csir net dec 2015 ls answer key

7/24/2019 Csir Net Dec 2015 Ls Answer Key

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Next Batch for CSIR NET Coaching Starts 10th& 11thJan 2016.

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CSIR NET

December 2015

ANSWER KEYPart A & B for Booklet B

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CSIR NET - LIFE SCIENCE DEC2015

Answer Key

Ans : 2

Explanation: SP = 10%of 100 + 100 = 110, Reselling price = 110 - 10%of 110 = 110 - 11 = 99, Hence

shopkeeper gets 110-99 = 11/- extra

Ans: 3

Explanation: Since, vessel is already filled with some volme of water (hence the graph will not

start with Origin, thus neglecting option 1 & 4) & its under filling process by a constant rate (Hence

option 2 is also neglected)


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Ans: 1

Explanation: since, x = 0, hence its a straight line at Y axis, (hence its a right angled triangle) Line 1 ,

y = x, Line 2 , y = 1-x, Equating line1 & 2 we get, x = 1-x, => x = 1/2 = 0.5, putting this value in line 2

equation, we get y = 0.5, Hence it is isoscales with 2 sides of same dimension.

Reference: http://gmatclub.com/forum/what-is-the-area-of-a-triangle-formed-by-lines-y-5-x-54110.html

Ans: 3

Ans: 1

Explanation: There are 20 small sqares & 10 big squares made by 4 small squares each


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Ans: 4

Explanation: Data is insufficient because by any means the person can't walk all these three

distances given in the answers in one hour.

Ans: 3

Explanation: time can never be decreased..


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Ans: 1

Explanation: At the top of the wheel velocity will be equal to 2v means 120 km/hr. At the bottom

of the wheel velocity will be -v at the top it will be v so add these vector quantities it will be equal

to 2v.

Ans: 1

Explanation: No effect will be observed, as soon as needle will come out the cell lipid bilayer will be

formed again.

Ans: 3


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Explanation: M=1000, D= 500, C=100, L= 50, V=5 , I=1. They are all roman numerals.

Reference:http://literacy.kent.edu/Minigrants/Cinci/romanchart.htm

Ans: 2

Explanation: porosity = 1- bulk density/density = 0.465 cc

Reference:http://lawr.ucdavis.edu/classes/SSC100/probsets/pset01.html

Ans: 4

Explanation:Area of square encolsed by each circle = (1/4) pi x r^2, (since 1/4th area of circle is

overlapping with sqaure), Since there are 4 squares, hence area of square under all 4 circles = 4 x

(1/4) pi x r^2 = pi x r^2, Since r = 1 unit, hence area under 4 circles = pi, Total area of square =

(side)^2 = 2^2 = 4, hence left out area = 4-pi
http://literacy.kent.edu/Minigrants/Cinci/romanchart.htmhttp://literacy.kent.edu/Minigrants/Cinci/romanchart.htmhttp://literacy.kent.edu/Minigrants/Cinci/romanchart.htmhttp://lawr.ucdavis.edu/classes/SSC100/probsets/pset01.htmlhttp://lawr.ucdavis.edu/classes/SSC100/probsets/pset01.htmlhttp://lawr.ucdavis.edu/classes/SSC100/probsets/pset01.htmlhttp://lawr.ucdavis.edu/classes/SSC100/probsets/pset01.htmlhttp://literacy.kent.edu/Minigrants/Cinci/romanchart.htm


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Ans: 4

Explanation: It will make triangle of height 3 meter, base 4 meter and hypotenuse 5 m.

Ans: 4

Explanation:A projector can have the same magnification but in any case resolution or the ability to

differentiate between closly placed objects will always be higher in case of microscope. So thats why

all other options are wrong, only fourth one is correct.

Reference:http://www.ncbi.nlm.nih.gov/books/NBK21629/

Ans: 4

Explanation: E. Six F. Five G. Four H. Seven Descending order gives H, E, F, G
http://www.ncbi.nlm.nih.gov/books/NBK21629/http://www.ncbi.nlm.nih.gov/books/NBK21629/http://www.ncbi.nlm.nih.gov/books/NBK21629/http://www.ncbi.nlm.nih.gov/books/NBK21629/


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Ans: 2

Explanation: Let l represent the length of one of the smaller rectangles, and let w represent

the width of one of the smaller rectangles, with w


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Ans: 4

Explanation: Dimension of box = 29x29x29 cm^3, Since, the thickness of the paper is 1 cm, hence

the next box which will be fitted in will have to be of dimension 27x27x27 cm^3 & similarly we

conclude that there will be 14 boxes which can be fitted inside.

