CSIR NET June 2014 Question paper + Answer Key ( Completely Solved )
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Answer: 2
Explanation: v=pi r^2 h Since the inside cylinders are identical, they have radius r/2
Answer: 2 and 4
Explanation: If rod is straight, 2r is maxium length which can be put inside, closer value is 17.3. but if
we consider that rod is bent, then it should be pi *r, so we can consider maximum value.
Answer: 4
Explanation: C=PI *D, THERE IS 1/2 CIRCUMFERENCE AND DEVIDED EQUALLY INTO 3 PARTS , SO IT IS 5 PI/ 3 MTR
Answer: 3
Explanation: GAIN TOTAL IS 200 RS FOR 1000 RS INVESTMENT, TAX IS 40 RS ON 200 RS GAIN. THE NET GAIN IS 160. IN TERMS OF PERCENTAGE IT IS 80
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Answer: 4
Explanation: Volume of cube = 4a^3 => Volume of cube of 20micron edge = (20 x 10^-6)^3 = 8000 x 10^-18, volume occupied by cube of edge of 2 micron = 8 x 10^-18, percentage of volume filled by fluid = [100 / 8000 x 10^-18 ] [(8000-8) x 10^-8] = 99.9%
Answer: 1
Explanation: Price of 2 carat = 8, price of 0.5 carat = price of 1 carat / 2 = 4/2 = 2, so price of 2.5 carat = 8+2 = 10
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Answer: 2
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Yes, You read it right.
Total 137 BioTecNika Students selected in CSIR NET December 2013
Toll Free Phone Help Line: 1800-200-3757 Address: Bangalore: #427,1st Floor, 27th Main, Sector-1, HSR Layout, Bangalore, Karnataka, India New Delhi + NCR : C-31, 1st Floor, Sector-2, NOIDA, Uttar Pradesh - 201301
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Answer: 2
Explanation: Total speed = 60 + 60 = 120 km/hr = 120 x 1000 / 60 x 60 = 200/60 m/s, Distance = 200m, hence, time = 6s
Answer: 3
Explanation: 2^17 x 3^2 x 5^14 x 7 = 2^5 x 2^5 x 2^5 x 2^2 x 3^2 x 5^4 x 5^4 x 5^4 x 5^2 x 7 = 5.04e+16
Answer: 4
Explanation: 1+1-2+3-4+5-6+7-8+9-10+11-12+13-14+15-16+17-18+19-20 = -9
Answer: 2
Explanation: 2+3=5, 2*3=6 3+4=7, 3*4=12, 4+5=9, 4*5=20
Answer: 2
Explanation: Let MP = x, 20% discount on MP = 0.2x so SP = x - 0.2x = 0.8x, Let CP = y, 20% profit on CP = 0.2y so actual CP = y - 0.2y = 0.8 y, Since, CP , y = 0.8x = Actual CP = 0.8 x 0.8X = 0.64x, Now MP + Actual CP / 2 = x + 0.64x / 2 = 0.82x > SP
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Answer: 4
Explanation: The reflection in the mirror will show X'YZ axis, since, we know that X' is on the opposite side of X axis, which matches with only one option
Answer: 4
Explanation: now A1 = 1 & d = 1, here n = 100, hence, An = 1 + 99 x 1 = 100, now A1 = 1 & d = 1, here n = 100, hence, An = 1 + 99 x 1 = 100
Answer: 4
Explanation: [1,15,12,6], [8,10,13,3], [14,4,7,9], [11,5,2,16], [6,7,10,11]
Answer: 3
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Explanation: Any leap year will repeat in exactly the same way every 28 years. So after 1965 wednesday will repeat on 1994
Answer: 1
Explanation: probability of being it a leap year is 1/4 beacause n=1 and leap year comes after every 4 year, so probability of being a non leap year is 3/4
Answer: 2
Explanation: present age of Bro1=x, present age of Bro2=y; 3 years ago, x-3-y+3 = 2---eq1;After 10 years, x+10+y+10= 2(x+y)---eq2; Solving 1 and 2, x=11, y=9; Age of elder bro 11.
Answer: 1
Explanation: 61 --> 16 = 4^2
52 --> 25 = 5^2
63 --> 36 = 6^2
94 --> 49 = 7^2
8^2 = 64 --> 46
18 --> 81 = 9^2
Answer: 2
Explanation: Ribosomes binds to two transmembrane glycoproteins on rough endoplasmic reticulam that are not found in Smooth ER. Ribophorins are mediating the attachment of ribosomes.
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Ans: 4
Ans: 1
Explanation: assembly of pre-replication (pre-RC) initiative complexes, helicase activation, and replisome loading. Cell cycle regulation by protein phosphorylation ensures that pre-RC assembly can only occur in G1 phase, whereas helicase activation and loading can only occur in S phase. Checkpoint regulation maintains high fidelity by stabilizing replication forks and preventing cell cycle progression during replication stress or damage.
