PYQ DEC 2012 - VedPrep

CSIR NET BIOLOGY (DEC - 2012)
Previous Year Question Paper with Solution.

21. Out of the following hydrogen bonding schemes shown by..., which one corresponds to the weakest hydrogen bond in a given solvent condition ?

(a) O – H ... O <

(b) N – H ... O <

(d) N – H ... N <

Ans. (d)

Sol. Weakest hydrogen bond will be formed when the electronegativity difference between the atom and H is the least. Hence, (d) will be the answer.

22. Which peptide bond (s) marked as a, b, c, d and e will be broken when the following oligopeptide is treated with trypsin at pH 7.0 ?

Lys a Arg b Pro c Lys d Arg e Gly

(a) a, b, d, e

(b) b, d, e

(d) d

Ans. (d)

Sol. Trypsin cleaves the peptide bond between the carboxyl group of arginine or the carboxyl group of lysine and the amino group of the adjacent amino acid.

23. In cellular respiration, which of the following processes occur only insdie mitochondria and not in the cytoplasm ?

(a) Glycolysis and pentose-phosphate pathway

(b) Glycolysis and the citric acid cycle

(d) Glycolysis and oxidative phosphorylation

Ans. (c)

Sol. The second stage of cellular respiration occurs in the mitochondrial matrix of cell and takes part through the citric acid cycle. Oxygen is essential in this stage. Acetyl CoA is the substrate of citric acid cycle. Two molecules of Acetyl CoA are produced in first stage of cellular respiration by breakdown of one molecule of glucose. In second stage, these two molecules of Acetyl CoA are oxidized to carbon dioxide with release of two molecules of ATP, six molecules of NADH and two molecules of FADH₂. One NADH provides three ATP and one FADH₂ provides two ATP via the electron transport chain. Therefore, in second stage of cellular respiration, two molecules of Aetyl CoA provide total 24 molecules of ATP.

24. An enzyme catalyzed reaction was measured in the presence and absence of an inhibitor. For an uncompetitive inhibition,

(a) only K_m is increased

(b) both K_m and V_max are decreased

(d) both K_m and V_max are not affected

Ans. (b)

Sol. On rearranging the equation K_max = K_m/(1 + I/K_ii) and V_max = V_m/(1+I/K_ii) we see that the apparent K_m and V_m do decrease as we predicted. K_ii is the inhibitor affects the intercept of the double reciprocal plot. Note that if I is zero, K_m and V_max unchanged.

25. The Gibbs free energy of binding of a ligand with a protein is determined using calorimetric measurements at 25°C. The value of thus determined is 1.36 kcal/mole. The binding constant for the ligand-protein association is

(a) 1.30 × 10^–12

(b) 0.10

(d) 0.97

Ans. (b)

Sol. = + RT ln K_b

0 = 1.36 kcal/mol + 2 × 10^–3 × 298 ln K_b

0 = 1.36 kcal/mol + 0.002 × 298 ln K_b

0 = 1.36 kcal/mol + 0.596 kcal/mol ln K_b

log K_b =

K_b = 0.10

26. A is converted to E by enzymes E_A, E_B, E_C, E_D. The k_m(M) values of the enzyme are 10^–2, 10^–4, 10^–5 and 10^–4, respectively. If all the substrates and products are persent at a concentration of 10^–4 M, and the enzymes have approximately the same V_max, the rate limitine step will be

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The larger the value of K_m, the smaller will be V₀ and therefore it will represent a slow step or rate limiting step.

27. KCl (100 mM) was entrapped inside large unilamellar vesicles. A diffusion potential across the bilayer can be generated by diluting with buffer containing

(a) 100 mM KCl and a protonophore

(b) 100 mM NaCl and a protonophore

(d) 100 mM NaCl and a K⁺ specific ionophore

Ans. (d)

Sol. Ionophores catalyze ion transport across hydrophobic membranes, such as liquid polymeric membranes (carrier-based ion selective electrodes) or lipid bilayers found in the living cells or synthetic vesicles (liposomes). We can create a perfectly K+-selective membrane by adding certain organic molecules, known as K+ionophores, to a planar lipid bilayer. Examples are valinomycin and gramicidin. These molecules have the ability to partition into bilayers and catalyze the diffusion of K+ across phospholipid membranes. Valinomycin and gramicidin act by different mechanisms, but both allow a current of K+ ions to flow across membranes.

28. An organisms that has peroxidase and superoxide dismutase but lacks catalase is most likely an

(a) aerotolerant anaerobe

(b) aerotolerant aerobe

(d) facultative anaerobe

Ans. (a)

Sol. Most aerotolerant anaerobes have superoxide dismutase and (non-catalase) peroxidase but don't have catalase. More specifically, they may use a NADH oxidase/NADH peroxidase (NOX/NPR) system or a glutathione peroxidase system. An example of an aerotolerant anaerobe is Cutibacterium acnes.

29. Blood group type A antigen is a complex oligosaccharide which differs from H antigen present in type O individuall by the presence of terminal

(a) glucose

(b) galactose

(d) fucose

Ans. (c)

Sol. All people synthesize a precursor carbohydrate, called the H antigen, which is attached to lipids or proteins on the outer surface of red blood cells. Specific enzymes synthesized by the ABO genes attach additional monosaccharides to the H antigen, and the completed carbohydrate determines that person's blood type. A and B blood differ from the O type in the presence of an additional sugar antigen (GalNAc and Gal, respectively) on the core H-antigen found on O-type RBCs.

30. Phosphatidyl serine (PS) is mostly located in the inner bilayer of plasma membrane of red blood cells (RBCs). You have to prove this fact about PS by an experiment. You are provided with PS-specific lytic enzymes (PSE) and other reagents needed. Identify the correct sequences of experiments to be carried out to settle this issue.

(a) RBCs inside out vesicles PSE thin layer chromatography (TLC)

(b) RBCs right side out vesicles TLC PSE

(d) RBCs PSE TLC inside out vesicles

Ans. (a)

Sol. The exposure of phosphatidylserine (PS) on the outer membrane leaflet of red blood cells (RBCs) serves as a signal for suicidal erythrocyte death or eryptosis, which may be of importance for cell clearance from blood circulation. It can also be detected using PS- specific lytic enzymes and other reagents.

31. In a given experiment the cells were labeled for 30 minutes with radioactive thymidine. The medium was then replaced with that containing unlabelled thymidine and the cells were grown for additional time. At different time points after replacement of medium the fraction of mitotic cells were analysed. Based on the results obtained, the figure below was drawn which shows the percentage of mitotic cells that are labeled as a function of time after brief incubation with radioactive thymidine.

Considering the above experiment, the following statements were made :

P. Cells in the S-phase of the cell cycle during the 30 minute labeling period contain radioactive DNA

Q. It takes about 3 hours before the first labeled mitotic cells appear

R. The cells enter the second round of mitosis at t₃₀ hours

S. The total length of the cells cycle about 17 hours with G₁ being more than 15 hours

Which of the combinations of the above statements is correct ?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (a)

Sol. This shows that cells in the S-phase of cell cycle during the 30 min. labelling period contain radioactive DNA and it takes about 3 hours before first labelled mitotic cells to appear.

32. ATP-driven pumps hydrolyze ATP to ADP and phiosphate and use the energy released to pump ions or solutes across a membrane. There are many classes of these pumps and representatives of each are found in all prokaryotic and eukaryotic cells. Which of the following statements about these pumps is not correct ?

(a) P-type pumps are multipass transmemebrane proteins which phosphorylate themselves during pumping and involve in ion transport.

(b) F-type pumps normally use the H⁺ gradient across the membrane to drive the synthesis of ATP.

(d) ABC transporters primarily pumps small molecules across cell membrane.

Ans. (c)

Sol. Vacuolar-type proton pumping ATPase (V-ATPase), initially identified in Saccharomyces cerevisiae and plant vacuoles, is a ubiquitous enzyme responsible for H+ (proton) transport across membranes and acidification of cellular compartments in animals.

33. Which of the following names is appropriate for the sequence 5'-G/ANNAUG-3' in a mammalian mRNA ?

(a) Shine-Dalgarno sequence

(b) Kozak sequence

(d) Translation termination site

Ans. (b)

Sol. Kozak sequence is the sequence 5'-G/ANNAUG-3' in a mammalian mRNA. It possess start codon.

34. During DNA replication, events at the replication fork require different types of enzymes having specialized functions except

(a) DNA polymerase III

(b) DNA gyrase

(d) DNA glycosylase

Ans. (d)

Sol. During DNA replication, events at the replication fork require different types of enzymes having specialized functions except DNA glycosylase. It participates in base excision repair.

35. With reference to Iac operon, what will be the phenotype of an E. coli strain having a genotype
?

(a) Constitutive for both -galactosidase and Lac permease

(b) Inducible for both -galactosidase and Lac permaease

(d) Constitutive for -galactosidase and inducible for Lac permease

Ans. (c)

Sol. With reference to Lac operon, phenotype of an E. coli strain having a genotype will be Inductible for -galactosidase and constitutive for Lac permease.

36. The specificity of tRNA recognition by a aminoacyl tRNA synthetase that is intrinsic to the tRNA molecule lies on

(a) acceptor stem

(b) acceptor stem and anticodon loop

(d) D-arm

Ans. (b)

Sol. The specificity of tRNA recognition by a aminoacyl tRNA synthetase that is intrinsic to the tRNA molecule lies on acceptor stem and anticodon loop.

37. Viral gene expression after T3 bacteriophage infection is controlled by

(a) repressor molecule

(b) slow injection of nucleic acid

(d) DNA polymerase

Ans. (b)

Sol. Viral gene expression after T3 bacteriophage infection is controlled by slow injection of nucleic acid.

38. A bacterial population has a plasmid with copy number 'n'. It was observed that on ann average in one out of 2^{(n – 1)} cell divisions, there was spontaneous plasmid curing. It was inferred from the observation that

P. Each cell divison does not have equal probability of plasmid curing.

Q. There is no evidence for any mechanisms of plasmid segregation in the two daughter cells.

R. Plasmid distribution to daughter cells is random.

S. Each plasmid has an equal chance of being in either of the two daughter cells.

Which of the combinations of above statements is true ?

(a) P and Q

(b) Q and S

(d) Q, R and S

Ans. (d)

Sol. In the condition mentioned in question following statements are true about plasmid curing :

Q. There is no evidence for any mechanisms of plasmid segregation in the two daughter cells.

R. Plasmid distribution to daughter cells is random.

S. Each plasmid has an equal chance of being in either of the two daughter cells.

39. Mutants of Iac Y(Y^–) gene of E. coli do not synthesize the lactose permease protein. The following statements refer to the behaviour of Iac Y^– mutants under different experimental conditions.

