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

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

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

Read the following instructions very carefully and strictly follow them:

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 are 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 candidate 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.
At which stage during evolution did human use hides to protect their bodies and buried their dead? [1]
(a) Homo habilis
(b) Neanderthal man
(c) Java man
(d) Homoerectus
Answer:
(b) Neanderthal man

The Neanderthal man with a brain size of 1400cc lived in near east and central Asia between 1,00,000-40,000 years back. They used hides to protect their body and buried their dead.

Question 2.
Given below are Column A with a list of certain Assisted Reproductive Technologies (ART) and in Column B the procedures followed during ART: [1]

Choose the option where ART correctly matches with the procedure.
(a) (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
(b) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)
(c) (A)-(iv), (B)-(iii), (C)-(i), (D)-(ii)
(d) (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)
Answer:
(d) (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)

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 for further development is another method attempted, intra cytoplasmic sperm injection (ICSI) is another specialised procedure to form an embryo in the laboratory in which a sperm is directly injected into the ovum.

The semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI – intra-uterine insemination) of the female. The zygote or early embryos (with upto 8 blastomeres) then be transferred into the fallopian tube (ZIFT-zygote intra fallopian transfer).

Question 3.
The decrease in the T-lymphocytes count in human blood will result in: [1]
(a) Decrease in antigens
(b) Decrease in antibodies
(c) Increase in antibodies
(d) Increase in antigens
Answer:
(b) Decrease in antibodies

The B-lymphocytes produce an army of proteins in response to pathogens into our blood to fight with them. These proteins are called antibodies. The T-cells themselves do not secrete antibodies but help B cells to produce them. Hence, decreases in number of T- lymphocytes automatically reduce the count of antibodies.

Note
If you have low number of lymphocytes, you are at higher risk of infections.

Question 4.
Given below is a sequence of bases in mRNA of a bacterial cell. Identify the amino acid that would be incorporated at codon position 3 and codon position 5 during the process of its translation. [1]
3′ AUCAGGUUUGUGAUGGUACGA 5′
(a) Phenylalanine, Methionine
(b) Cysteine, Glycine
(c) Alanine, Proline
(d) Serine, Valine
Answer:
(a) Phenylalanine, Methionine

In the given sequence of bases of bacterial cell, the 3 ^rd codon position is UUU which codes for Phenylalanine and the 5 ^th codon position is AUG that codes fòr Methionine.

Question 5.
A Tight one-to-one relationship between many species of fig tree and certain wasps is an example of – [1]
(a) Commensalism
(b) Parasitism
(c) Amensalism
(d) Mutualism
Select the pathogen mismatched with the symptoms of disease caused by it from the list given below : [1]
(a) Entamoeba histolytica : Constipation, abdominal pain.
(b) Epidermophyton : Dry scaly lesions on nail.
(c) Wuchereria bancrofti : Chronic inflammation of lymphatic vessels of lower limb.
(d) Haemophilus influenzae : Blockage of the intestinal passage.
Answer:
(d) The interaction confers benefits on both the interacting species. Such type of interaction is called Mutualism. One of the most common examples of mutualism is, the fèmale wasp uses the fruit not only as an oviposition (egg-laying) site hut uses the developing seeds within the fruit for nourishing its larvae. The wasp pollinates the fig inflorescence while searching for suitable egg-laying sites. In return for the favour of pollination the fig offers the wasp some of its developing seeds, as food for the developing wasp larvae.

Question 6.
The primary productivity in an ecosystem is expressed as: [1]
(a) gm ^-2 yr ^-1
(b) gm ^-2 yr
(c) K cal m ^-2 yr ^-1
(d) K cal m ^-2
Answer:
(d) K cal m ^-2

Bacteria like Streptococcus pneumoniae’ and Haemophilus influenzae are responsible for the disease pneumonia in humans which infects the alveoli (air filled sacs) of the lungs.

Question 7.
The primary productivity in an ecosystem is expressed as:
(a) gm ^-2 yr ^-1
(b) gm ^-2 yr
(c) K cal m ^-2 yr ^-1
(d) K cal m ^-2
Answer:
(a) gm ^-2 yr ^-1

The rate ofbiomass production is called productivity. It is expressed in terms of gm ^-1 yr ^-1 or (kcalm ^-2 )yr ^-1 . (1 mark)

Note
Productivity of an ecosystem can be divided into gross primary productivity and net primary productivity.

