Booklet Code - C · The one in a white shirt says: “I am a girl” (statement-I) The one in a...

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Booklet Code - C


June 18th 2019Special Edition

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Answer Key For CSIR NET June 2019

Booklet Code - C



3Phone 080-5099-7000 Toll Free 1800-1200-1818

CSIR NET JUNE 2019 Exam Question Paper + Answer Key

BOOKLET CODE C

PART A1 Of four agents Alpha Beta Gamma and Delta three have to be sent

together on a mission If Alpha and Beta cannot go together Beta and Gamma cannot go together and Gamma and Delta cannot go together then which of the following holds

1 Any three agents can be sent2 Alpha Delta and any one out of Beta and Gamma can be sent3 Beta Gamma and any one out of Alpha and Delta can be sent4 The mission is impossible

Answer 4

Explanation According to given information there can be only two possi-bilities Alpha Beta and Gamma or Alpha Beta and Delta But none of the options give any of these conditions

2 An open rectangular box is made by excluding the four identical cor-ners of a piece of paper as shown in the diagram and folding it along the dotted lines

The capacity of the box (in cm3) is

1 80002 10003 40004 6000

Answer 3

Explanation The capacity of the box is the volume of the cylinder length breadth heightlength of the box = 40cm - 10 cm -10 cm = 20 cmbreadth = 40 cm - 10cm - 10cm = 20 cmheight = 10 cmVolume = 202010 cc = 4000 cc

3 Which of the following is the largest 250 340 430 5201 2502 3403 4304 520

Answer 2

Explanation 2^50 3^40 4^30 5^20 can be written as(2^5)^10 = 32^10(3^4)^10 = 81^10(4^3)^10 = 64^10(5^2)^10 = 125^10So the largest number would ideally be = 5^20

4 A monkey climbs a tree to eat fruits The amount of energy gained from eating fruits and the energy spent in climbing on different branches have a relationship shown in the figure

The ratio of energy gained to the energy spent will be the maximum

1 At a point where the slope of the curve is the maximum2 At a point where the slope of the curve is the unity3 At a point on the curve where the tangent passes through the origin4 At the highest point on the curve

Answer 1

Explanation The ratio of energy gained to energy spent will be the maxi-mum at a point where the slope is the maximum because slope = yx = energy gained energy spent Therefore Option 1 is the correct answer

5 The length of the cylinder is measured 10 times yielding 10 distinct values For this set of values consider the following statements

A Five of these values will lie above the mean and five below itB Five of these values will lie above the median and five below itC At least one value will lie above the meanD At least one value will lie at the median

Which of the following statements are necessarily correct

1 B and C2 A and C3 B and D4 A C and D

Answer 1

Explanation Since there are 10 numbers the median value will be theaverage of the 5th and 6th number So B is correct and D is false

6 In the given circle O is the centre ㄥPAO = 400 ㄥPBQ = 300 and



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outer angle ㄥAOB = 2200 Then ㄥAQB = 2200

1 7002 8003 6004 110

Answer 1

Explanation For inscribed angles some theorems are applied1) According to Angle at centre theorem - an angle at the circumference of a

circle is half the angle at the centre subtended by the same arc Hence angle APB = 12 of inner angle AOBOuter angle AOB = 220 degInner angle AOB = 360 deg - 220 deg = 140 degtherefore angle APB = 12 x 140 deg = 70 deg

2) According to Angles subtended by the same arc theorem - angles subtend-ed by the same arc at the circumference are equalHence both the angles angle APB and angle AQB subtended by the same arc

AB should be equalangle APB = angle AQB = 70 deg

Reference httpswwwmathsisfuncomgeometrycircle-theoremshtml

7 A canal system is shown in the figure

Water flows from A to B through two channels Gates G1 and G2 are

operated independently to regulate the flow Probability of G1 to be open is 10 while that of G2 is 20 The probability that water will flow from A to B is

1 102 203 284 30

Answer 3

Explanation Given Probability of G1 open = 10 = 01 Probability of G2 open = 20 = 02

Therefore Probability of G1 closed = 90 = 09 Probability of G2 closed = 80 = 08

Possibilities of water moving from A to B is Only through G1 with G2 closed = 01 x 08 = 008Only through G2 with G1 closed = 02 x 09 = 018Both through G1 amp G2 = 01 x 02 = 002

Hence total probability = 008 + 018 + 002 = 028 = 28

Reference httpwwwnucengcaep714sept2009Chapter20420-20Probability20Tools20and20Techniquespdf

8 A long ream of paper of thickness t is rolled tightly As the roll becomes larger the length of the paper wrapped in one turn exceeds the length in the previous turn by

1 t2 2t3 πt4 2πt

Answer 4

Explanation Let the radius of the present roll be r (as shown in the image) so the total circumference (length of paper covering the entire turn) is 2πr If the paper has a thickness t then now the new radius for another turn in the roll will be (r+t) hence the new circumference will be 2π(r+t) Hence the differ-ence between the two will be 2π(r+t) - 2πr or 2πr + 2πt - 2πr = 2πt

9 Point A on a wheel of radius r touches the horizontal plane at point P It rolls without slipping till point A is at the highest position in the first turn What is the final distance AP

1 2r

2 rradic(1+π2)3 rradic(4+π2)4 2rradic(1+π2)

Answer 3

Explanation Let the opposite point of A be B at the final position (as shown in the image) Since the radius of the wheel is r the circumference of the wheel is 2πrIn order for the point A to appear at the highest position the wheel has to rotate half ie 2πr2 = πr which is the distance travelled from point P to BThe diameter of the wheel will be 2r which is the distance of A and BNow considering the distance A and P is asked this is a right angled triangle Hence to find the hypotenuse we use the formula H^2 = B^2 + Height^2 = (πr)^2 + (2r)^2 = π^2 x r^2 + 4r^2= r^2 ( π^2 + 4)Therefore H = radicr^2 ( π^2 + 4) or rradic(4 + π^2) Ans

10 In a bacterial cell a protein is synthesized at random location in the



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cytoplasm The protein has to reach one pole of the cell for its appropriate function The protein reaches the pole by

1 Chemical attraction2 Random movement3 Enzymatic action4 Attraction between opposite charges

Answer 1

Explanation Bacterial cell invariably inherits 2 poles-one old another new poles at the time of division Before cytokinesis the proteins gets localized at both the poles the other proteins located at the cytosol gets attracted by these protein gradient and distribute themselves accordingly

Reference httpswwwncbinlmnihgovpmcarticlesPMC3874780

11 A precious stone breaks into four pieces having weights in the pro-portion 1234 The value of such a stone is proportional to the square of its weight What is the percent loss in the value incurred due to breaking

1 02 303 704 80

Answer 3

Explanation Let the initial weight of the stone be 10x So initial value will be equal to 100x^2 After breaking the value will be equal to 30x^2 So percentage loss will be 100x^2 - 30x^2 So finally the loss is 70

12 Two runners starting together run on a circular path taking 6 and 8 minutes respectively to complete one round How many minutes later do they meet again for the first time on the start line assuming constant speeds

1 82 243 324 60

Answer 2

Explanation They will meet on the start line at an LCM of the individual time taken = 24First person would have covered 4 rounds while the second would have cov-ered 3 rounds of the circumference

13 The distribution of grades secured by students in a class is given in the table belowWhat is the least possible population of the class

1 22 43 84 10Answer 4

Explanation The least possible population of the class will be 10 So stu-dents getting grade A will be equal to 1 grade B 4 grade C 3 and grade D 2

14 The nine numbers x1x2x3x9 are in ascending order Their average m is strictly greater than all the first eight numbers Which of the follow-ing is true

1 Average (x1x2x9m) gt m and Average (x2x3x9) gt m2 Average (x1x2x9m) lt m and Average (x2x3x9) lt m3 Average (x1x2x9m) = m and Average (x2x3x9) gt m4 Average (x1x2x9m) lt m and Average (x2x3x9) = m

Answer 3

Explanation

15 Which among the following diagrams represents women mothers human beings

Answer 1

Explanation The outermost circle represents human beings then next circle is women and innermost is mother As all are human beings of which some are women and these women some are mothers

Reference httpswwwexamsbookcomlogical-venn-diagram-in-detail

16 A boy and a girl make the following statements of which at most one is correctThe one in a white shirt says ldquoI am a girlrdquo (statement-I)The one in a blue shirt says ldquoI am a boyrdquo (statement-II)Which of the following is the correct inference

1 Statement-I is correct but Statement-II is incorrect2 Statement-II is correct but Statement-I is incorrect3 Both the statements I and II are incorrect4 The correctness of the statements I and II cannot be ascertained

Answer 4

17 How many quadrilaterals does the following figure have


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1 172 18

3 194 20

Answer 3

18 12 balls 3 each of the colours red green blue and yellow are put in a box and mixed If 3 balls are picked at random without replacement the probability that all 3 balls are of the same colour is

1 142 1123 1364 155

Answer 4

Explanation The probability that 1st ball is of 1 colour = 3122nd ball also of same colour without replacement = 2113rd ball of same colour without replacement = 110

Total probability = P (A amp B amp C) 312 x 211 x 110 = 61320

Since there are 3 others colours (each having same no of balls mixed) All of the individual probabilities need to be added up The reason this has to be done is because they have not mentioned any particular colour of ball to be pickedie 61320 + 61320 + 61320 + 61320 = 241320 = 155 Ans

Referencehttpwwwprobabilityformulaorgprobability-without-replacementhtml

19 Some aliens observe that roosters call before sunrise every day Hav-ing no other information about roosters and sunrises which of the follow-ing inferences would NOT be valid

1 Rooster-call and sunrise may be independent cyclic events with the same periodicity2 Both may be triggered by a common cause3 Rooster-call may be causing the sunrise4 Sunrise cannot be the cause of rooster call as the rooster-call precedes

the sunrise

Answer 2

Explanation If it was triggered by a common cause it could have happened at other times also We have no concrete evidence of them being connected or them having a common cause

20 Twenty-one litres of water in a tank is to be divided into three equal parts using only 5 8 and 12 litre capacity cans The minimum number of transfers needed to achieve this is 1 3

2 43 54 7

Answer 4

Explanation In 7 steps 21 L of water can be divided into 3 equal parts ie 7 L using 5L 8L and 12 L cans Details of the steps are provided in the image

Reference httpswwwcareercupcomquestionid=11148437

PART B

21 The first step in glycogen breakdown releases glucose units as

1 glucose-6 phosphate2 glucose-1 phosphate3 glucose4 glucose and glucose-6 phosphate

Answer 2

Explanation Glycogen Phosphorylase catalyzes breakdown of glycogen into Glucose-1-Phosphate (G1P)

Reference pg- 562-564 Ch15-Principles of Metabolic Regulation Glu-cose and Glycogen Principles of Biochemistry Fourth Edition

22 The Na+K+ ATPase pump is found on the plasma membrane of most animal cells A mutation in the intrinsic phosphorylation site of the pump is most likely to affect

1 the outward movement of Na+ only2 inward movement of K+ only3 both the inward and outward movement of K+ and Na+4 has no effect on pump activity but affects its stability

Answer 3

Explanation The Na+K+-ATPase is an example of a P-type ion pump The ldquoPrdquo stands for phosphorylation indicating that during the pumping cycle the hydrolysis of ATP leads to the transfer of the released phosphate group to an aspartic acid residue of the transport protein which in turn causes an essential conformational change within the protein Conformational changes are neces-sary to change the affinity of the protein for the two cations that are transport-ed Consider the activity of the protein It must pick up sodium or potassium ions from a region of low concentration which means that the protein must have a relatively high affinity for the ions Then the protein must release the



8Phone 080-5099-7000 Toll Free 1800-1200-1818

ions on the other side of the membrane into a much greater concentration of each ion To do this the affinity of the protein for that ion must decrease Thus the affinity for each ion on the two sides of the membrane must be different This is achieved by phosphorylation which changes the shape of the protein molecule The change in shape of the protein also serves to expose the ion binding sites to different sides of the membrane so if any change in intrinsic phosphorylation site it will affect both inward and outward movement of Na and K ions

Reference Cell and Molecular Biology by Karp 6th edition pg 164

23 The site of the division plane during cytokinesis of animal cells is determined

1 by position of nucleus2 by the central spindle3 by the pre-prophase band4 randomly

Answer 2

Explanation The Microtubules of the Mitotic Spindle Determine the Plane of Animal Cell Division The part of the spindle that specifies the division plane varies depending on the cell type in some cells it is the astral microtu-bules in others it is the overlapping antiparallel microtubules in the central spindle

Reference httpswwwncbinlmnihgovbooksNBK26831

24 To prepare individual tissue cells from a primary culture the cell-cell and cell matrix interaction must be broken To achieve this one would NOT use

1 EDTA2 Trypsin3 Collagenase4 Separase

Answer 4

Explanation Apart from separase other enzymes contribute to breaking of cell-cell and cell - matrix interactions supported by following facts Proteases eg trypsin collagenase elastase and more recently dispase (a metallo- neu-tral-protease with a mild action) and enzymes which hydrolyse the extra-cel-lular matrix eg hyaluronidase and lysosyme are commonly used to separate cells Chelating agents (EDTA or citrate) are often included in order to remove the Ca 2+ and Mg 2+ ions that are essential for matrix stability and cell-matrix interactions but will not liberate cells alone EDTA act as a metal chelator which is added to trypsin solutions to enhance activityEDTA is added to re-move the calcium and magnesium from the cell surface which allows trypsin to hydrolyze specific peptide bonds The principle reason of using theEDTA along with trypsin is to remove cell to cell adhesion

Reference httpwwwicmsqmulacukflowcytometryusesmulticolouranal-ysiscellpreparationindexhtml

25 Which one of the following statements is NOT true about nucleoso-mal organization of core particle

1 The typical structure of DNA is altered in the middle of the core par-ticle2 In core particle DNA is organized as flat super helix with 165 turns

around the histone octamer3 While forming 30 nm fibres generally 6 nucleosomes per turn organ-

ize into a two-start helix4 The N-terminal histone tails in a core particle are strictly ordered and

exit from the nucleosomes between turns of the DNAAnswer 1

Explanation statement 2 and statement 3 are correct in terms of nucleosome structure statement 4 can be supported by following factThe complexity of the NCP surface is furthered by the histone N-terminal tails that protrude from the nucleosome surface either outside (H4 and H2A) or between (H3 and H2B) the DNA gyres making statement 1 incorrect (ie the right option)


26 During replication RNaseH removes all of the RNA primer except the ribonucleotide directly linked to the DNA end This is because

1 It can degrade RNA and DNA from their 5rsquo end2 It can only cleave bonds between two ribonucleotides3 It can degrade RNA and DNA from their 3rsquo end4 Activity of RNaseH is inhibited by the presence of duplex containing

both strands as DNA

Answer 2

Explanation RNase H removes all of the RNA primer except the ribonucle-otide directly linked to the DNA end This is because RNaseH can only cleave bonds between two ribonucleotides The final ribonucleotide is removed by a 5 exonuclease that degrades RNA or DNA from their 5 ends

Reference Molecular Biology of gene by Watson 7th edition Pg 271

27 Which one of the statements on protein conformation detailed below is INCORRECT

1 L-amino acids can occur in Type 1rsquo β-turns where Φ ψ are both pos-itive2 A peptide rich in proline is unlikely to adopt α-helical structure3 Proline residues have high propensity to occur in β-turns4 The dihedral angles Φ ψ of amino acids in unfolded proteins are ex-

clusively positive

Answer 4

Explanation Statement 1 is correct because Type I beta turns usually have 0 +30 +60 and +90 dihedral angles and whichever Ramachandran plots we consider are usually for L-Type amino acids only Statement 2 is correct be-cause prolines are helix breakers as these do not support H-bonding It also causes a bend in the structure Statement 3 is correct because because peptide bonds involving the imino nitrogen of proline readily assume the cis configu-ration a form that is particularly amenable to a tight turn So proline residues readily occur in beta turns Statement 4 should be wrong because we cannot randomly predict if the dihedral angle has to be positive or negative as the structure is unfolded

Reference httpswwwncbinlmnihgovpmcarticlesPMC2920885 httpwwwbiologyarizonaedubiochemistryproblem_setsaaProlinehtml

28 Choose the INCORRECT statement from the following statements made for an enzyme-catalyzed reaction

1 The kinetic properties of allosteric enzymes do not diverge from Michaelis-Menten behaviour 2 In feedback inhibition the product of a pathway inhibits an enzyme

of the pathway3 An antibody that binds tightly to the analog of the transition state in-

termediate of the reaction SP would promote formation of P when the analog is added to the reaction4 An enzyme with Kcat = 14 x 104 s-1 and Km = 9 x 10-5 M has activity

close to the diffusion controlled limit

Answer 1Explanation Statement 1 is incorrect because Allosteric enzymes show re-



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lationships between V0 and [S] that differ from Michaelis-Menten kinetics They do exhibit saturation with the substrate when [S] is sufficiently high but for some allosteric enzymes plots of V0 versus [S] produce a sigmoid saturation curve rather than the hyperbolic curve typical of non-regulatory enzymes On the sigmoid saturation curve we can find a value of [S] at which V0 is half-maximal but we cannot refer to it with the designation Km because the enzyme does not follow the hyperbolic MichaelisMenten relationship In-stead the symbol [S]05 or K05 is often used to represent the substrate concen-tration giving half-maximal velocity of the reaction catalyzed by an allosteric enzyme Statement 2 is correct because in feedback inhibition products inhibit either the first enzyme of a staight chain pathway or the first enzyme of the branch reaction Statement 3 is also correct because in abzymes If an antibody is developed to bind to a molecule thats structurally and electronically similar to the transition state of a given chemical reaction the developed antibody will bind to and stabilize the transition state just like a natural enzyme low-ering the activation energy of the reaction and thus catalyzing the reaction By raising an antibody to bind to a stable transition-state analog a new and unique type of enzyme is produced Statement 4 is correct because diffusion controlled limits range between 10^8 to 10^9 Ms and calculationg kcatKm values also give us value in this range

Reference httpswwwbarnardhealthusglucose-phosphatethe-ki-netic-properties-of-allosteric-enzymes-diverge-from-michaelisment-en-behaviorhtml httpwwwbiologydiscussioncomcell-biologygroup-ii-control-mechanisms-of-allosterism-with-diagram3459

29 On sequence analysis of a double stranded DNA the results showed the content of cytosine C was 20 What is the amount of A and T put together

1 202 303 504 60

Answer 4

Explanation Since C is 20 G will also be 20 Remaining 60 forms A+T This also justifies Chargaffs first parity rule

Reference httpfigcoxmiamiedu~cmallery150genechargaffhtm

30 Eukaryotic mRNAs are modified to possess a 5rsquo cap structure Which one of the following is an INCORRECT statement about the function of 5rsquo cap structure

1 It protects the mRNA from 5rsquo3rsquo exoribonuclease attack2 It facilitates splicing of the nascent transcripts3 It protects the transcript from degradation by RNAse III family enzymes4 It facilitates attachment to 40s subunit of ribosome

