CSIR UGC NET/JRF LIFE SCIENCE DECEMBER 2015 EXAM PAPER AND ANSWER KEY

last December CSIR UGC(Human Resource Development Group Council of Scientific & Industrial Research) organized NET/JRF EXAM on 20th December .This exam was held in three shifts A, B and C.

 For all booklets answer keys are different.

Here I am providing all three booklets official exam papers and answer keys. Candidates check there answer key according there booklet number.

CSIR UGC NET/JRF LIFE SCIENCE DEC. 2015 

Booklet A Question paper

Booklet B Question paper

Booklet C Question paper

DOWNLOAD ANSWER KEY FOR A B AND C BOOKLETS HERE-

ANSWER KEY

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 91 to 95

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86 TO 90 QUESTIONS WITH ANSWER OF CSIR UGC NET/JRF EXAMINATION

Q91.For successful fertilization in sea urchin,interaction between the surface of the egg and acrosomal proteins, specifically a 30.5 kDa protein called bindin, is necessary. The following factors could affect this interaction and prevent fertilization-
1.Removal of egg jelly polysaccharides.
2.Removal of bindin receptors on the egg vitelline membrane.
3.Removal of bindin receptors from the egg jelly.
4.Removal of bindin receptors from a single culster on the vitelline membrane.
Which one or the combination above statements is correct?
(A)1 and 4                                      (B)Only 2
(C)1 and 2                                      (D)Only 3
Answer-B
Description-Bindin is a 30.5kDa (Head of spermatozoa acrosomal protein) is a major protein of sea urchin required for fertilization(interaction with the surface of egg via specific receptor on egg surface)
Removal of bindin receptors on the egg vitelline memb. will affect this interaction and prevent fertilization.

Q92.A mutant embryo of Drosophila in which one of the major sex determining gene.sex lethal, can only undergo default splicing. was allowed to develop. The following statements are towards explaining the determination of sex of the embryo-
1.The embryo will develop into a male fly.
2.The embryo will develop into a female fly.
3.sex lethal gene product directly regulates sex specific alternate splicing of double sex RNA.
4. sex lethal  gene product regulates sex specific splicing of transformer RNA which in turn regulates splicing of double sex RNA.
The correct combination of above statements to explain sex determination of the given embryo is-
(A)1 and 3
(B) 1 and 4
(C)2 and 4
(D) 2 and 3
Answer-B
Description-Sxl-gene is sex lethal gene is a master switch gene for somatic cell determination in D. melanogaster.
In XX animals, Sxl become activated and imposes female development,while in XY animals. Sxl remains inactive and male development ensures.
Therefore,default splicing will inactivate Sxl protein product.Hence male develops.
Also sex lethal is an RNA splicing enzyme whose immediate target is transformer mRNA.Since, Sxl are not transcribed in males, its action on Transformer is restricted to females.
sex lethal acts positively in the functional splicing of transformer mRNA. Transformer is another splicing factor. Hence determines female development fate.

Q93.A two celled embryo made of blastomeres A and B. If the two blastomeres are experientially separated, the A blastomere generates all the cells it would normally make. However the B blastomere in isolation makes only a small fraction of cells it would normally make.Based on the above observations only, which one of the following conclusions is correct?
(A) A blastomere is autonomously specified while B blastomere is conditionally specified.
(B) A blastomere is conditionally specified while B blastomere is autonomously specified.
(C) Descendants of A blastomere are autonomously specified.
(D)Descendants of B blastomere can either be autonomously specified or conditionally specified.
ANSWER-A

94.A mutant was experimentally generated which has wings reduced to haltere like structure. The following statements are put forward regarding this phenotype-
1.ultrabithorex gene ectopically expressed in second thoracic segment.
2. antennapedia gene ectopically  expressed in second thoracic segment.
3.A homeotic mutation.
4. A mutation in gap gene.
The following combination of statements will be most appropriate explaining the molecular basis of mutant phenotype-
(A)1 and 2
(B)2 and 3
(C)3 and 4
(D)1 and 3
ANSWER-D
Description- Ultrabithorax (activated in absence of hunchback protein) Ubx gene found in insects.
In D.melanogaster it is expressed in the 3rd thoracic and 1st abdominal segments and repress wing formation. Ubx gene regulates the decisions regarding the no. of wings and legs.
Homeotic mutation causes tissues to alter their normal differentiation pattern producing integrated structures but in unusual locations.

