CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2011

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

  1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
  2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
  3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
  4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
  5.  Section D contains question number 23, Value Based Question of four marks.
  6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
  7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. Name the embryonic stage that gets implanted in the uterine wall of a human female.
Answer : The blastocyst gets implanted in the uterine wall of a human female.

Question.2. State the importance of biofortification.
Answer: To increase the amount of vitamin, minerals, protein, fat, Micronutrieflt and mineral content of the plants.

Question.3. Biotechnologists refer to Agrobacterium tumifaciens as a natural genetic engineer of plants. Give reasons to support the statement.
Answer : Agrobacterium tumifaciens is a pathogen of several dicot plants. It is able to deliver a piece of its DNA, called Ti (Tumor inducing) plasmid into normal plant cells where it gets incorporated in the host DNA & replicates along with it.

Question.4. How do algal blooms affect the life in water bodies ?
Answer: Algal blooms diminish water quality and deteriorate the water quality by depleting the oxygen content resulting asphyxiation and increase BOD. It also causes fish mortality and extremely toxic to human beings and animals.

Question.5. Name the common ancestor of the great apes and man.
Answer: Ramapithecus and Dryopithecus.

Question.6. Write a difference between net primary productivity and gross productivity.
Answer : Gross Primary (GPP) productivity : It is the total amount of energy captured or he total organic matter or biomass manufactured by the producet by the process of photosynthesis per unit time per unit area.
Net Primary productivity (NPP) : It is the amount of energy or biomass stored by the produces per unit areaper unit time. It is calculated by substracting the amount of energy utilised in respiration from the GPP.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-1

Question.7. Mention the contribution of genetic maps in human genome project.
Answer: Genetic and physical maps act as instruments for the completion of human genome project. Genetic maps were used to study polymorphism among the genes, use of RE (Restriction Endonuclease) to determine specific repetitive DNA sequence commonly present.

Question.8. Name the phase all organisms have to pass through before they can reproduce sexually.
Answer: (i) Juvenile phase (animals) or vegetative phase (plants)
(ii) Reproductive phase .

SECTION-B

Question.9. Name the emyme produced by Streptococcus bacterium. Explain its importance in medical sciences.
Answer : Streptokinase enzyme produced by Streptococcus bacterium. It is extremely useful in medical practice due to its ability to break down blood clots. .

Question.10. How is ‘Rosie’ considered different from a normal cow ?Explain.
Answer : The ‘Rosie’ is first, transgenic cow, Rosie, produced human protein-enriched milk. Milk contained the human gene alpha-lactalbumin and it was nutritionally a more balanced product for human babies than natural cow-milk.

Question.11. State the use of Biodiversity in modem agriculture.
Answer : Use of biodiversity in development of agriculture:-
(i) Modern agricultural technologies increases the productivity per unit area of land, and helps in the conservation and promotes farming of all wild and native varieties of plants.
(ii) Base for our agricultural food chain, development and safeguard of livestock’s etc.
(iii) The use of plant protection products is an important tool to control some of these invasive Which species and protect biodiversity, including that found on the farm.

Question.12. Write the full form of VNTR. How is VNTR different from “Probe”?
Answer : VNTR stands for Variable Number of Tandem Repeats. Probe is a small fragment of DNA or RNA used for identification of genes in biological system. VNTR is a small fragment of DNA containing tandemely repeated sequence, whose number and length vary among chromosome and’ individuals. A probe can be used for VNTR to find out the relationship between the people, to find out the criminals as to confirm parents of a child.

Question.13. Differentiate between benign and malignant tumours.
Answer : Difference between benign and malignant tumours :
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-2
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-3

Question.14. The above graph shows Species-Area relationship. Write the equation of the curve ‘a’ and explain.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-4
Answer: curve ‘a’represent the equation S=C\({ A }^{ Z }\) curve ‘b’ represent 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. The characteristic feature of curve is-
(i) Within a region richness of species increases with exploration of new areas in limit.
(ii) Straight line in the graph represent logarithmic value of species richness.
OR
Differentiate between in situ and ex situ approaches of conservation of biodiversity.
Answer: Difference between in situ and ex situ approaches of conservation of biodiversity
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-5

Question.15. The cell division involved.in gamete formation is not of the same type in different organisms. Justify.
Answer : Cell division results in the formation of gamete, heterogametic species produce male and female gametes. In Monera, Fungi, Algae and Bryophytes gametes are haploid (n), the gametes are produced by mitotic division. While organism such as pteridophytes, gymnosperms, angiosperms and animals including human beings, the parental body is diploid (2n), they undergo reduction division, to produce haploid gametes.

