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Series : SKS/1

Roll No.

Code No. 57/1/1Candidates must write the Code onthe title page of the answer-book.

Code number given on the right hand side of the question paper should be written on the title page ofthe answer-book by the candidate.

Please check that this question paper contains 30 questions.

Please write down the Serial Number of the questions before attempting it.

15 minutes time has been allotted to read this question paper. The question paper will be distributed at10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will notwrite any answer on the answer script during this period.

BIOLOGY

[Time allowed : 3 hours] [Maximum marks : 70]

General Instructions:

1. All questions are compulsory.

2. This questions paper consists of four Sections A, B, C and D. Section-A contains 8 questions of

one mark each, Section-B is of 10 questions of 2 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, an internal choice has been provided in one questions of

2 marks, one questions of 3 marks and one question of 5 marks weightage. A student has to

attempt only of the alternatives in such questions.

4. Wherever necessary, the diagrams drawn should be neat and properly labelled.

Studymate Solutions to CBSE Board Examination 2012-2013

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STUDYmate

SECTION-A

1. An anther with malfunctioning tapetum often fails to produce viable male gametophytes.

Give any one reason. [1]

Ans. Tapetum nourishes the developing pollen grain so a malfunctioning tapetum will fail to produce

viable male gamete.

2. Why sharing of injection needles between two individuals is not recommended? [1]

Ans. Sharing of injection needles may lead to transmission of HIV-infection.

3. Name the enzyme and state its property that is responsible for continuous and discontinuous

replication of the two strands of a DNA molecule. [1]

Ans. DNA dependent DNA polymerase. This enzyme catalyse polymerisation only in one direction,

i.e., 5' 3'. As a result on template strand with 3' 5' the replication is continuous while on

the template strand with polarity 5' 3' it is discontinuous.

4. Identify the examples of convergent evolution from the following: [1]

(i) Flippers of penguins and dolphins (ii) Eyes of octopus and mammals

(iii) Vertebrate brains

Ans. (i) Flippers of penguins and dolphins (ii) Eyes of octopus and mammals

5. Write the importance of MOET. [1]

Ans. In MOET cow is given hormones with FSH like activity. These hormones induce follicular

maturation and super-ovulatin which produces 6–8 eggs per cycle instead of one egg.

6. Why is the enzyme cellulase needed for isolating genetic material from plant cells and not

from the animal cells? [1]

Ans. Since plant cells have cell wall, so to digest it cellulase is required. Animals do not have cellwall so no cellulase required.

7. Name the type of biodiversity represented by the following: [1]

(a) 50,000 different strains of rice in India.

(b) Estuaries and alpine meadows in India.

Ans. (a) Genetic diversity

(b) Ecological diversity

8. Write the equation that helps in deriving the net primary productivity of an ecosystem. [1]

Ans. GPP – R = NPP [where GPP = Gross Primary Productivity, R = Respiratory losses]

SECTION-B

9. Geitonogamous flowering plants are genetically autogamous but functionally cross-pollinated.Justify. [2]

Ans. In Geitonogamous flowers the pollen grain comes from the same plant. It is being transferredto the stigma of another flower fo same plant. So it is functionally cross pollination by involving

pollinating agent but genetically it is autogamy.

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10. When and where do chorionic villi appear in humans? State their function. [2]

Ans. (a) Antibody - IgE

Chemical - Histamine and serotonin

(b) Drugs like anti histamine or adrenaline can be used for giving immediate relief.

11. In a cross between two tall pea plants some of the offsprings produced were dwarf. Show withthe help of Punett square how this is possible. [2]

Ans. Parent TT & Tt

Gametes T T, t

Punnet square

T T

T

t

TT(tall)

Tt (Tall)

TT(Tall)

tt (dwarf)

Tall : Dwarf

3 : 1

Gametes

12. A student on a school trip started sneezing and wheezing soon after reaching the hill station

for no explained reasons. But, on return to the plains, the symptoms disappeared. What issuch a response called? How does the body produce it? [2]

Ans. This response is called allergy. This immune response is produced for a substance called

allergen. Allergy is due to release of chemicals like histamine and serotonin from mastcells. The antibodies produced in these are IgE type.

