PYQ JUNE 2015 - VedPrep

CSIR NET BIOLOGY (JUNE - 2015)
Previous Year Question Paper with Solution.

21. In an alpha helical polypeptide, the backbone hydrogen bonds are between

(a) NH of n and CO of n + 4 amino acids

(b) CO of n and NH of n + 3 amino acids

(d) NH of n and CO of n + 3 amino acids

Ans. (c)

Sol. A helical state is one in which the phi and psi were roughly –60°, twisting repeatedly in the same direction.

The helical form models could be built with varying degrees of twist, but one model fit the atomic dimensions especially well.

Alpha helix : has 3.6 amino acids per turn of the helix, which places the C = O group of amino acid #1 exactly in the line with the H-N group of amino acid acid #5 (and C = O #2 with H-N #6).

So, for the C = O group of amino acid # n exactly in the line with H-N group of amino acid # (n + 4).

22. A 1% (w/v) solution of a sugar polymer is digested by an enzyme (20 µ.g, MW = 200, 000). The rate of monomer sugar (MW = 400) liberated was determined to have a maximal initial velocity of 10 mg formed/min. The turnover number (min^–1) will be:

(a) 5 × 10⁴

(b) 2.5 × 10^–2

(d) 2.5 × 10⁵

Ans. (d)

Sol.

23. Following are three single stranded DNA sequences that form secondary structures.

P. A T T G A G C G A T C A A T

Q. A T T G A G C G A T A T C A A T

R. A G G G A G C G A T C C C T

Based on their stability, which one is correct?

(a) P = Q = R

(b) R > P > Q

(d) Q > R > P

Ans. (b)

Sol. Due to presence of maximum palindromic sequence in R , ssDNA will form stable secondary structure.

24. In a 30-residue peptide, the dihedral angles have been determined by one or more methods. When their values are examined in the Ramchandran plot, it is

(a) not possible for values to be distributed in the helical as well as beta sheet region.

(b) possible that the values are all in the helical region although circular dichroism spectral studies indicate beta sheet conformation.

(d) not possible to conclude if the peptide is entirely helical or entirely in beta sheet conformation.

Ans. (c)

Sol. A Ramachandran plot is a way to visualize backbone dihedral angles psi (Ψ) against phi (Φ) of amino acid residues in protein structure. Proteins and most naturally occurring peptides are composed of amino acids in the L-configuration. However, D-amino acids have been detected in a variety of peptides synthesized in animal cells. In a 30-residues peptide it is possible to conclude than the peptide is composed of entirely D-amino acids.

25. Hydrogen bonds in proteins occur when two electronegative atoms compete for the same hydrogen atom Donor –H ......... Acceptor

The angle between donor and acceptor of a hydrogen bond was determined from large number of X-ray structures of proteins, as shown below:

Which one of the distribution of 'q' was observed from the proteins?

(a) Only Q

(b) Only P

(d) P and Q

Ans. (c)

Sol. R shows the correct 'q' distribution due to increase in number of hydrogen bonds with increase in theta angle.

26. In the accompanying figure, reaction kinetics of three proteins (a, b, c) is presented. Protein concentrations used to obtain this data are a – 1 mg/ml; b – 4 mg/ml and c – 2 mg/ml. If cataytic efficiency is defined as k_cat/K_m'

Which of the following statements is correct?

(a) b > c > a

(b) a > b > c

(d) c > a > b

Ans. (b)

Sol. These products then become involved in some other biological pathway that initiates certain functions of the human body. This is known as the catalytic efficiency of enzymes, which, by increasing the rates, results in a more efficient chemical reaction within a biological system. So, b will be the correct option.

27. When bacteria growing at 20°C are warmed at 37°C, they are most likely to synthesize membrane lipids with more

(a) short chain saturated fatty acids

(b) short chain unsaturated fatty acids

(d) long chain unsaturated fatty acids

Ans. (c)

Sol. At higher temperatures, the membrane needs to be made of fatty acids that are more viscous all fatty acids becomes less viscous at higher temps. Saturated long chain fatty acids are more viscous than unsaturated short chain fatty acids.

28. Cystic fibrosis transmembrane conductance regulator (CFTR) is known to control the transport of which ion?

(a) Ca⁺²

(b) Mg⁺²

(c)

(d) Cl^–

Ans. (d)

Sol. In the lung, the CFTR ion channel moves chloride ions from inside the cell to outside the cell. To get out of the cell, the chloride ions move through the center of the tube formed by the CFTR protein.

29. Beating of cilia is regulated by

(a) actin

(b) myosin

(d) nexin

Ans. (d)

Sol. Nexin was first identified in electron microscope images of axonemes as a linker that repeats every 96 nm between adjacent outer doublet microtubules. The nexin links are thought to be protease-sensitive structures that limit microtubule sliding during the ciliary beat cycle.

30. Given below are some statements about prokaryotic and eukaryotic mobile genetic elements or transponsons.

P. Most mobile genetic elements in bacteria transpose via an RNA intermediate.

Q. Most mobile genetic elements in bacteria are DNA

R. Mobile genetic elements eukaryotes are only retrotransposons.

S. Both, RNA and DNA transposons, are found in eukaryotes.

(a) P and R

(b) Q and R

(d) Q and S

Ans. (d)

Sol. Most mobile elements in bacteria transpose directly as DNA. In contrast, most mobile elements in eukaryotes are retrotransposons, but some eukaryotic transposons have been identified.

Transposons, also called transposable elements or jumping genes, are stretches of deo-xyribonucleic acid (DNA) that can move around an organism's chromosome. These "transpositions" occur at a very low frequency. A transposon can contain one gene or a set of genes and transposons are found in both eukaryotes and prokaryotes.

Prokaryote transposons may replicate DNA as well as cut and paste it. Transposons in eukaryotes do not replicate DNA. They move either by cutting and pasting or by creating a ribonucleic and (RNA) intermediate.

31. They key determinant of the plane of cytokinesis in mammalian cells is the position of

(a) chromosomes

(b) central spindle

(d) pre-prophase band

Ans. (b)

Sol. With contracitle ring assembly during prophase, a microtubule based structure termed the central spindle (or spindle midzone) forms when non-kinetochore microtubule fibres are bundled between the spindle poles. A number of different species including H. sapiens, D. melanogaster and C. elegans require the central spindle in order to efficiently undergo cytokinesis, although the specific phenotype described when it is absent varies from one species to the next. Seemingly vital for the formation of the central spindle and therefore efficient cytokinesis is a heterotetrameric protein complex called cent-ralspindlin.

Along with associated factors such as SPD-1 in C. elegans, central spindlin plays a role in bundling microtubules to form the spindle midzone during anaphase.

32. The mammalian oocyte prior to sperm entry is arrested at what stage of cell division?

(a) Prophase of mitosis

(b) Prophase of meiosis I

(d) Metaphase of meiosis II

Ans. (d)

Sol. Prior to fertilization, the meiotic cell cycle of the mammalian oocyte is arrested at metaphase II because of the prsence of active maturation promoting fator (MPF). MPF is a cell cycle modulator comprised of two subunits p34^cdc2 kinase and cyclin B1 and is responsible for inducing spindle assembly, chromatin condensation and nuclear envelope breakdown. During metaphase II arrest, MPF remains active due to the presence of 'cytostatic factor' which is comprised at least in part of the proteins Mos and Emil which prevent the degradation of cyclin B1. Shortly after sperm-oocyte fusion, elevation degradation and Mos undergoes uniquitin-dependent degradation by the proteasome. The resulting loss of active cytostatic factor and the inactivation of MPF allows entry of the female chromatin into anaphase II.

33. Which one of the following combinations must be present in a steroid receptor that is located in the cytoplasm?

(a) Nuclear Export Sequence (NES), leucine zipper

(b) NES, zinc finger motif

(d) Nuclear localization sequence (NLS), zinc finger motif

Ans. (d)

Sol. Steroid hormone receptors are found on the plasma membrane, in the cytosol and also in the nucleus of target cells. They are generally intracellular receptors (typically cytoplasmic) and initiate signal transduction for steroid hormones which lead to changees in gene expression over a time period of hours to days.

Steroid receptors of the nuclear receptor family are all transcription factors. Depending upon the tyre of receptor, they are either located in the cytosol and move to the cell nucleus upon activation or remain in the nucleus waiting for the steroid hormone to enter and activation them. This uptake into the nucleus is facilitated bynuclear localization signal (NLS) found in the hinge region of the receptor. A combinations of nuclear localization sequnces (NLS) and zinc finger motif is lcoated in the cytoplasm.

Secondary and teritary structure is distinct from that of classic zinc fingers.

34. Glycolipids and sphingomyelin are produced by the addition of sugars or phosphorylcholine to ceramide on cytosolic and luminal surfaces, respectively, of the Golgi apparatus. Finally, after such modifications, these molecules are located on the outer half of the plasma membrane. What key events are responsible for such localization?

(a) Membrane fusion only

(b) Action of Flippase and membrane fusion.

(d) Flip flop of these molecules in the Golgi membrane catalyzed by proton pump.

Ans. (b)

Sol. Flippases involved in creating membrane asymmetry are type-IV P-type ATPases (P4-ATPases) that catalyze the movement of specific phospholipid species from the extracellular leaflet to the cytosolic leaflet, whereas floppases are ABC-transporters that mediate the movement of phospholipids in the reverse direction. Membrane fusion is the process by which two initially separated lipid bilayers merge to form a single unity. It is a universal biological process in life, that is involved in many cellular events, e.g., in viral infection, fertilization, and intracellular trafficking.

35. When one isolates ribosomes from bacterial lysate, apart from 70S,50S and 30S ribosomal subunits, one also finds a small population of 100S, 130S and 150S sub-units. EDTA dissociates these larger ribosomal subunits into 50S and 30S, suggesting that they have both the subunits. Upon addition of cations, they reassociate into 70S, but none of the other forms could be detected. What is the reason for not obtaining the >70S forms?

(a) The effects of EDTA cannot be reversed by the addition of cations.

(b) 100S, 130S and 150S are modified form of ribosome that are irreversibly damaged by EDTA.

(d) They are obtained as an experimental artifact in preparations of ribosomes.

Ans. (c)

Sol. Statement c is correct.

36. Rec 8 is a meiosis specific cohesion that maintains centromeric cohesion between sister chromatids in meiosis I.

Which of the phenotypes listed below would you predict will be manifested in a rec8^– yeast?

(a) Only low viablility of dyads.

(b) Improper reduction division and low viability of tetrads.

(d) Low tetrad viability with no effect on reduction division.

Ans. (b)

Sol. Rec8 is a prominent component of the meiotic prophase chromosome axis that mediates sister chromatid cohesion, homologous recombination and chromosome synapsis. A detailed segregational analysis in the rec8-mutant revealed more spores disomic for chromosome III than in a wild-type strain. Aberrant segregations are caused by precocious segregation of sister chromatids at meiosis I, rather than by nondisjunction as a consequence of lack of crossovers. In situ hybridization further showed that the sister chromatids are separated prematurely during meiotic prophase. Moreover, the mutant forms aberrant linear elements and shows a shortened meiotic prophase.

37. When circular plasmids having a centromere sequence are transformed into yeast cells, they replicate and segregate in each cell division. However, if a linear chromosome is generated by cutting the plasmid at a single site with a restriction of the instability of the chromosome ends. What could be done so as to restore its stability and can be inherited?

(a) Methylation of adenine residues of the plasmid

(b) Complexing the plasmid ends with histone proteins.

(d) By incorporating acetylated histone proteins to the plasmid ends.

Ans. (c)

Sol. Recombination events occurring between different chromosome ends could lead to the dispersal and amplification of telomere-adjacent sequences.

