PYQ DEC 2016 - VedPrep

CSIR NET BIOLOGY (DEC - 2016)
Previous Year Question Paper with Solution.

21. Excess oxygen consumed after a vigorous exercise is

(a) to pump out lactic acid from muscle

(b) to increase the concentration of lactic acid in muscle

(d) to make ATP for gluconeogenesis.

Ans. (d)

Sol. After exercise has stopped, extra oxygen is required to metabolize lactic acid; to replenish ATP, phosphocreatine and glycogen and to pay back any oxygen that has been borrowed from helmoglobin, myoglobin (an iron containing substance similar to hemoglobin that is found in muscle fibres), air in the lungs and body fluid.

22. Choose the most, appropriate, pH at which the net charge is zero for the molecule from the data shown below:

(a) 2.02

(b) 2.91

(d) 6.87

Ans. (b)

Sol. This is an acidic amino acid, so here to calculate the PI value wer need to add pKa values of acidic side chains and which will be 2.02 + 3.80/2 = 2.91; at this pH this amino acid will have zero charge. At PI value of molecule has zero charge.

23. Choose the correct statement about peptides in the Ramchandan plot.

(a) Peptides that are unstuctured will have all the backbone dihedral angles in the disallowed regions.

(b) It is not possible to conclude whether peptide adopts entirely helix or entirely beta sheet conformation.

(d) The sequence of a peptide can be deduced.

Ans. (c)

Sol. Occurrence of beta turn can be deduced by Ramachandran plot. Unstructured peptide has some of its dihedral angles in the disallowed regions.

24. Equilibrium constant of a reaction is a ratio of product to substrate concentrations. The relation between and free energy change in a reaction is as follows:

Reaction A and Reaction B have values of 10 and 100, respectively. Which of the following statements is correct with respect to ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. As mentioned in the relation = –RT ln K_aq^', the larger the value of K_eq^', the smaller will be .

As the value of K^'_eq for reaction A is 10 and for reaction B it is 100, so K^'_eq(A) < K'_eq(B).

Hence, of A > of B.

25. From the following statements,

P. For a reaction to occur spontaneously the free energy change must be negative.

Q. The interaction between two nitrogen molecules in the gaseous state is pedominantly electrostatic.

R. By knowing bond energies, it is possible to deduce whether the bond is covalent bond or hydrogen bond.

S. Hydrophobic interactions are not important in a folded globular protein.

Pick the combination with all wrong statements.

(a) P and Q

(b) Q and R

(d) Q and S

Ans. (d)

Sol. The hydrophobic effect is considered to be the major driving force for the folding of globular proteins. It results in the burial of the hydrophobic residues in the core of the protein. As both nitrogen atoms have the same degree of electro-negativity, nitrogen gas (N2) is diatomic and non-polar. Therefore, nitrogen atoms stick together to form a liquid due to London dispersion forces.

26. In the biosynthesis of purine:

(a) All N atoms, C4 and C5 are from aspatic acid.

(b) N1 is from Aspartic acid; N3 and N9 are from Glutamine side-chain; N7, C4 and C5 are from Glycine.

(c) N1 is from Aspartic acid; N3 and N9 are from Glutamine side-chain; N9 from N attached to of Glutamine; N7, C4 and C5 are from Glycine.

(d) N1 is from Glutamine; N3 and N9 are from Glutamine side-chain; N9 from N attached to of Glutamine; N7, C4 and C5 are from Glycine.

Ans. (b)

Sol. Nitrogen N₁ comes from aspartic acid, N₃ and N₉ comes from of glutamine, N₇C₄ and C₅ comes from Glycine.

27. A researcher investigated a set of conditions for a protein with an isoelectic point of 6.5 and also binds to calcium. This protein was subjected to four indepdendent treatments : (i) pH 6.4, (ii) 10% glyceol, (iii) 10 mM Cacl₂, (iv) 40% ammonium sulphate. This was followed by centrifugation and estimation of the protein in the supenatant. The resuts ae depicted in the graph below:

Which of the following treatments best represents the results shown in the graph?

(a) a = ammonium sulphate, b = glycero, c = pH 6.4, d = CalCl₂

(b) a = CalCl₂, b = glycero, c = ammonium sulphate, d = pH 6.4

(d) a = CalCl₂, b = pH 6.4, c = glycero, d = ammonium sulphate

Ans. (d)

Sol. Protein near their PI value gets precipitated, as pH 6.4 is very near to its PI value so we will get minimum protein in the supermatant, which is either (b) or (d). Ammonium sulphate also precipitates the proteins, so minimum protein in the supernatant. CaCl₂ and glycerol will help in the solubilization they max protein will be present in the supernantant. So only correct is fourth option.

28. From the following statements:

1. Hydrogen, deuterium and tritium differ in the number of protons.

2. Hydrogen, deuterium and tritium differ in the number of neutrons.

3. Both deuterium and tritium are radioactive and decay to hydrogen and deuterium, respectively.

4. Tritium is radioactive and decays to helium.

5. Carbon-14 decays to nitrogen-14.

6. Carbon-14 decays to nitrogen-13.

Pick the combination with all correct statements:

(a) 1, 2 and 6

(b) 2, 4 and 5

(d) 3, 5 and 6

Ans. (b)

Sol. Hydrogen, Deuterium and Tritium differ in neutrons. Tritium is radioactive and it decays to form helium. Carbon-14 decays to Nitrogen-14.

29. The following are four statements on peptide/protein conformation:

P. Glycine has the largest area of conformationally allowed space in the Ramachandran plot of and .

Q. A 20-residue peptide that is acetylated at the N-terminus and amidated at the C-terminus has = –60° (±5), = –30° (±5) for all the residues. It can be concluded that conformation of the peptide is helix-turn strand.

R. The allowed values of , for amino acids in a protein ae not valid for short peptides.

S. A peptide Acetyl - A₁-A₂-A₃-A₄-CONH₂ (A₁-A₄ are amino acids) adopts a well defined -turn. The dihedral angles of A₂ and A₃ determine the type of -turn.

Pick the combination with all correct statements:

(a) P and Q

(b) Q and R

(d) R and S

Ans. (c)

Sol. Glycine is the most flexible amino acid so that's why it has the largest area in the Ramachandran plot. The peptide which has 20 residues and it acetylated and amidated at N and C terminals, phi value of –60 and psi value of –30 shows that it has helical conformation only, not helix turn strand. Theshort peptides are highly unstable structure so they form random structures, which can be present in disallowed regions as well. We can confirm type of beta turn with the help of dihedral angles.

30. A researcher was investigating the substrate of two different enzymes, X and Y, on the same substrate. Both the enzymes were subjected to treatment with either heat or an inhibitor which inhibits the enzyme activity. Following are the results obtained where a = inhibitor treatment, b = heat treatment and c = control. Which of the following statements is correct?

(a) Only protein X is specific for the substrate, S.

(b) Only protein Y is specific for the substrate, S.

(d) Both X and Y are non-specific for the substrate, S.

Ans. (c)

Sol. Specificity is defined as the ability of an enzyme to choose an exact substrate from a group of the same chemical molecules. Actually, specificity is a molecular recognition mechanism that works through complementarity in conformation and structure between the enzyme and the substrate. Graph shows that both X and Y are specific for substrate, S.

31. Which of the following is a food borne toxin?

(a) Tetanus toxin

(b) Botulinum toxin

(d) Diphtheria toxin

Ans. (b)

Sol. Botulinum toxin is a food borne toxin, generally. Cholera toxin is water born, some time it is food born but not generally.

32. Which one of the following describes the primary function of flippases?

(a) Help in increasing lipid-protein interaction in the outer leaflet of the bilayer.

(b) Move certain phospholipids from one leaflet of the membrane to another.

(d) Cause uncoupling of v-SNArES and t-SNArES after fusion of incoming vesicle with target membrane.

Ans. (b)

Sol. Flippases are the enzymic responsible for movement of proteins across lipid bilayer. It is required for maintaining asymmetric lipid bilayer structure. (REF-Moelcula Cell book).

33. Mitotic cyclin-CDK activity peaks in M phase. This is because

(a) mitotic cyclin is synthesized only in M phase.

(b) thershold level of mitotic cycin accumulates only in late G₂.

(d) the kinase subunit is activated by dephosphoylation only in M phase.

Ans. (d)

Sol. CDK1-Cyclin B synthesis occurs in late G2 phase but its activated by Cdc25P in M phase. Inhibitory phosphate group is attached on Try-15 or The-14 in mammalian CDK1 by Wee-1 kinases in late G2 phase. But during M phase transition Cdc25 Phosphatases removes inhibitory phosphate and CDK1 becomes activated in M-phase. (REF-Molecular Cell).

34. The gel to liquid crystalline phase transition temperature in phosphatidyl choline (PC) lipids composed of dioleoyl (DO), dipalmitoyl (DP), disteroyl (DS) and palmitoyl oleoyl (PO) fatty acids in increasing order will be

(a) DOPC > DPPC > POPC > DSPC

(b) DSPC > DPPC > POPC > DOPC

(d) POPC > DPPC > DOPC > DSPC

Ans. (b)

Sol. Gel is hard structure as compared to liquid crystallien phase; it means transition temperature will be higher for gel form. The saturated fatty acid containing lipids have high melting or transition temperature, so in gel form saturated and higher number of carbon containing fatty acids will be present. Stearic acid is saturated as well as it has 18 carbons, palmitic is 16 carbon and saturated, so the choline which has stearic acid at both positions will have higher melting temperature, than the one which has palmitic acid at both positions, than one palmitic and one oleaic acid (oleic acid is 18 carbon and mono unsaturated fatty acid) and lastly with oleic acid at both positions will have the minimum melting temperature.

35. Which of the following is not an example of transmembrane transport between different sub-cellular compartments?

(a) Transport from cytoplasm into the lumen of endoplasmic reticulum.

(b) Transport from endoplasmic reticulum to the Golgi complex.

(d) Transport from mitochondrial intermembrane space into the mitochondrial matrix.

Ans. (b)

Sol. Transport between ER and Golgi involves vesicular transport. Transport of molecules from cytosol to organelles involve trans-membrane transporters since they are topologically different. Movement between ER and Golgi involves vesicles movements. (Bruce Albertis-Molecular cell).

