*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in
Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com1Model Question Paper-01General Science
(A) & Life Sciences (B & C)Time: 3:00 Hrs. Max. Marks:
200This Test Booklet will contain 145 (20 Part A +50 Part B + 75
Part C) Multiple Choice Questions (MCQs). Candidates will be
required to answer 15 in part A, 35 in Part B and 25 questions in
Part C respectively (No. of questions to attempt may vary from exam
to exam). In case any candidate answers more than 15, 35 and 25
questions in Part A, B and C respectively only first 15, 35 and 25
questions in Parts A, B and C respectively will be evaluated.
Questions in Parts A and B carry two marks each and Part B
questions carry four marks each. There will be negative marking
@25% for each wrong answer. Below each question, four alternatives
or responses are given. Only one of these alternatives is the
CORRECT answer to the question.PART-A (General Science)1. A
shooting star is1. A small star moving away from the earth at a
very high speed2. A fast moving satellite that shines by sunlight3.
A heavenly object that shines because it is heated by the friction
of the earths atmosphere as it falls at a great speed4. A star of
an extremely high density2. By what process is heat transmitted
from the filament of an evacuated electric bulb to the glass?1.
Conduction2. Convection3. Radiation4. Heat cannot be transmitted
through a vacuum3. If the length of a heater coil is reduced by 10
percent of its original length, then the power consumed by the
heater will1. Increase over 10%2. Decreased by 10%3. Increase by
0.5%4. Decreased by 0.5%*MUDRA* Model Question Paper for NET-LS
& JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com24. The total output of photosynthesis
including the organic matter used up in respiration during the
period of measurement is called:1. Net primary productivity2. Gross
primary productivity3. Net community productivity4. Secondary
productivity5. Haemoglobin is 1. The coloring matter of leaves of
plants2. The coloring matter of blood3. A compound present in
milk4. A compound that transmits signals to the brain6. In solution
of HF and acetic acid,1. CH3COOH behaves as a base2. CH3COOH
remains unionized3. HF remains unionized4. HF behaves as a base7.
When a pencil is partly immersed in water in a beaker and held in a
slanting position, the immersed portion appears1. Bent towards the
bottom2. Bent towards the water surface3. Bent in a zigzag manner4.
As if it was not immersed8. A codon:1. Is a sequence of three bases
in a row2. Signals that three particular amino acids be
incorporated into a growing peptide3. Helps position in the new
amino acids correctly by hydrogen bonding with an anticodon of
tRNA4. May have more than one meaning depending upon its location
in the mRNA polymer*MUDRA* Model Question Paper for NET-LS &
JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com39. Which of the following are false
statements?I) Ozone is an allotropic form of oxygen.II) Phosphorus
is present in DNA and RNAIII) Halogens occur in free state in
nature.IV) The compounds of chlorine are used as cooling agents1.
I, IV 2. II, IV3. III, IV 4. I, II10. The incorrect gas is a
mixture of1. Silk Polyamide 2. Butane and propane3. Methane and
ethylene 4. Carbon dioxide and oxygen11. RuBisCO is made up of:1.
14 sub units 2. 15 sub unit3. 16 sub units 3. 18 sub unit 12. A
veena player compares with a tuning fork the fundamental frequency
generated by one of the strings of the veena and hears 4 beats per
sec. He then tightens the string a bit and hears only 3 beats per
second. Then1. The string has a higher frequency than the tuning
fork and must be tightened more2. The string has a lower frequency
than the tuning fork and must be tightened more to make the
frequencies equal3. The string has a lower frequency and its
tension has to be relaxed to make the frequencies equal4. The
string has a higher frequency than the fork and has to be loosened
for equality with fork13. A. It is advantageous to use microbial
enzymes in organic synthesis than mammalian enzymes.R.
Microorganisms can be made available. If needed, in much higher
amounts than the animals.1. Both A and R are correct and R is the
correct explanation of A 2. Both A and R are incorrect3. A is
correct and R is incorrect4. Both A and R are correct and R is not
the explanation of A*MUDRA* Model Question Paper for NET-LS &
JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com414. TATA box is found in:1. Promoter
region of gene rich in A, T bases2. Promoter region of gene rich in
G.C. bases3. Operator region of gene rich in A, T, bases4. Operator
region of gene rich in G.C. bases15. The species that will not give
positive test for CN ions is 1. NaCN 2. Ca(CN)23. CH3CN 4.
[Fe(H2O)4 Cl2]CN16. Metamorphic rocks originate from1. Igneous
rocks2. Sedimentary rocks3. Both igneous and sedimentary rocks4.
None of these17. What would be the output of following program1. 2
2. 3 3. 4 4. 618. On a PC, how much memory is available to
application software?1. 1024 Kilobytes 2. 760 Kilobytes3. 640
Kilobytes 3. 560 Kilobytes*MUDRA* Model Question Paper for NET-LS
& JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com519. A cube of side 1 cm is painted by
putting a lacquer of thickness , negligible compared to the side of
the cube. The volume of the painted cube is approximately 1. 1 +
cm3 2. 1 + 3 cm3 3. 1 + 33 cm3 4. 1 + 3 cm3 20. Which of the
following straight lines passes through the point (1,1)? 1. y = 2x
+ 3 2. 2y = x6 3. x = 1 4. x = y + 1PART-B (Life Sciences)21. Which
of the following is true for non-competitive inhibition?1. E+1=E1
2. ES + 1 = ESI 3. E+I=ET,ES+I=ESI 4. None of these22. Which of the
following hormones is not a secretion product of human placenta? 1.
Human chorionic gonadotropin2. Prolactin3. Estrogen4.
Progesterone23. DNA repair is carried out by which of the
polymerases in eukaryotes?1. and 2. and 3. and 4. and .24. The
world conservation was prepared by1. UNEP 2. UNEP and WWF3. IUCN,
UNEP and WWF 4. IUCN and WWF*MUDRA* Model Question Paper for NET-LS
& JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com625. DNA fingerprinting refers to1.
Molecular analysis of profiles of DNA samples2. Analysis of DNA
samples using imprinting devices3. Techniques used for molecular
analysis of different specimens of DNA4. Techniques used for
identification of fingerprints of individual26. Which one of the
following genes is defective in patients suffering from severe
combined immunodeficiency syndrome (SCID)?1. Cystic fibrosis
transmembrane conductor (CFTR)2. Adenosinedeaminase 3.
Ribonucleotide reductase4. a 2-microglobulm27. Mitochondria and
chloroplast carry out oxidative phosphorylation and
photophosphorylation, respectively, by means of 1. Conformational
coupling 2. Chemiosmotic coupling3. High energy intermediate
coupling4. Sliding filaments28. Two populations of land snails have
been effectively isolated from each other for a long period.
According to the biological species concept, which of the following
would demonstrate that the two populations have become separate
species?1. The two populations behave differently when subjected to
same dose of pesticides2. Sterile hybrids are produced when member
of the two populations are experimentally mated3. DNA nucleotide
sequence are different between two populations4. The two
populations have different electrophoretic pattern of proteins29.
Hydrolysis of starch is more efficiently catalysed by1. Sulphuric
acid 2. Diastase3. Both 4. None of these30. Duodenum has
characteristic Brunners gland which secrete two hormones called1.
Kinase, estrogen*MUDRA* Model Question Paper for NET-LS & JRF
of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com72. Secretin, cholecystokinin3. Prolactin,
parathormone4. Estradiol, progesterone 31. A pure or nearly pure
water contains a BOD of approximately1. 30 mg/L 2. 20 30 mg/L3. 0 3
mg/L 4. 10 12 mg/L32. Which of the following is an advantage of
confocal microscopy over conventional fluorescence microscopy?1.
The interaction of a laser beam with the cell surface allows the
imaging of individual macromolecules.2. The use of electrons
instead of light to image the specimen results in greatly increased
resolving power.3. Optical sections can be taken at different
depths in a specimen.4. Only scattered light enters the microscope
lens, making the object appear illuminated against a dark
background.33. Which of the following properties is common to all
cytoskeletal motor proteins (such as kinesins, dyneins, and
myosis)?1. An actin-binding domain 2. Two globular-head domains3.