13. Name two commonly used bioreactors. State the importance of using a bioreactor. [2]

Ans. Two commonly used bio reactors are

(a) Simple stirred tank bioreactor

(b) Sparged stirred - tank bioreactor

Importance of Bioreactor

Large volumes of cultures (100 - 1000 litres) can be processed. Also bioreactor provide optimal

conditions for getting the desired product by providing optimum growth conditions like temp,

pH, substrate, salts, vitamins and oxygen.

14. Write the function of adenosine diaminase enzyme. State the cause of ADA deficiency in

humans. Mention a possible permanent cure for a ADA deficiency patient. [2]

Ans. Adenosine deaminase enzyme is very important for functioning of immune system. ADA

deficiency is due to delection of the gene for adenosine deaminase.

Possible permanent cure is gene therapy where lymphocytes from the blood of the patientsare grown in a culture outside the body. Then functional ADA is introduced into these

lymphocytes which are then introduced into patient.

However it is important that gene isolated from marrow cells producing ADA is introduced

into cells at early embryonic stage for permanent cure.

15. Expand the following and mention one application of each: [2]

(i) PCR (ii) ELISA

OR

(a) Mention the difference in the mode of action of exonuclease and endonuclease.

(b) How does restriction endonuclease function?

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Ans. (i) PCR - Polymerase Chain Reaction

(ii) ELISA - Enzyme linked Immuno sorbent Assay.

OR

(a) Exonuclease - cuts the DNA at free ends.

Endonuclease - cuts within DNA segment

(b) Resticriction endonuclease, : cuts within the DNA, at specific sites called pallindromicsites.

e.g. Eco R1 cuts the DNA at following pallindromic sequences.

5' G A A3' C T T

T T C A A G 5'

3'

16. Name any two sources of e-Wastes and write two different ways for their disposal. [2]

Ans. Example of e-waste are: Irreparable computes and Electronic goods.

Two different ways for their disposals are

(i) Buried in landfills.

(ii) Incinerated.

17. Why the pyramid of energy is always upright? Explain. [2]

Ans. Pyramid of energy is upright because when energy flows from a trophic level to next trophiclevel, some energy is always lost as heat at each step. Finally it is lost to atmosphere and

never goes back to Sun.

18. Explain why very small animals are rarely found in polar region. [2]

Ans. Very small animals have large surface area with respect to volume, there is more chance

of heat loss, hence they are rarely found in polar region.

SECTION-C

19. Draw a diagram of the microscopic structure of human sperm. Label the following parts in it

and write their functions. [3]

(a) Acrosome (b) Nucleus

(c) Middle piece

Ans.

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20. With the help of any two suitable examples explain the effect of anthropogenic actions onorganic evolution. [3]

Ans. Effect of anthropogenic actions on organic evolutions.

(i) Industial melanism: It was due to smoke and soot coming out of man made industrieswhich caused a shift from white-winged moths on trees to dark winged or melanisedmoths.

Prior to industrialisation white coloured lichens covered the bark of trees which wasfavourable for white winged moths.

(ii) Use of herbicides and pesticides has resulted in selection of resistant varieties in veryshort time scale. Also development of microbes resistant to antibiotics in period ofmonths is due to anthropogenic actions.

21. (a) Why is human ABO blood group gene considered a good example of multiple alleles?[3]

(b) Work out a cross up to F1 generation only, between a mother with blood group A

(Homozygous) and the father with blood group B (Homozygous). Explain the pattern ofinheritance exhibited.

Ans. (a) ABO blood group is controlled by I gene.

I genes has 3 alleles IA, IB, i

Therefore ABO blood group is a good example of multiple allelism

(b) Genotype I IA A I IB B

(Mother) (Father)

I IA B

Progeny will have blood group AB.

– This shows codominance pattern of inheritance.

