CSIR-UGC (NET) LIFE SCIENCE ONLINE: August 2015

Wednesday, August 26, 2015

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^ra^-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

Tuesday, August 25, 2015

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

Monday, August 24, 2015

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

Tuesday, August 18, 2015

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

Thursday, August 13, 2015

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 11 TO 20

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 QUETION NO. 1 TO 10 WITH SOLUTION

Q11.Two platforms are separated horizontally by distance A and vertically by distance B.They are to be connected by a staircase having identical steps.If the minimum permissible step length is a,and a maximum permissible step height is b,the number of step the staircase can have is-
(a)≥B/b (b)≤A/a

(c)≥B/b and≤A/a (d)≤B/b and ≥A/a

Answer-C
Description-Max. permissible step length is a
Max. permissible step height is b
Max. number of step in terms of height=B/b
Max. number of steps in terms of length=A/a
so, Max. number of steps ≥B/b and≤A/a

Q12. Ajay,Bunty,Chinu, and Deb were agent, baker,compounder and designer,but not necessarily in the order.Deb told the baker that Chinu is on his way. Ajay is sitting across the designer and net to the compounder.The designer didn't say anything. What is each person's occupation?
(a)Ajay-compounder,Bunty-designer,Chinu-baker,Deb-agent
(b)Ajay-compounder,Bunty-baker,Chinu-agent,Deb-designer
(c)Ajay-baker,Bunty-agent,Chinu-designer,Deb-compounder
(d)Ajay-baker,Bunty-designer,Chinu-agent,Deb-compounder
Answer-D

Q13.Every month the price of a particular commodity falls in this order-
1024,640,400,250,....
What is the next value?
(a)156.25
(b)Approximately 39
(c)64
(d)40
Answer-A
Description-The percentage fall in price=640x100/1024 x100
=37.5
the per cant fall in price
37.5=price fall/present price x100
37.5=price fall/250x100
price fall=93.75
Future price=250-93.75
=156.25

Q14.We defina function f(N)=sum of digits of N,expressed as dimal number.e.g..f(137)=1+3+7=141.Evaluate f(2⁷3⁵5^{6 )-}
^{(a)10 (b)18}
^{(c)28 (d)11}
^Answer-B
^{Description-If f(137)=1+3+7=11 then,}
^f(2⁷3⁵5^{6 )=7+5+6=18}
read abut biomolecules on this link-
BIOMOLECULE

Q15.A certain day,which is x day before 17th August,is such that 50 days,it was 4x days since March 30th of the same year.What is x?
(a)18 (b)30
(c)22 (d)16
Answer-D

Q16.A mouse has to go from point A to B without retracing any part of the path,and never moving backwards.What is the total number of distinct paths that the mouse may take to go from A to B?

(a)11 (b)48
(c)72 (d)24
Answer-B
Description-Total number of distinct paths taken=2x4x3x2=48

Q17.The sum of first n natural numbers with one of them missed is 42.What is the number that was missed?
(a)1 (b)2
(c)3 (d)4
Answer-C
Description-Sum of n natural no.=n/2(2a+(n-1)d)
where, n=no. of natural no. added
a=Initial natural no.
d=difference between them
Sum=n/2(2x1+(n-1)1)
=n/2(n+1)
If the missed natural no. was x thus,
x+42=n/2(n+1)
Substituting the value of x which represents missed natural no.
If x=1
1+42=n/2(n+1)
86=n²+n

n²+n-86=0
No solution

If x=2
= 2+42=n/2(n+1)
= 88=n²+n
= n²+n-88=0
No Solution

If x=3
=3+42=n/2(n+1)
= 90=n²+n
=n²+n-90=0
=n=9

Ifx=4
=4+42=n/2(n+1)
=92=n²+n
=n²+n-92=0
No solution thus,as x=3 can predict that no. of natural no. are added together so option C is correct.

