l There are a total of 27 questions and four sections in the question paper. All questions are compulsory.

l Section A contains question number 1 to 5, very short-answer type questions of 1 mark each.

l Section B contains question number 6 to 12, short-answer type I questions of 2 marks each.

l Section C contains question number 13 to 24, short-answer type II questions of 3 marks each.

l Section D contains question number 25 to 27, long-answer type questions of 5 marks each.

l There is no overall choice in the question paper, however, an internal choice is provided in two questions of 1 mark, two

questions of 2 marks, four questions 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.

l Wherever necessary, the diagram drawn should be neat and properly labelled.

SET I

SECTION A

1. State from where do the signals forparturition originate in human females. [1]

2. Name the pattern of inheritance where F1

phenotype

(i) resembles only one of the two parents.[½]

(ii) does not resemble either of the two

parents and is in between the two. [½]

3. According to the Hardy-Weinbergprinciple, the allele frequency of apopulation remains constant.

How do you interpret the change offrequency of alleles in a population? [1]

Or Coelacanth was caught in South Africa.State the significance of discovery ofCoelacanth in the evolutionary history ofvertebrates. [1]

4. State the functions of mast cells in allergyresponse. [1]

Or State the function of interferons. [1]

5. What is the cell that receives arecombinant gene called? [1]

EXAMINATION

PAPER 2019BIOLOGY (ALL INDIA) Fully Solved

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General Instructions

SECTION B

6. Name a disorder a human suffers from as aresult of monosomy of the sexchromosome. Give the karyotype and writethe symptoms.

[2]

7. In the TS of a mature anther given below,identify ‘a’ and ‘b’ and mention theirfunctions. [1+1]

Or

What is cleistogamy? Write one advantageand one disadvantage of it to the plant.

[1+1]

8. State the role of thymus as a lymphoidorgan. Name the cells that are releasedfrom it and mention their function. [2]

9. ‘Artificial insemination helps overcomeseveral problems of normal mating in thecattle’s’. Do you agree? Support youranswer with any three reasons. [2]

10. Name and explain the interaction that isseen between clown fish and seaanemones. [2]

11. Write the relationship betweenproductivity, gross primary productivity,net primary productivity and secondaryproductivity. [2]

12. Justify the need for signing of ‘MontrealProtocol’ by the participating nations in1987. [2]

Or

Write the effective remedy found byAhmed Khan of Bengaluru for the efficientuse of the plastic waste generated by bigcities. [2]

SECTION C

13. Emasculation and bagging are the twoimportant steps carried during artificialhybridisation to obtain superior varieties ofdesired plants. Explain giving reasons, inwhich types of flowers and at what stagesare the two processes carried out. [3]

Or

State what is apomixis? Write itssignificance. How can it be commerciallyused? [3]

14. (i) Draw a sectional view of human ovary.

Label the following parts

(a) Primary follicle

(b) Secondary oocyte

(c) Graafian follicle

(d) Corpus luteum [2]

(ii) Name the hormones influencing folli-

cular development of corpus luteum.[1]

Or

(i) Draw an LS of pistil showing pollen tube

entering into the embryo sac. Label the

following

(a) Nucellus (b) Antipodals

(c) Synergids (d)Micropyle [2]

(ii) Write the functions of the following

(a) Synergids

(b) Micropyle [1]

15. Given below is the representation of arelevant part of amino acid composition ofthe β-chain of haemoglobin, related to theshape of human red blood cells.

a

b

Gene....CTC........GAG....

mRNA....GAG....

GluGluProThrLeuHisVal

1 2 3 4 5 6 7

HbA Peptide

(i) Is this representation of the sequence of

amino acids indicating a normal human

or a sufferer from a certain blood related

genetic disease? Give reason in support

of your answer. [1½]

(ii) Why is the disease referred to as a

Mendelian disorder? Explain. [1½]

Or

Name the kind of diseases/disorders andany two symptoms that are likely to occurin humans if

(i) mutation in the gene that codes for an

enzyme phenylalanine hydroxylase

occurs. [1½]

(ii) the karyotype is XXY. [1½]

16. Name the technique and the property ofplant cells that can help to growsomaclones of certain desired variety ofapple. Explain how somaclones of applecan be obtained in the lab so as to get thedesired variety on a large scale. [3]

17. Study the graph given below and answerthe questions that follows

(i) The curve ‘b’ is described by the

following equation

dN

dtrN

K N

K=

−

What does ‘K’ stand for in this equation?

Mention its significance.

(ii) Which one of the two curves is

considered a more realistic one for most

of the animal populations ?

(iii) Which curve would depict the population

of a species of deer if there are no

predators in the habitat? Why is it so? [3]

18. ‘A very small sample of tissue or even adrop of blood can help to determinepaternity’. Provide a scientific explanationto substantiate how it is possible. [3]

19. Explain the phenomena of dominance,multiple allelism and codominance takinghuman ABO blood group as an example.

[3]

20. Name the genus to which baculovirusesbelong. Describe their role in the integratedpest management programmes. [3]

21. Give reasons why

(i) DNA cannot pass into a host cell through

the cell membrane.

(ii) Proteases are added during isolation of

DNA for genetic engineering.