Ans: 2

Explanation: So after one hour bucket b will be having 1 liter and A will be having 1 liter. After one

and half hour A = 1.5 liter, B= 1/2 liter. After 2 and half hour, A= 0.5 liter, B= 1.5 liter. After 3 hours A=

1 liter, B= 1 liter. After 4 hours A= 0, B= 2 liter.


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Ans: 3

Explanation: Here colours are constant, levels are variable. So probability will be equal to 3^8.

Ans: 1

Explanation: Enzymes Accelerate Reactions by Facilitating the Formation of the Transition State


Ans: 1

Explanation: 1 molecule = moles * avagadro number = 1/6 * 10^-23
http://www.ncbi.nlm.nih.gov/books/NBK22431/http://www.ncbi.nlm.nih.gov/books/NBK22431/http://www.ncbi.nlm.nih.gov/books/NBK22431/http://www.ncbi.nlm.nih.gov/books/NBK22431/


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Ans: 2

Explanation: Ionic strength = number of ions produced after ionization, so in this case it will be

doubled = 2*0.2 M = 0.4 M

Ans: 1

Explanation: Glycophorin has a segment of 19 hydrophobic residues (residues 75 to 93) forms an

helix that traverses the membrane bilayer once.

Reference: Principles of Biochemistry by Leninger,fourth edition,page no 374

Ans: 3

Explanation: Since cell line is deficient in Salvage pathway so it will follow de novo pathway for purine

synthesis.In de novo pathway the amine group and nitrogen is given by all three amino acids.

Reference: Principles of Biochemistry by Leninger,fourth edition,page no 864-865


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Ans: 4

Explanation: Phosphorylation of Rb protein is the 1st step during G1-S phase transition.

Then Mitotic CDKs Promote Nuclear Envelope Breakdown by

phosphorylation of laminins. Polyubiquitination of securin cause the

separation of sister chromatids and finally the nuclear membrane and pore

complex reappear.

Reference:Molecular cell Biology by Lodish,7th edition page no-926-33

Ans: 4

Explanation: The natural environment of E.coli cells is the lower intestine of a warm-blooded animal.

Its optimal growth temperature is 37 0C. The doubling time or generation time for most E.coli strains

in a rich medium at 37 0C is between 20 to 40 minutes. As its given that the fresh round of

replication took 20 minutes it is understandable that the division process will take more time than


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20 minutes. Its highly likely that it will not take double the time it took to replicate so most probable

answer is 30 mins.

Reference:http://www.exptec.com/Expression%20Technologies/Bacteria%20growth%20media.htm

Ans: 3

Explanation: Puromycin is a well-known inhibitor of protein synthesis in vivo as well as in cell-free

systems.Its structure, analogous to aminoacyl-tRNA, leads to the premature release of unfinishedpolypeptide chains as polypeptidyl-puromycin derivatives.

Reference:http://www.pnas.org/content/70/12/3866.full.pdf

Ans: 3

Explanation: Coupling of reaction centers of ETC complexes occurs by combining Ubiquinones and

Cytochrome C.
http://www.exptec.com/Expression%20Technologies/Bacteria%20growth%20media.htmhttp://www.exptec.com/Expression%20Technologies/Bacteria%20growth%20media.htmhttp://www.exptec.com/Expression%20Technologies/Bacteria%20growth%20media.htmhttp://www.pnas.org/content/70/12/3866.full.pdfhttp://www.pnas.org/content/70/12/3866.full.pdfhttp://www.pnas.org/content/70/12/3866.full.pdfhttp://www.pnas.org/content/70/12/3866.full.pdfhttp://www.exptec.com/Expression%20Technologies/Bacteria%20growth%20media.htm


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Ans: 3

Explanation: The ability of a "vital" dye, acridine orange (AO), to intercalate into the DNA of living

cells was investigated by quantitative intensified fluorescence microscopy and digital imaging under

various conditions of dye concentration, excitation light intensity, and ionic concentration

Reference:http://www.ncbi.nlm.nih.gov/pubmed/2015848

Ans: 2

Explanation:Loading of the replicative DNA helicase at origins of replication is of central importance

in DNA replication. In eukaryotic cells, helicase loading is tightly restricted to the G1 phase of the cell

cycle.

