CBSE Class 12 Biology Question Paper (Outside Delhi 2023) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Outside Delhi 2023) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Outside Delhi 2023) with Solutions

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General Instructions:

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1. This question paper contains 33 questions. All questions are compulsory.
2. Question paper is divided into Five sections – section A, B, C, D and E.
3. In section A – question number 1 to 16 are Multiple Choice (MCQ) type questions carrying 1 mark each.
4. In section B – question number 17 to 21 are Very Short Answer (VSA) type questions carrying 2 marks each.
5. In section C – question number 22 to 28 are Short Answer (SA) type question carrying 3 marks each.
6. In section D – question number 29 to 30 arc case-based questions carrying 4 marks each. Each question has subparts with internal choice in one subpart.
7. In section E- question number 31 to 33 are Long Answer (LA) type questions carrying 5 marks each.
8. There is no overall choice. However, an internal choice has been provided in 1 question in Section B, 1 question in Section C, 2 questions in Section D and 1 question is Section E. A condidate has to attempt only one of the laternateives in such question.
9. Wherever necessary, neat and properly labelled diagrams should be drawn.

Section – A

Question 1.
A human male decides to adopt a surgical method for contraception. Identify the point in the diagram where a cut would be made and tied.

(a) Point S
(b) Point R
(c) Point Q
(d) Point P
Answer:
(c) Point Q

False

Note
Vasectomy is a form of male birth control that cuts the supply of sperm to your semen.

Question 2.
Which of the following structures is well-developed in a mature seed of black pepper’?
(a) Perisperm
(b) Thalamus
(c) Sepals
(d) Peduncle
Answer:
(a) Perisperm

True

Question 3.
Observe the fòllowing line diagram depicting the 28 days menstrual cycle of a healthy young woman.

Select the option of days on which this woman would be most and least fertile.

Answer:
(b) 10-17 day of menstrual cycle are most fertile days, because ovulation occurs in which the ovary releases an egg for fertilisation to occur but when this egg does not get fertilised, the endometrium lining sheds causing the release of the egg along with mucus, this occurs in the form of menstruation cycle. This occurs during 21 -28 days and the female is in the least fertile stage during 21-28 days.

Question 4.
Which on of the following was not present during the Mesozoic Era of the geological time scale?
(a) Ferns
(b) Horsetails
(c) Ginkgos
(d) Bryophytes
Answer:
(d) Bryophytes

False

Question 5.
Identify the element used by Hershey and Chase to label the protein in their experiment, from the following cptions:
(a) P ³²
(b) S ³²
(c) S ³⁵
(d) P ³⁵
Answer:
(c) S ³⁵

True

Question 6.
DNA profiles of the child and three individuals 1, 2 and 3 who claim to be the parents of the child are given below. Select the option that shows the correct actual parent/parents of the child.

Answer:
(d) Individual 1 exhibit largest number of common bands with the child X that are band 1,3,6 and 8 are common so individual 1 is expected to be the parent of child X

Question 7.
The given schematic illustration shows three steps ‘P’, ‘Q’ and ‘R’ of the polymerase chain reaction.

Which of the following statements are correct with reference to the illustration given above?
(i) Step T’ is showing denaturation at low temperature.
(ii) Step ’Q’ is a denaturation of DNA strand at high temperature, followed by annealing.
(iii) Step ‘R’ is the extension of DNA in presence of thermostable DNA polymerase.
(iv) Step ‘Q’ is extension with two sets of primers.
(a) (i) and (iii) only
(b) (ii) and (iii) only
(c) (ii) only
(d) (i) only
Answer:
(b) (ii) and (iii) only

False

Note
PCR is a laboratory technique for rapidly producing million to billion copies of a specific segment of DNA.

