PYQ JUNE 2017 - VedPrep

CSIR NET BIOLOGY (JUNE - 2017)
Previous Year Question Paper with Solution.

21. Denaturation of a highly helical protein having disulfide bridges and two phenyalanines can be monitored as a function of temperature by which one of the following techniques?

(a) Recording circular dichroism spectra at various temperatures.

(b) Monitoring the absobance at 214 nm at various temperatures.

(d) Monitoring the ratio of absorbance at 214 nm and at 250 nm at various temperatures.

Ans. (a)

Sol. Phenyl alanine absorbs maximally at 274 nm, so by measuring absorbance at 274 nm it is possible to look at the denautration profile of this pretein but not at 214 nm or 250 nm. Disulfide bonds cannot be broken by heating a protein up to 100 or 150°C. Circular dichroism spectra can be used to monitor the denaturation of this protein.

22. Glycerol is added to protein solutions to stablilize the preparations by

(a) increasing the viscosity of solution.

(b) stabilizing the pH.

(d) interacting and neutralising the surface charges on the proteins.

Ans. (c)

Sol. Glycerol stabilizes the protein solution preparations by acting as an amphiphilic interface between the hydrophobic surface of protein and polar solvent and due to the electrostatic interactions between protein in glycerol leads to preferential hydration that induces orientations of glycerol molecule at the protein surface. This leads to inhibition of protein aggression and shifting of native protein to more compact form.

23. Protein stability is represented as

Prior to development of sensitive calorimeters, thermodynamic parameters of processes were determined by following equation

and are standard changes in enthalpy and entropy, respectively.

Which one of the following statements is correct for estimating and ?

(a) Determining the ratio of folded and unfolded protein at 37°C.

(b) Plotting K_eq as a function of .

(d) Plotting K_eq against temperature.

Ans. (d)

Sol. K_eq = R (1/T) + /R, it can be observed that this equation exactly matches with y = mx + c. So, , and , can be measured by plotting K_eq versus 1/T, it will give us straight line curve with negative slope. Slope will define the /R value and intercept on y-axis will define /R values. Once we got and value, can be obtained by putting that in this equation = – T.

24. Rotenone is an inhibitor of the electron transport chain. The addition of rotenone to cells results in which of the following?

(a) Generation of mitochondrial reactive oxygen species and block in ATP generation.

(b) Block in ATP generation but no generation of reactive oxygen species.

(d) Permeabilization of the inner membrane to compounds which are usually not able to traverse the membrane.

Ans. (a)

Sol. Rotenone works by interfering with the electron transport chain in mitochondria. It inhibits the transfes chain in mitochondria. It inhibits the transfer of electrons from iron-sulphur centers in complex I to ubiquinone. This interferes with NADH during the formation of ATP. Complex I is unable to pass off its electron to CoQ creating a back-up of electrons within the mitochondrial matrix. Cellular oxygen is reduced to the radical, creating a reactive oxygen species, which can damage DNA and other components of the mitochondria.

25. Which one of the following statements is not true?

(a) Beta-oxidation of long chain fatty acids occurs in mitochondria.

(b) Fatty acid biosynthesis occurs in peroxisomes.

(d) Peroxisomes import their repertoire of proteins using sorting signals.

Ans. (b)

Sol. Fatty acid biosynthesis occurs in peroxisomes.

26. From the following statements:

1. In proteins the amino acids that can undergo oxidation are Cys and Met.

2. A tetrasaccharide composed of alternate L and D isomers will not be optically active.

3. The (kcal/mol) values for K_eq of 0.1, 0.01 and 0.001 are 1.36, 2.72 and 4.09, respectively. It can be concluded that the relationship between and K_eq is is parabolic.

4. The oxidation states of Fe in haemoglobin is +2. In cytochrome C, the oxidation states of Fe can be +2 or +3.

5. In DNA, the sugar and bases are planar.

6. High-energy bonds hydrolyze with large negative .

Choose the combination with only one wrong statement.

(a) 1, 5, 6

(b) 2, 3, 4

(d) 1, 2, 3

Ans. (a)

Sol. We need to choose the option in which only one option is wrong. A option is wrong because only cysteine amino acid can get oxidized or reduced not methionine. B statement is correct because a tetrasaccharide with two D-form and two with L-form can make a meso form, which will be optically inactive. C option is wrong because the relationship between an K_eq is straight line, all the log curves are straight line. The oxidation state of Fe in haemoglobin is 2+ and in cytochrome c it is +3, because in electron transport chain cytochrome c acts as a electron acceptor, if it will be present in 2+ state than it will not accept a electron. E and F statements are correct, because DNA is planer due to the sugar and bases arranged on the top of each other in DNA structure and strong bonds like covalent bonds hydrolyze with higher amount of .

27. Given below are statements related to protein structures:

P. The dihedral angles of an amino acid X in Acetyl-X-NMethyl amide in the Ramchandran plot, occur in very small but equal areas in the left and right quadrants. It can be concluded that X is not one of the 20-coded amino acids.

Q. The dihedral angles of a 20-residue peptide are represented in the Ramchandran plot. It is possible to conclude that the peptide does not have a proline.

R. Two proteins can have a simila fold even if they do not shae significant similarity in their pimay structure.

S. On denatuation of a potein by urea, the inteactions that wold be disrupted are ionic bonds and van der Waal's interaction but not disulfide bonds.

Choose the combination with all correct answes:

(a) P, Q, R

(b) P, R, S

(d) P, Q, S

Ans. (b)

Sol. On denaturation of a protein by urea, disulphide bonds will be reduced as urea is a reducing agent and thus disulphide bonds will be interrupted.

28. Various modifications of nucleotides occu in nucleic acids. Which of the following combinations contains at least one modification that does not occu in nucleic acids?

(a) N, N-dimethylguanosine, pseudouidine, 2'O-methyluridine.

(b) 2-thiouidine, dihydouridine, N-isopentenyladenine.

(d) dihydrouridine, 4-thiouridine, 2'O-methyluridine.

Ans. (c)

Sol. N, N-dimethylguanosine, pseudouridine, dihydrouridine, N-isopentyladenine, 5-methyldeoxycytosine, 2-Thiouridine are found in nucleic acids (either in RNA or DNA). 4-Thiouridine and 2'-amethyluridine are used for labelling or cross-linking nucleic acids.

29. Given below are statements that may or may not be correct?

1. Fructose 2, 6-bisphosphate is an allosteric inhibitor of phosphofruckokinase 1.

2. The TCA cycle intermediates, succinate and oxaloacetate can both be derived from amino acids.

3. A diet rich in cysteine can compensate for a methionine deficient diet in humans.

4. dTTP fo DNA synthesis can be obtained from UTP.

5. In the fatty acid biosynthetic pathway, the carbon atom from in the synthesis of malonyl CoA is not incorporated into palmitic acid.

Choose the option that repersents the combination of all the correct statements.

(a) 1, 2, 3 and 5

(b) 2, 4 and 5

(d) Only 2 and 3

Ans. (b)

Sol. C statement is not correct because methionine is an essential amino acid for plants as well as animals, so it can never be synthesized from cysteine amino acid. A statement is wrong because fructose 2, 6 biphosphate is an allosteric activator of PFK. Other statements are correct.

30. Three electron acceptors 'X', 'Y' and 'Z' have redox potential (E'₀) of +0.15 V, + 0.05 V and –0.1V, respectively. For a reaction B + 2H + 2e^– BH₂ E'₀= + 0.05 V

Which of these three electron acceptors are appropriate?

= free energy change; n = number of electrons; F = Faraday constant

(a) X and Y

(b) only X

(d) only Z

Ans. (a)

Sol. Electrosn flow spontaneously from low to higher redox potentials, so here +0.05 is just lower than that of the redox potential of 'X' that has redox potential of +0.15.

31. A serine protease was tested for its activity on the following peptide substrate of different lengths and sequences. The obtained kinetic parameters of the protease are shown along with the peptide.

Arrow denotes site of cleavage. Based on the above data, the following statements are made:

P. Catalytic efficiency (K_cat/K_m) increases with the size of the peptide.

Q. Amino acid at the hydrolytic cleavage position of the peptide is critical for binding of the peptide with the protease.

R. Catalytic efficiency decreases from three amino acid peptide to four amino acid peptide.

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

(a) P and Q

(b) P and R

(d) P, Q and R

Ans. (a)

Sol. As the size of peptide in increasing the K_cat value increases, means overall value of K_cat/K_m increases. As one can observe that by replacing alanine from C-terminal with valine K_m value increases drastically. So, this residue is important.

32. Metachromatic leukodystrophy (MLD) is caused by a deficiency of arylsulfatase A and affects the CNS. MLD is

(a) a lysosomal storage disorder

(b) a disease due to dysfunctional mitochondria

(d) caused by a defect in proteins of the nuclear envelope.

Ans. (a)

Sol. Lysosomal storage disorder.

33. Which one of the following pairs is not matched correctly?

(a) Glycocalx – adherence

(b) Fimbriae – motility

(d) Peptidoglycan – cell wall

Ans. (b)

Sol. Fimbriae is usually short hair like fibers. Pathogenic bacteria have adhesins on the fimbriae that allow them to attach to the target tissues of their host.

34. A membrane associated protein is composed of seven "-helices", with each helix containing 19 hydrophobic residues While treating the membrane with all kinds of proteases, a major portion of this protein remains intact. Treatment with high salt (till 1.5 M HaCI) and buffer with pH 5.0 failed to dissociate this protein from the membrane. Predict the most appropriate nature and orientation of this protein in the membrane.

(a) Peripheral glycoprotein.

(b) Integral protein with seven membrane spanning regions.

(d) Peripheral protein with both N and C-termlnal remain exposed to cytosollc surface of the cell membrane.

Ans. (b)

Sol. If it contains 19 hydrophobic amino acids in its one hydrophobic domain and this protein cannot be isolated with the help of 1.5 M-NACl it simply means it is a integral membrane protein. By cleaving it with lot of protease leaves major portion of the proteins intact means it contains several transmembrane domains, the major example of this kind of protein is OPCR or seven transmembrane helical proteins.

35. When the cholera toxin (protein of Mr 90,000 Da) gains access to the human intestinal tract, it binds tightly to specific receptors in the plasma membrane of the epithelial cells lining the small intestine, causing membrane bound adenylyl cyclase to undergo prolonged activation resulting in extensive loss of H₂O and Na⁺. Pretreatment of the epithelial cells with various phosphollpases and proteases failed to Inhibit the binding of cholera toxin to its receptor and the fluid loss but treatment with exoglycosidase, prior to binding, significantly reduces these effects. Which of the following molecule could be the receptor for this toxin?

