Download + 2 subjective question bank reproduction

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+ 2 SUBJECTIVE QUESTION BANK
REPRODUCTION
SECTION – A
I. Each question carries 1 mark.
1. The following statements describe the characters of wind-pollinated plants. Which one of these
statements is incorrect?
(i) The pollen grains are sticky
(ii) Stamens are well exposed
(iii) Flowers often have a single ovule
Ans. (i) The pollen grains are sticky.
2. Write the function of Nucleus of human sperm
Ans. Nucleus contains genetic material and has haploid set of chromosomes. It passes on paternal
characters to the offspring and helps in maintaining diploid chromosome number in species.
3. Name an organism where cell division in itself is a mode of reproduction.
Ans. Amoeba is an organism where cell division in itself is a mode of reproduction.
4. Name an alga that reproduces asexually through zoospores.
Ans. Chlamydomonas.
5. Name the phenomenon and one bird where the female gamete directly develops into a new
organism.
Ans. Parthenogenesis, Turkey
SECTION – B
II. Each question carries 2 marks.
1. What do you understand by amniocentesis? Why is there a statutory ban on this? Give reason.
Ans. It is a procedure in which amniotic fluid is taken from amniotic sac of the foetus to diagnose various
chromosomal and genetic disorders.
During this test sex of the baby is also revealed. This promotes the practice of female foeticide.
Hence, it is justified to ban the process to check female foeticide.
2. When and where do chorionic villi appear in humans? State their function.
Ans. After implantation, finger-like projections appear on the trophoblast called chorionic villi. The chorionic
villi and uterine tissue become inter digitated with each other and jointly form a structural and functional
unit between developing embryo (foetus) and maternal body called placenta. The placenta facilitates
the supply of oxygen, antibodies, hormones and nutrients to the embryo. It also helps in the elimination
of carbon dioxide and excretory wastes from foetus.
3. (a) How does cleistogamy ensure autogamy?
(b) State one advantage and one disadvantage of cleistogamy to the plant.
Ans. (a) Cleistogamous flowers do not open at all. In such flowers, the anthers and stigma lie close to each
other. When anthers dehisce in the flower buds, pollen grains come in contact with stigma to effect
pollination. So cleistogamous flowers are autogamous as there is no chance of cross pollen landing on
stigma.
(b) Advantage – Cleistogamous flowers produce assured seed-set even in the absence of pollinators.
Disadvantage – Genetic variations will not be exhibited by cleistogamous flower and hence the plants
will suffer from inbreeding depression.
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4. Explain the steps that ensure cross pollination in an autogamous flower.
Ans. (i) Removal of anthers from the flower bud in case of bisexual flower before the anther dehisces
(Emasculation).
(ii) Covering emasculated flowers with a bag of suitable size and generally made up of butter paper
(Bagging).
(iii) Dusting of desired pollen grain on stigma of bagged flower when the stigma of bagged flower
attains receptivity and then rebagging it.
5. Name all the haploid cells present in an unfertilized mature embryo-sac of a flowering plant.
Write the total number of cells in it.
Ans. 3 Antipodal cells, one large central cell having 2 polar nuclei, 1 egg cell and 2 synergids are the
haploid cells present in an unfertilized mature embryo sac of a flowering plant. Thus, it has 7-cells.
6. Differentiate between the two cells enclosed in a mature male gametophyte of an angiosperm.
Ans.
Generative cell
Vegetative cell
It is small and floats in the cytoplasm of the vegetative It is bigger, has abundant food material and a
cell. It is spindle shaped with dense cytoplasm and large irregularly shaped nucleus.
nucleus. It divides to form male gamete.
7. Explain the hormonal regulation of the process of spermatogenesis in humans.
Ans. Spermatogenesis starts at the age of puberty due to significant increase in the secretion of
gonadotropin releasing hormone (GnRH) from the hypothalamus. GnRH acts at the anterior pituitary
gland and stimulates secretion of two gonatodropins – luteinizing hormone (LH) and follicle stimulating
hormone (FSH). LH acts at the Leydig cells and stimulates secretion of testosterone – a principal
androgen. Testosterone is essential for sperm formation. Androgens in turn stimulates the process of
spermatogenesis. FSH acts on Sertoli cells and stimulates secretion of some factors (like ABP,
androgen binding protein) which help in the process of spermiogenesis. They bind to testosterone and
concentrate testosterone in seminiferous tubules.
SECTION – C
III. Each question carries 3 marks.
1.
Study the graph given above showing the levels of ovarian hormones during menstruation and
ovulation. Correlate the uterine events that take place according to the hormonal levels on:
(i) 6 – 15 days
(ii) 16 – 25 days
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(iii) 26 – 28 days (if the ovum is not fertilized)
Ans.
(i) Regeneration of endometrium and ovulation.
(ii) Uterus gets highly vascularised, spongy and is ready for embryo implantation
(iii) If ovum is not fertilised the endometrium disintegrates leading to menstruation.
2. Expand the following and explain any one of them.
(i) IVF
(ii) ZIFT
(iii) IUI
(iv) MTP
Ans.
(i) IVF – In Vitro Fertilisation
(ii) ZIFT - Zygote Intra Fallopian Transfer
(iii) IUI – Intra Uterine Insemination
(iv) MTP – Medical Termination of Pregnancy
(v) MTP (Medical termination of pregnancy) is a procedure done to get rid of unwanted pregnancies
before 20 weeks of pregnancy. It is also essential in certain cases where continuation of the
pregnancy could be harmful or even fatal either to the mother or to the foetus or both.
3. Draw the following diagrams related to human reproduction and label them:
(a) The zygote after the first cleavage division
(b) Morula stage
(c) Blastocyst stage (sectional view)
4. Suggest and explain any three Assisted Reproductive Technologies (ART) to an infertile couple.
Ans. Infertile couples could be assisted to have children through some special methods called assisted
reproductive technologies (ART).
(i) In vitro fertilisation (IVF) is the method in which the mature egg is removed from a woman’s ovary, kept
in laboratory culture dish and mixed with sperm from husband or donor male. The zygote or early
embryos (up to 8 blastomeres) could then be transferred into the fallopian tube of the female (ZIFT –
Zygote intra fallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT – intra
uterine transfer), to complete its further development. A normal baby can be born to such a mother.
Such a baby is called test tube baby.
(ii) Gamete intra fallopian transfer (GIFT) – An ovum is collected from a donor. It is mixed with 50,000
sperms and immediately transferred into the fallopian tube of another female who cannot produce one,
but can provide suitable environment for fertilisation and further development of embryo (invivo).
(iii) Intra-cytoplasmic sperm injection (ICSI) is a procedure to form an embryo in the laboratory in which a
sperm is directly injected into the ovum. Then the fertilized egg is implanted into woman’s uterus.
(iv) Artificial insemination (AI) technique is used in cases when male is unable to inseminate the female or
has very low sperm counts. In this technique, the semen collected from the husband or a healthy
donor is artificially introduced either into the vagina or into the uterus (IUI – intra uterine insemination)
of the female.
5. (a) Describe the endosperm development in coconut.
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(b) Why is tender coconut considered a healthy source of nutrition?
(c) How are pea seeds different from castor seeds with respect to endosperm?