36. During cell cycle progression from G₁ to S, cyclin D-CDK4 phosphorylates rb and reduces its affinity for E2F. E2F dissociates fom Rb and activates S-phase gene expression. Overexpression of protein 'A' arrests G₁ phases progression. Which of the following statements is true?

(a) 'A' inhibits rb-E2F interation.

(b) 'A' inhibits CDK4 activity.

(d) 'A' degrades Rb.

Ans. (b)

Sol. If A inhibits CDK4 it will ceases G₁ progression. CDK4 in the active form will be required for the progression of cell cycle form G₁ phase to S phase.

37. Cells in S-phase of the cell cycle were fused to cells in the following stages of cell cycle : (a) G₁ phases, (b) G₂ phase, (c) M phase. These cells were then grown in medium containing tritiated thymidine. Maximal amount of freshly labeled DNA is likely to be obtained S-phase cells fused with

(a) G₁ phase cells

(b) G₂ phase cells

(d) Both G₁ and G₂ phase cells

Ans. (a)

Sol. G₁ fusion with S phase nucleus will promote early entry of G₁ phase cells into S phase and therefore more radiolabelled DNA will be produced. Cell cycle progression occurs in the order of G₁-S-G₁-M. Cells in G₂ phase cannot come back to S phase if fused with S phase nucleus, since progression is unidirectional. (Molecular Cell).

38. Fluorescently tagged protein was used to study protein secretion in yeast. Fluorescence was obtained in:

P. the Golgi

Q. the secretory vesicles.

R. the rough ER.

Which of the following describes best the sequence in which these events occur?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Sequence is from ER-Golgi Secretory vesicles. Movement of secretory molecules occurs from RER-cisGN trans-GN Secretory vesicles and finally to the plasma membrane. (Molecular Cell).

39. In eukaryotes, a specific cyclin dependent kinase (CDK) activity is required of the activation of loaded helicases to initiate replication. On the contrary, this CDK activity inhibits the loading of helicases onto the oigin of replication. Considering the fact the during each cycle, there is only one opportunity of helicases to be loaded onto origins and only one opportunity of these loaded helicases to be activated, which one of the following graphs best depicts this CDK activity in G₁ and S phases of the cell cycle?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. In protozoan's when exon of one gene gets attached with exon of another gene or RNA the process is called trans-splicing.

40. In an experiment, red blood cells were subjected to lysis and any unbroken cells were removed by centrifugation at 600 g. The supernatant was taken and centrifuged at 100,000 g. The pellet was extracted with 5 M NaCl and again centrifuged at 100,000 g. Which of the following proteins would be present in the supernatant?

(a) Band 3

(b) Glycophorin

(d) Spectrin

Ans. (d)

Sol. Rest all three proteins aer membrane linked and Spectrin is a cytosolic protein. Membrane proteins are high density proteins are therefore will pellet down at low RPM during sub-cellular fractionation of RBC. Only cytosolic proteins will remain in the supernatant. (Molecular Cell).

41. rNA editing, a post-transcriptional process, is achieved with the help of guide NA (g-rNA). Which one of the following statements about the process is not tue?

(a) g-rNA dependent rNA editing happens in the kinetoplast DNA.

(b) g-rNA is involved in chemical modification of t-rNA.

(d) Sequences edited once may be re-edited using a second g-rNA.

Ans. (b)

Sol. Guiding RNA edit the mitochondrial DNA, which other way also called kinetoplast as well. This process involves insertion and deletion of uridines. Guiding RNA never modify the t-RNA. That's the wrong option.

42. Telomerase, an rNA-protein complex which completes the replication of telomeres during DNA synthesis, is a specialised

(a) rNA dependent DNA polymerase.

(b) DNA dependent DNA polymerase.

(d) RNA dependent RNA polymerase.

Ans. (a)

Sol. Telomerase synthesize only one strand, it already has its own RNA primer so we call it RNA dependent DNA polymerase.

43. Consider a short double-stranded linear DNA molecule of 10 complete turns with 10.5 bp/turn. The ends of the DNA molecule are sealed together to make a relaxed circle. The relaxed circle will have a linking number of

(a) 105

(b) 20.5

(d) 10.5

Ans. (c)

Sol. For Relaxed DNA the linking number is equal to the number of twist in DNA molecule. As number of turns is equal to 10 so the linking number will be 10. Lk = Tw + Wr.

44. Which of the following are not transcribed by RNA polymease II?

(a) mirNA and some snrNA

(b) mirNA and snorNA

(d) trNA and 5S-rNA

Ans. (d)

Sol. trNA and 5S-rNA are not tanscribed by RNA polymease II. These are transcribed by RNA polymerase III.

45. In order to ensure that only fully processed mature mRNA are allowed to exported to cytosol, pre-mrNAs associated with snrNPs are retained in the nucleus. To demonstrate this, an experiment was performed where a gene coding a per-mrNA with a single intron was mutated either at the 5' o 3' splice sites or both the splice sites.

Given below are a few possible outcomes:

P. Pre-mrNA having mutation at both the splice sites will be retained in the nucleus because of the presence of bound snRNPs.

Q. Pre-mrNA having mutation at both the splice sites will be exported to cytosol because of the absence of bound snrNPs.

r. Pre-mrNA mutated at either 3' o 5' splice sites will be retained in the nucleus because of the presence of bound snRNPs.

S. Pre-mrNA mutated at either 3' o 5' splice sites will be exported to cytosol because of the absence of bound snRNPs.

Choose the correct combination of the possible outcomes:

(a) Q and r

(b) P and S

(d) P and r

Ans. (a)

Sol. mRNA having snRNPs bound on it can't be transported to cytoplasm. So, we can say that Pre-mrNA having mutation at both the splice sites will be exported to cytosol because of the absence of bound snrNPs and Pre-mrNA mutated at either 3' o 5' splice sites will be retained in the nucleus because of the presence of bound snRNPs.

46. Telomerase, a protein-NA complex, has a special reverse transcriptase activity that completes replication of telomeres during DNA synthesis. Although it has many properties similar to DNA polymerase, some of them are also different. Which one of the following properties of telomerase is different from that of DNA polymerase?

(a) Telomerase requires a template to direct the addition of nucleotides.

(b) Telomerase can only extend a 3'-OH end of DNA

(d) Telomerase acts in a progressive manner.

Ans. (c)

Sol. Telomerase has its own RNA primer bound to it. It extends the DNA from 3' end only, so DNA polymerase requires templates to synthesize DNA. Telomerase as well as DNA polymerase both acts in the processive manner. DNA polymerase carry out the synthesis of both strands, means leading as well lagging strand but Telomerase only carry out the synthesis of leading strand not lagging strand. After telomerase synthesized the DNA on leading strand, DNA polymerase carry out the synthesis of lagging strand afterwards.

47. Polysome profiling of cells treated with three hypothetical translational inhibitors is shown in the plots below. These three inhibitors are

P. CHP - leaky inhibitor of translation.

Q. LTM - aest ribosome at the initiation codon.

r. PTM - inhibits ribosome scanning.

Match the polysome profiling to the inhibitor

(a) P-1; Q-2; r-3

(b) P-2; Q-3; r-1

(d) P-1; Q-3; r-2

Ans. (a)

Sol. As CHP is the leaky inhibitor of translation, there polysome formation will happen; it can be observed from (a) plot. LTM inhibits the translation at initiation codon it means, small and large sub-unit will get assembled but there will be no formation of polysome, which can be observed in (b) plot.

48. In mammals, CG rich sequences are usually methylated at C, which is a way for marking genes for silencing mammals. Which one of the following best explains it?

(a) Methylation of cytosine does not prevent the binding of rNA Pol II with the promote, so housekeeping genes are expressed.

(b) During housekeeping gene expression, the enzyme methyltransferase is temporaily silenced by miRNA, thus shutting down global methylation.

(c) Unlike within the coding region of a gene, CG rich sequences present in the promotes of active genes are usually not methylated.

(d) As soon as the cytosine is methylated in the promoter region, the enzymes of DNA repair pathways remove the methyl group, thereby ensuring gene expression.

Ans. (c)

Sol. Housekeeping genes are essential for the survival of the cell, so these genes will nto be methylated at any point of time.

49. Addition of the antibiotic cephalexin to growing E. coli cells lead to filamentation of the cells, followed by lysis. Cephalexin is an inhibitor of

(a) protein synthesis

(b) DNA synthesis

(d) rNA polymease

Ans. (c)

Sol. Cephalexin is responsible for the inhibition of peptidogycan synthesis. If proper cell wall is not formed than E. coli cells will get lysed.

50. In Typanosomes, a 35 base leader sequence is joined with several different tanscripts making functional mrNAs. The leader sequence is joined with the other RNAs by

(a) a specific rNA ligases

(b) the process of trans-splicing.

(d) a nucleophilic attack caused by a 2'OH of an internal a present in the leader sequence.

Ans. (b)

Sol. In Typanosomes, a 35 base leader sequence is joined with several different tanscripts making functional mrNAs. The leader sequence is joined with the other RNAs by the process of trans-splicing.

51. Mayfai genes (hypothetical) consists of a super family of transcription factors. They are found in 4 clusters in mammals; in 2 clusters in insect; and in a single cluster in an ancestor to insects. These data are consistent with all of the following explanations except:

(a) Two successive genome duplication events occurred between ancestral oganism and vertebrates.

(b) The first duplication may have taken place before divergence of vertebrates.

(d) Whole genome duplications could lead to such observations.

Ans. (c)

Sol. This question can be expalined through the following example. The developmental control genes containing an Antennapedia-type homeobox are clustered in insects and vertebrates. Subsequent duplication events generated a cluster of at least five homeobox genes in the last common ancestor of insects/vertebrate divergence two of the resulting genes had already undergone a further duplication, while the Abdomina-B precursor remained singular. Later in the evolution of vertebrates and insects, gene duplicatin in all three of the original classes.

52. In the below signalling cascade, which one of the following molecules is denoted by 'B'?

(a) STAT5

(b) SMAD6

(d) SMAD4

Ans. (d)

Sol. Activini/TGF-beta activates Type I and Type II receptor. Which in turn activates Smad 1 and Smad 3. Smad 4 is co-smad in TGF/Activin signalling which is always required for Smad-dimer formation. (Cooper-Cell).

