CBSE Class 12 Biology Question Paper (Delhi 2020) with Solutions

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CBSE Class 12 Biology Question Paper (Delhi 2020) with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

1. Question paper comprises five sections – A, B, C, D and E.
2. There are 27 questions in the question paper. All questions are compulsory.
3. Section A question number 1 to 5 are multiple choice questions, carrying one mark each.
4. Section B question number 6 to 12 are short answer questions type-I, carrying two marks each.
5. Section C question number 13 to 21 are short answer questions type-II, carrying three marks each.
6. Section D question number 22 to 24 are short answer questions type-III, carrying three marks each.
7. Section E question number 25 to 27 are long answer questions, carrying five marks each.
8. Answer should be brief and to the point also the above word limit be adhered to as far as possible.
9. There is no overall choice in the question paper. However, an internal choice has been provided in two questions of 1 mark, one question of 2 marks, two questions of 3 marks and three questions of 5 marks questions. Only one of the choices in such questions have to be attempted.
10. The diagram drawn should be neat proportionate and properly labelled, wherever necessary.
11. In addition to this, separate instructions are given with each section and question, wherever necessary.

Section A

Question 1.
Which one of the following part of the plant when put into the soil is likely to produce new offspring?
(a) Part of an internode
(b) A stem cutting with a node
(c) Part of a primary root
(d) A flower
Answer:
(b) A stem cutting with a node

Stem cuttings with leaf node when buried in soil tend to produce new offspring with new roots usually at the node. This process is called vegetative propagation in which plants parts like leaves, stem and roots are used to produce a new plant.

Question 2.
In a bacterium when RNA-polymerase binds to the promoter on a transcription unit during transcription, it
(a) terminates the process
(b) helps remove introns
(c) initiates the process
(d) inactivates the exons
Answer:
(c) initiates the process

When the RNA polymerase in bacteria get binds with the promoter in order to initiates the process of transcription. Promoter is a sequence of DNA that provides a binding site for RNA polymerase enzyme. It is located towards 5’-end (upstream) of a structure gene.

Question 3.
The hypothesis that “Life originated from pre-existing non-living organic molecules was proposed by
(a) Oparin and Haldane
(b) Louis Pastur
(c) S.L. Miller
(d) Hugo de Vries
Answer:
(a) Oparin and Haldane

“Life originated from pre-existing non-living organic molecules (RNA, protein and so on)” was proposed by Oparin and Haldane. He also stated that the formation of life was preceded by chemical evolution such as the formation of diverse organic molecules from inorganic constituents. The conditions on earth were-high temperature, volcanic storms, reducing atmosphere containing CH ₄ , NH ₃ and so on.

Question 4.
Mating of a superior male of a breed of a cattle to a superior female of another breed is called
(a) inbreeding
(b) outcrossing
(c) outbreeding
(d) crossbreeding
Answer:
(d) crossbreeding

Mating of a superior of a breed of cattle to a superior female of another breed is called cross-breeding. This breeding method allows the desirable qualities of two different breeds to be mated. The progeny hybrid animals are then used for commercial production. For example’ Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes and Marino rams.

OR

Large-holes in ‘Swiss-Cheese’ are due to
(a) Propionibacterium sharmanii
(b) Saccharomyces cerevisae
(c) Penicillium chrysogenum
(d) Acetobacter aceti
Answer:
(a) Propionibacterium sharmanii

The large holes in the ‘Swiss-cheese’ are due to the production of a large amount of carbon dioxide by bacterium Propionibacterium sharmanii.

Question 5.
Increased concentration of DDT in fish-eating birds is due to
(a) eutrophication
(b) bio-magnification
(c) cultural eutrophication
(d) accelerated eutrophication
Answer:
(b) bio-magnification

The increased concentration of DDT in fish-eating birds is due to biomagnifications. Biomagnifications refers to the increase in concentration of the toxicant at successive trophic levels. This takes place because a toxic substance is accumulated by an organism that cannot be metabolised or excreted.

OR

Species-Area relationship is represented on a log scale as
(a) hyperbola
(b) rectangular hyperbola
(c) linear
(d) inverted
Answer:
(b) rectangular hyperbola

The species-area relationship is represented on a log scale as rectangular hyperbola. Alexander von Humboldt observed that within a region species richness increased with increased explored area but only upto a limit. On a logarithm scale, the relationship is a straight line described by the equation:
log S = log C + Z log A
where S = species richness
A = Area
Z = slope of the line (regression coefficient)
C = Y-intercept

Section – B

Question 6.
State two advantages of an apomictic seed to a farmer.
Answer:
Apomixis is a form of asexual reproduction that mimics sexual reproduction. As some species of plants such as Asteraceae and grasses have evolved special mechanism in order to produce seeds without fertilisation.