Reference:http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660832/

Ans: 4

Explanation: phosphatidyl inositol is not a second messenger.IP3 and DAG these two second

messengers are from phosphatidyl inositol 4,5 bisphosphate
http://www.ncbi.nlm.nih.gov/pubmed/2015848http://www.ncbi.nlm.nih.gov/pubmed/2015848http://www.ncbi.nlm.nih.gov/pubmed/2015848http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660832/http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660832/http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660832/http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660832/http://www.ncbi.nlm.nih.gov/pubmed/2015848


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Ans: 4

Explanation: CD8 T cells recognize antigen that is expressed on the surface of class I MHC

molecules, and function mainly as cytotoxic T cells.

Reference: Kuby Immunology,7th edition page no 99

Ans: 4

Explanation: cadherin,selectin,immunoglobulin super family are all cell adhesion poteins.But laminin

is an extracellular matrix protein.Basal lamina the very good example of extracellular matrix is having

the protein laminin

Ans: 3


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Explanation: gain of function mutation will give rise to oncogene

Reference: Bruce albert,Cancer chapter

Ans: 2

Explanation:Alpha amanitin phalloidin poisonous mushroom. It will inhibit RNA polymerases. RNA

polymerase 1 is insensitive to alpha amanitin.RNA Pol2 is highly sensitive.While RNA pol3 is sensitive

in high concentration

Reference: Lubert Strayer sixth edition, transcription Chapter


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Ans: 4

Explanation: Stable isotopes of carbon is used to distinguish between C3,C4 and CAM plants.

These plants differs in their ability to assimilate different carbon isotopes.

Reference: (Plant physiology by Taiz and Zeiger, page 188- Added by Sangeeta),

https://books.google.co.in/books?id=GDIWBAAAQBAJ&pg=PA138&lpg=PA138&dq=why+carbon+12

+and+13+are+used+with+C4+and+C3+plants&source=bl&ots=3USsNboaz-

&sig=3YzgnYRYsujapcoMxZwmar21EtY&hl=en&sa=X&ved=0ahUKEwjJ-dCdperJAhXYj44KHfhdCOQQ6AEIMjAE#v=onepage&q&f=true

Ans: 2

Explanation: Class A will give carpels instead of sepals and stamens instead of petals,class A and B

will give petals, class B will give sepals instead of petals and carpels instead of stamen, Band C will

give stamens,Class C will have petals instead of stamen and C alone produces Carpels. B is mutated

then sepals sepals carpel carpel will be the result.

Reference: Taiz and Zeiger Plant physiology Pg 564

Ans: 1

Explanation: Ingression. The migration of individual cells from the surface layer into the interior of the

embryo.

Reference: Developmental Biology, 6th edition Scott F. Gilbert figure 8.6


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Ans: 3

Explanation:Aleurone layers in grains of long-day plants such as Barley, secretes more-amylase and had a higher responsiveness to GA3 which can be measured by

-amylase secretion. Storage of the grains increased both the percentage of

germination and the responsiveness of the aleurone to GA3.

Reference: (Plant physiology by Taiz and Zeiger, page 489- Added by Sangeeta),

http://plantsinaction.science.uq.edu.au/edition1/?q=content/2-1-3-sucrose-and-starch-synthesis

Ans: 1

Explanation: Phenyl alanine a precursor for phenolics and Alkaloids originated from Shikimic acid

Reference: (Plant physiology by Taiz and Zeiger page 291- Added by Sangeeta)

https://en.wikipedia.org/wiki/Shikimate_pathway


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Ans: 2

Explanation:Alveoli arise from endoderm only.

Reference:http://discovery.lifemapsc.com/library/review-of-medical-embryology/chapter-25-germ-

layers-and-their-derivatives

Ans: 2

Explanation: Chichkdevelopment of wings, feathers and claws is decided by regional specificity of

induction. In this case mesenchyma is going to play instructive role for the differentiation of epithelial

layer.


