CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2016

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.

SECTION-A

Question.1. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason.
Answer : It is because of Haplodiploidy which is a sex determination system in which males develop from unfertilized eggs and are haploid, and females develop from fertilized eggs and are diploid.

Question.2. Mention the role of‘genetic mother in MOET.
Answer : MOET is a programme for herd improvement to get more eggs. The genetic mother is available for another round of super ovulation in this technology.

Question.3. What is biopiracy ?
Answer : Biopiracy is the term used to refer to the use of bioresources without proper authorisation by the multinational companies and other organisations without compensatory payments to people concerned.

Question.4. Mention two advantages for preferring CNG over diesel as an automobile fuel.
Answer : Two advantages for preferring CNG over diesel are :
(1) It is a very cheap fuel.
(2) It is a greener fuel.

Question.5. Write the probable differences in eating habits of Homo habilis and Homo  erectus.
Answer : Difference in Eating habits of Homo habilis and Homo erectus are :
Homo habilis did not eat meat while Homo erectus ate meat.

SECTION-B

Question.6. A single pea plant in your kitchen garden produces pods with viable seeds, but the individual papaya plant does not. Explain.
Answer : In pea plant both male and female flowers are present on the same plant which prevents autogamy but not geitonogamy which results to produce viable seeds after self pollination. In papaya plant, male and female flowers are present on different plants i.e., each plant is either male or female which prevents both autogamy and geitonogany. Thus, a single papaya plant cannot produce yiable seeds.

Question.7. Following are the features of genetic codes. What does each one indicate ?
Stop codon; Unambigous codon; Degenerate codon; Universal codon.
Answer : Stop Codon : Not code for any amino acids. Unambiguous Codon: One codon codes for only on amino acids.
Degenerate Codon : Some amino acids are coded by more than one codon.
Universal Codon : It is same for all either bacteria or human.

Question.8. Suggest four important steps to produce a disease resistant plant through conventional plant breeding technology.
Answer : Important steps to produce a disease resistant plant through conventional plant breeding technology are :

1.  Collection of variability of germplasm.
2.  Evaluation and selection of parents.
3.  Cross hybridisation among the selected parents.
4. Selection and testing of superior recombinant.

9. Name a genus of baculovirus. Why are they considered good biocontrol agents ?
Answer: Nudeopolyhedrovirus is a genus baculovirus which “* are efficient bio-control agents. They are considered to be good bio-control agents because these viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications and show no negative inpacts on plants, mammals, birds or even non-target insects.

Question.10. Explain the relationship between CFC’s and Ozone in the stratosphere.
OR
Why are sacred groves highly protected ?
Answer : Relationship between CFC’s and Ozone :
When CFC’s are released into the stratosphere, they end up being broken up by the UV light, resulting in chlorine being released. This acts like a catalyst with the ozone molecules creating chlorine monoxide and molecular oxygen. Cl atoms are not consumed in the reaction. Hence whatever CFCs are added to the stratosphere, they have permanent and continuing affects on ozone level.
OR
Sacred Groves are relic forest patches traditionally protected by communities in reverence of a deity. Sacred Groves form important repositories of forest biodiversity. It also provide vital ecosystem services to local people are the last refuges for a large number of rare and threatened plants and animal species.

SECTION-C

Question.11. (a) Name the organic material exine of the pollen grain
is made up of. How is this material advantageous to pollen grain ?
(b) Still it is observed that it does not form a continuous layer around the pollen grain. Give reason.
(c) How are ‘pollen banks’ useful ?
OR
(a) Mention the problems that are taken care of by Reproduction and Child Health Care programme.
(b) What is amniocentesis and why there is a statutory ban on it ?
Answer : (a) The hard outer layer called exine is made up of sporopollenin which is one of the most resistant organic material. It can withstand high temperature, strong acids and alkalis. It cannot be degraded by any of the known enzymes. Hence, it acts as a shield and protects the pollen grain from getting damaged.
(b) Exine does not form a continuous layer around the pollen grain. Pollen grain exine has prominent aperture called germ pore where sporopollenin is absent. Germ pores serve as an oudet for the formation of pollen tube.
(c) Pollen grains at a large can be stored for years in liquid nitrogen (— 196°C). So, after this treatment they are stored in pollen banks. Such conserved pollen grains can be used in plant breeding programs.
OR
(a) The problems which are addressed through Repro¬duction and Child Health Care Programme are :
Creating awareness among people about the various reproduction related aspects and providing facilities for building up a reproductively healthy society.
(b) A foetal sex determination test based on the chromo¬somal pattern in the amniotic fluid surrouning the developing embryo is called amino- centesis. Statutory ban on aminocentesis is imposed because this test can be used for determining the sex. of foetus which is increasing female foeticides.

Question.12. What is a test cross ? How can it decipher the heterozygosity of a plant ?
Answer : If the progenies produced by. a test cross show 50% dominant trait and 50% recessive trait, then the cross in which the genotype of an unknown dominant phenotype can be determined by crossing it with an individual homozygous recessive phenotype for that trait is called test cross.
This cross determines whether the dominant character is coming from homozygous dominant genotype or heterozygous genotype, (eg. tallness coming from Tt) when Tt is cross with it, we obtain all Tt (tall) individuals in the program. Thus, test can be used to determine the heterozygosity of the plants.

Question.13. (a) What do *Y’ and ‘B’ stand for in ‘YAC’ and ‘BAC’used in Human Genome Project (HGP). Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the percentage of discovered genes whose functions are known as observed during HGP.
(c) Expand ‘SNPs’ identified by scientists in HGP.
Answer : (a) YAC stands for Yeast Artificial Chromosomes. BAC stands for Bacterial Artificial Chromosomes-BAC and YAC are specialized cloning vectors which are used in human genome project for cloning or amplification of human DNA fragents.
(b) Less than 2% of total genome codes for proteins in humans. Around approx. 30% of gene functions are known during HGP.
(c) SNP—Single Nucleotide Polymorphism.

Question.14. Differentiate between homology and analogy. Give one example of each.
Answer :

Question.15. (a) It is generally observed that the children who had suffered from chicken-pox in their childhood may not contract the same disease in their adulthood.’ Explain giving reasons the basis of such an immunity in an individual. Name this kind of immunity. (b) What are interferons ? Mention their role.
Answer : (a) The type of immunity is Passive Acquiredimmunity. Due to the development of memory-B cells in person’s body after primary exposure to the disease, the generated antibodies help to prevent the second recurrence of the disease in adulthood.
(b) Interferon : These are proteins made and released by host cells in response to the presence of pathogens such
as viruses, bacteria, parasites etc.Role: It inhibits viral infections and stimulates the entire immune system to fight disease. It also regulates many
kinds of cell functions.