Question 8.
Identify the fungus that ripens the famous ‘Roquefort’ cheese:
(a) Saccharomyees cerevisiae
(b) Propionibacterium sharmanii
(c) Monascus purpureus
(d) Penicillium notatum
Answer:
(b) Propionibacterium sharmanii

False

Question 9.
Select the options which is/are incorrect statement(s) with respect to T-lymphocytes in the human body.
(i) They are a type of white blood cells.
(ii) They are produced in bone marrow.
(iii) They remain active at all times in the body.
(iv) They mature in the bone marrow.
(a) (i) and (iv) only
(b) (iii) only
(c) (iv) only
(d) (iii) and (iv) only
Answer:
(d) (iii) and (iv) only

True

Question 10.
Which one among the following regions is not a hotspot of biodiversity?
(a) The Indo-Burma Region
(b) Jaintia Hills in Meghalaya
(c) The Western Ghats and Sri Lanka
(d) The Himalayas
Answer:
(b) Jaintia Hills in Meghalaya

True

Question 11.
Human settlement often leads to habitat loss which leads to fragmentation, forming smaller patches of habitats. Select the statements that describe how a small patch differs from a large patch of the same habitat.
(i) Invasive species will never be seen here.
(ii) Population of large animals decreases.
(iii) Biodiversity decreases.
(iv) Competition from surrounding habitats increases,
(a) (ii), (iii) and (iv) only
(b) (ii) and (iv) only
(c) (i) and (iii) only
(d) (i), (ii) and (iii) only
Answer:
(a) (ii), (iii) and (iv) only

False

1. Population of large animals decreases: This occurs because the large animals require larger territory to survive which is not possible at small patch. The population of large animals like herbivores or predators is likely to reduce.
2. Biodiversity decreases: Due to habitat loss and fragmentation of large patch into small patch the number of species is likely to reduce.
3. Competition from surrounding habitats increases: As the small patch developed after habitat loss and fragmentation there will be limited number of resources available so the population of small patch is likely to compete with members of the other neighboring patches for resources like food, water and shelter.

Question 12.
Identify the option that gives the correct type of evolution exhibited by the two animals shown, living in the same habitat in Australia.

(a) Convergent Evolution
(b) Disruptive Selection
(c) Divergent Evolution
(d) Homologous Ancestry
Answer:
(a) Convergent Evolution

True

For Questions number 13 to 16, two statements are given-one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 13.
Assertion (A): The Covid-19 virus has a shorter life-span and evolves into new strains at a fast speed.
Reason (R): RNA being unstable, mutates at a faster rate.
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

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Question 14.
Assertion (A): For DNA sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes.
Reason (R): Human genome is said to have approximately 3 × 10 ⁹ bp and the total estimated cost for sequencing is very high.
Answer:
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes this is due to the fact that there are technical limitations in sequencing very long strands of DNA in a single attempt. The entire human genome is said to have approximately 3 × 10 ⁹ base pairs. If the cost of sequencing is estimated to be about US $ 3 for each base pair, then the estimated cost of the entire project turns out to be 9 billion US dollars.

Question 15.
Assertion (A): Biologists are sure about how many prokaryotic species are living now.
Reason (R): The conventional taxonomic methods are not suitable for identifying microbial species.
Answer:
(d) Assertion (A) is false, but Reason (R) is true.

False

Question 16.
Assertion (A): Mary Mallon continued to spread typhoid for many years.
Reason (R): Salmonella typhi generally enters the small intestine through food and water contaminated with it.
Answer:
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

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Section – B

Question 17.
(a) Explain the process of the development of a male gametophyte in an angiosperm.
(b) Why is it called a male gametophyte? [2]
Answer:
(a) In angiosperms, the pollen grain is the male gametophyte. Maturation of the male gametophyte or pollen grain includes two mitotic divisions. First divisions form vegetative and generative cell and in the second mitotic division the generative cell forms basically two male gametes and their release occurs from a mature anther. Therefore, it has two male gametes and one vegetative cell. (2 marks)

b) In angiosperms, microspores indicate the male gametophyte . microspores divide mitotically resulting in two unequal type of cells . generative cell splits once again resulting in two male gametes or sperms .

Note
For fertilization to occur in angiosperms. pollen has to be transferred to stigma of a flower

Question 18.
(a) Name the two institutes which developed the technology of biogas production in India.
(b) Explain the main principle used in this technology. [2]
Answer:
(a) The technology of biogas production was developed in India mainly due to the efforts of Indian Agricultural Research Institute (IARI) and Khadi and Village Industries Commission (KVIC).

(b) Principle of Biogas: Biogas is produced by decomposition of organic matter.

• A mixture of farm waste is fed into the biogas plant. This waste is mixed to make a slurry.
• Slurry is then fed into the digester. Digester is an air-tight chamber and oxygen is not present in it.
• Anaerobic bacteria in digester carry out decomposition process of slurry. This results in production of biogas.
• Biogas is sent out through an outlet so that it can be suitably used.
• Decomposed matter is taken out and it can be used as manure.