Q95.Following are certain statements regarding the activities of homeotic genes of classes A,B and C involved in floral organ identity-
1.Activity of A alone specifies sepals.
2.Activity of B alone specifies petals.
3.Activity of B and C form stamens.
4.Activity of C alone specifies carpels.
Which one of the following combinations of above statements is correct?
(A)1,2 and 3
(B)1,2 and 4
(C)2,3 and 4
(D)1,3 and 4
ANSWER-D

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 86 to 90

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Q86. The following graph represents the expression of Tryptophan Synthetase (TS) in E.coli cels in absence or presence of Tryptophan in the medium-

If the two trp codons in the leader sequence of trp operon is mutated to ala, which of the following graph will best represent activity of TS in E.coli cells grown in the absence or presence of tryptophan?

(A)

(B)

(C)

(D)

ANSWER-B

Description-42-45 nucleotide coding leader sequence(trpL) contains an attenuator site.

Att-Transcription termination loop.(3:4)

Q87.The lifetime of a peptide bond in protein is very large,which statement below is incorrect with respect to stability of peptide bond?

(A)free energy of hydrolysis is negative

(B)The free energy of hydrolysis is positive and large

(C)The energy barrier to be crossed to go to the hydrolysed state in the large

(D)The peptide bond can be hydrolysed by6N HCl at 100 degree C.

ANSWER-B

Description-Because peptide bond hydrolysis is a spontaneous process. so it have highly negative value.

Q88. Following are certain statements related to euckryotic DNA replication-

1.The genome of multicellular animals contain many potential origins of replication.

2. During early development, when embryos are undergoing rapid cell divisions origin sites are uniformly activated.

3.'Pulse-chase' technique is used to label sites of DNA replication.

4.The rate of elongation of different DNA chains during genome replication varies drastically.

Which one of the following combinations of above statements are correct?

(A)1,2 and 3

(B)1,3 and 4

(C)2,3 and 4

(D)1,2 and 4

ANSWER-A

Description-Because the size of Eukaryotic genome is very large so,it contains multiple ori, uniformly transcribed at each ori, during early development stages.

'Pulse-chase' technique is also used to label the sites of DNA replication.

Method was used to determine the existence and function of Okazaki fragments(DNA replication)

Q89.A researcher wanted to immunize individuals of a particular area with viral infections. The researcher developed two different vaccine types(A and B) with the following properties-

1.When vaccine type A specific for a viral strain is administered to individuals, they develop strong neutralizing antibody response with very poor immunological memory.Hence it has to be administered in repetitive doses.

2.When vaccine type B specific for a viral strain is administered to individuals, they fail to develop circulating antibody response at the time of infection but they develop strong immunological memory.

If two viral strain V1(incubation period-2 days) and V2(incubation period-15days) are likely to infect the area which of the following vaccine combination would provide maximum immunization?

(A)V1 specific type A and V1 specific type B

(B)V1 specific type A and V2 specific type B

(C)V2 specific type A and V1 specific type B

(D)V2 specific type A and V2 specific type B

ANSWER-B

Description-Type A specific type to viral strain, produces strong neutralizing Abs but very poor immunological memory.

Hence repetitive booster doses are required.

But type B vaccine is generating strong memory.

Therefore to control V1 viral strain and V2, a combination of type A and B vaccine should be beneficial.

Q90.The expression of a hypothentical gene was analysed by Northen and Westenbolt hybridizations under control and induced condition.The results are summarized blow-

Expression of gene can be regulated by-

1.control at transcription initiation

2.alternative splicing

3.control the translation initiation

4.protein stability

Which of the above regulatory mechanisms can explain the observations show in the figures?

(A)Only 2

(B)Only 1 and 2

(C)Only 2 and 3

(D)1,2,3 and 4

ANSWER-D

Description-Northern bolt hybridization-RNA

                   Western bolt hybridization-Protiens

Expressions of gene can be regulated at any of following stages-A,B,C and D.

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CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 81 to 85

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 71 to 80

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CSIR UGC NET LIFE SCIENCE DEC. PAPER OUETION NO, 61 TO 70 WITH ANSWER

Q71. A gene producing red pigment was placed near centromeres  of fission yeast and thus subjected to position effect variegation and produced white colonies. A screen for mutants that increased the red pigment production was undertaken. Which of the following genes, when mutated, is likely to produce this genotype?
(a) Histone deacetylase 
(b) Histone acetylase
(c) RNA polymerase II
(d) TATA binding factor
Answer-A
Description- gene- red pigment
                    gene is placed near centromere: Inactivated due to position effect variegation. moved to centromere. Hence, produces white colonies. Normally deacetylase represses the gene expression but by the increased production of product could occur.

