1. Studymate Solutions to CBSE Board Examination 2011-2012
Series : SMA/1 Code No. 57/1/1
Candidates must write the Code on
Roll No. the title page of the answer-book.
 Please check that this question paper contains 6 printed pages.
 Code number given on the right hand side of the question paper should be written on the title
page of the answer-book by the candidate.
 Please check that this question paper contains 30 questions.
 Please write down the Serial Number of the questions before attempting it.
 15 minutes time has been allotted to read this question paper. The question paper will be distributed
at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and
will not write any answer on the answer script during this period.
BIOLOGY
[Time allowed : 3 hours] [Maximum marks : 70]
General Instructions:
1. All questions are compulsory.
2. This questions paper consists of four Sections A, B, C and D. Section-A contains 8 questions of
one mark each, Section-B is of 10 questions of 2 marks each, Section-C is of 9 questions of three
marks each and Section-D is of 3 questions of five marks each.
3. There is no overall choice. However, an internal choice has been provided in one questions of
2 marks, one questions of 3 marks and one question of 5 marks weightage. A student has to
attempt only of the alternatives in such questions.
4. Wherever necessary, the diagrams drawn should be neat and properly labelled.
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2. STUDYmate
SECTION-A
1. Mention the unique flowering phenomenon exhibited by Strobilanthus kunthiana
(Neelakuranaji). [1]
Ans. It flowers once in 12 years.
2. How does smoking tobacoo in human lead to oxygen deficiency in their body? [1]
Ans. Smoking increases the carbon monoxide content in blood and reduces the
concentration of oxygen bound to haeme group of haemoglobin.
3. A garden pea plant (A) Produced inflated yellow pod, and another plant (B) of the
same species produced constricted green pods. Identify the dominant traits. [1]
Ans. Dominant traits are Green and inflated.
4. Why is Eichhornia crassipes nicknamed as “Terror of Bengal”? [1]
Ans. Eichhornia crassipes is called terror ‘of Bengal’ because it grows very fast in standing
water. It drains oxygen from water thus leading to death of marine life.
5. Write the location and function of the sertoli cells in humans. [1]
Ans. Sertoli cells are located in the germinal epithelium of seminiferous tubules. They
provide nourishment to developing spermotozoa.
6. Name the following:
(a) The semi-dwarf variety of wheat which is high-yielding and disease-resistant.
(b) Any one inter-specific hybrid mammal. [1]
Ans. (a) Sonalika / Kalyan sova (b) Mule
7. Write the similarity between the wing of a butterfly and the wing of a bat. What do
you infer from the above with reference to evolution? [1]
Ans. Both butter fly and bat use their wings to fly. They are anatomically different. So it is
result of convergent evolution.
8. Write what do phytophagous insects feed on.
Ans. They feed on plants
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SECTION-B
9. Draw a neat labelled sketch of a replicating fork of DNA. [2]
Ans. 5 3
Template DNA
(parental strands)
continuous Discontinuous
5
synthesis 3 synthesis
3 Newly 5
5 synthesised 3
strands
10. Where is sporopollenin present in plants? State its significance with reference to its
chemical nature? [2]
Ans. Sporopollenin is present in exine layer of wall of pollen grain.
It is an organic polymer and is resistant to oxidation and leaching.
11. (a) Highlight the role of thymus as a lymphoid organ.
(b) Name the cells that are released from the above mentioned gland. Mention how
they help in immunity. [2]
Ans. (a) In thymus immature lymphocytes differnetiate into antigen – sensitive
lymphocytes. After maturation in thymus they migrate to secondary lymphoid
organs.
(b) The cells released from thymus are called T-lymphocytes. These lymphocytes
are responsible for cell mediated immunity which defends the body against
virus, fungi and some bacteria which has entered the hosts cells. Helper T cells
stimulate B-cells to produce antibodies and killer T-cells migrate to site of
infection.
12. Explain the work carried out by Cohen and Boyer that contributed immensely in
biotechnology. [2]
Ans. Stanley Cohen and Herbert Boyer constructed first artificial recombinant DNA. They
did this by isolating the antibiotic resistant gene by cutting out a piece of DNA from
a plasmid which was responsible for giving antibiotic resistance.
13. Why do clown fish and sea anemone pair up? What is this relationship called? [2]
Ans. This interaction is called commensalism. Here the clown fish lives in the tentacles of
sea anemone. The fish gets protected from predators which stay away from stinging
tantacles. However anemone does not get any benefit from clown fish.