Ans. (a) In coconut the endosperm development is of free nuclear type. The coconut water from tender
coconut is nothing but free nuclear endosperm and the surrounding white kennel is the cellular
endosperm. Primary endosperm nucleus undergoes repeated divisions and these nuclei are arranged
in the periphery, leaving a large central vacuole. Cytokinesis begins from periphery towards the centre
making it cellular at maturity. The number of free nuclei formed before cellularisation varies greatly.
(b) Tender coconut is considered a healthy source of nutrition as it is rich source of vitamins and minerals.
(c) In pea seeds the endosperm is completely consumed during development of embryo and they are called
non-endospermic seeds while castor seed has a massive endosperm. Such seeds are called
endospermic seeds.
6. Write the changes a fertilized ovule undergoes within the ovary in an angiosperm plant.
Ans. Immediately after fertilization the secondary nucleus starts developing into a triploid endosperm while
the zygote starts developing into embryo. The integuments form the seed coat. The zygote divides only
when certain amount of endosperm is formed. This is an adaptation to provide assured nutrition to the
developing embryo. The first division of zygote is transverse and produces hypobasal cell towards
micropyle and epibasal cell or embryo cell towards the chalaza.
The hypobasal and the remaining cells constitute the suspensor which pushes the embryo into the
endosperm to receive nutrition for the development of embryo. The first cell of the suspensor towards
the micropylar end becomes swollen and functions as a haustorium. The last cell of the suspensor at
the end adjoining the embryo is called as hypophysis which gives rise to the radicle.
The epibasal cell or embryo cell undergoes two vertical divisions and one transverse division to
form eight cells which are arranged in two tiers-epibasal (terminal) and hypobasal (near the suspensor).
The epibasal cell forms the two cotyledons and the plumule. The hypobasal cells produce the hypocotyl
except its tip.
Embryo depicts globular, heart shaped and ultimately torpedo shaped structures.
The various stages of development of embryo are depicted in diagrams drawn below:
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SECTION – D
IV. Each question carries 5 marks.
1.
(a) Draw a labelled diagram of sectional view of human ovary showing different stages of
oogenesis.
(b) Where is morula formed in humans? Explain the process of its development from the zygote.
Ans. (a)
Human ovary
(b) Morula is formed in Fallopian tube. The zygote divides mitotically in successive steps. In the first
division two blastomeres are formed, during second four and during third division eight blastomers are
formed. The eight to sixteen cell stage the embryo is called morula.
2. (a) Draw a labelled diagram of the sectional view of a typical anatropous ovule.
(b) Mention the fate of all the components of the embryo sac after fertilization.
Ans. (a)
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(b) Fate of components of the embryo sac –
 Antipodal and synergid degenerates.
 Egg cells form zygote (2n).
 Polar nuclei form primary endosperm nucleus (3n).
3. (a) Draw a diagrammatic sectional view of a mature anatropous ovule and label the following
parts in it:
(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 gains entry into the embryo sac.
(v) that attaches the ovule to the placenta.
(b) Describe the characteristic features of wind pollinated flowers.
Ans. (a)
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(b) Characters of wind pollinated flowers –
(1) Flowers are small, usually unisexual, inconspicuous, colourless, odourless and nectarless. They are
produced above the foliage.
(2) In case of unisexual flowers, male flowers are more abundant. In bisexual flowers, stamens are
produced in large number as compared to pistil.
(3) The anthers of wind pollinated flowers are extended and well-exposed so that the pollens are easily
dispersed into wind currents. They have large, often feathery stigma which may easily trap air-borne
pollen grains.
(4) Pollen grains are small, smooth, dry and produced in large quantities. In some plants pollen are
winged.
(5) Non-essential parts are either absent or reduced.
(6) The wind pollinated flowers often have a single ovule in each ovary and many flowers are packed
into an inflorescence. Example is corn cob in which silky tassel is nothing but the stigma and style
of many female flowers hanging in air to trap pollen grains.
4. (a) Draw a diagrammatic sectional view of the female reproductive system of human and label the
parts:
(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)
(b) Role of pituitary hormones in menstrual cycle – The secretion of gonadotropins (LH and FSH)
stimulates follicular development as well as secretion of estrogens by the growing ovarian follicles. Both
LH and FSH attain a peak level in the middle of the cycle (about 14th day). Rapid secretion of LH leading
to its maximum level during the mid-cycle called LH surge induces rupture of Graafian follicle and
thereby the release of ovum (ovulation). The ovulation is followed by the luteal phase during which the
remaining parts of Graffian follicle transform as the corpus luteum. LH maintains the corpus luteum as
well as secretion of progesterone by the corpus luteum.
Role of ovarian hormone - Estrogen and Progesterone are the two reproductive female hormones.
Estrogen secreted by Graafian follicles is a group of steroids responsible for the development of
secondary sexual characters in a mature woman. It also helps in the development of uterine wall during
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the follicular phase. Progesterone secreted by corpus luteum is essential for maintenance of the
endometrium where implantation of fertilised ovum takes place.
5. (a) Draw a labelled schematic diagram of the transverse section of a mature anther of an
angiosperm plant.
(b) Describe the characteristic features of an insect pollinated flower.
Ans. (a)
(b)
1. Majority of insect pollinated flowers are large, colourful, fragrant and rich in nectar.
2. When flowers are small, a number of flowers are clustered into an inflorescence to make them
conspicuous.
3. The flowers pollinated by flies and beetles secrete foul odours to attract these animals.
4. In many cases special markings occur on the petals for guiding the insect to nectar glands.
5. Some flowers provide safe place to insects for laying eggs.
6. Pollen grains are surrounded by yellow sticky substance called pollen kit.
7. Stigmas are often sticky.
8. Edible pollens are produced by many insect pollinated flowers.
6. Write two differences between spermatogenesis and oogenesis
Ans.
Spermatogenesis
Oogenesis
(i) Spermatogenesis is the process of formation of
haploid spermatozoa that takes place in testis
(ii) Starts at puberty.
(iii) One primary spermatocyte forms four functional
sperms.
(iv) Occurs through out the life of male after attaining
puberty.
(i) Oogensis is the formation of ovum which is
carried in ovary.
(ii) Starts in foetal life.
(iii) One primary oocyte forms one ovum only.
(iv) This process of oogenesis ceases after
menopause which usually occurs at 45 – 50
years of age.
7. (a) Draw a diagram of a mature embryo sac of an angiosperm and label the following part 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.
Ans. (a)
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b) On fertilization, egg cell forms zygote which further develops into an embryo while polar nuclei after
being fertilised develop into primary endosperm nucleus by the process of triple fusion. This then gives
rise to a nutritive triploid endosperm tissue.
GENETICS
SECTION – A
I. Each question carries 1 mark.
1. Name the enzyme and state its property that is responsible for continuous and discontinuous
replication of the two strands of a DNA molecule.
Ans. DNA-dependent. DNA polymerase is an enzyme involved in template directed synthesis of DNA from
deoxyribonucleotide triphosphates. This enzyme has the ability of polymerisation only in one direction,
that is 5’  3’
SECTION – B
II. Each question carries 2 marks.
1. A non-haemophilic couple was informed by their doctor that there is possibility of a haemophilic
child being born to them. Draw a checker board and find out the percentage of possibility of
such a child among the progeny.
Ans. The couple themselves are non-haemophilic. Thus, the mother must be a carrier and father must be
normal.