Question.16. (a) Write the two limitations of traditional breeding technique that led to promotion of micro propagation.
(b) Mention two advantages of micro propagation.
(c) Give two examples where it is commercially adopted.
Answer : (a) Two limitations of traditional breeding are :
(1) The required characteristics may not be present in the breeding population.
(2) The breeder does not know exactly what genes have been introduced to the new cultivars.
(b) Advantages of micro propagation are :
(1) Production of many plants that are clones of each other.
(2) It can be used to produce disease free plants.
(c) It is commercially adopted for :
(1) Banana, (2) Tomato

Question.17. (a) How do organic farmers control pests ? Give two examples.
(b) State the difference in their approach from that of conventional pest control methods.
Answer : (a) Organic farmers create a system where the insects are not eradicated, but are kept at manageable levels by a complex system within living and vibrant ecosystem.
Examples are :
(i) The ladybird and dragon flies are useful to get rid of aphids and mosquitoes respectively.
(ii) Bacillus thuringiensis (Bt) are used to control butterfly caterpillars.
(b) Organic farmer, works to create a system where insects that are sometimes called pests are not eradicated, but instead are kept at manageable levels by a complex system of checks and balanced within a living ecosystem, contrary to the conventional farming practices which often, use chemical methods to kill both useful and . harmful life from.

Question.18. (a) Name the selectable markets in the cloning vector pBR322 ? Mention the role they play.
(b) Why is the coding sequence of an enzyme b- galactosidase a preferred selectable marker in comparison to the ones named above ?
Answer : (a) Ampicillin, chloramphenicol are selectable markers in the cloning vector pBR322. Selectable marker, helps in identifying and eliminating non-transformants and Selectively permitting the growth of the transformants.
(b) Alternative selectable marker which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate. In this , a recombinant DNA is . inserted within the coding sequence of an enzyme, b galactosidase, which results into activation of the enzyme referred as insertional inactivation coding sequence for the enzyme b-galactosidase is preferred over antibiotic resistance genes because recombinants can be easily visualised.

Question.19. (a) Why must a cell be made ‘competent’ in biotechnology experiments ? How does calcium ion help in doing so ?
(b) State the role of ‘biolistic gun’ in biotechnology experiments.
Answer : (a) DNA being a hydrophilic molecule, can not pass through cell membranes, hence the cells should be made competent to accept the DNA molecules as competency is the ability of a cell to take up foreign DNA. Calcium ion helps in increasing the pore size in cell wall which enables the cell to take up the recombinant DNA.
(b) To introduce alien DNA into host cells, suitable for plants, cells are bombarded with high velocity micro¬particles of gold or tungsten coated with DNA molecules known as biolistic or gene gun play important role in biotechnology experiments.

Question.20. Explain enzyme-replacement therapy to treat adeno¬ sine deaminase deficiency. Mention two disadvantages of this procedure.
Answer: Adenosine deaminase (ADA) deficiency is a genetic disorder. In this disease, the gene coding for the enzyme ADA gets deleted leading to deficiency of ADA and problems in immune system. Adenosine deaminase (ADA) deficiency in patients can be treated by enzyme replacement therapy.
In this treatment, patients are regularly injected with the functional ADA enzyme.
Disadvantages of this procedure :
(i) It does not completely eradicate the disease.
(ii) Requirement of repeated doses of the enzyme
makes it expensive.

Question.21. Name and explain the type of interaction that exists in mycorrhizae and between cattle egret and cattle.
Answer : Mycorrhizae are associations between fungi and the roots of higher plants. The type of interaction that exists in mycorrhizae is mutualism in which both fungi and plants – are dependent on each other. Fungi absorb and transport essential nutrients to the plants and in turn plants provide the fungi with other energy carbohydrates.
The interaction that exist between cattle egret and cattle is known as commensalism. In this type of interaction, one specie is benefitted whereas the other is neither benefitted nor harmed. The cattle egrets (bird) always forage close to where the cattle are grazing because the cattle, as they move, stir up and flush out from the vegetation insects that otherwise might be difficult for the egrets to find and catch. Thus, the cattle is neither benefitted nor harmed but the egret is benefitted.

Question.22. Differentiate between primary and secondary succession. Provide one example of each.
Answer :

SECTION-D

Question.23. A large number of married couples the world over are childless. It is shocking to know that in India the female partner is often blamed for the couple being childless.
(a) Why in your opinion the female partner is often blamed for such situations in India ? Mention any two values that you as a biology student can promote to check this social evil.
(b) State any two reasons responsible for the cause of infertility.
(c) Suggest a technique that can help the couple to have a child where the problem is with male partner.
Answer : (a) Due to improper educational existence and lack of moral values and also along with the orthodox male dominant society nature in India. Females are blamed for infertility issues.
As a biology student:
(1) We must provide proper biological/Sex education at such , a basic stand which can be clear to every individual.
(2) General Health awareness programme must be scheduled to persons for their health related queries.
(b) Causes of Infertility in females :
(1) Ovulation disorders
(2) Problems in the uterus or fallopian tubes Causes of Infertility in Males :
(1) Low sperm count/Low sperm mobility
(2) Genetic abnormality
(c) Artificial Insemination (AI) is a technique that can help the couple to have a child where the problem is with male partner. In this technique, the semen collected either from the husband or a healthy donor is artifically introduced into the vagina or into the uterus of the female.
ICSI (Intra Cytoplamsic Sperm Injection) is another specialized procedure to form an embryo in the lab in which a sperm is directly injected into the ovum.