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14. (a) State the difference between meiocyte and gamete with respect to chromosome
number.
(b) Why is a whiptail lizard referred to as parthenogenetic? [2]
Ans. (a) Meiocyte (gamete mother cells) is diploid (2n), where as gamete is haploid (n).
(b) Whiptail lizard is said to be parthenogenetic because female gamete undergoes
development to form new organisms without fertilization.
15. Name the plant source of the drug popularly called “smack”. How does it affect the
body of the abuser? [2]
OR
Why is Rhizobium categorized as a ‘symbiotic bacterium’? How does it act as a
biofertiliser? [2]
Ans. It is obtained from Papver somniferum.
Smack is a stronger analgesic than morphine. It reduces heart beat, blood pressure
and increases blood sugar.
OR
Rhizobium lives in root nodules of leguminous plants. This association is mutually
beneficial. Rhizobium gets food and shelter and leguminous plant gets nitrogen in
return. Since Rhizobium is capable of fixing atmospheric nitrogen so it acts as
biofertiliser.
16. (a) State the role of DNA ligase in biotechnology.
(b) What happens when Meloidegyne incognitia consumes cells with RNAi gene?
[2]
Ans. (a) Role of DNA ligase in biotechnology is joining of DNA fragments end to end,
having same kind of sticky ends.
(b) If Meloidegyne incognitia consumes cells with RNA i gene silencing of specific
mRNA occurs due to a complementary ds RNA molecule formation that binds
to and prevents translation of mRNA (silencing) and thus causing death of the
nematode.
17. Some organisms suspend their metabolic activities to survive in unfavourable
conditions. Explain with the help of any four examples. [2]
Ans. Examples:
(a) In bacteria, fungi and lower plants various kinds of thick walled spore are formed
which help them to survive unfavourable conditions.
(b) In higher plants, seeds and some other vegetative reproductive propagules serve
as a means to tide over period of stress, besides helping in seed dispersal.
(c) Bears going into hibernation during winter.
(d) Snails and fishes go into aestivation to avoid summer related head problems
and dessication. OR
(e) Many zooplankton are known to enter diapause, a stage of suspended
development.
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18. (a) Name the Protozoan parasite that causes amoebic dysentery in humans.
(b) Mention two diagnostic symptoms of the disease.
(c) How is this disease transmitted to other? [2]
Ans. (a) Entamoeba histolytica.
(b) Symptoms of this disease include constipation, abdominal pain, cramps, stools
with excessive mucous and blood clots.
(c) Houseflies act as mechanical carriers and serve to transmit the parasite from
faeces of infected person to food and thus contaminating them.
SECTION-C
19. It is established that RNA is the first genetic material. Explain giving three reasons.
[3]
OR
(a) Name the enzyme responsible for the transcription of tRNA and the amino acid
the initiator tRNA gets linked with.
(b) Explain the role of initiator tRNA in initiation of protein synthesis. [3]
Ans. RNA was first genetic material because
(a) Many evidences suggest that essential life processes such as metabolism,
translations, splicing, etc. evolved around RNA.
(b) RNA used to act as genetic material as well as catalyst. Many important
biochemical reactions in living systems are catalysed by RNA.
(c) RNA being catalyst is highly unstable.
OR
(a) RNA polymerase in procaryotes and RNA polymerase III in Eukaryotes is
responsible for transcription of tRNA.
Initiator tRNA gets linked with methionine in eukaryotes and formylated
methionine in prokaryotes.
(b) Initiator tRNA combines with methionine in presence of amino acyl-tRNA
synthetas enzymes resulting in formation of charged tRNA. Now this initiator
tRNA combines with two subunits of ribosome and mRNA forming translation
initiation complex. First mRNA attaches to small subunit of ribosome and charged
initiator tRNA. The initiator tRNA joins the initiation codon AUG and signals the
start of translation. Now the large subunit of ribosome combines with small
subunit. Initiator tRNA lies at the P site of the ribosome.
20. State the theory of Biogenesis. How does Miller’s experiment support this theory?
[3]
Ans. Oparin and Haldane proposed that the first form of life could have come from pre-
existing non-living organic molecules (e.g. RNA, protein etc.).
According to them origin of live is abiogenesis first but biogenesis there after.