53. The secondary antibodies routinely used fo the detection of primary antibodies in western blotting experiment are

(a) anti-allotypic

(b) anti-idiotypic

(d) anti-praratypic

Ans. (c)

Sol. Secondary antibody are raised in different host. Therefore it is anti-isotypic. Isotypic antibodies have same constant region in same species. Therefore different species inherit different constant region of HC and LC. If primary antibody is raised in mouse secondary should of differnt host may be rabbit or goat. (Kuby).

54. Which of the following events will not usually leads to transformation of a normal cell into a cancer cell?

(a) Gain-of-function of oncogenes.

(b) Loss-of-function of tumor suppressors.

(d) Loss-of-function of pro-apoptosis related genes.

Ans. (c)

Sol. DNA repair enzymes are like tumor suppressor genes in which loss of function mutation result in cancer. Rest all options promoters cancer development.

55. In order to study the intracellular trafficking of protein 'A', it was tagged with GEP (A-GFP). Fluroescence microscopy showed that A-GFP co-localizes with LAMP1. In the presence of bafilomycin A, an inhibitor of H⁺-ATPase, A-GFP does not co-localize with LAMP1. Instead, it co-localizes with LC3 puncta. Which of the following statements is true?

(a) A-GFP tagets to the Er in the absence of bafilomycin A.

(b) Autophagy is required for tafficking of A-GFP to lysosomes.

(d) Bafilomycin A facilitates targeting of A-GFP to mitochondria.

Ans. (d)

Sol. LAMP₁ (lysosomal associated membrane protein) in present Lysosomes. LC₃ Puncta is present in Autophagosome. In autophagy, Autopahagosome fuse with Lysosomes and degrades its contents. The protein A-GFP should always colocalizes with LAMP₁, which is present in Lysosomes. Since, Bafilomycin inhibits formation of lysosome, the protein A-GFP will co-localizes with LC₃, which is present in L autopahosomes.

56. In animals, four separate families of cell-cell adhesion proteins are listed in Column A and their functional characteristics are given in Column B.

Column A Column B

P. Integrin 1. Lectins that mediate a vaiety of tansient, cell-cell adhesion

interactions in the blood steam.

Q. Cadherin 2. Contains extracelluar Ig-like domains and are mainly involved

in the fine turning of cell-cell adhesive interactions during

developement and regeneation.

R. Ig-superfamily 3. Mediates Ca²⁺ -depedndent strong homophilic cell-cell

adhesion.

S. Selectin 4. Transmembane cell adhesion proteins that acts as extracellular

matrix receptors.

Which of the following is the correct combination?

(a) P-1, Q-2, R-3, S-4

(b) P-2, Q-3, R-4, S-1

(d) P-4, Q-3, R-2, S-1

Ans. (d)

Sol. Integrins are the principal receptors used by animal cells to bind to the extracellular matrix. They are heterodimers and function as transmembrane linkers between the extracellular matrix and the actin cytoskeleton."Cadherins found in adherens junctions constitute the major family of transmembrane glycoproteins that mediate cell-cell adhesion by virtue of their ability to self-associate in a Ca2+-dependent manner."Ig- supetfamiy- Members of the Immunoglobulin superfamily include vascular and neural cell adhesions molecules (VCAM and NCAM), intercellular adhesion molecules (ICAM) and the Nectins and nectin-like (Necl) proteins."Selectins are a family of three cell adhesion molecules (L-, E-, and P-selectin) specialized in capturing leukocytes from the bloodstream to the blood vessel wall. This initial cell contact is followed by the selectin-mediated rolling of leukocytes on the endothelial cell surface.

57. Following are the list of some of the pathogens (column A) and the unique mechanism they employ for evading immune response (column B)

Column A Column B

P. Trypanosoma brucei 1. Capable of employing unusual genetic processes by

which they generate extensive variations in their variant

surface glycoproteins (VSG).

Q. Plasmodium falciparum 2. Capable of continualy undergoing maturational changes

in transformation to different forms which allow the

organism to change its surface molecules.

R. Haemophius influenzae 3. Capable of evading immune response by frequent

antigenic changes in its hemagglutinin and

neuaminidase glycoproteins.

Which of the following is the correct match between the organisms and the receptive mechanism to evade immune response?

(a) P-1, Q-2, R-3

(b) P-2, Q-3, R-1

(d) P-1, Q-3, R-2

Ans. (a)

Sol. Trypanosoma brucei parasites successfully evade the host immune system by periodically switching the dense coat of variant surface glycoprotein (VSG) at the cell surface. Each parasite expresses VSGs in a monoallelic fashion that is tightly regulated."Plasmodium falciparum, avoids detection—it changes a critical protein on its surface that it uses as one of several molecular "keys" to enter into a new red blood cell."Haemophilus influenzae uses Antigenic drift and shift to escape immunity. The gradual accumulation of mutations, mainly in the highly variable globular head region of HA, causes the influenza virus to escape recognition by virus neutralizing antibodies and allows it to cause seasonal epidemic outbreaks. This phenomenon is called antigenic drift.

58. A student treated cancer cells with an anti-cancer drug and performed western blot analysis. Which one of the following blots is the best representation of untreated control (C) and treated (T) samples?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Level of D1 should be less and p53 should be more in presence of anti-cancerous drugs. p53 is a tumor suppressor protein and therefore its level should increase upon administrating anti-cancerous drug. Cyclin D1 is G1 phase cyclin and required for cell cycle progression. If the level are reduced cells will get arrested in G1 phase and cancer will be inhibited. PARP is an enzyme required for DNA repair and therefore level should increase to inhibit cancer.

59. Two steroid hormone receptors X and Y both contain a ligand binding and a DNA binding domain. Using recombinant DNA technology, a modified hybrid receptor H is prepared such that it contains the ligand binding domain of X and DNA binding domain of Y. These sets of cells overexpressing receptors X, Y and H were then treated separately either with hormone X or with homone Y. Assuming that there is no crossreactivity, which one of the following graphs best represent the receptor-ligand binding in each case?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Hybrid H protein will only be able to find X hormone, because it has only X ligand binding domain. Third option shows high ligand bind in case of receptor X and H, both have binding sites for X. In case of Y hormone treatment, the ligand will only be able to bind to the Y receptor not X and H. Data based evaluation.

60. A protein X is kept in an inactive state in cytosol as complexed with protein Y. Under certain stress stimuli, Y gets phosphorylated resulting in its proteasomal degradation. X becomes free, tanslocates to nucleus and results in the transcription of a gene which causes cell death by apoptosis. Stress stimuli were given to following four different cases.

Case A : Protein Y has a mutation such that phosphorylation leading to proteasomal degradation does not occur.

Case B : Cells are transfected with a gene which encodes of a protein that inhibits the translocation of protein Y to the nucleus.

Case C : Cells are transfected only with empty vector used to transfect the gene for protein L.

Case D : Cells are treated with Z-VAD-FMK, a boad spectum caspase inhibito.

Which one of the following graphs best describes the apoptotic state of the cells in the above cases?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Case A says that if protein Y will not get phosphorylated it will not get detached from protein X, so X protein will not get translocated to nucleus and will not result in the transcription of gene which is responsible for the synthesis of apoptosis. Case B says that cells are transfected with a gene which codes for protein L which inhibits the translocation of Y, the % of apoptotic cells will be low. Empty vector cells won't synthesize protein L, it means protein Y will get translocated of the gene responsible for apoptosis will happen. Case D will inhibit all the caspases which are responsible for cell death, in this case also % of apoptotic cells will be low. Apoptosis will occur only in empty vector transfected cells. In option (b), (c) and (d) apoptosis will not occur.

61. Which of the following statements regarding B-cell receptor (BCr) and T-cell receptor (TCr) is not true?

(a) TCr is membrane bound does not appear as soluble from as does the BCr.

(b) Unlike BCr, most of the TCr are not specific for antigen alone but for antigen combined with MHC.

(d) The antigen binding interactions of BCr is much weaker than TCr.

Ans. (d)

Sol. BCR is MHC independent and therefore involves stronger interactions. (Kuby).

62. Apical ectodermal ridge induction is essential for tetrapod limb development. Which one of the following is not essential for the formation of the functional limb?

(a) Tbx genes and Wnt

(b) Androsterone

(d) Fibroblast growth factor

Ans. (b)

Sol. Androsterone is a type of chemical known as a "prohormone." These chemicals are changed by the body to other "muscle-building" hormones such as testosterone.

63. Which one of the following statements is wrong?

(a) The megasporocyte develops within the megasporangium of the ovule.

(b) Megasporocyte undergoes meiosis to produce for haploid megaspore.

(d) Female gametophyte is haploid.

Ans. (c)

Sol. In most angiosperms only one megaspore develops to produce mega-gametophyte.

64. Certain proteins or mrNA that are regionally localized within the unfertilized egg and regulate deveopment are called

(a) gene regulators

(b) morphometric determinants

(d) mosaic forming factors.

Ans. (c)

Sol. In drosophila, unfertilized egg contains certain maternal mRNAs which are located in different regions of egg and play an important role in the formation of anterior-posterior and dorsal-ventral symmetry in embryo. These maternal mRNAs are called cytoplasmic determinants.

65. Cell to cell communication is important in development of an organism. The ability of cells to respond to a specific inductive signal is called

(a) regional specificity of induction

(b) competence

(d) instructive interaction

Ans. (b)

Sol. Competence is ability to respond.

66. Consider the following events which occur during fertilization of sea chin eggs.

P. resact/speract are peptides released from the egg jelly and help in sperm attraction.

Q. Bindin, an acrosomal protein, interacts in a species-specific manner, with eggs.

r. A "respiratory burst" occurs during cross-linking of the fertilization envelope, where a calcium-dependent increase in oxygen levels is observed.

S. IP₃, which of the above statements is not true?

(a) Only r

(b) P and R

(d) Q and S

Ans.

