CBSE Class 12 Biology Question Paper 2020 Set 57-3 Solutions

Strictly Confidential: (For Internal and Restricted use only)
Senior School Certificate Examination-2020
Marking Scheme – BIOLOGY (SUBJECT CODE - 044)
(PAPER CODE – 57/3/1,2,3 )
General Instructions: -

1. You are aware that evaluation is the most important process in the actual and correct
assessment of the candidates. A small mistake in evaluation may lead to serious
problems which may affect the future of the candidates, education system and teaching
profession. To avoid mistakes, it is requested that before starting evaluation, you must
read and understand the spot evaluation guidelines carefully.Evaluation is a 10-12
days mission for all of us. Hence, it is necessary that you put in your best efforts
in this process.
2. Evaluation is to be done as per instructions provided in the Marking Scheme. It should
not be done according to one’s own interpretation or any other consideration. Marking
Scheme should be strictly adhered to and religiously followed. However, while
evaluating, answers which are based on latest information or knowledge and/or
are innovative, they may be assessed for their correctness otherwise and marks
be awarded to them.
3. The Head-Examiner must go through the first five answer books evaluated by each
evaluator on the first day, to ensure that evaluation has been carried out as per the
instructions given in the Marking Scheme. The remaining answer books meant for
evaluation shall be given only after ensuring that there is no significant variation in the
marking of individual evaluators.
4. Evaluators will mark( √ ) wherever answer is correct. For wrong answer ‘X”be marked.
Evaluators will not put right kind of mark while evaluating which gives an impression that
answer is correct and no marks are awarded. This is most common mistake which
evaluators are committing.

5. If a question has parts, please award marks on the right-hand side for each part. Marks
awarded for different parts of the question should then be totaled up and written in the
left-hand margin and encircled. This may be followed strictly.
6. If a question does not have any parts, marks must be awarded in the left-hand margin
and encircled. This may also be followed strictly.
7. If a student has attempted an extra question, answer of the question deserving more
marks should be retained and the other answer scored out.
8. No marks to be deducted for the cumulative effect of an error. It should be penalized
only once.
9. A full scale of marks 0-70 has to be used. Please do not hesitate to award full marks if
the answer deserves it.
10. Every examiner has to necessarily do evaluation work for full working hours i.e. 8 hours
every day and evaluate 20 answer books per day in main subjects and 25 answer books
per day in other subjects (Details are given in Spot Guidelines).
11. Ensure that you do not make the following common types of errors committed by the
Examiner in the past:-
 Leaving answer or part thereof unassessed in an answer book.
 Giving more marks for an answer than assigned to it.
 Wrong totaling of marks awarded on a reply.

 Wrong transfer of marks from the inside pages of the answer book to the title page.
 Wrong question wise totaling on the title page.
 Wrong totaling of marks of the two columns on the title page.
 Wrong grand total.
 Marks in words and figures not tallying.
 Wrong transfer of marks from the answer book to online award list.
 Answers marked as correct, but marks not awarded. (Ensure that the right tick mark
is correctly and clearly indicated. It should merely be a line. Same is with the X for
incorrect answer.)
 Half or a part of answer marked correct and the rest as wrong, but no marks
awarded.

12. While evaluating the answer books if the answer is found to be totally incorrect, it should
be marked as cross (X) and awarded zero (0)Marks.

13. Any unassessed portion, non-carrying over of marks to the title page, or totaling error
detected by the candidate shall damage the prestige of all the personnel engaged in the
evaluation work as also of the Board. Hence, in order to uphold the prestige of all
concerned, it is again reiterated that the instructions be followed meticulously and
judiciously.

14. The Examiners should acquaint themselves with the guidelines given in the Guidelines
for spot Evaluation before starting the actual evaluation.

15. Every Examiner shall also ensure that all the answers are evaluated, marks carried over
to the title page, correctly totaled and written in figures and words.

16. The Board permits candidates to obtain photocopy of the Answer Book on request in an
RTI application and also separately as a part of the re-evaluation process on payment of
the processing charges

Question Paper Code 57/3/1
SECTION A
Note : Choose the correct option from the choices given in each of the following questions :
1 Introduction of an alien DNA into a plant host cell is achieved by making them
(A) Competent with bivalent ions
(B) Using microinjections
(C) Using gene gun
(D) Using lysozymes and chitinase
Ans. C / Using gene gun =1
[1 Mark]
2. One of the ex situ conservation methods for endangered species is
(A) Biosphere reserves
(B) National parks
(C) Cryopreservation
(D) Wildlife sanctuaries
Ans. C / Cryopreservation = 1
[1 Mark]
OR
Ozone gas is continuously formed in the stratosphere by
(A) Action of UV rays on nascent oxygen
(B) Reaction of oxygen with water vapour
(C) Action of UV rays on molecular oxygen
(D) Action of UV rays on water vapour
Ans. C / Action of UV rays on molecular oxygen = 1
[1 Mark]
3. Intense lactation in mothers acts as a natural contraceptive due to the
(A) Suppression of gonadotropins
(B) Hypersecretion of gonadotropins
(C) Suppression of gametic transport
(D) Suppression of fertilization
Ans. A / Suppression of gonadotropins = 1
[1 Mark]
4. The principle of vaccination is based on the property of

3

(A) Specificity
(B) Diversity
(C) Memory
(D) Discrimination between ‘self’ and ‘non-self’
Ans. C / Memory =1 [1 Mark]

OR
Opioids act as

(A) Depressants
(B) Pain killers
(C) Euphoria providers
(D) Stimulants
Ans. A / Depressants =1 [1 Mark]
5. Nematode specific genes were introduced into the tobacco host plant by using the vector
(A) Plasmid
(B) Bacteriophage
(C) pBR 322
(D) Agrobacterium
Ans. D / Agrobacterium = 1 [1 Mark]
SECTION B
6. Spirulina is a rich source of proteins. Mention the two ways by which large scale culturing
of these microbes is possible.

Ans. Grown in waste water from potato processing plants (starch rich) /straw / molasses / animal manure/
sewage (any two) // bioreactor, using any two above mentioned materials = 1+1
[2 Marks]

7. How does EcoRI specifically act on DNA molecule ? Explain.
Ans. Inspects the length of a DNA sequence, finds its specific recognition sequence as
5' GAATTC 3'
, binds to the DNA, and cut each of the two strands of the double helix at
3' CTTAAG 5'
specific points in their sugar-phosphate backbones = ½ × 4

//

4

=½ =½
=½

=½

[2Marks]

8. (a) Explain the cause responsible in a human to have sex chromosomes as ‘XXY’
instead of ‘XX’ or ‘XY’.
(b) List any two ways such individuals are different from the normal being.
Ans. a) Failure of segregation of chromatids during cell division cycle/ nondisjunction/
aneuploidy, resulting in gain of an extra X chromosome in a male after fertization
=½×2
b) Development of breasts/ gynaecomastia , sterile individuals = ½ × 2
[2Marks]