Answer: 3
Explanation: IN THE EM STREAMS OF ELECTRONS ARE DEFLECTED BY AN ELECTROSTATIC OR ELECTROMAGNETIC FIELD
Answer: 3
Explanation: histone acetylation modification is primary event
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Answer: 3
Explanation: Tripsin is used to resuspend the adherent cell lines during Harvesting
Answer: 2
Answer: 3
Explanation: mass to charge ratioo m/z we have to consider.
Answer: 1
Explanation: 1 rad = 0.01 J/kg, 1 J/kg = 1/0.01 rad, hence, 14.9 J/kg = 14.9 / 0.01 = 1490 rad
Answer: 3
Explanation: in IEF, basis of separation is acids and base property, that ise due to positive and negative charge.
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Answer: 4
Answer: 3
Explanation: In Prokaryotes, the 50S (23S component) ribosome subunit contains the peptidyl transferase component and acts as a ribozyme.
Answer: 3
Explanation: Triple Response
Answer: 3
Answer: 3
Explanation: Granulomatus Reaction is involved in delayed type hypersensitivity reactions
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Answer: 1
Answer: 4
Ans: 3
Ans: 3
Exp: Since Acetylation is required to decondence the chromatin and make it ready for transcription. All other options are going to compact the DNA
Ans: 1
Exp: These proteins bind to the inner side of the nucleosomal DNA, altering the interaction between the DNA and the histone octamer. They bring about nucleosome remodelling
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Ans: 2
Exp: zahavi's handicap principle is regarding peacock tail.
Ans: 2
Exp: internal pathogen is always represented by MHC1 that is recognised by CD8+ lymphocytes.
Ans: 1
Exp: As axin and catenine are important regulator of Wnt pathway and axin is inhibitor, so in its presence catenin expression will not increase
Ans: 3
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Exp: guide strand is integrated into the RISC complex and is selected by the argonaute protein
Ans: 2
Ans: 4
Exp: the last O2 molecule bind to hemoglobin which is already in R stae with higher affinity
Ans: 4
Explanation: Synapsin I anchors synaptic vsicles to actin cytoskeleton
Answer: 1
Explanation: V1 receptor is involved in vasoconstriction. Since Ca insurge will help in Muscle contraction
Ans: 1
Explanation: Parental: AbC/AbC x aBc/aBc; F1: AbC/aBc x abc/abc (test cross); DCO: Abc/abc & aBC/abc; Comparing DCO with parental, order of gene acb
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Ans: 2
Ans: 4
Explanation: glycolate is transfered from chloroplast to peroxisome in C2 cyce
Answer: 3
Explanation: All other behaviours are shown by mother rats
Ans: 4
Explanation: Seed germination is inhibited by ABA in antagonism with gibberellin. ABA also prevents loss of seed dormancy
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Answer: 1
Explanation: Domain comprises of many kingdoms, it comes higher than kingdom.
Answer: 3
Explanation: Branchiostoma belongs to phylum chordata, Have fatenned body and soft notochord. Since phylum chordata comes under deuterosomes we can assign it to Branchiostoma. Deuterostomes are always enterocoelomates
Answer: 3
Explanation: Increase in population=645-600=45; growth rate=45/600=0.075; GR=BR-DR; DR=BR-GR; DR=0.125 - 0.075 = 0.05
Ans: 1
Explanation: Recessive allele e is masking the expression of B and b
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Ans: 2
Explanation: Plasmodium requires warm temperature for its development inside the vector mosquito
Answer: 4
Explanation: Big trees Bear fruits at a later age
Answer: 1
Population size,N=50; decline in variation=1/(2N)= 1/100 = 0.01
Answer: 1
Explanation: its due to random variable
Answer: 4
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Explanation: Long chain fatty acids are not involved in antigen specificity, most of times proteins are involved
Answer: 3
Explanation: "Two methods of producing transgenic mice are widely used:
transforming embryonic stem cells (ES cells)growing in tissue culture with the desiredDNA;
injecting the desired gene into thepronucleus of a fertilized mouse egg.
Answer: 3
Explanation: Non-Covalent interactions like Hydrogen bonding , hydrophobic interactions & vanderwaal forces are responsible for protein folding.
Answer: 3 ( Not sure, Still verifying )
Answer: 4
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Explanation: Radulla: structure used by molluscs for feeding, Clitellum: thick glandular, nonsegmented portion in Annelids in which eggs are deposited.
Answer: 2
Explanation: since C'-N has partial doulble bond character.
Answer: 2
Explanation: Non- polar aa are always buried , But the charged aa can or can not be burried.
Answer: 4
Explanation: Sphingolipids and chloresterol tend to self assemble into microdomains that keep floating within the artificial bilayer called as lipid rafts.
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Toll Free Phone Help Line: 1800-200-3757 Address: Bangalore: #427,1st Floor, 27th Main, Sector-1, HSR Layout, Bangalore, Karnataka, India New Delhi + NCR : C-31, 1st Floor, Sector-2, NOIDA, Uttar Pradesh - 201301
Yes, You read it right.Total 137 BioTecNika Students selected in CSIR NET December 2013