38. Lipid rafts are involved in signal transduction in cells. Rafts have composition different from rest of the membrane. Rafts were isolated and found to have cholesterol to spingolipid ratio of 2 : 1. The estimated size of the raft is 35 nm². If the surface areas of cholesterol is 40Å², and sphingolipid is 60 Å², how many cholesterol and sphingolipids are present in one raft?

(a) 50 cholesterol : 25 sphingolipid

(b) 200 cholesterol : 100 sphingolipid

(d) 20 cholesterol : 10 sphingolipid

Ans. (a)

Sol. Ratio of cholesterol to sphingolipid = , so, C = 2S. Total size of raft = 35 nm².

Surface area of cholesterol = 40 Å = 40 × 10^–20 m² and surface area of sphingolipid = 60 Å² = 60 × 10^–20 m²

Estimated size of the raft = 35 nm² = 35 × 10^–18 m²

Surface area of cholesterol + surface area of sphingolipid = Estimated size of the raft

C × 40 × 10^–20 + S × 60 × 10^–20 = 35 × 10^–18

40C + 60S = 3500, solving this equation with the help of C = 2S, we have

Sphingolipid (S) = 3500/140 = 25 and cholesterol (C) = 2S = 2 × 25 = 50.

39. In an attempt to study the transport of secretory vesicles containing insulin along microtubules in cultured pancreatic cells, how would treatment with "colcemid" affect the transport of these vesicles?

(a) Colcemid induces polymerization of microtubules, which in turn would activate vesicular transport.

(b) Polymerization of microtubules is inhibited by colcemid, which in turn would inhibit the transport of secretory vesicles.

(d) Colcemid activates t-SNARE proteins and in turn activates vesicular transport.

Ans. (b)

Sol. Colcemid depolymerises microtubules, limits microtubules formation and inactivate the spindle fiber mechanism during metaphase. Colcemid binds tubulin rapidly in comparison to colchicine which binds tubulin relatively slowly.

40. Chromatin condensation is driven by protein complexes called condensins which are members of a family of "structural maintenance of chromatin" (SMC) proteins that play a key role in the organization of eukaryotic chromosomes. Condensins along with another family of SMC proteins called cohesins significantly contribute to chromosome segregation during mitosis. If the cells are treated with an inhibitor of cdk1 phosphorylation immediately before the cells enter M phase, which of the following statements is most likely to be true?

(a) Sister chromatids are held together by condensins along the entire length of the chromosome.

(b) Sister chromatids are held together by cohesins along the entire length of the chromosome.

(d) Sister chromatids are held together by condensins and attached to each other only at the telomere.

Ans. (b)

Sol. Condensin regulates the number and distribution of double-strand breaks and crossovers, whereas cohesin is essential for the assembly of a structure called the axial element, which forms on meiotic chromosomes and is important for proper association of homologous chromosomes and for crossover recombination. Cyclin-dependent kinase 1 (Cdk1) is an archetypical kinase and a central regulator that drives cells through G2 phase and mitosis.

41. In bacteria, heat-shock response is primarily controlled by

(a)

(b)

(c)

(d)

Ans. (b)

Sol. The heat shock response control levels of chaperones and protesses to ensure a proper cellular environment for protein folding.

The first alternate sigma factor which plays a crucial role in regulating heat shock genes has been identified in E.coli and called sigma 32 (), this sigma factors is encoded by the rpoH gene and complexes with the RNAP core (E) to constitute E. E recognizes the classical heat shock promoters located upstream of all the heat shock genes belonging to the sigma-32 regulon such as dnaK, groEL and ion.

42. During each cycle of chain elongation in translation, how many conformational changes does the ribosome undergo that are coupled to GTP hydrolysis?

(a) Zero

(b) One

(d) Three

Ans. (c)

Sol. Translocations is one of the key events in translocation, requireing large-scale conformational changes in the ribosome, movements of two transfer RNAs (tRNAs) across a distance of more than 20 Å and the coupled movement of the messenger RNA (mRNA) by one codon, completing one cycle of peptide chain elongation. Translocation is catalyzed by elongation factor G (EF-G in bacteria), which hydrolyzes GTP in the process.

43. In type II splicing

(a) a `G-OH´ from outside makes a nucleophilic attack 5´–P of first base of intron.

(b) a free 2´–O of an internal adenosine makes a nuclephilic attack on 5´–P of first base of intron.

(d) the hydrolysis of last base of exon is carried out by U2/U4/U6.

Ans. (b)

Sol. The 3' to 5' proof-reading exonuclease works by scanning along directly behind as the DNA polymerase adds new nucleotides to the growing strand. If the last nucleotide added is mismatched, then the entire replication holoenzyme backs up, removes the last incorrect base and attempts to add the correct base again. The enzyme is "3' to 5'" because it scans in the opposite direction of DNA replication, which we learned must always be 5' to 3'.

44. Copying errors occuring during replication are corrected by the proof reading activity of DNA polymerase that recognize incorrect bases

(a) at the 5´ end of the growing chain and remove them by 5´–3´ exonuclease activity.

(b) at the 3´ end of the growing chain and remove them by 5´–3´ exonuclease activity.

(d) at the 5´ end of the growing chain and remove them by 5´–3´ exonuclease activity.

Ans. (c)

45. Hybrid dysgenesis in Drosophila is caused by P-elements. Which one of the following crosses between different cytotypes will lead to dysgenesis?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. P element movement within the fly is a maternally inherited trait. Movement is suppressed in flies with the P cytotype, but is permissible in the M cytotype. Furthermore, P-cytotype flies contain P elements, whereas M-cytotype elements do not. When movements does occur it happens in the germ line, that is those cell which produce the gametes. This movement of P elements creates many different changes in phenotype, but few are fatal. The progeny are described as being dysgenic which refers to biological deficiencies that are a result of the P element movement. Movement is supperssed in somatic cells where the effects would be more damaging. This condition of germ line abnormalities which generates mutations, chromosomal breakage and sterility is called P-M hybrid dysgenesis and is mediated by P element movement.

46. What will happen if DNA is labeled by nick translation while doing DNA foot-printing?

(a) Nick translation will facilitate better analysis because entire DNA will be labeled and proteins binding at any region of DNA can be demaracted with precision.

(b) This will allow arranging the DNA fragment in the desired order.

(d) The linear order of fragments from 5´ 3´ end of DNA cannot be arranged.

Ans. (d)

Sol. If DNA is labeled by nick translation while doing DNA foot-printing, the linear order of fragments from 5´ 3´ end of DNA cannot be arranged.

47. Eukaryotic DNA polymerase a has tightly associated primase activity but moderate processivity. DNA polymerase and are highly processive but lack primase activity. Given below are four statements about leading and lagging strand synthesis in eukaryotes. Which one is true?

(a) Both leading and lagging strands are synthesized by DNA polymerase . Moderate processivity is essential to maintain fidelity of replication.

(b) Entire leading and lagging strands are synthesized by and . Eukaryotic replication is primer independent process.

(d) Primers for both the strands are synthesized by DNA polymerase followed by "polymerase switching" with and .

Ans. (d)

Sol. In leading and lagging strength synthesis in eukaryotes, primers for both the strands are synthesized by DNA polymerase followed by "polymerase switching" with and .

48. In context to lac operon, if two bacterial strains P₁ and P₂ with the genotypes O^CI⁺Z^– and O⁺I^–Z⁺ respectivly, were used to produce merodiploid daughter strain D, which one of the following statements correctly predicts the expression of Z gene (-galactosidase activity) in all the three strains? (O⁺, I⁺ and Z⁺ denote the wild type allele of the respective genes)

(a) P₁-No expression; P₂- Constitutive expression; D-Inducible expression

(b) P₁-No expression; P₂- Constitutive expression; D-constitutive expression

(d) P₁-Inducible expression; P₂- Constitutive expression; D-Inducible expression

Ans. (a)

Sol.

49. The 3´ end of most eukaryotic mRNAs is defined by the addition of a polyA tail - a processing reaction called polyadenylation. The addition of polyA tail is carried out by the enzyme Poly(A) polymerase. Given below are few statements about this process:

1. Poly(A) polymerase is a template independent enzyme.

2. Poly(A) polymerase catalyses the addition of AMP from dATP to the 3´ end of mRNA.

3. Poly(A) polymerase is a RNA-template dependent enzyme.

4. Poly(A) polymerase catalyzes the addition of ADP from ATP to the 3´ end of mRNA.

5. Poly(A) polymerase catalyzes the addition of AMP from ATP to the 3´ end of mRNA.

6. Poly(A) polymerase catalyzes the addition of AMP from dADP to the 3´ end of mRNA.

Which of the following combination is true?

(a) 2 and 3

(b) 3 and 4

(d) 3 and 6

Ans. (c)

Sol. During poly A polymerisation at 3' end of mRNA poly A polymerase is a template independent enzyme and Poly A polymerase catalyzes the addition of AMP from ATP to the 3´ end of mRNA.

50. With an intention to identify the genes expressed in an organism at specific stage of development, mRNAs were isolated from the given organism, cDNAs were synthesized, cloned in a suitable vector and sequence. A few of the cDNA sequences showed no matches with the genomic DNA sequence. Further, it was observed that these sequences were U-rich and found to be instretches dispersed along the sequence. The following may bepossible reasons for appearance of such RNA:

1. Splicing

2. Alternate splicing

3. Trans-splicing

4. Guide RNA mediated introduction of Us involving endonuclease, terminal-U transferase and RNA ligase

5. Deaminations converting C to U

Which of the following is the most appropriate reason/s?

(a) 1 and 3

(b) 2 and 4

(d) 4 only

Ans. (d)

Sol. cDNA sequences may become unmatched A-T rich when guide RNA mediated introduction of Us involved endonuclease, terminal-U transferase and RNA ligase in mRNA.

51. In order to study the transcription factor TFIIH, it was cloned from a large number of human subjects. Surprisingly, the subjects having mutation in TFIIH, also showed defects in their DNA repair system. Given below are the explanations:

P. DNA damage is always associated with transcription inhibition.

Q. TFIIH has no role in DNA repair

R. In mammalian system, TFIIH plays an active role in transcription coupled DNA repair process.

S. Because of mutation in TFIIH, transcription initiation is inhibited and incompletely synthesized mRNAs remain attached to the template DNA leading to DNA damage.

Choose the correct answer.

(a) P and Q

(b) R only

(d) S only

Ans. (b)

Sol. In mammalian system, TFIIH plays an active role in transcription coupled DNA repair process.

52. Sting of a bee causes pain, redness and swelling. Melittin is a major peptide in bee venom. Melittin is a membrane binding peptide that is involved in activating phospholipases in the membrane. The possible target phospholipase that is activated by melittin is

(a) Phospholipase C to generate inositol phosphates.

(b) Phospholipase A₂ to generate arachidonic acid.

(d) Phospholipase A₁ to generate palmitic acid.

Ans. (b)

Sol. Phospholipase A2 comprises 10-12% of peptides and it is the most destructive component of apitoxin. It is an enzyme which degrades the phospholipids which cellular membranes are made of. It also causes decreased blood pressure and inhibits blood coagulation. Phospholipase A2 activates arachidonic acid which is metabolized in the cyclooxygenase-cycle to form prostaglandins. Prostaglandins regulate the body's inflammatory response.

53. What phenotype would you predict for a mutant mouse lacking one of the genes required for site-specific recombination inlymphocytes?

(a) Decrease in T cell counts

(b) Immunodeficient

(d) Increase in B cell counts

Ans. (b)

Sol. The mouse would be immunodeficient, lacking both B and T lymphocytes as a result of being unable to rearrange its immunoglobulin and T cell receptor genes.

54. Which one of the following events never activates the G-protein coupled receptor for sequestering Ca²⁺ release?

(a) Interaction of binding to sperm receptors.

(b) Activation of Frizzled by Wnt.

(d) DNA synthesis and nuclear envelope breakdown

Ans. (d)

Sol. GPCR is inhibited during DNA synthesis breakdown and nuclear envelope breakdown.