22. Describe the structure of a RNA polynuclotide chain having four different types of nucleotides.

[3]

Ans. RNA polynucleotide consists of namely

(i) Sugar = Ribose (ii) Phosphate group

(iii) Nitrogenous bases like

Adenine - A Uracil - U

Cytosine - C Guanine - G

23. Differentiate beteen inbreeding and outbreeding in cattle. State one advantage and one

disadvantage for each one of them. [3]

Ans. SN Inbreeding Outbreeding

1. Mating of more closely related

individuals

Mating of unrelated animals

2. Mating is between animals of same breed for 4-6 generations

If mating is between animals of same breed

them there should be no common ancestor

for 4-6 generations.

4. Mating is between same species. Mating can be between different species.

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Advantage of inbreeding

It increases homozygosity so is used for developing pure lines.

Advantage of out breeding

It produces hybrids with desirable characters like better lactation period and high milkproductions.

24. (a) Why are the fruit juices bought from market clearer as compared to those made at

home? [3]

(b) Name the bioactive molecules produced by Trichoderma polysporum and Monascus

purpureus.

Ans. (a) Fruit juices bought from market are clearer as compared to home because bottled juices

are clarified by the use of poctinases and proteases.

(b) Trichoderma polysporum produces Cyclosporin A.

Monosacus purpurens produces statins.

25. (a) Why are transgenic animals so called? [3]

(b) Explain the role of transgenic animals in

(i) Vaccine safety (ii) Biological products

with the help of an example each.

Ans. (a) Transgenic animals are so called because they have their DNA manipulated. As a

result they possess and express foreign gene.

(b) Role of in vaccine safety

Transgenic mice are being used to test the safety of the polio vaccine before they are

used on humans.

Biological products

Medicines required to treat many human diseases contain certain biological productswhich are expensive to make. These products can be produced by transgenic animals

by induction of genes in them which code for particular product e.g. human protein ( -

1 - antitrypsin)

26. How have human activities caused desertification? Explain. [3]

OR

How does algal bloom destroy the quality of a fresh water body? Explain. [3]

Ans. The development of the fertile top soil takes centuries. But it can be removed very easily dueto human activities like over- cultivation, unrestricted grazing, deforestation and poor

irrigation practices, resulting in arid patches of land. When large barren patches extend and

meet over time, a desert is created. Now-a-days, desertification is a major problem particularydue to increased urbanisation.

OR

Ans. Presence of large amounts of nutrients in water causes excessive growth of planktonic (free-

floating) algae, called an algal bloom which imparts a distinct colour to the water bodies. Algalblooms cause deterioration of the water quality and fish mortality. Some bloom forming algal

are extremely toxic to human beings and animals.

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27. Explain mutualism with the help of any two examples. How is it different from commensalism?[3]

Ans.. Mutalism is the interaction which confers benefits on both the interacting species. Lichensrepresent an intimate mutualistic relationship between a fungus and photosynthesisingalgae or cynobacteria. Similarly, the mycorrhizae are associations between fungi and roots

of higher plants. Fungi help the plant in the absorption of essential nutrients from the soilwhile the plant in turn provides the fungi with energy yielding carbohydrates.

• The interaction where one species is benefitted and the other is neither benefittednor harmed is called commensalism.

Species Species

A++

B+0

Name of interaction

MutualismCommensalism

SECTION-D

28. (a) Draw a diagrammatic sectional view of a mature anatropous ovule and label the following

parts in it: [5]

(i) that develops into seed coat.

(ii) that develops into an embryo after fertilization.

(iii) that develops into an endosperm in an albuminous seed.

(iv) through which the pollen tube grains entry into the embryo sac.

(v) that attaches the ovule to the placenta.

(b) Describe the characteristic features of wind pollinated flowers.

OR

(a) Draw a diagrammatic sectional view of the female reproductive system of human and

label the parts: [5]

(i) where the secondary oocytes develop

(ii) which helps in collection of ovum after ovulation

(iii) where fertilization occurs

(iv) where implantation of embryo occurs.

(b) Explain the role of pituitary and the ovarian hormones in menstrual cycle in human

females.

Ans.

(a) Egg = embryo

Polar nuclei +male gamete = endosperm

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(b) Characteristics of wind pollinated flowers:

(i) Large number of pollen grains are produced.

(ii) Pollen grains are light and non-sticky.

(iii) Well exposed stamens so that pollen grain are easily dispersed.

