PYQ JUNE 2013 - VedPrep

CSIR NET BIOLOGY (JUNE - 2013)
Previous Year Question Paper with Solution.

21. Which one of the following non-covalent interactions between the non-bonded atoms A and B is most sensitive to the distance between them ?

(a) A and B are permanent dipoles and are involved in hydrogen bonding.

(b) A and B are fully ionized and are involved in salt bridge formation.

(d) A and B are uncharged and attract each other.

Ans. (c)

Sol. The magnitude of van der waals interaction is calculated using following formula :

F or V =

In this equation, both repulsion and attraction forces are involved, in which r¹² factor indicates repulsion primarily between electron shells of two atoms while r⁶ factor attraction. Among these forces repulsive forces are most sensitive to distance between them. These forces are defined for uncharged atoms or molecules.

22. Which statement best describes the pK_a of amino groups in proteins ?

(a) pK_a of -amino group is higher than the pK_a of -amino group.

(b) pK_a of -amino group is lower than the pK_a of -amino group.

(d) pK_a of -amino group is higher than the pK_a of guanidine side chain of arginine.

Ans. (b)

Sol. Lysine, an essential amino acid, has a positively charged -amino group (a primary amine). Lysine is basically alanine with a propylamine substituent on the -carbon. The -amino group has a significantly higher pK_a (about 10.5 in polypeptides) than does the -amino group.

The amino group is highly reactive and often participates in a reactions at the active centers of enzymes. Proteins only have one amino groups. However, the higher pK_a renders the lysyl side chains effectively less nucleophilic. Specific environmental effects in enzyme active centers can lower the pK_a of the lysyl side chain such that it becomes reactive.

Note that the side chain has three methylene groups, so that even though the terminal amino group will be charged under physiological conditions, the side chain does have significant hydrophobic character. Lysines are often found buried with only the amino group exposed to solvent.

23. A protein has 30% alanine. If all the alanines are replaced by glycines,

(a) helical content will increase.

(b) -sheet content will increase.

(d) the alanine-substituted protein will be less structured than the parent protein.

Ans. (d)

Sol. In internal helical positions, alanine is regarded as the most stabilizing residue, whereas glycine, after proline, is the more destrabilizing.

24. The amino acid alanine has high propensity to occur in helical conformation. The circular dichrosim spectrum of an equimolar mixture of two 20-residue peptides, one composed of only L-alanine and the other only D-alanine is record in the region of 185-250 nm. Which one of the following will be observed ?

(a) No signal : as the chiroptical properties of the two peptides will cancel out.

(b) Bands with only negative ellipticity : as helix formed by the D-Ala peptides will cancel out.

(d) Bands with identical negative and positive ellipticity.

Ans. (a)

Sol. L-amino acids are used in protein synthesis, while D-amino acids are not. L-aminos rotate counterclockwise or left in a process known as levorotation, while D-amino acids rotate clockwise to the right, in what's known as dextrorotation. No signal will be observed as chiroptical properties of two peptides will cancel out.

25. The following small peptide substrates are used for determining elastase activity and the following data have been recorde.

Substrate K_m(mM) k_cat(s^–1)

P A P A G 4.02 26

P A P A A 1.51 37

P A P A F 0.64 18

The arrow indicates the cleavage site. From the above observations, it appears that :

1. PAPAF is digested most rapidly.

2. PAPAG is digested most rapidly.

3. A hydrophobic residue at the Cp-terminus seems to be favored.

4. A smaller residue at the C-terminus seems to be favored.

5. Elastase always requires a smaller residue at the N-terminus of cleavage site.

Which of the following is/are true?

(a) 1, 3 and 5

(b) 2, 4 and 5

(d) 4 and 5

Ans. (a)

Sol. A high Km means a lot of substrate must be present to saturate the enzyme, meaning the enzyme has low affinity for the substrate. On the other hand, a low Km means only a small amount of substrate is needed to saturate the enzyme, indicating a high affinity for substrate. So, PAPAF having least km value is digested more rapidly.

26. The apparent pH of a fluid is 7.45, where bicarbonate buffer is involved for maintaining its pH. Values of pK_a of carbonic acid are 6.15 and 10.45. The molar ratio or [conjugate base] : [acid] is

(Hint : antilog 1.3 = 20.0 and antilog 10^–3 = 1000)

(a) 1 : 20

(b) 20 : 1

(d) 1000 : 1

Ans. (b)

Sol.

27. A segment of B-DNA encodes an enzyme of molecular mass 50 kDa. The estimated length of this segment in µm would be

(a) 0.1547

(b) 0.1547 × 10^–3

(d) 0.4641 × 10^–3

Ans. (c)

Sol. Average molecular mass of an amino acid = 110 Da

Number of amino acid in enzyme = = 455

1 amino acid will be encoded by 3 nucleotides, then total number of nucleotides = 455 × 3 = 1365

Total number of base pairs in DNA = 1365

The distance between two neighboring base pairs is 0.34 nm = 0.00034 µm

Estimated length of segment of B-DNA = 1365 × 0.00034 µm = 0.4641 µm

28. In order to determine the primary structure of an octapeptide, amino acid composition was determined by acid hydrolysis (A). The intact oligopeptide was treated with carbkoxypeptidase (B), chymotrypsin (C), trypsin (D) and CNBr (E). The peptides were separated in each case and acid hydrolysis was carried out for B-E. Following results were obtained (the brackets represent mixtured of amino acids in each fragment) :

A. (2Ala, Arg, Lys, Met, Phe, 2Ser)

B. (Ala, Arg, Lys, Met, Phe, 2Ser) and Ala

C. (Ala, Arg, Phe, Ser), (Ala, Lys, Met, Ser)

D. (Ala, Arg), (Lys, Phe, Ser), (Ala, Met, Ser)

E. (Ala, Arg, Lys, Met, Phe, Ser), (Ala, Ser)

Which one is the correct sequence of the oligopeptide ?

(a) Arg-Ala-Ser-Lys-Met-Phe-Ser-Ala

(b) Arg-Ala-Ser-Lys-Phe-Met-Ser-Ala

(d) Ala-Arg-Phe-Ser-Lys-Met-Ser-Ala

Ans. (c)

Sol. Carboxypeptidase E (CPE) cleaves preferentially Arg or Lys at the C-terminal. Chymotrypsin preferentially cleaves at Trp, Tyr and Phe. Trypsin cleaves the peptide bond between the carboxyl group of arginine or the carboxyl group of lysine and the amino group of the adjacent amino acid. Cyanogen bromide (CNBr) cleaves at methionine (Met) residues.

29. What is the effect of 2, 4-dinitrophenol on mitochondria ?

(a) Blocks ATP synthesis without inhibiting electron transport by dissipating the proton gradient.

(b) Blocks electron transport and ATP synthesis by inhibiting ATP-ADP exchange across the inner mitochondrial membrane.

(d) Interacts directly with ATP synthase and inhibits its activity.

Ans. (a)

Sol. 2, 4-dinitrophenol (DNP) uncouples the mitochondria by shuttling H⁺ ions across the inner membrane, bypassing ATP synthase. DNP is a mobile ionophore, this means that it is a lipid soluble molecule that is capable of transporting ions across a biological membrane.

The way it does this is by shielding the charge of the ion within its hydrophobic exterior, thereby facilitating the transport of H⁺ ions across a biological membrane.

