S.No. Experiment
No
Topic
I. A – Fresh / Preserved Specimens
1. A1 Ecological adaptations of Hydrophyte
A2 Ecological adaptations of Xerophyte
A3 Ecological adaptations of Halophyte
A4 Ecological adaptations of Epiphyte
II.B Models/Photographs/Pictures
2. B1 E.Coli cloning vector (pBR322)
III. C - Solving the problems – Genetics and Ecology
3. C1 To verify Mendel’s Monohybrid cross
4. C2 Analysis of seed sample to study Mendelian Dihybrid Ratio
5. C3 Flow of energy and Ten percent law
6. C4 Determination of Population density and Percentage frequency by
Quadrat method
7. C5 Genetic - linkage map
IV. D - Experiments
8. D1 Dissect and display the Pollinia of Calotropis
9. D2 Study of Pollen germination on a slide
10. D3 Isolation of DNA from plant materials
I. Identify the given fresh / preserved specimen ‘A’ and give any two reasons
(Identification – 1, Any two reasons – 2) 3 marks
II. Identify the given model / photographs/ picture ‘B’ and give any two reasons
(Identification – 1, Any two reasons – 2) 3 marks
III. Analyse the given ecological / genetic problem ‘B’. Solve / Construct it by giving appropriate
reasons. (Identification – 1, Solve/ Construct– 2, Result-1) 4 mark
IV. Write the aim, procedure, observation and inference of the given experiment ‘D’.
(Aim – 1, Procedure – 2, Observation – 1, Inference – 1) 5 marks
Total: 15 mark
Record: 3 mark
Skil: 2 mark
Maximum marks: 20
I.A - Fresh / Preserved Specimens
Experiment No. A1 Ecological adaptations of Hydrophyte - Eichhornia (Water hyacinth)
Identification: The given specimen is identified as ecological adaptations of Hydrophyte - Eichhornia (Water
hyacinth)
Diagnostic Features:
1.Root system is poorly developed.
2.Root pockets are present.
3.Mechanical tissues are generally absent.
Experiment No. A2 Ecological adaptations of Xerophyte – Opuntia
Identification: The given specimen is identified as ecological adaptations of Xerophyte – Opuntia
Diagnostic Features:
1.The stem is flattened, green, thick and fleshly called phylloclade.
2.Mucilage is present which helps to retain the water.
3.Leaves are modified into spines.
Experiment No. A3 Ecological adaptation of Halophyte – Pneumatophores of Avicennia
Identification: The given specimen is identified as ecological adaptation of Halophyte – Pneumatophores of
Avicennia.
Diagnostic Features:
1.A special kind of negatively geotropic root called pneumatophores (respiratory roots) are present.
2.The leaves excrete salts through the salt glands.
Experiment No. A4 Ecological adaptation of Epiphyte – Vanda
Identification: The given specimen is identified as ecological adaptation of Epiphyte – Vanda.
Diagnostic Features:
1.It shows the number of joules transfer from one trophic level to next trophic level.
2.In this pyramid, the number of energy (joules) are gradually decreasing.
3.Hence, this energy pyramid is always upright.
II.B Models/Photographs/Pictures
Experiment No. B1 E.coli cloning vector (pBR 322)
Identification: The given picture is identified as E.coli cloning vector (pBR 322)
Diagnostic Features:
1.pBR 322 plasmid is a reconstructed plasmid.
2.It containing 4361 base pairs.
3.It is a widely used cloning vector.
III.C - Solving the Problems
Experiment No. C1 To verify Mendel’s Monohybrid cross
Identification: The given Genetical problems is identified as Mendel’s Monohybrid cross.
Phenotypic Ration: 3:1 (F1 is Yellow (64)
F2 is Yellow (48) and Green (16) so, 3:1 or 75% : 25%)
Genotypic Ration: 1:2:1 (F1 is Heterozygous Yellow (Yy-64)
F2 is Homozygous Yellow (YY-16),
Heterozygous Yellow (Yy-32),
Homozygous Green (yy-16) so, 1:2:1 or 25% : 50% : 25%)
Experiment No. C2 Analysis of seed sample to study Mendelian Dihybrid Cross
Identification: The given Genetical problems is identified as Mendel’s Dihybrid cross
Construction:
Result:
Phenotypic Ration: 9:3:3:1
Which is exactly the same as obtained by Mendel for a dihybrid cross. This indicates that the contrasting genes
for seed colour and seed shape show an independent assortment in the population of pea seeds.
