CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2014

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.



Question.1. Name the part of the flower which the tassels of the corn-cob represent.
Answer : Female reproductive parts-are: stigma and style.

Question.2. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Answer: The two contrasting traits with respect to seeds in pea plant that were studied by Mendel are :
(i) Seed shape: round and wrinkled.
(ii) Seed colour: yellow and green.

Question.3. Why is secondary immune response more intense than the primary immune response in human?
Answer: The primary immune response to antigen occurs on the first occasion and generate memory B and T cells with a high specificity for the inducing antigen. The secondary response, mediated by B cells with the help of T cells, quickly produces high-affinity and antigen-specific antibodies against pathogens.

Question.4. Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally?
Answer : Alien DNA requires specific sequence called recognition sites (sites where restriction enzymes cut DNA) to ligate itself with host chromosome. ‘ Recognition sites sequence should be close to the origin of replication (ori sequence where DNA replication starts). This site is necessary for the binding of DNA polymerase to start replication. As this site is not be present in alien DNA molecules, so an alien piece of DNA cannot replicate normally by attaching to any DNA.

Question.5. State the role of C peptide in human insulin.
Answer: Human insulin is produced as a pro-hormone. The 31 amino acid C-peptide of proinsulin is important for the biosynthesis of insulin. It helps in maintaining the level of active insulin.

Question.6. Name the enzymes that are used for the isolation of DNA from bacterial and fungal cells for recombination DNA technology.
Answer: Enzymes used to isolate DNA from bacteria: Lysozyme Enzymes used to isolate DNA from fungi: Chitinase

Question.7. State Gause’s Competitive Exclusion Principle.
Answer : The principle that when two species compete for the same resources within an environment, one of them will eventually outcompete and displace the other. The displaced species may become locally extinct, by either migration or death.

Question.8. Name the type of association that the genus Glomus exhibits with higher plants.
Answer : Mycorrhizal is the type of association.


Question.9. Why are the human testes located outside the abdominal cavity ? Name the pouch in which they are present.
Answer : Testes lie in the scrotum outside the abdominal cavity. It is so as it keeps the temperature l-3degree C(approx. 35°) less than that of the body (i.e about 37°C). A lesser temperature is required for the sustenance of sperm which is provided by the scrotum.

Question.10. In Snapdragon, a cross between true-breeding red flowered (RR) plants and true-breeding white flowered (rr) plants showed a progeny of plants with all pink flowers.
(a) The appearance of pink flowers is not known as blending.Why?
(b) What is this phenomenon known as ?
(a) The appearance of pink flower is not known as blending because it is due to partial influence of allele for white colour over the allele for red colour. On self-crossing the Fj plants, the F2 progeny are three types of plants-red flowered, pink flowered & white flowered in the ratio of 1 : 2 : 1. The occurrence of red & white flowered plants in F2 generation, indicate that the two alleles (red & white flower colour) and not blended but partially expressed as pink flower plants.
(b) The above example is known as incomplete dominance.

Question.11. With die help of one example, explain the phenomena of co-dominance and multiple allelism in human population.
Answer: A condition in which two different alleles for a genetic trait in a heterozygote are fully expressed thereby resulting in offspring with a phenotype that is neither dominant nor recessive.
When three or more alternative forms of a particular gene existing in a population, it is called multiple allelism.
Example : A typical example showing co-dominance is the ABO blood group system. For instance, a person having IA allele and IB allele will have a blood type AB because both the IA and IB alleles are co-dominant with each other.
ABO blood group is controlled by I gene. The gene I has 3 different alleles IA, IB and i. IA and IB produce two different types of sugars on the plasma membfane of red blood cells. The gene I does not produce any sugars. IA and IB are completely dominant over i. When IA and’lB’ are present together, they express their own type of sugars.

Question.12. Write the scientific name of the fruit-fly. Why did Morgan
prefer to work with fhiit-flies for’his experiments ? State any three reasons.
Linkage and crossing-over of genes are alternative of each other. Justify with the help of an example.
Answer : Scientific name of fruit-fly: Drosophila melanogaster. Morgan used fruitfly for his experiment because :
(i) The fruit-fly could be grown on a simple synthetic medium inside the laboratory
(ii) The life cycle of a fruit-fly is about only two weeks.
(iii) A single mating could produce a large number of progeny offsprings.
Answer : (i) There is some linkage between all genes located on the same chromosome. The linkage strength depends on the percentage of the distance between the two, But linkage can be easily broken by crossing over.
(ii) When genes located on the same chromosome, then . there is possibility of two situations, either a crossing
over between the two genes or no crossing between two genes.
(iii) Crossing over always occurs if genes are located very far from each other – 50% recombinants, 50% parental. Example :Morgan hybridized yellow-bodied, white-eyed females to brown-bodied, red eyed males and intercrossed their F progeny. He found that the genes for white and yellow were very lightly linked and showed only 13% recombinant while white and miniature wing showed 37.2% recombination.