Question 19.
Given below is a food web that involves nine organisms.

(a) Identify two producers and two carnivores shown in the food web.
(b) Is it possible to make an ecological pyramid depicting this food web? Give reason in support of your answer. [2]
Answer:
(a) According to the given food web, the producers are 1 and 2 whereas the two carnivores are 6 and 7.

True

Question 20.
(a) ‘Insertional inactivation’ is a method to detect recombinant DNA. Explain the method. [2]
OR
(b) Explain how recombinant DNA technology is used to detect a disease even before any clinical symptom appears. [2]
Answer:
(a) Insertional inactivation technique of recombinant DNA technology used to select bacteria that carry recombinant plasmids; a fragment of foreign DNA is inserted into a restriction site within a gene for antibiotic resistance, thus causing that gene to become nonfunctional. It is often used to identify recombinant vectors in gene cloning and in turn to distinguish a recombinant vector from a non-recombinant vector.

False

: (a) PCR, ELISA and autoradiography are some of the recombinant DNA technologies that can be used for the detection of disease even before any clinical symptom appears. (b) PCR, ELISA and autoradiography are some of the recombinant DNA technologies that can be used for the detection of disease even before any clinical symptom appears.

PCR is now routinely used to detect HIV in suspected AIDS patients. It is being used to detect mutations
in genes in suspected cancer patients too. It is a powerful technique to identify many other genetic disorders.

False

Question 21.
On August 22 in the year 2022,3358 fires were detected in the rainforests. Write one short-term and one long-term effect of this event on the biotic and abiotic components of the environment.
Answer:
One short term impact of forest fire:
On biotic component of the environment is mortality of large number of species causing biodiversity loss.
On abiotic component of the environment is the deterioration of the quality of air and water.
One long term impact of forest fire:
On biotic component of the environment is co-extinction of species.
On abiotic component of the environment will not support the survival of future species.

Section – C

Question 22.
Explain the following population interactions with the help of one example each: [3]
(a) Brood Parasitism
(b) Co-evolution of mutualists
Answer:
(a) Brood parasitism: in birds is a fascinating example of parasitism in which the parasitic bird lays its eggs in the nest of its host and lets the host incubate them. During the course of evolution, the eggs of the parasitic bird have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest. (3 marks)

plants need the help of animals for pollinating their flowers and dispersing their seeds . animals obviously have to be paid ‘fees’ for the services that plants expect from them . plants offer rewards or fees in the form of pollen and nectar for pollinators and juicy and nutritious fruits for seed dispersers .

animals that try to steal nectar without aiding in pollination . plant-animal interactions often involve co-evolution of the mutualists .

. - (FR) In many species of fig trees, there is a tight one-to-one relationship with the pollinator species of wasp . the female wasp uses the fruit not only as an oviposition site but also uses the developing seeds within the fruit for nourishing its larvae .

False

Question 23.
(a) Write the scientific name of the nematode that infests the tobacco plants and the part that it infests.
(b) How is Agrobacterium used to protect tobacco plant from this attack? [3]
Answer:
(a) A nematode Meloidegyne incognitia infects the roots of tobacco plants and causes a great reduction in yield.

nematode-specific genes were introduced into the host plant . introduction of DNA produced sense and anti-sense RNA in the host cells . dsRNA initiated RNAi and thus, silenced the specific mRNA of the nematode . transgenic plant thus got itself protected from the parasite .

Question 24.

(a) Name the category of drugs represented by the chemical structure given above.
(b) If the methyl group is substituted by acetyl group we get a bitter crystalline compound. Name the compound.
(c) Name the natural source of these compounds.
(d) State the harmful effects of this class of drugs on the human body. [3]
Answer:
(a) The given drug is morphine that belongs to the category of opioids. Opioids are the drugs, which bind to specific opioid receptors present in our central nervous system and gastrointestinal tract.

Note
Morphine are used as analgesics.

heroin is a depressant and slows down body functions . it is usually taken by snorting and injection .

(‘A mark’) Papaver somniferum is the natural source of these compounds.

d) The harmful effects of opioids include drowsiness, constipation, euphoria, nausea, vomiting and slowed breathing . a person using opioids over time can develop tolerance, physical dependence and opioid used disorder .