Question.16. Identify the type of the given ecological pyramid and give one . example each of pyramid of number and pyramid of biomass in such cases.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-6
Answer : The given ecological pyramid is the inverted pyramid.
Inverted pyramid of biomass: Lake: Phytoplankton —> Zooplankton —> fishes.
Inverted pyramid of number: Tree —> insects —> birds.

Question.17. Describe the Lactational Amenorrhea method of birth control.
Answer : Lactational Amenorrhea method (LAM) is considered a natural method of birth control based on the fact that higher lactation around the clock decreases menstruation and ovulation does not occur. Female will not get pregnant during the first six months after parturition.

Question.18. Name the type of bioreactor shown. Write the purpose for which it is used.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-7
Answer : The given bioreactor is the simple stirred tank bioreactor.
Its purpose is a large scale production of recombinant protein or enzymes, using microbial plants/animals/human cells. It is usually cylindrical or with a curved base to facilitate mixings of reactor content. The stirrer facilitates mixing and use available oxygen. It contains agitator system, an oxygen delivery system, a foam control system, a temperature control system, pH control system and sampling ports so that small volume of the culture can be withdrawn periodically.

SECTION – C

Question.19. Draw a labelled diagram of the reproductive system in a human female.
Answer:
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-8

Question.20. Branching descent and natural selection are the two key concepts of Darwinian Theory of Evolution. Explain each concept with the help of a suitable example.
Answer : Branching descent : Branching descent is the process of development of a new species from, single common descendant. New developed species became, geographically adapted to a new environment ultimately results in complete development of new species, e.g., Darwins Finches-varieties of Finches arose from grain eaters; Australian marsupials evolved from common marsupial.
Natural selection : Natural selection is a process in which better adapted organisms lead to better adaptation and survival while less adapted organisms gets eliminated in successive stages. Selected organisms reproduce and produce stable genetic quality to sustain the changes, e.g., white moth surviving before the industrial revolution and black moth surviving after the industrial revolution.

Question.21. Scientists have succeeded in recovering healthy sugarcane plants from a diseased one.
(a) Name the part of the plant used as explant by the scientists.
(b) Describe the procedure the scientists followed to recover the healthy plants.
(c) Name this technology used for crop improvement.
Answer: (a) Meristematic cells, apical and axillary
(b) Meristem tissue (sugarcane) is grown to nutrient medium leads in invitro.The tissue proliferates to form undifferentiated mass/callus. This callus formed is transferred to a medium containing growth hormones like auxins and cytokinins.
(c) Tissue culture or micropropagation.

Question.22. (i) Name the enzyme that catalyses the transcription of hnRNA.
(ii) Why does the hnRNA need to undergo changes ? List the changes hnRNA undergoes and where in the cell such changes take place.
Answer: (i) RNA polymerase II.
(ii) hnRNA has non-functional introns in between the functional exons. To remove these, it undergoes changes. The changes that hnRNA undergoes includes:
(a) Capping; Methyl guanosine triphosphate is added to 5′ end.
(b) Tailing: In which poly A tail is added at 3′ end.
(c) Splicing: Through which introns are removed and exons are joined.

Question.23. (i) Write the scientific names of the two species of filarial worms causing filariasis.
(ii) How do they affect the body of infected person(s) ?
(iii) How does the disease spread ?
Answer: (i) Filariasis is caused by organism called Wuchereria, two principal species belong to this category is Wuchereria bancrofti and Wuchereria malayi.
(ii) Filarial worm remain in the body for a long time and develops chronic inflammation. They inhabit lymphatic vessels of lower limbs resulting in the swelling of lower limbs and the disease is called elephantiasis or filariasis. Genital organ also gets affected resulting in deformities’ in its shape and size.
(iii) Transmission of infection generally occurs through bite of female mosquito vectors.

