostrils, from nostrils
air passes through pharynx (common
passage for food and air)→larynx (voice
box)→trachea (wind pipe)→bronchi (2
for each side lungs)→bronchioles→
alveoli (small sacs or pouches for
exchange of gases).
08The Total Lung Capacity (TLC) is
the total volume of air
accommodated in the lungs at the
end of a forced inspiration. This
includes [NEET (Oct.) 2020]
(a) RV, IC (Inspiratory Capacity), EC
(Expiratory Capacity) and ERV
(b) RV, ERV, IC and EC
(c) RV, ERV, VC (Vital Capacity) and FRC
(Functional Residual Capacity)
(d) RV (Residual Volume), ERV (Expiratory
Reserve Volume), TV (Tidal Volume)
and IRV (Inspiratory Reserve Volume)
Ans.(d)
The Total Lung Capacity (TLC) is the total
volume of air accommodated in the
lungs at the end of a forced inspiration.
This includes Residual Volume (RV),
Expiratory Reserve Volume (ERV), Tidal
Volume (TV) and Inspiratory Reserve
Volume (IRV).
TLC is also equals to vital capacity of
residual volume. Thus, option (d) is
correct.
09The maximum volume of air a
person can breathe in after a
forced expiration is known as
[NEET (Odisha) 2019]
(a) expiratory capacity
(b) vital capacity
(c) inspiratory capacity
(d) total lung capacity
Ans.(b)
Vital Capacity (VC) is the maximum
volume of air a person can breathe in
after a forced expiration. This
includes ERV, TV and IRV or the
maximum volume of air a person can
breathe out after a forced inspiration.
10Tidal Volume and Expiratory
Reserve Volume of an athlete is
500 mL and 1000 mL, respectively.
What will be his Expiratory Capacity
if the Residual Volume is 1200 mL?
[NEET (National) 2019]
(a) 1700 mL (b) 2200 mL
(c) 2700 mL (d) 1500 mL
Ans.(d)
The Expiratory Capacity of athlete will be
1500 mL.
It can be calculated as
Given, Tidal Volume (TV)=500mL
Expiratory Reserve Volume (ERV)=1000 mL
Expiratory Capacity=TV+ERV
= + =500 1000 1500mL
11Match the items given in Column I
with those in Column II and select
thecorrectoption given below
[NEET 2018]
Column I Column II
1. Tidal volume (i) 2500–3000 mL
2. Inspiratory
reserve volume
(ii) 1100–1200 mL
3. Expiratory
reserve volume
(iii) 500–550 mL
4. Residual volume (iv) 1000–1100 mL
Respiratory Volumes
and Capacities
TOPIC 3
Diaphragm
contracted
Ribs and
sternum
raised
Volume of
thorax
increased
Rib cage
Air entering lungs
Mechanism of inspiration
Mechanism of Breathing
TOPIC 2

172. 1 2 3 4
(a) i iv ii iii
(b) iii i iv ii
(c) iii ii i iv
(d) iv iii ii i
Ans.(b)
Tidal Volume(TV) is the volume of air
inspired or expired during normal breath.
It is about 500–550 mL.
Inspiratory Reserve Volume(IRV) is the
extra amount of air that can be inspired
directly after a normal inspiration. It is
about 2500–3000 mL.
Expiratory Reserve Volume(ERV) is
the extra amount of air that can be
expired forcibly after a normal
expiration. It is about 1000-1100 mL.
Residual Volume (RV)is the volume of
air which remains still in the lung after
the most forceful expiration. It is about
1100-1200 mL.
Therefore, option (b) is correct.
12Lungs are made up of air-filled
sacs, the alveoli. They do not
collapse even after forceful
expiration, because of[NEET 2017]
(a) Residual Volume (RV)
(b) Inspiratory Reserve Volume (IRV)
(c) Tidal Volume (TV)
(d) Expiratory Reserve Volume (ERV)
Ans.(a)
In lungs, even after the most forceful
expiration, some of the volume of air
remains. This volume is termed Residual
Volume (RV). Due to this, lungs do not
collapse even after the most forceful
expiration. RV is about 1100
mL -1200 mL.
13Lungs do not collapse between
breaths and some air always
remains in the lungs which can
never be expelled because
[NEET 2016, Phase II]
(a) there is a negative pressure in the
lungs
(b) there is a negative intrapleural
pressure pulling at the lung walls
(c) there is a positive intrapleural
pressure
(d) pressure in the lungs in higher than
the atmospheric pressure
Ans.(b)
Lungs do not collapse between breaths
and some air always remains in the lungs
which can be never expelled because
there is a negative intrapleural pressure
pulling at the lung walls. Alveoli are basic
functional unit of lungs.
The outer alveolar wall surface has cells
which secrete DPPC also called aslipid
surfactant.
The surfactant expands the alveoli due
to which the negative pressure inside
the alveoli increases. This prevents the
alveoli from collapsing. Since, in this
question both options (a) and (b) are
correct and option (b) provides more
appropriate explanation, therefore, it
must be chosen.
14Listed below are four respiratory
capacities( )1 4-and four jumbled
respiratory volumes of a normal
human adult
Respiratory
Respiratory
Capacities Volumes
1. Residual volume 2500 mL
2. Vital capacity 3500 mL
3. Inspiratory reserve volume 1200
mL
4. Inspiratory capacity 4500 mL
Which one of the following is the
correct matching of two capacities
and volumes?[CBSE AIPMT 2010]
(a) (2) 2500 mL, (3) 4500 mL
(b) (3) 1200 mL, (4) 2500 mL
(c) (4) 3500 mL, (1) 1200 mL
(d) (1) 4500 mL, (2) 3500 mL
Ans.(c)
Inspiratory Capacity(IC) is the maximum
amount of air that can be inspired after a
normal expiration,IC TV IRV= + . It is
3500 mL in adult male and 2400 mL in
adult female.
Residual Volume (RV) is the amount of
air remaining in the lungs after a
forced exhalation. Its average value is
1200 mL and 1100 mL in adult male and
female respectively.
15What is the vital capacity of our
lungs? [CBSE AIPMT 2008]
(a) Inspiratory reserve volume plus
tidal volume
(b) Total lung capacity minus expiratory
reserve volume
(c) Inspiratory reserve volume plus
expiratory reserve volume
(d) Total lung capacity minus residual
volume
Ans.(d)
Vital capacity is the sum of inspiratory
reserve volume, tidal volume and
expiratory reserve volume. It is about
4800 mL.
Total lung capacity is the sum of vital
capacity and residual volume, i.e., vital
capacity of our lungs is total lung
capacity minus residual volume.
Tidal volume is the amount of air which
normally passes into and out of the lungs
during each cycle of quite breathing. It is
about 800 mL in adult person.
Inspiratory reserve volume is the extra
volume of air that can be inhaled into
lungs during deepest possible
inspiration.
16The quantity 1500 mL in the
respiratory volumes of a normal
human adult refers to
[CBSE AIPMT 1996]
(a) maximum air that can be breathed in
and breathed out
(b) residual volume
(c) expiratory reserve volume
(d) total lung capacity
Ans.(b)
Residual air is the volume of air that
remains in the lungs after the most
forceful expiration. It equals to 1500 mL.
Residual air mostly occurs in alveoli.
Maximum air that can be breathed in
and breathed out is vital capacity
(3500–4500 mL) while the air that can
be expired over and above the tidal air
by most forceful expiration is the
expiratory reserve volume (1200 mL).
17Assertion(A) A person goes to high
altitude and experiences ‘altitude
sickness’ with symptoms like
breathing difficulty and heart
palpitations.
Reason(R) Due to low
atmospheric pressure at high
altitude, the body does not get
sufficient oxygen.
In the light of the above
statements, choose the correct
answer from the options given
below. [NEET 2021]
(a) Both A and R are true and R is the
correct explanation of A
(b) Both A and R are true, but R is not the
correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true
162 NEETChapterwise Topicwise Biology
Exchange and
Transport of Gases
TOPIC 4

173. Ans.(a)
Both A and R are true and R is the
correct explanation of A.
A person goes to the high altitude and
experiences altitude sickness like heavy
breathing and heart palpitation. It is due
to low atmospheric pressure at high
altitude, the body does not get sufficient
oxygen.
18Select the favourable conditions
required for the formation of
oxyhaemoglobin at the alveoli.
[NEET 2021]
(a) HighpO
2
, lowpCO
2
, less H
+
, lower
temperature
(b) LowpO
2
, highpCO
2
, more H
+
, higher
temperature
(c) HighpO
2
, highpCO
2
, less H
+
, higher
temperature
(d) LowpO
2
, lowpCO
2
, more H
+
, higher
temperature
Ans.(a)
The favourable conditions for the
formation of oxyhaemoglobin is highpO
2
,
lesser H
+
concentration and lower
temperature found in alveoli, whereas low
pO
2
, highH
+
concentration and high
temperature are favourable for
dissociation of oxygen from the
oxyhaemoglobin found in tissues.
19The partial pressures (in mm Hg) of
oxygen (O
2
) and carbon dioxide
(CO
2
) at alveoli (the site of
diffusion) are [NEET 2021]
(a)pO
2
=104 andpCO
2
=40
(b)pO
2
=40 andpCO
2
=45
(c)pO
2
=95 andpCO
2
= 40
(d)pO
2
=159 andpCO
2
=0.3
Ans.(a)
Partial pressures of oxygen( in mm Hg)
and carbon dioxide at alveoli arepO
2
=104andpCO
2
=40.
Respir
atory
gas
Atmos-
pheric
air
Alveoli
Blood
(deoxy
genated)
Blood
(oxyge
nated)
Tissue
O
2
159 104 40 95 40
CO
2
0.3 40 45 40 45
20Match the following columns and
select the correct option from the
codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. Pneumotaxic
centre
1. Alveoli
B.O
2
dissociation
curve
2. Pons region
of brain
C. Carbonic
anhydrase
3. Haemoglobin
D. Primary site of
exchange of gases
4. RBC
Codes
A B C D
(a) 1 3 2 4
(b) 2 3 4 1
(c) 3 2 4 1
(d) 4 1 3 2
Ans.(b)
Option (b) is correct match, which is as
follows. Pneumotaxic centre is present
in the pons region of the brain.
O
2
dissociation curve is useful in
studying the effect of factors likepCO
2
,
H
+
concentration, etc., on binding ofO
2
with haemoglobin.
Carbonic anhydrase is an enzyme
present on the surface of RBC.
Primary site of exchange of gases is the
alveoli of the lungs.
21Identify the wrong statement with
reference to transport of oxygen.
[NEET (Sep.) 2020]
(a) Partial pressure ofCO
2
can interfere
withO
2
binding with haemoglobin
(b) HigherH
+
concentration in alveoli
favours the formation of
oxyhaemoglobin
(c) LowpCO
2
in alveoli favours the
formation of oxyhaemoglobin
(d) Binding of oxygen with haemoglobin
is mainly related to partial pressure of
O
2
Ans.(b)
Statement in option (b) is incorrect with
reference to transport of oxygen. It can
be corrected as
In alveoli highpO
2
, lowpCO
2
, low H
+
concentration and lower temperature
are the factors that favour the formation
of oxyhaemoglobin.
22Reduction in pH of blood will
[NEET 2016, Phase I]
(a) reduce the blood supply to the brain
(b) decrease the affinity of haemoglobin
with oxygen
(c) release bicarbonate ions by the liver
(d) reduce the rate of heartbeat
Ans.(b)
Reduction in pH of blood, i.e. increase in
acidity favours the dissociation of
oxyhaemoglobin thereby giving upmore
O
2
. When this phenomenon occurs due
to increase inCO
2
concentration then it
is called Bohr effect.
23The partial pressure of oxygen in
the alveoli of the lungs is
[NEET 2016, Phase II]
(a) equal to that in the blood
(b) more than that in the blood
(c) less than that in the blood
(d) less than that of carbon dioxide
Ans.(b)
The partial pressure of oxygen( O )
2
pin
alveoli of lungs is 104 mm Hg, which is
more than that of blood in the blood
capillaries of lung alveoli (40 mm Hg).
This difference allows passive diffusion
ofO
2
from air filled in the lungs to the
blood vessels of lung alveoli.
24Approximately seventy percent of
carbon dioxide absorbed by the
blood will be transported to the
lungs [CBSE AIPMT 2014]
(a) as bicarbonate ions
(b) in the form of dissolved gas
molecules
(c) by binding to RBC
(d) as carbaminohaemoglobin
Ans.(a)
The largest fraction of carbon dioxide,
i.e. about 70% is converted to
bicarbonates(HCO )
3
−
and transported in
the plasma.
CO H O
2 2
Anhydrase
Carbonic
+a H CO
2 3a
Anhydrase
Carbonic
HCO H
3
− +
+
About−23%ofCO
2
is carried by
haemoglobin as carbaminohaemoglobin
CO Hb (haemoglobin) HbCO
2 3
+ s
Carbo amino haemoglobin
25People who have migrated from
the planes to an area adjoining
Rohtang Pass about six months
back [CBSE AIPMT 2012]
(a) have more RBCs and their
haemoglobin has a lower binding
affinity toO
2
(b) are not physically fit to play games
like football
Breathing and Exchange of Gases 163

174. (c) suffer from altitude sickness with
symptoms like nausea, fatigue, etc.
(d) have the usual RBC count but their
haemoglobin has very high binding
affinity toO
2
Ans.(a)
As a person moves up a hill thepO
2
and
total atmospheric pressure decreases.
Decrease inpO
2
due to the increasing
altitude, stimulates the Juxtaglomerular
cells of kidney to secrete erythropoietin
hormone which increases the number of
RBCs (polycythemia) to compensate the
supply ofO
2
. At higher altitude,
haemoglobin has lower binding affinity
toO
2
because the primary factor
responsible for binding ispO
2
which
decreases at higher altitude.
26Which two of the following changes
(1- )4usually tend to occur in the
plain dwellers when they move to
high altitudes (3,500 m or more)?
[CBSE AIPMT 2010]
1. Increase in red blood cell size
2. Increase in red blood cell
production
3. Increased breathing rate
4. Increase in thrombocyte count
Changes occurring are
[CBSE AIPMT 2010]
(a) 2 and 3 (b) 3 and 4
(c) 1 and 4 (d) 1 and 2
Ans.(a)
When a person moves to higher
altitudes, thepO
2
and total atmospheric
pressure decrease. Hypoxia stimulates
the Juxtaglomerular cells of the kidney
to release erythropoietin hormone which
stimulates erythropoesis in bone
marrow causing polycythemia (increase
in RBCs production). Hypoxia will also
increase breathing rate. Initially the size
of RBCs will also increase but with
increase in number of RBCs, the size of
RBCs becomes normal.
27The haemoglobin of a human
foetus [CBSE AIPMT 2008]
(a) has a lower affinity for oxygen than
that of the adult
(b) its affinity for oxygen is the same as
that of an adult
(c) has only 2 protein sub-units instead of 4
(d) has a higher affinity for oxygen than
that of an adult
Ans.(d)
Haemoglobin is the protein that makes
red blood corpuscles appear red, binds
easily and reversibly with oxygen. Normal
values for haemoglobin are 14–20 g/100
mL of blood in infants, 13–18 g/100 mL in
adult male and 12–16 g/100 mL in adult
females. Foetal red blood cells are not
sickle-shaped even in that destined to
have sickle-cell anaemia, i.e.
haemoglobin of foetus has higher
affinity of oxygen than that an adult.
28People living at sea level have
around 5 million RBC per cubic
millimeter of their blood whereas
those living at an altitude of
5400 metre have around 8 million.
This is because at high altitude
[CBSE AIPMT 2006]
(a) atmosphericO
2
level is less and
hence, more RBCs are needed to
absorb the required amount ofO
2
to
survive
(b) there is more UV radiation which
enhances RBC production
(c) people eat more nutritive food,
therefore, more RBCs are formed
(d) people get pollution-free air to
breathe and more oxygen is available
Ans.(a)
At high altitudes, the atmosphericO
2
level is less and hence, more RBCs are
needed to absorb the required amount
ofO
2
to survive. That is why, the people
living at sea level have around 5 million
RBC/mm
3
of their blood whereas, those
living at an altitude of 5400 meter have
around 8 million RBC/mm
3
of their blood.
29Blood analysis of a patient reveals
an unusually high quantity of
carboxyhaemoglobin content.
Which of the following conclusions
is most likely to be correct?
[CBSE AIPMT 2004]
(a) The patient has been inhaling polluted
air containing unusually high content
of carbon disulphide
(b) The patient has been inhaling polluted
air containing unusually high content
of chloroform
(c) The patient has been inhaling polluted
air containing unusually high content
of carbon dioxide
(d) The patient has been inhaling polluted
air containing unusually high content
of carbon monoxide
Ans.(d)
Inhalation of polluted air causes increase
in CO in the blood of a person. Carbon
monoxide forms a stable compound with
haemoglobin called carboxy-
haemoglobin as affinity of Hb for CO is
210 times greater than its affinity forO
2
.
In this form haemoglobin does not carry
oxygen resulting in death due to hypoxia.
Hb + CO Hb COa
Haemoglobin Carboxy haemoglobin
30WhenCO
2
concentration in blood
increases, breathing becomes
[CBSE AIPMT 2004]
(a)shallower and slow
(b) there is no effect on breathing
(c) slow and deep
(d) faster and deeper
Ans.(d)
WhenCO
2
concentration in blood
increases breathing becomes faster and
deeper. The effect of increasedCO
2
is to
decrease the affinity of haemoglobin for
O
2
.
Thus, due to Bohr’s effect theCO
2
released in respiring tissue accelerates
the delivery of oxygen by faster and
deeper breathing.
31The process of migration of
chloride ions from plasma to RBC
and of carbonate ions from RBC to
plasma is [CBSE AIPMT 1999]
(a) chloride shift
(b) ionic shift
(c) atomic shift
(d)Na
+
pump
Ans.(a)
To maintain electrostatic neutrality of
plasma, many chloride ions diffuse from
plasma into RBCs and bicarbonate ions
pass out.
The chloride content of RBCs increases
when oxygenated blood becomes
deoxygenated. This is termed as chloride
shift or Hamburger shift.
32The exchange of gases in the
alveoli of the lungs takes place by
[CBSE AIPMT 1998]
(a) simple diffusion
(b) osmosis
(c) active transport
(d) passive transport
164 NEETChapterwise Topicwise Biology

175. Ans.(a)
Oxygen diffuses from alveoli to
deoxygenated blood andCO
2
diffuses
from deoxygenated blood to alveoli by
simple diffusion. Diffusion is defined as,
the flow of the substance (gases) from a
region of their higher concentration to a
region of lower concentration.
33Which one of the following
statements about blood
constituents and transport of
respiratory gases is most
accurate? [CBSE AIPMT 1996]
(a) RBCs transport oxygen whereas
WBCs transportCO
2
(b) RBCs transport oxygen whereas
plasma transports onlyCO
2
(c) RBCs as well as WBCs transport both
oxygen andCO
2
(d) RBCs as well as plasma transport
both oxygen andCO
2
Ans.(d)
RBCs and plasma both transportO
2
and
CO
2
either in chemically bonded state or
in dissolved state.
34At high altitude, the RBCs in the
human blood will[CBSE AIPMT 1995]
(a) increase in size
(b) decrease in size
(c) increase in number
(d) decrease in number
Ans.(c)
At high altitude there is lowO
2
concentration, so RBCs increase in
number, i.e.O
2
supply can be
maintained to organs.
35The carbon dioxide is transported
viablood to lungs as
[CBSE AIPMT 1995]
(a) dissolved in blood plasma
(b) in the form of carbonic acid only
(c) in combination with haemoglobin only
(d) carbaminohaemoglobin and as
carbonic acid
Ans.(d)
Most of theCO
2
(70%) that dissolves in
plasma reacts with water forming
carbonic acid. This reaction occurs
rapidly inside RBCs because of the
presence of an enzyme carbonic
anhydrase. Moreover, above 23% of total
bloodCO
2
is transported by loosely
binding with haemoglobin forming an
unstable compound called
carbaminohaemoglobin.
36Although muchCO
2
is carried in
blood, yet blood does not become
acidic, because
[CBSE AIPMT 1995]
(a) it is absorbed by the leucocytes
(b) blood buffers play an important role in
CO
2
transport
(c) it combines with water to formH CO
2 3
which is neutralised byNa CO
2 3
(d) it is continuously diffused through
tissues and is not allowed to
accumulate
Ans.(b)
Buffer of the blood is sodium
bicarbonate which play an important
role inCO
2
transport. DuringCO
2
transportation, carbonic acid
dissociates intoH
+
andHCO
3
–
(bicarbonate ions). This bicarbonate
combines with sodium forming sodium
bicarbonate. Thus, concentration of
carbonic acid does not increase in blood
due to the presence of sodium and thus
blood does not become acidic. About
70% ofCO
2
released during cellular
respiration is transported by blood in the
form of sodium bicarbonate in plasma.
37Oxygen dissociation curve of
haemoglobin is[CBSE AIPMT 1994]
(a) sigmoid (b) hyperbolic
(c) linear (d) hypobolic
Ans.(a)
Oxygen haemoglobin dissociation curve
gives the relationship between the
saturation of haemoglobin and oxygen
tension. The curve obtained by plotting
the percent saturation of Hb against
time is sigmoid, at 38°C and pH 7.4.
Dissociation of oxyhaemoglobin can be
promoted by rise in the body
temperature and low pH (highCO
2
).
38Carbon dioxide is transported from
tissues to respiratory surface by
only [CBSE AIPMT 1993]
(a) plasma and erythrocytes
(b) plasma
(c) erythrocytes
(d) erythrocytes and leucocytes
Ans.(a)
Carbon dioxide is transported from
tissues to respiratory surface by only
plasma and erythrocytes. Carbon
dioxide(CO )
2
transportation by blood
is much easier than oxygen due to the
high solubility ofCO
2
in water. During
transport ofCO
2
, 7% ofCO
2
is
dissolved in plasma, 23% as
carbaminohaemoglobin and 70%
transported as bicarbonates (HCO
3
−
).
39Carbonic anhydrase occurs in
[CBSE AIPMT 1991]
(a) lymphocytes (b) blood plasma
(c) RBC (d) leucocytes
Ans.(c)
The erythrocyte (RBC) contains
sufficient amount of carbonic
anhydrase enzyme which catalyses
the reaction betweenCO
2
andH O
2
and
helps in transportation ofCO
2
from
tissues to the lungs.
40According to Central Pollution
Control Board (CPCB) what size (in
diameter) of particulate is
responsible for causing greater
harm to human health?
[NEET (Oct.) 2020]
(a) 3.5 micrometers
(b) 2.5 micrometers
(c) 4.0 micrometers
(d) 3.0 micrometers
Ans.(b)
According to Central Pollution Control
Board (CPCB), particulate size 2.5
micrometers or less in diameter (PM
2.5) are responsible for causing the
greatest harm to human health. These
fine particulates can be inhaled deep
into the lungs and can cause breathing
and respiratory symptoms, irritation,
inflammation and damage to the lungs
and premature deaths.
41Which of the following is an
occupational respiratory disorder?
[NEET 2018]
(a) Botulism (b) Silicosis
(c) Anthracis (d) Emphysema
Ans.(b)
Silicosisis an occupational respiratory
disorder which is caused due to
excessive inhalation of silica dust. It
usually affects the workers of grinding or
stone breaking industries. The long-term
exposure can cause lung fibrosis (or
stiffening), leading to breathing difficulties.
AnthracisorAnthraxis a bacterial
infection caused byBacillus anthracis.
Botulismis food poisoning infection
caused byClostridium botulinum.Its
symptoms include diarrhoea, vomiting,
abdominal distention, etc.
Breathing and Exchange of Gases 165
Disorders of
Respiratory System
TOPIC 5

176. 166 NEETChapterwise Topicwise Biology
Emphysemais a lung disease, that
damages the air sacs and causes
shortness of breathe. It may be caused
by smoking, deficiency of enzymes
alpha-1-antitrypsin and air pollution.
42Which one of the following options
correctly represents the lung
conditions in asthma and
emphysema, respectively?
[NEET 2018]
(a) Increased respiratory surface;
Inflammation of bronchioles
(b) Increased number of bronchioles;
Increased respiratory surface
(c) Inflammation of bronchioles;
Decreased respiratory surface
(d) Decreased respiratory surface;
Inflammation of bronchioles
Ans.(c)
Asthmais inflammation of bronchioles.
Its symptoms include wheezing,
coughing and difficulty in breathing
mainly during expiration.
Emphysemais an inflation or abnormal
distension of the bronchioles or alveolar
sacs of the lungs. Many of the septa
between the alveoli are destroyed and
much of the elastic tissue of the lungs is
replaced by connective tissue. As a
result alveolar septa collapse and the
surface area get greatly reduced.
43Name the chronic respiratory
disorder caused mainly by cigarette
smoking [NEET 2016, Phase I]
(a) asthma
(b) respiratory acidosis
(c) respiratory alkalosis
(d) emphysema
Ans.(d)
Emphysema is characterised by inflation
or distension of alveoli by dissolution of
wall of the two adjacent lung alveoli. It
generally occurs due to chronic
cigarette smoking.
44Name the pulmonary disease in
which alveolar surface area
involved in gas exchange is
drastically reduced due to damage
in the alveolar walls.
[CBSE AIPMT 2015]
(a) Pleurisy
(b) Emphysema
(c) Pneumonia
(d) Asthma
Ans.(b)
Empysema is a chronic respiratory
disease where there is over-inflation
of the air sacs (alveoli) in the lung,
causing a decrease in lung function
and often, breathlessness.
In this disease, the alveolar walls are
damaged leading to drastic reduction
in gas exchange.
45Which one of the following is the
correct statement for respiration in
humans? [CBSE AIPMT 2012]
(a) Cigarette smoking may lead to
inflammation of bronchi
(b) Neural signals from pneumotoxic
centre in pons region of brain can
increase the duration of inspiration
(c) Workers in grinding and stone
breaking industries may suffer, from
lung fibrosis
(d) About 90% of carbon dioxide(CO )
2
is
carried by haemoglobin as carbamino
haemoglobin
Ans.(c)
Irritating gases, fumes, dusts, etc.,
present in the work place result in
lung disorders.
This is because the defence
mechanism of the body cannot fully
cope with this situation of so much
dust.
Long exposure can give rise to
inflammation leading to fibrosis
(proliferation of fibrous tissue) and
thus causing serious lung damage.

177. 01Persons with ‘AB’ blood group are
called as "universal recipients". This
is due to [NEET 2021]
(a) absence of antigens-A and B on the
surface of RBCs
(b) absence of antigens-A and B in
plasma
(c) presence of antibodies, anti-A and
anti-B on RBCs
(d) absence of antibodies, anti-A and
anti-B in plasma
Ans.(d)
Person with AB blood group are universal
recipient because they do not have any
antibodies to anti-A and anti-B in their
blood hence they can receive blood from
a donor of any blood type.
During transfusion, plasma is matched
to avoid A and B antibodies in the
transfused plasma that will attack the
recipient's red blood cells. Person with
AB blood type does not contain A and B
antibodies in plasma. Therefore, they are
also called universal plasma donors.
02Which enzyme is responsible for
the conversion of inactive
fibrinogens to fibrins?[NEET 2021]
(a) Thrombin
(b) Renin
(c) Epinephrine
(d) Thrombokinase
Ans.(a)
Plasma contains prothrombin and
fibrinogen. During clotting, inactive
prothrombin converts to active thrombin.
Thrombin acts as a proteolytic enzyme
to convert soluble plasma protein
fibrinogen molecule produced from the
liver in the presence of vitamin-K to form
insoluble fibrin monomer.
03Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. Eosinophils 1. Immune response
B. Basophils 2. Phagocytosis
C. Neutrophils 3. Release histaminase
destructive enzymes
D. Lymphocytes 4. Release granules
containing histamine
A B C D
(a) 4 1 2 3
(b) 1 2 4 3
(c) 2 1 3 4
(d) 3 4 2 1
Ans.(d)
Option (d) is the correct. It can be
explained as follows
Eosinophils are associated with allergic
reactions and release histaminase,
destructive enzymes. Basophils secrete
histamine, serotonin, heparin, etc. and
are involved in inflammatory reactions.
Neutrophils are phagocytic cells.
Both B and T lymphocytes are
responsible for immune responses of
the body.
04Which of the following gastric cells
indirectly help in erythropoiesis?
[NEET 2018]
(a) Goblet cells (b) Mucous cells
(c) Chief cells (d) Parietal cells
Ans.(d)
Parietal cells(oxyntic cells) secrete
hydrochloric acid and castle intrinsic
factor. HCl converts iron (in diet) from
ferric to ferrous form which can be easily
absorbed and used during erythropoiesis
(formation of RBCs).
Castle intrinsic factor helps in absorbing
vitamin-B
12
and its deficiency causes
pernicious anaemia. The functions of
other cells are as follows
Mucous or Goblet cells secrete mucus
that lines the stomach and protects it
from the acid present in stomach.
Chief cells secrete gastric digestive
enzymes as proenzymes or zymogens.
05Match the items given in Column I
with those in Column II and select
thecorrectoption given below
[NEET 2018]
Column I Column II
1. Fibrinogen (i) Osmotic
balance
2. Globulin (ii) Blood clotting
3. Albumin (iii) Defence
mechanism
1 2 3
(a) (i) (iii) (ii)
(b) (i) (ii) (iii)
(c) (iii) (ii) (i)
(d) (ii) (iii) (i)
Ans.(d)
Fibrinogenis a soluble plasma protein
that is stimulated by thrombin and gets
converted into insoluble form fibrin. The
latter helps in the formation of blood clot
to seal the wound and stop bleeding.
BodyFluidsand
Circulation
18
Blood
TOPIC 1

178. Globulinsare simple proteins that form a
large fraction of blood serum proteins
involved in defence mechanism. There
are four main types of globulins that are
manufactured in liver, namely alpha-1,
alpha-2, beta and gamma.
Albuminis a plasma protein that is
manufactured by the liver. It helps in
maintaining osmotic pressure which
prevents the fluid-leakage out into the
tissues from the bloodstream.
06Adult human RBCs are enucleate.
Which of the following statement(s)
is/are most appropriate
explanation for this feature?
(I) They do not need to reproduce.
[NEET 2017]
(II) They are somatic cells.
(III) They do not metabolise.
(IV) All their internal space is
available for oxygen transport.
Codes
(a) Only (IV)
(b) Only (I)
(c) (I), (III) and (IV)
(d) (II) and (III)
Ans.(a)
The absence of nucleus in RBC is an
adaptation that allows it to contain more
haemoglobin and carry more oxygen by
providing empty space. This adaptation
also aids in effective diffusion of oxygen.
Concept EnhancerRBCs are initially
produced in bone marrow with a nucleus.
They, then undergo enucleation at
maturity, in which their nucleus is
removed.
07Name the blood cells, whose
reduction in number can cause
clotting disorder, leading to
excessive loss of blood from the
body. [NEET 2016, Phase II]
(a) Erythrocytes (b) Leucocytes
(c) Neutrophils (d) Thrombocytes
Ans.(d)
A reduction in number of thrombocytes
can lead to clotting disorders which will
result in excessive loss of blood from the
body. These are also called blood platelets.
08Serum differs from blood in
[NEET 2016, Phase II]
(a) lacking globulins
(b) lacking albumins
(c) lacking clotting factors
(d) lacking antibodies
Ans.(c)
When all clotting factors along with cells
are removed from plasma, it becomes
serum.
09Person with blood group AB is
considered as universal recipient
because he has[CBSE AIPMT 2014]
(a) Both A and B antigens on RBC but no
antibodies in the plasma
(b) Both A and B antibodies in the plasma
(c) No antigen on RBC and no antibody in
the plasma
(d) Both A and B antigens in the plasma
but no antibodies
Ans.(a)
Blood group AB is universal recipient
because the person with AB blood group
has both A and B antigens on RBC but no
antibodies in the plasma.Other blood
group and their genotypes are given
below
Blood
group
Antigen (s)
Present on
the RBC
Antibodies
Present in
Serum
Genotypes
A Antigen-A Anit-b AA/AO
B Anitigen-B Anti-a BB/BO
O
None
Anti-a and b O
10A certain road accident patient
with unknown blood group needs
immediate blood transfusion. His
one doctor friend at once offers his
blood. What was the blood group of
the donor? [CBSE AIPMT 2012]
(a) Blood group B (b) Blood group AB
(c) Blood group O (d) Blood group A
Ans.(c)
Blood group is tested by two types of
sera, i.e. anti-A (antibody-A) and anti-B
(antibody-B). Persons with blood group O
possess both antibodies in their plasma
but have no antigens in their RBCs. So,
RBCs of blood group ‘O’ do not show
clumping in any of the two sera. That’s
why, persons with blood group ‘O’ are
calleduniversal donorand they can
donate blood to a person with any type of
blood group.
11Which one of the following plasma
proteins is involved in the
coagulation of blood?
[CBSE AIPMT 2011]
(a) Serum amylase (b) A globulin
(c) Fibrinogen (d) An albumin
Ans.(c)
Fibrinogen (factor 1] is a soluble plasma
glycoprotein, synthesised by the liver. It is
converted by thrombin into fibrin during
blood coagulation. Fibrin is then
cross-linked by factor XIII to form a clot.
12A person with unknown blood
group under ABO system, has
suffered much blood loss in an
accident and needs immediate
blood transfusion. His one friend
who has a valid certificate of his
own blood type, offers for blood
donation without delay. What
would have been the type of blood
group of the donor friend?
[CBSE AIPMT 2011]
(a) Type AB (b) Type O
(c) Type A (d) Type B
Ans.(b)
Blood type ‘O’ has no antigen but both
types of antibodies ‘a’ and ‘b’. The
person with blood type ‘O’ is universal
donor.
13The most popularly known blood
grouping is the ABO grouping. It is
named ABO and not ABC because
‘O’ in it refers to having
[CBSE AIPMT 2009]
(a) other antigens besides A and B on
RBCs
(b) over dominance of this type on the
genes for A and B types
(c) one antibody only—either anti-A or
anti-B on the RBCs
(d) no antigens A and B on RBCs
Ans.(d)
Landsteiner divided human population
into four groups based on the presence
of antigens found in their RBCs. Each
group represented a blood group. Thus,
there are four types of blood groups A, B,
AB and O. Blood group ‘O’ does not
contain any antigen on RBCs, hence can
be given to any person, that’s why this
blood group is called universal donor.
14If you suspect major deficiency of
antibodies in a person, to which of
the following would you look for
confirmatory evidence?
[CBSE AIPMT 2007]
(a) Serum albumins
(b) Serum globulins
(c) Fibrinogen in the plasma
(d) Haemocytes
168 NEETChapterwise Topicwise Biology

179. Ans.(b)
Deficiency of antibodies can be
confirmed by serum globulins as
antibodies are also called
immunoglobulins and constitute the
gamma globulin part of blood proteins.
These are secreted by activated B-cells
or plasma cells.
15A drop of each of the following, is
placed separately on four slides.
Which of them will not coagulate?
[CBSE AIPMT 2007]
(a) Blood plasma
(b) Blood serum
(c) Sample from the thoracic duct of
lymphatic system
(d) Whole blood from pulmonary vein
Ans.(b)
Serum will not coagulate. Because
serum do not contain clotting factor,
RBCs or WBCs. It is blood plasma not
including the fibrinogens.
16Examination of blood of a person
suspected of having anaemia, shows
large, immature, nucleated
erythrocytes without haemoglobin.
Supplementing his diet with which of
the following, is likely to alleviate his
symptoms? [CBSE AIPMT 2006]
(a) Thiamine
(b) Folic acid and cobalamin
(c) Riboflavin
(d) Iron compounds
Ans.(d)
Anaemia refers to any condition in which
there is an abnormally low haemoglobin
concentration and/or blood cell count.
The most common cause is deficiency of
iron which is an essential element of
haemoglobin molecule. Thus, the iron
compounds in the diet will help to
alleviate the symptoms of anaemia.
17Antibodies in our body are complex
[CBSE AIPMT 2006]
(a) lipoproteins
(b) steroids
(c) prostaglandins
(d) glycoproteins
Ans.(d)
Antibodies are the proteins
(glycoproteins) called immunoglobulins.
These are produced by B-lymphocytes in
response to entry of a foreign substance
or antigen into the body. Lipoproteins
are the micellar complex of protein and
lipids.
Steroids are a group of lipids derived from
a saturated compound cyclopentano
perhydrophenanthrene which has a
nucleus of four rings.
Prostaglandin is a group of organic
compounds derived from essential fatty
acids and causing a range of
physiological effects in animals.
18Which of the following substances,
if introduced in the blood stream,
would cause coagulation, at the
site of its introduction?
[CBSE AIPMT 2005]
(a) Fibrinogen
(b) Prothrombin
(c) Heparin
(d) Thromboplastin
Ans.(d)
Lipoproteinaceous, thromboplastin is
released by the injured tissue which
causes blood clotting. In blood vessels,
thromboplastin do not release due to
which blood does not clot.
But external thromboplastin to blood will
cause blood clotting at the site of its
introduction due to the formation of
prothrombinase.
19You are required to draw blood
from a patient and to keep it in a
test tube for analysis of blood
corpuscles and plasma. You are
also provided with the following
four types of test tubes. Which of
them will you not use for the
purpose? [CBSE AIPMT 2004]
(a) Test-tube containing calcium
bicarbonate
(b) Chilled test tube
(c) Test tube containing heparin
(d) Test tube containing sodium oxalate
Ans.(a)
Clotting of collected blood can be
prevented by coating the test tubes with
silicon or adding chelating agents.
Citrate, oxalate, heparin and EDTA are
anticoagulants.
20What is correct regarding
leucocytes?[CBSE AIPMT 2000]
(a) These can squeeze out through (can
cross) the capillary walls
(b) These are enucleate
(c) Sudden fall in their number indicates
cancer
(d) These are produced in thymus
Ans.(a)
Most of the T and B-lymphocytes (types
of leukocytes) continuously circulate
between the blood and lymph. These
leave the blood stream, squeezing out
between specialised endothelial cuts
found in certain small vessels and enter
various tissues including all the lymph
nodes.
After percolating through a tissue, these
accumulate in small lymphatic vessels
which connect to a series of lymph
nodes, from where they ultimately enter
the main lymphatic vessel (thoracic duct)
which carries them back into the blood.
21What is correct for blood group ‘O’?
[CBSE AIPMT 2001, 1999]
(a) No antigens but both a and b
antibodies are present
(b) A antigen and b antibody
(c) Antigen and antibody both absent
(d) A, B antigens and a, b antibodies
Ans.(a)
In blood of ‘O’ group, no antigens are
present on red blood cells, but both
anti-a and anti-b antibodies are present
in plasma. Blood group A has antigen ‘A’
and antibody ‘b’.
Blood group B has antigen ‘B’ and
antibody ‘a’. Blood group AB has antigens
‘A’ and ‘B’ but no antibody in plasma.
22Haemoglobin is a type of
[CBSE AIPMT 1999]
(a) carbohydrate
(b) vitamin
(c) skin pigment
(d) respiratory pigment
Ans.(d)
Haemoglobin (Hb) is a proteinaceous
respiratory pigment made up of a protein
called globin with iron[Fe ]
2 +
containing
porphyrin as prosthetic group. It binds to
oxygen reversibly.
23Which is the principal cation in the
plasma of the blood?
[CBSE AIPMT 1999]
(a) Magnesium (b) Sodium
(c) Potassium (d) Calcium
Ans.(b)
The concentration ofNa
+
in plasma is
0.32% followed byK
+
(0.02%) and
magnesium (0.0025%).
The mineral ions likeNa
+
and others
present in the blood plasma play an
essential role in the maintenance of
osmotic pressure of the blood.
Body Fluids and Circulation 169

180. 24Which of the following is
agranulocyte?[CBSE AIPMT 1997]
(a) Lymphocyte (b) Eosinophil
(c) Basophil (d) Neutrophil
Ans.(a)
White Blood Corpuscles (WBC) or
leucocytes can be divided into two
groups on the presence/absence of
minute granules in the cytoplasm.
(a) Granulocytes whichcontain granules.
e.g. neutrophils, eosinophils and
basophils.
(b) Agranulocytes which do not contain
granules. lymphocytes, monocytes.
25The life span of human WBC is
approximately[CBSE AIPMT 1997]
(a) less than 10 days
(b) between 20-30 days
(c) between 2-3 months
(d) more than 4 months
Ans.(a)
Human WBC (or leukocytes) life span is
approximately less than 10 days.
Leukocytes constitute less than 1% of
the cells in human blood. They are large
in size than red blood cells. They have
nucleus but no haemoglobin.
26Which one of the following
vertebrate organs receives the
oxygenated blood only?
[CBSE AIPMT 1996]
(a) Gill (b) Lung
(c) Liver (d) Spleen
Ans.(d)
Only spleen and brain always receive
oxygenated blood. Gills and lungs are the
blood purifying organs, i.e. here blood
becomes oxygenated and these two
organs always receive deoxygenated
blood. Liver is a part of hepatic portal
system, in which hepatic portal vein
carries food laiden blood from
alimentary canal and associated glands
before finally returning to heart.
27Antigens are present
[CBSE AIPMT 1995]
(a) inside the nucleus
(b) on cell surface
(c) inside the cytoplasm
(d) on nuclear membrane
Ans.(b)
Antigens (Ag) are foreign particles
present on the surface of cell and when
introduced in the blood they initiate a
specific immune response against
themselves.
28Cells formed in bone marrow
include [CBSE AIPMT 1993]
(a) RBC
(b) RBC and leucocytes
(c) leucocytes
(d) lymphocytes
Ans.(b)
In the developing foetus, the
haemopoietic tissues are liver and
spleen, while after birth RBCs are mainly
produced in the bone marrow of the long
bones. The formation of leucocytes
occurs in the bone marrow, Peyer’s
patches, lymph nodes, thymus, spleen
etc. and it is called as leucopoiesis.
29Removal of calcium from freshly
collected blood would
[CBSE AIPMT 1989]
(a) cause delayed clotting
(b) prevent clotting
(c) cause immediate clotting
(d) prevent destruction of haemoglobin
Ans.(b)
Blood clotting starts when
prothrombinase in the presence of
Ca
2+
converts inactive prothrombin into
thrombin which in turn converts
dissolved fibrinogen protein into fine
thread-like fibrin. The network of fibrin
covers the wound in which blood
corpuscles get endulge and form clot. If
Ca
2+
is removed, it will prevent clotting.
30In the ABO system of blood groups,
if both antigens are present but no
antibody, the blood group of the
individual would be
[CBSE AIPMT 2004, 1991]
(a) B (b) O (c) AB (d) A
Ans.(c)
Blood
groups
Antigen on
RBC
Antibodies in
serum
A A anti-b
B B anti-a
AB A and B –
O – anti-a and anti-b
Hence, blood group AB has no antibodies
in serum but both antigens A and B.
31A person with blood group A
requires blood. The blood group
which can be given is
[CBSE AIPMT 1989]
(a) A and B (b) A and AB
(c) A and O (d) A, B, AB and O
Ans.(c)
Blood group A has A antigen and b
antibody and blood group O has no
antigens and both a and b antibodies so,
if a patient with blood group A needs
blood, both A and O blood group can be
given to him.
32Which one engulfs pathogens
rapidly? [CBSE AIPMT 1989]
(a) Acidophils (b) Monocytes
(c) Basophils (d) Neutrophils
Ans.(d)
Neutrophils are granulocytes, i.e.,
cytoplasm is filled with fine granules.
These granules are actually lysosome
and Golgi bodies. These are the chief
phagocytic cells of the body and engulf
the microbes by phagocytosis, so
neutrophils are also called soldiers of the
body.
33Child death may occur in the
marriage of
[CBSE AIPMT 1988, 2000]
(a)Rh
+
man andRh
+
woman
(b)Rh
+
man andRh
−
woman
(c)Rh
−
man andRh
−
woman
(d)Rh
−
man andRh
+
woman
Ans.(b)
Rhesus antibodies are formed in the
plasma ofRh
–
woman who have been
pregnanted withRh
+
babies, if the foetal
blood leaks across the placenta during
the birth the mother body starts
preparing antibodies against the
Rh-antigen. LaterRh
+
foetus would be at
risk and may suffer from haemolysis.
34RBCs do not occur in
[CBSE AIPMT 1988]
(a) frog (b) cow
(c) camel (d) cockroach
Ans.(d)
RBCs or red blood corpuscles are meant
for the transportation of respiratory
gases in the blood and these are present
in all vertebrates.
In cockroach haemolymph is present
which serves for the transportation of
nutrients, maintains hydrostatic
pressure and acts as a reservoir of
water.
35Breakdown product of
haemoglobin is[CBSE AIPMT 1988]
(a) bilirubin (b) iron
(c) biliverdin (d) calcium
170 NEETChapterwise Topicwise Biology

181. Ans.(a)
Haemoglobin of erythrocytes split off
into heme and globin. The core of iron in
heme is salvaged, bound to protein as
hemosiderin and stored for reuse.
The remaining of the heme group is
degraded to bilirubin a yellow pigment
that is released into the blood. Bilirubin
is picked up by liver cells which in turn
secrete it into the intestine where it is
metabolised to urobilinogen.
36One of the factors required for the
maturation of erythrocytes is
[CBSE AIPMT 1998]
(a) vitamin-D
(b) vitamin-A
(c) vitamin-B
12
(d) vitamin-C
Ans.(c)
Vitamin-B
12
(cobalamin) promotes DNA
synthesis, maturation of erythrocytes
and myelin formation. Vitamin-D
(calciferol), It is necessary for the
formation of healthy bones and teeth.
Vitamin-A (retinol), It is necessary for
proper body growth and night vision.
Vitamin-C (ascorbic acid), It helps in
wound healing, blood formation and
absorption of iron.
37The haemorrhagic disease of new
born is caused due to the
deficiency of[CBSE AIPMT 1995]
(a) vitamin-A
(b) vitamin-B
1
(c) vitamin-B
12
(d) vitamin-K
Ans.(d)
The haemorrhagic disease of new born is
caused due to the deficiency of
vitamin-K, characterised by delayed
blood clotting and haemorrhage (blood
loss).
38Vitamin-K is required for
[CBSE AIPMT 1993]
(a) formation of thromboplastin
(b) conversion of fibrinogen to fibrin
(c) conversion of prothrombin to
thrombin
(d) synthesis of prothrombin
Ans.(d)
Vitamin-K (phylloquinone) is required for
the synthesis of prothrombin necessary
for blood clotting. It is synthesised by
bacteria in the colon. Vitamin-K is
commonly called anti-haemorrhagic
vitamin. Its deficiency leads to bleeding,
i.e. no coagulation or clotting.
39The QRS complex in a standard
ECG represents[NEET (Sep.) 2020]
(a) depolarisation of auricles
(b) depolarisation of ventricles
(c) repolarisation of ventricles
(d) repolarisation of auricles
Ans.(b)
Option (b) is correct and can be explained
as in an ECG there occurs five
consecutive waves: P, Q, R, S and T.
P wave represents depolarisation of
atria and leads to contraction of both
atria.
QRS complex representsdepolarisation
of ventricleswhich leads to initiation of
ventricular contraction.
T wave represents return of ventricles
from excited to normal state.
40All the components of the nodal
tissue are autoexcitable. Why does
the SA node act as the normal
pacemaker?[NEET (Odisha) 2019]
(a) SA node has the lowest rate of
depolarisation
(b) SA node is the only component to
generate the threshold potential
(c) Only SA node can convey the action
potential to the other components
(d) SA node has the highest rate of
depolarisation
Ans(d)
The nodal musculature has the ability to
generate action potentials without any
external stimuli, i.e. it is autoexcitable.
However, the number of action
potentials that could be generated in a
minute vary at different parts of the
nodal system.
The SAN (Sino-Atrial Node) can generate
the maximum number of action
potentials, i.e 70-75 min, i.e. the highest
rate of depolarisation and is responsible
for initiating and maintaining the
rhythmic contractile activity of the
heart. Therefore, it is called pacemaker.
41A specialised nodal tissue
embedded in the lower corner of
the right atrium, close to
atrio-ventricular septum, delays
the spreading of impulses to heart
apex for about 0.1 sec. The delay
allows [NEET (Odisha) 2019]
(a) blood to enter aorta
(b) the ventricles to empty completely
(c) blood to enter pulmonary arteries
(d) the atria to empty completely
Ans.(d)
Atrio-Ventricular Node (AVN) present in
the lower corner of the right atrium,
delays the spreading of impulses to
heart ventricles for about 0.1 second.
This pause allows the atria to empty
completely into the ventricles before the
ventricles pump out the blood.
42What would be the heart rate of a
person if the cardiac output is 5 L,
blood volume in the ventricles at
the end of diastole is 100 mL and at
the end of ventricular systole is 50
mL? [NEET (National) 2019]
(a) 75 beats per minute
(b) 100 beats per minute
(c) 125 beats per minute
(d) 50 beats per minute
Ans.(b)
As per the given data, the heart rate of
the person would be 100 beats per minute.
It can be calculated as follows Given,
Cardiac output=5L (5000 mL) Blood
volume in ventricles at the end of diastole
=100 mL
Blood volume at the end of ventricular
systole
=50 mL
So, Stroke volume= − =100 50 50mL
Cardiac output=Stroke volume×Heart
rate, i.e.
5000mL=50 mL×Heart rate
Therefore, Heart rate=100 beats/min.
Body Fluids and Circulation 171
Structure of the Human
Circulatory System: Heart
TOPIC 2

182. 43Match the following columns.
[NEET (National) 2019]
Column I Column II
A. P - wave (i) Depolarisation of
ventricles
B. QRS complex (ii) Repolarisation of
ventricles
C. T-wave (iii) Coronary ischemia
D. Reduction in
the Size of
T-wave
(iv) Depolarisation of
atria
(v) Repolarisation of
atria
Select the correct option.
A B C D
(a) (iv) (i) (ii) (v)
(b) (ii) (i) (v) (iii)
(c) (ii) (iii) (v) (iv)
(d) (iv) (i) (ii) (iii)
Ans.(d)
(A)–(iv), (B)–(i), (C)–(ii), (D)–(iii)
In an Electrocardiograph (ECG), P-wave
represents the depolarisation of atria
which is caused by the activation of SA
node. QRS complex represents
depolarisation of ventricles which is
caused by the impulse of contraction from
AV node.
T-wave represents repolarisation of
ventricles and reduction in its size
signifies coronary ischemic, i.e. when
the heart muscles receive insufficient
oxygen as in arteriosclerotic heart
disease.
44Match the items given in Column I
with those in Column II and select
thecorrectoption given below
[NEET 2018]
Column I Column II
1. Tricuspid
valve
(i) Between left
atrium and left
ventricle
2. Bicuspid
valve
(ii) Between right
ventricle and
pulmonary artery
3. Semilunar
valve
(iii) Between right
atrium and right
ventricle
1 2 3
(a) (i) (ii) (iii)
(b) (i) (iii) (ii)
(c) (iii) (i) (ii)
(d) (ii) (i) (iii)
Ans.(c)
The atrioventricular opening between
the left atrium and left ventricle is
guarded by thebicuspid valve.It is also
called as mitral valve. The right
atrioventricular opening is guarded by
thetricuspid valve.It has three flaps.
Semilunar valveis found in right
ventricle and pulmonary artery.
Therefore, option (c) is correct.
45Doctors use stethoscope to hear
the sounds produced during each
cardiac cycle. The second sound is
heard when [CBSE AIPMT 2015]
(a) AV valves open up
(b) ventricular walls vibrate due to
gushing in of blood from atria
(c) semilunar valves close down after the
blood flows into vessels from
ventricles
(d) AV node receives signal from SA node
Ans.(c)
In healthy adults, there are two normal
heart sounds often described as lub and
dup. These are the first heart sound and
second heart sound produced by the
closing of the AV valves and semilunar
valves respectively.
46Which one of the following animals
has two separate circulatory
pathways? [CBSE AIPMT 2015]
(a) Frog (b) Lizard
(c) Whale (d) Shark
Ans.(c)
The circulatory system in which two
distinct and separate circulatory
pathways for blood flow are involved, is
called double circulatory system (also,
double-loop circulatory system). It
occurs in mammals and birds. Whale is a
mammal, so it shows above
characteristic.
47The diagram given here is the
standard ECG of a normal person.
The P-wave represents the
[NEET, CBSE AIPMT 2013, 2009]
(a) contraction of both the atria
(b) initiation of the ventricular
contraction
(c) beginning of the systole
(d) end of systole
Ans.(a)
In ECG, P-wave represents the
depolarisation of atria which leads to the
contraction of both atria. T-wave
represents the return of ventricles from
excited to normal state. The QRS
complex represents the depolarisation
of the ventricles which initiates
ventricular contraction. The contraction
starts shortly after Q and marks the
beginning of systole.
48‘Bundle of His’ is a part of which
one of the following organs in
humans? [CBSE AIPMT 2011]
(a) Heart (b) Kidney
(c) Pancreas (d) Brain
Ans.(a)
The bundle of His, are specialised
muscle fibres for electrical conduction
present in the heart which were named
after the Swiss cardiologist Wilhelm His,
Jr., who discovered them in 1893. These
are also known as AV bundle which is a
collection of heart muscle cells
49If due to some injury the chordae
tendinae of the tricuspid valve of
the human heart is partially
non-functional, what will be the
immediate effect?
[CBSE AIPMT 2010]
(a) The flow of blood into the aorta will be
slowed down
(b) The ‘pacemaker’ will stop working
(c) The blood will tend to flow back into
the left atrium
(d) The flow of blood into the pulmonary
artery will be reduced
Ans.(d)
If chordae tendinae of the tricuspid
valve became partially non-functional
due to the injury then the flow of blood
into the pulmonary artery will be
reduced. Because chordae tendinae,
arise from papillary muscles and insert
upon the flaps of tricuspid and bicuspid
valves and the valves in the heart allow
the flow of blood only in one direction,
i.e., from the atria to the ventricles and
from the ventricles to the pulmonary
artery or aorta.
172 NEETChapterwise Topicwise Biology
P
Q
R
T

183. 50In humans, blood passes from the
post caval to the diastolic right
atrium of heart due to
[CBSE AIPMT 2008]
(a) pushing open of the venous valves
(b) suction pull
(c) stimulation of the sino auricular node
(d) pressure difference between the post
caval and atrium
Ans.(d)
Due to the pressure difference between
the post caval and atrium, the blood
passes from the post caval to the
diastolic right atrium. Because the
action of heart includes contractions
and relaxations of the atria and
ventricles. The dynamics of blood flow in
blood vessels is no exception and blood
flows through the blood vessels along a
pressure gradient, always moving from
higher to lower pressure areas.
51Systemic heart refers to
[CBSE AIPMT 2003]
(a) enteric heart in lower vertebrates
(b) the two ventricles together in humans
(c) the heart that contracts under
stimulation from nervous system
(d) left auricle and left ventricle in higher
vertebrates
Ans.(a)
Systemic heart refers to enteric heart in
lower vertebrates. It pumps the blood to
different body parts and not to lungs.
52Bundle of His is a network of
[CBSE AIPMT 2003]
(a) nerve fibres distributed in ventricles
(b) nerve fibres found throughout the heart
(c) muscle fibres distributed throughout
the heart walls
(d) muscle fibres found only in the
ventricle wall
Ans.(d)
Bundle of His is a network of specialised
conducting muscle fibres (Purkinje
fibres) which transmit the impulse from
AV node to all parts of both the
ventricles.
53In mammals, histamine is secreted
by [CBSE AIPMT 1998]
(a) fibroblasts (b) histocytes
(c) lymphocytes (d) mast cells
Ans.(d)
Histamine is a potent vasodilator formed
by decarboxylation of the amino acid
histidine and released by mast cells in
response to appropriate antigens.
Mast cells are especially prevalent in the
connective tissue of the skin, respiratory
tract and in surrounding blood vessels.
54The correct route through which
pulse-making impulse travels in the
heart is [CBSE AIPMT 1995]
(a) AV node→Bundle of His→SA node
→Purkinje fibres→Heart muscles
(b) AV node→SA node→Purkinje fibres
→Bundle of His→Heart muscles
(c) SA node→Purkinje fibres→Bundle
of His→AV node→Heart muscles
(d) SA node→AV node→Bundle of His
→Purkinje fibres→Heart muscles
Ans.(d)
The correct route through which
pulse-making impulse travels in the
heart is : SA node→AV node→Bundle
of His→Purkinje fibres→Heart muscles.
55Pacemaker of heart is
[CBSE AIPMT 1994, 99, 2002, 04]
(a) AV node (b) bundle of His
(c) SA node (d) Purkinje fibres
Ans.(c)
SA node lies in the right wall of right
auricle below the opening of superior
vena cava. It is also called pacemaker as it
is first to originate the cardiac impulses
and determines the rate of heartbeat.
56Dup sound is produced during
closure of [CBSE AIPMT 1994]
(a) semilunar valves
(b) bicuspid valve
(c) tricuspid valve
(d) Both (b) and (c)
Ans.(a)
The period between the end of one
heartbeat to the end of next heartbeat is
called cardiac cycle. Cardiac cycle is
formed of three phases.
Atrial systole, ventricular systole and
joint diastole. During ventricular systole
closing of Auriculo ventricular (AV) valves
at the start of ventricular systole
produces first heart sound called ‘lubb’ or
systolic sound. During joint diastole
rapid closure of semilunar valves at the
beginning of ventricular diastole produces
the second heart sound called ‘dup’.
57Tricuspid valve is found in between
[CBSE AIPMT 1989]
(a) sinus venosus and right auricle
(b) right auricle and right ventricle
(c) left ventricle and left auricle
(d) ventricle and aorta
Ans.(b)
Right auricle opens in right ventricle
through a wide circular right
auriculoventricular aperture guarded by
right auriculoventricular valve which is
formed of three flaps called cusps, so it
is called tricuspid valve. It regulates the
unidirectional flow of blood from right
auricle to right ventricle.
58Blood pressure in the pulmonary
artery is [NEET 2016, Phase I]
(a) more than that in the carotid
(b) more than that in the pulmonary vein
(c) less than that in the vena cava
(d) same as that in the aorta
Ans.(b)
Blood pressure in different blood vessels :
Artery > Arteriole > Capillary > Venule >
Vein (vena cava). The pulmonary arteries
have thicker smooth muscle and
connective tissue then the pulmonary
veins to accomodate the higher
pressure and high rate of blood flow.
59In mammals, which blood vessel
would normally carry largest amount
of urea? [NEET 2016, Phase I]
(a) Dorsal Aorta
(b) Hepatic Vein
(c) Hepatic Portal Vein
(d) Renal Vein
Ans.(b)
Urea is synthesised in liver. So,
maximum amount of urea is present in
hepatic vein and minimum in renal vein.
60Figure shows schematic plan of
blood circulation in human with
labelsA-D. Identify the correct label
with given function[NEET 2013]
Body Fluids and Circulation 173
Structure of Human Circulatory
System Blood vessels
TOPIC 3
A
BC
D

184. (a)A–pulmonary vein–takes impure blood
from body parts,pO
2
60=mm Hg
(b)B–pulmonary artery takes blood from
heart to lungs,pO
2
90=mm Hg
(c)C–vena cava takes blood from body
parts to right auricle,pCO
2
45=mm Hg
(d)D–dorsal aorta takes blood from heart
to body parts,pO
2
95=mm Hg
Ans.(c)
The correct labelling of parts with their
respective functions is as follows.
A. Pulmonary
vein
takes oxygenated blood
from lung and carried it
to left auricle.
B. Dorsal aorta takes blood from heart to
body parts,pO
2
95=mm
Hg.
C. Vena cava takes blood from body
parts to right auricle
pCO
2
45=mm Hg
D. Pulmonary
artery
takes blood from heart to
lungs,pO
2
90=mm Hg
61Arteries are best defined as the
vessels which[CBSE AIPMT 2011]
(a) carry blood away from the heart to
different organs
(b) break up into capillaries which reunite
to form a vein
(c) carry blood from one visceral organ to
another visceral organs
(d) supply oxygenated blood to the
different organs
Ans.(a)
Arteries are blood vessels that carry
blood away from the heart towards
different organs. They generally contain
oxygenated blood (except pulmonary
artery which contains deoxygenated
blood). The blood flows in an artery under
alternate increased pressure and with
jerks.
62Which one of the following
statements is correct regarding
blood pressure?[CBSE AIPMT 2011]
(a) 100/55 mm Hg is considered an ideal
blood pressure
(b) 105/50 mm Hg makes one very active
(c) 190/110 mm Hg may harm vital organs
like brain and kidney
(d) 130/90 mm Hg is considered high and
requires treatment
Ans.(c)
Blood pressure [190/110 mm Hg) of an
individual is 140/90 [140 over 90] or
higher, it shows hypertension. High blood
pressure [190/110 mm Hg) leads to heart
diseases and also affects vital organs like
brain and kidney. Hypertension means the
blood pressure that is higher than normal
[120/80]. In this measurement, 120 mm
Hg (millimeter of mercury pressure) is the
normal systolic or pumping, pressure and
80 mm Hg is the normal diastolic or
resting pressure.
63Pulmonary artery is different from
pulmonary vein because it has
[CBSE AIPMT 2000]
(a) larger lumen
(b) thick muscular walls
(c) no endothelium
(d) valves
Ans.(b)
Arteries have thick walls, narrow lumen
but no valves. Endothelium is present in
both arteries and veins.
64The thickening of walls of arteries
is called [CBSE AIPMT 1999]
(a) arthritis (b) atherosclerosis
(c) anaeurysm (d) Both (a) and (c)
Ans.(b)
Atherosclerosis involves thickening of
inner walls of arteries due to deposition
of lipid (cholesterol) which prevents the
dilation of arteries.
65An adult human with average
health has systolic and diastolic
pressures as[CBSE AIPMT 1998]
(a) 80 mm Hg and 80 mm Hg
(b) 70 mm Hg and 120 mm Hg
(c) 120 mm Hg and 80 mm Hg
(d) 50 mm Hg and 80 mm Hg
Ans.(c)
In a normal human being, the systolic
and diastolic pressure are 120 mmHg and
80 mmHg respectively.
66Closed circulatory system occurs in
[CBSE AIPMT 1994]
(a) cockroach (b) tadpole/fish
(c) mosquito (d) house fly
Ans.(b)
Closed circulatory system is usually high
pressure system, in which blood flows in
closed tubular structures called blood
vessels (arteries, veins and capillaries). It
is found in most of annelids,
cephalopods, among the molluscs and all
vertebrates including human beings. In
this type of system there is no direct
contact between body tissues and blood.
This is more efficient as blood
circulation is completed in short period.
67Wall of blood capillary is formed of
[CBSE AIPMT 1993, 91]
(a) haemocytes (b) parietal cells
(c) endothelial cells (d) oxyntic cells
Ans.(c)
Each capillary is lined by a single layer of
flat cells, called endothelium. The
endothelium allows the exchange of
materials like the nutrients, respiratory
gases, waste products, hormones, etc
between the blood and surrounding
tissue cells through the tissue fluid.
68Splenic artery arises from
[CBSE AIPMT 1990]
(a) anterior mesenteric artery
(b) coeliac artery (or celiac artery)
(c) posterior mesenteric artery
(d) intestinal artery
Ans.(b)
Splenic artery is the blood vessel that
supplies oxygenated blood to the spleen.
It branches from the celiac artery and
follows a course superior to the
pancreas.
69A vein possesses a large lumen
because [CBSE AIPMT 1990]
(a) tunica media and tunica externa form
a single coat
(b) tunica interna and tunica media form
a single coat
(c) tunica interna, tunica media and
tunica externa are thin
(d) tunica media is a thin coat
Ans.(d)
The tunica media is comparatively thin in
the veins making a large lumen in veins.
Basically each artery and vein consists
of three layers, an inner lining of
squamous endothelium, the tunica
interna, a middle layer of smooth muscle
and elastic fibre, the tunica media and an
external layer of fibrous connective
tissue with collagen fibres, the tunica
externa.
70Arteries carry oxygenated blood
except [CBSE AIPMT 1989]
(a) pulmonary
(b) cardiac
(c) hepatic
(d) systemic
Ans.(a)
Right ventricle pumps deoxygenated
blood into pulmonary artery which
supplies it to the lungs where
oxygenation of blood takes place.
174 NEETChapterwise Topicwise Biology

185. 71Which of the following statements
is true for lymph?
[CBSE AIPMT 2002]
(a) WBC and serum
(b) All components of blood except RBCs
and some proteins
(c) RBCs, WBCs and plasma
(d) RBCs, proteins and platelets
Ans.(b)
Lymph is known as blood minus RBCs
and some proteins. The main site of
lymph formation is interstitial space and
normally the rate of lymph formation is
equal to the rate of its return to blood
stream.
72Which of the following is not main
function of lymph glands?
[CBSE AIPMT 1998]
(a) Forming WBC
(b) Forming antibodies
(c) Forming RBC
(d) Destroying bacteria
Ans.(c)
Cells of lymph nodes perform the
following functions. (a) produce
lymphocytes (b) synthesise antibodies (c)
destroy bacteria by phagocytosis.
73The lymph serves to
[CBSE AIPMT 1995]
(a) transport oxygen to the brain
(b) transport carbon dioxide to the lungs
(c) return the interstitial fluid to the
blood
(d) return the WBCs and RBCs to the
lymph nodes
Ans.(c)
Lymph acts as a middle man between
the blood and tissue cells. Lymph is a
transparent fluid derived from blood and
other tissues which accumulates in the
interstitial spaces as the interstitial fluid
and it passes on food andO
2
from blood
to tissue cells and handed ones
excretory wastes, hormones and
CO
2
from the body cells to the blood.
74Which of the following conditions
causes erythroblastosis foetalis?
[NEET (Oct.) 2020]
(a) Mother Rh+ve and foetus Rh−ve
(b) Mother Rh−ve and foetus Rh+ve
(c) Both mother and foetusRh−ve
(d) Both mother and foetusRh+ve
Ans.(b)
If mother is Rh−ve and foetus is Rh+ve
then there can occur a condition called
erythroblastosis foetalis. It is a special
case of Rh incompatibility in which
during the second pregnancy of Rh−ve
mother carrying Rh+ve foetus, the Rh
antibodies from mother (Rh−ve) can
leak into the blood of the foetus (Rh+ve)
and destroy the foetal RBCs
75Continuous bleeding from an
injured part of body is due to
deficiency of[CBSE AIPMT 2002]
(a) vitamin-A (b) vitamin-B
(c) vitamin-K (d) vitamin-E
Ans.(c)
Vitamin-K is required for clotting
process, it is required for the formation
of prothrombin in liver, the deficiency of
which leads to severe bleeding
disorders. Deficiency of vitamin-A
causes night blindness, xerophthalmia,
keratomalacia, retarded growth.
Deficiency of vitamin-B causes beri-beri
disease. Deficiency of vitamin-E causes
sterility.
76The blood cancer is known as
[CBSE AIPMT 1995]
(a) leukemia
(b) thrombosis
(c) haemolysis
(d) haemophilia
Ans.(a)
Blood cancer is known as leukemia
which is characterised by uncontrolled
division of leukocytes.
Body Fluids and Circulation 175
Disorders of
Circulatory System
TOPIC 5
Lymph and
Lymphatic System
TOPIC 4

186. 01Match the items in Column-I with
those in Column II
[NEET (Odisha) 2019]
Column I Column II
1. Podocytes i. Crystallised
oxalates
2. Protonephridia ii. Annelids
3. Nephridia iii. Amphioxus
4. Renal calculi iv. Filtration slits
Select the correct option from the
following
1 2 3 4
(a) (iii) (iv) (ii) (i)
(b) (iii) (ii) (iv) (i)
(c) (iv) (iii) (ii) (i)
(d) (iv) (ii) (iii) (i)
Ans.(c)
The correct matches are
1. Podocytes (iv) Filtration slit
2. Protonephridia (iii)Amphioxus
3. Nephridia (ii) Annelids
4. Renal calculi (i) Crystallised
oxalates
Podocytes are cells in Bowman’s capsule
in kidneys. They have filtration slits
through which the blood is filtered.
Protonephridia help in osmoregulation.
Nephridia in annelids help in
osmoregulation and excretion. Renal
calculi are kidney stones which mainly
consist of crystallised oxalates.
02Uricotelic mode of passing out
nitrogenous wastes is found in
[CBSE AIPMT 2011]
(a) birds and annelids
(b) amphibians and reptiles
(c) insects and amphibians
(d) reptiles and birds
Ans.(d)
Reptile, birds, land snails and insects
excrete nitrogenous waste as uric acid in
the form of pellet of paste with a
minimum loss of water and are called
uricotelic animals.
03Which one of the following is not a
part of a renal pyramid?
[CBSE AIPMT 2011]
(a) Convoluted tubules
(b) Collecting ducts
(c) Loop of Henle
(d) Peritubular capillaries
Ans.(a)
Convoluted tubule is the highly
convoluted segments of nephron in the
renal labyrinth of the kidney.
It is made up of the proximal tubule
leading from the Bowmans capsule to
the descending limb of Henle’s loop and
the distal tubule leading from the
ascending limb of Henle’s loop to a
collecting tubule.
04Uric acid is the chief nitrogenous
component of the excretory
products of[CBSE AIPMT 2009]
(a) man
(b) earthworm
(c) cockroach
(d) frog
Ans.(c)
Cockroach excretes uric acid as the
chief nitrogenous excretory product.
Man excretes urea as the chief
nitrogenous excretory product.
Earthworm excretes 40% urea, 20%
ammonia and 40% amino acids.
Frog excretes urea as the chief
nitrogenous product.
05Consider the following four
statements (A–D) about certain
desert animals such as kangaroo
rat
A. They have dark colour and high
rate of reproduction and excrete
solid urine.
B. They do not drink water, breathe
at a slow rate to conserve water
and have their body covered with
thick hairs.
C. They feed on dry seeds and do
not require drinking water.
D. They excrete very concentrated
urine and do not use water to
regulate body temperature.
Which two of the above
statements for such animals are
true? [CBSE AIPMT 2008]
(a) C and D (b) B and C
(c) C and A (d) A and B
Ans.(a)
Kangaroo rat is a desert rodent. Its body
is covered by hairs. Its urine is more than
20 times concentrated as its plasma.
This concentrated waste enables it to
live in dry or desert environment where
little water is available for him to drink.
ExcretoryProducts
andTheirElimination
19
Modes of Excretion
TOPIC 1

187. Excretory Products and Their Elimination 177
Most of its water is metabolically
produced from the oxidation of
carbohydrates, fats and proteins in the
seeds that it eat. The animal remains in
cool burrow during day time and the
respiratory moisture condensed in nasal
passages.
06In Ornithine cycle which one pair of
the following wastes are removed
from the blood?
[CBSE AIPMT 2005, 06]
(a)CO
2
and urea
(b)CO
2
and ammonia
(c) Ammonia and urea
(d) Urea and sodium salts
Ans.(b)
CO
2
and ammonia are the pair of wastes
removed from the blood in Ornithine
cycle. Urea is formed in Ornithine cycle
or urea cycle and urea is fomed of two
molecules of ammonia and one molecule
ofCO
2
. Urea cycle is represented as
follows
Most of the urea is produced in the liver.
The liver cells continuously release urea
into the blood and kidneys withdraw it
from the blood to excrete it in urine.
07Uricotelism is found in
[CBSE AIPMT 2004]
(a) mammals and birds
(b) fishes and freshwater protozoans
(c) birds, reptiles and insects
(d) frogs and toads
Ans.(c)
The animals which excrete mainly uric
acid are uricotelic and this phenomenon
is called uricotelism. Uric acid is
excreted by terrestrial reptiles (lizard,
snakes, etc) birds and insects to
conserve water in their body.
Frog and mammals excrete urea and so
they are called as ureotelic animals and
this phenomenon is known as
ureotelism.
08In living beings, ammonia is
converted into urea through
[CBSE AIPMT 2000]
(a) ornithine cycle
(b) citrulline cycle
(c) fumarine cycle
(d) arginine cycle
Ans.(a)
Ornithine combines with one molecule of
NH
3
andCO
2
to produce citrulline.
Citrulline combines with another
molecule ofNH
3
to form arginine.
Arginine is broken down into urea and
ornithine which repeats the cycle. This is
called Ornithine cycle or urea cycle or
Krebs-Henseleit cycle.
09The enteronephric nephridia of
earthworms are mainly concerned
with [CBSE AIPMT 2000]
(a) digestion
(b) respiration
(c) osmoregulation
(d) excretion of nitrogenous wastes
Ans.(d)
Enteronephridia is concerned with
excretion of nitrogenous waste. In
annelids, the nephridia are the excretory
organs. In earthworm, three types of
nephridia are found (a) septal (b)
pharyngeal and (c) integumentary.
The septal nephridia do not discharge
the excretory fluid to the exterior.
Instead, these pour it into the intestine.
Hence, these are also called
enteronephric nephridia.
10Aquatic reptiles are
[CBSE AIPMT 1999]
(a) ammonotelic
(b) ureotelic
(c) ureotelic in water
(d) ureotelic over land
Ans.(b)
Ureotelic animals include,Ascaris,
earthworm, cartilaginous fishes,
semiaquatic amphibians aquatic or
semiaquatic reptiles like turtles and
alligators.
11In ureotelic animals, urea is formed
by [CBSE AIPMT 1997]
(a) Ornithine cycle
(b) Cori cycle
(c) Krebs’ cycle
(d) EMP pathway
Ans.(a)
Urea is the main nitrogenous excretory
product of ureotelic animals. It is
produced by liver cells from deaminated
excess amino acidsviaurea cycle, also
called Ornithine cycle or Krebs-Henseleit
cycle.
12The kidney of an adult frog is
[CBSE AIPMT 1997]
(a) pronephros
(b) mesonephros
(c) metanephros
(d) opisthonephros
Ans.(b)
Mesonephric kidney consists of a large
number of tubules which develop
internal glomeruli enclosed in capsules
forming Malpighian bodies. In
amphibians, (e.g. frog) it is functional
both in embryo as well as adults.
13Uric acid is nitrogenous waste in
[CBSE AIPMT 1994]
(a) mammals and molluscs
(b) birds and lizards
(c) frog and cartilaginous fishes
(d) insects and bony fishes
Ans.(b)
Uric acid is least soluble nitrogenous
waste and 1 g of it needs only 10 mL of
water to be expelled out of body.
Another advantage of it is that it is least
toxic among all nitrogenous wastes and
can be retained in the body for longer
period, so it is of greater advantage to
animals which have limited access to
water like birds and lizards.
14Nitrogenous waste products are
eliminated mainly as
[CBSE AIPMT 1991]
(a) urea in tadpole and ammonia in adult
frog
(b) ammonia in tadpole and urea in adult
frog
(c) urea in both tadpole and adult frog
(d) urea in tadpole and uric acid in adult
frog
Ans.(b)
Ammonia is the main nitrogenous waste.
It is soluble in water and highly toxic. A
large amount of water is required for its
excretion. Tadpole is aquatic and lives in
plenty of water so, nitrogenous wastes
in tadpole are eliminated as ammonia.
Frog being amphibious, excretes its
nitrogenous wastes as urea.
OmithineArginine
Citrulline
Arginase
NH
3
CO
2
NH
3
H O
2 Urea
H O
2
H O
2
Urea cycle

188. 15Match the items given in Column I
with those in Column II and select
thecorrectoption given below
[NEET 2018]
Column-I
(Function)
Column-II
(Part of
Excretory
System)
1. Ultrafiltration i. Henle’s loop
2. Concentration
of urine
ii. Ureter
3. Transport of
urine
iii. Urinary
bladder
4. Storage of
urine
iv. Malpighian
corpuscle
v. Proximal
convoluted
tubule
1 2 3 4
(a) v iv i ii
(b) iv i ii iii
(c) iv v ii iii
(d) v iv i iii
Ans.(b)
Ultrafiltration orGlomerular filtrationis
carried out in the glomerular capillaries
found in Malpighian corpuscle. This
process is carried out under high
pressure.
Henle’s loop continuously absorbs the
water from glomerular filtrate, because
of the hyperosmolarity created by
counter-current mechanism. This helps
in theconcentration of urineand hence,
it becomes hypertonic.
Ureter are narrow, tubular structures
that convey ortransport urinefrom
kidney to urinary bladder.
Urinary bladder is pear-shaped,
muscular, sac-like structure that
temporarilystores urine.
16Figure shows human urinary
system with structures labelled
A-D. Select option, which correctly
identifies them and gives their
characteristics and/of functions
[NEET 2013]
(a) A−Adrenal gland—located at the
anterior part of kidney. Secrete
catecholamines, which stimulate
glycogen break down
(b) B–Pelvis—broad funnel shaped space
inner to hilum, directly connected to
loops of Henle
(c) C–Medulla—inner zone of kidney and
contains complete nephrons
(d) D–Cortex—outer part of kidney and do
not contain any part of nephrons
Ans.(a)
A–Adrenal gland it is correctly
mentioned. It is located at the anterior
part of kidney and secretes
catecholamines which stimulate
glycogen breakdown.
17The principal nitrogenous excretory
compound in humans is
synthesised[CBSE AIPMT 2010]
(a) in kidneys but eliminated mostly
through liver
(b) in kidneys as well as eliminated by
kidneys
(c) in liver and also eliminated by the
same through bile
(d) in the liver but eliminated mostly
through kidneys
Ans.(d)
In humans, the principal nitrogenous
excretory compound (i.e. urea) is
synthesised in liver by Ornithine cycle
and is eliminated mostly through kidney
as nitrogeneous excretory product. In
liver, one molecule ofCO
2
is activated by
biotin and combines with two molecules
ofNH
3
in the presence of carbamyl
phosphate synthetase enzyme and 2ATP
to form carbamyl phosphate and one
molecule ofH O
2
is released.
Carbamyl phosphate reacts with
Ornithine and forms Citrulline. Citrulline
combines with another molecule of
ammonia and form arginine that is broken
into urea and Ornithine in the presence of
an enzyme arginase and water.
2NH + CO NH CO NH
3 2 2 2
Arginase
→  
+ H O
2
18What will happen if the stretch
receptors of the urinary bladder
wall are totally removed?
[CBSE AIPMT 2009]
(a) Urine will not collect in the bladder
(b) Micturition will continue
(c) Urine will continue to collect normally
in the bladder
(d) There will be no micturition
Ans.(c)
If stretch receptors of urinary bladder
wall are totally removed, the urine will
continue to collect normally in the
bladder. The urinary bladder is a
pear-shaped, hollow muscular organ
situated in the pelvic cavity which is
made up of smooth and involuntary
muscles. The lumen of urinary bladder is
lined by transition epithelium which has
great power of stretching.
19Bowman’s glands are found in
[CBSE AIPMT 2006]
(a) olfactory epithelium
(b) external auditory canal
(c) cortical nephrons only
(d) juxtamedullary nephrons
Ans.(a)
Bowman’s glands (olfactory glands)
occur below the olfactory epithelia.
Their ducts open on the olfactory
epithelial surface. These glands secrete
watery mucus to protect and keep the
epithelium moist.
20Part not belonging to uriniferous
tubule is [CBSE AIPMT 1994]
(a) glomerulus
(b) Henle’s loop
(c) distal convoluted tubule
(d) collecting tubule
Ans.(a)
From the option given glomerulus do not
belong to uriniferous tubule. Each
nephron is about 6 cm long and is divided
into two parts : Bowman’s capsule and
nephric or uriniferous tubule. Glomerulus
is a group of about 50 capillaries. Its
capillary wall has numerous minute
pores, so the permeability of glomerular
membrane increases 100-500 times as
high as that of usual capillary.
178 NEETChapterwise Topicwise Biology
Human Excretory
System: Structure
TOPIC 2
A
B
C
D
Urinary bladder
Kidney

189. While Henle’s loop, DCT and collecting
tubules are part of uriniferous tubule.
21Proximal and distal convoluted
tubules are parts of
[CBSE AIPMT 1990]
(a) seminiferous tubules
(b) nephron
(c) oviduct
(d) vas deferens
Ans.(b)
Nephron is the structural and functional
unit of kidney. Nephrons are also called
renal tubules or uriniferous tubules.
Each nephron is formed of two parts.
(i) Bowman’s capsule and (ii) Nephric
tubule which is a long and coiled and is
formed of proximal convoluted tubule,
loop of Henle and distal convoluted
tubule.
22Brush border is characteristic of
[CBSE AIPMT 1990]
(a) neck of nephron
(b) collecting tube
(c) proximal convoluted tubule
(d) All of the above
Ans.(c)
Proximal convoluted tubule is present in
cortex and is convoluted. It is about
12-24 mm in length. It is lined by brush
bordered cuboidal epithelium with
numerous microvilli. These cells have
numerous mitochondria for active
transport.
23The increase in osmolarity from
outer to inner medullary
interstitium is maintained due to
[NEET (Oct.) 2020]
I. close proximity between Henle’s
loop and vasa recta
II. counter-current mechanism
III. selective secretion ofHCO
3
−
and
hydrogen ions in PCT.
IV. higher blood pressure in
glomerular capillaries
(a) Only II
(b) III and IV
(c) I, II and III
(d) I and II
Ans.(d)
Statements in option (d) are correct as
the close proximity between the Henle’s
loop and vasa recta as well as the
counter-current in them help in
maintaining an increasing osmolarity
towards the inner medullary interstitium,
i.e. from 300 mOsmolL
−1
in the cortex
(outside) to about 1200 mOsmolL
−1
in the
inner medulla (inside).
24Match the following parts of a
nephron with their function.
[NEET (Odisha) 2019]
Column I Column II
1. Descending
limb of Henle’s
loop
(i) Reabsorption
of salts only
2. Proximal
convoluted
tubule
(ii) Reabsorption
of water only
3. Ascending limb
of Henle’s loop
(iii) Conditional
reabsorption
of sodium ion
and water
4. Distal
convoluted
tubule
(iv) Reabsorption
of ion, water
and organic
nutrients
Select the correct option from the
following.
1 2 3 4
(a) (i) (iii) (ii) (iv)
(b) (ii) (iv) (i) (iii)
(c) (i) (iv) (ii) (iii)
(d) (iv) (i) (iii) (ii)
Ans.(b)
The correct matches are
Part of nephron Function
1. Descending limb of
Henle’s loop
(ii) Reabsorption of
water only
2. Proximal
convoluted tubule
(iv) Reabsorption of
ion, water and
organic nutrients
3. Ascending limb of
Henle’s loop
(i) Reabsorption of
salts
4. Distal convoluted
tubule
(iii) Conditional
reabsorption of
sodium ion and
water
25Which of the following statements
is correct? [NEET 2017]
(a) The ascending limb of loop of Henle is
impermeable to water
(b) The descending limb of loop of Henle
is impermeable to water
(c) The ascending limb of loop of Henle is
permeable to water
(d) The descending limb of loop of Henle
is permeable to electrolytes
Ans.(a)
The ascending limb of loop of Henle is
impermeable to water and permeable to
K , Cl and Na
+ +−
and partially permeable
to urea. Due to this, sodium, potassium,
calcium, magnesium and chloride are
reabsorbed here making the filterate
hypotonic.
26The part of nephron involved in
active reabsorption of sodium is
[NEET 2016, Phase II]
(a) distal convoluted tubule
(b) proximal convoluted tubule
(c) Bowman’s capsule
(d) descending limb of Henle’s loop
Ans.(b)
Proximal convoluted tubule is involved in
active reabsorption of sodium. The
majority (about 70%) of sodium is
reabsorbed here, into the cytosol of the
epithelial cells of the nephron.
27Which of the following causes an
increase in sodium reabsorption in
the distal convoluted tubule?
[CBSE AIPMT 2014]
(a) Increase in aldosterone levels
(b) Increase in antidiuretic hormone
levels
(c) Decrease in aldosterone levels
(d) Decrease in antidiuretic hormone
levels
Ans.(a)
Increase in aldosterone levels cause an
increase in sodium reabsorption in DCT.
It is secreted by outer layer of adrenal
gland when aldosterone is present in the
blood and all theNa
+
ions in the filterate
are reabsorbed. Retaining Na
+
raises the
osmotic pressure of the blood and
reduces the water loss from the body.
28The maximum amount of
electrolytes and water (70-80%)
from the glomerular filtrate is
reabsorbed in which part of the
nephron? [CBSE AIPMT 2012]
(a) Ascending limb of loop of Henle
(b) Distal convoluted tubule
(c) Proximal convoluted tubule
(d) Descending limb of loop of Henle
Excretory Products and Their Elimination 179
Human Excretory
System: Physiology
TOPIC 3

190. Ans.(c)
From the Bowman’s capsule, the
glomerular filtrate enters the proximal
convoluted tubule (PCT). PCT is
surrounded by a network of peritubular
capillaries and is the seat of
reabsorption. About 75% of glomerular
filtrate is normally reabsorbed in PCT
before reaching the loop of Henle.
The reabsorbed materials include
glucose, amino acids, vitamins,
hormones, sodium, potassium,
chlorides, phosphates, bicarbonates,
most of the water and some urea, etc.
29Which one of the following
correctly explains the function of a
specific part of a human nephron?
[CBSE AIPMT 2011]
(a) Henle’s loop — Most
reabsoroption of
the major
substances from
the glomerular
filtrate
(b) Distal
convoluted
tubule
— Reabsorption of
ions into the
surrounding blood
capillaries
(c) Afferent
arteriole
— Carries the blood
away from the
glomerulus
towards renal vein
(d) Podocytes — Create minute
spaces (slit pores)
for the filtration of
blood into the
Bowman’s capsule
Ans.(d)
Podocytes or visceral epithelial cells are
the cells in Bowman’s capsule in the
kidneys that wrap around the capillaries
of glomeruls. They create minute pores
(slit pores) for the filtration of blood into
the Bowman’s capsule.
30Injury to adrenal cortex is not likely
to affect the secretion of which one
of the following?[CBSE AIPMT 2010]
(a) Aldosterone
(b) Both androstenedione and
dehydroepi- androsterone
(c) Adrenalin
(d) Cortisol
Ans.(c)
The adrenal medulla synthesises two
hormones adrenaline (epinephrine) and
non-adrenaline (non-epinephrine).
These hormones are proteinaceous in
nature and derived from amino acid
tyrosine. Thus, injury to adrenal cortex
will not affect the secretion of
adrenaline.
31Which one of the following
statements in regard to the
excretion by the human kidneys is
correct? [CBSE AIPMT 2010]
(a) Descending limb of Loop of Henle is
impermeable to water
(b) Distal convoluted tubule is incapable
of reabsorbingHCO
3
(c) Nearly 99% of the glomerular filtrate
is reabsorbed by the renal tubules
(d) Ascending limb of loop of Henle is
impermeable to electrolytes
Ans.(c)
The plasma fluid that filters out from
glomerular capillaries into Bowman’s
capsule of nephron is called glomerular
filtrate. A comparison of the volume of
the filtrate formed per day (180 L/day)
with that of the urine released (1.5L),
suggests that nearly 99% of the filtrate
has to be reabsorbed by the renal
tubules in a process called reabsorption.
32The net pressure gradient that
causes the fluid to filter out of the
glomeruli into the capsule is
[CBSE AIPMT 2005]
(a) 20 mmHg
(b) 75 mmHg
(c) 30 mmHg
(d) 50 mmHg
Ans.(a)
Kidneys help in the formation of urine,
from the blood flowing through
glomerular capillaries. About 20% of
plasma fluid filters out into the
Bowman’s capsule through a thin
glomerular-capsular membrane due to a
net or effective filtration of about
10-15 mm Hg. So, it is the nearest option
(a) which is correct.
33If Henle’s loop were absent from
mammalian nephron, which of the
following is to be expected?
[CBSE AIPMT 2003]
(a) The urine will be more concentrated
(b) The urine will be more dilute
(c) There will be no urine formation
(d) There will be hardly any change in the
quality and quantity of urine formed
Ans.(b)
The main function of the Henle’s loop is
to absorb water from the tubular lumen,
thus making the urine concentrated. If
loop of Henle absent then the urine
becomes more dilute.
34The ability of the vertebrates to
produce concentrated
(hyperosmotic) urine usually
depends upon the
[CBSE AIPMT 2000]
(a) area of Bowman’s capsule epithelium
(b) length of Henle’s loop
(c) length of the proximal convoluted
tubule
(d) capillary network forming glomerulus
Ans.(b)
Length of Henle’s loop determined the
concentration of urine. Urine is
concentrated through counter current
mechanism which involves (a) the loop of
Henle (b) the vasa recta (c) nearby
collecting tubules and ducts, (d) the
interstitial fluid.
35The basic functional unit of human
kidney is [CBSE AIPMT 1997]
(a) nephron
(b) pyramid
(c) nephridia
(d) Henle’s loop
Ans.(a)
Each human kidney consists of about
one million structural and functional
units called nephrons. Each nephron is
mainly made up of two parts :
(i) Malpighian body
(ii) Renal tubule.
Nephridia are excretory tubules found in
the Platyhelminthes (flatworms) and
annelids (earthworms).
36Under normal conditions which one
is completely reabsorbed in the
renal tubule?[CBSE AIPMT 1991]
(a) Urea
(b) Uric acid
(c) Salts
(d) Glucose
Ans.(d)
Glucose is high threshold substance, i.e.,
it is totally or mostly reabsorbed from
the nephric filtrate in the blood
capillaries.
180 NEETChapterwise Topicwise Biology

191. Renal threshold, i.e., upper limit of
kidney to reabsorb such high threshold
substances of kidney for reabsorption of
glucose is about 180 mg/100 mL of
nephric filtrate. When blood sugar level
reaches beyond this, sugar also appears
in urine.
37Reabsorption of useful substances
from glomerular filtrate occurs in
[CBSE AIPMT 1989]
(a) collecting tube
(b) loop of Henle
(c) proximal convoluted tubule
(d) distal convoluted tubule
Ans.(c)
Proximal convoluted tubule is the main
site for the reabsorption of useful
substances from glomerular filtrate. In
PCT complete reabsorption by active
transport takes place for glucose.
PCT reabsorbs most of the amino acids
and vitamin-C, about 70% ofNa
+
and
nearly 75% ofK
+
and a large amount of
Ca
2+
from glomerular filtrate.
Cl
−
is reabsorbed by diffusion. 70% of
water from filtrate is also reabsorbed in
PCT by osmosis.
38Select the correct statement.
[NEET (Oct.) 2020]
(a) Atrial Natriuretic Factor increases
the blood pressure
(b) Angiotensin II is a powerful
vasodilator
(c) Counter-current pattern of blood flow
is not observed in vasa recta
(d) Reduction in Glomerular Filtration
Rate (GFR) activities JG cells to
release renin
Ans.(d)
Statement in option (d) is correct as a
reduction in glomerular filtration rate
activate the JG cells to release renin
which converts angiotensinogen in blood
to angiotensin-I. Other statements are
incorrect and can be corrected as
Atrial Natriuretic Factor (ANF) causes
vasodilation and thereby decreases the
blood pressure.
Angiotensin-II is a powerful
vasoconstrictor, it increases the GFR.
Counter-current pattern of blood flow is
observed in vasa recta.
39Which of the following factors is
responsible for the formation of
concentrated urine?
[NEET (National) 2019]
(a) Maintaining hyperosmolarity towards
inner medullary interstitium in the
kidneys
(b) Secretion of erythropoietin by
juxtaglomerular complex
(c) Hydrostatic pressure during
glomerular filtration
(d) Low levels of antidiuretic hormone
Ans.(a)
Maintaining hyperosmolarity towards
inner medullary interstitium in the
kidneys is the factor responsible for the
formation of concentrate urine because
it provides concentration gradient
necessary for water reabsorption in
renal tubules.
Erythropoietin secretion by juxtaglomerular
complex is related to red blood cell
production and hydrostatic pressure
during glomerular filtration to amount of
filtrate formed by kidneys.
High levels of antidiuretic hormone and
not low levels produce urine that is more
concentrated.
40A decrease in blood
pressure/volume will not cause the
release of [NEET 2017]
(a) renin
(b) atrial natriuretic factor
(c) aldosterone
(d) ADH
Ans.(b)
A decrease in blood pressure/volume
stimulates the hypothalamus to release
ADH (Anti Diuretic Hormone) as well as
JGA (Juxtaglomerular Apparatus) cells to
release renin. Renin byrenin
angiotensinmechanism activates the
adrenal cortex to release aldosterone.
Atrial Natriuretic Factor(ANF) is
produced by atria of heart during
increased blood pressure/volume.
It can cause vasodilation and there by,
decrease the blood pressure, therefore,
option (b) is correct
41Human urine is usually acidic
because [CBSE AIPMT 2015]
(a) the sodium transporter exchanges
one hydrogen ion for each sodium
ion, in peritubular capillaries
(b) excreted plasma proteins are acidic
(c) potassium and sodium exchange
generates acidity
(d) hydrogen ions are actively secreted
into the filtrate
Ans.(d)
The proximal tubule is the portion of
nephron of the kidney which leads from
the Bowman’s capsule to the loop of
Henle.
It regulates the pH of the filtrate by
exchanging hydrogen ions in the
interstitium for bicarbonate ions in the
filtrate. Due to theH
+
ions the human
urine is usually acidic. ThusH
+
are
actively secreted into the filtrate is the
correct option.
42Which one of the following
statements is correct with respect
to kidney function regulation?
[CBSE AIPMT 2011]
(a) Exposure to cold temperature
stimulates ADH release
(b) An increase in glomerular blood flow
stimulates formation of
angiotensin-II
(c) During summer when body loses lot
of water by evaporation, the release
of ADH is suppressed
(d) When someone drinks lot of water
ADH release is stopped
Ans.(d)
When someone drinks lot of water which
is not required by body, the
osmoregulation of the blood will
decrease. The decrease in osmoregulation
will inhibit the release of ADH.
43A person who is on a long hunger
strike and is surviving only on
water, will have[CBSE AIPMT 2007]
(a) more sodium in his urine
(b) less amino acids in his urine
(c) more glucose in his blood
(d) less urea in his urine
Ans.(d)
A person who is on a long hunger strike
and is surviving only on water, will have
less urea in his urine.
Excretory Products and Their Elimination 181
Regulation of
Kidney Function
TOPIC 4

192. As urea is an organic compound which is a
waste product produced during body
metabolism.
44Angiotensinogen is a protein
produced and secreted by
[CBSE AIPMT 2006]
(a) macula densa cells
(b) endothelial cells (cells lining the blood
vessels)
(c) liver cells
(d) Juxtaglomerular (JG) cells
Ans.(c)
Angiotensinogen is a plasma protein
produced and secreted by the liver cells.
Renin is secreted from juxtaglomerular
cells and acts enzymatically on
angiotensinogen to release 10 amino
acid contaning peptide angiotensin-I.
Macula densa is actually a plaque in wall
at the end of thick assending limb of
nephrons.
45A person is undergoing prolonged
fasting. His urine will be found to
contain abnormal quantities of
[CBSE AIPMT 2005]
(a) fats (b) ketones
(c) amino acids (d) glucose
Ans.(b)
If a person is undergoing prolonged
fasting, his urine will be found to contain
abnormal quantities of ketones. During
fasting energy is obtained by the
oxidation of reserved fats. As a result of
fatty acid oxidation large amount of
ketone bodies are produced such as
acetoacetate,β-hydroxybutyrate and
acetone.
46Glucose is taken back from
glomerular filtrate through
[CBSE AIPMT 1993]
(a) active transport
(b) passive transport
(c) osmosis
(d) diffusion
Ans.(a)
Glucose is taken back from glomerular
filtrate through active transport in
proximal convoluted tubule.
47Which one of the following is
correctly matched pair of the given
secretion and its primary role in
human physiology?
[CBSE AIPMT 2000]
(a) Sebum — Sexual attraction
(b) Sweat — Thermoregulation
(c) Saliva — Tasting food
(d) Tears — Excretion of salts
Ans.(b)
Sweat glands are coiled tubular glands
situated in the dermis and connected to
a sweat duct which open as pore on the
surface of the skin.
These secrete sweat which contains
0.1-0.4% of sodium chloride, sodium
acetate and urea.
Sweating occurs when the body
temperature increases. As sweat
evaporates from the skin surface,
energy is lost from the body as latent
heat of vaporation and in this way
sweat reduces body temperature.
48Presence of which of the following
conditions in urine are indicative of
diabetes mellitus?
[NEET (Sep.) 2020]
(a) Uremia and Renal calculi
(b) Ketonuria and Glycosuria
(c) Renal calculi and Hyperglycaemia
(d) Uremia and Ketonuria
Ans.(b)
Presence of glucose (glycosuria) and
ketone bodies (ketonuria) in urine are
indicative of diabetes mellitus. In
diabetes mellitus the body produces
excess ketones as an indication that it is
using an alternative source of energy. It
is seen more commonly in type 1
diabetes mellitus.
Presence of glucose indicates Type II
diabetes. In some cases, insulin cannot
transport blood sugar into the body’s
cells effectively.
This can also cause blood sugar to be
passed out in urine.
49Which of the following would help
in prevention of diuresis?
[NEET (Sep.) 2020]
(a) Reabsorption ofNa
+
and water from
renal tubules due to aldosterone
(b) Atrial natriuretic factor causes
vasoconstriction
(c) Decrease in the secretion of renin by
JG cells
(d) More water reabsorption due to
undersecretion of ADH
Ans.(a)
Adrenal cortex secretes
mineralocorticoids, i.e. aldosterone
which increase the reabsorption ofNa
+
and water from renal tubule thereby
preventing diuresis.
50Use of an artificial kidney during
hemodialysis may result in :
[NEET (National) 2019]
A. Nitrogenous waste build-up in
the body
B. Non-elimination of excess
potassium ions
C. Reduced absorption of calcium
ions from gastrointestinal tract
D. Reduced RBC production
Which of the following options is
the most appropriate?
(a) (B) and (C) are correct
(b) (C) and (D) are correct
(c) (A) and (D) are correct
(d) (A) and (B) are correct
Ans.(b)
Statements (C) and (D) are correct.
Use of an artificial kidney during
haemodialysis may result in reduced
182 NEETChapterwise Topicwise Biology
Disorders of
Excretory System
TOPIC 6
Accessory
Excretory Organs
TOPIC 5

193. Excretory Products and Their Elimination 183
RBC production and reduced absorption
of calcium ions from gastrointestinal
tract. The former would occur due to the
low level of erythropoietin hormone
secreted by juxtaglomerular cells.
The later would be caused due to the
elimination of calcium ions along with
phosphate ions during dialysis.
Statements (A) and (B) are incorrect
because dialysis helps in the removal of
nitrogenous waste and potassium ions
from the body.
51Match the items given in Column I
with those in Column II and select
thecorrectoption given below
[NEET 2018]
Column I Column II
1. Glycosuria (i) Accumulation of
uric acid in joints.
2. Gout (ii) Mass of
crystallised salts
within the kidney.
3. Renal
calculi
(iii) Inflammation in
glomeruli
4. Glomerular
nephritis
(iv) Presence of
glucose in urine.
1 2 3 4
(a) (ii) (iii) (i) (iv)
(b) (i) (ii) (iii) (iv)
(c) (iii) (i) (iv) (i)
(d) (iv) (i) (ii) (iii)
Ans.(d)
Increased level of glucose in blood which
may be caused due to untreated
diabetes mellitus results inglycosuria. In
this condition, glucose is present in the
urine.
Gout is a form of arthritis characterised
by severe pain and tenderness in joints.
It is caused due to the accumulation of
uric acid crystals in joints.
Renal calculiorkidney stonesare small
masses of crystalline salts within the
kidneys. These stones can be of calcium,
uric acid, struvite (magnesium
ammonium phosphate), etc.
Glomerular nephritisis the inflammation
of filtering unit, i.e. glomerulus of kidney.
It is also known asBright’s disease. It
may cause haematuria (blood in urine)
and proteinuria (proteins in blood).
52A condition of failure of kidney to
form urine is called
[CBSE AIPMT 1998]
(a) deamination
(b) entropy
(c) anuria
(d) None of these
Ans.(c)
The termsanuria, oligonuria, polynuria
and dysuria are used for absence of
urine, scanty urine, large amounts of
urine and painful urination respectively.
Deamination is the removal of an amino
(—NH )
2
group frequently from an amino
acid by transaminase enzymes.
53If kidneys fail to reabsorb water,
the effect on tissue would
[CBSE AIPMT 1994]
(a) remain unaffected
(b) shrink and shrivel
(c) absorb water from blood plasma
(d) take moreO
2
from blood
Ans.(b)
If kidney fails to reabsorb water the
concentration of urine will be low and
urination will be more frequent, a
condition called polyuria as a result, the
tissues of the body will be dehydrated
and shrink.

194. 01During muscular contraction, which
of the following events occur?
[NEET 2021]
I. ‘H’ zone disappears
II. ‘A’ band widens
III. ‘I’ band reduces in width
IV. Myosine hydrolyses ATP,
releasing the ADP and Pi
V. Z-lines attached to actins are
pulled inwards.
Choose the correct answer from
the options given below.
(a) I, III, IV and V
(b) I, II, III and IV
(c) II, III, IV and V
(d) II, IV, V and I
Ans.(a)
Statements I, III, IV, and V are correct
and statement II is incorrect and can be
corrected as During muscular
contraction by utilising energy from ATP
hydrolysis the myosin head binds to the
exposed active sites on actins to form a
cross bridge.
This pulls the attached actin filaments
towards the center of A band. This
movement contricts ‘A’ band.
02Calcium is important in skeletal
muscle contraction because it
[NEET 2018]
(a) detaches the myosin head from the
actin filament
(b) activates the myosin ATPase by
binding to it
(c) binds to troponin to remove the
masking of active sites on actin for
myosin
(d) prevents the formation of bonds
between the myosin cross bridges
and the actin filament
Ans.(c)
Calcium plays a key regulatory role in
muscle contraction.Ca
2+
ions bind to
troponin and changes its shape and
position. This in turn, alter the shape and
position of tropomyosin and hence, the
active sites on F-actin are exposed. Due
to this, myosin cross-bridges are able to
bind to these active sites and muscle
contraction occurs.
03Lack of relaxation between
successive stimuli in sustained
muscle contraction is known as
[NEET 2016, Phase I]
(a) fatigue (b) tetanus
(c) tonus (d) spasm
Ans.(b)
Sustained muscle contraction due to
repeated stimulus is known as tetanus.
This results due to muscle fatigue.
04Name the ion responsible for
unmasking of active sites for
myosin for cross-bridge activity
during muscle contraction.
[NEET 2016, Phase II]
(a) Calcium (b) Magnesium
(c) Sodium (d) Potassium
Ans.(a)
Ca
+ +
ions bind withTp
c
unit of troponin
which is responsible for masking of
active sites for myosin through
tropomyosin. The event initiates
cross-bridge activity during muscle
contraction mechanism.
05Smooth muscles are
[NEET 2016, Phase II]
(a) involuntary, fusiform, non-striated
(b) voluntary, multinucleate, cylindrical
(c) involuntary, cylindrical, striated
(d) voluntary, spindle-shaped,
uninucleate
Ans.(a)
Smooth muscles are involuntary,
fusiform and non-striated.
These muscles are located in the inner
walls of hollow visceral organs of the
body like alimentary canal, reproductive
tract, etc.
They do not exhibit any striation and are
smooth in appearance. Their activities
are under autonomic and hormonal
control and thus they are also known as
involuntary muscles.
Hence, option (a) is correct.
06The H-zone in the skeletal muscle
fibre is due to [NEET 2013]
(a) the absence of myofibrils in the
central portion of A-band
(b) the central gap between myosin
filaments in the A-band
(c) the central gap between actin
filaments extending through myosin
filaments in the A-band
(d) extension of myosin filaments in the
central portion of the A-band
Ans.(c)
H-zone in skeletal muscle is due to the
central gap between actin filaments
extending through myosin filaments in
the A-band. Alternate arrangement of
dark and light bands gives the striated
appearance to a skeletal muscle. At the
centre of A-band, a comparatively less
dark zone called H-zone is present.
Locomotion
andMovement
20
Skeletal Muscle
and its Contraction
TOPIC 1

195. Locomotion and Movement 185
In the centre of H-zone, M-line is
present, formed by the threats that
connect the myofilaments.
07Which one of the following is the
correct pairing of a body part and
the kind of muscle tissue that
moves it? [CBSE AIPMT 2009]
(a) Heart wall — Involuntary unstriated
muscle
(b) Biceps of
upper arm
— Smooth muscle fibres
(c) Abdominal
wall
— Smooth muscle
(d) Iris — Involuntary smooth
muscle
Ans.(d)
Smooth muscles are plain, non-striated,
involuntary or unstriped muscles due to
the absence of striations. These occur in
the walls of hollow internal organs, in
capsules of lymph glands, spleen, etc., in
iris and ciliary body of eyes, skin dermis,
penis and other accessory genitalia, etc.
08The contractile protein of skeletal
muscle involving ATPase activity is
[CBSE AIPMT 2006]
(a) myosin (b) a-actinin
(c) troponin (d) tropomyosin
Ans.(a)
The kinesin, myosin and dynein proteins
of skeletal muscle involve ATPase
activity. This activity cause the
contraction of skeletal muscles,
propelling action of cilia and flagella and
the intracellular transport of organelles.
09ATPase enzyme needed for muscle
contraction is located in
[CBSE AIPMT 2004]
(a) actinin (b) troponin
(c) myosin (d) actin
Ans.(c)
During muscular contraction myosin
cross bridges heads. Myosin is not only
an actin binding protein, it is also an
ATPase (an enzyme) which is activated
byCa
2+
andMg
2+
ions. Hence, ATPase
enzyme is located in myosin.
10Which statement is correct for
muscle contraction?
[CBSE AIPMT 2001]
(a) Length of H-zone decrease
(b) Length of A-band remains constant
(c) Length of I-band increases
(d) Length of two Z-line increases
Ans.(b)
WhenCa
+
ions combine with troponin
then in consequence muscle contraction
will initiates.
During contraction, the Z-lines come
closer together and the sarcomere
becomes shorter. The length of A-band
remains constant. I-bands shortens and
H-zone narrows.
11What is sarcomere?
[CBSE AIPMT 2001]
(a) Part between two H-lines
(b) Part between two A-lines
(c) Part between two I-bands
(d) Part between two Z-lines
Ans.(d)
Huxley reported the fine structure of
striated muscle fibre or myofibril. Each
myofibril is made up of A-bands (dark
band anisotropic) and I-bands = isotropic
(light bands isotropic). These two bands
are made up of myosin and actin
filament respectively. Each I-band is
divided into two equal halves by a thin,
fibrous and transverse zig-zag partition,
called Z-band or Z-disc or Krause’s
membrane.
The part of the myofibril between two
successive Z-lines functions as a
contractile unit called sarcomere.
12The functional unit of contractile
system in striated muscle is
[CBSE AIPMT 1998]
(a) myofibril (b) sarcomere
(c) Z-lines (d) cross bridges
Ans.(b)
Skeletal muscle is also called striated
muscle because the regular
arrangement of the myofilaments
creates as repeating pattern of light and
dark bands. Each repeating unit is a
sarcomere,the basic functional unit of
the muscle.
13Which of the following is the
contractile protein of a muscle?
[CBSE AIPMT 1998]
(a) Myosin (b) Tropomyosin
(c) Actin (d) Tubulin
Ans.(a)
Myosin-II, a two-headed tailed variety
of myosin is involved in muscle
contraction.
14Which ion is essential for muscle
contraction?[CBSE AIPMT 1994]
(a) Na (b) K (c) Ca (d) Cl
Ans.(c)
Calcium is an essential element required
for the contraction of muscles. Release
of calcium ions from sarcoplasmic
reticulum trigger the muscle contraction
process. In fact, calcium ions and ATP,
actin and myosin interact forming
actomyosin, which causes the muscles
to contract.
15Intercostal muscles occur in
[CBSE AIPMT 1988]
(a) abdomen (b) thigh
(c) ribs (d) diaphragm
Ans.(c)
Intercostal muscles, i.e. internal
intercostal and external intercostal are
attached with the ribs, these are the
main muscles for normal quite
breathing. External intercostal muscle
and diaphragm contract during
inspiration and relax during expiration.
16Match the List I with List II.
[NEET 2021]
List I List II
A. Scapula 1. Cartilaginous joints
B. Cranium 2. Flat bone
C. Sternum 3. Fibrous joints
D. Vertebral
column
4. Triangular flat bone
Choose the correct answer from
the options given below.
A B C D
(a) 1 3 2 4
(b) 2 3 4 1
(c) 4 2 3 1
(d) 4 3 2 1
Ans.(d)
(A)-(4), (B)-(3), (C)-(2), (D)-(1)
ScapulaIt is a flat, triangular shaped
bone. It is located at the upper thoracic
region on the dorsal surface of the ribcage.
CraniumIt is the part of the skull that
encloses the brain. They have
immovable fibrous joint.
SternumIt is the flat bone located in the
middle of chest. Also referred as
breastbone.
Vertebral columnIt is a series of
approximately 33 bones called
vertebrae. They have cartilaginous type
joints designed for weight bearing.
Skeletal System
TOPIC 2

196. 17Select the incorrectly matched pair
from following.[NEET (Oct.) 2020]
(a) Chondrocytes–Smooth muscle cells
(b) Neurons–Nerve cells
(c) Fibroblast–Areolar tissue
(d) Osteocytes–Bone cells
Ans.(a)
The incorrectly match pair is option (a). It
can be corrected as
Chondrocytes are cartilage cells. The
intercellular material of cartilage is
solid, pliable and resists compression.
Cells of this tissue are called
chondrocytes which are enclosed in
small cavities within the matrix
secreted by them.
18Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. Floating
ribs
1. Located between
second and seventh
ribs
B. Acromion 2. Head of the
humerus
C. Scapula 3. Clavicle
D. Glenoid
cavity
4. Do not connect with
the sternum
A B C D
(a) 1 3 2 4
(b) 3 2 4 1
(c) 4 3 1 2
(d) 2 4 1 3
Ans.(c)
The correct option is (c). It can be
explaed as follows
11th and 12th pairs of ribs are not
connected ventrally with sternum and are
therefore, called floating ribs.
Acromion is a flat expanded process of
spine of scapula. The lateral end of
clavicle articulates with acromion
process. Scapula is a flat triangular bone
in the dorsal part of the thorax between
2nd and the 7th rib.
Glenoid cavity of scapula articulates with
head of the humerus to form the
shoulder joint.
19Select the correct option.
[NEET (National) 2019]
(a) 11th and 12th pairs of ribs are
connected to the sternum with the
help of hyaline cartilage
(b) Each rib is a flat thin bone and all the
ribs are connected dorsally to the
thoracic vertebrae and ventrally to
the sternum
(c) There are seven pairs of
vertebrosternal, three pairs of
vertebrochondral and two pairs of
vertebral ribs
(d) 8th, 9th and 10th pairs of ribs
articulate directly with the sternum
Ans.(c)
The statement that there are seven pairs
of vertebrosternal, three pairs of
vertebrochondral and two pairs of
vertebral ribs is correct.
Rest statements are incorrect. Correct
information about the statements is as
follows Vertebrosternal ribs are first
seven pairs of true ribs which are
attached dorsally to thoracic vertebrae
and ventrally to the sternum.
Vertebrochondral ribs (8th, 9th, 10th
pairs) are false ribs which are not
attached to sternum directly. They are
attached to the seventh rib with the help
of hyaline cartilage.
Vertebral ribs are the last two pairs of
floating ribs which are attached to
vertebrae dorsally and are not attached
ventrally.
20Out of ‘X’ pairs of ribs in humans
only ‘Y’ pairs are true ribs. Select
the option that correctly
represents values ofXandYand
provides their explanation.
[NEET 2017]
(a)X Y= =12 7, True ribs are
attached dorsally to
vertebral column and
ventrally to the
sternum.
(b)X Y= =12 5, True ribs are
attached dorsally to
vertebral column and
sternum on the two
ends.
(c)X Y= =24 7, True ribs are dorsally
attached to vertebral
column, but are free
on ventral side.
(d)X Y= =24 12, True ribs are dorsally
attached to vertebral
column, but are free
on ventral side.
Ans.(a)
In the rib cage, the true ribs are those
which are attached to the sternum in the
front and vertebral column at back.
These are 7 in numbers. Although there
are total 12 ribs in the rib cage.
The 11th and 12th ribs are attached to the
vertebral column and keep floating in the
thoracic cavity, so are called floating
ribs.
21Which of the following is not a
function of the skeletal system?
[CBSE AIPMT 2015]
(a) Production of erythrocytes
(b) Storage of minerals
(c) Production of body heat
(d) Locomotion
Ans.(c)
Production of body heat is the function
of muscles. The contraction of muscle
produce heat, which keeps the body
warm during the winters.
22An acromion process is
characteristically found in the
[CBSE AIPMT 2005]
(a) pelvic girdle of mammals
(b) pectoral girdle of mammals
(c) skull of frog
(d) sperm of mammals
Ans.(b)
An acromion process is found in pectoral
girdle of mammals. Pectoral girdle
consists of scapula and clavicle. The
scapula consists of a sharp ridge, the
spine and a triangular body. The end of
the spine projects as a flattened and
expanded process called acromion.
23What will happen if ligaments are
cut or broken?[CBSE AIPMT 2002]
(a) Bones will move freely at joints
(b) No movement at joint
(c) Bone will become unfix
(d) Bone will become fixed
Ans.(c)
Ligaments are specialised connective
tissues which connect bones together;
hence, if they are cut or broken the bone
will become unfixed.
186 NEETChapterwise Topicwise Biology
1
2
3
4
5
6
7
8
9
10
True
ribs
False
ribs
11
12
Floating
ribs
Vertebral
column
Ribs
Sternum
Ribs and
rib cage

197. 24Which one of the following is a skull
bone? [CBSE AIPMT 2000]
(a) Atlas (b) Coracoid
(c) Arytenoid (d) Pterygoid
Ans.(d)
Pterygoid is a small bone articulated with
the palatine. In human it becomes the
pterygoid processes of the sphenoid bone.
25Ligament is a[CBSE AIPMT 1999]
(a) modified yellow elastic fibrous
tissue
(b) inelastic white fibrous tissue
(c) modified white fibrous tissue
(d) None of the above
Ans.(b)
Ligaments are made up of inelastic
white fibrous tissue and connect bones
at joints. It is also known as articular
ligament, articular larua or fibrous
ligament.
26Tendon is made up of
[CBSE AIPMT 1999]
(a) adipose tissue
(b) modified white fibrous tissue
(c) areolar tissue
(d) yellow fibrous connective tissue
Ans.(b)
The modified white fibres tissue form
cords called tendons which connect
muscles with the bones.
27Total number of bones in the hind
limb of man is[CBSE AIPMT 1998]
(a) 14 (b) 30
(c) 24 (d) 21
Ans.(b)
1 femur+1 fibula+1 tibia+1 patella+7
tarsals+5 meta tarsals+14 phalanges
make one hind limb of man. Total of 30
bones.
28The lower jaw in mammals is made
up of [CBSE AIPMT 1998]
(a) mandible (b) dentary
(c) maxilla (d) angulars
Ans.(b)
Dentary is tooth bearing membrane bone
of lower jaw of the vertebrates-one on
each side.
29The number of floating ribs, in the
human body, is[CBSE AIPMT 1995]
(a) 6 pairs (b) 5 pairs
(c) 3 pairs (d) 2 pairs
Ans.(d)
Usually, there are 12 pairs of ribs, but
occasionally these may be 11, 13 or even
14 pairs. The first seven pairs of ribs are
known as true ribs.
Pairs 8, 9, 10 are false ribs, they are
attached indirectly to sternum by means
of cartilages. Last two pairs (11 and 12) do
not reach to the sternum and are known
as ‘floating ribs’.
30Which is a part of pectoral girdle?
[CBSE AIPMT 1994]
(a) Glenoid cavity (b) Sternum
(c) Ileum (d) Acetabulum
Ans.(a)
A cavity known as glenoid cavity is
present at the tip of scapula and
coracoid process, for articulation of
head of humerus. Pectoral girdle
consists of a membranous bone called
clavicle and a large replacing bone called
shoulder blade or scapula coracoid.
31Long bones function in
[CBSE AIPMT 1993]
(a) support
(b) support, erythrocyte and leucocyte
synthesis
(c) support and erythrocyte synthesis
(d) erythrocyte formation
Ans.(b)
In higher animals, movements and
locomotion depend on the association of
skeletal muscles and skeletal system
(bones and joints). Bones function to
provide support as well as produce
erythrocytes and leucocytes in the bone
marrow.
32A deltoid ridge occurs in
[CBSE AIPMT 1990]
(a) radius
(b) ulna
(c) femur
(d) humerus
Ans.(d)
Humerus is the bone of forelimb. It bears
head, greater tuberosity and lesser
tuberosity. Proximal part of humerus
bears a slight ridge, which is called as
deltoid ridge.
33Number of cervical vertebrae in
camel is [CBSE AIPMT 1990]
(a) more than that of rabbit
(b) less than that of rabbit
(c) same as that of whale
(d) more than that of horse
Ans.(c)
Cervical vertebrae are seven in number,
constant in most of the mammals.
Whale, camel, giraffe, etc. have same
number of cervical vertebrae. However,
long neck in camel and giraffe is due to
more length of vertebrae.
34Extremities of long bones possess
cartilage [CBSE AIPMT 1989]
(a) calcified (b) fibrous
(c) elastic (d) hyaline
Ans.(d)
Hyaline cartilage is the most abundant
cartilage. It forms the cushions or pads
upon the articular surfaces at joints of
long bones, known as articular cartilage.
It forms the costal cartilage at the
ventral ends of ribs, and also helps to
form the nose, larynx, trachea, bronchi
and bronchial tubes.
35Match the following joints with the
bones involved[NEET (Odisha) 2019]
Column I Column II
1. Gliding
joint
(i) Between carpal
and metacarpal
of thumb
2. Hinge joint (ii) Between atlas
and axis
3. Pivot joint (iii) Between the
carpals
4. Saddle
joint
(iv) Between
humerus and ulna
Select the correct option from the
following
1 2 3 4
(a) (iii) (iv) (ii) (i)
(b) (iv) (i) (ii) (iii)
(c) (iv) (ii) (iii) (i)
(d) (i) (iii) (ii) (iv)
Ans.(a)
The correct matches are
1. Gliding joint–(iii) Between the
carpals
2. Hinge joint – (iv) Between humerus
and ulna
3. Pivot joint – (ii) Between atlas and
axis
4. Saddle joint – (i) Between carpal and
metacarpal of thumb
Locomotion and Movement 187
Joints
TOPIC 3

198. 188 NEETChapterwise Topicwise Biology
36The pivot joint between atlas and
axis is a type of [NEET 2017]
(a) fibrous joint (b) cartilaginous joint
(c) synovial joint (d) saddle joint
Ans.(c)
The joint between atlas and axis is a type
of synovial joint. A considerable movement
is allowed at all synovial joints. They are
surrounded by tubular articular capsule.
The capsule consists of two layers, i.e.
outer fibrous capsule and inner synovial
membrane, which secretes synovial fluid
lubricates and providing nourishment to
articular cartilage.
Concept Enhancer
Fibrous jointThey do not allow
movement because the bones are held
firmly by bundles of white fibrous tissue.
e.g. joint between the bones of skull.
Cartilaginous jointsThey allow slight
movement; because of the elastic pads
of fibrocartilage present between the
ends of the bones taking part in the
joints, e.g. pubic symphysis of pubis.
Saddle jointsThis is the type of synovial
joints. This type of joint is like ball and
socket joint, but not developed fully, e.g.
joint between carpal of hand.
37Which of the following joints would
allow no movement?
[CBSE AIPMT 2015]
(a) Fibrous joint
(b) Cartilaginous joint
(c) Synovial joint
(d) Ball and socket joint
Ans.(a)
Fibrous joints are immovable joints that
occur between the bones of the cranium
and in the tooth sockets. They do not
allow movement because the bones are
held firmly together by bundles of strong
white collagen fibres. The immovable
joints are often known as the sutures.
The periosteum that covers the bones
dips between them as a septum.
38Select the correct matching of the
type of the joint with the example
in human skeletal system
[CBSE AIPMT 2014]
Types of Joint Example
(a) Cartilaginous
joint
Between frontal
and parietal
(b) Pivot joint Between third and
fourth cervical
vertebrae
(c) Hinge joint Between humerus
and pectoral
girdle
(d) Gliding joint Between carpals
Ans.(d)
Gliding joint is a type of synovial joint,
found between carpal bones and tarsal
bones.
Cartilaginous joint between surfaces of
skull bones. Pivot joint is found between
the atlas and axis. Hinge joint is found in
elbow, knee ankle and interphalangeal
joints.
39The characteristics and an example
of a synovial joint in humans is
[NEET 2013]
Characteristics Examples
(a) Fluid cartilage
between two
bones, limited
movements
Knee joints
(b) Fluid filled
between two
joints, provides
cushion
Skull bones
(c) Fluid filled synovial
cavity between
two bones
Joint between
atlas and axis
(d) Lymph filled
between two
bones, limited
movement
Gliding joint
between
carpals
Ans.(c)
Joint between atlas and axis is pivot
joint, which is an example of synovial
joint characterised by the presence of a
fluid-filled synovial cavity between the
articulating surface of the two bones.
Knee joint (hinge joint) is a synovial joint
characterised by the presence of fluid
filled synovial cavity between the
articulating surfaces of the two bones.
Fluid cartilage between two bones
having limited movements is slightly
movable joint.
In freely movable joints fluid filled
between two joints provides cushion. In
gliding joint, the articulating bones can
slide upon one another.
40Elbow joint is an example of
[CBSE AIPMT 2009]
(a) pivot joint
(b) hinge joint
(c) gliding joint
(d) ball and socket joint
Ans.(b)
In hinge joint, the convex surface of one
bone fits into the concave surface of
another bone, e.g. knee, elbow and
interphalangeal joints.
Pivot joint, one bone is fixed and second
articulated, e.g. atlas and axial of skull
rotate with axis vertebra.
Gliding joints primarily permit
side-by-side and back-and-forth gliding
movements, e.g. intercarpal joints and
intertarsal joints.
In ball and socket joint, ball of one
bone articulates in socket of another
bone, e.g. head of humerus and
glenoid cavity of pectoral girdle.
41Which of the following pairs, is
correctly matched?
[CBSE AIPMT 2005]
(a) Hinge joint — Between
vertebrae
(b) Gliding joint — Between
zygapophyses of
the successive
vertebrae
(c) Cartilaginous
joint
— Skull bones
(d) Fibrous joint — Between
phalanges
Ans.(b)
Gliding joint is present between
zygapophyses of the successive
vertebrae. This joint permits sliding
movements of two bones over each
other.
42The joint found between sternum
and the ribs in humans is
[CBSE AIPMT 2000]
(a) angular joint (b) fibrous joint
(c) cartilaginous joint (d) gliding joint
Ans.(c)
Cartilaginous jointsare found between
the centra of vertebrae; at pubic
symphysis and between ribs and
sternum.
Fibrous jointsare also called as sutures
or immovable joints because in these
joints the adjoining bones cannot move
upon each other, e.g. coronal suture
between frontal and parietal bones of skull.
Ingliding joints, two bones can slide
upon each other, e.g. joints between
carpals in wrist.
In angular joints one bone is movable on
another bone in two planes side to side
and back and forth, e.g. wrist joints.