Advantage of Apomixis:
Apomixis reduces the cost of hybrid production and helps plant breeders to produce new varieties of seeds more quickly and more cheaply.

Question 7.
Explain when is a genetic code said to be
(a) Degenerate
(b) Universal
Answer:
(a) Degenerate: Code is said to be degenerate as some amino acids are coded by more than one codon.
(b) Universal: The code is said to be universal as a genetic code is same in all organisms. For example: from bacteria to human UUU would code for Phenylalanine (Phe).

Note
Codon is set of the three nucleotides that codes for a single amino acid. As genetic codon is triplet in nature and there are 64 codons. Out of which, 61 codons codes for amino acid and 3 are stop codons.

Question 8.
Differentiate between opioids and cannabinoids on the basis of their
(a) specific receptor site in human body.
(b) mode of action in human body.
Answer:
Difference between opioids and cannabinoids:

Pioneer species	Climax succession	Sere
1. It is the first biotic community develops in the bare area.	It is the final biotic community that develops in an area.	It is a transitional community that develops in an area during succession
2. Life span of organisms is short.	Life span of organism is long.	Life span of organism is long.
3. It consists of fewer small-sized organisms.	It consists of large and small sized organisms.	The size of individuals is small.
4. Growth is fast.	Growth is slow.	Growth is high.

Note
Opioids lire the drugs that are obtained from the latex of poppy plant called Papaver somniferum. Whereas cannabinoids are the group of chemicals that are naturally obtained from the inflorescence of the plant Cannabis saliva.

Question 9.
(a) Name the two techniques employed to meet the increasing demand of fish in the world.
(b) Name any two fresh water fishes.
OR
Describe the contributions of Alexander Fleming, Ernest Chain and Howard Florey in the field of microbiology.
Answer:
(a) To meet the increasing demands on fisheries different techniques such as aquaculture and pisciculture techniques are employed to increase production. By using these two techniques, both fresh-water and marine fish production will increases.
(b) Example of fresh water fishes are Catla, Rohu and common carp.

OR

Alexander Fleming while working on Staphylococci bacteria observed that a mould growing in one of his unwashed culture plates around which Staphylococci could not grow. During this, he found that it was due to a chemical produced by the mould and he named it as Penicillin after the Penicillium notatum. Ernest Chain and Howards Florey give its full potential to prove it as an effective antibiotic. This antibiotic was used for treatment of American soldiers that wounded in World War II. Fleming, Chain and Florey were awarded the Nobel Prize in 1945 for this discovery.

Question 10.
All cloning vectors do have a ‘selectable marker’. Describe its role in recombinant DNA-technology.
Answer:
The use of selectable marker in Recombinant DNA helps in the identification and elimination of non-transformants. It selectively permits the growth of the transformants.

Transformation refers to the process through which a piece of DNA is introduced in a host bacterium and the genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline or kanamycin are considered as useful selectable marker for E.coli.

Question 11.
Mention how have plants developed mechanical and chemical defence against herbivores to protect themselves with the help of one example of each.
Answer:
Unlike animals, plants cannot move away from their predators. So they developed a variety of mechanical and chemical defences against herbivores. Herbivores are predator. Some plants such as Acacia and Cactus develop specific structures like thorns for defence. Many plants produce and store chemicals that make herbivore sick when they are eaten as they inhibit feeding and digestion. This chemical also disrupts its reproduction or even kills it.

For example: A weed called Calotropis growing in abandoned fields produces a highly poisonous cardiac glycosides that causes cardiac arrest in herbivores.

Question 12.
Name and explain the processes earthworm and bacteria carry on detritus.
Answer:
Decomposition is a process in which the Earthworm and bacteria carry on detritus. Detritus is composed of dead remains of plants such as leave, bark, flowers and dead remains of animals involves fecal matter. All these material serves as a raw material for decomposition.