Question.24. Name the genus to which baculoviruses belong. Describe, their role in the integrated pest management programmes.
Answer: Nucleopolyhedrovirus.
Role : Baculoviruses is regarded as natural pest management microorganism. They control only species-specific pest, do not affect non-target organisms or beneficial insects are conserved, they thus aid in IPM problems and there is no negative impact on plants or other animals. These viruses can attack wide range of arthropod and some other insects.

Question.25. Unambiguous, universal and degenerate are some of the terms used for the genetic code. Explain the salient features of each one of them.
Answer : This small combination of amino acids sequence was named as genetic code or codon. Sir Har Gobind Khorana developed chemical method for synthesis of RNA molecule. Some of the important characteristic feature of genetic codes is-
(1) Unambiguous — One codon codes for one amino acid,e.g. AUG(Methionine).
(2) Universal – Codon and its corresponding amino acid are the same in all organisms, eg. Bacteria to human UUU ‘ codes for phenylalanine.
(3) Degenerate – Coding of some amino acids are done by more than one sets of codon therefore they are termed as degenerate, e.g. – UUU and UUC code or phenylalanine.

Question.26. Water is very essential for life. Write any three features both for plants and animals which enable them to survive in water scarce environment.
Answer : Features which enable them to survive in water scarce environments are :
Animal adaptation
(i) Kangaroo rats of North America maintain water requirement by internal fat oxidation.
(ii) Some animals have the ability to concentrate urine so that minimum amount of water is lost during excretion.
(iii) Some have burrowing nature to minimize water loss. Plant adaptation
(i) Desert plants develop thick cuticle and deeply placed stomata to decrease rate of transpiration.
(ii) Use of CAM photosynthetic pathway helps stomata to remain inactive or closed during day time.
(iii) Some desert plants develop spikes to replace leaf so that rate of photosynthesis is done by flat stems.
OR
How do organisms cope with stressful external environmental conditions which are localised or of short duration?
Answer : The following methods are employed by organisms to cope with stressful environmental conditions :
(i) Migrate temporarily from the stressful habitat to a hospitable area.
(ii) Suspend activities
(iii) Form thick walled spores
(iv) Form dormant seed
(v) Hibernate during winter
(vi) Planktons undergo diapause

Question.27. (i) State the consequence if the electrostatic precipitator of a thermal plant fails to function.
(ii) Mention any four methods by which the vehicular air pollution can be controlled.
Answer : (i) Particulate matter is considered very harmful, major step for the removal of particulate matter is the implementation of electrostatic precipitator. They are placed near the exhaust of thermal power plant. Precipitator has an electrode wire with thousands of volts, releasing electrons which get attached to the dust particles (negatively charged). On the other side collecting plates attract charged dust particle, where scrubber cleans gases like sulphur oxide.
(ii) Automobile are major source of air pollution, PCB have suggested that particulate size 2.5 micrometers or less in diameter (PM 2.5) are responsible for causing the greatest harm to human health. Four important methods that can be implemented to reduce vehicular pollution :
(a) Use of CNG, (b) Phasing out of old vehicles,
(c) Use of unleaded petrol, (d) Use of low sulphur fuel

SECTION-D

Question.28. Give reasons why:
(i) Most zygotes in angiosperms divide only after certain amount of endosperm is formed.
(ii) Groundnut seeds are exalbuminous and castor seeds are albuminous.
(iii) Micropyle remains as a small pore in the seed coat of a seed.
(iv) Integuments of an ovule harden and the water content is highly reduced, as the seed matures.
(v) Apple and cashew are not called true fruits.
Answer:
(i) To obtain nutrition from the endosperm for the developing embryo.
(ii) The Groundnut seeds rates are albuminous because the endosperm is completely consumed. Whereas, castor seeds rates albuminous because the endosperm persists.
(iii) For the entry of water and oxygen required for germination.
(iv) To protect the embryo and keep the seed viable, until favourable conditions return for germination.
(v) In apple and cashew, ovary does not take part in fruit formation, instead thalamus contributes to fruit formation.
OR
(a) Draw a labelled diagram of L.S. of an embryo of grass (any six labels).
(b) Give reason for each of the following:
(i) Anthers of angiosperm flowers are described as dithecous.
(ii) Hybrid seeds have to be produced year after year.
Answer:
(a)
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-9
(b) (i) Each anther of angiosperm is dithecous i.e bilobed in nature with two layered protection called theca (dithecous). (ii) Use of hybrid vegetable and crop is growing exponentially in present era. They have significantly increased productivity of the plants with higher nutritive value and because progeny shows segregation and do not maintain hybrid characters.

