12th std Bio-Botany EM Practical notes.pdf

12th std Bio-Botany EM Practical notes

1. 12th standard Bio-Botany Practical notes


Name:_______________
Reg.No:______________
Subject:______________
Class:_____Section_____
Date:______Batch______
Session:______Time_____











IDENTIFICATION
Q.No. Ex.No. Content
I Preparation and Demonstration of Slides
A1 T.S. of Mature Anther
II Fresh or preserved
specimens/Models/Photographs/Charts
B7 Pyramid of Biomass
III Solving the Problems
C5 Chromosomal Aberration - Deletion
IV Experiments
D2 Study of pH of different types of soil
V Economic Importance of Plants
E4 Henna Powder












































Q. No. Ex.No. Content
I A Preparation and Demonstration of Slides
1. A1 T.S of Anther Diagnostic Features
2. A2 L.S of an Angiospermic ovule
3. A3 T.S. of Nerium Leaf
II B Fresh or preserved specimens/Models/Photographs/Charts
4. B1 Wind Pollinated Flowers - Anemophily
B2 Insect Pollinated Flowers - Entemophily
5. B3 Dicot seed
6. B4 E.coli cloning vector (pBR 322)
7. B5 Plant Tissue Culture – Callus with plantlets
8. B6 Pyramid of Number
B7 Pyramid of Biomass
B8 Pyramid of Energy
III C Solving the Problems
9. C1 To verify the Mendel’s Monohybrid Cross
10. C2 Analysis of seed sample to study Mendelian Dihybrid Cross
11. C3 Flow of energy and Ten percent law
12. C4 Determination of Population density and Percentage frequency
by Quadrat method
13. C5 Chromosomal Aberration - Deletion
C6 Chromosomal Aberration - Duplication
C7 Chromosomal Aberration - Inversion
14. C8 Genetic - linkage map
IV D Experiments
15. D1 Study of Pollen germination on a slide
16. D2 Study of pH of different types of soil
17. D3 Isolation of DNA from plant materials
V E Economic Importance of Plants
18. E1 Sesame or Gingelly oil
E2 Rubber
E3 Flaked Rice or Aval
E4 Henna Powder
E5 Aloe Gel
2022-23G.P.S 1
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2. Bio-Botany Practical Question Paper


Time: 1:30 hr Marks: 10

I. Identify the given slide ‘A’ and give any two reasons. Draw a neat labeled diagram. 2 mark
(Identification – ½, Any two reasons – ½, Diagram – ½, Labelling – ½)

II. Identify the given specimen / model / photograph/ chart ‘B’ and give any two reasons
(Identification – ½, Any two reasons – ½) 1mark

III. Analyse the given ecological / genetic problem ‘C’. Solve it by giving appropriate reasons.
(Identification – ½, Solve/ Construct– ½, Reason/Observation, Inference/Answer – ½) 1½ mark

IV. Write the aim, procedure, observation and inference of the given experiment ‘D’.
(Aim – ½, Procedure – ½, Observation, Inference – ½) 1½ mark

V. Identify the economically important plant product ‘E’. Mention its Botanical name,
useful part and their uses.
(Identification and Botanical name – ½, Useful part – ½, uses – ½) 1½ mark


Total: 7½ mark
Record: 1½ mark
Skil: 1 mark
Maximum marks: 10























I - Preparation and Demonstration of Slides
Exercise 1 (A1):
T.S of Anther Diagnostic Features
Identification: The given slide A1 is identified as T.S of Anther
Diagnostic Features:
1. A mature anther is bilobed (dithecous) and the two lobes are joined by a connective.
2. Each anther lobe has two pollen chambers in which pollen grains are produced.








Exercise 2 (A2):
L.S of an Angiospermic ovule
Identification: The given slide A2 is identified as L.S of an Angiospermic ovule
Diagnostic Features:
1.Ovule or megasporangium is protected by one or two coverings called integuments.
2.The stalk of the ovule is called funicle.








Exercise 3 (A3):
T.S. of Nerium Leaf
Identification: The given slide A3 is identified as T.S. of Nerium Leaf
Diagnostic Features:
1.Presence of multilayered epidermis with thick cuticle.
2.Sunken stomata are present only in the lower epidermis.








G.P.S 2
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3. II - Fresh or preserved specimens / Models / Photographs / Charts

Exercise 4 (B1):
Wind Pollinated Flowers - Anemophily
Identification: The given B1 specimens/Models/Photographs /Charts is identified as
Wind Pollinated Flowers - Anemophily
Diagnostic Features:
1.The flowers are small, inconspicuous, colourless, odourless and nectarless.
2. Anthers and stigmas are commonly exerted.


Exercise 4 (B2):
Insect Pollinated Flowers – Entemophily
Identification: The given B2 specimens/Models/Photographs /Charts is identified as
Insect Pollinated Flowers - Entemophily
Diagnostic Features
1.The flowers are showy, brightly coloured and scented.
2.The flowers produce nectar or ediblepollen.


Exercise 5 (B3):
Dicot seed
Identification: The given B3 specimens/Models/Photographs /Charts is identified as Dicot seed
Diagnostic Features:
1. Seeds of gram have two cotyledons and an embryonal axis.
2. Each seed is covered by two seed coats (a) Testa – outer coat (b) Tegmen – inner coat.


Exercise 6 (B4):
E.coli cloning vector (pBR 322)
Identification: The given B4 specimens/Models/Photographs /Charts is identified as
E.coli cloning vector (pBR 322)
Diagnostic Features:
1.pBR 322 plasmid is a reconstructed plasmid containing 4361 base pairs and most widely used
as cloning vector.
2.In pBR, p denotes plasmid and B and R respectively the notes of scientists Boliver and
Rodriguez who developed the plasmid.