Sol. The peptide Resact/superact acts as a chemoattractant for sea urchin sperm. in the sea urchin, contact with egg jelly initiates the acrosomal reaction, which is a mediated by calcium. The acrosomal vesicles fuses with plasma membrane, release enzymes from the tip of the sperm that degrade the egg jelly. At this time, bindin is deposited on the surface of the acrosome reacted sperm. One of the hallmarks of fertilization is that it is species-specific : sperm from one species cannot fertilize eggs of another. Sea urchin eggs synthesize H₂O₂ in a "respiratory burst" at fertilization, as an extracellular oxidant for cross linking their protective surfce evelopes. The slow block to poly-spermy is sea urchin egg is mediated by PIP2 second messenger.

67. Following statements were given regarding the decisions taken during development of the mammalian embryos:

P. The pluripotency of the inner cell mass (ICM) is maintained by a core of three tanscription factors, Oct 4, Sox 2, Nanog.

Q. Prior to blastocyst formation each blastomere expresses both Cdx 2 and the Oct 4 tanscription factors and appears to be capable of becoming either ICM or trophoblast.

r. Both ICM and trophoblast cells synthesize transcription factor Cdx 2.

S. Oct 4 activates Cdx 2 expression enabling some cells to become trophoblast and other cells to become ICM.

Which of the above statements are true?

(a) P and Q

(b) P and r

(d) Q and r

Ans. (a)

Sol. Four genes, Oct4, Sox2, Nanog, and Tcf3, represent transcription factors crucial for the maintenance of pluripotency. These factors form a self-sustaining autoregulatory loop by binding to each other's promoter regions and activating their transcription. OCT-4 and CDX2 are essential for early development and gene expression involved in differentiation of ICM and TE lineages in bovine embryos.

68. Apoptosis during early development is important for proper formation of different structures. In C. elegans, apoptosis is accentuated by ced-3 and ced-4 genes, which in turn are negatively regulated by ced-9 and eventually EGL-1. When compared to mammals, functionally similar homologues have been identified. Accordingly, which of the following statement is not correct?

(a) CED-4 resembles Apaf-1.

(b) CED-9 resembles Bcl-XL

(d) CED-4 resembes caspase-9.

Ans. (d)

Sol. In C elegans ced-9 blocks the apoptotic protein similarly as Bcl-XI in mammals, so CED-9 is homolog of Bcl-XI, while CED-3 is act as the effector caspase like mammalian caspase-3. CED-4 is the initiator protein function to activate effector apoptotic protein CED-3 like in mammals Apaf-1 function to activate effector caspase 9. So, All given option are correct except option (d) as CED-o4 is homolog of mammalian Apaf-1.

69. In case of amphibians, the dorsal lip cells and their deivatives are called as "Spemann Mangold Organizer". Following statements related tothe "organizer" were made:

1. It includes the host's ventral tissue to change their fates to form a neural tube and dorsal mesodermal tissues.

2. It cannot organize the host and donor tissues into the secondary embryo.

3. It does not have the ability to self-differentiate into dorsal mesoderm.

4. It has the ability to initiate the movements of gastrulation.

5. Both -catenin and chordin are produced by the organizer.

Which of the above statements are correct?

(a) 1 and 4

(b) 4 and 5

(d) 2 and 3

Ans. (a)

Sol. Spemann referred to the dorsal lip cells and their derivatives (notochord, prechordal mesoderm) as the organizer because they induced the host's ventral tissues to change their fates to form a neural tube and dorsal mesodermal tissue (such as somites) and they organized host and donor tissues into a secondary embryo with clear anterior-posterior and dorsal ventral axes and have the ability to induce movements of gastrulation.

70. Diesch perfomed the famous pressure plate experiment involving intricate recombination with an 8-celled sea urchin embryo. The procedure reshuffled the nuclei that normally would have been in the egion destined to fom endodem into the persumptive ectodem egion. If segregation of nuclear determinants had occurred, the resulting embryo should have disordered. However Driesch obtained normal larvae from these embryos. The possible interpretations regarding the 8-celled sea urchin embryo are:

P. The prospective potency of an isolated blastomere is greater than its actual prospective fate.

Q. The prospective potency and the prospective fate of the blastomere were identical.

r. Sea-urchin embryo is a 'hamoniously equipotential system, because all of its potentially independent parts interacted together to form a single embryo.

S. regulative development occurs where location of the cell in the embryo determines its fate.

Which of the above interpretation is true?

(a) Only P

(b) Only S

(d) P, R and S

Ans. (d)

Sol. Driesch had established that the prospective potency of an isolated blastomere (from those cell types it was possible to form it) is greater than its probable fate (those cell types it would normally give rise to over the unchanged route of its development). Next, Driesch concluded that the sea urchin embryo is a "harmonious equi-potential system" because all of its potentially self-governing parts functioned together to form a singel organism. Third, he concluded that the fate of a nucleus depend solely on its location in the embryo.

71. Individual and overlapping expression of homeotic genes in adjacent whorls of a flower determine the pattern of floal organ development. In an Arabidopsis mutant, the floral organs ae distributed as follows:

Whorl 1 (outemost) - carpel

Whorl 2 - stamens

Whorl 3 - stamens

Whorl 4 (innermost) - carpel

Loss-of-function mutation in which one of the following genes would have caused the above pattern of floral organ development?

(a) APETALA 2

(b) APETALA 3

(d) AGAMOUS

Ans. (a)

Sol. The loss of function of AP2 in Arabidopsis has been previously demonstrated to cause a homeotic transition of sepals to carpels and petals to stamens.

72. Which of the following compounds is not a part of alkaloid class of secondary metabolites?

(a) Lignin

(b) Indole

(d) Pyrroidine

Ans. (a)

Sol. Indole, Tropane and pyrrolidone all are alkaloid class of molecules. Ligning is a phenolic compound.

73. Which one of the following best describes the symplast pathway of water flow from the epidermis to endodermis in a plant root?

(a) Water moves though cell walls and extracellular spaces without crossing any membrane.

(b) Water travels across the root cortex via the plasmodesmata.

(d) Transport across the tonoplast.

Ans. (b)

Sol. Symplast is living pathway of cells connected through plasmodesmata.

74. The herbicide, dichloophenyldimethyllurea, is an inhibitor of,

(a) Shikimate pathway for biosynthesis of aromatic amino acids.

(b) electron transport from P680 to P700

(d) electron transport from P700 to ferredoxin.

Ans. (b)

Sol. DCDU stops the electron transport from PS II to PS I, so PS I is P700 and PS II is P680. So third option is correct.

75. Which one of the following plant derived signalling molecule induces hyphal branching of arbuscular mycorrhizal fungi, a phenomenon that is observed at the initial stages of colonization by these fungi?

(a) Salicyclic acid

(b) Abscisic acid

(d) Systemin

Ans. (c)

Sol. Strigolactones promotes VAM.

76. 'A' is an inhibito of chlooplast function. The poduction of O₂ and the synthesis of ATP ae measued in illuminated chlooplasts befoe and afte addition of 'A' as shown below:

Which statement is correct?

(a) 'A' inhibits the reduction of NADP⁺.

(b) 'A' inhibits the proton gradient and the reduction of NADP⁺.

(d) 'A' inhibits neither the proton gradient nor the reduction of NADP⁺.

Ans. (c)

Sol. As form the plots it can be observed that there is no change in the O₂ production but there is inhibition of ATP synthesis, so this inhibitor is reducing or inhibiting proton gradient which is responsible for the ATP synthesis, but not the reduction of NADP⁺.

77. read the following statements related to plant-pathogen interaction:

P. Systemic acquired resistance is observed following infection by compatibe pathogen.

Q. Induced systemic resistance is activated following infection by compatibe pathogen.

r. A bacterial infection can induce effector triggered immunity (ETI) leading to hypersensitive response locally.

S. NPr1 monomes that ae released in cytosol due to salicylic acid accumulation is rapidly translocated to nucleus.

Which of the above statements is correct?

(a) P, Q and r

(b) P, r and S

(d) Q, r and S

Ans. (b)

Sol. ISR is caused by non-infective microbes.

78. Given below are statements describing various features of solute transport and photoassimilate translocation in plants.

P. Apoplastic phloem loading of sucrose happens between cells with no plasmodesmatal connections.

Q. Growing vegetative sinks (e.g. young leaves and roots) usually undergo symplastic phloem unloading.

r. Movement of water between phloem and xylem occurs only at the source and sink regions.

S. Symplastic loading of sugars into the phloem occrus in the absence of plasmodesmatal connections.

Select the option that gives a combination of correct statements:

(a) Only P and R

(b) Only Q and r

(d) Only P and Q

Ans. (d)

Sol. Apoplastic loading, sucrose produced in mesophyll cells enters the cell wall space (apoplast) and is taken up into the minor vein phloem by transporters. This is a thermodynamically active process that uses the proton gradient as an energy source. In symplastic loading, driven by a downhill concentration gradient from the mesophyll to the phloem and requiring high plasmodesmatal densities . It usually occurs in young leaves and roots.

79. Given below are names of phytohomones in column I and their associated features/effects/functions in column II.

Column I Column I

P. Auxin 1. Delayed leaf senescence

Q. Gibberellins 2. Epinastic bending of leaves

R. Cytokinin 3. Polar transpot

S. Ethylene 4. Removal of seed domancy

Select the correct set of combinations from the options given below:

(a) P-3, Q-2, R-4, S-1

(b) P-4, Q-3, R-1, S-2

(d) P-1, Q-4, R-3, S-2

Ans. (c)

Sol. Polar auxin transport is the regulated transport of the plant hormone auxin in plants. It is an active process, the hormone is transported in cell-to-cell manner and one of the main features of the transport is its asymmetry and directionality (polarity).

GA stimulates the seed germination. GA exerts its influence in two manners, first by increasing the growth potential of embryo and second by inducing hydrolytic enzymes."Cytokinin production slows down the process of senescence.

Ethylene - Leaf epinasty involves the downward bending of leaves as a result of disturbances in their growth, with a greater expansion in adaxial cells as compared to abaxial surface cells.

80. In photosynthetic electron transpot, electrons travel through carriers organized in the "Z-scheme". The following are indicated as directions of electron flow:

Which one of the combinations is correct?