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 50% daughter normal (XX)
 50% daughter carrier (XX)
 50% son normal (XY)
 50% son haemophilic (XY)
2. In a particular plant species majority of the plants bear purple flowers. Very few plants bear
white flowers. No intermediate colours are observed. If you are given a plant bearing purple
flowers, how would you ascertain that it is a pure breed for that trait? Explain.
Ans. To ascertain whether the plant is of pure breed for purple coloured flowers a test cross will be made.
In a test cross the given purple coloured plant will be crossed with homozygous plant with white flowers.
If all the flowers of the progeny are purple, the plant is homozygous dominant.
3 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.
Ans.
This type of inheritance is called incomplete dominance as red colour allele (R) is not completely
dominant over recessive (r) allele. So partial expression of both alleles in a heterozygote results in a
phenotype intermediate between those of two homozygotes, i.e., pink colour of flowers.
4. In a typical monohybrid cross the F2-population ratio is written as 3 : 1 for phenotype but
expressed as 1 : 2 : 1 for genotype. Explain with the help of an example.
Ans. This particular monohybrid cross is between a homozygous tall plant and a homozygous dwarf plant.
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Homozygous tall : heterozygous tall : Homozygous dwarf
5. Work out a cross to find the genotype of a tall pea plant. Name the type of cross.
Ans. To determine the genotype of a tall pea plant, a test cross is made in which the tall pea plant will be
crossed with a dwarf pea plant.
In case 1, the plant is heterozygous tall and the F1 ratio between tall and dwarf plants will be 1 : 1. In
case 2, the tall plant is homozygous tall and all the offsprings will be tall.
6. (a) Write the specific features of the genetic code AUG.
(b) Explain aminoacylation of the tRNA
Ans. (a) Genetic code AUG has dual functions. It codes for methionine (met) and it also acts as initiator
codon.
(b) Activation of amino acid in presence of ATP and its linkage to their cognate tRNA is called as charging
of tRNA or aminoacylation of tRNA.
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SECTION – C
III. Each question carries 3 marks.
1. What is the genetic basis for proof that codon is a triplet?
Ans. George Gamow, a physicist argued that since there are only 4 bases and if they have to code for all
the 20 amino acids, the code should be made of three nucleotides. This was a very bold proposition
because a permutation combination of 43(4 x 4x 4) would generate 64 codons. This proposition was
later proved by Khorana and Nirenberg.
If the codon consists of 4 letters (4 x 4), only sixteen codes are possible which is less than
the number of amino acids. As the ribosomes move on mRNA the codons are read in a contiguous
fashion.
2. What is satellite DNA in a genome? Explain their role in DNA fingerprinting.
Ans. Such DNA sequences which are repeated many a times and show a high degree of polymorphism and
also form bulk of DNA in a genome are called as satellite DNA.
DNA from every tissue from an individual shows the same degree of polymorphism and is
heritable. Hence, it is very useful in DNA fingerprinting.
3. Describe the structure of an RNA polynucleotide chain having four different types of nucleotides.
Ans.
RNA is a single stranded polynucleotide structure composed of ribose
sugar, phosphate and nitrogenous bases (adenine, guanine, cytosine and
uracil).
A nitrogenous base is linked to the first carbon of pentose sugar through an
N-glycosidic linkage to form a nucleoside. When a phosphate group is linked to
5'-OH of a nucleoside through phosphoester linkage, a corresponding nucleotide
is formed. Two nucleotides are linked through 3' – 5' phosphodiester linkage to
form a dinucleotide. More nucleotides can be joined in such a manner to form a
polynucleotide chain. A polymer thus formed has at one end a free phosphate
moiety at 5’-end of ribose sugar, which is referred to as 5'-end of polynucleotide
chain. Similarly, at the other end of the polymer the ribose has a free 3'-OH
group which is referred to as 3' end. The backbone of the chain is formed of
sugars and phosphates.
4. Why are human females rarely haemophilic? Explain. How do haemophilic patients suffer?
Ans. A woman has two X-chromosomes. If only one X-chromosome bears an allele for haemophilia, its
dominant allele present on the other X-chromosome will not allow the expression of recessive
haemophilic allele since haemophilia is a sex linked recessive disease. Such females will be normal but
are called carriers for haemophiliac gene. This disease occurs mostly in the males because they have
one X-chromosome (XY) only.
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Haemophilia is a disorder of the blood which prevents its clotting due to absence of antihaemophilic globulin. The patient will continue to bleed even from a minor skin cut and as a result of
continuous bleeding, he or she may die due to excessive blood loss.
5. In a maternity clinic, for some reasons the authorities are not able to hand over the two newborns to their respective real parents. Name and describe the technique that you would suggest
to sort out the matter.
Ans. To sort out the matter DNA finger printing of parents and new born are to be conducted to hand over
the new borns to their real biological parents.
DNA fingerprints can be carried out by taking DNA from extremely minute amounts of blood, semen,
hair bulb or any other cells of the body. The major steps are as follows:
(i) Isolation of DNA from nuclei of cells.
(ii) If the content is limited, DNA can be amplified using Polymerase Chain Reaction.
(iii) Digestion of DNA by restriction endonucleases that cuts them into fragments.
(iv) Separation of DNA fragments according to their size by gel electrophoresis.
(v) The separated fragments of DNA in the gel are copied on to a synthetic membranes such as nylon
or nitrocellulose by southern blotting technique.
(vi) Special DNA-probes are made in the laboratory and are made radioactive by labelling with
radioactive isotopes. The radioactive DNA-probes bind to the repetitive sequences on the nylon
membrane.
(vii) Detection of DNA fragments by autoradiography – An X-ray film is exposed to the nylon
membrane to mark the places where the radioactive DNA-probes have bound to the DNA fragments.
These places are marked as dark bands when X-ray film is developed. These dark bands on X-ray
film represent the DNA-fingerprints.
Bands of new born will have resemblance with their respective parents.
6. (a) Explain DNA polymorphism as the basis of genetic mapping of human genome.
(b) State the role of VNTR in DNA fingerprinting
Ans. (a) If an inheritable mutation is observed in a population at high frequency, it is referred to as DNA
polymorphism. The probability of such variations to be observed in non-coding DNA sequence would
be higher as mutations in these sequences may not have any immediate effect in an individuals
reproductive ability. These mutations keep on accumulating generation after generation and form one of
the basis of variability/polymorphism ranging from single nucleotide change to very large scale changes.
Such DNA segments are called repetitive DNA and show high degree of polymorphism and form the
basis of DNA fingerprinting.
(b) VNTR (Variable Number of Tandem Repeats) belongs to a class of satellite DNA referred to as minisatellite. A small DNA sequence is arranged tandemly in many copy numbers. The copy numbers
varies from chromosome to chromosome in an individual. The numbers of repeat show very high
degree of polymorphism. As a result the size of VNTR varies in size from 0.1 to 20 kb. Consequently,
after hybridisation with VNTR probe, the autoradiogram gives many bands of different sizes. These
bands give a characteristic pattern for an individual DNA. It differs from individual to individual in a
population except in the case of identical twins.
