We are listing out some of the CSIR December 2013 Exam Doubtful Questions and their answers?
If you have any doubtful questions, please use below comments section.
Finally, helpBIOTECH will collect all and submit to CSIR (latest by 13th March, 2014).
Please spend time and will help CSIR Aspirants….
Booklet A (Doubtful Questions/Answers).
100: Cells from an early frog blastula were removed from the animal pole and used to replace cells from the vegetal pole of the blastula. The following events may be expected.
B. Transplanted cells would develop as cells of the animal pole of the adult on the vegetal pole.
C. Region of the animal pole from where the cells were removed would be missing in the adult.
D. Remaining cells in the animal pole would compensate for the cells that were removed.
Which of the following are true?
1. B, C and D
2. A, B and D [CSIR Answer]
3. A, B and C
4. A, C and D
108: An action potential was generated on a nerve fiber by a threshold electrical stimulus. When a second stimulus was applied, no matter how strong it was, during the absolute refractory period of the action potential, the nerve fibre was unable to generate second action potential. Th is observation was explained in the following statements;
B. The critical number of sodium channels required to produce an action potential could not be recruited.
C. A large fraction of sodium channels was voltage inactivated.
D. The critical number ofpotassium channels required to produce an action potential could not be recruited.
Which one ofthe following is true
1. Only A
2. A and B
3. Only C
4. C and D [CSIR Answer]
127: An observation was made on a species experiencing three factors A, B and C in order to infer a density dependent population regulation by a factor. The following graph shows the relationship between the adverse effect of the factors in terms of number and population density.
1. A - Density independent; B - Density dependent; C - Inversely density dependent
2. A - Inversely density dependent; B = Density independent; C - Density dependent [CSIR Answer]
3. A - Density dependent; B - Inversely density independent; C - Density independent
4. A - Density dependent; B - Density independent; C - Inversely density dependent
130: Microevolution is the term used for changes in allele frequencies that occur over time. This occurs
B) within a community at genus level
C) due to appearance of new genes from infections
D) due to mutation, natural selection, gene flow and genetic drift
Which of the following combinations is NOT appropriate?
1. A and C
2. A and D [CSIR Answer]
3. B and C
4. B and D
145: An EEG was recorded and its power spectrum analyses were done in rats with implanted electrode for a long time. The power of the EEG waves decreased two months after electrode implantation.
A. Glial cells accumulate surrounding the exposed tips of electrodes.
B. Degeneration of neurons occur surrounding the electrode tips due to metal ion deposition.
C. Coating of electrodes are destroyed with time.
D. The microsocket becomes loose with time.
Which one of the following is true?
1. Only A
2. A and B [CSIR Answer]
3. Only C
4. C and D
More questions add soon (updated)
JRF – Around 100 Marks
LS (NET) - Around 90 Marks
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96 Comments
Qtn 108 correct answer is Only C
ReplyDeletehelp biotech team ,
ReplyDeleteplease send these doubtful questions to csir exam board and save a lot of marks for aspirants
sir what about hormone profile after mating clearly 4 graph is correct and 124 is facilitation tolerance model.138 should be abc becoz tdna is flanked by imperfect repeats and not perfect repeats and only right repeat is used for transfer
ReplyDeleteHelp biotech are you sure about the cut off? Because life sciences never had 100 marks as the cut off..
ReplyDeletesir, (helpbiotech) please provide correct answers of above doubtful questions with refrences/reasons/justifications so that we too can send these representations to csir. thank you
ReplyDeletefrom 2011,jrf cut off onaverage 95, i think its not more than 96
ReplyDeleteFor qtn no-108.. the correct answer is C
ReplyDeleteYes that's what I am trying to say..100 cut off is way too much.. please let us know is this a confirmed news?
ReplyDeletesir even 121 is doubtful because animals are multicellular also so more specific answer is asexual and sexual reproduction which is characteristic feature of plants only that is why biologists considered some sea weeds as plants
ReplyDeleteI evn found dec'13 paper bit tricky and tough. Dec'12 cutoff ws 96 though that paper ws easy as compared to dec'13.. Lets hope fr d best !!
ReplyDeleteFor Q108, the correct answer is "Only C" and for Q100 the correct answer is "
ReplyDelete2"
CSIR NET Life Sciences-Dec 2013
ReplyDeleteSET- A
Ques No: 130
Page 23
According to the CSIR key, the answer is 2 (i.e. A and D)
However, the correct answer should be 3 (i.e. B and C)
Please refer to the explanations from various sources:
1) Biology, 7th ed. Neil A. Campbell & Jane B. Reece:
Macroevolution: Evolutionary change above the species level, including the appearance of major evolutionary developments, such as flight, that we use to define higher taxa.