This mode of action provides less resistane of the H⁺ ion than as if it were to move through ATP synthase and thus it becomes the preferred route. Because of this, less ATP is produced inversely proportional to the concentration of DNP. Subsequently the potential energy that was stored in the concentration gradient is released as heat energy. If the concentration of DNP becomes too great, the cell will be unable to produce ATP and eventually die.

30. The gel to liquid crystalline transition temperature (T_m) of phospholipids is dependent on the fatty acid composition. Considering this, T_m of

(a) all the phospholipids will be identical.

(b) DPPC will be lowest and DOPC will be highest.

(d) DOPC will be lowest and DPPC will be highest.

Ans. (d)

Sol. T_m of phospholipids cannot be identical and also saturated phospholipid (DPPC) cannot have lower melting point than unsaturated phospholipid (DOPC). However standard values of T_m for DPPC is 41°C and for DOPC it is –17°C.

DPPC – Dipalmitoylphosphatidylcholine (16 carbons an no double bond)

DOPC – Dioleoyl-sn-glycero-3-phosphocholine (18 carbons and single double bond)

POPC – 1-palmitoyl-2-oleoyl-sn-glycero-3-phosphocholine

DMPC – 1, 2-dimyristoyl-sn-glycero-3-phosphocholine.

31. Which of the following pairs of subcellular compartments is likely to have same pH and electrolyte composition ?

(a) Cytosol and lysosomes.

(b) Cytosol and mitochondrial inter membane space.

(d) Mitochondrial matrix and inter membane space.

Ans. (b)

Sol. The intermembrane space (between the outer membrane and the inner membrane) is biochemically similar to the cytosol; however, this space also contains several enzymes that utilize ATP and several cell death promoting factors including apoptosis-inducing factor (AIF), capsases, and cytochrome C.

32. Regarding microtubule assembly and disassembly during cell division, which will be the most answer ?

(a) Once formed, kinetochore microtubules depolymerize at the plus ends throughout mitosis.

(b) Once formed, kinetochore microtubules polymerize at the plus ends throughout mitosis.

(d) Kinetochore microtubules polymerize at their minus ends up to cytokinesis, at which point they depolymerize.

Ans. (c)

Sol. At metaphase, polymerization at kinetochores occurs at the flux rate, whereas in anaphase, polymerization stops at kinetochores. Kinetochores then become tightly bound to the microtubule lattice ("park" state), and are pulled poleward by flux.

33. You are following the intracellular sorting of an integral plasma membrane protein in a living cell, in culture. You have decided to probe this protein by metabolic labeling technique with ³⁵S-methionine (pulse-chase technique). After one cycle of division, the cells were treated with a potent inhibitor of protein biosynthesis and processed for subcellular fractionation. In which of the following fractions will you expect the presence of this protein upon immunoprecipitation with a specific antibody ?

(a) Only cytoplasm

(b) Only plasma membrane

(d) Only secretory vesicle and endoplasmic reticulum

Ans. (c)

Sol. Pulse-chase analysis is a commonly used technique for studying the synthesis, processing and transport of proteins. Cultured cells expressing proteins of interest are allowed to take up radioactively labeled amino acids for a brief interval ("pulse"), during which all newly synthesized proteins incorporate the label. So, the protein will be present in both ER and PM.

34. The diploid genome of a species comprises 6.4 × 10⁹ bp and fits into a nucleus that is 6µm in diameter. If base pairs occur at intervals of 0.34 nm along the DNA helix, what is the total length of DNA in a resting cell ?

(a) 3.0 m

(b) 3.5 m

(d) 4.0 m

Ans. (c)

Sol. Total number of base pairs = 6.4 × 10⁹ bp

Distance between two consecutive base pair = 0.34 nm = 0.34 × 10^–9 m

Total length of DNA = 6.4 × 10⁹ × 0.34 × 10^–9 = 2.2 m

35. The principal pathway for transport of lysosomal hydrolases from the Golgi network (pH 6.6) to the late endosomes (pH 6.0) and the recycling of M6P (mannose 6 phosphate) receptors back to the Golgi depends on the pH difference between those two compartments. From what you know about M6P receptor binding and recycling and the pathways for delivery of material to lysosomes, predict what would hapen if the pH in late endosomes was raised to 6.6 ?

(a) M6P will bind to hydrolases but will not release the hydrolases in the late endosomes.

(b) M6P will bind to hydrolases and will release the hydrolases in the endosomes.

(c) At higher endosomal pH, the receptor would not release the hydrolase and could not be recylced back to the trans Golgi network.

(d) M6P will be degraded at higher pH.

Ans. (c)

Sol. The early endosome is acidified to pH ~6.2 by the V-ATPase, allowing the uncoupling of bound ligands from their receptors. By contrast, late endocytic organelles, late endosomes, and lysosomes are more acidic, with a luminal pH ~5.0–5.5, which is required for the full activation of lysosomal hydrolases. As a hallmark of endosomal maturation, the gradual acidification ensures the activation and inactivation of proteins and signaling molecules.

36. Phosphorylation of serines as well as methylation and acetylation of lysines in histones tails affect the stability of chromatin structure above the nucleosome level and have important consequences for gene expression. The resulting changes in charge are expected to affect the ability of the tails to interact with DNA because

(a) DNA is negatively charged.

(b) DNA-histone interaction is independent of net charge.

(d) Methylation and acetylation of lysine increases DNA-histone interaction.

Ans. (a)

Sol. Histone phosphorylation confers a negative charge to the histone, resulting in a more open chromatin conformation. It is therefore associated with gene expression and is involved in DNA damage repair and chromatin remodelling. DNA methylation inhibits gene expression in animal cells, probably by affecting chromatin structure. Biochemical studies suggest that this process may be mediated by methyl-specific binding proteins that recruit enzymatic machinery capable of locally altering histone modification.

37. Cells that grow and divide in a medium containing radioactive thymidine covalently incorporate the thymidien into their DNA during S phase. Consider a simple experiment in which cells are labeled by a brief (30 minutes) exposure to radioactive thymidine. The medium is then replaced with one containing unlabeled thymidine, and the cells grow and divide for some additional time. At different time points after replacement of the medium, cells are examined under a microscope. Cells in mitosis are easy to recognize by their condensed chromosomes and the fraction of mitotic cells that have radioactive DNA can be estimated by autoradiography and plotted as a function of time after the thymidine labeling as in the figure below :

The rise and and fall of the curve is because :

(a) initial rise of the curve corresponds to cells that were just finishing DNA replication when radioactive thymidine was added (S phase).

(b) the peak of the curve corresponds to cells in M phase.

(d) the fall in curve after 10 min indicates the cells exiting M phase.

Ans. (a)

Sol. S phase includes replication of both euchromatin and heterochromatin DNA, the radioactive thymidine will be incorporated in both types of DNA.

38. A rapidly growing bacterial species such as E. coli exhibits a typical phase of growth cycle in liquid nutrient broth (lag phase log phase stationary phase death phase). If a bacterial culture has a starting density of 10³ cells/ml has a lag time of 10 minutes and a generation time of 10 minutes, what will the cell density (cells/ml) be at 30 minutes ?

(a) 6.0 × 10³

(b) 2.0 × 10³

(d) 4.0 × 10³

Ans. (d)

Sol. Given, lag time = 10 minutes, generation time (T) = 10 minutes

B = number of bacteria at the beginning = 10³ cells/ml

Growth time of bacteria (t) = 30 min – 10 min = 20 min

Number of generations (n) = Growth time of bacteria/generation time = t/T = 20/10 = 2

As we know that, b = B × 2ⁿ = 10³× 2² = 4 × 10³ cells/ml; where, b = number of bacteria at the end.