Question.29. Describe the mechanism of pattern of inheritance of ABO blood groups in humans.
Answer: In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles IA, IB and i. Hence, referred to as multipleallelism. A person possesses any two of the three alleles. Z4 and IB dominate over i. But with each other, Z and Z4 are co-dominant.
Table: The Genetic Basis of Blood Groups in Human Population
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-10
OR
(a) Why is haemophilia generally observed in human males ? Explain the conditions under which a human female can be haemophilia
(b) A pregnant human female was advised to undergo M.T. E It was diagnosed by her doctor that the foetus she is carrying has developed from a zygote farmed by an XX- egg fertilized by Y-carrying sperm. Why was she advised to undergo M.T P.?
Answer : (a) Haemophilia is sex linked recessive disease. It is transmitted from unaffected female carrier to a male child with haemophilia. Y has no allele for this. If male inherits Xh from the mother, he will be haemophilic (with the genotype XhY). If female inherits XhXh, one from the carrier mother and one from her haemophilic father, then she can be haemophilic.
(b) In normal human being fusion of two gamete sperm and egg results in the development of zygote, with 23 pairs of chromosomes, or 46 chromosomes. Gamete develops after meiosis containing one set of chromosome so called haploid (22 autosome) and one sex chromosome. Fusion of sperm (male) and egg (female) gamete forms diploid zygote with 22 pairs of autosome and a pair of sex chromosome. Male (sperm) possess heterogametic sex chromosome X and Y, while female have only hohaogafnetic chromosome XX. Above given problem deals with trisomic condition, nondisjunction abnormality such as Klinefelter syndrome where males have an extra X chromosome. Genotype of which is XXY or sometime it may be XXYY, XXXY. She was advised to undergo MTP since the child will have the following problems :
(i) Male with feminine traits
(ii) Gynaecomastia
(iii) Underdeveloped testes
(iv) Sterile

Question.30. (i) Describe the characterisucs a cloning vector must possess,
(ii) Why DNA cannot pass through the cell membrane ? Explain. How is a bacterial cell made “competent” to take up recombinant DNA from the medium ?
Answer: (i) Characteristics features of cloning vector must have:
(a) Presence of selectable marker genes encoding for an antibiotic resistant or genes encoding for a-galactosidase.
(b) Cloning site or recognition site for convenient insertion and removal of plasmid DNA, by the use of RE.
(c) Easy propagation and maintenance of the clone.
(d) Ori or origin of replication
(ii) DNA is a hydrophilic molecule, therefore it cannot pass through the cell membrane. Bacterial cells are made competent by:
(a) Treating bacteria with specific concentration of divalent cation results in increase efficiency of DNA to move inside bacterium cell wall.
(b) Followed by the incubation of cell with recombinant DNA on ice, then place them briefly at 42°C (heat shock), and again back on ice.
OR
If a desired gene is identified in an organism for some experiments, explain the process of the following :
(i) Cutting this desired gene at specific location .
(ii) Synthesis of multiple copies of this desired gene Answer : (i) Cutting of the desired gene at specific location is attained by the implementation of restriction enzymes (RE). Firstly, the restriction endonucleases that recognise the palindromic nucleotide sequence of the desired gene is identified. These enzymes are specialized to cut the fragment of DNA at specific locations. It cuts each of the double helix . at a specific point which is a little away from the centre of the palindromic site. The cutting site is between the same two bases on the opposite strands. This results in over¬hanging single stranded stretches which act as sticky ends, (ii) Multiple copies of the desired gene is synthesised by polymerase chain reaction (PCR) method. In this method, the desired gene is synthesised in vitro. The double stranded DNA is denatured using high temperature of 95°C arid the strands are separated. Each separated strand acts as template. Two sets of chemically synthesised oligonucleotides (primers) and DNA polymerase are being used in vitro for the multiplication of desired gene. The thermostable Taq polymerase extends the primers using nucleotides provided in the reaction mixture.