(iii) Single cloning site is preferred in a

vector. [ ]1 3×

22. State the medicinal value and the bioactivemolecules produced by Penicillium

notatum, Monascus purpureus andTrichoderma polysporum. [ ]1 3×

23. Describe the roles of (i) high temperature,(ii) primers and (iii) bacterium– Thermus

aquaticus in carrying the process ofpolymerase chain reaction. [ ]1 3×

Or

How does β-galactosidase codingsequence act as a selectable marker? Whyis it a preferred selectable marker toantibiotic resistance genes? Explain. [3]

24. Answer the following questions based onMeselson and Stahl’s experiment onE. coli

(i) Write the name of the chemical substance

used as the only source of nitrogen in the

experiment. [½]

(ii) Why did they allow the synthesis of the

light and the heavy DNA molecules in

the organism? [1]

(iii) How did they distinguish the heavy DNA

molecules from the light DNA

molecules? Explain. [1]

(iv) Write the conclusion the scientists

arrived at, at the end of the experiment.[½]

Time ( )t

Po

pu

latio

nd

en

sity

()

N

Ka

b

SET II

(ONLY UNCOMMON QUESTIONS FROM SET I )

5. Write the specific point in the palindrome and the bond that is cut by Eco RI. [1]

6. Explain polygenic inheritance with the help of an example. [2]

8. Write the scientific name of the source from where cocaine is obtained. How does its use affectthe human body? [2]

Natality

( )B

Mortality

( )D

Immigration ( )I

Emigration ( )E

Population Density

( )N

–+

+

–

Ovum

A

B C

D

E

SECTION D

25. Describe the process of megasporogenesisup to fully developed embryo sacformation in an angiosperm.

[5]

Or

Given below is the diagram of a humanovum surrounded by a few sperms. Studythe diagram and answer the followingquestions.

(i) Which one of the sperms would reach the

ovum earlier? [1]

(ii) Identify ‘D’ and ‘E’. Mention the role of ‘E’.[1]

(iii) Mention what helps the entry of sperm

into the ovum and write the changes

occurring in the ovum during the process.[2]

(iv) Name the specific region in the female

reproductive system where the event

represented in the diagram takes place.[1]

26. Study the schematic representation of thegenes involved in the lac operon givenbelow and answer the questions thatfollows

p i p o z y a

(i) Identify and name the regulatory gene in

this operon. Explain its role in ‘switching

off ’ the operon. [2]

(ii) Why is the lac operon’s regulation

referred to as negative regulation? [1]

(iii) Name the inducer molecule and the

products of the genes ‘z’ and ‘y’ of the

operon. Write the functions of these gene

products. [2]

Or

(i) How does the Hardy-Weinberg equation

explain genetic equilibrium? [1]

(ii) Describe how this equilibrium is

disturbed that may lead to founder effect.[4]

27. (i) Study the flow chart given below and

complete the equation that follows by

identifying 1, 2, 3 and 4.

N Nt t+ = + + − +1 1 2 3 4[( ) ( )] (2½)

(ii) Mention the different ways by which the

population density of different species

can be measured. [2½]

Or

(i) ‘The pyramid of energy is always

upright.’ Explain. [2]

(ii) Explain with the help of labelled

diagrams, the difference between an

upright pyramid of biomass and an

inverted pyramid of biomass. [3]

10. How is an alien species invasion

considered as one of the cause of

biodiversity loss? Support your answer

with the help of an example. [2]

19. Effluent from the primary treatment of the

sewage is passed through large aeration

tanks for biological treatment.

Explain the complete process that follows

till the water is ready to be released into the

natural water bodies. [3]

20. Expand ‘BAC’ and ‘YAC’. What are they

and what is the purpose for which they are

used? [3]

21. (i) Mention the importance of gel-

electrophoresis in biotechnology. [1]

(ii) Explain the process of this technique. [2]

25. (i) Why and how must the ozone layer in the

stratosphere be protected? Explain. [2]

(ii) How do deforestation and greenhouse

gases negatively affect our environment?

Explain. [3]

Or Biomagnification and acceleratedeutrophication are both caused due toindiscriminate use of chemicals andirresponsible human activities. Do youagree? Support your answer withexplanation and an example of each. [5]

SET III

(ONLY UNCOMMON QUESTIONS FROM SET I AND II )

5. Write the two specific ‘cry’ genes thatencode the proteins which control cottonbollworms. [1]

6. Write the genotypes of the parents of achild suffering from thalassemia. State thecause of this disease. [2]

8. List any four common approaches for thetreatment of cancer. [2]

10. State Gause’s ‘Competitive ExclusionPrinciple’. How have the recent studiesmodified this principle? [2]

19. Explain the phenomenon of evolution bynatural selection as supported by thevariations observed in white winged anddark winged moth populations in Englandbetween 1850-1920. [3]

20. Explain ‘mutation breeding’ as a method ofdeveloping disease resistant crop plants.Give an example of a pulse crop where thistechnique has helped. [3]

21. Explain how ‘sticky ends’ are obtained in aDNA segment. Write their importance inDNA technology. [3]

27. (i) Work out the crosses so as to obtain the

phenotypic ratios given below :