Question 25.
(a) Darwin’s theory of Natural Selection is widely accepted but some limitations have been identified by modern biologists. Mention the limitations identified.
(b) Name and state the most accepted theory of evolution in modern times.
(c) Mention any two ways the limitations identified in Darwin’s theory of evolution are explained in modern biology. [3]
Answer:
(a)

1. The theory of evolution could not explain how and where variations have arisen.
2. It also could not explain how the variations are inherited.

deVries brought forth the idea of mutations – large difference arising suddenly in a population . deVries believed that it is mutation which causes evolution and not the minor variations (heritable) that Darwin talked about .

False

1. According to Darwin evolution is a gradual process caused by minor variation this was contradicted by de Vries who proposed that evolution is an outcome of mutations that appear suddenly. He called it saltation (single step large mutation).
2. According to Darwin evolution is a slow directional process whereas according to deVries evolution is caused by random chance events or mutations resulting in sudden evolution of characters in a single or few generation.

Question 26.
(a) (i) How many types of RNA polymerases are there in a eukaryote cell? Mention which one of them transcribes hnRNA.
(ii) Write the changes that hnRNA undergoes before it leaves the nucleus as mRNA. [3]
OR
(b) The length of DNA in any cell is far greater than the dimension of its nucleus. Explain how this enormous DNA is packaged in a eukaryotic cell. [3]
Answer:
(a) (i) Eukaryotes have three RNA polymerases which are structurally distinct complexes, though share certain subunits in common, and have a specific function and specific promoter sequence. RNA polymerase I synthesize preribosomal RNA (pre- rRNA), which contains the precursor for the 18S, 5.8S, and 28S rRNAs. RNA polymerase II is synthesized mRNAs and some specialized RNAs. RNA polymerase III makes tRNAs, the 5S rRNA, and some other small specialized RNAs. The enzyme RNA Polymerase II is responsible for the transcription of eukaryotic hnRNA. (VA marks)

ii) hnRNA is required to undergo spicing process before becoming functional mRNA in eukaryotes . all the non coding (introns) are removed and all the coding (exons) are bind together and forms functional mRNA .

in eukaryotes, the length of the chromatin material is large so it is required to be packed in a manner so that it can be condensed in a single nucleus . there is a set of positively charged, basic proteins called histones . Histones are rich in the basic amino acid residues lysine and arginine . both the amino acid residues carry positive charges in their side chains .

False

nucleosomes in chromatin are seen as ‘beads- on-string’ structure when viewed under electron microscope (EM)

False

Question 27.
Expand and explain the following techniques used in the Test Tube Baby programme: [3]
(a) GUT
(b) ZIFT
(c) IUI
Answer:
(a) GIFT: In this assisted reproductive technologies (ART) the transfer of an ovum collected from a donor into the fallopian tube (GIFT – gamete intra fallopian transfer) of another female who cannot produce one, but can provide suitable environment for fertilisation -and further development is another method attempted. (1 mark)

True

embryos with more than 8 blastomeres, into the uterus (IUT – intra uterine transfer) to complete its further development . embryos with more than 8 blastomeres, into the uterus (IUT – intra uterine transfer) to complete its further development .

Note
Assisted reproductive technology is used to treat inferurr-ty.

Question 28.
Trace the journey of a zygote from the isthmus of the fallopian tube up to its implantation in the uterus of a human female. Highlight the changes the zygote undergoes during the course of its journey up to implantation. [3]
Answer:
(a) The ovaries produce the female egg cells, called the ova or oocytes. The oocytes are then transported to the fallopian tube where fertilisation by a sperm may occur. Then the zygote travels down the fallopian tube, where it becomes a morula.

a) The fertilised egg then moves to the uterus, where the uterine lining has thickened . b) The fertilised egg then moves to the uterus, where the uterine lining has thickened .

c) Once fertilised in the ampulary region of the fallopian tube, the embryo the uterine cavity at the blastocyst stage after 5 days of fertilisation .

d) During implantation the embryo adheres to the wall of the uterus . (d) During implantation the embryo adheres to the wall of the uterus .

Section – D

False

Question 29.
The following pedigree chart shows the inheritance of a genetic disorder up to three generations of a family. Observe the chart and answer the questions that follow.

