CBSE Class 12 Biology Question Paper (Delhi 2017) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Delhi 2017) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Delhi 2017) with Solutions

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 questions number 1 to 5, very short-answer type questions of 1 mark each.
3. Section B contains questions number 6 to 10, short-answer type I questions of 2 marks each.
4. Section C contains questions number 11 to 22, short-answer type II questions of 3 marks each.
5. Section D contains question number 23, value based question of 4 marks.
6. Section E contains questions number 24 to 26, long-answer type questions of 5 marks each.
7. 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 questions of 5 marks. In these questions, an examinee is to attempt any one of the two given alternatives.

Section – A

Question 1.
Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers.
Answer:
Test cross is used for the determination of genotype of pea plant bearing violet flowers.

Note
Test cross helps the students to determine whether the genotype of the plant is heterozygous dominant (Ww) or homozygous dominant (WW).

Question 2.
State two postulates of Oparin and Haldane with reference to origin of life.
Answer:
Oparin and Haldane proposed that the first form of life could have been evolved from pre-existing non-living organic molecules such as RNA, protein and so on.

They also state that the formation of life was preceded by chemical evolution such as the formation of diverse organic molecules from inorganic constituents.

The conditions on the earth for survival were high temperature, volcanic storms and reducing atmosphere contains CK₄, NH₃ and so on.

Question 3.
A herd of cattle is showing reduced fertility and productivity. Provide one reason and one suggestion to overcome this problem.
Answer:
Reduced fertility and productivity in a herd of cattle is because of inbreeding depression. Inbreeding depression occurs because of inbreeding. This problem can be overcome by mating of selected animals with the unrelated superior animals of the same breed. This helps to restore the fertility and productivity.

Note
Inbreeding refers to the mating of more closely related individuals within the same breed for 4-6 generations.

Question 4.
What are Cry genes ? In which organism are they present ?
Answer:
Cry genes are present in bacterium Bacillus thuriengiensis. It codes for toxin called Bt toxin. The proteins encoded by the genes cryIAc and cryIIAb control the cotton bollworms whereas cryIAb controls corn borer.

Note
The Bt toxin protein exist as inactive protoxins but once an insect ingest the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that causes cell swelling and lysis and eventually cause death of the insects.

Question 5.
An electrostatic precipitator in a thermal power plant is not able to generate high voltage of several thousands. Write the ecological implication because of it.
Answer:
An electrostatic precipitator will not be able to remove particulate matter present in the exhaust of thermal power plants and because of this dust particles will be released into the air. This results in air pollution.

Section – B

Question 6.
A pollen grain in angiosperm at the time of dehiscence from an anther could be 2-celled or 3-celled. Explain. How are the cells placed within the pollen grain when shed at a 2-celled stage ?
Answer:
In many angiosperms, pollen grains are released in 2-celled stage while in other plant species, the generative cells are dividing into 2-male gametes and they will form 3-celled stage. When the pollen grain is shed at 2-celled stage then it has two unequal cells such as a bigger vegetative cell and smaller generative cell.

Question 7.
Differentiate between the genetic codes given below :
(a) Unambiguous and Universal
(b) Degenerate and Initiator
Answer:
(a) Difference between unambiguous and universal:

(b) Difference between degenerate and initiator

Note
In prokaryotes, GUG acts as a initiation codon and codes for amino acid Valine.

Question 8.
Mention one application for each of the following :
(a) Passive immunization
(b) Antihistamine
(c) Colostrum
(d) Cytokinin-barrier
Answer:
(a) Passive immunization: When readymade antibiotics are introduced into the body, then it is called passive immunization. Passive immunization provides quick immune response in the body.
(b) Anti-histamines: Anti-histamines are the chemicals that are given against allergic reactions.
(c) Colostrum: Colostrum is the yellow fluid produced during the initial days of lactation. It is rich in antibodies IgA. It provides innate immunity to the new baby.
(d) Cytokinin barrier: It involves interferons. Virus- infected cells secrete proteins called interferons that protect non-infected cells from further viral infections.

Question 9.
Name the microbes that help production of the following products commercially :
(a) Statin
(b) Citric acid
(c) Penicillin
(d) Butyric acid
Answer:
(a) Statin: It is produced by yeast called as Monascus purpureus. It is used for blood-cholesterol lowering agents.
(b) Citric acid: It is produced by fungus Aspergillus niger.
(c) Penicillin: Penicillin is an antibiotic that is produced by bacteria called Penicllium notatum.
(d) Butyric acid: It is produced by a bacterium called Clostridium butylicum.