Ans: 1

Explanation: h-channel is not a pacemaker potential of heart
http://discovery.lifemapsc.com/library/review-of-medical-embryology/chapter-25-germ-layers-and-their-derivativeshttp://discovery.lifemapsc.com/library/review-of-medical-embryology/chapter-25-germ-layers-and-their-derivativeshttp://discovery.lifemapsc.com/library/review-of-medical-embryology/chapter-25-germ-layers-and-their-derivativeshttp://discovery.lifemapsc.com/library/review-of-medical-embryology/chapter-25-germ-layers-and-their-derivativeshttp://www.ncbi.nlm.nih.gov/books/NBK9993/http://www.ncbi.nlm.nih.gov/books/NBK9993/http://www.ncbi.nlm.nih.gov/books/NBK9993/http://www.ncbi.nlm.nih.gov/books/NBK9993/http://discovery.lifemapsc.com/library/review-of-medical-embryology/chapter-25-germ-layers-and-their-derivativeshttp://discovery.lifemapsc.com/library/review-of-medical-embryology/chapter-25-germ-layers-and-their-derivatives


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Reference:http://circep.ahajournals.org/content/2/2/185.full

Ans: 2

Explanation: Preganglionic neurons of sympathetic nervous system secrete Acetylcholine as their

neurotransmitter

Reference:http://hal.bim.msu.edu/ome2/Content/Circulatory/Overview/Autonomic/Overview/page_1_text.html

Ans: 4

Explanation: Most of the triose phosphate synthesised in chloroplasts is converted to

either sucrose or starch. Starch accumulates in chloroplasts, but sucrose

is synthesised in the surrounding cytosol.

Reference: (Plant physiology by Taiz and Zeiger page 143- added by Sangeeta)http://plantsinaction.science.uq.edu.au/edition1/?q=content/2-1-3-sucrose-and-starch-synthesis
http://circep.ahajournals.org/content/2/2/185.fullhttp://circep.ahajournals.org/content/2/2/185.fullhttp://circep.ahajournals.org/content/2/2/185.fullhttp://hal.bim.msu.edu/ome2/Content/Circulatory/Overview/Autonomic/Overview/page_1_text.htmlhttp://hal.bim.msu.edu/ome2/Content/Circulatory/Overview/Autonomic/Overview/page_1_text.htmlhttp://hal.bim.msu.edu/ome2/Content/Circulatory/Overview/Autonomic/Overview/page_1_text.htmlhttp://circep.ahajournals.org/content/2/2/185.full


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Ans: 4

Reference: https://embryology.med.unsw.edu.au/embryology/index.php/Mouse_Estrous_Cycle,

http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0035538

Ans: 1

Explanation: Kussmaul breathing is a deep and labored breathing pattern often associated with

severe metabolic acidosis, particularly diabetic ketoacidosis (DKA) but also kidney failure. It is a form

of hyperventilation, which is any breathing pattern that reduces carbon dioxide in the blood due to

increased rate or depth of respiration.

Reference:https://en.wikipedia.org/wiki/Kussmaul_breathing
https://en.wikipedia.org/wiki/Kussmaul_breathinghttps://en.wikipedia.org/wiki/Kussmaul_breathinghttps://en.wikipedia.org/wiki/Kussmaul_breathinghttps://en.wikipedia.org/wiki/Kussmaul_breathing


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Ans: 1

Explanation: Pleotropic mutation is a mutation that has effects on several different characters. So the

best way to analyse the function of gene X will be by creating pleotropic mutants. Since the question

doesn't mention anything about the differential expression of the gene X under different temperature,

so we can't say conclusively that temperature sensitive mutants will help in analysing the gene's

function. As the mutation is lethal in haploids, so it will be lethal in recessive mutants as well,

therefore we can't see its function in recessive mutants. Same is the case with mutants with low

penetrance. So option 1 is most appropriate.

Reference:http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3558540/

Ans: 1

Explanation: Since the children of affected parents of generationn I are both normal, that means traitcannot be a recessive trait (because children of homozygous recessive parents are also homozygous
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3558540/http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3558540/http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3558540/http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3558540/


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recessive). So the trait here is a dominant one and both the parents of gen I are heterozygous for it.

Father affected with X-linked domiannt trait will always have affected daughters. Since that is not the

case here, so it is a Autosomal dominant trait. answer is option 1.