Q72. In order to prove that liposome can serve as a model membrane (mimicking cellular plasma membrane) and can be used as a target for complement- mediated immunolysis, an experiment as below is designed. To initiate such experiment, hepten- conjugated liposomes are made and loaded with umbelliferyl phosphate(UMP) ; hydrolysed product of UMP is umbelliferone and is fluorescent). Such loaded hapten conjugated liposomes in 10 mM Tris buffered saline,pH 7.4 were mixed with anti hapten antibodies and fresh guinea pigserum (as a source of complement) to induce immunolysis of lipsomal membrane. To quanyify only the membrane lysis component which of the assay sequences below is most appropriate?
(a) Mixture is ultra centrifuged and the supernatant reacted with alkaline phosphatase and fluorescence measured
(b) Mixture is sequentially reacted with phospholipase and alkaline phosphatase followed by fluorescence measurements
(c)Mixture is directly subjected to fluorescence measurement
(d) Mixture is treated with Triton X-100 and fluorescence measured
Answer-A
Description- If  the liposomal membrane is lysed, UMP fluorescence will be measured within the supernatant reacted with Alkaline phosphatase.

Q73.A null  mutation is created in a gene which is responsible for specific phosphorylation at 6th carbon position of mannose on acid hydrolases occurring in cis-Golgi. The following statements are given towards explaining the effect of this mutation-
(1) The lysosome will be devoid of lysosomal enzymes.
(2) lysosomal enzyme will be secreted out.
(3) Lysosomal enzymes will be get localized in cytoplasm.
which statement or combination of statements will explain the effect of mutation if the acid hydrolases in the mutant do not get degraded?
(a)1 and 3
(b) 2 and 3
(c)n 3 only
(d) 1 and 2
Answer-D
Description-Within cis-golgi an enzyme Glc-N-Ac phosphotransferase is responsible for adding a Glc-N-Ac-1-phosphate residue on the Mannose sugar through the formation of a phosphodiestet bond: Man-phosphate-Glc-N-Ac. Once formed the lysosomal enzymes are translocated to the trans golgi.

Q74.A newly identified sequence was experrimentally tested by in vitro transport assay using a radiolabelled protein containing the sequence to test import into mitochondria. Transport assay was done for a short time with or without membrane potential and after the assay, the mitochondria were either treated on not treated with proteinase K. At the end of the assay the mitochondria were pelleted and total protein of the pellet was isolated and separated on SDS-PAGE and autoradiographed. A representative auto-radiogram is shown below. Based on this experimental data, which of the following statements is not correct?
(A) The protiens goes into the matrix
(b)Not all the added protein was imported
(c) The protein requires membrane potential for import
(d) The protein is associated with the outer mitochondrial membrane
Answer-D
Description- The proteins inserted into mitochondria requires the membrane potential. The presence of proteinase K will be able to degrade the protein outside the mitochondria.

Q75. Acetyl-(Ala)18-CONH2 exists in alpha-helical conformation in solution. Most of the backbone dihedral angles(ᵠ and ᵩ) will be-
(A) -60 degree and -30 degree
(b) 60 AND 3. Degree
(c)-60 and -30 degree(50%) and 60 and 3. degree(50%)
(D)-80 and -120 degree
ANSWER-A


Q76. Enzyme parameters of four isoenzymes is given below-
          Isozyme                   Km micromolar                              Vmax
              A                                0.1                                               15
              B                                1.5                                               45
              C                                4.0                                               100
              D                                0.01                                             10
         These isoenzymes are localized in different tissues. In liver the substrate concentration is 0.2 micromolar. The liver isozyme is likely is to be-
(a)A                                        (b) B
(c) C                                       (d) D
Answer-A
Description- Km of an enzyme is the substrate concentration at which the reaction reaches half of its maximum.
At               [S] = Km;  V=Vmax/2
The Km of an enzyme tends to be similar to the maximum conc. of its substrate so the liver isozyme is likely to be A.

Q77. DNA is not hydrolysed by alkali whereas RNA is readily hydrolysed. This is due to-
(A) The double helix structure of DNA
(B) The presence of uridine in RNA
(C) Due to features observed in RNA such as stem-loop structures
(D) The presence of 2'-OH group in RNA
Answer-D


Q78. Two homologous proteins were isolate from a psychrophile(p) and a thermophile (T). The purified proteins were subjected to denaturation, protease digestion and circular dichroism (CD). Following observations were made-
1. The CD spectra of P and T proteins are identical
2.Their amino acid composition is 95% identical
3. T and P are equally susceptible to proteolysis in the presence or absence of reducing agent
4. T has higher midpoint of thermal denatu ration than P
The reason for enhanced stability in T is due to-
(A) Altered secondary structure
(B) Increased number of disulfieds in T
(C) Increase in water of hydration
(D) Increase in number of salt bridges
Answer-D
Description- The thermophile protein (T) is more stable as compared to psychrophile (P) as it possess large no. of salt bridges.

