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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
SOME ADDITIONAL QUESTIONS FOR +2 BOARD EXAMS
REPRODUCTION
SECTION – A
I. Each question carries 1 mark.
1. Give one example of an animal which exhibits Oestrous cycle.
Ans. Female Dog (bitch), Cow, Buffalo etc.
2. Give one example each of a fungus which reproduce by:
(a) budding
(b) conidia
Ans. (a) Budding : Yeast (Saccharomyces)
(b) Conidia : Penicillium
3. Name an IUD that you would recommend to promote the cervix hostility to the sperms.
Ans. Hormone releasing IUDs, make the uterus unsuitable for implantation and the cervix hostile to
sperms.
4. In which two of the following organisms is the fertilization external?
Bony fishes, ferns, frogs, birds
Ans. External fertilization. Bony fishes and Frogs.
SECTION – B
II. Each question carries 2 marks.
5. Differentiate between outbreeding and outcrossing.
Ans. Differentiate between outbreeding and outcrossing
Outbreeding
Outcrossing
(i) Refers to the breeding of unrelated animals, it (i) Refers to the mating animals within the
may include outcrossing or cross-breeding or
same breed but having no common
inter specific hybridisation.
ancestors upto 4-6 generations.
(ii) The offspring may be an out-cross or an (ii) The offspring is known as outcross.
intra/interspecific hybrid.
6. “Intra-cytoplasmic sperm injection” and “gamete intrafallopian transfer” are two assisted
reproductive technologies. How is one different from the other?
Ans. In “intra-cytoplasmic sperm injection” a single sperm is directly injected into ovum to form an embryo
in the laboratory. It is used in cases of oligospermia, abnormal sperm motility or morphology and in
cases where IVF has failed. In “gamete intrafallopian transfer ovum is collected from the female or the
donor female. The oocytes are mixed with 50,000 sperms and then transferred into the ampulla of the
fallopian tube of the affected female who cannot produce the ovum but can provide suitable
environment for fertilization and further development.
7. Banana crop is cultivated by farmers without sowing of seeds. Explain how the plant is
propagated.
Ans. In banana no seeds are produced as the fruits are parthenocarpic. Banana plants are cultivated by
planting pieces of the underground stem-the rhizome. At nodes, the axillary buds develop shoots and
roots.
8. Explain the events that occur during fertilization of an ovum in humans. How is it that only one
sperm enters the ovum?
Ans. The process of fusion of sperm with an ovum is called fertilization. Secretions of acrosome of sperm
help it to enter into the cytoplasm of ovum through zona pellucida and the plasma-membrane. It
induces meiotic division-II to form haploid ovum (ootid) and second polar body. The fusion of haploid
nucleus of sperm with that of ovum form the diploid zygote.
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
When the sperm comes in contact with, the zona pellucida layer of the ovum it induces changes in the
membrane which block the entry of additional sperms.
9. Write the effect of the high concentration of L.H. on a mature Graafian follicle.
Ans. High level of L.H. around the mid-cycle induces rupture of Graafian follicles to release the secondary
oocyte (ovum) from the ovary, a process called ovulation.
10. Why do algae and fungi shift to sexual mode of reproduction just before the onset of adverse
conditions?
Ans. In order to tide over the unfavourable conditions, these organisms take the course of sexual
reproduction. They form gametes which fuse to form zygote which develops a thick wall that is resistant
to desiccation and damage. It undergoes a period of rest before germination under favourable
conditions.
11. Name the organic materials the exine and intine of an angiosperm pollen grains are made up of.
Explain the role of exine.
Ans. Exine is made of highly resistant organic substance called sporopollenin and intine is
pectocellulosic in nature.
Exine is a protective layer. It checks the natural decay of pollen grain and also possesses proteins for
enzymatic and compatibility reactions.
12. List the post-fertilisation events in angiosperms.
Ans. Post-fertilisation events include all those events which occur in a flower, after double fertilization.
The major events are:
(i) Development of endosperm
(ii) Development of embryo
(iii) Maturation of ovule (s) into seed (s)
(iv) Maturation of ovary into fruit.
SECTION – C
III. Each question carries 3 marks.
13. How is apomixis different from parthenocarpy?
Ans. Apomixis is a form of asexual reproduction that mimics sexual reproduction, in which seeds are
produced without fertilization while parthenocarpy is the process of formation of fruits without
fertilization.
14. Describe endosperm development in angiosperm.
In angiosperms Endosperm development occurs after fertilization (vegetative) and precedes embryo
development. It is a food-laden tissue used for nutrition of developing embryo and its development is of
three types-nuclear endosperm, cellular endosperm and helobial endosperm. Nuclear endosperm is
the most common type. In this type the PEN divides successively to give rise to free nuclei. Later cell
wall is formed to make it cellular.
In coconut, the development of endosperm is nuclear-type. The coconut water remains as free nuclear
endosperm and the surrounding white kernel becomes cellular endosperm.
15. (a) Mature seeds of legumes are non-albuminous. Then, can it be assumed that double
fertilization does not occur in legumes? Explain your answer.
(b) List the differences between the embryos of dicot (pea) and monocot (grass family).
Ans. (a) No; double fertilization does occur in legumes and the endosperm is formed. However during the
development of embryo whole of endosperm is utilised and there are no remnants of endosperm in the
mature seeds. The seeds are non-endospermic or exalbuminous.
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
(b)
Dicot embryo
(i) Embryo is present between two cotyledons.
Monocot embryo
(i) Embryo is present at the lateral side of the
cotyledon.
(ii) Hypocotyl and epicotyl are distinct.
(ii) Hypocotyl and epicotyl not distinct.
(iii) Plumule and radicle are not covered by any (iii) Plumule and radicle are covered by coleoptile
sheath.
and coleorhiza respectively.
(iv) Embryo development include globular, heart (iv) No such stages during embryo development.
shaped and mature embryo stages.
16. Make a list of any three outbreeding devices that flowering plants have developed and explain
how they help to encourage cross-pollination?
Ans. Outbreeding Devices: To avoid inbreeding depression and promote outbreeding, the plants have
evolved certain devices/contrivances such as:
(i) Dicliny (unisexuality) : Flowers unisexual e.g., Papaya, Maize, Mulberry.
(ii) Dichogamy: Sex-organs mature at different times (a) Protandry – anthers mature earlier, e.g.,
Salvia, Sunflower (b) Protogyny – stigmas mature earlier, e.g., Gloriosa, Plantago, Mirabilis, etc.
(iii) Self-sterility/Self-incompatibility: Pollen grains do not germinate on the stigma of same flower, a
genetic mechanism to prevent inbreeding.
17. A woman has certain queries as listed below, before starting with contraceptive pills. Answer
them.
(a) What do contraceptive pills contain and how do they act as contraceptives?
(b) What schedule should be followed for taking these pills?
Ans.
(a) The contraceptive pills contain small doses of either progesterone or progesterone-oestrogen
combinations. They act by inhibiting ovulation and implantation, as well as alter the quality of cervical
mucus to prevent/retard entry of sperms.
(b) Pills have to be taken daily for 21 days starting within the first five days of menstrual cycle. After a gap
of 7 days, it has to be repeated in the same pattern till she desires to prevent conception.
18. Name and explain the role of inner and middle walls of the human uterus.
Ans. The walls of uterus are composed of three layers of tissues. The perimetrium is an outer thin covering
of peritoneum. The myometrium is a middle thick layer of smooth muscle fibres which show strong
contraction during delivery of the baby. The endometrium is inner glandular layer that lines the uterine
cavity and undergoes cyclic changes during the menstrual cycle.
SECTION – D
IV. Each question carries 5 marks.
19. Explain the different phases of menstrual cycle and correlate the phases with the different levels
of ovarian hormones in a human female.
Ans. Menstrual Cycle: The cycle of events occurring in the ovaries and uterus in the female primates is
called menstrual cycle. One ovum is released during the middle of this cycle.
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
Diagrammatic representation of various events during a menstrual cycle.
A menstrual cycle consists of:
(a) Menstruation: Menstrual flow occurs for 3 – 5 days due to breakdown of endometrium because of
low level of LH and progesterone. It occurs only if released ovum of previous cycle is not fertilised.
(b) Follicular phase: (Proliferative phase) Primary oocyte transforms into Graafian follicle and the
endometrium regenerates through proliferation. This occurs due to changes in levels of pituitary
and ovarian hormones. Level of LH and FSH gradually increases and stimulates follicular
development and secretion of estrogen by growing follicles. LH & FSH attain a peak in the middle
of the cycle. Maximum level of LH reaches at about 14th day and induces rupture of Graafian
follicle to release ovum containing secondary oocyte (ovulation).
(c) Luteal phase: Remnants of Graafian follicle forms corpus luteum, which secretes large amount of
progesterone, essential for maintenance of endometrium, a necessity for implantation and for
pregnancy to continue. During pregnancy all events of menstrual cycle stop and no menstruation
occurs. In the absence of fertilization, corpus luteum degenerates and progesterone levels fall.
This causes disintegration of endometrium, resulting in menstrual flow marking a new cycle.
20. Explain the events that occur, upto fertilization, when the compatible pollen grain lands on the
stigma.
Ans. After pollination, the compatible pollen grains land on the stigma. Germination of pollen grain is
connected with compatibility-incompatibility reaction in-between proteins present over the pollen grain
and stigma. The compatible pollen grains absorb water and nutrients from the stigmatic surface. It
results in the swelling of pollen grains. They germinate and form pollen tubes through one of the germ
pores. Pollen tube grows by obtaining nourishment from the interior of stigma and style. The contents
of the pollen grain shift into the pollen-tube with the tube or vegetative nucleus moving to its tip,
followed by generative nucleus two male gametes. Further, growth of pollen tube occurs towards its tip.
In solid style the pollen tube grows through transmitting tissue by digesting its cells, by secreting
enzymes and absorbing the food. In the ovary, the growth of pollen tube towards ovule is directed by
another tissue called obturator. The pollen tube enters the ovule either through its micropyle or chalaza
or the sides after piercing through integuments. After entering the ovule, the pollen tube is attracted
towards the micropylar end of the embryo sac by the chemicals secreted by synergids/help cells. The
pollen tube enters one of the synergids and burst open into it. Synergid gets destroyed under its
impact. The released contents include vegetative nucleus and (two male gametes, in plants which shed
pollen in the three celled condition) generative nucleus which divides to form two male gametes.
All these events from pollen deposition on stigma until pollen tube enters the ovule, are together
referred to as pollen-pistil interaction.
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
21.
(a) Explain the different ways apomictic seeds can develop. Give an example of each.
(b) Mention one advantage of apomictic seeds to farmers.
(c) Draw a labelled mature stage of a dicotyledonous embryo.
Ans. (a) Apomixis/agamospermy is a form of asexual reproduction that mimics sexual reproduction, in
which seeds are produced without fertilization. It may be:
(i) Recurrent – a diploid embryo-sac formed directly from nucellar cell (apospory) or diploid
megaspore mother cell without meiosis (diplospory). The diploid egg develops into diploid
embryo, called diploid parthenogenesis. Synergids or antipodal of diploid embryo-sac may
develop into embryo-diploid apogamy.