55. The main difference between normal and transformed cells are

(a) Immortality and contact inhibition

(b) apoptosis and tumour suppressor gene hyperfunction

(d) inactivation of oncogenes and shorter cell cycle duration

Ans. (a)

Sol. Cancer cells are sometimes described as transformed cells, because they can be created from normal cells. Once transformed, cancer cells display distinct growth properties. They often have markedly decreased requirements for external growth factors. Transformed cells, unlike normal cells, lack contact inhibition and continue to crowd, eventually piling up on each other. Normal cells usually will not grow unless they are attached to a firm surface (such as a petri dish). However, transformed cells are often anchor-age independent; that is, they continue to divide even when suspended in a soft agar gel. Normal cells have a limited life span in the laboratory; they many divide in a petri dish 10 or 50 times, but then they cease growing. Transformed cells usually are immortal in that they seem to have an unlimited life span and will continue to divide for years under appropriate conditions.

56. Collagens are the most abundant component of the extracellular matrix. In order to maintain normal physiological processes like wound healing, bond development, etc. Which one of the following type of enzymes is most important?

(a) Peptidases

(b) Proteases

(d) Lipases

Ans. (b)

Sol. Matrix metalloproteinases (MMPs) are members of the metzincin group of proteases which share the conserved zinc-binding motif in their catalytic active site. Due to the broad spectrum of their substance specificity. MMPs contribute to the homeostasis of many tissues and pariticipate in several physiological processes, such as bone remodelling, angiogenesis, immunity and wound healing. MMP activity is tightly controlled at the level of transcription, pro-peptide activation and inhibition by tissue inhibitors of MMPs.

57. A researcher would like to monitor changes in the level of a serum protein for which an antibody is available.

Which one of the following methods would be best suited for the purpose?

(a) Immunofluorescence microscopy

(b) Fluorescence in situ hybridization

(d) Fluorescence activated cell sorting

Ans. (c)

Sol. Enzyme linked immunosorbent assays (ELISAs) have served as important methods for clinical protein determinations. Detection limits (DLs) approach 1 kpg mL^–1 for some protein biomarkers, but ELISA is difficult to develop for point of care use and requires significant technical expertise. Classical ELISA methods suffer limitations in analysis time, sample size, equipment cost and measuring collections of proteins.

58. Certain chemokines are known to suppress HIV infection whereas proinflammatory cytokines are known to enhance infection. In order to explain these findings, control and chemokine receptor knock-out animals were treated with proinflammatory cytokines followed by HIV administration and then was assessed periodically. Which one of the graphical representation given below best explain the experimental results.

_________ control

.................. Chemokine receptor knockout animals

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Chemokine receptor knockout animals showed less viral load in blood than control ones.

59. A patient with breast cancer was given a dose of radiation along with chemotheraphy and was apparently cured of the tumor. After five years, a tumor was noticed in the patient's lungs, but the doctors confirmed that it was derived from cells of the mammary gland. The following possibilities were suggested by the doctor.

P. Bacterial infection, after radiation, led to development of the tumors in the lungs.

Q. Migration of residual chemo-resistant cells from the mammary gland resulted in tumors in the lungs.

R. Epithelial-to-mesenchymal transition had occured in the lungs.

S. Cells in the lungs were induced to become a tumor after chemotherapy, and form factors secreted by mammary cells.

Which of the following is correct?

(a) Q and S

(b) Only Q

(d) P and R

Ans. (b)

Sol. Despite the best care and significant progress made in treatment, cancer comes back. When this happens it is called a recurrence or relapse. The likely relapse occurs is that a few of the original cancer cells survived the initial treatment. When cancer cells resist the effects of drugs used for treatment, they can grow and reform tumors, a process known as recurrence or relapse. Sometimes resistance develops quickly, within a matter of weeks of starting treatment. In other cases, it develops months, or even years, later.

60. A researcher was studying a protein 'X' which has been observed to move across cells when an extracelular electrical stimulus is provided. An artificial peptide 'P' was prepared which resembles the structure of connexins and competitively inhibits connexon formation. Which one of the following statements will best explain the fate of protein 'X' if the cells are treated with peptide 'P' and the electrical stimulus is provided?

(a) X fails to move across cells due to improper formation of tight junctions.

(b) X fails to move across cells due to improper formation of gap junctions.

(d) X fails to move across cells due to improper formation of desmosomes.

Ans. (b)

Sol. Gap junctions allow the exchange of ions, second messengers, and small metabolites between adjacent cells and are formed by two unrelated protein families, the pannexins and connexins. Mutations in connexin genes cause a variety of genetic disorders, implicating a critical role in tissue homeostasis.

61. There are three substances A, B and C. Given below are the pattern of immunological responses in rabbits when (i) A is administered along with C, (ii) B is administered along with C and (iii) A is conjugated with B and administered along with C.

Which one of the following is the correct identification?

(a) A-protein, B-hapten, C-adjuvant

(b) A-hapten, B-protein, C-adjuvant

(d) A-hapten, B-adjuvant, C-protein

Ans. (b)

Sol. Hapten, also spelled haptene, small molecule that stimulates the production of antibody molecules only when conjugated to a larger molecule, called a carrier molecule. An adjuvant is an ingredient used in some vaccines that helps create a stronger immune response in people receiving the vaccine. In other words, adjuvants help vaccines work better.

62. Which of the following gives the correct human disease-causal microbe match for each?

Human disease Causal Microbe

P. Chronic gastritis 1. Borrelia burgdorferi

Q. Lyme disease 2. Helicobacter pylori

R. Scarlet fever 3. Rickettsia prowazekii

S. Typhus 4. Streptococcus pyogenes

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

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

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

Ans. (a)

Sol. Helicobacter pylori is a Gram-negative bacterium that selectively colonizes the human stomach. In 1983, Barry Marshall and Robin Warren first identified H. pylori within the gastric epithelium of patients with chronic gastritis.

Lyme disesase is a tick-transmitted infection casued by the spirochete Borrelia burgdorferi. B. burgdorferi enters the skin at the site of the tick bite. After 3 to 32 days, the organisms migrate locally in the skin the organisms migrate locally in the skin aroung the bite, spread via the lymphatics to cause regional adenopathy or disseminate in blood to organs or other skin sites.

Scarlet fever also called scarlathina, acute infectious disease caused by group A hemolytic streptococcal bacteria, in particular Streptococcus pyogenes. Scarlet fever can affect people of all ages, but it is most often seen in children. It is called scarlet fever because of the red skin rash that accompanies it.

Typhus is a bacterial disease spread by lice or flesas. Typhus is caused by 2 types of bacteria : Rickettsia typhi or Rickettsia prowazekii.

Rickettsia prowazekii causes epidemic typhus. It is spread by lice.

Brill-Zinsser disease is a mild form of epidermic typhus. It occurs when the bacteria becomes active again in a persome who was previously infected. It is more common in the elderly.

63. Hydra shows morphallactic regeneration and involves which one of the following signal transduction pathway in its axis formation?

(a) -catenin pathway

(b) Retinoic acid pathway

(d) Delta-Notch pathway

Ans. (a)

Sol. -Catenin signaling plays crucial roles in regenerative processes in eumetazoans. It also acts in regeneration and axial patterning in the simple freshwater polyp Hydra, whose morphallactic regenerative capacity is unparalleled in the animal kingdom.

64. The pluripotency of the inner cell mass in mammals is maintained by a core of three transcription factors namely,

(a) Oct 4, Sox 2 and Nanog

(b) Oct 4, Sox 2 and Cdx 2

(d) Oct 4, Cdx2 and Nanog

Ans. (a)

Sol. Pluriopotency is the central property of all ES cells. It refers to the ability to generate any type of cells in the body. Developmentally, zygotes are totipotent, capable of giving rise to a whole animal, including all the cell types.

The roles of Nanog, Oct-4 and Sox-2 have not been fully development and maintenance of pluripotency in ESCs.

65. Development of vulva in C. elegans is initiated by the induction of a small number of cells by short range signals from a single inducing cell. With reference to this, following statements were put forward.

P. When the anchor cell was ablated early in development, no vulva formed.

Q. In a dominant negative mutant of let-23, a primary vulva formed but the secondary vulva formation did not take place.

R. A cell adopting a primary fate inhibits adjacent cells from adopting the same fate by lateral inhibition involving LIN-39 and also induces the secondary fate in these cells.

S. A constitutive signal from the hypodermis inhibits the development of both the promary and secondary fates but it is overruled by the initial signal from the anchor cell.

Which of the above statement is true?

(a) P and Q

(b) P and R

(d) Q and S

Ans. (c)

Sol. The AC is the key organizer of vulval patterning and morphogenesis, and as a consequence, some mutations that affect the development of the AC also affect vulval development. Wild-type hermaphrodites have a single AC, generated after interactions between two equivalent cells. Statement P and S is correct.

66. What will happen if wingless RNAi is expressed in wingless expressing cells from the stage when this gene initiates its expression in a developing Drosophila embryo?

P. The enhanced expression of wingless thus caused will broaden the area of engrailed expression.

Q. Since wingless protein makes a long range gradient, its effect will not be seen in the same segment.

R. The posterior compartment of each future segment will get affected.

S. Since engrailed expression is initiated by pair rule genes, the posterior segment will not be affected.

Which one of the following will most appropriately answer the question?

(a) P and R

(b) Only R

(d) Only S

Ans. (b)

Sol. Its earliest identifiable role during development of Drosophila is in the embryonic segmentation cascade, wherein wingless functions as a segment polarity gene and serves to pattern each individual segment along the antero-posterior axis of the developing embryo. RNA interference (RNAi) is the process by which the expression of a target gene is effectively silenced or knocked down by the selective inactivation of its corresponding mRNA by double-stranded RNA (dsRNA). This will affect posterior compartment of future segment.

67. Formation of digits and sculpting the tetrapod limb requires death of specific cells in the limb in a programmed manner. Which one of the following interactions could explain proper limb formation?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The -catenin signaling pathway is an evolutionarily conserved mechanism that plays a preeminent role in maintaining cellular homeostasis. It regulates embryo development, cell proliferation and differentiation, apoptosis, and inflammation-associated cancer. Option a is correct.

68. Of the following signaling processes, which one is not involved in cellular movement or cytosekletal changes?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The MAPK/ERK pathway is a chain of proteins in the cell that communicates a signal from a receptor on the surface of the cell to the DNA in the nucleus of the cell.

69. Which one of the following is the correct combination?

(a) A – i, B – iv, C – ii

(b) A – iv, B – iii, C – i

(d) A – v, B – ii, C – iii

Ans. (c)

Sol. Option (c) is correct.

70. Which one of the following about development of sea urchhin embryos is true?

(a) Each blastomere of a 4-cell stage possesses a portion of the original anima-vegetal axis and if isolated and allowed to develop will form a complete but smaller sized larva.

(b) Each blastomere of a 8-cell stage has the capacity to form a complete embryo but by the 16-cell stage, blastomeres will develop according to their presumptive fate.

(d) After an intricate recombination at the 16-cell stage, the resulting embryo looses its ability to form a complete larva.

Ans. (a)

Sol. The four cells of the animal tier divide meridionally into eight blastomeres, each with the same volume. These cells are called mesomeres. The vegetal tier, however, undergoes an unequal equatorial cleavage to produce four large cells, the macromeres, and four smaller micromeres at the vegetal pole . If isolated and allowed to develop they will form a complete but small sized embryo.

71. Which one of the following compounds is generally translocated in the phloem?

(a) Sucrose

(b) D-Glucose

(d) D-Fructose

Ans. (a)

Sol. Photosynthates are produced in the mesophyll cells of photosynthesizing leaves. From there, they are translocated through the phloem where they are used or stored.