Reference:https://www.nlm.nih.gov/medlineplus/ency/article/002049.htm

Ans: 1

Explanation: Cross: AaBb x AaBb; in next generations following general genotypes will be seen- i)

A_B_:phenotype Green as A and B both present. Proportion 9/16 ii) A_bb: Phenotype yellow as A is

present alone, so only yellow intermediate is formed. Proportion 3/16. iii) aaB_: Phenotype white. as A

itself is absent, so the yellow intermediate itself can't be formed. Proportion 3/16. iv) aabb: since Both

A and B is absent, again phenotype will be white. Proportion 1/16. So the final proportion will be

Green (9): White (4): Yellow (3).

Reference:http://biolearnspot.blogspot.in/2013/11/recessive-epistasis.html
https://www.nlm.nih.gov/medlineplus/ency/article/002049.htmhttps://www.nlm.nih.gov/medlineplus/ency/article/002049.htmhttps://www.nlm.nih.gov/medlineplus/ency/article/002049.htmhttp://biolearnspot.blogspot.in/2013/11/recessive-epistasis.htmlhttp://biolearnspot.blogspot.in/2013/11/recessive-epistasis.htmlhttp://biolearnspot.blogspot.in/2013/11/recessive-epistasis.htmlhttp://biolearnspot.blogspot.in/2013/11/recessive-epistasis.htmlhttps://www.nlm.nih.gov/medlineplus/ency/article/002049.htm


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Ans: 3

Explanation: When crossing over occurs between gene locus and centromere in a heterozygous

individual, then the segregation of the 2 alleles will occur in second meiotic division i.e. anaphase II,

when the sister chromatids separate.

Reference:http://www.doctortee.com/dsu/tiftickjian/genetics/tetrad-analysis.html

Ans: 2

Explanation: evolution is a random process that takes place continuously but very slowly over the

years of time. not every evolutionary changes is beneficial. we can not moniter the process as ittakes million of years to take place and we can see only the resultant phenotype only if it survives.

For example, modern days birds have supposed to be evolved from reptiles and acheoptyrx is the

link organism we all know but no one has seen the intermediate form other than this.thus evolution

is continuous process of selection.

Reference:http://evolution.berkeley.edu/evolibrary/article/misconcep_02
http://www.doctortee.com/dsu/tiftickjian/genetics/tetrad-analysis.htmlhttp://www.doctortee.com/dsu/tiftickjian/genetics/tetrad-analysis.htmlhttp://www.doctortee.com/dsu/tiftickjian/genetics/tetrad-analysis.htmlhttp://evolution.berkeley.edu/evolibrary/article/misconcep_02http://evolution.berkeley.edu/evolibrary/article/misconcep_02http://evolution.berkeley.edu/evolibrary/article/misconcep_02http://evolution.berkeley.edu/evolibrary/article/misconcep_02http://www.doctortee.com/dsu/tiftickjian/genetics/tetrad-analysis.html


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Ans: 3

Explanation:Animals possessing body cavity/coelom- Annelida (Earthworm), the body cavity not

lined with mesoderm- pseudocoelom, Aschelminthes/Nematodes (Roundworm), animals without bodycavity- Acoelom- Playthelminthes (Flatworm)

Reference:http://schools.aglasem.com/14813

Ans: 4

Explanation: The IUCN Red List of Threatened Species has identified amphibians as being the most

threatened vertebrate group assessed so far, with around 41% at risk of extinction.

Reference: (http://cmsdocs.s3.amazonaws.com/summarystats/2015-

4_Summary_Stats_Page_Documents/2015_4_RL_Stats_Table_3a.pdf- Added by Sangeeta),

http://www.iucn.org/about/work/programmes/species/our_work/amphibians/
http://schools.aglasem.com/14813http://schools.aglasem.com/14813http://schools.aglasem.com/14813http://schools.aglasem.com/14813


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Ans: 4

Explanation: The primary greenhouse gases in Earth's atmosphere are water vapor,carbon dioxide,

methane, nitrous oxide, and ozone

Reference:https://en.wikipedia.org/wiki/Greenhouse_gas

Ans: 2

Explanation:Anthocerophyta are group of non-vascular plants charaterised by presence of greem

gametophyte with single large chloroplast and sporophyte with stomata. Rhizoids are presenet whichare usually unicellular.