Question 8.
Given below is the restriction site of a restriction endonuclease Pst-I and the cleavage sites on a DNA molecule. [1]

Choose the option that gives the correct resultant fragments by the action of the enzyme Pst-I.

Answer:
(d) Restriction endonucleases make cuts at specific positions within the DNA. Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands. Therefore resultant fragment will be

Question 9.
The IUCN Red Data List (2004) in the last 500 years documents the extinction of nearly 784 species including: [1]
(a) 330 invertebrates
(b) 338 invertebrates
(c) 359 invertebrates
(d) 362 invertebrates
Answer:
(c) 359 invertebrates

The IUCN Red List (2004) documents the extinction of 784 species (including 338 vertebrates, 359 invertebrates and 87 plants) in the last 500 years.

Question 10.
Given below are the list of the commercially important products and their source organisms. Select the option that gives the correct matches. [1]

Options:
(a) (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
(b) (A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)
(c) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)
(d) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
Answer:
(d) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

Statins produced by the yeast Monascuspurpureus have been commercialised as blood-cholesterol lowering agents.
Cyclosporin A, that is used as an immunosuppressive agent in organ-transplant patients, is produced by the fungus Trichoderma polysporum.
Penicillin was the first antibiotic to be discovered extracted from Penicillium notatum.
Streptokinase is used as a clot buster is obtained from bacteruim streptococcus.

Question 11.
Important attributes belonging to a population but not to an individual are : [1]
(i) Birth rate and death rate
(ii) Male and female
(iii) Birth and death
(iv) Sex-ratio
Select the correct option from the given options :
(a) (i) only
(b) (ii) only
(c) (ii) and (iii)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)

A population has certain attributes whereas, an individual organism does not. An individual may have births and deaths, but a population has birth rates and death rates. In a population these rates refer to per capita births and deaths. Another attribute characteristic of a population is sex ratio. An individual is either a male or a female but a population has a sex ratio.

Question 12.
Select the option that shows the correctly identified ‘U’, ‘X’, ‘Y’ and ‘Z’ in a developing dicot embryo.

(a) X – Plumule (2n), Y – Suspensor (n), Z – Cotyledon (2n), U – Radicle (2n).
(b) X – Plumule (2n), Y – Suspensor (2n), Z – Radicle (2n), U – Cotyledon (2n).
(c) X – Suspensor (2n), Y – Cotyledon (2n), Z – Radicle (2n), U – Plumule (2n).
(d) X – Cotyledon (2n), Y – Radicle (n), Z – Plumule (n), U – Suspensor (n).
Answer:
(c) X – Suspensor (2n), Y – Cotyledon (2n), Z – Radicle (2n), U – Plumule (2n).

(1 mark)

Note
The above figure represent stage in embryo develpment in a dicot.

Question Nos. 13 to 16 consists of two statements, Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Question 13.
Assertion (A) : Determining the sex of an unborn child followed by MTP is an illegal practice.
Reason (R) : Amniocentesis is a practice to test the presence of genetic disorders also. [1]
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Statutoryban on amniocentesis for sex-determination to legally check increasing menace of female foeticides, massive child immunisation, etc., are some programmes that have been mentioned in this connection. In aminocentesis some of the amniotic fluid of the developing foetus is taken to analyse the fetal cells and dissolved substances. This procedure is used to test for the presence of certain genetic disorders such as, Down syndrome, haemoplilia, sickle-cell anemia, etc. to determine the survivability of the foetus.

Question 14.
Assertion (A) : Synthetic oligonucleotide polymers are used during Annealing in a PCR.
Reason (R) : The primers bind to the double stranded DNA at their complementary regions. [1]
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

In polymerase chain reaction, multiple copies of the gene (or DNA) of interest is synthesised in vitro using two sets of primers (small chemically synthesised oligonucleotides that are complementary to the regions of DNA) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template.

Question 15.
Assertion (A) : Decomposition process is slower if detritus is rich in lignin and chitin.
Reason (R): Decomposition is largely an oxygen requiring process. [1]
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Decomposition is largely an oxygen-requiring process. The rate of decomposition is controlled by chemical composition of detritus and climatic factors. In a particular climatic condition, decomposition rate is slower if detritus is rich in lignin and chitin, and quicker, if detritus is rich in nitrogen and water-soluble substances like sugars.

