CBSE Class 12 Biology Question Paper (Delhi 2016) with Solutions

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CBSE Class 12 Biology Question Paper (Delhi 2016) 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 question number 1 to 5, Very Short Answer type questions of 1 mark each.
3. Section B contains question number 6 to 10, Short Answer type-I questions of 2 marks each.
4. Section C contains question 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 question 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.
According to de-Vries what is saltation ?
Answer:
According to Hugo de Varies saltation is a single step large mutation. He said that speciation is caused due to mutation. Saltation is responsible for speciation.

Note
Mutation is defined as the sudden and permanant alteration in the nucleotide sequence of DNA.

Question 2.
Excessive nutrients in a fresh water body cause fish mortality. Give two reasons.
Answer:
Fish mortality rate increases due to algal bloom that is caused due to the presence of large amount of nutrients in the water bodies. Excessive nutrients support the growth of algae in the water bodies that deteriorate the quality of water by depleting the dissolved oxygen in water.

Question 3.
Suggest the breeding method most suitable for animals that are below average in milk productivity.
Answer:
Outcrossing is the best known breeding method for animals that are below average in milk productivity. This method involves the mating of animals within the same breed that have no common ancestors on both side of their pedigree for 4-6 generations.

Question 4.
State a difference between a gene and an allele.
Answer:
The difference between gene and allele are as follows:

Question 5.
Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer:
DNA fragments are separated by using molecular biology technique called gel electrophoresis. In this technique, DNA molecules are separated on the basis of their size in an electric field. DNA is negatively charged so it moves towards a positively charged anode in an agarose gel matrix. (1 Mark)

Note
As, DNA fragments are separated according to their size through the pores of agarose gel so, smaller the size of fragment farther it moves.

Section – B

Question 6.
Explain the significance of meiocytes in a diploid organism.
Answer:
Meiocytes are specialised cells that are found in sexually reproductive organisms. Meiocytes undergo the process of meiosis in order to produce male and female gametes that carry one set of chromosomes. Thus, they help in maintaining the chromosome number in organisms.

Question 7.
Mention the kind of biodiversity of more than a thousand varieties of mangoes in India represent. How is it possible?
Answer:
In India, there are around 1000 different varieties of mangoes produced due to genetic diversity. Genetic diversity is produced due to difference in soil found in different regions. It also occur due to different agricultural practices as well as use of various horticulture techniques that are used in India.

Note
Genetic diversity is defined as the type of diversity in which the number and types of genes as well as chromosomes that are found in different species. It leads to variation in the genes and their alleles in the same species. It leads to speciation as well as evolution of new species.

Question 8.
List the events that reduce the Biological Oxygen Demand (BOD) of a primary effluent during sewage treatment.
Answer:
The primary effluent is passed into large aeration tanks just after the primary treatment. In the large aeration tanks, the primary effluent is constantly agitated mechanically and air is also pumped into it. This process supports the vigorous growth of useful microbes that consists of floes of bacteria and fungi. Such microbes consume large portion of organic matter to the effluent also reduce the biochemical oxygen demand of the primary effluent.

Question 9.
Discuss the role the enzyme DNA ligase plays during DNA replication.
Answer:
During DNA replication, the lagging strand of template DNA is discontinuous (5’ → 3’). The formed segments are short segments of replicated DNA (3 ’ → 5 ’) are called OKAZAKI fragments. The OKAZAKI fragments of DNA are joined together by DNA ligase enzyme.

Question 10.
Name the causative organism of the disease amoebiasis. List three symptoms of the disease.
OR
Identify ‘A’, ‘B’, ‘C’ and ‘D’ in the given table.

Answer:
Amoebiasis or amoebic dysentery is caused by Entamoeba histolytica.
Symptoms of Amoebiasis:

• Appearance of blood in the stool.
• Pain in the abdomen.
• Fever
• Diarrhoea

OR

Section – C

Question 11.
Why is breast-feeding recommended during the initial period of an infant’s growth ? Give reasons.
Answer:
Doctors recommend breast feeding during the initial period of infant’s growth to maintain the health of baby. The first yellow milk that comes out from the mammary gland of the mother just after parturition is called colostrum. It is rich in protein such as lactalbumin and lactoprotein. It also contains antibody IgA that provide innate immunity to the infant.