(a) P and Q

(b) Q and r

(d) P and S

Ans. (b)

Sol. On excitation by the absorption of a photon –P680* quickly transfers an electron to a close by pheophtin a. Pheophytin a is chlorophyll a molecule. The electron is then get transferred to a strongly bound plastoquinone at the QA site of the D2 subunit. The electron is then transferred to a transferable plastoquinone located at the QB site of the D1 subunit. The incoming of a second electron to the QB site with uptake of two protons from the stroma produces plastoquinol, PQH2. Plastoquinone (PQ) carry the electrons from PSII to the cytochrome bf complex. This complex functions as plastoquinol-plastocyanin oxido-reductase. The final stage of the light reactions is catalyzed by PSI. Upon excitation either by direct absorption of a photon or exciton transfer-P700* transferes an electron through a chlorophyll (A0) and a bound phllyoquinone (A1) to a set of 4Fe-4S clusters. From these clusters the electron is transferred to ferredoxin (Fd). The flow of electron occur from water to photosystem-II, from photosystem-II to photosystem-I and PS-I to NADP⁺ and arrangement is described as Z-scheme.

81. Phytochrome-mediated control of photomorphogeneis is linked to many other gene functions. The following statements are made on the mechanism of phytochrome action:

P. Phytochorme function requires COP1, an E3 ubiquitin ligase that brings about protein degradation.

Q. COP1 is slowly exported from the nucleus to the cytoplasm in the presence of light.

r. HY5 is targeted by COP1 for degradation in pesence of light.

S. HY5 is a tanscription factor involved in photo-morphogenetic response.

Which one of the following combinations is correct?

(a) P, Q and r

(b) Q, r and S

(d) P, r and S

Ans. (c)

Sol.

82. The C₄ carbon cycleis CO₂ concentrating mechanism evolved to reduce photorespiration. The following are stated as impotant features of C₄ pathway.

P. The leaves of C₄ plants have Kranz anatomy that distinguishes mesophyll and bundle sheath cells.

Q. In the peripheral mesophyll cells, atmospheric CO₂ is fixed by phosphoenolypyruvate carboxylase yielding a four-carbon acid.

r. In the inner layer of mesophyll, NAD-malic enzyme decarboxylates four-carbon acid and releases CO₂.

S. CO₂ is again re-fixed through Calvin cycle in the bundle sheath cells.

Which one of the following combinations is correct?

(a) Q, r and S

(b) P, Q and r

(d) P, r and S

Ans. (c)

Sol. Decarboxylation does not occur in mesophyll.

83. Vasopressin section does not increase with

(a) exercise

(b) an increase in extracellular fluid volume

(d) vomiting

Ans. (b)

Sol. Vasopressin or ADH helps in water conservation in the body and thus is secreted under conditions like sweating, diarrohea, vomitting etc. Vassopressin secretion is also affected by body posture with a progressive increas in secretion while standing. Moreover, a decrease in extracellular volume and not an increase will cause ADH secretion.

84. Which one of the following does not occur due to stimulation of baroreceptors?

(a) Bradycardia

(b) Hypotension

(d) Vasoconstriciton

Ans. (d)

Sol. Baroreceptors or pressoreceptors, which are located in the wall of each internal carotid artery at the carotid sinus and in the wall of the aortic arch. After the baroreceptor signals have entered the medulla, secondary signals eventually inhibit the vasoconstrictor centre of the medulla and excite the vagal centre. The net effects are vasodilatation of the veins and arterioles througout the peripheral circulatory system and decreased heart rate and strength of heart contraction.

85. Serum has essentially the same composition as plasma except that it lacks

(a) Albumin

(b) Stuart-Prower factor

(d) Hageman factor

Ans. (c)

Sol. Serum doesn't have fibrinogen, Clotting factor II, V and VIII. Factor VIII is called anti-hemophilic factor.

86. Which type of cells located in gastric glands is responsible for the release of histamine?

(a) Mucous neck cells

(b) Enterochromaffin-like cells

(d) Parietal cells

Ans. (b)

Sol. Histamine in the stomach occurs in endocrine cells (so called enterochrmaffin like ECL) cells. Local release of histamine is thought to be necessary for the stimulation of parietal cells. They are located basally in the oxyntic gland area, in the chief-cell-rich region.

87. If in a blood transfusion, type A donor blood is given to a recipient having type B blood, the red blood cells (rBCs) of donor blood would agglutinate but the recipients's rBCs would be least affected. These observations can be explained in the following statements:

P. Agglutinins in recipient's plasma caused agglutination by binding with type A agglutinogens.

Q. The agglutinins of donor blood was diluted in recipient's plasma resulting in low agglutination.

r. Low of titre of anti-A agglutinins is the cause of low agglutination of recipients's rBCs.

S. High agglutination of donor rBCs is the outcome of high titre of anti-B agglutinins.

Which of the above statement(s) is/are incorrect?

(a) Only P

(b) P and Q

(d) R and S

Ans. (d)

Sol. If in a blood transfusion, type A donor blood is given to a recipient having type B blood, the red blood cells (rBCs) of donor blood would agglutinate but the recipients's rBCs would be least affected. This can be explain when Agglutinins in recipient's plasma caused agglutination by binding with type A agglutinogens and The agglutinins of donor blood was diluted in recipient's plasma resulting in low agglutination.

88. The arteria pressure usually rises and false 4 to 6 mm Hg in a wave like manner causing 'respiratory waves'. The probable mechanism of these waves has been proposed in the following statement:

P. The more negative intrathoacic pressure during insipiration reduces the quantity of blood rerturning to the left side of the heat causing decerased cardiac output.

Q. The changes of intrathoracic pressure during respiration can excite vascular and atrial stretch receptors which affect heat and blood vessels.

r. The activity of medullray respiratory centeres can influence the vasometer centre.

S. The 'respiratory waves' are outcome of the oscillation of the central nervous system ischemic pressure control mechanism.

Which of the above statement(s) is/are not appropriate?

(a) Only P

(b) P and Q

(d) Only S

Ans. (d)

Sol. If the arteria pressure usually rises and false 4 to 6 mm Hg in a wave like manner causing 'respiratory waves'. The mechanism involve The more negative intrathoacic pressure during insipiration reduces the quantity of blood rerturning to the left side of the heat causing decerased cardiac output. The changes of intrathoracic pressure during respiration can excite vascular and atrial stretch receptors which affect heat and blood vessels and The activity of medullray respiratory centeres can influence the vasometer centre.

89. The uptake of nitrous oxide (N₂O) and cabon monoxide (CO) in the blood of lung alveolar capillary relative to their partial pressure and the transmit time of red blood cell in capillary is shown in the figure below:

The reasons of difference in the pattern of alveolar gas exchange of N₂O and CO have been proposed in the following statements:

P. N₂O does not chemically combine with proteins in blood but equilibrate between aveolar gas and blood.

Q. CO has high solubility in blood.

r. CO has high solubility in the alveolar capillary membrane.

S. The dispersion of N₂O between alveolar gas and blood is considered as diffusion limited.

Which of the above statement(s) is/are incorrect?

(a) Only P

(b) P and Q

(d) r and S

Ans. (d)

Sol. The pattern of alveolar gas exchange of N₂O and CO are different because N₂O does not chemically combine with proteins in blood but equilibrate between aveolar gas and blood and CO has high solubility in blood.

90. External pressure given on a mixed nerve cause loss of touch sensation while pain sensation remains relatively intact. On the other hand, application of local anaesthetics on the same nerve, induces loss of pain sensation keeping touch sensation least affected. These observations can be explained by the following statements:

P. External pressure causes loss of conduction of impulses in small diameter sensory nerve fibres.

Q. Local anaesthetics depress the conduction of impulses in large diameter sensory nerve fibres.

r. Touch-induced impulses are carried by fibre type A.

S. Fibre type C is responsible for pain sensation.

Which of the above statement(s) is/are incorrect?

(a) P and Q

(b) r and S

(d) Only S

Ans. (a)

Sol. External pressure given on a mixed nerve cause loss of touch sensation while pain sensation remains relatively intact. On the other hand, application of local anaesthetics on the same nerve, induces loss of pain sensation keeping touch sensation least affected. Here, Touch-induced impulses are carried by fibre type A. and Fibre type C is responsible for pain sensation.

91. The probable effects of lesion of left optic tract on the vision of human subject are given below. Identify the correct statement.

(a) Blindness is the left eye but the visual field of right intact.

(b) Blindness is the right eye but the visual fields of both the eyes.

(c) Blindness is the left half of the visual field of left eye and blindness in the right half of the visual field of the right eye.

(d) Blindness is the left half of the visual field of both the eyes.

Ans. (b)

Sol. Lession of left optic tract may cause, Blindness is the right eye but the visual fields of both the eyes.

92. The following diagram represents steroidogenic pathway in the Zona Glomerulosa of the adrenal cotex. What do A, B and C represent, respectively?

(a) sER, progesterone, 11 (OH) cortisol.

(b) Mitochondria, progesterone, corticosterone.

(d) Mitochondria, 3-pregnenolone, 11(OH) cortisol.

Ans. (b)

Sol. Cortical mitochondria contain various oxidases which alter cholesterol to progenolone. Further pregnenolone converted to progesterone. 11-Deoxy-corticosterone converted to corticosterone.

93. If non-disjunction occurs in meiosis I, which of the following scenario is likely to occur?

(a) Two gametes will be n + 1 and two will be n – 1.

(b) One gamete will be n + 1, two will be 'n' and one will be n – 1.

(d) Two gametes will be normal and two will be n + 1.

Ans. (a)

Sol. Nondisjunction is that a chromosome pair failed to separate during the meiotic division. This will create one daughter cell with an extra chromosome and another daughter cell with one less chromosomes. If the nondisjuction occurs during the first meiotic division (meiosis I), all the gametes derived will be abnormal.

94. Which of the following is true for cells harboing F' palsmid?

(a) Their F plasmid is non-functional.

(b) They exhibit increased rates of transfer of all chromosomal gene.

(d) They fail to survive as the chromosomal origin of replication is inactivated.

Ans. (c)

Sol. F' plasmids have extra chromosomal DNA, which gets wrongly excised. If any bacteria has fertility factors they are called merodiploids.