(iv) Large and feathery stigma to trap pollen grains.

OR

Ans. (a)

(b) Changes hapenning in ovary and uterus during menstrual cycle are caused by change

in the levels of pituitary and ovarian hormones.

After menstrual phase gonadotropin i.e., Luteinising hormone (LH) and follicularstimulating hormone (FSH) are released from pituitary. Their level gradually increases

during follicular phase and it stimulates development of follicles and secretion ofestrogen by growing follicles.

Both LH and FSH attain peak on 13th & 14th day.

Rapid secretion LH causes ovulation on day 14.

During luteal phases Graafian follicle changes to corpus luteum which secretesProgesterone Progesterone maintains endometrium which is necessary for

implantation.

29. Describe the asexual and sexual phases of life cycle of Plasmodium that causes malaria inhumans. [5]

OR

(a) What is plant breeding? List the two steps the classical plant breeding involves.

(b) How has the mutation breeding helpes in improving crop varieties? Give one examplewhere this technique has helped.

(c) How has the breeding programme helped in improving the public nutritional health?State two examples in support of your answer.

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Ans.

OR

Ans. (a) Plant breeding: It is manipulation of plant species so as to produce plants with desired

characters like better yield and disease resistance.

Classical plant breeding involves

(i) Hybridisation of pure lines.

(ii) Artificial selection for producing plant with desired characters of higher yield or

resistance to diseases.

(b) Mutational breeding- creates genetic variations through changes in the base

sequences within genes. This results in new character which is not found in parents.

Improved varieties can be produced by inducing mutations artificially in the plants

having desirable characters.

For example in mung bean resistance to yellow mosaic virus and powdery mildew was

induced by limitations.

(c) Breeding crops with higher levels of vitamins and minerals or higher proteins andhealthier fats has improved public health.

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× · × · × · × · ×

Breeding programmes have been designed with objectives of

(i) improved protein content and quality.

(ii) Oil content and quality

(iii) Vitamin content

(iv) Micronutrients and mineral contains

maize hybrids having double amount of lysine and tryopophan have been developed.Similarly IARI New Delhi have developed many vegetable crops rich in vitamins

and minerals like spinach, carrots, pumpkin etc.

30. A child by the family from Thalassemia is born to a normal couple. But the mother is being

blamed by the family for delivering a sick baby. [5]

(a) What is Thalassemia?

(b) How would you counsel the family not to blame the mother for delivering a child suffering

from this disease? Explain.

(c) List the values your counselling can propagate in the families.

Ans. (a) Thalassemia - is an autosomal reccessive blood disease which can occur due to deletion

of the genes controllign the formation of globin chains (commonly and ) of haemoglobin.

(b) Mother cannot be blamed for the disease as, it is an autosomal reccessive blood disorder.

The genes for the synthesis of globin chains are present on autosomes i.e.

(i) Formation of chain is controlled by 2 genes present on chromosome 16 and

(ii) Formation of -chain is controlled by one gene present on chromosome 11manifestation of the disease occurs when the progeny receives defective genes

from both the parents (as thalassaemia is expressed is homozygous recessive

condition only).

(c) Values that can be propogated in families are

(i) One should be aware of such autosomal diseases ad should get themselves checkedfor the same before marriage.

(ii) Manifestation of the disease occurs due to defective genetic set up of both theparent (so it is not only mother to be blamed.

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SECTION-A

5. Identify the examples of homologous structures from the following.

(i) Vertebrate hearts

(ii) Thorns in Bougainvillea and tendrils of Cucurbita

(iii) Food storage organs in sweet potato and potato.

Ans. (i) and (ii)

9. Describe the gene therapy procedure for an ADA-deficient patient.

Ans. In gene therapy, lymphocytes from the blood of patient are grown in a culture ouside the

body. A functional ADA CDNA (using reteroviral vector) is then introduced into theselymphocytes which are subsequently returned to the patient.

However as these cells are not immortal, patient requires periodic infusions of suchgenetically engineered lymphocytes.

14. (a) How does cleistogamy ensure autogamy?

(b) State one advantage and one disadvantage of cleistogamy to the plant.