Miller’s Experiment: That simple organic compounds could be formed in nature in
the manner explained above was experimentally demonstrated in 1953 by
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6. STUDYmate
Stanley L Miller. Miller designed a glass apparatus comprising a gas flask, a condenser,
and a liquid flask interconnected with tubes and fitted with sources of energy. The
apparatus simulated the conditions on the primitive earth, including a “reducing
atmosphere” and an “ocean”. He circulated in this apparatus a mixture of methane
(CH4), ammonia (NH3) and hydrogen (H2) and water vapour (H2O) at 800°C. These
gases were believed to prevail in the ancient atmosphere. He provided energy for the
interaction of the gases present in the mixture in the form of electric. The electric
sparks simulated lightning. Then the gases were condensed in a narrow tube and
passed through a liquid flask. Here, energy was provided as heat with an electric
heater. He kept the experiment working continuously for 18 days. A mixture of small
organic molecules was formed in the gas flask (atmosphere) and was carried by
condensation (rain) to the liquid flask (ocean). He found many simple organic
compounds which included amino acids, such as glycine, alanine and aspartic acid;
adenine and simple sugars such as ribose.
21. Name the two different categories of microbes naturally occurring in sewage water.
Explain their role in cleaning sewage water into usable water. [3]
Ans. Different microbes, occuring in sewage water are aerobic and anaerobic bacteria,
Protozoans, and filamentous fungi.
The primary effluent is passed into large aeration tanks where it is constainlly agitated.
This allows abundant growth of aerobic microbes (bacteria and filamentous fungi)
into floes which is a mesh like structure. The growth of these microbes reduces BOD
of effluents. Once the BOD is reduced significantly, then the effluent is passed into
settling tanks where the bacterial floes are allowed to sediment. This sediment is
called activated sludge. A small part of activated sludge is again introduced into large
tanks called anaerobic sludge digesters. Here anaerobic bacteria digest the bacterial
and fungi in the sludge. This digestion produces methane, H2S and CO2 gas. These
gases form biogas. The effluent from secondary treatment is them released into natural
water bodies.
22. Write the function of each one of the following:
(a) (Oviducal) Fimbriae (b) Coleoptile
(c) Oxytocin [3]
Ans. (a) (Oviducal) Fimbriae: It receives the ova which gets released on ovulation.
(b) Coleoptile: It is the protective covering of plumule present inside the monocot
seed.
(c) Oxytocin:
(i) It causes contraction of uterus at the time of child birth.
(ii) It also helps in production of milk during lactation period.
23. Name the genes responsible for making Bt cotton plants resistant to bollworm attack.
How do such plants attain resistance against bollworm attacks? Explain. [3]
Ans. The gene responsible for making Bt cotton plant resistant to bollworms is cryIAC and
cryIIAb these genes form protein crystals during a particular phase of their growth.
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7. STUDYmate
These crystals contain insecticidal protein. These crystals exist as inactive prototoxins
but once an insect ingest the inactive toxin it is converted into active form of toxin
inside alkaline conditions of gut of insect. This solubilizes the crystals. The active
toxin binds to the surface of midgnt epithelial cells and create pores that cause cell
swelling and lysis and causes death of the insect.
24. Study a part of the life cycle of malarial parasite given below. Answer the questions
that follows: [3]
(a) Mention the roles of ‘A’ in the life cycle of the malarial parasite.
(b) Name the event ‘C’ and the organ where this event occurs.
(c) Identify the organ ‘B’ and name the cells being released from it.
Ans. (a) A is female Anopheles mosquito that act as vector for Plasmodium.
(b) C = fertilization that takes place in mosquito intestine.
(c) B = salivary gland cells which are released sporozoites.
25. Given below is the representation of amino acid composition of the relevant translated
portion of -chain of haemoglobin, related to the shape of human red. [3]
Gene ..... CTC ....
..... GAG ....
mRNA ... GAG...
Val His Leu Thr Pro Glu Glu
1 2 3 4 5 6 7
HbA Peptide
(a) Is this representation indicating a normal human or a sufferer from certain
related genetic disease? Give reason in support of your answer.
(b) What difference would be noticed in the phenotype of the normal and the sufferer
related to this gene?
(c) Who are likely to suffer more from the defect related to the gene represented the
males, the females or both males and females equally? And why?
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Ans. (a) This representation indicates a normal human because in the respective amino
acid chain, Glutamic acid is present at the 6th position.
(b) In the sufferer who exhibit sickle cell trait, defect is caused by the substitution
of Glutamic acid (Glu) by Valine (Val) at the 6th position of Beta globin chain of
the haemoglobin.