(a) Phosphatidylcholine

(b) Sodium-potassium ATPase

(d) Chloride-bicarbonate exchanger

Ans. (c)

Sol. When cholera toxin is released from the bacteria in the infected intestine, it binds to the intestinal cells known as enterocytes through the interaction of the pentameric B subunit of the toxin with the GM1 ganglioside receptor on the intestinal cell, triggering endocytosis of the toxin.

36. -bungarotoxin binds to acetylcholine receptor (AChR) protein with high specificity and prevents the ion-channel opening. This interaction can be exploited to purify AChR from membrane using

(a) ion-exchange chromatography.

(b) gel filtration chromatography,

(d) density gradient centrifugation.

Ans. (c)

Sol. Affinity chromatography. Bungarotoxin binds to acetylcholine receptor and therefore receptor can be isolated using bungarotoxin linked affinity column.

37. In Schizosaccharomyces pombe, the recessive (cdc2r) and dominant (cdc2°) mutants have opposing phenotypes. wihile cd2° produces abnormally small cells, cdc2r produces abnormally long cells. Some possible explanations are given below.

P. cdc2° may lack interaction with WEE1.

Q. cdc2r may not interact with CDC13 kinase.

R. cdc2° may not interact with CDC25 phosphatase.

S. cdc2r cells may be deficient In interaction with either CDC25 or WEE1.

Which combination of the above statements is correct?

(a) P and Q only

(b) b P, R and S only

(d) P, Q and R only

Ans. (a)

Sol. Wee-1 kinase causes inhibitory phosphorylation on Cdc2 and the inhibitory phosphate group is removed by CDC25 phosphatases. No integration of Wee-1 leads to production of smaller cells as inhibitory phosphate will not be added. Non-interaction of CDC25 phosphatases or CDC13 causes no mitosis leading to large sized cell production.

38. In tay-Sach disease, accumulation of glycollplds occurs especially In nerve cells. These cells are greatly enlarged with swollen lipd-endosomes and the children with this disease die at a very early stage. Such condition occurs due to a specific defect in

(a) specific lysomomal enzyme that catalyzes a step In the breakdown of gangllosldes.

(b) sorting of an enzyme that adds a phosphate group at 6^th position of mannose In all acid hydrolases.

(d) v-SNARE molecules which cause abnormal vesicle tethering and docking and affect vesicle fusion with lysosomes.

Ans. (a)

Sol. Tay-Sachs disease is caused by the absence of a vital enzyme called hexoseminidase. A it is inherited from a person's parents in an autosomal recessive manner. Due to the absence of hexosaminidase A, which results in the build up of the toxin GM2 ganglioside within cells.

39. In eukaryotes, precursors of micro RNAs (miRNAs) and small Interfering RNAs (siRNAs) are usually synthesized by

(a) RNA Pol I and III, respectively.

(b) RNA Pol III and I, respectively.

(d) only RNA Pol II.

Ans. (d)

Sol. In eukaryotes miRNA and siRNA are synthesized by RNA polymerase II.

40. Aminoacyl tRNAs are escorted to the ribosome by the elongation factor

(a) EF-Ts

(b) Ep-G

(d) eEF'2

Ans. (c)

Sol. After bioding with GTP, EF-Tu transports aminoacyl-tRNA to the A-site of the ribosome.

41. Scientists usually find difficulty in identifying the exact transcription termination site in eukaryotes because

(a) immediately following termination of transcription, the 3'-end is polyadenylated.

(b) the 3'-end is generated by cleavage prior to actual termination of transcription.

(d) 3'-end of transcript is complexed with 5'-end for initiation of translation.

Ans. (b)

Sol. The transcription in eukaryotes even after the transcription terminator sites aer reached, RNA Pol II continues to synthesize at the 3' end, after that one exonuclease starts cleaving the extra RNA From 5' end to 3' End. So, that''s why it is difficult to identify the transcription termination site in eukaryotes.

42. In eukarvotic replication, priming of DNA synthesis and removal of RNA primer Is catalyzed by

(a) DNA Pol and PCNA, respectively.

(b) DNA Pol a and FEN1, respectively,

(d) DNA Pol z and PCNA, respectively.

Ans. (b)

Sol. DNA polymerase contains primase activity, initiates DNA synthesis. The RNA primers are excised by ribonuclease FEN-1, which degrades RNA present in RNA-DNA duplexes.

43. Which one of the following modification of proteins is co-translational?

(a) Palmitoylation

(b) Myristoylation

(d) Addition of cholesterol

Ans. (b)

Sol. Myristoylation is done by both means co-translationally and post-translationally.

44. The lambda and P22 phages are two related lambdoid bacteriophages. A recombinant lambda phage was derived from the wild type lambda phage by replacing its CI repressor gene and the CI binding sites with those from the P22 phage. Both the X™ and the were used independently to infect Escherichia cod strain over-producing CI repressor. Following outcomes were surmised

P. Infection with will lyse the E. coli used.

Q. Infection with will invariably establish lysogeny in the E. coli used.

R. Infection with will lyse the E. coli used.

S. Infection with will invariably establish lysogeny in the E. coli used.

Which combination of the above statements is correct?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (b)

Sol. When mutant phase (Lambda wt) infects E. coli, it takes up Lytic cycle as original cII and cIII forces the cell to take up lytic cycle before cII activates of expression. Thus, statement (iii) is correct. Wild-type lambda can make a choice of either taking up lytic or lysogeny cycle depending upon availability of resources. As only one combination of option satisfies our observation. Hence, option B is correct.

45. Chloramphenicol is a broad-spectrum antibiotic which inhibits protein synthesis in prokaryotes. Given below are a few statements regarding the mode of action of chloramphenicol.

P. Chloramphenicol inhibits the peptidyl transferase activity of ribosomes.

Q. Chloramphenicol can be used to treat moderate to severe infections, because mitochondrial ribosomes are not sensitive to chloramphenicol.

R. Chloramphenicol binds to one of the domains of 23S rRN(A)

S. Chloramphenicol competes for binding with the E-slte tRN(A)

which of the following options describes correctly the mechanism of action of chloramphenicol?

(a) Q and S only

(b) P and R only

(d) Q, R and S

Ans. (b)

Sol. In prokaryotes 23S rRNA catalyze the peptide bond formation. Chloramphenicol binds to the one end of the domains of 23S rRNA and inhibit the peptide bodn formation, thus inhibit the elongation step.

46. A merodiploid strain of E. coli with the genotype was constructed. The activity of -galactosidase enzyme was measured in this strain upon following treatments,

P. no induction.

Q. induction with n moles of IPTG.

R. induction with n moles of lactose.

S, induction with n moles of lactose in the presence of n moles of glucose.

Which one of the following graphs depicts the expected trends in -galactosidase activity under the four different conditions?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. IPTG is a very good inducer as compared to lactose. So, higher amount of galactosidase activity will be observed in case of IPTG as compared to lacotse. In case of lactose and glucose will show less amount of expression.

47. Two experiment were performed. In the first one, Okazaki fragments were prepared from a replicating cell of E. coli grown in the presence of ³²P. In the other, the two strands of E. coli chromosome were separated into a H strand and L strand, immobilized onto a nitrocellulose membrane and hybridized with the Okazaki fragments prepared in the first experiment. Which one of the following options correctly describes the observation?

(a) Okazaki fragments will hybridize to only H strand.

(b) Okazaki fragments will hybridize to only L strand.

(d) Because the H and L strands have been prepared from different cultures of E. coli, the Okazaki fragments will hybridize to neither.

Ans. (c)

Sol. Okazaki fragments will hybridize with only heavy (H) strands.

48. Eukaryotic mRNAs have an enzymatic appended cap structure consisting of a 7-methylguanosine residue joined to the initial 5' nucleotide of the transcripts. Given below are a few statements regarding capping.

P. Capping protects the mRNA from degradation by 5'-exoribonucleases.

Q. During capping, the a-phosphate is released from the 5'-end of the nascent mRN(A)

R. Phosphorylation mediated conformational change in carboxyl terminal domain (CTD) of RNA Pol II enables its binding with capping enzymes.

S. During capping, a 5'-5' triphosphate bond is formed between the p-phosphate of the nascent mRNA and a-phosphate of GTP. Which of the above statement(s) is/are incorrect?

(a) R only

(b) Q only

(d) R and S

Ans. (b)

Sol. During capping the and phosphate is released from the 5'-end of nascent mRNA.

49. In E.coli grown under untrient rich conditions, replcation of entire genome lakes takes about 40 min., yet it can divide every 20 min. This is no because

(a) while E.colii divides divides every 20 min, equal transfer of genetic material occurs only in the alternate rounds of cell divisions.

(b) a second round of genome replication begins before the completion of first round of replication, and by the timce cell is ready to divide, two coples of the genome are availabe.

(d) d-during cell division, only one of tho strands of tho genome whoso synthesis can be achieved in 20 mi transferred to tho daugher cell.

Ans. (b)

Sol. The total genome of bacteria replicate in 40 minutes but its generation time (cell division) is 20 minutes because the second round of genome replication begins before the completion of first round of replication. During this period, cell ready to divide and two copies of genome are available.

50. Which one of the following Is not a bacterial disease?

(a) Tuberculosis

(b) Typhoid

(d) Small pox

Ans. (d)

Sol. Smallpox was an infectious disease caused by either of two virus variants, Variola major and Variola minor.

51. The second messenger, which opens calcium Ion pores In endoplasmic reticulum and plasma membrane is

(a) diacYlglyceroi

(b) cAMP

(d) Inositol triphosphate

Ans. (d)

Sol. During PIP-2 signalling IP3 is produced as a secondary messenger which goes and bind to the IP-3 receptor on SER and causes opening of Ca ion channels.

52. Following are list of some proteins:

P. BCL-2

Q. BCL-XL

R. Al

S. BAX

Which of the protein(s) Is/are not antl-apoptotlc?

(a) S only

(b) R only

(d) Q and S only

Ans. (a)

Sol. BAX is the nuclear-encoded protein present in higher eukaryotes that are able to pierce the mitochondrial outer membrane to mediate cell death by apoptosis. In the given list of BAX is a pro-apoptotic protein, rest all are anti-apoptotic proteins.

53. Which one of the following cells generally does not secrete IFN-y?

(a) CD8⁺T-cells

(b) TH1 cells

(d) TH2 cells

Ans. (d)

Sol. IFN- is produced mainly by natural killer (NK) and natural killer T (NKT) cells as part of the innate immune response and by CD4 Th1 and CD8 cytotoxic T lymphocytes (CTL) effector T cells once antigen-specific immunity develops. Th2 cells upon activation secretes anti-inflammatory cytokines and opposes inflammation. INF is a pro-inflammatory molecule majorly secreted by Th1, Macrophages, Tc etc.