Detritivores such as earthworm breakdown detritus into smaller particles and this process is called fragmentation. Then, by the process of leaching, the water soluble inorganic nutrients go down into the soil norizon and get precipitated as unavailable salts. Bacterial and fungal enzymes degrade detritus into simpler inorganic substances by the process called as catabolism. During the process of decomposition, humiliation and mineralisation also occurs.

Note
Humification involves the priucss of accumulation of dark co-loured amorphous substance that is highly resistant to microbial action. Whereas humus is further degraded by some microbes and release of inorganic nutrients occurs and this proccss is called mineralisation.

Section – C

Question 13.
Explain three different modes of pollination that can occur in a chasmogamous flower.
OR
Explain the formation of placenta after implantation in a human female.
Answer:
The three types of pollination that takes place in a chasmogamous flower are as follows:
(a) Autogamy: In this type, the process of pollination is achieved within the same flower. It involves the transfer of pollen grains from the anther to the stigma of the same flower. Chasmogamous flowers are type of flowers that are similar to the flowers of other species with exposed anthers and stigma.

In chasmogamous flowers, the anthers and stigma lie close to each other. When anther dehisces in the flower buds, the pollen grain comes in contact with the stigma to effect pollination.

(b) Geitonogamy: It involves the transfer of pollen grains from the anther to the stigma of another flower of the same plant. It is functionally a type of cross-pollination that involves pollinating agents. But genetically it is similar to autogamy so the pollen grains come from the same plant.

(c) Xenogamy: It involves the transfer of pollen grains from anther to stigma of a different plant. This is the only type of pollination in which pollination brings genetically different types of pollen grains to the stigma.

Note
Pollination refers to the process of transfer of pollen grains (shed from the anther) to the stigma of a pistil.

OR

After implantation of foetus, a finger-like projections appears on the trophoblast called chorionic villi that is surrounded by the uterine tissue and maternal blood. The chorionic villi and uterine tissue become interdigitated with each other and they jointly give rise to a structural and functional unit between developing embryo as well as maternal body called as placenta. (3 Marks)

Note
Placenta helps to facilitate the supply of oxygen and nutrients to the developing embryo and also responsible for the removal of carbon dioxide as well as other excretory waste produced by the embryo.

Question 14.
State Mendel’s law of dominance. How did he deduce the law? Explain with the help of a suitable example.
Answer:
Mendal’s law of dominance states that characters are controlled by discrete unit called factors.
They occur in pair. In a dissimilar pair of factors one member of the pair dominates (dominant) over the other (recessive).

This law is used to explain the expression of only one of the parental characters in a monohybrid cross in the F ₁ generation. The expression of both alleles in the F ₂ generation. Fie also explained the proportion of 3:1 obtained at the F ₂ generation.

Diagrammatic Representation of an Example of Mendel’s law of Dominance:

Question 15.
What are ‘SNPs’? Where are they located in a human cell? State any two ways the discovery of SNPs can be of importance to humans.
Answer:
SNPs are single nucleotide polymorphism that occurs in humans. These are the most common form of genetic variation among people. Each SNPs represents a difference in a single nucleotide sequence of DNA. SNPs are located in a DNA between the genes such as promoters, exons, introns or 5’ and 3’ untranslated regions. Basically they are located within a gene or in a regulatory region near a gene.
They act as biological markers that helps scientist to locate the genes that are associated with diseases. It helps to measure the genetic variations among individuals.

Question 16.
(a) State what does the study of Fossils indicate.
(b) Rearrange the following group of plants according to their evolution from Palaeozoic to Cenozoic periods:
Rhynia; Arborescent Lycopods; Conifers; Dicotytedon.
Answer:
(a) The study of fossils is considered as the important evidence for evolution because it represents that the life on the earth was once different from the life that is found on earth today. The study of fossils is important because it helps to determine and study the physical structure of extinct organisms. With the help of fossils, one can determine that how long life has existed on earth and how the different plants as well as animals are evolved and are related to each other.

Note
Fossils are the dead remains impression or trace of an animal on plant of are preserved in Earth’s rust from a past geological age.

(b) Paleozoic – Rhynia
Arborescent – Lycopods
Mesozoic – Conifers
Dicotyledons

Question 17.
(a) Explain the mode of action of Cu ⁺⁺ releasing IUDs as a good contraceptive. How is hormone releasing IUD different from it?
(b) Why is ‘Saheli’ a preferred contraceptive by women (any two reasons)?
Answer:
(a) IUDs are Intra Uterine Devices (IUDs) increases phagocytosis of sperms within the uterus and the copper ions are released suppress sperm motility and the fertilising capacity of sperms.
Example of copper releasing IUDs are cuT, cu7, Multiload 375.
The hormone releasing IUDs make the uterus unsuitable for implantation and the cervix hostile to the sperms. IUDs are ideal contraceptives for the females who want to delay pregnancy.
Example of hormone releasing IUDs is Progestasert, LNG-20.