SECTION-E

Question.24. (a) Explain the menstrual phase in a human female. State the levels of overian and pituitary hormones during this phase.
(b) Why is follicular phase in the menstrual cycle also referred as proliferative phase ? Explain.
(c) Explain the events that occur in a graafian follicle at the time of ovulation and thereafter.
(d) Draw a graafian follicle and label antrum and secondary oocyte.
OR
(a) As a senior biology student you have been asked to demonstrate to the students of secondary level in your school, the procedure(s) that shall ensure cross pollination in a hermaphrodite flower. List the different steps that you would suggest and provide reasons for each one of them.
(b) Draw a diagram of a section of a megasporangium of an angiosperm and label funiculus, micropyle, embryosac and nucellus.
Answer : (a) In human females, menstruation is repeated at an average interval of about 28/29 days and the cycle of events starting from one menstruation till the next one is called the menstrual cycle. This cycle starts with the menstrual phase, when menstrual flow occurs and it lasts for 3-5 days. The menstrual flow results due to breakdown of endometrial lining of the uterus and its blood vessels which forms liquid that comes out through vagina.
During this phase the levels of estrogen and proges- terone are low.
(b) The proliferative phase is the part of the menstrual cycle during which follicles inside the ovaries develop and mature in preparation for ovulation. The levels of FSH.
increase in the bloodstream during the proliferation phase, stimulating the maturation of follicles. Each follicle contains an ovum, or egg. Although many follicles may grow and increase in size during this phase, only one will reach full growth and release the ovum at the time of ovulation.
(c) During the mid cycle Leutinizing hormone secretes to its maximum level which induces rupture of Graafian follicle and thereby the release of ovum (ovulation). The ovulation is followed by the luteal phase during which the remaining parts of the Graafian follicle transform as the corpus luteum which secretes large amounts of progesterone which is essential for maintenance of the endometrium. Such an endometrium is necessary for implantation of the fertilised ovum and other events of pregnancy.

OR
(a) The different steps that would suggest for cross pollination
in a hermaphrodite flower are :
(i) Removal of anthers frono the flower bud before the anther dehisces using a pair of’forceps is necessary referred as emasculation.
(ii) Emasculated flowers covered with a tag of suitable size to prevent contamination of its stigma called Bagging.
(iii) When the stigma of bagged flower attains receptivity, mature pollen grains collected from anther of the male parent are dusted on the stigma, and the flowers are rebagged.

Question.25. Describe Meselson and Stahl’s experiment that was carried in 1958 on E. coli. Write the conclusion they arrived at after the experiment.
OR
(a) Describe the process of transcription in bacteria.
(b) Explain the processing the hnRNA needs to
undergo before becoming functional mRNA eukaryotes.
Answer : The experiment was performed by Meselson and Stahl. The following steps were followed in the experiment.
coli was grown in a medium containing 1SNH4 Cl the heavy isotope 15N as the sole nitrogen source. This led to the incorporation of 15N into the newly synthesised which ultimately made the DNA heavy.This heavy DNA was separated from the normal DNA by density gradient centrifugation using cesium chloride as the gradient.
The cells were then transferred into a medium with 14N as the nitorgen source. Samples were taken from this medium and the DNA was extracted.
Observation:
Since E. coli divides every 20 minutes, the DNA extracted after 20 minutes in the experiment had a hybrid density. The DNA extracted after 40 minutes had equal amounts of hybrid and light densities.
Conclusion:
DNA extracted from the culture after another generation was composed of equal amounts of this hybrid DNA and of light DNA.
This implies that the newly synthesised DNA obtained one of its strands from the parent. Thus, replication was semiconservative.

OR
(a) Transcription has three steps : Initiation, elongation and termination.
Initiation : RNA polymerase binds to promoter and initiates transcription. It associates with initiation factor and alters the specificity of RNA polymerase to initiate the transcription.
Elongation : RNA polymerase uses nucleoside triphosphate as substrate, and polymerises in a template depended fashion following the rule of complementarity and facilitates opening of the helix and continues elongation.
Termination occurs when termination factor (rho) alters the specificity of RNA polymerase.
Only a short stretch of RNA remains bound to the enzyme. Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination for translation of transcription.

(b) The precursor of mRNA, i.e., hnRNA, contains both introns and exons. Introns are removed and exons are joined by a process called splicing. The remaining mRNA is processed in two ways :
Capping : An unusual nucleotide (methyl gunosine triphosphate) is added to the 5′-end of hnRNA.
Tailing : Adenylate residues (200-300) are added at 3′-end in a template independent manner.
When hnRNA is full processed, it is known as mRNA, which is transported out of the nucleous.

Question.26. (a) Name the two growth models that represent population growth and draw the respective growth curves they represent.
(b) State the basis for the difference in the shape of these curves.
(c) Which one of the curves represent the human population growth at present ? Do you think such a curve is sustainable ? Give reason in support of your answer.
OR
(a) Taking an example of a small pond, explain how the four components of an ecosystem function as a unit.
(b) Name the type of food chain that exists in a pond.
Answer. (a) Two growth models are :

1.  Exponential
2. Logistic

Exponential Growth :
N : Population size b : Birth rates
d : Death rates


(b) The difference in the shape of these curves is the amount of resources available for the given population. When resources are unlimited each species realizes its innate potential to grow in number and result in a J-shaped curve in exponential growth while in logistic growth no population has unlimited resources leads to competition for resources and show S-shaped curve.
(c) Logistic growth curve represents the human population growth at present because the number of human beings are increasing rapidly but the available resources are not increasing enough. Such a curve is not sustainable because at one point the human population would reach a place where resource there would not be finite resources for every one.
OR
(a) The four components of an ecosystem are :

1.  Productivity
2.  Decomposition
3. Energy flow
4. Nutrient cycling

A pond is a shallow water body in which all the four components of in ecosystem are well exhibited. The abiotic component is the water with all the dissolved inorganic and organic substances and the rich soil deposit at the bottom of the pond. The solar input, the cycle of temperature, day- length and other climatic conditions regulate the rate of function of the entire pond. The autotrophic components include the phytoplankton, some algae and the floating sub merged and marginal plants found at the edges. The consumers are represented by the zooplancton, the free swimming and bottom’dwelling forms. This system performs all the functions of any ecosystem and of bisophere as a whole i.e. conversion of inorganic into organic material with the help of the radiant energy of the sun by the autoprophs : Consumption of the autotrophs by heterotrophs; decomposition and mineralization of the dead matter to release them back for reuse by the autotrophs. There is unidirectional movement of energy towards the higher trophic levels and its dissipation and loss as heat to the environment.
(b) Aquatic food chain exists in a pond.

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.

SET-I

SECTION-A

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.

SECTION – B

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 ?
Answer:

(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.
OR
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.
Answer:

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.

SECTION-C

Question.19. Make a list of any three outbreeding devices that flowering plants have developed and explain how they help to encourage cross-pollination ?
OR
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.
OR
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.
Conclusions:

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.