The ability to bind to biological membranes 4. ATPase activity34.
Two varieties of maize averaging 48 and 72 inches in height,
respectively, are crossed. The F1 progeny is quite uniform
averaging 60 inches in height. Of the 500 F2 plants, the shortest 2
are 48 inches and the tallest 2 are 72 inches. What is the probable
number of polygenes involved in this trait?1. Four.2. Eight.3.
Sixteen.4. Thirty two.*MUDRA* Model Question Paper for NET-LS &
JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com835. The term zygotic induction refers
to1. Embryogenesis of the fertilized egg2. Process of
fertilization3. Prophage induction in a F ~ (F minus) recipient
bacteria after Hfr strain mediated conjugation4. Prophage entering
the lytic cycle after UV irradiation of a lysogen36. What would be
the effect of addition of 2,4 D on the production of berberine by
cell culture of Thalictrum minus.1. To stimulate growth and thereby
urease secondary metabolite production2. Stimulate
dedifferentiation and thereby decrease secondary metabolite
production3. Stimulate proliferation and reduce secondary
metabolite production4. None of the above37. Which of the following
techniques is NOT ideal for immobilizing cell free enzymes?1.
Physical entrapment by encapsulation2. Covalent chemical bonding to
surface carriers3. Physical bonding by flocculation4. Covalent
chemical bonding by cross linking the precipitate38. The full
length coding sequence of an eukaryotic gene was expressed in
bacteria and the protein was purified. However, in the functional
assay, no activity was detected for the purified protein. The
reason could be:1. The host bacteria produced an enzyme that
inhibited the activity of the expressed eukaryotic protein2. The
purified protein was contaminated with bacteria3. The host bacteria
did not produce the essential co factors4. No post translational
modification on the protein expressed in bacteria39. Which of the
following fluorescent probes is used to monitor the progress of
amplification in Real time PCR?1. SYBR green 2. Rhodamine3. FITC 4.
Cyan blue*MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com940. The linked characters will always
inherit together till they are1. Delinked due to segregation 2.
Masked by dominance 3. Mutated 4. Separated due to crossingover.41.
Nullisomy is the term used for the condition when an organism has1.
An additional chromosome 2. One chromosome less than normal3. A
complete set of chromosomes except one homologous pair4. None of
the above42. Which part of the human body is most affected by
chronic lead toxicity?1. Muscles and bones 2. Nervous system3.
Reproductive system 4. Vascular system43. Pyrology is the study
of1. Distribution of animals in relation to temperature2. Role of
fire in ecology3. Distribution of plants in forests in relation to
sunlight4. Variation in species distribution in relation to ambient
temperature44. When nonsense mutations occur in the reading frame
of mRNA, protein synthesis gets terminated at the nonsense mutation
to deliver a truncated polypeptide. However, in certain bacterial
strains, this does not happen; these bacterial cells are able to
synthesize full-length polypeptide. This phenomenon is due to:1.
Compensatory frame shift mutation that occurs elsewhere in the
mRNA2. Involvement of suppressor RNAs3. Polypeptide splicing at the
broken point4. Post-transcriptional editing of the nonsense
mutation45. The net ecosystem production (NEP) is a function of1.
Gross primary production (GPP) and is independent of autotrophic
respiration (Ra)*MUDRA* Model Question Paper for NET-LS & JRF
of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com102. GPP, Ra, heterotrophic respiration
(Rh) and can be either positive or negative3. GPP, Ra,
heterotrophic respiration (Rh) and can be positive4. Net primary
production (NPP)46. Which of the following accounts for the extent
of proliferation observed in the retrovirus-infected cells not
treated with growth factors?1. The v erbB product has a function
similar to that of activated EGF receptor.2. EGF binds to both the
EGF receptor and the v erbB product.3. The v erbB product activates
both the LCGF and the EGF receptors.4. The v erbB product
antagonizes the action of the EGF receptor.47. Which of the
following takes place during anaphase of mitosis in an animal cell?
1. Kinetochore microtubules elongate to push chromosomes towards
the Metaphase plate.2. The chromosomes align on the metaphase
plate.3. Sister chromatids remain attached to each other at the
centromere and move toward the pole as a unit.4. Polar microtubules
elongate and slide to push the spindle poles apart48. Larger
islands may have greater species diversity than smaller islands
because larger islands1. Are in the tropics2. Are farther from
continents than smaller islands are3. Have more habitats than
smaller islands do4. Have greater genetic drift than smaller
islands do49. Two of the premises that form the basis of Darwins
concept of natural selection are1. Ecotype and race2. heritability
and fitness3. uniformitarianism and catastrophism4. Geographic and
reproductive is isolation 50. Blood fibrinogen is converted into
fibrin during1. CO2 transport 2. Oxygen transport3. An immune
repose 4. Clot formation*MUDRA* Model Question Paper for NET-LS
& JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com1151. In calves that consume large
quantities of milk, the curdling of milk takes place due to 1. A
large variety of useful bacteria2. The high acid content of gastric
juice3. The action of rennin4. The action of pepsin52. Honey is
formed by1. Worker bees from juicy plants2. The action of digestive
juices of worker bees on nectar collected from flowers3. The
selective absorption of fructose by worker bees from juicy plants
and its processing in their guts4. Worker bees in the nectarines of
flowers53. Acetyl CoA, a common key compound in intermediary
metabolism, is produced in the body byA. Breakdown of glucose under
aerobic conditionsB. Anaerobic breakdown of glucoseC. oxidation of
fatty acidsD. Breakdown of anti ketogenic amino acids1. A, C 2. A,
B, D3. B, D 4. B, C, D54. A normal couple has five children, two of
whom suffer from a somewhat uncommon genetic disorder that has,
however, appeared occasionally in this familial line. What kind of
gene is involved in this case?1. Codominant 2. completely dominant
3. Completely recessive 4. Incompletely dominant55. Given below are
two statements, one labeled as Assertion (A) and the other as
Reason (R).Assertion (A): Sonora 64 is a dwarf variety of wheat
that played a significant role in the green revolution of India. It
is called so because:Reason (R): A scientist, P. Sonora, developed
it in 1964, through mutation breeding.In view of the above two
statements, which one of the following is correct?*MUDRA* Model
Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences
(Model Question Paper No. 1) www.mudralifesciences.com121. Both A
and R are true and R is the correct explanation of A2. Both A and R
are true but R is not the correct explanation of A3. A is true but
R is false4. A is false but R is true56. Ribosomes similar to those
of bacteria are found in1. Pancreatic endoplasmic reticulum 2.
Liver mitochondria 3. Skeletal muscle cytoplasm 4. Daisy
chloroplasts.57. Bateson used the terms coupling and repulsion for
linkage and crossing over. Name the correct parental or coupling
type along with its cross over or repulsion?1. Coupling AABB, aabb;
Repulsion AABB, AABB2. Coupling AABB, aaBB; Repulsion AABB, aaBB3.
Coupling aaBB, aabb; Repulsion AABB, aabb4. Coupling AABB, aabb;
Repulsion aaBB, aaBB58. The biggest drawback of using cost benefit
analysis in environmental impact assessment is 1. Its theoretical
approach2. Conversion of intangible aspects into monetary units3.
Choice of discount rate4. Rudimentary nature of the process59. What
is the disadvantage in using electrostatic precipitators for air
pollution control1. Small particles cant be removed2. High pressure
drop3. Problem in handling hot gases4. High initial cost60. In
addition to the requirement of macronutrients, plants also need
micronutrients.1. Fe, Mn, Cu, Zn, Mo, B and Cl2. Cs, Sr, I, Mn, Zn,
Ba and Cl*MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com133. C, H, O, S, P Ca and K4. C, Na, Mn,
Mg, Cu, Mo and Cl61. Isozymes can be characterized by1. The
different chemical reactions that they catalyze2. The differences
in their elution profile from a size-exclusion column3. Differences
in their amino acid sequences4. All of the above62. Sodium Dodecyl
Sulphate (SDS) is used while separating proteins by polyacrylamide
gel electrophoresis because1. It helps in solubilization of
proteins thereby making it easier to separate 2. It binds to
proteins and confers uniform negative charge density thereby making
them move during electrophoresis3. Decreases the surface tension of
the buffer used for electrophoresis4. Stabilizes the proteins63.