Q18.A2.2 m wide rectangular steel plate is corrugated as shown in the diagram.Each corrugation is a semicircle in cross section having a diameter of 7cm.What will be the widh of steel after it is cirrugated?

(a)1.4m (b)1.6m
(c)0.7m (d)1.1m
Answer-A

Q19.If N,E and T are distinct poitive integers such that NxExT=2013,then which of the followongs is the maximum possible sum of N,E and T?
(a)39 (b)2015
(c)675 (d)671
Answer-C

Q20.The areas of the inner circle and the shaded ring are equal.The radii r1 and r2 are related by-

(a)r1=r2 (b)r1=r2√2

(c)r1=r2√3 (d)r1=2r2

Answer-B

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CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 QUETION NO. 21 TO 30 WITH SOLUTION

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Wednesday, August 5, 2015

CSIR-UGC NET/JRF LIFE SCIENCE EXAM.DECEMBER 2014 SOLVED PAPER WITH DESCRIPTION Q.NO. 1 TO 10

LIFE SCIENCE

PART-A

Q1. What is the 94th term of the following sequence?
1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4
(a)8 (b)9
(c)10 (d)11
Answer-C
Description-The series follows the order,on the basis of repetition of terms thus the 94th term of the sequence is 10.

Q2.Which of the following number is a perfect square?
(a)1022121 (b)2042122
(c)3063126 (d)4083128
Answer-A
Description-√1̅0̅2̅2̅1̅2̅1̅̅̅=1011
As the square roots of all other options are in fraction so only the option 1 is the perfect square root.

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Q3.The equation M²-33n+1=0,where m &n are intergers,has-
(a)no solution
(b)exactly one solution
(c)exactly two solution
(d)infinitely many solution
Answer-A
Description-The equation contains two unknown variables 'm' and 'n' so cannot have any solution.

Q4. The following graphs depict variation in the value of dollar and Euro in terms of the rupee over six months.
which of the following statements is true?

(a)Value of Dollar and Euro rose steadily from January to June
(b)value of dollar and Euro rose by equal rate between January to march
(c)The rise in the value of dollar from April to May is three times the fall in Euro during the same period
(d)Values of Dollar and Euro rose equally between May and June
Answer-C
description-The rise in the value of Dollar is approximately 1.5 rupee which is thrice of the fall in the value of Euro i.e., of 0.5 rupees falls.

Q5.What is the maximum number of whole laddoos having diameter of 6cm that can be packed in a box whose inner dimensions are 24×18x17 cm3?
(a)24 (b)30
(c)33 (d)36
Answer-C
Description-maximum number of whole laddoos that can be packed is 33.

Q6. What is the next term in the following sequence?
7,11,13,17,19,23,29.....
(a)37 (b)35
(c)31 (d)33
Answer-c
description-The sequence as given in the question represents prime numbers and the next prime no. in the series is 31.

Q7. What is the area of the triangle bounded by the lines y=2x,y=-2x and y=6?
(a)36 (b)18
(c)12 (d)24
Answer-B
Description-Area of triangle=1/2xbasex height
=1/2x6x6
=18
you can also read about cell biology on this link-
CELL BIOLOGY
Q8.Three volumes of a hindi book.identical in shape and size,are next to each other in a shelf,all upright,so that their spines are visible,left to right:1,2 and 3.A worm starts eating outside front cover of volume 1.eats its way horizontally to the outside back cover of volume 3.What is the distance travelled by the worm,if each volume is 6cm thick?
(a)6cm
(b)12cm
(c)18cm
(d)a little more then 18cm
answer-A

Q9.A cubical piece of wood was filed to make it into the largest possible sphere. What fraction of the original volume was removed?
(a)more then 3/4
(b)1/2
(c)slightly less then 1/2
(d)slightly more then 1/2
Answer-B

CSIR UGC NET SYLLABUS

For 11 to 20 questions click here-
CSIR NET LIFE SCIENCE JUNE 2014 PAPER -
QUESTION NO. 21 TO 30 WITH SOLUTION