Question 16.
Assertion (A): In Thalassemia an abnormal myoglobin chain is synthesized due to a gene defect.
Reason (R): α -Thalassemia is controlled by genes HBA1 and HBA2 on chromosome 16. [1]
Answer:
(d) (A) is false, but (R) is true.

In thalassemia, the formation of abnormal haemoglobin molecules resulting into anaemia which is characteristic of the disease. A thalassemia is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16 of each parent and it is observed due to mutation or deletion of one or more of the four genes.

Section – B

Question 17.
The graph given below shows the number of primordial follicles per ovary in women at different ages. Study the graph and answer the questions that follow.

(a) What is the average age of the women at the onset of menopause?
(b) At what age are maximum primordial follicles present in the ovary, according to the given graph? [1 + 1 = 2]
Answer:
(a) The average age of women at the onset of menopause is 45. According to the given graph, perimenopausal women start at 45 and post menopaual women start at 50. So, average age will be 45+50/2 = 45 age.
(b) According to the given graph, the maximum primordial follicles are seen at between the age group range of 10 to 20 years.

Question 18.
“Cattle and goats do not browse the Calotropis plant.” Justify the statement giving reasons. [2]
Answer:
The plant Calotropis produces highly poisonous cardiac glycosides and that is why we never see any cattle browsing on this plant. A wide variety of chemical substances that we can extract from plants on a commercial scale (nicotine, caffeine, quinine, strychnine, opium, etc.,) are produced by them actually as defences against grazers and browsers.

Note
Menstrual cycle starts with a menstrual phase, when menstrualfiow occurs and it last for 3-5 days

Question 19.
By using Punnett square depict the genotypes and phenotypes of test crosses (where green pod colour (G) is dominant over yellow pod colour (g)) in Garden pea with unknown genotype. [2]
Answer:
The tests cross between green pod colour (GG) and yellow pod colour (gg) in garden pea.

Hence, the genotype and phenotype of this test cross is 1 : 1

Question 20.
(a) (i) Give an example of a genus of virus used as narrow spectrum insecticidal biocontrol agent.
(ii) How does its use serve as an aid in overall integrated pest management programme? [2]
OR
(b) Why a malignant tumour considered to be more damaging than a benign tumour? Explain. [2]
Answer:
(a) (i) Nucleopolyhedrovirus are excellent candidates for species-specific, narrow spectrum insecticidal applications.
(ii) Nucleopolyhedrovirus have been shown to have no negative impacts on plants, mammals, birds, and fish or even on non-target insects. This is especially desirable when beneficial insects are being conserved to aid in an overall integrated pest management (IPM) programme, or when an ecologically sensitive area is being treated.

OR

(b) Benign tumors normally remain confined to their original location and do not spread to other parts of the body and cause little damage. The malignant tumors, on the other hand are a mass of proliferating cells called neoplastic or tumor cells. These cells grow very rapidly, invading and damaging the surrounding normal tissues. As these cells actively divide and grow they also starve the normal cells by competing for vital nutrients. Cells sloughed from such tumors reach distant sites through blood, and wherever they get lodged in the body, they start a new tumor there.

Question 21.
(a) Write the scientific name of the source organism of the themostable DNA polymerase used in PCR.
(b) State the advantage of using Thermostable DNA polymerase. [2]
Answer:
(a) Thermus aquaticus is the source organism of the thermostable DNA polymerase used in PCR.

Note
PCR stand for Polymerase chain reaction. In this reaction multiple copies of gene of interest is synthezied in vitro.

(b) Advantages of using thermostable DNA polymerase is- It remains active during the high temperature induced denaturation of double stranded DNA.

Section – C

This section consists of one case followed by 6 questions. Besides this 6 more questions are given. Attempt any 10 questions from this section. The first 10 questions attempted would be evaluated.

Question 22.
Name and explain a surgical contraceptive method that can be adopted by the male partner of a couple. [3]
Answer:
Surgical methods, also called sterilisation, are generally advised for the male/female partner as a terminal method to prevent any more pregnancies. Surgical intervention blocks gamete transport and thereby prevents conception. Sterilisation procedure in the male is called ‘vasectomy’ and that in the female, tubectomy’.
In vasectomy, a small part of the vas deferens is cut or tied up through a small incision on the scrotum.