SECTION-D

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.
OR
(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.
(c)

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 ?
OR
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.
   

OR
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.
OR
(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. .

OR
(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.

SET-II

SECTION-A

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).

SECTION-B

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).

SECTION – C

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.

SECTION – D

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.
   

OR
“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 :

SET III

SECTION-A

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.

SECTION-B

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.

SECTION-C

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.

SECTION-D

Question.28. Explain the process of sewage water treatment before it can be discharged into natural water bodies. Why is this treatment essential ?
OR
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.

OR
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.
   

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.

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 :

Question.14. The above graph shows Species-Area relationship. Write the equation of the curve ‘a’ and explain.

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

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.

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.

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:

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)

(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

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)

(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)

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:

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.

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.

CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2013

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 an organism where cell division in itself is mode of reproduction.
Answer : The division of cell in itself is a mode of reproduction found in amoeba and paramoecium.

Question.2. When does a human body elicit a anamnestic response ?
Answer : When our body attacked by pathogens for the second time the memory cells which were formed during the first attack produces a highly intensified secondary or anamnestic response.

Question.3. Name any two disease the ‘Himgiri’ variety of wheat is resistant to.
Answer : The ‘Himgiri’ variety of wheat is resistant to leaf and stripe rust and hill bunt disease.

Question.4. State the role oi transposons in silencing of mRNA in eukaryotic cells.
Answer : Role of transposons : Silencing of a gene is done in order to prevent translation of mRNA, where transposons act as a complementary RNA that is used to stop translation.

Question.5. Why are green algae not likely to be found in the deepest  strata of the ocean ?
Answer : Green algae not likely to be found in the deepest strata of the ocean because deep inside the sea presence of sufficient light for photosynthesis and brackish water are not available so green algae are not present at this level, instead algae inhabits littoral zone of water.

Question.6. State what does‘standing crop’of a trophic level represent.
Answer : ‘Standing crop’ of a trophic level represent certain j mass of a living material at a particular time.

Question.7. Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic converters ?
Answer : The use of unleaded petrol- recommended for 1 motor vehicles equipped with catalytic converters because lead in petrol inactivates the catalysts which convert harmful pollutants (CO, unburnt hydrocarbons, nitric oxide) to lesser ,harmful pollutants (C02, HzO, N2).

Question.8. Name the type of biodiversity represented by the following :
(i) 1000 varieties of mangoes in India
(ii) Variations in terms of potency and concentration of reserpine in Rauwolfia vomitoria growing in different
regions of Himalayas.
Answer : (i) Genetic Biodiversity : (ii) Genetic Biodiversity

SECTION – B

Question.9. In angiosperm, zygote is diploid while primary endosperm cell is triploid. Explain.
Answer : (i) Fertilisation of haploid egg cell by one haploid male gamete to form diploid zygote also called syngamy.

(iii) Fertilisation of two (diploid) polar nuclei by the other haploid male gamete to form triploid primary endosperm nucleus also called triple fusion.

Question.10. A cross between red flower bearing plant and a white flower bearing plant of Antirrhinum produced all plants having pink flowers. Work out a cross to explain how this is possible.
Answer:

Phenotypic ratio: 1:2:1
Genotypic ratio: 1:2:1
R(Red) factor is not completely dominant over r(white) factor is incomplete dominance.

Question.11. List the two main propositions of Oparin and Haldane.
Answer : The two main propositions of Oparin and Haldane are : (i) The first form of life could have come from preexisting non-living organic molecules (e.g., RNA, protein etc.) i.e., first abiogenesis and biogenesis later.
(ii) The first form of life was preceded by chemical evolution i. e., formation of organic molecules from inorganic molecules like CH4, NH3 etc.

Question.12. Write the events that take place when a vaccine for any disease is introduced into the human body.
Answer : Vaccines are a non-virulent form of pathogens, when administered into the body, the body start making antibodies against the antigens present in the vaccine. , The vaccine also generate memory – B and T – cells, that recognize the pathogen quickly on. subsequent second exposure and wipe out the invaders with a massive production of antibodies.
OR
Why is a person with cuts and bruises following an accident administered tetanus antitoxin ? Give reasons. Answer : Tetanus antitoxins neutralize and provide passive immunity to the bacterial toxin. The antitoxin contain antibody against pathogen it attach and inactivate pathogen.

Question.13. Name the bacterium responsible for the large holes seen in ‘Swiss Cheese’. What are these holes due to ?
Answer : Propionibacterium sharmanii is responsible for the large holes seen in Swiss Cheese. The large holes are due to production of a large amount of C02 by a bacterium.

Question.14. Name the source of the DNA polymerase used in PCR technique. Mention why it is used.
Answer : Thermus aquaticus because it is heat stable DNA polymerase. Polymerase chain reaction (PCR) is a method in which the desired gene is synthesised in vitro in following steps:
(a) Denaturation : The double-stranded DNA is denatured ’ by applying high temperature of 95°C for 15 seconds. Each separated single stranded now acts as template for DNA synthesis.
(b) Annealing : Two sets of primers are added which anneal to the 3′ end of each separated strand. Primers act as initiators of replication.
(c) Extension : DNA polymerase extends the primers by adding nucleotides complementary to the template provided. in the reaction. A thermostable DNA polymerase (Taq polymerase) is used in the reaction which can tolerate the high temperature of the reaction. All these steps are repeated many times to obtain several copies of desired DNA.

Question.15. Write any four ways used to introduce a desired DNA segment into bacterial cell in recombinant technology experiment.
Answer : The four ways used to introduce a desired DNA segment into bacterial cell in recombinant. technology experiment as following:
(i) Chemical Method : Poration by divalent cation such as calcium.
(ii) Micro injection : Recombinant DNA is directly injected into the nucleus of an animal cell.
(iii) Biolistic or gene gun: Plant cells are bombarded with hi gh velocity micro-particles of gold or tungsten coated with DNA.
(iv) Disarmed pathogen vectors : when allowed to infect the cell, transfer the recombinant DNA into the host.

Question.16. Why is proinsulin so called ? How is insulin different from it?
Answer : Proinsulin is a protein molecule and like a pro¬enzyme. It contain an extra strech of C peptide so it need to be processed to become fully mature and functional hormone like insulin, it is a mature hormone and is produced by the beta cells. Proinsulin is different from insulin because it serves as a precursor hormone to insulin.