Microevolution: Evolutionary change below the species level; change in the genetic makeup of a population from generation to generation.
2) Evolution, Douglas J. Futuyma:
macroevolution: A vague term, usually meaning the evolution of substantial phenotypic changes, usually great enough to place the changed lineage and its descendants in a distinct genus or higher taxon.
microevolution: A vague term, usually referring to slight, short-term evolutionary changes within species.
3) Evolution: The History of Life on Earth (2009), Russ Hodge (Page 225):
macroevolution: evolutionary changes that happen over very long periods of time. This usually refers to the development of large new branches of life, such as vertebrates or mammals.
microevolution: evolutionary changes that happen on a small scale, often within a single species, such as a change in the frequency of a particular allele within just a few generations
4) Evolution: A Beginner’s Guide, Burton S. Guttman:
Biologists customarily divide the processes of evolution into three broad categories. Microevolution refers to changes that occur within a single species. Speciation means division of one species into two or more. And macroevolution refers to the larger changes in the variety of organisms that we see in the fossil record. We will begin with an overview of evolution as a whole.
5) Encyclopedia of Biology (2004), Don Rittner & Timothy L. McCabe, Ph.D. (Page 220):
macroevolution: Evolution that deals with large scale and complex changes such as the rise of species, mass extinctions, and evolutionary trends.
microevolution: The smallest scale of evolution; changes within a species; a change in allele or genotype frequencies over time.
can u explain how qustn No. 100 has option 2 right answer
ReplyDeletehormone profile after rat mating the correct ans is 4th but in csir ans it is 3rd which is a graph of normal mensuration cycle. but after mating the correct graph is 4th one.
ReplyDeleteanswer for q.no108 is only C .Reference is Medical physiology by guyton.But i am not sure csir team will correct their answers.
ReplyDeleteI am wondering why everyone is so confused about q.No.130.Answer given by csir is correct ie A and D.If anyone has doubt u can check wikipedia or ecology by peter stiling
June2013 Key too was highly flawed. But they did not review those questions, why would they do it this time. Also I doubt the cut offs according to my OMR copy and the csir key I was securing 2
ReplyDelete92 marks clearly not enough for jrf but I did not even make it to LS. CSIR should certainly start uploading Individual score cards as before.
i think the crt ans for Quest.127 is option 4
ReplyDeleteplz also check the ques from booklet code C ques no. 124...csir say the right ans is 2
ReplyDeleteplz also check ques of booklet code C.....ques no.42, 60, 145
ReplyDeletechetan...ull get NET -LS if ur score is 92.Because for december exams cut off for net is mostly 86 and for jrf is 96.
ReplyDeletefirst of all i thank helpbiotech for providing aspirants to share their thoughts on a common platform and supporting csir aspirants ...
ReplyDeletealong with the above question in booklet A in part B ,question no 58 the option should be either 2 or 1 but it cannot be 3 and if 2 is correct then option 1 is also correct it is printed as encoclomates in option 3 ..i didnt find anything as encoclomates in internet ..if it is there let me know and if we think that is printing mistake ..closest printing mistake is eucoelomates and if eucoelomates is the correct replacement for encoclomates then pseudocoelomates--encoclomates---protostomes----deutrostomes cannot be right ...and if according to csir 58 the question is right than question 51 in booklet A ..is based on same concept and they themselves have given a evolutionary tree than answer for that question is also wrong .. guys keep postion early doubts ..along with helpbiotech we can also represent the doubtful answers and can make a strong case for replacements of answers ..thank you ..
and i believe answers for question 100 and question 130 in part c in booklet A are wrong people who believe that csir key is wrong can they please check carefully and comment ...
in Q 130 all combination are wrong except B
ReplyDeleteQ 127 all the options are wrong
ReplyDeleteMicroevolution is the changes in allele frequencies that occur over time within a population. This change is due to four different processes: mutation, selection (natural and artificial), gene flow, and genetic drift.
ReplyDeletethis matter is from wikipedia , this says that micro evolution occurs with in populations and due to gnetic drift and gene flow.
by this the not appropriate answer is b and c (3).
Q 113 booklet A answer should be option 4
ReplyDeleteSir in set B and qurtion no.31 most apropriate spectral band for vegetation analysis using remote sensing platform is infrared and visible bcoz plant can reflect visible light efficiently so the answer is 2 that is IR and Visible
ReplyDeleteIN set B and quetion no 73 the most exact answer is 1 that is DIAUXIE bcoz two substrate r simultaneously used and in lag period cells produce the enzymes needed to metabolize second sugar
ReplyDeleteSET B qution no. 47 most suitable answer is 3 that is trunk and forelimb
ReplyDeleteFGF10 play a role in limb devlopment , anterio-posterio patterning if it overexpressed , prevent the formation of trunk and forelimb
qouestion no.106 should be c
ReplyDeletein term of vegetation analysis they all material can reflect visible light but for the vegetation there is a NDVI where only red and infrared is required so this ans is wrong.