Question 12.
Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross.
Answer:
Sickle cell anaemia is an example of autosomal recessive hereditary disorder. In this disorder, the erythrocytes become sickle shaped due to deficiency of oxygen. This disorder is caused due to the formation of abnormal haemoglobin-S. Genes for sickle cell erythrocytes is represented by Hbs whereas normal genes are represented by Hb^A.
Representation of cross of sickle cell anaemia:

Note
In case of sickle cell anaemia, abnormal haemoglobin Hbs is differ from normal haemoglobin HbA only by one amino acid at 6th amino acid of beta-globin chain of haemoglobin. The glutamic acid is replaced by valine due to substitution mutation of Thymine (T) by Adenine (A) at the second position of triplet codon such as CTC into CAC. The substitution mutation occurs at 11th chromosome. Codon CTC is transcribed into GAG that codes for amino acid glutamic acid. Whereas CAC is transcribed into GUG that codes amino acid valine.

Question 13.
Describe the experiment that helped Louis Pasteur to dismiss the theory of spontaneous generation of life.
Answer:
Louis Pasteur conducted the Swan-necked flask experiment to dismiss the theory of spontaneous generation of life which is as follows:

(a) Two swan-necked flasks are used that contain nutrient broth.
(b) The nutrient broths were made sterile by boiling in order to kill existing microbes present in the broth.
(c) One of the swan necked flask was broken just after sterilisation and kept open for sometime.
(d) The dust particles contain bacteria through air entered into the broken neck of flask.
(e) The broth present in broken flask became cloudy whereas broth of unbroken flask remained clear.
(f) The appearance of cloudiness of nutrient broth indicates the presence of microbial growth in the broken flask.
(g) This experiment concluded that, appearance of microbial life in the broken flask after sterilisation indicates that pre-existing life form.

Question 14.
Plant breeding technique has helped sugar industry in North India. Explain how.
Answer:
In North India, a variety of sugarcane such as Saccharum barberi was grown that had poor sugar content and low yield. Whereas another sugarcane variety Saccharum officinarum that had high sugar content and thicker stems. This variety of sugarcane does not grown in North India. So these two different varieties of sugarcane were crossed in order to obtain the desired qualities of sugarcane such as higher yield, high sugar content, thick stems and ability to grow in the belt of North India.

Question 15.
Suggest and describe a technique to obtain multiple copies of a gene of interest in vitro.
Answer:
With the help of recombinant DNA technology called Polymerase Chain Reaction (PCR) technique multiple copies of gene of interest are obtained Invitro. A single PCR amplification cycle involves three steps which are as follows:

(a) Denaturation: This is the first step of PCR, in which the target DNA is heated at high temperature such as 94-96°C. It facilitates the separation of two strands of DNA. Each separated strand of DNA acts as a template for synthesis of DNA.

(b) Annealing: This is the second step of PCR, in which two oligonucleotide primers are used to hybridize each single stranded template DNA. The sequence of primers is complementary to 3’ end of the template DNA strand.
This step of PCR occurs at low temperature 40- 60°C than denaturation. The annealing temperature depends upon the length and sequence of the primers.

(c) Extension: This is third and last step of PCR, in which enzyme TagDNA polymerase synthesizes the DNA between the primers. This step also requires dNTPS and Mg²⁺⁺. The optimum temperature for extension is 72°C.

Note
Enzyme used in PCR is a DNA polymerase such as Taq DNA polymerase. This enzyme is stable at high temperature as it is isolated from thermostable bacteria Thermus aquaticus.

Question 16.
What is a GMO ? List any five possible advantages of a GMO to a farmer.
Answer:
GMOs are Genetically modified organisms. They are defined as a living organisms whose genes are manipulated or altered by using recombinant DNA technology.

