CBSE Class 12 Biology Question Paper (Outside Delhi 2013) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Delhi 2013) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Delhi 2013) with Solutions

Time Allowed: 3 Hours
Maximum Marks : 70

General Instructions:

1. All questions are compulsory.
2. This question paper consist of four Sections A, B, C and D. Section A contains 8 questions of one mark each, Section B is of 10 questions of two marks each, Section C is of 9 questions of three marks each and Section D is of 3 questions of five marks each.
3. There is no overall choice. However, internal choice has been provided in one question of 2 marks, one question of 3 marks and two questions of 5 marks weightage. A student has to attempt only one of the alternatives in such questions.
4. Wherever necessary, the diagrams drawn should be neat and properly labelled.

Section – A

Question 1.
Name an organism where cell division in itself is a mode of reproduction.
Answer:
The cell division is itself is a mode of reproduction in Monera and Prostista. For example: Bacteria and amoeba.

Question 2.
When does a human body elicit an anamnestic response?
Answer:
Anamnestic response is the secondary immune response which is produced when the body encounters the same antigen which is entered previously in the body. As the body recognises the pathogen, immune system starts producing antibodies against the foreign antigens for subsequent encounter and this response is very intense.

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

Question 4.
State the role of transposons in silencing of mRNA in eukaryotic cells.
Answer:
Transposons are also called as mobile genetic elements and the process of mRNA silencing is used for the prevention of mRNA translation. In mRNA silencing, transposons are act as complementary RNA in order to stop the process of mRNA translation.

Note
Transposons are repetitive DNA sequences that have capability to move from one genome location to another genome location.

Question 5.
Why are green algae not likely to be found in the deepest strata of the ocean?
Answer:
The green algae acts as producer and they synthesise their own food by the process of photosynthesis. In the deepest strata such as in the benthic zone sunlight is not available. The green algae are not likely found in the deepest strata of the ocean because of the non-availability of sunlight for the process of photosynthesis.

Question 6.
State what does ‘standing crop’ of a trophic level represent.
Answer:
Standing crop at a trophic level is referred to the amount of biomass or mass of living material at a successive trophic level at a given time.

Question 7.
Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic converters?
Answer:
Motor vehicles equipped with catalytic converter should use unleaded petrol because lead in the petrol inactivates the catalyst. Catalytic converters may reduce the emission of harmful gasses from the automobiles.

Question 8.
Name the type of biodiversity represented by the following:
(i) 1000 varieties of mangoes of India.
(ii) Variations in terms of potency and concentration of reserpine in Rauwolfia vomitoria growing in different regions of Himalayas.
Answer:
(i) Genetic diversity refer to the total number of genetic characteristics in the genetic makeup of a species. 1000 varieties of mangoes in India is an example of genetic diversity. The vast genetic diversity in India is observed as it lies within tropical latitudes that provide a constant and predictable environment as well as availability of more solar energy results in higher productivity.

(ii) The genetic variation represented by the medicinal plant Rouwolfia vomitoria which is growing in different Himalayan ranges is because of the potency and concentration of the active chemical (reserpine) produced by this plant. It is an example of genetic diversity.

Section – B

Question 9.
In angiosperms, zygote is diploid while primary endosperm cell is triploid. Explain.
Answer:
The fusion of one haploid male gamete with haploid female gamete (egg cells) results in the formation of a diploid zygote and this process is called sexual reproduction. Whereas endosperm is formed when other male gamete move towards the two polar nuclei which is located in the central cell and its fusion with the two polar nuclei results in the formation of triploid primary endosperm nucleus (PEN).

This process is called triple fusion. The central cell after triple fusion becomes the primary endosperm cell (PEC) and develops into the endosperm whereas the zygote develops into an embryo.

Note
The process of double fertilisation involves two types offusions such as syngamy and triple fusion that takes place in the embryo sac of the flowering plants.

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

The phenotypic and genotypic ratio obtained in incomplete dominance is 1 : 2 : 1

Note
Incomplete dominance states that a form of gene interaction in which two alleles that controls a trait is dominant over each other and the progeny obtained is the intermediate of the two alleles.

Question 11.
List the two main propositions of Oparin and Haldane.
Answer:
Oparin of Russia and Haldane of England proposed the following:

1. The first form of life could come from pre-existing non-living organic molecules such as RNA, proteins, etc. and the formation of life was preceded by chemical evolution such as formation of diverse organic molecules from inorganic constituents.
2. The conditions on Earth were – high temperature, volcanic storms, reducing atmosphere containing CH ₄ , NH ₃ , and so on.