ReplyDeletejust check Remote sensing and GIS or NDVI
in set A Q 9 CSIR give ans 3 while the option 4 is best suitable because all the points are spreed allover
ReplyDeletein Q 127 set A all graph shows positive relation. in case of positive relation how one factor is negative and how one is independent. since all are wrong
ReplyDeletewhile in Q 130 micro-evolution is at species level not at genus and in two options they consider at genes level means both two not correct then how one is ans.
all questions came in under doubt has no single correct option. even they have two similar option or no one is correct. Vipin
Booklet B:
ReplyDelete96. Which of the following biotic provinces are part of Deccan Peninsula biogeographic zone of India?
Given answer: Option C:
Reason of Rejection: Chotta Nagpur Plateau is not under Deccan plateau rather it comprises eastern regions of India including Jharkhand, Odisha, West Bengal, Bihar and Chhattisgarh.
Right answer: Option D
csir life science dec 2013
ReplyDeleteBooklet code B:
Part B
Ques 30 correct answer should be 2
Ques 35 correct answer should be 4
Ques 47 correct answer should be 3
Part C
Ques 76 correct answer should be 3
Ques 84 correct answer should be 4
Booklet:B
ReplyDeleteQuestion No: 73
Answer given: 3 (Catabolite repression)
But it can also be diauxie
Hallo everyone....Can anyone tell e what will be answer of Q No 86 in booklet C? CSIR answer was facilitation inhibition...But, I think it should be facilitation tolerance..plz anyone confirm it..
ReplyDeletein BOOKLET C, for Q NO 92 i.e about microevolution, 3rd option i.e B and C is correct
ReplyDeleteBooklet-A, Q,113- option 3 is correct becoz female generally permit mating only after up regulation of uterine presser receptor & this up regulation occurs due to increase conc. of estrogen.Progesterone secretion occurs only after ovulation from corpus luteum. Mating does not surely indicate that ovulation has already been done (Tough it may occur after ovulation) bt it surely indicate that mating occurs as the female wanted it which was due to up regulation of uterine pressor receptor & that up regulation was obviously due to estrogen over secretion.
ReplyDeleteq.no :- 124, ecology
ReplyDeleteyes the matrix model was given by ecology is facilitation and tolerance, not inhibition,
according to GOTELLI, who's given models for succession, is tolerance, in tolence all will get linked and biderectional arrows.. plz copy this link and ovserve, who has doubts in this ecology question plz observe this link and try to post same link so that CSIR will consider majority,
http://www.anselm.edu/homepage/bpenney/teaching/BI320/Lectures/8Communitypatternsoverti.html
actually these diagrams are included in aresearch article which is not having free accession.
booklet code B
ReplyDeleteQ. 12 ANSWER MUST BE OPTION 3
For example we take only 3 circles A1, A2 and A3
Considering option 3
Here N=3
Ï€(N+1)N+N-1
Ï€ (3+1)3+3-1
Ï€ (4)3+2
12 π+2
Distance travelled around A1 circle = 2 π1
A1- A2 distance=1
Distance travelled around A2 circle= 2 π2
A2- A3 distance=1
Distance travelled around A2 circle= 2 π3
Total distance=2 π1+1+2 π2+1+2 π3=12 π+2
part c
ReplyDeleteset A
q.no-106
they'v given CRY1 AND CRY2 genes althogh perive blue light but not involved in phototropism, it's wrong,
Recently, Ahmad et al. (1998) published evidence that a
cry1cry2 double mutant was deficient in first positive phototropic
curvature and postulated a role for cry1 and cry2
in phototropism.
excat research says that All of the 3-d-old cry mutants, both single and double, showed reduced curvatures in comparison to the wild
type. This reduction may reflect either a real genetic difference
between the mutants and the wild type, implicating
the cry1 and cry2 photoreceptors in modulating the magnitude
of the phototropic response
plz all who have doubts follow this link address,
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC59300/
q:no:-74
ReplyDeleteset A
enzyme kinetics
km not only dissociation constent and even assoiation constent, both we have to consider,
Kcat is rate constent for es complex to substrate bound enzyme and product.
both are also true.
the best way to compare the catalytic efficiencies of different enzymes or the turnover of different substrate by the same enzyme is to compare the ratio of kcat/ km for two reactions. it's a constant for the conversion of E+S to E+P. (page no 263, lehniger)
explenation :-
kcat/km consider for enzyme to specific substarte, as km changes with substarte, and it's not indivedual property it's strcictly bound enzyme and substarte,
kcat may be indiviudally means can caluclate the efficiency, but km stictly depends upon subtrate, as km changes with substrate how we can expect kcat/ km is not been a property of free substrate and enzyme.