Question.17. Where would you expect more species biodiversity in tropics or in polar region ? Give reasons in support of your answer.
Answer : High species or biodiversity lies in tropical areas because tropics are :
(i) Undistributed habitats since millions of year in comparison to temperate and polar region which face frequent glaciation. It favours speciation, as speciation is product of time.
(ii) Less seasonal variation than polar areas.
(iii) High availability of solar radiatipns than polar area, which harbours more plant species.

Question.18. “It is possible that a species may occupy more than one trophic level in the same ecosystem at the same time.” Explain with the help of one example.
Answer : Yes, as the trophic level of a species represents the functional role of organism in energy flow which is determined by the food intake. The availability of food is depends on what the organism want to eat, so have more than one trophic level at a time.
Ex. Sparrow is a Primary consumer, when eating seeds where as Secondary consumer, when eating insects.

SECTION-C

Question.19. Explain the steps in the formation of an ovum from an oogonium in humans.
Answer : The steps in the formation of an ovum from an oogonium in humans involves in oogenesis process. It can be divided into three stages :
(a) Multiplication phase (b) Growth phase
(c) Maturation phase
(a) Multiplication phase : In this stage primordial germ cells or ovum mother cells are repeatedly divided by mitosis to form large number of diploid oogonia. This process completes in embryo stage of female in most higher animals.
(b) Growth phase : In this process oogonia grow in size and form primary oocytes. The growth phase is the longest phase oogenesis (except humans). During growth phase size of egg increases many times.
(c) Maturation phase : Oogenesis takes place in the ovaries. In contrast to males the initial steps in egg production occur prior to birth. By the time the foetus is 25 weeks old, all the oogonia that she will ever produce, are already formed by mitosis. Hundreds of these diploid cells develop into primary oocytes, begin the first steps of the first meiotic division, proceed up to diakinesis, and then stop any further development. The oocytes grows much larger and complete the meiosis I, forming a large secondary oocyte and a small polar body that receives very little amount of cytoplasm but one full set of chromosomes.

OR
Suggest and explain any three Assisted Reproductive Technologies (ART) to an infertile couple.
Answer : Three assisted reproductive technologies (ART) to an infertile couple are :
(i) In vitro fertilisation (IVF-fertilisation outside the body in almost similar conditions as that in the body) followed by embryo transfer (ET) is one of such methods. In this method, popularly known as,, test tube baby programme, ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions-in the laboratory. The zygote or early embryos (with upto 8. blastomeres) could then be transferred into the fallopian tube (ZIFT- zygote intra fallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT —intra uterine transfer), to complete its further development.
(ii) Embryos formed by in-vivo fertilization (fusion of gametes within the female) also could be used for such transfer to assist those females who cannot conceive. Transfer of an ovum collected from a donor into the fallopian tube (GIFT – gamete infra fallopian transfer) of another female who cannot produce one, but can provide suitable environment for fertilisation and further development is another method attempted.
(iii) Intra cytoplasmic sperm injection (ICSI) is another ‘ specialised procedure to form an embryo in the laboratory in which a sperm is directly injected into the ovum.
Infertility cases either due to inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates, could be corrected by Artificial Insemination (AI) technique. In this technique, the semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI-Intrauterine Insemination) of the female.

Question.20. Why a human females are rarely haemophilic ? Explain. How do haemophilic patients suffer ?
Answer : 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 tnother and one from her haemophilic father, then she can be haemophilic. Simple cut will results to increased bleeding time in haemophilie patients.

Question.21. In a maternity clinic, for some reasons the authorities are not able to hand over the two new-borns to their respective real parents. Name and describe the technique that you would suggest to sort out the matter.
Answer: DNA Fingerprinting or DNA test is the technique that suggested to describe the parental identification of these two new born babies in a maternity clinic.
The procedure of finger printing is as follows :
(i) Isolation of DNA,
(ii) Digestion of DNA by restriction endonucleases,
(iii) Separation of DNA fragments by electrophoresis,
(iv) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
(v) Hybridisation using labelled VNTR probe, and
(vi) Detection of hybridised DNA fragments by autoradio-graphy. Half of the band of child will resemble to father and half to mother.

Question.22. Explain the increase in the number of melanic (dark winged) moths in the urban areas of post – industrialisation period in England.
Answer: In England, before industrial revolution the environ-ment was unpolluted. The white-winged moths were more and lichens on the barks of trees were pale. The white-winged moths could easily camouflage, while the dark winged were spotted out by the birds for food. Hence, they could not * survive. After industrial revolution the lichens became dark (due to soot-deposit). This favoured the dark-winged moths while the white-winged were picked by birds. The population of the former which was naturally selected increased.

Question.23. Describe how biogas is generated from activated sludge. List the components of biogas.
Answer : Biogas .can be produced by anaerobic digestion or fermentation of biodegradable materials. Bio wastes are collected and slurry of dung is mixed, a floating cover is placed over the slurry. The slurry having gas outlet is placed which keeps on rising as the gas is produced in the tank due to microbial activity of methanogens like Methanobacterium. Anaerobic fermentation of waste biomass can be visualized in three stages:
(i) The facultative anaerobic microbes degrade the complex polymers to simple monomers by enzymatic action. The Polymer like cellulose, hemicellulose, proteins and lipids get degraded into monomers but lignins and inorganic salts are left as residue because they do not degrade.
(ii) In second stage, monomers are converted in to organic acids by microbial action under partially aerobic conditions which are finally converted to acetic acid.
(iii) In third stage acetic acid is oxidized in to methane by the activity of anaerobic methanogenic bacteria. These bacteria are commonly found in the anaerobic sludge during sewage treatment. In this whole process digestion of cellulose takes place at very slow rate so that it is the “rate limiting factor in biogas production.”

Question.24. Name the pest that destroys the cotton balls. Explain the role of Bacillus thuringiensis in protecting the cotton crop against the pest to increase the yield.
Answer : Cotton ballworm is die pest that destroy the cotton balls. Bt toxin protein is produced by a soil bacterium called Bacillus thuringiensis in inactive prototoxin and crystalline form. The prototoxin form does not kill the bacteria. It becomes active and toxic when it is consumed by insects such as lepidopterans (armyworm), coleopterans (beetles) and dipterans (flies/ mosquitoes) due to presence of alkaline pH in the gut. The activated toxin (delta endotoxins) binds to the epithelial cells in the midgut of an insect and creates pores that cause lyses and swelling, eventually killing the insect.