7. Explain co-dominance taking an example of human blood groups in the population.
Ans. When two alleles present together in an organism express their trait independently instead of showing
dominant recessive relationship, they are called co-dominant alleles and the phenomenon is called codominance. Example is ABO blood groups in humans. Four humans blood groups A, B, AB and O are
the phenotypes for this trait. The letter A and B refer to the two types of glycoproteins (antigens) coating
the Red Blood Cells (RBC) of persons with either type A or type B blood, respectively. Type AB blood
contains both the glycoproteins while type O blood contains neither.
Three different alleles IA; IB and i of a gene determine the phenotypes of the four blood groups.
Both IA and IB are dominant on recessive allele “i”. They show their independent dominant effect and
produce A and B blood types. When both IA & IB come together, they are both expressed independently
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in the phenotypes as AB blood group, and are thus said to be co-dominant. A person carries only two
out of these three alleles. When a person has two recessive alleles ‘ii’, it produces ‘O’ blood group.
SECTION – D
IV. Each question carries 5 marks.
1. (a) Write what DNA replication refers to.
(b) State the properties of DNA replication model.
(c) List any three enzymes involved in the process along with their functions.
Ans. (a) DNA replication refers to synthesis of DNA.
(b) The two strands of DNA separate and act as template for the synthesis of new complimentary
strands. After the completion of replication each DNA molecule would have one parental and
one newly synthesised strand i.e., DNA replication is semiconservative. It is continuous on
one strand and discontinuous on the other strand as replication occurs only in 5' – 3 ' direction.
The replication does not initiate randomly at any place in DNA. It starts at definite regions
called origin of replication. For long DNA molecules, since the two strands of DNA cannot be
separated in its entire length (due to very high energy requirement), the replication occurs
within a small opening of the DNA helix, referred to as replication fork.
(c) 1. DNA polymerase III – adds nucleotides
2. RNA primase – adds RNA primer
3. Helicase – breaks hydrogen bonds and separates two DNA strands
4. Gyrase – relaxes supercoiling
5. DNA ligase – joins Okazaki fragments
2. Inheritance patterns of flower colour in garden pea plant and snap dragon differ. Why is
the difference observed? Explain the difference with the help of crosses in their
inheritance patterns.
Ans. Inheritance pattern of flower colour in Pisum sativum (garden pea) follows law of dominance
that is out of the two alleles of the flower colour (gene) the dominant allele is expressed
(phenotypically) and the recessive allele is suppressed when both are present together in
heterozygous condition. The recessive trait is expressed only when the recessive allele is
present in homozygous condition.
In case of Antirrhinum (snap dragon) the flower colour shows incomplete dominance and all the F1
progeny is of pink colour. It is because the allele for red colour is not completely dominant over its
recessive allele. The law of dominance is not exhibited in this case.
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3. A child suffering from Thalassemia is born to a normal couple. But the mother is being blamed
by the family for delivering a sick baby.
(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 a group of inherited haematological disorders of reduced haemoglobin synthesis
caused by defects in one or more of the haemoglobin (Hb) chains. -Thalassemia is caused by reduced
or absent synthesis of –globin chain. -Thalassemia is caused by reduced or absent synthesis of globin chain. There is an increased production of gamma (γ) and delta (δ) chains. Globin chain
imbalance causes haemolysis and anemia. It is of two types – thalassemia minor and major (severe).
(b) The family will be counselled as follows:
(i) Thalassemia is a genetic disease which can be transmitted to the child from any of the parents as its
gene is located on a autosome and not on sex chromosome. Father and mother both are equally
responsible for transmission of this trait.
(ii) If only one parent is the carrier of thalassemia gene, then only some of the offsprings may inherit this
faulty gene. Since this gene is recessive in nature the child will have thalassemia minor. But in
case both the parents are carrier of this defective gene and the child becomes homozygous for this
gene, then he will suffer from thalassemia major.
(c) During the counselling programme, the values that can be propagated in the families are as follows:
(i) Mothers are not solely responsible for autosomal disorders.
(ii) In case they have a history of thalassemia in their family they need to go in for genetic counselling.
(iii) After knowing the scientific principle of this disorder they should seek proper scientific advise for
treatment.
4. (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.
Ans. (a) Human female has 44 autosomes + XX sex-chromosomes while male has 44 autosomes + XY sex
chromosomes. The males produce two types of sperms. 50% of the sperms carry X-chromosome and
the other 50% carry Y-chromosome. The females produce only one type of gametes (eggs or ovum).
All the eggs bear only one, i.e., ‘X’ type of chromosome. When the sperm bearing X-chromosome fuses
with egg (ovum) during fertilisation to form zygote, such zygote develops into a female baby having XX
sex-chromosomes. If sperm bearing Y-chromosome fuses with egg to form zygote, this zygote gives
rise to a male baby having XY sex-chromosome. Thus, in humans, there is 50 per cent chances of a
baby boy and 50 per cent chances of a baby girl to be borne. In human, it is the sperm which
determines the sex of the child.
(b) In humans we see male heterogamety as human males produce two types of sperms X and Y types.
Female heterogamety is seen in birds where female produces two types of egg Z and W while males
are homogametic which produce only Z type of sperms.
5. (a) Explain Mendel’s law of independent assortment by taking a suitable example.
(b) How did Morgan show the deviation in inheritance pattern in Drosophila with respect to this
law?
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Ans. (a) When a cross is made between plants (parents) differing in two or more pairs of contrasting
characters, the inheritance of one pair of contrasting character is independent of the other pair of
contrasting character. The two characters are transmitted independently in their own patterns. This law
can be explained with the help of a dihybrid cross. A cross between two parents which differ in two
pairs of contrasting characters is known as dihybrid cross. The cross can be made between true
breeding parent plants, one having yellow and round seeds (YYRR) and another plant having green and
wrinkled seeds (yyrr). This cross is shown as under:
9
: 3
:
3
: 1
Phenotypic ratio : round yellow : round green : wrinkled yellow : wrinkled green
To understand the law of Independent assortment, study the formation of gametes in F1 plant
(RrYy). Consider the segregation of one pair of genes R and r. 50% of gametes have the gene R and
other 50% have r. In addition each gamete should also have the allele Y or y. We can see segregation
of 50% R and 50% r is independent from the segregation of 50% Y and 50% y. Therefore, 50% of r
bearing gametes have Y and the other 50% y. Similarly, 50% of R bearing gametes have Y and other
50% have y. Thus four different kinds of gametes, i.e., RY, Ry, rY, ry are formed.
(b) Morgan’s experiment on Drosophila melanogaster (tiny fruit flies) Morgan hybridised yellow bodied –
white eyed female flies with brown bodied-red eyed males and inter-crossed their F1 progeny. He
observed that the two genes did not segregate independently of one another and the F2 ratio deviated
very significantly from 9 : 3 : 3 : 1 ratio. Morgan knew that the genes were located on the Xchromosome and saw quickly that when the two genes in a dihybrid cross were situated on the same
chromosome, the proportion of parental gene combinations were much higher than the non-parental
combinations. Morgan attributed this to linkage or physical association of genes on a chromosome.
The total number of genes present on one chromosome constitute a linkage-group. The genes are
called linked genes.
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EVOLUTION
SECTION – A
I. Each question carries 1 mark.
1. Write the formula to calculate allele frequency in future generations according to Hardy-Weinberg
genetic equilibrium.
Ans. (p + q)2 = p2 + 2pq +q2 = 1
2. Identify the examples of convergent evolution from the following:
(i) Flippers of penguins and dolphins.