Question.13. List of symptoms of Ascariasis. How does a healthy person acquire this infection?
Answer: Symptoms of Ascariasis include : Worms in stool, coughing up worms, loss of appetite, fever. Severe symptoms of Ascariasis include : Vomiting, shortness of breath, swelling of the abdomen, severe stomach pain, and intestinal blockage.
Mode of Transmission:
(1) It is transmitted by improper disposal of human stool containing the eggs of Ascaris.
(2) Healthy persons may get infection from contaminated water, vegetables, fruits, other food articles & fomites.

Question.14. Explain the significant role of the genus Nucleopolyhedrovirus in an ecological sensitive area.
Answer: The nucleopolyhedrovirus, a sub group of Bacu- loviruses is a virus. It affects insects, predominantly moths and butterflies and used as a biological control agent. It has been used as a pesticide for crops infested by insects specially arthropods. Though this virus is species specific, making it effective under certain circumstances and there no negative effect on plants, mammals, birds, fish or other is non-target insects.

Question.15. How does a restriction nucleases function ? Explain.
Answer : Restriction nucleases are of two different types- endonucleases cut at a specific position inside DNA strand. Exonuclease remove nucleotides from the end of a DNA. strand. Restriction endonucleases recognize short, usually palindromic (meaning the base sequence reads the same backwards and forwards), sequences of 4—8 bp and, in the presence of Mg2+, cleave the DNA within or in close proximity to the recognition sequence. For example, EcoRI digestion produces “sticky” ends.
Whereas, Smal restriction enzyme cleavage produces “blunt” ends:

Question.16. How have transgenic animals proved to be beneficial in:
(a) Production of biological products (b) Chemical safety testing.
Answer : (a) Production of biological products :
The transgenic farm mammal was produced, a sheep called ‘Rosie cow’, had a human gene that expressed high levels of the human protein alpha-1-antitrypsin. The protein, which missing in humans, can lead to a rare form of emphysema.
(b) Chemical safety testing Transgenic animals, toxicity-sensitive transgenic animals have been produced for chemical safety testing. Transgenic animals can also be used to test the identity and purity of human proteins used as drugs. A transgenic » animal that makes a human protein (e g human insulin) will recognise this substance as its own and will therefore not produce an immune response against it.

Question.17. Describe the mutual relationship between fig tree and wasp and comment on the phenomenon that operates in their relationship.
Answer : Mutual relationship : Fig tree and wasp shows mutualism between them. The interaction in which both the interacting species get benefit of each other called mutualism. Fig flower is pollinated only by wasp and not by any other species. Female wasp lays eggs inside the developing fruit and also uses the developing seeds within the fruit for nourishing its larvae. Co-evolution exists between their close specific tight relationship.

Question.18. Construct an age pyramid which reflects an expanding
growth status of human population.
Expanding pyramids of human population :

A population at any given time is composed of different age groups. These three groups include :

  1.  Pre-reproductive
  2. Reproductive
  3. Post-reproductive
  4.  If the age distribution (percent individuals of a given age or age group) is plotted for the population, the resulting structure is called an age pyramid.
  5.  In human – beings, the age pyramids show the age distribution of male and female in a combined diagram.
  6. In expanding pyramid, individuals in reproductive age group are more in number so the pyramid is expanding.


Question.19. Make a list of any three outbreeding devices that flowering plants have developed and explain how they help to encourage cross-pollination ?
Why are angiosperm anther called dithecous ? Describe the structure of its microsporangium.
Answer : The three outbreeding devices to encourage, cross-pollination
(i) Protoandry : The pollen grain and stigma of the flower mature at two different times, so that pollen release and stigma receptively are not simultaneous.
(ii) Protogyny : Mechanical barrier on the stigmatic surface of flowers, so that the anther and stigma cannot come in contact with each other of same flower.
(iii) Self incompatibility : The receptive stigma retard the growth of the pollen tube of fallen mature pollen grains of the same flower.
Angiosperm anther is bilobed. Each lobe has two theca (microsporangium) so it is known as dithecous. Structure of microsporangium: The transverse section of a typical microsporangium are circular in outline. The microsporangium surrounded by four separate wall layers: epidermis, endothecium, middle layers and tapetum. The innermost wall layer tapetum provide nourishment to developing pollen grains. Tapetum cells are multi-nuclei ‘ and have dense cytoplasm. The outer three wall layers perform the function of protection and help iif dehiscence of anther to release the pollen. When the anther is young, a group of compactly arranged homogeneous cells called the sporogenous tissue occupies the centre of each microsporangium.