Question 12.
Write the events that take place when a vaccine for any disease is introduced into the human body.
OR
Why is a person with cuts and bruises following an accident administered tetanus antitoxin? Give reasons.
Answer:
The principle of immunisation or vaccination is based on the property of ‘memory’ of the immune system. In vaccination, a preparation of antigenic proteins of pathogen or inactivated/ weakened pathogens (vaccine) is introduced into the body.

The antibodies that are produced in the body against these antigens would neutralise the pathogenic agents during actual infection and vaccines also generate memory B and T-cells that recognise the pathogen quickly on subsequent exposure and encounters the foreign antigen with a massive production of antibodies. (2 Marks)

Note
Immunization refers to the process in which a person is made immune or resistant to an infectious disease by the administration of a vaccine.

OR

Person with cuts and bruises following an accident is administered with tetanus antitoxin provides protection from the infection by bacteria Clostridium tetani. This bacterium is responsible for causing tetanus infection which can be fatal if left untreated. The bacterium enters into the body through cuts or wounds and is usually found in soil and manure.

So, at the time of accidents there is a higher risk of entering bacteria in the skin and hence tetanus antitoxin is administered in order to reduce the chances of infection by providing passive immunity to the bacterial toxin.

Note
The overall ability of a body to fight against disease – causing microorganisms by the immune system is called immunity.

Question 13.
Name the bacterium response for the large holes seen in “Swiss Cheese”. What are these holes due to?
Answer:
The production of large holes in ‘Swiss cheese’ is because of the production of a large amount of carbon dioxide by a bacterium named Propionibacterium sharmanii.

Question 14.
Name the source of the DNA polymerase used in PCR technique. Mention why it is used.
Answer:
DNA polymerase used in PCR is Taq DNA polymerase which is isolated from a bacterium Thermus aquaticus. This bacterium is found in hot springs and hydrothermal vents. The Taq polymerase remains active at high temperature during denaturation process of PCR.

Note
PCR stands for Polymerase Chain Reaction is a molecular biology technique used for the formation of large number of copies of samples produced in small quantities. PCR amplification is commonly used by medical and forensic applications.

Question 15.
Write any four ways used to introduce a desired DNA segment into a bacterial cell in recombinant technology experiments.
Answer:
The four ways used for the introduction of a desired DNA segment into bacterial cell in the recombinant DNA technology are as follows:

(i) Chemical method: In this process, the bacterial cells must be first made ‘competent’ to take up DNA. It is done by treating the bacterial cell with a specific concentration of a divalent cation such as calcium that increases the efficiency with which DNA enters the bacterium through pores in its cell wall. Recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock). Then, putting them back on ice. So, this enables the bacteria to take up the recombinant DNA.

(ii) Microinjection: In this method, recombinant DNA is directly injected into the nucleus of an animal cell.

(iii) Biolistics or gene gun: This method is suitable for plants, as cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA.

(iv) Disarmed pathogen: In this method, vector is allowed to infect the cell results in the transfer of the recombinant DNA into the host.

Question 16.
Why is proinsulin so called? How is inuslin different from it?
Answer:
In humans, insulin is synthesised as a prohormone that contains an extra stretch called the C peptide and it is not present in the mature insulin. As a mature insulin contains two short polypeptide chains such as Chain A and Chain B that are linked together by a disulphide bond. The prohormone is non-functional whereas mature insulin is functional.

Question 17.
Where would you expect more species biodiversity – in tropics or in polar regions? Give reasons in support of your answer.
Answer:
Tropics has more species diversity than polar regions because of the availability of sufficient sunlight that promotes higher productivity. Less seasonal variations are observed in tropic regions than polar Regions and tropics remain undisturbed from many years so they have a long evolutionary time for diversification of species.

Question 18.
“It is possible that a species may occupy more than one trophic level in the same ecosystem at the same time.” Explain with the help of one example.
Answer:
Yes, it is possible that a species may occupy more than one trophic level in the same ecosystem at the same time. For example, man is an omnivore as he can consume plants so they are called as a primary consumer and when man consumes animals such as goat and chicken which consume plants and become secondary consumer. In this way, a species occupying more than one trophic level in the same ecosystem at the same time.

Section – C

Question 19.
Explain the steps in the formation of an ovum from an oogonium in humans.
OR
Suggest and explain any three Assisted Reproducive Technologies (ART) to an infertile couple.
Answer:
The process of formation of a mature female gamete is called oogenesis. This process is initiated during the embryonic development stage when a couple of million gamete mother cells or oogonia are formed within each fetal ovary as no more oogonia are formed and added after birth.