The sucrose is actively transported against its concentration gradient into the phloem cells using the electrochemical potential of the proton gradient. This is coupled to the uptake of sucrose with a carrier protein called the sucrose-H⁺ symporter.

72. Which one of the following statements about LEAFY (LEY), a regulatory gene in Arabidopsis thaliana, is correct?

(a) LEAFY (LEY) is involved in floral meristem identity.

(b) LEAFY (LEY) is involved in leaf expansion.

(d) LEAFY (LEY) is responsible for far-red light mediated seedling growth.

Ans. (a)

Sol. Flowering plants produce floral meristems in response to intrinsic and exitrinsic flowering inductive signals. In Arabidopsis, the floral meristem identify genes LEAFLY (LFY) and APETALA1 (AP1) are activated to play a pivotal role in specifying floral meristems during floral transition.

The first step in flower development is the generation of a floral meristem by the inflorescence meristem.

LEAFY interacts with another floral control gene, APETALA1, to promote the transition from inflorescence to floral meristem.

73. Nitrogen gas is reduced to ammonia by nitrogen fixation method. In order to execute the process, which one of the following compounds is usually required?

(a) ATP

(b) GTP

(d) ADP

Ans. (a)

Sol. Nitrate in the natural environment is relatively rare. Microbes capable of usnig alternative nitrogen sources have an advantage and a subset of microbes is capable of obtaining the nitrogen they need from nitrogen gas. Nitrogen gas makes up about 79% of our atomosphere and is easily available. Molecular nitrogen is a stable unreactive gas with a triple bond between the two atoms and the reduction of it to ammonia is an energy expensive process. A large amount of ATP, protons and electrons are required to reduce just one molecule of nitrogen gas.

N₂ + 8H⁺ + 8e^– + 16ATP 2NH₃ + H₂ + 16ADP + 16P_i

The enzyme that catalyzes the reaction is called nitrogenase and is made of two separate protein components, dinitrogenase reductase and dinitrogenase.

74. The quantum yield of oxygen evolution during photosyntheiss drastically drops in far-red light. This effect is known as:

(a) far red drop

(b) red drop

(d) visible spectrum drop

Ans. (b)

Sol. Red Drop Effect : The quantum yield of oxygen evolution dramatically drops for far-red light wavelength of greater than 680 nm.

75. Which one of the following enzymes is not a part of pyruvate dehydrogenase enzyme complex in glycolysis pathway?

(a) Pyruvate dehydrogenase

(b) Dihydrolipoyl transferase

(d) Dihydrolipoyl oxidase

Ans. (d)

Sol. Pyruvate dehydrogenase is a multi-enzyme complex that uses three enzymes :

1. E₁ : Pyruvate dehydrogenase which uses thiamine pyrophosphate (TPP) as its prosthetic group.

2. E₂ : Dihydrolipoyl transacetylase which uses lipoamide and coenzyme A (also known as coASH) as its prosthetic groups.

3. E₃ : Dihydrolipoyl dehydrogenase which uses flavin adenine dinucleotide (FAD) and nicotinamide adenine dinucleotide (NAD⁺) as its cofactors.

Note : Prosthetic groups are molecules that are covalently bonded to an enzyme.

The net reaction of converting pyruvate into acetyl coA and CO₂ is :

2 pyruvate + 2NAD⁺ + 2coA 2acetyl coA + 2ANDH + 2CO₂

76. Dark-growth seedling display 'triple response' when exposed to ethylene. Which one of the following is not a part of 'triple response'?

(a) Decrease in epicotyl elongation

(b) Rapid unfolding and expansion of leaves.

(d) Horizontal growth of epicotyl

Ans. (b)

Sol. The "triple response" of dark-grown seedlings is a common bioassya for ethylene. Reduced epicotyl elongation and swelling by ethylene can also be induced in stems and petioles that are made to grow rapidly, either by growing them, horizontally in the dark or treating them with hormones, such as GA. The triple response is characterized by the inhibiton of hypocotyl and root elongation, a thickening of hypo-cotyl and an exaggerated apical hook.

77. Following are certain statements regarding seed development in plants:

P. During final phase of development, embryos of "orthodox" seeds become tolerant todesiccation, dehydrate losing up to 90% of water.

Q. Dormant seeds will germinate upon rehydration while quiescent seeds require additional treatments or signals for germination.

R. Precocious germination is germination of seeds without passing through the normal quiescent and/or dormant stage of development.

S. Abscisic acid is known to inhibit precocious germination.

Which of the following combination is correct?

(a) P, Q and R

(b) P, Q and S

(d) P, R and S

Ans. (d)

Sol. In plants, vivipary (precocious or premature germination) involves the germination of seeds while still on the parent plant. It is a widespread phenomenon in plants characterized by the lack of seed dormancy. Abscisic acid inhibit precocious germination.

78. The glycolysis and citric acid cycles are important pathway to generate energy in the cell. Given below are statements regarding the production of ATP.

P. Electrons released during the oxidative steps of glycolysis and citric acid cycle produce 10 molecules of NADH and 2 molecules of FADH₂ per molecule of glucose

Q. Electrons released during the oxidative steps of glycolysis and citric acid cycle produce 20 molecules of NADH and 4 molecules of FADH₂ per molecule of glucose

R. The coenzymes produced are oxidized by electron transfer chain.

S. The conversion of ADP and P_i to ATP takes place in the intermembrane space of mitochondria.

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

(a) P and Q

(b) Q and R

(d) P and R

Ans. (d)

Sol. The oxidation of NADH and FADH2 proceeds in a highly organized system known as electron transport chain (ETC) which consists of complexes of proteins and coenzymes. The ultimate electron acceptor is oxygen, which is converted to water.

79. Symbiotic nitrogen fixation in legume nodules involves complex interaction between Rhizobium and legum roots. This complex interaction is governed by

P. Integration of sym plasmid of Rhizobium in the root nuclear genome.

Q. Sensing of plant flavonoids by rhizobia.

R. Activation of nod genes in rhizobia.

S. Activation of NODULIN genes in legume roots.

Which of the following combinations is correct?

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (c)

Sol. Legumes are able to form a symbiotic relationship with nitrogen-fixing soil bacteria called rhizobia. The result of this symbiosis is to form nodules on the plant root, within which the bacteria can convert atmospheric nitrogen into ammonia that can be used by the plant. Nod factors (nodulation factors or NF), are signaling molecules produced by soil bacteria known as rhizobia in response to flavonoid exudation from plants under nitrogen limited conditions. Nod factors initiate the establishment of a symbiotic relationship between legumes and rhizobia by inducing nodulation.

80. Pyruvate dehydrogenase is subject to feedback inhibition by its products in glycolysis. Some of the chemical compounds which might be involved in the process, are listed below:

P. NADH

Q. FAD

R. Acetyl-CoA

S. Acetaldehyde

Which one of the following combinations of above chemical compounds is involved in feedback inhibition of pyruvate dehydrogenase?

(a) P and Q

(b) Q and R

(d) P and R

Ans. (d)

Sol. In vitro, the pyruvate dehydrogenase complex is sensitive to product inhibition by NADH and acetyl-coenzyme A (CoA).

81. Light is an important factor for plant growth and development. There are several photoreceptors in higher plants such as Arabidopsis thaliana involved in perception of various wavelengths oflight. Some statements are given below related to photoreceptors:

P. Red light photoreceptors are represented by a gene family.

Q. Phytochrome C is the most prominent photoreceptor to perceive red light.

R. Cryptochrome 1 and cryptochrome 2 have evolved from bacterial DNA photolyases.

S. Far-red light perceived by phytochrome D.

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

(a) P and Q

(b) Q and R

(d) P and R

Ans. (d)

Sol. The amino acid sequences of the cryptochromes are more similar to those of the class I photolyases than the class II photolyases, although they lack any apparent DNA repair activity. Cryptochromes bind similar chromophores and absorb in the blue light region, like the photolyases.

82. Carbohydrates synthesized by photosynthesis are converted into sucrose and transported via phloem to other parts of the plant. The following aspects are associated with sucrose uploading in phloem and its transport:

P. Both reducing and non-reducing sugars are transported efficiently through phloem.

Q. Sucrose uploading can be both symplastic and apoplastic.

R. The route of phloem uploading is mesophyll cells phoem parenchyma companion cells sieve tubes.

S. Transport in sieve tubes is as per the 'pressure-flow model'.

Which one of the following combinations is correct?

(a) P, Q and R

(b) Q, R and S

(d) S, Q and P

Ans. (b)

Sol. Phloem loading of sucrose is a crucial step that drives long-distance transport by elevating hydrostatic pressure in the phloem. Three phloem loading strategies have been identified, two active mechanisms, apoplastic loading via sucrose transporters and symplastic polymer trapping, and one passive mechanism. In apoplast-loading species, sucrose reaches phloem parenchyma cells through plasmodesmata. Sucrose is loaded and accumulates in the phloem by passing through the apoplast between the PP and the CC.

83. A 'Z' scheme describes electron transport in O₂ -evolving photosynthetic organisms. The direction of electron flow is presented in the following sequences:

P. P680* Pheophytin Q_A Q_B PC cytochrome b₆f P700

Q. P700* Phylloquinone FeS_A FeS_B FeS_X Fd

R. P680* Pheophytin Q_A Q_B cytochrome b₆fPC P700

S. P700* Phylloquinone FeS_x FeS_A FeS_B Fd

Which one of the following combinations is correct?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (c)

Sol. In the Z-scheme, electrons are removed from water (to the left) and then donated to the lower (non-excited) oxidized form of P680. Absorption of a photon excites P680 to P680*, which "jumps" to a more actively reducing species. P680* donates its electron to the quinone-cytochrome bf chain, with proton pumping.

84. Following are certain statements related to plants exposed todehydration stress:

P. When the water potential of the rhizosphere decreases due to water deficit, plants can continue to absorb water as long as plant water potential is lower that of soil water.

Q. The ratio of root to shoot growth increases in response to water deficit.

R. Plant cells tend to release solutes to lower water potential during periods of osmotic stress.

S. Abscisic acid is synthesized at a higher rate when leaves are dehydrated, and more ABA accumulates in the leaf apoplast.

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

(a) P, Q and R

(b) Q, R and S

(d) P, R and S

Ans. (c)

Sol. Water-deficit stress can be defined as a situation in which plant water potential and turgor are reduced enough to interface with normal functions. In response to water deficit ABA induces stomatal closure and the expression of numerous stress-responsive genes. By contrast, ABA promotes the accumulation of seed storage compounds during seed development and is also required for the induction and maintenance of seed dormancy.

85. The S wave of normal human ECG originates due to

(a) septal and left ventricular depolarization.

(b) late depolarization of the ventricular walls moving back toward the AV junction.

(d) repolarization of atrium.

Ans. (b)

Sol. The S wave of normal human ECG originates due to late depolarization of the ventricular walls moving back toward the AV junction.

86. In which of the following conditions is Basal Rate (BMR) the lowest?

(a) Awake and resting

(b) Prolonged starvation

(d) Higher environmental temperature

Ans. (b)

Sol. The BMR refers to the amount of energy you body needs to maintain homeostasis. Your BMR is largely determined by your total lean mass, especially muscle mass, because lean mass requires a lot of energy to maintain. Anything that reduces lean mass will reduce your BMR.

An anverage man has a BMR of around 7,100 kJ per day, while an average woman has a BMR of around 5,900 kJ per day. Energy expenditure is continuous, but the rate varies throughout the day. The rate of energy expenditure is usually lower in the early morning and it is lowest while we sleeping.

87. Which one of the following skeletal muscles of human body contains highest number of muscle fibre in a motor unit?

(a) Muscles of hand

(b) Extraocular muscles

(d) Muscles of face

Ans. (c)

Sol. Motor unit consists of a motor neuron and the group of skeletal muscle fibers which it innervates.