Note
IUDs is one of most widely accepted methods of contraception in India.

(b) Saheli-The new oral contraceptive pill fcr the females that contains a non-steroidal preparation. It is a ‘once a week’ pill with very few side effects and high contraceptive value. They inhibit ovulation and implantation as well as alter the quality of cervical mucus to prevent or retard the entry of sperms.

Question 18.
(a) Explain why bee-hives are setup on the farms for some of our crop-species. Name any two such crop species.
(b) List any three important steps to be kept in mind for successful bee keeping.
Answer:
(a) Bee-keeping or apiculture is the maintenance of hives of honeybees for the production of honey. Handling and collection of honey and of beeswax. Bees are the pollinators of many of our crop species such as sunflower, Brassica, apple and pear. Keeping the beehives in crop fields during flowering period increases pollination efficiency and improves the yield-beneficial both from the point of view of crop yield and honey yield.

(b) The following points are important for successful bee-keeping:

• Knowledge of the nature and habits of bees.
• Selection of location for keeping of beehives
• Catching and hiving of Swarms or group of bees,
• Management of beehives during different seasons.

Question 19.
Why GMOs are so called? List the different ways in which GMO plants have benefitted and have become useful to humans.
Answer:
GMOs are Genetically Modified Organisms that are produced by alteration in genes of plants, bacteria, fungi and animals. GM plants are useful in many ways such as:

• Genetic modification made crops more tolerant to abiotic stresses such as cold, drought, salt and heat.
• It also reduces the reliance of chemical pesticides such as pest-resistant crops.
• It helps to reduce post-harvest losses.
• It also increases the efficiency of mineral usage by plants
• GM enhanced nutritional value of foods such as vitamin A enriched rice.

Question 20.
Differentiate between “Pioneer-species”; “Climax- community” and “Seres”.
OR
Explain any three ways other than zoological parks, botanical gardens and wildlife safaries, by which threatened species of plants and animals are being conserved ‘ex situ’.
Answer:
Difference between pioneer species, climax succession and sere are as follows:

Categories	Opioids	Cannabinoids
Specific receptor site in human body	Opioid receptors present in the central nervous system and gastrointestinal tract of human beings.	Cannabinoid receptors present in the brain.
Mode of action in human body	Opioids such as heroin act as a depressant and slow down body functions.	Cannabinoids such as marijuana, hashish, charas and ganja effects on cardiovascular system of the body.

OR

Except from zoological parks, wildlife sanctuaries and botanical gardens, ex-situ conservation also involves a method for preservation of endangered or extinct species such as by

Cryopreservation: Gametes (sperms, eggs, tissues and embryo) of several endangered plants and animal species can be preserved by methods involving cryopreservation (-196°C). It can be fertilised In vitro followed by propagation through tissue culture methods.

Seeds of different genetic strains of commercially important plants are also kept for longer period in seed banks.

Question 21.
Explain ‘Integrated organic’ farming as successfully practiced by Ramesh C. Dagar, a farmer in Sonepat (Haryana).
Answer:
‘Integrated organic farming’ is a cyclical, zero-waste procedure in which the waste products from one process are cycled in as nutrients for other processes. This allows the maximum utilisation of resources and increases the efficiency of production. Ramesh Chandra Dagar, a farmer in Sonipat, Haryana is practising this farming. He involves bee-keeping, dairy management. water-harvesting, composting and agriculture in a chain of processes that support each other and also allow an extremely economical as well as sustainable venture.

In this type of farming, there is no need to use chemical fertilisers for crops such as cattle excreta (cow dung) are used as manure. Crop waste is also used to create compost that can be used as a natural fertiliser or can be used to generate natural gas for satisfying the energy needs of the farms.

Section – D

Question 22.
Study the diagram showing the entry of HIV into the human body and the processes that are followed:

(a) Name the human cell ‘A’ HIV enters into.
(b) Mention the genetic material ‘B’ HIV releases into the cell.
(c) Identify enzyme ‘C’.
Answer:
(a) The human cell ‘A’ is helper lymphocyte in which HIV virus enters.
(b) Viral RNA is introduced into the cell.
(c) Viral DNA is produced by reverse transcriptase enzyme and then the viral DNA is incorporates into the host genome.

