Download to view

RESOURCE MATERIAL
BIOLOGY
CLASS-XII 2015-16
RESOURCE PERSON: Dr Komal Sahi
Alka Pradhan
Mentored by
N Mandal
Principal, APS Sukna
QUESTION BANK /BIOLOGY/XII
Page 1
HOW TO PREPARE BIOLOGY FOR CLASS 12 BOARD EXAM
Biology is a very important subject for class 12 board exam and various medical
exams like AIPMT and AIIMS. In board exam, Biology carries 100 marks with 70
marks in theory and 30 marks in practical. Hence students aspiring to appear in
AIPMT, AIIMS and other state medical entrance exam must concentrate on Biology.
First of all, let us take a brief look at marks distribution of various chapters of
Biology. Theory paper of Biology carries 70 marks in the board exam.
Sr. No.
Unit
Marks Distribution
(1)
Reproduction
14
(2)
Genetics and Evolution
18
(3)
Biology in Human Welfare
14
(4)
Biotechnology and its Applications
10
(5)
Ecology and Environmental Issues
14
Looking at the above mentioned marks distribution it is wise to
prepare Reproduction, Biology in Human Welfare and Ecology units first as
these are simple and together carry 42 marks in the board exam. You need to learn
lots of topics and terms in these units hence these units should be completed
first. Genetics and Biotechnology are related to each other and hence these two
units require more concentration and should be prepared together.
Biology doesn’t only involve learning
Often students complain that in Biology they only need to learn various biological
terms which is quite boring. But chapters like Genetics and Biotechnology do not
require only learning work. You need to understand them well to score better in the
exam. If you succeed in understanding different topics of Genetics then it will
become the easiest chapter in Biology. Otherwise you will find it difficult to prepare
Genetics and Biotechnology.
How to learn difficult Biology terminology?
The best tip to learn the difficult biological terminology is to write them on paper
many times. By writing you can memorize any biological terms. Suppose you need
to learn the genus of a Baculovirus i.e. “Nucleopolyhedrovirus”. Undoubtedly it is
a big term but you can easily learn it by writing the term repeatedly on paper.
Importance of Diagrams in Biology
Diagram is very important in Biology. Even in medical exams like AIPMT and AIIMS,
many diagram based questions are asked these days. Hence you should fully
concentrate on various biological diagrams. There are many questions in Biology
QUESTION BANK /BIOLOGY/XII
Page 2
like- How Hershey and Chase proved that DNA is the genetic
material? Theoretically it is a big question but you can answer this question by
drawing diagram of the experiment conducted on Bacteriophage by Hershey and
Chase too. And you will get full marks for it in the board exam.
Importance of reading NCERT Books
NCERT is the best book for preparing Biology for class 12 board exam. If you will
take the help of various sidebooks for board exam preparation neglecting NCERT
book then it will only elongate your syllabus. But you can take the help of sidebooks
to understand different topics of NCERT. Try to read NCERT Biology book daily and
give equal emphasis on revising learned chapters. Proper revision is mandatory to
secure good marks in Biology. Try to solve objective questions of Biology more and
more to attempt 1 mark question in the board exam. You can take the help of
sidebooks for medical exam preparation. NCERT is really important as some
questions are asked directly from the book. Take Pre-Boards seriously. Though
your performance in it is no reflection of your 'to be' performance, but do prepare for
them. Practice few model test papers before exams. They also play importance in
your performance. Don’t overdo them, around 5 would be enough. Don’t worry and
relax. Try to avoid any kind of pressure, anxiety or frustration during preparation
day. Talking to friends and teacher is of great help.
It is very important for you to set a realistic schedule. Set targets that are achievable
and try to achieve these targets even if you have to burn the midnight oil. Class XII
can be very stressful as students have to prepare for both Board exams as well as
entrance. However, one must make a timetable and stick to it. Work hard for two
years and then you can reap its benefits forever. These are important if you want to
achieve your dream. Besides studying regularly, one should also listen to their
teachers and get all their doubts clear. Teachers play a very important role in your
success.
Class 12th is very important for your career. Hence, I would recommend that you
stay away from all possible distractions and stay focused towards your goal. Your
commitment to your goal and hard work will never go waste. Cracking Board exams
is no rocket science. Once you score well in your class XII and entrance exams,
your career will be set. Also try to listen to your teachers when they’re telling you
something. However, do not take unnecessary stress. Relax and give yourself
breaks from time to time.
So, Cheers!
All the best.
N Mandal
Principal
APS SUKNA
QUESTION BANK /BIOLOGY/XII
Page 3
CONTENTS
1. Reproduction in organism
2. Sexual reproduction in flowering plnts
3. Human reproduction
4. Principles of inheritance and variations
5. Heredity and variations
6. Molecular basis of inheritance
7. Evolution
8. Health and disease
9. Strategies of enhancement in food production.
10. Microbes in Human Welfare
11. Biotecnology: principles and processes
12. Biotechnology and its Applications
13. Organism and populations
14. Ecosystem
15. Biodiversity and conservation
16. Environmental issues
 sample question papers
QUESTION BANK /BIOLOGY/XII
6-11
12-21
22-32
33-36
37-72
73-98
99-113
114-126
127-134
135-144
145-164
165-172
173-179
180-184
185-190
191-198
199-217
Page 4
DESIGN OF QUESTION PAPER
BIOLOGY (Code - 044)
CLASS XII
UNIT-WISE BLUE PRINT
Unit
Marks
14
18
14
10
14
70
1. Reproduction
2. Genetics and Evolution
3. Biology and Human Welfare
4. Biotechnology and its Applications
5. Ecology and Environment
Total
CHAPTER-WISE BLUE PRINT
Sr.
No.
UNIT
VSA
SA-I
SA-II
VBQ
LA
TOTAL
1
Reproduction
1 (1)
1 (2)
2 (6)
-
1 (5)
5 (14)
2
Genetics and
Evolution
Biology and
Human
Welfare
1(1)
1(2)
2 (6)
1 (4)
1 (5)
6(8)
1 (1)
2 (4)
3 (9)
-
-
6 (14)
-
1 (2)
1 (3)
-
1 (5)
3 (10)
2 (2)
-
4 (12)
-
-
6 (14)
5 (5)
5 (10)
12 (36)
1 (4)
3 (15)
26 (70)
3
4
5
Biotechnology
and its
Applications
Ecology and
Environment
TOTAL
NOTE:
1. There is no overall choice.
2. However, an internal choice has been provided in one question of 2 marks, one
question of 3 marks and all the three questions of 5 marks weightage
QUESTION BANK /BIOLOGY/XII
Page 5
TYPOLOGY OF QUESTIONS
QUESTION BANK /BIOLOGY/XII
Page 6
Chapter-1.REPRODUCTION IN ORGANISMS
VERY SHORT ANSWER TYPE QUESTIONS (1marks)
1) No two individuals, especially in mammals (except monozygotic twins) look
alike. What distinguishes them from the rest?
Ans. Monozygotic twins develop from a single zygote. In rest of the individuals,
the zygote develops directly into an individual
2) A plant was introduced in India because of its beautiful flowers & shape of
Leaves. It propagates vegetatively at a phenomenal rate and spread all over the
water bodies and has become a threat. Which plant is being referred to?
Ans. Water hyacinth. It grows in abundance as result oxygen depletion occurs.
3) A few plants exhibit unusual flowering phenomena which flowers once in
their life time and die. Suggest an evidence of the same from monocotyledon.
Ans.Bamboo.
4) Identify A and B Label (1) and (2) in the given figure 1
Ans.A- Penicillium 1. ConidiaB-Sponge 2.Gemmules
5) The Nilgiris belt form the large blue stretches in the hilly areas of Kerala,
Karnataka, & Tamilnadu and attracted a large number of tourists. What reason
do you attribute to this phenomenal change of nature once in 12 years?
Ans.Its shows mass flowering of Strobilanthuskunthaiana.
6) Why estrous cycle is not seen in humans?
Ans.Man is an advanced animal, with more reproductive capacity.
QUESTION BANK /BIOLOGY/XII
Page 7
7) All papaya and date palm plants produce flowers yet only few papaya and
date palm seen to produce fruit. What could be the possible reason for the rest
not producing them?
Ans.Papaya and date are dioecious plants.
8) Often the number of male gametes produced in an organism is in large
number as compared to female. Why has nature taken up this developmental
disparity?
Ans.Male gametes are motile in nature and they are destroyed.
9) This figure shows fruit containing seed inside. Label S & P shown in fig.
Ans.S- Seed, P-Protective pericarp.
10) In nature for both plants & animals, hormones are responsible for transitions
between the three phases of reproduction. Which 3 phases are being referred to
here?
Ans.a) Pre-fertilization b) Post-fertilization c) Embryogenesis
11) Name an organism where cell division itself is a mode of reproduction.
Ans.Unicellular organisms.
12) Mention the characteristic features and a function of zoospores in some algae.
Ans.zoospores are motile,microscopic and thin-walled.They are asexual reproductive
structures.
13)Why can't man be oviparous? Justify the statement.
Ans. Man is a placental animal having adaptation to give birth.
14) Differentiate between oviparous and viviparous animals.
Ans.Oviparous-embryo develops outside the body in the female eg.reptiles.
Viviparous-embryo develops inside the body of the female eg.Humans.
15)Mention one difference between monoecious and dioecious plant.
QUESTION BANK /BIOLOGY/XII
Page 8
Ans.When it bears both male and female flowers on the same plant it is monoecious.if it
bears exclusively either male or female then it is dioecious.
16)How is continuity of a species maintained for generation?
Ans.continuity of a species maintained for generation through reproduction.
17) Offspring derived by asexual reproduction are called clones.justify.
Ans.they are genetically and morphologically identical among themselves as well as to the
parent.
18) Mention the characteristic feature and function in some algae.
Ans.zoospores are microscopic motile structures.they are asexual reproductive units.
19) Name the phase all organisms have to undergo before they can reproduce sexually.
Ans.juvenile phase.
20) All papaya plants bear flowers, but fruits are seen only in some.explain
Ans. Papaya is dioecious,the which produce male flowers do not bear fruits while the female
flower bearing plants give fruits.
21)Why is the offspring formed by asexual reproduction referred to as term clone?
Ans. Offspring are morphologically and genetically similar to their parents and they are
produced asa result of mitosis, i.e. ASEXUAL REPRODUCTION
22)What may happen if meiosis does not take place during gametogenesis?
Ans.Gametes will be diploid and they are not viable gametesSyngamy will not takesplace
between the diploid gametes
SHORT ANSWER TYPE QUESTIONS (2 marks)
23)Label A and B and write the type of the vegetative propagule.
A
B
Ans.A-Eyes B-Germinating eye bud Type-Tuber
QUESTION BANK /BIOLOGY/XII
Page 9
24)(i) Though ginger is found under the soil. Yet it is not a root, but stem.
Justify your answer with two reasons.
(ii) What are the specialized cells which undergo meiosis in the diploid
organisms,called as?
Ans.(i) It is a modified stem and has nodes, internodes and bud.
(ii) Cell division and cell differentiation.
25)Mention the site of zygote formation in the ovule of a flowering plant. What
happens to sepals, petals and stamens after fertilisation? State the fate of zygote, ovule
and ovary in these plants.
Ans.Embryo sac.
Sepals, Petals and Stamens dry and fall off. Zygote develops into embryo.
Ovule develops into seed and ovary into fruit
26) (i) Name the process of development of embryo from the zygote.
(ii) What are the two changes which the zygote undergoes during this
process?
Ans. i)Embryogenesis ii)cell differentiation and organogenesis.
27)Why do algae and fungi shift to sexual mode of reproduction just before the onset of
adverse conditions?
Ans.As asexual reproduction may produce large population that may not survive due to lack
of resources.sexual reproductions brings a variations which might help the individual to
adapt to the changed conditions and survive. Thus insuring continuity of species.
28) Why dogs and cats have oestrus cycle but human beings have menstrual
cycle, though all are mammals?
Ans.Dogs and cats are seasonal breeders having heat period during which
Ovulation takes place but human females have this cycle every month.
29) In bisexual flowers, why is the transfer of pollen grains easier than in
unisexual flowers? Name the specialized event in unisexual flowers which helps
in transfer of pollen.
Ans.Both the reproductive organs stamens and pistils are present in the same
flower close to each other but in unisexual flower they are present in two
different flowers. ii) Pollination
30) Can we refer the off -springs formed by asexual reproduction, as a clone?
If yes, why?
QUESTION BANK /BIOLOGY/XII
Page 10
Ans.Yes, since all the organisms are exactly similar to the parents
31) The posterior end of cockroach shows the following structures. Mention if it
is a male or female. Also label the part marked 'X'.
Ans.The figure indicates part of female cockroach, name of the part- ovary.
32) Arrange the following events in proper sequence:(a) Embryogenesis
(b) Fertilization
(c) Gametogenesis
(d) Zygote formation.
33) Unicellular organisms are immortal whereas multicellular are not justify.
Ans.Unicellular organisms reproduce by cell division there is no natural death for them and
hence they are considered immortal.In multicellular,reproduction occurs in specialeised
organs involving specialized cells.their body as whole dies due to aging and senescence.
34) State the difference between meiocyte and gamete with respect to chromosome
number.Why is whiptail lizard referred as parthenogenetic?
Ans. Meiocyte is a diploid cell,while a gamete is haploid.whiptail lizard referred as
parthenogeneticas it develops from female gamete without fertilization.
35) How does the floral pattern of mediterrnean orchid Ophrys guarantee cross
pollination?
Ans.orchidOphrys resembles the female wasp in colour,smell and appearance.the male
pollinators mistake them as female.Therefore in the process of pseudocopulation they
pollinate the flower.
QUESTION BANK /BIOLOGY/XII
Page 11
36)In oogamous organisms, female gamete is large and non-motile but the male gamete
is very small. Why such type of adjustment is there in higher organisms?
Ans.Female gamete is large and non motile, is an adaptation for storing more food which will
be required for the future development. The male gamete has to move to reach the
counterpart, so it has the machinery for its reaching and delivering the chromosomes. It is
therefore both the gametes have specialized themselves for their functions
37)What will happen if meiosis does not take place during gametogenesis?
Ans. Gametogenesis, fertilization, zygote embryogenesis.Gametes will be diploid.
38) coconut palm is nomoceious while date palm is dioecious.why are they so called?
Ans.
Cocnut is monoecious because both male and female flowers borne on the same plant.while
date palm bears exclusively either male or female flowers.
QUESTION BANK /BIOLOGY/XII
Page 12
Chapter-2.SEXUAL REPRODUCTION IN FLOWERING PLANTS
VERY SHORT ANSWER TYPE QUESTIONS (1marks)
1)Do pollen grains survive in adverse conditions?
Ans.Yes, sporopollenin is present in exine which is not digested by enzymes.
2) Non- albuminious seeds do not have endosperm, then from where do they take the
food during germination?
Ans.Theytake food from cotyledons
3) T.S. of anther shows four layers in the wall-epidermis, endothelium, tapetum and
middle layer. Arrange them from outermost toinnermost.
Ans.outer most- Epidermis,Middle layer-Endothecium,Inner most-Tapetum
4) Identify the figure given below and label the parts indicated in the figure.
Ans. A. vacuolesB. Nucleus,Development of pollen grain
5) Complete the flow chart.
Ans. (i) Pollination
(ii) Geitonogamy
6) If the number of chromosomes in the leaf cell of a flowering plant is 28, What
number would you expect in the embryo and endosperm?
Ans.Embryo – 28.Endosperm - 42
7) Why do pollen grains of valliesneria have a mucilaginous covering?
QUESTION BANK /BIOLOGY/XII
Page 13
Ans.Presence of mucilaginous covering to protect from desiccation in water.
8)Why are pollen grains produced in enormous quantity in Maize?
Ans.To ensure pollination because Maize is pollinated by wind
9)In same species of Asteraceae and grasses, seed are formed withoutfusion of gametes.
Mention the scientific term for such form of reproduction.
Ans.Apomixis
10) Name the type of flower which favours cross-pollination?
Ans.Chasmogamous which are similar to flowers of other species with exposed anthers and
stigma favours cross pollination.
11) Why is typical embryo sac referred to as monosporic?
Ans.Since one of the four megaspores formed develops into the embryo sac while the others
degenerate the embryo sac is monosporic.
12) Outer envelope of pollen grain made of a highly resistant substance. What
is that substance? At which particular point the substance is not present?
Ans. Sporopollenin; at germpore sporopollenin is absent.
SHORT ANSWER TYPE QUESTIONS (2 marks)
13) (a) "The microspore is haploid while that of microspore mother cell is
diploid"comment.
(b) How many male gametes and female gamets are produced by?
(i) Five microspore mother cell (ii) Five megaspore mother cell
Ans.(a) MMC under goes meiotic division to form microspores.
(b) (i) Male gametes-20
(ii) Female gamets-5
14) (a) what is the process shown in the diagram given below?
(b)Name the structure at (a) of the figure given below
QUESTION BANK /BIOLOGY/XII
Page 14
8) (a) Microsporogenesis
(b) Pollen grain tetrad.
(a) Microsporogenesis
(b) Pollen grain tetrad
15) Why do you think that the zygote is dormant for some time in a fertilized ovule?
Ans.Zygote divides only after the formation of endosperm because endosperm nourishes
the developing embryo.
16) What will be the fate of ovule if the synergids are absent in the embryo sac?
Ans.Synergidshavefiliform apparatus at the micropylar end which guides pollen tube
into the egg apparatus, otherwise pollen tube may not enter to embryo sac, nofertilization.
17) Your friend would like to cross-pollinate the bisexual flower. How can you guide
him to be successful in his experiment?
Ans.Emasculation followed by artificial pollination, i.e. Pollens of selected plants have
to be dusted on the stigma before bagging the flower.
18) What is geitonomgamy? State its differenceswith xenogamy.
Ans.Geitonogamy
1. Transfer of pollen grains from the another to stigma of another
2. Does not provide opportunity for gametic recombination.
Xenogamy
1.Transfer of Pollen grains from another flower of the same plant.
2.Provide opportunity for gametic recombination
19) A student wants to know the ploidy of coconut. After studying its different parts he
inferred the ploidy of the following parts. Check whether the student is correct.
QUESTION BANK /BIOLOGY/XII
Page 15
(a) Water inside the fruit-n
(b) White Kernal-2n
(c) Seed coat-n
(d) Embryo-3n
(e) Tepal-2n
Ans.(a) 34
(b) 34
(c) 24
(d) 24
(e) 24
20) Inangiospermic plant before formation of microspore sporogenous tissue
undergo cell division.
(a) Name the type of cell division.
(b) What would be the ploidy of the cells of tetrad?
Ans. (a) meiosis division (b) haploid.
21) Do plant breeders ensure cross pollination in economically important bisexual
flower?
Ans.it is important that only desired pollen grains are used for pollinating the selected
female parent so the stigma of the female parent has to be protected from contamination by
unwated pollen. Emasculation and bagging are the two techniques to achieve it.
22) Apple is a false fruit while banana is parthenocarpic fruit. explain .
Ans.apple is a false fruit because its thalamus also contributes to the formation of fruit other
then its ovary.while banana is parthenocarpic fruit because it develops from ovary without
fertilization.
23) Banana is parthenocarpic fruit whereas orange show polyembryony.how are they
different from each other with respect to seeds?
Ans.Parthenocarpic fruits are generally seedless or have non-viable seed.
In oranges, each seed has more than one embryo ie.polyembryony.
24) Differentiate between the two cells enclosed in a mature male gametophyte of an
angiosperm.
Ans.vegetative and generative cell.the vegetative cell is larger in size and have vacuolated
cytoplasm.the generative cell have thin dense cytoplasm with prominent nuclei that give rise
to two male gametes.
25) Name all the haploid cells present in an unfertilized mature embryo sac of flowering
plant.write the total number of cells in it.
QUESTION BANK /BIOLOGY/XII
Page 16
Ans. an unfertilized mature embryo sac is composed of 7 cells ie 7 celled and 8 nuclei.among
the 8 nuclei,6 are enclosed by cell walls and organized into cells which are haploid in
number-3 antipodal,2 synergids and 1 egg cell and a large central nuclei with two pollen
nuclei.
26) How does pollination take place in Vallisneria and Zostera?
Ans a) In Vallisneria, the female flowers reach the surface of water by the long stalk and the
male flowers or pollen grains are released on to the surface of water.
They are carried passively by water currents; some of them eventually reach the female
flowers and the stigma. In Zostera, female flowers remain submerged in water and the
pollen grains are released inside the water. Pollen grains in many such species are long,
ribbon like and they are carried passively inside the water; some of them reach the stigma
and achieve pollination
27) Differentiate between Autogamy, Geitonogamy and Xenogamy
Ans.Autogamy: Transfer of pollen grains from the anther to the stigma of the same
flower. No genetic variations seen.
Geitonogamy :Transfer of pollen grains from the anther to the stigma of another
flower of the sameplant. No genetic variatins seen
Xenogamy :Transfer of pollen grains from anther
to the stigma of a different plant.Genetic variations seen.
28) State one advantage and one disadvantage of cleistogamy.
Ans. Advantage:produces assured seeds even in absence of pollinators
Disadvantage:cleistogamous flowers are invariably autogamous so there is no changce of
cross pollination.
29)Explain any two devices by which autogamy is prevented in flowering plants.
Ans.i)anthers and stigma are placed in sucha a way that the pollen of the same flower cannot
fall on the stigma.
ii)self-incompatibility is a genetic process that prevents germination of pollen from the same
flowers on the stigma.
30) Name the cell from which the endosperm of coconut develops.Give the
characteristic features of endosperm of coconut.
Ans.cell formation occurs and the endosperm becomes cellular.The number of free nuclei
formed before cellularisation varies greatly.The coconut water is free nuclear endosperm.It is
made up of thousands of nuclei and the surrounding white kernel is the cellular endosperm.
31)What similarities do you find in seed bank and pollen bank? Write their importance.
QUESTION BANK /BIOLOGY/XII
Page 17
Ans.Pollen grain of different species of plants can be stored for years in liquid nitrogen
(196ºC) like seeds. Such type of storage is called pollen banks similar to seed banks, in crop
breeding programme
SHORT ANSWER TYPE QUESTIONS (3marks)
32) How does the megaspores mother cell develop into 7-celled and 8-nucleate embryo
sac in an angiosperm? Draw a labeled diagram of a mature embryo sac.
Ans.The megaspore mother cells undergoes mitosis to form two nuclei which migrate to
opposite poles forming a 2-nucleate embryo sac.Furthur mitotic division leads to the
formation of 4-nucleate followed by 8-nucleate stages of the embryo sac.In these mitostic
divisions,nuclear division is not followed by cell division.aftwer the 8 nucleate stage the cell
wall are laid down and a typical female gametophyte or embryo sac is formed.among the 8
nuclei,6 are enclosed by the cell wall and organized in to cells while the polar nuclei are
situated above the egg apparatus in the central cell.out of the 6,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 anti podal cells.thus a
typical angiosperm embryo sac after maturity is 7-celled and 8-nucleate.
33) Continued self pollination lead to inbreeding depression. List three devices,
which flowering plants have developed to discourage self pollination?
Ans.(a) Release of pollen and stigma receptivity is not synchronized in somespecies.
(b) Anther and stigma are at different position/heights in some plants.
(c) Self-incompatibility a genetic mechanism.
34) Draw the embryo sac of a flowering plants and label:
(a) (i) Central Cell (ii) Chalazal end (iii) Synergids
(b) Name the cell that develops into embryo sac and explain how this cell leads to
formation of embryo sac.
QUESTION BANK /BIOLOGY/XII
Page 18
(c) Mention the role played by various cells of embryo sac.
(d) Give the role of filiform apparatus.
Ans.
a) Refer to above answer.
b) Functional Megaspore, Refer text on page 27 NCERT book.
c) Egg : Fuses with male gamete to form zygote or future embryo
Synergid : Absorption of nutrient, attract and guides pollen tube.
Central Cell : After fusion with second male gamete forms Primary
endosperm cell which gives rise to Endosperm
d) Guides the entry of pollen tube.
35)(a) Draw a diagrammatic sectional view of a mature anatropousfollowing parts in it:
(i) That develops into a seed coat.
(ii) That develops into an embryo after
(iii) That develops into an endosperm in an albuminous seed.
(iv) Through which the pollen tube gains entry into the embryo sac.
(v) That attaches the ovule to the placenta.
(b) Describe the characteristics features of wind pollinated flowers.
Ans.
(a) . Refer to figure 2.8(c) page 26 NCERT book
(i) That develops into seed coat - Integument
(ii) That develops into an embryo after fertilization -Embryo sac
(iii) That develops into an endosperm in an albuminous seed- Nucellus
(iv) Through which the pollen tube gains entry into the embryo sac-Micropyle
(v) That attaches the ovule to the placenta-FunicleII | Bilog
(b) The characteristic feature of wind pollinated flowers are
(i) It shows complete inflorescence
(ii) It has well exposed stamens.
(iii) They have light seeds, with some kind of method for being carried on thewind.
(iv) The pollens are dry.
36)(a) Why is fertilization in an angiosperm referred to as double fertilization? Mention
the ploidy of the cells involved.
(b) Draw a neat labeled sketch of L.S of an endospermous monocot seed.
Ans.a) At the time of fertilization, pollen tube brings two male gametes and the processes
that take place in the embryo sac are(i) Syngamy- It is the fusion of both haploid male and female gamete to form a diploid
zygote which further develops to form embryo.
Male Gamete + Egg (female gamete) Zygote Embryo Syngamy is also called generative
fertilization. (ii) Triple fusion- The other male gamete moves to the antreal cell to fuse with
the two polar nuclei to form a triploid primary endosperm nucleus (PEN) which grows to
form primary endosperm cell (PEC) and which divide mitotically to from endosperm.
QUESTION BANK /BIOLOGY/XII
Page 19
Two polar nuclei + Male Gamete PEN PEC Endosperm Endosperm provides nutrition to the
developing embryo. Thus triple fusion is also called vegetative fertilization. Since this
process involves two types of fertilization, it is referred to as double fertilization.
37)Describe in sequence the events that lead to the development of a 3-celled pollen
grain from microspore mother cell in angiosperms.
Ans.Microspore mother cell , meiosis , 4 haploid microspores , tetrad , Each micorspore is a pollen
grain , with two layered wall - outer exine (made of sporopollenin) and inner intine (made up of
cellulose and pectin), pollen grain contain one larger vegetative cell , and a smaller generative cell ,
which divides mitotically , into two male gametes
38)What is Double fertilization? Why it is called so? Elaborate the term Syngamy?
Ans. Fertilization: Pollen grains germinate on stigma & pollen tube grows through style.
This pollen tube reaches micropyle & releases two male gametes into embryo-sac.
Fertilisation is the process of fusion of male & female gametes (n+n) to form a diploid(2n)
zygote.SYNGAMY: Fusion of one male gamete(n) with egg (n) so that Zygote(2n) is
produced.First Fusion is of two Polar Nuclei(n+n=2n).Second Fusion is of Male Gamete
which Fuses with the fusion product of the two polar nuclei(3n).Third Fusion fusion of male
gamete with egg cell.
DOUBLE FERTILIZATION: In an embryo sac, two type of fusions are seen,i)Fusion of
male gamete with egg – First fertilisation, ii)Fusion of fusion product of polar nuclei with
male gamete-Second Fertilisation. Hence there are two Fertilizations occurring hence this
phenomenon is known as double fertilization
39)Give Biological name of the following:
(i)
Three cells present at the chalazal end in the embryo sac.
(ii)
A small pore in the ovule through which the pollen tube enter.
(iii)
Wall of fruit having mesocarp , endocarp, epicarp.
(iv)
Two cells present on either side of egg cell in an embryo sac.
(v)
Mass of parental cells enclosed within the integument
QUESTION BANK /BIOLOGY/XII
Page 20
Ans.(i)Antipodals
(ii) Micropyle
(iii) Pericarp
(iv) Synergids
(v) Nucellus
40)What are the post-fertilisation changes?
Ans.POST FERTILISATION CHANGES:
STAGES OF EMBRYO DEVELOPMENT AFTER FERTILISATION:
1. Zygote divides by mitosis into suspensor & embryo cells.
2. Suspensor cell forms a globular basal cell which remains embedded in the endosperm & a
multicellular suspensor bearing the embryo.
3. Globular embryo becomes heart-shaped & then grows into mature embryo with radicle,
plumule & Cotyledons.
4.Primary endosperm nucleus – divides repeatedly to form endosperm, food for the embryo.
Further ature ovary becomes fruit and mature ovule becomes seed.
41)Explain the term out breeding devices?
Ans.Continuation of self-pollination results in breeding depression. Flowering plants have
developed many devices to discourage self-pollination & encourage cross-pollination these
are:
(i) Bearing unisexual flowers
(ii) Anther & stigma mature at different times
(iii) Anther & stigma placed at different positions
(iv) Self-incompatibility where pollen grains of a flower do not germinate on the stigma of
the same flower.
42.a)Identify the following figure:
b) Name x, y and z in the above figure.
c) Mention one characteristic feature and one function of 'z' in the above figure.
Ans.
QUESTION BANK /BIOLOGY/XII
Page 21
a) Microsporangium
b) x → endothecium
y → microspore mother cell
z → tapetum
c) Characteristic: dense cytoplasm/multi-nucleated [any one]
Function: nourishes the growing pollen grains
43. a) Identify the following figure:
b) Name the initial structure from which this structure has developed. Also write the
ploidy of the initial structure.
c) Draw and name the next mature stage and label the parts.
Ans.
a) Globular embryo
b) Zygote, 2n
c)
Heart-shaped embryo
QUESTION BANK /BIOLOGY/XII
Page 22
Chapter- 3-HUMAN REPRODUCTION
VERY SHORT ANSWER TYPE QUESTIONS (1marks)
1) Zygote undergoes mitosis to form 16 celled stage of embryo. What is it
known as?
Ans.morula
2) Name the important mammary gland secretions that help in resistance of the
new born baby.
Ans.Colostrum
3) Failure of testes to descend into scrotal sacs leads to sterility. Why?
Ans.High temperature of abdomen kills the spermatogenic tissue of the testes,so no sperm
are formed.
4) Both vaccine and colostrum produce immunity. Name type of immunityproduced by
these.
Ans..Vaccine .Active immunity Colostrum .Passive immunity.
5) Write the location and function of sertoli cells inhumans.
Ans.within the lining of seminiferous tubule of testis. They provide nutrition to the
developing sperm cells.
6) The changes the primary oocyte undergoes in the tertiary follicular stage in human
embryo.
Ans.primary oocyte grows in size,completes meiosis I forms larger cells,the secondary
oocyte and smaller cell,the polar body.
7)Mention the differences between spermiogenesis and spermiation.
Ans.Spermiogenesis:transformation of spermatids into spermatozoa.
Spermiation:the process of release of spermatozoz from sertoli cells in to the cavity of
seminiferous tubules.
8)Where is acrosome present in humans?write its functions.
Ans acrosome is ac ap like structure present at the top of the sperm head inuman males helps
the sperm in fertilizing the ovum by dissolving the wall of ovum facilitating sperm nucleus
enter the ovum.
QUESTION BANK /BIOLOGY/XII
Page 23
9)How is the sperm ensured into n ovum during fertilization in humans entry of one?
Ans.During fertilization a sperm comes in contact with the zona pellucid layer of the ovum
and induces changes in the membrane that blocks the entry of additional sperm.
10)What is function of trophoblast in human embryo?
Ans.Outer layer of blastocyst helps in attachment of it to the endometrium lining of the
uterus.
11)What stimulates the pituitary to release hormone responsible for parturition.name
the hormone.
Ans.‘foetal ejection reflex stimulates pituitary to release oxytocin hormone which is
responsible for partituition.
12)How does colostrums provide protection against diseases to new born infants.
Ans.It contains necessary antibodies IgA that provides protection against new born infants.
SHORT ANSWER TYPE QUESTIONS (2 marks )
13)Why does fertilisation take place in fallopian tube and not in uterus?
Ans The pH is not suitable for fertilization in uterus.
14) Fill in the boxes
Ans.a) Primary spermatocytesb) Spermatids
15) Which cell organelle is present in the neck of the sperm? What is itssignificance?
Ans. Mitochondria It produces energy for the movement of tail that facilitates sperm motility
essential for fertilization.
16) Failure of fertilization leads to menstruation. Explain.
QUESTION BANK /BIOLOGY/XII
Page 24
Ans. In the absence of fertilization corpus luteum degenerates. This causesdisintegration of
endometrium leading to menstruation
17)Give reason for the following :
(a) The first half of the menstrual cycle is called follicular phase as well
as proliferative phase.
(b) The second half of the menstrual cycle is called luteal phase as well
as secretory phase.
Ans.(a) During this phase, primary follicles transform into Graafian follicle underFSH
stimulation. Graafian follicles secrete estrogens with stimulateenlargement of Endometrium
of uterus.
(b) During this phase, Corpus luteum is fully formed and secretes largequantity of
Progestrone.
18) Where are leydig cells present?what is their role in reproduction.
Ans.In the interstitial spaces between the seminiferous tubules.it synthesizes and secretes
hormones androgens eg.testosterone.
19) Where are fimbriae located in human females.give their functions.
Ans.It is a funnel shaped edges of the fallopian tube or oviduct in human female reproductive
system.they help in collection of ovum or secondary oocyte after ovulation.
20) Differentiate between vas deferens and vas efferentia
Ans.vas deferens: it’s a tube like structure which conducts the spermatozoa from the
epididymis to penis.
vas efferentia: it connects the rete testis to epididymis.
21) Mention the sites of action of the hormone GnRH and FSH during spermatogenesis
in human males.Give one function of each hormones.
Ans.GnRH is hypothalamic hormone spermatogenesis starts at puberty due to its significant
increase.its increase level acts on interior pituitary gland to stimulate the secretion of FSH
and LH.LH acts on leydig cells to stimulate and synthesize and secretes androgens.FSH acts
on sertoli cells which stimulates the secretion of some factors that help in spermiogenesis.
22) Placenta acts as an endocrine tissue justify.
Ans.It secretes hormones like
a)human chorionic gonadotropin hCG
b)human placental lactogen hPL
c)oestrogen.
d)progesterone.
QUESTION BANK /BIOLOGY/XII
Page 25
23) Spermatogenesis in human is hormone regulated process.Justify.
Ans.gonadotropin releasing hormones is released from the hypothalamus during puberty.
It stimulates ant pituitary to secrete gonadotropins i.e LH and FSH or ICSH.
LH acts on leydig cells to stimulate the synthesis and secrete androgens
Androgens stimulates the process of spermatogenesis.
FSH acts on sertoli cells and stimulates them to secrete certain factors which are necessary
for the process of spermiogenesis.
24) Mention the target cell of LH in human males and females.Explain the effect and
changes which the hormone induces in each case.
Ans. Target cel of LH in males are leydig cells.
In females are mature follicles.
LH stimulates the leydig cells in males to secrete testosterone,which controls
spermatogenesis.In females LH stimulates ovulation,formation of corpus luteum..
25)Why placenta is called Functional Unit?
Ans. It is called so as it helps in the transfer of oxygen and nutrients to the embryo from
mother and removal of carbon di-oxide and excretory waste from the embryo.
26) Differentiate b/w blastulla and morulla.
Blastula: It is a hollow sphere of 32 or more cells formed by the rearrangement of
blastomeres.
Morulla: It is a solid sphere of 8- 16 cells blastomeres formed by cleavage of zygote
SHORT ANSWER TYPE QUESTIONS (3marks)
27) T.S. of mammalian testis revealing seminiferous tubules show different
types of cell.
(i) Name the two types of cells of germinal epithelium.
(ii) Name of cells scattered in connective tissue and lying betweenseminiferous tubules.
Differentiate between them on the basis of their functions.
Ans.(i) Germinal epithelium have two types of cell.
1. Spermatogonium.
2. Sertoli cells
(ii) Leydig cells or Interstitial cells.
Functions
Spermatogonium undergoes meiotic division leading to spermformation.
Sertolicell : Nourishes germ cells
Leydigcell :Synthesise and Secrete hormone androgen.
QUESTION BANK /BIOLOGY/XII
Page 26
28) How does the ovum ensure that only one sperm fertilizes it?
Ans. During fertilisation a sperm comes in contact with Zonapellucida layer of the
ovum& induces change in the membrane that block entry of additional sperms.
Thus it ensures single sperm entry.
29) Name the part of the female reproductive system where the embryo is
implanted. Mention the type of tissue by which it is made up of and give their
functions?
Ans. Uterus.Uterus is made of three tissue layersa) External thin membranous layer_Perimetrium
b) Middle thick layer of smooth muscle__Myometrium
c) Inner glandular layer__Endometrium.
Endometrial layer undergoes cyclic changes during menstrual cycle.
Myometrium exhibits strong contraction during delivery of the baby.
30) What is the fate of inner cell mass in the blastocyst? Mention their
significance.
Ans.The inner cell mass of embryo differentiates into
i) Outer layer----Ectoderm
ii) Inner layer----endoderm
iii) Middle layer----Mesoderm
These three layers give rise to all tissues (organs) in adults.
31) Label a,b,c in the following diagram.
4.a) Nipple b) Ampulla c) Fat
32)Differentiate between Endometrium and Myometrium.
Ans.
Endometrium
Myometrium
It is innermost glandular It is the middle thick layer of
QUESTION BANK /BIOLOGY/XII
Page 27
layer that lines the uterine smooth muscles
cavity
uterine wall.
of
the
Implantation occurs in this It is responsible for the
layer
uterine movement.
It does not undergo any
It undergo cyclic changes
cyclic changes during the
during the menstrual cycle
menstrual cycle
33)a diagram of microscopic structure of human sperm. label the following part in it
and write their function. a) Acrosome b) Nucleus c) middle Piece.
Ans.
34) Differentiate between:Spermatogenesis and oogenesis.
Spermatogenesis
Produces
(sperm).
male
Oogenesis
gametes Produces female gametes
(oocytes).
Occurs in the seminiferous
Occurs in the ovaries.
tubules (in testes).
Involves meiosis occurs
Involves meiosis –occurs
after
puberty
until
throughout life after puberty.
menopause.
Humans normally produce
May produce 400,000,000 per
one oocyte during each
day.
ovarian cycle.
Primary spermatocyte divide Primary otocyte divide
equally to form two similar unequally to form one
QUESTION BANK /BIOLOGY/XII
Page 28
secondary spermatocytes.
large secondary oocyte and
a small polar body.
An oogonium produces
One
spermatogonium
one functional ovum and 3
produces
4
functional
non
functional
polar
spermatozoa.
bodies.
35)What is the diffrence b/w spermatogenesis and spermiogenesis?
Spermatogenesis
Spermiogenesis
It is the process of formation of mature
spermatozoa in the testis
It is a process of transformation of spermatids
into spermatozoa
It involves meiotic and mitotic division
It does not involve any division.
It is controlled by hormone LH and androgen. It is controlled by hormone LH only.
36)What are the stages of maturation of ova?
Ans.
LONG ANSWER TYPE QUESTIONS (5 marks )
37) Give the term / reason
a) Mechanism responsible for parturition.
QUESTION BANK /BIOLOGY/XII
Page 29
b) Role of oxytocin during expulsion of the baby out of uterus
c) Why does zonapellucida layer block the entry of additional sperms?
d) Sperm cannot reach ovum without seminal plasma.
e) All copulations do not lead to fertilization and pregnancy.
Ansa) complete nuero-endocrine mechanism
b) Oxytocin acts on uterine muscle for stronger contraction
c) To ensure the fusion of one sperm
d) Seminal plasma is a liquid medium which helps the sperm to move &nourishes it.
e) The movement of sperm and ovum does not occur simultaneously manytimes.
38) Women are often blamed for giving birth to girl child in our society. What is
your view?
Ans. The belief is totally wrong.
It is the probability of X or Y chromosome combination that decides the sex of the
child.(Detail in page 52 of NCERT class xii (sex determination))Chromosome Y decides the
sex.
39) Furnish the technical term for the following:
a) Cushion of fatty tissue covered by skin and pubic hair in femaleexternal genitalia.
b) The finger like projections which collect ovum after ovulation
c) The middle thick layer/wall of uterus
d) Semen without sperm
e) The finger-like projections appearing on the trophoblast afterimplantation.
Ans.a) Mons pubis
b) Fimbriae
c) Myometrium
d) Seminal plasma
e) Chorionic villi
40)The following is the illustration of the sequence of ovarian events “a” to “i” in a
human female:
(a) Identify the figure that illustrates corpus luteum and name the pituitary hormone
that influences its formation.
(b) Specify the endocrine function of corpus luteum. How does it influence the uterus?
Why is it essential?
QUESTION BANK /BIOLOGY/XII
Page 30
(c) What is the difference between “d” and “e”?
(d) Draw a neat labeled sketch of Graafian follicle.
Ans.(a) Figure ‘f’ illustrates the corpus luteum and the pituitary hormones responsible for are
luteinizing hormone (LH) and follicle stimulating hormone (FSH).
(b) The corpus luteum secretes progesterone, which is a steroid hormone responsible for the
development of the endometrium and maintenance, respectively. The endometrium of the
uterus gets thickened and blood supply to the endometrium increases. It is essential for
implantation of the fertilized ovum and other events of pregnancy.
(c) ‘d’ illustrates the developing follicle whereas ‘e’ illustrates the mature Graafian follicle.
The difference is that primary follicles in the ovary grow to become a fully mature Graafian
follicle.
41)(a) Draw a diagrammatic sectional view of the female reproductive system of
human and label the parts
(i) where the secondary oocytes develop
(ii) which helps in collection of ovum after ovulation.
(iii) where fertilization occurs.
(iv) where implantation of embryo occurs.