(c) Both the males and females suffer equally because sickle cell anaemia is not a
sex linked disease. It is an autosomal disease and sickle shaped RBC will cause
equal deficiency of oxygen in both males and females.
26. By the end of 2002 the public transport of Delhi switched over to a new fuel. Name
the fuel. Why is this fuel considered better? Explain. [3]
Ans. Delhi Government shifted to CNG (Compressed Natural Gas). CNG is better because
it burns more efficiently unlike petrol or diesel, in automobiles and very little of it is
left unburnt. Also CNG is cheaper fuel them petrol and diesel.
27. Draw a schematic sketch of pBR322 plasmid and label the following in it:
(a) Any two restriction sites.
(b) Ori and rop genes
(c) An antibiotic resistant gene.
Ans. Cla I
Eco R I Hind III
Bam HI
Pvu I
Pst I Sal I
Ori Rop
Origin of Pvu II
replication
Diagrammatic representation of E.coli cloning vector pBR322 showing restriction sites
R R
(Hind III, Eco RI, Bam HI, Sal I, Pvu II, Pst I, Cla I), Ori V and (amp and tet ). Rop
codes for the proteins involved in the replication of the plasmid.
SECTION-D
28. Explain the carbon cycle with the help of a simplified model. [5]
OR
Explain how does: [5]
(a) a primary succession start on a bare rock and reach a climax community?
(b) the algal bloom eventually choke the water body in an industrial area?
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9. STUDYmate
Ans.
 71% carbon is found dissolved in oceans. This oceanic reservoir regulates the
amount of CO2 in the atmosphere.
 Fossil fuel also represent a reservois of carbon.
 Carbon cyclic occurs through atmosphere, ocean and through dead and living
organisms.
 A considerable amount of CO2 is fixed by process of photosynthesis.
 Carbon from plants moves thruogh food chains.
 Detritus food chain releases it to soil.
 Respiration and combustion of fossil fuels, burning of forest firewood and organic
debris releases CO2 in atmosphere.
 Decomposers pass substantial CO2 to pool by their processing of water materials
and dead organic matter of land or ocean.
 Some amount of fixed carbon is lost to sediment and removed from circulation.
 Human activities have significantly influenced carbon cycle. Deforestation and
massive burning of fossil fuel has significantly increased the rate of release of
CO2 into atmosphere.
OR
(a) Such a succession is called Xerarch succession.
In this succession,
(i) Pioneer specie is lichens which secrete acids to dissolve rock causing
weathering and soil formation.
(ii) Next seral stage will be bryophytes which can hold in the small amount of
soil.
(iii) Bryophytes are then succeeded by grasses.
(iv) Grasses eventually will pave way for bigger trees which will form stable
climax community. This remains stable as long as the environment remains
unchanged.
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(b) Effluent from industries contains large amounts of nutrients. This causes
excessive growth of free-floating algae causing algal bloom. Algae starts
consuming oxygen for its own consumption. This will decrease BOD of water
body. This reduced BOD eventually causes death of all aquatic life thus choking
the water body.
29. [5]
(i) Identify the figure that illustrates ovulation and mention the stage of oogenesis
it represents.
(ii) Name the ovarian hormone and the pituitary hormone that have caused the
above mention event.
(iii) Explain the changes that occur in the uterus simultaneously in anticipation.
(iv) Write the difference between ‘c’ and ‘h’.
(v) Draw a labelled sketch of the structure of a human ovum prior to fertilization.
OR
How does the megaspore mother cell develop into 7-celled, 8 nucleate embryo sac in
an angiosperm? Draw a labelled diagram of mature embryo sac. [5]
Ans. (i) (a) f = stage of ovulation
(b) Meiosis I of oogenesis is completed.
(ii) Ovarian hormone = estrogen
Pituitary hormone = Luetenizing hormone
(iii) Endometrium lining gets thickened and vascularized.
(iv) c = Developing graffian follicle
h = Degenerating corpus luteum
(v)
Human ovum prior to fertilization
OR
 The nucleus of the functional megaspore divides mitotically to form two nuclei
which move to the opposite poles, forming the 2-nucleate embryo sac. Two more
sequential mitotic nuclear divisions result in the formation of the 4-nucleate
and later the 8-nucleate stages of the embryo sac.
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 These mitotic divisions are strictly free nuclear, that is, nuclear divisions are
not followed immediately by cell wall formation. After the 8-nucleate stage, cell
walls are laid down leading to the organisation of the typical female gametophyte
or embryo sac.