9. Name and explain the technique that can be used in developing improved crop
varieties in plants bearing female flowers only.
Ans. Artificial hybridization = ½
The female flower buds are bagged before the flower open, when stigma becomes receptive
pollination is carried out using the desired pollen , and flower is rebagged (and fruits are allowed to
develop) =½×3

[2Marks]
OR
When are the non-flowering plants said to be homothallic and monoecious; and
heterothallic and dioecious ? Give an example of each.
Ans. Homothallic and monoecious : bisexual condition / having both male and female reproductive structures
on the same plant, eg. Chara and several fungi or any other suitable example = ½ × 2
Heterothallic and dioecious: unisexual condition / having either male or female reproductive structures
present in different plants, example Marchantia or any other suitable example =½×2
[2Marks]

5

10. Mention the kind of interaction mycorrhizae exhibit. How is Glomus in mycorrhizal
association beneficial to the plants ?
Ans. Symbiotic relationship / mutualism = ½

absorbs phosphorus from the soil, provides resistance to the root borne pathogens, tolerance to
salinity and drought, overall increase in the plant growth ( any three)= ½× 3
[2Marks]

11. Given below is the segment of a DNA strand. Write its complementary strand and the
RNA strand that can be transcribed from the DNAmolecule formed.
5' TAC CGT GAC GTC 3'
Ans. Complementary strand 3' ATG GCA CTG CAG 5' ( correct polarity) = ½ (correct
sequence) = ½
RNA strand 5' UAC CGU GAC GUC 3' ( correct polarity) = ½ (correct sequence) = ½
[1+ 1 = 2Marks]
12. Name the type of Ecological Pyramid that can exist as upright as well as inverted. Explain
how does it happen.
Ans. Pyramid of number = ½

Pyramid of number

A. Upright Pyramid B. Inverted Pyramid

=½×2

Grassland Ecosystem Tree Ecosystem (both example) = ½

//
Pyramid of Biomass = ½
A. Upright Pyramid B. Inverted Pyramid

6

=½×2

Tree Ecosystem Sea Ecosystem (both example) = ½

//
In the upright pyramid of number the producers / grass are more than the number of primary
consumer / secondary consumer = ½
Eg. grassland ecosystem = ½
In inverted pyramid of number the producer is less in number than the number of primary consumer
=½
Eg. Tree and insects feeding on it = ½
//
In the upright pyramid of biomass the producers / tree are more in biomass than the primary consumer
/ secondary consumer = ½
Eg. forest ecosystem = ½
In inverted pyramid of biomass the biomass of the producer/phytoplanktons are less than the
biomass of the zooplanktons / small standing crop of phytoplanktons supports
large standing crop of zooplanktons = ½
Eg. Pyramid of biomass in sea = ½
[2Marks]
SECTION C
13. ‘‘Cotton bollworms enjoy feeding on cotton plants, but get killed when feed on Bt cotton
plant.’’ Justify the statement.
Ans. Once a bollworm feeds on Bt cotton plant the inactive protoxin produced by Bacillus
thuringiensis, is converted into an active form of toxin, due to the alkaline pH of the gut which
solubilise the crystals, the activated toxin binds to the surface of its midgut epithelial cells, create
pores that cause cell swelling and lysis, and eventually cause death of the insect =½×6
[3 Marks]
OR
(a) Mention the cause of ADA deficiency in humans.

7

(b) How is gene therapy carried out to treat the patients suffering from this disease ?
(c) State the possibility of a permanent cure of this disease.
Ans. a) Caused due to the deletion of the gene for adenosine deaminase = ½
b) lymphocytes from the blood of the patient are grown in a culture medium outside the body,
a functional ADA cDNA (using a retroviral vector) is then introduced into theselymphocytes
which are subsequently returned to the patient, patient requires periodic infusion of such ge-
netically engineered lymphocytes = ½ × 3
c) if the gene isolated from bone marrow cells producing ADA, is introduced into cells at early
embryonic stages it could be a permanent cure = ½ + ½
[3 Marks]
14. (a) Differentiate between Intrauterine insemination and Intrauterine transfer.
(b) Mention one positive and one negative application of amniocentesis.
Ans. a) IUI - the semen collected either from the husband or a healthy donor is artificially
introduced either into the vagina or into the uterus of the female = 1
IUT - embryos with more than 8 blastomeres transferred into the uterus = 1
b) Positive application: used to test for the presence of certain genetic disorders /
Down’s syndrome / Klinefelter’s syndrome / Haemophilia / Sickle-cell anemia = ½

Negative application : used to detect sex of the foetus and leads to female foeticides = ½
[3 Marks]
15. Explain the solutions found by Ahmed Khan, a Bengaluru based plastic sack manu facturer,
after realising the problems created by plastic wastes.
Ans. Polyblend a fine powder was made from recycled modified plastic , this was mixed with the bitumen
to lay roads which proved to enhance the bitumen’s water repellant properties , and helped to
increase road life by the factor of three = 1 × 3
[3 Marks]

16. Mention the chemical nature of an antibody and name the type of cells they are produced
by. Write the difference between active and passive immune responses on the basis of
antibodies.
Ans. Made up of proteins / peptide , B-lymphocytes / B-cells = ½ + ½
Active immunity - due to exposure to antigens / pathogens / vaccination / immunisation leads to
production of antibodies by the individual, slow process = ½ + ½
Passive immunity - Ready-made antibodies are directly given to protect the body of an individual
against foreign agents, fast process/ provide immediate immunity = ½ + ½
[3 Marks]
OR
Name the cells that act as HIV factory in humans when infected by HIV. Explain the events
that occur in these infected cells.
Ans. Macrophages = ½

8

Once the virus enters the human body the virus / viral genome infects macrophages where its RNA
genome replicates , to form viral DNA , with the help of the enzyme reverse transcriptase, this viral
DNA gets incorporated into host cell’s DNA , and directs the infected cells to produce virus
particles = ½ × 5
[3 Marks]
17. (a) Why is the collection of white winged moths and dark winged moths made in England
between 1850 – 1920 considered a good example of natural selection ?
(b) ‘‘Evolution is based on chance events in nature and chance mutations in organ isms.’’
Justify the statement.
Ans. a) During Pre-industrialisation white-winged moths survived due to white coloured lichens on
trees, During post-industrialisation white-winged moths did not survive due to predation /
predators could spot the moth against contrasting back ground, then the dark-winged or
melanised moths survived, this showed that in a mixed population of white and dark winged
moths those who can adapt better will survive =½ × 4
b) excess use of herbicides /pesticides and antibiotics has resulted in selection of resistant
varieties that developed due to chance mutation (in much lesser time scale) = 1
[2 +1 = 3 Marks]
18. (a) Compare the mechanism of sex determination in humans with that of honey bees,
with respect to chromosome number.
(b) How is the gamete formation comparable in the above two cases ?
Ans. a) In honeybee union of a sperm and an egg develops as a female (queen or worker), and an
unfertilised egg develops as a male (drone) by means of parthenogenesis, the females are dip-
loid / having 32 chromosomes and males are haploid / having 16 chromosomes , in humans-
both male and female individuals have 23 pair of chromosomes / diploid / female 44+ XX and
males 44+ XY =½ × 4
b) In humans gametes are formed by meiosis, in honeybee female gametes are formed by meiosis
and male gametes by mitosis =½ × 2
[2 +1 = 3 Marks]
19. Differentiate between the pattern of inheritance in humans of the blood diseases, haemophilia
and thalassemia.
Ans. Haemophilia Thalassemia
1. Sex linked recessive disorder 1. Autosomal recessive disorder
2. heterozygous female carrier 2 Transmitted from both the carrier
may transmit it both to her son and parents
daughter
3. males are generally affected 3. both sexes can be affected
but female are rarely affected