Generation Total Number
of individuals
Genotypes Phenotypes
YY Yy yy
F1 64 - Heterozygous
Yellow
- Yellow
Total - 64 - 64
F2 64 Homozygous
Yellow
Heterozygous
Yellow
Homozygous
Green
Yellow and
Green
Total 16 32 16 64
Total Number of
seeds observed
No.of yellow
round seeds
No.of yellow
wrinkled seeds
No.of green
round seeds
No.of green
wrinkled seeds
160 90 30 30 10
Yellow round Yellow wrinkled Green round Green wrinkled
9 3 3 1
Experiment No. C3 Flow of energy and Ten percent law
Identification: The given Ecological problems is identified as Ten Percent Law of Energy
Construction:
i) T1 – Grass (Producer) = 30,000 J of Energy
ii) T2 – Rabbit (Primary Consumer) = 3000 J of Energy
iii) T3 – Snake ( Secondary Consumer) = 300 J of Energy
Result:
The third tropical level T3 – Snake (Secondary Consumer) receives 300 J of Energy.
Experiment No. C4
Determination of Population density and Percentage frequency by Quadrat method.
Identification: The given Ecological problems is identified as Determination of Population density and
Percentage frequency by Quadrat method.
Construction:
Total number of individuals in all quadrats studied
Population Density =
Total number of quadrats studied
Total number of quadrats in which species occurred
Percentage freguency = X 100
Total number of quadrats studied
Result:
S
.
N
o
Plant
species
No.of individuals per
quadrat
Total number
of individuals
in all the
quadrats
studied (N)
Total number of
quadrats in
which each
species
occurred (A)
Total
number of
quadrats
studied (B)
Populati
on
Density
(N/B)
Frequency
Percentage
(A/B) X
100 I II III IV
1 I 4 6 3 5 18 4 4 4.5 100%
2 II 2 3 1 0 06 3 4 1.5 75%
3 III 6 4 2 1 13 4 4 3.25 100%
4 IV 0 4 3 0 07 2 4 1.75 50%
5 V 0 0 4 0 04 1 4 1 25%
Experiment No. C5 Genetic - linkage map
Identification: The given problems is identified as Genetic linkage map
Construction:
Result:
1. The frequency of crossing over is directly proportional to the relative distance of the genes on the
chromosomes.
2. More crossing over = More distance between two genes and
Less crossing over = Less distance between the two genes.
3. In the above problem, the sequence of the genes on the linkage map is B, A, C.
IV.D – Experiments
Experiments No: D1 To dissect and display the pollinia of Calotropis
Aim Procedure Observation Inference
To dissect and
observe the
structure of
pollinia and
understand the
mechanism of
pollination.
i) Take a mature flower of
Calotropis and remove the calyx
and corolla.
ii) Pentangular stigmatic disc is
identified.
iii) At the angles of the
stigma pollinium is present.
The stamens of Calotropis
produce pollinium.
The structure of pollinium
of Calotropis is well
suited to achieve
pollination.
Experiment No: D2 Study of Pollen germination on a slide
Aim Procedure Observation Inference
To study the pollen
germination.
i) 1gm Sucrose + 100 ml
distilled water.
ii) add few pollen grains.
iii) after 5 minutes view the slide.
Different stages of
germination pollens are
observed.
Pollen tube containing
tube nucleus and two male
gametes.
Experiment No: D3 Isolation of DNA from plant materials
Aim Procedure Observation Inference
To isolate DNA
from available
plant materials.
i) Grinded plant material + water
+ Nacl.
ii) filter the solution.
iii) add liquid soap solution and
chilled ethanol.
DNA appears as white
precipitate of very fine
thread like structure.
DNA can be isolated from
the plant cell nucleus.