Ans fr ques no. 130 is correct I.e.option 2 because micro evolution is the term used for population at specie level only .
ReplyDeleteq.no:- 114
ReplyDeleteset A
csir given answer as A
but it should be d
bcoz each cell individually equal to any mother cell, r normal cell
before deviding it'l be 100% and after daughter cell is 50%
but when u consider two cell togeater also final each daughter cell should be individually 50% choice of indipenednet assortment on combinig 25% of each daughter cell.
cell r in meotic meta phase only so each cell we to cosider as either microspore mother cell and megaspore mother cell,
plz anyone form genetics background conform this..
@ ch dantu i totally agree with your answer for q.no 106
ReplyDeleteQ.NO :- 102
ReplyDeletecsir says hydrogen perxoide got cleaved by catalse as wrong but it write
asada halliwal pathway come into picture when ROS was formed from electrons accepted by oxygen, till now there is no sysytem saying than oxygen is alternative electron acceptor...
4 enzymes plays a role in scavenging free radicals which include catalase also
here oxygen is not alternative electron acceptor it's main electron acceptor in plants.
Peroxidase, CAT and APX appear to play an essential
protective role in the scavenging process when coordinated
with SOD activity. They scavenge H2O2 generated primary
through SOD action
ref link:-
Antioxidant defense responses: physiological plasticity in higher
plants under abiotic constraints
Cheruth Abdul Jaleel Æ Ksouri Riadh Æ Ragupathi Gopi Æ Paramasivam Manivannan Æ
Jallali Ine`s Æ Hameed Jasim Al-Juburi Æ Zhao Chang-Xing Æ Shao Hong-Bo Æ
Rajaram Panneerselvam
for question number 130 the correct answer should be b and c
ReplyDeleteBooklet code A:-) life science
ReplyDeletePart c question no. 80 answer should be diauxic instead of catabolite repression as provided by csir
Question 100 answer should be 4 the
Q-124 answer should be 1st 3rd is not possible according to the question
Q-127 answer should be 4th 1st can be possible in the case when the graph given is just opposite so 4 the is the right answer
I hope csir wil make correction in answer key and will not dissapoint to the csir aspirants who done well in the examination by filling the right answer......I also thanks to help biotech unit for this initiative it will be great help by or side if correction done by csir in answer key before publishing the result
Set A
ReplyDeleteQ.no. 38 answer should be 3
Becoz FGF 10 is capable of inducing AER in d competent ectoderm bet d dorsal and ventral side of d embryo .This junction is imp .In mutunt in which d limb bud is dorsalize & there is no dorsal ventral junction.AER fails to form and limb devlopment cease( ref Devlopmental bio by Gilbert page no. 451)
Certain FGF can induce trunk and hindlimb . Neuronal expression in d epiblast cell(chick mutunt is limbless)
Ref Dev bio by Gilbert page no. 311
I also agree with ch Dantu for q no 86 in booklet c...the second diagram of comunity succession clearly says that it is tolerance model..according to tolerance model, there are interactions between each and every species because a new species compete with the existing species for resources and eventually the 1st species dies and 2nd prevails...and sometime, 2nd spp can also face competition from the 1st spp which become more competitive with time and compete for the resouces and again they prevail...thus they are related in bidirectional ways...hope CSIR will correct their answers and help CSIR aspirants
ReplyDeleteDear Sir,
ReplyDeleteFor Q.80 the correct answer should be A& C both because Catabolite repression results into Diauxie growth.When there are two separate Carbon sources (for example glucose and lactose) for the growth E.coli will preferably choose glucose for the growth and during this time the genes for lactose operon will remain shut down. After the whole Gilucose sources has been exhausted from the culture media then it will choose lactose for the growth and lac operon will switch on.But this growth will happen after a lag period of time in during which cells produce the enzymes needed to metabolize the second sugar.In the case of Glucose : Lactose this ratio is 1:3 . So we will find two sequential and separate growth pattern. According to the answer as given by CSIR Catabolite repression is the correct mechanism but this will also results into Diauxic growth for three reasons .
First we have two or more than two carbon sources for growth so they can never be utilized simultaneosly and there must occur diauxic shift after a definite time period or lag period.
Second Catabolite repression allows bacteria to adapt quickly to a preferred (rapidly metabolisable) carbon and energy source first. This is usually achieved through inhibition of synthesis of enzymes involved in catabolism of carbon sources other than the preferred one.