When prospective neuroectoderm from an early amphibian gastrula is
transplanted in the prospective epidermal region of a recipient
(early gastrula) embryo, the donor tissue will give rise to:1.
neural tube.2. epidermis.3. neural tube and notochord.4. neural
tube and epidermis.64. A group of six cells called 'equivalence
group cells' divide to form the vulval structure in Caenorhabditis
elegans. They are called so because:1. they have similar fates
during development of vulva.2. all the six cells are competent to
form vulva and can replace each other under various experimental
conditions.3. they are all under the influence of the anchor cell,
signals from which initiate vulval development.4. they interact
with each other to form the vulval structure.*MUDRA* Model Question
Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model
Question Paper No. 1) www.mudralifesciences.com1465. In higher
plants, the red/far-red sensory photoreceptor, phytochrome, is a
light-regulated kinase. Which of the following classes of kinases
does it represent?1. Two-component sensor regulator (histidine
kinase).2. Two-component sensor regulator (serine/threonine
kinase).3. Leucine rich repeat (LRR) receptor kinase.4.
Calcium-dependent protein kinase.66. Absorption of UV radiation by
proteins and nucleic acids is due to transition of Electrons
between the1. Vibrational energy levels 2. Rotational energy
levels3. Nuclear energy levels4. Electronic energy levels67. A
myasthenia gravis patient develops muscle paralysis because1. the
nerve terminal at the neuromuscular junction fails to release
acetylcholine.2. although enough acetylcholine is released at the
neuromuscular junction, it is destroyed by acetylcholinesterase.3.
the patient develops immunity against his own acetylcholine
receptor.4. the patient develops antibody against his own
acetylcholine.68. During the course of prolonged starvation and
fasting, glucose or glycogen is synthesized from non -carbohydrate
precursors by the process of1. Glycogenesis 2. Glycolysis3.
Gluconeogenesis 4. Glycogenolysis69. Deletion of the leader
sequence of trp operon of E. coli would result in1. decreased
transcription of trp operon.2. increased transcription of trp
operon.3. no effect on transcription.4. decreased transcription of
trp operon in the presence of tryptophan.*MUDRA* Model Question
Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model
Question Paper No. 1) www.mudralifesciences.com1570. In the
endodermis of higher plants, the role of Casperian strip is to
control the water movement so that it flows1. between the cells.2.
through the plasma membrane.3. through the cell wall.4. through the
transfusion tissue.*****PART-C71. To approximate the actual
concentration of enzyme in a bacterial cell, assume that the cell
contains equal concentration of 1,000 different enzymes in solution
in the cytosol and that each protein has a molecular weight of
100000. Assume also that the bacterial cell is cylinder having a
diameter of 1m and height 2.0 m, that the cytosol (specific gravity
1.20) is 20% soluble protein by weight, and that the soluble
protein consists entirely of enzyme. Calculate the average
molecular concentration of each enzyme in this hypothetical cell.1)
2.4 x 10-5 M2) 1.4 x 10-6 M3) 1.4 x 10-4 M4) 2.4 x 10-6 M72. The
following is the pathway for the biosynthesis of molecule D
catalyzed by the enzyme E1, E2, and E3. A B C DE1 E2 E3The reaction
catalyzed by E1 in the absence of the other enzymes. You find that
the rate of the reaction decreases as you add increasing
concentration of D. What does this tell about the mechanism of
regulation of enzyme E11) Enzyme E3 is probably regulated by
feed-back inhibition2) Enzyme E1 is probably regulated by feed-back
inhibition3) Enzyme E1 is probably regulated by competitive
inhibition4) Enzyme E3 is probably regulated by competitive
inhibition*MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com1673. The stem of bamboo, a tropical
grass, can grow at phenomenal rate of 0.3m/day under optimal
condition. Given that the stems are composed almost entirely of
cellulose fibers oriented at the direction of growth, calculate the
number of sugar residues per second that must be added
enzymatically to growing cellulose chain to account for the growth
rate. Each D-glucose unit in the cellulose molecule is about 0.45nm
long.1) 5,800 residues2) 7,700 residues3) 10,700 residues4) 20,800
residues74. When a small amount of sodium dodecyl sulphate (Na+CH3
(CH2)11OSO-3) is dissolved in water, the detergent ions enter the
solution as a monomeric species. As more detergent is added, a
concentration is reached (the critical micelle concentration) at
which the monomers associated to form micelles. The critical
micelle concentration of SDS is 8.2mM. The micelle has an average
particle weight (the sum of molecular weight of the constituent
monomer) of 18,000. Calculate the number of detergent molecules in
the averaged micelle.1) 23 SDS molecules/micelle 2) 43 SDS
molecules/micelle3) 63 SDS molecules/micelle4) 83 SDS
molecules/micelle75. The sequence of monosaccharides including
position and configuration of glycosidic bonds in a glycoprotein is
to be determined. Which one of the following methods can be
employed?1. Glycoprotein removal of oligosaccharides by alkaline
hydrolysis nuclear magnetic resonance analysis of cleaved mixture
of oligosaccharides2. Two dimensional nuclear magnetic resonance
spectroscopic analysis of the glycoprotein3. Glycoprotein release
of oligosaccharides with endoglycosidases followed by purification
to separate oligosaccharides enzymatic hydrolysis of purified
oligosaccharides with specific glycosidases mass spectroscopic
analysis of smaller oligosaccharides4. Glycoprotein treat with
trypsin followed by MALDI analyses of tryptic peptides76. In
multipass transmembrane proteins the polypeptide chain passes back
and forth repeatedly across the lipid bilayer. It is thought that
an internal signal peptide serves as a start-transfer signal in
these proteins to initiate translocation, which continues until a
stop-transfer peptide is reached. In double-pass transmembrane
proteins, for example, the polypeptide is released into the bilayer
at this point. By recognizing the first appropriate hydrophobic
segment to emerge from the ribosome, the SRP sets the *MUDRA* Model
Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences
(Model Question Paper No. 1) www.mudralifesciences.com17reading
frame, if translocation is initiated, the next appropriate
hydrophobic segment will be recognized as a stop-transfer peptide,
causing the region of the polypeptide chain in between to be
threaded across the membrane. A similar scanning process continues
until all of the hydrophobic regions in the protein have been
inserted into the membrane. Because membrane proteins are always
inserted from the cytosolic side of the ER in this programmed
manner, all copies of the same polypeptide chain will have the same
orientation in the lipid bilayer. This generates:-1) a molecular
scaffold which help itself to pass through various barriers2) an
asymmetrical ER membrane in which the protein domains exposed on
one side are different from those domains exposed on the other.3) a
symmetrical ER membrane helps the protein domains exposed only one
side of the membrane4) Nothing happens77. The lipid-linked
precursor oligosaccharide is linked to the dolichol by a high
energy pyrophosphate bond, which provides the activation energy
that drives the glycosylation reaction. The entire oligosaccharide
is built up sugar by sugar on this membrane-bound lipid molecule
prior to its transfer to a protein. The sugars are first activated
in the cytosol by the formation of nucleotide-sugar intermediates,
which then donate their sugar (directly or indirectly) to the lipid
in an orderly sequence. All of the diversity of the linked
oligosaccharide structures on mature glycoproteins results from
later modification of the original precursor structure. This
oligosaccharide trimming or processing continues in the:1)
Endoplasmic reticulum2) Golgi complex3) Cytosol4) Vesicles78. The
proteins which take basic stain are known as the basic proteins.