(ii) Adventive embryony – embryos develop directly from diploid cells of nucellus or integuments,
e.g., Citrus, Opuntia beside the normal embryo produced from zygote.
The phenomenon of occurrence of more than one embryo in a seed is called polyembryony.
(b) If seeds collected from hybrids are sown, the plants in the progeny will segregate and do not maintain
hybrid characters. Thus, hybrid seeds have to be produced each year and fall very expensive for the
farmer. If these hybrids are made into apomicts by transfer of apomictic genes into the hybrid
varieties then there is no segregation of characters in the hybrid progeny. The farmers can keep on
using these hybrid seeds to raise crop year after year and will not need to buy seeds every year.
(c)
A typical dicot embryo.
22. Coconut palm is monoecious, while date palm is dioecious. Why are they so-called?
Ans. Coconut is a bisexual plant as both male flowers and female flowers are born on the same plant
while date palm is a dioecious or unisexual plant as the male flowers and female flowers are born on
separate plants.
23. (a) Draw a labelled diagrammatic view of human male reproductive system.
(b) Differentiate between:
(i) Vas deferens and vasa efferentia
(ii) Spermatogenesis and spermiogenesis
Ans. (a)
Diagrammatic view of male reproductive system
(part of testis is opened to show inner details)
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
(b) (i) Vasa efferentia are the ciliated ductules which carry sperms from the rete testis (formed by joining
of free end of seminiferous tubules) to the epididymis. While vasa deferentia carry sperms from the
epididymis, it ascends to the abdomen, receives a duct from the seminal vesicle and opens into
urethra as ejaculatory duct.
(ii) Spermatogenesis is a process by which primary spermatocytes undergo meiotic division to form
four equal haploid spermatids. Spermiogenesis is the process by which spermatids transform into
spermatozoa (sperms).
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
GENETICS
SECTION – A
I. Each question carries 1 mark.
24. Female of many birds has a pair of dissimilar ZW chromosomes, while the males possess a pair
of similar ZZ chromosomes. State whether true/false and why?
Ans. (a) Statement is a correct statement because in birds, the female is heteromorphic and shows
heterogamety.
25. What will happen if DNA replication is not followed by cell division in a eukaryotic cell?
Ans. It will result in polyploidy i.e., presence of more than two sets of chromosomes.
26. Name the material used as matrix in gel-electrophoresis and mention its role.
Ans. Agarose gel. It provides a sieving effect and helps in separation of DNA fragments according to their
size. The smaller the fragment size, the farther it moves.
27. What is ‘saltation’ according to de Vries?
Ans. According to de Vries, ‘Saltation’ means single step large mutations arising suddenly in a population
causing evolution.
28. A boy has been diagnosed with ADA-deficiency. Suggest any one possible treatment.
Ans. The possible treatment for ADA deficiency is enzyme replacement therapy in which functional ADA
is injected in to the patient.
29. What are ‘true breeding lines’ that are used to study inheritance pattern of traits in plants?
Ans. ‘True breeding lines’ are the pure, true-breeding homozygous plants which produce the phenotypically
similar plants on continued self-breeding e.g., a pure homozygous tall plant (TT).
30. Mention how does DNA polymorphism arise in a population?
Ans. DNA polymorphism are the variations at genetic level and arise due to mutations. The polymorphism
in repetitive DNA sequences can be used in DNA fingerprinting.
31. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic
experiments?
Ans. Repetitive DNA is separated from bulk DNA by Density Gradient Centrifugation.
SECTION – B
II. Each question carries 2 marks.
32. Name the disease that was first to get the gene therapy treatment. Write the cause of the
disease and the effect it has on the patient.
Ans. Name of disease : Adenosinedeaminase deficiency (ADA).
Cause: Deletion of the gene for adenosine deaminase
Effect: Immune system does not function due to non-functional T-lymphocytes
33. Show DNA replication with the help of a diagram only.
Ans.
Replicating Fork
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DR. ARVIND'S BIOLOGY CLASSES
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34. In Snapdragon, a cross between true-breeding red flowered (RR) plants and true breeding white
flowered (rr) plants showed a progeny of plants with all pink flowers.
(a) The appearance of pink flowers is not known as blending. Why?
(b) What is this phenomenon known as?
Ans. (a) The appearance of pink flowers in Snapdragon is not a case of blending because in F2 generation,
the red flowers and white flowers again make their reappearance.
(b) Incomplete dominance.
35. Write the scientific name of the fruit-fly. Why did Morgan prefer to work with fruit-flies for his
experiments. State any three reasons.
or
Linkage and crossing-over of genes are alternative of each other. Justify with the help of an
example.
Ans. Scientific name of fruit-fly: Drosophila melanogaster.
Reasons for selections:
(i) They could be grown on simple synthetic medium in laboratory.
(ii) They complete their life cycle in about two weeks.
(iii) A single mating could produce a large number of progeny flies.
(iv) Male and female flies are easily distinguishable.
(v) It has many types of hereditary variations that can be seen with low power microscopes.
or
Linkage: is the phenomenon of inheritance of a number of genes or factors together due to their
physical occurrence on the same chromosome. Linked genes do not show independent assortment
and do not give a 9 : 3 : 3 : 1 dihybrid phenotypic ratio and 1 : 1 : 1 : 1 as double test cross ratio. For
example in Drosophila, the cross between grey body and long wings flies with black body and vestigial
(short) wing flies shows a ratio of 3 : 1 parental phenotypes (3 grey body long wings and one black body
vestigial wing) instead e.g., 9 : 3 : 3 : 1 phenotypic ratio.
Cross-over is a recombination of genes due to the exchange of genetic material between two
homologous chromosomes. Beside parental combinations, recombinant types are also produced. For
example a cross between red-eyed normal winged and purple eyed vestigial wings produced red-eyed
vestigial winged and purple eyed normal winged flies as recombinant types beside the parental types.
Only 9.3% recombinant types were observed which is quite different from 50% recombinant in case of
independent assortment (1 : 1 : 1 : 1 dihybrid test cross ratio). Thus, complete linked genes show only
parental phenotypes while cross-over genes (incomplete linkage) beside parental phenotypes, also
show a little percentage of recombinant type. Thus, linkage and cross-over are showing alternatives of
each other.
36. State the difference between the structural genes in a Transcription Unit of Prokaryotes and
Eukaryotes.
Ans. Cistron is a segment of DNA coding for a polypeptide. In prokaryotes, the structural genes are
grouped together in the operon and are regulated by a common promoter and regulator gene. They
transcribe together to form more than one polypeptide. These structural genes are called polycistronic.
In eukaryotes, the structural genes are monocistronic. Each structural gene has exons (coding or
expressed sequences) and introns (sequence which do not appear in processed RNA) and are termed
split-gene. The eukaryote structural genes occur singly and each gene has its own promoter and
regulator gene.
37. Explain the two factors responsible for conferring stability to double helix structure of DNA.
Ans. DNA is more stable than RNA because (a) instead of 2’-OH group of sugar there is only H. The 2’-OH
group on every nucleotide of RNA is a reactive group and makes RNA labile and easily degradable.
(b) Thymine in place of uracil makes it more stable.
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
38. How does the gene ‘I’ control ABO blood groups in humans? Write the effect the gene has on
the structure of red blood cells.
Ans. ABO blood groups are controlled by the gene ‘I’. The gene is involved in the formation of sugar
polymers that protrude from the surface of RBCs. The gene has three alleles IA, IB and i. The allele IA
and IB produce slightly different form of sugar polymers while allele i does not produce any sugar. The
alleles IA and IB are completely dominant over the allele i but when IA and IB are co-dominant and both
express their own types of sugars.
39. (i) Name the scientist who suggested that the genetic code should be made of a combination of
three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion.
Ans. (i) George Gamow
(ii) DNA contains only four types of nitrogen bases while the number of amino acids used in protein
synthesis is 20. A singlet code (one amino acid specified by one nitrogen base) can specify only
4 amino acids (41), a doublet code only 16 (42) while a triplet code can specify up to 64 amino
acids (43). As there are 20 amino acids, a triplet code (three nitrogen bases for one amino acid)
can be operative.
SECTION – C
III. Each question carries 3 marks.
40. (a) What are the transcriptional products of RNA polymerase III?
(b) Differentiate between ‘Capping’ and ‘Tailing’.
(c) Expand hnRNA
Ans. (a) RNA polymerase-III transcribes t-RNA; 5srRNA and snRNA.
(b) In capping, an unusual nucleotide (methyl guanosine triphosphate) is added to the 5’ end of the hn–
RNA. ‘Tailing’ involves the addition of 200-300 adenylate residue at 3’ end.
The fully processed hn-RNA is now called m-RNA, ready for translation.
(c) hn-RNA = Heterogeneous Nuclear RNA
41. Give three reasons, write how Hardy-Weinberg equilibrium can be affected.
Ans. The three reasons which affect Hardy-Weinberg equilibrium are:
(i) Gene flow/migration – The movement of a section of population from one place to another results in
the addition of new alleles to the local gene pool of the host population. This is called gene migration.
Migration causes change in gene frequencies in original and the new population.
(ii) Genetic drift – The random changes in gene frequency in a population occurring by chance alone
rather than by natural selection is called genetic drift. The effects of genetic drift are more prominent
in small populations
(iii) Mutation – These are new sudden inheritable discontinuous variations produced in the organisms due
to permanent change in their genotypes. They are the source of variations and these variations may
accumulate and give rise to new species.
42. Morgan carried out several dihybrid crosses in Drosophila and found F2-ratios deviated very
significantly from the expected Mendelian ratio. Explain his findings with the help of an
example.
Ans. Based upon his breeding experiments on fruitfly Drosophila melanogaster, Morgan proposed
chromosome theory of linkage which states that linked genes occur on the same chromosome in a
linear sequence and maintain parental combination of genes except for occasional cross overs.
Linkage: The phenomenon of inheritance of a number of genes or factors due to their physical
occurrence on the same chromosome is called linkage.
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DR. ARVIND'S BIOLOGY CLASSES
(A Unit of Med-Xel Tutorials)
Linked genes occur on same chromosome while unlinked genes occur in different chromosomes.
Linked genes do not show independent assortment. They do not give a 9 : 3 : 3 : 1 dihybrid phenotypic
ratio and 1 : 1 : 1 : 1 as double test cross ratio. Linkage may be:
(i) complete
(ii) incomplete
Complete linkage
Parents:
Phenotypes
F1 :
Phenotypes
Test Cross :
Phenotypes
T.C. Progeny : Phenotype
Ratio
Red eyes, Normal Wings (♀) x Purple Eyes, Vestigial Wings (♂) (Pure)
Red Eyes, Normal Wings
Red Eyes, Normal Wings (Hybrids)
Red Eyes, Normal Wings (♂) x Purple Eyes, Vestigial Wings (♀)
Red Eyes, Normal Wings
Purple Eyes, Vestigial Wings
2
:
2
1
:
1
Recombinants:
Nil
Figure: Morgan’s Experiment on Drosophila A dihybrid test cross showing complete linkage.