Question 23.
Human Genome Project (HGP) was a mega project launched in the year 1990 with some important goals.
(a) Enlist any four prime goals of HGP.
(b) Name any one common non-human animal model organism which has also been sequenced thereafter. [3]
Answer:
(a) The prime goals of HGP are :

1. Identify all the approximately 20,000-25,000 genes in human DNA;
2. Determine the sequences of the 3 billion chemical base pairs that make up human DNA;
3. Store this information in databases;
4. Improve tools for data analysis.

(b) There are many non-human model organisms, such as bacteria, yeast, Caenorhabditis elegans (a free living non-pathogenic nematode), Drosophila (the fruit fly), plants (rice and Arabidopsis), etc., have also been sequenced.

Question 24.
Industrial melanism in England after 1850 is an excellent example of Natural selection. Explain how? [3]
Answer:
In a collection of moths made in 1850s, i.e., before industrialisation set in, it was observed that there were more white-winged moths on trees than dark-winged or melanised moths. However, in the collection carried out from the same area, but after industrialisation, i.e., in 1920, there were more dark-winged moths in the same area, i.e., the proportion was reversed.

The explanation put forth for this observation was that ‘predators will spot a moth against a contrasting background’. During post industrialisation period, the tree trunks became dark due to industrial smoke and soots. Under this condition the white-winged moth did not survived due to predators, dark-winged or melanised moth survived.

Before industrialisation set in, thick growth of almost white- coloured lichen covered the trees – in that background the white winged moth survived but the dark-coloured moth were picked out by predators. They will not grow in areas that are polluted. Hence, moths that were able to camouflage themselves, i.e., hide in the background, survived. This understanding is supported by the fact that in areas where industrialisation did not occur e.g., in rural areas, the count of melanic moths was low. This showed that in a mixed population, those that can better- adapt, survive and increase in population size. Remember that no variant is completely wiped out.

Question 25.
One of the major approaches of crop improvement programme is Artificial Hybridisation. Explain the steps involved in making sure that only the desired pollen grain pollinate the stigma of a bisexual flower by a plant breeder. [3]
Answer:
Artificial hybridisation is one of the major approaches of crop improvement programme. In such crossing experiments it is important to make sure that only the desired pollen grains are used for pollination and the stigma is protected from contamination (from unwanted pollen). This is achieved by emasculation and bagging techniques. If the female parent bears bisexual flowers, removal of anthers from the flower bud before the anther dehisces using a pair of forceps is necessary. This step is referred to as emasculation.

Emasculated flowers have to be covered with a bag of suitable size, generally made up of butter paper, to prevent contamination of its stigma with unwanted pollen. This process is called bagging. When the stigma of bagged flower attains receptivity, mature pollen grains collected from anthers of the male parent are dusted on the stigma, and the flowers are rebagged, and the fruits allowed to develop.

Question 26.
(a) “Plasmodium protozoan needs both a mosquito and a human host for its continuity.” Explain. [3]
OR
(b) We all must work towards maintaining good health because ‘health is wealth’. Enlist any six ways of achieving good health. [3]
Answer:
(a) In the life cycle of Plasmodium, Plasmodium enters the human body as sporozoites (infectious form) through the bite of infected female Anopheles mosquito. The parasites initially multiply within the liver cells and then attack the red blood cells (RBCs) resulting in their rupture. The rupture of RBCs is associated with release of a toxic substance, haemozoin, which is responsible for the chill and high fever recurring every three to four days.

When a female Anopheles mosquito bites an infected person, these parasites enter the mosquito’s body and undergo further development. The parasites multiply within them to form sporozoites that are stored in their salivary glands. When these mosquitoes bite a human, the sporozoites are introduced into his/ her body, thereby initiating the events mentioned above. Hence, malarial parasite requires two hosts – human and mosquitoes – to complete its life cycle.

Note
Different species of plasmodium (P vivax, P malaria and P falciparum) are responsible for causing malaria.

OR

(b) Health can be defined as a state of complete physical, mental and social well-being. When people are healthy, they are more efficient at work. This increases productivity and brings economic prosperity. Health also increases longevity of people and reduces infant and maternal mortality.

