Sex in plants

n 1.
Angiosperms bearing unisexual flowers are said to be either monoecious or dioecious.
Explain with the help of one example each.

Plant bearing flowers of both sexes, i.e. staminate and pistillate flowers called
monoecious, e.g. lea mays (maize).
When both sexes, i.e. staminate and pistillate flowers, are present on different plants;
these plants are called dioecious, e.g. Carica papaya (Papaya).

2.
These pictures show the gynoecium of (A) Papaver and (B) Michelia flowers. Write the
difference in the structure of their ovaries.

The gynoecium of Papaver is multicarpellary and syncarpous (pistils are fused

together), the ovary can be unilocular to multilocular. The gynoecium of Michelia is
multicarpellary and apocarpus (pistils are free) and the ovary is always unilocular.

3.
Name the parts of the flower which the tassels of corn cob represent.

The parts of the flower that represent the tassels of corn cob are stigma and style which
wave in the wind to trap pollen grains flowing with the wind.

5.
State the function of filiform apparatus found in mature embryo sac of an angiosperm.

The special cellular thickenings present in synergids at the micropylar tip called filiform
apparatus, found in mature embryo sac of an angiosperm help in guiding the entry of
pollen tubes up to the synergids.

6.
Give an example of a plant which came into India as a contaminant and is a cause of
pollen allergy.

Parthenium or Carrot grass is a major contaminant which came to India and caused
pollen allergy.
7.
A bilobed, dithecous anther has 100 microspore mother cells per microsporangium.
How many male gametophytes this anther can produce?
An anther is a four-sided (tetragonal) structure consisting of four microsporangia.
Each microsporangium has 100 microspore mother cells, so total number of microspore
mother cells in anther = 4 × 100 = 400 microspore mother cells. Meiosis in each
microspore mother cell produces 4 male gametes, so 400 cells will produce = 4 × 400 =
1600 male gametes.

8.
An anther with malfunctioning tapetum often fails to produce viable male gametophytes.
Give one reason. Or Write the function of tapetum in anthers.

The anther with malfunctioning tapetum cannot provide complete nutrition to the
developing microspores or male gametophytes. So, it fails to produce viable male
gametophyte.

9.
In the TS of a mature anther given below, identify ‘a’ and ‘b’ and mention their functions.

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

● Sporogenous tissue has cell which are potential Pollen Mother Cell (PMC) or
microspore mother cell and give rise to microspore tetrad after meiotic cell division.

● Tapetum nourishes the developing microspores or pollen grains.

10.
A pollen grain at the time of dehiscence from an anther could be 2-celled or 3-celled.
Explain, how are the cells placed within the pollen grain when shed at a 2-celled stage?

A pollen grain is partly germinated microspore representing the male gametophyte. It

divides by unequal mitotic division and forms two cells. Thus, each mature pollen grain
in angiosperms have a generative cell and a vegetative cell.

In about 60% of angiosperms, pollen grains are shed at this 2-celled stage. However, in
about 40% flowering plants, the generative cell may further divide mitotically to give rise
to two male gametes and pollen grains are shed at this 3-celled stage.
 11.
A mature embryo sac in a flowering plant may possess 7-cells, but 8-nuclei. Explain with
the help of a diagram only.

A typical angiospermic embryo sac is 8-nucleated and 7-celled.

12.
In a flowering plant, a microspore mother cell produces four male gametophytes while a
megaspore mother cell forms only one female gametophyte. Explain.

In flowering plants, microspore mother cells are found embedded in the spOrophytic
tissue of anther. These cells undergo meiosis and give rise to four microspores that
remain together in a microspore tetrad. After attaining maturity, these microspores
separate from each other and each microspore develops into a male gametophyte or
pollen grain.

On the other hand, megaspore mother cell develops in the ovary of a flower and divides
by meiotic division to produce four megaspores. From these, three degenerate while, the
one undergoes further development and mitotic divisions to produce female
gametophyte. Thus, in a flowering plant, a microspore mother cell produces four male
gametophytes while, megaspore mother cell produces one female gametophyte.

13.
Gynoecium of a flower may be apocarpous or syncarpous. Explain with the help of an
example each.

Gynoecium of a flower is called as apocarpous when the carpels are free, e.g. apocarpus
in Ranunculus. Whereas it is called syncarpous when the carpels are fused, e.g.
syncarpous in Petunia.

15.
Differentiate between the two cells enclosed in a mature male gametophyte of an
angiosperm.

Haploid pollen grains represent the male gametophyte. It contains two cells, i.e.
vegetative cell and generative cell.
The vegetative or tube cell is larger in size as compared to generative cell and have a
vacuolated cytoplasm. The generative cell on the other hand have thin dense cytoplasm
with prominent nuclei that give rise to two male gametes, while vegetative cell does not.

16.
How many haploid cells are present in mature female gametophyte of a flowering plant?
Name them.
An unfertilised embryo sac of angiosperm is composed of 7 cells, i.e. 7-celled and
8-nucleated. Among 8-nuclei, 6 are enclosed by cell walls and organised into cells, which
are haploid in number (3 antipodals, 2 synergids and 1 egg cell) and a large central cell
with 2 polar nuclei.

18.
Explain the function of germ pores.

Germ pores are prominent apertures of pollen grain exine where sporopollenin is
absent. These are the regions where intine comes out forming a pollen tube to release
male gamete in the embryo sac.

19.
Identify and label the parts in the given anatropous ovule.

A – Micropyle B – Outer integument

C – Inner integument D – Embryo sac

20.
How are pollens preserved in the pollen banks? Explain. How are such banks benefitting
our farmers? Write any two ways.

Pollen grains are stored for years in liquid nitrogen (-196°C) in pollen banks for later use
in plant breeding programmes. Plant breeding is a technique of manipulation of plant
species in order to create desired plant types that are better suited for cultivation, give
better yield and are disease resistant.
The objectives of such pollen banks include incorporation of certain traits or characters
into crop plants in order to enhance the food production such as
(i) Increased tolerance to environmental stresses such as salinity, extreme temperature,
drought, etc.
(ii) Resistance to pathogens like viruses, fungi and bacteria.

21.
(i) what organic material is exine made up of. Advantage of material?
(ii) exine does not form a continuous layer around the pollen grain. Give reason.
(iii) How are ‘pollen banks’ useful?

(i) The organic material of exine of pollen grain is sporopollenin. This is most resistant
biological material known so far. It protects pollen grains from damages.
(ii) Exine in pollen grain is not a continuous layer. It is thin at places and pollen tube
germinates by growth of intine through these thin parts of exine called germ pores.
(iii) Pollen banks are used to store pollen grains for short as well as long period of time
in viable conditions.

22.
Why are angiosperm anthers called dithecous?

Since, the angiosperm’s anthers are bilobed, each lobe bearing two thecae, they are
referred to as dithecous. Microsporangium appears circular in outline and is usually
surrounded by four wall layers. The outer three layers epidermis, endothecium and
middle layers are protective in function. They also help in dispersal of pollens by
dehiscing themselves. While, the innermost layer tapetum is nutritive in function and
nourishes the developing pollen grains. The centre of the microsporangium comprises of
compact sporogenous tissue. The cells of this sporogenous tissue undergo meiotic
divisions to form microspore tetrads, that further develop to form pollen grains.

24.
Explain the process of microsporogenesis in angiosperms.

The formation of microspores from a pollen mother cell through meiosis is called
microsporogenesis.

Microspores are arranged as tetrad. As the anthers mature and dehydrate they
dissociate from each other and develop into mature pollen grains. Pollen grains or the
male gametophytes are released by dehiscence of anther.

27.
Describe the structure of a 3-celled pollen grain of an angiosperm.

The 3-celled pollen grain structure in an angiosperm consists of two male gametes and
one vegetative cell. The vegetative cell is bigger, has abundant food reserve and a large
irregularly-shaped nucleus. The generative cell is small and floats in the cytoplasm of the
vegetative cell. It is spindle-shaped with dense cytoplasm and a nucleus. In over 60% of
angiosperms, pollen grains are shed at this 2-celled stage. In 40% of the angiospermic
species, the generative cell divides mitotically to give rise to the two male gametes
before pollen grains are shed (3-celled stage).
 28.
Describe the process of megasporogenesis up to fully developed embryo sac formation
in an angiosperm.

In angiosperms, the process of megasporogenesis starts inside the nucellus of the

ovule. During megasporogenesis, the Megaspore Mother Cell (MMC) undergoes meiosis
resulting in the production of four megaspores. Out of the four megaspores, only one is
functional while the other three degenerate.

The functional megaspore undergoes mitosis to form two nuclei, which migrate to
opposite poles, forming a 2-nucleate embryo sac.

Further, mitotic divisions lead to the formation of 4-nucleate and 8-nucleate stages of
the embryo sac. In these mitotic divisions, nuclear division is not followed by cell
division. After the 8-nucleate stage, cell walls are laid down and a typical female
gametophyte or embryo sac is formed.
Among the 8 nuclei, 6 are enclosed by cell walls and organised into cells, while the
remaining 2 nuclei (polar nuclei) are situated above the egg apparatus in a large central
cell.

Out of the six cells, three are grouped at the micropylar end and constitute the egg
apparatus. It is made up of two synergids and one egg cell. The other three cells are
located at the chalazal end and are called antipodals. Thus, a typical mature angiosperm
embryo sac is 8-nucleate and 7-celled.

30.
Trace pollen grain development from sporogenous tissue in the anther.

(i) Development of pollen grain from Pollen Mother Cell (PMC)

● Pollen mother cell or microspore mother cell undergoes meiosis to form

microspore tetrad or haploid microspores.

● As the anther matures, the microspores dissociate from the tetrad and develop
into pollen grains.

● Nucleus of the microspores undergoes mitosis to form a large vegetative cell and
small spindle-shaped generative cell.

● They develop a two-layered wall, the outer exine made of sporopollenin and the
inner intine made of cellulose and pectin.

● Usually the pollen grains are liberated at this 2-celled stage.

● In certain species, the generative cell divides mitotically to form two male gametes
and the pollen grains are 3-celled during liberation.

32.
(i) Describe the formation of mature female gametophyte within an ovule in
angiosperms.
(ii) Describe the structure of cell that guides the pollen tube to enter the embryo sac.

(i) The functional megaspore undergoes mitosis to form 2 nuclei, which migrate to
opposite poles, forming a 2-nucleate embryo sac. Further, mitotic divisions lead to the
formation of 4-nucleate and 8-nucleate stages of the embryo sac. In these mitotic
divisions, nuclear division is not followed by cell wall formation. After the 8-nucleate
stage, cell walls are laid down and a typical female gametophyte or embryo sac is
formed. Among the eight nuclei, six are enclosed by cell wall and organised into cells,
while the remaining two nuclei (polar nuclei) are situated above the egg apparatus in a
large central cell.

Out of the six cells, three are grouped at the micropylar end and constitute the egg
apparatus made up of two synergids and one egg cell. The other three cells are located
at the chalazal end and are called antipodals. Thus, a typical angiosperm embryo sac
after maturity is 8-nucleate and 7-celled.

(ii) The egg apparatus present towards the micropylar end, comprises of two synergids
and an egg cell.
These synergids possess special cellular thickenings at their micropylar tip and called
filiform apparatus. This filiform apparatus guides the pollen tube to enter into embryo
sac.

35.
Mention the characteristic features and function of tapetum.
Tapetum is the inner nourishing layer of microsporangial wall. The cells of tapetum have
dense cytoplasm and more than one nucleus. These cells nourish the developing pollen
grains.

Explain the following giving reasons

(a) Pollen grains are well-preserved as fossils.
(b) Pollen tablets are in use of people these days.

(a) The outer exine layer of pollen grain is highly resistant because of sporopollenin. It is
an organic material which can withstand harsh conditions, action of alkalis and acids.
No enzyme can degrade sporopollenin. Thus, pollen grains are well-preserved as fossils.

(b) Pollen grains are rich in nutrients. So, used by people as health tablets or food
supplements

36.
Write one advantage and one disadvantage of cleistogamy to flowering plants.

The advantage of cleistogamy is that it ensures pollination in the absence of pollinators.

Disadvantage of cleistogamy is that there is no chance of variation to occur.

37.
What is pollen-pistil interaction and how is it mediated?

Pollen-pistil interaction is a chain or group of events that take place from the falling of
pollen over the stigma to the formation of pollen tube and its entry into the ovule. It is
mediated by chemical components of pollen grain, interacting with that of pistil.

38.
Differentiate between xenogamy and geitonogamy.

Xenogamy is the transfer of pollen grains from anther of one flower to the stigma of
another flower of a different plant, while geitonogamy is the transfer of pollen grains
from anther of one flower to the stigma of another flower on same plant.

39.
Why do the pollen grains of Vallisneria have a mucilaginous covering?

As the pollination of Vallisneria takes place by means of water, the pollen grains are
covered by mucilaginous coating that protects them from damage and desiccation.
40.
What is cleistogamy? Write one advantage and one disadvantage of it, to the plant.

Cleistogamy is a type of self-pollination that occurs in a permanently closed flower.

Advantage and disadvantage of cleistogamy are as follows

● Advantage Cleistogamous flowers produce assured seed-set even in the absence

of pollinators.

● Disadvantage Cleistogamous flowers are invariably autogamous. So, there is no

chance of cross-pollination. Hence, less variations are generated in the progeny.

41.
You are conducting artificial hybridisation on papaya and potato. Which one of them
would require the step of emasculation and why ? However for both you will use the
process of bagging. Justify giving one reason.

Papaya produces unisexual flowers and potato produces bisexual flowers. Therefore, the
step of emasculation will be done on potato because emasculation is done on bisexual
flower to avoid self-pollination. But, bagging is done on unisexual flowers, so to dust
suitable pollen grains on the stigma when the stigma turns receptive and the flowers are
rebagged.

42.
Express the process of pollination in Vallisneria.

Vallisneria is a water pollinated plant. In this plant, the process of pollination involves
reaching of female flower at the surface of water by the long stalk and release of pollen
grains onto the surface of water. These pollen grains are carried by water currents to
reach the stigma eventually.

43.
A single pea plant produces pods with viable seeds, but the individual papaya plant does
not. Explain.

A single pea plant produces pods with viable seeds because the pea plant is
autogamous, i.e they have the ability of self-pollination. Whereas the individual papaya
plant is prevented from both autogamy and geitonogamy. In this plant, male and female
flowers are present on different plants, i.e. each plant is either male or female(diecious)

45.
List the different types of pollination depending upon the source of pollen grain.

Depending on the source of pollen grain, pollination can be classified into

● Autogamy; It is the transfer of pollen grain from anther to the stigma of the same
flower.
● Geitonogamy It is the transfer of pollen grains from anther of one flower to the
stigma of another flower on the same plant. Geitonogamy is functionally
cross-pollination involving pollinating agent, but genetically it is equivalent to autogamy
since the pollen grains come from the same plant.

● Xenogamy It is the transfer of pollen grains from anther to the stigma of different
plants of same species. It brings genetically different types of pollen grains to the
stigma.

46.
In angiosperms, zygote is diploid, while primary endosperm cell is triploid. Explain.

In angiosperms or flowering plants, zygote is diploid and primary endosperm nucleus is

triploid. It is because in these plants, one of the male gametes fuses with egg cell, which
results in the formation of zygote. So, zygote is diploid. While primary endosperm cell is
triploid because the nucleus of the second male gamete (n) fuses with the two haploid
polar nuclei or diploid secondary nucleus (2n) of the central cell to form a triploid
primary endosperm nucleus (3n). This process is referred to as triple fusion. The central
cell is now called primary endosperm cell.

49.
Why should a bisexual flower be emasculated and bagged prior to artificial pollination?

Emasculation in a bisexual flower is required to prevent contamination of the stigma

with self-pollen grains. Bagging is done to prevent contamination of the stigma of the
emasculated flower with any other unwanted pollen grains. That is why a bisexual flower
should be emasculated and bagged prior to artificial pollination. Emasculation is not
required in unisexual flowers.

50.
Write the functions of the following
(a) Synergids
(b) Micropyle

(a)Synergids These possess special cellular thickenings at their micropylar tip called
filiform apparatus. This filiform apparatus guides the pollen tube to enter embryo sac.
(b) Micropyle facilitates the entry of pollen tube and thus fertilisation.

51.
Emasculation and bagging are the two important steps carried during artificial
hybridisation to obtain superior varieties of desired plants. Explain giving reasons

If the plant bears bisexual flowers, emasculation and bagging are carried out before the
anther dehisces.
If the plant bears unisexual flowers, emasculation is not required. The female flower
buds are bagged before the opening of flowers are

It is the transfer of pollen grains from the anther to the stigma of another flower of same
plant.

nogamy gamy

he transfer of pollen grains from the anther he

t transfer of pollen grains from the anther to the
igma of another flower of same plant. a of different plants.

ollen grains are genetically similar to the plaollen grains are genetically different from the pla

52.
(i) Differentiate between geitonogamy and xenogamy.

(i) geitonogamy result in purelines, e.g. homozygous. They are genetically similar.
Xenogamy results in hybrids, e.g. heterozygous. They show variations in characters.

54.
(i) Can a plant flowering in Mumbai be pollinated by pollen grains of the same species
growing in New Delhi?
(ii) Draw the diagram of a pistil where pollination has successfully occurred. Label the
parts involved in reaching the male gametes to its desired destination.

(i) Yes, a plant flowering in Mumbai can be pollinated by pollen grains of the same
species growing in New Delhi. It is mainly because there are certain agents of pollination
that can carry pollen grains to long distance. Plants can use either abiotic or biotic
agents for pollination. Abiotic pollinators include wind and water while biotic pollinators
are insects, birds,
(ii) The parts involved in transferring the male gametes to its desired destination are
stigma, style, micropyle, filiform apparatus and synergids.

55.
What does an interaction between pollen grains and its compatible stigma result in after
pollination? List two steps in sequence that follow after the process.

When the pollen grains fall on the stigma, the pollen tube enters one of the synergids
and releases two male gametes.

● One of the male gametes moves towards the egg cell and fuses with it to complete
syngamy to form the zygote.

● The other male gamete fuses with the two polar nuclei and forms triploid Primary
Endosperm Nucleus (PEN). This is termed as triple fusion.

● Since, two kinds of fusion syngamy and triple fusion takes place, the process is
known as double fertilisation and is characteristics of flowering plants.

56.
As a senior biology student you have been asked to demonstrate to the students of
secondary level in your school, the procedure(s) that shall ensure cross-pollination in a
hermaphrodite flower. List the different steps that you would suggest and provide
reasons for each one of them. (All India 2016)

Cross-pollination is done to mix two desired characters of two different species of a

plant. For example, purple and white flower of a pea.

● Select two pea plants one with white and other with purple flower.

● Label them as male (white flowered) and female (purple flowered) plant.

● Cut anthers from purple flower with the help of scissors before their dehiscence to
avoid self-pollination and cover it with white paper bag.
● Now collect pollens from the white flower (male plant) with the help of brush.

● Dust the pollens on the stigma of female (purple fewer) flower.

● Cover it again with paper bag till seed formation.

57.
Make a list of any three outbreeding devices that flowering plants have developed and
explain how they help to encourage cross-pollination.

Hermaphrodites or bisexual flowers develop outbreeding devices to ensure

cross-pollination and avoid self-pollination. The three outbreeding devices that flowering
plants have developed to discourage self-pollination are
(i) Unisexuality ;Flowers are unisexual, so that self-pollination is not possible. The plants
may be monoecious (bearing both male and female flowers, e.g. maize) or dioecious
(bearing male and female flowers on different plants, e.g. mulberry, papaya).

(ii) Dichogamy Anthers and stigmas mature at different times in a bisexual flower for
preventing self-pollination.
(a) Protandry Anthers mature earlier than stigma of the same flower. The pollens thus
become available to stigmas of the older flowers, e.g. sunflower, Salvia.
(b) Protogyny Stigmas mature earlier, so that they get pollinated before the anthers of
the same flower develop pollen grains, e.g. Mirabilis jalapa,
Gloriosa, Plantago.

(iii)self-incompatibility. It is a genetic mechanism that prevents self-pollen from

fertilising the ovules by preventing pollen germination or pollen tube growth in the pistil.

58.
Explain the phenomenon of double fertilisation.

The phenomenon of double fertilisation occurs in following steps

● In an angiospermic plant, two male gametes are discharged by a pollen tube into
the cytoplasm of a synergid of the embryo sac.

● One of the male gametes fuses with the egg to form a zygote. This process is
called syngamy.

● Other male gamete fuses with the secondary nucleus to form the primary
endosperm nucleus, this process is called triple fusion.

● Since, there are two fusions (syngamy and triple fusion) inside an ovule during
fertilisation, it is known as double fertilisation.

59.
Write the differences between wind pollinated and insect pollinated flowers. Give an
example of each type.
The differences between wind pollinated and insect pollinated flowers are

pollinated flowers t pollinated flowers

e are small. are either large or grouped to form large clusters.

ly inconspicuous due to dull colouresence of bright colours in corolla, calyx or bracts to

ct insects.

are odourless and devoid of nectargly odoured and usually possess nectar or edible polle

ns are produced in large numbers. r pollen grains are produced.

rtica, Maize, Parthenium. ose, Snapdragon, Calotropis.

60.
what led to inbreeding depression and why?

Geitonogamy will lead to inbreeding depression because the pollen grains are genetically
similar resulting into inbreeding. Continuous inbreeding reduces fertility.

61.
(i) Write the characteristic features of anther, pollen and stigma of wind pollinated
flowers.
(ii) How do flowers reward their insect pollinator? Explain.

(i) In wind pollinated flowers,

● Anthers are well-exposed for easy dispersal of pollen grains.

● Pollen grains are light and non-sticky, so that they can be transported by wind
currents.

● Stigma is large and feathery to trap pollens.

(ii) Flower rewards their insect pollinators easily by offering

● Nectar and edible pollen grains.

● Safe place for insects to lay eggs by some flowers, e.g. Amorphophallus and
Yucca.

62.
(i) Describe any two devices in a flowering plant which prevent both autogamy and
geitonogamy.
(ii) Explain the events up to double fertilisation after the pollen tube enters one of the
synergids in an ovule of an angiosperm.

(i) The two devices that prevent both autogamy and geitonogamy in flowering plants are
as follows
(a) Self-incompatibility In some plants when pollen from same flower or other flower of
the same plant comes on the stigma, it is incapable of bringing about fertilisation.
It is due to the presence of similar self-sterile gene, e.g. tobacco, potato, etc. It prevents
autogamy and geitonogamy.

(b) Dioecy In several species such as papaya, male and female flowers are present on
different plants. Thus, each plant is either male or female. This condition also prevents
both autogamy and geitonogamy.

(ii) In the ovule, the pollen tube is attracted by secretions of synergids. Usually the
pollen tube enters the embryo sac by passing into one of the two synergids which starts
degenerating. The pollen tube bursts up by absorbing hydrolytic substances secreted by
degenerating synergids. It is followed by double fertilisation in flowering plants.

Double fertilisation It is the fusion of two male gametes to two different cells of the
same female gametophyte in order to produce two different structures.

● Syngamy of the egg nucleus with one male gamete is called syngamy. This fusion
results in the formation of diploid cell, the zygote.

● Triple fusion Along with syngamy, the other’male gamete moves towards the two
polar nuclei located in the central cell and fuses with them to produce a triploid Primary
Endosperm Mother (PEM) cell. In this way, fertilisation occurs in flowering plants.

64.
(i) Plan an experiment and prepare a flow chart to ensure that only desired seeds are
formed from the sets of pollen grains. Name the experiment (ii) importance of such
experiments.

(i) Artificial hybridisation is carried out to ensure that seeds are formed from the desired
set of pollen grains. This is done by emasculation and bagging.
The flow chart below shows the steps to be followed

(ii) Importance of such experiments are

(a) Creation of new genetic recombination with better qualities.
(b) Incorporation of a large number of desirable characters into a single variety.

65.
flowers may be monoecious, cleistogamous or show self-incompatibility. Describe
features and state which one s promotes inbreeding and outbreeding.

The characteristic features of flowers

(i) Monoecious flowers are unisexual, i.e. they have either the male reproductive or
female reproductive part in separate flowers, both produced on same plant. The flowers
(male and female) are separate. It prevents self¬pollination and promotes
cross-pollination.

(ii) Cleistogamous ; flowers in which anthers and stigmas lie close to each other and do
not open at all, even at maturity. These flowers are invariably autogamous and promote
inbreeding depression as there is no chance for cross-pollination at all.

(iii) Self-incompatible In angiospermic flowers, there is a genetic mechanism, wherein

the flowers prevent the self-pollens from fertilising the ovules or inhibit their
germination on stigma. This device or mechanism promotes outbreeding.

66.

Explain the events that occur, up to fertilisation, when the compatible pollen grain lands
on the stigma.

The events that occur when compatible pollen grains fall on stigma in the sequence are
as follows
(a) Pollen-pistil interaction ;Once the compatible pollen grains fall on stigma which is
receptive, it recognises and accepts the pollen with the aid of chemical components
interacting with pollen.

(b) Germination of pollen grain Once the pollen is recognised, it germinates on the
stigma of flower. The tube cell of pollen grain protrudes out through germ pores to form
a pollen tube. The generative cell divides to form two male gametes that are released
into the tube.

(c) Growth of pollen tube The pollen tube grows down through the tissues of stigma and
style and enters ovule, usually through micropyle. Inside ovule, the filiform apparatus
guides the pollen tube, carrying gametes to the egg cell.

(d) Double fertilisation After releasing the two male gametes into the synergids, one of
them fuses with egg to form a diploid zygote (syngamy) and other male gamete fuses
with 2 polar nuclei to form triploid primary endosperm cell (triple fusion). Because of the
occurrence of these two types of fusions, it is called double fertilisation.

67.
Why is fertilisation in an angiosperm referred to as double fertilisation? Mention the
ploidy of the cells involved.

In fertilisation (in angiosperm), two types of fusion occur, i.e. syngamy and triple fusion,
in the embryo sac. That is why it is called double fertilisation.
Ploidy of cells involved in double fertilisation Zygote is diploid (2n). It is formed as a
result of syngamy, i.e. fusion of two haploid gametes (male gamete + egg). Primary
endosperm nucleus (3M) is formed as a result of triple fusion, i.,e. fusion of two haploid
polar nuclei with male gamete.

69.
Geitonogamy is functionally a cross-pollination, but genetically similar to autogamy.
Explain.

Transfer of pollen grains from the anther to stigma of another flower of the same plant is
called geitonogamy. It is functionally cross-pollination as it involves a pollinating agent,
but genetically similar to autogamy since, the pollen grains come from the same plant
(genetically same parent).

Continued self-pollination results in inbreeding depression because majority of flowering

plants produce hermaphrodite flowers and pollen grains generally come in contact with
the stigma of same flower.

70.
Given below is a section of maize grain. Identify A and state its function.
A is endosperm. It provides nutrition to the developing embryo.

71.
Identify A in the figure showing a stage of embryo development in a dicot plant and

mention its function.

A is cotyledon. It is the storehouse of food.

72.
Mention the function of coleorhiza.

Coleorhiza is a protective sheath covering the young root of the embryo in plants of the
grass family.

73.
The meiocyte of rice has 24 chromosomes. Write the number of chromosomes in its
endosperm.

The meiocyte is a diploid cell and have 24 chromosomes. Thus, its haploid chromosome
number is 12.
Number of chromosomes in endosperm is
12 × 3 = 36

74.
Write the function of scutellum.

The cotyledon of embryo of grass family is called scutellum. It is an embryonic leaf.

75.
Banana is a true fruit but is also a parthenocarpic fruit. Give reason.
The fruit of banana is formed from the ovary, so it is a true fruit. It is a parthenocarpic
fruit because the ovary develops into fruit without fertilisation and is thus seedless.

76.
Why is apple referred to as a false fruit?

In apple, the thalamus also contributes to fruit formation. So, apples are called false
fruits.

77.
Name the mechanism responsible for the formation of seed without fertilisation in
angiosperms. Give an example

Apomixis is the mechanism responsible for the formation of seeds without fertilisation
in angiosperms, e.g. grasses.

78.
It is said apomixis is a type of asexual reproduction. Justify.

Through apomixis, viable seeds can be produced without fertilisation and zygote
formation through gametic fusion. This is not the case during sexual reproduction. In
sexual reproduction, seeds are produced through gametic fusion following fertilisation.
So, apomixis is called a type of asexual reproduction.

79.
Name the type of fruit apple is and why? Mention two other examples

Apple is categorised as false fruit because it does not develop from the ovary, but
thalamus. Cashew and strawberry belong to the same category.

80.
Write the difference between the tender coconut water and the thick, white kernel of a
mature coconut and their ploidy.

The tender coconut water represents the free nuclear endosperm while, the white kernel
is the cellular endosperm. The water and kernel of the endosperm are both triploid.

81.
Suggest two advantages to a farmer using apomictic seeds of hybrid varieties.

Two advantages of apomictic seeds to a farmer are as follows

● It lowers the cost of production.

● Apomictic seeds do not have to be produced every year.

82.
List the post-fertilisation events in angiosperms.

The post-fertilisation events in angiosperms include

● Endosperm and embryo development.

● Maturation of ovule into seed.

● Maturation of ovary into fruit.

84.
Write the fate of egg cell and polar nuclei after fertilisation.

The egg cell after fertilisation with one of the two male nuclei forms zygote which
further develops into an embryo. Polar nuclei fuse with other male nuclei to form primary
endosperm nucleus which further develops into endosperm. In case of coconut, the
endosperm is watery.

85.
Some seeds are ‘albuminous’, whereas others are perispermic. explain

Some angiospermic seeds are albuminous as they retain endosperm even after embryo
development, i.e. not completely consumed by embryo, e.g. wheat, maize, castor.
nucellus are persistent which is referred to as perisperm, e.g. black pepper and beet.

86.
Differentiate between albuminous and non-albuminous seeds, one example each.

minous seed albuminous seed

sperm is not completely consumed by thesperm is completely consumed by the developi

oping embryo, so a portion of it remains iyo before the maturation of seed, so there is no
eed. sperm left in the seed.

oconut, castor and maize. ea, bean and mustard.

87.
(i) Given below is a TS of an apple.
(i) In the given figure of TS of an apple,
A – Thalamus, B – Seed, C – Endocarp

89.
Differentiate between parthenocarpy and parthenogenesis. Give example of each.

The differences between parthenocarpy and parthenogenesis are

enocarpy enogenesis

he phenomenon of formation of fruits

he phenomenon in which the unfertilised female ga
ut fertilisation. m) develops into an adult.

urs in plants only. urs in plants and animals both.

produced by this procedure are of duals produced by parthenocarpy are usually weak
al type. e.g. Pineapple fruits. viable, e.g. Plants-Datura, Animals- drones (honeybe

90.
If the meiocyte of a maize plant contains 20 chromosomes. Write the number of
chromosomes in the endosperm and embyro of the maize grain and give reasons in
support of your .

If the meiocyte of a maize plant contains 20 chromosomes, then the number of

chromosomes in the endosperm and the embryo of maize grain will be 30 and 20,
respectively.
The meiocyte is a diploid cell and have 20 chromosomes. Thus, its haploid
chromosome is 10.
Number of chromosome in endosperm is 10 × 3 = 30.
Number of chromosomes in embryo is 2n,
2 × 10 = 20.

91.
Do you think apomixis can be compared with asexual reproduction?give reason.
Yes, apomixis can be compared with asexual reproduction. It is also called a form of
asexual reproduction that mimics sexual reproduction.
In apomixis, seeds are produced without fertilisation and zygote formation through
gametic fusion. So, it can be called as a form of asexual reproduction.

92.
How are parthenocarpic fruits produced by some and apomictic seeds by other?.

Parthenocarpic Seeds
Some fruits develop without undergoing fertilisation, these are called parthenocarpic
fruits, e.g. banana and this process of formation of fruit without fertilisation is called
parthenocarpy. It can be induced through the application of growth hormones and such
fruits are seedless.

Apomictic Seeds
It is a form of asexual reproduction that mimics sexual reproduction, but produces viable
seeds without fertilisation. It does not involve formation of zygote through the gametic
fusion. It occurs in some species of Asteraceae and grasses. In some species, the
diploid egg cell is formed without reduction division and develops into embryo without
fertilisation. It is an asexual reproduction in the absence of pollinators and takes place in
extreme environments.In some species like Citrus, some of the nucellar cells
surrounding the embryo sac start dividing and develop into embryo. It occurs in the
megaspore mother cell that does not undergo meiosis, thus produces diploid embryo
sac through mitotic divisions.

Farmers pick apomictic seeds in plants which do not reproduce sexually. So that farmers
use this method for the production of cloned seed.

93.
Significance of apomixis, How can apomixis be commercially used?

Significance

● It helps in fixing heterosis or hybrid vigour in plants permanently.

Rapid multiplication of genetically uniform individuals can be achieved without risk of
segregation.

Commercial use Apomixis is used in plant breeding. It increases the chance of

developing superior gene combinations and facilitates the rapid incorporation of
desirable traits.

94.
Double fertilisation is reported in both, castor and groundnut. However, the mature seeds
of groundnut are non-albuminous and castor are albuminous. Explain the
post-fertilisation events that are responsible for it.
Double fertilisation is reported in both castor and groundnut, but their mature seeds are
different in terms of endosperm. The primary endosperm nucleus formed after
fertilisation divides mitotically without cytokinesis to initiate the formation of
endosperm. At this stage, the endosperm is called free nuclear endosperm.

Then, cell wall formation occurs and the endosperm becomes cellular type. The number
of free nuclei formed before cellularisation varies greatly. Endosperm may be completely
utilised by the developing embryo before the maturation of seeds as in groundnut. Such
seeds are called non-albuminous or non-endospermic seeds.

When a portion of endosperm remains in seeds and is used up during seed germination,
such seeds are called albuminous or endospermic seeds, e.g. castor.

95.
Describe endosperm development in angiosperm.

Endosperm development takes place by three methods

(i) In nuclear type, which is a common method, the Primary Endosperm Nucleus (PEN)
undergoes repeated mitotic division without cytokinesis. At this stage, the endosperm is
called free nuclear endosperm.

(ii) In cellular type, cell wall formation occurs and the endosperm becomes cellular. The
number of free nuclei formed before cellularisation varies greatly, e.g. in coconut, the
water is free nuclear endosperm and surrounding white kernel is cellular endosperm.

(iii) In helobial type endosperm formation, one half of endosperm is nuclear type and
other half is cellular type.

96.
(i) How is apomixis different from parthenocarpy?
(ii) Describe any two modes by which apomictic seeds can be produced.

(i) main difference between apomixis and parthenocarpy is that seeds are formed in
former, while absent in later.

(ii) The two modes by which apomictic seeds can be produced are
(a) Agamospermy in which the seed or embryo is derived from diploid egg cell, formed
without meiosis and syngamy. This diploid egg cell develops into embryo without
undergoing fertilisation, e.g. apple, Rubus.
(b) Adventive embryony ;The method in which diploid cells surrounding the embryo sac,
e.g nucellus and integument protrude into the sac and develop into embryo. This may
also lead to the formation of more than one embryos in an embryo sac or ovule, leading
to condition called polyembryony, e.g. Citrus, Opuntia.

98.
(i) Describe the endosperm development in coconut.
(ii) Why is tender coconut considered as healthy source of nutrition?
(iii) How are pea seeds different from castor seeds with respect to endosperm?
(i) Coconut endosperm formation is nuclear type.
The primary endosperm nucleus undergoes free nuclear division without cell wall
formation.
(ii) Soft coconut is an endosperm. It is rich in nutrients like fats, proteins, carbohydrates,
minerals, vitamins, etc. Hence, it is considered as a healthy source of nutrition.
(iii) The seeds of pea are non-endospermic, while castor seeds are endospermic. The
endosperm in pea seeds is consumed completely during embryo development, but
endosperm is not utilised in castor seeds.

99.
Differentiate between perisperm and endosperm giving one example of each. (
Differences between perisperm and endosperm are

perm sperm

resents persistent remains of nucellus (of ovuleelops from Primary Endosperm Nucleus (PEN
seed.

part that belongs to seed. tains reserve food materials for developing se

sually dry. sually in fluid form or soft.

lack pepper. water of coconut.

101.
With the help of an example of each explain the following Apomixis, Parthenocarpy,
Polyembryony.

Apomixis The phenomenon in which seeds are produced without fertilisation is called
apomixis or agamospermy, e.g. grass.
Parthenocarpy It is a commercially important process in which seedless fruit is formed
without fertilisation, e.g. banana.
Polyembryony The occurrence of more than one embryo in a seed is known as
polyembryony, e.g. orange.

102.
(i) Explain any two ways by which apomictic seed can develop.
(ii) List one advantage and one disadvantage of an apomictic crop.

(i) Apomictic seed can develop in two ways as follows

(a) These are produced from segments of fruit (mango stem), male gametic content of
pollen (Cyperus) and other vegetative parts.
(b) In some species, the diploid egg cell is formed without reduction division and
develops into embryo without fertilisation.

(ii) Advantage of apomictic crop These are the products of asexual reproduction in the
absence of pollinators and takes place in extreme environment.
Disadvantage of apomictic crop During the production of apomictic crop, there is no
segregation of characters in the hybrid progeny.

103.
(i) When a seed of an orange is squeezed, many embryos, instead of one are observed.
Explain, how it is possible.
(ii) Are these embryos genetically similar or different? Comment.

(i) It is true that when we squeeze a seed of an orange, many embryos, instead of one
are observed. It occurs mainly due to a process named polyembryony. It is a
phenomenon of occurrence of more than one embryo in a seed. Polyembryony can be
spontaneous or induced experimentally. The polyembryony may arise by the following
reasons Formation of additional embryos from synergids or from antipodals and polar
nuclei (very rare). Some embryos can also be derived by the activation of some
sporophytic cells of ovule such as nucellus or integument.
Embryos can also be developed from an additional embryo sac in the same ovule.

(ii) The embryos formed as a result of polyembryony are genetically similar to one
another. However, the embryos arising from gametophytic tissues are similar to each
other, but not to their parents.

105.
Why does endosperm development precede embryo development in angiosperm seeds?
State the role of endosperm in mature albuminous seeds.

The embryo development starts only after a adaptation for assured nutrition of the
developing embryo therefore, endosperm development precedes embryo development.
The role of endosperm in mature albuminous seeds is storage of reserve food for
growing embryo.

106.
(i) Mature seeds of legumes are non-albuminous. Then, can it be assumed that double
fertilisation does occur in legumes? Explain your .
(ii) List the differences between the embryos of dicot (pea) and monocot (grass)
families.

(i) Seeds of legumes are non-albuminous that implies that endosperm in such seeds is
completely used up in providing nutrition to developing embryo. The endosperm is
formed as a result of triploid fusion, i.e. between a male gamete and two polar nuclei.
This makes it obvious that it cannot be formed in the absence of double fertilisation.
Therefore, though the seeds of legumes are non-albuminous, it clearly states the
occurrence of double fertilisation in them.

107.
(i) Why are seeds of some grasses called apomictic? Explain.

(i) The seeds of some grasses develop without fertilisation. It may be because a diploid
egg cell develops into an embryo directly (without undergoing meiosis and syngamy) or
some diploid cells of nucellus or integument surrounding the embryo sac, protrude
inside and develop into embryos. This phenomenon of developing embryo and seeds
without fertilisation is called apomixis and such seeds produced are referred to as
apomictic.

108.
Give reasons why?
(i) Most zygotes divide only after certain amount of endosperm is formed.
(ii) Groundnut seeds are exalbuminous and castor seeds are albuminous.
(iii) Micropyle remains as a small pore in the seed coat of a seed.
(iv) Integuments of an ovule hardens and the water content is highly reduced as the seed
matures.
(v) Apple and cashewnuts are not called true fruits.

(i) Zygotes in angiosperms mostly divide only after a certain amount of endosperm is
formed as an adaptation strategy to assure nutrition for the developing embryo.

(ii) Groundnut seeds are exalbuminous because the developing embryo utilises the
endosperm completely. So, there is no endosperm left in the seed.
Castor seeds are albuminous because endosperm is not completely used up by the
developing embryo. There is some amount of endosperm left in the seeds always.

(iii) Micropyle allows the entry of water and oxygen during seed germination.

(iv) During unfavourable conditions, seeds become dormant. The loss of water reduces
the metabolic activity of seeds and hardens the integuments.

(v) In these fruits, thalamus contributes in fruit formation. So, they are not called true
fruits.

109.
How are seeds advantageous to flowering plants?

Advantages of seeds to flowering plants are

● Provide protection to embryo in most delicate stage.

● Help in dispersal and spread in new habitats.

● Contain sufficient food reserves.

● Produce genetic variations.

● Seeds are related to pollination and fertilisation.

110.
Explain the development of the zygote into an embryo and of the primary endospermic
nucleus into an endosperm in a fertilised embryo sac of a dicot plant.

Embryo Development in Dicot Seed

● Embryo formation starts after a certain amount of endosperm is formed.

● Zygote divides by mitosis to form a proembryo.

● Formation of globular and heart-shaped embryo occurs, which finally becomes

mature embryo.

● In dicot plant, embryo consists of two cotyledons and an embryonal axis between
them.

● The portion of embryonal axis above the level of attachment of cotyledons is

epicotyl and terminates in the plumule.

● The portion of embryonal axis below the level of attachment of cotyledon is the
hypocotyl, it becomes radicle (root tip).

111.
Normally one embryo develops in one seed but when in orange seed many embryos of
different shape and size are seen. how has it happened.

In orange seed, embryos originate by adventive embryony from diploid cells of nucellus
or integuments and thus in orange seed many embryos of different sizes can be
observed.

112.
Name two end products of double fertilisation in angiosperms. How are they formed?
Write their fate during the development of seed.

The two end products of double fertilisation in angiosperms are diploid zygote and a
triploid Primary Endosperm Nucleus (PEN).
Diploid zygote is formed by the fusion of haploid gametes, i.e. male gamete and egg,
while another male gamete and two polar nuclei of central cell fuse to form triploid
primary endosperm nucleus.

During the development of seed, the zygote undergoes mitotic divisions to form a
mature embryo while, the primary endosperm cell gives rise to nutritive tissue called
endosperm, which provides nourishment to growing embryo.
 113.
List the parts of a typical dicot embryo.
Parts of a typical dicot embryo are
(a) Plumule (b) Radicle (c) Cotyledons (d) Hypocotyl

114.
Read the following statement and the s that follows.
‘A guaVa fruit has 200 viable seeds’.
(i) What are viable seeds?
(ii) Write the total number of
(a) Pollen grains
(b) Gametes are producing 200 viable guava seeds.
(iii) Prepare a flow chart to depict the post-pollination events leading to viable seed
production in a flowering plant. (Delhi 2017)

(i) Viable seeds are seeds having the ability to germinate when favourable conditions are
present.
(ii) (a) Number of pollen grains = 200
(b) Gametes are producing 200 viable guava seeds = 400
(iii) Post-pollination events include
(a) Double fertilisation (b) Formation of embryo (c) Formation of endosperm

115.
A flower has 520 ovules. However, it produces 480 viable seeds.
(i) What could have prevented the rest of the 40 ovules from maturing into viable seeds?
Explain giving a reason.
(iii) Why certain angiospermic seeds are albuminous, while others are exalbuminous

(i) Ovules can develop into viable seeds having the ability to germinate under suitable
environmental conditions, however, often ovules do not develop in viable seeds because
of the following reasons

● Excessive dry weather or high temperature.

● Damage to embryo.

● Starvation due to exhaustion of food in ovule.

(iii) In flowering plants, during the development of seeds, along with embryo, a nutritive
tissue called endosperm is also formed. During embryo development, endosperm
degenerates to release its content that helps in the growth of embryo. In some plants,
the endosperm is completely used up in embryo development.
In such seed, endosperm is absent, e.g. pea, gram, beans, etc. These seeds are called
exalbuminous seeds. On the other hand, in some plants, the endosperm tissue persists
in the viable seeds. These are called as albuminous seeds, e.g. castor.
 116.
(i) Explain the post-pollination events leading to seed production in angiosperms.
(ii) List the different types of pollination depending upon the source of pollen grain.

(i) Post-pollination events include pollen-pistil interaction, double fertilisation, formation

of endosperm, development of embryo, seed and fruit.

Development of an endosperm Endosperm development precedes embryo development.

The primary endosperm cell divides repeatedly to form a triploid endosperm tissue.

In most of the cases, the PEN undergoes successive nuclear divisions without
cytokinesis, to give rise to free nuclear endosperm. Subsequently, cell wall formation
starts from the periphery and the endosperm becomes completely cellular, e.g. coconut,
rice, maize, sunflower, etc.

Development of an embryo Embryo develops at the micropylar end of the embryo sac
where the zygote is situated. Most zygotes divide only after certain amount of
endosperm is formed.

The nutrition for the development of embryo is provided by the endosperm. The zygote
gives rise to proembryo and subsequently to the globular, heart-shaped and mature
embryo. Development of a seed Double fertilisation in angiosperms triggers the
transformation of ovule into a seed. Seeds are formed inside the fruits. Formation of a
fruit A fruit is formed as a result of cell division and differentiation in the ovary, which is
transformed into fruit as a result of stimuli received from pollination as well as from
developing seed.

(ii) Depending upon the source of pollen grains, pollination is of following three types
(a) Autogamy (Self-pollination) It is the kind of pollination achieved within the same
flower. The pollen from the anthers of a flower is transferred to the stigma of the same
flower, e.g. wheat, rice, pea, etc.

Autogamy is further classified as

● Cleistogamy In some plants, flowers never open up and the anthers dehisce inside
the closed flowers to ensure pollination.
Thus, cleistogamous flowers are invariably autogamous as there is no chance of
cross-pollination. These flowers produce assured seed sets even in the absence of
pollinators, e.g. Oxalis, Viola, etc.

● Homogamy In this method, both the anthers and the stigma of bisexual flowers
mature at the same time, e.g. Mirabilis.

(b) Geitonogamy The pollination where the pollen grains from the anther of a flower are
transferred to the stigma of another flower borne on the same plant but at different
branches. It usually occurs in plants, which show monoecious condition, e.g. Cucurbita.
It is functionally cross-pollination (involves a pollinating agent) but genetically it is
similar to autogamy (since pollen grains come from same plant).
(c) Xenogamy (Cross-pollination) It involves the transfer of pollen grains from the anther
of one plant to the stigma of another plant. This is the only type of pollination, which
brings genetically different types of pollen grains to the stigma during pollination, e.g.
papaya, maize, etc.

117.
(i) Explain the events after pollination leading to the formation of a seed in angiosperms.
(ii) Mention the ploidy levels of the cells of different parts of an albuminous seed.

(i) The following events take place between pollination and formation of a seed in
angiosperms
(a) Double fertilisation (b) Endosperm formation (c) Embryo development
(d) Seed formation ;Seeds are fertilised ovule that are developed inside a fruit. Each seed
consists of a seed coat, an embryonal axis and cotyledon.

(ii) The ploidy of the endosperm of an albuminous seed is 3n and ploidy of the embryo is
2n.

118.
A flower produces 240 viable seeds.
(i) minimum number of ovules present in per pollinated pistil?
(ii) How many microspore mother cells would be required?
(iii) How many pollen grains must have minimally pollinated the carpel?
(iv) How many male gametes would have used to produce these 200 viable seeds?
(v) How many megaspore mother cells were required in this process?

(i) The minimum number of ovules would also have been 240, as 240 viable seeds are
formed. After fertilisation, the ovule turns into seeds.
(ii) The minimum number of microspore mother cells in the above case would be 60.
This is because each microspore mother cell gives rise to 4 microspores.
Thus, to obtain 240, 60 cells are required.
(iii) The minimum number of pollen grains that must have been involved are 240, as 240
viable seeds are formed. This is because each pollen grain contains 2 male gametes, out
of which one fuses with the egg forming zygote that gives rise to seeds.
(iv) The number of male gametes involved would be 240. Each male gamete fuses with
one egg nucleus to form zygote that gives rise to seed.
(v) 240 megaspore mother cells were involved. They undergo meiotic division to form 4
haploid megaspores. Out of them, only 1 becomes functional megaspore, rest 3
degenerate.

119.
A flower produces 300 viable seeds.
the following s giving reasons
(i) How many ovules are involved?
(ii) How many megaspore mother cells are involved?
(iii) What is the minimum number of pollen grains that must land on stigma for
pollination?
(iv) How many male gametes are involved in the above case?
(v) How many microspore mother cells must have undergone reduction division prior to
dehiscence of anther in the above case? (Delhi 2015)

(i) The minimum number of ovules would be 300 as 300 viable seeds are formed.
(ii) 300 megaspore mother cells were involved.
(iii) The minimum number of pollen grains that must have been involved are 300.
(iv) The number of male gametes involved in seed formation are 300.
(v) Minimum 75 microspore mother cells.must have been involved to form 300 pollen
grains.

120.
Write the changes a fertilised ovule undergoes within the ovary in an angiosperm plant.

Changes taking place in a fertilised ovule within the ovary in an angiospermic plant are
Fertilised ovule – Seed
Funiculus – Stalk of ovule
Integument – Seed coat
(a) outer – Testa
(b) inner – Tegman
Polar nuclei – Endosperm
Nucellus – Utilised or remaining perisperm
Antipodal – Degenerate
Synergid – Degenerate
Egg – Embryo

121.
(i) Name the structures which the parts A and B shown in the diagram given below,
respectively develop into.

(ii) Explain the process of development which B undergoes in albuminous and

exalbuminous seeds. Give one example of each of these seeds.

(i) The part A is zygote which develops into the embryo. The part B is primary
endosperm nucleus which develops into the endosperm.
(ii) Endosperm Formation

● Primary endosperm cell divides repeatedly and forms triploid endosperm nucleus.
● Primary endosperm nucleus undergoes successive free nuclear divisions to give
rise to a number of free nuclei. At this stage, it is called free nuclear endosperm.

● Wall formation takes place from the periphery and proceeds towards the centre
and the endosperm becomes cellular.

● In albuminous seeds, some amount of endosperm persists in the mature seed as

the developing embryo does not consume it completely, e.g. wheat, maize.

● In exalbuminous seeds, the endosperm is completely consumed by the developing

embryo before seed maturation, e.g. in pea/groundnut.

122.
Give reason for each of the following
(a) Anthers of angiosperm flowers are described as dithecous.
(b) Hybrid seeds have to be produced year after year.

(a) A typical angiosperm anther is bilobed with each lobe having two thecae.
So, anther is called dithecous.
(b) Hybrid seeds show segregation of traits and do not maintain the hybrid character in
plants. So, they need to be produced every year and cannot be stored.

124.
(i) What is pollination? Give the different types of pollination.
(ii) How do flowers present submerged in water bodies adapt themselves for pollination?

(i) The transfer of pollen grains from the anther of the stamen to the receptive stigma of
a flower is called pollination. It is of two types, self-pollination and cross-pollination.
(ii) In submerged plants, the pollen grains are released in water. In such plants, pollen
grains are long, ribbon-like and are carried passively inside the water to stigma for
pollination.

125.
(i) What are false fruits? Give an example.
(ii) What is meant by parthenocarpy?
Name a fruit develops by this method.

(i) The fruit derived from ovary along with other accessory floral parts like thalamus is
called false fruit, e.g. apple, strawberry, etc.
(ii) In parthenocarpy, fruits develop without undergoing fertilisation. Such fruits are
seedless and called as parthenocarpic fruits, e.g. banana.

Inheritance

1.
Name the pattern of inheritance where F1phenotype
(i) resembles only one of the two parents.;Dominance
(ii) does not resemble either parents and is in between the two. ; Incomplete dominance

2.
RC Punnett developed a representation of a genetic cross called ‘Punnett Square’.
Mention the possible result this predicts of the genetic cross carried.

Punnett square helps to predict the probability of all the possible genotypes of offspring
in a genetic cross.

3.
Name the cross that would find the genotype of a pea plant bearing violet flowers.

To find the genotype of a pea plant bearing violet flowers, test cross would be carried
out in which the plant with dominant trait, i.e. violet flowers, will be crossed with its
recessive parent.

4.
State a difference between a gene and an allele.

A unit of inheritance which is passed down from parent to offspring through the
gametes over successive generations is known as gene. Genes consist of a pair of
contrasting forms for a character that are known as alleles.

5.
Give an example of polygenic trait in humans.
An example of a polygenic trait in humans is skin colour because it is under the control
of many genes.

6.
geneticist prefers to choose organisms with shorter life cycle. give reason.

A geneticist interested in studying variations and patterns of inheritance in living beings

prefers to choose organisms with shorter life cycle, because it enables the geneticist to
study many generations of the organism in a short time period.

8.
What are ‘true-breeding lines’ that are used to study inheritance pattern of traits in
plants.

True-breeding lines are those plants, which have undergone continuous self-pollination
and show stable trait inheritance and expression for several generations.

9.
How many phenotypes would you expect in F2-generation in a monohybrid cross
exhibiting codominance?
In codominance, alleles are able to express themselves independently when present
together. Thus, in a monohybrid cross there would be three kinds of phenotype in the
F2-generation showing codominance.

10.
Name the stage of cell division where segregation of an independent pair of
chromosomes occurs.
; During meiotic anaphase-I of cell division, the separation of independent pair of
chromosomes occurs.

Name the event during cell division cycle that results in the gain and loss of
chromosomes.
;
Non-disjunction of chromosomes during anaphase-I of meiosis results in the gain or
loss of chromosomes.

13.
In a dihybrid cross, when would the proportion of parental gene combinations be much
higher than non-parental types, as experimentally shown by Morgan and his group?

The proportion of parental gene combination is much higher than non-parental types,
when the two genes show linkage and are inherited together.

14.
Write possible genotypes Mendel got when he crossed F1 tall plant with a dwarf pea
plant.

Tt and tt (in ratio of 11) genotypes were obtained on crossing F1 tall plant with a dwarf
parent plant. It is a test cross.

18.
Mention the type of allele that expresses itself only in homozygous state in an organism.

Recessive allele expresses itself only in homozygous condition because in the presence
of a dominant allele its effect is masked.

19.
Pea flowers produce assured seed sets. Give a reason.

Pea flowers produce assured seed sets because they have cleistogamous flowers,
which undergo natural self-pollination.

21.
When does a geneticist need to carry a test cross?
A geneticist needs to carry a test cross when he/she wants to determine the genotype of
an organism, with a dominant phenotype trait, whether it is homozygous or
heterozygous.

How would you find the genotype of an organism exhibiting a dominant phenotype?