1. No synthesis of -galactosidase when Y^– cells are induced with lactose.

2. Synthesis of -galactosidase when cells are induced with lactose.

3. No synthesis of -galactosidase when cells are induced with IPTG.

4. Synthesis of -galactosidase when cells are induced with IPTG.

5. The cells induced with IPTG cannot grow in the presence of TONPG (TONPG is a compound, whose uptake is mediated by lactose permease and cleaved by -galactosidase to release a toxic compound).

6. Cells induced with IPTG can grow in the presence of TONPG.

Which combination of the above statements is correct ?

(a) 1, 4 and 6

(b) 2, 3 and 5

(d) 1, 3 and 5

Ans. (a)

Sol. The behaviour of Lac Y^– mutants under different experimental condition can be refered as

No synthesis of -galactosidase when Y^– cells are induced with lactose.

Synthesis of -galactosidase when cells are induced with IPTG.

Cells induced with IPTG can grow in the presence of TONPG.

40. The semiconservative nature of DNA replication was established by Meselson and Stahl in their classic experiment with bacteria. They grew bacteria in N¹⁵-NH₄Cl containing medium, washed and then incubated in fresh medium with N¹⁴-containing compounds and allowed to grow for three generations. CsCl density gradient centrifugation of isolated DNA established the nature of semiconservative DNA replication. The pictorial representation below shows the position of differentially labeled DNA in CsCl density gradient.

Had the DNA replication been conservative, what would have been the pattern ?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. From the description mentioned in question, we can conclude that Transcription of gene X is coupled to mRNA capping, Transcription elongation is coupled to splicing and phosphorylation of CTD is required for the recruitment of capping and splicing enzymes.

41. HeLa cell extract was used to study transcription of a gene X having six introns. RNA Pol II complex containing all associated proteins was isolated from actively transcribing system and subjected to proteome analysis. Results showed the presence of both splicing and capping enzymes in the complex. When transcription elongation was inhibited by flavopiridol, polymerase complex contained only capping enzymes. When phosphorylation of the CTD domain of Pol II was inhibited by a kinase inhibitor, the complex contained neither splicing nor capping enzymes. From these results, following conclusions were made :

P. Transcription of gene X is coupled to mRNA capping.

Q. Transcription elongation is coupled to splicing.

R. Phosphorylation of CTD is required for the recruitment of capping and splicing enzymes.

S. Both capping and splicing of mRNAs occur simulaneously.

Identify the correct set of conclusions :

(a) P, Q and R

(b) Q, R and S

(d) S, P and Q

Ans. (a)

Sol. As per the information in question pattern in option (c) is correct.

42. In bacteria, N-fomyl methionine is the first amino acid to be incorporated into a polypeptide chain. Accordingly, one would think that all bacterial proteins have a formyl group at their amino terminus and the first amino acid is methionine. However, this is not the case, because of the following possible reasons.

P. Deformylase removes the formyl group only during or after the synthesis of the polypeptides.

Q. Aminopeptidase removes onlyy the amino terminal methionine.

R. Aminopeptidase removes only the amino terminal methionine.

S. Deformylase removes the formyl group as well as amino terminal methionine and adds one or two amino acids to it.

Choose the combination of correct answers from the following :

(a) Q and R

(b) P and Q

(d) P and S

Ans. (c)

Sol. Bacterial proteins do not have formyl group at their amino terminus this is because deformylase removes the formyl group only during or after the synthesis of the polypeptides and amino peptidase removes only the amino terminal methionine.

43. Bacteriophage is a temperate phage. Immediately after infection, viral specific mRNAs for N and Cro proteins are expressed followed by early mRNAs. At the commitment phase, either lytic cycle starts with the expression of genes for head tail, and lytic proteins or lysogenisation cycle begins with the expression of repressor and integrase genes. During induction of lysogens both INT and XIS proteins are needed along with host factors. Out of the four processes below, some govern integration of viral genome and its excision.

P. Repression of transcription

Q. Retroregulation

R. Rearrangement of viral genome

S. Repression of translation

Identify the correct set of combination :

(a) P and Q

(b) Q and R

(d) S and P

Ans. (b)

Sol. During induction of lysogens both INT and XIS proteins are needed along with host factors. Retroregulation and Rearrangement of viral genome govern integration of viral genome and its excision.

44. In E. coli, recA gene is involved in recombination as well as repair and dnaB gene is involved in unwinding of DNA double strands during replication. Which of the following statement is/are correct about RecA and DnaB ?

P. Mutation in E. coli recA gene is lethal.

Q. E. coli with mutated dnaB gene does not survive.

R. DnaB after uncoling DNA double strands, prevents further reannealing at the separated strands.

S. recA gene is involved in SOS response and helps DNA repair.

The correct options are :

(a) Q and R

(b) P and Q

(d) P and R

Ans. (c)

Sol. DNA-B is helicase which is required for replication and without it E.coli can't survive. The recA gene product is a multifunctional enzymes that plays a role in homologous recombination, DNA repair and induction of the SOS response. In homologous recombination, the protein functions as a DNA-dependent ATPase, promoting synapsis, heteroduplex formation and strand exchange between homologous DNAs.

45. The challenges faced by aminoacyl tRNA synthetases in selecting the correct amino acid is more daunting than its recognition of the appropriate tRNA. In case of amino acids with similar structures like valine and isolecucine, this challenge is met by the enzyme possibly through its

P. Catalytic pocket

Q. Editing pocket

R. Anticodon loop

S. Acceptor arm

Choose the correct set from the following :

(a) P and Q

(b) P and R

(d) Q and R

Ans. (a)

Sol. The challenges faced by aminoacyl tRNA synthetases in selecting the correct amino acid is more daunting than its recognition of the appropriate tRNA. In case of amino acids with similar structures like valine and isolecucine, this challenge is met by the enzyme possibly through its Catalytic pocket, Editing pocket.

46. Locus control region (LCR) lies far upstream from the gene cluster and is required for the appropriate expression of each gene in the cluster. LCR regulates expression of globin genes in the cluster through the following ways.

P. LCR interacts with promoterrs of individual genes by DNA looping through DNA-binding proteins.

Q. The LCR-bound proteins attract chromatin-remodelling complexes including histon-modifying enzymes and components of the transcription machinery.

R. LCR acts as an enhancer for global regulation of gene cluster and does not regulate individual genes.

S. LCR participates in converting inactive chromatin to active chromatin around the gene cluster.

Choose the correct set of combinations :

(a) P and Q

(b) P and R

(d) Q and S

Ans. (a)

Sol. LCR regulates expression of globin genes in the cluster when LCR interacts with promoterrs of individual genes by DNA looping through DNA-binding proteins, and LCR-bound proteins attract chromatin-remodelling complexes including histon-modifying enzymes and components of the transcription machinery.

47. Acetylcholine receptor is an archetype for :

(a) Ligand gated ion channel

(b) ATPase dependent voltage-gated ion channel

(d) ATPase independent voltage gated ion channel

Ans. (a)

Sol. The nicotinic acetylcholine receptor (nAChR) is the archetypal ligand-gated ion channel. A model of the alpha7 homopentameric nAChR is described in which the pore-lining M2 helix bundle is treated atomistically and the remainder of the molecule is treated as a "low resolution" cylinder. The surface charge on the cylinder is derived from the distribution of charged amino acids in the amino acid sequence (excluding the M2 segments).

48. Which of the following factors is not true for the low levels of immune response in Plasmodium infection ?

(a) Different types of antigens are expressed at various stages of Plasmodium life cycle

(b) Most of the phases in the life cycle of Plasmodium are intracellular

(d) Plasmodium infection primarily destroys macrophages and dendritic cells

Ans. (d)

Sol. Anemia a serious clinical manifestation of malaria is due to increased destruction of both infected and uninfected red cells due to membrane alterations, as well as ineffective erythropoiesis.

49. Presence of the nuclear localization signal (NLS) in a steroid receptor indicates that the receptor resides

(a) on the nuclear membrane

(b) within the nucleus

(d) in the cytosol

Ans. (d)

Sol. The activated GR complex up-regulates the expression of anti-inflammatory proteins in the nucleus or represses the expression of pro-inflammatory proteins in the cytosol.

50. Which of the following is an intracellular anchor protein ?

(a) Vitronectin

(b) Vinculin

(d) Elastin

Ans. (b)

Sol. As intracellular protein of 130,000 molecular weight was recently isolated in this laboratory from chicken gizzard smooth muscle. By immunofluorescence observations of cultured chicken fibroblasts, it was shown to be concentrated on the ventral surfaces of the cells where they formed focal adhesions to the substratum [Geiger, B. (1979) Cell 18, 193-205]. Focal adhesions are sites where, inside the fibroblast, microfilament bundles are known to terminate at the cell membrane.

51. Out of the following matches of orcogenes with the proteins that each specifies, which one is incorrect ?

(a) erbA-thyroid hormone receptor

(b) erbB-epidermal growth factor receptor

(d) fos-platelet derived growth factor

Ans. (d)

Sol. Fructooligosaccharides (FOS) also sometimes called oligofructose or oligofructan, are oligosaccharide fructans, used as an alternative sweetener. FOS exhibits sweetness levels between 30 and 50 percent of sugar in commercially prepared syrups.

52. A nerve fibre cannot be stimulated during the absolute frefractory peirod of a previous stimulus because

(a) sodium permeability remains high

(b) sodium potassium pump does not operate

(d) voltage gated calcium channels remain closed

Ans.

Sol. The period of time after an action potential begins during which an excitable cell cannot generate another action potential in response to a normal threshold stimulus is called the refractory period. It can be absolute or relative. During the absolute refractory period, even a very strong stimulus cannot initiate a second action potential. This period coincides with the period of Na⁺ channel activation and inactivation. Inactivated Na⁺ channels cannot reopen; they first must return to the resting state. The relative refractory period is the time period during which a second action potential can be initiated, but only by a larger-than normal stimulus. It coincides with the period when the voltage gated K⁺ channels are still open after inactivated Na⁺ channels have returned to their resting state.

53. p24 is an important core protein of HIV. This protein is abundant during active replication of the virus. The serum of an HIV patient was examined for the presence of p24 and antibody against p24 for proper diagnosis of the infection stage. Match the clinical observations in column A with the inferences in column B.