Third Also when the answer is Catabolite Repression the repression molecule must be breakdown product of Catabolite that E.Coli consume first preferrably for example in the case of glucose It was thought that a breakdown product of glucose metabolism (a “catabolite” of glucose) might be involved in the repression, and so this repression of the lac operon (and many other operons for alternate energy sources) was called “catabolite repression” As mentioned in the answer option A
In the absence of glucose,lactose is used after a lag period of about three times as long as the lag period for glucose utilization in this case When glucose & lactose both are present E coli first consumes Glucose although lactose is also present but the lac operon isis shut down. Why? Lactose would be binding the repressor and inactivating it. So what is keeping the RNA polymerase from transcribing the LacZ and LacY genes?
It turns out that catabolite repression (the presence of glucose) is instead mediated by the absence of cyclic AMP
(In the absence of glucose, a bacterial “phosphotransferase” protein [“IIAGlc”] accumulates in its activated form, and begins to phosphorylate (and thereby activate) Adenylyl Cyclase. Adenylyl cyclase therefore begins synthesizing cAMP at a high rate. Levels of cAMP rise in the cell.)
Q. 9 Set A) Relation between height and weight according to the given diagram
ReplyDeleteAns. according to CSIR is option 3 (i.e. high heights does not imply proportionality with high weight)
Justification
According to this option high heights shows independency on weight means height and weights are independent that means option “4” is also right.
And according to that diagram the middle heights also shows low to high weight point spread through the graph, then we can also able to say the middle height also not dependent on weight.
Answer
The option 4) weight and height are independent characters
Dear Sir,
ReplyDeleteThe correct answer for the Q.No 108 should be Only C but CSIR answered C& D both correct for this. The explanation and argument behind option Only C is that Action Potential is brief all-or-none depolarization of a neuron’s plasma membrane and it is is a temporary reversal of the electrical potential along the membrane for a few milliseconds. Sodium gates and potassium gates open in the membrane to allow their respective ions to cross. Sodium and potassium ions reverse positions by passing through membrane protein channel gates that can be opened or closed to control ion passage. Sodium crosses first. At the height of the membrane potential reversal, potassium channels open to allow potassium ions to pass to the outside of the membrane. Potassium crosses second, resulting in changed ionic distributions, which must be reset by the continuously running sodium-potassium pump.
Opening Na+ channels, in contrast, will allow Na+ to flow into the cell and will depolarize the cell reducing the negative charge and even making it positive. When a stimulus depolarizes the membrane Na+ channels open, allowing Na+ to diffuse into the cell. After the Na+ have opened they close and K+ channels open. K+ flows out of the cell rapidly reversing the polarity of the cell and briefly undershooting the cell’s resting potential briefly before the resting potential is restored. During the undershoot a second action potential cannot be initiated and this time when the cell is insensitive to stimulation is called the refractory period.
Whenever we talk about action potential we mean to say change or reversal of the normal resting membrane potential (RMP) and this happens by local current of depolarization which propagates along the length of neuron by Opening Na+ channels and this will allow Na+ to flow into the cell and will depolarize the cell reducing the negative charge and even making it positive.
As in the Option D it is mentioned the critical number of Potassium ion channels could not be recruited, but recruitment of K ion or opening of K ions leads into hyperpolarisation of the membrane and Opening K+ channels will increase the flow of K+ out of the cell and hyperpolarize the cell making the potential more negative. Whenever a stimulus is given to a nerve occupying RFM a segment of the whole neuron undergoes depolarization whereas the previous segment is still under RFM and a local current propagates along the nerve impulse making the next segment to undergo depolarization and the previous segment which was previously under depolarization now acquires again RFM so whenever we open K ion channel it \may lead into hyper polarization but never depolarization and hence no action potential even a new stimulus is being given . Also during the refractory period recruitment of K ion channels or opening of K ion channels never lead into generation of Action potential ( as the statement is itself explained in the question)
Booklet c question no. 106 ans should be c
ReplyDeleteReference-
Lehninger, principles of biochemistry.
Topic-
VOLTAGE GATED ION CHANNELS PRODUCE NEURON ACTION POTENTIALS
sir plz chak boklet B (bilingual)
ReplyDeleteQ no. 58,right ans is 1. aromatc aminoacid only but csir given 3. both aminoacid and heme.
Qno. 73 right ans is 1 diauxie, but csir given 3. catabolic ripression.
Q no 76. right ans is 3. only c, but csit given 4. c&d both
Q no 84. right ans is 4 but csir given 3(hormon profile after mating)
sir pls send quari for these Q also if my ans is right.....
sir pls chack booklet b(bilingual) Q no 58, 73, 76, 84.....