There are various way to achieve a good health:

1. Balanced diet, personal hygiene and regular exercise are very important to maintain good health.
2. Yoga has been practised since time immemorial to achieve physical and mental health.
3. Awareness about diseases and their effect on different bodily function are also necessary for achieving good health.
4. Vaccinations (immunisation) against infectious diseases are necessary for achieving good health.
5. Proper disposal of wastes is important,
6. Control of vectors and maintenance of hygiene in food and water resources are necessary for achieving good health.

Question 27.
On spraying Bacillus thuringiensis on an infected cotton crop field the pests are killed by the toxin, however the toxin although produced by the bacteria does not affect it. Explain giving reason. [3]
Answer:
Some strains of Bacillus thuringiensis produce proteins that kill certain insects such as lepidopterans (tobacco budworm, armyworm), coleopterans (beetles) and dipterans (flies, mosquitoes)

B. thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. The Bt toxin protein exist as inactive protoxins but once an insect ingest the inactive toxin, it 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 midgut epithelial cells and creates pores that cause cell swelling and lysis and eventually cause death of the insect.

Question 28.
“Biodiversity plays a major role in many ecosystem services that nature provides.”
(a) Describe any two broadly utilatarian arguments to justify the given statement.
(b) State one ethical reason of conserving biodiversity. [3]
Answer:
(a) The broadly utilitarian argument says that biodiversity plays a major role in many ecosystem services that nature provides.
The two broadly utilitarian arguments are:

1. The fast dwindling Amazon forest is estimated to produce, through photosynthesis, 20 per cent of the total oxygen in the earth’s atmosphere.
2. Pollination (without which plants cannot give us fruits or seeds) is another service, ecosystems provide through pollinators layer – bees, bumblebees, birds and bats.

(b) The ethical argument for conserving biodiversity relates to what we owe to millions of plant, animal and microbe species with whom we share this planet.
Philosophically or spiritually, we need to realise that every species has an intrinsic value, even if it may not be of current or any economic value to us.

Section – D

This section consists of one case followed by 6 questions. Besides this 6 more questions are given. Attempt any 10 questions from this section. The first 10 questions attempted would be evaluated.

Q. Nos. 29 and 30 are case based questions. Each question has subparts with internal choice in one subpart.

Question 29.
When a microorganism invades a host, a definite sequence of events usually occur leading to infection and disease, causing suffering to the host. This process is called pathogenesis. Once a microorganism overcomes the defense system of the host, development of the disease follows a certain sequence of events as shown in the graph. Study the graph given below for the sequence of events leading to appearance of a disease and answer the questions that follow:

(a) In which period, according to the graph there are maximum chances of a person transmitting a disease/ infection and why? [1]

(b) Study the graph and write what is an incubation period. Name a sexually transmitted disease that can be easily transmitted during this period. Name the specific type of lymphocytes that are attacked by the pathogen of this disease. [2]
OR
(b) Draw a schematic labelled diagram of an antibody. [2]
(c) In which period, the number of immune cells forming antibodies will be the highest in a person suffering from pneumonia? [1]
Answer:
(a) In the period of illness, there are maximum chances of a person transmitting a disease because during this period the number of micro-organisms is highest.
(b) Incubation period (also known as the latent period or latency period) is the time elapsed between exposure to a pathogenic organism, a chemical, or radiation, and when symptoms and signs are first apparent.
AIDS (Acquired immune deficiency syndrome) can be easily transmitted during incubation period.
T lymphocytes (helper T cells) are attacked by the pathogen of HIV.

OR

(b)

(c) In the period of illness, the number of immune cells forming antibodies will be highest in a person suffering from pneumonia.
B-cells (B-lymphocytes) are the immune cells that produce antibodies.

Question 30.
The chromosome number is fixed for all normal organisms leading to species specification whereas any abnormality in the chromosome number of the fixed number of chromosomes both in male and female. In male it is ’44 + XY’ and in female it is ’44 + XX’. Thus the human male is heterogametic, in other words produces two different types of gametes one with ’22 + X’ chromosomes and the other with ’22 + Y’ chromosomes respectively: Human female, on the other hand is homogametic i.e. produces only one type of gamete with ’22 + X’ chromosomes only. Sometimes an error may occur during meiosis of cell cycle, where the sister chromatids fail to segregate called nondisjunction, leading to the production of abnormal gametes with altered chromosome number. On fertilisation such gametes develop into abnormal individuals.