Question.25. (a) Write the importance of measuring the size of a population in a habitat or an ecosystem.
(b) Explain with the help of an example how the percentage cover is a more meaningful measure of population size than more numbers.
Answer : (a) The importance of measuring the size of a population cover is more meaningful measure of population size than numbers. The size of the population tell us alot about its status in the habitat.
(b) Percentage cover is more meaningful measure of popula-tion size than more numbers because the’ relative abundance of a species is not only determined by number of individual w but by both i. e., the relative abundance in number and relative abundance in biomass.
Ex. In unit area the number of a grass species individuals or relative abundance in number is high but not in relative abundance of biomass. If the same area has one or two ficus benghalensis (Bargad) tree as it is very low in relative abundance in number while high in relative abundance of biomass.

Question.26. Differentiate between two different types of pyramids of biomass with the help of example of each.
Answer : Difference between two types of-pyramids of biomass:
Pyramid of Biomass is graphic representation of amount of biomass per unit area in the trophic levels with producers at the base & top carnivore at the apex. Biomass is maximum in producers. Only 10% of biomass is passed to next level. This is in accordance to the 10% law by Lindeman, (1942). Thus,.the biomass at higher trophic levels become smaller and smaller.
The pyramid of biomass is upright in grassland ecosystem.

For an aquatic system, the pyramid of biomass may be inverted or spindle-shaped. This is because the diatoms & other phytoplankton have a small standing crop, but they have a high annual productivity & high turn over rate. The reason or small standing crop is their short life span.

Question.27. (a) Describe the endosperm development in coconut.
(b) Why is tender coconut considered a healthy source of nutrition ?
(c) How are pea seeds different from castor seeds with respect to endosperm ?
Answer : Endosperm is ,a nutritive triploid tissue formed from mitotic divisions in primary endosperm nucleus (PEN). The cells of this tissue are filled with reserve food material & are used for the nutrition of the developing embryo.
In cococut, the type of endosperm formed is cellular endosperm, For this, the PEN divides many times and each division is followed by wall formation.

SECTION – D

Question.28 (a) Draw a L.S. of a pistil showing pollen tube entering the embryo-sac in an angiosperm and level any six parts other than stigma, style and ovary.
(b) Write the changes a fertilized ovule undergoes within the ovary in an angiosperm plant.
OR
(a) Draw a diagrammatic sectional view of a human seminiferous tubules and label sertoli cells, primary spermatocyte, spermatogonium and spermatozoa in it.
Answer: (a)

(b) Fertilised ovule forms seed which contains embryonic plant, reserve food and protective coat. A seed contains/ consists of two parts—seed contains/consists
(i) Seed Coat: It is the outer covering of the seed, functioning as a protective coat. It is formed from the integuments of the ovule, which hardens after fertilisation. The outer seed coat is called Testa & is formed from outer integument. The inner seed coat is called Tegumen is formed from inner integument. The micropyle remains as a small pore in the seed coat. It helps in the entry of oxygen & H20 into the seed during germination. Above the micropyle, hilum is visible as a depression, being remanant of the attachment point.

(ii) Embryo : The embryo is made up of embryonal axis Cotyledon/s and endosperm. Embryonal axis is the central axis, also called ‘Tigellum. One end of Tigellum bears Radicle (future root) & the other end bears plumule (future shoot). Cotyledon is present at the node of embryo axis. It is a fleshy structure, used for storage of food. Endosperm is the nutritive layer. Present inside the seed coat providing nutrition to growing embryo.

OR
(a) Spermatogenesis is under the control of endocrine hormones.

(i) Hypothalamus produces Gonadotropin Releasing Hormone (GnRH).
(ii) GnRH acts on anterior pituitary to produce gonadotropins,
ICSH (Interstitial Cell stimulating Hormone) & FSH. * ICSH acts on interstitial or leydig cells which produce testosterone.
(iii) FSH stimulates sertoli cells to develop ABP (Androgen Binding Protein) which help in concentrating testosteron in seminiferous tubules.
(iv) Excess testosterone inhibits LH/ICSH production by anterior pituitary & subsequendy GriRH production by Hypothalamus.
(v) Sertoli cells also produce a glycoprotein called Inhibition which suppresses FSH synthesis by anterior pituitary and GnRH by Hypothalamus.
This is called Negative Feedback control for release of testosterone.

Question.29. (a) Write the conclusion drawn by Griffith at the end of his experiment with streptococcus pneumonia.
(b) How did O. Avery, C. MacLeod and M. McCarty prove the DNA was the genetic material ?
Answer : (a) Griffith transformation experiment : Griffith performed his experiment in 1928 on Streptococcus pneumoneae bacteria which cause pneumonia in mice.
He used two strains of bacteria.
(i) Rough strain : Non capsulate non-virulent rough colonies on culture media.
(ii) Smooth strain : Capsulated virulent form smooth colonies on media.
Experiment: (i) Mice + Smooth strain bacteria —> Dead mice.
(ii) Mice + Rough strain bacteria —> Living mice.
(iii) Mice + Heat killed bacteria + Rough bacteria —> Dead Bacteria.
On the basis of third experiment he proposed that rough bacteria absorb some heat stable material from dead smooth bacteria and transformed into smooth bacteria which killed the mice.
(b) O Avery, C. Macleod and M. McCarty prove the DNA was the genetic material by purifying biochemicals (proteins, DNA, RNA etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed. They also discovered that protein – digesting enzymes (proteases) and RNA – digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material.
OR
(a) Explain the mechanism of sex-determination in humans.
(b) Differentiate between male heterogamety and female heterogamety with the help of an example of each.
Answer : (a) Sex determination in humans :
(i) The males have 22 pairs autosomes and A pair of XY-
chromosome.
(ii) The females have 22 pairs autosomes and a pair of XX- chrornosomes
(iii) In male, 50% of sperms carry X- chromosome and other 50% carry Y – chromosomes.
(iv) In females, all ova contains X-chromosomes.
(v) The sex of an individual is determined by the type of sperm fertilizing the ovum.
(vi) If the ovum is fertilized by Y-chromosome, the zygote (XY) develops info a male and if the ovum in fertilized by X-chromosome, Zygote (XX) develops into a female.