Question 23.
Following a road accident four injured persons were brought to a nearby clinic. The doctor immediately injected them with tetanus antitoxin.
(a) What is tetanus antitoxin?
(b) Why were the injured immediately injected with this antitoxin?
(c) Name the kind of immunity this injection provided.
Answer:
(a) Antitoxins are the preparation containing antibodies to the toxins. Tetanus antitoxins is preparation that contains antibodies that kill Clostridium tetnani. A causal organisms of tetanus.
(b) Tetanus antitoxin is administered to an injured person to neutralize the effect of toxin produced by Clostridium tetani.
(c) Tetanus antitoxin provides a short-term passive immunity to injured person. As it contains preformed antibodies against Clostridium tetani and produces a quick immune response in the patients.

Question 24.
“The population of a metro city experiences fluctuations in its population density over a period of time.”
(a) When does the population in a metro city tend to increase?
(b) When does the population in metro city tend to decline?
(c) If ‘N’ is the population density at the time’t’, write the population density at the time’t + 1′.
Answer:
(a) The size of a population changing with time and it depends on various factors such as availability of food, predation pressure, weather conditions. So the size of population in a metro city tends to increase when all the conditions are favourable. The size of population in a particular habitat fluctuates because of four basic processes such as natality, mortality, immigration and emigration. Out of which Natality (refers to the number of births during a given period of time in a population) and Immigration (refers to the number of individuals of the same species that have come into the habitat from somewhere else in a specific time period) tends to increase the population size in a given locality.

(b) Mortality (refers to the number of deaths in the population during a given period) and Emigration (refers to the number of individuals of a population who left the habitat and gone somewhere else during the time period) tends to decreases the population.

(c) If N is the population density at time t, then its density at time t + 1 is:
N _t+1 = N _t + [(B + I) – (D + E)]
Where B = Natality
I = Immigration
E = Emigration
D = Mortality

Section – E

Question 25.
(a) Describe the process of megasporogenesis, in an angiosperm.
(b) Draw a diagram of mature embryo sac of angiosperm, label its any six parts.
OR
(a) Where and how in the testes process of spermatogenesis occur in humans.
(b) Draw diagram of human sperm and label four parts.
Answer:
(a) The process of formation of megaspores from the megaspore mother cell is called megasporogenesis. The megaspore mother cell undergoes the process of meiosis and forms a four haploid tetrad megaspores. The chalazal megaspore remains functional whereas the other 3 will degenerate. So, the functional megaspore is the first cell of the female gametophyte. The cell enlarges and undergoes three free nuclear miiotic divisions.

So the first meiotic division produces two nucleate embryo sac and two nuclei shift to the two ends and again gets divide and forms four nucleate. In this way, eight nucleate structures is formed. One nucleus from each side moves to the middles and they are called polar nuclei. Then the remaining three nuclei form cells at the two ends, 3-celled egg apparatus at the micropylar end and three antipodal cells at the chalazal end.

(b) Diagrammatic Representation of mature embryo sac in angiosperm:

OR

(a) In the testis, the immature male germ cells or spermatogonia produce sperms by spermatogenesis that begins at puberty. The spermatogonia present on the inside wall of seminiferous tubules get multiply by mitotic division and increases in numbers. Each spermatogonium is diploid in nature as it contains 46 chromosomes.

Some of the spermatogonia are called primary spermatocytes that periodically undergo meiosis. A primary spermatocyte completes the first meiotic division results in the formation of two equal, haploid cells called secondary spermatocytes that contains only 23 chromosomes each.

The secondary spermatocytes undergo the second meiotic division to produce four equal, haploid spermatids. The spermatids are then transformed into spermatozoa or sperms by the process called spermiogenesis. After spermiogenesis, sperm heads become embedded in the sertoli cells and then are finally released from the seminiferous tubules by the process called spermiation.

(b) Diagrammatic Representation of human sperm:

Question 26.
(a) Why did T.H. Morgon select Drosophila melanogaster for his experiments?
(b) How did he disprove Mendelian dihybrid F2 phenotypic ratio of 9 : 3 : 3 : 1? Explain giving reasons.
OR
(a) List any four major goals of Human Genome project.
(b) Write any four ways the knowledge from HGP is of significance for humans.
(c) Expand BAC and mention its importance.
Answer:
(a) T.H Morgan selected Drosophila melanogaster for his experiments because he found that fruit flies are suitable for studies as they could be easily grown on simple synthetic medium in the laboratory.