SET-II

SECTION-A

Question.3. Why is it essential to have a ‘selectable marker’ in a cloning vector ?
Answer : Selectable markers are essential to identify and eliminate non-transformants, and selectively permiting the growth of the transformants.

SECTION – B

Question.9. Why are some molecules called bioactive molecules? Give two examples of such molecules.
Answer : Bioactive molecules are those molecules which are biologically active. Because microbes like bacteria or fungi are used in their production.
e.g., Citric acid produced by a fungus Aspergillus niger.
Acetic Acid produced by a bacteria Acetobacter aceti.

Question.13. List the two types of immunity a human baby is bom with. Explain the differences between the Jyvo types.
Answer : The two types of immunity are innate and passive/ acquired immunity. Innate immunity is a non-specific type of defense that provides a barriers to the entry of antigens. Passive immunity is a pathogen-specific type of defense that develops in response to encounter with pathogen. The foetus receives antibodies through the placenta.

Question.16. Explain the response of all communities to environment over time.
Answer : Environmental factors like temperature, water, light, soil, etc., may influence the members of communities in varying degrees. Organisms in response to these factors shall try to adapt according to their capacities. This change is orderly,’sequential and parallel with the changes in the physical environment. In this process, they may also try to maintain a constant internal environment through homeostasis or migrate to a less stressful environment or may even suspend activities till favourable conditions return. These changes lead to a community is in real equilibrium with the environment called the climax community.
(i) The cow is administered with FSH to induce follicular maturation and super-ovulation to produce 6 to 8 eggs.
(ii) The animal is either mated with an elite bull or artificially inseminated.
(iii) The fertilised eggs 8-32 cells stage are recovered non- surgically and transferred to surrogate mother where they develop into an improved variety.
(iv) This technology is being used to get high milk yielding females and to increase herd size in short time in cattle.

Question.23. (a) Name the causative agent of typhoid in humans.
(b) Name the test administered to confirm the disease.
(c) How does the pathogen gain entry into the human body?
Write the diagnostic symptoms and mention the body organ that gets affected in severe cases.
Answer : (a) The causative agent of typhoid in humans is Salmonella typhi.
(b) The test administerd to confirm the disease is Widal test.
(c) The pathogen gains entry via small intestine through contaminated food and water and migrate to other organ through blood.
The symptoms include sustained high fever (30°C to 40°C), weakness, stomach pain, constipation, headache, loss of appetite’.

Question.25. (a) Name the scientist who called t-RNA an adaptor molecule.
(b) Draw a clover leaf structure of t-RNA showing the following:
(i) tyrosine attached to its amino acid site.
(ii) anticodon for this amino acid in its correct site (codon for tyrosine is UCA).
(c) What does‘the actual structure of t-RNA look like ?
Answer : (a) Francis Crick (b)
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-11
(c)the actual structure of tRNA looks like inverted L.

SECTION-C

Question.21. Describe the technology that has successfully increased the herd size of cattle in a short time to meet the increasing demands of growing human population.
Answer: Multiple ovulation embryo transfer technology (MOET) has successfully increased the herd size of cattle.

SECTION-D

Question.30. (a) With the help of diagrams show the different steps in the formation of recombinant DNA by action of restriction endonuclease enzyme EcoRI.
Name the technique that is used for separating the fragments of DNA cut by restriction endonucleases.
Answer: (a)
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-12
Steps in formation of recombinant DNA by action of restriction endonuclease enzyme EcoBl
(b) Gel electrophoresis is used for separating the fragments of DNA cut by restriction endonucleases.
OR
(a) Name the source from which insulin was extracted earlier. Why is this insulin no more in use by diabetic people ?
(b) Explain the process of synthesis of insulin by Eli Lilly Company. Name the technique used by the company.
(c) How is the insulin produced by human body different from the insulin produced by the above mentioned company?
Answer: (a) Earlier, insulin was extracted from pancreas of slaughtered catde and pig. However, this insulin use caused some patients to develop an allergic reaction to this foreign protein.
(b) Eli Lilly used the following producer for insulin synthesis :
(i) Two DNA sequences corresponding to A and B chains of insulin were prepared.
(ii) These sequences were then introduced in plasmids of E. coil.
(iii) The two insulin chains are produced separately.
(iv) The two chains are extracted and combined with creating disulphide bonds to form the assembled mature molecule of insulin.
(c) The pro-hormone (like a pro-enzyme) synthesiszed in the human body has an extra stretch of C peptide.