(a) State what is aneuploidy. [1]
(b) If during spermatogenesis, the chromatids of sex chromosomes fail to segregate during meiosis, write only the different types of gametes with altered chromosome number that could possibly be produced. [1]
(c) A normal human sperm (22 + Y) fertilises an ovum with karyotype ’22 + XX’. Name the disorder the offspring thus produced would suffer from and write any two symptoms of the disorder. [2]
OR
(c) Name a best known and most common autosomal aneuploid abnormality in human and write any two symptoms. [2]
Answer:
(a) Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome, called aneuploidy.
(b) If sister chromatids tail to separate during meiosis II. the result is one gamete that lacks that chromosome, two normal gametes with one copy of the chromosome, and one gamete with two copies of the chromosome.
(c) Klinefelter’s Syndrome is genetic disorder is also caused due to the presence of an additional copy of X chromosome resulting into a karyotype of 47, XXY. Such an individual has overall masculine development, however, the feminine development (development of breast, i.e., Gynaecomastia) is also expressed. Such individuals are sterile.

OR

(c) Down’s Syndrome is autosomal aneuploid abnormality causes due to genetic disorder. It is the presence of an additional copy of the chromosome number 21 (trisomy of 21). The symptoms shows in individual is short statured with small round head, furrowed tongue and partially open mouth Palm is broad with characteristic palm crease. Physical, psychomotor and mental development is retarded.

Note
Down syndrome was first described by Langdon Dawy (1866).

Section – E

This section consists of one case followed by 6 questions. Besides this 6 more questions are given. Attempt any 10 questions from this section. The first 10 questions attempted would be evaluated.

Question 31.
(a) (i) Explain the monosporic development of embryosac in the ovule of an angiosperm. [3]
(ii) Draw a diagram of the mature embryo sac of an angiospermic ovule and label any four parts in it. [2]
OR
(b) (i) Explain the formation of placenta after the implantation in a human female. [3]
(ii) Draw a diagram showing human foetus within the uterus and label any four parts in it. [2]
Answer:
(a)
(i)In a majority of flowering plants, one of the megaspores is functional while the other three degenerate. Only the functional megaspore develops into the female gametophyte (embryo sac). This method of embryo sac formation from a single megaspore is termed monosporic development. The nucleus of the functional megaspore divides mitotically to form two nuclei which move to the opposite poles, forming the 2-nucleate embryo sac.

Two more sequential mitotic nuclear divisions result in the formation of the 4-nucleate and later the 8- nucleate stagesof the embryo sac. It is of interest to note that these mitotic divisions are strictly free nuclear, that is, nuclear divisions are not followed immediately by cell wall formation. After the 8- nucleate stage, cell walls are laid down leading to the organisation of the typical female gametophyte or embryo sac. Observe the distribution of cells inside the embryo sac. Six of the eight nuclei are surrounded by cell walls and organised into cells; the remaining two nuclei, called polar nuclei are situated below the egg apparatus in the large central cell.

(ii)

Diagrammatic representation of the mature embryo sac.

OR

(b)
(i) After implantation, finger-like projections appear on the trophoblast called chorionic villi which are surrounded by the uterine tissue and maternal blood. The chorionic villi and uterine tissue become interdigitated with each other and jointly form a structural and functional unit between developing embryo (foetus) and maternal body called placenta. The placenta facilitates the supply of oxygen and nutrients to the embryo and also removal of carbon dioxide and excretory/waste materials produced by the embryo.

The placenta is connected to the embryo through an umbilical cord which helps in the transport of substances to and from the embryo. Placenta also acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens, progestagens, etc.

(ii)

Question 32.
(a) Name and describe the steps involved in the technique widely used in forensics that serves as the basis of paternity testing in case of disputes. [5]
OR
(b) It is sometimes observed that the Fj progeny has a phenotype that does not resemble either of the two parents and has intermediate phenotype. Explain by taking a suitable example and working out the cross upto F ₂ progeny. [5]
Answer:
(a) DNA finger printing is a technique to identify a person on the basic of his / her DNA specificity
DNA of the human is almost the same for all individuals but very small amount differs from person to person and forensic science analysis this DNA to identify people.