Genotype of the dominant phenotype is determined by a test cross. In it, the F1 progeny
is crossed to its recessive parent. When F1 progeny (heterozygous) crossed with dwarf
plant, the monohybrid test cross ratio is 1 1. But, all tall plants are obtained when both
homozygous parents are crossed.

22.
Write the scientific name of the fruitfly. Why did Morgan prefer fruit flies for his
experiments? State three reasons.

The scientific name of fruitfly is Drosophila melanogaster.

TH Morgan preferred this organism for his study because of the following reasons

● It has fast and short life cycle.

● It has only four pairs of chromosomes.

● It reproduces quickly.

23.
Explain pleiotropy with the help of an example.

Pleiotropy is the phenomenon in which a single gene exhibits multiple phenotypic

expressions. The genes exhibiting pleiotropy are called pleiotropic genes. Pleiotropism
occurs mainly because of mutation in a particular gene, e.g. phenylketonuria which is a
disorder caused by mutation in the gene coding for the enzyme phenylalanine
hydroxylase. In the absence of this enzyme, phenylalanine is not converted into tyrosine
and accumulation of phenylalanine takes place. The affected individual shows hair and
skin pigmentation and mental problems.

24.
Why are F2 phenotypic and genotypic ratios are same in a cross between red flowered
snapdragon and white flowered snapdragon plants? Explain with the help of cross.

The given condition represents the case of incomplete dominance. In snapdragon, the
inheritance of flower colour shows incomplete dominance. Neither of the alleles of gene
for flower colour is completely dominant over the other and hybrid shows an
intermediate phenotype. Therefore, F2 phenotypic and genotypic ratios are same in a
cross between red flowered snapdragon and white flowered snapdragon plants.

It can be explained with the help of a cross given below

25.
With the help of example, explain codominance and multiple allelism in humans

In human population, the phenomena of codominance and multiple allelism can be

explained by the inheritance pattern of ABO blood groups which are controlled by three
alleles, i.e. IA,IB and I.

● Codominance IA and IB both are codominant as both of them express themselves

independently in blood group AB (IA IB). There is no mixing of the effects of two alleles
and the expressed phenotype is the combination of two phenotypes. They do not follow
Mendelian inheritance.

● Multiple allelism In this phenomenon, genes exist in more than two allelic forms or
combinations. For example, the gene for blood group exists in three allelic forms IA, IB
and i. These alleles are produced due to repeated mutation of the same gene in different
direction. They do not follow Mendelian pattern of inheritance.

26.
Linkage and crossing over are alternatives of each other. Justify with an example.

Linkage is the tendency of certain loci or alleles (genes) to be inherited together. While
crossing over in the segregation of genes. The former helps to preserve parental
characters in offsprings whearas the latter produces new combination of characters.
The genes on a chromosome either follow linkage path or crossing over to form the
gametes during gametogenesis in human. Therefore, linkage and crossing over of genes
are alternatives of each other.
 27.
In snapdragon, a cross between true-breeding red flowered (RR) plants and true-breeding
white flowered (rr) plants showed a progeny with all pink flowers.
(i) The appearance of pink flowers is not known as blending. Why? Blending is the mixing
of two colours, but in this example red and white colours appear independently at
cellular level. Thus, no blending occurs. The red and white colours reappear in
F2-generation.
(ii) What is this phenomenon known as? This phenomenon is known as incomplete
dominance.

28.
A cross was carried out between two pea plants showing the contrasting traits of height
of the plants. The results of the cross showed 50% parental characters.
(i) Work out the cross with the help of a Punnett square.
(ii) Name the type of the cross carried out.

(i) A Punnett square representing the cross given in the is as follows

(ii) The type of cross carried out here is a test cross, in which an individual with an
unknown dominant phenotype is crossed with a homozygous recessive for that trait.

29.
How does the gene i control ABO blood group in humans? Write the effect the gene has
on the structure of red blood cells.

In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three
alleles, i.e. IA, IB and i. A person possesses any two of the three alleles. IA and IB are
codominant and they both are dominant over i. These alleles help to determine the blood
group of a person. The plasma membrane of red blood cells has sugar polymers that
protrude out from its surface and the kind of sugar is regulated by the gene ‘I’ of ABO
blood group. The alleles IA and IB produce enzymes that produce A and B types of sugar
respectively on the surface of red blood cells, while allele i does not produce any sugar.

30.
Study the figures given below and the .
Identify in which of the crosses, the strength of linkage between the genes is higher. Give
reasons in support of your .

The strength of linkage is higher in the cross A than in cross B because linkage is higher
when two genes are present closely on the same chromosome than those genes which
are far apart. In cross B, the chances of crossing over or recombination are higher
because the genes are loosely linked.

34.
In a cross between two tall pea plants, some of the offsprings produced were dwarf.
Show with the help of Punnett square, how this is possible.

Tall plants may either have genotype TT or Tt. Two tall pea plants that produce some
dwarf plants among their progenies must be heterozygous with the genotype Tt. Plants
with genotype TT cannot produce dwarf offspring as they lack the allele for dwarfness
(t) and hence, cannot transfer it to the progeny. But, if both the parents are heterozygous
tall (Tt), 25% of F, progeny would contain ‘t’ allele in homozygous (tt) condition. It can be
represented using Punnett square as follows

35.
In a typical monohybrid cross, the F2 population ratio is written as 3 1 for phenotype, but
expressed as 1 2 1 for genotype. Explain with the help of an example.

In a monohybrid cross, when a pureline tall plant is crossed with a pureline dwarf plant,
following ratios are produced
The phenotypic ratio 3 1 represents that 1/4th of F2 plants are tall, whereas 1/4th plants
are dwarf. The genotypic ratio 1 2 1 represents 1 true breeding dominant, 2
heterozygous dominant and 1 true breeding recessive progeny.

36.
How is the phenotypic ratio of F2-generation in a dihybrid cross is different from
monohybrid cross?

In a monohybrid cross, the phenotypic ratio of F2-generation is 3 1, whereas in dihybrid

cross, the pheonotypic ratio of F2-generation is 9 3 3 1.

37.
In a dihybrid cross, white-eyed, yellow-bodied female Drosophila crossed with red-eyed,
brown-bodied male Drosophila produced in F2-generation 1.3% recombinants and 98.7%
progeny with parental type combinations. This observation of Morgan deviated from
Mendelian F2-phenotypic dihybrid ratio. Explain, giving reasons Morgan’s observation.

The results obtained were due to the linkage. It is the phenomenon in which two or more
linked genes are inherited together and their frequency of recombination in a test cross
progeny is less than the expected 50%. In Morgan’s experiment on Drosophila, the genes
for eye colour and body colour show linkage and do not allow crossing over during
gamete formation. Hence, parental type progeny is in greater ratio than that of
recombinants.

39.
Compare in any three ways the chromosomal theory of inheritance as proposed by
Sutton and Bovery with experimental results on pea plant presented by Mendel.

Through any of the given ways chromosomal theory of inheritance and experimental
results presented by Mendel can be compared

● In a diploid organism, the factors (genes) and chromosomes occur in pairs.

● Both chromosomes as well as genes segregate at the time of gamete formation
such that only one of each pair is transmitted to a gamete.

● A gamete contains only one chromosome of a type and only one of the two alleles
of a trait.

● The paired condition of both chromosomes as well as Mendelian factors is

restored during fertilisation.

40.
(i) Explain linkage and recombination as put forth by TH Morgan based on his
observations with Drosophila melanogaster crossing experiment.
(ii) Write the basis on which Alfred Sturtevant explained gene mapping.

(i) TH Morgan studied X-linked gene in Drosophila and saw that when the two genes in a
dihybrid cross were situated on the same chromosome, the proportion of parental gene
combinations were much higher than the non-parental type.
He attributed this due to the physical association or linkage of the two genes on a
chromosome and coined the term linkage. The term recombination describes the
generation of non-parental gene combination in offsprings.

(ii) Alfred Sturtevant explained gene mapping by using the frequency of recombination
between the gene pairs on the same chromosome as a measure of the distance
between genes and he mapped their position on the chromosome.

41.
What is a test cross? How can it decipher the heterozygosity of a plant?

This is a method devised by Mendel to determine the genotype of an organism. In this

cross, the organism with an unknown dominant genotype is crossed with the recessive
parent, instead of self-crossing. For example, in a monohybrid cross, between violet
colour flower (W) and white colour flower (w), the F1-hybrid was a violet colour flower. If
all the F1-progenies are of violet colour, then the dominant flower is homozygous and if
the progenies are in 1 1 ratio, then it is deciphered that dominant flower is heterozygous.

42.
How would you find genotype of a tall pea plant bearing white flowers? Explain with the
help of a cross. Name the type of cross you would use. (Delhi 2016)

We can find out the genotype of a plant by test cross by allowing it to cross with its
recessive parent. The tall plant may be either homozygous or heterozygous.
Case I Tall (homozygous) pea plant with white flowers crossed with dwarf pea plant with
white flowers.

If plant produces all tall plants with white flowers as offspring, then genotype of plant is
TTpp, i.e. homozygous tall plant with white flowers.
Case II Tall (heterozygous) pea plant with white flowers is crossed with dwarf pea plant
with white flowers. If plant produces both tall plant with white flowers and dwarf plant
with white flowers, then genotype of plant is Ttpp, i.e. heterozygous tall pea plant with
white flowers.

43.
Although Mendel published his work it remained unrecognised till 1900. Explain three
reasons.

The following are the three reasons that led to delay in accepting Mendel’s work.

● Lack of communication and less published work.

● His concept of factors (genes) as discrete units which did not blend with each
other was not accepted in the light of variations occurring continuously in nature.

● Mendel’s approach to explain biological phenomenon with the help of

mathematics was also not accepted.

44.
Explain the laws that Mendel derived from his monohybrid crosses.

From monohybrid crosses, Mendel derived law of dominance and law of segregation.
 45.
to find the genotype of pea plants bearing purple flowers Name and explain the cross
that will make it possible.

Test cross is a method devised by Mendel to determine the genotype of a plant with
dominant phenotype. In a test cross, the unknown dominant genotype is crossed with
recessive parent (white, WW in the given case).
(i) If the progeny consists of purple and white flowers in ratio of 11, the purple flower is a
hybrid with PW genotype. It can be seen from the given cross.

(ii) If the progeny obtained have all purple flowers, both parents are homozygous, i.e.
genotype of purple flower is PP.
It can be seen from the cross that follows

46.
During a monohybrid cross involving a tall pea plant with a dwarf pea plant, the offspring
were tall and dwarf in equal ratio. Work out a cross to show how it is possible.

In a monohybrid test cross, involving a heterozygous tall plant (Tt) and a pure dwarf
plant (tt), the progeny consists of tall and dwarf plants in the ratio of 1 1. This can be
shown as follows
47.
Explain with the help of a suitable example, the inheritance of a trait where two different
dominant alleles of a trait express themselves simultaneously in the progeny. Name this
kind of inheritance pattern.

Codominance is the inheritance of a trait where two different dominant alleles of a trait
express themselves simultaneously in the progeny.
For example, ABO blood groups in human population.

● Gene i for blood group exhibits three allelic forms, i.e. IA, IB and i.

● IA and IB produce RBC surface antigen A and B respectively, whereas i does not
produce any antigen.

● IA and IB are codominant alleles, and they both are dominant over ‘i’ which is a
recessive allele.

● In case IA and IB are present together, both express themselves equally and
produce both surface antigen A and B. The resultant offspring is of ‘AB’ blood type.

48.
Explain polygenic inheritance with the help of a suitable example.

Polygenic inheritance is an inheritance pattern controlled by three or more genes

(multiple genes) and the graded phenotypes are due to the additive or cumulative effect
of all the different genes of the trait, e.g. skin colour in human population shows
variation.

Skin colour in humans is produced by a pigment called melanin. The quantity of melanin
is due to three pairs of polygenes (A, B and C). If a black or very dark (AA BB CC) and
white or very light (aa bb cc) individuals marry each other, the offsprings or individuals of
F1-generation show intermediate colour and they are often called mulatto (Aa Bb Cc).

49.
Morgan carried out several dihybrid crosses in Drosophila and found F2 ratios deviated
very significantly from the expected Mendelian ratio. Explain his finding with the help of
an example.

Morgan’s studies on Drosophila were based on the genes that were located on the
X-chromosome. He found when the two genes in a dihybrid cross were situated on the
same chromosome, the proportion of parental gene combinations were much higher
than the non-parental type. Morgan stated this association as linkage to describe the
physical association of genes on a chromosome. Recombination is a term used to
describe the generation of non-parental gene combination. Morgan also found that
some genes were tightly linked (low recombination) and others were loosely linked (high
recombination). He concluded that in case of inheritance of linked genes, the phenotypic
ratio deviates from expected 9 3 3 1 ratio of Mendel’s dihybrid cross.
To prove his findings, Morgan hybridised yellow-bodied and white-eyed females with
brown-bodied and red-eyed males (wild type) and intercrossed their F1-progeny (cross
A). It was observed that the two genes did not segregate independently for each other
and the F2-ratio deviated significantly from 9 3 3 1 ratio.

In F2-generation, parental combinations were 98.7% and the recombinants were 1.3%. In
another cross (cross B), between white-bodied female fly with miniature wing and a
male fly with yellow body and normal wing, parental combinations were 62.8% and
recombinants were 37.2% in F2-generation. Thus, it was proved from the crosses that
the linkage between genes for yellow body and white eyes was stronger than the linkage
between the white body and miniature wing.

50.
(i) Write the conclusion Mendel arrived at on dominance of traits on the basis of
monohybrid crosses that he carried out in pea plants.
(ii) Explain why a recessive allele is unable to express itself in a heterozygous state.

(i) In a monohybrid cross in pea plant, Mendel found that only dominant trait expresses
itself in offspring no matter it is whether present in homozygous state (TT) or in
heterozygous state (Tt). Dominant trait does not require another similar allele to produce
its effect on the phenotype, in fact it has the ability to mask the effect of recessive allele.
Based on this observation, Mendel proposed the law of dominance.

(ii) Since, the characters are controlled by genes which occur in pair, when two alternate
forms of a trait or character (genes or alleles) are present in an organism, only one factor
expresses itself in F1-generation. This factor is dominant, while the other factor that
remains masked by the dominant factor is called as recessive. The recessive allele is
unable to express itself in a heterozygous state because it forms an incomplete or
defective polypeptide or enzyme, so its expression does not contain any effect. In
contrast, dominant allele can form complete polypeptide or enzyme to express its effect.

51.
In pea plants, the colour of the flower is either violet or white, whereas human skin
colour shows many gradation.
Explain giving reasons how it is possible.

In pea plants, the colour of the flower is either violet or white because the colour is
purely dependent on two alleles and the crossing between them, i.e. violet and white in
the parents generation, which results in F1-generation. Thus, the resultant offsprings are
either violet or white coloured. Hence, the inheritance pattern of flower colour in pea
plant follows the law of dominance.

But in case of humans, skin colour is produced by polygenes. Such inheritance is

controlled by one or more genes in which dominant alleles have cumulative effect with
each dominant allele expressing a part of the trait. The full trait is shown only when all
the dominant alleles are present together.
The quantity of human skin pigment melanin determines the human skin colour

53.
A pea plant with purple flowers was crossed with white flowers producing all plants with
only purple flowers. On selfing, these plants produced 482 plants with purple flowers and
162 with white flowers. What genetic mechanism accounts for these results? Explain.

Purple flowered plants White flowered plants are in the ratio of 482 162 which is
approximately equal to 3 1.

The genetic mechanism for these results is explained below

(i) Factors segregate from each other during gamete formation that remain together in a
parent.
(ii) A homozygous parent produces all gametes that are similar, while heterozyogous
parent produces two kinds of gametes in equal ratio.

54.
Work out a cross between true breeding red and white flowered dog flower plants
(snapdragon) upto F2 progeny. Explain the results of F1 and F2-generation.
.In F1-generation Pink flowered plants obtained. It is due to incomplete dominance.
In F2-generation Alleles of the hybrid (F1) segregate during gamete formation and the
parental characters reappear without any change. So, the phenotypic and genotypic
ratios in F2-generation are same.

55.
(i) Work out the crosses so as to obtain the phenotypic ratios given below
(a) 1 2 1 (in F2-)
(b) 3 1 (in F2)
(c) 1 1 (in F1)
(ii) Differentiate between pleiotropy and polygenic inheritance patterns.
(i) (a) 1 2 1 (in F2 ) is the phenotypic ratio of incomplete dominance.

(b) 3 1 (in F2-generation) is the phenotypic ratio of monohybrid cross.

(c) 1 1 (in F1/sub>-generation) is the phenotypic ratio of test cross..

(ii) Differences between pleiotropy and polygenic inheritance are as follows

ropy enic inheritance

e gene product confers many e phenotypic effect is under the control o

ological effects. genes.

enes involved are called pleiotropic enes involved are called polygenes.
s.

henylketonuria. uman skin colour.

56.
Differentiate between incomplete dominance and codominance. one example of each.

Differences between codominance and incomplete dominance are as follows

minance mplete dominance

he appearance of both parental he appearance of an intermediate phenotyp

otypes together in the offspring when
h is a combination of both parental alleles
is done between individuals with tw a cross is done between individuals with t
ent phenotypes. ent phenotypes.

parental alleles produce their effect t of the two parental alleles is intermediate
endently. ffspring.

parental alleles can be observed in thof parental alleles can be observed in the
ring. ring.
 ples include ABO blood group, etc. Sples include inheritance of flower colour in
ent genotypes of human ABO bloodogflower, etc. Here, the genotypic ratio of F
p show four phenotypes-A, B, AB and ns same as Mendel’s monohybrid cross, i.e
, but phenotypic ratio changes from 3 1 to

. 58.
State and explain the ‘law of independent assortment’ in a typical Mendelian dihyhrid
cross.

Law of Independent Assortment (Third law) is based on the inheritance of two genes, i.e.
dihybrid cross which states that when two pairs of contrasting traits are combined in a
hybrid, segregation of one pair of characters is independent of the other pair of
characters.

These factors randomly rearrange in the offspring producing both parental and new
combination of characters. It means inheritance of one character does not affect the
inheritance of another character and both characters assort independently during
gamete formation. The Punnett square can be used to understand the independent
segregation of the two pairs of genes during meiosis.
 59.
(i) What is polygenic inheritance? Explain with the help of a suitable example.
(ii) How are pleiotropy and Mendelian pattern of inheritance different from polygenic
pattern of inheritance?

(i) Polygenic Inheritance was proposed by Galton and he suggested that many
instances of continuous variations are heritable. In this case, a trait is controlled by three
or more genes and the graded phenotypes develop due to the additive or cumulative
effect of all the different genes of the trait, e.g. human skin colour, height, intelligence
etc.
Polygene is a gene where one dominant allele controls only a unit or partial quantitative
expression of a trait.

It also takes into account the influence of environment and is also called as quantitative
inheritance, as the character or phenotype can be quantified, like the amount of pigment,
intelligence in human beings and milk yield in animals. These characters have been
found to be determined by many genes and their effects have been cumulative. These
traits are called polygenic traits, e.g. human skin colour explains the phenomenon of
polygenic inheritance. Skin colour in human is produced by a pigment called melanin.
The quantity of melanin is due to three pairs of polygenes (A, B and C). If black or very
dark (AABBCC) and white or very light (aabbcc) individuals marry each other, the
offsprings or individuals of F,-generation show intermediate colour which are often
called mulatto (AaBbCc).

A total of eight allele combinations are possible in the gametes forming 27 distinct
genotypes distributed into 7 phenotypes, i.e. 1 very dark, 6 dark, 15 fairly dark, 20
intermediate, 15 fairly light, 6 light and 1 very light.

(ii) In pleiotropy, single gene product may produce multiple or more than one phenotypic
effect whereas in polygenic inheritance single phenotypic trait (human skin colour) is
controlled by 3 pairs of genes (A, B and C). In Mendelian inheritance, one gene controls
one phenotypic character (flower colour red or white). However, in polygenic inheritnace
one phenotype is controlled by more than one gene.

60.
(i) A couple with blood group A and B, respectively have a child with blood group O. Work
out a cross to show how it is possible and the probable blood groups that can be
expected in their other offspring.
(ii) Explain the genetic basis of blood groups in human population.

(i) Parents must be heterozygous since blood group ‘O’ appears in progeny. The progeny
can have all the four blood groups. A, B, AB and O. There are three alleles of the gene
controlling blood group character, i.e. IA, IB and i. IA and IB are dominant over i and
together they are codominant to each other.

(ii) ABO blood grouping in humans shows the phenomenon of codominance.

61.
(i) State and explain the law of segregation as proposed by Mendel in a monohybrid
cross.
(ii) Write the Mendelian F2 phenotypic ratio in a dihybrid cross. State the law that he
proposed on the basis of this ratio. How is this law different from the law of
segregation?

(i) Law of segregation states that the factors or alleles of a pair segregate from each
other during gamete formation in a way that a gamete receives only one of the two
factors. They do not show any blending. The pattern of inheritance can be understood
by crossing F1-hybrid (Tt heterozygous) of a tall plant with a dwarf plant.
Here, hybrid tall makes two types of gametes (T) and (t), while pure dwarf makes only
one type of gamete, i.e. (t). It is because only one allele can enter in a gamete for a
character.

(ii) (a) Mendelian F2 phenotypic ratio in a dihybrid cross is 9 3 3 1.

(b) Law of independent assortment was proposed on the basis of dihybrid cross. It
states that when two pairs of contrasting traits are combined in a hybrid, segregation of
one pair of characters is independent of the other pair of characters.

Differences between law of segregation and law of independent assortment are

of segregation of independent assortment

ased on the monohybrid cross. ased on the dihybrid cross.

 lains non-mixing of two alleles of a genlains non-mixing of characters during the
time of gamete formation. tance to the next generation.

62.
Give a genetic explanation for the following cross. When a tall pea plant with round
seeds was crossed with a dwarf pea plant with wrinkled seeds then all the individuals of
F1-population were tall with round seeds. However, selfing among F1 population led to a
9 3 3 1 phenotypic ratio. (All India 2015)

The cross given in the is a dihybrid cross, which explains the third law of Mendel, i.e. law
of independent assortment. In a dihybrid cross inheritance pattern of two traits are
considered simultaneously.
This law states that when two pairs of traits are combined in a hybrid, segregation of
one pair of character is independent of the other pair of characters at the time of gamete
formation. The alleles controlling these characters also get randomly rearranged in the
offspring producing both parental and new combinations of characters.

The ratio 9 3 3 1 was. obtained because the factors of height of plant and those for
shape of seeds have segregated independently and each gamete has one factor for
each of these two traits.

63.
A cross was carried out between a pea plant heterozygous for round and yellow seeds
with a pea plant having wrinkled and green seeds.
(i) Show the cross in a Punnett square.
(ii) Write the phenotype of the progeny of this cross.
(iii) What is this cross known as? State the purpose of conducting such a cross.
(ii) Phenotypic ratio Round yellow Round green Wrinkled yellow Wrinkled green = 1 1 1
1
Genotypic ratio 1 1 1 1

(iii) This cross is known as dihybrid test cross. The purpose of this cross is to determine
the relationship between different allelic pairs.

64.
Work out a monohybrid cross up to F1-generation between two pea plants and two
Antirrhinum plants both having contrasting traits with respect to colour of flower.
Comment on the pattern of inheritance in the crosses carried above.

Phenotypic ratio
Purple White
3 1
Genotypic ratio
PP Pp pp
1 2 1
Inheritance of flower colour in garden pea shows true dominance. In F1-generation,
dominant colour purple is expressed and in F2-generation, both dominant (purple) and
recessive (white) colours are expressed in the ratio of 3 1.

Inheritance pattern of flower colour in Antirrhium (snapdragon), In this phenomenon,

neither of the two alleles is completely dominant over the other and the hybrid is
intermediate between the two. Hence, red is homozygous dominant, white is
homozygous recessive, while hybrid is intermediate, i.e. pink.

65.
(i) Differentiate between dominance and co-dominance.
(ii) Explain codominance taking an example of human blood groups in the population.

(i) Difference between dominance and codominance is as follows

Dominance
Out of the two contrasting alleles of a gene, only one can produce effect in heterozygous
condition, e.g. trait of tallness in pea plants.
Codominance
Both the alternative forms of a gene can produce effect in heterozygous condition, e.g.
ABO blood grouping in humans.

66.
iHow did Morgan show the deviation in inheritance pattern in Drosophila with respect to
this law?

Morgan found that linkage is an exception to the law of independent assortment.

67.
What is the inheritance pattern observed in the size of starch grains and seed shape of
Pisum sativum. Work out the monohybrid cross showing the above traits. How does this
pattern of inheritance deviate from that of Mendelian law of dominance?

The starch synthesis in pea plants is controlled by a single gene. It has two alleles B and
b. BB homozygotes produce large starch grains as compared to that produced by bb
homozygotes.

After maturation, it was observed that BB seeds were round and bb were wrinkled. When
they were crossed, the resultant progeny have intermediate sized (Bb) seeds.
The cross involved is

Deviation from Mendel’s law of dominance If starch grain size is considered as the
phenotype, then, the alleles show incomplete dominance.
Thus, dominance is not an autonomous feature of a gene, it depends on gene product
and production of particular phenotype from this product.

68.
Differentiate between the following.
(i) Polygenic inheritance and pleiotropy.
(ii) Dominance, codominance and incomplete dominance.

(i) Differences between polygenic inheritance and. pleiotropy are as follows

enic inheritance opy

 e phenotypic effect is under the contro
e gene product confers many phenotypic
any genes. ts.

enes involved are called polygenes, e.enes involved are called pleiotropic gene
n skin colour. enylketonuria.

(ii) Differences between dominance, codominance and incomplete dominance are as

follows

ance minance plete dominance

relationship between he phenomenon of expression olso known as partial or

s of a single gene, in whithe alleles in heterozygous ic dominance where no
llele masks the phenotyption. In this, alleles do not show
e two contrasting alleles
ssion of another allele atnant-recessive relationship andrs is dominant, e.g.
ame gene locus, e.g. ble to express themselves mplete dominance in 4 ‘O
ss in pea plant. endently, e.g. ABO blood group plant.
ns.

Incomplete dominance
It is also known as partial or mosaic dominance where none of the two contrasting
alleles or factors is dominant, e.g. incomplete dominance in 4 ‘O’ clock plant.

69.
(i) List three different allelic forms of gene T in human. Explain the different phenotypic
expressions, controlled by these three forms.
(ii) A woman with blood group A marries a man with blood group O. Discuss the
possibilities of the inheritance of the blood group in the following starting with ‘yes’ or
‘no’ for each.
(a) They produce children with blood group A only.
(b) They produce children, some with 0 blood group and some with A blood group.
(i) In humans, the ABO blood groups are controlled by a gene called T. It has three
alleles. These are IA, IB and i..

(ii) (a) No, its not necessary as mother could have a genotype IAIA or IAi. If the genotype
is IA IA, all the offsprings would have A blood group, but in the second case, offsprings
can have either ‘A’ or ‘O’ blood group as their father has ‘O’ blood group.

(b) Yes, if the mother is of genotype IAi and father is ‘O’ (genotype ii), blood group of
some children can be ‘O’ and some can be ‘A’.

70.
(i) Explain monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work
out the cross up to F2-generation.
(ii) State the law of inheritance that can be derived from such a cross.
(iii) How is the phenotypic ratio of F2-generation different in a dihybrid cross?

(i) In a monohybrid cross, when homozygous dominant and homozygous recessive

parents are crossed, F1-hybrid would be heterozygous for the trait and would express the
dominant allele.

(ii) The laws of inheritance that can be derived from such a cross are
(a) Law of dominance
(b) Law of segregation

(iii) Phenotypic ratio in F2-generation.

In monohybrid cross – 31
In dihybrid cross – 9 3 3 1

71.
When a garden pea plant with violet flowers was crossed with another plant with white
flowers, 50% of the progeny bear violet flowers.
How does the inheritance pattern of flower colour in snapdragon differ from the above?
In this test cross, violet and pure white flowers when crossed produce violet and white
flowers.
Whereas in snapdragon, the F1-generation hybrid was pink coloured. F2-generation
consists of red, pink and white flowers in the ratio of 1 red 2 pink 1 white. This is due to
incomplete dominance. Similar type of cross can be made for plants with green and
yellow pods.

72.
Write the sex of a human having XXY chromosome With 22 pairs of autosomes. Name
the disorder this human suffers from.

The sex of a human having XXY chromosomes with 22 pairs of autosomes is male. The
disorder from which this human is suffering is Klinefelter’s syndrome.

73.
State the fate of a pair of autosome during gamete formation.

During gamete formation, the homologous pair of autosomes gets separated from each
other and moves to different gametes, so that each gamete receives haploid set of
chromosomes.

74.
Give an example of a human disorder that is caused due to a single gene mutation.

Sickle-cell anaemia is an example of a human disorder that is an caused due to a single

gene mutation.

75.
Give an example of a sex-linked recessive disorder in humans.

Haemophilia is a sex-linked recessive disorder in humans.

76.
example of an organism that exhibits haplo-diploidy sex-determination system.

Haplo-diploidy sex-determination system is seen in honeybees.

77.
Give one example of organism exhibiting female heterogamety

Female heterogamety is exhibited by birds. They have ZW-ZZ type of sex-determination

mechanism. In this type, the male is homogametic and the female is heterogametic.
Therefore, there are two types of eggs, i.e. with Z and with W and only one type of
sperms, i.e. each with Z.

When egg with Z-chromosome is fertilised by a sperm with Z-chromosome, a zygote

with ZZ-chromosomes (male) is formed. Similarly when egg with W-chromosome is
fertilised by a sperm with Z-chromosome, a zygote with ZW-chromosomes (female) is
formed.

78.
Identify the correct statement
(i) Females of many birds have a pair of dissimilar ZW-chromosomes, while the males
possess a pair of similar 2Z-chromosomes.
(ii) Females of many birds have a pair of similar ZZ-chromosomes, while the males
possess a pair of dissimilar ZW-chromosomes.

The correct statement is

(i) Females of many birds have a pair of dissimilar ZW-chromosomes, while the males
possess a pair of similar ZZ-chromosomes.

79.
Identify and write the correct statement
(i) In grasshopper males, two sex chromosomes are X and Y type.
(ii) In grasshopper males, there exist XO type of sex-determinants.

The correct statement Is

(ii) In grasshopper males, there exist XO type of sex-determinants.

80.
Identify and write the correct statement
(i) Drosophila male has one X and one Y-chromosome.
(ii) Drosophila male has two X-chromosomes.

The correct statement is

(i) Drosophila male has one X and one Y-chromosome.

81.
Why do normal red blood cells become
elongated and sickle-shaped in structure in a person suffering from sickle-cell anaemia?

In sickle-cell anaemia, substitution of valine in place of glutamic acid in the sixth

position of haemoglobin chain occurs. The mutant haemoglobin molecule undergoes
polymerisation under low oxygen tension causing the change in the shape of the RBC
from biconcave disc to the elongated sickle-like structure.

82.
Name one autosomal dominant and one autosomal recessive Mendelian disorder in
humans.

Autosomal dominant Mendelian disorder- Huntington’s disease.

Autosomal recessive Mendelian disorder-Sickle-cell anaemia.

83.
Write the genotype of
(i) an individual who is the carrier of sickle-cell anaemia gene, but apparently unaffected
and
(ii) an individual affected with the disease.

(i) HbA Hbs

(ii) Hbs Hbs

84.
Write the cause of Down’s syndrome in humans.

This is due to nondisjunction, i.e. failure of segregation of homologous chromosomes

during gamete formation resulting in the gain of an extra copy of chromosome number
21 (21 Trisomy).

85.
The son of a haemophilic man may not get this genetic disorder. Mention the reason.

The gene responsible for haemophilia is located on X-chromosome and males have only
single copy of X-chromosome with no alternate normal allele. The son gets
X-chromosome from mother only. So, if the female is normal, the son may not get this
genetic disorder from male (father).

86.
Write the genotypes of the parents of a child suffering from thalassemia. State the
cause of this disease.

The genotypes of the parents of a child suffering from thalassemia are as follows
Mother → ThB ThB+
Father → ThB ThB+
Thalassemia is an autosomal-linked recessive disease, which results in reduced rate of
synthesis of one of the globin chains of haemoglobin. The disease is controlled by a
single pair of allele, ThB and ThB+. ThB codes for normal B-protein and ThB+ codes for
an abnormal B-protein. Only homozygous individuals for ThB+ (ThB+ ThB+) show the
diseased phenotype.
87.
Name a disorder a human suffers from as a result of monosomy of the sex
chromosome. Give the karyotype and write the symptoms.

Monosomy of the sex chromosome in humans results in Turner’s syndrome.

The number of chromosomes is 45 (44+ XO)

Symptoms are as follows

● The affected females are sterile as ovaries are rudimentary.

● Other symptoms include lack of secondary sexual characters, short stature, etc

88.
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.
(i) Represent this cross in the form of a pedigree chart. Give the genotypes of the
daughter and her husband.
(ii) Write the conclusion you draw from the inheritance pattern of this disease.

Genotypes of daughter and her husband are XhX and XY, respectively.

(ii) Haemophilia is a sex-linked recessive disease. It is transmitted from the carrier

female to her sons. From the above pedigree chart, it can be observed that the disease is
being transmitted from the carrier female to her daughter (carrier) and son (affected).
The carrier daughter transmits this disease to the grandson. This pattern of inheritance
is called criss-cross inheritance.

89.
Is haemophilia in humans a sex-linked or autosomal disorder? Work out a cross in
support of your .

Haemophilia is a sex-linked disorder and more specifically it is X-linked disorder.

If normal parents give rise to a haemophilic child, then their genotypes should be Father
– XY (normal) Mother – XXh (carrier/heterozygous non-haemophilic)
 90.
A couple with normal vision bears a colourblind child. Work out the cross to show how it
is possible and mention the sex of the affected child.
Or
A colourblind child is born to a normal couple. Work out a cross to show how it is
possible. Mention the sex of this child.

Colour blindness is a sex-linked disease which results in the defect in either red or green
cone cells of eyes. The gene for this disorder is present on the X-chr. Hence, it is carried
by normal females that do not express the disease.

If a colourblind child is born to a normal couple, then the mother would be the carrier of
the disease. The following cross shows the inheritance of the disorder

Thus, the sex of colourblind child would be male.

91.
Differentiate between male and female heterogamety.

Differences between male and female heterogamety are

heterogamety le heterogamety

produces two different types of gametle produces two different types of

tes.
 ple human male produces gametes wiple female birds produce gametes with
X or Y-chromosomes. Z or W type chromosomes.

92.
Why is the possibility of a human female suffering from haemophilia rare? Explain.

Haemophilia is an X-linked recessive disease. Therefore, the females having haemophilic

allele on single X-chromosome do not produce haemophilic phenotype. Females suffer
from this disease only if the father is haemophilic and the mother is at least the carrier
of disease. Such condition is rare and hence, the human females are rarely haemophilic.

93.
Why is pedigree analysis done in the study of human genetics? State the conclusions
that can be drawn from it.

The study of inheritance of genetic traits in several generations of a human family in the
form of a family diagram is called pedigree analysis. Inheritance pattern of traits in
human beings cannot be studied by crosses. Thus, pedigree analysis is done. Based on
the pedigree studies, inheritance of a specific trait, abnormality or disease can be traced.

96.
Name a blood related autosomal Mendelian disorder. Why is it called Mendelian
disorder? How is the disorder transmitted from parents to offspring?

A blood related autosomal Mendelian genetic disorder is sickle-cell anaemia. It occurs

by the change of a single base pair in the gene, leading to the substitution of glutamic
acid by valine at the 6th position of p-globin chain of haemoglobin.

Since, its transmission follows Mendelian principles, it is called Mendelian disorder.

Inheritance pattern It is transmitted from parents to the offspring, when both the
partners are carriers (heterozygous) of the disease.

97.
Write the types of sex-determination mechanisms the following crosses show. Give an
example of each type.
(i) Female XX with male XO.
(ii) Female ZW with male ZZ.

(i) The type of sex-determination mechanism shown in female XX and males XO is called
male heterogamety. In this case, males are heterogametic with half of the male gametes
carrying X-chromosome while the other half being devoid of it, e.g. grasshopper.

(ii) The type of sex-determination mechanism is female heterogamety because female

produces two different types of gametes, i.e. Z and W while males are ZZ type, e.g. birds.
99.

Give reasons, which explain that haemophilia is (i) sex-linked (ii) caused by X-linked
gene.

(i) (a) It is X-linked trait which is transmitted from an unaffected carrier female to some
of the male offsprings.
(b) Females rarely become haemophilic as in this cases, the mother has to be at least a
carrier and father should be haemophilic.

(ii) Gene for haemophilia is X-linked or present on X-chromosome because male receives
this gene from mother.
(a) So, a carrier female transmits it to her son.
(b) Disease appears more in males because they have only one X-chromosome.

100.

haemoglobin of a normal person is given below

The codon for the sixth amino acid is GAG. The sixth codon GAG mutates to GAA as a
result of mutation ‘A’ and to GUG as a result of mutation ‘B’. Haemoglobin structure did
not change as a result of mutation ‘A’, whereas haemoglobin structure changed because
of mutation ‘B’ leading to sickle-shaped RBCs.
Explain giving reasons how. could mutation ‘B’ lead to sickle-shaped RBCs. Explain
giving reasons how could mutation.‘B’ change the haemoglobin structure or bring down
mutation and not mutation ‘A’.

In mutation A, the change in amino acid does not occur when codon GAG changes to
GAA because both GAG and GAA code for amino acid, glutamic acid. Thus, there is no
change in haemoglobin structure.

In mutation B, the change in amino acid occurs when codon GAG is changed to GUG
because GUG codes for valine, while the original codon GAG codes for glutamic acid.
Thus, there will be a change in the haemoglobin structure which would lead to sickle-cell
anaemia.

101.
Given below is the representation of a relevant part of amino acid composition of the
ß-chain of haemoglobin, related to the shape of human red blood cells.
(i) Is this representation of the sequence of amino acids indicating a normal human or a
sufferer from a certain blood related genetic disease? Give reason in support of your .
(ii) Why is the disease referred to as a Mendelian disorder? Explain. (All India 2019)
Or
Given below is the representation of amino acid composition of the relevant translated
portion of ß-chain of haemoglobin, related to the shape of human red blood cells.

(i) Is this representation indicating a normal human or a sufferer from certain related
genetic disease? Give reason in support of your .
(ii) What difference would be noticed in the phenotype of the normal and the sufferer
related to this gene?
(iii) Who are likely to suffer more from the defect related to the gene represented the
males, the females or both males and females equally? And why? (Delhi 2012)

(i) This representation indicates normal human because normal peptide HbA is with
glutamic acid (Glu) at sixth position of (1-globin chain of haemoglobin molecule.
Sickle-cell anaemia is referred to as Mendelian disorder because each parent
contributes one mutant allele to affected offspring and the recessive sickle trait is
expressed in homozygous condition only.

(ii) The normal individual has biconcave, disc-like RBCs, whereas the sufferer has
sickle-shaped RBCs.

(iii) It is an autosomal recessive disorder. So, both males and females suffer equally.

102.

) Name the kind of diseases/disorders that are likely to occur in humans if

(a) mutation in the gene that codes for an enzyme phenylalanine hydrolase occurs,
(b) there is an extra copy of chromosome 21,
(c) the karyotype is XXY.
(ii) Mention any one symptom of the diseases/disorders named above.
(i) The following types of diseases are likely to occur in human
(a) Phenylketonuria
(b) Down’s syndrome
(c) Klinefelter’s syndrome

(ii) Symptoms of above diseases are

(a) Phenylketonuria Accumulation of phenylketonuria causes mental retardation.
(b) Down’s syndrome Affected individuals are short statured with small round head.
(c) Klinefelter’s syndrome The individuals are sterile.

103.
Explain the mechanism of sex-determination in birds. How does it differ from that of
human beings?

Differences between XY and ZZ types of sex-determination are as follows

pe sex-determination pe sex-determination

mans) rds)

les are homogametic, while males are les are heterogametic, while males are
ogametic. ogametic.

le produces egg with X-chromosomes le produces eggs with either Z or

while male produces sperms with eitherromosomes, while male produces sperm
hromosomes Z-chromosomes.

ple humans, Drosophila ple fish, birds, etc.

104.
Explain how trisomy of 21st chromosome occurs in humans. List any four characteristic
features in an individual suffering from it.

An additional copy of chromosome number 21 (trisomy of chromosome number 21) in

humans results in Down’s syndrome.
Following characteristic symptoms are likely to develop in the child suffering from
Down’s syndrome

● Short statured with small, round head.

● Furrowed tongue and partially open mouth.

● Broad palm with characteristics palm crease.

● Slow mental, physical and psychomotor development.

105.
Both haemophilia and thalassemia are blood related disorders in humans. Write their
causes and the difference between the two. Name the category of genetic disorder they
both come under.

Haemophilia and thalassemia both are Mendelian or gene related human disorders.
Cause of Haemophilia It is caused due to the absence of antihaemophilic globulin and
plasma thromboplastin factor.
Cause of Thalassemia It is caused due to mutation or deletion of the genes controlling
the formation of globin chains (a and P) of haemoglobin.
Differences between haemophilia and thalassemia are as follows

mophilia ssemia

nce on single trait. nce on multiple traits.

tance is criss-cross. tance is straight from both the parents to all the
rings.

Category of genetic disorder

Haemophilia – Sex-linked disease
Thalassemia – Autosomal recessive disease

106.
Give the example of an autosomal recessive trait in humans. Explain its pattern of
inheritance with the help of a cross.

Sickle-cell anaemia is an autosomal recessive trait in human. The inheritance of

sickle-cell anaemia is shown in the cross given below
 107.
Which chromosomes carry the mutant genes causing thalassemia in humans? What are
the problems caused by these mutant genes?)

Thalassemia is an autosomal-linked recessive disease, which occurs due to either

mutation or deletion of genes, resulting in reduced rate of synthesis of one of the globin
chains of haemoglobin. It is associated with the mutation in chromosome 11 or 16
which code for p-chain and a – chain of haemoglobin, respectively. Haemoglobin is the
oxygen carrying component of the red blood cells. It consists of two different proteins,
an a and a p. If the body does not produce enough of either of these two proteins, the red
blood cells do not form properly and cannot carry sufficient oxygen.
This results in anaemia that begins in early childhood and lasts throughout the life.

108.
Name the phenomenon that leads to situations like ‘XO’ abnormality in humans. How do
humans with ‘XO’ abnormality suffer? Explain

XO abnormality or monosomy of
X-chromosome represents chromosomal disorder called Turner’s syndrome.

Non-disjunction is responsible for this chromosomal disorder. It is the phenomenon of

failure of segregation of the members of homologous pairs of the chromosomes. Such
disorder occurs due to the absence of one X-chromosome, i.e. 45 with XO (karyotype).
The affected individual has underdeveloped feminine characters. Females are sterile and
ovaries are rudimentary.

109.
(i) Sickle-cell anaemia in humans is a result of point mutation. Explain.
(ii) Write the genotypes of both the parents, who have produced a sickle-celled anaemic
offspring.
(i) Point mutation occurs due to change in a single base pair of DNA. Sickle-cell anaemia
occurs due to defect caused by the single base substitution at the sixth codon of the
P-chain of haemoglobin from GAG to GUG.

This causes substitution of glutamic acid by valine. The defective haemoglobin molecule
undergoes polymerisation under low oxygen tension causing sickle-shaped red blood
cells.

(ii) The parents must be heterozygotes, i.e. HbA Hbs and HbA Hbs to produce a
sickle-celled anaemic offspring.

110.
Name a disorder, give the karyotype and write the symptoms where a human male
suffers as a result of an additional X-chromosome.

It is a chromosomal disorder called Klinefelter’s syndrome, which occurs in males. The

affected human males have XXY sex chromosome (47 chromosomes).
Karyotype 44 + XY
Symptoms

● Development of feminine characters like breast development.

● Body hair sparse.

● Individual is sterile.

111.
(i) Why are grasshopper and Drosophila said to show male heterogamety? Explain.
(ii) Explain female heterogamety with the help of an exampl

(i) In male heterogamety, males produce two different types of gametes. In humans and
Drosophila, the males have one X and Y chromosome, whereas in grasshopper, the male
have only one X-chromosome (XO type). Thus, the males of these organisms show male
heterogamety as they produce two types of gametes
Either with or without X-chromosome
Some gametes with X-chromosome and some with Y-chromosome.

(ii) In some organisms, females produce two different types of gametes. This is termed
as female heterogamety. In birds and some reptiles, female has two different sex
chromosomes (one Z and one W-chromosome) whereas male has a pair of same
chromosome (a pair of Z-chromosome). (11/2)

112.
(i) Why are thalassemia and haemophilia categorised as Mendelian disorders? Write the
symptoms of these diseases. Explain their pattern of inheritance in humans.
(ii) Write the genotypes of the normal parents producing a haemophilic son.

(i) Thalassemia and haemophilia are categorised as Mendelian disorders because these
disorders are due to alteration in a single gene. Also, they are transmitted to offspring
through Mendelian principles of inheritance.
Symptoms and pattern of inheritance are given below

(a) Thalassemia It is an autosome-linked recessive blood disorder characterised by

defect in α, ß or δ-chain resulting in abnormal Hb molecule.
Symptom Anaemia Inheritance Two mutant alleles (one from each parent) must be
inherited for an individual to be affected, i.e. homozygous. Heterozygous are carriers and
may pass the mutant allele to their children.

(b) Haemophilia It is a sex-linked recessive disorder whose gene is located on

X-chromosome.
Symptom Prolonged clotting time and internal bleeding, even in a minor injury.

Inheritance The gene is present on X-chromosome, so it is inherited by males as they

have a single X-chromosome. Affected males are said to be hemizygous. Females have
2 X-chromosomes, thus their possibility of being affected is rare as the mother of such
female has to be at least carrier and father should be haemophilic.

(ii) Genotypes of the normal parents producing a haemophilic son are XCX (carrier
mother) and XY (father).

113.

About 8% of human male population suffers from colour blindness whereas only about
0.4% of human female population suffers from this disease. Write an explanation to
show how it is possible.
.

(ii) Colour blindness is an X-linked recessive trait which shows transmission from carrier
female progeny. In males, the defect can appear in heterozygous condition beacause
males possess only one X-chromosome whereas in females, it appears in homozygous
condition only, which is very rare. Thus, it is more common in males than females.

114.
(i) Explain the mechanism of sex-determination in humans.
(ii) Differentiate between male heterogamety and female heterogamety with the help of
an example of each.

A sex (male or female) is heterogametic, if it has two different sex chromosomes.

Male heterogamety The female has two X-chromosomes (XX), while male has only one
X-chromosome and at the time of gametogenesis the latter produces two types of
gametes, 50% with X-chromosome, while other 50% without X-chromosomes.
Heterogametic r tales are of two types

● XX-XY type, e.g. Drosophila, man, etc.

● XX-XO type, e.g. insects like grasshopper. Female heterogamety In some species
like birds,’fishes, etc. females are heterogametic, i.e. produce two types of gametes,
while males produce only one type of gamete.

Heterogametic females are of two types

● ZO-ZZ type, e.g. butterflies

● ZW-ZZ type, e.g. insects, fish, reptiles, birds, etc.

 115.
(i) Why is haemophilia generally observed in human males? Explain the conditions under
which a human female can be haemophilic.
(ii) A pregnant human female was advised to undergo MTP. It was diagnosed by her
doctor that the foetus she is carrying has developed from a zygote formed by an XX-egg
fertilised by Y carrying sperm. Why was she advised to undergo MTP?

(i) The genes for haemophilia arepresent on X-chromosome. A male has only one
X-chromosome and bears only one allele for the trait. He is hemizygous for the trait as
Y-chromosome does not have a corresponding allele. A female contains two
X-chromosomes. She has to be homozygous recessive to be haemophilic. It means her
father must be a sufferer and mother must be either a carrier or sufferer to carry forward
the disease.

(ii) The zygote will have XXY-chromosomes.

It will develop into a male with Klinefelter’s syndrome. Such males are sterile and show
feminine characters. That is why, female was advised to undergo MTP.

116.
(i) How does a chromosomal disorder differ from a Mendelian disorder?
(ii) Name any two chromosomal aberration-associated disorders.
(iii) List the characteristics of the disorders mentioned above that help in their diagnosis.

(i) Differences between chromosomal disorders and Mendelian disorders are

mosomal disorders elian disorders

e are caused due to the absence or excess of o

e are due to alteration in a single
ore chromosomes or abnormal arrangement of
more chromosomes.

are not transmitted as the affected individual iare transmitted to generations

e. gh Mendelian principles of
tance.

ple Down’s syndrome, Turner’s syndrome. ple colour blindness,

ylketonuria.

(ii) Chromosomal disorders Down’s syndrome and Klinefelter’s syndrome.

(iii) Down’s syndrome The affected individual is short statured with small round head,
furrowed tongue and partially open mouth. Physical, psychomotor and mental
development is also retarded.

Klinefelter’s syndrome It occurs in males, which show overall muscular development, but
feminine development also occurs. Such individuals are sterile.

117.
Explain the causes, inheritance pattern and symptoms of any two Mendelian genetic
disorders.

Two Mendelian genetic disorders are as follows

● Phenylketonuria It occurs due to defective allele on the autosome. It is inherited

from parent, who are heterozygous for the gene to the offspring.

The affected individual lacks an enzyme phenylalanine hydroxylase that converts the
amino acid phenylalanine into tyrosine. As a result, phenylalanine gets accumulated in
brain and body and gets converted into phenyl pyruvate and other derivatives.

Symptoms Accumulation of these compounds in the brain causes mental retardation.

They are also excreted in the urine due to poor absorption by kidney.

118.
Study the given pedigree chart showing the pattern of blood group inheritance in a
Family.

(i) Give the genotype of the following

(a) Parents
(b) The individual X in second generation.
(ii) State the possible blood groups of the individual Y in third generation.
(iii) Explain codominance with the inheritance of the blood group AB. (All India 2010C)

(i) (a) Parents A – IA IO

B- IB IO
(b) Xindividual-IB IO (B-type)

(ii) Individual Y-blood groups can be O or A.

(iii) IA and IB when stay together, show the phenomenon of codominance and express
themselves in the presence of each other.
In heterozygous hybrid, when both alternative alleles coexist, both the alleles show their
effect and result in the progeny with AB blood group. This is called codominance.
119.
Study the following pedigree chart of a family starting with mother with AB blood group
and father with 0 blood group.

(i) Mention the blood group as well as its genotype of the offspring numbered 1 in
generation II.
(ii) Write the possible blood groups as well as their genotypes of the offsprings
numbered 2 and 3 in genearation III.

(i) Offspring numbered 1 has blood group B with genotype IO IB.

(ii) Offspring 2 may have blood group A (IAIO) or blood group O (IOIO).
There can be two cases for offspring numbered 3 to know the possible blood groups as
well as their genotypes.
Case I When parent is homozygous for A, i.e. IA IA

120.
Naresh, a haemophilia patient, forbades his daughter to marry Mohan who is also a
haemophilia patient. Is Naresh right?
Naresh is right because his daughter can be carrier of haemophilia disease (an X-linked
disorder).
If she marries a haemophilic person, there are 75% chances that their offspring will also
suffer from haemophilia.

121.
Avantika’e doctor advised her to undergo MTP because through diagnostic tests she
came to know that her foetus has XO chromosomes. Do you think Avantika should
undergo MTP? Give reason.

(i) Yes, Avantika should undergo MTP.

(ii) The foetus contains XO chromosomes, i.e. one X-chromosome is lacking.
It means that the baby will be suffering from Turner’s syndrome. Therefore, MTP is
recommended.

122.
Why cannot O positive blood be transfused into o -ve body?
(ii) Human blood grouping is an example of?

(i) O+ blood group contains antigen for Rh factor. Rh– blood lacks the antigen, so if the
two blood mix, it will lead to clotting and thus, death of the patient.
(ii) Multiple allelism.

Molecular inheritance

1.
Write the dual purpose served by deoxyribonucleoside triphosphates in polymerisation.)

Deoxyribonucleoside Triphosphates (dNTPs) serve the dual purpose of

● acting as a substrate.

● providing energy (from two terminal phosphates).

2.
Why does hnRNA undergo splicing? Where does splicing occur in the cell?

hnRNA undergoes splicing to remove non-coding sequences, i.e. introns and joins exons.
Splicing occurs in the nucleus of the cell.

3.
Name the enzyme that transcribes hnRNA in eukaryotes.

The RNA polymerase-II transcribes the precursor of mRNA, i.e. the heterogeneous
nuclear RNA (hnRNA).
4.
Why is RNA more reactive in comparison to DNA?

The OH-group in the ribose of RNA at 2’end makes the molecule more reactive,
compared with DNA. RNA is also not stable under alkaline conditions.

5.
Name the negatively charged and positively charged components of a nucleosome.

Histone proteins forming an octamer is the positively charged component and the DNA
helix is the negatively charged component of nucleosome.

6.
What is cistron?

The segment of DNA coding for a polypeptide is known as cistron.

7.
Name the transcriptionally active region of chromatin in a nucleus.

Euchromatin (lightly stained) is the transcriptionally active region of chromatin in a

nucleus.

8.
What will happen if DNA replication is not followed by cell division in a eukaryotic cell?

If cell division is not followed after DNA replication then replicated chromosomes (DNA)
would not be distributed to daughter nuclei. A repeated replication of DNA without any
cell division results in the accumulation of DNA inside the cell.
This would increase the volume of the cell nucleus, thereby causing cell expansion.
Further, it will lead to polyploidy.

9.
Why is it not possible for an alien DNA to become part of chromosome anywhere along
its length and replicate normally?

An alien DNA cannot become a part of chromosome anywhere along its length and
replicate normally due to the absence of origin of replication where the replication
process is initiated.

10.
Name the enzyme and state its property that is responsible for continuous and
discontinuous replication of the two strands of a DNA molecule.

DNA-dependent DNA polymerase This enzyme can polymerise deoxynucleotides in 5′ →

3′ direction only.
Due to this, replication of DNA is continuous in one strand with polarity 3′ → 5’while
discontinuous in another polarity 5′ → 3′.

11.
Name the enzyme that joins the small fragment of DNA of a lagging strand during DNA
replication.

DNAligase.

12.
Which one of an intron and an exon is the reminiscent of antiquity?

Intron is considered to be the reminiscent of antiquity because these are non-coding

sequences.

