Jagran Josh Logo
  1. Home
  2. |  
  3. Board Exams|  

CBSE Class 12 Biology Solved Question Paper: 2016

Nov 2, 2017 15:24 IST
CBSE Class 12 Biology Solved Question Paper: 2016
CBSE Class 12 Biology Solved Question Paper: 2016

CBSE Solved Question Paper for Class 12th Biology 2016 board exam is available here. CBSE Class 12th Biology 2016 board exam was held on 21st March 2016. With this Solved Paper students can easily understand the level of questions which are asked in CBSE Class 12 Biology board exam. Solutions given in this Solved Question Paper are to the point and with these solutions, students can easily understand how to give proper answers in CBSE 12th Biology board exam to secure maximum marks.

Complete CBSE Class 12 Biology Solved Question Paper 2016 is given below

Solved Question Paper 2016

Class ‒ XII

Subject ‒ Biology (Theory)

All India

Time allowed: 3 hours                                                                           Maximum Marks: 70

General Instructions:

(i) There are a total of 26 questions and five sections in the question paper. All questions are compulsory.

(ii) Section A contains questions number 1 to 5, very short-answer type questions of 1 mark each.

(iii) Section B contains question number 6 to 10, short-answer type I questions of 2 marks each.

(iv) Section C contains question number 11 to 22, short-answer type II questions of 3 marks each.

(v) Section D contains question number 23, value based question of 4 marks.

(vi) Section E contains question number 24 to 26, long-answer type questions of 5 marks each.

(vii) There is no overall choice in the question paper, however, an internal choice is provided in one question of 2 marks, one question of 3 marks and all the three question of 5 marks. In these questions, an examinee is to attempt any one of the two given alternatives.

Section A

Question1. A male honeybee has 16 chromosomes while its female honeybee has 32 chromosomes. Give one reason.

Solution1. A male honeybee is produced by parthenogenesis and hence is haploid (n=16) while its female is produced by fertilized egg, the product of fusion of drone’s sperm (n=16) and female’s egg (n=16), and hence is diploid (2n=32).

CBSE Class 12 Biology Syllabus 2017 - 2018

Question2. Mention the role of genetic mother in MOET.

Solution2. Multiple Ovulation Embryo Transfer (MOET) includes administration of hormones to genetic mother to induce production of more number of mature eggs at a time, i.e. superovulation, which in turn are then transferred to surrogate mother.

Question3. What is biopiracy?

Solution3. Unauthorised use of biological resources without any compensatory payment done to the concerned people is termed as biopiracy.

Question4. Mention two advantages of preferring CNG over diesel as an automobile fuel.

Solution4. CNG is more efficient fuel than diesel as burning of CNG leaves very little residues.

2: Being less costly than diesel, it is less subjected to adulteration.

Question5. Write the probable differences in the eating habits of Homo habilis and Homo erectus.

Solution5. Homo habilis were not flesh eater and taking up “the meat eating” habit lead to appearance of meat eating Homo erectus with development of larger brain, small teeth and smaller stomach and gut.

Section B

Question6. A single pea plant in your kitchen garden produces pods with viable seeds, but individual papaya plant does not. Explain.

Solution6.

Pea plant is a dioecious plant and produce bisexual flowers bearing both stamen and pistel on the same flower. Therefore, self fertilization between pollen and ovule of same flowers produce viable seeds. Papaya is a monoecious plant and produce unisexual flowers which in turn facilitate only cross fertilization. Hence, single papaya plant cannot produce viable seeds.

Question7. Following are the features of genetic code. What does each one indicate? Stop codon, Unambiguous codon, Degenerate codon and Universal codon.

Solution7.

Stop codon: Codons that cause termination of protein synthesis and do not code for any amino acid; Example: UAA (ochre), UGA (opal) and UAG (amber).

Unambiguous codon: one codon for one amino acid only; Example: GAA codes for glutamine only.

Degenerate codon: multiple codons specify single amino acid; Example: UUU and UUC code for phenylalanine.

Universal codon: a particular codon specifies a particular amino acid in all the organisms. The mRNA codon "ACU" codes for threonine, irrespective of its source.

Question8. Suggest four important steps to produce disease resistant plant through conventional plant breeding technology.

Solution8.

1. Selection of parent plants with desired combination of characters.

2. Hybridization of selected parent plants to obtain progeny carrying the desired characters of both parent plants.

3. Selection of superior recombinant offspring.

4. Testing of selected recombinants followed by their release for commercial production.

Question9. Name a genus of baculovirus. Why they are considered good biocontrol agents?

Solution9.