Column A Column B

P. p24 is present in the serum 1. Viral latency

Q. Anti-p24 antibody is high in the serum 2. Progression of HIV from latency to lytic

stage

R. Anti-p24 antibody begins to decline 3. Early stage of infection

with corresponding increases in p24

Choose the correct matching

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

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

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

Ans. (c)

Sol. High levels of p24 are present in the blood serum of newly infected individuals during the short period between infection and seroconversion, making p24 antigen assays useful in diagnosing primary HIV infection. Presence of high Anti-p24 antibody in serum meaning the virus is in dormant stage (lysogenic phase) and decline in anti-p24 antibody with increase in p24 shows progression of HIV ( lytic phase).

54. After successive surgery and chemotherapy, the tumor of a beast patient subsided. However, after almost 5 years, the tumor relapsed in a more aggressive manner and did not respond to the conventional chemotheapy delivered earlier. The following postulations were made.

P. Chemoresistant cells were persisting within the tumor even after therapy.

Q. A population of quiescent cells existed, which under favourable conditions, transformed to new tumor cells.

R. High ABC (ATP-Binding Cassette)-transporter expressing cells perissted in the breast during chemotherapy.

S. Breast tumor cells which may have migrated to other tissues, returned to the breast immediately after chemotherapy was terminated.

Which of the above combination of statements is true ?

(a) P and S (b) P, Q and R (c) only Q (d) Q and S

Ans. (b)

Sol. Recurrent cancer may be more aggressive than the original cancer if it's already spread to other parts of the body, or if it's become resistant to chemotherapy. The sooner the cancer returns, the more serious it likely is. Cancers come back when small numbers of cancer cells can remain in the body after treatment. Quiescent cells may be present which under favourable conditions transformed to new tumor cells.

55. Epidermal growth factor (EGF) is needed for growth of almost all cells. EGF receptor is a transmembraned protein having an extracellular ligand-binding domain, a transmembrane domain, a transmembrane domain an a cytosolic domain of protein tyrosine kinase (PTK). Binding of EGF to the receptor activates PTK resulting in activation of transcription factors through intracellular transducers. In cell type A, much of the extracellular ligand-binding domain is deleted by proteases such that cytosolic domain of PTK becomes constitutively active whereas cell type B is having normal EGF receptor. What will be the best fit graph for the growth of the cultures of cell type A and B in complete medium in presence (+) and absence (–) of EGF ?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The epidermal growth factor receptor protein is involved in cell signaling pathways that control cell division and survival. Sometimes, mutations (changes) in the EGFR gene cause epidermal growth factor receptor proteins to be made in higher than normal amounts on some types of cancer cells. Epidermal growth factor (EGF) may bind to EGF receptor (EGFR) on cells to activate a variety of signal pathways including MAPK-Erk, PI3K-Akt, and STAT pathways. These pathways play important roles in the cell differentiation and proliferation. As in cell type A, cytosolic domain of PTK is constitutively active, number of cells will be highest. Whereas in Norma cell type B, no. of cells will be high when EGF is present and less when EGF is absent.

56. A particular type of cancer cell undergoes apoptosis by both extrinsic and intrinsic pathways when treated with a chemotherapeutic agent X. Caspase 8 and Caspase 9 arer the initiator caspases associated with extrinsic and intrinsic pathways respectively. Now, if caspase 9 is silenced in the cancer cell by shRNA transfection, what will be the best-fit graph for apoptosis scenairo in the cancer cell when treated with agent X ?

cell before transfection cell after transfection

(a)

(b)

(c)

(d)

Ans. (a)

Sol. As initiator caspase 9 is silenced, so intrinsic pathway of apoptosis will not be followed and cells will keep growing declining the apoptosis percentage. So, graph (a) best represent the apoptosis scenario.

57. Following are the experimental observations made on treatment of B-cells

P. Anti-immunoglobulin (anti-Ig) antibody treatment results in B-cells apoptosis.

Q. Anti-Ig plus CD40 ligand treatment results in B-cell proliferation.

R. Anti-Ig plus CD40 ligand plus IL-4 treatment results in B-cell proliferation and switching to IgG1.

S. Anti-Ig plus IL-40 treatment results in less B-cell Proliferation but switching to IgE.

From the above observation, which one of the following is the correct interpretation for the role of CD40 in B-cell function ?

(a) Induce death of B-cells

(b) Rescue B-cells from death and Ig class switch to IgG1

(d) Induce Ig class switching to both IgG1 and IgE and inhibit B-cell proliferation

Ans. (b)

Sol. CD40 signaling of B cells promotes germinal center (GC) formation, immunoglobulin (Ig) isotype switching, somatic hypermutation (SHM) of the Ig to enhance affinity for antigen, and finally the formation of long-lived plasma cells and memory B cells. So, CD40 rescue B-cells from death and Ig class switch to IgG1.

58. Mouse IgG is left either intact (left lane in A, B, C, D) or digested with papain or pepsin or treated with -mercaptoethanol (-ME) and run on non-reducing SDS-PAGE and stained with Coomassie blue. In a separate experiment, papain-digested aer immunoblotted with an anti-idiotypic monoclonal antibody. Following four profiles are attributed to each of these treatments.

Which one of the following possibilities is correct ?

(a) A (pepsin), B (papain), C (-ME), D (papain, followed by anti-idiotype immunoblot)

(b) A (pepsin), B (papain), C (papain, followed by anti-idiotype immunoblot), D (-ME)

(d) A (-ME), B (papain), C (pepsin), D (papain, followed by anti-idiotype immunoblot)

Ans. (a)

Sol. A : Pepsin

Digestion by the enzyme pepsin normally produces one F(ab')2 fragment and numerous small peptides of the Fc portion. The resulting F(ab')2 fragment is composed of two disulfide-connected Fab units. The Fc fragment is extensively degraded, and its small fragments can be separated from F(ab')2.

B : Papain

When IgG molecules are incubated with papain in the presence of a reducing agent, one or more peptide bonds in the hinge region are split, producing three fragments of similar size: two Fab fragment and one Fc fragment.

C : beta-mercaptoethanol

2-Mercaptoethanol alters the properties of the Fc fragment of the IgG antibody molecule by cleaving inter-H-chain-disulfide bonds. Treatment of the antibodies with 2-ME resulted in a significant reduction of their capacity to induce cytotoxicity of lymphocytes against the respective antigen-antibody complex.

D : Papain, followed by anti-idiotype immunoblot.

59. A potentially valuable therapeutic approach for killing tumor cells without affecting normal cells is the use of immunotoxins. Immunotoxins constitute monoclonal antibodies against tumor cells conjugated to lethal toxins. Which of the following molecular approaches do you think is not appropriate for generating tumor cells-specific immunotoxin that will not kill normal cells ?

(a) Cell surface receptor binding polypeptide chain of toxin molecule should be replaced by monoclonal antibodies against a particular tumor cell type

(b) Constant region Fc domain of tumor cell specific monoclonal antibody should be replaced by ligation of toxins

(d) Inhibiton polypeptide chain of toxin should be conjugated to F(ab)₂ domain of tumor cell specific monoclonal antibody

Ans. (c)

Sol. The variable region, composed of 110-130 amino acids, give the antibody its specificity for binding antigen. If that region is altered then the tumor cells-specific monoclonal antibody will lose its specificity and will kill normal cells.

60. Capacitation of sperms in human

(a) occurs during copulation

(b) occurs after the acrosome reaction

(d) takes place in the epididymis of testis

Ans. (c)

Sol. Capacitation is the sperm activation process in mammals that occurs in the female reproductive tract. In the female reproductive tract, some secretion destabilizes the acrosomal part and activates sperm to penetrate the egg and fertilize the ovum.

61. With respect to development of any organisms, "autonomous specification" would result in which type of development ?

(a) Regulative

(b) Mosaic

(d) Definitive

Ans. (b)

Sol. Autonomous specification gives rise to a pattern of development referred to as mosaic development, since the embryo appears to be constructed like a tile mosaic of independent self-differentiating parts.

62. The group of cells which generates the vascular tissues including the pericycle in roots of higher plants are called

(a) procambium

(b) protoderm

(d) apical meristem

Ans. (a)

Sol. The procambium is a meristematic tissue concerned with providing the primary tissues of the vascular system; the cambium proper is the continuous cylinder of meristematic cells responsible for producing the new vascular tissues in mature stems and roots including the pericycle in roots of higher plants.

63. If an embryo undergoes 13 cleavage divisions during embroyogenesis, then the size of the embryo compared to zygote

(a) increases 13 times

(b) increases in an exponential fashion

(d) remains almost the same

Ans. (d)

Sol. During the cleavage process, the zygote frequently restores the large cytoplasmic mass into a large number of small blastomeres. It pertains to cell division without the growth in size due to cells resuming to be maintained within the zona pellucida. The enormous volume of egg cytoplasm is divided into numerous smaller, nucleated cells. These cleavage-stage cells are called blastomeres

64. Flowers represent a complex array of functionally specialized structures that differ substantially from the vegetative platn body in form and cell types. Following are statements made regarding floral meristems.

P. Floral meristems can usually be distinguished from vegetative meristems by their larger size.

Q. The increase in the size of the meristem is largely a result of increased rate of cell division in central cells.

R. The increase in the size of the meristem is due to larger size of the cells, which in turn results from rapid cells expansion only.

S. A network of genes control floral merphogenesi in plants.

Which combination of the above statements is true ?

(a) P, Q and S

(b) P, Q and R

(d) P, R and S

Ans. (a)

Sol. Flowers contain the male and female sexual organs that are critical for plant reproduction and survival. Each individual flower is produced from a floral meristem that arises on the flank of the shoot apical meristem and consists of four organ types: sepals, petals, stamens, and carpels. Floral meristems are larger in size than the vegetative meristems. Increase in size of the floral meristem is due to larger size of the cells which in turn result from rapid cell expansion only.

65. Three embroys, X (wild type), Y (mutant for bicoid) and Z (mutant for nanos) were injected with bicoid mRNA in their posterior pole at early stage. What would be the phenotypes of the resulting embryos ?

(a) Embroy X will develop head on both anterior and posterior side, while embroys Y and Z will develop head on posterior side only

(b) Embroys X and Z will develop head on both anterior and posterior side, while embryo Y will develop head on posterior side only

(d) Embryo X will develop head on anterior side, embryo Y will develop no head, while embryo Z will develop head on anterior as well as posterior side

Ans. (b)

Sol. The Drosophila gene bicoid functions as the anterior body pattern organizer of Drosophila. Embryos lacking maternally expressed bicoid fail to develop anterior segments including head and thorax. The Bicoid protein activates hunchback gene transcription in the anterior part of the embryo, while the Nanos protein inhibits the translation of hunchback RNA in the posterior part of the embryo.

66. In C. elegans during embryogenesis, an anchor cell and 6 hypodermal vulval precursor cells (VPCs) get involved in forming the vulva. If 3 of the hypodermal VPCs are killed by a laser beam, a normal vulva is still formed. This could be due to the following possible reasons.