Reference:http://www.bryoecol.mtu.edu/chapters/2-8Anthocerotophyta.pdf
https://en.wikipedia.org/wiki/Greenhouse_gashttps://en.wikipedia.org/wiki/Greenhouse_gashttps://en.wikipedia.org/wiki/Greenhouse_gashttp://www.bryoecol.mtu.edu/chapters/2-8Anthocerotophyta.pdfhttp://www.bryoecol.mtu.edu/chapters/2-8Anthocerotophyta.pdfhttp://www.bryoecol.mtu.edu/chapters/2-8Anthocerotophyta.pdfhttp://www.bryoecol.mtu.edu/chapters/2-8Anthocerotophyta.pdfhttps://en.wikipedia.org/wiki/Greenhouse_gas


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Ans: 3

Explanation: Ovulate & microsporangiate cones on separate plants; fleshy-coated seeds

Reference:http://focusky.com/efwx/aerg

Ans: 4

Explanation: birds do not have a good sense of smell, so bird-pollinated flowers usually have little

odor.

Reference:www.life.umd.edu/CBMG/faculty/Moctezuma/.../Lec13_Pollination.ppt

Ans: 1
http://focusky.com/efwx/aerghttp://focusky.com/efwx/aerghttp://focusky.com/efwx/aerghttp://www.life.umd.edu/CBMG/faculty/Moctezuma/.../Lec13_Pollination.ppthttp://www.life.umd.edu/CBMG/faculty/Moctezuma/.../Lec13_Pollination.ppthttp://www.life.umd.edu/CBMG/faculty/Moctezuma/.../Lec13_Pollination.ppthttp://www.life.umd.edu/CBMG/faculty/Moctezuma/.../Lec13_Pollination.ppthttp://focusky.com/efwx/aerg


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Explanation: Exponential word problems almost always work off the growth / decay formula, A =

Pert, where "A" is the ending amount of whatever you're dealing with (money, bacteria growing in

a petri dish, radioactive decay of an element highlighting your X-ray), "P" is the beginning amount

of that same "whatever", "r" is the growth. dN/dT = rN is the logarithmic form of formula.

Reference: http://www.mpi-bremen.de/Binaries/Binary13037/Wachstumsversuch.pdf

http://study.com/academy/lesson/logistic-population-growth-equation-definition-graph.html

Ans: 1

Explanation: radioisotpes are used in non-clinical studies to assess drug

absorption,distribution,metabolism,and excretion(ADME). Some isotopes that are currently inuse

are H3,C14,P32, and I131but the C14 is the choice of most of the ADME studies.

Reference:cdn.intechopen.com/pdfs-wm/23691.pdf

Ans: 2

Explanation: Animals are monogamous because it was the only way they could guard their mates

and thus their breeding rights. Males stayed with one female to ensure their young were not killed
http://cdn.intechopen.com/pdfs-wm/23691.pdfhttp://cdn.intechopen.com/pdfs-wm/23691.pdfhttp://cdn.intechopen.com/pdfs-wm/23691.pdfhttp://cdn.intechopen.com/pdfs-wm/23691.pdf


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by another male. Its one thing to sire a litter of pups and another to ensure that they actually

survive to reproduce to carry on their genetic lineage.

Reference: http://healthland.time.com/2013/07/30/the-

reason-for-monogamy-researchers-disagree/

Ans: 4

Explanation: The neomycin phosphotransferase gene (nptII) conferring resistance to kanamycin and

neomycin is one of the antibiotic resistance genes commonly present in GM plants.

Reference:http://www.ncbi.nlm.nih.gov/pubmed/21722080

Ans: 4

Explanation: As the conc. of glycerol will be increased in aquous solution viscosity will also

be inccreased. By making a standard plot it can be easily determined.

Refer:http://www.jlr.org/content/5/3/314.full.pdf

Ans: 4
http://www.ncbi.nlm.nih.gov/pubmed/21722080http://www.ncbi.nlm.nih.gov/pubmed/21722080http://www.ncbi.nlm.nih.gov/pubmed/21722080http://www.jlr.org/content/5/3/314.full.pdfhttp://www.jlr.org/content/5/3/314.full.pdfhttp://www.jlr.org/content/5/3/314.full.pdfhttp://www.jlr.org/content/5/3/314.full.pdfhttp://www.ncbi.nlm.nih.gov/pubmed/21722080


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Explanation: Angiosperms evolved during the late Cretaceous Period, about 125-100 million years

ago.Angiosperms have developed flowers and fruit as ways to attract pollinators and protect their

seeds, respectively.