Ans. (a) Cleistogamy ensures autogamy as cleistogamous flowers do not open at all and hence

there is no chance of cross pollination as there is no chance of cross-pollen landing on

the stigma.

(b) Advantage

• Clestogamous flowers produce assured seed set even is absence of pollinators.

Disadvantage

• No variation is there in the progeny.

15. When and where do chorionic villi appear in humans? State their function.

Ans. Chorionic villi appear

1. After implantation

2. On trophoblast.

Function: The chorion villi and uterine tissue become interdigitated with each other andjoin to form, a structural and functional unit betwwen developing embryo (foetus and maternal

body called placenta).

18. (a) Explain how to find whether an E.coli bacterium has transformed or not when a

recombinant DNA bearing amplicillin resistant gene is transferred into it.

(b) What does the amplicilling resistant gene act as in the above case?

Ans. (a) If E. Coli has been transformed then these ‘transformed E.coli’ cells would be able to

grew on ampicillin medium.

(b) Ampicillin resistant gene acts as a selectable marker in this case.

Code No. 57/1/2

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTIONS ONLY

Series : SKS/1

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STUDYmate

SECTION-C

22. (a) Explain how to overcome inbreeding depression in cattle.

(b) List three advantages of inbreeding in ctalle.

(c) Name an improved breed of cattle.

Ans. (a) Inbreeding depression in Cattle is overcomed by mating the selected animals of breedingpopulation with unrelated superior animals of same breed, to restore fertility and yield.

(b) Advantages of Inbreeding in Cattle:

(i) Helps in evolving a pureline in any animal.

(ii) It exposes harmful recessive genes that are eliminated by selection.

(iii) Helps in accumulation of superior genes.

(c) Improved variety of Cattle is Hisardale – a new breed of sheep developed by crossingBikaneri ewes and Marino rams.

27. (a) Draw a diagram of the structure of a human ovum surrounded by corona radiata. Labelthe following parts:

(b) (i) Ovum

(ii) Plasma Membrane

(iii) Zona Pellucida

Ans. (a)

Ovum

Plasma membrane

Zonapellucida

(b) Function of Zona Pellucida

It completely surrounds the ovum and prevents polyspermy.

SECTION-D

30. (a) Draw a labelled schematic diagram of the transverse section of a mature anther of anangiosperm plant.

(b) Describe the characteristic features of an insect pollinated flower.

OR

(a) Describe the events of spermatogenesis with the help of a schematic representation.

(b) Write two differences between spermatogenesis and oogenesis.

Ans. (a)

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(b) Majority of Insect pollinated flowers are

1. Large

2. Colourful

3. Fragnant

4. Rich in nectar

5. Flowers pollinated by flies and beettles secrete foul odours to attract them.

OR

Ans. (a)

(i) Spermatogenesis starts at the age of puberty

(ii) Spermatogonia are present on the inside wall of semniferous tubules multiply bymitotic divisions and increase in numbers.

(iii) Each spermatogonium is diploid and contain 46 chromosomes

(iv) Some of spermatogomia called primary spermato cytes undergo meiosis I to

produce 2 equal haploid cells called - secondary spermatocytes with 23

chromosomes each

(v) Secondary spermatocyte undergo second meiotic division to produce 4 equal

haploid spermatids.

(vi) Spermatid later transform into spermatozoa.

(b) Spermatogenesis Oogenesis

(i) It occurs in testes It occurs in ovaries

(ii) A spermato gonium forms 4 spermatozoa An oogonium forms one ovum

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SECTION-A

6. “Sweet potato tubers and potato tubers are the result of convergent ecolution”. Justify the

statement.

Ans. Both sweet potato tuber and potato tuber have same function but anatomically sweet potato

tuber is root whereas potato tuber is stem so convergent evolution.

SECTION-B

13. Explain the steps that ensure cross pollination in an autogamous flower.

Ans. Following steps can ensure cross pollination in autogamous flowers

(i) Pollen grain and stigma receptivity should not be synchronized.

(ii) Anther and stigma are placed at different positions so that pollen cannot come incontact with stigma of the same flower.

(iii) Self incompatibility also prevents inbreeding.