95. Maternal inheritance of coiling of shell in snail (Limnaea peregra) is well established. The dextral coiling depends on dominant allele D and sinistral coiling depends upon recessive allele d. A female F₁ progeny of dextral (Dd) type is crossed with a male sinistral snail. What will be the ratio of heterozygous: homozygous individuals in its F₂ progeny?

(a) 3 : 1

(b) 1 : 1

(d) 1 : 2 : 1

Ans. (b)

Sol. The inheritance pattern for certain genes in which the genotype of the mother directly determines the phenotype of the offspring is called maternal type of inheritance. 1DD : 2Dd : 1dd all are dextral he ratio between heterozygous and homozygous will be 1:1.

96. Which of the following mutagen is most likely to result in a single amino acid change in a gene product?

(a) Acidine orange

(b) X-rays

(d) Ethidium bromide

Ans. (c)

Sol. EMS is alkylating agent.

97. Inversions are considered as cross-over suppressors because

(a) Homozygous inversions are lethal and thus they do not appear in the next generation.

(b) Inversion heterozygotes, i.e. one copy having normal chromosome and its homologue having inversion, does not allow crossing over to occur as they cannot pair at all.

(c) Due to inversion present, four chromosomes take part in the pairing and crossing over events and make the structure difficult for separation and gamete formation.

(d) The pairing and crossing overs do occur in inversion heterozygotes but the gametes having cross over poducts are lethal.

Ans. (d)

Sol. Inversions are considered as cross-over suppressors because the pairing and crossing overs do occur in inversion heterozygotes but the gametes having cross over poducts are lethal.

98. Three met^– E. coli mutant strains were isolated. To study the nature of mutation these mutant strains were treated with mutagens EMS or proflavins and scored for revertants. The results obtained are summarized below:

(+ stands for revertants of the original mutants and – stands for no revertants obtained)

Based on the above and typical effects of EMS and proflavin, what was the nature of the original mutation in each strain?

(a) P - Transversion; Q - Insertion or deletion of a single base; R- Deletion of multiple bases.

(b) P - Transition; Q - Transversion; R - Insertion o deletion of a single base

(d) P - Transition; Q - Insertion or deletion of a single base; R - Transversion

Ans. (c)

Sol. Proflavin is a intercalating agent so it results in insertion or deletion.

99. The following pedigree shows the inheritance pattern of a trait.

From the following select the possible mode of inheritance and the probability that the daughter in generation III will show the trait.

(a) X-linked recessive, probability is 1/2.

(b) X-linked recessive, probability is ¼

(d) Autosomal recessive, probability is 1/3

Ans. (c)

Sol. By observing pedigree chart the trait is autosomal recessive and its probability = 1 × 1 × ½ = ½

100. A pair of allele governs seed size in a crop plant. 'B' allele responsible for bold seed is dominant over 'b' allele controlling small seed. An experiment was carried out to test if an identified dominant DNA marker (5 kb band) is linked to alleles controlling seed size. A plant heterozygous of the marker and alleles was crossed to a small seeded plant lacking the 5 kb band. 100 progeny obtained from the cross were analyzed for the presence and absence of the DNA marker. The results are tabulated below:

Based on the above observations which one of the following conclusions is correct?

(a) The DNA make assorts independently of the phenotype.

(b) The 5 kb band is linked to the allele 'B'.

(d) The DNA make assorts independently with bold seed but is linked to small seed trait.

Ans. (a)

Sol. Test cross ratio is 1 : 1 : 1 : 1, so independent assortment.

101. The following scheme repersents deletions (1-4) in the rII locus of phage T₄ from a common reference point.

Deletion 1 ––––––––––––––––

Deletion 2 ––––––––––––––––––––––

Deletion 3 ––––––––––––––––––––––––––––

Deletion 4 –––––––––––––––––––––––––––––––––––

(The bars represent the extent of deletion in each case)

Four point mutations (P to S) are tested against four deletions for their ability (+) or inability (–) to give wild type (rII⁺). The results are summarized below.

Based on the above the predicted order of the point mutations is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Deletion 4 (largest deletion) result in the ability of point mutation c to produce wild type phenotype (c⁺) deletion 3-result in c⁺ and a⁺ deletion 2-result in c⁺, a⁺ and b⁺ deletion 1-c⁺, a⁺, b⁺ and d⁺ so the correct order will be d-b-a-c.

102. red hair is a recessive trait in humans. In a randomly mating population in Hardy-Weinberg equilibrium, approximately 9% of individuals are red-haired. What is the fequency of heterozygotes?

(a) 80%

(b) 49%

(d) 18%

Ans. (c)

Sol. For a population is genetic equilibrium : p + q = 1.0 (The sum of the frequencies of both alleles is 100%) (p + q)² = 1

so p² + 2pq + q² = 1

p² = frequency of AA (homozygous dominant)

2pq = frequency of Aa (heterozygous)

q² = frequency of aa (homozygous recessive)

here wer have q² = 9/100 = 0.09

q = 0.3p + 0.3 = 1;

thus 0.7 frequency of heterozygotes = 2pq = 2 × 0.3 × 0.7 = 0.42 or 42%

103. Interrupted mating experiments were performed using three different Hfr strains (1-3). The three strains have different combinations of selectable markers. The time of entry for markers for each strain is shown in the table below.

Using the above data, predict the correct sequence of markers on the E. coli chromosome.

(a) met-thr-str^r-phe-pro-pur^r-his

(b) pur^r-pro-his-met-thr-str^r-phe

(d) his-met-phe-thr-pro-str^r-pur^r

Ans. By the given time intervals between markers of all strains, we can calculate the distance between these markers which will be, met-thr-12, thr-str-8, str-phe-5, phe-pur-8, pur-pro-5, pro-his-10 and his-met-10 so by arrange the markers with respect to their distance wer will get the order of markers. Because these markers are in circular DNA so there transfer depend on mating time. Order will be, pur-pro-his-met-thr-str-phe.

Sol.

104. An alga having chlorophyll a, floridean starch as storage product and lacking flagellate cells belongs to the class:

(a) Phaeophyceae

(b) Chlorophyceae

(d) Xanthophyceae

Ans. (c)

Sol. Floridean starch is characteristic feature of red algae.

105. Which of the following is not true for monocots?

(a) Sieve tube members with companion cells.

(b) Vasculature atactostelic.

(d) Vascular cambium absent

Ans. (c)

Sol. Tricolpate pollens are a characteristic of eu-dicots.

106. A plot of dN/dt as a function of population density yields a

(a) rectangular hyperbola

(b) negative exponential curve

(d) bell-shaped curve

Ans. (d)

Sol. Plot of dN/dt or change in the population size, first increases with increase in the populations size, reaches to max at half carrying capacity after that decreases as the population size approach carrying capacity.

107. For a species having logistic growth, if K = 20,000 and r = 0.15, the maximum sustainable yield will be

(a) 450

(b) 1500

(d) 6000

Ans.

Sol. In majority of the ecosystems root biomass resides in the upper 1m of the soil. Grasslands allocate a large proportion of their biomass below ground, resulting in large root to shoot ratios. For forests the root : shoot ratios decreases because of the accumulation of standing above ground biomass, such as in the stems of forest trees.

108. Which of the following is a correct ranking of ecosystems based on the root: shoot ratio of plants?

(a) Tropical wet forest > Tropical dry forest > Temperate grassland > Tropical grassland

(b) Temperate grassland > Tropical grassland > Tropical wet forest > Tropical dry forest

(d) Temperate grassland > Tropical grassland > Tropical dry forest > Tropical wet forest

Ans. (d)

Sol. The root-to-shoot ratio is a measurement of the amount of plant tissues with supportive functions (roots) compared to the amount of plant tissue with growth function (shoots). A plant with a higher proportion of roots than a neighboring plant will be able to compete more successfully for soil nutrients. Root/shoot (R/S) ratios were highest for tundra, grasslands, and cold deserts (ranging from 4 to 7). Root/shoot ratio (RSR) The RSR ranged from 0.001 (in tropical rainforest) to 45.615 (in cool temperate grasses).

109. Individuals occupying a paticular habitat and adapted to it phenotypicaly but not genotypically aer known as

(a) Ecophenes

(b) Ecotypes

(d) Coenospecies

Ans. (a)

Sol. Morphological forms are ecophenes (ecads).

110. Peripatus is an interesting living animal having unjointed legs, nephridia, haemocoel, trachea, dorsal tubular heart, claws, jaws, continuous muscle layers in body wall. This is considered as a connecting link between

(a) Nematoda and Annelida: continues muscle layers in body wall, unjointed legs and nephridia being nematode characters while haemocoel, trachea and dorsal tubular heart being annelid characters.

(b) Annelida and Arthropoda: unjointed legs and nephridia being annelid characters while claws, jaws, haemocoel, trachea and dorsal tubula heat being arthropod character.

(c) Arthropoda and Mollusca: unjointed legs and nephridia being mollusca characters while claws, jaws, trachea and dorsal tubular heart being arthropod characters.

(d) Nematoda and Arthropoda: continuous muscle layers, unjointed legs and nephridia being nematode characters while claws, jaws, trachea and dorsal tubular heart being arthropod characters.

Ans. (b)

Sol. Peripatus is the living connecting link between arthropods and annelids. Its arthropods have claws, jaws, haemocoel, tracheae and dorsal tubular heart. The annelids have continuous muscle layers in the body wall, unjoined legs and nephridia.

111. The following schematic diagram represents secondary growth in the angiosperms.

Based on the above scheme, which of the following options represents the correct identity of cambia labelled as P, Q, R and S.

(a) P - Inter-fascicular, Q - Fascicular, R-Vascular, S - Cork

(b) P - Fascicular, Q - Inter-fascicular, R-Vascular, S - Cork

(d) P - Cork, Q - Fascicular, R- Inter-fascicular, S - Vascular

Ans. (c)

Sol. The vascular cambium produces secondary xylem and secondary phloem and the cork cambium (phellogen) produces cork cells.

112. The table below lists the major fungal groups and thei characteristics:

Fungal groups Characteristics

P. Ascomycota 1. Hyphae aseptate, coenocytic, asexual repoduction by sporan-giophoes.

Q. Chytids 2. Hyphae aseptate, coenocytic, asexual repoduction by zoospoes.

R. Glomeromycetes 3. Hyphae aseptate, coenocytic, no sexual spores

S. Zygomycetes 4. Hyphae septate or unicellular, asexual repoduction by conidia.

Which one of the following options represents the appropriate match between the fungal group and their characteristics?

