PYQ JUNE 2014 - VedPrep

CSIR NET BIOLOGY (JUNE - 2014)
Previous Year Question Paper with Solution.

21. The peptde (–C'O–NH–) is planer due to

(a) restriction around –C' bond

(b) restriction around C'–N bond

(d) hydrogen bonding between carbonyl oxygen and imino hydrogen of the peptide backbone

Ans. (b)

Sol. Peptide bonds have a planar, trans, configuration and undergo very little rotation or twisting around the amide bond that links the a-amino nitrogen of one amino acid to the carbonyl carbon of the next. This effect is due to amido–imido tautomerization.

22. Which one of the following is unfavorable for protein folding?

(a) Hydrophobic ineraction

(b) van dar Waals interaction

(d) Hydrogen bonding

Ans. (c)

Sol. An important goal of protein design is to enchance protein stability. The loss of conformational entropy upon folding is one of the main unfavourable features that needs to be overcome to reach the native state.

23. The maximum number of hydrogen bonds that can form between H₂N–NH₂ (hydrazine) and water is

(a) 2

(b) 1

(d) 4

Ans. (d)

Sol. Maximum number of hydrogen bonds that can be formed between NH₂-NH₂ is 4.

Hydrogen bond molecules in water and ice.

24. Which one of the following is the most appropriate statement regarding folded proteins?

(a) Charged amino acid side chains are always buried

(b) Charged amino acid side chains are seldom buried

(d) Tyrosine residues are always buried

Ans. (b)

Sol. In proteins with globular folds, hydrophobic amino acids tend to be interspersed along the primary sequence, rather than randomly distributed or clustered together. Charged amino acid side chains are also seldom buried.

25. The reaction of glutamate and ammonia to glutamine and water has a value of +14 kJ mol^–1 for . This is coupled with the ATP reaction (ATP + H₂O ADP + phosphate). The (kJ mol^–1) for the coupled reaction Gluamate + NH₃ + ATP Glutamine + ADP + Phosphate, under equilibrium condition is

(a) 16

(b) –44

(d) 44

Ans. (c)

Sol. Glutamate + NH₃ → Glutamine + water; = +14 kJ/mol

ATP + H₂O ADP + Phosphate; = –30 kJ/mol

Sum of above reaction will give us

Glutamate + NH₃ + ATP Glutamine + ADP + Phosphate

= +14 kJ/mol + +(–30) kJ/mol = –16kJ/mol

26. 10 nM acetate buffer (pH 4.00) is diluted one million times with distilled water (pH 7.00). pH of this diluted buffer is

(a) 4.00

(b) 7.04

(d) 6.96

[Help : log₁₀ 10^x = x; log₁₀ 1.10 = 0.04; log₁₀ 1.01 = 0.004]

Ans. (d)

Sol. 10 mM acetate buffer pH = 4

When diluted the pH = (10/1000)*4 = 0.04

pH of this diluted buffer = 7 – 0.04 = 6.96

27. The internal energy of a gas increases by 1J when it is compressed by a force of 1 Newton through 2 meteres. The heat change of the system is

(a) 1J

(b) –1 J

(d) –2 J

Ans. (b)

Sol. The heat change of the system = 1 – 2 × 1 = 1 – 2 = –1 J

28. for the base pairing of oligonucleotides (n = 5) at 300 K is –18 kJ mol^–1. What would be the approximate value of the equilibrium constant K?

(a) 100

(b) 10

(d) 1

Ans. (c)

Sol. = –2.303 RT log K

–18 kJ/mol = –2.303 × 8.314 × 10^–3 × 300 log K

log K = = 3.15

K_c 1000

29. Consider the structure less oligopeptide, R-G-P-S-T-K-M-P-E-Y-G-S-T-D-Q-S-N-W-H-F-R. The number of bonds that will be cleaved by trypsin and chymotrypsin treatments separately, are:

(a) 1, 2

(b) 2, 2

(d) 1, 3

Ans. (a)

Sol. Trypsin cleaves the peptide bond between the carboxyl group of arginine or the carboxyl group of lysine and the amino group of the adjacent amino acid. Chymotrypsin is a digestive enzyme component of pancreatic juice acting in the duodenum, where it performs proteolysis, the breakdown of proteins and polypeptides.

30. Which one of the following is requried to anchor the synaptic vesicles to the cytoskeletal proteins in the presynaptic nerve terminals?

(a) Syntaxin

(b) Synaptobrevin

(d) Synapsin

Ans. (d)

Sol. Synapsin I a synaptic vesicle-specific phosphoprotein, links synaptic vesicles to the cytoskeleton. The hydrophobic carboxy terminus of synapsin I is attached to the synaptic vesicles by binding to Ca²⁺/calmodulin-dependent protein kinase II (CaM kinase II). Other domains of synapsin I bind to cytoskeleton actin, spectrin and tubulin and thus anchor the vesicles to the cytoskeleton.

31. The dye used in Gram staining is

(a) Rhodamine

(b) Methylene blue

(d) Crystal voilet

Ans. (d)

Sol. Gram staining is a common technique used to differentiate two large groups of bacteria based on their different cell wall constituents. The Gram stain procedure distinguishes between Gram positive and Gram negative groups by colouring these cells red or violet. Gram positive bacteria stain violet due to the presence of a thick layer of peptidoglycan in their cell walls, which retains the crystal violet these cells are stained with. Alternatively Gram negative bacteria stain red, which is attributed to a thinner peptidoglycan wall, which does not retain the crystal violet during the decolouring process.

32. Molecules primarily responsible for the formation of lipid raft are

(a) phosphatidyl serine and phosphatidyl choline

(b) phosphatidyl inositol and cholesterol

(d) sphingolipids and cholesterol.

Ans. (d)

Sol. A lipid raft. Lipid rafts are small, specialized areas in membranes where some lipids (primarily sphingolipids and cholesterol) and proteins (green) are concentrated.

33. Membrane-bound and free ribosomes, are structurally identical, but differ only at a given gime in terms of association with

(a) acetylated proteins

(b) glycosylated proteins

(d) nascent proteins

Ans. (d)

Sol. Free ribosomes are in the cytoplasm, while attached ribosomes are anchored to the endoplasmic reticulum. Free ribosomes produce proteins in the cytosol, while attached ribosomes produce proteins that are inserted into the ER lumen. They both are attached to different nascent proteins.

34. You are studying a protein that inserts itself into a model membrane (liposomes) during a reconstitution process. The protein has an N-terminal, 18-amino acid hydrophilic segment that is located on the outside of the membrane, a 19-amino acid hydropobic segment flanked by negatively and positively charged amino acids, and a C-terminal domain that resides inside the lumen (as depicted below in the form of a carbon).

For proper reconstitution of the protein, which of the following strategies will be appropriate?

(a) Increase the number of negatively charged amino acids in the N-terminal

(b) Increase the number of positively charged amino acids in the C-terminal

(d) Increase the length of hydrophobic segment

Ans. (c)

Sol. Removal of positively charged amino acid from C-terminal will lead to proper reconstitution of protein.

35. Cyclins are the regulatory subunits and cyclin-dependent kinases (CDKs) are the catalytic subunits. Following diagram represents the involvement of cyclins and CDKs in various stages of cell cycle:

If we knockdown cyclin D in a cell by siRNA, which one of the following graphs correctly represents the level of CDK2 activity?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. If no cyclin is present, Cdk is inactive, and targets specific to the G1/S transition are not phosphorylated. Nothing happens, and S phase factors remain "off".

36. To test whether bacteria with enhanced toluene degradative abilities could be created for cold environment, a TOL (Toluene-degrading) plaasmid from a mesophilic bacterial strain was transferred b conjugation into a facultative psychrophile. he psychrophile was able to degrade salicylae (SAL) but not toluene. The recoombinant strain carried the introduced TOL plasmind and its own SAL plasmid. The results are as follows:

Identify A, B, C X and Y:

(a) A-mesophile, B-psychrophile, C-transformant, X-Toluene, Y-Salicylate

(b) A-memsophile, B-psychrophile, C-transformant, X-Salicylate, Y-Toluene

(d) A-transformant, B-psychrophile, C-mesophile, X-Toluene, Y-Salicylate

Ans. (b)

Sol. A mesophile is an organism that grows best in moderate temperature, neither too hot nor too cold, with an optimum growth range from 20 to 45°C. The optimum growth temperature for these organisms is 37°C. Psychrophiles or cryophiles are extremophilic organisms that are capable of growth and reproduction in low temperatures, ranging from –20°C to 20°C. They have an optimal growth temperature at 15 °C. The bacterial cells that take up the foreign DNA are called transformants, while the bacterial cells that do not take foreign DNA are called non-transformants. Option (b) is correct.

37. Chromosome organization becomes clearer from a series of biochemical, electron microscopic and X-ray crystallographic studies. When interphase chromatin is isoloated ni low salf buffer and observed under EM, 11 nm bead on string organization is seen. Interphase chromatin directly observed under EM shows 30 nm fibre. When histones are depleted from metaphase chromosome and visualized under EM, it shows a huge number of very large loops associated with scaffold.

Following interpretations can be made from these:

P. 11 nm fibre is formed when nucleosomes are brought closer by scaffold.

Q. 30 nm interphase chromatin is formed by zig-zag organization of nuclesomes of 11 nm fibre.

R. 30 nm fibre makes a solenoid packing to form the metaphase chromosome.

S. 30 nm fibre gets organized into loops due to SARs getting associated with scaffold proteins and coming closer.

The correct combination of interpretations is

(a) P and S

(b) P and R

(d) Q and S

Ans. (d)

Sol. Statement Q and S is true.

38. The tetrapeptide "KDEL" is well known as a retrieval signal of several newly synthesized proteins. This process is mediated through specific receptor-KDEL interaction. Any single amino acid change in this tetrapeptide is not allowed in terms if its binding with its receptros and its subsequent retention in specific organelle, whereas secretory proteins are devoid of such tetrapeptide. From this observation indicate the localization of the receptor of this tetrapeptide

(a) Plasma membrane

(b) Golgi

(d) Mitochondria

Ans. (c)

Sol. KDEL is a target peptide sequence in mammals and plants located on the C-terminal end of the amino acid structure of a protein. The KDEL sequence prevents a protein from being secreted from the endoplasmic reticulum (ER) and facilitates its return if it is accidentally exported.

39. The amount of each enzyme present in the chloroplast stoma is regulated by mechansms that control the concerted expression of nuclear and chloroplast genomes. Following are certain statements regarding the regulation of chloroplast enzymes:

P. Nucleus-encoded enzymes are translated on 70S ribosomes in the cytosol and subsequently transported into the plastid.

Q. Planstid encodedenzymes are translocated in the stroma on prokaryote-like 70S ribosomes.

R. Light modulates the expression of stromal enzymes encoded by the nuclear genome via specific photoreceptors.

S. The eight small subunits of rubisco is encoded in plastid.

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

(a) P and Q

(b) P and R

(d) R and S

Ans. (c)

Sol. Light acts as a regulatory signal by activating certain carbon assimilation enzymes, enabling the cells to switch between dark and light metabolism. Statement q and r is correct.

40. Patiebnts suffering from foamilial hyperchloesterolemia (FH) are mostly homozygous for the defective gene and have profoundly elevated levels of serum chloesterol. The reason may be that the gene for highly specific receptor for LDL is either defective or missing in these patients. In an experiment, cells were taken from both normal individual and homozygote (FH) subjects, incubated in buffer with ¹²⁵I-labeled LDL in presence or absence of excess unlabeled LDL for various time periods and then ¹²⁵I-labeled LDL bound to cells was measured. Which of the following is the best-fi graph for the above experiment?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Graph A represent correct experiment. Normal experiment shows high level of I-LDL. In FH case, both level of I-LDL and I-LDL is decreased.