P. Six hypodermal VPCs form equivalence grou of cells, out of which only 3 participate in vulva formation and 3 cells remain as reserve cells.

Q. When 3 hypodermal VPCs are killed, the 3 neighboring hypodermal non-VPCs get freshly recruited.

R. Anchor cell functions as an inducer which can induce epithelial cells of the gonad to get recruited to compensate for the loss.

S. Anchor cell acts as an inducer which can spatially induce only 3 hypodermal cells to the fomr th vulva.

Which combination of the above statements is correct ?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (d)

Sol. Generation of VPCs six vulval precursor cells (VPCs) are specified among the 11 cells, which are located in the ventral epidermis

Vulval precursor patterning a signal from the gonad and signalling among the VPCs specify three VPCs to generate vulval cells.

The anchor cell induces nearby epidermal precursor cells to generate vulval cells via an epidermal growth factor (EGF) signaling pathway. The precise pattern of vulval precursor cell fates involves the graded action of the EGF signaling and LIN-12 (Notch) mediated lateral signaling.

67. In sea urchins, a group of cells at the vegetal pole become specified as the large micromere cells. These cells are determined to become skeletogenic mesenchme cells that will leave the blastula epithelim to ingress into the blastocoel. This specification is controlled by the expression of Pmar1 which is a repressor of HesC. HesC represses the genes encoding transcription factors activating skeleton forming genes. The gene regulatory network is given below.

Activation Repression

Below column I lists the experiments carried with mRNA/antisense RNA of different genes injected into single-celled sea urchin embryo while column II lists the developmental outcomes. Match the following :

Column I (injection of) Column II (developmental outcomes)

P. mRNA of Pmar1 1. All cells will start ingressing into the

blastocoel

Q. mRNA of HesC 2. Skeleton mesenchyme will not be

formed

R. Antisense of Pmar1

S. Antisense of HesC

Which of the following combinations is correct ?

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

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

(d) P-2, Q-2, R-2, S-2

Ans. (c)

Sol. Specification of sea urchin embryo micromeres occurs early in cleavage, with the establishment of a well defined regulatory state. A gene encoding a transcriptional repressor, pmar1, is activated specifically in micromeres, where it represses transcription of a second repressor that is otherwise active globally. Thus, the micromere-specific control genes, which are the target of the second repressor, are expressed exclusively in this lineage.

68. Which of the following cellular communications shown below will override the process of normal development and lead to cancer ?

(a) Q and R

(b) P and R

(d) Q and S

Ans. (b)

Sol. Inhibition of apoptotic factors by anti- apoptotic factor ( Bcl-2) will lead to overgrowth of cells causing cancer.

69. In tadopoles, if the tail is amputated in can regenerate. However, if the tail is amputated and then exposed to retinoic acid, it develops limbs instead of regenerating the tail. This could be due to the following reasons :

P. Retinoic acid is a morphogen and induces genes responsibel for limb formation.

Q. Retinoic acid raises the positional values in that region for limb development to take place.

R. This is a random phenomenon and is not well understood.

S. Retinoic acid possibly acts as a mutagen and the phenotype observed is a result of seveal mutations.

Which combination of the above statements is true ?

(a) P and Q

(b) R and S

(d) Q and R

Ans. (a)

Sol. Retinoic acid was thought to play a key instructive role during limb bud initiation and subsequent patterning. New results argue instead that its role is permissive: retinoic acid is essential only to antagonize early axial Fgf signals that otherwise inhibit the limb field.

70. The chlorosis (yellowing) symptom of iron deficiency is influenced by

(a) sodium and potassium

(b) sodium and phosphorus

(d) potassium and phosphorus

Ans. (d)

Sol. Excesses of potassium, magnesium and phosphorus also contribute to chlorosis. When present in excess, these elements cause some trees, particularly oaks and maples, to take up inadequate amounts of the micronutrients iron and manganese. If iron or maganese deficiency is suspected, there are both long-term and short-term treatment the pH as well as the availability of nutrients that cause chlorosis. Stressors, such as temperature extremes, drought, poor drainage (which limits soil aeration) or restricted root growth, further limit nutrient uptake in plants sensitive to chlorosis.

71. A plant hormone that promotes the acquisition of desiccation tolerance in developing seed is

(a) ABA

(b) ethylene

(d) GA3

Ans. (a)

Sol. ABA plays important roles in many aspects of seed development, including accumulation of storage compounds, acquisition of desiccation tolerance, induction of seed dormancy and suppression of precocious germination.

72. Change in Ca²⁺ concentration can initiate various repsonses in plants. Which one of the following responses is not known to be initiated by change in Ca²⁺concentration ?

(a) Closure of stomata

(b) Reorentation of growth in pollen tubes

(d) Lateral root formation

Ans. (d)

Sol. The plant hormone auxin appears to play a critical role in lateral root initiation. Exogenous application of auxin stimulates lateral root formation and some auxin-resistant mutants have reduced numbers of lateral roots.

73. Water can move through the soil-plant-atmosphere continuum, only if water potential along that path

(a) decreases

(b) increases

(d) fluctates rapidly in eithe direction

Ans. (a)

Sol. Water potential is the measure of potential energy in water and drives the movement of water through plants. Plants use water potential to transport water to the leaves so that photosynthesis can take place. Roots have higher water potential than plant and atmosphere has the least water potential. So, movement of water is from soil- plant- atmosphere.

74. The citric acid cycle in respiration yields :

(a) 1 GTP, 3 NADH, 1 FADH₂, 2 CO₂

(b) 2 GTP, 2 FADH₂, 6 NADH, 2 CO₂

(d) 32 GTP, 2 NADH, 4 FADH₂, 4 CO₂

Ans. (a)

Sol. Citric Acid Cycle : The citric acid cycle can be summarized by either of the diagrams below. The diagram below occurs twice, once for each acetyl CoA.

Coenzyme A is removed when the two carbon compound is attached to a four-carbon compound producing a six-carbon compound (citrate).

Each citrate molecule undergoes a series of reactions that removes 2 carbon atoms which are released as CO₂. In addition, 3 NADH, 1 ATP and 1 FADH₂ are produced. In addition, the four-carbon compound that began the cycle is regenerated.

75. Following are statements related to the organization of the four major protein complexes of thylakoid membrane.

P. Photosystem II is located predominantly in the stacked regions of the thylakoid membrane.

Q. Photosystem I is found in the unstacked regions protruding into stroma.

R. Cytochrome b₆f complex is confined to stroma only.

S. ATP synthase is located in the unstacked regions protruding into stroma.

Which one of the following combinations of above statements is correct ?

(a) P, Q and R

(b) P, Q and S

(d) R, S and P

Ans. (b)

Sol. The light reactions of photosynthesis in green plants are mediated by four large protein complexes, embedded in the thylakoid membrane of the chloroplast. Photosystem I (PSI) and Photosystem II (PSII) are both organized into large supercomplexes with variable amounts of membrane-bound peripheral antenna complexes. PSI consists of a monomeric core complex with single copies of four different LHCI proteins and has binding sites for additional LHCI and/or LHCII complexes. PSII supercomples are dimeric and contain usually two to four copies of trimeric LHCII complexes. These supercomplexes have a further tendency to associate into mega complexes or into cyrstalline domains, of which several types have been characterized. Together with the specific lipid composistion, the structural features of the main protein complexes of the thylakoid membranes form the main trigger for the segregation of PSII and HCII from PSI and ATPase into stacked grana membranes. We suggest that the margins, the strongly folded regions of the membranes that connect the grana, are essentially protein-free and that protein-protein interactions in the lumen also determine the shape of the grana. We also discuss which mechanisms determine the stacking of the thylakoid membranes and how the supramolecular organization of the pigment-protein complexes in the thylakoid membrane and their flexibility may play roles in various regulatory mechanisms of green plant photosynthesis.

76. In plants, the energy of sunlight is a first absorbed by the pigments present in their leaf cells followed by the fixation of carbon through photosynthesis. Consider the following statements.

P. Chlorophyll a and b are abundant in green plants.

Q. Chlorophylls c and d are found in some protists and cyanobacteria.

R. Out of different types of bacteriochlorophyll, type a is the most widely distributed.

S. Out of different types of bacterriochlorphyll, type b is the most widely distributed.

Which one of the following combination of above statements is correct ?

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (a)

Sol. There are four types of chlorophyll: chlorophyll a, found in all higher plants, algae and cyanobacteria; chlorophyll b, found in higher plants and green algae; chlorophyll c, found in diatoms, dinoflagellates and brown algae; and chlorophyll d, found only in red algae. Bacteriochlorophyll a, with absorption peaks in vivo at 800–805 nm and around 870 nm, is especially widespread: it is found in many types of photosynthetic Proteobacteria, including anaerobic sulfur oxidizing bacteria.

77. Nitrate reductase is an important for nitrate assimilation. Given below are some statements on nitrate reductase enzyme :

P. Nitrate reductase of higher plants is composed of two identical subunits.

Q. One subunit of nitrate reductase contains three prosthetic groups.

R. One of the prosthetic groups attached to both subunits is heme.

S. One of the prosthetic groups complexed with ptein in magnesium.

Which one of the following combination of statements on nitrate reductase mentioned above is correct ?

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (a)

Sol. Nitrate reductase catalyzes NAD(P)H reduction of nitrate to nitrite. Nitrate reductase is a homodimeric enzyme with three electron-transferring prosthetic groups per subunit: flavin (FAD), heme and Mo cofactor (Moco). Moco consists of Mo covalently bound to two S atoms in the tricyclic molecule pterin.

78. A farmer growing a particular variety of grape platns in vineyard, observes the following :

P. Fruit size nromally remained small.

Q. Natural seed abortion.

R. Development of fungal infection as the pedicels are small in size due to which moisture is retained in the bunches of grapes.

Experts suggested spraying gibberellic acid during the fruit development. This treatment would help in getting rid of

(a) P, Q and R

(b) only P and Q

(d) only Q and R

Ans. (c)

Sol. Gibberellins can be used to end seed dormancy, promote flowering, increase fruit size. Plant growth regulator gibberellins (GAs) are responsible for elongation of stem leaf expansion, bolting, enzymeformation, parthenocarpy, etc. Their ability to cause increase in length of axis is used to make stalk of grapes lengthy, thereby allowing the grapes to grow larger and as certain better yield.

79. Light is perceived by various photoreceptors in plants. The photoreceptors predominantly work at specific wavelengths of light. Some of the following statements are related to the functions of plant photoreceptors.