(a) P-2, Q-3, R-1, S-4

(b) P-4, Q-2, R-3, S-1

(d) P-2, Q-4, R-3, S-1

Ans. (b)

Sol. Factual question. Ascomycetes reproduces asexually by conidia, hyphae septate or unicellular (yeast). Zygomycetes hyphae aseptate, coenocytic and asexual reproduction by sporoangiophorse.

113. As a biologist, you want to classify three taxa, A, B and C. You have the information on three taits p, q and r. The trait that is ancestral is counted '0' and the trait that is deived is counted as '1'. The distribution of traits found in three taxa is given below.

A B C

p. 1 1 0

q. 1 1 0

r. 0 1 1

Based on the above table, the following cladograms were drawn:

Based on trait distribution and the principle of parsimony, select the correct option.

(a) Both P and Q cladograms are possible

(b) Only Q cladogram is possible

(d) Only P cladogram is possible

Ans. (d)

Sol. Only first cladogram is possible, because p and q character ar matching between A and B species. A is close to B, so B can not be close C.

114. Given below are some pathogens and diseases of humans, animals and plants.

P. Bordetella pertusis 1. Lyme disease of humans

Q. Tiletia indica 2. Grain rot of rice

r. Borrelia burgdorferi 3. Karnal bunt of wheat

S. Anaplasma marginale 4. Whooping cough in humans

T. Burkholderia glumae 5. Heamolytic anemia in cattle

Which one of the following is the correct match between the pathogen and disease caused?

(a) P-4, Q-3, R-1, S-5, T-2

(b) P-4, Q-5, R-1, S-2, T-3

(d) P-2, Q-5, R-1, S-3, T-4

Ans. (a)

Sol. Factual question. Bordetella pertusis cause the Whooping cough, Tilletia indica causes Karnal bunt of wheat, Boirrelia burgorferi causes Lyme disease, Anaplasma cause hemolytic anemia in cattle, Burkholderia glumae Grain rot in rice.

115. Given below are statements pertaining to organisms belonging to three domains of life. Identify the incorrect statement.

(a) Unlike bacteria and eukarya, some archaeal membrane lipids contain long chain hydrocarbons connected to glycerol molecules by ether linkage.

(b) Peptidoglycans are absent in the cell wall of archaea.

(d) Mycoplasma, a group of low GC content, gram positive bacteria that lack cell wall, belong to the same family as the gram positively Mycobacteriaceae.

Ans. (d)

Sol. Mycoplasmataceae lack the cell wall, featuring some of the smallest genomes known and are "metabolically challenged", i.e. missing some essential pathways of free-living organisms. Many Mycoplasmataceae species are pathogenic in humans and animals.

116. You observed that two species of barnacles, species 1 and species 2, occupy upper and lower strata of intertidal rocks, respectively. Only when species 2 was removed by you from the lower strata, species 1 could occupy both the upper and lower strata. From the choices given below, what would be your inference from these observations?

(a) Upper strata of the intertidial rock is the realized niche of species 1.

(b) Upper strata of the intertidial rock is the fundamental niche of species 1.

(d) Species 1 can compete out species 2.

Ans. (a)

Sol. If species 1 is removed from upper strata, nthan species 2 occupy this area means for species 1 it is realized niche.

117. Match the correct local names of the temperate grasslands with their geographical range.

Geographical range Local name of the grassland

P. Asia 1. Pampas

Q. North America 2. Prairies

r. South America 3. Steppes

S. South Africa 4. Veldt

(a) P-3, Q-2, R-4, S-1

(b) P-3, Q-2, R-1, S-4

(d) P-2, Q-3, R-1, S-4

Ans. (b)

Sol. Correct option is (b)

118. Following is a hypothetical life table for a species.

Which one of the following is the correct net reproductive rate (R₀)?

(a) 0.0

(b) 0.3

(d) 1.5

Ans. (c)

Sol. Survivorship between 0-5 is 1 and fertility is 0 so R₀ = 0 (R₀ = l_xm_x)

For age between 5-10 the survivorship is 05 and fertility is 0.5 so R₀ = 0.25

For 10-15 it will be 0.3 and 0.5 R₀ = 0.15

For 15-20 it will be 0.2 and 1 so R₀ = 0.20

For 20-15 it will be 0.1 and 1 so R₀ = 0.1

Add all of them up Net reproductive rate R₀ = 0 + 0.25 + 0.15 + 0.20 + 0.1 = 0.7

R₀ = = 0 + 0.25 + 0.15 + 0.20 + 0.10 = 0.70

119. Which one of the following statements is true for the trends of Dissolved Oxygen (DO) and Biological Oxygen Demand (BOD) in a water stream receiving pollutants from a point soruce?

(a) In septic zone, both DO and BOD levels remain stationary.

(b) In recovery zone, both DO and BOD levels increase rapidly.

(d) In septic zone, DO level decreases and BOD level increases whereas in the recovery zone DO increases and BOD decreases.

Ans. (c)

Sol. Critical dissolved oxygen (D.O.) deficit occurs in Zone of Active Decomposition of the 'oxygen sag curve' in case of self-purification of natural streams. Clear water, presence of fishes. Dissolved oxygen level falls due to decomposition of Organic Matter in this Zone. BOD is a measure of the amount of oxygen required to remove waste organic matter from water in the process of decomposition by aerobic bacteria (those bacteria that live only in an environment containing oxygen). The increasing oxygen consumed in the decomposition process robs other aquatic organisms of the oxygen they need to live. Therefore, organisms that are more tolerant of lower dissolved oxygen levels may replace a diversity of natural water systems containing bacteria, which need oxygen (aerobic) to survive.

120. Following are the graphical representations of various hypotheses proposed for explaining the possible relationships between species richness (X-axis) and community services (Y-axis).

Which of the following options is the correct match between the graphical representations and the hypotheses?

(a) P: redundancy, Q : Keystone, R : rivet, S : Idiosyncratic.

(b) P: Idiosyncratic, Q : rivet, R : Keystone, S : redundancy.

(d) P: rivet , Q : Keystone, R : redundancy, S : Idiosyncratic.

Ans. (c)

Sol. According to Rivet hypothesis many species many get lost unnoticed. As per Redudancy hypothesis, some species can substitute each other in their roles for ecosystem functioning and a kind of saturation is reached if all roles are sufficently filled, additional species have no or only minor effects on the degree of ecosystem functioning. According to this hypothesis, each species is unique in its role. Thus ecosystem function changes when diversity changes but the magnitude because the roles of individual species are complex and varied. Thus if a keystone species is removed then ecosystem functioning will drop rapidly. A positive trajectory in case of keystone species implies that a keystone species contribute positively to ecosystem functioning.

121. Which of the following statement supports the concept of trade-off in the evolution of the life history traits?

(a) Level of parental care and clutch size are positively correlated.

(b) Animals maturing early tend to live longer.

(d) Allocation of higher energy for reproduction leads to higher population growth.

Ans. (c)

Sol. A trade-off in evolution means one trait cannot increase without a decrease in another (or vice versa).

122. Which of the following periods is known as "Age of fishes"?

(a) Devonian

(b) Jurassic

(d) Carboniferous

Ans. (a)

Sol. Age of Fishes. In Devonian time, from about 415 to 355 million years ago.

123. Which of the following is not a assumption of the Hardy-Weinberg model?

(a) Population mates at random with respect to the locus in question.

(b) Selection is not acting on the locus in question.

(d) The population is effectively infinite in size.

Ans. (c)

Sol. Hardy Weinberg principle states. If an infinitely large, random mating population is free from outside evolutionary forces (i.e. mutation, migration and natural selection) then the gene frequencies will not change over time.

124. Which of the following geological period is characterized by the fist appearance of mammals?

(a) Tertiary

(b) Creataceous

(d) Triassic

Ans. (d)

Sol. The oldest known haramiyids are from 208 million years ago in the Triassic.

125. In which of the following mating system there is likely to be no conflict of interest over the reproductive success between sexes?

(a) Polyandry

(b) Monogamy

(d) Polygamy

Ans. (b)

Sol. Monogamy is one to one mating.

126. Given below is a graphical representation of changes in morphological featuers over a period of geological time scale, where population A accumulates heritable morphological changes and give rise to distinct species B. Population B splits into a distinct species B₂. Which of the following lineages represent the pattern of speciation by cladogenesis?

(a) Lineage 1

(b) Both the lineages 1 and 2

(d) Neither of the lineage 1 and 2

Ans. (c)

Sol. Cladogenesis is the formation of a new group of organisms or higher taxon by evolutionary divergence from an ancestral form. As species A does not show divergence, it accumulated heritable morphological changes and converted into B. However, B2 is showing divergence from B. So, Lineage 2 represent pattern of speciation by cladogenesis.

127. The frequency of M-N blood types in a population of 6129 individuals is as follows:

The frequency of L^N allele in this population is

(a) 0.4605

(b) 0.2121

(d) 0.2911

Ans. (a)

Sol. Individuals with L^NL^N = 1303, so q² = 1303/6129 = 0.21 take under root of that you will get 0.46, which will be the frequency of L^N allele.

128. In a natural system, a species producing large numbers of offsprings, with little or no parental care, generally exhibits which one of the following kind of survivorship curves?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. In natural ecosystem if a species producing large amount of offspring with no parental care % survivorship will always gets decreased with time, only second option is correctly depicting it.

129. Which genes have been introduced in Bollagard II cotton to get resistance against cotton bollworm, tobacco budworm and pink bollworm?

(a) cry1Ab + cry1Ac

(b) cry1Ac + cry2Ab

(d) cry9C + cry2Ab

Ans. (b)

Sol. The Cry1ac and Cry2ab genes are encoded by an organism known as Bacillus thuringiensis. These genes code for a toxin protein that causes lysis in the gut of the bollworms. The protein acts as a toxin when ingested by the pests. The protein becomes activated after reaching the gut of pests.