The number of muscle fibers in a motor unit is also directly related to the size of its motor neuron. Small motor nerve fibers form small motor units and large motor nerve fibers form large motor units. Muscles of leg have highest number of muscle fibre in a motor unit.

88. Which one of the following viruses cause acute gastrointestinal illness due to combination of drinking water?

(a) Norovirus

(b) Poliovirus

(d) Filoviruses

Ans. (a, c)

Sol. Acute gastrointestnial illness (AGI) resulting from pathogens directly entering the piping of drinking water distribution system is insufficiently understood. Noroviruses (NVs) are a genus of the Caliciviridae family of viruses found in the 'used' water.

Norovirus cause a majority of the cases of foodborne gastroenteritis. Foodborne sources of gastrointestinal illness associated with noroviruses include cake frosting, drinking water, orange juice, salads, fresh produce, sandwiches, lettuce and European oysters. Waterborne outbreaks of norovirus-induced gastrenterities usually occur as a result of the contamination of well water and to a lesser extent, surface water by sewage.

89. An extraordinary sensory ability that elephants possess is :

(a) emission and detection of ultra high frequency sounds.

(b) emission and detection of ultra low frequency sounds.

(d) possession of ultraviolet vision

Ans. (b)

Sol. Elephants produce a broad range of sounds from very low frequency rumbles to higher frequency snorts, barks, roars, cries and other idiosyncratic calls. Asian elephants also produce chirps. The most frequently used category of calls, at least for African elephants, is the very low frequency rumble. Elephants can produce very gentle, soft sounds as well as extremely powerful sounds.

90. During physical exercise, the oxygen supply to the active muscles is increased, which has been explained by the following statements:

P. PO₂ declines and Pco₂ rises in the active muscles.

Q. The temperature is increased and pH is decreased in active muscles.

R. 2, 3-biphosphoglycerate is decreased in RBC and P₅₀ rises.

S. Metabolic accumulating in the active muscles increase the affinity of hemoglobin to oxygen.

Which one of the following combinations is not correct?

(a) P only

(b) P and Q

(d) R and S

Ans. (d)

Sol. During physical exercise, the oxygen supply to the active muscles is increased. This is achieved when 2, 3-biphosphoglycerate is decreased in RBC and P₅₀ rises and metabolic accumulating in the active muscles increase the affinity of hemoglobin to oxygen.

91. The fractional clearance of neutral and cationic dextran molecules of various sizes through kidneys of a rat is shown in the figures below. Which one of the following is correct?

__________ Neutral dextran ---------------- Cationic dextran

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Graph in option (b) correctly represents fractional apperance of neutral and cationic dextran molecules through kidney.

92. Estradiol synthesis follows a 2-cell-2-gonadotropin theory, where partial synthesis occurs in the granulosa cells and the rest in thetheca interna cells of the Graafian follicle. Which one of the following correctly represents estradiol synthesis secretion?

C — Cholesterol E — Estradiol

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Option (a) correctly represents estradiol synthesis and secretion.

93. The mechanism of sound localization in horizontal plane by the human auditory system can be explained by

P. The difference in time between the arrival of stimuus in two ears.

Q. The difference in phase of the sound waves on two ears.

R. The difference in tuning curves of two auditory nerves.

S. The activity of neurons in superior olivary nucleus, but not the neuronal activity of auditory cortex.

Which one of the following is not correct?

(a) P only

(b) P and Q

(d) R and S

Ans. (d)

Sol. Sound localization refers to a listener's ability to identify the location of origin of a detected sound in direction and distance. It may also refer to the methods inacoustical engineering to stimulate the placement of any auditiory cue in a virtual 3D space transient disparity. Any instantaneous difference between the arrival times of a sound wave at the two ears, including interaural differences in the onset time of the sound, the termination time and the time of any sudden change in pitch or intensity, but excluding ongoing or continuous phase delay. It is used in conjunction with phase delay as a cue for localizing sounds below about 1,000 hertz having wavelength greater than the diameter of the head rather than being reflected from or absorbed by it, thus preventing a sonic shadow from forming on the side furthest from the sound source.

94. In high altitude, hypoxia induces increased number of circulating red blood cells, which can be explained by the following changes:

P. The transcription factors, HIFs, are produced.

Q. Erythropoietin secretion is increased.

R. Myoglobin context is decreased.

S. Cytochrome oxidase is decreased.

Which one of the following is not correct?

(a) P only

(b) P and Q

(d) R and S

Ans. (d)

Sol. In high altitude, hypoxia induces increased number of circulating red blood cells but it does not require decrease in myoglobin content and cytochrome oxidase.

95. 'Segregation of alleles can occur at Anaphase I or at Anaphase Ii of meiosis'. With reference to this statement, which one of the following organism is an ideal model system for identifying stage of allelic segretation at meiosis?

(a) Neurospora crassa

(b) Saccharomyces cerevisiae

(d) Pisum sativum

Ans. (a)

Sol. Neurospora crassa is an ideal model system for identifying stage of allelic segretation at meiosis.

96. Genes A, B and C control three phenotypes which assort independently. A plant with the genotype Aa Bb Cc is selfed. What is the probability for progeny which shows the dominant phenotype for at least one of the phenotypes controlled by genes A, B and C?

(a) 1/64

(b) 27/64

(d) Cannot be predicted

Ans. (c)

Sol. Different types of gametes produced by AaBbCc :

The Aa produces two types of gametes A and a = 2 gametes

The Bb produces two types of gametes B and b = 2 gametes

The Cc produces two types of gametes C and c = 2 gametes

Total number of gametes = 2 × 2 × 2 = 8

It is given as the individual with AaBbCc genotype is selfed. Both parents will produce 8 gametes each and thus there will be 8 × 8 = 64 progenies. Among the 64 progenies one will be with homozygous dominant for all the allele (AABBCC) and one will be homozygous recessive for all the alleles (aabbcc). The rest of the individuals (62 numbers) will have at least one dominant character either in homozygous or heterozygous conditions. Probability for progeny which shows the dominant phenotype for at least one of the phenotypes controlled by genes A, B and C = 1 – 1/64 = 63/64.

97. In an experiment, clones of a plant is grown in a field. The plants were observed to be of different heights. When a graph was plotted for frequency of plants (Y-axis) against different heights (X-axis), a bell-shaped curve was obtained. From the above it can be concluded that the observed variation in height is due to

(a) it being a polygenic trait

(b) environmental effect

(d) influence of environmental on different genotypes

Ans. (b)

Sol. From the above it can be concluded that the observed variation in height is due to environmental effect.

98. A cross was made between Hfr met⁺ arg⁺ leu⁺ str^s × F^– met^– arg^– leu^– str^R, in which leu⁺ exconjugants are selected. If the linear organization of the genes are leu⁺ arg⁺ met⁺, which one of the following genotypes is expected to occur in the lowest frequency?

(a) leu⁺ arg^– met^–

(b) leu⁺ arg⁺ met^–

(d) leu⁺ arg^– met⁺

Ans. (d)

Sol. Leu⁺ and Met⁺ genes are far away possessing Arg⁺ in between them. Hence, Leu⁺ Arg^– Met⁺ would have lowest frequency.

99. Two homozygous individuals (P1 and P2) were genotyped using dominant DNA markers A and B, as shown below. The F1 progeny obtained was test crossed. The frequency of progeny with which different genotypes appear, is given below:

The following conclusions were made:

P. In the F1, markers A and B are linked and in coupling phase (cis).

Q. In the F1, markers A and B are linked and in coupling phase (trans).

R. The distance between A and B is 10 cM.

S. The distance between A and B is 15 cM.

Which of the above conclusions are correct?

(a) P and R

(b) P and S

(d) Q and S

Ans. (c)

Sol. In the F1, markers A and B are linked and in coupling phase (trans).

The distance between marker A and B =

100. A plant with red fruit is crossed to a plant with white fruit. The F1 progeny had red fruits. On selfing the F1, two kinds of progeny were observed, plants with red-fruits and those with white-fruits. To test whether it was a case of recessive epistatic interaction a chi-square test was performed. A value of 1.062 was obtained (chi-square valueat P_0.05 = 3.841 for degree of freedom = 1). The following statements were made:

P. The null hypothesis was that plants with red and white fruit will occur in 9 : 7 ratio.

Q. The null hypothesis was that plants with red and white fruits will occur in 1 : 1 ratio.

R. Based on the chi-square value, it is a case of recessive epistatic interaction.

S. Based on the chi-square value, it is not a case of recessive epistatic interaction.

Which of the combination of above statement is correct?

(a) P and R

(b) P and S

(d) Q and S

Ans. (a)

Sol. As pre chi-square test the null hypothesis was that plants with red and white fruit will occur in 9 : 7 ratio and based on the chi-square value, it is a case of recessive epistatic interaction.

101. Somatic recombination was caused by mild exposure to radiation on files heterozygous for a given allele during specific stages of development and the individuals were allowed to develop. Such individuals are likely to have

P. Clones of homozygous cells in heterozygous body.

Q. Site specific mutagenesis.

R. Twin spots, i.e. patches of mutants cells and homozygous wild-type cells in heterozygous body.

S. Tissue specific expression of the given allele.

Which of the following combination of answers will be most appropriate?

(a) P and Q

(b) Q and R

(d) R and S

Ans. (c)

Sol. Somatic recombination was caused by mild exposure to radiation on files heterozygous for a given allele during specific stages of development and the individuals were allowed to develop. Such individuals are likely to have clones of homozygous cells in heterozygous body and twin spots, i.e. patches of mutants cells and homozygous wild-type cells in heterozygous body.

102. The below pedigree shows the inheritance of a rare allele. The allele is

(a) X-linked recessive

(b) autosomal recessive

(d) autosomal recessive with incomplete penetrance

Ans. (c)

Sol. By observing the pedigree chart the allele is dominant with incomplete penetrance.

103. Sickle cell anemia is a recessive genetic disease caused due to a point mutation in the 6^th codon abolishing one of the MspIi endonuclease digestion site present in the β-globin gene. MspII digested DNA from a normal person gives two bands, 1150 bp and 200 bp, in β-globin gene. A family with a proband (based on the disease phenotype)

The following conclusions were down:

1. Son (I) is the proband and the given mutation is not present in Son (II).

2. The daughter is a carrier for the given mutation.

3. The gene is X-linked and thus Son (I) becomes the proband.

4. The father and daughter are affected.

5. A de novo mutation in the same site on normal allele has allowed appearance of disease phenotype in the proband.

Which of the following combination of conclusions will be the most appropriate for the figure given above?

(a) 1, 2 and 5

(b) 1,2 and 3

(d) 3 and 4

Ans. (a)

Sol. By observing MspII digestion pattern Son (I) is the proband and the given mutation is not present in Son (II), the daughter is a carrier for the given mutation and a de novo mutation in the same site on normal allele has allowed appearance of disease phenotype in the proband.

104. In eusocial insects, males develop from unfertilized eggs while females develop from fertilized eggs. The ultimate consequence of this difference is that

(a) in any colony there are always more males than females.

(b) a female is genetically more closely related to her sister than to her own offspring

(d) in any colony there are always more females than males.

Ans. (b)

Sol. Hymenoptera have a haplodiploid sex determination system (whereby females arise from fertilized diploid eggs and males arise from unfertilized haploid eggs). As a result female is genetically more closely related to her sister than to her own offspring.

105. The phylum in which the animals are bilaterally symmetrical in the larval stage and radially symmetrical in the adult stage is

(a) Coelenterate

(b) Nematoda

(d) Echinodermata

Ans. (d)

Sol. Echinodermal is the common name given to any member olf the Phylum Echinodermata of marine animals. The adults are recognizable by their (usually five-point) radial symmetry and include such well-known animals as starfish, sea urchins, sand dollars and sea cucumbers as well as the sea lilies or "stone lilies". Echinoderms evolved from animals with bilateral symmetry. Although adult echinoderms possess pentaradial or five-shaded symmetry, echinoderm larvae are ciliated, free-swimming organsims that organize in bilateral symmetry which makes them look like embryonic chordates.