199. 43Chronic auto immune disorder
affecting neuro muscular junction
leading to fatigue, weakening and
paralysis of skeletal muscle is
called as [NEET 2021]
(a) arthritis
(b) muscular dystrophy
(c) myasthenia gravis
(d) gout
Ans.(c)
Myasthenia gravis is defined as an
autoimmune neuromuscular disorder
that leads to fluctuating muscle
weakness and fatigue. The cause of this
disease is circulating antibodies that will
block acetylcholine receptors at the
post-synaptic neuromuscular junctions.
Other options can be explained as :
Muscular dystrophyis a disease
characterised by progressive
degeneration of muscle fibres without
the involvement of nervous system.
Goutis caused by two conditions that
includes one which occur due to
excessive formation of uric acid and
another by the inability to excrete it.
Uric acid deposits in the form of
monosodium salts.Arthritisis the
swelling and tenderness of one or more
number of joints. It is caused by injury,
abnormal metabolism, and genetic makeup.
44Match the following columns and
select the correct option from the
codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. Gout 1. Decreased levels
of estrogen
B. Osteoporosis 2. LowCa
2+
ions in
the blood
C. Tetany 3. Accumulation of
uric acid
crystals
D. Muscular
dystrophy
4. Autoimmune
disorder
Codes
A B C D
(a) 2 1 3 4
(b) 3 1 2 4
(c) 4 3 1 2
(d) 1 2 3 4
Ans.(b)
The option (b) is correct match which is
as follows
Gout is inflammation of joints due to
accumulation of uric acid crystals.
Osteoporosis occurs due to decreased
levels of oestrogen in females. It is an
age-related disorder characterised by
decreased bone mass and increased
chances of fractures.
Tetany is rapid spasms (wild
contractions) in muscle when there is low
Ca
2+
ions in the blood.
Muscular dystrophy is progressive
degeneration of skeletal muscle mostly
due to genetic disorder.
45Which of the following muscular
disorders is inherited?
[NEET (National) 2019]
(a) Muscular dystrophy
(b) Myasthenia gravis
(c) Botulism
(d) Tetany
Ans.(a)
Muscular dystrophy is an inherited
muscular disorder in which the skeletal
muscles degenerate progressively. It is
caused due to the absence of dystrophin
protein which helps to keep muscle cells
intact.
Myasthenia gravis is an autoimmune
neuromuscular disease that causes
paralysis of skeletal muscles.
Botulism is a type of food poisoning
caused by bacteriumClostridium
botulinum.
Tetany is an involuntary muscle
contraction caused due to the low
level of calcium in body.
46Osteoporosis, an age-related
disease of skeletal system, may
occur due to[NEET 2016, Phase II]
(a) immune disorder affecting
neuromuscular junction leading to
fatigue
(b) high concentration of Ca
+ +
and Na
+
(c) decreased level of oestrogen
(d) accumulation of uric acid leading to
inflammation of joints
Ans.(c)
Osteoporosis is caused by decreased
level of oestrogen. Oestrogen deficiency
causes both the early and late forms of
osteoporosis in post-menopausal women.
Osteoporosis is thinning or weaking of
bones which makes them fragile and
more likely to break.
Women have low oestrogen level when
they are transitioning through
menopause. Hence, option (c) is correct.
47Select the correct statement with
respect to locomotion in humans.
[NEET 2013]
(a) A decreased level of progesterone
causes osteoporosis in old people
(b) Accumulation of uric acid crystals in
joints causes their inflammation
(c) The vertebral column has 10 thoracic
vertebrae
(d) The joint between adjacent vertebrae
is a fibrous joint
Ans.(b)
Inflammation of joints due to the
accumulation of uric acid crystals is
gout. Fibrous joints are formed by the
flat skull bones, which fuse end-to-end
with the help of dense fibrous
connective tissues in the form of
sutures to form cranium. The vertebral
column is formed by 26 serially arranged
units called vertebrae. The less
secretion of progesterone causes
abortion as it basically supports pregnancy.
48Select the correct statement
regarding the specific disorder of
muscular or skeletal system.
[CBSE AIPMT 2012]
(a) Muscular dystrophy –Age related
shortening of muscles
(b) Osteoporosis –Decrease in bone mass
and higher chances of fractures with
advancing age
(c) Myasthenia gravis –Autoimmune
disorder which inhibits sliding of
myosin filaments
(d) Gout–Inflammation of joints due to
extra deposition of calcium
Ans.(b)
Osteoporosis is age related disease in
which bones loose minerals and fibres
from the matrix causing decreased bone
mass and higher chances of fractures
with advancing age.
Major causative factors of osteoporosis
are imbalance of hormones like
calcitonin of thyroid, parathormone of
parathyroids, sex hormones and
deficiencies of calcium and vitamins.
The disease may be classified as primary
type 1, primary type 2 or secondary.
The form of osteoporosis most common
in women after menopause is referred to
as primary type 1 or postmenopausal
osteoperosis.
Secondary osteoporosis may arise at any
age and affect men and women equally.
Locomotion and Movement 189
Disorders of Bones
and Muscles
TOPIC 4

200. 01Receptor sites for
neurotransmitters are present on
[NEET 2017]
(a) membrances of synaptic vesicles
(b) pre-synaptic membrane
(c) tips of axons
(d) post-synaptic membrane
Ans.(d)
The post-synaptic membrane of the
synapse of a neuron contains the
receptors for neurotransmitters.
02Myelin sheath is produced by
[NEET 2017]
(a) Schwann cells and Oligodendrocytes
(b) Astrocytes and Schwann cells
(c) Oligodendrocytes and Osteoclasts
(d) Osteoclasts and Astrocytes
Ans.(a)
The myelin sheath is a greatly extended
and modified plasma membrane
wrapped around the nerve axon in a
spiral fashion. It is originated from
Schwann cells in the peripheral nervous
system and oligodendroglial cells in the
central nervous system.
03Destruction of the anterior horn
cells of the spinal cord would result
in loss of [CBSE AIPMT 2015]
(a) sensory impulses
(b) voluntary motor impulses
(c) commissural impulses
(d) integrating impulses
Ans.(b)
Destruction of the anterior horn cells of
the spinal cord would result in loss of
voluntary motor impulses. It is because
the anterior horn cells (also called
anterior grey column), which is the front
column of grey matter in the spinal cord
contains motor neurons that affect the
axial muscles.
04How do parasympathetic neural
signals affect the working of the
heart? [CBSE AIPMT 2014]
(a) Reduce both heart rate and cardiac
output
(b) Heart rate is increased without
affecting the cardiac output
(c) Both heart rate and cardiac output
increase
(d) Heart rate decreases but cardiac
output increases
Ans.(a)
Parasympathetic neural signal reduces
both heart rate and cardiac output,
through the post ganglionic fibres.
These fibres are very short, and are
cholinergic in nature.
05When a neuron is in resting state,
i.e. not conducting any impulse, the
axonal membrane is
[CBSE AIPMT 2011]
(a) equally permeable to both Na
+
and K
+
ions
(b) impermeable to both Na
+
and K
+
ions
(c) comparatively more permeable to K
+
ions andnearly impermeableto Na
+
ions
(d) comparatively more permeable to Na
+
ions and nearly impermeable to K
+
ions
Ans.(c)
Neurons are excitable cells because
their membrane are in a polarised state.
Different types of selectively permeable
channels are present on the neural
membrane. When a neuron is not
conducting any impulse, or in the resting
stage, the axonal membrane is
comparatively more permeable to
potassium ion (K
+
) and nearly
impermeable to sodium ion (Na
+
).
06During the propagation of a nerve
impulse, the action potential
results from the movement of
[CBSE AIPMT 2008]
(a)K
+
ions from extracellular fluid to
intracellular fluid
(b)Na
+
ions from intracellular fluid to
extracellular fluid
(c)K
+
ions from intracellular fluid to
extracellular fluid
(d)Na
+
ions from extracellular fluid to
intracellular fluid
Ans.(a)
During the nerve impulse when a
stimulus of adequate strength is applied
to a polarised membrane, the
permeability of the membrane to Na
+
is
increased at the point of stimulation. As
NeuralControl
andCoordination
21
Structure of the
Nervous System
TOPIC 1
Axon
Axon
terminal
Synaptic
vesicles
Pre-synaptic
membrane
Synaptic cleft
Post-synaptic
membrane
Receptors for
neurotransmitter
Neurotransmitters
Synapse
Diagram showing axon terminal and
synapse

201. Neural Control and Coordination 191
a result the sodium ion channels permit
the influx ofNa
+
by diffusion into the
intracellular fluid from extracellular fluid.
07During the transmission of nerve
impulse through a nerve fibre, the
potential on the inner side of the
plasma membrane has which type of
electric charge?[CBSE AIPMT 2007]
(a) First negative, then positive and again
back to negative
(b) First positive, then negative and
continue to be negative
(c) First negative, then positive and
continue to be positive
(d) First positive, then negative and again
back to positive
Ans.(a)
During the transmission of nerve
impulse through a nerve fibre, the
potential on the inner side of the
plasma membrane has first become
negative charged, then positive and
again negative by repolarisation.
08Which one of the following pairs of
structures distinguishes a nerve
cell from other types of cell?
[CBSE AIPMT 2007]
(a) Perikaryon and dendrites
(b) Vacuoles and fibres
(c) Flagellum and medullary sheath
(d) Nucleus and mitochondria
Ans.(a)
A nerve cell consists of cell body
(perikaryon) containing the nucleus,
Nissl granules, dendrites and an axon.
These are specialised cells.
09Which of the following statements
is correct about node of Ranvier?
[CBSE AIPMT 2002]
(a) Axolemma is discontinuous
(b) Myelin sheath is discontinuous
(c) Both neurilemma and myelin sheath
are discontinuous
(d) Covered by myelin sheath
Ans.(b)
Neurons are the chief functional units of
the nervous system. An ordinary neuron
has a soma or cyton and a long thread
and, called as axon which is enclosed in a
multilayered myelin sheath. The myelin
sheath is interrupted at the spaces
between Schwann cells to form gaps.
These gaps are callednodes of Ranvier.
These nodes and the myelin sheath
create condition that speed up the nerve
impulses.
10Which of the following is regarded
as a unit of nervous tissue?
[CBSE AIPMT 1999]
(a) Myelin sheath (b) Axons
(c) Dendrites (d) Neurons
Ans.(d)
The nervous tissue is made up of nerve
cells (the repeating units) also called
neurons.
Each neuron has a cell body or cyton and
two kinds of cell processes
(a) Dendrons, come out from cyton.
(b) Axon, an elongated nerve fibre.
11The junction between the axon of
one neuron and the dendrite of the
next is called[CBSE AIPMT 1999]
(a) junction point (b) a synapse
(c) a joint (d) constant bridge
Ans.(b)
The end to end position of the axon of
one neuron and the dendrites of another
neuron is called thesynapse.
Most of the neurons do not actually
touch other neurons with which they
communicate, instead there is a minute
space. This separating gap is called the
synaptic cleft.
12Sympathetic nervous system
induces [CBSE AIPMT 1997]
(a) heartbeat
(b) secretion of digestive juices
(c) secretion of saliva
(d) All of the above
Ans.(a)
Medulla of brain has two regions
affecting heart rate (a) cardiac inhibitory
centre, (b) cardiac accelerator centre.
Sensory nerves originating from the
accelerator centre run parallel to the
spinal cord and enter the sino-atrial
node. Stimulation by these nerves, which
are part of sympathetic nervous system
cause an increase in heartbeat.
13In humans, visceral organs are
innervated by[CBSE AIPMT 1996]
(a) sympathetic nerves and are under
conscious control
(b) parasympathetic nerves and are
under conscious control
(c) Both (a) and (b)
(d) both sympathetic and
parasympathetic nerves but are not
under conscious control
Ans.(d)
Both sympathetic and parasympathetic
nerve fibres innervate visceral organs
and coordinate their activity
antagonistically, but this is not under
body's conscious control.
14Afferent nerve fibres carry
impulses from[CBSE AIPMT 1992]
(a) effector organs to CNS
(b) receptors to CNS
(c) CNS to receptors
(d) CNS to muscles
Ans.(b)
Afferent nerve fibres are formed of only
sensory nerve fibres, conduct nerve
impulses from sensory organs or
receptors to central nervous system to
produce sensation, e.g. optic nerve.
15One function of parasympathetic
nervous system is
[CBSE AIPMT 1990]
(a) contraction of hair muscles
(b) stimulation of sweat glands
(c) acceleration of heartbeat
(d) constriction of pupil
Ans.(d)
Parasympathetic nervous system
involves conservation of energy and
brings about relaxation, comfort,
pleasure etc, at the time of rest. Another
function of this system is that during
emergency or stress while SNS dilates
pupil for more light, PNS constricts the
pupil to its normal condition.
Whereas,sympathetic nervous
systeminvolves expenditure of energy
and increases the defence system of
body against adverse conditions, so, it
operates during stress, pain, fear and
anger.
Dendrites
Nucleus
Cell body
Axon
Nodes of
RanvierMyelin
sheaths
Schwann
cell
Neuron structure

202. 16Which of the following is
associated with decrease in
cardiac output?[NEET (Oct.) 2020]
(a) Sympathetic nerves
(b) Parasympathetic neural signals
(c) Pneumotaxic centre
(d) Adrenal medullary hormones
Ans.(c)
Parasympathetic neural signals (a
component of autonomic nervous
system) decreases the rate of heartbeat,
speed of conduction of action potential
and thereby the cardiac output.
17Which of the following statements
is not correct?[NEET (Odisha) 2019]
(a) An action potential in an axon does
not move backward because the
segment behind is in a refractory
phase
(b) Depolarisation of hair cells of cochlea
results in the opening of the
mechanically gated potassium-ion
channels
(c) Rods are very sensitive and
contribute to daylight vision
(d) In the knee-jerk reflex, stimulus is the
stretching of muscle and response is
its contraction
Ans.(c)
Option (c) is not correct because rods
and cones are photoreceptor cells in our
eyes. The rod cells contain a purple
pigment rhodopsin that is useful in night
vision or scotopic vision. Daylight
(photopic) vision and colour vision are
the functions of cones.
18Which part of the brain is
responsible for thermoregulation?
[NEET (National) 2019]
(a) Hypothalamus
(b) Corpus callosum
(c) Medulla oblongata
(d) Cerebrum
Ans.(a)
Hypothalamus is the thermoregulatory
centre in the brain and it maintains the
constant body temperature of 37°C. The
hypothalamus contains a number of
centres, which control body temperature.
Corpus callosum is the thick band of
nerve fibres that divide the cerebrum
into left and right hemispheres.
Medulla oblongata is the component of
hindbrain. It receives and integrates
signals from spinal cord and sends them
to cerebellum. Cerebrum is the large
part of the brain and consists of two
hemispheres.
19Stimulation of a muscle fibre by a
motor neuron occurs at
[CBSE AIPMT 2014]
(a) the neuromuscular junction
(b) the transverse tubules
(c) the myofibril
(d) the sarcoplasmic reticulum
Ans.(a)
Stimulation of a muscle fibre by a motor
neuron occurs at neuromuscular junction
(the area of contact between a nerve and
muscle fibre). It is also called motor-end
plate. At neuromuscular junction a
neuron activates a muscle to contract
during the excitation contraction
coupling of vertebrate skeletal muscles.
20Injury localised to the hypothalamus
would most likely disrupt
[CBSE AIPMT 2014]
(a) short term memory
(b) co-ordination during locomotion
(c) executive function, such as decision
making
(d) regulation of body temperature
Ans.(d)
The hypothalamus performs many
functions which are important for the
survival and enjoyment of life. It serves
as a link between ‘mind’ and ‘body’ and
between the nervous and endocrine
system. The hypothalamus is
responsible for hormone production.
The hormone produced by this area
govern body temperature thirst hunger,
sleep, circoction rhythm, mood sex drive
etc. This area of the brain also controls
the functioning of pituitary gland. Thus,
if any injury localised to the
hypothalamus it will disrupt the
complete regulation of body
temperature and other activities.
21A diagram showing axon terminal
and synapse is given. Identify
correctly at least two ofA-D
[NEET 2013]
(a)A–Receptor,C–Synaptic vesicles
(b)B–Synaptic connection,D–K
+
(c)A–Neurotransmitter,
B–Synaptic cleft
(d)C–Neurotransmitter,D–Ca
2+
Ans.(a)
A–Receptor,B–Synaptic cleft,
C–Synaptic vesicles,D–Ca
2+
.
22The human hindbrain comprises
three parts, one of which is
[CBSE AIPMT 2012]
(a) spinal cord
(b) corpus callosum
(c) cerebellum
(d) hypothalamus
Ans.(c)
The hindbrain generally has its anterior
roof enlarged to form a pair of cerebellar
hemispheres.
Its floor is thickened to form the pons
anteriorly and the medulla oblongata
posteriorly.
23The nerve centres which control
the body temperature and the urge
for eating are contained in
[CBSE AIPMT 2010]
(a) hypothalamus
(b) pons
(c) cerebellum
(d) thalamus
Ans.(a)
Hypothalamus is the part of the sides
and floor of the brain derived from the
forebrain. It lies at the base of thalamus.
The hypothalamus contains a number of
centres, which control body temperature,
urge for eating and drinking. It also
contains several groups of neurosecretory
cells, which secrete hormones called as
hypothalamic hormones.
192 NEETChapterwise Topicwise Biology
A
C
B
D
Physiology of the
Nervous System
and Reflex Arc
TOPIC 2

203. 24Which part of human brain is
concerned with the regulation of
body temperature?
[CBSE AIPMT 2009]
(a) Medulla oblongata
(b) Cerebellum
(c) Cerebrum
(d) Hypothalamus
Ans.(d)
In human brain,hypothalamusis a
centre for hunger, thirst, sweating,
sleep, fatigue, temperature, anger,
pleasure, love, hate and satisfaction.
Medulla oblongatais the centre for
heartbeat, respiration, digestion, blood
pressure, involuntary functions, and
urination etc.Cerebellumregulates
posture and balance.
Cerebrumis the centre for intelligence,
emotion, will power, memory,
consciousness, imagination, etc.
25Which one of the following
statements is correct?
[CBSE AIPMT 2006]
(a) Neurons regulate endocrine activity,
but notvice versa
(b) Endocrine glands regulate neural
activity and nervous system regulates
endocrine glands
(c) Neither hormones control neural
activity nor the neurons control
endocrine activity
(d) Endocrine glands regulate neural
activity, but notvice versa
Ans.(a)
The autonomous nervous system
regulates the secretion of glands
whereas, the glands do not regulate the
nervous system.
26Which one of the following do not
act as a neurotransmitter?
[CBSE AIPMT 2006]
(a) Acetylcholine (b) Epinephrine
(c) Norepinephrine (d) Cortisone
Ans.(d)
Cortisone does not act as a
neurtransmitter. Cortisone is a
corticosteroid that is itself biologically
inactive and is formed naturally in the
adrenal gland (adrenal cortex) from the
active hormone cortisol. Cortisol
promotes the synthesis and storage of
glucose and is important in the normal
response to stress, suppresses
inflammation and regulates deposition
of fat in body.
Acetylcholineis a neurotransmitter
secreted from the nerve endings.
Epinephrineandnorepinephrineare
secreted from the medulla of adrenal
gland and these also act as
neurotransmitter.
27One of the examples of the action
of the autonomous nervous system
is [CBSE AIPMT 2005]
(a) knee-jerk response
(b) pupillary reflex
(c) swallowing of food
(d) peristalsis of the intestine
Ans.(d)
Peristalsis of the intestine is related with
autonomous nervous system whereas,
knee-jerk response, pupillary reflex and
swallowing of food are related to reflex
action.
28Four healthy people in their
twenties got involved in injuries
resulting in damage and death of a
few cells of the following. Which of
the cells are least likely to be
replaced by new cells?
[CBSE AIPMT 2005]
(a) Osteocytes
(b) Malpighian layer of the skin
(c) Liver cells
(d) Neurons
Ans.(d)
Neuron cells are least likely to be
replaced by new cells. These cells are
specilalised to conduct electrochemical
current.
Nerve cells do not have the capability of
division as they are restricted at
G
0
-phase of the cell cycle.
29In the resting state of the neural
membrane, diffusion due to
concentration gradients, if allowed,
would drive[CBSE AIPMT 2004]
(a)K
+
into the cell
(b)K
+
andNa
+
out of the cell
(c)Na
+
into the cell
(d)Na
+
out of the cell
Ans.(c)
In the resting nerve fibre, the cytoplasm
inside the axon has a high concentration
ofK
+
and a low concentration ofNa
+
in
contrast to the fluid outside the axon.
Thus, if diffusion occurs then through
concentration gradientNa
+
enters the
fibre.
30What used to be described as
Nissl’s granules in a nerve cell are
now identified as
[CBSE AIPMT 2003]
(a) ribosomes (b) mitochondria
(c) cell metabolites (d) fat granules
Ans.(a)
Main cell body of neuron is called as
cyton or soma. It contains a large and
centrally located nucleus, mitochondria,
Golgi bodies, rough endoplasmic
reticulum, lysosomes, fat globules.
Besides these soma also contains Nissl’s
granules or neurofibrils. These are
masses of ribosomes and rough
endoplasmic reticulum and are engaged
in the process of protein synthesis.
31Which cells do not form layer and
remain structurally separate?
[CBSE AIPMT 2001]
(a) Epithelial cells
(b) Muscle cells
(c) Nerve cells
(d) Gland cells
Ans.(c)
Only nerve cells do not form layers, they
also remain structurally separate from
each other (though communicate with
each other through synapse).
Nerve cells or neurons are the cells
specialised to conduct an
electrochemical current, nerve tissue is
made up of these cells and supporting
cells.
32An action potential in the nerve
fibre is produced when positive and
negative charges on the outside
and the inside of the axon
membrane are reversed, because
[CBSE AIPMT 2000]
(a) more potassium ions enter the axon
as compared to sodium ions leaving it
(b) more sodium ions enter the axon as
compared to potassium ions leaving it
(c) all potassium ions leave the axon
(d) all sodium ions enter the axon
Ans.(b)
When a nerve fibre is stimulated, its
membrane becomes more permeable to
sodium ions. Hence, more sodium ions
enter the axon than potassium ions
leaving it. As a result, the positive and
negative charges on the outside and
inside of the membrane are reversed.
The membrane with reversed polarity is
called depolarised.
Neural Control and Coordination 193

204. 33Which cranial nerve has the highest
number of branches?
[CBSE AIPMT 1999]
(a) Facial nerve (b) Trigeminal
(c) Vagus nerve (d) None of these
Ans.(c)
Vagus nerve has five branches
(a) Superior laryngeal nerve
(b) Recurrent laryngeal nerve
(c) Cardiac nerve
(d) Pneumogastric nerve
(e) Depresser nerve
34The Nissl’s granules of nerve cell
are made up of[CBSE AIPMT 1997]
(a) ribosomes (b) protein
(c) DNA (d) RNA
Ans.(a)
Nissl’s granules (or Nissl’s bodies) are
the groups of ribosomes and rough
endoplasmic reticulum. These are
actively involved in the synthesis of
proteins.
35The roof of the cranium of frog is
formed by [CBSE AIPMT 1997]
(a) parasphenoid (b) alisphenoid
(c) frontoparietal (d) orbitosphenoid
Ans.(c)
Dorsal part of the cranium is formed of
two large and flat frontoparietals which
are articulated together by a mid dorsal
sagittal suture together and are collectively
called frontoparietal. Endo-frontoparietal
consists of a frontal bone (in front) and a
parietal bone (behind). But now it has
been proved that it is only the frontal
bone, the parietals are not present in
frog due to the absence of neck.
36The sympathetic nerves, in
mammals arise from
[CBSE AIPMT 1995]
(a) sacral nerves
(b) cervical nerves
(c) thoraco-lumbar nerves
(d) III, VII, IX and X cranial nerves
Ans.(c)
Sympathetic nerves arise from
thoracic and lumbar spinal segments.
37Respiratory centre is situated in
[CBSE AIPMT 1994, 99]
(a) cerebellum
(b) medulla oblongata
(c) hypothalamus
(d) cerebrum
Ans.(b)
Normally, the breathing process
(inspiration and expiration) is controlled
involuntarily by a breathing centre
located in themedulla oblongata.The
ventral portion of the breathing centre
(inspiratory centre) increases the rate
and depth of inspiration while the dorsal
and lateral portions of the centre
(expiratory centre) inhibit inspiration and
stimulate expiration.
38CNS is mostly made of
[CBSE AIPMT 1993]
(a) motor neurons and sensory neurons
(b) sensory neurons and association
neurons
(c) association neurons
(d) motor neurons and association
neurons
Ans.(c)
Central nervous system is mostly made
up of association neurons.
39Ivan Pavlov performed experiments
on
(a) simple reflexes
(b) conditioned reflexes
(c) cardiac reflexes
(d) origin of life
Ans.(b)
Conditional reflexes are those responses
which can be initiated to a stimulus
other than the one which normally
initiates that response.
Conditional reflexes were first
demonstrated by a Russian physiologist I
Pavlov (1929). He conducted‘Bell
experiment on dog’.He rang a bell
every time he offered food to a dog,
finally he noticed that merely ringing bell
can substitute sight or smell of food to
initiate salivation.
40Vagus nerve is
[CBSE AIPMT 1992, 97]
(a) X (b) IX
(c) VII (d) V
Ans.(a)
Vagus nerve is Xth cranial nerve. It is
mixed in nature having both sensory and
motor fibres.
41Third ventricle of brain is also
known as [CBSE AIPMT 1990]
(a) metacoel (b) rhinocoel
(c) paracoel (d) diacoel
Ans.(d)
Inside diencephalon there is a narrow
cavity called 3rd ventricle of brain or
diacoel,which is connected anteriorly
with lateral ventricles (also called
paracoel) of cerebral hemisphere (called
1st and 2nd ventricle) by a common
aperture calledforamen of Monro.
While it is connected posteriorly with 4th
ventricle of medulla oblongata through a
narrow longitudinal canal called
iter/aqueduct of Sylvius.
42Which of the following cranial
nerves can regulate heartbeat?
[CBSE AIPMT 1989]
(a) X (b) IX (c) VIII (d) VII
Ans.(a)
Xth cranial nerve is vagus or
pneumogastric nerve which originates
from lateral side of medulla oblongata
behind IX cranial nerve. It is a mixed
nerve, its sensory fibres innervate to
receptor present in the wall of visceral
organs. Whereas, its motor fibres
innervates to muscles in the wall of
visceral organs-like heart, alimentary
canal, trachea, lungs, kidneys, genital
tracts etc. It also regulates heartbeat.
43Match the following columns and
select the correct option from the
codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. Rods and
cones
1. Absence of
photoreceptor
cells
B. Blind spot 2. Cones are
densely packed
C. Fovea 3. Photoreceptor
cells
D. Iris 4. Visible coloured
portion of the eye
Codes
A B C D
(a) 3 1 2 4
(b) 2 3 1 4
(c) 3 4 2 1
(d) 2 4 3 1
Ans.(a)
The option (a) is correct match which is
as follows. Rods and cones are
photoreceptor cells of eye.
194 NEETChapterwise Topicwise Biology
Sensory Organs
TOPIC 3

205. Blind spot is the area where there is
absence of any photoreceptor cells in
the eye.
Fovea is the area in the eye where cones
are densely packed.
Iris is the visible coloured portion of the
eye.
44Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. Organ of
Corti
1. Connects middle
ear and pharynx
B. Cochlea 2. Coiled part of the
labyrinth
C. Eustachian
tube
3. Attached to the
oval window
D. Stapes 4. Located on the
basilar
membrane
A B C D
(a) 3 1 4 2
(b) 4 2 1 3
(c) 1 2 4 3
(d) 2 3 1 4
Ans.(b)
Option (b) is correct because organ of
Corti is located on the basilar
membrane. The coiled portion of the
labyrinth is called cochlea. The
Eustachian tube connects the middle
ear cavity with the pharynx. The middle
ear contains ossicle called stapes that is
attached to the oval window of the
cochlea.
45Which of the following receptors
are specifically responsible for
maintenance of the balance of
body and posture?
[NEET (Odisha) 2019]
(a) Basilar membrane and otoliths
(b) Hair cells and organ of Corti
(c) Tectorial membrane and macula
(d) Crista ampullaris and macula
Ans.(d)
The inner ear contains crista ampullaris
and macula as the specific receptors of
the vestibular apparatus responsible for
maintenance of balance of the body and
posture.
46Which of the following statements
is correct?[NEET (National) 2019]
(a) Cornea consists of dense connective
tissue of elastin and can repair itself
(b) Cornea is convex, transparent layer
which is highly vascularised
(c) Cornea consists of dense matrix of
collagen and is the most sensitive
portion of the eye
(d) Cornea is an external, transparent
and protective proteinaceous
covering of the eyeball
Ans.(c)
The statement that cornea consists of
dense matrix of collagen and is the most
sensitive portion of the eye is correct.
Rest statements are incorrect. The
correct information about the
statements is as follows
The outer layer of the wall of eyeball,
sclera, consists of a dense connective
tissue containing mainly collagen and
some elastic fibre. Cornea is convex,
transparent layer which is
non-vascularised. The cornea is the clear
part of eye’s protective covering.
47The transparent lens in the human
eye is held in its place by
[NEET 2018]
(a) smooth muscles attached to the iris
(b) ligaments attached to the iris
(c) ligaments attached to the ciliary body
(d) smooth muscles attached to the
ciliary body
Ans.(c)
The lens in the human eye is held in place
by thesuspensory ligamentsattached to
the ciliary body. The function of other
components are as follows
The smooth muscles attached to the
ciliary body helps to control the shape of
lens.
Smooth muscles of iris help in regulating
the diameter of pupil.
Pactinate ligament attached to iris is
involved in the drainage of aqueous
humor because it contains spaces
between the fibres.
48Photosensitive compound in
human eye is made up of
[NEET 2016, Phase I]
(a) opsin and retinal
(b) opsin and retinol
(c) transducin and retinene
(d) guanosine and retinol
Ans.(a)
Photosensitive pigment rhodopsin in
human eye is made up of opsin protein
and retinal [aldehyde form of vitamin-A
(retinol)]. These pigments are present in
the rod cells of retina layer of eye.
49Choose the correct statement.
[NEET 2016, Phase II]
(a) Nociceptors respond to changes in
pressure
(b) Meissner’s corpuscles are
thermoreceptors
(c) Photoreceptors in the human eye are
depolarised during darkness and
become hyperpolarised in response
to the light stimulus
(d) Receptors do not produce graded
potentials
Ans.(c)
The photosensitive compounds
(Rhodopsin) in the human eye is
composed of opsin (a protein) and retinal
(an aldehyde of vitamin–A, i.e. retinol).
It is present in the rod cells
(Photoreceptors). Light induces
dissociation of retinol, from opsin thus
changing the structure of opsin. This
creates potential differences in the
photoreceptors and they become
hyperpolarised.
However, during darkness rhodopsin is
resynthesised from opsin and retinine to
restore the dark vision and
photoreceptors are depolarised.
The correct form of other statements
are
(a) Nociceptors are sensory nerve
cells that respond to potentially
damaging chemical or mechanical
stimuli and send them to brain and
spinal cord.
(b) Meissner’s receptors are tactile
receptors receiving the stimuli of
pressure.
(d) Receptors always produce graded
potentials.
50In mammalian eye, the ‘fovea’ is the
center of the visual field, where
[CBSE AIPMT 2015]
(a) high density of cones occur, but has
no rods
(b) the optic nerve leaves the eye
(c) only rods are present
(d) more rods than cones are found
Ans.(a)
At the posterior pole of the eye lateral to
the blind spot, there is a yellowish
pigmented spot called macula lutea with
a central pit called the fovea. It is a
thinned-out portion of the retina where
only the cones are densely packed. It is
the point where the visual acuity
(resolution) is the highest.
Neural Control and Coordination 195

206. 51Which one of the following
statements is not correct?
[CBSE AIPMT 2014]
(a) Retinal is the light absorbing portion
of visual photopigments
(b) In retina the rods have the
photopigment rhodopsin, while cones
have three different photopigments
(c) Retinal is a derivative of vitamin-C
(d) Rhodopsin is the purplish red protein
present in rods only
Ans.(c)
Retinal pigment of epithelium shields the
retina from excess incoming light. It
supplies omega-3 fatty acid and glucose
to the retina. The former is used for
building photoreceptive by membranes
the latter used for energy retinal is
supplied by the visual vitamin-A cycle.
52PartsA, B, CandDof the human
eyes are shown in the diagram.
Select the option, which gives
correct identification along with its
functions/characteristics
[NEET 2013]
(a)A–Retina–contains
photoreceptors–rods and cones
(b)B–Blind spot–has only a few rods and
cones
(c)C–Aqueous chamber–reflects the
light, which does not pass through
the lens
(d)D–Choroidits anterior part forms
ciliary body
Ans.(a)
A–Retina—Contains photoreceptors -
rods and cones. The daylight vision is
function of cones and twilight vision is
related to rods.
B–Blind spot—Photoreceptor cells are
not present in this part.
C–Aqueous chamber contains a thin
watery fluid called aqueous humour.
D–Sclera is the external layer of eye
having dense connective tissue.
53Which part of the human ear plays
no role in hearing as such but is
otherwise very much required?
[CBSE AIPMT 2012]
(a) Eustachian tube
(b) Organ of Corti
(c) Vestibular apparatus
(d) Ear ossicles
Ans.(c)
The inner ear contains a complex system
called vestibular apparatus which is
located above the cochlea. It has no role
in hearing but is influenced by gravity
and movements. Its specific receptors
calledcristaandmaculaare responsible
for maintenance of balance of the body
and posture.
54The purplish red pigment rhodopsin
contained in the rods type of
photoreceptor cells of the human
eyes is a derivative of
[CBSE AIPMT 2011]
(a) vitamin-C (b) vitamin-D
(c) vitamin-A (d) vitamin-B
Ans.(c)
There are two types of photoreceptor
cells of retina, namely rods and cones.
The rods contain a purplish red protein
called therhodopsin(visual purple),
which is a derivative of vitamin-A.
55Cornea transplant in human is
almost never rejected. This is
because [CBSE AIPMT 2008]
(a) its cells are least penetrable by
bacteria
(b) it has no blood supply
(c) it is composed of enucleated cells
(d) it is a non-living layer
Ans.(b)
Cornea is a transparent portion that
forms the anterior one-sixth of the
eyeball. The cornea admits and helps to
focus light waves as they enter the eye.
It is avascular, i.e., has no blood supply
therefore, cornea transplant in human is
almost never rejected.
56Given below is a diagrammatic
cross section of a single loop of
human cochlea.[CBSE AIPMT 2008]
Which one of the following
options correctly represents the
names of three different parts?
(a)B—Tectorial membrane,
C—Perilymph,D— Secretory cells
(b)C—Endolymph,D—Sensory hair cells,
A— Serum
(c)D—Sensory hair cells,A—Endolymph,
B— Tectorial membrane
(d)A—Perilymph,B—Tectorial
membrane,C— Endolymph
Ans.(d)
Cochlea is the main hearing organ, which
is connected with saccule by short
ductus reunions leading from the
saccule. It consists of three fluid filled
chambers or canals the upper scala
vestibuli, lower scala tympani and middle
scala media. Both scala vestibuli and
scala tympani are filled with perilymph
(A) while scala media is filled with
endolymph.(C) The tectorial membrane
(B) overhangs the sensory hair in scala
media.
57Which one of the following is the
correct difference between rod
cells and cone cells of our retina?
[CBSE AIPMT 2008]
Rod cellsCone cells
(a) Visual
acuity
High Low
(b) Visual
pigment
contained
Iodopsin Rhodopsin
(c) Overall
function
Vision in
poor light
Colour vision
and detailed
vision in bright
light
(d) Distribution More
concentrate
d in centre
of retina
Evenly
distributed all
over retina
Ans.(a)
The rods contain the rhodopsin (visual
purple) pigment and enable the animals
to see in darkness. Therefore, present in
large number in nocturnal animals. The
cones contain the iodopsin (visual violet)
pigment and chiefly concerned with
distinction in colour and light vision
during day time.
196 NEETChapterwise Topicwise Biology
A B
C
D
A
Perilymph
Tectorial
membrane
Endolymph
Basilar
membrane
Organ of
Corti
(A)
(B)
(C)
(D)
Cochlea
A
B
C
D
Lens
Iris

207. Neural Control and Coordination 197
58Bowman’s glands are located in the
[CBSE AIPMT 2007]
(a) proximal end of uriniferous tubules
(b) anterior pituitary
(c) female reproductive system of
cockroach
(d) olfactory epithelium of our nose
Ans.(d)
Many olfactory glands (Bowman’s glands)
occur below the olfactory epithelium
that secrete mucus over the epithelium
to keep it moist.
59What is the intensity of sound in
normal conversation?
[CBSE AIPMT 2001]
(a) 10-20 dB (b) 35-60 dB
(c) 70-90 dB (d) 120-150 dB
Ans.(b)
The intensity of sound in normal
conversation is around 35-60 dB. The
word noise is orginated from the Latin
word nausea and is defined as unwanted
or unpleasant sound that causes
discomfort.
Intensity of some noise sources is as
follows
Source Intensity (dB)
Breathing 10
Broadcasting studio 20
Trickling cl’ock 30
Library 30-35
Normal conversation 35-60
Telephone 60
Office noise 60-80
Alarm clock 70-80
Traffic 50-90
Motor cycle 105
Jet fly (over 1000') 100-110
Train whistle (50') 110
Air craft (100') 110-120
Commercial jet air
craft (100')
120-140
Space rocket
(launching)
170-180
60Characteristic feature of human
cornea is that[CBSE AIPMT 2001]
(a) it is secreted by conjunctiva and
glandular tissue
(b) it is lacrimal gland which secretes tears
(c) blood circulation is absent in cornea
(d) in old age it become hard and white
layer deposits on it which causes the
cataract
Ans.(c)
Human eye is about 1 inch in diameter
and is covered and protected by the
sclera, which is made up of tough
connective tissue. The front of eye is
transparent thus allows the light to enter
the eye. This portion of the eye’s outer
layer is calledcornea.It lacks a blood
supply.
It derives nutrientsviaaqueous humour
from cell body. Cornea not only allows
light to enter the eye but also bend it as
well. This makes it a characteristic
feature of human eye, i.e. cornea.
61When we migrate from dark to
light, we fail to see for some time
but after a time visibility becomes
normal. It is an example of
[CBSE AIPMT 2001]
(a) accommodation
(b) adaptation
(c) mutation
(d) photoperiodism
Ans.(b)
It takes some time for rhodopsin to split
into scotopsin and retinal (bleaching) and
release of transmitter passing nerve
impulseviabipolar and ganglion cells to
the optic nerves. This is a case of
adaptation. It differs from
accommodation which is a reflex
mechanism by which the focus of the
eye change to make the images of
distant and near objects sharp on the
retina.
62In the chemistry of vision in
mammals, the photosensitive
substance is called
[CBSE AIPMT 1997]
(a) sclerotin (b) retinal
(c) rhodopsin (d) melanin
Ans.(c)
Rods of the retina contain the
light-sensitive pigment rhodopsin which
is formed by a combination of the
protein molecule called scotopsin and a
small light absorbing molecule called
retinene (retinal).
63In frog, ‘fenestra ovalis’ is
[CBSE AIPMT 1997]
(a) the opening in the auditory capsule
which separates the middle ear from
internal ear
(b) the air-filled cavity of the middle ear
(c) the communication between the
pharynx and the tympanic cavity
(d) the external opening of the tympanic
cavity which is covered by the
tympanic membrane
Ans.(a)
‘Fenestra ovalis’ is an oval aperture
through which tympanic cavity is
connected with auditory capsule which
houses the internal ear.
64Cornea transplantation is
outstandingly successful because
[CBSE AIPMT 1996]
(a) cornea is easy to preserve
(b) cornea is not linked up with blood
vascular and immune systems
(c) the technique involved is very simple
(d) cornea is easily available
Ans.(b)
Cornea is non-vascular, i.e. no blood
supply so, its transplantation is
outstandingly successful.
65Retina is most sensitive at
[CBSE AIPMT 1993]
(a) optic disc
(b) periphery
(c) macula lutea
(d) fovea centralis
Ans.(d)
Retina is most sensitive at fovea
centralis. It is with a depression,fovea
centralisin its middle-N it is the area of
most distinct day vision.Yellow spot or
macula lutea or area centralis is a small
area on retina which lies opposite to
optical axis of the lens. Only cones are
present in this area so, it most sensitive
to day light vision.
66Function of iris is to
[CBSE AIPMT 1993]
(a) move lens forward and backward
(b) refract light rays
(c) bring about movements of eyelids
(d) alter the size of pupil
Ans.(d)
Iris controls the amount of light that
reaches the photosensors at the back of
the eye. It consists of circular sphincters
and radial dilators. The function of iris is
to alter the size of pupil.
67Light rays entering the eye are
controlled by[CBSE AIPMT 1993]
(a) pupil
(b) iris
(c) cornea
(d) lens

208. 198 NEETChapterwise Topicwise Biology
Ans.(a)
Pupil is the black hole in the centre of
the iris. It is the area through which
light enters the eyeball, i.e. pupil is the
aperture that controls the light
entering into the eye.
68Iris is part of[CBSE AIPMT 1992]
(a) sclerotic
(b) choroid/uvula
(c) choroid and retina
(d) sclerotic and choroid
Ans.(c)
Iris consists of two layers, outer one lies
in continuation with choroid, while inner
one is in continuation with retina.
69Sensitive pigmented layer of eye is
[CBSE AIPMT 1989]
(a) cornea
(b) retina
(c) sclerotic
(d) iris
Ans.(b)
Retina consists of a pigmented layer and
a nervous tissue layer, first there is the
photoreceptor layer containing
photosensitive cells, the rods and cones.
Rod cells are sensitive towards light and
are used for vision in dim light, having no
ability to detect colour, whereas cones
are used for bright light vision with the
ability to make coloured images of the
object.
Next is the intermediate layer containing
short sensory bipolar neurons. Bipolar
cells inturn synapse with the retinal
ganglion cells, whose axons bundle
together as the optic nerve.
70Acute vision is present in
[CBSE AIPMT 1988]
(a) vulture (b) shark
(c) bat (d) frog
Ans.(a)
Acute vission is found in birds like
vulture.
71Snow blindness in Antarctic region
is due to [NEET (Sep.) 2020]
(a) inflammation of cornea due to high
dose of UV-B radiation
(b) high reflection of light from snow
(c) damage to retina caused by infrared
rays
(d) freezing of fluids in the eye by low
temperature
Ans.(a)
Snow blindness in Antarctic region is due
to inflammation of cornea due to high
dose of UV-B radiation. It is a painful,
temporary loss of vision due to
overexposure to the sun’s UV rays. It also
called ’photo keratitis’(photo = light,
keratitis= inflammation of the cornea).
72In a man, abducens nerve is
injured. Which one of the following
functions will be affected?
[CBSE AIPMT 2005]
(a) Movement of the eye ball
(b) Swallowing
(c) Movement of the tongue
(d) Movement of the neck
Ans.(a)
Abducens (abducent) nerve is a cranial
nerve which originates from the ventral
surface of medulla oblongata. It
innervates the lateral rectus muscle of
eyeball. It is a motor nerve and controls
the movements of the eyeball. Hence, if
abducens nerve is injured in a man,
movement of eyeball will be affected.
73Parkinson’s disease (characterised
by tremors and progressive rigidity
of limbs) is caused by degeneration
of brain neurons that are involved
in movement control and make use
of neurotransmitter
[CBSE AIPMT 2005]
(a) acetylcholine (b) norepinephrine
(c) dopamine (d) GABA
Ans.(c)
Abnormal release of neurotransmitter
dopamine leads to Parkinson's disease in
humans. It is caused by degeneration of
brain neuron that are involved in control
movement.
74Injury to vagus nerve in human is
not likely to affect
[CBSE AIPMT 2004]
(a) tongue movements
(b) gastrointestinal movements
(c) pancreatic secretion
(d) cardiac movements
Ans.(a)
Vagus nerve is a mixed cranial nerve
which is, controlling much of the gut,
ventilatory system and heart. It does not
affect tongue movements. Tongue
movement is controlled by
glossopharyngeal nerve.
75A person suffering from the
deficiency of the visual pigment
rhodopsin is advised to take more
[CBSE AIPMT 2000]
(a) radish and potato
(b) apple and grapes
(c) carrot and ripe papaya
(d) guava and ripe banana
Ans.(c)
Carrot and ripe papaya contain carotene
from which vitamin-A is synthesised.
Vitamin-A is necessary for the formation
of rhodopsin.
The visual pigments in vertebrate eyes
are located in the tips of specialised
sensory cells calledrod cellandcone
cells.Rod cells contain rhodopsin and
are responsible for black and white
vision.
Disorders of the Nervous
System and Sensory
Organs/System
TOPIC 4

209. 01Match the following columns and
select the correct option from the
codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. Pituitary
hormone
1. Steroid
B. Epinephrine 2. Neuropeptides
C. Endorphins 3. Peptides,
proteins
D. Cortisol 4. Biogenic
amines
A B C D
(a) 4 1 2 3
(b) 3 4 2 1
(c) 4 3 1 2
(d) 3 4 1 2
Ans.(b)
Option (b) is correct match which is as
follows
Pituitary hormones are chemically
peptides and proteins.
Epinephrine is a biogenic amine.
Endorphins are neuropeptides.
Cortisol is a steroid hormone.
02Hormones stored and released
from neurohypophysis are
[NEET (Oct.) 2020]
(a) thyroid stimulating hormone and
oxytocin
(b) oxytocin and vasopressin
(c) follicle stimulating hormone and
leutinizing hormone
(d) prolactin and vasopressin
Ans.(b)
Neurohypophysis (pars nervosa) is also
known as posterior pituitary which
stores and releases two hormones
called oxytocin and vasopressin, which
are actually synthesised by the
hypothalamus and are transported
axonally to neurohypophysis.
03GnRH, a hypothalamic hormone,
needed in reproduction, acts on
[NEET 2017]
(a) anterior pituitary gland and stimulates
secretion of LH and oxytocin
(b) anterior pituitary gland and
stimulates secretion of LH and FSH
(c) posterior pituitary gland and stimulates
secretion of oxytocin and FSH
(d) posterior pituitary gland and
stimulates secretion of LH and relaxin
Ans.(b)
GnRH is a hypothalamic hormone. It
stimulates the anterior lobe of pituitary
gland to secrete LH and FSH.
04Hypersecretion of growth hormone
in adults does not cause further
increase in height because
[NEET 2017]
(a) growth hormone becomes inactive in
adults
(b) epiphyseal plates close after
adolescence
(c) bones loose their sensitivity to growth
hormone in adults
(d) muscle fibres do not grow in size after
birth
Ans.(b)
Chronic hypersecretion of Growth
Hormone (GH) leads to gigantism or
acromegally depending on the age of the
individual.
If its hypersecretion occurs before the
ossification of epiphyseal plates, it
causes exaggerated and prolonged
growth in long bones. It results in
gigantism.
In adults, hypersecretion of GH leads to
accromegaly. No increase in height
occurs because of the ossified
epiphyseal plate.
05The amino acid, tryptophan is the
precursor for the synthesis of
[NEET 2016, Phase I]
(a) thyroxine and tri-iodothyronine
(b) oestrogen and progesterone
(c) cortisol and cortisone
(d) melatonin and serotonin
Ans.(d)
Melatonin and serotonin are derivatives
of tryptophan amino acid while thyroxine
and tri-iodothyronine are iodinated
tyrosine amino acid derivatives.
06Which of the following pairs of
hormones are not antagonistic
(having opposite effects) to each
other? [NEET 2016, Phase I]
(a) Insulin Glucagon
(b) Aldosterone Atrial Natriuretic
Factor
(c) Relaxin Inhibin
(d) Parathormone Calcitonin
ChemicalCoordination
andIntegration
22
Hormones and Their
Chemical Nature
TOPIC 1

210. Ans.(c)
Relaxin hormone which is secreted by
posterior pituitary gland relaxes the
pubic symphysis during parturition while
inhibin decreases the secretion of FSH
from anterior pituitary.
07Name a peptide hormone which
acts mainly on hepatocytes,
adipocytes and enhances cellular
glucose uptake and utilisation.
[NEET 2016, Phase II]
(a) Insulin (b) Glucagon
(c) Secretin (d) Gastrin
Ans.(a)
Insulin is the peptide hormone which
enhances the uptake of glucose molecules
by liver cells (hepatocytes) and fat cells
(adipocytes) for its cellular utilisation.
Such an activity of insulin brings down
the level of glucose in the blood.
08Which one of the following
hormones is not involved in sugar
metabolism? [CBSE AIPMT 2015]
(a) Cortisone (b) Aldosterone
(c) Insulin (d) Glucagon
Ans.(b)
Aldosterone is not involved in sugar
metabolism. It is a steroid hormone
(mineralocorticoid) produced by the
outer section (zona glomerulosa) of the
adrenal cortex in the adrenal gland. It
plays a central role in the regulation of
blood pressure mainly by acting on the
distal tubules and collecting ducts of the
nephron, increasing reabsorption of ions
and water in the kidney, to cause the
conservation of sodium, secretion of
potassium, increase in water retention
and decrease in blood pressure and
blood volume.
09Which one of the following
hormones though synthesised
elsewhere, is stored and released
by the master gland?
[CBSE AIPMT 2015]
(a) Antidiuretic hormone
(b) Luteinising hormone
(c) Prolactin
(d) Melanocyte stimulating hormone
Ans.(a)
Antidiuretic Hormone (ADH) or
vasopressin is a peptide hormone
synthesised in the hypothalamus, but
stored and released from the posterior
pituitary lobe.
10Identify the hormone with its
correct matching of source and
function. [CBSE AIPMT 2014]
(a) Oxytocin – Posterior pituitary, growth
and maintenance of mammary glands
(b) Melatonin – Pineal gland, regulates the
normal rhythm of sleepwake cycle
(c) Progesterone – Corpus luteum,
stimulation of growth and activities of
female secondary sex organs
(d) Atrial natriuretic factor – Ventricular
wall increases the blood pressure
Ans.(b)
Melatoninis a hormone present in
animals, plants and microbes. In animals
melatonin allows the regulation of
cicarcadian rhythms.
Oxytocinis a neurohypophysial hormone
which stimulates the muscle contraction
(smooth muscle) in the wall of uterus
during childbrith.
Progesteroneis a female hormone
produced by the corpus luteum after
ovulation.
This hormone maintain the wall of uterus
throughout the pregnancy.
ANF stimulates the secretion of Na and
H O
2
by the kidneys and helps in
regulating blood pressure.
11A person entering an empty room
suddenly finds a snake right in
front on opening the door. Which
one of the following is likely to
happen in his neurohormonal
control system?[CBSE AIPMT 2012]
(a) Sympathetic nervous system is
activated releasing epinephrine and
norepinephrine from adrenal medulla
(b) Neurotransmitters diffuse rapidly
across the cleft and transmit a nerve
impulse
(c) Hypothalamus activates the
parasympathetic division of brain
(d) Sympathetic nervous system is
activated releasing epinephrine and
norepinephrine from adrenal cortex
Ans.(a)
Epinephrine and nor-epinephrine are
secreted by adrenal medulla (under the
control of sympathetic nervous system)
in response to stress of any kind or
during emergency situations.
These are also calledemergency
hormones. Thus, they would be released
when the person enter an empty room
and suddenly finds a snake.
12Which one of the following pairs of
hormones are the examples of
those that can easily pass through
the cell membrane of the target
cell and bind to a receptor inside it
(mostly in the nucleus)?
[CBSE AIPMT 2012]
(a) Insulin and glucagon
(b) Thyroxin and insulin
(c) Somatostatin and oxytocin
(d) Cortisol and testosterone
Ans.(d)
Cortisol and testosterone are lipid
soluble hormones, which can directly
pass through the cell membrane of the
target cell and bind with interacellular
receptors.
13What is correct to say about the
hormone action in humans?
[CBSE AIPMT 2012]
(a) Glucagon is secreted byβ-cells of
islets of Langerhans and stimulates
glycogenolysis
(b) Secretion of thymosine is stimulated
with ageing
(c) In females, FSH first binds with
specific receptors on ovarian cell
membrane
(d) FSH stimulates the secretion of
oestrogen and progesterone
Ans.(c)
FSH hormone is one of the
gonadotropins secreted by anterior lobe
of pituitary. It is a proteinaceous
hormone, so binds with extra cellular or
membrane bound receptors.
14Which one of the following pairs is
incorrectly matched?
[CBSE AIPMT 2010]
(a) Glucagon — Beta cells (source)
(b) Somatostatin — Delta cells (source)
(c) Corpus luteum— Relaxin (secretion)
(d) Insulin —Diabetes mellitus (disease)
Ans.(a)
In pancreatic islets, alpha or A–cells
constitute about 15% of pancreatic islets
cells and secrete glucagon. Its molecule
consists of a single polypeptide chain of
29 amino acid residues. Glucagon
intensifies glycogenolysis, deamination
and gluconeogenesis and inhibits
glycogenesis in liver cells. It also
intensifies lipolysis in adipose tissue.
Thus, it is a promoter of catabolic
metabolism.
200 NEETChapterwise Topicwise Biology

211. 15Foetal ejection reflex in human
female is induced by
[CBSE AIPMT 2009]
(a) pressure exerted by amniotic fluid
(b) release of oxytocin from pituitary
(c) fully developed foetus and placenta
(d) differentiation of mammary glands
Ans.(b)
Oxytocin (child birth hormone) secreted
by neurohypophysis of pituitary gland
stimulates contraction of uterus
muscles. It stimulates labour pain for
child birth. When secretion of
progesterone hormone declines it will
result in making the end of pregnancy.
As the sensory impulse of increasing
labour pain reaches hypothalamus,
more and more oxytocin is released
from posterior pituitary under a
positive feedback regulation, it dilates
the cervix (vaginal stretching).
16The blood calcium level is lowered
by the deficiency of
[CBSE AIPMT 2008, 1999]
(a) parathormone (b) thyroxine
(c) calcitonin (d) Both (a) and (c)
Ans.(a)
The chief cells of the parathyroid
secrete parathormone. Its deficiency
causes the lowering of blood calcium
level. This increases the excitability of
nerves and muscles causing cramps and
convulsions. This causes parathyroid
tetany characterised by sustained
contractions of the muscles of larynx,
face, hands and feet.
Calcitoninis secreted when calcium
level is high in blood. It has an opposite
action to that of parathyroid hormone
and lowers the calcium level by
suppressing release of calcium ions
from the bones.
Thyroxineis secreted from the thyroid
gland. It regulates the metabolic rate of
the body and thus, maintain basal
metabolic rate, stimulate protein
synthesis and therefore, promote
growth of the body tissues.
17In human adult females, oxytocin
[CBSE AIPMT 2008]
(a) is secreted by anterior pituitary
(b) stimulates growth of mammary
glands
(c) stimulates pituitary to secrete
vasopressin
(d) causes strong uterine contractions
during parturition
Ans.(d)
Oxytocin hormone is secreted from the
posterior lobe of pituitary gland.
It promotes contraction of uterine
muscle during parturition and
contraction of myoepithelial cells of
lactating breast, sequeezing milk into
the large ducts behind the nipple.
Because of its role oxytocin is called
birth hormone.
18Feeling the tremors of an
earthquake a scared resident of
seventh floor of a multistoryed
building starts climbing down the
stairs rapidly. Which hormone
initiated this action?
[CBSE AIPMT 2007]
(a) Thyroxin
(b) Adrenaline
(c) Glucagon
(d) Gastrin
Ans.(b)
Adrenaline hormone is responsible for
this action, as adrenaline hormone is
known as 3F hormone, i.e. fright, flight
and fight.
19In the human female, menstruation
can be deferred by the
administration of
[CBSE AIPMT 2007]
(a) LH only
(b) combination of FSH and LH
(c) combination of estrogen and
progesterone
(d) FSH only
Ans.(c)
When the production of progesterone
and estrogen hormone stops in blood
this leads to shedding of the lining of
uterine endometrium. Therefore, by
supply of oestrogen and progesterone
the menstruation can be deferred.
20Compared to a bull a bullock is
docile because of
[CBSE AIPMT 2007]
(a) higher levels of thyroxin
(b) higher levels of cortisone
(c) lower levels of blood testosterone
(d) lower levels of
adrenaline/noradrenaline in its blood
Ans.(c)
Testes are degenerated due to which
testosterone level in blood is reduced.
This hormone promotes the growth of
many body tissues such as muscles.
21Withdrawl of which of the following
hormones is the immediate cause
of menstruation?
[CBSE AIPMT 2006]
(a) Oestrogen (b) FSH
(c) FSH-RH (d) Progesterone
Ans.(d)
Menstruation is caused by the reduction
of oestrogen and progesterone.
Hormone level, (especially
progesterone) at the end of monthly
ovarian cycle. The first effect is
decreased stimulation of the
endometrial cells by these two
hormones followed rapidly by involution
of the endometrium itself to about 65%
of its previous thickness.
22Which hormone causes dilation of
blood vessels, increased oxygen
consumption and glucogenesis?
[CBSE AIPMT 2006]
(a) ACTH (b) Insulin
(c) Adrenaline (d) Glucagon
Ans.(c)
Adrenaline (epinephrine) is a hormone
produced by adrenal medulla and is
secreted in great amounts during
emotional states. It elevates the glucose
level in blood stream (by glucogenesis)
which is accompanied by increase in
oxygen consumption, body temperature,
heat production. Adrenaline also causes
an increase in the flow of blood by
dilating the blood vessels.
Insulinregulates the glucose level in
blood.
ACTH(Adreno Corticotropic Hormone) is
secreted by anterior pituitary and
stimulates the adrenal cortex.
Glucagonis a polypeptide hormone
secreted by the alpha cells of islets of
Langerhans of pancreas. It also acts to
promote glycogenolysis.
23Sertoli cells are regulated by the
pituitary hormone known as
[CBSE AIPMT 2006]
(a) FSH (b) GH
(c) prolactin (d) LH
Ans.(a)
Sertoli cells are the cells that line the
seminiferous tubules in the testis. These
cells protect the spermatids and convey
nutrients to both the developing and
mature spermatozoa. Sertoli cells are
regulated byFSH(Follicle Stimulating
Hormone) as the FSH receptors are
confined to the Sertoli cells.
Chemical Coordination and Integration 201

212. FSH stimulates Sertoli cells to
produce androgen-binding protein
and inhibin and together with
testosterone, promotes the
proliferation of Sertoli cells.
24A steroid hormone which regulates
glucose metabolism is
[CBSE AIPMT 2006]
(a) cortisol
(b) corticosterone
(c) 11-deoxycorticosterone
(d) cortisone
Ans.(a)
Cortisol(a steroid hormone) is the
principal glucocorticoid hormone of
many mammals including humans
(corticosterone is more abundant in
some small mammals). It regulates the
glucose metabolism and promotes
gluconeogenesis, especially during
starvation and raises blood pressure.
Cortisoneis an inactive form of cortisol.
25Chemically hormones are
[CBSE AIPMT 2004]
(a) biogenic amines only
(b) proteins, steroids and biogenic
amines
(c) proteins only
(d) steroids only
Ans.(b)
Chemically hormones are of different
nature like protein hormones
(hypothalmic hormones), steroids (sex
hormones) and biogenic amines (like
thyroxin hormone).
26Which of the following hormones is
not a secretion product of human
placenta? [CBSE AIPMT 2004]
(a) Human chorionic gonadotropin
(b) Prolactin
(c) Oestrogen
(d) Progesterone
Ans.(b)
Prolactin is secreted by anterior pituitary
gland (not human placenta) which
stimulates mammary gland development
during pregnancy and lactation after
child birth.
Placenta is a connection between the
uterine wall of mother and the foetus. It
helps in exchange of material between
these two. Placenta secretes human
chorionic gonadotropin, oestrogen and
progesterone.
27Which one of the following
hormones is a modified amino
acid? [CBSE AIPMT 2004]
(a) Epinephrine
(b) Progesterone
(c) Prostaglandin
(d) Oestrogen
Ans.(a)
Epinephrine is synthesised from amino
acid tyrosine. While oestrogen and
progesterone are modified steroids and
prostaglandins are basically fat.
28Mainly which type of hormones
control the menstrual cycle in
human beings?[CBSE AIPMT 2002]
(a) FSH
(b) LH
(c) FSH, LH, estrogen
(d) Progesterone
Ans.(c)
Follicle Stimulating Hormone (FSH),
Luteinising Hormone (LH) and estrogen
all play an important role in controlling
the menstrual cycle in human females.
29When both ovaries are removed
from rat which hormone is
decreased in blood?
[CBSE AIPMT 2002]
(a) Oxytocin
(b) Prolactin
(c) Estrogen
(d) Gonadotropic releasing factor
Ans.(c)
If both the ovaries are removed from rat
then the blood plasma level of oestrogen
will be affected as it is produced by
theca interna cells of Graafian follicles.
Oestrogen regulates growth and
development of female accessary
reproductive organs, secondary sexual
characters and sexual behaviour.
30Melanin protects from
[CBSE AIPMT 2002]
(a) UV-rays
(b) visible rays
(c) infra-red rays
(d) X-rays
Ans.(a)
Melanin is a protective pigment
synthesised fromtyrosine. Melanocytes
under the influence of melanocyte
secreting hormone secrete melanin
which protects the body from harmful
effects of UV rays.
31Adrenaline directly affects
[CBSE AIPMT 2002]
(a) SA node
(b)β-cells of Langerhans
(c) dorsal root of spinal cord
(d) epithelial cells of stomach
Ans.(a)
The hormone adrenaline (epinephrine) is
secreted by adrenal medulla and directly
affects SA node thereby increasing heart
rate. This hormone is responsible for the
alarming reactions. It also increases
breathing and blood glucose level.
32Which steroid is used for
transformation?[CBSE AIPMT 2002]
(a) Cortisol (b) Cholesterol
(c) Testosterone (d) Progesterone
Ans.(b)
Cholesterol forms a major component of
animal cell membranes. Liposomes
(artificially created spheres surrounded
by a phospholipid bilayer like a
membrane) are used for transformation
(transgenics).
33Secretion of progesterone by
corpus luteum is initiated by
[CBSE AIPMT 1999]
(a) thyroxine (b) LH
(c) MSH (d) testosterone
Ans.(b)
LH (Luteinising Hormone), secreted by
anterior pituitary, stimulates the corpus
luteum to secrete the hormone
progesterone.
34Hormones thyroxine, adrenaline
and the pigment melanin are
formed from [CBSE AIPMT 1997]
(a) tryptophan (b) glycine
(c) tyrosine (d) proline
Ans.(c)
Tyrosine is the precursor of : (a)
hormone epinephrine (i.e. adrenaline)
and thyroid hormones, (b)
neurotransmitter dopamine, (c) melanin
(the black pigment of skin).
35Which one of the following
hormones stimulates the ‘let down’
(release) of milk from the mother’s
breasts when the baby is sucking?
[CBSE AIPMT 1995]
(a) Progesterone (b) Oxytocin
(c) Prolactin (d) Relaxin
202 NEETChapterwise Topicwise Biology

213. Ans.(b)
Oxytocin induces contraction of the
mammary gland muscles, helps in the
flow of milk from mammary glands to
mouth of the child, hence, called ‘milk
ejection hormone’.
36ADH or vasopressin is
[CBSE AIPMT 1991]
(a) enzyme that hydrolyses peptides
(b) hormone secreted by pituitary that
promotes reabsorption of water from
glomerular filtrate
(c) hormone that promotes
glycogenolysis
(d) energy rich compound connected
with muscle contraction
Ans.(b)
ADH or vasopressin is synthesised in
hypothalamus and stored and released
by neurohypophysis or posterior lobe of
pituitary gland. It controls the
permeability of wall of collecting tubules
and DCT of renal tubules to water, which
stimulates reabsorption of water so, it
controls the osmoregulation.
37Insulin is [CBSE AIPMT 1990]
(a) vitamin (b) lipid
(c) hormone (d) enzyme
Ans.(c)
Insulin is earliest known hormone. It is
also calledhypoglycemicor
antidiabetic factor, as it decreases
glucose level in blood and prevents
occurrence of diabetes. It is secreted
byβ-cells of the islets of Langerhans.
38Addition of a trace of thyroxine or
iodine in water containing tadpoles
will [CBSE AIPMT 1990]
(a) keep them in larval stage
(b) hasten their metamorphosis
(c) slow down their metamorphosis
(d) kill the tadpoles
Ans.(b)
Thyroxine controls tissue differentiation
and metamorphosis of tadpole larva into
frog.Gundernatch(1912) proved that
metamorphosis of tadpole into adult
frog is controlled by thyroxine. Addition
of thyroxine in water will hasten the
metamorphosis.
39Which hormone possesses anti
-insulin effect?[CBSE AIPMT 1988]
(a) Cortisol (b) Calcitonin
(c) Oxytocin (d) Aldosterone
Ans.(a)
Cortisol is a glucocorticoid, secreted by
adrenal cortex. It is primarily meant for
carbohydrate metabolism, which
increases the rate of gluconeogenesis
(conversion of proteins in liver into
sugars) and decreases peripheral
utilisation of glucose, thus it possess
anti-insulin effect.
40Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. Pituitary
gland
1. Grave’s disease
B. Thyroid
gland
2. Diabetes
mellitus
C. Adrenal
gland
3. Diabetes
insipidus
D. Pancreas 4. Addison’s
disease
A B C D
(a) 3 2 1 4
(b) 3 1 4 2
(c) 2 1 4 3
(d) 4 3 1 2
Ans.(b)
The correct option is (b). It can be
explained as follows
Grave’s disease occurs is due to excess
secretion of thyroid hormones (T3 and T4).
Diabetes mellitus is due to
hyposecretion of insulin fromβ-cells of
pancreas.
Diabetes insipidus is due to
hyposecretion of ADH from posterior
pituitary. Addison’s disease is due to
hyposecretion of hormone from adrenal
cortex.
41Select the correct statement.
[NEET (Sep.) 2020]
(a) Glucagon is associated with
hypoglycemia
(b) Insulin acts on pancreatic cells and
adipocytes
(c) Insulin is associated with
hyperglycemia
(d) Glucocorticoids stimulate
gluconeogenesis
Ans.(d)
Statement in option (d) is correct. Rest
statements can be corrected as
Glucagon is associated with
hyperglycemia. Insulin acts on
hepatocytes and adipocytes and is
associated with hypoglycemia.
42Which of the following conditions
will stimulate parathyroid gland to
release parathyroid hormone?
[NEET (Odisha) 2019]
(a) Fall in active vitamin-D levels
(b) Fall in bloodCa
+2
levels
(c) Fall in boneCa
+2
levels
(d) Rise in bloodCa
+2
levels
Ans.(b)
The fall in blood Ca
+2
levels is the
condition which will stimulate
parathyroid gland to release parathyroid
hormone. This hormone exerts its effects
on bones and kidneys. When calcium
levels are low, parathyroid hormone is
released by the parathyroid glands into
the blood and causes the bones to
release calcium and increase levels in
the bloodstream.
43Which of the following glucose
transporters is insulin-dependent?
[NEET (National) 2019]
(a) GLUT II (b) GLUT III
(c) GLUT IV (d) GLUT I
Ans.(c)
GLUT IV is an insulin dependent glucose
transporter. It helps in the transport of
glucose into muscles and adipose cells
under anabolic conditions. It permits the
facilitated diffusion of circulating
glucose down its concentration gradient
into muscle and fat cells. On the other
hand, GLUT I, II and III are insulin
independent glucose transporters.
44Artificial light, extended work-time
and reduced sleep-time disrupt the
activity of[NEET (National) 2019]
(a) thymus gland (b) pineal gland
(c) adrenal gland
(d) posterior pituitary gland
Ans.(b)
Artificial light, extended work time and
reduced sleep time disrupt the activity
of pineal gland. It is a small
pea-shaped gland in the brain. It
produces melatonin, which helps
maintain circadian rhythm and
regulate reproductive hormones.
Chemical Coordination and Integration 203
Human Endocrine
System: Major Glands
TOPIC 2

214. 45Which of the following hormones
can play a significant role in
osteoporosis? [NEET 2018]
(a) Estrogen and parathyroid hormone
(b) Progesterone and aldosterone
(c) Aldosterone and prolactin
(d) Parathyroid hormone and prolactin
Ans.(a)
Estrogen and parathyroid hormone can
play significant role in osteoporosis. It is
caused due to the deficiency of estrogen
and excessive activity of parathormones.
Estrogen helps to promote the activity
of osteoblast (helps in the formation of
bone cells) and inhibits osteoclast
(destruct the bones). On the other hand,
parathormone promotes the
mobilisation of calcium from bones into
blood hence causes demineralisation.
The other listed hormones also contribute
to osteoporosis but their effects are
insignificant or very less.e.g.,low level
of progesterone and aldosterone causes
bone loss whereas raised level of prolactin
have been linked with osteoporosis.
46Which of the following structures
or regions isincorrectlypaired with
its function? [NEET 2018]
(a) HypothalamusProduction of
releasing hormones
and regulation of
temperature, hunger
and thirst.
(b) Limbic systemConsists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(c)Medulla
oblongata
Controls respiration
and cardiovascular
reflexes.
(d)Corpus
callosum
Band of fibres
connecting left and
right cerebral
hemispheres.
Ans.(b)
Limbic systemconsists of four major
components namely hippocampus,
amygdala, septal nuclei and mammilary
bodies. It controls the emotional
behaviour, food habits and sex behaviour
of an organism. It is not involved in
controlling movements.
The rest three options are correctly
paired with their functions.
47The posterior pituitary gland is not
a ‘true’ endocrine gland because
[NEET 2016, Phase II]
(a) it is provided with a duct
(b) it only stores and releases hormones
(c) it is under the regulation of
hypothalamus
(d) it secretes enzymes
Ans.(b)
The posterior pituitary gland is not a
‘true’ endocrine gland because it only
stores and releases two hormones–
oxytocin and vasopressin, which are
actually synthesised by the hypothalamus
and are transported to the posterior
pituitary through hypophysial portal
system. Hence option (b) is correct.
48Fight or flight reactions cause
activation of[CBSE AIPMT 2014]
(a) the parathyroid glands, leading to
increased metabolic rate
(b) the kidney, leading to suppression of
reninangiotensin-aldosterone
pathway
(c) the adrenal medulla, leading to
increased secretion of epinephrine
and norepinephrine
(d) the pancreas leading to a reduction in
the blood sugar levels
Ans.(c)
Fight or flight reaction is a physiological
reaction that occurs in response to
emergency. Emergency hormone is
secreted byadrenal medullaof adrenal
gland. Adrenal medulla secretes
epinephrine and nor-epinephrine
commonly called catecholamines
hormones which stimulates sweating,
heartbeat and breathing rate.
These hormones also causes dilation of
coronary artery (supplying blood to heart
muscles) a bronchioles (for increasing
inspiratory volume) and pupil (for better
vision).
49Which of the following statement is
correct in relation to the endocrine
system? [NEET 2013]
(a) Adenohypophysis is under direct
neural regulation of the
hypothalamus
(b) Organs in the body like
gastrointestinal tract, heart, kidney
and liver do not produce any
hormones
(c) Non-nutrient chemicals produced by
the body in trace amount that act as
intercellular messenger are known as
hormones
(d) Releasing and inhibitory hormones
are produced by the pituitary gland
Ans.(c)
Hormones are non-nutrient chemicals,
which act as intracellular messengers and
are produced in trace amounts.
Endocrine cells are present in different
parts of the gastrointestinal tract,
e.g. gastrin, secretin, GIP.
Atrial wall of our heart secretes a
peptide hormone called ANF (Atrial
Natriuretic Factor), RH/IH are produced
by hypothalamus. Adenohypophysis is
not directly under neural control, it is
under the control of hypothalamic
hormones, brought by portal system.
50Select the answer which correctly
matches the endocrine gland with
the hormone it secretes and its
function/deficiency symptom
[NEET 2013]
Endocrine
gland
Hormone
Function/
deficiency
symptoms
(a) Anterior
pituitary
Oxytocin Stimulates
uterus
contraction
during child
birth
(b)Posterior
pituitary
Growth
Hormone
(GH)
Oversecretion
stimulates
abnormal
growth
(c) Thyroid
gland
ThyroxineLack of iodine
in diet results
in goitre
(d) Corpus
luteum
Testoster
one
Stimulates
spermatogene
sis
Ans.(c)
Lack of iodine in diet results in goitre.
Oxytocin is produced by
neurohypophysis, which stimulates
uterus contraction during child birth.
Anterior pituitary secretes Growth
Hormones (GH) its over secretion
stimulates abnormal growth.
Testosterone is secreted by Leydig cells
of testes in males.
51Match the source gland with its
respective hormone as well as the
function. [CBSE AIPMT 2011]
Source
gland
Hormone Function
(a)Posterior
pituitary
Vasopressin Stimulates
resorption of
water in the
distal tubules
in the nephron
204 NEETChapterwise Topicwise Biology

215. (b)Corpus
luteum
Oestrogen Supports
pregnancy
(c)Thyroid Thyroxine Regulates
blood calcium
level
(d)Anterior
pituitary
Oxytocin Contraction of
uterus muscles
during child
birth
Ans.(a)
The pituitary gland is located in a bony
cavity calledsella tursicaand is
attached to hypothalamus by a stalk. It is
divided anatomically into an
adenohypophysis and a neurohypophysis.
The neurohypophysis latter is also called
pars nervosa or posterior pituitary.
It stores and releases two hormone
calledoxytocinandvasopressin, which
are actually synthesised by the
hypothalamus and are transported
axonally to neurohypophysis.
Vasopressin acts mainly at the kidney
and stimulates reabsorption of water
and electrolytes by the distal tubules in
the nephron and thereby reducing the
loss of water through urine (diuresis).
Hence, it is also called as Anti-Diuretic
Hormone (ADH).
52Given ahead is an incomplete table
about certain hormones, their
source glands and one major effect
of each on the body in humans.
Identify the correct option for the
three blanks A, B and C.
[CBSE AIPMT 2011]
Gland Secretion Effect on body
A Oestrogen Maintenance of
secondary sexual
characters
Alpha cells
of islets of
Langerhans
B Raises blood
sugar level
Anterior
pituitary
C Over secretion
leads to
gigantism
A B C
(a)Placenta Insulin Vasopressin
(b)Ovary Insulin Calcitonin
(c)Placenta Glucagon Calcitonin
(d)Ovary Glucagon Growth
hormone
Ans.(d)
Hypersecretion of growth hormone
(somatotropic hormone) during adulthood
causes acromegaly. It is characterised by
bossing of frontal bone, prominent cheek,
coarse hair, hirsutism, enlarged hands
and feet.
53Which one of the following pairs of
organs includes only the endocrine
glands? [CBSE AIPMT 2008]
(a) Parathyroid and adrenal
(b) Pancreas and parathyroid
(c) Thymus and testes
(d) Adrenal and ovary
Ans.(a)
The endocrine glands have no ducts
and their secretions get absorbed into
the immediate surrounding blood
circulation to reach the specific
organs to initiate a particular
metabolic change, e.g. thyroid,
parathyroid, adrenal, thymus.
54A person is having problems with
calcium and phosphorus
metabolism in his body. Which one
of the following glands may not be
functioning properly?
[CBSE AIPMT 2007]
(a) Parathyroid (b) Parotid
(c) Pancreas (d) Thyroid
Ans.(a)
The parathormone secreted by
parathyroid gland regulates the calcium
and phosphate balance between the
blood and other tissues.
55Which of the following is an
accumulation and release centre of
neurohormones?
[CBSE AIPMT 2006]
(a) Posterior pituitary lobe
(b) Intermediate lobe of the pituitary
(c) Hypothalamus
(d) Anterior pituitary lobe
Ans.(c)
Almost all hormonal secretion by the
pituitary gland are controlled by
hormonal signal from hypothalamus.
The neurohormones are secreted and
accumulated by hypothalamus.
56Melatonin is secreted by
[CBSE AIPMT 2000]
(a) skin (b) thymus
(c) pituitary (d) pineal gland
Ans.(d)
The pineal gland (epiphysis) secretes
the hormone melatonin. It regulates
the working of gonads by inhibiting
gonadotropins and their effects. Its
production is inhibited by exposure of
animal to light.
57Melanocyte Stimulating Hormone
(MSH) is produced by
[CBSE AIPMT 2000]
(a) anterior pituitary
(b) posterior pituitary
(c) pars intermedia of pituitary
(d) parathyroid
Ans.(c)
The intermediate lobe (pars
intermedium) of the pituitary gland
secretes MSH which causes dispersal of
pigment granules in the pigment cells
(melanocytes) which give colour to the
skin.
In the frogs and tadpoles, MSH is
responsible for the generalised
blackening of the skin.
58A common scent-producing gland
among mammals is
[CBSE AIPMT 2000]
(a) anal gland
(b) prostate gland
(c) adrenal gland
(d) Bartholin’s gland
Ans.(a)
Scent or musk glands are located
around the face, feet or anus of many
mammals. These secrete pheromones
which may be involved in defence,
recognition and territorial behaviour.
59The gonadotropic hormones are
produced by [CBSE AIPMT 1999]
(a) interstitial cells of testes
(b) adrenal cortex
(c) adenohypophysis of pituitary
(d) posterior part of thyroid
Ans.(c)
The gonadotropic hormones (FSH and
ICSH=LH) are secreted by anterior lobe
of the pituitary (adenohypophysis).
Gonadotropic Hormone (GTH) controls
the function of gonads (ovary in
females and testes in males).
60The hormone which regulates the
basal metabolism in our body is
secreted from[CBSE AIPMT 1998]
(a) pituitary (b) thyroid
(c) adrenal cortex (d) pancreas
Ans.(b)
Thyroxine(T )
4
and tri-iodothyronin(T )
3
hormones are secreted by the thyroid
follicular cells. These hormones
maintain the Basal Metabolic Rate (BMR)
of the body.
Chemical Coordination and Integration 205

216. 61Occurrence of Leydig’s cells and
their secretion is
[CBSE AIPMT 1998, 93, 91]
(a) ovary and oestrogen
(b) liver and cholesterol
(c) pancreas and glucagon
(d) testis and testosterone
Ans.(d)
The endocrine part of testis is formed of
groups of cells, called interstitial cells or
Leydig’s cells,scattered in connective
tissue between the sperm producing
seminiferous tubules of the testis.
These cells are stimulated to produce
male sex hormones, called androgens by
ICSH of anterior pituitary. Testosterone
is main androgen and it is a steroid
hormone.
62Which of the following endocrine
gland stores its secretion in the
extracellular space before
discharging into the blood?
[CBSE AIPMT 1995]
(a) Pancreas
(b) Adrenal
(c) Testis
(d) Thyroid
Ans.(d)
Thyroid gland secretes thyroxine.
Thyroxine hormones are stored in the
lumen of the follicle, the extracellular
space.
63MSH of pars intermedia of middle
pituitary is responsible for
[CBSE AIPMT 1988]
(a) darkening of skin in lower vertebrates
(b) light colouration of skin in lower
vertebrates
(c) Both (a) and (b)
(d) darkening of skin in human beings
Ans.(a)
MSH controls the synthesis and
dispersal of melanin granules in the
chromatophores in the skin of fishes,
amphibians and some reptiles
(poikilothermal animals) so, it causes
darkening of skin. It is non-functional in
man.
64Erythropoietin hormone which
stimulates RBC formation is
produced by [NEET 2021]
(a) alpha cells of pancreas
(b) the cells of rostral adenohypophysis
(c) the cells of bone marrow
(d) juxtaglomerular cells of the kidney
Ans.(d)
The juxtaglomerular cells of kidney
produce peptide hormone called
erythropoietin which stimulates
(erythropoiesis). It stimulates the stem
cells of the bone marrow to increase red
blood cell production.
65Match the following hormones with
the respective disease.
[NEET (National) 2019]
A. Insulin (i) Addison's disease
B. Thyroxin (ii) Diabetes insipidus
C. Corticoids (iii) Acromegaly
D. Growth
hormone
(iv) Goitre
(v) Diabetes mellitus
Select the correct option.
A B C D
(a) (ii) (iv) (iii) (i)
(b) (v) (iv) (i) (iii)
(c) (ii) (iv) (i) (iii)
(d) (v) (i) (ii) (iii)
Ans.(b)
(A)–(v), (B)–(iv), (C)–(i), (D)–(iii)
The deficiency of insulin causes
diabetes mellitus which is characterised
by high blood sugar levels. Excess
production of thyroid hormone,
thyroxine causes goitre, deficiency of
corticoids causes Addison’s disease or
adrenal insufficiency. Hypersecretion of
growth hormones causes acromegaly in
adults. It is characterised by enlarged
hands and feet of affected person.
66Graves’ disease is caused due to
[NEET 2016, Phase II]
(a) hyposecretion of thyroid gland
(b) hypersecretion of thyroid gland
(c) hyposecretion of adrenal gland
(d) hypersecretion of adrenal gland
Ans.(b)
Graves’ disease or hyperthyroidism is
caused due to the secretion of excessive
amount of thyroid hormone by the
thyroid gland, i.e. hyperthyroidism.
67A pregnant female delivers a baby,
who suffers from stunted growth,
mental retardation low intelligence
quotient and abnormal skin. This is
the result of [NEET 2013]
(a) deficiency of iodine in diet
(b) low secretion of growth hormone
(c) cancer of the thyroid gland
(d) over secretion of pars distalis
Ans.(a)
Hypothyroidism during pregnancy
causes defective development and
maturation of the growing foetus leading
to stunted growth. Low secretion of GH
results in stunted growth resulting in
pituitary dwarfism.
Pars distalis or anterior pituitary
produces growth hormone. Prolactin
(PRL), Thyroid Stimulating Hormone
(TSH), Adrenocorticotrophic Hormone
(ACTH), Lutenising Hormone (LH) and
Follicle Stimulating Hormone (FSH).
68Toxic agents present in food which
interfere with thyroxin synthesis
lead to the development of
[CBSE AIPMT 2010]
(a) toxic goitre (b) cretinism
(c) simple goitre (d) thyrotoxicosis
Ans.(c)
Toxic agents in food which interfere with
thyroxin synthesis will lead to simple
goitre. Thyrotoxicosis and toxic goitre
are under the category of
hyperthyroidism.
69A health disorder that results from
the deficiency of thyroxin in adults
and characterised by
[CBSE AIPMT 2009]
I. a low metabolic rate
II. increase in body weight
III. tendency to retain water in
tissues is
(a) hypothyroidism (b) simple goitre
(c) myxoedema (d) cretinism
Ans.(c)
Myxoedema (Gulls disease) occurs due to
the deficiency of thyroxine in adults, it
causes low BMR (by 30-40%), low body
temperature, tendency to retain water in
tissues, reduced heart rate, pulse rate,
blood pressure and cardiac output, low
sugar and iodine level in blood, muscular
weakness and oedema (accumulation of
interstitial fluid that causes the facial
tissues to swell and look fluffy).
Decrease secretion of thyroxine
hormone from thyroid gland causes
hypothyroidism.Simple goitreis caused
by lower intake of iodine through diet.
Cretinismoccurs due to hyposecretion
of thyroxine during the growth years or
birth.
206 NEETChapterwise Topicwise Biology
Hormones and
Related Disorders
TOPIC 3

217. 70Which one of the following pairs
correctly matches a hormone with
a disease resulting from its
deficiency?[CBSE AIPMT 2004]
(a) Luteinising hormone−Failure of
ovulation
(b) Insulin−Diabetes insipidus
(c) Thyroxine−Tetany
(d) Parathyroid hormone−Diabetes
mellitus
Ans.(a)
Ovulation occurs under the influence of
LH and FSH of anterior pituitary gland.
Disease Deficiency
Diabetes mellitus Insulin
Tetany Parathormone
Diabetes insipidus ADH
71Acromegaly is caused by
[CBSE AIPMT 2002]
(a) excess of STH
(b) excess of thyroxin
(c) deficiency of thyroxin
(d) excess of adrenaline
Ans.(a)
Hypersecretion of growth hormone
(somatotropic hormone) during
adulthood causes acromegaly. It is
characterised by bossing of frontal
bone, prominent cheek, coarse hair,
hirsutism, enlarged hands and feet.
72Diabetes is due to
[CBSE AIPMT 1999]
(a) iodine deficiency
(b) hormonal deficiency
(c)Na
+
deficiency
(d) enzyme deficiency
Ans.(b)
Diabetes mellitus occurs due to the
deficiency of hormone insulin which is
secreted byislets of Langerhansof
pancreas. It is a peptide hormone.
73Which of the following radioactive
isotope is used in the detection of
thyroid cancer?
[CBSE AIPMT 1995, 2002]
(a) Iodine-131 (b) Carbon-14
(c) Uranium-238 (d) Phosphorus-32
Ans.(a)
Radioactive iodine, i.e. Iodine-131 is
administered to patients suffering from
thyroid cancer for its detection.
74IdentifyA,BandCin the
diagrammatic representation of
the mechanism of hormone action.
Select the correct option from the
following
[NEET (Odisha) 2019]
(a)A– Steroid hormone;
B–Hormone-receptor complex,
C– Protein
(b)A–Protein hormone,B–Receptor;
C–Cyclic AMP
(c)A– Steroid hormone;B-Receptor,
C-Second messenger
(d)A– Protein hormone;B– Cyclic AMP,
C– Homone-receptor complex
Ans.(b)
In the given mechanism of hormone
action, the labels are
A–Protein hormone
B–Receptor
C–Cyclic AMP
Hormones which interact with
membrane bound receptors do not enter
the target cell. These bind with
membrane proteins (receptors) and
generate second messengers which in
turn regulate cellular metabolism, e.g.
FSH on ovarian cell membrane.
75How does steroid hormone
influence the cellular activities?
[NEET (National) 2019]
(a) Binding to DNA and forming a
gene-hormone complex
(b) Activating cyclic AMP located on the
cell membrane
(c) Using aquaporin channels as second
messenger
(d) Changing the permeability of the cell
membrane
Ans.(a)
Steroid hormones are able to bind to
DNA and form a gene-hormone complex.
These hormones can easily cross the
lipid bilayer of the cell and do not require
secondary messengers for the same.
Steroid hormones bind to intracellular
receptors in the nucleus to form
hormone receptor complex which in turn
interact with the genome.
cAMP pathway, secondary messengers
and change in cell membrane
permeability is required in case of
proteinaceous hormones which cannot
pass through lipid bilayer on their own.
76According to the accepted concept
of hormone action, if receptor
molecules are removed from target
organs, then the target organ will
[CBSE AIPMT 1995]
(a) not respond to the hormone
(b) continue to respond to hormone
without any difference
(c) continue to respond to the hormone
but in the opposite way
(d) continue to respond to the hormone
but will require higher concentration
Ans.(a)
Hormones are released in normal blood
circulation but each hormone stimulates
only a specific target organ to initiate a
specific response. It is because of the
presence of specific receptor protein
only in the specific target cell. If these
receptor molecules are removed from
target organs, the organ will not respond
to hormone.
Chemical Coordination and Integration 207
A
B
Response 1
C
Physiological responses
Mechanism of
Hormonal Control
TOPIC 4

218. 01Genera likeSelaginellaandSalvinia
produces two kinds of spores.
Such plants are known as
[NEET 2021]
(a) homosorus (b) heterosorus
(c) homosporous (d) heterosporous
Ans.(d)
Heterosporous plants are the ones that
produce two diverse sorts of spores, i.e.
microspores and the megaspores. The
microspores and megaspores germinate
and give rise to male and female
gametophyte in these plants and are
retained on the parent sporophytes for
variable periods, e.g.Selaginellaand
Salvinia.
02Vegetative propagule inAgaveis
termed as [NEET (Oct.) 2020]
(a) rhizome (b) bulbil
(c) offset (d) eye
Ans.(b)
Vegetative propagule inAgaveis bulbil. It
develops as a small bud on the axial leaf
and after falling on the soil, it develop
into new plants.
03Strobili or cones are found in
[NEET (Sep.) 2020]
(a)Pteris (b) Marchantia
(c)Equisetum (d)Salvinia
Ans.(c)
Strobili or cones are found inEquisetum.
Equisetumspecies have rhizomes that
grow deep below the grounds surface.
Some species have cone-like structures
called strobili that produce and house
the plants spores for reproduction.
04Which of the following statements
is incorrect?[NEET (National) 2019]
(a)Clavicepsis a source of many
alkaloids and LSD
(b) Conidia are produced exogenously
and ascospores endogenously
(c) Yeasts have filamentous bodies with
long thread-like hyphae
(d) Morels and truffles are edible
delicacies
Ans.(c)
The statement ‘‘ yeasts have filamentous
bodies with long thread-like hyphae” is
incorrect. Correct information about the
statement is as follows.
Yeast is a unicellular sac fungus which
lacks filamentous structures or hyphae.
However, they may form short
temporary filamentous structure called
pseudomycelium. Rest statements are
correct.
05Offsets are produced by
[NEET 2018]
(a) parthenocarpy
(b) mitotic divisions
(c) meiotic divisions
(d) parthenogenesis
Ans.(b)
Offsetsare produced bymitotic
divisions. They are one internode long
runners that occur in some aquatic
plants. Breaking of offsets helps in
vegetative propagation. They give rise to
new plants, e.g.Eichhornia.Meiotic
divisionsoccur in only germ cells.
Parthenocarpyis the development of
seedless fruits.Parthenogenesiscan
be defined as the development of an egg
into a complete individual without
fertilisation.
06Which one of the following
statements is not correct?
[NEET 2016, Phase II]
(a) Offspring produced by the asexual
reproduction are called clone
(b) Microscopic, motile asexual
reproductive structures are called
zoospores
(c) In potato, banana and ginger, the
plantlets arise from the internodes
present in the modified stem
(d) Water hyacinth, growing in the
standing water, drains oxygen from
water that leads to the death of
fishes
Ans.(c)
Statement (c) is incorrect as in potato,
banana and ginger new plantlets always
arise from the nodes of the modified
stem.
Internodes are the area between the two
nodes.
Reproduction
inOrganisms
23
Asexual Reproduction
TOPIC 1
Leaves
Adventitious
roots
Offset
Offset of water hyacinth (Eichhornia)

219. 07Which of the following pairs is not
correctly match?
[CBSE AIPMT 2015]
Mode of reproductionExample
(a) Offset Water hyacinth
(b) Rhizome Banana
(c) Binary fissionSargassum
(d) Conidia Penicillium
Ans.(c)
The plant bodySargassumis a diploid
sporophyte. It does not multiply
asexually by means of spores. Instead it
reproduce by vegetative means,
i.e. fragmentation which is the only
known method of vegetative
reproduction in the free floating species
ofSargassum.
08Which one of the following is
correctly matched?
[CBSE AIPMT 2012]
(a) Onion – Bulb
(b) Ginger – Sucker
(c)Chlamydomonas– Conidia
(d) Yeast – Zoospores
Ans.(a)
The correctly matched pair is
onion-bulb. Onion is a simple tunicated
layered bulb while ginger is a straggling
rhizome having uniparous cyme
branching with sympodial axis. While
yeast reproduces by budding and
Chlamydomonasby zoospores.
09Which one of the following is
correctly matched?
[CBSE AIPMT 2012]
(a) Onion – Bulb
(b) Ginger – Sucker
(c)Chlamydomonas– Conidia
(d) Yeast – Zoospores
Ans.(a)
Onion is a simple tunicated layered bulb
while ginger is a straggling rhizome
having uniparous cyme branching with
sympodial axis.
10The ‘Eyes’ of the potato tuber are
[CBSE AIPMT 2011]
(a) flower buds
(b) shoot buds
(c) axillary buds
(d) root buds
Ans.(c)
Tuber is oval or spherical swollen
underground modified stem lacking
adventitious roots.
It possesses a number of spirally
arranged depression called eyes. Each
eye represents node and consists of 1-3
axillary buds in the axils of small scally
leaves.
11Vegetative propagtion inPistia
occurs by [CBSE AIPMT 2010]
(a) stolon (b) offset
(c) runner (d) sucker
Ans.(b)
Vegetative propagation inPistiaoccurs
by offset. Offsets are branches
originated from the mainstem and upper
portion of each branche is curved
bearing a group of leaves while lower
portion bearing roots. Each branch when
separate can grow independently
forming a new plant.
12Vegetative propagation in mint
occurs by [CBSE AIPMT 2009]
(a) offset (b) rhizome
(c) sucker (d) runner
Ans.(c)
Vegetative propagation in mint occurs
by sucker. These develop from the
nodes of underground stem and comes
up obliquely above the ground in the
form of shoots. It also takes place in
Chrysanthemum.
13In which one pair both the plants
can be vegetatively propagated by
leaf pieces?[CBSE AIPMT 2005]
(a)AgaveandKalanchoe
(b)BryophyllumandKalanchoe
(c)AsparagusandBryophyllum
(d)ChrysanthemumandAgave
Ans.(b)
Marginal notches inKalanchoeand
Bryophyllumpossess adventitious buds
in their leaves for vegetative propagation.
14During regeneration modification
of an organ to other organ is known
as [CBSE AIPMT 2001]
(a) morphogenesis
(b) epimorphosis
(c) morphallaxis
(d) acretopmaruy growth
Ans.(b)
Epimorphosis is the replacement of a
lost organ of the body by proliferating
new cells from the surface of the wound
or injured part.
Morphogenesis (Gr.morphe=form and
genesis=origin) is the growth, shaping
and arrangement of body parts
according to genetically predefined
patterns. The extent direction and rate
of morphogenesis depend on genetic
controls and environmental factors.
15A population of genetically identical
individuals, obtained from asexual
reproduction is[CBSE AIPMT 1993]
(a) callus
(b) clone
(c) deme
(d) aggregate
Ans.(b)
Clone refers to the population of
genetically identical individuals obtained
from asexual reproduction or produced
vegetatively from single organism. An
individual member of a clone is called
ramete.
16Vegetative reproduction ofAgave
occurs through[CBSE AIPMT 1991]
(a) rhizome (b) stolon
(c) bulbils (d) sucker
Ans.(c)
Vegetative propagation ofAgaveoccurs
by bulbils. These are condensed axillary
bud capable of giving rise to shoots, i.e.
independent plant.
17Which of the following plant is
monoecious? [NEET 2021]
(a)Carica papaya
(b)Chara
(c)Marchantia polymorpha
(d)Cycas circinalis
Ans.(b)
When male and female reproductive
structures are present on same plant it
is called monoecious condition, whereas
when both reproductive structures are
present on different plants then this
condition is called dioecious.
Reproduction in Organisms 209
Sexual Reproduction
TOPIC 2

220. 210 NEETChapterwise Topicwise Biology
Charais monoecious. It has oogonium
(female organ) and antheridium (male
organ) on the same plant.
Other options are incorrect because
Carica papaya(papayas)Marchantia
polymorphaemelandCycas circinalis
(gymnosperm) are all dioecious plants.
18Which one of the following flowers
only once in its lifetime?
[NEET 2018]
(a) Mango
(b) Jackfruit
(c) Bamboo species
(d) Papaya
Ans.(c)
Bambooplants are perennial,
monocarpic plants. They flower only
once in their lifetime, usually after
50–100 years. They produce large
number of fruits and die.Mango,
JackfruitandPapayaare polycarpic
plants, i.e. they flower repeatedly at
regular intervals every year.
19Which one of the following
generates new genetic
combinations leading to variation?
[NEET 2016, Phase II]
(a) Vegetative reproduction
(b) Parthenogenesis
(c) Sexual reproduction
(d) Nuclear polyembryony
Ans.(c)
The sexual reproduction brings about
variation through genetic
recombinations. First of all genetic
recombination occurs in
non–homologous chromosomes during
crossing over of pachytene stage in
meiotic cell division.
Secondly, the random union of gametes
also contributes in forming new
combinations of characters. Other
options (a), (b) and (d) do not contribute in
the formation of progenies with new
variations.
Vegetative reproductionoccurs
through vegetative parts of organisms
(plants). Parthenogenesis is formation of
new individual without fertilisation (e.g.
frog).
20Match column I with column II and
select the correct option using
the codes given below
[NEET 2016, Phase II]
Column I Column II
A. Pistils fused
together
1. Gametogenesis
B. Formation of
gametes
2. Pistillate
C. Hyphae of higher
ascomycetes
3. Syncarpous
D. Unisexual female
flower
4. Dikaryotic
Codes
A B C D
(a) 4 3 2 1
(b) 2 1 4 3
(c) 1 2 4 3
(d) 3 1 4 2
Ans.(d)
(a) Pistils fused together — Syncarpous
(b) Formation of gametes — Gametogenesis
(c) Hyphae of higher ascomycetes —
Dikaryotic
(d) Unisexual female flower—Pistillate
Concept EnhancerAlways remember,
in Biology the latin word‘‘Syn’’signifies
togetherness or fusion.
21Select the wrong statement.
[NEET 2013]
(a) Isogametes are similar in structure,
function and behaviour
(b) Anisogametes differ either in
structure, function and behaviour
(c) In oomycetes female gamete is
smaller and motile, while male
gamete is larger and non-motile
(d)Chlamydomonasexhibits both
isogamy and anisogamy andFucus
shows oogamy
Ans.(c)
Oomycetesinclude water moulds,
white rusts and downy mildews. In
these, female gamete is larger and
non-motile, whereas, male gamete is
smaller and motile. Isogametes are
found in algae likeUlothrix,
Chlamydomonas, Spirogyra, etc.,
which are similar in structure,
function and behaviour.
Anisogametes are found in
Chlamydomonasin which one gamete
is larger and non-motile and the other
one is motile and smaller.Oogamyis
the fusion of non-motile egg with
motile sperm. The gametes, differ
both morphologically as well as
physiologically. It occurs in
Chlamydomonas,Fucus, Chara, Volvox,
etc.
22Product of sexual reproduction
generally generates[NEET 2013]
(a) longer viability of seeds
(b) prolonged dormancy
(c) new genetic combination leading to
variation
(d) large biomass
Ans.(c)
Sexual reproduction leads to new
genetic combination leading to variation
in new products.
Longer viability of seeds, prolonged
dormancy and large biomass are not
related to sexual reproduction.
23Why is vivipary an undesirable
character for annual crop plants?
[CBSE AIPMT 2005]
(a) It reduces the vigour of the plant
(b) It adversely affects the fertility of the
plant
(c) The seeds exhibit long dormancy
(d) The seeds cannot be stored under
normal conditions for the next
season
Ans.(d)
Vivipary is the condition when seeds
germinate on the plant. It is an
undesirable character for annual crop
plants because germinated seeds
cannot be stored under normal
conditions for the next season.
24In oogamy, fertilisation involves
[CBSE AIPMT 2004]
(a) a small non-motile female gamete
and a large motile male gamete
(b) a large non-motile female gamete
and a small motile male gamete
(c) a large non-motile female gamete
and a small non-motile male gamete
(d) a large motile female gamete and a
small non-motile gamete

221. Ans.(b)
In oogamous type of sexual
reproduction, the female gamete
(ovum/egg) is big, Passive while male
gametes (spermatozoids) are smaller,
active and motile.
25The process of series of changes
from larva to adult after embryonic
development is called
[CBSE AIPMT 1999]
(a) regeneration
(b) metamorphosis
(c) growth
(d) ageing
Ans.(b)
Metamorphosis (meta=change,
morphe=form) is a process by which an
animal under goes a comparatively rapid
change from larval to adult form.
Regeneration is regrowth of the part of
body which has been removed due to the
injury or other causes. Growth is an
increase in dry mass of an organism.
Ageing is progressive deterioration in
activity of cell, tissues, organs, etc.
26A perennial plant differs from
biennial in [CBSE AIPMT 1995]
(a) having underground perennating
structure
(b) having asexual reproductive structures
(c) being tree species
(d) not dying after seasonal production
of flowers
Ans.(d)
Perennialplants have life span of more
than two years, these may be herbs,
shrubs or trees.Biennialsare those plants
which complete their life cycle in two
years. These plants are usually herbs.
27Syngamy means[CBSE AIPMT 1991]
(a) fusion of gametes
(b) fusion of cytoplasms
(c) fusion of two similar spores
(d) fusion of two dissimilar spores
Ans.(a)
Syngamy refers to the fusion of male and
female gametes or compatible gametes.
It is also known as fertilisation.
In seed plants, fertilisation/syngamy
occurs with the help of pollen tube and is
known as siphonogamy.
28New banana plants develop from
[CBSE AIPMT 1990]
(a) rhizome
(b) sucker
(c) stolon
(d) seed
Ans.(b)
New banana plants develop through
sucker, Sucker is the sub-aerial
modification of stem which originates
from the basal and underground portion
of main stem. It grows obliquely upwards
giving rise to leafy shoot or a new plant.
It also occurs in mint,Chrysanthemum,etc.
Reproduction in Organisms 211

222. 01The term used for transfer of
pollen grains from anthers of one
plant to stigma of a different plant
which, during pollination, brings
genetically different types of pollen
grains to stigma is[NEET 2021]
(a) xenogamy
(b) geitonogamy
(c) chasmogamy
(d) cleistogamy
Ans.(a)
Xenogamy is a cross pollination in which
pollen grains are transferred from anther
of one flower to the stigma of another
flower in order to get fertilised. This is
the only type of cross pollination which
during pollination brings genetically
different types of pollen grains to the
stigma, e.g. Sunflower.
Other options can be explained as:
Geitonogamyis a type of self-pollination
in which pollens are transferred from the
anther of one flower to the stigma of
another flower of the same plant, e.g.
Corn.
Chasmogamyis the condition in which
bisexual flowers have exposed anthers
and stigma. Both self-pollination and
cross pollination can occur in these
flowers, e.g.Catharanthus.
Cleistogamyoccurs in flowers that do
not open, their anther and stigma lie
close to each other thus, production of
seeds is a result of autogamy, e.g.
Arachis hypogaea.
02Diadelphous stamens are found in
[NEET 2021]
(a) China rose
(b)Citrus
(c) Pea
(d) China rose andCitrus
Ans.(c)
Diadelphous condition is a condition of
arrangement of filaments and stamen in
a flower, e.g. Pea. In this condition,
filaments of nine different stamens are
connected into one unit and the tenth
posterior stamen remains out of the
bundle as a stand part.
The androecium of pea flower is
diadelphous because the filaments of
the anther are united in two bundles. In
the case of pea, out of ten, nine stamens
form a staminal tube while one is free.
Thus, the correct answer is 'Pea' which
exhibit diadelphous condition.
China roseof Malvaceae family possess
numerous stamens. The filaments of
stamens are united in one group thus
forming a staminal tube around the style.
Such stamens are called monadelphous.
The polyadelphous stamen is seen in
Citrus,these have many small bunches
of the fused stamen.
03A typical angiosperm embryo sac at
maturity is [NEET 2021]
(a) 8-nucleate and 7-celled
(b) 7-nucleate and 8-celled
(c) 7-nucleate and 7-celled
(d) 8-nucleate and 8-celled
Ans.(a)
In dicotyledons, at the time of
fertilisation, the female gametophyte
develops from a single megaspore.
The megaspore will undergo three
successive mitotic divisions to form
eight nucleated embryo sac. Two nuclei
generated in the first mitotic division in
the megaspore will move to opposite
poles. These nuclei divided and redivide
at their ends to form eight nucleated
stage. Thus, each end have four nuclei,
out of which at micropylar end towards,
three nuclei differentiate into two
synergids and one egg cell, while at the
chalazal end, three nuclei differentiate
as antipodal cells.
The remaining two cells, one at
micropylar end and other at chalazal end
migrate the center and fuse.
Hence, a typical angiosperm embryo sac
at maturity is 7 celled and 8 nucleated
structure.
04In some members of which of the
following pairs of families, pollen
grains retain their viability for
months after release?[NEET 2021]
(a) Poaceae; Rosaceae
(b) Poaceae: Leguminosae
(c) Poaceae: Solanaceae
(d) Rosaceae : Leguminosae
Ans.(d)
In some members of Rosaceae,
Leguminosae and Solanaceae pollen
grains maintain viability for a month due
to sporopollenin.
The outer wall or exine of pollen grains
contains sporopollenin. It is one of the
most resistant organic compounds
known. It protects pollen grains from
external factors such as temperature,
acid, alkali, etc. because of
sporopollenin, pollen grains are
preserved as fossils.
SexualReproduction
inFloweringPlants
24
Pre-fertilisation
Structures and Events
TOPIC 1

223. Sexual Reproduction in Flowering Plants 213
05Which of the following is incorrect
for wind pollinated plants?
[NEET (Oct.) 2020]
(a) Well exposed stamens and stigma
(b) Many ovules in each ovary
(c) Flowers are small and not brightly
coloured
(d) Pollen grains are light and non-sticky
Ans.(b)
Option (b) is incorrect because wind
pollinated plants have single ovule in
each ovary. In case of wind pollination or
anemophily, many pollens are produced
because anemophily is highly wasteful
and non-directional process.
These pollens are also light-weighed,
small, dusty and dry. To catch the
wind-borne pollens, stigma is exposed
and hairy. Flowers are small and
inconspicuous, colourless and
nectarless.
06In water hyacinth and water lily,
pollination takes place by
[NEET (Sep.) 2020]
(a) water currents only
(b) wind and water
(c) insects and water
(d) insects or wind
Ans.(d)
In water hyacinth and water lily, the
flowers emerge above the level of water
and are thus pollinated by insects or wind.
Water lily and water hyacinth are in the
water but their stem part which is above
the thalamus is not in water. The pollen
grains are in the upper part of thalamus
so pollination cannot be done by water.
That is why it done by insects or wind.
07The plant part which consists of
two generations, one within the
other [NEET (Sep.) 2020]
I. Pollen grains inside the anther.
II. Germinated pollen grain with two
male gametes.
III. Seed inside the fruit.
IV. Embryo sac inside the ovule.
(a) I, II and III (b) III and IV
(c) I and IV (d) Only I
Ans.(c)
The plant part which consists of two
generations one within the other are
pollen grains inside the anther and
embryo sac inside the ovule.
This can be explained as :
In an ovule or megasporangia( )2nthere
occurs an embryo sac or female
gametophyte which consists of an egg
cell( ).nThis egg cell or female gamete is
the part of next generation which form
zygote after syngamy. Likewise an
anther( )2nis a male reproductive
structure of a plant. As the anther
mature and dehyrate, it form pollen
grainsviamicrosporogenesis. These
pollen grains represents the male
gametophyte comprising of male
gametes( ).nSo in this way there occurs
two generations one with the other.
08What type of pollination takes
place inVallisneria?
[NEET (Odisha) 2019]
(a) Pollination occurs in submerged
condition by water
(b) Flowers emerge above surface of
water and pollination occurs by
insects
(c) Flowers emerge above water surface
and pollen is carried by wind
(d) Male flowers are carried by water
currents to female flowers at the
surface of water
Ans.(d)
Vallisneriais a water pollinated plant. In
Vallisneria, the female flowers reach the
surface of water by the long stalk and
the male flowers or pollen grains are
released on to the surface of water. They
are carried passively by water currents
to female flowers at surface of water.
09In which of the following, both
autogamy and geitonogamy are
prevented?[NEET (Odisha) 2019]
(a) Wheat (b) Papaya
(c) Castor (d) Maize
Ans.(b)
Autogamy and geitonogamy both are
prevented in papaya plant. In papaya,
male and female flowers are present on
different plants that is each plant is
either male or female (dioecy).
10Pollen grains can be stored for
several years in liquid nitrogen
having temperature of[NEET 2018]
(a)− °196 C (b)− °80 C
(c)− °120 C (d)− °160 C
Ans.(a)
Pollen grains can be stored for several
years in liquid nitrogen having a
temperature of− °196 C. Pollen grains can
be later used in plant breeding
programmes.
11Which of the following has proved
helpful in preserving pollen as
fossils? [NEET 2018]
(a) Oil content
(b) Cellulosic intine
(c) Pollenkitt
(d) Sporopollenin
Ans.(d)
Sporopollenin has proved helpful in
preserving pollen as fossils. The
covering of pollen grain, sporoderm is
consist of two layers,viz.,exine and
intine.Exineis made of a highly
resistant fatty substance called
sporpollenin.
It could not be degraded by any enzyme.
It is not affected by high temperature,
strong acid or strong alkali. Thus, it keeps
the pollen grains well-preserved as fossils.
Pollenkittis a yellowish, viscous, sticky
and oily layer that covers exine of some
insect pollinated pollen grains. Intine of
pollen grains is made up of pectin and
cellulose.
12Winged pollen grains are present in
[NEET 2018]
(a) mango (b) Cycas
(c) mustard (d)Pinus
Ans.(d)
Winged pollen grains are present in
Pinus. These wings are spirally arranged
microsporophylls that arise from the
lateral side and help in pollination.
The sperms (pollen grains) ofCycasare
top-shaped. The pollen grains ofmango
are spheroidal, while that ofmustard
are prolate to subspheroidal.
Vacuole
Exine
Intine
Germ pore Generative
cell
Tube
Nucleus
Tube cell
A mature pollen grain of an angiosperm
Microspore
Prothallial cells
Central cell
Wings
Tetrad of
microspores
Single microspore
with air filled wings
Pinus: Pollen grain

224. Concept EnhancerInPinus,the
pollination is anemophilus. Pollen
remains suspended in the air for a long
time due to wings. It appears as yellow
dust. This is popularly called as
phenomenon of ‘sulphur shower’.
13Attractants and rewards are
required for [NEET 2017]
(a) anemophily (b) entomophily
(c) hydrophily (d) cleistogamy
Ans.(b)
Attractant and rewards are required
for entomophily (insect pollination).
Flowers produce specific odour and
nectar to attract the insect for
effective pollination. Entomophilous
flowers are large with bright colours.
14A dioecious flowering plant
prevents both [NEET 2017]
(a) autogamy and xenogamy
(b) autogamy and geitonogamy
(c) geitonogamy and xenogamy
(d) cleistogamy and xenogamy
Ans.(b)
Dioecious flowering plants contain
unisexual flower. In dioecious condition
two types of unisexual flowers occur on
different plants.
Hence, it does not favour autogamy
and geitonogamy because autogamy
takes place in bisexual flowers and
geitonogamy takes place between
different flower of the same plant.
15Functional megaspore in an
angiosperm develops into
[NEET 2017]
(a) ovule (b) endosperm
(c) embryo sac (d) embryo
Ans.(c)
In angiosperms, functional megaspore
develops into an embryo sac. The
functional megaspore is the first cell
of female gametophyte.
16Which one of the following
statements is not true?
[NEET 2016, Phase I]
(a) Exine of pollen grains is made up of
sporopollenin
(b) Pollen grains of many species cause
severe allergies
(c) Stored pollen in liquid nitrogen can be
used in the crop breeding
programmes
(d) Tapetum helps in the dehiscence of
anther
Ans.(d)
Tapetum is the inner layer of
microsporangium (anther) which
provides nourishment to developing
pollen grain after meiotic cell division.
17Proximal end of the filament of
stamen is attached to the
[NEET 2016, Phase I]
(a) connective
(b) placenta
(c) thalamus or petal
(d) anther
Ans.(c)
A typical stamen consist of anther and
filament. The proximal end of filament
is attached to thalamus or petal of the
flower whereas distal end bears
anther.
18Which of the following statements
is not correct?[NEET 2016, Phase I]
(a) Insects that consume pollen or nectar
without bringing about pollination are
called pollen nectar robbers
(b) Pollen germination and pollen tube
growth are regulated by chemical
components of pollen interacting
with those of the pistil
(c) Some reptiles have also been
reported as pollinators in some plant
species
(d) Pollen grains of many species can
germinate on the stigma of a flower,
but only one pollen tube of the same
species grows into the style
Ans.(d)
Pollen grains of different species are
incompatible, so they fail to germinate.
Only the pollen of the same species
germinate and can form pollen tube
which grows and finally dispatches male
gamete to embryo sac.
19In majority of angiosperms
[NEET 2016, Phase II]
(a) egg has a filiform apparatus
(b) there are numerous antipodal cells
(c) reduction division occurs in the
megaspore mother cells
(d) a small central cell is present in the
embryo sac
Ans.(c)
In most of the angiosperms megaspore
mother cell( )2ndivides meiotically to
produce 4 cells. Out of these
3 degenerate and one remains which
forms functional megaspore. This
divides mitotically and forms embryo sac
which contains following structures.
(i) One egg cell with 2 synergids
forming an egg apparatus.
(ii) There are 3 antipodal cells.
(iii) There are two central cells which
are seen as secondary nucleus (2n).
Egg does not have filiform apparatus. It
is the synergids which have special
cellular thickenings at the micropylar tip
which is called filiform apparatus. This
guides pollen tube into the synergids.
20Pollination in water hyacinth and
water lily is brought about by the
agency of [NEET 2016, Phase II]
(a) water
(b) insects or wind
(c) birds
(d) bats
Ans.(b)
Water hyacinth has single spike of 8-15
conspicuous attractive flowers which
attract bees and other insects. These
insects pollinate them. This plant is an
aquatic weed.
Water lily is also an aquatic plant with
large conspicuous coloured flowers.
Most of the species are bee pollinated
except few species where wind
pollination occurs. Insect pollinated
flowers have coloured petals, big size
and they offer fragrance and nectar to
attract insects which bring about
pollination in them.
21The ovule of an angiosperm is
technically equivalent to
[NEET 2016, Phase II]
(a) megasporangium
(b) megasporophyII
(c) megaspore mother cell
(d) megaspore
Ans.(a)
Ovule of an Angiosperm is equivalent to
megasporangium which consists of
2 synergids, 1 egg, 3 antipodal cells and a
secondary nucleus.
Megaspore mother cell( )2ngives rise
to ovule. Megasporophylls are sterile
structures on which ovules may be
present.
214 NEETChapterwise Topicwise Biology
Parietal cell
Megaspore
mother cell
Degenerating
megaspores
Functional
megaspore
Linear tetrad
of megaspores

225. 22Male gametophyte in angiosperms
produces [CBSE AIPMT 2015]
(a) two sperms and a vegetative cell
(b) single sperm and a vegetative cell
(c) single sperm and two vegetative cells
(d) three sperms
Ans.(a)
Pollen grain or male gametophyte in
angiosperms contain one generative
cell. The generative cell further divides
mitotically to form two male gametes
(sperms). So, a mature male
gametophyte in angiosperms contains
two sperms and one vegetative cell.
23Filiform apparatus is characteristic
feature of [CBSE AIPMT 2015]
(a) generative cell (b) nucellar embryo
(c) aleurone cell (d) synergids
Ans.(d)
Filiform apparatus are finger-like
projections present at the micropylar
end of synergids of embryo sac.
24In angiosperms, microsporogenesis
and megasporogenesis
[CBSE AIPMT 2015]
(a) occur in anther
(b) form gametes without further divisions
(c) involve meiosis
(d) occur in ovule
Ans.(c)
Both event microsporogenesis and
megasporogenesis involve the process
of meiosis which results in the formation
of haploid gametes from the microspore
or megaspore mother cells.
25Geitonogamy involves
[CBSE AIPMT 2014, 10, 94]
(a) fertilisation of a flower by the pollen
from, another flower of the same plant
(b) fertilisation of a flower by the pollen
from the same flower
(c) fertilisation of a flower by the pollen
from a flower of another plant in the
same population
(d) fertilisation of a flower by the pollen
from a flower of another plant
belonging to a distant population
Ans.(a)
Geitonogamy is a type of self pollination.
In other word geitonogamy is transfer of
pollen grain from the anther of one
flower to the stigma of another flower of
either same or genetically similar plant.
26Pollen tablets are available in the
market for [CBSE AIPMT 14]
(a)in vitrofertilisation
(b) breeding programmes
(c) supplementing food
(d)ex situconservation
Ans.(c)
Pollen grains are rich in nutrients. They
are taken as tablets and syrups to
improve health. Pollen consumption has
been claimed to enhance the
performance of athletes and race
horses.
27Function of filiform apparatus is to
[CBSE AIPMT 2014, 2008]
(a) recognise the suitable pollen at
stigma
(b) stimulate division of generative cell
(c) produce nectar
(d) guide the entry of pollen tube
Ans.(d)
Filiform apparatus of synergids secrete
some chemotropically active
substances, which direct the pollen tube
towards micropyle of ovule.
28Megasporangium is equivalent to
[NEET 2013]
(a) embryo sac
(b) fruit
(c) nucellus
(d) ovule
Ans.(d)
Megasporangium is equivalent to an
ovule. An ovule generally has a single
embryo sac formed from a megaspore
through reduction division. It is a small
structure attached to the placenta by
means of a stalk called funicle.
Each has one or two protective
envelopes called integuments. Nucellus
is a mass of cells enclosed with in the
integuments. ovule forms Megaspore
Mother Cell (MMC) by meiosis which
further forms megaspore.
Megaspore nucleus forms embryo sac.
Fruits develop from the ovary of flower,
other floral parts degenerate and fall off.
29Which one of the following
statement is correct?[NEET 2013]
(a) Hard outer layer of pollen is called
intine
(b) Sporogenous tissue is haploid
(c) Endothecium produces the
microspores
(d) Tapetum nourishes the developing
pollen
Ans.(d)
Tapetum is the innermost layer of
microsporangium. It nourishes the
pollen grains. The inner wall of pollen
grain is called intine. Endothecium is the
wall around the microsporangium, which
provide protection and help in
dehiscence of anther to release the
pollen. Sporogenous tissue is diploid. It
undergoes meiotic division to form
microspore tetrads.
30Advantage of cleistogamy is
[NEET 2013]
(a) higher genetic variability
(b) more vigorous offspring
(c) no dependence on pollinators
(d) vivipary
Ans.(c)
In cleistogamous flowers, anthers and
stigma lie close to each other. When
anthers dehisces in the flower buds,
pollen grains come in contact with the
stigma to effect pollination. Thus, these
flowers produce assured seed set even
in the absence of pollinators.
The higher genetic variability and more
vigorous offsprings are produced due to
variations obtained by sexual
reproduction. Vivipary relates to the
birth of young babies from mammals.
31Both, autogamy and geitonogamy
are prevented in[CBSE AIPMT 2012]
(a) papaya
(b) cucumber
(c) castor
(d) maize
Ans.(a)
Autogamy involves pollination within the
same flower, whilegeitonogamyinvolves
transfer of pollen grains from the anther
of one flower to the stigma of another
flower of the same plant.
Both the processes are prevented in
papaya because it is a dioecious plant
(i.e. male and female sex organs are born
on separate plants) and it always needs
cross-pollination.
Sexual Reproduction in Flowering Plants 215
Synergids ( )n
Egg cell ( )n
Vacuole
Secondary
nucleus (2 )n
Antipodal
cells ( )n
Egg apparatus
Structure of an ovule

226. 32An organic substance that can
withstand environmental extremes
and cannot be degraded by any
enzyme is[CBSE AIPMT 2012, 1994]
(a) cuticle
(b) sporopollenin
(c) lignin
(d) cellulose
Ans.(b)
Sporopollenin is a fatty substance
present in the exine of pollen grains. It is
resistant to microbial and chemical
decomposition and can withstand the
extreme environmental conditions. Due
to the presence of sporopollenin, pollen
grains are well preserved during
fossilisation.
33Even in absence of pollinating
agents seed-setting is assured in
[CBSE AIPMT 2012]
(a)Commelina (b)Zostera
(c)Salvia (d) fig
Ans.(a)
Commelina benghalensis bears aerial,
chasmogamous(stigma and anthers
exposed to pollinating agents), insect
pollinated flowers and underground
cleistogamousflowers. Cleistogamous
flowers are bisexual flowers which never
open, i.e. always remain closed. In such
flowers the anthers and stigma lie very
close to each other. When anthers dehisce
in the flower buds, pollen grains come in
contact with the stigma of the same
flower, i.e. autogamy occurs. So, these
flowers produce assured seed set, even in
the absence of pollinators.
34Filiform apparatus is a
characteristic feature of
[CBSE AIPMT 2011]
(a) egg (b) synergid
(c) zygote (d) suspensor
Ans.(b)
The synergid cell wall forms a highly
thickened structure called the filiform
apparatus at the micropylar end,
consisting of numerous finger-like
projections into the synergid cytoplasm.
It is believed to play a major role in pollen
tube guidance and reception.
35Wind pollination is common in
[CBSE AIPMT 2011]
(a) lilies
(b) grasses
(c) orchids
(d) legumes
Ans.(b)
Wind pollination (anemophily) is quite
common in grasses. Anemophilous
flowers are small and inconspicuous
with long and versatile stamens.
Pollen grains are dry, powdery, light
and non-sticky, e.g. maize, wheat,
sugarcane, bamboo,Pinusand papaya.
36In which one of the following
pollination is autogamous?
[CBSE AIPMT 2011]
(a) Xenogamy (b) Chasmogamy
(c) Cleistogamy (d) Geitonogamy
Ans.(c)
Bisexual flowers which remains always
closed are calledcleistogamousand
such condition of flowers is called
cleistogamy. In such flowers, the
anthers and stigma lie close to each
other. When anthers dehisces in the
flowers buds, pollen grains come in
contact with stigma to affect pollination.
Thus, cleistogamous flowers are
invariably autogamous.
37Wind pollinated flowers are
[CBSE AIPMT 2010]
(a) small, brightly coloured, producing
large number of pollen grains
(b) small, producing large number of dry
pollen grains
(c) large, producing abundant nectar and
pollen
(d) small, producing nectar and dry pollen
Ans.(b)
Pollination taking place by means of
wind is known as anemophily, e.g.Zea
mays, Cannabis, Pinus, etc.
Anemophilous flowers are small and
produces large number of pollen grains
which are dry, powdery light and non-sticky.
38Which one of the following plants is
monoecious?[CBSE AIPMT 2009]
(a)Marchantia (b)Pinus
(c)Cycas (d) Papaya
Ans.(b)
Pinusis monoecious as it bear male cone
as well as female cone on same tree, (but
on separate branches).
Marchantia, Cycasand papaya are
dioecious plants.
39Which one of the following is
resistant to enzyme action?
[CBSE AIPMT 2008]
(a) Cork (b) Wood fibre
(c) Pollen exine (d) Leaf cuticle
Ans.(c)
Pollen exine is resistant to enzyme
acition. The pollen wall is consisted of
two layers, the outer exine and inner
intine. The exine is chiefly made up of
sporopollenin, which is derived by the
oxidative polymerisation of carotenoids.
Sporopollenin is one of the most
resistant biological materials known.
Exine is thin in beginning but become
very thick with maturity.
40Unisexuality of flowers prevents
[CBSE AIPMT 2008]
(a) autogamy, but not geitonogamy
(b) both geitonogamy and xenogamy
(c) geitonogamy, but not xenogamy
(d) autogamy and geitonogamy
Ans.(a)
Unisexuality of flowers prevents
autogamy, but not geitonogamy.
Autogamy also called self-pollination
involves the transfer of pollen grains of
one flower to the stigma of same flower.
Self-pollination occurs only in bisexual or
hermaphrodite flowers.
Geitonogamy involves the transfer of
pollen grains from a male flower to stigma
of female flower of same plant. Thus,
geitonogamy operates only in monoecious
plants, i.e. plants having male and female
flowers on different places.
Allogamy or xenogamy also known as
cross-pollination involves the transfer of
pollen grains of male flower to the
stigma of genetically different female
flower. It takes placeviavarious
agencies like wind, water, insects, etc.
41Which one of the following pairs of
plant structures has haploid
number of chromosomes?
[CBSE AIPMT 2008]
(a) Megaspore mother cell and antipodal
cells
(b) Egg cell and antipodal cells
(c) Nucellus and antipodal cells
(d) Egg nucleus and secondary nucleus
Ans.(b)
Egg cell and antipodal cells have haploid
number of chromosomes. In
angiosperms a single diploid megaspore
mother cell matures within an ovule.
Through meiosis it gives rise to four
megaspores (haploid). In most plants
only one of these megaspore survive.
This functional megaspore divides three
times by mitosis and produces eight
haploid nuclei enclosed within a 7-celled
embryo sac.
216 NEETChapterwise Topicwise Biology