54. The table given below lists organisms (Column A) and characteristic features (Column B):

Column A Column B

P. Caulobacter 1. Multicellular fruiting body

Q. Myxobacteria 2. Endospore

R. Methylotroph 3. Non-free living, penicillin resistant

S. Bacillus subtills 4. Immortal stalk cells

T. Mycoplasma 5. Can use formate, cyanide and carbon monoxide as a source of

carbon

Choose the option that correctly matches organisms with their characteristic features,

(a) P-l, Q-5, R-4,-s-2, T-3

(b) P-4, Q-l, R-5,-s-2, T-3

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

Ans. (b)

Sol. One of the daughter cells of Caulobacter is a mobile "swarmer" cell that has a single flagellum at one cell pole. The myxobacteria produce fruiting bodies in starvation conditions, cells. Methylotrophs are microbes with the intriguing ability to utilize singel carbon (C₁) compounds as the sole energy source for their growth. Bacillus subtilis is a gram positive, rod shaped bacterium that forms a tough, protective endospore.

55. Toll-like receptors (TLR) present In mammalian macrophages are recognized by types of macromolecules that are not present In vertebrates but are present In certain groups of microbial pathogens. When these pathogens infect macrophages, TLR signalling Is stimulate(d) Following are the list of macromolecules In column A and type of TLR In column b.

Column A Column B

P. Lipopolysaccharide (LPS) 1. TLR3

Q. Flagellin 2. TLR4

R. Double stranded RNA 3. TLR5

S. Unmethylated CpG dinucleotides 4. TLR9

which of the following Is the best possible match of the pathogenic ligand

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

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

(d) p-3, Q-4, R-2, S-l

Ans. (c)

Sol. A B

i. Lipopolysaccharide (b) TLR4

ii. Flagellin (c) TLR5

iii. Double stranded RNA (a) TLR3

iv. Unmethlyated CpG dinucleotides (d) TLR9

56. Preventing the blocking action of patched protien leads to activation of Cos-2, which dissociates itself from microtubules, activates Ci/gli, which binds to CBP 9CREB-binding protien) and promontes transcripton of target genes. Which one of the following treatment of cells will mostly prevent CI/GLi activated transcription in the cells/

(a) Small molecules which target Frizzled.

(b) Azepine, an inhibitor of y-secretase.

(d) Cdk blockers, which negatively regulate TGF-induced growth.

Ans. (c)

Sol. Hedgehog signalling involves to membrane molecules. Patched and Smoothened where patched is a negative regulator of smoothened. Upon smoothened activation Ci protein induces gene expression. Drug like cyclopamine inhibits smoothened activity and these no gene expression occurs.

57. The second messenger cAMP, synthesised by adenylyl cyclase transduces a wide variety of physiological signals in various cell types in mammalian cells. Most of the diverse effects of cAMP are mediated through activation of protein kinase A (PKA), also called cAMP dependent protein kinase. Which of the following statements regarding PKA is not correct?

(a) Inactive PKA is a tetramer consisting of two regulatory (R) subunits and two catalytic (C) subunits.

(b) Each R subunit binds the active site in a catalytic domain and inhibits the activity of the catalytic subunits.

(d) Binding of cAMP to R subunit causes a conformational change resulting in binding to site other than catalytic site causing strengthening of binding to C subunit activating its kinase activity.

Ans. (d)

Sol. Inactive PKA is tetramer of 2 identical regulatory subunits and 2 catalytic subunits. 2 molecules of c-AMP interacts with regulatory subunits causing its dislocation fro catalytic subunits. Catalytic subunits become free and active.

58. Given below is a list of some proteins known to be associated with apoptosis, their subcellular localization (but not in correct order) and possible role in apoptosis.

Proteins Localization Role in apoptosis

A. Effector caspase (a) Cytosol 1. Promotes

B. Apaf-1 (b) Cytosol, mitochondria 2. Inhibits

C. Bax (c) Cytosol, nucleus

Choose the right combination which matches the proteins with their correct localization and role in apoptosis.

(a) A-a-2, B-b-2, C-c-1

(b) A-c-1, B-a-2, C-b-2

(d) A-c-1, B-a-1, C-b-1

Ans. (d)

Sol. Effector caspase, BaX and Apaf-1 all promotes apoptosis. Caspases are localise in the cytosol and nucleus. Apaf-1 for a complex with Cyt C in the cytosol and BaX is a pro-apoptotic proteins functional in mitochondria and cytosol.

59. Given below are a list of some extracellular matrix (ECM) proteins in column A and their characteristics in column B, but not in correct order.

Which one of the following is the most appropriate match?

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

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

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

Ans. (c)

Sol. Fibronectins are large proteins composed of two polypeptides of approximately 235 kD linked by disulfide bonds near their C-termini. "Laminins are a major constituent of the basement membrane which is an intricate meshwork of proteins separating the epithelium, mesothelium, and endothelium from connective tissue. Laminins are heterotrimeric proteins, composed of one a, one ß, and one ? chain. "Nidogens, formerly known as entactins, are a family of sulfated monomeric glycoproteins located in the basal lamina of parahoxozoans. Two nidogens have been identified in humans: nidogen-1 (NID1) and nidogen-2 (NID2)."The main amino acids that make collagen are proline, glycine and hydroxyproline. These amino acids group together to form protein fibrils in a triple helix structure. Your body also needs the proper amount of vitamin C, zinc, copper and manganese to make the triple helix.

60. The major histocompatibility complex (MHC) is referred to as the human leukocyte antigen (HLA) complex in humans and as the H-2 complex in mice. In an experiment, H-2^K mice were primed with the lymphoctic choriomeningitis virus (LCMV) to Induce cytotoxic T lymphocytes (CTLs) specific for the virus. Sleepen cells from this LCMV-primed mouse were then added to target cells of the same (H-2 ) or different H-2 haplotypes (H-2^b) that where intracellularly radiolabelled with s'Cr and either infected or not infected with LCMV. CTL mediated killing of target cells were then measured by the release of ⁵¹Cr into the culture supernatant (Cr release assay). In which of the following cells, ⁵¹Cr will be released into the culture supernatant?

(a) H–2^k target cells

(b) H–2^k infected target cells

(d) H–2^b LCMV infected target cells

Ans. (b)

Sol. Since H-2^k mice were infected with LCMV and therefore from spleen reactive Tc cells can be isolated whic hwill show reactivity to only MHC of H-2^k mic infected with LCMV. T cells show MHC dependent response.

61. Inwart movement of an expanding outer layer spreading over the internal surface during gastrulation is termed as

(a) invagination

(b) ingression

(d) delamination

Ans. (c)

Sol. Involution at or near the blastoderm margin occurs during gastrulation. This movement folds the blastoderm into two cellular layers, the epiblast and hypoblast, within a ring (the germ ring) around its entire circumference. It is inward movement of expanding outer layer spreading over internal surface during gastrulation.

62. The ability of cells to achieve their respective fates by interacting with other cells is known as

(a) autonomous specification

(b) conditional specification

(d) competence

Ans. (b)

Sol. Conditional specification is the ability of cells to achieve their respective fates by interactions with other cells.

63. The dorsal-most vegetal cells of the amphibian embryo that is capable of inducing the organizer is called as a Nieuwkoop centre and is marked by the presence of

(a) Chordin

(b) -catenin

(d) Nanos

Ans. (b)

Sol. In amphibians, the Nieuwkoop center – a primary inducing region – has a central role in the induction of dorsal mesodermal cells to form the Spemann organizer. In teleosts, such as the zebrafish, Danio rerio, the functional equivalent of the amphibian Spemann organizer is the dorsal shield. The Nieuwkoop center is a region of -catenin signaling. A very important advance on the nature of the Nieuwkoop center signal has been the discovery that -catenin provides the earliest asymmetry found in the Xenopus egg. The dorsal signal is mediated by the Wnt/-Catenin signal.

64. Which kind of cleavage is shown in mammals?

(a) Holoblastic rotational

(b) Meroblastic rotational

(d) Meroblastic radial

Ans. (a)

Sol. Mammals have holoblastic rotational cleavage, in which the plane of the first cleavage is parallel to the animal vegetetal axis, but the second cell division takes place across two planes at right angles to each other.

65. During embryo germination in a grass family an absorptive organ that forms interface between the embryo and the starchy endosperm tissue is called

(a) Coleorhiza

(b) Coleoptile

(d) Mesocotyl

Ans. (c)

Sol. The scutellum is thought to be a modified cotyledon. The single cotyledon of grasses is transformed into the absorptive scutellum which lies between the endosperm and the embryo axis.

66. The following are certain statements regarding stem cells:

P. All types of stem cells have the ability to give rise to a complete embryo.

Q. Muftipotent stem cells are those whose commitment is limited to a relatively small subset of all possible cell types.

R. Stem cell niches allow controlled self-renewal and also survival of the cells that leave the niche.

S. The pluripotency of the stem cells in an embryo is essentially maintained by Fgf8, Nanog and TGFp.

T. Adult ceHs may be reprogrammed to gain pluripotency by modifying the following genes: Oct 3/4, Sox2, c-myc, WK Which one of the following combinations of statements is correct?

(a) P, Q and S

(b) Q and T

(d) P, R and S

Ans. (b)

Sol. Only embryonic stem cells can be used for embryo formation. Multipotent stem cells differentiate to limited cells only. Cells leaving stem cell niche are not dependent on stem cell factors. For iPS technology all the factor Oct 4, Sox2, c-Myc and Klf4 is required.

67. When the 4 blastomere pairs of the 8-cell stage embryo are dissociated, each forms of the structures it would have formed had it remained in the embryo. However, the notochord and nervous system get specified only if different blastomers get the change to intract. Given below are certain interpretations formulated from the above results:

P. Each pair of blastomers forming respective structures indicate autonomous spefification.

Q. Each pair of blastomeres forming respective structures indicate conditional specification

R. The notochord and nervous system development Indicate autonomous specification

S. The notochord and nervous system development indicate conditional specification.

Which combination of interpretations Is most appropriate?

(a) P and R

(b) Q and S

(d) Q and R

Ans. (c)

Sol. Autonomous specification in the early tunicate embryo. When the four blastomere pairs of the 8-cell embryo are dissociated, each forms structures that is would have formed if it had remained in the embryo. But the notochord and nervous system are specified by cellular interactions.