(ii) Eyes of octopus and mammals
(iii) Vertebrate brains
Ans. (i) Flippers of penguins and dolphins and (ii) Eyes of octopus and mammals are analogous organs
which are a result of convergent evolution a process by which unrelated species become more similar in
order to survive in similar environmental conditions.
3. Identify the examples of homologous structures from the following:
(i) Vertebrate hearts.
(ii) Thorns in Bougainvillea and tendrils of Cucurbita.
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(iii) Food storage organs in sweet potato and potato
Ans. (i) Vertebrate hearts and (ii) Thorns in Bougainvillea and tendrils of cucurbita, are the examples of
homologous structures.
Homologous structures are result of divergent evolution and represent those structures which have
same structural plan and developmental origin but differ in function they perform.
SECTION – B
II. Each question carries 2 marks.
1. How do Darwin’s finches illustrate adaptive radiation?
Ans. Darwin observed that there were many varieties of finches (small black birds) on the same island. All
the varieties, he concluded, evolved on the island itself. All the varieties of finches were evolved from
the original seed eating variety. Many other forms with altered beaks arose, enabling them to become
insectivorous and vegetarian finches. This process of evolution in a given geographical area starting
from a point and literally radiating to other geographical habitats is an example of adaptive radiation.
2. List the two main propositions of Oparin and Haldane.
Ans. Oparin and Haldane proposed that “first form of life could have come from pre-existing non-living
organic molecules (e.g., RNA, protein etc.) and that formation of life was preceded by chemical
evolution i.e., formation of diverse organic molecules from inorganic constituents.
The conditions on earth were high temperature, volcanic storms, reducing atmosphere containing
CH4, NH3 etc.). Reducing atmosphere promoted increased synthesis of energy rich molecules and
abiotic synthesis of complex organic molecules took place.
3. Write the Oparian and Haldane hypothesis about the origin of life on Earth. How does meteorite
analysis favour this hypothesis?
Ans. Oparian and Haldane proposed that the first form of life could have come from pre-existing non-living
organic molecules (e.g., RNA, protein etc.) and that formation of life preceded by chemical evolution i.e.,
formation of diverse organic molecules from inorganic constituents. Meteorites are hypothesised to be
another source for early organic molecules as meteorite content contains some similar organic
components as found in early earth’s atmosphere. The stony meteorites are well known to contain rich
mix of organic compounds. This analysis of meteorite content indicates that similar processes are
occurring elsewhere in space.
SECTION – C
III. Each question carries 3 marks.
1.
With the help of any two suitable examples explain the effect of anthropogenic actions on
organic evolution.
Ans. Man has bred selected plants and animals for agriculture, horticulture, sport and security. Man has
domesticated many crops and animals. This intensive breeding programme has created breeds that
differ from other breeds like dog but still are of the same group. In nature such an evolution would have
taken millions of years.
2. Explain the increase in the numbers of melanic (dark winged) moths in the urban areas of postindustrialisation period in England.
Ans. In England, it was observed that (peppered moth) Biston betularia showed cryptic colouration with two
phenotypes, white (light winged) moths which outnumbered dark black winged moths. After
industrialisation the falling smoke particles darkened the background of tree trunk. Against a dark
background, the light coloured moths became conspicuous and were picked up by the predator birds.
Sooner the dark coloured moth outnumbered the light coloured moths and they had a much better
chance of survival under the new conditions created by industrial pollution. Before industrialisation set
in, thick growth of almost white-coloured lichen covered the tree. In that background the white winged
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moth survived but the dark-coloured moth were picked up by predators. Thus, industrial melanism
demonstrates the action of natural selection.
SECTION D
IV. Each question carries 5 marks.
1. How does the process of natural selection affect Hardy-Weinberg equilibrium? Explain. List the
other four factors that disturb the equilibrium.
Ans. Natural selection affects Hardy Weinberg Equilibrium principle which states that in sexually
reproducing organisms the allele frequency is stable and is constant from generation to generation
under certain conditions of stability. The gene pool, i.e., total genes and their alleles in a population
remains constant. This is called genetic equilibrium.
Natural selection is a process in which heritable variations enabling better survival are enabled to
reproduce and leave greater number of progeny. Natural selection can lead to:
(i) Stabilisation in which more individuals acquire mean character value.
(ii) Directional change in which more individuals acquire value other than the mean character value i.e.,
towards one of the two extremes of the distribution curve.
(iii) Disruption in which more individuals acquire peripheral character value at both ends or extreme
ends of the distribution curve.
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The other four factors that disturb the genetic equilibrium are:
(i) Gene migration/flow : When migration of a section of population to another place occurs, there is
change in gene frequencies.
(ii) Genetic drift: The random changes in gene frequency in a population occurring by chance alone
rather than by natural selection.
(iii) Mutation: Pre-existing advantageous mutations when selected will result in observation of new
phenotypes and over few generations this would result in speciation.
(iv) Recombination: Variations due to recombination during gametogenesis result in changed
frequency of genes and alleles in future generations.
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BIOLOGY IN HUMAN WELFARE
SECTION – A
I. Each question carries 1 mark.
1. Why sharing of injection needles between two individuals is not recommended?
Ans. Intravenous drug users that share needles are at high risk of contracting diseases like AIDS and
hepatitis. So sharing of needles is not recommended.
2. Write the importance of MOET.
Ans. Multiple Ovulation Embryo Transfer (MOET) is a technique adopted for herd improvement. It involves
super ovulation followed by embryo transfer. In this method a healthy high milk yielding breed of female
animal is injected with gonadotropic hormones like FSH and LH to release more eggs per cycle from its
ovaries.
3. Name an improved breed of cattle.
Ans. Jersey is an improved breed of cow.
4. When does a human body elicit an anamnestic response?
Ans. Anamnestic response (intensified secondary response) occurs during any subsequent encounter with
a pathogen (germ).
5. Name any two diseases the ‘Himgiri’ variety of wheat is resistant to.
Ans. Leaf and stripe rust and hill bunt.
6. List the components of biogas.
Ans. Methane, hydrogen and carbon dioxide.
7. “Pranay suffered from measles at the age of 10 years. There are rare chances of his getting
infected with the same disease for the rest of his life”. Give reason for the statement.
Ans. The first infection works as a vaccine and the immune system responds by forming memory-B and Tlymphocytes. These cells recognise the pathogen quickly on subsequent exposure and overwhelm the
invader with the massive production of antibodies.
SECTION – B
II. Each question carries 2 marks.
1. 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 is such a
response called? How does the body produce it?
Ans. Such a response is called allergy. Allergy is due to the release of chemicals like histamine and
serotonin from the mast cells.
2. A young boy when brought a pet dog home started to complain of watery eyes and running nose.
The symptoms disappeared when the boy was kept away from the pet.
(a) Name the type of antibody and the chemicals responsible for such a response in the boy.
(b) Mention the name of any one drug that could be given to the boy for immediate relief from such a
response.
Ans.
(a) Antibody IgE type and chemicals histamine and serotonin are responsible for watery eyes and running
nose which are because of an allergic response.
(b) Anti-histamine, adrenalin and steroids are drugs that could be given to the boy for immediate relief from
such a response.