ReplyDeleteand send quari for these Q also.....
Dear sir
ReplyDeleteMe too have doubt in booklet A question no. 80 its ans should be diauxic .
I also have doubt in ques 83 ,100,108,124
Any one pls clear my doubt on ques 83
a/c to me its ans should be 1
1) q.no:- 104
ReplyDeleteCSIR ans:- 1 (statements A,B,C)
1) Here with i submit my answer to this question as per Ahmed et al, Nature. 1998 Apr 16;392(6677):720-3; reported CRY1 & CRY2 genes are involved int he phototropism by percieving blue light; and as transgenic arabidopsis plants over express CRY1 & CRY2 shows enhanced phototrophic curvature; and hence cryptochrome is one of the photoreceptors mediating phototropism.
therfore i submit my answer for this question is statements A,B,D; i.e option 4.
here by i'm attaching supporting nature reference of their abstract.
2) q.no:- 100
CSIR ans:- 4 (statements A,C,D)
a) as per cheruth et al, 2009, acta physiol plant (31:427-436) molecular oxygen is the primary electron acceptor in the production of ROS, and hence option A which is saying that molecular oxygen is alternative electron acceptor is incorrect answer as per following research reports.
b) CAT is the site for where hydrogen peroxide is reduced by catalase to form water which is the main detoxification step in Asada- halliwal pathway as per Ron Mittler, 2002, trends in plant science, vol 7(31:427-436).
c) CAT and SOD are the most efficient antioxidant enzymes and their combined action neutralize the potentially dangerous O2 - and H2O2 to H2O (water) and molecular oxygen (O2) thus converting cellular damage (Scandalios 1997).
The correct as per the scientific documants and research publications I submit my answer to be correct as B,C,D are correct statements and option is 1= B,C,D
3) q.no:- 86
CSIR ans:- option 3 (facilitation and inhibition)
according Connell and Slatyer 1977, thi given second matrix is showing tolerance model of succession, on the basis of each species neither facilitate nor inhibits others species. so patch can change from unpredictably from one species to another species.
ref:-
Connell, Joseph H. and Ralph O. Slatyer. “Mechanisms of succession in natural communities and their role in community stability and organization.” The American Naturalist 111 (982) (Nov. - Dec. 1977): 1119-1144.
NICHOLAS J. GOTELLI A Primer of Ecology Fourth Edition, page no:-194
the correct answer as per scientific documents the 1st model represents facilitation and 2nd model represents tolerance, option no 1.
4) q.no:- 124
CSIR ans;- statement B,D :option 2
reference:- enzyme kinetics from a Principles of biochemistry Lehninger, 3rd edition, page no: 263, 264.
Km will be either association constent or dissociation constent based and the reaction direction.
Kcat/Km is also called specificty constant for E+S== E+P. on the basis neither substare nor enzyme is free state, and statement D is wrong. it's in complex/ intermediate state from ES ===EP.
The correct answer could be combination of statement A&C is option 1.
5) q.no:- 112
CSIR ans:- 1
reference:- principles of genetics, 8th edition, page:- 29
the principle of indipendent assortment in metaphase 1 of meotic division must be 100% in each mother cell, and the gamets will be 50%, choice to either of gamets i.e AB, Ab, ab, aB will be 25% each from two mother cell during independent assortment. while writing punnet square for each gamete use to mention 1/4 th 25% is based on this rule only. if Mother cell is representing with 50% intially while writing punnet square or law of independent assortment for each gamete it'l be1/8th or 12.5%.
mother cell= 100%
Daughter cell from single mother cell= 50%
the choice of indeendent assortment for all 4 gamets = 25%
hence correct answer could be option 4th.
6) q.no:- 97
csir ans:- 1 ( statements A,B,C)
reference:- text book of botany (an introduction to plant biology, 4th edition by James D. mauseth)
statement A conider to be incorrect because the bisexual flower always shows haplo diploid nature, hence the early development stages when gynecium develops andrecium will be immature vice verse, to avoid self pollination showing temporal separation of maturation leading to haplo diploid condition in same flower.
the correct answer could be statements B,C,D i.e option 2.
booklet code: A
ReplyDeletefor q.no. 100, the right answer is 4
for q.no. 108, the right answer is 3
for q.no. 113, the right answer is 4
booklet code: C
ReplyDeletefor Q-71, all the options are incorrect, it the peptide is cyclic then how trypsin will digest lysine of arginine from carboxyl group??