(both corresponding points to be written for credit) =1×3
[3 Marks]

9

20. Identify i, ii, iii, iv, v and vi in the following table :
No. Organism Bioactive molecules Use
1 Monascus purpureus i ii
2 iii iv Antibiotic
3 v Cyclosporin A v

Ans. i) statin
ii) cholesterol lowering agent
iii) Penicillium notatum
iv) Penicillin
v) Trichoderma polysporum
iv) Immunosuppressive agent = ½ × 6
[3 Marks]

21. (a) Write the scientific name of methanogen bacteria. Where are these bacteria gener-
ally found ? Explain their role in biogas production.
(b) Name the components of biogas.
Ans. a) Methanobacterium =½
found in the anaerobic sludge / rumen of cattle , they grow anaerobically on cellulosic material and
cause its breakdown (into methane CO2 and H2) = ½ × 2
b) methane , CO2 , H2 = ½ × 3
[1½+1 ½= 3 Marks]
SECTION -D
22. Given below is the diagram representing the observations made for separating DNA
fragments by Gel electrophoresis technique. Observe the illustration and answer the
questions that follow :

A B

10

(a) Why are the DNA fragments seen to be moving in the direction A  B ?
(b) Write the medium used on which DNA fragments separate.
(c) Mention how the separated DNA fragments can be visualised for further technical
use.
Ans. a) Because the DNA fragments are negatively charged =1
b) Agarose gel =1
c) After staining DNA with ethidium bromide, followed by exposure to UV rays =½ × 2
[3 Marks]
23. Study the transverse section of human ovary given below and answer the questions that
follow :

(a) Name the hormone that helps in the growth of A  B  C.
(b) Name the hormone secreted by A and B.
(c) State the role of the hormone produced by D.
Ans. a) Gonadotropins // FSH and LH = 1
b) Estrogen =1
c) Maintenance of uterine endometrium =1
[3 Marks]
24. Indiscriminate use of chemicals, pesticides and weedicides by humans are polluting our
water bodies, which in turn are harming the living organisms. Study the flow chart and
answer the questions based on it.

11

(a) Why does the concentration of DDT seem to be considerably highin the top con-
sumer ?
(b) How would the organisms at the highest level be affected ?
(c) Name the phenomenon observed.
Ans. a) DDT cannot be metabolised or excreted so gets accumulated by an organism , increase
in concentration at successive trophic levels =½ × 2
b) Disturb calcium metabolism in fish eating birds / causes thinning of egg shell ,
premature breaking of eggs leading to population decline =½ × 2
c) Biomagnification =1 [3 Marks]

SECTION - E
25. (a) According to ecologists, tropical regions in the world account for greater biological
diversity. Justify.
(b) Why are habitat loss and alien species invasion considered as the causes of biodiversity
loss ? Explain with the help of an example of each.

Ans. a) (i) have remained relatively undisturbed for millions of years / had a long evolutionary
time for species diversification =1
(ii) environment less seasonal / more constant and predictable / such constant environment
promotes niche specialization =1
(iii) more solar energy available in the tropics contributes to higher productivity and greater
diversity =1
b) Habitat loss
Amazon rain forest is being cut for cultivating soyabeans / degradation of habitat by pollution /
human activities leading to clearing of forests for commercial or tourism purpose =1 (any other
relevant example)
Alien species invasion -
The Nile perch introduced into Lake Victoria in East Africa led eventually to the extinction of an
ecologically unique assemblage of more than 200 species of cichlid fish in the lake / Recent illegal
introduction of the African catfish Clarias gariepinus for aquaculture purposes is posing a threat to
the indigenous catfishes in our rivers / carrot grass / lantana / water hyacinth causes threat to our
indigenous species =1
(any other relevant example)
[5 Marks]
OR
(a) What is an ecological succession ?
(b) Differentiate between primary and secondary succession. Why is secondary succession
faster than primary succession ? Explain with suitable examples.

12

(c) What are pioneer species ? Give examples of pioneer species in Xerarch and Hydrarch
successions respectively.
Ans. a) The gradual and fairly predictable change in the species composition of a given area is called
ecological succession =1
b) Primary succession Secondary succession
Starts in an area where Areas that somehow
no living organisms lost all the living organisms
ever existed / that existed there /
bare rock / newly created pond / abandoned farmlands /
reservoir / bare area flooded field area / burnt forest . =1
Secondary succession is faster since some soil or sediment is already present =1
eg. abondoned lands/ burnt or cut forests/ lands that have been flooded =½
c) The species that invade a bare area are called pioneer species = ½
Xerarch- lichens , Hydrarch- phytoplanktons =½ × 2
[5 Marks]
26. (a) Name the type of DNA that forms the basis of DNA fingerprinting and mention two
features of this DNA.
(b) Write the steps carried out in the process of DNA fingerprinting technique, and men-
tion its application.
Ans. a) Satellite DNA / repetitive DNA =½
These sequences normally do not code for any proteins, these sequence show high degree of poly-
morphism = ½ × 2
b) (i) isolation of DNA,
(ii) digestion of DNA by restriction endonucleases,
(iii) separation of DNA fragments by electrophoresis,
(iv) transferring (blotting) of separated DNA fragments to synthetic membranes such as
nitrocellulose or nylon,
(v) hybridisation using labelled VNTR probe,
(vi) detection of hybridised DNA fragments by autoradiography =½ x 6
Application - Forensic science / determining population and genetic diversities / paternity test = ½
[5 Marks]
OR
Explain the role of different genes in a lac operon, when in a ‘Switched On’ state.
Ans. The regulator gene produces repressor, inactivated in the presence of inducer, that is lactose, RNA
polymerase then gets access to the promoter gene and transcription proceeds =½ × 4
z gene codes for beta-galactosidase (ß-gal), responsible for the hydrolysis of the disaccharide
lactose into galactose and glucose =½ × 2

13

y gene codes for permease , which increases permeability of the cell to ß-galactosides /
lactose =½ × 2
a gene encodes enzyme transacetylase =1
//

[5 Marks]
27. (a) Draw a diagram of a fully developed embryo sac of an angiosperm. Label its chalazal
end and any other five parts within the embryo sac.
(b) Why does the development of an endosperm precede that of the embryo in
angiosperm ?
(c) Number of chromosomes in an onion plant cell is 16. Name the cells of the embryo
sac having 16 and 24 chromosomes formed after fertilisation.