The most important basic proteins of the nucleus are
nucleoprotamines and the nucleohistones. The nucleoprotamines are
simple and basic proteins having very low molecular weight (about
4000 daltons). The most abundant amino acid of these proteins has
the pH 10 to 11. The protamines usually remain bounded with the DNA
molecules by the salt linkage. The protamines occur in the
spermatozoa of the certain fishes. The nucleohistones have high
molecular weight, i.e., 10,000 to 1,00,000 daltons. The histone
proteins remain associated with the DNA by the ionic bonds and they
occur in the nuclei of most organisms. According to the composition
of the amino acids following types of histone proteins have been
recognized, as:1. Tyrosine, Threonine and Histidine2. Lysine,
Alanine and Serine3. Arginine, Lysine and Histidine.4. Proline,
Serine and Threonine79. Drosophila larvae are small, white,
wormlike creatures. When they first emerge from their eggs they are
about 1-2 mm in length, but rapidly grow to about 7-8 mm before
pupating. During the rapid larval growth stage, the larva's cells
need large quantities of proteins. Polytene chromosomes contain
many copies of *MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com18DNA molecules that have replicated
several times side by side to provide additional copies of DNA as
templates for transcription. The extra copies of DNA cause the
chromosomes to expand to an unusually large size, so they can
easily be visualized under a microscope, to confirm the site
specific expression of gene (transcription) which method would you
prefer?I) In situ hybridization using 3H containing probesII) In
vitro isolation of cromatin from polytene chromosomeIII) cDNA
analysis by cloning in bacterial cellIV) Southern Hybridization
method1) I only2) I and III only3) I and IV only4) II and III
only80. A man who had purple ears came to the attention of a human
geneticist. The human geneticist did a pedigree analysis and made
the following observations:In this family, purple ears proved to be
an inherited trait due to a single genetic locus. The man's mother
and one sister also had purple ears, but his father, his brother,
and two other sisters had normal ears. The man and his normal-eared
wife had seven children, including four boys and three girls. Two
girls and two boys had purple ears. The purple-ear trait is most
probably: 1. autosomal, dominant 2. autosomal, recessive 3.
sex-linked, dominant 4. sex-linked, recessive81. This antigen is
not found in man, however a great variety of other cells and
tissues could sensitize man to this antigen. Following an attack of
infectious mononucleosis, an infection by the Epstein-Barr virus,
antibodies are produced which react not only to the virus but also
to a completely unrelated antigen the sheep red cell. This is known
as the heterophile antibody response and forms the basis of the
Paul-Bunnel test for infectious mononucleosis. Name the antigen:1)
MHC antigens2) Monoclonal antibodies3) Unspecified antigens4)
Forssman antigen 82. During many important cell processes, many
proteins need to undergo degradation to culminate a part of the
process. For example, during cell cycle, cycling proteins need to
be degraded to allow the cells to exit *MUDRA* Model Question Paper
for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question
Paper No. 1) www.mudralifesciences.com19mitosis. This is achieved
by selective ubiquitination of cyclin followed by its degradation
by proteasomes. The specific protein factor that is involved in
this process is called Anaphase Promoting Complex (APC). APC is
possibly a protein which is known as1. E1 enzyme.2. E2 enzyme.3. E3
enzyme.4. Protease.83. Digestion of 4kb DNA molecule with EcoRI
yields two fragments of 1 kb and 3 kb each. Digestion of same
molecule with HindIII yields fragments of 1.5kb and 2.5kb. Finally
digestion with EcoRI and HindIII in combination yields fragments of
0.5 kb. 1 kb, and 2.5 kb. Select from the following the correct
restriction map indicating the position of the EcoRI and HindIII
cleavage sites.EcoRIHindIII1.0 kb 2.5kb2.5 kb4)EcoRIHindIII1.0 kb
2.5kb2.5 kb3)EcoRIHindIII1.5 kb 1.0kb2.5 kb2)EcoRIHindIII1.5 kb
1.0kb2.5 kb1)84. Upon studying a considerable number of different
crosses in Drosophila, Morgan reached the conclusion that all genes
of this fly were clustered into four linked groups corresponding to
the four pairs of chromosomes. Further studies revealed that
linkage is not absolute and it is broken frequently. It is broken
in prophase by a process called1. Recombination.2. Jumping of
genes.3. Integration.4. Mutation.*MUDRA* Model Question Paper for
NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper
No. 1) www.mudralifesciences.com2085. DNA repair, synthesis and
recombination are intimately connected and inter dependent. An
apparent commonality between processes of DNA replication and
repair in the enzymatically catalyzed synthesis of DNA
polynucleotide segments, which can be assembled with preexisting
polynucleotides, leading to repair or replication. Synthesis of
these polynucleotide segments is catalyzed by a group of enzymes
DNA-dependant DNA polymerases. In the case of E.coli, DNA
polymerase has been isolated in three distinctforms whereas five
main types of polymerase have been isolated from mammalian cells.
All the polymerases synthesize polynucleotides only in the 5' 3'
direction. If polynucleotide chains could be elongated in 3' 5'
direction, the hypothetical growing 5' terminus, rather than the
incoming nucleotide, would carry a triposphate that is unsuitable
for further elongation. The 3' 5 exonuclease activity is not
associated with all the polymerases and only present in:(A) All E.
coli DNA polymerases but not all mammalian polymerases.(B) Pol I,
Pol II, Pol III, Pol o, Pol |.(C) Pol I, Pol II, Pol o, Polc, Pol
(D) Pol I, Pol II, Polo, Pol c.The correct statements are1. (A) and
(B).2. (A), (B) and (C).3. (A) and (C).4. (A), (C) and (D).86.
Bacteriophage genetic circuit may be represented as follows:The
control of gene expression occurred during the phage infection may
be described as follows:(A) N and Q protein act as
antiterminator(B) CI acts only as repressor(C) CII act as a
retroregulator(D) CI and CII both act as positive and negative
regulatorWhich one of the statements are correct?1. (A), (B) and
(C).2. (B), (C) and (D).3. (A) and (D) only.4. (A), (C) and
(D).*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR
in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com2187. A type of gangrene develops in
necrotic (dead) tissue that has lost its blood supply. As a
consequence, spores of certain obligate anaerobes, can germinate
and vegetative cells can proliferate there. As they do, they
secrete hydrolytic enzymes and cytotoxins that kill and digest
surrounding host cells, expanding the necrotic area in which the
pathogen cells grow. Select the correct pathogen of this cause:1)
Streptococcal pharyngitis2) Staphylococcus aureus3) Clostridium
perfringens4) Bacillus anthracis88. Adrenergic hormones bind to a
family of G protein-linked receptors known as adrenergic receptors.
The individual members of this family differ mainly in their
preference for epinephrine or norepinephrine and in which G protein
is linked to the receptor. They can be broadly classified into -
and -adrenergic receptors. The -adrenergic receptors bind both
epinephrine and norepinephrine. The -adrenergic receptors bind
epinephrine much better than norepinephrine. A) - receptors are
located on the smooth muscles that regulate blood flow to visceral
organs. B) - receptors are found on smooth muscles associated with
arterioles that feed the heart smooth muscles of the bronchioles in
the lungs, and skeletal muscles.C) - receptors are located near the
medulla of KidneyD) - receptors are found on pyramids of KidneyFind
the appropriate locations of these receptors in effect to its
action:1) A Only2) B Only3) Both A and B4) A, B, C and D89. A young
dicot seedling (e.g soyabean) is subjected to gravity stimulus by
laying it horizontally on a surface the shoot bends upwards and
root bends downward. Indicate the reason.1. Redistribution of auxin
throughout the seedlings is responsible for stimulatory unequal
growth in shoots and roots.2. Redistribution of auxin in shoots
while cytokinine in roots is responsible for stimulatory unequal
growth.3. Redistribution of auxin in roots while cytokinine in
shoots is responsible for stimulatory unequal growth.4.
Redistribution of cytokinine throughout the seedlings is
responsible for stimulatory unequal growth in shoots and roots.90.
Quorum sensing is a type of decision-making process used by
decentralized groups to coordinate behavior. Many species of
bacteria use quorum sensing to coordinate their gene expression
according to the local density of their population. Similarly, some
social insects use quorum sensing to make collective decisions
about where to nest. In addition to its function in biological
systems, quorum sensing has several useful applications for
computing and robotics.Quorum sensing can function as a
decision-making process by:*MUDRA* Model Question Paper for NET-LS
& JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com221) assessing the number of other
components they interact with and a standard response once a
threshold number of components is detected.2) gene transcription to
be activated, the cell must encounter signaling molecules secreted
by other cells in its environment.3) regulating a host of different
processes, essentially serving as a simple communication network4)
All of these91. Plants grown in greenhouse at 25C when exposed
first to 35C for 6 hours and subsequently to 42C for 12 hours adapt
better to the high temperature (42C) in comparison to those
directly transferred to 42C for the same duration. What is the
phenomenon called and what is its main physiological basis?1.