Nucleopolyhedrovirus. Being species specific selective and narrow spectrum biocontrol agents, these viruses do not exert any harmful impact on crops, animals and human and non target insects. For example, Nucleopolyhedrovirus targets mainly moths and butterflies.

CBSE Class 12 Biology Question Paper 2017

Question10. Explain the relationship between CFC’s and ozone in stratosphere.

OR

Why sacred grooves are highly protected?

Solution10.

Exposure of CFCs to UV rays releases the chlorine atoms which in turn react with ozone molecules and release diatomic or molecular, oxygen. When a free oxygen atom reacts with this chlorine-oxygen compound, diatomic oxygen molecules and the chlorine is released. The released chlorine devastates more ozone molecules. Molecular oxygen cannot keep UV rays from reaching the Earth's surface like ozone molecules.

OR

Sacred grooves are highly protected by local communities as their religious sentiments are attached with those forest fragments. Example: grooves present in Gumpa forest of Arunachal Pradesh.

Question11.

(a) Name the organic material the exine of pollen grain is made up of. How is the material advantageous to pollen?

(b) Still it is observed that it does not for 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 Reproductive and child health care programme.
(b) What is amniocentesis and why there is a statutory ban on it?

Solution11.

a. Exine is the outer covering of pollen grain which is highly sculptured with sporopollenin; a chemically inert polymer of carotenoid ester which is highly resistant to degradation by any enzyme, as known so far.  It can withstand extremes of temperature and desiccation. It protects the pollen from any damage.

b. The regions of exine where sporopollenin is absent are known as germ pores. These pores serve as site for formation of pollen tube or germ tube after germination.

c. Pollen banks store the pollen grains of different crops in liquid nitrogen (-196 degree Celsius) and serve as reservoir of germplasm to be used for plant breeding programmes.

OR

a. The programme aims to create awareness about different aspects and problems concerned with reproduction. For the purpose, it helps to build and maintain reproductive health society by providing required facilities and support.

b. Use of amniotic fluid to detect presence of any genetic defect in unborn child is called amniocentesis; the fluid has free foetal cells which are cultured and tested for any genetic disorder. Since amniocentesis can also disclose sex of foetus, it has been misused by expecting parents abort the developing foetus if it is a female child. Sex determination by amniocentesis caused genetic imbalance by reducing number of females to males and thus its use for sex determination of foetus has been banned in India.

Question12. What is a test cross? How can it decipher the heterozygosity of a plant?

Solution12. A test cross is performed to determine the genotype of a dominant parent if it is a heterozygous- or homozygous-dominant. For the purpose, the dominant parent is crossed with homozygous recessive parent. Presence of recessive progeny confirms presence of recessive allele in otherwise dominant parent; it is heterozygous dominant. All dominant progeny confirms that the dominant parents is homozygous and does not carry the recessive allele. For example: a dihybrid test cross between TtRr and ttrr gives 1:1:1:1 ratio while a dihybrid test cross between TTRR and ttrr gives all TtRr progeny.

Question13. (a). What do “Y” and “B” stands for in YAC and BAC used in Human Genome Project (HGP)? Mention their role in the project.

(b). Write the percentage of total human genome that codes for proteins and the percent of discovered genes whose functions are known as obscured during HGP.

(c) Expand “SNPs” identified by scientists in HGP.

Question13.

a. Bacterial Artificial Chromosome (BAC) and Yeast Artificial Chromosome (YAC) are the artificial construct used as cloning vector in HGP for cloning of human DNA fragments.

b. Almost 2% of total human genes i.e. around 30,000 genes code for proteins and around 50% of total human genes carry out unknown functions.

c. Single Nucleotide Polymorphisms (SNPs)

Question14. Differentiate between analogy and homology. Give one example of each.

Solution14. Homologous organs are adapted to perform different functions and have different appearance but share common basic structural plan and origin. Homologous organs exhibit divergent evolution and the condition is termed as homology. Thorns of Bougainvillea and tendrils of Cucurbita are modified branches, present in axil of leaves modified for help the plant in climbing and protect the plants from predators respectively.

Analogous organs perform similar function and have similar appearance but have different basic structure and origin. They exhibit convergent evolution and the condition is termed as analogy. Wings of birds are modified forelimbs with feathery surface while wings of butterfly are sac like extensions of body wall; both modified for flying.

Question15. a. It is generally observed that children who have suffered from chicken pox in childhood may not contact the same disease in their adulthood. Explain giving reasons that basis of such immunity in an individual. Name this kind of immunity.

b. What are interferons? Mention their role.