227. One nucleus is located near the opening
of embryo sac in egg cell.
Two nuclei are located in a single cell in
the middle of embryo sac and are called
polar nuclei, two nuclei are contained in
cells called synergids that flank the egg
cell and other three nuclei are resided in
the cells called antipodals, located at the
end of embryo sac opposite the egg cell.
42Male gametes in angiosperms are
formed by the division of
[CBSE AIPMT 2007]
(a) microspore
(b) generative cell
(c) vegetative cell
(d) microspore mother cell
Ans.(b)
Male gametes in angiosperms are
formed by the division of generative cell.
Before pollination the pollen grain
cytoplasm divides in generative cell and
vegetative cell. The generative cell
divides to form two male gametes.
43Which one of the following is
surrounded by a callose wall?
[CBSE AIPMT 2007]
(a) Microspore mother cell
(b) Male gamete
(c) Egg
(d) Pollen grain
Ans.(a)
The wall of the pollen mother cell
(microspore mother cell) is deposited by
callose (β-1, 3-glucan). Callose plays a
significant role in reproductive biology of
angiosperms. It acts as a temporary wall
to prevent the product of meiosis from
cohesion and fusion and its dissolution
results in the release of free spores.
44The arrangement of the nuclei in a
normal embryo sac in the dicot
plants is [CBSE AIPMT 2006]
(a) 3 +2 + 3
(b) 2 + 3 + 3
(c) 3 + 3 + 2
(d) 2 + 4 + 2
Ans.(a)
In angiosperms (dicots), thePolygonum
type of embryo sac is most common. In
this embryo sac, the arrangement of the
nuclei is3 2 3+ +, i.e. 3 in antipodal cells,
2 as polar nuclei (which later fuse and
form a diploid secondary nucleus) and 3
in egg apparatus (2 in synergids and 1 in
egg cell).
45What would be the number of
chromosomes in the cells of the
aleuron layer in a plant species with
8 chromosomes in its synergids?
[CBSE AIPMT 2006]
(a) 24 (b) 32
(c) 8 (d) 16
Ans.(a)
Synergids are two, short-lived, haploid
cells lying close to the egg in mature
embryo sac of flowering plant ovule.
While the endosperm is a triploid tissue
formed after triple fusion.
Thus, if the synergids have 8
chromosomes, the aleurone layer (part
of endosperm) will have just triple of that
of chromosomes in the synergids, i.e. 24
chromosomes.
46Which one of the following
represents an ovule, where the
embryo sac becomes horse
shoe-shaped and the funiculus and
micropyle are close to each other?
[CBSE AIPMT 2005]
(a) Amphitropous
(b) Circinotropous
(c) Atropous
(d) Anatropous
Ans.(a)
In an amphitropous ovule, the embryo
sac becomes horse-shoe shaped and
the funiculus and micropyle are close to
each other. This type of ovule is found in
Alismaceae, Butomaceae families.
47Anthesis is a phenomenon which
refers to [CBSE AIPMT 2004]
(a) reception of pollen by stigma
(b) formation of pollen
(c) development of anther
(d) opening of flower bud
Ans.(d)
Anthesisis the opening of floral buds.
Reception of pollen by stigma is called
pollination. Formation of pollen is called
microsporogenesis.
48An ovule which becomes curved so
that the nucellus and embryo sac
lie at right angles to the funicle is
[CBSE AIPMT 2004]
(a) hemitropous
(b) campylotropous
(c) anatropous
(d) orthotropous
Ans.(a)
In hemitropous type, the ovule becomes
curved and nucellus and embryo sac lie
at right angles to the funicle, e.g.
Ranunculaceae and Primulaceae.
49In a flowering plant, archesporium
gives rise to[CBSE AIPMT 2003]
(a) only tapetum and sporogenous cells
(b) only the wall of the sporangium
(c) both wall and the sporogenous cells
(d) wall and the tapetum
Ans.(c)
The archesporial cells divide periclinally,
cutting off primary parietal layer
(forming wall later on) towards the outer
side and primary sporogenous cells
towards the inner side. The cells of the
primary parietal layes divide by
successive periclinal and anticlinal
division to form concentric layers of
pollen sac wall.
50Which type of association is found
in between entomophilous flower
and pollinating agent?
[CBSE AIPMT 2002]
(a) Mutualism (b) Commensalism
(c) Cooperation (d) Co-evolution
Ans.(a)
A plant and its pollinator have a
mutualistic relationship. The plant uses
its pollinator to ensure cross-pollination
while pollinator uses the plant as food.
51In angiosperms all the four
microspores of tetrad are covered
by a layer which is formed by
[CBSE AIPMT 2002]
(a) pectocellulose (b) callose
(c) cellulose (d) sporopollenin
Sexual Reproduction in Flowering Plants 217
Chalaza
Funicle
Embryo
sac
Nucellus
Integuments
Raphe
Amphitropous ovule
NucellusEmbryo sac
Micropyle
Funicle
Chalaza

228. Ans.(b)
The deposition of callose starts in pollen
mother cell as it enters meiosis and is
complete by the end of first meiotic
division. By the time tetrad are formed,
the common callose wall dissolves, even
then all the four microspores lie within a
common callose wall.
52What is the direction of micropyle
in anatropous ovule?
[CBSE AIPMT 2002]
(a) Upward (b) Downward
(c) Right (d) Left
Ans.(b)
Body of the anaptropus ovule gets
inverted and micropyle is on lower side
(downward). Further micropyle and
funiculus lie side by side and micropyle is
close to hilum.
53Anemophily type of pollination is
found in [CBSE AIPMT 2001]
(a)Salvia (b) bottle brush
(c)Vallisneria(d) coconut
Ans.(d)
Pollination through air is known as
anemophily e.g. coconut.
InSalvia,the pollination is taken place by
insects (entomophily).
InVallisneria,the pollination occurs
through water (hydrophily).
In bottle brush(Callistemon)the
pollination occurs through birds
(ornithophily).
54Eight nucleate embryo sacs are
[CBSE AIPMT 2000]
(a) always tetrasporic
(b) always monosporic
(c) always bisporic
(d) sometimes monosporic, sometimes
bisporic and sometimes tetrasporic
Ans.(d)
Megaspore is a haploid structure which
divides and gives rise to embryo sac
which is also called as female
gametophyte.
Eight-nucleate embryo sacs may be
monosporic (e.g.Polygonum) or bisporic
(e.g.Allium) or tetrasporic (e.g.Adoxa).
55Flowers showing ornithophily show
few characteristic like
[CBSE AIPMT 1999]
(a) blue flower with nectaries at base of
corolla
(b) red sweet scented flower with
nectaries
(c) bright red flower into thick
inflorescence
(d) white flowers with fragrance
Ans.(b)
Ornithophily is an allogamy performed by
birds, such as long beaked small birds
(sun birds, humming birds), crow, parrot,
bulbul, etc. Ornithophilous flowers are
large and showy cup-shaped with
abundant nectar or edible part, e.g.
Bombax, Agave, etc.
56How many pollen grains will be
formed after meiotic division in 10
microspore mother cells?
[CBSE AIPMT 1996]
(a) 10 (b) 20
(c) 40 (d) 80
Ans.(c)
One microspore mother cell form four
pollen grains after meiotic division, so,
10 microspore mother cells will form 40
pollen grains through dividing
meiotically.
57In an angiosperm, how many
microspore mother cells are
required to produce 100 pollen
grains? [CBSE AIPMT 1995]
(a) 25 (b) 50
(c) 75 (d) 100
Ans.(a)
In angiosperms, eachMicrospore
Mother Cell(MMC) undergoes meiosis to
produce four microspores which develop
into pollen grains. Thus, to produce 100
pollen grains, 25 micropore mother cells
are required.
58Chief pollinators of agricultural
crops are [CBSE AIPMT 1994]
(a) butterflies (b) bees
(c) moths (d) beetles
Ans.(b)
Bees are considered the most common
pollinators which pollinate about 80% of
the total insect pollinated flowers.
59Ovule is straight with funiculus,
embryo sac, chalaza and micropyle
lying on one straight line. It is
[CBSE AIPMT 1993]
(a) orthotropous
(b) anatropous
(c) campylotropous
(d) amphitropous
Ans.(a)
Orthotropous or atropous is the erect
ovule in which the body of ovule lies
straight and upright over the funicle.
Hilum, chalaza and micropyle occur on
one straight line, e.g.
family–Polygonaceae and Piperaceae.
60Number of meiotic divisions
required to produce 200/400 seeds
of pea would be[CBSE AIPMT 1993]
(a) 200/400 (b) 400/800
(c) 300/600 (d) 250/500
Ans.(d)
200 seeds of pea would be produced
from 200 pollen grains and 200 embryo
sacs. 200 pollen grains will be formed by
50 microspore mother cells, while 200
embryo sacs will be formed by 200
megaspore mother cells. Similarly 400
seeds of pea would be produced from
400 pollen grains and 400 embryo sacs.
400 pollen grains will be formed by 100
Microspore Mother Cell (MMC) and 400
embryo sacs fromed by 400 Megaspore
Mother Cell (MMC). Thus, number of
meiotic divisions required to produce
200/400 seeds would be 250/500.
61Meiosis is best observed in dividing
[CBSE AIPMT 1992]
(a) cells of apical meristem
(b) cells of lateral meristem
(c) microspores and anther wall
(d) microsporocytes
Ans.(d)
Microsporocytes or microspore mother
cells are diploid cells formed from
sporogenous cells in the anther. The
pollen/microspore mother cells undergo
meiosis and form tetrads of microspores
or pollen grains. The wall of pollen
mother cell degenerates and pollen or
microspores separate.
62Point out the odd one
[CBSE AIPMT 1991]
(a) nucellus (b) embryo sac
(c) micropyle (d) pollen grain
Ans.(d)
Nucellus is the parenchymatous
nutritive tissue of ovule, micropyle is the
pore present in the integuments at one
end of ovule and embryo sac represents
the female gametophyte. Thus, nucellus,
micropyle and embryo sac are part of
ovule, whereas, pollen grain/microspore
represents the immature male
gametophyte.
218 NEETChapterwise Topicwise Biology

229. 63Pollination occurs in
[CBSE AIPMT 1991]
(a) bryophytes and angiosperms
(b) pteridophytes and angiosperms
(c) angiosperms and gymnosperms
(d) angiosperms and fungi
Ans.(c)
Pollination is the transfer of pollen
grains from anther to stigma of same or
different flower. It is a characteristic
features of higher plants,
i.e. angiosperm and but also found in
few gymnosperms.
64Embryo sac occurs in
[CBSE AIPMT 1991]
(a) embryo
(b) axis part of embryo
(c) ovule
(d) endosperm
Ans.(c)
Embryo sac occurs in ovule. Megaspore
mother cell is developed inside the
nucellus of the ovule and by a meiotic
division it forms four megaspores, out of
which three degenerate. The functional
megaspore divides mitotically to form
embryo sac.
65Female gametophyte of
angiosperms is represented by
[CBSE AIPMT 1990]
(a) ovule
(b) megaspore mother cell
(c) embryo sac
(d) nucellus
Ans.(c)
In angiosperms, female gametophyte is
represented by embryo sac. Embryo sac
is produced by the functional megaspore
formed through meiosis of megaspore
mother cell in ovule.
The common type of embryo sac is
monosporicPolygonumtype, it is
7-celled and 8-nucleate structure
covered by a thin membrane formed of
megaspore wall.
66Male gametophyte of
angiosperms/monocots is
[CBSE AIPMT 1990]
(a) microsporangium
(b) nucellus
(c) microspore
(d) stamen
Ans.(c)
Microspore or pollen grain is an
immature male gametophyte and thus,
represents the first cell of gametophytic
generation in angiosperms. Microspores
are generally rounded, small,
uninucleate haploid cells produced as a
result of meiosis in microspore mother
cell inside the microsporangia. The
process of formation of microspore
inside the pollen sacs of anthers is
known as microsporogenesis.
67Male gametophyte of angiosperms
is shed as [CBSE AIPMT 1988]
(a) four-celled pollen grain
(b) three-celled pollen grain
(c) microspore mother cell
(d) anther
Ans.(b)
In angiosperms, pollen grains are
generally shed from the anther at
2-celled stage, i.e. one generative cell
and one vegetative cell. But, in some
angiosperms, the generative cell divides
prior to the dehiscence of anther and
shed at 3-celled stage, i.e. one
vegetative cell and two male gametes.
Double fertilisation was discovered by
Nawaschin(1898) inFritillariaandLilium.
It was confirmed byGuignard(1899).
68Double fertilisation is[NEET 2018]
(a) fusion of two male gametes with one
egg
(b) fusion of one male gamete with two
polar nuclei
(c) fusion of two male gametes of pollen
tube with two different eggs
(d) syngamy and triple fusion
Ans.(d)
Double fertilisationis the fusion of two
male gametes to two different cells of
the same female gametophyte. It
consists of following two events
(i)SyngamyFusion of the egg
nucleus with one male gamete is
called syngamy.
(ii)Triple fusionIt is the fusion of
second male gamete and central
cell.
Concept EnhancerSyngamy results in
the formation of diploid (2n) zygote.
Triple fusion involves three nuclei, i.e.
one of male gamete and two polar nuclei.
It gives rise to a triploid( )3nendosperm.
69Double fertilisation is exhibited by
[NEET 2017]
(a) gymnosperms (b) algae
(c) fungi (d) angiosperms
Ans.(d)
Double fertilisation is the characteristic
feature of angiosperms. In this type of
fertilisation one male gamete fuses with
an egg and forms zygote. The second
male gamete fuses with diploidsecondary
nucleus of central cell to form a triploid
primary endosperm nucleus.
70Through which cell of the embryo
sac, does the pollen tube enter the
embryo sac [CBSE AIPMT 2005]
(a) egg cell
(b) persistant synergid
(c) degenerated synergid
(d) central cell
Ans.(c)
Egg apparatus is present towards the
micropylar end of an ovule. Egg
apparatus has two lateral synergid cells
and one centrally located egg cell.
During entry of pollen tube within the
ovule synergid cells become disintegrate
and provide path for entry of pollen
tube within the chamber of embryo sac.
Sexual Reproduction in Flowering Plants 219
Pollen grains
Stigma
Pollen tubes
Conducting
tissue
Ovary
Ovule
Antipodals
Pollen tube
Secondary
nucleus(Polar nuclei)
Pollen tube tip
Obturator
Male
gametes
Fertilisation in an angiosperm
through porogamy
Double Fertilisation
TOPIC 2

230. 220 NEETChapterwise Topicwise Biology
71In angiosperms pollen tubes
liberate their male gametes into
the [CBSE AIPMT 2002]
(a) central cell (b) antipodal cell
(c) egg cell (d) synergid
Ans.(d)
The contents of the pollen tube are
discharged in the synergid from where
the first male gamete is transferred to
the egg cell, while the other male
gamete moves to the central cell
through cytoplasmic current.
72Double fertilisation leading to
initiation of endosperm in
angiosperms require
[CBSE AIPMT 2000]
(a) fusion of one polar nucleus and the
second male gamete only
(b) fusion of two polar nuclei and the
second male gamete
(c) fusion of four or more polar nuclei
and the second male gamete only
(d) all of the above kinds of fusion in
different angiosperms
Ans.(b)
In angiosperms the endosperm is a
special tissue which is formed as a result
of triple fusion. In triple fusion the
second male gamete fuses with
secondary nucleus (diploid nucleus
formed by fusion of two polar nuclei) to
form triploid primary endosperm
nucleus.
73Fertilisation involving carrying of
male gametes by pollen tube is
[CBSE AIPMT 1994]
(a) porogamy (b) siphonogamy
(c) chalazogamy (d) syngonogamy
Ans.(b)
In angiosperms, the male gametes are
carried by the pollen tube and such type
of fertilisation or fusion of male and
female gametes or syngamy is called
siphonogamy. It was discovered by
Strasburger(1884).
74Double fertilisation and triple
fusion were discovered by
[CBSE AIPMT 1993, 88]
(a) Hofmeister
(b) Nawaschin and Guignard
(c) Leeuwenhoek
(d) Strasburger
Ans.(b)
Double fertilisation was discovered by
Nawaschin(1898) inFritilariaandLilium.
It was confirmed byGuignard(1899).
Syngamy (one male gamete fuse with
egg cell to form zygote) and triple fusion
(second male gamete fuses nucleus to
with two polar nuclei or secondary
nucleus to form triploid primary
endosperm nucleus occurs
simultaneously in angiosperms and this
is called double fertilisation.
75Double fertilisation is fusion of
[CBSE AIPMT 1991]
(a) two eggs
(b) two eggs and polar nuclei with pollen
nuclei
(c) one male gamete with egg and other
with synergid
(d) one male gamete with egg and other
with secondary nucleus
Ans.(d)
Double fertilisation deals with the fusion
of one male gamete with egg to form
zygote and the other male gamete with
secondary nucleus to form Primary
Endosperm Nucleus (PEN).
76Entry of pollen tube through
micropyle is[CBSE AIPMT 1990]
(a) chalazogamy (b) mesogamy
(c) porogamy (d) pseudogamy
Ans.(c)
Porogamy refers to the most common
method by which the pollen tube enters
the ovule through micropyle, e.g. lily.
When pollen tube enters through
chalaza, it is known as chalazogamy and
when it enters laterally through
integuments, then such process is
known as mesogamy.
77Generative cell was destroyed by
laser but a normal pollen tube was
still formed because
[CBSE AIPMT 1989]
(a) vegetative cell is not damaged
(b) contents of killed generative cell
stimulate pollen growth
(c) laser beam stimulates growth of
pollen tube
(d) the region of emergence of pollen
tube is not harmed
Ans.(a)
Pollen grain or microspore divides
mitotically forming a larger tube or
vegetative cell and a small generative
cell. Since, vegetative cell gives rise to
pollen tube and generative cell divides to
form two male gametes, thus if a
generative cell is destroyed by laser, a
normal pollen tube will still form through
the vegetative cell which is undestroyed.
78Total number of meiotic divisions
required for forming 100
zygotes/100 grains of wheat are
[CBSE AIPMT 1988]
(a) 100 (b) 75
(c) 125 (d) 50
Ans.(c)
Total number of meiotic divisions
required for forming 100 zygote/100
grains of wheat are 125. Zygote is the
fusion product of pollen grains and egg.
100 zygotes require 100 pollen grains and
100 embryo sacs. 100 pollen grains are
produced from 25 microspore mother
cells while 100 embryo sacs are formed
from 100 functional megaspores which
as a result being produced by 100
megaspore mother cells, since three,
out of four megaspores degenerate in
each case.
79In some plants thalamus
contributes to fruit formation. Such
fruits are termed as
[NEET (Oct.) 2020]
(a) false fruits
(b) aggregate fruits
(c) true fruits
(d) parthenocarpic fruit
Ans.(a)
A fruit in which floral parts other than
ovary, like thalamus, base of sepals,
petals, etc., fuse with pericarp and
contribute in fruit formation is called
false fruit or pseudocarp. Apple,
mulberry and strawberry are false fruits.
80Which one of the following
statements regarding
post-fertilisation development in
flowering plants is incorrect?
[NEET (National) 2019]
(a) Zygote develops into embryo
(b) Central cell develops into endosperm
(c) Ovules develop into embryo sac
(d) Ovary develops into fruit
Ans.(c)
The statement that “ovules develop into
embryo sac’’ is incorrect. Correct
information about the statement is as
follows. During post-fertilisation event,
ovule develops into seed.
Post-fertilisation:
Structures and Events
TOPIC 3

231. On the other hand, embryo sac is a
multicellular structure which is derived
from the megaspore. Rest statements
regarding post-fertilisation development
in flowering plants are correct.
81Seed formation without
fertilisation in flowering plants
involves the process of
[NEET 2016, Phase I]
(a) budding
(b) somatic hybridisation
(c) apomixis
(d) sporulation
Ans.(c)
Apomixis is a special mechanism found
in flowering plants to produce seeds
without fertilisation. It is a type of
asexual reproduction which mimics the
sexual reproduction and is commonly
found inCitrusvarieties.
82The coconut water from tender
coconut represents
[NEET 2016, Phase I]
(a) fleshy mesocarp
(b) free-nuclear proembryo
(c) free-nuclear endosperm
(d) endocarp
Ans.(c)
Coconut milk represents free-nuclear
endosperm where the division of primary
endosperm nucleus is not followed by
formation of cell walls (cytokinesis) thus
all nucleus remain free in liquid form. It is
rich in plant hormone cytokinin.
83Cotyledon of maize grain is called
[NEET 2016, Phase I]
(a) coleorhiza (b) coleoptile
(c) scutellum (d) plumule
Ans.(c)
Large, shield shaped cotyledon of grass
family is called scutellum. Coleorhiza is a
sheath protecting the root of
germinating grass.
Coleoptile is a sheath covering emerging
shoot. Plumule is rudimentary shoot of
an embryo plant.
84The wheat grain has an embryo
with one large, shield-shaped
cotyledon known as
[CBSE AIPMT 2015]
(a) epiblast
(b) coleorrhiza
(c) scutellum
(d) coleoptile
Ans.(b)
The grain of wheat, maize or rice is a
caryopsis. The embryo in such grains lies
laterally near the base of the grain. The
embryo consists of an upper large, shield
shaped cotyledon known as scutellum.
The scutellum is closely pressed against
the endosperm and helps in the
translocation of nutrients from
endosperm to the growing embryo at the
time of germination and seedling
growth.
85Coconut water from a tender
coconut is [CBSE AIPMT 2015]
(a) immature embryo
(b) free nuclear endosperm
(c) innermost layers of the seed coat
(d) degenerated nucellus
Ans.(b)
The coconut water from tender coconut
that we are familiar with, is nothing but
free nuclear endosperm (made up of
thousands of nuclei) and the surrounding
white kernel is the cellular endosperm.
In the most common type of endosperm
development, the Primary Endosperm
Nucleus (PEN) undergoes successive
nuclear divisions to give rise to free
nuclei. This stage of endosperm
development is called free-nuclear
endosperm.
86Non-albuminous seed is produced
in [CBSE AIPMT 2014]
(a) maize (b) castor
(c) wheat (d) pea
Ans.(d)
Non-albuminous or non-endospermic
seeds are those in which cotyledons are
usually smaller and less developed like in
pea. Such seeds are seen when most or
all of the endosperm is used up much
before germination.
87Nucellar polyembryony is reported
in species of[CBSE AIPMT 2011]
(a)Gossypium (b)Triticum
(c)Brassica (d)Citrus
Ans.(d)
Nucellar polyembryony is reported in
species ofCitrus. Occurrence of more
than one embryo in a seed is referred to
as polyembryony. In manyCitrusand
mango varieties, some of the nucellar
cells surrounding the embryo sac start
dividing, protrude into the embryo sac
and develop into the embryos.
88Apomictic embryos inCitrusarise
from [CBSE AIPMT 2010]
(a) synergids
(b) maternal sporophytic tissue in ovule
(c) antipodal cells
(d) diploid egg
Ans.(b)
Apomictic embryos inCitrusarise from
maternal sporophytic tissue (e.g.
nucellus or integuments) in ovule.
Apomixis is the formation of new
individuals directly through asexual
reproduction without involving the
formation and fusion of gametes.
89Endosperm is consumed by
developing embryo in the seed of
[CBSE AIPMT 2008]
(a) coconut (b) castor
(c) pea (d) maize
Ans.(c)
Endosperm is consumed by developing
embryo is the seeds of pea (Pisum
sativum). These seed are called
non-endospermic seeds. Other
examples are gram, beans, orchids, etc.
90The scutellum observed in a grain
of wheat or maize is comparable to
which part of the seed in other
monocotyledons?
[CBSE AIPMT 2010, 06]
(a) Cotyledon
(b) Endosperm
(c) Aleurone layer
(d) Plumule
Ans.(a)
In wheat or maize (family–Poaceae), the
scutellum is thought to be a modified
cotyledon or seed leaf.
91In angiosperms, triple fusion is
required for the formation of
[CBSE AIPMT 1996]
(a) embryo (b) endosperm
(c) seed coat (d) fruit wall
Ans.(b)
In angiosperms, triple fusion is required
for the formation of endosperm. Triple
fusion refers to the vegetative
fertilisation, i.e. the fusion of nucleus of
a male gamete with the two polar nuclei
or the diploid secondary (fusion) nucleus.
Triple fusion converts central cell into
triploid primary endosperm cell which
forms the endosperm, a nutritive tissue.
Sexual Reproduction in Flowering Plants 221

232. 222 NEETChapterwise Topicwise Biology
92Study of formation, growth and
development of new individual
from an egg is[CBSE AIPMT 1993]
(a) apomixis
(b) embryology
(c) embryogeny
(d) cytology
Ans.(b)
Embryology (Gr.en– in;bryo– swell;
logos –study) is the study of formation
growth and development of embryo. It
includes sporogenesis, fertilisation and
embryogeny, i.e. the overall process
starting from formation of gametes,
fertilisation, zygote and embryo formation
and development of new individual.
93Nucellus embryo is
[CBSE AIPMT 1989]
(a) amphimictic haploid
(b) amphimictic diploid
(c) apomictic haploid
(d) apomictic diploid
Ans.(c)
Nucellus embryo is apomictic haploid.
Nucellus is a parenchymatous, haploid
nutritive tissue of the ovule of
phanerogams. The phenomenon of
formation of embryo or new individual
directly through asexual reproduction or
adventitiously from a cell other than egg,
i.e. nucellus, integuments, etc., is known
as apomixis. Nucellus embryo is thus,
known as apomictic embryo.
94In some plants, the female gamete
develops into embryo without
fertilisation. This phenomenon is
known as [NEET (National) 2019]
(a) parthenocarpy
(b) syngamy
(c) parthenogenesis
(d) autogamy
Ans.(c)
Parthenogenesis is the process by which
the female gamete develops into embryo
without fertilisation.
It is of two type-haploid and diploid. In
former, embryo develops from haploid
egg and in latter case, a diploid egg
develops into embryo.
Parthenocarpy is the development of
seedless fruits from an unfertilised egg
of the flower.
Syngamy is the fusion of male and
female gamete during sexual
reproduction in plants.
Autogamy is a type of self-pollination in
which a flower is pollinated by its own
pollen.
95Which one of the following fruits is
parthenocarpic?[CBSE AIPMT 2015]
(a) Brinjal (b) Apple
(c) Jackfruit (d) Banana
Ans.(d)
Parthenocarpy is a process of
developing fruits without involving
fertilisation/seed formation. Therefore,
the seedless varieties of economically
important fruits like orange, banana,
watermelon, lemon, etc. are produced
using this technique. This technique
involves inducing fruit formation by the
application of plant growth hormones
such as auxins.
96In a type of apomixis known as
adventive embryony, embryos
develop directly from the
[CBSE AIPMT 2005]
(a) nucellus or integuments
(b) zygote
(c) synergids or antipodals in an embryo
sac
(d) accessory embryo sac in the ovule
Ans.(a)
Apomixis does not involve gamete
formation and fertilisation as found in
amphimixis. Adventive embryo
formation is a type of apomixis in which
embryos are formed from diploid
integumental or nuceller cells.
97Adventive embryony inCitrusis
due to [CBSE AIPMT 2001]
(a) nucellus
(b) integuments
(c) zygotic embryo
(d) fertilised egg
Ans.(a)
Nucellus embryony occurs in
crassinucellate ovules (e.g.Citrus,
Opuntia). On the other hand
integumentary embryony occurs in
tenuinucellate ovules (e.g.Euonymus).
98Formation of gametophyte directly
from sporophyte without meiosis is
(a) apospory [CBSE AIPMT 1988]
(b) apogamy
(c) parthenogenesis
(d) amphimixis
Ans.(a)
Apospory is the formation of
gametophyte directly from sporophyte
without forming spores and meiosis. The
gametophytes formed through apospory
are usually diploid. Apospory leads to
polyploidy and hence, new species in
bryophytes and pteridophytes.
99Prothallus (gametophyte) gives rise
to fern plant (sporophyte) without
fertilisation. It is
[CBSE AIPMT 1988]
(a) apospory (b) apogamy
(c) parthenocarpy (d) parthenogenesis
Ans.(b)
Apogamy refers to the development of
sporophyte from gametophyte without
fertilisation.
In fern plant, prothallus (gametophyte)
gives rise to main plant body
(sporophyte) directly from somatic cell
without forming gametes. Sporophyte
thus formed is haploid in nature.
100Development of an organism from
female gamete/egg without
involving fertilisation is
[CBSE AIPMT 1989]
(a) adventitive embryony
(b) polyembryony
(c) parthenocarpy
(d) parthenogenesis
Ans.(d)
Parthenogenesis can be defined as
formation of embryo from an unfertilised
egg or female gamete. Parthenogenetic
plants are homozygous so, they have
more importance in plant breeding and
genetics.
Special Mechanism
of Reproduction
TOPIC 4

233. 01Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. Placenta 1. Androgens
B. Zona pellucida 2. Human Chorinoic
Gonadotropin (hCG)
C. Bulbourethral
glands
3. Layer of the ovum
D. Leydig cells 4. Lubrication of the
penis
A B C D
(a) 1 4 2 3
(b) 3 2 4 1
(c) 2 3 4 1
(d) 4 3 1 2
Ans.(c)
The correct option is (c). It can be
explained as follows.
Placenta secretes human chorionic
gonadotropin (hCG).
Zona pellucida is a primary egg
membrane secreted by the secondary
oocyte. The secretions of Bulbourethral
glands help in lubrication of the penis.
Leydig cells synthesise and secrete
testicular hormones called androgens.
02Select the correct sequence for
transport of sperm cells in male
reproductive system.
[NEET (National) 2019]
(a) Seminiferous tubules→Rete testis
→Vasa efferentia→Epididymis→
Vas deferens→Ejaculatory duct→
Urethra→Urethral meatus
(b) Seminiferous tubules→Vasa
efferentia→Epididymis→Inguinal
canal→Urethra
(c) Testis→Epididymis→Vasa
efferentia→Vas deferens→
Ejaculatory duct→Inguinal canal→
Urethra→Urethral meatus
(d) Testis→Epididymis→Vasa
efferentia→Rete testis→Inguinal
canal→Urethra
Ans.(a)
The correct sequence of sperm
transport in male reproductive system is
seminiferous tubules→Rete testis→
Vasa efferentia→Epididymis→Vas
deferens→Ejaculatory duct→Urethra
→Urethral meatus. The pathway of
sperm transport is shown in the diagram
below
03Which of the following depicts the
correct pathway of transport of
sperms? [NEET 2016, Phase II]
(a) Rete testis→Efferent ductules→
Epididymis→Vas deferens
(b) Rete testis→Epididymis→Efferent
ductules→Vas deferens
(c) Rete testis→Vas deferens→
Efferent ductules→Epididymis
(d) Efferent ductules→Rete testis→
Vas deferens→Epididymis
Ans.(a)
The sperms are produced in the
seminiferous tubules. The rete testis is
connected to these tubules at one end
and transfers sperms to vasa efferentia
(small tubular structures between rete
testis and epididymis).
The sperms reach to epididymis through
vasa efferentia where they are
temporarily stored for maturation and
then transferred to seminal vesicle
through vas deferens.
Thus, the correct route is
Rete testis→Efferent ductules→
Epididymis→vas deferens.
04The shared terminal duct of the
reproductive and urinary system in
the human male is
[CBSE AIPMT 2014]
(a) urethra (b) ureter
(c) vas deferens (d) vasa efferentia
Ans.(a)
In the human male, urethra provides an
exit for urine from the urinary bladder as
well as semen from vasa differentia
during ejaculation. Thus, it is also known
as urogenital duct.
In males, it is about 8 inches (20 cm) long
and opens at the end of the penis.
HumanReproduction
25
Human Reproductive
System
TOPIC 1
Ejaculatory
duct (6)
Vas
deferens (5)
Epididymis (4)
Urethra (7)
Urethral
meatus (8)
Seminiferous
tubules
(1)
Vasa efferentia (3)
Rete testis (2)
Enlarged view of testis

234. Vas deferens and vasa efferentia are the
male sex accessory ducts.
Ureters are the tubes that carry urine
from the kidney to urinary bladder.
05The Leydig cells as found in the
human body are the secretory
source of [CBSE AIPMT 2012]
(a) progesterone (b) intestinal mucus
(c) glucagon (d) androgens
Ans.(d)
Interstitial cellsorcells of Leydigare
present in the connective tissue lying in
between seminiferous tubules. These
cells secrete oestradiol-steroid
androgens, e.g. testosterone.
Androgens stimulate male characters,
influence male sexual behaviour (libido)
and regulate the development,
maturation and functions of male
accessory sex organs.
06If for some reason, the vasa
efferentia in the human
reproductive system get blocked,
the gametes will not be transported
from [CBSE AIPMT 2011]
(a) epididymis to vas deferens
(b) ovary to uterus
(c) vagina to uterus
(d) testes to epididymis
Ans.(d)
Vasa efferentia (ductuli efference) are
10-20 fine tubules which connect rete
testis with an epididymis (ductus
epididymis). The latter is a pair of ducts
from each testis which is formed by
union of its vasa efferentia. If the vasa
efferentia get blocked, the sperms will
not be transported from testes to
epididymis.
07The figure given below depicts a
diagrammatic sectional view of the
female reproductive system of
humans. Which one set of three
parts out of A-F have been
correctly identified?
[CBSE AIPMT 2011]
(a) C–Infundibulum, D–Fimbriae, E–Cervix
(b) D–Oviducal funnel, E–Uterus, F–Cervix
(c) A–Perimetrium, B–Myometrium,
C–Fallopian tube
(d) B–Endometrium, C–Infundibulum,
D–Fimbriae
Ans.(a)
The Fallopian tube is about 10-12 cm long
and extends from the periphery of each
ovary to the uterus. The part closer to
the ovary is funnel shaped and is called
infundibulum. The edges of the
infundibulum possess finger-like
projections calledfimbriae, which help
in collection of the ovum after ovulation.
The uterus opens into vagina through a
narrow cervix.
08The testes in humans are situated
outside the abdominal cavity inside
a pouch called scrotum. The
purpose served is for
[CBSE AIPMT 2011]
(a) escaping any possible compression
by the visceral organs
(b) providing more space for the growth
of epididymis
(c) providing a secondary sexual feature
for exhibiting the male sex
(d) maintaining the scrotal temperature
lower than the internal body
temperature
Ans.(d)
The testes in humans are situated
outside the abdominal cavity in scrotal
sacs. This is because the temperature of
scrotal sacs is 2-2.5°C which is less than
internal body temperature.
09Sertoli cells are found in
[CBSE AIPMT 2010]
(a) ovaries and secrete progesterone
(b) adrenal cortex and secrete
adrenaline
(c) seminiferous tubules and provide
nutrition to germ cells
(d) pancreas and secrete cholecystokinin
Ans.(c)
The epithelium of seminiferous tubule is
made up of two types of cells, i.e. Sertoli
cells and spermatogenic cells. Sertoli
cells are elongated and pyramidal which
partially envelope the spermatogenic
cells. These nourish spermatozoa, act as
nurse cells for differentiating
spermatozoa. These secrete a
glycoprotein hormone, called inhibin
which is involved in the negative feed
back control of sperm production.
10Vasa efferentia are the ductules
leading from[CBSE AIPMT 2010]
(a) testicular lobules to rete testis
(b) rete testis to vas deferens
(c) vas deferens to epididymis
(d) epididymis to urethra
Ans.(b)
Rete testis is connected to epididymis
by 12-20 fine tubules called vasa efferentia
or ductuli efferens. These collect sperms
from inside the testis and transfer them
to the epididymis. Vasa deferens arises
from caudal epididymis, conducts
sperms from epididymis to urethra.
11Seminal plasma in human males is
rich in [CBSE AIPMT 2010, 09]
(a) fructose and calcium
(b) glucose and calcium
(c) DNA and testosterone
(d) ribose and potassium
Ans.(a)
Seminal plasma is composed of the fluid
and sperms from the vas deferens
(about 10% of the total), fluid from the
seminal vesicles (almost 60 percent),
fluid from the prostate gland (about 30
percent) and small amount of mucous
glands, especially the bulbourethral
glands.
12The part of Fallopian tube closest
to the ovary is[CBSE AIPMT 2010]
(a) isthmus (b) infundibulum
(c) cervix (d) ampulla
Ans.(b)
The Fallopian tubes, uterus and vagina
constitute the female accessory ducts.
Each Fallopian tube extends from the
periphery of each ovary to the uterus.
The part closer to the ovary is
funnel-shaped infundibulum, which help
in collection of the ovum after ovulation.
13Given below is a diagrammatic
sketch of a portion of human male
reproductive system. Select the
correct set of the names of the
parts labelledA,B,C,D.
[CBSE AIPMT 2009]
224 NEETChapterwise Topicwise Biology
B
A
F
E
C
D
A
B
C
D

235. A B C D
(a) Ureter Prostate Seminal
vesicle
Bulbourethral
gland
(b) Vas
deferens
Seminal
vesicle
Prostate Bulbourethral
gland
(c) Vas
deferens
Seminal
vesicle
Bulboure-
thral gland
Prostate
(d) Ureter Seminal
vesicle
Prostate Bulbourethral
gland
Ans.(b)
Option (b) is correct,A=Vas deferens.
B=Seminal vesicle,C=Prostrate and
D=Bulbourethral gland.
14Bartholin’s glands are situated
[CBSE AIPMT 2003]
(a) on either side of vagina in humans
(b) on either side of vas deference in
humans
(c) on the sides of the head of some
amphibians
(d) at the reduced tail end of birds
Ans.(a)
Bartholin’s glands(Bulbovestibular
glands) are one pair, small sized glands
find just behind the labia minora, one on
either sides of vaginal orifice. These
lubricate the vagina during mating and
parturition by secretion of mucus.
15Location and secretion of Leydig’s
cells are [CBSE AIPMT 1991]
(a) liver — cholesterol
(b) ovary — estrogen
(c) testis — testosterone
(d) pancreas — glucagon
Ans.(c)
The endocrine part of testis is formed of
groups of cells, called interstitial cells or
Leydig’s cells, scattered in connective
tissue between the sperm producing
seminiferous tubules of the testis.
These cells are stimulated to produce
male sex hormones, called androgens
by ICSH of anterior pituitary.
Testosterone is main androgen and is
a steroid hormone.
16Select the correct option of haploid
cells from the following groups.
[NEET (Oct.) 2020]
(a) Primary oocyte, secondary oocyte,
spermatid
(b) Secondary spermatocyte, first polar
body, ovum
(c) Spermatogonia, primary spermatocyte,
spermatid
(d) Primary spermatocyte, secondary
spermatocyte, second polar body
Ans.(b)
Out of the given the option, the haploid
cells are secondary spermatocyte, first
polar body, ovum, secondary oocyte,
spermatids and second polar body.
The diploid cells are primary oocyte and
primary spermatocyte. Thus, option (b) is
correct.
17Match the following columns and
select the correct option from the
codes given belows
[NEET (Oct.) 2020]
Column I Column II
A. Ovary 1. Human chorionic
gonadotropin
B. Placenta 2. oestrogen and
progesterone
C. Corpus
luteum
3. Androgens
D. Leydig
cells
4. Progesterone
only
Codes
A B C D
(a) 4 3 2 1
(b) 1 2 3 4
(c) 1 3 2 4
(d) 2 1 4 3
Ans.(d)
The option (d) is the correct match which
is as follows
Ovary produces oestrogen and
progesterone.
Placenta produces Human Chorionic
Gonadotropin (hCG).
Corpus luteum produces progesterone
only.
Leydig cells produce androgens.
18Which of the following hormone
levels will cause release of ovum
(ovulation) from the Graafian
follicle? [NEET (Sep.) 2020]
(a) High concentration of progesterone
(b) Low concentration of LH
(c) Low concentration of FSH
(d) High concentration of oestrogen
Ans.(d)
High level of oestrogen will send positive
feedback to anterior pituitary for release
of ovum from Graafian follicle. FSH, LH
and oestrogen are at peak level during
mid of menstrual cycle (28 day cycle). LH
surge leads to ovulation.
19Meiotic division of the secondary
oocyte completed
[NEET (Sep.) 2020]
(a) at the time of copulation
(b) after zygote formation
(c) at the time of fusion of a sperm with
an ovum
(d) prior to ovulation
Ans.(c)
Meiotic division of secondary oocyte is
completed after the entry of sperm in
secondary oocyte which lead to the
formation of a large ovum and a tiny IInd
polar body.
20No new follicles develop in the
luteal phase of the menstrual cycle
because [NEET (Odisha) 2019]
(a) follicles do not remain in the ovary
after ovulation
(b) FSH levels are high in the luteal
phase
(c) LH levels are high in the luteal phase
(d) both FSH and LH levels are low in the
luteal phase
Ans.(d)
No new follicles develop in the luteal
phase of menstrual cycle. It is because
during this phase, Luteinising Hormone
(LH) and Follicle Stimulating Hormone
(FSH) levels decrease. Instead, the
already ruptured follicle closes after
releasing the egg and forms a corpus
luteum during luteal phase, which
produces progesterone.
21What is the fate of the male
gametes discharged in the
synergid?[NEET (National) 2019]
(a) All fuse with the egg
(b) One fuses with the egg, other(s)
fuse(s) with synergid nucleus
(c) One fuses with the egg and other
fuses with central cell nuclei
(d) One fuses with the egg other(s)
degenerate (s) in the synergid
Ans.(c)
Out of the male gametes discharged in
the synergid, one fuses with the egg and
other fuses with central cell nuclei. The
fusion between male gamete and egg is
Human Reproduction 225
Gametogenesis, Menstrual
Cycle and its Hormonal Control
TOPIC 2

236. called syngamy or true fertilisation
which forms zygote (2)n. The fusion
between male gamete and central cell
nuclei is called triple fusion and it results
in the formation of a triploid primary
endosperm nucleus (3)n.
22Extrusion of second polar body
from egg nucleus occurs
[NEET (National) 2019]
(a) after fertilisation
(b) before the entry of sperm into ovum
(c) simultaneously with first cleavage
(d) after the entry of sperm but before
fertilisation
Ans.(d)
Extrusion of second polar body from egg
nucleus occurs after the entry of sperm
but before fertilisation. The entry of
sperm into female egg causes the
breakdown of Metaphase Promoting
Factor (MPF) and turns on Anaphase
Promoting Factor (APF). Hence, the
secondary oocyte completes its meiotic
division after fertilisation and is said to
be activated.
23The difference between
spermiogenesis and spermiation is
[NEET 2018]
(a) In spermiogenesis, spermatozoa from
Sertoli cells are released into the
cavity of seminiferous tubules, while
in spermiation spermatozoa are
formed
(b) In spermiogenesis, spermatozoa are
formed, while in spermiation
spermatids are formed
(c) In spermiogenesis, spermatids are
formed, while in spermiation
spermatozoa are formed
(d) In spermiogenesis, spermatozoa are
formed, while in spermiation
spermatozoa are released from
Sertoli cells into the cavity of
seminiferous tubules
Ans.(d)
Spermiogenesisis the process of
transformation of spermatids (n) into
spermatozoa (n) or sperms. It involves
the differentiation phase in which one
spermatid develops into one
spermatozoan.
Spermiationinvolves the release of
sperms from seminiferous tubules
through Sertoli cells.
24Match the items given in Column I
with those in Column II and select
thecorrectoption given below.
[NEET 2018]
Column I Column II
1. Proliferative
phase
i. Breakdown of
endometrial
lining
2. Secretory
phase
ii. Follicular
phase
3. Menstruation iii. Luteal phase
1 2 3
(a) ii iii i
(b) i iii ii
(c) iii ii i
(d) iii i ii
Ans.(a)
Duringproliferative phase, the follicles
start growing in size under the influence
of Follicle stimulating Hormone (FSH).
Hence, this phase is also called follicular
phase.
Duringsecretory phase, corpus luteum
secretes progesterone that helps to
thicken the endometrial lining. Due to
the persistence of corpus luteum, this
phase is also called luteal phase.
Menstruationor bleeding occurs due to
the breakdown of endometrial lining in
the absence of pregnancy. During this
phase, corpus luteum regresses and
progesterone level decreases.
25A temporary endocrine gland in the
human body is [NEET 2017]
(a) pineal gland
(b) corpus cardiacum
(c) corpus luteum
(d) corpus allatum
Ans.(c)
Corpus luteum is a temporary endocrine
gland in the human body. It secretes
small amount of estradiol and significant
amount of progesterone hormone. In the
absence of fertilisation, the corpus
luteum degenerates.
26Select the incorrect statement.
[NEET 2016, Phase I]
(a) LH and FSH triggers ovulation in ovary
(b) LH and FSH decrease gradually during
the follicular phase
(c) LH triggers secretion of androgens
from the Leydig cells
(d) FSH stimulates the Sertoli cells which
help in spermiogenesis
Ans.(b)
In follicular phase of menstrual cycle, LH
and FSH increase gradually and
stimulate follicular development as well
as secretion of oestrogens by the
growing follicles.
27Identify the correct statement on
‘inhibin’. [NEET 2016, Phase I]
(a) Is produced by granulosa cells in
ovary and inhibits the secretion of
FSH
(b) Is produced by granulosa cells in
ovary and inhibits the secretion of LH
(c) Is produced by nurse cells in testes
and inhibits the secretion of LH
(d) Inhibits the secretion of LH, FSH and
prolactin
Ans.(a)
Inhibin is produced by granulosa cells of
ovarian follicles in the ovary and has
negative feedback effect on the
secretion of FSH.
28Changes in GnRH pulse frequency
in females is controlled by
circulating levels of
[NEET 2016, Phase I]
(a) oestrogen and inhibin
(b) progesterone only
(c) progesterone and inhibin
(d) oestrogen and progesterone
Ans.(d)
High levels of oestrogen and
progesterone give negative feedback to
hypothalamus for the release of GnRH.
Thus, inhibiting the gonadotropin
release.
29Which of the following layers in an
antral follicle is acellular?
[CBSE AIPMT 2015]
(a) Granulosa (b) Theca interna
(c) Stroma (d) Zona pellucida
Ans.(d)
Follicles that form an antrum during
maturation are called antral follicle or
Graafian follicle.
During the development of the follicle, a
glycoprotein polymer capsule called the
zona pellucida which is acellular, forms
around the oocyte, separating it from
the surrounding granulosa cells.
This layer remains with the oocyte after
ovulation, and contains enzymes that
catalyse with sperm to allow
penetration.
30Which of the following events in
not associated with ovulation in
human female?[CBSE AIPMT 2015]
(a) Decrease in oestradiol
(b) Full development of Graafian follicle
(c) Release of secondary oocyte
(d) LH surge
226 NEETChapterwise Topicwise Biology

237. Ans.(a)
Oestradiol levels fall after ovulation and
before menstruation while, its levels
peak prior to ovulation. Oestradiol are
not associated with ovulation. Decrease
in oestradiol level result in the cessation
of menstruation.
31The main function of mammalian
corpus luteum is to produce
[CBSE AIPMT 2014, 1995 ]
(a) oestrogen only
(b) progesterone
(c) human chorionic gonadotropin
(d) relaxin only
Ans.(b)
The main function of mammalian corpus
luteum is the secretion of progesterone
which is essential for the maintenance
of endometrium. Endometrium is
necessary for implantation of the
fertilised ovum and other events of
pregnancy.
Corpus luteum also secretes some
amount of estrogen to maintain
pregnancy.
hCG (human Chorionic Gonadotropin) is
secreted by placenta for maintaining the
corpus luteum.
Relaxin is secreted by corpus luteum
during the end of gestation period.
32What is the correct sequence of
sperm formation? [NEET 2013]
(a) Spermatid, Spermatocyte,
Spermatogonia, Spermatozoa
(b) Spermatogonia, Spermatocyte,
Spermatozoa, Spermatid
(c) Spermatogonia, Spermatozoa,
Spermatocyte, Spermatid
(d) Spermatogonia, Spermatocyte,
Spermatid, Spermatozoa
Ans.(d)
Spermatogonia→Spermatocyte→
Spermatid→Spermatozoa
Spermatogonia is present on the inside
wall of seminiferous tubule which undergo
mitotic division and increase their number.
Spermatocytes are some of the
spermatogonia, which periodically
undergo meiosis. The secondary
spermatocytes undergo the second
meiotic division to produce four, equal
haploid spermatids. The spermatids are
transformed into spermatozoa (sperm).
33Menstrual flow occurs due to lack
of [CBSE AIPMT 2012]
(a) progesterone (b) FSH
(c) oxytocin (d) vasopressin
Ans.(a)
Menstrual flow occurs due to the lack of
progesterone.
Progesterone is secreted by corpus
luteum and is essential for the
maintenance of endometrium. This
endometrium is responsible for
implanation of the festilised ovum, i.e.
pregnancy.
FSHStimulates gonadal activity and also
called as gonadotrophins.
OxytocinStimulates contraction in
uterus during childbirth.
VasopressinStimulate resorption of
water and electrolytes by the distal
tubules, also called as Antidiuretic
Hormone (ADH).
34Which one of the following
statements is false in respect of
viability of mammalian sperm?
[CBSE AIPMT 2012]
(a) Sperm is viable for only up to 24 hrs
(b) Survival of sperm depends on the pH
of the medium and is more active in
alkaline medium
(c) Viability of sperm is determined by its
motility
(d) Sperms must be concentrated in a
thick suspension
Ans.(a)
Viability of a sperm means the capability
of a sperm, to fertilise an egg.
Sperms are viable for 24 h to 48 h,
whereas the ovum is viable for only 24 h.
35Which one of the following
statements about human sperm is
correct? [CBSE AIPMT 2010]
(a) Acrosome has a conical pointed
structure used for piercing and
penetrating the egg, resulting in
fertilisation
(b) The sperm lysins in the acrosome
dissolve the egg envelope facilitating
fertilisation
(c) Acrosome serves as a sensory
structure leading the sperm towards
the ovum
(d) Acrosome serves no particular
function
Ans.(b)
Penetration of human sperm is a
chemical mechanism. In this, acrosome
of sperm undergoes acrosomal reaction
and releases certain sperm lysins, which
dissolve the egg envelope locally and
make the path for the penetration of
sperm. Sperm lysins are acidic proteins.
These sperm lysins contain a lytic
enzyme hyaluronidase (that dissolves
the hyaluronic acid polymers in the
intercellular spaces, which holds the
granulosa cells of corona radiata
together) corona penetrating enzyme
and acrosin.
36Which one of the following is the
most likely reason of not occurring
regular menstruation cycle in
females? [CBSE AIPMT 2009]
(a) Fertilisation of the ovum
(b) Maintenance of the hypertrophical
endometrial lining
(c) Maintenance of high concentration of
sex-hormones in the blood stream
(d) Retention of well-developed corpus
luteum
Ans.(a)
If fertilisation occurs and foetus is
implanted in the endometrium, the
trophoblast cells of the developing
placenta secrete a hormone human
Chorionic Gonadotrophic (hCG). This
hormone like LH maintains the corpus
luteum and the secretion of
progesterone and estradiol by it. These
two hormones check the breakdown of
the endometrium of the uterus. The
absence of menstrual bleeding is the
earliest sign of pregnancy.
37Which one of the following is the
correct matching of the events
occurring during menstrual cycle?
[CBSE AIPMT 2009]
(a)Ovulation LH and FSH attain
peak level and
sharp fall in the
secretion of
progesterone
(b)Proliferative
phase
Rapid regeneration
of
myometrium and
maturation of
Graafian
follicle
(c)Development
of corpus
luteum
Secretory phase and
increased secretion
of
progesterone
(d)Menstruation Breakdown of
myometrium and
ovum not fertilised
Ans.(b)
In secretory phase during ovulation, the
follicle breaks and collapes under the
continuous influence of Luteinising
Hormone (LH).
Human Reproduction 227

238. It begins to enlarge and forms a
yellowish structure, called the corpus
luteum. The corpus luteum plays an
important role in the preparation of the
endometrium for the implantation of the
fertilised egg by secreting oestrogens
and progesterone.
38Which one of the following
statement is incorrect about
menstruation?[CBSE AIPMT 2008]
(a) During normal menstruation about 40
mL blood is lost
(b) The menstrual fluid can easily clot
(c) At menopause in the female, there is
especially abrupt increase in
gonadotropic hormones
(d) The beginning of the cycle of
menstruation is called menarche
Ans.(b)
During normal menstruation
approximately 40 mL of blood and an
additional 35 mL of serous fluid are lost.
The menstrual fluid is normally
non-clotting because a fibrinolysin is
released alongwith necrotic endometrial
material.
39At the end of first meiotic division,
male germ cell differentiates into
[CBSE AIPMT 2008, 1994]
(a) secondary spermatocyte
(b) primary spermatocyte
(c) spermatogonium
(d) spermatid
Ans.(a)
First meiotic division takes place in
diploid primary spermatocyte and it
forms two haploid cells called
secondary spermatocytes.
40Which part of ovary in mammals
acts as an endocrine gland after
ovulation? [CBSE AIPMT 2007]
(a) Graafian follicle (b) Stroma
(c) Germinal epithelium
(d) Vitelline membrane
Ans.(a)
During ovulation, the mature follicle or
Graafian follicle bursts and the ovum is
released. This is named ascorpus
luteumwhich serves as a temporary
endocrine gland by releasing
progesterone and oestrogen.
41If mammalian ovum fails to get
fertilised, which one of the
following is unlikely?
(a) Corpus luteum will disintegrate
(b) Estrogen secretion further decreases
(c) Primary follicle starts developing
(d) Progesterone secretion rapidly
declines
Ans.(b)
If mammalian ovum fails to get fertilized,
the estrogen secretion does not
decrease further, while corpus luteum
will disintegrate. Primary follicle starts
developing and progesterone secretion
rapidly declines.
42Ovulation in the human female
normally takes place during the
menstrual cycle
[CBSE AIPMT 2004]
(a) at the mid secretory phase
(b) just before the end of the secretory
phase
(c) at the beginning of the proliferative
phase
(d) at the end of the proliferative phase
Ans.(d)
Ovulation takes place under the
influence of LH and FSH. It normally
takes place at the end of proliferativeie,
14
th
day or mid way during menstrual
cycle. The LH surge stimulates
completion of reduction division of
oocyte. Following ovulation, the Graafian
follicle changes to corpus luteum.
43Which set is similar?
[CBSE AIPMT 2001]
(a) Corpus luteum — Graafian follicle
(b) Sebum — Sweat
(c) Bundle of His — Pacemaker
(d) Vit-B
7
— Niacin
Ans.(a)
Corpus luteum is temporary endocrine
tissue developing from ruptured
Graafian follicle.
Sebum is an oily lipid containing
secretion of mammalian sebaceous
glands.
Sweat is an aqueous secretion of
mammalian sweat glands.
Bundle of His is a part of conducting
system of heart and pace-maker is
responsible for initiation of heart beat in
right auricle SA node.
Vitamin-B
5
is also known as niacin.
44Middle piece of mammalian sperm
possesses[CBSE AIPMT 1999, 91]
(a) mitochondria and centriole
(b) mitochondria only
(c) centriole only
(d) nucleus and mitochondria
Ans.(b)
The middle piece of human sperm
contains mitochondria which are coiled
around an axial filament called
mitochondrial spiral. These provide
energy for the movement of sperm.
45After ovulation, Graafian follicle
regresses into[CBSE AIPMT 1999]
(a) corpus luteum
(b) corpus callosum
(c) corpus albicans
(d) corpus artesia
Ans.(a)
Just after ovulation, LH (secreted by
anterior lobe of pituitary gland)
stimulates remaining ovarian follicles to
develop into corpus luteum.
The corpus luteum plays an important
role in the preparation of the
endometrium for the implantation of the
fertilised egg by secreting oestrogens
and progesterone.
46In 28 days human ovarian cycle,
ovulation occurs on
[CBSE AIPMT 1997, 94]
(a) 1 day
(b) 5 day
(c) 14 day
(d) 28 day
Ans.(c)
The first phase (proliferative phase) of
menstrual cycle ends on 14
th
day, the
ovarian follicles rupture and ovulation
occurs.
47Fertilisins are emitted by
[CBSE AIPMT 1997, 91]
(a) immature eggs
(b) mature eggs
(c) sperms
(d) polar bodies
Ans.(b)
Eggs secrete the chemical fertilisin,
which is made up of glycoprotein. It
interacts with the antifertilisin (protein
on sperm surface) of a sperm of same
species.
228 NEETChapterwise Topicwise Biology
Nucleus
AcrosomeCentriole
Mitochondria
2.5mm
Head
Neck
Tail
Middle
piece

239. 48Human eggs are
[CBSE AIPMT 1997, 89]
(a) alecithal
(b) microlecithal
(c) mesolecithal
(d) macrolecithal
Ans.(a)
The eggs almost free of yolk are called
alecithal, e.g. human eggs (ova). In such
eggs the cleavage pattern is holoblastic
(Gr.holo-whole;blastos-germ) that is the
cleavage extends completely through
the egg.
49Ovulation occurs under the
influence of[CBSE AIPMT 1994]
(a) LH
(b) FSH
(c) oestrogen
(d) progesterone
Ans.(a)
The Luteinising Hormone (LH) of anterior
pituitary regulates the ovulation from
the Graafian follicle. This LH surge
causes ovulation to occur.
50In telolecithal egg the yolk is found
[CBSE AIPMT 1993]
(a) all over the egg (b) on one side
(c) both the sides (d) at centre
Ans.(b)
Yolk is concentrated towards vegetal
pole. The nucleus and major part of
cytoplasm is displaced to animal pole as
in mesolecithal and macrolecithal eggs
of vertebrates.
51Freshly released human egg has
[CBSE AIPMT 1991]
(a) one Y-chromosome
(b) one X-chromosome
(c) two X-chromosomes
(d) Both (a) and (b)
Ans.(c)
Human female is homozygous, i.e.
produces same kind of gametes that
carry X-chromosome while human male
is heterozygous, i.e. produces unlike
gametes that carries either
X-chromosome or Y-chromosome.
52Egg is liberated from ovary in
[CBSE AIPMT 1989]
(a) secondary oocyte stage
(b) primary oocyte stage
(c) oogonial stage
(d) mature ovum stage
Ans.(a)
In most of the organisms including
human female the ovulation, i.e., release
of ovum from ovary occurs at secondary
oocyte stage in which meiosis-I has been
completed and first polar body has been
released.
53There is no DNA in
[CBSE AIPMT 2009]
(a) an enucleated ovum
(b) mature RBCs
(c) a mature spermatozoan
(d) hair root
Ans.(a)
The chromatin material inside the nucleus
is composed of DNA, some proteins and
RNA. Infact it is basically DNA—protein
complex. Thus, in an enucleated ovum,
DNA will be absent. The mature RBCs,
mature spermatozoan and root hair are
nucleated which contain DNA.
54Receptors for sperm binding in
mammals are present on
(a) corona radiata [NEET 2021]
(b) vitelline membrane
(c) perivitelline space
(d) zona pellucida
55Capacitation occurs in[NEET 2017]
(a) rete testis
(b) epididymis
(c) vas deferens
(d) female reproductive tract
Ans.(d)
Capacitation is a process, where the
spermatozoa acquire the capacity to
fertilise the eggs. It occurs in female
reproductive tract.
Concept Enhancer
Capacitation involves following changes
(i) Dilution of inhibitory factors of
semen.
(ii) Removal of cholestrol vesicles
covering sperm head and acrosome.
(iii) Increase in the permeability of
acrosome.
Ans.(d)
Receptors for sperm binding are present
on zona pellucida layer of ovum in
mammals.
A sperm comes in contact with zona
pellucida layer of ovum and induces
changes that blocks further entry of
other sperms.
It ensures entry of only one sperm inside
the ovum for fertilisation.
56Fertilisation in humans is
practically feasible only if
[NEET 2016, Phase I]
(a) the ovum and sperms are transported
simultaneously to ampullary - isthmic
junction of the Fallopian tube
(b) the ovum and sperms are transported
simultaneously to ampullary-isthmic
junction of the cervix
(c) the sperms are transported into
cervix within 48 hrs of release of
ovum in uterus
(d) the sperms are transported into
vagina just after the release of ovum
in Fallopian tube
Ans.(a)
Fertilisation in humans, is practically
feasible only if the sperms and ovum are
transported simultaneously at
ampullary-isthmic junction of Fallopian
tube.
57Ectopic pregnancies are referred
to as [CBSE AIPMT 2015]
(a) pregnancies with genetic abnormality
(b) implantation of embryo at site other
than uterus
(c) implantation of defective embryo in
the uterus
(d) pregnancies terminated due to the
hormonal imbalance
Ans.(b)
Ectopic pregnancy develops when an
embryo implants somewhere other than
the uterus, such as in one of Fallopian
tube. It is also known as eccysis or tubal
pregnancy.
58In human females, meiosis-II is not
completed until[CBSE AIPMT 2015]
(a) puberty
(b) fertilisation
(c) uterine implantation
(d) birth
Ans.(b)
In human females, meiosis II is not
completed until fertilisation. Secondary
oocyte is arrested in metaphase II stage
until sperm cell contacts plasma
membrane of the oocyte and completes
meiosis II resulting in release of ovum.
Human Reproduction 229
Fertilisation and
Implantation
TOPIC 3

240. 59Select the correct option
describing gonadotropin activity in
a normal pregnant female
[CBSE AIPMT 2014, 12]
(a) High level of FSH and LH stimulates
the thickening of endometrium
(b) High level of FSH and LH facilitate
implantation of the embryo
(c) High level of hCG stimulates the
synthesis of estrogen and
progesterone
(d) High level of hCG stimulates the
thickening of endometrium
Ans.(c)
hCG (human Chorionic Gonadotrophin)
secretion occurs about 48-72 hours after
implantation. Its level increases and
excess of hCG leaks into urine which is
the indication of pregnancy.
This hormone like LH stimulates the
corpus luteum to secrete high levels of
progesterone and some estrogen to
maintain pregnancy. There steroids are
required to maintain the development of
placenta, initiate the development of
mammary glands and inhibit ovulation.
60Which one of the following
statements about morula in
humans is correct?
[CBSE AIPMT 2010]
(a) It has almost equal quantity of
cytoplasm as an uncleaved zygote but
much more DNA
(b) It has far less cytoplasm as well as
less DNA than in an uncleaved zygote
(c) It has more or less equal quantity of
cytoplasm and DNA
(d) It has more cytoplasm and more DNA
than an uchleaved zygote
Ans.(a)
Cleavage divisions are mitotic division, in
which the single-celled zygote is
converted into a multicellular morula.
But during cleavage divisions, there is no
growth of resultant daughter cells
/blastomeres. So, the DNA content will
increase, but there is no increase or
insignificant increase in amount of
protoplasm.
61A change in the amount of yolk and
its distribution in the egg will
affect.[CBSE AIPMT 2009, 95, 93]
(a) formation of zygote
(b) pattern of cleavage
(c) number of blastomeres produced
(d) fertilisation
Ans.(b)
Patterns of cleavage are based on
amount of yolk and its distribution in the
eggs. Any change in amount and
distribution of yolk directly affects the
pattern of cleavage.
As holoblastic cleavage is found in
microlecithal, mesolecithal or
telolecithal types of eggs. Meroblastic
cleavage is found in macrolecithal and
highly telolecithal eggs of reptiles, birds
and monotreme mammals.
62Gray crescent is the area
[CBSE AIPMT 2005]
(a) at the point of entry of sperm into
ovum
(b) just opposite to the site of entry of
sperm into ovum
(c) at the animal pole
(d) at the vegetal pole
Ans.(b)
The gray crescent area is the area
just opposite to the entry of sperm into
ovum.
63What is true for cleavage?
[CBSE AIPMT 2002]
(a) Size of embryo increases
(b) Size of cells decreases
(c) Size of cells increases
(d) Size of embryo decreases
Ans.(b)
Cleavage is a succession of rapid cell
division during which the cells undergo
the S phase (DNA synthesis) and
M-phase (mitosis) of the cell cycle but
often virtually skip theG
1
andG
2
-phases.
Cleavage simply divides the cytoplasm of
larger cells into smaller cells, called
blastomeres.
64Cleavage in mammalian egg is
[CBSE AIPMT 2000]
(a) equal holoblastic
(b) unequal holoblastic
(c) superficial meroblastic
(d) discoidal meroblastic
Ans.(b)
The progressive division of cells during
embryonic growth is calledcleavage. It
is not accompanied by increase in the
overall size of embryo.
The first cleavage occurs in human
zygote about 30 hours after fertilisation.
It is holoblastic (complete) and gives rise
to the blastomeres-one of which is
slightly larger than the other.
65Blastopore is
[CBSE AIPMT 2000, 1992]
(a) opening of neural tube
(b) opening of gastrocoel
(c) future anterior end of embryo
(d) found in blastula
Ans.(b)
Blastopore is the opening of gastrocoel.
Gastrulation is a process of migration and
rearrangement of cells which are already
present in blastula, a new cavity is formed
named archenteron or gastrocoel which
open outside through blastopore.
66What is true about cleavage in
fertilised egg of human?
[CBSE AIPMT 1994]
(a) Meroblastic
(b) Starts when egg reaches uterus
(c) Starts in Fallopian tube
(d) It is identical to normal mitosis
Ans.(c)
In human, cleavage occurs in the
Fallopian tube (oviduct) during the
movement of zygote towards uterus.
67Acrosome reaction in sperm is
triggered by[CBSE AIPMT 1993]
(a) capacitation
(b) release of lysin
(c) influx ofNa
+
(d) release of fertilisin
Ans.(d)
Ovum secretes a chemical substance,
calledfertilisinwhich has number of
spermophilic sites on its surface where
the sperm of species specific type can
be bound by their antifertilisin site. This
fertilisin-antifertilisin reaction triggers
acrosome reaction The main aim of this
reaction is to thin out the number of
sperms to reduce the chances of
polyspermy.
68Extrusion of second polar body
from egg nucleus occurs
[CBSE AIPMT 1993]
(a) after entry of sperm but before
completion of fertilisation
(b) after completion of fertilisation
(c) before entry of sperm
(d) without any relation of sperm entry
Ans.(a)
The sperm entry stimulates the
secondary oocyte to undergo Meiotic II
division, which produces the ovum and
second polar body.
230 NEETChapterwise Topicwise Biology

241. 69Termination of gastrulation is
indicated by[CBSE AIPMT 1993]
(a) obliteration of blastocoel
(b) obliteration of archenteron
(c) closure of blastopore
(d) closure of neural tube
Ans.(a)
Termination of gastrulation is indicated
by obliteration of blastocoel.
70Meroblastic cleavage is a type of
division [CBSE AIPMT 1992]
(a) horizontal
(b) partial/parietal
(c) total
(d) spiral
Ans.(b)
In meroblastic cleavage zygote divides
partially. This type of cleavage is found in
mesolecithal and centrolecithal eggs like
eggs of reptiles, birds, insects and
egg-laying mammals.
71During cleavage, what is true about
cells? [CBSE AIPMT 1991]
(a) Nucleocytoplasmic ratio remains
unchanged
(b) Size does not increase
(c) There is less consumption of oxygen
(d) The division is like meiosis
Ans.(b)
Cleavage is the rapid mitotic cell division
of the zygote upto the completion of
blastula stage. Cleavage differs from the
ordinary mitosis in following respects:
(a) Interphase are short and do not
involve growth.
(b) Divisions are very rapid, nuclear
cytoplasmic ratio increases with
each division.
(c) There is a decrease in the size of
blastomeres (cells resulting from
cleavage), but size and shape of the
embryo do not change.
72How many sperms are formed from
a secondary spermatocyte?
[CBSE AIPMT 1990]
(a) 4 (b) 8
(c) 2 (d) 1
Ans.(c)
As it is clear from the figure that each
secondary spermatocyte, after second
meiosis give rise to two spermatids,
each of which develops into sperm.
73Which of the following secretes the
hormone, relaxin, during the later
phase of pregnancy?[NEET 2021]
(a) Graafian follicle (b) Corpus luteum
(c) Foetus (d) Uterus
Ans.(b)
Corpus luteum is formed in ovary after
the ovulation and degenerates if
pregnancy does not occur.
In later phase of pregnancy the corpus
luteum secretes relaxin hormone.Relaxin
dilates the cervix and helps in
parturition. Graafian follicle, uterus and
foetus has no role in relaxin secretion.
74Which of these is not an important
component of initiation of
parturition in humans?[NEET 2021]
(a) Increase in oestrogen and
progesterone ratio
(b) Synthesis of prostaglandins
(c) Release of oxytocin
(d) Release of prolactin
Ans.(d)
Prolactin hormone is not important for
initiation of parturition because it is
released after the parturition in order to
secrete the milk from the mammary
gland. Parturition is usually initiated by
the release of oxytocin hormone from
the maternal pituitary.
This hormone increases the
concentration of uterine muscles.
Prostaglandins are synthesised and
released in response to the oxytocin
hormone which induces stronger
contraction resulting in expulsion of baby.
During parturition there is an increase in
oestrogen and progesterone ratio in the
female body.
75In human beings, at the end of 12
weeks (first trimester) of
pregnancy, the following is
observed. [NEET (Oct.) 2020]
(a) Eyelids and eyelashes are formed
(b) Most of the major organ systems are
formed
(c) The head is covered with fine hair
(d) Movement of the foetus
Ans.(b)
In human beings, at the end of 12 weeks
(first trimester), most of the major organ
systems are formed, e.g. the limbs and
external genital organs gets
well-developed.
76Select the correct sequence of
events [NEET (Odisha) 2019]
(a) Gametogenesis→Gamete transfer
→Syngamy→Zygote→Cell division
(Cleavage)→Cell differentiation→
Organogenesis
(b) Gametogenesis→Gamete transfer
→Syngamy→Zygote→Cell divison
(Cleavage)→Organogenesis→Cell
differentiation
(c) Gametogenesis→Syngamy→
Gamete transfer→Zygote→Cell
division (Cleavage)→Cell
differentiation→Organogenesis
(d) Gametogenesis→Gamete transfer
→Syngamy→Zygote→Cell
differentiation→Cell division
(Cleavage)→Organogenesis
Ans.(a)
The correct sequence of events are
Gametogenesis (i.e. production of
gametes)→Gamete transfer (i.e.
movement of gamete at the site of
fertilisation)→Syngamy (i.e. the fusion
of gametes)→Zygote (i.e. a eukaryotic
cell formed by fertilisation of gametes)
→Cell division (cleavage)→Cell
differentiation (i.e the process where a
cell changes from one cell type to
another)→Organogenesis (i.e. the
process of formation of three germ
layers).
77Which of the following hormones is
responsible for both the milk ejection
reflex and the foetal ejection
reflex? [NEET (Odisha) 2019]
(a) Oestrogen
(b) Prolactin
(c) Oxytocin
(d) Relaxin
Human Reproduction 231
Germ cell (2 )n
Spermatogonia
Primary
spermatocyte (2 )n
Secondary
spermatocyte ( )n
Spermatid ( )n
Sperm
Spermatogenesis
Multiplicative
phase
Growth phase
Maturation
phase Meiosis -I
Meiosis -II
Embryogenesis, Parturition
and Lactation
TOPIC 4

242. Ans.(c)
Oxytocin hormone is responsible for
both, the milk ejection reflex and foetal
ejection reflex. It is a peptide hormone
normally produced in the hypothalamus
and released by the posterior pituitary
gland.
78Colostrum, the yellowish fluid,
secreted by mother during the
initial days of lactation is very
essential to impart immunity to the
new born infants because it
contains [NEET (National) 2019]
(a) monocytes
(b) macrophages
(c) immunoglobulin A
(d) natural killer cells
Ans.(c)
Colostrum, the yellowish fluid, secreted
by mother during the initial days of
lactation is very essential to impart
immunity to the newborn infants
because it contains immunoglobulin A.
The type of immunity provided by
colostrum is natural passive immunity.
As IgA is secreted in mother’s milk, it is
also called secretory immunoglobulin.
Monocytes are a type of white blood cell
having simple oval nucleus.
Macrophages are cells of the immune
system. These cells can engulf bacteria,
fungi, viruses and parasites.
Natural killer cells are lymphocytes and
are a component of innate immune
system.
79The amnion of mammalian embryo
is derived from [NEET 2018]
(a) mesoderm and trophoblast
(b) endoderm and mesoderm
(c) ectoderm and mesoderm
(d) ectoderm and endoderm
Ans.(c)
Amnion of mammalian embryo is derived
fromectodermandmesoderm. It is one
of the extraembryonic membrane which
is formed by the amniogenic cells of
ectodermal origin on inner side and
somatopleuric extraembryonic
mesoderm on outer side. This
membrane acts as a shock absorber for
the foetus, regulates foetal body
temperature and prevents desiccation.
The origin of other extraembryonic
membranes is as follows
ChorionTrophoectoderm and
mesoderm.
Allantois and Yolk sacOuter
mesoderm and inner endoderm.
80Hormones secreted by the
placenta to maintain pregnancy are
[NEET 2018]
(a) hCG, hPL, progestogens, estrogens
(b) hCG, hPL, estrogens, relaxin, oxytocin
(c) hCG, hPL, progestogens, prolactin
(d) hCG, progestogens, estrogens,
glucocorticoids
Ans.(a)
The hormones secreted by the placenta
to maintain pregnancy arehCG,hPL,
progestogensandestrogens. Placenta is
the intimate connection between the
foetus and uterine wall of the mother to
exchange the materials. It has endocrine
function and secretes the following
hormones
(i)Human Chorionic Gonadotropins
(hCG) It stimulates and maintains
the corpus luteum to secrete
progesterone until the end of
pregnancy.
(ii)Human Placental Lactogen(hPL) It
is also known Human Chorionic
Somatomam- motropin (HCS), it
stimulates the growth of mammary
glands during pregnancy.
(iii)Progesterone and estrogen
support foetal growth, maintain
pregnancy, inhibit uterine contractions,
etc.
On the other hand, the sources of other
hormones are as follows
OxytocinSecreted by posterior lobe of
pituitary gland during foetal ejection
reflex.GlucocorticoidSecreted by
adrenal gland of foetus to induce foetal
ejection reflex.
RelaxinSecreted by corpus luteum to
increase flexibility of pubis symphysis.
ProlactinSecreted by anterior lobe of
pituitary, helps in the secretion of milk.
81Match column I with column II and
select the correct option using the
codes given below.
[NEET 2016, Phase II]
Column I Column II
A. Mons pubis 1. Embryo formation
B. Antrum 2. Sperm
C. Trophecto
derm
3. Female external
genitalia
D. Nebenkern 4. Graafian follicle
Codes
A B C D
(a) 3 4 2 1
(b) 3 4 1 2
(c) 3 1 4 2
(d) 1 4 3 2
Ans.(b)
The correct match are
(a) Mona pubis–Femaleexternal genitalia
(b) Antrum–Graafian follicle
(c) Trophectoderm–Embryo
development
(d) Nebenkern–Sperm
Concept EnhancerNebenkern is a
mitochondrial structure present in the
sperm of certain insects.
82Several hormones like hCG, hPL,
oestrogen, progesterone are
produced by[NEET 2016, Phase II]
(a) ovary (b) placenta
(c) Fallopian tube (d) pituitary
Ans.(b)
Several hormones like–hCG, hPL,
oestrogen, progesterone are produced
by placenta. It is a structural and
functional connectivity between the
developing embryo (foetus) and the
maternal body. It is connected to embryo
through an umblical cord which helps in
transport of substances to and from the
embryo. Placenta also acts as an
endocrine tissue by producing the above
mentioned hormones.
83Which one of the following is not
the function of placenta? It
[NEET 2013]
(a) facilitates supply of oxygen and
nutrients to embryo
(b) secretes oestrogen
(c) facilitates removal of carbon dioxide
and waste material from embryo
(d) secretes oxytocin during parturition
Ans.(d)
Pituitary secretes oxytocin during
parturition. The functions of placenta
are supply of oxygen and nutrients to
embryo. It aslo secretes estrogen,
facilitates removal of carbon dioxide and
waste materials from embryo.
84Signals for parturition originate
from [CBSE AIPMT 2012, 10]
(a) both placenta as well as fully
developed foetus
(b) oxytocin released from maternal
pituitary
(c) placenta only
(d) fully developed foetus only
Ans.(a)
The process of delivery of the foetus
(childbirth) is called parturition which is
induced by a complex neuroendocrine
mechanism.
232 NEETChapterwise Topicwise Biology

243. The signals for parturition originate from
the fully developed foetus and the
placenta which induce mild uterine
contractions calledfoetal ejection
reflex.
This triggers release of oxytocin from
maternal pituitary. Oxytocin causes
stronger uterine contractions which in
turn stimulate further secretion of
oxytocin. The stimulatory reflex between
the uterine contraction and oxytocin
secretion continues resulting in stronger
and stronger contractions. This leads to
expulsion of the baby out of the uterus
through the birth canal, i.e. parturition.
85The first movements of the foetus
and appearance of hair on its head
are usually observed during which
month of pregancy?
[CBSE AIPMT 2010]
(a) Fourth month (b) Fifth month
(c) Sixth month (d) Third month
Ans.(b)
During development of foetus in human
by week 20, hair begin to grow including
eyebrows and eyelashes, fingerprints
develop. Fingernails and toe nails grow.
Firm hand grip. Between 16 and 20
weeks the first movements of the
foetus are observed.
86Which extra-embryonic membrane
in humans prevents desiccation of
the embryo inside the uterus?
[CBSE AIPMT 2008]
(a) Chorion (b) Allantois
(c) Yolk sac (d) Amnion
Ans.(d)
Amnion is an extra embryonic membrane
that surrounds embryo in reptiles, birds
and mammals. It provides a kind of
private aquarium to the embryo and
protects it from mechanical shock and
desiccation.
Chorion (serosa) is the outermost extra
embryonic membrane in reptiles, birds
and mammals. It surrounds the whole
embryonic system of embryo.
Yolk sac contains yolk in reptiles and
birds. In mammals yolk sac is also know
umbilical vesicle. It is connected to
enteron of embryo by a slender yolk stalk.
87During embryonic development,
the establishment of polarity along
anterior/ posterior, dorsal/ventral
or medial/lateral axis is called
[CBSE AIPMT 2003]
(a) anamorphosis
(b) pattern formation
(c) organiser phenomena
(d) axis formation
Ans.(d)
Embryonic axis are formed very early in
development and sometimes by the
polarity of the egg.
88Extra-embryonic membranes of
the mammalian embryo are derived
from [CBSE AIPMT 1994]
(a) inner cell mass
(b) trophoblast
(c) formative cells
(d) follicle cells
Ans.(b)
Extra-embryonic membranes are
formed outside the embryo from the
trophoblast in amniotes (reptiles, birds
and mammals) and perform specific
function. These are yolk sac, amnion,
allantois and chorion.
89Gonads develop from embryonic
[CBSE AIPMT 1990]
(a) ectoderm (b) endoderm
(c) mesoderm (d) Both (b) and (c)
Ans.(c)
Gonads develop from mesoderm. Beside
gonads mesoderm also forms muscles,
connective tissue, dermis of skin, bones
and cartilages, peritoneal layers,
coelom, circulatory system (heart, blood
vessels, blood, lymphatic system),
kidneys, ureters and adrenal cortex.
90Cells become variable in
morphology and function in
different regions of the embryo.
The process is[CBSE AIPMT 1989]
(a) differentiation
(b) metamorphosis
(c) organisation
(d) rearrangement
Ans.(a)
After formation of three primary germ
layers (i.e., ectoderm, mesoderm and
endoderm), cells of these three layers
become variable in morphology, shape,
size and more specified to form organs
so as to meet out the future functional
needs of the foetus, this process is
called differentiation.
Human Reproduction 233

244. 01Which one of the following is an
example of hormone releasing IUD?
[NEET 2021]
(a) Cu-T (b) LNG-20
(c) Cu-7 (d) Multiload-375
Ans.(b)
Levonorgestrel hormone is released
from LNG-20. It is highly effective for
contraception. The risk of unwanted
pregnancy is lower with LNG-20 because
it causes endometrial atrophy and alter
the stroma to inhibit the process of
implantation.
Lippes loop is an IUD impregnated with
barium sulphate. IUDs like CuT and
multiload-375 release copper that
suppress the motility of sperm thus
reducing its fertilising capacity. Thus,
Lippes loop, CuT and multiload-375 are
not hormone releasing IUDs.
02Match the List-I with List-II.
List I List II
A. Vaults 1. Entry of sperm through
cervix is blocked
B. IUDs 2. Removal of vas
deferens
C. Vasectomy 3. Phagocytosis of sperms
within the uterus
D. Tubectomy 4. Removal of Fallopian
tube
Choose the correct answer from
the options given below.
[NEET 2021]
A B C D
(a) 4 2 1 3
(b) 1 3 2 4
(c) 2 4 3 1
(d) 3 1 4 2
Ans.(b)
(A)-(1),(B)-(3),(C)-(2),(D)-(4)
Vaultsis a rubber dome, which fits over
the vaginal vault or cervix. They prevents
the entry of sperm into the uterus.
IUDsare the devices which are inserted
by the doctors or trained nurses into the
uterus of female. These IUDs increase
phagocytosis of sperm in uterus.
VasectomySurgical removal of the small
part of vas deferens. It is a form of male
birth control that inhibits to transport of
male gamete.
TubectomySurgical removal of the small
part of Fallopian tube to prevent the
sperm from reaching the egg to fertilise it.
03Progestogens alone or in
combination with oestrogens can
be used as a contraceptive in the
form of [NEET (Oct.) 2020]
(a) implants only
(b) injections only
(c) pills, injections and implants
(d) pills only
Ans.(c)
Progesteronse alone or in combination
with oestrogens can be used as a
contraceptive in the form of pills,
injections and implants under the skin.
They inhibit ovulation and implantation
of the zygote as well as alter the quality
of cervical mucus to prevent/retard
entry of sperms.
04Which of the following is a correct
statement?[NEET (Odisha) 2019]
(a) IUDs once inserted need not be
replaced
(b) IUDs are generally inserted by the
user herself
(c) IUDs increase phagocytosis of
sperms in the uterus
(d) IUDs suppress gametogenesis
Ans.(c)
Option (c) is correct as Intrauterine
Devices (IUDs) increase phagocytosis of
sperms within the uterus and the Cu ions
released suppress sperm motility and
fertilising capacity of sperms.
Other statements can be corrected as
IUDs can be removed as these are a
reversible contraception method.
IUDs are inserted by doctors or expert
nurses in the uterus through vagina.
IUDs do not affect gametogenesis.
05Select the hormone- releasing
Intra-Uterine Devices.
[NEET (National) 2019]
(a) Multiload 375, Progestasert
(b) Progestasert, LNG-20
(c) Lippes Loop, Multiload 375
(d) Vaults, LNG-20
Ans.(b)
Progestasert and LNG-20 are hormone
releasing Intrauterine Devices (IUDs).
These devices release small quantities
of hormone which suppresses
endometrial changes, cause an ovulation
and insufficient luteal activity.
06Which of the following
contraceptive methods do involve
a role of hormone?
[NEET (National) 2019]
(a) Barrier method, Lactational
amenorrhea, Pills
(b) Cu-T, Pills, Emergency
contraceptives
(c) Pills, Emergency contraceptives,
Barrier methods
(d) Lactational amenorrhea, Pills,
Emergency contraceptives
Ans.(d)
Lactational amenorrhea, pills and
emergency contraceptives provide
contraception due to the role of hormones.
ReproductiveHealth
26
Birth Control Measures
and Amniocentesis
TOPIC 1

245. Reproductive Health 235
In lactational amenorrhoea, high
prolactin level during active lactation
period decreases the gonadotropin level
in the blood.
Pills usually contain progesterone or
progesterogen- oestrogen combinations
which prevent ovulation. Emergency
contraceptives also contain
progesterone and estradiol preparation.
Other contraceptive methods include
Intrauterine Devices (IUDs) which release
copper and destroys the sperms. On the
other hand, barrier method represents a
physical method of contraception. Also,
copper-T (Cu-T) acts by releasing copper
and not any hormone.
07The function of copper ions in
copper releasing IUD’s is
[NEET 2017]
(a) they suppress sperm motility and
fertilising capacity of sperms
(b) they inhibit gametogenesis
(c) they make uterus unsuitable for
implantation
(d) they inhibit ovulation
Ans.(a)
An intra uterine device is a small, often
T-shaped birth control device that is
inserted into a woman’s uterus to
prevent pregnancy. The copper
releasing IUD’s primarily work by
disrupting sperm motility and damaging
sperm. Copper acts as a spermicide. It
can also alter the endometrial lining,
preventing implantation.
08Which of the following approaches
does not give the defined action of
contraceptive?[NEET 2016, Phase I]
(a) Intra uterine
devices
Increase
phagocytosis of
sperms, suppress
sperm motility
and fertilising
capacity of
sperms
(b) Hormonal
contraceptives
Prevent/ retard
entry of sperms,
prevent ovulation
and fertilisation
(c) Vasectomy Prevents
spermatogenesis
(d) Barrier
methods
Prevent
fertilisation
Ans.(c)
In vasectomy, a small part of the vas
deferens is removed or tied up through a
small incision on the scrotum in males.
Vasectomy blocks the gamete transport
and does not affect spermatogenesis.
09In context of amniocentesis, which
of the following statement is
incorrect? [NEET 2016, Phase I]
(a) It is used for prenatal
sex-determination
(b) It can be used for detection of down
syndrome
(c) It can be used for detection of cleft
palate
(d) It is usually done when a woman is
between 14-16 weeks pregnant
Ans.(c)
Cleft palate is a developmental
abnormality which may occur in the
developing foetus and so it can be
detected by sonography, not by
amniocentesis.
Amniocentesis is being misused for
foetal sex-determination test so it is
banned in India.
10Which of the following is incorrect
regarding vasectomy?
[NEET 2016, Phase II]
(a) No sperm occurs in seminal fluid
(b) No sperm occurs in epididymis
(c) Vasa deferentia is cut and tied
(d) Irreversible sterility
Ans.(b)
Epididymis is an accessory duct present
in the male reproductive system.
Sperms produced by testes are sent
here by vasa efferentia and are
temporarily stored here. These sperms
do not reach to seminal vesicle when vas
deferens is cut and tied in vasectomy.
11Which of the following is
hormone-releasing IUD?
[NEET 2016, Phase II]
(a) LNG-20 (b) Multiload-375
(c) Lippes loop (d) Cu-7
Ans.(a)
Hormone releasing IUD (Intra Uterine
Devices) is LNG-20. The IUD’s are ideal
contraceptive methods used by females
to prevent pregnancy. The hormone
releasing IUD’s make the uterus
unsuitable for implantation and the
cervix hostile to the sperms. Hence,
option (a) is correct.
Concept EnhancerLippes loop is a
non-medicated IUD whereas Cu-7 and
Multiload-375 are copper releasing IUDs.
12Tubectomy is a method of
sterilisation in which
[CBSE AIPMT 2014]
(a) small part of the Fallopian tube is
removed or tied up
(b) ovaries are removed surgically
(c) small part of vas deferens is removed
or tied up
(d) uterus is removed surgically
Ans.(a)
Tubectomy is a type of sterilisation.
During this procedure female’s, Fallopian
tubes are cut and tied up to block entry
of sperms into the ovary. In females, this
is permanent birth control.
13Which of the following is a
hormone releasing Intrauterine
Device (IUD)?[CBSE AIPMT 2014]
(a) Multiload 375 (b) LNG-20
(c) Cervical cap (d) Vault
Ans.(b)
LNG-20 is a hormone releasing IUD, that
makes the uterus unsuitable for
implantation and cervix hostile to sperm.
Intra uterine devices are plastic or metal
objects which are inserted by the
doctors into the uterus through vagina.
These are available as:
Non-medicated IUDs (i.e. lippes loop)
Copper relasing IUDs (e.g. multiload 375)
Hormone releasing (e.g. progestasert).
Cervical caps and vaults are made up of
rubber and are inserted into the female
reproductive tract to cover the cervix
before coitus. They prevent conception.
14Which of the following cannot be
detected in a developing foetus by
amniocentesis? [NEET 2013]
(a) Klinefelter’s syndrome
(b) Sex of the foetus
(c) Down’s syndrome
(d) Jaundice
Ans.(d)
Amniocentesis is a foetal sex
determination test based on the
chromosomal pattern in the amniotic
fluid surrounding the developing
embryo. Jaundice is a condition not
based on chromosomal pattern. It is a
disease related to liver dysfunctions.
Vas Deferens
cut and tied
Vasectomy

246. 15One of the legal methods of birth
control is [NEET 2013]
(a) abortion by taking an appropriate
medicine
(b) by abstaining from coitus from day
10-17 of the menstrual cycle
(c) by having coitus at the time of day
break
(d) by a premature ejaculation during
coitus
Ans.(b)
One of the legal method of birth control
is periodic abstinence in which couples
abstain from coitus from day 10 to 17 of
the menstrual cycle. Abortion by taking
medicine is not a legal method. A day
break coitus may increase the chances
of contraception.
16What is the figure given below
showing in particular?
[CBSE AIPMT 2012]
(a) Ovarian cancer (b) Uterine cancer
(c) Tubectomy (d) Vasectomy
Ans.(c)
The figure depicts tubectomy The
process of cutting and ligating both the
oviducts or Fallopian tubes of female is
calledtubectomy. It is very reliable
method of birth control, the approximate
failure rate of which is less than 1%.
17Which one of the following is the
most widely accepted method of
contraception in India, at present?
[CBSE AIPMT 2011]
(a) Cervical caps
(b) Tubectomy
(c) Diaphragms
(d) IUDs (Intra Uterine Devices)
Ans.(d)
At present, the most widely accepted
method of contraception in India is IUDs
(Intra Uterine Devices). These devices
are effective and popular. These devices
are inserted by doctors and expert
nurses in the uterus through vagina.
18The permissible use of the
technique amniocentesis is for
[CBSE AIPMT 2010]
(a) detecting sex of the unborn foetus
(b) artificial insemination
(c) transfer of embryo into the uterus of
a surrogate mother
(d) detecting any genetic abnormality
Ans.(d)
Amniocentesis is a technique for the
diagnosis of congenital abnormalities
before birth. By karyotypic studies of
somatic cells, abnormalities due to the
changes in chromosome number like
Down’s syndrome, Turner’s syndrome,
Klinefelter’s syndrome, etc, can be
determined.
It can be used as a pre-natal diagnostic
technique. These days, amniocentesis is
being misused also. Mothers even get their
normal foetus aborted if it is a female.
19Cu ions released from
copper-releasing Intra Uterine
Devices (IUDs)
[CBSE AIPMT 2010, 2000]
(a) make uterus unsuitable for
implantation
(b) increase phagocytosis of sperms
(c) suppress sperm motility
(d) prevent ovulation
Ans.(c)
Intra Uterine Device (IUD) is a small
device made up of copper, plastic or
stainless steel. It is inserted by a doctor
and experts nurses into the uterus
through vagina and can be left these for
long periods. This device needs to be
replaced after 3-5 yrs when Cu release
become scanty due to calcium
deposition. It supress the sperm mobility
and fertilising capacity of sperms.
20Consider the statement given
below regarding contraception and
answer as directed thereafter
[CBSE AIPMT 2008]
1. Medical Termination of
Pregnancy (MTP) during first
trimester is generally safe.
2. Generally chances of conception
are nil until mother breast-feeds
the infant up to two years.
3. Intra uterine devices like
copper-T are effective
contraceptives.
4. Contraception pills may be taken
upto one week after coitus to
prevent contraception.
Which two of the above
statements are correct?
(a) 1, 3 (b) 1, 2
(c) 2, 3 (d) 3, 4
Ans.(a)
Option I and III are correct as. Intra
uterine devices like copper-T are
effective contraceptives for birth
control. It suppresses sperm motility and
the fertilising capacity of the sperm.
Medical termination of pregnancy or
induced abortion is voluntary or
intentional termination of pregnancy
before full term of foetus. It is
comparatively safe up to 12 weeks (the
first trimester) of pregnancy.
21Given below are four methods
(A-D) and their modes of action
(i-iv) in achieving contraception.
Select their correct matching from
the four option that follow.
[CBSE AIPMT 2008]
Method Mode of Action
A. The pill (i) Prevents sperms
reaching cervix
B. Condom (ii) Prevents
implantation
C. Vasectomy (iii) Prevents
ovulation
D. Copper-T (iv) Semen contains
no sperms
A B C D
(a) (iii) (iv) (i) (ii)
(b) (ii) (iii) (i) (iv)
(c) (iii) (i) (iv) (ii)
(d) (iv) (i) (ii) (iii)
Ans.(c)
The contraceptive pills are hormones
either in combination or progesterone
only that primarily prevent release of
egg. It is convenient and highly effective,
significant non-contraceptive health
benefits such as protection against
ovarian and endometrial cancers.
Condom is thin rubber sheath for penis
that collects semen. It is easy to use,
effective and inexpensive.
Vasectomy is the cutting and tying off
the ductus deference so, that sperm
connot enter the ejaculate.
Copper-T is small plastic device placed
in the uterus that prevent fertilisation or
implantation.
22In a population, unrestricted
reproductive capacity is called
[CBSE AIPMT 2002]
(a) biotic potential
(b) fertility
(c) carrying capacity
(d) birth rate
236 NEETChapterwise Topicwise Biology

247. Ans.(a)
The biotic potential or innate capacity to
increase of a population refers to the
maximum rate of increase in the
population that can possibly occur under
ideal conditions (unlimited resources, no
hindrances).
23Progesterone, which is the most
important component of oral
contraceptive pils, prevents
pregnancy by[CBSE AIPMT 2000]
(a) preventing the formation of egg
(b) preventing the cleavage of the
fertilised egg
(c) creating unfavourable chemical
environment for the sperms to
survive in the famale reproductive
tract
(d) blocking ovulation
Ans.(d)
Most contraceptive pills contain the
hormones oestrogen and progesterone.
Progesterone, alongwith oestrogen,
disturbs the normal menstrual cycle to
prevent ovulation. It inhibits the pituitary
from secreting FSH and LH, which leads
to blocking of ovulation.
24Tablets to prevent contraception
contain [CBSE AIPMT 1999]
(a) progesterone (b) FSH
(c) LH (d) Both (b) and (c)
Ans.(a)
Contraceptive pills for women contain
female sex hormones oestrogen and
progesterone. These prevent
development of eggs and ovulation by
inhibiting secretion of FSH.
Some pills contain progesterone only in
such cases, ovulation may occur but
cervical mucus is thickness preventing
the entry of sperm.
25Amniocentesis is a process to
[CBSE AIPMT 1997]
(a) determine any disease in heart
(b) determine any hereditary disease in
the embryo
(c) know about the disease of brain
(d) All of the above
Ans.(b)
Cells of amniotic fluid are cultured to
increase their number and to obtain
dividing cells. The chromosomes of the
dividing cells can be examined for
abnormalities.
The cells and the fluid sample may also
be tested for products of faulty genes
(e.g. high level ofα-fetiprotein, i.e. AFP)
indicated high risk of birth defects.
26Human population growth in India
[CBSE AIPMT 1996]
(a) tends to follow a sigmoid curve as in
case of many other animal species
(b) tends to reach zero population
growth as in case of some animal
species
(c) can be reduced by permitting natural
calamities and enforcing birth control
measures
(d) can be regulated by following the
National programme of family
planning
Ans.(d)
Though India is very crowded and over
populated country but still human
population can be regulated by following
the National programme of family
planning.
27In India, human population is
heavily weighed towards the
younger age groups as a result of
[CBSE AIPMT 1995]
(a) short life span of many individuals
and low birth rate
(b) long life span of many individuals and
low birth rate
(c) short life span of many individuals
and high birth rate
(d) long life span of many individuals and
high birth rate
Ans.(c)
In India human population is heavily
weighed towards the younger age group
as a result of high birth rate and short
life span of individuals.
28Veneral diseases can spread
through [NEET 2021]
I. Using sterile needles.
II. Transfusion of blood from
infected person.
III. Infected mother to foetus.
IV. Kissing.
V. Inheritance.
Choose the correct answer from
the options given below
(a) I, II and III (b) II, III and IV
(c) II and III (d) I and III
Ans.(c)
Veneral disease or sexually transmitted
disease is defined as a medical condition
that can be passed from one person to
anotherviasexual contact, transfusion of
blood from infected person and by
infected mother to foetus.
Thus, option (c) is correct.
29Which of the following STDs are
not curable?[NEET (Oct.) 2020]
(a) Genital herpes, hepatitis-B, HIV
infection
(b) Chlamydiasis, Syphilis, genital warts
(c) HIV, gonorrhoea, trichomoniasis
(d) Gonorrhoea, trichomoniasis,
hepatitis-B
Ans.(a)
Except for hepatitis-B, genital herpes
and HIV infections other sexually
transmitted diseases are completely
curable if detected early and treated
properly. Diseases or infections which
are transmitted through sexual
intercourse are collectively called
Sexually Transmitted Diseases (STDs) or
Venereal Diseased (VD) or Reproductive
Tract Infections (RTI).
30Select the option including all
sexually transmitted diseases.
[NEET (Sep.) 2020]
(a) Gonorrhoea, Malaria, Genital herpes
(b) AIDS, Malaria, Filaria
(c) Cancer, AIDS, Syphilis
(d) Gonorrhoea, Syphilis, Genital herpes
Ans.(d)
Gonorrhoea, syphilis, genital herpes are
sexually transmitted diseases.
Gonorrhoea is caused by a bacterium
Neisseria gonorrhoeae. It is a sexually
transmitted disease (STD) that can infect
both men and women. It can cause
infections in the genitals, rectum and
throat.
Syphilis is caused by a bacterium
Treponema pallidum.It starts as painless
sore typically on the genitals, rectum or
mouth. Genital herpes is caused by a
virus Type-II Herpes simplex virus.
It causes herpetic sores, which are
painful bilsters (fluid-filled bumps) that
can break open and ooze fluid.
Reproductive Health 237
Sexually Transmitted
Diseases
TOPIC 2

248. 238 NEETChapterwise Topicwise Biology
31Which of the following sexually
transmitted diseases do not
specifically affect reproductive
organs? [NEET (Odisha) 2019]
(a) Genital warts and Hepatitis-B
(b) Syphilis and Genital herpes
(c) AIDS and Hepatitis-B
(d) Chlamydiasis and AIDS
Ans.(c)
AIDS and Hepatitis-B are sexually
transmitted diseases which do not
specifically affect reproductive organs.
AIDS affects the overall immune system
of the individual and Hepatitis-B affects
the liver. These are called STD because
these spread through unsafe sex or
unprotected sex.
32Match the following sexually
transmitted diseases (column I)
with their causative agent (column II)
and select the correct option.
[NEET 2017]
Column I Column II
(A) Gonorrhea 1. HIV
(B) Syphilis 2. Neisseria
(C) Genital Warts 3.Treponema
(D) AIDS 4. Human
Papilloma Virus
Codes
A B C D
(a) 2 3 4 1
(b) 3 4 1 2
(c) 4 2 3 1
(d) 4 3 2 1
Ans.(a)
Column I Column II
(A) Gonorrhea 1. HIV
(B) Syphilis 2. Neisseria
(C) Genital Warts 3. Treponema
(D) AIDS 4. Human
Papilloma Virus
33In which of the following
techniques, the embryos are
transferred to assist those females
who cannot conceive?
[NEET (Sep.) 2020]
(a) GIFT and ZIFT
(b) ICSI and ZIFT
(c) GIFT and ICSI
(d) ZIFT and IUT
Ans.(d)
Option (d) is the correct answer because
the techniques by which the embryos are
transferred to assist those females who
cannot conceive are ZIFT and IUT, i.e.
Zygote Intra Fallopian Transfer and Intra
Uterine Transfer respectively, both are
Embryo Transfer (ET) methods. Option
(a), (b) and (c) are incorrect because in
GIFT (Gamete Intra Fallopian Transfer),
gamete is transferred into the Fallopian
tube of female who cannot produce ova.
ICSI is Intra Cytoplasmic Sperm Injection
in which sperm is directly injected into
the ovum.
34In case of a couple, where the male
is having a very low sperm count,
which technique will be suitable for
fertilisation? [NEET 2017]
(a) Intrauterine Transfer
(b) Gamete Intracytoplasmic Fallopian
Transfer
(c) Artificial Insemination
(d) Intracytoplasmic Sperm Injection
Ans.(c)
Assisted Reproductive Technologies
(ARTs) is a general term referred to the
method used to achieve pregnancy by
artificial means or partial artificial means
and is primarily used in infertility
treatment. Artificial insemination is a
type of ARTs.
35Embryo with more than 16
blastomeres formed due toin vitro
fertilisation is transferred into
[NEET 2016, Phase II]
(a) uterus (b) Fallopian tube
(c) fimbriae (d) cervix
Ans.(a)
Embryo with more than 16 blastomeres
formed due toin vitrofertilisation is
transferred into the uterus to complete
its further development.
Concept EnhancerIntra Uterine
Transfer or IUT is an assisted
reproductive technology to solve the
infertility problems.
In this process, the implantation of
embryo takes place in the uterus where
it develops into a foetus which forms a
child. On completion of gestation, the
mother will give birth to a normal child.
36A childless couple can be assisted
to have a child through a technique
called GIFT. The full form of this
technique is[CBSE AIPMT 2015]
(a) Gamete Inseminated Fallopian
Transfer
(b) Gamete Intra Fallopian Transfer
(c) Gamete Internal Fertilisation and
Transfer
(d) Germ Cell Internal Fallopian Transfer
Ans.(b)
The full form of GIFT is “Gamete Intra
Fallopian Transfer”.
This method is used in females who
cannot produce ova but can provide
suitable environment for fertilisation and
further development of embryo in the
oviducts. In such cases, ovum from the
donar female is surgically removed and
is then introduced into the Fallopian
tube of such females. Such women then
accept sperms from her husband during
copulation.
37Assisted reproductive technology,
IVF involves transfer of
[CBSE AIPMT 2014]
(a) ovum into the Fallopian tube
(b) zygote into the Fallopian tube
(c) zygote into the uterus
(d) embryo with 16 balastomeres into the
Fallopian tube
Ans.(b)
In vitrofertilisation is a process in which
an egg is fertilised by sperm outside the
woman’s womb and the zygote up to 8
celled stage is implanted into the Fallopian
tube. IVF is a major treatment for
infertility. The process involves
hormonally controlled ovulatory process,
removing ova from the woman’s ovaries
and letting sperm fertilise them in fluid
medium.
The Zygote or the embryo up to 8-celled
stage is then transferred into the
patient’s Fallopian tube with the intent to
be successful. When the zygote is more
than 8 blastomere stage it is placed directly
into the uterus to establish pregnancy.
38Artificial insemination means
[NEET 2013]
(a) transfer of sperms of a healthy donor
to a test-tube containing ova
(b) transfer of sperms of husband to a
test-tube containing ova
(c) artificial introduction of sperms of a
healthy donor into the vagina
(d) introduction of sperms of healthy
donor directly into the ovary
Infertility and Assisted
Reproductive Technologies
TOPIC 3

249. Reproductive Health 239
Ans.(c)
Artificial Insemination (AI) means
artificial introduction of sperms of a
healthy male donor into the vagina of
female. Infertility due to the inability of
the male to inseminate the female or due
to very low sperm counts in the
ejaculates can be corrected by artificial
insemination.
39The test-tube baby programme
employs which one of the following
techniques? [CBSE AIPMT 2012]
(a) Intra cytoplasmic Sperm Injection (ICSI)
(b) Intra Uterine Insemination (IUI)
(c) Gamete Intra Fallopian Transfer (GIFT)
(d) Zygote Intra Fallopian Transfer (ZIFT)
Ans.(d)
The test-tube baby programme employs
the technique ofin vitrofertilisation (IVF)
and Zygote Intra Fallopian Transfer
(ZIFT) technique. ZIFT is a method used
to treat infertility in which an egg
fertilisedin vitro(outside the body) is
placed into a women’s Fallopian tube
(oviduct).
It is an assisted reproductive procedure
similar to IVF and embryo transfer, the
difference being that zygote proembryo
is transferred into the Fallopian tube
instead of the uterus. Because the
fertilised egg (zygote) is directly
transferred into the tubes the procedure
is also referred to astubal embryo
transfer.
40Medical Termination of Pregnancy
(MTP) is considered safe up to how
many weeks of pregnancy?
[CBSE AIPMT 2011]
(a) Eight weeks (b) Twelve weeks
(c) Eighteen weeks (d) Six weeks
Ans.(b)
Intentional or voluntary termination of
pregnancy before full term is called
Medical Termination of Pregnancy (MTP)
or induced abortion. MTPs are considered
relatively safe during the first trimester,
i.e. upto 12 weeks of pregnancy.
Second trimester abortions are much
more riskier.
41In vitrofertilisation is a technique
that involves transfer of which one
of the following into the Fallopian
tube? [CBSE AIPMT 2010]
(a) Embryo only, up to 8 cell stage
(b) Either zygote or early embryo up to 8
cell stage
(c) Embryo of 32 cell stage
(d) Zygote only
Ans.(b)
In vitrofertilisation (IVF) or test-tube
baby technique involves fertilising one or
more eggs outside the body and then
transferring the fertilised eggs known as
pre-embryos back into the uterus.
Zygote Intra Fallopian Transfer (ZIFT) is
an example of IVF. In this, the zygote or
early embryos up to 8 blastomeres are
transferred into the Fallopian tube. If the
embryo has more than 8 blastomeres
then it is transferred into uterus this is
known as IUT.
42Certain characteristic demographic
features of developing countries
are [CBSE AIPMT 2004]
(a) high fertility, low or rapidly falling
mortality rate, rapid population
growth and a very young age distribution
(b) high fertility, high density rapidly
rising mortality rate and a very young
age distribution
(c) high infant mortality, low fertility
uneven population growth and a very
young age distribution
(d) high mortality high density uneven
population growth and a very old age
distribution
Ans.(a)
The characteristic demographic
features of developing countries is high
fertility, low or rapidly falling mortality
rate, rapid population growth and a very
young age distribution.
In India it is all due to the reproductive
and child health care programmes
operating throughout the country
43Test-tube baby means a baby born
when [CBSE AIPMT 2003]
(a) the ovum is fertilised externally and
there after implanted in the uterus
(b) it develops from a non-fertilised egg
(c) it is developed in a test-tube
(d) it is developed through tissue culture
method
Ans.(a)
Patrick Stepote andRobert Edwardfirst
time developed test-tube baby
technique in 1978. In certain cases,
where normal fertilisation is not
possible, ovum from the female and the
sperm from the male are matedin vitro.
The zygote, later on is implanted in the
uterus where further development into
baby take place.
44Test-tube baby is one who
[CBSE AIPMT 1996]
(a) is born out of artificial insemination
(b) has undergone development in a
test-tube
(c) is born out of the technique of
fertilisationin vitro
(d) has been developed without
fertilisation
Ans.(c)
Test-tube baby technique is a recent
solution to infertility. In this techniquein
vitrofertilisation is done in which after
viewing through a pencil thin laproscope
a ripe egg is removed from a woman
ovary.
The egg is kept in laboratory culture dish
and mixed with sperm from future
father.
The fertilised egg undergoes cleavage in
the laboratory dish and when it reaches
the 8-celled stage, it is transferred into
the mothers uterus for implantation. A
normal baby can be born to such a
mother.