In Q-86, the answer should be facilitation model for fig1 and tolerance model for fig2
In Q-104, all the options are incorrect bcos the fourth statement says that phy1 percepts blue light which is incorrect. also the third statement which says that cry1 and cry2 are not involved in phototropism but they are involved in phototropism (in circadian rhythms).
In Q-106, the answer should be only C. Bcos during absolutely refractive period sodium channels are inactivated. And it is written in Alberts and Lodish both
in Q-28, ced-9 is pro-apoptotic and hence the correct answer should be option A.
CSIR NET Life Sciences-Dec 2013
ReplyDeleteSET- A
Ques No: 90
Page 15
"Q: A ligand recognizestwo different cell surface receptors, A and B, on the same cell type. Receptor A, after binding with ligands internalized along with the ligand whereas receptor B, after binding with the ligand, initiates tyrosine kinase activity of the intracellular domain. One particular disease may be resulted due to"
According to the CSIR key, the answer is 2 (i.e. B and C) only.
However, the option 3 can also be a correct answer (i.e. C and D)
because cytoplasmic domains of receptors are recognized by adaptor proteins of clathrin coats then only endocytosis is possible.
So disease can be due to defect in endocytosis also
Explanations /Reference:
Adaptins are components of the adaptor complexes which link clathrin to receptors in coated vesicles. Clathrin-associated protein complexes are believed to interact with the cytoplasmic tails of membrane proteins, leading to their selection and concentration. Gamma-adaptin is a subunit of the golgi adaptor. Alpha adaptin is a heterotetramer that regulates clathrin-bud formation. The carboxyl-terminal appendage of the alpha subunit regulates translocation of endocytic accessory proteins to the bud site.
1) Current Biology, Volume 5, Issue 10, October 1995, Pages 1168–1178
Role of the regulatory domain of the EGF-receptor cytoplasmic tail in selective binding of the clathrin-associated complex AP-2
2) Basic Neurochemistry: Principles of Molecular, Cellular, and Medical Neurobiology by Scott Brady, George Siegel, R. Wayne Albers, Donald Price, page no 122
All answers are CORRECT,..except 130..acoording to question, 1, 3, and 4 are NOT appropriate. if in question NOT is removed then only 2 will be d right answer. i think NOT is mistakenly typed in question.
ReplyDeletei'v posted 6 question with possible explanation all those question number is from set C BOOKLET,
ReplyDeletecsir dec 2013 life science
ReplyDeleteaccording to booklet code A
Qno-100= Ans should be 4th option instead of 2nd option
Explanation= option b can not be correct as cell was taken from animal pole of early frog means it is conditional rather then determined so the transplanted cell can not develop as cell of the animal pole of the adult on the vegetal pole
Region of animal pole from where the cell were removed would be miising in the adult this is correct because the cell from an early frog blastula were removed from the animal pole when it has been removed definitely it will show missing that region in adult so A&C is correct answer which is not present in any other option together except 4 th so A,C,D is the correct answer
QNo-124= Ans should be 1st instead of 3rd opion
Explanation= fig -1 represent facilitation model which is correct but the fig-2 represent tolerance model because in fig 2 reversible sign is present in every step of succession and highly interlinked as by definition inhibition model of succession suggest that the change in plant species dominance over time is caused by death which is not observed in fig 2.
As by definition of facilitation model of succession change in plant species dominance over time is caused by modification in the abiotic environment that are imposed by the developing community thus the entry and growth of the later species depend on earlier species which is observed in fig .1 not in fig 2
So for fig 2 both facilitation model and inhibition model is wrong
QNo-127= Ans should be 4th option instead of 2nd option
Explanation= Species experiencing factor A growing in adverse effect & reach to its maximum carrying capacity means they are highly competitive so they are K selected species and k selected spp show density dependent .
Species experiencing factor B showing no any adverse effect on its population density don’t have any d3etrimentaleffect on its population density and its population is continuously increasing therefor represent density independent
Species experiencing factor c showing initially its population density is inceased frequently but when slightly adverse effect increased or increasing then population density is not with that rate as it was previously increasing so it can be said the relationship is inversely density dependent so 4th is the correct option
QNo.80= Ans should be diauxie not catabolite repression………
CATABOLITE REPRESSION comes in account If both lactose and glucose are present, synthesis of β-galactosidase is not induced until all the glucose has been utilized. Thus, the cell conserves its metabolic machinery (that, for example, induces the lac enzymes) by utilizing any existing glucose before going through the steps of creating new machinery to metabolize the lactose. In the first observation, it was clearly explaned that in absence of glucose lactose is used which can only be explained by the DIAUXIE not by catabolite repression since the lag period between the utilization of the two carbon source can be better explained by this way. Rest all the observation explain the possible utilization of simple hydrocarbon than the complex one.