(b) Explain the role of pituitary and the ovarian hormones in menstrual cycle inhuman
females
Ans.(i) where the secondary oocytes develop- ovary
(ii) which helps in collection of ovum after ovulation- fimbriae
(iii) where fertilization occurs - fallopian tubes
(iv) where implantation of embryo occurs- uterus
(b) Menstrual cycle starts with menstrual flow caused due to the breakdown of the
endometrium of the uterus, which is followed by the follicular phase the role of
pituitary and the ovarian hormones in menstrual cycle are
(i) In this phase, the release of gonadotropins (leutnizing hormone and
follicle stimulating hormone) increases. Growing follicles produce
estrogen.
(ii) In the 14th day, LH and FSH rupture the Graffian follicles to release ovum.
(iii) The corpus luteum secretes large amounts of progesterone which is
essential for maintenance of endometrium
QUESTION BANK /BIOLOGY/XII
Page 31
42)Describe various stages of fertilisation which leads to implantation,with the help of a
well labelled diagram.
Ans.
43)What are the accessory glands associated with male reproductive system?
Ans. Three accessory glands provide fluids that lubricate the duct system and nourish the
sperm cells. They are the seminal vesicles, the prostate gland, and the bulbourethral glands
(Cowper glands).
Seminal vesicles-Seminal vesicles are sac-like structures attached to the vas deferens at one
side of the bladder. They produce a sticky, yellowish fluid that contains fructose. This fluid
provides sperm cells energy and aids in their motility.
Prostate gland-The prostate gland surrounds the ejaculatory ducts at the base of the urethra,
just below the bladder. The prostate gland is responsible for the production of semen, a liquid
mixture of sperm cells, prostate fluid and seminal fluid.
Bulbourethral gland-The bulbourethral glands, also called Cowper glands, are two small
glands located on the sides of the urethra just below the prostate gland. These glands produce
a clear, slippery fluid that empties directly into the urethra. It produces substances related to
nourishment of spermatozoa.
44) Give details of spermatogenesis and oogenesis.
Ans.
QUESTION BANK /BIOLOGY/XII
Page 32
45)a) Draw a diagram of the human female reproductive system and label on it:
i) site of fertilisation of ova
ii) three walls of uterus
iii) uterine fundus
b) Write a short note on the structure and function of mammary glands in
humanfemales.
Ans.Female reproductive system:
b) The mammary glands are paired structures (breasts) that
contain glandular tissue and variable amount of fat.The glandular tissue of each breast is
divided into 15-20mammary lobes containing clusters of cells called alveoli.
The cells of alveoli secrete milk, which is stored in the cavities
(lumens) of alveoli.The alveoli open into mammary tubules. The tubules of each
lobe join to form a mammary duct.Several mammary ducts join to form a wider mammary
ampullawhich is connected to lactiferous duct through which milk issucked out.
QUESTION BANK /BIOLOGY/XII
Page 33
Chapter- 4-REPRODUCTIVE HEALTH
VERY SHORT ANSWER TYPE QUESSTIONS (2 marks)
1) A large number of couples are said to be infertile. The couples could be
assisted to have children through certain special techniques. Name thetechniques.
Ans.Assisted reproductive technologies(ART)
2) At what stage Zygote can be introduced in the fallopian tube in Zygote Intra
Fallopian Transfer (Z.I.F.T)?
Ans.8-celled stage
3)A woman's husband is infertile. So the lady has decided to have baby by
taking sperms from sperm bank. Which technique will you suggest for her
pregnancy?
Ans.Intra cytoplasmic sperm injection(ICSI)
4) A newly married couple does not want to produce children at least for one
year and also not to use any contraceptives. Suggest a method to prevent
pregnancy.
Ans.Periodic abstinence or coitus interruptus
5) A doctor has been observed the chromosomal disorders in developing foetus
and advised the couple to undergo abortion. suggest the technique by which
doctor absorbed the chromosomal disorders.
Ans.Amniocentesis
6) What precautions a lady can take to prevent unwanted pregnancy?
i) Name the barrier
ii) Mention the composition of it.
Ans.Pills (Progestogen-estrogen combination.)orProgestogen
7)The fluid from which foetal cells are extracted for chromosomalanalysis.
Ans.Amniotic fluid.
8)Mention one positive and one negative applications of aminocentesis.
Ans.It can be used to diagonise any chromosomal abonormality or genetic disorder in foetus.
It is used to determine the sex of the soetus and leads to female infanticide.
QUESTION BANK /BIOLOGY/XII
Page 34
9)Why is tubectomy considered as contraceptivemethod.
Ans.a small part of fallopian tube is tied up to block the transportation of sperm so as to
prevent fretilisation.so it is considered as contraceptive method.
10)After a successful in vitro fertilisation, the fertilised egg begins to divide.Where is
this egg transferred before it reaches the 8-celled stage and whatis this technique
called?
Ans. Fallopian tube; Zygote intra fallopian transfer (ZIFT)
11) Give technical name of female used to bring up in vitro fertilized egg tomaturity.
Ans.Surrogate mother.Name the fluid from which foetal cells are extracted for chromosomal
analysis.
SHORT ANSWER TYPE QUESTIONS (2 marks )
12) During lactation chances of conception are almost zero.
(i) Give the reason.
(ii) Give the term used to describe the phenomenon.
Ans.a) Ovulation does not take place
b) Lactational amenorrhea.
13)What is 'Saheli'? Name the institute where it was it developed?
Ans .Oral contraceptive for women
CDRI - Central Drug Research Institute, Lucknow
14) When do implantation occurs?
Ans. The blastocyst is a liquid-filled ball of cells. Occurs around 5 – 8 days after conception.
Implantation in the endometrium occurs at this stage.
15)What are the surgical methods towards birth control?
Ans. Surgical Methods: These are more or less permanent methods of contraception.
Tubal Ligation: Both the female tubes are tied off and usually cut during tubal
ligation to prevent the sperm from reaching theovum during intercourse.
Vasectomy: The two tubes which carry sperm from the testes to the penis are thevas
deferens. Tying them off and cut.
Essure: Essure is a method in which small micro-inserts are placed at the mouth of the
fallopian tubes to cause scarring and block them. This prevents sperm from reaching the
ovum for fertilization.
QUESTION BANK /BIOLOGY/XII
Page 35
16) What are implants? How do they help in preventing fertilisation?
Ans.The structures which contain hormones like progesterone and estrogenand are placed
under the skin.
17) Differentiate b/w tubectomy and vasectomy.
Ans.
Tubectomy
Vasectomy
Method of sterilisaion in females
Method of sterilization in males
Fallopian tubes of both sides are cut and
tied.
Vas deference is cut and tied.
It prevents ova to reach the place of
fertilization.
It prevents sperms to reach the place of
fertilization.
18) Briefly explain two natural barriers for birth control.
Ans.Periodic abstinence .couple should avoid coitus from 10th to 17th day ofmenstrual cycle.
Coitus interruptus . Male partner withdraws his penis from the vagina justbefore ejaculation
of semen.
SHORT ANSWER TYPE QUESTIONS (3marks )
19) Howare non medicated IUD.S different from hormone releasing IUD.S? Give
Examples.
Ans.(a) Non medicated IUDs = Lippes loop, Copper releasing IUD.S ( CuT,
Multiload 375) These increase phagocytosis of sperms within uterus
and release copper ions which suppress sperm motility and fertilising
capacity of sperm.
(b) Hormone releasing IUDs .Progestasert, LNG.20 . These makes
uterus unsuitable for implantation and the cervix hostile to sperms
20)(a) Identify the given diagram. What it is used for?
Ans.Implants
Functions
i) They contain progestogens or progestogen. estrogen combination
ii) They inhibit ovulation and implantation of embryo to the uterine wall.
QUESTION BANK /BIOLOGY/XII
Page 36
21) Differentiate b/w GIFT and ICSI.
Ans.
GIFT
ICSI
Called as gamete intra fallopian
transfer(b).transfer of ovum from
a donor female to provide
suitable environment .
Intra
cytoplasmic
sperm transfer this is a
special technique to
prepare embryo in lab.
Sperm is not directly injected to The sperm is directly
ovum
injected to ovum.
Artificial insemination
No artificial insemination is done is done for infertile
couple
22) What are the various methods of IVF?
Ans. Zygote intrafallopian transfer (ZIFT) infertility treatment where blockage in the
fallopian tubes prevents the normal binding of sperm to the egg. Egg cells removed from
woman's ovaries, and in vitro fertilised. Resulting zygote placed into the fallopian tube by
laparoscopy. The procedure spin-off of the gamete intrafallopian transfer (GIFT) procedure..
First, the woman must take a fertility medication to stimulate egg production in the ovaries.
The doctor will monitor growth of ovarian follicles, once they are mature, woman will be
injected with human chorionic gonadotropins (hCG).After fertilization in laboratory resulting
early embryos or zygotes are placed into the woman's fallopian tubes using laparoscope.
Gamete intrafallopian transfer (GIFT) assisted reproductive technology against infertility.
Eggs removed from a woman's ovaries, placed in one of the Fallopian tubes, along with the
man's sperm. The technique, which was pioneered by endocrinologist Ricardo Asch, allows
fertilization to take place inside the woman's body. Takes, an average of four to six weeks to
complete a cycle of GIFT. First, the woman must take a fertility drug to stimulate egg
production in the ovaries. The doctor will monitor growth of ovarian follicles, once they
mature, woman will be injected with Human chorionic gonadotropin (hCG). The eggs will be
harvested approximately 36 hours later, mixed with the man's sperm, and placed back into
the woman's Fallopian tubes using laparoscope.
Intracytoplasmic Sperm Injection (ICSI): technique in which a single sperm injected into
the centre of the egg, in order to achieve fertilization. Sperm is collected from the male
partner by masturbation. Single healthy sperm then injected into the prepared ovum. The
advantage of this method is that only a single sperm is needed - even men with a very low
sperm count can become fathers with this treatment. Men found to be azoospermic, that is
with no sperm at all in the semen, sperm can be suctioned out of the vas deferens ( male
tubes). Sperm can also be liberated from the testes itself by careful testicular biopsy and
culture by a method called MESA - Microepididymal sperm aspiration.
QUESTION BANK /BIOLOGY/XII
Page 37
GENETICS AND EVOLUTION
Chapter -5.Principles of Inheritance and Variation
VERY SHORT ANSWER TYPE QUESTIONS (1 marks )
1)Write the percentage of the pea plants that would be heterozygous tall in the F2
generation, when tall heterozygous F1 pea plants are selfed.
Ans.50%
2)In a test cross progeny of pea plants, all were bearing violet flowers. Give the
genotypes of the parent pea plants.
Ans.VV and vv
3)Name the respective pattern of inheritance where F1 phenotype
(a) does not resemble either of the two parents and is in between the two.
(b) resembles only one of the two parents.
Ans.Incomplete dominance/Dominance
4)In a dihybrid cross, would the proportion of parental gene combination be much
higher than non-parental types, as experimentally shown by Morgan and his group?
When the two genes show linkage Write the possible genotypes Mendel got when he
crossed F1 tall pea plant with a dwarf pea plant.
Ans.Tt and tt (in the ratio 1.1)
5)Why, in a test cross, did Mendel cross a tall peplant with a dwarf pea plant only?
Ans.Dwarfness is a recessive trait which is expressed only in homozygous condition, So he
was sure of the genotype of the dwarf plant as tt.
6)Name the event during cell division cycle that results in the gain or l oss of
chromosomes.
Ans.Failure of segregation of members of homologous pairs of chromosomes (nondisjunction) during anaphase I of meiosis.
QUESTION BANK /BIOLOGY/XII
Page 38
7)Name an autosomal dominant and one autosomal rec essive Mendelian disorder in
humans.
Ans.Autosomal dominant disorder: Myotonic dystrophy.Recessive : Sickle cell anaemia.
8)Write the genotype of:
(i) an individual who is carrier of sickle-cell anaemia gene, but apparently unaffected
(ii) an individual affected with the disease.
Ans.
1. HbA HbS
2. HbS HbS
9)human being suffering from Down’s syndrome shows trisomy of 21st chromosome. M
ention the cause of this chromosomal abnormality.
Ans.It is due to non-disjunction (non-separation) of the 21st chromosomes during ova
formation and the fertilization of an ovum having two 21st chromosomes by a normal sperm.
There is an additional copy of the 21st chromosomesin such individuals.
10)What is heterogamety? Give an example of an organ ism showing it.
Ans.Heterogamety is the phenomenon in which an individual produces two types of gametes
with reference to sex-chromosomes. eg Human males and female fowls are heterogametic.
11)gene I that controls the ABO blood grouping in human beings has three alleles IA,
IB and i.
(a) How many different genotypes are likely to be present in the human population?
(b) Also, how many phenotypes are possibly present
Ans.Six
Four
12) Mention the type of allele that expresses itself only in homozygous state in an
organism.
Ans.Recessive allele.
SHORT ANSWER TYPE QUESTIONS (2marks )
13)A cross is made between different homozygous pea plants for contrasting flower
positions.
(a) Find out the position of flowers in F1 generation on the basis of genotypes.
(b) Work out the cross up to F2 generation.
(c) Compute the relative fraction of various genotypes in the F2 generation?
Ans.
(a) Axial position
QUESTION BANK /BIOLOGY/XII
Page 39
(b)
AA
x
aa
(Axial)
(Terminal)
A
a
P
F1
Aa (Axial)
Aa
x
Aa
A a
AA
Selfing
A a
Aa
AA = 1/4
F1
Aa
Aa = 1/2
aa
F2
aa = 1/4
14)In a typical monohybrid cross the F2- population ratio is written as 3 : 1 for
phenotype but expressed as 1 :2 : 1 for genotype. Explain with the help of an example.
Ans.In garden Pea plant
Parental
TT
x
Tall
Gametes
Dwarf
T
F1 generation
tt
t
Tt
Tall
Selfing
F2 generation
T
t
QUESTION BANK /BIOLOGY/XII
Page 40
T TT Tt
t
Tt
tt
genotype 1 :2 : 1-- one is homozygous dominant TTand 2 are heterozygous but dominant Tt
and one is homozygous recessive tt
phenotype 3 : 1 -- three are dominantand one shows recessive character (dwarf)
15)When a tall pea-plant was selfed, it produced one-fourth of its progeny as dwarf.
Explain with the help of a cross.
Ans.Since dwarfness (recessive trait) has appeared in the progeny, the tall pea plant must be
heterozygous for tallness, ie. its genotype is Tt
Selfing
:
Tall
Tt
Tall
x
Tt
Gametes
:
T, t
T, t
Progeny
:
T
t
TT
Tt
Tall
Tall
Tt
tt
Tall
Dwarf
T
t
The phenotypic ratio is 3 Tall : 1 dwarf.
-Since tallness is dominant over dwarfness, the homozygotes(TT) as well as the heterzygotes
(Tt) are tall.
-The recessive trait is expressed only under homozygous condition (tt) and hence appears
only in one-fourth of the progeny.
16)A cross between a red flower-bearing plant and a white flower-bearing of
Antirrhinum produced all plants having pink flowers. Work out a cross to ex plain how
this is possible.
Ans.Inheritance of flower colour in Antirrhinum majus :
Parents :
Red
QUESTION BANK /BIOLOGY/XII
White
Page 41
flowers
flowers
RR
Gametes :
x
rr
R
r
F1generation :
Rr-Pink flowers
- This character shows incomplete dominance , where neither of the two alleles of the gene
is completely dominant and the F1 hybrid is intermediatebetween the two.
17)In Snapdragon a cross between true-breeding red flowered (RR) plants and truebreeding white flowered (rr) plants showed a progeny of plants with all pink flowers.
(a) The appearance of pink flowers is not known as blending. Why?
(b) What is this phenomenon known as?
Ans.
(a) RR (red)
x
rr (white)
F1 = Rr (pink)
Rr x Rr (selfing)
F2
RR : Rr : rr
1
(red)
2
1
(pink) (white)
On selfing the pink flowers we get the parental forms – Red & White
(b) Incomplete dominance
18) Work out a cross to find the genotype of a tall plant. Name the type of cross.
In the test cross , an individual showing dominant phenotype is crossed to an individual with
recessive phenotype for the same trait.
(i) If the individual is homozygous dominant (Tall) TT, all the individuals in the progeny
would show the dominant phenotype ie tall.
Tall
TT
T
x
tt dwarf
t
QUESTION BANK /BIOLOGY/XII
Page 42
Tt ( Tall) Dominant phenotype
(ii) If the individual is heterozygous Tt tall, the progeny will show dominant phenotype Tall
and recessive phenotype dwarf in the ratio of 1 :1.
Tt
x
tt
T, t
t,t
Tt
tt
Dominant
Recessive
phenotype
phenotype
-Thus, the genotype of the individual withdominant phenotype can be ascertained.
19)How does a test-cross help in identifying the genotype of the organism? Explain.
Ans.In the test cross , an individual showing dominant phenotype would be crossed to an
individual with recessive phenotype for the same trait.
(i) If the individual is homozygous dominant, all the individuals in the progeny would show
the dominant phenotype.
AA
x
aa
A
a
Aa (Dominant phenotype)
(ii) If the individual is heterozygous , the progeny will show dominant phenotype and
recessive phenotype in the ratio of 1 :1.
Aa
x
aa
A, a
Aa
a
aa
Dominant
Recessive
QUESTION BANK /BIOLOGY/XII
Page 43
phenotype
phenotype
-Thus, the genotype of the individual with dominant phenotypecan be ascertained.50% of the
progeny are heterozygous
20)With the help of a Punnett square, find the percentage of heterozygous tall plants in
a F2 population involving a true-breeding tall and a true breeding dwarf pea plant.
Ans.
Parents
:
Pure tall-
Gametes :
x
TT
tt
T
t
F1generation :
Selfing
Gametes
Pure dwarf
Tt-Tall
:
Tt
x
Tt
T, t
T,t
F2generation :
T
t
TT
Tt
Tall
Tall
Tt
Tt
Tall
dwarf
T
t
50% of the progeny are heterozygous tall.
21) Tallness of pea plant is a dominant trait while dwarfness is the alternate recessive
trait. When a pureline tall is crossed with a pureline dwarf, what fraction of tall plants
in F2 shall be heterozygous? Give reasons
Ans.Refer above answer
F2 gen -phenotypic ratio :3 Tall : 1 Dwarf
QUESTION BANK /BIOLOGY/XII
Page 44
Genotype ratio : 1 TT :2 Tt :1 tt
-Two –third of the tall progeny is heterozygous.
-It is because the gene for tallness (T) is dominant and expresses itself in the heterozygous
condition,Tt.
22)Differentiate between multiple allelism and pleiotropy, with the help of an example
each.
Ans.
Multiple allelism
Pleiotropy
It is the phenomenon in which a gene
for a trait exists in more than two
allelic forms.
e.g. the gene for human blood groups
exists in three allelic forms. IA, IB i.
- It is the ability of a single gene to have
more than one phenotypic effect.
e.g. a single gene in garden pea controls
seed shape and size of starch grains.
SHORT ANSWER TYPE QUESTIONS (3marks )
23)A relevant portion of B-chain of haemo-globin of a normal human is given below:
Val
1
His
Leu
2
3
Thr
Pro
Gl u
u
Glu
4
5
6
7
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` Hemoglobin structure did
not change as a result of mutation ‘A` whereas hemoglobin structure changed due to
mutation 'B' leading to sickle-shaped RBCs. Explain giving reasons how could
mutation ‘B` change the haemoglobin structure and not mutation ’A`.
Ans.In mutation A, the change in the codon GAG to GAA does not change the amino acid
coded i.e. both GAG and GAA code for amino acid glutamic acid, hence there is no change
in heamoglobin structure In mutation B, the codon GAG is changed to GUG, where GUG
codes for valine while the original codon GAG codes for gultamic acid, hence there is change
in haemoglobin structure and it leads to sickle-cell anaemia
QUESTION BANK /BIOLOGY/XII
Page 45
24)Write the scientific name of the fruit-fly. Why did Morgan prefer to work with fruitflies for his experiments? State any three reasons.
Ans.Drosophila melanogaster Any 3 reasons (refer pg. 83 of NCERT text book)
25)Why is pedigree analysis done in the study of human genetics? State the conclusions
that can be drawn from it.
- Ans.Control crosses that can be performed in pea plant or some other organisms are not
possible in case of human beings
- Whether trait in question is dominant or recessive
- Whether trait is autosomal or sex linked .
26)Identify ‘a`, ‘b`, ‘c`, ‘d`, 'e' and ‘f` in the table given below:
No. Syndrome
1
Down`s
Cause
Trisomy of
21
Characteristics of
Sex Male/
affected individuals Female/Both
‘a`
(i)
‘b`
(ii)
2
‘c`
XXY
Overall masculine
development
‘d`
3
Turner`s
45 with XO
‘e`
‘f`
Ans.
No.
1
Syndrome
Down`s
Cause
Characteristics of
affected individuals
Sex Male/
Female/Both
Trisomy of
21
'a' (i) Short Statured
Both
Overall masculine
development
2
Klinefelter`s
XXY
3
Turner`s
45 with XO
(ii) Small round
head
(i) Sterile
Male
Female
(ii) lack of other
secondary sexual
characters
QUESTION BANK /BIOLOGY/XII
Page 46
27)Why are human females rarely haemophilic? Explain. How do haemophilic patients
suffer?
Ans.The gene for haemophilia is present on the X chromosome, i.e. sex –linked.
-The disorder is due to a recessive mutant allele; hence a female with XX sex chromosomes,
must be homozygous recessive to produce the disease.
-She must receive one of the defective alleles from her haemophilic father and the other Xchromosome with the defective allele from her mother, who is also haemophilic or at least a
carrier (heterozygous for the trait, XXh).
28)Recently a girl baby has been reported to suffer from hemophilia. How is it possible?
Explain with the help of a cross
Ans.The gene for haemophilia is present on the X chromosome ,i.e.sex –linked.
-The disorder is due to a recessive mutant allele; hence a female with XX sex chromosomes,
must be homozygous for the recessive gene to produce the disease.
-She must receive one of the defective alleles from her haemophilic father and the other Xchromosome with the defective allele from her mother, who is either haemophilic or a carrier
(heterozygous for the trait, XXh).
The cross is as follow:
Parents
:
Father
x
Mother
(haemophilic)
Gametes :
(carrier)
Xh Y
XXh
Xh , Y
X, Xh
Progeny :
Xh
X
Xh
Y
XXh
XhXh
Carrier
Haemophilic
Female
Female
XY
XhY
Normal Haemophilic
Male
Male
QUESTION BANK /BIOLOGY/XII
Page 47
29) Given below is the representation of amino acid composition of the relevant
translated portion of B-chain of haemoglobin, related to the shape of human red blood
cells.
(a) Is this representation indicating a normal human or a sufferer from certain related
genetic disease? Give reason in support of you answer.
(b) What difference would be noticed in the phenotype of the normal and the sufferer
related to this gene?
(c) 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?
Ans.
(a) The representation indicates a normal human : HbA is a normal peptide with glutamic acid at
the sixth position of Beta globin chain.
(b) The normal individual has biconcave, disc-like RBCs, whereas ,the sufferer has elongatedsickle –shaped RBCs
(c) Both males and females suffer equally , because it is autosomal recessive disorder.
30)Name the phenomenon that leads to situation like ‘XO` abnormality in humans.
How do humans with ‘XO` abnormalities suffer? Explain.
Ans.Aneuploidy - which is due to non disjunction of chromosomes
Non-disjunction is the phenomenon of failure of segregation of the members of
homologous pairs of chromosomes.
Absence of one of the X chromosomes ; 44 + XO, 2n = 45 chromosomes
The individual lacks secondary sexual characters.
Ovaries are rudimentary.
She is sterile.
31)Name a disorder, give the karyotype and write the symptoms a human suffers from
as a result of monosomy of sex-chromosomes.
QUESTION BANK /BIOLOGY/XII
Page 48
Ans.
Turner's syndrome
44 + XO
2n = 45
The individual lacks secondary sexual characters.
Ovaries are rudimentary.She is sterile.
32)Name the genetic disorder caused by trisomy of 21st chromosome in humans. Write
the diagnostic features of the disorder.
Ans.Down's Syndrome
Diagnostic features:
(i) Partially open mouth with furrowed tongue.
(ii) Short statured with small round head
(iii) broad palm with characteristic palm crease.
(iv) Physical, psychomotor and mental development is retarded.
Trisomy of 21st chromosome; 2n = 47
33)(a) Sickle-celled anaemia in humans is a result of point mutation. Explain.
(b) Write the genotypes of both the parents who have produced a Sickle-celled anaemia
offspring.
Ans. (a)Mutation arising due to a change in a single base pair of DNA is called point
mutation.
- the defect is caused by the single base substitution at the sixth codon of the beta chain of
hemoglobin from GAG to GUG, this leads to substitution of glutamic acid by valine.
- The mutant or defective hemoglobin molecule undergoes polymerization under low oxygen
tension causing sickle-shaped RBCs.
(b) The parents must be HbA HbS and
HBA HbS.
34)Name a disorder, give the karyotype and write the symptoms which a human male
suffers, as a result of an additional X- chromosome.
-
-
Ans.Klinefelter`s syndrome.
The individual has 22 pairs of autosomes and XXY sex chromosomes, i.e. 44 + XXY = 47
chromosomes
Symptoms:
The individual is a male.
He shows development of feminine characters like development of breasts Gynaecomastia.
Body hair is sparse.The individual is sterile.
35)During his studies on genes in Drosophila that were sex-linked, T.H Morgan found
F2 population phenotypic ratios deviated from the expected 9:3:3:1. Explain the
conclusion, he arrived at.
QUESTION BANK /BIOLOGY/XII
Page 49
-
-
Ans.Morgan`s Conclusion
The genes were located on the X- chromosomes.
In a dihybrid cross the genes did not segregate independently of each other and the ratio
deviated from 9:3:3:1( expected when the two genes are independent).
They found that when 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 gene
combinations, called recombinants.
Morgan attributed this to linkage, i.e. physical association between the genes on a
chromosome.
He also found that some genes were tightly linked and showed very low recombination,while
some genes were loosely linked and showed higher recombination frequency.
36)(i) Why are grasshopper and Drosophila said to show male heterogamety? Explain.
(ii) Explain female heterogamety with the help an example.
Ans.(a) Male heterogamety (XY) or (XO)
- A male grasshopper (XO) produces two types of gametes with reference to sex
chromosomes i.e. 50% of them with one X- chromosome and 50% of them with no Xchromosome.
- A male Drosophila (XY) produces 50% gametes with one X- chromosome and 50% of
them with one Y- chromosome.
- Since they produce two types of gametes with reference to sex chromosomes, they are said
to show male heterogamety.
Sex of the offspring is determined by the type of sperm that fertilizes the ovum.
(b) Female heterogamety (ZW)
- It is the phenomenon in which female of a species produce two types of gametes with
reference to sex- chromosomes
- It is seen in fowls, where a female has ZW sex chromosomes and produces 50% of ova with
one Z- chromosome and 50% of them with one W- chromosome.
- The sex of the offspring is determined by the type ovum fertilized.
37) Explain the mechanism of sex-determination in insects like Drosophila and
grasshopper.
-
Ans.Sex determination in Drosophila:
In Drosophila, sex determination is of XY type and both males and females have the same
number of chromosomes, i.e. 4 pairs
The males have three pairs of autosomes and XY- chromosomes.
The females have three pairs of autosomes and XX- chromosomes.
The males are heterogametic and produce two types of sperms, 50% of them having one Xchromosome and the other 50% with one Y- chromosome.
The females are homogametic and all ova contain one X- chromosome
QUESTION BANK /BIOLOGY/XII
Page 50
-
-
The sex of the insect is determined by the type of sperm fertilizing the ovum. The ovum
fertilized by the X-carrying sperm develops into a female, while that fertilized by the Ycarrying sperm develops into a male.
Sex determination in Grasshopper:
Males have one chromosome less than the females.
Sex determination in grasshopper is of XO type.
The males have only one X- chromosomes other than the autosomes.
The females have two X- chromosomes (XX) other than the autosomes.
Males produce two types of sperms, 50% of them with one X- chromosome and 50% of them
with no X- chromosome.
Ova fertilized by a sperm carrying one X- chromosome develop into females, while those
fertilized by a sperm, carrying no X- chromosome develop into males.
Who proposed chromosomal theory of inheritance? Point out any two similarities in the
behavior of chromosomes and genes.
Chromosomal theory of inheritance was proposed by Sutton and Boveri.
Similarities between genes and Chromosomes:
(i) Both genes and Chromosomes occur in pairs in normal diploid cells.
(ii) Both of them segregate during gamete formation and enter different gametes i.e. one
member into one gamete and the other member into another gamete.
(iii) Members, of each pair segregate independently of the members of the other pair(s).
38) Study the given pedigree chart showing the pattern of blood inheritance in a family.
A
AB
X
A
B
A
O
Y
a) Give the genotype of the following:
(i) Parents
(ii) The individual ‘X` in second generation
State the possible blood groups of the individual ‘Y` in the third generation.
( c) How does the inheritance of this blood group explain codominance.
Ans.
QUESTION BANK /BIOLOGY/XII
Page 51
a) (i) Father IAi and mother IBi
(ii) IAIA or IAi or ii or IAIB
(b) Y can be blood group A or O
(c ) -The gene for blood group trait has three alleles, IA, IB and i
IA and IB are not only dominant over i, but are also codominant i.e both of them express
themselves in the presence of each other, in producing the phenotype, blood group AB.
39)
Study the pedigree chart above, showing the inheritance pattern of blood groups in a
family and answer the following questions:
(a) Give the possible genotypes of the individuals 1 and 2.
(b) Which antigen or antigens will be present on the plasma membranes of the RBCs of
individuals 5 and 9?
( c) Give the genotypes of the individuals 3 and 4?
Ans.(a) Individual 1 – IB i
Individual 2 IA i
(b) Individual 5 – Both glycoprotein A and glycoprotein B.
Individual 9 - no glycoprotein
(c ) Individual 3 – IBi
Individual 4 – IAi
40)A non-haemophilic couple was informed by their doctor that there is a possibility of
a haemophilic child to be born to them. Explain the basis on which the doctor conveyed
this information. Give the genotypes and phenotypes of all the possible children who
could be born to them.
QUESTION BANK /BIOLOGY/XII
Page 52
Ans.The doctor must have used pedigree analysis / (studied the family history), which refers
to the analysis of distribution of traits in several generations of a family.
Since the non-haemophilic parents can give birth to a haemophilic child, their genotypes
should be:
Father : XY (normal)
Mother : XXh (carrier/heterozygous,non-haemopilic)
Parents : Father
Mother
x
XY
XXh
Gametes :
Progeny :
X
Y
X
x
Xhh
x
XXh
X
carrier, non-haemophilic
female
XY
XhY
Y
Normal male
Haemophilic Male
The progeny can consist of the following genotypes and phenotypes:
XX
Normal Female
XX
XXh
XY
Normal carrier, non- Normal Male
Female haemophilic
female
XhY
Haemophilic
Male
41)Study the give pedigree chart and answer the questions that follow.
(a) Is the trait recessive or dominant?
(b) Is the trait sex-linked or autosomal?
( c) Give the genotypes of the parents in generation I and of their third and fourth child
in generation II.
Ans.
(a) It is a recessive trait.
(b) It is an autosomal trait.
(c ) Generation I parents – Aa and Aa.
QUESTION BANK /BIOLOGY/XII
Page 53
Generation II – Third child –aa.
Fourth child -Aa or AA
42)Study the given pedigree chart and answer the questions that follow.
(a) Is the trait recessive or dominant?
(b) Is the trait sex-linked or autosomal?
( c) Give the genotypes of the parents shown in generation I and their third child shown
in generation II and the first grandchild shown in generation III.
Ans.a) The trait is dominant.
(b) It is autosomal.
( c) Genotypes of parents in generation I:
Female aa, male Aa
Genotype of third child in generation II : Aa.
Genotype of first grandchild in generation III: Aa.
LONG ANSWER TYPE QUESTIONS (5marks )
43)a) Differentiate between dominance and co-dominance.
b)Explain co-dominance taking example of human blood groups in the population.
a)Dominance – expression of only one characteristic in the heterozygous condition
Ans.(a) co-dominance- in a heterozygote both alleles express themselves independently
(b) Blood groups are controlled by 3 alleles of a single gene I (IA, IB, i) alleles IA and IB
are dominant over i which is recessive but IA and IB are co-dominant when occurring
together the blood group is AB.
44)(a) Why are colour blindness and thalassemia categorized as Mendelian disorders?
Write the symptoms of the disease seen in people suffering from them.
(b) 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.
Ans.a) Both are caused due to mutation / alteration in a single gene and follow Mendelian
inheritance.
Colour blindness – unable to discriminate between red and green colours .
Thalassemia – (formation of abnormal haemoglobin resulting in) Anaemia .
(b) it is due to a recessive mutation in the X chromosomes.Males have only one X
chromosome and females have two. Female will be colour blind only in a homozygous
recessive state/ both X chromosomes carry defective gene XcXc whereas male will be colour
blind if they are XcY/heterozygous .
QUESTION BANK /BIOLOGY/XII
Page 54
XXc
x
XcY
X Xc
XXc
Xc y
XXc X Y
Xc Xc
XY = ½
X Xc
Xc Y
carrier normal colour colour
blind
x
XX X Y
X Y
Xc X
Xc Y
=1
normal normal carrier colour
blind
blind
45 a) Why is human ABO blood group gene considered a good example of multiple
alleles?
b) Work out a cross up to F1 generation only, between a mother with blood group A
(homozygous) and the father with blood group B (homozygous). Explain the pattern of
inheritance exhibited.
Ans.(a) The gene controlling ABO blood group in human exists in three allelic forms, IA, IB
and i i.e. multiple alleles.
(b) Mother-homozygous A group -IAIA
Father –homozygous B group - IBIB
The progeny will be as follows:
Parents : Father
x
Mother
Gametes:
B group
A group
(IBIB)
(IAIA)
IB
IA
IAIB – AB Group
Progeny :
QUESTION BANK /BIOLOGY/XII
Page 55
The progeny will consist of individuals with only AB blood group.
The alleles IA and IB are both dominant over i
But these two alleles both express themselves when they are present together ie they are
codominant.
46)(a) Explain the mechanism of sex determination in humans.
(b) Differentiate between male heterogamety and female heterogamety with the help of
an example of each.
Ans.Humans
- A female (mother) is homogametic and produces ova, all containing one X-chromosome
and 22 autosomes. 22 + X
- A male (father) is heterogametic and produces sperms of two types – 50% of them carrying
one X- chromosome and the other 50% carrying Y- chromosome and 22 autosomes each.
- 22+X , 22+Y
- The sex of child is determined by the type of sperm fertilising the ovum.
- If the ovum is fertilised by a sperm carrying X-chromosome a female child results.
- If the ovum if fertilised by a sperm carrying Y-chromosome a male child is formed.
(b)
Male heterogamety
Female heterogamety
-It is the phenomenon in which the male
produces two types of sperms, with
reference to the sex-chromosome
- It is the phenomenon in which the female
produces two types of ova with reference to
the sex-chromosomes.
- The sex of the offspring is determined by
the type of sperm fertilizing the ovum
- The sex of the offspring is determined by
the type of ovum that is fertilised.
eg Man
eg Fowl
47)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 Mendeli an law of dominance
Ans.Inheritance of size of starch grains and seed-shape in Pisum sativum
In Pisum sativum, a single gene controls two phenotypes:
(i) the size of starch grains.
(ii) the seed-shape.
such a phenomenon, is called pleiotropy.
Starch is synthesized effectively by homozygote BB and the starch grains are
large.Homozygotes bb, have lesser efficiency for starch synthesis and hence the starch grains
produced are small.
QUESTION BANK /BIOLOGY/XII
Page 56
- After maturation, BB seeds are round and bb seeds are wrinkled.
Hetreozygotes, Bb, produce round seed, i.e. round seed-shape is dominant over wrinkled
seed-shape.
But the starch grains in them (Bb) are of intermediate size. This shows that the alleles show
incomplete dominance for the size of starch grains, though they show complete dominance
(round seeds) for seed shape.
- The cross is as follows;
Parents
:
Round
Wrinkled
Seed
Seeds
Large
Small
grains
grains
BB
bb
Gametes
:
B ,B
b, b
F1 generation :
Selfing
Gametes
F2 generation
Bb (Round seed and intermediate starch grain)
Bb
Bb
B
:
b
B
b
:
B
B BB
b
b
Bb
Round
Round
Seed
Seed
Large
Intermediate
grains
grains
Bb
bb
Round
Wrinkled
Seeds
Seed
Intermediate
Small grains
grains
The phenotypic and genotypic ratios are as follows:
QUESTION BANK /BIOLOGY/XII
Page 57
Seed
shape
3 round
Size
of
starch
grains
1 large :
1BB
1 BB
:
: 2Bb
1 wrinkled
: 1bb
2 intermediate : 1 small
:
2 Bb
:
1 bb
- The phenotypic ratio of size of starch grains has deviated from the Mendelian monohybrid
phenotypic ratio of 3 : 1 due to incomplete dominance.
48)a) State the cause and symptoms of Down’s syndrome. Name and explain the event
responsible for causing this syndrome.
(b) Hemophilia and Thalassemia are both examples of Mendelian disorders, but show
difference in the inheritance pattern. Explain how.
Ans.a) It is due to trisomy of 21st chromosome.
Trisomy of 21st chromosomes occurs due to non-disjunction, i.e. non-separation of the
homologous pair of 21st chromosomes at anaphase I of meiosis during ova formation and
ferlilisation of such an ovum having two 21st chromosomes with a normal sperm.
Symptoms of Down`s Syndrome:
(i) Partially open mouth with furrowed tongue.
(ii) Short statured with small round head
(iii) broad palm with characteristic palm crease.
(iv) Physical, psychomotor and mental developmentis retarded.
(b)
Haemophilia
Thalassemia
- Sex linked recessive disorder
- The gene for haemopilia
- is present on the
X-chromosome.
- A female passes the X-chromosome to the
male offspring while the male parent
parent to the female progeny.
- It appears more in male than in females.
- Expresses in homozygous condition in
females and in heterozygous condition in
males
Autosomal recessive disorder
The gene is present on the autosome.
- Since it is autosomal, both the parents can
pass it on to the male and female offspring
with equal chances.
- It occurs in equal frequency among male
and females.
- Expresses in homozygous condition in
both males and females
49)a) Explain a monohybrid cross, taking seed coat colour as a trait in Pisum sativum.
Work out the cross upto F2 generation.
QUESTION BANK /BIOLOGY/XII
Page 58
(b) State the laws of inheritance that can be derived from such a cross.
(c ) What is the phenotypic ratio in a dihybrid cross?
a) Monohybrid cross
- In Pisum sativum yellow seed are dominant over green seed.
- A cross between a true-breeding yellow-seeded plant and a true breeding green-seeded
plant is as follows:
Parents :
Yellow seeded
Green seeded
YY
x
yy
Gametes :
Y,Y
F1 generation
Selfing :
Yy
Gametes :
Y,y
F2 generation :
YY
Y
YY
Yellow seeded
y
Yy
yellow seeded
y,y
Yy(yellow seeded)
x
Yy
Y,y
yy
Yy
Yellow Seeded
yy
green seeded
The F2 phenotypic ratio is
3 yellow-seeded : 1 green-seeded
The F2 genotypic ratio is
1YY : 2 Yy : 1yy
(b) Mendel`s law from the above cross.
(i) Law of dominance
When individuals differing in a pair of contrasting character are crossed the character that
appears in the F1 hybrid is dominant (yellow seed coat colour) over the alternate form of the
trait that remains hidden(green seed coat colour) which is recessive.
(ii) Law of segregation:
- The two factors of the trait (Y and y) that remained together in the hybrid segregate during
gametogenesis and enter different gametes.
(c ) The phenotypic ratio of a dihybrid cross is 9 : 3: 3 : 1
50)ABO blood grouping in human population exhibits four possible phenotypes form
six different genotypes. Explain the different mechanisms of inheritance involved in
exhibiting the possibility of four phenotypes and six genotypes
Ans.Multiple allelism and codominance.
QUESTION BANK /BIOLOGY/XII
Page 59
-
-
The gene for blood group character exist in the three allelic forms IA, AB and i
It is the phenomoenon of multiple allelism as there are more than two allelic forms of a gene.
Any individual carries two of the three alleles.
The allele IA codes for glycoprotein A and the allele IB codes for glycoprotein B that are
found on the surface of RBCs. The allele i does not produce any glycoprotein.
The allele IA is dominant over i and IB is also dominant over i.
When the alleles IA and IB are together they are equally dominant and both glycoproteins A
and B are produced making the blood group AB; this phenomenon where both alleles are
equally dominant is known as codominance.
The blood group is determined by the glycoprotein (s) in the RBCs.
There are six genotypes and four phenotypes as given in table
Genotypes (s)
IAIA or IA i
IB IB or IB i
IA IB
ii
Dominance
Dominance
Codominance
Dominance
Blood group
A
B
AB
O
51) State and explain with the help of a cross, the law o f segregation as proposed by
Mendel.
Ans.Law of segregation:
- Law of segregation states that the members of the allelic pair that remained together in the
parent/hybrid segregate during gamete formation and enter different gametes.
- As a result gametes have only one allele for a trait and are pure for a character.
Parent :
Tall
Dwarf
plant x
plant
TT
tt
Gametes :
T
t
F1 generation
Self-pollination :
Gametes
:
F2 generation
Tt (Tall)
x
Tt
T,t
Tt
T,t
:
T
T
t
t
TT
Tall
Tt
Tall
Tt
Tall
tt
Dwarf
Tall plants :
Dwarf plants
3 : 1
In this case tallness is dominant and dwarfness is recessive.
QUESTION BANK /BIOLOGY/XII
Page 60
The F1 hybrid is tall (dominant character)
- The recessive character dwarfness remains hidden in the F1 but reappears in the F2 generation
without any change.
