Download CBSE XII BIO QS with key(2009-2015)

KENDRIYA VIDYALAYA CRPF AVADI CHENNAI-65
CBSE- QUESTIONS- BIOLOGY (chapter wise) 2009-2015
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
Chapter 1
REPRODUCTION IN ORGANISMS
Name the phase all organisms have to pass through before they can reproduce
sexually.
Ans :juvenile/growth =1
Mention the site where syngamy occurs in amphibians and reptiles respectively.
Ans: amphibians: outside the body/ external medium =1/2, in reptiles – inside the body
Name an organism where cell division itself is a mode of reproduction? 2013
ans]protists/monerans/amoeba/paramecium/any other
Name an alga that reproduces asexually through zoospores. Why are these
reproductive units so called?
ans : chlamydomonas , because they are motile = ½+1/2 =1
Name an organism where cell division itself is a mode of reproduction?
(1) ans]protists/monerans/amoeba/paramecium/any other
Name the phenomenon and one bird where the female gamete directly develops
into a new organism?
ans: parthenogenesis = ½
turkey =1/2
Name the vegetative propagules in the following(2014)
a) Agave
b) Bryophyllum
Ans. Agave – Bulbil =1/2
Bryophillum – leaf buds/ adventitious buds = 1/2
Why do moss plants produce very large number of male gametes? Provide
one reason. What are these gametes called?(2015)
To compensate the loss of male gametes during their transport ( to the non-motile female
gametes) through water/ to increase chances of fertilization, antherozoids=1+1
A liverwort plant is unable to complete its life- cycle in a dry environment.
State two reasons.
Ans : 1) They need water as medium of gamete transfer for fertilisation=1
2) For germination of spores =1
Chap 2
SEXUAL REPRODUCTION IN FLOWERING PLANTS
How do the pollen grains of vallisneria protect themselves?
Have mucilaginous covering ( to prevent it from getting wet)
Name the type of flower which favours cross pollination.
Chasmogamous.
Why is bagging of the emasculated flower essential during hybridization
experiments?
To prevent the entry of unwanted pollen// in order to have desired cross pollination
Pea flowers produce assured seed sets. Give a reason.
Pea flower is a cleistogamous flower/do not open=1
Why banana is considered a good example of parthenocarphy?
Formation of fruit without fertilization/no formation of seeds.
Name the part of the flower which the tassels of the corn –cob represent.(2014)
Ans: Style = ½ ,stigma = ½
Give an example of a plant which came into India as a contaminant and is a cause
XII BIO
Page 1
1
1
1
1
1
1
1
2
2
1
1
1
1
1
1
1
8
9
10
11
12
13
14
15
of pollen allergy.
Parthenium/carrot grass
State one advantage and one disadvantage of cleiostogamy.
Advantage – self pollination assured/seed production assured
Disadvantage – least variations/ leading to inbreeding depression.
In angiosperms, zygote is diploid while primary endosperm cell is triploid.
Explain?(2013)
ans: zygote is diploid because of the fusion / syngamy , of male and female gametes ,
Primary endosperm cell is triploid as one male gamete fuses,with two polar nuclei [
each having n no. of chromosomes ] = ½*4 .
Name all the haploid cells present in an unfertilized mature embryo – sac of
flowering plant. Write the total number of cells in it? 2013
ans: egg cells , synergids , antipodals = ½*3 = 1 ½ .seven cells = ½
Explain the function of each of the following
a) coleorhizae – protects the radical of monocot embryo = 1
b)germ pores – allow germination of pollen grains/formation of pollen tubes =1
Draw a vertical section of a maize grain and label i)
pericarp,ii)scutellum,iii)coleoptiles and iv) radical
Ans:
Or
Mention any two uses of seeds.
b)better adaptive strategies for dispersal to new habitats
they have food reserves for the seedling or embryo
the seed coat provides protection to the embryo
generate genetic variation
remain viable for a considerable period of time.(any two) =1x2
State what is apomixis? Comment on its significance. How can it be
commercially used? (2015)
Form of asexual reproduction producing seeds without fertilization/ type of asexual that
mimics sexual reproduction to form seeds without fertilization=1
Parental characters are maintained in the progeny/ offspring (as there is no meiosis/
segregation of characters)=1
If desired hybrid seeds are made apomictics. The farmers can keep on using the hybrid
seeds to raise new crops year after year=1
Differentiate between perisperm and endosperm giving one example each.
Perisperm – persistant nucleus =1/2,diploid=1/2,eg., black pepper/beet=1/2
Endosperm – nutritive tissue for embryo = ½,triploid = ½, ex: maize/rice =1/2
Fertilization is essential for production of seed, but in some angiosperms seeds
develop without fertilization.
XII BIO
Page 2
2
2
2
2
2
2
3
3
i)
16
17
18
19
Give an example of angiosperm that produces seeds without
fertilization.name the process.
ii)
Explain two ways by which seeds develop without fertilization.
ans: i) Members of Asteraceae/grasses/citrus/mango, apomixes = ½+1/2
ii)two ways-in some species the diploid egg cell is formed without reduction division
and develops into embryo without fertilization = 1
b) More often, as in many citrus and mango varieties some of the nucellar cells
surrounding the embryo sac start dividing, protrude into the embryo sac and develop into
the embryos.
a) Mention any four strategies adopted by flowering plants to prevent self
pollination.
b) Why is geitonogamy also referred to as genetical autogamy?
Ans: a)pollen release and stigma receptivity are at different times=1/2
Anther and stigma are placed at different positions=1/2
Self incompatibility/prevents pollen from fertilizing the ovule of the same
plant=1/2
Production of unisexual flowers=1/2
b)pollen grains come from the same plant=1
Make a list of any three out breeding devices that flowering plants have developed
and explain how they help to encourage cross – pollination. (2014)
Ans: (i) Time of pollen release and stigma receptivity are different (not synchronized),
self pollination prevented.
(ii) Anther and stigma are placed at different positions, so the pollen cannot come in
contact with the stigma of the same flower.
(iii) Self incapability, genetic mechanism (prevent the pollination on the stigma of the
same flower).
(iv) Production of unisexual flowers / dioecious plants, cross pollination ensured.
any three) = 1/2x6
OR
Why are angiosperm anthers called dithecous? Describe the structure of its
microsporangium. (2014)
Ans: (i) Anther bilobed, each lobe of anther has two theca.1/2+1/2
(ii) Microsporangium surrounded by four wall layers named as epidermis,
endothecium , middle layer and tapetum.1
(iii) In young anther a group of completely arranged homogenous cells called
sporogenous tissue occupies the centre of each microsporangium which produces
microspores / pollen grains.1
Why are some seeds referred to as apomictic seeds? Mention one
advantage and one disadvantage to a farmer who uses them.(2015)
Seeds produced without fertilization are referred to as apomictic =1
Advantage: desired characters retained in offspring (progeny)/ no segregation of
characters in offspring(progeny)/ assured seed production in absenceof
pollinators=1(any other relevant advantage)
Disadvantage: cannot control accumulation of deleterious genetic
mutation/usually restricted to narrow ecological niches/ lack ability to adapt to
changing environment = 1(any other relevant disadvantage)
a)Why is fertilization in angiosperm referred to as double fertilization? Mention
the ploidy of the cell involved.
b) Draw a neatly labeled sketch of L.S of an endospermous monocot seed.
XII BIO
Page 3
3
3
3
5
a)fertilization of haploid egg cell by one haploid male gamete to form diploid
zygote is called syngamy 1/2x3=11/2
fertilization of two (diploid) polar nuclei by the other haploid male gamete to
form triploid primary endosperm nucleus is called triple fusion =1/2x3=1 ½
b)
20
Any four labels including endosperm =2
a) Trace the development of embryo after syngamy in a dicot plant.
b) Endosperm development precedes embryo development. Explain
c) Draw a diagram of a mature dicot embryo and label cotyledons, plumule,
radical and hypocotyls in it.
explanation of the figure / above diagram can be considered//
the zygotes divides unequally to form two cells namely embryo cell and
suspensor cell, the smaller cell divides repeatedly to produce a row of 4-8 cells,
the terminal cell divides to produce a cluster of cells called the globular embryo/
proembryo and the remaining cells constitute the suspensor,
a few cells of the proembryo nearest of the suspensor develop into hypocotyls
and radical while other cells give rise to epicotlplumule and cotyledons = 1/2x4
b) provides nourishment to the developing embryo=1
c)
XII BIO
Page 4
5
hypocotyl
21
1/2x4=2
Give reasons why:1x5=5 marks
i) Most zygotes in angiosperms divide only after certain amount of endosperm is
formed.
ii)groundnut seeds are exalbuminous and castor seeds are albuminous
iii)micropyle remains as a small pore in the seed coat of a seed
iv) Integuments of an ovule harden and the water content is highly reduced, as the
seed matures.
v) Apple and cashew are not called true fruits.
Ans:i)to obtain from the endosperm for the developing embryo=1
ii)ground nut- endosperm completely consumed =1/2
castor – endosperm persists =1/2
iii)for the entry of water/oxygen,for germination=1/2=1/2=1
iv) protect the embryo/keep the seed viable,untilfavourable conditions return for
germination=1/2=1/2=1
v)ovary does not take part in fruit formation/thalamus contributes to fruit formation=1
or
a)draw a labeled diagram of L.S of an embryo of grass(any six labels)
c) Give reason for each of the following
i) Anthers of angiosperm flowers are described as dithecus.
ii) Hybrid seeds have to be produced year after year.
i)
each bilobed anther has two theca
ii)
progeny show segregation, do not maintain hybrid characters=1/2=1/2=1
XII BIO
Page 5
5
22
How does the pollen mother cell develop into a mature pollen grain? Illustrate the
stages with labeled diagrams.
PMC/Microspore mother cell undergoes meiosis, to form microspore tetrad by the
process called microsporogenesis, each microspore develops into pollen grain, each
pollen grain undergoes unequal mitotic division, and produces two cells – vegetative cell
and generative cell 1/2x5=2 ½
23
XII BIO
1/2x5=2 1/2
a) Trace the development of megaspore mother cell up to the formation of a
mature embryo-sac in flowering plant
b) draw a labeled diagram of the structure of mature dicot embryo
ans:MMCmegaspore mother cell undergoes meiosis to form four megaspores,one
of the megaspore is functional while other three degenerate, the nucleus of
functional megaspores divides thrice mitotically, to form 8 nucleate stage of the
embryo sac followed by the formation of cell walls, cells are arranged as one egg
cell and two synergids at micropylar end, 3 antipodals at chalazal end and two
polar nuclei at the centre =1/2x6
b
Page 6
5
5
24
5
Draw a L.S of a pistil showing pollen tube entering the embryo-sac in an
angiosperm and label any six other than stigma, style and ovary.
(b) Write the changes a fertilized ovule undergoes within the ovary in an
angiosperm plant.
Longitudinal section of a flower Showing growth of pollen tube
(Pollentube,antipodal, polar nuclei, eggcell, synergids, pollen, embryo sac any 6) =
1/2 × 6
(b) Zygote
embryo
Integument
seed coat
Micropyle remains
Synergids and antipodals degenerate
Ovule
seed (any 4) = ½ ×4
[3+2=5marks]
25
A) Describe the endosperm development in coconut.
B) Why is tender coconut considered a healthy source of nutrition?
C) How are pea seeds different from castor seeds with respect to endosperm?
Ans. A) the primary endosperm nuclei undergoes successive nuclear division to form
free nucie then cell wall formation occurs toward the periphery and endosperm becomes
cellular leaving free nuclear endosperm in central part.=1/2 +1/2=1
B) reserve food material=1
C)peas-endosperm used up-so no endosperm in mature seed ---in
Castor-endosperm remain in mature seed=1=(1+1+1=3 )
5
26
a) Explain the different ways apomictic seeds can develop. Give an example of
each.
b) Mention one advantage of apomictic seeds to farmers.
c) Draw a labeled mature stage of a dicotyledonous embryo. (2014)
Ans. (a) (i) Diploid egg cell is formed without reduction division and
embryo without fertilization, eg. Asteraceae/ grasses.
(ii) In citrus / mango, some of the diploid nucellar cells surrounding
the
embryo sac start dividing, protrude into embryo sac & develop into a embryo.
5
(b) No segregation of character in hybrid seeds, economically beneficial / desired
varieties are cultivated
XII BIO
Page 7
27
1
2
3
4
5
6
7
8
(c )
any four labeling = 1/2x4=2
a) Plan an experiment to and prepare a flow chart of the steps that you
would follow to ensure that the seeds are formed only from the desired sets
of pollen grains. Name the type of experiment that you carried out.
b) Write the importance of such experiments.(2015)
a) selection of flowers from desired plants- emasculationbagging  dusting
of the pollens on the stigma of the flowers that were bagged flower rebagged
fruit formed=1/2x6
artificial hybridisation=1
b)production of superior/ improved desired varieties of plants. =1
CHAP-3
HUMAN REPRODUCTION
Where are the leydig cells present?what is their role in reproduction?
Ans: between / outside the seminiferous tubules// interstitium of the testes=1
Role – stimulates and secretion of androgens =1
Name the embryonic stage that gets implanted in the uterine wall of a human
female.
Blastocyst/blastula
How is the entry of only one sperm and not many ensured into an ovum during
fertilization in humans?
On contact with zonapellucida, sperm induces changes in ovarian membrane. That
prevents the entry of others.
Where is acrosome present in human? Write its function.
On the sperm head-1/2, Has enzymes to dissolve the follicles of ovum / facilitate 1the
entry of sperm nucleus for fertilization/ help the sperm enter into the cytoplasm of the
ovum. 1
What stimulates pituitary to release the hormone responsible for parturition?
Name the hormone.
Ans:signal from the developed foetus and placenta induces mild uterine
contractions/foetal ejection reflex =1/2
Oxytocin =1/2
Explain the functions of umbilical cord.
Transports nutrients and respiratory gases and metabolic wastes to and from mother and
foetus =1 mark
Give reasons for the following
a) The human testes are located outside the abdominal cavity?
b)some organisms like honey-bees are called parthenogenetic animals
a) to maintain the temperature (2-2.5c) lower the normal body temperature essential for
spermatogenesis
1
b) they/drone/males develop from unfertilized eggs.
1
a) Where do the signals for parturition originate from the humans?
b) Why is it important to feed the newborn babies on colostrums?
Ans: a) from the fully developed foetus/placenta/foetal ejection reflex
b) contains IgA antibodies to (passively) immunize the baby
XII BIO
Page 8
5
1
1
1
1
1
1
2
2
9
10
11
12
13
14
15
Differentiate between major structural changes in human ovary during the
follicular and luteal phase of the menstrual cycle.
Ans: follicular phase: primary follicle matures to form graffian follicle=1
luteal phase: the ruptured graafian follicle changes into corpus luteum.=1
Why are humans testes located outside the abdominal cavity? Name the pouch in
which they are present? (2014)
Ans: Needs lower temperature (2 – 2.5⁰C) less than the body temperature for the
formation of sperms / for spermatogenesis = 1
Scrotal sac / Scrotum = 1
Write the location and functions of the following in human testes :
(a) Sertoli cells
(b) Ledig cells
(2014)
Ans: (a) location : Lined inside the seminiferous tubules 1/2
Functions: provide nutrition to the germ cells 1/2
(b) Location : Outside the seminiferous tubules / Interstitial space 1/2
Functions : Synthesise or secrete male hormones / Testicular hormones /
Androgens 1/2
Write the effect of high concentration of LH on a mature graafian follicle.(2014)
Rupture of graafian follicle, release of ovum/secondary oocytes/ovulation. = 1+1
Draw a labeled diagram of the reproductive system in a human female.
Ans:
2
Explain the steps in the formation of an ovum from an oogonium in humans.