The advantages of GMOs are as follows:

1. GMOs are resistant to diseases, pest and insects. So it reduces the use of harmful pesticides and other chemical fertilisers that harm the crops.
2. GMOs crops are more tolerant to abiotic stress such as cold, drought, heat and salt stress.
3. Such crops have high crop yield and nutritional value.
4. GMOs reduce post-harvest loss of crops.
5. GMOs increase the efficiency of mineral usage by plants and also prevents exhaustion of soil.

Note
Golden rice is genetically modified crop which is obtained by recombinant DNA technology. It contains good quantities of beta-carotene which is a principal source of vitamin A. Due to the presence of beta-carotene, the rice grain appear golden in colour.

Question 17.
During a school trip to ‘Rohtang Pass’, one of your classmate suddenly developed ‘altitude sickness’. But, she recovered after sometime.
(a) Mention one symptom to diagnose the sickness.
(b) What caused the sickness ?
(c) How could she recover by herself after sometime ?
Answer:
(a) Altitude sickness is also called mountain sickness. It is caused due to the presence of low oxygen pressure at higher altitude. Nausea and fatigue are the symptoms of altitude sickness. (1 Mark)
(b) Altitude sickness in person is caused when the body is not able to get sufficient oxygen due to atmospheric pressure.
(c) After some time her body is able to adapt the changing atmosphere. Her body is ready to compensate low oxygen pressure by producing more red blood cells. It also tends to decrease the bind affinity for haemoglobin by increasing breathing rate.

Question 18.
How has RNAi technique helped to prevent the infestation of roots in tobacco plants by a nematode Meloidegyne incognitia ?
Answer:
Meloidegyne incognitia is a nematode that causes infections in the root of tobacco plants. This reduces the yield of tobacco plants. In order to protect the tobacco plants from infection, a process called RNA interference occurs in all eukaryotic organisms as a method of cellular defense. The process of RNA interference involves mRNA silencing because of complementary dsRNA molecule that get binds to and prevents the translation of mRNA.

The complementary RNA is obtained due to the infection by viruses that contain a RNA genome or mobile genetic elements (transposons). It replicate via a RNA intermediate.

By using Agrobacterium vectors, nematode-specific gene were used to introduced into the host plant. After the introduction of DNA into the host, it produces both sense and anti-sense RNA in the host cells.

The two RNA’s are complementary to each other formed a double strand (dsRNA) that initiates the process of RNAi and silenced the specific mRNA of the nematode. After that, the nematode is not able to survive in a transgenic host expressing specific interfering RNA.
In this way, the transgenic plant got itself to be protected from the parasite.

Question 19.
“In a food-chain, a trophic level represents a functional level, not a species.” Explain.
OR
(a) Name any two places where it is essential to install electrostatic precipitators. Why it is required to do so?
(b) Mention one limitation of the electrostatic precipitator.
Answer:
In a trophic level, the position of species is determined by their function as well as mode of nutrition in a specific food chain. The species may occupy more than one trophic level within the same ecosystem. If the mode of nutrition of species changes results in the change in trophic level of species. One species can become primary level of consumer in one food chain and also become secondary level of consumer in another food chain.

Note
In ecology, food chain is defined as the sequence of transfer of energy and matter in the form offood from one organism to other. Food chain is divided into four levels such as producers (plants), primary consumers (herbivores), secondary consumers (carnivores) and tertiary consumers (carnivore).

OR

(a) Electrostatic precipitator is installed in automobiles and thermal power plants because a large amount of harmful gases are released from automobiles and thermal power plants. Electrostatic precipitators are used to remove particulate matter from their exhaust.
(b) One drawback of using electrostatic precipitators is that it cannot remove particulate matter having size less than 2.5 micrometer in diameter. Such particulate matters are harmful for health.