16. A student on a school picnic to a park on a windy day started sneezing and having difficulty

in breathing on reaching the park. The teacher enquired whether the student was allergic

to something.

(a) What is an allergy?

(b) Write the two unique characteristics of the system involved in the response observedin the student.

Ans. (a) The exaggerated response of the immune system to certain antigens present inthe environment is called allergy.

(b) The system involved is immune system. It unique characteristics are:

(i) It recognizes foreign antigens & responds to them.

(ii) It has memory so it remembers antigens

17. Why and how bacteria can be made ‘competent’?

Ans. Bacterial cell is made competent so that it can take up DNA. Taking up DNA is not easy asit is hydrophilic molecule and cannot pass through cell membrane. Bacterial cell can be

made competent by treating it with a specific concentration of a divalent cation such as

calcium as it increases the efficiency with which DNA can enter the bacteria through itthe pores of cell wall.

18. (a) Name the deficiency for which first clinical gene therapy was given.

(b) Mention the cause of and one cure for this deficiency.

Ans. (a) Adenosine deaminase (ADA) Deficiency

(b) It is caused due to deletion of the gene for adenosine de aminase. It can be cured by

bone marrow transplantation in some children or it can also be treated by enzymereplacement therapy.

Code No. 57/1/3

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTIONS ONLY

Series : SKS/1

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STUDYmate

21. (a) Differentiate between inbreeding and outbreeding.

(b) Explain inbreeding depression and how it can be overcome.

(c) Mention two advantages of inbreeding programme in cattle.

Ans. (a) • When breeding is between animals of the same breed it is called inbreeding.

• when crosses between different breeds are called outbreeding.

(b) Inbreeding exposes harmful recessive genes that are eliminated by selection. It alsohelps in accumulation of superior genes and elimination of less desirable genes. thereforethis approach where there is selection at each step, increases the productivity of inbredpopulation. However, continued inbreeding, expecially close inbreeding usually reducesfertility and even productivity. This is called inbreeding dipression.

(c) Two advantages of inbreeding programme in cattle are

(i) In breeding increases homozygosity.

(ii) A superior female, in the case of cattle is the cow or buffalo hthat produces moremilk per lactation, it also helps in accumulation of superior genes and eliminationof less desirable gens.

SECTION-D29. (a) Draw a diagram of a mature embryo sac of an angiosperm and label the following

parts in it:

(i) Filiform apparatus (ii) Synergids

(iii) Central cell (iv) Egg cell

(v) Polar nuclei (vi) Antipodals

(b) Write the fate of egg cell and polar nuclei after fertilization.

OR

(a) Describe the events of Oogenesis with the help of schematic representation.

(b) Write two differences between Oogenesis and Spermatogenesis.

Ans. (a)

(b) During fertilization, 1st male gamete fuses with the egg cell and forms the diploidzygote, while the second male gamete fuses with the polarnuclei forms the triploidendosperm.

(n) + (n) 2n Diploid Zygote

1st male Egg Cell

gamete

(n) + (2n) 3n Triploid endosperm

2nd male Central Cell

gamete having the polar nuclei

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OR

Ans. (a)

The process of formation of a nature female gamete is called oogenesis, which isdifferent from spermatogenesis. Oogenesis is initiated during the embryonic

development stage when a couple of million gamete mother cells (oogonia) are

formed within each fetal ovary, no more oogonia are formed and added after birth.These cells start division and enter into prophase 1 of the meiotic division and get

temporarily arrested of the stage, called primary oocytes. Each primary oocyte then

gets surrounded by a layer of granulosa cells and then called the primary follicle.The primary follicles get surrounded by more layers of granulose cells and a new

theca and called secondary follicle which is characterized by a fluid filled cavity

called antrum. Primary oocyte within the tertiary called antrum. Primary oocytewithin the tertiary follicle grows in size and completes its first meiotic division. It

is an unequal division resulting in the formation of a large haploid secondary oocyte

and a tiny first polar body. The tertiary follicle further changes into the maturefollicle or graffian follicle. Graffian follicle now ruptures to release the secondary

oocyte (ovum) from the ovary by the process called ovulation.

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