41. After activation of a promoter by the DNA binding activity of a transcription factor, a co-activator is recruited at the region targeted for transcription which in turn creates a binding site for a chromatin remodeling complex. Which one of the following activities of the co-activator is responsible for the recruitment of chromatin remodeling comlex?

(a) Histone deacetylase activity

(b) Histone methyl tansferase activity

(d) DNA methyl tansferase actvity

Ans. (c)

Sol. Histone acertyl transferase activity of the co-activator is responsible for the recruitment of chromatin remodeling complex.

42. Virus induced gene silencing (VIGS) is a process that takes advantages of the RNAi-mediated antiviral defence-mechanism. Which one of the follwing ultimately guides siRNA to degrade the target tanscript (mRNA)?

(a) dsRNA

(b) ssRNA

(d) dsRAN binding protein

Ans. (c)

Sol. RNA Induced Silencing (RIS) complex ultimately guides siRNA to degrade the target tanscript (mRNA).

43. Each origin of replcation s acivated only once. This is achieved because

(a) pre-replicative complex can only form in G1 and replication can only be initiated when pre-replication complex is disassembled at the beginning of S-phase.

(b) replication can only by initiated when pre-replication complex is intact.

(d) pre-replicative complex can only form n S-phase.

Ans. (a)

Sol. Each origin of replcation s acivated only once. This is achieved because pre-replicative complex can only form in G1 and replication can only be initiated when pre-replication complex is disassembled at the beginning of S-phase.

44. During protein synthesis in prokaryotes, the peptidyl transferase activity required for peptide bond formation is due to

(a) ribosomal protein L26

(b) 16S ribosomal RNA

(d) aminocyl tRNA

Ans. (c)

Sol. The new aminoacyl-tRNA binds to the A site and is aligned with the peptidyl-tRNA at the P site and a peptide bond is formed by the action of peptide bond is formed by the action of peptidyltranferase. Peptidyltransferase activity is catalyzed by 23S ribosomal RNA (a ribozyme). The formation of the peptide bond attaches the growing polypeptide to the newly bound aminoacyl-tRNA, which is now a peptidyl-tRNA.

45. Prior to transcription, chromatin changes from an inactive state to an active state by various factors in a stepwise manner. Which one of the follwing is involved in the initial step during activation of a chromatin loop?

(a) HMG 14

(b) Single standed DNA-binding protein

(d) Toposomerase I

Ans. (d)

Sol. Prior to transcription, chromatin changes from an inactive state to an active state by various factors in a stepwise manner. Toposomerase I is involved in the initial step during activation of a chromatin loop.

46. The homologous genetic recombination s DNA repair process referred to as recombination repair. Which one of the following statements is incorrect for recombination repair?

(a) DNA polymerase III stalls at the site of the damage

(b) DNA polymerase III leaves a gap n the daughter strand

(c) The gap is filled by recombination between complementary parent strand homologous to daughter strand and the gapped daughter strand.

(d) Homologous recombination process is catalyzed by topoisomerase II.

Ans. (d)

Sol. Topoisomerae II is involved in alteration of topology of DNA while replication but it is not involved in homologous recombination.

47. Histone acetylase and chromatin remodeling complezes are recruited to specific regions of chromatin by

(a) gene activator proteins

(b) specific promoter sequence

(d) dephosphorylation of chromatin remodeling complexes

Ans. (a)

Sol. Histone acetylase and chromatin remodeling complezes are recruited to specific regions of chromatin by gene activator proteins.

48. Aminoacyl-tRNA synthetases are very specific for aminoacylation of tRNAs with the correct cognate amino acids. However, there is possibility of a mismatch between the tRNA and its congnate amino acid. This error is corrected by the inherent proof-reading activity of the aminoacyl-tRNA synthetaase. In case of two very similar amino acids, namely valine and isoleucine, isoleucyl-tRNA synthetase employs the following possibl approaches for an error-free aminoacylation.

P. It removes an incorrect amino acid by hydrolyzing the aminoacyl-AMP linkage following the first reaction step.

Q. It is activated for proof-reading activity, leading to breakage of the bond between the wrong amino acid and tRNA.

R. It has an intrinsic ability to recognize the structural difference between amino acids leading to abortive elimination of the non-cognate amino-acid.

S. it gets sequestered ini thesecond step with the wrong amino acid, and that freezes the amino-acylation process.

Which of the following combinations is correct?

(a) P and Q

(b) P and S

(d) R and S

Ans. (a)

Sol. For error free aminoacylation, isoleucyl-tRNA synthetase removes an incorrect amino acid by hydrolyzing the aminoacyl-AMP linkage following the first reaction step and it is activated for proof-reading activity, leading to breakage of the bond between the wrong amino acid and tRNA

49. In order to identify the regulatory regions of a novel promoter sequence shown below. for 150 bp deletion constructus were made in a luciferase reporter system as indiacted above in boxes A to D. After transfection, the observed level of promoter activity (%) as analyzed by luciferase assay of all the constructs is indicated in the right of the figure. Identify the best correct combination of regions in the options given below that indicate the presence of a positive and a negative regulatory element respectively.

(a) B and D

(b) A and C

(d) A and B

Ans. (b)

Sol. By observing the deletion and activity in graph option (a) has postive regulatory elements while option (c) has negative regulatory elements.

50. The following statements are made on DNA replication

P. Replication fork is a branch point in a replication 'eye' or 'bubble'

Q. A replication bubble contains two replication forks

R. DNA replication is continuous according to the interpretation made by Okazaki

S. Multiple priming events are required for both leadng and lagging strands to initiate DNA snythesis.

Which one of the following is the correct combination?

(a) P and Q

(b) Q and R

(d) P and R

Ans. (a)

Sol. During replication, Replication fork is a branch point in a replication 'eye' or 'bubble' and A replication bubble contains two replication forks.

51. MicroRNAs (miRNAs) have recently been shown to play a significant role in the fine tuning of gene expression. Some miRNAs induce gene silecing by binding to mRNAs and inducing inhivition of translation.

P. miRNA is partially complementary to a region of target mRNA in the 3' UTR.

Q. miRNA always base pairs with mRNA around an AU-rich sequence

R. miRNA base pairs with mRNA through 6-7 nucleotides at ts 5' end referred to as "seed sequence" as well as few additional bases elsewhere.

S. miRNA is always partially complementary to the 5' UTR of the target mRNA.

Choose the correct option from the following :

(a) P and Q

(b) P and R

(d) P and S

Ans. (b)

Sol. miRNA is partially complementary to a region of target mRNA in the 3' UTR and miRNA base pairs with mRNA through 6-7 nucleotides at ts 5' end referred to as "seed sequence" as well as few additional bases elsewhere.

52. Attenuation is mechanism involved in the regulation of tryptophan operon iin E. coli. When tryptophan levels are high in the cell, region 2 of the trpL is blocked from pairing with region 3. This allows the pairing of region 3 and 4 leading to the formation of a rho-nidependent terminator. What would be the structure of the trpL region in E.coli cells where protein synthesis has been inhibited?

(a) Region 2 pairs with region 3 allowing transcription of the structrual genes

(b) Region 1 and 2 will pair, allowing 3 and 4 to pair leading to attenuation

(d) Region 2 and 3 will pair leadng to attenuation.

Ans. (b)

Sol. Region 1 and 2 will pair, allowing 3 and 4 to pair leading to attenuation. Hence, double secondary structure leads to inhibiton of protein synthesis in trp operon.

53. Mutation is a gene x in Arabidopsis thaliana results is more number of lateral root formation. Which one of the following is the correct statements?

(a) The gene product acts as a positive regulator of lateral root formation

(b) The gene product acts as a negative regulator of lateral root formation

(d) The gene product promotes replicaion for lateral root development.

Ans. (b)

Sol. LeMYC2 works as a negative regulator of blue light mediated photomorphogenesis. Further-more, LeMYC2 specifically binds to the G-box of LeRBCS-3A promoter. Over-expressing mutation in LeMYC2 has led to increased root length with more number of lateral roots.

54. Satellite RNAs (sat RNAs) are species of RNA ssociated with specific strains of some plant RNA viruses, although it is not necessary for their replication. Few statements are given below on sat-RNA.

P. Presence of sat-RNA leads to reducton in severity of disease symptoms

Q. Presence of sat-RNA leads to increase in serverity of disease symptoms

R. sat-RNA is consstitutively expressed like coat proteins and is independent of virus infection

S. sat-RNA is not constitutively expressed like coat proteins but is expressed only after virus infection

Which one of the following combinations of above statements regarding sat-RNA is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (d)

Sol. Presence of sat-RNA leads to reducton in severity of disease symptoms and sat-RNA is not constitutively expressed like coat proteins but is expressed only after virus infection

55. The proximal distal growth and ifferentiation o the tetraped limb bud are made possible by a series of interactions between the apical ectodermal ridge (AER) and limb bud mesenchyme directly beneath it. Some of the interactions performed in chick demonstrated the following results:

P. When the AER was removed at any time tmie of development, further development of distal limb skeletal elements ceased.

Q. When leg mesenchyme was placed directly beneath the wing AER, distal hindlimb structres developed at the end of the wing.

R. When limb mesenchyme was replaced by a non-limb mesenchyme beneath the AER, the limb still developed.

S. When an extra AER was grafted onto an existing limb bud, the development of the limb ceassed.

Which of the above combinations is correct?

(a) P and Q

(b) P and R

(d) Q and R

Ans. (a)

Sol. When the AER was removed at any time tmie of development, further development of distal limb skeletal elements ceased and when leg mesenchyme was placed directly beneath the wing AER, distal hindlimb structres developed at the end of the wing.

56. Which one of the followng graphs represents he relative expression of proteins of Wnt-activated signaling cascades involved during development of an embryo?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Subsequently, -catenin was identified as a GSK3 substrate: GSK3-mediated phosphorylation triggers -catenin destabilization 9, 10. This finding thus established a central role for GSK3 in Wnt--catenin signaling.

57. Vasopression acts on blood vessels leading to their constriction. Which one of the following signaling cascades will apply to the effect of vasopressin?

(a) V₁R Phosphatidylinositol hydrolysis Ca²⁺ increase

(b) V₁R Adenylate cyclase cAMP

(d) V₂R Adenylate cyclase cAMP

Ans. (a)

Sol. Vasopressin regulates water excretion from the kidney by increasing the osmotic water permeability of the renal collecting duct – an effect that is explained by coupling of the V2R with the Gs signaling pathway, which activates cAMP.

58. A pathogen like Mycobacterium, which colonizes inside thecells of the host, is likely to be least affected by which one of the following host immune defense mechanisms?

(a) Cell-mediated immune response

(b) CD⁴⁺ T lymphocytes

(d) humoral immune response

Ans. (d)

Sol. Mycobacteria are endowed with mechanisms through which they can evade the onslaught of host defense response. These mechanisms are including deminishing the ability of antigen presenting cells to present antigens to CD⁴⁺ T cells; production of suppressive cytokines; escape from fused phagosomes and including T cell apoptosis.

59. Which one of the following is a type of intercellular junction in animal cells?

(a) Middle lamella

(b) lasmodesmata

(d) Glycocalyx

Ans. (c)

Sol. A cell junction (or intercellular bridge) is a type of structure that exists within the tissue of some multicellular organisms, such as animals. Cell junctions are especially important in enabling communication between neighbouring cells via specialized proteins called communicating junctions. Cell junctions are also important in reducing stress placed upon cells.