13.
Name the specific components and the linkage between them that forms
deoxyadenosine.

The specific components of deoxyadenosine are adenine and deoxyribose. These are
linked by N-glycosidic linkage.

14.
Which one out of rho factor and sigma factor acts as an initiation factor during
transcription in a prokaryote?

Sigma factor acts as an initiation factor and binds to the promoter during transcription
in prokaryotes.

15.
Name the enzyme involved in continuous replication of DNA strand. Mention the polarity
of the template strand.

Enzyme involved in continuous replication of DNA strand is DNA polymerase.

Template strand has 3′ → 5′ polarity.

16.
Name the two basic amino acids that provide positive charges to histone proteins.

Lysine and arginine.

17.
If the base adenine constitutes 31% of an isolated DNA fragment, then what is the
expected percentage of the base cytosine in it?

According to Chargaff’s rule A + G = C+ T = 50%

∴ if A = 31% then T = 31%;
C + T = 50%
∴ C = 50% – 31% = 19%

18.
A structural gene has two DNA strands X and Y shown along side. Identify the template
strand.

‘X’ is template strand. It is because the template strand has the polarity in 3′ → 5′
direction.

19.
What is an origin of replication in a chromosome? State its function.

Replication does not initiate randomly at any place in DNA (chromosome). So, there is a
definite region in DNA (chromosome) termed as origin of replication. It is the place from
where DNA replication originates. It helps in propagation of a piece of DNA, during
recombinant DNA procedures.

20.
Although a prokaryotic cell has no defined nucleus, yet DNA is not scattered throughout
the cell. Explain.

Prokaryotes lack a defined nucleus. However, the DNA is not scattered throughout the
cell. This is due to the fact that the negatively charged DNA is held with some positively
charged proteins in nucleoid or the middle region. In this region, the DNA is organised in
large loops held by proteins.

21.

Draw a labelled diagram^of a nucleosome. Where is it found in a cell?

Nucleosome is a complex of negatively charged DNA. wrapped around the positively
charged histone octamer (unit of 8 molecules of histone). It is found in the nucleus of
the cell.
It is made up of four types of proteins; H2A, H2B, H3 and H4 occurring in pairs. The
histone proteins acquire positive charge due to abundance of amino acid residues, i.e.
lysine and arginines. Both these amino acids carry positive charges in their side chains.
A typical nucleosome consists of 200 base pairs of DNA helix.

22.
Discuss the role of DNA ligase during DNA replication.

DNA ligase facilitates the joining of Okazaki fragments in lagging DNA strands together
by catalysing the formation Of phosphodiester bond. It also plays a role in repairing
single-strand breaks in duplex DNA.

23.
In a typical nucleus, some regions of chromatin are stained light and others dark. Explain
why is it so and what is its significance.

In a typical nucleus, some regions of chromatin are stained light because of loose
packing of chromatin and some regions of chromatin are stained dark because,
chromatin is densely packed. Euchromatin is transcriptionally active chromatin (lightly
stained) while heterochromatin (darkly stained) is transcriptionally inactive.

24.
Explain the two factors conferring stability to double helix structure of DNA.

Two factors responsible for conferring stability to double helix structure of DNA are

● Stacking of one base pair over other.

● H-bonds between nitrogenous bases.

25.
difference between structural genes in a transcription unit of prokaryotes and
eukaryotes.

Prokaryotic structural genes are continuous and are without any non-coding region,
while eukaryotic structural genes contain exons (coding DNA) and introns (non-coding
DNA).

26.
Show DNA replication with the help of a diagram only.
The replication fork of DNA formed during DNA replication.

27.
A template strand is given below. Write down the corresponding coding strand and the
mRNA strand that can be formed, along with their polarity.
3′- ATGCATGCATGCATGCA TGCATGC-5′

For the given template strand 3- ATGCATGCAT GCATGCATGCATGC- 5′

Coding strand is 5′- TACGTACGTACGTACGTACG TACG – 3′
and mRNA strand is 5′- UACGUACGUACGUACGU ACGUACG – 3′

28.
Draw a schematic diagram of a part of double-stranded dinucleotide DNA chain having
all the four nitrogenous bases showing the correct polarity.

Schematic diagram of a double stranded dinucleotide DNA chain having all the four
nitrogenous bases (A, T, G, C) with polarity.

29.
State the functions of the following in a prokaryote
(i) tRNA (ii) rRNA

(i) tRNA helps in transferring amino acids to ribosome for the synthesis of polypeptide
chain.
(ii) rRNA plays structural and catalytic role during translation.

30.
Differentiate between a cistron and an exon.

Differences between cistron and exon are as follows

on

on is segments of DNA that possess bo

s are coding regions of a gene that code
and intron. ent proteins.

ers to the whole gene. ers to a part of gene.

31.
Differentiate between exons and introns.

Differences between exons and introns are as follows

s ns

ns of a gene which become part of ns of a gene which do not form part of mRNA
A.

for the different proteins and henceoved during the processing of mRNA because
d coding sequence. are non-coding sequences.

32.
State the dual role of deoxyribonucleoside triphosphates during DNA replication.

● The deoxyribonucleoside triphosphates are the building blocks for the DNA strand
(polynucleotide chain), i.e. they act as substrates.

● These also serve as energy source in the form of ATP and GTP because of the
presence of high energy terminal phosphate groups.
33.
the s based on the dinucleotide shown below.

(i) Name the type of sugar guanine base is attached to.

(ii) Name the linkage connecting the two nucleotides.
(iii) Identify the 3′ end of the dinucleotide. Give a reason for your .
(i) Pentose sugar or deoxyribose sugar.
(ii) Two nucleotides are linked through 3′-5′ phosphodiester linkage to form a
dinucleotide.
(iii) The ribose sugar has a free 3′ – OH group which is referred to as 3’end of the
polynucleotide chain.

34.
Make a labelled diagram of an RNA dinucleotide showing its 3′ → 5′ polarity.

RNA dinucleotide.

35.
the following s based on Meselson and Stahl’s experiment on E.coli.
(i) chemical substance used as the source of nitrogen.
(ii) Why did they synthesise light and heavy DNA in the organism?
(iii) How did they distinguish heavy DNA from light DNA?
(iv) conclusion arrived at.

(i) NH4Cl (Ammonium chloride) with N14 atom.

(ii) It is to show that after one generation E. coli with 15N -DNA in a medium of 14N, has
DNA of intermediate density between the light and heavy DNAs.
It shows that of the two strands, only one strand is synthesised newly, using the
14N-nitrogen source in the medium.
(iii) The heavy and light DNA can be differentiated by centrifugation in a cesium chloride
(CsCl) density gradient.
The 15N -DNA was heavier than 14N -DNA and the hybrid 15N – 14N -DNA was
intermediate between the two newly form DNA strands.
(iv) Scientists concluded that the DNA replication is semi-conservative, i.e. of the two
strands of DNA, one is the parental strand, while the other is newly synthesised.
36.
Explain the mechanism of DNA replication. Role of DNA-ligase in replication fork?

Replication in DNA strand occurs within a small opening of the DNA helix, known as
replication fork.
DNA polymerases catalyse polymerisation only in one direction, i.e. 5′ → 3′. It creates
additional complications at the replicating fork. Consequently, on one strand (template 3′
→ 5′), the replication is continuous. This is known as leading strand, while on the other
strand (template 5′ → 3′), it is discontinuous. This is known as lagging strand.
The discontinuously synthesised fragments called Okazaki fragments are later joined by
DNA ligase.

37.
Construct and label a transcription unit from which the RNA segment given below has
been transcribed. Write the complete name of the enzyme that transcribed this RNA.

The process of copying genetic information from one strand of DNA into RNA is called
‘transcription’. Transcription is catalysed by ‘DNA dependent RNA polymerase’.
The RNA molecule given in should be

As RNA have uracil in place of thymine.

For given RNA, the transcription unit will be

38.
Why is DNA molecule considered as a better hereditary material than RNA molecule?
DNA is considered a better genetic material when compared to RNA because of its high
stability and low reactivity. RNA being more reactive, is labile and easily degradable.
The high reactivity of RNA is contributed by the 2-OH group in the nucleotides. RNA as a
catalyst (e.g. ribozyme) is also more reactive. Uracil present in RNA makes it unstable
over DNA that contains thymine.

39.
Name the three polymerases found in eukaryotic cells and mention their functions.
Three types of RNA polymerases are found in eukaryotic cells and their functions are as
follows

● RNA polymerase-I transcribes rRNAs.

● RNA polymerase-II transcribes the precursor of mRNA called hnRNA.

● RNA polymerase-III transcribes tRNA, 5 SrRNA and snRNAs.

41.
Study the diagram given below
Name the linkage X, Y, Z and the respective molecules formed by them.

X N-glycosidic linkage. It forms nucleoside.

Y Phosphoester linkage. It forms nucleotide.
Z 3′-5’phosphodiester linkage. It forms polynucleotide.

42.
Describe the experiment that helped demonstrate the semi-conservative mode of DNA
replication.

Matthew Meselson and Franklin Stahl conducted an experiment with Escherichia coli
(1958) as follows

● They grew many generations of E. coli in a medium that contained 15NH4Cl (15N
is the heavy isotope of nitrogen) as the only source of nitrogen. The result was that 15N
was incorporated into the newly synthesised DNA. Upon centrifugation in a cesium
chloride (CsCl) density gradient, this heavy DNA molecule could be distinguished from
the normal DNA.

● The cells were then transferred into a medium containing normal 14NH4Cl.

● At definite time intervals, as the cells multiplied, samples were taken and the DNA
which remained as double-stranded helices were extracted.

● The samples were separated independently on CsCl gradient to measure the

densities of DNA.

● The DNA obtained from the culture, one generation after the transfer from 15N to
14N medium had a hybrid or intermediate density.
● The DNA obtained from the culture after another generation (generation II), was
composed of equal amounts of hybrid DNA and Tight’ DNA.

Thus, Meselson and Stahl concluded that the DNA replication is semi-conservative, i.e.
out of the two strands of DNA one is the parental
strand, while another is newly synthesised.

43.
(i) Differentiate between a template strand and coding strand of DNA.
(ii) Name the source of energy for the replication of DNA.

(i) Differences between template strand and coding strand are as follows

late strand g strand

3′- 5’polarity. 5′- 3’polarity.

s transcribed. s not get transcribed.

quence is complementary to mRNquence is same as mRNA except it contains

ed. ad of ‘U’.

(ii) The source of energy for the replication of DNA are the deoxyribonucleoside
triphosphates that have two terminal high energy phosphates.

44.
A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine
containing nucleotides. How many pyrimidine bases this DNA segment possesses?

According to Chargaff’s rule, ratio of purines to pyrimidines is equal,

i.e. A + G = C + T
The number of adenine (A) is equal to the number of thymine (T).
A = 240(given)
Therefore, T = 240
Also, the number of guanine (G) is equal to cytosine (C).
Thus, G + C = 1000 – [A + T]
G + C = 1000 – 480= 520
Hence, G = 260, C = 260
The number of pyrimidine bases,
i.e. C + T= 240 + 260 = 500

(ii) Diagrammatic sketch of portion of DNA segment

45.
With the help of a schematic diagram, explain the location and role of the following in a
transcription unit. Promoter, structural gene, terminator.

Structure of a transcription unit

The promoter and terminator flank the structural gene in a transcription unit. The
promoter is located towards 5’end (upstream) of the structural gene and it helps to
initiate transcription by binding with RNA polymerase. The terminator is located toward
3’end (downstream) of the coding strand and it usually defines the end of the process of
transcription. The structural gene is present in between promoter and terminator. It
codes for enzyme or protein for structural functions.

46.
(i) What are the transcriptional products of RNA polymerase-III?
(ii) Differentiate between capping and tailing.
(iii) Expand hnRNA.

(i) RNA polymerase-III is responsible for the transcription of tRNA, 5s rRNA and snRNAs
(small nuclear RNAs).
(ii) In capping process, an unusual nucleotide (methyl guanosine triphosphate) is added
to ‘5’ end of hnRNA.
In tailing process, 200-300 adenylate residues are added at 3′ end of hnRNA.
(iii) hnRNA is heterogeneous nuclear RNA.

47.
It is established that RNA is the first genetic material. Explain giving three reasons.

RNA is the first genetic material in cells because

● RNA is capable of both storing genetic information and catalysing chemical

reactions

● Essential life processes (such as metabolism, translation, splicing, etc.) were

evolved around RNA

● It shows the power of self-replication.

48.
(i) Construct a complete transcription unit with promoter and terminator on the basis of
hypothetical template strand given below.

(ii) Write the RNA strand transcribed from the above transcription unit along with its
polarity.

(i) Transcription unit

(ii) RNA strand transcribed from the above transcriptional unit

49.
List the salient features of double helix structure of DNA.
Salient features of DNA double helix are as follows
● It is made up of two polynucleotide chains containing the backbone of
sugar-phosphate from which the bases project inside.

● The two chains have anti-parallel polarity, i.e. one of them is 5′ → 3′, the other has
3′ → 5’polarity.

● The bases in two strands are paired through hydrogen bond (H-bonds) forming
base pairs (bp). Adenine pairs through two hydrogen bonds with thymine on opposite
strand and vice-versa. In the same way, guanine is bonded with cytosine through three
H-bonds. Due to which, purine always comes opposite to a pyrimidine.

● The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm
and there are roughly 10 bp in each turn. Consequently, the distance between base pair
in a helix is about 0.34 nm.

● The plane of one base pair stacks over the other in double helix. This confers
stability to the helical structure.

50.
How is hnRNA processed to form mRNA?

The precursor of mRNA transcribed by RNA polymerase-II is called heterogeneous

nuclear RNA (hnRNA). It undergoes following processing to form nascent rRNA.

● Splicing In this process, the non-coding introns are removed and coding sequences
called exons are joined in a definite order. This is required because primary transcript
contains both introns and exons.

● Capping Tailing

● The fully processed mRNA is released from the nucleus into cytoplasm for
translation.

51.
The base sequence in one of the strands of DNA is TAGCATGAT.
(i) Give the base sequence of the complementary strand.
(ii) How are these base pairs held together in a DNA molecule?
(iii) Explain the base complementarity rule. Name the scientist who framed this rule.

(i) ATCGTACTA.
(ii) Base pairs are held together by weak hydrogen bonds, adenine pairs with thymine by
two H-bonds and guanine pairs with cytosine through three H-bonds.
(iii) Base complementarity rule For a double-stranded DNA, this rule states that the ratio
between adenine and thymine and that between guanine and cytosine are constant and
equal to one. Erwin Chargaff framed this rule.

52.
Why do you see two different types of replicating Strands in the given DNA replication
fork? Explain. Name these strands.

Two different types of parent strands function as template strands.

On the template strand with 3′ → 5’polarity, the new strand is synthesised as a
continuous strand because the enzyme DNA polymerase can carry out polymerisation of
the nucleotides only in 5′ → 3′ direction. This is called continuous synthesis and the
strand is called leading strand.

On the template strand with 5′-» 3’polarity, the new strand is synthesised from the point
of replication fork in short stretches called Okazaki fragments and this strand is called
lagging strand. Okazaki fragments are later joined by DNA ligases to form a continuous
strand.

53.
Monocistronic structural genes in eukaryotes have interrupted coding sequences.
Explain. How are they different in prokaryotes?

Monocistronic structural genes in eukaryotes have interrupted coding sequences due to

the presence of introns, i.e. non-coding sequences.

Differences between monocistronic structural gene in prokaryotes and in eukaryotes are

as follows

tural gene in prokaryotes tural gene in eukaryotes

sts of only coding sequences. sts of both coding and non-coding sequenc

mation is continuous as only exonsmation is split due to the presence of introns

resent. een exons.

is no need of splicing. ng is required to make functional mRNA.

54.
Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase.
Hershey and Chase used 35S and 32P in their culture medium. These are radioactive
sulphur and phosphorus, respectively. These two components were used to detect
whether the genetic material is DNA or protein.

Role of 32P and 35S Viruses grown on medium with 32S. had non-radioactive genetic
material and radioactive protein as sulphur is not present in DNA but found in protein.
While those grown on 32P had radioactive genetic material because DNA contains
phosphorus but proteins does not contain it. Thus, it was established that DNA is the
genetic material.

55.
Describe the initiation process of transcription in bacteria.

Initiation process of transcription in bacteria RNA polymerase becomes associated

transiently to an initiation factor sigma (cr) and binds to specific sequence on DNA
called promoter to initiate transcription.
A single DNA-dependent RNA polymerase catalyses the transcription of all the three
types of RNA.

56.
Describe the elongation process of transcription in bacteria.

Elongation process of transcription in bacteria RNA polymerase facilitates opening of

the DNA helix after binding to promoter. It uses ribonucleoside triphosphates as
substrate and polymerises the nucleotides in a template dependent fashion following
complementarity. The process continues till RNA polymerase reaches the terminator
region on the DNA strand.

57.
Describe the termination process of transcription in bacteria.

Termination occurs when RNA polymerase reaches the terminator region and the
nascent RNA falls off. The RNA polymerase being transiently associated with
termination factor rho (p) also falls off the transcription unit.

58.
Draw a schematic representation of a dinucleotide. Label the following
(i) The component of a nucleotide
(ii) 5′ end
(iii) N-glycosidic linkage
(iv) Phosphodiester linkage

Schematic representation of a dinucleotide.

59.
Describe the role of RNA polymerases in transcription in bacteria and in eukaryotes.

In bacteria, there is a single DNA-dependent RNA polymerase, which catalyses the

formation of mRNA, tRNA and rRNA.
In eukaryotes, there are three types of RNA polymerases, which show division of labour.
These are as follows

● RNA polymerase-I transcribes rRNAs, (28S, 18S and 5.8S)

● RNA polymerase-II transcribes the precursor of mRNA called hnRNA.

● RNA polymerase-III transcribes tRNA, 5 SrRNA and snRNAS.

60.
(i) State the ‘central dogma’ as proposed by Francis Crick.
(ii) Explain, how the biochemical characterisation (nature) of ‘Transforming Principle’
was determined, which was not defined from Griffith’s experiments.

(i) Central Dogma It states that there is one way or unidirectional flow of information
from master copy DNA to working copy RNA (transcription) and from working copy RNA
to building plan of polypeptide chain (translation).

Central dogma of molecular biology was proposed by Crick (1958). It can also be written
as follows

Reverse transcription is an exception to central dogma. The discovery of ‘RNA

dependent DNA polymerase’ gave the concept of central dogma reverse or teminism. It
suggested that information flow can also take place from RNA to DNA.

During this, transcription or RNA synthesis occurs over DNA and translation or protein
synthesis occurs over ribosomes. In eukaryotes, these two processes are separated in
time and space. This protects DNA from respiratory enzymes and RNA’s degradation
from nucleases.

(ii) Various experiments were conducted to prove that DNA is the genetic material.
Griffith did his experiment and concluded that some substance causes transformation.
However, he did not tell the substance that had caused the transformation.

Biochemical Characterisation of Transforming Principle Three scientists Avery, MacLeod

and McCarty revealed that the chemical nature of the transforming substance was DNA.
They showed that DNA isolated from S-strain could itself confer the pathogenic
properties to R-strain.

This fact suggested that DNA possesses the genetic properties.

The outline of the experiment is given below
Thus, on the basis of their experiment, they concluded that DNA is the hereditary
material.

61.
(i) Why does DNA replication occur in replication forks and not in its entire length?
(ii) Why is DNA replication continuous and discontinuous in a replication fork?
(iii) State the importance of origin of replication in a replication fork.

(i) DNA replication occurs in small replication fork and not in its entire length because
whole DNA cannot be opened in one stretch due to high energy requirement.

(ii) DNA replication is continuous and discontinuous in a replication fork because the
enzyme DNA polymerase can carry out polymerisation of the nucleotides only in 5-3′
direction. On the template strand with 3-5′ polarity, DNA replication is continuous. On the
template strand with polarity 5′-3′, the DNA replication occurs in short stretches and is
called discontinuous.

(iii) Replication of DNA does not initiate randomly, and DNA polymerases on their own
cannot initiate replication. So, there is a need of specific sequence, called origin of
replication from which the replication starts. DNA polymerase binds to it and continues
the process.

62.
the following s based on Hershey and Chase’s experiments
(i) virus they worked with and why?
They worked with bacteriophage, i.e. viruses that infect bacteria. These viruses were
used because during infection they transfer their genetic material into bacteria.

(ii) Why did they use two types of culture media to grow viruses in? They used
two types of culture media, containing 35S and 32P, so as to compare that which one out
of DNA and proteins gets transferred from virus to bacteria and acts as genetic material.

(iii) What was the need for using a blender and later a centrifuge during their
experiments?
A blender and centrifuge was used to open up the bacterial cells and viral particles, so,
that genetic material could be exposed

(iv) State the conclusion drawn by them after the experiments.

Conclusion DNA is the genetic material.

How did Hershey and Chase established that DNA is transferred from virus to bacteria?
Alfred Hershey and Martha Chase (1952) established that DNA is transferred from
viruses that infect bacteria. In their experiment, bacteriophages (viruses that infect
bacteria) were used. They grew some viruses on a medium that contained radioactive
phosphorus and some others on medium containing radioactive sulphur. Viruses grown
in the presence of radioactive phosphorus contained radioactive DNA, but not
radioactive protein because DNA contains phosphorus, but protein does not. In the same
way, viruses grown on radioactive sulphur contained radioactive protein, but not
radioactive DNA because DNA does not contain sulphur. Radioactive phages were
allowed to attach to E.coli bacteria. As the infection proceeded, viral coats were
removed from the bacteria by agitating them in a blender. The virus particles were
seperated from the bacteria by spinning them in centrifuge.

Bacteria which were infected with viruses that had radioactive DNA were radioactive,
indicating that DNA was the material that passed from the virus to the bacteria. Bacteria
that were infected with viruses that had radioactive proteins were not radioactive. This
indicated that the proteins did not enter the bacteria from viruses. It proved that DNA is
the genetic material that is passed from virus to bacteria.

63.
Name the stage in the cell cycle where DNA replication occurs.;;During the synthetic
phase of interphase of cell cycle DNA duplicates or replicates.

Explain the mechanism of DNA replication. Highlight the role of enzymes in the process

Process of DNA Replication Replication is an energy expensive process,

deoxyribonucleoside triphosphate serves the dual purpose of

● acting as a substrate.

● providing energy (from two terminal phosphates).

Why is DNA replication said to be semi-conservative

It is said to be semi-conservative because in newly synthesised DNA, one strand is

parental and one is new, so it conserves the one strand.

What would happen if cell division is not followed after DNA replication?

DNA replication is followed by cell division. In case, the latter fails to occur, polyploidy
may occur.

(i) Draw a labelled diagram of a ‘replicating fork’ showing the polarity. Why does DNA
replication occur within such ‘forks’?In a long DNA molecule, replication takes place
within a small opening of the DNA helix, known as replication fork because whole DNA
does not open in one stretch due to high energy requirement. DNA dependent DNA
polymerases catalyse polymerisation only in one direction, i.e. 5′ → 3′. This brings
additional complications at the replication fork.

Name two enzymes involved in the process of DNA replication, along with their
properties.

DNA Ligase It facilitates the joining of DNA strands together by catalysing the formation
of phosphodiester bond. It plays a role in repairing single strand breaks in duplex
DNA.On one strand (template with polarity 3′ → 5′), replication is continuous, while on
the 5′ → 3′ strand, replication is discontinuous.
Thus, the fragments synthesised are discontinuous and later joined by the enzyme DNA
ligase.Helicase It unwinds the DNA strand, i.e. separates the two strands from one point
for the formation of a replication fork.DNA-Dependent DNA Polymerase It is the main
enzyme which uses a DNA template to catalyse the polymerisation of deoxynucleotides.
The average rate of polymerisation is 2000 bp (base pairs) per second approximately.
These enzymes are highly efficient as they have to catalyse polymerisation of a large
number of nucleotides in a very short time. These polymerases act very fast, catalyse
the replication process with high degree of accuracy as any mistake would result in
mutations.

The DNA polymerases cannot initiate replication process on their own and replication
does not initiate randomly anywhere in DNA.
It begins at definite regions in a DNA molecule known as origin of replication (Ori).

64.
List the criteria of a molecule that can act as genetic material must fulfil. Which one of
the criteria is best fulfilled by DNA or by RNA thus making one of them a better genetic
material than the other? Explain.

From the Hershey and Chase experiment, the fact was established that DNA acts as
genetic material. But later studies revealed that in some viruses (e.g. Tobacco Mosaic
Viruses, QB bacteriophage, etc.) RNA is the genetic material. Following are the criteria
that a molecule must fulfil to act as a genetic material.

● It- should be able to replicate.

● It should be chemically and structurally stable.

● It should provide the scope for slow changes (mutation), which are required for
evolution.

● It should be able to express itself in the form of ‘Mendelian characters’.

According to these criteria, both DNA and RNA have the ability to direct their
duplications (because of the rule of base pairing and complementarity).

So, both the nucleic acids (DNA and RNA) have the ability to direct their duplications,
whereas the other molecules in the living system, fail to fulfil first criteria itself, e.g.
protein. The most important criteria of genetic material is the stability as the genetic
material should not change with the different stages of life cycle, age or with change in
the physiology of an organism. Both DNA and RNA have the ability to mutate. Since,
RNA is unstable, it mutates at a faster rate. That is why, those viruses, which have RNA
genome and a shorter lifespan, undergo mutation and thus, evolve rapidly.

DNA is dependant on RNA for protein synthesis, whereas RNA directly codes for protein
synthesis. This proves that both RNA and DNA act as genetic material, but DNA being
more stable is preferred for the storage of genetic information.

65.
What is central dogma? Who proposed it?

Central Dogma ;Francis Crick proposed the central dogma in molecular biology, which
states that the genetic information flows from DNA → RNA → Proteins.

66.
(i) How are the following formed and involved in DNA packaging in the nucleus of a cell?
(a) Histone octamer
(b) Nucleosome
(c) Chromatin
(ii) Differentiate between euchromatin and heterochromatin. (All India 2016)

(i) (a) Histone Octamer ;Histones are the proteins that are rich in basic amino acids, i.e.
lysine and arginine. Both these amino acids carry positive charges in their side chains.
So, histones are a set of positively charged, basic proteins (histones + protamine). A
histone octamer is formed by the organisation of two molecules each of H2A, H2B, H3
and H4 histones so as to make a unit of 8 molecules. It helps to package DNA into
nucleosome.

(b) Nucleosome The negatively charged DNA is wrapped around the positively charged
histone octamer, forming a structure known as nucleosome. It forms the fundamental
repeating units of eukaryotic chromatin. It is used to pack the large eukaryotic genome
into the nucleus.

(c) Chromatin One nucleosome contains 200 base pairs of DNA helix, approximately.
Nucleosomes are the repeating unit of chromatin, which are thread-like stained
(coloured) bodies present in nucleus.

The nucleosomes in chromatin look like ‘beads-on-string’ when observed under an

electron microscope. The chromatin is further packed to form a solenoid structure of 30
mm diameter and further supercoiling tends to form a looped structure called chromatin
fibre and then chromatid. This further coils and condenses at metaphase stage to form
the chromosomes.

(ii) Differences between euchromatin and heterochromatin are as follows

omatin ochromatin

ghtly stained region. arkly stained region of the chromatin

mosome).

oosely coiled region and thus, hahe compactly coiled region and thus, has more
DNA.
 anscriptionally active and is anscriptionally inert and cannot be transcribed
cribed into mRNA. mRNA due to very tight coiling.

67.
(i) Describe the experiment which demonstrated the existence of transforming principle.
(ii) How was the biochemical nature of this transforming principle determined by Avery,
MacLeod and McCarty? (Foreign 2015)

(i) Transforming Principle Frederick Griffith in 1928, carried out a series of experiments
with Streptococcus pneumoniae (a bacterium that causes pneumonia).He observed that
when these bacteria (Streptococcus pneumoniae) were grown on a culture plate, some
of them produced smooth, shiny colonies (S-type), whereas the others produced rough
colonies (R-type).This difference in appearance of colonies (smooth/rough) was due to
the presence or absence of mucus (polysaccharide) coat.
In his experiments, he first infected two separate groups of mice as follows S-strain
(virulent strain) → Inject into mice → Mice die.

R-strain (non-virulent strain) → Inject into mice → Mice live.

S-strain (heat-killed) → Inject into mice → Mice live.
S-strain (heat-killed) + R-strain (live) → Inject into mice → Mice die.

Griffith concluded that the live R-strain, bacteria, had been transformed by the heat-killed
S-strain bacteria. He concluded that, there was some ‘transforming agent that was
transferred from the heat-killed S-strain, which helped the R-strain bacteria to synthesise
a smooth polysaccharide coat and thus, become virulent.He concluded that must be due
to the transfer of the genetic material. However, he was not able to define the
biochemical nature of genetic material from his experiments.

(ii) Contributions of MacLeod, McCarty and Avery worked to determine the chemical
nature of transforming principle of Griffith’s experiments. They discovered that protein
digesting enzymes (proteases) and RNA-digesting.enzymes (RNase) did not affect
transformation. Digestion with DNase did inhibit transformation suggesting that DNA
was responsible for the transformation.

68.
Explain the processing the /mRNA needs to undergo before becoming functional mRNA
in eukaryotes.

In eukaryotes, the primary transcript is often larger than the functional RNA, called
heterogeneous nuclear RNA or hnRNA.
Post-transcription processing is necessary to convert primary transcript of different
types of RNAs into functional RNAs.
The processing includes

● Cleavage Larger RNA precursors are cleaved to form smaller RNAs.

● Splicing Eukaryotic transcripts possess extra segments called introns which are
non-functional. They do not appear in mature or processed RNA. The functional coding
sequences are called exons. Splicing is the removal of introns and fusion of exons in a
definite order to form functional RNAs.

● Terminal additions (capping and tailing) Capping includes addition of an unusual

nucleotide to 5’ end of hrRNA. These unusual segments are CCA segment in tRNA and
cap nucleotides at 5’end of mRNA. In tailing, poly-A segment (200-300 residues) are
added at 3’end of mRNA. Cap is formed by the modification of GTP into 7-methyl
guanosine or 7 mG.

● Nucleotide modifications They are most common in tRNA-methylation (i.e.

methyl cytosine, methyl guanosine), deamination (e.g. inosine form adenine),
dihydrouracil, pseudouracil, etc.

70.
(i) Explain the role of DNA-dependent RNA polymerase in initiation, elongation and
termination during transcription in bacterial cell.
(ii) How is transcription a more complex process in eukaryotic cells? Explain.

(i) Role of DNA-dependent RNA polymerase

● RNA polymerase becomes associated transiently with initiation factor (σ) and
binds to the promoter site on DNA and initiates transcription.

● It uses the nucleoside triphosphate as substrates and polymerises them in a

template-dependent fashion following the base complementarity rule in the 5′ → 3′
direction.

● It also facilitates the opening of the DNA helix and continues the elongation
process.

● When the polymerase falls off a terminator region on the DNA, the nascent RNA
separates. This results in termination.

(ii) Reasons that transcription is more complex in eukaryotes are

(a) The three types of RNA polymerases in the nucleus show division of labour.

● RNA polymerase-I transcribes rRNAs (28S, 18S and 5.8S).

● RNA polymerase-II transcribes the precursor of mRNA, called hnRNA.

● RNA polymerase-III transcribes tRNA, 5 SrRNA and snRNAse.

(b) hnRNA contains both coding sequences called exons and non-coding sequences
called introns. So, it undergoes a process called splicing, in which the non-coding
sequences (introns) are removed and the coding sequences (exons) are joined together
in a defined order.

(c) In capping, unusual nucleotide, methyl guanosine triphosphate residues are added at
the 5’end of the hnRNA.

(d) In tailing, 200-300 adenylate residues are added at the 3’ end of the hnRNA.

71.
Mention the contributions of Maurice Wilkins and Rosalind Franklin

(i) (a) Maurice Wilkins and Rosalind Franklin They carried out X-ray diffraction studies on
the structure of DNA molecule.

73.
Name one amino acid, which is coded by only one codon.

Methionine.

74.
Give an example of a codon having dual function.

AUG is a codon with dual functions. It codes for the amino acid methionine (met) and
also acts as an initiator codon of polypeptide synthesis during protein synthesis.

75.
Write the two specific codons that a translational unit of mRNA is flanked by one on
either sides.

Two specific-codons that are flanked on either sides of mRNA in a translation unit are

● Start codon (AUG)

● Stop codon (UAG or UAA or UAA).

76.
How does a degenerate code differ from an unambiguous one?

When some amino acids are coded by more than one codon, the code is said to be
degenerate.
On the other hand, when one codon codes for only one amino acid, it is called
unambiguous.

77.
How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic
experiments?

Satellite DNA is separated from bulk genomic DNA by density-gradient centrifugation

technique.

78.
State which human chromosome has
(i) the maximum number of genes and
(ii) the one which has the least number of genes.

(i) Chromosome 1 (2968 genes).

(ii) Chromosome Y (231 genes).

79.
Which one of the two subunits of ribosome encounters an mRNA?

Smaller subunit.

80.
Differentiate between the following Inducer and repressor in lac operon.

Inducer It is a chemical which after coming in contact with repressor, changes it into
non-DNA binding state, so as to free the operator gene.
Repressor It is a regulator protein meant for blocking the operator gene.

81.
Mention the role of the codons AUG and UGA during protein synthesis.

AUG acts as initiation codon and codes for amino acid methionine.
UGA acts as stop/termination codon that signals termination of polypeptide synthesis.

82.
Mention the contribution of genetic maps in human genome project.

Genetic maps are used as a starting point in the sequencing of whole genome.

83.
Mention any two ways in which Single Nucleotide Polymorphisms (SNPs) identified in
human genome, can bring out revolutionary changes in biological and medical sciences.

● By tracing human history.

● By finding chromosomal locations for disease associated sequences.

84.
Differentiate between the genetic codes given below
(i) Unambiguous and universal
(ii) Degenerate and initiator

(i) The difference between unambiguous and universal genetic codes is as follows
Unambiguous
In genetic code, one codon codes for only one amino acid, hence it is unambiguous.

Universal
The genetic code is universal, i.e. each codon codes for same amino acid in all
organisms.

(ii) The difference between degenerate and initiator codes is as follows

nerate or

e amino acids are coded by more than one e codons act as start signal for
n, hence the code is degenerate. ation,

erine, leucine, arginine are encoded by 6 codoUG acts as initiator codon and it
s for methionine.

85.
Following are the features of genetic codes. What does each one indicate?
Stop codon, Unambiguous codon, Degenerate codon, Universal codon.

● Stop codon does not code for any amino acids, e.g. UGA.

● Unambiguous codon codes for only one . amino acid, e.g. CCG codes for proline.

● Degenerate codon Genetic code is described as degenerate when a single amino

acid is coded by more than one codon, e.g. serine is coded by 6 codons.

● Universal codons codes for the same amino acid in all organisms (except in
mitochondria and few protozoan).

86.
What is aminoacylation? State its significance.

Aminoacylation or charging of tRNA is a process in which amino acids get activated in

the presence of ATP and get linked to their cognate tRNA.
Significance of aminoacylation The charged tRNA carries amino acids to the site of
protein synthesis. If two charged tRNAs are brought close to each other, the formation of
peptide bond is favoured energetically.
87.
State the functions of ribozyme and release factor in protein synthesis.

Ribozyme in bacteria is 23S rRNA, that acts as an enzyme for the formation of a peptide
bond between two amino acids.
Release factor binds to the stop codon (UAA) to terminate translation and release the
complete polypeptide.

88.
How would lac operon operate in E. coli growing in a culture medium, where lactose is
present as source of sugar?

When lactose is present in a medium having E. coli, it will act as a substrate for enzyme
3-galactosidase and switches on the operon.
Hence, it is also termed as an inducer. It inactivates repressor by binding to it and
allows RNA polymerase access to the promoter.

89.
Where does peptide bond formation occur in a bacterial ribosome and how?

A peptide bond is formed between carboxyl group (-COOH) of amino acid at P-site and
amino group (-NH) of amino acid at A-site. It is formed by the enzyme peptidyl
transferase in a bacterial ribosome.

90.
(i) Name the scientist who suggested that the genetic code should he made of a
combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion.

(i) George Gamow suggested that the genetic code should be made of a combination of
three nucleotides.
(ii) George stated that a codon must be of three bases in order to code for 20 amino
acids, since there are only four bases (i.e. 43 or 4 × 4 × 4 = 64) which code for 20 amino
acids.

93.
Draw a schematic diagram of lac operon in its switched off position. Label the following
(i) Structural genes
(ii) Repressor bound to its correct positions
(iii) Promoter gene
(iv) Regulatory gene
94.
Write the full form of VNTR. How is VNTR different from Probe?

VNTR-Variable Number Tandem Repeat.

VNTR
It is a class of satellite DNA, where a small sequence is arranged tandemly in many copy
numbers. Probe
It is a radiolabelled VNTR, used for hybridisation with DNA segments in .

95.
(i) Differentiate between unambiguous and degenerate codons.

mbiguous nerate

mbiguity for a particular codon. For exampleis degenerate for a particular amino ac
s an ambiguous codon, it codes for glycine
s glutamic acid.

ticular codon will always code for the same mino acid is coded by more than one
o acid, where it is found. n, e.g. phenylalanine has two codons, i.
and UUC.

96.
‘A very small sample of tissue or even a drop of blood can help determine paternity’.
Provide a scientific explanation to substantiate how it is possible.

DNA fingerprinting is used to distinguish between individual of same species by using

their DNA sample. The DNA is isolated from the cells and further amplified to produce
many copies by using polymerase chain reaction. This amplified DNA is further
processed to detect the presence of similarities between the parent and child. Because,
the DNA from sample can be amplified to produce many copies in DNA fingerprinting,
only a very small sample of tissue or even a drop of blood can help determine paternity.

97.
Expand ‘BAC’ and ‘YAC’ what are they and what is the purpose for which they are used?

BAC — Bacterial Artificial Chromosome

YAC — Yeast Artificial Chromosome.
‘BAC’ and ‘YAC’ are used as vectors in cloning of DNA.

98.
(i) Expand VNTR and describe its role in DNA fingerprinting.
(ii) List any two applications of DNA fingerprinting technique. (2018)

(i) VNTR is Variable Number of Tandem Repeats. These are short nucleotide repeats in
the DNA. These are highly specific for individuals. No two individuals have the same
VNTR.

Role of VNTR in Fingerprinting VNTRs are used as probe markers in the identification of
DNA of different individuals because no two individuals can have the same VNTRs
(except in case of monozygotic twins).

(ii) Applications of DNA Fingerprinting

● DNA fingerprinting can identify the real genetic mother, father and offspring.

● DNA fingerprinting is very useful in the detection of crime and legal pursuits.

99.
(i) List the two methodologies which were involved in human genome project. Mention
how they were used.

(i) Two major methodologies involved in HGP are as follows

● Expressed Sequence Tags (ESTs) This method focuses on identifying all the
genes that are expressed as RNA.

● Sequence annotation This method involves the sequencing of whole set of

genome (that contained all coding and non-coding sequence) and then assigning
functions to the different regions in the sequence.

100.
In a maternity paternity issues. Name and describe the technique that you would
suggest to sort out the matter.

The technique that can help in the identification of victims is DNA fingerprinting which is
used to distinguish between individuals of same species by using their DNA sample. The
chemical structure of DNA is same in everyone (99.9%) except the order of base pairs,
i.e. only 0.1% of DNA makes every individual unique.
DNA fingerprinting exploits the highly variable tandem repeating sequences, i.e. VNTRs
for profiling. These VNTRs are highly conserved among members of the same species.

Technique
This technique is carried out in following steps

● DNA Isolation DNA is extracted from the cells in a high speed centrifuge.

● Amplification Many copies of the extracted DNA can be made by the use of
polymerase chain reaction.

● Digestion of DNA by restriction endonucleases.

● Separation of DNA fragments by electrophoresis.

● Blotting Transfer of separated DNA fragments to synthetic membranes (like nylon

or nitrocellulose).

● Hybridisation, with the help of a radio-labelled VNTR probe (small segments of

DNA which help to detect the presence of a gene in a long DNA sequence). These
probes target a specific nucleotide that is complementary to them.

● Autoradiography Detection of hybridised DNA fragments by autoradiography.

The presence of similarities between the victims and their relatives determines their
association on the basis of which dead bodies or newborns can be identified and
handed over to their families.

101.
percentage of the total human genome that codes for proteins and the percentage of
discovered genes whose functions are known as observed during HGP.

Less than 2% of the total human genome codes for protein, functions of 50 % of
discovered genes are not known.

103.
Write any three goals of human genome project.

The following are the goals of the human genome project

● Identify all the (approximately) 20000-25000 genes in human DNA.

● Determine the sequence of the 8 billion chemical base pairs that make up human
DNA.

● Improve tools for data analysis.

 104.
What is mutation? Explain with the help of an example how does a point mutation affect
the genetic code. Name another type of mutation.

A mutation is a sudden, stable, inheritable change in genetic material. A classical

example of point mutation is a change of single base pair in the gene for p-globin chain
that results in the change of amino acid residue glutamate to valine in a cell of a person
suffering from sickle-cell anaemia. It is a genetic disease.

The other type of mutation includes frameshift insertion or deletion mutation. The
insertion or deletion of one or more bases may change the reading frame from the point
of insertion or deletion. Insertion or deletion of three or its multiple bases insert or delete
one or multiple codon, hence one or multiple amino acid may be formed in polypeptide.

105.
Explain the significance of satellite DNA in DNA fingerprinting technique.

A DNA satellite is a region that consists of short DNA sequences repeated many times.
The variation between individuals in the lengths of their DNA satellites forms the basis
of DNA fingerprinting.DNA satellites are of two types, i.e. microsatellites and
minisatellites. Their characteristic that makes them useful for identification is that they
are highly polymorphic. The length of each satellite in DNA is inherited.The length of
satellite regions is highly variable between people. These form small peaks during
density gradient centrifugation and thus, are invaluable for identification purposes.

106.
Describe how the lac operon operates, both in the presence and the absence of an
inducer in E. coli.

Lac operon is made up of one regulatory gene i and three structural genes (z, y, a).
Its function in the presence and the absence of inducer is as follows
(i) When inducer (lactose) is absent, i gene regulates and produces repressor mRNA.
The repressor protein binds to the operator region of operon and as a result prevents
RNA polymerase to bind to the operon. The operon is switched off in this situation.

(ii) When inducer (lactose) is present, lactose acts as an inducer and binds to the
repressor. Thus, forming an inactive repressor. The repressor fails to bind the operator
region. The RNA polymerase binds to the operator and transcripts lac mRNA.

Lac mRNA is known to be polycistronic which produces all three enzymes, i.e.
P-galactosidase, permease and transacetylase required for the hydrolysis of lact >se.
Operon is switched on in this situation.

107.
What is a genetic code?
Genetic code is the sequence of three nucleotides present on tnRNA which codes for a
specific amino acid.

108.
Explain the process of translation

Translation is the process of polymerisation of amino acids to form a polypeptide with

the help of mRNA, tRNA and ribosomes and many enzymes involved in the process.

109.
(ii) Write the scientific importance of single nucleotide polymorphism identified in
human genome

(ii) Scientists have identified about 1.4 million locations, where single base DNA
differences or Single Nucleotide Polymorphisms (SNPs) occur in humans. Since, these
sequences have high degree of polymorphism they form the basis of DNA fingerprinting.

110.
(i) Name the scientist who postulated the presence of an adapter molecule that can
assist in protein synthesis, (ii) Describe its structure with the help of a diagram. Mention
its role in protein synthesis.

(i) Francis Crick proposed the presence of an adapter molecule, i.e. /RNA which could
read the code and bind to the specific amino acids, thus assisting in protein synthesis.
(ii) A clover leaf structure of tRNA

A tRNA- functions as carrier of amino acids and participates in protein synthesis.

111.
(i) Explain DNA polymorphism as the basis of genetic mapping of human genome.
(ii) State the role of VNTR in DNA fingerprinting.

(i) DNA polymorphism exhibited by certain repetitive DNA sequences is used to

construct genetic and physical maps of the genome which are used in human genome
project. (ii) Variable Number of Tandem Repeats (VNTRs) belong to a class of satellite
DNA called as minisatellite. VNTRs are used as probes in DNA fingerprinting.

113.
How are the structural genes activated in lac operon in E. coli ?
Lac operon consists of

● an operator, which controls all structural genes as a unit

● a regulatory gene (i gene)

● three structural genes (z, y, a), which code for enzymes and

● a promoter, where RNA polymerase binds for transcription.

The regulatory gene codes for a repressor protein that has high affinity for the operator
region and which prevents RNA polymerase from transcribing the structural genes, i.e.
the lac operon is switched off or inactive.

Lac operon in the presence of inducer When inducer (lactose) is added, it binds with
repressor protein and inactivates it. This allows RNA polymerase to have access to the
promoter and transcription proceeds via the activation of structural genes.

115.
How is the lac operon switched on in the bacteria? Mention the state of the operon when
lactose is digested.

In lac operon, when lactose is added, it enters the cell wall with the help of permease, a
small amount of which is already present in cell. Lactose binds itself to active repressor
and changes its structure. The repressor now fails to bind to the operator.

Then, RNA polymerase starts transcription of operon by binding to promoter site-P. All
the three enzymes for lactose metabolism are synthesised.
After some time, when whole of lactose is consumed, there is no inducer present to bind
to the repressor. Due to this, the repressor becomes active again, attaches itself to the
operator and finally switches off the operon.

116.
(i) Why is tRNA called an adapter?
(ii) Draw structure of tRNA. How does the actual structure of tRNA look like?

(i) tRNA binds to a specific amino acid and it also reads the codon of the amino acid
bound to it through its anticodon. So, it is called an adapter molecule.
In actual structure, the /RNA is a compact molecule, that looks like an inverted L.

117.
The following is the flow chart higlighting the steps in DNA fingerprinting technique.
Identify a, b, c, d, e and f.

a – Restriction endonuclease
b – Polyacrylamide gel
c – Nitrocellulose or Nylon
d – VNTR e – Hybridisation
e- Autoradiography

118.
Study the schematic representation of the genes involved in the lac operon given below
and the s that follows

(i) Identify and name the regulatory gene in this operon. Explain its role in switching off
the operon.
(ii) Why is lac operon’s regulation referred to as negative regulation?
(iii) Name the inducer molecule and the products of the genes z and y of the operon.
Write the function of these gene products.
(i) i gene-regulatory gene It codes for the repressor protein of the operon, which is
synthesised constitutively. The repressor has the affinity for the operator gene. It binds
to the operator and prevents the RNA polymerase from transcribing the structural genes.
(ii) When repressor binds to the operator, the operon is switched off and transcription is
stopped. So, it is called negative regulation.
(iii) Lactose is an inducer molecule.
Gene ‘Z’ codes for p-galactosidase, which is responsible for the hydrolysis of lactose into
galactose and glucose.
‘y’ gene codes for permease. It increases the permeability of the cell to lactose.

119.
(i) Write the contributions of the following scientists in deciphering the genetic code.
George Gamow, Har Gobind Khorana, Marshall Nirenberg, Severo Ochoa.
(ii) State the importance of genetic code in protein biosynthesis.

(i) George Gamow coined the term genetic code and argued that since there are only
four bases and if they have to code for 20 amino acids, the code should constitute a
contribution of bases.
He suggested that in order to code for all* the amino acids the genetic code should be
made up of 3 nucleotides.

Har Gobind Khorana developed a chemical method for the synthesis of RNA molecule
with defined base combinations (homopolymers and copolymers). Marshall Nirenberg
Put forward a cell-free system for protein synthesis that helps in deciphering the code.
Severo Ochoa Showed that the polynucleotide phosphorylase also helped in
polymerising RNA with defined sequences in a template independent manner (enzymatic
RNA synthesis).

(ii) The genetic code is the set of rules by which information encoded in genetic material
is translated into proteins by living cells. The genetic code is nearly universal language
that encodes directions for cells. Their arrangement as codons, stores the blueprint for
amino acid chain. This chain in turn forms proteins which regulate the biological process
in every living beings.

121.
Mention what would happen if lactose is withdrawn from the culture medium.

If lactose is withdrawn from the culture medium, the repressor of the operon is
synthesised from the i gene. This repressor protein binds to the operator region of
operon and prevents RNA polymerase from transcribing the operon.

122.
Write any two different levels at which regulation of gene expression could be exerted in
eukaryotes.

The different levels at which regulation of gene expression can be exerted in eukaryotes
are following
● Transcriptional level (formation of primary transcript).

● Processing level (regulation of splicing)

124.
Name the major types of RNAs and explain their role in the process of protein synthesis
in a prokaryote.

(i) Translation is the process of polymerisation of amino acids to form a polypeptide.

The role of mRNA, tRNA and ribosomes in the process is described below
(a) Role of mRNA The order and sequence of amino acids are defined by the sequence
of bases in the mRNA. It stores the genetic information from DNA. The amino acids are
joined by a bond, which is known as a peptide bond.
This process requires energy mRNA also possesses untranslated sequences called
Untranslated Regions (UTR) at both 5′ end and 3′ end. They help in efficient translation.

(b) Role of (RNA It acts as an adaptor molecule. Activation of amino acids occurs in the
presence of ATP and activated amino acids get linked to their cognate tRNA, i.e.
charging of fRNA or aminoacylation of tRNA.

If two such charged fRNAs are brought closer, the formation of peptide bond between
them would occur energetically in the presence of a catalyst.

(c) Role of /RNA It is formed in nucleolus and it forms 80% of total RNA present inside
the cell. It is also the most stable type of RNA. rRNA is associated with structural
organisation of ribosomes (rRNA forms about 60% of weight of ribosomes), which are
seats of protein synthesis.

(d) Role of ribosomes Initiation of polypeptide synthesis occurs in ribosomes (cellular

factory for protein synthesis).

● Ribosome consists of structural RNAs and about 80 different proteins.

● In its inactive state, it exists as two subunits, i.e. a large and a small subunit.

● When the small subunit encounters an mRNA, the process of translation of the
mRNA to protein begins.

● There are two sites in the large subunit, i.e. the P-site and A-site for subsequent
amino acids to bind to and thus, be close enough to each other for the formation of a
peptide bond.
The small subunit (with the fRNA) attaches to the large subunit in such a way that the
initiation codon (AUG) comes to the P-site.

127.
Given below is the schematic representation of lac operon of E.coli. Explain the
functioning of this operon when lactose is provided in the growth medium of the
bacteria.

Lactose in the lac operon regulates operon, by switching on and off. If lactose is present
in the medium, it acts as an inducer here and the substrate for the enzyme
p-galactosidase.
It binds to the repressor and forms an inactive repressor, which fails to bind to the
operator region of the operon.

The RNA polymerase thus, binds to the operator and transcribes lac-mKNA. Lac mRNA
produces all three enzymes, i.e. P-galactosidase, permease and transacetylase of
polycistronic structural gene present in E.coli.
Therefore, operon will be switched on in the presence of the lactose.

128.
(i) DNA polymorphism is the basis of DNA fingerprinting technique. Explain.
(ii) Mention the causes of DNA polymorphism.

(i) DNA polymorphism It is the occurrence of inheritable mutations at a frequency

greater than 0.01 in a population.

● Such variations often occur in non-coding sequences. They keep on accumulating

generation after generation.

● Types of polymorphism range from single nucleotide change to very large scale
changes.

● Single nucleotide polymorphism is used to diagnose disease related sequences of

DNA on the chromosome.

● Variable number of tandem repeats show a high degree of polymorphism.

(ii) DNA polymorphism occurs due to mutations.

130.
(i) Name the enzyme responsible for transcription of tRNA and the amino acid, the
initiator tRNA gets linked with.
(ii) Explain the role of initiator tRNA in initiation of protein synthesis.

(i) RNA polymerase transcribes tRNA and the amino acid, the initiator tRNA gets linked
with is methionine.
(ii) The initiator tRNA binds the amino acid methionine at its amino acid acceptor site. It
has anticodon loop, which has anticodon for methionine, i.e. UAC. It recognises the start
codon (AUG) at P-site and binds to it according to complementarity of bases.

131.
(ii) ‘Process of transcription and translation are coupled in prokaryotes, but not in
eukaryotes’. Explain.

(ii) In bacteria, both processes occur in cytoplasm as there is no nucleus. In eukaryotes,

transcription occurs in nucleus, while translation occurs in the cytoplasm.

Complexities in eukaryotic transcription Eukaryotes have additional complexities in gene

expression than prokaryotes as mentioned in gene expression below
(a) There are at least three RNA polymerases in the nucleus other than the RNA
polymerase in organelles. The RNA polymerase-I transcribes rRNAs (28S, 18S and 5.8S).
RNA polymerase-III is responsible for the transcription of tRNA, 5srRNA and snRNAs
(small nuclear RNAs). RNA polymerase-II transcribes precursor of mRNA, the
heterogeneous nuclear RNA (hnRNA).

(b) Another complexity is that, the primary transcripts contain both the exons and the
introns and are nonfunctional. Hence, it is subjected to a process called splicing.
In this process, introns are removed and exons are joined in a definite order.
(c) hnRNA undergoes additional processing called capping and tailing. In capping, an
unusual nucleotide is added to the 5′ end of ImRNA.

In tailing, adenylate residues (200-300) are added at 3′ end in a template. Capping and
tailing protect the mRNA from degradation by the activity of digestive enzymes present
in the cytoplasm. It is the fully processed hnRNA, now called mRNA, that is transported
out of the nucleus for translation process.

Complexities in translation in eukaryotes

● The mRNA formed in nucleus has to be transported to the cytoplasm.

● Transcription and translation cannot be coupled in eukaryotes.

132.
(i) Describe the structure and function of a tRNA molecule. Why is it referred to as an
adapter molecule?
(ii) Explain the process of splicing of hnRNA in a eukaryotic cell. (All India 2017)

(i) Structure of tRNA

● The secondary structure of tRNA looks like a clover-leaf.

● All tRNA molecules have a guanine residue at its 5’end.

● At its 3′ end an unpaired CCA sequence is present. Amino acids get attached to
this end during translation.

● tRNA has an anticodon loop, an amino acid acceptor arm, and a ribosome site.

● The anticodon loop has bases complementary to the code. Amino acid acceptor
end binds to amino acids.