Incomplete Linkage
The F1 females of above cross are crossed with homozygous recessive males. The ratio comes out to
be 9 : 1 : 1 : 8. The two genes did not segregate independently.
Phenotype
Progeny
Observed
Expected if
Expected if
Complete
Independent
Linkage
Assortment
Parental Types
(a) Red eyed, normal winged
1339
1420
710
(b) Purple eyed, vestigial winged
1195
1420
710
Recombinant Types
(a) Red eyed, vestigial winged
152
zero
710
(b) Purple eyed, normal winged
152
zero
710
Note: In Drosophila melanogaster-the genes for eye colour and wing size are present in the
X-chromosome. D. melanogaster has XX-XY sex mechanism. Thus, heterozygous female produces
two types of eggs, while the males produce only one type of sperms.
43. With the help of a schematic diagram, explain the location and the role of the following in a
transcription unit:
Promoter, Structural gene, Terminator.
Ans.
A transcription unit.
A transcription unit consists of three regions in the DNA:
(a) Promoter located towards 5’ end of coding strand and provides the recognising site and the binding
site for RNA-polymerase.
(b) Terminator: Located at 3’-end of coding strand and stops the process of transcription.
(c) Structural gene : Segment of DNA which transcribes RNA, it is flanked by the promoter and
terminator regions.
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DR. ARVIND'S BIOLOGY CLASSES
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44. Following a severe accident, many charred : disfigured bodies are recovered from the site
making the identification of the dead very difficult. Name and explain the technique that would
help the authorities to establish the identity of the dead to be able to hand over the dead to their
respective relatives.
Ans. Name of technique: DNA fingerprinting. DNA fingerprinting is a technique to identify a person on the
his/her DNA specificity. Each person has a unique DNA fingerprint.
Principle of DNA fingerprinting: 99.9% of the nucleotide bases are exactly similar in all human
beings. Only 0.1% of human genome represents the variability among human beings and it is this
difference that provides individuality to each human being.
The DNA of an individual has some specific sequence of nucleotides which are repeated many times
and are found at many places throughout the DNA. These are very specific for each person and vary in
number from person to person. These are the Variable Number Tandem Repeats or VNTR, also
called minisatellites. These sequences are inherited.
Technique for DNA Fingerprinting
Southern Blot
The Southern Blot is one way to analyze the genetic patterns which appear in a person’s DNA, it
involves the following steps Source of DNA: White blood corpuscles, blood, semen, saliva, vaginal swabs, skin cells, bone cells,
cells from hair root, etc. are used as a source of DNA for fingerprinting. The amount of DNA required
can be met by 1 microgram of tissue or 1,00,000 cells.
(a) Isolation and Extraction – isolation of DNA from DNA source cells in a high speed refrigerated
centrifuge.
(b) DNA Amplification – Many copies of extracted DNA are made by PCR-technique.
(c) Fragmentation/Digestion – of DNA by restriction endonuclease to produce VNTRs.
(d) Separation – of DNA fragments (VNTRs) by size through gel electrophoresis.
(e) Single stranded DNA – denaturing of VNTRs either by heating or chemically treating the fragments
in the gel.
(f) Southern Blotting – transferring (blotting) of separated single stranded DNA fragments to synthetic
membranes such a nitrocellulose or nylon.
(g) Hybridisation – using radioactive/labelled VNTR probe.
(h) Exposure to X-Ray films/Autoradiography – detection of hybridised DNA by autoradiography.
After hybridisation autoradiography shows many dark bands of different sizes. These bands give a
characteristic pattern for an individual DNA.
DNA prints or DNA profiles of dead bodies are compared with DNA prints of the close, blood relatives
(sisters, brothers, parents) to determine the identity of the dead bodies.
45. Explain polygenic inheritance with the help of a suitable example.
Ans. Polygenic inheritance is also called quantative inheritance. It is type of inheritance controlled by one
or more genes in which the dominant alleles have commutative/additive effect with each dominant allele
expressing a part of the trait. The full trait is shown only when all the dominant alleles are present.
The genes involved are called Polygenes. A few examples are cob length in maize, skin colour in
human beings, human intelligence, milk and meat yield in animals, height in human beings.
Intermediate types are quite common in polygenic inheritance. Skin colour in human beings is
controlled by three genes A, B and C. The dominant genes A, B and C form dark skin colour and the
recessive alleles a, b and c produce light skin colour. The genotype with all the dominant allele
(AABBCC) produces darkest skin colour and the genotype with recessive alleles (aabbcc) form lightest
skin colour. The different shades of skin colour from darkness to lightness are determined by the
number of each type of the alleles in the genotype of an individual.
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DR. ARVIND'S BIOLOGY CLASSES
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46. Why is pedigree analysis done in the study of human genetics? State the conclusions that can
be drawn from it.
Ans. A record of inheritance of certain genetic traits for two or more generations presented in the form of a
diagram or a family tree is called pedigree analysis. Pedigree analysis is done in the study of human
genetics as the use of Mendel’s laws for a study of problems of heredity in humans is very difficult
because of - (i) in humans the generation time is very long 20 years or more (ii) the number of offspring
per couple is also small (iii) Controlled crosses cannot be made in case of human beings. Hence
pedigrees are maintained.
Uses of Pedigree analysis –
(i) In human genetics, pedigree study provides a strong tool, which is used to trace the inheritance of a
specific trait, abnormality or disease.
(ii) It can be used to understand whether the trait in question is dominant, recessive or X-linked.
(iii) It is useful for genetic counsellors to advice couples about possibility of having children with genetic
defects.
47. Identify ‘a’, ‘b’, ‘c’, ‘d’, ‘e’ and ‘f’ in the table given below:
No.
Syndrome
Cause
Characteristics
of Sex
affected individuals
Male/Female/Both
1.
Down’s
Trisomy of 21
‘a’
‘b’
2.
‘c’
XXY
Overall
masculine ‘d’
development
3.
Turner’s
45 with XO
‘e’
‘f’
Ans. ‘a’
(i) Round face, broad forehead, partially open mouth, protruding tongue.
(ii) Short neck, flat hands with palm crease, furrowed tongue, mental retardation.
‘b’ = Both
‘c’ = Kleinfelter’s syndrome
‘d’ = Male
‘e’ = (i) Sterile females with rudimentary ovaries, small uterus, undeveloped breasts, short stature. (ii)
Webbed neck and abnormal intelligence, may not menstruate or ovulate.
‘f’ = Female
48. Mendel published his work on inheritance of characters in 1865, but it remained unrecognized
till 1900. Give three reasons for the delay in accepting his work.
Ans. Mendel’s work remained unrecognised for many years due to:
(i) Limited circulation of the “Proceedings of Brunn Natural Science Society” in which it was
published.
(ii) Mendel’s conclusions about heredity were ahead of his time.
(iii) Lack of aggressiveness in his personality.
(iv) He used statistical methods and mathematical logic which were unfamiliar to other biologists.
(v) The scientific world was being rocked at that time by Darwin’s theory of evolution.
49. A cross between a normal couple resulted in a son who was haemophilic and a normal
daughter. In course of time, when the daughter was married to a normal man, to their surprise,
the grandson was also haemophilic.
a) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and
her husband.
b) Write the conclusion you draw of the inheritance pattern of this disease
Ans. The gene for haemophilia is present on X-chromosome. A male always receives X-chromosome from
his mother. Since in the given cross, a haemophilic son and a normal daughter were born to a normal
couple, it indicates that their father is normal (XY), while the mother is a normal but carrier for
haemophilia (XXh). As the daughter is normal, her genotype may be XX or XX h. Further, when the
daughter was married to a normal man, she gave birth to a haemophilic son, indicating that the
daughter of normal couple was carrier (XXh) for the disease.
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(a) Pedigree chart
Genotype of daughter = XXh
Genotype of husband = XY
(b) It is a sex-linked inheritance (X-linked recessive). The mother passes her X-linked trait to both her
daughters and sons while the father passes his X-linked traits to only his daughters and not to his
sons.
SECTION – D
I. Each question carries 5 marks.
50. Describe the process of transcription in a bacterium
Ans. Transcription is the process of synthesis of RNA chain over DNA template. Only a segment of DNA
and only one of the two strands is copied into RNA. Transcription requires a DNA dependent enzyme
RNA polymerase. The transcription includes following steps:
(i) Activation of Ribonucleotides in the nucleoplasm through phosphorylation. These are ATP, GTP,
UTP and CTP.
(ii) A single RNA – polymerase binds to the promoter site of the DNA template. Chain opening of DNA
segment by uncoiling occurs from the site of polymerase binding.
(iii) Initiation of transcription occurs by the complementary pairing of free activated ribonucleotides with
the nitrogen bases of DNA template.
(iv) The RNA polymerase is only capable of catalysing the process of elongation. It associates
transiently with initiation factor () and termination factor (  ) to initiate and terminate the
transcription, respectively. Activated ribonucleotide triphosphate act as substrate and also provide
energy for polymerisation.
(v) As the RNA chain formation initiates, the sigma () factor separates.
RNA polymerase moves along the coding region of DNA template causing elongation of RNA
chain.
(vi) RNA synthesis stops as soon as the polymerase reaches the terminator region in association with
the Rho factor (  ). The RNA polymerase falls off and transcription is terminated.
Transcription in Bacteria
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51. (a) What is a genetic code?
(b) Explain the following:
Degenerate code ; Unambiguous code; Universal code; Initiator code
Ans. (a) The relationship between the sequence of amino acids in a polypeptide and nucleotide sequence
of DNA and m-RNA is called genetic code.
(b) (i) Degenerate code: Some amino acids are coded by more than one codons meaning that the code is
degenerate. The first two nitrogen bases are similar while third one is different and this has no effect
on coding. This third one is called Wobble position. Only Tryptophan (UGG) and Methionine
(AUG) are coded by single codons. All other amino acids are coded by 2-6 codons.
(ii) Unambiguous code : One codon will specify only one amino acid.
(iii) Universal code: The genetic code is applicable universally i.e., a codon specifies same amino
acid in all organisms.
(iv) Initiator code: Polypeptide synthesis is signalled by two initiation codons., AUG and GUG.
These have dual function as they also code for Methionine and Valine respectively.
52. Explain the process of translation.
Ans. Translation – The polymerisation of amino acids to form a polypeptide, under the control and
guidance of codons of m-RNA is called translation. The order and sequence of amino acids is
determined by the sequence of bases in the m-RNA.
Mechanism:
(i) Amino acids are activated by ATP to form amino acid-AMP complex.
(ii) The activated amino acid gets linked to OH– at 3’ end of specific t-RNA and the process is called
charging of t-RNA or amino acylation of t-RNA. The charged t-RNA moves to the site of protein
synthesis, the ribosome.
(iii) Translation begins when small subunit of ribosome links to m-RNA called initiation complex. The
small and large subunits of ribosomes must come together for protein synthesis. This is brought
about by mRNA by formation of initiation complex. Then the larger subunit joins them to form
active ribosome. Large subunit contains two sites, one for binding t-RNA-amino acid complex,
acceptor site (A-site) and second peptidyl-tRNA site, donor site (P-site).