3. Name the bioactive molecules produced by Trichoderma polysporum and Monascus purpureus.
Ans. Trichoderma polysporum produces Cyclosporin A while Monascus purpureus produces Statins.
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SECTION – C
III. Each question carries 3 marks.
1.
Differentiate between inbreeding and outbreeding in cattle. State one advantage and one
disadvantage for each one of them.
Inbreeding
Outbreeding
Inbreeding refers to the mating of more closely Outbreeding is the breeding of the unrelated
related individuals within the same breed for 4 to 6 organisns which may be between individuals of the
generation.
same breed but having no common ancestors for
4 – 6 generations (out-crossing) or between
different breeds (cross-breeding) or different
species (inter specific hybridisation).
Advantage – It helps in evolving a pure line in any Advantage – A single outcross helps to overcome
animal as it increases homozygosity. It helps in inbreeding depression. Cross breeding helps in
accumulation of superior genes and elimination of developing new animal breeds e.g. Hisardale a
less desirable genes.
new breed of sheep was developed in Punjab by
crossing Bikaneri ewe and Marino ram.
Disadvantage – Reduces fertility and productivity, Disadvantage – There is a possibility of
called inbreeding depression.
introduction of undesirable characters.
SECTION – D
IV. Each question carries 5 marks.
1. Describe the asexual and sexual phases of life cycle of Plasmodium that causes malaria in
humans. (Also draw the flow chart).
Ans. Four species of Plasmodium are known to cause different types of malaria in man. They are P-vivax,
P-ovale, P-malariae and P-falciparum. P-vivax is the most common infecting malarial parasite. The life
cycle of plasmodium starts in the stomach of female mosquito. The asexual phase occurs in human
body while sexual phase occurs in the body of female Anopheles. Plasmodium enters human body as
sporozoite through the bite of infected female Anopheles. The parasites initially multiply within the liver
cells and then attack the RBCs resulting in their rupture. The rupture of RBCs is associated with release
of toxic substance, haemozoin, which causes chills and fever. When a female Anopheles bites an
infected person, these parasites enter the mosquito’s body and undergo further development. The
parasites multiply in mosquito’s body to form sporozoites that are stored in their salivary glands. When
these mosquitoes bite a human, the sporozoites are introduced into human body.
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2. (a) What is plant breeding? List the two steps the classical plant breeding involves.
(b) How has the mutation breeding helped in improving crop varieties? Give one example where
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.
Ans. (a) Plant breeding is the purposeful manipulation of plant species in order to create desired plant
types that are better suited for cultivation, give better yield and are disease resistant. Classical plant
breeding involves –
(i) Crossing or hybridisation of pure lines.
(ii) Artificial selection to produce plants with desirable traits of higher yield, nutrition and resistance to
diseases.
(b) Mutation is a sudden and heritable change in the character of an organism. Genetic variations are
induced through changes in the chromosome number, chromosome structure or base sequence of
24
the concerned gene. It is possible to induce mutations artificially through use of chemicals or
radiations like gamma radiations. In mung bean, resistance to yellow mosaic virus and powdery
mildew were induced by mutations.
(c) Breeding crops with higher levels of vitamins and minerals, or higher protein and healthier fats – is
the most practical means to improve public health and is brought about by the process of bio
fortification.
Maize hybrids that have twice the amount of the amino acids, lysine and tryptophan, compared to
existing maize hybrid were developed. An iron-fortified rice variety containing over five times as much
iron as in commonly consumed varieties, has been developed. Wheat variety Atlas 66, having a high
protein content has been used as a donor for improving cultivated wheat. At the same time IARI has
also released several vegetable crops that are rich in vitamins and minerals.
3. 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.
Ans. (a) The person should not be compelled to leave the colony as AIDS is not a communicable disease
transmitted by air, water, touch or vectors. It only infects when a person indulges in unsafe sex with a
person having AIDS, by transfusion of contaminated blood and Blood products or by sharing
contaminated needles.
(b) Following possible preventive measures would be suggested to the residents:
 Avoid unprotected illicit sexual relations.
 Before getting blood transfusion, it is important to be aware whether that blood is HIV +ve or –ve.
 Use of barrier contraceptive devices like condoms.
 Avoid sharing needles and syringes.
 Avoid getting shaved at barbers’ shop who use the same razor for different people.
(c) AIDS is caused by the Human Immuno deficiency Virus (HIV)
Symptoms – Fever, diarrhoea and weight loss in initial stages. Because of continuous fall in immunity of
body, patient suffers from opportunistic infections such as T.B., pneumonia, toxoplasmosis, meningitis,
diarrhoea and tumours for which a normal person is immune.
4. Aditya participated in a group discussion in his school on “The ill effects of Tobacco on Human
Health”. In the evening he goes with his family for dinner and insists on sitting in the “Nonsmoking Area” to which his father (who is a heavy smoker) objects.
(a) In this situation who wins your support – Aditya’s concern for health and environment or his
father’s objection? Justify giving two reasons.
(b) Suggest any three effective propaganda campaigns for anti-tobacco awareness.
Ans.
(a) I will support Aditya’s concern for the health of his family. Both Active as well as Passive smoking is
injurious to health as smoking is associated with increased incidence of cancers of lungs, urinary
bladder, throat and oral cavity, bronchitis, emphysema/coronary heart disease/gastric ulcer,
hypertension and diabetes etc.
(b) (i) By printing statutory warning on cigarette packets.
(ii) As advertisement in mass media such as television, newspaper, internet etc.
(iii) Designating non-smoking Zones in public areas such as restaurants, airports etc.
25
BIOTECHNOLOGY
SECTION – A
I. Each question carries 1 mark.
1. Give the name of the carcinogenic dye which is used to stain gel to make the DNA visible under
UV light.
Ans. Ethidium bromide (Et. Br).
2. Which main technique and instrument is used to isolate DNA from a plant cell?
Ans. The technique is called centrifugation and the instrument used is called centrifuge.
3. How is the action of normal endonuclease enzymes different from that of restriction
endonucleases?
Ans. While normal endonuclease cuts the DNA at random positions, restriction endonucleases make cut in
DNA only at specific positions.
4. State the role of transposons in silencing of mRNA in eukaryotic cells.
Ans. Transposons or Jumping genes are source of complementary mRNA which is used for RNA
interference. This method involves silencing of a specific mRNA due to a complementary dsRNA
molecules that binds to and prevents translation of mRNA.
SECTION – B
II. Each question carries 2 marks.
1. Why is the introduction of genetically engineered lymphocytes into an ADA deficiency patient not
a permanent cure? Suggest a possible permanent cure?
Ans. The patient requires periodic infusion of such genetically engineered lymphocytes as these
lymphocytes are not immortal. If a gene coding for ADA is isolated from bone marrow cells and
introduced into the cells of foetus at early embryonic stages, it can lead to a permanent cure.
2. In which technique do we use Taq polymerase enzyme and why?
Ans. This enzyme is used in Polymerase Chain reaction (PCR). The reason for this is that it is a
thermostable DNA polymerase enzyme. It is isolated from bacteria Thermus aquaticus from hot water
springs and does not get denatured at high temperature during PCR. It works as a normal DNA
polymerase enzyme.
3. Name two commonly used bioreactors.
Ans. (i) Simple stirred-tank bioreactor.
(ii) Sparged stirred-tank bioreactor.