Q79. Binding of two ligands to their binding proteins were investigated. Following binding isotherms were obtained-
which of the following statement is correct?
(A) A is a obtained with an oligomeric protein and B is obtained with a monomeric protein
(B) B is obtained with protein with positive cooperativity
(C) A and B were obtained by the same protein at two different temperatures
(D) The profile B is not possible
ANSWER-B
Description- According to scatchard Plot, ligand binds specifically and non covalently with the receptor. A receptor  may have one or more binding site over itself.

Q80. Which of the following statement best described archaebacteria?
(A) Mostly autotrophic, cell wall contains peptidoglycan, 60S ribosomes, live in extreme environment
(B) Divide by fission not susceptible to lysozyme, live in extreme environments, mostly autotrophic
(C) Not susceptible to lysozyme, contain Golgi and linear chromosomes
(D) Chitinous cell wall, obligate aerobic, circular chromosomes
ANSWER-B
Description- The archaebacteria are the organisms that divide by fission, are not susceptible to lysozyme, can survive at extreme environmental conditioning and mostly these are autotrophic in nature.
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CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 81 to 85

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CSIR UGC/ JRF NET life science December exam paper Q.No. 71-80

Q81.Puromycin is an antibiotic used to inhibit protein  synthesis. Given below are few statements about the antibiotic.

1.It enters the E-site of the ribosome where it prevents the release of deacetylated tRNA after the action of peptidyl transferase.

2.It blocks the translocation process by binding to the translocation factor EF-G

3. Puromycin resembles the initiatior tRNA, tRNA_i^f-met 

^{4.It resembles the aminoacyl tRNA and binds to the A-site of the ribosome}

^{5.Puromycin inhibits only prokaryotic protein synthesis.}

^{6.Puromycin inhibits both prokaryotic and eukaryotic protein synthesis.}

^{Which of the above statements are true?}

^{(A)1 and 5}

^{(B) 2 only}

^{(C) 4 and 6}

^{(D) 3 and 5}

ANSWER-C

Description- Puromycin is a secondary metabolite of Streptomyces alboniger which is a structural analog of aminoacyl t-RNA and binds to A-site on the ribosome . Peptidyl transferase enzyme catalyse the covalent linkage of puromycin to growing peptide chain covalently ultimately inhibiting protein biosynthesis in P.K and E.K

Q82.Total RNA was isolated separately from cytosol and nuclei of human cells growing in a cell culture. Each sample was mixed with a purified denatured fragment of a DNA corresponding to a large intron of a house keeping gene incubated under renaturating condition. given below are the statements amde about the outcome of the experiment.

1.RNA isolated from nuclei will from RNA-DNA duplexes because of the presence of introns in the primary RNA.

2.Cytosolic RNAs usually will not from RNA-DNA duplexes.

3. Both cytosolic and nuclear RNA will not from RNA-DNA duplexes as transcription and splicing occur simultaneously.

4.cytosolic RNA will from RNA-DNA duplexes because unspliced cytosolic RNAs are exceptionally stable.

Which of the above statements are most likely to be true?

(A) only 3

(B) 1 and 4

(C) 1 and 2

(D) only 4

ANSWER-C

Description-hnRNA in nucleus do not undergo ant splicing.Hence contains intronic sequences along with exonic part. After splicing that takes places in Nucleus introns are removed and exons are ligated to from m RNA molecule, ultimately m RNA is transported into cytosol i.e., out of the nucleus.

Q83. If a proteasome inhibitor is added to synchronously cycling human cells in G2 phase which one of the following events is likely to happen?

(A) Induce re-replication of DNA

(B) Arrest cells in G2 phase

(C) Arrest cells in Anaphase

(D) Block chromatin condensation

ANSWER-C

Description-proteasome- multi subunit enzyme complex that play a central role in the regulation of proteins that control cell cycle progression and apoptosis, an imp. target of cancer therapy.in G2 phase cell growth continues and proteins are synthesized in preparation for mitosis.

the spindle- Assembly checkpoint monitors the alignment of chromosomes the mataphase spindle.

Q84. A promoter deletion study was done in order to determine the binding sites for a transcription factor on the promoter, which is activated on treatment with the drug 'X'. The following constructs were made-

Luciferase assay revealed the following results-

The following statements can be made-

1. Region between -1800 ans -1210 contains a binding site for the activator

2.Region between -868 and -1210 contains a binding site for a repressor

3.Region between -868 and -432 contains a binding site for a repressor

4. Region between -1210 ans -868 contains a binding site for the activator

Which of above are true?