Solution15. a. Exposure of B cell with chicken pox antigen and its binding to the antigen receptors stimulates B cell to undergo clonal expansion to produce plasma cells and memory cells. Upon second encounter of person with chicken pox antigen, the already present memory cells divide quickly to produce more lymphocytes which in turn quickly produce antibodies. This kind of immune response selectively targeted against particular antigen, to which body has previously been exposed, is known as acquired immunity and is responsible to prevent second time infection.

b. Interferons are carbohydrate containing proteins that are produced by virus infected fibroblasts and leukocytes. They have antiviral and anticancer properties and protect the nearby cell from getting infected by inhibiting replication of virus.

Question16.

a. Write the two limitations of conventional breeding techniques that led to promotion of micropropagation.

b. Mention two advantages of micropropagation.

c. Give two examples where it is commercially adopted.

Solution16. a. Two limitations of conventional breeding programme:

1. It is prone to seasonal fluctuation and does not allow production of plants throughout the year.

2. It is a slow process that fails to keep pace with high demand of commercial crops.

b. Two advantages of micropropagation are:

1. Micropropagation can produce large number of plants from small explant within a short period.

2. It can produce plants throughout the year, irrespective of the seasonal variations.

c. Micropropagation is commercially adopted for

1. Production of virus-free plants of sweet potato (Ipomea batatus) and yam (Discorea rotundata).

2. Commercial production of crop plants such as tomato and potato.

Question17.

a. How do organic farmers control pests? Give two examples.

b. State the difference between their approaches from that of the conventional pest control methods.

Solution17.

a. Organic farmers use biological controlling agents as insecticides and pesticides that are species specific, narrow spectrum agents which do not cause any harm to non target species. Bacillus thuringiensis (Bt) produces Bt toxin during sporulation. The toxin bind to receptors on midgut epithelial cells of gallworm, become inserted into the plasma membrane where they form pores leading to cell death through osmotic lysis; thus serving as biopesticide for gall worms. Trichoderma serve as very effective biocontrolling agent for soil borne plant disease management. It is a free-living fungus, and reduces pathogenic infections by different mechanisms like competition, antibiosis, hyphal interactions, mycoparasitism and enzyme secretion.

b. On the contrary to the complete destructive approach of conventional breeder of killing the pests, organic farmers avoid total eradication of pests as it can adversely affect the other organisms that are dependent on them. Rather they try to keep them under controlled number so as to avoid any possible harm to the crop.

Question18.

a. Name the selectable marker in the cloning vector PBR322? Mention the role they play.

b. Why is the coding sequence of an enzyme β galactosidase a preferred selectable marker in comparison to the ones named above?

Solution18.

a. Selectable markers are the antibiotics resistant genes present in cloning vectors. The cloning vector PBR322 has two antibiotic genes, the ampicillin and tetracycline resistance genes that serve as selectable markers to distinguish the transformed cells from the non transformed cells and the recombinant cells from the non recombinant ones.

b. The β galactosidase gene serves as alternate selectable marker and exhibit insertional inactivation i.e. the gene is inactivated upon insertion of foreign sequence into it. Growing bacteria on chromogenic substrate gives blue coloured colonies of non recombinants and colour less colonies of the recombinants. β galactosidase gene is preferred as selectable marker over antibiotic resistant genes due to easy visualization of recombinants making the process of identification less cumbersome.

Question19.

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.

Solution19.

a. Biotechnology experiments require the host cells to take up the foreign DNA sequences. Hence, the cells are made ‘competent’ to enable them for DNA uptake. One the method of making the cells competent includes treatment of cells with divalent cations (such as calcium ions) which in turn attracts both negatively charged DNA and negatively charged groups in the inner core of cell membrane allowing the DNA to pass through it and enter the cell.

b. Biolistic gun is the method of DNA transfer in plant cells wherein cells are shot with microscopic gold or tungsten particles coated with hundreds of copies of the genes to be transfected into the target cell.

Question20. Explain the enzyme replacement therapy to treat adenosine deaminase deficiency. Mention two disadvantages of this procedure.

Solution20.

Enzyme replacement therapy (ERT) is replacing the defective enzyme in usually by an intravenous (IV) infusion containing the enzyme. The process includes preparation of modified enzyme by conjugating polyethylene glycol with purified bovine ADA followed by periodic administration of this PEG-ADA conjugate by intramuscular injection.

Two disadvantages of this procedure:

1. It treats disease temporarily.

2: Periodic administration of enzyme makes the process very expensive.

Question21.

Name and explain the type of interactions that exist in mycorrhizae and between cattle egret and cattle.

Solution21.