Function of tRNA The function of tRNA is to align the required amino acids according to
the nucleotide sequence of mRNA. tRNA is also called the adapter molecule because on
one hand it can read the code and on the other hand it can bind to specific amino acid. It
acts as intermediate molecule between triplet code of mRNA and amino acid sequence
of polypeptide chain.
(ii) Splicing of hnRNA Eukaryotic transcripts possess extra segments called introns or
intervening sequences or non-coding sequences. They do not appear in mature or
processed RNA. The functional coding sequences are called exons. Splicing is removal
of introns and fusion of exons to form functional RNAs. A complex called spliceosome
is formed between 5’end (GU) and 3’end (AG) of intron. Energy is obtained from ATP. It
removes the introns. The adjacent exons are brought together. The ends are sealed by
RNA ligase.

133.
Where do transcription and translation occur in bacteria and eukaryotes, respectively?
Explain the complexities in transcription and translation in eukaryotes that are not seen
in bacteria.

In bacteria, both processes occur in cytoplasm as there is no nucleus.

In eukaryotes, transcription occurs in nucleus, while translation occurs in the cytoplasm.
Complexities in eukaryotic transcription Refer to No. 2. (ii)

Complexities in translation in eukaryotes are

● The TRNA formed in nucleus has to be transported to the cytoplasm.

● Transcription and translation cannot be coupled in eukaryotes.

134.
According to human genome project, about 99.9% nucleotide bases are exactly the
same in all humans. Do you think the discrimination of people on the basis of colour,
creed and religion is correct? Justify.

No discrimination of people on these grounds is not correct because all people have
same genetic material (i.e. DNA) and are similar in their makeup.

135.
An organism is able to survive on a culture medium, containing nutrient A, by the
enzyme-catalysed reactions.

A mutant organism failed to survive on this medium, but grew well when nutrient C was
added to it.
(i) Which gene of this mutant organism is defective?
(ii) What does such a condition indicate of?
(iii) Indicate the value expressed in this sequence of reactions.

(i) The gene q is defective.

(ii) It indicates that one gene controls the synthesis of one enzyme.
(iii) Each one of reaction has a role to play, when one fails, progress is not possible.

Human Reproduction
1.
Why are human testes located outside the abdominal cavity? Name the pouch

Human testes are located outside the abdominal cavity as it helps in maintaining low
temperature (2-2.5%) lower than body temperature) required for spermatogenesis.
Testes are enclosed in a pouch called scrotum.

2.
Write the location and functions of
(i) Sertoli cells
(ii) Leydig cells

(i) Location; Within the lining of seminiferous tubule of testis.

Function;provide nutrition to the developing sperms or germ cells.

(ii) Location;In the interstitial spaces between the seminiferous tubules. Function; They
synthesise and secrete male hormones, androgens, testosterone.

3.
function of the seminal vesicle.

Seminal vesicle produces an alkaline secretion containing prostaglandins, proteins and

fructose. The high fructose content provides energy to the spermatozoa. These
secretions form 60-70% of the fluid in the semen.

4.
List the parts of human oviduct through which ovum travels till it meets the sperm

The parts of human oviduct through which the ovum travels are given below;

● Fimbriae, finger-like projections;Collect the ovum from ovary after ovulation.

● Infundibulum;Ovum from fimbriae is guided into funnel-shaped infundibulum, part

of Fallopian tube.

● Ampulla A wider part of oviduct that leads ovum into isthmus.

● Isthmus; has narrow lumen which opens into uterus.

● In the junction of ampulla- isthmus, the ovum gets fertilised.

6.

(i) Identify ‘X’ and write its location.

(ii) Name the accessory gland ‘Y’ and its secretion.
(iii) Name and state the function of ‘Z’.

(i) .X-Testis. It is located outside the abdominal cavity within a pouch called scortum.
(ii) K-Seminal vesicle. It produces an alkaline secretion rich in fructose and constitutes
60% of the volume of the semen.
(iii) Z-Epididymis. It stores the sperms and secretes a fluid which helps in the maturation
of sperms.

7.
(ii) Differentiate between vas deferens and vasa efferentia.

(ii) Distinction between vas deferens and vasa efferentia

eferens efferentia

tube-like structure which conducts the nects the rete testis to the
matozoa from the epididymis to the penis. dymis.

arise from the rete testis. arise from the cauda epididymis.

are only 2 in number. vary from 15-20 in number.

eferens are thick. efferentia are fine.

lining has many stereocilia. lining bears many ciliated cells.

8.
Name and explain the role of inner and middle walls of human uterus.

The innermost wall of uterus is called endometrium.

Role of Endometrium
(i) It lines the uterine cavity and is glandular.
(ii) It undergoes cyclic changes during menstrual cycle.
The middle wall or layer of uterus is called myometrium.

(i) It is made up of thick layer of smooth muscles.

(ii) It shows strong contractions during the delivery of baby.

10.
When do the oogenesis and the spermatogenesis initiate in human females and males,
respectively?

Oogenesis initiates during foetal or embryonic stage in females, whereas

spermatogenesis in males starts at puberty.

11.
differences between spermiogenesis and spermiation.

miogenesis miation

he process of transforming m heads are embedded in Sertoli cells to obta

matids into mature spermatozoa orshment and finally released from seminiferou
ms through differentiation. es by the process called spermiation.

12.
Where is acrosome present? Write its function.

acrosome is present in the anterior portion of the head of sperm. Function ;Hydrolytic
enzymes or sperm lysins present in acrosome help in penetration of sperm into egg,
during fertilisation.

13.
List the changes the primary oocyte undergoes in the tertiary follicular stage in human
embryo.

The primary oocyte within the tertiary follicle grows in size. The fully grown primary
oocyte completes its first meiotic division to produce two daughter nuclei in which larger
haploid cell is called secondary oocyte and the tiny one is called first polar body. The
secondary oocyte retains bulk of nutrient rich cytoplasm of primary oocyte.

14.
Name the hormones influencing follicular development of corpus luteum.

The hormones influencing follicular development of corpus luteum are FSH, oestrogen
and progesterone.

15.
Explain the hormonal regulation of the process of spermatogenesis in humans.

● Gonadotropin Releasing Hormone (GnRH) is released significantly from the

hypothalamus during puberty.

● GnRH stimulates anterior pituitary to secrete gonadotropins, i.e. LH and FSH or

Interstitial Cell Stimulating Hormone (ICSH).

● Luteinising Hormone (LH) acts on Leydig cells to stimulate the synthesis and
secretion of androgens which then stimulate the process of spermatogenesis.

● Follicle Stimulating Hormone (FSH) acts on Sertoli cells and stimulates them to
secrete inhibin which then stimulates the process of spermiogenesis.

16.
What happens to corpus luteum in human female if the ovum is (i) fertilised, (ii) not
fertilised?

● In case of fertilisation, the corpus luteum continues secreting progesterone which

is required for the maintenance of endometrium during pregnancy.

● In the absence of fertilisation, the corpus luteum degenerates and gets converted
into corpus albicans. Deficiency of progesterone causes disintegration of the
endometrium leading to menstruation and thus, a new cycle starts.

17.
parts and functions that assist the sperm to reach and gain entry into the female
gamete.

The parts that assist sperm to reach and gain entry into female gamete are,
Tail Its wiggling movement helps the sperm to swim in female reproductive tract.
Middle piece It contains mitochondria that provide energy for sperm movement.
Head It contains acrosome loaded with sperm lysins. These help to dissolve the layers
of female gamete or egg.
 18.
effect of high concentrations of LH on a mature Graafian follicle.
The high concentration of LH (LH surge) induces rupture of Graafian follicle, which
results in the release of secondary oocyte hence, causing ovulation in females.

19.
Differentiate between spermatogenesis and spermiogenesis.

matogenesis miogenesis

process of formation of sperms from process of transformation of a circular

ture germ cells. matid to a motile spermatozoa.

ber of cells increases as each hange in number of cells as only one

matogonium produces four spermatids.matid develops into a spermatozoa.

20.
Draw and label head region of a human sperm.

21.

Identify A, B, C and D with reference to gametogenesis in humans

A – Leydig’s cell
B – Sertoli cell
C – Spermatogenesis
D – Spermiogenesis
22.
Name the labels A, B, C, D, E and F in seminiferous tubule.

A – Spermatogonium
B – Primary spermatocyte
C – Secondary spermatocyte
D – Spermatid
E – Spermatozoa
F – Sertoli cell

23.
Differentiate between menarche and menopause.

rche pause

ning of menstrual cycle in human age of menstrual cycle at the age of about
e at puberty. in human female.

of reproductive phase. f reproductive phase.

24.
Study the sectional view of human testis showing seminiferous tubules given below.

(i) Identify A, B and C.

(ii) Write the function of A and D.

(i) A-Spermatogonia
B-Interstitial cells
C-Spermatozoa.

(ii) A-Spermatogonia undergo meiosis to produce spermatozoa (sperms).

D-Sertoli cells provide nutrition to the germ cells.

25.
How and at what stage of menstrual cycle is corpus luteum formed? When does it
regress?

After ovulatory phase (ovulation), the luteal phase starts. The remaining parts of
ruptured Graafian follicle transform into corpus luteum in this phase. The corpus luteum
secretes large amount of progesterone which is essential for the maintenance of
endometrium.In the absence of fertilisation, the corpus luteum degenerates in the ovary
and gets converted into corpus albicans.

27.
(i) How many primary follicles are left in each ovary in a human female at puberty?

(i) A large number of primary follicles degenerate in females during the period from birth
to puberty by the process called follicular atresia. As a result, about 60000-80000
primary follicles are left in each ovary at puberty.

28.
Explain the events in a normal woman during her menstrual cycle on the following days
(i) Ovarian event from 13-15 days.
(ii) Ovarian hormones level from 16-23 days.
(iii) Uterine events from 24-29 days.

(i) In the ovarian event from 13-15 days, a immature ovum (egg cell) is released from the
Graafian follicle. Both LH and FSH attain maximum peak. FSH helps Graafian follicle to
attain maturity and LH helps in its rupture. Ovum covered by a number of layer and a
yellow fat layer forms corpus luteum. It releases (secretes) progesterone.

(ii) During menstrual cycle, the period level from 16-23 days is called luteal phase
(secretory phase). The corpus luteum secretes large amount of progesterone which is
essential for the maintenance of endometrium.

(iii) Uterine events from 24-29 days are under the influence of progesterone hormone. It
influences the maintenance of the endometrium for any pregnancy to occur. In the
absence of pregnancy, the corpus luteum degenerates and endometrium sheds off. It
causes the menstrual flow or bleeding.

29.
Explain the events in a normal menstrual cycle on (i)
Pituitary hormone levels from 12 days.
(ii) Uterine events from 13-15 days.
(iii) Ovarian events from 16-23 days.

(i) The period of 8-12 days after the onset of menstruation is the follicular phase. During
this phase, GnRH from hypothalamus stimulates anterior pituitary to release FSH and
LH. FSH stimulates the ovarian follicles to secrete oestrogen, which in turn stimulates
the proliferation of the endometrium of the uterine wall. This causes the endometrial
lining to thicken.

(ii) The uterine events between day 13 and 15 are governed by the high LH and FSH
levels. The endometrium is intact due to the effect of these gonadotropin hormones and
also prepares itself for pregnancy, if fertilisation occurs.

(iii) During 16-23 days, ruptured Graafian follicle gets converted into corpus luteum in the
ovary. It starts secreting progesterone which maintains the endometrium, necessary for
the implantation of fertilised ovum followed by other events of pregnancy.

31.
Explain the steps in the formation of an ovum from oogonium in humans.

In human females, primary oocytes are formed from the oogonia during the embryonic
developmental stages in the foetal ovaries.

● Oogonial cells start dividing and enter prophase-I of meiosis. They remain
suspended at this stage as primary oocytes.

● Each primary oocyte is surrounded by a layer of granulosa cells and becomes the
primary follicle.

● The primary follicle when surrounded by more layers of granulosa ceils, is called a
secondary follicle.

● Secondary follicle transforms into a tertiary follicle, with the development of a

fluid-filled cavity (antrum) around the primary oocyte

● Granulosa cells become organised into an outer theca externa and an inner theca
interna.

● primary oocyte completes meiosis-I and forms a larger haploid secondary oocyte
and a tiny first polar body.

● Tertiary follicle grows and becomes a mature follicle called Graafian follicle.

● Secondary oocyte secretes a new membrane called zona pellucida around it.

● At this stage, follicle ruptures to release the secondary oocyte, which moves into
the cytoplasm.
● Secondary oocyte completes meiosis-II only when a sperm enters its cytoplasm. It
forms a larger cell, the ootid and a small second polar body. This event occurs in the
ampulla of Fallopian tube.

32.
(i) Explain menstrual cycle in human females.
(ii) How can the scientific understanding of the menstrual cycle of human females help
as a contraceptive measure ?

(i) Menstrual Cycle; The inner lining of uterus called endometrium, grows and thickens
each month and prepares itself for the implantation of an embryo. If pregnancy does not
occur, the endometrium sheds off. The monthly development and shedding of the
functional layer of the uterus is called the menstrual phase and the monthly maturation
of an egg and its release is called the ovarian cycle. A typical menstrual cycle completes
in an average of about. 28 days. It starts at the age of 13 or 15 and continues till about
50 years of age. Menstrual cycle occurs in three major phases namely menstrual phase,
follicular phase and secretory phase.
(a) Menstrual Phase It lasts for 3-4 days. It occurs due to the breakdown of
endometrium lining cf uterus and blood vessels.
(b) Follicular or Proliferative Phase. regulated by hormones secreted by anterior pituitary
gland who stimulate the ovarian follicle to secrete oestrogens.(c) Secretory Phase or
Luteal phase The phase of menstrual cycle with possibility of fertilisation is the initial
luteal phase. It is marked by the presence of corpus luteum (yellow body). During
pregnancy, this yellow body secretes progesterone. In the absence of pregnancy, it
regresses to form corpus albicans and menstruation starts (menstrual phase). Due to
the hormones secreted by corpus luteum, i.e. oestrogen and progesterone, the release of
FSH and LH is inhibited. This prevents the development of new follicles.This phase lasts
for 14 days. Maintenance of endometrium by progesterone is necessary for
the implantation of embryo and pregnancy. During pregnancy, menstrual cycle stops due
to high level of progesterone.

(ii) The scientific understanding of the menstrual cycle of human females helps as a
contraceptive method. This method is known as periodic abstinence. In this method, the
couples avoid or abstain the coitus from day 10 to 17 of the menstrual cycle when
ovulation could be expected. The chances of fertilisation are very high during this period.

33.
(i) Arrange the following hormones in the sequence of their secretion in a pregnant
woman. hCG, LH, FSH, Relaxin
(ii) Mention their source and the function they perform.

(i) The sequence of secretion of the given hormones in a pregnant woman is

FSH → LH → hCG → Relaxin

one ce ion
 ior pituitary lobe ulates the growth of ovarian follicles in ovary.

ior pituitary lobe gers ovulation.

nta mulates the corpus luteum to secrete progesterone.

in nta litates parturition by softening the connective tissu

mphysis pubica.

34.
(i) Explain the menstrual phase in a human female. State the level of ovarian and
pituitary hormones during this phase.
(ii) Why is follicular phase in the menstrual cycle also referred as proliferative phase?
Explain.
(iii) Explain the events that occur in a Graafian follicle at the time of ovulation and
thereafter.
(iv) Draw a Graafian follicle and label antrum and secondary oocyte. (Delhi 2016)

(i) The reproductive cycle in female primates, e.g. monkeys, apes and human beings is
called menstrual cycle.
Menstrual phase starts from 3rd day and ends on 5th day of the menstrual cycle. It is
initiated due to the reduced secretion of progesterone and oestrogen from the
regressing corpus luteum in the ovary. The endometrium breaks down and blood along
with degenerated ovum constitutes the menstrual flow. The secretion of pituitary
hormones, i.e. FSH and LH is also reduced during this phase.

(ii) In follicular phase, primary follicles in the ovary grow under the influence of Follicle
Stimulating Hormone (FSH). It starts from 6th day and ends on 13 th or 14th day of 28
day cycle. FSH also stimulates the ovarian follicles to secrete oestrogen which in turn
stimulates endometrium to proliferate, so that it becomes thicker and highly
vascularised. Thus, it is also called proliferative stage of menstrual cycle.

(iii) Graafian Follicle at Ovulation

At the time of ovulation following events occur

● LH and FSH reach at their peak levels (about 14th-16th day of cycle).

● High level of LH induces Graafian follicle to rupture and the release of secondary
oocyte from it.
● After ovulation, the remaining cells of Graafian follicle are stimulated by LH to
develop corpus luteum (an endocrine gland which secrete progesterone hormone).

(iv) Diagrammatic sectional view of Graafian follicle

35.
(i) Explain the process of spermatogenesis in humans.
(ii) Draw a human sperm and label acrosome and middle piece. Mention their functions.
Outside (Delhi 2016C)

(i) Spermatogenesis is the production of sperms in males.

● In testis, the immature male germ cells (spermatogonia) produce sperms by

spermatogenesis. It begins at puberty due to significant increase in the secretion of
gonadotropins, i.e. luteinising hormone and follicle stimulating hormone under the
influence of Gonadotropin Releasing Hormone (GnRH) released from hypothalamus.

● Spermatogonia (sing, spermatogonium) present on the inside wall of seminiferous

tubules multiply by mitotic division and increases in numbers.

● Each spermatogonium is diploid and contains 46 chromosomes. Some of the

spermatogonia transform to primary spermatocytes.

● The primary spermatocyte undergoes meiosis -1 and forms two haploid secondary
spermatocytes containing 23 chromosomes each.

● The secondary spermatocytes undergo meiosis – II and form four equal sized
haploid spermatids.

● Spermatids transform into the spermatozoa by spermiogenesis.

● After spermiogenesis, the sperm heads get embedded in the Sertoli cells and
released from the seminiferous tubules via spermiation process.
(a) The acrosome is filled with hydrolytic enzymes that help the sperm to penetrate the
ovum.
(b) Middle piece possesses many mitochondria to produce energy for the movement of
tail to facilitate sperm motility.

36.
Explain the ovarian and uterine events that occur during a menstrual cycle in a human
female under the influence of pituitary and ovarian hormones, respectively.
The cycle of events starting from one menstruation till next in female primates is called
menstrual cycle. It comprises of four phases which are regulated by both pituitary (LH
and FSH) and ovarian (oestrogen and progesterone) hormones that affect ovaries and
uterus, respectively. The events occurring in a menstrual cycle are as follows

trual phase (from

ed by reduced secretion of LH, progesterone and oestrogen. Th
th day in a 28 dametrium breaks down and blood along with unfertilised ovum
) itutes menstrual flow.

ular phase (fromSH (Follicle Stimulating Hormone) secreted by anterior pituitary

3th day in a 28 dlates ovarian follicle to secrete oestrogens. These oestrogens
) late proliferation of uterine walls as a result of which
metrium gets thickened (due to rapid cell division and increase
e glands and blood vessels).

tory phase (14thary hormones, i.e. LH and FSH reach the highest level in middle
n 28 day cycle) ycle. Rapid secretion of LH causes ovulation thus, inducing the
re of Graafian follicle to release secondary oocyte and a polar
 l or secretory ituitary hormone LH stimulates the remaining cells of ovarian
e (from 15th-28thes to develop into corpus luteum. This corpus luteum secretes
n a 28 day cycle)amount of progesterone and maintains endometrium thickenin
e implantation of fertilised ovum during pregnancy. In the absen
tilisation, the hormone levels are reduced (LH and progesterone
ndometrium disintegrates leading to onset of another menstrua

For graphical illustration of menstrual cycle;

Refer to figure 3.J0 on page no. 55.

37.
Explain the development of a secondary oocyte (ovum) in a human female from the
embryonic stage up to its ovulation. Name the hormones involved in this process. (Delhi
2015)
Or
When and where are primary oocytes formed in a human female? Trace the development
of these oocytes till ovulation (in menstrual cycle). How do gonadotropins influence this
developmental process? (Delhi 2010)

Embryonic Stages Refer to No. 22.

Influence of Gonadotropins on Oogenesis

● Gonadotropins, i.e., LH and FSH stimulate follicular development and secretion of

oestrogen by the growing follicles.

● Both LH and FSH attain a peak level in the middle of the cycle (14th day).

● Rapid release of LH during mid-cycle causes ovulation.

● LH also stimulates the formation of corpus luteum from the ruptured follicle and
secretion of progesterone from corpus luteum.

38.
(i) How is ‘oogenesis’ markedly different from ‘spermatogenesis’ with respect to the
growth till puberty in the humans?
(ii) Draw a sectional view of human ovary and label the different follicular stages, ovum
and corpus luteum. (Delhi 2014)

(i) Oogenesis is markedly different from spermatogenesis in the following aspects

matogenesis nesis
 urs in males. starting from puberty till therts before birth during embryonic
lete life cycle. opment and occurs till menopause.

gle spermatogonium after second meioticgle oogonium, after second meiotic

on forms four haploid sperrftatids that on, produces one ovum and two
re to form four spermatozoa. unctional polar bodies.

rocess of spermatogenesis, i.e. second econd meiotic division of oogenesis

ic division completes in testes and releaseletes in Fallopian tube when sperm
re sperms. s the secondary oocyte.

(ii) Refer to figure 3.6 on page no. 53.

39.
Schematically represent and explain the events of spermatogenesis in humans. (Delhi
2014C)

Refer to No. 26 (i)

40.
(i) Draw a transverse section of a human ovary showing the sequential development of
different follicles up to the corpus luteum.
(ii) Comment on the corresponding ovarian and pituitary hormone levels during these
events. (Delhi 2014C)

(i) Refer to figure 3.6 on page no. 53.

(ii) Refer to No. 28.

41.
The following is the illustration of the sequence of ovarian events (A-I) in a human
female.

(i) Identify the figure that illustrates ovulation and mention the stage of oogenesis it
represents.
(ii) Name the ovarian hormone and the pituitary hormone that have caused the above
mentioned event.
(iii) Explain the changes that occur in the uterus simultaneously in anticipation.
(iv) Write the differences between C and H.
(v) Draw a labelled sketch of the structure of a human ovum prior to fertilisation. (Delhi
2012)

(i) Figure F illustrates ovulation. It represents the ovulatory stage of oogenesis.

(ii) Ovarian and pituitary hormones involved in causing ovulation are
Ovarian hormone Oestrogen.
Pituitary hormone LH and FSH.
(iii) In anticipation of receiving the fertilised egg, the endometrium of the uterus gets
thickened and also the blood supply to the endometrium increases.
(iv) In the figure, C stage represents the secondary follicle and the H stage represents
the degenerating corpus luteum.

ndary follicle us luteum

urrounded by layers of granulosa s of granulosa cells and theca cells are abse
and theca layer.

tains an oocyte s not contain oocyte as it is formed after the

se of secondary oocyte.

s not secrete retes progesterone.

(v) Structure of a human ovum

42.
The following is the illustration of the sequence of ovarian events A-I in a human female.
Identify the figure that illustrates corpus luteum and name the pituitary hormone that
influences its formation.
(ii) Specify the endocrine function of corpus luteum. How does it influence the uterus?
Why is it essential?
(iii) What is the difference between D and E?
(iv) Draw a neat and labelled sketch of Graafian follicle. (Delhi 2012)

(i) Gis corpus luteum. LH secreted by anterior pituitary’ influences its formation.
(ii) The endocrine function of corpus luteum is to secrete progesterone which is
essential for the maintenance of endometrium layer of uterus. Thickened endometrium
is necessary for the implantation of fertilised ovum and other events of pregnancy.
(iii) D is a developing secondary follicle containing primary oocyte and undifferentiated
thecal cells. E is a mature tertiary follicle that contains a secondary oocyte and
differentiated theca cells.
(iv) Structure of a Graafian follicle

43.
(i) Describe the stages of oogenesis in human females.
(ii) Draw a labelled diagram of a human ovum released after ovulation. (Delhi 2012)

(i) For oogenesis, Refer to No. 22.

(ii) Human ovum Refer to No. 32 (v).

44.
(i) Draw a diagrammatic labelled sectional view of a seminiferous tubule of a human.
(ii) Describe in sequence the process of spermatogenesis in humans.

(i) Sectional view of a seminiferous tubule

Refer to figure 3.3 on page no. 51.
(ii) Refer to No. 26 (i).
45.
State from where do the signals for parturition originate in human females. (All India
2019)

Foetal-ejection reflex generated by fully developed foetus and placenta stimulates

pituitary to release the hormone oxytocin responsible for parturition.

46.
Mention the function of zona pellucida. (Delhi 2015C, 2013)

During (ertilisation, a sperm comes in contact with the zona pellucida layer of ovum and
induces changes in the membrane and forms fertilisation envelope. It helps block the
entry of additional sperms. So, zona pellucida ensures that only one sperm fertilises an
ovum.

47.
How does the sperm penetrate through the zona pellucida in human ovum? (Delhi
2013C)

The sperm penetrates through the zona pellucida by releasing sperm lysins
(hyaluronidase and protease) present in the acrosome.

48.
How is the entry of only one sperm ensured into an ovum during fertilisation in humans?
(All India 2012)

Presence of zona pellucida layer ensures that only one sperm can fertilise an ovum.
Refer to No. 2.

49.
Write the function of oxytocin. (Delhi 2012)

Oxytocin secreted by posterior pituitary helps in parturition by inducing and enhancing

uterine muscle contractions.

50.
Mention the function of trophoblast in human embryo. (Delhi 2011)

Trophoblast is the outer layer of blastocyst which helps in the attachment of blastocyst
to the endometrium of the uterus.

51.
Name the embryonic stage that gets implanted in the uterine wall of a human female.
(All India 2011)
Blastocyst gets implanted in the uterine wall of human female.

52.
What stimulates pituitary to release the hormone responsible for parturition? Name the
hormone. (All India 2011)

Foetal-ejection reflex stimulates pituitary to release the hormone responsible for

parturition. This hormone is oxytocin.

53.
Why is breastfeeding recommended during the initial period of an infant’s growth? Give
reasons. (Delhi 2016)
Or
(i) Why is mother’s milk considered very essential for the healthy growth of infants?
(ii) What is the milk called that is produced in the initial days of lactation? (All India
2016)

The first milk which comes out from the mother’s mammary glands just after childbirth
is called colostrum. This is a thin, yellowish, opalescent fluid.

It is rich in proteins and energy along with several antibodies (IgA) that provide passive
immunity to the newborn. This immunity helps the newborn to attain resistance to many
infections, but it is low in fat and iron content. Hence, breastfeeding during the initial
period of infant growth is highly recommended by the doctors for bringing up a
disease-free, healthy baby.

54.
Where does fertilisation occur in humans? Explain the events that occur during this
process.

In humans, the fertilisation of ovum takes place in ampullary-isthmic junction of the

Fallopian tube.
The events that occur during the process of fertilisation are

● The sperm reaches the junction of ampulla and isthmus and comes in contact with
zona pellucida layer of ovum.

● Acrosome of sperm head releases sperm lysin enzyme that dissolves corona
radiata and digests zona pellucida layer to enter the cytoplasm of ovum.

● After the entry of sperm into the ovum, zona pellucida induces changes in
membrane and blocks the entry of additional sperms

● Entry of sperm stimulates the secondary oocyte to complete its suspended

second meiotic division thus, producing haploid egg or ovum and second polar body.

● Nucleus of sperm and of ovum fuse to form a diploid zygote.

55.
When and where do chorionic villi appear in humans? State their function.

After implantation in the uterus, finger-like projections called chorionic villi appear on the
trophoblast of blastocyst. It interdigitates with uterine tissue to form placenta that
transports oxygen and nutrients to foetus and remove waste products and carbon
dioxide produced by the foetus.

56.
(i) Where do the signals for parturition originate in humans?
(ii) Why is it important to feed the newborn babies on colostrum?

(i) The signals for parturition originate from the fully developed foetus and the placenta,
which induces mild uterine contraction called foetal-ejection reflex.
(ii) Colostrum contains necessary antibodies (IgA) that provide protection against
disease to infants.

57.
Explain the function of umbilical cord.

The placenta is connected to developing embryo through an umbilical cord, which helps
in the transport of substances like oxygen, nutrients, etc. to and from the developing
embryo. Blood in umbilical cord can be collected during delivery and stored to treat
various diseases in future because it is rich in stem cells.

58.
Mention the number of cells in the following stages.

yonic Stage ber of Cells

ocyst

(i) 1 (ii) 16 (iii) 64 (2)

59.
Name the hormones produced only during pregnancy in human female. Mention their
source organs. (Foreign 2011)
The hormones produced only during pregnancy in human female are human Chorionic
and placenta Gonadotropin (hCG), human Placental Lactogen (hPL) and relaxin.
The source of hCG and hPL is placenta and that of relaxin is ovary and placenta.

60.
Why is parturition called a neuroendocrine mechanism? (All India 2011C)

Process of parturition is induced by both neural system and endocrine system therefore,
it is called a neuroendocrine mechanism.
Vigorous contraction of the uterus at the end of pregnancy causes expulsion/delivery of
the foetus. The process of delivery of foetus (childbirth) is called parturition.
Relaxin hormone is secreted by the ovary and placenta to facilitate parturition by
softening the connective tissue of symphysis pubica. Foetus and the placenta induce
mild uterine contractions called foetal-ejection reflex.

This initiates the release of oxytocin hormone from the posterior pituitary. Oxytocin acts
on uterine muscles to cause stronger contractions, which in turn stimulate further
secretion of oxytocin.
This causes more stronger contractions leading to the expulsion of the baby out of the
uterus, through the birth canal. After the baby is delivered, the placenta is also expelled
out of the uterus.

61.
(ii) Name the embryonic stage that gets implanted in human females. Explain the
process of implantation.

(ii) Blastocyst gets implanted in the uterine wall.

62.
The given diagram shows a part of the human female reproductive system.

(i) Name the gamete cells that would be present in X’ if taken from a newborn baby.
(ii) Name ‘Y’ and write its function.
(iii) Name ‘Z’ and write the events that take place here.

(i) Primary oocytes are present in A, i.e. ovary of a newborn baby.

(ii) Y is fimbriae. It helps in the collection of ovum from ovary after ovulation.
(iii) Z is the ampullary-isthmic junction. It is the site of fertilisation in humans.

64.
(i) How is placenta formed in human female?
(ii) Name any two hormones which are secreted by it and are also present in a
non-pregnant woman.

(i) After implantation of blastocyst, the finger-like projections called chorionic villi,
appear on the trophoblast. They get surrounded by uterine tissue and maternal blood
which become interdigitated with each other to form placenta.
(ii) The two hormones secreted by placenta that are also present in a non-pregnant
woman are oestrogen and progesterone.

65.
Given below is the diagram of a human ovum surrounded by a few sperms. Study the
diagram and the following s.

(i) Which one of the sperms would reach the ovum earlier?
(ii) Identify ‘D and ‘E. Mention the role of ‘E.
(iii) Mention what helps the entry of sperm into the ovum and write the changes
occurring in the ovum during the process.
(iv) Name the specific region in the female reproductive system where the event
represented in the diagram takes place.

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

(ii) D-Cororia radiata, E-Zona pellucida-During fertilisation, a sperm comes in contact
with zona pellucida layer of the ovum and induces changes in the membrane that block
the entry of additional sperms. Hence, E helps to prevent polyspermy.
(iii) The hydrolytic secretions of the acrosome help the sperm to enter into the
cytoplasm of the ovum through the zona pellucida and the plasma membrane. This
induces the completion of the meiotic division of the secondary oocyte.
(iv) These events take place in the ampullary-isthmic junction of the Fallopian tube of
uterus.

66.
(i) Draw a diagram of the adult human female reproductive system and label the
different
(a) parts of Fallopian tube
(b) layers of uterus wall
 67.
Briefly explain the events of fertilisation and implantation in an adult human female.
(ii) Comment on the role of placenta as an endocrine gland. (Delhi 2016)

(i) Fertilisation is the process of fusion of a sperm with an ovum.

● The motile sperms move through the cervix’, enter the uterus and reach the
junction of the isthmus and ampulla (ampullary-isthmic junction) of the Fallopian tube.

● The ovum released from the ovary also reaches the ampullary-isthmic junction
where fertilisation takes place.

● Fertilisation can only occur if the ovum and sperms are transported
simultaneously to this junction. This explains why all copulations do not lead to
fertilisation and pregnancy.

● The sperm comes in contact with the zona pellucida of the ovum and releases
sperm lysins which induces changes in the egg membrane. It blocks the entry of the
other sperms into the egg. Thus, it ensures that only one sperm can fertilise an ovum.

● The secretions of the acrosome help the sperm to enter into the cytoplasm of the
ovum through the zona pellucida and the plasma membrane.

● This induces the completion of meiotic division of the secondary oocyte. The
secondary meiotic division results in the formation of a secondary polar body and a
haploid ovum (ootid).

● The haploid nucleus of the sperm and that of ovum fuse together to form a diploid
zygote.
Implantation The mitotic division starts as the zygote moves through the isthmus of the
oviduct towards the uterus called cleavage thus, forming 2, 4, 8, 16 daughter cells called
blastomeres.

● The embryo with 8-16 blastomeres is called morula. But, it is not larger than a
zygote.

● The morula continues to divide and transforms into blastocyst as it moves further
into the uterus.

● The blastomeres in the blastocyst gets arranged into an outer layer called
trophoblast and the inner group of cells attached to trophoblast called the inner cell
mass.

● The trophoblast layer then gets attached to the endometrium and the inner cell
mass differentiates into the embryo. After attachment, the uterine cells divide rapidly
and cover the blastocyst.

As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is
called implantation and it leads to pregnancy.

(ii) Placenta also acts as an endocrine tissue and produces several hormones like
human Chorionic Gonadotropin (hCG), human Placental Lactogen (hPL) oestrogen,
progesterone, etc.
The hCG stimulates and maintains corpus luteum to secrete progesterone. hPL
stimulates the growth of mammary glands during pregnancy. Relaxin helps in parturition
by softening the connective tissue of pubic symphysis.

68.
During the reproductive cycle of a human female when, where and how does placenta
develop? What is the function of placenta during pregnancy and embryo development?

Placenta is an organ that connects the developing foetus to the uterine wall of mother
for supporting pregnancy.
After implantation, finger-like projections appear on the trophoblast called chorionic villi,
which are surrounded by the uterine tissue and maternal blood. The chorionic villi and
the uterine tissue become interdigitated with each other and jointly form a structural and
functional unit between foetus and maternal body, i.e. placenta.

Functions of Placenta

● It facilitates the supply of oxygen and nutrients to the developing embryo.

● It also facilitates the removal of carbon dioxide and waste materials produced by
the foetus.

● The placenta is connected to the developing embryo through the umbilical cord,
which helps in the transport of substances to and from the developing embryo.

● Placenta also acts as an endocrine tissue and produces several hormones like
human Chorionic Gonadotropin (hCG), human Placental Lactogen (hPL), oestrogen,
progesterone, etc.

The production of these hormones is necessary to support foetal growth, metabolic

changes in the mother and maintenance of pregnancy.

69.
(iii) Name-and draw a labelled sectional view of the embryonic stage that gets
implanted. (Delhi 2010)

(iii) Blastocyst is the stage implanted in the uterus.

71.
(i) Mention the event that induces the completion of the meiotic division of the
secondary oocyte.
(ii) Trace the journey of the ovum from . the ovary, its fertilisation and further
development until the implantation of the embryo.

(i) Secondary oocyte completes meiosis-II only

when a sperm enters into its cytoplasm. It forms a larger cell, the ootid and a small cell,
the second polar body.

(ii) (a) The secondary oocyte is released by the rupture of the Graafian follicle in the
process called ovulation.
(b) It is moved into the Fallopian tube with the help of fimbriae.
(c) It reaches the ampullary-isthmic junction of the Fallopian tube where fertilisation
takes place.
(d) After fertilisation, cleavage starts in the zygote.
(e) Cleavage results in the formation of 2, 4, 8 and 16 daughter cells called blastomeres.
The embryo with 8-16 blastomeres is a solid spherical structure called morula.
(f) Morula continues to divide and blastomeres rearrange themselves as it moves further
into the uterus.
(g) As a result, blastocyst is formed, which contains trophoblast (outer layer) and inner
cell mass.
(h) The trophoblast attaches to endometrium and blastocyst gets embedded in it
(implantation).

72.
(i) Where does the secondary oocyte develop?
(ii) Which help in the collection of ovum after ovulation?
(iii) Where does fertilisation occur?
(iv) Where does implantation of embryo occur?

(i) The secondary oocyte develops in ovary.

(ii) Fimbriae help in the collection of ovum after ovulation.
(iii) In the junction of ampulla-isthmus, the ovum gets fertilised.
(iv) Implantation of embryo occurs in the cavity of uterus.

73.
Write the function of each one of the following
(i) Fimbriae (oviducal)
(ii) Coleoptile
(iii) Oxytocin

(i) Fimbriae are the feathery finger-like projections present at the end of Fallopian tubes.
They help to collect the ovum after its release from the ovary during ovulation.
(ii) Coleoptile is a conical sheath present in the monocot seeds and its function is to
protect the developing plumule.
(iii) Oxytocin is a hormone secreted by the posterior pituitary that stimulates the
contraction of uterine muscles during child birth (parturition).

74.
Write the function of each of the following
(i) Middle piece in human sperm
(ii) Tapetum in anthers
(iii) Luteinising hormone in human males

(i) Middle piece in human sperm contains several mitochondria which produce energy
for the motility of the sperm.
(ii) Tapetum in anthers It is the innermost layer of the anther. The main function of
tapetum is to provide nourishment to the developing pollen grains.
(iii) Luteinising hormone in human males stimulates the Leydig cells to produce
testosterone which is necessary to complete the process of spermatogenesis.

75.
Write the function of each of the following
(i) Seminal vesicle
(ii) Seutellum
(iii) Acrosome of human sperm

(i) Seminal vesicle During spermatogenesis, it secretes an alkaline fluid rich in fructose
and prostaglandins. High level of fructose provides energy to sperms.
(ii) Scutellum It is a tissue present in seed to absorb food from the adjacent endosperm
and develops into growing embryo.
(iii) Acrosome of human sperm It is the cap-like structure that is present at the tip of the
sperm (male gamete) in head region. The acrosome contains hydrolytic enzyme, which
helps the sperm to enter into the cytoplasm of the ovum and thus, helps in fertilisation.

76.
Give reasons for the following
(i) The human testes are located outside the abdominal cavity.
(ii) Some organisms like honeybees are called parthenogenetic animals.

(i) Testes are located outside the abdominal cavity within a pouch called scrotum. It
maintains low temperature of the testes (2-2.5°C lower than normal body temperature)
required for spermatogenesis.
(ii) In honeybees, the drone bees are developed parthenogenetically from an unfertilised
egg. Since, they develop from unfertilised diploid eggs (and do not undergo fertilisation),
they are called parthenogenetic animals.

77.
(i) Explain the following phases in the menstrual cycle of a human female.
(a) Menstrual phase
(b) Follicular phase
(c) Luteal phase
(ii) A proper understanding of menstrual cycle can help immensely in family planning. Do
you agree with the statement? Provide reasons

(i) (a) Menstrual phase The menstrual cycle starts with menstrual phase. It lasts for
about 3-5 days. The menstrual flow results due to the breakdown of endometrial lining of
the uterus and its blood vessels that gome out through vagina.
(b) Follicular phase It lasts till about 13th day of menstrual cycle. In this phase, the
primary follicles in the ovary grow to become a fully mature Graafian follicle. The
secretion of gonadotropins (LH and FSH) from anterior pituitary increases gradually
during the follicular phase. They stimulate follicular development as well as secretion of
oestrogen by the growing follicles.
(c) Luteal phase This phase lasts for about 10-14 days. In this phase, the ruptured
Graafian follicle transforms into corpus luteum. It secretes large amount of progesterone
which is essential to maintain endometrium.

(ii) Yes, a proper understanding of menstrual cycle can help in family planning as this
knowledge can be used to avoid the meeting of sperms and ovum. This is known as
periodic abstinence or rhythm method of birth control, i.e. temporary avoidance of sex.
In this method, a couple can avoid or abstain the coitus from day 10 to 17 of the
menstrual cycle because ovulation occurs during this period. The chances of fertilisation
are very high during this period.

78.

(i) One of the sperms is observed to penetrate ‘A’ of the ovum, as shown in the above
diagram. Name ‘A’.
(ii) How is the sperm able to do so?
(iii) Where exactly in the Fallopian tube does this occur?
(iv) Explain the events thereafter up to morula stage.

(i) A is zona pellucida.

(ii) Sperm is able to penetrate the egg with the help of sperm lysins contained in
acrosome.
(iii) Ampullary-isthmic junction.
(iv) As soon as the sperm enters the egg, meiotic division of secondary oocyte gets
completed. As a result a second polar body and a haploi ovum are formed. Then, haploid
sperm and ovum fuse together to form a diploid zygote which undergoes cleavage via
mitotic division. At 8-16 cell stage, the embryo is called morula.

79.
(ii) Enumerate the events in the ovary of a human female during
(a) follicular phase
(b) luteal phase of menstrual cycle

(ii) (a) Follicular phase The primary follicle in the ovary grows to become fully mature
Graafian follicle. Endometrium gets thickened and the follicular cells secrete oestrogen.
(b) Luteal phase The remaining part of ruptured Graafian follicle transforms into corpus
luteum that secretes progesterone.

80.
(i) Write the specific location and the functions of the following cells in human males
(c) Primary spermatocytes
(ii) Explain the role of any two accessory glands in human male reproductive system.
Primary spermatocytes are located inside the seminiferous tubule and are involved in
the formation of spermatozoa.
(ii) Accessory glands of human male reproductive system are
(a) Prostate and seminal vesicles Their secretions provide a fluid medium for the sperms
to swim towards the ovum. They provide nutrition to sperms.
(b) Bulbourethral glands Their secretion helps in the lubrication and neutralising the
acidity of urogenital tract.

81.
(i) When does oogenesis begin?
(ii) Differentiate between the location and function of Sertoli cells and Leydig cells.

(i) Oogenesis begins during the embryonic development stage when a million gamete
mother cells (oogonia) are formed within each foetal ovary.
(ii) (a) Sertoli cells are located on the inside lining of seminiferous tubule. These cells
provide nutrition to the germ cells.
(b) Leydig cells or interstitial cells are located in the regions outside the seminiferous
tubule called interstitial spaces. These cells synthesise and secrete testicular hormone
called androgens.

83.
What is implantation?

(i) The implantation is the embedding of the blastocyst in the endometrium of the
uterus.

84.
(i) What is the average duration of human pregnancy known as?
(ii) What is the term given to the process of delivery of foetus?

(i) The average duration of human pregnancy is called as gestation period. It is

approximately 270 days in humans.
(ii) The term given to the process of delivery of foetus is called parturition.

Sex in humans

Human Reproduction

1.
Why are human testes located outside the abdominal cavity? Name the pouch

Human testes are located outside the abdominal cavity as it helps in maintaining low
temperature (2-2.5%) lower than body temperature) required for spermatogenesis.
Testes are enclosed in a pouch called scrotum.
 2.
Write the location and functions of
(i) Sertoli cells
(ii) Leydig cells

(i) Location; Within the lining of seminiferous tubule of testis.

Function;provide nutrition to the developing sperms or germ cells.

(ii) Location;In the interstitial spaces between the seminiferous tubules. Function; They
synthesise and secrete male hormones, androgens, testosterone.

3.
function of the seminal vesicle.

Seminal vesicle produces an alkaline secretion containing prostaglandins, proteins and

fructose. The high fructose content provides energy to the spermatozoa. These
secretions form 60-70% of the fluid in the semen.

4.
List the parts of human oviduct through which ovum travels till it meets the sperm

The parts of human oviduct through which the ovum travels are given below;

● Fimbriae, finger-like projections;Collect the ovum from ovary after ovulation.

● Infundibulum;Ovum from fimbriae is guided into funnel-shaped infundibulum, part

of Fallopian tube.

● Ampulla A wider part of oviduct that leads ovum into isthmus.

● Isthmus; has narrow lumen which opens into uterus.

● In the junction of ampulla- isthmus, the ovum gets fertilised.

6.

(i) Identify ‘X’ and write its location.

(ii) Name the accessory gland ‘Y’ and its secretion.
(iii) Name and state the function of ‘Z’.
(i) .X-Testis. It is located outside the abdominal cavity within a pouch called scortum.
(ii) K-Seminal vesicle. It produces an alkaline secretion rich in fructose and constitutes
60% of the volume of the semen.
(iii) Z-Epididymis. It stores the sperms and secretes a fluid which helps in the maturation
of sperms.

7.
(ii) Differentiate between vas deferens and vasa efferentia.

(ii) Distinction between vas deferens and vasa efferentia

eferens efferentia

tube-like structure which conducts the nects the rete testis to the
matozoa from the epididymis to the penis. dymis.

arise from the rete testis. arise from the cauda epididymis.

are only 2 in number. vary from 15-20 in number.

eferens are thick. efferentia are fine.

lining has many stereocilia. lining bears many ciliated cells.

8.
Name and explain the role of inner and middle walls of human uterus.

The innermost wall of uterus is called endometrium.

Role of Endometrium
(i) It lines the uterine cavity and is glandular.
(ii) It undergoes cyclic changes during menstrual cycle.
The middle wall or layer of uterus is called myometrium.

(i) It is made up of thick layer of smooth muscles.

(ii) It shows strong contractions during the delivery of baby.

10.
When do the oogenesis and the spermatogenesis initiate in human females and males,
respectively?

Oogenesis initiates during foetal or embryonic stage in females, whereas

spermatogenesis in males starts at puberty.

11.
differences between spermiogenesis and spermiation.

miogenesis miation

he process of transforming m heads are embedded in Sertoli cells to obta

matids into mature spermatozoa orshment and finally released from seminiferou
ms through differentiation. es by the process called spermiation.

12.
Where is acrosome present? Write its function.

acrosome is present in the anterior portion of the head of sperm. Function ;Hydrolytic
enzymes or sperm lysins present in acrosome help in penetration of sperm into egg,
during fertilisation.

13.
List the changes the primary oocyte undergoes in the tertiary follicular stage in human
embryo.

The primary oocyte within the tertiary follicle grows in size. The fully grown primary
oocyte completes its first meiotic division to produce two daughter nuclei in which larger
haploid cell is called secondary oocyte and the tiny one is called first polar body. The
secondary oocyte retains bulk of nutrient rich cytoplasm of primary oocyte.

14.
Name the hormones influencing follicular development of corpus luteum.

The hormones influencing follicular development of corpus luteum are FSH, oestrogen
and progesterone.

15.
Explain the hormonal regulation of the process of spermatogenesis in humans.
● Gonadotropin Releasing Hormone (GnRH) is released significantly from the
hypothalamus during puberty.

● GnRH stimulates anterior pituitary to secrete gonadotropins, i.e. LH and FSH or

Interstitial Cell Stimulating Hormone (ICSH).

● Luteinising Hormone (LH) acts on Leydig cells to stimulate the synthesis and
secretion of androgens which then stimulate the process of spermatogenesis.

● Follicle Stimulating Hormone (FSH) acts on Sertoli cells and stimulates them to
secrete inhibin which then stimulates the process of spermiogenesis.

16.
What happens to corpus luteum in human female if the ovum is (i) fertilised, (ii) not
fertilised?

● In case of fertilisation, the corpus luteum continues secreting progesterone which

is required for the maintenance of endometrium during pregnancy.

● In the absence of fertilisation, the corpus luteum degenerates and gets converted
into corpus albicans. Deficiency of progesterone causes disintegration of the
endometrium leading to menstruation and thus, a new cycle starts.

17.
parts and functions that assist the sperm to reach and gain entry into the female
gamete.

The parts that assist sperm to reach and gain entry into female gamete are,
Tail Its wiggling movement helps the sperm to swim in female reproductive tract.
Middle piece It contains mitochondria that provide energy for sperm movement.
Head It contains acrosome loaded with sperm lysins. These help to dissolve the layers
of female gamete or egg.

18.
effect of high concentrations of LH on a mature Graafian follicle.
The high concentration of LH (LH surge) induces rupture of Graafian follicle, which
results in the release of secondary oocyte hence, causing ovulation in females.

19.
Differentiate between spermatogenesis and spermiogenesis.

matogenesis miogenesis
 process of formation of sperms from process of transformation of a circular
ture germ cells. matid to a motile spermatozoa.

ber of cells increases as each hange in number of cells as only one

matogonium produces four spermatids.matid develops into a spermatozoa.

20.
Draw and label head region of a human sperm.

21.

Identify A, B, C and D with reference to gametogenesis in humans

A – Leydig’s cell
B – Sertoli cell
C – Spermatogenesis
D – Spermiogenesis

22.
Name the labels A, B, C, D, E and F in seminiferous tubule.

A – Spermatogonium
B – Primary spermatocyte
C – Secondary spermatocyte
D – Spermatid
E – Spermatozoa
F – Sertoli cell

23.
Differentiate between menarche and menopause.

rche pause

ning of menstrual cycle in human age of menstrual cycle at the age of about
e at puberty. in human female.

of reproductive phase. f reproductive phase.

24.
Study the sectional view of human testis showing seminiferous tubules given below.

(i) Identify A, B and C.

(ii) Write the function of A and D.

(i) A-Spermatogonia
B-Interstitial cells
C-Spermatozoa.

(ii) A-Spermatogonia undergo meiosis to produce spermatozoa (sperms).

D-Sertoli cells provide nutrition to the germ cells.

25.
How and at what stage of menstrual cycle is corpus luteum formed? When does it
regress?

After ovulatory phase (ovulation), the luteal phase starts. The remaining parts of
ruptured Graafian follicle transform into corpus luteum in this phase. The corpus luteum
secretes large amount of progesterone which is essential for the maintenance of
endometrium.In the absence of fertilisation, the corpus luteum degenerates in the ovary
and gets converted into corpus albicans.

27.
(i) How many primary follicles are left in each ovary in a human female at puberty?

(i) A large number of primary follicles degenerate in females during the period from birth
to puberty by the process called follicular atresia. As a result, about 60000-80000
primary follicles are left in each ovary at puberty.

28.
Explain the events in a normal woman during her menstrual cycle on the following days
(i) Ovarian event from 13-15 days.
(ii) Ovarian hormones level from 16-23 days.
(iii) Uterine events from 24-29 days.

(i) In the ovarian event from 13-15 days, a immature ovum (egg cell) is released from the
Graafian follicle. Both LH and FSH attain maximum peak. FSH helps Graafian follicle to
attain maturity and LH helps in its rupture. Ovum covered by a number of layer and a
yellow fat layer forms corpus luteum. It releases (secretes) progesterone.

(ii) During menstrual cycle, the period level from 16-23 days is called luteal phase
(secretory phase). The corpus luteum secretes large amount of progesterone which is
essential for the maintenance of endometrium.

(iii) Uterine events from 24-29 days are under the influence of progesterone hormone. It
influences the maintenance of the endometrium for any pregnancy to occur. In the
absence of pregnancy, the corpus luteum degenerates and endometrium sheds off. It
causes the menstrual flow or bleeding.

29.
Explain the events in a normal menstrual cycle on
(i) Pituitary hormone levels from 12 days.
(ii) Uterine events from 13-15 days.
(iii) Ovarian events from 16-23 days.

(i) The period of 8-12 days after the onset of menstruation is the follicular phase. During
this phase, GnRH from hypothalamus stimulates anterior pituitary to release FSH and
LH. FSH stimulates the ovarian follicles to secrete oestrogen, which in turn stimulates
the proliferation of the endometrium of the uterine wall. This causes the endometrial
lining to thicken.

(ii) The uterine events between day 13 and 15 are governed by the high LH and FSH
levels. The endometrium is intact due to the effect of these gonadotropin hormones and
also prepares itself for pregnancy, if fertilisation occurs.
(iii) During 16-23 days, ruptured Graafian follicle gets converted into corpus luteum in the
ovary. It starts secreting progesterone which maintains the endometrium, necessary for
the implantation of fertilised ovum followed by other events of pregnancy.

31.
Explain the steps in the formation of an ovum from oogonium in humans.

In human females, primary oocytes are formed from the oogonia during the embryonic
developmental stages in the foetal ovaries.

● Oogonial cells start dividing and enter prophase-I of meiosis. They remain
suspended at this stage as primary oocytes.

● Each primary oocyte is surrounded by a layer of granulosa cells and becomes the
primary follicle.

● The primary follicle when surrounded by more layers of granulosa ceils, is called a
secondary follicle.

● Secondary follicle transforms into a tertiary follicle, with the development of a

fluid-filled cavity (antrum) around the primary oocyte

● Granulosa cells become organised into an outer theca externa and an inner theca
interna.

● primary oocyte completes meiosis-I and forms a larger haploid secondary oocyte
and a tiny first polar body.

● Tertiary follicle grows and becomes a mature follicle called Graafian follicle.

● Secondary oocyte secretes a new membrane called zona pellucida around it.

● At this stage, follicle ruptures to release the secondary oocyte, which moves into
the cytoplasm.

● Secondary oocyte completes meiosis-II only when a sperm enters its cytoplasm. It
forms a larger cell, the ootid and a small second polar body. This event occurs in the
ampulla of Fallopian tube.

32.
(i) Explain menstrual cycle in human females.
(ii) How can the scientific understanding of the menstrual cycle of human females help
as a contraceptive measure ?

(i) Menstrual Cycle; The inner lining of uterus called endometrium, grows and thickens
each month and prepares itself for the implantation of an embryo. If pregnancy does not
occur, the endometrium sheds off. The monthly development and shedding of the
functional layer of the uterus is called the menstrual phase and the monthly maturation
of an egg and its release is called the ovarian cycle. A typical menstrual cycle completes
in an average of about. 28 days. It starts at the age of 13 or 15 and continues till about
50 years of age. Menstrual cycle occurs in three major phases namely menstrual phase,
follicular phase and secretory phase.
(a) Menstrual Phase It lasts for 3-4 days. It occurs due to the breakdown of
endometrium lining cf uterus and blood vessels.
(b) Follicular or Proliferative Phase. regulated by hormones secreted by anterior pituitary
gland who stimulate the ovarian follicle to secrete oestrogens.
(c) Secretory Phase or Luteal phase The phase of menstrual cycle with possibility of
fertilisation is the initial luteal phase. It is marked by the presence of corpus luteum
(yellow body). During pregnancy, this yellow body secretes progesterone. In the absence
of pregnancy, it regresses to form corpus albicans and menstruation starts (menstrual
phase). Due to the hormones secreted by corpus luteum, i.e. oestrogen and
progesterone, the release of FSH and LH is inhibited. This prevents the development of
new follicles.This phase lasts for 14 days. Maintenance of endometrium by
progesterone is necessary for the implantation of embryo and pregnancy. During
pregnancy, menstrual cycle stops due to high level of progesterone.