(A) 1 and 3

(B) 2 and 3

(C)1 and 4

(D) 2 only

ANSWER-C

Description-Activator protein binding region lies upstream to the transcription start site of sense strand.Promoter region is defined as DNA sequence, where transcription of a gene by RNA pol. begins. Both RNA pol. & necessary transcription factors(T.F) binds to the promoter sequences.

Q85. A pharmacy student designed a drug to specifically target the receptors for retinoic acid in order to prevent stem cell differentiation. After in vitro trial, the investigator found that the drug seemed to be underwent differentiation and the drug seemed to be ineffective. The following reasons were given by the student-

1.The size of drug exceeded the size of molecules that could cross the membrane.

2.The drug was small in size but hydrophobic in nature.

3.the drug did not bind to its receptors.

Which one of the above cold be the probable reason for drug ineffectiveness?

(A) Only 3

(B)1 and 3

(C) 1,2 and 3

(D)Only 2

Answer-B

Description-Retinoic acid receptor is a type of nuclear receptor which can also act as a T.F. that is activated by both all trans retinoic acid. so if the drug small+ hydrophobic.

Therefore easily cross PM and nuclear membrane.

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 61 to 70

CSIR UGC NET LIFE SCIENCE EXAM. DEC.  QUESTION NO.51 TO 60 WITH SOLUTION

Q61. The degree of genetic relatedness between the offspring and their parents-
(a) higher than that between sister and brother
(b) lower than that between sister and brother
(c) the same as that between sister and brother
(d) dependent on the number of siblings
Answer-C

Q62. You want to purify a recombinant protein of your interest. You can use affinity chormatograpty of purity as you have nickel columns available in the laboratory. With what molecule will you tag the protein to purity using those columns?
(a) GST
(b) Histidine
(c) Histamine
(d) proline
Answer-B

Q63. During which geological period was there an explosive increase in the number of many marine invertebrate phyla?
(a) Ordovician
(b) Devonian
(c) Permian
(d) Cambrian
Answer-D
Q64. An example of the species interaction called commensalism is-
(a) nitrogen- fixing bacteria in association with legume plants roots
(b) microbes in living human gut
(c) female mosquito deriving nourishment from human blood
(d) orchid plant growing on the trunk of a mango tree
Answer-D

Q65. In which ecosystem is the detrital pathway of energy flow most important?
(a) Lakes
(b) Grasslands
(c) Tropical rain forests
(d) Oceans
Answer-C

Q66. What parameter, plotted on Y- axis against generation time, would yield the curve slown in the figure?
(a) Survivorship
(b) Body size
(c) Lifespan
(d) Intrinsic rate of increase
Answer-D

Q67. In an experiment of detect a new protein in fixed cells, no secondary antibody tagged with fluorescence dye is available. what should be the best choice out of the following to detect the proteins?
(a) Protein A-FITC
(b) Protien A- Sepharose
(c) Biotin FITC
(d) Avidin-FITC
Answer-A

Q68. Lower limits of detection by sensord is important. Which method of detection is more sensitive than glass electrode used for pH measurement?
(a) Absorption spectroscopy
(b) Refractive index
(c) Circular dichroism
(d)Fluorescence spectroscopy
Answer-D

Q69. If  a researcher intends to identify a specific brain area activity linked to a cognitive function in human subjects, which one of the following techniques should be used?
 (a) CAT
(b) MRI
(c) fMRI
(d) Patch clamp
Answer-C

Q70. Which of the following statement is incorrect for fluorescence in situ hybridization (FISH) technique?
(a) A fluorescence or confocal microscope is used for detection of signal
(b) A labelled sequence of nucleotides are used
(c) Specific fluorescence tagged antibodies are used
(d) A stringent washing step is essenyial to remove appearance of non specific signal
Answer-C

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 51 to 60

CSIR UGC NET LIFE SCIENCE EXAM. DEC.  QUESTION NO.41 TO 50 WITH SOLUTION

 Q51. An interrupted mating experiment was performed between Hfr Str^s a⁺  b⁺ c^{+  and} f^- Str^r a^- b^- c^{- s}    strains. The genotype of majority of streptomycin resistant exconjugant after 10,20 and 30 minutes of interrupted mating is given below-
The most probable gene order would be-
(a) a b c                      (b) c a b
(c) b a c                      (d) a c b
Answer-D

Q52.Which one of the following functions is not served by the plasma proteins?
(a) Blood clotting
(b) O2 transport
(c) Hormone binding and transport
(d) Buffering capacity of blood
Answer-B