Question 10.
List four benefits to human life by eliminating the use of CFCs.
OR
Suggest two practices giving one example of each, that help protect rare or threatened species.
Answer:
CFCs (Chlorofluorocarbons) are responsible for affecting the human life. As the chlorinated molecules from CFCs releases leads to cause pollution and ozone layer depletion. Reduction in the use of CFCs helps in following ways:

• It helps in the prevention of ozone layer depletion.
• It also helps in reduction of greenhouse effect.
• It also reduces El Nino effect or odd climatic changes.
• Reduction in CFCs will also help in the prevention of snow blindness and inflammation of cornea.

Note
CFCs are widely released into the atmosphere from refrigerators and is discharged in the lower part of atmosphere. Then it move upwards and reach atmosphere. Chlorine atoms are releasing into the atmosphere by the action of UV rays. It results in the ozone depletion.

OR

Practices that help to prevent rare or threatened species are as follows:

1. Ex-situ conservation: In this, the rare or threatened plants and animal species are taken out from their natural habitat and place them in special area where they can be protected and given special care.
2. Cryopreservation: Gametes (sperms, eggs, tissues and embryo) of several endangered plants and animal species can be preserved by methods involves cryopreservation (-196°C). It can be fertilised in Invitro followed by propagation through tissue culture methods.
3. In-situ conservation: This approach involves conservation and protection of entire ecosystem, in order to protect its biodiversity at each and every level of food chain. As if we save forest then we save the tiger. (2 Marks)

Note
Ecologically unique and biodiversity-rich regions in India are legally protected as biosphere reserves, national parks and sanctuaries.

Section – C

Question 11.
(a) Can a plant flowering in Mumbai be pollinated by pollen grains of the same species growing in New Delhi ? Provide explanations to your answer.
(b) Draw the diagram of a pistil where pollination has successfully occurred. Label the parts involved in reaching the male gametes to its desired destination.
Answer:
(a) Yes, it can be only possible by means of artificial hybridisation in which a pollen grain of one flower is artificially introduced on the stigma of another flower. But it does not involve self-incompatibility of flowers.

• In this, one flower is emasculated and bagged.
• After some time, the bag is removed and then desired pollen grains are introduced on its stigma.

Note
Emasculation refers to the removal of anthers from the floral bud before the anther dehiscence by using a pair of forceps. Whereas bagging refers to the covering of emasculated flower with a bag made of butter paper in order to prevent contamination of stigma with unwanted pollen.

(b) Diagrammatic representation of pistil:

Question 12.
Both Haemophilia and Thalassemia are blood related disorders in humans. Write their causes and the difference between the two. Name the category of genetic disorder they both come under.
Answer:

Question 13.
(a) List the two methodologies which were involved in human genome project. Mention how they were used.
(b) Expand ‘ YAC’ and mention what was it used for.
Answer:
(a) Human Genome Project was also called mega project that has approximately 3 × 10⁹ bp. Human Genome Project involves two major approaches such as ESTs and Sequence Annotation.

• Expressed sequence Tags: This approach aims at identifying all the genes that are expressed as RNA.
• Sequence Annotations: This approach involves sequencing of the entire set of genome either coding or non-coding sequences or later assigning different region with functions.

(b) YAC stands for Yeast Artificial Chromosome. It is used as cloning vectors for cloning DNA fragments in suitable host to carry out the process of DNA sequencing.

Question 14.
Write the characteristics of Ramapithecus, Dryopithecus and Neanderthal man.
Answer:

Question 15.
Name a human disease, its causal organism, symptoms (any three) and vector, spread by intake of water and food contaminated by human faecal matter.

OR

(a) Why is there a fear amongst the guardians that their adolescent wards may get trapped in drug/alcohol abuse?
(b) Explain ‘addiction’ and ‘dependence’ in respect of drug/alcohol abuse in youth.
Answer:
The infectious disease that is caused because of the intake of contaminated food and water by human faecal matter is Amoebic dysentery or Amoebiasis.
This disease is caused by a protozoa Entamoeba histolytica. Its symptoms involve constipation, abdominal pain and cramps, stools with excess mucous and blood clots. Houseflies act as carrier for the transmission of parasites from faeces of infected person to food and water.

OR

(a) The adolescent’s wards may get trapped in drug/ alcohol abuse involves following reasons:

• Peer pressure
• Curiosity and need to try for adventure, excitement and experiment.
• To escape from stress, frustration and depression.
• To overcome hardships of life.
• Unstable or unsupportive family structure.