In vertebrates, there are three major types of cell junctions :

Adherens junctions, desmosomes and hemidesmosomes (anchoring junctions)

Gap junctions (communicating junction)

Tight junctions (occluding junctions)

60. Which one of the following is not related to immediate hypersensitivity reactions?

(a) Mast cell degranulation results in histamine-mediated allergc reactions.

(b) Regnc antibodies trigger allergic reactions.

(d) Anaphylactic reaction is triggered primarily by IgE.

Ans. (c)

Sol. The traditional classification for hypersensitivity reactions is that of Gell and Coombs and is currently the most commonly known classification system. It divides are hyper-sensitivity reactions into the following 4 types :

Type I reaction (i.e., immediate hypersensitivity reactions) involve immunoglobulin E (IgE) mediated release of histamine and other mediators from mast cells and basophils.

Type II reactions (i.e., cytotoxic hypersensitivity reactions) involve immunoglobulin G or immunoglobulin M antibodies bound to cell surface antigens, with subsequent complement fixation.

Type III reactions (i.e., immune complex reactions) involve circulating antigen-antibody immune complexes that deposit in postcapillary venules with subsequent complement fixation.

Type IV reactions (i.e., dealyed hypersensitivity reactions, cell mediated immunity) are mediated by T cell rather than by antibodies.

61. ¹²⁵I-labelled diaminofluorene (DAF) is a well known photoactivatable hydrophobic probe of plasma membrane integral protein. To dertemine the approximate length and number of hydrophobic domain in any intergal membrane protein, a controlled experiment (following standard protocol) is carried out. In order to ascertain the aforesaid aspects indicate the correct combination of experimental protocols from the following choices:

(a) Intact membrane was allowed to ineract with DAF and unicorporated DAF measured.

(b) Intact membrane was allowed to interact with DAF, lysed and total protein precipitated with TCA and amount of radioactivity incorporated in thetotal proteins in the TCA precipitated fraction measured.

(c) Intact membrane was allowed to interact with DAF, then membrane was solubilized with detergentm, digested with proteases (limited proteolysis) run on SDS-PAGE followed by autoradiography.

(d) Intact membrane was allowed to interact with DAF followed by complete proteolysis, SDS-PAGE and finally autooradiography.

Ans. (c)

Sol. Option C is correct.

62. In a plaque-forming cell assay, antigen specific B-cell numbers are assessed. In this assay, antigen coated sheep red blood cells (SRBCs) are lysed by the hapten-sepcific B-cells by complement-mediated cytotoxicity. In an assay that tried to enumerate the TNP-specific B-cells elicited in TNP-KLH-primed mice, no plaques were formed despite the presence of antigen-specific anibody producing B-cells which of the following is not the reason for the absence of plaques?

(a) The source of complement has anti-TNP antibody.

(b) The SRBC were stored for too long.

(d) he complement had anti-KLH antibody.

Ans. (d)

Sol. PFC assays measure several aspects of the total cellular immune response induced in a given experiment, i.e. the product of clonal recruitment, clonal expansion and activation of enhanced antibody secretory rates.

63. Following sets of Plasmodium falciparum sporozoites

P. normal sporozoites

Q. sporozoites with mutation in C-terminal of circumsporozoite (CS) antigen

R. sporozoites with mutation in the N-terminal region of circumsporozoite (CS) antigenare injuected into 2 groups of mice, one normal (Group A) and one (Group B) where localized knockdown of heparan sulfate (receptor for CS antigen in liver cells) is achieved by injecting specific shRNA expressing lentiviral particles in the liver prior to infection. 15 days post infection, parasitemia is masured by counting infected RBC through Giemsa staining.

Which of the following groups will show maximum level of parasitemia?

(a) Group B with set P

(b) Group A with set Q

(d) Group B with set Q

Ans. (c)

Sol. An immunologically cryptic epitope of Plasmodium falciparum circumsporozoite protein facilitates liver cell recognition and induces protective antibodies that block liver cell invasion. The N-terminal region encompasses KLKQP motif, which is vital in the entry inside the hepatocytes, while the C-terminal region composes of a polymorphic Th2R and Th3R sub-regions. So group A with set R will show maximum level of parasitemia.

64. The following statements have been proposed for a cancer cell.

P. Binding of p53 with MDM2, a ubiquitin E3 ligase, is a precondition for cancer progression

Q. Phosphroylation of a tyrosine residue in the C-terminus of human c-Src is essential for cell invasion and motility.

R. Loss-of-function of both alleles of a tumor suppressor gene prevents metastasis

S. Dimerization of c-myc-Max leads to enhanced celll proliferation.

Which of the combinations of the above statements is correct?

(a) P and Q

(b) R and S

(d) Q and R

Ans. (c)

Sol. MDM2 directly binds to the transactivation domain of p53 and inhibits its transcriptional activity, causes the ubiquitination and proteasomal degradation of p53, and exports p53 out of the nucleus which promotes p53 degradation and inhibits its activity. MYC stimulates cell-cycle progression and the cellular proliferation through the regulation of genes related to cell-cycle control. MYC induces positive cell-cycle regulators such as several cyclins, CDKs and E2F transcription factors. P and S statement is correct.

65. Collagen is the most prevalent extgracellular matrix protein. Which of the follownig is not true for collagen?

(a) Collagen is compose4d of triple helix consisting of two α polypeptide chain and one B polypeptide chain wound around one another n a rope-like structure.

(b) Glycine accounts for almost one third of the amino acids within collagen molecule.

(d) Individual collagen polypeptide chains are synthesized on membrane-bound ribosomes with N-terminal signal sequences for directing them to ER lumen.

Ans. (a)

Sol. Collagen is protein molecules made up of amino acids. It provides structural support to the extracellular space of connective tissues. Due to its rigidity and resistance to stretching, it is the perfect matrix for skin, tendons, bones, and ligaments. The defining feature of collagen is an elegant structural motif in which three parallel polypeptide strands in a left-handed, polyproline II-type (PPII) helical conformation coil about each other with a one-residue stagger to form a right-handed triple helix.

66. CD4⁺ T-cells are co-cultured with macrophages in the presence of immobilized anti-CD3 antibody under four different conditions:

P. Interleukin (IL)-4 plus anti-IFN antibody

Q. IL-12 and anti-IL-4 antibody

R. Transforming growth factor (TGG)-β

S. TGF- and IL-6

for three rounds to induce T-helper cell differentiation identifiable by the cytokines they express predominantly. Which one of the following is the most likely combination of predominant cytokine expression in these cultures?

(a) (P) IL-4, (Q) IFN-, (R) IL-10, (S) IL-17

(b) (P) IFN-, (Q) IL-4, (R) IL-17, (S) IL-10

(d) (P) IL-17, (Q) IL-10, (R) IL-4, (S) IFN-

Ans. (a)

Sol. IL-10 functions as a potent B cell stimulator that enhances activation, proliferation, and differentiation of B cells. Studies have shown that transforming growth factor-beta (TGF-beta) and interleukin 6 (IL-6) are required for the lineage commitment of pathogenic IL-17-producing T helper cells (T(H)-17 cells). Interleukin-12 (IL-12) and IL-4 are known to differentially promote T helper (Th) cell differentiation. While IL-12 induces interferon- (IFN-) production and maturation of Th1 cells, IL-4 is thought to antagonize IL-12 and to favour Th2 development.

67. Four groups of one-day old female BALB/c neonates had received the following treatments:

P. Epidermal cells from C57BL/6 male plus anti- microglobulin antibody

Q. Epidermal cells from C57BL/6 male plus antibodies to microglobulin, CD40 ligand, CD80 and CD86.

R. Epidermal cells from C57BL/6 female plus anti-CD80 antibody.

S. Epidermal cells from C57BL/6 female mice

When these BALB/c neonates grew six weeks old, they received skin transplant from C57BL/6 male mice. Transplantation rejection time varied between these four groups. Starting from th fastest to the slowest rejection, which one of the following is the most likely order?

(a) S > R > P > Q

(b) P > Q > R > S

(d) S > Q > R > P

Ans. (a)

Sol. Correct order of rejection will be S = R > P > Q (fastest to slowest).

68. Upon prologed illumination, activited rhodopsindoens not activate transducin, hence the vision is imparired. This could be because of thefollowing explanations:

P. Most of the activated rhodopsin gets phosphorylated and is unable to activate transducin.

Q. Most of the activated rhodopsin gets dephosphorylated and is unable to activate transducin.

R. Arrestin further interacts with phosphorylated rhodopsin.

S. Arrestin further interacts with dephosphorylated rhodospin.

(a) P and R

(b) Q and S

(d) Q and S

Ans. (a)

Sol. Statement P and R is correct.

69. Sleeping sickness is disease caused by protozoan parasite and the following statements pertain in that disease:

1. The vactor for this disease is tsetse fly.

2. The vector for ths disease is Trypanosoma brucei.

3. The parasite's body is covered by a dense coat of variable surface glycoprotein (VSG)

4. There are several thousands of VSG genes, only one of them expressing at a time, which helps parasite in evadling host's immune response.

5. Several thousand copies of VSG genes express concurrently, paralyzing the host immune system.

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

(a) 2, 3 and 4

(b) 1, 3 and 4

(d) 2, 3 and 5

Ans. (b)

Sol. Trypanosoma brucei, the causative agent of African sleeping sickness, confounds antibodies by rapid antigenic variation. This pathogen can spontaneously modify its expression of its variable surface glycoprotein (VSG), the molecule that is normally the main target of humoral responses to this parasite. Trypanosome persistence in the mammal is due to antigenic variation, which involves changes in the identity of the variant surface glycoprotein (VSG) that forms a dense cell surface coat to shield invariant surface antigens from immune recognition.

70. In order to develop a vaccine against a regulatory T-cell-promoting but Th1 suppressing viral infection, four gropus (A-D) of mice were pried with either killed virus (A) or a virus-derived immune dominant peptide (B), or the same peptide but with two substitutions (C) or were left unprimed (D). Upon infection challenge, the order of increasing severity of infection was observed to be B > A > D > C. To explain the contrasting effects of these two peptides (B) and (C), their MHC-binding affinities were assessed but no difference was found. Which of the following possibilities most likely to explain their constrasting effects?

(a) The wild-type peptide (B) elicits T-cell expansion but the mutant peptide (C) fails.

(b) The wild-type peptide-MHC complex binds T-cell deletion but he mutant peptide does not.

(c) The wild-type peptide-MHC complex binds T-cell receptor with significantly higher affinity than the mutant peptide-MHC complex

(d) The wild-type peptide induces delection of T-Reg cells but increase. IFN production by T-cells, whereas the mutant peptide fails to induce thee effects.

Ans. (d)

Sol. Regulatory T cells (Tregs) are a specialized subpopulation of T cells that act to suppress immune response, thereby maintaining homeostasis and self-tolerance. It has been shown that Tregs are able to inhibit T cell proliferation and cytokine production and play a critical role in preventing autoimmunity. The pleiotropic cytokine IFN- is produced primarily by activated T cells and NK cells. The cellular effects of IFN- are mediated by its heterodimeric cell surface receptor IFN-R. The IFN-R is comprised of a- and ß-chains, both of which belong to the class II family of cytokine receptors. So, wild-type peptide induces deletion of T-reg cells but increase IFN gamma production by T-cells, whereas the mutant peptide fails to induce these effects.

71. During which one of the following stages of Arabidopsis embryogenesis, cell elogation throughout the embryonic axis and further development of the cotyledons occur?