Question.20. If implementation of better techniques and new strategies are required to provide more efficient care and assistance to people, then why is there a statutary ban on amniocentesis ? Write the use of this technique and give reason to justify the ban.
Answer: Amniocentesis is a prenatal technique of diagnosing the genetic & metabolic disorders of the foetus by taking out a small quantity of amniotic fluid. Amniotic fluid contains foetal cells, placental cell, foetal enzymes, proteins & other biochemicals. Foetal cells give information about the sex of the foet us and any abnormality in the chromosomes. It the foet us suffers from in curable genetic & metabolic disorders, then the foetus needs to be aborted through MTP.
But this very useful technique has been misused to know the sex of the developing foetus & destroying the same if the foetus is female. Therefore, the test has been banned except it few centres & this ban is justified.

Question.21. Why is pedigree analysis done in the study of human genetics ? State the conclusions that can be drawn from it.
Answer : Pedigree analysis is the study of family history about the inheritance of a particular trait. It can be used to draw the inheritance of a specific trait, abnormality or disease in humans because control crosses are not possible in case of human being.

  1. Identification of the recessive or dominant nature of a specific trait could be done by pedigree analysis.
  2.  The trait is linked to sex chromosome or autosomal can be find out by pedigree’ analysis, for example, haemophilia is a sex linked recessive disease. X-linked recessive trait shows transmission from carrier female to – male progeny.
  3.  The pattern of inheritance of Mendelian disorders can be traced in a family by pedigree analysis. For example, most common Mendelian disorders are haemophilia, cystic
    fibrosis, sickle cell anaemia, colour blindness, phenylketonuria, thalassemia, myotonic dystrophy (autosomal dominant trait), etc.

Question.22. Identify ‘a’, ‘b’, ‘c’, ‘d’, ‘e’, and ‘f’ in the table given below:
Answer: a. (i) Palm is broad with characteristics palm crease; short statured with small round head.
(ii) Physical, mental, psychomotor development is retarded,
(b) Both (c) Klinefelter’s syndrome (d) Male
(e) (i) Short stature and underdeveloped feminine character,
(ii) Such females are sterile as ovaries are rudimentary. They also do not have well developed secondary sexual characters.
(f) Female

Question.23. Community service department of your school plans a visit to a slum area near the school with an objective to educate the slum dwellers with respect to health and hygiene.
(a) Why is there a need to organise such visits ?
(b) Write the steps you will highlight, as a member of this department, in your interaction with them to enable them to lead a healthy life.
Answer: (a) The well-being of each human being depends on their environment. In slum areas individuals live in congested and insanitary conditions. In this type of conditions they are more susceptible to suffer from diseased condition. So there is a need to organize such visits to educate them about the importance of health and hygiene.
(b) Steps to enable slum dwellers guide to healthy life

  1.  Use of mosquito nets while sleeping, get wire mesh fixed to doors and windows, prevent water logging, regularly change water of water-coolers to avoid mosquito breeding.
  2.  Wash hands before eating and after toilet use, maintain the environment clean so that flies do not breed. Disinfect water by chlorine tablets if it is drawn from well or any other source.
  3. Clean toilets and use disinfectants regulary.
  4. Educate people about the benefit of vaccine which are available at the health centres such as DPT for diphtheria, pertusis (whooping cough) and tetanus, polio vaccine. MMR vaccine for measles, mumps, rubella.

Question.24. The following graph shows the species —area relationship. Answer the following questions as directed.
(a) Name the naturalist who studied the kind of relationship shown in the graph. Write the observations made by him.
(b) Write the situations as discovered by the ecologists when the value of ‘Z’ (slope of the line) lies between
(i) 0.1 and 0.2 (ii) 0.6 and 1.2
What does ‘Z’ stands for ?
(c) When would the slope of the line ‘b’ become steeper ?
Answer : (a) Species area relationship was studied by Alexander Von Humboldt. He made a observation that within a region, species richness increased with increasing explored area but only upto a limit.
(b) (i) Z = 0.1 to 0.2: the stope of regression lines are similar the slope of regression is steepel when are analyse the species area relationship among very large areas like entire countinent.
(ii) Z = 0.6 to 1.2 : for large area for example entire continent.
(c) The slope of the line b become steeper when species area relationship is analyzed in a very large area like the entire continents.

Question.25. Name and describe the technique that helps in separating the DNA fragments formed by the use of restriction endonuclease.
Answer: Agarose gel electrophoresis is used to separate DNA * fragments formed by restriction endonuclease.
Agarose gel electrophoresis : The DNA cleavage by restric¬tion endonucleases which results in DNA fragments. Electrophoresis is a technique used to separate and sometimes purify nucleic acids that differ in size, charge or conformation. As such, it is one of the most widely-used techniques in biochemistry and molecular biology. When DNA molecules are placed in an electric field. DNA molecules are negatively charged due to their phosphate backbone, and migrate toward the anode. The DNA fragments separate (resolve) according to their size through sieving effect’ provided by the agarose gel. Hence, the smaller the fragment size, the further it moves. The separated DNA fragments can be visualized only after staining the DNA with a compound # known as ethidium bromide followed by exposure to UV radiation (pure DNA fragments cannot be seen in the visible light and without staining). The bands are cut from the gel and extracted by using a convenient technique. This step is called elution. The eluted DNA fragments are then purified and used in constructing recombinant DNA by joining them with cloning vectors.