(b) Addiction: Addiction is defined as the psychological attachment to certain effects such as Euphoria or temporary feeling of well-being.
Dependence: Dependence is defined as the tendency of the body to show withdrawal syndrome or appearance of symptoms because of regular doses of drug or alcohol abuse is abruptly discontinued.

Question 16.
(a) Write the desirable characters a farmer looks for in his sugarcane crop.
(b) How did plant breeding techniques help north Indian farmers to develop cane with desired characters ?
Answer:
(a) The desirable characteristics a farmer wants to see in his sugarcane crop are:

• Higher yielding capacity
• Thicker stem
• Higher sugar content
• Ability to grow in North India.

(b) North Indian farmers developed sugarcane with desired characters by crossing two varieties of sugarcane such as Saccharum barberi that was originally grown in North India. It had low sugar content and yield. While Saccharum officinarum was originally grown in South India and had thicker stems and higher sugar content.

When these two species of sugarcane were successfully crossed desirable qualities such as higher yielding capacity, thicker stems, higher sugar content and ability to grow in the sugar cane areas of North India.

Question 17.
Secondary treatment of the sewage is also called Biological treatment. Justify this statement and explain the process.
Answer:
Secondary treatment of sewage is also called biological treatment because it involves biological organisms such as aerobic and anaerobic microbes and fungi for digestion of organic waste.

In this, the primary effluent is passed into the large aeration tanks and is constantly agitated mechanically. In this air is pumped and this allows the vigorous growth of useful aerobic microorganisms into floes. These microbes consume the maximum part of the organic matter in the effluent. This significantly reduces the biochemical oxygen demand (BOD) of the effluent. The sewage water is treated till the BOD is reduced.

Note
Floes are the masses of bacteria associated with fungal filaments to form a mesh like structures. BOD refers to the amount of oxygen consumed if all the organic matter in one litre of water were oxidised by bacteria.

Question 18.
(a) Explain the significance of ‘palindromic nucleotide sequence’ in the formation of recombinant DNA.
(b) Write the use of restriction endonuclease in the above process.
Answer:
(a) The palindromic sequences are the groups of letters that form the same word that is when both read forward and backward. For e.g “MALAYALAM”. Palindrome where the same word is read in both directions, the palindrome in DNA is a sequence of base pairs that reads the same on the two strands when orientation of reading is kept the same.

For example: The following sequences read the same on the two strands in 5’→3’ direction. The same sequence read in the 3’→5’ direction.
5’——GAATTC——3’
3’—–CTTAAG—–5’

(b) On finding the palindrome, the endonuclease binds to the DNA. It cuts the opposite strands of DNA, but between the same bases on both the strands and form ‘Sticky-ends’. This ‘sticky ends’ facilitates the action of enzyme DNA ligase and also helps in the formation of recombinant DNA.

Question 19.
Describe the roles of heat, primers and the bacterium Thermus aquaticus in the process of PCR.
Answer:
Role of heat in PCR:
Heat helps in the denaturation process in PCR. In this process, the dsDNA is heated in this process at very high temperature (94-96°C) results in the separation of two strands of DNA into single strands.

Role of primers in PCR:

Primers are the short synthetic single stranded DNA fragments that are complementary to DNA sequences that flank the target region of the DNA.

DNA polymerase enzyme extends the primers by using the nucleotides provided in the reaction and the genomic DNA as template. It helps in the extension of new chain.

Role of Bacterium Thermus aquaticus:

The bacteria Thermus aquaticus is thermostable bacteria. An enzyme Taq DNA polymerase is isolated from these thermostable bacteria. This enzyme remains active during high temperature during denaturation of double stranded DNA.

Question 20.
Explain the various steps involved in the production of artificial insulin.
Answer:
The various steps involved in the production of artificial insulin are as follows:

• The artificial insulin consists of two short polypeptide chains such as chain A and chain B.
• These two short polypeptide chains are linked together by disulphide bond.
• In mammals such as humans, insulin is synthesised as a prohormone that contains an extra stretch called the C peptide.
• This C peptide is not present in mature insulin and is removed during maturation into insulin.
• The two DNA sequences corresponding to A and B polypeptide chains of human insulin were prepared and these were introduced into E.coli in order to produce A and B chains separately, and these chains were extracted and then combined by creating disulphide bonds.