- This is because the factors T and t remained together in the hybrid but segregated during
gamete formation and entered different gametes.
Diploid condition is restored during fertilisation
52 a) Why is haemophilia generally observed in human males? Explain the conditions
under which a human female can be haemophilic.
(b) A pregnant human female was advised to undergo MTP. It was diagnosed by her
doctor that the foetus she was carrying had developed from a zygote formed by an XXegg fertilized by a Y-carrying sperm. Why was she advised to undergo MTP?
Ans.a) The gene for haemophila is present on the X-chromosome and is recessive.
- A male has only one X-chromosome and bears only one allele for the trait i. e. he is
hemizygous for the trait as Y-chromosome does not have a corresponding allele.
- A female has two X-chromosomes (one from each of her parents) so she has to be
homozygous recessive i.e. her father must be a sufferer and mother either a sufferer or a
carrier to develop the disease.
Female XXh (carrier) or( haemophilic) Xh Xh
Male - haemophilic XhY
b) The zygote will be XXY and develop into a male with Klinefelter`s syndrome.
- Such individuals are sterile.
- They show gynaecomastia and feminine character.
- To avoid such a hereditary disorder the womanwas advised MTP.
53)a) How does a chromosomal disorder differ from a Mendelian disorder?
(b) Name any two chromosomal aberration-associated disorders.
(c ) List characteristics of the disorders mentionedabove that help in their diagnosis.
Ans.
QUESTION BANK /BIOLOGY/XII
Page 61
Mendelian disorder
-Mendelian disorders are due to alteration
or mutation in a single gene.
- They are transmitted to the progeny in the
same way as Mendelian principles of
inheritance
- The pattern of inheritance can be traced in
a family tree by pedigree analysis.
e.g. Phenylketonuria, Colour blindness.
-
Chromosomal disorders
- Chromosomal disorders due to absence or
excess number of one or more
chromosome(s) (aneuploidy) or abnormal
arrangement of one/more chromosome(s).
- They are not transmitted as the afflicted
individual is sterile.
- The disease can be comfirmed by
karyotyping e.g. Down`s syndrome,
Turner`s syndrome.
b) Down`s syndrome, Turner`s syndrome.
Down's Syndrome
Diagnostic features:
(i) Partially open mouth with furrowed tongue.
(ii) Short statured with small round head
(iii) broad palm with characteristic palm crease.
(iv) Physical, psychomotor and mental development is retarded.
(v) Trisomy of 21st chromosome; 2n = 47
Turner's syndrome
Female
Absence of one of the X chromosomes ; 44 + XO chromosomes = 45
The individual lacks secondary sexual characters.
Ovaries are ruimentary.She is sterile.
54)Explain the causes, inheritance patterns and symptoms of any two Mendelian
genetic disorders
Ans.i) Sickle-cell anaemia: It is caused by a change of a single base pair in the gene on an
autosome, leading to substitution of glutamic acid by valine.
- It is transmitted from parents to the offspring when both the partners are carriers
(heterozygous)for the disease.
- Symptoms
The RBCs become elongated and sickle-shaped as the mutant haemoglobin undergoes
polymerization under low oxygen tension
Phenylketonuria:It is due to a defective allele on the autosome.
It is passed on from the parents who are heterozygous for the gene to the offspring.
Symptoms
The affected individual lacks an enzyme that converts the amino acid phenylalanine into
tyrosine. Consequently, phenylalanine gets accumulated and converted into phenylpyruvate
and other derivatives.
QUESTION BANK /BIOLOGY/XII
Page 62
- Accumulation of these compounds in the brain results in mental retardation. They are also
excreted in the urine.
55) Write the symptoms of heamophilia and sickle-sell anaemia in humans. Explain
how the inheritance pattern of the two disease differ fromeach other.
Ans.Symptoms of Haemophilia:Since a protein necessary for blood clotting is a not formed
the blood does not clot and there is non-stop bleeding in case of an injury in the afflicted
individual.
Symptoms of sickle-cell anaemia:The RBCs become sickle-shaped.The oxygentransport to the
tissues is impaired.
Haemophilia
Sickle-cell anaemia
-It is due to a defective allele present on the
X-chromosome i.e. it is a sex-linked
recessive disorder.
- More males than females are affected
- The female parent passes on the disorder
to male progeny but father never passes it
on the male progeny.
-Expresses in homozygous condition in
females and in heterozygous condition in
males
female XXh or Xh Xh
Male
- It is due to point mutation i.e. single base
pair change leading to a change in an
amino acid on an autosome i.e it is an
autosomal recessive disorder.
- Both males and females are affected
equally.
- The female parent passes on the disorder
to male or female progeny in equal
frequency and father also passes on the
disorder to male and female progeny
-Expresses in homozygous condition in
both males and females
XhY
56)A particular garden pea plant produces only violet flowers.
(a) Is it homozygous dominant for the trait or heterozygous?
(b) How would you ensure the genotype? Explainwith the help of crosses.
a) It must be homozygous dominant, since it produces only violet flowers.
(b) The plant must be crossed with a plant bearing white (recessive) flowers.
- If the progeny consists of plants all producing violet flowers the plant is homozygous
dominant (cross 1)
- If the progeny contains 50% violet-flowered plants and 50% white-flowered plants the plant
is heterozygous.(cross 2)
(i) Cross 1: Homozygous dominant
QUESTION BANK /BIOLOGY/XII
Page 63
a) It must be homozygous dominant, since it produces only violet flowers.
(b) The plant must be crossed with a plant bearing white (recessive) flowers.
- If the progeny consists of plants all producing violet flowers the plant is homozygous
dominant (cross 1)
- If the progeny contains 50% violet-flowered plants and 50% white-flowered plants the plant
is heterozygous.(cross 2)
(i) Cross 1: Homozygous dominant
Parents : Violet-flowered plants
x
White-flowered plants
VV
vv
Gametes :
V
v
Progeny :
Vv-violet-flowered
(ii) Cross 2: Heterozygous
Parents :
Violet-flowered plants
x
White-flowered plants
Vv
vv
Gametes :
V,
v
v,
v
Progeny :
V v , Vv, vv, vv
Vv
:
vv
1 Violet-flowered
1 White-flowered
57)Inheritance pattern of flower colour in garden pea plant and snapdragon differs.
Why is this difference observed? Explain showing the crosses upto F2 generation.
Ans.Inheritance of flower colour in garden pea:
Inheritance of flower colour in garden pea shows true dominance and the F1 hybrid expresses
one of the parental characters i.e. dominant trait and F2 generation shows both dominant and
recessive traits in the ratio of 3 : 1.
Parents
:
purple- flowered
x
white- flowered
PP
Gametes
:
pp
P
p
Pp
(purple-flowered)
QUESTION BANK /BIOLOGY/XII
Page 64
Selfing
:
Pp
Gametes :
x
Pp
P,p
F2 Progeny :
P,p
P
P
p
p
PP
Pp
Purpleflowered
Purpleflowered
Pp
Pp
PurpleFlowered
Whiteflowered
The phenotypic ratio is:
3 purple – flowered: 1 White-flowered.
The genotypic ratio is
1 pp : 2 Pp : 1 pp
Inheritance of flower colour in snapdragon:
- Inheritance of flower colour in snapdragon show incomplete dominance - a phenomenon in
which netiher of the two alleles is completely dominant over the other and the hybrid is
intermediate between the two.
(c)
(d)
(e)
(f)
(g)
(h)
(i)
F2 progeny
Parents :
Red flowered
RR
R
Gametes :
F1 generation :
Selfing
:
Gametes
:
:
(Rr Pink-flowered)
Rr
x
R ,r
R
R RR
QUESTION BANK /BIOLOGY/XII
x
white flowered
rr
r
Rr
R, r
r
Rr
Page 65
r
Redflowered
Pinkflowered
Rr
rr
Pink-
White
flowered
flowered
The phenotypic and genotypic ratios are the same.
Red-flowered
RR
Pink-flowered
:
Rr
1
White-flowered
:
2
rr
1
58)With the help of one example each, provide genetic explanation for the following
observations.
(i) F1 generation resembles both the parents.
(ii) F1 generation does not resembleeither of the parents.
Ans.(i) F1 generation resembles both the parents if there is codominance.
- codominance is the phenomenon in which two alleles of a gene are equally dominant and
both are expressed in the hybrid e.g. AB blood group of human, where alleles IA and IB are
codominant.
Mother-homozygous A group -IAIA
Father –homozygous B group - IBIB
The progeny will be as follows:
Parents
:
Father
x
Mother
B group
A group
(IBIB)
(IAIA)
QUESTION BANK /BIOLOGY/XII
Page 66
Gametes :
IB
IA
IAIB – AB
Group
Progeny :
The progeny will consist of individuals with onlyAB blood group.
ii) F1 generation does not resemble any of the parents, but is intermediate between the two in
case of incomplete dominance.
- Incomplete dominance is the phenomenon in which neither of the two alleles of a gene is
completely dominant over the other and hence the hybrid shows intermediate form of the
trait. eg flower colour in snapdragon
Parents :
Red flowered
x White flowered
RR
rr
Gametes :
R
r
Progeny :
Rr
(Pink-flowered)
59)a) A true breeding pea plant homozygous for axial violet flowers is crossed with
another pea plant with terminal white flowers(aavv). Work out the cross to show the
phenotypes and genotypes of F1 and F2 generations along with the ratios.
(b) State the law that Mendel proposed on the basis of such a cross
Ans.(a) Parents
:
Axial violet flowered
AAVV
Gametes :
AV
x
Terminal white flowered
aavv
av
F1 generation : AaVv – Axial-violet flowered
Selfing
:
AaVv
QUESTION BANK /BIOLOGY/XII
AaVv
Page 67
Gametes
:
AV,Av,aV,av
AV,Av,aV,av
F2generation :
AV
Av
aV
av
AV
Av
aV
av
AAVV
AAVv
AaVV
AaVv
Axial,violet- Axial,violet-
Axial,violet-
Axial,violet
flowered
flowered
flowered
Flowered
AAVv
AAvv
AaVv
Aavv
Axial
Axial
Axial
Axial
violet
white
violet
Violet
AaVV
AaVv
aaVV
aaVv
Axial
Axial
Terminal
Terminal
Violet
Violet
Violet
Violet
AaVv
Aavv
aaVv
aavv
Axial
Axial
Terminal
Terminal
violet
white
violet
white
(b) The phenotypic ratio of F2 generation:
Axial
violetflowered
9
:
Axial
WhiteFlowered
3
:
Terminal
violetflowered
3
:
Terminal
whiteflowered
1
-
This phenotypic ratio is the result of independent assortment i.e. segregation of alleles for
flower position independently of the alleles for flower colour.
- Phenotypic ratio = 9 : 3 ; 3 : 1
- genotypic ratio = AAVV : AAVv : AaVV : AaVv : AAvv : Aavv : aaVV : aaVv : aavv
1
: 2
: 2
: 4 : 1 : 2 :
1 : 2 : 1
(c) Law of independent assortment
QUESTION BANK /BIOLOGY/XII
Page 68
This law states that when two pairs of traits are combined in hybrid the factors of every
character segregate independentlyof the factors of other of characters.
60)Given below is a table showing the genotypes and the phenotypes of blood groups in the
human population? (i) Identify the genotypes (W) and (X) and the phenotypes (Y) and
(Z).
(ii) How is codominance different from incomplete dominance and dominance?
(iii) Name the pattern of inheritance exhibited by the phenotypes (Y) and (Z) in the
table.
Sr.No
Genotype
Phenotypic
1
(W)
A
2
IBIO
(Y)
3
IAIB
(Z)
4
(X)
O
Ans.(i) W = IAIA or IA I
X = ii
Y= B
Z = AB
(ii) Refer question No 91
(iii)Y : The allele IB is dominant over i and hence the phenotype is B –group.-- Dominance
Z : The alleles IA and IB are codominant and both express themselves and the phenotype is
AB - group – Codominance.
61)Using Punnett square, show the F2 result of a dihybrid cross, where the pure-bred
parents contrasting traits with reference to seed shape and seed colour in Pisum
sativum. Give the phenotypic ratio.
Ans.
QUESTION BANK /BIOLOGY/XII
Page 69
62)Let ‘Y` be the genotype symbol for dominant yellow seed colour, symbol ‘y` for
recessive green seed colour, symbol ‘R` for dominant round shape of seed and symbol
‘r` for recessive wrinkled seed shape in garden pea. Using these symbols, explain
Mendel’s law of independent assortment.
Ans.See answer above
Mendel's law of independent assortment states that when two pairs of traits are combined in a
hybrid, segregation of one pair of characters is independent of the other pair of characters.
Dihybrid cross - two Traits
1.Seed colour : Yellow YY, green yy
2. Seed shape : Round RR, wrinkled rr
Parents RRYY and rryy
Refer dihybrid cross shown above.
QUESTION BANK /BIOLOGY/XII
Page 70
63)(a) State the cause and symptoms of Down’s syndrome. Name and explain the event
responsible for causing this syndrome.
(b) Hemophilia and Thalassemia are both examples of Mendelian disorders, but show
difference in the inheritance pattern. Explain how.
Ans.(a) It is due to trisomy of 21st chromosome.
Trisomy of 21st chromosomes occurs due to non-disjunction, i.e. non-separation of the
homologous pair of 21st chromosomes at anaphase I of meiosis during ova formation and
ferlilisation of such an ovum having two 21st chromosomes with a normal sperm.
Symptoms of Down`s Syndrome:
(i) Partially open mouth with furrowed tongue.
(ii) Short statured with small round head
(iii) broad palm with characteristic palm crease.
(iv) Physical, psychomotor and mental development is retarded.
Haemophilia
Thalassemia
- Sex linked recessive disorder
-
Autosomal recessive disorder
- The gene for haemopilia is present on the
X-chromosome.
-
The gene is present on the autosome.
-
Since it is autosomal, both the parents
can pass it on to the male and female
offspring with equal chances.
-
It occurs in equal frequency among
male and females.
-
Expresses in homozygous condition in
both males and females
- A female passes the X-chromosome to
the male offspring while the male
parent parent to the female progeny.
- It appears more in male than in females.
- Expresses in homozygous condition in
females and in heterozygous condition
in males
64)(a) List the three different allelic forms of gene ‘I` in humans. Explain the different
phenotypic expressions, controlled by these three forms.
(b) A woman with blood group`A` marries a man with blood group `O` Discuss the
possibilities of the inheritance of the blood groups in the following, starting with `yes` or
`no` for each .
(i) They produce children with blood group `A` only.
(ii) They produce children some with `O` blood group and some with `A` blood group.
Ans.(a) Inheritance of human blood-group character
QUESTION BANK /BIOLOGY/XII
Page 71
- The gene for blood group character exists in three allelic forms, IA ,IB and i, it is a case of
multiple alleles
Blood group
Genotype (s)
A
IA IA or IA i
B
IB IB or IB i
AB
IA IB
O
ii
- The alleles IA and IB are codominant and express themselves as blood group AB, when
they are together.
- Blood group O is homozygous recessive.
- (b) (i) Yes when the woman is homozygous (IAIA) for blood group A the children can be
with only blood group A
Parents
:
Mother
Father
A-group
O group
IAIAii
Gametes :
IA
i
IA i(A-group)
Progeny :
OR
No when the woman is heterozygous for blood gp A(IA i ), chidren can have either A or O
blood gp.
ii) Yes when the woman is heterozygous (IA i ) for blood gp A then some children can have
A blood gp and some O
Parents
:
Mother
Father
A-group
O group
IA iii
Gametes
:
Progeny
:
i ,
IA
i
IA i ( A ) , ii (O )
65)(a) Explain the genetic basic of ABO blood groups in human population.
(b) How do ABO blood groups explain the phenomena for dominance and
codominance?
QUESTION BANK /BIOLOGY/XII
Page 72
-
Ans.The gene for blood group character exist in the three allelic forms IA, AB and i
It is the phenomoenon of multiple allelism as there are more than two allelic forms of a gene.
Any individual carries two of the three alleles.
The allele IA codes for glycoprotein A and the allele IB codes for glycoprotein B that are
found on the surface of RBCs: the allele i does not produce any glycoprotein.
DOMINANCE
The allele IA is dominant over i and IB is also dominant over i.
When alleles i and IA are present together we get blood gpA
When alleles i and IB are present together we get blood gpB
CODOMINANCE
When the alleles IA and IB are together they are equally dominant and both
glycoproteins A and B are produced making the blood group AB, this phenomenon where
both alleles when present together are equally dominant and both express themselves is
known as codominance.
The blood group is determined by the glycoprotein (s) in the RBCs.
There are six genotypes and four phenotypes as given in table
Genotypes (s)
Blood group
A A
A
I I or I i
A
B B
B
I I or I i
B
A B
I I
AB
ii
O
QUESTION BANK /BIOLOGY/XII
Page 73
Chapter – 6 Molecular Basis of Inheritance
VERY SHORT ANSWER TYPE QUESTIONS (1marks )
1)Name the enzyme and state its property that is responsible for continuous and
discontinuous replication of the two strands of a DNA molecule
Ans.( DNA dependent )DNA – polymerase
It polymerises the nucleotides only in 5`
3` direction.
2)How does the flow of genetic information in HIV deviate from the ‘central dogma`
proposed by Francis Crick?
Ans.HIV shows reverse transcription i.e. formation of DNA on RNA template.
3)Which one out of Rho factor and sigma factor, acts as initiation factor during
transcription in prokaryote?
Ans.Sigma factor.
4)Name the specific components and the linkages between them that form
deoxyguanosine.
Ans.The nitrogenous base guanine is linked to deoxyribose sugar by N glycosidic linkage.
5)which one is tailed with adenylate residue between 3' end and 5' end of hnRNA?
Ans.3` end.
6) two basic amino acids positive charge that provide to histone proteins.
Ans.Lysine and Arginine
7)Mention the role of the codons AUG and UGA during protein synthesis,
Ans.AUG codes for methionine, and it is theinitiation codon.
UGA is the termination codon that signals termination of polypeptide synthesis.
8)State which human chromosome has:
(i) the maximum number of genes, and
(ii) the one which has the least number of genes.
QUESTION BANK /BIOLOGY/XII
Page 74
Ans.(i) Chromosome 1
(ii) Y- chromosome.
9)If the base adenine constitutes 30 per cent of an isolated DNA fragment, then what is
the expected percentage of the base cytosine in it?
Ans.20 per cent
10)
A structural gene has two DNA strands X and Y shown aboveIdentify the template
strand.
Ans.Y.
11)Why is hnRNA required to undergo splicing?
Ans.Since hnRNA has both exons (coding sequences) and introns (non-coding sequences), it
has to undergo splicing to remove non coding introns
12)The two additional processings which hnRNA needs to undergo after splicing so as
to become functional.
Ans.Capping and tailing.
13)When and at what end does the 'tailing' of hnRNA take place?
Ans.Tailing takes place after splicing. It occurs at the 3' end.
14)Which ends do 'capping' and 'tailing' ' of hnRNA occur, respectively
Ans.Capping occurs at the 5' end. Tailing occurs at the 3' end
15) How is the length of DNA usually calculated?
Ans.Length of DNA is calculated by multiplying the total number of bp with distance
between two consecutive bp.
QUESTION BANK /BIOLOGY/XII
Page 75
16) Name the parts 'A' and B' of the transcription unit givenbelow
Ans.A-Promoter; B-coding strand
17) Name the components `a’ and `b’ in the nucleotide with a purine given below:
Ans.a. Phosphate; b Purine
18)Given below is a schematic representation of a lac operon in the absence of an
inducer. Identify ‘a` and ‘b` in it.
a- Repressor
b- repressor binds to the Operator region O.
19) What are ‘a` and ‘b` in the transcription unit represented below:
Ans.a-, Terminator.b- Promoter
QUESTION BANK /BIOLOGY/XII
Page 76
20) Retroviruses have no DNA. However the DNA of the infected host cell does possess
viral DNA. How is it possible?
Ans.Reverse transcription of viral RNA into DNA. Then integrates / incorporates with the
host DNA.
SHORT ANSWER TYPE QUESTIONS (2 marks)
21)What is the genetic basis for proof that codon is a triplet and is read in a contiguous
manner without punctuations?
Ans.Since there are only four bases which code for twenty amino acids the code should be
made up of three bases i.e. (4 x 4 x 4) = 64 codons a number more than the required.
22)What is satellite DNA in a genome? Explain their role in DNA fingerprinting
Ans.DNA sequences which are repeated many a times show a high degree of polymorphism
and form a bulk of DNA in a genome called as satellite DNA .DNA from every tissue from
an individual shows the same degree of polymorphism and is heritable hence very useful in
DNA fingerprinting.
23)Between a cistron and an exon.
Ans.
Cistron
- A cistron is a segment of DNA coding-for
a polypeptide.
- It includes both introns and exons.
Exon
- Exon is the coding sequence of a gene.
- Only exons appear in a mature or
processed mRNA.
24)How do histones acquire positive charge?
Ans.A protein acquires a charge depending on the abundance of amino acid residues with
charged side chains.Histones are rich in basic amino acid residues, lysines and arginines,
which carry positivecharges in their side chains. Hence histones are positively charged.
25)Difference between VNTR and probe.
Ans.
VNTR
- It is a class of satellite DNA, referred to
as mini satellite. A small DNA sequence
is arranged tandemly in many copy
QUESTION BANK /BIOLOGY/XII
Probe
- Probe is a radioactively labeled single
stranded DNA used for hybridization
with complementary DNA .
Page 77
numbers.
26) How is the translation of mRNA terminated? Explain.
Ans.The ribosome moves from codon to codon along the m RNA.When it reaches one of the
termination codons / stop codons (UAA, UAG, UGA), which does not code for any amino
acid and there is no tRNA molecule for it,A release factor binds to stop codon terminating
translation.The polypeptide synthesised is released from the ribosome.
27) Explain the dual function of AUG codon. Give the sequence of bases it is
transcribed from and its anticodon.
Ans.Dual function of AUG codon.
- It codes for amino acid, methionine.
- It functions as the initiation codon.
It is transcribed by TAC on DNA.
Its anticodon is UAC.
28)
Look at the above sequence and mention the events A , B and C
(b) What does central dogma state in molecular biology? How does it differ in some
viruses?
Ans.
a) A - Replication of DNA.
B - Transcription.
C - Translation.
(b) - The central dogma of molecular biology states that the genetic information flows from
DNA to RNA to proteins.
- In some viruses (Retroviruses), the flow of information is in reverse direction, i.e. from
RNA to DNA and is called reverse transcription.
29)Give one function each of histone protein and non-histone chromosome protein in an
eukaryotic cell.
QUESTION BANK /BIOLOGY/XII
Page 78
Ans.Negatively charged DNA is wrapped around positively charged histone octamer to form
nucleosome. Non-histone chromosomal proteins help in packaging of chromatinat higher
levels.
30)Draw a schematic diagram of a part of double-stranded dinucleotide DNA chain,
having all the four nitrogenous bases and showing the correct polarity.
Ans.
SHORT ANSWER TYPE QUESTIONS (3marks)
31)
(a) Write what DNA replication refers to.
(b) State the properties of DNA replication model.
(c) List any three enzymes involved in the process along with their functions.
Ans.
(a) DNA synthesis
(b) (i) Semi conservative
(ii)Semi discontinuous
(iii)Unidirectional
(iv)occurs in the replication fork
(v)Origin of replication
(c) DNA polymerase – adds nucleotides
RNA primase – synthesizes primer.
32)Describe the structure of an RNA poly-nucleotide chain having four different types
of nucleotides.
Ans.
- A nucleotide has three components —
QUESTION BANK /BIOLOGY/XII
Page 79
-
a nit rogenous base, a pentose (ribose) sugar and a phosphate group
There are two types of nitrogenous bases—purines and pyrimidines.
The purines are adenine and guanine
The pyrimidines are cytosine and uracil.
A nitrogenous base is linked to the ribose sugar through a N-glycosidic linkage, to form a
nucleoside (adenosine, guanosine, cytidine or uridine)
- When a phosphate group is attached to 5'-OH of a nucleoside, through a phosphoester
linkage, a nucleotide is formed
- Two nucleotides are linked through 3-5' phosphodiester linkage to form a dinucleotide
- When more nucleotides are joined in this manner, it becomes a polynucleotide
- A polynucleotide has at one end a free phosphate moiety at 5' end of ribose sugar; it is
referred to as the 5'-end of the polynucleotide
- The other end of the polynucleotide has a free 3'-OH group of ribose; it is called the 3'-end
ofthe polynucleotide chain
- The backbone of the polynucleotide chain is formed by the sugar and phosphates and the
nitrogenous bases project from the backbone
33)Given below are the sequences of ' nuleotides' in a particular mRNA and amino
acids coded by it.
UUU AUG UUC GAG UUA GUG UAA
Phe - Met - Phe - Glu - Leu – Val
Write the properties of genetic code that can be and that cannot be correlated from the
above given data
Ans.
i) UAA does not code for any amino acid; it is a termination /stop codon
ii) Genetic code is specific and unambiguous, - one codon codes for only one amino acid.
iii) Genetic code is degenerate, some amino acids are coded by more than codon, eg UUU
and UUC code for phenylalanine.
iv) Genetic code is read in a contiguous manner without any punctuation
v) The codon is a triplet
vi) The code is universal
vii) AUG has dual functions :it is initiation codon as well as codes for Methionine.
34)Correlatethe codons on mRNA strand with the amino acids of a polypeptide
translated
5'
AUG ACC UUU CAC UUC GUG UAA 3'- mRNA
Met -Thr- Phe - His - Phe- Val - Translated polypeptide
Infer any three properties of genetic code with examples from the above information.
Ans.
(i)Genetic code is specific and unambiguous, ie one codon codes for only one amino acid
only.
(ii) Genetic code is degenerate, as one amino acid is coded by more than codon, eg UUU and
UUC code for phenylalanine.
QUESTION BANK /BIOLOGY/XII
Page 80
(iii) Genetic code is read in a contiguous manner without any punctuation (any three)
(iv) The codon is a triplet.
35)It is established that RNA is the first genetic material. Explain giving three reasons.
Ans.RNA is the first genetic material because:
(i) RNA can directly code for the synthesis of proteins and hence can easily express the
character; it is the genetic material in many viruses.
(ii) RNA can also act as a catalyst; there are some important biochemical reactions in living
systems that are catalysed by RNA catalysts and not proteins
(iii) Many essential life processes such as metabolism, translation, splicing etc have evolved
around RNA.
36)(a) Name the enzyme responsible for the transcription of tRNA and the amino acid,
the initiator tRNA gets linked with.
(b) Explain the role of initiator tRNA in initiation of protein synthesis
Ans.
(a)RNA polymerase III transcribes tRNA. Methionine is the amino acid
(b) -The initiator tRNA binds to the amino acid, methionine, at its amino acid acceptor site .
- At its anticodon loop, it has the anticodon for methionine, i e UAC;
- it recognises the start codon (AUG) on the m RNA and binds to it following
complementarity of bases rule.
37)How are the structural genes inactivated in lac operon in E.coli? Explain
Ans.The lac operon consists of
(i) three structural genes (z, y and a) which code for enzymes.
(ii) an operator, which controls all the structural genes as a unit.
(iii) a regulatory gene or inhibitor gene and
(iv) a promoter, where RNA polymerase binds for transcription.
- The regulatory gene codes for a repressor protein (all the time constitutively).
In the absence of an inducer - lactose, the repressor protein has high affinity for the operator
and binds to the operator region and prevents RNA polymerase from transcribing the
structural genes, i.e.. the lac operon is switched off or inactive.
38)How are the structural genes activated in the lac operon in E.coli?
QUESTION BANK /BIOLOGY/XII
Page 81
Ans.The lac operon consists of
(i) three structural genes (z, y and a) which code for ß-galactosidase, permease and
transacetytase, respectively
(ii) an operator, which controls the structural genes as a unit
(iii) a regulatory gene (i.e, inhibitor gene) and
(iv) a promoter, where the RNA-polymerase binds for transcnption
- The regulatory gene codes for the repressor protein, all the time (constitutively). In the
absence of an inducer, lactose, the repressor binds to the operator to inactivate the operon
- In the presence of lactose (inducer)- lactose binds to the repressor and preventsit from
binding to the operator
- This allows RNA polymerase an access to the promoter end transcription starts, i.e lac
operon is activated
39)Why DNA is considered a better hereditary material than RNA
Ans.DNA is a better genetic material for the following reasons.
(i) The genetic material should be stable and should not change with age or change in
physiology. This stability is given to DNA by its
(a) double stranded nature
(b) presence of thymine
(c) deoxyribose sugar
(ii) RNA cannot be genetic material because the 2`-OH group of RNA nucleotides is a
reactive group that makes RNA labile and easily degradable.
(iii) RNA is also catalytic i.e. it is reactive.
40)How ishnRNA processed to form mRNA?
Ans.In Eukaryotes:
- The structural genes are split.
- They have coding sequences (exons) interspersed with non-coding sequences (introns).
- The Primary transcript of RNA hnRNA undergoes a process called splicing by which the
introns are removed and the exons are joined together.
- The hnRNA (precursor of mRNA) also undergoes capping and tailing to become mRNA.
- In capping methyl guanosine triphosphate is added to the 5` end of hnRNA
- In tailing adenylate residues (about200-300) are added at the 3` end.
The fully processed mRNA is released from the nucleus into the cytoplasm.
41)What is hnRNA? Explain the changes hnRNA undergoes during the processing to
form to mRNA
Ans.Heterogeneous nuclear RNA-- precursor of mRNA in eukaryotes
- It has coding sequences (exons) interspersed with non-coding sequences (introns).
- It undergoes a process called splicing by which the introns are removed and the exons are
joined together.
QUESTION BANK /BIOLOGY/XII
Page 82
- The hnRNA (precursor of mRNA) undergoes capping and tailing to become mRNA.
- In capping methyl guanosine triphosphate is added to the 5` end of hnRNA
- In tailing adenylate residues (about200-300) are added at the 3` end.
The fully processed mRNA is released from the nucleus into the cytoplasm.
42)State the conditions when 'genetic code' is said to be
(i) degenerate
(ii) unambiguous and specific
(iii) universal.
Ans.Genetic code is degenerate as some amino acids are coded by more than one codon, e.g
proline and glycine are coded by four codons each
- Genetic code is unambiguous and specific as each codon codes only for a particular amino
acid, e. g. GUG codes only for valine, UUU codes only for phenylalanine AUG codes only
for methionine.
Genetic code is universal as one codon codes for the same amino acid in all organisms, be it
a bacterium or human e.g. AUG codes for methionine in all organisms.
43)Explain the process of transcription in a bacterium.
Ans.Transcription unit in a bacterium consists of a promoter, structural genes and a
terminator
- In bacteria, there is a single RNA polymerase, which catalyses transcription of alt the three
types of RNAs (mRNA, tRNA and rRNA).
Initiation - RNA polymerase binds to the promoter
Uses nucleoside triphosphates as substrate and polymerises in a template depended fashion
following the rule of complementarity.
- It also facilitates opening of the double helical DNA and only the DNA strand with 3' to 5'
polarity is transcribed. as the enzyme can polymerise the nucleotides only in 5' to 3' direction
- Once the polymerase reaches the terminator sequence, the nascent RNA falls off as also
RNA polymerase .This results in termination of transcription.
44)The base sequence in one of the strands of DNA is TAGCATGAT.
(i)
(ii)
(iii)
(iv)
Give the base sequence of its complementary strand.
How are these base pairs held together in a DNA molecule?
Explain the base complementarity rule.
Name the scientist who framed this rule
Ans.(i)ATC GTA CTA
(ii) They are held together by weak hydrogen bonds between adenine and thymine and
guanine and cytosine.
(iii) A purine always pairs with a pyrimidine and also very specifically adenine pairs with
thymine and guanine with cytosine.
QUESTION BANK /BIOLOGY/XII
Page 83
(iv)Erwin Chargaff
45)Name the enzyme that catalyses the transcription of hnRNA.Why does the hnRNA
need to undergo changes? List the changes hnRNA undergoes and where in the cell
such changes take place.
Ans.RNA polymerase II catalyses transcription of hnRNA in eukaryotes
(ii) - Since eukaryotes have split gene arrangement the hnRNA has both coding sequences
(exons) and non-coding sequences (introns) and is non-functional; so it has to undergo
processing to make it functional.
- splicing, the process, in which introns are removed and exons are joined.
- capping, an unusual nucleotide, called methyl guanosine triposphate is added at the 5' end of
hnRNA
- tailing, about 200-300 adenylate residues are added at the 3' end in a template-independent
manner.
- take place in the nucleus.
46)Describe the initation process of transcription in bacteria.
Ans.Initiation process of transcription in bacteria:
- The RNA polymerase binds transiently to an initiation factor (Sigma a factor) and binds to a
specific sequence on the DNA, called promoter, to initiate transcription.
-
The DNA strand with 3` ------ 5`polarity acts as the template.
47) Describe the elongation process of transcription in bacteria.
Ans.Elongation process of Tanscription in bacteria.
- After binding to the promoter the RNA polymerase facilitates the opening of the DNA.
- It uses nucleoside triphosphates as substrate and polymerises the nucleotides in a templatedependent fashion following complementarity of bases in the 5`to 3` direction.
QUESTION BANK /BIOLOGY/XII
Page 84
The process continues till the RNA polymerase reaches the terminator region on the DNA
strand.
48)(a)List the structural genes involved in the digestion of lactose in E.coli.
(b)Highlight the function of any one.
(c)What triggers the transcription of these genes?
Ans.Structural genes of lacoperon
- There are three structural genes, z, y and a in lac operon
Gene z codes for ß-galactosidase (ß gal), that hydrolyses lactose into glucose and galactose
Gene y codes for permease that helps increase permeability of the cell.
- Gene a codes for transacetylase that makes lactose into its active form.
(b) -Presence of Lactose as an inducer
- It binds to the repressor coded by the regulatory (i) gene and prevents it from binding to the
operator.
- So, the RNA polymerase gains access to promoter and transcription of the structural genes
is triggered.
49)(a)Why is tRNA called an 'adapter'?
(b) Draw and label a secondary structure of tRNA. How does the actual structure of
tRNA look like?
Ans.Since tRNA on one hand binds to a specific amino acid and on the other hand reads the
code on mRNA, it is called an ‘adapter`.
(b)
-It actually looks like an inverted L.
50)How do initiation and termination of translation processes occur in bacteria? Where
are untranslated regions located in an mRNA? Mention their role.
Ans.Initiation of translation
- When the small subunit of ribosome binds to the mRNA the process of translation starts: in
bacteria the ribosome also acts as a catalyst (23S rRNA is ribozyme) for peptide bond
formation
- The ribosome binds to mRNA at the Start codon (AUG), that is recognised by the initiator
tRNA; it involves certain initiation factors.
QUESTION BANK /BIOLOGY/XII
Page 85
Termination of translation
-
When the ribosome reaches a termination codon, translation is terminated and the
polypeptide is released from the ribosome
Untranslated regions are present at the 5' end before the start codon and also at the 3' end
after the termination codon.
They are required for efficient translation
51)(a)What is this diagram representing?
(b) Name the parts a, b and c.
(c) In eukaryotes, the DNA molecules are organised within the nucleus. How is the
DNA molecule organized in a bacterial cell in absence of a nucleus?
Ans.It is a nucleosome.
(b) a – Histone octamer
b – DNA
c – H1 Histone.
(c) In prokaryotes the DNA is held with some positively-charged proteins to form a nucleoid
(in the cytoplasm).
52)What is satellite DNA in a genome? Explain their role in DNA fingerprinting.
Ans.DNA sequences which are repeated many times, show a high degree of polymorphism
and form a bulk of DNA in a genome, called satellite DNA.
DNA from every tissue from an individual shows the same degree of polymorphism and is
heritable, hence very useful in DNA fingerprinting.
53)The role of
35S
end 32P experiments conducted by Hershey and Chase.
Ans.
- The viruses/bacteriophages grown on radioactive sulphur (35S) contained, radioactive
protein but not radioactive DNA, because DNA does not contain sulphur.
- When these viruses were allowed to infect bacteria the bacteria did not contain
radioactivity, because proteins did not enter the bacteria; hence protein is not the genetic
material
QUESTION BANK /BIOLOGY/XII
Page 86
- The viruses grown on radioactive phosphorus (32P) contained radioactive DNA, because
DNA contains phosphorus and not proteins.
When these viruses were allowed to infect the bacteria, the bacteria were radioactive,
indicating that DNA is the genetic material that has passed from the virus into bacteria.
LONG ANSWER TYPE QUESTIONS (5marks)
54)Explain the process of protein synthesis from processed m-RNA.
Ans.For initiation the ribosome binds to the mature m-RNA at the start codon (AUG) that is
recognized by the initiator t – RNA. During elongation charged tRNA sequentially binds to
the appropriate codon in m-RNA with the anticodon present on tRNA. The ribosome moves
form one codon to another adding amino acids one after the other to form a polypeptide i.e.
translation. During termination the release factor binds to stop codon (UAA, UAG, UGA)
terminating translation and releasing the polypeptide chain.
55)a) Write the conclusion drawn by Griffith at the end of his experiment with
Streptococcus pneumoniae.
(b) How did Avery, C. MacLeod and M.McCarty prove that DNA was the genetic
material? Explain.
Ans.
a) Frederick Griffith (1928) performed the experiments on Bacterial transformation with
Streptococcus pneumoniae, the bacterium that causes pneumonia.
QUESTION BANK /BIOLOGY/XII
Page 87
- He concluded that the R-strain had somehow been transformed by the heat-killed S-strain
bacteria. Some transforming principle, transferred from the heat killed S strain had enabled
the R strain to synthesize a smooth polysaccharide coat and become virulent and this must be
due to the transfer of genetic material.
(b) Avery, MacLeod and McCarty.
- They purified biochemicals like proteins, DNA and RNA from the heat-killed S-cells.
- When these fractions were added individually to the culture of live R-cells, only DNA
was able to cause transformation of R-cells into S-cells.
- They also found that protein digesting enzymes and RNA- digesting enzymes did not affect
transformation indicating that transforming substance is not a protein or RNA.
- Digestion with DNase did inhibit transformation. This suggests that DNA caused
transformation and therefore DNA is the genetic material.
56)
a)Write the specific features of the genetic code AUG.
b) codes can be universal and degenerate. Write about them giving one example of each.
c) aminoacylation of the tRNA.
Ans.
(a) AUG – is starting codon, codes for methionine
(b) Universal – UUU codes for phenylalanine in all organismDegenerate – many codons
code for same amino acid –(any one eg) – UUU/ UUC both code for phenylalanine
(c) Amino acid is activated in the presence of ATP linked to their cognatetRNA.
57)Describe Frederick Griffith`s experiment on Streptococcus pneumoniae. Discuss the
conclusion he arrived at.
Ans.Frederick Griffith (1928) performed the experiments on Bacterial transformation with
Streptococcus pneumoniae, the bacterium that causes pneumonia.
- He observed two strains of this bacterium, one forming smooth shiny colonies with capsule
(S-type) and the other forming rough colonies without capsule (R-type).
- The S-type cells are virulent while the R-type cells are non-virulent
- When live S-type cells were injected into the mice, they suffered from pneumonia and died
- When live R-type cells were injected into the mice, the disease did not appear and the mice
survived.
- When heat-killed S-type cells were injected, the disease did not appear and mice lived.
- When heat-killed S-type cells were mixed with live R-type cells and injected into the mice,
the mice died of pneumonia and live S-type cells were isolated from the body of the dead
mice.
- He concluded that the R-strain had somehow been transformed by the heat-killed S-strain
bacteria into live S strain, which must be due to the transfer of genetic material. But he did
notspecify the biochemical nature of the genetic material.
58)a)Explain the experiment performed by Griffith on Streptococcus pneumoniae.What
did he conclude?
QUESTION BANK /BIOLOGY/XII
Page 88
b)Name the three scientists who followed up Griffith`s experiment.
c)What did they conclude and how?
Ans.- When heat-killed S-type cells were mixed with live R-type cells and injected into the
mice, the mice died of pneumonia and live S-type cells were isolated from the body of the
mice.
- He concluded that the R-strain had somehow been transformed by the heat-killed S-strain
bacteria, into S strain which must be due to the transfer of genetic material, the transforming
principle.
(b) Avery, MacLeod and McCarty.
c) - They purified biochemicalslike proteins, DNA and RNA from the heat-killed S-cells.
- When these fractions were added individually to the culture of live R-cells, only DNA
was able to cause transformation of R-cells into S-cells.
- They also found that protein digesting enzymes and RNA- digesting enzymes did not affect
transformation indicating that transforming substance was not a protein nor RNA.
- Digestion with DNase did inhibit transformation suggesting that DNA caused
transformation and is the genetic material.
59)Explain Mac Leod, McCarty and Avery`s work that followed Griffith`s experiment.
State the conclusion they arrived at.
Ans.They purified biochemicals like proteins, DNA and RNA from the heat-killed S-cells.
- When these fractions were added individually to the culture of live R-cells, only DNA
was able to cause transformation of R-cells into S-cells.
- They also found that protein digesting enzymes and RNA- digesting enzymes did not affect
transformation indicating that transforming substance is not a protein or RNA.
- Digestion with DNase did inhibit transformation this suggests that the DNA caused
transformation.
DNA is the genetic material.
60)
(a) Identify the structure shown above.
QUESTION BANK /BIOLOGY/XII
Page 89
(b) Redraw the structure as a replicating fork and label the parts.
(c) Write the source of energy for this replication and list the enzymes involved in this
process.
(d) Mention the difference in the synthesis based on the polarity of the two
template strands.
Ans.(a) It represents the parental / template DNA strands.
(b)
The deoxyribouncleoside triphosphates provide energy.
- DNA dependent DNA polymerase and DNA ligases are the enzymes involved.