Ans. Oogonia- prophase 1-(temporary arrest)-primary oocyte, primary oocytes complete
first meiosis, to form sec oocyte +first polar body, sec oocyte –completes meiosis,
ovum, +2 polar bodies=1/2 x6= 3.
a) When and how does placenta develop in human female?
b) How is the placenta connected to the embryo?
c) Placenta acts as an endocrine gland. Explain.
Ans: a) after implantation of the embryo placenta develop in the human female, the
chorionic villi, and uterine tissue, become interdigitated with each other and jointly form
a structural and functional unit between developing embryo(foetus) and maternal body
called placenta=1/2x4=2
b)through the umbilical cord=1
c)placenta acts as an endocrine tissue and produces several hormones like human
chorionicgonadotropin (hCG) human placental lactogen(hPL) /
estrogen/placental progestogens/
any two=1/2+1/2=1
in the maternal blood increased level of these hormones is essential for
supporting the foetal growth/metabolic changes in mother and maintenance of
pregnancy (any two effects) =1/2+1/2
3
XII BIO
Page 9
2
2
2
3
5
16
The following is the illustration 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?
c) What is the difference between “d” and “e”?
d) Draw a neat labeled sketch of graffian follicle.
5
a)’g’-1/2 lutenisingharmone LH-1/2
b)produces the harmoneprogesterone,causes proliferation of the endometriumwhich
gets highly vacularised,it is essential for the implantation of the fertilized ovum and
to maintain the same during the pregnancy.-1/2x3
c)”d” is the developing tertiary follicle-1/2
“e” is the graafian follicle-1/2
d)
17
Study the flow chart given below. Name the hormones involved at each stage and
explain their functions.
Hypothalamus
Pituitary
Ovary
Pregnancy
Ans: hypothalamus produces gonadotropins releasing hormone, which induces anterior
pituitary, to release LH, and FSH, to effect oogenesis, and ovulation, the graffian follicle
after shedding ovum becomes corpus luteum, which secretes progesterone, to maintain
endometrium, and implantation of the embryo = 1/2x10
XII BIO
Page 10
5
18
Draw a diagrammatic sectional view of a human seminiferous tubule, and label
sertoli cells, primary spermatocyte, spermatogonium and spermatozoa in it.
a) Explain the hormonal regulation of the process of spermatogenesis in
human.
(b) GnRH acts on/stimulates anterior pituitary(1/2)
Stimulates secretion of LH,FSH = ½
LH acts on Leydig cells = ½
Leydig cells produce androgen-stimulates spermatogenesis = ½
FSH acts on sertoli cells and stimulates secretion of some factors,= ½
These factors help in the process of spermiogenesis = ½ = ½ × 6 = 3
19
5
[2+3=5]
a) Draw a diagrammatic labeled sectional view of a seminiferous tubule of a
human.
b) Describe in sequence the process of spermatogenesis in humans.
5
b)at puberty spermatogonia undergo mitosis to form primary spermatocytes,each
undergoes first meiotic division to form two secondary spermatocytes, each
completes second meiotic division to form two spermatids, spermatid undergoes
spermiogenesis to form spermatozoa =1/2x4=2
// labeled schematic representation may be considered in lieu of explanation.
20
Describe the post zygotic events leading to implantation and placenta formation in
XII BIO
Page 11
5
21
22
humans. Mention any two functions of placenta.
Ans:zygote moves through oviduct and cleavage begins,formsblastomeres,at 8-16 cell
stage embryo is called morula,mitosis continues and embryo moves to uterus,
blastomeres form outer layer called trophoblast, the rest of the cells form mass of cells,
at blastocyst stage embryo gets implanted, trophoblast puts forth chorionic villi which
finall along with uterine tissue forms placenta=1/2x8
Functions of placenta:
Supplies oxygen and nutrients to the embryo
Removes co2 and excretory waste
Endocrinal function producing human chorionic gonadotropins / human placental
lactogen/estrogen/progestogens
(Any two)=1/2x2
(a) Where does fertilization occur in humans? Explain the events that occur during
this process. (2014)
(b A couple where both husband and wife are producing functional gametes, but
the wife is still unable to conceive, is seeking medical aid. Describe any one
method that you can suggest to this couple to become happy parents. (chap -4)
Ans. (a)
i.
Ampullary Isthmic junction in fallopian tube/ fallopian tube.
ii. The sperms come in contact with zona pellucida.
iii. Induces change in the membrane.
iv. Blocks entry of other sperms/ ensures only one sperm
fertilizes the ovum/ prevents polyspermy.
v. The secretion of acrosome helps the sperm to enter the c
cytoplasm.
vi. Entry of sperm induces completion of second meiotic
division forming ovum and 2nd polar body.
vii. The haploid nucleus of sperm and that of ovum fuses.
viii. Formation of diploid Zygote, fertilization completed.
(8x1//2 = 4)
b)
Methods IVF / ZIFT/ AI
IVF : Ova from wife and sperm from the hushband is collecte
It is induced to formed zygote under laboratory conditions.
//
ZIFT : Zygote or early embryo are then transferred to the fallopian tube (ZIFT)
or into uterus (IUT) to complete further development.
//
AI : Semen collected from the husband is artificially Introduced either into the
vagina or into the uterus (IUI) of the wife. [5 marks]
5
Describe the role of pituitary and ovarian hormones during the menstrual cycle in a 5
human female.
Pituitary hormones: (when levels of FSH is high) FSH, induces follicular growth,
secretion of estrogen by follicles.( when LH surge there in the mid of the cycle)
lutinising hormones/ LH, along with FSH leads to ovulation, and then formation of
corpus luteum
= 1/2x6
Ovarian hormones:
Estrogen , repair/ proliferation of endometrium,
Progesterone, maintains endometrium for implantation = 1/2x4
( low level of progesterone leads to menstrual flow)
Chap 4
REPRODUCTIVE HEALTH
XII BIO
Page 12
1
2
3
4
Why do some women use “saheli” pills?
Once a week pill/with very few side effects/high contraceptive value/prevents
pregnancy. any two 1+1
Describe the lactational amenorrhea method of birth control.
Ans:during intense lactation/breast feeding=1
No menstrual cycle.no ovulation =1
Explain the zygote intra fallopian transfer technique (ZIFT). How is intra uterine
transfer technique IUT different from it?
ZIFT – when the zygote at 8 celled stage is transferred to the fallopian tube =1
IUT – when the embryo more than 8 celled stage is implanted in the uterus =1
Suggest and explain any three assisted reproductive technologies (art) to an
infertile couple.
Ans. In vitro fertilization /INF(test tube baby ) –fertilization outside the body in
almost similar conditions as that in the body followed by embryo transfer//
ZIFT/zygote intra fallopian transfer , the zygote of early embryo can be transferred into
the fallopian tube and embryos with more than 8 blastomeres into the uterus//
GIFT/ gamete intra fallopian transfer , transfer of an ovum collected from a donor into
the fallopian tube//
ICSI/intra cytoplasmic sperm injection –sperm is directly injected to the ovum..
AI/artificial insemination, the semen collected either from the husband or a healthy
donor is artificially introduced into the uterus.
5
If implementation of better techniques and new strategies are required to provide
more efficient care and assistance to people, then why is there a statutory ban on
amniocentesis? Write the use of his technique and give reason to justify. (2014)
Ans: To legally check female foeticide / Misuse of sex determination technique 1
To detect the abnormal chromosomes / genetic disorder .1
Justify: Prevent (female foeticides) change in sex ratio of the population.1
6 .
A women has certain queries as listed below, before starting with contraceptive
pills.
Answer them.
(a) What do contraceptive pills contain and how do they act as
contraceptives?
(b) What schedule should be followed for taking these pills? (2014)
Ans : (a) progesterone / progesterone – estrogen combination;
Inhibit ovulation, inhibit implantation, alter quality of cervical mucus to
prevent or retard Entry of sperm.
(any two) = 1+1
(c) Taken daily for a period of 21 days starting within first five days of menstrual cycle
(to be repeated after a gap of 7 days) = 1
7
a) Name of two two copper releasing IUDs. (2014)
(b) Explains how they act as effective contraceptives in human females.
Ans: (a) Intra Uterine devices –CuT , Cu7, Multiload 375 (any two) ½+1/2=1
(b)Supress the sperm motility, suppresses fertilizing capacity of sperms, increase
phagocytosis of sperms with uterus. Any two1+1
8
Your school has been selected by the Department of Education to organize and host
an interschool seminar on “reproductive health – problems and practices”however, many parents are reluctant to permit their wards to attend it. Their
argument is that the topic is “too embarrassing”.
Put forth four arguments with appropriate reasons and explanation to justify the
topic to be very essential and timely.(2015)
XII BIO
Page 13
2
2
2
3
3
3
3
4
1
2
3
4
5
6
7
8
Ans. 1. The issue of puberty and adolescence need to be addressed effectively with the
respective age gp because many changes take place in the body during adolescence of
which they are supposed to be aware of =1
2. to bring in awareness about their reproductive health and its effect on their physical,
emotional and social being =1
3. to address the increase in sex abuse and sex crimes in our country =1
4. Myths and misconceptions related to reproduce issues =1(any other reasons may also
be accepted)
Chap 5 PRINCIPLES OF INHERITANCE AND VARIATION
Name the respective pattern of inheritance when F1 phenotype
a) doesn’t resemble either of the two parents and is in between the two
b)resembles only one of the two parents
ans: a) incomplete dominance
b) dominance
A garden pea plant produced axial white flowers. Another of the same species
produced terminal violet flowers. Identify the dominant traits.
Ans: axial, violet flowers.
Write the genotype of
i) An individual who is carrier of sickle cell anemia gene but apparently
unaffected
ii) An individual affected with the disease.
Ans: i) HbAHbS = ½
ii) HbSHbS = ½
[½+½ = 1mark
A human being suffering from Down’s syndrome shows trisomy 21st chromosome.
Mention the cause of this chromosomal abnormality.
Ans: Failure of segregation of chromatids during cell division/ presence of extra 21st
chromosome. = 1mark
Mention any two contrasting traits with respect to seeds in pea plant that were
studied by Mendel.(2014)
Ans: Round / wrinkled = ½ , Yellow/ Green = ½
Name the stage of cell division where aggregation of an independent pair of
chromosomes occurs.(2014)
Anaphase – 1of meiosis -1/ Anaphase-1
How many chromosomes do drones of honeybee possess? Name the type of
cell division involved in the production of sperms by them.(2015)
16, mitosis = 1/2+1/2
A relevant portion of b-chain of haemoglobin of a normal human is given below:
The codon for the sixth amino acid is GAG. The sixth codon GAG mutates to GAA
as a result of mutation ‘A’ and into GUG as a result of mutation ‘B’. Haemoglobin
structure did not change as a result of mutation ‘A’ where as haemoglobin changed
because of mutation ‘B’ leading to sickle shaped RBCs. Explain giving reason how
mutation ‘B’ could change the haemoglobin structure and not mutation ‘A’.
Ans: GAA also codes for glutamic acid and hence no change in RBC. = 1
GUG codes for Valine and so the sickle shaped RBCs. = 1[1+1= 2 marks]
XII BIO
Page 14
1
1
1
1
1
1
1
2
9
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 a help of an example
ans : in garden pea plant = ½
PARENTAL
TT * tt
TALL DWARF
↓
↓
GAMETES
T
t
↓
F1 GENERATION
Tt
SELFING
TALL
T
t
F2 GENERATION
10
11
12
13
T
TT
Tt
Tt
tt
t
genotype 1:2:1 one homozygous dominant
and two are
heterozygous dominant and one is homozygous recessive =1/2
phenotype 3:1
three are dominant and one shows recessive character =1/2
Why are human females rarely haemophilic? Explain. How do haemophilic
patients suffer?
Ans. Haemophilia is a sex linked recessive disorder, the female has xx sex
chromosomes and if one of the 2x is normal she remains a carrier and not
diseased=1+1=2
//only if mother is a carrier , father haemophilic =1+1=2
Non stop bleeding/, no clotting =1/2+1/2
With the help of one example, explain the phenomena of co- dominance and
multiple allelism in human population. (2014)
Ans: ABO blood group in human being is an example of multiple allelism = ½
Three allels for the gene I i.e. I , I , I = ½
When I and I are present together the blood group is AB = ½
Both A and B are expressed and is called co dominance = ½
In Snapdragon, A cross between true breeding red flower (RR) plants and true
breeding white flower (rr) plants showed by a Progeny of plants with all pink
flowers. (2014)
(a) The appearance of pink flowers is not known a s blending. Why?
(b) What is the phenomenon known as?
Ans: (a) R (dominant allele red colour ) is not completely dominant over r(recessive
allele white colour) / r maintains its originality and reappear in F₂ generation = 1
(b) Incomplete dominance = 1
Write the scientific name of the fruit fly. Why did Morgan prefer to work with
fruit-flies for his experiments? State any three reasons. (2014)
Ans: Drosophila melanogaster = ½
Grown in simple synthetic medium , complete the life cycle in two weeks/ short lie
cycle , single mating produce more progeny , dimorphism , many heritable variations /
easy to handle . (any three ) = 1 ½
OR
Linkage or crossing over of genes is alternatives of each other. Justify with the
help of an example. (2014)
Ans: In Drosophila a yellow bodied white eyed female was crossed with brown bodied
red eyed male, F₁ progeny produced and intercrossed the F₂ phenotypic ratio of
XII BIO
2
Page 15
2
2
2
2
14
15
16
17
18
Drosophila deviated significantly from Mendel’s 9:3:3:1, the genes for eye colour and
body colour are closely located on the ‘X’ chromosome showing linkage and therefore
inherited together, recombinants were formed due to crossing over but at low
percentage.
Why is pedigree analysis done in the study of human genetics? State the conclusions
that can be drawn from it. (2014)
Ans: (i) Control crosses are not possible in case of human beings. 1
(ii)Analysis of traits in several generations of a family / to pattern of inheritance /
whether the trait is dominant or recessive / sex linked or not.
(Any two) = 1+1
a) Explain the phenomena of multiple allelism and codominance taking ABO blood
group as an example.
One gene I has three alleles IA IB and i. hence multiple allelism. We inherit any two of
them. When the genotype is IAIB, the individual has AB blood group since both IA and IB
equally influence the formation of antigen A and B codominance.
b) What is the phenotype of the following?
(i) IAi
(ii) ii
Ans:a)A blood group
b)O blood group
Explain the mechanism of sex determination in insects like Drosophila and
grasshopper.
Ans: In Drosophila sex is determined by the type of sperm, with X or with Y; fertilizing
the ovum with X. In grasshopper sex is determined by the type of sperm, with X or
without sex chromosomes, fertilizing the ovum with egg. [½ * 6= 3marks]
Explain the sex determination mechanism in humans. How is it different in birds?
Ans: Human: Females produces all gametes with X chromosomes. If the egg fuses with
a sperm with X chromosome a daughter is born. If the egg fuses with the sperm with a Y
chromosome a son is born.
[½*3 = 1½ marks]
Birds: male produces all gametes with Z chromosomes. If the sperm fuses with egg with
Z chromosome a male is born. If the sperm fuses with egg with W chromosome a female
is born. [½*3 = 1½ marks]
[1½+1½= 3marks]
Recently a girl baby has been reported to suffer from hemophilia. How is it
possible? Explain with the help of a cross.
Ans:
XhY
Father is haemophilic
=1
19
3
3
3
3
XXh
mother is a carrier
=1
XhXh
Possibility of the daughter being haemophilic. =1
[1+1+1 = 3marks]
Identify ‘a’, ‘b’, ‘c’, ‘ d’ , ‘e’ , and ‘f’ in the table given below: (2014)
NO.
SYNDROME
CAUSE
CHARACTERI
STICS
OF AFFECTED
INDIVIDUAL
Down’s
Trisomy of 21
‘a’ (i) (ii)
1.