P. Phytochrome A predominantly perceives the red and far-red light.

Q. Phytochrome B predominantly perceives red light.

R. Cryptochromes regulate plant development.

S. Phototropins are involved in blue light perception and chloroplast movements.

Which one of the following combinations based on above statements is correct ?

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (b)

Sol. Phytochromes mainly perceive red (R)/far-red light (FR), while the cryptochromes and phototropins recognize UV-A and blue light . Cryptochromes regulate plant development and photomorphogenesis whereas phototropins are primarily implicated in photomovement responses such as phototropism and chloroplast relocation.

80. From the following statements :

P. Triose phosphate is utilized for the synthesis of both starch and sucrose.

Q. Triose phosphate is translocated to cytosol from chloroplast.

R. Triose phosphate is confined to chloroplast and is utilized for synthesis of starch only.

S. Triose phosphate is translocated from cytosol to chloroplast.

Which one of the following combinations is correct regarding starch and sucrose synthesis during day time ?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (a)

Sol. Triose phosphate is utilised for the synthesis of both starch and sucrose. Triose phosphate is trans located from cytosol to chloroplast. Triose phosphates are the principal product of photosynthesis. They are used within the chloroplast for starch synthesis, or translocated to the cytosol where they are used to fuel sucrose synthesis.

81. Shown below, is a graph representing the growth of different plant species subjected to salinity relative to that of unsalnized control. Which of the following statements is not true ?

(a) Plants in group IA are extreme halophytes while very sait sensitive species will be part of group III.

(b) Plants in group IA are very salt sensitive and extreme halophytes will be part of group III.

(d) Non-halophytes, which are salt tolerant but lacks salt glands will be a part of group II.

Ans. (b)

Sol. 1A are salt sensitive and group 3 are extreme halophytes. Halophytes live in salt marshes and are exposed to excessive salt as well as physiological drought.

82. Which one of the following is responsible for the ejection of milk from mammary glands in mammals ?

(a) Oxytocin

(b) Prolactin

(d) Melatonin

Ans. (a)

Sol. There are two hormones involved in the synthesis and release of milk : prolactin and oxytocin. Both are produced in a pea-sized gland attached to the abse of the brain called the pituitary gland and carried to the breasts by the blood.

Prolactin is produced by the anterior (front part) of the pituitary gland. It stimulates the cells in the breasts to synthesise milk. Prolactin synthesis is stimulated by several minutes of the infrant sucking at the breast.

Oxytocin is synthesised by the posterior (rear part) of the pituitary gland. It stimulates the release of the milk from the breast (also called milk 'let down'). Oxytocin production is also stimulated by sucking at the breast.

83. The T-wave of ECG indicates

(a) atrial depolarizatio

(b) ventricular depolarization

(d) atrial repolarization

Ans. (c)

Sol. The T wave represetns ventricular repolarization (i.e., conduction of the repolarization wave is slower than the wave of depolarization). Sometimes a small positive U wave may be seen follownig the T wave (not shown in figure at top of page). This wave represents the last remnants of ventricula repolarization. Inverted or prominent U waves indicates underlying pathology or conditions affecting repolarization.

84. An experimentalist stimulates a nerve fibre in the middle of an axon and records the following observation. Which one of the observation is correct ?

(a) Nerve impulse is travelling in a direction towards cell body.

(b) Nerve impulse is travelling in a direction towards telodendrons.

(d) Nerve impulse is not moving in either direction.

Ans. (c)

Sol. A nerve electrical impulse only travels in one direction. There are a few reasons why nerve impulses only travel in one direction, the most important being synaptic transport.

In order for a depolarization wave or "nerve impulse" to pass from cell to cell, there are what we call synaptic junctions. This means that the nerve cells are lined up head to tail all the way down a nerve track and are not connected, but have tiny gaps between them and the next cell. These tiny gaps are called synapses.

85. Desert animals have longer loop of Henie compared to that of humans. It may be due to the following reasons :

P. Long loop of Henle is associated with greater amount of vasopressin secretion.

Q. In long loop of Henle, the counter-current exchanger is more effective.

R. Long loop of Henle conserves more water.

S. Long loop of Henle stimulates production of angiotensin II.

Which of the above reasons(s) is/are correct ?

(a) P and Q

(b) Q and R

(d) Only S

Ans. (b)

Sol. Water is reabsorbed from descending loop of Henle if there is long loop of Henle, there will be more reabsorption of water. So, In long toop of Henle, the counter-current exchanger is more effective and Long loop of Henle conserves more water.

86. A boy eats a large serving of cheese having high amount of sodium. he hardly drinks any fluid. In splite of this, the water and electrolyte balance was maintained. Which one of the following explanation is correct ?

(a) His aldosterone was decreased and alcohol dehydrogenase (ADH) was increased.

(b) His aldosterone was increased and ADH was decreased.

(d) His sympathoadrenal system was stimulated.

Ans. (a)

Sol. Aldosterone acts mainly at the renal tubules and stimulates the reabsorption of Na⁺ and water and excretion of K⁺ and phosphate ions. Thus, aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure. Antidiuretic hormone (ADH) or vasopressin acts mainly at the kidney and stimulates reabsorption of water and thereby reduces loss of water through urine (diuresis).

87. The blood volume decreased when a mammal was bled rapidly. However, the cardiovascular changes resulting from hemorrhage could be minimized by the following compensatory mechanism :

P. Increased cerebral blood flow.

Q. Reduction of baroreceptor activity and stimulation of chemoreceptors.

R. Reabsorption of tissue fluid in blood.

S. Increased release of enkephalins and -endorphins.

Which of the above is/are correct ?

(a) P and Q

(b) Q and R

(d) only S

Ans. (b)

Sol. Juxta glomerular cells in efferent arteriol act as baroreceptor to check out blood pressure. In hemorrhage condition there is reduction in blood pressure due to blood loss and baro-receptor activity may cause more blood loss by increasing blood pressure. So, compensatory mechanism reduce baro-receptor activity and stimulate chemo-receptors and tissue fluid also get reabsorb in blood to compensate blood loss.

88. The stomach of a person was partially removed during surgery of a gasteric tumour. Despite taking a balnced diet, the person developed anemia. Following possible explanations were offered :

P. Lower gastric secretion inhibits folic acid absorption.

Q. Protein digestion was disturbed in partial gastrectomy.

R. Lower HCl secretion from stomach reduced iron absorption.

S. Lower secretion of intrinsic protein factor from stomach reduced VitB₁₂ absorption.

Which of the above explanations were correct ?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (c)

Sol. Lower HCl secretion from stomach reduced iron absorption and Lower secretion of intrinsic protein factor from stomach reduced VitB₁₂ absorption.

89. A cross was made between puer wild type males and brown eyed, curled winged females of D. melangaster. The F1 females were test crossed. The F2 progeny obtained was as follows :

Wild-type 200

Brown eyes, curled wings 150

Brown eyes, normal wings 30

Normal eyes, curled wings 20

Total 400

The genetic distance (cM) between brown eye and curled wing loci is

(a) 12.5

(b) 50

(d) 25

Ans. (a)

Sol. Genetic distance between brown eye and curled wing

90. The effect of nonsense mutation could be nullified by reversion as well as suppression. Which of the following processes will help to distinguish between the two kinds of revertants ?

(a) Complementation

(b) Transgenesis

(d) Recombination

Ans. (d)

Sol. Revertant produce by reversion would have corrected sequence which resembles to sequence prior to non-sense mutation whereas revertant with suppresion would have existing non-sense mutation. So, recombination will help to distinguish between the two kinds of revertants.

91. 2-aminopurine induces mutation by

(a) base pair change

(b) frameshift

(d) insertion

Ans. (a)

Sol. 2-aminopurine induces mutation by substitution i.e., base pair change.

92. In a transformation experiment, donor DNA from an E. coli strain with the genotype Z⁺ Y⁺ was used to transform a strain of genotype Z^– Y^–. The frequencies of transformed classes were

Z⁺ Y⁺ 200

Z⁺ Y^– 400

Z^– Y⁺ 400

Total 1000

What is the frequency (%) with which Y locus is co-transformed with the Z locus ?

(a) 1

(b) 20

(d) 40

Ans. (b)

Sol.

93. The RFLP pattern observed for two pure parental lines (P1 and P2) and their F1 progeny is represented below. Further, the P1 plant had red flowers while the P2 had white flowers. The F1 progeny was backcrossed to P2. The result obtained, showing the number of progeny with red and white flowers and their RFLP patterns is also represented below.

Which one of the following conclusions made is correct ?

(a) The DNA marker and the gene for the flower color are 10 cM apart.

(b) The marker and the gene for the flower color are 5 cM apart.

(d) The marker and the gene for the color segregate from one another.

Ans. (a)

Sol. By observing RFLP pattern we found that,

Total number of recombinant in cross between F1 and P2 = 5 + 5 = 10

So, distance between DNA marker and gene =

= = 10 cM

Hence, option (a) is correct

94. Wild-type T4 bacteriophage can grow on B and K strains of E. coli forming small plaques. rII mutants of T4 bacteriophage cannot grow on E. coli strain K (non-permissive host), but form large plaques on E. coli strain B (permissive host). The following two experiments were carried out :

Experiment I : E. coli K cells were simultaneously infected with two rII mutants (a^– and b^–). Several plaques with wild-type morphology were formed.

Experiment II : E. coli B cells were simultaneously infected with the same mutants as above. T4 phages were isolated from the resulting plaques and used to infect E. coli K cells. Few plaques with wild type morphology were formed.

Which one is the correct conclusion made regarding the rII mutants, a^– and b^– from the above experiments ?

(a) The mutations a^– and b^– belong to two different cistrons (experiment I) and there is no recombination between them (experiment II).

(b) The mutations a^– and b^– belong to two different cistrons (experiment I) and they recombined (experiment II).

(d) The mutations a^– and b^– belong to the same cistron (experiment I) and they did not recombine (experiment II).

Ans. (b)

Sol. The mutations a^– and b^– belong to two different cistrons (experiment I) and they recombined (experiment II).

95. The following pedigree represents inheritance of a trait in an extended family

What is the probable mode of inheritance and which individuals conclusively demonstrate this mode of inheritance ?

(a) Autosomal recessive, III-2, 3 and IV-1, 2 conclusively demonstrates the mode of inheritance.

(b) Autosomal recessive, I-1, 2 and II-2 conclusively demonstrates the mode of inheritance.

(d) X-linked recessive, II-3, 4 and 5 conclusively demonstrates the mode of inheritance.

Ans. (a)

Sol. By observing pedigree chart, we can come to the conclusion that trait is Autosomal recessive, III-2, 3 and IV-1, 2 conclusively demonstrates the mode of inheritance.