DNA polymerase polymerises the deoxyribonucleoside triphosphates in the 5`to 3` direction
into the polynucleotide chains.
- On the template strand with 5' to 3' polarity the polynucleotide strand is synthesized in short
stretches. Later these segments are joined together by DNA ligases
(d) On the template strand with 3`to 5` polarity DNA synthesis is continuous in the 5`to 3`
direction.
- On the template strand with 5`to 3` polarity DNA synthesis is discontinuous DNA strand is
synthesized as short segments also in the 5`to3` direction and the discontinuously synthesized
segments become joined by DNA ligase.
61)
Shown above is the electron micrograph (EM) picture of 'beads-on-string'.
(a)Identify and explain the detailed structure of a bead with the help of a labelled
diagram.
(b) Describe the packaging of ‘beads –on -string' in a eukaryotic cell.
Ans.
(a) The bead-like structures are the nucleosomes.
(b) There is a set of positively charged proteins, called histones, rich in basic ammo acid
residues, lysine and arginine
QUESTION BANK /BIOLOGY/XII
Page 90
(c) Histones are organised to form a unit of eight molecules, called histone octamer.
(d) The negatively charged DNA is wrapped around the positively charged histone
octamer, to form a nucleosome; a nucleosome contains 200 bp of DNA helix
b) The nucleosomes constitute the repeating units of chromatin, which appear as beads-onstring structure under an electron microscope.
- These are further packaged to form the chromatin fibres, which condense to form
chromosomes.
- The packaging of chromatin at higher levels requires additional set of proteins called nonhistone chromosomal (NHC) proteins.
62)How did Meselson and Stahl experimentally prove that DNA replication is semi conservative? Explain
Ans.It was to show that DNA replication is semi conservative.
- After completion of replication, each new/daughter molecule of DNA has one parental
strand and one newly synthesised strand, this is called semi-conservative replication of DNA.
Experiment
They grew E.coli, in a medium containing 15NH4Cl, for many generations until 15N was
QUESTION BANK /BIOLOGY/XII
Page 91
- incorporated into the two strands of newly formed E.coli cells;
- Then they transferred the cells into a medium with normal 14NH4Cl and took out samples
atvarious time intervals
- the 'heavy' DNA can be separated from the normal 14N-DNA by centrifugation in cesium
chloride (CsCl) density gradient
- They extracted the DNA and centrifuged it to measure the densities.
- The DNA extracted from cells after one generation after transfer from the 15N medium to
14
N medium (i.e. after about 20 minutes), had an intermediate/ hybrid density.
- The DNA extracted after two generations (i.e. after 40 minutes) consisted of equal amount
of 'light' DNA and 'hybrid' DNA.
This proves that after replication, each DNA molecule has one parental strand and one
newly-synthesised strand, i.e. replication is semi conservative
63)(a)Differentiate between repetitive and satellite DNA.
(b) How can satellite DNA be isolated? Explain.
(c) List two forensic applications of this technology.
Ans.
(a)
Repetitive DNA
Satellite DNA
- Repetitive DNA refers to the sequences of
DNA in which a small stretch of DNA is
repeated many times
- refers to those repetitive DNA sequences
- which do not code for proteins but form a
large part of human genome
- Satellite DNA can be classified as
micro satellites, mini satellites etc.
- Show high degree of polymorphism
(b) Satellite DNA is separated from the genomic DNA by density gradient centrifugation;
the satellite DNA forms smaller peaks, while the genomic DNA forms a major peak,
(c) (i) to identify a criminal from the crime site
(ii) It is the basis of paternity testing, in case of disputes.
(iii) in determining population and genetic diversities
64)(a) Describe the process of synthesis of fully functional mRNA in a eukaryotic cell.
(b) How is this process of mRNA synthesis different from that in prokaryotes?
Ans.
(a) Synthesis of mRNA in eukaryotes.
- In eukaryotes.
QUESTION BANK /BIOLOGY/XII
Page 92
RNA polymerase II transcribes the precursor of mRNA, called heterogeneous nuclear RNA
(hnRNA).
- This primary transcript contains both coding sequences (exons) and non-coding sequences
(introns) and is non-functional.
- So it has to undergo splicing, i.e.removal of introns and joining of exons.
- After splicing, the hnRNA has to undergo capping, i.e. the process in which an
unusual nucleotide, methyl guanosine triphosphate is added 5' end
Then it undergoes tailing, the process in which about 200-300 adenylate residues
are added at the 3' end of hnRNA.
This fully processed hnRNA is now called mRNA and is ready for release into
cytoplasm for translation.
(b)
mRNA synthesis in prokaryotes
-
There is a single RNA polymerase to
catalyse formation of all types of RNAs.
There is no need for splicing as
information is continuous.
There is no processing of RNA
transcript.
takes place in cytoplasm
mRNA synthesis in eukaryotes
- There are three RNA polymerases and
polymerase II catalyses synthesis of
mRNA
-There is need for splicing as information
is split.
- The hnRNA also undergoes capping
and tailing.
takes place in nucleus
65)Name the scientists who proved experimentally that DNA is the genetic material.
Describe their experiment.
-
-
Ans.Alfred Hershey and Martha Chase proved that DNA is the genetic material.
Experiment of Hershey and Chase
They used bacteriophages and E.coli
- They made two different preparations of the phage, in one, the DNA was made radioactive
with 32P and in the other, the protein coat was made radioactive with 35S.
These two phage preparations were allowed to infect the E. coli bacterial cells separately.
- Soon after infection, the cultures were gently agitated in a blender to separate the protein
coats of the virus from the bacterial cells
- The culture was then centrifuged to separate the viral particles and the bacterial cells.
- It was found that when the phage containing radioactive DNA was used to infect the
bacteria , radioactivity was found in the bacterial cells (in the sediment) indicating that the
DNA had been passed into the bacterial cell.
- Bacteria that were infected with viruses that had radioactive proteins were not radioactive
showing that proteins did not enter the bacteria from viruses
QUESTION BANK /BIOLOGY/XII
Page 93
-
Therefore, DNA is the genetic material that passed on from viruses to bacteria and not
proteins.Refer diagram given in above answer.
66)Describe with the help of labelled diagrammatic sketches, the experiments
conducted by Hershey and Chase. Write the inference drawn by them.
Ans.They proved that DNA is the genetic material
67)Answer the following questions based on Meselson and Stahl`s experiment
 Why did the scientists use 15NH4Cl and 14NH4Cl as sources of nitrogen in the
culture medium for growing E.coli?
 Name the molecule(s) that 15N got incorporated into.
 How did they distinguish between 15N labelled molecules from 14N ones?
 Mention the significance of taking the E.colisamples at definite time intervals for
Observations
 Write the observations made by them from the samples taken at the end of 20
minutes and 40 minutes respectively.Write the conclusion draw by them at the
end of their experiment.
Ans.
(a) They used 15NH4Cl and 14NH4Cl to grow E.coli so that the density of DNA of E.coli
grown in these media could be distinguished. 15N is the heavy isotope of Nitrogen.
(b) 15N was incorporated into the nitrogenous bases (adenine, guanine, cytosine and
thymine) of DNA
15
(c) The N-DNA can be distinguished from the 14N-DNA by centrifugation in cesium chloride
density gradient. 15N-DNA is heavier than 14N-DNA.
(d) They took samples at definite intervals that is after every generation as the DNA
replicated after every 20 min and extracted the DNA
(e) At the end of 20 minutes the DNA of E.coli was a hybrid (one heavy strand and one light
strand) and had intermediate density.At the end of 40 minutes the DNA extracted from
QUESTION BANK /BIOLOGY/XII
Page 94
the cells contained equal quantities of hybrid DNA and light DNA (both strands having
light 14N).
(f) They concluded that DNA replication is semi-conservative.
68)a) State the arrangement of different genes that in bacteria is referred to as `operon’
b) Draw a schematic labeled illustration of lac operon in a `switched on’ state.
c) Describe the role of lactose in lac operon.
Ans.
(a) The different genes in an operon are
(i) Structural genes
(ii) Operator gene,
(iii) Promoter gens and
(iv) Inhibitory/Regulatory gene
- The operator gene 'O' lies adjacent to the structural gene(s) 'z', 'y' , 'a' .On its other side,
lies the promoter gene 'p'
- Promoter gene on its other side has regulatory/inhibitor gene 'i' .
(b)
c)
- Lactose is the Role of lactose substrate for the enzyme -galactosidase
- It functions as the inducer and regulates the switching on and off of the operon
When lactose is present, it combines with the repressor protein, which otherwise has high
affinity for the operator.
-inactivates the repressor from binding to the operator and hence, transcription continues, i.e.
the operon is switched on.
69)
(a) Mention the contributions of the following scientists:
(i) Maurice Wilkins and Rosalind Frankin.
(ii) Erwin Chargaff.
(b) Draw a double-stranded dinucleotide chain with all the four nitrogen bases Label
the polarity and the components of the dinucleotide.
Ans.
a)
QUESTION BANK /BIOLOGY/XII
Page 95
(i) Maurice Wilkins and Rosalind Franklin provided the X-ray diffraction data of DNA.
(ii) Erwin Chargaff observed that for a double-stranded DNA, the ratios between adenine
and thymine and between guanine and cytosine are constant and gave the base pairing/
complementarity rule that A pairs with T and C with G.
b)
70)The schematic representation of the genes involved in the lac operon given below
and answer the question that follow
p
i
p
o
z
y
a
(i) Identify and name the regulatory gene in this operon. Explain its role in
`switching off’ 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 functions of these gene products.
Ans.
i) The ‘i` gene is the regulatory gene of the operon.
- It codes for the repressor protein of the operon, which is synthesised constitutively.
The repressor has affinity for the operator gene; it binds to the operator and prevents the
RNA polymerase from transcribing the structural genes.
(ii) When the product of regulator gene acts to stop transcription it is called negative
regulation. In the lac operon the repressor inhibits transcription therefore its regulation is
referred to as negative regulation.
(iii) Lactose is the inducer molecule.
- Gene ‘z' codes for -galactosidase, which is responsible for the hydrolysis of lactose into
galactose and glucose
- Gene ‘y` codes for permease, which increases the permeability of the cell to lactose.
71)
i) DNA polymorphism is the basis of DNA fingerprinting technique. Explain.
QUESTION BANK /BIOLOGY/XII
Page 96
(ii) Mention the causes of DNA polymorphism.
Ans.(i) DNA fingerprinting involves identifying differences in some specific regions in DNA
called as repetitive DNA.
These sequences show a high degree of polymorphism i.e. variation at genetic level, and
form the basis of DNA fingerprinting.
These sequences normally do not code for any proteins, but form a large portion of human
genome
Since DNA from every tissue from an individual show thesame degree of polymorphism,
they become a very important tool in forensic applications.
Polymorphisms are inheritable from parents to children, therefore, DNA fingerprinting is the
basis of paternity testing in case of disputes. Allelic sequence variation is described as DNA
polymorphism if more than one variant allele at a locus appears in human population with a
frequency greater than 0.01.
(If an inheritable mutation is observed in a population at high frequency it is referred to as
DNA polymorphism.)
- The probability of such variations is more in the non-coding sequences.These variations
keep accumulating generation after generation and form one of the basis of variability/
polymorphism
- There is a variety of types of polymorphisms ranging from single nucleotide change to very
large scale changes
ii) DNA polymorphism, i.e. variation at genetic level, arises due to mutations.
Mutations can arise in the somatic cells or the germ cells.
If a germ cell mutation does not seriously impair an individual's ability to have offspring who
can transmit the mutation, it can spread to the other members of the population through
sexual reproduction.
72)Study the flow chart given below and answer the question that follow:
(i)Name the organism and differentiate between its two strains R and S respectively.
(ii) Write the result A and B obtained in steps c and d respectively.
QUESTION BANK /BIOLOGY/XII
Page 97
(iii)Name the scientist who performed the steps (a) , (b) and ( c).
(iv) Write the specific conclusion draw form the step (d).
Ans.
Streptococcus pneumoniae.
Differences:
- R- cells are those cells that form rough colonies without a capsule and are non-virulent.
- S-cells those cells that form smooth colonies protected by a capsule and are virulent.
(ii)
A. Mice died
B. Mice lived
(iii)
Frederick Griffith.
When DNase is added to the medium, the DNA of the heat-killed cells gets digested and is
unable to carry out the transformation.This indicates that DNA caused the
transformation.They concluded that DNA is the hereditary material.
(i)
73)Name and describe the technique that helps in solving a case of paternity dispute
over the custody of a child by two different families.
Ans.
DNA-Fingerprinting
The steps/procedure of DNA-fingerprinting include the following.
(i) Isolation of DNA.
(ii) Digestion of DNA by restriction endonucleases
(iii)Separation of DNA fragments by electrophoresis.
(iv) Southern Blotting. Transferring (blotting) of separated DNA fragments to synthetic
membranes, such as nitrocellulose or nylon
(v)Hybridisation. using labeled VNTR probes
(vi) Detection of hybridized DNA fragments by autoradiography
74)
(a) State the Central Dogma of molecular biology. Who proposed it? Is it universally
applicable? Explain.
(b) List any four properties of a molecule to be able to act as a genetic material.
Ans.(a) The Central Dogma of molecular biology states that the genetic information flows
from DNA to RNA to protein, it is shown below:
QUESTION BANK /BIOLOGY/XII
Page 98
- It was proposed by Francis Crick.
- No, it cannot be considered universal, because in some viruses (retroviruses), the flow of
information is in reverse direction, i.e. from RNA to DNA; it is called reverse transcription.
- Such viruses have the enzyme reverse transcriptase.
b) Properties of Genetic Material
(i) The genetic material should be able to generate its replica ( replication ).
(ii) It should be chemically and structurally stable.
(iii) It should provide scope for slow changes (mutation) that are necessary for evolution.
(iv) It should be able to express itself in the form of 'Mendelian Characters'
75)Draw a labeled schematic structure of a transcription unit. Explain the function of
each component of the unit the process of transcription.
Ans.
Functions:
- The promoter is a DNA sequence that provides binding site for the RNA polymerase.
- The structural genes code for the enzymes/proteins and transcribe the mRNAs for the same.
- The terminator is a sequence of bases on DNA, that defines the end of transcription process.
- The strand of DNA with 3' —> 5' polarity acts as the template for transcription of mRNA.
- The strand of DNA with 5' —> 3' polarity is the coding strand; it does not code for RNA,
but all reference points regarding transcription are made with this strand.
QUESTION BANK /BIOLOGY/XII
Page 99
CHAPTER7.EVOLUTION
VERY SHORT ANSWER TYPE QUESTIONS (1 marks)
1)Write the formula to calculate allele frequency in future generations according to
Hardy – Weinberg genetic equilibrium.
Ans.(p + q)2 = p2 + 2pq + q2 = 1
2)Identify the examples of homologous structures from the following:
(a) Vertebrate hearts.
(b) Thorns in Bougainvillea and tendrils of Cucurbita.
(c) Food storage organs in sweet potato and potato.
Ans (a), and (b).
3)Write the similarity between the wing of a butterfly and the wing of a bat. What do
you infer from the above with reference to evolution?
Ans - They both help in flying.
They are analogous organs, which perform similar functions.
- Analogous organs are the result of convergent evolution.
4)Comment on the similarity between the wing of a cockroach and the wing of a bird.
What do you infer from the above with reference to evolution?
Ans They both help in flying.
They are analogous organs, which perform similar functions.
- Analogous organs are the result of convergent evolution.
5)Coelacanth was caught in 1938 in South Africa. Why is it very significant in the
evolutionary history of vertebrates?
Ans Coelacanth belongs to lobefins which evolved into the first amphibians; they are fish
with stout and strong fins that could move on land and go back to water.- connecting link
between fish and amphibians
- It was thought to be extinct.
6)When does a species become founders to cause founder effect?
Ans.When the change in allele frequency is so different in the new sample of population that
they become a different species. The original drifted population becomes founders and the
effect is called founder effect.
7)What does Hardy-Weinberg equation p2 + 2pq + q2 = 1, convey?
QUESTION BANK /BIOLOGY/XII
Page 100
Ans.It indicates genetic equilibrium, i.e. the allele frequencies in a population are stable and
remainconstant from generation to generation.
1.5 mya
2.0 mya
Java man
Homo habilis –
more – man – like
Australopithecines
– hunted
With stones
8)Study the ladder of human evolution given above and answer the following questions:
(a) Where did Australopithecines evolve?
(b) Write the scientific name of Java man.
Ans.(a) Australopithecines evolved in East African Grasslands.
(b) Homo erectus
9)Are the thorns of Bougainvillea and tendrils of Cucurbita homologous or analogous?
What type of evolution has brought such a similarity in them?
Ans.They are homologous.- Divergent evolution-has brought such a similarity.
10)Why are lichens regarded as pollution indicators?
Ans.Since, lichens do not grow in polluted areas; they are regarded as pollution indicators.
11)Name the placental mammals corresponding to the Australian 'spotted cuscus' and
'Tasmanian tiger cat', which have evolved as a result of convergent evolution.
Ans.Spotted cuscus - Lemur.
Tasmanian tiger cat - Bobcat.
12)Mention what caused evolution according to de Vries.
OR
What causes speciation according to Hugo de Vries?
Ans.According to de Vries, single-step large mutation (saltation) caused evolution/
speciation.
SHORT ANSWER TYPE QUESTIONS (2marks)
13)How do Darwin`s finches illustrate adaptive radiation?
Ans.Original stock of seed eating finches migrated to different habitats (of Galapagos Island)
and adapted to different feeding methods by altered beak structure and evolved into different
types of finches.
QUESTION BANK /BIOLOGY/XII
Page 101
14)List the two main propositions of Oparin and Haldane.
Ans.i)The first form of life could have come from pre-existing non-living organic molecules
like RNA, proteins, etc.
ii)Formation of life was preceded by chemical evolution that resulted in the formation of
diverse organic molecules from inorganic constituents
15)Name the scientist who had used the set-up shown below:
Write the purpose of 'a'in the set-up and the conclusion, the scientist arrived at.
Ans.S.L Miller.
- The electrodes (a) are used to create an electric discharge, similar to the lightning in the
primitive earth; it was to provide energy.
- He observed formation of amino acids; it proved the chemical evolution.
16)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?
1. Ans.They have evolved from a common ancestor.
2. Their forelimbs are the result of divergent evolution, i.e. the same structure developed
along different directions due to adaptations to different needs.
3. They are modified vertebrate forelimbs to perform different functions.
4. They are categorised as homologous organs.
17)How does palaeontological evidence support evolution of organisms on Earth?
Ans.Paleontology is the study of fossils. It indicates
(i)
the geological time period in which the organisms existed.
(ii)
that life forms varied over time and certain life forms are restricted to certain
geological time spans.
QUESTION BANK /BIOLOGY/XII
Page 102
(iii)
that new forms of life have arisen at different times in the history of earth.
18)Name the ancestors of man based on the features given below:
(a) Human-like meat eater with 900 cc brain lived in Java.
(b) More human with brain size 1400 cc, lived in Central Asia, used hides and buried
their dead.
(c)Human-like, vegetarian, with brain capacity between 650 cc and 800 cc.
(d) Man-like primate, that existed about 15 mya. Fossils found in Tanzania.
Ans(a) Homo erectus.
(b) Neanderthal man.
(c)Homo habilis.
(d) Ramapithecus
19) Categorise the following pairs of examples as convergent or divergent evolutions:
(a) Eyes of Octopus and mammals.
(b) Wings of butterfly and birds.
(c) Tuber of sweet potatoes and potato
(d) Thorns in Bougainvillea and tendrils in Cucurbits.
Ans.
a) Convergent evolution.
(b) Convergent evolution.
(c) Convergent evolution.
(d) Divergent evolution
20)Divergent evolution leads to homologous structures. Explain with the help of an
example.
Ans.Divergent evolution is the evolutionary process, where the same structure develops
along different directions in different groups of organisms as adaptations to different needs.
- the homologous structures are those which have a similar anatomical structure but are
modified differently to perform different functions, e.g. the thorn of Bougainvillea and
tendril of Cucurbita are homologous organs, as both of them are modified stems, which
perform different functions.
21) Convergent evolution leads to analogous structures. Explain with the help of an
example.
Ans.Convergent evolution is the evolutionary process, where anatomically different
structures in different groups of organisms evolve towards the same function, in similar
habitats.
- Such structures in different groups of organisms which perform similar function, but are
anatomically different, are called analogous organs.
QUESTION BANK /BIOLOGY/XII
Page 103
- The wings of a butterfly and those of bird are anatomically different, but perform the same
function (flying); hence they are said to be analogous organs.
22)Write Oparin and Haldane`s hypothesis about the origin of life on Earth. How does
meteorite analysis favour this hypothesis?
Ans.Life arose from pre existing non living organic molecules.Formation of life was
preceded by chemical evolution.Meteorite analysis revealed similar compounds to show that
similar processes occurred else when in space.
23)Convergent evolution leads to analogous structures. Explain with the help of an
example.
Ans.Convergent evolution is the evolutionary process, where anatomically different
structures in different groups of organisms evolve towards the same function, in similar
habitats.
-Such structures in different groups of organisms which perform similar function, but are
anatomically different, are called analogous organs.
- The wings of a butterfly and those of bird are anatomically different, but perform the same
function (flying); hence they are said to be analogous organs
24) Why are the wings of butterfly and birds said to be analogous organs? Name the
type of evolution, the analogous organs are a result of?
Ans.The wings of a butterfly and those of bird are anatomically different, but perform the
same function (flying); hence they are said to be analogous organs.
- Such structures in different groups of organisms which perform similar function, but are
anatomically different, are called analogous organs.
-Convergent evolution is the evolutionary process, where anatomically different structures in
different groups of organisms evolve towards the same function, in similar habitats and result
in analogous structures.
25)How is 'mutation' explained by Hugo de Vries different from the Darwinian
variations?
Ans.
Darwinian Concept
- According to Darwin, speciation occurs
with the accumulation of minor heritable
variations.
- Evolution was gradual and occurs
through a number of generations.
- Variations are small and directional.
de Vries Concept
- According to de Vries, mutations arising
suddenly in a population, are responsible for
speciation.
- Evolution occurs by a large single-step
mutation (saltation).
- Mutations are random and directionless.
26)Give the sum total of all the allelic frequencies in a stable population. List any two
factors which disturb the stability of a population. How does this disturbance affect the
population?
Ans.The sum total of all the allelic frequencies of a stable population is one.
The five factors which affect Hardy-Weinberg's equilibrium are as follows:
(i) Gene migration
- When some individuals of a population migrate to other populations or when certain
QUESTION BANK /BIOLOGY/XII
Page 104
-
individuals come into a population (ie. emigration and immigration), some genes are lost
in the first case and added in the second.
(ii) Genetic drift
Random changes in the allele frequencies of a population occurring only by chance,constitute
genetic drift.
- The change in allele frequency may become so drastically different that they form species.
(iii) Mutations
- Though mutations are random and occur at very slow rates, they are sufficient to create
considerable genetic variation for speciation to occur.
(iv) Genetic Recombination
- New combinations of genes occur due to crossing over in meiosis during gamete
formation.
(v) Natural selection
- It is the most critical evolutionary process, that leads to changes in allele frequencies and
promotes adaptation as a product of evolution.
(any two)
The disturbance in genetic equilibrium in a population would result in evolution.
27)How does ‘fitness` of a population help in evolution?
Ans.Fitness, according to Darwin, refers ultimately and only to reproductive fitness.
- Those who are better fit in an environment would outbreed those, who are less fit in that
environment;
- they would leave more reproductively fit progeny (with more fit individuals) than the
others.
- They will have better chances to survive and will be selected by nature (natural selection)
to reproduce and increase their population size
28) How is genetic drift different from gene migration? Explain.
Ans.
Genetic drift
Gene migration /Gene flow
- Random changes in the allele frequencies - It refers to the change in allelic frequencies
of a population, occurring only by chance, of a given population, when individuals
constitute genetic drift.
migrate into the population (immigration) or
from the population (emigration).
29)What does the comparison between the eyes of Octopus and those of mammals say
about their ancestry and evolution?
Ans.Eyes of Octopus and those of mammals are analogous structures, which have resulted
from convergent evolution.
-They have not evolved from common ancestors.
30) Why are the wings of butterfly and birds said to be analogous organs? Name the
type of evolution, the analogous organs are a result of?
Ans.The wings of a butterfly and those of bird are anatomically different, but perform
the same function (flying); hence they are said to be analogous organs.
Such structures in different groups of organisms which perform similar function, but are
anatomically different, are called analogous organs.
-Convergent evolution is the evolutionary process, where anatomically different structures in
different groups of organisms evolve towards the same function, in similar habitats and result
in analogous structures.
SHORT ANSWER TYPE QUESTIONS (3marks)
QUESTION BANK /BIOLOGY/XII
Page 105
31)Anthropogenic action can hasten evolution. Explain with the help of a suitable
example.
Ans.
- Anthropogenic actions, i.e. human activities have been found to hasten evolution.
Resistant organisms / cells are appearing in a time scale of months or years and not centuries.
e.g. 1. Use of DDT has resulted in evolution of DDT-resistant mosquitoes.
2. Evolution of antibiotic or drug-resistant microbes.
- When DDT was used for the first time, many mosquitoes died, but a few survived.
- These few mosquitoes showed resistance to DDT and reproduced in the presence of DDT.
- Most of the offspring were also resistant to DDT.
- Hence, the mosquito populations of today consist mainly of DDT-resistant mosquitoes and
hence, DDT is not effective.
- In the absence of DDT, these DDT-resistant mosquitoes had no advantage over those
mosquitoes, which were sensitive to DDT.
32)Explain the increase in the numbers of melanic (dark-winged) moths in the urban
areas of post-industrialisation period in England.
OR
How does industrial melanism support Darwin's theory of Natural Selection? Explain.
Ans.Industrial Melanism:
- In England, before industrial revolution, there were more white-winged or dull-grey moths
on the tree trunks, than the dark-winged or melanic moths.
- In the collection of moths, carried out in the same area after industrial revolution, there
were more dark-winged moths.
- The explanation given for this observation was that predators will spot a moth against a
contrasting background.
- Before industrialisation, there used to be a thick growth of the almost white-coloured
lichens on the tree trunks and in that background, the white-winged moths survived better;
the dark-coloured moths were easily spotted and picked up by their predators.
- During the post-industrialisation period, the trunks became dark with the industrial smoke
and soot.
Under this condition, white-winged moths did not survive as predators could easily spot
these, while dark-winged or melanic moths survived better.
- In this case, the moths which were able to camouflage and hide in the background survived
and increased their population size through reproduction.
-This shows natural selection-- those that can better adapt, survive and increase in number /
population size. However, no variant is completely wiped out.
This proves Darwin's theory of natural selection that is survival of the fittest.
33)State the theory of abiogenesis. How does Miller's experiment support this theory?
OR
State the views of Oparin and Haldane on evolution. How does S.L. Miller's experiment
support their views?
OR
(a) Explain the theory of abiogenesis.
(b) How did Miller demonstrate experimentally the chemical evolution that happened
three billion years ago?
Ans.- Theory of chemical evolution or abiogenesis was proposed by Oparin and Haldane.
- It states that the first form of life could have come from pre-existing non-living organic
molecules (like RNA, proteins, etc.) and that formation of life was preceded by chemical
evolution, i.e. formation of diverse organic molecules from inorganic constituents.
Miller's experiment:
QUESTION BANK /BIOLOGY/XII
Page 106
- He created conditions similar to the primitive earth in the laboratory.
- Electric discharge was produced (by using electrodes) in a closed flask, containing
methane, hydrogen, ammonia and water vapour.
- The temperature was kept at 800 °C
- After a week, he observed formation of amino acids.
- Such molecules must have reacted among themselves to form giant, self-replicating
molecules, the first form of life.
34)Evolution is a change in the gene frequencies in a population in response to changes
in the environment in a time scale of years and not centuries. Justify this statement with
reference to DDT. How does the theory of Hugo de Vries support this?
Ans.As the environment changes, the organisms which are better adapted to the changed
environment, would survive better and reproduce.
- When DDT was used, initially most of the mosquitoes died, but a few survived.
- These few mosquitoes reproduced and the offspring were also resistant to DDT.
- Today, the population of mosquitoes consists mainly of DDT-resistant mosquitoes
- The DDT-resistant mosquitoes have appeared in a time scale of months or years and not
centuries due to anthropogenic actions.
- Therefore, evolution is not a direct process but stochastic process based on chance events.
According to Hugo de Vries, evolution occurs due to mutations, i.e. large differences arising
suddenly in a population.
According to his theory a large, single-step mutation, -- saltation, must have been the cause
of DDT-resistant population of mosquitoes.
35) Study the schematic representation of evolutionary history of plant forms given
below and mention:
a) The plant form Ferns and Conifers are most related to.
(b) The nearest ancestors of flowering plants.
(c) The most primitive group of plants.
(d) Common ancestry Psilophyton provides to
(e) The common ancestors of Psilophyton and seed ferns.
QUESTION BANK /BIOLOGY/XII
Page 107
(f) The common ancestors of mosses and tracheophytes.
Ans.
a) Psilophyton
(b) Seed ferns
(c) Chlorophyte ancestors
(d) Ferns, Conifers, Seed ferns ,Flowering plants
(e) Tracheophyte ancestors
(f) Chlorophyte ancestors.
36)Branching descent and natural selection are the two key concepts of Darwinian
theory of evolution. Explain each concept with the help of a suitable example.
(i) Branching descent.
Ans.Homology is accounted for by the idea of branching descent
eg of homologous organs fore limb of man and fore limb of horse
.- Common ancestry
-Same structure developed along different directions due to adaptations to different needs
- Divergent evolution
(ii) Natural Selection:
eg Industrial melanism
- Nature selects those individuals who are fit in the environment
- Fitness, according to Darwin, refers to ultimately and only -reproductive fitness.
- Those individuals who are better adapted, reproduce in large number and the progeny
consists of more fit individuals, who are selected by nature, i.e. natural selection.
37)Convergent evolution and divergent evolution are the two concepts explaining
organic evolution. Explain each one with the help of an example.
Ans.Convergent evolution:
- It is the evolutionary process in which anatomically different organs in different groups of
organisms, evolve towards the same function
- Analogous organs are the result of convergent evolution, e.g. Sweet potato (root
modification) and potato (stem modification) are analogous, as they both are different
structures with the same function-storage of food
Divergent evolution:
- It is the evolutionary process in which the same structure develops in different directions in
different groups of organisms as adaptations to different needs.
- Homologous organs are the result of divergent evolution, e.g. human fore limbs are
homologous to the wings of a bird, as their basic structure (number and types of bones) is
similar but are modified differently to perform different functions
38)Given below is a diagrammatic representation of the experimental set-up usedfor his
experiment:
QUESTION BANK /BIOLOGY/XII
Page 108
Ans.(a)- The gases are methane, ammonia, hydrogen and water vapour.
- An electric discharge is created using electrodes.
- The Temperature was kept at 800 degrees C.
(b) Amino acids.
(c) life could have arisen from pre-existing non-living organic molecules and their formation
was preceded by chemical evolution formation of organic molecules from inorganic
constituents.ie. life arose from non living molecules.
39) Name and explain the principle, the given equation represents:
p2 + 2pq + q2 = 1
Ans.Hardy Weinberg principle
The given equation represents Hardy-Weinberg's (genetic) equilibrium.
This principle states that allele frequencies in a population are stable and remain constant
from generation to generation.
(a) The sum total of all the allelic frequencies is one.
- Individual frequencies are named p and q, which represent the frequency of allele A and a,
respectively, in a population of diploid organisms.
- The frequency of AA individuals is p2, while that of aa is q2 and Aa is 2pq; hence
p2 + 2pq + q2 = 1.
(b) When the frequency measured differs from the expected values, the difference indicates
the extent of evolutionary change.
- The disturbance in genetic equilibriumwould then be interpreted as resulting in evolution.
40(a) How does the Hardy Weinberg's expression (p2 + 2pq + q2 = 1) explain that
genetic equilibrium is maintained in a population?
(b) List any two factors that can disturb the genetic equilibrium.
Ans.
(a) refer above answer
(b) gene migration, genetic drift, mutation etc
41)
QUESTION BANK /BIOLOGY/XII
Page 109
(a) Recognize and explain the process by which Tasmanian wolf evolved.
(b) Give one example of an animal that has evolved along with Tasmanian wolf.
(c) Compare and contrast the two animals shown?
Ans.
(a) Adaptive radiation – The process of evolution of different species in a geographical area
starting from a point and literally radiating to other areas of geography (habitats)
(b) Tiger cat/ Banded ant eater/ Marsupial rat/ Kangaroo/ Wombat/ Bandicoot/ Koala/
Marsupial mole/ Sugar glider.
(c) Wolf is a placental mammal, whereas Tasmanian wolf is a marsupial mammal
42) Explain adaptive radiation and convergent evolution by taking examples of some
Australian marsupials and Australian placental mammals
OR
(a) What is adaptive radiation?
(b) Explain with the help of a suitable example, where adaptive radiation has occurred
to represent convergent evolution.
Ans.
- Adaptive radiation is the evolutionary process in which a common ancestor evolves into a
number of species, in a given geographical area, starting from a point and literally radiating
to other areas of geography.
- A number of Australian marsupials, each different from the other, evolved from an
ancestral stock, all within the Australian island continent.
- When more than one adaptive radiation appeared to have occurred in an isolated
geographical area, representing different habitats, it can be called convergent evolution.
- Placental mammals of Australia also exhibit adaptive radiation in evolving into varieties of
such placental mammals, each of which appears to be 'similar' to a corresponding marsupial,
Placental mammal
Australian marsupials
e.g.Bob Tasmanian tiger cat.
Lemur
Spotted cuscus.
Anteater
Numbat.
43)
(a) Write your observations on the variations seen in the Darwin's finches shown above.
(b) How did Darwin explain the existence of different varieties of finches on Galapagos
Islands?
Ans.(a) The variations in the beaks are adaptations to their food habit-flesh-eating,
insectivorous or seed/ fruit-eating.
QUESTION BANK /BIOLOGY/XII
Page 110
(b) - Darwin conjectured that all the varieties must have evolved on the same island itself.
- From the original seed-eating features, other forms with altered beaks arose, enabling
them to become insectivorous and vegetarian finches.
- This process of evolution of different species in a given geographical area, starting from
a point and literally radiating to other geographical areas, is called adaptive radiation.
44)
-
Ans.
a) Australia
(b) The phenomenon responsible for the evolution of such diverse species, is adaptive
radiation.
- Adaptive radiation is the evolutionary phenomenon, in which a common ancestral stock
evolves into a number of species in a given geographical area starting from a point and
literally radiating to other geographical areas or habitats.
(c) - When more than one adaptive radiation appeared to have occurred in an isolated
geographical area, it can be called convergent evolution.
Placental mammals of Australia have also exhibited adaptive radiation and evolved
into a variety of placental mammals, each of which appears to be similar' to a corresponding
marsupial, e.g. Bob cat and Lemur are the placental mammals, which appear similar to
Tasmanian tiger cat and spotted cuscus, respectively.
45)
a) Rearrange the following in an ascending order of evolutionary tree, reptiles,
salamander, lobefins, frogs.
(b) Name two reproductive characters that make reptiles more successful than
amphibians.
(a) Lobefins ---- Frogs ---- Salamander --- Reptiles.
(b) Reptiles are more successful than amphibians because:
QUESTION BANK /BIOLOGY/XII
Page 111
(i) fertilisation is internal.
(ii) they lay fertilised eggs, covered by a hard calcareous shell which do not dry up in the
sun.
56)How did Darwin's theory of natural selection explain the appearance of new forms
of life on earth?
Ans.Any population has built in variation for every character.
- Individuals with those characters which enable them to survive better would outbreed the
others, who are less adapted.
- Fitness, according to Darwin, is reproductive fitness, i.e. individuals who are better fit in an
environment leave more progeny than others.
- These progenies survive better and more and more fit individuals are added to the
population, ie. Natural Selection.
- The population now comes to possess more fit individuals, ie. nature selects the fit
individuals and over a long period of time, through a number of generations, the
population slowly becomes modified into a different form, or a new species
i.e Reproduction- over production of offspring
Competition - inter specific, intra specific, environmental
Struggle for existence
Survival of the fittest
Variations
New species
57) What is speciation? Explain the role of natural selection in speciation.
Ans.Formation of new species
Darwin`s theory of Natural Selection:
See above answer.
LONG ANSWER TYPE QUESTIONS (5marks)
58) List the various causes of variations in the progeny of a population.
(b) Describe the three different ways in which natural selection operates in nature, with
regard to organic evolution.
Ans.
(a) (i) Mutation,
(ii) genetic recombination during gametogenesis,
(iii) gene flow, and
(iv) genetic drift.
(v) natural selection
(b) Different ways in which natural selection results in different populations include the
following:
(i) Stabilisation—More individuals acquire mean character value, i.e. variation is much
reduced.
(ii) Directional change—More individuals acquire value other than the mean character
value.
(iii) Disruption—More individuals acquire peripheral character value at both ends of the
distribution curve.
59)
(a) How does Hardy-Weinberg equation explain genetic equilibrium?
(b) Describe how this equilibrium gets disturbed which may lead to founder effect.
QUESTION BANK /BIOLOGY/XII
Page 112
Ans.(a) Hardy-Weinberg Principle:
p2 + 2pq + q2 = 1
- It states that the allele frequencies in a population are stable and remain constant from
generation to generation; it is called genetic equilibrium.
- The sum total of all the allelic frequencies is one.
- Individual frequencies are named p and q, which represent the frequency of allele A and a,
respectively, in a population of diploid organisms.
- The frequency of AA individuals is p2, while that of aa is q2 and Aa is 2pq; hence
p2 + 2pq + q2 = 1.
(b) Genetic drift:
- Genetic drift refers to the changes in allele frequencies that occur only by chance events.
- Sometimes the change in allele frequency is so different in the new sample of population,
that they become a different species.
- The originally drifted population becomes the founder and such an effect is called founder
effect.
60)
a) Name the primates that lived about 15 million years ago. List their characteristic
features.
(b)(i) Where was the first man-like animal found?
(ii) Write the order in which Neanderthals, Homo habilisand Homo erectusappeared
on theearth. State the brain capacity of each one of them.
(iii) When did modern Homo sapiensappear on this planet?
Ans.
(a) Dryopithecus and Ramapithecus:
Characteristics:
(i) They were hairy.
(ii) They walked like gorillas and chimpanzees.
(iii) Ramapithecuswas more man-like while Dryopithecuswas more ape-like.
(b) (i) East African grasslands.
(ii) Homo habilis, Homo erectusand Neanderthals had brain capacities of 650-800 cc, 900 cc
and 1400 cc, respectively.
(iii) During the ice age between 75,000-10,000 years ago.
61)Fitness is the end-result of the ability to adapt and get selected by nature. Explain
with suitable example.
Ans.Fitness, according to Darwin, refers ultimately and only to reproductive fitness.
- Those individuals who are better fit in an environment leave more progeny than the others;
the population comes to possess more fit individuals, as nature selects for fitness.
- The so-called fitness is based on characteristics which are inherited; hence there must be a
genetic basis for getting selected and to evolve.
- In other words, adaptive ability is inherited and has a genetic basis; fitness is the ability to
adapt and get selected by nature.eg Industrial melanism.
62)(a) Natural selection operates when nature selects for fitness. Explain.
(b) The rate of appearance ofnew forms is linked to the life span of an organism.
Explain with the help of a suitable example.
Ans.Nature selects for fitness.- natural selection
QUESTION BANK /BIOLOGY/XII
Page 113
Refer earlier answer
(b) - The rate of appearance of new forms is linked to the lifespan of the species.
- Microbes have the ability to divide and multiply to produce millions of individuals within
hours.
- A colony of bacteria (say A) growing on a given medium, has built-in variation in terms of
ability to utilise a food component.
- A change in the composition of the medium will allow only those individuals who can
survive in the changed condition.
- In due course of time, the variant population (say B) would outgrow the population A and
appear as a new species; this would happen within a few days.
- This is also true for microbes to develop antibiotic-resistant varieties.
- Use of pesticides like DDT against mosquitoes has resulted in mosquito populations with
DDT-resistance within a few years.
- For a change to occur in a fish or fowl would take million years, as their lifespans are in
years.
63)(a) Explain taking one example of vertebrate anatomy that evolution of life has
occurred on earth.
(b) 'Nature selects for fitness'. Explain with suitable example.
-
Ans.(a)The forelimbs of cheetah, whales, bats and human have a similar anatomical structure
but have been modified differently to perform different functions.
All of them have the bones humerus, radius, carpals, metacarpals and phalanges.
It indicates divergent evolution where the same structure has developed along different
directions as adaptations to different needs.
Such structures are called homologous organs and homology indicates common ancestry and
divergent evolution
(b)Refer natural selection and industrial melanism.
QUESTION BANK /BIOLOGY/XII
Page 114
Chapter – 8 Human Health and Disease
VERY SHORT ANSWER TYPE QUESTIONS (1marks)
1)When does a human body elicit an anamnestic response?
Ans.When a human body encounters a pathogen for the second time, it elicits an anamnestic
response.
2)State the function of mast cells in allergic response.
Ans.Mast cells release chemicals tike serotonin and histamine which cause allergy.
3)How do interferons protect us?
Ans.Interferons protect the non-infected cells from viral infection.
4)Some allergens trigger sneezing and wheezing in human beings. What causes this type
of response by the body?
Ans.Pollen grains are allergens and cause allergic reactions in some people by causing
release of histamine and serotonin from the mast cells.
5)What is it that prevents a child to suffer from a disease he/she is vaccinated against?
Give one reason.
Ans.The vaccine contains antigens so the body develops antibodies that circulate in the body
fluids and neutralise the pathogen during actual infection.