Ans. a)

chalazal end

14

Chalazal end = ½, and any other five parts =½ × 5
b) the cells get filled with reserve food materials, used for providing the nutrition of
developing embryo / for assured nutrition of developing embryo =½ × 2
c) cells with 16 chromosome- zygote,
cells with 24 chromosome- endosperm = ½ × 2 [5 Marks]
OR
Describe the events that occur after fertilisation of an ovum till implantation in a hu-
man female.
Ans. The mitotic division called cleavage , starts as the zygote moves towards the uterus and forms 2-4 -
8 -16 daughter cells called blastomeres , and forms the first embryonic stage morula, the morula
continues to divide and transforms into blastocyst as it moves further into the uterus, the blas-
tomeres in the blastocyst are arranged into an outer layer called trophoblast, and an inner group of
cells attached to trophoblast called the inner cell mass, the trophoblast layer then gets attached to
the endometrium, and the inner cell mass gets differentiated as the embryo, after attachment the
uterine cells divide rapidly and covers the blastocyst, that gets embedded in the endometrium of the
uterus (this is called implantation ) = ½ × 10
[5 Marks]

15

Question Paper Code. 57/3/2
Note : Choose the correct option from the choices given in each of the following questions :
SECTION A
1. Nematode specific genes were introduced into the tobacco host plant by using the vector
(A)
Plasmid
(B) Bacteriophage
(C) pBR 322
(D) Agrobacterium
Ans. D / Agrobacterium = 1 [1 Mark]
2. The principle of vaccination is based on the property of
(A) Specificity
(B) Diversity
(C) Memory
(D) Discrimination between ‘self’ and ‘non-self’
Ans. C / Memory =1 [1 Mark]

OR
Opioids act as
(A) Depressants
(B) Pain killers
(C) Euphoria providers
(D) Stimulant
Ans. A / Depressants =1 [1 Mark]
3. After spermiogenesis, the sperm heads get embedded in which of the following cells ?
(A) Leydig cells
(B) Sertoli cells
(C) Germinal epithelium
(D) Seminal vesicle
Ans. (B) / Sertoli cells [1Mark]

4. Introduction of an alien DNA into a plant host cell is achieved by making them

(A) Competent with bivalent ions

16

(B) Using microinjections

(C) Using gene gun

(D) Using lysozymes and chitinase

Ans. C / Using gene gun =1
[1 Mark]

5. One of the ex situ conservation methods for endangered species is

(A) Biosphere reserves

(B) National parks

(C) Cryopreservation

(D) Wildlife sanctuaries

Ans. C / Cryopreservation = 1

[1 Mark]

OR

Ozone gas is continuously formed in the stratosphere by

(A) Action of UV rays on nascent oxygen

(B) Reaction of oxygen with water vapour

(C) Action of UV rays on molecular oxygen

(D) Action of UV rays on water vapour

Ans. C / Action of UV rays on molecular oxygen = 1

[1 Mark]

SECTION B

6. Name and explain the technique that can be used in developing improved crop varieties in
plants bearing female flowers only.

Ans. Artificial hybridization = ½
The female flower buds are bagged before the flower open, when stigma becomes receptive
pollination is carried out using the desired pollen , and flower is rebagged (and fruits are allowed to
develop) =½×3
[2 Marks]

17

OR
When are the non-flowering plants said to be homothallic and monoecious; and
heterothallic and dioecious ? Give an example of each.
Ans. Homothallic and monoecious : bisexual condition / having both male and female reproductive
structures on the same plant, eg. Chara and several fungi or any other suitable example = ½ × 2
Heterothallic and dioecious: unisexual condition / having either male or female reproductive struc-
tures present in different plants, example Marchantia or any other suitable example
=½×2

[2Marks]
7. (a) Explain the cause responsible in a human to have sex chromosomes as ‘XXY’ in
stead of ‘XX’ or ‘XY’.
(b) List any two ways such individuals are different from the normal being.
Ans. a) Failure of segregation of chromatids during cell division cycle/ nondisjunction/
aneuploidy , resulting in gain of an extra X chromosome in a male after fertization
=½×2
b) Development of breasts/ gynaecomastia , sterile individuals = ½ × 2
[2Marks]

8. Spirulina is a rich source of proteins. Mention the two ways by which large scale culturing
of these microbes is possible.
Ans. Grown in waste water from potato processing plants (starch rich) /straw / molasses / animal m a -
nure/ sewage (any two) // bioreactor, using any two above mentioned materials = 1+1
[2Marks]
9. How does EcoRI specifically act on DNA molecule ? Explain.

5' GAATTC 3'
Ans. Inspects the length of a DNA sequence, finds its specific recognition sequence as
3' CTTAAG 5'
, binds to the DNA , and cut each of the two strands of the double helix at specific points as sugar-
phosphate backbones = ½ x 4
//

=½ =½
=½

=½

[2Marks]

18

10. Name any two autotrophic microbes and state how they serve as biofertilizers.

Ans. Anabaena / Nostoc / Oscillatoria (any two) = ½ + ½ (any other relevant example) Fix
atmospheric nitrogen , add organic matter to the soil = ½ + ½

[2 Marks]

11. How is the study of fossils an evidence of evolution of life forms which have taken place on
the Earth ? Explain giving two reasons.

Ans. A study of fossils in different sedimentary layers indicate that different types of organisms existed in
different geological period, the life-forms varied over time and certain life forms are restricted to
certain geological time span = 1 x 2
[2 Marks]

12. How is the normal human body temperature of 37°C maintain during (i) Summer, and
(ii) Winter ? Explain.

Ans. (i) In summer we sweat (profusely), evaporation causes cooling = ½ × 2

(ii) We start to shiver a kind of exercise, which produces heat = ½ × 2

[2 Marks]

SECTION C

13. (a) Write the scientific name of methanogen bacteria. Where are these bacteria
generally found ? Explain their role in biogas production.

(b) Name the components of biogas.
Ans. a) Methanobacterium =½
found in the anaerobic sludge / rumen of cattle, they grow anaerobically on cellulosic material and
cause its breakdown (into methane CO2 and H2 ) = ½ × 2
b) methane , CO2 , H2 = ½ × 3
[3 Marks]
14. Explain double fertilization in an angiosperm.