Echinoderms exhibit secondary radial symmetry in portions of their body at some stage of life. This, however, is an adaptation to their sessile existence. They developed from other members of the Bilateria and exhibit bilateral symmetry in their larval stage. Many crinoids and some seastars exhibit symmetry in multiples of the basic five, with starfish such as Labidiaster annulatus known to possess up to fifty arms and sea-lily Comaster schlegelii having two hundred.

106. Which of the following fungal groups has septate hyphae and reproduces asexually by budding, conidia and fragmentation?

(a) Basidiomycota

(b) Zygomycetes

(d) Glomeromycota

Ans. (a)

Sol. Basidiomycota (club fungi) have septate hyphae and includes fungi that produce fruiting structures called mushrooms.

Sexual spores are basidiospores formed externally on a base pedestal called a basidium.

Asexual spores in some are conidiospores.

Examples : Cryptococcus neoformans Amanita phalloides (Death Angel, deathcap) Psilocybe spp. (P. cubensis, P. mexicana, P. tampanensis and P. atlantis).

107. The general relation between generation time (T) and population growth rate (r) is described by the equation

(a) In r = In a – b In T

(b) r = a – b T

(d) r = a + b T

Ans. (a)

Sol. (a) is correct option to show relation between generation time (T) and population growth rate (r).

108. Which of the following is not semelparous?

(a) Dracena

(b) Bamboo

(d) Mayfly

Ans. (a)

Sol. Semelparity and iteroparity refer to the reproductive strategy of an organism. A species is considered semelparous if it is characterized by a single reproductive episode before death and iteroparous if it is characterized by multiple reproductive cycles over the course of its lifetime. Semlaparous animals include many insects, including some species of butterflies, cicadas and mayflies, many arachnids and some molluscs. Long-lived semelparous plants include century plant (agave), Lobelia telekii and some species of bamboo.

109. The dynamics of any subpopulation within a metapopulation differs from that of a normal population in that the

(a) birth rates are lower than the death rates

(b) death rates are lower than the birth rates

(d) immigration and emigration rates are negligible

Ans. (c)

Sol. A metapopulation consists of a group of spatially separated populations of the same species which interact at some level. The interconnected local populations may differ in size and may be subject to different degrees of emigration, immigration gene flow, genetic drift, local adaptation and temporal persistence.

Average population size of a metapopulation depends on emigration and population growth rates and on dispersers mortality rate, which is directly related to isolation between patches : Avaerage population size increases with decreasing isolation due to decreasing mortality during dispersal. Average population size also increases with increasing fraction of occupied patches. Local population size increases with immigration and local population growth and decreases with emigration. Immigration significantly increases the growth rate of small local populations. Hanski's model assumes emigration to be density-independent. I question this assumption on the basis that a patch population becomes very dense, high density should induce emigration.

110. Which of the following is likely to contribute to the stability of an ecosystem?

(a) High number of specialists

(b) Fewer number of functional links

(d) Linear rather than reticulate food webs

Ans. (c)

Sol. Omnivores have double roles in ecosystems, being able to consume both plant food and animal food as their primary food source. They contributes more stability to ecosystem.

111. The following table shows selected characters used in analyzing the phylogenetic relationships of four plant taxa:

Taxa T₁, T₂, T₃ and T₄ are respectively:

(a) Ferns, Oaks, Pines, Hornworts

(b) Oaks, Pines, Hornworts, Ferns

(d) Ferns, Pines, Oaks, Hornworts

Ans. (a)

Sol. T1 : Similar to flowering plants, ferns have roots, stems and leaves. However, unlike flowering plants, ferns do not have flowers or seeds; instead, they usually reproduce sexually by tiny spores or sometimes can reproduce vegetatively, as exemplified by the walking fern.

T2 : Oaks have spirally arranged leaves, with rounded edges in many species; some have leaves with jagged edges or entire leaves with smooth margins. Many deciduous species do not drop dead leaves until spring. In spring, a single oak tree produces both male flowers (as catkins) and small female flowers.

T3 : Pine is a non-flowering tree i.e. it does not produce flowers. It reproduces from seeds that grow within pine cones instead of flowers.

T4 : Like all bryophytes, hornworts lack vascular tissue (xylem and phloem), and therefore do not have true roots, stems, or leaves. The hornwort plant body typically consists of a flattened, somewhat lobulated structure called a thallus, which is usually less than 0.8–1.6 inches (2 centimeters) in diameter.

112. Given below are the main characteristics of a few mammalian orders. Match the names of the animals with the characteristics of their orders:

P. Hooves with even number of toes on each 1. Tapir

foot, omnivorous

Q. Teeth consisting of many thin tubes 2. Lemur

cemented together, eats ants and termites

R. Opposable thumbs, forward facing eyes, 3. Aardvark

well developed cerebral cortex, omnivorous

S. Hooves with an odd number of toes on each 4. Pig

foot; herbivorous

Choose the correct combination.

(a) P–3, Q–4, R–1, S–2

(b) P–1, Q–4, R–2, S–3

(d) P–4, Q–3, R–1, S–2

Ans. (c)

Sol. Pigs are known to be even-toes ungulates which is a term that refers to a hoofed animal whose weight is spread evenly by more than one toe. Pigs are also known as hos and swine.

Pigs are omnivores meaninng that they eat both plants and animals. Pigs are scavengers by nature and will eat nearly anything that they come across from plants and fruit to dead insects and tree bark.

Tubulidentata – Teeth consisting of many thin tubes cementeed together; eats ants and termites (ex : aadrvarks).

Primates – Opposable thumbs; forward facing eyes; well-developed cerebral cortex; omnivorous (ex : lemurs, monkeys, chimpanzees, gorillas, humans). Perissodactyla possess hooves with an odd number of toes on each foot; herbivores. Ex : Horses, zebras, tapirs, rhinoceroses.

113. Which of the following options match the plant tissue type with its correct function in vascuar plants?

Tissue Function

P. Tracheids 1. Chief water-conducting element in gymnosperms

Q. Vessel elements 2. Chief water-conducting element in angiosperms

R. Sieve tube element 3. Food-conducting element in gymnosperms

S. Sieve cell 4. Food-conducting element in angiosperms

(a) P–1, Q–2, R–4, S–3

(b) P–2, Q–1, R–3, S–4

(d) P–1, Q–3, R–4, S–2

Ans. (a)

Sol.

114. Match the five (P – T) group of organisms with their correct taxonomic rank (1 – 5) given below :

Group Taxonomic rank

P. Crustacea 1. Order

Q. Hominidae 2. Domain

R. Dermaptera 3. Class

S. Ctenophora 4. Phylum

T. Archaea 5. Family

(a) P–3, Q–1, R–5, S–4, T–2

(b) P–1, Q–2, R–3, S–4, T–5

(d) P–3, Q–5, R–1, S–4, T–2

Ans. (d)

Sol. Crustaceans are a taxonomic class. It form a very large group of arthropods, ususally treated as a subphylum, which includes such familiar animals as crabs, lobsters, crayfish, shrimp, krill and barnacles.

Dermaptera comes from the Greek words derma, which means skin and ptera, which means wings. It has simple development, short wing covers-second pair not always developed, chewing mouth parts.

The Hominidae also konwn as great apes or hominids are a taxonomic family of primates that includes seven extant species in four genera : Pongo, the Bornean and Sumatran orangutan; Gorilla, the eastern and western gorilla; Plan, the common chimpanzee and bonobo; and Homo, the human.

Ctenophora is a phylum of animals that live in marine waters worldwide. Their most distinctive feature is the 'combs' groups of cilia which they use for swimming they are the largest animals that swim by means of cilia.

The Archaea constitute a domain or kingdom of single-celled microorganisms. These microbes are prokaryotes meaning that they have no cell nucleus or any other membrane-bound organelles in their cells.

115. For the aquaculture farming of Indian Major carps several techniques are used. Which one of the following techniques is not used for this purpose?

(a) Induced breeding

(b) Selective breeding

(d) Composite fish farming

Ans. (c)

Sol. With the accelerated human population explosion, demand for fish is also increasing rapidly. In Indian context, total fish production comes from the two sectors, marine and inland. Aquaculture is defined as the system of breeding and farmint of aquatic species in a controlled environment. Innovative aqua-culture involves the applications of new and more effective ways or solutions for aquaculture management. Here are some innovations which have played or are playing or will play a significant role in the development of aquaculture sector.

(a) Induced Breeding : Technological breakthrough in induced breeding of carps through hypophysation in 1957 by H. Chaudhuri and K.H. Alikunhi has revolutionised aquaculture in India. Since then carp breeding by hypophysation has been an established technology for seed production with an annual output of around 21,000 million fry. Based on this principle, various prime fish species were bred and endeavour is on to breed other commercially important fishes.

(b) Composite Fish Culture : Composite fish culture, with the aim of utilising all niches of water column was started, involving the three Indian major carps, viz. Catla, Rohu and Mrigal along with three exotic carps. Silver carp, Grass carp and common carp. It has been a mainstay in aquaculture industry since 1980s. Due to its immense popularity among fish farmers, today composite fish culture alone in contributing more than 80% of total aquaculture production of the country.

(c) Wastewater Aquaculture : Wastewater aquaculture was developed in Germany at the end of nineteenth century and independently in Calcutta in 1930. At present, there are more than 130 wastewater aquaculture units in India covering about 10,000 ha. Out of it, 80% are located in West Bengal. In this system, generally multiple stocking and multiple harvesting practices are followed for a period of 3-5 months with carp fingerlings and tilapia.

116. Which one of the following is most appropriate match for the protected areas of India?

Category Protected area

P. Biosphere reserve 1. Chambal

Q. National Park 2. Loktak

R. Ramsar site 3. Nanda Devi

S. Wildlife sanctuary 4. Rajaji

5. Sundarbans

(a) P–3, Q–4, R–2, S–1

(b) P–2, Q–4, R–3, S–1

(d) P–3, Q–2, R–5, S–4

Ans. (a)

Sol. Nanda Devi Biosphere Reserve is located in the northern part of west Himalaya in the biogeographical classification zone 2B. The Biosphere Reserve spreads over three districts of Uttarakahand-Chamoli in Garhwal and Bageshwar and Pithoragarh in Kumaun. The two core zones of the Biosphere reserve i.e., the Nanada Devi National Park and the Valley of Flowers National Park are inscribed as World Heritage sites in 2005.

Rajaji National Park is an Indian national park and Tiger Reserve that encompasses the Shivaliks, near the foothills of the Himalayas. It is spread over 820 km² and three districts of Uttarakhand : Haridwar, Dehradun and Pauri Garhwal.

Among the six Ramsar sites, Loktak Lake in Manipur has been identified and listed in 1990 and in 1993, the Montreux record categorized it as "where ecological changes have occurred, are occurring or likely to occur". The other five Ramsar sites include Chilka Lake (Orissa), Keoladeo National Park, Bharatpur in Rajashtan, Sambhar Lake (Rajasthan), Wular Lake in Kashmir and Harike in Punjab.

National Chambal Sanctuary, also called the National Chambal Ghharial Wildlife Sanctuary, is a 5,400 km² (2,100 sq mi) tristate protected area in northern India for the critically endangered gharial (small crocodiles), the red-crowned roof turtle and the endangered Ganges river dolphin. Located on the Chambal River near the tripoint of Rajasthan, Madhya Pradesh and Uttar Pradesh, it was first declared in Madhya Pradesh in 1978 and now constitutes a long narrow eco-reserve coadministered by the three states.