Q.no.38 ,59 ,80 and 130
ReplyDeleteCsir gives the wrng answers
what is expected date of result
ReplyDeleteSir, will they change the key or no????
ReplyDeletewe request you csir plz display the result after a proper judgemnt because there are lots of queries in your answer keys and every student has to face this so plzz..read all comments
ReplyDeleteI dont know what will happen this time...Will csir change d key or not? or they will just declare d result...I request to all of u v should raise our voice if csir will not take any action regarding the key...
ReplyDeleteWhat will be the cut off, and when will be the result ?
ReplyDeleteIf a candidate write wrong roll number in box but correct in bubbles, will the OMR sheet be cancelled?
ReplyDeleteWat nonsense lots of ans wrong
ReplyDeletein question of hormone profile option 4 is correct because in all option there is rise in LH so there in no question about it .after that when a female rat mates with male there is rise in progesteron level up to 3 days whether pregnancy is concieve or not .in case of pregnancy progesteron level further increases . even stimulating the vagina by a rod in case of female rat results in pseudopregnancy .
ReplyDeleteI applied for JRF, if I have a LS score will my name come in the result?
ReplyDeletesure but not your name only your roll number will come
Deleteif a candidate used black ballpoint pen for shading bubbles , but he used blue ballpoint pen for filling his name and centre name , will the omr sheet be canceled ?
ReplyDeleteLife Science : Booklet C :
ReplyDeleteQuestion Number 3 : Correct option is 3
Question Number 16 : Correct answer is R√3 because minimum distance between A and B is a straight line which pass through interior of the Earth.
Question Number 43 : Correct options are 1 and 2 both.
Question Number 86 : Correct option is 1
@ manjishta even i'v called them, they told me that it would be after april 15th only bcoz of arguments regarding key, it's almost all 5 months and almost all one semester for a students... who cares.....no schedule nothing....
ReplyDeleteFriends,
ReplyDeleteIn the CSIR question K1 of original MM equation is replaced by Km and K2 of original MM equation is replaced by Kcat.
Now discuss all the options.
Option A :
It is correct. Because in the given equation, Km is the association constant.
Option B :
It is wrong. Because in the given equation, Km is the association constant, not disassociation constant.
Option C :
It is wrong. Because Kcat is the rate constant for the chemical conversion of ES complex to free enzyme (not substrate bound enzyme) and product.
Option D :
It is wrong. Kcat/Km ratio refers to the ratio of (thus properties of) (E+P)(E+S) / (ES)(ES).
I think ques74 correct ans is 1
ReplyDeleteCSIR is going to declare results on april last week, had long conversation with them, and informed that lot of issues with the key, because of that they are delaying results, hope for the best...
ReplyDeleteif any one have doubts they can call CSIR exam unit numbers given in the website, so no need to check CSIR website daily...
Please let me know about ques 74 set A.
ReplyDeleteI got 100 mark in life sciences and 3 question those r Wong by csir will I qualify jrf
ReplyDeletedont worry guys csir first declared a modified answer key and then published a result .and result will declared next 3-4 days.......... so all the best
ReplyDeleteSir can you tell me where is the result.. Today is 29th of april..
ReplyDeletePlease dont tell a lie.
@ranjan pandey
ReplyDeletey can't u call csir people to ask about schedule, telephone numbers are there in csir website, u can call them, on monday i'v called them, it'l be after 4 days, no one is sure about results here, call them and inform us also...like u we r also eagerly waiting for results ..
Today is 30 april . Result is still not decleared . Next exam is coming . This wait may affect the preparation of next exam .
ReplyDeleteSHAMe on CSIR....seriously...ICAR NET was conducted frm 29th march to april 10th n today on 30th april we got our result ...but for CSIR xam on 22nd dec n still no results.............horrible terrible.......so disgstng yaar...........
ReplyDeleteThe only institution who conduct a next exam....before declaring the result of previous one?
ReplyDeleteA. CSIR
B.CSIR
C CSIR
D CSIR
7 more days as per CSIR officials informed,
ReplyDeletedeclared results with out key, whats happening, any one know about cutoff?
ReplyDeleteSir, r u taking any action against Q.127 of set A? Sir pls take an action as it is surely a wrong question.
ReplyDeletek when they are publishing the keys?
ReplyDeleteHELPBIOTECH plz come forward to support thousands of your students for their future. Even you know CSIR is working like an opaque system which is not to be so.Since year they had neither published cut off marks nor marks statement of last exam (i.e- june 2013). plz do something!!!
ReplyDeletereally nice post it will help in preparing for the coming CSIR net exam in june
ReplyDeleteIs dere any concession in the marks for sc st obc quota?
ReplyDelete