• They tend to complete their life cycle in about two weeks and a single mating could produce a large number of progeny flies.
• Sexes are clearly differentiated between male and female flies.
• They exhibit several types of hereditary variations that can be observed with low power microscope.

(b) Morgan carried out many dihybrid crosses in Drosophilla to study the genes that were sex-linked. These crosses were similar to the dihybrid cross performed by Mendel in peas. He hybridised yellowbodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F ₁ progent.
Then he observed that the two genes did not segregate independently to each other and the F ₂ ration deviated very significantly from the 9:3:3:1 ratio.

He observed that the genes were located on the X-chromosome and they saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type. According to Morgan, this is because of physical association or linkage of the two genes and coined the term linkage. It used to describe this physical association of genes on a chromosome and is called recombination. It is term used to describe the generation of non-parental gene combinations.

OR

(a) The important goals of Human Genome Project are as follows:

• Identify all the approximately 20,000-25000 genes in human DNA.
• Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
• Store this information in databases
• Improve tools for data analysis.

(b) The four ways by which HGP is significant for humans are as follows:

• Repeated sequences make up very portions of the human genome.
• Repetitive sequences are stretches of DNA sequences that are repeated several times sometimes hundred to thousand times.
• Scientist have identified about l .4 million locations where single base DNA differences called SNPS or single nucleotide polymorphism occur in humans.
• The human genome contains 3164.7 million nucleotide bases.

(c) BAC is bacterial artificial chromosome. BAC is a vector that is used for transforming DNA and cloning in bacteria especially in E.coli.

Question 27.
(a) Name the insect that attacks cotton crops and causes lot of damage to the crop. How has Bt cotton plants overcome this problem and saved the crop? Explain.
(b) Write the role of gene Cry IAb.
OR
(a) Explain the different steps carried out in Polymerase Chain Reaction, and the specific roles of the enzymes used.
(b) Mention application of PCR in the field of
(i) Biotechnology
(ii) Diagnostics
Answer:
(a) Cotton bollworms are the insect that attack and destroy the cotton crops. Some strains of Bacillus thuringiensis produces proteins that kill insects that the cotton crops. They forms protein crystals during a particular phase of their growth. These crystals contain a toxic called insecticidal protein.

This protein exists in an inactive form but once it is ingested by the insect, this protein is converted into an active form of toxin because of alkaline pH of the gut. It tends to solubilise the crystals. Then the activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling as well as lysis and eventually results in the death of the insect.

(b) The gene crylAb controls the growth of cotton com borer. (1 Mark)
OR
(a) With the help of recombinant DNA technology called Polymerase Chain Reaction technique (PCR) multiple copies of gene of interest are obtained In vitro. A single PCR amplification cycle involves three steps which are as follows:
(i) Denaturation: This is the first step of PCR, in which the target DNA is heated at high temperature such 94-96°C. It facilitates the separation of two strands of DNA. Each separated strand of DNA acts as a template for synthesis of DNA.

(ii) Annealing: This is the second step of PCR, in which two oligonucleotide primers are used to hybridize each single stranded template DNA. The sequence of primers is complementary to 3′ end of the template DNA strand.
This step of PCR occurs at low temperature 40- 60°C than denaturation. The annealing temperature depends upon the length and sequence of the primers.

(iii) Extension: This is third and last step of PCR, in which enzyme Taq DNA polymerase synthesizes the DNA between the primers. This step also requires dNTPS and Mg2++. The optimum temperature for extension is 72°C.

(b) (i) Application ofPCRin the field of biotechnology:
Polymerase Chain Reaction is a widely used method in the field of biotechnology as it helps in amplification of billions of copies of desired DNA sequence from small sample of DNA.
(ii) Application of PCR in the field of diagnostic:
Now days, PCR can be used in molecular diagnostic and biochemical analyses. Even with the help of small sequence of DNA, the genotypes can be determined. PCR helps in diagnosis of genetic disorder.

Note
Enzyme used in PCR is a DNA polymerase such as Taq polymerase. This enzyme is stable at high temperature as it is isolated from thermostable bacteria Thermus aquaticus.