Acquired thermo-tolerance because of the induction of
mutagensresulting into improved stability of all the proteins.2.
Induced thermo-tolerance because of the induction of heat shock
proteins.3. Induced thermo-tolerance because changes in RNA
polymerase II resulting in efficient and improved transcription.4.
Acquired thermo-tolerance because of efficient post translational
modification of proteins.92. Several atoms are assembled and held
together to form thousands of molecules which participate in the
building structural and functional biological systems. A bond is
any force which holds two atoms together. The formation of bond
between two atoms is due to some redistribution or regrouping of
electrons to form a more stable configuration. The regrouping of
electrons in the combining atoms may take place in which of the
following way(s):1) By a transfer of one or more electrons from one
atom to another2) By a sharing of one or more pairs of electrons
between the combining atoms3) By a combination of the two processes
of transfer and haring4) All of these93. In the 1920, while working
with Streptococcus pneumonia, the agent that causes pneumonia,
Griffith injected mice with different types of bacteria. For each
of the following bacteria types injected, indicate whether the mice
lived or died:A) Type IIRB) Type IIIRC) Heat Killed IIISD) Type IIR
+ Heat Killed IIIS*MUDRA* Model Question Paper for NET-LS & JRF
of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com23Results:A- LivedB- DiedC- LivedD-
DiedFind the appropriate event(s) from the above result1) Both A
and B2) Both A and B and C3) C only4) All A; B; C and D94. Random
copolymers were used in some of the experiments that revealed the
characteristics of the genetic code. For each of the following
ribonucleotide mixtures, give the expected codons and their
frequencies, and give the expected proportions of the aminoacids
that would be found in a polypeptide directed by the copolymer in a
cell-free Protein synthesizing system.1) 2 A : 6 C2) 4 G : 1 C3) 1
A : 3 U : 1 C4) 1 A : 1 U : 1 G : 1 C95. During development and
differentiation, there is a dynamic program of differential
expression of sets of genes. In bacteria, phage infections are
among the simplest examples of developmental process. Typically,
only a subset of the phage genome, offer referred to as immediate
early genes, are expressed in the host immediately after phage
infection. As time passes, early genes start to be expressed, and
the immediate early genes and bacterial genes are turned off. In
the final stage of phage infection, the early genes give way to
late genes. One of the simplest way it is achieved is through:(A)
expression of cascade of o factors(B) expression of new RNA
polymerases(C) expression of different holoenzymes(D) expression of
different transcription factorsThe correct reasons are1. (A), (D)2.
(A), (C), (D)3. (A), (B), (D)4. (A), (B), (C)*MUDRA* Model Question
Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model
Question Paper No. 1) www.mudralifesciences.com2496. During
receptor-mediated endocytosis, ligand first binds with cell surface
receptor, then traffic through Rab5 positive early endosomal
compartment. Finally, it moves to the Lamp1 positive lysosomes via
Rab7 positive late compartment. In order to understand the
trafficking of ligand A in epithelia cells, cells were allowed to
internalize ligand A for various period of times at 37C. Finally,
cells were stained with anti-ligand antibody and probed with
secondary antibody labelled with Alexa-Red fluorescence dye. Same
cellswere also co-stained with anti-Rab5, anti-Rab 7 or anti-Lamp1
antibody and probed with appropriate secondary antibody labelled
with Alexa-green fluorescence dyes. Cells were viewed in confocal
microscope and observations are (I) 5 min internalize ligand (Red)
in cells are colocalize with anti Rab5 antibody but not with
anti-Lamp1 antibody and (II) 90 min internalized ligand are
colocalized with anti-Lamp1 antibody but not with anti-Rab5
antibody. The following conclusions could be arrived at from the
above bservations.(A) Ligand A travels to early endosomal
compartment by 5 min.(B) Ligand A travels to early endosomal
compartment by about 90 min.(C) Ligand A travels to lysozome by
about 5 min.(D) Ligand A travels to early endosome by about 90
min.Identify the correct inferences.1. (A) and (B)2. (B) and (D)3.
(C) and (D)4. (D) and (C)97. Fur color in the babbit, a furry
little animal and popular pet, is determined by a pair of allele, B
and b. BB and Bb babbits are black, and bb babbits are white. A
farmer wants to breed babbits for sale. True-breeding white (bb)
female babbits breed poorly. The farmer purchase a pair of black
babbits, and these mate and produce six black and two white
offspring. The farmer immediately sells his white babbits, and then
he comes to consult you for a breeding strategy to produce more
white babbits. If he performed crosses between pairs of F1 black
babbits, what proportion of the F2 progeny would be white?1) 1/9 2)
2/93) 3/9 4) 6/998. Genes a and b are sex-linked and are located 7
mu apart of the X chromosome of Drosophilia. A female of genotype
a+b/ab+ is mated with a wild type (a+b+/Y). What is the probability
that one of her sons will wither a+ b+ or a b+ in phenotype?1) 0.25
%2) 0.50 %3) 0.75 %4) 1.00 %*MUDRA* Model Question Paper for NET-LS
& JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com2599. Two protein kinases, K1 and K2
function sequentially in regulating intracellular pathway in
response to extracellular signal. The following observations are
made:(i) Response is observed even in the absence of extracellular
signal when a mutation permanently activates K1.(ii) Response is
observed even in the absence of extracellular signal when K1
contains an activating mutation and K2 with inactivating
mutation.(iii) No response in the cells is detected even in the
presence of extracellular signal when both kinases are inactivated
by mutation.Which one of the following is correct?1. K1 activates
K22. K2 activates K13. K1 inhibits K24. K2 inhibits K1100. It is
claimed that the mean () arsenic concentration in the ground water
of village is 20g/L with = 3. In a random sample of 16 measurements
what values of arsenic concentration should lead to rejection of
the claim with 95% confidence?1. values lower then 18.53 and values
higher that 21.472. values higher than 21.473. values lower than
18.534. values lower than 18.07 and values higher than 21.93101.
Pattern baldness is more frequent in males than in females. This
appreciable difference in frequency is assumed to result
from-Y-linkage of this trait1) X-linked recessive with sex-limited
inheritance2) Sex-influenced autosomal inheritance3) Excessive
beer-drinking in males, consumption of gin being approximately
equal between the sexes.102. The structure of a protein is known
from X-ray diffraction studies which gave 30% o- helix, 50% |-sheet
and 20% random coil. Circular dichroism (CD) measurements gave 50%
o-helix, 40% |-sheet and 10% random coil. What could not be a
possible explanation for these observations.1. Protein structure in
the crystal is different from that in the solution.2. CD analysis
for structural components is not appropriate for this
protein.*MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com263. Contributions from other chromophores
also contribute to the CD spectrum of the protein4. Protein
contains high content of disulphide bonds.103. Which of the
following redox reaction would be expected to proceed as written?