- Also generate memory B cells and memory T which recognize pathogen quickly on
subsequent exposure and produce antibodies
6)A boy of ten years had chicken pox. He is not expected to have the same disease for
the rest of his life. Mention how it is possible.
Ans.He acquired active immunity against chicken pox.
His body developed antibodies during actual infection as well as memory B cells and
memory T cells.
7)What type of virus causes AIDS? Name its genetic material.
Ans.Retrovirus causes AIDS.
RNA is its genetic material.
8)Where are mucosal associated lymphoid tissues present in the human body and why?
QUESTION BANK /BIOLOGY/XII
Page 115
Ans.Mucosal associated lymphoid tissues are present within the lining of the respiratory,
digestiveand urinogenital tracts; they constitute about 50 per cent of the lymphoid tissue to
recognize and respond to foreign antigens.
10)What role do macrophages play in providing immunity to humans?
Ans.Macrophages phagocytose and destroy the microbes (pathogens) and provide protection
from diseases.
11)What causes swelling of the lower limbs in patients suffering from filariasis?
Ans.The filarial worms live for many years in the lymphatic vessels of lower limbs and cause
chronic inflammation there.
12)High fever, loss of appetite, stomach pain and constipation are some of the
symptoms seen in a patient. How would the doctor confirm that the patient is suffering
from typhoid and not amoebiasis?
By Widal test.
13)How do virus-infected cells provide innate immunity to healthy cells?
Virus – infected cells secrete interferons, which protect the other non-infected cells from
viral infection.
14)It was diagnosed by a specialist that the immune system of the body of a patient has
been suppressed. Name the disease the patient is suffering from and its causative agent.
- The patient is suffering from Acquired Immuno-Deficiency Syndrome (AIDS)
Human Immunodeficency Virus (HIV) causes the disease.
15)Why is secondary immune response more intense than the primary immune
response in humans.
Ans.Our body has memory of the first encounter with the same pathogens.
SHORT ANSWER TYPE QUESTIONS (2 marks)
16)A young boy on bringing a pet dog home, started to complain of watery eyes and
running nose. The symptoms disappeared when the boy was kept away from the pet.
(a) Name the type of antibody and the chemicals responsible for such a response in the
boy.
(b) Mention the name of any one drug that could be given to the boy for immediate
relief from such a response.
Ans.
QUESTION BANK /BIOLOGY/XII
Page 116
(a) IgE, the chemicals are serotonin and histamine.
(b) Adrenaline, antihistamine, steroids, (any one)
17)Write the events that take place when a vaccine for any disease is introduced into the
human body.
Ans.Active immunity is produced by injecting the microbes deliberately during
immunization.
- Antibodies are produced by the B-cells of our body in response to the antigens, they
neutralise the pathogenic agents during actual infection
- The vaccines also generate memory B-cells and T-cells, which recognise the same pathogen
on subsequent exposure and destroy them by a massive production of antibodies.
18)How does a vaccine for a particular disease immunise the human body against that
disease?
Ans.Active immunity is produced by injecting the microbes (vaccine) deliberately during
immunization for the particular disease.
produced by the B-cells of our body in response to the antigens; they neutralise the
pathogenic agents during actual infection.
- The vaccines also generate memory B-cells and T-cells, which recognise the same pathogen
on subsequent exposure and destroy them by a massive production of antibodies.
19)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?
Ans.Amoebiasis
Entamoeba histolytica.
Large intestine.
Houseflies.
20)Name the different types of cells that are responsible for producing the acquired
immune response in a human body. How do these cells respond when a pathogen enters
the body?
Ans.B-lymphocytes and T- lymphocytes
Both these types of cells multiply and produce a clone of cells.
B-lymphocytes produce an array of proteins, called antibodies
The T-lymphocytes stimulate the B-cells to produce antibodies.
T- cells mediate CMI response.
Memory B-cells and T-cells remain in the body, for secondary immune response, in case
of subsequent encounter by the pathogen.
21) (a) Highlight the role of thymus as a lymphoid organ.
(b) Name the cells that are released from the above mentioned gland.
QUESTION BANK /BIOLOGY/XII
Page 117
Mention how they help in immunity.
Ans.
(a) Thymus is a primary lymphoid organ, where immature lymphocytes differentiate into
antigen-sensitive T-lymphocytes,
(b) T-lymphocytes are released.
- They help B-cells to secrete antibodies.
- They are involved in cell-mediated immune response (in organ transplantation).
22)Name the plant source of the drug popularly called ‘smack'. How does it affect the
body of the abuser?
Ans. (a) Highlight the role of thymus as a lymphoid organ.
(b) Name the cells that are released from the above mentioned gland.
Mention how they help in immunity.
Smack is obtained from Papaver somniferum (Poppy)
- It binds to specific opioid receptors present in our central nervous system and
gastrointestinal tract.
It is a depressant and slows down the body functions
23) one plant and the addictive drug extracted from its latex. How does this drug affect
the human body?
Ans.Papaver somniferum (Poppy)Heroin
- It binds to specific opioid receptors present in our central nervous system and
gastrointestinal tract.
It is a depressant and slows down the body functions.
24)a)Name the group of viruses responsible for causing AIDS in humans. Why are these
viruses so named?
(b) List any two ways of transmission of HIV infection in humans other than sexual
contact.
Ans.
(a) Retroviruses.They have RNA as the genetic material and can make DNA on RNA
template with the help of enzyme reverse transcriptase hence they are called retroviruses .
(b) Transmission of HIV infection,
(i ) by transfusion of blood and blood products contaminated with HIV.
(ii) by sharing infected needles as in the case of intravenous drug abusers from the infected
mother to the foetus, through placenta.
25) Name and explain the two types of immune response in humans.
Ans.The two types of immune responses are:
(i) Primary immune response-The immune response that occurs as a result of the first
encounter of the individual with an antigen is called primary immune response of low
intensity.
QUESTION BANK /BIOLOGY/XII
Page 118
(ii) Secondary immune response-The immune response that occurs at the second and
subsequent encounters of the individual with the antigen, is called secondary immune
response highly intensified
26)Describe the role of lymph nodes in providing immunity.
Ans.Lymph nodes are secondary lymphoid organs
- The lymph nodes serve to trap the microorganisms or other antigens which have entered the
lymph and tissue fluid.
- These trapped antigens activate the lymphocytes present in the lymph nodes to cause
immune response.
27)Name the plant source of cocaine. How does it affect the human body?
Ans.Cocaine is obtained from the plant Erythroxylum coca.
- It interferes with the transport of neurotransmitter, dopamine.
- It has a stimulating action on the central nervous system and produces a sense of euphoria
and increased energy.Excess of it causes hallucination.
28)Name the different types of cells providing cellular barrier responsible for innate
immunity in humans.
Ans.The cellular barrier includes:
(i) Polymorpho nuclear leukocytes (PMNL)
(ii) Monocytes.
(iii) Natural killer type of lymphocytes.
(iv) Macrophages.
29)Name the two types of immunity in a human body Why are cell-mediated and
humoral immunities so called?
Ans.Innate immunity and acquired immunity.
- Cell-mediated immunity is so called because it involves specialized cells the Tlymphocytes.( as in organ transplant)
Humoral immunity is so called because it includes the antibodies, which are found
circulating in the body fluid the blood. (humors - body fluids.)
30)Identify A, D, e and F in the diagram of an antibody molecule given below:
QUESTION BANK /BIOLOGY/XII
Page 119
Ans.
A – Antigen binding site.
D – Light chain.
e – Heavy chain.
F – Disulphide bond/bridge.
31)What is colostrum? Why is it important to be given to the newborn infants?
Ans.Colostrum is the yellowish fluid / milk produced by the mammary glands of the mother
during the initial few days of lactation.
- It contains IgA antibodies that are necessary for developing resistance in the newborn
babies.
32)(a)Name the lymphoid organ in humans, where all the blood cells are produced.
(b) Where do the lymphocytes produced by the lymphoid organ mentioned above,
migrate and how do they affect immunity?
Ans.
a) Bone marrow.
(b) Some of the lymphocytes remain in the bone marrow and mature into B-lymphocytes.
-B-lymphocytes produce antibodies that neutralize the pathogen during actual infection.
- Some of the lymphocytes move to the thymus and mature into T-lymphocytes.
- T-lymphocytes stimulate B-cells to produce antibodies and some T – cells are involved in
cell – mediated immunity and are responsible for rejecting organ transplants.
33) Explain metastasis. Why is it fatal?
Ans.Metastasis is the property of neoplastic or cancerous cells by which the cancerous cells
sloughed off from a tumour reach distant sites through blood and establish a new tumour
wherever they get lodged.It is fatal because there is no specific location for the treatments
like radiation therapy or surgery
34) How do macrophages in the human body act as 'HIV factory'?
Ans.After entering the macrophage the RNA of the virus produces DNA catalysed by the
enzyme reverse transcriptase.
- This viral DNA gets incorporated into the DNA of the host cell or macrophage and directs
the host cell to produce virus particles.
- The macrophage continues to produce new viruses and thus acts as a HIV factory.
35) State the functions of primary and secondary lymphoid organs in humans.
Ans.
Benign Tumours
QUESTION BANK /BIOLOGY/XII
Malignant Tumours
Page 120
-
-
These are the tumours which remain
confined to their original location and do
not spread to other parts of the body.
These tumours cause little damage to the
body.
Metastasis is not shown.
-
-
These are the tumours whose neoplastic
cells separate and move to other sites.
Cells grow rapidly invade and damage
surrounding normal tissue
They cause more damage to the body.
Metastasis - they spread to other parts of
the body.
36)
(a) Explain the property that prevents normal cells from becoming cancerous.
(b) All normal cells have inherent characteristic of becoming cancerous. Explain
Ans.(a) Contact inhibition is the property shown by normal cells, by the virtue of which
contact with other cells inhibits their uncontrolled proliferation and growth.
(b) All normal cells have cellular oncogenes (c-onc) or proto-oncogenes; which when
activated under certain conditions, could lead to oncogenic transformation of cells. i.e they
become cancerous.
37) Due to undue peer pressure, a group of adolescents started using opioids
intravenously?What are the serious problems they might face in future?
Ans.They are more likely to get infections like AIDS and hepatitis B as the viruses of these
diseases are spread by the use of contaminated needles and syringes; both are chronic
infections and are fatal.
- Since, the tolerance levels of the receptors present in our body to these drugs increases, they
respond only to higher doses of drugs leading to greater intake and addiction - drug
dependence.
Opioids are depressants and slow down body functions
38) How do normal cells get transformed into cancerous neoplastic cells? Mention the
difference between viral oncogenes and cellular oncogenes.
Ans.The transformation of normal cells into cancerous neoplastic cells is induced by
physical, chemical or biological agents collectively called carcinogens.
Those genes of the viruses which cause cancer, are called viral oncogenes.Cellular oncogenes
are those genes present in normal cells, which when activated under certain conditions can
cause oncogenic transformation of the cell.
SHORT ANSWER TYPE QUESTIONS (3 Marks)
39)Name the cells HIV attacks first when it gains entry into a human body. How
doesthis virus replicate further to cause immunodeficiency in the body?
Ans. entering the body of a person, the virus enters the macrophages.
- The viral genome (RNA) undergoes reverse transcription to become viral DNA with the
help of reverse transcriptase.
QUESTION BANK /BIOLOGY/XII
Page 121
- The viral DNA gets incorporated into the DNA of the host cells and directs these cells to
produce virus particles.
- The macrophages continue to produce viral particles and function as HIV factory.
- Simultaneously HIV enters into helper T-lymphocytes and replicate to produce progeny
viruses.
- The progeny viruses released in the blood attack new helper T- lymphocytes.
- This process is repeated and there is a progressive decrease in the number of helper Tcellsin the body of the infected person leading to immunodeficiency in the body.
40)A woman has tested positive for AIDS. Name the pathogen that infected her How
does the pathogen weaken her immune system?
Ans.The pathogen is Human Immunodeficiency virus.
After entering the body of a person, the virus enters the macrophages.
- The viral genome (RNA) undergoes reverse transcription to become viral DNA with the
help of reverse transcriptase.
- The viral DNA gets incorporated into the DNA of the cells and directs these cells to
produce virus particles.
- The macrophages function as HIV factory and produce a number of HIV particles.
- These HIV particles infect the helper T-lymphocytes and replicate to produce progeny
viruses.
- The progeny viruses released in the blood attack new helper T-cells.
- This process is repeated and there is a progressive decrease in the number of helper T-cells
thereby weakening her immune system.
41)Name the stage of Plasmodium that is transmitted to human body by the vector.
Describe the life cycle of the parasite in humans.
Ans.Sporozoite stage
Life cycle of Plasmodium in human body:
- When an infected female Anopheles mosquito bites a person, the sporozoites are injected
into the body
- They reach the liver cells through blood.
- The parasite reproduces asexually in the liver cells and by the bursting of liver cells, new
sporozoites are released into the blood.
- They enter the red blood cells (RBCs) and reproduce asexually, burst the red blood cells,
and release haemozoin, which causes the cycles of fever, chill and shivering.
- The released parasite cells infect new red blood cells; some continue the asexual
reproduction and cause the cycles of fever.
Others enter the sexual stage and form gametocytes in the RBCs, which are picked by the
Anopheles mosquito, along with the blood meal.
42)(i) How and at what stage does Plasmodium enter into a human body?
(ii) Why does the victim show symptoms of high fever?
(iii) With the help of a flow chart only, show the stages of asexual reproduction in the
life cycle of the parasite in the infected human.
QUESTION BANK /BIOLOGY/XII
Page 122
Ans.(i)When an infected female Anopheles mosquito bites a person, the sporozoites are
injected into the body.
ii) The rupture of RBCs is associated with the release of a toxic substance, haemozoin, which
is responsible for the chill and fever recurring every three to four days.
iii)
43) Why lymph nodes and bone marrow are called lymphoid organs? Explain the
functions of each of them.
Ans.Lymph nodes are secondary lymphoid organs which provide the sites for interaction of
lymphocytes with antigen, which then proliferate to become effector cells.
Bone marrow is a primary lymphoid organ where immature lymphocytes differentiate and
proliferation of lymphocytes occur.
- Lymph nodes serve to trap the microorganisms or other antigens that enter the lymph and
tissue fluid; the lymphocytes in the lymph nodes are then activated by these antigens and
cause the immune response.
- Bone marrow is the primary lymphoid organ where all blood cells including lymphocytes
are produced. The B lymphocytes mature in the bone marrow.
44) Study a part of the life cycle of malarial parasite given below. Answer the questions
that follow.
QUESTION BANK /BIOLOGY/XII
Page 123
(a) Mention the roles of 'A' in the life cycle of the malarial parasite.
(b) Name the event 'C' and the organ where this event occurs.
(c) Identify the organ B' and name the cells being released from it.
Ans. (a)A is a female Anopheles mosquito.Host and vector. A part of the life cycle of the
parasite occurs in the body of female Anopheles mosquito( host); these mosquitoes act as
vectors and transmit the disease from patients to healthy individuals.
(b) 'C' is Fertilization, it occurs in the stomach of mosquito.
(c) 'B' is Salivary gland. Sporozoites are the cells released from it.
45)(a) Name the stage of Plasmodium that gains entry into the human body.
(b) Trace the stages of Plasmodium in the body of female Anopheles after its entry.
(c) Explain the cause of periodic recurrence of chill and high fever during malarial
attack in humans.
Ans. (a) Sporozoite (b) i) Anopheles mosquito picks up the gametocytes along with the
blood meal.
ii) Gametocytes reach the stomach of mosquito; fertilization occurs here.
iii)Zygote develops and produces sporozoites in the intestine.
iv) Sporozoites escape from the intestine and reach the salivary glands, where they remain
stored.
(c)The rupture of RBCs is associated with the release of a toxic substance called haemozoin,
which is responsible for the recurring chill and high fever.
46)Study the diagram showing replication of HIV in humans and answer the following
question accordingly:
(i)
(ii)
(iii)
(iv)
Write the chemical nature of the coat ‘A`
Name the enzyme ‘B` acting on ‘X` to produce molecule ‘C` Name ‘C`.
Mention the name of the host cell ‘D` the HIV attacks first when it enters
into the human body.
Name the two different cells the new viruses ‘E` subsequently attack.
QUESTION BANK /BIOLOGY/XII
Page 124
Ans.
(i) A – Protein coat.
(ii) B – Reverse transcriptase.
C – It is viral DNA.
(iii) Macrophage
(iv) Macrophages and helper T lymphocytes.
47) Explain the role of the following in providing defence against infection in human
body:
(i)
Histamine
(ii)
Interferons
(iii) B – cells.
(i) Histamines cause allergic reaction of the immune system against allergens.
(ii) Interferons protect the non-infected cells of our body from viral infections.
(iii) B – cells produce antibodies against antigens/pathogens
48)A person is suffering from Ascariasis. Mention the pathogen causing the disease and
an organ of the body affected, three symptoms and one mode of transmission of the
disease.
Ans.Ascariasis is caused by Ascaris lumbricoides
The intestine becomes blocked and damaged.
Symptoms
- Internal bleeding
- Muscular pain
- Anaemia
Fever
- Blockage of intestinal passage.
Transmission
Infection is acquired through contaminated water, fruits and vegetables
The eggs of the parasite from the faeces of infected person contaminate soil, water and
plants.
49)A person is suffering from ringworm disease. Mention the pathogen and the part of
the human body affected. Give two symptoms of the disease along with two modes of
transmission.
Ans. Ringworm is caused by Microsporum. Trichophyton and Epidermphyton.
- Skin, nails and scalp are affected.
Symptoms
- Dry, scaly lesions on body parts like skin, scalp and nails.
QUESTION BANK /BIOLOGY/XII
Page 125
- The lesions are accompanied by intense itching.
Transmission
- By contact with the articles (like towels, comb, clothes, etc.) used by the infected person.
From soil.
50) Name the type of immunity that is present at the time of birth in humans. Explain
any two ways by which it is accomplished.
Ans. Innate immunity is present at the time of birth in humans.
It is accomplished by providing different types of barriers:
(i) Physical barriers:
- Skin is the main barrier which prevents entry of micro-organisms.
Mucus coating on the epithelial lining of respiratory, gastro intestinal and other tracts helps in
trapping microbes.
(ii) Physiological barriers:
- Acid in the stomach.
- Saliva in the mouth, tears from eyes, prevent microbial growth
(iii) Cellular barriers:
- Leucocytes like neutrophils and monocytes.
- Natural killer cells in blood
- Macrophages in the tissues-all the above can phagocytose and destroy microbes.
(iv) Cytokine barriers:
Virus-infected cells secrete proteins called interferons, which protect the non-infected cells
from further viral infection.
LONG ANSWER TYPE QUESTIONS (5 Marks)
51)Your classmate complains of headache and cough to the doctor. The doctor confirms
that he is suffering from pneumonia and not just common cold. How did the doctor
reach such a conclusion? Mention any two precautions to be followed to prevent the
spread of this disease.
Ans. Doctor confirms pneumonia on the basis of the following symptoms – fever / chills/
grey – blue lips and finger nails (any two)
And not common cold as the following symptoms are not observed – Nasal congestion/ sore
throat/ hoarseness (any two)
PrecautionsCover the nose when near the patient. Do not share glasses and utensils/ articles used by the
infected person.
52) Malarial parasite ‘Plasmodium' completes its life cycle in two hosts. Draw its
complete life cycle and explain various stages it follows throughout its life.
QUESTION BANK /BIOLOGY/XII
Page 126
Ans.Life cycle from NCERT text book Biology class XII pg 148 = 2
Stages.
(a) The stage in which the parasite enters in the body of humans through saliva of mosquitosporozoite stage.
(b) Asexual reproduction of sporozoites in liver cells resulting into bursting of those cells
and releasing outside into the blood.
(c) Sporozoites infect RBCs cause them to burst, with release of haemozoin, indicated by
repeated cycles of fever. Released parasites also infect other RBCs.
(d) This is followed by sexual stage in RBCs. Gametocytes develop in the RBCs .
(e) Female mosquito takes up gametocytes with the blood of host. Fertilization and
development takes place in the stomach and intestine of mosquito.
(f) From intestine parasite (sporozoites) reach the salivary glands for storage from where
they reach the human body by the bite of the infected female Anopheles mosquito, and
that is how the cycle continues.
53) Describe the asexual and sexual phases of life cycle of Plasmodium that causes
malaria in humans.
Ans.
Asexual Phase
- It occurs in the human host.
- The sporozoites enter the human body through the bite of an infected female Anopheles
mosquito.
- They reach the liver through blood and multiply within the liver cells.
- Such liver cells burst and release the parasite into blood.
- The parasites enter RBCs, multiply in them and cause their rupture.
- The rupture of RBCs is associated with the release of a toxin, called haemozoin, which is
responsible for the chill/shivering and the high temperature (fever) of the body.
Sexual Phase
- It starts with the formation of gametocytes in the RBCs of man.
- The parasite then enters the female Anopheles mosquito along with the blood when it bites
infected person.
- The gametes fuse to form a zygote in the stomach.
- The zygote undergoes further development in intestine to form sporozoites.
- Sporozoites are transported to and stored in the salivary glands of the mosquito and are
transferred to a human body during the bite of the mosquito.
QUESTION BANK /BIOLOGY/XII
Page 127
Chapter –9 Strategies for Enhancement in Food Production
VERY SHORT ANSWER TYPE QUESTIONS (1marks)
1)Write the importance of MOET.
.High milk-yielding breeds of females and high meat-yielding bulls are bred to increase the
herd size in a short time through MOET.
2) Name the following:
(a) A semi-dwarf variety of wheat which is high-yielding and disease-resistant,
(b) Any one interspecific hybrid mammal.
Ans.(a) Sonalika or Kalyan Sona. (b) Mule
3) State the importance of biofortification.
Ans.Biofortification -breeding crops with higher levels of vitamins and minerals, or higher
protein and healthier fats, is the most practical means to improve public health.
4) Which of the following is the semi-dwarf wheat that is high-yielding and disease
resistant?Pusa Shubra, Kalyan Sona, Ratna.
Ans.Kalyan Sona
5) What is the economic value of Saccharum officinarum?
Ans.Saccharum officinarum is the sugarcane grown in South India which has thicker stems
and higher sugar content and yield.
6) Mention the strategy used to increase homozygosity in cattle for desired traits.
Ans.Inbreeding, i.e. mating of more closely related individuals of the same breed, for 4-6
generations.
7) What is the economic value of Spirulina?
Ans.Spirulina is employed in the commercial production of single cell protein.
Can be grown on waste matter thereby reducing environmental pollution.
8) Tissue culture experiment has been performed with a plant tissue infected with
TMV. Meristematic tissue produces healthy plant. Reason out the possibility of
obtaining such result.
Ans.Meristematic tissues are free of virus.
QUESTION BANK /BIOLOGY/XII
Page 128
9)Which of the following is the semi-dwarf wheat that is high-yielding and diseaseresistant?Pusa Shubra, Kalyan Sona, Ratna.
Ans.Kalyan Sona
10)Which of the following produces single cell proteins?
Sonalika, Spirulina, Saccharomyces.
Ans. Spirulina
SHORT ANSWER TYPE QUESTIONS (2marks)
11) (a) What is micropropagation? Why are the plants produced by micropropagation
called somaclones?
(b) Name the technique by which healthy plants can be recovered from the diseased
plants.
Ans.(a) Micropropagation is the process of producing plants (in large numbers) through
tissue culture.
- genetically identical
(b) Disease-free or healthy plants can be recovered from a diseased plant by meristem
culture. It is because the meristems are free of viruses even though the plant is infected with a
virus.
12)Identify A, B,C and D in the table given below:
Crop
Variety
Wheat
A
B
Pusa Subhra
Cowpea
Pusa Komal
Brassica
Karan Rai
Resistance to Disease
Leaf and stripe rust
Black rot
C
D
Ans.(A) Himgiri
(B) Cauliflower
(C) Bacterial blight
(D) White rust
13) Following are the steps in MOET programme for herd improvement in which a cow
has been administered hormones with FSH – like activity. Arrange steps A to D in their
correct sequence.
A – Transferred to a surrogate mother.
B – It is either mated with an elite bull or artificially inseminated.
C - Fertilised eggs at 32 cell stage are recovered non surgically.
D – It produces 6 – 8 eggs instead of one egg which they normally yield per cycle.
Ans.D
B
C
A
QUESTION BANK /BIOLOGY/XII
Page 129
14)Explain the advantages of animal inbreeding programme. Mention when inbreeding
depression would occur.
Ans.Advantages of inbreeding:
(i) Inbreeding helps to evolve a pure line in animals, as it increases homozygosity.
(ii) It exposes the harmful recessive alleles, which are eliminated by selection.
(iii) It helps in accumulation of superior genes and elimination of less desirable genes.
(iv) This approach increases productivity of the inbred population, as there is selection at
each step.
Continued inbreeding, especially close inbreeding usually reduces fertility and even
productivity. This is called inbreeding depression.
15) How are somaclones produced? How are they different from somatic hybrids?
Ans.Somaclones are produced through tissue culture / micropropagation
Somaclones
- Somaclones are produced by micropropagation, i.e. tissue culture.
- All the plants are genetically identical
among themselves and also to the plant
from which they are produced.
Somatic hybrids
- Somatic hybrids are produced by somatic
hybridization (protoplast fusion) and then
growing the hybrid protoplasts.
- The plants are genetically dissimilar
among' themselves and also from the
parent plants.
16) Demand for mushroom as food has led to its culturing on a large scale. Similarly it
is perceived that microbes too would become acceptable as food. Identify a microbe
which can be cultured as a food source and give the applicability of its culture in the
given context.
Ans.Spirulina
Produces large quantities of food rich in protein, minerals, fats, carbohydrate and vitamins or
Methylophilus methylotrophus
250gm of this microorganism produces 25 tonnes of protein per day
17) Study the steps given below:
a) Cow is administered with FSH hormone
b) ---------------(1)----------------------c) 6-8 eggs per cycle are derived
d) Artificially inseminated
e) Fertilized eggs at 8-32 cells are recovered
f) -----------------(2)-----------------i) Identify the events that take place at stages (1) and (2) respectively.
(ii) State the importance of the technology explained above.
Ans.
(i) 1. The hormone induces follicular maturation and super-ovulation,
QUESTION BANK /BIOLOGY/XII
Page 130
2. They are transferred to surrogate mothers.
(ii) This method has been used to increase the herd size in a short time of high milk-yielding
breeds of females and high quality meat-yielding bulls.
SHORT ANSWER TYPE QUESTIONS (3marks)
18) Differentiate between inbreeding and out-breeding in cattle. State one advantage
and one disadvantage for each one of them.
Ans.
Inbreeding
Out-breeding
- Inbreeding refers to the mating of more - Out- breeding refers to the breeding of
closely related individuals within the same unrelated animals of the same breed
breed for 4-6 generations.
with no common ancestors (out crossing), or between two different
breeds (cross - breeding) or different
species (inter specific hybridization).
Advantages:
- Inbreeding: It is used to evolve purelines in animals.
- Out-breeding: A single out-cross helps to overcome inbreeding depression.
Disadvantages:
- Inbreeding: Continuous inbreeding leads to inbreeding depression and reduction in
productivity.
Out-breeding: The success rate may be low and the desired combinations of characters may appear
in low frequency.
19) Name the prominent South Indian and North Indian species of sugarcane used for
cross-breeding. List the desired qualities of the hybrid cane obtained from this cross.
Ans.
- Saccharum barberi, is the North Indian sugar cane, with poor sugar content and yield.
- Saccharum officinarum, is the South Indian sugar cane, with thick stems and higher sugar
content.The hybrid can grow in North India too and has thick stem, high sugar content and
yield.
20)(a) Explain how to overcome inbreeding depression in cattle.
(b) List three advantages of inbreeding in cattle.
(c) Name an improved breed of cattle.
Ans. (a) Inbreeding depression can be overcome by out crossing.
(b) (i) Inbreeding is necessary to evolve pure lines in any animal, as it increases
homozygosity.
(ii) Inbreeding exposes the harmful recessive alleles, which become eliminated by selection.
(iii) It helps in accumulation of superior genes and elimination of the less desirable ones.
(c) Jersey.
QUESTION BANK /BIOLOGY/XII
Page 131
21)(i) Mention the property that enables the explants to regenerate into a new plant.
(ii) A banana herb is virus-infected. Describe the method that will help in obtaining
healthy banana plants from this diseased plant.
Ans.
(i) Totipotency
(ii) Tissue culture
Healthy banana plants can be obtained by meristem culture, as the meristems (apical or
axillary) are free of virus,
- An explant from the meristem of the banana plant is taken.
- It is grown in a test tube, under sterile conditions in special nutrient media.
- The nutrient medium should provide a carbon source like sucrose, inorganic salts,
vitamins, amino acids and growth regulators (auxins, cytokinins etc).
Disease free banana plantlets are obtained which can then be planted in the soil.
22) Scientists have succeeded in recovering healthy sugar cane plants from a diseased
one:
(a) Name the part of the plant used as explant by the scientists.
(b) Describe the procedure the scientists followed to recover the healthy plants.
(c)Name this technology used for crop improvement.
Ans.
a) Shoot tip (meristem).
(b) The explant/shoot tip is grown in a test tube under sterile conditions in special nutrient
media.
- The nutrient medium must provide a carbon source like sugar and also inorganic salts,
vitamins, amino acids and growth regulators such as auxins, cytokinins, etc.
- The plantlets formed are then shifted to the soil in the field.
(c)Tissue culture /Micro propagation.
23) Mention the property of plant cells that has helped them to grow into a new plant in
in vitro conditions. Explain the advantages of micropropagation.
Ans.Totipotency is the property by which each plant cell can grow into a new plant.
Advantages of micro propagation are as follows:
(i) By this method, thousands of plants can be raised in a short duration.
(ii) They are genetically identical (somaclones)
(iii) Recovery of healthy plants from diseased ones.
(iv) Many important food plants like tomato, banana and apple have been produced on
commercial scale using this technique.
24)Why is biofortification considered the most practical means to improve public
health? Explain taking three examples.
Ans.Biofortification refers to breeding crops with improved nutritional quality, i.e. with
higher levels of vitamins, minerals, proteins and healthier fats, e.g.
(i) Maize hybrids with high levels of amino acids, lysine and tryptophan.
QUESTION BANK /BIOLOGY/XII
Page 132
(ii) Wheat variety, with high protein content
(iii) Vitamin A-enriched carrots.
25)(a) Name the Indian scientist, whose efforts brought 'Green Revolution' in India.
(b) Mention the steps that are essentially carried out in developing a new genetic variety
of crop under plant breeding programme.
Ans.
a) Dr. M.S. Swaminathan
(b) The steps in developing a new genetic variety of crop are:
(i) Collection of genetic variability or germplasm.
(ii) Evaluation and selection of parents.
(iii) Cross-hybridization among the selected parents,
(iv) Selection and testing of superior recombinants or hybrids.
(v) Testing, release and commercialization of new cultivars.
26)Out crossing and cross-breeding are two breeding practices in animal husbandry.
How the two practices are different from each other and what advantage are they to
breeders? Explain.
Ans.
Out crossing
- Out crossing refers to the mating of animals within the same breed, but having no
common ancestor on either side of their
pedigree up to 4-6 generations.
- A single out cross helps to overcome
inbreeding depression.
Cross- breeding
- Cross – breeding is the mating of a
superior selected male of one breed with
the selected female of another breed.
- Cross – breeding allows combining of
desirable qualities from the two breeds.
The hybrid animals may be used for
commercial production or for developing
new breeds by inbreeding and selection.
27)Define totipotency of a cell. List the requirements, if the objective is to produce
somaclones of a tomato plant on commercial scale.
Ans.Totipotency is defined as the capacity of any cell to generate the whole plant from it.
Requirements ( for tissue culture):
(i) Explant
- Any part of the tomato plant taken out for growing a new plant.
ii) sterile/ aseptic conditions.
(iii) Nutrient media
- The nutrient medium must provide a carbon source such as sucrose, inorganic salts,
vitamins, amino acids growth regulators like auxins, cytokinins etc.
(iii) Suitable conditions of light and temperature
From such tissue culture a number of plants can be grown, such plants are genetically
identical and are called somaclones.
QUESTION BANK /BIOLOGY/XII
Page 133
28) Why is 'MOET' considered to be a successful programme in cattle breeding?
(b) What kind of male and female cattle are selected for this programme?
(c) Why is the cow administered with FSH-like hormones? Explain.
Ans(a) MOET is considered to be a successful programme in cattle breeding as the chances
of successful production of hybrid are high and the herd size is increased in a short time.
(b) A superior male that yields high quality meat i.e lean meat with less lipid and a superior
female that is high milk yielding are selected.
(c) It is to induce super ovulation i.e. 6-8 eggs instead of one per cycle
LONG ANSWER TYPE QUESTIONS (5marks)
29)
(a)What is plant breeding? List the two steps the classical plant breeding involves.
(b) How has the mutation breeding helped in improving crop varieties? Give one
example where this technique has helped.
(c)How has the breeding programme helped in improving the public nutritional health?
State two examples in support of your answer.
Ans.
(a) Plant breeding is the purposeful manipulation of plant species in order to create desired
plant types that are suited for cultivation, give better yield and are disease-resistant.
- The two steps in classical plant breeding are:
(i) Crossing or hybridization of pure lines.
(ii) Artificial selection to produce plants with desirable traits of higher yield, nutrition and
resistance to diseases.
(b) Mutation breeding involves inducing mutations artificially and selecting and using the
plants that have the desirable characters for further breeding.
e.g. Resistance to yellow mosaic virus and resistance to powdery mildew in mung bean was
induced by mutation breeding.
(c) Through biofortification - i.e. breeding crops with higher levels of vitamins and minerals,
or higher protein and healthier fats
e.g. Vitamin A-enriched spinach, Iron and calcium enriched spinach and bathua, iron
fortified rice etc.
30) How is a pure line in an animal raised? Explain.
Ans.
- In animals a pure line is raised by inbreeding, as inbreeding increases homozygosity.
- Inbreeding refers to the mating of more closely related individuals within the same breed
for 4-6 generations.
- In this process, superior males and females are identified within the same breed and mated
in pairs.The progeny obtained from such crosses are evaluated and superior males and
females are identified for further matings.
QUESTION BANK /BIOLOGY/XII
Page 134
31) With advancements in genetics, molecular biology and tissue culture new traits
have been incorporated into crop plants.Explain the main step in breeding a new
genetic variety of a crop.
-
Ans.
Collection of variability
Evaluation & selection of parents
Cross hybridization among the selected parents
Selection & testing of superior recombinants
Testing, release & commercialization of new cultivars.
32)a) State the objective of animal breeding.
(b) List the importance and limitation of inbreeding. How can the limitation be
overcome?
(c) Give an example of a new breed each of cattle and poultry
a) Animal breeding aims at increasing the yield of animals and improving the desirable
qualities of the produce .
(b) Importance
- increases homozygosity / pure line .
- accumulation of superior genes and elimination of less desirable genes.
- exposes harmful recessive genes .
- Limitation.
- inbreeding depression .
-can be overcome by mating with unrelated superior animal of the same breed.
(c) cattle - Jersey; poultry - Leghorn .
QUESTION BANK /BIOLOGY/XII
Page 135
Chapter 10- Microbes in Human Welfare
VERY SHORT ANSWER TYPE QUESTIONS (1 marks)
1)Name the nutrient that gets enhanced while curdling of milk by Lactobacillus.
Ans.Vitamin B12
2)Name the gas released and the process responsible for puffing up of the bread dough
when Saccharomyces cerevisiae is added to it.
Ans.The carbon dioxide released is responsible for the rising of dough.
- Fermentation is the process
3)What makes the nucleopolyhedrovirus a desirable biological control agent?
Ans.They are species-specific and narrow spectrum insecticides; they have no negative
impacts on other organisms.
4) Mention the role of cyanobacteria as a biofertiliser.
Ans.(i) Cyanobacteria can fix atmospheric nitrogen.
(ii) They also add organic matter to the soil to increase
5) Mention the information that the health workers derive by measuring BOD of a
water body.
Ans.
- The greater the BOD of a water body, more is its polluting potential.
- BOD is the measure of uptake of oxygen by the microbes in the water sample and indicates
the amount of organic matter present in the water.
6)Milk starts to coagulate when Lactic Acid Bacteria (LAB) is added to warm milk as a
starter. Mention any two other benefits which LAB provide.
Ans.(i) LAB improves the nutrient quality of curd by increasing the content of vitamin B12.
(ii) They also check the disease-causing microbes in our stomach.
7)Why is sewage water treated until the BOD is reduced? Give a reason.
Ans.Sewage water is treated to reduce the BOD, because higher the BOD of the water,
greater is its polluting potential.
8)Which of the following is a free-living bacterium that can fix nitrogen in the soil?
QUESTION BANK /BIOLOGY/XII
Page 136
Spirulina, Azospirillum, Sonalika.
Ans.Azospirillum
9)Which of the following is a cyanobacterium that can fix atmospheric nitrogen?
Azospirillum, Oscillatoria, Spirulina.
Ans.Oscillatoria.
10)Name the type of association that the genus Glomus exhibits with higher plants.
Ans.Mycorrhiza
SHORT ANSWER TYPE QUESTIONS (2 marks)
11)Name the bacterium responsible for the large holes in 'Swiss cheese'. What are these
holes due to?
Ans.Propionibacterium sharmanii
- Since this bacterium produces large amount of carbon dioxide during fermentation, large
holes are produced.
12)How does 'starter' added to milk help it set into curd?
Ans.The small amount of curd added to the milk contains millions of lactic acid bacteria.
They start multiplying at the suitable temperature.
- LAB produces acids that coagulate and partially digest the milk proteins; thus, milk is
converted into curd.
-Vitamin B12 content is increased.
13) List four advantages that a symbiotic mycorrhizal association provides to the host
plant.
Ans.
-The fungus absorbs phosphorus from the soil and passes it to the plant.
- Plants with mycorrhiza show resistance to root-borne pathogens.
- They show increased tolerance to salinity and drought.
- There is an overall increase in plant growth and development.
14)Why is Rhizobium categorised as a 'symbiotic bacterium?' How does it act as a
biofertiliser?
Ans.Rhizobium lives in the root nodules of leguminous plants and provides the plants with
nitrogenous nutrients and the plant provides shelter to the bacterium; since both are mutually
benefited, it is called symbiotic bacterium.
QUESTION BANK /BIOLOGY/XII
Page 137
- It fixes the atmospheric nitrogen into organic forms which are used by the plant.
15) Name the source of streptokinase. How does this bioactive molecule function in our
body?
Ans.Streptococcus
- Streptokinase functions as a clot-buster, for removing the clots from the blood vessels of
patients who have suffered myocardial infarction leading to heart attack.
16)Name the source of statin and state its action on the human body.
Ans.Monascus purpureus is the source of statin.
- Statins are used as blood cholesterol-lowering agents, as they competitively inhibit the
enzyme responsible for the synthesis of cholesterol.
17)Why are some molecules called bioactive molecules? Give two examples of such
molecules.
Ans. molecules produced by living organisms perform certain functions in the body of other
organisms and modify the metabolism; hence they are called bioactive molecules.
- Statins, cyclosporin A and streptokinase are bioactive molecules.
18)Name the enzyme produced by Streptococcus bacterium. Explain its importance in
medical sciences.
Ans.Streptokinase is the enzyme produced by Streptococcus.
- It is used as a clot-buster to remove clots from the blood vessels of patients who have
suffered myocardial infarction.
19)During the secondary treatment of primary effluents, how does a significant
decrease in BOD occur?
Ans.During secondary treatment, the aeration allows vigorous growth of useful aerobic
microbes into flocs, i.e. masses of bacterial cells in association with fungal filaments,
forming mesh-like structures.
-As they grow, the microbes consume a major part of the organic matter in the effluent;
Therefore, BOD is significantly reduced.
20) Identify A, B, C and D in the table given below:
Micro-organism
Product
A
Streptokinase
QUESTION BANK /BIOLOGY/XII
Biological
activity
Clot buster
Medical
ailment/Procedure
D
Page 138
Trichoderma
polysporum
B
C
Transplant
surgery
Ans.A - Streptococcus
B - Cyclosporin A
C - Immunosuppressive agent
D - Clot buster
21) Name the blank spaces a, b, c and d in the table given below:
Type of Microbe
Name
Fungus
Bacterium
a
Acetobacter aceti
Commercial
Product
Penicillin
b
c
Aspergillus niger
Citric acid
d
Ethanol
Ans.
Yeast
(a)
notatum
(b) Acetic acid
(c) Fungus
(d) Saccharomyces cerevisiae
Penicillium
22) Explain the process of secondary treatment given to the primary effluent up to the
point it shows significant change in the level of biological oxygen demand (BOD) in it.
Ans.The primary effluent is passed into large aeration tanks where it is constantly agitated,
mechanically pumping air into it, this allows vigorous growth of useful aerobic microbes into
flocs, these microbes consumes the major part of organic matter in the effluent (this
significantly reduces the BOD of the effluent)
23)Explain the significant role of the genus Nucleopolyhedrovirus in an ecological
sensitive area.
Ans.They are species – specific, narrow spectrum bio-control agents
They have no negative impacts on plants, mammals, birds, fish or non-target insects.
24)(a) Patients who have undergone myocardial infarction are given clot buster.
Mention the clot buster administered and its microbial source.
(b) A person recuperating from illness is advised to have curd regularly. Why?
Ans.(a) Streptokinase, Streptococcus
QUESTION BANK /BIOLOGY/XII
Page 139
(b) Curd contains Lactic Acid Bacteria which play beneficial role in checking disease
causing microbes/ It is a source of vitamin B12 too.
SHORT ANSWER TYPE QUESTIONS (3marks)
25)“Microbes play a dual role when used for sewage treatment as they not only help to
retrieve usable water but also generate fuel Explain.