Q53. Two plants with white flower are crossed. White flower arise due top recessive mutation all F1 progeny have red flowers. When the F1 plants are selfed, both red and white flowered progeny are observed. In what ratio will red- flowered plants and white flowered plants occur?
(a)1:1           (b) 3:1
(c)9:7           (d) 15:1
Answer-C

Q54. The population density of an insect species increases from 40 to 46 in one month. If the birth rate during that period is 0.4 what is the death rate?
(a)0.25           (b)0.15
(c)0.87           (d)0.40
Answer-A

Q55. Two 18-residue helical peptides A and B are enantiomers. They can be distinguished by-
(a) recording there MALDI mass spectrum
(b) hydrolysis followed by amino acid analysis
(c) sequencing by Edman's method
(d) examining their circular dichroism spectra
Answer-D

Q56. Schizocoelous coelom formation, mouth formation from embryonic blastopore, spiral and determinate cleavage are characteristics of-
(a) deuterostomes
(b) pseudocoelomates
(c) protists
(d) protostomes
Answer-D

Q57.Which species concept utilizes morphological and molecular characters to distinguish between species?
(a) Evolutionary
(b) Ecological
(c) Biological
(d) Phylogenetic
Answer-D

Q58. Worker bees, instead of themselves reproducing, help the queen reproduce. This behaviour is explained as an example of-
(a) kin selection
(b) group selection
(c) sexual selection
(d) natural selection
Answer-A

Q59.Which of the following is a correct match of the animal with its taxonomic group?
(a)Hirudinea-leech, Chelicerata-Horse shoe creb, Cestoda-Tapeworm Echinoidea-sea urchins, Earthworm Oligochaeta-Earthworm
(b)Hirudinea-Earthworm,Chelicerata-Horse shoe creb,Cestoda-Octopus, Echinoidea-Tapeworm,Cephalopoda-Earthworm,Oligochaeta-leech,
(c)Hirudinea-Tapeworm,Chelicerata-leech, Cestoda-Tapeworm,Echinoidea-Horse shoe creb,  Cephalopoda-Earthworm,Oligochaeta- Octopus,
(d)Hirudinea-leech, Chelicerata-Tapeworm,Cestoda-Earthworm,Echinoidea-sea urchins, Cephalopoda- Octopus,Oligochaeta-Horse shoe creb, 
Answer-A

Q60.The wings of insect and the wings of bats represent a case of-
(a) divergent evolution
(b) convergent evolution
(c) parallel evolution
(d) neutral evolution
Answer-B

CSIR UGC/JRF NET LIFE SCIENCE DEC. EXAM PAPER QUESTION NO. 61 TO 70 WITH SOLUTION

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 41 to 50

CSIR UGC NET LIFE SCIENCE EXAM. DEC.2014QUESTION NO. 30 TO 40 WITH SOLUTION

41. Light is the dominant enviromental signal that conrols stomatal movement in leaves of well- watered plants grown in natural environment.Which one of the following wavelengths of light is responsible for such regulation?
(a) Red light                                (b) Blue light
(c) Green light                             (d) Far red light
Answer-B

Q42.Which one of the following is not the main factor that contributes to water potential during plant growth under normal conditions?
(a) Solute potential
(b) Hydrostatic pressure
(c) Gravity
(d) Temperature
Answer-D

Q43.Which one of the following cell is the renal corpuscle can influence Glomerular filtration by its contraction?
(a) Podocytes
(b) Endothelial cells of glomerular capillaries
(c) Parietal epithelial cells of Bowman's capdule
(d) Mesangial cells
Answer-D

Q44. Production of excessive amount of corticotropin (ACTH) occurs in which one of the following-
(a) Grave's disease
(b) Cushing's syndrome
(c) Grieg's syndrome
(d) Alport's syndrome
Answer-B
Description- Excess production of corticotropin (ACTH)- Cushing syndrome.

Q45. The plant hormone indole-3-acyetic acid (IAA) is present in most plants. The structure of this hormone is related to which one of the following amino acid?
(a)Glutamic acid
(b) Asparatic acid
(c) Lysine
(d) Tryptophan
Answer-D

Q46. The type one glomus cells present in the carotid bodies contain granules which release some substances during hypoxia. Which one of the following is released in hypoxia?
(a) Serotonin  
(b) GABA
(c) Dopamine
(d) IL 8
Answer-C

Q47.Lndividuals with greater mass have a smaller surface area to volume ratio, which helps to conserve heat. This is known as-
(a) Leibig's rule 
(b) Cope's rule
(c) Gloger's rule
(d) Bergmann's rule
Answer-D