Answer 3

Explanation The cap structure is characteristic of all RNA polymerase transcripts and consists of an inverted 7-methyl guanosine linked via a 5lsquo-5lsquo triphosphate bridge to the first transcribed residue m7G(5rsquo)ppp(5rsquo)X The cap structure has been implicated in many aspects of RNA metabolism It provides resistance to 5rsquo-3lsquo exonucleases and contributes to a variety of cellular pro-cesses including pre-mRNA splicing polyadenylation RNA nuclear export and mRNA translation Capping occurs co-transcriptionally and is one of the earliest RNA modifications Decapping enzymes such as DCP2 with ad-ditional cofactors hydrolyze the 5rsquo cap exposing the mRNA to decay that is carried out by XRN1 a processive exoribonuclease that completely hydrolyz-es decapped (5rsquo monophosphorylated) RNA in the 5rsquorarr3rsquo direction Bacterial RNase III plays important roles in the processing and degradation of RNA transcripts


httpswwwncbinlmnihgovpmcarticlesPMC3742305 httpsfebsonlinelibrarywileycomdoipdf101111j1432-1033199700461x

31 Which one of the following does NOT belong to human antimicrobial proteins and peptides at epithelial surfaces forming part of innate immu-nity

1 Lactoferrin2 Defensin3 Calprotectin4 Vimentin

Answer 4

Explanation Lactoferrin is found primarily in mucosal secretions synthe-sized by epithelial cells Lactoferrin is considered a first-line defense protein involved in protection against a multitude of microbial infectionsβ-defensins are secreted by most leukocytes and epithelial cellsβ-Defensins are produced by the respiratory epithelium and the alveolar macrophage and secreted into the airway surface fluid Calprotectin is also an epithelial cell-derived antimi-crobial peptide Vimentin Vimentin is a type III intermediate filament protein within the cytosol but is also expressed on the host cell surface table given in images for reference

Reference httpswwwncbinlmnihgovpmcarticlesPMC2915836 httpswwwncbinlmnihgovpmcarticlesPMC59560 httpswwwncbinlmnihgovpmcarticlesPMC3816379 httpsmbioasmorgcontent91e00160-18 httpswwwncbinlmnihgovpmcarticlesPMC5040971 -vimentin

32 Which one of the following best describes death-upon-detachment

1 Necroptosis2 Anoikis3 Extravasation4 Metastasis

Answer 2

Explanation Anoikis is a form of programmed cell death that occurs in an-chorage-dependent cells when they detach from the surrounding extracellular matrix (ECM)

Reference httpslinkspringercomchap-ter101007978-1-59745-221-2_6

33 Fruit bats are known to harbour and spread several viruses that can infect other animals and humans Which one of the following viruses is NOT reported to spread by fruit bats

1 Ebola2 Nipah3 SARS4 HIV

Answer 4

Explanation In recent years bats have gained notoriety after being impli-cated in numerous (Emerging infectious diseases)EIDs Bat-borne viruses that can affect humans and have caused EIDs in humans fall into different families paramyxoviruses including Hendra viruses and Nipah viruses Ebola hemor-rhagic fever filoviruses Marburg hemorrhagic fever filoviruses and sudden acute respiratory syndrome-like coronaviruses (SARS-CoV)

Reference httpswwwintechopencombooksbatsbats-bat-borne-viruses-and-environmental-changes34 In a type 1 hypersensitivity-mediated asthamatic response which one



10Phone 080-5099-7000 Toll Free 1800-1200-1818

of the following is thought to contribute significantly to the prolonged bronchospasm and build-up of mucous seen in asthmatics

1 Thromboxane2 Leukotriene3 TGFβ4 Chondroitin

Answer 2

Explanation Leukotrienes play a key role in asthma in three ways caus-ing inflammation bronchoconstriction and mucus production The cysteinyl leukotrienes (LTC 4 LTD 4 and LTE 4 ) have been shown to be the most potent bronchoconstrictor in humans and are believed to play a crucial role in asthmatic airway obstruction Leukotrienes may attract white blood cells to the lungs increasing swelling of the lung lining Leukotrienes also increase mucus production and make it easier for fluids to accumulate (an important part of inflammation)

Reference httpwwwasthmaandallergycentercomarticleleukot-

rienes-role-in-asthma httpswwwncbinlmnihgovpmcarticlesPMC59560

35 In a human cell line a large fraction of double-strand DNA breaks are repaired by non-homologous end joining (NHEJ) An inhibitor of FLAP endonuclease will affect

1 Recruitment of DNA-dependant kinase2 Gap trimming3 DNA unwinding 4 Pair of micro-homology regions

Answer 2

Explanation Inhibitor of Flap endonuclease will affect gap trimming as FEN is responsible for removing the flap It is a structure-specific nuclease which possesses double -strand specific 5 to 3 exonuclease activity Mechanism of DSB microhomology-mediated alternative NHEJ involves recognition of microhomology by MRN complex and PARP1 followed by recruitment of FEN1 (Flap endonuclease) which can remove the flap Further recruitment of XRCC1ndashLIGASE III at the site helps in ligating the DNA ends leading to an intact DNA Further image is given for reference in image sheet

Reference httpswwwnaturecomarticlescddis201558 httpswwwncbinlmnihgovpmcarticlesPMC1190240

36 Sugar puckering in double stranded nucleic acids is exclusively

1 C-2rsquo endo in double stranded DNA2 C-3rsquo endo in double stranded DNA3 C-2rsquo endo in double stranded RNA4 C-3rsquo endo in hybrid duplex with one strand as DNA and the other as RNA

Answer 1

Explanation The sugar puckers in DNARNA structures are predominately in either C3-endo (A-DNA or RNA) or C2-endo (B-DNA) corresponding to A or B form conformation As the options given option 1 is the most appropri-ate among all as B-form of double stranded DNA is most accepted one

Reference jenalibleibniz-flideIMAGE_DNA_MODELShtml

37 Homeobox transcription factors (Hox proteins) play important roles in specifying whether a particular mesenchymal cell will become stylopod zeugopod or autopod Based on the expression patterns of these genes a model was proposed wherein these Hox genes specify the identity of a

limb region What would be the observed phenotype for homozygous for a HOXD13 mutation

1 No zeugopod formation2 Abnormalities of the hands and feet wherein the digits fuse3 Deformities in stylopods4 No femur or patella formation

Answer 2

Explanation Hox genes specify the identity of the limb region (as per ques-tion) so whenever there is a mutation in HOXD13 some or the other abnor-malities will be observed in the hands amp feet wherein the digits fuse Disease caused is called as Synpolydactyly 1 (SPD1) Limb malformation that shows a characteristic manifestation in both hands and feet

Reference httpswwwuniprotorguniprotP35453

38 Which one of the following described the function of silicon in plants

1 Constituent of amino acids2 Contributes to cell wall rigidity and elasticity3 Constituent of the photosynthesis reaction centre4 Maintenance of cell turgor and electro-neutrality

Answer 2

Explanation Silicon is deposited primarily in the endoplasmic reticulum cell wall and intercellular species as hydrated amorphous silica Silicon con-tributes to cell wall rigidity and strengthening besides increasing cell wall elasticity during extension of growthIn the primary cell walls silicon inter-acts with cell wall constituents such as pectins and polyphenols and these cross links increase cell wall elasticity during extension growth

Reference Plant Physiology by Gupta Pg 139 httpsbooksgooglecoinbooksid=iBBbPWBQdyYCamppg=PA139amplpg=PA139ampdq=-silicon+contributes+to+cell+wall+rigidity+and+elasticityampsource=-blampots=Vli4q_mboTampsig=ACfU3U1vs9evkMroiEHzyogaDilJeXX-e3wamphl=enampsa=Xampved=2ahUKEwjlxa_Cu-3iAhVGbn0KHdW_CK0Q6AEwA3oECAkQAQv=onepageampq=silicon20contributes20to20cell20wall20rigidity20and20elasticityampf=false

39 Most of the plant disease resistance (R) gene products contain

1 G-Box domains2 Transcription repression domains3 Leucine-rich repeats4 Enzymatic activities

Answer 3

Explanation Plant resistance gene analogs (RGAs) as resistance R gene candidates have conserved domains and motifs that play specific roles in path-ogens resistance Well-known RGAs are nucleotide binding site leucine rich repeats receptor like kinases and receptor like proteinsRGAs can be grouped as either nucleotide binding site leucine rich repeat (NBS-LRR) or transmem-brane leucine rich repeat (TM-LRR)


40 Out of several gibberellins identified in plants which one of the fol-lowing is NOT bioactive

1 GA12 GA33 GA44 GA5Answer 4


11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818

Explanation The bioactive GAs are GA1 GA3 GA4 and GA7

Reference httpswwwsciencedirectcomsciencearticlepii

B9780128115626000049

41 Nitrogenase a complex metal containing enzyme is involved in con-version of N2 to NH3 Which one of the following metals is NOT involved in the activity of nitrogenase

1 Molybdenum (Mo)2 Iron (Fe)3 Vanadium (V)4 Cobalt (Co)

Answer 4

Explanation All nitrogenases are two-component systems made up of Component I (also known as dinitrogenase) and Component II (also known as dinitrogenase reductase) Component I is a MoFe protein in molybdenum ni-trogenase a VFe protein in vanadium nitrogenase and a Fe protein in iron-on-ly nitrogenase Component II is a Fe protein that contains the Fe-S cluster which transfers electrons to Component I Component I contains 2 key metal clusters the P-cluster and the FeMo-cofactor (FeMo-co) Mo is replaced by V or Fe in vanadium nitrogenase and iron-only nitrogenase respectively During catalysis electrons flow from a pair of ATP molecules within Component II to the Fe-S cluster to the P-cluster and finally to the FeMo-co where reduction of N2 to NH3 takes place


42 Which one of the following agents cause relaxation of mesangial cells

1 Histamine2 Thrombaxane A23 Norepinephrine 4 Dopamine

Answer 4

Explanation Mesangial cell relaxation can be mediated by enhanced cAMP or cGMP generation Dopamine is a catecholamine known to increase c-AMP levels in mesangial cellsDopamine stimulates renin release from gran-ular cells via D1 receptors and cAMP generation

Reference httpswwwfasebjorgdoiabs101096fasebj143308611 httpswwwresearchgatenetpublication15199280_Effect_of_dopa-mine_on_ecto-5-nucleotidase_expression_in_human_glomerular_me-sangial_cells httpswwwsciencedirectcomtopicsveterinary-science-and-veteri-

nary-medicineafferent-arterioles

43 A patient comes to the hospital complaining of vomiting and diar-rhoea The doctor suggested that the patient take glucose and electrolyte solution orally Which one of the following membrane proteins is likely to be involved in rehydrating the patient

1 Cystic fibrosis transmembrane regulator (CFTR)2 Sodium glucose transporter protein 1 (SGLT1)3 Insulin receptor protein (IRP)4 Sucrase-isomaltase protein (SIP)

Answer 2

Explanation The SLC5A1 gene located at 22q123 encodes sodium-glu-cose cotransporter 1 (SGLT1) Intestinal SGLT1 actively transports glucose and water into enterocytes against their concentration gradients using energy

generated by sodium and electrical gradients Loss-of-function mutations of the SLC5A1 gene prevent glucose and galactose from being absorbed and re-sult in a build-up in the intestinal tract of glucose galactose and water leading to severe diarrhea and dehydration Oral rehydration takes advantage of glu-cose-coupled sodium transporta process for sodium absorption which remains relatively intact in infective diarrheas due to viruses or to enteropathogenic bacteria whether invasive or enterotoxigenic Glucose enhances sodium and secondarily water transport across the mucosa of the upper intestine

Reference httpswwwsciencedirectcomtopicsneurosciencesodium-gl-

cose-cotransporter-1httpsrehydrateorgorsorthtm

44 In certain plants the mechanism where timing of another dehiscence and stigma receptivity do not coincide to avoid self-pollination is called

1 dichogamy2 herkogamy3 monoecy4 dioecy

Answer 1

Explanation Dichogamy is generally thought to be a mechanism that pre-vents self-fertilization in flowering plants If we go through other options Herkogamy is again used to reduce sexual interference between male and fe-male function but it supply a spatial separation of the anthers amp stigma Mono-ceious amp Dioceious are the two different plant groups that includes single plant that bears both male and female flowers and distinct male amp females plants respectively


45 In Xenopus embryos β-catenin plays an important role in the DorsalVentral axis development What would you expect if the endogenous gly-cogen synthase kinase 3 (GSK3) is knocked out by a dominant-negative form of GSK3 in the ventral cells of the early embryo

1 Blocking of GSK3 on the ventral side has no effect A normal embryo will form2 The resulting embryo will only have ventral sides3 A second axis will form4 The dorsal fate is suppressed

Answer 3

Explanation β-catenin is necessary for forming the dorsal axis since ex-perimental depletion of β-catenin transcripts with antisense oligonucleotides results in the lack of dorsal structures Moreover the injection of exogenous β-catenin into the ventral side of the embryo produces a secondary axis β-cat-enin is part of the Wnt signal transduction pathway and is negatively regulated by the glycogen synthase kinase 3 GSK-3 also plays a critical role in axis formation by suppressing dorsal fates Activated GSK-3 blocks axis formation when added to the egg If endogenous GSK-3 is knocked out by a dominant negative protein in the ventral cells of the early embryo a second axis forms

Reference Scott F Gilbert Early Amphibian Development in Devel-opmental Biology 6th ed (Sunderland Sinauer Associates 2000) httpwwwncbinlmnihgovbooksNBK10113 Chapter 10 Page No-280

46 Assuming that the A B C and D genes are not linked the probability of a progeny being AaBBccDd from a cross between AaBBccDd and aaB-BccDD parents will be

1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3

Explanation GENOTYPE OF PARENTS ndash AABbccDd X aaBBccDD

Gametes from parent 1 ndash A parent 2 ndash a B or b B c c D or d D

Genotype for individual gene pairs in progeny ndash Aa BB Bb cc DDDd Probability of (AaBBccDd) Genotype of progeny - 1 x 12 x 1 x 12 = 14

(As genes are not linked individual probability as multiplied by product rule of probability for independent events)


47 The new born baby of a mother having blood group AB Rh+ and father having blood group O Rh- got mixed with other babies in the hos-pital The baby with which of the following blood groups is expected to be of the said couple

1 O Rh+2 O Rh-3 AB Rh-4 B Rh+

Answer 4

Explanation Genes from the mother = IA and IB Genes from the father = i and i Genotype of children = IAi IBi = A blood group and B blood group Rhe-sus antigen is represented by D Mother can have genotype DD or Dd Father has genotype dd So children can have either Dd or dd (the chance of Dd is higher = Rhesus positive) So the most probable child is B+ve

Reference httpswwwncbinlmnihgovbooksNBK2267httpswwwncbinlmnihgovpmcarticlesPMC2535884

48 What is required for Vitamin B12 absorption for small intestine

1 Cobalophilin2 Hephaestin3 Hepcidin4 Na+ -cotransporter

Answer 1

Explanation Cobalophilin is a Vitamin B12 binding protein that is secreted in saliva that binds the vitamin in the stomach when it is released from dietary proteins by the action of gratic acid and pepsin Going through the other op-tions Hephaestin is involved in the metabolism amp homeostasis maintenance of Fe amp possibly copper Hepcidin is a key regulator of the entry of Iron into the circulation in mammals amp is a central regulator of Fe homeostasis

Reference httpswwwncbinlmnihgovpmcarticlesPMC2855274 httpswwwncbinlmnihgovpubmed1056622 httpswwwsciencedirectcomtopicsneurosciencehephaestin

49 Which one is the correct sequence of events that takes place during phototransduction when light falls onto the retina

1 Closure of Na+ channels activation of transducin decreased re-lease of glutamate decrease in intracellular cGMP structural chang-es in rhodopsin

2 decreased release of glutamate structural changes in rhodopsin activation of transducin decrease in intracellular cGMP Closure of Na+ channels 3 structural changes in rhodopsin activation of transducin decrease

in intracellular cGMP Closure of Na+ channels decreased release of glutamate4 decrease in intracellular cGMP activation of transducin decreased

release of glutamate structural changes in rhodopsin Closure of Na+ channels

Answer 3

Explanation When the retinal moiety in the rhodopsin molecule absorbs a photon its configuration changes from the 11-cis isomer to all-trans retinal this change then triggers a series of alterations in the protein component of the molecule The changes lead in turn to the activation of an intracellular mes-senger called transducin which activates a phosphodiesterase that hydrolyzes cGMP All of these events take place within the disk membrane The hydrol-ysis by phosphodiesterase at the disk membrane lowers the concentration of cGMP throughout the outer segment and thus reduces the number of cGMP molecules that are available for binding to the channels in the surface of the outer segment membrane leading to channel closure


50 A Lod score of 3 represents a Recombination Frequency (RF) that is

1 3 times as likely as the hypothesis of no linkage2 30 times as likely as the hypothesis of no linkage3 100 times as likely as the hypothesis of no linkage4 1000 times as likely as the hypothesis of no linkage

Answer 4

Explanation LOD score is actually an acronym for log of the odds LOD LOD score actually refers to a numerical result when estimating whether two genes or a gene and a disease are linked to one another LOD scores are most often used to describe the data one gets out of family studies where you are looking at large families and an inheritance of traits or diseases within the families In a LOD score of 3 3 is translated roughly into about 1000-to-one odds that this gene really is linked to this disease as opposed to the alternative hypothesis which is unlinked

Reference httpswwwgenomegovgenetics-glossaryLOD-Score

51 Which one of the following is a fungal disease of plants

1 Cucumber mosaic 2 Fire blight of pear3 Crown gall 4 Apple scab

Answer 4

Explanation Apple scab or black spot is caused by the fungus Venturia in-aequalis It infects leaves shoots buds blossoms and fruit It occurs almost everywhere apples are grown and is the most serious and widespread disease of this crop especially important in regions with high rainfall and relative hu-midity during the growing season

Reference httpagriculturevicgovauagriculturepests-diseases-and-weeds

plant-diseasesfruit-and-nutspome-fruit-diseasesapple-scab httpwwwagrgccaresourcesproddocscipubpdfcarisse_scab_epdf

52 Which one of the following influenza A virus subtypes caused severe avian flu and was responsible for disease outbreak in the year 1997 in Hong Kong



14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4

Explanation Bird flu also called avian influenza a viral respiratory disease mainly of poultry and certain other bird species including migratory water-birds some imported pet birds and ostriches that can be transmitted directly to humans The first known cases in humans were reported in 1997 when an outbreak of avian influenza A virus subtype H5N1 in poultry in Hong Kong led to severe illness in 18 people one-third of whom died

Reference httpswwwncbinlmnihgovpmcarticlesPMC88966 httpswwwbritannicacomsciencebird-flu

53 Which one of the following DNA markers can be used to distinguish between a homozygote and a heterozygote

1 RAPD2 AFLP3 RFLP4 ISSR

Answer 3

Explanation Codominant markers are markers for which both alleles are expressed when co-occurring in an individual Therefore with codominant markers heterozygotes can be distinguished from homozygotes allowing the determination of genotypes and allele frequencies at loci

Reference httpwwwscielobrscielophpscript=sci_arttetamppid

=S1415-47572004000400019

54 Which one of the following statements is correct

1 ectomycorrhizal associations predominantly reduce phosphorus limi-tation and endomycorrhizal associations reduce both nitrogen and phos-phorus limitation2 endomycorrhizal associations predominantly reduce phosphorus lim-

itation and ectomycorrhizal associations reduce both nitrogen and phos-phorus limitation3 Ecto-and endo-mycorrhizal associations do not reduce nitrogen and

phosphorus limitation4 Ecto-and endo-mycorrhizal associations are able to reduce only phos-

phorus limitation

Answer 2

Explanation P deficiency shifts the priority of energy expenditure from growth development and reproduction to P acquisition a situation that ad-versely affects the plant productivity To overcome or minimize the effect of P shortage plants associate with AMF as obligate biotrophs but also oth-er beneficial root-colonizing fungi such as Piriformospora indicaArbuscu-lar mycorrihza is known as Endomycorrhiza Hence according to the above explanation statement 3 amp 1 is not correct In statement 2-ectomycorrhizal associations reduce phosphorus limitation-its true as they have the ability to move deep inside the cortical cells of rootOption 4 is also incorrect as ecto is involved in reduction of nitrogen limitation So here option 2 is correct