Question.26.State the function of a reservoir in a nutrient cycle. Explain the simplified model of carbon cycle in nature.
Answer : Function of a reservoir : To meet with the deficit which occurs due to imbalance in ttie rate of influx and efflux of nutrients.

Question.27. Since the origin of life on Earth, there were five episodes of > mass extinction of species.
(i) How is the ‘Sixth Extinction’, presently in progress, different from the previous episodes?
(ii) Who is mainly responsible for the ‘Sixth Extinction’ ?
(iii) List any four points that can help to overcome this disaster.
Answer : (i) The current extinction “sixth extinction” rates are estimated to be 100 to 1000 times faster than in pre human times.
(ii) Human activities in ecosystem are mainly responsible for sixth extinction.
Main reason for this extinction is :
(a) Habitat loss and fragmentation, (b) Over exploitation
(c) Alien species introduction (d) Co extinction
(iii) Afforestation: Creation of sacred groves in which all the trees and wild life are venerated and given total protection.
(iv) By preventing habitat loss : Zoological park, botanical gardens, wildlife sanctuaries can also help to overcove such extinction.
(v) By the use of Diverse species.
(vi) By in-situ conservation & ex-situ conservation.


Question.28. (a) Where does fertilization occur in humans ? Explain the events that occur during this process.
(b) A couple where both husband and wife are producing
functional gametes, but the wife is still unable to conceive, is seeking medical aid. Describe any one method that you can suggest to this couple to become happy parents.
Answer: (a) In humans fertilisation of male & female gamete occurs in the junction of ampulla and isthmus of fallopain tube. The various events which occur during the fusion of gametes are :
(A) Acrosomal reaction : As sperm comes in contact with the egg surface, it secretes/release enzyme Hyaluronidase which dissolves the corona radiata.
(ii) As sperm reaches Zona pellucida, the acrosome release Acrosin/zona lysin and dissolves zona pellucida.
(iii) Compatibility reaction also stimulates development of an outgrowth by the oocyte called Fertilisation cone. Egg cell has fertilizin protein and sperm has antifertilizing protein.
(B) Sperm Entry : Sperm head comes in contact with the fertilisation cone. Sperm & egg membrane dissolve at this point and components of head (nucleus), neck and middle piece of sperm enter the cytoplasm of egg. Tail of sperm is left out.
(C) Zona Reaction : The zona pellucida stiffens after ,, entry and does not allow any other sperm to enter. The phenomenon called Monospermy.
(D) Activation of oocyte to ovum : Egg is secondary ” oocyte stage undergo meiosis II by removal by MPF & development of APC/APF resulting in mature ovum/ ootid and second polar body.
(E) Karyogamy : It is the find stage of fertilisation. The sperm nucleus fuses with the egg nucleus. Nuclear envelops breakdown forming a spindle & thus form the zygote under laboratory condition. The zygote or early embryo is transferred in the fallopian tube.
(b) Couple able to produce functional gamete but unable to conceive can assist to have children through one of following techniques commonly “called as — Assisted Reproductive Technologies (ART).
In vitro fertilization followed by embryo transfer : Ova from the wife/donor and sperms from the husband/ donor male are collected and fused to form zygote under laboratory condition. The zygote or early embryo is transferred in the fallopian tube.
(a) Explain the different ways apomictic seeds can develop.
Give an example of each.
(b) Mention one advantage of apomictic seeds to farmers.
(c) Draw a labelled mature stage of a dicotyledonous embryo.
Answer : (a) Different ways apomictic seeds development;
(i) The diploid egg cell is formed without reduction division and develops into an embryo without fertilization. Example : Grasses/Asteracease.
(ii) Nuclear cells surrounding the embryo sac start dividing and protrude into the embryo sac and develop into the embryo, for example, citrus and mango. They have more than one embryo in a seed known as polyembryony.
(b) Advantage of apomictic seeds to farmers :
As apomictic seed formation does not involves meiosis and fertilization, they are genetically identical to their parents.
If the hybrid seeds become apomictic they will maintain their traits generation after generation. As does not involves meipsis so lack of segregation of characters & not involves fertilization so no recombination and trait will be maintained for several generations, so the farmers can use these apomictic seeds to raise new crop year after year.