117. Given below is a matrix of possible interactions (beneficial (+), harmful (–), neutral (0)) between species 1 and 2. The names of interactions, A, B, C and D, respectively, are

(a) Predation, competition, mutualism, commensalism

(b) Mutualism, competition, amensalism, commensalism

(d) Competition, mutualism, commensalism, predation

Ans. (c)

Sol. A. Competition is most typically considered the interaction of individuals that vie for a common resource that is in limited supply, but more generally can be defined as negative - negative interaction.

B. Predation is an interaction in which one organism, the predator, eats all or part of the body of another organism, the prey. (+,- interaction).

C. Mutualism is defined as an interaction between individuals of different species that results in positive (beneficial) effects on per capita reproduction and/or survival of the interacting populations (+,+ interaction).

D. Amensalism, association between organisms of two different species in which one is inhibited or destroyed and the other is unaffected (-,0 interaction).

118. Two lakes (I and II) with a similar trophic structure of phytoplankton-zooplankton-planktivorous fish food chain were chosen. To understand the 'top-down' effects, some piscivorous fish (those that feed on planktivorous fish) were introduced into lake I, making it a system with four trophic levels. Lake II was enriched by adding large quantities of nitrates and phosphates to study the 'bottom-up' effects over a period of time. Changes in the biomasses of each trophic level were measured. The expected major changes in the two lakes are

(a) In Lake I zooplankton biomass increases, phytoplankton biomass decreases. In Lake II both phytoplankton and planktivorous fish biomasses increases.

(b) In Lake I zooplankton biomass decreases, phytoplankton biomass increases. In Lake II both phytoplankton and planktivorous fish biomasses increases.

(c) In Lake I planktivorous fish and phytoplankton biomass decreases. In Lake II phytoplankton biomasses increases, planktivorous fish biomass decreases.

(d) In Lake I planktivorous fish and zooplankton biomass increases. In Lake II phytoplankton and planktivorous fish biomass increases.

Ans. (a)

Sol. In lake 1, 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.

In lake 2, both phytoplankton and planktivorous fish biomass increases.

119. Which of the following is true about the Digestion Efficiency (DE) (assimilation/consumption) and Ecological Efficiency (EE) (Production/consumption) of ectotherms and endotherms?

(a) Endotherms have a high DE and ectotherms have a high EE.

(b) Endotherms have a low DE and ectotherms have a high EE.

(d) Endotherms have a low DE and ectotherms have a high EE.

Ans. (a)

Sol. An endotherm is an organism that maintains its body at a metabolically favourable temperature, largely by the use of heat set free by its internal bodily functions instead of relying almost purely on ambient heat. Such internally generated heat is mainly an incidental product of the animal's routine metabolism, but under conditions of excessive cold or low activity an endotherm might apply special mechanisms adapted specifically to heat production. Endotherms have high digestion efficiency.

Ecotherms on the other hand, do not have efficient temperature regulation systems. In warm regions when the temperature is high, the ectotherms seek shade and low-temperature areas to prevent over-heating. Whereas, in cool regions they seek sunlight and expose themselves to the sunlight for warmth. For the same reason, they can also search out and stay near fire pits, campfires etc.

So when wer calculate the energy demands, the ectotherms need more energy as they regulate their own metabolisms to match the need, they have high ecological efficiency.

120. Which of the following is/are not valid explanation(s) for the observed pattern of species richness?

1. Older communities are more species-rich.

2. Large areas support more species.

3. Natural enemies promote reduced species richness at local level.

4. Communities in climatically similar habitats may themselves be similar in species richness.

5. Greater productivity permits existence of more species.

(a) 2, 3 and 4

(b) Only 3

(d) 1, 2 and 5

Ans. (b)

Sol. Natural enemies can help keep plant-feeding insects from attaining damaging population levels. It is important to understand how to manage ecosystems to take advantage of the services of natural enemies. Natural enemies reduce species richness at local level.

121. The following table shows the number of individuals of each species found in two communities:

(Hint: In values for 0.05, 0.10, 0.25 and 0.80 are –3.0, –2.3, –1.4 and –0.2, respectively)

The calculated Shannon diversity index (H) values for communities C1 and C2, respectively are

(a) 1.4 and 0.69

(b) 1.2 and 0.34

(d) 1.8 and 0.37

Ans. (a)

Sol. For community C1, Shannon diversity table will look like as :

H' = p_i ln p_i = –(–1.4) = 1.4

For community C2, Shannon diversity table will look like as

H' = – p_i ln p_i = –(–0.69) = 0.69

122. Compared to K-selection, r-selection favours

(a) rapid development, smaller body size and early, semelparous reproduction.

(b) rapid development, smaller body size and early, iteroparous reproduction.

(d) slow development, smaller body size and late, iteroparous reproduction.

Ans. (a)

Sol. The selection pressures that determine the reproductive stratege and therefore much of the life history of an organism can be understood in terms of r/K selection theory. r-selected organisms usually :

mature rapidly and have an early age of first reproduction.

have a relatively short lifespan.

have a large number of offspring at a time and few reproductive events or are semelparous.

have a high mortality rate and a low offspring survival rate.

have minimal parental care/investment.

123. The most commonly used molecular tool for phylogenetic analysis involves sequencing of

(a) mitochondrial DNA

(b) mitochondrial RNA

(d) nuclear DNA

Ans. (a, c)

Sol. Ribosomal RNA (rRNAs) are highly conserved among species through evolutionary history. In fact, rRNAs are present in all extant species examined to date and the functionally important rRNA-based peptidyl transferase center, which catalyzes peptide bond formation in the ribosome, has even been recovered through artificial selection from a random sequence pool. rRNA sequences have been widely used in phylogenetic studies.

In general, residues located on the surface of the rRNA molecule evolve faster. Additionally, different rRNA secondary structures evolve at different rates in different lineages.

124. According to which evolutionary theory, there are long periods without significant evolutionary changes interrupted by short episodes of rapid evolution?

(a) Punctuated equilibrium

(b) Saltation

(d) Neutrality

Ans. (a)

Sol. "Punctuated equilibrium" refers to a concept in evolutionary biology that is both controvesial and widely misunderstood. Both punctuated equilibrium and its alternatives have significant drawbacks, either in plausibility or evidence. Punctuated equilibrium seeks to reconcile the idea of natural evolution with the missiing links in the fossils recored. Punctuated equilibrium is an attempt to reconcile available evidence with the idea of naturalistic evolution. It is, in many ways, another example of reinterpreting facts in order to fit and ideology. Still, any willingness to modify evolutionary theory in light of evidence is good, since this can only lead closer and closer to the idea of a Creator God. Pursuit of these ideas might help close the philosphical gaps between atheistic naturalism and intelligent design-which might be the very reason punctuated equilibrium is so highly resisted in some academic circles.

125. In the evolutionary tree given below, terms A, B, C, D and E represent respectively

(a) Homo erectus, Homo heidelbergensis, Neanderthal, Denisovan and Homo sapiens.

(b) Homo heidelbergensis, Homo erectus, Denisovan, Neanderthal and Homo sapiens.

(d) Homo heidelbergensis, Homo sapiens, Denisovan, Neanderthal and Homo erectus.

Ans. (a)

Sol. A. The extinct ancient human Homo erectus is a species of firsts. It was the first of our relatives to have human-like body proportions, with shorter arms and longer legs relative to its torso.

C. Neanderthals made and used a diverse set of sophisticated tools, controlled fire, lived in shelters, made and wore clothing, were skilled hunters of large animals and also ate plant foods, and occasionally made symbolic or ornamental objects.

D. Alternatively, on arriving in Denisovan habitat, H. sapiens may have outcompeted or killed their cousins, or brought lethal diseases with them. Climatic events could have been crucial too.

126. Homing pigeons, when released at a place far away from their home, use earth's magnetic field or the sun as navigational cues and choose the right direction to fly. To test the hypotheses, two experiments were conducted. In experiment I, one group of pigeons (Test) were equipped with Helmholtz coils (which disrupt magnetic field direction), while the second group (control) were not. Both groups were released on sunny day. Experiment II used results, if the hypotheses is true, would be

(a) In experiment I, both control and test groups fly in the right direction, but in experiment II, only control group does.

(b) In both experiments, test groups fail to choose the right direction.

(d) In experiment I, both groups fly in the right direction, but in experiment II, both groups fail to choose the right direction.

Ans. (a)

Sol. In experiment 1, on sunny day test pigeons used sun for navigation. But on cloudy day , when sun was not visible only control group does using earth's magnetic field which was disrupted in test pigeons.

127. Brothers A and B have the same father but different mothers. B wants A to help him which involves both benefits (b) and costs (c) for A. If A incurs a cost of 30 'Darwinian fitness units' in that act, under what condition should he help B, following Hamilton's rule?

(a) Only if b > 30

(b) Only if b > 60

(d) Only if b > 240

Ans. (c)

Sol. Kin selection will favour the altruistic act, if rb – c > 0,

where,

c = cost incurred

b = benefit received by the recipients

r = coefficient of relatedness.

c = 30

r = 0.25

rb > c,

So, b > 120

128. In a population at Hardy-Weinberg equilibrium, the genotype frequencies are: (A₁A₁) = (0.59; f(A₁A₂) = 0.16; f(A₂A₂) = 0.25. What are the frequencies of the two alleles at this locus?

(a) A₁ = 0.59 ; A₂ = 0.41

(b) A₁ = 0.75 ; A₂ = 0.25

(d) A₁ = 0.55 ; A₂ = 0.44

Ans. (c)

Sol. If p and q represent allele frequencies for a gene with two alleles; the sum of p + q must equal 100% of the alleles (or gametes) in the population. In other words, p + q = 1.

If p², 2pq, q² represent genotype frequencies; p² corresponds to one homozygous genotype, 2 pq represents the hetrozygous genotype and q² is the other homozygous genotype. Determining allele and genotype frequencies can be done by the method involves converting the initial numbers of each genotype to frequencies and then doing all calculations as frequencies. In this case the frequency of the p allele = the frequency of the p/p homozygotes + 1/2 the frequency of the heterozygotes. The frequency of the q allele = the frequency of theq/q homozygotes + 1/2 the frequency of the heterozygotes.

Here

frequency of p/p homozygotes

f(A₁ A₁) = 0.59;

f(A₁A₂) = 0.16;

and

frequency of q/q homozygotes

f(A₂A₂) = 0.25

Frequency of the A₁ allele

= f(A₁ A₁) + f(A₁ A₂)

= 0.59 + × (0.16)

= 0.59 + 0.08

= 0.67

and frequency of the A₂ allele

= f(A₂A₂) + f(A₁A₂)

= 0.25 + × (0.16)

= 0.25 + 0.08 = 0.33

129. Following are the main types of defense employed by prey species against predators.

Types of defense Prey species

P. Chemical with aposematic coloration 1. Grasshoppers and seahorses

Q. Cryptic coloration 2. Hoverflies and wasps

R. Batesian mimicry 3. Bombardier beetles, ladybird beetes, many butterflies

S. Intimidation display 4. Frilled lizard, Porcupine fish

(a) P–1, Q–3, R–2, S–4

(b) P–4, Q–2, R–1, S–3

(d) P–2, Q–3, R–1, S–4

Ans. (c)

Sol. In Batesian mimicry, the mimic incurs a benefit at the cost of the model (see adaptive value). In Mullerian mimicry, the mimic is always the organism, and the model is the common signal among the mimic species that honestly indicates inedibility. An intimidation display with a means to threat are exhibited through: hair bristling, feather ruffling, raising skin folds and crest, teeth displaying, horn displaying, making sound, etc. Option (c) is correct.

130. Following is the list of some important events in the history of life and the names of the epochs of Cenozoic era.

Events Epochs

P. Angiosperm dominance increases; continued 1. Paleocene

radiation of most present day mammalian orders

Q. Major radiation of mammals, birds and 2. Pleistocene

pollinating insects

R. Origins of many primate groups 3. Oligocene

S. Origin of genus Homo 4. Pliocene

T. Appearance of bipedal human ancestors 5. Eocene

U. Continued radiation of mammals and 6. Miocene

angiosperms, earliest direct human ancestors.