68. The presence of -catenln in the nuclei of blastomeres in the dorsal portion of the amphibian embryo is one of the determinants for laying down the dorso-ventral axis. What will be the outcome of expressing a 2a negative form of GSK3 in the ventral cells of early embryo? expressing a dominant

(a) The dorsal cell will be ventralize

(b) A second axis will be formed

(d) The embryo will develop normally.

Ans. (b)

Sol. The importance of GSK3 in Xenopus development has been shown with a dominant-negative mutant, which stabilizes -catenin and induces and ectopic axis.

69. Extensive molecular genetic studies on miR156, miR172, SPL genes and AP2-like genes have yielded the following functional model on the juvenile adult reproductive transition in Arabidopsis:

Based on these results, the following schematic diagram has been proposed to predict the expression kinetics of these genetic factors. Which of the following combinations is most likely to be correct?

(a) a-miR156; b-SPL genes; c-miR172; d-AP2 like genes

(b) a-miR156; b-miR172; c-SPL genes; d-AP2 like genes

(d) a-miRlse; D-AP2 like genes; c-miR172; d-SPL genes

Ans. (d)

Sol. The correct option is (d).

70. Injection of noggin mRNA in cells that will become the future ventral side of a frog embryo mimics the effect of an organizer graft to the ventral side. This experiment demonstrates that

P. Noggin is a transcription factor.

Q. Noggin induces ventral fates.

R. Noggin is involved in organizer fate.

S. Noggin Is required to induce a secondary axis.

Which one of the following options represents correct combination of statements/s?

(a) P and R

(b) R and S

(d) Q and R

Ans. (b)

Sol. Noggin mRNA is first localized in the dorsal blastopore lip region and then becomes expressed in the notochord. Rescue of dorsal structures by Noggin protein occurs when Xenopus eggs are exposed to ultraviolet radiation, cortical rotation fails to occur and the embryos lack dorsal structures in a dosage-related fashion. Hence, statement C is correct. If too much noggin message is injected, the embryo produces dorsal anterior tissue at the expense of ventral and posterior tissue, becomin little more than a head. Hence, statement D is partially correct. As noggin is Sufficient to induce secondary axis, not required. So, only option possible is C and D.

71. Antennapedia complex n Drosophila contains five genes, lab, pb, dfd, scr and Antp and they express in parasegments 1 to 5, respectively in a non-overlapping manner. In the larva stages of development, the region of Antp (Antennapedia) expression corresponds to a part of second thoracic segment. A mutation in Antp is known as to casue transformation of antenna to leg-like structure. Below are certain statement made in respect to the functions of Antennapedia:

P. In the above describe Antp mutation, the gene ectopically express in the head region.

Q. One of the functiosn of Antp is to be repress that induce antenna development.

R. Antp express in thorax and forms a concentration gradient in the posterior-anterior direction, thus affecting head development.

S. Homozygous resseive mutation of Antp is expected to transform the leg to antenna in the second thoracic segment. which combination of the above statements corrrectly describe the function of Antennapedia?

A homozygous recessive mutation of Antp is expected to transform the leg to antenna in the second thora Which combination of the above statements correctly describes the function of Antennapedia?

(a) P, Q and R

(b) Q and R

(d) P, Q and S

Ans. (d)

Sol. Antennapedia complex in Drosophila have five genes lab (labial), pb (proboscipedia), dfd (deformed), scr (sex combs reduced), Antp (Antennapedia). Lab, pb and dfd expressed in head region, scr express in T₁ segment and participate in leg formation within it. Antp express in T₂ segment and participate in legs and wings formation within it. So, the expression of Antp not affect the head development.

72. The energy-rich fuel molecules produced in the TCA cycle are

(a) 2 GTP, 2 NADH and 1 FADH₂

(b) 1 GTP, 2 NADH and 2 FADH₂

(d) 2 GTP and 3 NADH

Ans. (c)

Sol. Per turn of TCA cycle produced 1 GTP, 3 NADH and 1 FADH₂.

73. The following statements are made regarding secondary metabolites of plants:

P. All secondary metabolites are constitutively produced in all cells of a plant during its entire life

Q. They serve as signals to help the plant survive adverse conditions.

R. They may be volatile compounds.

S. They contribute to flower colour.

Which one of the following options represents a combination of correct statements?

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (b)

Sol. Secondary metabolities are the compounds which are derived by pathways from primary metabolism, and are not essential to sustain the life of cells. These compounds do not have a continuous production. Secondary, metabolities are produced during non-growth phase of cells.

74. For which one of the plant hormone biosynthetic pathways, 1-aminocyclopropane-l-carboxylic acid is an intermediate?

(a) Abscisicacid

(b) Brassinosteroid

(d) Gibberellic acid

Ans. (c)

Sol. ACC plays an important role in the biosynthesis of the plant hormone ethylene. Ethylene ACC is an intermediate in the ethylene hormone biosynthesis during YANG cycle.

75. In a study, it was found that K⁺ ion concentration in root cells of a pea plant was ~75 times greater than that of the nutrient medium in which the plant was grown. This indicated that K⁺ ions were absorbed from the medium

(a) because the plant was grown continuously in the dark.

(b) by an active, energy-dependent process.

(d) through plasmodesmatal connections between the epidermis and the medium.

Ans. (b)

Sol. The K⁺ ion concentration in root cells of a pea plant = 75 times greater than that the nutrient medium in which the plant was grown. So, the uptake of K⁺ ion is an active process which require energy currency.

76. A researcher wanted to study light reaction during photosynthesis by blocking photosynthetic electron flow using the herbicide, dichlorophenyl-dimethylurea (DCMU) and paraquat. The researcher listed the following observations:

P. Both DCMU and paraquat the electron flow in Photosystem II.

Q. Both DCMU and parqquat block the electron flow in Photosystem I.

R. DCMU blocks electron flow Photosystem I while parquuat blocks in photosystem II.

S. DCMU blocks electron flow in Photosystem II while paraquat blocks in Photosystem I.

Which of the following combinations of the above statement is incorrect?

(a) P, Q and R

(b) P, Q and S

(d) Q, R and S

Ans. (a)

Sol. DCMU act by blocking electron flow at the quinone acceptors of photosystem II (block the electron flow to Q_R). Paraquet act by accepting electrons from the early acceptors of photosystem I.

77. Following are a few statements regarding water potential in plants:

P. Solute concentration and pressure potential contribute to water potential of a cell in a given state.

Q. When a flaccid cell Is placed In a solution that has a water potential less negative than the intracellular water potential, water will move from solution into the cell.

R. When a flaccid cell is placed in a solution that has a water potential less negative than the interacellular water potential, water will move out from cell into the solution.

S. Water potential of a plant cell under werver water stress is always ngative as compared to that of unstressed cells.

Which combination of the above statements is correct?

(a) P and Q

(b) Q and R

(d) R and S

Ans. (a)

Sol. Chemical potential is the free energy of the molecule in the system. In terms of water it become water potential ().

Solute concentration and pressure potential contribute to water potential of a plant cell in a given state. So; = .

Pure water has value is zero (0). So water potential of in solution will be negative.

Water move from highg to low . The of solution is less negative then the intracellular . So water move from solution into the cell.

78. The following scheme shows the flowering status of a plant species and the photoperiod regimes In which it is grown (L denotes light period; D denotes dark period).

(a) The species is a short day plant; length of the dark phase determines flowering status.

(b) The species is a long day plant; length of the dark phase determines flowering status.

(d) The species is a long day plant; length of the light phase determines flowering status.

Ans. (a)

Sol. Short day plant produce flower when light period is less then critical day length. According to the question, species is the short day plant; length of the dark phase determines flowering status.

79. In a photoresponse experiment, imbibed seeds were kept under the following light regimes and their germination status was noted as follows:

D Darkness; R: Red light; FR: Far-red light

In an independent biochemical experiment, it was demonstrated that the red light photoreceptor phytochrome interconverted between two forms. P and P', by red or far-red light.

keeping these information in mind, which of the following combination of conclusion is correct?

(a) Red light convertes P to P'; P' promotes seed germination.

(b) Far-red light P to P'; P' promotes seed germination

(d) Far-red ligth convertes P' P promotes seed germination.

Ans. (a)

Sol. Red light promotes seed germination because it converts P (inactive form of phytochrome) to P' (active form of phytochrome).

80. Following are a few statement reagardin the structure of terpenes:

P. Isopentenyl diphosphate and farnesyl diphosphate are monoterpene and sesquiterpene

Q. Squalene and geranyl diphosphate are triterpene and monoterpene, respectively.

R. Dimethylallyl diphosphate and geranylgeranyl diphosphate have 10 and 20 carbons respectively.

S. Diterpenes have 20 carbons, whereas sesquiterpenes have 15 carbons.

Which combination of the above statement is correct?

(a) P and Q

(b) Q and S

(d) R and S

Ans. (b)

Sol. Squalene is a natural 30-carbon organic compound, so it is interpene. Geranyl diphosphate contain 10 carbon, so, it is monoterpene. Diterpenes have 20 carbons, whereas sesquiterpenes 15 carbons.

81. Consider the following facts regarding the control of shoot apical meristem (SAM) size in Arabido

P. Loss of the CLAVATA1 (CLV1) gene leads to bigger SAM.

Q. Loss of the C LAV ATA 3 (CLV3) gene leads to bigger SAM.

R. Loss of the WUSCHEL (WUS) gene leads to smaller SAM.

S. Loss of both CLV1 and WUS leads to smaller SAM.

T. Loss of both CLV3 and WUS leads to smaller SAM.

U. Loss of both CLV1 and CLV3 leads to bigger SAM.

V. Overexpression of CLV3 leads to smaller SAM.

W. Overexpression of CLV3 in the loss of function mutant of CLV1 leads to bigger SAM.

Based on the above information, which of the following genetic pathways describes the relationship among CLV1, CLV3 AND WUS most appropriately?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. In the subapical zone of developing embryo WUS gene activity high that maintain meristem activity of the SAM and also activate CLV3 gene, CLV3 diffuse from cytosolic area to apoplastic area and bind with its specific receptor CLV1 and 2. After binding of CLV3 to CLV1 the activity of CLV1 is activated. Active CLV1 activate ROP, KAPP that ultimately inactive the WUS protein. It is also essential for node and internode development.

82. Filtration slits are formed by

(a) podocytes c mesangial cells

(b) endothelial cells of capillary

(d) lacis cells

Ans. (a)

Sol. Podocytes are the foot projections wrap around the capillaries and leave slits between them. Blood is filtered through these slits, each known as a filtration slit or slit diaphragm.