Question.29. (a) Describe the various steps of Griffiths experiment that led to the conclusion of the ‘Transforming Principle’.
(b) How did the chemical nature of the ‘Transforming Principle’ get established ?
Describe how the lac operon operates, both in the presence and absence of inducer in E.coli.
Answer : (a) Transformation is change of genetic material of an organism by obtaining genes from other organism (dead relative). Fredrick Griffith a British bacteriologist in 1928, carried out experiments on ‘Transforming principle’. He worked with strains of streptococcus pneurnonical. These are two strains of this bacteria :
(i) Virulent or S. Strain, which produces smooth colony and has the capacity to cause the disease (pneurnonical).
(ii) Non-virulent or R-strain, which produces rough colony and does not cause pneumonia.
Griffith’s experiment was carried out as follows:

  1. R-type (strain) of live bacteria injected in the mice No disease observed in mice.
    Mice + R-strain (live) ——-> No disease in mile
  2.  Live S-strain of bacteria injected in the mice-mice has occurence of pneumonia & dies. .-
    Mice + S-strain (live) ——–> Disease seen & mice dies
  3.  Heat-killed s-strain bacteria injected in mice.
    No disease observed. Mice survives
    Mice + s-strain (Heat-killed) No disease Mice Survives
  4.  Mice injected with a mixture of heat killed s-strain of bacteria and live R-strain. Mice dies of pneumonia.
    Mice + s-strain (heat killed) ——–> Disease occurs + R-strain (live) Mice dies
    On observing the blood of mice, it showed presence of both R-strain & S-strain live bacteria. Occurrence of live s-strain was possible only th rough a change transformation of R-strain into s-strain through transfer of biochemical substance.

(b) Oswald TAvery, Collin Macleod & Maclyn McCarty in 1944 established the chemical nature of transforming principle.

  1. The heat killed s-strain of bacteria and separated their’ components – DNA, proteins & Carbohydrates
  2. The DNA component was segregated into two. One with hydrolysing enzyme DNA-ase & the other without it.
  3. They then mixed these components of s-strain with live R-strain in Separate culture media.
  4.  There was no change in three culture having additions of neat- killed s-stiain carbohydrates, proteins & DNA (with DNAase).
  5. But the fourth culture medium having neat-killed s-strain DNA without DNAase showed presence oflive s-strain bacteria.
    It was concluded that the live S-strain bacteria must have been formed from R-strain with the help of DNA of S-strain. Thus, DNA is the genetic material was established.

In E.Coli, the breakdown of lactose requires three enzymes. These enzymes are synthesized together in a coordinated manner & the unit is known as lac operon. Since the addition of lactose itself stimulates the production of lactose itself stimulates the production of required enzymes. It is also referred as Inducible system. It gets switched off in normal conditions.
The genes involved in lac operon are as follows :

  1.  Strcutural Genes : These genes code for the proteins needed by the cell which include enzymes or other proteins having structural functions. Lac operon has three structural genes :
    (a) Lap z : Gene coding for enzyme b-galactosidase for splitting lactose into glucose & galactose.
    (b) Lacy y : Gene coding for enzyme permease/Galactoside permease which is required for entry of lactose.
    (c) Lac a ; Gene coding for enzyme Transacteylase/ Galactoside acetylase.
    The three structural genes of lac operon produce a single polycistronic mRNA.
  2. Operator gene (O): It gives passage to RNA polymerase when the structural genes are to express themselves. Normally, it is covered by a repressor & is in off position.
  3.  Promoter gene (p) : This, gene is the recognition centre/ initiation point for RNA polymerase of the operon.
  4.  Regulator gene (i) : It is also called inhibitory gene. It produces a repressor protein that binds the operator gene, when the substrate (lactose) is not available, so as to keep it non-functional. It prevents the passage of RNA polymerase from promoter to structural gene.
  5.  Repressor (p) : It is a small portion formed by regulator gene which binds to operator gene & blocks the passage of RNA polymerase towards the structural genes. It has two allosteric sites, one for attaching to operator gene & second for binding to the inducer.
  6.  Inducer : It is a chemical which attaches to repressor, and changes the shape of operator binding sites so that the repressor rem ains no more attached to the operator. Mechanism of Lac Operon :
    (I) In the absence of induce (lactose) : The lac operon is generally off which is ensured by the formation of repressor by the regulator gene which blocks the operator gene. Thus, there is no transcription & no enzymes are produced. Operon is switched aff.
    (II) On the other hand, when inducer (lactose) is added, the repression protein (produced by gene i) gets bound is removed from the operator. RNA Polymerase is now allowed to act & the transcription of lac genes take place. The operon is now switched ON. All the three genes are transcribed to form a single mRNA strand. It is a polycistronic mRNA. This process continues till the inducer .is consumed. Once inducer finishes, the repressor again blind the operator gene & switches OFF the operon.