(a) 1 : 2 : 1 (in F2 -generation) [1]

(b) 3 : 1 (in F2 -generation) [1]

(c) 1 : 1 (in F1 -generation) [1]

(ii) Differentiate between pleiotropy and

polygenic inheritance patterns. [2]

Or (i) Describe the process of amino acylation.[2½]

(ii) ‘Process of transcription and translation

are coupled in prokaryotes, but not in

eukaryotes’. Explain. [2½]

SET I

1. Foetal-ejection reflex stimulates pituitary to release the

hormone responsible for parturition. This hormone is

oxytocin. (1)

2. (i) Dominance (½)

(ii) Incomplete dominance (½)

3. The change of frequency of alleles will be interpreted as

resulting in evolution. (1)

Or

The discovery of Coelacanth (lobefins), first amphibians

is significant as they prove that amphibians have evolved

from fish-like organisms. Lobefins were the ancestors of

modern day frogs and salamanders. (1)

4. Allergy occurs when chemicals like histamine and

serotonin are released from the mast cells. These

chemicals cause inflammatory reactions in the body.(1)

OrInterferons are a type of cytokine barriers released from

virus infected cells. They protect non- infected cells from

the other viral infected cells. (1)

5. Host cell is the cell that receives a recombinant gene.(1)

6. Monosomy of the sex chromosome in humans results in

Turner’s syndrome.

The number of chromosomes is 45 (44 XO)+ . (1)

Symptoms are as follows

The affected females are sterile as ovaries are

rudimentary.

Other symptoms include lack of secondary sexual

characters, short stature, etc. (1)

7. In the given figure, a is sporogenous tissue and b is

tapetum.

(i) Sporogenous tissue has cells which are potential

Pollen Mother Cell (PMC) or microspore mother cell

and give rise to microspore tetrad. (1)

(ii) Tapetum nourishes the developing microspores or

pollen grains. (1)

Or

Cleistogamy is a type of self-pollination that occurs in a

permanently closed flower.

Advantage and disadvantage of cleistogamy are as

follows

(i) Advantage Cleistogamous flowers produce

assured seed-set even in the absence of pollinators.(1)

(ii) Disadvantage Cleistogamous flowers are invariably

autogamous. So, there is no chance of

cross-pollination. Hence, less variations are

generated in the progeny. (1)

8. The thymus is a specialised primary lymphoid organ of

the immune system. Within the thymus, T-lymphocytes

mature.

T-lymphocytes form Cell-Mediated Immune System

(CMIS). They defend against pathogens including

protists and fungi that enter the body. (2)

9. Yes, I do agree. Artificial insemination helps overcome

several problems of normal mating in the cattle. The

reasons to support the statement are as follows

(i) Artificial insemination helps to overcome problems

related to normal mating.

(ii) Semen can be transported very easily in frozen form

from one place to where female is housed and thus

desirable matings are carried out.

(iii) Semen can be stored and frozen for the later use.(2)

10. Clown fish maintains commensalism with the sea

anemone. In this interaction, one species is benefitted

and the other is neither harmed nor benefitted. Sea

anemone has stinging tentacles that provide protection

to clown fish from predators. The anemone does not

appear to derive any benefit from the clown fish. (2)

11. The rate of biomass production is called productivity in

term of gm−2 yr−1 for an ecosystem. It can be divided in

Gross Primary Productivity (GPP) and Net Primary

Productivity (NPP). GPP of an ecosystem is the rate of

production of organic matter during photosynthesis.

NPP is the available biomass for the consumption of

heterotrophs (herbivores and decomposers).

Secondary productivity is defined as the rate of formation

of new organic matter by consumers. (2)

12. Montreal Protocol was signed at Montreal (Canada) in

1987 (effective in 1989) to control the emission of ozone

depleting substances like CFCs and aerosols.

The balance of ozone production and degradation has

been disrupted due to enhancement of ozone

degradation by Chlorofluorocarbons (CFCs). Once

CFCs are added to the stratosphere, they have

permanent and continuous effects on ozone levels. (2)

Or

Ahmad Khan developed a fine powder called polyblend

of recycled modified plastic. This mixture is mixed with

the bitumen and used to lay roads.

It enhanced the bitumen’s water repellant properties and

helped to increase road life by a factor of three. By this

way, the problems created by plastic waste was solved

and plastic manufacturers thought of producing

polyblend. (2)

EXPLANATIONS

13. If the plant bears bisexual flowers, emasculation and

bagging are carried out before the anther dehisces.

If the plant bears unisexual flowers, emasculation is not

required. The female flower buds are bagged before the

opening of flowers. (3)

Or Apomixis is the special mechanism to produce seeds

without fertilisation. It is a form of asexual reproduction

that mimics sexual reproduction. (1)

Significance

(i) It helps in fixing heterosis or hybrid vigour in plants

permanently.