(i) Is the disease sex-linked or autosomal as per the chart? Give reasons in support of your answer. [1]
(ii) Is it a recessive or a dominant disorder? [1]
(iii) Write the genotypes of the individuals ‘C’, ‘D’ and ‘H’. [1]
(iv) (a) Efthe female D’ marries a normal man, what Will be the probability of their daughter being a sufferer of this disease? [1]
OR
(iv) (b) If the mother ‘B’ is a carrier of the disease, what will be the probability of their daughter being a sufferer of this disease? [1]
Answer:
(i) According to the given pedigree chart, among the parents in the generation I A is affected which is according to pedigree symbol is a male. The disease is passed from the male to the progeny (male children) in generation II and III so the disease is Y-linked or sex linked.

the Y chromosome is one of the two sex chromosomes in each of a male’s cells . in Y-linked inheritance, a variant can only be passed from father to son . therefore, the given case is an example of Y-linked dominant disorder .

iii) According to the given pedigree chart, the genotypes of individuals ‘C’, ‘D’ and ‘H’ are XX, XX and XX respectively because these are females and mother is not the carrier of the disease in the generation I.

to: (iv) (a) XX (female D) X XY (normal man) then the probability of a daughter being a sufferer of the disease is zero. This is because the given pedigree chart is indicative of X linked disorder so male progeny is likely to be affected by it and also the mother is not the carrier of the disorder so the female progeny will not be the carrier or affected by the disease.

iv) (b) If X (mother B) is the carrier of the disease then the probability of their daughter being the sufferer of the disease is zero . in this case one of the daughter can be the carrier of the disease but will remain unaffected from the disease because the given disease is Y linked dominant trait which will not be transferred in females .

Question 30.
The diagram shows the life cycle of a pathogenic protozoan.

(i) Name the parasitic stage that is being transferred from host ‘X’ to host ‘Y’. [1]
(ii) Write the changes the parasite undergoes in the liver. [1]
(iii) Write the changes the parasite undergoes when it enters the RBC. [1]
(iv) (a) Trace the changes the parasite undergoes when the host ‘X’ takes its blood meal from infected host Y. [1]
OR
(iv) (b) At which stage during the life cycle of the pathogen does the host ‘Y’ experience the symptoms of the disease? Name the disease and the toxic substance responsible for these symptoms. [1]
Answer:
(i) Sporozoite is the parasitic stage that is being transferred from host X (mosquito) to host Y (human).

ii) The parasite reproduce asexually in the liver cells of the human host . it leads to bursting of cells and the parasite spreads into the bloodstream .

paraphrase: (iii) Parasite when enters into the red blood cells reproduces asexually causing bursting of the red blood cells .

a) Host X is mosquito which will feed on host Y to obtain its blood meal . b) Host X will take up gametocytes with blood meal .

when the parasite enters into the red blood cells of the host it reproduces asexually . when the red blood cells get burst this causes cycles of fever and other symptoms . the disease is malaria and the rupture of RBCs is associated with release of a toxic substance, haemozoin .

Section – E

Question 31.
(a) Protein synthesis requires the services of all three types of RNAs, namely t-RNA, m-RNA and r-RNA. Explain the role of each of them during the process of protein synthesis in prokaryotes. [5]
OR
(b) A homozygous tall pea plant with green seeds is crossed with a homozygous dwarf pea plant with yellow seeds.
(i) Write the possible phenotype and genotype of F ₁ generation.
(ii) State the laws of Mendel that are proved true by the F ₁ generation.
(iii) Mention the F ₂ phenotypic ratio along with their possible phenotypes.
(iv) Write the genotypes of the male and female gametes produced by F ₁ progeny. [5]
Answer:
(a) Theroleoft-RNA,m-RNAandr-RNAduringtheprocess of protein synthesis in prokaryotes is as follows: t-RNA: tRNA functions as an adapter molecule during the translation process. It was earlier known as soluble RNAor sRNA. As an adapter, it links the amino acids to nucleic acids. It carries the amino acid to be added in the peptide chain and also deciphers the codon for the same in the mRNA molecule.

True

True

Note
There are many types of RNA found in eukaryotic and prokaryotic cells. Three main types of RNA are mRNA, tRNA, rRNA.

(b) (i) According to the cross:
TTGG (tall green) X ttgg(dwarf yellow)
F ₁ generation progeny with have tall green (TtGg) genotype and tall green phenotype.