Ans. After entering one of the synergids the pollen tube releases the two male gametes , one of the
male gametes moves towards the egg cell , fuses with its nucleus thus completing the syn-
gamy and forms a diploid zygote , the other male gamete moves towards the two polar n u -
clei in the central cell and triple fusion occur , resulting in formation of primary endosperm
nucleus, syngamy and triple fusion occuring in the embryo sac is double fertilization = ½ × 6

[3 Marks]

19

15. Differentiate between the pattern of inheritance in humans of the blood diseases, haemophilia
and thalassemia.
Ans. Haemophilia Thalassemia
1. Sex linked recessive disorder 1. Autosomal recessive disorder
2. heterozygous female carrier 2. Transmitted from both the carrier
may transmit it both to her son and parents
daughter
3. males are generally affected 3. both sexes can be affected
but female are rarely affected

(both corresponding points to be written for credit) =1×3

[3 Marks]
16. (a) Compare the mechanism of sex determination in humans with that of honeybees,
with respect to chromosome number.
(b) How is the gamete formation comparable in the above two cases ?
Ans. a) In honeybee union of a sperm and an egg develops as a female (queen or worker), and an
unfertilised egg develops as a male (drone) by means of parthenogenesis, the females are dip-
loid / having 32 chromosomes and males are haploid / having 16 chromosomes , in humans-
both male and female individuals have 23 pair of chromosomes / diploid / female 44+ XX and
males 44+ XY =½ × 4
b) In humans gametes are formed by meiosis, in honeybee female gametes are formed by meiosis
and male gametes by mitosis = ½ × 2
[3 Marks]

17. Differentiate between Dominance, Incomplete dominance and Co-dominance with
the help of a suitable example of each.

Ans. Dominance Incomplete Dominance Co-dominance

In heterozygous In heterozygous condition In heterozygous condition
condition one allele is not completely both the alleles are equally
one allele of the trait dominating the other expressed
dominates the other and an intermediate phenotype

one is expressed.

Eg. In pea heterozygous Eg.In Heterozygous Eg. In heterozygous IA IB

tall plant / Tt condition- T Rr condition of snapdragon condition both alleles IA

20

expresses itself the pink colour is expressed. and IB expressed as blood
and supresses t. group AB = ½×6

[3Marks]

18. Mention the chemical nature of an antibody and name the type of cells they are produced
by. Write the difference between active and passive immune responses on the basis of
antibodies.
Ans. Made up of proteins / peptide, B-lymphocytes = ½ + ½
Active immunity - due to exposure to antigens / pathogens / vaccination / immunisation
leads to production of antibodies by the individual, slow process = ½ + ½
Passive immunity - Ready-made antibodies are directly given to protect the body of an
individual against foreign agents, fast process/ provide immediate immunity = ½ + ½
[3 Marks]
OR
Name the cells that act as HIV factory in humans when infected by HIV. Explain the
events that occur in these infected cells.
Ans. Macrophages = ½
Once the virus enters the human body the virus / viral genome infects macrophages where its RNA
genome replicates , to form viral DNA, with the help of the enzyme reverse transcriptase, this viral
DNA gets incorporated into host cell’s DNA, and directs the infected cells to produce virus
particles = ½ × 5
[3 Marks]

19. Explain the solutions found by Ahmed Khan, a Bengaluru based plastic sack manufacturer,
after realising the problems created by plastic wastes.
Ans. Polyblend a fine powder was made from recycled modified plastic, this was mixed with the bitumen
to lay roads which proved to enhance the bitumen’s water repellant properties, and helped to
increase road life by the factor of three = 1 × 3
[3 Marks]
20. Explain the effect of the sewage discharges on the characteristic (quality) of a river.
Ans. Due to discharge of sewage there is an increase in the activity of the micro-organisms involved i n
degradation of the organic wastes, sharp decline in dissolved oxygen occurs in the downstream ,
increased biochemical oxygen demand (causes mortality of fish and other aquatic
creatures) = 1 + 1 + 1

[3 Marks]
21. ‘‘Cotton bollworms enjoy feeding on cotton plants, but get killed when feed on Bt cotton
plant.’’ Justify the statement.

21

Ans. Once a bollworm feeds on Bt cotton plant the inactive protoxin produced by Bacillus thuringiensis,
is converted into an active form of toxin, due to the alkaline pH of the gut which solubilise the
crystals, the activated toxin binds to the surface of its midgut epithelial cells, create pores that cause
cell swelling and lysis, and eventually cause death of the insect =½x6
[3 Marks]
OR
(a) Mention the cause of ADA deficiency in humans.
(b) How is gene therapy carried out to treat the patients suffering from this disease ?
(c) State the possibility of a permanent cure of this disease.
Ans. a) Caused due to the deletion of the gene for adenosine deaminase = ½
b) lymphocytes from the blood of the patient are grown in a culture medium outside the
body, a functional ADA cDNA (using a retroviral vector) is then introduced into these
lymphocytes which are subsequently returned to the patient, patient requires periodic
infusion of such genetically engineered lymphocytes = ½ × 3
c) if the gene isolated from bone marrow cells producing ADA, is introduced into cells at
early embryonic stages it could be a permanent cure = ½ + ½
[3 Marks]

SECTION D
22. Indiscriminate use of chemicals, pesticides and weedicides by humans are polluting our
water bodies, which in turn are harming the living organisms. Study the flow chart and
answer the questions based on it.

(a) Why does the concentration of DDT seem to be considerably high in the top con-
sumer ?
(b) How would the organisms at the highest level be affected ?
(c) Name the phenomenon observed.
Ans. a) DDT cannot be metabolised or excreted so gets accumulated by an organism , increase
in concentration at successive trophic levels =½ × 2
b) Disturb calcium metabolism in fish eating birds / causes thinning of egg shell ,
premature breaking of eggs leading to population decline =½ × 2
c) Biomagnification =1 [3 Marks]
23. Given below is the diagram representing the observations made for separating DNA
fragments by Gel electrophoresis technique. Observe the illustration and answer the
questions that follow :

22

(a) Why are the DNA fragments seen to be moving in the direction A  B ?
(b) Write the medium used on which DNA fragments separate.
(c) Mention how the separated DNA fragments can be visualised for further technical use.
Ans. a) Because the DNA fragments are negatively charged =1
b) Agarose gel =1
c) After staining DNA with ethidium bromide, followed by exposure to UV rays =½ × 2
[3 Marks]

24. Study the transverse section of human ovary given below and answer the questions that
follow :

.

a) Name the hormone that helps in the growth of A  B  C.
(b) Name the hormone secreted by A and B.
(c) State the role of the hormone produced by D.
Ans. a) Gonadotropins // FSH and LH = 1
b) Estrogen =1
c) Maintenance of uterine endometrium =1
[3 Marks]

23

SECTION E
25. Name the type of DNA that forms the basis of DNA fingerprinting and mentiontwo features of
this DNA. Write the steps carried out in the process of DNA fingerprinting technique, and
mention its application.
Ans. a) Satellite DNA / repetitive DNA =½
These sequences normally do not code for any proteins, these sequence show high degree of
polymorphism = ½ × 2
b) (i) isolation of DNA,
(ii) digestion of DNA by restriction endonucleases,
(iii) separation of DNA fragments by electrophoresis,
(iv) transferring (blotting) of separated DNA fragments to synthetic membranes such
as nitrocellulose or nylon,
(v) hybridisation using labelled VNTR probe,
(vi) detection of hybridised DNA fragments by autoradiography =½ x 6
Application - Forensic science / determining population and genetic diversities / paternity
test = ½
[5Marks]
OR
Explain the role of different genes in a lac operon, when in a ‘Switched On’ state.
Ans. The regulator gene produces repressor, inactivated in the presence of inducer, that is lactose,
RNA polymerase then gets access to the promoter gene and transcription proceeds =½ × 4
z gene codes for beta-galactosidase (ß-gal), responsible for the hydrolysis of the disaccharide lactose
into galactose and glucose = ½ × 2
y gene codes for permease, which increases permeability of the cell to ß-galactosides /
lactose =½ × 2
a gene encodes enzyme transacetylase =1

//

24

=½ x 10
[5 Marks]

26. (a) According to ecologists, tropical regions in the world account for (greater biologi-
cal diversity. Justify.
(b) Why are habitat loss and alien species invasion considered as the causes of biodiversity
loss ? Explain with the help of an example of each.

