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Which was the toughest part in CSIR NET exam held on 23-12-2012
Part A
52%
Part B
13%
Part C
22%
Entire Question paper
13%
Total votes: 188







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Comments
2012 December CSIR
I checked csirhrdg site just now for results...
But the result has still not declared!!
CSIR NET EXAM
When is the result gona come out?
CSIR NET
It was really good...bt some questions required more time to solve
whts the cut off??
whts the cut off??
paper was easy
paper was easy
thank you
thank you
how many marks
plez let me know how many marks you got?
==123 not sure..
according to IFAS JODHPUR KEY.
haaaaaaaaaa!!!!!!!!!!!!!!
hey.. i got jrfffff:):) hope u also.
too cool! Congrats!! I dint!
too cool! Congrats!!
I dint! Writing again! :)
congrats
congrats buddy
code a 21) 4 25) 4 32) 2 33)
code a
21) 4
25) 4
32) 2
33) 4
90) 4
reason:
Charged Residues of the Rotor Protein FliG Essential
for Torque Generation in the Flagellar Motor of
Escherichia coli
Scott A. Lloyd and David F. Blair*
81) 1
CODE A 21.4 25 4 32
CODE A
21.4
25 4
32 2
agreed!!
qsn 33) can you justify how option 4th is correct??
becuase i think option 2nd and 3rd are also correct
90) option 3 is also correct becz if we mutate the proton ATPase, bacteria will become nonmotile...
(((((nonmotile mutations of fliM and fliN, but not fliG, can be cured by overexpression.))))
question no. 81:::: the receptors for P53 get desensitized when the level of p53 gose beyond its limit, further it will lead to cancer.
for 90th
90) We called proton ATPase means ATP will degrade or synthesize . in the case of flagella rotation it wont happen ,so we wont called proton ATPase . In the case of mitochondria or chloroplast or bacterial membrane where ATP synthesis that subunit only we called proton ATPase .
and also i found many journals mutation of FLI G will show non motile ones.
for eg: Charged Residues of the Rotor Protein FliG Essential for Torque Generation in the Flagellar Motor of Escherichia coli
Scott A. Lloyd and David F. Blair*
The FliG protein of Escherichia coli is essential for assembly and function
of the flagellar motor. Certain mutations in FliG give a non-motile, or
Motÿ, phenotype, in which flagella are assembled but do not rotate. Mutations
with this property are clustered in a C-terminal segment of FliG
that is stable when expressed alone, and thus probably constitutes an
independently folded domain. Previously, we suggested that this domain
forms the rotor portion of the active site for torque generation in the
motor. In this work, we have used a mutational approach to identify the
amino acid residues in the C-terminal domain of FliG that are most important
for motor function. Site-directed mutagenesis was used to replace
each of the conserved residues in this domain with alanine, and the
effects on motor function were measured. Because charged residues have
often been suggested to have important roles in torque generation, conserved
charged residues were changed individually and in all pairwise
combinations. The results show that three charged residues of FliG,
Arg279, Asp286 and Asp287, are directly involved in torque generation.
Mutations in these residues cause motility defects that suggest that they
function jointly, in an active site whose most important property is a
speci®c arrangement of charges. Two other charged residues, Lys262 and
Arg295, may also be involved in torque generation, but are less critical
than Arg279, Asp286 or Asp287. Uncharged residues of the FliG motility
domain do not appear to have direct roles in torque generation, although
some are needed for the stability of the protein or for normal clockwise/
counter-clockwise switching. The Motÿ mutations of ¯iG isolated
previously by random mutagenesis do not alter the putative active-site
residues, but render the proteins abnormally susceptible to proteolysis,
suggesting signi®cantly altered conformations or reduced stabilities.
90th qsn
ATP generated
by glycolysis powers a membrane H-ATPase that maintains the protonmotive
force.
FliG, FliM, and FliN are also referred to as the “switch complex,” since many
mutations of fliG, fliM, and fliN lead to defects in switching (in control of the
direction of rotation) (32, 33).
An electron micrographic analysis of basal-body structures found in nonmotile missense mutations of fliG, fliM, and fliN indicates loss of the C-ring, the components of which (FliM andFliN) can be recovered in the cytoplasm
And nonmotile mutations of
fliM and fliN, but not fliG, can be cured by overexpression.
receptor 4 p53 means the
receptor 4 p53 means the element on DNA where p53 binds,,,,
@kaur simran
dat means ans 4 81 is 2?????????????
81th
we need to remember 2nd option is not ubiquitinases it is deubiquitinases.
if p53 deubiquitinases more it ll degrade MDM2 protein( mdm 2 is ubiquitine ligase ) so p53 level going to increase . if p53 is more it inhibit cyclin and cdks so 3rd and 4th option also correct .one more if cdk and cyclins inactivate it wont phosphorilate Rb. so 1st option also correct
the question is cancer with normal p53 gene effect
according to me 1st option
according to me 1st option can never be the answer...
beacuse non-phosphorylatable form of Rb will never release the elongation factor...??
how this will lead to cancer?? explain?
see!!! p53 is going to act on
see!!! p53 is going to act on DNA element, which will lead to the production of p21 which will further inhibit G1/S- Cdks and phosphorylation of Rb protien..
but what if that DNA element gets desensitized beacuse of the excess OF P53----->> whole process will stop ---->> leading to CANCER..