250. 58.
01The production of gametes by the
parents, formation of zygote, theF
1
andF
2
plants, can be understood
from a diagram called[NEET 2021]
(a) Bullet square (b) Punch square
(c) Punnett square (d) Net squ are
Ans.(c)
Punnett square is a tool that helps to
show all possible allelic combinations of
gametes in a cross of parents with
known genotypes in order to predict the
probability of their offspring possessing
certain sets of alleles. For a cross
involving two genes, a Punnett square is
still a good strategy.
02The number of contrasting
characters studied by Mendel for his
experiments was[NEET (Oct.) 2020]
(a) 14 (b) 4 (c) 2 (d) 7
Ans.(d)
Mendel conducted breeding experiments
on garden pea by selecting seven pairs
of contrasting characters. Luckily all
these characters were related as
dominant and recessive and none of
them showed linakge. The seven pairs of
contrasting characters in pea plant were
Characters of pea plant
Character
Contrasting Trait
Dominant Recessive
Stem height Tall Dwarf
Flower colour Violet White
Flower position Axial Terminal
Pod shape Full Constricted
Pod colour Green Yellow
Seed shape Round Wrinkled
Seed colour Yellow Green
03How many true breeding pea plant
varieties did Mendel select as pairs,
which were similar except in one
character with contrasting traits?
[NEET (Sep.) 2020]
(a) 2 (b) 14 (c) 8 (d) 4
Ans.(b)
Mendel (father of genetics) selected 14
true-breeding pea plant varieties, in
pairs, which were similar except for one
character with contrasting traits. A true
breeding line refers to the plant that has
undergone continuous self-pollination
and showed stable trait inheritance and
expression for several generations.
04Identify the wrong statement with
reference to the gene ‘I’ that
controls ABO blood groups.
[NEET (Sep.) 2020]
(a) A person will have only two of the
three alleles
(b) WhenI
A
andI
B
are present together,
they express same type of sugar
(c) Allele ‘i’ does not produce any sugar
(d) The gene (I) has three alleles
Ans.(b)
WhenI
A
andI
B
are present together, they
express same type of sugar is wrong
statement with reference to the gene ‘I’
that controls ABO blood group because
I
A
and I
B
are completely dominant overI
O
,
but whenI
A
andI
B
are present together,
they both express their own types of sugar
and thus behaving as codominant alleles.
05The production of gametes by the
parents, the formation of zygotes, the
F
1
andF
2
plants, can be understood
using [NEET (Odisha) 2019]
(a) piediagram (b) apyramiddiagram
(c) Punnett square (d) Venn diagram
Ans.(c)
The production of gametes by the
parents, the formation of zygotes, the
F
1
andF
2
plants can be understood
from a diagram called Punnett square.
It was developed by a British
geneticist, Reginald C. Punnett. It is a
graphical representation to calculate
the probability of all possible genotypes
of offspring in a genetic cross.
06In a marriage between male with
blood group A and female with
blood group B, the progeny had
either blood group AB or B. What
could be the possible genotype of
parents? [NEET (Odisha) 2019]
(a)I i
A
(Male) :I I
B B
(Female)
(b)I I
A A
(Male) :I I
B B
(Female)
(c)I I
A A
(Male) :I i
B
(Female)
(d)I i
A
(Male) :I i
B
(Female)
Ans.(a)
The possible genotype of parents having
progeny with either blood group AB or B
isI i
A
(male):I I
B B
(female)
07InAntirrhinum(Snapdragon), a red
flower was crossed with a white
flower and inF
1
-generation, pink
flowers were obtained.
When pink flowers were selfed, the
F
2
-generation showed white, red
and pink flowers.Choose the
PrinciplesofInheritance
andVariation
27
Mendelism
TOPIC 1
I
A I
B
I I
A B
I
B
iI I
A B
I
B
i
Blood
group
B
i
´
AB
Blood
group
AB
Blood
group
B
Blood
group
I
B

251. Principles of Inheritance and Variation 241
incorrect statement from the
following[NEET (National) 2019]
(a) Pink colour in F
1
is due to incomplete
dominance
(b) Ratio of F
2
is
1
4
(Red) :
2
4
(Pink) :
1
4
(White)
(c) Law of segregation does not apply in
this experiment
(d) This experiment does not follow the
principle of dominance
Ans.(c)
The statement that ‘law of segregation
does not apply in this experiment’ is
incorrect because the law of segregation
applies universally. The reappearence of
parental (red and white) flowers in the
F
2
-generation also confirms, that law of
segregation applies in this experiment.
Rest statements are correct.
08A gene locus has two alleles A, a. If
the frequency of dominant allele A
is 0.4, then what will be the
frequency of homozygous dominant,
heterozygous and homozygous
recessive individuals in the
population?[NEET (National) 2019]
(a) 0.16(AA); 0.24(Aa); 0.36 (aa)
(b) 0.16(AA); 0.48(Aa); 0.36 (aa)
(c) 0.16(AA); 0.36(Aa); 0.48 (aa)
(d) 0.36(AA); 0.48(Aa); 0.16 (aa)
Ans.(b)
The frequency of homozygous dominant,
heterozygous and homozygous
recessive individuals would be 0.16 (AA);
0.48 (Aa); 0.36 (aa). The frequencies are
calculated as follows:
Frequency of dominant allele (p)=0.4
(given)
Frequency of recessive allele (q)
= − =1 0 4 0 6. .
Frequency of homozygous dominant
individuals (AA)=p
2
=(0.4)=0.16
Frequency of heterozygous individual
(Aa)=2pq
= =2 0 4 0 6 0 48( . ) ( . ) .
Frequency of homozygous recessive
individual (aa)
= = =q
2
( . ) .0 6 0 36
2
09Select thecorrectstatement.
[NEET 2018]
(a) Spliceosomes take part in translation
(b) Punnett square was developed by a
British scientist
(c) Franklin Stahl coined the term
‘linkage’
(d) Transduction was discovered by S.
Altman.
Ans.(b)
Punnett Squareis a checker-board used
to show the result of a cross between
two organisms. The checker board was
devised by a British geneticist,Regnald
Punnett(1927). It depicts both genotypes
and phenotypes of the progeny.
Franklin StahlwithMatthew Meselson
proved thesemi-conservativereplicationof
DNA.Spliceosomeis formed during
post-transcriptional changes in
eukaryotes. It is a complex, formed
between′5end (GU) and′3end (AG) of
intron to remove it.
Transductionis a method of sexual
reproduction in bacteria. It involves the
transfer of foreign genes by means of
viruses. It was discovered byZinderand
his teacherLederberg(1952) in
Salmonella typhimurium.
10Which of the following
characteristics represents
‘Inheritance of blood groups’ in
humans? [NEET 2018]
1. Dominance
2. Codominance
3. Multiple allele
4. Incomplete dominance
5. Polygenic inheritance
(a) 2, 4 and 5 (b) 1, 2 and 3
(c) 2, 3 and 5 (d) 1, 3 and 5
Ans.(b)
Dominance, codominanceandmultiple
allelesare the characteristics that
represent ‘inheritance of blood groups’ in
humans. ABO blood groups are
determined by the gene I. There are
multi ple(three)alleles;I , I
A B
andI
0
of
this gene. AlleleI
A
andI
B
aredominant
overI
0
. However, whenI
A
andI
B
alleles
are present together, they show
codominance.
Therefore, option (b) is correct.
11Which one from those given below
is the period of Mendel’s
hybridisation experiments?
[NEET 2017]
(a) 1856 - 1863
(b) 1840 - 1850
(c) 1857 - 1869
(d) 1870 - 1877
Ans.(a)
Mendel was a great Mathematician and
was Austrian Monk. He became
interested in genetics and conducted
experiments in pea plant (Pisum
sativum). He hybridised the contrasting
characters of the plant and conducted
his experiments for more than 10 years
between 1856-1863; this experimental
data was published in 1865.
12Among the following characters,
which one was not considered by
Mendel in his experiments on pea?
[NEET 2017]
(a) Stem – Tall or Dwarf
(b) Trichomes – Glandular or
Non-glandular
(c) Seed – Green or Yellow
(d) Pod – Inflated or Constricted
Ans.(b)
Trichomes are the epidermal tissues
structure. When epidermal cells become
glandular hair, it is called trichome. This
character was not amongst the seven
characters of pea, which mendel
selected for his hybridisation
experiments.
13The genotypes of a husband and
wife areI I
A B
andI i
A
. Among the
blood types of their children, how
many different genotypes and
phenotypes are possible?
[NEET 2017]
(a) 3 genotypes ; 3 phenotypes
(b) 3 genotypes ; 4 phenotypes
(c) 4 genotypes ; 3 phenotypes
(d) 4 genotypes ; 4 phenotypes
(a) Glandular (b) Stellate
(c) Urticating (d) Stinging
Different types of trichomes

252. Ans.(c)
A cross between two individuals, one
with AB blood group and other with A
blood group will produce four genotypes
and three phenotypes.
Parents Male Female
Phenotype AB A
Genotype I I
A B
I i
A
Gametes I ,I
A B
I ,i
A
X
I
A
I
B
I
A
I I
A A
(A)
I I
A B
(AB)
i I i
A
(A)
I i
B
(B)
Offsprings Genotypes4(I I , I I , I i, I i)
A A A B A B
Phenotypes3 (A, B, AB)
14A true breeding plant is
[NEET 2016, Phase II]
(a) one that is able to breed on its own
(b) produced due to cross-pollination
among unrelated plants
(c) near homozygous and produces
offspring of its own kind
(d) always homozygous recessive in its
genetic constitution
Ans.(c)
A true breeding plant (pureline) has
homozygous genes for a character (e.g.
TT for tall or tt for dwarf). It always
produces offsprings which are true
(pure) for its characters.
15Match the terms in column I with
their description in column II and
choose the correct option.
[NEET 2016, Phase I]
Column I Column II
A. Dominance 1. Many genes
govern a single
character
B. Codominance 2. In a heterozygous
organism only one
allele expresses
itself
C. Pleiotropy 3. In a heterozygous
organism both
alleles express
themselves fully
D. Polygenic
inheritance
4. A single gene
influences many
characters
Code
A B C D
(a) 2 3 4 1
(b) 4 1 2 3
(c) 4 3 1 2
(d) 2 1 4 3
Ans.(a)
Dominance—Expression of only one
allele in a heterozygous organism.
Codominance—Side by side full expression
of both alleles.F
1
resembles both
parents.
Pleiotropy– Single gene can exhibit
multiple phenotypic expression, e.g.
Phenylketonuria.
Polygenicinheritance—Many genes
govern a single character, e.g. Human
skin colour.
16A tall true breeding garden pea
plant is crossed with a dwarf true
breeding garden pea plant. When
the F
1
plants were selfed the
resulting genotypes were in the
ratio of [NEET 2016, Phase I]
(a) 1 : 2 : 1 :: Tall heterozygous : Tall
homozygous : Dwarf
(b) 3 : 1 :: Tall : Dwarf
(c) 3 : 1 :: Dwarf : Tall
(d) 1 : 2 : 1 :: Tall homozygous : Tall
heterozygous : Dwarf
Ans. (d)
Parents – TT × tt
(Tall) Tt (Dwarf)
F
1
-generation (Heterozygous tall)
K-Pollen→
T
On selfing
t
Egg
T
TT
(Tall)
Tt
(Tall)
F
2
-genera
tion
t
Tt
(Tall)
tt
(Dwarf)
Phenotypic ratio=3 1:[Tall : Dwarf]
Genotypic ratio⇒1 : 2 : 1
[Homozygous tall : Heterozygous tall :
Dwarf]
17A gene showing codominance has
[CBSE AIPMT 2015]
(a) one allele dominant on the other
(b) alleles tightly linked on the same
chromosome
(c) alleles that are recessive to each other
(d) Both alleles independently expressed
in the heterozygote
Ans.(d)
A gene shows codominance when both
alleles in heterozygous condition,
express their traits independently
instead of showing dominant-recessive
relationship and such alleles are called
codominant alleles.
18In his classic experiments on pea
plants, Mendel did not use
[CBSE AIPMT 2015]
(a) seed colour (b) pod length
(c) seed shape (d) flower position
Ans.(b)
Pod length was not considered by
Mendel in his experiments. For his
experiments, Mendel choose seven
characters of pea plants which are
1. Seed colour
2. Seed shape
3. Flower colour
4. Pod colour
5. Pod shape
6. Flower position and
7. Plant height
19A pleiotropic gene
[CBSE AIPMT 2015]
(a) is expressed only in primitive plants
(b) is a gene evolved during Pliocene
(c) controls a trait only in combination
with another gene
(d) controls multiple traits in an individual
Ans.(d)
Pleiotropic gene is a gene that controls
multiple traits is an individual. It is also
called polyphenic gene, e.g.
phenylketonuria causing multiple
adverse effects due to the mutation in a
single gene coding for enzyme
phenylalanine hydroxylase.
20Fruit colour in squash is an
example of [CBSE AIPMT 2014]
(a) recessive epistasis
(b) dominant epistasis
(c) complementary genes
(d) inhibitory genes
Ans.(b)
Fruit colour in squash is an example of
dominant epistasis in which the
dominant gene (epistatic gene) masks
the effect of other gene (recessive
hypostatic gene).
Squash fruit appear white due to the
epistatic effect of ‘W’ allele (white colour)
over ‘G’ allele (green colour).
242 NEETChapterwise Topicwise Biology

253. 21If two persons with ‘AB’ blood group
marry and have sufficiently large
number of children, these children
could be classified as ‘A’ blood
group : ‘AB’ blood group : ‘B’ blood
group in 1 : 2 : 1 ratio. Modern
technique of protein
electrophoresis reveals presence
of both ‘A’ and ‘B’ type proteins in
‘AB’ blood group individuals. This is
an example of [NEET 2013]
(a) codominance
(b) incomplete dominance
(c) partial dominance
(d) complete dominance
Ans.(a)
Incodominanceboth alleles of a pair
express themselves fully inF
1
hybrid. It is
contrary to the situation seen in
incomplete dominance, where traits
express themselves only partially. This is
not the example of partial dominance or
complete dominance
AB→ I I
A B
Genotype
→Antigen A
+Antigen B→ Codominance
22Which Mendelian idea is depicted
by a cross in which the
F
1
-generation resembles both the
parents? [NEET 2013]
(a) Incomplete dominance
(b) Law of dominance
(c) Inheritance of one gene
(d) Codominance
Ans.(d)
In codominance, both alleles of a pair
express themselves fully inF
1
hybrid, so,
it resembles both the parents. In
incomplete dominance, the two genes of
allelomorphic pair are not related as
dominant or recessive, but each of them
express itself partially. Law of
dominance states that when a cross is
made between two homozygous
individuals considering contrasting trait
of simple character then the trait that
appear inF
1
hybrids is called dominant.
Inheritance of one gene is based on
crossing between single traits.
23F
2
-generation in a Mendelian cross
showed that both genotypic and
phenotypic ratios are same as 1 : 2
: 1. It represents a case of
[CBSE AIPMT 2012]
(a) codominance
(b) dihybrid cross
(c) monohybrid cross with complete
dominance
(d) monohybrid cross with incomplete
dominance
Ans.(d)
Monohybrid cross with incomplete
dominance shows both genotypic and
phenotypic ratio as same (1 : 2 : 1).
Gen
Genotypic ratio – 1 (AA) : 2 (Aa) : 1 (aa)
Phenotypic ratio – 1 (Red) : 2 (Pink) : 1(white)
23ABO blood groups in humans are
controlled by the gene I. It has
three alleles−I , I
A B
and i. Since
there are three different alleles, six
different genotypes are possible.
How many phenotypes can occur?
[CBSE AIPMT 2010]
(a) Three (b) One
(c) Four (d) Two
Ans.(c)
The ABO blood group system has at least
6 genotypes. On the basis of presence or
absence of antigens and antibodies, four
blood groups (phenotypes) have been
differentiated—A, B, AB and O blood
groups. In ABO blood group system,
inheritance of grouping is controlled by a
single autosomal gene on chromosome 9
with three major alleles A, B and O (
I , I and I
A B O
).
24Which one of the following cannot
be explained on the basis of
Mendel’s Law of Dominance?
[CBSE AIPMT 2010]
(a) The discrete unit controlling a
particular character is called a factor
(b) Out of one pair of factors one is
dominant, and the other recessive
(c) Alleles do not show any blending and
both the characters recover as such
in F
2
-generation
(d) Factors occur in pairs
Ans.(c)
Out of the following statement (c) is
incorrect because the law of dominance
does not occur universally. After Mendel
several cases were recorded by
scientists, where a clear deviation from
law of dominance was seen.
Such a deviation may be seen in the form
of incomplete dominance or blending
inheritance and co-dominance.
Where,F
1
hybrids exhibited a mixture or
blending of character of two parents, the
case is considered as that of incomplete
dominance or blending inheritance. It
simply means that two genes of
allelomorphic pair are not related as
dominant or recessive, but each of them
expresses itself partially. In the case of 4
O’clock plant, when plants with red
flowers are crossed with plants having
white flower, theF
1
hybrids bear pink
flower. When these pink flowers are self
pollinated, they develop red, pink and
white flowers in the ratio of 1 : 2 : 1
respectively.
25The genotype of a plant showing
the dominant phenotype can be
determined by[CBSE AIPMT 2010]
(a) test cross
(b) dihybrid cross
(c) pedigree analysis
(d) back cross
Ans.(a)
In genetics, a test cross, first introduced
by Gregor J Mendel, is used to determine
if an individual exhibiting a dominant
trait is homozygous or heterozygous for
that trait. More simply, test cross
determines the genotype of an individual
with a dominant phenotype. In some
sources, the test cross is defined as
being a type of back cross between the
recessive homozygote andF
1
-generation.
26In pea plants, yellow seeds are
dominant to green. If a
heterozygous yellow seeded plant
is crossed with a green seeded
plant, what ratio of yellow and
green seeded plants would you
expect inF
1
-generation?
[CBSE AIPMT 2007]
(a) 50 : 50 (b) 9 : 1 (c) 1 : 3 (d) 3 : 1
Principles of Inheritance and Variation 243
AA aa
Red White
A a
Aa
Pink
aA
a
A
A
a
Aa
Pink
AA
Red
Aa
Pink
aa
White
Parents
Gametes
F -Generation
1
Gametes
F -generation
2

254. Ans.(a)
In the given cross the ratio is 50 : 50 of
yellow and green seeded plants in
F
1
-generation.
27A common test to find the
genotype of a hybrid is by
[CBSE AIPMT 2007]
(a) crossing of oneF
2
progeny with male
parent
(b) crossing of oneF
2
progeny with
female parent
(c) studying the sexual behaviour of
F
1
-progenies
(d) crossing of oneF
1
progeny with male
parent
Ans.(d)
Test cross, i.e. crossing ofF
1
-progeny to
the recessive parent is used to find the
genotype of the progeny.
28A human male produces sperms
with the genotypes AB, Ab, aB and
ab pertaining to two diallelic
characters in equal proportions.
What is the corresponding
genotype of this person?
[CBSE AIPMT 2007]
(a) AaBb (b) AaBB
(c) AABb (d) AABB
Ans.(a)
The corresponding genotype of person
will be AaBb.
29Test cross involves[CBSE AIPMT
2006]
(a) crossing between two genotypes with
recessive trait
(b) crossing between twoF
1
hybrids
(c) crossing theF
1
hybrid with a double
recessive genotype
(d) crossing between two genotypes with
dominant trait
Ans.(c)
The test cross involves the crossing ofF
1
hybrid with a double recessive genotypic
parent. By test cross, the heterozygosity
and homozygosity of the organism can
be tested.
Thus, the offspring will be 100%
dominant, if the individual which crossed
with recessive parent, i.e. (tt) was
homozygous dominant and ratio will be
50% dominant and 50% recessive if the
individual was heterozygous dominant.
In dihybrid test cross, ratio will be 1:1:1:1.
30How many different kinds of
gametes will be produced by a
plant having the genotype
AABbCC ?
[CBSE AIPMT 2006]
(a) Three (b) Four
(c) Nine (d) Two
Ans.(d)
The types of gametes produced by a
plant depend upon the number of
heterozygous pair.
Number of types of gametes=2
n
n=number of heterozygous pair2 2
1
=
The gametes are – ABC and AbC.
31In Mendel’s experiments with
garden pea, round seed shape (RR)
was dominant over wrinkled seeds
(rr), yellow cotyledon (YY) was
dominant over green cotyledon (yy).
What are the expected phenotypes
in theF
2
-generation of the cross
RRYY × rryy ?[CBSE AIPMT 2006]
(a) Only round seeds with green
cotyledons
(b) Only wrinkled seeds with yellow
cotyledons
(c) Only wrinkled seeds with green
cotyledons
(d) Round seeds with yellow cotyledons
and wrinkled seeds with yellow
cotyledons
Ans.(d)
When a cross (dihybrid) is made between
plants bearing round yellow (RRYY) and
wrinkled green (rryy) seeds, all the plants
inF
1
-generation are with yellow round
seeds (showing the genotype RrYy). The
phenotype inF
2
will be as follows
32Phenotype of an organism is the
result of
[CBSE AIPMT 2006]
(a) mutations and linkages
(b) cytoplasmic effects and nutrition
(c) environmental changes and sexual
dimorphism
(d) genotype and environmental
interactions
Ans.(d)
Phenotype is the observable
characteristics or the total appearance
of an organism. It is determined by its
genes, the relationships between the
alleles and by the interaction during
development between its genetic
constitution (genotype) and the
environment.
33In order to find out the different
types of gametes produced by a
pea plant having the genotype
AaBb, it should be crossed to a
plant with the genotype
[CBSE AIPMT 2005]
(a) aaBB
(b) AaBb
(c) AABB
(d) aabb
Ans.(d)
In the given question AaBb should be
crossed with aabb. Scientists perform
test cross to find out the different types
of gametes or the genotype of an
unknown individual. Test cross is
performed always between theF
1
heterozygous plants and pure recessive
(homozygous) parent plant. So, in the
given case AaBb should be crossed with
aabb.
34In a plant, red fruit (R) is dominant
over yellow fruit (r) and tallness (T)
is dominant over shortness (t). If a
plant with RRTt genotype is
crossed with a plant that is rrtt
[CBSE AIPMT 2004]
(a) 25% will be tall with red fruit
(b) 50% will be tall with red fruit
(c) 75% will be tall with red fruit
(d) all of the offspring will tall with red
fruit
Ans.(a)
In the given experimnet 50% will be tall
with red fruits. It can be explanied as
244 NEETChapterwise Topicwise Biology
AB
ab
aB
Ab
B
b
A a
Genotype
of gametes
(yellow) (green)
Yy yy
Y y y y
Yy Yy yy yy
Yellow
50%
Green
50%

255. Conclusion :
1. All plants are red.
2. 50% are red tall.
3. 50% are red dwarf.
35A male human is heterozygous for
autosomal genes A and B and is
also hemizygous for haemophilic
gene h. What proportion of his
sperms will be abh?
[CBSE AIPMT 2004]
(a)
1
8
(b)
1
32
(c)
1
16
(d)
1
4
Ans.(a)
The genotype of human male in question
must be Aa BbX Y.
h
Hence2 2 2 8× × =types of gametes
would be formed.AB X
h
,ABY,aB X
h
.
aBY, Ab X , AbY, ab X
h h
, abY.
Hence, 1/8 proportion of his sperms
would be abh.
36Two crosses between the same
pair of genotypes or phenotypes in
which the sources of the gametes
are reversed in one cross, is known
as [CBSE AIPMT 2003]
(a) dihybrid cross (b) reverse cross
(c) test cross (d) reciprocal cross
Ans.(d)
Since genotypes/phenotypes of both
parents are same—only sources of
gametes are reversed, these crosses
are called reciprocal crosses.
37The genes controlling the seven
pea characters studied by Mendel
are now known to be located on
how many different chromosomes?
[CBSE AIPMT 2003]
(a) Five (b) Four
(c) Seven (d) Six
Ans.(b)
As a result of studies by S Blixt, it is now
known that Mendel’s seven selected
characters are located on four
chromosomes. Of these, two characters
are located on chromosome number 1,
three of chromosome number 4 and one
each on chromosome number 5 and 7.
38Which one of the following traits of
garden pea studied by Mendel was
a recessive feature?
[CBSE AIPMT 2003]
(a) Green pod colour
(b) Round seed shape
(c) Axial flower position
(d) Green seed colour
Ans.(d)
Green seed colour was a recessive
character in Mendel’s experiment. When
a pair of contrasting characters are
crossed together thenF
1
-generation has
only one type of character. This
expressed character is known as
dominant character while the character
which could not express inF
1
-generation
is known as recessive character. In pea
plants, tallness, round seed, yellow seed,
purple flower, green pod, inflated pod
and axial flower are dominant over
dwarfness, wrinkled seed, green seed,
white flower, yellow pod, constricted
pod and terminal flower respectively.
39A plant ofF
1
-generation has
genotype ‘AABbCC’. On selfing of
this plant, the phenotypic ratio in
F
2
-generation will be
[CBSE AIPMT 2002]
(a) 3 : 1
(b) 1 : 1
(c) 9 : 3 : 3 : 1
(d) 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1
Ans.(a)
Since, AABbCC contains only one
heterozygous allelic pair ‘Bb’, the cross
would behave as monohybrid cross
leading to 3 : 1 phenotypic ratio inF
2
.
Phenotypic ratio is 3 : 1.
40In his experiment, Mendel obtained
wrinkled pea. The wrinkling was due
to deposition of sugar instead of
starch. This happened due to the
enzyme [CBSE AIPMT 2001]
(a) amylase
(b) invertase
(c) diastase
(d) absence of starch-branching enzyme
Ans.(d)
In round seeds (RR/Rr) Starch-Branching
Enzyme-1 (SBE-1) is found but is absent
from wrinkled seeds. In rr seeds, a small
DNA segment interferes with SBE-1
activity so that starch is not formed and
free sugar gets accumulated. Resulting
high osmotic pressure in rr seeds leads
to shrinkage and wrinkling.
41Which of these do not follow
independent assortment?
[CBSE AIPMT 2001]
(a) Genes on non-homologous
chromosomes and absence of linkage
(b) Genes on homologous chromosomes
(c) Linked genes on same chromosome
(d) Unlinked genes on same chromosome
Ans.(c)
Genes which are ‘linked’ and are situated
on same chromosome, cannot ‘separate’
during gametes formation and hence,
cannot assort independently.
42Two non-allelic genes produce the
new phenotype when present
together but fail to do so
independently, it is called
(a) epistasis[CBSE AIPMT 2001]
(b) polygene
(c) non-complementary gene
(d) complementary gene
Ans.(d)
In complementary genes, two separate
pair of genes interact to produce the
phenotype in such a way that neither of
the dominant genes is expressive unless
the other one is absent.
43Ratio of complementary genes is
[CBSE AIPMT 2001]
(a) 9 : 3 : 4
(b) 12 : 3 : 1
(c) 9 : 3 : 3 : 4
(d) 9 : 7
Ans.(d)
In case of complementary genes, the
ratio of 9 : 7 is obtained in
Principles of Inheritance and Variation 245
AABbCC AABbCC
ABC AbC ABC
Selfing
AbC
ABC
AbCAABbCC
AABBCC AABbCC
AAbbCC
ABC AbC
F2
F1
Gametes
-generation
-generation
RRTt
(red tall)
rrtt
(yellow dwarf)
RT Rt rt Gametes
RrTt
Rrtt
RrTt
Rrtt
rt rt F -generation
1
RT
Rt
+
rt

256. F
2
-generation. This was first discovered
byBatesonandPunnett.
Complementary genes are those genes
which express themself when present
together. None of these two get
expressed yourself when present alone.
44Which one of the following
characters studied by Mendel in
garden pea was found to be
dominant? [CBSE AIPMT 2000]
(a) Green seed colour
(b) Terminal flower position
(c) Green pod colour
(d) Wrinkled seed
Ans.(c)
Green pod colour was a dominant
character in Mendel’s garden pea
experiment.
45Hybridisation betweenTt tt×gives
rise to the progeny of ratio
[CBSE AIPMT 1999]
(a) 1 : 1 (b) 1 : 2 : 1
(c) 1 : 2 (d) 4 : 1
Ans.(a)
Offsprings with genotypes Tt
(heterozygous tall) and tt (homozygous
dwarf) are produced in the ratio of 1 : 1.
46How many types of genetically
different gametes will be produced
by a heterozygous plant having
genotype AABbCc?
[CBSE AIPMT 1998]
(a) Two (b) Four
(c) Six (d) Nine
Ans.(b)
First pair AA is homozygous hence, it will
contribute only one type of gene to
gametes, Bb will yield two types of
gametes—B and b similarly, Cc will yield
two types of gametes—C and c. Hence,
1 2 2 4× × =types of gametes would be
produced having the genotypes
ABC, ABc, AbC and Abc.
47If Mendel had studied the seven
traits using a plant with 12
chromosomes instead of 14, in
what way would his interpretation
have been different?
[CBSE AIPMT 1998]
(a) He would have mapped the
chromosome
(b) He would have discovered blending or
incomplete dominance
(c) He would not have discovered the law
of independent assortment
(d) He would have discovered sex-linkage
Ans.(c)
If Mendel would have studied seven
traits in 12 chromosomes instead of 14
he would not have discovered
independent assortment.
48When a single gene influences
more than one traits it is called
[CBSE AIPMT 1998]
(a) pleiotropy (b) epistasis
(c) pseudodominance
(d) None of these
Ans.(a)
Pleiotropy is the condition in which a
single gene influences more than one
traits, e.g. gene for single cell
produces anaemia as well as
resistance to malaria.
49Alleles that produce independent
effects in their heterozygous
condition are called
[CBSE AIPMT 1996]
(a) codominant alleles
(b) epistatic alleles
(c) complementary alleles
(d) supplementary alleles
Ans.(a)
Codominant alleles produce
independent effects in their
heterozygous condition. Both the
genes produce their independent
effect.
50In a dihybrid cross AABB × aabb,
F
2
progeny of AABB, AABb, AaBB
and AaBb occurs in the ratio of
[CBSE AIPMT 1994]
(a) 1 : 1 : 1 : 1 (b) 9 : 3 : 3 : 1
(c) 1 : 2 : 2 : 1 (d) 1 : 2 : 2 : 4
Ans.(d)
In a dihybrid cross the genotypic ratio
can be represented as
follows—assuming a cross between
AABB and aabb
AABB × aabb — Parents
AB↓ab — Gametes
AaBb — F
1
-generation
Selfing — F
2
-generation
1BB — 1 AABB
1 AA 2 Bb — 2 AABb
1 bb — 1 Aabb
1 BB — 2AaBB
2 Aa 2Bb — 4AaBb
1 bb — 2Aabb
1 BB — 1 aaBB
1 aa 2Bb — 2aaBb
1bb — aabb
So, the ratio of AABB, AABb, AaBB and
AaBb will be 1 : 2 : 2 : 4.
51A cross between pure tall pea plant
with green pods and dwarf pea
plant with yellow pods will produce
dwarfF
2
plants out of 16
[CBSE AIPMT 1994]
(a) 9 (b) 3
(c) 4 (d) 1
Ans.(c)
This is an example of dihybrid cross
because two characters are being studied
simultaneously in which dwarfness is a
recessive characters, so in
F
2
-generation 4 plants will be dwarf out
of 16.
52A child of blood group O cannot
have parents of blood groups
[CBSE AIPMT 1994]
(a) AB and AB/O
(b) A and B
(c) B and B
(d) O and O
Ans.(a)
Blood group of the child is determined by
allelic genes, i.e.I I
A B
andI
O
in whichI
A
andI
B
are dominant overI
O
, so if one of
the parent either mother or father is of
blood group AB, then she or he will have
both genes, i.eI
A
andI
B
and child of
such parent cannot have blood group O.
53Which of the following is suitable
for experiment on linkage?
[CBSE AIPMT 1993]
(a) aaBB × aaBB (b) AABB × aabb
(c) AaBb × AaBb (d) AAbb × AaBB
246 NEETChapterwise Topicwise Biology

257. Ans.(b)
AABB×aabb is suitable for experiment
on linkage. Linkage may be defined as
the tendency of two genes of the same
chromosome to remain together in the
process of inheritance.
54Mendel studied inheritance of seven
pairs of traits in pea which can have
21 possible combinations. If you are
told that in one of these
combinations, independent
assortment is not observed in later
studies, your reaction will be
[CBSE AIPMT 1993]
(a) independent assortment principle
may be wrong
(b) Mendel might not have studied all the
combinations
(c) it is impossible
(d) later studies may be wrong
Ans.(b)
Mendel’s law of independent assortment
is applicable to only those genes which
are located on different chromosomes,
so if in one of the combination
independent assortment is not observed
that means Mendel might not have
studied all the combinations.
55A polygenic inheritance in human
beings is
[CBSE AIPMT 1993, 99, 2006, 07]
(a) skin colour
(b) phenylketonuria
(c) colour blindness
(d) sickle-cell anaemia
Ans.(a)
Human skin colour is controlled by
polygenic effect atleast by three
separate genes. Skin colour is
determined by cumulative genes and
this hypothesis was designed by
Devenport and Devenport in 1910.
56An organism with two identical
alleles is [CBSE AIPMT 1992]
(a) dominant
(b) hybrid
(c) heterozygous
(d) homozygous
Ans.(d)
An organism with two identical alleles
is homozygous, e.g. rr, tt, RR, TT, etc.
57Segregation of Mendelian factors
(no linkage, no crossing over)
occurs during[CBSE AIPMT 1992]
(a) anaphase-I
(b) anaphase-II
(c) diplotene
(d) metaphase-I
Ans.(a)
At the end of anaphase-I, two groups of
chromosomes (one at each pole) are
produced. Each such group is having half
the original number of chromosomes
present in the parent nucleus. So,
anaphase-I results in the reduction of
chromosome number to half and
segregation of Mendelian factors.
58An allele is dominant if it is
expressed in
[CBSE AIPMT 1992, 2002]
(a) both homozygous and heterozygous
states
(b) second generation
(c) heterozygous combination
(d) homozygous combination
Ans.(a)
Dominant allele expresses itself both in
homozygous and in heterozygous states.
59A gene pair hides the effect of
another. The phenomenon is
[CBSE AIPMT 1992, 95, 99]
(a) epistasis (b) dominance
(c) mutation (d) None of these
Ans.(a)
Epistasis is the phenomenon by which
a gene suppresses the phenotypic
expression of a non-allelic gene.
60RR (red)Antirrhinumis crossed
with WW (white) one. Offspring RW
are pink. This is an example of
[CBSE AIPMT 1991]
(a) dominant-recessive
(b) incomplete dominance
(c) hybrid
(d) supplementary genes
Ans.(b)
The pink colour of the offspring is an
example of incomplete dominance. In
which the expression of the characters
inF
1
individual is intermediate of the
factors as found in homozygous state.
61The allele which is unable to
express its effect in the presence
of another is called
[CBSE AIPMT 1991]
(a) codominant
(b) supplementary
(c) complementary
(d) recessive
Ans.(d)
In heterozygous condition where both
the contrasting alleles are present only
one allele is able to express, called
dominant, while other which remain
suppressed is called recessive.
62The contrasting pairs of factors in
Mendelian crosses are called
(a) multiple alleles[CBSE AIPMT 1991]
(b) allelomorphs
(c) alloloci
(d) paramorphs
Ans.(b)
Two allelomorphs are the different
forms of a gene which are responsible
for different expression of same
characters, e.g. for colour of flower is R
and r.
63First geneticist/father of genetics
was [CBSE AIPMT 1991]
(a) De Vries
(b) Mendel
(c) Darwin
(d) Morgan
Ans.(b)
Mendel for his great contribution in
genetics is now known as father of
genetics.
64Mendel’s last law is
[CBSE AIPMT 1991]
(a) segregation
(b) dominance
(c) independent assortment
(d) polygenic inheritance
Ans.(c)
Mendel’s law of independent assortment
is related with inheritance of two or
more genes at one time. The distribution
of genes in the gametes and in the
progeny of subsequent generation is
independent of each other.
Principles of Inheritance and Variation 247

258. 65A dihybrid condition is
[CBSE AIPMT 1991]
(a) tt Rr (b) Tt rr
(c) tt rr (d) Tt Rr
Ans.(d)
Dihybrid condition means simultaneous
transmission of two pairs of genes. So,
dihybrid condition is TtRr.
66Blue eye colour is recessive to
brown eye colour. A brown eyed
man whose mother was blue eyed
marries a blue eyed women. The
children shall be[CBSE AIPMT 1991]
(a) both blue eyed and brown eyed 1 : 1
(b) all brown eyed
(c) all blue eyed
(d) blue eyed and brown eyed 3 : 1
Ans.(a)
A brown eyed man, whose mother was
blue eyed must have the genotype Bb
where B represents brown eye colour
and b represents blue eye colour. When
a man of such genotype will marry to a
blue eyed woman, the children shall be
67Multiple alleles control inheritance
of [CBSE AIPMT 1991]
(a) phenylketonuria
(b) colour blindness
(c) sickle-cell anaemia
(d) blood groups
Ans.(d)
Multiple alleles control inheritance of
blood groups in human. A, B, O blood
group system is found in humans where
the alleleI
A
for A antigen is codominant
with the alleleI
B
for the B antigen.
BothI
A
andI
B
are completely dominant
to the alleleI
O
.
Hierarchy of dominance relationships is
symbolised as( )I I I
A B O
= > .
68Which contribute to the success of
Mendel? [CBSE AIPMT 1988]
(a) Qualitative analysis of data
(b) Observation of distinct inherited traits
(c) His knowledge of Biology
(d) Consideration of one character at one
time
Ans.(d)
Mendel took one trait at a time for his
experiments, this contributed a lot to his
success.
69Chromosomal theory of inheritance
was proposed by[NEET (Oct.) 2020]
(a) Sutton and Boveri
(b) Bateson and Punnett
(c) TH Morgan
(d) Watson and Crick
Ans.(a)
Chromosomal theory of inheritance was
proposed by Sutton and Boveri
independently in 1902.
The two workers found a close similarity
between the transmission of hereditary
traits and behaviour of chromosomes
while passing from one generation to the
next through gametes.
70Experimental verification of the
chromosomal theory of inheritance
was done by [NEET (Sep.) 2020]
(a) Sutton (b) Boveri
(c) Morgan (d) Mendel
Ans.(c)
Experimental verification of the
chromosomal theory of inheritance was
done by Thomas Hunt Morgan. Sutton
and Boveri proposed chromosomal
theory of inheritance but it was
experimentally verified by TH Morgan.
According to this theory, genes are the
units of heredity and are found in the
chromosomes.
71Crossing over takes place between
which chromatids and in which stage
of the cell cycle?
[NEET (Odisha) 2019]
(a) Non-sister chromatids of
non-homologous chromosomes at
zygotene stage of prophase I
(b) Non-sister chromatids of
homologous chromosomes at
pachytene stage of prophase I
(c) Non-sister chromatids of
homologous chromosomes at
zygotene stage of prophase I
(d) Non-sister chromatids of
non-homologous chromosomes at
pachytene stage of prophase I
Ans.(b)
Crossing over takes place between
non-sister chromatids of homologous
chromosomes at pachytene stage of
prophase-I. This stage of prophase-I in
meiosis is characterised by the
appearance of recombination nodules,
the site at which crossing over occurs
between non-sister chromatids of the
homologous chromosomes.
72What map unit (Centimorgan) is
adopted in the construction of
genetic maps?
[NEET (National) 2019]
(a) A unit of distance between two
expressed genes representing 100%
cross over
(b) A unit of distance between genes on
chromosomes, representing 1% cross
over
(c) A unit of distance between genes on
chromosomes, representing 50%
cross over
(d) A unit of distance between two
expressed genes representing 10%
cross over
Ans.(b)
In the construction of genetic maps,
map unit or centimorgan is a unit or
distance between genes on
chromosomes, representing 1%
crossover.
i. e. 1 map unit=1% crossover
Hence, the genetic distance between
genes is based on average number of
cross over frequency between them.
73The frequency of recombination
between gene pairs on the same
chromosome as a measure of the
distance between genes was
explained by[NEET (National) 2019]
(a) Gregor J Mendel
(b) Alfred Sturtevant
(c) Sutton-Boveri
(d) TH Morgan
Ans.(b)
Alfred Sturtevant was the first to explain
the concept of chromosomal mapping. It
is drawn on the basis of recombination
frequency between gene pairs on the
same chromosome. This frequency is
directly proportional to the distance
between these two genes.
It can be used to determine the exact
location of a gene on the chromosome.
248 NEETChapterwise Topicwise Biology
(brown eye)
Bb
(blue eye)
bb
Bb
(brown eyed
children)
Parents
Genotype
of
progenies
B b b b
(blue eyed
children)
Bbbb bb
Gametes
Chromosomal Theory of
Inheritance and Linkage
TOPIC 2

259. 74In a test cross involvingF
1
dihybrid
flies, more parental-type offspring
were produced than the
recombinant type offspring. This
indicates [NEET 2016, Phase I]
(a) chromosomes failed to separate
during meiosis
(b) the two genes are linked and present
on the same chromosome
(c) both of the characters are controlled
by more than one gene
(d) the two genes are located on two
different chromosomes
Ans.(b)
When two genes in a dihybrid cross are
situated on the same chromosome, the
proportion of parental gene combinations
are much higher than the non-parental
or recombinant type as linked genes are
inherited together in offspring.
75The mechanism that causes a gene
to move from one linkage group to
another is called
[NEET 2016, Phase II]
(a) inversion (b) duplication
(c) translocation (d) crossing over
Ans.(c)
Translocation is the process causing a
gene to move from one linkage group to
another. It is the separation of a
chromosome segment and its union to a
non-homologous chromosome. It is of
two types–simpleandreciprocal. In
simple translocation one chromosome
shows deletion or deficiency while a
nonhomologous chromosome comes to
have an additional segment. In reciprocal
translocation two non- homologous
chromosomes exchange segments
between themselves to create new
linkagegroupsinboththechromosomes.
Hence, option (c) is correct.
76The term “linkage” was coined by
[CBSE AIPMT 2015]
(a) TH Morgan (b) T Boveri
(c) G Mendel (d) W Sutton
Ans.(a)
The term linkage was coined by TH
Morgan. He carried out several dihybrid
crosses inDrosophilato study genes that
were sex-linked. He described the physical
association of genes on a chromosome.
77Which of the following statements
is not true of two genes that show
50% recombination frequency?
[NEET 2013]
(a) The genes may be on different
chromosomes
(b) The genes are tightly linked
(c) The genes show independent
assortment
(d) If the genes are present on the same
chromosome, they undergo more
than one crossovers in every meiosis
Ans.(b)
Out of the given statements (b) is incorrect
because the tightly linked genes on
chromosomes show 100% parental types
and 0% recombinants. Two genes that
undergo independent assortment
indicated by a recombinant frequency of
50% are either on non- homologous
chromosomes or located far apart in a
single chromosome.
As the distance between two genes
increases, crossover frequency
increases. More recombinant gametes,
fewer parental gametes.
78Select the correct statement from
the ones given below with respect
to dihybrid cross.
[CBSE AIPMT 2010]
(a) Tightly linked genes on the same
chromosome show higher
recombinations
(b) Genes far apart on the same
chromosome show very few
recombinations
(c) Genes loosely linked on the same
chromosome show similar
recombinations as the tightly linked
ones
(d) Tightly linked genes on the same
chromosome show very few
recombinations
Ans.(d)
Morgan and his group found that when
genes were grouped on the same
chromosome, some genes were very
tightly linked (showed very low
recombination), while others were
loosely linked (showed higher
recombination).
Recombination is a process of
rearrangement of genes during meiosis
so that a gamete contains a haploid
genotype with a new gene combination.
79Two genes R and Y are located very
close on the chromosomal linkage
map of maize plant. When RRYY
and rryy genotypes are hybridised,
thenF
2
segregation will show
[CBSE AIPMT 2007]
(a) higher number of the recombinant
types
(b) segregation in the expected 9 : 3 : 3 : 1
ratio
(c) segregation in 3 : 1 ratio
(d) higher number of the parental types
Ans.(d)
Higher number of the parental types
formed when RRYY and rryy genotypes
are hybridised giving the condition that R
and Y genes are closely linked.
Law of independent assortment does
not applicable when the gene of
different character occupy the same
homologous chromosome i.e. are linked
gene.
80In which mode of inheritance do
you expect more maternal
influence among the offspring?
[CBSE AIPMT 2006]
(a) Autosomal
(b) Cytoplasmic
(c) Y-linked
(d) X-linked
Ans.(b)
The more maternal influence can be
expected in the cytoplasmic inheritance,
i.e. the inheritance of genes contained in
the cytoplasm of a cell, rather than the
nucleus.
The reason is that the female
reproductive cell or the egg has a large
amount of cytoplasm containing many
such organelles which contain their own
genes and can reproduce independently,
e.g. mitochondria and chloroplast and
which are consequently incorporated
into the cytoplasm of all the cells of the
embryo.
The male reproductive cells (sperm or
pollen) consist almost solely of a
nucleus. Cytoplasmic organelles are
thus, not inherited from the male parent.
This is why, the cytoplasmic inheritance
is also called maternal inheritance.
A gene located in the X-chromosome is
said to be X-linked and its inheritance is
called X-linked inheritance. In this, a
male transmits his X-chromosome only
to his daughters while a female
transmits one of her X-chromosomes to
the offspring of both sexes.
81The recessive genes located on
X-chromosome in humans are
always [CBSE AIPMT 2004]
(a) lethal
(b) sublethal
(c) expressed in males
(d) expressed in females
Principles of Inheritance and Variation 249

260. Ans.(c)
The recessive genes located on
X-chromosome in humans are always
expressed in males because a female
may be homozygous or heterozygous
while male is always hemizygous (i.e.
only one allele is present). Haemophilia,
colour blindness are some human diseases
which are frequently found in males.
82Extranuclear inheritance is a
consequence of presence of genes
in [CBSE AIPMT 2004]
(a) mitochondria and chloroplasts
(b) endoplasmic reticulum and
mitochondria
(c) ribosomes and chloroplast
(d) lysosomes and ribosomes
Ans.(a)
Extranuclear or extrachromosomal or
cytoplasmic or organellar inheritance is
a consequence of presence of genes in
mitochondria and chloroplast.
Extrachromosomal units function either
independently or in collaboration with
nuclear genetic system.
83Lack of independent assortment of
two genes A and B in fruit fly
Drosophilais due to
[CBSE AIPMT 2004]
(a) repulsion (b) recombination
(c) linkage (d) crossing over
Ans.(c)
TH Morgan (1910) explained the lack of
independent assortment inDrosophila
due to the linkage. When genes closely
present adhere or link together in a
group and transmitted as a single unit,
the phenomenon is called linkage. It
stops the process of independent
assortment. Incomplete linkage is
broken down due to the crossing over.
84Pattern baldness, moustaches and
beard in human males are
examples of[CBSE AIPMT 2003]
(a) sex differentiating traits
(b) sex determining traits
(c) sex linked traits
(d) sex limited traits
Ans.(d)
Sex limited traits are those which are
limited to one sex only. Moustaches,
beard are found in human males only. It
was suggested on the basis of statistical
analysis that premature baldness is
controlled by a dominant gene, which
expresses only in the presence of a
certain level of male hormone (androgen).
85The linkage map of X-chromosome
of fruit fly has 66 units, with yellow
body gene (y) at one end and
bobbed hair (b) gene at the other
end. The recombination frequency
between these two genes (y and b)
should be [CBSE AIPMT 2003]
(a)≤50% (b) 100%
(c) 66% (d) >50%
Ans.(b)
The actual distance between two genes
is said to be equivalent to the
percentage of crossing over between
these two genes. Since, the two genes
lie at the ends of the chromosome, there
are 100% chances of their segregation
during crossing over.
86When a cluster of genes show
linkage behaviour they
[CBSE AIPMT 2003]
(a) do not show independent assortment
(b) induce cell division
(c) do not show a chromosome map
(d) show recombination during meiosis
Ans.(a)
Linked genes do not show independent
assortment because they are located on
the same chromosome. But genes which
are located on the same chromosomes
(called linked genes) do not assort
independently. Such type of genes are
called linked genes and this
phenomenon is called as linkage.
87In recent years, DNA sequences
(nucleotide sequence) ofmtDNA
and Y-chromosomes were
considered for the study of human
evolution, because
[CBSE AIPMT 2003]
(a) their structure is known in greater
detail
(b) they can be studied from the samples
of fossil remains
(c) they are small and therefore, easy to
study
(d) they are uniparental in origin and do
not take part in recombination
Ans.(d)
Wilsonand Sarich choose mitochondrial
DNA (mtDNA) for the study of maternal
line inheritance. While Y-chromosomes
were considered for the study of human
evolution particularly male domain. It is
possible because they are uniparental in
origin and do not take part in
recombination.
88Genetic map is one that
[CBSE AIPMT 2003]
(a) shows the stages during the cell
division
(b) shows the distribution of various
species in a region
(c) establishes sites of the genes on a
chromosome
(d) establishes the various stages in
gene evolution
Ans.(c)
Genetic map is a diagram which shows
the relative position of genes on a
chromosome.Strutevantin 1911
prepared the first genetic map of two
chromosomes of fruit fly.
89Genes for cytoplasmic male
sterility in plants are generally
located in [CBSE AIPMT 2003]
(a) nuclear genome
(b) cytosol
(c) chloroplast genome
(d) mitochondrial genome
Ans.(d)
Mitochondria are originated from
pre-existing mitochondria. These are
semi-autonomous, living, organelles
present in all eukaryotic cells. These
contain DNA (mtDNA). The available
evidences show that the genes located
inmtDNA control the cytoplasmic male
sterility.
90There are three genes a, b, c,
percentage of crossing over
between a and b is 20%, b and c is
28% and a and c is 8%. What is the
sequence of genes on
chromosome?[CBSE AIPMT 2002]
(a) b, a, c (b) a, b, c
(c) a, c, b (d) None of these
Ans.(a)
According to the given question the
sequence of gene son chromosome are
b, a, c.
91A and B genes are linked. What
shall be the genotype of progeny in
a cross between AB/ab and ab/ab?
[CBSE AIPMT 2001]
(a) AAbb and aabb
(b) AaBb and aabb
(c) AABB and aabb
(d) None of these
250 NEETChapterwise Topicwise Biology
b a c
28%
20% 8%

261. Ans.(b)
Linked genes occur on the same
chromosome and do not separate during
inheritance (complete linkage).
92Extranuclear inheritance occurs in
[CBSE AIPMT 2001]
(a) KillerParamecium
(b) KillerAmoeba
(c)Euglena
(d)Hydra
Ans.(a)
Extranuclear inheritance or cytoplasmic
inheritance is the inheritance of the
characters of only one parent (generally
the female parent). e.g. some strains of
Parameciumcalled killer strain.
93Crossing over in diploid organism is
responsible for[CBSE AIPMT 1998]
(a) dominance of genes
(b) linkage between genes
(c) segregation of alleles
(d) recombination of linked alleles
Ans.(d)
Crossing over in diploid organism is
responsible for recombination of linked
alleles.
94The polytene chromosomes were
discovered for the first time in
[CBSE AIPMT 1995]
(a)Drosophila
(b)Chironomus
(c)Musca nebulo
(d)Musca domestica
Ans.(b)
Polytene chromosomes were first time
discovered by the Italian cytologist
EG Balbiani(1881) in the salivary gland
cells ofChironomuslarva.
95Two dominant non-allelic genes are
50 map units apart. The linkage is
[CBSE AIPMT 1993]
(a)cistype
(b)transtype
(c) complete
(d) absent/incomplete
Ans.(d)
Chromosome mapping is based on two
genetic principles
(a) The frequency of crossing over
between two genes is directly
proportional to the distance
between them in the chromosome.
(b) Genes are arranged in a linear order
in the chromosome.
50 map unit distance between the genes
is quite enough to change thecis
arrangment of dominant genes intotrans.
So, there is no fixed linkage present.
96Two linked genes a and b show
20% recombination. The
individuals of a dihybrid cross
between ++/++ × ab/ab shall show
gametes [CBSE AIPMT 1989]
(a) ++80 : ab : 20
(b) ++50 : ab : 50
(c) ++ 40 : ab 40 : + a 10 : + b : 10
(d) ++ 30 : ab 30 : + a 20 : + b : 20
Ans.(c)
The gametes of a dihybrid cross
between ++/++×ab/ab will be
++ 40 : ab40 : +a10 : + b : 10
97Crossing over in diploid organism is
responsible for[CBSE AIPMT 1998]
(a) dominance of genes
(b) linkage between genes
(c) segregation of alleles
(d) recombination of linked alleles
Ans.(d)
The genes present on the same
chromosome do not always remain
together. These usually get separated
and recombine with genes present on
homologous chromosomes to form new
combinations (recombinants).
98A fruit fly heterozygous for
sex-linked genes, is mated with
normal female fruit fly. Male
specific chromosome will enter
egg cell in the proportion
[CBSE AIPMT 1997]
(a) 1 : 1 (b) 2 : 1 (c) 3 : 1 (d) 7 : 1
Ans.(a)
Genes which are present on sex
chromosomes are called sex linked
genes. MaleDrosophilacontains XY sex
chromosome, while female contains
XX-chromosomes. During gamete
formation male produces 50% male
specific gametes and 50% female
specific gametes while female produces
only one type of gametes, i.e. female
specific. As male produces two types of
gametes in equal proportion. There is an
equal opportunity to getting a male or
female offspring.
99After crossing two plants, the
progenies are found to be male
sterile. This phenomenon is found
to be maternally inherited and is
due to some genes which are
present in [CBSE AIPMT 1997]
(a) nucleus (b) chloroplast
(c) mitochondria (d) cytoplasm
Ans.(c)
Factors responsible for cytoplasmic
male sterility are located in
mitochondrial DNA. Mitochondria are
found only in eukaryotic cells.
They contain a single circular double
stranded DNA molecule (mtDNA) and
mitochondria of female parent are
transferred to progeny during fertilisation.
100When two genetic loci produce
identical phenotypes incisand
transposition, they are considered
to be [CBSE AIPMT 1995]
(a) pseudoalleles (b) different genes
(c)multiplealleles (d)partsofsamegene
Ans.(a)
Pseudoalleles are closely placed genes
producing related phenotypic effect
which is distinguishable only through a
rare crossing over, e.g. dominant star and
recessive asteroid eye traits inDrosophila.
101Genes located on Y-chromosome
are [CBSE AIPMT 1994]
(a) mutant genes (b) sex-linked genes
(c) autosomal genes (d) holandric genes
Ans.(d)
Holandric genes are those that occur on
the Y-chromosome only they are not
expressed in females. These genes are
directly transmitted from father to son.
Hairy ears (hypertrichosis) in man is
inherited through genes on
Y-chromosomes.
102Mr. Kapoor has Bb autosomal gene
pair and d allele sex-linked. What
shall be proportion of Bd in
sperms? [CBSE AIPMT 1993]
(a) 0 (b) 1/2 (c) 1/4 (d) 1/8
Ans.(c)
Mr. Kapoor will have the genotype Bb, d,
so 1/4th of the sperms will have Bd.
Principles of Inheritance and Variation 251
AB
ab
ab ab
AaBb
AB ab ab ab
AaBb
aabbaabb
+
×
Genotype is AaBb and aabb

262. 103When a certain character is
inherited only through female
parent, it probably represents
[CBSE AIPMT 1992]
(a) multiple plastid inheritance
(b) cytoplasmic inheritance
(c) incomplete dominance
(d) Mendelian nuclear inheritance
Ans.(b)
The amount of nuclear hereditary
material contributed by the two sexes is
almost equal but the cytoplasm in egg is
always much more than that of the
sperm, so when a certain character is
inherited only through female parent it
represents cytoplasmic inheritance.
104Out of 8 ascospores formed in
Neurosporathe arrangement is 2a
: 4a : 2a showing[CBSE AIPMT 1992]
(a) no crossing over
(b) some meiosis
(c) second generation division
(d) first generation division
Ans.(c)
InNeurosporaafter crossing over
between the gene and centromere, the
paired arrangement of ascospores is AA
aaaa AA or 2a : 4a : 2a. This is known as
second division segregation.
105Match the items of Column I with
Column II.[NEET (Odisha) 2019]
Column I Column II
1. XX-XO method of
sex-determination
i. Turner’s
syndrome
2. XX-XY method of
sex-determination
ii. Female
heterogamety
3. Karyotype-45 iii. Grasshopper
4. ZW-ZZ method of
sex- determination
iv. Female
homogamety
Select the correct option from the
following
1 2 3 4
(a) (ii) (iv) (i) (iii)
(b) (i) (iv) (ii) (iii)
(c) (iii) (iv) (i) (ii)
(d) (iv) (ii) (i) (iii)
Ans.(c)
The correct matches are
1. XX-XO method of sex-determination
is seen in (iii) Grasshopper, where
males have only one X-chromosome
besides autosomes and females
have a pair of X-chromosome.
2. XX-XY method of
sex-determination-(iv) Female
homogamety as seen in human
beings.
3. Karyotype-45 (i) Turner’s syndrome
with 45+XO.
4. ZW-ZZ method of
sex-determination-(ii) Female
heterogamety as seen in birds.
106Select the incorrect statement.
[NEET (National) 2019]
(a) In male grasshoppers, 50% of sperms
have no sex-chromosome
(b) In domesticated fowls, sex of
progeny depends on the type of
sperm rather than egg
(c) Human males have one of their sex
chromosomes much shorter than the
other
(d) Male fruitfly is heterogametic
Ans.(b)
The statement ‘‘in domesticated fowls,
sex depends on the type of sperm rather
than egg’’ is incorrect. The correct form
of statement is, in domesticated fowls,
sex of progeny depends on the type of
egg rather than type of sperm. In fowls,
the females are heterogametic and
produce two type of eggs containing
either (A+Z) or (A+W) chromosomes. The
males are homogametic and produce
only one type of sperm containing (A+Z)
chromosomes. Rest statements are
correct.
107Which one of the following pairs is
wronglymatched? [NEET 2018]
(a) XO type
sex-determination
– Grasshopper
(b) ABO blood grouping – Codominance
(c) Starch synthesis in
pea
– Multiple alleles
(d) TH Morgan – Linkage
Ans.(c)
In the given pairs, option (c) is wrongly
matched. Starch synthesis in pea is an
example of pleiotropy. A pleiotropic gene
is a single gene which produces many or
multiple unrelated phenotypes. Rest of
the pairs are correctly matched.
Concept EnhancerThe gene for starch
synthesis in pea seeds has two alleles B
and b. In BB genotype, large starch
grains are produced. After maturation
the seeds are round. In bb homozygous
condition, smaller starch grains are
produced and mature seeds are
wrinkled. Bb heterozygotes form round
seeds so that B seems to be dominant
allele. However, Bb seeds have starch
grains of intermediate size, showing
incomplete dominance.
108Which one of the following
conditions correctly describes the
manner of determining the sex in
the given example?
[CBSE AIPMT 2011]
(a) XO type of sex chromosomes
determine male sex in grasshopper
(b) XO condition in humans as found in
Turner syndrome, determines female
sex
(c) Homozygous sex chromosomes (XX)
produce male inDrosophila
(d) Homozygous sex chromosomes (ZZ)
determine female sex in birds
Ans.(a)
Grasshopper is an example of XO type of
sex determination in which the males
have only one X-chromosome besides
the autosomes, whereas females have a
pair of X-chromosomes.
109InDrosophila,the sex is determined
by [CBSE AIPMT 2003]
(a) the ratio of pairs of X-chromosomes
to the pairs of autosomes
(b) whether the egg is fertilised or
develops parthenogenetically
(c) the ratio of number of
X-chromosomes to the set of
autosomes
(d) X and Y-chromosomes
Ans.(a)
Calvin Bridges demonstrated that in
Drosophila,the sex is determined by
ratio of the number of X-chromosomes
to the sets of autosomes. According to
genic balance theory of
sex-determination, Y-chromosome of
Drosophiladoes not important for the
determination of sex.
110Drosophilaflies with XXY genotype
are females, but human beings with
such genotype are abnormal males.
It shows that[CBSE AIPMT 2000]
(a) Y-chromosome is essential for sex
determination inDrosophila
(b) Y-chromosome is female
determinating inDrosophila
(c) Y-chromosome is male determining
in human beings
(d) Y-chromosome has no role in sex
determination either inDrosophilaor
in human beings
252 NEETChapterwise Topicwise Biology
Sex-Determination
TOPIC 3

263. Ans.(c)
In human beings, the presence of a
Y-chromosome is required for the
development of a male sex phenotype.
X-chromosomes present in any number
(XXX, XXXX) in the absence of a
Y-chromosome gives rise to female.
Presence of even a single
Y-chromosome confers maleness. In
Drosophila,Y-chromosome plays no
significant role in sex determination. In
Drosophilamale determines were
located on autosomes. One X and two
autosomes produce male while two X (X,
X) and two autosomes produce female.
111Foetal sex can be determined by
examining cells from the amniotic
fluid by looking for
[CBSE AIPMT 1997]
(a) Barr bodies (b) autosomes
(c) chiasmata (d) kinetochore
Ans.(a)
Females have XX-chromosome,
presence of Barr body indicates female
child while absence indicates male.
Amniotic fluid contains living cells flaked
off from the skin of baby or amnion
(derived from zygote and identical to
foetus cells).
The non-dividing cells are examined. One
X-chromosome always appears in the
active state. If another is present, it is
seen in a resting state as tightly coiled
dark staining body (Barr body).
112Genetic identity of a human male is
determined by[CBSE AIPMT 1997]
(a) autosome (b) nucleolus
(c) sex chromosome
(d) cell organelles
Ans.(c)
Genetic identity of a human male is
determined by sex chromosomes.
As sex of a child is determined at the
time of fertilization. If male gamete
containing X-chromosome fertilizes
the ovum, the child would be female, if
Y-chromosome does it, the child
would be male.
113An individual exhibiting both male
and female sexual characteristics
in the body is known as
[CBSE AIPMT 1996]
(a) hermaphrodite (b) intersex
(c) gynandromorph (d) bisexual
Ans.(c)
Gynandromorph is typically male in
certain portions of the body and typically
female in others.
The cases of gynandromorphism has
been reported in man.Drosophila,
silkworm, bees, butterflies, beetles, etc
114Sex is determined in human beings
[CBSE AIPMT 1993]
(a) by ovum
(b) at the time of fertilisation
(c) 40 days after fertilisation
(d) seventh to eight week when genitals
differentiate in foetus
Ans.(b)
In human beings male produces two
types of sperms. 50% of them carry
X-chromosome and 50% have
Y-chromosome. Sex in human beings is
determined at the time of fertilisation
because sex of baby depends upon
which sperm fertilises the ovum.
115Diploid chromosome number in
humans is [CBSE AIPMT 1989]
(a) 46 (b) 44 (c) 48 (d) 42
Ans.(a)
There are 23 pairs of chromosomes in
humans in which 22 pairs of autosome
and one pair of sex chromosome is
present thus, total number of diploid
chromosome is 23 pairs = 46
chromosomes.
116A family of five daughters only is
expecting sixth issue. The chance
of its beings a son is
[CBSE AIPMT 1988]
(a) Zero (b) 25% (c) 50% (d) 100%
Ans.(c)
Chances of a baby to be either boy or girl
is always 50%. because in human beings
there are 22 pairs of autosome and one
pair of sex chromosome. Female is
homozygous while male is heterozygous
and genetically responsible for sex of the
child. Sperms are of two types, i.e.
sperms having X-chromosome
responsible for producing a girl and
sperms having Y-chromosome
responsible for producing a boy.
117Now-a-days, it is possible to detect
the mutated gene causing cancer
by allowing radioactive probe to
hybridise its complimentary DNA in
a clone of cells, followed by its
detection using autoradiography
because [NEET 2021]
(a) mutated gene partially appears on a
photographic film
(b) mutated gene completely and clearly
appears on a photographic film
(c) mutated gene does not appear on a
photographic film as the probe has no
complementarity with it
(d) mutated gene does not appear on
photographic film as the probe has
complementarity with it
Ans.(c)
Autoradiography is an imaging technique
that uses radioactive sources contained
within the exposed sample.
The single-stranded DNA or RNA,
tagged with a radioactive molecule
(probe) is allowed to hybridise to its
complementary DNA in a clone of cells
followed by detection using
autoradiography. The clone having the
mutated gene will hence not appear on
the photographic film, because the
probe will not have complementarity
with the mutated gene.
118Under which of the following
conditions will there be no change
in the reading frame of following
mRNA?
5′AACAGCGGUGCUAUU3’
[NEET (National) 2019]
(a) Deletion of G from 5th position
(b) Insertion of A and G at 4th and 5th
positions, respectively
(c) Deletion of GGU from 7th, 8th and 9th
positions
(d) Insertion of G at 5th position
Ans.(c)
The reading frame of givenmRNA will not
change even after the deletion of GGU
from 7th, 8th and 9th positions.
However, the amino acid glycine will not
be formed at third position in this case
which is being coded by GGU. In rest of
the cases, insertion or deletion of one or
two nucleotide bases would result in the
complete alteration in the reading frame
ofmRNA.
Principles of Inheritance and Variation 253
Father Mother
X Y X X
XX XX XY XY
Fertilisation
Sons
Parents
Gametes
XY XX
Daughters
Mutations
TOPIC 4

264. 119A cell at telophase stage is
observed by a student in a plant
brought from the field. He tells his
teacher that this cell is not like
other cells at telophase stage.
There is no formation of cell plate
and thus the cell is containing more
number of chromosomes as
compared to other dividing cells.
This would result in
[NEET 2016, Phase I]
(a) polyploidy
(b) somaclonal variation
(c) polyteny
(d) aneuploidy
Ans.(a)
Polyploid cells have a chromosome
number that is more than double the
haploid number, e.g.Triticum aestivum
(wheat) is a hexaploid( )6n.
120Point mutation involves
[CBSE AIPMT 2009]
(a) insertion
(b) change in single base pair
(c) duplication
(d) deletion
Ans.(b)
The point mutations involve alterations
in the structure of gene by altering the
structure of DNA, i.e. change in single
base pair.
Point mutations are of two types,
i.e. base pair substitution and frameshift
substitution.
Insertion is the addition of one or more
nitrogenous bases to a nucleotide chain.
Duplication is the presence of one block
of genes more than once in a haploid
component.
Deletion is the removal of one or more
nitrogenous bases from a nucleotide chain.
121Select the incorrect statement
from the following
[CBSE AIPMT 2009]
(a) linkage is an exception to the
principle of independent assortment
in heredity
(b) galactosemia is an inborn error of
metabolism
(c) small population size results in
random genetic drift in a population
(d) baldness is a sex limited trait
Ans.(d)
Out of the given statements (d) is
incorrect because baldness is not a
sex-limited trait.
122In a mutational event, when
adenine is replaced by guanine, it is
the case of[CBSE AIPMT 2004]
(a) frameshift mutation
(b) transcription
(c) transition
(d) transversion
Ans.(c)
In case of transition, purine base is
replaced by another purine (e.g.
A G3) and pyrimidine is replaced by
another pyrimidine (e.g.C T3) and
vice versa. In case of transversion purine
is replaced by a pyrimidine andvice
versa.
123One of the parents of a cross has
mutation in its mitochondria. In
that cross, that parent is taken as a
male. During segregation of
F
2
-progenies that mutation is
found in [CBSE AIPMT 2004]
(a) one-third of the progenies
(b) none of the progenies
(c) all of the progenies
(d) fifty per cent of the progenies
Ans.(b)
In the present case the male parent (not
female) had mutation in mitochondria,
there are negligible chances of the
mutation being inherited.
It is the female reproductive cell which
usually carries more cytoplasm and
cytoplasmic organelles than the male
cell and hence, naturally would be
expected to influence non-Mendelian
traits.
124Which of the following discoveries
resulted in a Nobel Prize?
[CBSE AIPMT 2003]
(a) Recombination of linked genes
(b) Genetic engineering
(c) X-rays induce sex-linked recessive
lethal mutations
(d) Cytoplasmic inheritance
Ans.(c)
HJ Muller was awarded Nobel Prize in
1946 for his discovery of the production
of mutations by X-ray radiation.
125Change in the sequence of
nucleotide in DNA is called as
[CBSE AIPMT 2002]
(a) mutagen (b) mutation
(c) recombination (d) translation
Ans.(b)
Change in the sequence of nucleotide in
DNA is called as mutation. The point
mutations involve minor changes in the
genetic material, while macro mutations
involve large segments of
chromosomes. ‘Frameshift mutations’
includes the addition or deletion of
nucleotide (not involving 3 base pairs) so
that the reading frame of the RNA is
shifted to left or right during translation.
126Male XX and female XY sometime
occur due to[CBSE AIPMT 2001]
(a) deletion
(b) transfer of segments in X and
Y-chromosomes
(c) aneuploidy
(d) hormonal imbalance
Ans.(d)
Hormonal inbalance may lead to
development of male characters in
female orvice versa.Deletion is the
removal of one or few nitrogenous bases
from a nucleotide chain. Aneuploidy is a
chromosomal aberration in which
certain chromosomes are present in
extra copies or certain are deficient in
number.
127During organ differentiation in
Drosophila,an organ is modified to
another organ (such as wings may
be replaced by legs). Genes
responsible for such
metamorphosis are called
[CBSE AIPMT 2000]
(a) double dominant genes
(b) plastid genes
(c) complementary genes
(d) homeotic genes
Ans.(d)
Homeotic genes are control genes which
either by getting expressed or by
remaining silent during development,
influence the differentiation of organs.
These have been found in insects, one
nematode and some plants. A DNA
sequence called homeobox, present in
these genes, is involved in specification
of organs.
254 NEETChapterwise Topicwise Biology

265. A mutation that causes a body part to
develop in appropriate position in an
organism, is called homeotic mutation,
e.g. inDrosophila, such mutation may
cause legs to develop on the head in
place of antennae.
128Mutation generally produces
[CBSE AIPMT 2000]
(a) recessive genes (b) lethal genes
(c) polygenes (d) dominant genes
Ans.(a)
Mutations generally produce recessive
genes. Mutation is a sudden heritable
change in the characteristics of an
organism. The individual which shows
these heritable changes is known as
mutant.
129Which of the following is the main
category of mutation?
[CBSE AIPMT 1999]
(a) Somatic mutation
(b) Genetic mutation
(c) Zygotic mutation
(d) All of these
Ans.(b)
Mutation is a sudden heritable change in
genes structure of an organism. The
term genetic mutation covers somatic
mutation as well as germinal mutation
(occurring during reproduction).
130Albinism is known to be due to an
autosomal recessive mutation. The
first child of a couple with normal
skin pigmentation was an albino.
What is the probability that their
second child will also be an albino?
[CBSE AIPMT 1998]
(a) 100% (b) 25%
(c) 50% (d) 75%
Ans.(b)
Since, albinism is a recessive character,
a child will be albino only if it is
homozygous for albinism genes. Since,
parents have normal skin, it means they
are heterozygous.
As a result of cross between two
heterozygous parents, 25% of the
children will be homozygous recessive.
The nature of the second child is not
affected in any way by the nature of the
first child because both are independent
events.
131Which base is responsible for
hotspots for spontaneous point
mutations? [CBSE AIPMT 1998]
(a) Guanine (b) Adenine
(c) 5-bromouracil (d) 5-methylcytosine
Ans.(d)
5-methylcytosine residues occur at the
position of each hot spot. The term ‘hot
spots’ was used by Benzer for the sites
which are more mutable than other
sites.
132Loss of an X-chromosome in a
particular cell, during its
development, results into
[CBSE AIPMT 1998]
(a) diploid individual
(b) triploid individual
(c) gynandromorphs
(d) Both (a) and (b)
Ans.(c)
Gynandromorphs are those individuals in
which one part of the body is female
while another part is male. It occurs due
to the irregularity in mitosis at the first
cleavage of the zygote.
One of the X-chromosomes of an XX
(female) zygote lags in the spindle, one
daughter nucleus receives only one
X-chromosomes, while the other
receives two X-chromosomes. A mosaic
body pattern is thus established which is
known as gynandromorph.
133The formation of multivalents at
meiosis in diploid organism is due
to [CBSE AIPMT 1998]
(a) monosomy (b) inversion
(c) deletion
(d) reciprocal translocation
Ans.(d)
The formation of multivalents at meiosis
in diploid organism is due to the
reciprocal translocation.
134A mutation at one base of the first
codon of a gene produces a
non-functional protein. Such a
mutation is referred as
[CBSE AIPMT 1997]
(a) frameshift mutation
(b) mis-sense mutation
(c) non-sense mutation
(d) reverse mutation
Ans.(b)
If mutation at one base of the first codon
of a gene takes place then all the
subsequent codons will be out of
register (genetic code).
This results into the formation of
mis-sense protein which is formed due
to the mutation in the first base of first
codon and thus called missense
mutation.
135Different mutations referrable to
the same locus of chromosome
give rise to[CBSE AIPMT 1997]
(a) pseudoalleles (b) polygenes
(c) oncogenes (d) multiple alleles
Ans.(d)
The phenomenon of multiple allelism is
the simultaneous occurrence of more
than two alleles (multiple alleles) at a
given gene locus. Any mutation
occurring within a gene (at the same
locus) will give rise to a new form or new
allele of that gene.
136HJ Muller was awarded Nobel Prize
for his [CBSE AIPMT 1996]
(a) discovery that chemicals can induce
gene mutations
(b) discovery that ionizing radiations can
induce gene mutations
(c) work on gene mapping inDrosophila
(d) efforts to prevent the use of nuclear
weapons
Ans.(a)
HJ Muller was awarded Nobel Prize for
discovering that ionizing radiations can
induce gene mutations.
137The most striking example of point
mutation is found in a disease
called [CBSE AIPMT 1995]
(a) thalassemia
(b) night blindness
(c) Down’s syndrome
(d) sickle-cell anaemia
Ans.(d)
Sickle-cell anaemia is an example of
point mutation. It is a genetic disease
reported from negroes due to a
molecular mutation of geneHb
A
on
chromosome 11 which producesβ-chain
of adult haemoglobin.
In this disease nucleotide triplet CTC is
changed to CAC, these changes takes
place at a particular point of chromosome,
so they are called as point mutation.
138Out of A==T, G≡≡C pairing, bases
of DNA may exist in alternate
valency state owing to
arrangement called
[CBSE AIPMT 1994]
Principles of Inheritance and Variation 255

266. (a) analogue substitution
(b) tautomerisational mutation
(c) frameshift mutation
(d) point mutation
Ans.(b)
Tautomerisation occurs through
rearrangement of electrons and protons
of the molecules.
Due to this the purines and pyrimidines
in DNA and RNA may exist in several
alternate forms or tautomers.
139A normal green male maize is
crossed with albino female. The
progeny is albino because
[CBSE AIPMT 1989]
(a) trait for albinism is dominant
(b) the albinos have biochemical to destroy
plastids derived from green male
(c) plastids are inherited from female
parent
(d) green plastids of male must have
mutated
Ans.(c)
Besides nucleus some genes are also
present in the cytoplasm of the female
parent, these genes are called
plasmogenes.
In the given example the progeny is
albino because of inheritance of plastids
from female parent.
140Haploids are able to express both
recessive and dominant
alleles/mutations because there
are [CBSE AIPMT 1988]
(a) many alleles for each gene
(b) two alleles for each gene
(c) only one allele for each gene in the
individual
(d) only one allele in a gene
Ans.(c)
In haploids there is only one allele for
each gene in the individual, that’s why
haploids are better for mutation work
because in them all mutations whether
dominant or recessive are expressed.
141In a cross between a male and
female, both heterozygous for
sickle-cell anaemia gene, what
percentage of the progeny will be
diseased? [NEET 2021]
(a) 50% (b) 75% (c) 25% (d) 100%
Ans.(c)
The genotype of both male and female,
heterozygous for sickle-cell anaemia
gene can be represented asHb Hb
A S
Thus,
Thus, the percentage of diseased
progeny will be 25%
142The best example for pleiotropy is
[NEET (Oct.) 2020]
(a) skin colour
(b) phenylketonuria
(c) colour blindness
(d) ABO blood group
Ans.(b)
The best example for pleiotropy is
phenylketonuria which occurs in
humans. In pleiotropy, a single gene can
exhibit multiple phenotypic expressions.
This gene is called pleiotropic gene. In
phenylketonuria, a single gene mutation
that codes for enzyme phenylalanine
hydroxylase is seen. This manifests itself
through phenotypic expression
characterised by mental retardation and
a reduction in hair and skin
pigmentation.
143Select the correct match.
[NEET (Sep.) 2020]
(a) Phenylketonuria – Autosomal
dominant trait
(b) Sickle-cell
anaemia
– Autosomal
recessive trait,
chromosome-11
(c) Thalassemia – X-linked
(d) Haemophilia – Y-linked
Ans.(b)
Option (b) is correct whereas option
(a),(c) and (d) are incorrect because
Phenylketonuriais an autosomal
recessive disorder.Due to this disorder,
enzyme phenylalanine hydroxylase is not
synthesised. This enzyme is required for
conversion of phenylalanine (PA) into
tyrosine. If enzyme production stops,
the concentration of PA in body tissues
increases,this accumulated PA gets
converted into phenyl pyruvic acid which
is responsible for damaging the brain.
Thalassemiais an autosomal recessive
disorder. Thalassemia are of two types
depending upon which protein chain of
haemoglobin gets synthesised or not
synthesised in a defective manner.
Haemophiliais an X- linked disorder in
which the clotting time is delayed.
144In which genetic condition, each
cell in the affected person, has
three sex chromosomes XXY?
[NEET (Odisha) 2019]
(a) Thalassemia
(b) Klinefelter’s syndrome
(c) Phenylketonuria
(d) Turner’s syndrome
Ans.(b)
Klinefelter’s syndrome is a genetic
condition in which each cell in the
affected person has three sex
chromosomes XXY.
It is caused due to the presence of an
additional copy of X-chromosome
resulting into a karyotype of 47, XXY.
Such individuals are sterile.
145What is the genetic disorder in
which an individual has an overall
masculine development
gynaecomastia and is sterile ?
[NEET (National) 2019]
(a) Klinefelter’s syndrome
(b) Edward syndrome
(c) Down’s syndrome
(d) Turner’s syndrome
Ans.(a)
In Klinefelter’s syndrome, individual has
overall masculine development,
gynaecomastia and is sterile. This
condition is represented as 44 +X(47)
due to the presence of an extra
X-chromosome in males.
Edward syndrome is 18 trisomy and it
causes severe developmental delay.
Down’s syndrome is 21-trisomy and it is
identified as Mongolism due to the short
stature of affected individual.
Turner’s syndrome is characterised by a
missing X-chromosome in females. It
causes sterility in females.
146Thalassemia and sickle-cell
anaemia are caused due to a
problem in globin molecule
synthesis. Select the correct
statement. [NEET 2017]
256 NEETChapterwise Topicwise Biology
Genetic Disorders
TOPIC 5
Hb
A
Hb
S
Hb
A
Hb
S
Hb
A
Hb
S
Hb
A
Hb
S
Hb Hb
A A
unaffected
Hb Hb
A S
unaffected
and
carrier
Hb Hb
A S
unaffected
and
carrier
Hb Hb
S S
affected
Parents
Gametes
F progeny
1
×

267. (a) Both are due to a qualitative defect in
global chain synthesis
(b) Both are due to a quantitative defect
in globin chain synthesis
(c) Thalassemia is due to less synthesis
of globin molecules
(d) Sickle-cells anaemia is due to a
quantitative problem of globin
molecules
Ans.(c)
Thalassemia is a autosomal recessive
disease, which occurs due to mutation in
genes. This results in reduced rate of
synthesis of the globin chains of
haemoglobin. Anaemia is the main
feature of this disease. There are two
forms of Thalassemia, i.e.α-thalassemia
(production of affectedα-globin chain,
which is governed by genes on 16th
chromosome), andβ-thalassemia
(production of affectedβ-chain, which is
governed by a gene on 11th
chromosomes).
Concept EnhancerThalassemia differs
from sickle-cell anaemia in that the
former is a quantitative problem of
synthesising few globin molecules, while
the later is a qualitative problem of
synthesising an incorrectly functioning
globin.
147A disease caused by an autosomal
primary non-disjunction is
[NEET 2017]
(a) down’s syndrome
(b) klinefelter’s syndrome
(c) turner’s syndrome
(d) sickle-cell anemia
Ans.(a)
Non-disjunction is the failure of
chromosomes to disjoin or separate and
move away to opposite poles.
Non-disjunction of 21st chromosome
during oogenesis is the cause of down’s
syndrome. It occurs due to the presence
of an additional copy of chromosome no.
21 (trisomy of 21st chromosome) is
humans.
148Which of the following most
appropriately describes
haemophilia?[NEET 2016, Phase I]
(a) X-linked recessive gene disorder
(b) Chromosomal disorder
(c) Dominant gene disorder
(d) Recessive gene disorder
Ans.(a)
Haemophilia is X-linked recessive gene
disorder. It is a blood clotting disorder
and shows criss-cross inheritance.
In this, characters from father are
transmitted to daughter and from
mother to son.
149Pick out the correct statements.
[NEET 2016, Phase I]
I. Haemophilia is a sex-linked
recessive disease.
II. Down's syndrome is due to
aneuploidy.
III. Phenylketonuria is an autosomal
recessive gene disorder.
IV. Sickle-cell anaemia is an X -
linked recessive gene disorder.
(a) II and IV are correct
(b) I, III and IV are correct
(c) I, II and III are correct
(d) I and IV are correct
Ans.(c)
Sickle-cell anaemia is an autosomal
recessive gene disorder in which
sickle-celled RBCs are formed instead of
normal ones. They carry very less
content ofO
2
as their haemoglobin is
malformed. The person suffering from
this disease show symptoms of
anaemia.
150If a colourblind man marries a
woman who is homozygous for
normal colour vision, the
probability of their son being
colourblind is[NEET 2016, Phase II]
(a) 0 (b) 0.5
(c) 0.75 (d) 1
Ans.(a)
If a colourblind man marries a woman
who is homozygous for normal colour
vision, the probability of their son
being colourblind is zero. Colour
blindness is a recessive sex-linked trait
in which the eye fails to distinguish
between red and green colour. In
females, colour blindness appears only
when both sex chromosomes carry
recessive gene( ).X X
c c
However, in
human males, the defect appears due to
single recessive gene(X Y)
c
because Y
chromosome does not carry gene for
colour vision.
This disease shows criss-cross
inheritance.
Thus, there is zero probability of son
being colourblind. Hence, option (a) is
correct.
151In the following human pedigree,
the filled symbols represent the
affected individuals. Identify the
type of given pedigree.
[CBSE AIPMT 2015]
(a) Autosomal dominant
(b) X-linked recessive
(c) Autosomal recessive
(d) X-linked dominant
Ans.(c)
The given pedigree shows the autosomal
recessive disorder.
In this disorder, the individual inherit two
mutated genes, one from each parent.
This disorder is usually passed on by two
carriers. Health is rarely affected, but
individual have one mutated gene
(recessive gene) and one normal gene
(dominant gene) for the condition.
The carriers have a 25% chance of
having an unaffected child with normal
genes, 50% chance of having an
unaffected child who also is a carrier and
a 25% chance of having an affected child
with recessive genes.
152A colourblind man marries a
woman with normal sight who has
no history of colour blindness in her
family. What is the probability of
their grandson being colour blind?
[CBSE AIPMT 2015]
(a) 0.5 (b) 1
(c) Nil (d) 0.25
Principles of Inheritance and Variation 257
MotherFather
SonDaughter
X X
XX
Ova
X
c
Y
Sperms
X
c
Y
X
c
Y
X X
c
X X
c
X Y
X Y
X
X
Colourblind manNormal woman
Parents
Gametes
Offsprings : 2 Carrier girls; 2 Normal boys
I.
II.
III.
IV.

268. Ans.(d)
When a colourblind man( )X Y
C
marries to
a woman with normal sight (XX) who has
no family history of colour blindness, all
of their sons will be normal pure and all
of their daughters will be carriers as
shown below
So in the next generation, the children of
all of their son will be normal in all
conditions (except the case in which the
wife involved is not carrier neither
colourblind). For carrier daughters.
(i) If they many to a normal man50%of
their grandsons will become
colourblind as
(ii) If carrier daughter marries to a
colourblind man50%of their
grandson will be colourblind along
with50%of the grand daughter
while rest50%of the grand
daughters will be carriers as
So in both the above cases the result
shows 50% of grand sons will be
colourblind which in terms of over all
progenies (son + daughters) comes as
25% thus confirming the probability as
0.25.
153A man whose father was colour
blind marries a woman, who had a
colour blind mother and normal
father. What percentage of male
children of this couple will be
colour blind?[CBSE AIPMT 2014]
(a) 25% (b) 0%
(c) 50% (d) 75%
Ans.(a)
As colour blindness is a sex-linked
recessive genetic disorder, for it is
present at X-chromosome. Thus,
according the gene to the situation given
in the question, a man whose father was
colour blind (will be, i.e. XY normal)
marries a woman whose mother was
colourblind and father was normal ( i.e.
this woman will be a carrier) according to
the cross given below
Thus, when marriage will happen
between a normal man and a carrier
woman, in that case percentage of a
male child to be colourblind is 25% (this
can be easily observed from the cross
given below)
154A human female with Turner’s
syndrome [CBSE AIPMT 2014]
(a) has 45 chromosomes with XO
(b) has one additional X-chromosome
(c) exhibits male characters
(d) is able to produce children with
normal husband
Ans.(a)
A human female with Turner syndrome
has the absence of one of the
X-chromosome, i.e. 45 with XO (or 44 +
XO).
Turner syndrome is a chromosomal
condition that affects the development
in females.
The most common feature of Turner
syndrome is short stature, which
become evident by the age of 5. An early
loss of ovarian function is also very
common.
The ovaries develop normally at first, but
egg cells usually die prematurely and
most ovarian tissue degenerates before
birth.
155If both parents are carriers for
thalassaemia, which is an
autosomal recessive disorder, what
are the chances of pregnancy
resulting in an affected child?
[NEET 2013]
(a) No chance
(b) 50%
(c) 25%
(d) 100%
Ans.(c)
In the given question since both parents
carry a haemoglobinopathy trait of
thalassemia the risk is 25% for each
pregnancy for an affected child.
R r
r RR Rr
r Rr rr
RR–Unaffected (25%)
Rr–Carrier (50%)
rr–Affected (25%)
So, the chances of pregnancy resulting
in an affected child is 25%.
156The incorrect statement with
regard to haemophilia is
[NEET 2013]
(a) it is a sex-linked disease
(b) it is a recessive disease
(c) it is a dominant disease
(d) a single protein involved in the
clotting of blood is affected
Ans.(c)
Out of the following statement (c) is
incorrect because haemophilia is a sex
linked recessive disease. In this disease,
a single protein that is a part of the
cascade of protein involved in the
clotting of blood is affected.
The heterozygous female for
haemophilia may transmit the disease to
sons.
258 NEETChapterwise Topicwise Biology
X
C
Y
XY
XY
X
X
+
Normal woman
XX
+
XY
All carrier
daughters
All normal
son
XX
C
XX
C
Colourblind man
X Y
XX XY
X Y
c
X
+
XX
c
+
XY
Colourblind
son
X
C
X X
c
Carrier womanNormal man
X
c
Y
XX
c
XY
X X
c c
X Y
c
X
X
c
+
Colourblind
Carrier Normal
Colourblind
XX
c
+
X Y
c
Carrier woman Colourblind man
Daughters Son
X X
c c
X X, X X,
c c
X Y, X Y
c c
50% carrier
daughter
50% colourblind
son
X (XY)
X X
c
X XY
X X, X X,
c
XY,X Y
c
Normal
daughter
Carrier
daughter
Normal
son
Colourblind
son
25% each

269. 157A normal-visioned man whose
father was colour blind, marries a
woman whose father was also
colour blind. They have their first
child as a daughter. What are the
chances that this child would be
colour blind?[CBSE AIPMT 2012]
(a) 100% (b) 0%
(c) 25% (d) 50%
Ans.(b)
Colour blindness is an X-linked disease.
So, woman whose father was colour
blind will be carrier for the disease.
So, possibility of a colour blind daughter
(i.e.,X X
c c
inF
1
-generation is 0%.
158Which one of the following symbols
and its representation, used in
human pedigree analysis is
correct? [CBSE AIPMT 2010]
Ans.(a)
The given symbol show the correct
representation in human pedigree
analysis
159Study the pedigree chart given
below.
What does it show?
[CBSE AIPMT 2009]
(a) Inheritance of a sex-linked inborn
error of metabolism like phenylketonuria
(b) Inheritance of a condition like
phenylketonuria as an autosomal
recessive trait
(c) The pedigree chart is wrong as this is
not possible
(d) Inheritance of a recessive sex-linked
disease like haemophilia
Ans.(d)
In the given pedigree chart, squares are
representing males and circles females.
InF
1
-generation, 1-male and 1-female are
diseased and in next generation only
male is diseased. This shows the
inheritance of a recessive sex-linked
disease like haemophilia.
160Which one of the following
condition in humans is correctly
matched with its chromosomal
abnormality/linkage?
[CBSE AIPMT 2008]
(a) Klinefelter’s syndrome— 44
autosomes + XXY
(b) Colour blindness — Y-linked
(c) Erythroblastosis foetalis— X-linked
(d) Down syndrome— 44 autosomes + XO
Ans.(a)
Klinefelter’s syndrome is represented by
44 autosomes + XXY. When an abnormal
egg with XX chromosome is fertilised by
a sperm carrying Y-chromosome a
zygote having XXY sex chromosomes is
formed.
The resulting young one is an abnormal
sterile male.
The 44 autosome + XO condition is due
to Turner’s syndrome. Such females are
sterile and have short stature, webbed
neck, broad shield-shaped chest, etc.
Down’s syndrome is due to the trisomy of
21st pair of chromosome.
It is characterised by moderate mental
retardation, large tongue, short stature,
stubby fingers, an enlarged liver and
spleen.
Colour blindness is an X-linked disease.
The person suffering from this disease
can’t differentiate between red and
green colours.
Erythroblastosis foetalis is caused due
to Rh factor in a child born due to
marriage betweenRh
+
man andRh
−
woman.
161A man and a woman, who do not
show any apparent signs of a
certain inherited disease, have
seven children (2 daughters and 5
sons). Three of the sons suffer
from the given disease but none of
the daughters are affected. Which
of the following mode of
inheritance do you suggest for this
disease ? [CBSE AIPMT 2005]
(a) Autosomal dominant
(b) Sex-linked dominant
(c) Sex-limited recessive
(d) Sex-linked recessive
Ans.(d)
In given problem, disease is the result of
sex-linked recessive genes. As neither
man nor woman shows signs of disease
it means woman would be carrier for
disease. In their children none of the
daughters suffer from disease, while the
sons were sufferred, it means daughters
are also carrier (i.e. X-linked recessive).
Suppose, genotype of man= X Y
Genotype of woman X X
d
=
(d-disease causing gene)
For each delivery the probability for each
combination is 25%.
So, among seven children 2 normal
daughter, 3 diseased sons and 2 normal
sons are possible.
162Haemophilia is more commonly
seen in human males than in
human females because
[CBSE AIPMT 2005]
(a) this disease is due to an X-linked
dominant mutation
(b) a greater proportion of girls die in
infancy
(c) this disease is due to an X-linked
recessive mutation
(d) this disease is due to a Y-linked
recessive mutation
Principles of Inheritance and Variation 259
X Y XX
c
XXXX
c
XYX Y
c
(normal man) (carrier woman)
Gametes
F -Generation
1
XY × X Y
c
Parents
(normal
daughter)
(carrier
daughter)
(normal
son)
(colourblind
son)
(a) = Mating between relatives
(b) = Unaffected male
(c) = Unaffected female
(d) = Male affected
=Mating between relatives.
Man
XY
X Y
Woman
X X
d
X
d X
X X
d
X X
(carrier but
normal in
appearance
daughter)
(normal
daughter)
(diseased
son)
X Y
d
(normal
son)
X Y

270. Ans.(c)
Haemophilia, a hereditary (recessive
X-linked) disease is caused due to fault
in genes controlling the factor VIII and IX,
located on X-chromosome. The male
carries only one X-chromosome, other
sex chromosome carries no genes for
blood clotting, so the condition is usually
seen only in males where only one faulty
chromosome is needed.
While a female with one faulty
X-chromosome will be carrier. So, in
females two faulty X-chromosomes are
needed to cause the disease.
163A woman with normal vision, but
whose father was colour blind,
marries a colour blind man.
Suppose that the fourth child of
this couple was a boy. This boy
[CBSE AIPMT 2005]
(a) must have normal colour vision
(b) will be partially colour blind since he is
heterozygous for the colour blind
mutant allele
(c) must be colour blind
(d) may be colour blind or may be of
normal vision
Ans.(d)
A woman, whose father was colour blind,
will be carrier for colour blind trait.
Marriage of this woman with a colour
blind man will result into following
possibilities.
164A normal woman whose father was
colour blind is married to a normal
man. The sons would be
[CBSE AIPMT 2004]
(a) 75% colour blind
(b) 50% colour blind
(c) all normal
(d) all colour blind
Ans.(b)
The genotype of normal woman with
colour blind father=XX
c
The genotype of normal man=XY
∴50% of the sons would be colour blind.
Refer Ans 32.
165Down’s syndrome is caused by an
extra copy of chromosome number
21. What percentage of offspring
produced by an affected mother
and a normal father would be
affected by this disorder?
[CBSE AIPMT 2003]
(a) 50% (b) 25% (c) 100% (d) 75%
Ans.(a)
In the given question 50% of offsprings
would be affected by this disorder
because Down’s syndrome is the result
of trisomy, in which chromosome pair
number 21st contains an extra copy of
chromosome (2A + 1). Affected mother
will produce 50% normal egg cells and
rest 50% eggs are of abnormal type.
166Which one of the following
conditions though harmful in itself,
is also a potential saviour from a
mosquito borne infectious disease?
[CBSE AIPMT 2003]
(a) Pernicious anaemia
(b) Leukaemia
(c) Thalassemia
(d) Sickle-cell anaemia
Ans.(d)
Sickle-cell anaemia is a genetic disorder
in which abnormal haemoglobin is
formed because valine replaces
glutamic acid at the sixth position in
β-chain of haemoglobin, but the persons
having this disease do not suffer from
malaria as the parasite fails to thrive in
sickle-shaped RBCs.
167If a diploid cell is treated with
colchicine then it becomes
[CBSE AIPMT 2002]
(a) triploid (b) tetraploid
(c) diploid (d) monoploid
Ans.(b)
Cholchicum autumnaleprovides an
alkaloid called colchicine which is used
in plant breeding for doubling the
chromosome number. Treatment with
0.1% colchicine inhibits spindle
formation so that chromatids fail to
separate during anaphase.
168Which of the following is the
example of sex-linked disease?
[CBSE AIPMT 2002]
(a) AIDS (b) Colour blindness
(c) Syphilis (d) Gonorrhoea
Ans.(b)
Colour blindness and haemophilia are well
known examples of sex-linked diseases.
169Pleiotropic gene is
[CBSE AIPMT 2002]
(a) haemophilia
(b) thalassemia
(c) sickle-cell anaemia
(d) colour blindness
Ans.(c)
The ability of a gene to affect an
organism in many ways is called
pleiotropy (Gr.Pleion– more) and that
gene is called as pleiotropic gene, e.g.
individuals heterozygous for the
sickle-cell anaemia(Hb Hb )
A S
are
resistant to malaria.
170Number of Barr bodies in XXXX
female [CBSE AIPMT 2001]
(a) 1 (b) 2 (c) 3 (d) 4
Ans.(c)
The number of Barr bodies in XXXX
female are 3. Barr body is a
condensed mass of chromatin found in
the nuclei of placental mammals which
contain one or more X-chromosomes, so
named after its discoverer Murray Barr.
The number of Barr bodies is one less
than the number of X-chromosomes
present.
171Haemophilic man marries a normal
woman. Their offspring will be
[CBSE AIPMT 1999]
(a) all boys haemophilic
(b) all normal
(c) all girls haemophilic
(d) all haemophilic
Ans.(b)
Haemophilia is also a sex-linked
recessive disease (like colour blindness).
260 NEETChapterwise Topicwise Biology
x X x Y
(carrier
woman)
(colourblind
man)
x x x X X Yx Y
(colourblind
daughter)
(carrier
daughter)
(colourblind
son)
(normal
son)
X X
c
XY
X
c
X X Y
X X
c
X Y
c
XX XY
(normal woman) × (normal man)
(carrier
daughter)
(colourblind
son)
(normal
daughter)
(normal
son)

271. None of the children would suffer from
haemophilia, though girls would be
carriers of the disease.
Results : All daughters are carrier while
all sons are normal.
172A woman with two genes (one on
each X-chromosome) for
haemophilia and one gene for
colour blindness on the
X-chromosomes marries a normal
man. How will the progeny be?
[CBSE AIPMT 1998]
(a) All sons and daughters haemophilic
and colour blind
(b) Haemophilic and colour blind
daughters
(c) 50% haemophilic colour blind sons
and 50% haemophilic sons
(d) 50% haemophilic daughters and 50%
colour blind daughters
Ans.(c)
Haemophilia and colour blindness both
are recessive X-linked traits. They
express in males when present in single
copy (heterozygous) but in females they
express only when present in
homozygous condition.
Results
(a) 50% sons are colour blinds and
haemophilic.
(b) 50% sons are haemophilic only.
(c) 50% daughters are carrier for
colour blindness and haemophilia.
(d) 50% daughters are carrier for
haemophilia only.
173Mental retardation in man,
associated with sex chromosomal
abnormality is usually due to
[CBSE AIPMT 1998]
(a) reduction in X-complement
(b) increase in X-complement
(c) moderate increase in Y-complement
(d) large increase in Y-complement
Ans.(b)
Sterile males with undeveloped testes,
mental retardation, etc. are due to
increase in their X-complement which
takes place in a disorder called.
Klinefelter’s syndrome. These are
formed by union of an XX egg and a
normal Y sperm or normal X egg and
abnormal XY sperm. The individual thus
has 47 chromosomes (44 + XXY).
174A man with a certain disease
marries a normal woman. They
have eight children (3 daughters
and 5 sons). All the daughters
suffer from their father’s disease
but none of the sons are affected.
Which of the following mode of
inheritance do you suggest for this
disease?[CBSE AIPMT 1996, 2002]
(a) Sex-linked recessive
(b) Sex-linked dominant
(c) Autosome dominant
(d) Sex-limited recessive
Ans.(b)
Daughters have 2 X-chromosomes one
of them is from father and other comes
from mother, in this case all the
daughters are suffering from the fathers
disease hence, X-chromosome of father
must be carrying a dominant trait, i.e.
inheritance pattern is sex-linked dominant.
175A person with 47 chromosomes
due to an additional Y-chromosome
suffers from a condition called
[CBSE AIPMT 1996, 97]
(a) Down’s syndrome
(b) Super female
(c) Turner’s syndrome
(d) Klinefelter’s syndrome
Ans.(d)
HF Klinefelter first described this
condition in 1942. The chromosome number
is 2n= 47 with the formula 44A + XXY.
Phenotypically these individuals are
males, but they can show some female
secondary sexual characteristics and are
usually sterile.
176A woman with albinic father
marries an albinic man. The
proportion of her progeny is
[CBSE AIPMT 1994]
(a) 2 normal : 1 albinic
(b) all normal
(c) all albinic
(d) 1 normal : 1 albinic
Ans.(d)
Daughter of an albinic father will be
carrier of this disease, when such
woman gets married to an albinic man,
50% of her progeny will be normal and
50% will be albinic.
177A colourblind woman marries a
normal visioned male. In the
offspring [CBSE AIPMT 1994]
(a) both son and daughter are colour
blind
(b) all daughters are colour blind
(c) all sons are normal
(d) all sons are colour blind
Ans.(d)
When a colour blind woman marries a
normal visioned male, daughters will be
carrier for this disease and all sons will
be colour blind, this can be represented
as follows
178Of both normal parents, the chance
of a male child becoming colour
blind are [CBSE AIPMT 1993]
(a) no
(b) possible only when all the four grand
parents had normal vision
(c) possible only when father’s mother
was colour blind
(d) possible only when mother’s father
was colour blind
Principles of Inheritance and Variation 261
hc
X
h
X
(Woman)
XY
(Man)
hc
X
h
X X YGametes
Parents
X Y
hc
X
h
X
hc
X X
hc
X Y
h
XX
h
XY
+
(haemophilic man)
XX
(normal woman)
h
X
X
X X
h
X
Y
h
XX
h
X X
XY XY
Gametes
Fertilisation
h
X Y
Y
Parents
(Carrier) (Carrier)
+
´
X X
c
X X
c
(carrier
daughter)
(colourblind
son)
Parents
Progenies
Gametes
(carrier
daughter)
(colourblind
son)
X Y
c
X Y
c
X
c
XX
c
Y
X
c
XX
c
Y

272. Ans.(d)
Colour blindness is a X-linked recessive
disease and the chance of a male child
becoming colour blind of a normal
parents is only when mother’s father was
colour blind.
This is criss-cross inheritance in which
genes are transferred to a child from his
maternal grandfather through his
mother.
179Of a normal couple, half the sons
are haemophilic while half the
daughters are carriers. The gene is
located on [CBSE AIPMT 1993]
(a) X-chromosome of father
(b) Y-chromosome of father
(c) one X-chromosome of mother
(d) both the X-chromosomes of mother
Ans.(c)
Haemophilia is a sex-linked disease.
Gene of this disease is located on
X-chromosome. In the given case where
half the sons are haemophilic and half
the daughters are carriers this is
possible only when the gene responsible
for haemophilia is located on one
X-chromosome of mother.
Daughters
50% = Carrier, 50% = Normal
Sons
50% = Haemophilic, 50% = Normal
180In human beings 45
chromosomes/single X/XO
abnormality causes
[CBSE AIPMT 1992]
(a) Down’s syndrome
(b) Klinefelter’s syndrome
(c) Turner’s syndrome
(d) Edward’s syndrome
Ans.(c)
In human beings 45 chromosomes/XO
abnormality causes Turner’s syndrome.
Henery H Turnerfirst described this
condition in 1938. Chromosomal formula
is 44+ XO.
Phenotypically these individuals are
females but ovaries are rudimentary and
always sterile.
181A colourblind mother and normal
father would have
[CBSE AIPMT 1992, 99, 2006]
(a) colour blind sons and normal/carrier
daughters
(b) colour blind sons and daughters
(c) all colour blind
(d) all normal
Ans.(a)
A colourblind mother and normal father
would have colour blind sons and carrier
daughters. Daughters will be normal
phenotypically but they will be carrier
genotypically.
182Down’s syndrome is due to
[CBSE AIPMT 1992, 2000, 02, 03]
(a) crossing over
(b) linkage
(c) sex-linked inheritance
(d) non-disjunction of chromosomes
Ans.(d)
Non-disjunction of 21st chromosome
during oogenesis is the cause of Down’s
syndrome. It is also called mongolian
syndrome.
183A colourblind girl is rare because
she will be born only when
[CBSE AIPMT 1991]
(a) her mother and maternal grandfather
were colourblind
(b) her father and maternal grandfather
were colourblind
(c) her mother is colour blind and father
has normal vision
(d) parents have normal vision but grand
parents were colourblind
Ans.(b)
For a girl to be colourblind, the genotype
of her father should beX Y
C
and of her
mother eitherX X
C
orX X
C C
, whereX
C
represents colour blind gene.
In the given options this is only possible
when her father and maternal
grandfather were colour blind.
184Which one is a hereditary disease?
[CBSE AIPMT 1990]
(a) Cataract (b) Leprosy
(c) Blindness (d) Phenylketonuria
Ans.(d)
Phenylketonuria was discovered by the
Norwegian physician A Fooling in 1934,
an autosomal recessive mutation of
gene on chromosome 12.
Phenylketonuria results when there is a
deficiency of liver enzyme phenylalanine
hydroxylase that converts phenylalanine
into tyrosine. Increased phenylalanine in
the blood interferes with brain
development, muscles and cartilage of
the legs may be defective and the
patients cannot walk properly.
185Haemophilia is more common in
males because it is a
[CBSE AIPMT 1990]
(a) recessive character carried by
Y-chromosome
(b) dominant character carried by
Y-chromosome
(c) dominant trait carried by
X-chromosome
(d) recessive trait carried by
X-chromosome
Ans.(d)
Haemophilia is a disorder, which is
sex-linked (X-chromosome) recessive
condition. Males have only one
X-chromosome, so this disease appears
in them more then female as they have
2X-chromosomes.
186Both husband and wife have
normal vision though their fathers
were colour blind. The probability
of their daughter becoming colour-
blind is [CBSE AIPMT 1990]
(a) 0% (b) 25% (c) 50% (d) 75%
Ans.(a)
In this case when the fathers of both
husband and wife were colour blind and
they have normal vision, husband have
normal vision while wife is carrier of this
disease. Daughter of such parents will be
carrier but there is no chance of her to
be colour blind.
187In Down’s syndrome of a male
child, the sex complement is
[CBSE AIPMT 1990]
(a) XO (b) XY (c) XX (d) XXY
Ans.(b)
XY is the sex complement of a male child
with Down’s syndrome. The cause of
Down’s syndrome is non-disjunction of
21st chromosome during oogenesis. This
chromosomal abnormality is related with
autosome so, the sex complement of a
male child in this syndrome will be XY.
262 NEETChapterwise Topicwise Biology
(carrier
mother)
(normal
father)
X
h
X X Y
X X
h
X Y
h
XYXX
(carrier
daughter)
(haemophilic
son)
(normal
daughter)
(normal
son)
Parents
Progenies
(colour blind
mother)
(normal
father)
X
c
X
c
X Y
X X
c
X X
c
X Y
c
X Y
c
(Carrier
daughters)
(Colour blind
sons)

273. 01Which one of the following
statements about histones is
wrong? [NEET 2021]
(a) Histones are organised to form a unit
of
8 molecules.
(b) The pH of histones is slightly acidic
(c) Histones are rich in amino acids
lysine and arginine
(d) Histones carry positive charge in the
side chain
Ans. (b)
Statement in option (a), (c) and (d) are
correct. Histone proteins are composed
of basic amino acid. These proteins
attach to the DNA to form the
nucleosome. Histones are organised to
form a unit of eight molecules called as
histone octamer.
Histones are rich in basic amino acid
residues lysine and arginine. It carries
positive charges in their side chains and
negatively charge DNA wrap around it.
Statement in option (b) is incorrect and
can be corrected as
The pH of histones is slightly basic.
02Which is the ‘‘only enzyme’’ that has
“capability” to catalyse initiation,
elongation and termination in the
process of transcription in
prokaryotes? [NEET 2021]
(a) DNA-dependent DNA polymerase
(b) DNA-dependent RNA polymerase
(c) DNA ligase
(d) DNase
Ans.(b)
Prokaryotes utilize one RNA
polymerase for transcription of all
types of RNA. The enzyme RNA
polymerase is needed for RNA
formation from DNA, i.e. DNA
dependent RNA polymerase. It occurs
in the cytoplasm of prokaryotic cells.
RNA polymerase is the only enzyme
which, has the capability to catalyse
all initiation, elongation and
termination in prokaryotes.
03Which of the following RNAs is not
required for the synthesis of
protein ? [NEET 2021]
(a)mRNA (b) tRNA
(c)rRNA (d) siRNA
Ans. (d)
siRNA mainly protect the cell from
exogenousmRNA attacks. It degrades
the growingmRNA and stop gene
expression. It is highly specific and
reduces the synthesis of particular
proteins by reducing the translation of
specific messenger RNAs. Hence,siRNA
is not required for protein synthesis but
is used to reduce its synthesis.
WhereasrRNA,mRNA andtRNA are
required for synthesis of protein.
04lf adenine makes 30% of the DNA
molecule, what will be the
percentage of thymine, guanine
and cytosine in it?[NEET 2021]
(a) T : 20, G : 30, C : 20
(b) T : 20, G : 20, C : 30
(c) T : 30, G : 20, C : 20
(d) T : 20, G : 25, C : 25
Ans. (c)
Chargaff rule - In DNA there is always
equality in quantity between the bases A
and T and between the bases G and C.
According to Chargaff rule
(A)+(G)+(C)+(T)=100%
A=30% therefore T is also 30%
Therefore G+C =100% – 60% = 40%
Hence, G =20% and C=20%
05Complete the flow chart on central
dogma [NEET 2021]
(a) (a)-Replication: (b)-Transcription;
(c)-Transduction: (d)-Protein
(b) (a)-Translation: (b)-Replication;
(c)-Transcription: (d)-Transduction
(c) (a)-Replication: (b)-Transcription;
(c)-Translation (d)-Protein
(d) (a)-Transduction; (b)-Translation
(c)-Replication; (d)-Protein
Ans. (c)
Central dogma of molecular biology was
proposed by Francis Crick which states
that the genetic information flows from
DNA→RNA→Protein.
Here,a b c, ,anddare
a-Replication,b-Transcription,
c-Translation,d-Protein
06E. colihas only4 6 10
6
.×base pairs
and completes the process of
replicaton within 18 minutes, then
the average rate of polymerisation
is approximately[NEET (Oct.) 2020]
(a) 2000 bp/s (b) 3000 bp/s
(c) 4000 bp/s (d) 1000 bp/s
Ans. (c)
E.colihas46 10
6
.×base pairs.
MolecularBasis
ofInheritance
28
The Genetic Material-
DNA and RNA
TOPIC 1

274. It completes replication process in 18
minutes i.e.18 60×seconds.
Rate of polymerisation=
×
×
4.6 10 bp
18 60 s
6
=
×
×
=
×46 10
18 6
46 10
108
5 4
.
=
460000
108
=42591bp/s
or approximately 4000 bp/sec
Thus, the correct option is (c).
07Name the enzyme that facilitates
opening of DNA helix during
transcription.[NEET (Sep.) 2020]
(a) DNA helicase
(b) DNA polymerase
(c) RNA polymerase
(d) DNA ligase
Ans. (c)
The correct option is (c) because
RNA polymerase facilitate opening of
DNA helix during transcription. RNA
polymerase is the main transcription
enzyme. Transcription begins when RNA
polymerase binds to a promoter
sequence near the beginning of a gene.
DNA helicases function in other cellular
processes where double-stranded DNA
must be separated, including DNA repair
and transcription.
DNA ligases helps in joining breaks in the
phosphodiester backbone of DNA that
occur during replication. DNA
polymerase does not function during
transcription.
08Who coined the term ‘Kinetin’?
[NEET (Oct.) 2020]
(a) Skoog and Miller
(b) Darwin
(c) Went
(d) Kurosawa
Ans. (a)
The term kinetin was coined by Skoog
and Miller in 1955. Chemically kinetin is
6-furfuryl aminopurine. It was the first
cytokinin to be discovered from the
degraded auto claved herring sperm
DNA. Kinetin does not occur naturally.
09In the polynucleotide chain of DNA,
a nitrogenous base is linked to
the –OH of [NEET (Oct.) 2020]
(a)2′C pentose sugar
(b)3′C pentose sugar
(c)5′C pentose sugar
(d) 1′C pentose sugar
Ans. (d)
In a DNA polynucleotide chain, a
nitrogenous base is linked to the hydroxy
( OH) of1 C′pentose sugar. It is
represented in the structure given below
10The term ‘Nuclein’ for the genetic
material was used by
[NEET (Oct.) 2020]
(a) Franklin (b) Meischer
(c) Chargaff (d) Mendel
Ans. (b)
The nucleic acid was first reported by
Friedrich Miescher in 1869 from the
nuclei of pus cells and was named
nuclein.
11Which of the following statements
is correct? [NEET (Sep.) 2020]
(a) Adenine pairs with thymine
through one H-bond
(b) Adenine pairs with thymine through
three H-bonds
(c) Adenine does not pair with thymine
(d) Adenine pairs with thymine through
two H-bonds
Ans. (d)
The statement in option (d) is correct
because
Adenine pairs with thymine through two
H-bonds, i.e. A = T and the cytosine pairs
with guanine by three hydrogen bonds .
Between the G-C base pairs there are 3
hydrogen bonds which makes this bond
pair stronger than the A-T base pair.
12If the distance between two
consecutive base pairs is 0.34 nm
and the total number of base pairs
of a DNA double helix in a typical
mammalian cell is6 6 10
9
.×bp ,
then the length of the DNA is
approximately[NEET (Sep.) 2020]
(a) 2.5 meters (b) 2.2 meters
(c) 2.7 meters (d) 2.0 meters
Ans. (b)
The distance between two consecutive
base pairs is 0.34 nm(0.34 10 m).
9
×
−
The
length of DNA double helix in a typical
mammalian cell can be calculated by
multiplying the total number of bp with
distance between the two consecutive
bp, i.e.6.6 10 bp
9
× 0.34 10
9
× ×
−
m/bp
=2.2metres (the length of DNA). Thus,
option (b) is correct.
13In RNAi, the genes are silenced
using [NEET (Odisha) 2019]
(a)dsRNA
(b)ssDNA
(c)ssRNA
(d)dsDNA
Ans. (a)
In RNAi, the genes are silenced using
dsRNA. RNA interference (RNAi) takes
place in all eukaryotic organisms as a
method of cellular defence. This
method involves silencing of a specific
mRNA due to a complementarydsRNA
molecule that binds to and prevents
translation of themRNA (silencing).
14What initiation and termination
factors are involved in transcription
in eukaryotes?[NEET (Odisha) 2019]
(a)σandρ, respectively
(b)αandβ, respectively
(c)βandγ, respectively
(d)αandσ, respectively
Ans.(a)
This question is not correct because out
of the given initiation and termination
factors, none is involved in transcription
in eukaryotes. Only option (a) gives
initiation and termination factors which
are involved in transcription. These
factors (σandρ) initiate and terminate
transcription in prokaryotes (not in
eukaryotes). Initiation and termination
factors involved in transcription in
eukaryotes are General Transcription
Factors (TF IIA - TF II H) and
Transcription Termination Factor-1
(TTF-1), respectively.
15Which scientist experimentally
proved that DNA is the sole genetic
material in bacteriophage?
[NEET (Odisha) 2019]
(a) Beadle and Tatum
(b) Meselson and Stahl
(c) Hershey and Chase
(d) Jacob and Monod
264 NEETChapterwise Topicwise Biology
HOPO
O
O
–
4' 1'
3' 2'
OHH
O
N
NH
2
N
N
N
Phosphate
Deoxyribose
sugar
Nitrogenous
base
5'

275. Ans. (c)
Alfred Hershey and Martha Chase (1952)
experimentally proved that DNA is the
sole genetic material in bacteriophage.
On the other hand, Beadle and Tatum
(1940s) experimentally showed one
gene-one enzyme hypothesis using
Neurospora. Meselson and Stahl first
showed that DNA replicates
semiconservatively through experiments
onE.coli.Jacob and Monod were first to
explainlacoperon.
16What will be the sequence ofmRNA
produced by the following stretch
of DNA?
3'ATGCATGCATGCATG5'
TEMPLATE STRAND
5'TACGTACGTACGTAC3' CODING
STRAND [NEET (Odisha) 2019]
(a) 3'- AUGCAUGCAUGCAUG 5'
(b) 5'- UACGUACGUACGUAC 3'
(c) 3'- UACGUACGUACGUAC 5'
(d) 5'- AUGCAUGCAUGCAUG 3'
Ans. (b)
ThemRNA will be complementary to the
DNA strand, but in RNA, uracil will be
present in place of thymine. If the
template strand is 3′- A T G C A T G C A T
G C A T G - 5 then the base sequence of
mRNA for the given DNA strand will be 5′
- U A C G U A C G U A C G U A C - 3’.
17Which of the following nucleic acids
is present in an organism having
70S ribosomes only?
[NEET (Odisha) 2019]
(a) Single-stranded DNA with protein
coat
(b) Double-stranded circular naked DNA
(c) Double-stranded DNA enclosed in
nuclear membrane
(d) Double-stranded circular DNA with
histone proteins
Ans. (b)
Double-stranded circular naked DNA
type of nucleic acid is present in an
organism having 70S ribosomes.
These are present in prokaryotic
organisms or cells. All prokaryotic cells
have a single double-stranded (double
helix), circular DNA molecule for their
genetic material. This DNA is attached to
the inner cell membrane where the DNA
replicating machinery is located. The
DNA is ‘‘naked’’, it does not have proteins
associated with it as eukaryotic DNA
does.
18Purines found both in DNA and RNA
are [NEET (National) 2019]
(a) adenine and guanine
(b) guanine and cytosine
(c) cytosine and thymine
(d) adenine and thymine
Ans.(a)
Adenine and guanine are the purines
which are found both in DNA and RNA.
Cytosine and thymine are the pyrimidines
which are found in DNA. In case of RNA,
thymine is replaced by uracil.
19The experimental proof for
semiconservative replication of
DNA was first shown in a
[NEET 2018]
(a) plant (b) bacterium
(c) fungus (d) virus
Ans. (b)
The experimental proof for
semiconservative replication of DNA was
first shown in a bacterium,Escherichia
coli.It was discovered by Meselson and
Stahl (1958).
Interpretation of results of experiment
of Meselson and Stahl (1958) to prove
semi-conservative replication of DNA
In this mode of replication, one strand of
parent DNA is conserved in the progeny
while the second is freshly synthesised.
Meselson and Stahl proved this by using
heavy isotope of Nitrogen( )N
15
.
20The final proof for DNA as the
genetic material came from the
experiments of [NEET 2017]
(a) Griffith
(b) Hershey and Chase
(c) Avery, MacLeod and McCarty
(d) Hargobind Khorana
Ans. (b)
The final proof that DNA is the genetic
material came from the experiments of
Alfred Hershey and Martha Chase (1952).
Griffith’s experiment proved the
existance of genetic material while
Avery, MacLeod and McCarty worked to
determine the biochemical nature of
transforming principle.
Concept EnhancerHershey and Chase
during their experiment, grew viruses in
two mediums, one containing
32
Pand
other
35
S, when these were allowed to
infect bacteria, they observed that
viruses containing
32
PDNA were
radioactive while those with
35
Sprotein
were not radioactive. Hence, DNA not
protein coat entered bacterial cells from
viruses.
21The association of histone H1 with
a nucleosome indicates[NEET 2017]
(a) transcription is occurring
(b) DNA replication is occurring
(c) the DNA is condensed into chromatin
fibre
(d) the DNA double helix is exposed
Ans. (c)
The association of H
1
histone with
nucleosome indicates that DNA remains
in its condensed form.
Concept EnhancerIn eukaryotes, DNA
packaging is carried out with the help of
histone proteins. Nucleosome is the unit
of compaction. Its core consists of four
pairs of histones(H A, H B, H and H )
2 2 3 4
.
The linker DNA, consisting of H
1
histone
connects two adjacent nucleosomes.
They together constitute
chromatosome. It gives rise to a
chromatin fibre after further
condensation.
Molecular Basis of Inheritance 265
Parent DNA molecule
15
N
14
N
15
N
15
N
14
N
14
N 14
N
15
N
First replication
H Histone
1
DNA
Octamer of Histones
A nucleosome

276. 22Spliceosomes are not found in cells
of [NEET 2017]
(a) plants
(b) fungi
(c) animals
(d) bacteria
Ans. (d)
Spliceosome is a large molecular
complex found in nucleus of eukaryotic
cells of plants, animals and fungi, etc. It
is assembled fromsnRNAs and protein
complexes that plays an important role
in splicing of introns. Spliceosome is
absent in cells of bacteria.
23DNA replication in bacteria occurs
[NEET 2017]
(a) during S-phase
(b) within nucleolus
(c) prior to fission
(d) just before transcription
Ans. (c)
Bacteria lack a cell nucleus. Due to their
primitive nature they lack a well marked
S-phase. In bacteria DNA replication
occurs before fission.
Concept Enhancer: Bacterial cell cycle
is divided into theB C,andDperiods. The
Bperiod extends from the end of cell
division to the beginning of DNA
replication. DNA replication occurs
during theCperiod. TheDperiod refers
to the stage between the end of DNA
replication and the division of bacterial
cell into two daughter cells.
24Which of the following RNAs should
be most abundant in animals cell?
[NEET 2017]
(a)rRNA (b) tRNA
(c)mRNA (d) miRNA
Ans. (a)
There are three main types of RNA, i.e.
rRNA,tRNA andmRNA.rRNA is the most
abundant form of RNA; because it is
responsible for coding and protein
synthesis in the cell and associated with
ribosomes.mRNA provides the template
for translation.tRNA brings amino acids
and reads the genetic code.
25A complex of ribosomes attached
to a single strand of RNA is known
as [NEET 2016, Phase I]
(a) polymer
(b) polypeptide
(c) okazaki fragment
(d) polysome
Ans. (d)
In prokaryotes, several ribosomes may
attach to singlemRNA and form a chain
called polyribosomes or polysomes.
26DNA-dependent RNA polymerase
catalyses transcription on one
strand of the DNA which is called
the [NEET 2016, Phase II]
(a) template strand (b) coding strand
(c) alpha strand (d) anti strand
Ans.(a)
DNAdependent RNA polymerase
catalyses transcription on one strand of
the DNA called atemplate strand. A
template can be considered as one of
those strands of DNA which decodes its
information directly through RNA
polymerase.
This information is then restored within
the RNA molecule and transferred
outside the nucleus for protein synthesis
within the cytoplasm.
27A molecule that can act as a
genetic material must fulfill the
traits given below, except
[NEET 2016, Phase II]
(a) it should be able to express itself in
the form of ‘Mendelian characters’
(b) it should be able to generate its
replica
(c) it should be unstable structurally and
chemically
(d) it should provide the scope for slow
changes that are required for
evolution
Ans. (c)
A molecule that can act as a genetic
material must be unstable structurally
and chemically.
The criteria that a molecule must fulfil to
act as a genetic material are as following
(i) It should be able to replicate.
(ii) It should be chemically and
structurally stable.
(iii) It should provide the scope for slow
changes, i.e. mutations which are
required for evolution.
(iv) It should be able to express itself in
the form of ‘Mendelian characters’.
28Which of the followingrRNAs act
as structural RNA as well as
ribozyme in bacteria?
[NEET 2016, Phase II]
(a) 5srRNA (b) 18 srRNA
(c) 23srRNA (d) 5 ⋅8srRNA
Ans. (c)
Bacterial cells use their 23srRNA as an
enzyme during protein synthesis. This is
the only non-proteinaceous enzyme
known so far.
29Taylor conducted the experiments
to prove semiconservative mode of
chromosome replication on
[NEET 2016, Phase II]
(a)Vinca rosea
(b)Vicia faba
(c)Drosophila melanogaster
(d)E. coli
Ans. (b)
The use of radioactive thymidine to
detect the semiconservative mode of
replication of newly synthesised DNA in
the chromosomes was performed on
Vicia fababy Taylor and colleagues in
1958. This experiment proved that the
DNA in chromosomes replicates
semiconservatively. Hence, the option (b)
is correct.
30The equivalent of a structural gene
is [NEET 2016, Phase II]
(a) muton (b) cistron
(c) operon (d) recon
Ans. (b)
Cistron is the segment of DNA which
determines the synthesis of complete
polypeptide. Thus, it is considered as
equivalent to a structural gene.
Therefore, option (b) is correct and
others are incorrect.
Concept EnhancerEukaryotic structural
gene is monocistronic whereas
prokaryotic structural gene is
polycistronic.
Muton Smallest unit of DNA in which
mutation occurs.
Operon Functional unit of genomic DNA
containing a cluster of genes under
control of single promoter.
Recon Smallest unit of DNA for
recombination.
266 NEETChapterwise Topicwise Biology
BperiodCperiod Dperiod
Chromosome
replication
Cell
division
Bacterial cell cycle

277. 31Which one of the following is not
applicable to RNA?
[CBSE AIPMT 2015]
(a) Complementary base pairing
(b)5′phosphoryl and3′hydroxyl ends
(c) Heterocyclic nitrogenous bases
(d) Chargaff’s rule
Ans. (d)
Chargaff’s rule is not applicable to RNA.
He is the generalisations formulated
about DNA structure. The rule states
that DNA from any cell of all organisms
should have a 1 : 1 ratio (base pair rule) of
pyrimidine and purine bases, i.e. the
amount of guanine is equal to cytosine
and the amount of adenine is equal to
thymine. Further complementary base
pairing is sometimes, visible in RNA as
well (in doubled stranded RNAs of
viruses) hence option (a) is not taken into
consideration.
32Identify the correct order of
organisation of genetic material
from largest to smallest.
[CBSE AIPMT 2015]
(a) Chromosome, gene, genome,
nucleotide
(b) Genome, chromosome, nucleotide,
gene
(c) Genome, chromosome, gene,
nucleotide
(d) Chromosome, genome, nucleotide,
gene
Ans. (c)
The correct order of organisation of
genetic material from largest to smallest
is as follows
Genome, chromosome, gene,
nucleotide.
GenomeIt is the total genetic material
of an individual.
ChromosomeIt is a packed and
organised structure containing most of
the DNA of a living organism.
Gene It is a segment of DNA that
encodes for a protein.
Nucleotide It is one of the structural
components, or building blocks, of DNA
and RNA.
33Transformation was discovered by
[CBSE AIPMT 2014]
(a) Meselson and Stahl
(b) Hershey and Chase
(c) Griffith
(d) Watson and Crick
Ans. (c)
Transformation was discovered by F
Griffith [1928]. He isolated the DNA as
genetic material that inherit the genetic
information between two generations by
using two strain ofPneumococcus
bacteria which infect mice. i.e. a type III
S(smooth) and type II R (rough) strain.
34Removal of RNA polymerase-III
from nucleoplasm will affect the
synthesis of[CBSE AIPMT 2012]
(a)tRNA (b) hnRNA
(c)mRNA (d) rRNA
Ans. (a)
RNA polymerase III transcribestRNA,
thereforetRNA synthesis will be
affected. RNA polymerase-II synthesises
mRNA while, RNA polymerase-I
synthesisrRNA in eukaryotes.
35Which one of the following is not a
part of a transcription unit in DNA?
[CBSE AIPMT 2012]
(a) The inducer
(b) A terminator
(c) A promoter
(d) The structural gene
Ans. (a)
Transcription unit consists of
promoter, structural gene and
terminator. The inducer
(lactose/allolactose) is not a
component of transcription unit.
36Ribosomal RNA is actively
synthesised in[CBSE AIPMT 2012]
(a) lysosomes
(b) nucleolus
(c) nucleoplasm
(d) ribosomes
Ans. (b)
Nucleolus is the centre for synthesis of
ribosomal RNA (rRNA). Ribosomal
proteins migrate to the nucleolus from
their assembly sites in the cytoplasm
and are packaged into ribonucleoproteins.
These return to the cytoplasm where
they become mature ribosome particles.
37Removal of introns and joining of
exons in a defined order during
transcription is called
[CBSE AIPMT 2012]
(a) looping (b) inducing
(c) slicing (d) splicing
Ans. (d)
The primary transcript from a typical
eukaryotic gene contains introns as well
as exons. During RNA splicing, introns
are removed and exons are joined in a
defined order, to produce functional
RNA.
38What are the structures called that
give an appearance as ‘beads on
string’ in the chromosomes when
viewed under electron
microscope? [CBSE AIPMT 2011]
(a) Genes (b) Nucleotides
(c) Nucleosomes (d) Base pairs
Ans. (c)
Nucleosome appear as "beads-on-string"
in the chromosones. Nucleosome is
sub-microscopic sub-unit of chromatin
which is formed by wrapping of DNA over
a core of histone proteins. The term was
coined by Oudet,et. al[1975). It is oblate
structure with a length of 10 nm and a
thickness of 5-5.7 nm. Its core is called
nu-body. The latter is formed of four
pairs of histone molecules-H A
2
,H B
2
,H
3
andH
4
. DNA makes 1.75 turns over the
octamer to form a nucleosome.
Two adjacent nucleosomes are
connected by a short segment of
unbound DNA called linker DNA. A fifth
type of histone calledH
1
is attached
over the linker DNA.
39Whose experiments cracked the
DNA and discovered unequivocally
that a genetic code is a triplet?
[CBSE AIPMT 2009]
(a) Nirenberg and Matthaei
(b) Hershey and Chase
(c) Morgan and Sturtevant
(d) Beadle and Tatum
Ans. (a)
The existence of triplet code was simply
an assumption till 1961, when Nirenberg
and Matthaei proved its existence by
experiments. They were able to
synthesise artificialmRNA, which
contained only one nitrogenous base, i.e.
uracil. This synthetic poly U sequence
was then placed in a cell free system
containing protein synthesising enzymes
(extracted from bacteriumE. coli) and
20 amino acids together with necessary
ATP. During the process, a small
polypeptide molecule was produced,
which was formed by the linking of
phenylalanine.
Molecular Basis of Inheritance 267

278. This suggested that UUU is the code for
phenyl alanine. Nirenberg got Nobel
Prize for his contributions.
40Polysome is formed by
[CBSE AIPMT 2008]
(a) several ribosomes attached to a
singlemRNA
(b) many ribosomes attached to a strand
of endoplasmic reticulum
(c) a ribosome with several subunits
(d) ribosomes attached to each other in a
linear arrangement
Ans. (a)
The group of ribosomes together with
the singlemRNA molecules, they are
translating is called polysome. They are
formed by several ribosomes attached
to a singlemRNA.
In eukaryotic cells the ribosomes are
attached to rough endoplasmic
reticulum by ribophorin protein. Electron
microscopy reveals that membranes of
homogenised endoplasmic reticulum
disrupt to form closed vesicles called
microsomes. Microsomes derived from
rough endoplasmic reticulum are
studied with ribosomes and are called
rough ribosomes.
41The Okazaki fragments in DNA
chain growth[CBSE AIPMT 2007]
(a) result in transcription
(b) polymerise in the 3′to 5′direction and
forms replication fork
(c) prove semi-conservative nature of
DNA replication
(d) polymerise in the 5′to 3′direction and
explain 3′to 5′DNA replication
Ans. (a)
The Okazaki fragments in DNA chain
growth polymerise in the 5′to 3′
direction. The replicated DNA results in
transcription.
42The length of DNA molecule greatly
exceeds the dimensions of the
nucleus in eukaryotic cells. How is
this DNA accommodated?
[CBSE AIPMT 2007]
(a) Deletion of non-essential genes
(b) Super-coiling in nucleosomes
(c) DNAse digestion
(d) Through elimination of repetitive DNA
Ans. (b)
In eukaryotic cells, DNA is
accommodated by super-coiling in
nucleosomes.
43Molecular basis of organ
differentiation depends on the
modulation in transcription by
[CBSE AIPMT 2007]
(a) RNA polymerase
(b) ribosome
(c) transcription factor
(d) anticodon
Ans. (c)
Transcription factor is molecular basis
of organ differentiation.
44Telomere repetitive DNA
sequences control the function of
eukaryotic chromosomes because
they [CBSE AIPMT 2007]
(a) act as replicons
(b) are RNA transcription initiator
(c) help chromosome pairing
(d) prevent chromosome loss
Ans. (d)
Telomeres, i.e. the ends of
chromosome, have repetitive DNA
sequences and are stable and
resistant to exonuclease digestion
hence, prevent chromosome loss.
45Which one of the following makes
use of RNA as a template to
synthesise DNA?
[CBSE AIPMT 2005]
(a) Reverse transcriptase
(b) DNA dependant RNA polymerase
(c) DNA polymerase
(d) RNA polymerase
Ans. (a)
In 1970 H Temin and D Baltimore
independently discovered the enzyme
reverse transcriptase. This enzyme uses
RNA as template for the synthesis of
cDNA (complementary DNA).
46Which one of the following
hydrolyses internal phosphodiester
bonds in a polynucleotide chain?
[CBSE AIPMT 2005]
(a) Lipase (b) Exonuclease
(c) Endonuclease (d) Protease
Ans. (c)
Endonuclease hydrolyses internal
phosphodiester bonds in a
polynucleotide chain.
47During transcription holoenzyme
RNA polymerase binds to a DNA
sequence and the DNA assumes a
saddle like structure at that point.
What is that sequence called?
[CBSE AIPMT 2005]
(a) CAAT box (b) GGTT box
(c) AAAT box (d) TATA box
Ans. (d)
TATA box is present in eukaryotic
promoter region. It has a resemblance
with Pribnow box of prokaryotes. TATA
box was identified by Dr. Hogness and
so, it is also called as Hogness box. It is a
7 bp long region located 20 bp upstream
to the start point.
During the process of transcription the
RNA polymerase (a holoenzyme which
has a core unit and a sigma factor for
proper initiation of transcription) binds
to TATA box due to which DNA assumes
a saddle like structure at this place.
48Telomerase is an enzyme which is a
[CBSE AIPMT 2005]
(a) repetitive DNA
(b) RNA
(c) simple protein
(d) ribonucleoprotein
Ans. (d)
Ends of an eukaryotic chromosome are
known as telomeres.
Telomerase, which is a special
ribonucleoprotein molecule (enzymatic
in nature) is responsible for the
synthesis of these telomeres.
49During replication of a bacterial
chromosome DNA synthesis starts
from a replication origin site and
[CBSE AIPMT 2004]
(a) RNA primers are involved
(b) is facilitated by telomerase
(c) moves in one direction of the site
(d) moves in bi-directional way
Ans. (a)
The events for initiation of DNA
replication in prokaryotes may be
classified into (a) pre-priming (occurring
only at the origin); (b) priming (recurring
with the initiation of each Okazaki
fragment during elongation phase.
Unwinding of DNA is followed by the
synthesis of RNA primers by RNA
primase.
50The telomeres of eukaryotic
chromosomes consist of short
sequences of[CBSE AIPMT 2004]
(a) thymine rich repeats
(b) cytosine rich repeats
(c) adenine rich repeats
(d) guanine rich repeats
268 NEETChapterwise Topicwise Biology