Exercise 7 (B5):
Plant tissue culture – Callus with plantlets
Identification: The given B5 specimens/Models/Photographs /Charts is identified as
Plant tissue culture – Callus with plantlets
Diagnostic Features:
1.The callus is an unorganized mass of undiff erentiated tissue.
2.Roots and shoots are diff erentiated from the callus.









Exercise 8 (B6):
Pyramid of numbers
Identification: The given B6 specimens/Models/Photographs /Charts is identified as
Pyramid of numbers
Diagnostic Features:
1. The number of organism that are present in successive trophic levels of an ecosystem is shown
in the pyramid of numbers of a grassland ecosystem.
2.Therefore, pyramid of number in grassland ecosystem is always upright.


Exercise 8 (B7):
Pyramid of biomass
Identification: The given B7 specimens/Models/Photographs /Charts is identified as
Pyramid of biomass
Diagnostic Features:
1.Pyramid of biomass represents the total biomass or standing crop (dry weight) of organisms in
each trophic level at a particular time.
2.Therefore, here the pyramid of biomass is always inverted in shape.


Exercise 8 (B8):
Pyramid of energy
Identification: The given B8 specimens/Models/Photographs /Charts is identified as
Pyramid of energy
Diagnostic Features:
1.Pyramid of energy represents the number of joules transferred from one trophic level to next.
2. Therefore pyramid of energy is always upright.
















G.P.S 3
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4. III - Solving the Problems
Exercise 9 (C1):
To verify Mendel’s Monohybrid cross
Identification: The given C1 Genetical problems is identified as Mendel’s Monohybrid cross

Construction:

















Answer:







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%)









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


Exercise 10 (C2):
Analysis of seed sample to study Mendelian Dihybrid Cross
Identification: The given C2 Genetical problems is identified as Mendel’s Dihybrid cross

Construction:

















Answer:




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.

Exercise 11 (C3):
Flow of energy and Ten percent law
Identification: The given C3 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
Answer:
The third tropical level T3 – Snake (Secondary Consumer) receives 300 J of Energy.

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 G.P.S 4
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5. Exercise 12 (C4):
Determination of Population density and Percentage frequency by Quadrat method.
Identification: The given C4 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 individuals in all the quadrats studied
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%

Exercise 13 (C5):
Chromosomal Aberration – Deletion
Identification: The given C5 problems is identified as Chromosomal Aberration – Deletion
Reasons:
1. The deletion of the chromosomal segement A and D.
2. When there is a loss of a segment of the genetic material in a chromosome it is called deletion.
Significance:
Most of the deletions lead to death of an organism.

Exercise 13 (C6):
Chromosomal Aberration – Duplication
Identification: The given C6 problems is identified as Chromosomal Aberration – Duplication
Reasons:
1. When a segment of a chromosome is present more than once in a chromosome, then it is called duplication
2. The order of the genes in a chromosome is A,B,C,D,E,F,G,H and I. Due to aberration, the genes B and C
are duplicated and the sequence of genes becomes A,B,C,B,C,D,E,F,G,H and I.
Significance:
Some duplications are useful in the evolution of the organism.








Exercise 13 (C7):
Chromosomal Aberration – Inversion
Identification: The given C7 problems is identified as Chromosomal Aberration – Inversion
Reasons:
1. When the order of genes in a chromosomal segment is reversed due to rotation by an angle of 180°,
it is called inversion.
2. The order of genes in a chromosome is A,B,C,D,E,F,G,H and I. Due to aberration,
the sequence of genes become A,D,C,B,E, F,G,H and I.
Significance:
Sometimes inversion is responsible for evolution of the organism.

Exercise 14 (C8):
Genetic - linkage map
Identification: The given C8 problems is identified as Genetic linkage map

Construction:





Reasons:
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







G.P.S 5
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6. IV - Experiments
Exercise 15 (D1):
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.


Exercise 16 (D2):
Study of pH of different types of soil


Aim

Procedure
Observation
Inference soil sample pH value
To study pH of
different types of
soil.
i) 1gm soil + 100 ml
distilled water.
ii) filter the solution.
iii) dip a small piece of pH paper.
1.River soil 7.0 pH value is required for
plant growth 2.Pond soil 5.5
3.Lake soil 7.5



Exercise 17 (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.







V - Economic Importance of Plants

Exercise 18:
Economically important plant products:

S.No Identification Botanical Name Useful Parts Uses
18.
(E1)
Gingelly oil Sesamum indicum Seeds 1.Used for culinary purpose.
2.Used to make soap, paints, lubricant.

S.No Identification Botanical Name Useful Parts Uses
18.
(E2)
Rubber Hevea brasiliensis Latex 1.Used to make erasers, footwear, rubber bands.
2.Used to make gloves and balloons.

S.No Identification Botanical Name Useful Parts Uses
18.
(E3)
Flaked Rice
(Aval)
Oryza sativa Seeds 1.Used as a breakfast.
2.Used as a snacks.

S.No Identification Botanical Name Useful Parts Uses
18.
(E4)
Henna
powder
Lawsonia inermis Leaves 1.Used as a hair dye.
2.Used to colouring skin and nails.

S.No Identification Botanical Name Useful Parts Uses
18.
(E5)
Aloe gel Aloe vera Leaves 1.Used as a skin tonic.
2.Used to make shampoos and shaving creams.











G.P.S 6
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7. G.P.S 7
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8. G.P.S 8
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9. G.P.S 9
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10. G.P.S 10
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11. G.P.S 11
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12. G.P.S 12
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13. G.P.S 13
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14. G.P.S 14
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15. G.P.S 15
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16. Sesame / Gingelly oil







Rubber G.P.S 16
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17. Flaked Rice







Henna Powder

















Aloe Gel G.P.S 17
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