(a) Globular stage

(b) Torpedo stage

(d) Mature stage

Ans. (b)

Sol. Torpedo stage : Differentiation of the shoot apical meristem (SAM) less obvious than the RAM. As the cotyledon lobes begin to grow, a group of cells between them is seen to divide infrequently. Later, the cotyledons and subtending tissues begin expressing genes for storage proteins. The group of quiescent cells does not express these genes and the lower boundary of the non-expressing cells in at a prominent cell wall corresponding to the O' line.

72. Which one of the following development processes in animals is more dependent on cellular movements?

(a) Pattern formation

(b) Morphogenesis

(d) Growth

Ans. (b)

Sol. Morphogenesis (from the Greek morphe^{^} shape and genesis creation, literally "beginning of the shape") is the biological process that causes an organisms to develop its shape. It is one of three fundamental aspects of developmental biology along with the control of cell growth and cellular differentiation

73. Engrailed expression in Drosophila defines

(a) anterior margin of the segment

(b) anterior compartment of each segment

(d) posterior compartment of each segment.

Ans. (d)

Sol. In the model organism Drosophila melanogaster engrailed acts as a segment-polarity gene in early embryonic development. It is initially expressed in stages 8-11 of development in 14 isolated bands of cells along the embryo's anterior posterior axis. The cells expressing engrailed define the anterior most region of each parasegment. Once proper segments form, engrailed expressing cells are found in the posterior most region of each segment.

74. Torpedo is a trans-membrane receptor on follicle cells that binds with Gurken protein located in the presumptive dorsal surface of the oocytes and inhibits a cascade leading to nuclear localzation of the Dorsal protein. In an experiment, Drosophila germ line chimeras were made by interchanging pole cells (germ line precursors) between wild type embryos and embryos from mother homozygous for a mutation of torpedo gene. These transsplants produced:

(i) Wild-type females whose egg came from mutant mother, and

(ii) tropedo deficient females whose egg came from wild-type mother.

The possible outcome of this experiment can be:

1. Tropedo deficient eggs developed in wild-type ovary produced normal embryos.

2. Wild-type eggs developed in Torpedo deficient ovary produced ventralized embryos.

3. Torpedo deficient eggs developed in wild-type ovary produced ventralized embryos.

4. Dorsal protein enters in the nuclei of dorsal side of embryos which came from wild-type eggs developed in Tropedo deficient ovary.

Which of the above combinations is correct?

(a) 1, 2 and 4

(b) 2, 3 and 5

(d) 1, 3 and 5

Ans. (a)

Sol. The torpedo mutant eggs were able to produce normal embryos if they developed within a wild-type ovary. Thus, unlike the gurken gene product, the wild-type torpedo gene is needed in the follicle cells, not in the egg itself. Statement 1,2 and 4 is correct.

75. At the 2-celled stage of Caenorhabditis elegans development the blastomeres were experimentally separated and allowed to proceed in development. One of the blastomere (P1) developed generating all types of cells it would normally make.

The following conclusion could be drawn:

P. The determination of both P1 and AB was autonomous.

Q. The determinaion of boht P1 and AB was conditional.

R. The determination of P1 was autonomous and AB was conditional.

S. Both asymmetric division and cell-cell interactions specify cell fate in early development.

Which of the above combinations is correct?

(a) P and R

(b) Q and S

(d) R and S

Ans. (d)

Sol. In the first cell division, the cleavage furrow is located asymmetrically along the anterior-posterior axis of the egg, closer to what will be the posterior pole. It forms a founder cell (AB) and a stem cell (P1). In the early divisions of the C. elegans zygote, one daughter cell becomes a founder cell (producing differentiated descendants) and the other becomes a stem cell (producing other founder cells and the germ line).

76. During lens formation in the Xenopus, the following statements have been proposed:

P. Lens induction can be achieved in the absence of optic vesicle after priming of head ectoderm by the anterior neural plate.

Q. The optic vesicle can induce the presumptive trunk ectoderm of form the lens.

R. Only the head ectoderm can respond to direct signals from the optic vesicle to from the lens.

S. The anterior neural plate primes the head ectoderm via BMP4 and Fgf8 prior to signals from the optic vesicle.

Which of the above combinations is correct?

(a) R and S

(b) Q and S

(d) P and R

Ans. (a)

Sol. The lens placode invaginates, forming a pit that pinches off from the surface ectoderm to form the lens vesicle. The surface ectoderm reforms a continuous layer that will become the corneal epithelium. Chick explant assays provide evidence that at gastrula stages, BMP activity inhibits neural fate and regulates the initial specification of lens placodal cells. Consistently, Bmp4 knockout mouse embryos lack morphological lens placodes. FGF and BMP signals act in the neural plate border region in an opposing manner, to restrict neural and epidermal cell fate, respectively.

77. During the operations of C₂ oxidative photosynthetic cycle, which of the following metabolites is transported from chloroplast to peroxisome?

(a) Glycerate

(b) Serine

(d) Glycolate

Ans. (d)

Sol.

78. Arabidopsis thaliana seeds were planted on Murashige Skoog (MS) plates with or without a hormone added to the medium. Seeds wee found to germinate late in the hormone containing MS plates as compared to MS plates without hormone. Identify the hormone.

(a) Jasmonic acid

(b) Cytokinin

(d) Abscisis acid

Ans. (d)

Sol. Abscisic acid (ABA) reversibly arrests embryo development at the brink of radicle growth initiation, inhibiting the water uptake which accompanies embryo growth. Seeds which have been kept dormant by ABA for several days will, after removal of the hormone, rapidly take up water and continue the germination process.

79. Ethylene signaling pathway is important for fruit ripening. Which one of the following responses is routinely used to identify ethylene signaling pathway components?

(a) Cotyledon expansion response

(b) Lateral root formation response

(d) Flowering time response

Ans. (c)

Sol. The ability of screen mutant in Arabidopsis has made possible the ioslation of a wide range of ethylene-response mutants. Such screens have in particular used the 'triple response' as an indicator of an intact ethylene detection systems. The tirple responses is a set of morphological changes to etiolated (dark grown) seedlings that normally occur in response to ethylene, including short and thick hypocotyls, short root and exaggerated apical hooks.

80. Aquaporins are a class of proteins that are relatively abundant in plant membranes. Following are certain statements regarding the properties of aquaporins:

P. Aquaporins from water channels in membrane

Q. Some aquporins also transport uncharged molecules such as NH₃.

R. The activity of aquaporins is not regulated by phosphorylation.

S. The activity of aquaporin is regulated by calcium concentration and reactive oxygen species.

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

(a) P, Q and S

(b) Q, R and S

(d) P, Q and R

Ans. (a)

Sol. The primary function of most aquaporins is to transport water across cell membranes in response to osmotic gradients created by active solute transport. Statement P, Q and S is correct.

81. Following statements are related to oxidative phosphorylation.

P. redox reactions of electron tansport chain coupled with ATP synthesis are collectively called oxidative phosphorylation.

Q. Three major processes : glycosis, oxidative pentose phosphate pathway and citric acid cycle are related to oxidative phosphorylation

R. Electron transport proteins are bound to outer of the two mitochondrial membranes.

S. In the electron transport chain electrons are transferred to oxygen from NADH.

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

(a) P and S

(b) Q and R

(d) P and R

Ans. (a)

Sol. Oxidative phosphorylation is the process by which ATP synthesis is coupled to the movement of electrons through the mitochondrial electron transport chain and the associated consumption of oxygen. During oxidative phosphorylation, electrons derived from NADH and FADH₂ combine with O₂, and the energy released from these oxidation/ reduction reactions is used to drive the synthesis of ATP from ADP.

82. In terpene biosynthesis pathways, three acetyl-CoA are joined together stepwise to form mevalonic acid. Which one of the following three steps is required by mevalonic acid to form isopentenyl diphosphate or isopentenyl pysrophosphate (IPP)?

(a) Pyrophosphorylation, decarboxylation and dehydration

(b) Alkylation, pyrophosphorylation and decarboxylation

(d) Phosphorylation, carboxylation and methylation.

Ans. (a)

Sol. By cyclization and other changes, this compound will finally result in cholesterol. Three phases will be considered in the biosynthetic process: (1) acetate to mevalonate, (2) mevalonate to squalene, and (3) squalene to cholesterol. Option a is correct.

83. Nitrogen fixation is basically a process of converting nitrogen gas into ammonia (NH₃). One of the key enzymes in the process is "nitrogenase". The production and activity of nitrogenase is very highly regulated as highlighted below:

P. Nitrogen fixation through nitrogenase is an energetically expensive process.

Q. Nitrogenase encoding gene is under a constitutive promoter.

R. Nitrogenase is highly sensitive to oxygen.

S. Endogenous availability of the cofactor of nitrogenase enzyme is very low.

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

(a) P and Q

(b) P and R

(d) Q and S

Ans. (b)

Sol. Nitrogenase (Nase) is an enzyme that fixes atmospheric nitrogen (N₂) into ammonia. Though abundantly present in the atmosphere, most organisms cannot utilize N₂ directly, and must instead take it in through other forms, like ammonia or nitrate. Nitrogenase is a complex, bacterial enzyme that catalyzes the ATP-dependent reduction of dinitrogen (N₂) to ammonia (NH₃). In its most prevalent form, it consists of two proteins, the catalytic molybdenum-iron protein (MoFeP) and its specific reductase, the iron protein (FeP).

84. One hemoglobin molecule containing four groups can bind with four O₂. The reactions of Hb and O₂ are shown below. Which one of the following reactions is fastest?

(a) Hb₄ + O₂ Hb₄O₂

(b) Hb₄O₂ + O₂ Hb₄O₄

(d) Hb₄O₆ + O₂ Hb₄O₈

Ans. (d)

Sol. Oxygen on entering the red cell, undergoes the following reactions with the Hb, which is called oxygenation

1. Hb₄ + O₂ Hb₄O₂

2. Hb₄O₂ + O₂ Hb₄O₄

3. Hb₄O₄ + O₂ Hb₄O₆

4. Hb₄O₆ + O₂ Hb₄O₈

The affinity of Hb to O₂ in the 3rd and 4th stages is much more than the Hb present in the 1st stage. This is the reasong for the sigmoid shape of oxygen dissociation curve.

In the last case any molecule which is stable must be either Hb₄ or Hb₄O₈; the intermediate molecules may be formed, but if so they are less stable than those in reduced and fully oxidised molecules and therefore cannot remain in their presence.

85. Respiration can be inhibited voluntarily for some time. The point at which respiration cannot be voluntarily inhibited is known as breaking point. Following explanations are offered for the breaking point.

P. J-receptors stimulate respiratory centers.

Q. Hering-Breuer reflex operates.

R. The rise of arterial pCO₂ stimulates the respiratory centre

S. The fall of arterial pO₂ stimulates the respiratory centre.

Which of the above combination is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (c)

Sol. Respiration can be inhibited voluntarily for some time. The point at which respiration cannot be voluntarily inhibited is known as breaking point. At breaking point, the rise of arterial pCO₂ and fall of arterial pO₂ stimulate the respiratory center.

86. A mechanical pressure was exerted on a specific location of a peripheral nerve of a mammal. The touch or pain receptors were stimulated from the skin surface innervated by the same nerve. The action potential generated by touch receptor stimulation was blocked beyond the point of mechanical pressure. But the action potential generated by pain receptro stimulation passes through the point of mechanical presssure.