(ii) Rapid multiplication of genetically uniform

individuals can be achieved without risk of

segregation. (1)

Commercial use Apomixis is used in plant breeding. It

increases the chance of developing superior gene

combinations and facilitates the rapid incorporation of

desirable traits. (1)

14. (i)

(2)

(ii) The hormones influencing follicular development of

corpus luteum are FSH, oestrogen and

progesterone. (1)

Or

(i)

(2)

(ii) (a) Synergids These possess special cellular

thickenings at their micropylar tip called filiform

apparatus. This filiform apparatus guides the

pollen tube to enter embryo sac. (½)

(b) Micropyle It facilitates the entry of pollen tube

and thus fertilisation. (½)

15. (i) This representation indicates normal human

because normal peptide HbA is with glutamic acid

(Glu) at sixth position of β-globin chain of

haemoglobin molecule. Sickle-cell anaemia is

referred to as Mendelian disorder because each

parent contributes one mutant allele to affected

offspring and the recessive sickle trait is expressed in

homozygous condition only. (1½)

(ii) Since, this disease transmission follows Mendelian

principles, it is called Mendelian disorder.

Inheritance pattern It is transmitted from parents to

the offspring, when both the partners are carriers

(heterozygous) of the disease. In this disease, RBCs

become sickle-shaped. (1½)

Or (i) Phenylketonuria It is an inborn error of metabolism,

which is inherited as an autosomal recessive trait.n The disease is caused due to the deficiency of an

enzyme phenylalanine hydroxylase that converts theamino acid phenylalanine into tyrosine. In thedeficiency of this enzyme, phenylalanine is notconverted into tyrosine.

n The phenylalanine is accumulated and getsconverted into phenyl-pyruvic acid and otherderivatives.

n Their accumulation in brain results in mentalretardation.

n It is also excreted through urine because of its poorabsorption by kidney. (1½)

(ii) Klinefelter’s syndrome It is caused due to the

presence of an additional copy of X-chromosome

(XXY), resulting into 47 chromosomes. So, it is an

example of trisomy of sex chromosome. It is

represented as ( )2 1n + .n Male individuals have masculine development, but

feminine characters (development of breast, i.e.gynaecomastia) also occurs.

n The individuals are sterile. (1½)

16. Micropropagation is the technique of tissue culture that

can help to grow somaclones. All plants produced by

tissue culture are genetically identical to the original plant

from where they are cultured, so they are called

somaclones.

Somaclones cultured from explants are grown in following

steps

(i) The explant is grown in a petridish/test tube under

sterile conditions on a special nutrient medium.

(ii) The medium must contain nutrient sources like

sugar, inorganic salts, vitamins, minerals, amino

acids, growth regulators like auxins and cytokinins.

(iii) The medium should be replaced regularly to restore

nutrients in it.

(iv) Different combination of plant hormone induces

growth and differentiation.

Blood vessels

Primary follicle

Tertiary follicleshowing antrum

Graafian follicle

Secondary oocyte(Ovulation)Corpus luteum

Secondary follicle

Diagrammatic sectional view of ovary showing

different stages

Stigma

Pollen tube

Chalazal end

Antipodals

Secondary nucleus

Synergids

Filiform apparatus

Male gametes

Micropyle

Hilum

Embryo sacNucellus

Integument

LS of pistil

(v) Cells differentiate into roots and shoots and plantlets

are formed.

(vi) These plantlets are transferred to soil or pots and

grown to maturity. (3)

17. (i) ‘K ’ stands for ‘carrying factor’. The carrying capacity

signifies the limit of habitat, i.e. limited resources in a

given habitat to support growth up to a certain level

beyond which no further growth can take palce.(1)

(ii) The curve ‘b’ is considered a more realistic one for

most of the animal populations. It is so, becasue in

the curve ‘b’, the sources of food and space are

limited and it supports the growth curve of animal

populations. (1)

(iii) Since, there are no predators in the habitat, the

population of a species of deer will show exponential

growth curve. (1)

18. DNA fingerprinting is the basis of paternity testing in case

of disputes. This technique is used to distinguish

between individual of same species by using their DNA

sample.

The DNA is isolated from the cells and further amplified

to produce many copies by using polymerase chain

reaction. This amplified DNA is further processed to

detect the presence of similarities between the parent

and child.

Because, the DNA from sample can be amplified to

produce many copies in DNA fingerprinting, only a very

small sample of tissue or even a drop of blood can help

determine paternity. (3)

19. Codominance It is a phenomenon in which two alleles

are able to express themselves independently when

present together, i.e. both alleles are codominant.

These alleles are called codominant alleles. The

offsprings show resemblance to both the parents.

(i) A common example of codominance is ABO blood

group in humans, which is controlled by the gene I.

(ii) The gene for blood group exists in three allelic forms

I , IA B and i. However, an individual contains any two

of the three I alleles.

(iii) IA and IB produce RBC surface antigens A and B,

respectively, whereas ‘I’ does not produce any

antigen.

(iv) IA and IB both are dominant alleles, whereas ‘i’ is the

recessive allele.

(v) When IA and IB are present together, both express

equally and produce both the surface antigens A and

B, respectively.

(vi) These three different alleles may produce six

different genotypes of human ABO blood group that

may show four phenotypes, i.e. A, B, AB and O.