Q48. Which one of the following is not a characteristic property of carotenoids?
(a) They possess complex porphyrin ring
(b) They are integral constituent of thylakoid membrane
(c) They are also called accessory pigments
(d) They protect plants from damages caused by light
Answer-A

Q49.5-Bromouracil is a base analog that can cause mutation when incorporated into DNA. Which of the following is the most likely change that 5- Bromouracil induces-
(a) T: A to C:G
(B) T: A to A:T
(c) G:C to T:A
(d) C:G to A: T
Answer-A

Q50. The following pedigree shoes the inheritance of a common phenotype controlled by an autosomal recessive allele. The probability of carriers in the population is 1/3-

Which is the probability that a child from parents II-3 and II-4 will show the phenotype?
(a)1/16
(b) 1/18
(c) 1/36
(d) 3/16
Answer-B

CSIR UGC NET / JRF LIFE SCIENCE DEC.2014 EXAM PAPER QUESTION NO. 51 TO 60

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 31 to 40

Read 21 to 30 question answer on this link-
CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 QUESTION NO. 21 TO 30 WITH SOLUTION

Q.31 A patient with ER+/PR+ breast cancer was cured with a drug 'T', whereas a second patient did not respond to 'T'. Which one of the following is the best therapy that you should suggest for the second patient?
(a) Surgery,the followed by HER-2/neu targeted drugs
(b) A drug that targets triple negative (ER-/PR-/HER-2-) breast cancer.
(c)Radiation,followed by drug 'T'
(d)Surgery,followed by radiation only
Answer-B
Description-1st patient                               2nd patient
                    ER+/PR+ Breast cancer         No response to drug 'T'
          PR=Progestrone receptor
          ER=Estrogen receptor
         HER-2=Human epidermal growth factor receptor
           Therefore, a drug, which targets ER-,PR-,HER- breast cancer will be successful and effective drug for both the patients.

Q32.If you run a pentavalent IgM through SDS-polyacrylamide gel electrophoresis,bow many bands you are supposed to get by Western blottng using alkaline phosphatase conjugated secondary antibody?
(a)Five                      (b)Four
(c)Three                    (d)One
Answer-D
Description-SDS causes multimeric protein yo break into single subunits.  

Q33.The splitting or migration or one sheet of cells into two sheets as seen during hypoblast formation in bird embroygenesis is termed as-
(a)delamination                   (b) ingression
(c)involution                       (d)invagination
 Answer-A
Description-Splitting or migration of sheet of cells on to 2 sheets of cells delamination.

Q34. Which of the following statements about meiosis is not true?
(a) Kinetochores of sister chromatids attech to opposite poles in meiosis 1
(b) Kinetochores of sister chromatids attech to opposite poles in meiosis II
(c) Chiasma is formed in prophase I
(d) Homologous chromosomes are segregated in meiosis I
Answer-A
Description- Kinetochores of Non sister chromatids attach to opposite poles in meiosis I.

Q35.In chloroplast, the site of coupled oxidation-reduction reactions is the-
(a)outer membrane                      (b) inner membrane
(c)thylakoid space                       (d) stromal space
Answer-C

Description- In chloroplast site of coupled oxidation reduction is thylakoid spaces.

Q36.Lens formation requires sequenitial events whereby the anterior neural plate signals the anterior ectoderm to promote secretion of Pax 6, which renders the anterior ectoderm more receptive to secretions from the optic vesicle. The above can be best explained by which of the following phenomenon?
(a) Instructive interactions only
(b) epithelial-Mesenchymal interaction
(c) Permissive interactions
(d) Induction and competence
Answer-D

Q37.The group of cells of amphibian blastula capable of inducing the organizer is called as-
(a) Hensen's node
(b) Nieuwkoop centre
(c) Dorsal blastopore lip
(d) Hypoblast
Answer-B
Description- Group of cells in amphibians blastula
                                             ↓
                                    Inducing organizer

Q38. Glycosaminoglycans are usually linked to proteins to form proteoglycans Which of the following is not a proteoglyean?
(a) Hyaluronan                      (b) Aggrecan
(c) Betaglycan                       (d) Syndecan-I

Answer-A

Description- Hyaluronan is a Glycosaminoglycan and not a proteoglycan.