Ans. a) (i) have remained relatively undisturbed for millions of years / had a long evolu-
tionary time for species diversification =1
(ii) environment less seasonal / more constant and predictable / such constant envi-
ronment promotes niche specialization =1
(iii) more solar energy available in the tropics contributes to higher productivity and
greater diversity =1
b) Habitat loss
Amazon rain forest is being cut for cultivating soyabeans / degradation of habitat by pollution /
human activities leading to clearing of forests for commercial or tourism purpose =1 (any other
relevant example)
Alien species invasion-
The Nile perch introduced into Lake Victoria in East Africa led eventually to the extinction
of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake / Recent
illegal introduction of the African catfish Clarias gariepinus for aquaculture purposes is posing a
threat to the indigenous catfishes in our rivers / water hyacinth / lantana / carrot grass / causes threat
to our indigenous species =1
(any other relevant example)
[5 Marks]

25

OR
(a) What is an ecological succession ?
(b) Differentiate between primary and secondary succession. Why is secondary
succession faster than primary succession ? Explain with suitable examples.
(c) What are pioneer species ? Give examples of pioneer species in Xerarch and
Hydrarch successions respectively.
Ans. a) The gradual and fairly predictable change in the species composition of a given area
is called ecological succession =1
b) Primary succession Secondary succession
Starts in an area where Areas that somehow
no living organisms lost all the living organisms
ever existed / that existed there /
bare rock / newly created pond / abandoned farmlands /
reservoir / bare area flooded field area / burnt forest . =1
Secondary succession is faster since some soil or sediment is already present =1
eg. abondoned lands/ burnt or cut forests/ lands that have been flooded =½
c) The species that invade a bare area are called pioneer species = ½
Xerarch- lichens , Hydrarch- phytoplanktons =½ × 2
[5 Marks]

27. Where does fertilization occur in the oviduct of a human female ?

Describe the process of fertilization.

Ans. • Ampullary isthmus junction / ampullary region of oviduct =1
• Sperm comes in contact with the zona pellucida layer of the ovum , induces changes
in the membrane, blocks the entry of additional sperms/ it ensures entry of one sperm,
secretion of acrosome help the sperm enter into the cytoplasm of the ovum , induces the
completion of the meiotic division of the secondary oocyte, the unequal second meiotic
division results in the formation of a smaller second polar body and a larger haploid
ovum (ootid), the haploid nucleus of the sperm and that of the ovum fuse together, to form the
diploid zygote = ½× 8

[5 Marks]
OR

(a) Where does microsporogenesis occur in an angiosperm ? Describe the process of
microsporogenesis.
(b) Draw a labelled diagram of the two-celled male gametophyte of an angiosperm.
How is the three-celled male gametophyte different from it ?

26

(a) Microsporangium / sporogenous tissue / PMC = ½
Each cell of the sporogenous tissue acts as microspore mother cell, undergoes meiosis forming
tetrad of haploid microspore, as the anthers mature the tetrad dissociates and develop into pollen
grains = ½×3

(b)

(any five) = ½× 5

Three celled male gametophyte has one vegetative cell and two male gametes (instead of a generative
cell ) = ½

[5 Marks]

27

Question Paper Code. 57/3/3
SECTION A
Note : Choose the correct option from the choices given in each of the following questions :
1. Hormones released in human females only during pregnancy are
(A) hCG, hPL, Progesterone
(B) Relaxin, hCG, hPL
(C) hCG, hPL, Oxytocin
(D) hPL, Thyroxine, hCG
Ans. B / Relaxin, hCG, hPL
[1 Marks]
2. Nematode specific genes were introduced into the tobacco host plant by using the vector
(A) Plasmid
(B) Bacteriophage
(C) pBR 322
(D) Agrobacterium
Ans. D / Agrobacterium = 1
[1 Mark]
3. The principle of vaccination is based on the property of
(A) Specificity
(B) Diversity
(C) Memory
(D) Discrimination between ‘self’ and ‘non-self’
Ans. C / Memory =1 [1 Mark]

OR
Opioids act as

(A) Depressants
(B) Pain killers
(C) Euphoria providers
(D) Stimulants
Ans. A / Depressants =1 [1 Mark]
4. One of the ex situ conservation methods for endangered species is

28

(A) Biosphere reserves
(B) National parks
(C) Cryopreservation
(D) Wildlife sanctuaries
Ans. C / Cryopreservation = 1
[1 Mark]
OR
Ozone gas is continuously formed in the stratosphere by
(A) Action of UV rays on nascent oxygen
(B) Reaction of oxygen with water vapour
(C) Action of UV rays on molecular oxygen
(D) Action of UV rays on water vapour
Ans. C / Action of UV rays on molecular oxygen = 1
[1 Mark]
5. Introduction of an alien DNA into a plant host cell is achieved by making them
(A) Competent with bivalent ions
(B) Using microinjections
(C) Using gene gun
(D) Using lysozymes and chitinase
Ans Ans. C / Using gene gun =1
[1 Mark]
SECTION B
6. Why are humming birds not found in the Polar region ?
Ans. Have large surface area relative to their volume , tend to loose body heat very fast when there is
cold outside, they have to expend much energy , to generate body heat through
metabolism. = ½ × 4

[2Marks]
7. Spirulina is a rich source of proteins. Mention the two ways by which large scale culturing
of these microbes is possible.
Ans. Grown in waste water from potato processing plants (starch rich) /straw / molasses / animal manure/
sewage (any two) // bioreactor, using any two above mentioned materials = 1+1
[2 Marks]
8. State what are biofertilizers. Name any three sources of biofertilizers.
Ans. Organism that enrich the nutrient quality of the soil = ½
Bacteria, fungus and cyanobacteria = ½ × 3

[2 Marks]
29

9. How does EcoRI specifically act on DNA molecule ? Explain.
Ans. Inspects the length of a DNA sequence, finds its specific recognition sequence as
5' GAATTC 3'
, binds to the DNA, and cut each of the two strands of the double helix at
3' CTTAAG 5'
specific points as shown in their sugar-phosphate backbones = ½ x 4
[2Marks]

10. Name and explain the technique that can be used in developing improved
crop varieties in plants bearing female flowers only.
Ans. Artificial hybridization = ½
The female flower buds are bagged before the flower open, when stigma becomes receptive polli-
nation is carried out using the desired pollen , and flower is rebagged (and fruits are allowed to
develop) =½×3