Genetic Basis of Blood Groups in Human Population

Allele fromparent 1

Allele from

parent 2

Genotype of

offspring

Blood types

of offspring

IA

IA

I IA A A

IA

IB

I IA B AB

IA i I

A i A

IB

IA

I IA B AB

IB

IB

I IB B B

IB i I

B i B

i i ii O

Multiple Allelism When more than two alleles are

present for a character then, the condition is known as

multiple allelism. It can be explained by ABO blood

grouping. In this case, more than two, i.e. three alleles

are governing the same character. Multiple alleles can be

found only when population studies are made since, an

individual can have only two alleles. (3)

20. Baculoviruses are pathogens that attack insects and

other arthropods. Majority of baculoviruses which are

used as biocontrol agents belong to the genus–

Nucleopolyhedrovirus.

These are species specific, narrow spectrum

insecticides. They do not harm plants, mammals, birds,

fish and other non-target insects.

Baculoviruses are beneficial in Integrated Pest

Management (IPM) programme, in which beneficial

insects are conserved. (3)

21. (i) Hydrophobic molecules can diffuse through the lipid

bilayer of the plasma membrane and not hydrophilic

molecules. DNA being hydrophilic with the

sugar-phosphate backbone, cannot pass through

the cell membrane. (1)

(ii) Proteases, catalyse the breakdown of proteins

present in the solution to its component amino acid.

If the proteins are not removed from DNA preparation

then, they could interfere with any downstream

treatment of DNA (such as action of restriction

endonuclease, DNA ligase, etc). (1)

(iii) Single cloning sites are preferred as the presence of

more than one recognition sites within the vector that

will generate several fragments leading to

complication in gene cloning. (1)

22. (i) Penicillium notatum is used for the production of

penicillin. Significance of this antibiotic is that it

contributed towards the welfare of the human

society; they have helped in controlling various

harmful diseases. (1)

(ii) Statins are produced by yeast Monascus purpureus.

These have been commercialised as blood

cholesterol lowering agents. (1)

(iii) Cyclosporin-A is produced by the fungus

Trichoderma polysporum and it is used as an

immunosuppressive agent in case of organ

transplants. (1)

23. (i) Role of Heat In PCR (in vitro), the DNA strands are

separated by heating at 95°C for two minutes.

Heating causes the H-bonds between bases of two

strands to get broken leading to unwinding. (1)

(ii) Role of Primers Primers are short lengths of DNA

about 20bp long that are required to start DNA

polymerisation in PCR. The primers hybridise to their

complementary sequence on the DNA strands at

40-50°C temperature and help in DNA

polymerisation. (1)

(iii) Role of Thermus aquaticus An enzyme called Taq

polymerase is isolated from Thermus aquaticus.

Since, this bacteria thrives in temperature as high as

95°C, this enzyme can also tolerate high temperature

without undergoing denaturation. Therefore, this

enzyme is used in PCR instead of normal DNA

polymerase. (1)

Or

Selectable Marker It helps in identifying or selecting

transformants and eliminating non-transformants and

selectively permit the growth of the transformants.

β-galactosidase acts as a selectable marker by inducing

the property of insertional inactivation in transformed

cells. In this process, the recombinants and

non-recombinants are differentiated on the basic of

colour production in the presence of chromogenic

substrate.

A recombinant DNA is inserted within the coding

sequence of an enzyme β-galactosidase which results

into inactivation of the enzyme.

Therefore, the bacterial colonies having inserted

plasmid, show no colouration while, those without

plasmid show blue colour. (3)

24. (i) NH Cl4 (Ammonium chloride) with N14 atom. (1/2)

(ii) It is to show that after one generation, E. coli with 15N

-DNA in a medium of 14N, has DNA of intermediate

density between the light and heavy DNAs.

It shows that of the two strands, only one strand is

synthesised newly, using the 14N-nitrogen source in

the medium. (1)

(iii) The heavy and light DNA molecules can be

differentiated by centrifugation in a cesium chloride

(CsCl) density gradient.

The 15N -DNA was heavier than 14N -DNA and the

hybrid 15 14N N− -DNA was intermediate between the

two newly formed DNA strands. (1)

(iv) Scientists concluded that the DNA replication is

semiconservative, i.e. of the two strands of DNA, one

is the parental strand, while the other is newly

synthesised. (1/2)

25. The functional megaspore undergoes mitosis to form

two nuclei, which migrate to opposite poles, forming a

2-nucleate embryo sac. Further, mitotic divisions lead to

the formation of 4-nucleate and 8-nucleate stages of the

embryo sac. In these mitotic divisions, nuclear division is

not followed by cell division. After the 8-nucleate stage,

cell walls are laid down and a typical female

gametophyte or embryo sac is formed.

Among the eight nuclei, six are enclosed by cell walls

and organised into cells, while the remaining two nuclei

(polar nuclei) are situated above the egg apparatus in a

large central cell.

Out of the six cells, three are grouped at the micropylar

end and constitute the egg apparatus. It is made up of

two synergids and one egg cell. The other three cells are

located at the chalazal end and are called antipodals.

Thus, a typical angiosperm embryo sac after maturity is

8-nucleate and 7-celled. (5)

Or (i) The sperm ‘A’ would reach the ovum earlier. (1)

(ii) D – Corona radiata, E–Zona pellucida-During

fertilisation, a sperm comes in contact with zona

pellucida layer of the ovum and induces changes in

the membrane that blocks the entry of additional

sperms. (1)

(iii) The secretions of the acrosome help the sperm enter

into the cytoplasm of the ovum through the zona

pellucida and the plasma membrane. This induces

the completion of the meiotic division of the

secondary oocyte. (2)

(iv) This event takes place in the Fallopian tube of uterus.(1)

26. (i) i gene — regulatory gene.