Mycorrhizae is association between fungi and roots of higher plants wherein fungal hyphae decompose the organic matter present in soil, increase the surface area of roots of the plant for water and mineral absorption while the plant provides the fungus with organic nutrients. Type of association wherein both the organisms are benefitted from each other in one or other ways is called as mutualistic association.

Cattle egret the birds found in proximity to grazing cattle. Movement of cattle in grass makes the insects of grass to come out which are then easy target for egret bird. Here, the bird is being benefitted from the association and the cattle is neither being harmed nor benefitted; this type of association is called as commensalism.

Question22.

Differentiate between primary and secondary succession. Provide one example of each.

Solution22.

The succession that takes place in the area never occupied by any life form ever is termed as primary succession. It is a very slow process as the soil is not fertile. Succession in cooled down lava is primary succession. The succession occurring the area where the previous life forms were destroyed is called as secondary succession. Being previously occupied by life forms, soil is fertile initially making the process fast as compared to the primary succession. Succession in burnt forest is secondary succession.

Section D

Question23. A large number of couples the world over the 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 the situation in India? Mention any two values that as a biology student can promote the check this social evil.

b. State any two reason 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.

Solution23.

a. The females in India are blamed for infertility owing to lack of awareness about various reasons for infertility. As a biology student, I will

1. Aware them about the possible causes of infertility in both male and female; it is not solely caused by problems with female partners.

2. I will suggest them to go for proper diagnosis first to know the reason of infertility and not to blame the females always for the same.

b. Infertility is caused by congenital diseases and drug intake.

c. Artificial insemination includes artificial insertion of sperms into female vagina to overcome the fertility barriers caused by inability of male in penis erection or low sperm count in semen.

Section E

Question24. a. Explain the menstrual phase in a human female. State the levels of pituitary hormones during this phase.

b. Why is follicular phase in menstrual cycle also referred as proliferative phase? Explain.

c. Explain the events that occur in 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 demisntrate to the student of secondary level in your school, the procedure that shall ensure cross pollination in a hermaphrodite flower. List the different steps that you would suggest and provide reason for each of them.

b. Draw a diagram of a section of a megasporangium of an angiosperm and label funiculus, micropyle, embryosac and nucellus.

Solution24. a. Menstrual cycle is the monthly series of events that involve ovary and uterus. The follicular phase (day 1 to 13) of menstrual cycle is marked by secretion of FSH by anterior pituitary gland. It stimulates growth, development and hormonal secretion of Graafian follicle and ovum maturation. Developing follicles secrete estrogen (ovarian hormone). LH is also secreted by anterior pituitary which inhibits secretion of FSH, stimulate maturation and rupture of Graafian follicle and release of ovum (ovulation) on the 14th day of cycle. Ovulation is followed by luteal phase (day 15-28) and is marked by secretion of estrogen and progesterone by corpus luteum. Negative feedback by progesterone inhibits synthesis of pituitary hormone LH.

b. Follicular phase in ovary stimulates proliferative phase in uterus. It is marked by uterine endometrium thickening, under influence of estrogens as secreted by developing follicles, to prepare it for implantation; hence the name proliferative phase.

c. LH surge between 12 to 14 days of cycle stimulates rupture of Graafian follicle and release of ovum i.e. ovulation. Ovulation is followed by rupturing of Graafian follicle with a clot inside called as corpus hemoragicum. Absorption of clot and other changes transforms it into corpus luteum which secretes estrogen and progesterone during first four months of pregnancy.

d.

Graafian follicle

OR

a. A hermaphrodite flower has both pollen and stigma; to prevent self pollination following techniques can be used:

Emasculation: removal of anthers from flowers to prevent self pollination is called as emasculation.

Bagging: cover the stigma with polycovers to ensure cross pollination of stigma and to prevent self pollination.

Dusting of pollens from desired plant: followed the bagging, when stigma becomes receptive, pollens from the selected plants are dusted over it and the stigma is rebagged to check any contamination.

b.

Question25. 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. Explant the processing the hnRNA needs to undergo before becoming functional mRNA in eukaryotes.

Solution25. Meselson-Stahl showed semi conservative mechanism of DNA replication in which a DNA strand serves as a template for the synthesis of a new strand and produces two new DNA molecules, each with one new strand and one parental strand. They cultured E. coli cells initially in a medium containing ammonium salts prepared with heavy nitrogen (15N) to label all cellular DNA. Then cells were transferred to a medium containing the normal light isotope (14N). DNA was analyzed by equilibrium density-gradient centrifugation in periodically collected sample to separate heavy-heavy (H-H), light-light (L-L), and heavy-light (H-L) duplexes into distinct bands.