Translation
(iv) Polypeptide formation involves 3 stages:
Initiation; Elongation and Termination of amino acid chain. During initiation there is formation of a
complex chain comprising of an m-RNA; a tRNA bearng the first amino acid of the polypeptide
and the two subunits of ribosome. Synthesis always starts with same aa, methionine.
(v) During chain elongation charged t-RNAs sequentially bind to the appropriate codons in m-RNA by
forming complementary base pairs with t-RNAs anticodon. Ribosome moves from codon to codon
along m-RNA in 5’-3’ direction. Amino acids are added one by one, translated into polypeptide
dictated by m-RNA.
(vi) A releasing factor binds to stop codon, terminates translation and releases polypeptide from the
ribosome.
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53. “DNA replication is semi-conservative”. Name the scientists who proposed it and who proved
it. How was it proved experimentally? Explain.
or
Colourblindness in humans is a sex-linked trait. Explain with the help of a cross.
Ans.Mathew Messelson and Franklin Stahl’s experiment demonstrated that DNA replication is semiconservative. Meaning that the two daughter DNA molecules produced would have one parental strand
and the other newly synthesised strand.
(i) Mathew Messelson and Franklin Stahl 1958 grew E. coli in a medium containing 15NH4Cl (15N is the
heavy isotope of nitrogen) as the only nitrogen source for many generations. The result was that
15
N was incorporated into newly synthesised DNA (as well as other nitrogen containing
compounds). This heavy DNA molecule could be distinguished from the normal DNA by
centrifugation in a caesium chloride (CsCl) density gradient.
(ii) They transferred the cells into a medium with normal 14NH4Cl and took samples at various definite
time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded
helices. The various samples were separated independently as CsCl gradients to measure the
densities of DNA.
(iii) Thus, the DNA that was extracted from the culture, one generation after the transfer from 15N to 14N
medium (that is after 20 minutes; E. coli divides in 20 minutes) had a hybrid or intermediate density.
DNA extracted from the culture after another generation (that is after 40 minutes, II generation) was
composed of equal amounts of hybrid DNA and of light DNA.
Messelson and Stahl’s Experiment.
or
Ans. (a) Red-Green Colour Blindness/Colour Blindness: It is a recessive sex-linked defect in which
human beings are unable to distinguish red and green colours. This defect is due to mutation in certain
genes present in the X-chromosome. It occurs in about 8 per cent of males and only about 0.4 per cent
of females because the genes that lead to red-green colour blindness are on the X-chromosome.
Genotype of a colour blind woman is XCXC, that of colour blind man is XCYand carrier woman with
normal vision XXC. Males cannot be carriers.
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54. How did the chemical nature of the ‘Transforming Principle’ get established?
or
Describe how the lac operon operates, both in the presence and absence of an inducer in E.coli.
Ans. Avery, Macleod, McCarty discovered that DNA from the heat killed S-strain caused the living Rstrain bacteria to become transformed into living S-type. DNA is the transforming principle. They found
proteases and RNAases did not affect transformation while DNAases inhibited transformation. Thus,
they concluded that DNA is the hereditary material.
or
Lac-operon, in bacteria a transcriptionally regulated system was proposed by Jacob and Monod. The
polycistronic structural genes are regulated by a common promoter and regulator gene.
The lac operon
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DR. ARVIND'S BIOLOGY CLASSES
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There are three structural genes z, y and a. z-codes for beta-galactosidase which hydrolyses lactose
into galactose and glucose, y-gene codes for permease which increases permeability of the cell to
lactose and the a-gene codes for transacetylase. All three of these gene products are required for
lactose metabolism.
Lactose in the cell, regulates switching on or off of the operon and is termed inducer.
When lactose is present then the repressor protein synthesised by i-gene combines with inducer
(lactose) to form inactive repressor which does not bind to operator gene and this allows the RNA
polymerase access to the promoter gene and transcription proceeds. In the absence of lactose,
repressor binds to the operator gene which is switched off and RNA polymerase cannot move from
promoter gene and thus there is no transcription. This type of regulation of lac operon by repressor is
called negative regulation. The repressor of operon from i-gene is synthesised all the time, hence, igene is a constitutive gene.
55. (a) Explain the process of DNA replication with the help of a schematic diagram.
(b) In which phase of the cell cycle does replication occur in Eukaryotes? What would happen if
cell-division is not followed after DNA replication?
Ans. (a)
(i) Mechanism of DNA replication – DNA strands separate and act as template. DNA polymerase
enzyme catalyses the polymerisation of deoxyribonucleotides.
(ii) Activate deoxyribonucleotide triphosphates act as substrate and also provide energy for
polymerisation reaction.
(iii) Replication occurs within a small opening of DNA-helix called replication fork. DNA-polymerase
polymerises nucleotides only in one direction 5’  3’.
(iv) At replication fork, replication is continuous in template with polarity 3’  5’ (leading strand) while
in the other template with polarity 5’  3’, replication is discontinuous (lagging strand) due to
exposure of small stretches of template at one time. Discontinuously synthesised okazaki
fragments are later joined by the enzyme DNA ligase.
Replicating Fork
(v) Replication does not initiate randomly at any place in the DNA. Origin of replication are definite
regions in E.coli DNA where the replication originates. During recombinant DNA procedures the
origin of replication is provided by the vectors thus facilitating the propagation of the DNA.
(b) The replication of DNA occurs in S-phase (synthetic phase) of cell cycle in Eukaryotes. If cell division
does not follow the doubling of DNA content then duplicate set of chromosomes/genes will be retained
in the single cell. It results in auto polyploidy.
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DR. ARVIND'S BIOLOGY CLASSES
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EVOLUTION
SECTION – A
I. Each question carries 1 mark.
56. Why are analogous structures a result of convergent evolution?
Ans. The analogous structures are superficially similar and perform similar functions but are different in
their structural details and origin and are present in different groups of organisms. They are a result of
convergent evolution e.g., wings of an insect and wings of bird.
SECTION – B
II. Each question carries 2 marks.
57. Rearrange the following in increasing order of evolution:
Gnetales; Ferns; Zosterophyllum; Ginkgo
Ans. Increasing order of evolution
Zosterophyllum  Ferns  Ginkgos  Gnetales
SECTION – C
I. Each question carries 3 marks.
58. According the Darwinian theory, the rate of appearance of new forms is linked to their life
cycles. Explain.
Ans. The variations, indirectly the rate of appearance of new forms is linked to the life cycle or the life span.
The individuals which have a short life span will show a high degree of variation over a fixed period of
time (even days in case of microbes) in comparison to the individuals with a long life cycle of life span.
In case of microbes which have the ability to multiply and become millions of individuals within hours
and in due course of time, this variant population outgrows the others and appears as a new species
within few days. The same event in fish or fowl would take million of years as the life span of these
animals are in years.
59. Cite an example where more than one adaptive radiations have occurred in an isolated
geographical area. Name the type of evolution your example depicts and state why it is so
named.
Ans. When more than one adaptive radiation appeared to have occurred in an isolated geographical area
(representing different habitats) it is termed as convergent evolution. Placental mammals in Australia
also exhibit adaptive radiation in evolving into varieties of such placental mammals each of which
appears to be ‘similar’ to a corresponding marsupial (e.g., Placental wolf and Tasmanian wolfmarsupial).
SECTION – D
IV. Each question carries 5 marks.
60. (a) Explain Darwinian theory of evolution with the help of one suitable example. State the two
key concepts of the theory.
(b) Mention any three characteristics of Neanderthal man that lived in near east and central Asia.
Ans. (a) Darwin’s theory of natural selection is based on certain facts such as:
Factors of the Theory
1) Rapid Multiplication: All animals and plants tend to multiply in geometrical progression. In other
words, a population, that doubles its number in a first year, can increase to four times in a second year
and to eight times in a third year, and so on. Each pair of mice produces dozens of young ones, insects
lay hundred of eggs and plants produce thousands of seeds.
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DR. ARVIND'S BIOLOGY CLASSES
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In spite of the enormous reproductive potential of organisms under natural conditions, the
number of individuals of each species remains nearly constant over long periods of time. This is
because the great majority of potential offsprings die.
2) Limited Space and Food: The carrying capacity of environment, that is, the number of individuals it
can support, does not allow its population to grow beyond the limit and equilibrium is reached. The
population fluctuates around this equilibrium.
3) Struggle for Existence: Because of excessive multiplication by the parents and limited space and food
supply, there starts a severe competition among the offspring for their requirements. The struggle for
existence is three-fold for every individual:(i) Intraspecific Struggle: This is the struggle between the individuals of the same species. This is
the keenest form of struggle as the needs of the individuals of the same species are identical.
(ii) Interspecific Struggle: This is the struggle between the individuals of different species. This
struggle is illustrated by the efforts of a snake for catching a rat, and of the rat for escape.
(iii) Environment Struggle: This is the struggle of the animals with the changes in environmental
factors, such as heat, cold, famine, light, etc.
4) Variation: Every individual varies in some respects, such as size, shape, structure and behaviour, from
others of its species.
5) Natural Selection (Survival of the Fittest): In the struggle for existence, the individuals which have
more favourable variations will enjoy a competitive advantage over others and will survive and
reproduce. Individuals with unfavourable variations have a selective disadvantage and are rejected by
nature.
This sorting out of the individuals with useful variations has been called natural selection by Darwin
and survival of the fittest by Herbert Spencer.
The individuals which survive in the struggle for existence, grow into adults and produce young
ones.
6) Inheritance of Useful Variations: The individuals pass on their useful variations to the next
generation. Darwin did not differentiate between continuous and discontinuous variations. He held that
any variations, which are favourable to its possessor, could be inherited. His inheritable variations
included even the acquired characters. In this respect, Darwin agreed with Lamarck.
7) Formation of New Species: In each generation, new favourable variations appear, and supplement
the favourable variations inherited from the parents. After a number of generations, the variations
become so many and so prominent that their possessors turn into the members of a new species. New
species, thus, arise by gradual modification of the older ones.
Branching descent and natural selection are said to be the two key concepts of
Darwin’s theory of Evolution.
Mimicry and Protective coloration found in certain animals are very good examples of Natural
Selection.
Branching descent and natural selection are the two key concepts of natural selection. Branching
descent means appearance of new forms, linked to life cycle or lifespan while natural selection is
based on those variations which are heritable and make resource utilisation better by those better
adapted to the habitat. These better adapted ones reproduce and leave more progeny. Over a period
of time, new forms appear to arise.
(b) Characteristics of Neaderthal man:
(i) The cranial capacity was 1400 cubic centimetres.
(ii) Lived about 1,00,000 – 40,000 years back.
(iii) Used hides to protect their body and buried their dead.
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DR. ARVIND'S BIOLOGY CLASSES
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BIOLOGY IN HUMAN WELFARE
SECTION – A
I. Each question carries 1 mark.
61. State one reason for adding blue-green algae to the agricultural soil.
Ans. Blue-green algae serve as an important biofertilizer, as they fix atmospheric nitrogen and add organic
matter to the soil.
62. Identify the two correct statements from the following:
(i) Apiculture means apical meristem culture
(ii) Spinach is iron-enriched.