83. Which one of the following vitamins is nor absorbed in the small intestine by Na⁺-co-transporters?

(a) Thiamine

(b) Riboflavin

(d) Ascorbic acid

Ans. (c)

Sol. Vitamin C (ascorbic acid) not absorbed in the small intestine by Na⁺ co-transporter.

84. Which one of the following is not formed after post-translational processing of preproglucagon?

(a) Glicentin

(b) p-lipotropin

(d) Oxyntomodulin

Ans. (b)

Sol. Lipotropin is a hormone produced by the cleavage of pro-opiomelanocortion (POMC). The anterior pituitary gland produces the pro-hormone POMC, which is then cleaved again to form adrenocorticotropin (ACTH) and -lipotropin (-LPH), Proglucagon is processed differently in the islet cells and the intestinal endocrine L cells to release either glucagon or glucagon-like peptide P-(7-37), GLPI-(7-37), peptide hormones with opposing actions in vivo. The kexin-like prohormone convertase, PC2 (SPC2), is responsible for generating the typical -cell pattern of proglucagon processing, giving rise to glucagon and leaving unprocessed the entire C-terminal half-molecule known as major proglucagon fragment or MPGF. PC3 (SPC3), the major neuroendocrine prophormone convertase, reproduces the intestinal L cell processing phenotype, in which MPGF is processed to release two glucagon-related peptides, GLP1 and GLP2, while the glucagon-containing N-terminal half-molecule (glicentin) is only partially processed to oxyntomodulin and small amounts of glucagon.

85. Which one of the following Is the most powerful buffer system of blood?

(a) Bicarbonate

(b) Phosphate

(d) Haemoglobin

Ans. (a)

Sol. The major buffer system in the ECF is the CO₂-bicarbonate buffer system. This is responsible for about 80% of extracellular bufferring. It is the most important ECF buffer for metabolic acids.

86. In kidney, Na⁺ is reabsorbed across the second half of proximal tubule due to positive transepithelial voltage (i.e. tubu.ar fluid becomes positive relative to blood) and by other mechanisms. The following proposed statements could explain the development of this positive transepithelial voltage.

P. CI^– concentrat on gradient in the second half of the proximal tubule favours diffusion of CI^– from tubular lumen to mtercellular space v.a a paracellular route, which generates the positive transepithelial voltage.

Q. The Na⁺ – H⁺ antiporters in the second half of proximal tubules create the positive transepithelial voltage.

R. The Na'-glucose symporters operating in the proximal part of renal tubules are responsible for this positive transepithelial voltage.

S. The positive transep.thelial voltage is created by the operation of INa⁺ – 1K⁺ – 2CI^– symporter in the proximal tubules.

Select the option with correct statements.

(a) Only P

(b) Q and R

(d) Only S

Ans. (a)

Sol. Trans epithelial voltage Cl^– concentration gradient. CI^– concentrat on gradient in the second half of the proximal tubule favours diffusion of CI^– from tubular lumen to mtercellular space v.a a paracellular route, which generates the positive transepithelial voltage.

87. An action potential of a nerve fibre is described by different components including after-hyperpolarization. The mechanism of generation of this after-hyperpolarization has been proposed in the following statements:

P. The increased conductance of Na⁺ has returned to the base line level but the conductance of K⁺ remains elevated during after-hyperpolarization phase.

Q. The membrane potential is pulled even closer to the K⁺ equilibrium potential at the after-hyperpolarization phase.

R. The conductance of Na⁺ is increased before any change of K⁺ conductance during after-hyperpolarization phase.

S. At the after-hyperpolarization phase, the membrane potential is driven closer to Na⁺ equilibrium optential. Choose the option with both correct statements:

(a) P and Q

(b) Q and R

(d) P and S

Ans. (a)

Sol. The increased conductance of Na⁺ has returned to the base line level but the conductance of K⁺ remains elevated during after-hyperpolarization phase. and The membrane potential is pulled even closer to the K⁺ equilibrium potential at the after-hyperpolarization phase.

88. In an experiment on healthy young men, the muscarinic receptor antagonist, atropine was administered to one group (Group A) while the -adrenergic receptor antagonist, propranolol was administered to another group (Group B) in four increasing doses of equal concentration for both the drugs. The effects of these two drugs on the heart rate are shown below:

On the basis of these observations, an investigator proposed the following statements:

P. Atropine and propranolol block sympathetic and parasympathetic effects on the heart, respectively.

Q. As the change of heart rate is more in Group A than in Group B, the sympathetic tone usually predominates in healthy resting individuals.

R. Atropine and propranolol block parasympathetic and sympathetic effects on the heart, respectively.

S. as substantia, changes occur in the heart rate with atropine, the parasympathetic tone is predominant in healthy resting individuals. Select the option with incorrect statements.

(a) Only P

(b) P and Q

(d) P and S

Ans. (b)

Sol. Atropine-parasympathetic stimulation via muscarinic receptor.

89. The P₅₀ value of haemoglobin for oxygen is increased during exercise. The mechanism of this change is described in the following proposed statements.

P. Increased CO₂ production by muscles elevated PCO₂ of blood which affects P₅₀ value.

Q. The affinity of haemoglobin for oxygen increases as 2,3-bisphosphoglycerate (2,3-BPG) level is elevate(d)

R. Increased body temperature shifts the oxyhaemoglobin dissociation curve to the left.

S. The decreased pH of blood reduces the affinity of haemoglobin for oxygen. Which of the above statements is/are incorrect?

(a) Only P

(b) Q and R

(d) P and S

Ans. (b)

Sol. The P₅₀ value of haemoglobin for oxygen is increased during exercise. During this changes Increased CO₂ production by muscles elevated PCO₂ of blood which affects P₅₀ value and The decreased pH of blood reduces the affinity of haemoglobin for oxygen.

90. There is evidence that following pyrogenic stimuli, cytokines produced by the CNS cause fever, possibly by local release of prostaglandins. Accordingly, the following statements have been proposed:

P. Cytokines act independently and directly on thermoregulatory centres.

Q. Intrahypothalamic injection of prostaglandin receptor agonists will prevent fever.

R. Antipyretic effect of aspirin is exerted on the hypothalamus to prevent prostaglandin synthesis

S. Aspirin blocks infections and eventually prevents fever.

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

(a) P and S

(b) Q and R

(d) P and R

Ans. (d)

Sol. An agonist is a chemical that binds to a receptor and activates, the receptor to produce a biological response, so (b) is incorrect. IL-1 affect the activity of the hypothalamus, the thermoregulatory center, which IL-1 is called an endogenous pyrogen. Aspirin has antipyretic (fever-reducing) activity, Aspirin acts on the hypothalamus, a small gland at the base of the brain that helps regulate body temperature, by inhibiting the productions of the prostaglandins that stimulate the hypothalamus to elevate the body's temperature set-point.

91. Following are some statements about the mechanism of stimulation of receptors for touch, pain, vision and warmth that may or may not be correct.

P. The touch receptor does not require any voltage gated cation channel for its activation.

Q. Ught causes closing of Na⁺ channels in the outer segments of rods and cones.

R. Pain sensation is caused by opening of Na⁺ or Na⁺/Ca⁺⁺ channels in free sensory nerve endings.

S. The warmth receptor is activated by non-selective anion channels.

Choose the option with both statements as correct?

(a) P and R

(b) Q and R

(d) P and S

Ans. (b)

Sol. Light induces rohodopsin activity causing closure of c-GMP induced Na ion channels and making the membrane slightly hyper polarised.

Pain mediated signalling involves Ca ion response.

92. In Drosophila melanogaster males, homologous chromosomes pair and segregate during meiosis but crossing over does not occur. At which stage of meiosis does segregation of 2 alleles of a gene take place in these individuals?

(a) Zygotene

(b) Diakinesis

(d) Anaphase II

Ans. (c)

Sol. Anaphase-I no crossing over leads to segregation of alleles at anaphase-I.

93. A recessive Inherited disease is expressed only in individuals of blood group O and not expressed In blood groups A, B or A(B) Alleles controlling the disease and blood group are Independently inherited. A normal woman blood group A and her normal husband with blood group B already had one child with the disease. The women is pregenent for second time. What is the probability that the second child will also have the disease?

(a) 1/2

(b) 1/4

(d) 1/64

Ans. (c)

Sol. It is already mentioned that the couple has a child with blood group O and the child is diseased. Thus, the parents are heterozygotes for the diesased allele (as parents are normal and the disease is recessively inherited) and for the blood group. The probability of O blood group therefore can be calculated by IA i × iB i = ii (probability of O blood group is thus 1/4). Similarly, for the disease to be inherited the individuals should have both the recessive alleles, thus Dd × Dd = dd (probability to get the disease is 1/4). Therefore, probability that the second child has the diesease = P(the child is blood group O) × P(the child has both the recessive alleles). Thus, the P(diseased child) = 1/4 × 1/4 = 1/16.

94. A lac^– culture of E coli was mutagenlse(d) On what media would one spread the mutagenised lac⁺ cells?

(a) Minimal media + lactose.

(b) Rich media + lactose.

(d) Rich media + IPTG + X-Gal.

Ans. (a)

Sol. To identify the lac⁺ mutants we need to provide the cell rich media, an inducer in the form of IPTG and the substrate X-gal which is a colourless substrate. If the galactosidase enzyme is present in the fully functional state, galactosidase will convert this substrate to a blue product. So, lac^– cells will give white colonies and lac⁺ cells will give blue colonies.

95. The following Is a schematic representation of a hypothetical pathway involved in formation of eye color in an insect species.

Genes A and B are linked and have a map distance of 10 cM. Females with genotypes a⁺ a b⁺ b are test crossed. Further, in these females, the two genes are linked in c is a⁺ and b⁺ represent wild type alleles, while a and b are null alleles. The progeny of the test cross have individuals with four different eye colours. What is the expected ratio of individuals with eye color Red: Vermillion: Brown: White in the progeny?

(a) 9 : 3 : 3 : 1

(b) 1 : 1 : 1 : 1

(d) 1 : 9 : 9 : 1

Ans. (c)

Sol. a⁺ab⁺b × aabb

a⁺ab⁺b 45 red

aabb 45 white

aab⁺b 5 vermilion

a⁺abb 5 brown

96. In the following diagram, segments A and C are copies of 10 base pair repeat DNA sequences, flanking a unique stretch shown as (B) A and C are in inverted orientation as indicated by arrows. Intramolecular recombination between A and C leads to which event:

(a) The complete region encompassing A to C will be inverted.

(b) Only A and B will be inverted.

(d) Only regions A and C will be inverted.