4. Expand the term “ELISA” and mention its application:
Ans. ELISA – Enzyme Linked Immuno-Sorbent Assay. It is used to detect infection by pathogens such as
AIDS by presence of antigens or by detecting the antibodies synthesised against pathogens.
5. (a) Mention the difference in the mode of action of exonuclease and endonuclease.
(b) How does restriction endonuclease function?
Ans. (a) Exonucleases remove nucleotides from the ends of DNA whereas, endonucleases make cuts at
specific positions within DNA.
(b) Restriction endonuclease, inspects the length of DNA sequence, recognises specific palindromic DNA
sequence and binds to it. Then it excises each of the two strands of the double helix at specific points in
sugar – phosphate backbones. It cuts the DNA strand a little away from the centre of palindromic sites,
but between the same two bases on the opposite strands. For example, EcoRI cuts the DNA between
bases G and A only when the sequence GAATTC is present in the DNA.
6. Describe the gene therapy procedure for an ADA-deficient patient.
Ans. Gene therapy is an in vivo method that allows correction of a gene defect that has been diagnosed in
a child embryo.
26
(1) As a first step towards gene therapy, lymphocytes from the blood of the patient are grown in a
culture outside the body.
(2) A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which
are subsequently returned to the patient.
(3) The patient requires periodic infusion of such genetically engineered lymphocytes as these cells are
not immortal. However, if the gene isolated from bone marrow cells producing ADA is introduced
into cells at early embryonic stage, it could be a permanent cure.
7. (a) Explain how to find whether an E. coli bacterium has been transformed or not when a
recombinant DNA bearing ampicillin resistant gene is transferred into it.
(b) What does the ampicillin resistant gene act as in the above case?
Ans. (a) When an E. Coli bacterium is transformed with a recombinant DNA having ampicillin resistant gene
in its plasmid, the recombinant plasmid will lose tetracycline resistance due to the insertion of foreign
DNA. But it can still be selected out from non-recombinant ones by plating the transformants on
ampicillin containing medium. The transformants growing on ampicillin containing medium are then
transferred on a medium containing tetracycline. The recombinants will grow in ampicillin containing
medium but not on that containing tetracycline. But non-recombinants will grow on the medium
containing both the antibiotics.
(b) Ampicillin resistant gene acts as a selectable marker and helps in selecting the transformants.
8. Why and how bacteria can be made ‘competent’?
Ans. As DNA is hydrophilic molecule, it cannot pass through cell membrane. In order to force bacteria to
take up the plasmid, the bacterial cell is made ‘competent’ to take up DNA. This is done by treating
them with specific concentration of a divalent cation such as calcium which increases the efficiency with
which DNA enters the bacterium through pores in its cell wall.
9. Write any four ways used to introduce a desired DNA segment into a bacterial cell in recombinant
technology experiments.
Ans. A desired DNA segment is introduced by using following four methods after making the bacterial cell
competent:
(i) Recombinant DNA can be forced into competent cells by incubating the cells with recombinant DNA
on ice followed by placing them briefly at 42C (heat shock) and then putting them back on ice. This
enables the bacteria to take up recombinant DNA.
(ii) Micro-injection during which recombinant DNA is directly injected into nucleus of an animal cell.
(iii) Gene/Particle Gun – A method suitable for plants. Cells are bombarded with high velocity microparticles of gold or tungsten coated with DNA in a method known as biolistics or gene gun.
(iv) ‘Disarmed pathogens’ vectors which when allowed to infect the cell transfer the recombinant DNA into
the host.
(v) Electroporation – A method which employs electric current to create transient pores in the cell
membrane of the host through which foreign DNA enters the cells.
10. Why is proinsulin so called? How is insulin different from it?
Ans. In mammals including humans insulin is synthesized as a prohormone. Proinsulin contains an extra
stretch called the C-peptide, (in addition to Chains A and B), which is absent in mature insulin and is
removed during maturation into insulin.
Insulin on other hand consists of two short polypeptide chains: Chain A and Chain B which are
linked together by disulphide bridges.
11. (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) This disorders is caused due to the deletion of gene which codes for enzyme ADA.
It can be cured by using any one of the following methods –
(i) Bone marrow transplantation.
(ii) Enzyme replacement therapy in which functional ADA is given to the patient by injection.
(iii) Gene therapy.
27
SECTION – C
III. Each question carries 3 marks.
1. A vector is engineered with three features which facilitate its cloning within the host cell. List the
three features and explain each one of them.
Ans. The following are the three features by which a vector is engineered to facilitate its cloning within the
host cell:
i) Origin of replication (Ori) – This is a sequence from where replication starts and any piece of DNA
when linked to this sequence can be made to replicate within the host cell. This sequence is also
responsible for controlling the copy number of the linked DNA.
ii) Selectable marker – It helps in identifying and eliminating non-transformants and selectively
permitting the growth of the transformants.
iii) Cloning sites also known as recognition sites. To this site the alien DNA is linked.
2. Explain the role of transgenic animals in (i) Vaccine safety and (ii) Biological products with the
help of an example of each.
Ans.
(i) Transgenic mice are being developed for use in testing the safety of the vaccines before they are
used on humans. If found to be successful & reliable, they could replace the use of monkeys to test
the safety of batches of the vaccine. Transgenic mice are being used to test the safety of the polio
vaccine.
(ii) Transgenic animals that produce useful biological products can be created by the introduction of the
portion of DNA (or genes) which code for a particular product such as human protein -1-antitrypsin
used to treat emphysema. Similar attempts are being made for treatment of phenylketonuria and
cystic fibrosis.
28
ECOLOGY
SECTION – A
I. Each question carries 1 mark.
1 ‘Eutrophication’ is the natural ageing of a lake. Mention another feature which defines this term.
Ans. The other term used are
(i) Nutrient enrichment
(ii) Depletion of dissolved oxygen in water.
2. Name the type of biodiversity represented by the following:
 Estuaries and alpine meadows in India
Ans. Ecological diversity
3. Write the equation that helps in deriving the net primary productivity of an ecosystem.
Ans. The rate at which energy/biomass/organic matter is accumulated in the organism excluding the energy
it uses for the process of respiration.
GPP – R = NPP
GPP = Gross Primary Productivity, R = Respiration rate i.e., consumption of food during respiration and
NPP = Net Primary Productivity
4. Why are green algae not likely to be found in the deepest strata of the ocean?
Ans. The environment in deep oceans is perpetually dark and its inhabitants do not receive sunlight. The
green algae cannot manufacture food. So green algae are not likely to be found in the deepest strata of
the ocean.
5. Name the type of biodiversity represented by the following:
(i) 1000 varieties of mangoes in India.
(ii) Variations in terms of potency and concentration of reserpine in Rauwolfia vomitoria growing
in different regions of Himalayas.
Ans. (i) Genetic diversity (As it refers to variation of genes within species).
(ii) Genetic diversity (As it refers to variation of genes within species).
6. Write the basis on which an organism occupies a space in its community/natural surroundings.
Ans. Organisms occupy a place in the natural surrounding or in a community according to their feeding and
relationship with other organisms.
7. What is a detritus food chain made up of? How do they meet their energy and nutritional
requirements?
Ans. The detritus food chain begins with dead organic matter. It is made of decomposers which are
heterotrophic organisms mainly fungi and bacteria. Decomposers secrete digestive enzymes that
breakdown dead material into simple, inorganic material which are subsequently absorbed by them.