The steps involve in DNA finger printing are:

1. Isolation of DNA,
2. Digestion of DNA by restriction endonucleases,
3. Separation of DNA fragments by electrophoresis.
4. Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
5. Hybridisation using labelled VNTR probe. and
6. Detection of hybridised DNA fragments by autoradiography.

OR

(b) If the F ₁ progeny has a phenotype that does not resemble either of the two parents, this could due to incomplete dominance. For example, Incomplete dominance is seen in cross-pollination experiments between red and white snapdragon plants. The allele that produces the red color (R) is not completely expressed over the recessive allele that produces the white color (r). The resulting offspring are pink.

Question 33.
(a) Bioreactors are the containment vehicles of any biotechnology-based production process. For large scale production and for economic reasons the final success of biotechnological process depends on the efficiency of the bioreactor. [5]
Answer the following questions w.r.t. the given paragraph:
(i) List the operational guidelines that must be adhered to so as to achieve optimisation of the bioreactor system. Enlist any four.
(ii) Mention the phase of the growth we refer to in the statement “Optimisation of growth and metabolic activity of the cells”.
(iii) Is the biological product formed in the bioreactor suitable for the intended use immediate? Give reason in support of your answer.

OR

(b) (i) ‘EcoRI’ has played very significant role in r-DNA technology.
(I) Explain the convention for naming EcoRI.
(II) Write the recognition site and the cleavage sites of this restriction endonuclease. [3]
(ii) What are the protruding and hanging stretches of DNA produced by these restriction enzymes called? Describe their role in formation of r-DNA. [2]
Answer:
(a)

(i) The four operational guidelines that must be adhered to for achieving optimization of the bioreactor system:

1. Monitoring and control of key parameters: In order to achieve optimal growth and productivity of the biological system, it is essential to monitor and control key parameters such as temperature, pH, dissolved oxygen, and nutrient levels. This can be achieved by using sensors and control systems that can adjust these parameters as needed.

2. Optimization of mixing and aeration: Proper mixing and aeration are essential for ensuring uniform distribution of nutrients and oxygen to the biological system. The mixing and aeration rate should be optimized based on the specific requirements of the system, as well as the characteristics of the bioreactor being used.

3. Minimizing contamination: Contamination can have a significant impact on the productivity of the biological system, so it is essential to minimize the risk of contamination by maintaining aseptic conditions throughout the process. This can be achieved by using sterile equipment, air filtration systems, and other appropriate measures.

4. Maintenance and cleaning of the bioreactor: Regular maintenance and cleaning of the bioreactor are essential for ensuring optimal performance and longevity of the system. This includes routine inspection of the system, cleaning of all components, and replacement of any worn or damaged parts as needed.

(ii) The cells can also be multiplied in a continuous culture system wherein the used medium is drained out from one side while fresh medium is added from the other to maintain the cells in their physiologically most active log/exponential phase.

(iii) After completion of the biosynthetic stage, the product has to be subjected through a series of processes before it is ready for marketing as a finished product. The processes include separation and purification, which are collectively referred to as downstream processing. The product has to be formulated with suitable preservatives. Such formulation has to undergo thorough clinical trials as in case of drugs. Strict quality control testing for each product is also required. The downstream processing and quality control testing vary from product to product.

OR

(b) (i)
(I) EcoRI comes from Escherichia coli RY 13. In EcoRI, the letter ‘R’ is derived from the name of strain. Roman numbers following the names indicate the order in which the enzymes were isolated from that strain of bacteria.

(ii) Restriction endonucleases are used in genetic engineering to form recombinant molecules of DNA,which are composed of DNA from different sources/Genomes. When cut by the same restriction enzyme, the resultant DNA fragments have the same kind of sticky-ends’ and these can be joined together (end-to-end) using DNA Ligases.

(II) Restriction endonuclease Cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands. This leaves single stranded portions at the ends. There are overhanging stretches called sticky ends on each strand. These are named so because they form hydrogen bonds with their complementary cut counterparts. This stickiness of the ends facilitates the action of the enzyme DNA ligase.

Note
DNA ligase is a DNA-joining enzyme. In DNA cloning, restriction enzymes and DNA ligase are used to insert genes and other pieces of DNA into plasmids.