Question 20.
Prior to a sports event blood & urine samples of sportspersons are collected for drug tests.
(a) Why is there a need to conduct such tests ?
(b) Name the drugs the authorities usually look for.
(c) Write the generic names of two plants from which these drugs are obtained.
Answer:
(a) There is a need to conduct blood and urine test for sports persons because some sports persons take narcotics analgesics, anabolic steroids, diuretics and several hormones in order to increase their muscle strength and performance. These tests help to determine whether a sports person has taken any drugs and it also ensures a fair play.
(b) The sports authorities are usually looking for narcotic drugs such as analgesics, diuretics and hormones such as gonadotropins and steroids, opiates such as and cannabinoids.
(c) The generic names of two plants are given below:

• Smack or heroin is obtained from the Papaver somniferum.
• Ganja and marijuana is obtained from the Cannabis saliva.

Question 21.
Describe the experiment that helped demonstrate the semiconservative mode of DNA replication.
Answer:
The semiconservative nature of DNA replication was experimentally proved by Matthew Meselson and Franklin Stahl in 1958 by using heavy nitrogen (¹⁵N) in E.coli. It involves following steps such as:
(1) They grow E.coli in a medium containing ¹⁵NH₄Cl. As nitrogen serves as the only source for many generations. ¹⁵N is the heavy isotope of nitrogen. So, the ¹⁵N was incorporated into newly synthesized DNA.

(2) This heavy DNA molecule is distinguished from normal DNA molecule by cesium chloride (CsCl) density gradient centrifugation. Then they transfer the cells into another medium containing normal ¹⁴NH₄Cl. They took the samples at different time interval as the cells multiplied and then DNA was extracted from the cells. In this way, different samples were separated independently on CsCl gradients in order to measure the densities of DNA.

(3) The DNA was extracted from the culture medium one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid or intermediate densities. The DNA extracted from the culture medium after another generation was composed of equal amounts of this hybrid and of light DNA.

Diagrammatic representation of semiconservative DNA replication:

Note
An experiment was performed by Taylor and his colleagues in 1958 to experimentally prove that the DNA in chromosomes also replicate semiconservatively. This experiment involves radioactive thymidine in order to detect the distribution of newly synthesised DNA in the chromosomes. This experiment was performed on Vicia faba (faba beans).

Question 22.
Given below is a list of six micro-organisms. State their usefulness to humans.
(a) Nucleopolyhedrovirus
(b) Saccharomyces cerevisiae
(c) Monascus purpureus
(d) Trichoderma polysporum
(e) Penicillium notatum
(f) Propionibacterium sharmanii
Answer:
(a) Nucleopolyhedovirus : Nucleopolyhedovirus is a biological control agent that is used as a species- specific insecticide.
(b) Saccharomyces cerevisiae : Saccharomyces cerevisiae is also called baker’s yeast. It is used in baking and beverage industry. It is used for making breads, south Indian cuisine, cakes and in beverages it is used for making alcohol.
(c) Monascus parpureus : It is yeast that is used for the production of a blood-cholesterol lowering agents called as statins.
(d) Trichoderma polysporum: It is fungus used for the formation of bioactive molecule called cyclosporine A. It is used as an immuno suppressive agent in organ-transplant patients. (‘A Mark)
(e) Penicillium notatum: It is a bacteria used for the production of antibiotics.
(f) Propionibacterium sharmanii: It is a bacterium used for the production of “Swiss cheese”. The appearance of large holes in the ‘Swiss cheese” is because of the production of carbon dioxide by bacteria.

Section – D

Question 23.
Reproductive and Child Healthcare (RCH) programmes are currently in operation. One of the major tasks of these programmes is to create awareness amongst people about the wide range of reproduction related aspects. As this is important and essential for building a reproductively healthy society.
(a) “Providing sex education in schools is one of the ways to meet this goal.” Give four points in support of your opinion regarding this statement.
(b) List any two ‘indicators’ that indicate a reproductively healthy society.
Answer:
(a) Sex education is one of the best ways in order to create a reproductively healthy society. It also helps people in following ways:

• Sex education helps to provide proper knowledge to curious adolescents. It helps them to prevent from being them misguided.
• It helps to create awareness about sexually transmitted disease and its prevention ways.
• Sex education provides proper knowledge about reproductive organs and other changes related to puberty to the adolescents.