Question. 30. With advancements in genetics, molecular biology and tissue culture, new traits have been incorporated into crop plants.
Explain the main steps in breeding a new genetic variety of a crop.
(a) State the objective of animal breeding.
(b) List the importance and limitations of inbreeding. How can the limitations be overcome ?
(c) Give an example of a new breed each of cattle and poultry.
Answer : Different steps in breeding a new crop variety.

  1. Collection of variability: Genetic variability is essential for breeding program. If genetic variability is not present than new variety can not be develops thus it is pre requisite condition for breeding. The collection of all the different alleles for all genes in a given crop is called germplasm collection.
  2.  Evaluation and selection of parent: Different germplasm is evaluated for desired trait and plants having the desired character are selected as parent. The selected plants are multiplied and pure lines obtained, which are used for hybridization.
  3.  Cross hybridization among the selected plant : The selected plants are hybridized to combine the character of two different parents.
  4.  Selection and testing of superior recombinants : On the basis of presence of desired character in hybrid, superior recombinants are selected. Plants are then self pollinated for several generations.
  5. Testing, release and commercialization of new cultivars: These new recombinant are evaluated for their yield and different agro climatic condition (such as quality and disease resistance) for several years along with best available local check variety. If these lines are superior than local check then they are released for commercial cultivation. .

(a) Objective of animal breeding: To increase the yield of animal & improving the desirable qualities of product.
(b) Importance of Inbreeding :

  1.  Superior male & superior female of same breed are identified for mating.
  2.  To evolve a pure line of animal.
  3. Exposes harmful recessive gene that are eliminated by selection.
  4.  Also accumulates superior genes & elimination of less desirable gene.
  5.  Increases the productivity of inbreed population. Limitation of Inbreeding :
    Continued inbreeding specially closed inbreeding . usually reduces fertility & productivity. This is called as inbreeding depression.

(c) New breed of cattle —> Hisardale, and New breed of poultry —> Leghorn.



Question.1. Why is Gambusia introduced into drains and ponds ?
Answer: Gambusia introduced into drains and ponds because they feed on mosquito larvae.

Question.7. Why are analogous structures a result of conveigent evolution ?
Answer : Analogous structures are not anatomically similar though they perform similar functions so they are a result of convergent evolution.

Question.8. Name the vegetative propagules in the following:
(a) Agave (b) Bryophyllum
Answer : (a) Agave : Bulbils.
(b) Bryophyllum : nodes (leaves).


Question.11. State the difference between the structural gene in a Transcription Unit of Prokaryotes and Eukaryotes.
Answer : Prokaryote structural genes consist of only exons (functional) while eukaryotes consists of both introns and exons. Introns are removed by the process of splicing before translation.
Prokaryotes are having polycistronic and continuous structural genes while eukaryotes have monocistronic and split.

13. Write the location and function of the following in human testes:
(a) Sertoli Cells (b) Leydig Cells 
Answer : (a) Sertoli Cells The sertoli cells are located within the seminiferous tubules. Their task is the creation of a hemato-testicular barrier and the nourishment of the spermatozoa.
(b) Leydig Cells: Leydig cells, also known as iriterstitial cells of Leydig, are found adjacent to the seminiferous tubules in the testicle. They produce testosterone in the presence of luteinizing hormone (LH).


Question.21. A woman has certain queries as listed below, before starting with contraceptive pills. Answer them.
(a) What do contraceptive pills contain and how do they acts as contraceptive ?
(b) What schedule should be followed for taking these pills ?
Answer : (a) Contraceptive pills contains progesterone and estrogen combination. This disrupts hormone patterns needed for pregnancy and affects the ovaries and the development of the uterine lining, making pregnancy less likely. They prevent ovulation (the egg leaving the ovary and moving into the fallopian tube). They block the hormones
needed for the egg to be able to be fertilized. They may affect the lining of the uterus and thus alters sperm transport, i which prevents sperm from reaching the egg to fertilize it.
(b) The pills have to be taken daily for a period of 21 days starting from the fifth days of menstrual cycle to the 25 th day. After a gap of 7 days (during which menstruation occurs), it has to be repeated in the same pattern till the female desires to prevent conception.

Question.24. Two types of aquatic organisms in a lake show specific growth pattems as shown below, in a brief period of time. The lake is adjacent to an agricultural land extensively supplied with fertilizers.
Answer the questions based on the fact given above :
(i) Name the organisms depicting the pattern A and B.
(ii) State the reason for the growth pattern seen in A.
(iii) Write the effects of the growth patterns seen above.
Answer : (i) A—>Planktonic Algae (free floating); B—>Fish
(ii) The reason for the growth pattern in ATPresence of large amount of nutrients in fertilizers in water causes excessive growth of planktonic (free – floating) algae known as Algal Bloom which consumes a lot of Oxygen and nutrients. As a result there is a sharp decline in the dissolve oxygen in the lake.
(iii) The increase in BOD (Biochemical Oxygen Demand) due to algal bloom which causes deterioration of the water quality which results in fish mortality. Some bloom forming
algae are extremely toxic to human beings and animals also.