Following explantions were offered for these observations:

P. The large diameter 'A' fibres were affected by mechanical pressure

Q. The small diameter 'C' fibre were not affected by mechanical pressure

R. The intermediate diameter 'B' fibres were affected by mechanical pressure

S. The large diameter 'A' fibres were not affected by mechanical pressure

Which of the above combinations is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (a)

Sol. If the mechanical pressure was exerted on a specific location of a peripheral nerve of a mammal. The touch or pain receptors were stimulated from the skin surface innervated by the same nerve. The action potential generated by touch receptor stimulation was blocked beyond the point of mechanical pressure. But the action potential generated by pain receptor stimulation passes through the point of mechanical presssure. This can be possible if the large diameter 'A' fibres were affected by mechanical pressure or the small diameter 'C' fibre were not affected by mechanical pressure.

87. When a skeletal muscle with an intact nerve supply is stretched, the muscle contracts and the tone increases.

The following explanations are offered for this ovbservation:

P. Golgi tendon organ ws timulated by the stretching of muscle.

Q. -motor neurons were excited by the stimulated afferent nerve fibres from the stretched muscle.

R. Muscle spindle was stimulated by strething of muscle

S. -motor neurons were excited by the stimulated afferent nerve fibres from the stretched muscle.

Which of the combination is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (c)

Sol. During muscle contraction, Muscle spindle was stimulated by strething of muscle and -motor neurons were excited by the stimulated afferent nerve fibres from the stretched muscle.

88. An elderly person suffering from calcium deficiency was advised to take calcium rich food and to supplement the diet with vitamin D. The absorption of calcium in the intestine was increased with the suplementation of vitamin D. Following explanations were offered for his increased calcium absorption by vitamin D :

P. The synthesis of calbindin-D_9K and calbindin-D_28K in enterocytes was stimulated.

Q. The number of Ca²⁺ -ATPase molecules in enterocytes was stimulated.

R. The synthesis of divalent metal transporters 1 (DMT1) in the eenterocyres was stimulated.

S. The number of hephaestin in the enterocyres was increased.

Which of the above combinations is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (a)

Sol. Vitamin D increases calcium absorption by the synthesis of calbindin-D_9K and calbindin-D_28K in enterocytes was stimulated and the number of Ca²⁺-ATPase molecules in enterocytes was stimulated.

89. In an experiment involving mapping of 3 genes (a, b, and c) in Drosophila a three point test cross is carried out. The parental cross was AAbbCC × aaBBcc. The genotypes of the double crossvers are : Aabbcc and aaBbCc. Based on this, determine the order of the genes.

(a) a c b

(b) c a b

(d) b a c

Ans. (a)

Sol. Parental genotype = AAbbCC × aaBBcc

Double crossover genotype = Aabbcc and aaBbCc

The order of genes will be (a c b).

90. A black labrador homozygous for the dominant alleles (BBEE) is crossed with a yellow labrador homozygous for the recessive alleles (bbee). On intercrossing the F1, the F2 progeny was obtained in the following ratio :

9 black : 3 bron : 4 yellow

This is an example of

(a) recessive epistasis where allele e is epistatic B and b

(b) dominant epistasis where allele E is epistatic to B and b

(d) complementary epistasis where allele b is epistatic.

Ans. (a)

Sol. By observing F2 progeny ratio 9 black : 3 brown : 4 yellow, it is obvious this is an example of recessive epistasis where allele e is epistatic B and b.

91. In the following example, 3 indepndently assorting genes are known ot govern coat color in mice. The genotype of few of the coat colors is given below:

Agouti : A – B – C

Black : aa B – C –

Albino : – – – – cc

What will be the expected frequency of albinos in the F2 progeny from crosses of pure black with albino of the genotype AAbbcc?

(a) 1/4

(b) 1/16

(d) 9/64

Ans. (a)

Sol. On crossing pure black color mice (aaBBCC) with albino colored mice (AAbbcc),

aaBBCC × AAbbcc

Gametes aBC Abc

F1 progeny AaBbCc

F1 crossing AaBbCc × AaBbCc

Genotype for albino coat color will be 1/4.

92. The distance between bacterial genes as determined from interrupted mating experiments are measured in units of

(a) cM

(b) minutes

(d) micrometers

Ans. (b)

Sol. The distance between genes was measured as time units (minutes) between the two genes in terms of their appearance as recombinants in the recipient. The genetic map of E. coli was constructed in just this way, with map units in minutes, timed from an arbitrarily located origin.

93. The inheritance pattern of a common trait which shows complete penetrance is shown below. Based on the below pedigree, fill in the blanks from the options given. "The trait is [A]. The probability that a child from the marriage of individual III-1 and III-2 will show the trait is [B] considering that the individual III-1 is a carrier of the trait."

(a) A–Y-linkded, B–0

(b) A–Y-linkded, B–1/2

(d) A–Autosomal B–1/6

Ans. (c)

Sol. By observing the pedigree chart and information mentioned in question the trait will be autosomal and probability for child from marriage of III-1 and III-2 will be 1/8.

94. The following statements were made regarding chromosome pairing (shown in the figure below) and subsequent segregation during meiosis I in the reciprocal translocation heterozygote:

P. Three ways of segregation in Anaphase I would be : adjacent 1 (vertically in the above figure), and alternate.

Q. Gametes resulting from adjacent 1 and adjacen 2 segregation will be non-viable because of deletions and duplications of several genes.

R. All gametes resulting from alternate segregation will be viable as they will carry both normal chromosomes or both chromosomess having trnaslocations in the two poles, respectively.

S. A dicentic and an acentric chromosome will be generated following alternate segregation.

Which of the following combination of statements will most appropriately explain the consequence?

(a) P, Q and S

(b) P, Q and S

(d) Only P and R

Ans. (b)

Sol. Segregation during meiosis I in the reciprocal translocation heterozygote mentioned in question explains that three ways of segregation in Anaphase I would be : adjacent 1 (vertically in the above figure), and alternate, gametes resulting from adjacent 1 and adjacen 2 segregation will be non-viable because of deletions and duplications of several genes and all gametes resulting from alternate segregation will be viable as they will carry both normal chromosomes or both chromosomess having trnaslocations in the two poles, respectively.

95. The following table summarizes the result of a cross between two strains of Neurospora having the alleles D and d, respectively. The table shows the different patterns of octad arrangement and the number of ascus observed of each type.

Total = 300

Based on the above, fill in the blanks from the options given below,

"The first two columns are from meiosis with no crossover between locus D and [A]. The patterns for these two columns represent [B] segregation pattern. The distance between the locus D and D centromere is [C] map units."

A B C

(a) d allele first division 10

(b) centromere first division 10

(d) centromere second division 10

Ans. (b)

Sol. The distance between the locus D and the centromere

A is centromere, B represents first division and distance (C) 10 cM.

96. A series of cell lines was created by fusing mouse and human somatic cells. In mouse-human somatic cell hybrids, human chromosomes tend to get lost before becoming a stable cell line. Some hybrid cell lines may carry human chromosome deletions. Each cell line wa examined for the presence of chromosomes and for the production of an enzyme. The following results were obtained :

Which segment of the chromosome has the gene encoding for the enzyme?

(a) 1p

(b) 5p

(d) 4p

Ans. (b)

Sol. By examining the chromosomal division, the gene encoding for the enzyme must be on 5p.

97. A hypothetical gene encodes a protein with the following amino acid sequence:

Phe-Pro-Thr-Ala-Val-Arg-Ser

A mutation of single nucleotide alters the amino acid sequence to

Phe-Leu-Leu-Leu-Leu-Val

A second single nucleotide mutation occurs in same gene restoring back the amino acid sequence ot the original. The following statements were made regarding the nature and location of the first mutation and that of the intragenic suppressor mutation:

P. The first mutation is a deletion in the second codon.

Q. The first mutation is an insertion in the second codon.

R. The intragenic suppressor mutation is an insertion in the second codon.

S. The intragenic suppressor mutation is deletion in the third codon

Which combination of the above statements is correct?

(a) P and R

(b) P and S

(d) Q and S

Ans. (a)

Sol. The first mutation is a deletion in the second codon. and the intragenic suppressor mutation is an insertion in the second codon.

98. The organs radula and clitellum are found in

(a) Coelenterata and Echinodermata, respectively

(b) Echinodermata and Coelenterata, respectively

(d) Mollusca and Annelida, respectively

Ans. (d)

Sol. The radula is an anatomical structure that is used by molluscs for feeding, sometimes compared rather inaccuraely to a tongue. The clitellum is a thickened glandular and non-segmented section of the body wall near the head in earthworms and leeches, that secretes aviscid sac in which the eggs are deposited. This organ is used in sexual reproduction of some annelids. The clitellum becomes apparent in mature annelids, but may be hard to locate visually in younger annelids. In leeches, it appears seasonally. Its colour is usually slightly lighter than that of the body of the annelid. Occasionally, living segments of the worm will be shed with the clitellum.

99. Branchiostoma is a

(a) deuterostome and schizocoelmate

(b) protosome and schizocoelomate

(d) protostome and entercoelomate

Ans. (c)

Sol. Enterocoelomate, any animal in which the mesoderm-lined body cavity (coelom) arises in the embryonic stage as an outpocketing of the developing gut (enteron). This form of development, found in echnioderms (e.g., starfishes, sea urchins) and a few other invertebrate phyla and inchordates (e.g., fishes, amphibians, reptiles, birds, mammals) has been viewed as evidence of the common ancestry of echinoderms and chordates. Enterocoelomates are grouped together and are often referred to as deuterostomes. Branchiostoma is a deuterostome and enterocoelomate.

100. Which of the following is a correct hierarchial sequence for classifying a living organism?

(a) Domain-kingdom-phylum-Class-Order-Family-Genus-Species

(b) Kingdom-Domain-Phylum-Class-Order-Family-Genus-Species

(d) Kingdom-Domain-Phylum-Order-Class-Family-Genus-Species.

Ans. (a)

Sol. Domain Eukarya

Organisms that have nuclei and other membrane enclosed organelles.

Kingdom Plantae

Terrestrial, multicellular, photosynthetic organisms

Phylum Anthophyta

Vascular plants with flowers, fruits and seeds

Class Monocathyledones

Monocots: Flowering plants with one seed leaf (cotyldeon) and flower parts in threes

Order Commelinales

Monocots with reduced flower parts, elongated leves and dry one-seeded fruits.

Family Poaceae

Grasses with hollow stems; fruits is a grain and abundant endosperm in seed

Genus Zea

Tall annual grass with separate female and male flowers.

Species Zee mays

Corn

101. The present global warming trend is expected to result in an increased incidence of malaria in temperate countries.

The supposed underlying mechanism is that

(a) higher temperatures make temperate country people more vulnerable to diseases

(b) malarial parasite grows better at higher temperatures

(d) anti-malaria drugs are less effective in temperate countries

Ans. (c)

Sol. Mosquitoes are more likely to bite when the weather is warm and sunny. This is because they like to breed in warm, humid areas. Mosquitoes are cold-blooded insects. They prefer to live in areas that are around 70-80 degrees Fahrenheit.

102. The most important reproductive stategies of big trees in a forest are

(a) earlier age at first reproduction and production of large number of small seeds.

(b) earlier age at first reproduction and production of a small number of large seeds

(d) later age at first reproduction and production of a small number of large seeds.

Ans. (d)

Sol. The tradeoff between offspring size and number is ubiquitous and manifestly similar in plants and animals despite fundamental differences between the evolutionary histories of these two major life forms. Fecundity (offspring number) primarily affects parental fitness, while offspring size underpins the fitness of parents and offspring. Big trees in forest produce later in age and produce small number of large seeds.