Ans. (c)

Sol. As A and C are in inverted orientation as indicated by arrows in diagram, intramolecular recombination between A and C leads to the condition where only B will be inverted.

97. Somatic cell hybridization is used to assign a gene to a particular chromosome. When two cell lines from two different species are fused, they form a heterokaryon which tends to lose chromosomes as they divide, preferentially from one species. A panel of cell lines was created from mouse-monkey somatic cell fusions. Each line was examined for the presence

On the basis of these results, which chromosome has the gene that codes for the given enzyme?

(a) Chromosome 10

(b) Chromosome 7

(d) Chromosome 5

Ans. (b)

Sol. Chromosome 7 of monkey is required for enzyme synthesis.

98. A phenotypically normal fruit fly was crossed to another fly whose phenotype was not recorded of the progeny, 3/8 were wild type, 3/8 had ebony body color, 1/8 had vestigial wings and 1/8 had ebony body color vestigial wings. Ebony body color and vestigial wings are recessive characters and their genes are located on two different autosomes. Based on this information which one of the following is the likely genotype of the pare different

(a) ee vgvg and e⁺ e⁺ vg⁺ vg

(b) ee vg⁺vg and e⁺ e vg⁺ vg

(d) e⁺e vg⁺vg and e⁺ e vg⁺ vg

Ans. (b)

Sol. The genotype of parents should be ee vg⁺vg and e⁺ e vg⁺ vg.

99. In normal individuals, there are three Mstll restriction sites, two flanking the -globin gene and one within the gene. In individuals affected by a disease, a single nucleotide polymorphism in the -globin gene abolishes the internal MstI recognition site. The RFLP pattern for this locus, obtained by hybridization using a prob internal to the flanking Mstll sites, from three siblings of a family is shown below:

Based on the above profile, what is the nature of the genetic disorder?

(a) X-linked recessive

(b) Autosomal dominant

(d) X-linked dominant

Ans. (c)

Sol. The normal son showing the three band, in this it can be observed only when the son is heterozygous to this trait, the length between two MstII sites flanking the globin gene is 1.35 kb. The daughter is showing two bands and thats only possible when she contains two copies of wild type alleles. The affected son showing only one band which can be observed when the individual have two copies of mutated allele.

100. In a transduction experiment using a⁺ b⁺ c⁺ genotype as a donor and a^– b^– c^– as the recipient, a⁺ transductants were selected and screened for b and c. The data obtained are shown below:

The cotransduction frequencies for a⁺ b⁺ and b⁺ c⁺, respectively, are:

(a) 17% and 12%

(b) 22% and 9%

(d) 17% and 9%

Ans. (b)

Sol.

101. As per the cladistic taxonomy, Archosaurs are a group of diapsid amniotes which include extinct dinosaurs. The living representatives of the group consist of

(a) Anurans and Aves

(b) Aves and Crocodilia

(d) Osteichthyes and Squamata

Ans. (b)

Sol. The archosaurs (group of diapsid amniotes) were the direct ancestors of the dinosaurs. The evolution of the archosaurs is a very significant event in the history of life and one branch led to Aves while the other to Crocodilia.

102. If you want to divide a human body into dorsal and ventral sections, what plane will you use?

(a) Coronal

(b) Abdominopelvic

(d) Sagittal

Ans. (a)

Sol. The Coronal Plane/Frontal Plane, is a vertical plane running from side to side which divides the body or any of its parts into anterior/ventral and posterior/dorsal portions.

103. Which one of the following bryophyte has multicellular rhizoids and its cells mostly contain numerous chloroplasts

(a) Anthoceros

(b) Sphagnum

(d) Marchantla

Ans. (b)

Sol. Sphagnum consists of many multicellular rhizoids with oblique septa and has many chloroplasts per cell.

104. Which of the following Is not true for the Anammox bacteria?

(a) They convert nitrate and ammonium Into dlnltrogen.

(b) They are responsible for 30-50% of the dlnltrogen gas produced In the ocean.

(d) Membranes of these bacteria contain laddcran lipids.

Ans. (a)

Sol. Anammox bacteria convert nitrite (not nitrate) and ammonium into dinitrogen.

105. To understand prey-predator relationship, Didinium (predator) and Paramecium (prey) were use(d) Paramecium population was grown with sand sediment as hiding place or refuge. To this population, Dldlnlum was Introduced only once. What would happen to the prey population In the course of time?

(a) The population will steadily decrease and vanish.

(b) The population will Initially Increase and then stabilize.

(d) The population will steadily Increase.

Ans. (c)

Sol. First prey were eliminated from the clear-fluid medium by the predators and then afterwards the predators starved to death in the absence of prey. The sand sediment served as a refuge for the prey from where they emerged later, increased its number via reproduction and finally reached the stable equilibrium.

106. Which one of the following Is nor correct?

(a) island ecosystems are less prone to biological invasion because of their distance from mainland.

(b) Invasive species have greater phenotypic plasticity compared to native species.

(d) At a large scale, diversity rich ecosystems are generally more prone to invasion.

Ans. (a)

Sol. Diverse rich ecosystem should be more resistant to invasion. The reasons being the more diverse the community, the fewer are the resources left available for new colonists thereby decreasing invasion success.

107. Which one of the following is in the correct decreasing order for the major reservoirs of carbon on Earth?

(a) Terrestrial soils > Terrestrial vegetation > Atmospheric CO₂ > Large lake sediments.

(b) Terrestrial soil > Large lake sediments > Terrestrial vegetation > Atmospheric CO₂.

(d) Large lake sediments > Terrestrial soils > Atmospheric CO₂ > Terrestrial vegetation.

Ans. (d)

Sol. Terrestiral soils hold the bulk of the carbon that is processed by microbes which through decomposition release the carbon back into the atmosphere as carbon dioxide followed by larger take sediments, then the terrestrial biota and least in the atmosphere.

108. In an experiment to determine the number of rats in a field, 80 rats were initially captured, marked and released. After one month, 100 rats were captured in the same field, of which 20 were previously marked ones. Based on the above observation, estimated population size of the rats in the field will be

(a) 160

(b) 200

(d) 1600

Ans. (c)

Sol.

Here, n₁ = 80

m₂ = 20,

n₂ = 100

Hence, N = = 400.

109. What do mayflies, Pacific salmon (Oncorhynchus spp.) and annual grain crops have in common? They are all

(a) semelparous

(b) iteroparous

(d) viviparous

Ans. (a)

Sol. Semelparous species focus all reproductive efforts on a single large event with very high reproductive rates per event.

110. In the following columns, certain terms and their descriptions are given in random order.

Column I Column II

P. Protostome 1. A fluid filled cavity lying inside the external body wall bathing the internal organs

Q. Deuterostome 2. Mouth forming from the blastopore

R. Pseudocoely 3. Coelom formed by splitting the mesodermal tissue

S. Schizocoely 4. Mouth forming from a second opening other than blastopore

T. Enterocoely 5. Coelom formed from pouches pinched off from the digestive tract

Which of the following combination gives correct match for the terms in column I from column II?

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

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

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

Ans. (d)

Sol. Protosomes are those organisms in which the oral end of the animal develops from the blastopore while in Deuterostomes the oral end of the animal develops from a second opening on the dorsal surface of the animal and the blastopore becomes the anus. Mesoderm of epithelial origin forms by cyagination of pouches from the archenteron and coincides with coelom formation through enterocoely. Embryotic mesenchyme may either be returned as mesenchymal mesodermal tissue in the adults or reorganized into coelomic epithelium by undergoing a mesenchymal to epithelial transition, a process referred to as schizocoely. Pseudocoely is when a body cavity is not a product of gastrulation and is not lined with a well-defined mesodermal membrnae. The mesoderm is present, as small pouches in between ectoderm and endoderm.

111. Given below are some statements on vertebrates. Which one of the following statements Is Incorrect?

(a) Muscular post-anal tail and pharyngeal slits are derived characters in vertebrates like notochord arid dorsal hollow nerve cord.

(b) Like echinoderms, vertebrates are deuterosomes

(c) Presence of two or more sets of HOX genes in living vertebrates distinguish them from cepha,ochordates urochordates which have only one set.

(d) Since adult hagnshes and lampreys lack vertebra, column, they are categorized outside class Vertebrata, but are retained under "chordata" along with Cepha,ochordates and urochordates.

Ans. (d)

Sol. Indeed the characters mentioned in (A) are of vertebrates but vertebrates are classified in chordates and on the other hand in option (D) there is an recent update regarding Hagfishes and lampreys i.e., they are placed outside the subphylum Vertebrate of the phylum Chordata.

112. A comparison of Bacteria, Archaea and Eukarya with respect to a few characteristics is given below:

Which of the following combinations present a correct comparison of characteristics in the table above?

(a) P, Q, R and T

(b) P, R and S

(d) P, Q and T

Ans. (d)

Sol. In bacteria formylated methionine is the initiator amino acid for protein synthesis while in archaea and eukaya methionin is the initiator amino acid.

Bacteria contains single copy of covalently closed circular double stranded DNA without histone while the genome of archaea and eukarya packed with histone proteins.

Bacterial genome free from intron but some genes of archaea contain introns and most genes of eukarya contain introns.

113. The table given below provides a list of diseases and causal organisms.

Disease Casual organism

P. Sleeping sickness in humans 1. Trypanosoma cruzi

Q. Chagas disease in humans 2. Trypanosoma brucei

R. Blast disease of rice 3. Magnaporthe graminis

S. Powdery mildew of grasses 4. Magnaporthe oryzae

5. Blumeria oryzae

6. Blumeria graminis

Which of the following options represent the correct match between disease and the causal organism?

(a) P-l, Q-2, R-5, S-6

(b) P-2, Q-l, R-3, S-5

(d) P-2, Q-l, R-4, S-6

Ans. (d)

Sol. Sleeping sickness, also known as African trypanosomiasis, is caused by Trypanosoma brucei whereas Chagas disease is caused by Trypanosoma cruzi. Blast disease of rice is caused by Magnaporthe oryzae and Powdery mildew of grasses is caused by Blumeria graminis.

114. The table given below lists species and conservation status.

Species Conservation Status

P. White bellied Heron 1. Critically endangered

Q. Ganges river dolphin 2. Endangered

R. Gaur 3. Vulnerable

S. Clouded leopard

Which one of the following is the correct pairing between Indian animal species and their conservation status?

(a) P-l, Q-l, R-2, S-3

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

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

Ans. (c)

Sol. White bellied Heron is classified as Critically Endangered because it has an extremely small and rapidly declining population. Gangetic river dolphin is listed under the Endangered category. Gaur and Clouded leopard together are scheduled under Vulnerable category.