(b) There are two types of sex determining mechanisms, i.e., XO type and XY type. But in both cases males produce two different types of gametes,
(i) Either with or without X-chromosome or,
(ii) Some gametes with X-chromosome and some with Y-chromosome.
Such types of sex determination mechanism is designated to be an example of male heterogamety. In some other organisms, e.g., birds a different mechanism of sex determination is observed. In this case the total number of chromosome is same in both males and females. But two . different types of gametes in terms of the sex chromosomes, are produced by females, i.e., female heterogamety. The two different sex chromosomes of a female bird has been designated to be the Z and W chromosomes. In these organisms the females have one Z and one W chromosome, ‘ whereas males have a pair of Z-chromosomes besides the autosomes.

Question.30. A person in your colony has recently been diagnosed with AIDS. People/Residents in the colony want him to leave the colony for the fear spread of AIDS.
(a) Write your views on the situation, living reasons.
(b) List the possible preventive measures that you would suggest to the residents of your locality in a meeting organized by you so that they understand the situation.
(c) Write the symptoms and the causative agent of AIDS.
Answer: (a) AIDS is infectious but not contagious it does not spread by shaking hand and use of common utensils so there is no need of fear to live with AIDS patient.
(b) Making blood (from blood banks) safe from HIV, ensuring the use of only disposable needles and syringes in public and private hospitals and clinics, free distribution of condoms, controlling drug abuse, advocating safe sex and promoting regular check-ups for HIV in susceptible populations, are some such steps taken up.
(c) Symptoms and the causative agent of AIDS :
AIDS is caused by the Human Immuno Deficiency Virus (HIV), a member of a group of viruses called retrovirus, which have an envelope enclosing the RNA genome. These target the T lymphocytes, due to which the person starts suffering from infections that could have been otherwise overcome such as those due to bacteria especially Mycobacterium, viruses, fungi and even parasites like Toxoplasma. The patient becomes so immuno-deficient that he/she is unable to protect himself/herself against these infections.

SET-II

SECTION-A

Question.3. Write the basis on which an organism occupies a space in its community/natural surroundings.
Answer: It is based on the feeding relationships of that organism with other organisms and source of their nutrition or food.

SECTION-B

Question.8. Name an algae that reproduces asexually through zoospores. Why are these reproductive units so called.
Answer : Algae that reproduces asexually .through zoospores is “Chlamydomonas”. They are microscopic and motile, due to flagella so known as zoospores.

SECTION-C 

Question.12. “Stability of a community depends on its species richness”. Write how did David Tilman show this experimentally.
Answer : (i) David Tilman s long term ecosystem experiment – shows that plots with more species show less year-to-year variation in total biomass.
(ii) He also showed in his experiment that, increased diversity contributed to higher productivity.

Question.14. Name the haploid cells present in an unfertilized mature embryosac of a flowering plant. Vfite the total number of cells in it.
Answer : In an unfertilized mature embryo sac of a flowering
k plant 6 haploid cells are present, three antipodal cells, two synergids and one egg cell.

Question.15. In a typical monohybrid cross the F2-population ratio is written as 3 : 1 for phenotype but expressed as 1 : 2 : 1 for genotype. Explain with the help of an example.
Answer:

Genotype 1 : 2 : 1- One is homozygous dominant and 2 are heterozygous but dominant and one is homozygous recessive .

Question.16.Mention the contribution of S. L. Miller’s experiment on Origin of Life.
Answer : Miller carried out the experiment as a proof of oparin-Haldane hypothesis. The experiment was as follows :
(i) Miller sealed a minture of water vapour (H20), NH3, CH4 8H2 in a spark chamber, which was provided with electrodes to provide electrical discharges.
(ii) The electrical sparks were of 75,000 volts and the ratio of CH4 : NH3 : H2 as 2 : 1 : 2 and water vapour at 800°C.
(iii) The spark chamber was connected to another falsk with arrangement for boiling H20 for evaporation.
(iv) The other end of spark chamber was connected to a condenser for condensation and collection of equeous solution.
(v) A trap was connected with flask for boiling HzO.
(vi) The apparatus was a controlled one, without any energy source.
(vii) After 18 days, he analysed the products after cooling (hem and proposed that HGN was formed from methane and NH3 and reacted with other compounds of gas to born amino acids like Alanine, Glycine and As partic Acid.
He gare the conclusion that similar synthesis could have occured in the prinitire atmsopheric condition.

Question.20. (a) Explain “birth rate” in a population by taking a suitable example.
(b) Write the other two chlracteristics which only a population shows but an individual cannot.
Answer : (a) It is the average number of new individuals added per unit population per year due to births, hatchings and germinations. If in a pool there were 20 lotus plants last year and plants reproduced and gave rise to 8 new plants, the total population is now 28. Birth rate will be equal to 8/20=0.4 offspring per lotus per year (b) (i) Death rate : per capita death
(ii) Sex Ratio : An individual is either a male or female, but a population has a sex ratio, e.g. 60% of the population of females and 40% of males.

SECTION-D

Question.25. A burglar in a huff forgot to wipe off his blood – stains from the place of crime where he was involved in a theft. Name the technique which can help in identifying the burglar from the blood stains. Describe the technique.
Answer: The technique which help in identifying the burglar from the blood stain is DNA fingerprinting. The procedure of DNA fingerprinting is as follows :
(i) Isolation of DNA : Isolation of DNA from the blood stain .
(ii) Cutting, sizing, and sorting : Special enzymes called restriction enzymes are used to cut the DNA at specific places. The DNA pieces are sorted according to size by a sieving technique called electrophoresis. This technique is the biotechnology equivalent of screening sand through progressively finer mesh screens to determine particle sizes. The DNA fragments contain VNTRs (Variable Nun her of Tandem Repeats)
(iii) Transfer of DNA to nylon : The distribution of DNA pieces is transferred to a nylon sheet by placing the sheet on the gel and soaking them overnight.
(iv) Probing : Adding radioactive or coloured probes to the nylon sheet produces a pattern called the DNA fingerprint. Each probe typically sticks in only one or two specific places on the nylon sheet.
(v) Radiograph : The -final DNA fingerprint is built by using several probes (5-10 or more) simultaneously. These places are marked as dark bands when X-ray film is developed. This process is called auto radiography.