(b) The indicators that represent a reproductively healthy society are as follows:

• Increase medical facilities for all sex related problems.
• On time detection and better cure of sexually transmitted diseases.

Section – E

Question 24.
(a) Explain the post-pollination events leading to seed production in angiosperms.
(b) List the different types of pollination depending upon the source of pollen grain.
OR
(a) Briefly explain the events of fertilisation and implantation in an adult human female.
(b) Comment on the role of placenta as an endocrine gland.
Answer:
(a) Pollination is a defined as the process of transfer of pollen grains from anthers to stigma. It involves followings steps such as:

• When the pollen grains fall on the stigma, pollen tube is formed and it enters one of the synergids and also releases two male gametes.
• One of the male gametes moves towards the egg and fuse to form a zygote.
• While the other male gamete fuses with polar nuclei and forms a primary endosperm nucleus. This process is termed as triple fusion.
• The central cell becomes the primary endosperm cell after the process of triple endosperm. The primary endosperm nucleus forms endosperm whereas zygote is further developed into the embryo.
• A seed refers to the fertilised ovules that are further inside a fruit.
• The integuments of the ovules are hardened to form the seed coat whereas the micropyle facilitates the entry of oxygen and water into the seeds.

(b) There are three different types of pollinations such as:

(i) Autogamy: This type of pollination requires transfer of pollen grains from anther to stigma of same flower. In this, the anther and stigma lie close to each other so that self-pollination can occur. Some plants such as Oxalis, Commeline and viola produces two types of flowers such as Chasmogamous flowers: Such flowers are similar to the flowers of other species with exposed anthers and stigma.
Cleistogamous flowers: Such flowers do not open at all. The anthers and stigma lic close to each other.

(ii) Geitonogamy: In this type of pollination, the transfer of pollen grain from anther to the stigma of another flower of same plant. It involves cross-pollination through pollinating agents. But genetically it is similar to autogamy as the pollen grains come from the same plant.

(iii) Xenogamy: This type of pollination involves transfer of pollen grains from anther to stigma of a different plant. In this, pollination brings genetically different types of pollen grains to the stigma.

Note
Plants use one biotic agent such as animals and two abiotic agent such as wind and water for pollination.

OR

(a) Fertilisation: The process of fertilisation is defined as the process of fusion of a sperm with an ovum.

• Fertilisation occurs only if the ovum and sperms are transported simultaneously to the ampullary- isthmic junction.
• During the process of fertilisation, a sperm comes in contact with the zona pellucida layer of the ovum and also induces the changes in the membrane in order to block the entry of additional sperms.
• Sperm secretion help it to enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane.
• It induces the completion of the meiotic division of the secondary oocyte.
• The second meiotic division unequally results in the formation of a second polar body and a haploid ovum.
• Zygote is formed by the fusion of haploid nucleus of the sperms and the ovum.

Implantation:

• The mitotic division starts when the zygote moves through the isthmus of the oviduct called cleavage towards the uterus. It forms 2, 4, 8, 16 daughter cells called blastomeres.
• Then, the embryo with 8 to 16 blastomeres stages is called morula.
• It continues to divide and then transforms into blastocyst as it moves further into the uterus.
• The blastomeres present in the blastocyst are arranged into an outer layer called trophoblast whereas the inner groups of cells are attached to the trophoblast called inner cell mass.
• After this, the trophoblast layer gets attached to the endometrium whereas the inner cell mass gets differentiated into the embryo.
• After attachment, the uterine cells rapidly cover the blastocyst.
• Now the blastocyst embedded in the endometrium of the uterus and this process is called Implantation that results in pregnancy.