103. Some key characterisstics of the four classes fo phylum Mollusca are listed below :

P. They have two lateral (left and right) shells (valves) hinged together dorsally; they do not have distinct head or radula; they disperse from place to place largely as larvae.

Q. They generally creep on their foot; the heads of most of thi group have a pair to tentacles with eyes at the end; during embryological development, they undergo torsion.

R. They have oval bodies with overlapping calcareous plates; underneath the plates, the body is not segmented; they creep along useing a broad, flat food surrounded by a groove or mantle cavity in which the gills are arranged.

S. They have highly developed nervous system; most members of this class have closed circulatory systems.

The correct match of the above characteristics with the classes of Mollusca is

(a) P–Polyplacophora, Q–Bivalvia, R–Gastropoda, S–Cephalopods

(b) P–Cephalopoda, Q–Polyplacophora, R–Bivalvia, S–Gastropoda

(d) P–Gastropods, Q–Bivalvia, R–Cephalopods, S–Polyplacophora

Ans. (c)

Sol. P. Bivalvia : Clams, cockles, mussels, oysters, scallops, and shipworms are bivalves. Most are completely enclosed by the shell, the two valves of which are joined by an elastic ligament, and by two sheets of tissue called the mantle. Bivalves have no head.

Q. Gastropods typically have a large foot with a flat sole for crawling, a single coiled shell that covers the soft body, and a head that bears a pair of eyes and tentacles. However, they are so diverse that some forms lack shells, while animals in one genus have shells with two halves, like bivalves.

R. Polyplacophora - The main characteristics of Polyplacophora include (1) elongate or oval, dorsoventrally flattened, bilaterally symmetrical, marine; (2) with dorsal shell of eight plates embedded in a tough mantle; (3) mantle-edge stiffened (called the girdle); (4) large, muscular, ventral foot (girdle and foot can act as suction cup.

S. Cephalopods are bilaterally symmetrical and typically have a highly developed centralized nervous system. Their image-forming eyes are similar in structure to vertebrate eyes, and their heads are armed with tentacles that have rows of round suction disks.

104. The list below inclues names of animal phyla and classes.

1. Enchinodermata 2. Cephalopoda 3. Annelida 4. Mollusca

5. Hirudinea 6. Asteroidea 7. Arthropoda 8. Crustacea

For a leech and lobster, the correct classification of phylum and class, respectively, is

(a) Leech: Phylum – 4, Class – 2; Loster : Phylum – 1, Class–8.

(b) Leech: Phylum – 3, Class – 2; Loster : Phylum – 4, Class–3.

(d) Leech: Phylum – 1, Class – 7; Loster : Phylum – 3, Class–6.

Ans. (c)

Sol. Leeches are placed within the phylum Annelida either in the order Hirudinea, class Clitellata. "Lobsters are classified in the phyllum Arthropoda and class Crustacea.

105. Which of the following set of observations is true with refernce to comparison of aquatic (A) and terrestrial (T) ecosystems?

(a) Number of tropihc levels is mor in A than in T. Productivity/Biomass ratio is higher in T than in A. Herbivore assimilation efficiency is higher in A than in T.

(b) Number of trophic levels is more in T than in A. Productivity/Biomass ratio is greater in A than in T. Herbivore assimilation efficiency is higher in T than in A.

(c) Number of trophic lelvels is more in T than in A. Productivity/Biomass ratio is higher in T than in A. Herbivore assimilation efficiency is higher in T than in A.

(d) Number of trophic levels is more in A than in T. Productivity/Biomass ratio is greater in A than in T. Herbivorwe assimilation efficiency is higher in A than in T.

Ans. (d)

Sol. There is not enough energy to support another trophic level after four up to six energy transfers. Aquatic ecosystems can support more trophic levels compare to the land ecosystems because of the higher efficiency. Productivity/biomass ratio is greater in aquatic than in terrestrial.

106. The following statements are related to excreion in invertebrates:

P. Flame cells are found in molluscs and jelly fish.

Q. Nephridia an Malpighian tubules convert ammonia to urea for water conservation

R. Green glands are found in flatworms and help in the excreta elimination

S. Excretory canals in nematodoes carry waste materials to excretory pores in the body wall.

Choose the correct answer.

(a) Only R

(b) P and R

(d) Q and S

Ans. (d)

Sol. Flame cells and nephridia in worms perform excretory functions and maintain osmotic balance. Some insects have evolved Malpighian tubules to excrete wastes and maintain osmotic balance. Nephridia and Malphigian tubules convert amminia to urea for water conservation. An additional excretory structure has evolved in the roundworms. Excretory canals located on both sides of the intestine facilitate waste disposal by carriage of material to an excretory pore in the body wall.

107. Fish species X and Y feed on mayfly nymphs in their stream habitat. In a laboratory experiment, the predation intensity of X and Y on their prey was tested under dark (D) and light (L) conditions. Thus, the experimental protocol inclueded four aquraia -LX, LY, DX and DY. In each aquarium containing 100 mayfly nymphs, one fish was introduced and allowed to feed for 30 minutes. Then the fish was removed and the number of mayfly nymphs left uneaten in each aquarium was counted. The results are shown graphically below:

The most significient conclusion from the results is:

(a) X is a visual predator, but has less predation impact on the prey than Y.

(b) X is visual predator and has greater predation impact on the prey than Y.

(d) Y is a visual predator but has less predation impact on the prey than X.

Ans. (b)

Sol. Data tells that X is visual predator and has greater predation impact on prey than Y. Because the no. Of left uneaten is high for X in dark and is same for Y in both light and dark . X has greater predation impact.

108. There are three species of frogs–A, B and C. Species A does not provide parental care for its eggs and larvae. Species B is subjected to predation by a predator that selectively feeds only on small-sized larvae. Species C faces progressively decreasing opportunities for breedling with increasing age. Assuming that resources available for reproduction are imilar for A, B and C, which of the following strategies would have been favored?

(a) A should produce large number of small-sized offspring; B should produce a small number or large-sized offspring; C should breed earlier in life.

(b) Species A and B should produce a small number of large-sized offspring and C should breed earlier in life.

(d) Speciess A should produce a small number of large-sized offspring; B should produce a large number of small-sized offspring and C should breed earlier in life with a small clutch size.

Ans. (a)

Sol. Statement A is true that species A should produce large no. of small sized offsprings in order to increase the survival chances of its eggs and larvae. B should produce small no. or large- sized offspring in order to avoid predation. C should breed early in life.

109. A small lake has three trophic levels-phytoplankton (autotrophs), zooplankton (herbivore) and planktivo-rous fish (primary carnivore). Into this lake, a population of piscivorous fish (secondary carnivore) was introduced to study the 'Top-down' effects. What is the expected long-term consequence of such an introduction to phytoplankton and zooplankton trophic levels?

(a) Zooplankton biomass will increase and phytoplankton biomass will decrease

(b) Zooplankton biomass will decrease and phytoplankton biomass will increase

(d) The biomasses of both zooplankton and phytoplankton will decrease.

Ans. (a)

Sol. Secondary carnivore will feed on primary carnivore and will reduce their number. In return number of herbivores (zooplankton biomass) will increase due to lack of predator (primary carnivore). This will ultimately lead to less autotrophs (phytoplankton biomass) due to increase in number of herbivores. This is an example of 'Top-down' effect.

110. The three graphs (A, B, C) show population growth (N) patterns in relation to N or ime (t).

Which of the following in correct with reference to the Y-axis label and the type of population growth?

(a) A: Y-axis: N, exponential growth

B: Y-axis:dN/dt, logistic growth

C: Y-axis: In(N), exponential growth

(b) A: Y-axis: dN/dt, exponential growth

B: Y-axis:In/(N), logistic growth

C: Y-axis: N, exponential growth

B: Y-axis:dN/dt, logistic growth

C: Y-axis: N, exponential growth

(d) A: Y-axis: dN/dt, exponential growth

B: Y-axis: In(N), logistic growth

C: Y-axis: N, exponential growth

Ans. (a)

Sol. Exponential growth models are more useful to predict investment returns when the rate of growth is steady. A growth curve is a visual depiction of the growth of a phenomenon, with the x-axis typically representing time and the y-axis exponential growth.

The logistic growth curve represents the logistic population growth rate. The Y-axis shows the total population increase or decreases going up and down on the graph. dN/dT. Exponential growth curve. Y-axis represent ln(N).

111. Following are the characteristics of specie hat make them more or less prone to extinction:

Which of the following is the correct combination of characteristics that makes the species more prone to extinction?

(a) a d f g j l n

(b) a c f h i k m

(d) b c f h j k m

Ans. (a)

Sol. Many rare and/or endemic species exhibit one or more of the following attributes which make them especially prone to extinction:

(1) narrow (and single) geographical range,

(2) only one or a few populations (rare),

(3) small population size and little genetic variability,

(4) over-exploitation by people,

(5) feed at high trophic level,

(6) have short life span.

112. Three Indian animals – Cormorant, Lion-tailed Macaque and Gerbil and are to be matched with the ecosystem they inhabit-Wetland (A), Desert (B), Deciduous forest (C). or Rain Forest (D). Which of the following is the correct match of each animal with its habitat?

(a) Cormorant – D; Lion-Tailed macaque – C; Gerbil – B

(b) Cormmorant – A; Lion-Tailed macaque – C; Gerbil – D

(d) Cormorant – B; Lion-Tailed macaque-C; Gerbil – D

Ans. (c)

Sol. The Indian Cormorant is a bird of larger freshwater wetlands and seems to do quite well even along mangroves and estuaries.

Lion-tailed macaques are the flagship species of the North Western Ghats Moist Deciduous Forests ecoregion, located in the Indian Tropical Coastal Forests (IM2) bioregion.

The Indian desert jird or Indian desert gerbil (Meriones hurrianae) is a species of jird found mainly in the Thar Desert in India.

113. Following is the diagram of thre idealized survivoship curves of animals. Find the correct match between the group of animals and the respective survivorship curves.

(a) Marine pelagic fish and large mammals – III and I, respectively.

(b) Marine pelagic fish and large mammals – I and III, respectively.

(d) Marine pelagic fish and some birds – I and III, respectively.

Ans. (a)

Sol. The Type I curve, illustrated by the large mammals, tracks organisms that tend to live long lives (low death rate and high survivorship rate); toward the end of their life expectancies, however, there is a dramatic increase in the death rate. Trees, marine invertebrates, and most fish have a Type III survivorship curve. In a Type III curve, very few organisms survive their younger years. However, the lucky ones that make it through youth are likely to have pretty long lives after that.

114. In rats, after the delivery of the offspring mother shows the following behaviors. Which one of the following behaviours is not maternal?

(a) Licking the pups

(b) Huddling above the pups to access the ventrum

(d) Bringing back to the nest pups that wander away from it.

Ans. (c)

Sol. Lordosis is a reflex action that causes many non-primate female mammals to adopt a body position that is often crucial to reproductive behavior. The posture moves the pelvic tilt in an anterior direction, with the posterior pelvis rising up, the bottom angling backward and the front angling downward.

115. The population size of a bird increased from 600 tp 645 in one year. If the per capita birth rate of this population is 0.125, what is its per capita death rate?

(a) 0.25

(b) 0.15

(d) 0.02

Ans. (c)

Sol. As wer know that, dN/dT = r × N; where, r is the instantaneous rate of increase and N is the initial population.