(ii) The scientific understanding of the menstrual cycle of human females helps as a
contraceptive method. This method is known as periodic abstinence. In this method, the
couples avoid or abstain the coitus from day 10 to 17 of the menstrual cycle when
ovulation could be expected. The chances of fertilisation are very high during this period.

33.
(i) Arrange the following hormones in the sequence of their secretion in a pregnant
woman. hCG, LH, FSH, Relaxin
(ii) Mention their source and the function they perform.

(i) The sequence of secretion of the given hormones in a pregnant woman is

FSH → LH → hCG → Relaxin

one ce ion

ior pituitary lobe ulates the growth of ovarian follicles in ovary.

ior pituitary lobe gers ovulation.

nta mulates the corpus luteum to secrete progesterone.

 in nta litates parturition by softening the connective tissu
mphysis pubica.

34.
(i) Explain the menstrual phase in a human female. State the level of ovarian and
pituitary hormones during this phase.
(ii) Why is follicular phase in the menstrual cycle also referred as proliferative phase?
Explain.
(iii) Explain the events that occur in a Graafian follicle at the time of ovulation and
thereafter.
(iv) Draw a Graafian follicle and label antrum and secondary oocyte.

(i) The reproductive cycle in female primates, e.g. monkeys, apes and human beings is
called menstrual cycle.
Menstrual phase starts from 3rd day and ends on 5th day of the menstrual cycle. It is
initiated due to the reduced secretion of progesterone and oestrogen from the
regressing corpus luteum in the ovary. The endometrium breaks down and blood along
with degenerated ovum constitutes the menstrual flow. The secretion of pituitary
hormones, i.e. FSH and LH is also reduced during this phase.

(ii) In follicular phase, primary follicles in the ovary grow under the influence of Follicle
Stimulating Hormone (FSH). It starts from 6th day and ends on 13 th or 14th day of 28
day cycle. FSH also stimulates the ovarian follicles to secrete oestrogen which in turn
stimulates endometrium to proliferate, so that it becomes thicker and highly
vascularised. Thus, it is also called proliferative stage of menstrual cycle.

(iii) Graafian Follicle at Ovulation

At the time of ovulation following events occur

● LH and FSH reach at their peak levels (about 14th-16th day of cycle).

● High level of LH induces Graafian follicle to rupture and the release of secondary
oocyte from it.

● After ovulation, the remaining cells of Graafian follicle are stimulated by LH to

develop corpus luteum (an endocrine gland which secrete progesterone hormone).
(iv) Diagrammatic sectional view of Graafian follicle

35.
(i) Explain the process of spermatogenesis in humans.
(ii) Draw a human sperm and label acrosome and middle piece. Mention their functions.
Outside (Delhi 2016C)

(i) Spermatogenesis is the production of sperms in males.

● In testis, the immature male germ cells (spermatogonia) produce sperms by

spermatogenesis. It begins at puberty due to significant increase in the secretion of
gonadotropins, i.e. luteinising hormone and follicle stimulating hormone under the
influence of Gonadotropin Releasing Hormone (GnRH) released from hypothalamus.

● Spermatogonia (sing, spermatogonium) present on the inside wall of seminiferous

tubules multiply by mitotic division and increases in numbers.

● Each spermatogonium is diploid and contains 46 chromosomes. Some of the

spermatogonia transform to primary spermatocytes.

● The primary spermatocyte undergoes meiosis -1 and forms two haploid secondary
spermatocytes containing 23 chromosomes each.

● The secondary spermatocytes undergo meiosis – II and form four equal sized
haploid spermatids.

● Spermatids transform into the spermatozoa by spermiogenesis.

● After spermiogenesis, the sperm heads get embedded in the Sertoli cells and
released from the seminiferous tubules via spermiation process.
(a) The acrosome is filled with hydrolytic enzymes that help the sperm to penetrate the
ovum.
(b) Middle piece possesses many mitochondria to produce energy for the movement of
tail to facilitate sperm motility.

36.
Explain the ovarian and uterine events that occur during a menstrual cycle in a human
female under the influence of pituitary and ovarian hormones, respectively.
The cycle of events starting from one menstruation till next in female primates is called
menstrual cycle. It comprises of four phases which are regulated by both pituitary (LH
and FSH) and ovarian (oestrogen and progesterone) hormones that affect ovaries and
uterus, respectively. The events occurring in a menstrual cycle are as follows

Menstrual phase Initiated by reduced secretion of LH, progesterone and

(from 3rd-5th day oestrogen. The endometrium breaks down and blood along with
in a 28 day cycle) unfertilised ovum constitutes menstrual flow.

Follicular phase The FSH (Follicle Stimulating Hormone) secreted by anterior

(from 6th-13th day pituitary stimulates ovarian follicle to secrete oestrogens. These
in a 28 day cycle) oestrogens stimulate proliferation of uterine walls as a result of
which endometrium gets thickened (due to rapid cell division and
increase in uterine glands and blood vessels).

Ovulatory phase Pituitary hormones, i.e. LH and FSH reach the highest level in
(14th day in 28 day middle of the cycle. Rapid secretion of LH causes ovulation thus,
cycle) inducing the rupture of Graafian follicle to release secondary
oocyte and a polar body.
 Luteal or secretory The pituitary hormone LH stimulates the remaining cells of
phase (from ovarian follicles to develop into corpus luteum. This corpus
15th-28th day in a luteum secretes large amount of progesterone and maintains
28 day cycle) endometrium thickening for the implantation of fertilised ovum
during pregnancy. In the absence of fertilisation, the hormone
levels are reduced (LH and progesterone) and endometrium
disintegrates leading to onset of another menstrual cycle.

37.
Explain the development of a secondary oocyte (ovum) in a human female from the
embryonic stage up to its ovulation. Name the hormones involved in this process.

● Gonadotropins, i.e., LH and FSH stimulate follicular development and secretion of

oestrogen by the growing follicles.

● Both LH and FSH attain a peak level in the middle of the cycle (14th day).

● Rapid release of LH during mid-cycle causes ovulation.

● LH also stimulates the formation of corpus luteum from the ruptured follicle and
secretion of progesterone from corpus luteum.

38.
(i) How is ‘oogenesis’ markedly different from ‘spermatogenesis’ with respect to the
growth till puberty in the humans?

(i) Oogenesis is markedly different from spermatogenesis in the following aspects

Spermatogenesis Oogenesis

It occurs in males. starting from puberty It starts before birth during

till the complete life cycle. embryonic development and occurs
till menopause.

A single spermatogonium after second A single oogonium, after second

meiotic division forms four haploid meiotic division, produces one
spermatids that mature to form four ovum and two non-functional polar
spermatozoa. bodies.
 The process of spermatogenesis, i.e. The second meiotic division of
second meiotic division completes in oogenesis completes in Fallopian
testes and releases mature sperms. tube when sperm enters the
secondary oocyte.

41.
The following is the illustration of the sequence of ovarian events (A-I) in a human
female.

(i) Identify the figure that illustrates ovulation and mention the stage of oogenesis it
represents.
(ii) Name the ovarian hormone and the pituitary hormone that have caused the above
mentioned event.
(iii) Explain the changes that occur in the uterus simultaneously in anticipation.
(iv) Write the differences between C and H.
(v) Draw a labelled sketch of the structure of a human ovum prior to fertilisation.

(i) Figure F illustrates ovulation. It represents the ovulatory stage of oogenesis.

(ii) Ovarian and pituitary hormones involved in causing ovulation are
Ovarian hormone Oestrogen.
Pituitary hormone LH and FSH.
(iii) In anticipation of receiving the fertilised egg, the endometrium of the uterus gets
thickened and also the blood supply to the endometrium increases.
(iv) In the figure, C stage represents the secondary follicle and the H stage represents
the degenerating corpus luteum.

Secondary follicle Corpus luteum

It is surrounded by layers of Layers of granulosa cells and theca cells

granulosa cells and theca layer. are absent.

It contains an oocyte It does not contain oocyte as it is formed

after the release of secondary oocyte.
 It does not secrete It secretes progesterone.

(v) Structure of a human ovum

42.
The following is the illustration of the sequence of ovarian events A-I in a human female.

Identify the figure that illustrates corpus luteum and name the pituitary hormone that
influences its formation.
(ii) Specify the endocrine function of corpus luteum. How does it influence the uterus?
Why is it essential?
(iii) What is the difference between D and E?
(iv) Draw a neat and labelled sketch of Graafian follicle. (Delhi 2012)

(i) Gis corpus luteum. LH secreted by anterior pituitary’ influences its formation.
(ii) The endocrine function of corpus luteum is to secrete progesterone which is
essential for the maintenance of endometrium layer of uterus. Thickened endometrium
is necessary for the implantation of fertilised ovum and other events of pregnancy.
(iii) D is a developing secondary follicle containing primary oocyte and undifferentiated
thecal cells. E is a mature tertiary follicle that contains a secondary oocyte and
differentiated theca cells.
(iv) Structure of a Graafian follicle

45.
State from where do the signals for parturition originate in human females.
Foetal-ejection reflex generated by fully developed foetus and placenta stimulates
pituitary to release the hormone oxytocin responsible for parturition.

46.
Mention the function of zona pellucida.

During fertilisation, a sperm comes in contact with the zona pellucida layer of ovum and
induces changes in the membrane and forms fertilisation envelope. It helps block the
entry of additional sperms. So, zona pellucida ensures that only one sperm fertilises an
ovum.

47.
How does the sperm penetrate through the zona pellucida in human ovum?

The sperm penetrates through the zona pellucida by releasing sperm lysins
(hyaluronidase and protease) present in the acrosome.

48.
How is the entry of only one sperm ensured into an ovum during fertilisation?

Presence of zona pellucida layer ensures that only one sperm can fertilise an ovum.

49.
Write the function of oxytocin.

Oxytocin secreted by posterior pituitary helps in parturition by inducing and enhancing

uterine muscle contractions.

50.
Mention the function of trophoblast in human embryo.

Trophoblast is the outer layer of blastocyst which helps in the attachment of blastocyst
to the endometrium of the uterus.

51.
Name the embryonic stage that gets implanted in the uterine wall of a human female.

Blastocyst gets implanted in the uterine wall of human female.

52.
What stimulates pituitary to release the hormone responsible for parturition? Name the
hormone.

Foetal-ejection reflex stimulates pituitary to release the hormone responsible for

parturition. This hormone is oxytocin.
 53.
Why is breastfeeding recommended during the initial period of an infant’s growth? Give
reasons.

The first milk which comes out from the mother’s mammary glands just after childbirth
is called colostrum. This is a thin, yellowish, opalescent fluid.It is rich in proteins and
energy along with several antibodies (IgA) that provide passive immunity to the
newborn. This immunity helps the newborn to attain resistance to many infections, but it
is low in fat and iron content. Hence, breastfeeding during the initial period of infant
growth is highly recommended by the doctors for bringing up a disease-free, healthy
baby.

54.
Where does fertilisation occur in humans? Explain the events that occur during this
process.

the fertilisation of ovum takes place in ampullary-isthmic junction of the Fallopian tube.
The events that occur during the process of fertilisation are

● The sperm reaches the junction of ampulla and isthmus and comes in contact with
zona pellucida layer of ovum.

● Acrosome of sperm head releases sperm lysin enzyme that dissolves corona
radiata and digests zona pellucida layer to enter the cytoplasm of ovum.

● After the entry of sperm into the ovum, zona pellucida induces changes in
membrane and blocks the entry of additional sperms

● Entry of sperm stimulates the secondary oocyte to complete its suspended

second meiotic division thus, producing haploid egg or ovum and second polar body.

● Nucleus of sperm and of ovum fuse to form a diploid zygote.

55.
When and where do chorionic villi appear in humans? State their function.

After implantation in the uterus, finger-like projections called chorionic villi appear on the
trophoblast of blastocyst. It interdigitates with uterine tissue to form placenta that
transports oxygen and nutrients to foetus and remove waste products and carbon
dioxide produced by the foetus.

56.
(i) Where do the signals for parturition originate in humans?
(ii) Why is it important to feed the newborn babies on colostrum?

(i) The signals for parturition originate from the fully developed foetus and the placenta,
which induces mild uterine contraction called foetal-ejection reflex.
(ii) Colostrum contains necessary antibodies (IgA) that provide protection against
disease to infants.

57.
Explain the function of umbilical cord.

The placenta is connected to developing embryo through an umbilical cord, which helps
in the transport of substances like oxygen, nutrients, etc. to and from the developing
embryo. Blood in umbilical cord can be collected during delivery and stored to treat
various diseases in future because it is rich in stem cells.

58.
Mention the number of cells in the following stages.

yonic Stage ber of Cells

ocyst

(i) 1 (ii) 16 (iii) 64 (2)

59.
Name the hormones produced only during pregnancy in human female. Mention their
source organs.

The hormones produced only during pregnancy in human female are human Chorionic
and placenta Gonadotropin (hCG), human Placental Lactogen (hPL) and relaxin.
The source of hCG and hPL is placenta and that of relaxin is ovary and placenta.

60.
Why is parturition called a neuroendocrine mechanism?

Process of parturition is induced by both neural system and endocrine system therefore,
it is called a neuroendocrine mechanism.
Vigorous contraction of the uterus at the end of pregnancy causes expulsion/delivery of
the foetus. The process of delivery of foetus (childbirth) is called parturition.
Relaxin hormone is secreted by the ovary and placenta to facilitate parturition by
softening the connective tissue of symphysis pubica. Foetus and the placenta induce
mild uterine contractions called foetal-ejection reflex. This initiates the release of
oxytocin hormone from the posterior pituitary. Oxytocin acts on uterine muscles to
cause stronger contractions, which in turn stimulate further secretion of oxytocin.
This causes more stronger contractions leading to the expulsion of the baby out of the
uterus, through the birth canal. After the baby is delivered, the placenta is also expelled
out of the uterus.

61.
(ii) Name the embryonic stage that gets implanted in human females. Explain the
process of implantation.

(ii) Blastocyst gets implanted in the uterine wall.

62.
The given diagram shows a part of the human female reproductive system.

(i) Name the gamete cells that would be present in X’ if taken from a newborn baby.
(ii) Name ‘Y’ and write its function.
(iii) Name ‘Z’ and write the events that take place here.

(i) Primary oocytes are present in A, i.e. ovary of a newborn baby.

(ii) Y is fimbriae. It helps in the collection of ovum from ovary after ovulation.
(iii) Z is the ampullary-isthmic junction. It is the site of fertilisation in humans.

64.
(i) How is placenta formed in human female?
(ii) Name any two hormones which are secreted by it and are also present in a
non-pregnant woman.

(i) After implantation of blastocyst, the finger-like projections called chorionic villi,
appear on the trophoblast. They get surrounded by uterine tissue and maternal blood
which become interdigitated with each other to form placenta.
(ii) The two hormones secreted by placenta that are also present in a non-pregnant
woman are oestrogen and progesterone.

65.
Given below is the diagram of a human ovum surrounded by a few sperms. Study the
diagram and the following s.
(ii) Identify ‘D and ‘E. Mention the role of ‘E.
(iii) Mention what helps the entry of sperm into the ovum and write the changes
occurring in the ovum during the process.
(iv) Name the specific region in the female reproductive system where the event
represented in the diagram takes place.

(ii) D-Cororia radiata, E-Zona pellucida-During fertilisation, a sperm comes in contact

with zona pellucida layer of the ovum and induces changes in the membrane that block
the entry of additional sperms. Hence, E helps to prevent polyspermy.
(iii) The hydrolytic secretions of the acrosome help the sperm to enter into the
cytoplasm of the ovum through the zona pellucida and the plasma membrane. This
induces the completion of the meiotic division of the secondary oocyte.
(iv) These events take place in the ampullary-isthmic junction of the Fallopian tube of
uterus.

66.
(i) Draw a diagram of the adult human female reproductive system and label the
different
(a) parts of Fallopian tube
(b) layers of uterus wall

67.
Briefly explain the events of fertilisation and implantation in an adult human female.
(ii) Comment on the role of placenta as an endocrine gland.

(i) Fertilisation is the process of fusion of a sperm with an ovum.

● The motile sperms move through the cervix’, enter the uterus and reach the
junction of the isthmus and ampulla (ampullary-isthmic junction) of the Fallopian tube.
● The ovum released from the ovary also reaches the ampullary-isthmic junction
where fertilisation takes place.

● Fertilisation can only occur if the ovum and sperms are transported
simultaneously to this junction. This explains why all copulations do not lead to
fertilisation and pregnancy.

● The sperm comes in contact with the zona pellucida of the ovum and releases
sperm lysins which induces changes in the egg membrane. It blocks the entry of the
other sperms into the egg. Thus, it ensures that only one sperm can fertilise an ovum.

● The secretions of the acrosome help the sperm to enter into the cytoplasm of the
ovum through the zona pellucida and the plasma membrane.

● This induces the completion of meiotic division of the secondary oocyte. The
secondary meiotic division results in the formation of a secondary polar body and a
haploid ovum (ootid).

● The haploid nucleus of the sperm and that of ovum fuse together to form a diploid
zygote.

Implantation The mitotic division starts as the zygote moves through the isthmus of the
oviduct towards the uterus called cleavage thus, forming 2, 4, 8, 16 daughter cells called
blastomeres.

● The embryo with 8-16 blastomeres is called morula. But, it is not larger than a
zygote.

● The morula continues to divide and transforms into blastocyst as it moves further
into the uterus.

● The blastomeres in the blastocyst gets arranged into an outer layer called
trophoblast and the inner group of cells attached to trophoblast called the inner cell
mass.

● The trophoblast layer then gets attached to the endometrium and the inner cell
mass differentiates into the embryo. After attachment, the uterine cells divide rapidly
and cover the blastocyst.
As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is
called implantation and it leads to pregnancy.

(ii) Placenta also acts as an endocrine tissue and produces several hormones like
human Chorionic Gonadotropin (hCG), human Placental Lactogen (hPL) oestrogen,
progesterone, etc.
The hCG stimulates and maintains corpus luteum to secrete progesterone. hPL
stimulates the growth of mammary glands during pregnancy. Relaxin helps in parturition
by softening the connective tissue of pubic symphysis.

68.
During the reproductive cycle of a human female when, where and how does placenta
develop? What is the function of placenta during pregnancy and embryo development?

Placenta is an organ that connects the developing foetus to the uterine wall of mother
for supporting pregnancy.

After implantation, finger-like projections appear on the trophoblast called chorionic villi,
which are surrounded by the uterine tissue and maternal blood. The chorionic villi and
the uterine tissue become interdigitated with each other and jointly form a structural and
functional unit between foetus and maternal body, i.e. placenta.

Functions of Placenta

● It facilitates the supply of oxygen and nutrients to the developing embryo.

● It also facilitates the removal of carbon dioxide and waste materials produced by
the foetus.

● The placenta is connected to the developing embryo through the umbilical cord,
which helps in the transport of substances to and from the developing embryo.

● Placenta also acts as an endocrine tissue and produces several hormones like
human Chorionic Gonadotropin (hCG), human Placental Lactogen (hPL), oestrogen,
progesterone, etc.

The production of these hormones is necessary to support foetal growth, metabolic

changes in the mother and maintenance of pregnancy.

69.
(iii) Name-and draw a labelled sectional view of the embryonic stage that gets
implanted. (Delhi 2010)
(iii) Blastocyst is the stage implanted in the uterus.

71.
(i) Mention the event that induces the completion of the meiotic division of the
secondary oocyte.
(ii) Trace the journey of the ovum from . the ovary, its fertilisation and further
development until the implantation of the embryo.

(i) Secondary oocyte completes meiosis-II only

when a sperm enters into its cytoplasm. It forms a larger cell, the ootid and a small cell,
the second polar body.

(ii) (a) The secondary oocyte is released by the rupture of the Graafian follicle in the
process called ovulation.
(b) It is moved into the Fallopian tube with the help of fimbriae.
(c) It reaches the ampullary-isthmic junction of the Fallopian tube where fertilisation
takes place.
(d) After fertilisation, cleavage starts in the zygote.
(e) Cleavage results in the formation of 2, 4, 8 and 16 daughter cells called blastomeres.
The embryo with 8-16 blastomeres is a solid spherical structure called morula.
(f) Morula continues to divide and blastomeres rearrange themselves as it moves further
into the uterus.
(g) As a result, blastocyst is formed, which contains trophoblast (outer layer) and inner
cell mass.
(h) The trophoblast attaches to endometrium and blastocyst gets embedded in it
(implantation).

72.
(i) Where does the secondary oocyte develop?
(ii) Which help in the collection of ovum after ovulation?
(iii) Where does fertilisation occur?
(iv) Where does implantation of embryo occur?

(i) The secondary oocyte develops in ovary.

(ii) Fimbriae help in the collection of ovum after ovulation.
(iii) In the junction of ampulla-isthmus, the ovum gets fertilised.
(iv) Implantation of embryo occurs in the cavity of uterus.

73.
Write the function of each one of the following
(i) Fimbriae (oviducal)
(ii) Coleoptile
(iii) Oxytocin

(i) Fimbriae are the feathery finger-like projections present at the end of Fallopian tubes.
They help to collect the ovum after its release from the ovary during ovulation.
(ii) Coleoptile is a conical sheath present in the monocot seeds and its function is to
protect the developing plumule.
(iii) Oxytocin is a hormone secreted by the posterior pituitary that stimulates the
contraction of uterine muscles during child birth (parturition).

74.
Write the function of each of the following
(i) Middle piece in human sperm
(ii) Tapetum in anthers
(iii) Luteinising hormone in human males

(i) Middle piece in human sperm contains several mitochondria which produce energy
for the motility of the sperm.
(ii) Tapetum in anthers It is the innermost layer of the anther. The main function of
tapetum is to provide nourishment to the developing pollen grains.
(iii) Luteinising hormone in human males stimulates the Leydig cells to produce
testosterone which is necessary to complete the process of spermatogenesis.

75.
Write the function of each of the following
(i) Seminal vesicle
(ii) Seutellum
(iii) Acrosome of human sperm

(i) Seminal vesicle During spermatogenesis, it secretes an alkaline fluid rich in fructose
and prostaglandins. High level of fructose provides energy to sperms.
(ii) Scutellum It is a tissue present in seed to absorb food from the adjacent endosperm
and develops into growing embryo.
(iii) Acrosome of human sperm It is the cap-like structure that is present at the tip of the
sperm (male gamete) in head region. The acrosome contains hydrolytic enzyme, which
helps the sperm to enter into the cytoplasm of the ovum and thus, helps in fertilisation.

76.
Give reasons for the following
(i) The human testes are located outside the abdominal cavity.
(ii) Some organisms like honeybees are called parthenogenetic animals.

(i) Testes are located outside the abdominal cavity within a pouch called scrotum. It
maintains low temperature of the testes (2-2.5°C lower than normal body temperature)
required for spermatogenesis.
(ii) In honeybees, the drone bees are developed parthenogenetically from an unfertilised
egg. Since, they develop from unfertilised diploid eggs (and do not undergo fertilisation),
they are called parthenogenetic animals.

77.
(i) Explain the following phases in the menstrual cycle of a human female.
(a) Menstrual phase
(b) Follicular phase
(c) Luteal phase
(ii) A proper understanding of menstrual cycle can help immensely in family planning. Do
you agree with the statement? Provide reasons

(i) (a) Menstrual phase The menstrual cycle starts with menstrual phase. It lasts for
about 3-5 days. The menstrual flow results due to the breakdown of endometrial lining of
the uterus and its blood vessels that gome out through vagina.
(b) Follicular phase It lasts till about 13th day of menstrual cycle. In this phase, the
primary follicles in the ovary grow to become a fully mature Graafian follicle. The
secretion of gonadotropins (LH and FSH) from anterior pituitary increases gradually
during the follicular phase. They stimulate follicular development as well as secretion of
oestrogen by the growing follicles.
(c) Luteal phase This phase lasts for about 10-14 days. In this phase, the ruptured
Graafian follicle transforms into corpus luteum. It secretes large amount of progesterone
which is essential to maintain endometrium.

(ii) Yes, a proper understanding of menstrual cycle can help in family planning as this
knowledge can be used to avoid the meeting of sperms and ovum. This is known as
periodic abstinence or rhythm method of birth control, i.e. temporary avoidance of sex.
In this method, a couple can avoid or abstain the coitus from day 10 to 17 of the
menstrual cycle because ovulation occurs during this period. The chances of fertilisation
are very high during this period.

78.

(i) One of the sperms is observed to penetrate ‘A’ of the ovum, as shown in the above
diagram. Name ‘A’.
(ii) How is the sperm able to do so?
(iii) Where exactly in the Fallopian tube does this occur?
(iv) Explain the events thereafter up to morula stage.

(i) A is zona pellucida.

(ii) Sperm is able to penetrate the egg with the help of sperm lysins contained in
acrosome.
(iii) Ampullary-isthmic junction.
(iv) As soon as the sperm enters the egg, meiotic division of secondary oocyte gets
completed. As a result a second polar body and a haploi ovum are formed. Then, haploid
sperm and ovum fuse together to form a diploid zygote which undergoes cleavage via
mitotic division. At 8-16 cell stage, the embryo is called morula.

79.
(ii) Enumerate the events in the ovary of a human female during
(a) follicular phase
(b) luteal phase of menstrual cycle

(ii) (a) Follicular phase The primary follicle in the ovary grows to become fully mature
Graafian follicle. Endometrium gets thickened and the follicular cells secrete oestrogen.
(b) Luteal phase The remaining part of ruptured Graafian follicle transforms into corpus
luteum that secretes progesterone.

80.
(i) Write the specific location and the functions of the following cells in human males
(c) Primary spermatocytes
(ii) Explain the role of any two accessory glands in human male reproductive system.

Primary spermatocytes are located inside the seminiferous tubule and are involved in
the formation of spermatozoa.
(ii) Accessory glands of human male reproductive system are
(a) Prostate and seminal vesicles Their secretions provide a fluid medium for the sperms
to swim towards the ovum. They provide nutrition to sperms.
(b) Bulbourethral glands Their secretion helps in the lubrication and neutralising the
acidity of urogenital tract.

81.
(i) When does oogenesis begin?
(ii) Differentiate between the location and function of Sertoli cells and Leydig cells.

(i) Oogenesis begins during the embryonic development stage when a million gamete
mother cells (oogonia) are formed within each foetal ovary.
(ii) (a) Sertoli cells are located on the inside lining of seminiferous tubule. These cells
provide nutrition to the germ cells.
(b) Leydig cells or interstitial cells are located in the regions outside the seminiferous
tubule called interstitial spaces. These cells synthesise and secrete testicular hormone
called androgens.

83.
What is implantation?

(i) The implantation is the embedding of the blastocyst in the endometrium of the
uterus.
 84.
(i) What is the average duration of human pregnancy known as?
(ii) What is the term given to the process of delivery of foetus?

(i) The average duration of human pregnancy is called as gestation period. It is

approximately 270 days in humans.
(ii) The term given to the process of delivery of foetus is called parturition.

Reproductive Health

1.
Our government has intentionally imposed strict conditions for MTP in our country.
Justify giving a reason. (Delhi 2017)

MTP or induced abortion is the termination of pregnancy due to certain medical reasons.
Government of India legalised MTP in 1971 with strict conditions to avoid its misuse, i.e.
to check indiscriminate and illegal female foeticides.

2.
Name an IUD that you would recommend to promote the cervix hostility to sperms.
(Delhi 2014C)

The hormone releasing IUDs, e.g. progestasert and LNG-20 are recommended to
promote the cervix hostility to sperms.

3.
State one reason, why breastfeeding the baby acts as a natural contraceptive for the
mother. (Delhi 2014C)

Lactation or Breastfeeding the baby delays the onset or return of menstruation and
ovulation cycle due to interference of hormone prolactin. Therefore, the chances of
conception are nil during this period, i.e. up to six months. Hence, breastfeeding the baby
may act as a natural contraceptive (lactational amenorrhea) for mother. (1)

4.
Mention one positive and one negative application of amniocentesis. (Delhi 2010)

Applications of amniocentesis are

● Positive,application It can be used to diagnose any chromosomal abnormality or

genetic disorder in foetus.

● Negative application It can be used to determine the sex of foetus and lead to
female foeticide.
5.
Why is tubectomy considered a contraceptive method? (Foreign 2010)

In tubectomy, a small part of Fallopian tube or oviduct is cut and tied up to block the
passage of ova from ovary to the site of fertilisation in Fallopian tube. It prevents
fertilisation. So, it is considered as a contraceptive method.

6.
Mention the problems that are taken care of by Reproduction and Child Healthcare
Programme. (All India 2016)

Reproduction and Child Healthcare (RCH) programmes cover wide range of reproduction
related areas. They include

● Creating awareness among people about various reproduction related aspects.

● Support for building up a reproductively healthy society by providing increased

medical facilities, better postnatal care, better detection and cure of diseases like STDs,
etc.

7.
What is amniocentesis? How is it misused? (Delhi 2014C)
Or
Why there is a statutory ban on amniocentesis? Why is this technique so named? (Delhi
2012C)
Or
What is amniocentesis? Why has the government imposed a statutory ban inspite of its
importance in medical field? (Foreign 2010)

Amniocentesis is a prenatal diagnostic test. It is named so, because it is based on the

chromosomal pattern of the cells in the amniotic fluid that surrounds the developing
foetus in the womb.
It is misused to detect the sex of pre-born child that leads to female foeticide. Hence,
there is statutory ban on amniocentesis.

8.
What do oral pills contain and how do they act as effective contraceptives? (Delhi
2014C)

Oral contraceptives or pills contain either progestogens or progesterone-oestrogen

combinations. They function as contraceptives by

● inhibiting ovulation.

● inhibiting implantation.
● altering the quality of cervical mucus to prevent the motility of sperms in female
reproductive tract.

9.
Why is Cu-T considered a good contraceptive device to space children? (Delhi 2011)

Copper-T (Cu-T) is an Intra Uterine Device (IUD) that is inserted by experts and it serves
as an effective contraceptive in the following ways

● Increases phagocytosis of sperms within the uterus.

● Copper ions released by Cu-T suppress the motility of sperms and their fertilising
ability.

10.
Name an oral pill used as a contraceptive by human females. Explain, how does it
prevent pregnancy? (Delhi 2011)
Or
Why is Saheli a well-accepted contraceptive pill? (Foreign 2010)

‘Saheli’ is a non-steroidal contraceptive pill used by females to space children.

Saheli inhibits ovulation and implantation. It alters the quality of cervical mucus to
prevent the entry of sperms into cervix.

11.
Describe the lactational amenorrhea method of birth control. (All India 2011)

Lactation amenorrhea refers to the absence of menstruation during the period of intense
lactation following parturition. It is a birth control method because

● ovulation and other events of menstrual cycle are stopped at this time.

● as long as the mother breastfeeds her child, chances of conception are nil
because of the suppressed gonadotropin activity. However, this method is generally
reliable upto only six months after delivery.

12.
At the time of Independence, the population of India was 350 million, which exploded to
over 1 billion by May 2000.
List any two reasons for this rise in population and any two steps taken by the
government to check this population explosion. (Foreign 2011)

Reasons for rise in population include

● All-round development in various fields and increased health facilities along with
better living conditions.

● Reduced maternal and infant mortality rate.

Two major steps taken by the government to check this population growth are

● People are educated and support the idea of small family by using various
contraceptive methods.

● There is statutory raising of marriageable age of females to 18 years and of males

to 21 years.

13.
How do copper and hormone releasing IUDs act as contraceptives? Explain. (All India
2010)

Copper and hormone releasing IUDs act as contraceptives because

● Copper IUDs (Cu-T, Cu-7) release Cu ions, which suppress sperm motility and the
fertilising capacity of sperms.

● Hormone releasing IUDs (progestasert, LNG-20)

(a) inhibit ovulation.

(b) make the cervix unreceptive to sperms.
(c) make the uterus unsuitable for implantation.

(iii) They both increase the phagocytosis of sperms within the uterus.

14.
(i) List any four characteristics of an ideal contraceptive.
(ii) Name two intrauterine contraceptive devices that affect the motility of sperms. (All
India 2016)

(i) An ideal contraceptive must have the following four characteristics

● It must be safe and user friendly.

● It must be easily available.

● It must be reversible with little or no side effects.

● It must not interfere with the sexual drive, desire or sexual act of the user.

(ii) Cu-T and Cu-7 are two examples of IUDs that affect the motility of sperms.

15.
Name two hormones that are constituents of contraceptive pills. Why do they have high
and effective contraceptive value? Name a commonly prescribed non-steroidal oral pill.
(All India 2014)
The two hormones that are the constituents of oral pills are

● progesterone

● oestrogen

They inhibit ovulation and fertilisation and also modify the quality of cervical mucus to
prevent or retard the entry of sperms. Hence, they have high and effective contraceptive
value. Saheli is the most commonly prescribed new oral contraceptive pill for females. It
contains a non-steroidal preparation called centchroman. It is once a week pill with few
side effects and high contraceptive value. It was developed by CDRI (Centra! Drug
Research Institute), Lucknow.

16.
If implementation of better techniques and new strategies are required to provide more
efficient care and assistance to people, then why is there a statutory ban on
amniocentesis? Write the use of this technique and give reason to justify the ban. (All
India 2014)

Though implementation of better techniques and new strategies is required to provide

more efficient care and assistance to people still, there is a statutory ban on
amniocentesis. Amniocentesis helps to determine any chromosomal abnormalities or
genetic disorders, sex of foetus and foetal infections, etc., by using minute amount of
amniotic fluid surrounding the foetus.

This prenatal diagnostic test is particularly useful for those women who are at increased
risk or have genetic disorders or chromosomal problems. However, this is also misused
to determine the sex of foetus which had ultimately lead to increased female foeticides.
Therefore, government has imposed a statutory ban to prevent its further misuse and to
balance the unequal sex-ratio prevailing in human population. (3)

17.
A woman has certain queries as listed below, before starting with contraceptive pills.
them.
(i) What do contraceptive pills contain and how do they act as contraceptives?
(ii) What schedule should be followed for taking these pills? (Delhi 2014)

(i) Refer to No. 8. (2)

(ii) The oral contraceptive pills are to be taken daily for 21 days, preferably within the first
five days of menstrual cycle. After the onset of menstruation cycle, i.e. 5-7 days, the
process is to be repeated in the same pattern (again for 21 days). This schedule is to be
followed till the woman wants to avoid conception. (1)

18.
(i) Name any two copper releasing IUDs.
(ii) Explain, how do they act as effective contraceptives in human females. (All India
2014)

(i) The copper releasing IUDs are Cu-T, Cu-7 and multiload.-375.
(ii) Refer to No. 9.

19.
Name and explain the surgical method advised to human males and females as a mean
of birth control. Mention its one advantage and one disadvantage. (Foreign 2014)

The surgical or sterilisation methods advised to human males and females as effective
means of birth control are

● Vasectomy (In males) A sterilisation method in which a small portion of vas

deferens is removed or tied up through a cut or incision on scrotum, thus blocking the
transport of sperms from the testes to the copulatory organ.

● Tubectomy (In females) A sterilisation method in which small part of Fallopian

tube is removed or tied up through incision in abdomen or through vagina. It blocks the
passage of ova from ovary to the site of fertilisation.

The advantage of these two sterilisation methods in both human males and females is
that it is a very effective method for preventing conception as it blocks the transport of
gametes. The disadvantage of this method is that this surgical procedure cannot be
reversed, so it is helpful for only those who already have children and do not want to
extend their family further.

20.
How do ‘implants’ act as an effective method of contraception in human females?
Mention its one advantage over contraceptive pills. (Delhi 2012)

Subcutaneous ‘implants’ contain synthetic progesterone and are placed under the skin.
They are an effective contraceptive method as they check ovulation and thicken cervical
mucus to prevent sperm transport.

‘Subcutaneous implants’ are more advantageous than contraceptive pills as they are
long lasting and once implanted, they are effective for up to 5 years.

21.
Your school has been selected by the Department of Education to organise and host an
interschool seminar on ‘Reproductive Health Problems and Practices’. However, many
parents are reluctant to permit their wards to attend it. Their argument is that the topic is
‘too embarrassing’.
Put forth four arguments with apporpriate reasons and explanation to justify the topic to
be very essential and timely. (All India 2015)

Parents should encourage their children to attend such seminar as they will get right
information regarding myths and misconceptions about sex related aspects. Following
are the four points to justify this topic to be essential

● Awareness of problems due to uncontrolled population growth, social evils like sex
abuse and sex related crimes, etc., need to be created so that children should think and
take up necessary steps to prevent them and thereby build up a reproductively healthy
society.

● Large group of school students comprises of adolescents who have attained

puberty. Therefore, these seminars are necessary to provide medical help and care for
reproduction related problems like menstrual problems, infertility, pregnancy, delivery,
contraception, abortions, etc.

● Knowledge about Sexually Transmitted Diseases (STDs) is essential as children

should be aware that unprotected sex with multiple partners results in the transmission
of sex related problems.

● Increasing population is a major problem of India which is directly related with

reproductive health. Children should be aware of family planning programmes such as
Reproductive and Child Healthcare (RCH) programmes.

22.
Name any two assisted reproductive technologies that help infertile couples to have
children. (Delhi 2012C)

The assisted reproductive technology methods that can help infertile couple to have
children are

● Zygote Intra Fallopian Transfer (ZIFT) and

● Artificial Insemination (AI) Technique.

23.
Expand GIFT and ICSI. (All India 2012C)

GIFT – Gamete Intra Fallopian Transfer.

ICSI – Intra Cytoplasmic Sperm Injection.

24.
After a brief medical examination a healthy couple came to know that both of them are
unable to produce functional gametes and should look for an ‘ART’ (Assisted
Reproductive Technique). Name the ‘ART’ and the procedure involved that you can
suggest to them to help them bear a child. (Delhi 2015)

The ART that would help the couple to bear a child is IVF (In Vitro Fertilisation) or Test
tube baby programme. In this process, ova from wife/donor female and sperms from the
husband/donor male are collected and fused to form zygote in the laboratory under
same conditions as in the body. This is in vitro fertilisation (fertilisation outside the
body).

Zygote or early embryo is transferred into Fallopian tube or uterus for further
development. This is called Embryo Transfer (ET). It can be Zygote Intra Fallopian
Transfer (ZIFT) or Intra Uterine Transfer (IUT).

25.
An infertile couple is advised to adopt test tube baby programme. Describe two principal
procedures adopted for such technologies. (Delhi 2015)
Or
Explain the Zygote Intra Fallopian
Transfer Technique (ZIFT). How is Intra Uterine Transfer (IUT) Technique different from
it? (All India 2010)

ZIFT (Zygote Intra Fallopian Transfer) is the technique in which zygote or early embryo
with up to 8 blastomeres is transferred into the Fallopian tube of female.
On the other hand in IUT, embryo with more than 8 blastomeres is transferred into the
uterus.
These are the two principal procedures adopted for test tube baby programme.

26.
A childless couple has agreed for a test tube baby programme. List only the basic steps
the procedure would involve to conceive the baby. (Delhi 2015C)
Or
(i) Give any two reasons for infertility among young couple.
(ii) Test tube baby programme is a boon to such couples. Explain the steps followed in
the procedure. (All India 2010C)

(i) The reasons of infertility in young people can be physical, congenital diseases, use of
drugs, immunological or even psychological factors.
(ii) In test tube programme,
(a) Ova from the wife or a donor female and the sperms from the husband or a donor
male are allowed to fuse under simulated conditions in the laboratory. It is called in vitro
fertilisation.
(b) Embryo is then transferred into the uterus or Fallopian tube for further development.

The process of embryo transfer is done in following ways Zygote or embryo up to 8

blastomeres is 1 transferred into Fallopian tube (ZIFT). Embryo with more than 8
blastomeres is transferred into uterus (IUT).

27.
‘Intra Cytoplasmic Sperm Injection (ICSI)’ and ‘Gamete Intra Fallopian Transfer (GIFT)’
are two assisted reproductive technologies. How is one different from the other? All
(India 2014C)
In Gamete Intra Fallopian Transfer (GIFT), the ovum from a healthy donor female is
transferred to a female, who cannot produce ova. However, she can provide suitable
environment for fertilisation and embryo development (in vivo fertilisation).

In Intra Cytoplasmic Sperm Injection (ICSI), the fertilisation is done in vitro by injecting
sperms directly into the ovum from a donor female, under simulated conditions. The
embryo is thus, formed in laboratory and which is later transferred to the uterus or
Fallopian tube for further development.

28.
Why is ZIFT a boon to childless couples? Explain the procedure. (Delhi 2013C)

ZIFT (Zygote Intra Fallopian Transfer) is a boon to childless couples as it helps them to
become parents. In this technique, ova from wife/donor female and sperms from
husband/donor male are fused to form zygote in laboratory. Zygote is allowed to divide
up to 8 blastomeres stage and it is at this stage, a zygote or early embryo is transferred
into the Fallopian tube. Implantation takes place in the uterus where further development
takes place.

29.
State any four methods to overcome infertility in human couples. (Delhi 2011C)

Following are the four methods to overcome infertility problems in human couples

● Test tube baby programme In this method, the fusion of ovum and sperm is done
outside the body of a woman (in vitro fertilisation) to form zygote which divides to form
embryo. The embryo is then implanted in the uterus where it develops into a foetus and
then into the child.

● Intra Cytoplasmic Sperm Injection (ICSI) In this technique, embryo is formed in the
laboratory by directly injecting the sperm into the ovum followed by embryo transfer.

● Artificial Insemination Technique (AIT) Semen (containing sperms) from husband

or donor is artificially introduced into the vagina hr uterus (IUI).

● Gamete Intra Fallopian Transfer (GIFT) Sperm and unfertilised ova are transferred
into the Fallopian tube of the female and they are allowed to fuse naturally.

30.
A couple where both husband and wife are producing functional gametes, but the wife is
still unable to conceive, is seeking medical aid. Describe any one method that you can
suggest to this couple to become happy parents. (All India 2014)

In case, if both husband and wife are producing functional gametes, but wife is not able
to conceive, the IVF technique can be employed to bless them with child.
Method of in vitro fertilisation is given below
● Gametes from both husband and wife are collected, i.e. sperm and ova. These are
fused to form zygote under laboratory conditions. As the fertilisation takes place outside
the female body, it is referred to as in vitro fertilisation.

● The zygote or embryo is then either transferred to Fallopian tubes (if up to 8

blastomeres), i.e. ZIFT or to the uterus (more than 8 blastomeres), i.e. IUT.

31.
Suggest and explain any three Assisted Reproductive Technologies (ART) to an infertile
couple. (All India 2013)

Refer to No. 3 and 4.

32.
(i) Explain one application of each one of the following (Delhi 2019)
(a) Amniocentesis
(b) Lactational amenorrhea
(c) ZIFT
(ii) Prepare a poster for the school programme depicting the objectives of ‘Reproductive
and Child Healthcare Programme’.

(i) (a) Amniocentesis The benefits of amniocentesis include the diagnosis of

chromosomal abnormalities and developmental disorders of foetus.
(b) Lactational amenorrhea It is the absence of menstruation during the period of
intense lactation following parturition. Because ovulation does not occur in this period,
the chances of conception are nill.
(c) ZIFT (Zygote Intra Fallopian Transfer) is related to embryo transfer in the test tube
programme. In this technique, the zygote or embryo up to 8 blastomeres is transferred
into the Fallopian tube.

33.
Reproductive and Child Healthcare (RCH) programmes are currently in operation. One of
the major tasks of these programmes is to create awareness amongst people about the
wide range of reproduction related aspects. As this is important and essential for
building a reproductively healthy society.
(i) ‘Providing sex education in schools is one of the way to meet this goal’. Give four
points in support of your opinion regarding this statement.
(ii) List any two ‘indicators’ that indicate a reproductively healthy society. (Delhi 2016)

(i) (a) Introducing sex education in schools is a good step for providing useful
information to the adolescents, so as to discourage them from believing in myths and
misconceptions about sex related issues.
(b) Better awareness about sex related matters.
(c) Better detection and cure of STDs.
(d) Awareness of problems due to uncontrolled population growth.

(ii) (a) Increased number of medically assisted deliveries and better postnatal care
leading to decreased maternal and infant mortality rates.
(b) Awareness of problems due to uncontrolled population growth, social evils like sex
abuse and sex related crimes, etc., should be created to enable people to think and take
necessary steps to prevent them.

34.
A large number of married couples in the world are childless. It is shocking to know that
in India, the female partner is often blamed for the couple being childless.
(i) Why in your opinion the female partner is often blamed for such situations in India?
Mention any two values that you as a biology student can promote to check this social
evil.
(ii) State any two reasons responsible for the cause of infertility.
(iii) Suggest a technique that can help the couple to have a child where the problem is
with the male partner. (All India 2016)

(i) There is a common myth prevailing in our society that inability of a couple to have
child is due to the infertility of female partner only, it is because female carries the child
in her womb. Being a biology student we should create awareness among people that
both male and female partners equally contribute for the birthy of a child. This is
because baby is formed from zygote that is formed by the fusion of both male and
female gametes. Hence, infertility in either male or female can be responsible for the
failure of conception. Infertility in both males and females can be easily cured as there
are so many specialised infertility clinics which provide treatments to childless couples.
In case treatment is not possible, the couples can be assisted to have children through
certain special techniques called the assisted reproductive technologies.

(ii) Causes of infertility could be

● Sexually transmitted diseases both in males and females.

● Some physiological problems in females/males, so gametes (sperm/ova) are not
produced.

(iii) IVF (In Vitro Fertilisation) and Artificial Insemination (AI) can be done if the sperm
count of male is low.

Biotech

2.Write the specific point in the palindrome and the bond that is cut by Eco Rl.

palindromic sequence, i.e. GAATTC. The type of bond broken by Eco RI is

phosphodiester bond between the G and A bases of the palindrome. This site is known
as restriction site.

4.
Mention the use of gel electrophoresis in biotechnology experiments.

Gel electrophoresis is used to separate the fragments of DNA that were cut by R.Es.

5.
Name the technique that is used to alter the chemistry of genetic material (DNA, RNA) to
obtain desired result.

The technique used to alter the chemistry of genetic material to obtain desired result is
called genetic engineering.

6.Why is it not possible for an alien DNA to become part of a chromosome anywhere
along its length and replicate normally?)

The alien DNA itself cannot multiply and replicate but requires a specific sequence for
initiating its replication called origin of replication. in a chromosome ori acts as the
starting point of replication as they it aids in binding of DNA polymerase.

7.Mention the type of host cells suitable for the gene guns to introduce an alien DNA.
Plant host cells are suitable for the gene guns to introduce an alien DNA.

8.
Write the two components of first artificial rDNA molecule constructed by Cohen and
Boyer.
The two components of first artificial rDNA molecule constructed by Cohen and Boyer
are

● Antibiotic resistance gene

● Plasmid of Salmonella typhimurium

9.
Name the host cells in which microinjection technique is used to introduce an alien DNA.
The microinjection technique is carried out in animal cell to inject alien DNA into the
nucleus.

10.
Name the material used as matrix in gel electrophoresis and mention its role.

The material used as matrix in gel electrophoresis is agarose.

This agarose gel acts as a sieve to separate the DNA fragments according to their size.

11.
Write any four ways used to introduce a desired DNA segment into a bacterial cell in
recombinant technology experiments.

Ways to introduce desired DNA into bacterial cell are

● microinjection

● disarmed pathogen vectors

● treatment of host cell by bivalent cation such as calcium

● biolistic or gene gun

12.
How can retroviruses be used in experiments in spite of them being disease causing?

Retroviruses can be used in biotechnology experiments after being disarmed, i.e.

removal of virulent gene so that it is unable to cause infection in hosts they are
transferred to.

13.
State what happens when an alien gene is ligated at Pvu I site of pBR322 plasmid.

An alien gene ligated at Pvu I site of pBR322 plasmid cause the transformant cell to
loose the ampicillin-resistance as ampR gene becomes non-functional. Thus, the
recombinant does not grow in the presence of ampicillin.

14.
Why is ‘plasmid’ an important tool in biotechnology experiments?

Plasmid have the ability to replicate within bacterial cells independently of chromosomal
DNA. They have high copy number, therefore an alien DNA ligated to it, will have equal
copy number as that of plasmids. So, it is used as a vector in gene cloning experiments
and thus acts as an important tool in biotechnology.

15.
State what happens when an alien gene is ligated at Sal I site of pBR322 plasmid.
When an alien gene is ligated at Sal I site of tetracycline resistance gene in the vector
pBR322, the recombinant plasmid lose its tetracycline resistance.

16.Mention the uses of cloning vector.

● Helps in linking the foreign/alien DNA with the host’s DNA.

● Helps in the selection of recombinants from the non-recombinants.

17.Agrobacterium tumefaciens is the natural genetic engineer of plants. Give reasons

Agrobacterium tumefaciens is a pathogen of dicot plants. It is used as a natural genetic

engineer because it delivers a piece of its DNA (called T-DNA) to transform normal plant
cells into tumour cells. It direct the tumour cells to synthesise the chemicals required by
the pathogen.

18.Why is it essential to have a selectable marker in a cloning vector?

Selectable marker in cloning vector helps in identifying and selecting the recombinants
and eliminating the non-recombinants.

19.How is the action of exonuclease different from that of endonuclease?

Exonucleases ;Remove nucleotides from the ends of DNA. Endonucleases;Cut DNA at

specific points.

20.What is the host called that produces a foreign gene product? What is this product?

Transgenic organisms or genetically modified organisms are the host that produces a
foreign gene product. Recombinant proteins are the products formed by these host
cells.

21.ß-galactosides enzyme is considered a better selectable marker. Justify .

Coding sequence of p-galactosidase is a better maker, as the recombinants and

non-recombinants are differentiated on the basis of their ability to produce colour in the
presence of a chromogenic substrate, while the selection of recombinants due to
inactivation of antibiotic resistant gene is a tedious and time taking process to grow
them simultaneously on two antibiotics containing media.

22.How does a restriction nuclease function?.

Restriction nucleases function by inspecting the length of DNA sequence and then
binding to specific recognition sequences and cutting the strands at sugar phosphate
backbones.
These nucleases are of two types depending on their mode of action
● Restriction exonucleases cut sequences at terminal ends of DNA.

● R.Es, e.g. Eco RI, cut between the two bases of recognition sequence.

23.Write the role of ori and restriction site in a cloning vector pBR322.

Ori is a sequence of DNA from where replication starts. Any piece of DNA that needs to
replicate in the host cell has to be linked to it. Restriction site is the recognition site
made of palindromic sequence for restriction enzymes.

24.how was Agrobacterium tumefaciens made a cloning vector for plant cells.

Agrobacterium infects plant tissues by transferring its plasmid T-DNA to the plant
genome. This property was exploited to transfer desired gene to a particular plant. The
desired gene is inserted in the plasmid T-DNA and the engineered Agrobacterium is
allowed to infect that particular plant. Hence, it acts as a natural cloning vector.

25.Explain with the help of a suitable example the naming of a R.E.

Naming of R.E involves the following rules

● The first letter of the enzyme comes from the genus and next two letters from
species of the prokaryotic cell from where enzymes are extracted.

● The Roman numbers following the name shows the order in which the enzymes
were isolated from the bacterial strain. For example, Eco RI is derived from Escherichia
coli RY 13, Hind II from Haemophilus influenzae Rd, etc.

26.How are sticky ends formed on a DNA strand? Why are they called so?

Sticky ends on DNA are formed by the action of enzymes R.Es. These enzymes cut the
strand of DNA a little away from the centre of the palindrome sequence between the
same two bases on both the strands.This results in single-stranded stretches on both
the complementary strands at their ends. These overhanging stretches are called sticky
ends as they form hydrogen bonds with the complementary base pair sequences.

27.How is insertional inactivation of an enzyme used as a Selectable marker to

differentiate recombinants from non-recombinants?

The insertional activation of an enzyme, e.g. ß-galactosidase occurs by inserting the

desired gene in the coding region of enzyme. It results in inactivation of ß-galactosidase
gene in recombinants. Due to this, the recombinant or transformed hosts are unable to
produce any colour when grown on chromogenic substrate. Thus, ß-galactosidase acts
as a selectable marker to differentiate recombinants from non-recombinants.

28.Explain palindromic nucleotide sequence with the help of a suitable example.

The palindromic nucleotide sequence is the sequence of base pairs in DNA that reads
the same on both the complementary strands of DNA, with same orientation of reading.
5-GAATTC-3′
3-CTTAAG-5′

29.Why are molecular scissors called so? Write their use.

Molecular scissors are so called because they cut DNA at specific sequences between
base pairs. Molecular scissors or restriction enzymes cut DNA at desired sequences and
generate sticky ends that facilitate the cut DNA to join with host genome or vector DNA.
They play an important role in genetic engineering because these enzymes can cut the
desired gene and introduce into vectors for expression.

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

Since, DNA molecules are hydrophilic, they cannot pass through cell membranes. For
rDNA to be integrated into vector or host genome, it is necessary for the DNA to be
inserted in the cell. Therefore, making the host cells competent is necessary in
biotechnology experiments.The two ways by which cells can be made competent to take
up DNA are

● Chemical action; The host cell is treated with a specific concentration of divalent
cation, i.e. calcium increases the pore size in the cell membrane. DNA is then incubated
with treated bacterial cell at 42°C, thereby increasing the efficiency of DNA to enter it
through pores in cell wall.

● Heat shock treatment;Incubating the cells with rDNA on ice, followed by brief
treatment of heat at 42°C and again putting them back on ice.

31.How is an exonuclease functionally different from an endonuclease? Give an

example of any two endonucleases other than Sal I.

Exonucleases cleaves base pairs of DNA at their terminal ends and act on single-strand
of DNA or gaps in double stranded DNA. endonuclease cleaves DNA at any point except
the terminal ends and can cut on one strand on both the strands of double-stranded
DNA, e.g. Eco R1 and Hind II.

32.Explain the work carried out by Cohen and Boyer.

Stanley Cohen and Herbert Boyer constructed the first artificial rDNA molecule.They
isolated the antibiotic-resistance gene by cutting out a piece of DNA from a plasmid with
the help of restriction enzyme and linked it to a native plasmid of Salmonella
typhimurium with the help of DNA ligase.