8. How are standing crop and biomass related to each other?
Ans. The mass of living material in a trophic level at a particular time is called standing crop. Standing crop
is measured as the mass of living organism (biomass) or the number in a unit area. Fresh weight and
dry weight are measurement units for biomass of a species.
SECTION – B
II. Each question carries 2 marks.
1. Differentiate between a detrivore and a decomposer giving an example of each.
29
Ans. Animals feeding on dead decaying organic material (detritus) are called detritivores. Example –
Earthworm and Free living nematode.
Decomposers are micro-organisms which break down complex organic matter into simpler inorganic
matter. Example – Bacteria and Fungi.
2. The gradual and predictable change in the species composition of a given area is called
ecological succession. What do you understand with the pioneer and climax community in this
context?
Ans. The species that invade a bare area are called the pioneer species. In primary succession on rocks
the pioneer community are usually lichens and in a lake they are phytoplanktons.
The last sustainable community which is in equilibrium with the environment is called climax community
i.e., forests.
3. Write two different ways for the disposal of electronic wastes (e-wastes).
Ans. (a) The waste can be recycled and metals like copper, iron, silicon, nickel and gold are recovered.
(b) Can be buried in landfills or incinerated
4. Why the pyramid of energy is always upright? Explain.
Ans. Pyramid of energy is always upright, can never be inverted because when energy flows from a
particular trophic level to the next trophic level, some energy is always lost as heat as well as for
performing various activities at each step.
5. Explain why very small animals are rarely found in polar region.
Ans. Since small animals have a larger surface area relative to their volume, they tend to lose body heat
very fast in cold environment. So, they have to expand much energy to generate body heat through
metabolism. This is the main reason for the absence of very small animals in polar regions.
6. “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.
Ans. Yes, it is possible that a given species may occupy more than one tropic level in the same ecosystem
at some time. For example a sparrow is a primary consumer when it eats seeds, fruits and a secondary
consumer when it eats insect and worms.
7. “Stability of a community depends on its species richness”. Write how did David Tilman show
this experimentally.
Ans. David Tilman’s long-term experiments using outdoor plot provide some tentative answer for the
phrase. He found that plots with more species showed less year-to-year variation in total biomass. He
also showed that in his experiments, increased diversity contributed to higher productivity.
SECTION – C
III. Each question carries 3 marks.
1. Interspecific interactions of two species of any population may be beneficial, detrimental or
neutral. Explain each of them with the help of suitable examples.
Ans. Interspecific beneficial interaction is mutualism. This interaction confers benefits on both the
interacting species. Example: lichen. In lichens the fungal partner provides shelter and helps in
absorbing nutrients while the algal partners provide food to fungal partner that it synthesizes.
Parasitism is a detrimental interspecific relationship. It ensures free lodging and meals to the parasite
while the host is at a loss. Example : leech. It sucks blood thus harming the host.
Commensalism is an interspecific interaction between two species in which one species benefits and
the other is neither harmed nor benefited. Example : Orchid growing as an epiphyte on a mango branch
30
and barnacles growing on back of the whale are benefited and the other is neither harmed nor
benefited.
2. Water is very essential for life. List any three features that enable plants and animals to survive
in water scarce environment.
Ans. Features of plants:
(i) Waxy cuticle.
(ii) Sunken stomata
(iii) Succulent leaves and stem to store water
(iv) Capable to complete life cycle in short period
(v) Deep tap roots
(vi) Presence of C4 pathway.
(vii) Deciduous leaves
Features of animals:
(i) Body covered with scales
(ii) Excrete uric acid (uricotelic).
(iii) Thick skin
(iv) No sweating
(v) Deposition of fat in sub epidermal layer
(vi) Burrowing nature
(vii) Excretion of concentrated urine
3. (a) What would be the consequence of failure of the electrostatic precipitator of a thermal plant?
(b) Mention any four methods by which vehicular air pollution can be controlled.
Ans. (i) 99 per cent particulate present in the exhaust from a thermal power plant will be released in
atmosphere due to failure of its electrostatic precipitator, resulting in increase in particulate pollution.
(ii)  Use of CNG
 Use of unleaded petrol
 Use of low sulphur fuel
 Use of catalytic converters
 Phasing out of old vehicles
 Application of stringent pollution level norms for vehicles.
4. How have human activities caused desertification? Explain.
Ans. Over-cultivation, unrestricted grazing, deforestation and poor irrigation practices, have resulted in arid
patches of land. When large barren patches extend and meet over time, a desert is created.
Internationally it has been recognised that desertification is a major problem now-a-days, particularly
due to increased urbanisation.
5. How does algal bloom destroy the quality of a fresh water body? Explain.
Ans. When an excess of fertilisers and pesticides containing nitrate or phosphate are used in the
agricultural field then some of this is carried to water bodies (lakes). The presence of nitrate and
phosphate salts overstimulate the growth of algae, causing unsightly scum and unpleasant odours and,
robbing the water of dissolved oxygen vital to other aquatic life. Toxins released by bloom infested
water body inhibits the growth of other algae and aquatic animals die due to toxicity or lack of oxygen.
Decomposition of aquatic flora and fauna further depletes the oxygen content of water body which
ultimately causes the lake to choke to death.
6. (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 mere numbers.
Ans. (a) The size of the population tells us a lot about its status in the habitat. Whatever ecological
processes we wish to investigate in population be it be outcome of competition with another species, the
impact of a predator or the effect of a pesticide application, we always evaluate them in terms of
population size.
(b) Although total number is generally the most appropriate measure of population density, it is in some
cases meaningless or difficult to determine. If in an area, there are 200 parthenium plants but only a
single huge banyan tree with a large canopy, stating that population density of banyan tree is low related
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to that of Parthenium, amounts to underestimating the enormous role of the banyan tree in the
community. In such cases the per cent cover is more meaningful measure of the population size.
7. Differentiate between two different types of pyramids of biomass with the help of one example of
each.
Ans. Pyramid of biomass deals with the relationship between the dry weight of organisms at different
trophic levels. It can be upright or inverted in nature.
Upright pyramid of biomass of a grassland shows a sharp decrease in biomass at higher trophic levels.
In oceans the pyramid of biomass is inverted in which a small standing crop of phytoplankton
supports large standing crop of zooplankton in volume of at any one time.
8. (a) Explain “birth rate” in a population by taking a suitable example.
(b) Write the other two characteristics which only a population shows but an individual cannot.
Ans. (a) Birth rate refers to the number of births taking place during a given period in the population that are
added to the initial density.
If in a pond there are 20 lotus plants last year and through reproduction 8 new plants are added, taking
the current population to 28, we calculate the birth rate as 8/20 = 0.4 offspring per lotus per year.
(b) Sex ratio and death rate are the two characteristics which cannot be studied at individual level but only
at population level.
9. (a) Explain “death rate” in a population by taking a suitable example.
(b) Write the other two characteristics which only a population shows but an individual cannot.
Ans. (a) Number of deaths taking place in the population during a given period. Example, if 4 individuals
out of 40 deer died in a specific time period, say a week, the death rate in the population of deer during
that period is 4/40 = 0.1 individual per deer per week
(b) Birth rate and sex ratio are the two characteristics which only a population shows but an individual
cannot.