96. Following is the diagram of a pacentric inversion heterozygote ABCDEFG/ABCFEDCG involved in recombination during meiosis I :

The consequence of this recombination will be teh formation of

P. a dicentric and an acentric chromosome in meiosis I as the chiasmata gets terminated.

Q. no dicentric or acentric chromosome but appearance of deletion and duplication in both the chromosomes

R. all non-viable gametes.

S. non-viable gametes from crossover products.

Which of the above statements are correct ?

(a) P and Q

(b) P and R

(d) Q and R

Ans. (c)

Sol. The presence of a chiasmata at meiosis within paracentrically inverted segments results in a dicentric chromosome and an acentric fragment, which cannot be regularly transmitted.

97. An E. coli strain has metB1 (90 min) and leuA5 (2 min) mutations. It also has strA7 (73 min) mutation and Tn5 transposon which confers streptomycin and kanamycin resistance, respectively, inserted in its chromosome. The mutant strain was crossed with an Hfr strain that is streptomycin sensitive and has a hisG2 mutation (44 min) that makes it require histidine. After incubation for 100 min, the cells were plated on minimal plate supplemented with leucine, histidine and streptomycin to select the metB marker. After purifying 100 of the Met⁺ transconjugants, one finds that 15 are His^–, 2 are Leu⁺ and 12 are kanamycin sensitive. Teh unselected markers are

P. metB1 and leuA5 mutation

Q. leuA5 and Tn5 insertion mutation

Which of the above statements is correct and what is the position of transposon insertion ?

(a) P and before 73 min

(b) Q and before 44 min

(d) P and before 44 min

Ans.

Sol. As crossing over is occuring between a normal homologous chromosome and homologous chromosome with inversion. So, The consequence of this recombination will be the formation of a dicentric and an acentric chromosome in meiosis I as the chiasmata gets terminated and non-viable gametes from crossover products.

98. A chemist synthesized three new chemical compounds, M1, M2 and M3. The compounds were tested for their mutagenic potential and were found to be highly mutagenic. Tests were made to characterize the nature of mutations by allowing the reversion with other mutagenes. The following results were obtained :

Which one of the following conclusions drawn regarding the nature of muations by the compounds is correct ?

(a) M1-transversion, M2-insertion, M3-deletion

(b) M1-transition, M2-transversion, M3-insertion

(d) M1-transversion, M2-transition, M3-insertion

Ans. (d)

Sol. Mutagens like 2-aminopurine and nitrous acid can produce or reverse transtition mutation whereas acridine orange which is an intercalating agent can cause addition/insertion or deletion of nucleotides.

99. The 'Tribe' refers to a taxonomic group recognized between the ranks

(a) genus and species

(b) family and genus

(d) class and order

Ans. (b)

Sol. In biology, a tribe is a taxonomic rank between family and genus. It lies below ranks with names derived from family, such as subfamily. It is sometimes subdivided into subtribes.

100. A plant species has been described for the first time by author 'x'. Later, the species has been transferred to some other genus by author 'y'. Then the author citation for the new combination will be

(a) x et y

(b) x ex y

(d) (y) x

Ans. (c)

Sol. In botanical nomenclature, author citation is the way of citing the person or group of people who validly published a botanical name, i.e. who first published the name 'x'. In case if the species is transferred to some other genus by author 'y', then author citation will be- (x)y

101. The group which is no longer considered under fungi is

(a) Ascomycetes

(b) Basidiomycetes

(d) Oomycetes

Ans. (d)

Sol. Fungi and Oomycetes are the two most important groups of eukaryotic plant pathogens. Fungi form a separate kingdom and are evolutionarily related to animals. Oomycetes are classified in the kingdom Protoctista and are related to heterokont, biflagellate, golden-brown algae.

102. The following table shows survival and fertility data for a seasonally breeding species.

Based on above data, net reproductive rate (R₀) of the species will be

(a) 1

(b) 5

(d) 20

Ans. (c)

Sol.

103. Which of the following is not a characteristic of late successional forest plant species ?

(a) Large seed size, high root to shoot ratio

(b) Long seed dispersal distance, long seed viability

(d) Low light saturation intensity, high efficiency at low light

Ans. (b)

Sol. Long-distance dispersal is a characteristically extreme event of propagule movement in any plant or animal population, typically occurring with an extremely low probability but potentially reaching an extremely long distance. So, late successional forest plant species don't have long speed dispersal distance and long seed viability.

104. Which of the following organisms do not possess the ability to fix nitrogen ?

(a) Organisms specialized for high altitude

(b) Marine plankton

(d) Acidophilic organisms

Ans. (c)

Sol. Nitrogen Fixation by Free-Living Heterotrophs.

Many heterotrophic bacteria live in the soil and fix significant levels of nitrogen without the direct interaction with other organisms. Examples of this type of nitrogen-fixing bacteria include species of Azotobacter, Bacillus, Clostridium, and Klebsiella. Eukaryotic organisms are only able to obtain fixed nitrogen through their symbiotic interactions with nitrogen-fixing prokaryotes. These symbioses involve a variety of host organisms, including animals, plants, fungi and protists.

105. Which of the following green house gases has got highest atmospheric life time ?

(a) CO₂

(b) CH₄

(d) CFCs

Ans. (d)

Sol. The longest lifetime in the atomsphere would have to be CFC-13, CCIF3 which has a half life of 640 years. One you didn't ask about Sulphur hexafluoride, SF6 is actually the longest at 3,000 years half-life.

106. Four Cnidarians with the following characteristics were observed :

P. Asexual polyps and sexual medusae; solitary or colonial; both freshwater and marine

Q. Polyp stage reduced or absent, medusae with velum; solitary; all marine

R. Polyp stage reduced, bell shaped medusae; solitary; all marine

S. All polyps, no medusae; solitary or colonial; all marine

They can be identified to their respective classes

(a) P-Scyphozoa, Q-Anthozoa, R-Cubozoa, S-Hydrozoa

(b) P-Hydrozoa, Q-Scyphozoa, R-Cubozoa, S-Anthozoa

(d) P-Cubozoa, Q-Scyphozoa, R-Anthozoa, S-Hydrozoa

Ans. (b)

Sol. Hydrozoa are a taxonomic class with some solitary and some colonial animals. They have both sexual (medusae) and asexual (polyp) forms. They are found in both fresh water and saline water. Therefore, option (b) is correct.

107. Identify the proteobacteria based on the key given below :

i. Causes disease in humans (ii)

i. Do not cause diseases in humans (iii)

ii. An obligate intracellular parasite (A)

ii. Not an obligate intracellular parasite (B)

iii. Live in insects (C)

iii. Do not live in insects (iv)

iv. Chemoautotrophic (D)

iv. Not Chemautototrophic (v)

v. Plant pathogen (E)

v. Not a plant pathhogen (vi)

vi. Fix nitrogen (viii)

vi. Do not fix nitrogen (F)

vii. Associated with legumes (G)

vii. Not associated with legumes (H)

(a) A-Rickettsia; B-Brucella; C-Wolbachia; D-Nitrobacter; E-Agrobacterium; F-Acetobacter; G-Rhizobium; H-Azospirillum

(b) A-Rickettsia; B-Wolbachia; C-Brucella; D-Nitrobacter; E-Acetobacter; F-Agrobacterium; G-Rhizobium; H-Azospirillum

(d) A-Rickettsia; B-Brucella; C-Wolbachia; D-Nitrobacter; E-Acetobacter; F-Agrobacterium; G-Azospirillum; H-Rhizobium

Ans. (a)

Sol. Rickettsia are gram negative bacteria. They are obligate intracellular parasite which are non-motile. Brucella is not an obligate intracellular parasite. Nolbachia is a genus of intracellular bacteria which lives in insects.

Nitrobacter are not shaped, gram negative, chemoautotrophic bacteria.

Agrobacterium is a gram negative bacteria that uses to cause tumor in plants.

108. Which of the following phylogenetic trees appropriately uses principle of parsiomony ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. The principle of parsimony in phylogeny states that the tree with fewest common ancestors is the most likely.

109. Associate the forest/vegetation type with the plants :

1. Grass land

2. Subalpine forest

3. Shola forest

4. Subtropical pine forest

5. Tropical thorn forest

6. Tropical dry deciduous forest

7. Tropical semievergreen forest

8. Tropical wet evergreen forest

(a) 1-Ilex, 2-Dichanthium, 3-Abies, 4-Pinus, 5-Acacia, 6-Angogeissus, 7-Cinnamomum, 8-Dipterocarpus.

(b) 1-Dichanthium, 2-Abies, 3-Ilex, 4-Pinus, 5-Acacia, 6-Angogeissus, 7-Cinnamomum, 8-Dipterocarpus.

(d) 1-Angogeissus, 2-Dichanthium, 3-Ilex, 4-Pinus, 5-Acacia, 6-Abies, 7-Cinnamomum, 8-Dipterocarpus.

Ans. (b)

Sol. Dichanthium is found in Grassland. Pinus is an gymnosperm found in subtropical pine forest. Acacia is found in Tropical thorn forest.

110. Identify the characters shown in the diagram depicting phylogentic relationships among major groups of ferns and ferm allies.

(a) A-Roots absent, B-Sporangiophores, C-Vertical, interrupted annulus, D-Heterospory, E-Leaves scale like, F-Elaters

(b) A-Roots absent, B-Leaves scale like, C-Sporangiophores, D-Elasters, E-Heterospory, F-Vertical, interrupted annulus

(d) A-Heterospory, B-Roots absent, C-Elaters, D-Sporangiophores, E-Leaves scale like, F-Vertical, interrupted annulus

Ans. (b)

Sol. Roots are absent in the members of psilotaceal and they are characterized by the presence of scale like leaves. Heterospory i.e., the presence of two different types of spores is found in Marsileaceal

111. The following table shows the summary of characters between two taxa based on presence (1) and absence (0) data

Which of the following represents Jaccard's coefficient and Simple matching coefficient respectively ?

(a) 0.8, 0.5

(b) 0.6, 0.5

(d) 0.5, 0.6

Ans. (d)

Sol.

Where, a = number of characters occurring in both samples.

b = number of characters occurring in A but not B.

c = number of characters occurring in B but not A.

d = number of characters absent from both samples.

Jaccard's coefficient (S_J) =

Simple matching coefficient (S_SM) =

112. Possible explanations for the age related decline in primary productiviey of trees are :

P. As trees grow longer with age, they have more tissues that respire and loose energy and proportionately less leaf area to photosynthesize.

Q. Nutrient limitation by nitrogen due to reduced rate of woody litter decomposition as forest ages.

R. As trees become larger, water transport to the top canopy leaves becomes limited because of increased hydraulic resistance. This results in reduced stomatal conductance and reduction in photosynthetic rate.