Ans.Microbes naturally present in the sewage are employed in the secondary treatment of the
sewage.
- The effluent from the primary treatment is passed into large aeration tanks.
- This allows the rapid growth of aerobic microbes into flocs which consume the
organic matter of the sewage and reduce the BOD.
- Then the effluent is passed into a settling tank where the flocs are allowed to
sediment forming the activated sludge.
- Major part of this activated sludge is pumped into anaerobic sludge digesters where
the anaerobic bacteria digest microbes in the activated sludge.
- During this digestion bacteria produce a mixture of gases like methane, hydrogen
sulphide and carbon dioxide which form the biogas and can be used as a source of
energy. The effluent is generally released into rivers and streams.
26)What does BOD of a water body stand for? How is it related to water pollution?
Ans. – Biological Oxygen Demand - it refers to the amount of oxygen that would be
consumed if all the organic matter in one liter of water was oxidized by bacteria. The BOD
test measures the rate of uptake of oxygen by microorganisms in a sample of water and thus
indirectly BOD is a measure of the organic matter present in the water. The greater the BOD
of water more is its polluting potential.
27)Describe how biogas is generated from activated sludge. List the components of
biogas.
Ans.The activated sludge is pumped into large tanks called anaerobic sludge digesters,
which have other kinds of bacteria which grow anaerobically, digest bacteria and fungi in the
sludge, this causes the gases to be generated, the components are CO2, methane, H2S
28)Name the two different categories of microbes naturally occurring in sewage water.
Explain their role in cleaning sewage water into usable water.
Ans.- Aerobic microbes and anaerobic microbes are the two types of microbes naturally
occurring in sewage water.
(i) Aerobic microbes
- When the primary effluent is passed into aeration tanks and agitated along with the
pumping air into it, the aerobic microbes show vigorous growth to form flocs.
QUESTION BANK /BIOLOGY/XII
Page 140
- During their growth, the microbes consume a major part of the organic matter in the
effluent and significantly reduce the BOD of the effluent.
(ii) Anaerobic microbes
- The effluent from aeration tank is passed into settling tanks, where the flocs sediment to
form the activated sludge.
- When the activated sludge is pumped into anaerobic sludge digesters, the anaerobic bacteria
digest the bacteria and fungi in the sludge and produce gases like methane, carbon dioxide
and hydrogen sulphide and the water is fit to be released into a natural water body.
29)
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.
Ans.- The bio-wastes are collected in the concrete tank which is 10 - 15 feet deep and a
slurry of cow dung is also fed into it.
- A floating cover is placed on the slurry, which keeps rising as the gas is produced in the
tank by the activity of the microbes.
- The biogas plant has an outlet that is connected to a pipe to supply biogas to nearby places
to be used for cooking and lighting.
- The spent slurry is removed through another outlet and may be used as fertilizer.
a - Sludge
b - Gas holder
c - Dung + water
30)(a) Why do farmers prefer biofertilisers to chemical fertilisers these days? Explain.
(b) How do Anabaena and mycorrhiza act as biofertilisers?
Ans.
Biofertilisers
- Biofertilisers are organisms that enrich the
nutrient quality of the soil.
QUESTION BANK /BIOLOGY/XII
Chemical Fertilisers
- Fertilisers are synthetic chemicals added to
soil to supply specific nutrient(s) that is/are
Page 141
- They provide almost all the nutrients, but
slowly.
- They do not degrade the soil quality.
- Their culturing does not cause environmental pollution.
lacking.
- They supply only the specific nutrient(s),
but quickly.
- They degrade the soil quality.
- Their manufacture and use causes
environmental pollution.
b) Anabaena acts as biofertiliser because
(i) It fixes atmospheric nitrogen and enriches soil fertility.
(ii) it also adds organic matter to the soil ( it is an autotrophic microbe).
Mycorrhizae are the symbiotic associations between certain fungi and roots of higher plants.
Mycorrhizae act as biofertiliser because
- The fungus absorbs phosphorus from the soil and passes it to the plant.
- Plants with mycorrhiza show resistance to root-borne pathogens.
- They show increased tolerance to salinity and drought.
- There is an overall increase in plant growth and development.
31)Name the genus to which baculoviruses belong. Describe their role in integrated pest
management programmes.
Ans.Baculoviruses belong to the genus Nucleopolyhedrovirus.
- They are the pathogens which attack insects and other arthropods and serve as biocontrol
agents.
- They are species-specific, narrow spectrum insecticides, which have no negative impact on
plants, birds, mammals, fish, and even non-target insects.
- This is desirable because beneficial insects are conserved to aid in integrated pest
management (IPM) programmes.
32)Explain the different step involved in sewage treatment before it can be released into
natural water bodies
Ans.The treatment of sewage involves two steps:
(i) Primary treatment and (ii) Secondary treatment.
(i) Primary treatment
Primary treatment is a physical process of removal of small and large particles through
filtration and sedimentation.
- The first step is to remove the floating debris by sequential filtration.
- Then the grit such as soil and small pebbles are removed by sedimentation.
-All solids that settle down form primary sludge and the supernatant forms the effluent.
- The effluent from the primary settling tank is taken for secondary treatment.
(ii) Secondary Treatment
-It is a biological process that employs the heterotrophic bacteria naturally present in the
sewage.
- The effluent from the primary treatment is passed into large aeration tanks, where it is
constantly agitated and air is pumped into it.
QUESTION BANK /BIOLOGY/XII
Page 142
This allows the rapid growth of aerobic bacteria into flocs, which consume the organic matter
of the sewage and reduce the BOD.
- The effluent is passed into a settling tank, where the flocs are allowed to sediment forming
the activated sludge.
A small part of the activated sludge is pumped back into aeration tank as inoculum.
- The remaining major part of the sludge is pumped into anaerobic sludge digesters, where
the anaerobic bacteria digest the bacteria and fungi in the sludge and produce a mixture of
methane, hydrogen sulphide and carbon dioxide which forms biogas.
- The effluent from this is released into the natural water bodies such as rivers and streams.
33)a) How is activated sludge formed during sewage treatment?
(b) This sludge can be used as an inoculum or as a source of biogas. Explain.
(a) Formation of activated sludge:
- The primary effluent is passed into large aeration tanks; it is constantly agitated and air is
pumped into it
- Consequently, the useful aerobic microbes grow vigorously and form flocs, which are
masses of bacteria associated with fungal filaments.
- These microbes consume sufficient quantities of the organic matter and thereby reduce the
biological oxygen demand (BOD).
- Once the BOD is reduced, the effluent is passed into settling tanks, where the flocs
sediment to form activated sludge.
(b) A small amount of activated sludge is pumped back into the aeration tank as inoculum to
grow into flocs and consume the organic matter to reduce BOD.
The major part of activated sludge is pumped into anaerobic sludge digesters, where the
anaerobic bacteria digest the bacteria and fungi of the flocs; during this process gases like
methane, hydrogen sulphide and carbon dioxide are formed, which constitute the biogas.
34)Identify a, b, c, d, e and f in the table given below:
a)
Organism
Bioactive molecule
Use
1. Monascus purpureus
(yeast)
2. c
a
b
d
Antibiotic
Cyclosporin A
f
3. e
Statin
(b) It is a blood-cholesterol lowering agent
(c)Penicillium notatum
(d) Penicillin
(e) Trichoderma polysporum
(f) It is used as an immuno-suppressive agent in organ-transplant patients.
35)An organic farmer relies on natural predation for controlling plant pests and
diseases. Justify giving reasons why this is considered to be a holistic approach.
QUESTION BANK /BIOLOGY/XII
Page 143
Ans
- 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.
- The organic farmer works to create a system where the insects (called pests) are not
eradicated, but kept at manageable levels by a complex system of checks and balance within
a living and vibrant ecosystem.
- According to the organic farmer, the eradication of the creatures, called pests, is not only
possible but also undesirable, because many beneficial predatory and parasitic insects cannot
survive without them.
- Such a use of biocontrol measures reduces the use of chemical pesticides and thereby
pollution.
36)a) Baculoviruses are excellent candidates for integrated pest management in an
ecologically sensitive area. Explain giving two reasons.
(b) What is organic farming? Why is it suggested to switch over to organic farming ?
Ans.(a) They are species – specific, narrow spectrum bio-control agents.
They have no negative impacts on plants, mammals, birds, fish or non-target insects
(b) Organic farming refers to the use of biofertilisers and biopesticides.
Organic farming is advised as it does not have the following harmful effects caused by
chemical fertilizers and pesticides:
(i) The chemical fertilisers cause pollution of environment and in the long run make the soil
unsuitable for cultivation.
(ii) The chemical pesticides are toxic and destroy a large number of useful insects along with
the pests.
(iii) They also enter the food chain and cause harmful effects.
37) How are flocs produced in the secondary treatment of the sewage? Explain their
role.
Ans.When air is pumped into the effluent in the aeration tanks and constantly agitated the
aerobic microbes grow vigorously and form flocs.
They consume a major part of the organic matter from the primary sludge and reduce the
BOD of the sewage.
In the settling tanks, the flocs sediment to form activated sludge which is used as an
inoculum and also to produce biogas.
38) Why should biological control of pests and pathogens be preferred to the
conventional use of chemical pesticides? Explain how the following microbes act as
biocontrol agents:
(a)
Bacillus thuringiensis
(b)
Nucleopolyhedrovirus.
Ans Biological control of pests and pathogens must be preferred to the use of chemical
pesticides because
(i) The chemicals are toxic and extremely harmful to human beings and other animals
QUESTION BANK /BIOLOGY/XII
Page 144
(ii) The chemicals also cause pollution of soil, groundwater and our agricultural products.
(iii) They are not species specific.
(a) Bacillus thuringiensis
- They are available in sachets as dried spores, which have to be mixed with water and
sprayed onto vulnerable plants.
- When they are eaten by the larvae of specific insects, the toxin is released in the gut of the
larvae and kills the larvae.
- Through genetic engineering, the gene coding for the toxic protein is introduced into crop.
39)State the medicinal value and the bioactive molecule produced by Streptococcus,
Monascus and Trichoderma.
Ans.Streptococcus : Streptokinase ,clot buster/ remove clot from the blood vessels
Monascus : Statin, blood cholesterol lowering agent/ it inhibits the enzymes responsible for
synthesis of cholesterol
Trichoderma , cyclosporin A , immunosppressive agent used in organ transplantation.
40)Cow dung and water is mixed and the slurry is fed into the biogas plant for digestion
by microbes. The person performing the process shares that there is no need to provide
inoculum for it. Why? What is the role of microbes at the source? Under which
condition will they be most active and effective?
Ans.- They are already present in the cow dung
- The source is the Rumen of cattle where they help in
- Breakdown of cellulose
- Under Anaerobic condition
41)What are methanogens? How do they help to generate biogas?
Ans.Anaerobic , methane producing bacteria
Methanogens generate biogas, when they act on cellulose rich biowaste (anaerobically)
QUESTION BANK /BIOLOGY/XII
Page 145
Chapter – 11.Biotechnology : Principles and Processes
VERY SHORT ANSWER TYPE QUESTIONS (1 marks)
1)Why is the enzyme cellulase needed for isolating genetic material from plant cells and
not from animal cells?
Ans.Cellulase is used to digest the cell wall (cellulose) of plant cells; animal cells have no
cell wall and hence, it is not needed.
2)How does an alien DNA gain entry into a plant cell by 'biolistic' method?
The cells are bombarded with high velocity micro-particles of gold or tungsten, coated with
the alien DNA.
3)How is Agrobacterium tumefaciens able to transform a normal plant cell into a
tumour?
Ans.When it infects a plant cell, it delivers a part of its DNA, called T-DNA (or Tumourinducing Ti plasmid) into the plant cell and transforms it into a tumour cell
4)How can retroviruses be used efficiently in biotechnology experiments in spite of them
being disease causing?
Ans.Retroviruses are disarmed (disease-causing gene is removed/inactivated) and used as
vectors to deliver the recombinant/alien DNA into animal cells.
5)State what happens when an alien gene is ligated at Pvu I site of pBR322 plasmid.
Ans.The transformant loses ampicillin-resistance.
6)Why is 'plasmid' an important tool in biotechnology experiments?
Ans.Since plasmids can replicate within the bacterial cell independently of the genomic
DNA, any alien DNA ligated to it will also multiply, ie. it is used as a vector as well as in
gene cloning
7)Name the specific sequence of DNA in a plasmid that the 'gene of interest' ligates
with, to enable it to replicate.
AnsOrigin of replication (Ori).
8)Mention the source of thermostable DNA ploymerase.
Ans.Thermus aquaticus.
QUESTION BANK /BIOLOGY/XII
Page 146
9)How can bacterial DNA be released from the bacterial cell for biotechnology
experiments?
Ans.The bacterial cell has to be treated with lysozyme, to digest the cell wall to release DNA.
10)Why is it essential to have a 'selectable marker' in a cloning vector?
Ans.It helps in identifying the recombinants from the non-transformants transformants/
/recombinants.
11)Why do DNA-fragments move towards the anode during gel electrophoresis?
Ans.DNA-fragments are negatively charged and hence, move towards the anode.
12)In the year 1963, two enzymes responsible for restricting the growth of
bacteriophage in E.coli were isolated. How did the enzymes act to restrict the growth of
the bacteriophage?
Ans.- One of them added methyl groups to DNA.
- The other (restriction endonuclease) cut at specific points within the DNA.
13)What is the role of ethidium bromide during agarose-gel electrophoresis of DNA
fragments?
Ans.The gel is stained by ethidium bromide, to view the separated DNA bands when exposed
to UV light.
14)Give the name of the carcinogenic dye which is used to stain gel to make the DNA
visible in UV light.
Ans.Ethidium bromide (EtBr)
15)Which main technique and instrument is used to isolate DNA form a plant cell?
Ans.Centrifugation and centrifuge
16)How is the action of normal endonuclease enzymes different from that of restriction
endonucleases?
Ans.Normal Endonuclease: makes cut at random position within a DNA
Restriction endonuclease : makes cuts only at specific position within a DNA
QUESTION BANK /BIOLOGY/XII
Page 147
17)Why is not possible for an alien DNA to become part of a chromosome anywhere
along its length and replicate normally?
Ans.Has to be linked with the origin of replication.
18)Name the enzymes that are used for the isolation of DNA from bacterial and fungal
cells for recombinant DNA technology.
Ans.Lysozyme for bacterial cells ,
chitinase for fungal cells .
SHORT ANSWER TYPE QUESTIONS (2marks)
19) Name two commonly used bioreactors. State the importance of using a bioreactor.
Ans.i) Simple stirred-tank bioreactor and
(ii) Sparged stirred-tank bioreactor
- Large volumes of culture can be processed in bioreactor for commercial production of
the product.
Bioreactors provide optimal conditions for achieving the desired product in large quantities.
20) (a) Mention the difference in the mode of action of exonuclease and endonuclease.
(b) How does restriction endonuclease function?
Ans. (a) - Exonucleases remove the nucleotides from the ends of the DNA.
- Endonucleases make cuts at specific positions within the DNA
(b) Restriction endonuclease functions by 'inspecting' the length of the DNA sequence and
find the specific recognition sequence (pallindromic nucleotide sequences), where it binds to
and cuts the two DNA strands at specific points in the sugar-phosphate backbone.
21) (a) Explain how to find whether an E.coli bacterium has transformed or not, when a
recombinant DNA bearing ampicillin-resistant gene is transferred into it.
(b) What does the ampicillin-resistant gene act as, in the above case?
Ans. (a) The transformant can be selected out from the non-transformants by plating them
on ampicillin - containing medium. - The transformants will grow in it, while the nontransformants will not grow
(b) It acts as a selectable marker.
22) Name the source of the DNA-polymerase used in PCR technique. Mention why it is used.
Ans.Taq polymerase is obtained from the bacterium Thermus aquaticus.
- It is a thermostable enzyme that can withstand the high temperature used in the denaturation and
separation of the two strands of DNA during PCR; hence it can be used for a number of cycles of
amplification.
QUESTION BANK /BIOLOGY/XII
Page 148
23) Write any four ways used to introduce a desired DNA segment into a bacterial cell in
recombinant DNA technology experiments.
Ans.i) Microinjection
(ii) Biolistics /gene gun
(iii) Making the cell competent (Heat-shock method )
(iv) Using 'disarmed' pathogen vectors
24) Explain the work carried out by Cohen and Boyer that contributed immensely to
biotechnology.
Ans.Cohen and Boyer were the first to construct a recombinant DNA (rDNA) molecule.
- They isolated the antibiotic-resistance gene by cutting out a piece of DNA with the help of
a restriction enzyme and linked it with a native plasmid (vector) of Salmonella typhimurium
with the help of DNA ligase.
- This rDNA was introduced into Escherichia coli for multiplication to make many copies of
antibiotic gene ( cloning).
25) a) A recombinant vector with a gene of interest inserted within the gene of agalactosidase enzyme is introduced into a bacterium. Explain the method that would
help in selection of recombinant colonies from non-recombinant ones.
(b) Why is this method of selection referred to as 'insertional inactivation'?
Ans.a) - The method is 'insertional inactivation.'
The recombinants can be differentiated from the non-recombinants 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 the chromogenic substrate in the medium.
(b) Since, the insert inactivates the enzyme, a-galactosidase, this method is called insertional
inactivation.
26)How can the following be made possible for biotechnology experiments?
(a) Isolation of DNA from bacterial cell.
(b)Reintroduction of the recombinant DNA into a bacterial
Ans. (a) - The bacterial cells are treated with lysozyme, to remove the cell wall.
- The proteins associated with the DNA are removed by treatment with proteases and the
RNA by treatment with ribonuclease .
- Similarly other molecules (if any) are removed by appropriate treatments.
- The purified DNA is precipitated by the addition of chilled ethanol
(b) - The recipient bacterial cell is made 'competent' to take up the recombinant DNA by
treatment with a specific concentration of calcium ions.
- The rDNA is forced into the cells by incubating the cells with DNA on ice, followed by
placing them briefly at 42 °C (heat shock) and then putting back on ice.
27) List the key tools used in recombinant DNA technology.
QUESTION BANK /BIOLOGY/XII
Page 149
Ans.The key tools include restriction enzymes, polymerase enzymes, ligases, vectors and the
host organism.
28) Explain the role of Ti plasmids in biotechnology.
Ans.Agrobacterium tumifaciens, a pathogen of several dicot plants is able to deliver a
piece of its DNA known as T -DNA to transform normal plant cells into a tumour.
- The tumour inducing Ti plasmid of Agrobacterium has now been modified into a cloning
vector which is no more pathogenic to plants but can still deliver genes of our interest to a
variety of plants.
29) How are recombinant vectors created? Why only one type of restriction
endonuclease is required for creating one recombinant vector?
Ans.The source DNA ( gene of interest) and the vector DNA are cut with a particular
restriction enzyme
- The alien DNA is then linked with the plasmid (vector) DNA using a ligase to form the
recombinant vector.
Since 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 desired source so that the sticky ends of both form complementary
hydrogen bonds which facilitates action of ligases.
30) Study the diagram given below and answer the following the questions:
(a) Why have DNA fragments in band 'D' moved farther away in comparison to those
in band 'C?
(b) Identify the anode end in the diagram.
(c)How are these DNA fragments visualised?
Ans. (a) DNA fragments in band D are smaller in size than those of band C.
QUESTION BANK /BIOLOGY/XII
Page 150
- The DNA fragments separate according to their size through the sieving effect provided by
the gel; hence the smaller fragments move farther away than the larger ones.
(b) Anode towards B end.
(c)The gel containing DNA fragments is stained with ethidium bromide and exposed to UV
radiation; orange-coloured bands (of DNA) become visible.
31) Explain giving reasons why an alien piece of DNA needs to be integrated to a
specific sequence of host DNA for its cloning.
Ans. The specific DNA sequence, where the replication of DNA is initiated, is called origin
of replication (Ori).
- Ori also controls the copy number of the linked DNA . Therefore, for the multiplication of
the alien DNA in the host, it has to be integrated to the origin of replication (Ori).
32) A recombinant DNA is formed when sticky ends of vector DNA and foreign DNA
join. Explain how the sticky ends are formed and get joined.
Ans.
- The same restriction enzyme is used for cutting the vector DNA as well as foreign DNA.
- Sticky ends are formed when a restriction enzyme cuts the strands of DNA a little away
from the centre of the palindromic sequence it recognizes.
- When cut by the same restriction enzyme, the resultant DNA fragments have the same
kind of 'sticky ends' and these can be joined together end to end using DNA ligases.
Since 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 foreign DNA, so that the sticky ends of both form
complementary hydrogen bonds which facilitates action of ligases.
33) How are the DNA fragments separated by gel electrophoresis visualised and
separated for use in constructing recombinant DNA?
Ans.
- The separated DNA fragments are stained with ethidium bromide.
- Then by exposure to UV-radiation, the separated DNAs become visible as orangecoloured bands.
The separated bands of DNA are cut out from the agarose gel and DNA is extracted from
these gel pieces by the process known as elution.
34) Explain the action of the restriction endonuclease EcoRI.
Ans.
- Restriction endonuclease, EcoRI cuts the DNA strands a little away from the centre of the
palindromic sequence, but between the same two bases G and A on the two strands as
shown below.
5`----G A A T T C----3`
QUESTION BANK /BIOLOGY/XII
Page 151
3`----C T T A A G----5`
- This leaves single-stranded portions, called sticky ends, overhanging at the end of each
strand.
- Since, the stickiness facilitates the action of DNA ligase, they easily form hydrogen bonds
with their complementary counterparts.Fig 11.1 , pg 196, NCERT text book)
35)A and B are the two different cloning vectors in two different bacterial colonies
cultured in chromogenic substrate. Bacterial colonies with cloning vector A were
colourless, whereas those with B were blue-coloured. Explain giving reasons the cause
of the difference in colour that appeared.
Ans.Bacterial colonies with cloning vector A are colourless, as they are recombinants with
the insert.
- Presence of the insert has resulted in insertional inactivation of the enzyme and hence, do
not produce any colour.
Bacterial colonies with cloning vector B are non-recombinants, i.e. have no insert and
produce colour as the enzyme is produced and is active.
36)What are recombinant proteins? How do bioreactors help in their production?
Ans.Isolation of DNA form bacterial cells:
- The bacterial cells are treated with enzyme, lysozyme to break the cells open to
release DNA along with RNA and proteins.
- RNA is removed by treatment with ribonucleases and proteins are removed by
treatment with proteases
Other molecules can be removed with appropriate treatment and purified DNA precipitates
out after addition of chilled alcohol.
37) Name the natural source of agarose. Mention one role of agarose in biotechnology.
Ans.Agarose is obtained from sea weeds.
It is used in gel electrophoresis to separate the DNA fragments according to their size
through the sieving effect provide by the agarose gel.
38) Study the linking of DNA fragments shown above.
(i) Name 'a' DNA and 'b' DNA.
QUESTION BANK /BIOLOGY/XII
Page 152
(ii) Name the restriction enzyme that recognises this palindrome.
(iii) Name the enzyme that can link these two DNA fragments.
Ans. (i) a – Vector/plasmid DNA
b- Foreign DNA.
(ii) EcoRI.
(iii) DNA ligase.
39) In which technique do we use Taq polymerase enzyme and why?
Ans.Polymerase chain reaction, because it is a thermostable DNA polymerase ,does not get
denatured at high temperature during PCR and works as normal DNA polymerase enzyme.
40)A vector is engineered with three features which facilitate its cloning within the host
cell. List the three features and explain each one of them.
Ans.
(i)
Origin of replication / ori site – From here the replication starts (and any piece of
DNA when linked can be made to replicate within the host cell) It also controls
copy number.
(ii)
At least two selectable markers – Helps in identifying and eliminating non
transformants.
Cloning sites / recognition sites – for the commonly used restriction enzymes. The foreign
DNA links to this region of the plasmid
41)Where and why do we use Taq polymerase enzyme when it works exactly as DNA
polymerase?
Ans.In PCR because it is a thermostable DNA polymerase enzyme isolated from bacteria
Thermus aquaticus and it does not get denatured at high temperature which is required during
PCR and works as normal DNA polymerase enzyme (whereas the normal DNA polymerase
gets denatured at high temperature).
42)Name the source of the DNA polymerase used in PCR technique. Mention why it is
used.
Ans.Bacterium Thermus aquaticus, thermostable/does not denature under high heat
43)How does a restriction nuclease function? Explain.
Ans.Exonuclease remove nucleotides from ends of DNA .Endonuclease – cuts at specific
positions within DNA.
SHORT ANSWER TYPE QUESTIONS (3marks)
QUESTION BANK /BIOLOGY/XII
Page 153
44) Name and describe the technique that helps in separation and isolation of DNA
fragments.
Ans. - Gel electrophoresis
- Since, DNA fragments are negatively charged; they move towards the anode under the
electric field through a medium / matrix (agarose).
The DNA fragments separate/resolve according to their size due to the sieving effect of
agarose gel.
- The separated fragments can be viewed by staining the DNA with ethidium bromide
followed by exposure to UV radiation.
- Elution is the process in which the separated bands of DNA are cut out from the gel and
extracted.
45) Explain the basis on which the gel electrophoresis technique works. Write any two
ways the products obtained through this technique can be utilised.
Ans. Since, DNA fragments are negatively charged; they are made to move towards the
anode under an electric field through a medium / matrix (agarose).
(i) The DNA fragments may be used to construct recombinant DNA by joining them with
cloning vectors.
(ii) The desired DNA fragments may be amplified i.e. used for making multiple copies by
polymerase chain reaction (PCR).
46) Explain in sequence the process of amplification of a gene of interest using
polymerase chain reaction.
Ans. Polymerase Chain Reaction
- In this reaction multiple copies of the desired gene or segment of DNA are
synthesized in vitro.
- Denaturation : high temperature treatment is used for denaturation and separation of
the two strands of DNA.
- Annealing : using two sets of primers ( chemically synthesized oligonucleotides that
are complementary to the regions of DNA of the two strands ) and the enzyme DNA
polymerase (Taq polymerase).
- Extension : The enzyme DNA polymerase extends the primers using the nucleotides
provided in the reaction and the genomic DNA as the template.
- Amplification : For repeated amplification to be achieved a thermostable DNA
polymerase (Taq polymerase) extracted from the bacterium, Thermus aquaticus is
employed. It remains active during the high temperature treatment used for
denaturation and separation of the two strands.
- The three steps in the PCR are shown in the figure below.
QUESTION BANK /BIOLOGY/XII
Page 154
47) Draw a schematic sketch of pBR322 plasmid and label the following in it:
(a) Any two restriction sites.
(b) Ori and rop genes,
(c) An antibiotic resistant gene.
Ans. Refer to fig 11.4 in the NCERT text book pg 199.
48) (a) Identify A and B illustrations in the following:
(i)
QUESTION BANK /BIOLOGY/XII
Page 155
(ii)
b) Write the term given to A and C and why?
c) Expand PCR. Mention its importance in biotechnology
Ans.(a) (i) A is recognition/restriction site (GAATTC) which the restriction enzyme, EcoRI,
recognises.
(ii) B is the Rop gene that codes for the protein involved in the replication of plasmid.
(b) A and C are called palindromes as the sequence of base pairs reads the same on the two
strands when the orientation of reading is kept same.
(c) Polymerase chain reaction.
- In this reaction multiple copies of the gene (or DNA) of interest can be synthesised in vitro.
49)(a) Why are restriction endonucleases so called?
(b) What is a palindromic nucleotide sequence? How do restriction endonucleases act
on palindromic sites, to create 'sticky ends'?
Ans. (a) Restriction endonucleases are called so because they restrict the growth of
bacteriophage by cutting their DNA at specific sequences.
(b) A palindrome in DNA is the sequence of base pairs that reads the same on the two strands
of DNA when the orientation of reading is kept the same.
Each restriction endonuclease recognizes a specific nucleotide sequence in the DNA.
- Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome
site but between the same two bases on both the strands.
- This creates single-stranded stretches overhanging at the ends of the palindrome; they are
called 'sticky ends'.
50) How are the following used in biotechnology?
(a) Plasmid DNA
(b) Recognition sequence
(c) Gel electrophoresis
Ans.(a) Plasmid DNA
- It is used for constructing recombinant DNA by ligating the gene of interest with it, it is
used as the cloning vector.
(b) Recognition Sequences
- These are the palindromic sequences of base pairs in DNA which are recognized by specific
restriction enzymes and where a restriction enzyme cuts the DNA.
QUESTION BANK /BIOLOGY/XII
Page 156
(c) Gel electrophoresis
- It is a technique used to separate the DNA fragments which separate/resolve according to
their size through the sieving effect of the gel under an electric field.
51)(a) What are ‘molecular scissors`? Given one example.
(b) Explain their role in recombinant DNA technology
Ans.(a) Molecular scissors are the restriction endonucleases, which cut the two strands of
DNA at specific points in their sugar-phosphate backbone, e.g EcoRI.
(b) - Each restriction endonuclease recognises a specific palindromic sequence in the DNA.
- It binds to the DNA and cuts each of the two strands of the double stranded DNA at specific
points in their sugar-phosphate backbones.
- The restriction enzymes cut the strand of DNA a little away from the centre of palindrome
site, but between the same two bases on the two strands.
- This creates single-stranded portions overhanging at the end of each strand; they are called
sticky ends and this stickiness facilitates the action of the enzyme DNA ligase.
- When cut by the same restriction endonuclease, the DNA fragments from the two sources
will have same kind of sticky ends and they form hydrogen bonds with their complementary
cut counterparts and joined together (end-to-end) using DNA ligases.
52)Why is Agrobacterium tumefaciens a good cloning vector? Explain.
Ans.Agrobacterium tumefaciens is a pathogen of several dicot plants.
- It is able to deliver a piece of DNA, called T-DNA into the plant cells and transform them
into tumour cells; it dictates the host cells to synthesise its nutrients.
- The Tumour inducing (Ti) plasmid of this bacterium is modified and made non-pathogenic
( disarmed).
- Though it is non-pathogenic, it still has the capacity to deliver its Ti plasmid to the plants
and the genes of interest can be ligated to it.
53)Explain the importance of (a) ori,
(b) ampR and
(c) rop in the E.coli vector shown below
Ans.
(a) Ori
- Ori is a sequence of bases on DNA, from where replication of DNA starts.
QUESTION BANK /BIOLOGY/XII
Page 157
- Any piece of alien DNA linked to this sequence, can be made to replicate within the host
cells; it is also responsible for controlling the copy number of linked DNA.
(b) ampR
- It is a selectable marker on the cloning vector pBR322.
- If a segment of alien DNA is ligated at the PvuI or PstI site of ampR gene, the recombinant
will lose its ampicillin resistance (insertional inactivation); thus, the recombinants can be
selected out from the non-recombinants.
(c) Rop
- Rop codes for the proteins involved in the replication of the plasmid.
54)DNA being hydrophilic cannot pass through the cell membrane of a host cell.
Explain how the recombinant DNA gets introduced into the host cell to transform the
latter.
Ans.
Introduction of rDNA into host cell
(i) Heat shock method
The cell is first made competent by treating it with a specific concentration of divalent cation
such as calcium which increase the efficiency with which DNA enters the host cell through
pores in its cell wall.
rDNA is forced into the competent cell by incubating the cell with rDNA on ice followed by
placing them briefly at 42 °C (heat shock) and then putting them back on ice. This enables
the host cell to take up rDNA
(ii) Microinjection
- In this method, the recombinant DNA is directly injected into the nucleus of an animal cell.
(iii) Gene gun/Biolistics
- In this method, used for plant cells, the cells are bombarded with high velocity
microparticles of gold or tungsten coated with DNA
(iv) By using disarmed pathogens.
55)
(a) Identify the selectable markers a and d in the diagram of E.coli vector shown above.
(b) How is the coding sequence of a-galactosidase considered a better marker than the
ones
identified by you in the diagram? Explain.
Ans.
(a) a = ampR
QUESTION BANK /BIOLOGY/XII
Page 158
d = tetR,
(b) - a-galactosidase marker differentiates the recombinants from the non-recombinants
based on their ability to produce colour in the presence of a chromogenic substrate.
- When a recombinant or an alien DNA is inserted into the coding sequence of the enzyme
a - galactosidase, there occurs insertional inactivation of the enzyme. Consequently no
colour develops in the colonies with an insert (recombinants)
- However, a blue colour develops with the chromogenic substrate, when the bacterial
colonies do not have an insert (non-recombinants)
Selection of recombinants due to inactivation of antibiotic-resistance is cumbersome because
it involves simultaneous plating of two plates having different antibiotics (tetracycline and
ampicillin).
56)A vector is engineered with three features, which facilitate its cloning within the host
cell. List the three features and explain each one of them.
Ans.Features of Cloning vectors:
(i) Origin of Replication (Ori)
- This is the sequence of DNA from where replication starts.
- Any piece of alien/foreign DNA is linked to it to replicate within host cell;
- it also decides the copy number of the linked DNA
(ii) Selectable marker
- A marker is a gene, which helps in selecting the host cells, which are
transformants/recombinants from the non-recombinant/ transformants, e.g. ampicillin and
tetracycline resistant genes in E.coli.
(iii) Cloning site
In order to link the alien DNA the vector should have very few, preferably one recognition
site for the commonly used restriction enzymes.
57)
(a) What does this diagram depict?
(b) What is meant by largest and smallest in the picture?
(c) Name the compound used to visualize them.
(d) Define elution.
Ans.(a) It depicts gel electrophoresis
(b) DNA bands/fragments
The DNA fragments separate according to their size through the sieving effect provided by
agarose gel. Hence, the smaller the fragment size, the farther it moves.
(c) Ethidium bromide.
QUESTION BANK /BIOLOGY/XII
Page 159
(d) Elution is the process in which the separated bands of DNA are cut from the agarose gel
and extracted from the gel piece.
58)An interesting property of restriction enzymes is molecular cutting and pasting.
Restriction enzymes typically recognize a symmetrical sequence of DNA.
Notice that the top strand is the same as the bottom strand, but reads backwards.
When the enzyme cuts the strands between G and A it leaves overhanging chains:
a. What is this symmetrical sequence of DNA known as?
b. What is the significance of these overhanging chains?
c. Name the restriction enzyme that cuts the strand between G and A.
Ans. a. Palindromic nucleotide sequence.
b. These overhanging single-stranded stretches are called sticky-ends. They from hydrogen
bonds easily with their complementary cut counterparts and this stickiness of the ends
facilitates the action of the enzyme DNA ligase.
c. EcoRI
59)Name the particular technique in Biotechnology whose steps are shown in the figure.
Use the figure to summarise the technique in three steps
QUESTION BANK /BIOLOGY/XII
Page 160
Ans.Recombinant DNA technology
The steps are:
(i) Isolation of the desired DNA segment
(using restriction enzymes).
(ii) Joining the DNA segment with the plasmid DNA
(using DNA-ligase)
(iii) Introduction of the recombinant DNA into a host cell.
60)Name the pest that destroys the cotton bolls. Explain the role of Bacillus
thuringiensis in protecting the cotton crop against the pest to increase the yield.
Ans.Cotton Boll worm.
Bacillus thuringienisis has Bt toxin genes, due to which it produces toxic proteins that kill the
pest, when the pest ingests it, the toxin gets activated due to alkaline pH of the gut, specific
Bt toxin genes were isolated from Bacillus thuringenisis and incorporated into the cotton
plants to make them pest resistant
61)Name and describe the technique that helps in separating the DNA fragments
formed by the use of restriction endonuclease.
Ans.Gel electrophoresis, DNA fragments negatively charged, separated by forcing them to
move towards anode, under electric field, through a medium / matrix (agarose), DNA
fragments separate according to their size through sieving effect of agarose gel.
62)
(a) Mark the positive and negative terminals.
(b) What is the charge carried by DNA molecule and how does it help in its separation?
(c) How the separated DNA fragments are finally isolated?
Ans.
(a) Positive terminal - ‘B`
Negative terminal - ‘A`
QUESTION BANK /BIOLOGY/XII
Page 161
(b) DNA being negatively charged, moves towards the positive electrode (anode) = 1
(c) By elution – separated bands of DNA are cut out from the agarose gel and extracted
from the gel piece
63)CryIAb is introduced in a plant to control infestation by corn borer.
(a) Name the resultant plant after successful insertion of the desired gene.
(b) Summarize the action of the gene introduced.
Ans.(a) Bt corn
(b) CryIAb / Bt toxin gene codes for crystal protein; the Bt toxin protein exists as an inactive
protein but once an insect ingests it gets converted into an active form due to the alkaline pH
of the gut which solubilizes the crystal. The activated toxin binds to the surface of mid gut
and creates pores that cause swelling, lysis and eventually death of the insect.
64)(a) In pBR322, foreign DNA has to be introduced in tetR region. From the restriction
enzymes given below which one should be used and why:
PvuI, EcoRI, BamHI
(b) Give reasons why the other two enzymes cannot be used.
Ans.
(a) Bam HI should be used as restriction site for the enzyme is present in tetR region.
(b) PvuI will not be used as restriction site for this enzyme is present in ampR region (not in
tetR)
EcoRI will not be used as restriction site for this enzyme is not present in selectable marker
tetR
LONG ANSWER TYPE QUESTIONS (5marks)
65)If a desired gene is identified in an organism for some experiments, explain the
process of the following:
(i) Cutting this desired gene at a specific location.
(ii) Synthesis of multiple copies of this desired gene.
Ans.i) – Cutting of desired gene at specific location is done by incubating the DNA with
specific restriction endonucleases.
- Each restriction endonuclease recognises a specific palindromic sequence in the DNA.
- It binds to the DNA and cuts each of the two strands of the double stranded DNA at specific
points in their sugar-phosphate backbones.
- The restriction enzymes cut the strand of DNA a little away from the centre of palindrome
site, but between the same two bases on the two strands.
- This creates single-stranded portions overhanging at the end of each strand; they are called
sticky ends and this stickiness facilitates the action of the enzyme DNA ligase.
(ii) Synthesis of multiple copies of the desired gene is carried out by Polymerase Chain
Reaction (PCR)
QUESTION BANK /BIOLOGY/XII
Page 162
-
-
In this reaction multiple copies of the desired gene or segment of DNA is synthesized
in vitro.
Denaturation : high temperature treatment is used for denaturation and separation of
the two strands of DNA.
Annealing : using two sets of primers ( chemically synthesized oligonucleotides that
are complementary to the regions of DNA of the two strands ) and the enzyme DNA
polymerase.
Extension : The enzyme DNA polymerase extends the primers using the nucleotides
provided in the reaction and the genomic DNA as the template.
Amplification : For repeated amplification to be achieved a thermostable DNA
polymerase (Taq polymerase) extracted from the bacterium, Thermus aquaticus is
employed. It remains active during the high temperature treatment used for
denaturation and separation of the two strands.
66)
(a) Why are engineered vectors preferred by biotechnologists for transferring the
desired
genes into another organism?
(b) Explain how 'ori, 'selectable markers' and 'cloning sites' facilitate cloning into a
vector.
Ans.(a) Engineered vectors are preferred because they help easy linking of foreign DNA and
selection of recombinants from non-recombinants.
(b) Ori
- It is sequence of bases on DNA from where replication starts- any piece of DNA when
linked to this sequence can be made replicate within the host cells.
- This sequence also controls the copy number.
Selectable marker
- The selectable marker helps in identifying and eliminating non-transformants or non
recombinants and permits selectively the growth of only recombinants.
Cloning sites
- Cloning sites are necessary to link the alien DNA. Single recognition sites of commonly
used restriction enzymes are preferred.
- When an alien DNA is introduced into the coding sequence of an enzyme or an antibiotic
resistance gene there is insertional inactivation- the enzyme is inactivated or the antibiotic
resistance is lost and hence recombinants can be selected from the non-recombinants.
67)What is a bioreactor? Draw a labelled diagram of a sparged stirred-tank bioreactor
its functioning.
Ans.Bioreactors can be considered as vessels in which raw materials are biologically
converted into specific products using microbial plant or human cells.
- A bioreactor provides optimal conditions for achieving the desired product by providing
optimum growth conditions, pH, substrate, salts, vitamins, oxygen, etc
- The commonly used bioreactors are of stirring type.
QUESTION BANK /BIOLOGY/XII
Page 163
- Air can be bubbled through the reactor / sterile air bubbles are sparged.
The stirrer facilitates the even-mixing and oxygen availability throughout the bioreactor.
Ref to fig 11.7(b) in the NCERT text book pg 204
68)(a) Mention the role of vectors in recombinant DNA technology. Give any two
examples.
(b) With the help of diagrammatic representation only, show the steps of recombinant
DNA technology.
Ans.(a) Role of Vectors
- The vectors have the ability to replicate within the bacterial cells independent of the control
of chromosomal DNA.
- If an alien piece of DNA is linked to the vector it can be made to multiply its number equal
to the copy number of the vector.
- Vectors are also used in the selection of recombinants from non-recombinants.
- e.g. plasmids and bacteriophages are the most commonly used vectors.
(b)
QUESTION BANK /BIOLOGY/XII
Page 164
QUESTION BANK /BIOLOGY/XII
Page 165
Chapter12-BIOTECHNOLOGY AND ITS APPLICATIONS
VERY SHORT ANSWER TYPE QUESTIONS (1 marks)
1)Name any two techniques that serve the purpose of early diagnosis of some
bacterial/viral human diseases.
Ans.(i) Enzyme-linked immunosorbent assay (ELISA) test.
(ii) Polymerase Chain Reaction (PCR)
2)Mention the source organism of the gene cryIAc and its target pest.
Ans Bacillus thuringiensis
Cotton bollworm is its target pest.
3)How does dsRNA gain entry into eukaryotic cell to cause RNA interference?
Ans.(i) Infection by viruses with RNA genome.
(ii) Transposons, the mobile genetic elements, that can replicate via an RNA intermediate.