 Six of the eight nuclei are surrounded by cell walls and organised into cells; the
remaining two nuclei, called polar nuclei are situated below the egg apparatus
in the large central cell.
 There is a characteristic distribution of the cells within the embryo sac. Three
cells are grouped together at the micropylar end and constitute the egg apparatus.
The egg apparatus, in turn, consists of two synergids and one egg cell.
 The synergids have special cellular thickenings at the micropylar tip called filiform
apparatus, which play an important role in guiding the pollen tubes into the
synergid.
 Three cells are at the chalazal end and are called the antipodals. The large
central cell, as mentioned earlier, has two polar nuclie. Thus, a typical angiosperm
embryo sac, at maturity, though 8-nucleate is 7-celled.
30. What is the inheritance pattern observed in the size of starch grains and seed shape
of Pisum sativum? Workout the monohybrid cross showing the above traits. How
does this pattern of inheritance deviate from that of Mendelian law of dominance?
[5]
OR
State the aim and describe Messelson and Stahl’s experiment. [5]
Ans. A round seed has well developed starch grains whereas wrinkled seeds don’t have
starch grains. So, a cross between two will give to an intermediate situations.
Suppose R is allele for round shape and r is allele for wrinkled.
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Round Wrinkled
RR rr Parents
R r gametes
Rr F1 (intermediate)
R r Gametes
R r Round intermediate wrinkled
R RR Rr 1 : 2 : 1
r Rr rr
This is an example of incomplete dominance. Whereas in Mendelian inheritance we
have complete dominance.
OR
Matthew Meselson and Franklin Stahl performed the following experiment in 1958:
(i) They grew E. coli in a medium containing 15NH4Cl (15N is the heavy isotope of
nitrogen) as the only nitrogen source for many generations. The result was that
15
N was incorporated into newly synthesised DNA (as well as other nitrogen
containing compounds). This heavy DNA molecule could be distinguished from
the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient
(Please note that 15N is not a radioactive isotope, and it can be separated from
14N only based on densities).
(ii) Then they transferred the cells into a medium with normal 14NH4Cl and took
samples at various definite time intervals as the cells multiplied, and extracted
the DNA that remained as double-stranded helices. The various samples were
separated independently on CsCl gradients to measure the densities of DNA.
(iii) Thus, the DNA that was extracted from the culture one generation after the
transfer from 15N to 14N medium [that is after 20 minutes; E. coli divides in 20
minutes] had a hybrid or intermediate density. DNA extracted from the culture
after another generation [that is after 40 minutes, II generation] was composed
of equal amounts of this hybrid DNA and of ‘light’ DNA.
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13. STUDYmate
Studymate Solutions to CBSE Board Examination 2011-2012
Series : SMA/1 Code No. 57/1/2
UNCOMMON QUESTIONS ONLY
SECTION-A
1. Cucurbits and papaya plants bear staminate and pistillate flowers. Mention the
categories they are put under separately on the basis of the type of flowers they bear.
Ans. Cucurbits are monoecious and papaya is Dioecious.
4. What is the interaction called between cuscuta and shoe flower blush?
Ans. Parasitism
5. When do the oogenesis and the spermatogenesis initiate in human females and males
respectively?
Ans. Oogenesis is initiated during the embryonic development, spermatogenesis begins at
puberty.
7. State the significance of the study of fossils in evolution.
Ans. Study of fossils indicates the geological period in which various life forms were arisen.
The calculation of geological period can be done via radioactive dating.
SECTION-B
13. Draw a schematic diagram of a part of double stranded dinucleotide DNA chain having
all the four nitrogenous bases and showing the correct polarity.
Ans.
14. Name the parasite that causes filariasis in human. Mention its two diagnostic
symptoms. How is this disease transmitted to others?
Ans. Wuchereria bancrofti and Wuchereria malayi causes filariasis in humans.
The two diagnostic symptoms are
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1. Chronic inflammation of the organs in which they live for many years.
2. The genital organs are also often affected resulting in gross deformities.
Pathogens are transmitted to a healthy person through the bite by the female mosquito
vectors.
15. Name the source of streptokinase. How does this bio-reactor molecule function in
our body?
OR
How do mycorrhizae act as biofertilizers? Explain. Name a genus of fungi that forms
a mycorrhizal association with plants.
Ans. Streptokinase produced by the bacterium streptococcus and modified by genetic
engineering is used as a clot buster for removing clots from the blood vessels of
patients who have undergone myocardial infection leading to heart attack.