XII BIO
3
3
SEX
MALE
/FEMALE/BOT
H
‘b’
Page 16
20
2.
‘c’
3.
Turner’s
45 with OX
‘d’
Overall
masculine
development
‘e’ (i) (ii)
‘f’
Ans: a. short statured / small round head /furrowed tongue / partially open mouth /
palm is broad / physical development retarded / psychomotor development retarded /
mental development retarded.
(any two) = ½
b. both / male and female = ½
c. klinefelter’s syndrome = ½
d. male = ½
e. sterile ovaries / rudimentary ovaries , lack of secondary sexual characters . = ½
f. female = ½(1/2x6=3)
During a monohybrid cross involving a tall pea plant with a dwarf pea
plant, the offspring populations were tall and dwarf in equal ratio. Work
out a cross to show how it is possible. (2015)
Parent
Tt (tall) X
tt (dwarf) -1
Gamete
Tt
21
XXY
T
t
t
X
Tt
t
tt
tt
3
-1
-1/2
F1 generation Tt, Tt = tall (50%) tt,tt dwarf (50%)
=½
Note: (similar cross shown in a Punnet square to be accepted)
Two independent monohybrid crosses were carried out involving a tall pea 3
plant with a dwarf pea plant. In the first cross, the offspring population had
equal number of tall and dwarf plants, whereas in the second cross it was
different. Work out the cross, and explain giving reasons for the difference
in the offspring populations.
Ans:
2nd cross
1 st cross
Parental generation
gametes
T
Tt x
t
tt
t
parental generation TT X
t
T
T
tt
t
t
F1
22
- Tt
: tt
F1 Tt
:
Tt
1:1 (50% Tall: 50% dwarf)
(Tall – 100%)
Reason – in the first cross the tall parent plant is heterozygous for the trait, in
second cross tall parent plant is homozygous for the trait = ½+ ½.
The F2 progeny of a monohybrid cross showed phenotypic and genotypic
ratio as 1:2:1, unlike that of Mendel’s monohybrid F2 ratio.with the help of
suitable example, work out a cross and explain how it is possible.
Ans: mirabilis jalapa/four O’clock plant/ Antirrhinum (majus)/ snapdragon
flower/dog flower = ½
XII BIO
Page 17
3
P generation Red (RR)
Gametes
white (rr) =1/2
R
r
F1 generation
Rr all pink =1/2
F2 generation
Gametes
R
R
r
r
RR
Rr
Rr
rr
phenotypic ratio red (1) : pink (2): white (1) =1/2
genotypic ratio : RR (1); Rr (2): rr(1)
in heterozygous condition a single dominant gene is not sufficient to produce red
colour, therefore it is a case of incomplete dominance = ½+1/2
23
a) Explain a monohybrid cross taking the seed coat colour as a trait in Pisum
sativum. Work out the cross upto F2 generation.
b) State the laws of inheritance that can be derived from such a cross.
c) How is the phenotypic ratio of F2 generation different in a dihybrid cross?
Ans: monohybrid cross
YY
Yellow
yy
green
Yy
Yellow
F2 selfing
Y
y
5
Y
YY
yellow
Yy
yellow
F1 =1/2
y
Yy
yellow
yy
green
Phenotypic ratio =3:1
Genotypic ratio = 1:2:1
24
Law of dominance
In a contrasting pair of factors one member of the pair dominates the other.
Law of segregation
Factors or allele of a pair segregate from each other such that a gamete receives
only one of the two factors.
c) phenotypic ratio of F2 in monohybrid cross is 3:1 whereas in a dihybrid cross the
phenotypic ratio is 9:3:3:1.
Describe the pattern of mechanism of inheritance of ABO blood group in human.
XII BIO
Page 18
5
25
(OR)
(a) Why is haemophilia generally observed in human males? Explain the condition
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 is carrying has developed from a zygote formed by an XX- egg
fertilized by Y-carrying sperm. Why was she advised to undergo MTP?
Ans: Human blood group is determined by glycoprotein/antigen A, and glycoprotein/
antigen B. = 1
The alleles are IA, IB and i– Hence referred to as multiple allelism.= 1
IAIA,IA i - A group
IBIB,IB i - B group
IA IB
- AB group
ii - O group = 1
In the case of A, B and O – Law of dominance is the pattern of inheritance as IA/ IB
dominant over i. = 1
In AB group both the alleles IA and IB express- it is the case of Co-dominance. = 1
[1+1+1+1+1 = 5marks]
(OR)
Ans: (a) Haemophilia is caused due to the recessive, gene on X chromosome. =½+½
Y has no allele for this/if a male is XhY then he is haemophilic/if a male inherits Xh from
the mother he is haemophilic(with a genotype XhY) = 1
If female inherits XhXh, one from the carrier mother and other from a haemophilic
father(then she can be haemophilic)= 1
(b) Embryo has (trisomy of sex chromosome) XXY/Klinefelter’s syndrome. =1
Advised MTP since the child will have the following problem:
Male with feminine traits/ like gynaecomastia/ underdeveloped tetes/sterile. = 1
[1+1+1+1+1 = 5marks]
a) A true breeding homozygous pea plant with green pods and axial flowers as
dominant characters is crossed with a recessive homozygous pea plant with yellow
pods and terminal flowers. Work out the cross up to F2 generation giving the
phenotypic ratio of F1 and F2 generation respectively.
(b) State the Mendelian principle which can be derived from such a cross and not
from monohybrid cross.
Ans: (a) Parent: GGAA
Green axial
Gamete:
GA
F1 -
ggaa = ½
yellow terminal
ga
GgAa GgAa (hybrid) green axial = ½
Selfing
XII BIO
Page 19
5
F2: [½+½+½+½+½=2½marks]
GA
Ga
gA
ga
GA
GGAA
Green
axial
GGAa
GgAA
Green axial Green
axial
GgAa
Green axial
Ga
GGAa
GGaa
Green
terminal
Ggaa
Green
terminal
Green
axial
GgAa
Green
axial
gA
GgAA
Green
axial
GgAa
ggAA
Green axial yellow
axial
ggAa
yellow
axial
ga
GgAa
Ggaa
Green
terminal
ggaa
Yellow
terminal
Green
axial
ggAa
yellow
axial
Phenotypes: green : green : yellow : yellow
axial terminal axial terminal
Phenotypes: 9 : 3
: 3
:
1
ratio
26
(b) Law of independent assortment, when two pairs of traits are combined in a
hybrid segregation of one pair of character is independent of the other pair of
character.= 1[½+½+2½+1=5marks]
(a) State the law of independent assortment.
(b) using punnet square demonstrate the law of independent assortment in a
dihybrid cross involving two heterozygous parents.
Ans: (a) When two pairs of traits are combined in a hybrid segregation of one pair of
character is independent of the other pair of character. = 1
(b) Parent
XII BIO
Page 20
5
27
Both parental type and recombinant types are observed to show that genes for the colour
and genes for the shape of seeds segregate independently during gametes formation. =1
[1+3+1= 5marks]
A particular garden pea plant produces only violet flowers.
a) Is it homozygous dominant for the trait or heterozygous?
b) How would you ensure its genotype? explain with the help of crosses.
Ans: a) it could be homozygous dominant , heterozygous dominant =1/2+1/2=1
b) by test cross = ½
if parent homozygous:
if the parent heterozygous
VV x
vv =1/2
Vv
x
vv
=½
(violet) (white)
(violet)
(white)
5
Gamete
V
v
=1/2
V
Vv =1/2
(All violet)
100%
28
v
v
Vv
(violet)
50%
=1/2
vv
(white)
50%
=1/2
=1/2
(1/2x10=5)
5
(a) You are given tall pea plants with yellow seeds whose genotypes are unknown.
How would you find the genotype of these plants? Explain with the help of crosses.
(b) Identify a, b and c in the table given below:
S.No
Pattern of Inheritance
Monohybrid F1, phenotypic expression
.
1.
Co dominance
a
2.
XII BIO
b
The progeny resembles only one of the
parent
Page 21
3.
Incomplete dominance
c
Ans: (a) Test cross =½
TTYY X ttyy
Cross I
TY
=
ty
TtYy (tall yellow) =
Cross II
TY
TtYy
TtYy
Ty
Ttyy
1
tY
ttYy
:
X
1
:
TtYy
=
ty
ty
ttyy
+
1 : 1
(b) a = Equal expression for both
b = Dominance
c = Intermediate = *3
[3 +1
29
30
= 5marks]
a)Difference between dominance and Co-dominance ,
(b)Explain co-dominance taking an example of human blood groups in the
population.
(a) Dominance-expression of only one characteristic in the heterozygous condition,
co-dominance in a heterozygote both alleles themselves independently. 1+1 =2
(b) Blood groups are controlled by 3 alleles of a single gene I(IA,lB,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
Write the conclusion drawn by Griffith at the end of his experiment with
Streptococcus pneumonia.
How did O. Avery, C. MacLeod and M. McCarty prove that DNA was the genetic
material? Explain
a) Transformation of R strain by the heat killed S strain indicated that some
transforming principle / genetic material have been transferred from the heat killed S
strain to the R strain, to make it virulent.
b) They purified, bio chemical i.e. proteins DNARNA etc, from heat killed S cells:
protease + and R Nase did not affect transformation, D Nase inhibited transformation,
XII BIO
Page 22
5
5
31
32
33
DNA alone from S caused transformation
Explain the mechanism of sex- determination in humans.
Differentiate between male heterogamety and female heterogamety with the help of
an example of each.
Ans: In humans female is XX and male is YY
Male produces 2 types of gametes (sperms ) – X and Y, female produces only
one type of gamete (ovum) –X, equal changes of X ova being fertilized by an X sperm
or Y sperm, resulting in XX female or XY male.
MALE HETEROGAMETY: example: XY grasshopper XO/ drosophila XY /human
XY, male produces 2 type of gametes.
FEMALE HETEROGAMETY: example: birds ZW, female produces 2 types of
gametes.
Work out a typical Mendelian dihybrid cross and state the law that he
derived from it. (2014)
Ans. Law of Independent Assortment ½, : it states that when two pairs
traits are combined in a hybrid, segregation of one pair of character is
independent of thee other pair of characters.=1
a) Why are thalassemia and haemophilia categorized as Mendelian
disorders? Write the symptoms of these diseases. Explain their pattern of
inheritance in humans.
b) Write the genotypes of the normal parents producing a hemophilic son.
a) both are caused due to alteration / mutation, in a single gene and follow
Mendelian pattern of inheritance = 1/2x2
symptoms:
thalassemia-anaemia (caused due to defective/abnormal Hb)
haemophilia – non stop bleeding even in minor injury= 1/2x2
pattern of inheritance:
thalassemia autosomal recessive inheritance pattern , inherited from
XII BIO
Page 23
5
5
5
34
heterozygous/ parent carrier= 1/2x2
haemophilia –X linked recessive inheritance, inherited from a haemophilic father/
carrier mother ( females are rarely haemophilic)=1/2x2
c) XhX Mother = ½
d) XY –Father = 1/2
a) Why are colorblindness and thalassemia categorized as Mendelian
disorders? Write the symptoms of these diseases seen in people
suffering from them.
b) About 8% of human male population suffers from colourblindness
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 single gene, and follow
Mendelian inheritance = 1/2x2
Colourblindness – unable to discriminate between red and green colours =1
Thallassemia – ( formation of abnormal haemoglobin resulting in ) Anaemia =1
c) It is due to recessive mutation in the X chromosomes = ½
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 the defective gene/ XcXc, whereas male will be colour blind if they are
XcY/ heterozygous=1/2X3
//
XXc x XcY
XXc
x XY =1/2
X
XXc
Xc
Xc
XcXc
Y
XY
X Xc
XcY
Carrier color blind normal color blind
X
Y
XX
X cX
XY
normal
carrier
normal
XcY =1
color blind
i)
a) How are Mendelian inheritance, polygenic inheritance and
pleiotropy different from each other?
b) Explain polygenic inheritance pattern with help of suitable example.
Ans:
Mendelian inheritance
Polygenic
Pleiotropy
inheritance
One gene controls one
Two or more genes One gene controls the
trait/character/phenotype influence the
expression of more than
expression of one
one
trait/ character/
trait/character/phenotype
phenotype
35
b)Human height/skin colour are examples of polygenic inheritance, height trait is
controlled by at least three gene pairs, additive effect of each allele contributes to
the phenotypic expression of the trait, more the dominant alleles more
pronounced is the phenotypic expression/ more the recessive alleles less
XII BIO
5
Page 24
5
pronounced is the phenotypic expression= 1/2x4
(skin colour may be accepted in place of height)
1
2
3
4
5
6
7
Chap 6
MOLECULAR BASIS OF INHERITANCE
Mention the contribution of genetic maps in human project?
Ans: Sequence of gene, DNA finger printing, tracing human history, and
chromosomal location for disease associated sequence. (Any one) = 1mark
Name the enzyme involved in the continuous replication of DNA strand. Mention
the polarity of template strand.
Ans: (DNA dependent) DNA polymerase. 3’ -> 5’ = 1mark
When and at what end the ‘tailing’ of hnRNA does takes place?
Ans: during conversion of hnRNA into functional mRNA, at 3’ end. = 1marks
State the role of transposons in silencing of m RNA in eukaryotic cells?2013
ans] transposons or mobile genetics elements in viruses are the sources of the
Complementary ds RNAs, that in turn binds/silences specific m RNA/causes RNA
interference of the parasite=1/2+1/2.
What is a cistron? (2015)
A segment of DNA , coding for a polypeptide=1/2+1/2
Write the full form of VNTR. How is VNTR different from ‘Probe’?
Ans: VNTR: Variable Number of Tandem Repeats. =1 mark
Probe- is labeled/radioactive (single stranded hybridized DNA fragments). = 1 mark
Explain the dual function of AUG codon. Give the sequence of bases it is
transcribed from and its anticodon.
Ans: Codes for Methonine and is an initiation codon. = 1mark
The sequence of bases from which it is transcribed is TAC. = ½
Its anticodon is UAC. = ½
[1 + ½ + ½ = 2marks]
8
9
1
1
1
1
1
2
2
2
(a ) Name the molecule ‘X’ synthesized by ‘i’ gene. How does this molecule
get inactivated?
(b) Which one of the structural genes codes for β-galactosidase?
(c) When will the transcription of this gene stop?
Ans: (a) Repressor = ½
Lactose (inducer) binds with the repressor molecule. = ½
(b) Z gene. = ½
(c) When all the lactose molecule are consumed/ repressor becomes free to bind
with the operator. = ½
[½+½+½+½ = 2marks]
Study the given portion of double stranded polynucleotide chain carefully. Identify
a, b, c and the 5’ end of the chain.
XII BIO
Page 25
2
10
11
12
13
Ans: a = hydrogen bond = ½
b = Adenine/ Thymine = ½
d = 5’ = ½
c = deoxyribose sugar = ½
[½ + ½ + ½ + ½ + ½ = 2marks]
State the difference between the structural genes in a Transcription Unit of
Prokaryotes and Eukaryotes. (2014)
Ans: Prokarotes
Eukaryotes
1. Polycistronic
Monocistronic
2. NO split genes / not interrupted Coding sequence
Split genes / interrupted
Coding sequences / exon and intron
4x1/2=2
Explain the two factors responsible for conferring stability to double helix structure
of DNA.
Presence of H bonds, the plane of one base pair stacks over the other, complementarity,
presence of thymine in place of uracil. Any two = 1+1
(i) Name the enzyme that catalyses the transcription of hnRNA.
(ii) Why does the hnRNA need to undergo change? List the changes hnRNA
undergoes and where in the cell such changes take place.
Ans: (i) RNA polymerase II = 1mark
(ii) Has (non functional) introns. = ½
(Methyl guanosine tri-phosphate is added to 5’ end) capping, tailing (Poly A
tail at 3’ end added), splicing (introns are removed and exons are joined). = ½*3 [1+2=
3mark]
i) Name the scientist who called t-RNA an adapter.