SET -III

SECTION-A

Question.1. What stimulates pituitary to release the hormone responsible for parturition ? Name the hormone.
Answer: The signal from the fully developed foetus and placenta or the foetal ejection reflex induces mild uterine contraction for parturition. The responsible hormone is oxytocin.

Question.2. Pollinating species of wasps show mutuajism with specific fig plants. Mention the benefits the female wasps derive from the fig trees from such an interaction.
Answer :The female wasp uses the fruit as oviposition and developing seeds for nourishing its larvae.

SECTION-B

Question.9. A relevant portion of P – chain of haemoglobin of a normal human is given below:
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-13
The codon for the sixth amino acid is GAG. The sixth codon GAG mutates to GAA as a result of mutation A’ and into GUG as a result of mutation ‘B’. Haemoglobin structure did not change as a result of mutation A’ whereas haemoglobin structure changed because of mutation ‘B’ leading to sickle shaped RBCs. Explain giving reasons how could mutation ‘B’ change the haemoglobin structure and not mutation A’.
Answer: Due to mutation A, GAG mutates to GAA. Both GAG and GAA code for glutamic acid and hence there is no change in RBCs. Whereas GUG formed due to mutation ‘B’ codes for valine and so the RBCs become sickle-shaped.

Question.10. Biopiracy should be prevented. State why and how.
Answer: Biopiracy is unauthorized exploitation of bioresources of developing or under-developed countries. There has been growing realization of the injustice, inadequate compensation and benefit sharing between developed and developing countries. Hence, it should be prevented. It can be prevented by developing laws to obtain proper authorization and by paying compensatory benefits.

Question.13. Why is tobacco smoking associated with rise in blood pressure f and emphysema (oxygen deficiency in the body) ? Explain.
Answer : Tobacco has nicotine, an alkaloid, that stimulates the release of adrenaline and noradrenaline which raise blood pressure and increase the heart rate. Smoking tobacco releases carbon monoxide which reduces the concentration of haem-bound oxygen. This causes emphysema.

Question.18. What is polyblend ? Why did the plastic manufacturers think of producing it ? Write its usefulness.
Answer : Polyblend is a fine powder of recycled modified plastic. Polyblend was produced to recycle plastic waste. It was developed by Ahmed Khan, a plastic manufacturer to solve the ever increasing problem and accumulation of waste. Polyblend can be used to lay roads that have increased road life. When blended with bitumen, it enhances the bitumens water repellent properties and increase the life of road.

SECTION-C

Question.23.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2011-14
Study the diagram showing replication of HIV in humans and answer the following questions accordingly:
(i) Write the chemical nature of the coat ‘A’.
(ii) Name the enzyme ‘B’ acting on ‘X’ to produce molecule ‘C’. Name ‘C’.
(iii) Mention the name of the host cell ‘D’. the HIV attacks first when it enters into the human body.
(Iv) Name the two different cells the new viruses ‘E’ subsequendy attack.
Answer: (i) Coat A’ is made up of protein.
(ii) The enzyme ‘B’ is reverse transcriptase, ‘C’ is viral DNA.
(iii) The host cell ‘D’ is macrophage.
(iv) The new viruses ‘E’ subsequendy attack macrophages and helper T-lymphocytes.

Question.25. Answer the following questions based on Messlson and Stahl’s experiment:
(a) Write the name of the chemical substance used as a source of nitrogen in the experiment by them.
(b) Why did the scientists synthesise the light and the heavy DNA molecules in the organism used in the experiment ?
(c) How did the scientists make it possible to distinguish the heavy DNA molecule from the light DNA molecule ? Explain.
(d) Write the conclusion the scientists arrived at after completing the experiment.
Answer: (a) Ammonium chloride (NH4CI).
(b) To check if DNA replication was semi-conservative.
(c) The heavy and light DNA molecules were distinguished by centrifugation in a Cesium chloride density gradient.
(d) The scientists concluded that DNA replicates semi- conservatively.