33.(i) A recombinant vector with a gene of interest inserted within the gene of
α-galactosidase enzyme is introduced into a bacterium. Explain the method that would
help in selection of recombinant colonies from non-recombinant colonies.
(ii) Why is this method of selection referred to as insertional inactivation?
(i) The recombinant colonies can be differentiated by their inability to produce colour in
the presence of a chromogenic substrate. The recombinants do not produce any colour,
while the non-recombinants produce a blue colour with chromogenic substrate in the
medium. It occurs because of the presence of α-galactosidase in former and its absence
in latter cells. (ii) The
enzyme α-galactosidase becomes inactivated on insertion of rDNA, within the coding
sequence of enzyme. Thus, the method is called insertional inactivation.

34.State the role of UV-light and ethidium bromide during gel electrophoresis of DNA
fragments.

DNA fragments are observed only after staining with ethidium bromide followed by their
exposure to UV radiation. This gives bright orange colour to DNA fragments.

36.List the key tools used in rDNA technology.

restriction enz, polymerases, ligases, cloning vectors and competent host organism or
cells.

37. Explain the role of Ti plasmids in biotechnology.

The Ti-plasmid isolated from Agrobacterium is responsible for the natural

transformation of plant cells into tumours. So, it is modified into a non-pathogenic vector
but still is able to deliver the DNA.This disarmed plasmid of Agrobacterium is used as a
vector for the transformation of plant cells.

38.How are recombinant vectors created? Why is only one type of R.E required for
creating one recombinant vector?

Creation of recombinant vectors;Vector DNA is cut at a particular restriction site by a

restriction enzyme. The alien DNA is then linked with the vector DNA using ligase to form
the recombinant vector.A restriction enzyme recognises and cuts the DNA at a particular
sequence called recognition site. The same restriction enzyme is used for cutting the
DNA segment from both the vector and the other source, so as to produce same sticky
ends in both DNA molecules to facilitate their joining.

39.Study the diagram given below and the following s.

(i) Why have DNA fragments in band D moved farther away in comparison to those in
band C?
(ii) Identify the anode end in the diagram.
(iii) How are these DNA fragments visualised?

(i) In band D, DNA fragments are smaller than those on band C. The fragments separate
according to their size through the sieving effect provided by the gel. So, the smaller
fragments move farther away than the larger ones.
(ii) B is anode end in the diagram as DNA fragments are moving towards this end.
(iii) Gel is stained with ethidium bromide and exposed to UV. Orange bands become
visible.

40.Explain how the sticky ends are formed and get joined?
The sticky ends are joined via complementary factor of two polynucleotide strands
bases and by the action of enzyme, DNA ligase.

41.Explain the action of Eco RI.

R.E Eco RI cuts the DNA strands a little away from the centre of the palindromic
sequence, but between the same two bases on the two strands, i.e. G and A, on both the
strands. The site of action of enzyme in palindrome sequence GAATTC is called
recognition site.

(i) Due to this, single-stranded portions called sticky ends, overhang at the end of each
strand.
(ii) Because of the stickiness, they easily form hydrogen bonds with their complementary
counterparts.

42.
How are the DNA fragments separated by gel electrophoresis visualised and separated
for use in constructing rDNA?

The separated fragments are stained with ethidium bromide. By the exposure to UV, the
fragments become orange bands.The separated bands of DNA are cut from agarose gel
and DNA is extracted from the gel in a process called elution.

43.Expand ‘BAC’ and ‘YAC’. What are they and what is the purpose for which they are
used?

‘BAC’ stands for Bacterial Artificial Chromosome and ‘YAC’ stands for Yeast Artificial
Chromosome. These are vectors used in cloning DNA. For sequencing the total DNA
from cell, the DNA is isolated and converted into smaller fragments. DNA fragments are
cloned in suitable host using specialised vectors, such as BAC and YAC. Fragments of
DNA are then sequenced by automated DNA sequences

Role of the sticky ends; The sticky ends are produced from hydrogen bonds with their
complementary cut counterparts. The stickiness of the ends facilitates the action of the
enzyme DNA ligase which helps to region the cut DNA.
 45.(i) Mention the importance of gel-electrophoresis in biotechnology.
(ii) Explain the process of this technique.

(i) DNA fragments formed by R.Es are separated by gel-electrophoresis.

(ii) DNA fragments are negatively charged molecules. Thus, they move towards the
positive charged anode under electric field through the medium. The smaller fragments
move farther in the gel as compared to larger fragments due to the sieving effect of gel.

46.How does p-galactosidase coding sequence act as a selectable marker? Why is it a

preferred selectable marker to antibiotic resistance genes? Explain.

Selectable marker helps in identifying or selecting transformants and eliminating

non-transformants and selectively permit the growth of the transformants.
ß-galactosidase acts as a selectable marker by inducing the property of insertional
inactivation in transformed cells. In this process the recombinants and
non-recombinants are differentiated on the basic of colour production in the presence of
chromogenic substrate. A rDNA is inserted within the coding sequence of an enzyme
ß-galactosidase which results into inactivation of the enzyme. Therefore, the bacterial
colonies having inserted plasmid, show no colouration while, those without plasmid
show blue colour.

47.
Give reason why
(i) DNA cannot pass into a host cell through the cell membrane.
(ii) Proteases are added during isolation of DNA for genetic engineering.
(iii) Single cloning site is preferred in a vector.

(i) hydrophilic molecules like DNA(sugar-phosphate backbone) cant diffuse through the
lipid bilayer of the plasma membrane.
(ii) Proteases catalyse the breakdown of proteins to its amino acid. If the proteins are
not removed from DNA, they interfere with downstream treatment (such as action of R.E,
DNA ligase, etc).
(iii) Single cloning sites are preferred as the the cloning of a sequence at more than one
recognition sites within the vector would generate several fragments leading to
complication in gene cloning.

48.Describe the formation of rDNA by the action of Eco RI.

 49.Explain the roles in rDNA technology.
(i) Restriction enzymes (ii) Plasmids

(i) The restriction enzymes are known as molecular scissors. These enzymes belong to
a larger group of enzymes called nucleases, which are of following two types

● Exonucleases Those, who remove nucleotides from the ends of the DNA (either 5′
or 3′) in one strand of duplex.

● Endonucleases Those, who make cuts at specific position within the DNA. Each
R.E functions by ‘inspecting’ the length of a DNA sequence.

Role of Restriction Enzyme in rDNA Technology The R.Es are used to cut plasmid DNA
as well as foreign DNA at desired sites. The foreign DNA is then inserted into plasmid
DNA and the plasmid takes the foreign DNA into the desired host organism.
Example The first discovered restriction enzyme is Hind II. It was isolated from
Haemophilus influenzae. It always cuts DNA at 5 GT (Pyrimidine T Or C (Purine A or G)
AC3′ and 3’CA (Purine A or G) (Pyrimidine T or C) TG5′.
It produces DNA segments with blunt ends.

(ii) Plasmids These are extrachromosomal, self-replicating, double-stranded, closed and

circular DNA molecules. It is found only in bacteria and few yeast cells.
Role of Plasmid in DNA Recombinant Technology These are used as vectors to carry the
desired gene (foreign genes) into the desired host organisms.
For example, pBR322 is widely used plasmid vector. In its name P signifies plasmid, B is
from Boliver and R is from Rodriguez. Boliver and Rodriguez were two scientists who
developed pBR322 in 1977. This plasmid has genes for resistance against ampicillin and
tatracycline. They have restriction sites for enzymes like PvuI, Pst I, Sal I, Bam HI.
 50.Explain the role(s) of the following in biotechnology
(i) R.E
(ii) Gel-electrophoresis
(iii) Selectable markers in pBR322

(i) R.Es These are the bacterial enzymes that cut dsDNA into fragments after recognising
and binding to the specific nucleotide sequences, known as recognition site. These
enzymes are used to form recombinant molecules of DNA, composed of DNA from
different sources.

(ii) Gel-electrophoresis is the technique which allows the separation and visualisation of
fragments of DNA on an agarose gel matrix. Since, the DNA fragments are negatively
charged molecules, they separate and move towards the anode (+ ve) under the
influence of an electric field. DNA fragments are separated on the basis of their size
through the sieving effect provided by the gel.

(iii) Selectable markers in pBR322 help in identification and selection of transformants.

pBR322, an E. coli cloning vector has two antibiotic resistance genes, i.e. for ampicillin
and tetracycline, which act as selectable marker. When a foreign DNA is ligated at the
site of tetracycline resistance (tetR) gene in pBR322, the recombinant plasmid lose
tetracycline resistance due to insertional inactivation of foreign DNA, but can still be
selected out from non-recombinants by placing the transformants on ampicillin
containing medium. The transformants growing on ampicillin containing medium are
then transferred on tetracycline containing medium. The recombinants grow on
ampicillin containing medium but not on tetracycline one whereas non-recombinants
grow on both the media.

51.
(i) significance of palindromic nucleotide sequences in rDNA.
(ii) use of R.E in the above process.

(i) Palindromic nucleotide sequences in the DNA are group of letters that form the same
words when read both forward and backward. For example, the following sequence
reads the same on the two strands in 5′ → 3’direction as well as 3′ → 5’direction.
5’— GAATTC —3′
3’—CTTAAG —5′
Significance These sequences act as recognition sites which are recognised by specific
R.Es.

(ii) Use of R.E During rDNA formation, these enzymes recognise and make a cut at
specific positions within the DNA and vector. Due to this function, R.Es are also called as
molecular scissors.

52.
(i) Name the selectable markers in pBR322. Mention the role
(ii) Why is the coding sequence of an enzyme p-galactosidase a preferred selectable
marker in comparison to the ones named above?
(i) Selectable markers in cloning vector pBR322 are ampicillin and tetracycline antibiotic
resistance gene. They help in the selection of transformants and eliminating the
non-transformants.
(ii) The selection of recombinants due to inactivation of antibiotics is a difficult process
and requires simultaneous plating on two plates having different antibiotics. Thus,
enzyme p-galactosidase is preferred as a selectable marker as it allows to differentiate
non-recombinants from recombinants easily by insertional inactivations technique.

53.(ii) State the role of ‘biolistic gun’ in biotechnology experiments. (All India 2016)
(ii) Biolistic guns or gene guns are used to bombard rDNA loaded on gold or tungsten
particles with high velocity into host cells. In this way, the rDNA is delivered to the
desired host cells.

54.How does Agrobacterium tumefaciens act as a suitable vector in the

biotechnological experiments? Site an example where it has been successfully used as a
vector.

Agrobacterium has been used as vector to introduce a gene from Tobacco Mosaic Virus
(TMV) into tobacco plants.

56.(i) Why was a bacterium used in the first artificial rDNA molecule?
(ii) Name the scientists who accomplished this and how. (Delhi 2016C)

(i) A bacterium Salmonella typhimurium was used in the first instance of construction of
artificial rDNA molecule because of the possibilities of linking a gene encoding antibiotic
resistance with a native plasmid of the bacterium. This was made possible by the
availability of restriction enzymes and the enzyme DNA ligase.

(ii) Stanley Cohen and Herbert Boyer accomplished this in 1972 by isolating the
antibiotic resistance gene by cutting out a piece of DNA from a plasmid which was
responsible for conferring antibiotic resistance. The cutting of DNA at specific sites was
possible with the availability of restriction enzymes. The cut DNA was linked with the
plasmid DNA using the enzyme DNA ligase. The plasmid DNA acts as vector to transfer
the piece of DNA attached to it.

57.
State the functions of the following in the cloning vector pBR322
(ii) rop and
(iii) Hind III sites (All India 2015C)

(ii) rop in pBR322 encodes for protein involved in plasmid replication.

(iii) Hind III is a restriction site in pBR322, where Hind III endonuclease makes a cut for
the introduction of foreign DNA segment.

58.
Name and explain the technique used for separating DNA fragments and making them
available for biotechnology experiments. (Foreign 2015; All India 2014)

DNA fragments formed by the use of R.Es are separated by gel electrophoresis.
(i) DNA fragments are negatively charged molecules. Thus, they move towards the
positive charged anode under electric field through the gel medium.
(ii) DNA fragments separate according to their size due to sieving effect of agarose gel.
(iii) The separated DNA fragments can be viewed by staining the DNA with ethidium
bromide followed by exposure to UV radiation.
(iv) The separated bands of DNA are cut and extracted from gel piece. This is known as
elution.

61.
What are ‘cloning sites’ in a cloning vector? Explain their role. Name any two such sites
in pBR322. (All India 2014C)

The cloning sites contain the specific unique recognition sequence for a particular
restriction enzyme, so as to link the foreign DNA with the vector DNA and thus, create a
rDNA molecule.

These sites are important for joining the DNA fragments of vector and alien DNA.
Multiple recognition sequences for a particular restriction enzyme within a DNA or
vector complicate the process of gene cloning. The two cloning sites in pBR322 are Bam
HI for tetracycline resistnt gene and Pvu I for ampicillin resistant genes.

62.
.
(ii) Write any two ways the products obtained through this technique can be utilised.
(Delhi 2013C)

(ii) Products obtained via gel electrophoresis can be utilised in following ways
(a) To construct a rDNA molecule by joining them with cloning vector.
(b) For amplification of desired segment using Polymerase Chain Reaction (PCR).

63.
How are the following used in biotechnology ?
(i) Plasmid DNA
(ii) Recognition sequence
(iii) Gel electrophoresis (All India 2011c)

(i) Plasmid DNA It is used for constructing rDNA, by ligating the alien piece of DNA with
it. It is used as a cloning vector and helps in the selection of recombinants from
non-recombinants.
(ii) Recognition sequences These are specific base sequences of DNA, where restriction
enzyme cuts the DNA. They are utilised to extract the desired gene or fragments from
DNA molecules.
(iii) Gel electrophoresis It is a technique, used to separate the DNA fragments according
to their size through seiviirg effect of the gel.

64.
(i) Name the organism in which the vector shown is inserted to get the copies of the
desired gene.
(ii) Mention the area labelled in the vector responsible for controlling the copy number of
the inserted gene.
(iii) Name and explain the role of a selectable marker in the vector shown. (All India
2010)

(i) Escherichia coli(E.coli)

(ii) Ori in the vector is responsible for controlling the copy number of inserted gene.
(iii) The genes encoding resistance to antibiotics like tetR resistant to tetracycline and
ampR resistant to ampicillin are used as selectable markers. If a foreign DN A is ligated
at the Bam HI site of tetracycline resistance gene, the recombinant plasmids will lose the
tetracycline resistance due to insertional inactivation but transformants can be selected
by growing them on ampicillin containing medium. The selectable markers help in
identifying and eliminating non-transformants and selectively permitting the growth of
transformants.

66.
(i) Name the technique used for separation of DNA fragments.

(i) Gel electrophoresis.

67.
Unless the vector and source DNA are cut, fragments separated and joined, the desired
recombinant vector molecule cannot be created.
(i) How are the desirable DNA sequences cut?
(iii) How are the resultant fragments joined to the vector DNA molecule? (Delhi 2015C)

(i) The desirable DNA sequences are cut by using resriction endonuclease enzyme.
These enzymes cut at specific site in palindromic sequence between same two bases
on both the strands.
(iii) The resulting fragments are joined together with vector DNA by DNA ligase enzyme.
It forms phosphodiester bonds between them.
 68.
(i) Describe the characteristics a cloning vector must possess.
(ii) Why DNA cannot pass through the cell membrane? Explain. How is a bacterial cell
made competent to take up rDNA from the medium? (All India 2011)

(i) Features which facilitate cloning of vector are

(a) Origin of replication (ori)

● This is the sequence of DNA from where replication starts.

● Any piece of alien or foreign DNA linked to it is made to replicate within host cell. It
also determine the copy number of the linked DNA.

(b) Selectable jmarker is a marker gene, which helps in selecting the transformants or
recombinants from the non-recombinant ones, e.g. ampicillin and tetracycline resistant
genes in E. coli.

(c) Cloning site is a unique recognition site in a vector to link the foreign DNA. The
presence of a particular cloning or recognition site helps the particular restriction
enzyme to cut the vector DNA.

(d) Small size of the vector The small size facilitates the introduction of the DNA into the
host easily.

70.
Name the enzymes that are used for the isolation of DNA from bacterial and fungal cells
for rDNA technology. (All India 2014; Foreign 2014)

The enzymes used for the isolation of DNA from bacterial cells is lysozyme and those
from fungal cells is chitinase.

71.
How can bacterial DNA be released from the bacterial cell for biotechnology
experiments? (Delhi 2o11)

Bacterial cells are treated with lysozyme which digest their cell wall for releasing DNA.

72.
Why is the enzyme cellulase used for isolating genetic material from plant cells hut not
for animal cells? (Delhi 2010)

Cellulase is used for digesting the cellulosic cell wall of plant cells. Animal cells do not
contain cell wall, so cellulase is not required.

73.
How is a continuous culture system maintained in bioreactors and why? (Delhi 2019)

The cells can be multiplied in a continuous culture system. In this, the used medium is
drained out from one side, while fresh medium is added from the other side to maintain
the cells in their physiologically most active (log/exponential) phase. This type of
culturing method produces a larger biomass leading to higher yields of desired products.

74.
Name the source of the DNA polymerase used in PCR technique. Mention why is it used?
(All India 2013, 2012; Foreign 2011)

The DNA polymerase used in PCR is Taq polymerase which is extracted from Thermus
aquaticus. It is a thermostable enzyme that can withstand high temperature used in the
denaturation and step for the separation of DNA strands. Hence, it can be used for a
number of cycles of DNA amplification without being denatured.

75.
Name the type of bioreactor shown. Write the purpose for which it is used. (All India
2011)

In the given figure, simple stirred-tank bioreactor is shown.

Bioreactors are used to produce large quantities of the desired gene products.

76.
What is genetic engineering? List the steps in rDNA technology. (All India 2011)

Genetic engineering is the process of artificial synthesis, isolation, modification,

combination, addition and repair of genetic material as needed.
Steps of rDNA technology involve

● Isolation of genetic material

● Cutting of DNA at specific locations

● Amplification of gene of interest using PCR

● Preparation and insertion of rDNA into host cell

● Obtaining desirable gene product.

77.
(i) Mention the number of primers required in each cycle of Polymerase Chain Reaction
(PCR). Write the role of primers and DNA polymerase in PCR.

(i) Two sets of primers are required in each cycle of polymerase chain reaction. Primers
are required for the addition of nucleotides to make multiple copies of the DNA of
interest. The enzyme DNA polymerase extends the primers by using the nucleotide
provided.

78.
A schematic representation of Polymerase Chain Reaction (PCR) upto the extension
stage is given below. the s that follows.

(i) Name the process A.

(ii) Identify B.
(iii) Identify C and mention its importance in PCR. (Foreign 2010)

(i) A-Denaturation of the double-stranded DNA

(ii) B-Primers
(iii) C-DNA polymerase or Taq polymerase

Importance in PCR
It extends the primers using the nucleotides provided in the reaction medium and the
genomic DNA as the template. Taq polymerase is thermostable enzyme and it can
withstand the high temperature used in denaturation step.

79.
Any rDNA with a desired gene is required in billion copies for commercial use. How is the
amplification done? Explain. (Delhi 2010C)

Amplification of rDNA gene is done using Polymerase Chain Reaction (PCR).

It is carried out in the following steps

● Denaturation The double-stranded DNA is denatured by applying high temperature

of 95°C for 15 seconds. Each separated strand acts as a template.

● Annealing Two sets of primers are added, which anneal to the 3′ end of each
separated strand.
● Extension Taq polymerase extends the primers by adding nucleotides
complementary to the template provided in the reaction. Taq polymerase is used in the
reaction because it can tolerate high temperature. All these steps are repeated many
times to get several copies of the desired DNA.

80.
Describe the roles of (i) high temperature, (ii) primers and (iii) bacterium.
Thermus aquaticus in carrying the process of polymerase chain reaction. (All India 2019)

Role of Heat In PCR (in vitro), the DNA strands are separated by heating at 95°C for two
minutes. Heating causes the breakage of H-bonds between bases of two strands leading
to its unwinding.

Role of Primers Primers are short lengths of DNA of about 20bp long that are required to
start DNA polymerisation in PCR. The primers hybridise to their complementary
sequence on the DNA strands at 40-50°C temperature and help in DNA polymerisation.

Role of Thermus aquaticus An enzyme called Taq polymerase is isolated from Thermus
aquaticus. Since, this bacteria thrives in temperature as high as 95°C, this enzyme can
also tolerate high temperature without undergoing denaturation. Therefore, this enzyme
is used in PCR instead of normal DNA polymerase.

83.
(i) How has the development of bioreactor helped in biotechnology?
(ii) Name the most commonly used bioreactor and describe its working. (2018)

(i) Bioreactors These are the large volume vessels approximately (100-1000 L) in which
raw materials are biologically converted into specific products, individual enzymes, etc.,
using microbial, plant, animal or human cells. A bioreactor provides the optimal
conditions for achieving the desired product by providing optimum growth conditions
like temperature, pH, substrate, salts, vitamins and oxygen. The cells can also be
multiplied in a continuous culture system.

In this, the used medium is drained out from one side while fresh medium is added from
the other side to maintain the cells in their physiologically most active log/exponential
phase. This type of culturing method produces a larger biomass leading to higher yields
of desired protein. Thus, it plays a very important role particularly in traditional
biotechnology.

(ii) The most commonly used bioreactor is simple stirred tank bioreactor. It is usually
cylindrical or with a curved base to facilitate the mixing of the reactor contents. The
stirrer activity facilitates even mixing and availability of 02 throughout the bioreactor.
Alternatively air can also be bubbles into the medium. It has an agitator system, a
temperature control system, a pH control system and sample ports so that small
samples can be withdrawn periodically.
 85.
Write the steps you would suggest to be undertaken to obtain a foreign-gene product.
(Delhi 2017)

To obtain a foreign-gene product following steps should be undertaken

● Identification of DNA with desirable genes.

● Introduction of the identified DNA into suitable host to form rDNA (rDNA).

● Maintenance of introduced DNA in particular host and gene cloning.

● Transfer of the DNA (gene transfer) to its progeny.

● Selection of the recombinants from non-recombinants.

● Expression of gene of interest by culturing recombinant cells.

● Culturing of cells in bioreactors for large scale production of desired gene product.
(3)

86.
Suggest and describe a technique to obtain multiple copies of a gene of interest in vitro.
(All India 2016)

Polymerase Chain Reaction (PCR) is a technique to obtain multiple copies of a gene of

interest in vitro. This technique amplifies DNA through a simple enzymatic reaction. This
technique was developed by Kary Mullis in 1965.
The basic requirements of a PCR are the following

● DNA template

● Primers

● Enzyme-Taq polymerase

88.
Draw a labelled sketch of sparged stirred-tank bioreactor. Write its application. (Delhi
2015)

Sparged stirred-tank bioreactor

Application These bioreactors are used to produce large quantities of products,
enzymes, etc., using microbial, plant, animal or human cells. (1)

89.
Rearrange the following in the correct sequence to accomplish an important
biotechnological reaction
(i) In vitro synthesis of copies of DNA of interest
(ii) Chemically synthesised oligonucleotides
(iii) Enzyme DNA polymerase
(iv) Complementary region of DNA
(v) Genomic DNA template
(vi) Nucleotides provided
(vii) Primers
(viii) Thermostable DNA polymerase (from Thermus aquaticus)
(ix) Denaturation of dsDNA. (All India 2015)

The correct sequence of reactions are

90.
Many copies of a specific gene of interest are required to study the detailed sequencing
of bases in it. Name and explain the process that can help in developing large number of
copies of this gene of interest. (Foreign 2015)

PCR is the process that can help in developing large number of copies of a gene of
interest.

91.
(i) What is a bioreactor? How does it work?
(ii) Name two commonly used bioreactors. (Delhi 2014c)

(i) Bioreactors are large vessels in which raw materials are biologically converted into
specific products by microbes, plant and animal cells or human cells. The bioreactors
work by providing optimal conditions to process the culture as well as the production of
desired product by maintaining optimum pH, temperature, oxygen and other grpwth
conditions required.

(ii) The two commonly used bioreactors are

● Simple stirred-tank bioreactors.

● Sparged stirred-tank bioreactors.

92.
What is a bioreactor used for? Name a commonly used bioreactor and any two of its
components. (All India 2014C)

The components of a commonly used stirred-tank bioreactor are

● Inlet for sterile air or oxygen

● Agitator system

● Temperature control system

● pH control system

● Foam control system

● Sampling ports

94.
(ii) State the purpose of such an amplified DNA sequence. (AII India 2015C)

(ii) Applications of PCR technique

● In diagnosis of pathogens for specifc infections, like HTV

● To identify mutations in organisms

● In DNA fingerprinting

● Prenatal diagnosis

● Gene therapy

96.
Write the function of the following in biotechnology. (Outside Delhi 2016C)
(i) Polymerase chain reaction technique.
(ii) R.Es.
(iii) Bacterium Thermus aquaticus.

(i) Polymerase chain reaction technique is used to prepare multiple copies of a gene of
interest in vitro using two sets of primers and the enzyme DNA polymerase.
(ii) R.Es are enzymes that make cut at specific positions within DNA.
(iii) Bacterium Thermus aquaticus produces an enzyme Taq polymerase which is heat
stable, i.e. resistant to denaturation by heat. The enzyme is used to amplify a specific
DNA fragment in PCR technique.

97.
Explain the basis on which the gel electrophoresis technique works. Write any two ways
the products obtained through this technique can be utilised. (Delhi 2013C)

Gel electrophoresis technique works on the principle of separation of DNA fragment on

the basis of their size and electric charge.
Since, DNA is negatively charged molecule so, they can be separated according to their
size when they towards anode under an electric field through a medium or matrix
(commonly used is agarose). Shorter molecules move faster towards anode and migrate
farther than the longer one.

The products obtained through this technique can be utilised in the following ways

● Construction of rDNA by joining with cloning vectors.

● Used in making multiple copies of same DNA by using PCR (Polymerase Chain
Reaction).

98.
How can the following be made possible for biotechnology experiments?
(i) Isolation of DNA from bacterial cell.
(ii) Reintroduction of the rDNA into a bacterial cell. (Foreign 2012)

(i) Isolation of DNA from bacterial cell can be done by

● treating the bacterial cells with enzymes such as lysozyme to remove cell wall.
● RNA associated with DNA can be removed by the treatment with ribonuclease,
whereas protein can be removed by treatment with protease. Similarly other molecules
(if any) are removed by appropriate treatment.

(ii) Reintroduction of the rDNA into bacterial cell can be done by the following methods

● The recipient bacterial cell is made ‘competent’ to take up the rDNA its by the
treatment with a specific increased concentration of calcium ions.

● The rDNA is forced into the cells by heat shock treatment, i.e. by incubating the
cells with rDNA on ice followed by placing them at 42°C (heat shock) and then again
putting them back on ice. This enables bacteria to take up rDNA.

99.
(i) Identify A and B illustrations in the following

(i) Write the term given to A and C and why?

(iii) Expand PCR. Mention its importance in biotechnology. (Delhi 2011)

(i) (a) A is recognition or restriction site (GAATTC), which is recognised by restriction

enzyme Eco RI.
(b) B is rop gene protein involved in the replication of plasmid coded by this gene.

(ii) A and C are called palindromes. These are sequence of base pairs that reads same
on the two strands of DNA, when orientation of reading is kept same.

(iii) PCR is Polymerase Chain Reaction in which multiple copies of the gene of interest
can be synthesised in vitro. Thus, PCR can be utilised to amplify a single gene or
fragment into thousands of copies to be used in cloning experiments.

100.
Industrial production of biologically important recombinant products utilises bioreactors.
Sohan was aware of this information but was curious to know about the process
involved. He asked his teacher about this, who explained the process in detail to him.
(i) What are bioreactors? Name the most common type of bioreactor utilised by
industries.
(ii) What is the sequence of events after completion of the biosynthetic phase in
bioreactor ?
(i) Bioreactore are large volume vessels in which raw materials are biologically converted
into specific products. The most common type of bioreactors are of stirring type.
(ii) The sequence of events after the completion of biosynthetic phase is called as
downstream processing. It involves (i) separation and (ii) purification of products
Obtained.

101.
DNA amplification has become an extensively employed technique in various fields. Arun
got curious about this process and asked his father, a molecular scientist to explain his
queries.
(i) What is meant by DNA amplification and how is it achieved?
(ii) Write down the basic steps involved in this process.
(iii) What values do you observe in Arun?

(i) DNA amplification refers to the production of multiple copies of gene of interest from
a single copy.
It is achieved by using PCR, i.e. Polymerase Chain Reaction.

(ii) Steps involved in PCR are

● Denaturation of DNA strands at 95°C.

● Annealing with primers at 45°C

● Extension using Taq polymerase and dNTPs.

(iii) Values shown by Arun are inquisitive mind, intelligence and awareness towards
scientific developments

Applications

1.
Write the two specific ‘cry’ genes that encode the proteins which control cotton boll
worms. (All India 2019)

cry IAc and cry IIAb control cotton bollworm and cry I Ab controls corn borer.

2.
Mention the chemical change that proinsulin undergoes, to be able to act as mature
insulin. (2018)
The C-peptide present in proinsulin is removed during its maturation.

3.
What are cry genes? In which organisms are they present? (All India 2017)

‘cry genes’ are genes found in Bacillus thuringiensis, a bacterium. These genes encode
for protein crystals that contain a toxic insecticidal protein called Bt toxin.

4.
Suggest any two possible treatments that can be given to patient exhibiting adenosine
deaminase deficiency.
Bone marrow transplantation.

Gene therapy (not permanent cure).

5.
Why do children cured by enzyme replacement therapy for adenosine deaminase
deficiency need periodic treatment? (All India 2015)

Children cured by enzyme replacement therapy for adenosine deaminase deficiency

need periodic treatment because in such type of treatment genetically engineered
lymphocytes are used and these cells are mortal.

6.
State the role of C-peptide in human insulin. (All India 2014)

The C-peptide is an extra stretch of the peptides that connect the A and B-polypeptide
chains of insulin in prohormone. During processing to release mature and functional
insulin, this C-peptide is removed.

7.
State the role of transposons in silencing of mRNA in eukaryotic cells. (All India 2012)

Mobile genetic elements, i.e. transposons are the possible source of RNA interference
(RNAi) gene which is further involved in the silencing of the specific mftNA and prevents
translation.

8.
Name a molecular diagnostic technique to detect the presence of a pathogen in its early
stage of infection. (Delhi 2010)

Techniques that serve the purpose of early diagnosis of some bacterial/viral human
diseases are as follows

● Polymerase Chain Reaction (PCR).

● DNA recombinant technology

● ELISA

9.
How does dsRNA gain entry into eukaryotic cell to cause RNA interference? (Delhi
2011C)

dsRNA gains entry into eukaryotic cell either through

● infection by virus having RNA genome.

● mobile genetic elements (transposons) that replicate via an RNA intermediate.

10.
Name the source organism of the gene cry IAc and its target pest. (Foreign 2011)

Source of gene cry IAc is Bacillus thuringiensis and its target pest is cotton bollworm.

11.
Name the source used to produce hapatitis-B vaccine using rDNA technology. (Delhi
2015C)

The source of hepatitis-B vaccine is yeast. Recombinant DNA technology has allowed
the production of antigenic polypeptides of pathogen in bacteria or yeast. Vaccines
produced using this approach allow large scale production and hence greater availability
for immunisation.

12.
Write the functions of
(i) cry IAc gene
(ii) RNA interference (RNAi) (Outside Delhi 2015C)

(i) cry IAc codes for toxic insecticidal protein as inactive protoxins in Bacillus
thuringiensis. This toxin kills the cotton bollworm.
(ii) RNA interference is associated with silencing of specific ntRNA and is a method of
cellular defence in eukaryotes.

13.
What is gene therapy? Name the first clinical case in which it was used.

Gene therapy is a corrective therapy or technique of genetic engineering that is used to

replace a faulty or non-functional gene with a normal healthy functional gene.The first
clinical gene therapy was given to a 4 year old girl with ADA (Adenosine Deaminase)
deficiency in 1990. It is caused due to the deletion of the gene coding for ADA, which
adversely . affects the functioning of imnune system.

14.
Why does Bt toxin not kill the bacterium that produces it, but kills the insect that ingests
it?

Bt toxin does not kill bacteria because in bacteria it exists in inactive state.
When Bt toxin is ingested by an insect, it gets converted into its active form due to the
alkaline pH of the gut.
The activated toxin binds to the surface of the epithelial cells of the midgut and creates
pores. Water enters through these pores and causes swelling and lysis of cells in insect
body.

15.
Explain how Eli Lilly, an American company, produced insulin by recombinant DNA
technology. (Foreign 2014)

Steps involved in insulin production by Eli Lilly company are as follows

● DNA sequences corresponding to the two polypeptide, A and B-chains of insulin

were synthesised in vitro.

● They were introduced into plasmid DNA of E. coli.

● This bacterium was cloned under suitable conditions.

● The transgene was expressed in the form of polypeptides A and B, secreted into
the medium.

● They were extracted and combined by creating disulphide bridge to form human
insulin.

16.What do ‘cry genes’ in Bacillus thuringiensis code for? State its importance for cotton
crop.
‘cry genes’ in Bacillus thuringiensis code for toxic insecticidal proteins called Cry
proteins which are encoded by different forms of a gene called cry gene, e.g. cry IAc and
cry II Ab control the cotton bollworm whereas cry IAb controls corn borer.Cry proteins
when expressed in cotton crops through genetic engineering confer pest resistance
against cotton bollworms and prevent damage. As the larva of these insects when feed
upon cotton plant parts, the toxin gets activated in their gut, lysing their cells and leads
to deah thus, making them pest resistant.

17.Why is proinsulin so called? How is proinsulin different from functional insulin in

humans? (All India 2013, 2012C)

Human insulin when initially synthesised in human body consists of three peptide
chains-A, B and C. The C-peptide is an extra stretch of amino acids joining the A and
B-chains. This is called proinsulin or prohormone. It undergoes processing or splicing to
release the functional mature insulin that can carry out its normal functions.
During processing, the C-peptide is removed. Only A and B-chains contribute to form the
functional insulin.

18.
Write any two ways how genetically modified plants are found to be useful? (All India
2014C)

The genetically modified plants are found to be useful as they

● reduce or minimise the use of agrochemicals, i.e. fertilisers, insecticides,

herbicides, etc.

● reduce post-harvest losses and enhance nutritional value of crop.

19.
(i) State the role of DNA ligase in biotechnology.
(ii) What happens when Meloidogyne incognita consumes cells with RNAi gene? (Delhi
2012)

(i) DNA ligase enzyme is used to join two DNA fragments from their ends.
(ii) When Meloidogyne incognita (parasite) consumes cells with RNA/ gene, parasite
cannot survive and the infection is prevented. It is mainly because introduced DNA
forms both sense and anti-sense RNA. These two strands being complementary to each
other form dsRNA. This dsRNA binds and prevents translation of nematode OTRNA.
Thus, the wRNA of nematode is silenced and the parasite cannot survive there. This
produces Meloidogyne incognita resistant tobacco plants.

20.
(ii) Name the vector used for transferring ADA-DNA into the recipient cells in humans.
Name the recipient cells.

(ii) Retroviral vector is used to transfer ADA-DNA into the recipient cells of human, i.e.
lymphocytes.

21Explain the process of RNA interference.

The strategy to prevent nematode infestation in tobacco plant is RNA interference.
It involves silencing of a specific mRNA due to a complementary dsRNA which binds and
prevents translation.

22.
Explain how a hereditary disease can be corrected. Give an example of the first
successful attempt made towards correction of such disease. (Delhi 2011)

Hereditary disease can be corrected by gene therapy. It is a collection of methods that

allow correction or replacement of defective genes. The first gene therapy was given in
1990 to a 4 year old girl with Adenosine Deaminase (ADA) deficiency. It is caused due to
the deletion of gene for adenosine deaminase.
The treatment involves following steps

● Lymphocytes from the blood of patient are grown on culture outside the body.

● A functional ADA, cDNA (using a retroviral vector) is then introduced into these
lymphocytes.

● Such genetically engineered lymphocytes are returned to the blood of patient.

● Periodic infusion of such genetically engineered lymphocytes is required by the

patient.

23.
How does recombinant DNA technology help in detecting the presence of mutant gene
in cancer patients? (All India 2o11c)

A single-stranded DNA or RNA, tagged with a radioactive molecule (probe) is allowed to

hybridise with its complementary DNA in a clone of cells followed by detection using
autoradiography. The clone having the mutated gene will not appear on photographic
film, because probe will not be complementary with mutated gene thus, helpful in
defecting the presence of mutated gene in cancer patients.

24.
Why is the introduction of genetically engineered lymphocytes into an ADA deficiency
patient not a permanent cure? Suggest a possible permanent cure. (Delhi 2010)

The genetically engineered lymphocytes have a definite lifespan. Hence, the patient
requires periodic infusion of genetically engineered lymphocytes, so the cure is not
permanent. The cure can be permanent, if the gene isolated from marrow cells
producing ADA is introduced into the cells at early embryonic stages.

25.
How did Eli Lilly synthesise the human insulin? Mention one difference between this
insulin and the one produced by the human pancreas. (All india 2010)
(i) For insulin synthesis, Refer to No. 15.
(ii) Differences between insulin produced by rDNA and insulin produced by human
pancreas are as follows

n produced by rDNA n produced by human pancreas

A and B-polypeptides. three polypeptides. A, B and C-chains before maturi

d the prohormone.

ctly synthesises mature ergoes processing to form mature and functional

one. one.

26.
How is Bt cotton made to attain resistance against bollworm? (Delhi 2010C)

Bt toxin genes cry IAc and cry IIAb control cotton bollworms. These genes are isolated
from the bacterium and are incorporated into cotton plants. For action of Bt toxin,

27.
How has the use of Agrobacterium as vectors helped in controlling Meloidogyne
incognita infestation in tobacco plants? Explain in correct sequence. (2018)

Several nematodes infect a wide variety of plants and animals including human beings.
A nematode Meloidogyne incognita infects the roots of tobacco plants which reduces
the production of tobacco.
The strategy adopted to prevent this infection is based on the process of RNA
interference (RNAi). RNA; mechanism takes place in all eukaryotic organisms as a
method of cellular defence.

This method involves the following steps

● Silencing of a specific mRNA due to the complementary dsRNA molecule that

binds to and prevents translation of the mRNA.

● Agrobacterium vectors are used to introduce nematode specific genes into the
host plant. It produces both sense and antisense RNA in the host cells.

● These two RNAs are complementary to each other and form a double-stranded
RNA (dsRNA) that initiates RNAi and hence, silence the specific mRNA of the nematode.

● The parasite cannot survive in a transgenic host, therefore the transgenic plant
gets itself protected from the parasite.
28.
Explain the various steps involved in the production of artificial insulin.

An American company Eli Lilly produced insulin via recombinant DNA technology in
1983.
Insulin production by using recombinant DNA technology is shown in flow chart below

29.
Why do lepidopterans die when they feed on Bt cotton plant?

Bt cotton plants are the transgenic plants that express a Bacillus thuringiensis gene
called cry gene. This gene, encodes for protein crystals having insecticidal properties
against insects of group Lepidoptera, Diptera and Coleoptera. Inside the bacterium,
these proteins remain inactive and do not harm the bacteria. However, these inactive
crystals can get activated in the alkaline pH of the gut of insects upon ingestion.After
activation, these crystals can bind to the receptors present on the membranes of gut
epithelial cells. Due to this binding, the membrane swells and pores are created on
them. These pores lead to bursting of cell and soon the lepidopteran dies due to
starvation.

31.
What is GMO? List any five possible advantages of a GMO to a farmer. (All India 2018)

The plants, bacteria, fungi and animals whose genes have been altered are called
Genetically Modified Organisms (GMOs). GM plants are useful in many ways.
Genetic modification has done the following changes to the phenotypic expression of
the plants,

● Crops become more tolerant to abiotic stresses like cold, drought, salt and heat.

● Dependence on chemical pesticides has reduced, i.e. pest-resistant crops.

● Helped to reduce post-harvest losses.

● Efficiency of mineral usage increased in plants, preventing early exhaustion of
fertility of soil.

● Enhanced nutritional value of food, e.g. vitamin-A enriched rice.

32.
How has the study of biotechnology helped in developing pest-resistant cotton crop?

Biotechnology had hepled in generating plant varieties against the pests that destroy the
cotton balls, i.e. cotton bollworms and cotton borer.
Bt cotton is created by using some strains of a bacterium. Bacillus thuringiensis (Bt is
short form).

● This bacterium produces protein that kills certain insects such as lepidopterans
(tobacco budworm and armyworm), coleopterans (beetles) and dipterans (flies and
mosquitoes).

● Bacillus thuringiensis forms protein crystals during a particular phase of their

growth. These crystals contain a toxic insecticidal protein.

● Bt toxin protein exists as inactive protoxin in bacteria, but once an insect ingests
this inactive toxin, it is converted into an active form due to the alkaline pH of the gut,
which solubilises the crystals.

● The activated toxin binds to the surface of midgut epithelial cells and creates
pores that cause cell swelling and lysis leading to death of insect.

● Specific Bt toxin genes were isolated from Bacillus thuringiensis and incorporated
into several crop plants.

● Most Bt toxins are insect-group specific. The toxin is coded by a gene named cry.
For example, the proteins encoded by the genes cry IAc and cry IIAb control the cotton
bollworms and cry IAb controls corn borer.

33.
Why is molecular diagnosis preferred over conventional methods? Name any two
techniques giving one use of each. (Delhi 2016C)

Using conventional methods of diagnosis (serum and urine analysis), early detection of
diseases is not possible. To overcome this problem, some molecular diagnosis
techniques were developed that provide early detection of diseases. These are as
follows
(i) Polymerase Chain Reaction (PCR) helps in early detection of diseases or pathogens
by the amplification of their nucleic acid.
PCR can amplify nucleic acids of pathogens even when their concentration is very low.
This technique can be used for detecting HIV in suspected AIDS patients, genetic
mutations in suspected cancer patients and in identifying genetic disorders.
(ii) Enzyme Linked Immunosorbent Assay (ELISA) is based on the principle of
antigen-antibody interaction. Infection by pathogen can be detected by the presence of
antigens (proteins, glycoproteins, etc.) or by detecting the antibodies synthesised
against the pathogen.

34.How has RNAi technique helped to prevent the infestation of roots in tobacco plants
by a nematode Meloidogync incognita?

A nematode Meloidogyne incognita infects the roots of tobacco plants, which reduces
the production of tobacco. The strategy adopted to prevent this infection is based on the
process of RNA interference (RNAf). RNAi takes place in all eukaryotic organisms as a
method of cellular defence. This method involves silencing of a specific mRNA due to
complementary rfsRNA molecule that binds to and prevents translation of mRNA.
Thus, resulting into the death of the nematode.

35.Explain enzyme replacement therapy to treat adenosine deaminase deficiency.

Mention two disadvantages of this procedure. (All India 2016)

In enzyme replacement therapy, the patient is given functional ADA (Adenosine

Deaminase) by injection.
Disadvantages

● The patient does not completely recover from the disease.

● It needs periodic injections of the enzyme to the patients.

36.
People are quite apprehensive to use GM crops. Give three arguments in support of GM
crops so as to convince the people in favour of such crops. (Outside Delhi 2016C)

The GMOs are the plants, animals, bacteria, etc. whose genes have been altered by
genetic manipulation. They were created because

● The agrochemicals used in agriculture are too costly.

● Excess use of chemicals in field adversely affects our environment by causing

pollution. GMOs are beneficial in many ways.

Advantages of GMOs are as follows

● Tolerance against abiotic stresses, such as cold, drought, salt, heat.

● Reduce dependence on chemical pesticides.

● Reduce post-harvest losses.

● Increased efficiency of mineral usage by plants.

 37.
(i) Explain the effect of deletion of the gene for ADA in an individual.
(ii) How does the gene therapy help in this case?

(i) Deletion of the gene for ADA in an individual leads to ADA deficiency disorder.
Adenosine Deaminase (ADA) enzyme is crucial for immune system to function.
(ii) Gene therapy is helpful in the treatment of ADA deficiency

38.
Describe any three potential applications of genetically modified plants. (All India 2015)

Potential applications of genetically modified plants are

● Nutritional enhancement, e.g. vitamin-A enriched rice.

● Stress tolerant crops became more tolerant to abiotic stresses such as cold,
drought, etc.

● Creation of tailor made plants by using GM plants to supply alternative resources

of industries in the form of starches, biofuels, etc.

39.
Name the host plant and its part that Meloidogyne incognita infects. Explain the role of
Agrobacterium in the production of dsRNA in the host plant.

The nematode Meloidogyne incognita infects the roots of tobacco plants.

The Agrobacterium is used as vectors carrying nematode specific genes to be
introduced in host plant. These genes when expressed inside host plant produce sense
and anti-sense RNA strands, complementary to nematode’s functional mRNA. This
binding results in the formation of double-stranded RNA and inhibiting or silencing the
translation of RNA specified. This process is called RNA interference.

40.
How does RNA interference help in developing resistance in tobacco plant against
nematode infection? (Delhi 2010)

● Infestation of tobacco plant with the nematode can be stopped by using RNA
interference (RNAi) process.

● Vector used for introducing the nematode specific gene in tobacco plant is
Agrobacterium.

41.
(ii) Why does this nematode die on eating such a GM plant? (Delhi 2010C)

(ii) Due to the RNAi process, specific mRNA of nematode is silenced. The result is that
the parasite could not survive after eating such GM or transgenic plant (host),
expressing, v specific interfering RNA.

43.
(i) Name the source from which insulin was extracted earlier. Why is this insulin no more
in use by diabetic people?

(i) Insulin was extracted earlier from pancreas of slaughtered pigs and cattle animals.
Insulin obtained from these sources caused some allergy or some other reactions to the
foreign protein.

44.
What is biopiracy? (Delhi 2017)

Biopiracy refers to the use of bioresources by multinational companies and other

organisations without proper authorisation from people and countries concerned.

45.
What are transgenic animals? Give an example. (All India 2016)

Animals with manipulated DNA to possess and express the characters of an extra
(foreign) gene are known as transgenic animals, e.g. transgenic
rats, rabbits, pigs, etc.

46.
Mention two objectives of setting up GEAC by our government. (All India 2015)

GEAC is an organisation setup by the Indian government to make decisions pertaining

with genetic research and experiments. The two objectives of setting up Genetic
Engineering Approval Committee (GEAC) by the Indian Government are

● To make decisions regarding the validity of GM research.

● To make decisions regarding the safety of introducing GM organisms for public

services.

47.
What is biopiracy? State the initiative taken by the Indian Parliament against it. (Delhi
2014)

The Indian Parliament has cleared a second amendment of Indian Patents Bill as an
initiative step against biopiracy. This bill considers issues including patent terms,
emergency provisions as well as research and development initiative.

48.
How have transgenic animals proved to be beneficial in
(i) production of biological products?
(ii) chemical safety testing? (Delhi 2014, 2013)

(i) The transgenic animals have been proved beneficial in the production of biological
products like human protein a-1 antitrypsin (by coding genes for that protein only), for
the treatment of emphysema and production of human protein (a-lactalbumin) enriched
milk by transgenic cow, i.e. Rosie. This milk was more nutritionally balanced for. human
babies than natural cow’s milk.
(ii) Transgenic animals are studied for testing toxicity of drugs and other chemicals, as
they carry genes that make them more sensitive to toxic substances.

49.
Mention any four benefits derived from transgenic animals. (Delhi 2012C)

Benefits derived from transgenic animals are as follows

● They are specially made to serve as models for studying human diseases.

● They produce useful biological products that can be created by the introduction of
a portion of gene, which codes for a particular product such as human protein,
a-1-antitrypsin is used to treat emphysema.

● Transgenic mice are being developed for testing the safety of vaccine before they
are used in humans.

● They can be used to study the effects of certain toxic substances.

50.
How is ‘Rosie’ considered different from a normal cow? Explain.
The transgenic cow Rosie, produced human protein-enriched milk (2.4 gm/L). It
contained the human a-lactalbumin and was nutritionally more balanced for human
babies than natural cow’s milk.

51l
(ii) Name the transgenic animal having the largest number amongst all the existing
transgenic animals.
(iii) Mention any three purposes for which these animals are produced. (2018 C)

(ii) Transgenic mice have the largest number (over 95%) amongst all the existing
transgenic animals.
(iii) For purpose of producing transgenic animals.

52.
Biopiracy should be prevented. State why and how. (All India 2011)

Biopiracy should be prevented because

● The countries and people concerned are not given adequate compensatory
payment.

● The countries/people also lose their right to grow and use breeding experiments
to improve the other varieties of the same species.
It may be prevented by implementing specific laws that take into consideration all the
biopatents and biopiracy related issues.

53.
Aditya’s father requested to the farmer’s in his village to stop using chemical fertilisers in
crop fields. Was he correct? Give reason.

Yes, agrochemicals like fertilisers and pesticides have harmful effects on the health of
living beings and the environment.

54.
(i) What is meant by GMO? Can the production of GMOs be considered unethical? Give
reason.

(i) GMO refers to Genetically Modified Organism. Yes, manipulation of living organism
without showing concern about them is considered unethical.

55.
(i) Why did the family declined the use of insulin injection despite its necessity ?
(ii) Name the process by which insulin is produced now-a-days and the organism
involved.

(i) Earlier the insulin was produced by slaughtering of animals like pigs. They also cause
allergic reactions in patients. The family declined because they were against such
process and products.
(ii) The process is recombinant DNA technology and organism used is E.coli.

Evolution

1.
State the two outcomes of the experiments conducted by Louis Pasteur on origin of life.

Louis Pasteur’s experiments demonstrated that life comes only from pre-existing life. He
showed that in swan-neck pre-sterilised flasks, life did not evolved from ‘killed yeast’
while in another flask open to air, new living organisms arose from ‘killed yeast.’

2.
State two postulates of Oparin and Haldane’s theory with reference to the origin of life.

Oparin and Haldane proposed the following postulates with reference to origin of life.
● The first form of life came from pre-existing non-living organic molecules.

● The conditions on earth favouring chemical evolution were high temperature,

volcanic storms and reducing atmosphere.

3.
Write the hypothetical proposals put forth by Oparin and Haldane.

Oparin and Haldane proposed the theory of chemical evolution. According to them, life
originated from pre-existing non-living organic molecules and the formation of life was
preceded by chemical evolution.

4.
Why are analogous structures a result of convergent evolution?

When two species have structures that are similar in function but differ in origin and
anatomy, they are called analogous structures. These structures develop in different
species which move from different areas to a common habitat where they adapt
themselves accordingly, therefore it is called convergent evolution.

5.
Name the evolution resulting in wings of butterfly and bird. What are they called?

Convergent evolution has resulted in the development of structures like wings of

butterfly and birds. Such structures are called analogous organs.

6.
Write the term used for resemblance of varieties of placental mammals to
corresponding marsupials in Australia.

Adaptive radiation occurring through parallel evolution results in the resemblance of

placental mammals to marsupials in Australia.

7.
Identify the examples of convergent evolution
(i) Flippers of penguins and dolphins
(ii) Eyes of Octopus and mammals
(iii) Vertebrate brains

(i) and (ii) are the examples of analogous organs representing convergent evolution.
Vertebrate brains are the example of divergent evolution.

8.
homologous structures

Homologous organs are derived through divergent evolution thus, indicating common
ancestry.Examples of homology are
(i) Vertebrate heart and brain.
(ii) In plants, thorns and tendrils of Bougainvillea and Cucurbita represent homology.
On the other hand, food storage organs, i.e. tubers in sweet potato and potato are
analogous organs.

9.
significance of the study of fossils in evolution

Fossils help us to know the morphology of the organisms present in the past and relate
them to present organisms for understanding evolution. We can trace the time at which
the organism existed.

10.
significance of biochemical similarities among diverse organisms in evolution.

Similarities in biochemicals such as DNA, help in deriving the line of evolution.

Organisms with more similar DNA sequences are considered close relatives that might
have evolved from the same ancestor.

11.
Write the similarity between the wing of a butterfly and the wing of a bat with reference
to evolution?

Similarity between the wings of butterfly , bat ,cockroach and bird or flippers of dolphin
and penguins is that they perform similar functions but are dissimilar in their structure
and development.They are analogous organs. With reference to evolution, it can be
inferred that these are formed as a result of convergent evolution.

12.
Name the scientist who disproved spontaneous generation theory.

Louis Pasteur disproved the spontaneous generation theory through his swan-neck
flask experiment..

13.
(i) homologous or analogous organs
(a) Sweet potato and potato.
(b) Eye of Octopus and eye of mammals.
(c) Thorns of Bougainvillea and tendrils of Cucurbita.
(d) Forelimbs of bat and whale.
(ii) State the kind of evolution they represent.

(i) The given pairs are identified as

(a) Analogous organs.
(b) Analogous organs.
(c) Homologous organs.
(d) Homologous organs.
(ii) (c) and (d) represent divergent evolution while (a) and (b) represent convergent
evolution.

(i) Select the homologous structures from the combinations given below
(a) Forelimbs of whale and bat
(b) Tuber of potato and sweet potato
(c) Eyes of Octopus and mammals.
(d) Thorns of Bougainvillea and tendrils of Cucurbita.
(ii) State the kind of evolution they represent.
(i) (a) Forelimbs of whales and bats and (d) thorns of Bougainvillea and tendrils of
Curcubita are homologous organs.
(ii) Both these structures represent divergent evolution, i.e. sharing common ancestry,
organs with same fundamental structure but different functions.

Write about the ancestry and evolution of bat, horse and human on the basis of a
comparative study of their forelimbs. What are these limbs categorised as?

All mammals, i.e. whales, cheetah, bat and human share similarities in the pattern of
bones of forelimbs. These forelimbs though perform different functions ,have similar
anatomical structure, i.e. all of them have humerus, radius, ulna, carpals, metacarpals
and phalanges in their forelimbs.

Hence, the same structure is developed along different directions due to the adaptations
to different needs in these animals. This is called divergent evolution and these
structures are homologous.

14.
Explain with the help of an example the type of evolution homology is based on.

Divergent evolution is a process, where the same structure develops along different
directions in different organisms due to adaptations to different needs. Divergent
evolution leads to the development of homologous structures, as they all have similar
anatomical structure and origin, but perform different functions.Examples, the thorn of
Bougainvillaea and tendrils of Cucurbita are homologous organs as both of them are
modified axillary buds, which perform different functions.

15.
(i) Explain adaptive radiation with the help of suitable example.
(ii) example where more than one adaptive radiation have occurred in an isolated area.
Name the type of evolution your example depicts and state why is it so named?

(i) The process of evolution of different species in a given geographical area starting
from a point and literally radiating to other areas or habitat is called adaptive radiation,
e.g. alterations in beaks of finches on Galapagos Islands.