(ii) The law of dominance can be applied in the F ₁ generation. The one which is expressed in the F ₁ generation is called the dominant trait and the one which is suppressed is called a recessive trait. In simple words, the law of dominance states that recessive traits are always dominated or masked by the dominant trait.

(iii) Phenotypic ratio in the F ₂ generation will be tall green: tall yellow: dwarf green: dwarf yellow and the genotypic ratio will be 9:3:3:1.

(iv) The genotypes of the male and female gametes produced by F ₁ progeny will 1:2:2:4:1:2:1:2:1.

Question 32.
(a) Answer the following questions with respect to recombinant DNA technology: [5]
(i) Why is plasmid considered to be an important tool in rDNA technology? From where can plasmids be isolated? (Any two sources)
(ii) Explain the role of’ori’ and selectable marker in a cloning vector.
(iii) “r-DNA technology cannot proceed without restriction endonuclease.” Justify.
OR
(b) Answer the following questions based on Bt-crops:
(i) Why do farmers prefer to grow. Bt cotton crop than genetically unmodified cotton crops?
(ii) Name any two insects that are killed by Bt toxin.
(iii) Explain the mechanism by which Bt toxin kills the insects but not the bacterium which possesses the toxin.
Answer:
(a) (i) Plasmids used in genetic engineering are called vectors. They are used to transfer genes from one organism to another. Plasmid is helps in linking of foreign DNA and selection of recombinants from nonrecombinants. Some plasmids may have only one or two copies per cell whereas others may have 15-100 copies per cell. Their numbers can go even higher. The two sources of plasmid are E.coli and Agrobacterium tumifaciens.

(i) Origin of replication (ori): This is a sequence from where replication starts and is responsible for controlling the copy number of the linked DNA. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin support high copy number. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin support high copy number.

True

Falk: restriction endonuclease cuts double stranded DNA very precisely . it was used to restrict the growth of viruses when it was discovered in bacteria . Falk: restriction endonuclease cuts double stranded DNA at specific recognition sites .

specific Bt toxin genes were isolated from Bacillus thuringiensis . cryllAb controls the growth of corn borer infecting the healthy cotton plants . the genetically unmodified cotton crops will remain unprotected with the corn borer .

(ii) Cotton bollworm and corn borer are the two insects that are killed by Bt toxin.

Bt toxin is produced by bacteria Bacillus thuriengiensis . the endotoxin that accumulates in the bacterium is an inactive precursor . it gets activated only in the alkaline gut of insect .

Question 33.
(a) (i) Describe the arrangement of nuclei and cells in a mature embryo sac of a typical angiosperm.
(ii) Explain the devices the flowering plants have developed to prevent the following types of pollination: [5]
(1) Prevents both autogamy and geitonogamy
(2) Prevents autogamy, but not geitonogamy
OR
(b) (i) Write the specific location of the following in the testis in humans:
(1) Sertoli cells
(2) Leydig cells
(ii) Explain the coordination between Gonadotropins, Leydig cells and Sertoli cells and their role in spermatogenesis. [5]
Answer:
(a) (i) In a typical embryo sac of an angiosperm there are seven cells- one central cell, two synergids, one egg cell, and three antipodals. The egg apparatus comprising a group of three cells (two synergids and one egg cell) is found at the micropylar end. Three antipodal cells are located at the chalazal end. Six of the eight nuclei are enclosed by cell walls, whereas, the remaining two nuclei (polar nuclei) are located in the central cell. Hence, a typical angiosperm embryo sac at maturity is 8-nucleate and 7-celled.

(i) (ii) (iii) (iv) (v) (v) (v) (iv) (v) (iv) (iv) (v) (iv) (v) (iv) (v) (iv) (v) (iv) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v)

True

Note
Geitonogamy is a functionally cross pollination involving a pollinating agent.

(b) (i) (ii) (ii) (iii) (iii) (iii) (iv) (v) (v) (v) (iv) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (v) (

(1) Sertoli cells are present in the seminiferous tubules of the male gonads in the testes.
(2) Leydig cells are present in the intertubular/interstitial space.

luteinising hormone (LH) acts at the Leydig cells and stimulates synthesis and secretion of androgens . follicle stimulating hormone (FSH) acts on the Sertoli cells and stimulates secretion of some factors which help in the process of spermiogenesis .

True

False

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