279. Ans. (d)
Telomeres have been shown to have
unique structures that include short
nucleotide sequences present as
tandemly repeated units. In eukaryotes
the telomeres terminate with a
single-stranded DNA [12-16 nucleotides
long) rich in guanine.
51During transcription, the nucleotide
sequence of the DNA strand that is
being coded is ATACG, then the
nucleotide sequence in themRNA
would be [CBSE AIPMT 2004]
(a) TATGC (b) TCTGG
(c) UAUGC (d) UATGG
Ans. (c)
If DNA hasATACGnucleotide sequence
then themRNA would contain UAUGC
sequence. The formation ofmRNA from
DNA is termed as transcription. This
process takes place in the nucleus
(eukaryotes) or in the cytoplasm
(prokaryotes).
The base sequence ofmRNA is
complementary copy of the template
DNA strand.
52Which form of RNA has a structure
resembling clover leaf?
[CBSE AIPMT 2004]
(a)rRNA (b) hnRNA
(c)mRNA (d) tRNA
Ans. (d)
The basic plan of the structure of
tRNA assumes the pattern of a clover
leaf. The structures of differenttRNAs
for almost all amino acids are now
available and all of these fit the clover
leaf model. ThetRNA structure can be
decomposed into its primary structure
and its secondary structure (usually
seen as clover leaf structure) and
tertiary structure.
53During transcription, the DNA site
at which RNA polymerase binds is
called [CBSE AIPMT 2003]
(a) receptor (b) enhancer
(c) promoter (d) regulator
Ans. (c)
Promoter is the nucleotide sequence
to which RNA polymerase binds and
initiates transcription. Formation of a
single stranded linear chain of
complementary RNA (mRNA) on the
template strand of DNA in nucleus
(eukaryotes) or in cytoplasm
(prokaryotes) is known as transcription.
54Chromosomes in a bacterial cell
can be 1-3 in number and
[CBSE AIPMT 2003]
(a) can be either circular or linear, but
never both within the same cell
(b) can be circular as well as linear within
the same cell
(c) are always circular
(d) are always linear
Ans. (c)
Bacterial chromosomes are circular DNA
molecules.
55Exon part ofmRNAs have code for
[CBSE AIPMT 2002]
(a) protein (b) lipid
(c) carbohydrate (d) phospholipid
Ans. (a)
Exon part ofmRNA consists of codons
for protein synthesis. Exon is the stretch
of bases which codes for amino acids,
while the non-coding stretches of bases
are called intron.
56Which of the following reunites the
exon segments after RNA splicing?
[CBSE AIPMT 2002]
(a) RNA polymerase (b) RNA primase
(c) RNA ligase (d) RNA protease
Ans. (c)
RNA ligase reunites the exon segments
after RNA splicing.
57Which statements is correct for
bacterial transduction?
[CBSE AIPMT 2002]
(a) Transfer of some genes from one
bacteria to another bacteria through
virus
(b) Transfer of genes from one bacteria
to another bacteria by conjugation
(c) Bacteria obtained its DNA directly
(d) Bacteria obtained DNA from other
external source
Ans. (a)
Transduction involves the picking up of
DNA by bacteriophage from one
bacterial cell and carrying it to another
where, the DNA fragment may get
incorporated into the bacterial host’s
genome.
58In a DNA percentage of thymine is
20. What is the percentage of
guanine? [CBSE AIPMT 2002]
(a) 20% (b) 40%
(c) 30% (d) 60%
Ans. (c)
Total DNA [100] = A + T + C + G
A = 20% (given)
A = T (base pairing rule)
100 = 20 + 20 + C + G
C + G = 100 – 40 = 60
C = G = 30 (C = G)
59Sequence of which of the following
is used to know the phylogeny?
[CBSE AIPMT 2002]
(a)mRNA (b) rRNA
(c)tRNA (d) DNA
Ans. (b)
The genes forrRNAs tend to be highly
conserved and, are therefore, often
employed for phylogenetic studies.
60In which directionmRNA is
synthesised on DNA template?
[CBSE AIPMT 2001]
(a)5 3′ → ′ (b)3 5′ → ′
(c) Both (a) and (b) (d) Any of above
Ans. (a)
5 3′→ ′is the direction of synthesis of
mRNA on DNA template.
61Gene and cistron words are
sometimes used synonymously
because [CBSE AIPMT 2001]
(a) one cistron contains many genes
(b) one gene contains many cistrons
(c) one gene contains one cistron
(d) one gene contains no cistron
Ans. (c)
Cistron is generally accepted as a
synonym for gene. Gene (Gr.genos=
birth, race) is the basic unit of heredity. It
is a sequence of nucleotides on a
chromosome that encodes a polypeptide
or RNA molecule and so, determines the
nature of individual’s inherited traits.
Cistron is a segment of DNA that codes
for one polypeptide.
62E. coliabout to replicate was
placed in a medium containing
radioactive thymidine for five
minutes. Then it was made to
replicate in a normal medium.
Which of the following observation
shall be correct?[CBSE AIPMT 2001]
(a) Both the strands of DNA will be
radioactive
(b) One strand radioactive
(c) Each strand half radioactive
(d) None is radioactive
Molecular Basis of Inheritance 269

280. Ans. (b)
Since, DNA replication is
semiconservative, the newly synthesised
strand of DNA would be normal while the
strand obtained from parent molecule
would be radioactive. In the given
expreiment
63Due to discovery of which of the
following in 1980’s the evolution
was termed as RNA world?
[CBSE AIPMT 2001]
(a)mRNA,tRNA,rRNA synthesise
proteins
(b) In some viruses, RNA is genetic
material
(c) Some RNAs have enzymatic property
(d) RNA is not found in all cells
Ans. (c)
Ribozymes are catalytically active RNA
molecules discovered in 1980’s. These
are self-splicing introns indicating their
possible role as intermediates in the
evolution of biological systems from
abiotic substances.
64During replication of DNA, its two
strands separate. Each of these
serves as a template for the
formation of new strands. Such
type of replication is called
[CBSE AIPMT 2000]
(a) non-conservative
(b) semi-conservative
(c) flexible
(d) conservative
Ans. (b)
Since, each daughter DNA molecule
contains one strand of the parent DNA
double helix (only one strand synthesised
afresh) the process of replication is
called semi-conservative. Mathew
Meselson and Franklin Stahl in 1958
proved experimentally that DNA
replication is semi-conservative.
65One of the similarities between
DNA and RNA is that both
[CBSE AIPMT 2000]
(a) are polymers of nucleotides
(b) are capable of replicating
(c) have similar sugars
(d) have similar pyrimidine bases
Ans.(a)
DNA Composition RNA Composition
Deoxyribose sugar. Ribose sugar.
Adenine, guanine
(both purine) and
cytosine, thymine
(both pyrimidine),
nitrogenous bases.
Adenine, guanine
(both purine) and
uracil, cytosine
(both pyrimidine)
nitrogenous bases.
Phosphate
molecules.
Phosphate
molecules.
It is polymer of
nucleotides.
It is also polymer
of nucleotides.
It is capable to
replicate in all
cases because it
functions as
heredity material.
It is formed from
DNA by the
process of
transcription only
in few cases (RNA
viruses) it
functions as
heredity material.
66ThePneumococcusexperiment
proves that [CBSE AIPMT 1999]
(a) DNA is the genetic material
(b) RNA sometime controls the
production of DNA and proteins
(c) bacteria undergo binary fission
(d) bacteria do not reproduce sexually
Ans. (a)
ThePneumococcusexperiment proves
that DNA is the genetic material as
Frederick Griffith (1928) found that
‘something’ passed from heat-killed
encapsulated forms ofPneumococcusto
live non-capsulated forms which caused
them to develop capsules and become
virulent. Averyet al,(1944) found this
transforming agent (hence, genetic
material) to be DNA.
67DNA elements, which can switch
their position, are called
[CBSE AIPMT 1998]
(a) exons (b) introns
(c) cistrons (d) transposons
Ans. (d)
Transposons are genetic elements
varying from 750 base pairs to 40 kilo
base pairs in length and can move from a
site in one genome to another site in the
same or in a different genome.
68Genes are packaged into a
bacterial chromo- some by
[CBSE AIPMT 1997]
(a) histones
(b) basic protein
(c) acidic protein
(d) actin
Ans. (b)
Bacteria are prokaryotic organisms.
Polyamines (basic proteins) like
spermidine and cadaverine (instead of
histones) are associated with DNA
packaging in bacteria.
69The hereditary material present in
the bacteriumE.coliis
[CBSE AIPMT 1997]
(a) single stranded RNA
(b) double stranded RNA
(c) single stranded DNA
(d) double stranded DNA
Ans. (d)
Bacterial chromosome is single, circular
double stranded DNA molecule.
70An enzyme that joins the ends of
two strands of nucleic acid is a
[CBSE AIPMT 1996, 2002]
(a) polymerase (b) synthetase
(c) helicase (d) ligase
Ans. (d)
Ligase enzyme joins the ends of two
strands of nucleic acid.
71Okazaki fragments are seen during
[CBSE AIPMT 1996]
(a) transcription (b) translation
(c) replication (d) transduction
Ans. (c)
During DNA replication in lagging strand
DNA fragments are formed in small
pieces these are called Okazaki
fragments.
72In split genes, the coding sequence
are called [CBSE AIPMT 1995]
(a) introns (b) operons
(c) exons (d) cistrons
Ans. (c)
In split genes coding region is called
exons. In higher organisms (eukaryotes)
gene is not continuous, within a single
gene there may be four or five silent
regions. These regions are called introns
(which do not transcribemRNA). The
remaining part is called as exons
(transcribemRNA).
270 NEETChapterwise Topicwise Biology
Both strands of DNA
contain radioactive
thymidine
E. coliafter 5
minutes from a
thymidine
radioactive
medium Replication
(semi-conservative)
Normal
DNA strand
(new
strand)
Old strand
containing
radioactive
thymidine
Old strand
containing
radioactive
thymidine
Normal
DNA strand
(new
strand)

281. 73Protein helping in opening of DNA
double helix in front of replications
fork is
(a) DNA gyrase
(b) DNA polymerase-I
(c) DNA ligase
(d) topoisomeras
Ans. (a)
DNA gyrase helps in opening of DNA
double helix in front of replication fork.
74Reverse transcriptase is
[CBSE AIPMT 1994]
(a) RNA dependent RNA polymerase
(b) DNA dependent RNA polymerase
(c) DNA dependent DNA polymerase
(d) RNA dependent DNA polymerase
Ans. (d)
Reverse transcriptase is RNA dependent
DNA polymerase. H Temin and D
Baltimore discovered reverse
transcription. Reverse transcriptase has
modified central dogma of molecular
biology as RNA→DNA→RNA→Protein.
75Nucleosome core is made of
[CBSE AIPMT 1993]
(a) H1, H2A, H2B and H3
(b) H1, H2A, H2B and H4
(c) H1, H2A, H2B, H3 and H4
(d) H2A, H2B, H3 and H4
Ans. (d)
A nucleosome is an octamer of histone
proteins and has a core of 8 molecules of
histone proteins (two each of H2A, H2B,
H3 and H4] wrapped by two turns of DNA.
76A DNA with unequal nitrogen bases
would most probably be
[CBSE AIPMT 1993]
(a) single stranded
(b) double stranded
(c) triple stranded
(d) four stranded
Ans. (a)
A single stranded DNA do not possess its
complementary base pairs so it would
have unequal nitrogen bases.
77During DNA replication, the strands
separate by
[CBSE AIPMT 1993]
(a) DNA polymerase
(b) topoisomerase
(c) unwindase/helicase
(d) gyrase
Ans. (c)
Unwinding of DNA helix is caused by
enzyme helicase.
78Who proved that DNA is basic
genetic material?
[CBSE AIPMT 1993]
(a) Griffith
(b) Watson
(c) Boveri and Sutton
(d) Hershey and Chase
Ans. (d)
Hershey and Chase [1952] proved that
DNA is basic genetic material.
79Nucleotide arrangement in DNA
can be seen by
[CBSE AIPMT 1993]
(a) X-ray crystallography
(b) electron microscope
(c) ultracentrifuge
(d) light microscope
Ans. (a)
Astbury by his X-ray diffraction studies
suggested 3-D configuration for DNA
molecules which was confirmed by
Wilkins and Franklin in 1952 and then in
1953 Watson and Crick designed the
model of DNA molecule.
80The transforming principle of
Pneumococcusas found out by
Avery, Mac Leod and McCarty was
[CBSE AIPMT 1993]
(a)mRNA
(b) DNA
(c) protein
(d) polysaccharide
Ans. (b)
Avery, MacLeod and McCarty [1944]
showed the significance of DNA in
hereditary transmission in bacteria
Pneumococcus. They discovered the
biochemical nature of gene.
81Experimental material in the study
of DNA replication has been
[CBSE AIPMT 1992]
(a)Escherichia coli
(b)Neurospora crassa
(c)Pneumococcus
(d)Drosophila melanogaster
Ans. (a)
Meselson andStahl[1958] proved
experimentally that inE. coliDNA is
replicated by semi-conservative
manner.
82Escherichia colifully labelled with
N
15
is allowed to grow inN
14
medium. The two strands of DNA
molecule of the first generation
bacteria have[CBSE AIPMT 1992]
(a) different density and do not resemble
parent DNA
(b) different density but resemble parent
DNA
(c) same density and resemble parent DNA
(d) same density but do not resemble
parent DNA
Ans. (b)
WhenE. colifully labelled withN
15
is
allowed to grow inN
14
medium, then after
first generation of replication one of the
two strands would haveN
15
and the other
strand would haveN
14
. The resulting
molecule would have a density which is
intermediate betweenN
15
DNA andN
14
DNA.
These two molecules of DNA will be
similar but not same in density.
83In RNA, thymine is replaced by
[CBSE AIPMT 1992]
(a) adenine
(b) guanine
(c) cytosine
(d) Uracil
Ans. (d)
DNA consists of nitrogenous bases,
adenine, guanine, cytosine and thymine,
whereas in RNA thymine is replaced by
uracil. The other nitrogeneous bases, i.e.
adenine, guanine, cytosine are present
both in RNA and DNA.
84A nucleotide is formed of
[CBSE AIPMT 1991]
(a) purine, pyrimidine and phosphate
(b) purine, sugar and phosphate
(c) nitrogen base, sugar and phosphate
(d) pyrimidine, sugar and phosphate
Ans. (c)
Nucleotide is the basic unit of nucleic
acids (DNA and RNA). It is composed of
nucleoside (nitrogeneous base +
pentose sugar) and phosphate group.
85The process of transfer of genetic
information from DNA to
RNA/formation of RNA from DNA is
[CBSE AIPMT 1991]
(a) transversion
(b) transcription
(c) translation
(d) translocation
Molecular Basis of Inheritance 271

282. Ans. (b)
The transfer of genetic information from
DNA to RNA (mRNA) is known as
transcription. Both the strands of DNA
do not transcribe RNA but only one of
them does it which is called as template
strand.
86Which is not consistent with
double helical structure of DNA?
[CBSE AIPMT 1990]
(a) A==T, C≡≡G
(b) Density of DNA decreases on heating
(c) A + T/C + G is not constant
(d) Both (a) and (b)
Ans. (c)
According toErwin Chargaff, the base
ratio A + T/G + C may vary from one
species to another, but is constant for a
given species. It is rarely equal to one
and varies from 0.4 and 1.9.
87DNA replication is
[CBSE AIPMT 1989]
(a) conservative and discontinuous
(b) semi-conservative and
semidiscontinuous
(c) semi-conservative and discontinuous
(d) conservative
Ans. (b)
DNA replication is semi-conservative
that means DNA formed after replication
contains one strand of its parent DNA
and this was proved by Meselson and
Stahl [1958].
During replication the strand formed in
leading strand is continuous, while the
strand formed in lagging strand is
discontinuous in the small pieces
(Okazaki fragments).
88Which one contains four pyrimidine
bases? [CBSE AIPMT 1994]
(a) GATCAATGC (b) GCUAGACAA
(c) UAGCGGUAA (d) TGCCTAACG
Ans. (a)
Pyrimidines are 6-membered nitrogen
bases that contain nitrogen at 1 and 3
positions, e.g. cytosine (C), thymine (T),
uracil (U).
89A segment of DNA has 120 adenine
and 120 cytosine bases. The total
number of nucleotides present in
the segment is[CBSE AIPMT 1991]
(a) 120 (b) 240
(c) 60 (d) 480
Ans. (d)
According to Chargaff’s rule, molar
amount of adenine is equal to that of
thymine and cytosine equals to guanine,
A + G = T + C. So, a segment of DNA with
120 adenine base and 120 cytosine base
will have same number of each thymine
and guanine base (as, A = T and C = G),
i.e. 120 thymine bases, 120 guanine
bases, thus a total of 480 nucleotides.
90Statement IThe condon ‘AUG’
codes for methionine and
phenylalanine.
Statement II‘AAA’ and ‘AAG’ both
codons code for the amino acid
lysine.
In the light of the above
statements, choose the correct
answer from the options given
below. [NEET 2021]
(a) Both statement I and statement II are
true
(b) Both statement I and statement ll are
false
(c) Statement I is true, but statement Il is
false
(d) Statement I is false, but statement II
is true
Ans. (d)
Statement I is false, but statement II is
true and can be corrected as
The codon AUG only codes for
methionine. As the codons are universal.
From bacteria to mammals AUG only
codes for methionine.
Some amino acids are coded by more
than one codon, hence the code is
degenerate. AAA and AAG both codons
code for the amino acid lysine.
91Identify the correct statement.
[NEET 2021]
(a) In capping, methyl guanosine
triphosphate is added to the 3 end of
hnRNA
(b) RNA polymerase binds withRho
factor to terminate the process of
transcription in bacteria
(c) The coding strand in a transcription
unit is copied to anmRNA
(d) Split gene arrangement is
characteristic of prokaryotes
Ans. (b)
Statement in option (b) is correct and
other statements can be corrected as
A heterogeneous nuclear RNA orhnRNA
is a primarymRNA transcript that is
localised in the nucleus. Capping is a
process in which at the5′end ofhnRNA,
a cap of 7-methly guanosine is added.
The template strand is a transcription
unit is coped to amRNA.
Split gene arrangement is
characteristic of eukaryotes.
92What is the role of RNA
polymerase-III in the process of
transcription in eukaryotes?
[NEET 2021]
(a) TranscribesrRNAs (28S, 18S and
5.8S)
(b) TranscribestRNA (5srRNA and
snRNA)
(c) Transcribes precursor ofmRNA
(d) Transcribes onlysnRNAs
Ans. (b)
In eukaryotes, at least three classes of
RNA polymerases (Pol I-III) are required
for the cellular RNA synthesis. In
eukaryote cells, RNA polymerase III
(also called Pol III) transcribes DNA to
synthesise ribosomal 5SrRNA,tRNA
and other small RNAs. The genes
transcribed by RNA Pol III fall in the
category of "housekeeping" genes
whose expression is required in all cell
types and most environmental
conditions.
93Identify the statement which is
incorrect. [NEET (Oct.) 2020]
(a) Sulphur is an integral part of
cysteine
(b) Glycine is an example of lipids
(c) Lecithin contains phosphorus atom
in its structure
(d) Tyrosine possesses aromatic ring in
its structure
Ans. (b)
Statement (b) is incorrect. It can be
corrected as Glycine is an example of
amino acid. It is a neutral amino acid
that contains only one amino group and
one carboxylic group with non-cyclic
hydrocarbon chain.
94The first phase of translation is
[NEET (Sep.) 2020]
(a) recognition of DNA molecule
(b) aminoacylation oftRNA
(c) recognition of an anti-codon
(d) binding ofmRNA to ribosome
272 NEETChapterwise Topicwise Biology
Genetic Code and
Gene Expression
TOPIC 2

283. Ans. (b)
The first phase of translation is
aminoacylation oftRNA, i.e. activation of
amino acids and the formation of AA-tRNA
complex. In the presence of an enzyme
tRNA synthetase, the amino acid (AA)
molecule is activated and then each
amino acid is attached to the specific
tRNA molecule at 3’ / CCA end to form
aminoacyl-tRNA complex. The reaction
needs ATP. This process is thus called
charging oftRNA or aminoacylation of
tRNA.
95The specific palindromic sequence
which is recognised byEcoRI is
[NEET (Sep.) 2020]
(a) 5'- GGAACC- 3'
(b) 5' - CTTAAG - 3' 3' - CCTTGG - 5'
3' - GAATTC - 5'
(c) 5' - GGATCC - 3'
(d) 5' - GAATTC - 3' 3'- CCTAGG - 5'
3' - CTTAAG - 5'
Ans. (d)
The correct option is (d) because the
specific palindromic sequence which is
recognised byEcoRI is5′-GAATTC-3′
3′-CTTAAG-5′.
A palindromic sequence is a sequence
made up of nucleic acids within double
helix of DNA or RNA that is the same when
read from 5’ to 3’ on one strand and 3’ to
5’ on the other, complementary strand.
96The sequence that controls the
copy number of the linked DNA in
the vector, is termed
(a)Orisite
(b) palindromic sequence
(c) recognition site
(d) selectable marker
Ans. (a)
The sequence that controls the copy
number of linked DNA in the vector is
called asOrisite. Origin of replication is a
sequence from where replication starts
and any foreign DNA is linked to this region.
Orisite is also responsible for controlling
copy number of linked DNA.
Therefore, if any person wants to
produce many copies of the target DNA
he/she should clone in a vector whose
Orisite supports high copy number.
97From the following, identify the
correct combination of salient
features of Genetic code.
[NEET (Odisha) 2019]
(a) Universal, Non-ambiguous,
Overlapping
(b) Degenerate, Overlapping, Commaless
(c) Universal, Ambiguous, Degenerate
(d) Degenerate, Non-overlapping,
Non-ambiguous
Ans. (d)
The correct combination of salient
features of Genetic code is degenerate,
Non-overlapping, Non-ambiguous.
These are explained as one codon codes
for only one amino acid, hence genetic
code is unambiguous and specific. Some
amino acids are coded by more than one
codon, hence the code is degenerate.
The codon is read inmRNA in a
contiguous fashion. There are no
punctuations and overlapping.
98Which of the following features of
genetic code does allow bacteria to
produce human insulin by
recombinant DNA technology?
[NEET (National) 2019]
(a) Genetic code is redundant
(b) Genetic code is nearly universal
(c) Genetic code is specific
(d) Genetic code is not ambiguous
Ans. (b)
Bacteria is able to produce human
insulin because genetic code is nearly
universal in all organisms. For example,
the codon AGG specifies amino acid
Arginine in bacteria, animals and plants.
But there are also some exceptions to
it, e.g. in mitochondria, stop codon
UGA specifies amino acid tryptophan.
99Match the following RNA
polymerases with their transcribed
products
1. RNA polymerase I i.tRNA
2. RNA polymerase II ii.rRNA
3. RNA polymerase III iii.hnRNA
Select the correct option from the
following [NEET (Odisha) 2019]
1 2 3
(a) (i) (iii) (ii)
(b) (i) (ii) (iii)
(c) (ii) (iii) (i)
(d) (iii) (ii) (i)
Ans. (c)
The correct matches are
1. RNA polymerse I
transcribes
(ii)rRNAs (28S,18S
and 5.8S)
2. RNA polymerase
II
(iii)hnRNA
3. RNA polymerase
III
(i)tRNA
100Expressed Sequence Tags (ESTs)
refers to[NEET (National) 2019]
(a) polypeptide expression
(b) DNA polymorphism
(c) novel DNA sequences
(d) genes expressed as RNA
Ans. (d)
Expressed Sequence Tags (EST) refers
to the genes expressed as RNA. These
are the DNA sequences that are
expressed asmRNA for protein
synthesis.
101AGGTATCGCAT is a sequence from
the coding strand of a gene. What
will be the corresponding sequence
of the transcribedmRNA?
[NEET 2018]
(a) ACCUAUGCGAU
(b) AGGTUTCGCAT
(c) AGGUAUCGCAU
(d) UCCAUAGCGUA
Ans. (c)
Coding strand is the one that codes for
mRNA. It has same nucleotide sequence
as that ofmRNA except thymine (T) is
replaced by uracil (U) inmRNA. Hence,
the corresponding sequence of
transcribedmRNA by template or
non-coding strand (complementary to
RNA) is AGGUAUCGCAU.
102If there are 999 bases in an RNA
that codes for a protein with 333
amino acids and the base at
position 901 is deleted such that
the length of the RNA becomes 998
bases, how many codons will be
altered ? [NEET 2017]
(a) 1
(b) 11
(c) 33
(d) 333
Ans. (c)
33 codons will be altered if the 901st
base is deleted and RNA has only 998
bases instead of 999 bases.
Total bases present in RNA=999
Bases left after deletion of 901st base in
RNA
= −999 901
=98
Number of codon present in98 33=
(Approximately as three codons code for
one amino acid).
Molecular Basis of Inheritance 273

284. 103Which one of the following is the
starter codon?[NEET 2016, Phase I]
(a) UGA (b) UAA
(c) UAG (d) AUG
Ans. (d)
AUG is the start codon. It also codes for
amino acid called methionine which is
the first amino acid in a polypeptide
chain. UAA, UAG and UGA are stop
codons and are meant for termination of
polypeptide chain during protein
synthesis.
104The diagram shows an important
concept in the genetic implication
of DNA. Fill in the blanksAtoC.
[NEET 2013]
(a) A–transcription, B–replication,
C–James Watson
(b) A–translation, B–transcription,
C–Erwin Chargaff
(c) A–transcription, B–translation,
C–Francis Crick
(d) A–translation, B–extension,
C–Rosalind Franklin
Ans. (c)
Central dogma is
DNA RNA Protein
Transcription Translation
→ →
A B
m
→
C
Francis Crick
105If one strand of DNA has the
nitrogenous base sequence as
ATCTG, what would be the
complementary RNA strand
sequence? [CBSE AIPMT 2012]
(a) TTAGU (b) UAGAC
(c) AACTG (d) ATCGU
Ans. (b)
If one strand of DNA has the
nitrogenous base sequence as ATCTG,
the complementary sequence of
mRNA will be UAGAC.
106What is not true for genetic code?
[CBSE AIPMT 2009]
(a) A codon inmRNA is read in a
non-contiguous fashion
(b) It is nearly universal
(c) It is degenerate
(d) It is unambiguous
Ans. (a)
The general features of genetic code are
(i) The genetic code is written in linear
form, using the ribonucleotide
bases that compose mRNA
molecule as letters.
(ii) Each word of codon consists of
three letters, i.e., the codon is
triplet.
(iii) The genetic code inside the cell
mediumis saidto benon-ambiguous.
(iv) The code is degenerate, i.e. a given
amino acid can be specified by
more than one codons.
(v) The codon contains ‘start’ and ‘stop’
signals.
(vi) The code is said to be commaless.
(vii) The code is non-overlapping.
107Which one of the following pairs of
codons is correctly matched with
their function or the signal for the
particular amino acid?
[CBSE AIPMT 2008]
(a) GUU, GCU — Alanine
(b) UAG, UGA — Stop
(c) AUG, ACG — Start/methionine
(d) UUA, UCA — Leucine
Ans. (b)
The group of nucleotides that specifies
one amino acid is a code word or codon.
The nucleotides ofmRNA are arranged
as a linear sequence of codons, each
codon consisting of three successive
nitrogenous bases.
Three codons UAG, UAA and UGA are the
termination codons. They do not code
for any of the amino acids.
In most organisms AUG codon is the
start or initiation codon, i.e. the
polypeptide chain starts either with
methionine or N-formylmethionine.
Leucine — UUA, UUG, CUU, CUC,
CUA, CUG.
Alanine — GUC, GCC, GCA, GCG.
GUU — Valine
UCA — Serine.
108One gene-one enzyme relationship
was established for the first time in
[CBSE AIPMT 2007]
(a)Neurospora crassa
(b)Salmonella typhimurium
(c)Escherichia coli
(d)Diplococcus pneumoniae
Ans. (a)
One gene-one enzyme relationship was
initially proposed by Beadle and Tatum
based on their experiments conducted
onNeurospora crassa.They were
awarded by Nobel Prize in 1958.
109A sequential expression of a set of
human genes occurs when a
steroid molecule binds to the
[CBSE AIPMT 2007]
(a) transfer RNA (b) messenger RNA
(c) DNA sequence (d) ribosome
Ans. (c)
The steroid hormone receptor protein
complex activate transcription of target
gene by binding to sepecific DNA
sequence.
110One gene-one enzyme hypothesis
was postulated by
[CBSE AIPMT 2006]
(a) R Franklin
(b) Hershey and Chase
(c) A Garrod
(d) Beadle and Tatum
Ans. (d)
‘One gene-one enzyme’ hypothesis was
given byBeadleandTatum[1948] which
states that particular gene controls the
synthesis of specific enzyme. Later, it
was modified to ‘one gene-one
polypeptide hypothesis’ by Yanofsky,et.
al,[1965).
111Amino acid sequence, in protein
synthesis is decided by the
sequence of[CBSE AIPMT 2006]
(a)tRNA (b) mRNA
(c)cDNA (d) rRNA
Ans. (b)
In the process of protein synthesis, the
messenger RNA (mRNA) is responsible for
carrying the genetic code transcribed from
DNA to specialised sites within the cell
(called ribosomes) where the information
is translated into protein. The sequence of
amino acids in a particular protein is
determined by the sequence of
nucleotides inmRNA. Sequence oftRNA,
cDNA orrRNA do not decide the amino
acid sequence in protein synthesis.
112After a mutation at genetic locus
the character of an organism
changes due to the change in
[CBSE AIPMT 2004]
(a) protein structure
(b) DNA replication
(c) protein synthesis pattern
(d) RNA transcription pattern
274 NEETChapterwise Topicwise Biology
DNA RNA Proteinm
A B
Proposed by
C

285. Ans. (a)
Normally, genetic information flows
from →DNA mRNA→protein. Hence,
any change in nucleotides due to the
mutation, would result in change in the
structure of protein/enzyme which
might result in some change in the
organism.
113The following ratio is generally
constant for a given species
[CBSE AIPMT 2004]
(a)A + G/C + T
(b)T + C/G + A
(c)G + C/ A + T
(d)A + C/ T + G
Ans. (c)
The base ratio A + T / G + C may vary
from one species to another, but is
constant for a given species. It is
rarely equal to one and varies between
0.4 and 1.9.
114In the genetic code dictionary, how
many codons are used to code for
all the 20 essential amino acids?
[CBSE AIPMT 2003]
(a) 61
(b) 60
(c) 20
(d) 64
Ans. (a)
Out of 64 codons three (UAA, UAG, UGA)
are chain terminating codons the
translating mechanism is not able to
read these codons and 61 codons are
used to code all the 20 essential amino
acids.
115What would happen if in a gene
encoding a polypeptide of 50 amino
acids, 25
th
codon (UAU) is mutated
to UAA? [CBSE AIPMT 2003]
(a) A polypeptide of 49 amino acids will
be formed
(b) A polypeptide of 25 amino acids will
be formed
(c) A polypeptide of 24 amino acids will
be formed
(d) Two polypeptides of 24 and 25 amino
acids will be formed
Ans. (c)
UAA is the ‘stop’ codon hence,
polypeptide chain will not grow after 24
th
amino acid. In the absence of new
initiating codon rest of codons will not
be able to translate.
116Degeneration of a genetic code is
attributed to the
[CBSE AIPMT 2003]
(a) entire codon
(b) third member of a codon
(c) first member of a codon
(d) second member of a codon
Ans. (b)
It has seen variously that onetRNA
molecule codes for more than one amino
acid molecules. This is possible due to
the improper pairing of third codon with
the first anticodon oftRNA.
117During translation initiation in
prokaryotes, a GTP molecule is
needed in [CBSE AIPMT 2003]
(a) association of 30S,mRNA with formyl
mettRNA
(b) association of 50S subunit of
ribosome with initiation complex
(c) formation of formyl mettRNA
(d) binding of 30S subunit of ribosome
withmRNA
Ans. (a)
During the process of translation an
initial complex is formed between
mRNA, 30S ribosomal sub-unit and
methionyltRNA. This complex is formed
due to the association ofIF , IF , IF
1 2 3
initiation factors and GTP molecule.
118Which one of the following triplet
codes, is correctly matched with its
specificity for an amino acid in
protein synthesis or as ‘start’ or
‘stop’ codon?[CBSE AIPMT 2003]
(a) UGU—Leucine (b) UAC—Tyrosine
(c) UCG—Start (d) UUU—Stop
Ans. (b)
UGU → Cistine
UAC → Tyrosine
UCG → Serine
UUU → Phenylalanine
UAG, UGA, UAA→ Stop codons
UAG → Start codon.
119‘Signal hypothesis’ for the
biosynthesis of secretory type of
proteins was proposed by
[CBSE AIPMT 2000]
(a) Camillo Golgi
(b) Blobel and Sabatini
(c) Baltimore
(d) Sheeler and Bianchi
Ans. (b)
A variety of proteins are synthesised on
ribosomes. However, these have
different destinations. David Sabatini
and G. Blobel proposed ‘signal sequence’
hypothesis according to which a short
amino acid sequence at the amino
terminus of a newly synthesised
polypeptide directs a protein to its
appropriate sequence.
120The transfer RNA molecule in 3D
appears [CBSE AIPMT 2000]
(a) L-shaped (b) E-shaped
(c) Y-shaped (d) S-shaped
Ans. (a)
Kimet al,(1973) suggested L shaped
model oftRNA by X-ray diffraction while
studying phenyl alaninetRNA of yeast. L
shape structure oftRNA is a
3-dimentional (3D) structure of 20 Å
thickness.
121Which is not involved in protein
synthesis? [CBSE AIPMT 1994]
(a) Transcription (b) Initiation
(c) Elongation (d) Termination
Ans. (a)
Transcription is the synthesis of RNA on
DNA template. It is not involved in
protein synthesis (translation).
122In DNA when AGCT occurs, their
association is as per which of the
following pair?[CBSE AIPMT 1999]
(a) ACGT
(b) AGCT
(c) ATGC
(d) All of these
Ans. (c)
In DNA AGCT is associated with pair
ATGC because in a DNA molecule, the
purine adenine in either chain is
associated with the pyrimidine
thymidine on the other. Similarly, purine
guanine in either chain is associated with
pyrimidine cytosine on the other.
123Protein synthesis in an animal cell
takes place [CBSE AIPMT 1997]
(a) only in the cytoplasm
(b) in the nucleolus as well as in the
cytoplasm
(c) in the cytoplasm as well as in
mitochondria
(d) only on ribosomes attached to a
nucleus
Molecular Basis of Inheritance 275

286. Ans. (c)
Protein synthesis is a complex process it
essentially involves DNA for the
synthesis ofmRNA (transcription) which
contains information for the synthesis of
proteins (translation). The process of
translation takes place on ribosomes
which are found in cytoplasm (in
attached form on ER) and in
mitochondria (in the free form).
124The RNA that picks up specific
amino acid from amino acid pool in
the cytoplasm to ribosome during
protein synthesis is called
[CBSE AIPMT 1997]
(a)mRNA (b) tRNA
(c)rRNA (d) RNA
Ans. (b)
tRNA (soluble RNA =sRNA) is a 70-75
nucleotide long molecule. 80% of this
RNA is double helical, one end of this
molecule has G and other C—C—A
sequences.
The clover leaf model [2D) oftRNA was
given by R Holley [1968] and Kimet al,
[1973] suggested ‘L’ shaped model (3D) of
tRNA by X-ray diffraction while studying
phenyl alaninetRNA of yeast.
Each amino acid had its own specific
tRNA molecule which transfers it from
cytoplasm to the ribosome.
125The codons causing chain
termination are[CBSE AIPMT 1997]
(a) TAG, TAA, TGA (b) GAT, AAT, AGT
(c) AGT, TAG, UGA (d) UAA, UAG, UGA
Ans. (d)
UAA, UAG and UGA act as stop codons
(terminator codons) because these are
not translated into amino acid. UAA is
called ochre, UAG as amber and UGA as
opal.
126The translation termination triplet
is [CBSE AIPMT 1996]
(a) UAU (b) UAA
(c) UAC (d) UGC
Ans. (b)
Termination codons are three in number
they are UAA (ochre) UAG (amber) and
UGA (opal).
127If the sequence of bases in DNA is
ATTCGATG, then the sequence of
bases in its transcript will be
[CBSE AIPMT 1995]
(a) CAUCGAAU (b) UAAGCUAC
(c) GUAGCUUA (d) AUUCGAUG
Ans. (b)
Transcription is the process of synthesis
ofmRNA on DNA template by the
complementary bases. As thymine is
replaced by uracil in RNA so, the
sequence of bases will be UAAGCUAC.
128Anticodon is an unpaired triplet of
bases in an exposed position of
[CBSE AIPMT 1995]
(a)mRNA (b) rRNA
(c)tRNA (d) sRNA
Ans. (c)
tRNA possess anticodon stem which
includes five paired bases. The
anticodon loop consists of 7 unpaired
bases. The third, fourth and fifth of
which form anticodon. This anticodon
permits temporary complementary
pairing with three bases onmRNA.
129DNA template sequence of
CTGATAGC is transcribed over
mRNA as [CBSE AIPMT 1994]
(a) GUCTUTCG
(b) GACUAUCG
(c) GAUTATUG
(d) UACTATCU
Ans. (b)
During transcription complementary
mRNA is formed on DNA template in
which T is replaced by U. So, the
sequence will be GACUAUCG.
130The number of base substitution
possible in amino acid codons is
[CBSE AIPMT 1994]
(a) 261 (b) 264
(c) 535 (d) 549
Ans. (d)
There are 64 codons out of which 61
codes for amino acid. Each codon
possess 3 bases which can undergo
transition and transversion, so the
number of base substitution possible in
amino acid codons is61 3 549
2
× = .
131Initiation codon of protein
synthesis (in eukaryotes) is
[CBSE AIPMT 1993, 94, 99, 2000]
(a) GUA (b) GCA
(c) CCA (d) AUG
Ans. (d)
At5′end ofmRNA where protein
synthesis starts codon AUG is present.
So, AUG is called as initiating or starting
codon or start signal.
132The process of translation is
[CBSE AIPMT 1993]
(a) ribosome synthesis
(b) protein synthesis
(c) DNA synthesis
(d) RNA synthesis
Ans. (b)
Translation is the process of protein
synthesis in which the triplet base
sequences ofmRNA molecules is
converted into a specific sequences of
amino acids in a polypeptide chain, this
occurs on ribosomes.
133Because most of the amino acids
are represented by more than one
codon, the genetic code is
[CBSE AIPMT 1993, 2002]
(a) overlapping (b) wobbling
(c) degenerate (d) generate
Ans. (c)
Degeneracy means lack of specificity.
Presence of more than one meaningful
codons for an amino acid is called
degeneracy, e.g. methionine and
tryptophan has single code for each. The
maximum number of codons for an
amino acid is six, e.g. serine, arginine
and leucine. Degeneracy provides a
protection against mutation.
134Khorana first deciphered the triplet
codons of [CBSE AIPMT 1992]
(a) serine and isoleucine
(b) threonine and histidine
(c) tyrosine and tryptophan
(d) phenylalanine and methionine
Ans. (b)
Dr. Hargobind Khorana deciphered first
triplet codon of threonine and histidine.
135In the genetic dictionary, there are
64 codons as[CBSE AIPMT 1990]
(a) 64 amino acids are to be coded
(b) 64 types oftRNAs are present
(c) there are 44 non-sense codons and
20 sense codons
(d) genetic code is triplet
Ans. (d)
It has been found that a sequence of 3
consecutive bases in a DNA molecule
codes for one specific amino acid. So,
genetic code is a triplet code and there
are 64 triplets which are called codons
( )4 4 4 64× × = of nitrogen bases for
protein synthesis.
276 NEETChapterwise Topicwise Biology

287. 136Genetic code consists of
[CBSE AIPMT 1988]
(a) adenine and guanine
(b) cytosine and uracil
(c) cytosine and guanine
(d) All of the above
Ans. (d)
The sequence of nitrogen bases on the
mRNA which store information for
linking the amino acids in a definite
sequence during synthesis of proteins is
calledgenetic code.These nitrogen
bases include adenine, guanine, cytosine
and uracil
137In the process of transcription in
eukaryotes, the RNA polymerase I
transcribes[NEET (Odisha) 2019]
(a)mRNA with additional processing,
capping and tailing
(b)tRNA, 5srRNA andsnRNAs
(c)rRNAs-28 S, 18 S and 5.8 S
(d) precursor ofmRNA,hnRNA
Ans. (c)
In the process of transcription (i.e.
copying of genetic information from one
strand of the DNA into RNA) in
eukaryotes, the RNA polymerase I
transcribesrRNA - 28S, 18S and 5.8S. On
the other hand,tRNA, 5srRNA and
snRNAs are transcribed by RNA
polymerase III. RNA polymerase II
transcribes precursor ofmRNA,hnRNA.
138Match the following genes of the
Lacoperon with their respective
products [NEET (National) 2019]
A.igene (i)β-galactosidase
B.zgene (ii) Permease
C.agene (iii) Repressor
D.ygene (iv) Transacetylase
Select the correct option.
A B C D
(a) (iii) (i) (ii) (iv)
(b) (iii) (i) (iv) (ii)
(c) (iii) (iv) (i) (ii)
(d) (i) (iii) (ii) (iv)
Ans. (b)
(A)–(iii), (B)–(i), (C)–(iv), (D)–(ii)
In aLacoperon,igene is a regulator
gene which produces a repressor that
binds to operator gene and stops its
functioning.z, yandaare the three
structural genes in thelacoperon ofE.
coli.zgene producesβ-galactosidase for
hydrolysing galactoside.ygene
produces permease for allowing the
entry of lactose from outside.
Agene produces transacetylase which
helps to transfer an acetyl group from
acetyl Co-A to beta- galactoside.
139Select thecorrectmatch.
[NEET 2018]
(a) Matthew Meselson
and F. Stahl
:Pisum sativum
(b) Alfred Hershey and
Martha Chase
: TMV
(c) Alec Jeffreys :Streptococcus
pneumoniae
(d) Francois Jacob
and Jacques
Monod
:Lacoperon
Ans. (d)
Jacob and Monod(1916) discovered the
lacoperon. An operon is a part of
genetic material or DNA which acts as a
single regulated unit. It possesses one or
more structural genes, an operator
gene, a promoter gene, a regulator gene,
a repressor gene and an inducer or
corepressor.
Matthew MeselsonandF Stahl
discovered the semi-conservative mode
of DNA replication inE. coli.Alfred
HersheyandMartha ChaseuseT
2
Bacteiriophage in their experiments to
infectE. coliand proved that DNA is the
genetic material.Alec Jeffreys(1984)
invented the DNA fingerprinting
technique. This technique determines
nucleotide sequences of certain areas of
DNA which are unique to each individual.
140All of the following are parts of an
operon except [NEET 2018]
(a) an enhancer
(b) structural genes
(c) an operator
(d) a promoter
Ans.(a)
Exceptenhancer, all the given
components are parts of an operon.
Enhancer sequences are present in
eukaryotes that, when bound by specific
proteins or transcription factors,
enhance the transcription of an
associated gene.
On the other hand, operon is a regulatory
unit of DNA containing a cluster of genes
in prokaryotes.
141Which of the following is required
as inducer(s) for the expression of
lacoperon? [NEET 2016, Phase I]
(a) galactose
(b) lactose
(c) lactose and galactose
(d) glucose
Ans. (b)
Lacoperon is an inducible operon.
Lactose is the substrate for the enzyme
β-galactosidase and it also regulates
switching on and off of the operon.
Hence, it is termed as inducer.
142Which one of the following is
wrongly matched?
[CBSE AIPMT 2014]
(a) Transcription – Writing information
from DNA totRNA
(b) Translation–Using information in
mRNA to make protein
(c) Repressor protein – Binds to operator
to stop enzyme synthesis
(d) Operon – Structural genes, operator
and promoter
Ans. (a)
Statement (a) is wrongly matched
because transcription is a process of
mRNA synthesis from a DNA template. It
involves three main events, i.e. initiation
(binding of RNA polymerase to as DNA),
elongation (development of a short
stretch of DNA) and termination
(recognition of the transcription
termination sequence and the release of
RNA polymerase).
143Which enzyme/s will be produced
in a cell in which there is a
non-sense mutation in thelac
Y-gene? [NEET 2013]
(a)β-galactosidase
(b) Lactose permease
(c) Transacetylase
(d) Lactose permease and
transacetylase
Ans.(a)
β-galactosidase is a structural gene,
which carry codes for the synthesis of
protein. Mutation in thelacY gene ofE.
colineeds residues of cytoplasmic
enzymeβ-galactosidase. Lactose
permease is a membrane protein, which
is a major facilitator superfamily.
Transacetylase is an enzyme transferring
acetyl groups from one compound to
another.
Molecular Basis of Inheritance 277
Regulation of
Gene Expression
TOPIC 3

288. 144Select the two statements out of
the four (I-IV) given below aboutlac
operon.
I. Glucose or galactose may bind
with the repressor and inactivate
it.
II. In the absence of lactose, the
repressor binds with the
operator region.
III. The z-gene codes for permease.
IV. This was elucidated by Francois
Jacob and Jacques Monod.
The correct statements are
[CBSE AIPMT 2010]
(a) I and III (b) I and III
(c) II and IV (d) I and II
Ans. (c)
Statement II and IV are true aboutlac
operon. In prokaryotes, a hypothesis was
given in 1961 to explain the protein
synthesis regulation. This hypothesis
was given by F Jacob and J Monod and
for this they were awarded Nobel Prize in
1965, the hypothesis was known by the
name of Operon Model.
The operator gene is the segment of
DNA, which exercise a control over
transcriptions. In the absence of
lactose, the repressor binds with the
operator gene.
145Differentiation of organs and
tissues in a developing organism is
associated with[CBSE AIPMT 2007]
(a) developmental mutations
(b) differential expression of genes
(c) lethal mutations
(d) deletion of genes
Ans. (b)
Differentiation of organs and tissues in a
developing organism is associated with
differential expression of genes. In
regulation of gene expression the
chromosomal proteins play important
role.
The chromosomal proteins are of two
types, histones and non-histones. The
regulation of gene expression involves
an interaction between histones and
non-histones.
146What does‘lac’refer to in what we
call thelacoperon?
[CBSE AIPMT 2003]
(a) Lac insect
(b) The number, 1,00,000
(c) Lactose
(d) Lactase
Ans. (c)
Lacoperon refers to the DNA sequence
in the genome of the bacteriumE. coli
encoding enzymes involved in lactose
uptake and metabolism.
147InE.coli,during lactose
metabolism repressor binds to
[CBSE AIPMT 2002]
(a) regulator gene (b) operator gene
(c) structural gene (d) promoter gene
Ans. (b)
InLacoperon, the repressor protein
combines with the operator gene to
express its functioning.
148Jacob and Monod studied lactose
metabolism inE.coliand proposed
Operon concept. Operon concept
applicable for[CBSE AIPMT 2002]
(a) all prokaryotes
(b) all prokaryotes and some eukaryotes
(c) all prokaryotes and all eukaryotes
(d) all prokaryotes and some protozoans
Ans. (b)
Jacob and Monod’s operon concept is
basically a theory of gene expression in
prokaryotes— though it is of some value
in the explanation of eukaryotic gene
expression.
149In negative operon
[CBSE AIPMT 2001]
(a) co-repressor binds with repressor
(b) co-repressor does not bind with
repressor
(c) co-repressor binds with inducer
(d)cAMP has negative effect onlac
operon
Ans.(a)
In negative operon co-repressor binds
with repressor to form repressor
co-repressor complex which further
binds with operator. Since, the product
of the regulator (the repressor) acts by
shutting off the transcription of
structural genes, it is referred to as a
negative control system.
150Genes that are involved in turning
on or off the transcription of a set
of structural genes are called
[CBSE AIPMT 1998]
(a) polymorphic genes
(b) operator genes
(c) reductant genes
(d) regulatory genes
Ans. (d)
The switching on and off of an operator
is controlled by respressor protein which
is coded by the regulator gene R.
151The wild typeE. colicells are
growing in normal medium with
glucose. They are transferred to a
medium containing only lactose as
sugar. Which of the following
changes takes place?
[CBSE AIPMT 1995]
(a) Thelacoperon is repressed
(b) All operons are induced
(c) Thelacoperon is induced
(d)E. colicells stop dividing
Ans. (c)
Inducible genes are the genes which
remain inactive or repressed in a cell and
can be activated when a certain
substrate is to be metabolised. It has
been seen when lactose is added to the
medium ofE.colithe operon is induced
and synthesis of enzymes required for
degradation of lactose to glucose and
galactose starts.
152InEscherichia coli lac Operonis
induced by [CBSE AIPMT 1994]
(a) lactose
(b) promoter gene
(c)β-galactosidase
(d) I-gene
Ans. (a)
InE. coli lacoperon is induced by adding
lactose sugar to the culture.
153Binding of specific protein on
regulatory DNA sequence can be
studied by means of
[CBSE AIPMT 1993]
(a) ultra centrifugation
(b) electron microscope
(c) light microscope
(d) X-rays crystallography
Ans.(d)
X-ray crystallography is an important
technique in molecular biology to
analyse the structure and orientation of
molecules. It is used to find out 3-D
positions of atoms in the molecules of
DNA, RNA and proteins, binding of
specific protein on regulatory DNA
sequences, 3-D structure of
haemoglobin, insulin, DNA, proteins,
collagen fibre, muscle and actin protein.
278 NEETChapterwise Topicwise Biology

289. 154DNA fingerprinting involves
identifying differences in some
specific regions in DNA sequence,
called as [NEET 2021]
(a) satellite DNA
(b) repetitive DNA
(c) single nucleotides
(d) polymorphic DNA
Ans. (b)
Repetitive DNA are DNA sequences that
are repeated in the genome. These
sequences do not code for protein. One
class termed highly repetitive DNA
consists of short sequences, 5-100
nucleotides, repeated thousands of
times in a single stretch and includes
satellite DNA.
Other options can be explained as:
The density of DNA is a function of its
base and sequence, and satellite DNA
with its highly repetitive DNA has a
reduced or a characteristic density
compared to the rest of the genome.
Single nucleotide polymorphisms,
frequently called SNPs (pronounced
‘snips’), are the most common type of
genetic variation among people. Each
SNP represents a difference in a single
DNA building block, called a nucleotide.
DNA polymorphisms are the different
DNA sequences among individuals,
groups, or populations. Polymorphism at
the DNA level includes a wide range of
variations from single base pair change,
many base pairs, and repeated
sequences. DNA polymorphisms are
endless, and more discoveries continue
at a rapid rate. These are called as
polymorphic DNA.
155DNA strands on a gel stained with
ethidium bromide when viewed
under UV radiation, appear as
[NEET 2021]
(a) yellow bands
(b) bright orange bands
(c) dark red bands
(d) bright blue bands
Ans. (b)
To make the DNA visible in the gel,
ethidium bromide is added to the gel
solution and the buffer. This positively
charged polycyclic aromatic compound
binds to DNA by inserting itself between
the basepairs (intercalation). The DNA
fragments when exposed to ultraviolet
light appear as orange colour bands, due
to the large increase in fluorescence of
the ethidium bromide upon binding to
the DNA.
156Which is the basis of genetic
mapping of human genome as well
as DNA fingerprinting?
[NEET (Oct.) 2020]
(a) Polymorphism in DNA sequence
(b) Single nucleotide polymorphism
(c) Polymorphism inhnRNA sequence
(d) Polymorphism in RNA sequence
Ans. (a)
Polymorphism in DNA sequence is the
basis of genetic mapping of human
genome as well as DNA fingerprinting.
Polymorphism simply means variation at
genetic level which arises due to mutations.
157Which of the following is not
required for any of the techniques
of DNA fingerprinting available at
present? [NEET 2016, Phase I]
(a) Zinc finger analysis
(b) Restriction enzymes
(c) DNA-DNA hybridisation
(d) Polymerase chain reaction
Ans.(a)
A zinc finger is a small protein structural
motif that is characterised by the
coordination of one or more Zn ions in
order to stabilise the folds.
158Satellite DNA is important because
it [CBSE AIPMT 2015]
(a) codes for proteins needed in cell
cycle
(b) shows high degree of polymorphism
in population and also the same
degree of polymorphism in an
individual, which is heritable from
parents to children
(c) does not code for proteins and is
same in all members of the
population
(d) codes for enzymes needed for DNA
replication
Ans. (b)
Satellite DNA forms the minor peak after
centrifugation of DNA. These are
repetitive DNA sequences that do not
code for any protein. They show high
degree of polymorphism and heritable
from parents to children, thus form the
basis of DNA fingerprinting.
159DNA fingerprinting refers to
[CBSE AIPMT 2004]
(a) molecular analysis or profiles of DNA
samples
(b) analysis of DNA samples using
imprinting device
(c) techniques used for molecular
analysis of different specimens of
DNA
(d) techniques used for identification of
finger-prints of individuals
Ans. (a)
DNA fingerprinting refers to molecular
analysis of DNA samples. Alec Jeffreys
[1985, 86] discovered this technique for
the first time.
160Nucleus of a donor embryonal
cell/somatic cell is transferred to
an enucleated egg cell. Then after
the formation of organism, what
shall be true?[CBSE AIPMT 2002]
(a) Organism will have extra-nuclear
genes of the donor cell
(b) Organism will have extra-nuclear
genes of recipient cell
(c) Organism will have extra-nuclear
genes of both donor and recipient cell
(d) Organism will have nuclear genes of
recipient cell
Ans. (b)
The organism will have extranuclear
genes of recipient cell. Since, the
recipient cell has already been
enucleated (its nucleus is removed), the
organism developing from it would have
the nuclear genes of donor cell.
161The basis for DNA fingerprinting is
[CBSE AIPMT 1996]
(a) occurrence of Restriction Fragment
Length Polymorphism (RFLP)
(b) phenotypic differences between
individuals
(c) availability of cloned DNA
(d) knowledge of human karyotype
Ans. (a)
The basis of DNA fingerprinting is the
occurrence of restriction fragment
length polymorphism which are
distributed throughout human genome.
DNA fingerprinting was developed by a
British geneticistProf. Alec Jeffreysin
1984.
The chromosomes of every human cell
contain short, highly repeated 15
nucleotide segments called
‘mini-satellites’orvariable number
tendem repeatsscattered through their
DNA.
Molecular Basis of Inheritance 279
Human Genome Project
and DNA Fingerprinting
TOPIC 4

290. 01After about how many years of
formation of earth, life appeared on
this planet?[NEET (Oct.) 2020]
(a) 500 billion years
(b) 50 million years
(c) 500 million years
(d) 50 billion years
Ans. (c)
During the course of evolution, earth
probably formed about 4.5 billion years
ago. After its formation, life appeared
about 500 million years ago in oceans.
The first living forms were believed to be
heterotrophs and used to derive
nutrition from external sources.
02From his experiments, SL Miller
produced amino acids by mixing
the following in a closed flask.
[NEET (Sep.) 2020]
(a)CH ,H ,NH
32 4
and water vapour at 800°C
(b)CH ,H ,NH
4 2 3
and water vapour at 600°C
(c)CH ,H ,NH
32 3
and water vapour at 600°C
(d)CH ,H ,NH
4 2 3
and water vapour at 800°C
Ans. (d)
The correct option is (d) because in 1953,
SL Miller, created electric discharge in a
closed flask containing CH
4
, H
2
, NH
3
and
water vapours at 800°C.
03Variations caused by mutation, as
proposed by Hugo de Vries are
[NEET (National) 2019]
(a) random and directionless
(b) small and directional
(c) small and directionless
(d) random and directional
Ans.(a)
Hugo de Vries proposed that the variations
caused by mutation are random and
directionless. These are the sudden,
heritable changes in the genetic material
and these variations constitute the raw
material for evolution. He also proposed
that mutations play a key role in speciation
and used the term saltation for single
step large mutations.
04According to Hugo de Vries, the
mechanism of evolution is
[NEET 2018]
(a) phenotypic variations
(b) saltation
(c) multiple step mutations
(d) minor mutations
Ans. (b)
According to Hugo de Vries, the
mechanism of evolution issaltation.
Hugo de Vries (1901) proposed mutation
theory of evolution and stated that
evolution is a jerky process in which new
species are evolved due to discontinous
sudden variations or saltation. These are
the single step large mutations
occurring in population.
05Following are the two statements
regarding the origin of life
[NEET 2016, Phase I]
I. The earliest organisms that
appeared on the earth were
non-green and presumably
anaerobes.
II. The first autotrophic organisms
were the chemoautotrophs that
never released oxygen.
Of the above statements which
one of the following options is
correct?
(a) II is correct but I is false
(b) Both I and II are correct
(c) Both I and II are false
(d) I is correct but II is false
Ans. (b)
The earliest organisms that appeared on
earth were anaerobic chemoautotrophs.
Chemoautotrophs were the first
autotrophic organisms. They were
unable to perform photolysis of water
and never released oxygen, e.g. sulphur
bacteria.
06Which of the following is the
correct sequence of events in the
origin of life?[NEET 2016, Phase II]
I. Formation of protobionts.
II. Synthesis of organic monomers.
III. Synthesis of organic polymers.
IV. Formation of DNA-based genetic
systems.
(a) I, II, III, IV (b) I, III, II, IV
(c) II, III, I, IV (d) II, III, IV, I
Ans. (c)
The sequential manner of events of
origin of life is as follows
II. Synthesis of organic monomers.
↓
III. Synthesis of organic polymers.
↓
I. Formation of protobionts.
↓
IV. Formation of DNA-based genetic
systems.
Thus, option (c) is correct.
07Which one of the following is
incorrect about the characteristics
of protobionts (coacervates and
microspheres) as envisaged in the
abiogenic origin of life?
[CBSE AIPMT 2008]
(a) They were able to reproduce
(b) They could separate combinations of
molecules from the surroundings
Evolution
29
Origin of Life
TOPIC 1

291. Evolution 281
(c) They were partially isolated from the
surroundings
(d) They could maintain an internal
environment
Ans. (d)
Homeostasis is keeping the internal
environment of the body constant. It is
necessary for normal life processes.
Microspheres are molecular aggregates
of proteinoids. Oparin and Sydney Fox
held that large organic molecules
synthesised abiotically on primitive earth
formed large colloidal aggregates due to
the intermolecular attraction.
These colloidal particles were called
coacervates. Oparin called giant
nucleoproteinoid molecules as
protobionts. These reproduce either by
budding or binary fission but do not
exhibit homeostasis.
08Which one of the following pair of
items correctly belongs to the
category of organs mentioned
against it?[CBSE AIPMT 2008]
(a)Thorn of
Bougainvilleaand
tendrils ofCucurbita
— Analogous
organs
(b) Nictitating
membrane and blind
spot in human eye
— Vestigial
organs
(c)Nephridia of
earthworm and
Malpighian tubules of
cockroach
— Excretory
organs
(d)Wings of honey bee
and wings of crow
— Homologou
s organs
Ans.(c)
In annelids likeNereis,earthworm, leech,
etc., the tubular coiled structures called
nephridia are excretory organs. In
phylum–Arthropoda, insects centipedes,
millipedes and arachnides possess
Malpighian tubules as their principal
excretory organ.Analogous organshave
almost similar appearance and perform
the same function but develop in totally
different groups and are totally different
in their basic structure and
developmental origin, e.g. wings of
butterfly, birds, bats.
The homologous organs have common
origin, perform different type of
functions and have different
appearance, e.g. thorns ofBougainvillea
and tendrils ofCucurbita.
Vestigial organs are useless remnants,
which might have been large and
functional in the ancestors, e.g.
nictitating membrane, vermiform
appendix, etc.
09The concept of chemical evolution
is based on[CBSE AIPMT 2007]
(a) crystalisation of chemicals
(b) interaction of water, air and clay
under intense heat
(c) effect of solar radiation on chemicals
(d) possible origin of life by combination
of chemicals under suitable
environmental conditions
Ans. (d)
The concept of chemical evolution is
based on possible origin of life by
combination of chemicals under suitable
environmental conditions.
10Evolutionary history of an organism
is known as[CBSE AIPMT 2006]
(a) ancestry
(b) palaeontology
(c) ontogeny
(d) phylogeny
Ans. (d)
Phylogeny (Gr.Phylon−tribe or race;
geneia= origin) is the origin and
diversification of any taxon or the
evolutionary history of its origin and
diversification. It is usually represented
as a diagrammatic phylogenetic tree
(that traces putative evolutionary
relationships), i.e., dendrogram.
Palaentology is the study of fossils.
Ontogeny is the whole course of an
individual’s development and life history.
11Which one of the following amino
acid was not found to be
synthesised in Miller’s experiment?
[CBSE AIPMT 2006]
(a) Aspartic acid (b) Glutamic acid
(c) Alanine (d) Glycine
Ans. (b)
Miller and Urey were the two scientists
who recreated the conditions of
primitive earth in laboratory and
abiotically synthesised amino acids and
bases. They synthesised glycine,
aspartic acid and alanine in abundant
quantities while, glutamic acid could not
be synthesised in their experiment.
12Which one of the following
experiments suggests that
simplest living organisms could not
have originated spontaneously from
non-living matter?
[CBSE AIPMT 2005]
(a) Larvae could appear in decaying
organic matter
(b) Microbes did not appear in stored
meat
(c) Microbes appeared from unsterilised
organic matter
(d) Meat was not spoiled, when heated
and kept sealed in a vessel
Ans. (d)
Pasteur performed experiments in which
he took sterilised (by boiling) yeast and
sugar solution in a long naked flask, then
he bent the neck of the flask like a neck
of swan. After one month he observed
that no life appeared in flask solution
because the curved flask neck acts as a
filter. He later on broke down the neck
and observed the solution. He found that
many microorganisms were originated in
solution.
13According to Oparin, which one of
the following was not present in
the primitive atmosphere of the
earth? [CBSE AIPMT 2004]
(a) Methane (b) Oxygen
(c) Hydrogen (d) Water vapour
Ans. (b)
According to Oparin the atmosphere of
primitive earth was reducing because H
atoms were most numerous end most
reactive. Large quantities ofH , N
2 2
,
water vapour,CO
2
,CH
4
andNH
3
were
present, but free oxygen was not
present in significant amount.
14Identify the correct sequence in
which the following substances
have appeared during the course of
evolution of life on earth
[CBSE AIPMT 1996]
(a) glucose, amino acids, nucleic acids,
proteins
(b) ammonia, amino acids, proteins,
nucleic acids
(c) water, amino acids, nucleic acids,
enzymes
(d) amino acids, ammonia, phosphates,
nucleic acids
Ans. (b)
Primitive atmosphere was reducing type
(without free oxygen). Hydrogen atoms
combined with oxygen forming water
and with nitrogen, forming ammonia.
Water and ammonia were probably the
first compound molecules on primitive
earth. The primitive atmosphere
contained gases likeCO
2
, CO, N,H
2
, etc.
and methane (CH
4
) was the first organic
compound formed in primitive
atmosphere.
In the primitive atmosphere, electric
discharge, ATP and solar energy
provided the source of energy for
polymerization reactions of organic

292. synthesis which lead to the synthesis of
amino acids, that joined to form
polypeptides and proteins. Simple sugar
units combined to form polysaccharides;
fatty acids and glycerol to form fats;
sugars, nitrogenous bases and
phosphates combined into nucleotides
which polymerised into nucleic acids in
the ancient oceans.
15The first organisms were
[CBSE AIPMT 1992]
(a) chemoautotrophs
(b) chemoheterotrophs
(c) autotrophs
(d) eukaryotes
Ans. (b)
The first cells were almost certainly
heterotrophs, obtaining energy and
nutrients from organic molecules in their
environment. The prokar- yotes evolved
before the eukaryotes, the earliest
prokaryotes must have been
chemoheterotrophs.
16Which was absent in the
atmosphere at the time of origin of
life? [CBSE AIPMT 1991]
(a)NH
3
(b)H
2
(c)O
2
(d)CH
4
Ans. (c)
The atmosphere of earth at the time of
origin of life was without free oxygen
atoms. The primitive atmosphere chiefly
consisted of methane, ammonia, water
vapour, hydrogen gas, nitrogen gas, and
some carbon monoxide.
Hydrogen atoms were most numerous
and most reactive in primitive
atmosphere. They combined with all
available oxygen atoms and formed
water.
17‘Origin of species’ was written by
[CBSE AIPMT 1989]
(a) Oparin
(b) Weismann
(c) Lamarck
(d) Darwin
Ans. (d)
Darwin andWallacepublished a joint
paper titled‘Origin of species’in 1858.
Later in 1859, Darwin published his
detailed theory in his book titled ‘Origin
of species by means of Natural
selection’.
According to Darwin, variations are
progressive factors for evolution.
18Evolution is [CBSE AIPMT 1989]
(a) progressive development of a race
(b) history and development of race
along with variations
(c) history of race
(d) development of race
Ans. (b)
Evolution is defined as history and
development of race with variations. The
term ‘evolution’ in Biology means gradual
changes. It is an excellent working
hypothesis to approach the problems of
diversity of organisms.
19First life on earth was
[CBSE AIPMT 2001]
(a) cyanobacteria
(b) chemoheterotrophs
(c) autotrophs
(d) photoautotrophs
Ans. (b)
First living beings were formed in the
environment of sea having abundant
organic molecules. They absorbed the
organic materials for the sake of
nutrition and hence, were
chemoheterotrophs.
20Match the List-I with List-II.
List-I List-II
A. Adaptive
radiation
1. Selection of
resistant radiation
varieties due to
excessive use of
herbicides and
pesticides
B. Convergent
evolution
2. Bones of forelimbs
in man and whale
C. Divergent
evolution
3. Wings of butterfly
and bird
D. Evolution by
anthropogeni
c action
4. Darwin Finches
Choose the correct answer from
the options given below.
[NEET 2021]
A B C D
(a) 4 3 2 1
(b) 3 2 1 4
(c) 2 1 4 3
(d) 1 4 3 2
Ans. (a)
(A)-(4), (B)-(3), (C)-(2), (D)-(1)
Adaptive radiationis a change that occur
in organism by adapting according to the
environment, e.g. Darwin finches.
Convergent evolutionis a process where
distant species develop similar structures,
e.g. wings of butterfly and birds.
Divergent evolutionis a process by
which an inbreeding species diverges
into two descendant species, e.g. bones
of forelimbs in man and whales.
Evolution by anthropogenic action
means evolution occurring due to human
activities, e.g. selection of resistant
varieties due to excessive use of
herbicides and pesticides.
21Embryological support for
evolution was proposed by
[NEET (Oct.) 2020]
(a) Ernst Heckel
(b) Karl Ernst von Baer
(c) Charles Darwin
(d) Alfred Wallace
Ans.(a)
Embryological support for evolution was
proposed by Ernst Haeckel. He proposed
biogenetic law in year 1864. According to
this law ‘structure of ancient origin
develops earlier than structure of newer
origin. In other words, it states
“Ontogeny repeats phylogeny”, i.e.
development of structures in an
organism follow the same sequence as
they evolved in his ancestors.
22The phenomenon of evolution of
different species in a given
geographical area starting from a
point and spreading to other
habitats is called[NEET (Oct.) 2020]
(a) saltation
(b) co-evolution
(c) natural selection
(d) adaptive radiation
Ans. (d)
The process of evolution of different
species in a given geographical area
starting from a point and literally
radiating to other areas of geography
(habitats) is called adaptive radiation.
23Embryological support for
evolution was disapproved by
[NEET (Sep.) 2020]
(a) Alfred Wallace
(b) Charles Darwin
(c) Oparin
(d) Karl Ernst von Baer
282 NEETChapterwise Topicwise Biology
Evidences of Evolution
TOPIC 2

293. Ans. (d)
Embryological support for evolution was
disapproved by Karl Ernst von Baer. He
observed the pattern of embryonic
development in different species.
24Which of the following refer to
correct example(s) of organisms
which have evolved due to changes
in environment brought about by
anthropogenic action?
[NEET (Sep.) 2020]
I. Darwin‘s Finches of Galapagos
islands.
II. Herbicide resistant weeds.
III. Drug resistant eukaryotes.
IV. Man-created breeds of
domesticated animals like dogs.
(a) I and III (b) II, III and IV
(c) Only IV (d) Only I
Ans. (b)
The correct option is (b) because
Herbicide resistant weeds, drug
resistant eukaryotes and man-created
breeds of domesticated animals like
dogs are examples of
evolution by anthropogenic action.
Darwin’s Finches of Galapagos islands
are examples of natural selection,
adaptive radiation and founder’s effect.
25Flippers of penguins and dolphins
are examples of[NEET (Sep.) 2020]
(a) convergent evolution
(b) industrial melanism
(c) natural selection
(d) adaptive radiation
Ans.(a)
Flippers of penguins and dolphins are
examples of convergent evolution.They
have similar function (helps in swimming)
but different origin so, they are also
called analogous organs. Penguin and
dolphins are not closely related to each
other but evolved similar traits (flippers)
which represent convergent evolution.
Hence analogous organs are a result of
convergent evolution.
26In Australia, marsupials and
placental mammals have evolved to
share many similar characteristics.
This type of evolution may be
referred to as[NEET (Odisha) 2019]
(a) adaptive radiation
(b) divergent evolution
(c) cyclical evolution
(d) convergent evolution
Ans. (d)
In Australia, marsupials and placental
mammals have evolved to share many
similar characteristics. This type of
evolution is referred to as convergent
evolution.
Convergent evolution is the independent
evolution of similar features in species
of different lineages. For example, a
number of marsupials, each different
from the other evolved from an ancestral
stock, but all within the Australian island
continent. Also, marsupials in Australia
resemble placental mammals in the rest
of the world, they evolved in isolation
after Australia separated from other
continents.
27In a species, the weight of newborn
ranges from 2 to 5 kg. 97% of the
newborn with an average weight
between 3 to 3.3 kg survive
whereas 99% of the infants born
with weights from 2 to 2.5 kg or 4.5
to 5 kg die. Which type of selection
process is taking place?
[NEET (National) 2021]
(a) Stabilising selection
(b) Disruptive selection
(c) Cyclical selection
(d) Directional selection
Ans.(a)
The given data represents stabilising
selection. It eliminates individuals from
both ends of a phenotypic distribution
and hence maintains the same
distribution average. In the given
situation, most of the newborn of
average weight 3-3.3 kg survive. Babies
having more or less weight had low
survival rate. Disruptive selection
favours both extremes of continuous
variation. Directional selection favours
one extreme of continuous variation.
Cyclical selection is regarded as a
source of polymorphism.
28Among the following sets of
examples for divergent evolution,
select theincorrectoption. :
[NEET 2018]
(a) Brain of bat, man and cheetah
(b) Heart of bat, man and cheetah
(c) Forelimbs of man, bat and cheetah
(d) Eye ofOctopus,bat and man
Ans. (d)
Divergent evolutionresults in
homologous structures. These organs
have the same fundamental structure
but are different in functions. Structural
homology is seen in brain, heart and
forelimbs of man, bat and cheetah.
Eyes ofOctopus, bat and manare
examples of analogous organs which
showconvergent evolution.Therefore,
option (d) is incorrect.
29The similarity of bone structure in
the forelimbs of many vertebrates is
an example of [NEET 2018]
(a) convergent evolution
(b) analogy
(c) homology
(d) adaptive radiation
Ans. (c)
The similarity of bone structure in the
forelimbs of many vertebrates is an
example ofhomology.The homologous
organs have the same fundamental
structure but are adapted to perform
different functions,e.g.forelimbs of
man, cheetah, whale and bat.
Analogous organsshowconvergent
evolution.These organs have similar
functions but are different in their
structural details and origin.
Development of different functional
structures from a common ancestral
form is calledadaptive radiation.
30Analogous structures are a result
of [NEET 2016, Phase I]
(a) convergent evolution
(b) shared ancestry
(c) stabilising selection
(d) divergent evolution
Ans.(a)
Analogous structures are a result of
convergent evolution. When organisms
with completely different organisation,
living in the same habitat come to
possess superficial resemblance, this is
known as convergent evolution.
31Which of the following structures is
homologous to the wing of a bird?
[NEET 2016, Phase I]
(a) Wing of a moth
(b) Hind limb of rabbit
(c) Flipper of whale
(d) Dorsal fin of a shark
Ans. (c)
Wings of bird and flipper of whale are
modified fore limbs of the two
organisms so have same origin wings
help in flying and flippers help in
swimming, but this perform the different
functions.
Evolution 283

294. 32The wings of a bird and the wings
of an insect are[CBSE AIPMT 2015]
(a) homologous structures and represent
divergent evolution
(b) analogous structures and represent
convergent evolution
(c) phylogenetic structures and
represent divergent evolution
(d) homologous structures and represent
convergent evolution
Ans. (b)
The wings of a bird and the wings of an
insect are analogous structures and
represent convergent evolution.
Analogous organs have the same
function and are superficially alike only.
However their fundamental structures
are quite different in morphology,
anatomy and embryonic origin. Analogy
is an example of convergent evolution.
33Forelimbs of cat, lizard used in
walking forelimbs of whale used in
swimming and forelimbs of bats
used in flying are an example of
[CBSE AIPMT 2014]
(a) analogous organs
(b) adaptive radiation
(c) homologous organs
(d) convergent evolution
Ans. (c)
Homologous organs are those organs
which have a common fundamental
anatomical plan and similar embryonic
origin but perform varied functions.
Forelimbs of cat, lizard used in walking,
forelimbs of whale used in swimming
and forelimbs of bats used in flying are
the example for homologous organs. All
are the examples of modified forelimbs,
with the same type of bones. They have
become different due to the adaptations
to different habitat.
34Which one of the following are
analogous structures?
[CBSE AIPMT 2014]
(a) Wings of bat and wings of pigeon
(b) Gills of prawn and lungs of man
(c) Thorns ofBougainvilleaand tendrils of
Cucurbita
(d) Flippers of dolphin and legs of horse
Ans. (a)
Analogous organs are the structures of
different species having similar or
corresponding functions but different
structure. They do not belong to the
same evolutionary origin.
Wings of bat are skin folds stretched
mainly between elongated fingers but
the wing of birds are feathers covering
all along the arm. They look similar
because they have a common use for
flying, but their origin is not common.
35The eyes ofOctopusand eyes of
cat show different patterns of
structure, yet they perform similar
function. This is an example of
[NEET 2013]
(a) homologous organs that have evolved
due to convergent evolution
(b) homologous organs that have evolved
due to divergent evolution
(c) analogous organs that have evolved
due to convergent evolution
(d) analogous organs that have evolved
due to divergent evolution
Ans. (c)
The analogous organs are not
anatomically similar structures though
they perform similar functions. Hence,
analogous structures are a result of
convergent evolution, i.e. different
structures evolving for the same
function and hence, having similarity.
Therefore, the eyes ofOctopusand eyes
of cat are examples of analogous organs,
though they are different in structure
but similar in function. Homologous
organs develop along different
directions due to the adaptations to
various needs. This is divergent
evolution and the structures are
homologous.
36The process by which organisms
with different evolutionary history
evolve similar phenotypic
adaptations in response to a
common environmental challenge,
is called [NEET 2013]
(a) natural selection
(b) convergent evolution
(c) non-random evolution
(d) adaptive radiation
Ans. (b)
Convergent evolution occurs in
unrelated group of organisms. It is the
development of similar functional
structures but in unrelated groups. The
process of evolution of different species
in a given geographical area starting
from a point and literally radiating to
other areas of geography is called
adaptive radiation. Natural selection is
the basis of evolution.
37Evolution of different species in a
given area starting from a point and
spreading to other geographical
areas is known as
[CBSE AIPMT 2012]
(a) adaptive radiation
(b) natural selection
(c) migration
(d) divergent evolution
Ans. (a)
The diversification of an ancestral group
into two or more species in different
habitats is called divergent evolution.
When this involves large number of
species to occupy different ritches, this
is called adaptive radiation. Adaptive
radiation is the process of evolution of
different species in a given geographical
area starting from a species of animals
or plants and literally radiating to other
areas of geography (habitats). Darwin’s
finches represent one of best examples
of this phenomenon.
38Which one of the following options
gives one correct example each of
convergent evolution and divergent
evolution? [CBSE AIPMT 2012]
Convergent
evolution
Divergent
evolution
(a) Eyes ofOctopus
and mammals
Bones of
forelimbs of
vertebrates
(b) Thorns of
Bougainvilleaand
tendrils of
Cucurbita
Wings of
butterflies and
birds
(c) Bones of
forelimbs of
vertebrates
Wings of butterfly
and birds
(d) Thorns of
Bougainvilleaand
tendrils of
Cucurbita
Eyes ofOctopus
and mammals
Ans. (a)
Convergent evolution involves the
independent development of similar
structures in organisms that are not
directly related. It is represented by
analogous organs, e.g. eyes ofOctopus
and mammals, wings of insects and
birds. Indivergent evolution, same basic
organ becomes adapted by specialisation
to perform different functions.
It is represented byhomologous organs,
e.g. Bones of forelimbs of vertebrates
(like seal’s flipper, bat’s wing, cat’s paw
horse’s front leg and human hand),
thorns ofBougainvilleaand tendrils of
Cucurbita.
284 NEETChapterwise Topicwise Biology

295. 39Peripatusis a connecting link
between [CBSE AIPMT 2009]
(a) Ctenophora and Platyhelminthes
(b) Mollusca and Echinodermata
(c) Annelida and Arthropoda
(d) Coelenterata and Porifera
Ans. (c)
Peripatusis connecting link between
Annelida and Arthropoda. It is a living
fossil that has similarities to both
arthropods (such as absence of external
segmentation, unjoined legs, the
presence of cuticle, etc) and annelides
(internal segmentation, eyes and
segmentally arranged nephridia etc.).
40Darwin’s finches are an excellent
example of[CBSE AIPMT 2010, 08]
(a) adaptive radiation
(b) seasonal migration
(c) brood parasitism
(d) connecting links
Ans. (a)
Adaptive radiation represents evolution
of new forms in several directions from
the common ancestral type. In 1831,
Darwin got an opportunity to travel by
HMS Beagle for a voyage of world
exploration. Beagle sailed to the
Galapagos Islands, here Darwin found a
living laboratory of evolution.
The common birds of Galapagos Islands,
the finches were markedly different
from the finches of main land. The
closely related species of finches had
beaks of different shapes and sizes and
adapted for feeding on completely
different diet showing adaptive radiation.
The transitional fossil forms which show
characteristic of two different groups of
living animals are called connecting
links, e.g.Archaeopteryx, Seymouria,etc.
41Thorn ofBougainvilleaand tendril
ofCucurbitaare examples of
[CBSE AIPMT 2008]
(a) analogous organs
(b) homologous organs
(c) vestigial organs
(d) retrogressive evolution
Ans. (b)
Thorns ofBougainvilleaand tendrils of
Cucurbitaare homologous organs. These
are modified branches and are axillary in
position. It means axillary branches in
Bougainvilleaare modified into thorns
for protection from burrowing animals
and inCucurbitathey are modified into
tendrils for climbing.
The analogous organs have almost
similar appearance and perform the
same function but these develop in
totally different groups and are totally
different in their basic structure and
developmental origin. The phyllode of
Ruscusor cladode ofAsparagusare
analogous to leaves of other plants.
The vestigial or rudimentary organs
are useless remnants of structures or
organs, which might have been large
and functional in ancestors, e.g. cutin
covered stomata on the stem of cacti
plants.
42Which one of the following
statements is correct?
[CBSE AIPMT 2007]
(a) Stem cells are specialised cells
(b) There is no evidence of the existence
of gills during embryogenesis of
mammals
(c) All plant and animal cells are
totipotent
(d) Ontogeny repeats phylogeny
Ans. (d)
Recapitulation theory or Biogenetic law
states thatontogeny(development of
embryo) is recapitulation of phylogeny
(ancestral sequence).
43What is common to whale, seal and
shark? [CBSE AIPMT 2007]
(a) Seasonal migration
(b) Thick subcutaneous fat
(c) Convergent evolution
(d) Homeothermy
Ans. (c)
Distantly related animals (as whale, seal
and shark) inhabiting similar habitats
often develop similar morphological
features that make them look similar.
This is termed adaptive convergence of
convergent evolution.
44When two species of different
geneology come to resemble each
other as a result of adaptation, the
phenomenon is termed
[CBSE AIPMT 2007]
(a) divergent evolution
(b) micro-evolution
(c) co-evolution
(d) convergent evolution
Ans. (d)
In convergent evolution lineages show
similar morphology under the
influence of similar environmental
factors.
45The finches of Galapagos islands
provide an evidence in favour of
[CBSE AIPMT 2007]
(a) special creation
(b) evolution due to mutation
(c) retrogressive evolution
(d) biogeographical evolution
Ans. (d)
Darwin’s finches of Galapagos islands
had common ancestors later on
whose beaks modified according to
their feeding habits. These provide
evidence of geographical distribution.
46An important evidence in favour of
organic evolution is the occurrence
of [CBSE AIPMT 2006]
(a) analogous and vestigial organs
(b) homologous organs only
(c) homologous and analogous organs
(d) homologous and vestigial organs
Ans. (d)
An important evidence in favour of
organic evolutionis the occurrence of
homologous and vestigial organs.
Homologous organsare those which
have the common origin and are built on
the same fundamental pattern but they
perform different functions and have
different appearances, e.g. whale’s
flipper, bat’s wings, cat’s paws, horse’s
front legs, bird’s wings, ox’s front legs
and human hands.
Vestigial organsin animals are those
having no function now, in them, but had
important functions in their ancestors.
Analogous organs are quite different
in fundamental structure and
embryonic origin but perform the
same function. The study of
analogous organs illustrates the
occurrence ofconvergent evolution.
47Which of the following is the
relatively most accurate method
for dating of fossils?
[CBSE AIPMT 2005]
(a) Radio-carbon method
(b) Potassium-argon method
(c) Electron-spin resonance method
(d) Uranium-lead method
Ans. (c)
Electron spin resonance method is the
most accurate method for dating of
fossils. It measures number of charges
occupying deep traps in crystal band
gap.
Evolution 285

296. 48Using imprints from a plate with
complete medium and carrying
bacterial colonies, you can select
streptomycin resistant mutants
and prove that such mutations do
not originate as adaptation. These
imprints need to be used
[CBSE AIPMT 2005]
(a) on plates with and without
streptomycin
(b) on plates with minimal medium
(c) only on plates with streptomycin
(d) only on plates without streptomycin
Ans. (c)
Plates having streptomycin allow to
propagate only those bacteria which are
resistant to the antibiotic. While those
plates in which streptomycin is absent,
both resistant and non-resistant
bacteria can grow normally.
49Presence of gills in the tadpole of
frog indicates that
[CBSE AIPMT 2004]
(a) fishes were amphibian in the past
(b) fishes evolved from frog-like
ancestors
(c) frogs will have gills in future
(d) frogs evolved from gilled ancestors
Ans. (d)
According tobiogenetic lawofErnst
Haeckel(1866)ontogeny repeats
phylogeny.Ontogeny is the life history of
an organism while, phylogeny is the
evolutionary history of the race of that
organism. In other words we can say an
organism repeats its ancestral history
during its development.
Hence, resemblance of amphibian to fish
is seen in most systems of the body as
both are cold blooded, both respire by
gills (as tadpole of frog), both usually lay
eggs in water leading to the conclusion
that amphibians have originated from
fishes.
50What kind of evidence suggested
that man is more closely related
with chimpanzee than with other
hominoid apes?[CBSE AIPMT 2004]
(a) Evidence from DNA from sex
chromosomes only
(b) Comparison of chromosomes
morphology only
(c) Evidence from fossil remains and the
fossil mitochondrial DNA alone
(d) Evidence from DNA extracted from
sex chromosomes, autosomes and
mitochondria
Ans. (d)
Chimpanzee is more closely related to
man than other hominoids. It is evidenced
by chromosome banding pattern, DNA
extracted from sex chromosomes,
autosomes and mitochondria. Molecular
clock based on mitochondrial DNA are
used to date recent events because this
DNA mutates 5-10 times faster than
nuclear DNA. Some similarities between
human and chimpanzee are
(a) DNA matching shows human
similarity with chimpanzee.
(b) There is little difference in banding
pattern in chromosomes 3 and 6 in
human and chimpanzee.
(c) Serum test indicates maximum
homology between human and
chimpanzee.
51Age of fossils in the past was
generally determined by
radio-carbon method and other
methods involving radioactive
elements found in the rocks. More
precise methods, which were used
recently and led to the revision of
the evolutionary periods for
different groups of organisms,
include [CBSE AIPMT 2004]
(a) study of carbohydrates/proteins in
fossils
(b) study of the conditions of
fossilisation
(c) Electron Spin Resonance (ESR) and
fossil DNA
(d) study of carbohydrates/proteins in
rocks
Ans. (c)
Electron Spin Resonance (ESR)
measures number of charges occupying
deep traps in crystal band gap. The basic
principle of ESR is same as those for
luminescence, i.e. electrons become
trapped and stored as a result of ionising
radiations e.g. dating of tooth enamel.
52Convergent evolution is illustrated
by [CBSE AIPMT 2003]
(a) dogfish and whale
(b) rat and dog
(c) bacterium and protozoan
(d) starfish and cuttle fish
Ans. (a)
Convergent evolution is the acquisition
of same or similar characters by
distantly related lines of descent.
Dogfish (pisces) and whale (mammals)
have acquired aquatic characters
though distantly related.
53Which one of the following
describes correctly the
homologous structures?
[CBSE AIPMT 2003]
(a) Organs appearing only in embryonic
stage and disappearing later in the
adult
(b) Organs with anatomical similarities,
but performing different functions
(c) Organs with anatomical
dissimilarities, but performing same
functions
(d) Organs that have no function now, but
had an important function in
ancestors
Ans. (b)
Homologous structures are similar in
origin but similar or dissimilar in
function, such as pectoral fins of fish
and forelimbs of horse are similar in
structure but different in functions.
54Which of the following pair is
homologous organ?
[CBSE AIPMT 2002]
(a) Wings of birds and locust
(b) Wings of birds (sparrow) and pectoral
fins of fish
(c) Wings of bat and butterfly
(d) Legs of frog and cockroach
Ans. (b)
Homologous organs are the organs
which have the same origin and similar
basic structure but may differ in external
appearance and function, wings of birds
and pectoral fins of fish are examples of
the same. Analogous organs are those
organs which are anatomically different
but functionally same.
55According to fossils discovered up
to present time origin and
evolution of man was started from
[CBSE AIPMT 2002]
(a) France (b) Java
(c) Africa (d) China
Ans. (c)
The first Hominid (ancestor from whom
humans evolved) arose at a time when a
change in weather led to the reduction in
the size of the African forests favouring
bipedalism.
56Sequence of which of the following
is used to know the phylogeny?
[CBSE AIPMT 2002]
(a)mRNA (b) rRNA
(c)tRNA (d) DNA
286 NEETChapterwise Topicwise Biology

297. Ans. (b)
The genes forrRNAs tend to be highly
conserved and are, therefore, often
employed for phylogenetic studies.
57Reason of diversity in living being is
(a) mutation [CBSE AIPMT 2001]
(b) gradual change
(c) long term evolutionary change
(d) short term evolutionary change
Ans.(c)
Though mutation provides the source of
variation, the diversity in living beings is
due to the natural selection of variations
and consequent evolutionary change
over a long period of time.
58Similarities in organisms with
different genotype indicates
[CBSE AIPMT 2001]
(a) micro-evolution
(b) macro-evolution
(c) convergent evolution
(d) divergent evolution
Ans. (c)
Increase in resemblance over time of
different evolutionary lineages (in one or
more phenotypic characters) thereby
increasing their phenetic similarities is
called convergence or convergent
evolution.
59Half-life period ofC
14
is about
[CBSE AIPMT 2001]
(a) 500 yr (b) 5730 yr
(c) 50 yr (d) 5 10
4
×yr
Ans. (b)
C
14
takes about 5730 year for half the
material to decay.
60Darwin’s finches provide an
excellent evidence in favour of
evolution. This evidence comes
from the field of
[CBSE AIPMT 2000]
(a) Biogeography (b) Anatomy
(c) Embryology (d) Palaeontology
Ans. (a)
Biogeography is the study of the
geographical distribution of life forms on
earth. Darwin undertook a voyage on the
ship HMS Beagle. The ship traversed the
Southern hemisphere where life is most
abundant and varied.
Along the way, Darwin found different
forms of life very different from those in
England. As he sailed southward along
the South America, he found that similar
species replaced each other. He thought
that related species could have been
modified according the environment.
His views got confirmed on Galapagos
islands (small group of Volcanic islands
of the Western coasts of South
America). Darwin found different
modified forms of finches which seemed
to have descended from mainland
finches as a result of the natural
selection.
61Which one of the following pair has
homologous organs?
[CBSE AIPMT 1999]
(a) Pectoral fins of a fish and forelimbs of
a horse
(b) Wings of a bat and wings of
cockroach
(c) Air sac of fish and lungs of frog
(d) Wings of a bird and wings of a
butterfly
Ans. (a)
Homologous organs are those organs
which are originally and anatomically
similar but functionally different.
The forelimbs of vertebrates are built on
same pentadactyl plan, though they may
have different functions, e.g. in birds
these are modified for flying.
62The age of the fossil of
Dryopithecuson the geological time
scale is [CBSE AIPMT 1998]
(a)5 10
6
×yr back
(b)25 10
6
×yr back
(c)50 10
6
×yr back
(d)75 10
6
×yr back
Ans. (b)
Dryopithecuslived about 20-25 million
years ago.Dryopithecushad the
combined characters of great apes, old
world monkeys and man. The main
structural characteristics of
Dryopithecusare broad jaws, large
canines, semi-erect walking, 5 cusped
molars and absence of brow ridges.
63Which one of the following is a
living fossil?[CBSE AIPMT 1997]
(a)Pinus longifolia(b)Dalbergia sissoo
(c)Mirabilis jalapa(d)Ginkgo biloba
Ans. (d)
Ginkgo bilobais believed to be the oldest
living seed plant. Its fossils have been
found in rocks as old as Triassic. It still
survives with little change over this long
period of time while other members of
its group have become extinct.
64The homologous organs are those
that show similarity in
[CBSE AIPMT 1995]
(a) size
(b) origin
(c) function
(d) appearance
Ans. (b)
Homologous structures are those which
have the same embryonic origin and basic
structure, though they may or may not
perform the same function.
65Homologous organs are
[CBSE AIPMT 1994]
(a) wings of insects and bat
(b) gills of fish and lungs of rabbit
(c) pectoral fins of fish and fore limbs of
horse
(d) wings of grasshopper and crow
Ans. (c)
Homologous structures are similar in
origin but similar or dissimilar in
function, as pectoral fins of fish and
forelimbs of horse are similar in
structure but different in function.
66Evolutionary convergence is
development of
[CBSE AIPMT 1993, 96]
(a) common set of characters in group of
different ancestry
(b) dissimilar characters in closely
related groups
(c) common set of characters in closely
related groups
(d) random mating
Ans. (a)
Convergent evolution or adaptive
convergence refers to the development
of similar adaptive functional structures
in unrelated groups of organisms, e.g.
wings of insect, bird and bat; Australian
marsupials and placental mammals.
67Study of fossils is
[CBSE AIPMT 1991]
(a) palaeontology
(b) herpetology
(c) saurology
(d) organic evolution
Ans. (a)
Palaeontology is the study of fossils
(remains or impressions of ancient
forms) and their distribution in rocks of
various ages. Study of animal fossils is
known as plaeozoology and study of
plant fossils is known as palaeobotany.
Evolution 287

298. 68Humming birds and hawk illustrate
[CBSE AIPMT 1988]
(a) convergent evolution
(b) homology
(c) adaptive radiation
(d) parallel evolution
Ans. (c)
Adaptive radiation or divergent evolution
refers to the formation of different
structures from a common ancestral
form, e.g. wings of humming birds and
hawk, fore limbs of horse, bat and
human beings.
69Genetic drift operates in
[NEET 2016, Phase II]
(a) small isolated population
(b) large isolated population
(c) non-reproductive population
(d) slow reproductive population
Ans. (a)
The genetic drift is a drastic change in
allele frequency when population size is
very small. Its effects are more marked
in a small isolated population.
70Industrial melanism is an example
of [CBSE AIPMT 2015]
(a) Neo Darwinism (b) Natural selection
(c) Mutation (d) Neo Lamarckism
Ans. (b)
Within a period of years in industrial or
polluted areas, the dark species has
almost replaced the light species. This is
called industrial melanism because of its
association with the sooty atmosphere
of industry. It is an example of natural
selection.
71Variation in gene frequencies within
populations can occur by chance
rather than by natural selection.
This is referred to as[NEET 2013]
(a) genetic flow
(b) genetic drift
(c) random mating
(d) genetic load
Ans. (b)
Genetic drift is a random change in allele
frequencies over the generations. It is
brought by chance alone. Gene flow or
gene migration is the physical
movement of alleles into and out of a
population.
72In the case of peppered moth
(Biston betularia), the
black-coloured form became
dominant over the light-coloured
form in England during industrial
revolution. This is an example of
[CBSE AIPMT 2009]
(a) natural selection whereby the darker
forms were selected
(b) appearance of the darker coloured
individuals due to very poor sunlight
(c) protective mimicry
(d) inheritance of darker colour
character acquired due to the darker
environment
Ans. (a)
The given case is an example of natural
selection. As a result of struggle for
existence only those organisms could
survive, which have favourable
variations to adapt to the environmental
conditions. With so many variations in
population of species the struggle for
existence results in survival of the
fittest.
The survival of the fittest is the result of
selection and proliferation of only those
organisms, which were most suitably
adapted to the environment and most
successful in mating, i.e. natural
selection.
73Which one of following scientists
name is correctly matched with the
theory put forth by him?
[CBSE AIPMT 2008]
(a) Weismann–Theory of continuity of
germplasm
(b) Pasteur–Inheritance of acquired
characters
(c) De vries–Natural selection
(d) Mendel–Theory of pangenesis
Ans. (a)
The correct theory is the Weismam
Theory of continuity of germplasm. Rest
of the theories were proposed by
Scientist Theory
Pasteur–Germ theory of disease
Charles Darwin–Theory of natural
selection
Lamarck–Inheritance of accquired
characters
Hugo de Vries–Mutation theory
Mendel–Laws of inheritance
Darwin–Theory of pangenesis
74Select the correct statement from
the following given options.
[CBSE AIPMT 2007]
(a) Darwinian variations are small and
directionless
(b) Fitness is the end result of the ability
to adapt and gets selected by nature
(c) All mammals except whales and
camels have seven cervical vertebrae
(d) Mutations are random and directional
Ans. (b)
Fitness (survival of the fittest) is a result
of selection and proliferation of only
those organisms which were most
suitably adapted to the environment and
get selected by nature.
75De Vries gave his mutation theory
on organic evolution while working
on [CBSE AIPMT 2005]
(a)Pisum sativum
(b)Drosophila melanogaster
(c)Oenothera lamarckiana
(d)Althea rosea
Ans. (c)
Hugo de Vries(1848-1935) proposed
mutation theory for the formation of new
species. He came to conclude this
theory while working onOenothera
lamarckianaevening primrose.
According to him, new species are not
formed by continuous variations but by
sudden appearance of variations which
he assigned as mutations. He stated that
these mutations are heritable and
persist in successive generations.
76Which one of the following
phenomena supports Darwin’s
concept of natural selection in
organic evolution?
[CBSE AIPMT 2005]
(a) Development of transgenic animals
(b) Production of ‘Dolly’, the sheep by
cloning
(c) Prevalence of pesticide resistant
insects
(d) Development of organs from ‘stem
cells’ for organ transplantation
Ans. (c)
According toDarwin’s concept of
natural selectionthe organisms which
are provided with favourable variations
would survive because, they are the
fittest to face their surrounding, while
the organisms which are unfit for
surrounding variations are destroyed.
Prevalence of pesticide resistant insects
is due to the adaptability of these
insects for the changes in environment
(due to use of pesticides).
288 NEETChapterwise Topicwise Biology
Theories of Evolution
TOPIC 3

299. 77Darwin in his ‘Natural Selection
Theory’ did not believe in any role
of which one of the following in
organic evolution?
[CBSE AIPMT 2003]
(a) Discontinuous variations
(b) Parasites and predators as natural
enemies
(c) Survival of the fittest
(d) Struggle for existence
Ans. (a)
Natural theory of Darwin did not believe
in any role of discontinuous variations.
Darwin called the variations as ‘sports’.
While, Hugo de Vries used the term
mutation to these variations. These
variations are sudden heritable changes
which can occur in any stage of
development.
78In a random mating population in
equilibrium, which of the following
brings about a change in gene
frequency in a non-directional
manner? [CBSE AIPMT 2003]
(a) Migration (b) Mutation
(c) Random drift (d) Selection
Ans.(b)
Given certain conditions, the allele
frequencies remain constant from
generation to generation. Under these
conditions, a population would be in
equilibrium and there will be no
evolutionary change. However, many
evolutionary changes usually occur
following the appearance of new alleles
and source of this is mutation.
79Random genetic drift in a
population probably results from
[CBSE AIPMT 2003]
(a) large population size
(b) highly genetically variable individuals
(c) interbreeding within this population
(d) constant low mutation rate
Ans. (b)
Genetic drift or Sewall Wright effect is
statically significant change in
population gene frequencies resulting by
chance and not from natural selection,
emigration or immigration. In simple
words, random loss of alleles is known as
genetic drift.
80Which one of the following
sequences was proposed by
Darwin and Wallace for organic
evolution? [CBSE AIPMT 2003]
(a) Variations, natural selection,
overproduction, constancy of
population size
(b) Overproduction, variations,
constancy of population size, natural
selection
(c) Variations, constancy of population
size, overproduction, natural
selection
(d) Overproduction, constancy of
population size, variations, natural
selection
Ans. (b)
Though living organisms tend to multiply
geometrically, the number of individuals
of a species tend to remain constant
over a long period of time. Out of
heterogenous population (due to the
variation) best adapted individuals are
selected by nature.
81Genetic drift operates in
[CBSE AIPMT 2002]
(a) small isolated population
(b) large isolated population
(c) fast reproductive population
(d) slow reproductive population
Ans. (a)
Genetic drift refers to changes in allele
frequencies of a gene pool due to
chance. Though it operates both in large
and small populations, it is expected to
be significant only in small populations,
where alleles may become extinct or get
fixed by chance alone.
82Reason of diversity in living being
is [CBSE AIPMT 2001]
(a) mutation
(b) gradual change
(c) long term evolutionary change
(d) short term evolutionary change
Ans. (c)
Though mutation provides the source of
variation, the diversity in living beings is
due to the natural selection of variations
and consequent evolutionary change
over a long periods of time.
83Darwin’s theory of pangenesis
shows similarity with theory of
inheritance of acquired characters
then what will be correct according
to it? [CBSE AIPMT 2001]
(a) Useful organs become strong and
developed while useless organs
become extinct. These organs help in
struggle for survival
(b) Size of organs increase with ageing
(c) Development of organs is due to will
power
(d) There should be some physical basis
of inheritance
Ans. (d)
According to both the views, something
is passed from parent to offspring which
causes development of specific
characters, i.e. all that has been
acquired by the organism during its life
time is preserved by generation and
transmitted to offsprings in form of
pangenes is or gemmules.
84The presence of gill slits, in the
embryos of all vertebrates,
supports the theory of
[CBSE AIPMT 1995]
(a) biogenesis
(b) recapitulation
(c) metamorphosis
(d) organic evolution
Ans. (a)
‘The theory of recapitulation’ or
‘Biogenetic law’, propounded byHaeckel
states that an individual organism in its
development (ontogeny) tends to repeat
the stages passed through by its
ancestors (phylogeny). During the life
history of frog, tadpole larva resembles
fishes, the ancestors of Amphibia. The
presence of gill clefts in vertebrate
embryo provides a strong evidence in
support of organic evolution.
85Which one does not favour
Lamarckian concept of inheritance
of acquired characters?
[CBSE AIPMT 1994]
(a) Lack of pigment in cave dwellers
(b) Absence of limbs in snakes
(c) Presence of webbed toes in aquatic
birds
(d) Melanization of peppered moth in
industrial areas
Ans. (d)
Lamarck believed in direct influence of
environment on the individual.
Lamarckian postulate of use and disuse
of organs is supported by rudimentary
eyes of cave dwellers, webbed feet of
swimming bird, elongated limbless body
of snake, vestigial organs of living
animals, etc.Biston betularia(peppered
moth) shows industrial melanism and
demonstrates natural selection.
Evolution 289

300. 86Frequency of a character increases
when it is [CBSE AIPMT 1994]
(a) recessive (b) dominant
(c) inheritable (d) adaptable
Ans. (d)
According to Darwin, the variations are
continuous and those which are helpful
in the adaptations of an organism
towards its surrounding would be passed
on to the next generation. In the
offsprings, these modifications become
more pronounced if they are exposed to
similar stress of the environment as
faced by their parents.
87Theory of natural selection dwells
on [CBSE AIPMT 1993]
(a) role of environment in evolution
(b) natural selection acting on favourable
variations
(c) changes in gene complex resulting in
heritable variations
(d) None of the above
Ans. (b)
According to Darwin’s theory of natural
selection, in the struggle for existence,
only those individuals survive which
possess the most useful variations.
Useful variations present in the surviving
individuals are passed on to the next
generation. Next generation repeats the
process of development of variations
and natural selection.
88Weismann cut off tails of mice
generation after generation but
tails neither disappeared nor
shortened showing that
[CBSE AIPMT 1993]
(a) Darwin was correct
(b) tail is an essential organ
(c) mutation theory is wrong
(d) Lamarckism was wrong in inheritance
of acquired characters
Ans. (d)
Lamarckian theory of inheritance of
acquired characters was discarded by
Weismann, while experimenting on mice.
He cut the tails of mice for 80 generations
but no single mouse was born without tail.
Weismann formulated his famous
‘Theory of continuity of germplasm’,
according to which only those
characters, whatever be their origin and
nature, that could affect germplasm are
heritable and others are not.
89Basic principles of embryonic
development were pronounced by
[CBSE AIPMT 1990]
(a) Von Baer (b) Weismann
(c) Haeckel (d) Morgan
Ans.(a)
The basic laws or principles of
embryonic development were given by
von Baer in 1828. According to his theory
during embryonic development
generalised features brain, spinal cord
etc. appeared earlier than special
features like hair. Haeckel (1866)
propounded ‘The theory of recapitulation
or Biogenetic law’.
90‘Continuity of germplasm’ theory
was given by[CBSE AIPMT 1989]
(a) De Vries (b) Weismann
(c) Darwin (d) Lamarck
Ans. (b)
August Weismann, a German scientist
formulated his famous ‘Theory of
continuity of germplasm’ in 1886, on
experimenting with mice. His theory
states that inheritance in multicellular
organisms takes place by means of germ
cells i.e. egg and sperm cells only.
91Theory of inheritance of acquired
characters was given by
[CBSE AIPMT 1989]
(a) Wallace (b) Lamarck
(c) Darwin (d) De Vries
Ans. (b)
Jean Baptiste de Lamarck (1744–1829)
was the greatest of French naturalists,
who for the first time suggested a
complete theory of evolution.
Lamarckian theory is also known as
‘Theory of inheritance of acquired
characters’ or ‘Theory of use and disuse
of organs’.
92The factor that leads to Founder
effect in a population is[NEET 2021]
(a) natural selection
(b) genetic recombination
(c) mutation
(d) genetic drift
Ans. (d)
The factor that leads to Founder effect
in a population isgenetic drift. It is the
change in the frequency of an existing
gene variant (allele) in a population due to
random sampling of organisms. Genetic
drift can cause big losses of genetic
variation for small population. The
Founder effect is when a few individuals
in a population colonize a new location
that is separated from the old population.
This also greatly reduces the population
size, as well as reduces the genetic
variability of the population.
Other options can be explained as
Natural selectionis the process through
which populations of living organisms
adapt and change.
Genetic recombination, in genetics,
primary mechanism through which
variation is introduced into populations.
Mutationis sudden heritable change.
93A population of a species invades a
new area. Which of the following
conditions will lead to adaptive
radiation?[NEET (Odisha) 2019]
(a) Area with large number of habitats
having very low food supply
(b) Area with a single type of vacant
habitat
(c) Area with many types of vacant
habitats
(d) Area with many habitats occupied by
a large number of species
Ans. (c)
Option (c) is correct.
The adaptive radiation occurs when a
population of a species invades a new
area. It is because it provides organisms
of a population a new habitat with plenty
of niche spaces. Darwin’s finches
represent one of the best examples of
this phenomenon.
94Which of the following represents
order of ‘Horse’? [NEET 2017]
(a) Equidae
(b) Perissodactyla
(c) Caballus
(d) Ferus
Ans. (b)
Order being the higher taxon of
classification exhibit the few similar
characters of families. The order of
horse is perissodactyla.
290 NEETChapterwise Topicwise Biology
Speciation and Isolation
TOPIC 4

301. Concept EnhancerThe order
perissodactyla includes odd toed
mammals. For example,Equus asinus
(donkey),Rhinoceros indicus(the great
one horned rhinoceros).
95In Hardy-Weinberg equation, the
frequency of heterozygous
individual is represented by
[NEET 2016, Phase II]
(a)p
2
(b) 2pq
(c)pq (d)q
2
Ans. (b)
Hardy-Weinberg principlestates that
the allele frequencies in a population
are stable and remain constant from
generation to generation. This can be
expressed by the equation :
p pq q
2 2
2 1+ + =
or ( )p q+ =
2
1
where,p
2
represents frequency
of homozygous dominant genotype,2pq
represents the frequency of
heterozygous genotype andq
2
represents the frequency of
homozygous recessive genotype.
Hence, option (b) is correct.
96The tendency of population to
remain in genetic equilibrium may be
disturbed by [NEET 2013]
(a) random mating
(b) lack of migration
(c) lack of mutations
(d) lack of random mating
Ans. (d)
According to Hardy-Weinberg principle,
allele frequencies in a population are
stable and is constant from generation
to generation only if the following
conditions are met
(i) There is no mutation, no gene flow
and all mating is random.
(ii) All genotypes reproduce equally
well (i.e., no natural selection). But
all there conditions are rarely met in
nature.
97According to Darwin, the organic
evolution is due to[NEET 2013]
(a) intraspecific competition
(b) interspecific competition
(c) competition within closely related
species
(d) reduced feeding efficiency in one
species due to the presence of
interfering species
Ans. (b)
Darwin stated that the organic evolution
is due to the interspecific competition. It
is the competition between members of
different species. Intraspecific
competition occurs amongst members
of the same species for obtaining
optimum amounts of their food, shelter,
mate, water, light, etc.
Closely related species if compete
cannot cause evolution. Reduced
feeding efficiency in one species due to
the presence of interfering species is
due to the struggle for existence.
98Adaptive radiation refers to
[CBSE AIPMT 2007]
(a) adaptations due to geographical
isolation
(b) evolution of different species from a
common ancestor
(c) migration of members of a species to
different geographical areas
(d) power of adaptation in an individual to
a variety of environments
Ans. (b)
Adaptive radiation is the development of
different functional structures from a
common ancestral form.
99Industrial melanism as observed in
peppered moth proves that
[CBSE AIPMT 2007]
(a) the true black melanic forms arise by
a recurring random mutation
(b) the melanic form of the moth has no
selective advantage over lighter form
in industrial area
(c) the lighter form moth has no
selective advantage either in polluted
industrial area or non-polluted area
(d) melanism is a pollution generated
feature
Ans. (a)
Industrial melanism is a term used to
describe the evolutionary process in
which darker individuals come to
predominate over lighter individuals
since the industrial revolution as a result
of natural selection. In 1848 a black form
of the moth was recorded in Manchester
and by 1895, 98% of the peppered moth
population in Manchester was black. This
black ‘melanic’ form arose by a recurring
random mutation, but its phenotypic
appearance had a strong selective
advantage in industrial areas.
100One of the important
consequences of geographical
isolation is[CBSE AIPMT 2007]
(a) no change in the isolated fauna
(b) preventing speciation
(c) speciation through reproductive
isolation
(d) random creation of new species
Ans. (c)
Speciation is the formation of new
species and the development of species
diversity occurs when gene flow within
the common pool is interrupted by an
isolating mechanism. The isolation can
occur through geographical separation
of population, known asallopatric
speciation.
101Which one of the following is not a
living fossil?[CBSE AIPMT 2006]
(a) King crab (b)Sphenodon
(c)Archaeopteryx(d)Peripatus
Ans. (c)
Archaeopteryxlithographica is not a
living fossil. It is a fossil bird that lived in
Jurassis period about 180 million years
ago. Its fossil displays the characters of
both reptiles (e.g. long tail, bones non
pneumatic, claw, presence of weak
sternum, free caudal vertebra, etc) and
birds (e.g. presence of feathers,
modified jaws, etc).
102Jurassic period of the Mesozoic era
is characterised by
[CBSE AIPMT 2006]
(a) radiation of reptiles and origin of
mammal-like reptiles
(b) dinosaurs become extinct and
angiosperms appear
(c) flowering plants and first dinosaurs
appear
(d) gymnosperms are dominant plants
and first birds appear
Ans. (d)
Jurassic period is the second geological
period of Mesozoic era. In this period,
thegymnospermswere dominant and
the plants included ferns, cycads,
Ginkgo, bushes and conifers. Among
animals, important invertebrates
included ammonites, corals,
brachiopods, bivalves and echinoids.
Reptiles dominated the vertebrates and
thefirst flying reptilesthe pterosaurs
appeared.
Thefirst primitive bird,Archaeopteryx,
also made its appearance.
Evolution 291

302. 103Industrial melanism is an example
of [CBSE AIPMT 2003]
(a) defensive adaptation of skin against
ultraviolet radiations
(b) drug resistance
(c) darkening of skin due to smoke from
industries
(d) protective resemblance with the
surroundings
Ans. (c)
Industrial melanism is an example of
directional selection. Changing
environment leading to changes in the
phenotypic/genotypic constitution of a
population.
104In which condition the gene ratio
remains constant for any species?
[CBSE AIPMT 2002]
(a) Sexual selection
(b) Random mating
(c) Mutation
(d) Gene flow
Ans. (b)
According to Hardy–Weinberg theorem,
the mixing of alleles at meiosis and their
subsequent recombination will not alter
the frequencies of alleles in the future
generations provided the mating within
the population is random.
105In which era reptiles were
dominant? [CBSE AIPMT 2002]
(a) Coenozoic era
(b) Mesozoic era
(c) Palaeozoic era
(d) Archaeozoic era
Ans. (b)
Mesozoic era began about 24.8 crore
year ago and lasted for about 18.3 crore
year. It is also known as the ‘Age of
reptiles’.
Main events in of this era are as follows:
(i) Gymnosperms dominate
landscape, first dinosaurs and
mammals.
(ii) Gymnosperms continue as
dominant plants, dinosaures
dominant, first birds.
(iii) Flowering plants (angiosperms)
appear, dinosaurs and many groups
of organisms become extinct at
end of period.
106Occurrence of endemic species in
South-America and Australia is due
to [CBSE AIPMT 2001]
(a) these species have been extinct from
other regions
(b) continental separation
(c) there is no terrestrial route to these
places
(d) retrogressive evolution
Ans. (b)
Seas separating the continents form
barriers to free intercontinental
movement causing evolution of
organisms independently in these
continents leading to endemism.
Endemic species are those species
which are found in a restricted area of
the world.
107Genetic drift operates only in
[CBSE AIPMT 1998]
(a) smaller populations
(b) larger populations
(c) Mendelian populations
(d) island populations
Ans. (a)
Genetic drift may be of significance in
small populations only, where alleles
may easily get extinct by chance alone.
108In general, in the developmental
history of a mammalian heart, it is
observed that it passes through a
two-chambered fish-like heart,
three-chambered frog-like heart
and finally to four-chambered
stage. To which hypothesis can this
above cited statement be
approximated?[CBSE AIPMT 1998]
(a) Hardy-Weinberg law
(b) Lamarck’s principle
(c) Biogenetic law
(d) Mendelian
Ans. (c)
Biogenetic law was propounded byErnst
Haeckelin 1866. According to this,
during its development an animal passes
through ancestral adult stages.
109‘Golden age of dinosaurs’/Age of
reptiles was[CBSE AIPMT 1994]
(a) Mesozoic
(b) Coenozoic
(c) Palaeozoic
(d) Psychozoic
Ans. (a)
Mesozoic era is the age of reptiles.
During this era, the dinosaurs achieved
an enormous size and were most
abundant, Mesozoic era is also known as
golden age of dinosaurs.
110The earliest fossil form in the
phylogeny of horse is
[CBSE AIPMT 1994]
(a)Merychippus
(b)Mesohippus
(c)Eohippus
(d)Equusprinciples
Ans. (c)
Hoofed animals like-horse originated in
Ecoene epoch in North America. First
horse-like animals from which the
modern horseEquusevolved was
Hyracotherium(old) nameEohippus). The
fossil record is most complete in horse.
111Two geographical regions
separated by high mountains are
[CBSE AIPMT 1994]
(a) Oriental and Australian
(b) Palaearctic and Oriental
(c) Nearctic and Palaearctic
(d) Neotropical and Ethiopian
Ans. (b)
Dr. PL Scalater (1858) proposed first time
the division of the world into six realms
or biogeographic regions according to
the distribution of birds. In 1876 AR
Wallace, adopted it for all the animals,
these regions include i.e. Palaearctic,
Ethiopian, Oriental, Australian, Nearctic
and Neotropical. Palaearctic and
Oriental realms are separated by high
mountain ranges, i.e. Himalayas.
112Genetic drift is change of
[CBSE AIPMT 1993]
(a) gene frequency in same generation
(b) appearance of recessive genes
(c) gene frequency from one generation
to next
(d) None of the above
Ans.(c)
Genetic drift is defined as any random
change, either directed or undirected in
gene frequency in a population.
113Correct order is[CBSE AIPMT 1991]
(a) Palaeozoic→Archaeozoic→
Coenozoic
(b) Archaeozoic→Palaeozoic→
Proterozoic
(c) Palaeozoic→Mesozoic→Coenozoic
(d) Mesozoic→Archaeozoic→
Proterozoic
Ans. (c)
Era is a division of geological time that
includes one or more periods. It follows
the order–Palaeozoic, Mesozoic and
292 NEETChapterwise Topicwise Biology

303. Coenozoic. Palaeozoic and Mesozoic
eras are sub-divided into periods but
Coenozoic is divided into periods and
epochs. Mesozoic era is the Age of
reptiles, and Coenozoic era is called as
‘Age of mammals and birds’.
114Parallelism is[CBSE AIPMT 1990]
(a) adaptive divergence
(b) adaptive divergence of widely
separated species
(c) adaptive convergence of widely
different species
(d) adaptive convergence of closely
related groups
Ans. (d)
When convergent evolution is found in
closely related species, it is called
‘Parallel evolution’ (or parallelism), e.g.
development of running habit in deer [2
toed) and horse (1 toed) with two vestigial
splint bones.
115A hominid fossil discovered in Java
in 1891, now extinct having cranial
capacity of about 900 cc was
[NEET (Oct.) 2020]
(a)Homo erectus
(b) Neanderthal man
(c)Homo sapiens
(d)Australopithecus
Ans. (a)
A Hominid fossil discovered in Java in
1891 now extinct having cranial capacity
of about
900 cc wasHomo erectus. It probably ate
meat.
116Which of the following statements
is correct about the origin and
evolution of men?
[NEET (Odisha) 2019]
(a) Agriculture came around 50,000
years back
(b) TheDryopithecusandRamapithecus
primates existing 15 million years ago,
walked like men
(c)Homo habilisprobably ate meat
(d) Neanderthal men lived in Asia
between 1,00,000 and 40,000 years
back
Ans. (d)
Statement (d) is correct about the origin
and evolution of men.
Neanderthal men lived in Asia between
1,00,000 and 40,000 years back. Other
statements are corrected as follows
Agriculture came around 10,000 years
back
About 15 mya , primates called
DryopithecusandRamapithecuswere
existing. They were hairy and walked like
gorillas and chimpanzees.
Homo habilisprobably did not eat meat
whileHomo erectusprobably ate meat.
117Match the hominids with their
correct brain size
[NEET (National) 2019]
A.Homo habilis (i) 900 cc
B.Homo neanderthalensis
(ii) 1350 cc
C.Homo erectus (iii) 650-800 cc
D.Homo sapiens (iv) 1400 cc
Select the correct option.
A B C D
(a) (iii) (ii) (i) (iv)
(b) (iii) (iv) (i) (ii)
(c) (iv) (iii) (i) (ii)
(d) (iii) (i) (iv) (ii)
Ans.(b)
(A)–(iii) (B)–(iv), (C)–(i) (D)–(ii)
The brain capacity ofHomo habilisor the
tool maker was 650-800 cc.Homo
neanderthalensis,who existed in the late
Pleistocene period had the brain size of
about 1400 cc.Homo erectuswere the
first to walk upright and stand erect.
Their brain size was about 900 cc.Homo
sapiensor the living modern man has the
brain size of about 1350 cc, which is
lesser than their immediate ancestors,
Cro-Magnon man.
118The chronological order of human
evolution from early to the recent
is
(a)Australopithecus→Ramapithecus→
Homo habilis→Homo erectus
(b)Ramapithecus→Australopithecus→
Homo habilis→Homo erectus
(c)Ramapithecus→Homo habilis→
Australopithecus→Homo erectus
(d)Australopithecus→Homo habilis→
Ramapithecus→Homo erectus
Ans. (b)
The fossils ofRamapithecuswas
discovered by Edward Levis from
Pliocene rocks of Shivalik hills in India.
They were present in Pliocene era nearly
14-15 million years ago (mya).
Australopithecus was first ape man
found in AfricanPlioceneera rocks
present5 mya.
Homo habilis were believed to be
present in East Africa 2 mya in Pliocene
era.
Homo erectus appeared nearly 1.5 mya
with cranial capacity of 800-1300 cc.
Thus, the correct sequence is
Ramapithecus→Australopithecus→
Homo habilis→Homo erectus
119What was the most significant
trend in the evolution of modern
man (Homo sapiens) from his
ancestors?[CBSE AIPMT 2012, 11]
(a) Shortening of jaws
(b) Binocular vision
(c) Increasing brain capacity
(d) Upright posture
Ans. (c)
The most significant trend in the
evolution of modern man is the
increased brain capacity from the
ancestors. The first human being was
the hominid, calledHomo habilis. The
brain capacities were between 650-800
cc. They probably did not eat meat.
Fossils discovered in Java in 1891
revealed the next stage, i.e.Homo
erectus.Homo erectushad a large brain
and probably ate meat.
The Neanderthal man with a brain size of
1400 cc lived in near east and central
Asia between 1,00,00-40,000 year back.
They used hides to protect their body
and burried their dead.Homo sapiens
arose in Africa and moved across
continents and developed into distinct
races. During ice age between
75,000-10,000 years back modernHomo
sapiensarose.
120The extinct human who lived
100000 to 40000 years ago, in
Europe, Asia and parts of Africa,
with short stature, heavy eye
brows, retreating fore heads, large
jaws with heavy teeth, stocky
bodies, a lumbering gait and
stooped posture was
[CBSE AIPMT 2012]
(a)Homo habilis
(b) Neanderthal human
(c) Cro-magnon humans
(d)Ramapithecus
Evolution 293
Human Evolution
TOPIC 5

304. Ans. (b)
Neanderthal man with a brain size of
1400cc lived in near East and Central
Asia, Europe and North Africa between
100000 to 40000 years back. It had
slightly prognathus face, sloping
forehead, eye brow ridges, smaller or no
chin, large receding jaws, thick-boned
skull and high domed head. They used
hides to protect their body and buried
their dead.
121Among the human ancestors the
brain size was more than 1000 cc in
[CBSE AIPMT 2007]
(a)Homo neanderthalensis
(b)Homo erectus
(c)Ramapithecus
(d)Homo habilis
Ans. (a)
The cranial capacity ofHomo
neanderthalensiswas about 1450 cc,
roughly to that of modern man.
122There are two opposing views
about origin of modern man.
According to one viewHomo
erectusin Asia were the ancestors
of modern man. A study of
variation of DNA however,
suggested African origin of modern
man. What kind of observation on
DNA variation could suggest this?
[CBSE AIPMT 2005]
(a) Greater variation in Asia than in Africa
(b) Greater variation in Africa than in Asia
(c) Similar variation in Africa and Asia
(d) Variation only in Asia and no variation
in Africa
Ans. (a)
There are two models about origin of
modern man (Homo sapiens sapiens), i.e.
(a)Multiregional modelAccording to
this view, modern humans evolved
in many parts of the world from
regional descendants of
Homo erectuswho dispersed from
Africa between 1 and 2 million years
ago.
(b)Monogenesis modelAccording to
this view only African descendants
ofHomo erectusgave rise to
modern humans.
In late 1980s Rebecca Cann and other
geneticists supported this view on the
basis of DNA of living humans. They
compared the mitochondrial DNA
(mtDNA) of a multiathenic sample of
more than 100 people representing four
continents. The greater the difference
between themtDNAs of two peoples, the
longer ago thatmtDNAs diverged from a
common source.
By using bioinformatics, they concluded
that the divergence ofmtDNA of
Africans from common source began
just 200000 year ago, much too late to
represent the dispersal ofHomo erectus.
Thus, there are greater variation in Asia
than in Africa.
123In recent years, DNA sequences
(nucleotide sequence) ofmtDNA
and Y-chromosomes were
considered for the study of human
evolution, because
[CBSE AIPMT 2003]
(a) they can be studied from the samples
of fossil remains
(b) they are small and, therefore, easy to
study
(c) they are uniparental in origin and do
not take part in recombination
(d) their structure is known in greater
detail
Ans. (c)
WilsonandSarichchoose mitochondrial
DNA (mtDNA) for the study of maternal
line inheritance. While, Y-chromosomes
were considered for the study of human
evolution particularly male domain. It is
possible because they are uniparental in
origin and do not take part in
recombination.
124Which of the following is correct
order of evolutionary history of
man? [CBSE AIPMT 2001]
(a) Peking man,Homo sapiens,
Neanderthal, Cro-magnon
(b) Peking man, Neanderthal,Homo
sapiens,Cro-magnon
(c) Peking man, Heidalberg man,
Neanderthal, Cro-magnon
(d) Peking man, Neanderthal,Homo
sapiens,Heidalberg man
Ans. (c)
Peking man→Heidalberg man→
Neanderthal man→Cro-magnon man is
the correct order/sequence of evolution
man.
125Which of following is closest
relative of man?[CBSE AIPMT 2001]
(a) Chimpanzee
(b) Gorilla
(c) Orangutan
(d) Gibbon
Ans. (a)
Banding patterns of human
chromosome number 3 and 6 are
remarkably similar to that of chimpanzee
indicating common origin for both.
126Homo sapiensevolved during
[CBSE AIPMT 2000]
(a) Pleistocene (b) Oligocene
(c) Pliocene (d) Miocene
Ans. (a)
The family–Hominidae includes humans
of today. These are the most intelligent
of the hominoids. They are distinguished
from the other families of hominoids in
that they arebipedal, i.e. they walk
upright on two legs. Hominids appeared
during Pliocene but modern human
(Homo sapiens) are believed to have
arisen about 1,50,000 year back during
Pleistocene.
127Which one of the following features
is closely related with the evolution
of humans? [CBSE AIPMT 2000]
(a) Loss of tail
(b) Shortening of jaws
(c) Binocular vision
(d) Flat nails
Ans. (b)
Evolution of modern man is
accompanied by the reduction in length
of jaw bones and teeth lines becoming
semi-circular instead of U-shaped.
128Which of the following primate is
the closest relative of humans?
[CBSE AIPMT 2000]
(a) Rhesus monkey (b) Orangutan
(c) Gorilla (d) Gibbon
Ans. (c)
Chimpanzees and gorillas are our closest
relatives of humans among the living
primates.
129Which is not a vestigial part in
humans? [CBSE AIPMT 2000]
(a) Segmental muscles of abdomen
(b) Finger nails
(c) Third molar
(d) Coccyx
Ans. (b)
Finger nails are not a vestigeal part in
humans. Structures or organs which are
present in an organism in a diminished
size but no longer useful are called
vestigial organ.
294 NEETChapterwise Topicwise Biology

305. Evolution 295
About 90 vestigial organs have been
reported from human body. These
include coccyx (tail bone), third molar
(wisdom tooth) and segmental muscles
of abdomen.
130Which one of the following
statements is correct?
[CBSE AIPMT 1998]
(a) Cro-magnon man’s fossil has been
found in Ethiopia
(b)Homo erectusis the ancestor of man
(c) Neanderthal man is the direct
ancestor ofHomo sapiens
(d) Australopithecus is the real ancestor
of modern man
Ans. (b)
Homo erectus(0.25–1.7 million years,
middle Pleistocene man) is known as
erect man, who walked erect over legs.
Homo sapiens(late Pleistocene man, 5
lakh year) is parent species to which
modern man belongs. It evolved from
Homo erectusin Africa.Homo sapiens
sapiensis the modern man, evolved
about 25000 year back but spread to
various parts of the world about
10000–11000 year ago.
131Common origin of man and
chimpanzee is best shown by
[CBSE AIPMT 1997]
(a) banding pattern in chromosomes
number 3 and 6
(b) cranial capacity
(c) binocular vision
(d) dental formula
Ans.(a)
A comparative study of the banding of
chromosomes of man and the great apes
has shown that the total amount of DNA
in human diploid cells and that of apes is
more or less similar. The banding pattern
of human chromosome numbers 3 and 6
are compared with those of particular
autosomes in the chimpanzee, which
shows a common origin. More over,
blood protein test also proves that man
is most closely related to great apes
(chimpanzee and gorilla).
132Which one of the following
statements about fossil human
species is correct ?
[CBSE AIPMT 1997]
(a) Fossils ofHomo neanderthalensis
have been found recently in South
America
(b) Neanderthal man and Cro-magnon
man did exist for sometime together
(c)Australopithecusfossils have been
found in Australia
(d)Homo erectuswas preceded byHomo
habilis
Ans. (d)
Homo erectuswas a large hominid. It had
a length of about 150 to 170 cm (5 to 5.5
feet) with a low but distinct forehead
strong browridge and a brain capacity of
about 1000 cc.Homo erectuswas social
and living in tribes of 20 to 50 people and
thus had a successful and complex
culture. It became widespread
throughout the tropical and temperate
old world.
133Which one of the following sets
includes only the vestigial
structures in man?
[CBSE AIPMT 1996]
(a) Body hair olecranon process, coccyx,
patella
(b) Wisdom teeth, mammary glands,
coccyx, patella
(c) Coccyx, nictitating membrane,
vermiform appendix, ear muscles
(d) Coccyx, body hair, ear ossicles,
vermiform appendix
Ans. (c)
Vestigial organs are non-functional,
degenerate and rudimentary organs that
correspond to fully developed and
functional organs of related organisms,
e.g. nictitating membranes, muscles of
ear pinna and third molar (wisdom tooth),
mammary gland in males, coccyx (caudal
vertebrae), vermiform appendix etc.
134Which one of the following is
regarded as the direct ancestor of
modern man? [CBSE AIPMT 1996]
(a)Homo erectus
(b)Ramapithecus
(c)Homo habilis
(d) Cro-magnon man
Ans. (d)
Homo sapiens sapiensis the modern
man evolved about 25000 years ago. The
direct ancestor of the living man was the
Cro-magnon (Homo sapiens fossilis),
discovered in 1868 from Cro-magnon
rocks of France by ‘Mac Gregor’ from
Holocene epoch. Cro-magnon was
emerged about 34000 years ago. It is
believed that Cro-magnon man was
somewhat more intelligent and cultured
than the man of today.
135Which one is irrelevant to evolution
of man? [CBSE AIPMT 1994]
(a) Perfection of hand for tool making
(b) Change of diet from hard nuts/roots to
soft food
(c) Increased ability to communicate or
develop community behaviour
(d) Loss of tail
Ans. (d)
Evolution simply means an orderly
change from one condition to another. It
is often called descent with
modification. Tail provides an evidence
of atavism or reversion, i.e.
reappearance of certain ancestral
characters which had disappeared or
were reduced. Tail is absent in man, but
occasionally early embryo of man
possesses an external tail (vestigial tail
vertebrae). In adults the tail is
represented by a string of caudal
vertebrae, which constitute the coccyx
(tail bone).