39. Plasmids are self replicating small circular DNA elements in bacterila cells that can be said to have a stable symbiotic existence with the host cell. They often carry genes useful to the host. Which of the following is a potential threat to the evolution and stability of the symbiotic coexistence ?

(a) 'Copy-up' mutations that increase the rate of plasmid replication per host cell cycle.

(b) Reversible integration of plasmid DNA into the host DNA.

(d) Spontaneous curing of plasmids in a small proportion of host cells.

Ans. (d)

Sol. Plasmid curing is the process of obviating the plasmid encoded functions such as antibiotic resistance, virulence, degradation of aromatic compounds, etc. in bacteria. Several plasmid curing agents have been reported in literature, however, no plasmid curing agent can eliminate all plasmids from different hosts.

40. Complex eukaryotic cells may have evolved from simpler prokaryotic cells because complexity of organization increases the

(a) growth rate

(b) efficiency of energy utilization

(d) ability to attain larger size

Ans. (d)

Sol. Prokaryotic cells are known to be much less complex than eukaryotic cells since eurkaryotic cells are usually considered to be present at a later time of evolution. It is likely that Eukaryotic cells have evolved from prokaryotic cells.

These differences in complexity can be seen at the cellular level. The DNA of prokaryotes is a circular and attached to the plasma membrane, while eukaryotic DNA is packed into chromosome bundles. Prokaryote's DNA is contained in nucleoid which is not surrounded by nuclear membrane. Eukaryotic DNA is more complex where it has histone (protein that winds the DNA into a more compact form) and nonhistone proteins in chromosomes.

Chromosomes are contained in the nucleus with a nuclear envelope (a defining feature of eukaryotic cells).

41. You have created a fusion between the trp operon, which encodes the enzymes for tryptophan biosynthesis, under the regulatory control of the Iac operator. Under which of the following conditions will tryptophan synthase be induced in the strain that carries the chimeric operator fused operons ?

(a) Only when both lactose and glucose are absent.

(b) Only when both lactose and glucose are present.

(d) Only when lactose is present and glucose is absent.

Ans. (d)

Sol. If trp operon have regulatory elements of lac operon, tryptophan biosynthesis will occur only when lactose is present and glucose is absent.

42. Origin of replication usually contains

(a) GC rich sequences

(b) both AT and GC rich sequences

(d) AT rich sequences

Ans. (d)

Sol. Repeated sequences are commonly present in the sites for DNA replication initiation in bacterial, archeal and eukaryotic replicons. Those motifs are usually the binding places for replication initiation proteins or replication initiation proteins or replication regulatory factors. In prokaryotic replication origins, the most abundant repeated sequences are DnaA boxes which are the binding sites for chromosomal replication initiation protein DnaA, interons which bind plasmid or phage DNA replication initiators, defined motifs for site-specific DNA methylation and 13-nucleotide-long motifs of a not too well-characterized function, which are present within a specific region of replication origin containing higher than average content of adenine and thymine residues.

In this review, we specify methods allowing identification of a replication origin, basing on the localization of an AT-rich region and the arrangement of the origin's structural elements. We describe the regulatory of the position and structure of the AT-rich regions in bacterial chromosomes and plasmids.

The importance of 13-nucleotide long repeats present at the AT-rich region, as well as other motifs overlapping them, was pointed out to be essential for DNA replication initiation including origin opening, helicase loading and replication complex assembly. We also summarize the role of AT-rich region repeated sequences for DNA replication regulation.

43. -subunit of E. coli RNA polymerase does not

(a) initiate transcription and fall off during elongation.

(b) increase affinity of the core enzyme to the promoter.

(d) ensures specificity of transcription by interacting with the one enzyme.

Ans. (c)

Sol. Every molecule of RNA polymerase holoenzyme contains exactly one sigma factor subunit, which in the model bacterium E. coli is one of those listed below. The numebr of sigma factors varies between bacterial species. E. coli has seven sigma factors. RNA polymerase holoenzyme complex consists of core RNA polymerase and a sigma factor executes transcription of a DNA template strand. One initiation of RNA transcription is cooplete, the sigma factor can leave the complex.

It long has been thought that the factor obligatorily leaves the core enzyme once it has initiated transcription, allowing the free to link to another core enzyme and initiate transcription at another site. Thus the cycles from one core to another. However, does not obligatorily leave the core. Instead, the changes its binding with the core during initiation and elongation. Therefore, the cycles between a strongly bound state during initiation and a weakly bound state during elongation.

44. The cap binding protein (eIF4E), which is involved in the global regulation of translation, is highly regulated in eukaryotic cells. In an experiment, a researcher transfected mammalian cells with (eIF4E) gene for its overexpression. Due to this, the cells will undergo

(a) apoptosis

(b) neoplastic transformation

(d) differentiation

Ans. (b)

Sol. The cap binding protein (eIF4E), which is involved in the global regulation of translation, is highly regulated in eukaryotic cells. In an experiment, a researcher transfected mammalian cells with (eIF4E) gene for its overexpression. Due to this, the cells will undergo neoplastic transformation.

45. Bacteriophage T4 infects E. coli and injects its DNA inside the cell. The transcription of viral genes occurs in three stages : immediate early, early and late. All the promoters on viral genome are available, but the control takes place at the level of

(a) promoter strength

(b) modification of host RNA polymerase

(d) turnover rate of RNA synthesis

Ans. (b)

Sol. T4 early promoter strength probed in vivo with unribosylated and ADP-ribosylated E. coli RNA polymerase : a mutation analysis.

46. In order to study the role of telomeres in DNA replication, genetically engineered mice were prepared, where the gene for telomease RNA was knocked out. When cells from these knockout mice were taken and cultured in vitro, they proliferated even after 100 cells divisions which is quite unlikely in the case of human cells. Which of the following is the correct reason ?

(a) Human and mice are fundamentally different with respect to their requirements for telomerase enzyme in the context of DNA replication.

(b) In vitro, mice DNA becomes circular due to end chromosome fusion and does not require telomerase for DNA and replication.

(d) In vitro, mice DNA replication does not require the removal of RNA primers.

Ans. (c)

Sol. Knockout mice proliferated even after 100 cell divisions because Mice have very long stretch of telomere DNA sequence compared to that of human.

47. You are working an in vitro eukaryotic transciption system, which produced both caped and uncapped mRNAs. You incubated these mRNAs with mammalian cell nuclear extract and then quantified the different products as shown below. Which of the following graphs correctly represents the expected result ?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Graph mentioned in option (a) is correct.

48. A non-enzymatic viral protein X was found to be inducing a cellular gene promoter acitivity. Although no in vitro DNA binding activity could be identified with X protein, it was found to be co-recruited on the cellular promoter along with a cellular transcription factor in vivo. Which one of the following statements seems to be the best interpretation of the above findings ?

(a) X is a DNA-binding protein.

(b) X physically interacts with the transcription factor.

(d) X is a chaperone.

Ans. (b)

Sol. As per the information mentioned in question, X physically interacts with the transcription factor.