(Assume standard conditions and the presence of appropriate
enzymes; E0values are shown below.)Half-reaction E0` (V)Fumarate +
2H+ + 2e- succinate 0.031Oxaloacetate + 2H+ + 2e- malate
-0.166Pyruvate + 2H+ + 2e- lactate -0.185Acetaldehyde + 2H+ +2e-
ethanol -0.197NAD+ + H+ + 2e- NADH -0.320Acetoacetat + 2H+
+2e--hydroxybutyrate -0.3461. Malate + NAD+ oxalaceteate + NADH +
H+2. Acetoacetate + NADH + H+ -hydroxybutyrate + NAD+3. Pyruvate +
-hydroxybutyrate lactate + acetoacetate4. Malate + pyruvate
oxaloacetate + lactate104. Which of the following sequences of
events occurs when E. coli are released from catabolite repression
by transfer to low-glucose medium?1. cAMP levels rise, cAMP binds
to CAP, cAMP-CAP complex binds to a site on DNA and activates
transcription.2. cAMP levels rise, cAMP binds to CAP, cAMP-CAP
complex binds to a site on DNA and represses transcription.3. cAMP
levels rise, cAMP binds to CAP, cAMP-CAP complex is removed from a
site on DNA and activates transcription.4. cAMP levels fall, cAMP
is removed from CAP, CAP binds to a site on DNA and activates
transcription.105. Opsonisation of a bacterium is a process by
which specific antibody binds with the surface molecule of the
bacteria. In an experimental condition, macrophage were infected
with either WT: Mycobacteria or with opsonised: Mycobacteria for 2
hrs at 37C. Subsequently, cells were washed and further incubated
for 24 *MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com27hrs at 37C. Finally, bacterial load in
macrophages were determined by colony forming unit (CFU). Which of
the following observation is true?1. WT: Mycobacteria inhibits its
transport to the lysosomes and survive in macrophages.2. Opsonised:
Mycobacteria inhibits its transport to the lysosomes and survive in
macrophages.3. WT: Mycobacteria are targeted to the lysosomes and
killed in macrophages.4. Opsonised: Mycobateria are targeted to the
lysosomes and survive in macrophages.106. The molar absorption
coefficient (extinction coefficient) of NADH at 340 nanometers is
6,220 liters per mole per centimeter, whereas that of NAD at 340
nanometers is 0. What absorbance will be observed when light at 340
nanometers passes through a 1- centimeter cuvette containing 10-
micromolar NADH and 10- micromolar NAD?1. 0.031 2. 0.0623. 0.124 4.
0.31107. During vertebrate limb development, a specialised
ectodermal structure, called Apical Ectodermal Ridge (AER), forms
at the dorso-ventral ectodermal boundary at the distal tip of the
developing limb bud.The following experimental facts about the AER
is available:(A) FGF 2, 4, and 8 are expressed in the AER(B)
Removal of the AER causes cessation of limb growth(C) Removal of
AER along with implantation of beads soaked in FGF 8 or(D) FGF 4 or
FGF 2 protein rescues the AER removal phenotype and gives rise to
normal limbWhich of the following statements cannot be made based
on the above facts?1. FGF 2, 4, and 8 are secreted proteins.2. FGF
2, 4, and 8 are necessary and sufficient for AER function3. FGF 2,
4, and 8 are sufficient for AER function4. FGF 2, 4, and 8 have
largely redundant functions108. A solution contains DNA polymerase
I, Mg2+ salts of dATP, dGTP, dCTP, and dTTP, and an appropriate
buffer. Which of the following DNA molecules would serve as a
template for DNA synthesis when added to this solution?1. A
single-stranded closed circle2. A single-stranded closed circle
base-paired to a shorter linear strand with a 3`-terminal
hydroxyl.3. A single-stranded closed circle base-paired to a
shorter linear strand with a 3`-terminal phosphate.4. A
double-stranded closed circle*MUDRA* Model Question Paper for
NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper
No. 1) www.mudralifesciences.com28109. When bacteriophage lambda
infects a sensitive bacterium, one of the first messenger RNA
species synthesized is very short, beginning at a site PL and
extending just through an adjacent gene N. After the appearance of
the gene N protein, messages become much longer, still beginning at
PL but extending far beyond gene N. The N gene encodesa) An
antiterminator activator for a promoter beyond gene Nb) A new sigma
factor acting on a promoter beyond gene Nc) An activator for a
promoter beyond agene Nd) An antirepressor that removes a protein
repressor bund at gene N110. In many different contexts of cell
differentiation, two distinct cell populations emerge from a
uniform cell population. This process is referred to as lateral
inhibition. Which one of the following must not be true about
lateral inhibition?(A) lateral inhibition results from morphogen
action(B) lateral inhibition requires direct cell cell contacts(C)
lateral inhibition requires reciprocal signalling between
twoneighbouring cells(D) lateral inhibition is preceded by
stochastic changes in gene expression in two neighbouring cells1.
(D)2. (A) and (D)3. (B) and (C)4. (A)111. In agamous mutant (flower
within flower phenotype) which of the following statements is
valid?1. Class A genes are expressed in the first two whorls, Class
B genes are expressed in the second and third whorls and Class C
genes are expressed in the third and fourth whorls.2. Class A genes
are not expressed. Class B and C genes are expressed in all the
whorls.3. Class A genes are not expressed. Class B genes are
expressed in the second and the third whorlsand Class C genes are
expressed in all the whorls.4. Class A genes are expressed in all
the whorls. Class B genes are expressed in the second and the third
whorls.112. During wing development in the chicken embryo, the
digit pattern (2-3-4) is thought to be controlled by a morphogen
concentration gradient that originates in the posterior of the
young wing bud as indicated in the diagrams above. An agar plug
soaked in retinoic acid (RA) can mimic the action of the morphogen.
Which of the following digit patterns would be expected to result
if an agar plug soaked in retinoic acid were paced in the anterior
of a developing wing bud?*MUDRA* Model Question Paper for NET-LS
& JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com291. 2 3 4 only 4. 4 3 2 only3. 2 3 4 4 3
2 5. 4 3 2 2 3 4113. Pyruvate kinase transfers a phosphate group
from phosphenolpyruvate to ADP, forming pyruvate and ATP. The
reaction catalyzed by this enzyme is essentially irreversible.
Which of the following is the best explanation for the irreversible
nature of this reaction?1. The binding of pyruvate to the active
site is very weak relative to the binding of phosphoenol
pyruvate.2. The reaction is coupled to the pyruvate dehydrogenase
reaction.3. The hydrolysis of ATP is highly favorable.4. The change
in free energy (G) for the overall reaction is large and
negative.114. Aspartate kinase is a key enzyme in the lysine
amino-acid biosynthesis in plants. With an objective of increasing
the lysine content in maize seeds, maize plants were transformed
with E.coli aspartate kinase with a strong seed specific plant
promoter. Resulting transgenic plants were found to express the
transgene; however, the content of lysine did not increase. Which
of the following option best explain the possible reason?1.
Bacterial proteins are not stable in plants.2. Bacterial proteins
are not properly folded in plants.3. Proper post-translational
modification did not take place in plants.4. Lysine causes feed
back inhibition of aspartate kinase*MUDRA* Model Question Paper for
NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper
No. 1) www.mudralifesciences.com30115. A response was observed when
a specific site in a rat brain was stimulated by passing electrical
pulses through indwelling electrode implanted surgically. In
another experiment in another rat, a cannula was surgically
implanted instead of the electrode and stimulated the area by
injecting excitatory neurotransmitter. However, the result of the
two experiments did not match. The possibilities of variations in
the results could be due to:1. animal variations only.2.
stimulation of cell bodies or nerve fibers only.3. difference in
anatomical brain areas only.4. variations in all the reasons
mentioned in 1, 2 and 3.116. Match the following1. Reflection a)
Light passing from one medium to another is bent at an angle2.
Transmission b) Light bounces back off surface of an object3.
Refraction c) Light is captured inside of object4. Absorption d)
Light passes directly through object5. Fluorescence e) Light
changes to a different wavelength1. 1-b; 2-d; 3-a; 4-c; 5-e 2. 1-d;
2-b; 3-c; 4-a; 5-e3. 1-e; 2-c; 3-a; 4-d; 5-a 4. None of these117.
Identify the position of each of the following on the accompanying
graph:1. Organisms divide at their most rapid rate2. New cells are
produced at same rate as old cells die3. Lag phase*MUDRA* Model
Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences
(Model Question Paper No. 1) www.mudralifesciences.com314. Log
phase5. Many cells undergo involution and deathChoose correct one
of these:1. 1-c; 2-b; 3-a; 4-b; 5-d 2. 1-b; 2-a; 3-b; 4-c; 5-d3.