Reference httpswwwncbinlmnihgovpmcarticlesPMC4608361 httpsjournalsplosorgplosonearticleid=101371journalpone0090841

55 Which one of the following statements is correct for the process of

speciation

1 Allopatric speciation occurs between adjacent populations2 Parapatric speciation may occur between geographically separated

populations3 Sympatric speciation occurs within one continuously distributed pop-

ulation4 Sympatric speciation occurs when continuously distributed popula-

tions are fragmented

Answer 3

Explanation Sympatric speciation does not require large-scale geographic distance to reduce gene flow between parts of a population Merely exploiting a new niche may automatically reduce gene flow with individuals exploiting the other niche This may occasionally happen when for example herbivo-rous insects try out a new host plantSympatric speciation is the evolution-ary process whereby species are formed from a single ancestral species while inhabiting the same geographic area In contrast to allopatric speciation the distribution ranges of species which evolve through sympatry may be identical or they may only overlap Rather than geographic distance prompting a reduc-tion of gene flow between populations sympatry occurs when members of one population make use of a new niche Hence this population is fragmented due to the choices of different niche

Reference httpsevolutionberkeleyeduevolibraryarticle0_0_0speciation-

modes_02 httpsbiologydictionarynetspeciation

56 Which one of the following does NOT contribute to microevolution-ary change

1 Mutation2 Random mating3 Genetic drift4 Natural Selection

Answer 2

Explanation Microevolution is simply a change in gene frequency within a population Evolution at this scale can be observed over short periods of time mdash for example between one generation and the next the frequency of a gene for pesticide resistance in a population of crop pests increases Such a change might come about because natural selection favored the gene because the population received new immigrants carrying the gene because some non-resistant genes mutated to the resistant version or because of random genetic drift from one generation to the nextMutation migration genetic drift and natural selection are all processes that can directly affect gene frequencies in a population

Reference httpsevolutionberkeleyeduevolibraryarticleevo_39

httplegacyhopkinsvillekctcseduinstructorsJason-ArnoldVLIModule3EvolutionModule3Evolution4html

57 According to Hamiltonrsquos rule altruistic behaviour can evolve when rb ＞ c where b is the extra benefit gained by the recipient as a result of the altruistic act c is the cost to the actor arising from performing the altruistic act and r is the relatedness between the

1 individual performing the altruistic act and the offspring of the recip-ient2 individual performing the altruistic act and the recipient3 recipient and the offspring of the individual performing the altruistic

act4 individual performing the altruistic act and the members of its popu-

lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818

Explanation The probability that the altruist and the recipient share a gene is called the coefficient of relatedness ( r ) The rule states that a social behav-iour will be favoured by natural selection if and only if rb gt c where lsquobrsquo is the effect of the behaviour on the reproductive success of others lsquocrsquo is the effect on onersquos own reproductive success and lsquorrsquo is the coefficient of relatedness which measures the statistical association between the genes of the actor who performs the behaviour and the genes of the recipient affected by it

Reference httpwwwflyfishingdevoncouksalmonyear2social_behavioursocial_

behaviourhtmhomilton_kin_selection_theory httpphilosophyofbrainscom201710312-relatedness-altruismaspx

58 Analyses of the nucleotide sequences of ribosomal RNA (rRNA) are particularly useful for evolutionary studies of living organisms because of the following reasons EXCEPT

1 rRNA is evolutionarily ancient2 No free-living organism lacks rRNA3 rRNA since critical for translation can undergo lateral transfer

amongst distant species4 rRNA has evolved slowly over geological time

Answer 3

Explanation rRNA gene sequences have become the gold standard for mi-crobial identification and the inference of deep evolutionary relationships rRNA genes occur in all cells and organelles and the rRNA genes are the most conservative large sequences in nature For instance the eukaryotic and bac-terial SSU rRNA genes typically have about 50 identity over the alignable lengths of their SSU rRNAs The rRNA genes have not undergone significant lateral transfer and the structural properties of the rRNA provide for optimi-zation of alignments Moreover at this time the SSU rRNA databases are the main source of information on environmental microbial diversity

Reference httpswwwncbinlmnihgovpmcarticlesPMC2786576httpsmicrobenetsimple-guidesfact-sheet-rrna-in-evolutionary-stud-

ies-and-environmental-sampling

59 Which one of the following is the most appropriate definition of lsquoGene Pyramidingrsquo in plants

1 Introducing different genes for resistance to a specific pest in different genotypes2 Introducing a single gene for resistance to a particular pest in different

genotypes3 Introducing different genes for resistance to a specific pest in a single

genotype4 Introducing a single gene for resistance to multiple pests in different

genotypes

Answer 3

Explanation Gene pyramiding is defined as a method aimed at assembling multiple desirable genes from multiple parents from a single genotype Ac-cording to this definition Option 3rd is the right answer


60 Depicted below is a phylogenetic tree of selected taxa

Based on the above which one of the following statements is correct

1 Group X is monophyletic and Group Y is polyphyletic2 Group X is paraphyletic and Group Y is monophyletic3 Both Group X and Y are monophyletic

4 Group X is monophyletic and Group Y is paraphyletic

Answer 4

Explanation A paraphyletic group includes the common ancestor but is missing some of the descendants a classic example being my own kingdom of choice Protista Reptiles are often used to demonstrate that as well since they gave rise to the birds and mammals that are not classified within the reptiles A group that includes all the descendants of a given ancestor along with their last common ancestor is called a clade A clade is sometimes referred to as a monophyletic group

Reference httpswwwdigitalatlasofancientlifeorglearnsystematics

phylogeneticstrees-classificationhttpsschaechterasmblogorgschaechter201010of-terms-in-biolo-

gy-monophyletic-paraphyletichtml Biology by Campbell (page num-ber-542)

61 Which of the following describes the identification features of non-poisonous snakes

1 Cylindrical tail and small belly scales2 Cylindrical tail broad transverse belly scales and 4th infralabial scale

is the largest3 Flat tail broad transverse scales and 3rd supralabial scale touches eye

and nose4 Cylindrical tail broad transverse belly scales and a loreal pit between

eye and nostril

Answer 1

Explanation Non-poisonous snakes have cylindrical tail and small belly scales

Reference Chordate Zoology by PS Verma Pg-411

62 A road is constructed through a wet tropical forest following which the population of a species of forest butterfly declines Which of the fol-lowing is NOT a possible explanation for the road causing a decline in the forest butterfly population

1 Road facilitates immigration of gap-loving species which compete with the forest species2 Road facilitates increased movement of individuals of the forest but-

terfly within the forest which reduces genetic diversity3 Road internally fragments the habitat and negatively affects impor-

tant micro-habitat conditions for the forest butterfly4 Road facilitates invasion by non-native plants that displace native host

and nectar plants of the forest butterfly

Answer 1

Explanation Roads are a major contributor to habitat fragmentation be-cause they divide large landscapes into smaller patches and convert interior habitat into edge habitatHabitat fragmentation is the process by which habitat loss results in the division of large continuous habitats into smaller more isolated habitat fragments Habitat fragmentation is one of the most important



16Phone 080-5099-7000 Toll Free 1800-1200-1818

processes contributing to population decline biodiversity loss and alteration of community structure and ecosystem

Reference httpwwwsafepassagecoalitionorgresourcesHabitat20Fragmentationpdf httpspublicationscsiroaurprdownloadpid=c-siroEP101968ampdsid=DS1

63 Which one of the following statements regarding normal distribution is NOT correct

1 It is symmetric around the mean2 It is symmetric around the median3 It is symmetric around the variance4 It is symmetric around the mode

Answer 3

Explanation Despite the different shapes all forms of the normal distribu-tion have the following characteristic properties Theyrsquore all symmetric The normal distribution cannot model skewed distributions The mean median and mode are all equal ie represented by the same number Half of the pop-ulation is less than the mean and half is greater than the mean The Empirical Rule allows you to determine the proportion of values that fall within certain distances from the mean

Reference httpsstatisticsbyjimcombasicsnormal-distribution

64 Tropical regions may have more species diversity because of the fol-lowing possible reasons EXCEPT

1 tropical regions have had more time to diversify under relatively sta-ble climatic conditions than temperate regions2 tropical regions have high spatial heterogeneity3 greater biological competition in the tropics leads to narrower niches 4 lower predation intensity in the tropics allows survival of more prey

species

Answer 4

Explanation The competition hypothesis is based on the idea that compe-tition is the most important factor of evolution in the tropics whereas natural selection at higher latitudes is controlled mainly by physical factors such as drought and cold (Dobzhansky 1950) Such catastrophic mortality factors are said to be rare in the tropics and thus competition for resources becomes keen-er and niches become smaller resulting in a greater opportunity for new spe-cies to evolve The predation hypothesis contradicts the competition hypoth-esis It claims that there are more predators andor parasites in the tropics and that these hold down individual prey populations enough to lower the level of competi tion between and among them The lowered level of competition then allows the addition and co-existence of new intermediate prey type which in turn support new predators in the system However the predation hypothesis does not explain why there are more predators andor parasites in the tropics to begin with

Reference httpswwwecologycenterustropical-foreststheories-to-explain-high-diver-sity-in-the-tropicshtml

65 A multimeric protein when run on an SDS gel showed 2 bands at 20 kDA and 40 kDA However when the protein was run on a native gel it showed a single band at 120 kDA The native form of the protein would be

1 homotrimer2 heterotetramer3 heterodimer4 heterotrimerAnswer 2

Explanation In SDS PAGE SDS (denaturing agent) cleaves the 120kd mul-timeric protein into four subunits 2 X 20 kd + 2 X 40kd However Native PAGE doesnt separate the subunits of proteins so it gives 120KD as total weight Here the four subunits have different weights So the native form of protein would be Heterotetramer

Reference httpwwwJBCorgcontent275117749full

66 A solution NADH and NAD+ both at 01mM concentration If NADH has a molar extinction coefficient of 6220 and that of NAD+ is negligible the optical density measured in a cuvette of 5 mm path length will be

1 0622 00623 0314 0031

Answer 4

Explanation Formula of OD= molar coefficientConcentrationLength The OD of NAD and NADH will be added then the answer is 0031

Reference httpswwwncbinlmnihgovpmcarticlesPMC2143013pdf8563639

pdf

67 Orientation of a closed DNA fragment (gene) in a plasmid vector can be checked by

1 PCR using two gene-specific primers2 Restriction digestion with an enzyme that has a single restriction site

within the cloned gene and none in the vector3 PCR using a combination of one gene-specific primer and one vec-

tor-specific primer4 Restriction digestion with an enzyme that has two restriction sites

within the vector sequence and none in the cloned gene

Answer 3

Explanation The usual PCR technique for fragment orientation involves one primer specific of the vector (upstream or downstream of the cloning site) and on primer within the cloning sequence If the primer set is correctly orient-ed PCR will give a unique band with a specific size according to the distance between primers If not properly oriented there will be no amplification For digestion principle is the same using one restriction enzyme specific of the vector (a unique site close to the cloning site) and one restriction specific of the insert but not lying right in the middle of it (it may be the same restriction enzyme specific of the vector)

Reference httpsgenomecshlporgcontent36338fullpdf

68 The emission maximum of tryptophan fluorescence in a protein is ~ 335 nm This suggests that tryptophan

1 is in a hydrophobic environment2 occurs in a helical segment3 has proximal cysteine residues4is oxidized

Answer

Explanation

Reference

69 To test the impact of cCAMP on protein kinase A conformation in



17Phone 080-5099-7000 Toll Free 1800-1200-1818

cells an investigator made FRET biosensor by fusing two fluorescent proteins at the N- and C-terminus of protein kinase A In the absence of cCAMP in the cellular milieu no FRET signal was detected However upon cCAMP addition a strong emission at 530 nm was observed What could be the best configuration of fluorophores that were used by the in-vestigator

1 Green fluorescent protein (GFP) and Red fluorescent protein (RFP)2 CYAN fluorescent protein (CFP) and Yellow fluorescent protein (YFP)3 Yellow fluorescent protein (YFP) and Red fluorescent protein (RFP)4 Red fluorescent protein (RFP) and CYAN fluorescent protein (CFP)

Answer 13

Explanation Since the emission is occurring in the presence of cAMP that means these two ends are coming nearby in the presence of cAMP Now in order to show fluorescence these two must have overlapping excitation and emission spectra from donors and recipients As a rule of thumb larger spec-tral overlap will yield a longer R0 therefore your FRET is going to be sen-sitive to longer distances RFP has excitation maximum of 558 nm and the emission maximum is 583 nm GFP from A victoria has a excitation peak at a wavelength of 395 nm and a minor one at 475 nm Its emission peak is at 509 nm So these two do not match either ways CFP has excitation of 433452 nm and emission of 475505 nm YFP has excitation of 516 nm and emission of 529 nm So these two also do not match either ways Comparing RFP and CFP again do not match However maximum coordination can occur between YFP and RFP where yellow one can excite YP as donor at 516 nm it will cause emission at 529 nm absorbing that RFP will get excited an will emit at 584 which can be calculated The YFP-RFP donor-acceptor FRET couple has been employed to study the dimerization of the human thyrotropin (TSH) receptor (Latif et al 2001) However most commonly used couples had been GFP-RFP and CFP-RFP out of which GFP-RFP is considered better So first option could also be correct

Reference httpwwwchmbrisacukmotmGFPGFPhhtmhttpswwwthermofishercominenhomelife-sciencecell-analysis

fluorophoresred-fluorescent-proteinhtmlhttpswwwncbinlmnihgovpmcarticlesPMC5038762httpwwwvisitechcoukassetsimaging-of-fluorescent-proteinspdf

70 In bioremediation by microorganisms detailed below choose the IN-CORRECT option

1 The organic contaminants provide a source of carbon 2 The bacteria do not get net energy by degrading contaminants 3 Bacteria can produce oxidized or reduced species that can cause met-

als to precipitate4 Bacteria act on contaminants by aerobic and anaerobic respirationAnswer 2

Explanation Microbes degrade contaminants because in the process they gain energy that allows them to grow and reproduce Microbes get energy from the contaminants by breaking chemical bonds and transferring electrons from the contaminants to an electron acceptor such as oxygen They invest the energy along with some electrons and carbon from the contaminant to produce more cells

Reference httpswwwnapeduread2131chapter418

PART C

71 Thermodynamics of protein folding is depicted as a free energy fun-nel below

Given below are regions in the diagram (Column X) and their representation (Column Y)

Choose the option that shows all correct matches

1 A - (ii) B - (iii) C - (iv) D - (i)2 A - (i) B - (ii) C - (iii) D - (iv)3 A - (iii) B - (iv) C - (ii) D - (i)4 A - (iv) B - (i) C - (ii) D - (iii)

Answer 1

Explanation This figure is of molten globule concept showing different intermediates observed in the folding of globular proteins It is very easy to identify D here it denotes the NATIVE STRUCTURE If you know only this match you can easily omit option 2 amp 4 Concentrating on options 1 amp 3 now Nothing is common in them except D-(i) match As it can be seen in the figure that enrgy is increasing from bottom to top and Entropy is highest at the top A can be matched to (ii) Hence if you through the options 1amp 3 Option 1 is correct as A-ii D-i

Reference httpswwwpnasorgcontent951911037

72 Table below shows the list of organelles (Column A) the signals (Col-umn B) that target proteins to the organelle


1 a - (ii) b - (iii) c - (iv) d - (i)2 a - (ii) b - (iv) c - (iii) d - (i)3 a - (iv) b - (iii) c - (i) d - (ii)4 a - (iv) b - (iii) c - (ii) d - (i)Answer 3



18Phone 080-5099-7000 Toll Free 1800-1200-1818

Explanation Lysosomes Lysosomes are composed of soluble and trans-membrane proteins that are targeted to lysosomes in a signal-dependent manner The majority of soluble acid hydrolases are modified with mannose 6-phosphate (M6P) residues allowing their recognition by M6P receptors in the Golgi complex and ensuing transport to the endosomallysosomal system Mitochondria Most mitochondrial proteins are encoded by the nuclear ge-nome and translocated into the mitochondria Like the proteins destined for other subcellular organelles the mitochondrially targeted proteins possess tar-geting signals within their primary or secondary structure that direct them to the organelle with the assistance of elaborate protein translocating and folding machines Mitochondrial preproteins that carry the cleavable N-terminal mi-tochondrial targeting sequence (MTS) This signal is classically characterized as an N-terminal motif predicted to form an amphipathic helix that is 15ndash70 residues in length and enriched in positively charged basic residues Nucleus Nuclear localization signals have since been identified in many other Most of these sequences like that of T antigen are short stretches rich in basic amino acid residues (lysine and arginine) Both the Lys-Arg and Lys-Lys-Lys-Lys sequences are required for nuclear targeting but the ten amino acids between these sequences can be mutated without affecting nuclear localization Per-oxisomes Peroxisomes are ubiquitous membrane-bound organelles involved in fatty acid oxidation and respiration Proteins destined for peroxisomes are synthesized exclusively on free polysomes and are post-translationally im-ported into the organelle without any detectable modifications Recent work has shown that anumber of mammalian peroxisomal proteins and one insect peroxisomal protein firefly luciferase are targeted to peroxisomes of CV-1 cells by a tripeptide sequence at the carboxyl termini of these proteins This tripeptide has the generalized structure SerAlaCys-LysArgHis-Leu and has been termed the peroxisomal targeting signal or PTS

Reference httpswwwsciencedirectcomsciencearticleabspiiS1096719211006524httpwwwjbcorgcontent2852014871fullpdfhttpswwwncbinlmnihgovbooksNBK9927httpwwwjbcorgcontent2663423197fullpdf

73 Following statements are made about chromatin remodelling in hu-man cells

A Local Chromatin conformation may play more important role than the local DNA sequence of the promoterB Histones in nucleosome can undergo many different covalent modifi-

cations which in turn alter the chromatin architecture locallyC Chromatin remodelling is a developmentally regulated passive pro-

cess which does not require ATP D Several histone variants exists which replace the standard histones in

specific types of chromatin

Select the option which has the best combination of all correct answers

1 A C D2 A B C3 A B D4 B C D

Answer 2

Explanation Chromatin remodellers helps in remodellingRemodelling is done by ATP hydrolysisThus called ATP Dependent Chromatin remodellars So in the given question the point C was incorrect

Reference httpswwwnaturecomarticles1204332

74 In an experiment the student has infected mammalian host cell with cytoplasmic RNA virus The virus growth was monitored by measuring the intracellular viral RNA at different time intervals It was observed that viral RNA titre progressively went down with time particularly 12 hours post infection Following are a few possibilities which can explain this observation