115. The following is a list of reproductive strucutre found n vascular and non-vascular plants.

P. Archegonia

Q. Megaspore

R. Capsule

S. Fern frond

T. Pollen

U. Corolla

. The net reproductive rate (R₀) is 1.5 for a given population. If N_t, the population of females at generation t, is 500, then what will be the population of females after four generation (N_{t + 4})?

(a) 11.25.000

(b) 2531.250

(d) 3796.875

Ans. (b)

Sol. Archegonia, megaspore and pollen grain associated with the gametophytic life cycle.

116. Two species, M and N, occupy the same habitat. Given below is a 'state-space' graph in which the abundance of species M is Plotted on the X-axis and abljndance of species N is plotted on the Y-axis. For each species, the zero-growth isocline is plotted.

– zero-growth isocline for species M

– zero-growth isocline for species N

K_H = carrying capacity of the habitat for species M in absence of species N

N_N = carrying capacity of the habitat for species N in absence of species M

= per capita effect of species N on M

= per capita effect of species M on N

Based on the above plot some deductions are made. Which one of the following statements is incorrect?

(a) At point A, populations of both the species M and N increase.

(b) At point B, population of species M increase while that of species N decreases,

(d) Ultimately species N will be eliminated.

Ans. (b)

Sol. The population of females after four generations can be calculated using the following formula :

N_t = N_o(R_o)^t where N_t = Population at time

"t", R_o = Net reproductive rate and t = time period

N_{(t + 4)} = 500(1.5)⁴

or 500(5.0625) = 2531.25.

117. Which one of the following statement is not correct?

(a) Herbivores enhance the productivity of a productive ecosystem and reduce the productivity of an unproductive ecosystem.

(b) Detritus based food chains are longer in more productive ecosystems.

(d) Production efficiency of carnivores is higher than herbivores.

Ans. (c)

Sol. Since the point B lies beyond the zero growth isocline for species N but within the zero isocline of species M, so the population of species N will rather decrease there and that of species M will increase.

118. Following are the descriptions used by conservation biologists for characterizing species/groups In a community:

P. Species with a dlsproportlonally large effect on its environment relative to its abundance.

Q. Species defining a trait or characteristics of the environment.

R. Species whose conservation leads to direct protection of other species.

S. Species which is instantly recognizable and used as the focus of a broader conservation effort.

Which of the following combination correctly identifies these species/groups?

(a) P-Keystone species, Q-Indicator species, R-Flagship species, S-Umbrella species

(b) P-Keystone species, Q-Indicator species, R-Umbrella species, S-Flagship species

(d) P-Umbrella species, Q-Indicator species, R-Keystone species, S-Flagship species

Ans. (c)

Sol. Consumption efficiency is the proportion of production that is ingested by next trophic level. It is lesser for herbivores of grassland ecosystem as they do not consume that entire plant body and leave behind some of the plant parts.

119. As per national air quality standard for India, which one of the following options gives correct concentration ljm (ugm-3, annual) of various gaseous air pollutants for a residential area?

(a) SO₂-100, NO₂-40, O₃-40, CO-50

(b) SO₂-50, NO₂-40, O₃-1OO, CO-O₂

(d) SO₂-50, NO₂-100, O₃-40, CO-O₂

Ans. (b)

Sol. Keystone species are those species whose impacts on its community or ecosystem are large and greater than would be expected from its relative abundance or total biomass. Umbrella species are generally large species that cover large areas in their daily or seasonal requirements. Flagship species are usually large and charismatic species that have wide appeal and thus draw attention to a conservation objective. Indicator species defines a trait or characteristic of the environment.

120. A plant is visited by bats during the night and sunbirds during the day. Given this information, which of the following characters best match this plant?

(a) The plant is a herb with saucer shaped white flowers.

(b) The plant is a shrub with tubular, red, diurnal flowers.

(d) The plant is a grass with white coloured fragrant, spikeiets.

Ans. (b)

Sol. For residential areas, the ranges are : for SO₂(50-80 µg/m³), for NO (40-80 µg/m³), for O₃ (100-180 µg/m³) and for CO (02-04 µg/m³) respectively.

121. The Western honey bee (Apis mellifera) collects nectar and pollen from flowers. The following are few hypotheses proposed to explain this behaviour in (A) mellifera:

P. In the past, those individuals that fed on nectar and pollen left more descendants than those who preferred only nectar or only pollen.

Q. The sensory stimulus from taste receptors in the honey bees lead to a positive reinforcement to look for more of the same food.

R. The honey bee's nervous system is predisposed to like the sweet taste.

S. The ancestor of honey bee was dependent on some sugar and protein rich diet and the honey bees have inherited the same taste perception.

Which of the following combination of ultimate hypotheses best explains the bee's feeding behaviour?

(a) P and Q

(b) Q and R

(d) Q and S

Ans. (c)

Sol. Sunbird feeds on flowers which are orange or red in colour having tubular structure. On the other hand, bats have long tongue which helps them to gather nectar from tubular flower.

122. A species of grass grows around a mine area having patches of heavy metal contaminated soil. Some of the populations of the species grew selectively on the soil that was contaminated with heavy metals. Over a period of time, though the tolerant and non-tolerant grass populations were continuously distributed and not separated by geographical barriers, they eventually evolved different flowering time and became different species. What kind of speciation would you call this?

(a) Allopatric speciation

(b) Sympatric speciation

(d) Bottleneck effect

Ans. (c)

Sol. Pollen is a bee's main source of protein and is necessary for queen vitality and brood production. Without sufficient pollen quality and quantity, a colony can start to decline in population, production and become susceptible to diseases not normally considered severe to colony healthy Bees see on the UV spectrum and use it to locate food.

123. The correct order of periods from Palaeozoic to Mesozoic era is

(a) Tnassic Jurassic Cretaceous Cambrian Ordovician Silurian Devonian Carboniferous Permian.

(b) Palaeocene Eocene Oligocene Miocene Pliocene Pleistocene Holocene.

(d) Pliocene Eocene Oligocene Silurian Devonian Carboniferous Triassic Jurassic Cretaceous.

Ans. (c)

Sol. Speciation occurs in overlapping areas called parapatric speciation.

124. Flufftalls In mainland Asia show high variation in tall colour. However, the far out Pacific Island, the flufftails show very little variation in tall colour. This variation tin tall coloour can be explained by all the following except.

(a) founder effect

(b) homologous evolution

(d) frequency dependent selection

Ans. (c)

Sol. Palezoic era is divided into the periods—Cambrian, Ordovician, Silurian, Devonian, Carboniferous and Permian from past to recent respectively. The Mesozoic era is divided into three periods—Triassic, Jurassic and Cretaceous from past to recent respectively.

125. Column A lists names of evolutionary biologists and Column B lists descri descriptions of evolutionary mechanisms proposed by them in random order.

Column A Column B

P. Jean-Baptist Lamarck 1. Variation at the molecular level is selectively neutral

Q. Charles Darwin 2. Inheritance of acquired characters

R. Motoo Kimura 3. Differential reproduction of genotypes

S. Seawall Wright 4. Changes In allele frequency due to random genetic drift

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

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

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

Ans. (b)

Sol. In genetic drift, the small population can show genetic variation through mutation and which can lead to the variation in tail colour. Founder effect also provides the explanation of gain or loss of variation in new population established from a large population. Frequency dependent selection can also lead to the loss of gain of tail colour variation. If the population are settled far a part from each other, there can never be homologous evolution between them which can lead to loss or gain of variation in colour.

126. Following diagrams represent various ways in which a character may evolve. Which of the following Is the correct definition for the character evolution patterns shown below?

(a) P-Autapomorphy, Q-Synapomorphy, R-Homoplasy

(b) P-Autapomorphy, Q-Homoplasy, R-Synapomorphy

(d) P-Synapomorphy, Q-Homoplasy, R-Autapomorphy

Ans. (b)

Sol. Jean-Baptiste Lamarck proposed the theory of inheritance of acquired characters. Whereas Charles Darwin proposed the idea of Natural selection. In modern evolutionary genetics, natural selection is defined as the differential reproduction of genotypes.

According to neutral theory of molecular evolution proposed by Motoo Kimura, most of the variation within and between species is not caused by natural selection but by genetic drift of mutant alleles that are neutral. Sewall Wright gave the principles of gene frequency fluctuations in small propulations : smaller the population, larger will be the fluctuation in gene frequency.

127. To understand the singing behaviour in song birds, the following three characters were measured as shown in the graph:

A. Territoriality rate

B. Female fertility rate

C. Song rate

Which one of the following conclusions is most appropriate?

(a) Male birds sing as a display of strength to rivals and to attract females.

(b) Male birds sing to display parental care behaviour.

(d) Male birds sing only to deter other male rivals from competing for territories.

Ans. (d)

Sol. A. Synapomorpy refers to derived characteristic of a clade. A characteristic present in an ancestral species and shared exclusively by its evolutionary descendants.

B. Homoplasy is a character shared by a set of species but not present in their common ancestor.

C. Autapomorphy is a distinctive feature, known as a derived trait that is unique to a given taxon. This trait is found only in one taxon, but not found in any others or outgroups taxa, not even those most closely related to the focal taxon.

128. Which one of the following statements regarding "Endosymblot, hypothesis of origin of eukaryotes" Incorrect?

(a) Mitochondria arose from an -proteobacter.um and p.astlds arose from cyanobactera.

(b) The event of engu.fment of a photosynthetlc cyanobacterium by a host cell was primitive to engulfment of an -proteobacterlum during the eukaryotlc origin.

(c) Protlsts cMorarachnlophytes, HMy evolved when heterotrophic outvote engulfed a green alga, exemplifying secondary endosymblosls,

(d) One of the membranes of the engu,fed double-membraned cyanobacter.a was lost In some of the hosts that eventually led to red and green algae descendants.

Ans. (a)

Sol. During the mating season in song birds, male bird reach the forest and sing the song to indicate self strength and attract the female birds.

129. In circadian rhythm studies, following may be possible generalizations for the effect of light entertainment to the day/light night cycle.

P. Shorter exposures have a greater effect than longer exposures.

Q. Bright light exposures have a greater effect than dim light.

R. Intermittent light exposures have a greater effect than consistent exposures.

S. Dim light can affect entrapment relative to darkness.

Which combination of the above statements is correct?

(a) Q and R only

(b) Q and S only

(d) p, q and S

Ans. (b)

Sol. There are no evidences of whether engulfment of photosynthetic cynobacterium was primitve than the engulfment of the proteobacterium by eukaryiotic cell origin.