Which of the above is/are correct ?

(a) P, Q and R

(b) Only P

(d) Only Q and R

Ans. (d)

Sol. Option (d) is correct explanation.

113. Species characteristics that make them more prone to extinction are listed below :

P. High degree of specialization

Q. High sexual dimorphism

R. High trophic status

S. Short life span

Which of the following is the correct combination ?

(a) P, Q and R

(b) P, R and S

(d) Q, R and S

Ans. (b)

Sol. General, the species with following characteristics are more susceptible to extinction :

Large body size.

Small population size.

Low reproductive rate.

Feeding at high trophic levels in the food chain.

Fixed migratory routes.

Narrow range of distribution.

High specialization.

Poor dispersal rate.

114. Identify the pollinators for the flowers with following pollination syndromes

P. Flowers dull colored, located away from foliage, floral parts turgid.

Q. Flowers bright red, crowded, trugid, nectar watery and sucrose rich.

R. Flowers white with pleasant odor, corolla tube long, night blooming.

(a) P-Bird, Q-Bat, R-Butterfly

(b) P-Bat, Q-Bird, R-Moth

(d) P-Bird, Q-Bat, R-Carrion fly

Ans. (b)

Sol. Bat species pollinate flower which are dull in colour, located away from the foliage. Birds prefer bright color of flower usually red which are rich in sucrose.

115. Based on the information given in the table below, which combination is correct ?

(a) A₁– B₃ – C₃

(b) A₂– B₁ – C₄

(d) A₄– B₄ – C₂

Ans. (c)

Sol. Takin, also called chamois is found in Mediterranean Ibex is a species of wild goat which are found in Indo-chinese.

Gibbon is found in Indo-Malayan.

116. Which of the following combinations is good for setting up a nature reserve :

Linked reserves

Large, compact shape

High edge to area ratio

Surrounding area of same ecosystems

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (d)

Sol. Option (d) is correct.

117. If the number of new species evolving is directly proportional to the number of existing species and the probability of extinction of any species is inversely proportional to the number of existing species, the number of species present at a time during evolution will follow a curve given by :

(a)

(b)

(c)

(d)

Ans. (c)

Sol. It will show hyperbola curved shaped graph.

118. Character similarity that can be misinterpreted as common descent is called

(a) symplesiomorphy

(b) synapomrophy

(d) homoplasy

Ans. (d)

Sol. In cladistics, a homoplasy or homoplastic character is a trait (genetic, morphological etc.) that is shared by two or more taxa because of convergence, parallelism or reversion.

119. Which of the following evolutionary processes played an important role in the evolution of complex immune system ?

(a) Reproductive isolation

(b) Adaptive radiation

(d) Co-evolution

Ans. (d)

Sol. Coevolution, the process of reciprocal evolutionary change that occurs between pairs of species or among groups of species as they interact with one another. The activity of each species that participates in the interaction applies selection pressure on the others. Co-evolution leads to evolution of complex immune system.

120. In some species of new world monkeys, only one female reproduces in a group. One or more younger females have suppressed reproduction and assist the reproductive female. This is an example of

(a) sexual selection

(b) group selection

(d) reciprocal altruism

Ans. (c)

Sol. Kin selection occurs when an animal engages in self-sacrificial behaviour that benefits the genetic fitness of its relatives. The theory of kin selection is one of the foundations of the modern study of social behaviour.

121. In bird species where both parents contribute equally to parental care, generally

(a) males are larger than females

(b) females are more colourful than males

(d) both sexes are morphologically similar

Ans. (d)

Sol. Parental care refers to the level of investment provided by the mother and the father to ensure development and survival of their offspring. In most birds, parents invest profoundly in their offspring as a mutual effort, making a majority of them socially monogamous for the duration of the breeding season.

122. The idea that an altruistic gene will be favoured if r > C/B, where r is the coefficient of relatedness, B is the benefit to the recipient of the altruism, and C is the cost incurred to the donor, is known as

(a) red queen hypothesis

(b) handicap principle

(d) competitive exclusion principle

Ans. (c)

Sol. Hamilton's rule asserts that a trait is favored by natural selection if the benefit to others, B, multiplied by relatedness, R, exceeds the cost to self, C. Specifically, Hamilton's rule states that the change in average trait value in a population is proportional to BR-C.

123. If the relationship between life time reproductive success and body size for size for males and females of a species as shown in figure below. The species is most likely to evolve

(a) sexual dimorphism

(b) asexual reproduction

(d) obligate monogamy

Ans. (a)

Sol. The phenomenon of sexual dimorphism is a direct product of evolution by natural selection, in that the struggle for reproductive success drives many male and female organisms down different evolutionary paths. This can produce forms of dimorphism which, on the face of it, would actually seem to disadvantage organisms. Sexual body size dimorphism is a difference in size between the two sexes, usually measured as a ratio of the male to female body weight. In most hominoids, the male is larger than the female.

124. Following tree represents phylogenetic relationships among species of a moth family. Circles represent species having eye spots on the wings. Other species do not have eye spots.

The following inferences were made by different researchers :

P. Eye spots were present in the ancestors and some species lost them.

Q. Eye spots were not present in ancestors.

R. Eye spots were lost more than once in evolution of the family.

S. Eye spots were gained only once while evolving from ancestor without them.

Which of the infereces are correct ?

(a) P and Q

(b) R and S

(d) Q and S

Ans. (c)

Sol. P and R statement is correct.

125. Wolbachia are obligate intracellular bacteria, many different strains of which are abundantly present in insects. They induce mating incompatibility in host, i.e. males infected with one strain can only fertilize females infected with the same strain. No other pathological effects are observed in host. A possible evolutionary consequence of this phenomenon would be :

(a) Extinction of many insect species.

(b) Termination of sexual reproduction in many insect species.

(d) Reproductive isolation leading to rapid speciation in insects.

Ans. (d)

Sol. Reproductive isolation is a collection of mechanisms, behaviors, and physiological processes that prevent the members of two different species that cross or mate from producing offspring, or which ensure that any offspring that may be produced is not fertile. Reproductive isolation is weak in the early stages of speciation, but changes to strong or complete in later stages of speciation.

126. Twenty small populations of a species, each polymorphic for a given locus (T, t) were bred in captivity. In 10 of them the population size was kept constant by random removal of individuals, while other 10 were allowed to incerase their population size. After several generations it was observed that in 7 of the size restricted populations only T was present, in the remaining 3 only t was present. In the growing populations 8 retained their polymorphism any in 2 only t was observed. This experiment illustrates

(a) genetic drift which is more likely in large populations

(b) genetic drift which is more likely in small populations

(d) density dependent selection against t

Ans. (b)

Sol. Typically, genetic drift occurs in small populations, where infrequently occurring alleles face a greater chance of being lost. Once it begins, genetic drift will continue until the involved allele is either lost by a population or until it is the only allele present in a population at a particular locus.

127. Some important events in the history of life on Earth are given below :

1. First vertebrates (jawless fishes); first plants.

2. Forest of ferns and conifers; amphibians arise; insects radiate.

3. Conifers dominant; dinosaurs arise; insects radiate.

4. Flowering plants appear; climax of dinosaurs followed by extinction.

5. Radiation of flowering plants, most modern mammalian orders represented.

6. Ice Ages, Modern humans appear.

Match the above withthe geological time periods and choose the correct combination,

(a) 1-Silurian; 2-Permian; 3-Triassic; 4-Jurassic; 5-Cretaceous; 6-Tertiary

(b) 1-Ordovician; 2-Carboniferous; 3-Triassic; 4-Cretaceous; 5-Tertiary; 6-Quaternary

(d) 1-Devonian; 2-Permian; 3-Triassic; 4-Cretaceous; 5-Tertiary; 6-Quaternary

Ans. (b)

Sol. Option (b) is correct.

128. Use of doubled haploid in plant breeding helps to

(a) reduce generation time while interogressing recessive traits

(b) reduce generation time while interogressing dominant traits

(d) interogress transgenic traits

Ans. (a)

Sol. Doubled haploid (DH) technology is a powerful tool in plant breeding to reduce the time and costs needed to produce pure lines, the cornerstone of hybrid seed production.

129. For sustained expression of a transgene in the successive generation of a cell line in culture, the ideal gene transfer can be obtained using

(a) lentiviral vector

(b) adenoviral vector

(d) only transgenic DNA

Ans. (a)

Sol. Cystic fibrosis (CF) is a chronic autosomic recessive syndrome, caused by mutations in the CF Transmembrane Conductance Regulator (CFTR) gene, a chloride channel expressed on the apical side of the airway epithelial cells. The lack of CFTR activity brings a dysregulated exchange of ions and water through the airway epithelium, one of the main aspects of CF lung disease pathophysiology. Lentiviral (LV) vectors, of the Retroviridae family, show interesting properties for CF gene therapy, since they integrate into the host genome and allow long-lasting gene expression. Proof-of-principle theat LV vectors can transduce the airway epithelium and correct the basic electrophysiological defect in CF mice has been given.

130. Desulphovibrio desulfuricans (A) and Pseudomonas species (B) are involved in mrecury bioremediation. Which of the statements below is correct ?

(a) A converts methyl mercury to mercuric ion, B converts mercury to methyl mercury.

(b) A converts mercury to methyl mercury, B converts methyl mercury to mercuric ion.

(d) A converts methyyl mercury to mercuric ion, B converts mercury to mercuric ion.

Ans. (c)

Sol. Two strains of Desulfovibrio desulfuricans, one known to synthesize monomethylmercury from ionic mercury, were grown to determine methylmercury toxicity and for comparison with an anaerobic strain of Clostridium pasteurianum, a H2 producer, and with the broad-spectrum mercury-resistant Pseudomonas putida strain FB-1, capable of degrading 1 µg of methylmercury to methane and elemental mercury in 2 h.

131. Optical density of a 400 base pair long 1 ml DNA soultion was foundn to be 0.052. How many DNA molecules are present in the solution ? [1 base pair = 650 dalton, optical density of 1.0 D corresponds to 50 µg DNA/ml]

(a) 6.023 × 10¹²

(b) 6.023 × 10¹⁸

(d) 5.2 × 10¹³

Ans. (a)

Sol. Optical density (OD) of one corresponds to 50 µg/ml of dsDNA

Therefore, OD of 0.052 corresponds to 50 × 0.052 = 2.6 µg/ml of dsDNA or 2.6 × 10^–6 g.

Average molecular weight of 1 nucleotide base pair is 650 Da.

Molecular weight of 400 bp = 650 × 400 = 2,60,000 Da.

The number of moles of a given substance is the ratio of its given weight (in grams) and molecular weight (in dalton).