306. 01Match the List I with List II.
[NEET 2021]
List I List II
A. Filariasis 1. Haemophilus
influenzae
B. Amoebiasis 2. Trichophyton
C. Pneumonia 3. Wuchereria
bancrofti
D. Ringworm 4. Entamoeba
histolytica
Choose the correct answer from
the options given below.
A B C D
(a) 4 1 3 2
(b) 3 4 1 2
(c) 1 2 4 3
(d) 2 3 1 4
Ans.(b)
(A)-(3), (B)-(4), (C )-(1) ,(D)-(2)
Filariasisor elephantiasis is caused by
filarial worm known asWuchereria
bancrofti. It affect the lymphatic vessels
of lower limbs resulting in gross
deformities.
Amoebiasisis a protozoan disease
caused byEntamoeba histolyticawhich
parasite the large intestine causing
constipation, abdominal pain, etc.
Pneumoniais a bacterial disease caused
by,Haemophilus influenzaeand
Sterptococcus pneumoniaebacteria.
These bacteria infect the alveoli leading
to several problems in respiration.
Ringwormis a fungal disease.
Trichophyton,Microsporumare
responsible for this disease, resulting in
appearance of dry scaly lesions on
various parts as the symptoms of this
disease.
02Match the following columns and
select the correct option from the
codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. Typhoid 1. Haemophilus
influenzae
B. Malaria 2. Wuchereria
bancrofti
C. Pneumonia 3.Plasmodium vivax
D. Filariasis 4.Salmonella typhi
Codes
A B C D
(a) 4 3 1 2
(b) 3 4 2 1
(c) 1 3 2 4
(d) 1 2 4 3
Ans.(a)
Option (a) is the correct match which is
as follows
Typhoid is causedSalmonella typhi.
Malaria is caused byPlasmodium vivax.
Pneumonia is caused byHaemophilus
influenzae.
Filariasis is caused byWuchereria
bancrofti.
03Match the following diseases with
the causative organism and select
the correct option.
[NEET (Sep.) 2020]
Column I Column II
A. Typhoid 1. Wuchereria
B. Pneumonia 2. Plasmodium
C. Filariasis 3. Salmonella
D. Malaria 4. Haemophilus
A B C D
(a) 3 4 1 2
(b) 2 1 3 4
(c) 4 1 2 3
(d) 1 3 2 4
Ans.(a)
Typhoidis caused by bacterium
Salmonella typhi. Typhoid fever is a type
of enteric fever. It spreads by drinking
water contaminated with the faeces of
an infected person. Fever that starts low
and increases daily, possibly reaching as
high as 104.9 F (40.5 C), muscle aches,
sweating, loss of appetite and weight
loss, abdominal pain and diarrhoea or
constipation are the symptoms of
typhoid.
Pneumoniais caused by bacterium
Haemophilus influenzae.It is a small,
Gram-negative, facultative anaerobic
organism which causes infection in the
upper respiratory tract. The bacteria are
usually transmitted by droplets in the air
from a sneeze, cough or close
conversation with an infected person.
Filariasisis a parasitic disease caused by
infectious worms calledWuchereria.
HumanHealth
andDiseases
30
Common Human Diseases
TOPIC 1

307. These spread by blood-feeding insects
such as black flies and mosquitoes.
Malariais a life-threatening disease
caused byPlasmodiumparasite. It’s
typically transmitted through the bite of
an infectedAnophelesmosquito.
Infected mosquitoes carry the
Plasmodiumspecies. After bite, the
parasite is released into the bloodstream
where it matures and begin to infect
RBCs resulting in symptoms that occur
in cycles that last two to three days at a
time.
04The infectious stage ofPlasmodium
that enters the human body is
[NEET (Sep.) 2020]
(a) sporozoites
(b) female gametocytes
(c) male gametocytes
(d) trophozoites
Ans.(a)
The infectious stage ofPlasmodiumthat
enters the human body is sporozoites,
present in salivary gland ofAnopheles
mosquito. The sporozoites grow and
multiply in the liver to become
merozoites. These merozoites invade
the erythrocytes (RBCs) to form
trophozoites, schizonts and
gametocytes, during which the symptoms
of malaria are produced.
05Identify the correct pair representing
the causative agent of typhoid fever
and the confirmatory test for
typhoid. [NEET (National) 2019]
(a)Streptococcus pneumoniae/ Widal
test
(b)Salmonella typhi/ Anthrone test
(c)Salmonella typhi/ Widal test
(d)Plasmodium vivax/ UTI test
Ans.(c)
Typhoid fever is caused by the bacterium
Salmonella typhiand widal test is the
confirmatory test for typhoid, which is
based on antigen antibody reaction.
Typhoid fever or enteric fever has the
incubation period of 1 to 2 weeks and it is
usually transmitted through
contaminated food and water.
06In which disease does mosquito
transmitted pathogen cause
chronic inflammation of lymphatic
vessels? [NEET 2018]
(a) Ringworm disease
(b) Ascariasis
(c) Elephantiasis
(d) Amoebiasis
Ans.(c)
Elephantiasisis a helminthic disease
caused byWuchereria bancrofti.The
infestation is transmitted by female
Culexmosquitoes from one individual to
the others. The worms live in the
lymphatic system.
Ascariasisis caused byAscaris
lumbricoides.It is an endoparasite of
the small intestine of human beings.
Amoebiasisis caused byEntamoeba
histolytica. It lives in the large intestine
of humans.Ringwormis a fungal skin
disease.
07Which of the following sets of
diseases is caused by bacteria?
[NEET 2016, Phase II]
(a) Cholera and tetanus
(b) Typhoid and smallpox
(c) Tetanus and mumps
(d) Herpes and influenza
Ans.(a)
Cholera and tetanus are diseases
caused by bacteria. Cholera is caused
by a bacteriumVibrio choleraeand
tetanus is caused by a bacterium
Clostridium tetani.Mumps, influenza,
herpes and smallpox are viral diseases.
Typhoid is a bacterial disease but it is
not paired with a bacterial disease.
Hence, option (a) is correct.
08Asthma may be attributed to
[NEET 2016, Phase I]
(a) allergic reaction of the mast cells in
the lungs
(b) inflammation of the trachea
(c) accumulation of fluid in the lungs
(d) bacterial infection of the lungs
Ans.(a)
Asthma is an allergic reaction
characterised by spasm of bronchi
muscles because of effect of histamine
released by mast cells.
09Which of the following diseases is
caused by a protozoan?
[CBSE AIPMT 2015]
(a) Syphilis
(b) Influenza
(c) Babesiosis
(d) Blastomycosis
Ans.(c)
Babesiosis is a malaria-like parasitic
disease caused by infection withBabesia
bigemina, a genus of protozoa
piroplasms.
Syphilis —
Treponema pallidum
(bacterium)
Influenza —
Influenza virus
Blastomycosis
—
Blastomyces
dermatitidis(fungus)
10Infection ofAscarisusually occurs
by [NEET 2013]
(a) drinking water containing egg of
Ascaris
(b) eating imperfectly cooked pork
(c) tse-tse fly
(d) mosquito bite
Ans.(a)
Infection ofAscarisoccurs in a healthy
person due to the contaminated water,
vegetables, (raw or uncooked) fruits, etc.
Mosquitos bite causes malaria due to the
entry toPlasmodiumparasite into the
blood by mosquito. Eating imperfectly
cooked pork causes trichinosis disease
(parasitic disease). Tse-tse fly causes
trypanosomiasis, an infection of the
central nervous system.
11Widal test is carried out to test
[CBSE AIPMT 2012, 10]
(a) malaria
(b) diabetes mellitus
(c) HIV/AIDS
(d) typhoid fever
Ans.(d)
Widal test is one of the most reliable
diagnostic tests for typhoid fever in
developing countries since its
introduction (over 100 years ago). This
test demonstrates the presence of
somatic (O) and flagellar (H) agglutinins
toSalmonella typhiin the patients blood
serum using suspensions of O and H
antigens. Antigens ofS. paratyphiA and
S. paratyphiB are included in most
commercial kits.
12Common cold differs from
pneumonia in, that
[CBSE AIPMT 2012]
(a) pneumonia is a communicable
disease, whereas the common cold is
a nutritional deficiency disease
(b) pneumonia can be prevented by a live
attenuated bacterial vaccine,
whereas the common cold has no
effective vaccine
(c) pneumonia is caused by a virus, while
the common cold is caused by the
bacteriumHaemophilus influenzae
(d) pneumonia pathogen infects alveoli
whereas the common cold affects
nose and respiratory passage but not
the lungs
Human Health and Diseases 297

308. Ans.(d)
Pneumonia is caused by the bacteria
Diplococcus pneumoniaewhich infects
the alveoli of lungs. It generally spreads
through the sputum of patient. Fever,
cold and difficulty in breathing are some
common symptoms of pneumonia. It can
be treated by the antibiotics.
Common cold is caused by a variety of
viruses, most commonly by rhinovirus
(RNA virus). It spreads through droplet
infection. It affects the upper respiratory
tract but not the lungs. Nasal and
bronchial irritation, sneezing and
coughing are some common symptoms
of cold.
13Ringworm in humans is caused by
[CBSE AIPMT 2010]
(a) bacteria (b) fungi
(c) nematodes (d) viruses
Ans.(b)
Ringworm refers to fungal infections
occurring on the surface of the skin.
Although the world is full of yeasts,
moulds and fungi, only a few cause skin
problems. These agents are called the
dermatophytes. Some common
dermatophytic fungi areTrichophyton
rubrum, T. tensurans, T. interdigitale, T.
mentagrophytes, Microsporum, Canis,
Albugo candidsandEpidermophyton
floccosum.
14Which of the following is a pair of
viral diseases?[CBSE AIPMT 2009]
(a) Ringworm, AIDS
(b) Common cold, AIDS
(c) Dysentery, common cold
(d) Typhoid, tuberculosis
Ans.(b)
Common cold and AIDS are viral
diseases, occur due to the rhino virus
and Human Immunodeficiency Virus
(HIV) respectively. Viral diseases can not
be treated by the use of antibiotics.
15Match the disease in column I with
the appropriate items
(pathogen/prevention/ treatment)
in column II
[CBSE AIPMT 2008]
Column I Column II
A. Amoebiasis (i)Treponema
pallidum
B. Diphtheria (ii)Use only sterilised
food and water
C. Cholera (iii) DPT vaccine
D. Syphilis (iv) Use oral
rehydration
therapy
(a) A–(i), B–(ii), C–(iii), D–(iv)
(b) A–(ii), B–(iv), C–(i), D–(iii)
(c) A–(ii), B–(i), C–(iii), D–(iv)
(d) A–(ii), B–(iii), C–(iv), D–(i)
Ans.(d)
Amoebiasis is caused byEntamoeba
histolytica.Prevention of infection is
entirely a matter of hygiene, both
personal as well as municipal. Their
prevention include use of properly
cooked food and sterilized water.
Diphtheria is caused byCorynebacterium
diphtheriae.The symptoms are fever,
sore throat, severe damage to heart,
nerve cell and adrenal glands. The
vaccine DPT is used for diphtheria,
pertussis and tetanus.
Cholera is caused byVibrio cholerae,a
Gram negative bacterium. It spreads by
faecal contamination. The dehydration
and loss of mineral salts can cause
death. It is treated by use of oral
rehydration therapy.
Syphilis is caused byTreponema
pallidum,a spirochaete and spread by
sexual contact and is STD.
16Sickle-cell anaemia has not been
eliminated from the African
population because
[CBSE AIPMT 2006]
(a) it is controlled by recessive genes
(b) it is not a fatal disease
(c) it provides immunity against malaria
(d) it is controlled by dominant genes
Ans.(c)
Sickle-cell anaemia (in which RBCs
become sickle-shaped and stiff) is a
genetic disorder that is autosomal and
linked to a recessive allele. It has not
been eliminated from the African
population because it provides immunity
against malaria. People who are
heterozygous for sickle cell allele are
much less susceptible for falciparum
malaria which is one of the main causes
of illness and death in them. Thus, the
sickle cell allele is maintained at high
levels in populations where falciparum
malaria is common.
17Both sickle-cell anaemia and
Huntington’s chorea are
[CBSE AIPMT 2006]
(a) bacteria-related diseases
(b) congenital disorders
(c) pollutant-induced disorders
(d) virus-related diseases
Ans.(b)
Both sickle-cell anaemia and
Huntington’s chorea arecongenital
genetic disorders.
Sickle-cell anaemiawas first reported
by James Herrick (1904). In this disease
the patient’s haemoglobin level reduced
to half of the normal and the RBCs
become sickle-shaped. A single
mutation in a gene cause sickle-cell
anaemia.
Huntington’s choreais caused by
autosomal mutation which is
dominant. The gene is present on
chromosome number 4.
18Which one of the following is not
correctly matched?
[CBSE AIPMT 2004]
(a)Glossina palpalis— Sleeping sickness
(b)Culex pipiens— Filariasis
(c)Aedes aegypti— Yellow fever
(d)Anopheles culicifacies
— Leishmaniasis
Ans.(d)
Leishmaniasis or kala-azar is caused by
a protozoan,Leishmania donovani. It is
spread by sand fly. It is also known as
dum-dum fever. It’s control includes
eradication of vector, and use of
antibiotics.
19Salmonellais related with
[CBSE AIPMT 2001]
(a) typhoid (b) polio
(c) TB (d) tetanus
Ans.(a)
Salmonella typhicauses typhoid fever.
The incubation period is about two
weeks. The patient first suffers from
high fever of 40°C and continual
headache. Polio, TB and tetanus are
caused by polio virus,Mycobacterium
tuberculosisandClostridium tetani
respectively. Polio is being eradicated by
polio vaccine. TB and tetanus can be
cured by antibiotics.
20Which of these is most infectious
disease? [CBSE AIPMT 2001]
(a) Hepatitis-B (b) AIDS
(c) Cough and cold (d) Malaria
Ans.(a)
Hepatitis may be transmittedviablood
transfusions, contaminated equipment,
unsterile needles (of drug addicts), or
any body secretion like saliva, sweat,
semen, breast milk, urine, faeces.
Infection to healthy persons is prevented
by proper vaccinations of hepatitis
specially hepatitis-B.
298 NEETChapterwise Topicwise Biology

309. 21Bovine spongiform encephalopathy
is a bovine disease. To which of the
following human diseases it is
related? [CBSE AIPMT 2000]
(a) Kala-azar
(b) Encephalitis
(c) Cerebral spondylitis
(d) Creutzfeldt Jacob disease
Ans.(d)
Bovine Spongiform Encephalopathy
(BSE) is a fatal brain disease known to
exists in beef and other dairy cattle in
UK, also known as mad cow disease. It is
believed to be caused by prions.
Creutzfeldt-Jacob Disease (CJD) is a
slow degenerative disease among
human affecting central nervous system
with dysfunction and degeneration of
the brain. Some scientists have
suggested that a few people in Britain
might have contracted CJD by eating
BSE-infected beef.
22A patient suffering from cholera is
given saline drip because
[CBSE AIPMT 1996, 2000]
(a)Cl
−
ions are important component of
blood plasma
(b)Na
+
ions help to retain water in the
body
(c)Na
+
ions are important in transport of
substances across membrane
(d)Cl
−
ions help in the formation of HCl in
stomach for digestion
Ans.(b)
Severe diarrhoea, vomiting, watery
stools are the chief symptoms of
cholera. All these lead to dehydration.
The toxin secreted byVibrio cholerae
causes a continuous activation of
adenylate cyclase of intestinal epithelial
cells.
The resultant high concentration of
cAMP triggers continual secretion of
Cl, HCO
3
− −
and water into the lumen of
the intestine. Administration of saline
not only supports the sodium–potassium
pump through which water in cell is
restored, but glucose is also symported
along with sodium.
23Koch’s postulates are not
applicable to[CBSE AIPMT 1999]
(a) cholera
(b) leprosy
(c) TB
(d) diphtheria
Ans.(b)
To apply Koch’s postulates, we have to
culture the suspected causal organism
in vitro. Mycobacterium lepraecannot be
culturedin vitro.Hence, Koch’s
postulates are not applicable to leprosy
because its incubation period is 2-5
years.
Cholera is caused byVibrio cholerae.
TB is caused byMycobacterium
tuberculosis.
Diphtheria is caused byMycobacterium
diphtheriae.
24Typhoid fever is caused by
[CBSE AIPMT 1998]
(a)Giardia (b)Salmonella
(c)Shigella (d) Escherichia
Ans.(b)
Salmonella typhicauses typhoid fever in
human beings. It is characterised by
constant fever due to the infection of
intestine.Giardiais a flagellate
protozoan,lambliaspecies of this
protozoan causes disease giardiasis, a
prolonged diarrhoeal disease of humans.
Bacterial genusShigellacauses
shigellosis orbacillary dysentery.
Escherichia coliis a facultative
anaerobes, found in the intestine of
human beings.
25Botulism caused byClostridium
botulinumaffects the
[CBSE AIPMT 1998]
(a) spleen
(b) intestine
(c) lymph glands
(d) neuromuscular junction
Ans.(d)
Clostridium botulinumbacterium causes
food poisoning (botulism).Clostridiumis
an obligate anaerobic
endospore-forming Gram positive,
rod-shaped bacterium. This bacterium
produces an exotoxin which is highly
toxic for the synaptic ends of the nerves
where it blocks the release of
acetylcholine. Later is a chemical
necessary for the transmission of nerve
impulse across the synapses.
26Diphtheria is caused by
[CBSE AIPMT 1997]
(a) poisons released by living bacterial
cells into the host tissue
(b) poisons released from dead bacterial
cells into the host tissue
(c) poisons released by virus into the
host tissues
(d) excessive immune response by the
host’s body
Ans.(a)
Toxins released fromCorynebacterium
diphtheriaecause diphtheria. Actually,
bacterial cells do not contain gene for
toxin production, i.e. a phage carries the
gene for it. Only those lysogenised cell of
C.diphtheriaewhich carryβ-phage, can
produce the toxin and cause diphtheria.
27Which of the following disease is
now considered nearly eradicated
from India? [CBSE AIPMT 1997]
(a) Smallpox (b) Polio myelitis
(c) Plague (d) Kala-azar
Ans.(a)
Small-pox is an acute highly
communicable viral disease. It is caused
by virus namedVariolavirus. Now, it is
eradicated from world including India by
the mass polio vaccination compaign
undertaken by government of India.
28Which of the following pair of
diseases is caused by virus?
[CBSE AIPMT 1996]
(a) Rabies, mumps
(b) Cholera, tuberculosis
(c) Typhoid, tetanus
(d) AIDS, syphilis
Ans.(a)
Rabies (hydrophobia) is caused by a virus
named as rabies virus. It is a lethal
disease. Mumps is an infectious disease
causing fever, difficulty in opening the
mouth and painful swelling of the parotid
glands which lie just below the lobe of
the ear. It is caused by a paramyxovirus.
29In which one of the following pairs
of diseases both are caused by
viruses? [CBSE AIPMT 1996]
(a) Tetanus and typhoid
(b) Whooping cough and sleeping
sickness
(c) Syphilis and AIDS
(d) Measles and rabies
Ans.(d)
Measles and rabies are viral diseases.
Disease Pathogen
Measles Rubeolavirus
Rabies Rabies virus.
Human Health and Diseases 299

310. 30If all ponds and puddles are
destroyed, the organism likely to be
destroyed is[CBSE AIPMT 1993]
(a)Leishmania (b)Trypanosoma
(c)Ascaris (d)Plasmodium
Ans.(d)
Anophelesis the host of malarial parasite
Plasmodiumis known to occur most
favourably in stagnant water, ditches,
ponds, moist and damp places.
Destruction of all the ponds and puddles,
i.e. the breeding places of larva and
pupae will cause destruction in the
number ofAnophelesandPlasmodium.
31Give the correct matching of
causative agent/germ and disease
[CBSE AIPMT 1993]
(a)Anopheles— malaria
(b)Leishmania— sleeping sickness
(c)Glossina— kala-azar
(d)Wuchereria— filariasis
Ans.(d)
Wuchereria bancrofticauses filariasis or
elephantiasis.
32The part of life cycle of malarial
parasitePlasmodium vivax, that is
passed in femaleAnophelesis
[CBSE AIPMT 1992]
(a) sexual cycle
(b) pre-erythrocytic schisogony
(c) exo-erythrocytic schisogony
(d) post-erythrocytic schisogony
Ans.(a)
Sexual phase in the life cycle of
Plasmodiumoccurs in the gut of
mosquito. Sexual phase involves the
gametocytes, megagametocytes
(female) and microgametocytes (male)
which reach the stomach of female
Anophelesmosquito by sucking human
blood.
33Who discoveredPlasmodiumin
RBCs of human beings?
[CBSE AIPMT 1991]
(a) Ronald Ross (b) Mendel
(c) Laveran (d) Stephen
Ans.(c)
In 1880,Charles Laverandiscovered
Plasmodium,the causative agent of
malaria in RBCs of human beings. In
1897,Ronald Rossdiscovered oocytes
ofPlasmodiumin the stomach of
mosquito.
34Malignant tertian malarial is caused
by [CBSE AIPMT 1991]
(a)Plasmodium falciparum
(b)P. vivax
(c)P. ovale
(d)P. malariae
Ans.(a)
Malignant tertian malaria is caused by
the malarial parasite,Plasmodium
falciparum,whereas,P. vivaxcauses
tertian malaria and benign tertian
malaria;P. ovalecauses mild tertian
malaria andP. malariaecauses Quartan
malaria.
35In hot summer and cold winter, the
number of malaria cases as well as
Anophelesdeclines, reappearance
of malaria in humid warm conditions
is due to [CBSE AIPMT 1990]
(a) surviving malarial parasites in human
carriers
(b) surviving sporozoites in surviving
mosquitoes
(c) monkeys
(d) mosquito larvae in permanent waters
Ans.(d)
The reappearance of malaria in humid
warm conditions is due to the mosquito
larvae in permanent waters.
36Amoebiasis is prevented by
[CBSE AIPMT 1990]
(a) eating balanced food
(b) eating plenty of fruits
(c) drinking boiled water
(d) using mosquito nets
Ans.(c)
Amoebiasis or amoebic dysentery is
caused by protozoan parasite
Entamoeba histolyticathat resides in the
upper part of large intestine. It spreads
through contaminated water and food
containing adult form (trophozoite) or
cyst ofEntamoeba.Trophozoite
damages intestinal wall by enzyme
histolysin, reaches blood capillaries and
feed on RBCs, bacteria, tissue debris,
resulting in abdominal pain, acidic
motions with mucus and blood. The
disease can be prevented by drinking
boiled and clean water and intake of
fresh and hygienic food.
37The vector for sleeping sickness is
[CBSE AIPMT 1989]
(a) house fly (b) tse-tse fly
(c) sand fly (d) fruit fly
Ans.(b)
Trypanosoma gambienseis the causative
agent of African sleeping sickness. Its
primary host is man and the secondary
(intermediate) host or vector is tse-tse
fly (Glossina palpalis).
38The infective state of malarial
parasitePlasmodiumthat enters
human body is[CBSE AIPMT 1989]
(a) merozoite
(b) sporozoite
(c) trophozoite
(d) minuta form
Ans.(b)
Sporozoites are small, spindle-shaped,
uninucleate organisms present in the
salivary glands of the mosquito.
Sporozoites represent the infective
stage, which along with saliva inoculates
into the blood stream of human and
undergo schizogony.
39Malaria fever coincides with
liberation of[CBSE AIPMT 1989]
(a) cryptomerozoites
(b) metacryptomerozoites
(c) merozoites
(d) trophozoites
Ans.(c)
Merozoites are the progeny of
sporozoites, formed in the liver of
human. These are produced several days
after the initial infection, which enter the
blood stream and infect erythrocytes.
40The Adenosine deaminase
deficiency results into[NEET 2021]
(a) dysfunction of immune system
(b) Parkinson’s disease
(c) digestive disorder
(d) Addison’s disease
Ans.(a)
Adenosine deaminase is an enzyme. The
deficiency of this particular enzyme
results in severe combined immuno
deficiency (SCID).
During the deficiency of adenosine
deaminase the patient lacks functional
T- lymphocytes and thus the immune
system does not work properly.
300 NEETChapterwise Topicwise Biology
Immunity
TOPIC 2

311. 41The yellowish fluid ‘colostrum’
secreted by mammary glands of
mother during the initial days of
lactation has abundant antibodies
(IgA) to protect the infant. This type
of immunity is called as
[NEET (Oct.) 2020]
(a) passive immunity
(b) active immunity
(c) acquired immunity
(d) autoimmunity
Ans.(a)
Passive immunity is when readymade
antibodies are directly given to protect
the body against foreign agents.
For example, the yellowish fluid
colostrum secreted by mother during
the initial days of lactation has abundant
antibodies (IgA) to protect the infant.
Also the foetus receives some
antibodies from their mother through
the placenta during pregnancy.
42Identify the wrong statement with
reference to immunity.
[NEET (Sep.) 2020]
(a) When readymade antibodies are
directly given, it is called ‘passive
immunity’
(b) Active immunity is quick and gives full
response
(c) Foetus receives some antibodies
from mother, it is an example for
passive immunity
(d) When exposed to antigen (living or
dead) antibodies are produced in the
host’s body. It is called ‘active
immunity’
Ans.(b)
The statement in option is (b) incorrect
because active immunity is slow and
takes time to give its full effective
response in comparison to passive
immunity where pre-formed antibodies
are administered.
43Humans have acquired immune
system that produces antibodies to
neutralise pathogens. Still innate
immune system is present at the
time of birth because it
[NEET (Odisha) 2019]
(a) is very specific and uses different
macrophages
(b) produces memory cells for mounting
fast secondary response
(c) has natural killer cells which can
phagocytose and destroy microbes
(d) provides passive immunity
Ans.(c)
Innate immunity is non-specific type of
defence that is present at the time of
birth because it has natural killer cells
which can phagocytose and destroy
microbes (cellular barriers). Other forms
of innate immunity are physical barriers,
physiological and cytokine barriers.
44Which of the following diseases is
an autoimmune disorder?
[NEET (Odisha) 2019]
(a) Myasthenia gravis
(b) Arthritis
(c) Osteoporosis
(d) Gout
Ans.(a)
Myasthenia gravis is a chronic
autoimmune neuromuscular disorder
that causes weakness in the skeletal
muscles. This is responsible for
breathing and moving parts of the body
including the arms and legs.
45Concanavalin A is
[NEET (National) 2019]
(a) an essential oil
(b) a lectin
(c) a pigment
(d) an alkaloid
Ans.(b)
Concanavalin A is a lectin or a
carbohydrate binding protein. It is a
T-cell mitogen that can activate the
immune system, recruit lymphocytes
and elicit cytokine production. It can also
induce programmed cell deathvia
mitochondria-mediated apoptosis.
46Which of the following immune
responses is responsible for
rejection of kidney graft?
[NEET (National) 2019]
(a) Humoral immune response
(b) Inflammatory immune response
(c) Cell-mediated immune response
(d) Auto-immune response
Ans.(c)
Cell-mediated immune response is
responsible for the rejection of kidney
graft. Cell-mediated immune response is
conferred by sensitised T-lymphocytes
and here, antibodies are not produced.
T-cells confer a long term memory and
they are able to discriminate between
self and non-self. These cells sometimes
consider graft as non-self and attack the
same which causes its rejection.
47Which of the following isnotan
autoimmune disease?[NEET 2018]
(a) Alzheimer’s disease
(b) Rheumatoid arthritis
(c) Psoriasis
(d) Vitiligo
Ans.(a)
Alzheimer’s diseaseis not an
automimmune disease. It is caused due
to the destruction of vast number of
neurons in theHippocampus. It occurs
due to a combination of genetic factors,
environmental or lifestyle factors and
the ageing process. There is loss of
neurotransmitter acetylcholine.
Individuals with this disease have trouble
remembering recent events.
Rheumatoid arthritis, vitiligo and
psoriasis all are autoimmune
diseases. Inrheumatoid arthritis,
antibodies are produced against the
synovial membrane and cartilage.
Vitiligo causeswhite patches on skin
whilepsoriasiscauses itch-skin.
48MALT constitutes about .................
per cent of the lymphoid tissue in
human body. [NEET 2017]
(a)50% (b) 20%
(c) 70% (d) 10%
Ans.(a)
MALT is mucosa associated lymphoid
tissue located within the linning of the
major tracts including respiratory,
digestive and urinogenital tracts. It is
nearly 50% of the total lymphoid tissue
in the human body.
49Transplantation of tissues/organs
fails often due to non-acceptance
by the patient’s body. Which type of
immune-response is responsible
for such rejections ?[NEET 2017]
(a) Autoimmune response
(b) Cell-mediated immune response
(c) Hormonal immune response
(d) Physiological immune response
Ans.(b)
Transplantation of tissue/organs may
fail, when that tissue is rejected by the
recipient’s immune system leading to its
destruction. Tissue rejection is a
function of cell-mediated immune
response that involves T-cells.These
cells have the ability to distinguish
between self and non-self. After the
recognition of non-self tissue, the killer
T-cells induces apoptosis of the target
tissue.
Human Health and Diseases 301

312. 50In higher vertebrates, the immune
system can distinguish self-cells
and non-self. If this property is lost
due to genetic abnormality and it
attacks self-cells, then it leads to
[NEET 2016, Phase I]
(a) graft rejection
(b) auto-immune disease
(c) active immunity
(d) allergic response
Ans.(b)
In auto-immune disease, the immune
cells are unable to distinguish between
self-cells and non-self cells and attack
self-cells which may lead to
auto-immune disorders like interstitial
lung disease in humans.
51Antivenom injection contains
preformed antibodies while polio
drops that are administered into
the body contain
[NEET 2016, Phase I]
(a) harvested antibodies
(b) gamma globulin
(c) attenuated pathogens
(d) activated pathogens
Ans.(c)
Oral polio vaccine consists of attenuated
pathogens. Attenuated pathogens are
living microorganisms or viruses
cultured under adverse condition,
leading to loss of their virulance. But
these organisms have the ability to
induce protective immunity. The oral
vaccine of polio contains three live polio
strains in attenuated forms.
52If you suspect major deficiency of
antibodies in a person, to which of
the following would you look for
confirmatory evidence?
[CBSE AIPMT 2015]
(a) Fibrinogin in plasma
(b) Serum albumins
(c) Haemocytes
(d) Serum globulins
Ans.(d)
Globulin is one of the protein found in
serum and it includes proteins, enzymes,
complement and immunoglobulins
(antibody).
That’s why, if major deficiency of
antibodies is suspected in a person, the
globulins in serum is tested as the
confirmatory evidence.
53Grafted kidney may be rejected in a
patient due to[CBSE AIPMT 2015]
(a) humoral immune response
(b) cell-mediated immune response
(c) passive immune response
(d) innate immune response
Ans.(b)
Grafted kidney may be rejected in a
patient due to the cell-mediated immune
response that is mediated by
T-lymphocytes. The body is able to
differentiate ‘self’ and ‘non-self’.
Therefore, tissue matching, blood group
matching are essential before
undertaking any graft/transplant and
even after this the patient has to take
immuno-suppressants all his/her life.
54Which of the following
immunoglobulins does constitute
the largest percentage in human
milk? [CBSE AIPMT 2015]
(a) IgD (b) IgM
(c) IgA (d) IgG
Ans.(c)
All types of immunoglobulin are found in
human milk. Out of these secretory IgA,
a type of immunoglobulin that protects
the ears, nose, throat and the
gastrointestinal tract, is found in largest
amount.
55The cell-mediated immunity inside
the human body is carried out by
[NEET 2013]
(a) T-lymphocytes (b) B-lymphocytes
(c) thrombocytes (d) erythrocytes
Ans.(a)
T-lymphocyte receptors can recognise
only antigen that bound to cell
membrane proteins. These lymphocytes
mediate CMI (Cell Mediated Immunity).
B-lymphocytes are the major effector
molecules of humoral immunity.
Erythrocytes are red blood cells.
Thrombocytes or platelets secrete
factors that are involved in vascular
repair.
56Which one of the following
statements is correct with respect
to immunity?[CBSE AIPMT 2012]
(a) Preformed antibodies need to be
injected to treat the bite by a viper
snake
(b) The antibodies against smallpox
pathogen are produced by
T-lymphocytes
(c) Antibodies are protein molecules,
each of which has four light chains
(d) Rejection of a kidney graft is the
function of B-lymphocytes
Ans.(a)
In artificially acquired passive immunity,
preformed antibody in an immune serum
of some other animal is introduced into
the body. As the antivenum used to treat
snake bites. In this case, the body does
not produce any antibodies. Antibody is a
protein molecule having two light chain
and two heavy chain. B-cells recognise
and bind antigens and may differentiate
to memory cell or plasma cells (produce
antibody). T-cells causes transplant
rejection.
57In which one of the following
options the two examples are
correctly matched with their
particular type of immunity?
[CBSE AIPMT 2012]
Examples
Type of
immunity
(a) Polymorphonuclear
leukocytes and
monocytes
Cellular
barriers
(b) Anti-tetanus and
anti-snake bite
injections
Active
immunity
(c) Saliva in mouth and
tears in eyes
Physical
barriers
(d) Mucus coating of
epithelium lining the
urinogenital tract
and the HCl in
stomach
Physiological
barriers
Ans.(a)
Phagocytosis is an important feature of
cellular innate immunity, performed by
cells called phagocytes that engulf or eat
pathogens or foreign particles. Common
examples of these phagocytes are
monocytes, macrophages, neutrophil
granulocytes (often referred to as
polymorphonuclear leukocytes or PMN
or PML, because of the varying shapes of
nucleus), tissue dendritic cells, mast
cells etc. Anti-tetanus and anti snake
bite injections are examples of passive
immunity.
58Consider the following four
statements (I-IV) regarding kidney
transplant and select the two
correct ones out of these.
I. Even if a kidney transplant is
proper the recipient may need to
302 NEETChapterwise Topicwise Biology

313. take immuno-suppresants for a
long time.
II. The cell-mediated immune
response is responsible for the
graft rejection.
III. The B-lymphocytes are
responsible for rejection of the
graft.
IV. The acceptance or rejection of a
kidney transplant depends on
specific interferons.
The two correct statements are
[CBSE AIPMT 2010]
(a) II and III (b) III and IV
(c) I and III (d) I and II
Ans.(d)
Both statements I and II are correct.
59A person likely to develop tetanus
is immunised by administering
(a) dead germs[CBSE AIPMT 2009]
(b) preformed antibodies
(c) wide spectrum antibiotics
(d) weakened germs
Ans.(b)
In passive immunity, the antibodies are
produced in some other organisms (e.g.
horse, rabbit, mouse) in response to the
given antigen. These antibodies are then
injected into the human body at the time
of need. This is known as inoculation,
e.g. persons infected by tetanus
(Clostridium tetani), rabies virus and
Salmonellathe sufficient amount of
antibodies, are given to enhance passive
immunity at the time of need.
60Globulins contained in human blood
plasma are primarily involved in
[CBSE AIPMT 2009]
(a) defence mechanisms of body
(b) osmotic balance of body fluids
(c) oxygen transport in the blood
(d) clotting of blood
Ans.(a)
Globulins are soluble in salt solutions of
strong acids and bases and insoluble in
pure water and moderately
concentrated salt solutions. These are
coagulated by heat. Globulins contained
in human blood plasma are primarily
involved in defense mechanisms of the
body. Some examples are i.e. Rabies
immune globulin, RhO(D) immune
globulin, specific immune globulin,
tetanus immune globulin, etc.
61The letter T in T-lymphocyte refers
to [CBSE AIPMT 2009]
(a) thyroid (b) thalamus
(c) tonsil (d) thymus
Ans.(d)
T-refers to thymus which is
haemopoietic as well as an endocrine
gland. Thymus is the ‘seedbed’ of thymic
lymphocytes (T-lymphocytes). Certain
stem cells, originating in yolk sac and
liver in early embryo, but only in bone
marrow in late embryo, migrate into the
thymus and proliferate to form a large
number of lymphocytes.
Thyroidis an endocrine gland.
Thalamusis the part of fore brain in
vertebrate lies above the hypothalamus.
Tonsilis a mass of lymphoid tissue,
several of which are situated at the
back of the mouth and throat in higher
vertebrates.
62Use of anti-histamines and steroids
give a quick relief from
[CBSE AIPMT 2009]
(a) allergy
(b) nausea
(c) cough
(d) headache
Ans.(a)
Allergy is the hypersensitive reaction of
a person to some foreign substances
coming in contact with or entering the
body. The common allergns are dust,
pollen mould, spores, fabricates,
bacteria, etc. During allergic reaction,
there is increased release of histamine
from mast cells. Use of anti-histamines
and steroids give a quick relief from
allergy.
63To which type of barriers under
innate immunity, do the saliva in
the mouth and the tears from the
eyes, belong?[CBSE AIPMT 2008]
(a) Cytokine barriers
(b) Cellular barriers
(c) Physiological barriers
(d) Physical barriers
Ans.(c)
Innate immunity (inborn) is the
resistance to infection, which an
individual possesses by virtue of his/her
genetic and constitutional make up.
Thus it comprises all those defence
elements with which an individual is born
and, which are always available to
protect a living body. Physiological
barriers like body temperature, pH of the
body fluid, and various body secretions
(saliva, tears) prevent growth of many
disease causing micro-organisms.
Skin is the physical barrier of the body.
Its outer tough layer the stratum
corneum prevents the entry of
bacteria and viruses.
64If you suspect major deficiency of
antibodies in a person, to which of
the following would you look for
confirmatory evidence?
[CBSE AIPMT 2007]
(a) Serum albumins
(b) Serum globulins
(c) Fibrinogen in plasma
(d) Haemocytes
Ans.(b)
Antibodies also called immunoglobulins
constitute the gamma globulin which
are the part of blood proteins. These are
secreted by activated B-cells or plasma
cells.
65Increased asthmatic attacks in
certain seasons are related to
[CBSE AIPMT 2007]
(a) hot and humid environment
(b) eating fruits preserved in tin
containers
(c) inhalation of seasonal pollen
(d) low temperature
Ans.(c)
Asthma is a respiratory disorder. It is
caused by foreign allergens and dust
particles present in the air passing
through the respiratory system, the
pollen grains present in air can cause
asthmatic attacks in certain seasons as
are produced in large number in that
particular seasons.
66What is true about T-lymphocytes
in mammals?[CBSE AIPMT 2004]
(a) They scavenge damaged cells and
cellular debris
(b) These are produced in thyroid
(c) There are three main types–cytotoxic
T-cells, helper T-cells and suppressor
T-cells
(d) These originate in lymphoid tissues
Ans.(c)
The function of T-cells is to provide
immunity (cellular type) and not to
scavenge damaged cells and cell debris.
These are produced in bone marrow and
get matured in thymus gland. Hence, the
only true statement is that there are
three types of T-cells, i.e. cytotoxic,
helper and suppressor.
Human Health and Diseases 303

314. 67The term ‘antibiotic’ was coined by
[CBSE AIPMT 2003]
(a) Selman Waksman
(b) Alexander Fleming
(c) Edward Jenner
(d) Louis Pasteur
Ans.(a)
The term ‘antibiotics’ was first time used
bySA Waksmanin 1945. Antibiotics are
the substances which are produced by
microorganisms such as fungi or
bacteria. These substances are harmful
to the growth of other microorganisms,
example of some of the antibiotics are
penicillin, streptomycin,
chloramphenicol, etc.
68Interferons are synthesised in
response to[CBSE AIPMT 2001]
(a)Mycoplasma(b) bacteria
(c) viruses (d) fungi
Ans.(c)
Cells infected by virus produce
interferons (an antiviral protein) which is
antiviral. It spreads to neighbouring cells
and makes them resistant to virus
infections by inhibiting viral growth.
69Small proteins produced by
vertebrate cells naturally in
response to viral infections and
which inhibit mutliplication of
viruses are called
[CBSE AIPMT 2000]
(a) immunoglobulins
(b) interferons
(c) antitoxins
(d) lipoproteins
Ans.(b)
Interferons (INFs) are a group of three
vertebrate glycoproteins( , , ).α β γOut of
these, two(αandβ)are produced within
viral infected cells. Interferon induces,
among adjacent cells, as antiviral state
by inducing synthesis of the enzymes
which inhibit the viral production cycle.
Thus, inhibiting multiplication of virus in
the body.
70If a person shows production of
interferons in his body, the
chances are that he has got an
infection of[CBSE AIPMT 1997]
(a) typhoid
(b) measles
(c) tetanus
(d) malaria
Ans.(b)
Interferons are proteins produced by a
cell infected by a virus and provide
protection to other healthy cells against
infection by viruses. Measles is also viral
disease. It is caused byparamyxovirus
(RNA virus). Interferon was discovered in
1957 byIssacsandLindenmann.
Typhoid and tetanus are bacterial
diseases and malaria is a protozoan
disease.
71Passive immunity was discovered
by [CBSE AIPMT 1996]
(a) Edward Jenner
(b) Emil von Behring
(c) Robert Koch
(d) Louis Pasteur
Ans.(a)
Passive immunity was first discovered
by Edward Jenner against chickenpox.
In passive immunity readymade
antibodies (γ-globulins) obtained from
human or animal serum, who already had
recovered from an infectious disease,
are injected into human body to develop
immunity.
It is used against measeles, rubella,
mumps, diphtheria, tetanus, snake
venom, scarlet fever, rabies and
Salmonellaand many other bacterial
infection.
72Hypersensitivity to an allergen is
associated with[CBSE AIPMT 1996]
(a) aberrant functioning of the immune
mechanism
(b) increase in ambient temperature
(c) age of the individual
(d) food habits
Ans.(a)
Allergy is hypersensitivity or
inappropriate over reaction or aberrant
functioning of the immune system.
73Which of the following diseases is
due to an allergic reaction?
[CBSE AIPMT 1995]
(a) Goitre (b) Skin cancer
(c) Hay fever (d) Enteric fever
Ans.(c)
Allergy also known as hypersensitivity, is
an inappropriate over-reaction of the
immune system. Hay fever is an allergic
reaction, antigens for such response are
pollens grains, dust and SPM in the
polluted air.
Symptoms of hay fever includes
closure of bronchial tubes that results
in difficulty in normal breathing, skin
rashes and eosinophilia.
74Cells involved in immune
mechanism are[CBSE AIPMT 1993]
(a) erythrocytes (b) lymphocytes
(c) eosinophils (d) thrombocytes
Ans.(b)
Lymphocytes are agranulocytes and
they play a key role in immunological
reactions. Lymphocytes are of two types
(i) B-lymphocytes–function in the
form of immunity called antibody
mediated immunity (humoral
immunity).
(ii) T-lymphocytes–function in
cell-mediated immunity (cellular
immunity).
75Small proteins produced by
vertebrate cells naturally in
response to viral infections and
which inhibit multiplication of
viruses are called
[CBSE AIPMT 2000]
(a) immunoglobulins
(b) interferons
(c) antitoxins
(d) lipoproteins
Ans.(b)
Interferons (INFs) are a group of three
vertebrate glycoproteins( , , ).α β γOut of
these, two(αandβ)are produced within
virally infected cells. Interferon act as
antiviral protein by inducing synthesis of
the enzymes which inhibit the viral
production cycle.
So, interferons are inhibitors of virus
particles.
76The antibodies are
[CBSE AIPMT 1999]
(a) germs
(b) carbohydrates
(c) proteins
(d) lipids
Ans.(c)
Antibodies are glycoproteins and are
secreted by mature vertebrate plasma
cells which are modified form of B-cells.
These selectively bind to epitopes of
antigens and clumping them
(agglutination) prior to phagocytic
engulfment.
304 NEETChapterwise Topicwise Biology

315. 77In mammals, histamine is secreted
by [CBSE AIPMT 1998]
(a) fibroblasts (b) histocytes
(c) lymphocytes (d) mast cells
Ans.(d)
Histamine is a potent vasodilator formed
by decarboxylation of the amino acid
histidine and released by mast cells in
response to appropriate antigens.
Mast cells are especially prevalent in the
connective tissue of the skin, respiratory
tract and in surrounding blood vessels.
78Interferons are[CBSE AIPMT 1996]
(a) antiviral proteins
(b) antibacterial proteins
(c) anticancer proteins
(d) complex proteins
Ans.(a)
Interferons(IFNs) are anti-viral,
regulatory glycoproteins, produced in
virus infected cells for defence. They are
non-antigenic protein of molecular
weight 20000 daltons; discovered by
Issacs and Lindemann (1957). These IFNs
induce formation of certain enzymes
that suppress viral multiplication in host
cell and protect host from further viral
reinfection.
79For effective treatment of the
disease, early diagnosis and
understanding its pathophysiology
is very important. Which of the
following molecular diagnostic
technique is very useful for early
detection? [NEET 2021]
(a) Western Blotting Technique
(b) Southern Blotting Technique
(c) ELISA Technique
(d) Hybridisation Technique
Ans.(c)
ELISA stands for Enzyme Linked
Immunosorbent Assay.
It is a technique to detect the presence of
antigens in biological samples. It is a very
effective molecular diagnostic technique
used for early detection. In this tectnique
the antibodies in the sample binds to the
specific antigen for the disease which is
to be detected.
There are different type of ELISA test,
that includes
Direct ELISA
Indirect ELISA
Sandwich ELISA
80Which of the following statements
is not true for cancer cells in
relation to mutations?
[NEET 2016, Phase I]
(a) Mutations destroy telomerase
inhibitor
(b) Mutations inactivate the cell control
(c) Mutations inhibit production of
telomerase
(d) Mutations in proto-oncogenes
accelerate the cell cycle
Ans.(c)
Cancerous cells have high telomerase
activity. The maintenance of telomere
stability is required for the long term
proliferation of tumors. This makes
telomerase a target not only for cancer
diagnosis but also for the development
of novel anti-cancer therapeutic agents,
e.g. telomerase inhibitors are used in
cancer treatment.
81Which of the following is correct
regarding AIDS causative agent
HIV? [NEET 2016, Phase II]
(a) HIV is enveloped virus containing one
molecule of single-stranded RNA and
one molecule of reverse
transcriptase
(b) HIV is enveloped virus that contains
two identical molecules of
single-stranded RNA and two
molecules of reverse transcriptase
(c) HIV is unenveloped retrovirus
(d) HIV does not escape but attacks the
acquired immune response
Ans.(b)
Statement (b) is correct. The correct
form of other statements are
(a) HIV is a virus containingssRNA and
reverse transcriptase enzyme
enveloped by protein coat.
(c) HIV is enveloped retrovirus.
(d) HIV escapes the immune cells and
attacks helper T-cells of immune
system.
82At which stage of HIV infection
does one usually show symptoms
of AIDS? [CBSE AIPMT 2014, 11]
(a) Within 15 days of sexual contact with
an infected person
(b) When the infected retro virus enters
host cells
(c) When HIV damages large number of
helper T-lymphocytes
(d) When the viral DNA is produced by
reverse transcriptase
Ans.(c)
T-lymphocyte receptors can recognise
only antigen that bound to cell
membrane proteins. These lymphocytes
mediate CMI (cell mediated immunity).
B-lymphocytes are the major effector
molecules of humoral immunity.
Erythrocytes are red blood cells.
Thrombocytes or platelets secrete
factors, that are involved in vascular
repair.
83Which one of the following is not a
property of cancerous cells,
whereas the remaining three are?
[CBSE AIPMT 2012]
(a) They compete with normal cells for
vital nutrients
(b) They do not remain confined in the
area of formation
(c) They divide in an uncontrolled manner
(d) They show contact inhibition
Ans.(d)
Contact inhibition involves major
histocompatibility complex and is the
natural process of arresting cell growth
when two or more cells come in contact
with each other. It is a property of
normal cells. Cancer cells divide in
uncontrolled manner and do not show
contact inhibition.
84A certain patient is suspected to be
suffering from acquired immuno
deficiency syndrome. Which
diagnostic technique will you
recommend for its detection?
[CBSE AIPMT 2011]
(a) MRI (b) Ultra sound
(c) WIDAL (d) ELISA
Ans.(d)
ELISA (Enzyme Linked Immuno Sorbent
Assay), also known as an Enzyme
Immuno Assay (EIA), is a biochemical
technique used mainly in immunology to
detect the presence of an antibody or an
antigen in a sample.
It is a useful tool for determining serum
antibody concentrations (such as with
the HIV test). The ELISA was the first
screening test widely used for HIV
because of its high sensitivity as it
detects antibodies at very low
concentrations.
Human Health and Diseases 305
Cancer and AIDS
TOPIC 3

316. 85Which one of the following
statements is correct with respect
to AIDS? [CBSE AIPMT 2010]
(a) The HIV can be transmitted through
eating food together with an infected
person
(b) Drug addicts are least susceptible to
HIV infection
(c) AIDS patients are being fully cured
cent per cent with proper care and
nutrition
(d) The causative HIV retrovirus enters
helper T-lymphocytes thus, reducing
their numbers
Ans.(d)
In AIDS patients, the virus attacks on
CD T - cells
4
+ (helper T-lymphocytes
responsible for the coordination of the
entire immune system), infecting and
killing them until none of them are left in
blood. Without these crucial immune
system cells, the body cannot fight
against invading bacteria or viruses
which leads to weaker immune system
and gradually the body of the HIV
positive persons become house of
infections leading to multiple problems.
86Which one of the following
statements is correct?
[CBSE AIPMT 2009]
(a) Patients, who had undergone surgery
are given cannabinoids to relieve pain
(b) Benign tumours show the property of
metastasis
(c) Heroin accelerates body functions
(d) Malignant tumours may exhibit
metastasis
Ans.(d)
Malignant tumour first grows slowly. No
symptoms are noticed. This stage is
called the latent stage. The tumour later
grows quickly. The cancer cells go
beyond adjacent tissue and enter the
blood and lymph. Once this happens,
they migrate to many other sites in the
body, where the cancer cells continue to
divide. It is called as metastasis. Only
malignant tumours are properly
designated as cancer.
87Carcinoma refers to
[CBSE AIPMT 2003]
(a) malignant tumours of the colon
(b) benign tumours of the connective
tissue
(c) malignant tumours of the connective
tissue
(d) malignant tumours of the skin or
mucous membrane
Ans.(d)
Carcinoma is a malignant or metastatic
tumour. It can extend to the
neighbouring cells, this process is called
as metastasis. These tumours are
generally located in epithelial tissue and
glands.
e.g. Breast cancer, skin cancer, stomach
cancer, lungs cancer, pancreas cancer,
etc.
88ELISA is used to detect viruses
where the key reagent is
[CBSE AIPMT 2003]
(a) DNA probe
(b) RNase
(c) alkaline phosphatase
(d) catalase
Ans.(c)
The Enzyme Linked Immuno Sorbent
Assay (ELISA), also known as the Enzyme
Immuno Assay (EIA) has become a widely
used serological technique for,
detection of AIDS in the blood serum of
HIV infected person. The enzymes used
for labelling in ELISA include horse
radish peroxidase, alkaline phosphatase,
β-galactosidase, lactoperoxidase, etc.
89Cancerous cells can easily be
destroyed by radiation due to
[CBSE AIPMT 2002]
(a) rapid cell division
(b) lack of nutrition
(c) fast mutation
(d) lack of oxygen
Ans.(a)
The ability of radiations to kill cells is
highest in the tissue with highest
number of dividing cells. Tumour cells
proliferate rapidly. Hence, tumours are
killed more rapidly by radiations.
90Reason of lung cancer is
[CBSE AIPMT 2001]
(a) coal mining
(b) calcium fluoride
(c) cement factory
(d) bauxite mining
Ans.(c)
Cancer is an uncontrolled growth and
division of certain body tissues. Lung
cancer is a cancer of epithelial tissue of
lungs. It is mainly (95%) caused by
smoking and can be found in both male
and female. It may also occur in the
people working in cement factory.
91Human Immunodeficiency Virus
(HIV) has a protein coat and a
genetic material which is
[CBSE AIPMT 1998]
(a) single stranded DNA
(b) single stranded RNA
(c) double stranded RNA
(d) double stranded DNA
Ans.(b)
AIDS (Acquired Immunodeficiency
Syndrome) was first reported in USA in
1981. It is caused by HIV (Human
Immunodeficiency Virus). HIV is the
member of retroviruses.
Later are so named because they
contain an enzyme reverse
transcriptase, which mediates the
formation of DNA from RNA. The genetic
material of HIV is single stranded RNA
(ssRNA).
92Hybridoma cells are
[CBSE AIPMT 1999]
(a) product of spore formation in
bacteria
(b) hybrid cells resulting from myeloma
cells
(c) nervous cells of frog
(d) only cells having oncogenes
Ans.(b)
A myeloma is a type of cancer
associated with abnormal production of
irregular antibodies. It occurs in
antibody-producing cells that have lost
their normal control. Clones of the hybrid
cell resulting from artificial fusion of a
normal antibody producing B-cell with
myeloma cell are called hybridomas.
93Which of the following symptoms
indicate radiation sickness ?
[CBSE AIPMT 1997]
(a) Red and ulcerated skin
(b) Nausea and anaemia
(c) Nausea and loss of hair
(d) Ulcerated skin, nausea and loss of
hair
Ans.(d)
Even lower doses of radiations cause
serious damages like skin burns, nausea,
loss of hairs and nails, change in blood
cell count, prolonged exposure causes
formation of tumours, cancer. High dose
(lethal radiation exposer) may cause
instant death.
306 NEETChapterwise Topicwise Biology

317. 94Which of the following will be
curable in next two decades?
[CBSE AIPMT 1997]
(a) tuberculosis (b) cancer
(c) polio myelitis (d) None of these
Ans.(b)
Cancer may be curable in next two
decades. The completion of the human
genome is causing profound changes in
thinking and direction of biomedical
research. Cancer is caused by
malfunctioning of genes, either through
activation of cancer causing oncogens
(proto-oncogenes) or through
inactivation of tumor suppressor genes.
By comparing the active genes in the
tumor to that of normal cells, the genes
causing the cancer can be determined.
Side by side there is a huge progress in
the field of genetic engineering and
biotechnology. All these aspects give us
hope that cancer may be curable in next
two decades.
95Retroviruses are implicated as a
cause for cancer in humans
because they[CBSE AIPMT 1996]
(a) carry gene for reverse
transcriptase
(b) may carry cellular protooncogenes in
their genome
(c) may carryv-oncogenes in their
genome
(d) carry single stranded RNA as their
genetic material
Ans.(b)
Retroviruses are implicated as a cause
of cancer in humans because they may
carry cellular proto-onchogenes in their
genome, when these proto-oncogenes
gets converted into oncogenes due to
some physical, chemical or biological
agents they cause cancer.
96ldentify the incorrect pair.
[NEET 2021]
(a) Alkaloids - Codeine
(b) Toxin - Abrin
(c) Lectins - Concanavalin-A
(d) Drugs - Ricin
Ans.(d)
Match pair in option (d) is incorrect and
can be corrected as:
Ricin and abrin are potent biological
toxins that are derived from plant
sources, e.g. castor beans. These toxins
inhibits protein synthesis in body leading
to cell death.
97Coca alkaloid or cocaine is
obtained from[NEET (Odisha) 2019]
(a)Papaver somniferum
(b)Atropa belladonna
(c)Erythroxylum coca
(d)Datura
Ans.(c)
Coca alkaloid or cocaine is obtained
from coca plantErythroxylum coca,
native to
South America. It interferes with the
transport of the neurotransmitter
dopamine.
98Drug called ‘Heroin’ is synthesised
by [NEET (National) 2019]
(a) acetylation of morphine
(b) glycosylation of morphine
(c) nitration of morphine
(d) methylation of morphine
Ans.(a)
Drug ‘Heroin’ is synthesised by the
acetylation of morphine.
Chemically heroin is diacetymorphine
and commonly it is called smack. It is an
opium derivative which is used as
medicine. Excessive use of it causes
addiction.
99Which is the particular type of drug
that is obtained from the plant
whose one flowering branch is
shown below?[CBSE AIPMT 2014]
(a) Hallucinogen
(b) Depressant
(c) Stimulant
(d) Pain-killer
Ans.(a)
The plant shown in the picture isDatura
which produce natural hallucinogens.
This kind of drug induce behavioural
abnormalities by changing thoughts,
feelings perceptions without any actual
sensory stimulus.
100Which one of the following fungi
contains hallucinogens?
[CBSE AIPMT 2014]
(a)Morchella esculenta
(b) Amanita muscaria
(c)Neurosporasp.
(d)Ustilagosp.
Ans.(b)
Amanite muscaria,is a fungus which is
known for its hallucinogenic
properties.
101Cirrhosis of liver is caused by the
chronic intake of
(a) opium [CBSE AIPMT 2012]
(b) alcohol
(c) tobacco (chewing)
(d) cocaine
Ans.(b)
The chronic intake of alcohol may be
fatal for the individual. On intake, a part
of alcohol is changed to acetaldehyde
which causes hangover. Acetaldehyde
stimulates formation of fat which is
deposited on artery walls (causing
coronary ailments) and in the liver
(causing fatty liver syndrome). Gradually,
the liver hardens and dries up as its cells
are replaced by fibrous tissue.
This kind of liver degeneration is called
liver cirrhosis (Laennec’s cirrhosis).
Excessive use of alcohol may also lead
liver failure, liver cell carcinoma, etc.
102Select the correct statement from
the ones given below.
[CBSE AIPMT 2010]
(a) Barbiturates when given to criminals
make them tell the truth
(b) Morphine is often given to persons
who have under gone surgery as a
pain killer
(c) Chewing tobacco lowers blood
pressure and heart rate
(d) Cocaine is given to patients after
surgery as it stimulates recovery
Ans.(b)
Serturner, a pharmacist isolated the
active principle of opium in 1806 and
named it morphine. Morphine is a
phenanthrene opioid receptor , its main
effect is binding to and activating the
μ-opioid receptors in the central nervous
system. In clinical settings, morphine
exerts its principal pharmacological
effect on the central nervous system
and gastrointestinal tract. Its primary
actions of therapeutic value are
analgesic and sedation at low doses.
Human Health and Diseases 307
Drugs and Alcohol Abuse
TOPIC 4

318. 308 NEETChapterwise Topicwise Biology
103Which one of the following is the
correct statement regarding the
particular psychotropic drug
specified? [CBSE AIPMT 2008]
(a) Hashish causes alter thought
perceptions and hallucinations
(b) Opium stimulates nervous system
and causes hallucinations
(c) Morphine leads to delusions and
disturbed emotions
(d) Barbiturates cause relaxation and
temporary euphoria
Ans.(a)
Charas is the dried resinous extract from
the flowering tops and leaves of
Cannabis sativa. In some countries, it is
calledhashish. It is a hallucinogen, which
alters a person’s thoughts, feelings and
perceptions.
104A person showing unpredictable
moods, outbursts of emotion,
quarrelsome behaviour and
conflicts with others is suffering
from [CBSE AIPMT 2006]
(a) schizophrenia
(b) Borderline Personality Disorder (BPD)
(c) mood disorders
(d) addictive disorders
Ans.(a)
Schizophreniais a group of severe
mental disorders that have common
symptoms as hallucinations, delusions,
blunted emotions, disordered thinking. It
can be caused by excessive dopamine
production, alternation of neuropeptide
and decreased frontal lobe activities.
Recovery is possible with regular use of
chloropromazine along with pychosocial
therapy.
105Which one of the following
depresses brain activity and
produces feelings of calmness,
relaxation and drowsiness?
[CBSE AIPMT 2005]
(a) Valium (b) Morphine
(c) Hashish (d) Amphetamines
Ans.(a)
Valium depresses brain activity and
produces feeling of calmness, relaxation
and drowsiness. It indirectly affects
central nervous system on long term
usage.
106Which one of the following pairs is
not correctly matched?
[CBSE AIPMT 2004]
(a)Streptomyces— Antibiotic
(b)Serratia — Drug addiction
(c)Spirulina— Single cell protein
(d)Rhizobium — Biofertiliser
Ans.(b)
Serratiais a harmful human pathogen
which has been known to cause urinary
tract infections, wound infections,
pneumonia.
107Which of the following is an opiate
narcotic? [CBSE AIPMT 1997]
(a) Barbiturates (b) Morphine
(c) Amphetamines (d) LSD
Ans.(b)
Narcotics are a group of pain killers
which includes heroin, morphine, codein
and methadone. All of these drugs are
derived fromopium. Later is a gummy
resin like substance found in unripe pods
of poppy plants.
108Nicotine acts as a stimulant,
because it nimics the effect of
[CBSE AIPMT 1995]
(a) thyroxine (b) acetylcholine
(c) testosterone (d) dopamine
Ans.(b)
Acetylcholine is rapidly acting excitatory
small sized neurotransmitter. Nicotine
and acetylcholine both have same
receptors and so the both have the same
effect
109The alkaloid ajmalicine is obtaine
from [CBSE AIPMT 1995]
(a)Atropa (b) Papaver
(c)Curcuma (d) Sarpgandha
Ans.(d)
Ajmalicine is obtained from Sarpgandha.
Its botanical name isRauwolfia
serpentinafamily—Apocyanaceae.
110Opiate narcotic is
[CBSE AIPMT 1993]
(a) bhang (b) charas
(c) heroin (d) nicotine
Ans.(c)
Opium is derived from unripe seed pods
of the poppy plantPapaver somniferum.
Opium has an analgesic effect and may
also reduce anxiety and tension, lowers
the blood pressure and breathing rate.
Opium and its derivatives including
morphine, codeineandheroinare
among the drugs collectively known as
narcotic drugs.
Narcotics induce addiction if used
repeatedly and heroin is most dangerous
narcotic.
111Analgesic drugs[CBSE AIPMT 1990]
(a) form tissues (b) relieve pain
(c) relieve fatigue (d) cause pain
Ans.(b)
Analgesic drugs are those drugs which
relieve us from pain. It acts on the
peripheral and central nervous system
which reversibly eliminate sensation of
pain.

319. 01Which of the following is not a step
in Multiple Ovulation Embryo
Transfer Technology (MOET)?
[NEET 2021]
(a) Cow is administered hormone having
LH like activity for super ovulation
(b) Cow yields about 6-8 eggs at a time
(c) Cow is fertilised by artificial
insemination
(d) Fertilised eggs are transferred to
surrogate mothers at 8-32 cell stage
Ans.(a)
Multiple Ovulation Embryo Transfer
Technology (MOET) is a programme for
herd improvement.
In this method, cow is administered
hormone, having FSH (not LH) like
activity to induce follicular maturation
and super ovulation.
02Inbreeding depression is
[NEET (Oct.) 2020]
(a) reduced motility and immunity due to
close inbreeding.
(b) decreased productivity due to mating
of superior male and inferior female
(c) decrease in body mass of progeny
due to continued close inbreeding
(d) reduced fertility and productivity due
to continued close inbreeding
Ans.(d)
Inbreeding depression is continued
inbreeding, especially close breeding
which reduces fertility and even
productivity in animals. This problem is
usually overcome by outbreeding.
03By which method was a new breed
‘Hisardale’ of sheep formed by
using Bikaneri ewes and Marino
rams? [NEET (Sep.) 2020]
(a) Mutational breeding
(b) Cross breeding
(c) Inbreeding
(d) Outcrossing
Ans.(b)
Hisardale is a new breed of sheep
developed in Punjab by crossing Bikaneri
ewes and Marino rams. In
cross-breeding, superior male of one
breed is mated with superior female of
another breed.
04Select the incorrect statement
regarding inbreeding.
[NEET (Odisha) 2019]
(a) Inbreeding helps in the elimination of
deleterious alleles from the
population
(b) Inbreeding is necessary to evolve a
pureline in any animal
(c) Continued inbreeding reduces
fertility and leads to inbreeding
depression
(d) Inbreeding depression cannot be
overcome by outcrossing
Ans.(d)
The incorrect statement regarding
inbreeding is the option (d). It is because
continuous inbreeding among cattle
causes inbreeding depression. It
decreases the fertility and even
productivity of an animal.
It can be overcome by applying
outbreeding in which mating is done
between different breeds or individuals
of the same breed but having no
common ancestors. Outbreeding
includes outcrossing, cross-breeding
and interspecific hybridisation.
05Mad cow disease in cattle is caused
by an organism which has
[NEET (Odisha) 2019]
(a) inert crystalline structure
(b) abnormally folded protein
(c) free RNA without protein coat
(d) free DNA without protein coat
Ans.(b)
Mad cow disease in cattle is caused by
prions which are abnormally folded
proteins. It is also known as Bovine
Spongiform Encephalopathy (BSE). It is a
progressive neurological disorder of
cattle.
06Which of the following statements
about methanogens is not correct?
[NEET (Odisha) 2019]
(a) They can be used to produce biogas
(b) They are found in the rumen of cattle
and their excreta
(c) They grow aerobically and breakdown
cellulose rich food
(d) They produce methane gas
Strategiesfor
Enhancement
inFoodProduction
31
Animal Husbandry
TOPIC 1

320. Ans.(c)
Statement (c) is incorrect. Correct
information about the statement is as
follows
Certain bacteria, which grow
anaerobically on cellulosic material,
produce large amount of methane along
withCO
2
andH
2
. These bacteria are
collectively called methanogens and one
such example isMethanobacterium.
Rest statements are correct.
07Homozygous purelines in cattle can
be obtained by [NEET 2017]
(a) mating of related individuals of same
breed
(b) mating of unrelated individuals of
same breed
(c) mating of individuals of different
breed
(d) mating of individuals of different
species
Ans.(a)
When closely related species of same
organisms are crossed continuously for
few successive generations, it results in
accumulation of recessive characters,
thus homozygous purelines are
obtained.
08Artificial selection to obtain cows
yielding high milk output
represents [NEET 2017]
(a) stabilising selection as it stabilises
this character in the population
(b) directional as it pushes the mean of
the character in one direction
(c) disruptive as it splits the population
into two, one yielding higher output
and the other lower output
(d) stabilising followed by disruptive as
stabilises the population of produce
higher yielding cows
Ans.(b)
The directional selection leads to
change in the phenotypic characters of a
population in one direction. In the case
of artificial selection, it is intentionally
done to increase the milk production, so
directional selection operates.
09Interspecific hybridisation is the
mating of [NEET 2016, Phase II]
(a) animals within same breed without
having common ancestors
(b) two different related species
(c) superior males and females of
different breeds
(d) more closely related individuals
within same breed for
4-6 generations
Ans.(b)
The interspecific hybridisation is the
mating or cross between two different
related species belonging to same genus.
10Among the following edible fishes,
which one is a marine fish having
rich source of omega-3 fatty acids?
[NEET 2016, Phase II]
(a) Mystus (b) Mangur
(c) Mrigala (d) Mackerel
Ans.(d)
Mackerel is a marine fish having rich
quantity of omega-3 fatty acid.
11Outbreeding is an important
strategy of animal husbandry
because it [CBSE AIPMT 2015]
(a) help in accumulation of superior
genes
(b) is useful in producing purelines of
animals
(c) is useful in overcoming inbreeding
depression
(d) exposes harmful recessive genes that
are eliminated by selection
Ans.(c)
The breeding of stocks or individuals
that are not closely related is called
outbreeding. It is an important strategy
of animal husbandry because it is useful
in overcoming inbreeding depression.
Inbreeding depression is the condition in
which the fertility and the productivity of
animals is reduced due to the continuous
breeding in same species.
12In cloning of cattle a fertilised egg
is taken out of the mother’s womb
and [CBSE AIPMT 2007]
(a) in the eight cell stage, cells are
separated and cultured until small
embryos are formed which are
implanted into the womb of other
cows
(b) in the eight cell stage the individual
cells are separated under electrical
field for further development in
culture media
(c) from this up to eight identical twins
can be produced
(d) the egg is divided into 4 pairs of cells
which are implanted into the womb of
other cows
Ans.(a)
During cloning of a cattle a fertilised egg
is taken out of the mother’s womb and in
the eight cell stage, cells are separated
and cultured until small embryos are
formed which are implanted into the
womb of other cows.
13Compared to a bull a bullock is
docile because of
[CBSE AIPMT 2007]
(a) higher levels of cortisone
(b) lower levels of blood testosterone
(c) lower levels of
adrenaline/noradrenaline in its blood
(d) higher levels of thyroxine
Ans.(b)
Compared to a bull a bullock is docile
because of lower levels of blood
testosterone.
A bullock is a castrated bull. Bulls are
castrated to make them more meek and
docile.
Castration is the removal or destruction
of one or both testicles and results in
sterility, decreased sexual desire and
inhibition of secondary sex
characteristics. It is performed for the
purpose of improving the quality of meat
and decreasing the aggressiveness of
farm animals; in pet animals it prevents
unwanted mating behaviour,
reproduction and wandering.
14Which one of the following is a viral
disease of poultry?
[CBSE AIPMT 2007]
(a) Coryza
(b) New castle disease
(c) Pasteurellosis
(d) Salmonellosis
310 NEETChapterwise Topicwise Biology
Peak shifts in
one direction
Diagrammatic
representation of the
Number of individuals
with phenotype
Phenotypes favoured
by artificial selection
Medium sized
Individuals
are favoured

321. Ans.(b)
New castle disease is a viral disease of
poultry. It is a highly contagious zoonotic
bird disease affecting many domestic
and wild avian species. Its effects are
most notable in domestic poultry due to
their high susceptibility and the potential
for severe impacts of an epidemic on the
poultry industries. It is endemic to many
countries.
15Which one of the following pair is
mismatched?[CBSE AIPMT 2007]
(a)Pila globosa— Pearl
(b)Apis indica— Honey
(c)Kenia lacca— Lac
(d)Bombyx mori— Silk
Ans.(a)
Out of the following the option (a) is
mismatched because pearl is obtained
from pearl oyster (Pinctada vulgaris)
while, honey fromApis indica, lac from
Kenia laccaand silk fromBombyx mori.
16The world’s highly prized wool
yielding ‘Pashmina’ breed is
[CBSE AIPMT 2005]
(a) goat
(b) sheep
(c) goat-sheep cross
(d) Kashmir sheep–Afghan sheep cross
Ans.(a)
Pashmina refers to a type of Kashmir
wool and textiles made from it. This wool
comes from a special breed of goat
indigenous to high altitudes of the
Himalayan mountains. The Himalayan
mountain goat, sheds its winter coat
every spring and the fleece is caught on
thorn bushes. One goat sheds
approximately 3-8 ounces of the wool
fibre.
17Honey is [CBSE AIPMT 1997]
(a) acidic
(b) neutral
(c) alkaline
(d) basic after some days
Ans.(a)
Honey is acidic as its pH is 2.5-4.0.Honey
is a byproduct of bee keeping. It is sweet
in taste and white to black in colour.
Smell of honey varies according to juices
collected from different flowers.
18High milk yielding varieties of cows
are obtained by[CBSE AIPMT 1997]
(a) super ovulation
(b) artificial insemination
(c) use of surrogate mother
(d) All of the above
Ans.(a)
In superovulation, a high milk yielding
cow is induced to shed 4-6 eggs every
6-8 weeks (instead of 20-21 days).
The superovulated donor is artificially
inseminated with semen from a quality
bull. the embryos developing from the
eggs so fertilised are flushed out. These
good quality embryos are now
transferred to surrogate mother for
delivery.
19Pebrine is a disease of
[CBSE AIPMT 1997]
(a) honeybee (b) fish
(c) silkworm (d)lacinsect
Ans.(c)
Pebrine disease is one of the most
damaging disease of silkworm. It is
caused byNosema bombycis nageli. The
other diseases of silk worm areFlacherie
which is an infectious viral disease
marked by body flaccidity and digestive
disorders.
Muscaridine, which is a fungal disease
caused bySpicariaorBotrytis.
20Pasteurisation of milk involve
heating for [CBSE AIPMT 1996]
(a) 60 min at about 90°C
(b) 30 min at about 50°C
(c) 30 min at about 65°C
(d) 60 min at 100°C
Ans.(c)
Pasteurisation of milk involves heating
of milk at 60-70°C for about 30 min so as
to kill the pathogens.
21The earliest animal to have been
domesticated by man was most
likely the [CBSE AIPMT 1996]
(a) horse (b) cow
(c) dog (d) pig
Ans.(c)
Dog was one of the earliest animals to be
domesticated by man.
22The long-term prospects for a truly
human civilisation depend in a large
measure on [CBSE AIPMT 1996]
(a) the ability of humanity to moderate its
fecundity
(b) increasing the food production
(c) colonisation of under populated areas
(d) control of human diseases
Ans.(d)
The long term prospects for a truly
human civilisation depend on a large
measure on control of human disease.
23The silkworm silk is the product of
[CBSE AIPMT 1995]
(a) cuticle of the larva
(b) cuticle of the adult
(c) salivary gland of the larva
(d) salivary gland of the adult
Ans.(c)
Caterpillar larva ofBombyx morisecretes
liquid silk from its salivary glands.
24Which of the following is not an
objective of biofortification in
crops? [NEET 2021]
(a) Improve protein content
(b) Improve resistance to diseases
(c) Improve vitamin content
(d) Improve micronutrient and mineral
content
Ans.(b)
Biofortification is the method developed
to produce crops with high level of
vitamins, proteins and minerals to
improve public health.
Improving resistance to disease is not the
objective of biofortification hence option (b)
is correct.
25Match the List -I with List - II.
List-I List-II
A. Protoplast fusion 1. Totipotency
B. Plant tissue
culture
2. Pomato
C. Meristem culture 3. Somaclones
D. Micropropagation 4. Virus free
plants
Choose tho correct answer from
the options given below.
[NEET 2021]
A B C D
(a) 3 4 2 1
(b) 2 1 4 3
(c) 3 4 1 2
(d) 4 3 2 1
Ans.(b)
(A)-(2), (B)-(1), (C)-(4), (D)-(3)
Pomatois the potato and tomato hybrid.
Pomato can be formed by somatic
hybridisation.
The mechanism by which two separate
species of plant protoplasts fuse
together to form hybrid is known as
somatic hybridisation.
Strategies for Enhancement in Food Production 311
Plant Breeding
TOPIC 2

322. Totipotencyis the basis of tissue
culture. Cells from growing root tips or
shoot tips or any other growing part can
be grown in nutrient medium under
sterilised condition. Afterwards, these
cells divides and grow into a mass of
tissue called callus.
Meristem cultureis one of the most
widely used methods for virus
elimination from infected plants and
production of virus-free plants. Apical
meristem culture is a proven means of
clonal propagation and also for
eliminating viruses from infected plants.
The plants raised through
micropropagationare calledsomaclones
because they are genetically identical to
the original plant from which they were
grown. It helps in producing plants that
are disease and pest resistant.
26Mutations in plant cells can be
induced by [NEET 2021]
(a) kinetin
(b) infrared rays
(c) gamma rays
(d) zeatin
Ans.(c)
Mutation is defined as the process by
which genetic changes are createdvia
changes in the sequences of bases
present within genes. This results in the
formation of a new trait or character not
found in the parental type.
Thus, it is possible to induce mutations
with the help of various chemicals or
radiations such as gamma radiations
artificially and then selecting and using
plants with desirable traits as a source in
breeding. This process is termed as
mutational breeding.
27In mung bean, resistance to yellow
mosaic, virus and powdery mildew
were brought about by
[NEET (Odisha) 2019]
(a) mutation breeding
(b) biofortification
(c) tissue culture
(d) hybridisation and selection
Ans.(a)
In mung bean, resistance to yellow
mosaic virus and powdery mildew were
induced by mutation breeding. Mutation
breeding is the process of exposing
seeds to chemicals or radiation in order
to generate mutants with desirable traits
to be bred with other cultivars.
28Which of the following is true for
Golden rice? [NEET 2018]
(a) It is pest resistant, with a gene from
Bacillus thuringiensis
(b) It is drought tolerant, developed
usingAgrobacteriumvector
(c) It has yellow grains, because of a
gene introduced from a primitive
variety of rice
(d) It is vitamin-A enriched, with a gene
from daffodil
Ans.(d)
Statement that Golden rice is vitamin-A
enriched, with a gene from daffodil is
true. Golden rice is genetically
engineered variety of rice to
biosynthesiseβ-carotene which is a
precursor of vitamin-A. It containspsy
gene (phytoene synthase) which is derived
from daffodil. Other statements are not
true for golden rice. The correct
information about the statements is as
follows The grains of golden rice appear
yellow due to high level ofβ-carotene in
it. Golden rice is neither drought tolerant
nor pest resistant.
29What triggers activation of protoxin
to activeBttoxin ofBacillus
thuringiensisin bollworm?
[NEET (National) 2019]
(a) Moist surface of midgut
(b) Alkaline pH of gut
(c) Acidic pH of stomach
(d) Body temperature
Ans.(b)
Alkaline pH of gut triggers activation of
protoxin to activeBttoxin ofBacillus
thuringiensisin bollworm. The inactive
protoxins contain toxic insecticidal
protein crystals.
When the alkaline pH of insect gut
solubilises the crystals, the activated
toxin binds to the epithelial cells of the
midgut and creates pores. It causes the
cell to swell and burst, eventually
causing the death of insect.
30Select the incorrect statement.
[NEET (National) 2019]
(a) Inbreeding is essential to evolve
purelines in any animal
(b) Inbreeding selects harmful recessive
genes that reduce fertility and
productivity
(c) Inbreeding helps in accumulation of
superior genes and elimination of
undesirable genes
(d) Inbreeding increases homozygosity
Ans.(b)
Statement that inbreeding selects
harmful recessive genes that reduce
fertility and productivity is incorrect.
The correct information regarding the
statement is as follows
Inbreeding does not select harmful
recessive genes. It exposes harmful
recessive genes that are eliminated by
selection and thus reduces fertility and
productivity. This is called inbreeding
depression.
Statements in other options are correct.
31A system of rotating crops with
legume or grass pasture to improve
soil structure and fertility is called
[NEET 2016, Phase I]
(a) contour farming
(b) strip farming
(c) shifting agriculture
(d) ley farming
Ans.(d)
Ley farming is a system of rotating crops
with legumes or grass pasture in order to
improve soil structure and fertility and
also to disrupt pest and disease life
cycles.
32In plant breeding programmes, the
entire collection (of plants/seeds)
having all the diverse alleles for all
genes in a given crop is called
[NEET 2013]
(a) selection of superior recombinants
(b) cross-hybridisation among the
selected parents
(c) evaluation and selection of parents
(d) germplasm collection
Ans.(d)
It is called germplasm collection.
Selection of superior recombinants
means selection of the best plant from
the whole lot by visual examination and
collecting their seeds for growing in
field. Cross hybridisation is the method
of combining the characters of different
plants together. The selection and
evaluation are the main steps of
hybridisation.
33Consider the following four
statements (I-IV) and select the
option which includes all the
correct ones only.
I. Single cellSpirulinacan produce
large quantities of food rich in
protein, minerals, vitamins, etc.
312 NEETChapterwise Topicwise Biology

323. II. Body weight-wise the
microorganismMethylophilus
methylotrophusmay be able to
produce several times more
proteins than the cows per day.
III. Common button mushrooms are
a very rich source of vitamin-C.
IV. A rice variety has been
developed which is very rich in
calcium. [CBSE AIPMT 2012]
(a) Statements III and IV
(b) Statements I, III and IV
(c) Statements II, III and IV
(d) Statements I and II
Ans.(d)
Out of the given statements (I) and (II) are
correct as single cellSpirulinacan
produce large quantities of food rich in
protein, minerals, vitamins, etc. And
body weight-wise the microorganism
Methylophilus methy lotrophusmay be
able to produce several times more
proteins than cows per day.
34‘Jaya’ and ‘Ratna’ developed for
green revolution in India are the
varieties of[CBSE AIPMT 2011]
(a) maize (b) rice
(c) wheat (d) bajra
Ans.(b)
‘Jaya’ and ‘Ratna’ are better yielding
semi-dwarf varieties of rice developed in
India for green revolution.
35‘Himgiri’ developed by hybridisation
and selection for disease
resistance against rust pathogens
is a variety of[CBSE AIPMT 2011]
(a) chilli (b) maize
(c) sugarcane (d) wheat
Ans.(d)
‘Himgiri’ is a variety of wheat. It is
resistant to leaf and stripe rust, hill bunt
diseases.
36Consider the following statements
(I-IV) about organic farming.
[CBSE AIPMT 2011]
I. Utilises genetically modified
crops likeBtcotton.
II. Uses only naturally produced
inputs like compost.
III. Does not use pesticides and
urea.
IV. Produces vegetables rich in
vitamins and minerals.
Which of the above statements
are correct?
(a) II, III and IV
(b) III and IV
(c) II and III
(d) I and II
Ans.(c)
Out of the given statements (II) and (III)
are correct as organic farming is the
form of agriculture that relies on
techniques such as crop rotation, green
manure, compost and biological pest
control to maintain soil productivity and
control pest on a farm. Organic farming
excludes or strictly limits the use of
manufactured fertilisers, pesticides
(which include herbicides, insecticides
and fungicides), plant growth regulators
such as hormones, food additives and
genetically modified organisms.
37Breeding of crops with high levels
of minerals, vitamins and proteins
is called [CBSE AIPMT 2010]
(a) somatic hybridisation
(b) biofortification
(c) biomagnification
(d) micropropagation
Ans.(b)
Breeding of crops with higher levels of
vitamins and minerals or higher protein
and healthier fats is called
biofortification. This is the most
practical aspect to improve the health of
the people.
38Which one of the following is linked
to the discovery of Bordeaux
mixture as a popular fungicide?
[CBSE AIPMT 2008]
(a) Loose smut of wheat
(b) Black rust of wheat
(c) Bacterial leaf blight of rice
(d) Downy mildew of grapes
Ans.(d)
Bordeaux mixture is the first inorganic
fungicide which was developed by RMA
Millardet (1882) against downy mildew
(Plasmopara viticola) of grape-vine at the
University of Bordeaux. It consists of
copper (II) sulphate(CuSO )
4
slaked lime
and water.
39Consider the following four
measures (1-4) that could be taken
to successfully grow chickpea in an
area where bacterial blight disease
is common
I. spray with Bordeaux mixture.
II. control of the insect vector of
the disease pathogen.
III. use of only disease-free seeds.
IV. use of varieties resistant to the
disease.
Which two of the above measures
can control the disease?
[CBSE AIPMT 2008]
(a) III and IV (b) I and IV
(c) II and III (d) I and II
Ans.(b)
Bacterial blight of chickpea is caused by
bacteriumXanthomonas campestris. The
stems and the leaves of infected plant
give blighted or burnt up appearance.
Control measures includes roguing,
3-year crop rotation, disease-free
seeds, spray of copper fungicides
(Bordeaux mixture) and antibiotics
besides sowing disease resistant
varieties.
40Farmers in a particular region were
concerned that pre-mature
yellowing of leaves of a pulse crop
might cause decrease in the yield.
Which treatment could be most
beneficial to obtain maximum seed
yield? [CBSE AIPMT 2006]
(a) Frequent irrigation of the crop
(b) Treatment of the plants with
cytokinins along with a small dose of
nitrogenous fertiliser
(c) Removal of all yellow leaves and
spraying the remaining green leaves
with 2,4,5-trichlorophenoxy acetic
acid
(d) Application of iron and magnesium to
promote synthesis of chlorophyll
Ans.(d)
If a pulse crop possesses premature
yellowing of leaves and decrease in yield
an application ofmagnesiumandironto
promote synthesis of chlorophyll may
become most beneficial to overcome
the problem and to obtain maximum
seed yield.
Magnesiumis an important part of ring
structure of chlorophyll molecule and its
deficiency causes chlorosis and
premature leaf abscission.
Inirondeficiency also, the leaves
become chlorotic because iron is
required for the synthesis of some of the
chlorophyll protein complexes in the
chloroplast.
Strategies for Enhancement in Food Production 313

324. 41In maize, hybrid vigour is exploited
by [CBSE AIPMT 2006]
(a) inducing mutations
(b) bombarding the protoplast with DNA
(c) crossing of two inbred parental lines
(d) harvesting seeds from the most
productive plants
Ans.(c)
Hybrid vigour has been commercially
exploited in different commercial crops
like maize, sorghum, bajra, tomato,
sugarbeet. Hybridisation or crossing of
two unrelated individuals or parental
lines leads to hybrid vigour or heterosis.
It refers to the superiority of the hybrid
over its parents. The changes in the
progeny or hybrid can be seen with
naked eyes. The main steps include, i.e.
selection of parents, selfing of parents,
emasculation, bagging, crossing of
desired and selected parents and finally
seed setting and harvesting.
42Golden rice is a transgenic crop of
the future with the following
improved trait
[CBSE AIPMT 2006, 05]
(a) high lysine (essential amino acid)
content
(b) insect resistance
(c) high protein content
(d) high vitamin-A content
Ans.(d)
Generally seeds of rice do not have
vitamin-A, but golden rice which is
developed through genetic engineering
has the high vitamin-A content.
43Crop plants grown in monoculture
are [CBSE AIPMT 2006]
(a) low in yield
(b) free from intraspecific competition
(c) characterised by poor root system
(d) highly prone to pests
Ans.(d)
The crop plants grown in monoculture
are highly prone to pests and thus, it
carries the risk of an entire crop being
destroyed with the appearance of a
single pest species or disease.
44Triticale,the first man-made cereal
crop, has been obtained by
crossing wheat with
[CBSE AIPMT 2006]
(a) rye (b) pearl millet
(c) sugarcane (d) barley
Ans.(a)
Triticaleis the first man-made cereal
crop. It has been obtained by crossing
wheat (Triticumsp.) with rye(Secale
cerele).
45Three crops that contribute
maximum to global food grain
production are[CBSE AIPMT 2005]
(a) wheat, rice and maize
(b) wheat, maize and sorghum
(c) rice, maize and sorghum
(d) wheat, rice and barley
Ans.(a)
Wheat, rice and maize belong to family–
Poaceae or Gramineae. The main fruit
type of these crops is caryopsis in which
fruit wall is fused with seed coat. These
crops are cultivated in all over the world
and are contributed maximum in global
food grain production.
46The name of Norman Borlaug is
associated with[CBSE AIPMT 2005]
(a) Green revolution
(b) Yellow revolution
(c) White revolution
(d) Blue revolution
Ans.(a)
Norman Borlaug is associated with
green revolution. The green revolution
means an increase in the production of
crops particularly cereals, such as
wheat, rice and maize.
47Why is vivipary an undesirable
character for annual crop plants?
[CBSE AIPMT 2005]
(a) It reduces the vigour of plant
(b) The seeds cannot be stored under
normal conditions for the next season
(c) The seeds exhibit long dormancy
(d) It adversely affects the fertility of the
plant
Ans.(b)
Vivipary is the condition when seeds are
germinated on the plant. It is an
undesirable character for annual crop
plants because germinated seeds
cannot be stored under normal
conditions for the next season.
48Which of the following is generally
used for induced mutagenesis in
crop plants?[CBSE AIPMT 2005]
(a) X-rays
(b) UV (260 nm)
(c) Gamma rays (from cobalt 60)
(d) Alpha particles
Ans.(c)
Cobalt 60 is the synthetic radioactive
isotope of cobalt. Gamma rays are
produced when an unstable atomic
nucleus like cobalt-60 releases energy to
gain stability. Sharbati Sonora and Pusa
Lerma are the two important varieties of
wheat that are produced by gamma rays
treatment of Sonora-64 and Lerma
Rojo-64 which are Mexican dwarf wheat
varieties.
49India’s wheat yield revolution in the
1960s was possible primarily due to
[CBSE AIPMT 2004]
(a) hybrid seeds
(b) increased chlorophyll content
(c) mutations resulting in plant height
reduction
(d) quantitative trait mutations
Ans.(c)
India’s wheat yield revolution in 1960s
was possible primarily due to the
mutations resulting in plant height
reduction. In 1963, ICAR introduced many
dwarf selections from CIMMYT, including
those developed by Norman Borlaug
using Norin-10 as the source of dwarfing
genes.
50Which of the following crops have
been brought to India from New
world? [CBSE AIPMT 2002]
(a) Cashewnut, potato, rubber
(b) Mango, tea
(c) Tea, rubber, mango
(d) Coffee
Ans.(a)
Cashewnut, potato, rubber are crops
that have been brought to India from
New World.
51What is the best pH of the soil for
cultivation of plants?
[CBSE AIPMT 2001]
(a) 3.4–5.4 (b) 6.5–7.5
(c) 4.5–8.5 (d) 5.5–6.5
314 NEETChapterwise Topicwise Biology
Triticum turgidum × Secale cerele
AA BB RR
AB R
ABR
AA BB RR
Triticale
(allohexaploid)
Colchicine
(chromosome doubling)
Gametes

325. Ans.(c)
Most cultivated soils have pH ranges
between 4.5-8.5. However most plants
grow best in soils with a neutral or
slightly acidic pH.
52Before the European invaders
which vegetable was/were absent
in India? [CBSE AIPMT 2001]
(a) Potato and tomato
(b) Simla mirch and brinjal
(c) Maize and chichinda
(d) Bitter gourd
Ans.(a)
Potato and tomato originated in the
New World. They were absent in India
previously.
53Which statement is correct about
centre of origin of plants?
[CBSE AIPMT 2001]
(a) More diversity in varieties
(b) Frequency of dominant gene is more
(c) Climatic conditions more favourable
(d) None of the above
Ans.(a)
The centres of origin of plants are more
appropriately called the centres of
diversity. These are the areas of
maximum diversity of these species.
54One of the most important reason
why wild plants should thrive is that
these are good sources of
[CBSE AIPMT 2000]
(a) unsaturated edible oils
(b) highly nutritive animals feed
(c) genes for resistance to diseases and
pests
(d) rare and highly sought after fruits of
medical importance
Ans.(c)
Wild plants have to survive without
getting any protection and for this, they
evolve various strategies/characters
which are exploited by plant breeders
like diseases resistance.
55The new varieties of plants are
produced by [CBSE AIPMT 1999]
(a) selection and hybridisation
(b) selection and introduction
(c) mutation and selection
(d) introduction and mutation
Ans.(a)
The new varieties of plants are produced
by selection and hybridisation. In
hybridisation, two or more plants of
unlike genotype are crossed together to
get offsprings with new desirable
combinations of characters, as a result
of genetic recombination.
56The reason why vegetatively
reproducing crop plants are best
suited for maintaining hybrid vigour
is that [CBSE AIPMT 1998]
(a) they can be easily propagated
(b) they have a longer life span
(c) they are more resistant to disease
(d) once a desired hybrid is produced,
there are no chances of losing it
Ans.(d)
Vegetative reproduction is the process
of multiplication in a small part or
portion of the plant body which
functions as a propagule and develops
into a new individual.
Thus, vegetative reproduction does not
involve meiosis; hence, recombination
and no loss of heterozygosity.
57Which one among the following
chemicals is used for causing
defoliation of forest trees?
[CBSE AIPMT 1998]
(a) Amo-1618
(b) Phosphon-D
(c) Malic hydrazide
(d) 2, 4-D
Ans.(d)
2, 4-D (2, 4 dichlorophenoxy acetic acid)
is an auxin hormone. It over stimulates
the growth activities of the cells of the
root due to which roots get destroyed
and thus plants finally destroyed. 2, 4-D
is used as a defoliant for broad leaves
dicots (mainly weeds).
58Which plant will loss its economic
value, if its fruits are produced by
induced parthenocarpy?
[CBSE AIPMT 1997]
(a) Grape
(b) Pomegranate
(c) Orange
(d) Banana
Ans.(b)
Testa is the edible part in pomegranate.
It is not formed if fruits are produced by
parthenocarpy (no seeds will be formed).
59Of the world’s top five crops (in
terms of annual production)
[CBSE AIPMT 1997]
(a) three belong to Poaceae (Gramineae),
one to Leguminosae, one to
Solanaceae
(b) four belong to Poaceae, one to
Leguminosae
(c) four belong to Poaceae, one to
Solanaceae
(d) All five belong to Poaceae
Ans.(c)
Top five crops of today are
wheat-Triticum aestivum(Poaceae),
corn–Zea mays(Poaceae), rice–Oryza
sativa(Poaceae), potato–Solanum
tuberosum(Solanaceae) and
barley–Hordeum vulgare(Poaceae).
60Most of our crop plants are
[CBSE AIPMT 1994]
(a) autopolyploid in origin
(b) allopolyploid in origin
(c) mixed genotypic in origin
(d) heterozygous in origin
Ans.(a)
Most of our crop plants are
autopolyploid in origin.
61Haploid plants are preferred over
diploids for mutation study
because in haploids
[CBSE AIPMT 1993]
(a) recessive mutation express
immediately
(b) induction of mutations is easier
(c) culturing is easier
(d) dominant mutation express
immediately
Ans.(a)
Haploid plants are preferred over
diploids for mutation study because of
their clear/obvious expression. In nature
mutations are generally recessive. In
case of a diploid it is difficult to trace out
the recessive mutation as its dominant
gene is present on other chromosome.
62In crop improvement programme,
haploids are important because
they [CBSE AIPMT 1989]
(a) require one half of nutrients
(b) are helpful in study of meiosis
(c) grow better under adverse conditions
(d) form perfect homozygous
Ans.(d)
Haploids are important in crop
improvement programme because they
produce a pureline and form perfect
homozygous.
Strategies for Enhancement in Food Production 315

326. 63To obtain virus-free healthy plants
from a diseased one by tissue
culture technique, which part/parts
of the diseased plant will be taken?
[CBSE AIPMT 2014]
(a) Apical meristem only
(b) Palisade parenchyma
(c) Both apical and axillary meristems
(d) Epidermis only
Ans.(c)
Both apical and axillary meristems are
free of virus for the healthy plant
cultivation because of strong interferon
activity in this region.
These tissues form a protective
impermeable covering around
themselves, which is non-penetrable by
any pathogen. Hence, these tissue are
used in the production of disease free
plants by tissue culture.
64Somaclones are obtained by
[CBSE AIPMT 2009]
(a) plant breeding
(b) irradiation
(c) genetic engineering
(d) tissue culture
Ans.(d)
Somaclones are obtained bytissue
culture. The plant regenerated from cell
and tissue cultures shows heritable
variation for both qualitative and
quantitative traits.Plant breedingis the
branch of biology, which is concerned
with developing varieties superior to
existing ones.
Irradiation means exposure to any form
of radiation.
Genetic engineeringis defined as the
manipulation of genes by man.
65In order to obtain virus-free plants
through tissue culture the best
method is [CBSE AIPMT 2006]
(a) meristem culture
(b) protoplast culture
(c) embryo rescue
(d) anther culture
Ans.(a)
Virus free plants can be developed by
using meristem as explant in tissue
culture. In infected plants, virus
concentration decreases as they
approach the apical meristem.
It is because the cells of apical meristem
undergo rapid mitotic divisions and virus
cannot divide so fast and thus it cannot
keep pace with the fast dividing
meristematic cells.
So, apical meristems are generally virus
free or they contain very low
concentration of virus. Before meristem
culture, viruses associated with
meristem are eliminated byin vitroheat
treatment (thermotherapy).Moreland
Maetin(1952) successfully obtained virus
freeDahliaplants through meristem
culture of infected plants.
66Haploid plant cultures are got from
[CBSE AIPMT 1994]
(a) leaves
(b) root tip
(c) pollen grain
(d) buds
Ans.(c)
Haploid plant cultures are obtained from
pollen grains as they are haploid while
leaves, root tip and buds are diploid.
316 NEETChapterwise Topicwise Biology
Plant Tissue Culture
TOPIC 3

327. 01Yeast is used in the production of
[CBSE AIPMT 2012]
(a) citric acid and lactic acid
(b) lipase and pectinase
(c) bread and beer
(d) cheese and butter
Ans.(c)
Saccharomyces cerevisiae is known as
baker’s yeastandSaccharomyces
ellipsoidensis calledwine yeast. These
are used in baking and brewing industry
respectively.
02Match the List-I with List-II.
[NEET 2021]
List-I List-II
A.Aspergillus niger1. Acetic Acid
B.Acetobacter aceti2. Lactic Acid
C.Clostridium
butylicum
3. Citric Acid
D.Lactobacillus 4. Butyric Acid
Choose the correct answer from
the options given below.
A B C D A B C D
(a) 3 1 4 2 (b) 1 2 3 4
(c) 2 3 1 4 (d) 4 2 1 3
Ans.(a)
(A)-(3) ,(B)-(1),(C)-(4),(D)-(2)
Citric acid is a principal organic acid
present in citrus fruits. To meet the
increasing demand, it is produced from
carbohydrate feedstock by fermentation
with the fungusAspergillus niger.
Acetobacter acetiuses sugars and
alcohols for its carbon source and turns
them into their acetic acid.
Clostridium butyricumare Gram-positive
bacteria that helps in the production of
butyric acid.
Lactobacillusis a genus of
Gram-positive, facultative an aerobic,
rod-shaped, non-spore forming
bacteria. They convert sugars to lactic
acid.
03Cyclosporin*A used as
immunosuppression agent, is
produced from[NEET (Oct.) 2020]
(a)Monascus purpureus
(b)Saccharomyces cerevisiae
(c)Penicillium notatum
(d)Trichoderma polysporum
Ans.(d)
Bioactive molecule, cyclosporin-A is
used as an immunosuppressive agent in
organ transplant patients. It is produced
by the fungusTrichoderma polysporum.
Monascus purpureusis a yeast which
helps to produce statins (blood
cholesterol lowering agent).
Saccharomyces cerevisaeis a yeast
which produces ethanol.Penicillium
notatumis a yeast which produces
pencillin (an antibiotics).
04Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A.Clostridium
butylicum
1. Cyclosporin-A
B.Trichoderma
polysporum
2. Butyric acid
C.Monascus
purpureus
3. Citric acid
D.Aspergillus niger4. Blood
cholesterol
lowering
agent
A B C D
(a) 2 1 4 3
(b) 1 2 4 3
(c) 4 3 2 1
(d) 3 4 2 1
Ans.(a)
The correct match is option (a). It can be
explained as follows
Butyric acid is produced byClostridium
butylicum. It is a strictly anaerobic
endosporeforming Gram-positive
bacteria.
Cyclosporin A is produced by the fungus
Trichoderma polysporumthat is used as
an immunosuppressive agent in organ
transplant patients.
Statins produced by the yeastMonascus
purpureushave been commercialised as
blood cholesterol lowering agent.
Citric acid is produced by fungus
Aspergillus niger.
Microbesin
HumanWalfare
32
Microbes in
Household Products
TOPIC 1
Microbes in
Industrial Products
TOPIC 2

328. 05Match the following organisms with
the products they produce
[NEET (National) 2019]
A.Lactobacillus(i) Cheese
B.Saccharomyces
cerevisiae
(ii) Curd
C.Aspergillus niger(iii) Citric acid
D.Acetobacter aceti(iv) Bread
(v) Acetic acid
Select the correct option.
A B C D
(a) (ii) (iv) (iii) (v)
(b) (iii) (iv) (v) (i)
(c) (ii) (i) (iii) (v)
(d) (ii) (iv) (v) (iii)
Ans.(a)
(A)–(ii), (B)–(iv), (C)–(iii), (D)–(v)
Lactobacillusbacteria helps in the
production of curd. The yeast,
Saccharomyces cerevisiaehelps in bread
making. The fungus,Aspergillus Nigeris
used for citric acid production. The
bacteriaAcetobacter acetiis used in the
production of acetic acid.
06Which of the following is correctly
matched for the product produced
by them? [NEET 2017]
(a)Acetobacter aceti: Antibiotics
(b)Methanobacterium: Lactic acid
(c)Penicillium notatum: Acetic acid
(d)Saccharomyces cerevisiae: Ethanol
Ans.(d)
Acetobacter acetiproduces acetic acid.
Methanobacteriumproduces methane.
Penicillium notatumproduces penicillin.
Saccharomyces cerevsiaeproduces
ethanol.
07Which of the following is wrongly
matched in the given table?
[NEET 2016, Phase I]
Microbe Product Application
(a)Monascus
purpureus
Statins Lowering of
blood
cholesterol
(b)Streptoco-
ccus
Streptoki-
nase
Removal of
clot from
blood vessel
(c)Clostridium
butylicum
Lipase Removal of oil
stains
(d)Trichoderma
polysporum
Cyclosporin–AImmunosuppr
ess-ive drug
Ans.(c)
Butyric acid is produced by fermentive
activity of the bacteria calledClostridium
butylicum.It does not produce lipase.
Lipase is obtained fromCandida
albicans.
08Match column I with column II and
select the correct option using the
codes given below
[NEET 2016, Phase II]
Column I Column II
A. Citric acid 1.Trichoderma
B. Cyclosporin 2.Clostridium
C. Statins 3. Aspergillus
D. Butyric acid 4.Monascus
Codes
A B C D
(a) 3 1 2 4
(b) 3 1 4 2
(c) 1 4 2 3
(d) 3 4 1 2
Ans.(b)
The correct match are
(a) Citric acid —Aspergillus
(b) Cyclosporin —Trichoderma
(c) Statins — Monascus
(d) Butyric acid —Clostridium
Thus option (b) is correct.
09Match the following list of microbes
and their importance.
[CBSE AIPMT 2015]
A.Saccharomyces
cerevisiae
1. Production of
immuno
suppressive agents
B.Monascus
purpureus
2. Ripening of Swiss
cheese
C.Trichoderma
polysporum
3. Commercial
production of ethanol
D.Propionibacteriu
m sharmanii
4. Production of
blood-cholesterol
lowering agents
Codes
A B C D
(a) 3 4 1 2
(b) 4 3 2 1
(c) 4 2 1 3
(d) 3 1 4 2
Ans.(a)
Correct match is
Column I Column II
A.Saccharomyces
cerevisae
3. Commercial
production of
ethanol
B.Monascus
purpureus
4. Production of
blood-cholesterol
lowering agents
C.Trichoderma
polysporum
1. Production of
immuno
suppressive
agents
D.Propionibacteriu
m shermanii
2.Ripening of Swiss
cheese
10The most abundant prokaryotes
helpful to human in making curd
from milk and in production of
antibiotics are the ones
categorised as[CBSE AIPMT 2012]
(a) cyanobacteria
(b) archaebacteria
(c) chemosynthetic autotrophs
(d) heterotrophic bacteria
Ans.(d)
Heterotrophic bacteriaare most
abundant in nature. Many of them have a
significant impact on human affairs.
These are helpful in making curd from
milk (e.g.Lactobacillussp.), production
of antibiotics (e.g.Streptomycessp.) and
fixing nitrogen in legume roots (e.g.
Rhizobiumsp).
11Monascus purpureusis a yeast used
commercially in the production of
[CBSE AIPMT 2012]
(a) ethanol
(b) streptokinase for removing clots
from the blood vessels
(c) citric acid
(d) blood cholesterol lowering statins
Ans.(d)
Monascus purpureus is a yeast used in
the production of statins which are
blood cholesterol lowering agents.
12A patient brought to a hospital with
myocardial infarction is normally
immediately given
[CBSE AIPMT 2012]
(a) penicillin
(b) streptokinase
(c) cyclosporin-A
(d) statins
Ans.(b)
Streptokinase (SK), a protein secreted by
several species ofStreptococcican bind
and activate human plasminogen. It is
used as an effective and inexpensive
thrombolysis medicationin some cases
ofmyocardial infarctionand pulmonary
embolism.
318 NEETChapterwise Topicwise Biology

329. 13Ethanol is commercially produced
through a particular species of
[CBSE AIPMT 2011]
(a)Clostridium(b)Trichoderma
(c)Aspergillus(d)Saccharomyces
Ans.(d)
Ethanol is commercially produced
through a particular species of yeast
called asSaccharomyces cerevisiae.
14The most common substrate used
in distilleries for the production of
ethanol is [CBSE AIPMT 2011]
(a) soya meal (b) ground gram
(c) molasses (d) corn meal
Ans.(c)
Molasses is a viscous byproduct of the
processing of sugarcane, grapes or
sugarbeets in sugar. It is the most
common substrate used in distilleries for
the production of ethanol. It can be used
as the base material for fermentation
into rum. In Australia, molasses is
fermented to produce ethanol for use as
an alternative fuel in motor vehicles.
15Which of the microrganism is used
for production of citric acid in
industries? [CBSE AIPMT 1998]
(a)Lactobacillus bulgaris
(b)Penicillium citrinum
(c)Aspergillus niger
(d)Rhizopus nigricans
Ans.(c)
Aspergillus nigeris the microorganism
used for production of citric acid in
industries. Citric acid has an
extraordinary range of uses.
It gives tartness and flavour to the
foods. It is an antioxidant and pH
adjuster in many foods and dairy
products, it often serves as an
emulsifier.
16Which one thing is not true about
antibiotics?[CBSE AIPMT 1996]
(a) The term ‘antibiotic’ was coined by
Selman Waksman in 1942
(b) first antibiotic was discovered by
Alexander Flemming
(c) Each antibiotic is effective only
against one particular kind of germ
(d) Some persons can be allergic to a
particular antibiotic
Ans.(c)
The statement (c) is wrong regarding to
antibiotics because antibiotics are
divided into two categories depending
upon their effect
(i)Broad spectrum antibioticsThey
have ability to act on several
pathogenic species differing from
each others in structure and
composition of cell wall.
(ii)Specific antibioticsThey act on a
few similar type of pathogens.
17Yeast (Saccharomyces cerevisiae)
is used in the industrial production
of [CBSE AIPMT 1998]
(a) citric acid (b) tetracycline
(c) ethanol (d) butanol
Ans.(c)
Yeast contains an enzyme zymase which
catalyse the fermentation of sugar to
form ethyl alcohol (ethanol) andCO
2
.
18Which one of the following
microorganisms is used for
production of citric acid in
industries? [CBSE AIPMT 1998]
(a)Penicillium citrinum
(b)Aspergillus niger
(c)Rhizopus nigricans
(d)Lactobacillus bulgaricus
Ans.(b)
Citric acid is commercially prepared is by
fermentation of sugar withA. niger.
Citric acid has an extraordinary range of
uses. It gives tartness and flavour to the
foods. It is an antioxidant and pH adjuster
in many foods and dairy products, it often
serves as an emulsifier.
19Match the following columns and
select the correct option from the
codes given below.
[NEET (Oct.) 2020]
Column I Column II
A. Dragonflies 1. Biocontrol agents of
several plant
pathogens
B.Bacillus
thuringiensis
2. Get rid of Aphids and
mosquitoes
C.Glomus 3. Narrow spectrum
Insecticidal
applications
D. Baculoviruses 4. Biocontrol agents of
lepidopteran plant
pests
5. Absorb phosphorus
from soil
Codes
A B C D
(a) 3 5 4 1
(b) 2 1 3 4
(c) 2 3 4 5
(d) 2 4 5 3
Ans.(d)
Option (d) is the correct match which is
as follows
Dragonflies help to get rid of Aphids and
mosquitoes.
Baccillus thuringiensisacts as biocontrol
agent for lepidopteran and plant pests.
Glomusis a fungus which forms
mycorrhiza to absorb phosphorus from
soil.
Baculoviruses are used in narrow
spectrum insecticidal applications.
20A biocontrol agent to be a part of
an integrated pest management
should be [NEET (Odisha) 2019]
(a) species-specific and symbiotic
(b) free-living and broad spectrum
(c) narrow spectrum and symbiotic
(d) species-specific and inactive on
non-target organisms
Ans.(d)
A biocontrol agent to be a part of an
Integrated Pest Management (IPM)
programme should be species-specific
and inactive or have no negative impacts
on non-target organisms like plants,
mammals, birds, fish and even on other
non-target insects. It should kill only
targeted insects/pests (organisms).
21Which of the following can be used
as a biocontrol agent in the
treatment of plant disease?
[NEET (National) 2019]
(a)Chlorella (b)Anabaena
(c)Lactobacillus (d) Trichoderma
Ans.(d)
Trichodermacan be used as a biocontrol
agent in the treatment of plant disease.
It is a filamentous soil fungus having
mycoparasitic activity. On the other
hand,Anabaenahelps in
nitrogen-fixation,Lactobacillushelps in
the production of organic acid, e.g. lactic
acid andChlorellais a single cell protein
which acts as food supplement.
22Select the correct group of
biocontrol agents.
[NEET (National) 2019]
(a)Trichoderma, Baculovirus, Bacillus
thuringiensis
(b)Oscillatoria, Rhizobium, Trichoderma
Microbes in Human Welfare 319
Microbes as
Biocontrol Agents
TOPIC 3

330. (c)Nostoc, Azospirillum,
Nucleopolyhedrovirus
(d)Bacillus thuringiensis,Tobacco
mosaic virus, Aphids
Ans.(a)
The correct group of biocontrol agents is
Trichoderma,BaculovirusandBacillus
thuringiensis.Baculovirusare pathogens
that attack insects and other
arthropods. Most ofBaculovirusesused
as biocontrol agent belong to the genus
Nucleopolyhedrovirus.
Trichodermais extensively used against
pathogenic fungi which causes soil
borne diseases.
Bacillus thuringiensissecretes toxin
crystals which kill the insect larvae. On
the other hand,Rhizobium,Nostoc,
AzospirillumandOscillatoriaare used as
biofertilisers. Tobacco mosaic virus is a
pathogen and aphids are pests that harm
crop plants.
23Which one of the following is an
example of carrying out biological
control of pests/diseases using
microbes? [CBSE AIPMT 2012]
(a)Trichodermasp. against certain plant
pathogens
(b) Nucleopolyhedrovirus against white
rust inBrassica
(c)Btcotton to increase cotton yield
(d) Lady bird beetle against aphids in
mustard
Ans.(c)
Out of the given statement ‘c’ is correct
example becauseBtcotton is
Genetically Modified (GM) cotton which
has an incorporated gene extracted
from the bacteriumBacillus thuringiensis.
This gene codes forBttoxin in plant
tissues which is harmful only to a small
fraction of insects, most notably the
larvae of lepidopterans, moths,
butterflies, beetles, flies, etc, and
harmless to other forms of life. So, it is
used as biological control of
pests/diseases.
24A common biocontrol agent for the
control of plant diseases is
[CBSE AIPMT 2010]
(a) Baculovirus
(b)Bacillus thuringiensis
(c)Glomus
(d)Trichoderma
Ans.(d)
Trichoderma is a genus of fungi that is
present in all soils. Several strains of
Trichodermahave been developed as
biocontrol agents against fungal
diseases of plants. The various
mechanisms include antibiosis,
parasitism, inducing host-plant
resistance and competition.
Most biocontrol agents are from the
speciesT. harzianum, T. virideand
T. hamatum.The biocontrol agent
generally grows in its natural habitat on
the root surface and so, affects root
disease in particular but can also be
effective against foliar diseases.
25The bacteriumBacillus thuringiensis
is widely used in contemporary
biology as a/an
[CBSE AIPMT 2009]
(a) indicator of water pollution
(b) insecticide
(c) agent for production of dairy products
(d) source of industrial enzyme
Ans.(b)
Bacillus thuringiensisis used as an
insecticide. It is a Gram-positive, soil
dwelling bacterium, also occurs naturally
in the gut of caterpillars of various types
of moths and butterflies.
During sporulation,B. thuringiensis
forms crystals of proteinaceous
insecticidalδ-endotoxins (crytoxins),
which are encoded bycrygenes.
It was determined that thecrygenes are
harbored in the plasmids ofB.
thuringiensisstrains.Crytoxins have
specific activities against species of the
order–Lepidoptera (moths and
butterflies), Diptera (flies and
mosquitoes) and Coleoptera (beetles).
Thus,B. thuringiensisserves as an
important reservoir ofcrytoxins andcry
genes for the production of biological
insecticides and insect resistant
genetically modified crops.
26What is true aboutBttoxin?
[CBSE AIPMT 2009]
(a) The inactive protoxin gets converted
into active form in the insect gut
(b)Btprotein exists as active toxin in the
Bacillus
(c) The activated toxin enters the ovaries
of the pest to sterilise it and thus,
prevent its multiplication
(d) The concernedBacillushas antitoxins
Ans.(a)
Bacillus thuringiensistoxin is an inactive
protoxin, which gets converted into
active form in the insect gut. It works as
an insecticide.
27Which of the following is not used
as a biopesticide?
[CBSE AIPMT 2009]
(a)Bacillus thuringiensis
(b)Trichoderma harzianum
(c) Nuclear Polyhedrosis Virus (NPV)
(d)Xanthomonas campestris
Ans.(d)
The bacteriumXanthomonas campestris
is the causative agent of plant disease,
black rot of cabbage.
Bacillus thuringiensis,T. harzianumand
NPV are biopesticides.
28Main objective of production/use of
herbicide resistant GM crops is to
[CBSE AIPMT 2008]
(a) eliminate weeds from the field
without the use of manual labour
(b) eliminate weeds from the field
without the use of herbicides
(c) encourage eco-friendly herbicides
(d) reduce herbicide accumulation in
food particles for health safety
Ans.(d)
The main objective of production/use of
herbicide resistant Genetically Modified
(GM) crops is to reduce herbicide
accumulation in food articles for health
safety.
The chemical substances, which are
used to kill or repel pest are called
pesticides. The chemical substance
which are used to destroy weeds are
called herbicides.
29Cry-Iendotoxins obtained from
Bacillus thuringiensisare effective
against [CBSE AIPMT 2008]
(a) mosquitoes (b) flies
(c) nematodes (d) bollworms
Ans.(b)
Cry-I endotoxins obtained fromBacillus
thuringiensisare effective against flies
(insects). Thecrygene ofBacillus
thuringiensisproduces a protein, which
forms crystalline inclusion in the
bacterial spores. These crystal proteins
are responsible for the insecticidal
activities of the bacterial strains.
30Which one of the following proved
effective for biological control of
nemato diseases in plants?
[CBSE AIPMT 2008]
(a)Pisolithus tinctorius
(b)Pseudomyces lilacinus
(c)Gliocladium virens
(d)Paecilomyces lilacinus
320 NEETChapterwise Topicwise Biology

331. Ans.(d)
Paecilomyces lilacinus has proved
effective for biological control of
nematodal disease in plants. It is easily
producedin vitro. It attack the eggs of
several nematode species and highly
effective treatment of plant matter, e.g.
seed tuber.
31Which one of the following proved
effective for biological control of
nematode diseases in plants?
[CBSE AIPMT 2008]
(a)Glicoladium virens
(b)Paecilomces lalacinus
(c)Pisolithus tinctorius
(d)Pseudomonas cepacia
Ans.(b)
Paecilomyces lilacinusis a fungus which
principally infects eggs of root knot
nematode (Meloidogyne sp.) and cyst
nematodes (GoboderaandHeterodera sp.)
It has been considered to have greatest
potential for application as a biocontrol
agent in sub-tropical and tropical
agricultural soils.
32The most likely reason for the
development of resistance against
pesticides in insect damaging a
crop is [CBSE AIPMT 2004]
(a) random mutations
(b) genetic recombinations
(c) directed mutations
(d) acquired heritable changes
Ans.(a)
The most likely reason for the
development of resistance against
pesticides in insect damaging a crop is
random mutations, because
environmental stress, i.e. pesticides
does not cause direct changes in
genome, instead, it simply selects rather
persisting mutations which result in
phenotypes that are better adapted to
the new environment in certain
pesticides.
33Biological control component is
central to advanced agricultural
production. Which of the following
is used as a third generation
pesticide? [CBSE AIPMT 1998]
(a) Pathogens
(b) Pheromones
(c) Insect repellents
(d) Insect hormone analogues
Ans.(b)
Insect hormones, i.e. pheromones, are
third generation pesticides.
Pheromones are the chemical
substances which when released into an
animal’s surroundings, influence the
behaviour or development of other
individuals of the same species.
Inorganic substances, oils, plant
extracts used as insecticides are called
first generation pesticides and synthetic
organic compounds as second
generation pesticides.
34What is agent orange?
[CBSE AIPMT 1998]
(a) A biodegradable insecticide
(b) A weedicide containing dioxin
(c) Colour used in fluorescent lamp
(d) A hazardous chemical used in
luminous paints
Ans.(b)
Agent orange is a weedicide containing
dioxin. It is so, called because of
distinctive orange stripe on its
packaging, combines equal parts of 2,
4-D and 2, 4, 5-T was later on found to
contain a highly poisonous chemical
dioxin as impurity.
35Suppression of reproduction of one
type of organism by utilising some
features of its biology or physiology
to destroy it or by use of another
organism is known as
[CBSE AIPMT 1996]
(a) competition
(b) predation
(c) biological control
(d) physiological control
Ans.(c)
Biological control is the suppression of
reproduction of one type of organism by
utilising some features of its biology or
physiology to destroy it or by use of
another organism.
36One of the major difficulties in the
biological control of insect pests is
the [CBSE AIPMT 1995]
(a) practical difficulty of introducing the
predator to specific areas
(b) method is less effective as compared
with the use of insecticides
(c) predator does not always survive
when transferred to a new
environment
(d) the predator develops a preference to
other diets and may itself become a
pest
Ans.(d)
The major difficulties in the biological
control of insects pests is that the
predator develops a preference to other
diets and may itself become a pest.
Biological control is mainly refers to the
introduction of living organisms which
destroy other harmful organisms.
37Which of the following in sewage
treatment removes suspended
solids ? [NEET 2017]
(a) Tertiary treatment
(b) Secondary treatment
(c) Primary treatment
(d) Sludge treatment
Ans.(c)
In sewage treatment, suspended solids
are removed during primary treatment. It
is also known as physical treatment.
It consists of shredding, churning,
screening and sedimentation.
Sequential filtration removes floating
and large suspended solids.
38Biochemical Oxygen Demand (BOD)
may not be a good index for
pollution in water bodies receiving
effluents from[NEET 2016, Phase II]
(a) domestic sewage
(b) dairy industry
(c) petroleum industry
(d) sugar industry
Ans.(c)
Biochemical Oxygen Demand (BOD) is not
a good index for checking the pollution
levels of water bodies receiving
effluents from petroleum industry. This
is because such effluents contain waste
which is non-biodegradable.
39What gases are produced in
anaerobic sludge digesters?
[CBSE AIPMT 2014]
(a) Methane andCO
2
only
(b) Methane, hydrogen sulphide andCO
2
(c) Methane, hydrogen sulphide andCO
2
(d) Hydrogen sulphide andCO
2
Ans.(b)
Methane, hydrogen sulphide andCO
2
are
gases that are produced in anaerobic
sludge digesters. These gases are
Microbes in Human Welfare 321
Microbes in Biofuels
and Sewage Treatment
TOPIC 4