130. A T₀ transgenic plant containing a transgene for herbicide resistance shows two bands on Southern blot analysis using a probe that is internal to the restriction sites used for genomic DNA digestion. However, it segregates in a 3:1 ratio for herbicide resistance: sensitivity in the T, progeny obtained by self pollination. Which one of the following statements is correct?

(a) The T₀ plant is a single-copy event.

(b) The T₀ plant is a double-copy event and the two transgene copies are tightly linked.

(d) The T₀ plant contains two unlinked copies of the transgene, both of which are truncated versions of the herbicide resistance gene.

Ans. (b)

Sol. Circadian rhythms are physical, mental and behavioural changes that follow a roughly 24-hour cycle, responding primarily to light and darkness in an organism's environment.

131. Which one of the following statements regarding crop improvement programs using molecular breeding approaches is incorrect?

(a) Allelic diversity for traits of interest should be available in the naturally occurring crossable germplasm.

(b) The gene/s of interest cannot be derived from a sexually incompatible organism.

(d) The crop plant should necessarily have an optimized robust system for production of doubled haploids.

Ans. (b)

Sol. It has two linked transgene copy in the T₀ plants.

132. Membrane potential in mitochondria is critical for oxidative phosphorylation and is monitored by

(a) patch clamping.

(b) measuring internal sodium ions after lysing the mitochondria.

(d) measuring the consumption of ATP.

Ans. (d)

Sol. The crop improvement using molecular breeding approaches can be achieved through if allelic diversity is available between two naturally occurring in corssable germplasm. The genes of interest cannot be derived from the sexually incompatible organism. The availablility of markers and linkage maps helps in the crop improvement. For the crop development is not neccessary that the crop plant must have robust system for the production double haploids, but is not necessary for sure.

133. The pH of a solution is 7.4 ± 0.02 where 0.02 is standard deviation obtained from eight measurements. If more measurements were carried out, the % of samples whose pH would fall between pH 7.38 and 7.42 is

(a) 996

(b) 95.4

(d) 99.8

Ans. (c)

Sol. Patch clamping is a technique used for measuring membrane potential.

134. In order to separate red and white blood cells, which of the following methods can be used?

(a) Ion-exchange chromatography and FACS.

(b) Hydrophobic chromatography and density gradient centrifugation.

(d) Hydrophobic chromatography and FACS.

Ans. (c)

Sol. If the variation in pH differ in between 7.38 to 7.42 than almost 50% molecules will have a closer value to 7.38-7.4 and 50% will have closer value to 7.4 to 7.42. So, 99.8% values will lie in between 7.38 to 7.42.

135. In order to check whether a protein has been phosphorylated during treatment with a drug, you would perform

(a) Southern hybridization

(b) Western blot analysis

(d) RELP

Ans. (c)

Sol. For separatign RBC's from WBC's most common technique used in FACS in which cell specific antibodies are used. RBC's are less dense and therefore can be separated using density gradient centrifugation also. Ligand affinity chromatography separation is based on unique interaction between the target analyte and a ligand, which is coupled covalently to a resin.

136. Given below are four statements regarding genetic transformation of plants in the laboratory.

P. Plants incapable of sexual reproduction cannot be transformed by agrobacterium tumefaciens.

Q. Integration of transgene in organellar (chloroplast) genome occurs primarily by homologous recombination.

R. An enhancer trap construct used in Agrobacterium-mediated transformation would contain a functional coding sequence of a reporter gene and a minimal promoter.

S. A T₀ transgenic plant containing two unlinked copies of a selection marker gene (hpr) and one copy of the passenger gene (gfp) would segregate in a 1:1 ratio for hygromycin resistance: sensitivity in the backcrossed progeny grown on selection media.

Which one of the combinations of above statements are correct?

(a) P and S

(b) Q and R

(d) Q and S

Ans. (b)

Sol. Phosphorylated protein can be identified by using Western blot analysis.

137. Agrobacterium mediated transformation was used to generate transgenic plants using a construct with a selection marker gene "X" and a passenger gene "Y". Expression levels of "Y" protein In eight independent transgenic plants are given below.

The following could represent probable reasons for the observed variability in transgene expression levels.

P. Position effects on passenger gene.

Q. Transgene silencing of the marker gene.

R. Variation in copy number of passenger gene.

S. mRNA instability of marker gene.

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

(a) P and R

(b) R and S

(d) P and Q

Ans. (c)

Sol. Maltose binding protein – Amylose

Streptavldin- Nickel

Glutathione S-transferase - Biotin

Flag-flag- Glutathione

6-Histidine tag - specific monoclonal antibody.

138. In a breeding experiment, two homozygous parental lines (P_I and P₂) were crossed to produce F₁hybrids. Due to an experimental error, seeds of these hybrids got mixed up with the seeds of two other germplasm lines (P₃ and P₄) and hybrid and hybrid seeds derived from them. A marker-based fingerprinting exercise was performed randomly selected seeds (F1-F6) from the mixed material and the four parental lines. Results of this analysis are shown below:

Based on the above data which one of the following options represents the correct set of parents and their F₁ progeny?

(a) PI X P2 = F3

(b) P3 X P4 = F2

(d) P3 X P4 = F6

Ans. (b)

Sol. Chloroplast transformation requires double homologous recombination. Therefore, two chloroplast DNA segments are used as franking sequences in chloroplast vectors to insert transgene cassette into the intergenic spacer region, without disrupting any functional genes. Hence, statement B is correct. Regulatory elements that impart a tissue-specific, stage-specific and/or environmental-stimuli-specific expression to the transgene are being identified and cloned using T-DNA based vectors that have been designed to identify and clone such regulatory sequences. The general principle behind this approach is to integrate a reporter gene that either lacks a promoter (gene/promoter trap) or carries only a minimal promoter (enchancer trap), at random sites in the wild type genome. A reporter gene cassette containing a minimal promoter (enhancer trap) close to the end of the insertion element can be cis activated when inserted close to a transcriptional enhancer that will drive the expression of the reporter gene.

139. The nuclear magnetic Resonance (ID and 2D) spectrum of a 30-residue peptide were recorded at 25°(C) The following observations were made.

P. The NH and C'H resonances were well resolved.

Q. The NOESY spectra showed extensive N-N₁₊₁ connectivities.

R. The NH resonances showed slow exchange with deuterium.

The spectra indicates that the peptide adopts

(a) helical conformations

(b) anti-parallel p-strand conformations

(d) p-turn conformation with four amino acids participating in the turn. Rest of the amino acids are unstructured.

Ans. (a)

Sol. Position effects and copy number of the transgene/passenger gene will be responsible for varied expression of passenger gene in different plant species. Expression of Y gene is not linked to X gene in the question.

140. Given below are a set of statistical methods/parameters (Column A) and their potential applications/utility in biological research (Column B), in a random manner.

Column A Column B

P. Variance 1. Measures strength of association between two variables.

Q. Correlation coefficient 2. Prediction of value of a dependent variable based on known value of an associated variable.

R. Regression analysis 3. Calculation of deviation between observed and expected values.

S. Chi-square analysis 4. Calculate the spread of a distribution.

Which of the following options is a correct match of entries In column A and B?

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

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

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

Ans. (c)

Sol. P1 × P2 = F1

P1 and P2 are the parents of F1 hybrid. We need to see the DNA bands of both parent that should be present in the progeny.

141. In an experiment designed to clone a PCR-amplified fragment in a cloning vector digested with Xhol (CTCGAG) and Smal (CCCGGG), which one of the following combinations of restriction enzymes can be used in the PCR primer to generate compatible ends for cloning?

('' ndicates the site of cleavage within the recognition sequence)

(a) Xbal (TCTAGA) and Spel (AlCTAGT)

(b) EcoRI (GAATTC) and Smal (CCCGGG)

(d) Hlndlll (AGCTT) and PvuII (CAGCTG)

Ans. (a)

Sol. This kind of NMR spectra will be observed only in case of anti parallel beta conformation, there hydrogen bonds are connected between N_i of one strand with N_{i + 1} residue of other strand.

142. A researcher was working with three proteins. A, B and C which may have potentail roles in gene expression. In order to validate the hypothesis, EMSA (electrophorebc mobility shift assay) was performed. The purified protiens were allowed to bind with a labelled DNA and the result obtained after shown below.

The following interpretations were made:

P. Protein A possesses the DNA binding motif.

Q. Protein B possesses the DNA binding motif.

R. Protein B binds to DNA-protein A complex.

S. Protein C binds to DNA only when protein A is bound.

Choose the correct combination of interpretations.

(a) P and S

(b) P and R

(d) R and S

Ans. (c)

Sol. Regression results in prediction of an independent variable based known value of an independent variable, Chi-square evaluates the deviation between observed and expected values. Variance is a measure of distribution, Correlation is a measure of association between two variables.

143. Point group symmetry operations such as inversion and mirror plane are not applicable to protein crystals. This is because

(a) protein molecules assemble in highly ordered fashion.

(b) protein molecules have handedness.

(d) hydrogen atoms in proteins diffract weakly.

Ans. (c)

Sol. SalI generates compatible ends which can be ligated to XhoI treated plasmid. As other enzyme generates blunt end it can be easily ligated.

144. Several fusion construct were developed to purify heterologous protein in E. coli. The table below lists fusion partners and llgands.

Partner Ligand

P. Maltose binding protein 1. Specific monoclonal antibody

Q. Streptavldin 2. Nickel

R. Glutathione-S-transferase 3. Glutathione

S. Flag-tag 4. Amylose

T. 6-Histidine tag 5. Biotin

Which one of the following Is the correct match of the fusion partner with the ligand?

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

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

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

Ans. (b)

Sol. In the gel picture only protein A can directly bind to the DNA. Protein B can interact to the DNA which is bound with A, can be seen in the upper-shift of the band.

145. In the following diagram, segments A and C are copies of 10 base pair repeat DNA sequences, flanking a unique stretch shown as (B) A and C are in inverted orientation as indicated by arrows. Intramolecular recombination between A and C leads to which event:

(a) The complete region encompassing A to C will be inverted.

(b) Only A and B will be inverted.

(d) Only regions A and C will be inverted.

Ans. (b)

Sol. Protein molecules shows handedness that's why form very soft crystals. That's why point group symmetry operations are not applicable to protein molecules.

CSIR NET BIOLOGY (JUNE - 2017)
Previous Year Question Paper with Solution.

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CSIR NET BIOLOGY (JUNE - 2017) Previous Year Question Paper with Solution.

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CSIR NET BIOLOGY (JUNE - 2017)
Previous Year Question Paper with Solution.