49. During elongation step of protein synthesis, translocation moves the mRNA and the peptidyl t-RNA by one codon through the ribosome. Translocation in E. coli involves GTP and EF-G. However, in vitro translocation can take place independent of GTP and EF-G. Based on these observations, the following hypotheses can be made

P. The molecular mechanism of translocation in vitro is completely different from that in vivo.

Q. Translocation activity is independent of GTP hydrolysis.

R. Translocation activity is completely dependent on GTP and EF-G.

S. Translocation activity is inherent in ribosomes, however, the rate of translocation in vivo is enhanced significantly in presence of GTP adn EF-G.

Which one of the following combinations is correct ?

(a) Only S

(b) P and R

(d) R and S

Ans. (a)

Sol. In vitro translocation can take place independent of GTP and EF-G because Translocation activity is inherent in ribosomes, however, the rate of translocation in vivo is enhanced significantly in presence of GTP adn EF-G.

50. DNA methylation plays an important role in transcription regulation in vertebrates. There is an inverse correlation between the level of DNA methylation in the vicinity of a gene and its transcription rate, whereas there is a direct correlation between histone acetylation and increased transcription. -thalassemia is a common genetic impairment of hemoglobin -chain synthesis in humans. If these patients can synthesizee hemoglobin-F instead of hemoglobin -chain in its place, they would be notably benefited. Administratio of 5-azacytidine to -thalassemia patients increases hemoglobin-F level in erythrocytes and thus benefit the patients. Which one of the following statements about 5-azacytidine is not correct ?

(a) Cells exposed to 5-azacytidine incorporate it into DNA in place of cytidine.

(b) 5-azacytidine decreases DNA methylation.

(d) 5-azacytidine does not promote gene expression.

Ans. (c)

Sol. Histone acetylation promote gene expression. 5-azacytidine does not promote gene expression. Hence, option (c) is incorrect.

51. Gram negative bacteria, Klebsiella pneumoniae, upon infecting humans, results in severe septic shock after a few hours of infection. Which of the following is not true for this type of infection ?

(a) Cell wall endotoxins cause overproduction of cytokines.

(b) Septic shock can be treated by anti- antibodies.

(d) Recombinant receptor antagonist can be used for the treatment of septic shock.

Ans. (c)

Sol. Klebsiella pneumoniae is a rare sepsis-causing bacteria, but it is well known for its severe outcomes with high mortality. Bacteremia caused by K. pneumoniae is seen more, and with a poorer prognosis. Current evidence does not support the use of human recombinant activated protein C in adults or children with severe sepsis or septic shock; moreover, there is an increased risk of bleeding associated with its use.

52. Which of the following is not associated with insulin action ?

(a) Increased glucose transport.

(b) Increased glycogen formation.

(d) Decreased rate of gluconeogenesis.

Ans. (c)

Sol. Insulin inhibited glycerol release by adipose tissue in all groups of rats with fructose in the medium. In fed rats a barely significant insulin inhibition occurred also in the presence of glucose and in fasted refed rats it was independent of whether a hexose was present in the medium or not.

The following effects of insulin on tissue of fasted-refed rats incubated in the absence of hexose wer observed : Inhibition of glycerol release with a dose effect relationship between 10 and 250 µ units insulin per milliliter; partial inhibiton of glycogen breakdown during incubation; diminished lactic and pyruvic acid release and a rise of the lactic to pyruvic acie ratio. These same affects were also exerted by theanilipolytic drug 5-methylpyrazole-3-carboxylic acid, which does not stimulate glucose uptake.

53. When adenoma is converted to metastatic adenocarcinoma, which of the following combination of proteins is almost certainly to be degraded ?

(a) Type IV collagen and laminin.

(b) Fibronectin and integrin.

(d) Elastin and selectin.

Ans. (a)

Sol. In metastasis, cancer cells break away from where they first formed and form new tumors in other parts of the body. Type IV collagen is the main component of the BM and provides a scaffold for assembly and mechanical stability, but is also important in cell adhesion, migration, survival, proliferation, and differentiation. It is a type of collagen found primarily in the skin within the basement membrane zone. It is degraded when adenoma is converted into metastatic adenocarcinoma.

54. Which of the following is considered to be a combined B- and T-cell deficiency ?

(a) Ataxia-telangiectasia

(b) Swiss type agammaglobulinemia

(d) Bruton's agammaglobulinemia

Ans. (b)

Sol. Combined B-cell and T-cell immunodeficiencies or SCID, is a group of medical disorders that are the result of genetic defects in both cellular and humoral immunity. The defects in humoral and cellular immunity have an early clinical presentation and if untreated, result in a fatal outcome in the first few years of life. This article focuses only on SCID disorders and outlines recent advances in therapeutics options for patients.

The profound degree of immune comprise in SCID leads to infections with bacterial, viral and fungal pathogens that cause significant morbidity and eventually mortality and eventually mortality in patients. Swiss type agammaglobulinemia is a type of combined immunodeficiency syndromes.

55. In the dark, rods show a large inward "dark" current which is suppressed by a flash of light. Which one of the following statements, explaining the effect of light, is true ?

(a) Sodium channels in the outer segment of rods are closed.

(b) Cytoplasmic cGMP concentration increases.

(d) Transducin dissociates from -arrestin.

Ans. (a)

Sol. Cyclic GMP is central to visual excitation in vertebrate retinal rod cells. Sodium channels in the plasma membrane of the outer segment are kept open in the dark by a high level of cGMP. Light closes these channels by activating an enzymatic cascade that leads to the rapid hydrolysis of cGMP.

56. Four groups of mice were studied for the factor required for mast cell generation : IL-3-deficient, GM-CSF-deficient, G-CSF-deficient and erythropoietin-deficient. In which mice, mast cell generation is most likely to be deficient ?

(a) IL-3-deficient

(b) GM-CSF-deficient

(d) Erythropoietin-deficient

Ans. (a)

Sol. IL-3 termed multi colony-stimulating-factor (multi-CSF) or mast cell growth factor (MCGF) is a haematopoietic growth factor which stimulates the formation of colonies for erythroid, megakaryocytic, granulocytic and monocytic lineages.

57. In cells having G protein coupled receptor, inhibition of protein kinase A by siRNA technology led to diminished transcription of androgen binding protein (ABP) and CREB protein. Addition of cAMP, which is a second messenger, will lead to

(a) increased transcription of ABP.

(b) increased phosphorylation of CREB protein.

(d) increased GTPase activity of subunit.

Ans. (c)

Sol. Signal transduction via G-protein coupled receptors (GPCRs) relies upon the production of cAMP and other signaling cascades. A given receptor and agonist pair, produce multiple effects upon cellular physiology which can be opposite in different cell types. Adenosine 3',5'-cyclic monophosphate (cAMP) is a nucleotide that acts as a key second messenger in numerous signal transduction pathways. cAMP regulates various cellular functions, including cell growth and differentiation, gene transcription and protein expression.

58. Binding of a ligand to a cell-surface receptor activates an intracellular signal transduction pathway through the sequential activiation of four protein kinases. In the human cell line A, these kinases are held in a signalling complex by a scaffolding protein whereas in another cell line B, these kinases are freely diffusible. Which one of the following possibilities do you think is not correct ?

(a) Speed of signal tarnsduction will be higher in cell A.

(b) Possibility of cross-linking with other signal transduction pathways will be lesser in cell A.

(d) Potency of spreading signal through other signaling pathways will be higher in cell B.

Ans. (c)

Sol. Protein kinases are an important class of intracellular enzymes that play a crucial role in most signal transduction cascades, from controlling cell growth and proliferation to the initiation and regulation of immunological responses. If they are held in signalling complex by scaffolding protein then the signal amplification will be higher.