A The virus infection triggered upregulation of miRNAs that might have downregulated the host factor critical for viral RNA replicationB The virus might encode miRNAs that regulate (inhibits) its own rep-

licationC One of the viral proteins inhibits replication of the viral RNA to re-

strict rapid proliferation D Viral RNA goes to nucleus with time and thus not detectable in the

cytoplasm 12 hour post infection

Which of the following options has all correct statements

1 A B and C2 A C and D3 A B and D4 B C and D

Answer 1

Explanation Statement C is correct as seen incase where Nuclear Factor NF45 Interacts with Viral Proteins of Infectious Bursal Disease Virus and In-hibits Viral Replication so there is a possibility that viral proteins can inhibit viral replication when associated with some cellular protein studies of vi-rus-encoded miRNAs predominantly from the double-stranded DNA Herpes-viridae family have shown that virus-encoded miRNAs target viral and host transcripts facilitate host immune response evasion and share targets with en-dogenous host miRNAs Statement D is incorrect as Cytoplasmic RNA viruses replicatetranscribe their genomes resulting in high levels of cytoplasmic viral RNA and small regulatory RNAs Statement A and B seems correct as there is evidence that Most times host miRNAs play a role in viral life-cycles and promote infection through complex regulatory pathways miRNAs can also be encoded by a viral genome and be expressed in the host cell Viral miRNAs can share common sequences with host miRNAs or have totally different se-quences They can regulate a variety of biological processes involved in viral infection including apoptosis evasion of the immune response or modulation of viral life-cycle phases

ReferencehttpswwwncbinlmnihgovpmcarticlesPMC5430039httpsjviasmorgcontent842010592 httpswwwncbinlmnihgovpmcarticlesPMC3685532

75 Irrespective of the chromosomal configuration a single X-chromo-some remains active in all diploid human somatic cell lines Which one of the following mechanisms best accounts for the above phenomenon

A A maternally inherited X-chromosome is developmentally pro-grammed to remain active by avoiding DNA methylationB Chromosome specific expression and binding of rox1 to one of the

X-chromosomes protects it from Xist mediated silencingC The T-six gene produces just enough of the Xist antisense RNA to

block one Xic locus D A cell produces just enough of the blocking factor to block one Xic

locus

Answer 3

Explanation After 7th round of replication only X-inactivation occursTill 7th round XIST and TSIX RNA production equal leading to RNA silencing After 7th round TSIX production lessened and XIST gene inactivated due to which 85 ie the somatic determinants gets inactivated while remaining 15v of the gene ie sex determining is active


76 In eukaryotic cells replication initiation from a replication origin occurs only once per cell cycle and S-phase CDKs play a vital role in the regulation of DNA replication In budding yeast a protein complex known as the origin recognition (ORC) is associated with DNA replication origin

Bangalore | Noida | Pune | Kolkata | Chennai | Online19

Phone 080-5099-7000 Toll Free 1800-1200-1818


during G1 however origins fire only once at the beginning of S-phase DNA replication does not start in G1 because

A MCM helicases are inactive in G1B Spindle checkpoint is active in G1C DNA polymerase is not recruited in G1 D ORC and initiation factors Cdc6 and Cdt1 do not recruit MCM heli-

case to the site of replication initiation in G1

Which of the above statements are correct

1 A and B2 A and C3 B and C4 B and D

Answer 2

Explanation Initiation of eukaryotic DNA replication is divided into two steps First step is formation of pre-initiation complex which occurs in G1 phase This involves the binding of protein ORC to the origin along with other proteins like cdc6 and cdt1 which finally recruit the helicase MCM Second step is activation of pre-replicative complex by the S cyclin-CDK via phos-phorylation This activates MCM which starts the DNA unwinding by helicase activity Activation also recruits DNA polymerases- DNA pol alpha delta and epsilon to initiate DNA replication So ideally the synthesis of DNA starts in S phase

Reference Molecular Biology of the Gene by Watson 5th Ed Chapter- 8 page no -224-227

77 Measurements of the rate of actin treadmilling in vivo show that it can be several times higher that can be achieved with pure actin in vitro The treamilling in vitro can be enhanced by providing

A profiling that binds G-actin on the site opposite the nucleotide binding cleftB cofiling binds specifically to the ADP containing F-actin and destabi-

lizes the actin filamentC buffer with ATP and low levels of cations D buffer with ADP and low levels of cations


1 A and D2 A and C3 C and D4 A and B

Answer 4

Explanation In the cell treadmilling of actin is regulated and accelerated by orders of magnitude by various actin binding proteins (ABPs) that prefer-entially recognize one of the nucleotide-related conformational states of actin Intriguingly it was proposed that ATP hydrolysis and Pi release do not intro-duce significant gradient of filament instability under crowded conditions of cellular environment and that actin filament treadmilling in the cell is defined primarily by regulatory ABPs that selectively recognize and intensify specif-ic nucleotide-dependent conformational transitions Some examples of ABPs preference for nucleotide specific states of actin include the much better bind-ing of cofilin and profilin to ADP- and ATP-G-actins respectively Cofilinrsquos inhibition of nucleotide exchange on G-actin is consistent with its proposed shifting of the actinrsquos cleft to a closed state On the other hand the acceleration of nucleotide exchange by profilin is in line with its documented opening of the nucleotide cleft on G-actin Profilin promotes assembly of actin filaments by acting as a nucleotide-exchange factor Profilin is the only actin-binding protein that allows the exchange of ATP for ADP When G-actin is complexed with other proteins ATP or ADP is trapped in the ATP-binding cleft of actin However because profilin binds to G-actin opposite to the ATP-binding cleft it can recharge ADP-actin monomers released from a filament thereby re-

plenishing the pool of ATP-actin Cofilin A second group of proteins which bind to actin filaments control the length of actin filaments by breaking them into shorter fragments Because the actin concentration in a cell favors the formation of filaments the breakdown of existing actin filaments and filament networks requires the assistance of severing proteins such as gelsolin and co-filin The actin-depolymerizing factor (ADF)cofilin protein family consists of small actin-binding proteins of 13ndash19thinspkDa that play central roles in acceler-ating actin turnover by disassembling actin filaments (F-actin) ADFcofilins bind to the side of the actin filament (F-actin) and preferentially interact with ADP-F-actin rather than ADP-Pi-F-actin or ATP-F-actin After binding F-ac-tin ADFcofilin severs the filament714 Severing mainly occurs at the point-ed end (P-end) of an ADFcofilin cluster ADFcofilin can also accelerate de-polymerization of the filament at basic pH (78 or 80) under certain conditions

ReferencehttpswwwncbinlmnihgovpmcarticlesPMC3670783httpswwwncbinlmnihgovbooksNBK21594httpswwwnaturecomarticless41467-018-04290-w

78 Equal volumes of pH 40 and pH 100 solutions are mixed What will be the approximate pH of the final solution

1 702 503 604 40

Answer 4

Explanation

pH of Solution A = 4 Therefore concentration of [H+] ion in solution A = 10^ndash4 M pH of Solution B = 10 Therefore Concentration of [H+] ion concentration of solution B = 10^ndash10

MOn mixing one litre of each solution total volume = 1L + 1L = 2L Amount of H+ ions in 1L of Solution A= Concentration times volume V = 10^ndash4 mol times 1L Amount of H+ ions in 1L of solution B = 10^ndash10 mol times 1L there4 Total amount of H+ ions in the solution formed by mixing solutions A and

B is (10^-10 mol + 10^ndash4 mol) This amount is present in 2L solution So total H+ = 10^-4 + 10^-10 2 = 05 x 10^-4 (very small values are ignored) = 5 x 10^-5 MpH = - log (5 x 10^-5)pH = - 07 + 5pH = 43The other way (if we do not ignore)

So total H+ = 10^-4 + 10^-10 2 = 10^-4 [1 + 10^-6] 2 = 05 x 10^-4 [1 + 10^-6] or 05 x 10^-4 [1 + 0000001] = 05 x 10^-4 + 00000005 x 10^-4 = 05000005 x 10^-4 = 50 x 10^-5 pH = - log (5 x 10^-5) pH = - 07 + 5 pH = 43

Reference httpsasklearncbseintcalculate-the-ph-of-a-solution-formed-by-mixing-equal-volumes-of-two-solutions13787

79 The inborn error of amino acid metabolism alkaptonuria is due to the lack of one of the following enzymes

1 Fumarylacetoacetate hydrolase2 Α-ketoacid decarboxylase


Phone 080-5099-7000 Toll Free 1800-1200-1818


3 Homogentisate oxidase4 p-hydroxyphenylpyruvate dehydroxylase

Answer 3

Explanation Alkaptonuria is an autosomal recessive disorder caused by a deficiency of the enzyme homogentisate 12-dioxygenase This enzyme defi-ciency results in increased levels of homogentisic acid a product of tyrosine and phenylalanine metabolism

Referencehttpswwwsciencedirectcomtopicsbiochemistry-genetics-and-molec-

ular-biologyhomogentisate-1-2-dioxygenase

80 The structure of a protein with 100 residues was determined by X-ray analysis at atomic resolution and NMR spectroscopy The following ob-servations are possible

A The dihedral angles determined from the X-ray structure and NMR will be identicalB The dihedral angles determined from the X-ray structure will be more

accurateC β-turns can be determined only by NMR D β-turns can be more accurately determined from the X-ray structure

Indicate the combination with ALL correct answers

1 A and C2 B and D3 B and C4 A and D

Answer 1

Explanation X-ray structure is merely a ldquosnapshotrdquo representing an ensem-ble average of a rigid macromolecule state and sometimes it is useful to utilize other techniques better suited for measuring protein dynamics such as NMR Hence statement B is incorrect So option 2 amp 3 are omitted straight away In the remaining options 1 amp 4-A statement is present that means it is correct check out for C amp D statements It is known that secondary structures are de-termined by NMR so option C is correct CD (circular dichroism) is used for accurate prediction of secondary structures not X-ray crystallography

Reference httpswwwncbinlmnihgovpmcarticlesPMC3138648 httpswwwncbinlmnihgovpubmed21989967 httpswwwpnasorgcontentpnas11224E3095fullpdf

81 The different arms in the tRNA structure are shown in Column A The specific signatures associated with the different arms are shown in Column B

Choose the correct matches from the following

1 A - (ii) B - (iv) C - (i) D - (iii)2 A - (i) B - (iii) C - (iv) D - (ii)3 A - (iv) B - (i) C - (ii) D - (iii)4 A - (ii) B - (iii) C - (iv) D - (i)

Answer 4

Explanation (1) The 3prime-end terminates into 5rsquoCCA3prime sequence that is al-ways unpaired (2)The terminal A residue is the site at which the amino acid is bound covalently(3) Then comes the first loop containing 7 unpaired bases This loop is designated as ldquoTΨ C looprdquo because it always contains a sequence 5prime ribothymidine- pseudouridine- cytidine 3prime This loop is involved in binding to ribosomes (4) After the ldquo5prime -T Ψ C-3prime looprdquo in the 5prime direction there occurs a loop of variable size called the extra loop or the ldquolumprdquo The lump may contain 3 to 21 bases (5) The third loop contains 7 unpaired bases and it has

the ldquoanticodonrdquo Anticodon consists of 3 bases At the 3prime -end of the anticodon there is a purine (A or G) while at the 5prime -end there is always uracil (U) At the time of protein synthesis anticodon pairs with its complementary ldquocodonrdquo on mRNA (6) The fourth loop is larger than others and contains 8-12 unpaired bases It is designated as ldquoD-looprdquo because it is rich in dihydrouridine (UH2) The enzyme aminoacyl synthetase binds to this loop[Refer Image]

Reference httpwwwbiologydiscussioncomacidsnucleic-acidtrna-mean-ing-structure-and-initiator-transfer-nucleic-acids35879

82 Some enzymes that serve as transient carriers of specific chemical-groups are shown below Choose the combination with all correct matches

1 A - (i) B - (ii) C - (iii) D - (iv)2 A - (ii) B - (i) C - (iv) D - (iii)3 A - (iii) B - (iv) C - (ii) D - (i)4 A - (iv) B - (iii) C - (i) D - (ii)

Answer 2

Explanation Coenzyme A serve as transient carriers of acyl group FAD is a second electron carrier used by a cell during cellular respirationPyridoxal phosphate functions as a carrier of amino groups and as an electron sink by facilitating dissociation of the α-hydrogen of the amino acid NAD helps in the transfer of hydride ions

Reference httpswwwmikeblaberorgoldwineBCH4053Lecture33Lecture33htm

83 In a laboratory experiment it was observed that both lsquoVirus Arsquo and lsquoVirus Brsquo could infect a mammalian host cell when infected individual-ly Interestingly if the cells were first infected with lsquoVirus Arsquo (with large MOI) Virus B failed to infect the same cell If the Virus B (with large MOI) is added first followed by Virus A both the virus can infect the cells However infection with lsquoVirus Arsquo was found to be in lesser extent Con-sidering X and Y are the receptorsco-receptors which may be involved for the virus entry following are few possibilities that can explain the observation

A lsquoVirus Arsquo uses lsquoXrsquo as receptor and lsquoYrsquo as coreceptorB lsquoVirus Brsquo uses exclusively lsquoYrsquo as receptor for entryC Both lsquoVirus Arsquo and lsquoVirus Brsquo need X receptor



21Phone 080-5099-7000 Toll Free 1800-1200-1818

Choose the option with all correct statements

1 A B and C2 A and B3 B and C4 A and C

Answer 4

Explanation X should be a receptor for both virus A and B as when MOI of virus A is more B cannot infect and in other case when MOI of B is more A can infect but in lesser extent so it means it has other receptor also so statement A and C are correct B is incorrect because if B was correct then when virus A infects with large MOI then B should have caused infection


84 While testing the effect of several potent anti-cancer compounds on cycling human oral cancer cells a student observed that a major per-centage of cells showed dose-dependant cell death after 12 hours of drug treatment However the remaining cells repopulated the culture dish once the compounds were removed and the cells were cultured in complete me-dium The student made the following assumptions

A Not all cells were equally affected by the compounds as they were not synchronized before treatmentB The compound selectively killed cells which were in G0 phaseC The cancer stem cells were impervious to the effects of the compounds

and therefore repopulated culture D The cancer cells differentiated into a mesenchymal phenotype and

grew in fresh culture medium containing inhibitors of epithelial-to-mes-enchymal transition (EMT)

Which one of the following combination of assumptions would best justify the results

1 B and C2 A and C3 B and D4 A and B

Answer 2

Explanation Depending on their mechanism of action the efficacy of chemotherapy drugs may be influenced markedly by the time of exposure (phase-specific or time-dependent drugs) or by the dose that can be adminis-tered (phase-nonspecific or dose-dependent drugs) The efficacy of phase-spe-cific anticancer drugs is time-dependent as only a fraction of tumor cells are in appropriate cell cycle phase for chemotherapy-mediated killing at any given time Since the question mentions a dose-dependent cell death there is no question that the drug will be specific to any phase Hence statement B cannot be the correct assumptionAlso statement C is correct Despite the recent advances in treatment modal-

ities nonndash small cell lung cancer remains to be one of the leading causes of mortality with a 5-year survival rate of 15This is owed to the increasingly emergent concept of cancer stem cells (CSCs) a subpopulation of the tumor mass alleged to be impervious to radiation and chemotherapeutic treatments making them responsible for recurrence of the disease in cancer patients

Reference httpswwwncbinlmnihgovpmcarticlesPMC3118128 httpsjournalssagepubcomdoipdf1011771010428317695915

85 When 8-cell embryo of tunicates is separated into 4 blastomere pairs and allowed to grow independently in culture medium then each blas-tomere pair can form most of the cell types however cells for nervous system are not developed The following statements are formed from the above observationsA Nervous system development demonstrated autonomous specifica-

tionB The other tissue types are formed due to conditional specificationC All the tissue types except nervous tissues that developed demon-

strated autonomous specificationD Nervous system development demonstrated conditional specification

The correct combination of statements that explains the above result is

1 A and B2 B and C3 C and D4 A and D

Answer 3

Explanation Early tunicate cells are specified autonomously each cell acquiring a specific type of cytoplasm that will determine its fateWhen the 8-cell embryo is separated into its four doublets (the right and left sides being equivalent) mosaic determination is the rule The animal posterior pair of blastomeres gives rise to the ectoderm and the vegetal posterior pair pro-duces endoderm mesenchyme and muscle tissue just as expected from the fate mapConditional specification however also plays an important role in tunicate development One example of this process involves the development of neural cells The nerve-producing cells are generated from both the animal and vegetal anterior cells yet neither the anterior or posterior cell of each half can produce them alone

Reference Scott F Gilbert DevelopmentalBiology 8th ed Chapter-3 pg-54-57Ch-8pg-238-241

86 In a strain of Ecoli a fusion between lac and trp operon took place and the new locus structure is shown below The strain lacks the wild-type trp operon

Given below are some of the potential scenarios

A Tryptophan will be synthesized in a medium containing lactose and tryptophan B Tryptophan synthesis will be repressed in a medium containing glu-

coseC Tryptophan synthesis will take place only in the absence of sufficient

tryptophan in the medium

Choose the option that correctly describes the behaviour of the fusion operon

1 A and B2 A and C3 C only4 B and D

Answer 1

Explanation The operon created is a fused operon with the regulatory genes of the lac operon and structural genes of trp operon So expression of trp structural genes will be under the regulation of lac regulatory genes It is no longer under the regulation of tryptophan presence or absence Expression of the lac operon occurs when lactose is present and glucose is absent Presence of glucose suppresses the lac operon by catabolite repression

Reference

231+ CSIR NET Dec 2018 amp GATE 2019 Toppers Are From Biotecnika



22Phone 080-5099-7000 Toll Free 1800-1200-1818

Molecular Biology of Gene by Watson 5th Ed Chapter- 16 page no -488-494

87 Following statements have been made about removal of supercoiling produced by DNA unwinding at the replication fork

A Accumulation of supercoils is the result of DNA helicase activity dur-ing unwinding of DNAB Problem of DNA supercoiling is valid only for circular chromosomes

of bacteria and not for the linear chromosomesC Supercoiling of DNA is removed by topoisomerases by breaking ei-

ther one or both strands of DNA on the unreplicated DNA in front of replication forkD Both topoisomerase I and topoisomerase II can remove positive su-

percoiling during replication

Which one of the following options has all correct statements

1 A B C2 A B D3 A C D4 B C D

Answer 3

Explanation DNA supercoils are generated during replication because of the DNA helicase eliminating the base pair between the two strands Supercoiling occurs in both circular and linear DNA Topoisomerases are used to resolves the supercoils Type I topoisomerases cleave one strand of DNA and Type II topoisomerases cleave both the strands to resolve the super-coils Both Type I and II can resolve positive supercoils

Reference Molecular Biology of Gene by Watson 5th Ed Chapter- 8 page no -198-199

httpwwwebiacukinterproentryIPR013499

88 Phosphorylation of the α-subunit of elF2 at Ser 51 position in Saccha-romyces cerevisiae leads to sequestration of elF2B a guanosine exchange factor This phenomenon is

1 known to activate translation of capped mRNAs in the cytosol2 known to activate translation of many key mRNAs possessing short

ORFs (uORFs) in the mRNA sequence that precedes the main ORF3 an essential requirement for translation of IRES containing mRNAs 4 an essential requirement for the transport of mature mRNAs out of

the nucleus

Answer 3

Explanation Many cellular IRES-containing mRNAs (eg cat-1 N-myc s-Src etc) were shown to be insensitive or much less sensitive than mRNAs without IRESs to the inhibition of protein synthesis caused by eIF2 phospho-rylation Phosphorylation of the alpha subunit of eIF2 (eIF2α) is a common consequence of many stress conditions (eg ER stress nutrient limitation and many others)This reduces the activity of the eIF2bullGTPbullMet-tRNAiMet terna-ry complex and thereby suppresses the overall rate of cap-dependent protein synthesis