(ii) Draw a clover leaf structure of t-RNA showing the following:
(a) Tyrosine attached to its amino acid site.
(b) Anticodon for this amino acid in its correct site
(iii) What does the actual structure of t-RNA look like?
Ans: (i) Francis Crick. = 1
(ii)
XII BIO
Page 26
2
2
3
3
14
15
(iii) tRNA looks like inverted L. = 1 [1+1+1 = 3marks]
Answer the following based on Meselson and Stahl’s experiment:
(a) Write the name of chemical substance used as a source of nitrogen in the
exp. by them.
(b) Why did the scientists synthesise the light and the heavy DNA molecule
in the organism used in the exp.
(c) How did the scientist make it possible to distinguish the heavy DNA
molecule from the light DNA molecule? Explain.
(d) Write the conclusion the scientist arrived at after completing the exp.
Ans: (a) NH4Cl = 1
(b) To check whether DNA replication was semi-conservative. = ½
(c) Distinguish them by centrifugation in the cesium chloride density gradient. =
½
(d) DNA replicated semi conservatively. = 1
[1+1+1=3marks]
In a series of experiment with Streptococcus and mice F. Griffith concluded that Rstrain bacteria had been transformed. Explain.
Ans: S strain bacteria when injected – mice die; R- mice live; heat killed S – mice
live; heat killed S + live R – mice die; recovered living S from dead mice; R strain
bacteria have been transformed to S strain by the genetic material of heat killed S
strain.
16
3
3
3
(a) What is the diagram representing?
(b) Name the parts a, b and c.
(c) In the eukaryotes the DNA molecules are organized within the nucleus.
Ans: (a) Nuclesome. = ½
(b) a- histone octamer = ½ ; b- DNA = ½; c- H1 histone. = ½
(c)In a bacteria DNA is nucleoid, is organized in large loops held by proteins. =
1mark
XII BIO
Page 27
17
18
19
20
21
[½ + ½ + ½ + ½ + 1 = 3marks]
What is the satellite genome? Explain their role in DNA fingerprinting.
Ans: DNA sequences which are repeated many times, show a high dergree of
polymorphism and form a bulk of DNA in a genome, called satellite DNA. = 1½
marks
DNA from every tissue from an individual shows the same degree of polymorphism
and is heritable, hence very useful in DNA fingerprinting. 1½ marks
[1½ +
1½ = 3marks]
a) Draw a schematic diagram of the structure of a transcription unit and show the
following in it:
(i)
Direction in which the transcription occurs
(ii)
Polarity of the two strands involved
(iii)
Template strand
(iv)
Terminator gene
(b) Mention the function of promoter gene in transcription.
1. (Ans: (a)
3
3
2marks
(b) Promoter – provides binding site for RNA polymerase/ initiates transcription. =1
marks.
[2+1 = 3marks]
(a) In human genome which one of the chromosomes has the most genes and which 3
one has the fewest?
(b) Scientists have identified about 1.4 million single nucleotides
polymorphism in human genome. How is the information of their existence
going to help the scientist?
Ans: (a) Chromosome 1 has most genes and Y has fewest. =1 mark
(b) This information promises to revolutionize the processes of finding chromosomal
locations for disease associated sequence. = 1marks
c) Tracing human history. = 1marks
[1 +1+1 = 3marks]
3
List the salient features of DNA double helix structures.
Ans: DNA helix made up of two polynucleotide chains, each constituted by sugar
phosphate bases, the chains are antiparellel in polarity (5’-----.> 3’ and 3’ ….>
5’), the bases are linked with H bonds, adenine pairs with thymine with two H
bonds while guanine pairs with cytosine with three H bonds, coiling of the chain
are in right handed fashion, pitch of the helix is 3.4 nm and there are 10bps per
turn, the plane of one base pair stacks over the other in a double helix. ( any six)
1/2x6 = 3
3
How are the structural genes activated in the lac operon in E.coli?
Lactose acts as the inducer , binds with repressor protein, frees operator gene, RNA
polymerase freely moves over the structural genes, transcribing Lac mRNA, which in
turn produces the enzyme responsible for digestion of lactose.= 1/2x6=3
XII BIO
Page 28
22
In a maternity clinic, for some reasons the authorities are not available are not able
to hand over the two new-born to their respective real parents. Name and describe
the technique that you would suggest to sort out the matter.
Ans. DNA finger printing=1
The steps include:
Isolation of DNA , Digestion of DNA with restriction endonuclease, separation of DNA
fragments by gel electrophoresis- southern blotting, hybridization by using lalabeled
VNTR probe- visualization BY AUTO RADIOGRAPHY = ½ x4
3
23
Explain DNA polymorphism as the basis of genetic mapping of human genome.
State the role of VNTR in DNA fingerprinting?
Ans. Polymorphism is variation at the genetic level arising due to mutation = ½ If an
inheritable mutation is observed at high frequency it is a DNA polymorphism = ½
3
Information on polymorphism of restriction endonuclease recognition sites and some
repetitive DNA sequences known as microsatellites’ were used for gene mapping= ½ A
satellite DNA that shows a very high degree of polymorphism is called
VNTR.(minisatellite) = 1
b) VNTR is used as probe.
24
25
26
[3 marks]
a) write the specific features of genetic code AUG .
(b) Genetic code can be universal and degenerate .write about them, giving one
example of each.
(c) Explain aminoacylation of the tRNA.
Ans. a) AUG-starting codon, codes for methionine ½ + ½ = 1
(b) Universal-UUU codes for phenylalanine, in all organism
Degenerate –many codons may code for same amino acid eg.(from table)---UUU/UUcboth for phenylalanine
(c) Amino acid is activated in the presence of ATP = 1 , linked to their
cognate tRNA=1
Explain the significance of satellite DNA fingerprinting technique.
1) They do not code for any proteins
2) They farm large part of human genome
3) They show high degree of polymorphism/ specific to each individual
A burglar in a huff forgot to wipe off his blood – strains from the place of crime
where he was involved in a theft and fight. Name the technique which can help in
identifying the burglar from the blood stains. Describe the technique?
Ans. DNA finger printing=1
The steps includes –
Isolation of DNA , Digestion of DNA with restriction endonuclease, separation of DNA
fragments by gel electrophoresis-Southern blotting, Hybridization by using a labeled
VNTR probe- visualization by auto radiography = ½ x4 [ ½ x 6=3]
“A very small sample of tissue or even a drop of blood can help to
determine paternity”. Provide a scientific explanation to substantiate the
statement.
Ans: 1. DNA from all cells of an individual shows the same degree of
polymorphism and therefore becomes a useful identification tool = 1
2.polymorphs are heritable and the child inherits 50% of the chromosome
from each parent =1
XII BIO
Page 29
3
3
3
27
28
29
3.with the help of PCR the small amount of DNA from blood can be
amplified and be used in DNA finger printing the paternity =1
Note: (1 mark only for DNA finger printing illustration )
(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) polycistronic structural gene/ three structural genes adjacent to, an operator, a
promoter, and a regulator. = 1 ½
(b)
=3
Lactose is the inducer that inactivates repressor, this allows RNA polymerase access to
the promoter, to initiate transcription of the natural gene or switch on also there= 1
How did Alfred Hershey and Martha Chase arrive at the conclusion that DNA is
the genetic material?
Ans: (refer the book for elaborate diagram)
= 5marks
(a) How did Griffith explain the transformation of R strain (non - virulent) bacteria
into S strain (virulent)?
(b) Explain how Macleod, McCarty and Avery determined the biochemical
nature of the molecule responsible for transforming R strain bacteria into S
strain bacteria.
Ans: (a)
XII BIO
Page 30
5
5
5
= 2marks
(b) They discovered that protein digesting enzymes and RNA digesting enzymes did not
affect transformation so the transforming substance was not a protein or RNA. = 1
Digestion with DNases did inhibit transformation. = 1
They conclude that DNA is the hereditary material/ by purifying the biochemical using
enzymes like proteases, RNases and DNases. = 1
[2+1+1+1=
5marks]
30
31
Describe Frederick Griffiths experiment on Streptococcus pneumonia. Discuss the
conclusion he arrived at.
Griffith observed two types of streptococcus pneumonia- smooth shiny colony
called Stype virulent with capsule, the other R type rough colony, non virulent
½+1/2=1
Experiment:
Live S type-- injected into mice- mice died= ½
Live R type-- injected into mice- no infection= ½
Heat killed type- injected into mice - no infection=1/2
Heat killed + live R type- injected into mice- mice died= ½
Griffith concluded that the genetic material of heat killed S type could transform
R type into virulent S type =1/2
He concluded that R strain bacteria had been transformed by the heat killed Sstrain bacteria=1
(5 marks)
a) Describe the various of Griffith’s experiment that led to the
conclusion of the ‘Transforming principle’.
(b)
How did the chemical nature of the ‘Transforming principle’
get established? (2014)
Ans.
(a)
Streptococcus pneumonia
S - Strain ‘inject into mice’ mice die.
R - Strain ‘inject into mice’ mice alive.
S - Strain ( heat killed) ‘inject into mice’ mice alive.
R - Strain (alive) + S (heat killed) strain ‘inject into mice’ mice
die.
R – Strain transformed into virulent.
(b)
Purified biochemicals (protein , DNA, RNA) from heat killed
S strain Treated with protease – did not affect transformation.
Treated with RNase – did not affect transformation.
Treated with DNase – transformation affected.(10X1/2=5)
OR
Describe how the lac operon operates, both in the presence and absence of
an inducer in E.coli. (2014)
Ans.
i. structural gene zya
ii. operator
XII BIO
Page 31
5
5
iii. i
iv. repressor
v. binding
vi. operon shut
vii. inducer
viii. inducer + binding
ix. operator free
x. enzymes / operator. (10X1/2=5)
32
5
Describe the Hershey and Chase experiment. Write the conclusion drawn by the
scientists after their experiment. (2014)
Ans. The Hershey – Chase experiment
Virus/ phage labelled with S -35, virus/ phase labelled with P -32 ,
Separately infected E.coli / bacteria growing in two different culture
media. infection proceeds/ for few generations, viral coats
removed from
the bacteria by agitating/blending, virus particles
separated from
bacteria by centrifugation/spinning ,radioactive S 35 of the viral coat
was detected in the supernatant, as coat did not enter the cell, radioactive P-32 was
detected in the bacterial cell.
Conclusion : DNA passes from virus to bacteria not the protein ,DNA is the
genetic material. ½+1/2
XII BIO
Page 32
33
34
5
How do m-RNA, t- RNA and ribosome help in the process of translation?
mRNA provides a template, with codons for specific amino acids to be linked to form a
polypeptide/ protein = ½+ ½
tRNA brings amino acid to the ribosomes, reads the genetic code with the help of its anti
– codons, initiator tRNA is responsible for starting polypeptide formation in the
ribosomes, tRNAs are specific for each amino acid = 1/2x4
Ribosome- (cellular factories for protein synthesis) its smaller sub unit binds with
mRNA to initiate protein synthesis at the start codon/AUG, in its larger sub unit there are
two sites present which brings two amino acids close to each other helping them to form
peptide bond, ribosome move from codon to codon along mRNA, amino acids are added
one by one to formpolypeptide/protein= 1/2x4
5
Explain the process of transcription in prokaryotes. How is the process
different in eukaryotes?
Ans: initiation : DNA dependent RNA polymerase associates with the
initiation factor /σ factor , and binds to the promoter site of DNA thus
initiates transcription= 1/2X3
Elongation : the RNA polymerase using nucleoside tri phosphates , polymerises in
a template dependent fashion in 5’ to 3’ direction following the rule of
complimentarity = 1/2X3
Termination : the enzyme associates with rho (ρ) factor at the terminator region
and both the enzyme and the newly formed /nascent RNA fall off from the DNA=
½X2
NOTE: ( Self explanatory diagrams with correct labeling may be accepted)
Difference:
i)
There are three different types of RNA polymerases in the nucleus of
eukaryotes(polymerizing the three different types of RNAmolecules)
but only one in prokaryotes
Primary transcripts (hnRNA/precussor mRNA) undergoes splicing capping and
tailing to give rise to functional RNA/mRNA(that moves out of the nucleus).the
processing is absent in prokaryotes= 1/2X 2
XII BIO
Page 33
a) Draw a labeled diagram of a “replicating fork” showing the polarity.
Why does DNA replication occur within such ‘fork’?
b) Name two enzymes involved in the process of DNA replication, along
with their properties.
35
a)
5’
5
3’
Template DNA (parental
strands)
3’
continuous
synthesis
3’
5’
5’
newly synthesized
strands
discontinuous synthesis
5’
3’
Correct diagram with polarity of parental strands and any other 3 labels = 1/2x4
Since two strands of DNA cannot be separated in its entire length due to very high
energy requirement/ high amount of energy is required to break the hydrogen
bonds holding the two strands the replication occurs in small opening of DNA
strands called the Replication fork. =1
b)1)DNA dependent DNA polymerase, adds nucleotides only in 5’ to 3’
directions/are very fast =1/2x2
2)DNAligase , joins the discontinuously synthesized DNA fragments during
1
2
3
4
5
replication = 1/2X2
Chap 7
EVOLUTION
Name the common ancestor of the great apes and man? (1 mark )
Ans: Dryopithecus/Ramapithecus
State the significance of coelacanth in evolution.
Ancestor of amphibians
=1
Comment on the similarity between the wings of a cockroach and the wing of a
bird. What do you infer from the above, with reference to evolution?
Similar in function/ analogous organs = ½ convergent evolution = 1/2
Why are analogous structures a result of convergent evolution? (2014)
Analogous structures are not anatomically similar/ do not have common ancestors and
evolving for similar function in the same habitat = ½+1/2
Branching descent and natural selection are the two key concepts of Darwinian
Theory of Evolution. Explain each concept with the help of suitable examples.
Ans: Branching descent
Different species descending from the common ancestor-get adapted in different
habitats = ½ e.g. Darwin’s finches- varieties of finches arose from grain eaters/
Australian marsupials- evolved
Natural selection:
A process in which heritable Variations enable better survival of a apices to reproduce in
large number=1/2
XII BIO
Page 34
1
1
1
1
2
6
7
8
9
10
e.g. White moth surviving before the industrial revolution and black moth surviving
after industrial revolution/Long necked giraffe survived/DDT resistant mosquito
survive(Any suitable example ) = ½+ 1/2 = 1.
(a) How does the Hardy Weinberg’s expression (P2 +2pq+q2 = 1 ) explain that
genetic equilibrium is maintained in the population?
(b)
List any two factors that can disturb the genetic equilibrium.
Ans.(a)
Gene frequencies in a population are stable, constant from generation, until
some change in frequency happens, due to some factor= ½ x ¼
(b)
Gene migration/ gene flow/gene drift/mutation/genetic
recombination/natural selection (Any two) = ½ x 2
According to Hardy- Weinberg’s principle the allele frequency of a population
remains constant. How do you interpret the change of frequency of alleles in a
population?
Ans : It indicates gene migration/ gene flow/gene drift/mutation/genetic
recombination/natural selection leading to evolution
(a) Select the homologous structures from the combinations given below:
(i) Forelimbs of whales and bats
(ii) Tuber of potato and sweet potato
(iii) Eyes of octopus and mammals
(iv) Thorns of Bougainvillea and tendrils of Cucurbita
(a) State the kind of evolution they represent.
i)
Forelimbs of whales and bats = ½
iii)
Thorns of Bougainvillea and tendrils of cucurbits= ½
b)divergent evolution=1
(a) Select the analogous structures from the combinations given below:
(i) Forelimbs of whales and bats
(ii) Eyes of octopus and mammals
(iii) Tuber of potato and sweet potato
(iv) Thorns of Bougainvillea and tendrils of Cucurbita
b)State the kind of evolution they represent.