OR
Mycorrhizae are associations between fungi and the roots of higher plants. The fungi
help the plant in the absorption of essential nutrients from the soil while the plant in
turn provides the fungi with energy yielding carbohydrates.
Boletus is the soil fungus that forms a mycorrhizal association with plants.
SECTION-C
19. Write the function of each of the following
(a) Middle piece in human sperm.
(b) Tapetum in anthers.
(c) Luteinizing hormone in human males.
Ans. (a) The middle piece possesses numerous mitochondria, which produces energy
for the movement of tail that facilitate sperm motility essential for fertilisation.
(b) Microsporangium are surrounded by 4 wall layers, epidermis, endothecium,
middle layer and tapetum. Outer 3 wall layers performs the function of protection
and help in dehiscence of anther to release the pollen. The innermost wall layer
is the tapetum, it nourishes the developing pollen grains. Cells of the tapetum
possess dense cytoplasm and generally have more than one nucleus.
(c) Luteinizing hormone acts at the Leydig cells and stimulates the synthesis and
secretion of androgens.
26. How does an algal bloom cause eutrophication of a water body? Name the weed that
can grow in such a eutrophic lake.
Ans. Presence of large amounts of nutrrients in waters also causes execessive growth of
planktonic algae called an algal bloom which imparrts a distinct colour to the wate
rbodies. Algal blooms causes deterioration of the water quality and fish mortality.
Eutrophication is the natural aging of a lake by nutrient enrichment of its water.
Streams draining into the lake increases nutrients such as nitrogen and phosphorous
which encourage the growth of aquatic organisms. As the lake’s fertility increases,
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15. STUDYmate
plant and animal life burgeons, and organic remains begin to be deposited on the
lake bottom. Over the centuries as silt and organic debris pile up, the lake grows
shallower and warmer, with warm-water organisms supplanting those that thrive in
a cold environment marsh plants take root in the shallows and begin to fill in the
original lake basin. Eventually, the lake gives way to large masses of floating plants,
finally converting into land.
Eichhornia crassipes water hyacinth can grow in entrophic lake.
28. (a) Draw a ‘pyramid of numbers’ of a situation where a large population of insects
feed upon a very big tree. The insects in turn, are eaten by small birds which in
turn are fed upon by big birds.
(b) Differentiate giving reasons, between the pyramid of biomass of the above
situation and the pyramid of numbers that you have drawn.
OR
(a) What are the two types of desirable approaches to conserve biodiversity? Explain
with examples bringing out the difference between the two types.
(b) What is the association between the bumble bee and its favourite orchid ophrys?
How would extinction or change of one would affect the other?
Ans. (a) Pyramid of Number
Large birds
Small birds
Insects
Tree
This is spindle shaped pyramid.
Pyramid of biomass
Large birds
Small birds
Insects
Tree
This is irregular shaped pyramid.
(b) In case of pyramid of number for first two steps number increases but them it
decreases.
Whereas in pyramid of biomass there is no clear trend.
OR
(a) Two types of desirable approaches to conserve biodiversity are:
(i) In-situ conservation: In-situ conservation is the most appropriate method
to maintain species of wild animals and plants in their natural habitats.
This approach includes protection of total ecosystems through a network
of protected areas. These are the biogeographical areas where biological
diversity along with natural/cultural resources are protected, maintained
and managed.
The common natural habitats (protected areas) that have been set for in-
situ conservation of wild animals and plants include –
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16. STUDYmate
(I) National parks
(II) Wild life sanctuaries
(III) Biosphere reserves
(IV) Several wetlands, mangroves and coral reefs
(V) Sacred grooves and lakes.
(ii) Ex-situ conservation: Ex-situ conservation includes the following:
(I) Sacred plants and home gardens.
(II) Seed banks, field gene banks, cryopreservation.
Botanical gardens, Arborata, Zoological gardens, Aquaria.
All these approaches help to conserve species and population diversity
outside the natural habitats.
In-situ Ex-situ
1. It is the most appropriate 1. In this approach, threatened
method to maintain species of animals and plants are not
wild animals and plants in their preserved in their natural
natural habitat. habitat and placed in special
setting, where they can be
protected and special care.
2. No such technique can be used 2. Gametes of threatened species
for In-situ conservation. can be preserved in viable and
fertile condition for long periods
using cryopreservation
techniques.
3. In this method organism 3. Reproduction in captivity often
alongwith its entire habitat is slows down and may not give
preserved. Chances of desired results.
fertilization and propagation
are higher.
4. Eg. National Park, Wild Life 4. Eg. Seed banks.
Sancturies.
(b) Association between the bumble bee and its favourite orchid ophrys is mutualism.
Orchids show a bewildering diversity of floral patterns many of which have
evolved to attract the right pollinator insect (bees and bumblebees) and ensure
guaranteed pollination by it. If the female bee’s colour patterns change even
slightly for any reason during evolution, pollination success will be reduced
unless the orchid flower co-evolves to maintain the resemblance of its petal to
the female bee.
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17. STUDYmate
Studymate Solutions to CBSE Board Examination 2011-2012
Series : SMA/1 Code No. 57/1/3
UNCOMMON QUESTIONS ONLY
SECTION-A
1. Mention the difference between spermiogenesis and spermiation. [1]
Ans. Spermiogenesis: Transformation of spermatids into spermatozoa.
Spermiation: The process of release of spermatozoa from Sertoli cells into the cavity
of sperminiferous tubules.
3. What is an interaction called when an orchid grows on a mango plant? [1]
Ans. Commensalism.
4. Write the names of two semi-dwarf and high yiuelding rice varieties developed in
India after 1966. [1]
Ans. Jaya/Ratna.
6. Mention the unique feature with respect to flowering and fruiting in bamboo species.
[1]
Ans. Bamboo species flower only once in their lifetime.
8. State the significance of biochemical similarities amongst diverse organism in
evolution. [1]
Ans. Biochemical similarities point to the same shared ancestory as structural similarities
among diverse organisms.
SECTION-B
15. Mention the importance of Lactic acid bacteria to humans other than setting milk
into curd. [2]
OR
How do methanagens help in producing biogas? [2]
Ans. (a) Lactic acid bacteria play a very beneficial role in checking disease causing
microbes.
(b) Methanogens such as Methanobacterium act on excreta of cattle and grow
anaerobically, producing large amount of methane along with CO 2 and H2.
Methane is the main predominant gas of bio gas.
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18. STUDYmate
SECTION-C
19. (a) Construct a complete transcription unit with promotor and terminator on the
basis of the hypothetical template strand given below:
A T GCA T GCA T A C
(b) Write the RNA strand transcribed from the above transcription unit along with
its polarity. [3]
OR
How are the structural genes inactivated in lac operon in E. Coli? Explain. [3]
Ans. (a)
A T G C A T G C A T A C
3 5
Promoter Terminator
5 3
T A C G T A C G T A T G
(b) RNA strand transcribed from above transcription unit.
5 3
UA C GU A CGU A UG
(c) In the absence of lactose, no inactivation of repressor occurs and hence the
repressor (which is constitutively produced from i gene) binds to the operator
region of the operon; thus preventing RNA polymerase from transcribing the
operon thus inactivating of the production of structural genes in E.coli occurs.
20. Write the function of each of the following: [3]
(a) Seminal vesicle (b) Scutellum
(c) Acrosome of human sperm
Ans. (a) Seminal vesicle: Constitute seminal plasma which is rich in fructose calcium
and enzymes.
(b) Scutellum: Is the papery cotyledon of the monocot seed and acts as a passage
for movement of nutrients from the endosperm to the developing embryo.
(c) Acrosome of human sperm: Contains hydrolytic enzymes that help in
penetration of egg during fertilization.
25. (a) Why are the colourful polysterene and plastic packagings used for protecting
the food, considered an environmental menace?
(b) Write about the remedy found for the efficient use of plastic waste by Ahmed
Khan of Bangalore. [3]
Ans. (a) Colourful polysterene and plastic packaging used for protecting the food are
considered to be environmental menace as plastic is non-biodegradable and its
recycling process is very costly and includes manual participation thus exposing
workers to toxic substances produced during recycling process.
(b) The remedy found for efficient use of plastic waste by Ahmed Khan of Bangalore
was making of Polyblend, a fine powder of recycled modified plastic. This mixture
is mixed with the bitumen, and is used to lay roads.
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19. STUDYmate
In collaboration with R.V. College of Engineering and Banglore city Corporation,
Ahmed Khan proved that blends of Polyblend and Bitumen, when used to lay
roads enhanced the bitumen’s water repellant properties and helped to increase
road life by a factor of three.
SECTION-D
30. Name the scientists who proved experimentally that DNA is the genetic material.
Describe their experiment. [5]
OR
(a) List the three different allelic forms of gene ‘I’ in humans. Explain the different
phenotypic expressions, controlled by these three forms.
(b) A woman with blood group ‘A’ marries a man with blood group ‘O’. Discuss the
possibilities of the inheritance of the blood groups in the folllowing starting
with ‘yes’ or ‘no’ for each:
(i) They produce children with blood group ‘A’ only.
(ii) They produce children some with ‘O’ blood group and some with ‘A’ blood
group. [5]
Ans. The two scientists that proove DNA is the genetic material are:
(i) Griffith Transformation Experiment
(a) In 1928, Frederick Griffith, in a series of experiments with Streptococcus
pneumoniae (bacterium responsible for pneumonia), witnessed a miraculous
transformation in the bacteria.
(b) When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a
culture plate, some produce smooth shiny colonies (S) while others produce
rough colonies (R). This is because the S strain bacteria have a mucous
(polysaccharide) coat, while R strain does not. Mice infected with the S
strain (virulent) die from pneumonia infection but mice infected with the R
strain do not develop pneumonia.
S strain  Inject into mice  Mice die
R strain  Inject into mice  Mice live
Griffith was able to kill bacteria by heating them. He observed that heat-
killed S strain bacteria injected into mice did not kill them.
S strain (heat-killed)  Inject into mice  Mice live
When he injected a mixture of heat-killed S and live R bacteria, the mice
died.
S strain (heat-killed) + R strain (live)  Inject into mice  Mice die
Moreover, he recovered living S bacteria from the dead mice.
(c) He concluded that the R strain bacteria had somehow been transformed
by the heat-killed S strain bacteria. Some ‘transforming principle’,
transferred from the heat-killed S strain, had enabled the R strain to
synthesise a smooth polysaccharide coat and become virulent. This must
be due to the transfer of the genetic material.
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20. STUDYmate
(d) Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty
(1933-44), the genetic material was thought to be a protein. They worked
to determine the biochemical nature of ‘transforming principle’ in Griffith’s
experiment.
(e) They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed
S cells to see which ones could transform live R cells into S cells. They
discovered that DNA alone from S bacteria caused R bacteria to become
transformed.
(f) They also discovered that protein-digesting enzymes (proteases) and RNA-
digesting enzymes (RNases) did not affect transformation, so the
transforming substance was not a protein or RNA. Digestion with DNase
did inhibit transformation, suggesting that the DNA caused the
transformation. They concluded that DNA is the hereditary material.
(ii) Hershey and Chase Experiment
(a) DNA is the genetic material came from the experiments of Alfred Hershey
and Martha Chase (1952). They worked with viruses that infect bacteria
called bacteriophages.
(b) The bacteriophage attaches to the bacteria and its genetic material then
enters the bacterial cell. The bacterial cell treats the viral genetic material
as if it was its own and subsequently manufactures more virus particles.
Hershey and Chase worked to discover whether it was protein or DNA
from the viruses that entered the bacteria.
(c) They grew some viruses on a medium that contained radioactive
phosphorus and some others on medium that contained radioactive sulfur.
Viruses grown in the presence of radioactive phosphorus contained
radioactive DNA but not radioactive protein because DNA contains
phosphorus but protein does not. Similarly, viruses grown on radioactive
sulfur contained radioactive protein but not radioactive DNA because DNA
does not contain sulfur.
(d) Radioactive phages were allowed to attach to E. coli bacteria. Then, as the
infection proceeded, the viral coats were removed from the bacteria by
agitating them in a blender. The virus particles were separated from the
bacteria by spinning them in a centrifuge.
(e) Bacteria which was infected with viruses that had radioactive DNA were
radioactive, indicating that DNA was the material that passed from the
virus to the bacteria. Bacteria that were infected with viruses that had
radioactive proteins were not radioactive. This indicates that proteins did
not enter the bacteria from the viruses. DNA is therefore the genetic material
that is passed from virus to bacteria.
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21. STUDYmate
OR
(a) Three different allelic forms of ‘I’ gene are
A
(i) i (ii) I
B
(iii) I
Genotype Blood Group
ii O
A A A
I I;I i A
B B B
I I ;I i B
A B
I I AB
(b) Possibility I
A A
I I ii
(woman) (man)
A A
I I
i IA i IA i
A A
i I i I i
Yes. A Blood group only
Possibility II
A
I i ii
(woman) (man)
A
I i
i IA i ii
A
i I i ii
Yes. 50% A Blood Group
50% O Blood Group
×·×·×·×·×
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