This equation can be written as, dN/dt × 1/N = r ...(1)

Instanteous rate of increase = Per capita birth rate – Per capita death rate

r = b – d

From equation (1), dN/dt × 1/N = r, we get, 45/600 = 0.075 = r

According to the question,

r = b – d

0.075 = 0.125 – d

Hence, d = 0.125 – 0.075 = 0.05

116. The long feather tail of a peacock is quoted as an example supporting

(a) Hamilton's rule

(b) Zahavi's handicap principle

(d) Haldane's rule

Ans. (b)

Sol. The Handicap Principle put forward a novel idea to explain several previously baffling aspects of animal behaviour, including the famous tail of the peacock. The principles relies on three chief tenets : (a) Animals communicate with each other through signals; (b) these signals, in order to be effective, must be honest and (c) honest signals are expensive, i.e., the animal producing an honest signal in the animal world as we understand it, it is not enough to be beautiful; for a phenotype to be selected, it must be 'fit' or have a higher ability to transmit the trait to the next generation.

117. In very small populations, genetic variation is often lost through genetic drift. If the population size of a mammal on an isolated island is 50, what percentage of its genetic variation is lost every generation?

(a) 0.01

(b) 0.5

(d) 0.05

Ans. (a)

Sol. The rate at which genetic variation will be lost within a population as a result of genetic drift is,

118. Assume that in terms of 'genetic fitness' the 'benefit' of performing an altruistic act to a relative is 500 units and the 'cost' involved is 150 units. Following Hamiton's rule, the act shold be performed if the relative is a

(a) only brother

(b) nephew or niece

(d) only step-sister

Ans. (a)

Sol. Kin selection will be favoured the altruistic act, if rB – C > 0; where, C = cost incurred, B = benefit received by the recipients, r = coefficient of relatedness.

The coefficient of relatedness must be 0.5 or more. According to the Hamilton's rule, the altruistic act will only favour if the relative is brother.

0.5 × 500 – 150 > 0

or, 100 > 0

It means that, the altruistic act will favour if the relative is brother.

119. The following situations might lead to the evolution of monogyny in birds :

P. Male has to assist the female in rearing the offspring

Q. Male guards the female against other male trying to mate with her

R. One male may not produce enough sperm required to fertilize all the eggs produced by the female.

Which of the above is/are correct?

(a) Only P

(b) Only Q

(d) P and R

Ans. (c)

Sol. Monogamy may evolve when the cost of acquiring mates is very high, when females have the ability to restrict male behavior, or when offspring survival requires more intensive care than can be provided by a single animal. Statement P and Q is correct.

120. The diagram below depicts a simplified Tree of Life with three domains and one of the domains including Whittaker's three major kingdoms. Which of the following is the correct naming of the numbered boxes?

(a) 1–Bacteria 2–Archaea 3–Eukarya 4–Fungi

(b) 1–Archaea 2–Bacteria 3–Eukarya 4–Plants

(d) 1–Archaea 2–Bacteria 3–Eukarya 4–Fungi

Ans. (a)

Sol. Recent evidence indicates that Archaea and Eukarya are more closely related to each other than either is to Bacteria. Option (a) is correct.

121. A few events in the history of life on earth are given below :

1. Radiation of mamals and birds; Flourishing of insects and angiosperms.

2. Primitive plants and fungi colonize land; Diversification of echinoderms

3. Seed plants appear; Fishes and trilobites abundant; Earliest amphibians and insects

4. Earliest birds and Angiosperms appear; gymnosperms dominant

5. Mass marine extinctionsl Reptiles radiate; Amphibians decline

Which of the following is a correct match of the above events with the geological period during which they had occurred?

(a) 1: Ordovician; 2; Tertiary; 3: Permian; 4: Silurian; 5: Devonian; 6: Jurassic.

(b) 1: Permian; 2; Devonian; 3: Silurian; 4: Ordovician; 5: Tertiary; 6: Jurassic.

(d) 1: Permian; 2; Devonian; 3: Jurassic. 4: Tertiary; 5: Silurian; 6: Ordovician;

Ans. (c)

Sol. Ordovician Period (490 to 443 mya)—origin of plants, marine alga abundant.

Silurian Period (443 to 417 mya)—diversity of jawless fishes, first jawed fishes, cononization of land by vascular plants and arthropods.

Devonian Period (417 to 354 mya)—diversification of bony fishes, first amphibians, insects.

Permian Period (290 to 248 mya)—extinction of many marine and terrestrial organisms, radiation of reptiles, origin of mammal like reptiles, most modern insects.

Jurassic Period (206 to 144 mya)—gymnosperms continue as dominant plants, dinosaurs dominant.

Tertiary Period (65 to 1.8 mya)

Paleocene Epoch (65 to 54.8 mya)—major radiation of mammals, birds, pollinating insects.

Eocene Epoch (54.8 to 33.7 mya)—angiosperm dominance increases, origins of most modern mammalian orders.

Oligocene Epoch (33.7 to 23.8 mya)—origins of many primate groups, including the apes.

Miocene Epoch (23.8 to 5.3 mya)—continued radiation of mammals and angiosperms.

Pliocend Epoch (5.3 to 1.8 mya)—apelike ancestors of humans appear.

122. Four different species concepts are given below :

P. Species separate based on their use of different ecological niches and their presence in different habitats and environments.

Q. Differences in physical characteristics or molecular characteristics are used to distinguish speices.

R. Species are distnict if they are reproductively isolated.

S. Phylogenetic trees and analyses of ancestry serve to differentiate species.

Which of the following gives the correct names of the above concepts?

(a) P : Biological; Q: Phylogentic; R: Evolutionary; S: Ecological

(b) P: Ecological; Q: Phylogentic; R: Biological; S: Evolutionary

(d) P: Phylogenetic; Q: Evolutionary; R: Ecological; S: Biological

Ans. (b)

Sol. Ecological species concept : A species is a set of organisms exploiting (or adapted to) a single niche or A species is a lineage or a closely related set of lineages, which occupies an adaptive zone minimally different from that of any other lineage in its range and which evolves separately from all lineages outside its range.

Phylogenetic species concept : A species is the smallest diagnosable cluster of individual organisms within which there is a parental pattern of ancestry and descent. Or A species is an irreducible (basal) cluster of organisms, diagnosable distinct from other such clusters and within which there is a parental pattern of ancestry and descent.

Biological species concept : Species are groups of actually or potentially interbreeding natural populations, which are reproductively isolated from other such groups. Or A species is a reproductive community of populations (reproductively isolated from others) that occupies a specific niche in nature.

Evolutionary species concept : A species is a lineage (an ancestral-descendant sequence of populations) evolving separately from others and with its own unitary evolutionary roles and tendencies or Evolutionary species concept : A species is a single lineage of ancestor-descendant populations which maintain its identify from other such lineages and which has it own evolutionary tendencies and historical fate.

123. Among the following antigens specific to a pathogen, which one is most likely to be ineligible as a vaccine with long lasting host protective effect?

(a) A cell surface protein

(b) An enzyme involved in pathogen metabolism.

(d) A long chain fatty acid.

Ans. (d)

Sol. Cells of the immune system have a high content of long-chain polyunsaturated fatty acids (LCPUFAs) in their membranes. Therefore, acquisition of different amounts of LCPUFAs by immune cells as they are developing could influence immune maturation and function and this could have a lasting effect on immune competence and risk of diseases involving immune dysfunction.

124. Which one of the following techniques is generally used to produce transgenic animals?

(a) Processed mRNA containing only exons are introduced into blastocyst stage embryo

(b) Entire foreign nucleus is introduced in enucleated unfertilized egg.

(d) cDNA of desired gene is introduced into animal embroys and implanted in a foster mother.

Ans. (c)

Sol. The three principal methods used for the creation of transgenic animals are DNA microinjection, embryonic stem cell-mediated gene transfer and retrovirus-mediated gene transfer.

125. Which one of the following statements is not correct for propagation and maintenance of mammalian cells in vitro?

(a) Transformed cell lines do need external supply of serum to grow

(b) The cells that are obtained directly from the organism is a primary culture

(d) HEPES buffer is generally used to maintain pH of the culture media.

Ans. (c)

Sol. When added to a cell culture, trypsin breaks down the proteins that enable the cells to adhere to the vessel. Trypsinization is often used to pass cells to a new vessel. When the trypsinization process is complete the cells will be in suspension and appear rounded.

126. In isoelectric focusing experiments, proteins are separated on the basis of their

(a) relative content of only positively charged residues

(b) relative content of only charged residues

(d) mass to charge ratios.

Ans. (c)

Sol. Proteins can also be separted electrophoretically on the basis of their relative contents of acidic and basic residues. The isoelectric point (pI) of a protein is the pH at which its net charge is zero. At this pH, its electrophoretic mobility is zero because z in equation 1 is equal to zero. This method of separating proteins accordiign to their isoelectric point is called isoelctric focusinig. The pH gradient in the gel is formed first by subjecting a mixture of polyampholytes (small multicharged polymers) having many pI values of electrophoresis. Isoelectric focusing can readily resolve proteins that differ in pI by as little as 0.01, which means that proteins differing by one net charge can be separated.

127. Which one of the following molecular marker types uses combination of both restriction enzyme and PCR techniques?

(a) SSR

(b) AFLP

(d) RAPD

Ans. (b)

Sol. AFLP : Amplified Fragment Length Polymorphism; a molecualr marker generated by a combination of restriction digestion and PCR amplification.

128. An any moving straight, upon encountering an obstacle, may turn either right or left continue moving. To test the hypothesis that the direction chosed by the ant is random, the most appropriate statistical test is

(a) Student's t-test

(b) of independence

(d) correlation test

Ans. (c)

Sol. The chi-square goodness of fit test is a hypothesis test. It allows you to draw conclusions about the distribution of a population based on a sample. Using the chi-square goodness of fit test, you can test whether the goodness of fit is "good enough" to conclude that the population follows the distribution.

129. A bioinformatics tool used to find the sequence similarity in the subunits of hemoglobin is

(a) FASTA

(b) BLAST

(d) PSI : PLOT

Ans. (b)

Sol. BLAST stands for Basic Local Alignment Search Tool. BLAST is a tools that is used to find the seqyuences homologous to a particular sequence. BLAST compares all the sequences in the database with one tht is searched for and provides many hits which are usually arranged in the increasing order of the scored obtained.

130. Electron microscopes have much resolution than any type of light microscope because

(a) of their higher magnification

(b) the lenses used are of much higher quality

(d) the images are viewed on screen rather than directly using an eye-piece or ocular lens.

Ans. (c)

Sol. An elctron microscope is a microscope that uses accelerated electrons as a source of illumination. Because the wavelength of an electron can be up to 100,000 times shorter than that of visible light photons, the electrons microscope has a higher resolving power than a light microscope and can reveal the structure of smaller objects. A tarnsmission electrons microscope can achieve better than 50 pm resolution and magnifications of up to about 10,000,000x whereas most light microscope are limited by diffraction to about 200 nm resoultion and useful magnifications below 2000s.

131. The average energy absorbed by 10 gm of tissue from ³²p radiation is 14.9 Jkg^–1. The average does in rads is

(a) 1490

(b) 1.49

(d) 1.49

Ans. (a)

Sol. 1 rad (radiation absorbed dose) 1 erg/g = 0.0.1 J/Kg

So, 1 J/Kg = 100 rad

Therefore, 14.9 J/Kg = 14.9 × 100 = 1490 rads

132. The electrospray ionization spectrum of a highly purified protein shows multiple closely spaced peaks. This most likely arises due to

(a) the presence of multiple conformations

(b) degradation of the protein during recording of the spectrum

(d) extensive aggregation of the protein

Ans. (c)

Sol. Electrospray ionisation is where voltage is applied, causing each particle to gain an H⁺ ion. This converts the sample into a gas made up of positive ions. Ionisation can also occur using electron impact ionisation. This is where an electron gun is used to fire high energy electrons at the particles.