Ans : (ii) Eyes of octopus and mammals= ½
(iii) Tuber of potato and sweet potato= ½
b) convergent evolution=1
(a) Select two pairs from the following which exhibit divergent evolution:
b)give reasons for your answer.
(i) Forelimbs of cheetah and mammals
(ii) flippers of dolphins and penguins
(iii) wings of butterflies and birds
(iv) Forelimbs of whales and mammals
Ans : a) i) & iv)= ½+½=1
b)Having similar anatomical structure/ origin, but performing different
functions=1/2+1/2=1
XII BIO
Page 35
2
2
2
2
2
11
12
13
14
15
16
3
(a) Write your observation 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) From the original seed eating features, many other forms with altered beaks
arose, enabling them to become insectivorous and vegetarian finches= ½ x 3
(b) The process of evolution of different species in a given geographical area
starting from one point, and literally radiating to other areas of geography(habitats), is
called adaptive radiation= ½ x3
How does industrial melanism support Darwin’s theory of natural selection?
Explain.
Before industrial revolution the environment was unpolluted the lichens on the barks of
trees – pale white winged moths could easily camouflage, while the dark winged were
spotted out by the birds for food – hence they could not survive. After industrial
revolution the lichens became dark. This favoured the dark winged moths while the
white winged were picked by birds the population of the former increases (naturally
selected) =1/2x6=3
Mention the contribution of S.L.Millers experiment on origin of life?2013
ans]created similar environment[high temp,volcanic stroms, reducing atm] in a lab yhat
existed before the life originated, by creating electric discharge in a closed flask
containing ch4,nh3 and water vapour at 800`c, observed formation of amino acids,
proved theory on chemical evolution of living molecules coming from on living organic
molecules = ½*4=2
diagram=1 ½ + ½
Explain the increase in the numbers of melanic (dark winged) moths in the urban
areas of post- industrialization period in England.
Ans. during the post industrialization period the tree trunks became dark due to smoke
and soot, predators spot albinic/ white moths against dark background, so dark moths
survive, due to Camouflage/ hide against the dark background =1+ ½ + 1 + ½ =3
(a) Explain adaptive radiation with the help of suitable examples. (2014)
(b) Cite an example where more than one adaptive radiation have occurred in an
isolated geographical area .Name the type of evolution your example depicts and
state why it is so named?
Ans: (a) Darwin finches / black birds (on Galapagos islands ) , evolved from
original seed eating features , into insectivorous and vegetarian features in different
habitat / islands 1/2x3=1+ 1/2
(b) Australian marsupials and placental mammals.1/2
Convergent evolution, more than one adaptive radiation occurred in isolated
geographical area.= ½+1/2
What does the following equation represent? Explain.
XII BIO
Page 36
3
3
3
3
3
1
2
3
4
5
6
7
8
9
10
11
p2 +2pq+q2=1
Hardy weinberg’s Principle/ allele frequencies in a population are stable and is
constant from generation to generation, 1 represents stable allelic frequency in a
population, indicating no evolution occuring, p2 frequency of homozygous
dominant/AA, 2pq frequency of heterozygous/Aa, q2 frequency of homozygous
recessive/aa=1/2x6
Note : ( if AA, Aa,aa have been indicated using any other alphabet correctly , that
can be accepted)
CHAP 8
HUMAN HEALTH AND DISEASE
Name the type of cells the AIDS virus enters into after getting in the human body.
Ans: macrophage/ Helper-T-cells. = 1mark]
A boy of 10years had chicken pox. He is not expected to have the same disease for
the rest of his life. Mention how is it possible.
Ans: Antibodies produced during the first infection; result in the memory of the first
encounter to protest the body in future.
[½+½ = 1mark
What type of virus causes AIDS? Name its genome material.
Ans: Retrovirus, RNA [½+½ = 1mark]
State two different roles of spleen in the human body.
Spleen stores lymphocytes, it filters microbes, acts as a reservoir to store erythrocytes(
any two) ½=1/2=1
when does human body elicit an anamnestic response?2013
ans]second encounter with pathogen/secondary response
Why is secondary immune response more intense than the primary immune
response in humans?(2014)
Ans : Body will have memory of the First Encounter / Presence of antibodies developed
during primary immune response .
Why is Gambusia introduced into drains and ponds?(2014)
To feed on mosquitoes’ larvae/ to eliminate the vectors responsible for causing malaria.
Retroviruses have no DNA. However, the DNA of the infected host cell does
possess viral DNA. How is it possible?(2015)
Reverse transcription of viral RNA into viral DNA, then integrates/ incorporates
with the host DNA = ½+1/2
List the two types of immunity a human baby is born with. Explain the difference
between the two types.
Ans: Innate immunity and Acquired/ Passive immunity. =1
Innate immunity: Non-specific type of defense, providing barriers to the entry of
antigens. = ½
Acquired immunity: specific type of defense, foetus receiving the antibodies from the
mother through the placenta. = ½
[1+½+½= 2marks]
Why is tobacco smoking associated with rise in blood pressure and emphysema?
Explain.
Ans: tobacco has nicotine that stimulates the release of adrenaline/non adrenaline to
raise blood pressure. = 1
Smoking tobacco releases CO which reduces the concentration of haem bound oxygen.
This causes emphysema. = 1
[1+1= 2marks]
Define the term ‘health’. Mention any two ways of maintaining it.
(OR)
XII BIO
Page 37
1
1
1
1
1
1
1
1
2
2
2
12
13
14
15
16
17
18
Why does a doctor administer tetanus antitoxin and not a tetanus vaccine to a child
injured in a roadside accident with a bleeding wound? Explain.
Ans: State of complete physical, mental and social well being. =1
Balance diet, personal hygiene, regular exercise. = ½+½
[1+½+½ = 2marks]
(OR)
Tetanus is a deadly disease requiring a quick immune response, so preformed
antibodies are injected directly. [1+1 = 2marks]
Name the opioid drug and its source plant. How does the drug affect the human
body?
Ans: Heroin (smack)/ morphine = ½
Poppy plant/Papaversomniferum. = ½
Binds with opioid receptor in CNS/ gastro intestinal tract, and slows down the system/
depressant. =½+½
[½+½+1= 2marks]
(a)Name the lymphoid organ in human 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. =1
(b) Secondary lymphoid organ/spleen/lymph nodes/tonsils/ Payer’s patches of small
intestine/appendix. =½
They trap the microbes/they activate the lymphocytes/they trap the antigens. =½
[1+½+½ = 2marks]
List the specific symptoms of typhoid. Name its causative agent.
Ans: Sustained high fever(300C to 400C), weakness, stomach pain, constipation,
headache, loss of appetite.(any three)= 1½
Causative agent: Salmonella typhi= ½
[1½+½ = 2marks]
Name the type of cells the AIDS virus first enters into after getting inside the
human body. Explain the sequence of events that the virus undergoes within these
cells to increase their progeny.
Ans: Macrophage/ Helper-T-cells, the virus enters the macrophages where RNs
genome of the virus replicates to form viral DNA, with the help of the enzyme reverse
transcriptase, this viral DNA gets incorporated into host cell’s DNA and directs the
infected cells to provide virus particles. [½*4 = 2marks]
Name on plant and the addictive drug extracted from the latex. How does this drug
affect the human body?
Ans: Poppy/ Papaver somniferum= ½
Opioids/smacks/heroin/morphine/diacetylmorphine = ½
It is a depressant. = 1
[½+½+1 = 2marks]
Name the group of viruses responsible for causing AIDS in humans. why are these
viruses named so?
a) List any two ways of transmission of HIV infection in humans, other than
sexual contact.
a)retrovirus,( have RNA genome) have reverse transcriptase// carry on
RNA…>DNA ….>RNA =1/2+1/2=1
b)infected blood transfusions, sharing syringes/ needles, children born to HIV
mothers.(any two) = 1/2+1/2=1
Name any two organisms that are responsible for ringworm in humans. mention
two diagnosticsymptoms.name the specific parts of the human body where these
organisms thrive and explain why.
Microsporum/ trichophyton/ epidermophyton = ½
XII BIO
Page 38
2
2
2
2
2
2
2
19
20
21
22
23
24
25
26
27
Dry/ scaly lesions on skin/nails/scalp/intence itching (any two) =1/2
Body- groin/ between toes,thrive better in heat/ moisture/perspiration = ½+1/2=1
Name the plant source of ganja. How does it affect the body of the abuser?
Cannabis sativa/hemp plant =1
Damages cardiovascular system =1
Name the two special types of lymphocytes in humans. How do they differ in their
roles in immune response?
B lymphocytes and T lymphocytes =1
B-cells produce pathogen specific antibodies/humoral immune response =1/2
T cells helps the B-cells to produce antibodies responsible for immediate immunity =1/2
Name the bacterium that causes typhoid. Mention two diagnostic symptoms. How is
the disease transmitted to others?
Salmonella typhi =1/2
Constipation, stomach pain, headache, weakness, loss of appetite, high fever (any two) =
½+1/2=1the disease is transmitted through contaminated food/water
What is metastasis? Is it fatal? Why?
Cells sloughed off from tumours reach distant sites through blood, and wherever they get
lodged in the body they start a new tumor there,this property is called metastasis
=1/2+1/2,
It is fatal because it causes malignant tumours/cancer =1
Write the events that take place when a vaccine for any disease is introduced into
the human body?
ans: the anti bodies are produced in the body against the anti gens of the disease , which
would neutralze the antigens during actual infection , the also cause the production of the
b and t cells , that recognise the pathogen quickly on subsequent exposure = ½*4.
Why is a person with cuts and bruises following an accident administered tetanus
anti toxin? Give reason?
ans : tetanus caused by deadly bacterium, quicker response required , so preformed anti
bodies anti toxin is administered to neutralize the effect of the bacterial toxin =
List the symptoms of Ascariasis .How does a healthy person acquire this infection?
(2014)
Ans: Internal bleeding, muscular pain, anemia, blockage of intestinal passage.
(any three) =1 ½
Intake of water, vegetables / fruits / foods contaminated with eggs of the parasite. =1/2
(a) Name the causative agent of typhoid in human.
(b) Name the test administered to confirm the disease.
(c) How does the pathogen gain the entry into the human body? Write the
diagnostic symptoms and mention the body organ that gets affected in severe cases.
Ans: (a) Salmonella typhi = ½
(b) Widal test = ½
(c) Through contaminated food and water =½
Sustained high fever (300C to 400C), weakness, stomach pain, constipation,
headache, loss of appetite. (any two) = ½+½
Intestine gets affected in severe cases. =½
[½+½½+1+½ = 3marks]
Study the diagram showing replication of HIV in humans and answer the following
question:
XII BIO
Page 39
2
2
2
2
2
2
2
3
3
28
29
(i) Write the chemical nature of the coat ‘A’
(ii) Name the enzyme ‘B’ acting on ‘X’ to produce molecular ‘C’.
(iii) Mention the name of the host cell ‘D’ the HIV attacks first when it enters
into the human body.
(iv) Name the two different cells the new viruses ‘E’ subsequently attacks.
Ans: (i) Protein = ½
(ii) ‘B’ –Reverse transcriptase = ½
‘C’ – viral DNA = ½
(iii) D – macrophage = ½
(iv) Macrophage, helper T lymphocyte = ½+½
[½+1+½+1= 3marks]
Community Service department of your school plans visit to a slum area near the
school with an objective to educate the slum dwellers with respect to health and
hygiene. (2014)
(a) Why is there a need to organize such visits?
(b) Write the steps you will highlight, as a member of this department, in your
interaction with them to enable them to lead a healthy life.
Ans: (i) To create awareness about disease and their effects on the body / about
immunization / health and hygiene.=1
(ii) Disposal of waste
(iii)Control vectors.
Hygienic food and water/ fresh drinking water / balanced diet / regular exercise /
yoga.(any four) = 1/2x4=2
A heavily bleeding and bruised road accident victim was brought to a
nursing home. The doctor immediately gave him an injection to protect
him against a deadly disease?
(a) Write what did the doctor injected into the patient’s body.
(b) How do you think this injection would protect the patient against the
disease?
(c) Name the disease against which this injection was given and the kind of
XII BIO
Page 40
3
3
30
31
32
33
34
immunity it provides?
a) tetanus anti toxins/ tetanus taxoid=1
b) the preformed antibody injected, act on the pathogen immediately to provide
protection=1/2X2
c) tetanus, passive immunity=1/2X2
Mention the name of the casual organism, symptoms and the mode of transmission
of the disease Amoebiasis.
Ans: Entamoebahistolytica=1
Constipation/abdominal pain/cramps/mucous stool/stool with blood clot. (Any two)
=½+½
Houseflies’ carrier from faeces to person via food products/ contaminated water.
=1[1+1+1= 3marks]
(a) Why do the symptoms of malaria not appear immediately after the entry of
sporozoites into the human body when bitten by female Anopheles? Explain.
(b) Give the scientific name of the malaria parasite that causes malignant
malaria in human.
Ans: (a) The symptoms of fever appear only when RBCs burst, releasing haemozoin
(along with the multiple no. of the parasite), but prior to this the parasite has to
complete an asexual cycle in liver cells, no symptoms appears in the infected person
between the period the parasite enters the body till RBCs release haemozoin. =½*4
(b) Plasmodium falciparum =1
[2+1 = 3marks]
a) Write the scientific names of the two species of filarial worms causing filariasis.
(b) How do they affect the body of infected person?
(c) How does the disease spread?
Ans: (a) Wuchereriabancrofti, Wuchereriamalayi= ½+½
(b) Inflammation of the lymphatic vessels of the lower limbs/ inflammation of the
genital organs/gross deformities of the lower limbs/deformities of the genital organs.
(Any two) =½+½
(c) Through the bite of female (Culex) mosquito. 1 [1+1+1 = 3marks]
Trace the life cycle of malarial parasite in the human body when bitten by an
infected female anopheles.
Sporozoites of plasmodium gets into human blood through the bite of female anopheles
mosquito, sporozoites reproduce asexually in liver cells, then they get into red blood
cells, there they reproduce asexually and infect more blood cells, after a while they
change into gametocytes, wait to be picked up by the mosquitoes = 1/2x6=3
A person in your colony has recently been diagnosed with AIDS. People/residents in
the colony want him to leave the colony for the fear of spread of AIDS.
(a) Write your view on the situation, giving reasons.
(b)List the possible preventive measures that you would suggest to the residents of
your locality in a meeting organised by you so that they understand the situation.
(c) Write the symptoms and the causative agent of AIDS.
Ans. (a) be allowed to stay
½
Reason-not contagious/infectious through touching/hugging/eating together etc. ½
(b) Educate the residents about the cause of spread of this disease emphasising that
i.
It is caused by sexual contacts with infected persons = ½
ii.
Through infected blood transfusion = ½
iii. Through infected needles = ½
Therefore the possible preventive measure could be –
To avoid the above situations/ Use condoms /Avoid disposable syringes/Avoid multiple
partners = ½
XII BIO
Page 41
3
3
3
3
5
35
1
2
3
4
5
6
(c) HIV / Human immune deficiency virus / a retrovirus = 1
(if only virus mentioned then give ½ mark )
Symptoms – about of fever, diarrhoea, weak and tired.(Any two) ½ + ½
[1+2+2=5marks]
Explain the process of replication of a retrovirus after it gains entry
into the human body. (2014)
Ans . Virus enters and infect the normal cell, viral RNA forms viral DNA with
the enzyme reverse transcriptase, viral DNA incorporates into host genome, new
viral RNA is produced by the infected cell, new viruses are produced which infect
other cells. 1x5=5
CHAP 9 STRATEGIES FOR ENHANCEMENT IN FOOD PRODUCTION
Write the name of the following
a) The most common species of bees suitable for apiculture.
b) An improved breed of chicken.
a)Apisindica/ Apismellifera/ Apisdorsata=1/2
b)leghorn/Rhode island red/ Minorca =1/2
State the use of biodiversity in modern agriculture.