Reference:https://www.boundless.com/biology/textbooks/boundless-biology-textbook/seed-plants-

26/evolution-of-seed-plants-158/evolution-of-angiosperms-620-11841/

Ans: 1

Explanation: The most common reason for failure of clinical trials of gene therapy trials is due to

inefficient gene transfer that also creates issued for regulation of gene expression and every

succeeding steps.

Reference:https://www.ndsu.edu/pubweb/~mcclean/plsc431/students99/oberoi.htm

Ans: 1

Explanation: Aromatic amino acids gives intrinsic fluoroscence in ultraviolet region (280 nm).

Intrinsic fluoroscence is due to presence of benzene or aromaticc rings in these amino acids.

Reference:

http://dwb.unl.edu/Teacher/NSF/C08/C08Links/pps99.cryst.bbk.ac.uk/projects/gmocz/fluor.htm
https://www.boundless.com/biology/textbooks/boundless-biology-textbook/seed-plants-26/evolution-of-seed-plants-158/evolution-of-angiosperms-620-11841/https://www.boundless.com/biology/textbooks/boundless-biology-textbook/seed-plants-26/evolution-of-seed-plants-158/evolution-of-angiosperms-620-11841/https://www.boundless.com/biology/textbooks/boundless-biology-textbook/seed-plants-26/evolution-of-seed-plants-158/evolution-of-angiosperms-620-11841/https://www.boundless.com/biology/textbooks/boundless-biology-textbook/seed-plants-26/evolution-of-seed-plants-158/evolution-of-angiosperms-620-11841/https://www.ndsu.edu/pubweb/~mcclean/plsc431/students99/oberoi.htmhttps://www.ndsu.edu/pubweb/~mcclean/plsc431/students99/oberoi.htmhttps://www.ndsu.edu/pubweb/~mcclean/plsc431/students99/oberoi.htmhttp://dwb.unl.edu/Teacher/NSF/C08/C08Links/pps99.cryst.bbk.ac.uk/projects/gmocz/fluor.htmhttp://dwb.unl.edu/Teacher/NSF/C08/C08Links/pps99.cryst.bbk.ac.uk/projects/gmocz/fluor.htmhttp://dwb.unl.edu/Teacher/NSF/C08/C08Links/pps99.cryst.bbk.ac.uk/projects/gmocz/fluor.htmhttps://www.ndsu.edu/pubweb/~mcclean/plsc431/students99/oberoi.htmhttps://www.boundless.com/biology/textbooks/boundless-biology-textbook/seed-plants-26/evolution-of-seed-plants-158/evolution-of-angiosperms-620-11841/https://www.boundless.com/biology/textbooks/boundless-biology-textbook/seed-plants-26/evolution-of-seed-plants-158/evolution-of-angiosperms-620-11841/


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Ans: 3

Explanation: Reprobing a Western blot save times and conserves sample while allowing

optimization to be peformed as needed. Reprobing also allows the same blot to be probed for

different target proteins.

Reference:https://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-

learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-

blots.html

Ans: 2

Explanation: The classic cloning vectors are plasmids, phages, and cosmids, which are limited to the

size insert they can accommodate, taking up to 10, 20, and 45 kb, respectively (Table 8.2).

Reference:http://www.blackwellpublishing.com/allison/docs/sample_ch8.pdf
https://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.htmlhttps://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.htmlhttps://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.htmlhttps://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.htmlhttps://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.htmlhttp://www.blackwellpublishing.com/allison/docs/sample_ch8.pdfhttp://www.blackwellpublishing.com/allison/docs/sample_ch8.pdfhttp://www.blackwellpublishing.com/allison/docs/sample_ch8.pdfhttp://www.blackwellpublishing.com/allison/docs/sample_ch8.pdfhttps://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.htmlhttps://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.htmlhttps://www.thermofisher.com/in/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/stripping-reprobing-western-blots.html


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Ans: 4

Explanation: transcriptome databases aid in the identification of genes that are differentially

expressed in distinct cell populations.

Reference:https://en.wikipedia.org/wiki/Transcriptome

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