It codes for the repressor protein of the operon,

which is synthesised constitutively.

The repressor has the affinity for the operator gene. It

binds to the operator and prevents the RNA

polymerase from transcribing the structural genes.(2)

(ii) When repressor binds to the operator, the operon is

switched off and transcription is stopped. So, it is

called negative regulation. (1)

Diagrammatic representation of the

mature embryo sac

(iii) Lactose is an inducer molecule.

Gene z codes for β-galactosidase, which is

responsible for the hydrolysis of lactose into

galactose and glucose.

y gene codes for permease. It increases the

permeability of the cell to lactose. (2)

Or

(i) According to Hardy-Weinberg principle, the allele

frequencies in a population are stable and is

constant from generation to generation. (1)

(ii) The four factors that affect genetic equilibrium aren Gene migrationn Genetic driftn Mutation and recombinationn Natural selection

Whenever the gene migration occurs multiple times,

it leads to some changes, that may sometimes result

in change in allele frequency (at random or by

chance).

This difference in allele frequency leads to a new

sample of population in such a way that they evolve

into a different species. Such populations are called

founders and the effect generated is called founder

effect. (4)

27. (i) If N is the population density at time t, then its density

at time t 1+ is

N N B I D Et t+ = + + − +1 [( ) ( )]

The population density will increase if the number of

births and the number of immigrants, i.e. (B+I) is

more than the number of death and the number of

emigrants, i.e. (D+E). (2½)

(ii) Three ways of measuring population density of a

habitat are as follows

(a) Per cent cover for trees with larger canopy.

(b) Number of fishes caught per trap.

(c) Pug marks or faecal pellets for tiger census.(2½)

Or

(i) Pyramid of energy is always upright because when

energy flows from a particular trophic level to the next

trophic level, some energy is always lost as heat at

each step.

Each bar in the energy pyramid indicates the amount

of energy present at each trophic level in a given

time. (2)

(ii) (a) The pyramid of biomass in a sea ecosystem is

inverted. Because, the sum total of the weight of

phytoplankton (producer) is far less than a few

fishes feeding at higher trophic levels.

(1)

(b) Pyramid of biomass in a forest ecosystem is

upright because producers are more in biomass

than primary consumers. Primary consumers are

more than secondary consumers and secondary

consumers are more than tertiary consumers

(top).

(2)

SET II5. Restriction endonuclease Eco RI cuts the DNA at the

sequence known as palindromic sequence, i.e.

5′ GAATTC 3′3′ CTTAAG 5′The bonds that are cut by Eco RI are phosphodiester

bonds. (1)

6. The concept of polygenic inheritance was given by

Galton in 1833. In this, traits are controlled by three or

more genes (multiple genes). Such traits are called

polygenic traits.

The phenotype is produced as a result of participation of

several genes and is also influenced by the environment.

It is also called quantitative inheritance as the character

or phenotype can be quantified. For example, human

skin colour is produced by a pigment melanin.

The quantity of melanin is due to three pairs of polygenes

(A, B and C). If a black or very dark (AA BB CC) and white

or very light (aa bb cc) individuals marry each other, the

offspring shows intermediate colour who are often called

mulatto (Aa Bb Cc). A total of eight allele combinations

are possible in the gametes forming 27 distinct

genotypes. (2)

8. Cocaine is derived from the leaves and young branches

of a South American plant called Erythroxylum coca.

Its mode of intake is sniffing and snorting.

It is a strong stimulant. When taken in high doses, it

causes headache, convulsions, hallucination and death

due to cardiovascular or respiratory failure. (2)

10. Alien species become invasive, compete with the native

species and cause extinction of indigenous species.

70

Dry weight (kgm )-2Trophic level

20

4

SC

PC

PP

Sea ecosytem

Trophic level Dry weight (kgm )-2

1.5

11

37

809

TC

SC

PC

PP

Forest ecosystem

(i) Introduction of Nile perch into Lake Victoria leads to

extinction of more than 200 species of cichlid fish in

that lake. (1)

(ii) Carrot grass (Parthenium) and Lantana introduced in

our country have become invasive and cause

environmental damage. They pose a threat to the

native species of plants in our forests. (1)

19. Secondary Treatment (Biological Treatment)

(i) Primary effluent is passed into large aeration tanks

with constant mechanical agitation and air supply.

(ii) This allows vigorous growth of useful aerobic

microbes into flocs (masses of bacteria associated

with fungal filaments to form mesh-like structures).

(iii) These microbes consume major part of organic

matter in the effluent, while growing and hence,

reduce the Biochemical Oxygen Demand (BOD) of

the effluent.

(iv) When BOD of the sewage gets reduced, the effluent

is passed into settling tank.

(v) The bacterial flocs are allowed to settle in the tank

and the sediment is called activated sludge.

(vi) A small amount of activated sludge is pumped back

into the aeration tank to serve as the inoculum.