Ans: A source of hybrid, GM plants, biopesticides, organic farming, biofertilisers. (Any
two) = 1 mark
Mention the strategy used to increase homozygosity in cattle for desired traits.
Ans: Inbreeding. = 1 mark
State the importance of biofortification.
Ans: Breeding of crops for improvement of nutritional quality. = ½
Higher levels of proteins/vitamins/minerals/healthier fats. (Any one) = ½ [½+½ = 1
Name any two diseases the himgiri variety of wheat is resistant to?
ans] leaf and stripe rust = ½
hill bunt = ½.
Write an alternate source of protein for animal and human nutrition.(2014)
XII BIO
Page 42
5
1
1
1
1
1
1
7
8
9
2
10
11
12
13
Single cell protein/ spirulina
(a) Why are the plants raised through micro propagation termed as
somaclones?
(b) Mention two advantages of this technique.
A) genetically identical =1
b) large no of plants in short duration, virus free plants = ½+1/2
MOET programme has helped in increasing the herd size of the desired verity of
cattle. List the steps involved in conducting the programme.
Ans: - Cow is administered hormone like FSH, super ovulation. = ½+½
Mated with selected bull or artificially inseminated. =½
Fertilized eggs are recovered and transferred to surrogate mother. = ½
[½+½+½+½ = 2marks]
Enumerate any six essentials of good, effective dairy farm management
practices.
Ans: selection of high yielding and diseases resistant breeds, well housed,
adequate water supply, maintained disease free, feeding in a scientific manner,
regular visits by veterinary doctors, regular inspection and record keeping,
cleanliness and hygiene while milking and transport (any six) = 1/2x6
What is the programme called that is involved in improving success rate of
production of desired hybrid and herd size of cattle?
b) Explain the method used for carrying this programme for cows.
Ans: multiple ovulation embryo transfer method/MOET =1/2
B) High milk yielding cow administered with FSH …..> 6-8 eggs produced-
inseminated artificially -- fertilized eggs recovered nonsurgically at 32cell stage --
transferred to surrogate mother for further growth = 1/2x5=2 1/2
Describe the technology that has successfully increased the herd size of cattle in a
short time to meet the increasing demand of growing human population.
Ans: Multiple ovulation embryo transfer technology (MOET). = ½
Cow is administered with FSH, to induce follicular maturation and super ovulation, to
produce 6 to 8 eggs, mated or artificially inseminated, fertilized eggs recovered nonsurgically, transferred to surrogate mother. = 2½
[½+2½ = 3 marks]
Scientist has succeeded in recovering healthy sugar plants from a diseased one.
(a)
Name the part of the plant used as explants by the scientist.
(b)
Describe the procedure the scientists followed to recover the healthy
plants.
(c)
Name this technology used for crop improvement.
Ans: (a) Meristem (apical, axillary) = ½
(b)Explant/ Virus free meristem is growing in nutrient medium, under aseptic
condition; tissue proliferates to form undifferentiated mass or callus, transferred to a
medium containing auxins and cytokinins. = 2 marks
(c) Tissue culture or micro propagation. = ½
[½+2+½ = 3marks]
(a) Mention the property that enables the explants to regenerate into a new plant.
(b) A banana herb is virus infected. Describe the method that will help in
obtaining healthy banana plants from this diseased plant.
Ans: (a) Totipotency. = 1
(b) Extract the disease free meristem, in vitro culture to get virus free plants. 1+1
[1+1+1 = 3 marks]
With advancement in genetics, molecular biology and tissue Culture,
new traits have been incorporated into crop plants.
XII BIO
Page 43
2
2
3
3
3
3
3
5
Explain the main steps in breeding a new variety of crop. (2014)
Ans. i. Collection of variability / germplasm collection, collection and
prevention of all different wil varieties, species and relatives of
Cultivated species / entire collection of plants. 1/2+1/2
ii. Evaluation and selection of parents to identify plant with desirable
combination of character/ pure lines are created. 1/2+1/2
iii. Cross hybridization among selected parents, cross hybridizing the two
parents to produce hybrids. 1/2+1/2
iv. Selection and testing of superior recombinants, selection among the
progeny of the hybrids that have desired character Combinations. Superior to both
the parents / self pollinated for several generations. 1/2+1/2
v. Testing, release and commercialization of new cultivars, newly selected
lines are evaluated for yield/ other agronomic traits of quality / disease resistance in
research feels followed by testing the material in farmers fields. 1/2+1/2
OR
1
2
3
4
5
a) State the objective of animal breeding.
b) List the importance and limitation of inbreeding. How the limitations can be
overcome.
(c ) Give an example of new breed each of cattle and poultry. (2014)
Ans.
(a)
Increase the yield of animal and improving the desirable
qualities of the products.
(b)
Importance :
Increases homozygosity / to evolve purline / expose harmful
recessive genes / help in accumulation of superior genes /
elimination of less desirable genes. ( any four)
Limitation
Reduces fertility and productivity / inbreeding depression.It can be overcome by
out breeding / cross breeding / out Cross / interspecific hybridization / selected animals
is to be bred with unrelated superior animals of the same breed.
(c ) Jersey / Hisardale - a new breed by crossing Bikaneri ewes and
Mirano rams (cattle) and Leghorn (poultry).
10-MICROBES IN HUMAN WELFARE
1-Which one of the following is the baker’s yeast used in fermentation?
1.
SaccharumBarberi
2.
Saccharomyces Cerevisiae
3
Sonalika
Ans:-2.Saccharomyces Cerevisiae
Mention the information that the health workers derive by measuring BOD of a
water body.
A measure of organic waste matter present in the water, greater the BOD more is its
polluting potential=1/2+1/2
Mention the role of cyanobacteria as a biofertilizer.
(a biological organism) that fixes atmospheric nitrogen =1
Name the type of association that the genus Globus exhibits with higher
plants.(2014)
Ans: Symbiosis / Mycorrhizae / Mutualism.
Why are some molecules called bioactive molecules? Give two examples of such
molecules?
Ans:-Microbes/bacteria fungus are used in their production = I
XII BIO
Page 44
1
1
1
1
2
6
7
8
9
10
11
12
13
14
e.g. Citric acid - acetic acid
Butyric acid
- lactic acid
Ethanol
- lipase
Streptokinase
- cyclosporine A(Any two)=1 = 1 =2
Many members of the genus Glomus form mycorrhiza.
a) What are called mycorrhizae?
Ans:-a) Fungal associations with roots of higher plants.
b) What are the other benefits shown by the plants having such association other
than increase in plant growth and development?
Ans:-b)Resistance to root borne pathogens ,tolerance to salinity and drought
Name the source of cyclosporine-A. How does this bioactive molecule function in
our body?
Trichodermapolysporum =1
Immunosuppressant , used in organs transplant patient = ½+1/2
Name the bacterium responsible for the large holes seen in “ swiss cheese” what are
these holes due to ?
ans: Propionibacteriumsharmanii , production of large amount of co2 = 1+1
Explain the different steps involved during primary treatment phase of
sewage.(2015)
Physical removal of particles (large and small),by filtration and sedimentation, forming
primary sludge/ sedimented solids, forming effluent (supernatant) for secondary
treatment=1/2x4
Explain the significant role of the genus Nucleopolyhedrovirus in an ecological
sensitive area. (2014)
Ans: Species specific, narrow spectrum , insecticidal application (IPM )n, no negative
impact on plants / mammals /birds / fish / even non target insects 1/2x4
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)= 1/2x4
Explain the function of “anaerobic sludge digester” in a sewage treatment
plant.
Anaerobic sludge digester has anerobic bacteria that digest the aerobic bacteria
and fungi present in the sludge =1
During the digestion these bacteria produce mixture of gases such as methane
and H2S and CO2/ biogas =1
Name the genus to which baculo viruses belong. Describe their role in the
integrated pest management program?
Nucleopolyhedrovirus =1
They control only species specific pest, do not affect non target organisms/beneficial
insects are conserved/they aid in IPM problems/ no negative impact on plants or other
animals = 1+1=2
Mention the product and its use produced by each of the microbes listed below:
(i) Streptococcus
(ii)Lactobacillus
XII BIO
Page 45
2
2
2
2
2
2
2
3
3
15
16
17
18
19
1
2
3
(iii)Saccharomyces cerevisiae
Ans:(i)Streptokinase, clot buster/dissolves clot
from blood vessels= ½ + ½
(ii)Lactic acid, coagulates milk/partial digestion of milk proteins= ½ + ½
(iii)Ethyl alcohol + CO2, ferments dough to make bread/idli= ½ + ½
How does activated sludge get produced during sewage treatment?
a) Explain how this sludge is used in biogas production.
Ans:a)when BOD of sewage water is reduced, the effluent passes to settling tank where,
bacterial flocs are allowed to sediment and this sediment is activated sludge=1/2x3
b) the major part of sludge is pumped into anaerobic sludge digester, here other
anaerobic bacteria digest the flocs,during this digestion biogas is produced=1/2x3
Describe how bio gas is generated from activated sludge. List the components of
biogas?
Ans. The activated sludge is pumped into large tanks, called anaerobic sludge digesters,
have other kinds of bacteria which grow anaerobically digest bacteria and fungi in the
sludge , this causes the gas to be generated the components are CO2 ,CH4 ,H2S (any 2
gases= ½ marks) =1/2x6=3
What are methanogens? How do they help to generate biogas?
Ans: anaerobic methane producing bacteria= 1/2x2
Methanogens generate biogas, when act on cellulose rich biowaste(anaerobically)
=1+1
State the medicinal value and the bioactive molecule produced by
Streptococcus, Monascus and Trichoderma.
Ans: Streptococcus: streptokinase, clot buster/ removed clot from the blood
vessels = ½+ ½
Monascus :statin, blood cholesterol lowering agent/ it inhibits the enzymes
responsible for synthesis of cholesterol = ½+1/2
Trichoderma: cyclosporine A, immunosuppressive agents used in organ
transplantation = ½+1/2
Explain the process of sewage water treatment before it can discharge into
natural water bodies. Why is the treatment essential? (2014)
Ans. Primary treatment – physical removal of particles, filtration/
sequential for grit/ soil and small pebbles, settled solids form primary sludge and
supernatant forms is effluent.1/2x4=2
Secondary treatment / biological treatment – Effluent passed in aeration tank and
agitated mechanically, air is pumped, vigorous growth of aerobic microbes
cosuming organic matter, BOD reduced. 1/2x4=2
Essential to control pollution in natural water bodies, to check Water borne
diseases/ pathodenic organism.1/2+1/2
11 PRINCIPLES OF BIOTECHNOLOGY
Biotechnologists refer to Agrobacterium tumefaciens as a natural genetic engineer of
plants .Give reasons to support the statement.
Ans: Can transfer gene naturally/ Can deliver a piece of TDNA/ has tumour inducing
plasmid.
3
3
3
5
1
1
How is the action of exonuclease different from that of endonuclease?
Exonuclease: removes nucleotides from the ends of DNA molecules=1/2
Endonuclease: makes cut at specific position within a DNA=1/2
Why is it not possible for an alien DNA to become a part of a chromosome
anywhere along its length and replicate normally?
Ans: Alien DNA must be linked to ori / origin of replication / site to start replication.
XII BIO
3
Page 46
1
4
5
6
7
8
9
10
11
12
13
Name the enzymes that are used for the isolation of DNA from bacterial and
fungal cells for recombinant DNA technology.
Ans: Bacteria: Lysozyme = ½ , fungi: chitinase = 1/2
What are recombinant proteins? How do bioreactors help in their production?
Ans: Any protein produced by genetically altered gene in a host .Bioreactors can be
thought of as vessels in which raw biologically converted into specific products./ A
bioreactor provides the optimal conditions for achieving the desired product by
providing optimum growth conditions.
A recombinant vector with a gene of interest inserted within the gene of α
galactosidase enzyme, is introduced into a bacterium.Explain the method that
would help in selection of recombinant colonies from non-recombinant ones
b) Why is this method of selection referred to as insertional inactivation?
Ans: bacteria is grown in a medium with chromogenic substrate, colonies formed show
blue colour – no recombinants =1/2x3=1 ½
b)gene for the enzyme is inactivated by insertion = 1/2
1
5’
GAATTC
3’
3’
CTTAAG
5’ is the palindrome of a restriction enzyme. Name the enzyme and
indicate the site where it cuts this palindrome sequence
Eco R 1 ; Between G and A
Identify the type of Bioreactor. Write one purpose for which it is used.
2
a) Simple stirred tank bioreactor
Used to produce large volumes(100-1000 litres)of specific products from transgenic animals
Mention any four methods by which the rDNA can be transferred in to the host.
A) Heat shock/ Traditional method
B) Microinjection
C) Gene gun/ Biolistics
Disarmed pathogen
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 =
1+1.
Write any four ways used to introduce a desired DNA segment into a bacterial cell
in recombinant technology experiments?
ans : desired DNA segment is inserted into cloning vector and bacterial , cell can be
made to take it up after making them competent by treating them with specific
concentration of divalent cations such as calcium = 2
How does a restriction nuclease function? Explain. (2014)
Ans: Restriction nuclease cut DNA at specific sites = 1
Exonuclease cuts DNA at the ends, endonuclease cuts at specific position with DNA .
Restriction endonuclease cuts the DNA at specific pallindromic sequence.1/2+1/2
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: Restriction enzymes cut the DNA sequence a little away from the centre of the
XII BIO
Page 47
2
2
2
2
2
2
2
3
14
15
16
palindrome site but between the same two bases on the opposite strands, leaving single
stranded portions at the ends. These overhanging stretches are called sticky ends on
each strand. They form hydrogen bonds with the complementary cut counterparts,
facilitates the action of ligase enzymes to join the foreign and vector DNA strands.
Name and explain the techniques used in th separation and isolation of DNA
fragments to be used in recombinant DNA technology
.
Ans: Gel Electrophoresis.=1/2
DNA fragments of the agarose gel are negatively charged molecules
and they
move towards the anode. The separated DNA fragments can be visualised after staining
with ethidium bromide followed by exposure to UV radiation,
Separated fragments
are extracted from the gel by elution.=1/2x5=21/2
(I) Mention the number of primers required in each cycle of polymerase chain
reaction(PCR). Write the role of primers and DNA polymerase in PCR.
(II) Give the characteristic feature and source organism of the DNA polymerase
used in PCR.
Ans: (I) 2 sets of primers. DNA polymerisation/extends the primers using the
nucleotides
. (II) Thermostable/ remain active during high temp induced denaturation of DNA.
Thermusaquaticus.
How is the amplification of a gene sample of interest carried out using PCR
polymerase chain reaction?
ds DNA is denatured at high temperature to unzip them, annealing using two sets of
primers, amplification in the direction of 5’--3’ using Taq polymerase. this enzyme is
thermo stable, 1 billion times amplified in 30 cycles =1/2x6=3
labeled illustration to be evaluated in lieu of the explanation 1/2x6=3
3
3
3
17
Plasmid is a boon to biotechnology. Justify this statement quoting the production of human
insulin as an example.
Plasmid can be taken out from bacteria, tailored,can be used to insert a required gene, inserted
into the bacteria,allowed to replicate in bacteria, to produce insulin=1/2x6
3
18
What are recombinant proteins? How do bioreactors help in their production?
Ans: Any protein produced by genetically altered gene in a host .Bioreactors can be
thought of as vessels in which raw biologically converted into specific products./ A
bioreactor provides the optimal conditions for achieving the desired product by
providing optimum growth conditions.