332. produced during biogas production by
the activity of bacteria called
methanogens.
40During sewage treatment, biogases
are produced, which include
[NEET 2013]
(a) methane, hydrogen sulphide and
carbon dioxide
(b) methane, oxygen and hydrogen
sulphide
(c) hydrogen sulphide, methane and
sulphur dioxide
(d) hydrogen sulphide, nitrogen and
methane
Ans.(a)
During sewage treatment biogas is
produced which include methane,
hydrogen sulphide and carbon dioxide.
Biogas is a mixture of gases (mainly
methane) produced by the microbial
activity and which may be used as fuel.
41Which of the following is mainly
produced by the activity of
anaerobic bacteria on sewage?
[CBSE AIPMT 2011]
(a) Propane (b) Mustard gas
(c) Marsh gas (d) Laughing gas
Ans.(c)
Marsh gas (methane) is mainly produced
by the activity of anaerobic bacteria on
sewage.
42Organisms called methanogens are
most abundant in a
[CBSE AIPMT 2011]
(a) cattle yard (b) polluted stream
(c) hot spring (d) sulphur rock
Ans.(a)
Organisms called Methanogens are most
abundant in a cattle yard. Methanogens
are present in the gut of several
ruminants animals such as cows and
buffaloes and they are responsible for
the production of methane (biogas) from
the dung of these animals.
43Secondary sewage treatment is
mainly a [CBSE AIPMT 2011]
(a) mechanical process
(b) chemical process
(c) biological process
(d) physical process
Ans.(c)
In secondary or biological treatment of
municipal waste, the organic matter is
decomposed with the help of microbes.
Decomposition of organic matter occurs
by one of the three methods-water
hyacinth pond, trickling filter method
and activated sludge method. After
decomposition the treated water is
sterilised through chlorination and
recycled.
44Select the correct statement from
the following[CBSE AIPMT 2010]
(a) Biogas is produced by the activity of
aerobic bacteria on animal waste
(b)Methanobacteriumis an aerobic
bacterium found in rumen of cattle
(c) Biogas, commonly called gobar gas, is
pure methane
(d) Activated sludge-sediment in
settlement tanks of sewage
treatment plant is a right source of
aerobic bacteria
Ans.(d)
Out of the following statements (d) is
correct because activated sludge is a
process dealing with the treatment of
sewage and industrial waste waters.
Atmospheric air or pure oxygen is
introduced to a mixture of primary
treated or screened sewage (or
industrial waste water) combined with
organisms to develop a biological flock,
which reduces the organic content of
the sewage. Sediment in settlement
tanks of sewage treatment plant is a rich
source of aerobic bacteria because small
amounts are used as inoculum in
secondary treatment or biological
treatment stage of sewage treatment.
45Which one of the following is being
utilised as a source of bio-diesel in
the Indian countryside?
[CBSE AIPMT 2007]
(a)Euphorbia (b) Beet root
(c) Sugarcane (d)Pongamia
Ans.(a)
Euphorbiais being utilised as a source of
biodiesel in the Indian countryside.
Some plants likeEuphorbia, Asclepias,
Capiaferaaccumulate hydrocarbons in
the form of latex which are used as
biodiesel.
46During anaerobic digestion of
organic waste, such as in
producing biogas, which one of the
following is left undegraded?
[CBSE AIPMT 2003]
(a) Hemicellulose
(b) Cellulose
(c) Lipids
(d) Lignin
Ans.(d)
During biogas formation, digestion of
cellulose is slow (rate-limiting) and most
of the lignin is not decomposed. After
cellulose lignin is the most abundant
plant polymer. It forms 20 to 30% of the
wood of the tree. Lignin is a complex
polymeric molecule, made up of phenyl
propanoid units. Cellulose is a large
chained polymer of glucose molecules
which are linked with each other by
glycosidic bonds. Hemicellulose are
branched polymers of glucose, xylose,
galactose, mannose and arabinose.
47Among the following pairs of
microbes, which pair has both the
microbes that can be used as
biofertilisers?[NEET (Odisha) 2019]
(a)AspergillusandRhizopus
(b)RhizobiumandRhizopus
(c) Cyanobacteria andRhizobium
(d)Aspergillusand Cyanobacteria
Ans.(c)
Biofertilisers are organisms that enrich
the nutrient quality of the soil.
For example, Cyanobacteria and
Rhizobium. AspergillusandRhizopusare
not used as biofertilisers.
48Which one of the following
microbes forms symbiotic
association with plants and helps
them in their nutrition?
[CBSE AIPMT 2012, 11]
(a)Azotobacter(b)Aspergillus
(c)Glomus (d)Trichoderma
Ans.(c)
Several species ofGlomus, including
G. aggregatum, are cultured and sold as
mycorrhizal inoculant for agricultrural
soils. Being endomycorrhiza, it helps the
plants in the absorption of nutrients
especially phosphorus from soil.
49A nitrogen-fixing microbe associated
withAzollain rice fields is
[CBSE AIPMT 2012]
(a)Spirulina (b)Anabaena
(c)Frankia (d)Tolypothrix
Ans.(b)
Anabaena azollaeis a free-living nitrogen
fixing blue-green alga or
322 NEETChapterwise Topicwise Biology
Microbes as Biofertilisers
TOPIC 5

333. cyanobacterium but it may also live
symbiotically in the leaf cavities of
Azolla, an aquatic, free floating,
freshwater pteridophyte (fern). This
cyanobacterium has nitrogenase
enzyme, therefore can fix nitrogen. If
this fern is grown in paddy (rice) fields, a
remarkable 50% increase in yield can be
noticed.
50An organism used as a biofertiliser
for raising soyabean crop is
[CBSE AIPMT 2011]
(a)Azospirillum(b)Rhizobium
(c)Nostoc (d)Azotobacter
Ans.(b)
Rhizobium leguminosarumis a symbiotic
bacteria found in root nodules of
legume. This bacterium has nitrogen
fixingnifgene. Soyabean is a legume.
Thus,Rhizobiumis used as a biofertiliser
for raising soyabean crop.
51Which one of the following is not a
biofertiliser?[CBSE AIPMT 2011]
(a)Rhizobium (b)Nostoc
(c) Mycorrhiza (d)Agrobacterium
Ans.(d)
Out of the following onlyAgrobacterium
is not a biofertiliser. It is a
Gram-negative bacterium that causes
tumours in plants. It is well known for its
ability to transfer DNA between itself
and plants, and for this reason it has
become an important tool for genetic
engineering.
A tumefacienscauses crown-gall
disease in plants. It has Ti-plasmid.
52The free-living, anaerobic
nitrogen-fixer is[CBSE AIPMT 2010]
(a)Beijerinckia(b)Rhodospirillum
(c)Rhizobium (d)Azotobacter
Ans.(b)
Rhodospirillumis a free-living, anaerobic,
nitrogen fixer. BothBeijerinckiaand
Azotobacterare free-living,
nitrogen-fixing, aerobic microbes.
Rhizobiumis a symbiotic, nitrogen fixer.
53The common nitrogen-fixer in
paddy fields is[CBSE AIPMT 2010]
(a)Rhizobium (b)Azospirillum
(c)Oscillatoria(d)Frankia
Ans.(b)
Azospirillumis a nitrogen fixing
bacterium in paddy fields. It is very
useful soil and root bacterium. It is an
associative symbioticN
2
-fixing bacteria.
When it is added to the soil, it multiplies
in millions and can supply 20-40 kg of
nitrogen per hectare per season.
54Which one of the following is not
used in organic farming?
[CBSE AIPMT 2010]
(a)Oscillatoria(b) Snail
(c)Glomus (d)Earthworm
Ans.(b)
Except snail, all these are used in
organic farming-
Glomus – Endomycorrhiza
Oscillatoria– BGA
Earthworm– Vermicompost
55Which one of the following
statements is correct?
[CBSE AIPMT 2007]
(a) Extensive use of chemical fertilisers
may lead to eutrophication of nearby
water bodies
(b) BothAzotobacterandRhizobiumfix
atmospheric nitrogen in root nodules
of plants
(c) Cyanobacteria such asAnabaenaand
Nostocare important mobilisers of
phosphates and potassium for plant
nutrition in soil
(d) At present it is not possible to grow
maize without chemical fertilisers
Ans.(a)
Out of the following statement (a) is
correct as eutrophication is caused by
run off water from fertilised fields,
sub-urban lawns, feed lots and
detergent rich sewage. It is phenomenon
of nutrient enrichment of a water body.
56A free-living nitrogen-fixing
cyanobacterium which can also
form symbiotic association with
the water fernAzollais
[CBSE AIPMT 2004]
(a)Tolypothrix(b)Chlorella
(c)Nostoc (d)Anabaena
Ans.(d)
Anabaenais a free-living nitrogen fixing
cyanobacterium which can form
symbiotic association with water fern
Azolla.
57Which of the following plants are
used as green manure in crop fields
and in sandy soils?
[CBSE AIPMT 2003]
(a)Saccharum munjaandLantana
camara
(b)Dichanthium annulatumandAzolla
nilotica
(c)Crotalaria junceaandAlhagi
comelorum
(d)Calotropis proceraandPhyllanthus
niruri
Ans.(c)
Crotalaria juncea(sunnhemp) andAlhagi
camelorumare among the plants which
are used as green manures in India.
These green manures help the soil
through increasement of area, water
holding capacities and fertility.
58The aquatic fern, which is an
excellent biofertiliser is
[CBSE AIPMT 2001, 1999]
(a)Azolla (b)Pteridium
(c)Salvinia (d)Marselia
Ans.(a)
Azollaleaves harbourAnabaenacolonies
which fix atmospheric nitrogen. The
nitrogen richAzollais used as
biofertiliser.
59Farmers have reported over 50%
higher yields of rice by sing which
of the following biofertiliser?
[CBSE AIPMT 2000, 99, 98]
(a) Mycorrhiza
(b)Azolla pinnata
(c) Cyanobacteria
(d) Legume-Rhizobium symbiosis
Ans.(b)
Anabaena azollae, a cyanobacterium
living in the cavities of fernAzolla, fixes
atmospheric nitrogen and releases it
into the leaf cavity of the fern. Farmers
have reported over 50% higher yields by
usingAzolla pinnata.
60Which of the following is
non-symbiotic biofertiliser?
[CBSE AIPMT 1998]
(a) VAM
(b)Azotobacter
(c)Anabaena
(d)Rhizobium
Ans.(b)
Free-living (non-symbiotic) bacteria like
AzotobacterandBacillus polymyxafix
atmospheric nitrogen and make it
available to crop plants.VAM(Vasicular
Arbuscular Mycorrhizae) is an
endosymbiosis between fungi and roots
of higher plants.
Anabaenais a cyanobacterium
(blue-green algae) which live solitary or
in association with other plant and can
fix atmosphericN
2
.
Rhizobiumbacterium makes symbiotic
association with leguminous plants.
Microbes in Human Welfare 323

334. 01Plasmid pBR322 has Pst I
restriction enzyme site within gene
amp
R
that confers ampicillin
resistance. If this enzyme is used
for inserting a gene for
β-galactoside production and the
recombinant plasmid is inserted in
anE.colistrain [NEET 2021]
(a) it will not be able to confer ampicillin
resistance to the host cell
(b) the transformed cells will have the
ability to resist ampicillin as well as
produceβ-galactoside
(c) it will lead to lysis of host cell
(d) it will be able to produce a novel
protein with dual ability
Ans.(a)
In plasmid vector pBR322, two unique
restriction sites PstI arid PvuI are
located within the amp
R
gene and
BamHl,SaII, etc., are located within the
tet
R
gene. The presence of restriction
sites within the marker genes tet
R
and
amp
R
permits an easy selection for cells
transformed with the I recombinant
pBR322. When restriction enzyme
BamHl orSaII is used, the DNA insert is
placed within the gene tet
R
making it
non-functional.
If this enzyme is used for inserting a
gene forβ-galactoside production and
the recombinant plasmid is inserted in
anE.colistrain, it will not be able to
confer ampicillin resistance to the host
cell.
02A specific recognition sequence
identified by endonucleases to
make cuts at specific positions
within the DNA is [NEET 2021]
(a) degenerate primer sequence
(b) Okazaki sequences
(c) palindromic nucleotide sequence
(d) poly(A) tail sequence
Ans.(c)
Palindromic nucleotide sequence is a
specific recognition sequence in a
double-standard DNA and RNA molecules
that is identified by endonucleases to
make cuts at specific positions. The
sequence is the same when one strand is
read from left to right andtheother
strand is read from right to left.
03Match the following techniques or
instruments with their usage.
[NEET (Oct.) 2020]
Column I Column II
A. Bioreactor (i) Separation of DNA
fragments
B. Electroph-or
esis
(ii) Production of large
quantities of products
C. PCR (iii) Detection of
pathogen, based on
antigen-antibody
reaction
D. ELISA (iv) Amplification of
nucleic acids
Select the correct option.
A B C D
(a) (iii) (ii) (iv) (i)
(b) (ii) (i) (iv) (iii)
(c) (iv) (iii) (ii) (i)
(d) (ii) (i) (iii) (iv)
Ans.(b)
Option (b) is correct match which is as
follows
Bioreactors are used for the industrial
scale production of products.
Electrophoresis helps in the separation
of DNA fragments based on their size.
PCR helps to amplify or generate large
number of copies of nucleic acids.
ELISA helps in the detection of pathogens
based on the principle of antigen-antibody
interaction.
04First discovered restriction
endonuclease that always cuts DNA
molecule at a particular point by
recognising a specific sequence of
six base pairs is[NEET (Oct.) 2020]
(a)EcoRI
(b) Adenosine deaminase
(c) Thermostable DNA polymerase
(d)Hind II
Ans.(d)
Hind II was the first discovered
endonuclease. It was isolated by Smith
Wilcox and Kelley (1968) from
Haemophilus influenzaebacterium. It
always cuts bacterium. DNA at particular
point by recognising a specific sequence
of six base pairs. It is known as the
recognition sequence forHind II and
reads as 5’-GTC GAC-3’, 3’-CAG CTG-5’.
Biotechnology:
PrinciplesandProcesses
33
Tools of Recombinant
DNA Technology
TOPIC 1

335. Biotechnology: Principles and Processes 325
05Choose the correct pair from the
following. [NEET (Sep.) 2020]
(a) Polymerases Break the DNA into
fragments
(b) Nucleases Separate the two
strands of DNA
(c) Exonucleases Make cuts at specific
positions within DNA
(d) Ligases Join the two DNA
molecules
Ans.(d)
The correct pair is option (d). Rest option
can be corrected as
The main function of DNA polymerase is
to synthesise DNA from
deoxyribonucleotides, the building
blocks of DNA.
Nucleases hydrolyse the phosphodiester
bonds of DNA and RNA.
Exonucleases are a broad class of
enzymes that cleave off nucleotides one
at a time from the 3’ or 5’ ends of DNA
and RNA chains.
06Identify the wrong statement with
regard to restriction enzymes.
[NEET (Sep.) 2020]
(a) They cut the strand of DNA at
palindromic sites
(b) They are useful in genetic engineering
(c) Sticky ends can be joined by using
DNA ligases
(d) Each restriction enzyme functions by
inspecting the length of a DNA
sequence
Ans.(c)
Statement in option (c) is incorrect. It
can be explained as follows.
Restriction endonucleases make cuts at
specific positions within the DNA known
as palindromic sites. They function by
inspecting the length of a DNA
sequence. They are used in genetic
engineering to form recombinant
molecules of DNA.
DNA ligases join the DNA fragments.
07A selectable marker is used to
[NEET (Odisha) 2019]
(a) help in eliminating the
non-transformants, so that the
transformants can be regenerated
(b) identify the gene for a desired trait in
an alien organism
(c) select a suitable vector for
transformation in a specific crop
(d) mark a gene on a chromosome for
isolation using restriction enzyme
Ans.(a)
To facilitate cloning into a vector, the
vector requires a selectable marker,
which helps in identifying and
eliminating non-transformants and
selectively permitting the growth of the
transformants.
08The two antibiotic resistant genes
on vector pBR322 are for
[NEET (Odisha) 2019]
(a) Ampicillin and Tetracycline
(b) Ampicillin and Chloramphenicol
(c) Chloramphenicol and Tetracycline
(d) Tetracycline and Kanamycin
Ans.(a)
The two antibiotic resistance gene on
E.colicloning vector pBR322 are for
ampicillin and tetracycline. Cloning
vectors are DNA molecules that carry a
foreign DNA segment and replicate
inside host cell. Plasmid inE.coliis a
cloning vector.
09Match the following enzymes with
their functions
Column I Column II
1. Restriction
endonuclease
i. Joins the DNA
fragments
2. Restriction
exonuclease
ii. Extends primers on
genomic DNA
template
3. DNA ligase iii. Cuts DNA at
specific position
4.Taq
polymerase
iv. Removes
nucleotides from
the ends of DNA
Select the correct option from the
following.[NEET (Odisha) 2019]
1 2 3 4
(a) (iii) (i) (iv) (ii)
(b) (iii) (iv) (i) (ii)
(c) (iv) (iii) (i) (ii)
(d) (ii) (iv) (i) (iii)
Ans.(b)
The correct matches are
1. Restriction
endonuclease
(iii) Cuts DNA at
specific site
2. Restriction
exonuclease
(iv) Removes
nucleotides from
the ends of DNA
3. DNA ligase (i) Joins the DNA
fragments
4.Taqpolymerase (ii) Extends primars
on genomic DNA
template.
10Following statements describe the
characteristics of the enzyme
Restriction Endonuclease. Identify
the incorrect statement.
[NEET (National) 2019]
(a) The enzyme binds DNA at specific
sites and cuts only one of the two
strands
(b) The enzyme cuts the
sugar-phosphate backbone at
specific sites on each strand
(c) The enzyme recognises a specific
palindromic nucleotide sequence in
the DNA
(d) The enzyme cuts DNA molecule at
identified position within the DNA
Ans.(a)
The statement about restriction
enzymes that the enzyme binds DNA at
specific sites and cuts only one of the
two strands is incorrect.
These enzymes cut both the strands of
DNA helix at specific sites in their sugar
phosphate backbone. The sequences
being recognised by restriction enzymes
are called palindromic sequences which
have same reading frame in both 5’→3’
and 3’→5’ directions. Rest statements
are correct.
11Which of the following is commonly
used as a vector for introducing a
DNA fragment in human
lymphocytes? [NEET 2018]
(a)λphage (b) Ti-plasmid
(c) Retrovirus (d) pBR 322
Ans.(c)
Usually aretrovirusis used as a vector
for introducing a DNA fragment in human
cells. They are used as vector in gene
therapy to introduce the desired gene so
as to replace the functioning of a
defected gene,. e.g. Severe Combined
Immune Deficiency (SCID) is caused due
to defect in the gene for the enzyme
adenosine deaminase.
In gene therapy against it, lymphocytes
are extracted from the bone marrow of
the patient. These are grown in a culture
outside the body. A functional ADA
cDNA, using a retroviral vector, is then
introduced into these lymphocytes.
These are reinjected into the patient’s
bone marrow.
λ-phageallows cloning of DNA
fragments upto
23 Kb lengths.Ti-plasmidis usually used
for plants.pBR-322is an artificial
cloning vector, usually used for bacteria.

336. 12A gene, whose expression helps to
identify transformed cells is known
as [NEET 2017]
(a) selectable marker
(b) vector
(c) plasmid
(d) structural gene
Ans.(a)
A gene whose expression helps to
identify transformed cell is known as
selectable marker. Usually, the genes
encoding resistance to antibiotics, such
as tetracycline, amphicillin, etc. are
useful selectable markers for, e.g.E.
coli.Concept Enhancer PlasmidpBR
322
has two resistance genes; ampicillin
resistance (amp
R
) and tetracyclin
resistance (tet
R
). These are considered
as useful selectable markers.
13The DNA fragments separated on
an agarose gel can be visualised
after staining with[NEET 2017]
(a) bromophenol blue
(b) acetocarmine
(c) aniline blue
(d) ethidium bromide
Ans.(d)
The DNA fragments separated on an
agarose gel can be visualised after
staining with ethidium bromide. It is
intercalating agent and a fluorescent
agent. The stained DNA fragments are
seen as bright orange coloured bands
under UV-light.
Thinking process Intercalation is the
insertion of molecules between the
planar bases of DNA. This process is
used as a method for analysing DNA.
Intercalation occurs, when ligands of an
appropriate size and chemical nature fit
themselves in between base pairs of
DNA. These ligands are mostly
polycyclic, aromatic and planar and
therefore often make good nucleic acid
stains. Intensively studied DNA
intercalator include ethidium bromide,
proflavine, etc.
14Which of the following is not a
feature of the plasmids?
[NEET 2016, Phase I]
(a) Circular structure
(b) Transferable
(c) Single-stranded
(d) Independent replication
Ans.(c)
Plasmid is extrachromosomal, double
stranded, circular DNA, found in
bacterial cells and some yeasts.
Discovery of plasmid has led to the
revolution in biotechnological research.
15Which of the following is a
restriction endonuclease?
[NEET 2016, Phase I]
(a) Protease (b) DNase I
(c) RNase (d) Hind II
Ans.(d)
Hind II is a restriction endonuclease.
Restriction endonucleases are enzymes
used for cutting of DNA at specific
locations.
HindII was the first restriction
endonuclease isolated by Smith Wilcox
and Kelley in 1968. It was found that it
always cut DNA molecules at a particular
point by recognising a specific sequence
of six base pairs.
16Which of the following restriction
enzymes produces blunt ends?
[NEET 2016, Phase II]
(a)SalI (b) EcoRV
(c)Xho (d)Hind III
Ans.(b)
EcoRV is a type II restriction
endonuclease isolated from strains ofE.
coli. It creates blunt ends. The enzyme
recognises the palindromic 6-base DNA
sequence and makes a cut at vertical
line. The blunt ends are formed as
↓
5′G – A – T – A – T – C3′
3′C – T – A – T – A – G5′
↑
Coverage site
↓
EcoRV
5
3
′ − −
′ − −



G A T
C T A
A T C
T A G
− − ′
− − ′



3
5
Blunt ends
17A foreign DNA and plasmid cut by
the same restriction endonuclease
can be joined to form a
recombinant plasmid using
[NEET 2016, Phase II]
(a)EcoRI (b)Taqpolymerase
(c) polymerase-III (d) ligase
Ans.(d)
DNA ligases (genetic gum) are used in
recombinant DNA technology to join two
individual fragments of double-stranded
DNA by forming phosphodiester
bonds between them to produce a
recombinant DNA (plasmid).
18The cutting of DNA at specific
locations became possible with the
discovery of[CBSE AIPMT 2015]
(a) restriction enzymes
(b) probes
(c) selectable markers
(d) ligases
Ans.(a)
Restriction enzymes are DNA cutting
enzymes found in bacteria. A restriction
enzyme recognises and cuts DNA only at
a particular sequence of nucleotides.
For example, the bacteriumHaemophilus
aegyptiusproduces an enzyme named
Hae III that cuts DNA wherever, it
encounters the sequence.
′5–G G C C–′3
′3–C C G G–′5
19The DNA molecule to which the
gene of interest is integrated for
cloning is called[CBSE AIPMT 2015]
(a) transformer (b) vector
(c) template (d) carrier
Ans.(b)
The DNA molecule to which the gene of
interest is integrated for cloning is called
vector. It is a DNA molecule used as a
vehicle to artificially carry foreign
genetic material into another cell, where
it can be replicated and/for expressed. A
vector containing foreign DNA is termed
as recombinant DNA.
20Which vector can clone only a small
fragment of DNA?
[CBSE AIPMT 2014]
(a) Bacterial artificial chromosome
(b) Yeast artificial chromosome
(c) Plasmid
(d) Cosmid
Ans.(c)
Plasmid is a small fragment of DNA
(about 10 Kbp size) that is physically
separate from and can replicate freely of
chromosomal DNA within a cell. It can
clone small DNA fragments.
Cosmid—45 Kbp
BAC—300-350 Kbp
YAC—1 Mbp/1,000 Kbp-2,500 Kbp)
21Commonly used vectors for human
genome sequencing are
[CBSE AIPMT 2014]
(a) T-DNA
(b) BAC and YAC
(c) Expression vectors
(d) T/A cloning vectors
326 NEETChapterwise Topicwise Biology

337. Ans.(b)
Commonly used vector for human genome
sequencing are BAC (Bacterial Artificial
Chromosome) and YAC. BAC is a DNA
construct, based on a functional fertility
plasmid (F plasmid) used for transforming
and cloning in bacteria (E. coli) and YAC are
genetically engineered chromosomes
derived from the DNA of the yeast,
(Saccharomyces cerevisiae) which is then
ligated into a bacterial plasma.
22The colonies of recombinant
bacteria appear white in contrast
to blue colonies of
non-recombinant bacteria because
of [NEET 2013]
(a) Non-recombinant bacteria containing
β-galactosidase
(b) Insertional inactivation of
α-galactosidase in non-recombinant
bacteria
(c) Insertional inactivation of
α-galactosidase in recombinant
bacteria
(d) Inactivation of glycosidase enzyme in
recombinant bacteria
Ans.(c)
The colonies of recombinant bacteria
appear white in contrast to blue colonies
of non-recombinant bacteria because of
insertional inactivation of alpha
galactosidase in recombinant bacteria.
Alpha galactosidase is a glycoside
hydrolase enzyme that hydrolyse the
terminal alpha galactosyl moieties from
glycolipids and glycoprotein. It is
encoded by the GLA gene.
β-galactosidase is an exoglycosidase,
which hydrolyses theβ-glycosidic bond
formed between a galactose and its
organic moiety.
23The given figure is the
diagrammatic representation of
theE. colivector pBR322. Which
one of the given options correctly
identifies its certain component(s)?
[CBSE AIPMT 2012]
(a)ori–original restriction enzyme
(b) rop–reduced osmotic pressure
(c)Hind III,EcoRI–selectable markers
(d)amp
R
,tet
R
–antibiotic resistance
genes
Ans.(d)
amp
R
andtet
R
are the antibiotic
resistant genes.Orirepresents the site
of origin of replication,roprepresents
the proteins that take part in the
replication of plasmid.HindIII,EcoRI are
the recognition sites of restriction
endonucleases.
24A single strand of nucleic acid
tagged with a radioactive molecule
is called [CBSE AIPMT 2012]
(a) vector
(b) selectable marker
(c) plasmid
(d) probe
Ans.(d)
Probes are 15-30 bases long radioactive
labelled oligonucleotides (RNA or DNA)
used to detect complementary
nucleotide sequences, used for disease
diagnosis, etc.
25For transformation, microparticles
coated with DNA to be bombarded
with gene gun are made up of
[CBSE AIPMT 2012]
(a) silver or platinum
(b) platinum or zinc
(c) silicon or platinum
(d) gold or tungsten
Ans.(d)
Biolistics or gene gun is a direct or
vectorless way used to introduce alien
DNA into host cells. In this method of
gene transfer, high velocity
micro-particles of gold or tungsten,
coated with DNA are bombarded on the
plant cells.
26Which one of the following is a
case of wrong matching?
[CBSE AIPMT 2012]
(a) Somatic hybridisation–Fusion of two
diverse cells
(b) Vector DNA – Site fortRNA synthesis
(c) Micropropagation –In vitroproduction
of plants in large numbers
(d) Callus – Unorganised mass of cells
produced in tissue culture
Ans.(b)
Out of the following the statement (b) is
wrong because a vector is a DNA
molecule used as a vehicle to transfer
foreign genetic material into desired
cell.
ThetRNA is synthesised in the nucleus
on a DNA tempelate. Only 0.025% of
total DNA content codes fortRNA.
27Given below is a sample of portion
of DNA strand giving the base
sequence on the opposite strands.
What is so, special shown in it?
[CBSE AIPMT 2011]
5′— GAATTC —3′
3′— CTTAAG —5′
(a) Replication completed
(b) Deletion mutation
(c) Start codon at the5′end
(d) Palindromic sequence of base pairs
Ans.(d)
Palindromic DNAis a base sequence of
DNA, which reads the same forward or
backward. It has similar sequence in
both the strands. Different types of
palindromic sequences are recognised
by restriction endonucleases.
28There is a restriction endonuclease
calledEcoRI. What does ‘co’ part in
it stand for?[CBSE AIPMT 2011]
(a) Colon (b) Coelom
(c) Coenzyme (d)Coli
Ans.(d)
Restriction endonuclease recognises a
particular palindromic sequence and
degrades the same. It was so, called
because it restricted the growth of
bacteriophage in the bacterium
(e.g.E. coli). The convention for naming
these enzymes is the first letter of the
name comes from the bacterial genus;
the second two letters come from the
species, and the fourth letter from
strain, e.g.EcoRI comes from
Escherichia coliRY13. Roman numbers
following the names indicate the order in
which the enzymes were isolated.
29Which one of the following is used
as vector for cloning genes into
higher organisms?
[CBSE AIPMT 2010]
(a) Baculovirus
(b)Salmonella typhimurium
(c)Rhizopus nigricans
(d) Retrovirus
Ans.(d)
Retroviruses are RNA containing animal
viruses that replicate through a DNA
intermediate. Retroviruses in animals
have the ability to transform normal cells
into cancerous cells. A better
Biotechnology: Principles and Processes 327
EcoRI
ClaI
Hind III
BamHI
SalI
PvuII
PstI
PvuI
amp
R
tet
R
pBR322
ori
rop

338. understanding of the act of delivering
genes by pathogen in these eukaryotic
hosts has generated knowledge to
transform these tools of pathogen into
useful vectors for delivering genes of
interest of humans. Retroviruses have
been disarmed and are now used to
deliver desirable genes into animal cells.
30Restriction endonucleases are
enzymes which
[CBSE AIPMT 2010, 06, 02, 01, 98, 95]
(a) make cuts at specific positions within
the DNA molecule
(b) recognise a specific nucleotide
sequence for binding of DNA ligase
(c) restrict the action of the enzyme DNA
polymerase
(d) remove nucleotides from the ends of
the DNA molecule
Ans.(a)
Restriction endonuclease recognises a
specific DNA base sequence
(recognition sequence or recognition
site, restriction sequence or restriction
site) and cleaves both the strands of
DNA at or near that site. The enzyme
cuts the DNA, generating restriction
fragments with overhanging ends or
blunt ends.
31The linking of antibiotic resistance
gene with the plasmid vector
became possible with
[CBSE AIPMT 2008]
(a) DNA polymerase
(b) exonucleases
(c) DNA ligase
(d) endonucleases
Ans.(c)
The linking of antibiotic resistance gene
with the plasmid vector became possible
with the enzyme DNA ligase, which acts
on cut DNA molecules and joins their
ends. This makes a new combination of
circular autonomously replication DNA
createdin vitroand is known as
recombinant DNA.
32Which one of the following is
commonly used in transfer of
foreign DNA into crop plants?
[CBSE AIPMT 2009]
(a)Trichoderma harzianum
(b)Meloidogyne incognita
(c)Agrobacterium tumefaciens
(d)Penicillium expansum
Ans.(c)
The uptake of foreign DNA or transgenes
by plant cells is called transformation.
A variety of techniques have been used
to introduce transgenes into plant cells,
these can be grouped into the following
two categories (i)Agrobacterium-
mediated and (ii) direct gene transfers.
Agrobacterium tumefaciensmediated
transformation eliminates the need for
regeneration from tissue explants.
33Restriction endonucleases
[CBSE AIPMT 2004]
(a) are present in mammalian cells for
degradation of DNA when the cell dies
(b) are used in genetic engineering for
ligating two DNA molecules
(c) are used forin vitroDNA synthesis
(d) are synthesised by bacteria as part of
their defense mechanism
Ans.(d)
Restriction endonucleases are found in
bacteria and are synthesised by bacteria
as a part of their defense mechanism.
They function in restricting the
multiplication of viruses in bacterial cells
by cutting up the genetic material of
invading virus.
34Manipulation of DNA in genetic
engineering became possible due
to the discovery of
[CBSE AIPMT 2003]
(a) restriction endonuclease
(b) DNA ligase
(c) transcriptase
(d) primase
Ans.(a)
Manipulation of DNA in genetic
engineering became possible due to the
discovery of restriction endonuclease.
These are isolated from bacterial cells,
and are tools for molecular biologists.
Several hundred restriction enzymes are
now known, each with a specific
sequence requirement dictating where it
will cut DNA. Therefore, digesting DNA
with a restriction enzyme creates a
characteristic set of fragments, which
can be isolated by electrophoresis and
subsequently analysed.
35A mutant strain ofT
4
-bacteriophage
R-II, fails to lyse theE. colibut when
two strains R-II
x
and R-II
y
are mixed
then they lyse theE. coli. What may
be the possible reason?
[CBSE AIPMT 2002]
(a) Bacteriophage transforms in wild
(b) It is not mutated
(c) Both strains have similar cistrons
(d) Both strains have different cistrons
Ans.(d)
The possible reason is that both strains
have different cistron because the
enzymes required for lysingE. colicould
not be synthesised by the mutant strain.
Two different strains had cistrons for
synthesising different enzymes which
acted together.
36In bacteria, plasmid is
[CBSE AIPMT 2002]
(a) extrachromosomal material
(b) main DNA
(c) non-functional DNA
(d) repetitive gene
Ans.(a)
Plasmid is a piece of circular DNA
molecule (mostly in bacteria but in yeast
also) which is not part of the normal
chromosomal DNA of a cell, and is
capable of replicating independently.
37Plasmid is[CBSE AIPMT 2000, 01]
(a) fragment of DNA which acts as vector
(b) fragment which joins two genes
(c)mRNA which acts as carrier
(d) autotrophic fragment
Ans.(a)
A plasmid is a piece of DNA, mostly in
bacteria (but also in yeast) not forming a
part of normal chromosomal DNA of a
cell, but capable of replicating
independently of it. These often act as
vehicles for gene transfer.
38Plasmids are suitable vectors for
gene cloning becauseu
[CBSE AIPMT 2000]
(a) these are small circular DNA
molecules which can integrate with
host chromosomal DNA
(b) these are small circular DNA
molecules with their own replication
origin site
(c) these can shuttle between
prokaryotic and eukaryotic cells
(d) these often carry antibiotic
resistance genes
Ans.(b)
Plasmids replicate autonomously. These
carry a signal situated at their replication
origin which determines how many
copies are to be made and this number
can be artificially increased for cloning a
given gene.
328 NEETChapterwise Topicwise Biology

339. Biotechnology: Principles and Processes 329
39The process of replication in
plasmid DNA, other than initiation,
is controlled by[CBSE AIPMT 1999]
(a) mitochondrial gene
(b) bacterial gene
(c) plasmid gene
(d) None of the above
Ans.(b)
The process of replication in plasmid
DNA, other than initiation, is controlled
by bacterial gene.
40Which of the following is related to
genetic engineering?
[CBSE AIPMT 1999]
(a) Mutation (b) Plasmid
(c) Plastid (d) Heterosis
Ans.(b)
Plasmids are used as vectors in genetic
engineering.
41Recombinant DNA is obtained by
cleaving the pro-DNA by
(a) primase [CBSE AIPMT 1998]
(b) exonucleases
(c) ligase
(d) restriction endonuclease
Ans.(d)
Recombinant DNA is obtained by
cleaving the pro-DNA by restriction
endonucleases. They can cleave DNA at
specific base sequences called
restriction sites.
42Genetic engineering is possible,
because [CBSE AIPMT 1998]
(a) the phenomenon of transduction in
bacteria is well understood
(b) we can see DNA by electron
microscope
(c) We can cut DNA at specific sites by
endonucleases like DNAs-I
(d) restriction endonucleases purified
from bacteria can be usedin vitro
Ans.(d)
Genetic engineering is the manipulation
of genetic material of an organism using
enzyme restriction endonuclease.
NathansandSmith(1970) isolated the
first restriction endonuclease.Jackson,
SymonsandPaul Berg(1972)
successfully generated recombinant
DNA moleculesin vitro.
43Two bacteria found to be very
useful in genetic engineering
experiments are[CBSE AIPMT 1998]
(a)NitrosomonasandKlebsiella
(b)EscherichiaandAgrobacterium
(c)NitrobacterandAzotobacter
(d)RhizobiumandDiplococcus
Ans.(b)
The most important tool in genetic
engineering of plants has been the Ti
plasmid of soil bacterium,Agrobacterium
tumefaciens.E. colihas also been
extensively used for genetic engineering
in animals, like in production of humulin,
somatotropin, etc.
44During the process of gene
amplification using PCR, if very
high temperature is not maintained
in the beginning, then which of the
following steps of PCR will be
affected first? [NEET 2021]
(a) Annealing (b) Extension
(c) Denaturation (d) Ligation
Ans.(c)
Denaturationis first step of PCR that
involves seperation of double-stranded
DNA. The DNA is subjected to heating at
high temperature (95ºC). This leads to
breaking of hydrogen bonds between
nucleotides and formation of
single-stranded DNA. Thus, if high
temperature is not maintained,
denaturation will be affected.
Annealingis second step of PCR which
involves annealing of primer to DNA
strands. In this step, DNA must be
cooled to 50ºC.
Extension, is third step in whichtaq
polymerase enzyme extends the primers
thus, completing replication of rest of
DNA.Ligationis the binding of amplified
sequence of interest.
45Which of the following is not an
application of PCR (Polymerase
Chain Reaction)? [NEET 2021]
(a) Molecular diagnosis
(b) Gene amplification
(c) Purification of isolated protein
(d) Detection of gene mutation
Ans.(c)
Polymerase Chain Reaction or PCR, is a
technique to make many copies of a
specific DNA regionin vitro(in a test tube
rather than an organism).
Following are the applications of PCR
The amplification of gene fragments
(Gene amplification).
The modification of DNA fragments.
The sensitive detection of pathogenic
microorganisms, if desired followed by
an accurate genotyping. (Molecular
diagnosis)
DNA analysis of archaeological
specimens.
Proof-reading PCR (PR-PCR) is designed
to detect known mutations within
genomic DNA.
46During the purification process for
recombinant DNA technology,
addition of chilled ethanol
precipitates out [NEET 2021]
(a) RNA (b) DNA
(c) histones (d) polysaccharides
Ans.(b)
The role of chilled ethanol and
monovalent cation is to remove the
solvation (solvent interface of any
chemical compound or biomolecule)
shell surrounding the DNA and
permitting the precipitation of the DNA
in pellet form.
Ethanol has a lower dielectric constant
than water, making it to promote ionic
bond formation the Na
+
(from the salt)
and the PO
3
−
(from the DNA backbone),
further, causing the DNA to precipitate.
47Which of the íollowing is a correct
sequence of steps in a PCR
(Polymerase Chain Reaction) ?
[NEET 2021]
(a) Denaturation, Annealing, Extension
(b) Denaturation, Extension, Annealing
(c) Extension, Denaturation, Annealing
(d) Annealing, Denaturation, Extension
Ans.(a)
PCR stands for Polymerase Chain
Reaction. It is a technique in which
multiple copies of gene of interest is
synthesised using two sets of primers
and the enzyme DNA polymerase. The
correct sequence of steps in a PCR
(Polymerase Chain Reaction) are
DenaturationIn which the
double-stranded template DNA is heated
at 95°C to separate it into two single
strands.
AnnealingIn which the temperature is
lowered to 50°C which enables the DNA
primers to attach to the template DNA.
Extension/ExtendingIn which the
temperature is raised and the new
Processes of Recombinant
DNA Technology
TOPIC 2

340. strand of DNA is made by thetaq
polymerase enzyme.
These three stages are repeated 20-40
times, doubling the number of DNA
copies each time.
48Select the correct statement from
the following.[NEET (Oct.) 2020]
(a) Gel electrophoresis is used for
amplification of a DNA segment
(b) The polymerase enzyme joins the
gene of interest and the vector DNA
(c) Restriction enzyme digestions are
performed by incubating purified DNA
molecules with the restriction
enzymes of optimum conditions
(d) PCR is used for isolation and
separation of gene of interest
Ans.(c)
Statement in option (c) is correct as
restriction enzyme digestions are
performed by incubating purified DNA
molecules with the restriction enzymes
of optimum conditions. Other statements
are incorrect and can be corrected as
Gel eletrophoresis is used for separation
and isolation of DNA fragments.
The polymerase enzyme uses DNA
templates to catalyse polymerisation of
deoxynucleotides.
PCR is Polymerase Chain Reaction, it is
used for amplification of a DNA
segment.
49In a mixture, DNA fragments are
separated by[NEET (Oct.) 2020]
(a) bioprocess engineering
(b) restriction digestion
(c) electrophoresis
(d) polymerase chain reaction
Ans.(c)
Electrophoresis is a technique used for
the separation of substances of
different ionic properties. It is used in
RDT to separate the DNA fragments that
are being cut by restriction
endonuclease. In this technique, DNA
fragments are loaded on agarose gel and
then electric field is applied. Due to
negatively charged, DNA fragments move
towards anodes (+vecharge).
The smaller fragments move farther
away as compared to larger fragments
and thus, these get separated.
50In recombinant DNA technology
antibiotics are used
[NEET Oct.) 2020]
(a) to keep medium bacteria-free
(b) to detect alien DNA
(c) to impart disease-resistance to the
host plant
(d) as selectable markers
Ans.(d)
In recombinant DNA technology,
antibiotics are used as selectable
markers, which help in identifying and
eliminating non-transformants and
selectively permitting the growth of the
transformants. Normally, the genes
encoding resistance to antibiotics such
as chloramphenicol, ampicillin,
tetracycline or kanamycin, etc., are
considered useful selectable markers
forE.coli.
51Match the following columns and
select the correct option.
[NEET (Sep.) 2020]
Column I Column II
A.Btcotton 1. Gene therapy
B.Adenosine deaminase
deficiency
2. Cellular
defence
C. RNAi 3. Detection of
HIV infection
D. PCR 4. Bacillus
thuringiensis
A B C D
(a) 3 2 1 4
(b) 2 3 4 1
(c) 1 2 3 4
(d) 4 1 2 3
Ans.(d)
The correct option is (d). It can be
explained as follows.
InBtcotton the specificBttoxin gene
was isolated fromBacillus thuringiensis.
The first clinical gene therapy was given
in 1990 to a 4-year old girl with adenosine
deaminase (ADA) deficiency.
RNAi (RNA interference) takes place in all
eukaryotic organisms as a method of
cellular defence.
PCR is now routinely used to detect HIV
in suspected AIDS patients.
52In gel electrophoresis, separated
DNA fragments can be visualised
with the help of[NEET (Sep.) 2020]
(a) ethidium bromide in UV radiation
(b) acetocarmine in UV radiation
(c) ethidium bromide in infrared radiation
(d) acetocarmine in bright blue light
Ans.(a)
In gel electrophoresis, separated DNA
fragments can be visualised with the
help of ethidium bromide in UV radiation
because DNA fragments cannot be seen
in visible light without staining.So, they
are stained with ethidium bromide and
made observable through UV radiation
as bright orange coloured bands.The
bands are cut out of agarose gel and
extracted. The purified DNA fragments
are then used in constructing
recombinant DNAs by attaching them to
cloning vectors.
53Given below are four statements
pertaining to separation of DNA
fragments using gel
electrophoresis. Identify the
incorrect statements.
1. DNA is negatively charged
molecule and so it is loaded on
gel towards the anode terminal.
2. DNA fragments travel along the
surface of the gel whose
concentration does not affect
movement of DNA.
3. Smaller the size of DNA
fragment larger is the distance
it travels through it.
4. Pure DNA can be visualised
directly by exposing
UV-radiation.
Select the correct option from the
following [NEET Odisha) 2019]
(a) 1, 3 and 4 (b) 1, 2 and 3
(c) 2, 3 and 4 (d) 1, 2 and 4
Ans.(d)
Statements (1), (2) and (4) are incorrect
because DNA fragments are negatively
charged molecules they can be
separated by forcing them to move
towards the anode under an electric field
through a medium/matrix. The
concentration of gel does affect the
resolution of DNA separation.
The separated DNA fragments can be
visualised only after staining the DNA
with a compound known as ethidium
bromide followed by exposure to UV
radiation. Only statement 3 is correct.
The DNA fragments spearate (resolve)
according to their size through sieving
effect provided by the agarose gel.
Hence, the smaller the fragment size,
the farther it moves.
54The correct order of steps in
Polymerase Chain Reaction (PCR) is
[NEET 2018]
(a) Denaturation, Extension, Annealing
(b) Annealing, Extension, Denaturation
(c) Extension, Denaturation, Annealing
(d) Denaturation, Annealing, Extension
330 NEETChapterwise Topicwise Biology

341. Biotechnology: Principles and Processes 331
Ans.(d)
The Polymerase Chain Reaction (PCR)
involves three basic steps; denaturation,
annealing and extension. In the
denaturationstep, DNA is heated at high
temperature(94°Cto96°C)to separate
the two strands. In the next step
(annealing), the two oligo-nucleotide
primers anneal to each single-stranded
template DNA.
This step is carried out at a lower
temperature
(40°Cto60°C). The final step is
extension, whereinTaqDNA
polymerase synthesises the
DNA region between the primers, using
dNTPs (deoxynucleoside triphosphates)
andMg
2 +
ions.
55The process of separation and
purification of expressed protein
before marketing is called
[NEET 2017]
(a) upstream processing
(b) downstream processing
(c) bioprocessing
(d) postproduction processing.
Ans.(b)
The process of separation and
purification of expressed protein before
marketing is called downstream
processing.
In this process, a whole range of
biochemical separation and purification
techniques are used such as drying,
chromatography, solvent extraction and
distillation. After purification quality
control testings are done.
56What is the criterion for DNA
fragments movement on agarose
gel during gel electrophoresis?
[NEET 2017]
(a) The larger the fragment size, the
farther it moves
(b) The smaller the fragment size, the
farther it moves
(c) Positively charged fragments move to
farther end
(d) Negatively charged fragments do not
move
Ans.(b)
Gel electrophoresis is used for the
separation of molecules of similar
electric charge on the basis of their size.
Hence, smaller the DNA fragment size
the farther it moves.
Thinking Process Agarose gel matrix
functions as sieve. Smaller DNA
fragments easily move and larger
fragments take time to move in gel
matrix.
57TheTaqpolymerase enzyme is
obtained from[NEET 2016, Phase I]
(a)Thiobacillus ferroxidans
(b)Bacillus subtilis
(c)Pseudomonas subtilis
(d)Thermus aquaticus
Ans.(d)
Taqpolymerase is a thermostable DNA
polymerase obtained fromThermus
aquaticus. Thermus aquaticusis a
bacterium that lives in hot springs and
hydrothermal vents.
58Which of the following is not a
component of downstream
processing?[NEET 2016, Phase II]
(a) Separation
(b) Purification
(c) Preservation
(d) Expression
Ans.(d)
Downstream process is the process of
separation and purification of products
synthesised in bioreactors. After the
completion of biosynthetic stage, the
product has to be subjected through a
series of processes before it is ready for
marketing as a finished product.
The processes include separation and
purification, collectively referred to as
downstream processing. The product is
then formulated with suitable
preservatives. Hence, option (d) is
incorrect and all other options are
correct.
59Stirred-tank bioreactors have been
designed for[NEET 2016, Phase II]
(a) purification of product
(b) addition of preservatives to the
product
(c) availability of oxygen throughout the
process
(d) ensuring anaerobic conditions in the
culture vessel
Ans.(c)
A stirred tank bioreactor is usually
cylindrical, possessing a stirrer which
facilitates even mixing of the reactor
contents and oxygen availability through
out the bioreactor.
The bioreactor has an agitator system,
an oxygen delivery system, foam control,
pH and temperature control systems.
Hence, option (c) is correct.
60An analysis of chromosomal DNA
using the Southern hybridisation
technique does not use
[CBSE AIPMT 2014]
(a) electrophoresis (b) blotting
(c) autoradiography (d) PCR
Ans.(d)
Southern hybridisation is a technique
used in molecular biology for detection
of a specific DNA sequence in DNA
samples in which except PCR we use all
three methods such as electrophoresis,
blotting and autoradiography. PCR is the
method used for the amplification of
DNA sample.In vitroclonal propagation
is characterised by PCR and RAPD.
Denaturation
Annealing
of primers
Primers
dsDNA
3¢
5¢
3¢
3¢
5¢
5¢
3¢
5¢
3¢
3¢ 5¢
5¢
Region to be amplified
DNA polymerase
( polymerase)
+ deoxynucleotides
Taq
3¢
5¢
3¢
5¢
5¢
3¢
5¢
3¢
Extension
of primers
30 cycles
Amplified
(~1 billion times)
Heat
Polymerase Chain Reaction (PCR) : Each cycle
has three steps : (i) Denaturation; (ii)
Primer annealing; and (iii) Extension of
primers
Motor
Foam beaker
Flat bladed
impeller
Culture
broth
Sterile air
Acid/base
for pH control
Steam for
sterilisation
(a) Simple stirred-tank bioreactor
Gas
entrainment
Increased
surface
area for
oxygen
Bubbles\
dramatically
increase the
oxygen transfer
area
(b) Sparged stirred-tank bioreactor through
which sterile air bubbles are sparged

342. 332 NEETChapterwise Topicwise Biology
61In vitroclonal propagation in plants
is characterised by
[CBSE AIPMT 2014]
(a) PCR and RAPD
(b) Northern blotting
(c) electrophoresis and HPLC
(d) microscopy
Ans.(a)
RAPD (Random Amplified Polymorphic
DNA) is a type of PCR reaction, but the
segments of DNA that are amplified are
random. Often, PCR is used to amplify a
known DNA sequence likein vitroclonal
propagation in plants.
62DNA fragments generated by the
restriction endonucleases in a
chemical reaction can be
separated by [NEET 2013]
(a) centrifugation
(b) polymerase chain reaction
(c) electrophoresis
(d) restriction mapping
Ans.(c)
DNA fragments generated by the
restriction endonucleases in a chemical
reaction can be separated by
electrophoresis. The polymerase chain
reaction is simply DNA replication in a
test-tube. Restriction mapping is the
process of obtaining structural
information on a piece of DNA by the use
of restriction enzymes,
e.g. endonucleases that recognise
specific 4-8 base regions of DNA.
63PCR and restriction fragment
length polymorphism are the
methods for [CBSE AIPMT 2012]
(a) study of enzymes
(b) genetic transformation
(c) DNA sequencing
(d) genetic fingerprinting
Ans.(d)
PCR and RFLP are methods used for
genetic fingerprinting. As Restriction
Fragment Length Polymorphism (RFLP) is
the basis of genetic (or DNA)
fingerprinting and is useful in identifying
individuals from their semen, blood or
tissues or any other DNA sample and
resolution of parent hood disputes.
Polymerase Chain Reaction (PCR) is also
useful in genetic fingerprinting as it can
amplify the DNA sample even if available
in a very small amount.
64Which one is a true statement
regarding DNA polymerase used in
PCR? [CBSE AIPMT 2012]
(a) It is used to ligate introduced DNA in
recipient cells
(b) It serves as a selectable marker
(c) It is isolated from a virus
(d) It remains active at high temperature
Ans.(d)
Polymerase Chain Reaction (PCR) is used
to amplify a DNA segment or to
synthesisein vitrothe multiple copies of
gene (or DNA) of interest, using two sets
of primers and the enzyme DNA
polymerase. This enzyme is isolated
from a bacteriumThermus aquaticusand
it remains active during the high
temperature but high temperature
induced denaturation of double stranded
DNA.
65Agarose extracted from sea weeds
is used in [CBSE AIPMT 2011]
(a) spectrophotometry
(b) tissue culture
(c) PCR
(d) gel electrophoresis
Ans.(d)
For gel electrophoresis the commonly
used matrix is agarose which is a natural
polymer extracted from seaweeds (e.g.
Gelidium, Gracilaria, Gigartina, etc.). Gel
electrophoresis is a technique to
separate fragments of DNA. Since, DNA
fragments are negatively charged
molecules they can be separated by
forcing them to move towards the anode
under an electric field through a
medium/matrix.
66Stirred-tank bioreactors have been
designed for
[CBSE AIPMT 2010]
(a) addition of preservatives to the
product
(b) purification of the product
(c) ensuring anaerobic conditions in the
culture vessel
(d) availability of oxygen throughout the
process
Ans.(d)
The most common type of aerobic
bioreactor in use today is the
stirred-tank bioreactor, which may
feature a specific internal configuration
designed to provide a specific
circulation pattern. The stirred tank
bioreactor have been designed for
availability of oxygen through out the
processes.
67Polyethylene glycol method is used
for [CBSE AIPMT 2009]
(a) biodiesel production
(b) seedless fruit production
(c) energy production from sewage
(d) gene transfer without a vector
Ans.(d)
Polyethylene glycol method is used for
gene transfer without a vector.
Introduction of DNA into plant cells
without the involvement of a biological
agent and leading to stable
transformation is known as direct gene
transfer. There are various methods for
direct direct gene transfer, one of which
is chemical method.
Certain chemicals, e.g. PEG
(Polyethylene Glycol), polyvinyl alcohol
and calcium phosphate enhance the
uptake of DNA by plant protoplast. PEG
and calcium phosphate are thought to
precipitate the DNA onto the outer
surface of plasmalemma and the
precipitate is taken up by the
endocytosis.
68Gel electrophoresis is used for
[CBSE AIPMT 2008]
(a) construction of recombinant DNA by
joining with cloning vectors
(b) isolation of DNA molecules
(c) cutting of DNA into fragments
(d) separation of DNA fragments
according to their size
Ans.(d)
When genomic DNA extracted from any
tissue of a plant or animal species is
digested with a restriction enzyme, it is
cleaved into segments. The segments of
different sizes can be separated through
gel electrophoresis. Gel electrophoresis
involves movement of fragments or
molecules from a well created on one
edge of the gel.

343. 01Match the organism with its use in
biotechnology.[NEET (Sep.) 2020]
Column I Column II
A.Bacillus
thuringiensis
1. Cloning vector
B.Thermus
aquaticus
2. Construction of
firstrDNA
molecule
C.Agrobacterium
tumefaciens
3. DNA polymerase
D.Salmonella
typhimurium
4. Cry proteins
Select the correct option.
A B C D
(a) 4 3 1 2
(b) 3 2 4 1
(c) 3 4 1 2
(d) 2 4 3 1
Ans.(a)
The correct option is (a). It can be
explained as follows
Bacillus thuringiensisis a source of Cry
proteins.
Thermus aquaticusis a source of
thermostable DNA polymerase (Taq
polymerase) used in PCR.
Agrobacterium tumefaciensis a cloning
vector.
The construction of 1st recombinant
DNA molecule was performed using
native plasmid ofSalmonella
typhimurium.
02RNA interference is used for which
of the following purposes in the
field of biotechnology?
[NEET (Oct.) 2020]
(a) to develop a plant tolerant to abiotic
stresses
(b) to develop a pest resistant plant
against infestation by nematode
(c) to enhance the mineral usage by the
plant
(d) to reduce post harvest losses
Ans.(b)
RNA interference is used to develop a
pest resistant plant against infestation
by nematode. RNA interference takes
place in all eukaryotic organisms as a
method of cellular defence. This method
involves silencing of a specificmRNA
due to a complementarydsRNA
molecule that binds to and prevents
translation of themRNA (silencing).
03Btcotton variety that was
developed by the introduction of
toxin gene ofBacillus thuringiensis
(Bt) is resistant to
[NEET (Sep.) 2020]
(a) fungal diseases
(b) plant nematodes
(c) insect predators
(d) insect pests
Ans.(d)
The correct option is (d). It can be
explained as follows
Btcotton is resistant to cotton bollworm
(insect pest).cryI Ac andcryII Ab genes
have been introduced in cotton to
protect it from cotton bollworm. This
makesBtcotton as biopestic.
04If an agricultural field is liberally
irrigated for a prolonged period of
time, it is likely to face problem of
[NEET (Odisha) 2019]
(a) metal toxicity (b) alkalinity
(c) acidity (d) salinity
Ans.(d)
Salinity of soil is a problem faced by
farmers if they liberally irrigate
agricultural field for a prolonged period
of time.
Irrigation salinity is the accumulation of
salts in the topsoil under irrigation. It is
caused by over irrigation of agricultural
land, inefficient water use, poor drainage
and the irrigation of unsuitable and leaky
soils.
05Which part of the tobacco plant is
infected byMeloidegyne
incognitia? [NEET 2016, Phase II]
(a) Leaf (b) Stem
(c) Root (d) Flower
Ans.(c)
Meloidegyne incognitiais a nematode,
which infects roots of tobacco plant and
results in root knot disease.
06Golden rice is a genetically modified
crop plant where the incorporated
gene is meant for biosynthesis of
[CBSE AIPMT 2015]
(a) vitamin-B (b) vitamin-C
(c) omega 3 (d) vitamin-A
Ans.(d)
Golden rice is a variety of rice produced
through genetic engineering to
biosynthesiseβ-carotene, a precursor of
vitamin-A, in the edible parts of rice.
Biotechnologyand
ItsApplications
34
Applications of
Biotechnology
in Agriculture
TOPIC 1

344. 07The introduction oftDNA into
plants involves[CBSE AIPMT 2015]
(a) infection of the plant by
Agrobacterium tumefaciens
(b) altering the pH of soil, heat-shocking
the plants
(c) exposing the plants to cold for a brief
period
(d) allowing the plant roots to stand in
water
Ans.(a)
Agrobacterium tumefaciens,a pathogen
of several dicot plants is able to deliver
a piece of DNA known astDNA to
transform normal plant cells into a
tumour and direct these tumour cells to
produce the chemicals required by the
pathogen.
08Which of the followingBtcrops is
being grown in India by the
farmers? [NEET 2013]
(a) Maize (b) Cotton
(c) Brinjal (d) Soyabean
Ans.(b)
In IndiaBtcotton is grown by farmers
extensively.Bttoxin is produced by a
bacterium calledBacillus thuringiensis
(Bt). Examples ofBtcrops areBtcotton,
Btcorn, rice, tomato, potato and
soyabean, etc.
09Consumption of which one of the
following foods can prevent the
kind of blindness associated with
vitamin-A deficiency?
[CBSE AIPMT 2012]
(a)Flavr savrtomato
(b) Canolla
(c) Golden rice
(d)Btbrinjal
Ans.(c)
Golden rice is a variety ofOryza sativa
(rice) produced through genetic
engineering to biosynthesise
beta-carotene, a precursor of vitamin-A
in the edible part of rice
(i.e. endosperm). The research that led
to golden rice was conducted with the
goal of helping children who suffer from
vitamin-A deficiency. Because many
children in countries where there is a
dietary deficiency in vitamin-A rely on
rice as a staple food, the genetic
modification of rice to produce the
vitamin-A precursor beta-carotene is
seen as a simple and less expensive
alternative to vitamin supplements.
10The process of RNA interference
has been used in the development of
plants resistant to
[CBSE AIPMT 2011]
(a) nematodes (b) fungi
(c) viruses (d) insects
Ans.(a)
The process of RNA interference (RNAi)
has been used in the development of
plants resistant to nematodes like
Meloidogyne incognitia, which infects the
roots of tobacco plants and causes a
great reduction in yield. RNA
interference takes place in all eukaryotic
organisms as a method of cellular
defense. This method involves silencing
of a specificmRNA due to a
complementarydsRNA molecule that
binds to and prevents translation of the
mRNA (silencing).
11Continuous addition of sugars in
‘fed batch’ fermentation is done to
[CBSE AIPMT 2011]
(a) produce methane
(b) obtain antibiotics
(c) purify enzymes
(d) degrade sewage
Ans.(c)
Continuous addition of sugars in fed
batch fermentation is done to purify
enzymes. A fed batch is a
biotechnological batch process which is
based on feeding of a growth limiting
nutrient substrate to a culture.
12The Genetically Modified (GM)
brinjal in India has been developed
for [CBSE AIPMT 2010]
(a) insect-resistance
(b) enhancing self life
(c) enhancing mineral content
(d) drought-resistance
Ans.(a)
The genetically modified brinjal orBt
brinjal has the samecry lacgene from
Bacillus thuringiensisas cotton. The
gene is supposed to make the plant
tolerant to the shoot and fruit borer
insect, which attacks it throughout its
life cycle.
13Some of the characteristics ofBt
cotton are [CBSE AIPMT 2010]
(a) long fibre and resistance to aphids
(b) medium yield, long fibre and
resistance to beetle pests
(c) high yield and production of toxic
protein crystals which kill dipteran
pests
(d) high yield and resistance to bollworms
Ans.(c)
Bacillus thuringiensisforms crystals
containing a toxic insecticidal protein.Bt
toxin protein exists as inactive protein
but once an insect ingests the inactive
toxin, it is converted into an active form
of toxin due to the alkaline pH of the gut,
which solubulise the crystals.
The activated toxin binds to the surface
of midgut epithelial cells and creates
pores that cause cell swelling and lysis
and eventually cause death of the insect.
SpecificBttoxin genes were isolated
fromB. thuringiensisand incorporated
into the several plants such as cotton.
The toxin is coded by a gene namedcry.
There are a number of them, e.g. the
proteins encoded by the genescryI Ac
andcryII Ab control the cotton
bollworms, that ofcryI Ab control corn
borers.
14A transgenic food crop which may
help in solving the problem of night
blindness in developing countries is
[CBSE AIPMT 2008]
(a)Btsoyabean
(b) golden rice
(c)flavr savrtomatoes
(d) starlink maize
Ans.(b)
Vitamin-A deficiency causes night
blindness among children. Vitamin-A
deficiency often occurs where rice is the
staple food since, rice grain does not
containβ-carotene. Three transgenes
providing phytoene synthase, phytoene
desaturase, beta carotene desaturase
and lycopene cyclase activities were
transferred into rice byAgrobacterium
mediated transformation. The resulting
transgenic rice popularly called goldern
rice contains good quantities of
β-carotene.
Flavr savris genetically modified
tomato, which remains fresh and
retain their flavour much longer than
normal tomato due to the blocking of
synthesis of fruit softening enzyme
polygalacturonase.
15Introduction of food plants
developed by genetic engineering
is not desirable because
(a) economy of developing countries may
suffer
334 NEETChapterwise Topicwise Biology

345. (b) these products are less tasty as
compared to the already existing
products
(c) this method is costly
(d) there is danger of introduction
viruses and toxins with introduced
crop
Ans.(a)
It is difficult for developing countries to
keep up and maintain genetically
engineered crops. Therefore, introduction
of food plants by genetic engineering is
not desirable for developing countries.
16When gene targetting involving
gene amplification is attempted in
an individual’s tissue to treat
disease, it is known as[NEET 2021]
(a) biopiracy
(b) gene therapy
(c) molecular diagnosis
(d) safety testing
Ans.(b)
Gene therapy is a type of treatment
designed to modify the expression of an
individual's genes or to correct abnormal
genes to treat a disease. Gene
amplification is common in cancer cells,
and some amplified genes may cause
cancer cells to grow or become resistant
to anticancer drugs. The presence of
gene amplification can have a
prognostic and a diagnostic value and
can help in orienting therapy in specific
tumour types.
17With regard to insulin choose the
correct options.
I. C-peptide is not present in
mature insulin.
II. The insulin produced byrDNA
technology has C-peptide.
III. The pro-insulin has C-peptide.
IV. A-peptide and B-peptide of
insulin are interconnected by
disulphide bridges.
Choose the correct answer from
the options given below.
[NEET 2021]
(a) II and IV (b) II and III
(c) I, III and IV (d) I and IV
Ans.(c)
Statement I, III and IV are correct.
Insulin consists of two short polypeptide
chains A and B which are linked together
by disulphide bridges. In mammals
insulin is synthesised as pro hormone
which contain an extra stretch called the
C, peptide also called pro-insulin.
C-peptide is removed during maturation
into insulin and is not present in mature
insulin.
Statement II is incorrect and be
corrected as
It is challenging to produce insulin by
usingrDNA techniques because it is
difficult to assemble insulin into its
mature form usingrDNA technique.
Thus, C- peptide is absent in insulin
produced byrDNA technology.
18Which of the following statements
is not correct?[NEET (Sep.) 2020]
(a) The proinsulin has an extra peptide
called C-peptide
(b) The functional insulin has A and B
chains linked together by hydrogen
bonds
(c) Genetically engineered insulin is
produced inE-coli
(d) In man, insulin is synthesised as a
proinsulin
Ans.(b)
Statement in option (b) is incorrect
because
Insulin is composed of two peptide
chains referred to as the A chain and B
chain. A and B chains are linked together
by two disulphide bonds, and an
additional disulphide is formed within
the A chain. Insulin molecules have a
tendency to form dimers in solution due
to hydrogen-bonding between the
C-termini of B chains.
19In India, the organisation
responsible for assessing the
safety of introducing genetically
modified organisms for public use
is [NEET 2018]
(a) Research Committee on Genetic
Manipulation (RCGM)
(b) Council for Scientific and Industrial
Research (CSIR)
(c) Indian Council of Medical Research
(ICMR)
(d) Genetic Engineering Appraisal
Committee (GEAC)
Ans.(d)
In India,Genetic Engineering Approval
Committee,i.e.,GEAC(NCERT) is
responsible for assessing the safety of
introducing genetically modified
organisms for public use. GEAC comes
under the Ministry of Environment and
Forests (MOE & F) while theReview
Committee on Genetic Manipulation
(RCGM) comes under Department of
Biotechnology.
TheCouncil of Scientific and Industrial
Research(CSIR) is the largest research
and development organisation in India.
The Indian Council of Medical Research
(ICMR) is the apex body in India for the
formulation, coordination and promotion
of biomedical research.
NoteThe name of GEAC is changed to
Genetic Engineering Appraisal
Committeefrom Genetic Engineering
Approval Committee in 2010.
20.The two polypeptides of human
insulin are linked together by
[NEET 2016, Phase I]
(a) phosphodiester bonds
(b) covalent bonds
(c) disulphide bridges
(d) hydrogen bonds
Ans.(c)
In humans, insulin is produced byβ-cells
of pancreas. It is synthesised as
prohormone in which two polypeptides
are synthesised with an extra stretch of
'C' polypeptide. During maturation extra
stretch of 'C' polypeptide is separated
and two polypeptide chains (A and B) are
linked together by disulphide linkages
(bridges).
21Which kind of therapy was given in
1990 to a four-year-old girl with
Adenosine Deaminase (ADA)
deficiency?[NEET 2016, Phase II]
(a) Gene therapy
(b) Chemotherapy
(c) Immunotherapy
(d) Radiation therapy
Ans.(a)
The first clinical gene therapy was given
in 1990 to a 4 years old girl with
Adenosine Deaminase (ADA) deficiency.
ADA deficiencyis a disorder caused due
to the deletion of the gene for adenosine
deaminase. Gene threapy is the
technique of genetic engineering that
allows correction of a gene defect that
has been diagnosed in child/embryo. In
this therapy a normal healthy functional
gene is inserted and the faulty gene is
replaced.
Biotechnology and Its Applications 335
Applications of
Biotechnology in Medicine
TOPIC 2

346. 22The first human hormone produced
by recombinant DNA technology is
[CBSE AIPMT 2014]
(a) insulin (b) estrogen
(c) thyroxin (d) progesterone
Ans.(a)
The first human hormone produced by
recombinant DNA technology is insulin.
It is peptide hormone, which controls the
level of blood sugar. It is formed by
joining of two polypeptide chain by
disulphide bonds.
23Human insulin is being
commercially produced from a
transgenic species of
[CBSE AIPMT 2008]
(a)Rhizobium (b)Saccharomyces
(c)Escherichia(d)Mycobacterium
Ans.(c)
In 1983 an American company Eli Lily
produced the first genetically
engineered insulin by first synthesising
two DNA sequences corresponding toα
andβinsulin chains. The two DNA
sequences or genes were made to fuse
with plasmids ofEscherichia coliand
later allowed to form insulin chains.
24Production of a human protein in
bacteria by genetic engineering is
possible because
[CBSE AIPMT 2005]
(a) bacterial cell can carry out the RNA
splicing reactions
(b) the human chromosome can
replicate in bacterial cell
(c) the mechanism of gene regulation is
identical in humans and bacteria
(d) the genetic code is universal
Ans.(d)
Production of human protein in bacteria
by genetic engineering is possible
because the genetic code is universal as
a codon codes for the same amino acid
in all the organisms.
25The genetic defect-Adenosine
Deaminase (ADA) deficinecy may
be cured permanently by
[CBSE AIPMT 2009]
(a) periodic infusion of genetically
engineered lymphocytes having
functional ADA cDNA
(b) administering adenosine deaminase
activators
(c) introducing bone marrow cells
producing ADA into cells at early
embryonic stages
(d) enzyme replacement therapy
Ans.(a)
Severe Combined Immuno Deficiency
(SCID) caused by Adenosine Deaminase
Deficiency (ADA) is the first genetic
disorder to be treated with gene therapy.
T-cell directed gene transfer was useful
in the treatment of ADA-SCID, whereas
the retroviral- mediated gene transfer to
haematopoietic stem cells was
insufficient for achievement of clinical
benefits.
26ELISA is used to detect viruses
where the key reagent is
[CBSE AIPMT 2004, 03]
(a) alkaline phosphatase
(b) catalase
(c) DNA probe
(d) RNase
Ans.(a)
Alkaline phosphate is the key reagent
used during detection of virus in ELISA
test. The test work by detecting
antibodies/substances or protein which
are produced in blood when virus is
present. The reagents are used to
provide antibody-antigen complex in a
specialised ELISA plate.
27The laws and rules to prevent
unauthorised exploitation of
bioresources are termed as
(a) biopatenting[NEET (Oct.) 2020]
(b) bioethics
(c) bioengineering
(d) biopiracy
Ans.(b)
Bioethics are sets of laws and rules to
prevent unauthorised exploitation of
bioresources.
Biopiracy is the use of bioresources by
multinational companies and other
organisations without proper
authorisation from the countries and
people concerned without
compensatory payment. Biopatenting is
patent granted on biological products,
organisms processes or bioresources.
Bioengineering is the application of the
life sciences, physical sciences,
mathematics and engineering principles
to define and solve problems in biology,
medicine, health care and other fields.
28Exploitation of bioresources of a
nation by multinational companies
without authorisation from the
concerned country is referred to as
[NEET (Odisha) 2019]
(a) bioweapon (b) biopiracy
(c) bioethics (d) biowar
Ans.(b)
Exploitation of bioresources of a nation
by multinational companies without
authorisation from concerned country is
referred to as biopiracy. For example, a
patent granted in USA covers the entire
basmati rice germplasm of our country.
29Exploration of molecular, genetic
and species level diversity for novel
products of economic importance
is known as[NEET (Odisha) 2019]
(a) biopiracy (b) bioenergetics
(c) bioremediation (d) bioprospecting
Ans.(d)
Exploration of molecular, genetic and
species level diversity for novel products
of economic importance is known as
bioprospecting. Biopiracy refers to
illegal use of bioresources.
Bioremediation is use of organisms to
clean up contamination during waste
treatment. Bioenergetics deals with
energy metabolism in living organisms.
30A ‘new’ variety of rice was patented
by a foreign company, though such
varieties have been present in India
for a long time. This is related to
[NEET 2018]
(a) Lerma Rojo
(b) Sharbati Sonora
(c) Co-667
(d) Basmati
Ans.(d)
In 1997, an American company got patent
rights on Basmati rice through the US
Patent and Trademark office. This ‘new’
variety of Basmati had actually been
derived from Indian farmer’s varieties.
This ‘new’ variety was produced by
crossing Indian Basmati with semi-dwarf
varieties.
Lerma RojoandSharbati Sonoraare high
yielding varieties of wheat.Co-667is a
variety of soyabean.
31Use of bioresources by
multinational companies and
organisations without authorisation
336 NEETChapterwise Topicwise Biology
Transgenic Organisms
and Ethical Issuees
TOPIC 3

347. from the concerned country and its
people is called [NEET 2018]
(a) biodegradation
(b) biopiracy
(c) bio-infringement
(d) bioexploitation
Ans.(b)
Biopiracyis referred to the use of
bioresources by multinational
companies and other organisations
without proper authorisation from the
countries and people concerned without
compensatory payment. Bio-infringement
is the commission of a prohibited act
with respect to a patented invention
without permission from the patent
holder. Bio-exploitation means taking
advantage of biological resources of
other country without permission.
Biodegradation is biological breakdown of
organic material by bacteria, fungi, etc.
32Maximum number of existing
transgenic animals is of
[CBSE AIPMT 2011]
(a) fish (b) mice
(c) cow (d) pig
Ans.(b)
Over 95% of all existing transgenic
animals are mice. Animals that have
their DNA manipulated to possess and
express an extra (foreign) gene are
known as transgenic animals, e.g. rats,
rabbits, pig, sheep, cows, fish, etc.
33Genetic engineering has been
successfully used for producing
[CBSE AIPMT 2010]
(a) transgenic mice for testing safety of
polio vaccine before use in humans
(b) transgenic models for studying new
treatments for certain cardiac
diseases
(c) transgenic cow-Rosie which produces
high fat milk for making ghee
(d) animals like bulls for farm work as
they have super power
Ans.(a)
Genetic engineering has been
successfully used for producing
transgenic mice which are being
developed for use in testing the safety of
vaccines before they are used on
humans. Transgenic mice are being used
for testing toxicity of drugs. Transgenic
animals are made to carry genes, which
make them more sensitive to toxic
substances than non-transgenic
animals.
They are then exposed to the toxic
substances and the effects studied.
Toxicity testing in such animals will allow
us to obtain results in less time.
34An improved variety of transgenic
basmati rice[CBSE AIPMT 2010]
(a) does not require chemical fertilisers
and growth hormones
(b) gives high yield and is rich in
vitamin-A
(c) is completely resistant to all insect
pests and diseases of paddy
(d) gives high yield but has no
characteristic aroma
Ans.(b)
Golden rice (transgenic basmati rice) is a
variety ofOryza sativaproduced through
genetic engineering to biosynthesise
beta-carotene, a precursor of
provitamin-A in the edible parts of rice.
The research that led to golden rice was
conducted with the goal of helping
children who suffer from vitamin-A
deficiency in poor countries. Golden rice
has been bred to be especially
disease-resistant, resulting in better
crop yields.
35Transgenic plants are the ones
[CBSE AIPMT 2009]
(a) generated by introducing foreign DNA
into a cell and regenerating a plant
from that cell
(b) produced after protoplast fusion in
artificial medium
(c) grown in artificial medium after
hybridisation in the field
(d) produced by a somatic embryo in
artificial medium
Ans.(a)
The plants obtained through genetic
engineering contain a gene or genes
usually from an unrelated organism,
such genes are called transgenes and
the plants containing transgenes are
known as transgenic plants. These
plants are often called as genetically
modified or GM crops, e.g.Flavr savr
tomatoes, golden rice. Plants are made
transgenic for identification, expressing
the gene activity in time, to produce
several chemicals like fatty acids,
sugars, cellulose, rubber, etc.
36In transgenics, expression of
transgene in target tissue is
determined by[CBSE AIPMT 2004]
(a) enhancer (b) transgene
(c) promoter (d) reporter
Ans.(d)
In transgenics, expression of transgene
in target tissue is determined by
reporter.
Reporter genes or screenable genes are
a number of marker genes which are
commonly used in plant transformation,
e.g. cat, lux, npt II, gus, etc.
37The Ti plasmid, is often used for
making transgenic plants. This
plasmid is found in
[CBSE AIPMT 2004]
(a)Azotobacter
(b)Rhizobiumof the roots of leguminous
plants
(c)Agrobacterium
(d) Yeast as a 2μm plasmid
Ans.(c)
A segment (T-DNA) of tumour-inducing
plasmid (T
1
) can be transformed from the
bacteriumAgrobacteriumto plant cells
at a wound site. This Ti-plasmid is often
for developing transgenic plants.
38Producing a giant mouse in the
laboratory was possible through
[CBSE AIPMT 2000]
(a) gene mutation
(b) gene manipulation
(c) gene synthesis
(d) gene duplication
Ans.(b)
Transgenic mice (of much larger size
than the normal) have been produced in
the laboratory by gene manipulation
(transfer of genes to fertilised eggs as
well as to stem cells).
39The first successfully cloned
mammals (animal) that gained
worldwide publicity was
[CBSE AIPMT 2000]
(a) Molly (a sheep) (b) Polly (a sheep)
(c) Chance (a bull) (d) Dolly (a sheep)
Ans.(d)
Dolly (a sheep) was the first successfully
cloned mammal that gained worldwide
publicity.
40The transgenic animals are those
which have [CBSE AIPMT 1995]
(a) foreign DNA in some of its cells
(b) foreign DNA in all its cells
(c) foreign RNA in all its cells
(d) DNA and RNA both in the cells
Ans.(b)
Transgenic animals have foreign DNA in
all its cells. The animals which carry
foreign genes are called transgenic
animals. The foreign genes are inserted
into the genome of animals using
recombinant DNA technology or gene
manipulation.
Biotechnology and Its Applications 337

348. 01Niche is [NEET 2018]
(a) the range of temperature that the
organism needs to live
(b) the physical space where an
organism lives
(c) all the biological factors in the
organism’s environment
(d) the functional role played by an
organism where it lives
Ans.(d)
Nicheis an ecological component of
habitat which is delimited by
functioning of an organism. A species
may live in more than one niche in
diferent stages of its life cycle.
02Which one of the following pairs is
mismatched?[CBSE AIPMT 2005]
(a) Savanna — Acaciatrees
(b) Prairie — Epiphytes
(c) Tundra — Permafrost
(d) Coniferous forest — Evergreen
trees
Ans.(b)
Prairies contain tall grasses and shrubs
(a) Savanna — Acaciatrees
(b) Tundra —Permafrost
(c) Coniferous forest—Evergreen trees
03Keystone species in an ecosystem
are those which[CBSE AIPMT 1997]
(a) are present in maximum number
(b) are most frequent
(c) attain a large biomass
(d) contribute to ecosystem properties
Ans.(d)
Keystone species of an ecosystem is a
species that exerts an important
regulatory effect on other species in
the community, i.e. contributes to
ecosystems properties. It maintains
higher species diversity in a community
by reducing the densities of strong
competitors.
04Niche of a species in an
ecosystem refers to its
[CBSE AIPMT 1996]
(a) function at its place of occurrence
(b) place of its occurrence
(c) competitive ability
(d) centre of origin
Ans.(a)
Niche word is used for the functional
role of species or population that it
plays in its ecosystem.
05Tropical forests occur in India
[CBSE AIPMT 1994]
(a) Jammu and Kashmir
(b) Rajasthan
(c) Kerala and Assam
(d) The forests do not occur in India
Ans.(c)
Tropical forest occur in India in Assam,
Western Ghats and Western Himalayas.
Major vegetation made of sal (Shorea
robusta), shrubs, grasslands and desert
community in the regions of decreasing
rainfall.
06Xeric environment is characterised
by [CBSE AIPMT 1994]
(a) precipitation
(b) low atmospheric humidity
(c) extreme of temperature
(d) high rate of vapourisation
Ans.(b)
Xeric environment is characterised by
low atmospheric humidity. This
environment is found in deserts.
07Study of inter-relationships
between living organisms and
their environment is
[CBSE AIPMT 1993]
(a) Ecology
(b) Ecosystem
(c) Phytogeography
(d) Ethology
Ans.(a)
Study of inter relationship between
living organisms and their environment
is called as Ecology. Ecology may be
called environmental biology.
08The sum total of the population of
the same kind of organisms
constitute [CBSE AIPMT 1993]
(a) colony
(b) genus
(c) community
(d) species
Ans.(d)
Species is a group of organisms similar
in structure, function and behaviour.
09What is true for individuals of
same species?
[CBSE AIPMT 2002]
(a) Live in same niche
(b) Live in same habitat
(c) Interbreeding
(d) Live in different habitats
Organismsand
Population
35
Ecology
TOPIC 1

349. Organisms and Population 339
Ans.(c)
According to the biological concept of
species.Speciesis a group of
organisms which can interbreed freely
in nature and produce fertile offsprings.
10Which part of the world has high
density of organisms?
[CBSE AIPMT 1999]
(a) Deciduous forests
(b) Grasslands
(c) Savannas
(d) Tropical rain forests
Ans.(d)
The tropical rain forests are most
diverse and highly dense with maximum
productivity (approximately 12000 kcal/
m
2
/yr).
11An association of individuals of
different species living in the
same habitat and having
functional interactions is
[CBSE AIPMT 2015]
(a) ecological niche
(b) biotic community
(c) ecosystem
(d) population
Ans.(b)
An association of individuals of
different species living in the same
habitat and having functional
interaction is called biotic community
(biocoenosis). The biotic community is
dominated by one of the character. It
determines the nature of community. It
can be biotic or abiotic.
12Competition for light, nutrients
and space is most severe
between [CBSE AIPMT 1988]
(a) closely related organisms growing
in different niches
(b) closely related organisms growing
in the same area/niche
(c) distantly related organisms
growing in the same habitat
(d) distantly related organisms
growing in different niches
Ans.(b)
Niche word was used for the first time
byJoseph Grinnel(1917). Niche means
functional role of an organism in an
ecosystem. Competition becomes most
severe between the closely related
organisms which share same niche.
13.Match the organisms in Column I
with habitats in Column II
[NEET (Odisha) 2019]
Column I Column II
1. Halophiles i. Hot springs
2. Thermoacidophiles ii. Aquatic
environment
3. Methanogens iii. Guts of
ruminants
4. Cyanobacteria iv. Salty area
Select the correct option from the
following
1 2 3 4
(a) (iv) (i) (iii) (ii)
(b) (i) (ii) (iii) (iv)
(c) (iii) (iv) (ii) (i)
(d) (ii) (iv) (iii) (i)
Ans.(a)
The correct match of organisms with
their habitats are
Halophiles live in salty areas
Thermoacidophiles live in hot springs
Methanogens live in guts of ruminants
Cyanobacteria live in aquatic
environment
14It is much easier for a small
animal to run uphill than for a large
animal, because
[NEET 2016, Phase I]
(a) smaller animals have a higher
metabolic rate
(b) small animals have a lower O
2
requirement
(c) the efficiency of muscles in large
animals is less than in the small
animals
(d) it is easier to carry a small body
weight
Ans.(a)
Basal metabolic rate is inversely
proportional to body size. So, smaller
animals have a higher metabolic rate,
thus have quick and more energy
required to go up the hills.
15The figure given below is a
diagrammatic representation of
response of organisms to abiotic
factors. What do (i), (ii) and (iii)
represent respectively?
[CBSE AIPMT 2010]
(i) (ii) (iii)
(a) Conformer Regulator Partial
regulator
(b) Regulator Partial
regulator
Conformer
(c) Partial
regulator
regulator Conformer
(d) Regulator Conformer Partial
regulator
Ans.(d)
In the given diagrammatic
representation of response of
organisms to abiotic factors
(i)RegulatorSome organisms are
able to maintain homeostasis by
physiological (sometimes
behavioural also) means which
ensures constant body
temperature, constant osmotic
concentration, etc. They are
known as regulators.
(ii)ConformerMost animals and
plants cannot maintain a constant
internal environment. Their body
temperature changes with the
ambient temperature. These
animals and plants are simply
called conformer.
(iii)Partial regulatorDuring the course
of evolution, the costs and
benefits of maintaining a constant
internal environment are taken
into consideration. Some species
have evolved the ability to regulate
but only over a limited range of
environmental conditions, beyond
which they simply conform. They
are partial regulators.
16Consider the following four
statements (I-IV) about certain
desert animals such as kangaroo
rat.
I. They have dark colour and high
rate of reproduction and
excrete solid urine.
(ii)
(i)
(iii)
Internal level
External level
Ecological Factors
TOPIC 2

350. II. They do not drink water,
breathe at a slow rate to
conserve water and have their
body covered with thick hairs.
III. They feed on dry seeds and do
not require drinking water.
IV. They excrete very
concentrated urine and do not
use water to regulate body
temperature.
Which two of the above
statements for such animals are
true? [CBSE AIPMT 2008]
(a) III and I (b) I and II
(c) III and II (d) II and III
Ans.(c)
Kangaroo rat feeds on dry seeds. It
seldom drinks water. The requirement
of water is met by food (10%) and
metabolic water (90%). Water loss is
prevented by living in burrows during
the day, concentration of urine and
solidification of faeces. It has a thick
coat to minimise evaporative
desiccation.
17.Annual migration does not occur
in the case of[CBSE AIPMT 2006]
(a) salmon (b) siberian crane
(c) salamander (d) arctic fern
Ans.(c)
Salamander is semiterrestrial lizard-like
tailed amphibian that lives under
stones, logs and inside cervices. They
show hibernation not annual migration.
Salmon areanadromous, i.e. they
spend their adult lives at sea but return
to freshwater to spawn. The pacific
species is legandry : after migrating
down stream as a smolt a sockeye
salmon ranges many hundreds of mile
over the pacific for nearly four year and
then returns to spawn in the head
waters of its parent stream.
Migration is characteristic feature of
birds.Arctic terntravels about 1100
miles during winter and returns back
during summer.
18In which one of the following pair
is the specific characteristic of
soil not correctly matched?
[CBSE AIPMT 2004]
(a) Laterite — Contains aluminium
compound
(b) Terra rossa — Most suitable for
roses
(c) Chernozems — Richest soil in the
world
(d) Black soil — Rich in calcium
carbonate
Ans.(d)
Black soil is dark or dark brown in
colour. It is formed from basaltic rock
under semi-arid condition. Black soil is
logically known asreguror black cotton
soil. Black soil is deficient in nitrogen
and phosphorus and rich in potash and
lime and not in calcium carbonate.
19In which one of the following
habitats does the diurnal
temperature of soil surface vary
most? [CBSE AIPMT 2004]
(a) Shrubland (b) Forest
(c) Desert (d) Grassland
Ans.(c)
Deserts have a very hot days and very
cold nights. Due to the bare plant cover,
the soil of desert is much more exposed
to these fluctuations as compared to
that of other areas. During day time, the
soil becomes hot and in night it
frequently, becomes cool.
20Diffuse porous woods are
characteristic of plants growing in
[CBSE AIPMT 2003]
(a) temperate climate
(b) tropics
(c) alpine region
(d) cold winter regions
Ans.(b)
In tropics, there is no sharp distinction
between the seasons, hence, there is
not much difference in the activity of
cambium. In a diffused porous wood,
the large sized vessels are distributed
through spring wood and autumn wood,
e.g.Syzygium cumini.
21Special kinds of roots called
pneumatophores are
characteristics of the plants
growing in[CBSE AIPMT 2000]
(a) sandy soils
(b) saline soils
(c) marshy places and salt lakes
(d) dryland regions
Ans.(c)
Pneumatophores are specialised roots
which grow vertically upwards into the
air from roots embedded in the mud.
Since, they are loosely constructed,
these make gaseous exchange possible
for submerged roots. These are found
in plants growing in marshes or saline
swamps.
22Temperature changes in the
environment affect most of the
animals which are
[CBSE AIPMT 1999]
(a) homeothermic
(b) aquatic
(c) poikilothermic
(d) desert living
Ans.(c)
Poikilothermy (cold bloodedness) is a
condition of any animal whose body
temperature fluctuates considerably
with that of its environment.
Homeothermy, on the other hand, is the
quality of maintaining a constant body
temperature.
23Extremities, tail and ear are
relatively shorter in animals living
in cooler regions as compared to
those inhabiting warmer zones.
This is [CBSE AIPMT 1996]
(a) Bergman’s rule
(b) Jordan’s rule
(c) Gloger’s rule
(d) Allen’s rule
Ans.(d)
According to Allen’s rule, extremities,
tail and ear are relatively shorter in
animals living in cooler regions as
compared to those inhabiting warmer
zones.
24Desert plants are generally
[CBSE AIPMT 1995]
(a) viviparous (b) succulent
(c) herbaceous (d) heterophyllus
Ans.(b)
Desert plants are generally succulents
or fleshy xerophytes, They are referred
as drought resisting xerophytes, e.g.
Opuntia, Bryophyllum, Euphorbia,
Mesembryanthemum(ice plant).
25Sunken stomata is the characteristic
feature of [CBSE AIPMT 1995]
(a) hydrophyte (b) mesophyte
(c) xerophyte (d) halophyte
Ans.(c)
Sunken stomata is the characteristic
feature of xerophytes, these stomata
are found generally on the lower
surface of leaves.
340 NEETChapterwise Topicwise Biology

351. 26Which of the following does not
have stomata?[CBSE AIPMT 1995]
(a) Hydrophytes
(b) Mesophytes
(c) Xerophytes
(d) Submerged hydrophytes
Ans.(d)
Sub-merged hydrophytes are those
plants which live completely inside the
water, so there is no need of
transpiration that’s why these plants do
not have stomata, e.g.Utricularia,
Ceratophyllum.
27Animals that can tolerate a
narrow range of salinity are
[CBSE AIPMT 1994]
(a) stenohaline (b) euryhaline
(c) anadromous (d) catadromous
Ans.(a)
Animals that can tolerate only a small
range of salinity are stenohaline.
28Soil best suited for plant growth is
[CBSE AIPMT 1993]
(a) clay (b) loamy
(c) sandy (d) gravel
Ans.(b)
Loamy soil containing about 1 part clay,
2 parts silt and 2 parts sand (20% clay,
40% silt and 40% sand) is best for plant
growth because it possesses good
aeration, sufficient nutritive salts and
good water retaining capacity.
29Soil particles determine its
(a) texture [CBSE AIPMT 1992]
(b) field capacity
(c) water holding capacity
(d) soil flora
Ans.(a)
Soil particles determine its texture. The
behaviour of water in the ground is
influenced by the type of soil present.
Soils are classified according to their
particle size as follows:
(i) Gravel – 2 mm – 75 mm
(ii) Sand – 0.05 mm – 2 mm
(iii) Silt – 0.002 mm – 0.05 mm
(iv) Clay – less than 0.002 mm
30A fertile agricultural soil appears
dark coloured at the surface as
compared to soil one metre down.
The reason for colour of top soil is
[CBSE AIPMT 1992]
(a) more moisture
(b) rich in organic matter
(c) rich in iron, calcium and magnesium
(d) recent formation
Ans.(b)
Dark colour of soil is due to
accumulation of leached organic
substances and organic matter which
serves as a reservoir of nutrients and
water in the soil, aids in reducing
compacting and surface crusting and
increases water infiltration into the soil.
31River water deposits
[CBSE AIPMT 1992]
(a) loamy soil (b) alluvial soil
(c) laterite soil (d) sandy soil
Ans.(b)
River water deposits are found in
alluvial soil. It is rich in nutrients and
may contain heavy metals. These soils
are formed when streams and rivers
slow their velocity and suspended soil
particles gets deposited on the river
bed.
32Deep black soil is productive due
to high proportion of
[CBSE AIPMT 1991]
(a) sand and zinc
(b) gravel and calcium
(c) clay and humus
(d) silt and earthworm
Ans.(c)
Black soil is productive due to the high
proportion of clay and humus, because
most of the minerals are present in it.
33CAM helps the plants in
[CBSE AIPMT 2011]
(a) secondary growth
(b) disease resistance
(c) reproduction
(d) conserving water
Ans.(d)
CAMplants are mostly succulent
xerophytes. The stomata in these
plants remain closed during the day.
This helps to check the transpiration. In
this way, water is conserved.
34Two different species cannot live
for long duration in the same
niche or habitat. This law is
[CBSE AIPMT 2002]
(a) Allen’s law
(b) Gause’s hypothesis
(c) Dollo’s rule
(d) Weismann’s theory
Ans.(b)
The principle of competitive exclusion
was postulated by Soviet ecologistG F
Gause. It states that if two species are
competing with one another for the
same limited resources, then one of the
species will be able to use that resource
more efficiently than the other and the
former will, therefore, eventually
eliminate the latter locally.
35In which of the following plant
sunken stomata are found?
[CBSE AIPMT 2001]
(a)Nerium (b)Hydrilla
(c) Mango (d) Guava
Ans.(a)
Presence of sunken stomata is an
adaptive feature of xerophytic plants.
These stomata are partially covered by
hairs and cuticle. Sunken stomata are
found inNeriumto check the
transpiration.
Mango is a mesophytic plant.
Hydrillais a hydrophytic plant.
Guava is also a mesophytic plant.
36Inspite of interspecific
competition in nature, which
mechanism the competing
species might have evolved for
their survival ?
[NEET 2021]
(a) Resource partitioning
(b) Competitive release
(c) Mutualism
(d) Predation
Ans.(a)
Resource partitioning is the
phenomenon where division of limited
resources occurs by species to help
avoid interspecific competition in an
ecological niche. If two species
compete for the same resource, they
could avoid competition by choosing,
for instance, different times for feeding
or different foraging patterns. Thus, we
can say inspite of interspecific
competition in nature, competing
species evolved a mechanism called
resource partitioning for their survival.
Other options can be explained as:
Competitive releaseoccurs when one
or two species competing for the same
resource disappears, thereby allowing
Organisms and Population 341
Population
TOPIC 3

352. the remaining competitor to utilize the
resource more fully than it could in the
presence of the first species.
Mutualismis the interaction that confers
benefits on both the interacting species.
Lichens represent an intimate mutualistic
relationship between a fungus and
photosynthesising algae or cyanobacteria.
Both the species benefit in mutualism and
both lose in competition in their
interactions with each other. In both
parasitism andpredationonly one
species benefits (parasite and predator,
respectively) and the interaction.
37In the exponential growth
equationN N e
t
rt
=
0
,erepresents
[NEET 2021]
(a) the base of number logarithms
(b) the base of exponential logarithms
(c) the base of natural logarithms
(d) the base of geometric logarithms
Ans.(c)
When resources in the habitat are
unlimited, each species has the ability
to realise fully its innate potential to
grow in number, as Darwin observed
while developing his theory of natural
selection. Then the population grows in
an exponential or geometric fashion.
The integral form of the exponential
growth can be represented by equation
as
N
t
=N
ert
0
where,
N
t
= Population density after timet
N
0
= Population density at time zero
r= intrinsic rate of natural increase
e= the base of natural logarithms.
In exponential growth, a population's
per capita (per individual) growth rate
stays the same regardless of population
size, making the population grow faster
and faster as it gets larger.
38Amensalism can be represented
as [NEET 2021]
(a) species A( )−: species B (0)
(b) species A (+) : species B (+)
(c) species A (−) : species B (−)
(d) species A (+) : species B (0)
Ans.(a)
Amensalism is the relationship between
two organisms, where one is hurt.
A prime example of amensalism is
penicillin killing bacteria. The bread
mouldPenicilliumsecretes penicillin
that ultimately kills bacteria. In this
contact between two organisms, one is
destroyed or inhibited, and other
remains unaffected. Hence, it is
represented as Species A (−); Species B
(0).
39Match the items inColumn Iwith
those inColumn II.
[NEET (Oct.) 2020]
Column I Column II
A.Herbivores-Plants (i) Commensalism
B.Mycorrhiza-Plants (ii) Mutualism
C.Sheep-Cattle (iii) Predation
D.Orchid-Tree (iv) Competition
Select the correct option.
A B C D
(a) (iv), (ii), (i), (iii)
(b) (iii), (ii), (iv), (i)
(c) (ii), (i), (iii), (iv)
(d) (i), (iii), (iv), (ii)
Ans.(b)
Option (b) is correct match, which is as
follows
The relationship between herbivores
and plants is prey-predator type in which
herbivores are predators and plants are
the prey.
In mycorrhiza plants association, both
species are benefited and thus it
represents mutualism. The sheep and
cattle show competition for common
resources like food, i.e. grass.
Orchid are epiphytes which grow on
trees and derive nutrients from it. In
this process, trees are neither harmed
nor helped. Thus, it is type
commensalism relationship.
40The impact of immigration on
population density is
[NEET (Oct.) 2020]
(a) negative
(b) Both positive and negative
(c) neutralised by natality
(d) positive
Ans.(d)
Population density is the member of
individuals present per unit area or
volume at a given time. It is calculated
by the formulaD N /S=, whereD=
Density,N =Total number of individuals
andS =Number of units of space. Since
immigration increases the number of
individuals in an area, population
density increase. Thus, immigration has
positive impact on population density.
41According to Alexander von
Humboldt [NEET (Oct.) 2020]
(a) species richness decreases with
increasing area of exploration
(b) species richness increases with
increasing area, but only up to
limit
(c) there is no relationship between
species richness and area
explored
(d) species richness goes on
increasing with increasing area of
exploration
Ans.(b)
Alexander von Humboldt was a German
naturalist and geographer. He proposed
that within a region, species richness
increases with increasing explored
area, but only up to a limit.
Accordingly, the relation between
species richness and area for a wide
variety of taxa turn out to be
rectangular hyperbola. The relationship
appears as a straight line on logarithmic
scale and described by the equation :
log log logS C Z A= =
Where,S–Species richness,A–Area
Z–Regression coefficient,
C–Y-Intercept.
42Which of the following is not an
attribute of a population?
[NEET (Sep.) 2020]
(a) Natality
(b) Mortality
(c) Species interaction
(d) Sex ratio
Ans.(c)
Species interaction is not an attribute
of a population. Rest Natality (Birth
rate), Mortality (Death rate) and Sex radio
are population attributes.
43Secondary metabolites such as
nicotine, strychnine and caffeine
are produced by plants for their
[NEET (Sep.) 2020]
(a) growth response
(b) defence action
(c) effect on reproduction
(d) nutritive value
Ans.(b)
A wide variety of chemical substances
(i.e. secondary metabolites) that we
extract from plants on a commercial
scale (nicotine, caffeine, quinine,
strychnine, opium, etc) are produced by
them (plants) as defence against
grazers and browsers.
342 NEETChapterwise Topicwise Biology

353. Organisms and Population 343
44Between which among the
following, the relationship is not
an example of commensalism?
[NEET (Odisha) 2019]
(a) Orchid and the tree on which it
grows
(b) Cattle egret and grazing cattle
(c) Sea anemone and clown fish
(d) Female wasp and fig species
Ans.(d)
Among the given examples, relationship
between wasp and fig species does not
show commensalism. In this
relationship, one species derives the
benefit and other neither harmed nor
benefitted.
Wasp and fig tree show mutualism.
Here fig flower is pollinated by wasp and
wasp lays its egg into fruit and leaves
them there for development.
Other options show examples of
commensalism.
45Carnivorous animals-lions and
leopards, occupy the same niche
but lions predate mostly larger
animals and leopards take smaller
ones. This mechanism of
competition is referred to as
[NEET (Odisha) 2019]
(a) character displacement
(b) altruism
(c) resource partitioning
(d) competitive exclusion
Ans.(c)
Carnivorous animals, lions and leopard,
occupy the same niche but lion
predates mostly larger animals and
leopard takes smaller ones. This is
called resource partitioning. It is a
mechanism in which there is the
division of limited resources by species
to help avoid competition in an
ecological niche. In any environment,
organisms compete for limited
resources, so organisms and different
species have to find ways to coexist
with one another. That is why lions
predate mostly larger animals and
leopards take smaller ones.
46Which of the following statements
is correct?[NEET (Odisha) 2019]
(a) Lichens do not grow in polluted
areas
(b) Algal component of lichens is
called mycobiont
(c) Fungal component of lichens is
called phycobiont
(d) Lichens are not good pollution
indicators
Ans.(a)
(a) Statement (a) is correct. Lichens
do not grow in polluted area. Rest
statements are incorrect.
The correct forms of the
statements are as follows
(b) Algal component of lichens is
called phycobiont.
(c) Fungal component of lichens is
called mycobiont.
(d) Lichens are good pollution
indicators.
47Pinusseed cannot germinate and
establish without fungal
association. This is because
[NEET (National) 2019]
(a) it has obligate association with
mycorrhizae
(b) it has very hard seed coat
(c) its seeds contain inhibitors that
prevent germination
(d) its embryo is immature
Ans.(a)
Pinus has an obligate association with
mycorrhizae due to which thePinus
seeds are unable to germinate and
establish in the absence of fungal
partner.
Fungus or mycorrhizae help thePinus
roots to absorb water and minerals by
increasing their surface area. In turn,
the fungus derives food from the plant.
48Match Column I with Column II.
Column I Column II
A. Saprophyte (i) Symbiotic
association of
fungi with plant
roots
B. Parasite (ii) Decomposition of
dead organic
materials
C. Lichens (iii) Living on living
plants or animals
D. Mycorrhiza (iv) Symbiotic
association of
algae and fungi
Choose the correct answer from
the option given below:
[NEET (National) 2019]
A B C D
(a) (iii) (ii) (i) (iv)
(b) (ii) (i) (iii) (iv)
(c) (ii) (iii) (iv) (i)
(d) (i) (ii) (iii) (iv)
Ans.(c)
A)–(ii), (B)–(iii) , (C)–(iv) (D)–(i)
Saprophytes are decomposers which
help in the decomposition of dead
organic material, e.g.Agaricus. Parasites
are the entitites which live on other living
plants or animals and derive nutrition
from them, e.g. tapeworm in humans.
Lichens represent symbiotic
association between algae and fungi.
Mycorrhiza is symbiotic association of
fungi and plant roots.
49Which one of the following plants
shows a very close relationship
with a species of moth, where
none of the two can complete its
life cycle without the other?
[NEET 2018]
(a) Banana (b) Yucca
(c)Hydrilla (d)Viola
Ans.(b)
Yucca gloriosahas developed an
obligate symbiotic relationship with
Pronubayuccasellamoth. The moth
cannot complete its life cycle with the
association ofYuccaflowers and in turn
Yuccahas no other pollinator.
Hydrilla is a hydrophilous plant while
Violais an entomophilous plant.
Bananasare usually parthenocarpic
fruits. Therefore, they do not require
pollination.
Concept EnhancerThe female moth
visits theYuccaflowers at night and
collects pollen in the form of balls. The
moth, then inserts its ovipositor into
ovary of the flower to lay eggs. The
temperature of the ovary is suitable for
hatching ofPronuba’seggs and works
as an incubator. After that, it climbs to
the top of the style and pushes the
pollen ball into stylar canal. Thus,
pollination occurs. Some seeds are
eaten by larvae which escape after
piercing the ovary wall.
50Natality refers to[NEET 2018]
(a) number of individuals leaving the
habitat
(b) birth rate
(c) death rate
(d) number of individuals entering a
habitat

354. Ans.(b)
Natalityisbirth rate. It refers to the
number of births during a given period
in the population that are added to the
initial density.
Death rateis termed as mortality. It
refers to the number of deaths in the
population during a given period.
Immigrationis the number of
individuals of the same species that
have come into the habitat, on the
other handemigrationis the number of
individuals of the population who left
the habitat.
51Which one of the following
population interactions is widely
used in medical science for the
production of antibiotics?
[NEET 2018]
(a) Parasitism (b) Mutualism
(c) Commensalism (d) Amensalism
Ans.(d)
Amensalism is widely used in medical
science for the production of
antibiotics.
It involves, the secretion of chemicals
called allochemics by one microbial
group to harm other microbes, e.g.,
Penicilliumsecretes chemicals to
inhibit the growth ofStaphylococcus
bacteria. These chemicals can be used
in medical science for the production of
antibiotics. On the other hand, no such
chemicals are secreted in parasitism,
mutualism and commensalism.
52In a growing population of a
country, [NEET 2018]
(a) reproductive and
pre-reproductive individuals are
equal in number
(b) reproductive individuals are less
than the post-reproductive
individuals
(c) pre-reproductive individuals are
more than the reproductive
individuals
(d) pre-reproductive individuals are
less than the reproductive
individuals
Ans.(c)
In a growing population, younger
population (or pre-reproductive
individuals) size is larger than that of
reproductive individuals. Such
population is represented by a
triangular-shaped age pyramid.
Whereas, the equal number of
reproductive and pre-reproductive
individuals represents a stable
population and the age pyramid is
bell-shaped.
Less number of pre-reproductive
individuals than reproductive
individuals represents declining
population and age pyramid appears
urn-shaped. The similar case is seen
when reproductive individuals are less
than the post-reproductive individuals.
53Asymptote in a logistic growth
curve is obtained, when
[NEET 2017]
(a) The value of ‘r’ approaches zero
(b)K N=
(c)K N>
(d)K N<
Ans.(b)
WhenK N=in a logististics growth
curve, it is asymptote.
It means a population growing in a
habitat with limited resources show
initially a lag phase, followed by phase
of acceleration and deceleration and
finally anasymptote, i.e. when the
population density( )Nreaches the
carrying capacity( )K
Population growth curve is logistic,
when responses are limiting the
growth,
hereKis carrying capacity andNis
population density.
54Mycorrhizae are the example of
[NEET 2017]
(a) fungistasis (b) amensalism
(c) antibiosis (d) mutualism
Ans.(d)
Mutualismis an association of two
species in, which both species are
benefitted.
Mycorrhizais a mutualistic relationship
between fungal hyphae and roots of
higher plants. The fungus helps in
mineral nutrition absorption for the
plants with, which they are associated
and obtained in turn, nutrients from
plants. Concept Enhancer
Amensalismis an interaction between
different species, in which one species
is harmed and other is neither
benefitted nor harmed, e.g.Penicillium.
AntibiosisIt is an antagonistic
association between two or more
organism, in which one is adversely
affected, e.g. antibiosis includes the
relationship between antibiotic and
bacteria.
Fungistasis inhibits the growth of fungi.
55Gause’s principle of competitive
exclusion states that
[NEET 2016, Phase I]
(a) Competition for the same
resources excludes species
having different food preferences
(b) No two species can occupy the
same niche indefinitely for the
same limiting resources
(c) Larger organisms exclude smaller
ones through competition
(d) More abundant species will
exclude the less abundant species
through competition
Ans.(b)
Gause’s principle of competitive
exclusion states that no two species
can occupy the same niche indefinitely
for the same limiting resources
56When does the growth rate of a
population following the logistic
model equal zero? The logistic
model is given asdN/dt=rN(l-N/K)
[NEET 2016, Phase I]
(a) whenNnears the carrying
capacity of the habitat
(b) whenN/Kequals zero
(c) when death rate is greater than
birth rate
(d) whenN/Kis exactly one
Ans.(d)
In logistic growth model population
growth equation is described as
dN
dt
rN
K N
K
=
−





where,N=Population density at timet
r=Intrinsic rate of natural increase
K=Carrying capacity
when,
N
K
=1then
K N
K
−
=0
Therefore,
dN
dt
=0
344 NEETChapterwise Topicwise Biology
dN
dt
=rN
K–N
K
K
Time (t)
Population density ( )
N

355. 57The principle of competitive
exclusion was stated by
[NEET 2016, Phase II]
(a) C Darwin
(b) GF Gause
(c) MacArthur
(d) Verhulst and Pearl
Ans.(b)
The principle of competitive exclusion
is stated by GF Gause. He studied the
effects of interspecific competition
between two closely related species.
He stated that two species competing
for the same food resource cannot
coexist at the same place as highest
degree of competitiveness exists
between them.
58If ‘+’ sign is assigned to beneficial
interaction, ‘–’ sign to detrimental
and ‘0’ sign to neutral interaction,
then the population interaction
represented by ‘+’ ‘–’ refers to
[NEET 2016, Phase II]
(a) mutualism
(b) amensalism
(c) commensalism
(d) parasitism
Ans.(d)
Parasitism is a relationship between
two living organisms of different
species in which one organism, i.e.
parasite obtains its food directly from
the host. In this relationship the
parasite is benefitted (+) and the host is
harmed (–). So, this type of population
interaction is represented by ‘+’ ‘–’.
59Which of the following is correct
forr-selected species?
[NEET 2016, Phase II]
(a) Large number of progeny with
small size
(b) Large number of progeny with
large size
(c) Small number of progeny with
small size
(d) Small number of progeny with
large size
Ans.(a)
r-selected are the species having the
ability to produce large number of
progenies (offsprings) with small size.
The population growth of these species
is a function of biotic potential. Hence,
option (a) is correct.
60In which of the following
interactions both partners are
adversely affected?
[CBSE AIPMT 2015]
(a) Competition (b) Predation
(c) Parasitism (d) Mutualism
Ans.(a)
Competition is a negative interaction
that occurs among organisms
whenever two or more organisms
require the same limited resource.
Population Interactions
Species A Species B Name of Interaction
+ + Mutualism
− − Competition
+ − Predation
+ − Parasitism
61A biologist studied the population
of rats in a barn. He found that
the average natality was 250,
average mortality 240,
immigration 20 and emigration
30. The net increase in population
is [NEET 2013]
(a) 10 (b) 15
(c) 05 (d) zero
Ans.(d)
A population has birth rates and death
rates. The rates are expressed as
change in numbers (increase or
decrease) with respect to members of
the population.
In this case, the net increases in
population will be zero. Because Birth
rate (B)+Immigration (I)−Death rate (D)
+Emmigration (I)=Density of
population.
Therefore, Density
= + − +[ ] [ ]250 20 240 30 =0
62A sedentary sea anemone gets
attached to the shell lining of
hermit crab. The association is
[NEET 2013]
(a) ectoparasitism (b) symbiosis
(c) commensalism (d) amensalism
Ans.(b)
This type of mutualism to called
symbiosis. In this type, the sea
anemone grows on the back of the
hermit crab. It protects the crab with
the help of its nematocysts. In
ectoparasitism an ectoparasite live on
the outside of host, e.g. human body
louse. In this interaction, the parasite
gets the benefit at the expense of the
host. Commensalism is an association
between organisms in which one or both
the species are benefitted and neither
species is harmed. In amensalism one
species is harmed, whereas the other is
unaffected.
63What type of human population is
represented by the following age
pyramid?
[CBSE AIPMT 2011]
(a) Vanishing population
(b) Stable population
(c) Declining population
(d) Expanding population
Ans.(c)
An age pyramid is a graphic
representation of proportion of various
age groups of a population with
pre-reproductive at the base,
reproductive in the middle and post
reproductive at the top. For human
population, the age pyramids show age
distribution of males and females in a
combined diagram. The shape of the
age pyramids reflects the growth status
of the population. In a declining
population the shape of pyramid is
urn-shaped.
64Consider the following four
conditions (I-IV) and select a
correct pair of them as adaptation
to environment in desert lizards.
Conditions [CBSE AIPMT 2011]
I. Burrowing in soil to escape
high temperature.
II. Losing heat rapidly from the
body during high temperature.
Organisms and Population 345
Density
–+
+
Natality Mortality
Immigration
Emigration
Post-reproductive
Reproductive
Pre-reproductive

356. III. Bask in sun when temperature
is low.
IV. Insulating body due to thick
fatty dermis.
(a) III, IV (b) I, III (c) II, IV (d) I, II
Ans.(b)
Desert lizards bask in the sun and
absorb heat when their body
temperature drops below the comfort
zone, but move into shade when the
ambient temperature starts increasing.
Some species are capable of burrowing
into the soil to hide and escape from
the above-ground heat.
65Which one of the following is
categorised as a parasite in true
sense? [CBSE AIPMT 2011]
(a) The femaleAnophelesbites and
sucks blood from humans
(b) Human foetus developing inside
the uterus draws nourishment
from the mother
(c) Head louse living on the human
scalp as well as laying eggs on
human hair
(d) The cuckoo (koel) lays its eggs in
crow’s nest
Ans.(c)
Human head louse (Pediculus) lives
among hair and surface of human body
feeding on blood. It spreads diseases
like typhus. It is a true parasite.
66Which one of the following is one
of the characteristics of a
biological community?
[CBSE AIPMT 2010]
(a) Stratification (b) Natality
(c) Mortality (d) Sex-ratio
Ans.(a)
Organisms are not uniformly distributed
throughout a community. They usually
occur in definite zones. This spatial
arrangement of populations is called
stratification which is characteristic of
biological community. Natality,
mortality, age structure and sex-ratio
are the basic characteristics of a
population.
67A country with a high rate of
population growth took measures
to reduce it. The figure below
shows age sex pyramids of
populations. A and B twenty years
apart. Select the correct
interpretation about them
[CBSE AIPMT 2009]
(a) ‘A’ is more recent and shows slight
reduction in the growth rate
(b) ‘B’ is earlier pyramid and shows
stabilised growth rate
(c) ‘B’ is more recent showing that
population is very young
(d) ‘A’ is the earlier pyramid and no
change has occurred in the
growth rate
Ans.(a)
Interpretation ‘a’ is correct for the given
figures. As ‘A’ is more recent and shows
slight reduction in the growth rate.
68A high density of elephant
population in an area can result in
[CBSE AIPMT 2007]
(a) mutualism
(b) intraspecific competition
(c) interspecific competition
(d) predation on one another
Ans.(b)
Intraspecific competition is an
important density dependent factor
regulating populations. Intraspecific
competition occurs between the
members of same population.
69The population of an insect
species shows an explosive
increase in numbers during rainy
season followed by its
disappearance at the end of the
season. What does this show?
[CBSE AIPMT 2007]
(a) S-shaped or sigmoid growth of
this insect
(b) The food plants mature and die at
the end of the rainy season
(c) Its population growth curve is of
J-type
(d) The population of its predators
increases enormously
Ans.(c)
Its population growth curve is J-shaped
in which density increases rapidly in
exponential fashion and then stops
abruptly as environmental resistance or
another limiting factor becomes
effective more or less suddenly.
70Geometric representation of age
structure is a characteristic of
[CBSE AIPMT 2007]
(a) biotic community
(b) population
(c) landscape
(d) ecosystem
Ans.(b)
Geometric representation of age
structure is a characteristic of
population. In most populations,
individuals are of different ages. The
proportion of individuals in each age
group is called age structure of that
population.
71If the mean and the median
pertaining to a certain character of
a population are of the same value,
the following is most likely to occur
[CBSE AIPMT 2007]
(a) normal distribution
(b) bi-modal distribution
(c) T-shaped curve
(d) skewed curve
Ans.(a)
For a normal distribution the mean,
median and mode are actually
equivalent.
72The formula for exponential
population growth is
[CBSE AIPMT 2006]
(a)dt/dN=rN
(b)dN/rN=dt
(c)rN/dN=dt
(d)dN/dt=rN
346 NEETChapterwise Topicwise Biology
Males Females
‘B’
15 12 9 6 3 0 3 6 9 12 15
Age
70+
60-69
50-59
40-49
30-39
20-29
10-19
0-9
Age (in years)
Males Females
‘A’
15 12 9 6 3 0 3 6 9 12 15
Age
70+
60-69
50-59
40-49
30-39
20-29
10-19
0-9
Age (in years)

357. Ans.(d)
J-shaped form of population growth is
mathematically described by an
equationofexponentialorgeometric
increase,which is as follows :
dN
dt
rN=
where,d=rate of change
t=time
N=number of females at a particular
time
r=biotic potential of each female
(Ncan also be considered as the total
population andras the biotic potential
of each individual).
73Niche overlap indicates
[CBSE AIPMT 2006]
(a) active cooperation between two
species
(b) two different parasites on the
same host
(c) sharing of one or more resources
between the two species
(d) mutualism between two species
Ans.(b)
Niche overlap is a measure of the
association of two or more species.
This indicate their similar habitat
requirement and may also indicate
competition if trophic niche/spatial
niche is same and food/space is
limiting, e.g. two different parasites on
the same host.
74Praying mentis is a good example
of [CBSE AIPMT 2006]
(a) Mullerian mimicry
(b) warning colouration
(c) social insects
(d) camouflage
Ans.(c)
Praying mantis(Mantis religiosa)is a
large social insect. It has small
triangular head, a long prothorax and an
abdomen consisting of ten segments.
The wings are well developed and the
pincer-like forelegs are modified for
grasping prey. It usually inhabits
plantation areas. It destroys certain
harmful insects so, it is useful.
75Animals have the innate ability to
escape from predation. Examples
for the same are given below.
Select the incorrect example
[CBSE AIPMT 2005]
(a) enlargement of body size by
swallowing air in puffer fish
(b) melanism in moths
(c) poison fangs in snakes
(d) colour change inChamaeleon
Ans.(c)
Animals resist predation by cryptic
colouration, deceptive marking,
behavioural defenses and the
possession of mechanical or chemical
defenses.
Example
Enlargement of body size by swallowing
air in puffer fish.
Melanism in moths.
Colour change inChamaeleon.
76Mycorrhiza is an example of
[CBSE AIPMT 2003]
(a) endoparasitism
(b) decomposers
(c) symbiotic relationship
(d) ectoparasitism
Ans.(c)
Mycorrhiza is a result of symbiosis
between the roots of higher plants
and fungi. In this association, plants
provide space and prepared food
material to fungi in exchange of this,
fungi help in absorption of minerals
and water to plants.
77The semilog of per minute
growing bacteria is plotted against
time. What will be the shape of
graph? [CBSE AIPMT 2002]
(a) Sigmoid
(b) Hyperbola
(c) Ascending straight line
(d) Descending straight line
Ans.(c)
Semilog of per minute growing
bacterium when plotted against time,
would yield ascending straight line.
78Choose the correct sequence of
stages of growth curve for
bacteria [CBSE AIPMT 2002]
(a) lag, log, stationary, decline phase
(b) lag, log, stationary phase
(c) stationary, lag, log, decline phase
(d) decline, lag, log phase
Ans.(a)
When microbes are grown in a closed
system or batch culture, the resulting
growth curve has usually four phases :
(a) lag phase
(b) exponential (log phase)
(c) stationary phase
(d) death phase
79Which type of association is found
in between entomophilous flower
and pollinating agent?
[CBSE AIPMT 2002]
(a) Mutualism (b) Commensalism
(c) Cooperation (d) Co-evolution
Ans.(a)
A plant and its pollinator have a
mutualistic relationship. The plant uses
its pollinator to ensure cross pollination
while pollinator uses the plant as food.
80Which of the following is a correct
pair? [CBSE AIPMT 2002]
(a)Cuscuta — Parasite
(b)Dischidia— Insectivorous
(c)Opuntia — Predator
(d)Capsella— Hydrophyte
Ans.(a)
Cuscuta, commonly known as dodder or
amarbel, is a parasitic plant.
81Two different species cannot live
for long duration in the same
niche or habitat. This law is
[CBSE AIPMT 2002]
Organisms and Population 347
4
8
16
32
64
128
256
1 2 3 4 5 6
Time (in minute)
Number of bacteria
7 8 9 10
Lag
phase
Log of
exponential
growth phase
Stationary
phase
Death or
logarithmic
decline phase
50 10
Time (h)
Lag of numbers of bacteria

358. 348 NEETChapterwise Topicwise Biology
(a) Allen’s law
(b) Mendel’s law
(c) Gause’s competitive exclusion
principle
(d) Weismann’s theory
Ans.(c)
The principle of competitive exclusion
was postulated by Soviet ecologistGF
Gause. It states that if two species are
competiting with one another for the
same limited resource, then one of the
species will be able to use that resource
more efficiently than the other and the
former will, therefore, eventually
eliminate the latter locally.
82An orchid resembling the female
of an insect, so as to be able to
get pollinated is due to the
phenomenon of
(a) mimicry [CBSE AIPMT 1998]
(b) pseudocopulation
(c) pseudopollination
(d) pseudoparthenocarpy
Ans.(b)
For its pollination, the orchidOphyrys
speculumhas picked on the most
selective attraction in the entire animal
kingdom. It is pollinated by a hairy wasp,
Colpa aurea. The wasp has a fixed habit
whereby its males leave the burrows for
above ground existence about four
weeks before the females emerge for
the open-air mating.
The orcind opens its flowers about the
same time the males appear and they
possess an appearance and odour
similar to those possessed by the
famele wasps.
The inexperienced males mistake the
Ophrysflowers for their female
counterparts and land to perform the
act of pseudocopulation. The insect
repeats the act with a number of orchid
flowers and carries pollinia from one
flower to another. This insect-plant
relationship is beneficial only to the
plant.
83The concept that population
tends to increase geometrically
while food supply increases
arithmetically was put forward by
[CBSE AIPMT 1995]
(a) Stuart Mill
(b) Adam Smith
(c) Charles Darwin
(d) Thomas Malthus
Ans.(d)
Malthusin his‘Essay on the principle of
population’(1798), pointed out that
population tends to increase in
geometric progression while food
supply increases only in arithmetic
progression.
84Which of the following pair is
correctly matched?
[CBSE AIPMT 1995]
(a) Uricotelism—Aquatic habitat
(b) Parasitism—Intra-specific
relationship
(c) Excessive perspiration—Xeric
adaptation
(d) Stream lined body—Aquatic
adaptation
Ans.(d)
The correct pair is stream lined body to
aquatic adaptation which helps these
animals in swimming.
85A mutually beneficial association
necessary for survival of both
partners are[CBSE AIPMT 91, 88]
(a) mutualism/symbiosis
(b) commensalism
(c) amensalism
(d) Both (a) and (b)
Ans.(a)
Mutualism is an association between
two organisms of different species
where both are benefitted but cannot
live separately (both favour the growth
and survival of each other and their
association is obligatory), e.g. pollination
by animals, dispersal of fruits and seeds
by animals, lichens, etc.
86The relation between algae and
fungi in a lichen is
[CBSE AIPMT 1989]
(a) symbiosis
(b) parasitism
(c) commensalism
(d) protocooperation
Ans.(a)
The relation between algae and fungi in
a lichen is symbiosis. In lichen fungi is
for water intake and algae is
photosynthetic and prepares food.
87Phenomenon when organisms
resembling others for escaping
from enemies is
[CBSE AIPMT 1988]
(a) adaptation (b) mimicry
(c) homology (d) analogy
Ans.(b)
Mimicry is the phenomenon of
resemblance of one species with
another. It is a means of adaptation and
protection against predation. The
species which is copied is called a
model, while the animal which copies
other is known as mimic, e.g. viceroy
butterfly mimics toxic monarch
butterfly.
88Association between sucker fish
(Remora) and shark is
[CBSE AIPMT 1988]
(a) commensalism
(b) symbiosis
(c) predation
(d) parasitism
Ans.(a)
Remora (Echeneis) has modified its
dorsal fin into a sucker. It attaches to
the body of sharks, whales, etc. This
type of association is known as
commensalism, in which one partner
gets benefits while other is not harmed.