59. Cells undergo apoptosis by the two distinct and inter-connected pathways : extrinsic and intrinsic. Extrinsic pathway is activated by extracellular ligand binding to cell surface death receptors. Whenever an apoptotic stimulus activates intrinsic pathway, the pro-apoptotic Bax and Bak proteins become activated and induce the release of cytochrome C from mitochondria leading to caspase cascade activation resulting in apoptosis. In cell A, cytochrome C is introduced by microinjection whereas in cell B, cytochrome C is introduced by microinjection but Bax and Bak are inactivated. What will be the most appropriate apoptotic response type in both cells ?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. During apoptosis, large pores form in the outer membranes of mitochondria. These pores are generated by two proteins—Bax and Bak—and they enable the mitochondrion to release proteins that activate processes involved in apoptosis. Pores also form in the mitochondrial membrane during necrosis. As caspase cascade activation is carried out by cyt c (which is active in both the cells), so equal % of apoptosis will be noticed in both the cells.

60. Dendritic cells (DC) from BALB/c mice were treated with IL-10 or with IFN-. Similarly, dendritic cells from -microglobulin deficient mice were also treated with IL-10 or with IFN-. These cells were co-culutred with CD8⁺ T-cells from hen egg lysozyme (HEL)-specific T-cell receptor transgenic mice in presence of the HEL peptide. Fve days later, CD8⁺ T-cells were assayed for target cell lysis. Which one of the following combinations will have the highest target cytotoxicity ?

(a) DC (BALB/c)^{IL – 10} × CD8⁺ T.

(b)

(c)

(d)

Ans. (b)

Sol. Dendritic cells (DCs) play a central role in the regulation of the balance between CD8 T cell immunity vs. tolerance to tumor antigens. Cross-priming, a process which DCs activate CD8 T cells by cross-presenting exogenous antigens, plays a critical role in generating anti-tumor CD8 T cell immunity.

61. The part of the embryo from which the ectoderm, mesoderm and endoderm are formed in chick is known as

(a) primitive streak

(b) hypoblast

(d) cytotrophoblast

Ans. (c)

Sol. The epiblast is capable of forming all three germ layers (ectoderm, mesoderm and endoderm) during gastrulation. Epiblast cells migrate to the primitive streak and invaginate into a space between the epiblast and the hypoblast. It is a cross section through the cranial region of the primitive streak at 15 days, illustrating and invagination of epiblast cells. The invaginating epiblast cells displace the hypoblast to create the definitive endoderm. Once the definitive endoderm is established, migrating epiblast cells also form the intraem-bryonic mesoderm. The remaining epiblast cells, which do not migrate through the primitive streak, remain in the epiblast to form the ecoderm. Hence, the epiblast gives rise to all three germ layers in the embryo.

62. Which protein secreted by the organizer induces neural tissue formation by inhibiting Bone Morphogenetic Protein ?

(a) -catenin

(b) Noggin

(d) Dishevelled

Ans. (b)

Sol. This protein is involved in the development of many body tissues, including nerve tissue, muscles, and bones. Noggin interacts with members of a group of proteins called bone morphogenetic proteins by inhibiting them. It is secreted by Spemann organizer.

63. The homologue of -catenin in Drosophila is

(a) Fushi tarazu

(b) Engrailed

(d) Cubitus interruptus

Ans. (c)

Sol. The armadillo repeat family of proteins is defined by the existence in length (42 amino acids) and spacing (typically end-to-end with no interventing sequences), though not necessarily highly conserved in sequence. Rather, it is the structure of the predominantly hydrophobic core of the the arm motif that is conserved; repeats of this domain together form a positively charged groove that mediates protein-protein interactions.

This is so-called 'arm motif' was originally identified in the protein product of the Drosophila segment polarity gene known as armadillo.

It was subsequently determined that the vertebrate homolgoues of armadillo, -catenin and plakoglobin and the Src tyrosine kinase substrate p120 catenin, also contain multiple arm repeats.

Together, these four proteins initially defined the armadillo repeat family of proteins. Although the function of these four proteins all coincidentally releate to cell-cell adhesion, it is now clear that members ofthe armadillo repeat family have widely varied roles in cellular physiology.

64. Which of the floral whorls is affected in apetala 3/pistillata (ap3/pi) mutants ?

(a) Sepals and petals

(b) Petals and stamens

(d) Sepals and stamens

Ans. (b)

Sol. Mutations in either AP3 or PI result in alterations in the 2nd and 3rd whorls, such that petals are replaced by sepals, and stamens are replaced by carpels.

65. Polyspermy results when two or more sperms fertilize an egg. It is usually lethal since it results in blastomeres with different numbers and types of chromosomes. Many species therefore, have two blocks to polyspermy : the fast block and the slow block.

In the case of sea urchins :

P. the fast block is immediate and causes the egg membrane resting potential to rise which does not allow the sperm to fuse with the egg and is mediated by an influx of sodium ions.

Q. the fast block is immediate and causes the egg membrane resting potential to rise which does not allow the sperm to fuse with the egg and is mediated by an efflux of sodium ions.

R. the slow block of cortical granule reaction is mediated by calcium ions.

S. the slow block of cortial granule reaction is mediated by potassium ions.

Which of the above statements are true ?

(a) P and R

(b) P and S

(d) Q and s

Ans. (a)

Sol. The fast block to polyspermy is transient, since the membrane potential of the sea urchin egg remains positive for only about a minute. This brief potential shift is not sufficient to prevent polyspermy, which can still occur if the sperm bound to the vitelline envelope are not somehow removed. The slow block of polyspermy is physical and is mediated by calcium ions. - Release of calcium ions propagates from the point of sperm entry, causing the cortical granules to fuse with the egg cell membrane.

66. Which of the inferences (P-S) given below would you draw from the following tissue transplantation experiments performed with the early and late gastrula stages of the newt ?

P. Cells of early newt gastrula exhibit conditional development.

Q. Cells of early newt gastrula exhibit autonomous development.

R. Cells of late newt gastrula exhibit conditional development.

S. Cells of late gastrula exhibit autonomous development.

The correct inferences are :

(a) P and S

(b) Q and R

(d) S only

Ans. (a)

Sol. In early gastrula the targeted tissue undergo conditional development because the ultimate fate depend on location in embryo. In late gastrulation, the transported tissue exhibits autonomous development independent of their final location after transplantation. Their fate is already determined.

67. Segmentation genes in Drosophila are divided into three groups (gap, pair rule and segment polarity) based on their mutant phenotype. Below are some of the major genes expressed in a sequential manner (with respect to the groups) affecting segmentation pattern.

P. Hairy paired tailless patched.

Q. Hunchback even-skipped fushi tarazu wingless.

R. Odd-skipped giant paired wingless.

S. Tailless hairy fushi tarazu gooseberry.

Which of the above sequence(s) of genes expressed from early to late embryo is/are correct ?

(a) S only

(b) P and Q

(d) Q and S

Ans. (d)

Sol. Q and S are true for sequence of genes expressed from early to late embryo.

68. Human chorionic gonadotropin (hCG) is known to facilitate attachment of blastocyst to uterus. In women with mutation in hcG gene, biologically inactive hcG was formed but implantation occurred. When hCG was immunoneutralized in the uterus of normal woman, implantation failed. This suggest that for implantation in humans :

(a) Biologically active circulating hCG is not required.

(b) Blastocyst can produce the required hCG, which helps locally in uterine attachement.

(d) Extra-embryonic tissue is not responsible for the attachment of embryo to uterus.

Ans. (b)

Sol. HCG is the first known hormonal signal of the conceptus. Its mRNA is transcribed as early as the 8-cell stage, while the blastocyst expresses HCG before its implantation. HCG is increasingly produced after implantation by the syncytiotrophoblast.

69. During reproductive development in plants :

P. Male and female gametes are produced as a result of two mitotic divisions after meiosis.

Q. Vegetative cells in pollen contribute to pollen development.

R. Antipodals provide nourishment to developing embryos.

S. Pollen tube ruptures and releases both the male gametes in one of the degenerated synergids.

Which of the above statements are true ?

(a) P and Q

(b) Q and S

(d) P and S

Ans. (b)

Sol. The known function of the vegetative cells in pollen grains is the extension of the pollen tube for the transportation of two sperm cells into the embryo sac for the process of fertilisation. In other words, the vegetative cells, following pollination, produce pollen tubes by serving as the pollen tube cell.

70. In an experiment, the cells that would normally become the middle segment of a Drosophila leg were removed from the leg forming area of the larva and were placed in the tip of the fly's antenna. Based on the "French flag" analogy for the operation of a gradient of positional information, which of the following statements is true ?

(a) The transplanted cells retain their committed status as leg cells, but respond to the positional information of their environment by becoming leg tip cells i.e. claws.

(b) The transplanted cells are determined as leg cells and therefore would form a complete limb instead of an antenna.

(c) The transplanted cells would intermingle with the cells present in the new environment and develop accordingly to give rise to an antenna.

(d) The transplanted cells retain their committed status as leg cells and would develop to form a chimeric structure having proximal region made of antenna and the distal region ending in a complete leg.

Ans. (a)

Sol. The transformation of antenna to leg is a classical model for understanding segmental fate decisions in Drosophila. Option (a) is correct statement.

71. During fertilization in mammals, sperm-egg interaction is mediated by zona pellucida (ZP) membrane proteins and their receptors present in sperm membrane. ZP3 has been identified to be the principle ZP protein whose post-translational modification is important for sperm-egg interaction. In a competitive inhibition assay the sperm is saturated with either active ZP3 or its modified forms, before studying sperm-egg-interaction. Which of the following experiments will not inhibit sprm-egg-interaction ?

(a) Saturate sperm with ZP3 protein prior to use.

(b) Deglycosylate the ZP3 protein and use it for saturation of sperm.

(d) Dephosphorylate the ZP3 protein and use it for saturation of sperm.

Ans. (b)

Sol. Presumably, the under-glycosylated ZP3 would fail to competitively inhibit sperm-egg binding, since chemically deglycosylated ZP3 no longer inhibits binding.

72. Which one of the following statements is incorrect about the role of oxidative pentose phosphate pathway in plant metabolism ?

(a) Generation of NADPH required to drive biosynthetic reactions.

(b) Production of pentose phosphate for the synthesis of nucleic acids.

(d) Productio of NADH to generate ATP.

Ans. (d)

Sol. This pathway, called the Pentose Phosphate Pathway, is special because no energy in the form of ATP, or adenosine triphosphate, is produced or used up in this pathway.

73. During photosynthetic carbon reduction cycle in green leaves, net production of one molecule of glyceraldehyde3-phosphate requires one of the following combinations of energy equivalents

(a) 9 NADPH and 6 ATP

(b) 3 NADPH and 9 ATP

(d) 6 NADPH and 9 ATP

Ans. (d)

Sol. For one turn: ATP used = 3 molecules, NADPH used = 2 molecules.

Thus for three cycles :

ATP used = 3 × 3 = 9 molecules

NADPH used = 2 × 3 = 6 molecules

74. Which one of the following essential micronutrients is associated with urease enzyme found in higher plants ?

(a) Nickel

(b) Molybdenum

(d) Copper

Ans. (a)

Sol. Nickel (Ni) is one of the essential micronutrients for higher plants and its known function is being the metal component of urease.

The effects of various Ni levels on urease activity in maize plnats grown in two nutrient media containing urea of ammnoium nitrate as two separate nitrogen sources were investigated. Ni, the most recently discovered essential element, is unique among plant nutrients in that its metabolic function was determined well before it was determined that its deficiency could disrupt plant growth. Subsequent to the discovery of its essentiality in the laboratory, Ni deficiency has now been observed under field condition in several perennial species.

75. Plants are able to percieve light through various photoreceptors and downstream genes. Which one of the following genes is not involved in light perception ?

(a) PIF3

(b) NPR1

(d) CRY 3

Ans. (b)

Sol. NPR1 (NONEXPRESSER OF PR GENES1) functions as a master regulator of the plant hormone salicylic acid (SA) signaling and plays an essential role in plant immunity.

76. If an Arabidopsis plant, mutated in lycopene biosynthetic pathway is grown in sunny tropical climate in the presence of oxygen :

(a) It would accumulate higher biomass due to higher rate of photosynthesis.

(b) There will not be any influence of this mutation on the rate of photosynthesis and plant growth.

(d) The leaves would be bluish purple in color because of higher accumulation of xanthophylls.

Ans. (c)

Sol. Plant carotenoids constitute the basic structural units of the photosynthetic apparatus. They serve a variety of critical functions such as light harvestation, and photoprotection of plants (light stress adaptation).

77. According to the current model of alternative oxidase regulation, the following factors cause induction of alternative oxidase :

P. Significant increase in the ubiquitin pool in the cytosol.

Q. Presence of -keto acids (like pyruvate and glyoxylate).

R. Cold stress.

S. Increase in cytosolic ATP concentration.

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

(a) P and S

(b) Q and R

(d) P and R

Ans. (b)

Sol. Alternative oxidase (AOX) is one of the terminal oxidases of the plant mitochondrial electron transport chain. AOX acts as a means to relax the highly coupled and tensed electron transport process in mitochondria thus providing and maintaining the much needed metabolic homeostasis by directly reducing oxygen to water. In response to stress AOX mediates a retrograde signaling pathway which in turn regulates gene expression both transcriptionally and post transcriptionally.

78. The oxidative pentose phosphate pathway provides the reducing equivalents for nitrite reduction in plastids (leucoplasts) of non-green tissues. Which one of the following statements would be correct for the above mentioned pathway ?

(a) Glutamate synthesized from NH₄⁺ is translocated from cytosol to leucoplast.

(b) -ketoglutarate is translocated from cytosol to leucoplast.

(d) Triose phosphate is translocated from cytosol to leucoplast.

Ans. (b)

Sol. Statement (b) is correct.

79. Perception of blue light in plants causes

(a) inhibition of cell elongation and stimulation of stomatal opening.

(b) stimulation of cell elongation and inhibition of stomatal opening.

(d) inhibition of cell elongation.

Ans. (a)

Sol. Blue light is involved in many plant responses, including phototropism initiation, stomata opening and hypocotyl elongation when the seedling emerges from the soil. Although Charles Darwin first documented plant responses to blue light in the 1800s, it was not until the 1980s that research began to identify the pigment responsible. Arabidopsis was used extensively in the study of the genetic basis of phototropism, chloroplast alignment and stomatal aperture and other blue light-influenced processes. These traits respond to blue light, which is preceived by three different blue light receptors : cryptochromes, phototropin light receptors and zeaxanthin.

Cryptochromes are necessary for seedling elongation just after emergece. Photoropin is a protein kinase involved in plant curvature during phototropic responses and zeaxanthin is a carotenoid which appears to be involved in stomatal openings in response to blue light.

80. Which one of the following pairs of precursor amino acidand alkaloid is correct ?

(a) 'Ornithine aspartate-nicotine' and 'tryptophan-quinine'.

(b) 'Ornithine-nicotine' and 'tyrosine-orphine'

(d) 'Ornithine-quinine' and 'ornithine apsartate-nicotine'.

Ans. (a)

Sol. Statement (a) is correct.

81. Following are few statements regarding water potential of soil.

P. The osmotic potential of soil water is generally negligible, except in saline soils.

Q. The osmotic potential of saline soil is always more than zero.

R. In dry soils the hydrostatic pressure of soil water potential is always positive.

S. Gravitational potential of soil water is always proportional to height of the tree.

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

(a) P and R

(b) Q and S

(d) S and P

Ans. (d)

Sol. The gravitational potential of soil water at each point is determined by the elevation of the point relative to some arbitrary reference level. If the point in question is above the reference, g is positive. If the point in question is below the reference, g is negative. Osmotic potential (o) is the decrease in the water potential which is due to the presence of solutes. Osmotic potential affects soil water flow whenever there is a gradient in solute concentration within the soil.

82. Typical morphological defects are routinely used in genetic screens to identify novel genes in signal transduction pathways. Which one of the following morphology has been used to decipher the ethylene signaling pathways ?

(a) Light grown morphology of seedling.

(b) Triple response morphology of seedling.

(d) Morphology of true leaves.

Ans. (b)

Sol. Exposure of plants to ethylene results in drastic morphological changes. Seedlings germinated in the dark in the presence of saturating concentrations of ethylene display a characteristic phenotype known as the triple response. This phenotype is robust and easy to score.

83. What would be the outcome if the theca interna cells were destroyed in a Graafian follicle ?

(a) Immediate formation of corpus albicans.

(b) Increased progesterone synthesis in the granulosa cells.

(d) Formation of corpus hemorrhagicum.

Ans. (c)

Sol. If the theca interna cells were destroyed in a Graafian follicle there is decreased estrogen synthesis in the granulosa cells because theca interna stimulate granulosa cells for estrogen synthesis.

84. The size of red blood cells (RBC) in venous blood is greater than that of arterial blood. The increased size of red blood cell in the venous blood is due to

(a) the increased permeability of red blood cell (RBC) membrane.

(b) the decreased osmotic pressure in plasma.

(d) the dissociation of cytoskeletal proteins in RBC.

Ans. (c)

Sol. The size of red blood cells (RBC) in venous blood is greater than that of arterial blood. The increased size of red blood cell in the venous blood is due to the increased osmotic pressure in RBC. It leads to inflow of aqueous content from blood plasma to RBC.

85. In bone marrow, stem cells are committed to different lineages. Factors that stimulate the colonies of these different lineages are interleukin-3 (multi-CSF), granulocyte-macrophage colony stimulating factor (GM-CSF) and granulocyte or macrophage colony stimulating factor (G-CSF or M-CSF). In a mouse deficient in GM-CSF, the number of hematopoietic cells will be altered. Which one of the following is correct ?

(a) Mast cells will be normal in number while granulocytes and macrophages will be deficient in number.

(b) Granulocytes count will be normal but not of macrophages.

(d) Mice will be deficient in all the three cell types.

Ans. (a)

Sol. In bone marrow, stem cells are committed to different lineages. Factors that stimulate the colonies of these different lineages are interleukin-3 (multi-CSF), granulocyte-macrophage colony stimulating factor (GM-CSF) and granulocyte or macrophage colony stimulating factor (G-CSF or M-CSF). In a mouse deficient in GM-CSF, the number of hematopoietic cells will be altered. In this condition, Mast cells will be normal in number while granulocytes and macrophages will be deficient in number.

86. An individual was suffering from digestive complication. It was observed that the individual had dehydrated gastrointestinal tract. When an advanced investigation was done, was found to have defects in the following :

P. cystic fibrosis transmembrane conductance regulator protein.

Q. glucose transporter protein.

R. Na⁺/K⁺ ATPase

S. Ca²⁺ ATPase

Which of the above could be the cause for such a digestive disorder ?

(a) P only

(b) Q and R

(d) S only

Ans. (a)

Sol. An individual was suffering from digestive complication. It was observed that the individual had dehydrated gastrointestinal tract. When an advanced investigation was done, was found to have cystic fibrosis transmembrane conductance regulator protein.

87. The action potential was recorded intracellularly from a squid giant axon bathed in two types of fluid such as sea water and artificial sea water having lower concentration of sodium ions while maintaining the same osmotic pressure with choline chloride. The nature of action potential was different in the two bathing fluids. Which of the following results in most likely ?

(a) The resting transmembrane potential was not changed but the amplitude of action potential was increased with lower sodium concentration in the bathing fluid.

(b) The amplitude of action potential was gradually decreased with reduction of sodium concentration in bathing fluid but the duration of action potential was prolonged.

(c) The resting transmembrane potential was decreased and the amplitude of action potential was also decreased with lower sodium conentration in the bathing fluid.

(d) The amplitude of action potential was not changed with reduction of sodium concentration in the bathing fluid but the duration of action potential was prolonged.

Ans. (b)

Sol. The action potential was recorded intracellularly from a squid giant axon bathed in two types of fluid such as sea water and artificial sea water having lower concentration of sodium ions while maintaining the same osmotic pressure with choline chloride. The nature of action potential was different in the two bathing fluids. The amplitude of action potential was gradually decreased with reduction of sodium concentration in bathing fluid but the duration of action potential was prolonged.

88. Three forms of dextrans namely neutral, polyanionic and polycationic having different molecular radii were injected separately in three groups of rats. The concentrations of dextrans in glomerular filtrate were measured to determine the filterability of the dextrans. The possible outcomes could be as follows :

P. The dextrans having smaller diameter have greater filterability than larger dextrans.

Q. Neutral dextrans were filtered more than poly-cationic and polyanionic dextrans.

R. Polycationic dextrans were filtered more than neutral and polyanionic dextrans.

S. Polyanionic dextrans were filtered more than neutral and polycationic dextrans.

Which one of the following combinations is correct ?

(a) P only

(b) Q only

(d) Q and S

Ans. (c)

Sol. Three forms of dextrans namely neutral, polyanionic and polycationic having different molecular radii were injected separately in three groups of rats. The concentrations of dextrans in glomerular filtrate were measured to determine the filterability of the dextrans. The dextrans having smaller diameter have greater filterability than larger dextrans and Polycationic dextrans were filtered more than neutral and polyanionic dextrans.

89. A chromosomes aberration leads to change in the order of genes in a genetic map but does not alter its linkage groups. This is due to

(a) translocation

(b) recombination

(d) inversion

Ans. (d)

Sol. Single crossovers within the inversion loop lead to recombinant gametes that are imbalanced for genetic material OUTSIDE the loop. One recombinant product is duplicated for the region at outside the loop at one end of the chromosome and simultaneously deleted for the material outside the loop at the other end of the chromosome. The reciprocal recombinant gamete has the reciprocal imbalance.