(iii) Green revolution has resulted in improved pulse-yields.
(iv) Aphids cannot infest rapeseed mustard.
Ans. (ii) Spinach is iron-enriched.
(iv) Aphids cannot infest rapeseed mustard. (Pusa Gaurav variety of
rapeseed mustard).
Hint: Apiculture means rearing of honey bees. Green revolution mainly concerns with development
of high yielding varieties of wheat and rice.
63. Why is secondary immune response more intense than the primary immune response in
humans?
Ans. It is due to the presence of memory cells that were produced during the primary response. The
response occurs more rapidly and is more intense.
64. Name any two types of cells which act as ‘Cellular barriers’ to provide Innate Immunity in
humans.
Ans. Polymorpho-nuclear leucocytes-(neutrophils), monocytes, natural killer cells (a type of lymphocytes)
in the blood and macrophages in tissues.
SECTION – B
II. Each question carries 2 marks.
65. Name two groups of organisms which constitute ‘flocs’. Write their influence on the level of
BOD during biological treatment of sewage.
Ans. Bacteria + Fungal hyphae
These microbes decompose aerobically the major part of the organic matter in the primary effluent.
Thus, it helps in reducing the BOD (biological oxygen demand) of the effluent.
66. Name an allergen and write the response of the human body when exposed to it.
Ans. Name of allergen = Dust or pollen in air. When the human body is exposed to the dust particles
in the air, they act as antigens to evoke an exaggerated response of immune system. It involves
the formation of I g E antibodies and release of histamine and serotonin from mast cells. These
chemicals produce symptoms of allergy such as sneezing, watery eyes, running nose and difficulty
in breathing etc.
67. How are morphine and heroin related? Mention the effect each one of them has on the human
body.
Ans. Morphine, a principal natural opium alkaloid extracted from the latex of poppy plant. It is a strong
analgesic. Heroin (diacetylmorphine hydrochloride) is formed from morphine by acetylation with acetic
acid and is semisynthetic opiate. It is 3 times more potent than morphine.
Morphine is a very effective sedative and painkiller and has a calming effect. It is very useful in patients
who have undergone surgery. It can release ADH and reduce urine output. Constipation is a
prominent feature of morphine action.
Heroin is more euphorient and is highly addictive. It is usually taken by snorting or injection. It is a
depressant and slows down body functions. It induces drowsiness and lethargy. Its aftereffects are,
indigestion, reduce vision, loss of weight, sterility and total loss of interest in work.
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DR. ARVIND'S BIOLOGY CLASSES
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68. Why is ‘starter’ added to set the milk into curd? Explain.
Ans. A small amount of curd, added to fresh milk contains million of Lactic Acid Bacteria as inoculum or
starter which multiply at suitable temperature thus converting milk into curd. They also improve its
nutritional quality by increasing vitamin B12. During growth, these bacteria produce enzymes that
convert milk sugar into acids that coagulate and partially digest the milk proteins to form curd.
69. Write the scientific name of the source plant of the drugs, marijuana and hashish and mention
their effect on the human body.
Ans. Name of source plant : Cannabis sativa.
Marijuana and hashish are group of chemicals called cannabinoids. They interact with cannabinoids
receptors present in the brain. They are taken by inhalation and oral ingestion. They have prominent
effect on cardiovascular system of the body. They bring about a state of well-being, euphoria,
excitement, sometimes uncontrolled laughter and dilation of pupil of eye. They are Psychedelic drugs
(Hallucinogens).
70. Name the drug obtained from Erythroxylum coca and write its effect on the human body.
Ans. Name of Drug – Cocaine (commonly called coke or crack) Effect : It interferes with the transport
of neurotransmitter Dopamine in the CNS. It has a potent stimulating action on central nervous system,
produces a sense of euphoria and increased energy. Excessive dosage causes hallucinations. It is an
addictive drug.
71. List the symptoms of Ascariasis. How does a healthy person acquire this infection?
Ans. Ascariasis is a helminthic disease caused by common roundworm – Ascaris lumbricoides. The
symptoms of disease include abdominal pain and cramps, blockage of intestinal passage, anaemia, internal
bleeding, nausea and headache. Disease spreads through raw vegetables, fruits and water contaminated
with the eggs of the parasite which are excreted along with faeces of an infected person.
SECTION – C
III. Each question carries 3 marks.
72. Suggest and describe a technique through which a virus-free healthy plant can be obtained from
a diseased sugarcane plant.
Ans. Micropropagation is the method of producing thousands of plants through tissue culture in a short
duration.
The production of virus-free healthy plants from a diseased sugarcane plant is done by using
meristem-culture technique. In a diseased plant, the meristems are always free from the virus even if
the plant is infested with a virus. Here, the apical and axillary buds are excised and used as Explant.
These are free from virus. They are inoculated in the culture medium, to produce plantlets. Then,
these plantlets are transferred into the soil where they grow into mature virus-free healthy plants and
are thus established in the field.
73. Draw a labelled sketch of a typical biogas plant.
or
(a) Name the causative organisms for the following diseases:
(i) Elephantiasis
(ii) Ringworm
(iii) Amoebiasis
(b) How can public hygiene help control such diseases?
Ans.
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DR. ARVIND'S BIOLOGY CLASSES
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A biogas plant.
or
(a) Causative organisms:
(i) Elephantiasis : Wuchreria bancrofti
(ii) Ringwom : Microsporum, Trichophyton, Epidermophyton
(iii) Amoebiasis : Entamoeba histolytica
(b) Public hygiene plays a great role in control of these diseases. Public hygiene like proper disposal of
domestic waste and sewage, control of vectors, supply of clean and pure drinking water, personal
healthy habits like washing hands before eating, proper washing of fruits and vegetables, wearing
clean clothes, not sharing personal used items are some important steps to help control these
diseases.
74. (a) Name the tropical sugar cane variety grown in South India. How has it helped in improving
the sugar cane quality grown in North India?
(b) Identify ‘a’, ‘b’ and ‘c’ in the following table:
No.
1.
2.
3.
Crop
Brassica
Flat bean
(c)
Variety
Pusa Gaurav
Pusa Sem 2
Pusa Sem 3
Pusa Sawani; Pusa A – 4
Insect Pests
(a)
(b)
Shoot and fruit borer
Ans. (a) Saccharum officinarum
Saccharum barberi originally growing in N. India, had poor sugar content and yield. S. officinarum
(tropical cane) grown in S. India has thicker stem and higher sugar contents, but could not grow
well in North India. A cross was made between these two species and the hybrid variety,
combined the desirable qualities of high yield, thick stems, high sugar and ability to grow in the
sugarcane areas of North India.
(b) a = Aphids
b = Jassids, aphids and fruit borer
c = Okra (Bhindi)
75. Why are beehives kept in crop field during flowering period? Name any two crop fields where
this is practised.
Ans. Bees are the most common pollinators of many crops such as sunflower, Brassica, apple, pear, etc.
Keeping beehives in crop fields during flowering season increases pollination efficiency and improves
both the crop yield and honey yield.
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DR. ARVIND'S BIOLOGY CLASSES
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SECTION – D
IV. Each question carries 5 marks.
76. With advancements in genetics, molecular biology and tissue culture, new traits have been
incorporated into crop plants.
Explain the main steps in breeding a new genetic variety of a crop.
or
(a) State the objective of animal breeding.
(b) List the importance and limitations of inbreeding. How can the limitations be overcome?
(c) Give an example of a new breed each of cattle and poultry.
Ans. Main steps involved in breeding a new genetic variety of a crop:
(a) Collection of variability – The basic need of a breeding programme involves collection and
preservation of all the different wild varieties, species and relatives of cultivated species and their
evaluation for their characteristics. The entire collection in the form of plants or seeds (having
diverse alleles for all genes) in a given crop is called germplasm collection.
(b) Evaluation and selection of parents – Germplasm is evaluated and plants with desirable
combination of characters are identified. These selected plants are multiplied and used in the
process of hybridisation. Pure lines are created wherever desirable and possible.
(c) Cross-hybridisation among the selected parents – We need to Cross hybridise the two parents
to produce hybrids that genetically combine the desired characters in one plant. It is a time
consuming and tedious process involving pollination and bagging. The rate of success is very low,
usually only one in few hundred to a thousand crosses shows desirable combinations.
(d) Selection and testing of superior recombinants – Selection among the progeny of hybrids for the
plants with desired character combinations. These selected superior plants are self-pollinated for
several generations till they reach a state of uniformity homozygosity so that characters will not
segregate in progeny.
(e) Testings release and commercialisation of new cultivars – Newly selected lines are evaluated
for desirable characters by growing these in the research fields and recording their performance
under ideal fertiliser application, irrigation and other crop management practices. It is followed by
testing the materials in farmer’s field for at least three growing seasons at different agroclimatic
zones. The material is evaluated in comparison to the best available local crop cultivar – a check or
reference cultivar.
(f) Release of new variety – The material evaluated is selected, certified and released as a new
variety.
or
(a) Animal breeding aims at:
(i) Increasing the yield of animals
(ii) Improving the desirable qualities of the produce
(b) (i) Inbreeding increases homozygosity, which is necessary to evolve a pure line in any animal.
(ii) Inbreeding exposes harmful recessive genes that are eliminated by selection.
(iii) It helps in accumulation of superior genes and elimination of less desirable genes.
(iv) Where there is selection at every step, there is increase in productivity of the inbred population.
However, continued inbreeding, especially close inbreeding, causes inbreeding depression
which reduces fertility and leads to low productivity.
This disadvantage of inbreeding can be overcome by outcrossing i.e., mating selected animals
of breeding population with unrelated superior animals of the same breed to restore fertility and
yield.
(c) (i) Cattle breed: Jersey
(ii) Poultry breed: White Leghorn.
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DR. ARVIND'S BIOLOGY CLASSES
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77. Explain the process of replication of a retrovirus after it gains entry into the human body.
Ans. AIDS virus is a retrovirus. It is a spherical virus that contains two single stranded RNA filaments,
associated with a reverse transcriptase enzyme which helps in reverse transcription i.e., formation of
DNA from RNA.
Process of replication:
(i) After entering into the body of a person the virus enters the macrophages. Here the RNA genome of
the virus replicates to form viral DNA with the help of enzyme reverse transcriptase.
(ii) This viral DNA gets integrated / incorporated into the host cell DNA and uses the raw material from the
host cells and the host cell machinery to produce virus particles.
(iii) The macrophages continue to produce virus particles and act like a HIV factory.
(iv) Simultaneously the virus enters the helper T-lymphocytes (TH), replicates and forms progeny viruses.
The progeny viruses released in the blood, attack the other helper T-lymphocytes, resulting in
progressive decrease in the number of helper T-lymphocytes in the infected person. Due to the
decrease in number of T-lymphocytes, the person becomes susceptible to infections like
mycobacterium, viruses, fungi and even parasites like Toxoplasma.
Note: Infected cell can survive while viruses are being replicated
and released replication of retrovirus
78. (a) Name the technology that has helped the scientists to propagate on large-scale the desired
crops in short duration. List the steps carried out to propagate the crops by the said technique.