[2Marks]
OR
When are the non-flowering plants said to be homothallic and
monoecious; and heterothallic and dioecious ? Give an example of each.
Ans. Homothallic and monoecious : bisexual condition / having both male and female reproductive
structures on the same plant, eg. Chara and several fungi or any other suitable example = ½ × 2
Heterothallic and dioecious: unisexual condition / having either male or female reproductive structures
present in different plants, example Marchantia or any other suitable
example = ½ × 2
[2Marks]
11. (a) Explain the cause responsible in a human to have sex chromosomes as ‘XXY’ in
stead of ‘XX’ or ‘XY’.
(b) List any two ways such individuals are different from the normal being.
Ans. a) Failure of segregation of chromatids during cell division cycle/ nondisjunction/
aneuploidy, resulting in gain of an extra X chromosome in a male after fertization
=½×2
b) Development of breasts/ gynaecomastia , sterile individuals = ½ × 2
[2Marks]
12. How is the mechanism of evolution explained by de Vries ?
Ans. Large difference arising suddenly in a population, because of mutation, in a single step / sal-
tation, random and directionless = ½ × 4

[2Marks]
SECTION C

30

13. (a) Write the scientific name of methanogen bacteria. Where are these bacteria gener-
ally found ? Explain their role in biogas production.
(b) Name the components of biogas
Ans. a) Methanobacterium =½
found in the anaerobic sludge / rumen of cattle, they grow anaerobically on cellulosic mate-
rial and cause its breakdown (into methane CO2 and H2 )
=½×2
b) methane , CO2 , H2 = ½ × 3
[3 Marks]
14. Majority of angiosperms have hermaphrodite flowers, but self-pollination is discouraged
by them. Explain any three outbreeding devices that they have developed to achieve this ?
Ans. i) Pollen release and stigma receptivity not synchronised,either pollen released before the stigma
is receptive or otherwise.
ii) Anther and stigma are placed at different position , stigma cannot come in contact with the
pollen of the same flower.
iii) In some plant flowers are unisexual , male flower and female flower born on different plant.
iv) Self incompatibility , it is a genetic mechanism and prevents self pollen from fertilising
the ovule by inhibiting pollen germination or pollen tube growth in the pistil (any three)
(each point to be ) = ½ × 2

[3Marks]
15. Differentiate between the pattern of inheritance in humans of the blood diseases, haemophilia
and thalassemia.
Ans. Haemophilia Thalassemia
1. Sex linked recessive disorder 1. Autosomal recessive disorder
2. heterozygous female carrier 2. Transmitted from both the carrier
may transmit it both to her son and parents
daughter
3. males are generally affected 3. both sexes can be affected
but female are rarely affected

(both corresponding points to be written for credit) =1×3

[3 Marks]
16. ‘‘Cotton bollworms enjoy feeding on cotton plants, but get killed when feed on Bt cotton
plant.’’ Justify the statement.
Ans. Once a bollworm feeds on Bt cotton plant the inactive protoxin produced by Bacillus thuringiensis,
is converted into an active form of toxin, due to the alkaline pH of the gut which solubilise the

31

crystals, the activated toxin binds to the surface of its midgut epithelial cells, create pores that cause
cell swelling and lysis, and eventually cause death of the insect = ½ x 6
[3 Marks]
OR
(a) Mention the cause of ADA deficiency in humans.
(b) How is gene therapy carried out to treat the patients suffering from this disease ?
(c) State the possibility of a permanent cure of this disease.
Ans. a) Caused due to the deletion of the gene for adenosine deaminase = ½
b) lymphocytes from the blood of the patient are grown in a culture medium outside the
body, a functional ADA cDNA (using a retroviral vector) is then introduced into these
lymphocytes which are subsequently returned to the patient, patient requires periodic
infusion of such genetically engineered lymphocytes = ½ × 3
c) if the gene isolated from bone marrow cells producing ADA, is introduced into cells at
early embryonic stages it could be a permanent cure = ½ + ½
[3 Marks]
17. (a) Compare the mechanism of sex determination in humans with that of honey bees,
with respect to chromosome number.
(b) How is the gamete formation comparable in the above two cases ?
Ans. Ans. a) In honeybee union of a sperm and an egg develops as a female (queen or worker),
and an unfertilised egg develops as a male (drone) by means of parthenogenesis, the females
are diploid / having 32 chromosomes and males are haploid / having 16 chromosomes , in
humans- both male and female in dividuals have 23 pair of chromosomes / diploid / female
44+ XX and males 44+ XY =½ × 4
b) In humans gametes are formed by meiosis, in honeybee female gametes are formed by
meiosis and male gametes by mitosis =½ × 2
[3 Marks]
18. You are asked to find the genotype of a pea plant bearing violet flowers, growing in your
school garden. Give your explanation to get the solution.
Ans. Carry a test cross / cross the violet flower bearing plant with the white flower bearing plant /
cross the violet flower bearing plant with the homozygous recessive plant =1
If the progeny has 100% violet flowers , the unknown genotype is homozygous dominant
=½+½
If the progeny has 50% violet flower and 50% white flower , then the unknown genotype is
heterozygous dominant = ½ + ½

//

32

[3Marks]
19. Mention the chemical nature of an antibody and name the type of cells they are pro-
duced by. Write the difference between active and passive immune responses on the
basis of antibodies.
Ans. Made up of proteins / peptide, B-lymphocytes = ½ + ½
Active immunity - due to exposure to antigens / pathogens / vaccination / immunisation
leads to production of antibodies by the individual, slow process = ½ + ½
Passive immunity - Ready-made antibodies are directly given to protect the body of an individual
against foreign agents, fast process/ provide immediate immunity = ½ + ½
[3 Marks]
OR
Name the cells that act as HIV factory in humans when infected by HIV. Explain the
events that occur in these infected cells.
Ans. Macrophages = ½
Once the virus enters the human body the virus / viral genome infects macrophages where its RNA
genome replicates , to form viral DNA, with the help of the enzyme reverse transcriptase, this viral
DNA gets incorporated into host cell’s DNA, and directs the infected cells to produce virus
particles = ½ × 5
[3 Marks]
20. Explain the solutions found by Ahmed Khan, a Bengaluru based plastic sack manufacturer,
after realising the problems created by plastic wastes.
Ans. Polyblend a fine powder was made from recycled modified plastic, this was mixed with the bitu-

33

men to lay roads which proved to enhance the bitumen’s water repellant properties,and helped to
increase road life by the factor of three = 1 × 3
[3 Marks]

21. (a) Name the process that makes the detritus become part of the soil in the nutrient
cycle.
(b) Write the factors responsible for controlling the rate of this process.
(c) Mention the condition when the rate would be faster.
Ans. a) Decomposition =½
b) Rate of decomposition is controlled by the chemical composition of the detritus , cli
matic condition where the detritus is present = ½ + ½
c) rate of decomposition is faster when detritus is rich in nitrogen ,water soluble sub-
stances (sugar) , warm and moist environment =½ × 3