(b) Placenta plays an essential role during pregnancy as it facilitates the supply of oxygen and nutrients to the embryo. It also helps in the removal of carbon dioxide and other metabolic waste produced by the embryo. It also acts as an endocrine tissue and produces hormones such as human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens, and progestrogens. Relaxin hormone is also secreted by the ovary during the later phase of pregnancy.

Question 25.
(a) How are the following formed and involved in DNA packaging in a nucleus of a cell ?
(i) Histone octomer
(ii) Nucleosome
(iii) Chromatin
(b) Differentiate between Euchromatin and Heterochromatin.
OR
Explain the role of lactose as an inducer in a lac operon.
Answer:
(a)

(i) Histone octomer: The histones are positively charged basic proteins. They are rich in basic amino acid such as lysine and arginines. They are organised to form a unit of eight molecules called a histone octomer.

(ii) Nucleosome: The negatively charged DNA molecule is wrapped around the positively charged histone octomer in order to form a structure called nucleosome. In a typical nucleosome, 200 bp of DNA helix are present.

(iii) Chromatin: The nucleosomes are unit together to form a chromatin. The nucleosome appears like beads-on-strings on the chromatin. It is packed to form chromatin fibres that further coil and condense at the metaphasic stage of cell division in order to form chromosome. It involves nonhistone proteins for packaging called non-histone chromosomal protein (NHC).

(b) The difference between euchromatin and hetero chromatin are as follows:

OR

Lactose acts as the substrate for enzyme beta-galactosidase. This enzyme regulates the switching on and off of 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 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.

Question 26.
(a) Why should we conserve biodiversity ? How can we do it ?
(b) Explain the importance of biodiversity hot-spots and sacred groves.
OR
(a) Represent diagrammatically three kinds of age- pyramids for human populations.
(b) How does an age pyramid for human population at given point of time helps the policy-makers in planning for future.
Answer:
(a) There is a need to conserve biodiversity because of the following reasons:

• Commercially important products such as food, timber and other essential industrial products are obtained from nature.
• Oxygen production and pollination is totally dependent on nature.
• There is need to conserve the endangered species and protect the biodiversity for our future generations.

Biodiversity can be conserved in two ways:

1. In-situ conservation: It involves the conservation of plants and animals species in their natural habitat. For this, biodiversity hotspots are being identified and protected. It involves wildlife sanctuaries, national parks and biosphere reserves.
2. Ex-situ conservation: The threatened and endangered species are taken out of their natural habitats and kept in special setting like zoological gardens and wildlife parks.
3. Cryopreservation: The gametes of endangered plants and animals are kept viable by preserving them at a very low temperature (-196°C) in a liquid nitrogen.

(b) Importance of biodiversity hotspots:
Biodiversity hotspots are regions that contain high level of species richness and higher degree of endemism. Such areas are very important because the total number of biodiversity hotspots in the world is 34 and hotspots can reduce the mass extinction by approximately 30%. It involves biodiversity regions such as Western Ghats and Sri-Lanka, Indo-Burma and Himalaya covers high biodiversity regions.

Importance of sacred groves:

Sacred groves are the forest regions that involve all trees and wildlife species. It provide protection to all plants and wildlife species.

In India, the sacred groves involve:

• Western Ghat regions of Karnataka and Maharashtra
• Khasi and Jaintia Hills in Meghalaya
• Aravalli Hills in Rajasthan
• Sarguja, Chanda and Bastar areas of Madhya Pradesh. (2 Marks)

OR

(a) There are three types of age distribution pyramids such as expanding, stable and declining. A population is composed of individuals of different age groups.
Diagrammatic representation of age pyramids for human population:

(b) The structure of age pyramid helps in the determination of growth status of the population. It involves three types of structure that represents that whether the population is expanding, stable or declining. The structure of age pyramids for human population emphasis on providing food to population, development of proper health care facilities and so on.
The age pyramids can help the policy makers to plan the healthcare programs, education policies, infras-tructure and employment policies etc.