359. 01Which one of the following
processes during decomposition
is correctly described?[NEET 2013]
(a) Fragmentation-Carried out by
organisms such as earthworm
(b) Humification-Leads to the
accumulation of a dark coloured
substance humus, which
undergoes microbial action at a
very fast rate
(c) Catabolism-Last step in the
decomposition under fully
anaerobic condition
(d) Leaching-Water soluble inorganic
nutrients rise to the top layers of
soil
Ans.(a)
Fragmentation is one of the steps
during decomposition, in which detritus
is converted into small fragments.
Humification leads to dark coloured
amorphous substance called humus
that is highly resistent to microbial
action and undergoes decomposition at
an extremely slow rate. Catabolism is
the set of metabolic pathways that
breaks down molecules into smaller
units to release energy. Leaching refers
to the loss of water soluble plant
nutrients from the soil due to the rain
and irrigation.
02Which ecosystem has the
maximum biomass? [NEET 2017]
(a) Forest ecosystem
(b) Grassland ecosystem
(c) Pond ecosystem
(d) Lake ecosystem
Ans.(a)
Biomass refers to the amount of living
organic matter. Forest ecosystem have
the maximum biomass, because it
includes organisms of all trophic levels
as compared to pond, lake or grassland
ecosystem. In forest ecosystems
productivity is also high that
contributes to maximum biomass.
03Presence of plants arranged into
well defined vertical layers
depending on their height can
best seen best in[NEET 2017]
(a) tropical savannah
(b) tropical rain forest
(c) grassland
(d) temperate forest
Ans.(b)
Tropical rain forests show
stratification. It can be defined as the
grouping of plants into two or more well
defined layers depending upon their
height. These layers are called strata or
storeys. There storeys consist of
respectively very tall emergent trees,
tall trees, small trees, a shrub layer and
a ground layer of ferns, mosses and
herbs.
04The term ecosystem was coined
by
[NEET 2016, Phase I]
(a) AG Tansley
(b) E Haeckel
(c) E Warming
(d) EP Odum
Ans.(a)
The term ecosystem was coined by AG
Tansley in 1935. Ecosystem is a self
regulated and self sustaining structural
and functional unit of nature. It consists
of living beings and their physical
environment.
05Which one of the following is a
characteristic feature of cropland
ecosystem?[NEET 2016, Phase II]
(a) Least genetic diversity
(b) The absence of weeds
(c) Ecological succession
(d) The absence of soil organisms
Ans.(a)
Cropland ecosystem is largest
anthropogenic ecosystem characterised
by less diversity and high productivity.
06The primary producers of the
deep-sea hydrothermal vent
ecosystem are
[NEET 2016, Phase II]
(a) green algae
(b) chemosynthetic bacteria
(c) blue-green algae
(d) coral reefs
Ans.(b)
The primary producers of the deep-sea
hydrothermal vent ecosystem are
archaebacteria.
These have chemosynthetic mode of
nutrition. Thus option (b), i.e.
chemosynthetic bacteria is the correct
option.
07Which one of the following is not a
functional unit of an ecosystem?
[CBSE AIPMT 2012]
(a) Energy flow (b) Decomposition
(c) Productivity (d) Stratification
Ans.(d)
Stratification is not the functional unit
of ecosystem. Vertical distribution of
different species occupying different
levels is called stratification. It
represent the structural unit of an
ecosystem. For example, trees occupy
top vertical strata or layer of a forest,
shrubs the second and herbs and
grasses occupy the bottom layers.
Ecosystem
36
Ecosystem and its
Components
TOPIC 1

360. 08Quercusspecies are the dominant
component in[CBSE AIPMT 2008]
(a) temperature deciduous forests
(b) alpine forests
(c) scrub forests
(d) tropical rain forests
Ans.(a)
Temperate deciduous forests grow in
continental climates with summer
rainfall and severe winters. They are
dominated by broad leaved deciduous
trees likeQuercus virginiana, magnolias,
bays and hallies as well as such tropical
species asFicus aureaandLysiloma.
09The slow rate of decomposition of
fallen logs in nature is due to their
[CBSE AIPMT 2008]
(a) low moisture content
(b) poor nitrogen content
(c) anaerobic environment around
them
(d) low cellulose content
Ans.(a)
The slow rate of decomposition of
fallen logs in nature is due to their low
moisture content.
The cellulose is in high amount in fallen
logs.
The environment arround the fallen logs
is aerobic, i.e.O
2
is present.
10An ecosystem which can be easily
damaged but can recover after
some time if damaging effect
stops, will be having
[CBSE AIPMT 2004]
(a) low stability and high resilience
(b) high stability and low resilience
(c) low stability and low resilience
(d) high stability and high resilience
Ans.(a)
Stability is the power of a system to be
in their state against unfavourable
factor. Resilience is the capability of
regaining its original shape or position
after being deformed. Hence, it has low
stability and high resilience.
11Which of the following is the most
stable ecosystem?
[CBSE AIPMT 1995]
(a) Forest
(b) Desert
(c) Mountain
(d) Ocean
Ans.(d)
Oceanic biome or ecosystem occupies
more than two-thirds of the earth’s
surface. This is the most stable
ecosystem.
12What is true of ecosystem?
[CBSE AIPMT 1998]
(a) Primary consumers are least
dependent upon producers
(b) Primary consumers out-number
producers
(c) Producers are more than primary
consumers
(d) Secondary consumers are the
largest and most powerful
Ans.(c)
In an ecosystem producers (green
plants) are always more than primary
consumers (herbivores).
13In an ecosystem, which one shows
one-way passage
[CBSE AIPMT 1998]
(a) free energy (b) carbon
(c) nitrogen (d) potassium
Ans.(a)
The flow of energy in any ecosystem is
unidirectional. The only source of
energy is sunlight. It gets trapped by
producers then it flows from herbivores
to carnivores or consumers at different
trophic level.
14Decomposers are organisms that
[CBSE AIPMT 1994]
(a) Elaborate chemical substances,
causing death of tissues
(b) operate in living body and
simplifying organic substances of
cells step by step
(c) attack and kill plants as well as
animals
(d) operate in relay terms, simplifying
step by step the organic
constituents of dead body
Ans.(d)
Decomposers are the organisms,
normally a fungus or bacterium, that
digest organic material by secreting
digestive enzymes into the
environment, in the process liberating
nutrients into the environment. These
are also known as microconsumers,
reducers or scavengers, as converting
complex organic constituents of dead
bodies of plants, animals, human
wastes into simple soluble forms.
15In the equation GPP−R=NPP, R
represents [NEET 2021]
(a) radiant energy
(b) retardation factor
(c) environment factor
(d) respiration losses
Ans.(d)
Net primary productivity is the available
biomass for the consumption to
heterotrophs. Gross primary
productivity is the rate of production of
organic matter during photosynthesis.
The overall productivity of a system can
be found in an equation, where the Net
Primary Productivity or NPP, is equal to
the Gross Primary Productivity or GPP,
minus the Carbon respiration
(respiration losses) or R.
The formula is the NPP = GPP−R.
16The rate of decomposition is
faster in the ecosystem due to
following factors except
[NEET (Oct.) 2020]
(a) detritus rich in sugars
(b) warm and moist environment
(c) presence of aerobic soil microbes
(d) detritus richer in lignin and chitin
Ans.(d)
The rate of decomposition is faster in
the ecosystem if detritus is rich in
nitrogen and water soluble substances
like sugars. Warm and moist
environment also favour decomposition
like wise high temperature and
presence of aerobic soil microbes also
helps in decomposition. But if detritus
is rich in lignin and chitin
decomposition rate gets slower. Thus,
option (d) is correct.
17Which of the following statements
is incorrect?[NEET (Oct.) 2020]
(a) Biomass decreases from first to
fourth trophic level
(b) Energy content gradually
increases from first to fourth
trophic level
(c) Number of individuals decreases
from first trophic level to fourth
trophic level
(d) Energy content gradually
decreases from first to fourth
trophic level
350 NEETChapterwise Topicwise Biology
Productivity and
Energy Flow
TOPIC 2

361. Ans.(b)
Statement (b) is incorrect. It can be
corrected as The energy content
decreases from the first (producer) to
fourth (consumer) level. At each level
about 90% of energy is lost and only
10% is passed to next level.
18Match the trophic levels with their
correct species examples in
grassland ecosystem.
[NEET (Sep.) 2020]
Column I Column II
A. Fourth trophic level 1. Crow
B. Second trophic level 2. Vulture
C. First trophic level 3. Rabbit
D. Third trophic level 4. Grass
Select the correct option.
A B C D
(a) 3 2 1 4
(b) 4 3 2 1
(c) 1 2 3 4
(d) 2 3 4 1
Ans.(d)
The correct match is option (d) as in
grassland ecosystem grass is the
producer (Ist Trophic level). Rabbit is
the primary consumer (IInd Trophic
level). Crow is secondary consumer
(IIIrd Trophic level). Vulture is tertiary
consumer (IVth Trophic level). The
primary consumers eat the producers.
Secondary consumers eat the primary
consumers, and so on. Grassland
ecosystem is a terrestrial ecosystem. It
includes various trophic levels.
Grass→Rabbit→Crow→Vulture
19In relation to gross primary
productivity and net primary
productivity of an ecosystem,
which one of the following
statements is correct?
[NEET (Sep.) 2020]
(a) Gross primary productivity is
always more than net primary
productivity
(b) Gross primary productivity and net
primary productivity are one and
same
(c) There is no relationship between
gross primary productivity and net
primary productivity
(d) Gross primary productivity is
always less than net primary
productivity
Ans.(a)
Gross primary productivity of an
ecosystem is the rate of production of
organic matter during photosynthesis.
Net primary productivity is
GPP – respiration. Hence, gross primary
productivity is always more than NPP.
20Most animals that in deep oceanic
water are [CBSE AIPMT 2015]
(a) primary consumers
(b) secondary consumers
(c) tertiary consumers
(d) detritivores
Ans.(d)
Most animals that live in deep oceanic
waters and called benthos are
scavengers ordetritivores. These
organisms include crustaceans,
polychaetes and some microorganisms.
21If 20 J of energy is trapped at
producer level, then how much
energy will be available to peacock
as food in the following chain?
Plant→Mice→Snake→Peacock
[CBSE AIPMT 2014]
(a) 0.02 J (b) 0.002 J
(c) 0.2 J (d) 0.0002 J
Ans.(c)
According to 10% law of energy flow by
Raynold Lindeman. The total amount of
energy that can be transferred to the
next trophic level is the 10% hence,
peacock will receive 0.02 J of energy as
top consumer.
Energy received by other organisms are
Plant→20 J
Mice→20 10 2× =%J
Snake→ × =2 10 0 2% .J
22Secondary productivity is rate of
formation of new organic matter
by [NEET 2013]
(a) producer (b) parasite
(c) consumer (d) decomposer
Ans.(c)
Secondary productivity is the rate of
formation of new organic matter by
consumers. Primary productivity
depends on the producers inhabiting a
particular area. Decomposers
breakdown complex organic matter into
inorganic substances like carbon
dioxide, water and nutrients. Parasitic
species feed on the body of other
organism.
23Identify the likely orgnaisms I, II, III
and IV in the food web shown
below. [CBSE AIPMT 2012]
I II III IV
(a) Deer Rabbit Frog Rat
(b) Dog Squirrel Bat Deer
(c) Rat Dog Tortoise Crow
(d) Squirrel Cat Rat Pigeon
Ans.(a)
In the given food web option (a) is
correct as producers utilise the radiant
energy of sun which is transformed to
chemical form during photosynthesis.
Thus, green plants occupy the first
trophic level. The herbivores constitute
the secondary trophic level and the
carnivores the third trophic level. Deer
is herbivores, rabbit and rat are also
herbivores but frog eats the
grasshoppers. Also deer is been eaten
by lion.
24Identify the possible link ‘A’ in the
following food chain
[CBSE AIPMT 2012]
Plant→Insect→Frog→‘A’→
Eagle
(a) Rabbit (b) Wolf
(c) Cobra (d) Parrot
Ans.(c)
The given food chain should be
Plant→Insect→Frog→Cobra→
Eagle
25Mass of living matter at a trophic
level in an area at any time is
called [CBSE AIPMT 2011]
(a) standing crop
(b) detritus
(c) humus
(d) standing state
Ecosystem 351
Vegetation/Seeds
Grass hopperMice(I)
(IV)
(II)
Sparrow
(III)
Garden
lizard
Foxes
Owis
Lion
Snake
Hawks

362. Ans.(a)
Standing crop is the total amount of
living matter in a specified population at
a particular time, expressed as biomass
(standing biomass) or its equivalent in
terms of energy. The standing crop may
vary at different times of the year for
example in a population of deciduous
trees between summer and winter.
26Of the total incident solar radiation
the proportion of PAR is
[CBSE AIPMT 2011]
(a) about 70% (b) about 60%
(c) less than 50% (d) more than 80%
Ans.(c)
The source of energy in all ecosystem is
solar energy. About 50% of the solar
energy incident over earth is present in
PAR (Photosynthetically active
Radiation). About 1-5% of incident solar
radiation or 2-10% of PAR is captured
by the photosynthetic organisms in the
synthesis of organic matter (gross
primary productivity).
Roughly 20% of it is consumed in
respiration so that net capture of
energy (net primary productivity) is
0.8-4% of incident radiation or 1.6-8%
of PAR.
27The biomass available for
consumption by the herbivores
and the decomposers is called
[CBSE AIPMT 2010]
(a) net primary productivity
(b) secondary productivity
(c) standing crop
(d) gross primary productivity
Ans.(a)
Net primary productivity is equal to the
rate of organic matter created by
photosynthesis minus the rate of
respiration and other losses. It is the
biomass available for consumption by
the herbivores and the decomposers.
28The correct sequence of plants in
a hydrosere is[CBSE AIPMT 2009]
(a) Oak→Lantana→Scirpus→Pistia
→Hydrilla→Volvox
(b)Volvox→Hydrilla→Pistia→
Scirpus→Lantana→Oak
(c) Pistia→Volvox→Scirpus→
Hydrilla→Oak→Lantana
(d) Oak→Lantana→Volvox→
Hydrilla→Pistia→Scirpus
Ans.(b)
The various stages in a hydrosere are
well studied in ponds, pools or lakes.
The various stages of hydrosere are:
(i)Phytoplankton stage, e.g. some
blue-green algae, green algae
(Volvox), diatoms and bacteria, etc.
(ii)Rooted submerged stage,e.g.
Hydrilla,Vallisneria, etc.
(iii)Floating stage,e.g.Nelumbo,
Nymphaea, etc. Some free floating
species arePistia, Azolla,Lemna,
etc.
(iv)Red-swamp stage, e.g. species of
Scirpus,Typha, etc.
(v)Sedge-meadow stage,e.g. species
of Cyperaceae and Gramineae.
(vi)Woodland stage,e.g.Lantana,
Salix, Populus,etc.
(vii)Forest stage,e.g. Tropical rain
forests, mixed forests ofAlmus,
Acer, Quercus(oak), tropical
deciduous forests.
29Which one of the following types
of organisms occupy more than
one trophic level in a pond
ecosystem?[CBSE AIPMT 2009]
(a) Phytoplankton (b) Fish
(c) Zooplankton (d) Frog
Ans.(b)
In a pond ecosystem, fishes occupy
more than one trophic levels. Small
fishes act as secondary consumer.
They feed on primary consumer. Large
fishes act as Tertiary consumer. They
feed on smaller fish.
30Consider the following statements
concerning food chains.
[CBSE AIPMT 2008]
I. Removal of 80% tigers from an
area resulted in greatly
increased growth of vegetation
II. Removal of most of the
carnivores resulted in an
increased population of deers
III. The length of food chains is
generally limited to 3-4 trophic
levels due to energy loss
IV. The length of food chains may
vary from 2 to 8 trophic levels
Which of the two above
statements are correct?
(a) I and II (b) II and III
(c) III aqnd IV (d) I and IV
Ans.(a)
Statements II and III are correct. A
simple food chain consists of
producers, herbivores and carnivores.
The length of food chain is generally
limited to 3–4 trophic levels due to the
energy loss. In grazing food chain the
producers (i.e. plants) are eaten by
herbivores (i.e. rabbit, deer, cow, etc)
and the herbivores are eaten by
carnivores. Therefore, the removal of
most of the carnivores resulted in an
increased population of deers.
31Which of the following ecosystem
types has the highest annual net
primary productivity?
[CBSE AIPMT 2007]
(a) Tropical rain forest
(b) Tropical deciduous forest
(c) Temperate evergreen forest
(d) Temperate deciduous forest
Ans.(a)
Productivity of tropical rain forest is
highest. The tropical rain forest cover
300,000 km
2
area. They contain more
than 50% of total flora and fauna of the
world.
32Which of the following is expected
to have the highest value
(gm /m /yr
2
) in a grassland
ecosystem?[CBSE AIPMT 2004]
(a) Secondary Production (SP)
(b) Tertiary Production (TP)
(c) Gross Production (GP)
(d) Net Production (NP)
Ans.(c)
The rate of total capture of energy or
the rate of total production of organic
material is gross primary productivity
while the balance or biomass remaining
after meeting the cost of respiration of
producers is net primary productivity.
Hence, gross productivity has highest
value in grassland ecosystem.
33Bamboo plant is growing in a far
forest then what will be the
trophic level of it?
[CBSE AIPMT 2002]
(a) First trophic level (T
1
)
(b) Second trophic level (T
2
)
(c) Third trophic level (T
3
)
(d) Fourth trophic level (T
4
)
352 NEETChapterwise Topicwise Biology

363. Ans.(a)
Plants, being photosynthetic, occupy
first trophic level(T )
1
in the food chain.
A trophic level is a step in the flow of
energy through an ecosystem, such as
the step at which plants manufacture
food or the step at which carnivores
feed on other animals.
34The transfer of energy from one
trophic level to another is
governed by the 2nd law of
thermodynamics. The average
efficiency of energy transfer from
herbivores to carnivores is
[CBSE AIPMT 1999, 96]
(a) 5% (b) 10%
(c) 25% (d) 50%
Ans.(b)
According to 10% law ofLindeman, only
10% of energy is transferred from one
trophic level to another, i.e. from
herbivores to carnivores.
35The rate at which light energy is
converted into chemical energy of
organic molecules is the
ecosystem’s
[CBSE AIPMT 1998]
(a) net primary productivity
(b) gross secondary productivity
(c) net secondary productivity
(d) gross primary productivity
Ans.(d)
Plants, found in an ecosystem are
known as producers, because they can
prepare food for themself by the
process of photosynthesis. The energy
fixed by the autotrophs during
photosynthesis gets incorporated into
organic compounds. The rate at which
organic molecules are formed in a
green plant (or a population of green
plants) is called gross primary
productivity.
36Which of the following ecosystem
has the highest gross primary
productivity?
[CBSE AIPMT 1997]
(a) Grasslands
(b) Coral reefs
(c) Mangroves
(d) Equatorial rain forest
Ans.(b)
TheGross primary productivityof an
ecosystem is the total amount of
organic matter produced by it’s
producers, i.e. autotrophs. Coral reefs
have maximum and deserts have lowest
productivity.
37In a biotic community, the primary
consumers are[CBSE AIPMT 1995]
(a) carnivores (b) omnivores
(c) detritivores (d) herbivores
Ans.(d)
In a biotic community, the primary
consumers or first order consumers are
herbivores, they feed on producers.
They are also calledkey industry
animalsbecause they convert plant
material into animal material, e.g. rat,
deer, rabbit, cattle, goat, sheep, insects
etc.
38If we completely remove the
decomposers from an ecosystem,
its functioning will be adversely
affected, because
[CBSE AIPMT 1995]
(a) energy flow will be blocked
(b) herbivores will not receive solar
energy
(c) mineral movement will be blocked
(d) rate of decomposition will be very
high
Ans.(c)
Decomposers like fungi, bacteria and
Actinomycetes are also called
mineralisers as they release minerals
trapped in organic matter. Thus, they
help in recycling of minerals, so if we
completely remove decomposers the
mineral movement will be blocked.
39In grass-deer-tiger food chain,
grass biogass is one tonne. The
tiger biomass shall be
[CBSE AIPMT 1994]
(a) 100 kg (b) 10 kg
(c) 200 kg (d) 1 kg
Ans.(b)
According to 10% law ofLindemann,if
1 tonne (1000 kg) biomass is present in
grass, only 10% of it means 100 kg will
go into deer and in tiger the biomass
will be only 10 kg, i.e. 10% of deer’s
biomass.
40Second most important trophic
level in a lake is[CBSE AIPMT 1994]
(a) zooplankton (b) phytoplankton
(c) benthos (d) neuston
Ans.(a)
In a lake ecosystem, the first trophic
level is occupied by phytoplankton and
then in second trophic level there are
zooplanktons which are primary
consumers.
41Food chain in which
microorganisms breakdown the
food formed by primary producers
are [CBSE AIPMT 1991]
(a) parasitic food chain
(b) detritus food chain
(c) consumer food chain
(d) predator food chain
Ans.(b)
Detritus food chain goes from dead
organic matter to detritivores
protozoan, bacteria, fungi and then to
organisms feeding on detritivores, e.g.
insect larva, nematodes. This food
chain is also called as saprophytic food
chain.
42Pick up the correct food chain.
[CBSE AIPMT 1991]
(a) Grass→Chameleon→Insect
→Bird
(b) Grass→Fox→Rabbit→Bird
(c) Phytoplankton→Zooplankton
→Fish
(d) Fallen leaves→Bacteria→Insect
larvae
Ans.(c)
The correct food chain is:
Phytoplankton→Zooplankton→Fish
i.e. Producers→Primary consumer
→Secondary consumer
43Upper part of sea/aquatic
ecosystem contains
[CBSE AIPMT 1998]
(a) plankton (b) nekton
(c) Both (a) and (b) (d) benthos
Ecosystem 353
Heterotrophs
SnakeRabbit HawkGreen
plants
Carnivores
3rd
trophic
level
4th
trophic
level
HerbivoresAutotrophs
1st
trophic
level
2nd
trophic
level

364. Ans.(a)
Planktons are passively floating
organisms living in the surface layers of
water due to absence of locomotory
organs, they are of two types:
Phytoplankton (photosynthetic
plankton) and zooplankton.
While nektons are actively floating
organisms and benthos are found in
the bottom and are usually sessile.
44Which of the following statement
is not correct? [NEET 2021]
(a) Pyramid of biomass in sea is
generally inverted
(b) Pyramid of biomass in sea is
generally upright
(c) Pyramid of energy is always
upright
(d) Pyramid of numbers in a grassland
ecosystem is upright
Ans.(b)
Pyramid of biomass in a sea is generally
inverted because the primary
producers (phytoplanktons) have a
lower biomass than that of succeeding
zooplanktons, which further have a
lower biomass than that of succeeding
small fishes and so on.
Pyramid of energy is the only pyramid
that can never be inverted and is always
upright. This is because some amount
of energy in the form of heat is always
lost to the environment at every trophic
level of the food chain.
In a grassland ecosystem, the number
of producers is always maximum,
followed by reducing number of
organisms at second trophic level, third
trophic level and other higher level (if
present). Thus, the pyramid of number
in grassland is upright.
45Which of the following ecological
pyramids is generally inverted?
[NEET (National) 2019]
(a) Pyramid of energy
(b) Pyramid of biomass in a forest
(c) Pyramid of biomass in a sea
(d) Pyramid of numbers in grassland
Ans.(c)
Pyramid of biomass in sea is
generally inverted because the
biomass of a trophic level depends
upon reproductive potential and
longevity of its members.
In a sea, the biomass of
phytoplanktons is usually lesser than
that of zooplanktons while the
biomass of carnivores is greater than
small carnivores and zooplanktons.
On the other hand, pyramid of energy
is always upright. Pyramid of biomass
in terrestrial ecosystems (forests,
grasslands) is also upright.
46What type of ecological pyramid
would be obtained with the
following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g[NEET 2018]
(a) Upright pyramid of numbers
(b) Pyramid of energy
(c) Inverted pyramid of biomass
(d) Upright pyramid of biomass
Ans.(c)
An inverted pyramid of biomass will be
obtained from the given data. The
biomass is continuously decreasing
from secondary consumer (120 g) to
primary consumer (60 g) to primary
producer (10 g).
Therefore, upright pyramid of biomass
cannot be obtained. The data is given in
the form of biomass, therefore pyramid
of number and energy cannot be
obtained. Further, pyramid of energy is
always upright.
47During ecological succession
[CBSE AIPMT 2015]
(a) the gradual and predictable
change in species composition
occurs in a given area
(b) the establishment of a new biotic
community is very fast in its
primary phase
(c) the numbers and types of animals
remain constant
(d) the changes lead to a community
that is in near equilibrium with the
environment and is called pioneer
community
Ans.(a)
The gradual and fairly predictable
change in the species composition of a
given area is called ecological
succession. During succession some
species colonise an area and their
populations become more numerous,
whereas populations of other species
decline and even disappear.
48Match the following and select the
correct option.[CBSE AIPMT 2014]
Column I Column II
A. Earthworm 1. Pioneer
species
B. Succession 2. Detritivore
C. Ecosystem
service
3. Natality
D. Population
growth
4. Pollination
Codes
A B C D A B C D
(a) 1 2 3 4 (b) 4 1 3 2
(c) 3 2 4 1 (d) 2 1 4 3
Ans.(d)
The species that invade a base area in
succession is called pioneer species
and earthworm is a detritivore.
Ecosystem services are the products of
ecosystem process, e.g. biodiversity
maintenance, crop pollination, etc. and
natality is the term used for population
growth or birth rate in population
ecology.
49Given below is an imaginary
pyramid of numbers. What could
be one of the possibilities about
certain organisms at some of the
different levels?
[CBSE AIPMT 2012]
(a) Level PC is insects and level SC is
small insectivorous birds
(b) Level PP is phytoplanktons in sea
and Whale on top level TC
354 NEETChapterwise Topicwise Biology
Secondary consumer
(120g)
Primary consumer
(60g)
Primary
producer
(10g)
Inverted pyramid of biomass
Ecological Pyramids
and Succession
TOPIC 3
TC
SC
PC
PP
10
50
500
1

365. (c) Level one PP is pipal trees and the
level SC is sheep
(d) Level PC is rats and level SC is
cats
Ans.(a)
The given figure shows spindle-shaped
pyramid of number in single tree
ecosystem. Here, a single large sized
tree provides food to a large number of
herbivores which support a few
carnivores and the later are eaten by
small number of top carnivores. So,
here PP is used for producer, i.e. single
tree, PC is Primary Consumers, i.e.
large number of insects, SC is
Secondary Consumers, i.e. small
insectivorous birds and TC is Top
Consumers which may be eagles or
falcon, etc.
50The upright pyramid of number is
absent in [CBSE AIPMT 2012]
(a) pond
(b) forest
(c) lake
(d) grassland
Ans.(b)
Pyramid of number represents the
number of individuals per unit area at
various trophic levels. It is always
upright in grassland, pond and lake
ecosystems. But in forest or single tree
ecosystem, it is spindle-shaped and of
parasitic food chain is considered then
it will be an inverted pyramid.
51The second stage of hydrosere is
occupied by plants like
[CBSE AIPMT 2012]
(a)Azolla
(b)Typha
(c)Salix
(d)Vallisneria
Ans.(d)
Vallisneriais at second stage of
hydrosere. It starts orginating in a pond
with colonisation of some
phytoplanktons which forms the
pioneer plant community. The stages
are
Ist – Bacteria, blue-green algae and
algae
IInd–Hydrilla,Potamogetonand
Vallisneria
IIIrd–Nelumbo,Nymphaea,Trapa,Azolla
andWolffia
IVth –TyphaandSagitaria
Vth–JuncusandCyperus
VIth–SalixandPopulus,Almus.
52Which one of the following
statements is correct for
secondary succession?
[CBSE AIPMT 2011]
(a) It occurs on a deforested site
(b) It follows primary succession
(c) It is similar to primary succession
except that it has a relatively fast
pace
(d) It begins on a bare rock
Ans.(a)
Secondary succession of subsere is
ecological succession that takes place
in a recently denuded area which still
contains a lot of organic debris,
remains and propagules of previous
living organisms. It is more common
and caused by baring of an area due to
the forest fires, deforestation,
excessive overgrazing, landslides,
earthquakes, repeated floods, etc.
53Which one of the following
statements for pyramid of energy
is incorrect whereas the remaining
three are correct?
[CBSE AIPMT 2011]
(a) It show energy content of different
trophic level of organisms
(b) It is inverted in shape
(c) It is upright in shape
(d) Its base is broad
Ans.(b)
Pyramid of energy is graphic
representation of energy per unit area
sequence-wise in various rising trophic
levels with producers at the base and
top carnivores at the apex. Pyramid of
energy is upright in all cases. It is also
more accurate than other types of
ecological pyramids.
54Which one of the following is not
used for constructing of ecological
pyramids? [CBSE AIPMT 2006]
(a) Dry weight
(b) Number of individuals
(c) Rate of energy flow
(d) Fresh weight
Ans.(d)
Ecological pyramids are the graphical
representation of the trophic structure
and function at successive trophic
levels. Ecological pyramids are of three
general types, listed as here under:
(a)Pyramid of numbers, showing the
number of organisms at each level.
(b)Pyramid of biomass, showing the
total dry weight of living matter.
(c)Pyramid of energy, showing the
rate of energy flow/productivity at
successive trophic levels.
Thus, fresh weight is not used for the
construction of ecological pyramids.
55The greatest biomass of
autotrophs in the world’s oceans is
that of [CBSE AIPMT 2000]
(a) benthic brown algae, coastal red
algae and dephnids
(b) benthic diatoms and marine
viruses
(c) sea grasses and slime molds
(d) free-floating micro-algae,
cyanobaceria and nanoplankton
Ans.(d)
The greatest biomass of autotrophs in
the world’s ocean is that of free floating
micro-algae, cyanobacteria and
nanoplankton. Phytoplanktons, diatoms
and dinoflagellates are the dominant
producers in the world’s oceans.
56In a terrestrial ecosystem such as
forest, maximum energy is in
which trophic level?
[CBSE AIPMT 1998]
(a)T
1
(b)T
2
(c)T
3
(d)T
4
Ans.(a)
There is 90% loss of energy at every
trophic level. Therefore, maximum
energy is atT
1
level.
57In a food chain, the largest
population is that of
[CBSE AIPMT 1996, 1994]
(a) decomposers
(b) producers
(c) primary consumers
(d) tertiary consumers
Ans.(b)
Producers are present in largest
number in any food chain.
58The nature of climax community
ultimately depends on
[CBSE AIPMT 1996]
(a) climate
(b) bed rock
(c) soil organisms
(d) pool of available nutrients
Ecosystem 355

366. Ans.(a)
The climax community ultimately
depends on the climate such as rain
forest in moist tropical area and mixed
coniferous or deciduous forest in
temperate area.
59The primary succession refers to
the development of communities
on a [CBSE AIPMT 1995]
(a) fleshly cleared crop field
(b) forest clearing after devastating
fire
(c) pond, freshly filled with water after
a dry phase
(d) newly-exposed habitat with no
record of earlier vegetation
Ans.(d)
Primary succession is the succession in
a totally barren area with no record of
earlier vegetation. It takes long time of
1000 year or more.
60The pyramid which cannot be
inverted in a stable ecosystem is
that of [CBSE AIPMT 1994]
(a) biomass (b) number
(c) energy (d) All of these
Ans.(c)
Pyramid of energy is graphic
representation of amount of energy
trapped per unit time and area in
different trophic levels of a food chain
with producers forming the base and
top carnivores or consumers the tip. It
is alwaysuprightin shape.
61Pyramid of number deals with
number of
[CBSE AIPMT 1993]
(a) species in an area
(b) individuals in a community
(c) individuals in a trophic level
(d) sub-species in a community
Ans.(c)
Pyramid of number is a graphic
representation of the number of
organisms per unit area of various
trophic levels. It deals with the number
of individuals in a trophic level. It deals
with the number of individuals in a
trophic level.
62Pyramid of number in a pond
ecosystem is
[CBSE AIPMT 1993, 1991, 1990]
(a) irregular (b) inverted
(c) upright (d) spindle-shaped
Ans.(c)
Pyramid of number in a pond ecosystem
is upright or erect, in which producers
are maximum in number and top
consumers are least in number.
63The amount of nutrients, such as
carbon, nitrogen, phosphorus and
calcium present in the soil at any
given time, is referred as
[NEET 2021]
(a) climax
(b) Climax community
(c) standing state
(d) standing crop
Ans.(c)
Standing state is the amount of
biogenetic nutrients present at any
specific time in the ecosystem. The
whole living matter is composed of
nutrients Like carbon, nitrogen and so
on.
Other options can be explained as:
Climax, in ecology, is the final stage of
biotic succession attainable by a plant
community in an area under the
environmental conditions present at a
particular time.
A climax community is one that has
reached the stable stage. When
extensive and well-defined, the climax
community is called a biome. Examples
are tundra, grassland, desert, and the
deciduous, coniferous and tropical rain
forests.
Each trophic level contains certain
mass of living matter at a specific time
is called standing crop.
64Which of the following statements
is incorrect regarding the
phosphorus cycle?
[NEET (Oct.) 2020]
(a) Phosphates are the major form of
phosphorus reservoir
(b) Phosphorus solubilising bacteria
facilitate the release of
phosphorus from organic remains
(c) There is appreciable respiratory
release of phosphorus into
atmosphere
(d) It is sedimentary cycle
Ans.(c)
Staement (c) is incorrect and can be
corrected as Phosphorus cycle is a type
of sedimentary cycle, i.e. it main
reservoirs are soil and rocks. It is mainly
found as phosphates in rocks. During
this cycle phosphorus solublising
bacteria likePseudomonas,
Acetobacter, etc., help to release
phosphorus from organic remains. A
large amount of phosphate is lost in sea
by sedimentation. There is no
respiratory release of phosphorus into
atmosphere because phosphorus is an
inorganic nutrient that does not take
part in respiration.
65In which of the following both
pairs have correct combination?
[CBSE AIPMT 2015]
(a) Gaseous
nutrient cycle
Carbon and
nitrogen
Sedimentary
nutrient
cycle
Sulphur and
phosphorus
(b) Gaseous
nutrient cycle
Carbon and
sulphur
Sedimentary
nutrient
cycle
Nitrogen and
phosphorus
(c) Gaseous
nutrient cycle
Nitrogen and
sulphur
Sedimentary
nutrient
cycle
Carbon and
phosphorus
(d) Gaseous
nutrient cycle
Sulphur and
phosphorus
Sedimentary
nutrient
cycle
Carbon and
nitrogen
Ans.(a)
The biogeochemical cycles are of two
types, i.e. gaseous cycles, in which the
reservoir for the nutrient elements is in
the atmosphere (air) or hydrosphere
(water). The four most abundant
elements in the living systems, i.e.
hydrogen, carbon, oxygen and nitrogen
have predominantly gaseous cycles.
In sedimentary cycles, the reservoir for
the nutrient elements is in the
sediments of the earth. Elements, such
as phosphorus, sulphur, potassium and
calcium have sedimentary cycles.
66Given below is a simplified model
of phosphorus cycling in a
terrestrial ecosystem with four
blanks (A-D). Identify the blanks
[CBSE AIPMT 2014]
356 NEETChapterwise Topicwise Biology
Nutrient Cycling
TOPIC 4

367. A B C D
(a) Rock
minerals
Detritus Litter fall Producers
(b) Litter fall Producers Rock
minerals
Detritus
(c) Detritus Rock
minerals
Producer Litter fall
(d) Producers Litter fall Rock
minerals
Detritus
Ans.(c)
67Natural reservoir of phosphorus is
[NEET 2013]
(a) rock
(b) fossils
(c) sea water
(d) animal bones
Ans.(a)
The natural reservoir of phosphorus is
in phosphate rocks. Phosphate is
added in small amount into the cycling
pool through weathering of rocks.
Phosphate circulate in the abiotic
environment in lithosphere as well as
hydrosphere.
68Which one of the following is not a
gaseous biogeochemical cycle in
ecosystem? [CBSE AIPMT 2012]
(a) Sulphur cycle
(b) Phosphorus cycle
(c) Nitrogen cycle
(d) Carbon cycle
Ans.(b)
Phosphorus cycle is a sedimentary
biogeochemical cycle. It describes the
movement of phosphorus through the
lithosphere, hydrosphere and biosphere
and the main reservoir pool is
lithosphere.
Atmosphere does not play any
significant role in the movement of
phosphorus because phosphorus and
phosphorus based compounds are
usually solids at the typical ranges of
temperature and pressure found on
earth. The production of phosphine gas
occurs only in specialised, local
conditions.
69About 70% of total global carbon
is found in[CBSE AIPMT 2008]
(a) grasslands
(b) agroecosystems
(c) oceans
(d) forests
Ans.(c)
Sea water contains 50 times moreCO
2
than air. i.e. about 70% of total global
carbon is found in oceans. This is in
form of carbonates and bicarbonates.
The atmosphere is the source ofCO
2
,
which is utilised by plants in
photosynthesis reduced to form carbon
compounds. The total mass of carbon
currently in the atmosphere is about
7 10
17
×g, i.e. 700000 million tonnes.
Oceans regulate theCO
2
content in the
atmosphere and thus, play a very
important role.
70Which one of the following pair is
mismatched?[CBSE AIPMT 2005]
(a) Biomass burning– Release ofCO
2
(b) Fossil fuel burning – Release of
CO
2
(c) Nuclear power – Radioactive
wastes
(d) Solar energy – Green house effect
Ans.(d)
Solar energy is not responsible for
green-house effect instead it is a
source of energy for the plants and
animals. Green plants prepare their
food by the use of this solar energy.CO
2
gas is mainly responsible for
greenhouse effect. Excess of this gas
forms a thick layer around the earth and
prevents re-radiation of earth climate
entering sun rays to atmosphere. Thus,
functions like the glass pannels of a
green house (or the glass windows of a
motor car). This is thus, called
greenhouse effect.
71If by radiation all nitrogenase
enzymes are inactivated, then
there will be no
[CBSE AIPMT 2004]
(a) fixation of nitrogen in legumes
(b) fixation of atmospheric nitrogen
(c) conversion from nitrate to nitrite
in legumes
(d) conversion from ammonium to
nitrate in soil
Ans.(c)
The rate of total capture of energy or
the rate of total production of organic
material is gross primary productivity
while the balance or biomass remaining
after meeting the cost of respiration of
producers is net primary productivity.
Hence, gross productivity has highest
value in grassland ecosystem.
72Which of the following pair is a
sedimentary type of
biogeochemical cycle?
[CBSE AIPMT 1995]
(a) Oxygen and nitrogen
(b) Phosphorous and sulphur
(c) Phosphorous and nitrogen
(d) Phosphorus and carbon dioxide
Ans.(b)
In sedimentary cycle the reservoir for
the elements is in the sediments of
earth (lithosphere), e.g. phosphorus,
calcium, magnesium, sulphur.
73The main role of bacteria in the
carbon cycle involves
[CBSE AIPMT 1998]
(a) photosynthesis
(b) chemosynthesis
(c) digestion or breakdown of organic
compounds
(d) assimilation of nitrogenous
compounds
Ans.(c)
Huge amount of plants, animals and
human wastes are decomposed by
bacteria and fungi present in
environment and large quantity ofCO
2
necessary for photosynthesis is
released into the atmosphere.
74BulkCO
2
-fixation occurs in
[CBSE AIPMT 1994]
(a) crop plants
(b) oceans
(c) tropical rain forests
(d) temperature forests
Ans.(b)
BulkCO
2
-fixation occurs in oceans. The
productivity of ocean ecosystem is very
high, here phytoplanktons, e.g. diatoms
are the greatest producers.
Ecosystem 357
Consumers Producer
inorganic
Phosphates
Soil solution
Detritus
By action of
organism
Liter fall
Phosphate
are released
in soil
Uptake
dissolved
inorganic
particles
Rock minerals
Phosphorus cycle

368. 01Which of the following regions of
the globe exhibits highest species
diversity? [NEET (Sep.) 2020]
(a) Madagascar
(b) Himalayas
(c) Amazon forests
(d) Western Ghats of India
Ans.(c)
The Amazonian rain forest in South
America has the greatest biodiversity
on earth. Rainforests have high
biodiversity because they are abundant
in nutrients, energy and have a
favorable climate for the biodiversity to
prosper.
02Decline in the population of Indian
native fishes due to introduction
ofClarias gariepinusin river
Yamuna can be categoriesd as
[NEET (Odisha) 2019]
(a) co-extinction
(b) habitat fragmentation
(c) overexploitation
(d) allien species invasion
Ans.(d)
Decline in the population of Indian
native fishes due to introduction of
Clarias gariepinusin river Yamuna can
be categorised as alien species
invasions. It is posing a threat to the
indigenous catfishes in our rivers and
causing a loss of biodiversity.
03Which of the following is the most
important cause for animals and
plants being driven to extinction?
[NEET (National) 2019]
(a) Drought and floods
(b) Economic exploitation
(c) Alien species invasion
(d) Habitat loss and fragmentation
Ans.(d)
Habitat loss and fragmentation is the
most important cause for animals and
plants being driven to extinction. Due to
habitat fragmentation and loss, a 14%
of the earth’ land surface, rainforest has
shrunk to only 6% in last few years.
Fragmentation and loss of large
habitats due to various human activities
badly affects mammals and birds
requiring large territories. Certain
animals with migratory habits are also
affected by habitat loss and
fragmentation. Thus, their populations
are driven towards decline and
extinction.
On the other hand, droughts and floods,
economic exploitation and alien species
invasion affect only a small part of
population at a time.
04Which is the National Aquatic
Animal of India?[NEET 2016, Phase I]
(a) River dolphin (b) Blue whale
(c) Seahorse (d) Gangetic shark
Ans.(a)
River dolphin is the National Aquatic
Animal of India. This mammal
exclusively reside in freshwater or
brackish water.
05Plants such asProsopis,Acacia
andCapparisrepresent examples
of tropical
[CBSE AIPMT 1998]
(a) grasslands
(b) thorn forests
(c) deciduous forests
(d) evergreen forests
Ans.(b)
Flora of thorn forests includeProsopis
cineraria, Acacia senagal, Capparis
decidua, Salvadora oleoides, Asparagus
racemosusandEphedra foliata.
06Alexander von Humbolt described
for the first time
[NEET 2017]
(a) ecological biodiversity
(b) law of limiting factor
(c) species area relationships
(d) population growth equation
Ans.(c)
Alexander von Humbolt was a great
German naturalist and geographer. He
did extensive explorations in the
wilderness of South American forests.
He established species area
relationship.
He observed that with in a region,
species richness increase with
increasing explored area, but upto a
certain limit. Infact, the relationship
between species richness and area for
a variety of taxa turns out to be a
rectangular hyperbola as shown in
figure below
Biodiversityand
ItsConservation
37
Biodiversity
TOPIC 1

369. Biodiversity and Its Conservation 359
Note that on log scale the relationship
becomes linear.
On a logarithmic scale, the relationship
is a straight line described by the
equation.
log log logS C Z A= +
where,
S=Species richness
A=Area
Z=Slope of the line (regression
coefficient)
C=Y-intercept
07Which of the following is the most
important cause of animals and
plants being driven to extinction?
[NEET 2016, Phase I]
(a) Alien species invasion
(b) Habitat loss and fragmentation
(c) Co-extinctions
(d) Over-exploitation
Ans.(b)
There are four major causes of
biodiversity loss in which most
important cause driving animals and
plants to extinction is habitat loss and
fragmentation.
08Red List contains data or
information on
[NEET 2016, Phase II]
(a) all economically important plants
(b) plants whose products are in
international trade
(c) threatened species
(d) marine vertebrates only
Ans.(c)
IUCN maintains aRed Data Book
which is a catalogue of the taxa facing
the risk of extinction. The threatened
species are the one which are at more
risk of become extinct. The list of these
species is called Red List.
09Which of the following is correctly
matched? [NEET 2016, Phase II]
(a) Aerenchyma —Opuntia
(b) Age pyramid — Biome
(c)Parthenium hysterophorus
— Threat tobiodiversity
(d) Stratification — Population
Ans.(c)
Parthenium hysterophorus(carrot grass)
is an alien species introduced
inadvertantly for some economic use,
turned invasive causing decline or
extinction of the indigenous species.
The other options are wrong because
(a) Aerenchyma is found in aquatic
plants (Vallisneria,Hydrilla), but
Opuntiais a xerophytic plant.
(b) Biome is total sum of all
ecosystem present in the planet
earth whereas age pyramid is the
graphical representation of age of
organisms of one population at a
specific time.
(d) Stratification is related with
different layers of vegetation in an
ecosystem (like forest/grass land)
and population is a term referred
to a group of same kind of
organisms which can freely
interbreed.
10Given below is the representation
of the extent of global diversity of
invertebrates. What groups the
four portions (A-D) represent
respectively?
[CBSE AIPMT 2014]
A B C D
(a) Insects Crustac-
eans
Other
animal
Groups
Molluscs
(b) Crusta-
cean
Insects Molluscs Other
animal
groups
(c) Molluscs Other
animal
groups
Crustac-
eans
Insects
(d) Insects Molluscs Crustac-
eans
Other
animal
groups
Ans.(c)
Phylum–Arthropoda is the largest
phylum of the animal kingdom with its
largest group, i.e. insecta (A). The
second largest population is of
phylum–Mollusca (B). The third one is
occupied by crustaceans (C). All other
animals combinally indicate the (D)
portion of pie chart.
11A species facing extremely high
risk of extinction in the immediate
future is called
[CBSE AIPMT 2014]
(a) vulnerable
(b) endemic
(c) critically endangered
(d) extinct
Ans.(c)
The extremely highest risk category
assigned by the IUCN Red List for wild
species is critically endangered
species. These are those species that
are facing a very high risk of extinction
in near future. There are currently 2129
animals and 1821 plants which have
been recorded in this category.
12The organisation which publishes
the Red List of species is
[CBSE AIPMT 2014]
(a) ICFRE (b) IUCN
(c) UNED (d) WWF
Ans.(b)
IUCN is International Union for
Conservation of Nature which publishes
the Red List of threatened species or
Red Data List which assesses the
conservation status of species. It is an
international organisation founded in
October, 1948. Its head quarter is in
Gland, Switzerland.
13Which of the following represent
maximum number of species
among global biodiversity?
[NEET 2013]
(a) Algae (b) Lichens
(c) Fungi (d) Mosses and ferns
Ans.(c)
B
C
D
A
Fungi
Mosses
Ferns and allies
Angiosperm
Lichens
Algae
Log =log
+ log
S
C Z A
S=CA
Area
Species richness
log-log scale
2
Showing species area relationship

370. The following figure shows global
biodiversity offungi, mosses, ferns,
algae, lichens and angiosperms.
Therefore, from the above figure it is
clear that fungi has maximum number
of species among the global biodiversity.
14Maximum nutritional diversity is
found in the group
[CBSE AIPMT 2012]
(a) Fungi (b) Animalia
(c) Monera (d) Plantae
Ans.(c)
Maximum nutritional diversity is shown
by the members of kingdom—Monera.
Some of them are autotrophic, (e.g.
photosynthetic autotrophic or
chemosynthetic autotrophic) while the
vast majority are heterotrophs (e.g.
saprotrophic or parasitic). Ecologically,
these may be producers or decomposers.
15The highest number of species in
the world is represented by
[CBSE AIPMT 2012]
(a) fungi (b) mosses
(c) algae (d) lichens
Ans.(a)
Fungi represent the highest number of
species in the world. Around 1,00,000
species of fungi have been formally
described by taxonomists but the global
biodiversity of kingdom fungi is not fully
understood.
16Which one of the following shows
maximum genetic diversity in India?
[CBSE AIPMT 2011, 2009]
(a) Rice (b) Maize
(c) Mango (d) Groundnut
Ans.(a)
There are an estimated 2,00,000
varieties of rice in India alone. The
diversity of rice in India is one of the
richest in the world. Basmati rice has
27 documented varieties grown in India.
17Which one of the following have
the highest number of species in
nature? [CBSE AIPMT 2011]
(a) Insects (b) Birds
(c) Angiosperms (d) Fungi
Ans.(a)
More than 70 per cent of all the species
recorded are animals. Among animals,
insects are the most species rich
taxonomic group, making more than
70 per cent of the total. It means out of
every 10 animals on this planet, 7 are
insects.
18Which one of the following expanded
forms of the following acronyms is
correct? [CBSE AIPMT 2011]
(a)UNEP — United Nations
Environmental Policy
(b)EPA — Environmental
Pollution Agency
(c)IUCN — International Union for
Conservation of
Nature and Natural
Resources
(d)IPCC — International Panel for
Climate Change
Ans.(c)
IUCN or IUCNNR (International Union
for Conservation of Nature and Natural
Resources) is now known as WCU
(World Conservation Union). Its
headquarter is at Gland. Switzerland. It
studies the threat to biodiversity in all
parts of the world by gathering
information about the geographical
distribution, population size and
population changes of various taxa. It
prepares a Red List or Red Data Book
categorising different organisms
belonging to different categories.
19ICBN stands for
[CBSE AIPMT 2007]
(a) Indian Congress of Biological
Names
(b) International Code of Botanical
Nomenclature
(c) International Congress of
Biological Names
(d) Indian Code of Botanical
Nomenaclature
Ans.(b)
ICBN stands for International Code of
Botanical Nomenclature. A body of
rules and recommendations governing
botanical names.
20Which one of the following pairs of
organisms are exotic species
introduced in India?
[CBSE AIPMT 2007]
(a)Ficus religiosa, Lantana camara
(b)Lantana camara,water hyacinth
(c) Water hyacinth,Prosopis cineraria
(d) Nile perch,Ficus religiosa
Ans.(c)
A species of organism that is not native
to a locality where it is flourishing and
have been moved there from its natural
range by humans or other agents is
called exotic species, e.g. water
hyacinth,Prosopis cineraria, etc.
21Which of the following pairs of an
animal and a plant represents
endangered organisms in India?
[CBSE AIPMT 2006]
(a)Bentinckia nicobaricaand red panda
(b) Tamarind and rhesus monkey
(c)Cinchonaand leopard
(d) Banyan and black buck
Ans.(a)
A plantBentinckia
condapanna/nicobarica(member of
familyArecaceae) and the animal, red
panda (an arboreal mammal) both have
been declared endangered in India.
High altitude area of Arunachal Pradesh
have formed Pangchen Red Panda
conservation alliance to recover
population of the endangered animal.
22What is a keystone species?
[CBSE AIPMT 2004]
(a) A species which makes up only a
small proportion of the total
biomass of a community, yet has a
huge impact on the community’s
organisation and survival
(b) A common species that has plenty
of biomass, yet has a fairly low
impact on the community’s
organisation
(c) A rare species that has minimal
impact on the biomass and on
other species in the community
(d) A dominant species that
constitutes a large proportion of
the biomass and which affects
many other species
Ans.(a)
Species having much greater influence
on community characteristics, relative
to their low abundance or biomass are
keystone species, removal of these
cause serious disruption in functioning
of community, e.g. in tropical forests,
figs are keystone species.
23According to IUCN Red List, what
is the status of red panda (Athurus
fulgens)?[CBSE AIPMT 2005, 2004]
(a) Vulnerable species
(b) Critically endangered species
(c) Extinct species
(d) Endangered species
Ans.(d)
Endangered species are those species
whose population have been reduced to
a critical level.
So, they are near to extinction in near
future. Approximately 300 species and
360 NEETChapterwise Topicwise Biology

371. sub-species of mammals are
considered as endangered by the
International Union for theConservation
of Nature and Natural Resources (IUCN).
Red pandais an endangered species
because it is facing a very high risk of
extinction in near future.
24Biodiversity act of India was
passed by the Parliament in the
year [CBSE AIPMT 2005, 2002]
(a) 1996 (b) 1992
(c) 2002 (d) 2000
Ans.(c)
According to the US office of
technology assessment (1987)
‘biological diversity is the variety
among living organisms and the
ecological complexes in which they
occur’. Biodiversity act of India was
passed by the Parliament in the year
of 2002.
25Which endangered animal is the
source of the world’s finest,
lightest, warmest and most
expensive wool —the shahtoosh?
[CBSE AIPMT 2003]
(a) Kashmiri goat (b) Chiru
(c) Nilgai (d) Cheetal
Ans.(c)
According to the US office of
technology assessment (1987)
‘biological diversity is the variety among
living organisms and the ecological
complexes in which they occur’.
Biodiversity act of India was passed by
the Parliament in the year of 2002.
26Which group of vertebrates
comprises the highest number of
endangered species?
[CBSE AIPMT 2003]
(a) Reptiles (b) Birds
(c) Mammals (d) Fishes
Ans.(c)
Approximately 69 mammalian species
and 40 bird species are threatened in
India alone.
27The endangered largest living
lemur Idri idri is inhabitant of
[CBSE AIPMT 2000]
(a) Madagascar (b) Mauritius
(c) Sri Lanka (d) India
Ans.(a)
The Lemurs are the inhabitants of
Madagascar and the Comoro Islands.
28Which of the following is mainly
responsible for extinction of
wildlife? [CBSE AIPMT 1999]
(a) Destruction of habitats
(b) Pollution of air and water
(c) Hunting for flesh
(d) All of the above
Ans.(d)
Destruction of habitats deprives wildlife
of their best places where they could
flourish, indiscriminate killing of wild
animals has greatly reduced their
population and pollution adversely
affects their life cycles.
29Which of the following is the main
factor of desertification?
[CBSE AIPMT 1995]
(a) Tourism
(b) Irrigated agriculture
(c) Over grazing
(d) All of these
Ans.(c)
The process of formation of desert is
desertification. Main factor of
desertification is over grazing of that
area by the herbivorous animals.
Cutting down of trees may be other
cause for desertification as it leads to
loss of wild life.
30If the forest cover is reduced to
half, what is most likely to happen
on a long basis?[CBSE AIPMT 1996]
(a) Tribals living in these areas will
starve to death
(b) Cattle in these and adjoining areas
will die due to lack of fodder
(c) Large areas will become deserts
(d) Crop breeding programmes will
suffer due to a reduced availability
of variety of germplasm
Ans.(c)
If the forest cover is reduced to half than
it will lead to desertification (formation
of desert) of that area in long term.
31Wildlife is destroyed most when
[CBSE AIPMT 1994]
(a) there is lack of proper care
(b) mass scale hunting for foreign trade
(c) its natural habitat is destroyed
(d) natural calamity
Ans.(c)
Destruction of habitat (including falling
of trees) exposes wildlife to a number of
adverse factors leading to diminishing
of their numbers.
32Which animal has become extinct
from India?[CBSE AIPMT 1994]
(a) Snow leopard (b)Hippopotamus
(c) Wolf (d) Cheetah
Ans.(d)
Cheetah has become extinct from India
in 1930. Asiatic cheetah are now found
in Iran and scientists are working to
breed them and bring them back to
Indian sub-continent.
33Species diversity increases as one
proceeds from[CBSE AIPMT 1994]
(a) high altitude to low altitude and
high latitude to low latitude
(b) low altitude to high altitude and
high latitude to low latitude
(c) low altitude to high altitude and
low latitude to high latitude
(d) high altitude to low altitude and
low latitude to high latitude
Ans.(a)
Species diversity increases from high
altitude or latitude to low altitude or
latitude due to the increase in
temperature and seasonal variability in
the concerned areas.
34American water plant that has
become a troublesome water weed
in India is[CBSE AIPMT 1993]
(a)Cyperus rotundus
(b)Eichhornia crassipes
(c)Trapa latifolia
(d)Trapa bispinosa
Ans.(b)
Eichhornia crassipesis a native of
America and is exotic species to India.
It has become a trouble some weed in
many aquatic habitats causing
eutrophication and many other
ecological problems.
35In the following in each set a
conservation approach and an
example of method of conservation
are given
(1)In situconservation–Biosphere
reserve
(2)Ex situconservation–Sacred
groves
Biodiversity and Its Conservation 361
Biodiversity Conservation
TOPIC 2

372. (3)In situconservation–Seed bank
(4)Ex situconservation–
Cryopreservation
Select the option with correct
match of approach and method.
[NEET (Oct.) 2020]
(a) (1) and (3) (b) (1) and (4)
(c) (2) and (4) (d) (1) and (2)
Ans.(b)
Option (b) is correct. It can be explained
asIn situconservation is on site
conservation technique, i.e. the species
are protected in their natural habitats
only. In this approach the important
components of biological diversity are
protected and managed through
protected areas, biosphere reserves
and sacred forests or lakes.
On the other hand,ex situconservation
involves the conservation of selected
rare flora or fauna in places outside
their natural habitat.
It includes off site collection and gene
banks. In gene banks, germplasm,
tissue or embryos are stored in
preserved conditions using the
technologies like orchards, tissue
culture and cryopreservation.
36Which one of the following is not a
method ofin situconservation of
biodiversity?
[NEET (National) 2019]
(a) Wildlife sanctuary
(b) Botanical garden
(c) Sacred grove
(d) Biosphere reserve
Ans.(b)
Botonical garden is not a method ofin
situconservation of biodiversity. It is a
type ofex situor off site conservation in
which rare plants are conserved in
places outside their natural habitat.
Rest all are methods ofin situ
conservation of biodiversity.
37The Earth Summit held in Rio de
Janeiro in 1992 was called
[NEET (National) 2019]
(a) for conservation of biodiversity
and sustainable utilisation of its
benefits
(b) to assess threat posed to native
species by invasive weed species
(c) for immediate steps to
discontinue the use of CFCs that
were damaging the ozone layer
(d) to reduceCO
2
emissions and
global warming
Ans.(a)
The Earth Summit held in Rio de
Janeiro in 1992 was called for the
conservation of biodiversity and
sustainable utilisation of its benefits. In
this summit, more than 130 nations
signed a convention on Biodiversity and
Climate Change. Canada was the key
player in the development of this
convention.
38All of the following are included in
ex-situconservation except
[NEET 2018]
(a) botanical gardens
(b) sacred groves
(c) wildlife safari parks
(d) seed banks
Ans.(b)
Sacred grovesis a mode ofin situ
conservation in which forest fragments
of varying size are protected by
religious communities. It helps to
protect the biota of that area on site.
On the other hand, botanical gardens,
seed banks and wildlife safari parks are
the examples ofex situconservation in
which the biota is protected outside its
natural habitat.
39Which one of the following is
related toEx situconservation of
threatened animals and plants?
[NEET 2017]
(a) Wildlife safari parks
(b) Biodiversity hotspots
(c) Amazon rainforest
(d) Himalayan region
Ans.(a)
Wildlife safari parks are used for theex
situconservation of threatened animals
and plants. They are taken out from
their natural habitat and placed in
special settings (wildlife safari park,
zoo). Here they are protected and given
special care.
40The region of biosphere reserve,
which is legally protected and
where no human activity is
allowed is known as[NEET 2017]
(a) core zone
(b) buffer zone
(c) transition zone
(d) restoration zone
Ans.(a)
Biosphere reserve consists of three
zones, i.e. core, buffer and transition
zone.Core zonecomprises an
undisturbed and legally protected
ecosystem, where no human activity is
allowed. Buffer zone surrounds the
core area and transitions zone is the
outermost area of the biosphere
reserve, where active cooperation
between reserve management and the
local people occur.
41Joint Forest Management
Concept was introduced in India
during [NEET 2016, Phase I]
(a) 1970s (b) 1980s
(c) 1990s (d) 1960s
Ans.(b)
Joint Forest Management Concept was
introduced in India during 1980s by the
Government of India to work closely
with the local communities for
protection and management of forests.
42Which of the following national parks
is home to the famous musk deer
or hangul?[NEET 2016, Phase II]
(a) Keibul Lamjao National Park,
Manipur
(b) Bandhavgarh National Park,
Madhya Pradesh
(c) Eaglenest Wildlife Sancturay,
Arunachal Pradesh
(d) Dachigam National Park, Jammu
and Kashmir
Ans.(d)
Dachigam National Park, situated at
Jammu and Kashmir is famous for
conservation of musk deer.
AtKeibul Lamjao National Park,
Manipur, brown deer (Sangai) is
protected. AtBandhavgarh National
Park,MP, tiger is protected.
Eaglenest Wildlife Sanctuary,
Arunachal Pradesh, protects elephants
and red panda.
43How many hotspots of biodiversity
in the world have been identified
till date by Norman Myers?
[NEET 2016, Phase II]
(a) 17 (b) 25
(c) 34 (d) 43
362 NEETChapterwise Topicwise Biology
Transition zone
Buffer zone
Core area
Human settlement
Zonation in terrestrial biosphere

373. Ans.(c)
The total number of biodiversity
hotspots in the world are 34 till date.
These are the areas of high endemism
and high level of species richness.
44The species confined to a
particular region and not found
elsewhere is termed as
[CBSE AIPMT 2015]
(a) keystone (b) alien
(c) endemic (d) rare
Ans.(c)
The species confined to a particular
region and not found elsewhere is
termed as endemic.
Their conservation requires peculiar
specific efforts due to their
unavailability in other parts of world.
45Just as a person moving from
Delhi to Shimla to escape the heat
for the duration of hot summer,
thousands of migratory birds from
Siberia and other extremely cold
Northern regions move to
[CBSE AIPMT 2014]
(a) Western Ghat
(b) Meghalaya
(c) Corbett National Park
(d) Keolado National Park
Ans.(d)
Every year in the season of winter the
famous Indian National Park, Keolado
National Park situated in Bharatpur,
Rajasthan, host thousands of migratory
birds which come from Siberia (Atlantic
Ocean) Central America and other
extremely cold Northern region.
October-February is the best time to
observe these migratory birds. Most of
them stay till march including the
Siberian crane.
46An example ofex situconservation
is
[CBSE AIPMT 2014, 2010]
(a) National Park
(b) Seed Bank
(c) Wildlife Sanctuary
(d) Sacred Grove
Ans.(b)
Ex situor off-site conservation is the
process of protecting endangered
species of plants and animals outside
their natural habitat. This involves the
conservation of genetic resources like
seeds in seed banks.
47Which one of the following is not
used forex situplant
conservation? [NEET 2013]
(a) Field gene banks
(b) Seed banks
(c) Shifting cultivation
(d) Botanical gardens
Ans.(c)
Shifting cultivationresults into
deforestation. Botanical gardens have
collection of living plants for reference.
Seed banksstore seeds as a source of
germplasm, in case seed reserves
elsewhere are destroyed.Field gene
banksare a type of biorepository which
preserve the genetic material.
48Which one of the following areas
in India, is a hot spot of
biodiversity?[CBSE AIPMT 2012]
(a) Eastern Ghats (b) Gangetic plain
(c) Sunderbans (d) Western Ghats
Ans.(d)
Hotspots are areas that are extremely
rich in species diversity, in its natural
habitat, have high endemism and are
under constant threat. In India, two
hotspots are found extending into
neighbouring countries. The Western
Ghats/Sri Lanka and the Indo-Burman
region (covering the Eastern Himalayas
which is also known as cradle of
speciation).
49Select the correct statement
about biodiversity.
[CBSE AIPMT 2012]
(a) The desert areas of Rajasthan and
Gujarat have a very high level of
desert animal species as well as
numerous rare animals
(b) Large scale planting ofBtcotton
has no adverse effect on
biodiversity
(c) Western Ghats have a very high
degree of species richness and
endemism
(d) Conservation of biodiversity is just
a fad pursued by the developed
countries
Ans.(c)
Western Ghats occur along the western
coast of India in Maharashtra,
Karnataka, Tamil Nadu and Kerala.
There is high degree of endemism as
well as richness of species of flowering
plants, amphibians, reptiles, some
mammals and butterflies.
50A collection of plants and seeds
having diverse alleles of all the
genes of a crop is called
[CBSE AIPMT 2011]
(a) germplasm (b) gene library
(c) genome (d) herbarium
Ans.(a)
A germplasm is a collection of genetic
resources for an organisms. For plants,
the germplasm may be stored as a seed
collection. It includes diverse alleles of
all the genes of organism occurring in
nature.
51Tiger is not a resident, in which
one of the following national
parks? [CBSE AIPMT 2009]
(a) Ranthambhor
(b) Sunderbans
(c) Gir
(d) Jim Corbett
Ans.(c)
Gir National Park(Gujarat) is not
concerned with tiger. The animals
found in Gir National Park are Asiatic
lion, panther, striped hyaena, sambar,
nilgai, cheetal, four-horned antelope
and chinkara.
Ranthambhor National Park
Sunderbans andJim Corbett
National Park(Uttarakhand) are tiger
reserves.
52Which one of the following is not
observed in biodiversity hot spots?
[CBSE AIPMT 2008]
(a) Endemism
(b) Accelerated species loss
(c) Lesser interspecific competition
(d) Species richness
Ans.(c)
Hotspots are the areas that are
extremely rich in species diversity, have
high endemism and are under constant
threat. The key criteria for determining
a hotspot are number of endemic
species and degree of threat which is
measured in terms of habitat loss 34.
hot spots had been identified globally
with an approximate area of 1.4%.
Among these hot spots 2 are found in
India, i.e. Western Ghat and Eastern
Himalayas. These areas are particularly
rich in floral wealth and endemism not
only in flowering plants but also in
reptiles, amphibians, swallo tailed
butterflies and some mammals.
Biodiversity and Its Conservation 363

374. 53World Summit on Sustainable
Development (2002) was held in
[CBSE AIPMT 2008]
(a) Brazil
(b) Sweden
(c) Argentina
(d) South Africa
Ans.(a)
In 1992, world leaders convened an
EarthSummitinRio de Janeiro, Brazil,
in search of international agreements
that could help to save the world from
pollution, poverty and the waste of
resources. Another Earth Summit was
convened from 26
th
August-4
th
September 2002 in Johannesburg,
South Africa.
54One of endangered species of
Indian medicinal plants is that of
[CBSE AIPMT 2007]
(a)Podophyllum
(b)Ocimum
(c) garlic
(d)Nepenthes
Ans.(a)
Podophyllumis an Indian endangered
plant of family–Berberidaceae. Its dried
roots and rhizomes are used in chronic
constipation and tumurous growth.
55Identify the odd combination of
the habitat and the particular
animal concerned.
[CBSE AIPMT 2007]
(a) Dachigam National Park
— Snow leopard
(b) Sunderbans — Bengal tiger
(c) Periyar — Elephant
(d) Rann of Kutch — Wild ass
Ans.(a)
Dachigam National Park is situated near
Dal Lake in Jammu and Kashmir. It is
known for conservation of the most
endangered Hangul or Kashmir stag in
paramount.
56Which one of the following is not
included underin situconservation?
[CBSE AIPMT 2006]
(a) Sanctuary
(b) Botanical gardens
(c) Biosphere reserve
(d) National park
Ans.(b)
Botanical gardens(i.e. man-made
areas that maintain living plant
collections representing a large number
of species, genera and families) are the
means ofex situconservation (i.e.
conservation outside the natural
habitats).
In situconservation involves the
conservation of genetic resources
through their maintenance within
natural ecosystems in which they
occur.
It includes National Parks, Sanctuaries,
Biosphere Reserves, Natural Reserves,
Natural Monuments, Cultural
Landscapes, etc.
57Which of the following is
considered a hotspot of
biodiversity in India?
[CBSE AIPMT 2006]
(a) Western ghats
(b) Indo-Gangetic plain
(c) Eastern ghats
(d) Aravalli hills
Ans.(a)
Hotspotsare the areas with high
density of diversity or megadiversity
which are also the most threatened
once. Today, the number of hot spots
identified by ecologists are 34, of which
two hot spots are present in India, i.e.
Western Ghats and North-East
Himalayas.
Western Ghats occur along the Western
Coast of India for a distance of about
1600 km in Maharashtra, Karnataka,
Tamil Nadu and Kerala extending over
to Sri Lanka.
58One of the most important
function of botanical garden is
that [CBSE AIPMT 2005]
(a) one can observe tropical plants
there
(b) they allowex situconservation of
germplasm
(c) they provide the natural habitat for
wild Life
(d) they provide a beautiful area for
recreation
Ans.(b)
A botanical garden is a garden
dedicated to the collection, cultivation
and display of a wide range of plants
with their botanical names.Ex situ
conservation means conservation of
plants or animals in the artificial
habitats which are quite similar to the
normal habitats of these organisms. In
this way botanical gardens provideex
situconservation of germplasm.
59In your opinion which is the most
effective way to conserve the plant
diversity of an area?
[CBSE AIPMT 2004]
(a) By tissue culture method
(b) By creating biosphere reserve
(c) By creating botanical gardens
(d) By developing seed banks
Ans.(b)
Biosphere reserve is anin situ
conservation method. Hence, it is the
most effective way among the four
above for preserving genetic diversity
by protecting wild population,
traditional life style and domesticated
plant genetic resource.
60Species restricted to a given area are
called [CBSE AIPMT 1998]
(a) sibling
(b) endemic
(c) sympatric
(d) allopatric
Ans.(b)
Species restricted to small areas are
called endemic; approximately 29% of
dicots in the Himalayas are endemic.
61MAB stands for[CBSE AIPMT 1997]
(a) Man And Biology programme
(b) Man And Biosphere programme
(c) Mammals And Biosphere
(d) Mammals And Biology programme
Ans.(b)
Man And Biosphere (MAB)programme
was formally launched by UNESCO in
1971. It is an interdisciplinary
programme of research and training
with emphasis on ecological approach
to the study of inter-relationship
between man and his environment.
364 NEETChapterwise Topicwise Biology

375. Biodiversity and Its Conservation 365
62A number of natural reserves have
been created to conserve specific
wildlife species. Identify the
correct combination from the
following [CBSE AIPMT 1996]
(a) Gir forest — Tiger
(b) Kaziranga — Elephants
(c) Rann of Kutch — Wild ass
(d) Manas Wildlife Sanctuary
— Musk deer
Ans.(c)
Rann of Kutch is situated in Gujarat and
provides protection mainly to wild ass,
whereas musk deers are mainly
protected in Kedarnath sanctuary. Gir is
famous for Asiatic lions. Kaziranga is
famous for one-horned rhinoceros.
63Which of the following is the
correct matching pair of a
sanctuary and its main protected
wild animal?[CBSE AIPMT 1995]
(a) Gir — Lion
(b) Sariska — Tiger
(c) Sunderban — Rhino
(d) Kaziranga — Musk deer
Ans.(b)
Sariskais a wildlife sanctuary and is
situated in Alwar, Rajasthan. Tiger is
main protected animal in Sariska as it
was selected as a Tiger Reserve in
Project Tiger (1973). Gir is a National
Park, associated with lions.
64National Park associated with
rhinoceros is[CBSE AIPMT 1994]
(a) Kaziranga
(b) Ranthambore
(c) Corbett
(d) Valley of flowers
Ans.(a)
Kaziranga National Park is associated
with rhinoceros. It is situated in
Golaghat and Nagaon districts of
Assam. This National Park is famous for
one- horned rhinoceros of India.
65Ranthambore National Park is
situated in[CBSE AIPMT 1994]
(a) Maharashtra
(b) Rajasthan
(c) Gujarat
(d) UP
Ans.(b)
Ranthambore National Park is situated
in Rajasthan.
This is one of the tiger reserves
established in 1973. Under the project
Tiger initiative taken by Indian
Government.

376. g369
01Air (Prevention and Control of
Pollution) Act was amended in 1987
to include among pollutants
[NEET (Oct.) 2020]
(a) vehicular exhaust
(b) allergy causing pollen
(c) noise
(d) particulates of size 2.5 micrometer or
below
Ans. (c)
Air prevention and control of pollution
acts was amended in 1987 to include
noise among pollutants. This act is
meant for preserving quality of air,
controlling air and noise pollution and
prevent their detrimental effects on
human health and health of other
biological entities.
02Due to increasing air-borne
allergens and pollutants, many
people in urban areas are suffering
from respiratory disorder causing
wheezing due to
[NEET (National) 2019]
(a) inflammation of bronchi and
bronchioles
(b) proliferation of fibrous tissues and
damage of the alveolar walls
(c) reduction in the secretion of
surfactants by pneumocytes
(d) benign growth on mucous lining of
nasal cavity
Ans.(a)
Wheezing occurs due to the
inflammation of bronchi and
bronchioles. It is one of the most
significant feature of asthma in which
people face difficulty in breathing. It is
usually caused due to increasing air
borne allergens and pollutants.
The allergens stimulate the release of
histamine from the mast cells which in
turn contracts the smooth muscles of
bronchioles.
03Which of the following protocols
did aim for reducing emission of
chlorofluorocarbons into the
atmosphere?[NEET (National) 2019]
(a) Kyoto Protocol
(b) Gothenburg Protocol
(c) Geneva Protocol
(d) Montreal Protocol
Ans.(d)
Montreal Protocol aimed to reduce the
emission of chlorofluorocarbons into
atmosphere because it has the
deleterious effects on stratospheric
ozone. This protocol was signed in
Montreal, Canada in 1987. Kyoto Protocol
aimed to reduce the emission of
CO , NO
2 2
and methane.
It was signed by 160 countries in a
convention held in Kyoto, Japan in 1997.
Geneva Protocol is a treaty to prohibit
the use of chemical or biological
weapons in international armed
conflicts.
Gothenberg Protocol is a multipollutant
protocol which focuses to reduce
eutrophications, acidification, emission
standards forSO
2
, etc.
04Which of the following is a
secondary pollutant?[NEET 2018]
(a)SO
2
(b)CO
2
(c) CO (d) O
3
Ans.(d)
Ozone( )O
3
is a secondary pollutant as it
is formed by the reaction amongst the
primary pollutants. On the other hand,
SO
2
is a primary pollutant. These
pollutants persist in the environment in
the form they are passed into it.COis
qualitative pollutant.
It is considered as pollutant only when
its concentration reaches beyond a
threshold value in the environment.CO
2
is a quantitative as well as a primary
pollutant.
05Which one of the following
statements is not valid for
aerosols ? [NEET 2017]
(a) They are harmful to human health
(b) They alter rainfall and monsoon
patterns
(c) They cause increased agricultural
productivity
(d) They have negative impact on
agricultural land
Ans.(c)
Aerosol refers to the suspended
particulate matter of less than 1μm size.
These are kind of air pollutants that are
suspended in our atmosphere. They have
a measurable effect on climate change as
they can modify the amount of energy
reflected by clouds.
As a result, they can change the
atmospheric circulation patterns and
affect agriculture negatively. These also
affect human health by causing breathing
problems.
06Which of the following are most
suitable indicators ofSO
2
pollution
in the environment?
[CBSE AIPMT 2015]
(a) Lichens
(b) Conifers
(c) Algae
(d) Fungi
Ans.(a)
Lichens are useful bioindicators for air
pollution, especially sulphur dioxide
pollution, since they derive their water
and essential nutrients mainly from
the atmosphere rather than from soil.
EnvironmentalIssues
38
Air Pollution and Its Control
TOPIC 1

377. Environmental Issues 367
07Acid rain is caused by increase in
the atmospheric concentration of
[CBSE AIPMT 2015]
(a)SO
2
andNO
2
(b)SO
3
and CO
(c)CO
2
and CO (d)O
3
and dust
Ans.(a)
Acid rain is caused by increase in the
atmospheric concentration ofSO
2
and
NO
2
. These mix with water vapour to
form sulphuric acid(H SO )
2 4
and nitric
acid(HNO )
3
respectively which falls on
earth in the form of acid rain.
08The UN conference of Parties on
climate change in the year 2012
was held at [CBSE AIPMT 2015]
(a) Durban (b) Doha
(c) Lima (d) Warsaw
Ans.(b)
The UN conference of Parties (COP-18)
was the UN framework convention on
climate change (UNFCCC) was held in Doha,
Qatar from 26th Nov. to 8th Dec, 2012.
09A scrubber in the exhaust of a
chemical industrial plant removes
[CBSE AIPMT 2014]
(a) Gases like sulphur dioxide
(b) Particulate matter of the size 5
micrometer or above
(c) Gases like ozone and methane
(d) Particulate matter of the size 2.5
micrometer or less
Ans.(a)
Scrubber is an electrostatic precipitator
in which the dirty air is cleaned by
capturing the gas likeSO
2
and other
oxides in water/lime spray(CaCO
3
). The
calcium in lime stone combines
chemically with the sulphur to produce
calcium sulphate(CaSO
4
), which is
separately collected. The detailed
mechanism is shown in the figure below:
10The Air Prevention and Control of
Pollution Act came into force in
[NEET 2013]
(a) 1975 (b) 1981
(c) 1985 (d) 1990
Ans.(b)
Air Prevention and Control of Pollution
Protection Act come into force in 1981 to
control and prevent air pollution. It was
amended in 1987. Environmental
Protection Act in 1986 and water
(Prevention and Control of Pollution) act
in 1974.
11Kyoto Protocol was endorsed at
[NEET 2013]
(a) CoP-3
(b) CoP-5
(c) CoP-6
(d) CoP-4
Ans.(a)
Kyoto Protocol is an international
agreement linked to United nations
framework convention on climate
change. It held at Kyoto, Japan in 1997
and entered into force on 16 February,
2005. The developed countries agreed to
specific targets for cutting their
emissions of greenhouse gases.
A general framework was defined for
this, with specifics to be detailed over
the next few years. This became known
as the Kyoto Protocol.
12Which one of the following
statements is wrong in case of
Bhopal gas tragedy?
[CBSE AIPMT 2011]
(a) Thousands of human being died
(b) adioactive fall out engulfed Bhopal
(c) It took place in the night of
December 23, 1984
(d) Methyl isocyanate gas leakage took
place
Ans.(d)
Bhopal gas gragedy (Bhopal disaster)
one of the world’s worst industrial
catastrophes. It occurred on the night of
December 23, 1984 at the Union Carbide
India Limited (UCIL) pesticide plant in
Bhopal, Madhya Pradesh.
A leak of methyl isocyanate gas and
other chemicals from the plant resulted
in the exposure of hundreds of
thousands of people.
The official immediate death toll was
2,259 and the government of Madhya
Pradesh has confirmed a total of 3,787
deaths related to the gas releases.
13dB is a standard abbreviation used
for the quantitative expression of
[CBSE AIPMT 2010]
(a) the density of bacteria in a medium
(b) a particular pollutant
(c) the dominantBacillusin a culture
(d) a certain pesticide
Ans.(b)
Noise pollution is a physical form of
pollution that affects the receiver
directly affecting the nervous system
which result into various disorders in
humans. dB (decibel) is a standard
abbreviation used for the quantitative
expression of noise. Noise or pollutant
sound has a value of 80 dB and above, it
harms hearing system. The WHO has
fixed 45 dB as the safe noise level for a
city.
14Steps taken by the Government of
India to control air pollution include
[CBSE AIPMT 2009]
(a) compulsory mixing of 20% ethyl
alcohol with petrol and 20%
biodiesel with diesel
(b) compulsory PUC (Pollution Under
Control) certification of petrol driven
vehicles, which tests for carbon
monoxide and hydrocarbons
(c) permission to use only pure diesel
with a maximum of 500 ppm sulphur
as fuel for vehicles
(d) use of non-polluting Compressed
Natural Gas (CNG) only as fuel by all
buses and trucks
Ans.(b)
Government of India have taken many
steps to control air pollution. Out of
which one includes compulsory PUC
(Pollution Under Control) certification of
petrol driven vehicles, which test for
carbon monoxide and hydrocarbons
emissions of the vehicles.
15According to Central Pollution
Control Board (CPCB), which
particulate size in diameter (in
micrometres) of the air pollutants
is responsible for greatest harm to
human health?[CBSE AIPMT 2008]
(a) 2.5 or less (b) 1.5 or less
(c) 1.0 or less (d) 5.2 or 2.5
Ans.(a)
According to CPCB, air pollutants of size
2.5 or less (in micrometres) diameter are
harmful to human health. It is the main
cause of respiratory disorders in
polluted cities like Delhi.
Water
Liquid in
Clean
Gas/Mist out
Dirt
Gas In
Slurry/Waste Out
Dirt
Gas In

378. 16In a coal fired power plant,
electrostatic precipitators are
installed to control emission of
[CBSE AIPMT 2007]
(a)SO
2
(b)NO
x
(c) SPM (d) CO
Ans.(c)
The electrostatic precipitors are
installed to control emission of
Suspended Particulate Matter (SPM) as it
is cause of various respiratory disorders
in humans.
17Lead concentration in blood is
considered alarming if it is
[CBSE AIPMT 2004]
(a) 20μg/100 mL (b) 30μg/100 mL
(c) 4-6μg/100 mL (d) 10μg/100 mL
Ans.(b)
The concentration of lead in blood
averages about 25μg/100mL. Increase
to70μg /100 mLis generally associated
with clinical symptoms. Hence, a level of
30μg /100 mLis considered alarming.
The chief sources of Pb to water are the
effluents of lead and lead processing
industries.
18In 1984, the Bhopal gas tragedy
took place because methyl
isocyanate [CBSE AIPMT 2004]
(a) reacted with DDT
(b) reacted with ammonia
(c) reacted withCO
2
(d) reacted with water
Ans.(d)
Bhopal gas tragedy occurred (23 Dec.
1984) when MIC (Methyl Isocyanate)
reacted with water in a tank, an
exothermic chemical reaction started
and produced a lot of heat. As a result,
the safety valve of tank burst because of
increase in pressure and gave rise to a
heavy gas which rapidly killed the people
around.
19What is the intensity of sound in
normal conversation?
[CBSE AIPMT 2001]
(a) 0-20 dB (b) 30-60 dB
(c) 70-90 dB (d) 120-150 dB
Ans.(b)
The word noise is taken from the Latin
word nausea and is defined as unwanted
or unpleasant sound that causes
discomfort.
Intensity of some noise sources is as
follows:
Source Intensity (dB)
Breathing 10
Broadcasting studio 20
Trickling clock 30
Library 30-35
Normal conversation 35-60
Telephone 60
Office noise 60-80
Alarm clock 70-80
Traffic 50-90
Motor cycle 105
Jet fly (over 1000') 100-110
Train whistle (50') 110
Air craft (100') 110-120
Commercial jet air craft
(100')
120-140
Space rocket
(launching)
170-180
20Which of the following is pollution
related disorder?[CBSE AIPMT 1999]
(a) Fluorosis (b) Leprosy
(c) Pneumonicosis (d) Silicosis
Ans.(d)
Silicosisis a pollution related disorder. It
is caused by inhalation of dust
containing free silica or silicon dioxide
especially by workers engaged in mining,
pottery, ceramic industry, sand blasting,
building and construction industries.
Fluorosisis caused due to deficiency of
fluoride.
Leprosyis caused byMycobacterium
leprae.
Pneumonicosisis caused byDiplococcus
pneumoniae.
21Which of the following is a
secondary pollutant?
[CBSE AIPMT 1999]
(a) Aerosol (b) CO
(c) PAN (d) CO
2
Ans.(c)
PAN (Peroxy Acetyl Nitrate) is secondary
pollutant. Pollutants formed by the
chemical interaction of primary
pollutants with atmospheric gas and
moisture, often catalysed by sunlight are
called secondary pollutants.
22In 1984, Bhopal gas tragedy was
caused due to the leakage of
[CBSE AIPMT 1999]
(a) potassium isocyanate
(b) sodium monoxide
(c) sodium thiocyanate
(d) methyl isocyanate
Ans.(d)
Methyl isocyanate gas, used as raw
material for synthesising carbonyl,
caused Bhopal gas tragedy in 1984.
23Carbon monoxide is a pollutant
because [CBSE AIPMT 1998]
(a) it reacts withO
2
(b) it inhibits glycolysis
(c) it reacts with haemoglobin
(d) it makes nervous system inactive
Ans.(c)
Carbon monoxide, (CO) when inhaled,
combines with blood haemoglobin to
form carboxy haemoglobin at a rate 210
times faster than the oxygen forms
oxyhaemoglobin. Thus, respiration is
impaired.
24TheCO
2
content by volume, in the
atmospheric air is about
[CBSE AIPMT 1997]
(a) 0.0314% (b) 0.34%
(c) 3.34% (d) 4%
Ans.(a)
CO
2
constitutes 0.0314% of the
atmosphere. Producers useCO
2
along
with energy from sun and make carbon
compounds such as glucose during the
process of photosynthesis. Consumers
use these compounds as energy source.
25The Taj Mahal is threatened due to
the effect of[CBSE AIPMT 1995]
(a) oxygen
(b) hydrogen
(c) chlorine
(d) sulphur dioxide
Ans.(d)
Taj Mahal of Agra is affected by gases
discharged from oil refinery in Mathura
which consists ofSO
2
,H S
2
and nitrogen
oxide.SO
2
corrodes metals, equipment
and damages buildings marble.
26Sound becomes hazardous noise
pollution at level[CBSE AIPMT 1994]
(a) above 30 dB (b) above 80 dB
(c) above 100 dB (d) above 120 dB
Ans.(b)
Noise pollution is measured in decibels.
Noise level up to 64 dB is well tolerated.
Noise above 80 dB causes discomfort in
man. WHO recommends an industrial
noise limit of 75 dB.
368 NEETChapterwise Topicwise Biology

379. 27Atmosphere of big/metropolitan
cities is polluted most by
[CBSE AIPMT 1994]
(a) automobile exhausts
(b) pesticide residue
(c) household waste
(d) radioactive fall-out
Ans.(a)
Automobile exhausts are the largest
source of air pollution in big cities.
Automobiles release carbon monoxide
(77.2%), hydrocarbons (13.7%) and
nitrogen oxide (7.7%).
28Major aerosol pollutant in jet plane
emission is [CBSE AIPMT 1990]
(a) sulphur dioxide
(b) carbon monoxide
(c) methane
(d) chlorofluoro-carbons
Ans.(d)
Aerosols are the chemicals released in
air with force. Jet aeroplanes are
important source of aerosol in upper
atmosphere. Aerosols contain CFCs
(Chlorofluoro Carbons).
29Which one is not a pollutant
normally? [CBSE AIPMT 1988]
(a) Hydrocarbons
(b) Carbon dioxide
(c) Carbon monoxide
(d) Sulphur dioxide
Ans.(b)
CO
2
is an essential component of the air
and its concentration is 0.03% but when
CO
2
concentration goes above this limit,
it acts as a pollutant.
30Acid rains are produced by
[CBSE AIPMT 1988]
(a) excessNO
2
andSO
2
from burning
fossil fuels
(b) excess production ofNH
3
by industry
and coal gas
(c) excess release of carbon monoxide
by incomplete combustion
(d) excess formation ofCO
2
by
combustion and animal respiration
Ans.(a)
SO
2
andNO
2
are the gases responsible
for acid rains.SO
2
mainly comes from
burning of coal, fossil fuel, in the form of
smoke from industries.
31Which of the following is put into
anaerobic sludge digester for
further sewage treatment?
[NEET (Sep.) 2020]
(a) Floating debris
(b) Effluents of primary treatment
(c) Activated sludge
(d) Primary sludge
Ans.(c)
Activated sludge is put into anaerobic
sludge digester for further sewage
treatment. It contains biological flocs
that contain bacteria and protozoan for
further digestion of organic wastes
under aerobic conditions.
32Which one of the following
equipments is essentially required
for growing microbes on a large
scale, for industrial production of
enzymes? [NEET (National) 2019]
(a) Sludge digester (b) Industrial oven
(c) Bioreactor (d) BOD incubator
Ans.(c)
Bioreactors are required for growing
microbes on large scale for the industrial
production of enzymes. These large
vessels provide biologically active
environment. On the other hand, sludge
digesters are used to decompose
organic solid waste under aerobic
conditions. BOD incubators are used to
maintain the temperature for tissue
culture growth, bacterial cultures, etc.
33Match the items given in Column I
with those in Column II and select
the correct option given below.
[NEET 2018]
Column-I Column-II
1. Eutrophication i. UV-B radiation
2. Sanitary landfill ii. Deforestation
3. Snow blindness iii. Nutrient
enrichment
4. Jhum cultivation iv. Waste disposal
1 2 3 4
(a) (iii) (iv) (i) (ii)
(b) (i) (iii) (iv) (ii)
(c) (ii) (i) (ii) (iv)
(d) (i) (ii) (iv) (iii)
Ans.(a)
Eutrophication is the nutrient
enrichment of water bodies containing
excessive population of phytoplanktons.
Sanitary landfillis a method of solid waste
disposal in which the waste material is
burried in the pits dug on the ground and
later they get covered by soil.
Snow blindnessis caused due to UV-B
radiations exposure. These radiations
can reach the earth surface due to the
depletion of ozone layer.
InJhum cultivation, land is cultivated
temporarily and then abandoned so that,
it can revert to its natural vegetation. It
is a long term process and usually leads
to deforestation.
34A river with an inflow of domestic
sewage rich in organic waste may
result in [NEET 2016, Phase I]
(a) increased population of aquatic food
web organisms
(b) an increased production of fish due
to biodegradable nutrients
(c) death of fish due to lack of oxygen
(d) drying of the river very soon due to
algal bloom
Ans.(c)
A river with an inflow of domestic
sewage rich in organic waste will reduce
the dissolved oxygen (DO). The organic
waste will increase biological oxygen
demand of the river thus depleting theO
2
content and may result in death of fish
due to lack of oxygen.
35The highest DDT concentration in
aquatic food chain shall occur in
[NEET 2016, Phase II]
(a) phytoplankton
(b) seagull
(c) crab
(d) eel
Ans.(b)
DDT is a toxic substance which gets
concentrated subsequently in a food
chain of an aquatic ecosystem in the
following manner
Phytoplanktons→Eel→Crab→
Seagull
(0.04 ppm) (0.5 ppm) (2 ppm) (25 ppm)
Thus, the animal or organism acquiring
the highest position in a food chain
would have the highest DDT
concentration (here seagull). This
process is known as biological
magnification or biomagnification.
Environmental Issues 369
Water Pollution
and Its Control
TOPIC 2

380. 36A lake which isrich in organic waste
may result in[NEET 2016, Phase II]
(a) increased population of aquatic
organisms due to minerals
(b) drying of the lake due to algal bloom
(c) increased population of fish due to
lots of nutrients
(d) mortality of fish due to lack of
oxygen
Ans.(d)
When much of organic matter is present
in lake, lots of microbial activity takes
place in its decomposition process. So,
demand for oxygen increases. This
increasedO
2
demand depletes the
dissolved oxygen in water at faster rates.
This adversely affects the living conditions
for other organisms like fishes, etc.
ultimately causing their death.
37Increase in concentration of the
toxicant at successive tropic levels
is known as [CBSE AIPMT 2015]
(a) biomagnification
(b) biodeterioration
(c) biotransformation
(d) biogeochemical cycling
Ans.(a)
Biomagnification is the sequence of
processes in an ecosystem by which
higher concentrations of a particular
toxicant, such as the pesticide or heavy
metal are reached in higher organisms of
the food chain, generally through a
series of prey-predator relationships.
38Eutrophication of water bodies
leading to killing of fishes is mainly
due to non-availability of
[CBSE AIPMT 2015]
(a) food
(b) light
(c) essential minerals
(d) oxygen
Ans.(c)
Eutrophication is a process where water
bodies receive excess nutrients that
stimulate excessive plant growth. This
can lead to overcrowding and
competition for sunlight, space and
oxygen. This condition creates the
lacking of essential nutrients for fishes
due to which they die.
39In an area where DDT had been
used extensively, the population of
birds declined significantly because
[CBSE AIPMT 2012]
(a) birds stopped laying eggs
(b) earthworms in the area got
eradicated
(c) cobras were feeding exclusively on
birds
(d) many of the birds laid eggs, that did
not hatch
Ans.(d)
DDT, its breakdown products and other
chlorinated hydrocarbon pesticides pose
serious threat to birds. These persistant
poisons enter the food chain and they
accumulate in the fatty tissues of
organisms at lower trophic level and
then tend to concentrate in considerably
toxic amount as they move up through
the food chain. This is called
biomagnification or bioconcentration. It
weakens the calcarious egg shell of birds
as it becomes very thin.
40Measuring Biochemical Oxygen
Demand (BOD) is a method used for
[CBSE AIPMT 2012]
(a) estimating the amount of organic
matter in sewage water
(b) working out the efficiency of oil
driven automobile engines
(c) measuring the activity of
Saccharomyces cerevisaein
producing curd on a commercial
scale
(d) working out the efficiency of RBCs
about their capacity to carry oxygen
Ans.(a)
Decomposition of organic matter by
microbial activity depends on oxygen
availability in water.
The degree of impurity of water due to
the organic matter is measured in terms
of BOD (Biochemical Oxygen Demand) or
BOD is the oxygen in milligrams required
for five days in one litre of water at 20°C
for the microorganisms to metabolise
organic waste.
41The domestic sewage in large
cities
[CBSE AIPMT 2012]
(a) has a high BOD as it contains both
aerobic and anaerobic bacteria
(b) is processed by aerobic and then
anaerobic bacteria in the secondary
treatment in Sewage Treatment
Plants (STPs)
(c) when treated in STPs does not really
require the aeration step as the
sewage contains adequate oxygen
(d) has very high amounts of suspended
solids and dissolved salts
Ans.(b)
Sewage is waste water having food
residue, animal and human excreta,
detergents, discharges from commercial
and industrial establishments.
The domestic sewage is processed first
by aerobic and then by anaerobic
bacteria in secondary treatment in
Sewage Treatment Plant (STPs) and then
recycled for further use.
42DDT residues are rapidly passed
through food chain causing
biomagnification because DDT is
(a) lipo soluble[CBSE AIPMT 2009]
(b) moderately toxic
(c) non-toxic to aquatic animals
(d) water soluble
Ans.(a)
Many of the pesticides, such as DDT,
aldrin and dieldrin are accumulated in
the environment.
They are fat soluble and generally
non-biodegradable. They get
incorporated into the food chain and
ultimately deposited in the fatty tissues
of animals and humans.
In the food chain, because of their build
up, they get magnified in the higher
trophic levels called biological
magnification. The phenomenon of
biological magnification is also reported
for certain other pollutants such as,
heavy metals, e.g. mercury, copper and
radioactive substances as strontium-90.
43Biochemical Oxygen Demand (BOD)
in a river water[CBSE AIPMT 2009]
(a) remains unchanged when algal
bloom occurs
(b) has no relationship with
concentration of oxygen in the water
(c) gives a measure ofSalmonellain the
water
(d) increases when sewage gets mixed
with river water
Ans.(d)
When sewage gets mixed with river
water, BOD will increase.Biochemical
OxygenDemand (BOD) is the amount of
oxygen used for biochemical oxidation of
organic mater by microorganisms in a
unit volume of water. Polluted water has
high BOD.
44A lake near a village suffered heavy
mortality of fishes within a few
days. Consider the following
reasons for this
370 NEETChapterwise Topicwise Biology

381. I. Lots of urea and phosphate
fertilizer were used in the crops
in the vicinity.
II. The area was sprayed with DDT
by an aircraft.
III. The lake water turned green and
stinky.
IV. Phytoplankton populations in the
lake declined initially thereby
greatly reducing photosynthesis.
Which two of the above were the
main causes of fish mortality in
the lake? [CBSE AIPMT 2008]
(a) II, III (b) III, IV
(c) I, III (d) I, II
Ans.(d)
A lake near a village suffered heavy
mortality of fishes within a few days,
because lots of urea and phosphate
fertiliser were used in the crops in the
vicinity and the area was sprayed with
DDT by an aircraft. Inorganic phosphorus
and nitrogen are responsible for the
growth of algae.
In polluted water, these increase due to
which algae increase greatly at the
surface of water forming water bloom.
Due to the death of these algae their
organic matter gets decomposed due to
which oxygen gets depleted and aquatic
animals die.
45A lake near a village suffered heavy
mortality of fishes within a few
days. Consider the following
reasons for this
I. Lots of urea and phosphate
fertiliser were used in the crops
in the vicinity.
II. The area was sprayed with DDT
by an aircraft.
III. The lake water turned green and
stinky.
IV. Phytoplankton populations in the
lake declined initially thereby
greatly reducing photosynthesis.
Which two of the above were the
main causes of fish mortality in
the lake? [CBSE AIPMT 2008]
(a) II and III
(b) III and IV
(c) I and III
(d) I and II
Ans.(d)
A lake near a village suffered heavy
mortality of fishes within a few days,
because lots of urea and phosphate
fertiliser were used in the crops in the
vicinity and the area was sprayed with
DDT by an aircraft.
Inorganic phosphorus and nitrogen are
responsible for the growth of algae. In
polluted water these increase due to
which algae increase greatly at the
surface of water forming water bloom.
Due to death of these algae their organic
matter gets decomposed due to which
oxygen gets depleted and aquatic animal
dies.
46Which one of the following is not a
bioindicator of water pollution?
[CBSE AIPMT 2007]
(a) Sludge worms
(b) Blood worms
(c) Stone flies
(d) Sewage fungus
Ans.(c)
Among the following stone flies are not
bioindicators of pollution. Stone flies
(e.gPerlasp.) belong to
order–Plecoptera of class–Insecta,
which has the terrestrial mandibulates.
47In which one of the following, the
BOD (Biochemical Oxygen Demand)
of sewage (S), distillery effluent
(DE), paper mill effluent (PE) and
sugar mill effluent (SE) have been
arranged in ascending order?
[CBSE AIPMT 2007]
(a) SE < S < PE < DE
(b) SE < PE < S < DE
(c) PE < S < SE < DE
(d) S < DE < PE < SE
Ans.(d)
The ascending order of BOD is Sewage
(S) < Distillary Effluent (DE) < Paper Mill
Effluent (PE) < Sugar Mill Effluent (SE).
48Photochemical smog pollution does
not contain[CBSE AIPMT 2006]
(a) ozone
(b) nitrogen dioxide
(c) carbon dioxide
(d) PAN (Peroxy Acyl Nitrate)
Ans.(c)
Photochemical smog is highly oxidising
pollutant comprising largely of ozone
(O )
3
, oxides of nitrogen(NO )
x
, hydrogen
peroxide(H O )
2 2
, organic peroxides,
Peroxy Acetyl Nitrate (PAN) and Peroxy
Benzyl Nitrate (PBN) but not carbon
dioxide(CO )
2
.
49Limit of BOD prescribed by Central
Pollution Control Board for the
discharge of industrial and
municipal waste water into natural
surface water, is
[CBSE AIPMT 2006]
(a) < 3.0 ppm (b) < 10 ppm
(c) < 100 ppm (d) < 30 ppm
Ans.(b)
The Central Pollution Control Board
prescribed the BOD limit for the
discharge of industrial and municipal
waste water as < 10 ppm.
50Which of the following is not used
for disinfection of drinking water?
[CBSE AIPMT 2005]
(a) Phenyl (b) Chloramine
(c) Chlorine (d) Ozone
Ans.(a)
Phenyl is not used for disinfection of
drinking water as it is a poisonous
substance. Water is chlorinated for
getting disinfected.
51Fluoride pollution mainly affects
[CBSE AIPMT 2004]
(a) teeth (b) kidney
(c) brain (d) heart
Ans.(a)
Prolonged intake of fluoride polluted
water causes stiffning of bone and joints
particularly spinal cord. Due to the
affinity with calcium, fluoride stores in
bones which causes mottling of teeth,
bone pains and outward bending of legs
from the knees. This is known asknock
knee syndrome.
52Which of the following is absent in
polluted water ?
[CBSE AIPMT 2002]
(a)Hydrilla (b) Water hyacinth
(c) Larva of stone fly (d) Blue-green algae
Ans.(c)
Stone flies are exopterygote insects with
aquatic nymphs; long antennae, biting
mouth parts and weak flight. Adults have
the tendency to feed on lichens and
unicellular algae. These are good
pollution indicators.
53What is BOD?[CBSE AIPMT 2001]
(a) The amount ofO
2
utilised by
organisms in water
(b) The amount ofO
2
utilised by
microorganisms for decomposition
(c) The total amount ofO
2
present in water
(d) All of the above
Environmental Issues 371

382. Ans.(b)
Biological Oxygen Demand(BOD) is the
amount ofO
2
required for biological
oxidation of organic matter present in
polluted water by microorganisms in any
unit volume of water.
54DDT is [CBSE AIPMT 1999]
(a) a non-degradable pollutant
(b) an antibiotic
(c) a biodegradable pollutant
(d) not a pollutant
Ans.(a)
DDT is non-degradable pollulant which
accumulates in the tissues. Its
concentration is estimated to be 13-31
ppm in body fat of Indians, which has
been accumulated in them through
biomagnification of DDT through food
chain.
55A sewage treatment process in
which a portion of the decomposer
bacteria present in the waste is
recycled into the beginning of the
process, is called
[CBSE AIPMT 1998]
(a) cyclic treatment
(b) primary treatment
(c) activated sludge treatment
(d) tertiary treatment
Ans.(c)
The word activated sludge system is
derived from the practice of adding
aerobic heterotrophic bacteria to the
incoming sewage, from a previous
batch. This sludge inoculum contains
large numbers of metabolising bacteria,
together with yeasts, molds and
Protozoa. An especially important
ingredient of the sludge are species of
Zoogloeabacteria, which form flocculant
masses (floc) in the aeration tanks.
The activity of these aerobic
micro-organisms oxidises much of the
effluent’s organic matter into carbon
dioxide and water. When the aeration
phase is completed, the floc (secondary
sludge) is allowed to settle to the
bottom, just as the primary sludge
settles in primary treatment.
56Which one of the following
organism is used as indicator of
water quality?[CBSE AIPMT 1998]
(a)Beggiatoa
(b)Chlorella
(c)Azospirillum
(d)Escherichia
Ans.(d)
A variety of the enteric group of bacteria
(facultative, aerobic) reside in the human
large intestine (e.g.E.coli). Therefore,
their presence in water supply indicates
that water supply has been
contaminated by sewage.
57Phosphate pollution is mainly
caused by [CBSE AIPMT 1997]
(a) phosphate rock only
(b) agricultural fertilisers only
(c) sewage and phosphate rocks
(d) sewage and agricultural fertilisers
Ans.(d)
Man has been releasing large quantities
of phosphorus into the biosphere in the
form of agricultural fertilisers
(superphosphates) and synthetic
detergents.
58Sewage drained into water bodies
kill fishes because
[CBSE AIPMT 1996]
(a) excessive carbon dioxide is added
to water
(b) it gives off a bad smell
(c) it removes the food eaten by fish
(d) it increases competition with fishes
for dissolved oxygen
Ans.(d)
Sewage provides food for decomposers.
Phosphorus and nitrogen compounds
present in sewage result into excessive
growth of algae on water surface, which
is also called algal bloom, this all results
in decrease in oxygen, which is
dangerous for aquatic life.
59In Minamata Bay of Japan, the
animals which remained free from
Minamata disease, are
[CBSE AIPMT 1995]
(a) pigs (b) rabbits
(c) dogs d) cats
Ans.(b)
Rabbits remained free from Minamata
disease as they are herbivores and this
disease is caused due to mercury
pollution in water.
60When huge amount of sewage is
dumped into a river, its BOD will
[CBSE AIPMT 1995]
(a) increase
(b) decrease
(c) sharply decrease
(d) remain unchanged
Ans.(d)
Taj Mahal of Agra is affected by gases
discharged from oil refinery in Mathura
which consists ofSO
2
,H S
2
and nitrogen
oxide.SO
2
corrodes metals, equipment
and damages buildings marble.
61Highest DDT deposition shall occur
in [CBSE AIPMT 1994]
(a) phytoplankton
(b) sea gull/birds
(c) crab
(d) eel
Ans.(b)
DDT concentration increases in amount
with rise in trophic level because they
accumulate in fat, this is
biomagnification. In the given options
sea gull/birds are the top consumers so
DDT concentration will be highest in
them.
62Disease caused by eating fish
found in water contaminated with
industrial waste having mercury is
[CBSE AIPMT 1994]
(a) Minamata disease
(b) Bright’s disease
(c) Hashimoto’s disease
(d) Osteosclerosis
Ans.(a)
Consumption of fishes poisoned with
mercury causes deformity called
Minamata disease, which is
characterised by diarrhoea, impairment
of senses, haemolysis, meningitis and
death.
63Domestic waste constitutes
[CBSE AIPMT 1991]
(a) non-biodegradable pollution
(b) biodegradable pollution
(c) effluents
(d) air pollution
Ans.(b)
Domestic waste constitutes
biodegradable pollutants, such
pollutants are naturally present organic
compounds which can be broken down
by microorganisms and can be recycled,
e.g. sewage.
64The maximum biomagnification
would be in which of the following
in case of aquatic ecosystem?
[CBSE AIPMT 1999]
(a) Fishes (b) Phytoplanktons
(c) Birds (d) Zooplanktons
372 NEETChapterwise Topicwise Biology

383. Ans.(a)
Non-degradable chemicals enter the
food chain and their concentration goes
up as they move up in the food chain.
This phenomenon is called
biomagnification. Naturally in a food
chain, Phytoplankton→Zooplankton
→Fishes→Birds, it would be highest in
fishes.
65Animals/organisms floating on the
surface of water are
[CBSE AIPMT 1988]
(a) plankton (b) pelagic
(c) benthos (d) neritic
Ans.(a)
Organisms floating on the surface of
water are planktons, these are of two
types
(a)ZooplanktonIf the floating
organisms are animals.
(b)PhytoplanktonIf the floating
organism are plants.
66Which of the following is an
innovative remedy for plastic
waste? [NEET (Odisha) 2019]
(a) Burning in the absence of oxygen
(b) Burrying 500 m deep below soil surface
(c) Polyblend
(d) Electrostatic precipitator
Ans.(c)
Polyblend is an innovative remedy for
plastic waste. Polyblend is a fine powder
of recycled modified plastic which when
mixed with bitumen, can be used to lay
roads.
67Which of these following methods
is the most suitable for disposal of
nuclear waste?
[NEET (National) 2019]
(a) Bury the waste under Antarctic
ice-cover
(b) Dump the waste within rocks under
deep ocean
(c) Bury the waste within rocks deep
below the Earth’s surface
(d) Shoot the waste into space
Ans.(c)
Nuclear waste is usually disposed by
burying it within rocks deep below the
earth’s surface. Nuclear waste disposal
is extremely hazardous. Before burrying
the waste, it is sealed in large containers
so as to reduce its radiation effects.
68Relative Biological Effectiveness
(RBE) is usually referred to
damages caused by
[CBSE AIPMT 2000]
(a) low temperature
(b) high temperature
(c) encephalitis
(d) radiation
Ans.(d)
RBE (Relative Biological Effectiveness)
is a comparison of the dose of the
radiation being studied with the dose of
standard radiation producing the same
effect.
69If by radiation all nitrogenase
enzymes are inactivated, then
there will be no[CBSE AIPMT 2004]
(a) fixation of nitrogen in legumes
(b) fixation of atmospheric nitrogen
(c) conversion from nitrate to nitrite in
legumes
(d) conversion from ammonium to
nitrate in soil
Ans.(a)
The enzyme nitrogenase is required for
the process of biological nitrogen
fixation only. Fixation of atmospheric
nitrogen occur through other route also.
Neither nitrification (conversion of
ammonium to nitrate) nor conversion of
nitrate to nitrite require nitrogenase.
70Relative Biological Effectiveness
(RBE) usually refers to the damages
caused by [CBSE AIPMT 2000]
(a) low temperature
(b) high temperature
(c) radiation
(d) pollution
Ans.(c)
RBE (Relative Biological Effectiveness)
is a comparison of the dose of radiation
being studied with the dose of standard
radiation producing the same effect.
71The supersonic jets cause pollution
by the thinning of
[CBSE AIPMT 1998]
(a)CO
2
layer
(b)SO
2
layer
(c)O
2
layer
(d)O
3
layer
Ans.(d)
Ozone layer is found in the stratosphere.
It protects us from the harmful UV
radiations coming from the sun. The
supersonic aircrafts flying at
stratospheric heights cause major
disturbances in ozone level due to
release of CFCs which react withO
3
present there, thus cause thinning ofO
3
layer.
72The worst environmental hazards
were created by accidents in
nuclear power plant and MIC gas
tragedy respectively in
[CBSE AIPMT 1996]
(a) Russia in 1990 and Bhopal in 1986
(b) Ukrain in 1988 and USA in 1984
(c) Bhopal in 1984 and Russia in 1990
(d) Ukrain in 1986 and Bhopal in 1984
Ans.(d)
Worst environmental hazards were in
Ukrain in 1986 and MIC gas tragedy in
Bhopal in 1984.
73Most hazardous metal pollutant of
automobile exhausts is
[CBSE AIPMT 1992]
(a) mercury (b) cadmium
(c) lead (d) copper
Ans.(c)
Lead particles are found in the smoke
from automobiles which cause nervous
disorder in man.
74Pedology is science of
[CBSE AIPMT 1991]
(a) earth (b) soil
(c) diseases (d) pollution
Ans.(b)
Pedology is the study of soil and soil
properties.
75Which of the following pairs of
gases is mainly responsible for
greenhouse effect?
[NEET (National) 2019]
(a) Oxygen and Nitrogen
(b) Nitrogen and Sulphur dioxide
(c) Carbon dioxide and Methane
(d) Ozone and Ammonia
Environmental Issues 373
Soil and Radioactive
Pollution
TOPIC 3
Greenhouse Effect
and Global Warming
TOPIC 4

384. Ans.(c)
Greenhouse effect is mainly contributed
by carbon dioxide (60%) and methane
(20%) along with nitrous oxide, nitrogen
dioxide and chlorofluorocarbons.
Greenhouse effect results in the rise in
temperature of earth because
greenhouse gases has the ability to trap
the heat of solar radiations.
76Global warming can be controlled
by [NEET 2013]
(a) Reducing deforestation, cutting
down use of fossil fuel
(b) Reducing reforestation, increasing
the use of fossil fuel
(c) Increasing deforestation, slowing
down the growth of human
population
(d) Increasing deforestation, reducing
efficiency of energy usage
Ans.(a)
Global warming can be controlled by
reducing deforestation, cutting down
use of fossil fuel, which results into
reduction in the concentration of one of
the greenhouse gas, i.e.CO
2
in the
atmosphere. The other ways of reducing
global warming are slowing down the
growth of human population, improving
efficiency of energy usage and encouraging
eco-friendly sustainable development.
77Which one of the following pairs of
gases are the major cause of
‘Greenhouse effect?
(a)CO
2
and CO [CBSE AIPMT 2011]
(b)CFCsandSO
2
(c)CO
2
andN O
2
(d)CO
2
andO
3
Ans.(c)
The phenomenon of keeping the earth’s
surface warm is due to the presence of
certain gases in the atmosphere that are
called greenhouse gases.
The name is based after a similar
warmer interior in a glass-enclosed
greenhouse where glass panels,CO
2
and
water vapour allow the solar radiations
to enter but prevent the escape of long
wave heat radiations.CO
2
andN O
2
are
the major cause of ‘greenhouse effect’.
CO
2
contributes 60% of total global
warming.N O
2
contributes 6% to
greenhouse effect.
78The two gases making highest
relative contribution to the
greenhouse gases are
[CBSE AIPMT 2010]
(a)CO
2
andCH
4
(b)CH
4
andNO
2
(c) CFCs andN O
2
(d)CO
2
andN O
2
Ans.(a)
The greenhouse effect is a naturally
occurring phenomenon that is
responsible for warming of earth’s
surface and atmosphere.CO
2
(60%) and
CH
4
(20%) are commonly known as
greenhouse gases because they are
responsible for the greenhouse effect,
that ultimately leads to global warming.
79Which one of the following is the
correct percentage of the two (out
of the total of four) greenhouse
gases that contribute to the total
global warming?[CBSE AIPMT 2008]
(a) CFCs 14%,CH
4
20%
(b)CO
2
40%, CFCs 30%
(c)N O
2
6%,CO
2
86%
(d)CH
4
20%,N O
2
18%
Ans.(a)
The percentage of various greenhouse
gases that contribute to total global
warming areCO
2
(warming effect 60%),
CH
4
(20%), CFCs (14%) and nitrous oxide
N O
2
(6%).
80Which one of the following pair is
mismatched?[CBSE AIPMT 2005]
(a) Biomass burning — Release ofCO
2
(b) Fossil fuel burning — Release ofCO
2
(c) Nuclear power — Radioactive wastes
(d) Solar energy — Greenhouse effect
Ans.(d)
From the following pair (d) is
mismatched because solar energy does
not cause greenhouse effect.
81Greenhouse effect refers to
[CBSE AIPMT 1999]
(a) production of cereals
(b) cooling of earth
(c) trapping of UV rays
(d) warming of earth
Ans.(d)
Greenhouse effects refer to warming of
earth due to increase inCO , CFCs, SO
2 2
and other substances that disturbe the
balance between the amount of energy
received and that reflected back into the
space.
82Warm ocean surge of the Peru
current recurring every 5 to 8 year
or so in the East Pacific of South
America is widely known as
[CBSE AIPMT 1998]
(a) Magnox (b) Gull stream
(c) El Nino (d) Aye Aye
Ans.(c)
EL Nino is a warm ocean surge of Peru
current (flowing North from Antarctica
along the West coast of South America
to South Equador, the West). It recurs
every 5-8 year or so, in the East Pacific
of South America.
83Which important greenhouse gas,
other than methane, is being
produced from the agricultural
fields? [CBSE AIPMT 1998]
(a) Arsine (b) Sulphur dioxide
(c) Ammonia (d) Nitrous oxide
Ans.(d)
In addition toCO
2
, some other gases also
contribute to greenhouse effect. These
include ozone, CFCs, nitrous oxideN O
2
and even methane(CH )
4
.
Nitrous oxide is produced by denitrifying
bacteria acting on artificial fertilisers
applied to poorly aerated soils.
84If there was noCO
2
in the earth’s
atmosphere the temperature of
earth’s surface would be
[CBSE AIPMT 1998]
(a) same as present
(b) less than the present
(c) higher than the present
(d) dependent on the amount of oxygen
in the atmosphere
Ans.(b)
CO
2
layer around earth surface acts as
insulator and does not allow heat of the
earth to escape into space thus keeping
the earth warm. It is an important
constituent of green house gases
present in earth’s atmosphere. The
percentage ofCO
2
in causing global
warming is 60%.
85The major contributor of
green-house gases to the
atmosphere is[CBSE AIPMT 1996]
(a) Russia
(b) USA
(c) Germany
(d) Brazil
Ans.(b)
The major contributor of greenhouse
gases to the atmosphere is USA.
86Green-house effect is warming due
to [CBSE AIPMT 1991]
(a) infra-red rays reaching earth
(b) moisture layer in atmosphere
374 NEETChapterwise Topicwise Biology

385. Environmental Issues 375
(c) increase in temperature due to
increase in carbon dioxide
concentration of atmosphere
(d) ozone layer of atmosphere
Ans.(c)
Increase inCO
2
concentration forms a
cover around the earth, this cover
inhibits the earth’s radiation to go out of
earth’s environment thus, increases
global temperature.
This increased temperature results in
melting of snow from hill tops and
poles.
87Dobson units are used to measure
thickness of [NEET 2021]
(a) CFCs (b) stratosphere
(c) ozone (d) troposphere
Ans.(c)
Ozone found in stratosphere is known as
good ozone or ozone layer. It acts as a
shield absorbing ultraviolet radiation
from the sun. The thickness of ozone in a
column of air from the ground to the top
of the atmosphere is measured in terms
of Dobson units.
88Montreal protocol was signed in
1987 for control of
[NEET (Sep.) 2020]
(a ) emission of ozone depleting
substances
(b) release of green house gases
(c) disposal ofe-wastes
(d) transport of genetically modified
organisms from one country to
another
Ans.(a)
Montreal protocol was signed in 1987 for
control emission of ozone depleting
substances. It is a global agreement to
protect the stratospheric ozone layer by
phasing out the production and
consumption of ozone-depleting
substances (ODS).
89Which of the following statements
about ozone is correct?
[NEET (Odisha) 2019]
(a) Tropospheric ozone protects us from
UV- radiations
(b) Stratospheric ozone is ‘bad’
(c) Tropospheric ozone is ‘good’
(d) Stratospheric ozone protects us
from UV- radiations
Ans.(d)
Statement (d) is correct. Stratospheric
ozone protects us from UV radiations of
the sun. Correct information about
incorrect statements is as follows.
Good ozone is found in the upper part of
the atmosphere, i.e. stratosphere. Bad
ozone is formed in the lower atmosphere
(troposphere) that harms plants and
animals.
90World Ozone Day is celebrated on
[NEET 2018]
(a) 16th September (b) 21st April
(c) 5th June (d) 22nd April
Ans.(a)
‘World Ozone Day’ is celebrated on 16th
September to controlO
3
depletion.
Ozone layer is a fragile shield of gas that
protects earth from harmful UV-rays.
On 21st April the Civil Service Day and
National Yellow Bat Day is celebrated.
5th Juneof every year is celebrated as
World Environment Day. Earth Day is an
annual event, celebrated on22nd April
of every year.
91In stratosphere, which one of the
following elements acts as a
catalyst in degradation of ozone
and release of molecular oxygen?
[NEET 2018]
(a) Fe (b) Cl
(c) Carbon (d) Oxygen
Ans.(b)
In stratosphere, Cl acts as a catalyst in
the degradation of ozone and release of
molecular oxygen. It is released by
action of UV rays on chlorofluorocarbon.
Chlorine reacts with ozone in a series of
chain reaction, converting it into oxygen.
One active chlorine can destroy 5000
molecules of ozone in one month.
CFCl CFCl + Cl
3
UV -C
2
→
CFCl CFCl + Cl
2
UV -C
→
Cl + O ClO + O
3 2
→
ClO + O Cl + 2O
3 2
→
Iron (Fe), carbon (C) and oxygen (O) are
not Ozone Depleting Substances (ODS).
92Depletion of which gas in the
atmosphere can lead to an
increased incidence of skin
cancers
[NEET 2016, Phase I]
(a) ozone (b) ammonia
(c) methane (d) nitrous oxide
Ans.(a)
Ozone is found in the upper part of the
atmosphere called stratosphere and it
acts as a shield absorbing ultraviolet
radiation from sun. So its depletion can
lead to incidence of skin cancers caused
by harmful solar radiations which will
reach earth in the absence ofO
3
layer.
93The zone of atmosphere in which
the ozone layer is present is called
[CBSE AIPMT 2014]
(a) ionosphere (b) mesosphere
(c) stratosphere (d) troposphere
Ans.(c)
The ozone layer is mainly found in the
lower portion of stratosphere, i.e.
approximately 20-30 km above the
earth, though its thickness varies
seasonally and geographically, ozone
layer acts as a shield absorbing UV rays
from the sun.
94Which one of the following is a wrong
statement? [CBSE AIPMT 2012]
(a) Most of the forests have been lost in
tropical areas
(b) Ozone in upper part of atmosphere is
harmful to animals
(c) Greenhouse effect is a natural
phenomenon
(d) Eutrophication is a natural
phenomenon in freshwater bodies
Ans.(b)
Ozone (O
3
)is an isotope of oxygen which
exists in so called ozone layer at a height
of about 15-60 km in the middle and
upper stratosphere and lower
mesosphere. This ozone layer absorbs
UV-rays of longer wavelength and
protects life on Earth from damaging
effects of these radiations.
Ozone in the lower atmosphere
(troposphere) is regarded as a pollutant.
95Montreal protocol aims at
[CBSE AIPMT 2009]
(a) reduction of ozone depleting
substances
(b) biodiversity conservation
(c) control of water pollution
(d) control ofCO
2
emission
Ans.(a)
In August 1989, 44 countries and EEC
ratified the Montreal Protocol, which
provides a mechanism to review the
efficiency of control measures. They
also agreed to phase out the halogens
and to control and reduce other Ozone
Ozone and Its Depletion
TOPIC 5

386. 376 NEETChapterwise Topicwise Biology
Depleting Substances (ODSs). In a policy
statement called Helsinki Declaration,
the attending nations agreed to phase
out the production and consumption of
controlled CFCs as soon as possible but
not later than the year 2007.
96Global agreement in specific
control strategies to reduce the
release of ozone depleting
substances, was adopted by
[CBSE AIPMT 2009]
(a) Rio de Janerio Conference
(b) Montreal Protocol
(c) Kyoto Protocol
(d) Vienna Convention
Ans.(b)
The Montreal Protocol is a landmark
international agreement designed to
protect the stratospheric ozone layer. by
reducing the release of ozone depleting
substances. The treaty was originally
signed in 1987 and substantially amended
in 1990 and 1992. The Montreal protocol
stipulates that the production and
consumption of compounds that deplete
ozone in the stratosphere—
chlorofluorocarbons (CFCs), halogens,
carbon tetrachloride and methyl
chloroform—are to be phased out by
2000 (2005 for methyl chloroform).
97Montreal protocol, which calls for
appropriate action to protect the
ozone layer from human activities
was passed in the year
[CBSE AIPMT 2006]
(a) 1986
(b) 1987
(c) 1988
(d) 1985
Ans.(b)
In 1987, twenty seven industrialised
countries signed theMontreal protocol
for reduction in production and release
of CFCs chlorofluorocarbons depleting
ozone layer, into the atmosphere. It was
followed by increasingly stringent
amendments in London in 1990 and in
Copenhagen in 1992.
98Identify the correctly matched pair.
[CBSE AIPMT 2004]
(a) Montreal protocol — Global warming
(b) Kyoto protocol — Climate change
(c) Ramsar convention — Ground water
pollution
(d) Basal convention — Biodiversity
conservation
Ans.(b)
Kyoto protocol deals with climate
changes while Montreal protocol deals
with ozone depletion.
99In coming years, skin related
disorders will be more common
due to [CBSE AIPMT 1997]
(a) air pollution
(b) use of detergents
(c) water pollution
(d) depletion of ozone layer
Ans.(d)
Ozone layer is found in the stratosphere
of atmosphere. It absorbs harmful
ultraviolet rays coming from the sun.
In coming years, when the ozone layer
may become thinner, ultra-violet
radiations may reach earth directly to
cause cancer, especially skin cancer.
100Formation of ozone hole is
maximum over
[CBSE AIPMT 1997]
(a) India
(b) Antarctica
(c) Europe
(d) Africa
Ans.(b)
An ozone hole (thinning) has been
formed over Antarctica as a result of
damage to ozone.
Most ozone is formed in the
stratosphere over the equator and
spread by winds around the globe.
Icy particles in polar stratospheric
clouds catalyse the release of chlorine
(from CFC) which destroys ozone. The
thinning in ozone is maximum because in
winter there is exceptionally cold.
101Ultraviolet radiations from sunlight
cause a reaction which produces
[CBSE AIPMT 1993]
(a)O
3
(b)SO
2
(c) CO (d)CH
4
Ans.(a)
Ozone is produced by action of
ultraviolet and other high energy
radiations on oxygen resulting in
splitting and interaction of oxygen
molecules3O
2
→2O
3
.
This ozone forms a layer that is also
called protective umbrella at 23 km from
earth surface.
102Polyblend, a fine powder of
recycled modified plastic, has
proved to be a good material for
[NEET (National) 2019]
(a) use as a fertiliser
(b) construction of roads
(c) making tubes and pipes
(d) making plastic sacks
Ans.(b)
Polyblend has proved to be a good
material for the constructions of roads.
It is a fine powder of recycled plastic and
it is mixed with bitumen to lay roads. The
first polyblend road was laid in Bangaluru
by the effort of Ahmed Khan.
103Chipko movement was launched
for the protection of
[CBSE AIPMT 2009]
(a) grasslands (b) forests
(c) livestock (d) wet lands
Ans.(b)
In 1973, the Chipko Movement (Chipko
means to hug or stick to the trunk of the
tree) was launched byChandi Prashad
Bhatt and Sunder Lal Bahugunaagainst
large scale falling of trees by timber
contractors in the Uttarakhand hills. The
starting point wasChambolidistrict of
Garhwalregion in Uttarakhand.
Case Studies
Related to Pollution
TOPIC 6