[3Marks]

SECTION D
22. Study the transverse section of human ovary given below and answer the questions that
follow :

(a) Name the hormone that helps in the growth of A  B  C.
(b) Name the hormone secreted by A and B.
(c) State the role of the hormone produced by D
Ans. a) Gonadotropins // FSH and LH = 1
b) Estrogen =1
c) Maintenance of uterine endometrium =1
[3 Marks]
23 . Indiscriminate use of chemicals, pesticides and weedicides by humans are polluting our
water bodies, which in turn are harming the living organisms. Study the flow chart and an-

34

swer the questions based on it

(a) Why does the concentration of DDT seem to be considerably high in the top con
sumer ?
(b) How would the organisms at the highest level be affected ?
(c) Name the phenomenon observed
Ans. a) DDT cannot be metabolised or excreted so gets accumulated by an organism , increase
in concentration at successive trophic levels =½ × 2
b) Disturb calcium metabolism in fish eating birds / causes thinning of egg shell ,
premature breaking of eggs leading to population decline =½ × 2
c) Biomagnification =1 [3 Marks]

24. Given below is the diagram representing the observations made for separating DNA
fragments by Gel electrophoresis technique. Observe the illustration and answer the
questions that follow :

(a) Why are the DNA fragments seen to be moving in the direction A  B ?
(b) Write the medium used on which DNA fragments separate.
(c) Mention how the separated DNA fragments can be visualised for further technical
use.
Ans. a) Because the DNA fragments are negatively charged =1
b) Agarose gel =1
c) After staining DNA with ethidium bromide, followed by exposure to UV rays =½ × 2
[3 Marks]

35

SECTION E
25. (a) When does the process of spermatogenesis begin in human males ? Describe the
process of spermatogenesis.
(b) Explain the role of gonadotropin in this process.
Ans. a) At puberty =½
The spermatogonia (sing. spermatogonium) present on the inside wall of seminiferous tu-
bules multiply by mitotic division and increase in numbers, some of the spermatogonia (2n)
called primary spermatocytes periodically undergo meiosis, a primary spermatocyte com-
pletes the first meiotic division (reduction division) leading to formation of two equal hap-
loid cells called secondary spermatocytes, the secondary spermatocytes undergo the second
meiotic division to produce four equal haploid spermatids, the spermatids are transformed
into spermatozoa (sperms) by the process called spermiogenesis = ½ × 5
//

(½ marks to be awarded for ploidy)
b) LH acts at the Leydig cells and stimulates synthesis and secretion of androgens / testoster-
one, which stimulates the process of spermatogenesis , FSH acts on the Sertoli cells , and stimulates
secretion of some factors which help in the process of spermiogenesis = ½ ×4
[5 Marks]

36

OR
Where does megasporogenesis occur in the ovule of an angiosperm ? Describe the pro-
cess up to the development of a mature embryo sac.
Ans. a) Occurs in the nucellus (towards the micropylar end) = ½
A single megaspore mother cell (MMC) undergoes meiotic division, results in the production of four
megaspores, one of the megaspores is functional while the other three degenerate, this type of
megasporogenesis is monosporic, the nucleus of the functional megaspore divides mitotically to form
the 8-nucleate stages of the embryo sac, six of the eight nuclei are surrounded by cell walls, the
remaining two polar nuclei remain in the large central cell, the three cells are towards the micropylar
end are the two synergids and one egg cell, three cells at the chalazal end are called the antipodals (a
typical angiosperm embryo sac at maturity is 8-nucleate and 7-celled) = ½ × 9
//

=½×9

[5 Marks]
26. (a) According to ecologists, tropical regions in the world account for greater biological
diversity. Justify.
(b) Why are habitat loss and alien species invasion considered as the causes of biodiversity
loss ? Explain with the help of an example of each.

37

Ans. a) (i) have remained relatively undisturbed for millions of years / had a long evolutionary
time for species diversification =1
(ii) environment less seasonal / more constant and predictable / such constant envi-
ronment promotes niche specialization =1
(iii) more solar energy available in the tropics contributes to higher productivity and
greater diversity =1
b) habitat loss
Amazon rain forest is being cut for cultivating soyabeans / degradation of habitat by pollution /
human activities leading to clearing of forests for commercial or tourism purpose =1 (any other
relevant example)
Alien species invasion-
The Nile perch introduced into Lake Victoria in East Africa led eventually to the extinction
of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake / Recent
illegal introduction of the African catfish Clarias gariepinus for aquaculture purposes is posing a
threat to the indigenous catfishes in our rivers / lantana / water hyacinth / carrot grass / causes threat
of our indigenous species =1
(any other relevant example)
[5 Marks]
OR
(a) What is an ecological succession ?
(b) Differentiate between primary and secondary succession. Why is secondary
succession faster than primary succession ? Explain with suitable examples.
(c) What are pioneer species ? Give examples of pioneer species in Xerarch and
Hydrarch successions respectively.
Ans. a) The gradual and fairly predictable change in the species composition of a given area
is called ecological succession =1
b) Primary succession Secondary succession
Starts in an area where Areas that somehow
no living organisms lost all the living organisms
ever existed / that existed there /
bare rock / newly created pond / abandoned farmlands /
reservoir / bare area flooded field area / burnt forest =1
Secondary succession is faster since some soil or sediment is already present =1
eg. abandoned lands/ burnt or cut forests/ lands that have been flooded =½
c) The species that invade a bare area are called pioneer species = ½
Xerarch- lichens , Hydrarch- phytoplanktons =½ × 2
[5 Marks]
27. (a) Name the type of DNA that forms the basis of DNA fingerprinting and mention
two features of this DNA.
(b) Write the steps carried out in the process of DNA fingerprinting technique, and men-

38

tion its application.
Ans. a) Satellite DNA / repetitive DNA =½
These sequences normally do not code for any proteins, these sequence show high degree of
polymorphism = ½ × 2
b) (i) isolation of DNA,
(ii) digestion of DNA by restriction endonucleases,
(iii) separation of DNA fragments by electrophoresis,
(iv) transferring (blotting) of separated DNA fragments to synthetic membranes such
as nitrocellulose or nylon,
(v) hybridisation using labelled VNTR probe,
(vi) detection of hybridised DNA fragments by autoradiography =½ x 6
Application - Forensic science / determining population and genetic diversities / paternity
test = ½
[5 Marks]
OR
Explain the role of different genes in a lac operon, when in a ‘Switched On’ state.
Ans. The regulator gene produces repressor, inactivated in the presence of inducer, that is lactose,
RNA polymerase then gets access to the promoter gene and transcription proceeds =½ × 4
z gene codes for beta-galactosidase (ß-gal), responsible for the hydrolysis of the disaccharide
lactose into galactose and glucose =½ × 2
y gene codes for permease, which increases permeability of the cell to ß-galactosides /
lactose = ½ × 2
a gene encodes enzyme transacetylase =1

//

39

[½ × 10 = 5 Marks]

40