Such cells start division and enter into prophase-I of the meiotic division. The cells get temporarily arrested at that stage called Primary oocytes.

Each primary oocyte then gets surrounded by a layer of granulosa cells and then called Primary follicle and the primary follicles get surrounded by more layers of granulosa cells and a new theca. This is called secondary follicle. Then the secondary follicle transform into a tertiary follicle which is characterised by a fluid filled cavity called Antrum.

The theca layer is organised into an inner theca internal and an outer theca externa. The primary oocyte within the tertiary follicle grows in size and completes its first meiotic division.

It is an unequal division results in the formation of a large haploid secondary oocyte and a tiny first polar body. The tertiary follicle changes into the mature follicle or Graafian follicle. The secondary oocyte forms a new membiane called Zona pellucida.

The Graafian follicle then ruptures to release the secondary oocyte (ovum) from the ovary by the process called Ovulation.

Diagrammatic representation of steps involved in the process of oogenesis:

OR

The ARTs suggested to infertile couples are as follows:

(i) In vitro fertilisation (IVF) technique involves collection of ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under stimulated conditions in the laboratory.

(ii) Zygote intra fallopian transfer (ZIFT) involves the transfer of zygote or early embryos upto 8 blastomeres into the fallopian tube of the female.

(iii) Gamete intra fallopian transfer (GIFT) involves the transfer of an ovum which is collected from a donor into the fallopian tube of another female who cannot produce egg but can provide suitable environment for fertilisation and its further development.

Note
ARTs (Assistea reproductive technology) is used for treatment of infertility in couples who are unable to produce offsprings. This method involves fusion of male gamete (sperm) and female gamete (egg) outside the body.

Question 20.
Why are human females rarely haemophilic?
Explain. How do haemophilic patients suffer?
Answer:
Haemophilia is a sex linked recessive disorder which is transmitted from unaffected carrier female to male progeny. This disorder is more common among males than females. Humans have 22 pairs of autosomal chromosomes and one pair of sex chromosome. There are 46 chromosomes in humans. Females have XX chromosome while males have X and Y chromosome. So, male offspring inherit X-chromosome from their mother and Y-chromosome from their father.

Males only have one X-chromosome if the X-chromosome carries mutation and this is the case of reason that males are suffering from haemophilia, While in females, as they have two X chromosomes, and this is a recessive disorder so females are only the carrier of this disease and can pass this disorder to male offsprings.

Genetic cross for haemophilia:

Haemophilia is caused because of the absence of blood clotting factor VIII (haemophilia-A) and IX (haemophilia B) and impairs the ability of the body to control blood clotting and coagulation even by a simple cut.

Question 21.
In a maternity clinic, for some reasons the authorities are not able to hand over the two new-borns to their respective real parents. Name and describe the technique that you would suggest to sort out the matter.
Answer:
DNA fingerprinting technique is used to describe the parental identification of two newborns in a maternity clinic.
The steps involved in the process of DNA fingerprinting are as follows:

• Isolation of DNA from the new borns and parents.
• DNA digestion by restriction endonucleases
• DNA fragments are separated by electrophoresis technique.
• Separated DNA fragments are transferred (blotting) to synthetic membranes such as nitrocellulose or nylon membrane.
• Hybridisation using labelled VNTR probe and,
• Hybridised DNA fragments are detected by using autoradiography.

Note
The DNA fingerprinting technique was developed by Alec Jeffreys and he used a satellite DNA as probe that represents a very high degree of polymorphism called VNTR (Variable Number of Tandem Repeats). VNTRs are repetitive units of 10-60 bp and show a higher degree of polymorphism as these base pair sequences are different in different individuals.

Question 22.
Explain the increase in the numbers of melanic (dark winged) moths in the urban areas of post-industrialization period in England.
Answer:
During post-industrialisation period, the tree trunks became dark due to industrial smoke and soots. So under this condition, the white winged moth did not survived due to predators, dark-winged or melanised moth survived. Whereas before industrialisation, thick growth of almost white-coloured lichen covered the trees-in that background the white winged moth survived but the dark-coloured moth were picked out by predators.

Question 23.
Describe how biogas is generated from activated sludge. List the components of biogas.
Answer:
Cow dung and other organic wastes are converted into slurry with 90% of water. A small quantity of activated sludge is pumped back into aeration tanks in order to serve as the inoculums. The remaining part of the sludge is pumped into large tanks called anaerobic sludge digesters. During the digestion, mixture of gases is produced called biogas by bacteria. This mixture involves methane, carbon dioxide and hydrogen gas.