Q39.Which one of the following statements regarding seed germination of a wild type plant is not correct?
(a) Low ABA and high bioachtive GA can break seed dormancy
(b) Light accompanied with high temperature can break seed            dormancy
(c) GA induce synthesis of hydrolytic enzymes in cereal grains
(d) Degradation of carbohydrates and storage proteins provide nourishment and energy to support seeding growth

Answer-B

Description- Light accomplished with increase in temp. can break seed dormancy

Q40.Some T lymphocytes respond to antigenic stimulation by synthesizing a growth factor that causes T cell proliferation thereby increasing the responsive T lymphocytes resulting in amplification of the immune response. This is an example of-
(a) endocrine signalling
(b) paracrine signalling
(c) autocrine signalling
(d) cyclic signalling

Answer-C

Read next 41 to 50 question answer on this link-
CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 PAPER QUESTION NO. 41 TO 50 WITH SOLUTION

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 21 to 30

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 PAPER QUESTION NO. 11 TO 20 WITH SOLUTION
21. Reaction products inhibit catalysis in enzymes by- 
(a) covalently binding to the enzyme
(b) altering the enzyme structure
(c) occupying the active site
(d) from a complex with the substrate
Answer-C
Description-If the reaction product is accumulated as it is not utilized within the chemical reaction pathway so will occupy the active site of the enzyme hence inhibiting it.

Q22. Chirality of DNA is due to-
(a)the bases
(b)base stacking
(c) hydrogen bonds between bases
(d) deoxyribosome
 Answer-D
Description- Chirality refers to the ability of a molecules to rotate the plane polarized light.
 For a compound to be chiral it must possess all the valency groups to be different from each other.

Q23. Which of the following statements regarding membrane transport is false?
(a)Poler and charged solutes will not cross cell membranes effectively without specific protein carriers
(b)Each protein carrier will only bind and transport one (or a few very similar) type of solute
(c) Sugars such as glucose are always transported by active transport rather then by facilitated diffusion carriers
(d)Ions are typically transported by special proteins that from membrane channels
Answer-C
Description- Sugar such as glucose are transported through facilitated diffusion mediated by glucose transporter (GLUT), thus statement 3 is false.

Q24.What will happen if histones are depleted from a metaphase chromosome and viewd under a transmission electron microscope?
(a)30mm chromatin fibres will be observed
(b) 10nm chromatin fibres will be observed
(c) A scaffold and a huge number of loops of DNA fibres will be observed
(d) A huge number of loops of DNA fibres without scaffold will be obsereved
Answer-C
Description-Scaffold proteins along with histones are responsible for maintaining the structure integrity of the DNA. If histones are lost or depleted then,DNA will be observed in the form of large no. of loops.

Q25.In proteins, hydrogen bonds from as follows- Donor (D)-H....Acceptor if the angle between D-H and A is-
(a)<90 Degree               (b)180 Degree
(c)180 Degree               (d)120 Degree
Answer-B
Description- The most stable hydrogen bonds are formed when the Doner-Hydrogen and acceptor are in front of each other i.e. at an angle of 180 Degree

READ 1 TO 10 QUESTION ANSWER ON THIS LINK-
1-10 QUESTION ANSWER WITH SOLUTION

Q26.Leader sequence in some of the protozoan parasites is transcribed elsewhere in the parasite genome and gets joined with several transcripts to make the functional RNA. The joining of the two transcripts occur by the process of-
(a) alternate splicing         (b) trans splicing
(c)ligation                         (d)RNA editing
Answer-B


Q27. Small nuclear RNAs used to process and chemically modify rRNAs are called-
(a)Sca RNAs           (b)Si RNA s
(c)Sno RNAs           (d)Sn RNAs
Answer-C
Description-Sno RNA (small nucleolar RNA) participate in the processing and modification of r RNA 

Q28.Protein motive force during oxidative phosphorylation is generated in mitochondria by-
(a) Exchanging protons for sodium ions
(b)pumping protons out into intermembrane space
(c)pumping hydroxyl ions into the mitochondria
(d)hydrolysis of ATP
Answer-B
Description- The proton motive force during oxidative phosphorylation is generated in mitochondria by the movement of [H+] across the inner mitochondrial membrane mediated by electron transport system.

Q29. During replication the RNA primer is degraded by the 5'-3' exonuclease activity of-
(a)RNAase H1(ribonuclease H1)
(b) FEN-1 (flap endonuclease 1)
(c) Topoisomererase II B
(d) DNA polymerase ɣ
Answer-B
Description- During DNA Replication FEN-1 degades RNA primer by its 5'-3' exo nuclease activity.

Q30.Which one of the following statements about eukaryotic translation is not true?
(a)ribosome binding site on m RNA is called Kozak consensus sequences
(b)initator tRNA is tRNAi^f-met  
(c)initator amino acid is nethionine
(d)translocation factor is eEF2
Answer-B
Description-During E.K. translation initiator t-RNA is tRNA met and not tRNAi^f-met  

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 CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 PAPER QUESTION NO. 31 TO 40 WITH SOLUTION