Name and describe the technique that helps in separating the DNA Fragments
formed by the use of restriction endonuclease. (2014)
Ans. Gel electrophoresis =1/2
DNA are –vely charged, forced to move towards anode, electric field in agarose
gel matrix, separate according to their size / sieving effect, Smaller fragments move
faster and further than the larger.=1/2x5
3
Name the organism in which the vector shown 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.
iii) name and explain the role of a selectable marker in the vector shown.
3
19
20
XII BIO
Page 48
3
21
22
23
Escherichia coli/ e.coli=1
ii) ori=1
iii) ampR is the marker gene that helps in identification and elimination of the non
transformant growing in ampicillin medium/ selectively permitting the growth of the
transformant resistant to ampicillin=1
//tetR is the marker gene that helps in identification and elimination of the non
transformant growing in tetracycline medium/ selectively permitting the growth of the
transformant resistant to tetracycline=1
(A) Name the source of Taq polymerase. Explain the advantage of its use in
biotechnology
(B) Expand the name of the enzyme ADA. Why is this enzyme essential in the
human body? Suggest a gene therapy for its deficiency.
Ans: (A) Thermusaquaticus
It is a Thermostable DNA polymerase, does not get denatured and remain active during
PCR.
(B) Adenosine deaminase, this enzyme is essential for immune system to function`, its
deficiency can be cured by gene therapy, lymphocytes from the patients are extracted
and cultured , functional ADA cDNA are introduced into lymphocyte using a vector,
lymphocytes are returned to the patient.
If a desired gene is identified in an organism for some experiments, explain the
process of the following:
(I) Cutting this desired gene at specific location.
Ans:
- Identifying the restriction nuclease that recognisesthe palindromic nucleotide
sequence of the desired gene. - The restriction endonuclease inspects the DNA
sequences- finds and recognises the site.- Cuts of the double helix at the specific point- a
little away from the centre of the palindromic site- between the same two bases on the
opposite strand. - Makes the overhanging stretch single stranded portion
as a sticky end. 2
(II) Synthesis of multiple copies of this desired gene.
Ans: - By PCR/ Polymerase Chain Reaction. - Desired gene is synthesised in vitro. DNA is denatured. - Annealed using two sets of primers.
- Thermostable Taqpolymerase extends the primers using nucleotides - Amplified
fragments are ligated. 3
(I) Describe the characteristics a cloning vector must possesss.
ANS: - Should have ori/ origin of relication.1/2
- Has selectable marker, genes encoding for an antibiotic resistance/ genes encoding for
alpha- galactosidase.1
- Has cloning site/ recognition site, for the restriction enzyme to recognise. 1
ii)Why DNA cannot pass through the cell membrane? Explain. How is a bacterial
XII BIO
Page 49
5
5
5
cell made ‘competent’ to take up recombinant DNA from the medium?
Ans: DNA is a hydrophilic molecule.1
Bacterial cell is made competent by treating with specific concentration of ca
ion/divalent ions incubating them on ice heat shock for a short period and placing it back
on ice again.1 ½
24
a) With the help of diagrams show the different steps in the formation of recombinant
DNA by action of restriction endonuclease enzyme Eco RI.
b) Name the technique that is used for separating the fragments of DNA cut by restriction
endonucleases.
5
b)Gel electrophoresis=1
Chap 12
1
2
3
4
5
6
BIOTECHNOLOGY AND ITS APPLICATIONS
Name the genetically engineered human protein produced to treat emphysema.
α – 1- antitrypsin
Recently, Indian Parliament passed the Indian Patent Bill. What is the necessity of such
bills?
Prevents unauthorized exploitation of bioresources and traditional knowledge
1
State the role of C-peptide in human insulin.
Ans: C-peptide (extra stretch of polypeptide ) which makes the insulin inactive /
proinsulin is inactive because it contain C-peptide.
Why do children cured by enzyme-replacement therapy for adenosine
deaminase deficiency need periodic treatment?(2015)
As this therapy does not cure the disease completely=1
State the cause of adenosine deaminase enzyme deficiency.
Deletion of gene for adenosine deaminase.
Suggest any two possible treatments that can be given to a patient
exhibiting adenosine deaminase deficiency.
Ans:1)enzymes replacement therapy (in which functional ADA is injected)
2) bone marrow transplantation
1
XII BIO
Page 50
1
1
1
1
7
8
9
10
11
12
13
14
15
16
3) gene therapy/ culturing the lymphocytes followed by introduction of functional
ADAcDNA into it & returning it into the patient’s body. (any two) = ½+1/2
You have developed a GM organism which government organisation will approach
you to obtain clearance for its mass production? Why is such a body necessary?
Give two reasons.
GEAC, It may give undesirable effects, so the validity and safety has to be checked.
A) Why is the bacterium Thermus aquaticus useful in r-DNA technology? B)
Explain its role in PCR
A) Resistant to high temperature, B)amplification of the segment of DNA
a) Mention the cause and body system affected by ADA deficiency in humans.
b) Name the vector used for transferring ADA-DNA into the recipient cells in
humans. Name the recipient cells.
a) defective gene into producing ADA, immune system is affected =1/2+1/2
b) a retroviral vector is used, recipient cells are lymphocytes = ½+1/2
How ‘Rosie’ is considered different from a normal cow? Explain.
Rosie is a transgenic cow=1
Rosie produced human enriched milk, containing human alpha lactalbumin=1/2+1/2=1
How have transgenic animals proved to be beneficial in: (2014)
(a) Production of biological products
(b) Chemical safety testing
Ans: (a) (Rosie – transgenic cow) produced human protein / alpha lactalbumin
enriched milk, alpha -1 antitrypsin used to treat emphysema.1/2+1/2
(d) (Toxicity testing) – more sensitive to toxic substances, results obtained in
less time. 1/2+1/2
Biopiracy should be prevented. State why and how.
Biopiracy is an unauthorised exploitation of bio resources of developing/ under
developed countries- hence to be prevented=1
By developing laws to obtain proper authorisation/pay compensatory benefits=1
Why is proinsulin so called ?How is insulin different from it?
ans : proinsulin is an inactive form of insulin , containing an extra stretch called
c peptide , insulin is made up of only 2 short poly peptide chains a and b linked by
disulphide bridges , is functional = ½*4
a) Tobacco plants are damaged severely when infested with Meloidegyne
incognitia.name and explain the strategy that is adapted to stop this infestation.
b) Name the vector used for introducing the nematode specific gene in tobacco
plant.
a) Nematode specific gene introduced into host plant, produced ds RNA, RNAi initiated,
specific mRNA of the nematode silenced and parasite dies. =1/2x4=2
b)Agrobacterium tumefaciens =1
Name the insect pest that is killed by the products of cry IAcgene.explain how the
gene makes the plant resistant to the insect pest.
Boll worm=1
The gene produces crystals of insecticidal protein which is inactive protoxin,when the
boll worm eats the protoxin the alkaline pH of the gut activates it, activated protoxin
binds to the midgut epithelial cells, creates pores/causes swelling/causes lysis/kill the
worm=1/2x4
Name the source from which insulin was extracted earlier. Why is this insulin no
more in use by diabetic people? Name the technique used by Eli Lilly Company to
produce insulin
Pancreas of slaughtered cattle, pigs.
XII BIO
Page 51
2
2
2
2
2
2
2
3
3
3
17
18
19
20
21
Develops allergy.
rDNA technique
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 worms, Bacillus thuringiensis has Bt cotton genes, due to which it
produces toxic proteins that kills the pest, when the pest injects it and then toxic gets
activated due to alkaline pH of the gut, specific Bt cotton genes were isolated from
Bacillus thuringiensis and incorporated into the cotton plants to make them pest
resistant. 1/2x6 =3
How did an American company, Eli Lilly use the knowledge of r-DNA
technology to produce human insulin?
Ans: two chains of DNA sequence corresponding to A&B chains of human insulin
prepared, introduced them into plasmids of E.coli to produce separate A&
Bchains,A& B chains extracted combined by creating disulphide bonds= 1x3
Describe any three potential applications of genetically modified plants.
More tolerant to abiotic stress, less dependence on chemical pesticides, reduces
post harvest losses, increase efficiency of mineral usage by plants, enhance
nutritional value of food. Eg, Vitamin A enriched rice (any three)= 1+1+1
Name the source and the types of cry genes isolated from it for incorporation into
crops by biotechnologists. Explain how these genes brought beneficial changes in
the genetically modified crops.
Ans:- source; bacillus thuringiensis=1
Types cry genes; cry I Ac / cry II Ab / cry I Ab (any2) =1
The gene produces an insecticidal protein / toxin that kills the worm =1
Rearrange the following in the correct sequence to accomplish an
important biotechnological reactions: (2015)
a) In vitro synthesis of copies of DNA of interest
b) Chemically synthesized oligonucleotide
c) Enzyme DNA polymerase
d) Complementary region of DNA
e) Genomic DNA template
f) Nucleotides provided
g) Primers
h) Thermostable DNA – polymerase( from thermos aquaticus)
i) Denaturation of ds DNA
j) Ans: correct sequence is ie b/g  g/b c/h h/c  f d a
1
//
ai e 
1
3
3
3
1
1
a) Name the source from which insulin was extracted earlier. Why is this insulin no
more in use by diabetic people?
b) Explain process of synthesis of insulin by Eli Lilly company. Name the technique
used by the company.
XII BIO
3
b/g  g/b c/h  h/c  f d
1
1
( no marks for the wrong sequence)
22
3
Page 52
5
23
1
2
3
4
c) How is the insulin produced by human body different from the insulin produced
by the above mentioned company?
a)pancreas of slaughtered cattle and pig=1/2
b) Eli Lilly used the following procedure
prepared two DNA sequences corresponding to A and B chain of insulin,introduced
them in plasmids of E.coli, insulin chains are produced separately, extracted and
combined by creating disulphide bonds(assembled mature molecule of insulin),
=1/2x4=2
using rDNA technique=1
c)the prohormone produced in the human body has an extra stretch of C peptide=1
How is transgenic tobacco plant protected against Meloidegyneincognitia? Explain
the procedure.
Using the technique RNA interference(RNAi) transgenic tobacco plant is protected
against Meloidegyne incognitia ,using Agrobacterium, as the vectors, nematode specific
genes were introduced into the host plant(introduction of ds RNA)produces both sense,&
anti sense RNA, these 2 RNAs form ds RNA ,silences specific mRNA of nematode, no
protein synthesis/no translation, hence nematode cannot survive in tobacco plant=1/2x10
Chap 13 ORGANISM AND POPULATIONS
State Gause’s Competitive Exclusion Principle .(2014)
Ans: Two closely related species competing for same resources cannot coexist
indefinitely (the inferior will be eliminated).
Explain brood parasitism with the help of an example.
Koel is a parasitic bird has evolved the technique of laying eggs in the nest of a crow =1,
its eggs bear resemblance to those of crow =1+1=2
Describe the mutual relationship between fig tree and wasp and comment on the
phenomenon that operates in their relationship. (2014)
Ans: Wasp – helps in pollination / pollinator (specific)
Oviposition / seeds and ovules used for nourishing larva.(any two) = ½+1/2
Co evolution exists between their close specific tight relationships = 1
Construct an age pyramid which reflects an expanding growth status of human
population. (2014)
5
1
2
2
2
Ans: Post – reproductive (1/2)
Reproductive (1/2)
Pre- reproductive (1/2)
EXPANDING
5
6
Construction of pyramid = 1/2
What is mutualism? Mention any two examples where the organisms
involved are commercially exploited in agriculture.(2015)
Interaction between two species in which both are benefitted=1
i)
Rhizobium in the roots (nodules) of legumes =1/2
ii)
Mycorrhiza/ glomus with the roots of higher plants=1/2
a) List any three ways of measuring population density of a habitat.
b) Mention the essential information that can be obtained by studying the
population density of an organism.
a) by physical counting, percent cover or total biomass, from relative density, counting
XII BIO
Page 53
2
3
7
8
9
10
pug marks, counting faecal pellets (any three) 1/2x3
b) Status of habitat, whether competition for survival exists or not, whether population is
increasing or declining, natality, mortality, emigration, immigration.
The following graph represents the organismic response to certain environmental
condition (e.g. temperature)
Which one of these,’ a’ or ‘b’ depicts conformers?
What does the other line graph depict?
How do these organisms differ from each other with reference to homeostasis?
Mention the category to which humans belong.
Ans: i) a = conformers = ½
ii)response of the regulators =1/2
iii)maintain homeostasis by physiological means, others either migrate, or suspend
activities = 1/2x3
iv)regulators =1/2
Water is very essential for life. write any three features both for plants and animals
which enable them to survive in water scarce environment
Ans: plants: ephemeral mode(complete life cycle in short period)/deep tap
roots/deciduous leaves/waxy cuticle/sunken stomata/ succulence to store water/C4
pathway of photosynthesis(any three)= 1/2x3
Animals:no sweating/uricotelic/deposition of fat in sub epidermal layer/burrowing
nature/thick skin/body covered with scales (any three) 1/2x3=11/2
How do organisms cope with stressful external environmental conditions which are
localized or short duration?
Migrate temporarily from the stressful habitat to a hospitable area/ suspended
activities/form thick walled spores/form dormant seeds/hibernate during winter/aestivate
during summer/planktons diapauses (any six)=1/2x6=3
a) Write the importance of measuring the size of a population in a habitat or an
ecosystem
b) Explain with the help of ex. how the percentage cover is a more meaningful
measure of population size than mere no?
Ans. Status in the habitat / outcome of competition with other species /impact of
predator / increase or decrease in population size (any 2) =1
b) ex. Banyan tree and parthenium plants =1
on comparing one banyan tree and 200 parthenium plants population of banyan
tree in terms of no. is much low as compared to 200 parthenium plants , but a
single banyan tree provide larger cover in comparison to 200 parthenium plants
XII BIO
Page 54
3
3
3
3
11
12
13
14
15
(therefore biomass or percentage cover is more meaningful measure of
population size in this ex.)=1/2x2=1
A) Explain “birth rate” in population by taking suitable ex.
B) Write 2 characteristics which only a population shows but an individual cannot.
Ans. Birth rate-no. of births per capita per year, ex. in a pond initial population of lotus
is 20 and 8 new plants added in one year so birth rate=8/20 =0.4 offspring per organism
per year=1+1
b)sex ratio , age distribution , population density, population growth (any2)= ½ +
½
[3marks]
a)State how the constant internal environment is beneficial to organisms.
b)explain any two alternatives by which organisms can overcome stressful external
conditions.(2014)
Ans: a)permits all biochemical reactions to proceeds with maximal efficiency, enhances
fitness of species. = ½+1/2
b)i )regulation, maintaining internal environment by maintaining constant body
temperature/ osmotic concentration. = ½+1/2
ii)suspend(conform), by suspending metabolic activities through
hibernation/aestivation/diapauses. = ½+1/2
iii)migration, organisms migrate temporarily to more hospitable areas.= ½+1/2
(any two) = 1/2x4=2
How do snails, seeds, bears, zooplanktons, fungi and bacteria adapt to
condition unfavorable for their survival?
Ans: snail – aestivation =1/2
Seeds – dormancy/ suspended metabolic activities = ½
Bear – Hibernation = ½
Zooplankton –diapause / suspended development =1/2
Fungi – spore/ zygospore = ½
Bacteria – cyst/spore =1/2
Explain co – evolution with reference to parasites and their hosts. Mention
any four special adaptive features evolved in parasites for their parasitic
mode of life.
Ans: if the host evolves special mechanism for rejecting or resisting the parasite,
the parasite has to (simultaneously) evolve/co- evolve the mechanism to counter
act and neutralize them. =1
a. Parasitic adaptation in animals.
i)Loss of unnecessary sense organs in animals
ii)Presence of adhesive organs/suckers
iii)Loss of digestive system
iv)High reproductive capacity
v)Resemblance of eggs in the case of brood parasitism
b. parasitic adaptation in plants
vi.) Haustoria in cuscuta
vii.) Loss of chlorophyll
viii) loss of leaves/foliage
note: ( any four adaptations with correct reference Animal or Plant can be allotted
marks).
a) List the different attributes that a population has and not an
individual organism.