(c) How are somatic hybrids obtained?
or
(a) Cancer is one of the most dreaded diseases of humans. Explain ‘Contact inhibition’ and
‘Metastasis’ with respect to the disease.
(b) Name the group of genes which have been identified in normal cells that could lead to
cancer and how they do so?
(c) Name any two techniques which are useful to detect cancers of internal organs.
(d) Why are cancer patients often given -interferon as part of the treatment?
Ans. (a) Micropropagation – a method of producing thousands of plants through tissue culture technique.
Technique of Plant Tissue Culture: The basic technique of plant tissue culture involves the following
steps.
1. Preparation of suitable Nutrient Medium: Suitable nutrient medium (depending on type of plant
tissues or cell that are used for culture) is prepared and transferred into suitable containers.
2. Selection of explants:
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3. Sterilisation of explants: Surface sterilization of the explants by disinfectants (e.g., sodium
hypochlorite) and then washing the explants with sterile distilled water is essential
4. Inoculation : Inoculation (transfer) of the explants into the suitable nutrient medium which is sterilized
in culture vessels under sterile conditions is done.
5. Incubations: Growing the culture in the growth chamber or plant tissue culture room, having the
appropriate physical condition.
6. Regeneration: Regeneration of plants from cultured plant tissues is carried out.
7. Hardening: Hardening is gradual exposure of plantlets to an environmental condition.
8. Plantlet Transfer: After hardening plantlets transferred to the greenhouse or field conditions.
(b) Somatic hybridisation is the process of fusing protoplast of somatic cells of two different
varieties/species of plants on suitable culture/nutrient medium under aseptic conditions. It includes:
(i) Isolation of protoplasts by Mechanical method using plasmolysis technique or by Enzymatic
digestion of cell wall using pectinase & cellulase.
(ii) Purification of naked protoplast by sedimentation & washing.
(iii) Fusion of isolated protoplast from two varieties of plants each having a desirable character to form
hybrid protoplast by either spontaneous fusion or induced fusion
(iv) Hybrid protoplast is further grown to form a somatic hybrid and can be induced to regenerate into a
whole plant.
or
(a) Contact inhibition is a property shown by normal cells in which contact of one cell with its
surrounding cells inhibits their uncontrolled growth. In cancer the cells lose this property of contact
inhibition and undergo unlimited and uncontrolled mitotic divisions. They invade and destroy the
surrounding tissues and form a growth or mass of cells called tumour or neoplasm. Metastasis is
a condition where the cells slough off from the primary tumour and reach distant sites through blood
or lymph, there they lodge and give rise to a secondary tumour. This is a characteristic of malignant
tumours.
(b) Cellular oncogenes (c-onc): Several genes called cellular oncogenes or proto-oncogenes are
present in normal cells. Under certain conditions (radiations, chemical carcinogens, tumour viruses
and physical irritants) they become activated, change into oncogenes. The product of these
oncogenes now do not respond to normal regulatory signals thus they transform the normal cells
into cancerous cells causing tumours
(c) (i) Magnetic resonance imaging (MRI)
(ii) Computed Tomography (CT)
(d) -Interferons come under the category of immunotherapy for cancer treatment. Tumour cells are
known to avoid detection and destruction by immune system. Therefore, patients are given
substances called biological response modifiers such as -interferon which activates the immune
system and helps in destroying the tumour.
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DR. ARVIND'S BIOLOGY CLASSES
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BIOTECHNOLOGY
SECTION – A
I. Each question carries 1 mark.
79. Why is it not possible for an alien DNA to become part of a chromosome anywhere along its
length and replicate normally?
Ans. Alien DNA after becoming a part of chromosome anywhere cannot replicate normally because it
needs a special sequence of bases constituting Ori, to start replication within host cells.
80. State the role of C-peptide in human insulin.
Ans. C-peptide, an extra-stretch helps the insulin prohormone to become fully mature and functional
hormone.
81. Name the enzymes that are used for the isolation of DNA from bacterial and fungal cells for
recombinant DNA technology.
Ans. The enzyme lysozyme in bacteria and the enzyme chitinase in fungi.
82. Mention the type of host cells suitable for the gene guns to introduce an alien DNA.
Ans. Suitable host cells belong to plants for gene guns.
SECTION – B
II. Each question carries 2 marks.
83. Why is making cells competent essential for biotechnology experiments? List any two ways by
which this can be achieved.
Ans. DNA is a hydrophilic molecule and cannot pass through the cell membranes. The cells are made
competent in order to take up the DNA present in their surrounding. Only then cell becomes
transformed with recombinant DNA.
Two ways in which the cell can be made ‘competent’ (i) For transformation bacterial cells are treated with a specific concentration of a divalent cation like
calcium which increases the efficiency with which DNA enters the bacterium through pores in its
cell wall.
(ii) The cells and the recombinant DNA are incubated on ice, followed by placing them briefly at 42C
(heat shock) & then putting them back into ice. This enables the bacteria to take up recombinant
DNA.
84. Human insulin when synthesised in the body needs to be processed before it can act. Explain
giving reasons.
Ans.
(i) Insulin consists of two short polypeptide chains, chain A and chain B, which are linked together by
disulphide bonds.
(ii) In mammals including humans insulin is synthesised as a prohormone, which contains an extrastretch called C-peptide, and it needs to be processed before it becomes a fully mature and
functional hormone.
(iii) The C-peptide is not present in the mature insulin and is removed during maturation of proinsulin.
85. How have transgenic animals proved to be beneficial in:
(a) Production of biological products
(b) Chemical safety testing
Ans.(a) Production of biological products: Transgenic animals have helped in (i) the mass production of
safe and more effective recombinant therapeutic drugs. (ii) Production of recombinant vaccines
including whole protein vaccines, polypeptide vaccines, DNA vaccines etc. (iii) Production of  – I –
antitrypsin, - lactoalbumin from transgenic cows (iv) Production of synthetic insulin (Humulin).
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(b) Chemical Safety testing : GMOs more sensitive to toxic substances in the environment or to
toxicity of drugs which are developed. They help us to study effects and to obtain results in less time,
e.g., polio vaccine is being tested for safety on transgenic mice and monkeys, transgenic bacteria
degrade oil spills, destroy dangerous chemicals. Pseudomonas is used in bioremediation i.e.,
depollution of environment.
86. Explain with the help of a suitable example the naming of a restriction endonuclease.
Ans. Naming of restriction enzyme is as follow:
(i) The first letter of the name comes from genus and the next two letters from the name of the species
of the bacterium from which they are isolated.
(ii) The next letter comes from the strain of the bacterium.
(iii) The roman number following these four letters indicate the order in which enzymes were isolated
from that strain of the bacterium e.g.,
(a) Eco R – I is isolated from Escherichia coli RY 13
(b) Hind-II is from Haemophilus influenzae
87. State how has Agrobacterium tumifaciens been made a useful cloning vector to transfer DNA to
plant cells.
Ans. Agrobacterium tumifaciens a pathogen of several dicot plants is able to deliver a piece of DNA known
as ‘T-DNA’ to transform normal plant cells into a tumour and directs these tumour cells to produce the
chemicals required by the pathogen.
The tumour inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified into a
cloning vector which is no more pathogenic to the plants but is still able to use the mechanism to deliver
genes of our interest into a variety of plants. So, once a gene or a DNA fragment has been ligated into
this cloning vector, it can easily be transferred into the plant host where it multiplies and produces the
desired product/result.
88. How are ‘sticky ends’ formed on a DNA strand? Why are they so-called?
Ans. Restriction enzymes inspect the length of DNA to recognise specific sequence. Then it binds to DNA
and cuts each of the two strands of double helix at specific points in their sugar-phosphate backbones
on the opposite strands producing single stranded overhangs at each end. These overhanging
stretches on each strand are called ‘sticky ends’. These ends form hydrogen bonds with their
complementary cut counterparts. The stickiness of the ends facilitates the action of enzyme DNA
ligase, to join two DNA fragments, one of vector and the other of foreign DNA to form a recombinant
DNA.
89. What is gene therapy? Name the first clinical case where it was used.
Ans. Gene therapy is a collection of methods that allow correction of a gene defect that has been
diagnosed in a child/embryo. It requires a mechanism to repair the defective gene or to substitute the
defective one with a healthy gene copy. Gene therapy may be germ line gene therapy or somatic gene
therapy.
(i) In this method genes are inserted into a person’s cells and tissues to treat a disease. Retroviral
vectors can be used to transform the cells in culture and then introduce them into the body.
(ii) It involves the delivery of a normal gene into the individual or embryo to take over the function of
and compensate for the non-functional gene.
The first clinical gene therapy was given in 1990 to a 4-year old girl deficient in enzyme adenosine
deaminase (ADA) crucial for functioning of immune system. A functional ADA cDNA (using a retro
viral vector) was introduced into the lymphocytes taken from the patient and then after culture these
genetically engineered lymphocytes were reintroduced into the patient.
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SECTION – C
III. Each question carries 3 marks.
90. How are Baculoviruses and Bacillus thuringiensis used as bio-control agents? Why are they
preferred over readily available chemical pesticides?
Ans. Baculoviruses – They are pathogens belonging to genus Nucleopolyhedroviruses, which attack
against insects and other arthropods. They are species specific and have narrow spectrum insecticidal
applications. They have no impact on plants, mammals, birds, fish or even on non-target insects. Thus
are good bio control agents.
Bacillus thuringiensis a bacterium, is a microbial bio-control agent, introduced into the crop plants in
order to control butterfly caterpillars. They are mixed with water and sprayed on susceptible plants
such as Brassica and fruit trees where they are eaten by insect larvae. The larvae get killed by toxins
released in their gut.
They are preferred over chemical pesticides as they are inexpensive eco-friendly and do not pollute
environment. They help in conserving the beneficial insects and help in an overall integrated pest
management programme.
91. What is a bioreactor used for? Name a commonly used bioreactor and any two of its
components.
Ans.Bioreactors are used for processing large volume of culture for obtaining the product of interest in
large quantities. In bioreactors the raw materials are biologically converted into specific products e.g.,
enzymes
A commonly used bioreactor is the stirred-tank reactor. It is cylindrical with a curved base.
The bioreactor has an Agitator System, an Oxygen delivery system and a foam control system. It also
has a temperature control system, pH control system and sampling ports so that small volumes of
culture can be withdrawn periodically.
92. Name the host plant and its part that Meloidogyne incognita infects. Explain the role of
Agrobacterium in the production of ds-RNA in the host plant.
Ans. The nematode Meloidegyne incognitia infects the roots of tobacco plants and reduces the yield.
Using Agrobacterium vectors, nematode specific genes were introduced into host plant. The
introduction of DNA was such that it produced both sense and antisense RNA in the host cells. These
two RNAs being complementary to each other formed a ds-RNA that initiated RNAi and thus silenced
the specific m-RNA of the nematode. As a result the parasite could not survive in a transgenic host
expressing specific interfering RNA. So the transgenic plant got itself protected from the parasite.
93. What are ‘cloning sites’ in a cloning vector? Explain their role. Name any two such sites in
pBR322.