Reference httpswwwncbinlmnihgovpmcarticlesPMC3048795 httpswwwncbinlmnihgovpmcarticlesPMC87228

89 Ecoli mutants isolated from a genetic screen showed following classes of mutation

A Point mutations in lacIB Deletions immediately downstream of the transcription start site of

the lacZYA mRNA

C Duplications of part or whole of lac YD Duplications of part or whole of lac A

Choose the option which is likely to result in contitutent expression of the lac operon

1 Both A and B2 Both B and C3 Both C and D4 Only A

Answer 4

Explanation Lac operon is inducible operon due to repressor protein coded by lacI which binds to operator and blocks the gene expression In case repres-sor becomes inactive it fails to bind to the operator and thus not block the gene expression Point mutation in lacI inactivates repressor as a result the operator is never blocked and gene is expressed constitutively


90 For Eschericia coli chromosomal DNA replication which one of the following statements is true

A DNA polymerase I is the main polymerase required for DNA replica-tionB DNA polymerase I though identified originally by Kornberg as the

one responsible for replication is not important for the DNA replication processC Requirement of DNA polymerase I is in the context of removal of

RNA primer needed for DNA synthesis and then fill in the same with DNA equivalentD DNA polymerase I is the primary enzyme for error prone DNA syn-

thesis in response to SOS

Answer 3

Explanation DNA pol I is responsible for primer removal Principle poly-merase for DNA synthesis is DNA Pol III For SOS repair special polymerase DNA pol IV and V are needed

Reference Biochemistry by Voet amp Voet 4th Ed Chapter- 30 and page no1178-1182

91 Following observations were made about variations among genomes of eukaryotic organisms

A Single nucleotide polymorphisms are the numerically most abundant type of genetic variantsB Both interspersed and tandem repeated sequences can show poly-

morphic variationC Mitotic recombination between mispaired repeats causes change in

copy number and generates minisatellite diversity in populationD Smaller variable segments in the genome can be identified by paired

end mapping technique

Select the option with all correct statements

1 A B C2 A C D3 B C D4 A B D

Answer 4

Explanation SNP are the most common type of genetic variation among people They occur almost once in every 1000 nucleotides on average which means there are roughly 4-5 million SNPs in a person genome (1) Both in-


23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818

terspersed and tandem repeated sequences can show polymorphic variation (2) Smaller variable segments in the genome can be identified by paired end mapping technique (3) Here in given option AB and D is correct

Referencehttpghrnlmnihgovprimergenomicresearchsnphttpspdfssemanticscholarorge914794a0aefb2176fc384d0e70a03b-577f1a5d5pdfHuman Molecular Genetics Page 410 Chapter 13

92 There are a number of specific T cell surface molecules involved in various functions of adaptive immune response Column X represents a list of T cell surface molecules and Column Y with the possible functional characteristics

Which of the following option has all correct matches

1 A - (i) B - (ii) C - (iii) D - (iv)2 A - (iv) B - (i) C - (ii) D - (iii)3 A - (iii) B - (iv) C - (i) D - (ii)4 A - (iv) B - (iii) C - (ii) D - (i)

Answer 4

Explanation T cell receptor consists of two polypeptides alpha and beta and in some cases gamma and deltaCD28 binds to B7 and acts costimulation for activation of T cellCD8 expressed on Tc cells binds to MHC class I and makes its activation self

MHC restrictedCD154 or CD40 L is responsible for binding to CD40 which helps in B cell

activation and differentiation

Reference Immunology by Kuby 5th Ed page no 37 202 233 260

httpswwwncbinlmnihgovpubmed10856699

93 In an experiment it was observed that a protein was upregulated in the cancer tissues (compared to control issues) that showed correlation with disease progression Following are a few possibilities which can ex-plain the above observation

A A mutation could be located in the 3rsquoUTR of the corresponding mRNA at a miRNA binding siteB A mutation changes the conformation of the protein resulting in its

better stabilityC A mutation in the corresponding mRNA promotes ribosome read-

through of the termination codon resulting in increased synthesis of the proteinD A mutation in the corresponding mRNA increased the stability of the

RNA due to change in secondary structure

Which one of the following combinations represents the most likely ex-planations

1 A B and C2 B C and D3 C D and A4 A B and D

Answer 4

Explanation If the termination codon is read through by the ribosome be-cause of the mutation it will result in a different protein not increased synthe-sis Hence this cannot be an explanation for upregulated synthesis Evidence for statement A to be a possible explanation The let-7 family in-

cludes 12 miRNA members which are thought to play important roles as tu-mor suppressors Let-7 binds to the 3-UTRs of key oncogenes including RAS and MYC and inhibits their expression Hence if this 3-UTR site is mutated in the corresponding mRNA then miRNA cannot bind and oncogenes expres-sion will be upregulated Evidence for statement B and D to be a possible explanation The cytoplas-

mic level of a messenger RNA and hence protein depends not only upon its rates of synthesis processing and transport but its decay rate as well Se-quence elements within an mRNA together with the protein andor small non-coding RNA factors that bind these elements dictate its decay rate Genetic alterations in mRNA stability can lead to various diseases including cancer heart disease and immune disorders Hence a mutation in the corresponding mRNA may make it more stable for increased translation

Reference httpcancerimmunolresaacrjournalsorgcontentcanimmea ly201902152326-6066CIR-18-0331fullpdf httpswwwresearchgatenetpublication233610363_Molecular_Mech-anisms_Regulating_mRNA_Stability_Physiological_and_Pathological_Significance

94 The extracellular matrix contains a number of non-collagen proteins that typically have multiple domains each with specific binding sites for other matrix molecules and cell surface receptors These proteins there-fore contribute to both organizing the matrix and helping cells attach to it The most well characterized matrix protein of this kind is fibronectin Which one of the following characteristics is NOT TRUE for fibronectin

A It is a large glycoprotein found in all vertebrates and important for many cell-matrix interactionsB It is composed of three polypeptides that are disulfide bonded into a

crosslink structureC In human genome there is only one fibronectin gene containing about

50 exons but the transcripts can be spliced in different ways to produce many different fibronectin isoformsD Fibronectin binds to integrin through an RGD motif Even very short

peptide containing RGD sequence can inhibit attachment of cells to fi-bronectin matrix

Answer 2

Explanation Fibronectin is the principal adhesion protein of connective tissues Fibronectin is a dimeric glycoprotein consisting of two polypeptide chains each containing nearly 2500 amino acids Fibronectin are proteins that connect cells with collagen fibers in the ECM allowing cells to move through the ECM So the given option 2 is incorrect

Reference httpsnptelacincourses102103012module6lec64html

95 Following are the list of some cellular receptors (Column X) and with



25Phone 080-5099-7000 Toll Free 1800-1200-1818

possible functional characteristics (Column Y)

Which one of the following is the correct match

1 A - (i) B - (ii) C - (iii)2 A - (ii) B - (iii) C - (i)3 A - (iii) B - (ii) C - (i)4 A - (i) B - (iii) C - (ii)

Answer 3

Explanation Asialoglycoprotein receptor is a transmembrane protein that plays a critical role in serum glycoprotein homeostasis by mediating the en-docytosis and lysosomal degradation of glycoproteins with exposed terminal galactose or N-acetylgalactosamine residues

The transferrin receptor is a membrane glycoprotein whose only clearly de-fined function is to mediate cellular uptake of iron from a plasma glycoprotein transferrin Iron uptake from transferrin involves the binding of transferrin to the transferrin receptor internalization of transferrin within an endocytic ves-icle by receptor-mediated endocytosis and the release of iron from the protein by a decrease in endosomal pH

Steroid receptors are a class of molecules that function as both signal transducers and transcription factors

Referencehttpswwwncbinlmnihgovgene432httpswwwncbinlmnihgovpubmed10582342httpswwwncbinlmnihgovpubmed2194782

96 Given below are statements related to different aspects of plant growth and development

A Leaf longevity is increased in ethylene-insensitive mutants etr1-1 and ein2 of ArabidopsisB Programmed cell death (PCD) is responsible for the formulation of

prickles thorns and spines in plants

C Senescence and PCD occur only in the development of vegetative tis-sues and does not occur in reproductive tissuesD Redifferentiation of organelles is an integral component during initial

stages of senescence in plants

Which one of the following represents the combination of all correct statements

1 A C and D2 B and C3 A B and D4 C and A

Answer 4

Explanation Go through options one by one Statement A is absolutely cor-rect hence option 2 is omitted Statement B is not true as thorns are modified sharp-pointed stems occurring at the base of a leaf or ends of twigs Spines (example Barberry cacti) are modified leaves or parts of leaves and are gener-ally thinner and shorter than thorns Prickles (example rose stem) are slender outgrowths of the plant outer layer (epidermis) they are not formed through PCD Hence this statement is wrong According to this we can omit option 3 also We are left with option 1 and 4 They both have A and C in common lets check with D statement Statement D is incorrect as Dedifferentiation is associated with Senescence Hence option 4 is the correct answer

Reference httpsacademicoupcomjxbarticle554062147585866 httpsanswersingenesisorgbiologyplantsthorns-and-thistles httpswwwncbinlmnihgovpmcarticlesPMC4844402

97 The following demonstrates proposed functions of different genes which determine the decision to become either trophoblast or inner cell mass (ICM) blastomere during early mammalian development

Based on the above figure which one of the following assumptions is

correct

1 The interplay between Cdx2 and Oct4 can influence the formation of ICM2 The ICM would form even if expression of Oct4 was inhibited3 YAP and TEAD4 are upstream components of Cdx2 and can be inhib-

ited by Nanog4 The expression of Stat3 is optional for maintaining pluripotency of the

ICM

Answer 1

Explanation Oct4 is a key component of the pluripotency regulatory net-work and its reciprocal interaction with Cdx2 has been shown to be a de-terminant of either the self-renewal of embryonic stem cells (ESCs) or their differentiation into trophoblast The internal population of cells termed the inner cell mass (ICM) is protected from differentiation by expression of the POU domain transcription factor Oct4 (also known as Oct3 Oct34 and Pou5f1) Following zygotic deletion of Oct4 the blastocyst eventually differ-entiates into trophectoderm STAT3 first identified as a transcription factor for the IL-6 family of cytokines was subsequently found to be crucial for ESC pluripotencyBy eliminating Stat3 in mouse oocytes and embryos we found STAT3 to be essential for the maintenance of ICM lineages but not for



26Phone 080-5099-7000 Toll Free 1800-1200-1818

blastocyst formation or TE maintenanceSTAT3 has an essential role in ICM lineage specification and maintenance and pluripotent stem cell identityThe nuclearcytoplasmic distribution of TAZYAP defines the first cell fate choice in the mouse embryo ndash the decision of embryonic cells to become either troph-ectoderm (TE) or inner cell mass (ICM)Nuclear TAZYAP control the activity of TEAD transcription factors to direct a TE-specific transcriptional program that includes the induction of Cdx2 In line with this deletion of Tead4 results in the loss of Cdx2 expression giving rise to embryos that fail to establish the TE Deletion of both Taz and Yap also results in cell fate specification defects but with embryos dying at the morula stage prior to the specification of TE or ICMReference httpswwwncbinlmnihgovpmcarticlesPMC4230828httpsdevbiologistsorgcontent14151001httpgenesdevcshlporgcontent27121378fullhtmlhttpsdevbiologistsorgcontent1418161

98 Jasmonate is known to inhibit root growth while auxin facilitates root growth Upon infection with pathogenic bacteria that produce coronatine we may expect the following in plants

A Upregulation of CO1-1 gene and inhibition of root growthB Upregulation of Aux1-1 gene and inhibition of root growthC Inhibition of Aux1-1 gene and promotion of root growthD Inhibition of CO1-1 gene and promotion of root growth

Which one of the following is correct

1 A B and C2 Only A3 Only B4 Only C

Answer 2

Explanation The expression of the JA receptor CORONATINE INSENSI-TIVE1 (COI1) and the key JA signaling regulator MYC2 is up-regulated in response to Al stress in the root tips Hence upregulation of COI-1 gene will occur over here which will further inhibit root growth Statement A is correct Either of the options 1 or 2 is correct This process together with COI1-me-diated Al-induced root growth inhibition under Al stress was controlled by ethylene but not auxin Option 1 includes both the regulatory mechanisms upregulation as well as inhib ition of Aux-I which is not possible Hence we can say that only A statement is correct or option 2 is correct

Reference httpwwwplantphysiolorgcontent17321420

99 Following are certain statements with regard to plant respiration

A Metabolism of glucose into pyruvate through glycolysis generates NADH and NADPHB Metabolism of glucose through oxidative pentose phosphate cycle

does not produce NADPHC Cyanide forms a complex with haem iron of cytochrome oxidase lead-

ing to prevention of change in valency which in turn stops electron trans-port in the respiratory chainD Alternative oxidase is insensitive to cyanide and has higher Km than

that of cytochrome oxidase

Which one of the following combinations is correct

1 A B and C2 B C and D3 B and D4 A C and D

Answer 4

Explanation Statement A is correct In glycolysis end product is pyruvate in addition to the generation of ATP and NADH not NADPH According to this

statement either of options 1 or 4 is correct Statement B is incorrect as PPP produces NADPH Cyanide acts an inhibitor in ETC it forms a complex with Fe of cytochrome oxidase which stops Electron transport Statement C is also correct Alternative oxidase obviously has a higher affinity than cytochrome oxidase and it is insensitive to cyanide too Statement D is also correct As statements AC and D are correct-- Here option 4 is the right answer

Reference httpswwwsciencedirectcomtopicsbiochemistry-genetics-and-molec-

ular-biologypentose-phosphate-pathway


A cry1 binds to COP1 and SPA1 complex leading to degradation of HY5B cry1 binds to COP1 and SPA1 complex and prevents degradation of

HY5C CCT is overexpressed and the plants are kept in darkD CCT is overexpressed and the plants are kept in light

Which of the following combinations of above statements will result in photo-morphogenesis

1 Only A2 Only B3 A B and C4 B C and D

Answer 2

Explanation The basic domainLeu zipper transcription factor ELONGAT-ED HYPOCOTYL5 (HY5) is a key photomorphogenesis-promoting factor downstream of COP1 and is destabilized by COP1 in darkness CRY binding inhibits COP1 activity Another factor that functions in phytochrome signal transduction SPA1 also interacts with COP1 Inhibition of COP1 may then allow accumulation of HY5 and the response to lightCOP1 is a protein that inhibits photomorphogenesis as it contains E3 ubiqui-

tin ligases and is involved in targeting proteins for 26 S proteosome-mediated degradation Usually COP1 leads to degradation of HY5 but when cry1 binds to COP1 its activity is inhibited means it prevents degradation of HY5-state-ment B is supporting that C amp D statements are contradictory hence only statement B is correct and option 2 is the right answer here


101 After absorbing light chlorophyll molecules in green plants exist in singlet and triplet states Following are certain statements on singlet and triplet states of chlorophyll molecules

A Singlet state is short lived compared to triplet stateB Singlet state is long lived compared to triplet stateC Singlet state contains electrons with anti-parallel spins while triplet

state has electrons with parallel spinsD Singlet state contains electrons with parallel spins while triplet state

has electrons with anti-parallel spins


1 A and B2 B and C3 A and C4 B and D

Answer 3

Explanation Excited electionic states are due to the promotion of an electron from tne ground state distribution changlng the electron density configurationto one of higher energy The excited singlet state is generally short-lived (10-

8-10-9 sec) witn the valence electrons having opposite spins An excited tri-plet state is longer-lived (milliseconds cr more) wlth the two electrons having



27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms

Reference httpspdfssemanticscholarorgce890c9b26ec4b2c3bf-d1e35243b2c55af94f282pdf

102 Only members of the plant kingdom and many bacteria have capa-bility of biological nitrogen reduction In this regard following statements are given

A Nitrogen is normally taken by the plant in their fully oxidized form but needs to be reduced before incorporation in organic moleculesB Conversion of oxidized nitrogen into reduced nitrogen needs energy

in the form of NAD(P)+C The metal associated with the enzyme nitrate reductase is Magnesi-

umD Nitrate reduction takes place in the cytoplasm whereas nitrite reduc-

tion takes place in chloroplast matrix

Which one of the following combinations of the above statements is cor-rect

1 A and C2 A B and C3 B and D4 A and D

Answer 4

Explanation Statement A is correct as Nitrogen is taken in the form of Nitric oxide and Nitrous oxide which is further reduced by the process of Nitogen metabolism amp assimilation According to this option 3 is incorrect Nitrate reductase are molybdoenzymes that reduce nitrate (NOminus3) to nitrite (NOminus2) it does not involve Mg rather includes FAD Mo amp Heme as co-factors State-ment C is also incorrect As you can see in the options- only option 4 is having A without C Hence option 4 is correct You can cross check also D statemnet is correct as Nitrite reduction takes place in chloroplast matrix and nitrate re-duction in cytoplasm

Reference httpwwwjbcorgcontent2071341fullpdf httpsplantcellbiologymastersgrkrajorghtmlPlant_Cell_Biochemistry_

And_Metabolism4-Nitrogen_Metabolismhtm


A May promote pathogen virulenceB May elicit avirulence responseC May suppress defense responseD May promote plant growth


1 A B and D2 A C and D3 A B and C4 B C and D

Answer 3

Explanation Pathogens attempting to invade plants have to overcome the first level of immunity in the prospective host and do so by developing ef-fector proteins The first purpose of effector proteins is to overcome this in-nate basic resistance of plants - Statement C is correct Effector proteins and PAMPs do not always use the terms pathogenicity virulence and avirulence appropriately and at the same time they do not trouble to give a clear new defi-nitions of these phenomena Currently the substances (effectors) that some

phytopathogen secretes into the host plant cells are taken to be either virulence or avirulence factors or sometimes also pathogenicity factors Effector mol-ecules are likewise primarily termed avirulence factors but they are also fac-tors that contribute to pathogen virulence - Statement A is correct Avirulence factors elicit avirulent responses in the host plant eg a response consisting in host cell necrosis at the site of inoculation (HR) - Statement B is correct

Reference httpwwwfupressnetindexphppmarticleview12077

104 During wing development in chick if Apical Ectodermal Ridge (AER) is removed the limb development ceases on the other hand plac-ing leg mesenchyme directly beneath the wing AER distal hindlimb struc-tures develop at the end of the wing and if limb mesenchyme is replaced by non-limb mesenchyme beneath AER regresses This may demonstrate that

A the limb mesenchyme cells induce and sustain AERB the mesenchyme cells specify the type wing or limbC the AER is responsible for specifying the type wing or limbD the AER is responsible for sustained outgrowth and development of

the limbE the AER does not specify the type wing or limb

Which combination of above statements is demonstrated by the experi-ment

1 A B C and D only2 A B D and E only3 C D and E only4 A and E only

Answer 2

Explanation The proximal-distal growth and differentiation of the limb bud is made possible by a series of interactions between the limb bud mesenchyme and the AER Although the mesenchyme cells induce and sustain the AER and determine the type of limb to be formed the AER is responsible for the sustained outgrowth and development of the limb The AER keeps the mesen-chyme cells directly beneath it in a state of mitotic proliferation and prevents them from forming cartilageStatement A B and D and E are correct according to the explanation Hence option 2 is correct here