Observation and conclusion: Since E coli cell takes 20 min to complete one round of cell cycle, the DNA extracted after 20 min had a hybrid density while that extracted after 40 min exhibited equal amounts of hybrid and light density; it was concluded that each new DNA molecule has one parental and one new strand i.e. the semi conservative mode of replication.

OR

a. Transcription occurs in three steps namely: initiation, elongation and termination.

Initiation of bacterial transcription includes binding of RNA polymerase to promoter; binding of RNA polymerase enzyme with sigma factor locate the enzyme to promoter sequence by destabilising all non specific interactions between the enzyme and DNA till it encounters the promoter sequence. It is followed by unwinding of DNA duplex.

Elongation of transcription includes dissociation of sigma factor and polymerisation of nucleoside triphosphates.

Rho dependent termination of transcription: The rho protein binds with the RNA at rho utilization elements (rut, CA rich sequence) and moves, in the 5’ to 3’ direction, towards the transcription complex that is paused at a termination site. Rho independent termination of transcription occurs by formation of hairpin structure by GC rich termination sequence.

b. hnRNA undergoes processing by three processes, namely, splicing, capping and tailing.

Splicing is removal of introns from the hnRNA and ligation of exons to form a continuous sequence specifying a functional polypeptide.

Capping of eukaryotic mRNAs refers to addition of 7-Methylguanosine to the 5’ end of almost all eukaryotic mRNAs in an unusual 5’,5’- triphosphate linkage. Tailing refers to cleavage of 3’ end and addition of 80 to 250 A residues to create a poly(A) tail.

Question26. 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 curve represents 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.

Solution26: a. Two types of growth models are:

1: Exponential growth model: In presence of unlimited resources, there is exponential growth of population. Let’s assume that N= population size, b= birth rate, d= death rate; increase or decrease in N during time t is:

dN/dt = (b ‒ d) × N

if (b ‒ d) = r ; r = intrinsic rate of natural increase

then dN/dt= r N

or Nt = N ert

N0 = Initial population density at time 0

Nt = Population density at time t

r = intrinsic rate of natural increase

e =base of natural log

2:  Logistic growth model: limited natural resources pose competition among individuals and survival of fittest only leads to logistic population growth. It is characterized by initial lag phase followed by increase/decrease and then asymptote, when the population obtains its carrying capacity (K). A curve having N plotted against t gives a sigmoid shape for logistic growth.

dN/dt= rN (K‒N/K)

b. “J” shaped curve of exponential growth model represents unlimited population growth in presence of unlimited natural resources. The “S” shaped curved of logistic growth model represents presence of limited growth resources causing decrease in population growth under scarcity of resources and levelling off the population growth when it reached the carrying capacity thereby giving a “S” shaped curve.

c. The present human population growth represents logistic growth model as limited natural resources cannot keep pace with increasing population growth. It is not sustainable as it would reach a point having scarcity of resources. We need to look for alternative resources to fulfil the demands of increasing population.

OR

a. Functioning of an ecosystem is studied by considering productivity, decomposition, nutrient cycling and energy flow.

Phytoplanktons are the producers in pond ecosystem that capture the sunlight and synthesize organic nutrients. The rate at which these producers convert the energy into organic nutrients is referred to as primary productivity. Hence, phytoplanktons impart primary productivity to pond ecosystem.

Energy Flow: Consumers of pond ecosystem are zooplanktons (primary consumers), nectic animals (secondary consumers) and benthic animals (tertiary consumers). Primary consumers depend on producers for nutrients while the secondary consumers feed on primary consumers and tertiary consumers on the secondary ones. Thus, energy flows from producers to consumers.

Decomposition and nutrient cycling: parasites and saprophytes constitute the decomposers of pond ecosystem and feed on dead and decaying organic matters of plants and animals. Decomposition releases nutrients in ecosystem which in turn serve as raw material for producers to carry out synthesis of organic matter (primary productivity).

b. Food chain in pond ecosystem starts with plants which are then eaten by primary consumers (zooplanktons) and primary consumers by secondary consumers. Type of food chain in which energy flows from autotrophic organisms to the heterotrophic organisms is referred to as grazing type food chain. Hence pond ecosystem exhibits grazing type of food chain.

 

Post Comment

Latest Videos

Register to get FREE updates

    All Fields Mandatory
  • (Ex:9123456789)
  • Please Select Your Interest
  • Please specify

  • By clicking on Submit button, you agree to our terms of use
    ajax-loader
  • A verifcation code has been sent to
    your mobile number

    Please enter the verification code below

Newsletter Signup
Follow us on
X

Register to view Complete PDF