Ans. In order to link the alien DNA, the vector needs to have very few, preferably single cloning or
recognition sites for the commonly used restriction enzymes. The ligation of alien DNA is carried out at
a restriction site specific for the enzyme used.
pBR 322 Vector, has EcoRI and Bam H-I as two cloning sites. The other sites are Hind III, Sal-I,
Pvu II, Pst-I, Cla-I.
94. Name and describe the technique that helps in separating the DNA fragments formed by the use
of restriction endonuclease.
Ans. Separation and isolation of DNA fragments is carried out by the technique of gel-electrophoresis.
(a) DNA fragments being negatively charged, can be separated by forcing them to move towards anode
under an electric field through a medium/matrix (agarose). DNA fragments separate (resolve)
according to their size through sieving effect provided by the agarose gel. The smaller the fragment
size, the farther it moves towards the anode.
(b) DNA fragments can be visualised by staining DNA with ethidium bromide followed by exposure to
UV radiations. Bright orange colour bands of DNA become prominent in the gel. The separated
bands of DNA are cut out from gel and extracted from the gel piece. This step is known as elution.
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Cathode
Anode
A typical agarose gel electrophoresis showing migration of undigested (lane 1)
and digested set of DNA fragments (lane 2 to 4).
(c) Purified DNA fragments are used for reconstructing recombinant DNA by joining them with cloning
vectors.
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DR. ARVIND'S BIOLOGY CLASSES
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ECOLOGY
SECTION – A
I. Each question carries 1 mark.
95. Write the level of biodiversity represented by a mangrove. Give another example falling in the
same level.
Ans. Ecological diversity. The other examples are deserts, rainforests, coral reefs, wetlands, estuaries,
alpine meadows etc.
96. Name the two gases contributing maximum to the greenhouse effect.
Ans. Methane (20%) and carbon dioxide (60%)
97. Where is good ozone present? Why is it called so?
Ans. Good ozone is present in stratosphere because it does not allow the harmful UV radiations to enter
the earth’s atmosphere.
98. Mention any two reasons why the primary productivity varies in different types of ecosystems.
Ans. Reasons: Variation in the (i) availability of environmental factors e.g., light, water (ii) availability of
nutrient in the soil in different types of ecosystems.
99. State Gause’s Competitive Exclusion Principle.
Ans. It states that two closely related species competing for the same resources cannot coexist indefinitely
and the competitive inferior one will be eliminated eventually. This may be true if the resources are
limited but not otherwise.
100. Name the type of association that the genus Glomus exhibits with higher plants.
Ans. Mutualism
101. Why is Gambusia introduced into drains and ponds?
Ans. Gambusia fish preys upon the larvae of the mosquitoes, found in drains and ponds. It is an effective
method of Biocontrol of malaria.
102. Give an example of a plant which came into India as a contaminant and is a cause of pollen
allergy.
Ans. Parthenium sp. (congress grass or carrot grass)
103. Write the name of the organism that is referred to as the ‘Terror of Bengal’.
Ans. Eichhornia crassipes (water hyacinth), the world’s most problematic weed is named as ‘Terror of
Bengal’.
104. How is ‘stratification’ represented in a forest ecosystem?
Ans. Vertical distribution of different species occupying different levels is called stratification. For example,
trees occupy top vertical strata or layer of a forest, shrubs the second and herbs and grasses occupy
the bottom layer.
105. Give an example of an organism that enters ‘diapause’ and why.
Ans. Some animals undergo a stage of suspended development called diapause to avoid unfavourable
conditions in lakes and ponds. Example : Zooplankton.
106. State the cause of Accelerated Eutrophication.
Ans. Pollutants from human activities like effluents from the industries and homes and run off fertilisers
from agricultural lands can accelerate the process of aging of a lake called Accelerated or Cultural
Eutrophication.
107. Name the two intermediate hosts which the human liver fluke depends on to complete its life
cycle so as to facilitate parasitization of its primary host.
Ans. Two intermediate hosts of human liver fluke are (i) snail, and (ii) a fish
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DR. ARVIND'S BIOLOGY CLASSES
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SECTION – B
II. Each question carries 2 marks.
108. Provide two reasons that make the count of prokaryotic species difficult.
Ans. The count of prokaryote species is difficult, due to some problems such as:
a) Conventional taxonomic methods are not suitable/sufficient for identifying these microbial species.
b) Many of these species cannot be cultured under laboratory conditions.
c) If we consider Biochemical or molecular criteria for delineating species for this group then their
diversity would run into millions.
109. ‘Fish mortality increases with influx of nutrients in a freshwater body. Write two reasons. How
will the influx of nutrients affect the BOD level of this water body?
Ans. Due to the influx of nutrients in freshwater body, the algal bloom causes drastic decrease in the
dissolved oxygen content of the water.
(i) Decreased oxygen level kills aquatic animal such as fish.
(ii) Increase in the organic load by the death of submerged plants and aquatic animals leads to
anaerobic conditions that produce toxic secondary pollutants that can kill fish.
Due to organic loading the BOD level of water body increases further.
110. List four causes of biodiversity loss
or
Name two metals used in a catalytic converter. How do they help in keeping the environment
clean?
Ans. Causes of biodiversity loss: The four major causes described as ‘The Evil Quartet’ are:
(i) Habitat loss and fragmentation
(ii) Over-exploitation
(iii) Invasion of alien species
(iv) Co-extinctions
or
Name of metals : (i) Platinum-palladium
(ii) Rhodium
These metals are used as catalytic converters, fitted in automobiles for reducing emissions of harmful
poisonous gases by (i) Converting the unburnt hydrocarbons into CO2 and H2O (ii) Converting CO and
nitric oxides to CO2 and nitrogen gas respectively.
111. List any four parasitic adaptations in a parasite.
Ans. Parasites have special adaptations in accordance with their lifestyle such as (a) Loss of unnecessary
sense organs (b) Presence of adhesive organs or suckers to cling to the host (c) loss of digestive
system and (d) high reproductive capacity.
112. Describe the mutual relationship between fig tree and wasp and comment on the phenomenon
that operates in their relationship.
Ans. The relation between fig tree and wasp is of Mutualism. A given species of fig can be pollinated only
by its partner wasp species and no other species. The female wasp uses the fruit not only as oviposition
(egg laying) site but uses the developing seeds within the fruit for nourishing its larvae. The wasp
pollinates the fig inflorescence while searching for suitable egg-laying sites. In return for the favour of
pollination, the fig offers the wasp some of its developing seeds, as food for the developing wasp
larvae.
113. Construct an age pyramid which reflects a stable growth status of human population.
Ans. Age pyramid of human population showing stable growth.
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114. Apart from being part of the food chain, predators play other important roles. Mention any two
such roles supported by examples.
Ans. Other than acting as conduits for energy transfer across trophic levels, predators play the following
roles - (a) They keep prey population under control which otherwise could achieve very high population
densities and cause instability in ecosystem. (b) They help in maintaining a species diversity in a
community by reducing the intensity of competition among competing prey species.
SECTION – C
III. Each question carries 3 marks.
115. Draw and explain expanding age pyramid of human population. Why is it so called?
Ans.
Expanding age pyramid is a graphic representation of a young or growing population which appear like a
triangle. The population has a very high proportion of pre-reproductive individuals. The proportional
number of reproductive individuals is moderate while post-reproductive individuals are proportionately
fewer. Because of very large number of pre-reproductive individuals, more and more individuals enter
the reproductive phase and this rapidly increases the size of population.
116. The following graph shows the species – area relationship. Answer the following questions as
directed.
(a) Name the naturalist who studied the kind of relationship shows in the graph. Write the
observations made by him.
(b) Write the situations as discovered by the ecologists when the value of ‘Z’ (slope of the line)
lies between
(i) 0.1 and 0.2
(ii) 0.6 and 1.2
What does ‘Z’ stand for?
(c) When would the slope of the line ‘b’ become steeper?
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DR. ARVIND'S BIOLOGY CLASSES
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Ans. (a) Alexander von Humboldt, a German naturalist and geographer studied this relationship. He
observed that within a region species richness increased with increasing explored area, but only upto a
limit. The relationship between species richness and area exhibits rectangular hyperbola for a wide
variety of taxa whether they are birds, bats, freshwater fishes or flowering plants. The value of Z lies in
the range of 0.1 to 0.2 regardless of the taxonomic group or region.
(b) (i) When the species area relationship is considered for a limited region.
(ii) The species area relationship is considered for every large area like the entire continents. ‘Z’
stands for slope of line or Regression coefficient.
(d) The slope of line ‘b’ becomes steeper with a value of 1.15 as in case of frugivorous birds and
mammals of tropical forests of different continents.
117. State the function of a reservoir in a nutrient cycle. Explain the simplified model of carbon
cycle in nature.
Ans. Reservoir is an abiotic component of ecosystem from which the nutrients move into the living biotic
component and are recycled during nutrient cycling. The function of the reservoir is to meet the defecit
which occurs due to imbalance in the rate of influx and efflux.
The reservoir for gaseous type of nutrient cycles (e.g., nitrogen, carbon cycles) exist in atmosphere and
for sedimentary cycles (e.g., sulphur and phosphorus cycles) the reservoir is located in earth’s crust.
Carbon cycle:
(i) Carbon is one of the important nutrient present in composition (organic compounds) of living
organism and constitutes 49% of the dry weight of organism, next only to water.
(ii) Oceanic reservoir of carbon (71% of total global carbon) regulates the amount of CO2 in the
atmosphere.
(iii) Atmosphere contains only 1% total global carbon.
(iv) Carbon cycling occurs through atmosphere, oceans, living and dead organisms.
(v) About 4 x 1013 kg of carbon is fixed in the biosphere through photosynthesis annually.
(vi) A great amount of carbon is returned to the atmosphere through the respiratory activities of
producers and consumers.
(vii) Some amount of fixed carbon is lost in the form of sediments (limestone, dolomite) and removed
from circulation.
(viii) Carbon dioxide is also returned to the atmosphere through:
Burning of fossil fuel; Fuel wood; Forest fire; Volcanic activities; Combustion of organic matter;
Decomposers also contribute to CO2 pool by processing the waste materials and dead organic matter;
Human activities such as deforestation and massive burning of fuel for energy and transport have
significantly increased the rate of release of CO2 into the atmosphere (greenhouse effect).
118. Since the origin of life on Earth, there were five episodes of mass extinction of species.
(i) How is the ‘Sixth Extinction’, presently in progress different from the previous episodes?
(ii) Who is mainly responsible for the ‘Sixth Extinction’?
(iii) List any four points that can help to overcome this disaster.
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Ans. (i) The earlier five episodes of mass extinction of species were natural and prehuman times. The
mass extinction occurred due to catastrophes, a number of times in geological history. But the sixth
extinction presently in progress, is due to human activities and is 100-1000 times faster than the
previous ones.
(ii) Destructive human activities and human interference in nature.
(iii) Approaches for conservation:
(a) Identification and protection of certain biodiversity rich regions called hot spots.
(b) Establishment of National Parks, wildlife sanctuaries and biosphere reserves.
(c) Establishment of zoological parks, botanical gardens, wildlife safari parks.
(d) Preservation of germ plasm.