Diagrammatic representation of artificial insulin:

Question 21.
(a) “Organisms may be conformers or regulators.” Explain this statement and give one example of each, (b) Why are there more conformers than regulators in the animal world ?
Answer:
(a) Conformers: Conformers are those organisms that are not able to maintain a constant internal
body temperature with the external environmental conditions. They change body temperature and osmotic concentration with the changing external environmental. For example: all plants and fishes.

Regulators: Regulators are those organisms that are able to maintain homeostasis is according to external environment. They have ability to maintain their constant body temperature and osmotic concentration. For example: Birds and mammals.

(b) There are more conformers than regulators in the animal world because they lack the ability to maintain the constant internal body temperature with the changing external environment.

Question 22.
Describe the inter-relationship between productivity, gross primary productivity and net productivity.
Answer:
Productivity: Productivity is defined as the rate of biomass production.

Gross primary productivity: Gross primary productivity is defined as the rate of organic matter production during photosynthesis.

Net primary productivity: Net primary productivity involves the gross-productivity minus respiratory losses (R).
So, NPP = GPP-R
As per their definitions, all the terms are interrelated to each other.

Section – D

Question 23.
It is commonly observed that parents feel embarrassed to discuss freely with their adolescent children about sexuality and reproduction. The result of this parental inhibition is that the children go astray sometimes.
(a) Explain the reasons that you feel are behind such embarrassment amongst some parents to freely discuss such issues with their growing children.
(b) By taking one example of a local plant and animal, how would you help these parents to overcome such inhibitions about reproduction and sexuality ?
Answer:
(a) The important reasons that parents feel embarrassed to discuss freely with their adolescent children about sexuality and reproduction are illiteracy, conservative attitude, myths and misconceptions among parents. They feel shy to discuss such issues with their children freely. But it is responsibility of every parent to give right information to their children about sexuality, reproduction, adolescence changes and sexual practices so that their children will never be misleaded.

(b) By an example of male honey bee and orchid ophrys flower, it is evident that sexual attraction is a natural phenomenon, the honey bee is attracted to a ophrys flower and assumes its one petal as its female partner & pseudo copulates with it. So it is a natural phenomenon and parents should talk regarding this matter to the children.

Section – E

Question 24.
(a) When a seed of an orange is squeezed, many embryos, instead of one are observed. Explain how it is possible, (b) Are these embryos genetically similar or different ? Comment.
OR
(a) Explain the following phases in the menstrual cycle of a human female:
(i) Menstrual phase
(ii) Follicular phase
(iii) Luteal phase
(b) A proper understanding of menstrual cycle can help immensely in family planning. Do you agree with the statement? Provide reasons for your answer.
Answer:
(a) The occurrence of more than one embryo in a seed in oranges is because of polyembryony. In orange, the nucellar cells, synergids and integument cells are developed into a number of embryos of different sizes. For Example: Citrus.

Note
Sometimes the formation of more than one egg in an emhryo silicon lead to polyembryony.

(b) Parental characters are maintained in the embryos formed as a result of polyembryony and hence they are genetically similar. As, in this process there is no segregation of characters in the progeny.

OR

(a) The menstrual cycle involves following phases such as:

1. Menstrual phase: This phase takes place when released ovum is not fertilised. This phase occurs within the first days of cycle where menstrual flow occurs because of the breakdown of endometrial lining of the uterus.
2. Follicular phase: This phase occurs within the 5^th-14^th day of the cycle where the primary follicles grow to become a fully mature Graafian follicle, endometrium of uterus regenerates and Graafian follicle ruptures to release ova as ovulation occurs on 14th day.
3. Luteal phase: This phase occurs within the 15^th – 28^th day. In this phase, the remaining parts of the Graafian follicle transform into Corpus luteum and secretion of progesterone occurs that is essential for the maintenance of endometrium.

(b) Yes, I agree with the statement that taking appropriate precautions between 10^th -17^th day of menstrual cycle when the chances of fertilisation are high.

Question 25.
(a) Compare, giving reasons, the J-shaped and S-shaped models of population growth of a species.
(b) Explain “fitness of a species” as mentioned by Darwin.
OR
(a) What is an ecological pyramid ? Compare the pyramids of energy, biomass and numbers.
(b) Write any two limitations of ecological pyramids.
Answer:
(a) The difference between J shaped-growth curve and S shaped-growth curve are as follows:

(b) According to Darwin, “Fitness of a species” means a specific species is reproductively fit. When the availability of resources are limited, competition takes place between the species and it only favours the survival of the fittest one who reproduce to leave more progeny. (2 Marks)