133. Following are certain statements regarding the use of Agrobacterium in plant transformation

P. A. tumefaciens causes crown gall disease and A. rhizogenes causes hairy root disease

Q. Region A in Ti plasmid is responsible for replication

R. Region D in Ti plasmid is responsible for virulence

S. Oncogenic (onc) region in T-DNA is responsible for unusual amino acid synthesis0

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

(a) P and Q

(b) R and S

(d) Q and S

Ans. (c)

Sol. Agrobacterium tumefaciens is a pathogenic bacterium that causes crown gall disease, a plant tumor affecting a wide range of plant species. Agrobacterium rhizogenes bacterium is a Gram-negative soil bacterium, causes hairy root disease in plants and it occurs in many dicotyledonous plants. VirD1 and VirD2 are involved in the processing of T-DNA during conjugation to produce the T-strand; this is the single-stranded DNA molecule that is transported to the host plant cell. During the processing, VirD1 will act as a topoisomerase to unwind the DNA strands. VirD2, a relaxase, will then nick one of the DNA strands and remain bound to the DNA as it is transferred to the recipient cell. Within the recipient cell, VirD2 will also work together with VirE2 to direct the transferred DNA to the recipient cell's nucleus. There are suggestions that VirD2 may be phosphorylated and dephosphorylated by different proteins, affecting its ability to deliver DNA. Conversely, little is known about VirD3, and mutational analyses have not provided any support for its role in the virulence of Agrobacterium. Finally, VirD4 is a crucial part of the conjugation process, serving as a coupling factor that recognizes and transfers the T-strand to the transport channel.

134. Cultured animal cells were transfected with expression vector encoding either -galactosidase (-gal) alone or expressing a fusion protein of -gal glucocorticoid receptor (GR). After transferction, cells were kept in presence or absence of Dexamethasone. Immunofluorescence with a labeled antibody specific for -gal was used to detect the expressed protein in cytoplasm or nucleus of transfected cells. Possible results of the experiments are:

1. Expression of -gal alnone in the cytoplasm in both absence or presence of Dexamethasone.

2. Expression of -gal GR in the cytoplasm in the avbsence of Dexamethasone.

3. Expression of -gal alone in the nucleus both in the presence or absence of Dexamethasone.

4. Expression of -gal GR in the nucleus in presence of Dexamethasone

5. Expression of -gal alone in both cytoplasm and nucleus in presence or absence of Dexamethasone.

6. Expression of -gal-GR in both cytoplasm and nucleus in presence of Dexamethasone.

Choose the correct combination of results from the following options

(a) 2, 3 and 4

(b) 1, 2 and 4

(d) 1, 2 and 6

Ans. (b)

Sol. Statement 1,2 and 4 is correct.

135. Marker-assisted selection (MAS) defined as selection based on molecular markers should have some important criterion for platn breeding activities. Some statements about these criteria are mentioned below:

P. Marker should co-segregae with the desired trait of interest.

Q. Marker should not co-segregate with the desired trait of interest.

R. Marker should be un-linked with the desired trait of interest

S. Marker is used for indirect selection of a genetic determinant or determinants of a trait of interest.

Which on of the above combinations is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (d)

Sol. Marker-assisted selection (MAS) is the process of using morphological, biochemical, or DNA markers as indirect selection criteria for selecting agriculturally important traits in crop breeding. This process is used to improve the effectiveness or efficiency of selection for the traits of interest in breeding programs. DNA markers can be used to detect the presence of allelic variation in the genes underlying these traits. By using DNA markers to assist in plant breeding, efficiency and precision could be greatly increased.

136. In a transgenic mice line, lox p sites are introduced in the target gene A in the following manner

This transgenic mice line was mated with another transgenic mice line where Cre recombinase is expressed only in B-cells. What will be the expression profile of gene A in Cre/lox recombinant mice?

(a) Gene will not be expressed in B-cells, as orientation of exon 1 will be invested by Cre.

(b) Gene will not be expressed in B-cells, as exon 2 will be deleted by Cre.

(d) Gene will not be expressed in B-cells as orientation of exon 2 will be inverted.

Ans. (b)

Sol. Cre-Lox recombination is a site-specific recombinase technology, used to carry out deletions, insertions, translocations and inversions at specific sites in the DNA of cells.

137. A small fraction of clear cellular lystate was run on an isoelectric focusing gel (IEF) to purify particular protein. which showed a number or sharp bands corresponding to different pI valves. The protein of interest has a pI of 5.2. Therefore, the band corresponding to pI 5.2 was cut, eluted with appropriate buffer and subjected to SDS-PAGE, which showed 3 distinct bands. Which one of the following inferences cannot be drawn from the above observations?

(a) Several different proteins having same pI may be present at the single band on IEF gel.

(b) SDS-PAGE showed 3 distinct bands which may represent molecular mass of different proteins.

(d) As the IEF gel showed a distinct band corresponding to pI 5.2, which is the pI of the protein of interest, the protein is composed of a single subunit.

Ans. (d)

Sol. IEF works the principle on the property of individual protein molecule to stop movement at a certain pI under an electric field. But once the electric field is removed the molecules start to diffuse. IEF finds its application in proteomics. The basic of proteomics is a multi-dimensional separation of protein molecules. As the protein is composed of 3 subunits , statement d is false.

138. The diagram below represens a 2 kb insert successfully introduced in between two BamHI sites of a 3.8 kb vector in desired orientation. The HindIII site on the insert and EcoRI site on the vector is also indicated. If the insert was introduced in the opposite orientation, which one of the following statements is incorrect?

(a) Digestion with EcoRI will linerize the 5.8 kb plasmid.

(b) Digestion with BamHI will yield one 3.8 and one 2.0 kb fragment

(d) Double digestion with EcoRI and HindIII will produce 2.1 kb and 3.7 kb fragmens.

Ans. (c)

Sol. Double digestion with EcoR1 and Hind3 will produce 1.7kb and 4.1kb fragments is incorrect statement as it will produce 2.1kb and 3.7kb fragments.

139. Given below are some of the methods used to assess evolutionary phylogenetic relationships among plant taxa.

P. 16S rRNA sequence

Q. Mitochondrial microsatellite

R. Biochemical characterization

S. Morphology

Which two of the above methods can best reveal the evolutionary phylogenetic relationships?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (a)

Sol. The 16S rRNA gene is used extensively in bacterial phylogenetics, in species delineation, and now widely in microbiome studies. However, the gene suffers from intragenomic heterogeneity, and reports of recombination and an unreliable phylogenetic signal are accumulating. These DNA sequences are typically non-coding. The number of repeated segments within a microsatellite sequence often varies among people, which makes them useful as polymorphic markers for studying inheritance patterns in families or for creating a DNA fingerprint from crime scene samples.

140. The hydrogen atoms in the (delta) methylene group of lysine will give the following splitting pattern in the ¹HNMR spectra of lysine.

(a) Triplet of triplets

(b) Quintet

(d) riplet of a doublet

Ans. (a)

Sol. The hydrogen atoms in the (delta) methylene group of lysine will give Triplet of triplets splitting pattern in the ¹HNMR spectra of lysine.

141. Combination of molecular markers with their classification based on either dominant or co-dominant types are shown below:

P. SSR and RFLP: co-dominant

Q. SSR and RAPD: co-dominant

R. RAPD and RFLP: dominant

S. AFLP and RAPD: dominant.

Which one of the following is the correct combination?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (d)

Sol. Dominant :

RAPD : Randomly Amplified Polymorphic DNA; a molecular marker based on the differential PCR amplification of a sample of DNAs from short oligonucleotide sequences.

AFLP : Amplified fragment Length Polymorphism; a molecular marker generated by a combination of restriction digestion and PCR amplification.

Co-dominant :

RFLP : Restriction Fragment Length Polymorphism; a molecular marker based on the differential hybridization of cloned DNA to DNA fragments in a sample of restriction enzyme digested DNAs; the marker is specific to a single clone/restriction enzyme combination.

Comparison of the most broadly used marker systems

142. Cytochrome-c has only one tryptophan residue (W) which is buried. The protein in cacodylate buffer (pH 6.0) is excited at 280 nm, and its emission spectrum measured in the range of 300-450 nm. The same measurement was repeated on the protein in the buffer containing 6M guanidine hydrochloride. It was observed that there is an increase in the intensity of the emission spectrum of the guanidine hydrochloride treated cytochrome-c. The most probable reason for this increase is

(a) W is near a hydrophobic patch present in the unfolded protein

(b) W is near the heme in the netive protein

(d) W is in a polar pocket in the native protein

Ans. (b)

Sol. Increase in emission spectrum of guanidine hydrochloride treated cytochrome c is because tryptophan is near heme in native protein.

143. Cytochrome-c has only one tryptophan residue (W) which is buried. The protein in cacodylate buffer (pH 6.0) is excited at 280 nm, and its emission spectrum measured in the range of 300-450 nm. The same measurement was repeated on the protein in the buffer containing 6M guanidine hydrochloride. It was observed that there is an increase in the intensity of the emission spectrum of the guanidine hydrochloride treated cytochrome-c. The most probable reason for this increase is

(a) W is near a hydrophobic patch present in the unfolded protein

(b) W is near the heme in the netive protein

(d) W is in a polar pocket in the native protein

Ans. (c)

Sol. Group A : When individuals with intermediate phenotypes are favoured and extreme phenotypes are selected against, the selection is said to be stabilizing.

Group B : Disruptive selection, also called diversifying selection, describes changes in population genetics in which extreme values for a trait are favored over intermediate values. In this case, the variance of the trait increases and the population is divided into two distinct groups.

Group C : Directional selection occurs when individuals homozygous for one allele have a fitness greater than that of individuals with other genotypes and individuals homozygous for the other allele have a fitness less than that of individuals with other genotypes.

144. A protein is composed of leucine, isoleucine, alanine, glycine, proline, one lysine, one arginine and two cysteines connected by a disulfide bond. Conformational analysis indicates that the protein has elements of helix and beta structure. The protein is most likely

(a) a non-specific protease

(b) not an enzyme

(d) a filppase

Ans. (b)

Sol. Enzymes are proteins comprised of amino acids linked together in one or more polypeptide chains. This sequence of amino acids in a polypeptide chain is called the primary structure. This, in turn, determines the three-dimensional structure of the enzyme, including the shape of the active site. Chemically, enzymes are generally globular proteins. (Some RNA molecules called ribozymes can also be enzymes. These are usually found in the nuclear region of cells and catalyze the splitting of RNA molecules). Enzymes are catalysts that breakdown or synthesize more complex chemical compounds.

145. A 6.4 kb plasmid DNA has two restriction endonuclease sites, HindIII and EcoRI. Complete double digestion of the plasmid with both the enztymes yields two fragments of 3.1 and 3.3 kb. In order to study DNA repair process, a G:T mismatch was introduced in one strand of HindIII site and the damaged plasmid was incubated in a reconstituted repair system containing all the factors and enzymes required for repair. If the efficiency of the repair system is 50%, which one of the following band patterns on agarose gel will be obtained after treating the repaired plasmid with both HindIII and EcoRI?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. D represent correct band pattern of agarose gel obtained after G:T mismatch in Hind3 site. One thick band will be present of complete plasmid DNA.

CSIR NET BIOLOGY (JUNE - 2014)
Previous Year Question Paper with Solution.

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CSIR NET BIOLOGY (JUNE - 2014)
Previous Year Question Paper with Solution.