Note
The bacteria that produce methane are called methanogens and one commonly known bacterium is Methanobacterium. These bacteria are found in the rumen of cattles and help in the breakdown of cellulose and provide essential nutrition to the cattle.

Question 24.
Name the pest that destroys the cotton bolls.
Explain the role of Bacillus thuringiensis in protecting the cotton crop against the pest to increase the yield.
Answer:
Cotton bollworms destroy the cotton bolls. The strains of Bacillus thuringiensis produce proteins that kill several insects such as lepidopterans, coleopterans and dipterans. Bacillus thuringiensis forms protein crystals contain a toxic called insecticidal protein. This toxin protein exists in an inactive form protoxins but once an insect ingest the inactive toxin, then 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 the midgut epithelial cells and creates pores that cause cell swelling and lysis results in death of the insect. The toxin is coded by a gene name cry. So, the protein encoded by the genes cryIAB controls com borer.

Question 25.
(a) Write the importance of measuring the size of a population in a habitat or an ecosystem.
(b) Explain with the help of an example how the percentage cover is a more meaningful measure of population size than more numbers.
Answer:
(a) Measurement of population in a habitat help in the determination of relative abundance of a particular species and their effect of the species on the available resources of habitat.

(b) In some cases it is very difficult to count the number of organisms for determination of population of an area. For example, in a laboratory culture of bacteria or counting number of trees in a dense forest is impossible as well as consumes lots of time. In such cases, calculation of percentage of the area which is covered by a particular organism than to sit or by counting them by number for the determination of population.

Question 26.
Differentiate between two different types of pyramids of biomass with the help of one example of each.
Answer:
The two types of pyramid of biomass are as follows:
(i)

Pyramid of biomass is upright and it occurs in grasslands where the biomass of the producers is much higher than that of primary consumer. This indicates that there is sharp decrease in biomass at higher trophic level.

(ii)

Pyramid of biomass is inverted for an aquatic ecosystems. Biomass of producers is much lower than that of primary consumers. This inverted pyramid indicates that a small standing crop of phytoplankton supports large standing crop of zooplankton.

Question 27.
(a) Describe the endosperm development in coconut.
(b) Why is tender coconut considered a healthy source of nutrition?
(c) How are pea seeds different from castor seeds with respect to endosperm?
Answer:
(a) The process of triple fusion involves fusion of male gamete with two polar nuclei which is located in the central cell results in the formation of a triploid primary endosperm nucleus (PEN). The central cell after triple fusion becomes the primary endosperm cell (PEC) and develops into the endosperm.

(b) The water of tender coconut is healthy source of nutrition because it is rich in nutrients such as proteins, carbohydrates, minerals, vitamins and fibre. It can also effectively replenish lost electrolytes in the body.

(c) Mature seeds are of two types such as non-albuminous and albuminous. Pea is a type of non-albuminous seeds that have no residual endosperm as it is completely consumed during embryo development. Whereas castor is a type of albuminous seeds that retain a part of endosperm as it is not completely used during embryo development.

Section – D

Question 28.
(a) Draw a L.S. of a pistil showing pollen tube entering the embryo-sac in an angiosperm and label any six parts other than stigma, style and ovary.
(b) Write the changes a fertilized ovule undergoes within the ovary in an angiosperm plant.
OR
(a) Draw a diagrammatic sectional view of a human seminiferous tubule, and label Sertoli cells, primary spermatocyte, spermatogonium and spermatozoa in it.
(b) Explain the hormonal regulation of the process of spermatogenesis in humans.
Answer:
(a) Diagrammatic representation of L.S of flower showing growth of a pollen tube:

(b) When the unfertilized ovule passes through double fertilization and then fertilised ovule is formed which further develops into seed. In this process, the hilum and funiculus are present. Outer integument is developed into texta whereas inner integument is developed into tegman. The chalaza and micropyle are also present while nucellus is absent. In embryo sac, synergids and antipodal cell are degenerate. The central cell then develops into endosperm and egg is developed into the zygote and then into the embryo.

Note
Hilum is the region where the body of the ovule fuses with funicle thus, hilum represents the junction between ovule and funicle.

OR

(a) Diagrammatic representation of sectional view of a seminiferous tubule

(b) The hormonal regulation of the process of spermatogenesis in humans are as follows:

• The process of spermatogenesis starts at the age of puberty due to significant increase in the secretion of gonadotrophin releasing hormone (GnRH).
• The increase in the level of GnRH acts at the anterior pituitary gland and also stimulates the secretion of two gonadotropins such as luteinising hormone (LH) and follicle stimulating hormone (FSH).
• LH acts at the Leydig cells and also stimulates the synthesis and secretion of androgens. Androgen then stimulates the process of spermatogenesis.
• FSH acts on the Sertoli cells and also stimulates the secretion of some factors that help in the process of spermiogenesis.