Question.26. Explain, giving three reasons, why tropics show greatest levels of species diversity.
Answer :

  1. Tropical latitude have remained relatively undisturbed for million of years so they have greatest level of species diversity.
  2.  Tropical environment are less seasonal, relatively more constant and predictable. Such constant environment promotes niche specialization and lead to a greater species diversity.
  3. There is more solar energy available in the tropics which contributes to higher productivity, so indirectly contribute to greater diversity.


Question.28. Describe the Hershey and chase experiment. Write the conclusion drawn by the scientists after their experiment.
Answer : Experiments by Hershey and Chase in the 1950’s using the bacteriophage T2 and E. coli cells demonstrated that DNA is the genetic material of the bacteriophage.

  1. Hershey and Chase conducted their experiments on the T2 phage, a virus whose structure had recendy been shown by electron microscopy.
  2.  The phage consists of a protein shell containing its genetic material. The phage infects a bacterium by attaching to its outer membrane and injecting its genetic material and leaving its empty shell attached to the bacterium.
  3.  In their first set of experiments, Hershey andChase labeled the DNA of phages with radioactive Phosphorus-32 (the element phosphorus is present in DNA but not  present in any of the 20 amino acids from which proteins are made).
  4. They allowed the phages to infect E. coli (Escherichia coli), observed that the transfer of P32 labeled phage DNA into the cytoplasm of the bacterium.
  5.  In their second set of experiments, they labeled the phages with radioactive Sulfur-35 (Sulfur is present in the amino acids cysteine and methionine, but not in DNA).
  6.  Following infection of E. coli they then sheared the viral protein shells off of infected cells using a high – speed blender and separated the cells and viral coats by using a centrifuge.
  7.  After separation, the radioactive S35 tracer was observed , id the protein sheik, but not in the infected bacteria,
    supporting the hypothesis that the genetic material which infects the bacteria was DNA and not protein.

Conclusion :

  1.  Hershey and chase concluded that DNA,not protein was the genetic material. They determined that a protective protein coat formed around the bacteriophage, but that the internal DNA is conferred its ability to produce progeny inside a bacteria.
  2.  They showed that, in growth, protein has no function, while DNA has some function. Only 20% of the P32 (radioactive) remained outside the cell and it was incorporated with DNA in the cell’s genetic material. All of S35(radioactive) in the protein coats remained outside the cell, it was not incorporated into the cell, and protein was not the genetic material.

“Work out a typical Mendelian dihybrid cross and state the law that he derived from it.
Mendelian Dihybrid Cross : a cross between two parents that differ by two pairs of alleles (AABBXaabb) and he derived it from law of independent assortment.
The phenotypes and general genotypes from this cross can be represented in the following manner :



Question.2. Name the stage of cell division where segregation of an independent pair of chromosomes occurs.
Answer: The stage of cell division in which the segregation of an independent pair of chromosomes occurs Anaphase of meiosis I.

Question.3. Write an alternate source of protein for animal and human nutrition.
Answer: Single cell protein is an alternative source of protein for animal and human nutrition.

Question.4 Give an example of a plant which came into India as a contaminant and is a cause of pollen allergy.
Answer : Plant came into India as a contaminant and is a cause of pollen allergy is Parthenium.


Question.16. Explain the two factors responsible for conferring stability to double helix structure of DNA.
Answer: Factors responsible for conferring stability to double helix structure of DNA:

  1.  Presence of Hydrogen bond in between base pair stack * confers stability to DNA.
  2. Presence of thymine at the plage of uracil gives more stability to DNA.

Question.18. Write the effect of the high concentration of L.H. on a mature Graafian follicle.
Answer : High levels of Luteinizing Hormone (LH) induces rupture of mature Graafian follicle and causes release of ovum known as ovulation.


Question.24. (a) Explain adaptive radiation with the help of suitable example.
(b) Cite an example where more than one adaptive radiations have occurred in an isolated geographical area. Name the type of evolution your example depicts and state why it is so named.
Answer : (a) Adaptive radiation or divergent evolution: Different species are evolved in a given geographical area starting from the single point and literally radiating to other habitats in that area.
Ex : In the Australian region, marsupials each different from the other evolved from an ancestral stock, but all within the Australian island continent when more than one adaptive radiation appeared to have occurred in an isolated geographical area, one can call this convergent evolution.
(b) Convergent evolution : Convergent evolution is the process by which unrelated or distandy related organisms evolve similar body forms, coloration, organs, and adaptations. Natural selection can result in evolutionary convergence under several different circumstances. Species can converge in sympatry, as in mimicry complexes among insects, especially butterflies.
Ex : Marsupial fauna of Australia and the placental mammals of the Old World. The two lineages are clades—that is, they each share a common ancestor that belongs to their own group, and are more closely related to one another than to any other clade— but very similar forms evolved in each isolated population.