(ii)Australian Marsupials are an example where there is more than one adaptive radiation
in an isolated geographical area. There are many marsupials which are evolved from an
ancestral stock within the isolated Australian island but all of these are adapted to
different habitats.example; Tasmanian wolf *marsupial* and a placental wolf *placental*

This depicts convergent evolution as these marsupials show development of similar

adaptive functional structures in unrelated groups of organisms.

16.
What was proposed by Oparin and Haldane on origin of life? How did SL Miller’s
experiment support their proposal?

Oparin and Haldane proposed that life originated on earth spontaneously from non-living
matter, i.e. organic molecules.

SL Miller’s experiment provided experimental evidence for chemical evolution. He

created conditions similar to primitive atmosphere in the laboratory such as high
temperature, reducing atmosphere consisting of CH4, NH3, etc. When he created an
electric discharge in the flask containing all the above stated components at 800°C,
organic molecules, e.g. amino acids were formed. Results indicated that the first
non-cellular forms of life were created about 3 bya. This supports the hypothesis that life
originated from organic matter.

17.
List the two main propositions of Oparin and Haldane.

Two main propositions of Oparin and Haldane were-

Proposition 1 The Earth’s early atmosphere was reducing, meaning it had a lack of
oxygen and was rich in gases like methane, ammonia, and hydrogen. This environment
was conducive to the formation of organic compounds from inorganic precursors.-
Proposition 2 These organic compounds eventually led to the formation of simple life
forms through a series of chemical reactions, which can be described as a naturalistic
or chemical evolution process.

18.
How do palaeontological evidences support evolution of organisms on earth?

Palaeontology is the study of past life based on fossil records. The study of fossils
reveals the type of life forms occurring in the past and highlights the course of evolution
of living organisms. The distribution of fossils in the sedimentary rocks of different ages
supports the concept of evolution. From the fossil records it has been concluded that
evolution has taken place from the simple to complex forms in a gradual manner.
19.
Write the Oparin and Haldane’s hypothesis about the origin of life on earth. How does
meteorite analysis favour this hypothesis?

Oparin-Haldane theory states that origin of life is the result of a long series of
physiochemical changes, brought about first by chemical evolution and then by
biological evolution. Analysis of meteorites also revealed the presence of similar
compounds as found in the primitive atmosphere, indicating the occurrence of similar
processes elsewhere in space.

20.
Convergent evolution leads to analogous structures. Explain with example.

Convergent evolution is a process of evolution, where anatomically dissimilar structures

in different organisms perform similar functions. It leads to the formation of analogous
structures in different organisms as they perform similar function, but are anatomically
different.

Examples, potato (stem modification) and sweet potato (root modification), flippers of
penguins and dolphins.

22.
(i) Differentiate between analogous and homologous structures.
(ii) Select and write analogous structures from the list given below.
(a) Wings of butterfly and birds
(b) Vertebrate hearts
(c) Tendrils of Bougainvillea and Cucurbita
(d) Tubers of sweet potato and potato. (2018)

(i) Differences between analogous structure and homologous structure are as follows

Analogous structures Homologous structures

These have different basic plan and These have similar basic plan and
origin. origin.

These are adapted to perform same These are adapted to perform

functions. different functions.
 These confirm convergent evolution, These confirm divergent evolution,
e.g. eye of Octopus and man. e.g. limbs of man and whale.

(ii) Analogous organs

(a) Wings of butterfly and birds.
(b) Tubers of sweet potato and potato.

23.
How do homologous organs represent divergent evolution? Explain with the help of a
suitable example.

Homologous organs as divergent evolution Homology is the relation among the organs
of different groups of organisms, that show similarity in the basic structure and
embryonic development, but have different functions. Homology in organs indicates
common ancestry. It is based on divergent evolution. When due to different needs, some
structures develop differently, the condition is called divergent evolution. This results in
the formation of homologous organs. Examples of homology in plants and animals ar as
follows

24.
Differentiate between homology and analogy. Give one example of each.

Differences between homology and analogy are as follows

Homology/Divergent evolution Analogy/Convergent evolution

 Homology is based on divergent Analogy is based on convergent
evolution. evolution.

Structures are anatomically similar but Structures are anatomically

functionally different. different but functionally similar.

e.g. in animals, forelimbs of whales, bats e.g. in animals, wings of butterfly

and cheetah. In plants, thorns of and birds. In plants, tubers of
Bougainvillea and tendrils of Cucurbita. sweet potato and potato.

25.
Describe Louis Pasteur’s experiment to dismiss the theory of spontaneous generation

Theory of spontaneous generation states that the life originated from dead, decaying or
rolling matters like storm, dead animals, etc.
Louis Pasteur rejected the theory of spontaneous generation and demonstrated that life
had evolved from pre-existing life. In his experiment, he kept killed yeast cells in pre
sterilised flask and in another flask open into air. The life did not evolved in the former,
but new living organisms evolved in the another flask.

27.
Explain adaptive radiation with the help of a suitable example.

Adaptive radiation is the process of evolution of different species in a given geographical

area starting from a point and radiating to other habitats.
Darwin went to Galapagos Island and observed that there were many varieties of finches
in the same island. All the varieties evolved on the island itself. Darwin suggested that
after originating from a common ancestral seed eating stock, the finches must have
radiated to different geographical areas and undergone adaptive changes in their beaks,
thus enabling some to become insectivorous while the other remained herbivore and ate
seeds.
Or
Many Australian marsupials, each different from the other, e.g. kangaroo, sugar glider,
etc., evolved from a common ancestral stock, but all within the Australian Island
continent. When more than one adaptive radiation occur in an isolated geographical
area, it is a convergent evolution. Australian placental mammals show adaptive
radiation in evolving into varieties of such placental mammals, each one of which
appears similar to a corresponding marsupial, e.g. placental wolf and Tasmanian wolf,
anteater and numbat, flying squirrel and flying phalanger, etc.
28.
Explain the interpretation of Charles Darwin’s finches on Galapagos Islands.

(i) Darwin found the variations in the beaks of small black birds on Galapagos Island due
to their adaptation to different food habits. Darwin explained
(a) All the varieties must have evolved within the same island itself. The original finches
were seed-eating. From them, some arose with altered beaks as insectivorous and some
as vegetarian finches.
(b) This process of evolution of different species in a given geographical area starting
from a point and radiating to other habitats is called adaptive radiation.

30.
Given below is a diagrammatic representation of the experimental setup used by SL
Miller for his experiment.

(i) gases contained and the conditions in the flask A.

(ii) type of organic molecule he collected in the water at B.
(iii) conclusion he arrived at.

(i) Gases were methane, ammonia, hydrogen and water vapour. In ‘A’ flask electric
discharge was created using electrodes.
(ii) The organic molecules collected in water at ‘B’ were amino acids.
(iii) He concluded that life could have come from pre-existing non-living organic
molecules and their formation was preceded by chemical evolution.

31.
State the views of Oparin and Haldane on evolution. How does SL Miller’s experiment
support this

The theory of biogenesis was proposed by Oparin and Haldane. It states that life could
have come from pre-existing non-living organic molecules (e.g. RNA, protein, etc.) and
the formation of life forms was preceded by chemical evolution, i.e. formation of diverse
organic molecules from inorganic constituents.
In 1953, Urey and Miller conducted an experiment to prove this theory. They created the
conditions of primitive earth, i.e. high temperature, volcanic storms, reducing
atmosphere containingCH4, NH3, etc., at laboratory scale. They then stimulated electric
discharge in a closed flask containing CH4,H2, NH3 and water vapour at 800°C. They
observed the formation of amino acids.

In similar experiments, they observed the formation of sugars, nitrogen bases, pigments
and fats. These small organic molecules are the building blocks for proteins and other
components. Hence, this experiment supported that life has came from pre-existing
non-living organic molecules.

33.
Anthropogenic actions hasten evolution. Explain with the help of suitable example.

Human activities, i.e. anthropogenic actions are found to enhance evolution.

example,
(i) Excessive use of DDT as a fertiliser in crops resulted in the evolution of DDT resistant
mosquitoes.

● When DDT was used first time, many mosquitoes died, but few survived.

● Survived mosquitoes showed resistance to DDT and reproduced even in the

presence of DDT.

● Offsprings produced by these mosquitoes were also resistant to DDT.

● Hence, DDT is not effective on mosquito population today.

(ii) Similarly, evolution of antibiotic resistant microbes has occurred due to the overuse
of antibiotics.

34.
How is analogy and homology considered as an evidence in support of evolution?
Homology and analogy show the similarities and differences among the organisms of
today and those existed years ago. These evidences come from the comparative study
of external and internal structure.

These can be determined by the following types Homology in organs indicates common
ancestry. It is based on divergent evolution. When due to different needs, some
morphologically similar structures develop differently, to perform different functions, the
condition is called divergent evolution. This results in the formation of homologous
organs.
Analogy had developed due to the convergent evolution where different structures
evolved for the same function and have morphologically dissimilar structures. These are
called analogous organs.

35.
(i) List any four evidences of evolution, (ii) Explain, any one of the evidences that helps to
understand, the concept of evolution.

(i) Evidences of evolution are derived from

● Palaeontology (Fossils)

● Comparative anatomy and morphology, i.e. homology and analogy

● Biochemical/Physiology

● Biogeography

● Embryology

(ii) Comparative anatomy and morphological evidences show the similarities and
differences among the organisms of today and those that existed years ago.

The evidences come from comparative study of external and internal structure.
I. (a) The organs with same structural design and origin, but different functions are
called homologous organs.
Examples are forelimbs of some animals like whales, bats and cheetah have similar
anatomical structure, such as humerus, radius, ulna, carpals, metacarpals and
phalanges.
(b) Homology in organ indicates common ancestry.
(c) Other examples of homology are vertebrate heart or brain. In plants also, thorns and
tendrils of Bougainvillea and Cucurbita represent homology.
(d) Homology is based on divergent evolution. The same structures developed along
different directions due to adaptations to different needs. The condition is called
divergent evolution.

II. (a) Organs which are anatomically different, but functionally similar are called
analogous organs.
For example, wings of butterfly and birds. In both, wings perform same function, but they
have different origin and structure.
(b) Analogy refers to a situation exactly opposite to homology.
(c) Analogous organs are a result of convergent evolution. It is the evolution in which
different structures evolve for same function and hence, have similarity. It can be said
that above organisms had different structures, but they came in the same environment
and evolved to perform same function.
(d) Other examples of analogy are eyes of Octopus and mammals; flippers of penguins
and dolphins.In plants, sweet potato (root modification) and potato (stem modification)
are analogous organs.

36.
How does the study of fossils help to understand evolution?

The fossils are the remains of past organisms preserved in sedimentary rocks.
Palaeontology is the study of fossils.

● Rocks form sediments and a cross-section of earth’s crust indicate the

arrangement of sediments one over the other during the long history of earth.

● Different aged rock sediments contain fossils of different life forms, who died
during the formation of the particular sediment.
Fossils which were present in a specific area explain the presence of that organism in
that area only.

● Some organisms appear similar to modern organisms. They represent extinct

organisms like dinosaurs.

● A study of fossils in different sedimentary layers indicates the geological period in

which they existed.
Fossils which are obtained from old rocks are of simple type, while which were obtained
from new rocks are of complex type.

● The study showed that life forms varied over time and certain life forms are
restricted to certain geological time scale. Hence, new forms of life have evolved at
different times in the history of earth. Thus, palaeontological evidences help in detailed
study of progress of evolution from old to new forms.
37.
According to the Hardy-Weinberg principle, the allele frequency of a population remains
constant. How do you interpret the change of frequency of alleles in a population?

According to Hardy-Weinberg principle, the change in frequency of alleles in a population

shows the extent of evolutionary change.

38.
State the significance of Coelacanth in evolution.

The discovery of Coelacanth (lobefins), the first amphibian is significant as it proved that
amphibians have evolved from fish-like organisms. Lobefins were the ancestors of
modern day frogs and salamanders.

39.
How did Charles Darwin express fitness?

According to Darwin, fitness of an individual is the ability of an organism to reproduce

successfully and leave a large number of progenies under a particular set of selection
pressures.

40.
Write the names of the following
(i) A 15 mya primate that was ape-like
(ii) A 2 mya primate that lived in East African grasslands.

(i) Dryopithecus (ii) Australopithecus

41.
What role does an individual organism play as per Darwin’s theory of natural selection?

According to the Darwin’s theory of natural selection, the role of an individual organism
is to pass on the necessary variations, changes or mutations from present generation to
the next generation, that has been selected by the nature.

42.
Write the probable differences in eating habits of Homo habilis and Homo erectus.

The probable differences in eating habit of Homo habilis and Homo erectus are as
follows
Homo habilis did not eat meat. Homo erectus probably ate meat.

43.
According to Hugo de Vries what is saltation?

Mutation theory of Hugo de Vries states that the evolution occurs due to single-step
large mutations occurring in a population. This is called saltation and it leads to new
species formation or speciation.

44.
reason for the increased dark coloured moths coinciding with the loss of lichens during
industrialisation

The increase in dark population of moths was due to industrial melanism. After
industrialisation, dark-winged moths became more than white-winged moths. This is
because tree trunks covered by lichens became dark due to the air pollution during
industrialisation. White-winged moths fail to camouflage and thus, decreased in number,
whereas dark-winged moths were able to escape predation.

45.
Write the basis of origin of variations in organisms as described by Hugo de Vries.

Mutations are the basis of origin of variations in an organism according to Hugo de

Vries.

46.
Name the common ancestor of the great apes and man.

Dryopithecus is the common ancestor of great apes and man.

47.
Mention how is mutation theory of Hugo de Vries different from Darwin’s theory of
natural selection.

Hugo de Vries theory It states that evolution occurs due to single step large mutations
called saltation, whereas Darwin’s theory states that the speciation occurs gradually
through a number of generations, with the accumulation of minor variations.

48.
List the two characteristics of mutation that help in explaining evolution.

According to mutation theory of evolution

1. Mutation are random, inheritable and appear in all conceivable directions.

2. Same type of mutations can appear in number of individuals of a species.

49.
When does a species become founder to cause founder effect? .

Founders effect occurs due to the change in allele frequency of a population. When the
change in the allele frequency is very different in the new sample of population, so that
they become a different species. The original drifted population becomes founder and
the effect is called founder effect.

50.

Study the ladder of human evolution given above and the following s.
(i) Where did Australopithecus evolve?
(ii) Write the scientific name of Java man.

(i) Australopithecus evolved in East African grasslands.

(ii) Java man -Homo erectus.

52.
how did Hardy-Weinberg explain that in a population frequency of alleles of a gene is to
remain the same through generations ?

Hardy-Weinberg’s principle states that allele frequencies in a population are stable. They
remain constant from generation to generation. The gene pool also remains constant.
This is called genetic equilibrium.

Thus, according to this principle, the sum total of all the allelic frequencies in a
population is always 1. Suppose in a diploid individual, p and q represent the frequency
of allele A and allele a, respectively. The probability that an allele A with a frequency of P
appears on both the chromosomes of a diploid organism in the p². Similarly of aa is q²,
of Aa is 2pq. Hence, p² + 2pq + q² = 1.

The difference measured in the expected values of frequencies, indicates the extent of
evolutionary change.

53.
Mention the evolutionary significance of the following organisms
(i) Shrews
(ii) Lobefins
(iii) Homo habilis
(iv) Homo erectus (Delhi 2017)

The evolutionary significance of the given organisms are as follows

(i) Shrews They are the first mammals. These were long tailed, insectivorous,
squirrel-like organisms. They gave rise to primitive primates. For example, leones and
tarsiers at the beginning of the Palaeocene era.
(ii) Lobe Fins They are the first amphibians. Modem day frogs and salamanders have
evolved from them.
(iii) Homo habilis The first human-like primates who lived in Africa about 2 mya. They
had brain capacity of 700 cc. They are also called as handy man as they were first and
the most skillful tool makers.
(iv) Homo erectus They appeared after Homo habilis, about 1.7 million years ago. They
had large brain capacities, i.e. 800-1100 cc and were omnivores.

54.
Name the first human-like hominid. Mention his food habit and brain capacity.

Homo habilis were the first human-like hominid. They probably did not consume meat
and their brain capacity was about 650-850cc.

55.
Explain how natural selection operates in nature by taking an example of white-winged
and dark-winged moths of England.

In England, prior to industrialisation, the tree trunks were covered with white lichens
hence, white moths could survive and were protected from predators due to white
colour. On the other hand, black moths (a dark-winged moths) could be easily identified
due to their dark colour and declined in number due to predation.

However, as industrialisation progressed, the lichens were replaced by soot and dust
particles and dark coloured moths were benefited due to camouflage, while
white-winged moths could be easily eaten up by the predators being easily identifiable.
Thus, only the dark-winged moths who were able to fit and survive, i.e. adapted well in
conditions, reproduced well in nature. Thus, natural selection operates in nature by
selecting the fittest characters of organisms.

56.
Rearrange the following in increasing order of evolution
Gnetales; Ferns; Zosterophyllum; Ginkgo.

The increasing order of evolution in plants is as follows Zcsterophyllum – Ferns –

Ginkgo – Gnetales

57.
Name the ancestors of a man based on the features given below.
(i) Human-like, meat-eater with 900 cc brain, lived in Java.
(ii) More human-like with brain size 1400 cc, lived in Central Asia, used hides and buried
their dead.
(iii) Human-like, vegetarian, with brain capacity between 650-800 cc.
(iv) Man-like primate, that existed about 15 my a. Fossils found in Tanzania. (All India
2013C)

(i) Homo erectus

(ii) Homo sapienes neanderthalensis
(iii) Homo habilis
(iv) Ramapithecus
 59.
(i) Write two differences between Homo erectus and Homo habilis.
(ii) Rearrange the following from early to late geological periods
Carboniferous, Silurian, Jurassic.

Homo erectus Homo habilis

Origin period is 1.5 mva Origin period is 1.2-1.5 mya

Brain capacity 900 cc meat First human like beings, brain capacity 650-800
eater, Fossils found in Java. cc. herbivorous and fossils found in Hast Africa.

(ii) The correct sequence from early to late geological period is Silurian → Carboniferous
→ jurassic.

60.
How can Hardy-Weinberg equilibrium be affected? Explain giving three reasons.

Factors which affect Hardy-Weinberg equilibrium are

● Gene migration; Due to migration, new genes or alleles are added to the population
and are lost from the old population thus, changing the frequencies of population.
Migration when happens multiple times, is termed as gene flow.

● Genetic drift; Changes occurring in frequencies by chance. Due to changes in

allele frequency in new population, some different species are formed. This is called
founder effect and the original population is called founder.

● Mutations These occur randomly and at a very slow rate. They lead to new
phenotypes and due to considerable genetic variations, speciation occurs.

● Recombination During gametogenesis, crossing over between homologous

chromosomes leads to new combinations of genes. It occurs during meiosis.

61.
Write the characteristics of Ramapithecus, Dryopithecus and Neanderthal man.

Ramapithecus

● 14-15 million years ago during late Miocene to Pliocene.

● walked erect on hindlegs.

● similar to ape, which lived on the tree tops, but also walked on the ground.

● They ate hard nuts and seeds like modem man. Their jaws and teeth were similar
to humans.

Dryopithecus

● 25 million years ago during Miocene period.

● Legs and heels indicate semi-erect posture and Knuckle walker.

● Dryopithecus was arboreal and herbivorous, who ate soft fruits and leaves.

● Dryopithecus had large canines and incisors.

Neanderthal man

● Neanderthal man existed in the late Pleistocene period.

● Neanderthal walked upright with bipedal movement.

● The face was slightly prognathous and had low brows, receding jaws and high
domed heads.

● The cranial capacity of Neanderthal man was about 1300-1600cc and of average
1450cc. Their jaws were deep with no chin and skull bones were thick.

62.
p² + 2pq + q² = 1. Explain this algebraic equation on the basis of Hardy-Weinberg’s
principle.

The equation p² + 2pq + q² = 1, mathematically represents the Hardy-Weinberg’s

principle. It is used to calculate the genetic variations among a population at
equilibrium.
Principle It states that allele frequencies in a population are stable and remain constant
from generation to generation.
In this equation,
p – frequency of allele A
q – frequency of allele a
p² – frequency of AA (homozygous) individuals in a population
q² – frequency of aa (homozygous) individuals
2pq – frequency of Aa (heterozygous) individuals

Also, the sum total of all the allelic frequencies is equal to 1. If the p and q allele
frequencies are known, then the frequencies of three genotypes can be calculated using
the Hardy-Weinberg’s equation. This equation can be used to measure the differences in
frequencies of observed genotype measured from the frequencies predicted by the
equation. The disturbance in genetic equilibrium results in evolution, thus the presence
of any difference indicates the extent of evolutionary change.
 64.
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.

(i) Sixth extinction is different from previous episodes in the following ways

● It takes place rapidly, due to the reduction in number of species per unit area per
unit time.

● In contrast to previous episodes which were naturally driven, sixth extinction is

accelerated by human activities such as deforestation, industrialisation, etc.

(ii) Human activities that ultimately lead to global warming and disruption of
environmental and ecological balance are responsible for sixth extinction.

(iii) The four measures that can be implemented to overcome this disaster are

● Aforestation.

● Reduction in overexploitation of natural resources.

● Conservation of species and their natural habitats to minimise their losses.

● Create awareness among people regarding global warming and its consequences.

65.
three ways in which natural selection operates on different traits in nature

The three different ways by which natural selection can affect the frequency of a
heritable trait in a population are
(i) Stabilisation It results in more number of individuals acquiring the mean character
value, i.e. variation is much reduced.
(ii) Directional change It results in more individuals acquiring value other than mean
character value, i.e. the peak shifts towards one direction.
(iii) Disruption ; more individuals acquire peripheral character value at both ends of the
distribution curve, i.e. two peaks are formed at periphery.

66.
According to Darwinian theory, the rate of new forms is linked to their life cycles. Explain.

Darwin’s theory states that the fitness of an organism is measured by its reproductive
ability. The appearance of new forms is linked to the lifespan of an organism. The
greater lifespan of an individual indicates that the more it can reproduce and hence,
greater new forms would appear. This can be observed in the development of
dark-winged moths due to industrial melanism.

67
Study the schematic representation of evolutionary history of plant forms given below
and mention

(i) The plant forms ferns and conifers are most related to.
(ii) The nearest ancestors of flowering plants.
(iii) The most primitive group of plants.
(iv) Common ancestry of Psilophyton provides to.
(v) The common ancestor of Psilophyton and seed ferns.
(vi) The common ancestors of mosses and tracheophytes.

(i) Psilophyton
(ii) Seed ferns,
(iii) Chlorophyte ancestor
(iv) Ferns, conifers and seeds ferns
(v) Tracheophyte ancestor
(vi) Chlorophyte ancestor.

68.
Branching descent and natural selection are the two key concepts of Darwinian theory of
evolution. Explain

The two key concepts of Darwinian theory of natural selection are as follows

Branching Descent

Members of a population vary in characteristics, even though they look superficially

similar. Most of these variations are heritable.Accumulation of variations over a period
of time through a number of generations leads to change in population characteristics.

e.g. Evolution of marsupials of Australia derived from a common ancestor.

Natural selection

Nature selects those individuals who are fit in the environment. Fitness according to
Darwin is reproductive fitness.Those who adapt better to the habitat reproduce more and
their progeny consists of more fit individuals, who are selected by nature, e.g. Industrial
melanism.

69.
(i) How does the Hardy-Weinberg’s expression (p² + 2pq + q² = 1), explain that genetic
equilibrium is maintained in a population?
(ii) List any two factors that can disturb the genetic equilibrium.

(i) The expression states that the sum total of all the allele frequencies is one. Suppose
there are two alleles ‘A’ and ‘a’ in a population. Their frequencies are p and q,
respectively. The frequency of ‘AA’ individual in a population is p². It can be explained
that the probability that an allele ‘A’ with a frequency p appears on both the
chromosomes of a diploid individual is simply the product of the probabilities, i.e. p². In
the same way, the frequency for aa is q² and for Aa it is 2 pq. p² + 2 pq + q² = 1 where, p²
represents the frequency of homozygous dominant genotype, 2 pq represents the
frequency of the heterozygous genotype and q² represents the frequency of
homozygous recessive.
(ii) Genetic equilibrium is disturbed by gene migration, genetic drift, mutation and gene
recombination during gamete formation.

70.
(i) Describe Hardy-Weinberg principle
(ii) List any four factors, which affect genetic equilibrium.
(iii) Describe founder effect.

(i) According to Hardy-Weinberg principle, the allele frequencies in a population are

stable and are constant from generation to generation.

(ii) The four factors that affect genetic equilibrium are

● Gene migration

● Genetic drift

● Mutation and recombination

● Natural selection

(iii) Whenever the gene migration occurs multiple times, it leads to some changes, that
may sometimes result in change in allele frequency (at random or by chance).
This difference in allele frequency leads to a new sample of population in such a way
that they evolve into a different species. Such populations are called founders and the
effect generated is called founder effect.

71.
(i) Explain Darwinian theory of evolution. State the two key concepts of theory.
(ii) Mention any three characteristics of Neanderthal man that lived in near East and
Central Asia.

(i) Darwinian theory of evolution/ Natural Selection

● All the populations have built in variations for each character, which help them to
adapt better to the environment.

● The characteristics, which enable some populations to survive better in natural

conditions (climate, food, physical factors) would outbreed others (survival of the
fittest).

● The population, which better fits in an environment is selected by nature and

survives more (natural selection).

● Adaptability is inherited and fitness is the end result of ability to adapt and get
selected by nature.

The two key concepts of Darwinian theory are

● Branching descent and

● Natural selection

(ii) The three characteristics of Neanderthal man that lived in near East and Central Asia
are

● Walked upright with bipedal movement.

● Cranial capacity was around 1300-1600 cc.

● Face slightly prognathous and jaws were deep with no chin.

72.
How does the process of natural selection affect Hardy-Weinberg equilibrium? Explain.
List the other four factors that disturb the equilibrium.

Natural selection affects Hardy-Weinberg equilibrium in the following ways

● It is a process in which heritable variations help in the survival of an organism,

enabling it to reproduce and give rise to a large number of offsprings.

● There may be change in the frequency of genes and alleles in the future
generations.

● It leads to the formation of new species. Hardy-Weinberg law states that allelic
frequencies in a population are stable and remain constant from generation to
generation but natural selection allows only one allele to adapt to changing conditions.

74.
(i) Name the primates that lived about 15 million years ago. List their characteristic
features.
(ii) (a) Where was the man-like animal found?
(b) Write the order in which Neanderthals, Homo habilis and Homo erectus appeared on
the earth. State the brain capacity of each of them.
(c) When did modern man Homo sapiens appear on this planet?

(i) Dryopithecus and Ramapithecus are the primates that lived about 15 million years
ago.
Their characteristics are

● They were hairy.

● They walked like gorillas and chimpanzees. Dryopithecus was more ape-like, while
Ramapithecus was more man-like.

(ii) (a) The man-like animal was found in East African grasslands.
(b) They appeared in the following order Homo habilis → Homo erectus → Neanderthal.
They had brain capacities of 650-800 cc, 900 cc and 1400 cc, respectively.
(c) During ice age between 75000-10000 years ago, modern man Homo sapiens
appeared

75.
Explain the salient features of Hugo de Vries theory of mutation. How is Darwin’s theory
of natural selection different from it?

Hugo de Vries explained that new species arise from pre-existing ones in a single
generation by a sudden appearance of marked differences called mutations. He believed
that it is mutation, which causes evolution.

The differences between de Vries theory and Darwin theory are as follows

de Vries theory Darwin theory

Evolution resulted from mutation. Evolution resulted from variations.

Evolution was sudden. Evolution was gradual.

Mutations are random nd directionless. Variations are small and directional.

77.
(i) Natural selection operates when nature-selects for fitness. Explain,

(i) The members of a population vary in characteristics even though they look similar.
The population usually increases exponentially but the natural resources are limited
leading to more competition. The individuals, which are fit and can adapt themselves are
able to survive. They grow, reproduce and survive. This is called natural selection as
stated by Darwin.

78.
(i) How do the observations made during moth collection in pre and post-industrialised
era in England to support evolution by natural selection?
(ii) Explain the phenomenon that is well-represented by Darwin’s finches other than
natural selection.

(i) Natural selection is the key concept of Darwin’s theory of evolution which was
explained by Charles Robert Darwin. According to this theory, population of all
organisms exhibits variations in characteristics, which help them to adapt better to
environment. It means that individuals of a population are never same.
Some of these characteristics, enable individuals to survive better in natural conditions
and reproduce. This is called as the survival of the fittest. The organisms which adapt
well in the environment are selected by nature and thus, survive more in nature.

This is called natural selection. From the description given below, we can figure out that
how the observations made during moth collection in pre and post-industrialised era in
England supported the idea of evolution by natural selection.

Industrial melanism There are two varieties of moth, white-winged and dark-winged.
(a) Before industrial revolution in England, white-winged moths were more in number
than dark-winged moths, because there was less pollution, which led to light trunk of
trees due to the presence of lichen on them. So, on light background white-winged
moths were not visible, while dark-winged moths could be eaten by predators very easily.

(b) After industrialisation, dark-winged moths became more than white-winged moths.
This is because during industrialisation, tree trunks covered by white lichens became
dark due to air pollution (dust and soot particles). So, now white-winged moths could be
detected easily. Due to this, white-winged moths could be easily eaten up by the
predators as they fail to camouflage. Whereas dark-winged moths escape predation. So,
nature selected only those moths which were better suited. However, none of them
eradicated completely.

(ii) The phenomenon represented by Darwin’s finches other than natural selection is
adaptive radiation.
Adaptive Radiation HF Osborn (1898) developed the concept of adaptive radiation or
divergent evolution. It involves the development of different functional structures from a
common ancestral form.

When a group of organisms shares a homologous structures, which are specialised to

perform a variety of different functions, it shows adaptive radiation. This represents the
evolution of new forms in several directions from the common ancestral type
(divergence).

The significance of adaptive radiation is that it suggests the existence of divergent

evolution based on the modification of homologous structures. The examples of
divergent evolution are as follows
(a) Darwin’s finches of Galapagos Islands had common ancestors. Later on, their beaks
modified according to their feeding habits.
(b) Australian marsupials and limbs of mammals are also examples of adaptive
radiation.

79.
(ii) Name the scientist who influenced Darwin and how?

(ii) Darwin was influenced by Thomus Robert Malthus, a British economist. He put
forward a theory of human population growth. He wrote the book ‘An Essay of the
Principles of Population’.
81.
statement; forelimbs of man and cheetah are structurally similar, though they perform
different functions. Give an example of such organs from the plant world.

Such organs are called homologous organs. Thorns of Bougainvillea and tendrils of
Cucurbita are homologous organs in plants.

82.
whether or not life is still originating on the earth today.
(i) As a student of biology what is your to the above problem and why?
(ii) What would have been the energy source for origin of life on the earth?

(i) There is no origin of life on earth today because the atmosphere is oxidising and any
new molecule if formed will get oxidised.
(ii) Sunlight (UV rays) and lighting electrical discharge would act as the energy source for
origin of life on earth.

83.
(i) Name the book in which Lamarckism was explained.
(ii) What are the three postulates of this theory?

(i) Philosophic Zoologique in 1809.

(ii) three postulates of this theory are

● New needs in respect to changing environment.

● Acquisition of characters.

● Inheritance of acquired characters.

Human Health and Disease

1.
State the functions of mast cells in allergy response. (All India 2019)

The function of mast cell is allergy response is that it releases histamin which cause
inflammatory reactions in the body.

2.
State the function of interferons. (All India 2019)

Interferons protect non-infected cells from the other viral infected cells by releasing
cytokine barriers.
 3.
How do monocytes act as a cellular barrier in humans to provide innate immunity?
(2018)

Monocytes phagocytose and destroy microbes present in the blood to provide innate
immunity.

4.
How do cytokine barriers provide immunity in humans? (2018)

Cytokine barriers include interferons, which are secreted by virus infected cells. They
provide protection to non-infected cells from further same viral infection.

5.
Name two diseases whose spread can be controlled by the eradication of Aedes
mosquito. (2018)

Dengue fever and chikungunya are two diseases, whose spread can be controlled by the
eradication of Aedes mosquito.

6.
Suggest a method to ensure an anamnestic response in humans. (Delhi 2017)

Anamnestic response is the quick and intense response, which occurs when an
individual encounters a particular antigen for the second time. It is aided by the memory
of primary response stored in B-cells. By the administration of a dead or attenuated
pathogen in the body of person, a secondary immune response can be ensured due to
the presence of memory cells in the body.

7.
Retroviruses have no DNA. However, the DNA of the infected host cell does possess viral
DNA. How is it possible? (All India 2015)

Retroviruses have RNA as their genetic material. After getting into the body of a person,
the virus enters the macrophages. Here, RNA is replicated to form viral DNA by using
enzyme reverse transcriptase. The viral DNA now, gets incorporated into the host cell’s
DNA and directs the infected cells to produce viruses.

8.
Indiscriminate diagnostic practices using X-rays, etc., should be avoided. Give one
reason. (Delhi 2015)

Indiscriminate diagnostic practices using X-rays should be avoided, because there are
several potential risks from exposure to ionising radiations, e.g. cancer can be developed
in later stages of life, various tissues can also get affected leading to cataract, hair loss,
etc.
 9.
In what way is monocyte a cellular barrier with reference to immunity? (Delhi 2015)
Or
Name any two types of .cells that act as ‘cellular barriers’ to provide innate immunity in
humans. (Delhi 2014)

Cellular barriers include certain types of leucocytes of our body such as

polymorphonuclear leucocytes, monocytes and natural killer in the blood as well as
macrophages in tissues. These can phagocytose and destroy microbes and provide
innate immunity to humans.

10.
How do cytokinin barriers help in evading viral infections? (Delhi 2015)
Or
How do interferons protect us? (All India 2012)

Cytokinins are virus infected cells, which secrete proteins called interferons. They
protect non-infected cells from further viral infection by inhibiting their replication and
making cell resistant to further infection.

11.
Why is Gambusia introduced into drains and ponds? (All India 2014)

Gambusia is a fish that feed upon mosquito larvae. Thus, it is introduced in drains and
ponds to destroy disease vectors.

12.
Why is secondary immune response more intense than the primary immune response in
humans? (All India 2014)
Or
When does a human body elicit an anamnestic response? (All India 2013; Delhi 2011C)

The secondary or anamnestic immune response is based on the memory of primary

response, i.e. first encounter with an antigen. Due to this, the second generated immune
response is more fast and have higher affinity for antigen and therefore, it is more
intense than primary immune response in humans.

13.
How does haemozoin affect the human body when released in blood during malarial
infection? (Foreign 2014)

The release of toxic haemozoin by the ruptured RBCs during malarial infection results in
recurrence of high fever and chill every 3-4 days.

14.
What is an autoimmune disease? Give an example. (Foreign 2014)
The abnormal response of an immune system, in which it fails to recognise ‘self’ and
‘non-self’ and starts destroying its own cells and molecules is called autoimmune
disease. Rheumatoid arthritis is an example of autoimmune disease, which destroys
articular cartilage and fusing bones.

15.
Why sharing of injection needles between two individuals is not recommended? (Delhi
2013)

Sharing of injection needles may act as a mode of transmission of certain diseases

including AIDS. Thus, it is not recommended.

16.
State two different roles of spleen in the human body? (All India 2012)

The two different roles of spleen in human body are

● Spleen acts as a filter to trap blood-borne microorganisms.

● It is a large reservoir of erythrocytes.

17.
Why do pollen grains of some flower trigger sneezing in some people? (Foreign 2012)

Pollen grains are allergens that cause allergy in some people due to release of chemicals
like histamine and serotonin from mast cells. These trigger the inflammatory responses
in body, e.g. sneezing, wheezing, etc.

18.
What is it that prevents a child to suffer from a disease he/she is vaccinated against?
Give one reason. (Delhi 2010)

Due to vaccination body produces antibodies in large numbers. It protects the child by
neutralising the pathogenic agents during infection. The vaccine also generates memory
B-cells and T-cells that can recognise pathogens on subsequent exposure and produce
intense immune response.

19.
How does malaria differ from chikungunya with reference to their vectors? (All India
2010C)

Malaria is spread by female Anopheles mosquito, whereas chikungunya is spread by

female Aedes mosquito.

20.
Malaria, typhoid, pneumonia and amoebiasis are some of the human infectious
diseases. Which one of these is transmitted through mechanical carriers? (Foreign
2010)

Amoebiasis is transmitted through mechanical carrier, i.e. houseflies.

21.
Differentiate between the roles of B-lymphocytes and T-lymphocytes in generating
immune responses. (Delhi 2019)

For B and T-lymphocytes, Refer to page no. 210.

22.
Principle of vaccination is based on the property of ‘memory’ of the immune system.
Taking one suitable example, justify the statement. (Delhi 2019)

For vaccination, Refer to page no. 211.

23.
Why is the structure of an antibody molecule represented as HgLa ? Name any two types
of antibodies produced in a human body. (2018C)

For structure of antibody molecule. Refer to page no. 210-211.

24.
Mention one application for each of the following
(i) Passive immunisation
(ii) Antihistamine
(iii) Colostrum
(iv) Cytokinin-barrier (All India 2017)

Applications of given components are as follows

(i) Passive immunisation provides a faster immune response.
(ii) Antihistamine is used to reduce the symptoms of allergy, such as sneezing, watery
eyes, rashes, running nose, etc.
(iii) Colostrum consists of antibodies (e.g. IgA) that provide immunity to an infant
against infections.
(iv) Cytokinin barrier produces interferons and protects non-infected cells from further
viral infection.

25.
Name the cells HIV (Human Immunodeficiency Virus) gains entry into after infecting the
human body. Explain the events that occur in these cells. (All India 2016)
Or
How do macrophages in the human body act as HIV factory? (All India 2010)
Or
Name the cells that act as HIV factory in humans when infected by HIV. Explain the
events that occur in the infected cell.
After infecting the human body, the HIV gains entry into macrophages.
Events occurring in these cells are as follows

● RNA genome of the virus replicates to form viral DNA by enzyme reverse
transcriptase.

● Viral DNA gets incorporated into the macrophage DNA and directs the infected
cells to produce new viruses.

● Macrophages continue to produce virus particles and thus called HTV factory.

26.
Name the causative organism of the disease amoebiasis. List three symptoms of the
disease. (All India 2016)
Or
(i) Name the protozoan parasite that causes amoebic dysentery in humans.
(ii) Mention two diagnostic symptoms of the disease.
(iii) How is this disease transmitted to others? (All India 2016, Delhi 2012)

The disease amoebic dysentery or amoebiasis is caused by an intestinal parasite,

Entamoeba histolytica, which is found in the large intestine of human. Transmitting
agent is housefly, which acts as mechanical carrier. It transmits the parasite from faeces
of infected person to the food.
Symptoms include constipation, abdominal pain and cramps, stools with excess
mucous and blood clots.

27.
(i) Name any two causative organisms responsible for ringworm.
(ii) State any two symptoms of the disease. (Delhi 2016C)

(i) Causative organisms responsible for ringworm are Microsporum, Epidermophyton

and Trichophyton (all fungi).

(ii) Two symptoms of the disease are

● Intense itching

● Appearance of dry, scaly lesions on various body parts.

28.
(i) Name any two helminths which are known pathogenic to human.
(ii) List any two symptoms of the diseases caused by any one of them. (Delhi 2016)
Or
List the symptoms of ascariasis. How does a healthy person acquire this infection? (All
India 2014)

(i) Helminths worms, which pathogenic to human are

● Ascaris, the roundworm.

● Wuchereria, the filarial worm.

(ii) Ascaris is an intestinal parasite that causes ascariasis. Symptoms of ascariasis are
internal bleeding, muscular pain, fever, anaemia and blockage in intestinal passage.
A healthy person acquires this infection through consumption of water, vegetables or
fruits contaminated with the eggs of parasite Ascaris.

29.
Name any two secondary lymphoid organs in a human body and state the function of
any of them. (Delhi 2016)

Secondary lymphoid organs provide the site for interaction of lymphocytes with the
antigen, which then proliferate to become effector cells, e.g. spleen, tonsils and lymph
nodes.
Lymph nodes help to trap antigens entering the tissue fluid, whereas spleen trap the
blood-borne microbes.

30.
How are oncogenic viruses different from proto-oncogenes?

Oncogenic viruses The cancer causing viruses which have viral genes or oncogenes are
called, oncogenic viruses.
Proto-oncogenes These genes are present in normal cells, which when activated under
certain conditions could lead to oncogenic transformation of the cells, thus leading to
cancer.

31.
(i) Which organ of the human body is initially affected when bitten by an infected female
Anopheles! Name the stage of the parasite that infects this organ.
(ii) Explain the events that are responsible for chill and high fever in the patient. (Delhi
2016C)

(i) Liver is initially affected by the sporozoites stage dud to the bite of female Anopheles.
(ii) Rupture of RBC and release of haemozoin is responsible for chill and high fever.

32.
Name an allergen and write the response of human body when exposed to it. (Delhi
2014C)

The allergen can be pollen grains, spores or dust particles. When the allergens are
inhaled or enter body system, they stimulate body to produce IgE antibodies and trigger
an antiallergic reaction. The chemicals such as histamine and serotonin are released
from mast cells in response to allergen, thereby causing dilation of blood vessels. These
chemicals also elicit inflammatory response that may result in sneezing, watery eyes,
running nose, etc.
33.
Differentiate between active and passive immunity. (Delhi 2014C)

Differences between active and passive immunity are as follows

e immunity ve immunity

ops when body’s own cells producops when antibodies produced in other
odies in response to infection or isms are injected or administered into a
ne. n to counteract antigen.

in response, but long lasting effectdes immediate relief, but short lived,

mmunity developed by vaccination.

njection of tatanus.

34.
How does a vaccine for a particular disease immunise the human body against that
disease? (Delhi 2013C)

During vaccination for a particular disease, an antigen or antigenic protein or weak

pathogen, which is in inactive form is introduced into the body to induce mild immune
response.

The vaccine generates antibodies that neutralises the toxin/pathogen and produces
memory-B or T-cells, which recognise the pathogen in the subsequent encounters and
produce antibodies.

35.
Why is a person with cuts and bruises following an accident administered tetanus
antitoxin? Give reasons. (All India 2013)
Or
Why does a doctor administer tetanus antitoxin and not a tetanus vaccine to a child
injured in a roadside accident with a bleeding wound? All India 2010

A person with cuts and bruises following an accident is administered tetanus antitoxin,
because this toxin contains performed antibodies against the pathogen Clostridium
tetani. This inactivates the pathogen and provide passive immunity. Also, tetanus
antitoxin provide instant response unlike tetanus vaccine, which take time to develop
immunity.

36.
A patient showed symptoms of sustained high fever, stomach pain and constipation, but
no blood clot in stools. Name the disease and its pathogen. Write the diagnostic test for
the disease. How does the disease get transmitted? (Delhi 2013C)

The symptoms, such as constant high fever, stomach pain and constipation, weakness
and headache are shown in typhoid.
Its causative’agent is a bacterium called Salmonella typhi. Widal test is used for its
diagnosis. Typhoid is transmitted through contaminated food and water.

37.
A student on a school picnic to a park on a windy day started sneezing and having
difficulty in breathing on reaching the park. The teacher enquired whether the student
was allergic to something.
(i) What is an allergy?
(ii) Write the two unique characteristics of the system involved in the response observed
in the student. (Delhi 2013)

(i) Allergy is a hypersensitive response to a foreign substance coming in contact with or

entering the body, e.g, sneezing, watery eyes, etc.
(ii) It is due to the release of histamine and serotonin by the mast cells.

38.
A young boy when brought a pet dog home started to complain of watery eyes and
running nose. The symptoms disappeared when the boy was kept away from the pet.
(i) Name the type of antibody and the chemicals responsible for such a response in the
boy.
(ii) Mention the name of any one drug that could be given to the boy for immediate relief
from such a response. (Delhi 2013)

(i) In case of allergy, IgE antibodies are involved. Histamine and serotonin are
responsible for such responses.
(ii) Antihistamine could be given for immediate relief.

39.
(i) Highlight the role of thymus as a lymphoid organs.
(ii) Name the cells that are released from the above mentioned gland. Mention, how they
help in immunity? (Delhi 2012)

(i) Thymus is a primary lymphoid organs of the immune system. Here, immature
lymphocytes get differentiated into antigen-sensitive T-lymphocytes.
(ii) T-lymphocytes are released from thymus, after their maturation get completed. They
themselves do not produce antibodies, but help B-cells to produce them. They are also
responsible for Cell Mediated Immune (CMI) response.

40.
Name the parasite that causes filariasis in humans. Mention its two diagnostic
symptoms. How is this transmitted to others? (Delhi 2012)

Wuchereria (W. bancrofti and W. malayi) is the filarial worm that causes filariasis in
humans.
Diagnostic Symptoms

● The presence of nematodes in lymph vessels causes collection of fluid. It may

lead, to swelling in arms, breasts, legs and genital region.

● Inflammation of lower limbs result in deformities.

It is transmitted to a healthy person through the bite of the female mosquito vector,
Culex.

41.
Name and explain the two types of immune responses in humans. (All India 2012)

Immune responses are of two types

● Primary response The reaction of the body’s immune system to the first attack of
microbe (antigen) is called primary immune response. It is slow and less intense.

● Secondary response The reaction of the body’s immune system to any subsequent
infection of the same microbe is termed as secondary immune response. It is fast and
intense.

42.
Name the two special types of lymphocytes in humans. How do they differ in their roles
in immune response? (All India 2012)

Two special types of lymphocytes in humans are

● B-lymphocytes or B-cells

● T-lymphocytes or T-cells

Differences between B and T-lymphocytes are as follows

phocytes phocytes
 produce antibodies against antigen. stimulate B-cells to produce antibodies.

do not respond to organ transplant. react to organ transplant.

43.
(i) Name the group of virus responsible for causing AIDS in humans. Why are these virus
so, named?
(ii) List any two ways of transmission of HIV infection in humans other than sexual
contact? All India 2012

(i) Retrovirus is the group of viruses causing AIDS in humans. They contain RNA as
genetic material and with the help of enzyme reverse transcriptase, they make viral DNA
using RNA as a template. Thus, they are called retrovirus.

(ii) (a) By sharing infected needles.

(b) By transfusion of blood contaminated with HIV.

44.
Why is an antibody represented as H2L2? (Foreign 2012)

Antibody is represented as H2I2 because each antibody molecule has four peptide
chains, i.e. two small light (L) chains and two longer heavy (H) chains.

45.
Name the different types of cell providing cellular barrier responsible for innate immunity
in humans. (Foreign 2012)

For cellular barrier as innate immunity, Refer to page no. 209.

46.
List any two emergent circumstances, when a medical doctor would recommend
injection of a preformed antibody into the body of a patient and why? (Delhi 2011C)

In case of snake bite and in tetanus infection. Preformed antibodies help in providing
quick immune response.

47.
List the two types of immunity a human baby inborn with. Explain the differences
between the two types. (All India 2011)
Two types of immunity with which human baby is born include.

● Innate immunity It is inherited type of immunity and it protects an organism from

birth throughout the life. It is not specific to particular pathogen and consists of four
types of barriers namely physical, physiological, cellular and cytokine.

● Natural passive immunity It is passively transferred from mother to foetus through

placenta, e.g. IgG antibodies can cross placental barrier to reach the foetus.

48.
How is an allergic reaction caused by an allergen? Name the drug that can reduce the
symptoms of allergy? (All India 2011C)

An allergic reaction is caused by allergens as these can produce IgE type of antibodies.
These antibodies causes the release of histamine and serotonin like chemicals from
mast cells, which cause allergic reactions. The use of drugs like antihistamine,
adrenaline and steroids quickly reduce the symptoms of allergy.

49.
Name the two types of immunity in a human body. Why are cell-mediated and humoral
immunities so called? (Delhi 2011)

Types of immunity system in humans are

● Innate immunity and acquired immunity are two main types of immunities in
human body.

● Cells mediated immunity is so called as it is mediated by specialised cells, the

T-lymphocytes that recognise self and non-self cells. Humoral immunity is so called
because it is mediated by antibodies, which are found circulating in body fluid (humor)
e.g. blood.

50.
Write the scientific names of the causal organisms of elephantiasis and ringworm in
humans. Mention the body parts affected by them. (Delhi 2011)

● Elephantiasis is caused by Wuchereria bancrofti and W. malayi. These affect lower

limbs and genital organs.

● Ringworm is caused by Microsporum, Trichophyton and Epidermophyton. They

affect the skin, nails and scalp.

51.
Identify A, D, E and F in the diagram of an antibody molecule given below, (Delhi 2011)
A-Antigen binding region,
D-Light chain
E-Heavy chain
F-Disulphide bond/bridge

52.
Name the host and the site, where the following occur in the life cycle of a malarial
parasite.
(i) Formation of gametocytes.
(ii) Fusion of gametocytes. (Delhi 2010)

(i) Formation of gametocytes occurs in the erythrocytes (RBCs) of human beings.

(ii) Fusion of gametocytes occurs in the intestine of mosquito.

53.
Define the term health. Mention any two ways of maintaining it. (All India 2010)

Health can be defined as a state of complete physical, mental and social well-being. It
can be maintained by taking balanced diet, maintaining personal hygiene, regular
exercise/yoga, vaccination against infectious diseases, etc. (1+1)

54.
Identify A, B, C and D in the following table. (Foreign 2010)

e of the human diseasese of the causal bacterial virusfic organ or its part affecte

oid onella typhi

mon cold
 monia tococcus pneumoniae

A – Small intestine
B – Rhinovirus
C – Nose, respiratory passage
D- Alveoli of lungs.

55.
The barriers in the innate immunity are given in the following table. Identify A, B, C and D.
(Delhi 2010C)

s of barrier ers

cal A

ological he eye

eron

ar D

A-Epithelium lining
B-Tears
C-Cytokinin
D-Polymorphonuclear leucocytes.

56.
(i) How does a vaccine affect immunity?
(ii) How can we get immunisation against tetanus? (All India 2010)

(i) In vaccination, a preparation of antigenic proteins of pathogen or inactivated/

weakened pathogen (vaccine) are introduced into the body. The antibodies produced in
the body against these antigens would neutralise the pathogenic agents during actual
infection. The vaccines also generate memory-B and T-cells.
(ii) Preformed antibodies for tetanus are directly injected to acquire quick immune
response. This is called passive immunisation against tetanus.
57.
Why do normal cells not show cancerous growth? (All India 2010)

Normal cells do not show cancerous growth as

● Their growth and division are regulated by certain regulatory mechanisms.

● They show the property of contact inhibition, by virtue of which contact with other
cells inhibit their uncontrolled growth.

58.
State the effect of carcinogens on human body. Name the carcinogenic ionising and
non-ionising radiations. Mention their carcinogenic effects. (All India 2010C)

Carcinogens can transform normal cell into cancerous cells.

Carcinogenic ionising radiations are X-rays and gamma rays. Carcinogenic non-ionising
radiations are UV-rays.
These radiations cause changes in base sequences, i.e. mutations that lead to
transformation of normal cells into cancerous cell.

59.
(i) Explain the property that prevents normal cells from becoming cancerous.
(ii) All normal cells have inherent characteristic of becoming cancerous. Explain.

(i) For concerous cells Refer to No. 57.

(ii) All normal cells have ceUular oncogenes (c-onc) or proto-oncogenes. When activated
under certain conditions, such genes could lead to oncogenic transformation of cells, i.e.
they may become cancerous.

60.
Name a human disease, its causal organism, symptoms (any three) and vector spread by
intake of water and food contaminated by human faecal matter. (All India 2017)
Or
A patient is down with amoebiasis. List the symptoms that confirm this infection. Name
the causative pathogen. (Delhi 2015c)

Amoebiasis is a disease spread by intake of water and food contaminated by human

faecal matter. Causal organism Entamoeba histolytica. Symptoms

● Abdominal pain

● Constipation with cramps

● Faeces with excess mucus Vector or carrier of pathogen is housefly.

 61.
(i) What precaution(s) would you recommend to a patient requiring repeated blood
transfusion?
(ii) If the advise is not followed by the patient there is an apprehension that the patient
might contract a disease that would destroy the immune system of his/her body. Explain
with the help of schematic diagram only how the immune system would get affected
and destroyed. (Delhi 2017)

(i) Repeated blood transfusion may result in contracting diseases like AIDS. The
recipient must ensure that the donor’s blood is being screened for HIV and other
pathogens. Also, he should make sure that doctors are using fresh needles.

(ii) In the absence of such measures, the patient can get infected by HIV (Human
Immunodeficiency Virus), which causes AIDS. It is a threatening disorder that weakens
the immune system by attacking helper T-cells in the body. A schematic diagram
showing the cycle of proliferation and effects of retrovirus (HIV) in infected person is as
follows

62.
(i) It is generally observed that the children who had suffered from chickenpox in their
childhood may not contract the same disease in their adulthood.
Explain giving reasons the basis of such an immunity in an individual. Name this kind of
immunity.
(ii) What are interferons? Mention their role. (Foreign 2016)

(i) A child who had suffered from chickenpox in childhood may not contract disease in
his/her adulthood. It is because during the first encounter with pathogen (chickenpox)
specific antibodies (by humoral immune response) are produced to counter the attack.
During this attack, memory cells are also produced. Due to this, on subsequent exposure
to the same pathogen, the immune response is more rapid and intense. That is why,
second exposure to the chickenpox does not cause disease. It is known as acquired
immunity of the body.

(ii) Interferons are special kind of proteins secreted by virus infected cells. These protect
the healthy cells from the virus attacks.

63.
Certain attributes of innate immunity are given in the table below. Identify A, B, C, D, E
and F respectively in it. (Delhi 2016C)

s of barrier ple of the barrier ion

nt microbial growth

morpho nuclear leucocytes

ytokine

(i) A – Physiological barriers

B – Lysozyme in saliva

(ii) C – Cellular barriers

D – Fhagocytose and destroy microbes

(iii) E – Interferons
F – Prevention of viral infections

64.
State the three characteristics of acquired immunity. List the different ways by which it
can be attained by humans. (Delhi 2016C)

The characteristics of acquired immunity are

● It is pathogen specific.
● It is characterised by memory.

● Responses can be characterised as a primary response (of low intensity) and

secondary response.

Acquired immunity can be attained by humans in the following ways

● Active immunity The antibodies are produced in the host body as a response to
foreign entitities, i.e. on living microbes or other proteins. The onset of response is slow.

