CBSE previous Year Solved Papers Class 12 Biology Outside Delhi 2016
Time allowed : 3 hours Maximum Marks: 70
General Instructions :
- There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
- Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
- Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
- Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
- Section D contains question number 23, Value Based Question of four marks.
- Section E contains question number 24 to 26, Long Answer type questions of five marks each.
- There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.
Question.1. 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.
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 :
- Collection of variability of germplasm.
- Evaluation and selection of parents.
- Cross hybridisation among the selected parents.
- 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.
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.
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.
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 ?
(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.
(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.
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.
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.
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.
(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.
(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.
(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.
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.
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.
(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.
(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 :
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.
(a) The four components of an ecosystem are :
- Energy flow
- 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.