(vii) The remaining major part of the sludge is pumped

into large tanks called anaerobic sludge digesters.

(viii) In anaerobic sludge digesters, other kinds of

bacteria, which grow anaerobically, digest the

bacteria and fungi in the sludge. During this process,

bacteria produce a mixture of gases, such as

methane, hydrogen sulphide and the carbon

dioxide, which form biogas (It can be used as source

of energy as it is inflammable).

(ix) The effluent from secondary treatment is generally

released into natural water bodies like rivers and

streams. (3)

20. ‘BAC’ stands for Bacterial Artificial Chromosome and

‘YAC’ stands for Yeast Artificial chromosome. These are

vectors used in cloning DNA.

For sequencing, the total DNA from cell, the DNA is

isolated and converted into relatively smaller size as

fragments. DNA fragments are cloned in suitable host

using specialised vectors, such as BAC and YAC.

Fragments of DNA are then sequenced by automated

DNA sequences. (3)

21. (i) DNA fragments formed by the use of restriction

endonucleases are separated by gel electrophoresis.(1)

(ii) The process of gel electrophoresis is as follows

(a) DNA fragments are negatively charged

molecules. Thus, they move towards the anode

under electric field through the medium.

(b) DNA fragments separate according to their size

due to seiving effect of agarose gel.

(c) The separated DNA fragments can be viewed by

staining the DNA with ethidium bromide followed

by exposure to UV radiation.

(d) The separated bands of DNA are cut and

extracted from gel piece. This is known as

elution. (2)

25. (i) The ozone layer in the stratosphere must be

protected, because the ozone in the stratosphere is

good ozone that acts as a shield and absorbs

ultraviolet radiations from the sun. UV rays are

extremely harmful to living organisms.

Ozone gas is continuously formed by the action of

UV rays on molecular oxygen and also degraded into

molecular oxygen in the stratosphere.

In order to protect the ozone layer, there should be a

balance between production and degradation of

ozone in the stratosphere.

In the recent years, the balance has been disrupted

due to enhancement of ozone degradation by

Chlorofluorocarbons (CFCs). So, the emission of

ozone degrading substances like CFCs need to be

controlled to protect the ozone layer in stratosphere.(2)

(ii) Effects of Deforestation Various effects of

deforestation are as follows

(a) Increased levels of CO2 concentration in the

atmosphere.

(b) Loss of biodiversity due to habitat destruction.

(c) Disturbs hydrologic cycle.

(d) Leads to soil erosion.

(e) Desertification also occurs in extreme cases.

Effects of Greenhouse Gases

(a) Greenhouse gases include CO2, CFCs, etc. Their

high level in the atmosphere allows the heat waves to

reach earth, but prevents their escape and earth

becomes warm. This increased temperature in last

three decades had caused changes in precipitation

patterns.

(b) The increased temperature also causes EL-Nino

effect, i.e. melting of the polar ice caps and

Himalayen snow. (3)

Or I agree that biomagnification and accelerated

eutrophication are both caused due to the indiscriminate

uses of chemicals and irresponsible human activities.

Biological magnification can be defined as

an increase in concentration of toxicants at

successive trophic levels in aquatic food chain.

l Mercury and DDT are well-known cause of biological

magnification.

l Toxic materials cannot be metabolised or excreted.

Therefore, they get accumulated in an organism and pass

on to higher trophic levels.

l DDT accumulates in birds and disturbs calcium

metabolism, which results in thinning of egg shell and

SET III5. cry I Ac and cry II Ab control cotton bollworm. (1)

6. The genotypes of the parents of a child suffering from

thalassemia are as follows

Mother → Th ThB B+ , Father → Th ThB B+

Thalassemia is an autosomal-linked recessive disease,

which results in reduced rate of synthesis of one of the

globin chains of haemoglobin. The disease is controlled

by a single pair of allele, ThB and ThB+ .

ThB codes for normal B-protein and ThB+ codes for an

abnormal B-protein. Only homozygous individuals for

ThB+ ( )Th ThB B+ + show the diseased phenotype. (2)

8. Treatment of Cancer It involves following methods

(i) Surgery Tumours are removed by surgery to check

further spread of cancer cells.

(ii) Radiation therapy Tumour cells are irradiated by

lethal doses of radiation, such as cobalt 60 while

taking care to protect the surrounding normal cells.

(iii) Chemotherapy Several chemotherapeutic drugs are

used to kill cancer cells. But, their side effects like hair

loss, anaemia, etc., are also reported. Most cancers

are treated by the combination of the above three.

(iv) Immunotherapy Biological modifiers like α-inter-

ferons are used to activate the immune system and

help in destroying the tumour. (2)

10. Gause’s competitive exclusion principle states that two

closely related species competing for the same

resources cannot coexist indefinitely and the

competitively inferior one will be eliminated eventually

(true if resources are limiting).

This principle has been modified to ‘complete

competitors cannot coexist’ one of the primary ways

niche-sharing species can coexist is the

competition-colonisation trade-off.

In simple terms, species that are better competitors will

be specialists, whereas species that are better colonisers

are more likely to be generalists. (2)

19. In England, prior to industrialisation, the tree trunks were

covered with white lichens hence, white moths could

survive and were protected from predators due to white

colour.