105 The following statements regarding the generation of dorsalventral axis in Drosophila was made

A Gurken protein moves along with the oocyte nucleus and signals folli-cle cells to adopt the ventral fateB Maternal deficiencies of either the gurken or torpedo gene cause ven-

tralization of the embryoC Gurken is active only in the oocyte and Torpedo is active only in the

somatic follicle cellsD the Pipe protein is made in the dorsal follicle cellsE the highest concentration of Dorsal is in the dorsal cell nuclei which

becomes the mesoderm

Which one of the following combination of the above statements is true

1 A and E2 C and D3 B and C4 B and E

Answer 3

Explanation The nucleus of the oocyte travels to what will become the dor-sal side of the embryo The gurken genes of the oocyte synthesize mRNA that becomes localized between the oocyte nucleus and the cell membrane where it is translated into Gurken protein The Gurken signal is received by the Tor-



28Phone 080-5099-7000 Toll Free 1800-1200-1818

pedo receptor protein made by the follicle cells Given the short diffusibility of the signal only the follicle cells closest to the oocyte nucleus (ie the dor-sal follicle cells) receive the Torpedo signal which causes the follicle cells to take on a characteristic dorsal follicle morphology and inhibits the synthesis of Pipe protein Therefore Pipe protein is made only by the ventral follicle cellsThe ventral region at a slightly later stage of development Pipe modifies an unknown protein (x) and allows it to be secreted from the ventral follicle cells Nudel protein interacts with this modified factor to split the products of the gastrulation defective and snake genes to create an active enzyme that will split the inactive Easter zymogen into an active Easter protease The Easter protease splits the Spatzle protein into a form that can bind to the Toll receptor (which is found throughout the embryonic cell membrane) This protease ac-tivity of Easter is strictly limited by the protease inhibitor found in the periv-itelline space Thus only the ventral cells receive the Toll signal This signal separates the Cactus protein from the Dorsal protein allowing Dorsal to be translocated into the nuclei and ventralize the cells Thus only ventral cells of the embryo has the dorsal protein located inside the nucleus

Reference Scott F Gilbert DevelopmentalBiology 8th edChapter-9 pg-262

106 A specialized area of teratogenesis involves the misregulation of the endocrine system Which one of the following statements regarding endo-crine disruptors is true

1 They can act as antagonist and inhibit the binding of a hormone to its receptors or block the synthesis of a hormone to its receptors2 They do not affect the synthesis elimination or transportation of a

hormone in the body 3 They do not mimic the effect of natural hormones4 Low dose of exposure of endocrine disruptors is not sufficient to pro-

duce significant disabilities later in life

Answer 1

Explanation Endocrine disruptors can

Mimic or partly mimic naturally occurring hormones in the body like estro-gens (the female sex hormone) androgens (the male sex hormone) and thy-roid hormones potentially producing overstimulationBind to a receptor within a cell and block the endogenous hormone from

binding The normal signal then fails to occur and the body fails to respond properly Examples of chemicals that block or antagonize hormones are an-ti-estrogens and anti-androgensInterfere or block the way natural hormones or their receptors are made or

controlled for example by altering their metabolism in the liver

Reference httpswwwniehsnihgovhealthtopicsagentsendocrineindexcfm

107 During prolonged illumination rhodopsin is desensitized which leads to the termination of visual response The associated response (Col-umn A) and their effects (Column B) are given below Which one of the matched combinations is correct

1 a - (iv) b - (ii) c - (i) d - (iii)

2 a - (ii) b - (iii) c - (iv) d - (i)3 a - (iii) b - (i) c - (ii) d - (iv)4 a - (ii) b - (iv) c - (iii) d - (i)

Answer 3

Explanation Photoactivated rhodopsin interacts not only with transducin but with two more proteins a protein kinase that specifically phosphorylates R (in contrast to dark-adapted rhodopsin) at multiple sites and an abundant soluble protein of 48 KDal (called 48 K-protein S-antigen or arrestin) that specifically binds to phosphorylated R Phosphorylation partially suppresses the ability of R to catalyze transducin-mediated phosphodiesterase activation even in the absence of arrestin Binding of arrestin to the phosphorylated R potentiates this inhibitory effect most probably because arrestin competes with transducin for binding on the phosphorylated R Phosphorylation in conjunction with arrestin binding therefore appears to be a mechanism that terminates the active state of the receptor R

Reference httpswwwncbinlmnihgovpubmed3040978

108 Esterus cycle in rats is controlled by pituitary and gonadal hor-mones While treating a set of rats with Vitamin D a student accidental-ly injected the rats with an inhibitor of 17α-hydroxypregnenolone and checked vaginal smears for 10 consecutive days

Which one of the following observations is correct

1 The smears showed well formed nucleated epithelial cells throughout the period2 The smears initially showed normal estrus stage but eventually en-

tered a prolonged diestrus stage3 The smears showed leukocytes and few epithelial cells4 The cells showed metestrus for 3 days and then returned to the proes-

trus stage

Answer 2

Explanation Consistent with the C57Bl6J mice WT mice shows 4ndash5 day cycle with four stages proestrous estrous diestrous and metaestrous1 Proestrous-It is the phase at which estrogen is lowest and gradually in-

creasesThe vaginal smear reveals a discharge of nucleated epithelial cells2 Estrous-This is the phase of highest estrogen level after which estrogen

declines It is a phase after which the epithelium of the vagina sloughsoff having reached its maximum thickness which shows cornified cells3 Metaestrous-A phase of declining concentration of estrogen With de-

clining concentrations epithelial cells along with many leukocytes slough off from the vagina 4 Diestrous-With low levels of E2 the vagina shows similar profile as me-

taestrous but with fewer leukocytic cellsvitamin D3 deficiency delays puberty and causes prolonged estrous cycles

characterized by extended periods of diestrus and reduced frequency of proes-trus and estrus17-Hydroxypregnenolone is important for the synthesis of es-trogenInhibiting 17-HOP results in estrogen deficiencyOvariectomy(lack of estrogen) in adult female mice has been shown to result

in extended diestrous phase similar to Vitamin D3 deficiencyThereby combining the above statements option 2 holds true

Reference httpswwwncbinlmnihgovpmcarticlesPMC2755182httpswwwncbinlmnihgovpmcarticlesPMC3431429httpssci-hubtw101016jphysbeh200911003

109 Centromere positions can be mapped in linear tetrads in some fun-gi A cross was made between two strains a b and arsquo brsquo and 100 linear tetrads were analyzed The genes a and b are located on two arms of the chromosome The tetrads were divided into 5 classes as shown belowBased on the above observation the following conclusions were drawn

A Class 1 is a result of a cross over between lsquoarsquo and the centromere



29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818

B Class 2 is a result of a double crossover involving 3 strands between lsquoarsquo and the centromereC Class 2 is a result of a double crossover between lsquoarsquo - centromere and

lsquobrsquo - centromere involving three strandsD Class 4 is a result of a double crossover involving all the 4 strands

Which one of the following options represents all correct statements

1 A and B2 A and C3 B and D4 C and D

Answer 2

Explanation If we analyze the arrangement in tetrads in class 1 the a is in 2222 ratio and b is in 44 ratio Hence the first class of tetrads is definitely a crossover product between a and the centromere Hence statement A is abso-lutely correct DCO between gene and centromere involving two strands can form 44 ratio but in statement B since three strand is given it is incorrect In class 5 both a and b are in 242 ratio which is crossover product between a and centromere and b and centromere hence a DCO involving 3 strands So statement C is correct Class 4 is PD that could result from DCO between two strands not all 4 so D is incorrect Hence correct combination is option 2


110 Two yellow mice with straight hair were crossed and the following progeny was obtained

frac12 yellow straight hair⅙ yellow fuzzy hairfrac14 gray straight hair112 gray fuzzy hair

In order to provide genetic explanation for the results and assign geno-types to the parents and progeny of this cross the following statements were given

A The 6231 ratio obtained here indicates recessive epistasisB The cross concerns two independent characteristics - body colour and

type of hairC The deviation of dihybrid ratio from 9331 to 6231 may be due to

one of the genes being a recessive lethalD The lethal allele is associated with straight hair

The most appropriate combination of statements to provide genetic ex-planation for this result is

1 B and C2 A only3 B C and D4 A C and D

Answer 2

Explanation As the total no of progeny is 6+2+3+1 = 12 definitely 4 prog-eny has died hence lethal alleles are involved in the cross The ratio is not as that of recessive epistasis 934 hence statement A is incorrect B is correct as we are not getting disproportionate number of parental and recombinant prog-

eny indicating genes exhibiting independent assortment As we look into the ratio there is decrease in the number of progeny yellow straight (6 instead of 9) and yellow fuzzy(2 instead of 3) All four progeny who have YY do not sur-vive hence mutation is dominant lethal not recessive D is incorrect because all progeny with straight hair survive Reference Page-370 iGenetics a molecular approach Peter J Russell

-- 3rd ed

111 A family was examined for a given trait which is represented in the pedigree shown below Further the degree of expression of trait is highly variable among members of the family some are only slightly affected while others developed severes symptoms at an early age

The following statements are made to explain the pattern of inheritance

shown in the pedigree

A X-linked dominant mutationB X-linked recessive mutationC Mitochondrial inheritanceD Variable expression can be due to heteroplasty

The best possible explanation for this inheritance is

1 A and D2 C and D3 B only4 A only

Answer 2

Explanation In the pedigree it is clearly shown that whenever mother is affected 100 of the progenies are affected irrespective of their sexes and also affected father is not inheriting the trait to any progeny This is definitely because of Mitochondrial inheritance as it is only inherited from the mother through egg cytoplasm Since the amount of defective mitochondria might be unequally distributed during oogenesis a process called as heteroplasmy it affects the degree of expression of the trait So both statement C and D are correct

Reference httpswwwsciencedirectcomtopicsmedicine-and-dentistrymater-

nal-inheritance

112 The location of six deletions (shown as solid line underneath the chromosome) has been mapped to the Drosophila chromosome as shown in the diagram given below Recessive mutations a b c de f and g are known to be located in the



31Phone 080-5099-7000 Toll Free 1800-1200-1818

region of deletions but the order of mutations on the chromosome is not known When flies homozygous for the recessive mutations are crossed with flies homozygous for deletion the following results were obtained where the letter lsquomrsquo represents a mutant phenotype and lsquo+rsquo represents the wild type

The relative order of the seven mutant genes on chromosome is

1 b c e a f g d2 b c d f g e a3 b c d e a f g4 c d e a g f b

Answer 3

Explanation The key principle here is that point mutations can recombine with deletions that do not extend past the mutation but they cannot recombine to yield wild-type phages with deletions that do extend past the mutation In a simpler way we can find the sequence by observing the length of mutation and which gene fuctions are getting affected by that and then comparing the overlapping regions we can predict the correct sequence In this question if we focus on the last gene in the map it is affected by only deletion 5 In the table function of g gene is affected by only deletion 5 So g must be last in the sequence Similarlyby overlapping the deletions and comparing the loss of funtion of genes we can predict the sequence which is option 3 here The correct order is bcdeafg


113 In the following pedigree individuals with shaded circle or shaded square show presence of a recessive autosomal trait

The calculated risk of occurrence of this trait for III-1 is

1 122 143 184 13

Answer 4

Explanation The trait given is autosomal recessive In this pedigree the manrsquos (II-3) parents must have both been heterozygotes Aa because they pro-duced affected aa child So among expected progeny of manrsquos (II-3) parents

(AA(14) Aa(12)aa(14) 23 probability that the man (II-3) is a carrier of the trait The woman II-4 is affected by the trait so she must have genotype aa So expected progeny genotype is Aa or aa So overall probability of the son to have the disease is 23 x 12 = 13


114 During the course of vertebrate evolution the jaw bones got mod-ified into three ear ossicles in mammals Which one of the following is a correct match of ear ossicle and its ancestral jaw bone

1 Stapes - Articular Incus - Hyomandibular Malleus - Quadrate 2 Stapes - Quadrate Incus - Articular Malleus - Hyomandibular3 Stapes - Quadrate Incus - Hyomandibular Malleus - Articular4 Stapes - Hyomandibular Incus - Quadrate Malleus - Articular

Answer 4

Explanation From fossil data comparative anatomy and developmental biology it is clear that the bones in the mammalian middle ear the malleus and incus are homologous to the quadrate and articular which form the ar-ticulation for the upper and lower jaws in non-mammalian jawed vertebrates The evolution of the stapes from the hyomandibula was an earlier and distinct event (1 2)

Reference httpswwwncbinlmnihgovpmcarticlesPMC3552421 httpwwwsfucabiologycoursesbisc316outlinesjawlecturehtml

115 The CO2 dissociation curves of oxygenated and deoxygenated blood are given along with dissolved CO2 below Following are the statements deduced from the curves above andor

based on the knowledge about CO2 transport which may or may not be correct

A The deoxygenated haemoglobin has greater affinity for CO2 than ox-ygenated haemoglobinB The deoxygenated haemoglobin does not bind with free H+ ions re-

leased during the formation of HCO3- from CO2 C The haemoglobin saturation with O2 has no effect on CO2 dissocia-

tion curveD O2 and CO2 bind to haemoglobin at different sites

Which one of the following options represents a combination of all cor-rect statements

1 A and B 2 B and C3 C and D4 A and D

Answer 4Explanation Carbon dioxide combines rapidly to the terminal uncharged



32Phone 080-5099-7000 Toll Free 1800-1200-1818

amino groups (R-NH2) to form carbamino compoundsHemoglobin is made up of four symmetrical subunits and four heme groups Iron associated with the heme binds oxygen so statemnet D is correctDeoxygenated hemoglobin has a higher affinity for CO2 because it is a better proton acceptor than oxy-genated hemoglobin so statemnet A is correct Therefore when hemoglobin is deoxygenated (ie at tissues) there is a right shift of the carbonic acid-bi-carbonate buffer equation to produce H+ which in turn increases the amount of CO2 which can be carried by the blood back to the lungs to be exhaled hydrogen ions bind easily to reduced haemoglobin which is made available when oxygen is released so statement B is incorrect

Reference httpsacademicoupcombjaedarticle56207331369httpswwwopenanesthesiaorghaldane_effect

116 Given below are the intervalsduration of electrocardiogram of a human subject (Column A) and the event in heart during the process (Column B)

Which one of the following options is a correct match of entries in Col-

umn A and B

1 a - (i) b - (iv) c - (ii) d - (iii)2 a - (ii) b - (iii) c - (i) d - (iv)3 a - (iv) b - (ii) c - (iii) d - (i)4 a - (iii) b - (i) c - (iv) d - (ii)

Answer 2

Explanation The QRS complex represents rapid ventricular depolarization as the action potential spreads through ventricular contractile fibers The SndashT segment which begins at the end of the S wave and ends at the beginning of the T wave represents the time when the ventricular contractile fibers are depolarized during the plateau phase of the action potential PR includes both atria and ventricular conduction The P wave represents atrial depolarization which spreads from the SA node through contractile fibers in both atria The second wave called the QRS complex begins as a downward deflection con-tinues as a largeupright triangular wave and ends as a downward wave The QRS complex

represents rapid ventricular depolarization The QndashT interval extends from the start of the QRS complex to the end of the T wave It is the time from the be-ginning of ventricular depolarization to the end of ventricular repolarization

Reference Principles of Anatomy and Physiologyby G J Tortora Pg 707-708

117 The pathway of synthesis of aldosterone in zona glomerulosa along with the intracellular locations is shown below

The enzymes below are required for different steps of aldosterone

Which one of the following options represents correct matched for A B and C

1 A - (i) B - (ii) C - (iii)2 A - (iii) B - (i) C - (ii)3 A - (ii) B - (iii) C - (i)4 A - (ii) B - (i) C - (iii)

Answer 3

Explanation

Reference httpsjasnasnjournalsorgcontent1581993

118 Given below is a figure of proopiomelanocortin (POMC) polypep-tide and its cleavage products ( marked as A B C D) which have differ-ent hormonal activities The names of the cleaved products obtained from POMC are shown in the table below the diagram Which one of the following options represents A B C and D correctly

1 A - (i) B - (ii) C - (iii) D - (iv)2 A - (ii) B - (iii) C - (i) D - (iv)3 A - (i) B - (iv) C - (iii) D - (ii)4 A - (iii) B - (ii) C - (iv) D - (i)

Answer 1Explanation The POMC gene is expressed in both the anterior and interme-



33Phone 080-5099-7000 Toll Free 1800-1200-1818

diate lobes of the pituitary gland The primary protein product of the POMC gene is a 285 amino acid precursor that can undergo differential processing to yield at least 8 peptides dependent upon the location of synthesis and the stim-ulus leading to their production Cleavage sites are indicated by the numbers 1 to 7 and consist of the sequences Arg-Lys Lys-Arg or Lys-Lys Adrenocorti-cotropic hormone (ACTH) [1-39 aa] and b-lipotropin [42-134aa] are products generated in the corticotrophic cells of the anterior pituitary under the control of corticotropin releasing hormone (CRH) Alpha-melanocyte stimulating hor-mone (a-MSH) [1-13aa] corticotropin-like intermediate lobe peptide (CLIP) g-lipotropin and b-endorphin [ 104-134aa] are products generated in the inter-mediate lobe of the pituitary under the control of dopamine a- b- and g-MSH are collectively referred to as melanotropin or intermedin The numbers in parentheses of hormone indicate the amino acids of POMC present in each

Reference Reference Module in Neuroscience and Biobehaviour-al Psychology Encyclopedia of Neurosciene (2009) Pages-1139-1141 Doi101016B978-008045046-901197-9

119 The table given below lists types of plant communities and types of growth forms Which of the following is the best match for the plant communities with most dominant growth form generally present in that community

1 i - D ii - A iii - B iv - C

2 i - C ii - A iii - D iv - D3 i - B ii - C iii - D iv - C4 i - C ii - A iii - D iv - C

Answer 4

Explanation Generally the biological spectra are worked out and compared with Raunkiaerrsquos normal spectrum The percentage of phanerophytes in differ-ent floras ranges from zero to over 74 in the tropical rain forest The higher percentage of phanerophytes is indicative of phanerophytic climate High per-centage of chamaephytes (above 50) indicates an extremely cold climateHigh percentage of hemicryptophytes in temperate forest vegetation indicates the conditions favourable for the development of extensive grassland

Reference httpwwwbiologydiscussioncomecologyphysiogno-my-life-form-classes-and-phytoclimatic-spectrum6772

120 Following are some of the generalizations regarding energy flow in an ecosystem

A Assimilation efficiency of carnivores is higher than herbivoresB Consumption efficiency of aquatic herbivores is higher than terrestri-

al herbivoresC Vertebrates have higher production efficiencies than invertebratesD Trophic-level transfer efficiency is higher in terrestrial food chains

than in marine

Based on the above select the correct option

1 Only A and C2 Only A and B3 A B and C

4 A C and D

Answer 2

Explanation Assimilation efficiency depends on the quality of the food source and the physiology of the consumerUnassimilated food returns as fec-es Carnivores have higher assimilation efficiency (about 80 percent) than do terrestrial herbivores (5 to 20 percent)Thus statement A is correct Production Efficiency = Net productionassimilationInvertebrates are better than verte-brates at this and cold blooded species do better than warm blooded species Vertebrates spend a huge amount of energy obtaining food and the basal met-abolic rate of warm blooded organisms uses a lot of energyHence statement C is wrong

Reference httpsglobalchangeumicheduglobalchange1currentlec-turesklingenergyflowenergyflowhtml