OR

Ecological pyramid is defined as the relationship between producers and consumers in an ecosystem
that can be graphically represented in the form of a pyramid.
Different types of ecological pyramids are as follows:

(i) Pyramid of number: It is defined as the relationship between producers and consumers in an ecosystem that can be presented in the form of a pyramid in terms of number.
Diagrammatic Representation of Pyramid of number:

(ii) Pyramid of biomass: It is defined as the relationship between producers and consumers in an ecosystem that can be represented in the form of a pyramid in terms of biomass. It can be upright or inverted.
Diagrammatic Representation of Pyramid of Biomass:

(iii) Pyramid of energy: It is defined as the relationship between producer and consumers in an ecosystem that can be represented in the form of pyramid in terms of flow of energy. It is always upright as the energy is lost as heat at each step.
Diagrammatic Representation of Pyramid of energy:

(b) The limitations of ecological pyramids are as follows:

• It levels takes into account the same species belonging to two or more trophic levels.
• Ecological pyramids assume a simple food chain that never existed in nature.

Question 26.
(a) Describe the structure and function of a t-RNA molecule. Why is it referred to as an adapter molecule?
(b) Explain the process of splicing of hn-RNA in a eukaryotic cell.
OR
Write the different components of a lac-operon in E.coli. Explain its expression while in an ‘open’ state.
Answer:
(a) Structure of t-RNA:

• The structure of t-RNA looks like a clover-leaf but its 3-D structure is inverted L-shaped
• It has an anticodon loop that has bases complementary to the code.
• It also contains amino acid acceptor end to which it gets bind with an amino acids.
• They are specific for each amino acid.
• The T-loop of t-RNA helps in binding to ribosome.
• D-Ioop help in binding of amino acyl synthetase.
• It also contains a variable loop.

Diagrammatic Representation of structure of t-RNA:

Function of t-RNA:

• It has anticodon loop that has bases complementary to the code.
• It also contains an amino acid acceptor end to which it gets binds with amino acid.
• t-RNA are specific for each amino.
• It is an adaptor molecule because it reads the code and on the another hand it binds with the specific amino acid.

(b) The primary transcript contains both coding and non-coding sequences called as exon and introns. So, the non-coding sequences or introns are removed by the process of splicing and exons are joined together in a specific manner. Primary transcript after splicing is called hnRNA. It undergoes two additional processes called capping and tailing.

In capping, an unusual nucicotide called methyl guanosine triphosphate is added to the 5’-end of hnRNA whereas in tailing, adenylate residues contains 200-300 bp are added at the 3’-end in a template independent manner. The fully processed hnRNA or primary transcript is called mRNA which is further transported into the nucleus for the process of translation.

OR

Lac operon was proposed by Jacob and Monad in 1961. It contains following components such as:

1.  Structural gene: There are three types of structural genes that codes for different enzymes and facilitates the process of transcription in the presence of inducer (lactose).
   • The z gene codes for enzyme beta-galactosidase that regulates the switching on and is responsible for the hydrolysis of disaccharide, lactose into its monomeric unit’s glucose.
   • y gene codes for enzyme permease that increases the permeability of the cell to beta-galactosides.
   • a gene codes for enzyme Iransacetylase.
2. Promoter: It is the sequence of DNA at which the RNA polyrnerase enzyme get binds and initiates the
   process of transcription.
3. Operator: It is sequence of DNA that is adjacent to promoter.
4. Regulator gene: A gene that codes for repressor protein and binds with the operator and because of it
   operon is switched “off”.
5. Inducer: Lactose is inducer that helps in switching “on” of operon.

Lactose acts as the substrate for enzyme beta-galactosidase. This enzyme regulates the switching on and off the operon because of this it is termed as inducer. So, in the absence of glucose (carbon source). if lactose is added in the growth medium of the bacteria. The lactose is transported into the cells by the action of permease enzyme that increases permeability of the cell to beta-galactosides.

Lactose induces the operon in following manner:

• In a lac operon, the repressor protein is synthesised from the i gene.
• This repressor protein gets bind with the operator region of the operon and prevents RNA polymerase enzyme from transcribing the operon.
• In the absence of lactose, the repressor gene produces repressor protein and get binds with the operator gene. It prevents the RNA polymerase enzyme to get binds with the operon.

Diagrammatic Representation of Lac operon in the absence of lactose:

• In the presence of lactose as an inducer, the repressor protein is inactivated. It allows RNA polymerase enzyme to activate the promoter and initiates the process of transcription by structural genes.

Diagrammatic Representation of operon in the presence of inducer: