Csir Net June 2015 Ques-Answer

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CSIR NET JUNE 2015 Question Paper + Answer Key


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PART A

Ans: 1

Explanation:

The

base

consist

of

4X4,

the

next layer would be 3X3 (as 0.5 has to be

step), next would be 2X2, 1X1,

Total=16+9+4+1=30

Ans: 4

Explanation: Explanation: Either B can lie or A

Ans: 3

Explanation: n=(360/theta)‐1, theta=60, n=5

Ans: 4

Explanation: Let the the third angle in the

traingle containing the x be y (seond angle

would be 180‐130=50

Fom the other triangle, 180‐y+50=130

==>y=100. If y=100, x=30.

Ans: 4

Explanation: The area would be maximum at

equator as the radius is maximum.


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Ans: 2

Explanation: Total distance=4+3=7, Speed =7,

Time=1

Ans: 4

Explanation: if m+n+mn = 118 then mn = 118

‐ (m+n). when 22 is substituted for m+n value

of mn

will

be

96.

96

can

be

written

as

16

x 6

that is value of m and n. hence 16 + 6 is 22. At

the same time 16 + 6 + (16*6) will be 118.

Ans: 1

Explanation: Let the price be x, after discount

it would be 0.9x. A loss of 90% incurred, so

final s.p=0.9*0.9x=0.81x

0.81x=729. Or x=900

Ans: 1

Explanation: X‐4 cannot be equal to x^2 ‐4^2,

it should be (x‐4)^2

Ans: 2

Explanation: We can select 11 players from 40

by 40 C 11 ways. We can choose 1 captain out

of 11 in 11 C 1 ways.Totally the number of

ways

11

C

1

x

40

C

11

=

11

x

40

C

11(11

C

1

=

11!/10! 1! = 11)


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Ans: 4

Explanation: The maximum diversity is

possible when all the bases occur in the same

proportion

Ans: 4

Explanation: Percentage increase in

2010=20/18*100:

Percentage increase in 2002=15/7*100:

Ans: 4

Explanation: 10000/3=3333

Ans: 2

Explanation: Hidden numbers are A: Eleven,

B: Nine, C: Eight, D: Ten. So according arrange

in ascending order C, B, D and A


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Ans: 4

Explanation: If we consider the southern

hemisphere, there is a ring near the South

Pole that has a circumference of one mile. So

what if we were standing at any point one

mile north

of

this

ring?

If we

walked

one

mile

south, we would be on the ring. Then one

mile east would bring us back to same point

on the ring (since it’s circumference is one

mile). One mile north from that point would

bring us back to the point were we started

from.

Ans: 2

Explanation: The answer most likely is option

2. Surface area changes by 5 % Volume of

sphere = 4/3 x pi x r^3 Since 7% is lost (10% of

70%), new volume is 0.93 x earlier volume So

the radii is also in the proportion r^3 = 0.93 x

R^3 (considering

r to

be

earlier

radius

and

R

new radius after loss of water) r = 0.976 R

Surface area = 4 x pi x r^2 So it will change by

a factor of (0.976)^2 = 0.952 So the change is

approx 5%

Ans: 1

Explanation: If the batch of 10 marbles we

moved from B to A contained k

black ones and 10 − k white ones, then A, at

the end, has k black marbles, and

B has 10 − (10 − k) = k white marbles. Thus the

two numbers are the same.

Ans: 1

Explanation: The distance between A and B

=pi R=3.14 R. As the speed is in ratio of 2:1

and hence

the

distance

covered

would

be

in

ratio of 2:1. i.e R from A and 2R from B


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Ans: 1

Explanation: aN when reversed is Na that is

sodium hence S. similarly eF = Fe = iron = I, gH

= Hg = mercurry and nS = tin hence T

Ans: 2

Explanation: mark a point O (center) on AB.

Join OC and OD. Hence OC, OD, OB will be

radius of circle = x. Let OP = y. Since OB is x OP

can be written as x ‐ 1. Apply pythorarus

theorem to triangle OCP. x^2 = 4+ (x‐1)^2. By

solvinf the equation x will be 2.5 which is the

radius of the circle.


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Biotecnika | CSIR NET Answer Keys June 2 5

1 www.biotecnika.org Ph: 1800‐200‐3757

Ans: 3

Explanation: Long chain saturated fatty acids would be

responsible for decreasing the fluidity of the plasma

membrane. As

the

bacteria

would

be

shifted

from

lower

to higher temperature therefore, it would be requiring

long chain saturated fatty acids for its survival.

Ref: Chapter 4: Structure and Function of Plasma

membrane, Page 134, Cell and Molecular Biology by

Karp (6th Edition)

Ans: 1

Explanation: Sucrose is trans located via phloem in

plants

Ref: https://en.wikipedia.org/wiki/Phloem

Ans: 4

Explanation: Most transposans in bacteria are DNA and

Eukaryotes have both DNA and RNA

Ref : https://en.wikipedia.org/wiki/Transposable_element

http://www.slideshare.net/ambicaparthasarathi/transposa

ble-elements

Ans: 1

Explanation: LEAFY is a floral meristem identity gene

Ref: http://www.ncbi.nlm.nih.gov/books/NBK10122/


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Ans: 1

Explanation: The process is coupled to the hydrolysis of

16 equivalents of ATP and is accompanied by the co‐

formation of one molecule of H2

Ref : https://en.wikipedia.org/wiki/Nitrogen_fixation

Ans: 2

Explanation: Wavelengths beyond 700nm are

apparently

of

insufficient

energy

to

drive

any

part

of

photosynthesis. So a huge drop in efficiency has been

noticed at 700nm.It is a part of the emerson effect.

Ref : Book: Gupta and Gupta, Plant physiology 2005

Ans: 3

Explanation: Book: Gupta and Gupta, Plant physiology

2005

Ref :

http://chemwiki.ucdavis.edu/Biological_Chemistry/Protei

ns/Protein_Structure, Lehninger Principles of

Biochemistry, chapter 4

Ans: 2

Explanation: The S wave is the first negative deflection

after the R wave. It represents the late ventricular

depolarisation.

Ref: http://www.cardionetics.com/conduction-

system#lateventriculardepolarisation

Ans: 4

Explanation: The transport of chloride ions helps

control the movement of water in tissues, which is

necessary for the production of thin, freely flowing

mucus

Ref: http://ghr.nlm.nih.gov/gene/CFTR


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Ans: 2

Explanation: The 2'OH of th adenosine will attack the

5'splice site

Ref : https://en.wikipedia.org/wiki/Group_II_intron

Ans: 4

Ref :

http://chemwiki.ucdavis.edu/Biological_Chemistry/Metab

olism/Pyruvate_Dehydrogenase_Complex

Ans: 2

Explanation: Immunodeficient. Mutation in genes

required for site specific recombination results in

reduced or absence of mature T and B cells. Immature

cells are still present resulting in immune deficient state

Ref: https://en.wikipedia.org/

wiki/Severe_combined_immunodeficiency

Ans: 4

Explanation: Kcat is the turnover number calculated as

Kcat = Vmax/Et where, Vmax = max velocity; Et = total

enzyme concentration. In the question we should

convert the concentration terms to molarity. Vmax = 10

*10^‐3/400 = 0.25 * 10^‐4; Et = 20 * 10^‐6/200000 =

10^‐10. Kcat = 0.25 * 10^‐4/10^‐10 = 2.5 * 10^5/min

Ans: 4

Explanation: Beating of cilia is carried through nexin ‐

dyenin microtubule activity .All the other options are

related to microfilaments.

Ref: http://jcs.biologists.org/content/123/4/519.full.pdf


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Ans: 4

Explanation: All others would result in increased

concentration of Ca2+ except for DNA synthesis and

nuclear envelope breakdown which might occur due to

activation of a cAMP dependent kinase.

Ref : https://books.google.co.in/books?id=dmokq_M-

gm8C&pg=PA337&lpg=PA337&dq=g+protein+coupled+r

eceptors+%2B+sperm&source=bl&ots=hrQEV2_OHH&s

ig=BpgomZxHHAfS2ymOC4GW-

5D7trc&hl=en&sa=X&ei=aG2GVYrnAsKfugS3u4OQAg&

ved=0CEIQ6AEwBA#v=onepage&q=g%20protein%20co

upled%20receptors%20%2B%20sperm&f=false

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3228701/

Ans: 1

Explanation: The pathway which plays an important

role in axis formation during morphallactic regenration

in Hydra would be beta catenin/Wnt pathway

Ref :

http://www.eecs.harvard.edu/~rad/courses/cs266/papers

/wolpert-chap13.pdf

Ans: 1

Explanation: immortality and contact inhibition.

transformed cells have loss of tumor suppressor activity

and activation of tumor suppressor genes

Ref:

http://library.med.utah.edu/WebPath/NEOHTML/NEOPL

105.html

Ans: 2

Explanation: The DNA sequence c has the highest

amount of GC content aomng the three and hence

would have maximum stability

Ans: 2


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Explanation: Polar and Kinetochore microtubules

present in the center of the cell are responsible for

determining the plane of formation of contractile ring

Ref : http://www.ncbi.nlm.nih.gov/books/NBK26831/

Ans: 2

Explanation: In triple response of seedlings when

exposed to ethylene, thickening and shortening of

hypocotyl with pronounced apical hook occurs.

Ref: https://en.wikipedia.org/wiki/Ethylene,

http://www.sccs.swarthmore.edu/users/08/wnekoba/Stud

ent5Handout.pdf

Ans: 1

Explanation: Inside cells, however, do not turn on the

Cdx2 gene, and express high levels of Oct4, Nanog, and

Sox2, These

genes

suppress

Cdx2

and

the

inside

cells

maintain pluripotency generate the ICM and eventually

the rest of the embryo proper

Ref : https://en.wikipedia.org/wiki/Inner_cell_mass

Ans: 3

Explanation: As the proof reading activity is carried out

at the 3; growing end and it done by 3'‐5'exonuclease activity

Ref : https://en.wikipedia.org/wiki/Proofreading_(biology)

Ans: 4

Explanation: Secondary oocyte is arrested at

Metaphase II of Meiosis II till the time sperm enters the

secondary oocyte which results in completion of 2nd

meiotic division of formation of ovum.

Ref : Chapter 14.3: Meiosis, Page 595, Cell and

Molecular Biology Karp (6th edition)


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Ans: 3

Ref :

https://books.google.co.in/books?id=KR1GlQs__9wC&p

g=PA257&lpg=PA257&dq=basal+metabolic+rate+would

+be+lower+during+prolonged+starvation,+sleep,+high+e

nvironmental+temperature&source=bl&ots=K2UYsT9slv

&sig=wedZnLXHhuhHVk1qIj8wKzX-

Au0&hl=en&sa=X&ei=rLiGVdCFIJeGuATMoY7wCg&ved

=0CB0Q6AEwAA#v=onepage&q=basal%20metabolic%

20rate%20would%20be%20lower%20during%20prolong

ed%20starvation%2C%20sleep%2C%20high%20enviro

nmental%20temperature&f=false

Ans: 1

Explanation: As there should be inverse relation

between r and T. So that can be justified only by option

1.

Option 2 denotes a direct relation between r and T

which is not possible. Option 3 and 4 are also not

possible since according to them if generation time is increasing then population growth constant should also

be increasing which is again not true.

Ref: https://en.wikipedia.org/wiki/Generation_time

http://vlab.amrita.edu/?sub=3&brch=65&sim=174&cnt=1

Ans: 3

Explanation: Steroid receptor requires a NLS signal and

the DNA binding domain is Zinc fingers

Ref :

https://en.wikipedia.org/wiki/Steroid_hormone_receptor

Ans: 3

Explanation: thigh muscles can have a thousand fibers

in each unit,

Ref: https://en.wikipedia.org/wiki/Motor_unit

Ans: 1

Explanation: The ordered tetrad is observed in N. Crasa

Ref : Griffith: Page 142


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Ans: 3

Explanation: rRNA and are used in reconstructing

phylogenies, due to the slow rates of evolution of this

region of the gene

Ref : https://en.wikipedia.org/wiki/16S_ribosomal_RNA,

http://eebweb.arizona.edu/blast/rna_lecture.pdf

Ans: 2

Explanation: Procollagen peptidase helps in the

processing of procollagen to collagen

Ref : http://www.medscape.com/viewarticle/423231_2,

https://en.wikipedia.org/wiki/Procollagen_peptidase

Ans: 2

Explanation: σ32 (RpoH) ‐ the heat shock sigma factor,

it is turned on when the bacteria are exposed to heat.

Ref : https://en.wikipedia.org/wiki/Sigma_factor

Ans: 1

Explanation: Semelparous organisms generally

reproduce only once in their lifetime producing many

offsprings at once. Cicada and Mayfly exhibit

semelparity along with bamboo which is a long lived

semelparous plant. Hence, option 1 would be correct.

Ref : Smith and Smith 2015, Elements of ecology, https://en.wikipedia.org/wiki/Semelparity_and_iteropa

rity

Ans: 1

Explanation: As only the elongation factors undergo for

conformational changes, ribosome do not undergo for

conformational changes during the GTP hydrolysis


Ans: 1

Explanation: Uv spectroscopy can be used to quantitate

proteins using Tyrosine and tryptophan content. rest of

the methods cannot be effectively used for quantitation

Ref : http://www.labome.com/method/Protein-

Quantitation.html


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Ans: 3

Explanation: As compared to normal population,

individuals of sub‐population within a metapopulation

are dynamic in nature which is determined by their

continuous immigration and emmigration.

Ref : Chapter 8: Properties of Population, Page 159,

Smith: Elements of Ecology (8th Edition)

Ans: 3

Explanation: The probability of having no dominant

phenotype=1/64 (tripple homozygous recessive)

Therefore the probability of having atleast one

dominant=1‐(1/64)=63/64

Ans: 3

Explanation: If high number of specialists would be

present then they would be dependent on a particular

type of food only. Similarly fewer number of functional

links would result in destabilization as high functional

diversity is required for stabilization. If an ecosystem is

having linear food web then the interdependence of

organisms on one other would be very strong while if

reticular food web is present then the interdependence

of organisms would be weak. Hence, linear food web

(food chain) would tend to decrease stability. While,

omnivores which are generalists would most likely be

increasing ecosystem stability since they can feed on

both plants and animals.

Ref : https://en.wikipedia.org/?title=Food_chain

Ans: 3

Explanation: T test is used in case of high variability

Ref : http://www.socialresearchmethods.net/kb/stat_t.php

Ans: 2

Explanation: Hybrid dysgenesis refers to the high rate of

mutation in germ line cells of Drosophila strains

resulting from a cross of males with autonomous P


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elements (P Strain/P cytotype) and females that lack P

elements (M Strain/M cytotype)

Ref: https://en.wikipedia.org/wiki/P_element

Ans: 4

Explanation: The protocol of DNA footprinting involves

one of the ends to be labelled so that after digestion

with DNaseI the distance from the end can be observed

by autoradiography. But of we incorporate label by nick

translation, this will allow the label to present across

the length of the DNA to be analyzed and we can not

accurately determined the length of the DNA fragments

Ref :

http://cshprotocols.cshlp.org/content/2013/5/pdb.prot074

328.short

Ans: 3

Explanation: As serum is outside the cell and hence

ELISA would be used

Ref :

http://labmed.ascpjournals.org/content/34/1/58.full.pdf

Ans: 1

Explanation: Noro virus and Rota virus both cause acute gastro intestinal illness. But Rota virus is more

associated with pediatric acute gastroenteritis. Where

ac Noro virus causes gastroenteritis in all age groups. It

is more contagious compared to Rota virus and is often

called as winter vomiting bug.

Ref : http://www.healthline.com/health/viral-

gastroenteritis#Overview1,

https://en.wikipedia.org/wiki/Norovirus

Ans: 3

Explanation: In the question taking path length as 1 cm.

A = ecl. The combined extinction coeff needs to be

considered whic is e = e trp * no. of trp + e tyr * no. of

tyr. In the question since both tyr and trp are 1 each. So

the combined molar extinction coeff is simply the sum

of both i.e 3000 + 1500 = 4500. Putting values in eqn,

0.90 = 4500 * c x 1; c = 0.90/4500 = 0.0002 M or 0.2 mM


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Ans: 1

Explanation: In punctuated equilibrium the species do

not evolve for a long period of time and would evolve

very rapidly for a short period of time.

Ref : https://en.wikipedia.org/wiki/Punctuated_equilibrium

Ans: 2

Explanation: The hearing capabilities of elephant is

superior compared to human beings. They use infra

sonics more frequently for hearing. they detect sounds

as low as 14 to 16 hz (human low range: 20 hz) and as

high as 12,000 hz (human high range: 20,000)

Ref: http://seaworld.org/animal-info/animal-

infobooks/elephants/senses/

Ans: 4

Explanation: As the mean remains same but the

variance decreases and hence this is an example of

stabilizing selection

Ref : https://en.wikipedia.org/wiki/Stabilizing_selection

Ans: 1

Explanation: Only Basidiomycotes have septate hyphae

which reproduce asexually by budding, conidia and

fragmentation.e.g. yeast. In zygomycetes, hyphae is

asepatate. It only forms septa where gametes are

formed or to wall off dead hyphae. Chytrids is also a

group of fungi which are asepatate and they reproduce

through zoospores. The Glomeromycota have generally

coenocytic (occasionally sparsely septate) mycelia and

reproduce asexually through blastic development of the

hyphal tip to produce spores called Glomerospores).

Ref: Sharma OP Textbook of fungi ,

https://books.google.co.in/books?id=4fcGAZBbZHAC&p

rintsec=frontcover&dq=op+sharma&hl=en&sa=X&ei=x5

OHVee8LI‐ jugS‐

6ZywCw&ved=0CB0Q6AEwAA#v=onepage&q=op%20sh

arma&f=false


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Ans: 1&4

Explanation: Bell‐shaped is a characteristics of

polygenic inheritance and almost all the polygenic inheritance is effected by environment

Ref:

http://highered.mheducation.com/sites/007246268x/stud

ent_view0/chapter7/index.html

Ans: 2

Explanation: Females have about 3/4 similarity with

their sisters, however, have only 1/2 similarities with

their daughters

Ans: 4

Ref : http://www.tutorvista.com/content/biology/biology-

iii/animal-kingdom/phylum-echinodermata.php

Ans: 2

Explanation: According to the question, glycolipids

would be facing the cytosolic side while sphingomyelin

would be facing the luminal side of Golgi. Now as the

vesicle would be budding from the Golgi, the luminal

portion of Golgi would be becoming the extracellular

portion of the plasma membrane while the cytoplasmic

portion would remain the cytoplasmic portion of plasma membrane. so, sphingomyelin would be facing the

extracellular side but for glycolipids to face the

extracellular side, we will be requiring flippase. So,

flippase along with membrane fusion between vesicle

and plasma membrane would be involved.

Ref : Chapter 8.3: The Endoplasmic Reticulum, Page

279, Cell and Molecular Biology by Karp (6th Edition)


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Ans: 3

Explanation: Telomeres are ends of linear

chromosomes. they are specific structures that protect

the inside relevant DNA region from getting digested by

nucleases and also prevent from chromsome fusions.

plasmids like pYAC system use this technology of having

telomeres at the ends to make the plasmid more stable.

Ref: http://www.els.net/WileyCDA/ElsArticle/refId-

a0000379.html

Ans: 1

Explanation:

The

surface

area

of

the

raft

is

35

nm2

which is equal to 3500 A2. Cholesterol to sphingolipid

ratio is 2:1. So if we have y residues of sphingolipid

there are 2y molecules of cholesterol. The equation will

be 2y * 40 +y * 60 =3500, this gives y = 25

(sphingolipids) and cholesterol as 50

Ans: 2

Explanation: Polymerization of microtubles is inhibited

by colcemid, which in turn would inhibit the transport

of secretory vesicles.


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Ref : Intracellular Trafficking of Proteins by Clifford J

Steer, https://en.wikipedia.org/wiki/Colcemid

Ans: 2

Explanation: PLA2 are commonly found in mammalian

tissues as well as insect and snake venom.[2]Venom

from both snakes and insects is largely composed of

melittin, which is a stimulant of PLA2. Due to the

increased presence

and

activity

of PLA2

resulting

from

a

snake or insect bite, arachidonic acid is released from

the phospholipid membrane disproportionately. As a

result, inflammation and pain occur at the site.

Ref: https://en.wikipedia.org/wiki/Phospholipase_A2

Ans: 4

Explanation: Once primase has created the RNA

primer, Pol α starts replication elongating the primer

with ~20 nucleotides.[18] Due to their high processivity,

Pol ε and Pol δ take over the leading and lagging strand

synthesis from Pol α respectively.

Ref: https://en.wikipedia.org/?title=DNA_polymerase


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Ans: 4

Explanation: Option A is correct as orthodox seeds are

long lived seeds. they can be successfully dried to

moisture contents as low as 5% without injury. Option C

and D are also correct as precocious germination is

defined as that kind of germination wherein the seed

germinates without undergoing all the four stages of

germination,

i.e.,

globular,

heart

shape,

torpedo

shape,

and cotyledonary stage. ABA is known to inhibit

precocious germination. Option B says that dormant

seeds would germinate upon rehydration. this is

incorrect. they might germinate or might not as

germination of seeds is dependent not only on water

but also on oxygen and temperature.

Ref: https://en.wikipedia.org/?title=Germination,

Copeland et al. Principles of seed science and

tecnhnology

Ans: 4

Explanation: oxidation of glucose involves glycolysis

during which 2 NADH are produced by the action of

glyceraldehyde ‐3‐ dehydrogenase, coversion of

pyruvate to acetyl coA by PDH complex produces 2

NADH, oxidation of acetyl coA by citric acid cycle

produces 6 NADH molecules and 2 FADH2 molecules.

These reduced co‐enzymes are oxidized back by

electron transfer chain.

Ref : Ref: Lehninger's Book of biochemistry Chapter no.

14, 16 and 19


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Ans:

Explanation:

Ans: 1

Explanation: It has been observed that 90% of N‐H‐‐‐O

bonds in proteins lie between 140 and 180°, and that

they are centred around 158°C. There fore graph b

seems more appropriate

Ref :

http://www.cryst.bbk.ac.uk/PPS2/projects/day/TDayDiss/

HBonds.html

Ans: 1

Explanation: P1 has genotype Oc I+ Z‐, this strain even

though the operator carries constitutive mutation

phenotype, will still not result in beta‐gal production as

the structural genes Z‐ is defective. Strain P2 phenotype

is O+ I‐ Z+, here repressor is defective, so this will result

in constitutive expression of beta‐gal as there is no

active repressor to block the operator region.

Merodiploid will be represented as Oc I+ Z‐/O+ I‐ Z+.

Here, Lac I is a trans acting factor, so in the merodiploid

strain, which has both I+ and I‐ copy, will essentially

behave as I+. However Operator is a cis‐acting factor so merodiploid Oc/O+ will behave as O+ as functional Z

gene is present on O+ chromosome. Therefore the

expression of beta‐gal will be inducible.

Ref:

http://dwb4.unl.edu/Chem/CHEM869N/CHEM869NLinks

/www.acsu.buffalo.edu/~jbarnard/LacOperonII.html


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Ans: 1

Explanation: breast cancer categorized as ER +ve are

more prone to distant reccurence and are associated

with dormancy. The main cause being Cells that are

chemo‐resistant get disseminated to lungs induce the

expression of Periostin (POSTN) in fibroblasts, which

reciprocally increases WNT signaling and promotes

tumor‐initiating potential and tumor outgrowth.

Similarly, lung micrometastases can also produce

tenascin C (TNC), an ECM protein, that facilitates

maintenance of tumor initiating potential by activating

WNT and Notch pathways in cancer cells.

Ref :

http://clincancerres.aacrjournals.org/content/19/23/6389.

full

Ans: 2

Explanation: Calculate Kcat for each of the protein as

Vmax/Protein concentration. Km is calculated from the

graph as substrate conc at which 1/2 Vmax is obtained.

this value for all the protein comes to be 1 uM. So the

Catalytic efficiency depends on Kcat alone.

Ref: For protein A: Kcat = 20/1 = 20; For protein B: Kcat

= 30/4 = 7.5; For protein C: Kcat = 12.5/2 = 6.25; So

catalytic effciency is in the order a>b>c


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Ans: 3

Explanation: Transcription termination in eukaryotes is

less understood but involves cleavage of the new

transcript followed by template‐independent addition

of adenines at its new 3' end, in a process called

polyadenylation. When the RNA is cleaved,

polyadenylation starts, catalysed by polyadenylate

polymerase. Polyadenylate polymerase builds the

poly(A) tail by adding adenosine monophosphate units

from adenosine triphosphate to the RNA, cleaving off

pyrophosphate

Ref :

https://en.wikipedia.org/wiki/Transcription_(genetics),

https://en.wikipedia.org/wiki/Polyadenylation

Ans: 3

Explanation: RNA editing through the addition and

deletion of uracil has been found in kinetoplasts from

the mitochondria of Trypanosoma bruce The

mechanism of the editosome involves an endonucleolytic cut at the mismatch point between the

guide RNA and the unedited transcript. The next step is

catalyzed by one of the enzymes in the complex, a

terminal U‐transferase, which adds Us from UTP at the

3’ end of the mRNA. Here U is found in the stretches

dispersed along the sequence, so deamination of C

resulting in U will also happen. Further trans splicing

follwed by demaination could result in sequences which

doesn't match genomic DNA.

Ref: https://en.wikipedia.org/wiki/RNA_editing,

https://en.wikipedia.org/wiki/Trans-splicing


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Ans: 1

Explanation: viral load will be increased in the test

group because of knockout of chemokine receptors. Plus due to further addition of proinflammatory

cytokines and HIV viral load will further increase. In the

control group, viral load will be comparitively less due

to presence of chemokine receptors.

Ans: 3

Explanation: Under such specific conditions in

microorganisms during stationary phase there is

presence of 100 S ribosome which do not have

translational activity but formation of such ribosomes

requires specific proteins

Ref : http://onlinelibrary.wiley.com/doi/10.1111/j.1365-

2443.2009.01364.x/full


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Ans: 2

Explanation: TFIIH, have helicase and ATPase activities

and help create the transcription bubble, and TFIIH is

involved in nucleotide excision repair.

Ref:

https://en.wikipedia.org/wiki/Transcription_factor_II_H Ans: 3

Explanation: Since all the vulval precursor cells are

under the control of anchor cell. so if anchor cell is

removed then it would result in no vulva formation.

Also, signals from hypodermis would be inhibiting the

development of primary and secondary vulva. However,

these signals are overcomed by the signals from anchor

cell (LIN 3)

Ref:

https://books.google.co.in/books?id=WbO6BwAAQBAJ&pg=PA252&lpg=PA252&dq=a+constitutive+fate+from+h

ypodermis+inhibits+the+development+of+both+primary+

and+secondary+fates&source=bl&ots=9nIjfsSVMS&sig=

LtfnaLKk8lKs6YIeHuHpnTbQfw4&hl=en&sa=X&ei=P4O

GVczrN8emuQTcoo2YCg&ved=0CB0Q6AEwAA#v=one

page&q=a%20constitutive%20fate%20from%20hypoder

mis%20inhibits%20the%20development%20of%20both

%20primary%20and%20secondary%20fates&f=false


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Ans: 2

Explanation: Since

cohesion

is mutated,

therefore

sister

chromatid will not be joined together and hence

improper reduction division in meiosis will result into

low viability of tetrads

Ans: 4

Explanation: The frequency depends upon the order of

the genes. Most frequent Leu+ arg‐met‐ followed by

leu+ arg+

met

‐and

least

frequent

leu+arg+met+

Ref : Figure 8.15: Page 214. Pierce Benjamin

Ans: 3

Explanation: As B, C, and D are involved in the steps of

nitogen fixation

Ref : https://en.wikipedia.org/wiki/Nod_factor

Ans: 2


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Explanation: Wingless is expressed in the posterior

region of each segment or in the anterior portion of

each parasegment. If RNAi of wingless is present then it

will not allow wingless to be expressed because of

which the posterior compartment of each future

segment will get affected.

Ans: 2

Explanation: Connexons are gap junction proteins. Gap

junctions are a specialized intercellular connection

between a multitude of animal cell‐types.They directly

connect the cytoplasm of two cells, which allows

various molecules, ions and electrical impulses to

directly pass through a regulated gate between cells. So

when the inhibior P is added, gap junctions are not

properly formed, thus the protein P can not move

between the cells

Ref : https://en.wikipedia.org/wiki/Gap_junction,

https://en.wikipedia.org/wiki/Connexin

Ans: 1

Explanation: Till 4 cell stage, since the cells contain part

of the animal and vegetal axis, therefore, if isolated they

would be giving rise to the entire embryo. However, at

the 8 cell stage, since the blastomeres have been

formed as a result of equational division, thereby the

animal and vegetal exis gets divided. So these

blastomeres would not be able to give rise to the intact pluteus larvae.



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Ans: 1

Explanation: In picture 1. no cellular signalling is

involved , It depicts cellular migration via

polymerisation of actin filaments along the leading end

of the cell

Ref :

https://en.wikipedia.org/wiki/Cell_migration,

http://www.ncbi.nlm.nih.gov/books/NBK9961/

Ans: 4

Explanation: Pyruvate dehydrogenase is inhibited when

the following ratios are increased: NADH/NAD+ and

acetyl‐CoA/CoA.

Ref :

https://en.wikipedia.org/wiki/Pyruvate_dehydrogenase_c

omplex,http://watcut.uwaterloo.ca/webnotes/Metabolism/pdhReg

ulation.html


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Ans: 3

Ref : https://en.wikipedia.org/wiki/Ingression_(biology)

Ans: 2

Explanation: The cells cannot enter the M phase

because of the inhibition of cdk 1 phosphorylation. At

this stage sister chromatids will be held together by

cohesins throught the length of the chromosome.

Ref:

https://books.google.co.in/books?id=a2YdBQAAQBAJ&p

g=PA212&lpg=PA212&dq=sister+chromatids+are+held+

together+by+condensins+and+attached+to+each+other

+only+at+the+centromere&source=bl&ots=-

wo82Yv1Cf&sig=3gzyO00T3j-

qQzgMBwr_561Wirw&hl=en&sa=X&ei=b5-

GVebSDJepuwTSk7ioBA&ved=0CFgQ6AEwCQ#v=one

page&q=sister%20chromatids%20are%20held%20toget

her%20by%20condensins%20and%20attached%20to%

20each%20other%20only%20at%20the%20centromere

&f=false

Ans:

Explanation:


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Ans: 1

Explanation: A is a protein, B is hapten and C isadjuvant.

Hence in first graph there is no anti A production. in the

second graph the adjuvant has enhanced the

immunogenicity of B (hapten) and there is increased

antbody production. In the third graph there is hapten protein conjugation plus enhanced immunogenicity

when administered with an adjuvant. Hence there in

production of more of anti B compared to anti A

Ref : https://en.wikipedia.org/?title=Hapten,

Ans: 4

Explanation: Phytochrome C is not the most prominent

photoreceptor of red light. In rice and Arabidopsis,

PHYTOCHROME C (PHYC) requires other phytochromes

for stability and function. option D is false, as Far red light is preceived by phy A. Statements A and C are true.

Ref : https://en.wikipedia.org/wiki/Cryptochrome,

http://www.pnas.org/content/111/28/10037


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Ans: 3

Explanation: As the parents have only one band

therefore F1 will have the allleles in trans. The

recombinants are 10 (5+5)/ Therefore distance =10cM

Ans: 2

Explanation: Nonreducing sugars are less likely to react

with other substances along the way.

Sucrose can be uploaded by either symplastic and apoplastic route

Route: mesophyll cells‐>phloem parenchyma‐

>companion‐>sieve

Transport is via pressure flow model

Ref:

https://books.google.co.in/books?id=e8BxM8FOTv8C&p

g=PA97&lpg=PA97&dq=sucrose+uploading&source=bl&

ots=j2J6PbgqhB&sig=xMDzs-d1-

feyiDmXWJnOSLGqN64&hl=en&sa=X&ei=l5SGVeW1L4

mNuASGj6bQDQ&ved=0CB4Q6AEwAA#v=onepage&q

=sucrose%20uploading&f=false

Ans: 3


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Explanation: Pheopytin‐>QA‐>QB‐>b6f ‐>PC‐>P700

P700‐>PQ ‐>FeSx‐>FeSA‐>FeSB‐>FD

Ref : Figure 7.21: Page 125. Taiz and Zinger

Ans: 4

Explanation: Changes in 2,3 BPG concentration are not

related to biochemical changes during physical

excercise. It only decreases the oxygen affnity of Hb at

high altitudes. Metabolites like lactic acid that

accumulate after excercise they decrease the pH thus

leading to decrease in affinity of Hb to oxygen (right

shift). Thus the statement D is also incorrect as it says

that accumulation of metabolites lead to increase in

affinity of Hb to oxygen

Ref :

http://ceaccp.oxfordjournals.org/content/4/6/185.full.pdf,http://link.springer.com/article/10.1007%2FBF00511232

Ans: 4

Explanation: Recessive epistasis ratio is 9:3:4, moreover

in chi squared test we assume null hypothesis to be true. Null hypothesis will be that plants with red and

white fruits will occur in 1: 1ratio. While interpreting the

results, If the calculated chi squared is less than the

critical value we accept the null hypothesis. Only when

this value is greater we reject it. In the question,

Observed Chi sq value is less than the critical value so

we accept the null hypothesis. Therefore statements B

and D are true

Ref :

http://www.ndsu.edu/pubweb/~mcclean/plsc431/mendel/

mendel4.htm, https://en.wikipedia.org/wiki/Chi-squared_test


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AnS: 2

Explanation : Since the glomerulus is negatively

charged, therefore, fractional clearance of cationinc

dextrans would be more as compared to the fractional

clearance of neutral dextrans which should be the of

the same molecular size. With increase in Stokes Radius or the molecular size, the fractional clearance keeps on

decreasing gradually. Hence, this is justified by graph 2.

Ref :

https://www.google.co.in/search?q=the+fractional+cleara

nce+of+cationic+and+neutral+dextrans+from+kidney&bi

w=1137&bih=752&source=lnms&tbm=isch&sa=X&ei=IK

yGVbiFE86VuATv3IL4Bg&ved=0CAYQ_AUoAQ&dpr=0.

9#imgrc=m5LJZ2fzb5FblM%253A%3BswP4yps_qBZ54

M%3Bhttp%253A%252F%252Fd6igaq6njxgjh.cloudfront

.net%252Fcontent%252Fphysrev%252F88%252F2%25

2F451%252FF11.large.jpg%3Bhttp%253A%252F%252F

physrev.physiology.org%252Fcontent%252F88%252F2%252F451%3B1280%3B770

Chapter 26: Urine Formation by the Kidneys, Page 317,

Textbook of Medical Physiology by Guyton (11th Edition)

Ans: 1

Explanation:

https://en.wikipedia.org/wiki/Loktak_Lake

https://en.wikipedia.org/?title=Nanda_Devi_National_Par

k

https://en.wikipedia.org/wiki/Rajaji_National_Park

https://en.wikipedia.org/wiki/National_Chambal_Sanctua

ry


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Ans: 3

Explanation: Statement C is false. Plant would

accumulate a larger amount of solute in it's cell rather

than relasing.

Ref : https://books.google.co.in/books?id=zDIaumF3MCYC&p

g=PA153&lpg=PA153&dq=plant+cells+tend+to+release

+solute+during+osmotic+stress&source=bl&ots=7GoP-

x2I6k&sig=rxIj3EjneB0zkAQiG8sF4asytxA&hl=en&sa=X

&ei=1Z2GVZzFK8ifugTA2ZHgBg&ved=0CCIQ6AEwAA#

v=onepage&q=plant%20cells%20tend%20to%20release

%20solute%20during%20osmotic%20stress&f=false

Ans: 4

Explanation: Somatic hybridization would cause, other

wise heterozygous individual to transform into

homozygous and hence patches of mutants cells and

homozygous wild type. Some of the tissue might end up

with only one kind of alleles and hence would cause

tissue specific expression of the allele.

Ans: 3

Explanation: Could not be X linked recessive as for the

trait to appear in daughter, it has to be present in


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father.

Could not be autosomal recessive as the trait is becuase

of rare allele and person coming from outside the family

would not have the allele (male in the second

generatioe) and hence his offspring would not have the

trait, which is not the case.

Autosomal dominant with incomplete penetrance, can

explain the pedigree. the female in the first generation

would be heterozygous and she would pass this

dominant allele to her daughter, but the trait is not

seen because of incomplete penetrance. In the next

generation it would be again expressed.

Ans: 1

Explanation: androstenedione is synthesized from

cholesterol which in turn gets converted to estrone or

estradiol directly or gets converted to testosterone first

and then to estradiol. Since LHR is present on Theca

Interna cells and FSHR on Granulosa cells. Therefore,

option 1 would be correct.

Ref : http://lsresearch.thomsonreuters.com/maps/849/

http://omicsonline.org/estradiol-synthesis-and-

metabolism-and-risk-of-ovarian-cancer-in-older-women-

taking-prescribed-or-plant-derived-estrogen-

supplementation-2157-7536.S12-003.pdf

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3327826/

Ans: 1

Explanation: Statements A, B and E are correct. Based

on the MspII digestion profile, we can observe that the

mother is normal phenotype (since she has both active

restriction site clear from the band pattern), father is a

carrier (since the digestion profile shows the presence

of bands coresponding to both presence as well as


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absence of the site‐heterozygous for the RFLP allele),

Son (I) is expressing the phenotype so he is the

proband. Son (II) again has normal phenotype. Daughter

also is a carrier for the RFLP allele.

Ans: 1

Explanation: Only oaks have flowers and rest 3 don’t

have any flowers. Ferns are peteridophytes, pine is a

gymnosperm and hornwort is a bryophyte

Ans: 4

Explanation: sound localiztion is the ability of an

induvidual to to first locate the origin of sound. Hence it

depends on time between the arrival of stimulus in 2

ears as the ear closer to the sound source percieves it

earlier compared to the other. At the same time th

phase also plays a role in sound localization. when

sound comes from one side the ear on the other side

percieves the sound shadow. the tuning curves of

auditory nerves help in sensing sound but not

localization and the tuning curves of auditory nerves

have many over laps. hence option c is false. The

superior olivatiry nucleus and the auditory cortex play a

role in sound localization hence option D is also wrong

Ans: 3

Ref : https://en.wikipedia.org/?title=Aardvark, https://books.google.co.in/books?id=4abcBwAAQBAJ&

pg=PA739&lpg=PA739&dq=opposite+thumbs,+forward

+facing+eye+well+developed+cerebral+cortex,+omnivor

ous&source=bl&ots=hN7618F2rv&sig=CCBSlve4ayMOc9

Pkvko‐

FaD9jVM&hl=en&sa=X&ei=QcCGVfvdJMSjugTzrbL4Bw&

ved=0CDgQ6AEwBQ#v=onepage&q=opposite%20thum


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bs%2C%20forward%20facing%20eye%20well%20develo

ped%20cerebral%20cortex%2C%20omnivorous&f=false,

https://en.wikipedia.org/wiki/Even‐toed_ungulate

Ans: 3

Explanation: In case of A, as both the species are

negatively affected so it would be competition. In B, one

of the species is benefitted another species is negatively

affected so it is a type of predation. In C, both the

species are positively aff ected, so it is mutualism. In D,

one of the species is negatively affected while other is

not affected at all so this would be ammensalism

Ans: 4

Explanation: transcription factors and erythropoitin

secretion and cytochrome oxidase levels and myoglobin

content are elevated in hypoxia. hence C and d are

wrong

Ref : http://www.ncbi.nlm.nih.gov/pubmed/11035279,

http://jeb.biologists.org/content/215/5/806.full


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Ans:

Explanation:

Ans: 1

Explanation: Since endotherms lose much of their energy as heat and in respiration, therefore they would

be having low digestion and ecological efficiency as

compared with ectotherms. This is because ectotherms

do not need to maintain their body temperature thus

they would be more efficient in deriving maximum

benefit from the consumed food

Ref : https://books.google.co.in/books?id=EwPwPHDM-

YsC&pg=PA175&lpg=PA175&dq=assimilation+and+ecol

ogical+efficiency+of+endotherms+and+ectotherms&sour

ce=bl&ots=AIFyH6cPYU&sig=qlv_BsweGmEIU39yCW_

L6SXYR_Q&hl=en&sa=X&ei=BnKGVbSMMpSIuwSTlYKQBQ&ved=0CCgQ6AEwAQ#v=onepage&q=assimilation

%20and%20ecological%20efficiency%20of%20endother

ms%20and%20ectotherms&f=false


44/54



Ans: 2

Explanation: Species richness is highest in area of

intermediate disturbances. This is because very

frequent disturbance eliminates sensitive species,

whereas very infrequent disturbance allows time for

superior competitors to eliminate species that cannot

compete. Hence, all the other statements are correct

except for statement C.

Ref:

http://www.biology.uco.edu/personalpages/CButler/BI

O3543/lectures/Week_10_Thursday.pdf

http://www.eoearth.org/view/article/156216/

Ans: 1

Explanation: Shannon Diversity index = ‐ summation of

(pi) x (ln pi)

pi = No. of organisms in a species/ Total no. of

organisms of all the species

Community 1

pi of A = 25/100 = 0.25

Same would be the pi of B, C and D

Shannon Diversity Index for Community 1 = ‐4 x (0.25) x

(‐1.4) = 1.4

Community 2

pi of A = 80/100 = 0.8

pi of B = 5/100 = 0.05

pi of C = 5/100 = 0.05

pi of D = 10/100 = 0.1

Shannon Diversity Index of Community 2 = ‐ (0.8) x (‐

0.2) + 2 x (0.05) x (‐3.) + (0.10) (‐2.3) = 0.69


45/54



Ans:

Explanation:

Ans: 1

Explanation: Lyme disease, also known as Lyme

borreliosis, is an infectious disease caused bybacteria of

the Borrelia type. H. pylori causes chronic gastritis,

Typhus fevers are caused by the rickettsiae bacteria and

transmitted by arthropod. Scarlet fever from

streptococcus pyogenes

Ref :

https://en.wikipedia.org/?title=Lyme_disease,

http://www.healthline.com/health/gastritis‐

chronic#Causes2,

http://www.healthline.com/health/typhus#Overview1,

https://en.wikipedia.org/wiki/Scarlet_fever

Ans: 1


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Explanation: In experiment I, both the pigeons will be

able to find the rightdirection on basis of sunlight. While

in second experiment the control group will be able to

find the direction by using magnetic field while the test

one will not.

Ans: 1

Explanation: R‐selected species have smaller body size,

reproduce only once in their lifetime producing many

offsprings (semelparous reproduction) and develop

rapidly

Ref : Part VII: Population Ecology, Page 510, Biology by

Raven (6th Edition)

Ans: 3

Explanation: rB>C. Here C=30 and r=0.25 (for step

siblings)

So B>30/0.25 or B>120

Ans: 1

Explanation: tracheids ‐chief water conducting element

in gymnosperms, vessel elements are chief water‐

conducting element in angiosperms, Sieve cell is the

food conducting element in gymnosperms and Sieve

tube element is the food conducting element in

angiosperms

Ref :

https://books.google.co.in/books?id=kGYdBQAAQBAJ&

pg=PA557&dq=chief+water+conducting+elements+in+g

ymnosperms&hl=en&sa=X&ei=nbiGVd_WCsa9ugTZ14P

wDQ&ved=0CBwQ6AEwAA#v=onepage&q=chief%20wa

ter%20conducting%20elements%20in%20gymnosperms

&f=false


48/54



Ans: 4

Explanation: Crustacea is a class of Arthropoda.

Hominidae is a family. Dermaptera is an order in

Arthropoda. Ctenophora is a phylum while Archae is a

Domain

Ref :

https://www.google.co.in/?gfe_rd=cr&ei=RGaGVYX5N4v

C8gezvKsI&gws_rd=ssl#q=dermaptera

https://en.wikipedia.org/wiki/Hominidae

https://en.wikipedia.org/?title=Crustacean

https://en.wikipedia.org/wiki/Ctenophora

http://www.ucmp.berkeley.edu/archaea/archaea.html

Ans: 3

Explanation: Frequency of A= homozygous frequency

+1/2*heterozygous

=0.59+1/2 (0.16)=0.67. Frequency of 'a'=1‐0.67=0.33

Ans:

Explanation:

Ans: 3


49/54



Explanation: The procedure is for Western blotting: C‐

A‐B‐D

Ans: 3

Explanation: Cre‐Lox recombination is known as a site‐

specific recombinase technology, and is widely used to

carry out deletions, insertions, translocations and

inversions at specific sites in the DNA of cells.The

system consists of a single enzyme, Cre recombinase,

that recombines a pair of short target sequences called

the Lox sequences. Placing Lox sequences appropriately

allows genes to be activated, repressed, or exchanged

for other genes

Ref : http://cre.jax.org/introduction.html

Ans: 3

Explanation: Inbreeding causes decrease in growth rate,

fecundity, and survival

Ref:

http://www.fao.org/docrep/006/x3840e/X3840E07.htm

Ans: 2

Explanation: PTMs on serine can be either

phosphorylation or glycosylation among the 4 options

given. Ubiquitination occurs on Lysine and ADP‐

ribsoylation occurs on Arginine. Among phosphorylation

or glycosylation, both will cause increase in molecular

weight leading to a slowly moving protein. However,

phosphorylation ac bring about change in the isoelectric

point of the protein as phosphoryl group is a charged

molecule and thus will affect the ionization of the

protein.

Ref : https://en.wikipedia.org/wiki/Post‐

translational_modification


50/54



Ans: 3

Explanation: Since the cells were resistant to ganciclovir

this means that the region of DNA construct between

the homology regions will not carry TK gene (thymidine

kinase). This is because ganciclovir along with TK is a

suicidal combination. TK converts it into a toxic product

that is generally used for elemination of TK+ cells. But

the knock out mice was resistant to it so it means it did

not carry a TK genes. These cells were sensitive to G418,

this is an antibiotic similar to gentamicin and resistance

is conferred by neo resistance genes. Since the cells

were sensitive to G418, so the region between

homology also doesn't have neo gene. So option 3

seems to be the construct that doesn't contain either

neo or TK gene

Ref : https://en.wikipedia.org/wiki/G418,

https://en.wikipedia.org/wiki/Thymidine_kinase

Ans: 2

Explanation: more stringent washing will remove non‐

sepcific hybridization mainly. More stringent the wash

is, the partially complementary sequence will be pulled

apart. More stringency is brought about by low salt and

high temeperature. the role of salt is to shield the

negative charges on the DNA thus stablizing the hybrid.

Ref :

http://www.cliffsnotes.com/sciences/biology/biochemistry

-ii/molecular-cloning-of-dna/dna-hybridization,

https://lifescience.roche.com/shop/PrintView?langId=-

1&storeId=15016&articleId=66008


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Ans: 3

Explanation:

https://books.google.co.in/books?id=4abcBwAAQBAJ&p

g=PA527&lpg=PA527&dq=angiosperm+dominance+incr

eases,+continued+radiation+of+most+present+day+ma

mmalian+order&source=bl&ots=hN7617J5tu&sig=AaGh

hGJjnk6UwfA6pnR22l8_lNQ&hl=en&sa=X&ei=EaqGVY_

2FZeUuASd9LiwCg&ved=0CB0Q6AEwAA#v=onepage&

q=angiosperm%20dominance%20increases%2C%20co

ntinued%20radiation%20of%20most%20present%20day

%20mammalian%20order&f=false

Ans: 4

Explanation: According to the question, statement 1, 2

and 3 all can be justified. So, option 4 would be correct.


52/54



Ans: 1

Explanation: Improving resolution means having a

lower value of D, which can be achieved by lower value of lambda or higher value of N.

Ans: 3

Explanation: The 95% of the data would be in between

u+ 2 sigma and u‐2sigma:

u=200 and sigma=24. Therefore null hypothesis will be

rejected with 95% confidence limit above

200+(2*24)=248

Ans: 2

Explanation: Auxin 2,4‐D is critical for Somatic

embryogenesis, maltose is more successful than

sucrose.

Ref: Book‐ Somatic embryogenesis in woody plants Vol

C, Jain et al. 2000,

https://books.google.co.in/books?id=HdboCAAAQBAJ&

pg=PA159&lpg=PA159&dq=plant+growth+medium+sucr

ose+%26+maltose&source=bl&ots=zyMWj9_2_a&sig=5‐

AR8_k8OaZoxfKLtwPFXJ6KWb0&hl=en&sa=X&ei=sdqGV

YKABNSRuASV5L‐

oCQ&ved=0CC4Q6AEwAg#v=onepage&q=plant%20gro

wth%20medium%20sucrose%20%26%20maltose&f=fals

e


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Ans: 4

Explanation: Since in case of Retrovirus, gene for Factor

IX would be integrated in non‐dividing cells, therefore,

these will keep on expressing Factor IX. In Adeno‐

associated virus, the gene for Factor IX is integrated in

non‐dividing cells so it would be expressed for the time

till non‐dividing cells survive. In adenovirus, the gene for

Factor IX is not integrated and only transfected in both

dividing and non‐dividing cells, so they would be

showing only a transient expression of factor IX.

Ans: 1

Explanation: Sulphonylureas inhibit ALS (acetolactate

synthase) by binding to the enzyme. Glyphosate's mode

of action is to inhibit a plant enzyme involved in the

synthesis of the aromatic amino acids

Ref : https://en.wikipedia.org/?title=Glyphosate,

http://www.jstor.org/stable/4045526?seq=1#page_sca

n_tab_contents


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