XII BIO
Page 55
3
3
3
3
5
1
2
3
4
5
6
b) What is population density? Explain any three different ways the
population density can be measured, with the help of an example
each.
a)Attributes of population
Birth rate, death rate, sex ratio, age pyramids/ age distribution (anytwo)= 1/2x2
b)Population densitynumber of individuals per unit area at a given time/period =1
1.Bio mass/ % cover, e.g Hundred Parthenium Plants and 1 huge banyan tree=
1/2X2
2.Relative Density, e.g Number of fish caught per trap from a lake = 1/2x2
3.Numbers , e.g Human population = 1/2x 2
4.Indirect estimation, e.g without actually counting / seeing them e.g tiger census
based on pugmarks and fecal pellets = 1/2X2(any three)
Chap 14
ECOSYSTEM
Write a difference between net primary productivity and gross productivity.
Gross productivity – rate of production of organic matter during photosynthesis = ½
Net primary productivity – available biomass for the consumption to heterotrophs/ GPPR=NPP=1/2
Why are green algae not likely to be found in the deepest strata of the ocean?
Ans. no light/wavelength of light at the bottom is unsuitable.
State what does ‘standing crop’ of a trophic level represent?
Ans. mass of living material [biomass], at a particular time =1/2+1/2.
State what does ‘standing crop’ of a trophic level represent?
ans. mass of living material [biomass] ,at a particular time =1/2+1/2.
List the features that make a stable biological community.
Should not show too much variations in productivity from year to year
Must be either resistant or resilient to occasional disturbance( natural/man made)
Must also be resistant to invasion by alien species.(any two)1+1
Name the pioneer and the climax species in a water body. Mention the changes
observed in the biomass and the biodiversity of the successive seral communities
developing in the water body.
Ans : pioneer species; phytoplanktons =1/2
Climax species; forest/trees = ½
There will be gradual increase in the biomass = ½
Free floating angiosperms/ rooted hydrophytes/sedges/grasses=1/2
7
8
1
1
1
1
2
2
2
Identify the type of the given ecological pyramid and give one example each of
pyramid of number and pyramid of biomass in each case.
Ans: inverted pyramid =1
Inverted pyramid of biomass in a lake – phytoplankton--- zooplankton-- fish1/2
Inverted pyramid of number – tree-- insects -- birds = 1/2
It is possible that a species may occupy more than one trophic level in the same
Ecosystem at the same time “. Explain with the help of one example?
ans : sparrow , primary consumer when eating seeds extra , secondary consumer when
XII BIO
Page 56
2
9
eating worms etc [ any other omnivore accepted ] =1+1/2 ½
Construct a pyramid of biomass starting with phytoplankton. Label three trophic
levels. Is the pyramid upright or inverted? Why?
3
fish/crustaceans/other=1/2
zooplanktons
=1/2
phytoplanktons
=1/2
10
11
12
the pyramid of biomass is generally inverted, because the biomass of fishes, and other
aquatic animals far exceeds that of phytoplanktons =1/2x3
Differentiate between two different types of pyramids of biomass with the help of
one ex of each?
Ans.In an upright pyramid the biomass decreases with each trophic level because the
biomass of the producers is more than that of the consumers=1 ex. forest ecosystem = ½
In inverted pyramid the biomass of the producers (phytoplankton) is less than that of the
consumers (fish) =1 ex. aquatic ecosystem = ½ [1+ ½ +1 ½ =3MARKS
State the function of a reservoir in a nutrient cycle. Explain the simplified model of
carbon cycle in nature. (2014)
Ans. Function : To meet the deficit which occurs due to imbalance in the
rate of influx & efflux. =1
1/2x4
‘It is often said that the pyramid of energy is always upright. On the other hand,
the pyramid of biomass can be both upright and inverted.” Explain with the help of
examples and sketches.
ANS:TC (Tertiary consumer
10 J (10%aavailable)
SC(secondary consumer
PC( primary consumer)
PP (primary producer
100 J (10%aavailable)
1000J(10%available)
1,000,000 J of sunlight
Upright pyramid of energy: e.g of any Grassland food chain depicting energy transfer at
each trophic level=1+1
XII BIO
Page 57
3
3
5
Trophic level
Upright Pyramid of biomass: e.g grassland food chain- grass rabbit foxtiger (any
other relevant example) = 1 for diagram= ½ for example
Note: ( if only two trophic levels are drawn with dry weight mentioned correctly can be
accepted)
PC
21
PP
4
1
2
3
4
5
Inverted pyramid of biomass: e.g aquatic ecosystem where small standing crop of
phytoplanktons support large standing crop of zooplanktons = 1 for diagram+ ½ for
example
CHAP 15
BIODIVERSITY AND CONSERVATION
Name the unlabeled areas ‘a’ and ‘b’ of the pie chart representing the biodiversity
of plants showing their proportionate number of species of major taxa.
A= fungi, b = angiosperm=1/2+1/2
Write the basis on which an organism occupies a space in its community / natural
surroundings?
ans : feeding relationships with other organisms food habits / trophic
India has more than 50000 strains of rice. Mention the level of biodiversity it
represents.
Genetic biodiversity
Name the type of biodiversity represented by the following;
[1] 1000 varieties of mangoes in India.
[2] Variations in terms of potency and concentration of reserpine in Rauwolfia
vomitoria growing in different regions of Himalayas.
ans: genetics diversity, in both / in 1 and 2 = ½ + ½ .
State the biodiversity in modern agriculture.
A source of hybrids, GM plants, biopesticides, organic farming, bio fertilizer, improved
varieties of plants, disease resistance plants or any other relevant use.(any two) 1
6
XII BIO
1
1
1
1
2
2
Page 58
7
8
9
10
11
The above graph shows species – area relationship. Write the equation of the curve
‘a’ and explain.
Ans: S=CAZ=1
I)
within the region, species richness increases with increasing explored area
but only up to a limit = ½
II)
relationship between species richness and area for a wide variety of taxa turns
out to be rectangular hyperbola =1/2
2
Where would you expect more species biodiversity –in tropics or in polar regions?
Give reasons in support of your answer?
ans : tropics = 1 ,
Reasons: undisturbed , less seasonal/more constant / predictable , more solar energy[any
two ] ½+1/2 .
2
“Stability of a community depends on its species richness”. Write how did David
Tillman show this experimentally?
ans : plots with more species show less year to year variation in total biomass , increased
diversity contributed to higher productivity = 1+1=2
Differentiate between in situ and ex situ approaches of conservation of biodiversity. 2
In situ
ex situ
Protection of endangered species – protection of endangered species by removing
Of plants and animals =1/2
them from the natural habitat=1/2
by protecting the natural habitat/
by placing under special care =1/2
ecosystem=1/2
List any four techniques where the principle of ex-situ conservation of
biodiversity has been employed.(2015)
Cryopreservation, in vitro fertilisation, micro propagation/ tissue culture, sperm
bank/ seed bank/ gene bank = 1/2x4
Alien species are highly invasive & are a threat to indigenous species. Substantiate
this statement with 3 examples.
Nile perch introduced into lake Victoria in east Africa led to the extinction of cichlid
fish.
Parthenium/lantana/Eichhornia invasive plants & poses a threat to indigenous species.
Introduction of African catfish aquaculture is threat to Indian catfishes.
(any three) = 1x3=3
XII BIO
Page 59
2
3
12
13
Taking one example of each of habitat loss & fragmentation, explain how the two
are responsible for biodiversity loss.
b) Explain two different ways of biodiversity conservation.
a)habitat loss: Amazon rain forest destroyed for soya bean cultivation/ for growing grass
land for grazing cattle/ colonization of pacific islands – extinction of 2000 species of
native birds =1
Fragmentation: b human activity – migratory birds & animals are affected.
b)exsitu:1/2 threatened organisms are taken out from the natural habitat and placed in
special setting with care and protected =1/2 eg: zoological park/ botanical garden =1/2
in-situ: ½ , threatened organisms are conserved in their natural habitat = ½, eg. national
parks/ biosphere reserves = 1/2
The following graph shows the species - area relationship. Answer the following
question as directed.
(2014)
3
3
(a) Name the naturalist who studied the kind relationship shown in the graph
Write the observation made by him.
(b) Write the situations discovered by the ecologists when the value of ‘Z’ ( slope of
the line) lies
(i)
(ii)
0.1 and 0.2
0.6 and 1.2
What does ‘Z’ stand for?
(c) When would the slope of the line ‘b’ become steeper?
Ans: (a) Alexander Von Humboldt. =1/2
Within a region species richness increased with increasing explored area but only
up to a limit. (1/2)
(b) i. the slopes of regression lines are similar / unaffected distribution in an area /
normal range. (1/2)
ii. The slope regression is steeper when we analyse the species area
relationship among very large areas like entire continent. (1/2)
Z (slope of the line) regression co-efficient =1/2
c) If species richness is more/0.62- 1.2= 1/2
14 Since the origin of life on the earth, there were five episodes of mass
extinction of species. (2014)
i)
How is the ‘Sixth Extinction’, presently in progress, different
from the previous episodes?
ii)
Who is mainly responsible for the ‘Sixth Extinction’’?
iii) List any four points that can help to overcome this disaster.
Ans. i) The rates are faster/accelerated/current species extinction
rate are estimated to be 100-1000 times faster than in
the pre-human times. =1/2
ii) Human activities. = ½
XII BIO
Page 60
3
15
1
2
3
4
5
6
7
8
9
iii ) a . Prevention habit loss and fragmentation
b. Checking overexploitation
c. Preventing alien species invasion
d. Preventing co- extinction
e. Conservation /Prevention of species.
(any four) = 4x1/2 = 2
[3 marks]
Explain, giving three reasons, why tropics show greatest levels of species
diversity.(2014)
I)
Tropical latitude has remained relatively undisturbed, have a long
evolutionary time for species diversification. = ½+1/2
II)
Less seasonal variation, constant and predictable environmental condition,
promotes niche specialization for greater species diversity. = ½+1/2
III)
More availability of solar energy, contributes to higher productivity. = ½+1/2
Chap 16
ENVIRONMENTAL ISSUES
Why is it desirable to use unleaded petrol in vehicles fitted with catalytic
converters?
Lead in petrol inactivates the catalyst, harmful pollutants are converted to lesser harmful
pollutants (Co2, H2O) =1/2+1/2=1
How do algal blooms affect the life in water bodies?
Ans: pollutes water/deterioration of the water quality/oxygen depletion/water becomes
toxic/BOD increases =1/2
Fish mortality/death of aquatic organism=1/2
How is snow blindness caused in humans?
High dose of UV-B radiation, inflammation of cornea=1/2+1/2
Why is the use of unleaded petrol recommended for motor vehicles equipped with
catalytic converters?
ans. lead in the petrol in activates the catalyst in the [which catalyses the conversion of
unburnt hydrocarbons into Co2 and H2o.
List two advantages of the use of unleaded petrol in automobiles as
fuel.(2015)
i)
Allows the catalytic convertor to remain active=1/2
ii)
Ii) reduces air pollution=1/2
How did Ahmed khan, plastic manufacturer from Bangalore, solve the ever
increasing problem of accumulating plastic waste?
Collected waste plastics- recycled- powdered to form polyblend, blended with bitumen,
used in road laying, increased road life by a factor of three/ more durable = 1/2x4=2
Explain the cause of algal bloom in water body. How does it affect an ecosystem?
Effluents from home/industries/chemical fertilizers/agriculture/sewage, bring or add
nutrients to the water body which enhances algal growth = ½+1/2
Effects – reduces oxygen content/BOD increases, deterioration of water quality, affects
all aquatic life forms, toxicity of water increases (any two0=1/2+1/2
Explain any three measures which will control vehicular air pollution in Indian
cities.
Use of CNG as the fuel, use of unleaded petrol, low sulphur petrol/ diesel, use of
catalytic converter in the vehicles, phasing out of old vehicles(any three) = 1x3
Ornithologists observed decline in the bird population in an area near a lake after
the setting of an industrial unit in the same area. Explain the cause responsible for
the decline observed.
Toxicants like DDT entered the trophic levels of the food chain, accumulated as it can
XII BIO
Page 61
3
1
1
1
1
1
2
2
3
3
10
neither be metabolized nor be excreted, toxicants disturb Ca metabolism, result in
thinning of egg shells, premature breaking of eggs, population declined=1/2x6=3
Particulate and gaseous pollutants along with harmless gases are released from the
thermal power plants.
i)name any two harmless gases released.
ii)name the most widely used device of removing particulate pollutants from the air
Explain how the device is used.
Ans:i)niyrogen,oxygen = ½+1/2
ii)electrostatic precipitator = ½
electrode wires at several thousand volts produce corona
3
release electrons attach dust particles & make it negatively charge =1/2
collected plates grounded & attract charged dust particles =1/2
i) State the consequence if the electrostatic precipitator of a thermal plant fails to
function.
ii) Mention any four methods by which the vehicular air pollution can be
controlled.
i)Particulate matter will pollute the air=1
ii) use of CNG/phasing out of old vehicles/use of unleaded petrol/use of low sulphur
fuel/use of catalytic converters/application of stringent pollution level norms (any four)
1+2=3
12 Two types of aquatic organisms in a lake show specific growth patterns as shown
below, in a brief period of time. The lake is adjacent to an agricultural land
extensively supplied with fertilizers. (2014)
11
3
3
A
B
13
TIME
Answer the questions based on the facts given below:
I)
Name the organisms depicting the the patterns A and B
II)
State the reason for the pattern
iv) Write the effect of the growth patterns seen above.
Ans: i) A- algae/ planktonic algae= ½
B – fish/aquatic animals = ½
ii)due to excessive loading of nutrients/fertilizers from adjacent
agriculture land resulting in increase in nutrients = 1
iii)decrease in dissolved oxygen(DO), increase in BOD, fish mortality,
unpleasant odour/Eutrophication. Any two ½+1/2
3
With the help of flow chart, show the phenomenon of biomagnifications of DDT in
aquatic food chains.
5 stages ½ mark each (1/2x5) the flow chart should show arrows in correct direction with
XII BIO
Page 62
increasing levels of DDT = (1/2)
Fish eating birds
DDT 5 ppm
Large fish(DDT
2PPM)
Small fish (DDT
0.5PPM)
Zooplankton
DDT 0.04PPM
Water (DDT
0.003 PPM)
14
With the help of a flow chart exhibit the events of eutrophication.
Ans: water in young lake is cold clear to support life
3
With time water is enriched with nutrients such as nitrogen and phosphorus by
streams draining into it.
As lake’s fertility increases plant and animal life increase/ proliferates
Organic matter begins to be deposited at the bottom of the lake Silt and organic
debris pile up and makes the lake shallower and warmer
Marsh plants develop roots and begin to fill the original lake basin
15
Eventually the lake gives way to large masses of floating plants finally converting
it into land ( natural aging)=1/2x6
5
a) What depletes ozone in stratosphere? How does it affect human life?
b) Explain biomagnifications of DDT in an aquatic food chain. How does it affect
the bird population?
a)chlorofluorocarbons (CFCs)
=1
UV (B) damages DNA causing mutation, skin cancer, inflammation of cornea, cataract,
aging of skin, snow blindness
(any two)
½+1/2=1
b)if DDT leaches from the agricultural field gets into the water body, it gets into the food
chain-- zoo planktons -- small fish -- large fish -- any fish eating bird//
concentration of DDT increases, along the food chain, reaching a high level in the top
carnivore bird=
1+ 1/2
DDT concentration disturbs ca++ metabolism- egg shells become thin --- premature
XII BIO
Page 63
breaking resulting in decline in bird population
XII BIO
1 + 1/2
Page 64