Number of moles of DNA = = 1 × 10^–11

1 mole contains of 6.023 × 10²³ molecules.

1 × 10^–11 moles will contain 6.023 × 10²³ × 1 × 10^–11 = 6.023 × 10¹² molecules.

132. In which of the following techniques does molecular fragmentation offer clues to the covalent chemical structures of biomolecules ?

(a) MALDI-TOF MS mass spectrometry

(b) MALDI-TOF MS/MS mass spectrometry

(d) LC-coupled ESI-TOF MS mass spectrometry

Ans. (b)

Sol. Matrix assisted laser desorption/ionisation-time of flight mass spectrometry (MALDI-TOF MS) is a relatively novel technique in which a co-precipitate of an UV-light absorbing matrix and a biomolecule is irradiated by a nanosecond laser pulse. Most of the laser energy is absorbed by the matrix, which prevents unwanted fragmentation of the biomolecule. The mass accuracy of MALDI-TOF MS will be sufficient to characterise proteins (after tryptic digestion) from completely sequenced genomes (e.g. methoanogens, yeast). The use of MALDI-TOF MS for typing of single nucleotide polymorphism using single nucleotide polymorphism using single nucleotide primer extension has made important progress.

133. The movement of a single cell was required to be continually monitored during development. This cells was marked with a reporter gene. To visualize this movement one would use

(a) phase contrast microscopy

(b) bright field microscopy

(d) atomic force microscopy

Ans. (c)

Sol. Fluorescent microscopy is often used to image specific features of small specimens such as microbes. It is also used to visually enhance 3-D features at small scales. This can be accomplished by attaching fluorescent tags to anti-bodies that in turn attach to targeted features, or by staining in a less specific manner.

134. During transgenesis, the location of the genes and their number integrated into the genome of the transgeneic animal are random. It is often necessary to determine the copy number of genes and their tissue-specific transcription. The following are the possible methods used for the determination.

P. Polymerase Chain Reaction (PCR)

Q. Southern blot hybridization

R. Reverse Transcriptase PCR

S. Western blot

Choose the correct set of combinations.

(a) P and Q

(b) Q and R

(d) P and S

Ans. (b)

Sol. Southern blotting is a hybridization technique for identification of particular size of DNA from the mixture of other similar molecules. This technique is based on the principle of separation of DNA fragments by gel electrophoresis and identified by labelled probe hybridization. Reverse transcription polymerase chain reaction is a laboratory technique combining reverse transcription of RNA into DNA and amplification of specific DNA targets using polymerase chain reaction.

135. Microbes produce either primary or secondary metabolites during fermentation. A metabolite production curve is shown below :

The following statements refer to the above figure ;

P. A primary metaboolite has a production curve that lags behind the line showing cell growth.

Q. A primary metabolite is produced after the trophophase is completed.

R. A secondary metabolite is produced mainly during idiophase.

S. The curve shows the production of penicillin from mold.

Which of the above statements are correct ?

(a) P and Q

(b) R and S

(d) Q and S

Ans. (b)

Sol. As per the figure mentioned in question, A secondary metabolite is produced mainly during idiophase and the curve shows the production of penicillin from mold.

136. Agrobacterium tumefaciens, also known as natural genetic engineers, causes crown-gall diseases in platns. However, when the same bacteria are used to raise transgenic platns with improved agronomic traits, no such tumor (disease) is observed. This is due to :

P. vir D₂ gene is mutated in Ti plasmid.

Q. Disamed Ti plasmid is generally used.

R. Heat-shock during transformation destroys virulence.

S. Oncogenes have been removed.

Which one of the following combination of above statements is correct ?

(a) P and R

(b) P and S

(d) Q and S

Ans. (d)

Sol. Agrobacterium tumefaciens, also known as natural genetic engineers, causes crown-gall diseases in platns. However, when the same bacteria are used to raise transgenic plants with improved agronomic traits, no such tumor (disease) is observed. This is due to Disarmed Ti plasmid is generally used and Heat-shock during transformation destroys virulence.

137. A student wrote following statements regarding comparison of Restriction Fragment Length Polymorphism (RFLP), Random Amplified Polymorphic DNA (RAPD), Amplified Fragment Length Polymorphism (AFLP) and Simple Sequence Repeats (SSRs) tehcniques used for generating molecular markers in plants :

P. All these tehcniques can be used for fingerprinting.

Q. Detection of allelic variation can be achieved only by RFLP and SSRs.

R. Use of radiosotopes is required for all the techniques.

S. Polymerase chain reaction is required for all the techniques.

Which one of the following combination of above statements are correct ?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (a)

Sol. All these tehcniques can be used for fingerprinting and Detection of allelic variation can be achieved only by RFLP and SSRs.

138. The molecular mass of a protein determined by gel filtration is 120 kDa. When its mass is determined by SDS-PAGE with the without -mercaptoethanol, it is only 60 kDa. What is the most probable explanation for these observations ?

(a) Protein is a dimer in which two identical chains are cross-linked by disulphide bond(s).

(b) Protein is a monomer of molecular mass 60 kDa but it is only 60 kDa but it is excluded from the gel matrix due to strong repulsion between the gel matrix and the protein.

(d) Protein is a monomer but it is nicked into half its size by SDS.

Ans. (c)

Sol. The molecular mass of a protein determined by gel filtration is 120 kDa. When its mass is determined by SDS-PAGE with the without -mercaptoethanol, it is only 60 kDa. This is becasue protein is most likely to be composed of two sub-units having identical molecular mass.

139. In a census for a lake fish, 10 individuals were marked and released. In second sampling after a few days 15 individuals were caught, of which 5 individuals were found marked. The estimated popultion of the fish in the lake will be

(a) 20

(b) 30

(d) 35

Ans. (b)

Sol.

Where, N = Total population size

M = Marked initially

R = Marked recaptures

T = Total in second sample

140. In order to clone an eukaryotic gene is pBR322 plasmid vector, the desired DNA fragment was produced by PstI cleavage and incubated with PstI digested pBR322 (PstI cleavage site lies within the ampicillin resistant gene) and ligated. Mixture of ligated cells were used to transform E. coli and plasmid containing bacteria were selected by their growth in tetracycline containing medium. Which type of plasmid/s will be found ?

(a) Circular PBR322 plasmid containing the target gene and resistant to only tetracycline.

(b) Circular pBR322 plasmid containing the target gene and resistant to tetracycline only and recircularized pBR322 plasmid resistant to both ampicillin and tetracycline.

(c) Circular pBR322 palsmid containing the target gene and resistant to only tetracycline, recircularized pBR322 resistant to both ampicillin and tetracycline and concatemerized pBR322 resistant to both ampicillin and tetracycline.

(d) Circular pBR322 plasimd containing the target gene and resistant to both ampicillin and tetracycline.

Ans. (c)

Sol. In order to clone an eukaryotic gene is pBR322 plasmid vector, the desired DNA fragment was produced by PstI cleavage and incubated with PstI digested pBR322 (PstI cleavage site lies within the ampicillin resistant gene) and ligated. Mixture of ligated cells were used to transform E. coli and plasmid containing bacteria were selected by their growth in tetracycline containing medium. In this conditions, Circular pBR322 palsmid containing the target gene and resistant to only tetracycline, recircularized pBR322 resistant to both ampicillin and tetracycline and concatemerized pBR322 resistant to both ampicillin and tetracycline.

141. During apoptosis, phosphatidyl serine (PS) usually present in the inner leaflet of the plasma membrane flips to the outer membrane. Annexin V is a protein that binds to PS. Using this as a tool, we identify the apoptotic cells from nectrotic and normal cells populations by FACs using FITC-tagged Annexin V. Propidium iodine (PI) is used to stain the nucleus which generally necrotic and late apoptotic cells. In which area of the plot you should get early apoptotic cells by FACS analysis ?

(a) Quadrant I

(b) Quadrant II

(d) Quadrant IV

Ans. (c)

Sol. Early prokaryotic cells would be found in quadrant III.

142. The muscle tone was increased after electrolytic lesin of the caudate nucleus in a cat. The muscle tone decreased within seven days. The following explanations were given by the researcher

P. The functional recovery was due to plastic changes of nervous system.

Q. The brain tissue surrounding the lesioned area was non-functional due to circulatory insufficiency immediately after surgery which led to greater functional loss.

R. The circulatory status in surrounding tissue recovered with time resulting in partial functional recovery.

S. The degenerating nerve fibres were regenerated which underline functional recovery.

Which one of the following is correct ?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (c)

Sol. The muscle tone was increased after electrolytic lesin of the caudate nucleus in a cat. The muscle tone decreased within seven days. This occurs because the circulatory status in surrounding tissue recovered with time resulting in partial functional recovery and the degenerating nerve fibres were regenerated which underline functional recovery.

143. A fluorophore when transferred from solvent A to solvent B results in an increase in the number of vibrational states in the ground state without any change in the man energies of either the ground or excited state. What would be the change seen in the fluorophore's emission spectrum ?

(a) An increase in emission intensity.

(b) An increase in emission bandwidth.

(d) A decrease in emission wavelength.

Ans. (b)

Sol. A fluorophore when transferred from solvent A to solvent B results in an increase in the number of vibrational states in the ground state without any change in the man energies of either the ground or excited state. This would result in An increase in emission bandwidth in the fluorophore's emission spectrum.

144. You wish to localize a given gene product at subcellular levels following immunofluorescence staining. Routine microscopy could not resolve whether the gene product is localized inside the nuclueus or on the nuclear membarne. Which of the following will resolve the unambiguously ?

P. Selecting fo cell followed by phase contrast microscopy.

Q. A simulation of 3D picture following confocal microscopy.

R. Optical selectioning and observing each section.

S. Freeze fracturing followed by Scanning Electron Microscopy.

(a) P and Q

(b) Q and R

(d) P and R

Ans. (b)

Sol. If you wish to localize a given gene product at subcellular levels following immunofluorescence staining. It can be done by a simulation of 3D picture following confocal microscopy and optical selectioning and observing each section.

145. The Triver-Willard hypothesis states that the physiological state of a female can bias the sex ratio of offspring. In an experiment in the bird species a group of females were fed a diet 30% lower in calories than the control females. After allowing both the groups to mate and breed freely, the offspring of control 1 group were 22 males and 18 females. The diet restricted females laid a 40 eggs. What would be the minimum deviation from the control to conclude that they have significantly female biased offspring sex ratio ? (Chi sq [0.05] df = 1 is 3.84)

(a) 18 male 22 female

(b) 20 male 20 female

(d) 10 male 30 female

Ans. (c)

Sol. The correct option is (c)

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Previous Year Question Paper with Solution.