1-b; 2-c; 3-a; 4-b; 5-d 4. None of these118. In order for chemical
reaction to proceed inside living cells, the energy of activation
must be supplied. Usually most reactions require more energy than
the cell has. How then do these reactions proceed?1. Cells produce
copious amounts of energy in the form of ATP2. Cellular enzymes
lower the activation energy of reactions3. Cells borrow energy from
adjacent cells4. The ribosomes actively synthesize proteins for
energy conservation119. In study of the biosynthesis of a
particular secretory glycoprotein, the first step was to
fractionate a crude RNA extract using an oligo-dT column. The RNA
bound to the column was eluted and translated in vitro in the
presence of [3H] leucine and [14C] mannose. An antibody specific
for the secretory glycoprotein was added and the resulting
immunoprecipitate was analyzed by sodium dodecyl sulfate (SDS)
polyacrylamide gel electrophoresis. Distribution of radioactivity
as a function of position in the gel was then analyzed as shown in
Figure 1. The in vitro translation experiment was repeated in the
presence of rough microsomes, which were then solubilized.
Immunoprescipitation and electrophoresis were performed as in the
precvious experiment. The results are shown in figure.*MUDRA* Model
Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences
(Model Question Paper No. 1) www.mudralifesciences.com32Why is the
oligo dT coumn used in this experiment?1) Only hnRNA will bind to
ligo-dT2) Poly RNA is purified by the procedure3) Ribosomal RNA is
bound to the column.4) Intact, rather than partially hydrolyzed,
RNA is retained on the column.120. In a haploid organism, the loci
A/a and D/d are 8 map units apart. In a cross Ad X aD, what will be
the proportion of each of the following progeny classes: (a) Ad (b)
Recombinants.1. 92%, 8%2. 46%, 8%3. 92%, 4%4. 46%, 4%121. Which of
the following illustrations explain the correct pairing preceeding
recombination between a chromosome (ABC-DEFG/ABC-DEFG) and its
inverted homologue (ABC-DGFE/ABC-DGFE). The dotin genotype
represent the centromere.122. The specific activity of the DNA
fragments in Figure 2 is defined as 32P disintegrations per minute
per microgram of DNA. Which of the following best describes the
relative specific activities of the fragments in lane1?1. Fragment
A has the highest specific activity, followed by B, C, D and E.2.
Fragment E has the highest specific activity, followed by D, C, B
and A.3. The specific activity depends on the order in which the
fragments were replicated.4. All fragments have the same specific
activity.*MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com33123. In a family, father is homozygous
dominant (AA) for a gene A and his wife is homozygous for its
ressive allele (aa) showing albino phenotype. It was surprising
that their child showed the albino phenotype. Which of the
following phenomenon can explain the phenotype?1. Nondisjunction2.
Uniparental Disomy3. Gene conversion4. All of the above124. The
protein 1-antitrypsin (1AT) inhibits the action of the proteolytic
enzyme elastase in lung tissue. A mutation has been described in
which the only change is the substitution of an Arg residue for a
Met residue in 1AT. The Altered 1AT does not inhibit elastase but
has the new property of inhibiting the blood coagulation protein
thrombin. The sequence around the active site of 1AT, altered 1AT,
and the natural thrombin inhibitor (antithrombin) are given below.
1At Met Ser Ile Pro Pro GluAltered 1AT Arg Ser Ile Pro Pro
GluAntithrombin Arg Ser Leu Asn Pro AsnWhich of the following would
be the best method to separate 1AT from altered 1AT?1. Size
exclusion Chromatography2. Ion exchange chromatography3. Thin layer
chromatography4. Sucrose gradient centrifugation125. Complete the
following sentence using options given below the sentence as a, b,
c, d and e.Species are critically endangered when it is not
endangered but is facing _________risk of extinction in the wild in
the _____________future.[(a) high; (b) very high; (c) extremely
high;(d) near; (e) immediate]1. a, a2. b, c3. c, e4. c, d126. An
experimenter generates library of plasmids containing 10-15
kilobase (kb) inserts from the genome of a bacterium by partially
digesting the bacterial genomic DNA with EcoRI and cloning the
resulting fragments into the EcoRI site of a Plasmid vector. The
experimenter must then identify the plasmids containing the pur B
gene. To do this 5 of the plasmids from the library were digested
with EcoRI and the *MUDRA* Model Question Paper for NET-LS &
JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com34digests were separated by gel
electrophoresis. In a second experiment, the same 5 plasmids were
analyzed by PCR using primers derived from sequences internal to
pur B and electrophoresis was performed on the PCR products. Both
gels were stained with ethidium bromide to visualize the DNA.Which
of the following methods would NOT be a useful alternative to using
PCR to determine which plasmids contain purB?1. Testing for
complementation of purB auxotroph2. Sequencing the inserts3.
Hybridizing the plasmids with a probe complementary to purB4.
Foot-printing with DNase127. Which one of the following trait set
characterizes best a r selected species?1. Usually a type III
survivorship curve, short life span and density dependent
mortality2. Usually a type I survivorship curve, short life span
and density dependent mortality3. Usually a type I survivorship
curve, long life span and density independent mortality4. Usually a
type III survivorship curve, short life span and density
independent mortality128. Muscle cells were cells incubated in the
presence of O2 and then quickly made anoxic. The concentrations of
various metabolites were measured immediately following the removal
of O2. The results are shown in the figure below.*MUDRA* Model
Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences
(Model Question Paper No. 1) www.mudralifesciences.com35The change
in the glucose 6-phosphate concentration can be explained best by
which of the following?1. Increased synthesis of glycogen2.
Increased conversion to free glucose3. Increased rate of
glycolysis4. Decreased synthesis of glucose 6-phsphate129.
Researchers studying the regulation of a hormone-responsive gene
isolated 750 base pairs of DNA immediately preceding the start site
of transcription (+1). They demonstrated that if these sequences
are cloned upstream of the bacterial chloramphenicol
acetyltransferase (CAT) gene and the DNA then introduced into
mammalian cells. CAT enzyme activity increases in response to
hormone treatment. To define the sequences involved in the
regulation of this gene, they made a series of deletions containing
various lengths of the 5`regulatory sequences. They cloned these
truncated DNA fragments upstream of the CAT gene as shown in the
figure below, introduced the constructs into mammalian cells, and
assayed for CAT enzyme activity in the absence (-) and presence (+)
of hormone. The figure below gives the results of a representative
experiment.Assuming that there is a single hormones-responsive
regulatory element in the gene, that element is located between1.
742 and 638 2. 638 and 424 3. 424 and 315 4. 315 and 116*MUDRA*
Model Question Paper for NET-LS & JRF of UGC-CSIR in Life
Sciences (Model Question Paper No. 1)
www.mudralifesciences.com36130. Three important biological
parameters generation time, population growth rate (r) and
metabolic rate per gram body weight are a function of the organisms
body size. Which of the curves (a) or (b) represents the correct
relation of each of the parameters to body size?1. Generation time
(a); population growth rate (b); metabolic rate/g bw (a)2.
Generation time (a); population growth rate (b); metabolic rate/g
bw (b)3. Generation time (b); population growth rate (a); metabolic
rate/g bw (a)4. Generation time (b); population growth rate (b),
metabolic rate rate/g bw (a)131. The female of a species of insects
lays about 300 eggs in June-July. Half of them hatch successfully
(equal proportion of males and females) by October. Forty percent,
of the larvae form pupae by January, and adults emerge from one
third of the pupae by March. Mating takes place during May, and 20%
of the adult insects manage to mate successfully. Thereafter, all
the adults die after the females have laid eggs in June-July. There
are no sex specific differences in survival, mortality, successful
completion of developmental stages and mating success. If 10
fertilized females are released in a very large enclosure in
June-July 9,001 how many eggs are likely to b laid during June-July
2005?1. 48000 2. 60002. 36000 3. 24000132. Eukaryotic cell
membranes were analyzed for hormone receptors. A membrane
preparation was incubated with radiolabeled hormone (3H-hormone)
for five minutes. A similar incubation of membranes and 3H-hormone
contained, in addition, a 1,000 fold excess of unlabeled hormone.