● Passive immunity The readymade antibodies are directly introduced to protect the
body against foreign agents.

65.
How are primary and secondary immune responses carried out in the human body?
Explain. (Delhi 2016C)

The primary response is the first response of immune system to a newly introduced
foreign agent, while a second intensified immune response to same foreign agent is the
secondary or anamnestic response. Immune responses are produced by two types of
lymphocytes

● B-Iymphocytes These produce an army of proteins in response to pathogens, i.e.

antibodies in the blood. The different types of antibodies secreted are IgA, IgM, IgE and
IgG. This response generated via antibodies of the immune system is also called the
humoral immune response.

● T-lymphocytes are the mediators of the cell-mediated immunity (CMI). The

cell-mediated immune response is responsible for the graft rejection.

66.
(i) HIV and Hepatitis-B are STDs. Mention the two other ways by which they can be
transmitted to a healthy person.
(ii) Why is early detection of STD essential? What can it lead to otherwise? Explain. (Delhi
2016C)

(i) STDs like AIDS and hepatitis-B can be transmitted to a healthy person in the following
ways

● Unsafe blood transfusion.

● From infected mother to foetus throught lactation.

(ii) Early detection of STD is essential for timely cure. Otherwise this can lead to Pelvic
Inflammatory Diseases (PIDs), abortions or even cancer of the reproductive tract. (IVi)

67.
A youth in his twenties met with an accident and succumbed to the injuries. His parents
agreed to donate his organs. List any two essential clinical steps to be undertaken
before any organ transplant. Why is the transplant rejected sometimes? What views
would you share with your health club members to promote organ donation? (Delhi
2015C)

Organ transplantation involves the removal of damaged/injured tissues or organs from

the body of a person and their substitution by similar tissues/organs from a donor.
Tissue matching, blood group matching are essential clinical steps before undertaking
any graft/transplant. Transplantation may result in the rejection of transplanted organs
as the immune system recognises the protein in the transplanted tissue or organs as
foreign and initiates cellular immunity. We should raise and promote awareness about
organ donation, about need of organ and tissue donors. There are lakhs of people
waiting for organ donation and many people die daily while waiting for transplant.
Organs and tissues from one donor can save upto 40-50% lives. So, we should
encourage and get registered for organ donation to save many lives.

68.
(i) State what happens in the human body when malarial parasites infected RBCs burst
to release the parasites in the blood.
(ii) Mention the specific sites in the host body, where production of
(a) sporzoites and
(b) gametocytes takes place in the life cycle of the malarial parasites. (Delhi 2015C)

(i) When the RBCs infected with malarial parasites burst open, they release a toxic
substance called haemozoin, which is responsible for the chill and high fever recurring
every three to four days.
(ii) (a) Sporozoites They are produced in the gut (inside oocyte on the surface of
stomach) of the female Anopheles mosquitoes.
(b) Gametocytes RBCs of human body.

69.
What is the functional difference between B-and T-cells. (Delhi 2015)

B-and T-cells are lymphocytes involved in immune responses generated by the host’s
body. Functional differences between B- Cells and T-Cells are as follows

phocytes phocytes

are formed and mature in bone marrow.are formed in bone marrow, but
ration occurs in thymus gland.
 produce antibody against antigen. The directly attach the antigen or direct
ne response produced is called humoras to produce antibody. They produce
ody mediated immunity. mediated immune response.

do not respond to organ transplantationrespond to organ transplantation.

70.
Mention any two human diseases caused by round worms. Name their causative agents
and their mode of transmission into the human body. (All India 2015C)

Roundworms are nematodes, that cause helminthic disease in humans.

Two human diseases caused by roundworms are

● Ascariasis It is caused by intestinal endoparasite of humans, Ascaris lumbricoides.

Infection occurs through contaminated vegetables, fruits and water.

● Filariasis It is caused by filarial worms, Wuchereria bancrofti and W. malayi. It is

transmitted by the bite of female Culex mosquito.

71.
At what stage is Plasmodium picked up by the female Anopheles? Describe the life cycle
of the parasite in this insect. (Delhi 2015C)

Female Anopheles pick up Plasmodium at gametocyte stage from human body.

Inside mosquito’s body, the gametocytes develop to form male and female gametes.
These gametes fuse and form zygote, which further divides to form many sporozoites in
the intestine of mosquito. These sporozoites later move into salivary glands. See text on
page no. 206.

72.
(i) Differentiate between benign and malignant tumours.
(ii) Why is colostrum a boon to the newborn baby? (Delhi 2015C)

(i) Differences between benign and malignant tumours are as follows

 n tumour nant tumour

ur remains confined to the umour spreads to other organs of the body. Rat
ted organ. The rate of tumour mour growth is usually rapid.
h is usually slow.

e tumours cause limited damage tumours have neoplastic cells that migrate to
sites of the body and start a new tumour
ever they land. This property is called metastasi

on-cancerous. ancerous.

(ii) The milk produced during the initial few days of lactation is called colostrum, which
contains IgA antibodies. These are essential to develop resistance in newborn babies as
they provide passive immunity. Thus, breast- feeding during the initial period of infant
growth is recommended by doctors for bringing up a healthy baby.

73.
A heavily bleeding and bruised road accident victim was brought to a nursing home. The
doctor immediately gave him an injection to protect him against a deadly disease.
(i) Write what did the doctor inject into the patient’s body?
(ii) How do you think this injection would protect the patient against the disease?
(iii) Name the disease against which this injection was given and the kind of immunity it
provides. (All India 2015)

(i) The doctor must have injected the tetanus vaccine into the patient.
(ii) The vaccine injection stimulates the body to make antibodies against the tetanus
toxin.
(iii) The disease is tetanus, which is caused by bacterium Clostridium tetani. It may enter
the skin through a cut or puncture wound. Once bacteria is under the skin, it makes a
toxin that causes severe and painful muscle spasms, which can even be fatal. The
injection provides passive immunity.

74.
Community service department of your school plans a visit to a slum near the school
with an objective to educate the slum dwellers with respect to health and hygiene,
(i) Why is there a need to organise such visits?
(ii) Write the steps you will highlight, as a member of this department, in your
interactions with them to enable them to lead a healthy life. All Indio 2Q14

(i) The community service department of schools plans a visit to a slum is educate them
about health, hygiene and nutrition. These, people are’always at risk of acquiring
infections due to poor hygiene. Therefore, there is always a need to organise visits to
slums so, as to educate and create awareness among them regarding the importance of
hygiene.

(ii) The points to be highlighted while interacting with the slum people may be

● Importance of cleanliness and hygiene of body as well as surroundings.

● Awareness and prevention of infectious diseases.

● Use of public facilities, i.e. toiletries.

● Consumption of properly cooked and hygienic food and water.

● Administration of vaccines to newborn children, so as to prevent diseases.

75.
(i) Name and explain giving reason, the type of immunity provided to the newborn by the
colostrum and vaccinations.
(ii) Name the type of antibody
(a) present in colostrum.
(b) produced in response to allergens in human body. (Foreign 2014)

(i) The immunity provided to the newborn by colostrum and vaccinations is called
passive immunity.

This is because both in colostrum and vaccines the antibodies conferred are not
produced by own body, but are rather transferred passively to recipient’s body. Such as
IgA antibodies pass through milk (colostrum) to infants and provides passive immunity
against infection.

(ii) (a) The type of antibody present in colostrum is IgA.

(b) IgE is produced in response to allergens in human body.

76.
(i) Name the causative organisms for the following diseases
(a) Elephantiasis
(b) Ringworm
(c) Amoebiasis
(ii) How can public hygiene help to control such diseases? Delhi 2014c

(i) The causative agents or organisms for the following diseases are
(a) Elephantiasis- Wuchereria bancrofti
(b) Ringworm- Microsporum
(c) Amoebiasis-Entamoeba histolytica

(ii) Maintenance of public hygiene includes

● keeping body and surroundings clean.

● consumption of clean drinking water, fruits and vegetables, etc.

● proper disposal of waste and excreta.

● regular cleaning and disinfection of tanks and other water reservoirs, etc.

● all the above measures help to control the increase in vectors of infectious
diseases and their breeding places. Thus, there would be reduced chances of
transmission of infectious diseases.

77.
Name the cells HIV attacks first, when it gains entry into a human body. How does this
virus replicate further to cause immunodeficiency in the body? (Delhi 2013C, 2010; All
India 2D10C)
Or
Trace the events occur in human body to cause immunodeficiency, when HIV gains entry
into the body. (Delhi 2011)

The HIV virus attacks the macrophages first in human body. The further replication of
virus causes immuno deficiency in the following way

● RNA is replicated to form viral DNA by the enzyme reverse transcriptase.

● Viral DNA gets incorporated into the host cell’s DNA and directs the infected cells
to produce viruses.

● Macrophages continue to produce virus particles and function as HIV factories.

● The virus particles enter helper T-lymphocytes in the blood, where they continue to
replicate and produce viral progenies.

● The number of helper T-lymphocytes progressively decreases in the body of the

infected person.

● With the decrease in number of T-cells, the immunity also decreases. The person is
unable to produce any immune response even against common bacteria like
Mycobacterium, parasites like Toxoplasma, viruses and fungi.

78.
Trace the life cycle of malarial parasite in human body, when bitten by infected female
Anopheles. (All India 2012)
Or
Trace the life cycle of Plasmodium in humans from the stage of entry until it is picked up
by the female Anopheles. (All India 2010)

Life Cycle of Malarial Parasite (Plasmodium) in Human Body

79.
Study a part of the life cycle of malarial parasite given below. the s that follows.

(i) Mention the role of A in the life cycle of the malarial parasite.
(ii) Name the event C and the organ where this event occurs.
(iii) Identify the organ B and name the cells being released from it. (Delhi 2012)

(i) A is female Anopheles mosquito, these mosquitoes act as vectors and transmit the
disease from infected to healthy individuals.
(ii) The event C is fertilisation. It occurs in the intestinal wall of mosquito.
(iii) B is salivary glands of mosquito, sporozoites cells are released from it.

80.
Study the diagram showing replication of HIV in humans and the following s
accordingly.

(i) Write the chemical nature of the coat A.

(ii) Name the enzyme B acting on X to produce molecule C. Name C.
(iii) Mention the name of the host cell D the HIV attacks first, when it enters into the
human body.
(iv) Name the two different cells the new viruses E subsequently attack. (All India 2011)

(i) A – Protein coat

(ii) B – Reverse transcriptase, A-viral RNA, C – Viral DNA
(iii) D – Macrophages (animal or human cell)
(iv) E – Macrophages and helper T-cells.

81.
(i) Name the causative agent of typhoid in humans.
(ii) Name the test administered to confirm the disease.
(iii) How does the pathogen gain entry into the human body? Write the diagnostic
symptoms and mention the body organ that gets affected in severe cases? All India
2011

(i) Salmonella typhi.

(ii) Widal test.
(iii) Pathogens enter the human body through contaminated food and water.
Diagnostic symptoms high fever, weakness, stomach pain. The body organ affected is
small intestine.

82.
An antibody molecule is represented as H2L2. Explain. (Delhi 2010.)

For Antibody H2L2, Refer to No. 44.

Refer to figure 8.2. on page no. 210.

83.
(i) All human beings have cellular oncogenes, but only few suffer from cancer disease.
Give reasons.
(ii) How is a malignant tumour different from a benign tumour? (Foreign 2010)

(i) Ail cells have cellular oncogenes (c-onc) or proto-oncogene, but only few suffer from
cancer disease because these genes code for certain growth factors. Under certain
conditions, they get activated and lead to oncogenic transformation causing cancer. This
transformation is induced by physical, chemical and biological factors called
carcinogens.

(ii) For differences between benign and malignant tumours, Refer to No. 72 (i).

84.
Under polio prevention programme, infants in India were given polio vaccines on a large
scale at regular intervals to eradicate polio from the country.
(i) What is a vaccine? Explain, how does it impart immunity to the child against the
disease.
(ii) With the help of an example each, differentiate between active and passive immunity.
(Foreign 2015)

(i) Vaccine is a preparation of inactivated or weakened pathogen of polio virus or protein

that is injected into a person to provide protection against disease. Refer to No. 34.
(ii) Refer to No. 33.

85.
(i) Cancer is one of the most dreaded diseases. Explain ‘contact inhibition’ and
‘metastasis’ with respect to disease.
(ii) Name the group of genes that have been identified in normal cells that could lead to
cancer. How do these genes cause cancer?
(iii) Name any two techniques that are useful in detecting cancers of internal organs.
(iv) Why are cancer patients often given a-interferon as part of the treatment? (Delhi
2014)

(i) Contact inhibition is the property exhibited by normal cells. It prevents their
uncontrolled proliferation when they are in contact with other neighbouring cells. But
cancerous cells seem to have lost this property and continue to divide despite being in
contact with other cells, which leads to masses of cells called tumours.
Metastasis is the property exhibited by malignant tumours which grows rapidly, invades
neighbouring tissues and is capable of reaching distant sites through blood and lymph
thus, spreading malignant tumours to other organs or parts of body. These two
properties make ‘cancer’ one of the dreaded diseases.

(ii) The group of genes called cellular oncogenes or proto-oncogenes in normal cells
could lead to cancer. These genes are present in inactivated or suppressed form. Some
factors, i.e. physical, chemical or biological called carcinogens are capable of activating
these oncogenes and thus, transforming normal cells into cancerous one.

(iii) The two techniques useful in detecting cancers of internal organs, are CT (Computed
Tomography) and MRI (Magnetic Resonance Imaging).

(iv) As tumour cells are capable of avoiding recognition and destruction by immune
system, the cancer patients are given a-interferons, which are biological response
modifiers. It helps in activating the immune system and destroy tumours.

86.
(i) Name and explain any four lymphoid organs present in humans.
(ii) Categorise the named lymphoid organs as primary or secondary lymphoid organs,
giving reasons. (Foreign 2014)

(i) The four lymphoid organs are

● Bone marrow Major lymphoid organs as both B and T-lymphocytes are formed
here and p-lymphocytes mature here only.

● Thymus T-lymphocytes mature in thymus and they are responsible for cell
mediated iinmupe response.

● Spleen Bean-shaped organ comprising of single mass of lymphoid tissues. In

foetal stage, it produces all type of blood cells but only lymphocytes are produced in
adult stage.

● Lymph nodes These are small solid structures composed of lymphoid tissue. They
produce lymphocytes and plasma cells and also act as filters for lymph.

(ii) The above described lymphoid organs, such as bone marrow and thymus can be
grouped under primary lymphoid organs, because these act as organs where both B and
T-lymphocytes mature and acquire their antigenic specificity. Whereas the spleen and
lymph nodes are considered as secondary lymphoid organs where the lymphocytes
undergo proliferation and differentiation. These are the site of acquired immune
response to antigens and formation of effector cells.

87.
A person in your colony has recently been diagnosed with AIDS. People/residents in the
colony want him to leave the colony for the fear of spread of AIDS.
(i) Write your view on the situation, giving reasons.
(ii) List the possible preventive measures that you would suggest to the residents of your
locality in a meeting organised by you so that they understand the situation.
(iii) Write the symptoms and the causative agent of AIDS. (All India 2013)

(i) AIDS is not contagious, i.e it does not spread by shaking hand, talking and use of
common utensils. So, there is no need of fear to live with the AIDS patient.
(ii) Some preventive and safe steps to be suggested are

● Taking out HIV affected blood from blood bank, ensuring the use of only
disposable needles and syringes in all public and private hospitals and clinics.

● Free distribution of condoms in public.

● Advocating safe sex and promoting regular check-up for HIV in population.

(iii) AIDS is caused by Human Immunodeficiency Virus (HIV), a retrovirus. This virus
attacks on T-helper cells, thus destroying the immune system.
The common symptoms of AIDS are weakness, fever, weight loss, regular illness, etc.

88.
Describe the asexual and sexual phases of life cycle of Plasmodium that causes malaria
in humans. (Delhi 2013)

For life cycle of Plasmodium, Refer to page no. 206.

89.
Mention the useful as well as the harmful drug obtained from the latex of poppy plant.
(Delhi 2013)

Morphine is obtained from latex of poppy plant. It is useful as an analgesic. Heroin

formed after acetylation of morphine is harmful as it is a depressant.

90.
How does smoking tobacco in human lead to oxygen deficiency in their body? (Delhi
2012)

Smoking increases carbon monoxide (CO) content in blood and reduces the
concentration of haem-bound oxygen. This causes oxygen deficiency in the body. (1)

91.
(i) Name the source plant of heroin drug. How is it obtained from the plant? (2018C)
(ii) Write the effects of heroin on the human body.20ia c

(i) Heroin is obtained from Papaver somniferum. It is extracted from the latex of the
plant.
(ii) Heroin is a depressant and slows down body function.
 92.
Why are adolescents especially advised not to smoke? How does smoking affect the
functioning of the body? (Outside Delhi 2016C)

Adolescents are advised not to smoke for the following reasons

● Smoking paves way for hard drugs.

● Smoking is associated with increased incidences of cancers of lung, throat and

bronchitis and emphysema. It also increases carbon monoxide content in blood and
reduces the concentration of haem-bound oxygen.

93.
Name two drugs obtained from poppy plant. ‘These drugs are medically useful, but are
often abused’. Taking the mentioned examples justify by giving reasons. (Delhi 2016 C)
Or
How are morphine and heroin related? Mention the effect each one of them has on the
human body? (All India 2014C)

Both morphine and heroin are extracted from the latex of plant Papaver somniferum.
Heroin is actually obtained by the acetylation of morphine. Thus, both heroine and
morphine are related.
Morphine acts as an effective sedative and pain-killer while heroin acts as depressant
and slows down body functions.

94.
What happens to an individual when a regular dose of drugs/alcohol is abruptly
discontinued? What characteristics manifest in the individual under such a situation?
(Outside Delhi 2016C)
Or
What is ‘withdrawal syndrome’? List any two symptoms it is characterised by. (Foreign
2014)

If the regular dose of drug or alcohol in an addicted person is discontinued abruptly, the
body exhibits characteristic and unpleasant symptoms called ‘withdrawal syndrome’.
The ‘withdrawal syndrome’ is characterised by symptoms like anxiety, nausea and
excessive sweating.

95.
Write the scientific name of the source plant of the drugs-marijuana and hashish and
mention their effects on human body. (Delhi 2014C)
Or
Name the plant source of ganja. How does it affect the body of the abuser? (All Indio
2012)
The scientific name of source plant of drugs marijuana, hashish and ganja is Cannabis
sativa. These drugs usually affect the cardiovascular system of human body.

96.
Name the plant source of the drug popularly called smack. How does it affect the body
of the abuser? (Delhi 2012)
Or
Name the opioid drug and its source plant. How does the drug affect the human body?
(All India 2010)

Smack is obtained from Papaver somniferum (poppy plant).

Drug’s affects

● It binds to specific opioid receptors present in our central nervous system and
gastro-intestinal tract.

● It is a depressant that slows down the body functions.

97.
Identify A, B, C and D in the following table. (Delhi 2012C)

tific name of the source plant ful effects/Human body part affect

ver somniferum essant/ slows body function

abis sativa abinoids

roxylum coca

A- Heroine
B – Cardiovascular system
C – Cocaine
D – Central nervous system

98.
Why is tobacco smoking associated with rise in blood pressure and emphysema (oxygen
deficiency in the body)? Explain. (All India 2011)

The nicotine present in tobacco stimulates

adrenal glands to secrete adrenaline and nor-adrenaline. Both these hormones increase
blood pressure and heart rates. ID
Smoking is associated with increased incidence of lung cancers. It increases carbon
monoxide level of the blood, which competes with oxygen for transport. As the
concentration of haem-bound oxygen decreases, there is oxygen deficiency in the body.
It also increases the craving for hard drugs.

99.
Why is there a fear amongst the guardians that their adolescent wards may get trapped
in drug/alcohol abuse? (All India 2017)

There is always a fear amongst guardians that their adolescents may get trapped in
drug/alcohol abuse due to following reasons

● Adolescence is accompanied by several biological and behavioural changes. It is a

vulnerable phase of mental and psychological development of an individual in which an
individual may get addicted to alcohol/drugs very easily.

● In this age, the first use of drugs or alcohol may be out of curiosity or
experimentation, which later on turns to addiction.

● Adolescents usually take drugs due to social pressure, need of adventure,

excitement to avoid stress, depression and frustration.

100.
Explain ‘addiction’ and ‘dependence’ in respect of drug/alcohol abuse in youth. (All India
2017)

Addiction is a psychological attachment to certain effects such as euphoria and a

temporary feeling of well-being associated with drugs and alcohol. These drive people to
take them without need or even when it becomes self-destructive. In the absence of any
guidance or counselling, the person gets addicted and becomes dependent on their use.

Dependence on drug/alcohol is the tendency of the body to manifest a characteristic

and unpleasant withdrawal syndrome, if regular dose of drugs/alcohol is discontinued
abruptly. Withdrawal symptoms are characterised by anxiety, shakiness, nausea and
sweating. Sometimes, it can be so severe that they may be life threatening.

101.
Prior a sports event blood and urine samples of sports persons are collected for drug
tests.
(i) Why is there a need to conduct such tests?
(ii) Name the drugs the authorities usually look for.
(iii) Write the generic names of two plants from which these drugs are obtained. (Delhi
2016)

(i) It is necessary to conduct these tests as sports-persons often take drugs to increase
their performance.
(ii) Cocaine and morphines are the drugs the authorities usually look for.
(iii) Morphine is extracted from the latex of poppy plant Papaver somniferum. Cocaine is
obtained from the coca plant Erythroxylum coca.

102.
A team of students are preparing to particfpate in the interschool sports meet. During a
practice session you find some vials with labels of certain cannabinoids.
(i) Will you report to the authorities? Why?
(ii) Name a plant from which such chemicals are obtained.
(iii) Write the effect of these chemicals on human body. (Delhi 2015)

(i) Yes, I will report it to the authorities, because cannabinoids are classified as drugs
and taking them without medical supervision is illegal.
(ii) Cannabinoids are obtained from various parts of plant Cannabis sativa.
(iii) Effect The cannabinoids interact with cannabinoid receptors in the brain and affect
the cardiovascular system of the body.

103.
Do you support ‘dope test’ being conducted on sports persons participating in a
prestigious athletic meet? Give three reasons in support of your . (All India 2014C)

Yes, the ‘dope test’ should be conducted on sports persons participating in a prestigious
athletic meet. This is done to find out if any participant had taken any kind of
performance enhancing drugs.
The use of drugs in sports should be banned as

● they increase muscle strength.

● promote aggressiveness.

● increase athletic performance.

Because of above reasons, use of such drugs, e.g. steroids, analgesics, diuretics should
be banned for participants as it would be unfair on the part of other participants (not
consuming such drugs).

104.
‘Prevention is better than cure’ is an apt slogan to safeguard adolescents from drug
abuse. List any 6 steps that could be taken in this regard. (All India 2013C)

Six steps that could be taken to prevent adolescents from drug alcohol abuse are as
follows

● A child should not be pushed unduly to perform beyond his/her limits in studies,
sports or any other activities.

● Educating and counselling him/her to face problems and stresses and accept
disappointments and failures as part of life.
● Parents and teachers can identify the danger signs and take appropriate steps to
diagnose the malady and underlying causes.

● Help should be taken from qualified psychologists and psychiatrists.

● Parents and teacher should become more supportive.

● Help of close friends and relatives can also be taken.

105.
Write the source and the effect on the human body of the following drugs
(i) Morphine
(ii) Cocaine
(iii) Marijuana (Delhi 2011)

(i) Morphine is obtained from the latex of Papaver somniferum. It is a depressant, which
slows down the body functions.
(ii) Cocaine is obtained from Erythroxylum coca. It is a stimulant and produces a sense
of euphoria and increased energy.
(iii) Marijuana is obtained from the inflorescence of Cannabis sativa. It affects
cardiovascular system of the body.

106.
(i) Name the drug used
(a) as an effective sedative and pain killer.
(b) for helping patients to cope with mental illness like depression, but often misused.
(ii) How does moderate and high dosage of cocaine affect the human body? (Foreign
2011)

(i) (a) Morphine is an effective sedative and pain-killer.

(b) Lysergic Acid Diethylamides (LAD) or barbiturates are often misused.

(ii) Moderate dose of cocaine have a stimulating action on central nervous system. It
produces a sense of euphoria and increased energy. High dosage of cocaine causes
hallucinations.

107.
Municipal corporation has deputed personals to check mosquito breeding in your
school. Which places they should check for mosquitoes and name two diseases which
are spread by them.

They should check water tanks, flower pots, etc. These are the places where mosquitoes
breed. Mosquitoes spread dengue and malaria.

108.
It is commonly observed that parents feel embarrassed to discuss freely with their
adolescent children about sexuality and reproduction. The result of this parental
inhibition is that the children go astray sometimes.
(i) Explain the reasons that you feel are behind such embarrassment amongst some
parents to freely discuss such issues with their growing children.
(ii) By taking one example of a local plant and animal, how would you help these parents
to overcome such inhibitions about reproduction and sexuality? (All India 2017)

(i) The reasons behind the embarrassment amongst some parents to freely discuss
sexuality and reproduction related issues are as follows

● Communication gap is a big reason for the same. The parents feel that talking
about such issues will have a negative impact on children.

● In India, sex-related issues are considered as taboos, so people feel awkward while
talking about them.

● Social beliefs are also responsible for this. Most parents think that there has to be
a V line of respect between parents and their children. Parents usually think that their
child is too young to discuss over this topic.

(ii) To overcome this inhibition, parents can make children understand about sexuality
via scientific perspective.
For example, in order to tell them about sexuality parents can take the example of
cucurbit and papaya. In cucurbits, both male and female reproductive structures are
present on the same plant, i.e, they are bisexual, while in papaya both male and female
reproductive structures are present on different plants, i.e. they are unisexual. Similarly,
in animals, earthworm is bisexual or hermaphrodite while cockroaches are unisexual.
The concept of reproduction can be taught in same way by citing the examples of
asexual reproduction in lower animals and sexual reproduction in higher animals.

109.
Modern life style in big cities and towns is surely making the life more easy and
comfortable for people. On the contrary, many more health issues and problems are on
the rise and one of them is allergic reactions.
(i) Write any four steps you would suggest to minimise the cause of the above allergic
responses.
(ii) List any two allergens. How does the human body respond to them? Explain. (Delhi
2014)

(i) Allergy is a hypersensitive reaction of the immune system to certain antigens present
in the environment. The following steps can be used to minimise the allergic reactions

● Provide a less protected environment in early childhood.

● Avoiding exposure to allergens like mites in dust, pollens, animal dander, etc.

● Use of drugs; like antihistamine, adrenaline and steroids.

● Exposing the patient to small doses of allergens and studying possible reactions.

(ii) Mist in dust and pollens are allergens. The human body responds to them by
producing IgE antibodies and releasing chemicals like histamine and serotonin from
mast cells.

110.
Peer pressure plays a negative role in triggering smoking habits in adolescents. As a
school captain list any two activities you would like to organise with the help of senior
students of your school and any other two activities you would like your school
authorities to organise for the students to tackle this problem. Explain how these
activities will help in doing so. (Delhi 2015)

As a captain, we will organise craft competition to create the awareness among the
students. Secondly, we will prepare some students to deliver debate during morning
assembly to make it more effective and with the help of school authorities, we will
prepare hoardings and put them Up on walls. We will also distribute some brochure
amongst the students including the lecture of principal during morning assembly, we can
also arrange a play on any auspicious occasion.

111.
You have attended a birthday party hosted by one of your classmates. You found some
guests at the party sitting in a corner making a lot of noise and consuming ‘something’.
After a while one of the boys from the group started screaming, behaving abnormally
and sweating profusely.
(i) Would you inform your parents/school authorities? Yes/No? Give reason in support of
your .
(ii) Prepare a note to be circulated amongst the schoolmates about the sources and
dangers of any two drugs.
(iii) Write any two ways that you will suggest to your school principal, so as to promote
awareness amongst the youth against the use of these drugs. (Foreign 2015)

(i) Yes, I will inform the school authorities, because such kind of behaviour may lead to
terrible consequences in future. It may lead to addiction to drugs. (1)
(ii) Following are two drugs that are most commonly available

s ces er

ne plant age to blood vessel, increased heart rate, even death

 uana abis sativa al retardation, lung infection, lung cancer.

(iii) Ways to promote awareness amongst the youth against the use of these drugs are
as follows

● All students must be inspired to adopt a healthy life style.

● There should be a counsellor who must talk to students about their problems and
situations that force them to adopt wrong habits.

112.
An active member of an awareness group conducts regular programmes to sensitise
public against alcoholism amongst youth as a serious health hazard in his locality.
Identify the values this member of the group is trying to propagate amongst the people
in his locality. (Value Based , Delhi 2013C)

Member of an awareness group is trying to aware public, commonly youth about the
harmful impacts of alcohol.
He wants to tell people that alcohol has several ill effects which affect the body of the
individual in many ways as follows

● Alcohol affects the foetus in case of pregnancy.

● It leads to reckless behaviour, vandalism and violence.

● It causes aggressiveness, rebellious behaviour and depression.

● Fatigue, isolation, fluctuations in weight are the other ill effects.

● Change in sleeping pattern, loss of appetite and lack of personal hygiene is also
seen in addicts.

Microbes in Human Welfare

1.
What are ‘floes’, formed during secondary treatment of sewage ? (Delhi 2019)

Floes are masses of bacteria held together by slime and fungal filaments to form
mesh-like structures. They are used during the secondary sewage treatment in the
aeration tank to increase the rate of decomposition.

2.
Why do we add an inoculum of curd to milk for curdling it? (Delhi 2015C)
Or
Why is ‘starter’ added to set the milk into curd? Explain. (All India 2014C)
Or
Name the nutrient that gets enhanced while curdling of milk by Lactobacillus? (All India
2014C)

When a small amount of curd as starter or inoculum is added to fresh milk, millions of
Lactic Acid Bacteria (LAB) present in the starter grow in milk and convert it into curd.
During this process, acids are produced by LAB that coagulate and partially digest the
milk proteins (casein). LAB increase vitamin-B12 content along with other vitamins in the
curd.

3.
How is lactic acid bacteria beneficial to us other than helping in curdling the milk? (All
India 2015C)

Two benefits of LAB are given below

● They improve the nutrient quality of curd by increasing the vitamin-B12 content.

● LAB also check the growth of disease causing microbes in the stomach.

4.
Give the scientific name of the source organisms from which the first antibiotic was
produced. (Foreign 2014)

The scientific name of the source organism, i.e. mould from which first antibiotic was
produced is Penicillium notatum.

5.
Name the gas released and the process responsible for puffing up of the bread dough
when Saccharomyces cerevisiae is added to it. (All India 2013c)

Saccharomyces cerevisiae (baker’s yeast) when added to dough causes its fermentation
and releases CO2 gas which is responsible for the puffed appearance of dough.

6.
Which of the following is the baker’s yeast used in fermentation ? Saccharum barberi,
Saccharomyces cerevisiae and Sonalika. (All India 2012,2011,2009)

Saccharomyces cerevisiae is the baker’s yeast used in fermentation.

7.
Write the scientific name of the microbe used for fermenting malted cereals and fruit
juices. (Delhi 2011)

Saccharomyces cerevisiae also called brewer’s yeast, is the microbe used for fermenting
malted cereals and fruit juices.
8.
Mention the information that the health workers derive by measuring BOD of a water
body. (All India 2010)

Biological Oxygen Demand or BOD value indicates the quantity of organic matter present
in the water. Higher the BOD of water body, more is its polluting potential and vice-versa.

9.
Why is sewage water treated until the BOD is reduced? Give a reason. (Delhi 2010C)

The higher the BOD of sewage water, more is its polluting potential. So, the sewage
water is treated, till its BOD-is reduced which further indicates the reduction in the
organic matter present in it.

10.
Distinguish between the roles of flocks and anaerobic sludge digesters in sewage
treatments. (Delhi 2016)

In sewage treatment, floes consume major part of the organic matter, converting it into
microbial biomass and releasing lot of minerals.
It reduces the BOD of sewage, while in anaerobic sludge digesters, many anaerobic
bacteria are present, which digest the organic mass. During this digestion, methane, CO2
etc. are produced.

11.
List the events that reduce the Biological Oxygen Demand (BOD) of a primary effluent
during sewage treatment. (Delhi 2016)

To reduce the BOD of primary effluent during sewage treatment, it is passed into large
aeration tanks with constant mechanical agitation and air supply. This allows vigorous
growth of aerobic microbes into floes which consume major part of organic matter in
the effluent. Hence, BOD of effluent is reduced.

12.
Explain the different steps involved during primary treatment phase of sewage. (All India
2015)

Primary treatment of sewage involves the physical removal of large and small particles
from sewage through filtration and sedimentation.
The steps involved in this process are

● Floating debris is removed by sequential filtration by passing through wire mesh

screens.

● After this, the grit (soil and small pebbles) is removed by sedimentation in settling
tanks. The sediment is called primary sludge and the supernatant forms the primary
effluent.
● The effluent is then taken for the secondary treatment.

13.
Mention a product of human welfare obtained with the help of each one of the following
microbes.
(i) LAB
(ii) Saccharomyces cerevisiae
(iii) Propionibacterium shermanii
(iv) Aspergillus niger (Delhi 2015)

be uct of human welfare

ccharomyces cerevisiae and cakes

ropionibacterium shermanii s cheese

spergillus niger acid

14.
Bottled fruit juices are clearer as compared to those made at home. Explain. (Foreign
2015)

Bottled fruit juices are clearer as compared to those made at home because, in these
juices pectinase enzyme is added, which digests the pectin and other fibres present in
juices.

15.
Name two groups of organisms which constitute ‘floes’. Write their influence on the level
of BOD during biological treatment of sewage. (All India 2014C)

The groups of organisms that constitute mesh-like structures called ‘floes’ are bacteria
and fungi.
These bacterial masses associated with fungal filaments called floes, consume the
major part of organic matter present in effluent, thereby reducing the BOD of the waste
significantly during biological or secondary treatment of sewage.
 16.
Name the bacterium responsible for the large holes seen in Swiss cheese. What are
these holes due to? (All India 2013)

Swiss cheese is produced by the bacterium Propionibacterium shermanii. The large

holes in Swiss cheese are due to the large amount of CO2 production.

17.
Name source of streptokinase. How does this bioactive molecule function in our body?
(Delhi 2012)
Or
Name the enzyme produced by Streptococcus bacterium. Explain its importance in
medical sciences. (All India 2011)

Streptokinase enzyme is produced by the bacterium Streptococcus. It is modified by

genetic engineering and is used as a clot buster for removing clots from the blood
vessels of patients who have suffered from myocardial infarction.

18.
Mention the importance of lactic acid bacteria to humans other than setting milk into
curd. (Delhi 2012)

● Lactic Acid Bacteria (LAB) are used to produce an acid called lactic acid that is an
important industrial product. It is also used in beverages, meat products, confectionary,
dairy products, etc.

● It checks the disease causing microbes in the stomach.

19.
Name the source of cyclosporin-A. How does this bioactive molecule function in our
body? (All India 2012.)
Or
Give the scientific name of the microbes from which cyclosporin-A and statin are
obtained. Write one medical use of each one of these drugs. (Foreign 2011)

Cyclosporin-A is produced by the fungus Trichoderma polysporum. It is used as an

immunosuppressive agent in organ-transplant patients as it suppresses the activation of
T-cells in body.

20.
Name the source of statin and state its action on the human body. (Foreign 2012)

Statin is produced by yeast Monascus purpureus. It is used as blood cholesterol

lowering agent.
It acts by competitively inhibiting the enzyme responsible for the synthesis of
cholesterol.
21.
Why are some molecules called bioactive molecules? Give two examples of such
molecules. (All India 2011)

Bioactive molecules are produced from microbes that are useful to other living
organisms in modifying their metabolism, e.g. streptokinase, cyclosporin-A, statins, etc.

23.
State the medicinal value and the bioactive molecules produced by Penicillium notatum,
Monascus perpureus and Trichoderma polysporum. (All India 2019,2015)

be tive molecule cinal value

illium illin as antibiotic against many fungal and bacteria

ses in humans and animals.

scus purpureus s as blood cholesterol lowering agent. It acts by

etitively inhibiting the enzyme responsible for
esis of cholesterol.

oderma polysporusporin-A as immuno suppressive agent in organ

plant patients.

24.
The three microbes are listed below. Name the product produced by each one of them
and mention their use. (2018C)
(i) Aspergillus niger
(ii) Trichoderma polysporum
(iii) Monascus purpureus

(i) Citric acid

(ii) Cyclosporin-A
(iii) Statins.

25.
Secondary treatment of the sewage is also called biological treatment. Justify this
statement and explain the process. (All India 2017)

The secondary treatment of sewage is also called biological treatment because in this
treatment, sewage is subjected to biodegradation. It means that it involves the
participation of microorganisms. The process of secondary treatment involves following
steps

● Primary effluent is passed into large aeration tanks with constant mechanical
agitation and air supply. This allows vigorous growth of useful aerobic microbes into
floes (masses of bacteria and fungi filaments).

● These microbes consume major part of organic matter in the effluent, while
growing. This reduces the BOD of the effluent.

● When BOD of sewage gets reduced, it is passed into settling tank. The bacterial
floes settle in tank and the sediment is called activated sludge. A, small amount of
activated sludge is pumped back into the aeration tank to serve as inoculum.

● The remaining major part of the sludge is pumped into large tanks called
anaerobic sludge digesters, where other kinds of bacteria, which grow anaerobically,
digest the bacteria and the fungi in the sludge.

During this process, bacteria produce a mixture of gases, such as methane, hydrogen
sulphide and the carbon dioxide, which form biogas. The effluent from secondary
treatment is generally released into natural water bodies. It helps to reduce water
pollution and water borne diseases.

26.
Describe how do ‘floes’ and ‘activated sludge’ help in sewage treatment. (Delhi 2017)

Roles of ‘floes’ and ‘activated sludge’ in sewage treatment are as follows

● Floes These are masses of bacteria held together by slime and fungal filaments to
form mesh-like structures. These are used during the secondary sewage treatment in the
aeration tank to increase the rate of decomposition.

● The microbes digest a lot of organic matter, converting it into microbial biomass
and releasing a lot of minerals. As a result, BOD of the sewage reduces. As the BOD of
waste is reduced to 10-15% of raw sewage, it is passed into settling tank. In these tanks,
floes are allowed to undergo sedimentation.

● Activated sludge The sediment of settling tank is called’activated sludge. A part of

it is used as inoculum in aeration tanks.

● The remaining part is passed into a large tank called anaerobic sludge digester. In
these tanks, anaerobic microbes are present that digest the organic mass as well as
aerobic microbes of activated sludge. The remaining sludge is used as manure or
compost.
27.
Make a list of three household products along with the names of the microorganisms
producing them. (All India 2016)

● Curd Lactobacillus

● Bread Saccharomyces cerevisiae

● Swiss cheese Propionibacterium shermanii.

28.
Determination of Biological Oxygen Demand (BOD) can help in suggesting the quality of
a water body. Explain. (Delhi 2015)

BOD is the amount of dissolved oxygen required for the microbial breakdown of
biodegradable organic matter. Aerobic organisms use a lot of oxygen and as a result,
there is a sharp decline in Dissolved Oxygen (DO) in the water body. This can cause
death of fishes and other aquatic species.

Determination of BOD is thus, an important parameter in determining the quality of a

water body. The presence of more organic waste increases the microbial activity thus,
decreasing the DO. BOD is higher in polluted water and lesser in clean water.

30.
Identify A, B, C, D, E and F in the table given below (Foreign 2014)

tific name of the organism uct produced n human welfare

tococcus tokinase modified

sporin-A

scus purpureus

bacillus milk into curd

The codes are identified as

A- Clot buster in patients who underwent myocardial infarction.
B- Trichoderma polysporum
C- Immunosuppressive agent in organ transplantation D-Statins
E – Blood cholesterol lowering agents
F – Lactic acid

32.
Identify A, B, C, D, E and F in the table given below (Delhi 2010C)

nism tive molecule

scus purpureus (yeast)

otic

sporin-A

A-Statins
B-They are used as blood cholesterol lowering agents,
C-Penicillium notatum
D-Penicillin
E-Trichoderma polysporum
F-Uscd as an immunosuppressive agent in organ transplant patients.

33.
Mention the product and its use, produced by each of the microbes listed below
(i) Streptococcus
(ii) Lactobacillus
(iii) Saccharomyces cerevisiae (All India 2010)

(i) Streptococcus Product is streptokinase. It is used as a clot buster for removing the
clots from the blood vessels of patients suffering from myocardial infarction.
(ii) Lactobacillus Product is lactic acid. It is used to convert milk into curd and improves
nutrient quality of curd by enriching it with vitamin-B12
(iii) Saccharomyces cerevisiae Product is ethanol and it is used in making bread and
beverages.

35.
List the events that lead to biogas production from waste water whose BOD has been
reduced significantly.
The gases from biogas are used as a source of energy because it is inflammable.

36.
(i) Name the category of microbes naturally occurring in sewage and making it less
polluted during the treatment.
(i) The category of microbes naturally occurring in sewage and making it less polluted
are bacteria and fungi, wherein masses of bacteria get associated with filaments of
fungi to form mesh-like structure called floes.

37.
Write any two places yhere methanogens can be found. (Delhi 2019)

● Rice fields

● Ruminants alimentary canal.

38.
Name the type of association that the genus Glomus exhibits with higher plants. (All
India 2014)

The genus-Glomus exhibits symbiotic association with higher plants called mycorrhiza.

39.
State one reason for adding blue-green algae to the agricultural soil. (Delhi 2014c)

Blue-green algae are added to agricultural soil because they add organic matter to the
soil and also increase its fertility.

40.
What makes the Nucleopolyhedrovirus a desirable biological control agent? (All India
2013C; 2012C)
Or
What is the significance of Nucleopolyhedrovirus in pest management?

Nucleopolyhedrovirus, a genus of baculoviruses is useful in controlling many insects and

other arthropods. They are species specific narrow spectrum bioinsecticide with no side
effects on plants, mammals, birds, fish and non-target insects. Therefore, they serve as
an important component of integrated pest management programme in dealing with
ecological sensitive areas. These properties are useful in organic farming.

41.
Mention the role of cyanobacteria as biofertilisers. (All India 2012)

Role of cyanobacteria as biofertilisers Cyanobacteria fix atmospheric nitrogen and

increase the organic matter of the soil through their photosynthetic activity.

42.
Mention two advantages of adding blue-green algae to paddy fields. (All India 2011)
In the paddy fields, cyanobacteria such as blue-green algae fix atmospheric nitrogen to
enrich the nitrogen content of soil. Therefore, the entire need of nitrogen to rice crop can
be supplied by blue-green algae, leading to increase in yield.

43.
Name any one symbiont, which serves as biofertiliser. Mention it’s specific role. (All India
2010C)

Rhizobium is a symbiotic bacteria that serves as biofertiliser.

The bacteria fix the atmospheric nitrogen into organic forms, which is used by the plants
as nutrients.

44.
Your advice is sought to improve the nitrogen content of the soil to be used for
cultivation of a non-leguminous terrestrial crop.
(i) Recommend two microbes that can enrich the soil with nitrogen.
(ii) Why do leguminous crops require such enrichment of the soil ? (2018)

(i) Azospirillum
Azotobacter

(ii) The leguminous plants have nodules in their roots. These root nodules are formed by
the symbiotic association of Rhizobium.

These bacteria fix atmospheric nitrogen and convert it into organic form. This organic
form of nitrogen is used later on by the plants nutrient. Therefore, they do not require
enrichment of the soil.

45.
How does the application of the fungal genus, Glomus, to the agricultural farm increase
the farm output? (Delhi 2017)

The application of Glomus to agricultural field increases the farm output by increasing
the nutrient availability to the crops. Glomus develops symbiotic association with the
roots of plants, called mycorrhiza. It absorbs phosphorus from the soil and passes it to
the plant it is associated with.
In return, it derives sugars from the host plant cells for its survival.
Thus, it acts as a biofertiliser. This association has other advantages also, like

● Resistance to root borne pathogens

● Tolerance to salinity and drought

● Increase in plant growth and development.

46.
How does the application of cyanobacteria help to improve agricultural output? (Delhi
2017)

Role of cyanobacteria in improving agricultural output Cyanobacteria are autotrophic

microbes found in aquatic and terrestrial environments. Most of these fix atmospheric
nitrogen, e.g. Anabaena, Nostoc, Oscillatoria, etc.
In paddy fields, cyanobacteria serve as an important biofertiliser as they enrich the
nitrogen content in the soil. They also add organic matter to the soil, thus increasing the
fertility. Thus, application of cyanobacteria helps in improving agricultural output.

47.
Name a genus of baculovirus. Why are they considered good biocontrol agents? (All
India 2016)

A genus of baculoviruses is Nucleopolyhedrovirus.

48.
What are methanogens? Name the animals in which methanogens occur and the role
they play there. (Delhi 2014)

Methanogens are the groups of anaerobic bacteria, that produce large amount of
methane.
Methanogens are found in the rumen of cattle and intestine of humans.
The methanogens present in the intestine of animals and humans act on cellulosic part
of food and digest them, thereby releasing methane along with CO2 and H2.

51.
How do plants benefit from having mycorrhizal symbiotic association?

Fungi form symbiotic association with plants, which is called mycorrhiza. The fungal
symbiont in these associations absorbs phosphorus from soil and passes it to the plant.

It also provides resistance to root borne pathogens and increases plant growth. Thus, it
acts as a biofertiliser. The fungi belonging to the genus -Glomus form mycorrhizal
associations with plants.

52.
How do methanogens help in producing biogas? (Delhi 2012)

Methanogens grow anaerobically on cellulosi material and produce large amount of

methane along with CO2 and H2. Since, biogas is a mixture of methane and CO2,
methanogens help in its production.

53.
Why is Rhizobium categorised as a symbiotic bacterium? How does it act as
biofertilisers? (Deihi 2012)
The nodules on the roots of leguminous plants are formed by Rhizobium bacteria for
their survival. These bacteria fix atmospheric nitrogen into organic form, which is used
by the plant as nutrients.

Since, Rhizobium forms symbiotic association with leguminous plants, these are
considered as symbiotic bacteria.
Rhizobiiim fixes the atmospheric nitrogen into organic form, i.e. nitrates which can be
utilised by the plant as nutrient. So, it is used as biofertilisers.

54.
Name a free-living and a symbiotic bacterium that serves as biofertiliser. Why are they
called so? (All India 2010C)

Azotobacter is a free-living bacteria serving as a biofertiliser. These bacteria absorb

free-nitrogen from the soil, air and convert it into salts of -nitrogen compounds and
enrich the soil nutrients.
Rhizobium is a symbiotic bacteria that lives in the root nodules of legumes and fixes
atmospheric nitrogen info organic compounds and enrich the soil nutrients.

57.
(i) Organic farmers prefer biological control of diseases and pests to the use of
chemicals for the same purpose. Justify.
(ii) Give an example of a bacterium, a fungus and an insect that are used as biocontrol
agents.

(i) Organic farmers do not use any chemical for raising crops. They simply depend on
biological control methods to control insects and pests. This way they avoid deleterious
effects of chemicals on food products as well as on the environment. These chemicals
get accumulated in food chain and ecosystem whereas biological control methods are
safe as they do not harm any form of life.

(ii) Examples of biological control agents are as follows

● Bacterium Bacillus thuringiensis for the control of cotton bollworm.

● Fungus Trichoderma species for controlling fungal soil borne diseases like
damping off of vegetables.

● Insects Ladybird beetle for the control of aphids.

59.
How are baculoviruses and Bacillus thuringiensis used as biocontrol agents? Why are
they preferred over readily available chemical pesticides? (All India 2014C)

Bacillus thuringiensis as Biocontrol Agent

● Through genetic engineering, the gene coding for the toxic protein is introduced
into crop plants, which makes them resistant to insect pests.

● When they are eaten by the larvae, the toxin becomes active in the gut of larvae
and kills the larvae.

● They are available in sachets as dried spores, which have to be mixed with water
and sprayed onto vulnerable plants.

Baculoviruses uses (Nucleopolyhedrovirus) as biocontrol agents

Biological control of pests and pathogens must be preferred over conventional use of
chemical pesticides because

● the chemicals cause pollution to soil, ground water and agricultural products.

● the chemicals are toxic and harmful to both human beings and animals.

● overuse of chemical fertilisers makes soil infertile.

● chemicals kill harmful as well as useful organisms indiscriminately.

60.
How is the Bt cotton plant created as a GM plant? How is it protected against bollworm
infestation? (Delhi 2013C)

Bt cotton is created by using some strains of a bacterium known as Bacillus

thuringiensis. This bacterium produces protein that kills certain insects such as
Lepidopterans, Coleopterans and Dipterans. Bt gene is cloned from this bacteria and had
been expressed in cotton plant to provide resistance to the insects without the need for
insecticides.

It is protected against corn borer disease by encoding Cry protein with the gene cry IAb.
These genes produce Bt toxins which are released in the gut of insects, who feed upon
them, thus killing them and protecting the plant from the pests.

61.

The diagram above is that of a typical biogas plant. Explain the sequence of events
occurring in a biogas plant.
Identify A, B and C. (Delhi 2011)
The sequence of events occurring in a biogas plant are as follows

● The biogas plant tank is fed with a slurry of dung.

● A floating cover is placed over the slurry, which keeps on rising as the gas is
produced in the tank due to the microbial activity.

● Methanogens like Methanobacterium grows anaerobically on cellulosic plant

material in cow dung to produce large amount of methane, CO2 and H2.

● The plant has an outlet, which is connected by a pipe to supply biogas in nearby
houses.

● The spent slurry is removed through another outlet and used as biofertilisers.
A-Sludge tank
B-Gas holder
C-Dung water

62.
(i) Why do farmers prefer biofertilisers to chemical fertilisers these days? Explain.
(ii) How do Anabaena and mycorrhiza act as biofertilisers? (Delhi 2011)

(i) A farmer relies on biofertilisers then chemical fertilisers because

● Chemical fertilisers significantly increase the soil pollution and reduce quality of
soil, cause water pollution, when it drains into nearby water bodies, after rain.

● Overuse of chemical fertiliser makes the soil unfit for raising any crop.

(ii) Anabaena fixes atmospheric nitrogen, thus enriching the nitrogen content of the soil,
as well as the organic matter.

In mycorrhiza, the fungal symbiont absorbs phosphorus from the soil and passes it to
the plant and provides resistance to root borne diseases. Since, they fulfil the nitrogen
and phosphorus requirement, they act as biofertilisers.

63.
Name the genus to which baculoviruses belong. Describe their role in the. Integrated
Pest Management (IPM) programme. (Delhi 2011; Foreign 2011)

Baculoviruses belong to the genus-Nucleopolyhedrovirus.

Their role in IPM is as follows

● Baculoviruses are pathogens that attack insects and other arthropods.

● The majority of baculoviruses act as biocontrol agents.

● These viruses are excellent for species-specific, narrow spectrum insecticidal
applications.

● They do not show negative impacts on plants, mammals, bird, fish or even
non-target insects. Therefore, they play an important role as biocontrol agent.

64.
An organic farmer relies on natural predation for controlling pests and diseases. Justify
by giving reasons, why this is considered to be a holistic approach? (Foreign 2010)

Organic farming is a holistic approach that seeks to develop an understanding of the

webs of interaction among the myriads of organisms that form the flora and fauna of the
field.

● An organic farmer works to create a system, where the insects are not eradicated,
but kept at manageable level by a complex system of checks and balance within a living
and vibrant ecosystem.

● Organic farming states that the eradication of pests is not only possible, but also
undesirable, because many beneficial predatory and parasitic insects cannot survive
without them.

● This use of biocontrol methods reduces the use of chemical pesticides and
thereby pollution.

65.
What are biofertilisers? Describe their role in agriculture. Why are they preferred to
chemical fertilisers? (Foreign 2015)

Biofertilisers are the living organisms that promote the growth of plants by replenishing
the nutrients in the soil. These include bacteria, fungi and cyanobacteria.

Role of Biofertlisers in Agriculture Some biofertilisers such as Rhizobium bacteria live in

symbiotic association with plants. They live within the root nodules of leguminous
plants. These bacteria fix atmospheric nitrogen and enrich the nitrogen content of soil.

Fungi such as Glomus forms symbiotic association with plants (mycorrhiza) by

absorbing phosphorus and passing it to plants. Cyanobacteria such as Nostoc and
Anabaena fix atmospheric nitrogen and act as biofertilisers especially in paddy fields.

Biofertilisers are preferred over chemical fertilisers. Refer to No. 26 (i).

66.
Name the microbes that help in the production of the following products commercially.
(i) Statin
(ii) Citric acid
(iii) Penicillin
(iv) Butyric acid

(i) Statin – Monascus purpureus

(ii) Citric acid – Aspergillus niger
(iii) Penicillin – Penicillium notatum
(iv) Butyric acid – Clostridium butylicum

67.
Choose any four microbes, from the following which are suited for organic farming
which is in great demand these days for various reasons. Mention one application of
each one chosen.
Mycorrhiza, Monascus, Anabaena, Rhizobium, Methanobacterium, Trichoderma.

The four microbes that can be chosen for organic farming are

● Rhizobium The nodules on the roots of leguminous plants are formed by the
symbiotic association of Rhizobium bacteria.
These bacteria fix atmospheric nitrogen into its organic form, which is used by the plants
as nutrient.

● Mycorrhiza (Glomus) Many members of genus-Glomus form symbiotic association

with plants called mycorrhiza. The fungal symbiont absorbs phosphorus from soil and
passes it to the plant. Mycorrhiza shows resistance to root-borne pathogens, tolerance
to salinity and drought and helps in overall increase in plant growth and development.

● Anabaena It is a cyanobacteria that is used as a biofertiliser. It fixes atmospheric

nitrogen.

● Trichoderma It is a free-living fungi that is very common in the root ecosystems. It

is a very effective biocontrol agent of several plant pathogens.

68.
(i) How do organic farmers control pests? Give two examples.
(ii) State the difference in their approach from that of conventional pest control
methods.

(i) The organic farmers control pests by the use of insect pests resistant varieties. The
two examples are

● The Pusa Gaurav variety of Brassica is resistant to aphids.

● Pusa Sawani variety of okra is resistant to shoot and fruit borer.

(ii) The use of resistant variety is safer to control the pests as it does not involve
chemical pesticides which are used in conventional method of controlling pests. Thus, it
is environmental friendly method and reduces soil pollution.