4)Name the cry genes that control cotton bollworm and corn borer, respectively.
Ans .cryIAc and cryIIAb control cotton bollworm cryIAb controls corn borer.
5)What was the speciality of the milk produced by the transgenic cow, Rosie?
Ans.It contains human protein, alpha-lactalbumin (2.4 g/l) that is nutritionally more balanced
for human babies than normal cow-milk.
6)How does silencing of specific mRNA prevent parasitic infestation?
Ans.The nematode cannot live in the transgenic host that expresses the specific RNA
interference as it prevents translation of mRNA.
7)How are tobacco plants benefitted, when nematode-specific genes are introduced into
them using certain vectors? Name the vectors used.
Ans.The tobacco plants become resistant to the nematode, as the nematode cannot survive in
the transgenic host expressing RNA interference.
- The vector is Agrobacterium tumefaciens.
8)State the role of transposons in silencing of mRNA in eukaryotic cells.
Ans.Transposons or mobile genetic elements in viruses are the source of the complementary
ds RNA that in turn binds/silences specific mRNA/causes RNA interference of the parasite.
QUESTION BANK /BIOLOGY/XII
Page 166
9) the function of adenosine deaminase enzyme. State the cause of ADA deficiency in
humans. Mention a possible permanent cure for an ADA deficiency
Ans.- Adenosine deaminase enzyme is necessary for the proper functioning of our immune
system.
- ADA deficiency is caused by the deletion of the gene for ADA.
- If the normal gene producing ADA can be isolated from marrow cells and introduced into
cells at early embryonic stages, it could be a permanent cure.
10)Expand the following and mention one application of each:
(a) PCR
(b) ELISA
Ans. (a) Polymerase Chain Reaction (PCR)
(b) Ezyme-linked Immunosorbent Assay (ELISA)
- PCR is used to detect HIV infection.
- PCR is also used to detect mutations in genes in suspected cancer patients.
- ELISA is used for the detection of any infection by a pathogen.
(any one)
11)Why is pro insulin so called? How is insulin different from it?
Ans.- Pro insulin is the pro hormone which needs to be processed before it becomes a fully
mature and functional hormone- insulin.
- Pro insulin has three polypeptide chains (A, B and C), is inactive, whereas insulin has only
two (A and B) polypeptide chains and is active.
12)Biopiracy should be prevented. State why and how.
Ans.Biopiracy refers to the use of bio-resources by multinational companies and other
organizations without proper authorization from the countries and people concerned and
without compensatory payment.
It should be prevented because
(i) The countries/people concerned are not given adequate compensatory payment.
(ii) The countries/people also lose their right to grow and use them in breeding experiments
to improve other varieties of the same species.
(iii) The countries/ people do not get any benefits.
Biopiracy can be prevented by developing laws to prevent the unauthorised exploitation of
their bioresouces and traditional knowledge.
13)How is recombinant DNA technology helping in detecting the presence of mutant
genes in cancer patients?
Ans.A single-stranded DNA or RNA is tagged with a radioactive molecule and used as a
probe.
QUESTION BANK /BIOLOGY/XII
Page 167
- It is allowed to hybridize with its complementary DNA in a clone of cells, followed by
detection using autoradiography.
The clone having the mutant gene will not appear on the photographic film, because the
probe will not be complementary with the mutated gene.
14) the steps undertaken during gene therapy to treat an ADA-deficient patient.
Ans.Gene therapy for ADA-deficiency:
- The lymphocytes from the blood of the patient are grown in culture in vitro.
- A functional ADA cDNA is introduced into lymphocytes using a retroviral vector.
- The lymphocytes are introduced back into the blood of the patient.
- But a patient requires periodic infusion of such genetically-engineered lymphocytes, as
lymphocytes have a particular life span.
If the functional gene (producing ADA) isolated from marrow cells is introduced into the
cells of early embryonic stages, it could be a permanent cure.
15)Why do the toxic insecticidal proteins secreted by Bacillus thuringiensis kill the
insect and not the bacteria itself?
Ans.Bacillus thuringiensis because it exists as an inactive protoxin.
- When Bt toxin is ingested by an insect it is converted into its active form when exposed to
the alkaline pH in the gut.
- The activated toxin binds to the surface of the epithelial cells of the midgut and creates
pores.that causes their swelling and lysis and eventually death of the insect.
16) any four advantages of genetically modified organisms
Ans.Advantages of GMOs:
(i) Genetic modification has made the crops more tolerant to abiotic stresses like cold, heat,
drought, salinity, etc.
(ii) It has reduced the dependence of crops on chemical pesticides as they are made pestresistant.
(iii) Post-harvest losses are much reduced.
(iv) These plants have increased efficiency of mineral usage and hence, the early exhaustion
of soil fertility is prevented.
(v) Food produced from GM crops has enhanced nutritional.
17)State the principle on which ELISA technique is based. How does it help in early
detection of a disease?
Ans.ELISA technique is based on the principle of antigen-antibody interaction.
- Infection by a pathogen can be detected either by the presence of antigens (like
glycoproteins or proteins) or by the antibodies synthesised by the body against the particular
pathogen.
18)Why is the introduction of genetically engineered lymphocytes into an ADA
deficiency patient not a permanent cure? Suggest a possible permanent
QUESTION BANK /BIOLOGY/XII
Page 168
Ans.Lymphocytes are not immortal but are short lived; hence the patient requires periodic
infusion of such genetically engineered lymphocytes; however if a gene producing ADA is
isolated from marrow cells and introduced into the cell at early embryonic stages it could be
a permanent cure.
19)Why is pro-insulin so called? How is insulin different from it?
Ans.Pro insulin is an inactive from of insulin, containing an extra stretch called C peptide,
insulin is made up of only 2 short polypeptide chains A and B linked by disulphide bridges,
is functional.
20)How have transgenic animals proved to be beneficial in
(a) Production of biological products
(b) Chemical safety testing
Ans. (a) Production of biological products.Rosie the first transgenic cow produced milk that
contained human protein, a-lactalbumin (2.4 g per liter) which is nutritionally more balanced
for human babies.
(b) Chemical safety testing.
Transgenic animals are made that carry genes which make them more sensitive to toxic
substances than non transgenic animals. They are then exposed to the toxic substances and
the effects studied.
SHORT ANSWER TYPE QUESTIONS (3marks)
21)How is Bt-cotton plant created as a GM plant? How is it protected against bollworm
infestation?
Ans.Specific Bt toxin genes cryIAc and cryIIAb which control cotton bollworms were
isolated from Bacillus thuringiensis and incorporated into cotton plants.
- The transgenic / GM cotton plants - Bt cotton plants produce the recombinant proteins,
called Cry proteins, encoded by the genes, during a particular phase of their growth.
- These crystals of insecticidal proteins, are toxic to cotton bollworm.
- These proteins exist as inactive protoxins, but get converted into an active form of toxin, in
the gut of insects due to the alkaline pH.
- The active toxin binds to the surface of midgut epithelium and creates pores that cause
swelling and lysis of cells; this leads to the death of insects as the plants are resistant to
bollworm.
22)(a) Explain the effect of deletion of the gene for ADA in an individual.
(b) How does gene therapy help in the case?
Ans. (a) Deletion of the gene for ADA leads to ADA deficiency in the individual.
- Since this enzyme is crucial for the immune system to function, it leads to weakening of the
immune system.
QUESTION BANK /BIOLOGY/XII
Page 169
(b) In gene therapy a functional ADA gene is introduced into the patient with the help of
rDNA technology so that the patient can synthesize the enzyme -adenosine deaminase.
- if the functional DNA can be introduced into the cells at early embryonic stages, it becomes
a permanent cure.
23)Plasmid is a boon to biotechnology. Justify this statement quoting the production of
human insulin as an example.
Ans.Plasmid is an autonomously replicating, extra-chromosomal, circular DNA found in
bacterial cells. As they can replicate within bacterial cells independent of the control of
chromosomal DNA ,they are used as vectos in rDNA technology
Production of Insulin:
- Eli Lilly, an American company prepared two DNA sequences in vitro coding for
the polypeptides A and B of insulin.
- These DNA sequences were introduced into the plasmids of Escherichia coli to
produce insulin chains.
- Chains A and B were produced separately, extracted and combined by creating
disulfide bonds to form human insulin.
24)Name the source and the types of cry genes isolated from it for incorporation into
crops by biotechnologists. Explain how these genes have brought beneficial change in
the genetically modified crops.
Ans.The source organism is Bacillus thuringiensis.
- The following types of cry genes are isolated from it:
cryIAc, cryIIAb, cryIAb
The proteins encoded by the genes cryIAc, cryIIAb control cotton bollworms, that of cryIAb
controls corn borer.
-These genes have been incorporated into crops to make GM plants like Bt cotton, Bt corn
etc
- The cry genes code for toxic insecticidal protein Bt toxin.
- The Bt toxin exists as inactive pro toxin and gets converted into active form (toxin) in the
alkaline pH of the gut of the insect.
- The activated toxin binds to the epithelial cells lining the surface of the midgut, and creates
pores that cause swelling and lysis of cells; this leads to the death of the insect.
- Thus, the genetically modified crop are resistant to insect pests without use of chemical
insecticides and crop yield is increased
25)How did an American company, Eli Lilly use knowledge of r-DNA technology to
produce human insulin?
Ans.Two chains of DNA sequences corresponding to A & B chains of human insulin
prepared,
introduced them into plasmids of E.coli to produce separate A & B chains, A & B chains
extracted, combined by creating disulphide bonds.
26)Rearrange the following in the correct sequence to accomplish an important
biotechnological reaction:
QUESTION BANK /BIOLOGY/XII
Page 170
(a) In vitro synthesis of copies of DNA of interest
(b) Chemically synthesized oligonucleotides
(c) Enzyme DNA-polymerase
(d) Complementary region of DNA
(e) Genomic DNA template
(f) Nucleotides provided
(g) Primers
(h) Thermostable DNA-polymerase (from Thermus aquaticus)
(i) Denaturation of ds-DNA
Ans.i---e---b/g---g/b---c/h---h/c---f---d---a
OR
a---i---e---b/g---g/b---c/h---h/c---f---d
27)Two of the steps involved in producing nematode resistant tobacco plants based on
the process of RNAi are mentioned below. Write the missing steps in its proper
sequence.
(a) ------------------------------------------------------(b) Using Agrobacterium as vector introduce it into tobacco
(c) -------------------------------------------------------(d) -------------------------------------------------------(e) Initiates RNA interference
(f) -------------------------------------------------------(g) ------------------------------------------------------(h) ------------------------------------------------------Ans.Isolation of nematode-specific genes.
(c) Production of sense and anti-sense RNA in the host
(d) Formation of ds RNA(as the sense and anti-sense RNA are complementary to each other)
(f) Silencing of specific mRNA of the nematode.
(g) Parasite could not survive in transgenic host expressing the specific interfering RNA.
(h) The transgenic plant is therefore protected from the parasite.
LONG ANSWER TYPE QUESTIONS (5marks)
28)(a) Name the source from which insulin was extracted earlier. Why is this insulin no
more in use by diabetic people?
(b) Explain the process of synthesis of insulin by Eli Lilly company.
Name the technique used by the company.
(c) How is the insulin produced by human body different from the insulin produced by
the above mentioned company?
QUESTION BANK /BIOLOGY/XII
Page 171
Ans. (a) - Pancreas of slaughtered pigs and cattle was the source of insulin.
- Insulin from these sources caused some patients to develop allergy or other types of
reactions to the foreign protein; hence no more used by diabetics.
(b) - Production of human insulin using rDNA was carried out by Eli Lilly, an American
company.
- It prepared two sequences of DNA that code for the A and B chains of insulin.
- These two DNA sequences were introduced into the plasmids of Escherichia coli to
produce insulin chains.
- Chains A and B were produced separately.
- They were extracted and combined by creating disulphide bridges to form human insulin.
(c) In humans insulin is synthesized as a pro hormone having three polypeptide chains A, B
and C and has to be processed to become functional.
- The insulin developed by Eli Lilly was functional
29)Name the process involved in the production of nematode-resistant tobacco plants,
using genetic engineering. Explain the strategy adopted to adopted to develop such
plants.
Ans.The process involved in the production of nematode-resistant tobacco plants is RNA
interference.
The process RNA interference (RNAi) involves silencing of a specific mRNA.
It takes place in all eukaryotic organisms as a method of cellular defence.
- The nematode-specific genes were introduced into the host plant by the use of
Agrobacterium vectors.
- The introduction of DNA was such that it produced both sense and anti-sense RNA in the
host cells.
- These two RNAs being complementary to each other formed a double-stranded RNA,which
binds to the mRNA and prevents its translation
- initiated RNA interference
and silenced the specific mRNA of the host.
- The parasite could not survive in a transgenic host that expresses the specific interfering
double-stranded RNA;
- The transgenic plant, therefore, got itself protected from the nematode ( parasite).
30)One of the main objectives of biotechnology is to minimise the use of insecticides on
cultivated crops. Explain with the help of a suitable example, how insect-resistant crops
have been developed using techniques of biotechnology.
Ans.Insect-resistant crops:
- .The bacterium Bacillus thuringiensis produces insecticidal proteins, called Bt toxins,
which kill the larvae of some insects.
- The Bt toxin gene has been isolated and cloned in bacteria.
QUESTION BANK /BIOLOGY/XII
Page 172
- Such genes are introduced into crop plants, which could be expressed in plants to provide
resistance to the insects, i.e. they function as biopesticides; so there is no need for
insecticides.
- Bt toxins are insect group-specific and coded by cry genes; the proteins encoded by genes
cryIAc and cryIIAb control cotton bollworms while cryIAb controls corn borer.
- eg Bt cotton.Specific Bt toxin genes cryIAc and cryIIAb which control cotton bollworms
were isolated from Bacillus thuringiensis and incorporated into cotton plants.
- The transgenic / GM cotton plants produce the recombinant proteins, called Cry proteins,
encoded by the genes, during a particular phase of their growth.
- These crystals of insecticidal proteins, are toxic to cotton bollworm.
- These proteins exist as inactive pro toxins, but get converted into an active form of toxin, in
the gut of insects due to the alkaline pH.
- - The active toxin binds to the surface of mid gut epithelium and creates pores that cause
swelling and lysis of cells; this leads to the death of insects as the transgenic plants are
resistant to bollworm.
31)What is ADA deficiency? Describe three methods to cure it.
Ans.ADA deficiency is deficiency of enzyme adenosine deaminase, which is crucial for the
normal functioning of our immune system. Deletion of the gene for ADA causes ADAdeficiency. In some children ADA deficiency can be cured by:
(i) Bone marrow transplantation.
(ii) Enzyme replacement therapy in which functional ADA is given to the patient by
injection. Both these methods are not completely curative.
(iii)Gene therapy- Lymphocytes from the blood of the patient are grown in a culture. A
functional ADA cDNA is introduced into these lymphocytes which are subsequently returned
to the patient. The permanent cure is done by introducing ADA cDNA into cells at
embryonic stages.
32)What are transgenic animals? Explain any four ways in which such animals can be
beneficial to humans.
Ans.Transgenic animals are those animals which have had their DNA manipulated to possess
and express an extra foreign gene.
They are beneficial for the following purposes.
(i) Transgenic animals are designed to allow the study of how genes are regulated and how
they affect the normal functions of the body and its development, e.g. study of complex
factors involved in growth such as insulin - like growth factor. By introducing genes from
other species that alter the formation of this factor and studying the biological effects that
result, information is obtained about the biological role of the factor in the body.
(ii) For production of useful biological products e.g. by introduction of the portion of DNA
gene which codes for human protein a-1-antitrypsin used for treating emphysema.
(iii) Transgenic animals are being developed to test the safety of vaccines, before they are
used on humans e.g. polio vaccine has been tested on transgenic mice.
(iv) Transgenic animals with more sensitivity to toxic substances are being developed to test
the toxicity of drugs.
QUESTION BANK /BIOLOGY/XII
Page 173
Chapter13-ORGANISMS AND POPULATIONS
VERY SHORT ANSWER TYPE QUESTIONS (1 marks)
1) Most living organisms cannot survive at temperatures above 45°c. How are
some microbes able to live in habitats with temperature beyond 100°c?
Ans. Microbes posses physiological and biochemical adaptations which allow them
to live in habitats with temperature exceeding. 100°c. There are branched chain
lipids in the cell membrane and special resistant enzymes which deal with high
temperatures.
2) An orchid plant is growing on a mango tree. How do you describe the
interaction between the orchid and the mango tree?
Ans.Commensalisms, In this interaction the orchid is benefitted on attaining the
Supportwhere as ,mango tree is neither benefitted nor harmed.
3) People living in higher altitudes have higher RBC count. Give reasons
Ans.Due to low atmospheric pressure at higher attitudes body compensates low
oxygen availability by increasing red blood cell production, decreasing the
binding affinity of hemoglobin and by increasing breathing rate.
4) Cattle or goats never graze on weeds of calotropis. Give reasons.
Ans.The plant produces highly poisonous cardiac glycosides.
5) Two closely related species competing for the same resources cannot co-exist
indefinitely.State the principle which supports this phenomenon.
Ans.Gause's competitive exclusion principle
6) What type of growth status the following pyramid represents ?
Ans.Declining population
7) "Abingdon tortoise in Galapagos islands became extinct within a decade after
goats were introduced in that island". Can you cite the possible reason for the
same?
QUESTION BANK /BIOLOGY/XII
Page 174
Ans.Goatshave greater browsing efficiency and hence the tortoises died of lack of
food.
8) State Allen's rule.
Ans.Mammals from colder climates generally have shorter ears and
limbs to minimise heat loss
9) What is the unit of productivity in terms of energy and weight respectively?
Ans.Energy → [Kcal m-2]yr-1
Weight → [g m-2]yr-1
10)Fresh water animals are unable to survive for long in sea water. Give
reason.
Ans. Due to osmotic problems.
11) Define diapause. Which organisms exhibit it?
Ans.A stage of suspended development, zooplanktons.
12) Calculate the death rate if 6 individuals in a laboratory population of 60
fruit flies died during a particular week.
Ans .6/60 =0.1 individuals per fruitfly per week.
13) In biological control method, one living organism is used against anotherto check its
uncontrolled growth. Which kind of population interaction isinvolved in this?
Ans. Predation.
14) An organism has to overcome stressful condition for a limited period of
time. Which strategies can it adopt to do so?
Ans. (i) Migration
(ii) Suspension of active life by hibernation/aestivation/spore formation.
15) Write what do phytophagous insects feed on?
Ans. Plant sap and other parts of plant
SHORT ANSWER TYPE QUESTIONS (2marks)
16)Mention the attributes which a population has but not an individualorganism.
QUESTION BANK /BIOLOGY/XII
Page 175
Ans.Birth rate, Death rate, Sex ratio, age groups.
17)Explain why very small animals are rarely found in polar region.
Ans.Since small animals have a larger surface area relative toto lose body heat very fast
when it is cold outside; then they have to expend muchto generate body heat through
metabolism. This is the main reason whyvery small animals are rarely found in Polar
Regions.
II | Biology
18) Study the graphical representation shown below and mention the
conditions responsible for the curves “a” and “b” respectively.
Ans. Curve 'a' represents exponential growth where the resources are not limiting
the growth.
b) Curve 'b' represents logistic growth where the resources are limiting the
growth.
19)
a) Which type of growth curve does it represent?
b) What do the notations represent in the above equation?
Ans.a) Logistic growth curve
b) N= population density at-time 't'
r=Intrinsic rate of increase
K= carrying capacity
20) Kangaroo rats can survive in the absence of an external source of water. How
do they adapt themselves to such conditions?
Ans. Kangaroo rat is capable of meeting all its water requirement through its
internal fat oxidation in which water is a by -product. It has the ability to
QUESTION BANK /BIOLOGY/XII
Page 176
concentrate urine.
21) If a marine fish is placed in fresh water aquarium, will the fish be able to
survive?Why?
Ans. No, it will not survive in fresh water aquarium because of osmotic problem it
would face.
22) "Snow leopards are not found in Kerala forests and tuna fish are rarely found
beyond tropical latitude in the ocean". Study the above two cases and states the
possible reasons for the same.
Ans. Change in temperature from their established habitats affects the kinetics of
the enzymes and through it the basal metabolism, activity and other
physiological functions of the organism
23)1920’s, in Australia, the prickly pear cactus caused havoc by spreading rapidly over
millions of hectares of rangeland. How it was brought In the early under control. What
term is used for such methods of controlling the prey?
Ans. cactus was brought under control only after a cactus feeding predator a moth from its
natural habitat was introduced in the country. It is called ‘Biological Control’.
24)Generally thermo regulation is energetically expensive for many organisms. So,
many smaller animals are rarely found in extreme hot or cold climatic condition. Why?
Write two specific reasons.
Ans.When it is cold out side, the animals tend to lose body heat quickly if they have large
surface area relative to their volume. They expend more energy to generate body heat by
metabolism. Small animals have limited reserves of energy, so they cannot survive.
25) Differentiate between stenohaline and euryhaline organisms.
Ans.Euryhaline : Organisms tolerant in wide range of salinities.
Stenohaline : Organisms tolerant to narrow range of salinities.
26)Mention the attributes which a population has but not an individualorganism.
Ans.Birth rate, Death rate, Sex ratio, age groups.
27) List four features which enable the Xeric plants to survive in the desertconditions.
Ans. (i) thick cuticle
(ii) Stomata in deep pits
(iii) Stomata closed during day time
(iv) leaves reduced to spines (CAM photosynthetic pathway).
QUESTION BANK /BIOLOGY/XII
Page 177
28)At high altitudes places like Manali or Mansarover we suffer from altitudes sickness
this is because in low atmospheric pressure, body does not get enough oxygen, but
gradually the problem is over. How did our body solve this problem?
Ans.By increasing RBC count, By decreasing binding capacity of hemoglobin, By increasing
breathing rate.
29)What is Gause Competitive Exclusion Principle?
Ans.Gauss Competitive Exclusion Principle -Two closely related species competing for
thesame resource cannot co-exist indefinitely and the competitively inferior one will be
eliminated eventually.
SHORT ANSWER TYPE QUESTIONS (3marks)
30) Mr. Ram on a trip to Rohtang Pass Suddenly experienced heart Palpitations,
Nausea, fatigue etc on reaching the destination. Suggest the reasons for his
sudden deterioration of health and also state Whether his body will withstand
this problem if he stays there for long and how?
Ans.Atmospheric pressure in Rohtang pass, which is at high altitude, is low and
hence the body does not get enough oxygen. Ram is suffering from altitude
sickness.
If he stays for long the following change will occur in the body and he will
Becomeacclimatised to the conditions.
a)RBC production increases
b) Breathing rate increases
c) Binding capacity of hemoglobin decreases
31) Anand on a visit through an under the ocean aquarium found that many sea
anemones are attached to hermit crab shells, sucker fisher attached to the ventral
surface of sharks and clown fish living among the sea anemones. He wondered
whether all these associations are of the same type; can you help him to arrive at
the correct conclusion.
Ans.a) Relation between sea anemones and hermit crab is mutualism since seaanemones
protects the hermit crab and crab provides bits of food to seaanemone, thus both benefitted.
b) Relation between shark and suckerfish is commensalisms because onlysucker fish gets
food and is benefitted while shark is neither harmed norbenefitted.
c) Relation between sea anemone and clown fish is also commensalism sinceonly the fish
gets protection from predators.
32)What is brood parasitism? Give an example. What adaptation has evolved in this
phenomenon?
QUESTION BANK /BIOLOGY/XII
Page 178
Ans.One species lays eggs in the nest of another bird, lets the host incubate them. e.g.
Cuckoo lays eggs in the nest of a crow. The Eggs of the parasite resemble the eggs of the
host in colour, size. Reduce chances of the host bird detecting the foreign eggs and ejecting
them from nest.
33)How does Ophrys get pollinated by bees?
Ans.
1. Sexual deceit.
2. One petal resembles female.
3. Male pseudocoupulates with the flower.
4. Pollen grain transferred from one flower to another.
34) How does the shape of age pyramid reflect the growth status of apopulation?
Ans. Shape of pyramids reflects growth statusof the population (a) growing (b) Stable (c)
declining.Refer page 227, Fig. 13.4, NCERT book, Biology - XII
35) Darwin showed that even a slow growing animal like elephant could reachenormous
number in absence of checks. With the help of yourunderstanding of growth models,
explain when is this possible? Why is thisnotion unrealistic?
Ans. Possible if the growth model is Exponential, i.e., having unlimited resources.Its an
unrealistic situation because resources are limited. Hence, it follows logistic growth model.
36).How will you measure population density in following cases?
(i) fish in a lake
(ii) tiger census in a national park
(iii) single huge banyan tree with large canopy.
(a) fish caught per trap.
(b) number per unit area
(c) percentage cover in biomass
(a) fish caught per trap.
(b) number per unit area
(c) percentage cover in biomass.
37) Species facing competition might evolve mechanism that promotes coexistence
rather than exclusion. Justify this statement in light of Gause.s
competitive exclusion principle, citing suitable examples
Ans.Gauss Competitive Exclusion Principle -Two closely related species competing for
thesame resource cannot co-exist indefinitely and the competitively inferior one will be
eliminated eventually.Mechanisms is resourcepartitioning. E.g., experiment of Mac Arthur
on Warblers (Refer page 325,NCERT book, Biology - XII).
38)How organisms respond to abiotic factors?
QUESTION BANK /BIOLOGY/XII
Page 179
Ans.Regulation
# Organisms maintain homeostasis achieved by physiological and behavioral means.
# Thermo regulation and osmo regulation.
Conformation
#cannot maintain constant internal environment.
# Body temperature and osmotic concentration of body changes with ambient temperature
and concentration of medium.
Migration
# Organism moves away temporarily to another habitat in stressful condition. e.g.- Migratory
birds.
Suspension
# Organisms suspend their metabolic activities during stressful condition
# Resume their function at the return of favorable conditions. E.g. Hibernation of Frog,
Reptiles, Polar Bear etc
# Aestivation in Snail and Fish.
# Seed dormancy.
QUESTION BANK /BIOLOGY/XII
Page 180
Chapter 14.ECOSYSTEM
VERY SHORT ANSWER TYPE QUESTIONS (1marks)
1)All the primary productivity is not available to the herbivore,give one reason.
Ans. A considerable part is utilized by the plants in respiration while some are lost as heat
into the environment.
2) Detritus contribute to the biogeochemical cycles, how?
Ans. By the decomposition of detritus, the simple minerals are released into theatmosphere &
from there it come back to the earth.
3) Can temperature regulate the rate of decomposition how?
Ans. High temperature favours decomposition and low temperature inhibitsdecomposition.
4) The detritus food chain and grazing food chain differ. How?
Ans.Detritus food chain begins from the dead and decaying matter while grazingfood chain
starts from the green plants(Producers).
5) As succession proceeds the numbers and types of animals and decomposers also
change. How?
Ans. Vegetation changes in turn change the food and shelter for various types ofanimals. As
a result the foresaid changes happen.
6) In burnt out Forests and flooded lands succession takes place faster. Why?
Ans. In burnt out forests and flooded lands some soil or sediment is present. Thereis no need
for soil to be formed.
7) Sedimentary cycle is quite different from a gaseous cycle with respect to itsreservoir.
Bring out the difference.
Ans. The reservoir of gaseous cycle exists in the atmosphere and for thesedimentary cycle it
is located in earth's crust.
8) Write the equation that helps in deriving the net primary productivity of an
Ecosystem
Ans.The equation of net primary productivity of an ecosystem is
NPP= GPP-RL
QUESTION BANK /BIOLOGY/XII
Page 181
Where, NPP- net primary productivity
GPP- gross primary productivity
RL- respiratory losses
9)Decomposition is faster if deteritus is rich in nitrogen and water soluble
substance like sugars. When is the decomposition process slower?
Ans.Its slower if detritus is rich in lignin and chitin.
10)If we count the number of insects on a tree and number of small birdsdepending on
those insects as also the number of larger birdseating thesmaller, what kind of pyramid
of number would we get?
Ans. Inverted Pyramid of Number.
11)Which metabolic process causes a reduction in the Gross PrimaryProductivity?
Ans.Respiration.
SHORT ANSWER TYPE QUESTIONS (2marks)
12)"Flow of energy is unidirectional but nutrient flow is in a cycle" Give reason.
Ans.Energy flow is always from the sun to 'producers' and to the different trophiclevels. so it
is unidirectional. But the nutrients are moving from the living tonon-living and vice-versa.
13) "Decomposition is an oxygen requiring process" comment.
Ans. Detritus is rich in nitrogen and sugars. For oxidation of nitrogen and sugarsoxygen is
required by a class of aerobic microbes.
14) Give an example of an ecological pyramid which is always upright. Justify your
answer.
Ans.Pyramid of energy is always upright and can never be inverted, becausewhen energy
flows from a trophic level to the next trophic level someenergy is always lost as heat at each
step.
15)What is the effect on decomposition rate if :
a) Detritus is rich in lignin and chitin
b) Detritus is rich is nitrogen and sugars
Ans. a) Decomposition rate is slower.b) Decomposition rate is faster.
16) What are the limitations of ecological pyramids?
QUESTION BANK /BIOLOGY/XII
Page 182
Ans.i) Does not take into account same species belonging to two or more trophic levels.
(ii) Assumes simple food chain, does not accommodate food web.
(iii) Saprophytes have not been given any place in ecological pyramids.
17) Given below is the primary hydrarch succession. Bring out the missing sere stages
in the process.
Ans.A-Submerged plant stage, B-Reed swamp stage ,C-scrub stage, D-Forest
18) Given below is a simplified model of phosphorus cycle. Write down thenatural
reservoir of phosphorus and also the processes that put in phosphorus tothe soil.
Ans.
Ans.A. Rock minerals B. Weathering C. Decomposition
QUESTION BANK /BIOLOGY/XII
Page 183
19) Give an example of an ecological pyramid which is always upright. Justifyyour
answer.
Ans. Pyramid of energy is always upright and can never be inverted, becausewhen energy
flows from a trophic level to the next trophic level some energy is always lost as heat at each
step.
20) Differentiate between primary succession and secondary succession.Which one
occurs faster?
Ans. Primary Succession : A process that starts where no living organisms arethere.
Secondary succession : A process that starts in areas which have lostall the living organisms
that existed there.
21) Gaseous nutrient cycle and sedimentary nutrient cycles have theirreservoir. Name
them. Why is a reservoir necessary?
Ans. Reservoir for Gaseous nutrient cycle: Atmosphere; for sedimentary nutrient cycle:
Earths crust. Reservoir is needed to meet with the deficitwhich occurs due to imbalance in
the rate of influx and efflux.
22) Why is the length of a food chain in an ecosystem generally limited to 3-4 trophic
levels?
Ans.As 90% energy is lost in the form of heat from one trophic level to another, residual
energy decreases drastically within 2-3 trophic levels.
23)What are the differences between detritus and grazing food chains?
Ans.a) Begins with Detritus-dead and decaying organic matter.
b) Grazing-Begins with Living green plants.
24) Mention two factors by which productivity is limited in an aquatic ecosystem.
Ans a) Light-decreases with increasing water depth.
b) Nutrient –Limiting factor in Deep Ocean
25) Why Phosphorus cycle is called an imperfect cycle, while Carbon and Nitrogen
cycle are perfect ?
Ans.In phosphorus cycle, the bulk of material remain in the relatively inactive and immobile
reservoir , while in carbon cycle , material remains is in circulation.
QUESTION BANK /BIOLOGY/XII
Page 184
SHORT ANSWER TYPE QUESTIONS (3marks)
26) Both carbon and phosphorus cycles are biogeochemical cycles but they differ in
threeaspects. List them.
Ans.
1) CARBON CYCLE –
1. Reservoir exists in atmosphere.
2. Considerable inputs of carbon through rain fall.
2) PHOSPHORUS CYCLE
1.Reservoir exists in earth's crust
2.Negligible inputs of phosphorusthrough rainfall.
27) Ecosystems should carry a hefty price tag for its various services. Enlist six of
them.
Ans.1. Purify air and water
2. Mitigate droughts and floods
3. cycle nutrients
4. Generate fertile soils
5. Provide wide life habitat
6. Pollinate flowers
7. Provide aesthetic, cultural and spiritual values
28) Detrivores like earthworm are involved in the process of decomposition of dead
plants and animals. Describe the different steps involved in the process of
decomposition.
Ans.The dead remains of plants and animals called detritus undergo decomposition and are
converted into simpler substances. The steps of this process are :
(i) Fragmentation:Breakdown of detritus into smaller pieces by detrivoures like earthworm.
(ii) Leaching:Water soluble inorganic nutrients go down into soil horizon and get precipitated
as unavailable salts.
(iii) Catabolism:Bacterial and fungal enzymes degrade detritus into simpler inorganic
substances.
(iv) Humification :It leads to accumulation of dark coloured amorphous substance called
humus which is highly resistant to microbial action so decomposes at slow rate and is rich in
nutrients.
(v) Mineralisation :Humus is further degraded by some microbes andrelease of inorganic
nutrients occurs.
29) Give account of factors affecting the rate of decomposition.
Ans.a) Climatic factor – i) temp ii) soil
b) Chemical quality of detritus Higher temp and moist condition – high rate of decomposition
Dry soil, High temp – Low rate
QUESTION BANK /BIOLOGY/XII
Page 185
Chapter -15. BIODIVERSITY
VERY SHORT ANSWER TYPE QUESTIONS (1marks)
1) a) India has more than 50000 different strains of rice and 1000 varieties ofmangoes.
b) Western Ghats have a greater Amphibian diversity than the Eastern Ghats.What do
you infer from the above two statements?
Ans. a) Genetic diversity b) Species diversity
2)What trend is observed in respect of species diversity when we move fromequator to
poles?
Ans. In general, species diversity decreases as we move away from the equator owards poles
3)Why is tropical environment able to support greater species diversity?
Ans. Tropical latitudes have remained undisturbed for millions of years and had a long
evolutionary history for species diversification.Thus it supports greater diversity.
4)Name the unlabelled areas 'a' and 'b' of the pie chart representing the global
biodiversity of invertebrates showing their proportionate number of species of major
taxa.
Ans.a-insectsb-molluscs.
5)Mention one application of pollen bank. Howare pollen stored in the bank?
Pollen banks are used to store pollens in liquid nitrogen(-196 C) for a very long period of
time in viable conditions and these pollens can be used in various crop breeding program me.
Cryopreservation
SHORT ANSWER TYPE QUESTIONS (2marks)
6) Identify a,b,c and d in the above pie diagram showing global biodiversity ofplants.
QUESTION BANK /BIOLOGY/XII
Page 186
Ans.a) Angiosperms b) Algae c) Fungi d) Mosses
7) Columbia located near the equator has nearly 1400 species of birds while new york at
41˚ N has 105 species and Greenland at 71˚ N has only 56 species. Comment.
Ans. Latitudinal gradients in biodiversity.
Biodiversity decreases as latitude increases .
Biodiversity decreases as one move from tropics to equator.
8) Would the extinction of one insect pollinator affect the ecosystem? Explain.
Ans.Yes,It may lead to co-extinction of species.In the case of a co-evolved plant-Pollinator
Mutualism where extinction of one invariably leads to the extinction of the other.
9) The species diversity of plants is much less than that of animals. What could be the
reason for this and figure out how animals achieved greater diversifications?
Ans.Animals have greater mobility .Animals migrate to different environmental conditions
and undergo adaptation.
10) Seeds of different genetic strains are kept for long periods in seed banks. Explain
the conservative strategy involved in this process.
Ans.Ex-situ conservation .They are preserved in viable and fertile condition for long periods
using cryopreservation techniques.
11) Amazon forests are regarded are "Lungs of the Planet". Why?
Ans.Amazon forests have the greatest biodiversity in the world and harbour a large number
of plant species which release large amount of oxygen into the atmosphere.
12) National parks come under 'in situ' conservation while Zoological parks
under'exsitu' conservation.Comment.
Ans.National parks- in situ- It is the practice of protecting the endangered species
in their natural habitats either by protecting or by defending the species from
predators.
QUESTION BANK /BIOLOGY/XII
Page 187
Zoological parks- ex situ- the threatened species are taken out from their natural
habitat and placed in special setting and given protection and special care.
13) There are 34 biodiversity hot spots in the world. What are the criteria of
selection of such hot spots?
Ans. To qualify as a biodiversity hotspot, a region must meet two strict criteria:
a)Contain at least 1,500 species of vascular plants as endemics (species found nowhere else
on Earth).
b)Have lost at least 70 percent of its original habitat.
14) Lantana &Eichornia are examples of two weeds. How do they affect the ecosystem?
Ans.Theseare examples of Alien species invasions. They threaten the indigenous species &
lead to their extinction.
15)Name the sociobioligist who popularized the term biodiversity. identify the levels of
biodiversity in India represented bya)Diversity amongst amphibians in eastern and western ghats.
b)50,000 strains of rice in India.
c)Presence of deserts,mangroves and coral reefs of India.
Ans.Edward Wilson.
Species diversity
Genetic diversity
Ecological diversity.
16)White Bengal tigers are protected in zoological parks. Tiger reserves are maintained
in western ghats.How do these two approaches differ from each other?mention
theadvantages of each one.
Ans.White Bengal tigers are protected in zoological parks this is called ex situ conservation
while tiger reserves are maintained in western ghats is called in situ conservation.
In situ Conservation:
1. It is conservation of endangered species in their natural habitats.
2. The endangered species are protected from predators.
3. The depleting resources are augmented.
4. The population recovers in natural environment.
Ex situ Conservation:
1. It is conservation of endangered species outside their natural habitats.
QUESTION BANK /BIOLOGY/XII
Page 188
2. The endangered species are protected from all adverse factors.
3. They are kept under human supervision and provided all the essentials.
4. Offspring produced in captive breeding are released in natural habitat for acclimatization.
17)Biodiversity decreases as one move from equator towards pole. Justify this statement
with a suitable example.
Ans.Equator has tropical environment and hence harbour more species e.g. Columbia near
the equator has 1400 species of birds , while New York at 410 N has 105 Species.
18)What is the scientific term for measurement of species diversity if
(i) number of species per unit area are measured.
(ii) relative abundance with which each species is represented in an area is measured.
Ans.(i) Species richness
(ii) Species evenness.
19)The tropics (between 23.5 0 N to 23.50 S) harbours more species than temperate and
polar regions. Explain the probable reasons for difference in biodiversity between
tropical and temperate regions.
Ans.Tropical have remained undisturbed for millions of years, temperate are subjected to
frequent graciations , tropical environment in less seasonal , more constant
&Predictable.suitable areas for species diversity.
SHORT ANSWER TYPE QUESTIONS (3marks)
20)Explain rivet popper hypothesis.name the ecologist who proposed it.
Ans.Hypothesis proposed by Paulehrlich.airplane/ecosystem joined by thousand
rivets/species.every passenger travelling in it starts to take a rivet home causing species to
become extinct.it may not affect the planes safety initially but once more and more rivets are
removed the plane becomes dangerously weak.critical loss of a rivet may seriously damage
the flight like removing a key species.
21)Hot spots are the regions of exceptionally high biodiversity. But they have become
regions of accidental habitat loss too. Name the three hot spots of our country. Why are
they called hot spot?
Ans.WestermGhats ,Indo-Burma; Himalaya arbiodiversity hot spots as they show
(i) High level of species richness
QUESTION BANK /BIOLOGY/XII
Page 189
(ii) High degree of endemism
22)What is so special about tropics that might account for their greater biological
diversity?
Ans a) Speciation is a function of time, unlike temperate regions subjectedo frequent
glaciations in the past, tropical latitude have remainedrelatively undisturbed for million of
years and thus had longevolutionary time for species diversification.
b) Tropical environment are less seasonal, more constant andpredictable.
c) More solar energy awailable in the tropics contributing to highproductivity leading to
greater diversity.
23) Why is the sobriquet .The Evil Quartet. used in context of biodiversity?Name the
members of this quartet. Why do we grieve for the genes whena species is lost
Ans.The .Evil Quartet.is used as a sobriquet to refer to the cause of loss of
biodiversity :
(i) Habitat loss and fragmentation:When large habitats are brokenup into smaller
fragments due to various human activities, the animalsrequiring large territories (elephants,
birds etc.) are badly affected andtheir populations decline.
(ii) Over-exploitation:When need of a resource becomes greed. e.g.,over exploitation of
passenger pigeon led to its extinction. Also marinefish is at brink of being endangered due to
over exploitations.
(iii) Alien species invasion:Intentional or non-Intentional introductionof a species to a
nearby area may disturb the harmony of existingspecies. e.g., Eichhornia after introduction
posed a big threat to thenative species.
(iv) Co-extinction:Extinction of one species invariably leads to extinctionof another when
they are associated with each other in an obligatoryway. e.g., when host species is extinct,
obligate parasites dependenton it also die.
(v) We grieve for the loss of genes, because the wild forms are hardy and more resistant to
pathogen attack and can be beneficial in cropbreeding programmes.
24) Biologists are not sure about how many prokaryotic species there might be.Give
reasons.
Ans.i)The conventional taxonomic methods are not sufficient for identifying these
microbial species
ii)Many of the species cannot be cultured under laboratory conditions.
iii) Biochemical and molecular biology techniques would put their diversityinto millions.
25) In an experiment, the slope of regression (z) is 0.2 and in another experiment
the value obtained is 1.2. Explain the two situations in respect of species
arearelationships.
Ans. 0.2 is obtained in studies regardless of the taxonomic group or the region
QUESTION BANK /BIOLOGY/XII
Page 190
1.2 is obtained if species area relationship is analysed among very large areaslike the entire
continents.
26) Would Western Ghats ecosystems be less functional if one of its tree frogspecies is
lost forever? Substantiate your answer in the light of hypothesisproposed by Paul
Ehrlich.