121 In an experiment to show that biogeochemical cycles interact nitro-gen fixing vines (Galactia sp) were grown in plots under normal levels of CO2 (Control) and under artificially elevated atmospheric CO2 (Experi-mental) Effect of elevated CO2 levels on nitrogen fixation was measured over a period of 7 years (Plot A) and the concentrations of iron and mo-lybdenum in the leaves of these plants were quantified at the end of the study (Plot B)

Which one of the following inferences CANNOT be made from the

above experiment

1 Decreasing rate of N-fixation correlates with decreased levels of leaf iron and molybdenum two micronutrients essential for N-fixation2 An initial exposure to elevated CO2 increased N-fixation by these

plants3 There is a continuous decrease in N-fixation due to elevated CO2 treat-

ment4 Plants exposed to continuous elevated levels of CO2 had lower levels

of iron and molybdenum in their leaves

Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818

Explanation Nitrogen-fixing plant species may respond more positively to elevated atmospheric carbon dioxide concentrations ([CO2]) than other spe-cies because of their ability to maintain a high internal nutrient supply Elevat-ed CO2 increased nodule size and number specific nitrogenase activity and plant N content and consequently increased biomass andor seed yield in leg-umes Also iron and molybdenon are essential cofactors required for the en-zyme nitrogenase which fixes nitrogen Hence if we observe plot A and B the correct statements are 1 and 4 as elevated CO2 enhances nitrogen fixation and hence utilization of these micronutrients increases Now lets compare state-ment 2 and 3 statement 2 makes more justice to the graph as initial exposure to high CO2 is increasing nitrogen fixation While in statement 3- in the 6th and 7th year it is becoming more or less stable hence statement 3 is incorrect

Referencehttpswwwfrontiersinorgarticles103389fpls201701546full

122 Match the following invasive plants to the likely habitats in which they are expected to occur

1 A - ii B - i C - iii2 A - i B - iii C - ii3 A - iii B - ii C - ii4 A - iii B - i C - ii

Answer 3

Explanation Lantana camara is found predominantly in tropical and sub-tropical environments but also capable of growing in warmer temperate and semi-arid regions whereas Eichhornia crassipes is a free floating aquatic plant that has invaded aquatic areas throughout the eastern and southern por-tions of the United States Eichhornia crassipes invades lakes ponds rivers marshes and other types of wetland habitats Puosopls juliflora is called as mesquite which usually invade Arid and semi-arid habitats

Reference httpswikibugwoodorgEichhornia_crassipes httpskeyserverlucidcentralorgweedsdatamediaHtmllantana_ca-marahtml httpswwwcabiorgiscdatasheet43942

123 Incorporating additional ecological factors into the Lotka-Voltera predator-prey model can change the predator isocline Given below are three state-space graphs (A-C) representing modification of predator iso-cline due to the ecological factors listed below (i-iii)

(i) Victim abundance acting as predator carrying capacity(ii) Availability of alternate prey (victim) population(iii) Predator carrying capacity determined by factors other than victim

abundance

Which one of the following options represents all correct matches of the state-space graphs with their ecological factors

1 A - (ii) B - (iii) C - (i)2 A - (ii) B - (i) C - (iii)3 A - (iii) B - (ii) C - (i)4 A - (i) B - (ii) C - (iii)

Answer 2

Explanation For the first statement given (i)-B as If prey abundances are not strongly positively correlated thenas one prey species becomes scarce the predator can continue to feed and

increase its population sizeIn this Unstable equilibriumpredators drive prey to extinction Predator carrying capacity predator can

no longer drive prey to extinction Stable coexistence as there are alternate prey options present for the predators (A-ii) In the third statement as the slope stays constant hence predator is independent of that of the presence of prey and therefore predator carrying capacity is dependent upon factors other than victim abundance

Reference httpwwwlifeillinoiseduib453453lec8predation2pdf

124 Birds in a population show two foraging phenotypes A and B Birds of phenotype A search attack and capture prey while birds of phenotype B steal prey from birds of phenotype A A and B are maintained in the population through negative frequency dependent selection The graph below shows the fitness of A (broken line) and B (solid line) at different relative frequencies of A (frequency of B = 1 - frequency of A)

Which of the following statements does the graph support

1 A outcompetes B at equilibrium A goes to fixation2 B outcompetes A at equilibrium B goes to fixation3 A and B are both maintained in the population the equilibrium fre-

quencies are A = 06 B = 04 4 A and B are both maintained in the population the equilibrium fre-

quencies are A = 09 B = 01



35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3

Explanation As you can clearly see from the graph that at equilibrium ie at the point when A and B meets at that point frequency of A=06 hence B=04 ie ( Frequency of B= 1-Frequency of A) hence 3rd is the right answerNega-tive frequency-dependent selection is a powerful type of balancing selection that maintains many natural polymorphisms but it is also commonly misin-terpreted

Reference httpswwwfrontiersinorgarticles103389fevo201800010full

125 Antelopes are proposed to form groups to reduce the risk of pre-dation A researcher measured the predation of individuals in groups of different sizes She found that per capita mortality risk decreased with in-creasing group size for males (solid line) but remained unchanged for fe-males (dashed line) Furthermore males in all groups experienced greater per capita mortality risk than females Identify the graph below that best depicts the above findings

Answer 4

Explanation per capita mortality risk decreased with increasing group size of males - so the solid line must have a decreasing slope from a higher valueper capita mortality risk was constant with increasing group size of females - so the dashed line must have constant slope So graph 2 is rejectedNow males in ALL GROUPS must have greater per capita mortality risk

than females - so the solid line must never have a y axis value below the dashed lineSo graph 4 is correct (In graph 3 it is seen that some males are having lower

mortality risk than females)

Reference httpscourseslumenlearningcomphysicschapter2-8-graphical-anal-

ysis-of-one-dimensional-motion

126 The prominent mammal species found in four different protected areas are listed below

Area A Tiger Wild dog Leopard Elephant

Area B Common langur Barking deer Wild dog ElephantArea C Tiger Indian rhinoceros Pygmy hog Wild pigArea D Blackbuck Indian gazelle Hyena Indian wolf

The area with the greatest phylogenetic diversity is

1 A2 B3 C4 D

Answer 2

Explanation All the four animals belong to 4 different mammalian orders thus exhibiting maximum diversity Common langur- Primates Barking deer-Artiodactyla Wild dog- Carnivora Elephant- Proboscidea

Reference httpswwwmammalsofindiaorg

127 The Hardy-Weinberg principle states that allele frequencies in a population will remain constant over generations if certain assumptions are met

A Random matingB Mate choiceC Small population sizeD Large population sizeE Lack of mutationsF Directional selection

Which of the above factors will cause changes in allele frequencies over generations

1 A D and F2 B D and F3 A C and E4 B C and F

Answer 4

Explanation Five assumptions of a Hardy-Weinberg population1) No Selection2) No Mutation3) No Gene Flow-No migration between populations to change allele fre-

quencies4) Infinite Population is necessary because the larger the population size is

the harder to change the allele frequency5) Random Mating which is pretty much what it says Mating being based on

nothing other than pure chance making it randomHence any deviation from these assumptions will cause allele frequencies to change over time

Reference httpwallacegeneticsugaedugroupsevol3000wikiadcc5HardyWeinberg_Principlehtml

128 Given below are few traits and related functions Match the above given traits to their most likely function

1 i - C ii - D iii - B iv - A



36Phone 080-5099-7000 Toll Free 1800-1200-1818

2 i - D ii - B iii - A iv - C3 i - B ii - D iii - A iv - C4 i - C ii - A iii - D iv - B

Answer 3

Explanation The term aposematism is commonly used as a synonym for warning coloration (ie something that is aposematic is warningly colored) The word literally means away signal Aposematism is the combination of a conspicuous signal and an unprofitable trait in a prey species(i-B)lsquoBask-ingrsquo is the most conspicuous thermoregulatory behavior in many animals that enable them to enhance physiological performance(ii-D) The cooperative hunting helps in acquiring food for examples species like Hyena Lions they hunt in packs so as to be efficient in acquiring food in short time (iii-A)Using the responses of territory owners to playback to infer the territorial function of acoustic signals is common practiceBIRD song is generally considered to have one or both of two main functions attracting a mate and proclaiming a territory The great tit (Parus major) is a typical songbird in that song is pri-marily (iv-B)

Reference httpswwwencyclopediacomhistorylatin-america-and-carib-beanguyana-historyaposematism httpswwwsciencedirectcomsciencearticlepiiS2351989417301889 httpswwwcbsumnedusitescbsumnedufilespublicdownloadsEvolution_of_Cooperative_Huntingpdf httpswwwncbinlmnihgovpmcarticlesPMC2662740 httpswwwnaturecomarti-cles271539a0

129 Following is a diagrammatic representation of human evolutionary tree

In the above diagram A B C and D respectively represent

1 Denisovan Homo habilis Homo erectus Homo neanderthalensis2 Homo habilis Homo erectus Homo neanderthalensis Denisovan3 Homo erectus Homo habilis Homo neanderthalensis Denisovan4 Homo erectus Denisovan Homo neanderthalensis Homo habilis

Answer 2

Explanation he famous paleoanthropologist named Louis Leakey along with his team discovered Homo habilis (meaning handy man) in 1964 Homo habilis was the most ancient species of Homo ever found In Asia in 1891 Eu-gene Dubois (also a paleoanthropologist) discovered the first fossil of Homo erectus (meaning upright man) which appeared 18 million years ago This fossil received several names The best known are Pithecanthropus (ape-man) and Sinanthropus (Chinese-man) Homo erectus appeared in East Africa and migrated to Asia where they carved refined tools from stoneeveral Homo species emerged following H erectus and quite a few coexisted for some time The best known one is Homo neanderthalensis usually called Neanderthals and they were known as the European branch originating from two lineages that diverged around 400000 years ago with the second branch (lineage) Homo sapiens known as the African branchDenisovans are a recent addition to the human tree In 2010 the first specimen was discovered in the Denisova cave in south-western Siberia Very little information is known on their behav-ior They deserve further studies due to their interactions with Neandertals and other Homo species

Reference httpwwwpasttimeorg201401epsode-10-field-guide-

the-hobbit-an-unexpected-discovery httpskidsfrontiersinorgarti-cle103389frym201900022

130 Given below are the names of the animals in column X and the ac-cessory respiratory organs in teleost fishes (column Y)

The correct match of the animals with the accessory respiratory organs they have are

1 A - (iv) B - (iii) C - (ii) D - (i)2 A - (iii) B - (iv) C - (i) D - (ii)3 A - (i) B - (iv) C - (ii) D - (iii)4 A - (ii) B - (i) C - (iii) D - (iv)

Answer 3

Explanation Anabas (lnclian climbing perch) often comes to water Sur-face and gulp down fresh air for aerial respiration lt has two spacious su-prabranchial cavities as dorsal outgrowths of the gill chambers Each of these cavities contains an accessory respiratory organ called labyrinthine organ It is formed by much folded concentric bony plates which develop from the first epibranchial bone This organ is covered by vascular mucous membrane Fresh air reaches the suprabranchial chamber through the mouth and after gas exchange the expiratory air is expelled through the opercular opening In Clarias (Indian cat fish) the wall of the gill chamber on either side evaginates to form a pair of suprabranchial cavities These cavities con-tain highly branched and tree-like accessory air breathing organs called dendritic or arborescent organExchange of respiratory gases occurs through the highly vascularized mucous membrane covering these organs Channa (Ophiocephalus) etc forms sac-like outgrowth called diverticulum which is lined by vascular epithelium Sometimes the sac extends above the gill pouchAir is drawn into the diverticulum and oxygen and carbon dioxide are exchanged Amphiopnous possesses a small and smooth diverticulumwhich opens through midventral gill slits and respiration occurs through air sacs

Reference Page-206 3rd edition Modern Textbook of Zoology Verte-brates RL Kotpal Rastogi Publictions ISBN 81-7133-891-7

131 The terms expressing some of the developmental events or specific body structures are given in column X and the names of animals that are associated with them in column Y

The correct match of the terms in Column X with the name of animals

in Column Y is

1 A - (i) B - (ii) C - (iii) D - (iv)2 A - (iv) B - (ii) C - (iii) D - (i)3 A - (iii) B - (i) C - (ii) D - (iv)4 A - (ii) B - (iv) C - (i) D - (iii)Answer 2



37Phone 080-5099-7000 Toll Free 1800-1200-1818

Explanation Torsion is the rotation of the visceral mass and foot 180 degrees with respect to the head and foot and is a unique synapomorphy of modern gas-tropods Torsion occurs during development in all gastropods usually in the late veliger stage In Obelia both hydroid and medusoid phases are diploid Thus infact there is no true alternation of generations in Obelia The medusa is only a modified zooid that swims freely This free swimming medusa is very useful to Obelia because it causes wide dispersal of gametes Secondly the gametes carried in the gonads of the medusae do not actually originate there These originate in the ectoderm of blastostyle develop and later migrate into medusae and enter the gonads This it is impossible to differentiate between sexual and asexual generations in Obelia The alternation of fixed (hydroid) and free swimming (medusae) phases is only an example of polymorphism The term metagenesis is used for such alternation of fixed and free phases (Alternation between two diploid phases is known as metagenesis) In Taenia the gravid proglottids are detached and pass through host faecal matter by a process called apolysis Pedicellariae of sea stars are minute whitish jaw like structures found on both body surfaces in association with spines

Reference Page-283355 800814 10th edition Modern Textbook of Zo-ology Invertebrates RL Kotpal Rastogi Publictions ISBN 978-81-7133-903-7

132 The table lists characteristic anatomical features and names of plants Choose the option that correctly matches plant with their characteristic

feature

1 i - C ii - B iii - D2 i - A ii - C iii - D3 i - C ii - A iii - B4 i - A ii - B iii - D

Answer 4

Explanation Siphonostele the pith is surrounded by the vascular tissue The concentric inner phloem cylinaer surrounds the central pith Next to the inner phloem is the concentric xylem cylinder which is immediately surrounded by outer phloem cylinder (eg in Marsilea)(B-ii) Protostele is found not only in ancient plants like Rhynia and Psilophyton but also in stems of Lycopodium some ferns and in many kinds of dicotyledonous roots(A-i) According to Brebner (1902) there is one more modification of the siphonostele known as eustele Here the vascular system consists of a ring of collateral or bicollat-eral vascular bundles situated on the periphery of the pith In such steles the inter-fascicular areas and the leaf gaps are not distinguished from each other very clearly The example of this type is Equisetum(D-iii)

Reference httpwwwbiologydiscussioncomplantsvascular-plantsevolution-of-stelar-forms-in-vascular-plants-botany69221httpwwwbiologydiscussioncomplant-tissuestissue-systemstelar-system-of-plant-definition-and-types-with-diagrams20313

133 The table given below provides a list of female gametophyte features and plant genera

Which one of the following options correctly matches the plant genera to female gametophyte features

1 i - D ii - C iii - A iv - B2 i - D ii - B iii - A iv - C3 i - A ii - B iii - D iv - C4 i - D ii - B iii - C iv - A

Answer 2

Explanation Embryo sacs may be mono-bi- or tetrasporicas well as they can be 4-8- or 16-nucleate Allium type is bisporic 8-nucleate in which case the egg is removed from the megaspore mother cell by four divisions Oeno-thera type is monosporic 4-nucleate when only four divisions intervene be-tween the megaspore mother cell and the egg Peperomia is tetrasporic 16-nu-cleate in which case the egg is removed from the megaspore mother cell by four divisions Polygonum is monosporic 8-nucleate in which case five divi-sions intervene between the megaspore mother cell and the egg

Reference httprepositoryiasacin31369131369pdf

134 Following are some generalizations related to wood anatomy of higher plants

A The axial system of conifer woods consist mainly or entirely of trac-heidsB The rays of conifers typically contain only parenchyma cellsC The rays of angiosperms typically contain both sclerenchyma cells

and tracheidsD Angiosperm wood may be either diffuse-porous or ring-porous


1 A and B only2 A and D only3 B and C only4 C and D only

Answer 2

Explanation In conifers which include pines spruces and firs this sys-tem is composed mainly of tracheids with some parenchyma cells In angi-osperms which include oaks ashes and elms the axial system contains in addition to tracheids and parenchyma vessel members and fibers (1)Vascular tissue of angiosperm is organized into discrete strands called vascular bun-dles each containing xylem and phloem Storage parenchyma and fibres are generally present and sclereids rarely present (2) Angiosperms wood may be either diffuse porous or ring porous (3) A and D options are the only correct statement when compared to other options given in the questions So it is con-cluded that the correct answer is 2

Reference httplifeofplantblogspotcom201012woodhtmlhttpswwwbritannicacomplantangiospermOrganization-of-the-vas-

cular-tissue httpssteurhhomexs4allnlengloofeloofhthtml

135 A field ecologist gathers following data (abundance values) in order



38Phone 080-5099-7000 Toll Free 1800-1200-1818

to study diversity of species in four plant communities

Based on the above observations the ecologist draws following conclu-

sions

A Plant communities C1 and C4 strong similarity with each otherB Plant communities C1 and C4 as well as communities C2 and C3 show

strong similarity with each otherC Plant community C1 is most diverseD Plant community C4 is most diverse

Which of the following statements is correct regarding above conclu-sions

1 All the conclusions are correct2 Only conclusions A and D are correct3 Only conclusions A and C are correct4 Only conclusions B and D are correct

Answer 2

Explanation An index of community similarity that is based on the relative abundance of species within the communities being compared is called as the percent similarity ( PS ) Based on percentage similarity C1 and C2 show strongest similarity The maximum values of the diversity index (Shannon in-dex) occurs when all species are present in equal numbers So here most even distribution of species is seen in C4 so it is the most diverse community among all

Reference Page-323332 Elements of Ecology Eighth Edition Thomas M Smith ISBN 978-0-321-73607-9

136 The blood plasma proteins (albumin and globulins) from a healthy person were separated by electrophoresis as shown above The diagnosis of acute inflammation can be done based on one of the following obser-vations

1 Increase in both α-1 and α-2 decrease in albumin2 Increase in albumin decrease in α-1 α-2 and szlig3 Increase in albumin and decrease in Ɣ-globulin4 Only decrease in albumin

Answer 1

Explanation Analbuminaemia or analbuminemia is a genetically inherited metabolic defect characterised by an impaired synthesis of serum albumin Hypoalbuminemia can result from decreased albumin production defective synthesis because of hepatocyte damage deficient intake of amino acids in-creased losses of albumin via GI or renal processes and most commonly acute or chronic inflammation Both alpha 1 and alpha 2 globulins increase during acute inflammation

Reference httpswwwncbinlmnihgovpubmed3789133 httpsemedicinemedscapecomarticle166724-overview httpswwwucsfben-ioffchildrensorgtests003540html

137 Following statements were given regarding factors influencing var-iation in expression levels of transgene in transgenic plants

A Difference in restriction enzyme sites within the T-DNAB Difference in copy number of the transgeneC Variation in site of integration of the T-DNA within the plant genomeD Presence of multiple promoters within the T-DNA region

Which one of the following options represents a combination of state-ments that would NOT lead to variations in transgene expression levels in transgenic plants generated using the same T-DNAbinary vector

1 A and C only2 B only3 C and D only4 A and D onlyAnswer 4

Explanation The routine generation of single-copy transgenic events is therefore a major goal for agricultural biotechnology Integration of transfect-ed DNA into the plant genome usually takes place by a process of non-homol-ogous or illegitimate recombination This random integration of transgenes but also variations in copy number and the configuration of the transgene which may result in gene silencing have been implicated to explain differenc-es in gene expression levels between individual transformants