Which one of the following is the correct match of events with the epochs?

(a) P–5, Q–2, R–1, S–3, T–4, U–6

(b) P–6, Q–1, R–2, S–4, T–3, U–5

(d) P–4, Q–1, R–2, S–3, T–5, U–6

Ans. (c)

Sol. The Eocene Epoch saw the replacement of older mammalian orders by modern ones."Mammals began a rapid diversification during this period. After the Cretaceous–Paleogene extinction event, which saw the demise of the non-avian dinosaurs, mammals began to evolve from a few small and generalized forms into most of the modern varieties we see today."Oligocene- origin of primate groups. Major changes during the Oligocene included a global expansion of grasslands, and a regression of tropical broad leaf forests to the equatorial belt."Pliocene- appearance of bipedal human ancestors"Eocene- angiosperm dominance increases"Miocene- continued radiation of mammals and angiosperms.

131. Which one of the methods listed below is the most sensitive label-free quantification method for proteins?

(a) UV spectroscopy

(b) Infra-red spectroscopy

(d) ¹³C content of protein

Ans. (a)

Sol. Simple UV Absorption spectrophotometry can determine the quantity of proteins in the sample by using the maximum absorption at 280 nm. The absorption at 260 nm correlates with the concentration of nucleotides.

132. Two groups (Control, Treated) are to be compared to test the effect of a treatment. Since individual variability is high in both groups, the appropriate statistical test to use is

(a) analysis of variance

(b) Kendal's test

(d) Mann-Whitney U-test

Ans. (d)

Sol. In statistics, the Mann-Whitney U test [also called the Mann-Whitney-Wilcoxon (MWW), Wilcoxon rank sum test (WRS) or Wilicoxon-Mann Whitney test] of the null hypothesis that two samples come from the same pouplation against an alternative hypothesis, especially that a particular population tends to have larger values than the other.

It can be applied on unknown distributions contrary to t-test which has to be applied only on normal distributions and it is nearly as efficient as the t-test on normal distributions.

133. A protein has one tryptophanand one tyrosine inits sequence. Assume molar extinction coefficients at 280 nm of tryptophan and tyrosine as 3000 and 1500^–1 and cm^–1, respectively. What would be the molar concentration of that protein if its absorption at 280 nm is 0.90?

(a) 2 mM

(b) 0.4 mM

(d) 0.02 mM

Ans. (c)

Sol. Molar extinction coefficient of tryptophan = 3000 M^–1 cm^–1. Molar extinction coefficient of tyrosine = 1500 M^–1 cm^–1

Total extinction coefficient = 1 × 3000 M^–1 cm^–1 + 1 × 1500 M^–1 cm^–1 = 4500 M^–1 cm^–1

As we know that, A = × c × l

Protein concentration (in molarity) =

134. The mean (µ) and standard deviation of body size in a Drosophila population are 8.5 and 2.2 mm, respectively. Under natural selection over many generations the µ and of body size change to 8.5 and 0.8 mm, respectively.

The type of natural selection responsible for the change is called

(a) directional

(b) neutral

(d) stabilizing

Ans. (d)

Sol. Stabilizing selection is the opposite of disruptive selection. Instead of favoring individuals with extreme phenotypes, it favors the intermediate variants. Stabilizing selection tends to remove the more severe phenotypes, resulting in the reproductive success of the norm or average phenotypes.

135. In transgenic mice, the orientation and location of the loxP sites determine whether Cre recombinase induces a deletion, an inversion or a chromosomal translocation. If a researcher wants to put the loxP sites in such a manner that only inversion will take place, which one of the following construct best justfies their intention? Gene X is the target gene.

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

136. A student, while constructing knock-out mice, isolated mouse embryonic stem cells and introduced an engineered DNA into the cells. However, none of the mice were transgenic. On checking the cells containing DNA construct, he found that he had made a mistake in constructing the DNA since the cells were resistant to ganciclovir but sensitive to G418. Which one of the following, constructs had he designed?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Through homologous recombination, an engineered mutation can be directed to a designated genetic locus. In this manner, a potentially important genomic clone can directly be utilized to create a mutation into a selected gene. Option c shows correct construct design because neomycin resistant gene and tk gene is still intact.

137. Somatic embryogenesis is an important exercise in micropropagation and genetic engineering of plants. The following steps are considered as critical for achieving somatic embryogenesis:

P. Reducing the concentration of sucrose in the medium by half.

Q. Addition of the hormone, 2, 4–dichlorophenoxyacetic acid, to induce somatic embryos.

R. Reduce agar concentration to 0.6% (w/v)

S. Use maltose in place of sucrose as a carbon source.

Which one of the following combinations is correct?

(a) P and R

(b) Q and S

(d) R and S

Ans. (b)

Sol. Somatic embryogenesis is an artificial process in which a plant or embryo is derived from a single somatic cell. Somatic embryos are formed from plant cells that are not normally involved in the development of embryos, i.e. ordinary plant tissue. It has been known that auxins mainly 2,4-dichlorophenoxyacetic acid (2,4-D) is required for somatic embryogenesis induction and embryo multiplication to scale-up the number of embryos which can be potentially produced by indirect somatic embryogenesis.

138. Factor IX is essential for blood clotting. Deficiency of Factor IX could be corrected by delivering factor IX gene using viral vectors. In an experiment the gene for Factor IX was cloned appropriately into Adenovirus (AV), Adeno associated virus (AAV) and Retrovirus (RV). Retrovirus integrates the gene into the dividing cells. AAV integrates partially into non-dividing cells. AV does not integrate the gene but transfects both dividing and non-dividing cells.

Following expression profile for factor IX was observed when the three vectors were injected intra-muscularly into three groups of experimental mice.

Which one of the following outcomes is correct?

(a) P by RV; Q by AV; R by AAV

(b) P by AV; Q by AAV; R by RV

(d) P by AV; Q by RV; R by AAV

Ans. (d)

Sol. D is correct outcome where, P is shown by Adenovirus, Q by retrovirus and R by Adeno associated virus.

139. The following genes have been genetically engineered to develop herbicide resistance in plants.

P. Resistance to glyphosate using the 5-enolpyruvyl shikimate-3 phosphate synthase gene.

Q. Bialaphos resistance using the bar gene.

R. Sulfonyl urea resistance using the bar gene.

S. Atrazine resistance using the glutathione S-transferase gene.

In which of the above two cases the mechanism is based on abolition of herbicide binding to the enzyme?

(a) P and S

(b) Q and R

(d) P and R

Ans. (d)

Sol. The enzyme 5-enolpyruvylshikimate-3-phosphate synthase (EPSPS) catalyzes the penultimate step of the shikimate pathway, and is the target of the broad-spectrum herbicide glyphosate. Kinetic analysis of the cloned EPSPS from Staphylococcus aureus revealed that this enzyme exerts a high tolerance to glyphosate, while maintaining a high affinity for its substrate phosphoenolpyruvate. Gene reintroduction experiments have confirmed that these amino acid substitutions are responsible for the herbicide resistance phenotypes. Transgenic plants carrying these genes are highly resistant to sulfonylurea herbicide applications.

140. In a confirmatory test for HIV, one or more viral antigens are detected in the blood of patients. Following are the steps to be performed for the experiment:

P. Transfer of viral antigens to nitrocellulose paper.

Q. Incubation with the bugffer containing antibodies specific for viral antigens.

R. Separation of viral antigens by SDS-PAGE.

S. Detection of bands by enzyme-linked secondary antibody.

Identify the correct sequence steps to be performed for the experiment.

(a) P Q R S

(b) Q R S P

(d) R Q P S

Ans. (c)

Sol. Correct sequence of steps will be :

R. Seperation of viral antigens by SDS-PAGE.

P. Transfer of viral antigens to nitrocellulose paper.

Q. Incubation with buffer containing antibodies specific viral antigens.

S. Detection of bands by enzyme - linked secondary antibody.

141. A protein undergoes post-translational modification. In an experiment to identify the nature of modification, following experimental results were obtained.

P. Protein moved more slowly in an SDS-PAGE.

Q. Isoelectric focusing [IEF] showed that there was no change inthe pI.

R. Mass spectrometric analysis showed that the modification was on serine.

The modification that the protein undegoes is likely to be

(a) phosphorylation

(b) glycosylation

(d) ADP-ribosylation

Ans. (b)

Sol. Post-translational modification (PTM) is the covalent and generally enzymatic modification of proteins following protein biosynthesis. Glycosylation, the attachment of sugar moieties to proteins, is a post-translational modification (PTM) that provides greater proteomic diversity than other PTMs. O-linked glycosylation is the attachment of a sugar molecule to the oxygen atom of serine (Ser) or threonine (Thr) residues in a protein.

142. A gene from genomic library is screened by using hybridization technique. After hybridizing the probe, usually a stringent washing step is given. The following statements are given to explain the stringent washing step:

1. Stringent washing takes care of removing unincorporated and non-specifically hybridized probe molecules.

2. Stringent washing is done is solution having high salt concentration and lower temperature.

3. Stringent washing is done in solution having low salt concentration keeping higher temperature.

4. Salt present in washing solution supports hybrids to stay inact by shielding the interference of water molecules.

5. Salt react with DNA molecules and allows easy dissociation of hybrids.

6. Stability of hybrid is directly proportional to the temperature.

Which combination of the above statements is mot appropriate for stringent washing step?

(a) 1, 2 and 4

(b) 1, 3 and 4

(d) 3, 5 and 6

Ans. (b)

Sol. After the incubation of the blot with the probe, washing steps are necessary to remove excess probe (sticking non-specifically to the membrane). The stringency of the washings is important for reasons mentioned above. Stringent washing is: high temperature, low salt."A commonly used washing solution is SSC (Saline Sodium Citrate, a salty sodium citrate solution, a mixture of NaCitrate and NaCl. Statement 1,3 and 4 is correct.

143. In response to a drug, changes in protein levels were examined in a cell line. A pulse chase experiment was performed using 35S labeled methionine. In comparison to untreated samples, the following observations were made: few minutes after stimulation, protein X accumulates and this was followed by reduction in protein Y and Z.

The correct interpretation of these observations would be:

(a) Protein X is a protease which degrades Y and Z.

(b) Protein X is a transcriptional repressor that controls expression of Y and Z.

(d) Information is not sufficient to distinguish between the three possibilities stated above.

Ans. (d)

Sol. A pulse-chase analysis is a method for examining a cellular process occurring over time by successively exposing the cells to a labeled compound and then to the same compound in an unlabeled form."Information is not sufficient to distinguish between three possibilities stated above.

144. The most important property of any microscope is its resolution (D) and can be calculated from the formula ; where D is minimum distance between two distinguishable objects, is the wavelength of incident light, is the angular aperture and N is the refractive index of the medium. Given below are several suggestions to improve the resolution of a microscope:

P. decrease the wavelength of incident light

Q. increase the wavelength of incident light

R. use oil which has a higher refractive index

S. use oil because of its lower refractive index

(a) P and R

(b) Only Q

(d) Q and S

Ans. (a)

Sol. An improved resolution means that lower value of D. In order to keep the value of 'D' small, should be small, and lens should be immersion oil lens, where immersion oil has been placed at the interface between the objective front lens and the speciman covered with cover slip.

145. The mean and standard deviation of serum cholesterol in a population of senior citizens are assumed to be 200 and 24 mg/dl, respectively. In a random sample of 36 senior citizens, what value of cholestero (to the nearest whole number) should lead to rejection of the null hypothesis at 95% confidence level?

(a) Above 224

(b) Above 248

(d) Below 192 and above 208

Ans. (d)

Sol. For random sample, a confidence interval for a population mean is given by,

For rejection of the hypothesis, = sample mean