On the other hand, black moths (a dark-winged moths)

could be easily identified due to their dark colour and

declined in number due to predation.

However, as industrialisation progressed, the lichens

were replaced by soot and dust particles and dark

coloured moths were benefitted due to camouflage,

while white-winged moths could be easily eaten up by the

predators being easily identifiable.

Thus, only the dark-winged moths who were able to fit

and survive, i.e. adapted well in conditions, reproduced

well in nature. Thus, natural selection operates in nature

by selecting the fittest characters of organisms. (3)

20. Mutation breeding involves inducing mutations

artificially by using chemicals or radiations (e.g. gamma

radiation).(1)

It is carried out in the following steps

n Inducing mutations in plants by various means.

n Screening the plant for resistance.

n Selecting the desirable plant for multiplication or for

breeding. (1)

In mung bean, resistance to yellow mosaic virus and

powdery mildew were introduced by mutations. (1)

21. Sticky ends on DNA are formed by the action of

enzymes restriction endonucleases.

These enzymes cut the strand of DNA a little away from

the centre of the palindrome sequence between the

same two bases on both the strands.

This results in single-stranded stretches on both the

complementary strands at their ends. These

overhanging stretches are called sticky ends as they

form hydrogen bonds with the complementary base pair

sequences.

their premature breaking. This results in decline of bird

population.

Eutrophication can be defined as the natural ageing of a

lake by nutrient enrichment of its water.

Process of Eutrophication

l Water in young lake is cold and clear to support little life.

l Gradually with time, it gets enriched with nutrients such as

nitrogen and phosphorus by streams draining in it.

l Due to this, aquatic life (plants and animals) flourish in

lake.

l Organic remains deposit at the bottom of the lake and

with time, water becomes warmer. This warm water

supports even more growth of plants.

Eventually, floating plants develop in the lake, finally

converting it into land. Natural Eutrophication generally

takes 1000 of years or more. But, human activities have

accelerated this natural process and as a result many

lakes are already extinct.

It is caused due to dumping of nutrient rich and thermal

waste into lakes. This is known as accelerated or cultural

eutrophication.

Example, Excessive growth of water hyacinth

(Eichhornia crassipes). It is the most problematic aquatic

weed, also called Terror of Bengal.

It grows abundantly in eutrophic water bodies and

imbalances water ecosystem. (5)

Role of the sticky ends These sticky ends produced

from hydrogen bonds with their complementary cut

counterparts. This stickiness of the ends facilitates the

action of the enzyme DNA ligase. (3)

27. (i) (a) 1 : 2 : 1 (in F2-generation) is the phenotypic ratio

of incomplete dominance.

(1)

(b) 3 : 1 (in F2-generation) is the phenotypic ratio of

monohybrid cross.

(1)

(c) 1 : 1 (in F1-generation) is the phenotypic ratio of

test cross.

In this example, the progeny consists of tall and dwarf

plants in the ratio of 1 1: . Thus, monohybrid test cross

ratio is 1 1: . (1)

(ii) Differences between pleiotropy and polygenic

inheritance are as follows

Pleiotropy Polygenic inheritance

Single gene productconfers manyphysiological effects.

Single phenotypic effectis under the control ofmany genes.

The genes involvedare called pleiotropicgenes.

The genes involved arecalled polygenes.

e.g. phenylketonuria. e.g. human skin colour.

(2)Or

(i) Aminoacylation or charging of tRNA is a process in which

amino acids get activated in the presence of ATP and get

linked to their cognate tRNA.

Significance of aminoacylation The charged tRNA

carries amino acids to the site of protein synthesis. If two

charged tRNAs are brought close to each other, the

formation of peptide bond is favoured energetically.(2½)

(ii) In bacteria, both processes occur in cytoplasm as there

is no nucleus.

In eukaryotes, transcription occurs in nucleus, while

translation occurs in the cytoplasm.

(a) Complexities in eukaryotic transcription

Eukaryotes have additional complexities in gene

expression than prokaryotes as mentioned in gene

expression below

n There are at least three RNA polymerases in the

nucleus other than the RNA polymerase in

organelles. The RNA polymerase-I transcribes

rRNAs (28S, 18S and 5.8S). RNA polymerase-III is

responsible for the transcription of tRNA, 5srRNA

and snRNAs (small nuclear RNAs).

n RNA polymerase-II transcribes precursor of

mRNA, the heterogeneous nuclear RNA (hnRNA).

(b) Complexities in translation in eukaryotes

n The mRNA formed in nucleus has to be transportedto the cytoplasm.

n Transcription and translation cannot be coupled ineukaryotes. (2½)

(Red)

RR

R

(White)

rr

r

Rr Rr

F -generation1Rr

R

R

r

r

F -generation2

Phenotypic ratio

Genotypic ratio

Red :RR :

Pink

Rr

Pink :

Rr :

White

rr

: :2

All pink

1 1

(Pink)

Rr(Red)RR

(Pink)

Rr

(White)

rr

Gametes

Parents

(Hybrid tall)

Tt

t

(Pure dwarf)

tt

tT

Tt tt

Tall Dwarf

Gametes

Parents

Progeny