Question.28. (a) Write the specific features of the genetic code AUG.
(b) Genetic codes can be universal and degenerative. Write about them, giving one example of each.
(c) Explain aminoacylation of tRNA.
Answer: (a) Features of the genetic code AUG are as follows:
(i) AUG has dual function.
(ii) AUG codes for Methionine and acts a initiator codon.
(b) (i) Universal: From bacteria to human UUU would code for Phenylalanine. Some exceptions are there in mitochondrial codons and in some protozoans.
(ii) Degenerate : Some amino acids are coded by more than one codon. E.g. Phenylalanine is coded by UUU and UUC.
(c) Aminoacylation of tRNA : (i) In the presence of enzyme Amino-acyl-tRNA synthetase from DHU loop of tRNA, specific amino acid binds with ATP.

(ii) The AA-AMP-E complex formed in first step reacts with specific tRNA. Thus, amino acid is transferred to tRNA. The enzyme & AMP are then released.

OR
(a) Differentiate between dominance and co-dominance.
(b) Explain co-dominance taking an example of human blood groups in population.
Answer:


(b) In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles IA, IBand i Hence, referred to as multiple allelism. A person possesses any two of the three alleles. IA and IB dominate over i. But with each other, IAand IB are co-dominant.

SET-III

SECTION-A

Question.1. What is detritus food chain made up of ? How do they meet their energy and nutritional requirements ?
Answer : Detritus food chain made up of decomposers which are heterotrophic organisms, mainly fungi and bacteria. They get energy and nutrients by decomposing dead organic matter or detritus.

Question.3. Name the phenomenon and one bird where the female gamete directly develops into a new organism.
Answer : Phenomenon : Parthenogenesis Bird : Turkey

SECTION-B

Question.10. What is meant by “alien species” invasion ? Name one plant one animal alien species that are a threat to our Indian species.
Answer : An alien species whose introduction does or is likely to pose threat to the survival of many native species and cause their extinction.
Plants ; Parthenium, Lantana, Eicchomia Animal: African catfish, Clarias gariepinus

SECTION-C

Question.14. Work out a cross to find the genotype of a tall pea plant. Name the type of cross.
Answer : (i) Test cross is used to find out the genotype of any trait. In this cross F hybrid of pure tall plant and a pure dwarf plant is crossed with a dwarf plant, e.g., Tt x tt Plants of F, Tt (Tall) x tt (Recessive)

The progeny consists of tall and dwarf plants in the ratio of 1:1
(ii) If the dominant plant are homozygous i.e. TT, then the progeny will have all tall plants :

Question.15. Write the Oparin and Haldanes hypothesis about the origin of life on earth. How does meteorite analysis favour this hypothesis.
Answer : Oparin-Haldane Hypothesis : Alexander I Oparin (1894-1980) a Russian biochemist &J.B.S. Haldane (1892-,1964), a British Scientist, put forward the concept that the first living organism evolved from non-living material. According to Oparin & Haldane (1929). Spontaneous generation of early molecules might have taken place through a series of chemical reactions from the earth’s primordial soup in a reducing atmosphere. The compounds of soup could be expected to react with one another producing a variety of chemical substances like amino acids, sugars, N2 bases. These precursor molecules then combined resulting in the appearance of proteins, polysacharides & nucleic acids. Energy required for these reactions was provided by UV radiations, cosmic rays, electric discharges etc.

Question.19 (i) Explain DNA polymorphism as the basis of genetic mapping of human genome.
(ii) State the role of VNTR in DNA fingerprinting.
Answer : (i) Polymorphism (variation at genetic level) are tlie result of mutation. Variation in allele sequence is DNA
polymorphism i.e., if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01. In other term we can say that if an inheritable mutation is observed in a population at high frequency, it is referred as DNA polymorphism. There is a variety of different types of polymorphism ranging from single nucleotide change to very large scale changes. For evolution and specification, such polymorphism are important. DNA polymorphism is also used in genetic and physical maps on the human genome. In genetic mapping, information on polymorphism of restriction endonuclease recognition sites are used.
(ii) Role of VNTR in DNA Finger printing : Radiolabelled VNTR (Variable Number of Tandem Repeat) is used as a probe in DNA finger printing. A tandem repeat is a short sequence of DNA that is repeated in a head-to-tail fashion at a specific chromosomal locus. Tandem repeats are interspersed throughout the human genome. Some sequences are found at only one site – a single locus – in the human genome. For many tandem repeats, the number of repeated units vary between individuals. Such loci are termed VNTRs.

SECTION-D

Question.30. How does the process of natural selection affect Hardy- Weinberg equilibrium ? Explain. List other four factors that disturb the equilibrium.
Answer : Population or Mendelian population is a group of individuals present in ageographical area which share acommon pool. Gene frequency is the percentage of an allele in relation to the total alleles of a gene in an interbreeding population. All the genes & their alleles together constitute gene pool. Normally the alleles tend to maintain an equilibrium with reference to one another over the generations. It is referred as genetic equilibrium. Such a population is referred as non¬evolving populaiton. G.H. Harady & withelm weinberg proposed a principle independently in 1908, about the genetic structure of a non-evolving population. It is known as Hardy-weinberg equilibrium. The allelic frequencies in non-evolving population are stable and remain constant from generation to generation.
The Hardy-Weinberg Equilibrium can change by process of natural selection. Natural or survival of the fittest is a major factor that adds variation in the population, change the gene frequies of the gene pool leading to evolution & formation of a new distinctive gene pool. It operates through differential or non-random reproduction. If same type of selection continues for a few generation; the gene popl will undergo change the alleles having the advanage of surviving. Thus it leads to change of gene pool.
OR
(a) Explain Mendel’s Law of independent assortment by taking a suitable example.
(b) How did Morgan show the deviation in inheritance pattern in Drosophila with respect to this law.
Answer : (a) Dihybrid cross is based on Law of Independent Assortment. This law states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.
The image shows a true-breeding plant with the dominant traits of green pod colour (GG) and yellow seed colour (YY) being cross-pollinated with a true-breeding plant with yellow pod colour (gg) and green seeds (yy). The resulting offspring are all heterozygous for green pod colour and yellow seeds (GgYy). If the offspring are allowed to self pollinate, a 9 : 3 : 3 : 1 ratio will be seen in the next generation. About 9 plants will have green pods and yellow seeds, 3 will have green pods and green seeds, 3 will have yellow pods and yellow seeds and 1 will have a yellow pod and green seeds.

(b) Morgan and his group observed that when the two genes in a dihybrid cross were located on the same chromosome, the proportion of parental gene combinations in the progeny were much higher than the non-parental or new combinations -(recombinants) of genes.