Question.25. (a) Name any two copper releasing IUDs.
(b) Explain how do they acts as effective contraceptives in human females.
Answer : (a) Two copper releasing IUDs are CuT, Cu7, Multiload 375.
(b) CuT is a method of intrauterine devices (IUTs). These devices are administered by the doctor in the uterus through vagina. The CuT is a copper releasing device which increases phagocytosis of sperms within the uterus and the cu ions released suppress sperm motility and fertilizing capacity of the sperm. So they acts as effective contraceptives in human females.

Question.27. (a) State how the constant internal environment is beneficial to organisms.
(b) Explain any two alternatives by which organisms can overcome stressful external conditions.
Answer : (a) The constant internal environment is beneficial to organisms because there is a continuous interaction between the organisms and the environment. An organism, fully adapted to environmental conditions in which they survive, grows and reproduces. The environment can be defined as the total of all physical and biotic conditions which influence the responses of organism.
(b) The two alternatives by which organisms can overcome stressful external conditions are :
(i) Direct Factors (ii) Indirect Factors
(i) , Direct factors : These factors influence the organisms
directly e.g., light, temperature, humidity and soil nutrients etc.
(ii) , Indirect factors: These factors affect organisms indirectly
by modifying other factors e.g., soil organisms, wind etc.


Question.28. Explain the process of sewage water treatment before it can be discharged into natural water bodies. Why is this treatment essential ?
Explain the process of replication of a retrovirus after it gains entry into the human body.
Answer : There are agronomic and economic benefits of wastewater used in agriculture. Irrigation with wastewater can increase the available water supply or release better quality supplies for alternative uses.
Sewage treatment generally involves three stages, called primary, secondary and tertiary treatment.

  1.  Primary Treatment : In primary treatment, the incoming flow is slowed in large tanks which allow the dirt, gravel, and other heavier components of the waste stream to settle out. Grease, oil, and other floatables are also removed here. Rotating arms simultaneously remove the settled solids from the bottom and the separated floatables from the top. Both pollutants are pumped into large heated holding silos, called digesters.
  2.  Secondary Treatment: This treatment removes dissolved and suspended biological matter. Secondary treatment is typically performed by indigenous, water-borne micro-organisms in a managed habitat. It may require a separation process to remove the micro-organisms from the treated water prior to discharge or tertiary treatment.
  3.  Tertiary Treatment : It is sometimes disinfected chemically or physically (for example, by lagoons and microfiltration) prior to discharge into a stream, river or it can be used for the irrigation of a green way or park. If it is sufficiently clean, it can also be used for groundwater recharge or agricultural purposes.
    This treatment is essential because the sewage water contains large amount of pathogenic microbes, organic matters.

HIV multiplies in human body first in macrophages during inhibition period & later in Helper T-cells during which symptoms of AID oppear.
(I) Cycle in Macrophages :

  1.  After gaining entry into the human body, the HIV passes to all parts through blood & other body fluids.
  2.  When it comes in contact with macrophage, the gp 120 spilce of virus binds with CD4 receptor of the macrophage.
  3.  A conformational change aids the virus to attach to another co-receptor called CCR5.
  4.  This triggers change in cell membrane of the macrophage which then endocytose HIV.
  5.  Once inside, it sheds the protective cover is shed. This frees the RNA along with reverse transcriptase in the cytoplasm of macrophage.
  6.  It synthnesizes copy of DNA from which a complement DNA is produced.
  7. The double stranded DNA attaches to host DNA in the form of provirus. It then directs the host cell machinery to form genomic RNA & mRNA.
  8.  mRNA synthesizes viral proteins including reverse transcriptase. Genomic RNA & viral proteins are packet together to form the virus. In this way, several copies of virus are formed.
  9.  These viruses then bud out of the macrophages by the process of exocytosis. These then invade new macrophages to further replicate.

The HIV undergoes similar cycle of replication in the Helper T-cells.

  1. The virus first attaches to CD4 receptor hy it gp 120. The complex then comes in contact with conceptor GXCR4.
  2.  The virus then passes into the cytoplasm ofT-lymphocytes through endocytosis.
  3. Inside the cytoplasm of T-cell, the virus coat is shed the naked RNA alonwith copy DNA & then complement DNA, which then gets integrated to host DNA as provirus.
  4.  The provirus directs the synthesis of two types of RNA- genomic & mRNA. mRNA forms vital proteins (including reverse transcriptase). Genomic RNA &viral proteins are packed together to form the virion.
  5. The virion comes in contactwith the surface oflymphocyte, raptures its cell membrane & come out. These further infect healthy cells.
  6.  Thus, number of T-cells decline, compromising the immune system of the body.