119. Two types of aquatic organisms in a lake show specific growth patterns as shown below, in a
brief period of time. The lake is adjacent to an agricultural land extensively supplied with
fertilisers.
Answer the questions based on the facts given above.
(i) Name the organisms depicting the patterns A and B.
(ii) State the reason for the growth pattern seen in A.
(iii) Write the effects of the growth patterns seen above.
Ans.
(i) Pattern A = Planktonic (free-floating) algae (algal bloom).
Pattern B = Aquatic animals such as fishes.
(ii) The excess growth of algae is due to nutrient enrichment of lake particularly with nitrogen and
phosphorus. It is due to the run off of fertilizers from the fields. This phenomenon is called
accelerated eutrophication.
(iii) Algal bloom and floating plants cut off light from submerged plants, which die, cause organic loading
and resulting in drastic decrease in oxygen level leading to death of aquatic animals. This further adds
to organic loading which leads to putrefaction and further decreases the dissolved oxygen. In this way,
a lake is literally choked to death.
120. Explain, giving three reasons, why tropics show greatest levels of species diversity.
Ans.
(i) Temperate regions were subjected to frequent glaciations in the past, the tropics have remained
undisturbed and hence have a greater species diversity.
(ii) As compared to temperate regions tropical environments are less seasonal, relatively more stable,
constant and predictable. Such constant environments have promoted niche specialisation and
greater species diversity.
(iii) There is more input of solar radiations available in the tropical regions, more available resources
and this contributes directly to more productivity, population sizes and thus to greater species
diversity.
121. (i) What is primary productivity? Why does it vary in different types of ecosystems?
(ii) State the relation between gross and net primary productivity.
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DR. ARVIND'S BIOLOGY CLASSES
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Ans.
(i) Primary production/productivity is defined as the amount of biomass or organic matter produced per
unit area of a time period by plants during photosynthesis. It is expressed in terms of weight (gm2) or
energy (kcal m2). Primary productivity depends on:
(a) environmental factors
(b) availability of nutrients
(c) photosynthetic capacity of plants, (plant species inhabiting a particular area)
(ii) (a) Gross Primary Productivity (GPP) – The rate of production of organic matter in an ecosystem
during photosynthesis.
(b) Net Primary Productivity : A major part of GPP is utilised by the plants in the respiration. The
remaining available biomass for consumption by heterotrophs (herbivores and decomposers)
constitutes the net primary productivity.
NPP = GPP – R
Where R represents respiratory losses.
122. What are Methanogens? Name the animals they are present in and the role they play there.
Ans. Methanogens are the bacteria which are grown anaerobically on cellulose material and produce large
amount of methane along with CO2 and H2O. Methanobacterium is the common methanogen found in
the anaerobic sludge during sewage treatment. These bacteria are also present in the rumen (a part of
the stomach) of cattle. In rumen, these bacteria help in the breakdown of cellulose present in the food
of cattle. Hence, they play an important role in digestion in these animals.
123. There are many animals that have become extinct in the wild but continue to be maintained in
Zoological parks.
(i) What type of biodiversity conservation is observed in this case?
(ii) Explain any other two ways which help in this type of conservation.
Ans. (i) Ex-situ conservation is observed in Zoological parks.
(ii) Other methods:
(a) Preservation of gametes by cryopreservation techniques
(b) Wildlife safaris
(c) Botanical gardens
(d) Seed Banks
SECTION – D
IV. Each question carries 5 marks.
dN
 rN represents. What does “r”
dt
represent in the equation? Write its importance in population growth.
(b) Explain the principle of carrying capacity by using population Verhulst-Pearl logistic growth
curve.
Or
(a) With suitable examples, explain the energy flow through different trophic levels. What does
each bar in this pyramid represent?
(b) Write any two limitations of ecological pyramids.
Ans. (a) The equation dN/dt = rN represents the Exponential Growth Model for population growth.
The “r” in the equation is called “intrinsic rate of natural increase” of population. It helps in assessing
impacts of any biotic or abiotic factor on population growth.
(b) In nature, a given habitat has enough resources to support a maximum possible number, beyond which
further growth is not possible. This limit is natures carrying capacity (K) for that species in that habitat.
The limited resources lead to competition between the individuals. Eventually the fittest individuals will
survive and reproduce.
124. (a) Name the population growth pattern the equation
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DR. ARVIND'S BIOLOGY CLASSES
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A population growing in a habitat with limited resources shows initially a lag phase, followed by a phase
of rapid growth called exponential or log phase, then there is a phase where growth slows down and
finally the population density reaches an equilibrium or stagnant phase. At this time the population size
has reached the point of maximum carrying capacity of the habitat/environment. A plot of N in relation
to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic
growth.
or
a)
An ideal pyramid of energy. Observe that primary producers
convert only 1% of the energy in the sunlight available to them into NPP.
In most ecosystems, the pyramid of energy is always upright i.e., producers have more energy than
the herbivores and herbivores have more energy than carnivores. The energy at the lower trophic
level is always more than that at a higher level. As energy (food) flows from a particular trophic
level to the next trophic level, some energy is always lost as heat at each step. The transfer of
energy follows 10 percent law – only 10 per cent of the energy is transferred to each trophic level
from the lower trophic level. This restricts the number of trophic levels in a food chain.
For example, in a grazing food chain, the plants – grass and trees capture solar energy and fix it
through the process of photosynthesis. From the producers, the energy is transferred into
herbivores-the rabbit, cow, insects, horse, birds, deer etc. From herbivores, the energy is
transferred to carnivores who feed on the flesh of herbivores. The carnivores include primary
carnivores (hyena, fox, jackal, eagle etc.) and top carnivores (lion, leopard, tiger etc.). As the level
of energy decreases, the number of organisms at higher trophic levels also decrease. Each bar in
the energy pyramid indicates the amount of energy present at each trophic level in a given time or
annualy per unit area.
(b) There are certain limitations of ecological pyramids.
(a) It does not take into account the same species operating at two or more trophic levels.
(b) They assume that food chains are simple whereas in nature simple food chains do not occur.
Instead food webs are present.
(c) It does not accommodate a food-web.
(d) Detrivores and Decomposers are not given any place in ecological pyramids, though they play a
vital role in an ecosystem.
125. (a) Draw a simplified model of phosphorus cycling in a terrestrial ecosystem.
(b) Write the importance of such cycles in ecosystems
Or
(a) Explain the narrowly utilitarian, broadly utilitarian and ethical arguments in favour of
conservation of biodiversity.
(b) How is designation of certain areas as hotspots a step towards biodiversity conservation?
Name any two hotspots in India.
Ans. (a)
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DR. ARVIND'S BIOLOGY CLASSES
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A simplified model of phosphorus cycling in a terrestrial ecosystem.
(b) Importance: Organisms need a constant supply of energy (food) and nutrients to grow, reproduce
and regulate various body function. The amount of nutrients (i.e., C, H, P, S, N, Ca, etc.) present in the
soil at any given time, varies in different kinds of ecosystems and also on seasonal basis. Nutrient
cycling comprises of movements of nutrient elements through various components of an ecosystem i.e.,
abiotic  biotic  abiotic. They are never lost and are recycled time and again indefinitely.
or
(a) Need for conservation of Biodiversity: Reasons for conservation of biodiversity can be grouped
into three categories:
(i) narrowly utilitarian
(ii) broadly utilitarian and
(iii) ethical
(i) Narrowly Utilitarian: The reasons for conserving biodiversity are obvious because of their:
(a) direct economic benefits such as (1) food (cereals, pulses, fruits) (2) firewood (3) fibre (4)
construction material (5) products of medicinal importance (6) industrial products (tannins,
gums, lubricants, dyes, resins, perfumes).
(b) More than 25% of the drugs are derived from plants.
(c) 25,000 species of plants are used as traditional medicines by native people.
(ii) Broadly Utilitarian: Biodiversity plays a major role in providing ecosystem services that nature
provides and which cannot be given a price tag are:
(a) Production of Oxygen
(b) Pollination of flowers by bees, bumblebees, birds and bats, etc. resulting in the formation of
fruits and seeds
(c) Aesthetic pleasures like bird watching, walking through the thick forests etc.
(iii) Ethical:
(a) We share this planet with millions of plants, animals and microbe species. Every species has an
intrinsic value even if it is not of any economic value to us.
(b) We have an essential duty to care for their well-being and pass on the biological legacy in a
proper form to our future generations.
(b) Hot spots – On a global basis, conservationists have identified certain biodiversity rich regions, called
hot spots which need to be protected. These regions have - (1) Very high level of species richness
(2) High degree of endemism (species confined to a particular area and not found anywhere else). (3)
Accelerated habitat loss.
Hot-spots of India
(i) Western ghats and Sri Lanka region
(ii) Indo Burma region
(iii) Himalaya region
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DR. ARVIND'S BIOLOGY CLASSES
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126. Explain the process of sewage water treatment before it can be discharged into natural water
bodies. Why is this treatment essential?
Ans. Sewage treatment: Municipal waste water, commonly called sewage contains food residue, animal
and human excreta, detergents and discharges from commercial and industrial establishments. It also
has a number of pathogens like cholera, typhoid, dysentery causing pathogens.
(i) There is great need to treat sewage before it is discharged into natural water bodies like rivers and
streams because discharge of untreated sewage into water bodies will increase the BOD (Biological
Oxygen Demand) due to consumption of O2 by decomposers. This will cause death of aquatic
organisms due to lack of oxygen. Moreover, it will lead to spread of waterborne diseases.
(ii) Sewage is treated in sewage treatment plants (STPs). Treatment of waste water is done by
heterotrophic microbes naturally present in sewage. It involves two steps:
(a) Primary treatment: In this step, there is physical removal of large and small particles from the
sewage through filtration and sedimentation.
(i) The first step is the removal of floating debris by sequential filtration.
(ii) Then it is passed into grit chamber to remove soil and small pebbles by sedimentation. All
solids that settle down form the primary sludge and the supernatant forms the effluent.
(iii) The effluent is taken for secondary treatment.
(b) Secondary treatment or Biological treatment: It involves the following steps:
(i) Primary effluent is passed into large aeration tanks, where it is constantly agitated mechanically
and air is pumped into it.
This allows vigorous growth of useful aerobic microbes into flocs (masses of bacteria
associated with fungal filaments to form mesh like structures).
The microbes decompose the major part of the organic matter in the effluent and reduce the
BOD (biological oxygen demand) of the effluent.
BOD refers to the amount of oxygen that would be consumed if all the organic matter in one litre
of water were oxidised by bacteria.
(ii) When the BOD of sewage is reduced significantly the effluent is passed into a settling tank
where the bacterial flocs are allowed to sediment forming the activated sludge.
A small part of activated sludge is pumped back into the aeration tank to serve as the inoculum.
The remaining major part of activated sludge is pumped into large tanks called anaerobic
sludge digesters.
(iii) Here, other types of anaerobic bacteria digest the bacteria and fungi in the sludge producing
methane, hydrogen Sulphide and carbon dioxide, i.e., biogas, which can be used as source of
energy as it is inflammable.
The effluent from secondary treatment is generally released into rivers and streams.