Reference httpswwwncbinlmnihgovpmcarticlesPMC2832237 httpsacademicoupcomnararticle261127291132539

138 The above figure shows the fluorescence emission spectra of three different proteins Protein (X) Protein (Y) and Protein (Z) excited at 280nm

Which one of the following statements gives the correct interpretation

1 Proteins (Y) and (Z) have tryptophan while protein (X) only has phe-nylalanine2 Protein (X) has only tyrosine and protein (Y) has tryptophan on the

surface while protein (Z) has tryptophan buried inside3 Protein (X) has tryptophan buried inside while proteins (Y) and (Z)

have tryptophan on the surface4 Protein (X) has only tyrosine and protein (Y) has tryptophan buried

and protein (Z) has tryptophan on the surface

Answer 4Explanation The most common use of Trp fluorescence λmax information



39Phone 080-5099-7000 Toll Free 1800-1200-1818

is to assign a Trp as buried and in a ldquonon-polarrdquo environment if λmax is lt ˜330 nm if λmax is longer than 330 nm the Trp is assigned a ldquopolarrdquo environment which almost always is intended to imply solvent exposure

Reference httpswwwsciencedirectcomsciencearticlepiiS0006349501761838fig1

139 Specimens for light microscopy are commonly fixed with a solution containing chemicals that crosslinkdenature cellular constituents Com-monly used fixatives such as formaldehyde and methanol could act in var-ious ways as described below

A Formaldehyde crosslinks amino groups on adjacent molecules and stabilizes protein-protein and protein-nucleic acid interactionsB Methanol acts as a denaturing fixative and acts by reducing the solu-

bility of protein molecules by disrupting hydrophobic interactionsC Formaldehyde crosslinks lipid tails in biological membranesD Methanol acts on nucleic acids crosslinks nucleic acids with proteins

and thus stabilizes protein-nucleic acid interactions

Which one of the following combinations represents all correct state-ments

1 A and C2 B and C3 B and D4 A and B

Answer 4

Explanation Ethanol (CH3CH2OH) and methanol (CH3OH) are consid-ered to be coagulants that denature proteins They replace water in the tissue environment disrupting hydrophobic and hydrogen bonding thus exposing the internal hydrophobic groups of proteins and altering their tertiary structure and their solubility in water Methanol is commonly used as a fixative for blood films and 95 ethanol is used as a fixative for cytology smears but both alcohols are usually combined with other reagents when used as fixatives for tissue specimens Hence B is correctFormaldehyde is an aldehyde which mildly oxidize reduced carbon or nitro-

gen It creates bonds between proteins lipids or between lipids and proteins Formaldehyde crosslinks protein and DNA Hence statement A is correct

Reference httpswwwleicabiosystemscompathologyleadersfixation-and-fixatives-3-fixing-agents-other-than-the-common-aldehydes

140 Run-off transcription assays were performed to establish the spec-ificity of three novel sigma factors for their promoters Results of the ex-periments are shown below

Following inferences were made from these results

A σA initiates transcription from P2 and σB from P1B σC can initiate transcription from both promotersC σB prevents initiation of transcription from P2D σA initiates transcription from P1

Choose the option that correctly interprets the results

1 A B and C only2 A and B only3 C and D only4 B C and D only

Answer 2

Explanation A run-off transcription assay is an assay in molecular biology which is conducted in vitro to identify the position of the transcription start site (1 base pair upstream) of a specific promoter along with its accuracy and rate of in vitro transcription To perform a run-off transcription assay a gene of interest including the promoter is cloned into a plasmid The plasmid is di-gested at a known restriction enzyme cut site downstream from the transcrip-tion start site such that the expected mRNA run-off product would be easily separated by gel electrophoresis To initiate transcription radiolabeled UTP the other nucleotides and RNA polymerase are added to the linearized DNA Transcription continues until the RNA polymerase reaches the end of the DNA where it simply ldquoruns offrdquo the DNA template resulting in an mRNA fragment of a defined length This fragment can then be separated by gel electrophore-sis alongside size standards and autoradiographed The corresponding size of the band will represent the size of the mRNA from the restriction enzyme cut site to the transcription start site (+1) Which means σA has transcribed from promoter P2 since band is visible whereas σB has transcribed P1 since band is visible only in P1 σc band is visible in both P1 and P2 so it can initiate transcription at both promoters

Reference httpswwwrevolvycompageRun252Doff-transcription

141 Based upon phenotypic observation it was concluded that an un-known gene responsible for an agronomically important trait is present in a particular plant In order to identify the gene a researcher proposes to use the following strategies

A PCR amplification of the geneB Map based cloning of the geneC Subtractive DNA hybridizationD Genome sequencingE Develop molecular markers linked to the trait

Which one of the following options is most suitable for identifying the unknown gene

1 A and C2 B and E3 C only4 A and D

Answer 2

Explanation An unknown gene responsible for a certain character can be identified through map based cloning and molecular markers For example With the construction of a linkage map to a genomic region linked to vernal-ization response in carrot the region was mapped in a linkage group with 78 molecular markers and flanking markers were at 069 and 079 cM This map could be used to develop molecular markers linked to the trait and also in the future to start physical mapping and sequencing using a carrot BAC library

Reference httprepositoriosdigitalesmincytgobarvufindRecordBDUNCU_8c905b45929767664fcbb78ce030785a

142 A T0 transgenic plant contains two unlinked copies of the T-DNA of which one is functional and the other is silenced Segregation of the transgenic to non-transgenic phenotype would occur in a ratio in prog-eny obtained by backcrossing and in a ratio in F1 progeny obtained by self-pollination

Fill in the blanks with the correct combination of (i) and (ii) from the



40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below

1 (i) - 31 and (ii) - 1512 (i) - 11 and (ii) - 313 (i) - 31 and (ii) - 314 (i) - 11 and (ii) - 151

Answer 2

Explanation there are 2 copies of T-DNA the which is functional lets take it as D and the non-functional to be d now since this cell contains 2copis it is Dd when Dd is backcrossed with either of the parent cell having either Dd the ratio will come as 11 when Dd is selfed the ratio comes as 31 it will never come as 151 because in that case both the genes are dominant which is not the case stated here

Reference httpswwwbritannicacomsciencebackcrosshttpswwwbritannicacomscienceplant-breedingBreeding-self-polli-

nated-species

143 Column X lists two diseases and Column Y lists name of proteins which are commonly used for routine clinical diagnosis of these diseases

Find out the correct combination

1 A - ii B - i A - iv B - iii2 A - iii B - iv A - ii B - i3 A - i B - ii A - iii B - iv4 A - iv B - iii A - i B - iiAnswer 1

Explanation In the early 1970s the World Health Organization (WHO) had defined the term myocardial infarction by the presence of 2 of the 3 following characteristics i) Symptoms of acute ischemia (chest pain) ii) development of Q waves in electrocardiogram (ECG) and iii) increase of enzymes in the blood [combination of total creatine kinase (CK) CK-myocardial band (MB) aspar-tate aminotransferase (AST) and lactate dehydrogenase (LDH)] However in 1999 the Joint European Society of Cardiology and the American College of Cardiology Committee jointly proposed the new definition for myocardial in-farction emphasizing the importance of sensitive and serological biomarkers for the diagnosis of acute myocardial infarction (AMI) and introduced cardiac troponins (cTn) as the gold standard Serum amylase remains the most com-monly used biochemical marker for the diagnosis of acute pancreatitis along-side pancreatic lipase However we have new markers developed


144 Given below are the steps to assess the population size of grasshop-pers in a given area

A lsquonrsquo individuals are collected randomly from the study area in a de-fined period of timeB The captured individuals are counted marked and released at the site

of collection Next day individuals are captured from the same site for same length of time Number of marked (nm) and unmarked (nu) individ-uals are separated and countedC This capture-release and recapture is continued till one gets 100

marked individuals

D The size of the population (N) is estimated as followsE The size of the population (N) is estimated as follows

The most appropriate combination of steps for estimating population

size using mark-recapture method is

1 A B and D2 A B and E3 A B C and D4 A B C and EAnswer 1

Explanation n= collected randomly first and marked In the next catch nm=marked and nu=unmarked hence total=(nm+nu) Total population is N Hence if nm are number of individual marked from the total population (nm+nu) Therefore the previous marked ien were from how much total pop-ulation N Just by simple proportionality theory the answer comes similar to statement D

Reference httpwww2nauedulrm22lessonsmark_recapturemark_recapturehtml httpswwwresourceaholiccom201610capturerecap-turehtml

145 A neurophysiologist was interested in using the patch-clamp tech-nique Following statements are related to this technique

A Intracellular movement of ion channelsB Post-translational modification of the ion channel proteinC Ligand that controls the opening or closing of ion channelsD Change in current flow in a single ion channel

Which one of the following combinations will be achievable using the patch-clamp technique

1 A and B2 B and C3 C and D4 D and A

Answer 3

Explanation Every cell expresses ion channels but the most common cells to study with patch-clamp techniques include neurons muscle fibers cardio-myocytes and oocytes overexpressing single ion channels To evaluate single ion channel conductance a microelectrode forms a high resistance seal with the cellular membrane and a patch of cell membrane containing the ion chan-nel of interest is removed Alternatively while the microelectrode is sealed to the cell membrane this small patch can be ruptured giving the electrode electrical access to the whole cell Voltage is then applied forming a voltage clamp and membrane current is measured Current clamp can also be used to measure changes in membrane voltage called membrane potential Voltage or current change within cell membranes can be altered by applying compounds to block or open channels These techniques enable researchers to understand how ion channels behave both in normal and disease states and how different drugs ions or other analytes can modify these conditions

Reference httpswwwmoleculardevicescomapplicationspatch-clamp-electrophysiologygref



41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



2Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer Key For CSIR NET June 2019

Booklet Code - C



3Phone 080-5099-7000 Toll Free 1800-1200-1818


BOOKLET CODE C




Answer 4




1 80002 10003 40004 6000

Answer 3



Answer 2





Answer 1






Answer 1





4Phone 080-5099-7000 Toll Free 1800-1200-1818


1 7002 8003 6004 110

Answer 1









1 102 203 284 30

Answer 3







1 t2 2t3 πt4 2πt

Answer 4



1 2r


Answer 3





5Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 1




1 02 303 704 80

Answer 3



1 82 243 324 60

Answer 2



1 22 43 84 10Answer 4




Answer 3

Explanation


Answer 1





Answer 4



6Phone 080-5099-7000 Toll Free 1800-1200-1818



7Phone 080-5099-7000 Toll Free 1800-1200-1818

1 172 18

3 194 20

Answer 3


1 142 1123 1364 155

Answer 4







the sunrise

Answer 2



2 43 54 7

Answer 4



PART B



Answer 2





Answer 3




8Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2





Answer 4












both strands as DNA

Answer 2





clusively positive

Answer 4











9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






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11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





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Answer 1





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1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








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Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



3Phone 080-5099-7000 Toll Free 1800-1200-1818


BOOKLET CODE C




Answer 4




1 80002 10003 40004 6000

Answer 3



Answer 2





Answer 1






Answer 1





4Phone 080-5099-7000 Toll Free 1800-1200-1818


1 7002 8003 6004 110

Answer 1









1 102 203 284 30

Answer 3







1 t2 2t3 πt4 2πt

Answer 4



1 2r


Answer 3





5Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 1




1 02 303 704 80

Answer 3



1 82 243 324 60

Answer 2



1 22 43 84 10Answer 4




Answer 3

Explanation


Answer 1





Answer 4



6Phone 080-5099-7000 Toll Free 1800-1200-1818



7Phone 080-5099-7000 Toll Free 1800-1200-1818

1 172 18

3 194 20

Answer 3


1 142 1123 1364 155

Answer 4







the sunrise

Answer 2



2 43 54 7

Answer 4



PART B



Answer 2





Answer 3




8Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2





Answer 4












both strands as DNA

Answer 2





clusively positive

Answer 4











9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



4Phone 080-5099-7000 Toll Free 1800-1200-1818


1 7002 8003 6004 110

Answer 1









1 102 203 284 30

Answer 3







1 t2 2t3 πt4 2πt

Answer 4



1 2r


Answer 3





5Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 1




1 02 303 704 80

Answer 3



1 82 243 324 60

Answer 2



1 22 43 84 10Answer 4




Answer 3

Explanation


Answer 1





Answer 4



6Phone 080-5099-7000 Toll Free 1800-1200-1818



7Phone 080-5099-7000 Toll Free 1800-1200-1818

1 172 18

3 194 20

Answer 3


1 142 1123 1364 155

Answer 4







the sunrise

Answer 2



2 43 54 7

Answer 4



PART B



Answer 2





Answer 3




8Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2





Answer 4












both strands as DNA

Answer 2





clusively positive

Answer 4











9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818


Referencehttpghrnlmnihgovprimergenomicresearchsnphttpspdfssemanticscholarorge914794a0aefb2176fc384d0e70a03b-577f1a5d5pdfHuman Molecular Genetics 10 Chapter 13




Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



5Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 1




1 02 303 704 80

Answer 3



1 82 243 324 60

Answer 2



1 22 43 84 10Answer 4




Answer 3

Explanation


Answer 1





Answer 4



6Phone 080-5099-7000 Toll Free 1800-1200-1818



7Phone 080-5099-7000 Toll Free 1800-1200-1818

1 172 18

3 194 20

Answer 3


1 142 1123 1364 155

Answer 4







the sunrise

Answer 2



2 43 54 7

Answer 4



PART B



Answer 2





Answer 3




8Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2





Answer 4












both strands as DNA

Answer 2





clusively positive

Answer 4











9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg


6Phone 080-5099-7000 Toll Free 1800-1200-1818



7Phone 080-5099-7000 Toll Free 1800-1200-1818

1 172 18

3 194 20

Answer 3


1 142 1123 1364 155

Answer 4







the sunrise

Answer 2



2 43 54 7

Answer 4



PART B



Answer 2





Answer 3




8Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2





Answer 4












both strands as DNA

Answer 2





clusively positive

Answer 4











9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



7Phone 080-5099-7000 Toll Free 1800-1200-1818

1 172 18

3 194 20

Answer 3


1 142 1123 1364 155

Answer 4







the sunrise

Answer 2



2 43 54 7

Answer 4



PART B



Answer 2





Answer 3




8Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2





Answer 4












both strands as DNA

Answer 2





clusively positive

Answer 4











9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



8Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2





Answer 4












both strands as DNA

Answer 2





clusively positive

Answer 4











9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



9Phone 080-5099-7000 Toll Free 1800-1200-1818




1 202 303 504 60

Answer 4





Answer 3






Answer 4





Answer 2





Answer 4





10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



10Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 2






Answer 2





Answer 1






Answer 2





Answer 2





Answer 3






11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg


11Phone 080-5099-7000 Toll Free 1800-1200-1818



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



12Phone 080-5099-7000 Toll Free 1800-1200-1818



B9780128115626000049



Answer 4





Answer 4






Answer 2







Answer 1





Answer 3




1 4322 3163 14



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



13Phone 080-5099-7000 Toll Free 1800-1200-1818

4 332

Answer 3








Answer 4





Answer 1








Answer 3





Answer 4





Answer 4







14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



14Phone 080-5099-7000 Toll Free 1800-1200-1818

1 H1N12 H7N73 H3N24 H5N1

Answer 4





Answer 3



=S1415-47572004000400019





phorus limitation

Answer 2




speciation





Answer 3






Answer 2







lationAnswer 2



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



15Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 3








genotypes

Answer 3







Answer 4









eye and nostril

Answer 1








Answer 1




16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



16Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





species

Answer 4








1 0622 00623 0314 0031

Answer 4



pdf






Answer 3





Answer

Explanation

Reference




17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



17Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 13









PART C





Answer 1








18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



18Phone 080-5099-7000 Toll Free 1800-1200-1818










Answer 2










Answer 1







locus

Answer 3





Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg


Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 2








Answer 4





1 702 503 604 40

Answer 4

Explanation









Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg


Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 3









Answer 1






Answer 4






Answer 2







21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



21Phone 080-5099-7000 Toll Free 1800-1200-1818



Answer 4









Answer 2









Answer 3










Answer 1


Reference




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg




22Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 3







the nucleus

Answer 3





the lacZYA mRNA




Answer 4








Answer 3










Answer 4



23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg


23Phone 080-5099-7000 Toll Free 1800-1200-1818



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



24Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4













Answer 4










Answer 2






25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



25Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3












Answer 4





correct



ICM

Answer 1




26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



26Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 2










Answer 4










Answer 2










Answer 3





27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



27Phone 080-5099-7000 Toll Free 1800-1200-1818

parallel spms









Answer 4








Answer 3









Answer 2










Answer 3




28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



28Phone 080-5099-7000 Toll Free 1800-1200-1818







Answer 1







1 a - (iv) b - (ii) c - (i) d - (iii)


Answer 3







trus stage

Answer 2













29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



29Phone 080-5099-7000 Toll Free 1800-1200-1818



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



30Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 2











Answer 2



-- 3rd ed







Answer 2



nal-inheritance




31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



31Phone 080-5099-7000 Toll Free 1800-1200-1818




Answer 3





1 122 143 184 13

Answer 4






Answer 4













32Phone 080-5099-7000 Toll Free 1800-1200-1818





umn A and B


Answer 2








Answer 3

Explanation







33Phone 080-5099-7000 Toll Free 1800-1200-1818






Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












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Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










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Answer 3






Answer 2







Answer 3





in Column Y is




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feature


Answer 4






Answer 2








Answer 2







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sions





Answer 2





Answer 1


















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Answer 4









Answer 2







Answer 2







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options given below


Answer 2



nated-species









marked individuals











Answer 3





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biotecnikaorg



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umn A and B


Answer 2








Answer 3

Explanation







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Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












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Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







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sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



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biotecnikaorg



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Answer 4






than in marine



4 A C and D

Answer 2





above experiment





Answer 3



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







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sions





Answer 2





Answer 1


















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Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



33Phone 080-5099-7000 Toll Free 1800-1200-1818



34Phone 080-5099-7000 Toll Free 1800-1200-1818





Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







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sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



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biotecnikaorg



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Answer 3





abundance



Answer 2












35Phone 080-5099-7000 Toll Free 1800-1200-1818

Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







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sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





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biotecnikaorg



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Answer 3




Answer 4










1 A2 B3 C4 D

Answer 2







Answer 4










36Phone 080-5099-7000 Toll Free 1800-1200-1818


Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







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sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



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biotecnikaorg



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Answer 3






Answer 2







Answer 3





in Column Y is




37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







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sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



37Phone 080-5099-7000 Toll Free 1800-1200-1818




feature


Answer 4






Answer 2








Answer 2







38Phone 080-5099-7000 Toll Free 1800-1200-1818



sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



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sions





Answer 2





Answer 1


















39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



39Phone 080-5099-7000 Toll Free 1800-1200-1818









Answer 4









Answer 2







Answer 2







40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



40Phone 080-5099-7000 Toll Free 1800-1200-1818

options given below


Answer 2



nated-species









marked individuals











Answer 3





41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



41Phone 080-5099-7000 Toll Free 1800-1200-1818



42Phone 080-5099-7000 Toll Free 1800-1200-1818

biotecnikaorg



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biotecnikaorg

Booklet Code - C · The one in a white shirt says: “I am a girl” (statement-I) The one in a...

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Transcript of Booklet Code - C · The one in a white shirt says: “I am a girl” (statement-I) The one in a...