Ans.According to the hypothesis proposed by Paul Ehrlich the "rivet popperhypothesis" each
species is essential in the balance of nature. If one is lost thatmuch imbalance is caused in the
ecosystem.
27)How do sanctuaries differs from national parks in terms of conservation?
Wildlife Sanctuary
A wildlife sanctuary is a declared protected area, where very limited human activity is
allowed. The ownership of this type of protected area could lie in the hands of either a
government or in any private organization or personthe hunting of animals is completely
prohibited. The trees can not be cut down for any purpose; especially the clearing of the
forest for agriculture is completely banned. The general public could use it up to a certain
extent so that the sanctuary is useful for them also. People can collect firewood, fruits,
medicinal plants…etc in small scale from a wildlife sanctuary.
National Park
A national park has a defined boundary, through which no person can get into the park
without an approval. Only an approved person can enter into a national park, either via
paying a visitor ticket or an approved letter from the governing body (mostly the
government). The visitors can only observe the park inside a vehicle that routes through
defined trails and they can not get out the vehicle for any reason unless there is an approved
place for visitors. Photographs are allowed but research and educational work can only be
done with a prior permission. The park can not be used for any reason viz. firewood, timber,
fruits…etc.
28) How to conserve biodiversity?
Ans.In-situ Conservation– Threatened /endangered plants and animals are provided
with urgent measures to save from extinction within their natural habitat( in wildlife
sanctuaries, national parks & biosphere reserves, sacred groves /lakes-i.e. in protected
areas).
Biodiversity hotspots – regions with very high levels of species richness and endemism.
Norman Myers developed the concept of hotspots in 1998 to designate priority areas
for insitu conservation. They are the most threatened reservoir of biodiversity on earth.
In India 2 hotspots are there,e.g.Western ghats, and the Eastern Himalayas.
Ex-situ Conservation –Threatened animals & plants aretaken out from their natural
habitat & placed in a setting where they can be protected and given care as in botanical
gardens, zoological gardens, seed/pollen/gene banks.
QUESTION BANK /BIOLOGY/XII
Page 191
Chapter16. ENVIRONMENTAL ISSUES
VERY SHORT ANSWER TYPE QUESTIONS ( 1marks)
1)Why should the velocity of air between the plates of an electrostaticprecipitator be
low?
Ans.To allow the dust to fall.
2)Mention the information that the health workers derive by measuring BOD of awater
body.
Ans.The information that the health workers derive by measuring BOD of a water body is
more BOD more is its polluting potential.
3) The birds egg shells become thinner in an area where there is an excess application of
pesticides. Comment.
Ans. High concentration of DDT disturb calcium metabolism in birds whichcauses thinning
of egg shell.
4) Motor vehicles with catalytic converter should use unleaded petrol. Why?
Ans. Lead inactivates catalyst.
5)Why is a scrubber used? Which spray is used on exhaust gases passing through a
scrubber?
Ans.To remove gases like sulphur dioxide. Spray of water or lime is used
6)The central pollution control board observed that a thermal power plant isemitting
80% of its particulate matter in the exhaust due to poor maintenance?What could be
the cause for this increase in % of particular matter?
Ans. Poor maintenance of the electrostatic Precipitator
SHORT ANSWER TYPE QUESTIONS (2marks)
7)Jhum cultivation has been in practice from earlier days, but its consideredmore
problematic these days. Why?
Ans. Enough time gap is not being given for the natural process of recoveryof land from the
effect of cultivation.
8) Electrostatic precipitator can remove over 99% particulate matter presentin exhaust
from a thermal power plant. How?
QUESTION BANK /BIOLOGY/XII
Page 192
Ans.Electrode wire at thousand volts, produce corona to release electrons,electrons attach to
dust particles giving them net negative charge,charged dust particles attracted/collected by
collecting plates which aregrounded.
9)Vehicles are fitted with catalytic converters. Give reason.
Ans. Catalytic converter has platinum palladium and rhodium as catalyst to reduce
emission of poisonous gases. As the exhaust passes through the converterunburnt
hydrocarbon are converted into CO2 and water and carbon monoxide;oxide are changed in to
CO2 and nitrogen gas.
10) It was observed that some of the aquatic birds’ population has been declining.On
analysis of the water in their habitat, high concentration of DDT was found.
a) What caused the decline is birds’ population?
b) What is this phenomenon known as?
Ans. a) DDT in the body interferes with the calcium metabolism, resulting inthinning of egg
shell & their premature breaking ultimately leading to a declinein bird population
b) Biomagnifications.
11) A man had a B. class Benz which has a very good catalytic converter. On hisway the
petrol was exhausted & he stopped at a petrol pump. The person at thepetrol pump had
only leaded petrol. Out of necessity the man had to fill leadedpetrol in his car.
a) What is its impact on the catalytic activity of the converter?
b) Explain the impact on the environment.
Ans. a) Lead in the petrol inactivates the catalyst .
b)Unburnt hydrocarbons are released which increases the air pollution.
12) A person has inhaled air having particulate pollutants of size less than
2.5micrometers in diameter. What could be the impact of their pollutants on
hisrespiratory system?
Ans. Particulate pollutants inhaled deep into the lungs cause irritation,inflammation, damage
to the lungs& premature death.
13) Mention the green house gases & their proportion in the pie diagram givenbelow.
QUESTION BANK /BIOLOGY/XII
Page 193
Ans. a = CO2 = 60%.b = methane = 20%.c = CFC'S = 14. D= N2O = 06%.
14) A farmer saw water in a pond turned green & with dead fishes.
a) What is the reason for the death of fishes?
b) Name the phenomenonthat leads tothe death of fishes.
Ans.a) Algal bloom resulting in depletion of oxygen.b) Eutrophication
15) CFCs are widely used as refrigerants. Then why is it suggested to reduce its
emission a far as possible? Explain.
Ans. CFCs are discharged in the lower part of the atmosphere move upward and reach the
stratosphere.In the stratosphere UV rays act on them releasing Cl atoms. Cl acts as aa catalyst
and degrades ozone releasing molecular oxygen.
16)
In the above graph what does A & B depict?
Ans.A-BOD .B-Dissolved oxygen
17) Identify the wrong pair/s and correct it.
a) Chernobyl incident-Radioactive waste.
b) Snow blindness cataract- High dose of CFC
c) Chipko movement- Save trees
d) Polyblend-Solves air pollution.
Ans.
b)-snow blindness cataract-High dose of UV-B
d)-Polyblend-solve plastic pollution.
18) Dumping waste in landfills is not a real solution for disposing wastes. Why?
QUESTION BANK /BIOLOGY/XII
Page 194
Ans a) The amount of garbage generation specially in the metros has increased somuch that
these sites are getting filled too.
a) Also there is a damage of seepage of chemicals etc from these land fillspolluting the
underground water resources
19)India in last two decades has seen a boom in automobile production and use.
(a) What has been its environmental impact?
(b) What steps are being taken so that the rapid increase in the number of automobiles
does not become a curse?
Ans.
(a) Enhance air pollution
(b) (i) Proper maintenance of automobiles along with use of lead-free petrol or diesel .
(ii) Use of catalytic converter for reducing emissionof poisonous gases .
(iii) Use of CNG in place of petrol, diesel
20)What are the advantages of CNG?
Ans.Advantages of CNG:(a) CNG burns most efficiently.
(b) Cheaper Cannot be siphoned.
(c) Cannot be adulterated.
21)Why are grains and vegetables produced in organic farms supposed to be better for
human consumption than those produced in farms where chemical fertilizers, pesticides
etc. has been used?
Ans.Chemical fertilizers / pesticides --non -biodegradable , enter into food chain through
vegetables / grains , causes biomagnification in humans . In organic farms manure is used,
whose produce causes no harm to human health.
SHORT ANSWER TYPE QUESTIONS ( 3marks)
22) Many coastal areas may get submerged due to the environmental changestaking
place at the present rate. Give the cause, and state two measures to checkit.
Ans.
1) Global warming due to the increase in conc. of green house gases
It can be checked by.
i) Growing more trees (afforestation)
ii) Reduce the use of fossil Fuel
iii) Prevent deforestation
23)People have been actively participating in the efforts for the conservation of forests.
(i) Name the award instituted in respect of Amrita Devi to promote suchefforts.
(ii) Name the movement launched to protect the trees by hugging them.
(iii) Name the step Government of India has undertaken in 1980.s to work
QUESTION BANK /BIOLOGY/XII
Page 195
closely with the local communities for protecting and managing forests.
Ans.
(i) Amrita Devi Bishnoi Wildlife Protection Award.
(ii) Chipko movement
(iii) Joint Forest Management (JFM).
24) Integrated organic forming is a cyclical zero waste procedure. Justify.
Ans. Integrated organic farming is a cyclical zero waste procedurea) Waste products from one process are cycled in as nutrients for theother process.
b) There is no need for chemical fertilizers as cattle excreta are used asmanure.
c) Crop waste is used to create compost, which is used as manure to
generate electricity
25) Ecosan toilets are a hygienic, efficient and cost effective solution to thedisposal of
human wastes. Justify.
Ans. Human excreta are also recycled into manure which reduces the use ofchemical
fertilizers.
26)Explain bio magnification of DDT in an aquatic food chain. How does it affect
the bird population?
Ans.DDT that stands for DichloroDiphenylTricholoro Ethane is the most commonly known
pesticide. DDT is helpful in killing pests like mosquitoes, fleas, house flies and bugs.
On a long run, DDT has proven to be harmful to both man and environment.
*DDT sprayed in farms get settled in the soil. It percolated into the deep layers of the soil
with water used in irrigating the field. Slowly, it reaches ground water.
*When the agriculture waste is dumped into water bodies, DDT enters water resources.
*It is a non-biodegradable substance that gets accumulated in the food chains. DDT being
toxic and interrupting the food chain by accumulation causes biomagnification.
* It is absorbed by aquatic organisms like small planktons.
*When these planktons are consumd by fish, DDT gets aaccumulated in the body of fishes
that feed on them.
*Fishes in turn pass on DDT to the birds that prey on it. In birds, DDT prevents formation of
egg shell, leading to breaking of eggs even before they can hatch. Ultimately, the organisms
which occupy the highest trophic level suffer the most.
Planktons ----->Small fish -----------> Large fish----------->Bird
(Producer) (Primary consumer) (Secondary consumer) (Tertiary consumer)
27)During the past century, lakes and ponds in many parts of the earth have been
gradually converted into land.
(a) What is the phenomenon called?
(b) How does human activity accelerate this phenomenon?
(c) What measures can we take to control this?
QUESTION BANK /BIOLOGY/XII
Page 196
Ans.
a) Accelerated eutrophication
(b) Sewage from homes, agricultural waste and Industrial wastes containing nutrients
like nitrogen , phosphorus etc.
(c) Domestic sewage and agricultural wastes should be treated before releasing into
river / avoid to release in river.
LONG ANSWER TYPE QUESTIONS (5 marks)
28)
a)Why are catalytic convertors recommended for vehicles?
b)Why should such vehicles use only unleaded petrol.
c)Why is CNG preferred to diesel as a fuel in vehicles.
Ans.a)A catalytic converter is a device that uses a catalyst to convert three harmful
compounds in car exhaust into harmless compounds.The three harmful compounds
are:Hydrocarbons carbon monoxide and nitrogen oxides.
In a catalytic converter, the catalyst (in the form of platinum and palladium) is coated onto a
ceramic beads that are housed in a muffler-like package attached to the exhaust pipe. The
catalyst helps to convert carbon monoxide into carbon dioxide. It converts the hydrocarbons
into carbon dioxide and water. It also converts the nitrogen oxides back into nitrogen and
oxygen.
b)Unleaded petrol inactivates the catalyst.
c)Reasons for CNG being better fuel
 CNG does not contain any lead, thereby eliminating fouling of spark plugs
 CNG fuel systems are sealed, preventing fuel losses from spills or evaporation.
 Increased life of lubricating oils, as CNG does not contaminate and dilute the
crankcase oil.
 Being a gaseous fuel, CNG mixes easily and evenly in air.
 Cheaper and cant be siphoned off by thieves and adulterated like petrol.
29) Observe the following figure carefully and answer the questions that follow
a) Label the parts A, B, C & D
b) What are the steps involved in the removal of particulate matter?
QUESTION BANK /BIOLOGY/XII
Page 197
c) How is it different from scrubber?
Ans.
A)a- collection plate b- Dirty airc-discharge corona d-Clean air .
B) i) The electrode wires are maintained in several thousand volts, which
producescorona that releases electron.
ii) Those electrons attach a dust giving negative charge.
iii) The collecting plates attract the charged dust particles
C) The electrostatic precipitator removes the particulate matter where as the
scrubber removes the gases like SO2.
30) Pollutant released due to human activities (like effluents from industriesand homes)
can radically accelerate the ageing process of the water body.
(a) Explain how does this process occurs during natural ageing of lake.
(b) Give the term used for accelerated ageing of water bodies. Also givethe term used
for the natural ageing of lake.
Ans (a) The phenomeon is eutrophication. More nutrients in water, aquatic lifeincreases
organic remains deposited on lake bottom, lake growsshallower and warmer, gradually
transforms into land due to depositionof silt and organic debris.
(b) Cultural or Accelerated eutrophication.Natural ageing is Eutrophication.
QUESTION BANK /BIOLOGY/XII
Page 198
SAMPLE QUESTION PAPER 1
CLASS - XII
SUBJECT – BIOLOGY
TIME: 3 Hrs
M.M.:70
General Instructions:
i
All questions are compulsory.
ii
The question paper consists of five sections A, B, C, D and E. Section-A contains
5questions of 1 mark each, Section -B is of 5 questions of 2 marks each, Section –
C has12 questions of 3 marks each, Section- D is of 1 question of 4marks
whereas Section E hasthree questions of five marks each.
iii
There is no overall choice. However an internal choice has been provided in one
question of 2 marks, one question of three marks and all the questions of 5
weightage. A student has to attempt only one of the alternatives in such
questions.
iv
Wherever necessary, the diagrams drawn should be neat and properly labelled.
SECTION – A
1.How many pairs of chromosomes does a male Drosophila fly have? Which one of
these bears the gene for eye colour ?
2. An exotic prickly pear cactus when brought into Australia in 1920s, became a
menace later. Mention how was this invasive cactus brought under control.
3. Which part of the blastula is destined to form the germ layers of the developing
embryo in humans?
4. Name the placental mammals corresponding to the Australian Spotted Cuscus
and Tasmanian tiger cat which have evolved as a result of convergent evolution.
5.How is the net primary productivity of an ecosystem derived?
SECTION – B
6.Name the region on the earth called the “lungs of the planet”. Mention giving reasons,
the activities which are being carried out in this region now.
7.Expand hn RNA.Name the enzyme that transcribes it.It is a precursor. What does it
form later?
8.Name the types of immunity provided by vaccines and colostrum in humans. Mention
one difference between them.
OR
8.What is contact inhibition in the normal cells of the body? What are the consequences
when this property is lost by these cells?
9.Write the specific palindromic nucleotide sequence in DNA that is
recognized by EcoRI.What does EcoRI stand for?
QUESTION BANK /BIOLOGY/XII
Page 199
10.Name the hormones produced only during pregnancy in a human female. Mention
their source organs.
SECTION – C
11.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?
12.What is CNG ?Why is it preferred to diesel or petrol?How does a catalytic converter
fitted in an automobile, help in reducing emission of poisonous gases. Explain.
13.Give the sequence of events followed in “Multiple Ovulation Embryo Transfer
Technology Programme” for increasing the herd size in cattle.
14.Why is it difficult for DNA to pass through cell membranes? How is a bacterial cell
made ‘competent’ to take up DNA ( plasmid) ?Explain any other method by which an
alien DNA can gain entry into a cell to form a recombinant DNA.
15.The events of the menstrual cycle are represented below. Answer the questions
following the diagram.
State the levels of FSH, LH and Progesterone simply by mentioning high or low, around
13th and 14th day and 21st to 23 rd day.In which of the above mentioned phases does
the egg travel to the fallopian tube?Why is there no menstruation upon fertilisation?
16.The figure given below shows relative contributions of various green-house gases to
the total global warming.
i)Name the gases (a) and (b)
QUESTION BANK /BIOLOGY/XII
Page 200
ii) Explain how increase in green-house gases in earth’s atmosphere leads to melting of
ice caps.
iii)Why is the ozone hole a threat to mankind?
OR
i)In which ecosystem is the pyramid of biomass inverted?
ii)Why is it inverted? Explain.
iii)Name the type of pyramid that is always upright. Give reasons.
17.iWhat are the symptoms of haemophilia?
ii)A haemophilic son is born to a normal couple. Explain the mechanism of this
inheritance.What is the probability of a haemophilic daughter being born to this
couple?
18.i)Name the mode of reproduction by which Plasmodium multiplies in the human
body and where it does so.
ii)Shivering and high fever are the common symptoms of malaria in
humans.Explain the cause of these symptoms.
19.See the figure given below.
AB and CD represent two strands of a DNA molecule.When this molecule undergoes
replication, forming a replication fork between A and C
i Name the template strands for replication.
iiUsing which strand as the template, will there be continuous synthesis of a
complementary DNA strand?
iii Complementary to which strand will Okazaki segments get synthesized /
discontinuous synthesis will occur.
ivWhat are template strands and Okazaki fragments?
v) In which direction is a new strand synthesised?
20.iMature seeds of legumes are non albuminous. Then can it be assumed
that double fertilization does not occur in legumes? Explain your answer.
iiList the differences between embryos of pea (dicot) and grass (monocot).
21.You have been deputed by your school principal to train local villagers in the use
of a biogas plant. With the help of a labelled diagram explain the various
parts of the biogas plant.
22.What are ‘flocs?’ State their role in effluent treatment and their ultimate fate in
sewage treatment tank.
SECTION – D
23.The human population is increasing. They require more space to live and dispose the
garbage. Man’s activities have converted villages and other areas into concrete jungles
leading to loss of biodiversity.
iWhat are the effects of biodiversity loss? Give three points.
QUESTION BANK /BIOLOGY/XII
Page 201
iiWhat does the sobriquet 'The Evil Quartet' denote?'
SECTION – E
24.
i) Draw the embryo sac of a flowering plant and label (a) central cell (b) Chalazal end of
the embryo sac (c) synergids.
ii) Name the cell that develops into the embryo sac and explain how this cell leads to the
formation of embryo sac . Also mention the role played by the various cells of the
embryo sac.
OR
i)Draw a labeled diagram of the microscopic structure of a human sperm.
ii)Trace the development of spermatozoa from the primary spermatocytes human
testes.
25.How is Bt cotton plant produced? Explain the mechanism by which the plant
is able to resist the infestation by cotton bollworms.
OR
i)Name the source of Taq polymerase. Explain the advantage of its use in
biotechnology.
ii)Expand the name of the enzyme ADA. Why is this enzyme essential in the human
body? Suggest a gene therapy for its deficiency.
26.Study the following carefully and explain why mutation (A) did not cause any sickle
cell anemia in spite of a change in the molecular structure of the gene which codes for
Haemoglobin, whereas a similar mutation (B) caused sickle cell anaemia.
(The question is based on the properties of the genetic code. C = codon, a = amino acid,
Hb = Haemoglobin)
Codons for Hb : C1-C2-C3-C4-C5-GAA-GAA-C8 .....................
Amino acids in Hb : a1-a2-a3-a4-a5-Glutamic acid –Glutamic acida8.............(Normal Haemoglobin)
Mutation (A) : C1-C2-C3-C4-C5-GAG-GAA-C8 .....................
a1-a2-a3-a4-a5-Glutamic acid-Glutamic acid-a8 ....................
(Normal Haemoglobin)
Mutation (B) : C1-C2-C3-C4-C5-GUG-GAA-C8 .....................
a1-a2-a3-a4-a5-Valine-Glutamic acid -a8 ....................
(Sickle cell Haemoglobin)
OR
26.One chromosome contains one molecule of DNA. In eukaryotes the length of the
DNA molecule is enormously large. Explain how such a long molecule fits into the tiny
chromosomes seen at Metaphase.
QUESTION BANK /BIOLOGY/XII
Page 202
SAMPLE QUESTION PAPER II
TIME: 3 Hrs
M.M.:70
CLASS - XII
SUBJECT - BIOLOGY
General Instructions:
i All questions are compulsory.
ii The question paper consists of five sections A, B, C, D and E. Section-A contains
5questions of 1 mark each, Section -B is of 5 questions of 2 marks each, Section –C
has12 questions of 3 marks each, Section- D is of 1 question of 4marks whereas Section
E hasthree questions of five marks each.
iiiThere is no overall choice. However an internal choice has been provided in one
question of 2 marks, one question of three marks and all the questions of 5
weightage. A student has to attempt only one of the alternatives in such
questions.
iv Wherever necessary, the diagrams drawn should be neat and properly labelled.
SECTION – A
1,Why do predators avoid the Monarch butterfly? How does the butterfly develop this
protective feature?
2.Why is the use of unleaded petrol recommended for motor vehicles equipped with
catalytic converter?
3.Name the fungus used in organ transplant treatment and write the product of the
organism.
4.What does ‘R’ represent in the given equation for productivity in an ecosystem?
GPP – R = NPP
Name the two intermediate hosts which the human liver fluke depends on to complete
its life cycle so as to facilitate parasitization of its primary host.
SECTION – B
6.i) Name the organism in which the vector shown below is inserted to get the
copies of the desired gene.
ii)Mention the area labeled in the vector responsible for controlling the copy number of
the inserted gene.
QUESTION BANK /BIOLOGY/XII
Page 203
OR
Name the insect pest that is killed by the products of cryIAc gene.Explain how the gene
makes the plant resistant to the insect pest.
7.Identify a, b, c, d in the following table with reference to birth control.
Method
Example
a
Diaphragm
Surgical
b
c
Saheli
IUD
d
8.Why have certain regions been declared as biodiversity ‘hot spots’ by
environmentalists of the world? Name any two hot spot regions of India.
9.Read the sequence of nucleotides in the given segment of m-RNA and the respective
amino acid sequence in the polypeptide chain.
m-RNA
: AUGUUUAUGCCUGUUUCUUAA
Polypeptide : met – phe – met – proline – valine – serine
iProvide the codons for valine and proline respectively.
iiWrite the nucleotide sequence of the DNA strand from which this m-RNA was
transcribed.
iiiWhat does the last codon of this RNA stand for?
10.Name the scientist who suggested that the genetic code should be made of a
combination of three nucleotides.the basis on which he arrived at this conclusion.
SECTION – C
11.iWhen does oogenesis begin in a human female?
iiDifferentiate between the location and function of Sertoli cells and Leydig
12.
iName the organic materials the exine and intine of an angiosperm pollen
grain are made up of.
iiExplain the role of exine?
iiiHow many microspore mother cells would be required to produce one
hundred pollen grains in a pollen sac and why?
QUESTION BANK /BIOLOGY/XII
cells.
Page 204
13.Explain the process of artificial hybridization if the female parent bears bisexual
flowers.
14.Name two groups of organisms which constitute flocs. Write their influence on the
level of BOD during biological treatment of sewage. What is BOD?
15.Draw a neat labeled diagram of a typical dicot embryo and label plumule, cotyledons
and radicle.
16.
i)What is this diagram representing?
ii)Name the parts a, b and c.
iii)In the eukaryotes the DNA molecules are organized within the nucleus. How is the
DNA molecule organized in a bacterial cell in absence of a nucleus.
17.i)Draw a schematic representation of a transcription unit showing the
polarity of both thestrands. Label the promoter gene and the template
strand.
ii)Mention the condition when the template strand becomes the coding strand.
iii)Give the function of the promoter gene.
18.“A population has been exhibiting genetic equilibrium”. Answer the following
questions with respect to the above statement.
i Explain the above statement.
ii Name the underlying principle.
iii List any two factors which would upset the genetic equilibrium of the
population.
iv Take up any one such factor and explain how the gene pool will change
due to that factor.
19.The steps in a programme are:
- Collection of germplasm
- Crossbreeding the selected parents
- Selecting superior recombinant progeny
- Testing, releasing and marketing new cultivars.
iWhat is this programme related to?
iiName two special qualities as basis of selection of the progeny.
iiiWhat was the outcome of the programme?
ivWhat is the popular term given to this outcome?
vName the Indian scientist responsible for this programme.
20.Two of the steps involved in producing nematode resistant tobacco plants based
QUESTION BANK /BIOLOGY/XII
Page 205
on the process of RNAi are mentioned below. Write the missing steps in proper
sequence.
a.
b. Using Agrobacterium as vector introduce it into tobacco
c.
d.
e. Initiates RNA interference
f.
g.
h.
OR
20.In a bacterial culture some of the colonies produced blue colour in the presence of a
chromogenic substrate and some did not due to the presence or absence of an insert
(rDNA) in the coding sequence of a-galactosidase.
i Mention the mechanism and the steps involved in the above experiment.
ii How is it advantageous over simultaneous plating on two plates having
different antibiotics?
21.Name any two organisms that are responsible for ringworm in humans. Mention
two diagnostic symptoms of ringworm. Name the specific parts of the human body
where these organisms thrive and explain why?
22.Name a drug used
ias an effective sedative and pain killer.
iifor helping patients to cope with mental illnesses like depression, butoften
misused.
iiiHow does the moderate and high dosage of cocaine affect the human body?
SECTION – D
23.
In the growth curve given below
i What is the status of food and space in the curves a and b?
ii In the absence of the predators, which curve a or b would appropriately depict
prey population?
iii Write the equation used to describe curve b.
iv Time has been shown on X –axis and there is a parallel dotted line above it. Give
the significance of this line.
QUESTION BANK /BIOLOGY/XII
Page 206
SECTION – E
24.When a garden pea plant with green pods was cross-pollinated with another plant with
yellow pods, 50% of the progeny bore green pods.
i Work out the cross to illustrate this.
ii How do you refer to this type of cross?
iii Why is such a cross done?
iv Sex determination in human beings is an example of male heterogamety. Why is
it so called?
OR
Study the flow chart given below and answer the questions that follow:
a)
S strain
into mice
mice die
b)
R strain
into mice
c) Heat killed S strain + live R strain
mice live
into mice ______A________
d) Heat killed S strain + DNase + live R strain
mice_____B______
into
i
Name the organism and differentiate between the two strains R and S
respectively.
ii
Write the results A and B obtained in steps c and d respectively.
iii
Name the scientist who performed steps a, b and c.
iv
Write the specific conclusion drawn from step d.
25 .i.
With an example explain how biotechnology has been applied in each of
the following:
a
In curing Diabetes Mellitus.
b
In raising pest resistant plants.
c
In producing more nutritionally balanced milk.
ii.
Do you think it is ethical to manipulate organisms for human benefit?
Why?
OR
Name any two cloning vectors. Describe the features required to facilitate
cloning into a vector.
26.
Explain how the xerarch succession progress from xeric to mesic conditions and
forms a stable climax community.
5
OR
i
Expand CFC.
ii
CFCs are a part of greenhouse gases. Name the other gases.
iii
Explain the major harm caused by these gases.
iv
Mention the consequences of degradation of O3
QUESTION BANK /BIOLOGY/XII
Page 207
SAMPLE QUESTION PAPER III
CLASS - XII
TIME: 3 Hrs
SUBJECT - BIOLOGY
M.M.:70
General Instructions:
i All questions are compulsory.
ii The question paper consists of five sections A, B, C, D and E. Section-A contains 5
questions of 1 mark each, Section -B is of 5 questions of 2 marks each, Section -C has
12 questions of 3 marks each, Section- D is of 1 question of 4marks whereas Section E
hasthree questions of five marks each.
iiiThere is no overall choice. However an internal choice has been provided in one
question of 2 marks, one question of three marks and all the questions of 5 marks
weightage. A student has to attempt only one of the alternatives in such questions.
iv Wherever necessary, the diagrams drawn should be neat and properly labelled.
SECTION – A
1. Give reasons why measurement of bio-mass in terms of dry weight is more accurate
than fresh weight.
2. Why do certain genes tend to be inherited together in a cell at the time of cell
division?
3. Why do blood pressure and heart rate increase after tobacco consumption?
4. Provide one sentence information about plasmid with respect to its (a) Chemical
nature (b) Its duplication
5. How do histones acquire positive charge?
SECTION – B
6. What are the endocrine functions of the placenta?
7. In a cross – between true breeding red flowered snap – dragon and white flowered
snap dragon plant, the F2 phenotypic and genotypic ratio are same. Explain the result
with the help of a cross.
8.How is a detritivore different from a decomposer? Give one example for each.
9. What is Ti plasmid? Name the organism where it is found. How does it help in
genetic engineering?
10. The species diversity of plants (22%) is much less than that of animals (72%). What
could be the explanation to how animals achieved greater diversification?
Or
QUESTION BANK /BIOLOGY/XII
Page 208
Among the ecosystem services are control of floods and soil erosion. How is this achived
by the biotic ecoponents of the ecosystem?
SECTION – C
11. What is meant by each of the following: (i) Primary follicle (ii) Secondary follicle
(iii) Tertiary follicle Or (a) Name the structures which the parts ‘A’ and B shown in the
diagram below respectively develop into (b) Explain the process of development which
‘B’ undergoes in alvuminous and exalbuminous seeds. Give one example of each of
these seeds.
12. (i) Mention the property that enables the explants to regenerate into a new-plant. (ii)
A banana herb is virus – infected, Describe the method that will help in obtaining
healthy banana plant from this diseased plant.
13. How do automobiles fitted with catalytic converters reduce air pollution? Suggest
the best fuel for such vehicles.
14. An engineered vector for r. DNA technology should have three essential features.
What are they? Explain
15. What will happen to an ecosystem if:- (a) all producers are removed; (b) All
organisms of herbivore level are eliminated and; (c) All top carnivore population is
removed.
16. Describe the technique by which genetic disorder in a developing foetus can be
detected? How do people often misuse this technique
Or
(a) How do plants overcome inbreeding depression? (b) How can seeds be produced
without fertilization.
17. (a) List two essential roles of ribosome during translation. (b) Differentiate between
exons and introns.
18. (a) How does our body adapt to low oxygen availability at high attitudes? (b) Why
are small animals rarely found in polar regions
19. a) Give a schematic representation of spermatogenesis in humans. (b) Meiotic
division during oogenesis is different from that in spermatogenesis. Explain. (c) What is
spermiogenesis?
20. Three Codons on mRNA are not recognized by t RNA. What are they ? What is the
general term used for them? What is their role in protein synthesis?
21. What are Algal blooms ? How are they formed ? Give two harmful effects of
formation of algal blooms in a water body.
22. What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances
of having a child with Down’s syndrome increases if the age of the mother exceeds forty
years?
23. Sukesh is an adolescent who accidentally got into quagmire of drugs. His parents
took him to a counsellor. Imagine that you are a counselor, how will you tackle the
situation? Why do you think that adolescents easily get into the quagmire of drugs?
SECTION – D
QUESTION BANK /BIOLOGY/XII
Page 209
24. (a) Briefly describe the methodologies inhuman genome project? (b) Which human
chromosome was sequenced last? (c) Name any 4 non human model organisms whose
DNA is sequenced.
Or
24.Answer the following questions based an messelson and stans experiment. (a) Why
did the scientists used 15NH4Cl and 14NH4Cl as sources of Nitrogen in the culture
medium for growing 15N got incorporated into? (b) Name the molecules that 15N got
incorporated into? (c) How did they distinguish between 15N labeled molecules from
15N ones? (d) Mention the significance of taking E-coli samples at definite time
intervals for observations? (e) Write the conclusion drawn by them at the end of their
experiment.
25.In peas, tallness is dominant over dwarfness, and red colour of flowers isdominant
over the white colour. When a tall plant bearing red flowers waspollinated by a dwarf
plant bearing white flowers, the different phenotypicgroups were obtained in the
progeny in numbers mentioned against them.
Tall, Red = 138
Tall, White = 132
Dwarf, Red = 136
Dwarf, White = 128
Mention the genotypes of the two parents and of the types of four offspring.
Or
25.Represent schematically the life cycle of malarial parasite.
26. Draw a diagrammatic sketch of biogas plant, and label its various components.
Or
26.The following diagrams are the age pyramids of different populations.Comment on
the status of these populations.
QUESTION BANK /BIOLOGY/XII
Page 210
CLASS - XII
TIME: 3 Hrs
SAMPLE QUESTION PAPER IV
SUBJECT - BIOLOGY
M.M.:70
General Instructions:
i All questions are compulsory.
ii The question paper consists of five sections A, B, C, D and E. Section-A contains 5
questions of 1 mark each, Section -B is of 5 questions of 2 marks each, Section -C has
12 questions of 3 marks each, Section- D is of 1 question of 4marks whereas Section E
hasthree questions of five marks each.
iiiThere is no overall choice. However an internal choice has been provided in one
question of 2 marks, one question of three marks and all the questions of 5 marks
weightage. A student has to attempt only one of the alternatives in such questions.
iv Wherever necessary, the diagrams drawn should be neat and properly labelled.
SECTION – A
1. What is Red data book? Give one endangered species listed in red data book.
2. Which end does the tRNA join with m RNA during the translation?
3. What is SCP?
4. It was diagnosed by a specialist that the immune system of the body of a patient
has been suppressed. Name the disease the patient is suffering from and its
causative agent.
5. If a double stranded DNA has 10% of cytosines, calculate the percent of adenine
in the DNA.
SECTION – B
6. (a) Who 1st proposed semi-conservative replication of DNA ?
(b) Which organism is used in this experiment?
(c) What is the result of 2nd and 3rd generation?
7. What are restriction enzymes? Name one restriction enzyme and one enzyme used
for joining alien DNA and vector DNA.
8. What do you mean by autoimmunity? Give one problem of
autoimmunity disorder.
OR
8. What is apomixis? Give the importance of apomixis in hybrid seed industry.
9. What are the major causes of species losses in a geographical region? Give the
name of one alien species of animal which is responsible of biodiversity losses.
10. What are bacteriophage vectors? Name the two phage vectors that are commonly
used.
SECTION –C
11. What is DNA fingerprinting? Mention its application.
12. Define decomposition and describe the processes and products of decomposition.
13. (a) Show the succession on bare rocks by diagram only.
QUESTION BANK /BIOLOGY/XII
Page 211
(b) A biosphere reserve has different zones.
(i) Label the guidelines marked 1 and 3.
(iii) What is the function of the zone marked 2?
14. What is biological magnification? How does DDT as a water pollutant undergo
biological magnification?
15. In the medium where E. coil was growing, lactose was added, which in induced the
lac-operon. But why does lac-operon shut down after sometime after, addition of lactose
in the medium?
16. In Antirrhinum majus:
RR is phenotypically red, rr is white and Rr is pink. Mention the phenotype and
the ratio in F1generation of the following crosses :
RR × Rr
rr × RR
Rr × Rr
rr × Rr
Name the other plant which shows similar type of inheritance. Give the scientific
name.
OR
16What is meant by R-cells and S-cells with which Frederick Griffith carried
out this experiments on Diplococcus pneumoniae? What did heprove from
these experiments.
17. What is an age pyramid? What do they show for human population? Represent
diagrammatically the different shapes of age pyramids and what shapes each of
them represent?
18. (i) Explain antibiotic resistance observed in bacteria in light of Darwinian
selection theory.
(ii) Describe one example of adaptive radiation.
19. Where do transcription and translation occur inside a living cell? Briefly describe
the threesteps involved in the process of translations.
20. (a) Why we should avoid antipyretic in mild fever. Give another term use for
endogenous pyrogen released by macrophages.
(b) What is the cause of Tertian malaria and Quartan malaria.Which is the most
serious parasite of malaria in India. What is Relapse malaria?
21. What is polymerase chain reaction? Describe the basic requirements of a PCR
reaction.
22. What is an age pyramid? What do they show for human population? Represent
diagrammatically the different shapes of age pyramids and what shapes each of them
represent?
SECTION – D
23. Rohit meets an accident. Iqbal his schoolmate takes him to hospital where Rohit
(AB bloodgroup) needs blood transfusion. Iqbal also has AB blood group and is willing
to donate his bloodbut Rohit’s mother object by saying “Iqbal belongs to different
community so has different typeof blood.” In your opinion Rohit’s mother is wrong or
right? Give your opinion by explaining theallelic composition of blood group AB.
QUESTION BANK /BIOLOGY/XII
Page 212
SECTION – E
24. What is menstrual cycle? Represent the various events during a menstrual cycle in a
mammalian female with hormonal and uterine events.
OR
24.a. In humans, males are heterogametic and females are homogametic. Explain. Are
there any examples where males are homogametic and females heterogametic?
b. Who determines the sex of an unborn child? Mention whether temperature has a role
in sex determination
25. What are transgenic bacteria? Illustrate using any one example.
OR
25.a) What are cry proteins? Name an organism that produces it. How has man
exploited this protein t his benefit?
(b) What is gene therapy? Illustrate using the example of adenosine deaminase
(ADA) deficiency.
26. T.H. Morgan while going on a walk, found a fruit covered with flies. He took the
flies to their laboratory. He along with his students performed experiment for
several generations. They surprised to see some of characters do not obey
Mendelian principal of independent assortment.
(i) Write common name of the flies and also its scientific name.
(ii) The tendency of two characters to remain inherited together for different
generations is called as ..............
(iii) Tendency of two characters to stay separately for different generation
is............
(iv) Draw the diagram of physical basis of this type of inheritance.
OR
26.Colour blindness is a sex linked disease. It is due to X-chromosome. Normal
parents have three daughter, all normal and one son colour blind. What is the
reason for it. Show the inheritance of a sex linked recessive case of human being.
QUESTION BANK /BIOLOGY/XII
Page 213
CLASS - XII
TIME: 3 Hrs
SAMPLE QUESTION PAPER V
SUBJECT - BIOLOGY
M.M.:70
General Instructions:
(i) All questions are compulsory.
(ii) This question paper consists of five Sections A, B, C, D and E. Section A contains 5
questions of onemark each, Section B is of 5 questions of two marks each, Section C is
of 12 questions of threemarks each. Section D is of 1 questions of four marks each and
Section E is of 3 questions of fivemarks each.
(iii) There is no overall choice. However, an internal choice has been provided in one
question of 2marks, one question of 3 marks and all the three questions of 5 marks
weightage. A student has toattempt only one of the alternatives in such questions.
(iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.
SECTION – A
1. Is it syngamy or triple fusion that leads to formation of zygote in an angiosperm?
Mention their ploidy.
2. Write the scientific name of the plant used by Mendel for his experiments.
3. Write the full form of SCID.
4. Define BOD.
5. What is anthropogenic pollution?
SECTION – B
6. Black pepper seeds have both perisperm and endosperm. What is the difference
between these two structures?
7. Why is colour blindness trait more common with the males? Also explain conditions
when a female will suffer from colour blindness trait.
8. Name the microsporangia and megasporangia of an angiosperm. After fertilization
what does megasporangia transform into?
9. Write the antigens and antibodies present in the blood of persons belonging to A, B,
AB and O blood groups?
10. Describe photochemical smog in brief? How does it lead to snow blindness?
OR
10. the cause of algal blooms in a water body. What will be its effect on other organisms
living the same area?
SECTION –C
11. a) Why do scientists prefer to perform experiments with fruit flies and bacteria?
b) What is the need of pedigree analysis for human beings?
12. a)What is ozone hole? Write the factors causing it.
b) What is good ozone and bad ozone?
QUESTION BANK /BIOLOGY/XII
Page 214
13.Write the principle with diagram on which an electrostatic precipitator works.
or
Explain ways to manage the hospital and electronic wastes in your surroundings.
14. Describe spermatogenesis under the following heads (draw diagrams to support
your answer):
a) formation of spermatocytes
b) formation of immature sperms
c) spermiogenesis and spermiation
15. If following is the sequence of the template strand of structural gene:
CAATAGCCTAGAGAT then find out the following :
a) the sequence of mRNA formed after transcription along with its polarity
b) the sequence of bases on the coding strand of DNA.
16. a)On what grounds is CNG a better fuel than diesel and petrol?
b) Why should vehicles fitted with catalytic converter use only unleaded petrol?
or
What is evil quartet? How has it become a threat for biodiversity?
17. What are palindromes? How are they significant to biotechnological processes?
18. How are microbes useful to a sewage treatment plant? Explain.
19. Through a single example explain the concepts of multiple allelism and co
dominance.
20. In the light of Darwin’s evolutionary theory explain :
a) branching descent in Australian marsupials
b) antibiotic resistance in bacteria
21. Discuss the various levels of biodiversity that are prevalent in the ecosystem. Give
suitable examples to support your answer.
22. What are cry genes and Cry proteins? Write scientific name of organism working as
their source. How these proteins have benefitted the human beings?
SECTION –D
23. Scientists are developing new strategies for enhancement of food production.
However is simply increasing the quantity of food a permanent solution for the growing
population of India or something else also needs to be taken care of ?
SECTION – E
24. a) Where are corpus luteum and corpus albicans present? Do they have any specific
function?
b) Draw a well labeled diagram of mature graafian follicle.
c) Describe the endocrine functions of human placenta.
or
a) Explain in detail all the events that occur during menstrual cycle.
b) Whatis menarche and menopause?
25. Describe the detailed structure of a eukaryotic chromosome with reference to
packaging of DNA within that chromosome. How is the packaging different from a
prokaryote?
QUESTION BANK /BIOLOGY/XII
Page 215
or
Why is the genetic code a triplet? What are non sense codons in genetic code? Explain
features of genetic code like degenerate, unambiguous, universal with suitable examples.
26. a)Define gene therapy. Site the example of ADA deficiency and its cure with
reference to gene therapy.
b) Write the positive and negative effects of developing transgenic crops.
or
What are biopiracy and biopatenting? How are they interrelated? Give any example of
biopiracy.How biopiracy can be harmful to the host country?
QUESTION BANK /BIOLOGY/XII
Page 216