@simran good observation
can you share the reference please...
deubiquitinase mean it will
deubiquitinase mean it will remove the ubiquitin molecules( which usually mark p53 for its degradation).
hence the level of the p53 will increase beyond its normal level....
and it will lead to desensitization....
me agree wth kaur simran
most probable ans for 81 is 2. @j reddy how 1st option bcome correct??
according to me yess!!!
according to me yess!!!
it was moderate...but C
it was moderate...but C section is always tough...but there also u can answer around 10 questons easily...try to do less negative questions in part B..and main thing is the choice of questions in part C..
@j reddy
y 81 is 1?? explain plzzzzz
doubt
csir dec 2011
code a
wt is the answer for 6?
Following night only. because
Following night only. because stars rarely or very slowly change their position, after six months from same place on earth will observe different stars view. Ans 1
After six monts position of
After six monts position of earth will be changed and it will observe many other new stars.
ohhhhhhhhhhh friends qust no.3
ans is 3................:(:(:( its sure.........
Math Professors of three
Math Professors of three different institutes observing answer 4 correct.
@rahul
the quest can be taken in 2 diff ways
1. which of the above are wrong statements? then the ans is 4 $5.
2. which of the above is the wrong step? then ans is 3to4. bcoz statement 3 is correct .and 4 is wrong. ie the mistake occur btn 3 and 4. rt? ie the wrong step is from 3 to 4.... clear ( in quest -which is wrong step)
think about it.. part a for lifescience and maths are same.... all of their answer key tell it is 3.
when will the result of dec
when will the result of dec 2011 net will be published? anyone have any idea..
Cutoff marks
Is there a separate cutoff for part A,B and C?? What is the expected cutoff for JRF+LS?? I expect my marks will be 80...Is there any hope of me clearing it?
hope 4 LS
hope 4 LS
friends, qust no.114
what the ans 1/8 or 1/4? pls give explanation
Answer
According to question the next children with blood group O and normal vision must be a FEMALE. So that only female progeny can be considered i.e. total 8 females out of these 2 females are with O blood group and normal vision. Correct Answer is 2/8= 1/4.
qust no.114
1/8 for sure...
the probability of getting blood grp O is 1/4
and probability of getting normal female is 1
and proababilty of getting female is 1/2
so its 1/4*1*1/2 =1/8
:) @ kaur Simran
AGAIN CONFUSED. IAGREE WITH U AND I HAVE MARKED 1/8.
BUT THE TWIST IN QUEST IS
"CHANCE OF NEST FEMALE CHILD HAS THIS TRAIT"
NOT"THE CHANCE OF THE PARENT HAS A FEMALE CHILD WITH THIS TRAIT"
ie no need to consider the chance of getting a female child(1/2).
so 1/4*1=1/4.
OMG
Oh god...!! now i am really afraid...lol...
@ bhasura
thanku...:)
CODE A Question 10th: CO2
CODE A
Question 10th: CO2 levels in earth
kindly tell me the correct answer.. i i think it was highest in the very early atm of the earth.. when there wr no plants n no photoautotrophs....
http://sbvor.blogspot.in/2000/01/atmospheric-co2-over-time.html
kindly clear my
doubts
.thanku...:)
Co2 level was very high in
Co2 level was very high in atmosphere of primitive Earth.
sorry quest no 128 ans is 3
the explanation is same...
The term iteroparity comes
The term iteroparity comes from the Latin itero, to repeat, and pario, to beget. An example of an iteroparous organism is a human—though many people may choose only to have one child, humans are biologically capable of having offspring many times over the course of their lives. Iteroparous vertebrates include all birds, most reptiles, virtually all mammals, and most fish. Among invertebrates, most mollusca and many insects (for example, mosquitoes and cockroaches) are iteroparous. Most perennial plants are iteroparous.
copyrite~ wikipedia.org
THIS LAST LINE is not letting me believe that the answer is 3rd.....
and the best strategy will be, that it will reproduce early and more then once...!!!!
still not satisfied:(
y confusion? that line is
y confusion? that line is correct. iteroparouscan be r selected or k selected.
in our qustion the sps is smaaaaallll and live only 20 days so they are r selected and semelparous, (not iteroparouscan) (all molluscs most of insects live more than 20 days so they not the species in question)
semelparous =only reproducing once before they die. Semelparous organisms may beshort-lived,like annual crops.
okay that means my ten marks
okay that means my ten marks r gone 5 for this and 5 for cross(genetics)
OH god...:(
@kaur simran quest 128
qust no 128 i think ans is 1.
tis type org is r selected (small size of organism short life expectancy many offspring are produced)
and in r selected sps there are
1.semelparous, only reproducing once before they die. Semelparous organisms may be short-lived,like annual crops.( in qust 128 organism is small<10cm, short lived 20 days so i think it is a semelparous and reproduce only once)
However, some semelparous organisms are relatively long-lived, such as the African flowering plant Lobelia telekii which spends up to several decadesaninflorescence that blooms only once before the plant dies,
[2. iteroparous, reproducing more than once in a lifetime. However, iteroparous organisms can be more r-selected than K-selected, such as a sparrow, which gives birth to several chicks per year but lives few years, (but in quest only 20 days)
i think it is clear now. for more refer wikipedia, google books.
@kaur simran
hi
qust 10.. ans 1. very early in earth. no doubt. its sure.
ANY DATE ABT RESULT OF JRF
ANY DATE ABT RESULT OF JRF 2012 N SAMPLE PAPER FORS FOR JRF