130. The tetanus vaccine given to humans in the case of a deep cut is a

(a) DNA vaccine

(b) recombinant vector vaccine

(d) toxoid vaccine

Ans. (d)

Sol. Tetanus vaccine is a toxoid type.

131. Different leads are used to record ECG of humans. Which one of the following is not unipolar leads?

(a) Augmented limb leads

(b) V₁ and V₂ leads

(d) VR and VL leads

Ans. (c)

Sol. Standard limb lead is not unipolar lead, rest all are unipolar.

132. The presence and distribution of specific mRNAs within a cell can be detected by

(a) Northern blot analysis

(b) Nase protection assay

(d) real-time PCR

Ans. (c)

Sol. Within the cell one has to do in situ hybridisation. In in-situ hybridisation we try to see the molecular interactions directly. Rest methods are outside the cellular systems or in-vitro methods involving molecular biology.

133. Which one of the following analytical techniques does not involve an optical measurement?

(a) ELISA

(b) Microarray

(d) Differential Scanning Calorimetry

Ans. (d)

Sol. DSC is a thermo-analytical technique in which the differences in the amount of heat required to increase the temperature of a sample and reference is measured as a function of temperature. Rest all techniques uses optical detectors.

134. An optical measurement of a protein is taken both before and after digestion of the protein by a protease. In which of the following spectroscopic measurements the signal change, i.e. before versus after protease treatment, could be the maximum?

(a) Absorbance at 280 nm

(b) Circular dichroism

(d) Fluorescence value

Ans. (b)

Sol. Circular dichroism (CD) spectroscopy is a form of light absorption spectroscopy that measures the difference in absorbance of right- and left-circularly polarized light (rather than the commonly used absorbance of isotropic light) by a substance. So, CD measures signal change of protein before and after protease treatment.

135. The electrospray ionization spectrum of a mixture of two peptides show peaks with m/z values 301, 401, 501, and 601. The molecular weights of the peptides are

(a) 1200 and 1250

(b) 1200 and 1500

(d) 1250 and 1350

Ans. (b)

Sol. ESI gives multiple charge species so peptide molecular weight can vary from its original molecular weight. A mass spectrum given M/Z ration where M is mass and Z is charge. So 301, 401 and 601 are representing one peptide 1200 M and 501 is representing 1500 one.

136. Following are plots representing biological rhythms at different time points depicted as SR = sunrise, N = Noon, SS = sunset, MN = Midnight. Which of the following plot(s) represent the ultradian biological rhythm(s)?

(a) Plot Q

(b) Plots P and r

(d) Plot S

Ans. (b)

Sol. In ultradian rhythm, the cycle has a period shorter than a day but longer than an hour. Examples of ultradian rhythms are blood circulation, pulse, heart rate, thermoregulation, blinking, micturition, appetite, arousal, etc. So, plot P and R represent ultradian biological rhythm as it is changing during daytime.

137. A population of non-poisonous butterflies have the same color pattern as some highly poisonous butterflies. Assume that the population of non-poisonous butterflies is higher than the population of poisonous butterflies. Given this, what will be the impact of this mimicry on the fitness of the population of the poisonous butterflies in the presence of the predator?

(a) It will lower the fitness, that is, fitness of the mimic is negatively frequency-dependent.

(b) It will increase the fitness, that is, fitness of the mimic is positively frequency-dependent.

(d) It will increase the fitness, that is, fitness of the mimic is negatively frequency-dependent.

Ans. (a)

Sol. In this mimicry, the mimic benefits but the model may actually find its survival threatened, specially if the harmless mimic becomes too common.

138. Agrobacterium Ti plasmid vectors are used to generate transgenic plants. The following are example of vir gene encoded proteins that are important for the transfer of T-DNA into plants.

P. Vir E, a single-stranded DNA binding protein.

Q. Vir D2 that generates T-strands

R. Vir A that senses plant phenolic compounds

S. Vir F which directs T-complex proteins for destruction in proteasomes.

Which one of the following combinations of proteins function inside the plant cells?

(a) Ony P and

(b) P, Q and R

(d) P, Q and S

Ans. (d)

Sol. VirA functions only in bacterial cells. VirD2 is functional in both bacteria and plant cell, VirE is solely functional in plant cells and VirF is functional in plant cells only.

139. A single copy homozygous transgenic plant containing the transgene 'A' for fungal was subsequently retransformed with another gene 'B' for confering resistance to salt-stress. The selection marker genes used for both the transformation experiments were different. Transgeneic plants obtained following the re-transformation experiment were screened for salt-stress resistance and single copy events were identified by Southern hybridization. These single copy events were self-pollinated. In the event of two T-DNAs (containing the A and B transgenes) getting integrated in unlinked locations in all the transgeneic plants, the phenotypic ratios among the T₁ progeny would be:

(a) 3 (Fungal resistant + Salt-stress resistant) : 1 (Fungal resistant).

(b) 1 (Fungal resistant) : 2 (Fungal resistant + Salt-stress resistant) : 1 (Salt-stress resistant).

(d) 1 (Fungal resistant) : 1 (Salt-stress resistant) : 1 (Fungal resistant + Salt-stress resistant)

Ans. (a)

Sol. A specific segment (T-DNA) of the resident Ti plasmid is transferred from this bacterium into plant cells and integrated into the plant cell genome. Statement a is true due to integration in unlinked locations in all transgenic plants.

140. A reseacher is investigating structural changes in a protein by following tryptophan fluroescence and by circular dichroism (CD). Fluroescence and CD spectra of a pure protein were obtained in the absence of any treatment (P), in the presence of 0.5 M Urea (Q), upon adding acrylamide, a quenche of tryptophan (R) and upon heating (S). The data are shown below:

Which one of the following statements is correct?

(a) CD is more sensitive to stuctual changes than fluorescence.

(b) Fluorescence is more sensitive to structural changes than CD.

(d) Acrylamide alters the secondary structure of the protein.

Ans. (b)

Sol. As it can be easily observed from the plot that structural changes occurring in different conditions are showing more changes in fluorescence spectra so fluorescence is more sensitive to structural changes.

141. Polynucleotide kinase (PNK) is frequently used for radiolabelling DNA or RNA by phosphorylating 5'-end of non-phosphorylated polynucleotide chains. Which of the following statement PNK is not true?

(a) PNK catalyzes the transfer of -phosphate from ATP to 5'-end of polyncleotide chains (DNA or rNA).

(b) PNK has 3'-phosphatase activity

(d) PNK is a T4 bacteriophage-encoded enzyme.

Ans. (a)

Sol. PNK catalyses transfers of gamma phosphate. PNK enzyme only uses ATP as a phosphate donor and catalyses transfer/exchange of phosphate group at the 5' end of DNA.

142. A gene encoding for protein X was cloned in an expression vector under the T7 NA polymerase promoter and lac opeator. Cells were induced by the addition of 1 nM IPTG at 37°C for 6 h. Cells were lysed and fractionated into insoluble bodies and cell-free supernatant by centrifugation. Protein X is present in the insoluble bodies. Which one of the following strategies would you use to express potein X in the soluble fraction (cell-free supenatant)?

(a) Increase the duration of induction with 1 mM IPTG.

(b) Grow cells at low temperature after induction with 1 mM IPTG.

(d) Grow cells at higher temperatue after induction with 1mM IPTG.

Ans. (b)

Sol. Growing cells under low temperature will stabilize the protein. At low temperature protein will not aggregate and will be produced in its native state thus will not go to pellet or incusion bodies.

143. Engineering of metabolic pathways in plants can be achieved by introduction and overexpression of appopiate candidate gene(s) using tansgenic technology. The figure given below epresents a biochemical pathway in plants where a precursor molecule 'A' is converted into products 'T' and 'X' through a series of enzymatic reactions. Enzymes 1-5 are involved in this pathway. Scientists attempted to increase the level of 'X' by introducing an additional copy of the gene for enzyme '5' under transcriptional contol of a strong constitutive promoter. However, the developed transgenic plants did not display a proportionate increase in the level of 'X'.

The following statements were proposed for explaining the above results:

P. Enzyme '4' has greater affinity for D than enzyme '3'.

Q. Feedback inhibition of enzyme '5' by compound X.

r. Substrate limitation fo enzyme '5'.

Which of the above statements could represent probable reasons for not obtaining a proportionate increase in the amount of X in the transgenic plants?

(a) Only r

(b) Only P and Q

(d) P, Q and R

Ans. (d)

Sol. As enzyme 4 has greater affinity for D, so it will convert more amount of D into T and little amount will get converted into E by 3 enzyme. It will limit the availability of E for enzyme 5 to convert it into X. If enzyme 5 gets inhibited by small amount of X again the overexpression of this enzyme wont have any effect on the production of X.

144. You are inserting a gene of 2 kb length into a vector of 3 kb to make a GST fusion protein. The gene is being inserted at the EcorI site and the insert has a HindIII site 500bp downstream of the first codon. You are screening for the clone with the correct orientation by restriction digestion of the plasmid using HindIII plus BamHI (H+B) and HindIII plus PstI (H+P). The map of the relevant region of the vector is shown below:

Given below is the pattern following restriction digestion of plasmid isolated from four independent clones (P, Q, R or S).

Which of the plasmids shown above represents the cone in the correct orientation?

(a) P

(b) Q

(d) S

Ans. (c)

Sol.

Based on gel picture

145. Fluorescence recovery after photobleaching (FRAP) is a method to estimate the diffusion of molecules in a membrane. Fluorescently labeled molecules such as:

1. a receptor tagged with green molecules such as

2. a receptor labelled with GFP which interacts with cytoskeleton.

3. a labeled lipid

4. a labeled protein that binds to the membrane surface.

The following data were obtained:

Which one of the combinations is correct?

(a) P =1, Q = 2

(b) Q = 3, P = 4

(d) S = 2, Q = 1

Ans. (b)

Sol. In FRAP (fluorescence recovery after photobleaching), fastest recovery after photobleaching will be observed in samples that have highest mobility. Among various molecules present in membrane lipids are known to have highest mobility therefore, recovery will be fastest in lipids this is appropriately represented by graph Q. Graph R and S represent slow recovery, (in fact R did not recover completely), therefore S may represent receptor that interacts with cytoskeleton.

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Previous Year Question Paper with Solution.