In both cases, the unbound hormone was removed by washing the
preparation and the amount of radioactivity remaining in the
membrane preparation was determined. The following results were
obtained.cpmonly hormone H000 , 53 cpmHormone Unlabeled Excess plus
hormone H500 , 13 Which of the following statements is most likely
true concerning the binding of 3H-hormone? 1. The total amount of
radioactivity bound in the absence of unlabeled hormone represents
the amount bound by the receptors.2. Most of the label is bound
nonspecidfically.3. The unlabeled hormone competes with 3H-hormone
for binding to the receptors.4. The cpm bound in the absence of
unlabeled hormone minus the cpm bound in the presence of unlabeled
hormone is a measure of nonspecific binding.*MUDRA* Model Question
Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model
Question Paper No. 1) www.mudralifesciences.com37133. In the
process of nitrification by organisms, the respective bacteria A
and B in the following reaction are:NH4 +1/2 O2 A NO2- +2 H+
H2ONO2-+1/2 O2 B NO31. Azotobacter, Nitrobacter2. Nitrobacter,
Azotobacter3. Nitrosomonas, Nitrobacter4. Nitrobacter,
Nitrosomonas134. Inspite of its two-fold cost, sexual reproduction
is the most dominant mode of reproduction among the living
organisms. Which of the following reasons might account for
this?(A) Sexual reproduction generates genetic heterogeneity
through recombination(B) Sexual reproduction helps in purging
deleterious mutations(C) Sexual reproduction evolved to stay
evolutionarily ahead of fast evolving internal parasites.1. (A)
only2. (A) and (B)3. (C) only4. (A), (B) and (C)135. The
filamentous alga Cladophora is illuminated with light dispersed by
a prism. As shown in the diagram above, aerobic bacteria
Pseudomonas included in the medium congregate where the alga is
illuminated with light at 656 nanometers and at 486 nanometers.
Pseudomonas do not congregate in this manner if the Cladophora is
removed from the medium. Which of the following is the most likely
explanation fro the bacterial movement?1. Bacteria have a
phototropic response 2. Starch made by photosynthesis is Secreted
from the alga in regions illuminated by light at 656 nm and 486
nm.3. O2 evolved by photosynthesis in the regions illuminated by
light at 656 nm and at 486 nm attracts bacteria.4. Alga and
bacteria have a symbiotic relationship; the bacteria need O2
suppolied by the alga and the alga requires CO2 supplied by the
bacteria.*MUDRA* Model Question Paper for NET-LS & JRF of
UGC-CSIR in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com38136. It is found that people with the
genetic disease called sickle cell anaemia are resistant to
malaria. Which of the following best describes the underlying
mechanism?1. Frequency-dependent selection2. Superiority of
heterozygotes3. Transient polymorphism4. Balanced polymorphism137.
The autoradiograms above (after electrophoresis and Southern
blotting) show human DNA digested with a specific restriction
enzyme and probed with labeled rRNA. In the autoradiogram on the
lert, the probe was 28S rRNA; at the right, the probe was 18S rRNA.
If the arrows in the following maps locate the recognition sites
the restriction enzyme, which map best explains the results shown
above?1)2)3)4)138. A severe winter storm kills many chickadees. An
investigation comparing the body size of dead birds with that of
survivors reveals that the dead birds included mainly the largest
and smallest members of the population. This winter storm
exemplifies1. Kin selection 2. Stabilizing selection3. Directional
selection 4. Balancing selection139. If a given gene in a randomly
mating population has three alleles a, b and c in the ratio of 0.5,
0.2 and 0.3 respectively, what is the expected frequency of
genotypes ab and bc in the population at equilibrium?1. 0.1 and
0.062. 0.2 and 0.153. 0.2 and 0.124. 0.04 and 0.09*MUDRA* Model
Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences
(Model Question Paper No. 1) www.mudralifesciences.com39140. One
summer the moose population on Isle Royale was unusually high, and
park naturalists noticed signs of malnutrition among the adults.
The wolf population was fairly low, near 20. That winter, for the
first time in many years, a substantial number of seemingly healthy
adult moose as well as calves and crippled animals were killed and
eaten by wolves. This description is part of a general situation in
which the wolf and moose populations1. Are maintained in a stable
equilibrium, from year to year2. Are simultaneously becoming
extinct 3. Fluctuate out of phase with each other4. Fluctuate
independently of each other141. Animal cell cultures are frequently
used for production of therapeutic proteins. NIH3T3 (a fibroblast
cell line) and CHO (Chinese hamster ovarian cell line) are some of
the popular cell lines used. Choose the best combination of cell
line (for transfection) and starting material for purification of
human growth hormone, a secretary protein1. NIH3T3 cell pellet2.
CHO and cell pellet3. NIH3T3 and culture medium4. CHO and culture
medium142. It is now thought that the atmosphere of the primitive
Earth was composed largely of carbon dioxide, nitrogen, and water
vapor. The composition of certain iron-containing minerals suggests
that the carbondioxide began to be replaced by oxygen about 2
billion years ago. Which of the following is the best explanation
for the change in atmospheric composition?1. Ozone produced in the
upper atmosphere by ultraviolet light broke down to oxygen.2.
Photosynthesis was established in primitive bacteria.3. Oxygen was
present in volcanic gases and slowly accumulated with time.4. Water
was broken down into oxygen and hydrogen by lightning
discharges.143. Some individuals suffer form an autosomal recessive
disorder known as alpha1-antitrypsinn deficiency. The recessive
homozygote for this disorder lacks the enzyme that ordinarily
degrades trypsin. The genotype of the recessive disorder may be
designated as aa. Normal individuals are either either homozygous
dominant (AA) or heterozygous carriers (Aa). Enzymatic tests and
colorimetric analysis reveal that a given individual possesses one
of three distinct levels of alpha1-antitrypsin enzyme activity. The
data below show the levels of trypsin inhibited per milliliter of
serum (which contains the antitrypsin enzyme) in three different
groups.Groups I. The general populationGroups II. Several families
in which some members have alpha1-antitrypin deficiency*MUDRA*
Model Question Paper for NET-LS & JRF of UGC-CSIR in Life
Sciences (Model Question Paper No. 1)
www.mudralifesciences.com40Groups III. Patients with
alpha1-antitrypsin deficiencyWhen the serum of an adult woman was
tested colorimetrically, a level of 0.5 milligram of trypsin
inhibited per milliliter of serum was found. This can be
interpreted as indicating which of the following?1. She is
homozygous dominant2. She is an asymptomatic heterozygous
carrier.3. She will exhibit alpha1-antitrypsin deficiency.4. She
will always transmit the abnormal recessive gene to her
offspring.144. In replicated experiments, plots are planted with
flax seeds in densities of 60, 1,440, and 3,600 per square meter,
respectively. The factors of soil, water, and light are similar for
each plot. Dry weights of the mature plants are
obtained.[DISTRIBUTION OF DRY WEIGHTS OF INDIVIDUALS IN POPULATIONS
OF FLAXPLANTS SOWN AT DIFFERENT DENSITIES]The experiments are most
likely performed to test for 1. Interspecific competition 2.
Intraspecific competition3. Competitive exclusion 4. Founder
effect*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR
in Life Sciences (Model Question Paper No. 1)
www.mudralifesciences.com41145. Following are some statements about
Agrobacterium plant interactions(A) Agrobacterium transfers a part
of its chromosome into plant cell.(B) Agrobacterium transfers a
part of one of its plasmid DNA into plant cell.(C) All the
virulence genes of Agrobacterium are inducible.(D) All the
virulence genes of Agrobacterium are functional only inside the
bacterial cells.(E) Some of the virulence genes of Agrobacterium
are inducible.(F) Some of the virulence genes of Agrobacterium are
functional both in bacterial and plant cells.Which of the following
combination of statements is true?1. (A), (C) and (D)2. (B), (E)
and (F)3. (C), (D) and (E)4. (B), (E) and (F)*MUDRA* Model Question
Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model
Question Paper No. 1) www.mudralifesciences.com42ANSWER SHEET FOR
MODEL QUESTION PAPER No.1(Must be printed on 100 gsm or Card
Paper)Name of the
Candidate:______________________________________________________Address:____________________________________________________________________________________________________________________________________Roll
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