Question 29.
(a) Write the conclusion drawn by Griffith at the end of his experiment with Streptococcus pneumoniae.
(b) How did O. Avery, C MacLeod and M. McCarty prove that DNA was the genetic material? Explain.
OR
(a) Explain the mechanism of sex-determination in humans.
(b) Differentiate between male heterogamety and female heterogamety with the help of an example of each.
Answer:
(a) Griffith concluded that the R-strain bacteria had somehow been transformed by the heat-killed S strain bacteria. Some ‘transforming principle,’ transferred from the heat killed S-strain that had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent. This is because of the transfer of the genetic material.

(b) The biochemical characterisation of Transforming principle was determined by Oswald Avery, Colin Macleod and Maclyn McCarty. Prior it was thought that the genetic material was protein.

The following steps are involved in his experiment:

• They purified biochemicals such as proteins, DNA and RNA from the heat-killed S cells to determine which one could transform live R cells into S cells.
• They discovered that DNA alone from S bacteria caused R bacteria to become transformed.
• They also discovered that protein-digesting enzymes such as proteases and RNA-digesting enzymes such as RNases did not affect the transformation.
• Hence, it was proved that the transforming substance was not protein and RNA.
• Digestion with DNase did inhibit transformation and the DNA caused the transformation.
  So, they concluded that DNA is the hereditary material.

OR

(a) The chromosome pattern in the human female is XX and that in the male is XY. So, all the haploid gametes produced by the female (ova) have the sex chromosome X while in the male gametes (sperms) the sex chromosome could be either X or Y. Hence, 50 % of the sperms carry the X chromosome whereas the other 50 % carry the Y. After the fusion of the male and female gametes the zygote carries either XX or XY depends on whether the sperm carrying X or Y chromosome fertilized the ovum.

Zygote carrying XX would develop into a female baby and XY would become male baby. It is concluded that the sex of a child is dependent on father not on mother.

(b) The difference between male and female heterogamety are as follows:
Male heterogamety: Male heterogamety in human males is XY while males of insects such as grasshopper and bugs is XO.
Female heterogamety: Female heterogamety is observed in some species of birds, fishes and insects. Females of butterfly and moth consists of ZO sex chromosomes and females of fish, reptiles and birds consist of ZW sex chromosome.

Question 30.
A person in your colony has recently been diagnosed with AIDS. People/residents in the colony want him to leave the colony for the fear of spread of AIDS.
(a) Write your view on the situation, giving reasons.
(b) List the possible preventive measures that you would suggest to the residents of your locality in a meeting organised by you so that they understand the situation.
(c) Write the symptoms and the causative agent of AIDS.
Answer:
(a) Recently AIDS is diagnosed in our area and it is a life- threatening disease. So it is our prime responsibility to create awareness among the people in order to combat the situation. There is no need to isolate the person who is suffering from AIDS from the society.
The infected person needs help and sympathy instead of being shunned by society. We must have educated people about AIDS.

(b) In a society meeting, we must have to educate people about the mode of infection of AIDS as it cannot be spread through direct contact with the infected person. It can be spread by following reasons:

• Sexual intercourse with multiple partners
• Transfusion of contaminated blood and blood products
• By sharing of infected needle as in case of intravenous drug abusers
• From infected mother to her foetus through placenta.

Preventive measures involves:

• Use of condom during sexual intercourse with the infected person.
• roper disposal of needles after use
• Use clean needle
• Pregnant women take medical care immediately
• Avoid sharing of sharp objects or needles

(c) AIDS is Acquired Immunodeficiency Syndrome and is caused by the Human Immuno deficiency Virus (HIV), which is a group of viruses called retrovirus. This virus has an envelope that encloses the RNA genome.

Symptoms: The infected person is suffering from a bouts of fever, diarrhoea and weight loss. Due to decrease in the number of helper T lymphocytes, the person starts suffering from infections caused by Mycobacterium, viruses, fungi and even parasites like Toxoplasma.

The infected person becomes immune deficient as the patient is unable to protect themselves from such infections.

Note
AIDS cart be diagnosed by Enzyme Linked Immunosorbent assay (ELISA). It can be treated by using anti-retroviral drugs which is partially effective and such drugs only prolong the life of the patient but cannot prevent death and are inevitable.