All gametes have the expected amount of DNA for everything within the inversion loop. If the centromere is in the inversion (pericentric) then each meiotic product has a centromere. If the centromere is outside of the inversion loop then the meiotic products that result from a single cross over with in the loop are imbalanced for the centromere as well as the surrouding genes one recombinant product will be dicentric and the other will be acentric.

90. The concept of recon was proposed by Seymour Benzer by studying recombination between

(a) lysis mutants of bacteriophage T4.

(b) white eye mutantsof Drosophila melangaster.

(d) auxotrophic mutants of Escherichia coli.

Ans. (a)

Sol. The concept of recon was proposed by Seymour Benzer by studying recombination between lysis mutants of bacteriophage T4.

91. Aspartic acid (Asp) is specified by the codon GAU and GAC. After mutation, Asp is changed to Alanine represented by GCX, where X may be A, U, C or G. The reversion of the mutation could only be done with reactive oxygen species. The nature of the mutation is considered to be

(a) transition

(b) transversion

(d) depurinztion

Ans. (b)

Sol. Aspartic acid (Asp) is specified by the codon GAU and GAC. After mutation, Asp is changed to Alanine represented by GCX, where X may be A, U, C or G. The reversion of the mutation could only be done with reactive oxygen species. The nature of the mutation is considered to be transversion.

92. A cross is made between two plants with white flowers. All the F1 progeny had red coloured flower. This is because of

(a) complementation

(b) recombination

(d) reversion

Ans. (a)

Sol. A cross is made between two plants with white flowers. All the F1 progeny had red coloured flower. This is because of complementation.

93. A novel enzyme was identified in humans. The following approaches are available to identify the chromosome on which the gene encoding the enzyme is present :

P. Suppress the activity of enzyme by RNAi.

Q. Identify polymorphism in the population and carry out pedigree analysis to study its inheritance.

R. Purify the enzyme, deciphre its amino acid sequence, predict its DNA sequence and search for its presence in the available human genome sequence.

S. Create chromosome addition lines by making somatic hybrids between human and mouse cells, identify lines showing the enzymes activity and the human chromosome present in it.

Which of the above approaches can be used ?

(a) P or Q

(b) Q or R

(d) P or R

Ans. (c)

Sol. To identify the chromosome on which the gene encoding the enzyme is present, we should purify the enzyme, deciphre its amino acid sequence, predict its DNA sequence and search for its presence in the available human genome sequence. We can also create chromosome addition lines by making somatic hybrids between human and mouse cells, identify lines showing the enzymes activity and the human chromosome present in it.

94. In an experiment on transposition in an eukaryotic system, an intron was cloned within a transposable element and allowed to transpose from a plasmid to genomic DNA. The intron was found to be absent in the transposable element in its new location. It is

(a) not a case of transposition.

(b) a case of replicative mode of transposition.

(d) a retroposon.

Ans. (d)

Sol. Retroposon : A transposition of DNA sequences that does not occur in DNA itself but rather when mRNA is transcribed back into the genomic DNA. Or it is A class of genetic elements that include retroviruses and transposons that have an intermediate RNA stage. A transposon that was created by reverse transcription of an RNA molecule.

95. In a plant species, a segregating line (one that contains both homozygotes and heterozygotes at a locus) can be made homozygous by repeated selfing for several generations. What is the level of remaining heterozygosity after three generations of selfing, if the level of heterozygosity in generation '0' is denoted as 1 ?

(a) 0.5

(b) 0.25

(d) 0.0625

Ans. (c)

Sol. Of the ensemble of all heterozygous loci, then after one generation of selfing, only 1/2 will still be heterozygous; after two generations, 1/4; after three, 1/8. In the n^th generation.

Het_n =

where Het_n is the proportion of heterozygous loci in the n^th generation and Het₀ is the proportion in the 0 generation.

96. Given below is the result of a complementation test for six independent mutants (1 to 6).

'+' represents complementation; '0' represents non-complementation.

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

(a) The mutations can be ordered in a single cistron as 1-3-5-2-4-6.

(b) All mutations belong to a single cistron, but their order cannot be determined.

(c) There are three cistrons, mutations 1, 3 and 6 represent one cistron, 4 and 5 represent the second cistrons and 2 represents the third cistrons.

(d) There are three linkage groups, mutations 1, 3 and 6 represent linkage group A, 4 and 5 represent linkage group B, and 6 represents linkage group C.

Ans. (c)

Sol. By observing results of complementation test, we can conclude that there are three cistrons, mutations 1, 3 and 6 represent one cistron, 4 and 5 represent the second cistrons and 2 represents the third cistrons.

97. In a hospital three babies were mixed up. The blood group of the babies were A, B and AB. In order to identify the parents of the babies, the blood groups of the parents were determined. The results obtained were :

Parent set 1 - A and AB

Parent set 2 - AB and O

Parent set 3 - B and AB

Which of the following conclusions can be definitively made ?

(a) The baby with blood group A is the child of the parent set 2.

(b) The baby with blood group AB is the child of the parent set 1.

(d) The parentage of none of the babies can be determined from the given information.

Ans. (d)

Sol. By observing the result of blood group of parents set we can say that the parentage of none of the babies can be determined from the given information.

98. There are two mutant plants, One shows taller phenotype than wild-type, whereas the other has the same height as the wild-type. When these two mutations were brought in together by genetic crosses, the double mutant displayed even taller phenotype than the tall mutant plants. This genetic interaction is called

(a) antagonistic interaction

(b) additive interaction

(d) suppressive interaction

Ans. (c)

Sol. Synergy occurs when the contribution of two mutations to the phenotype of a double mutant exceeds the expectations from the additive effects of the individual mutations. The molecular characterization of genes involved in synergistic interactions has revealed that synergy mainly results from mutations in fructionally related genes. Recent research in Arabidopsis thaliana has advanced our understanding of the scenarios resulting in synergistic phenotypes.

Those involving homologous loci usually result from various levels of functional redundancy. Some of these loci fail to complement each other or become dose-dependent in certain multiple mutant combinations, suggesting that the effects of haploinsufficiency and redundancy can combine.

Synergy involving non-homologous genes probably reflects the topology of the regulatory or metabolic networks and can arise when pathways that converge at a node are disrupted. The Hub genes provide a remarkable example, these genes have an extraordinary number of connections and mutations that interact with many unrelated pathways.

99. Cladistic classification is based on

(a) sequential order in which branches arise from a phylogenetic tree.

(b) the order of sequence divergence.

(d) cellular organization and cytoskeleton.

Ans. (a)

Sol. Cladistics shows the evolutionary relationship of recent ancestors and descendants or the pathway of evolution from the very start. It can be represented with the help of a cladogram. Phylogenetic trees are based on clades; hence cladistics is a part of the Phylogeny.

100. Tautonym is an informal taxonimic designation used for animals referring to

(a) same name for genus and species.

(b) same name for species and subspecies.

(d) the name of the author for the species.

Ans. (a)

Sol. A tautonym is a binomial name in which the genus and the species are given the same name.

101. A marine biologist dug up a small animal from the ocean floor. The animal was uniformly segmented with short, stiff appendages and soft, flexible skin. It had a complete digestive system and an open circulatory system but no exoskeleton. Based on this description, the animal appears to be a

(a) lancelet

(b) roundworm

(d) crustacean

Ans. (b)

Sol. Roundworms (nematodes) are bilaterally symmetrical, worm-like organisms that are surrounded by a strong, flexible noncellular layer called a cuticle. Their body plan is simple. The cuticle is secreted by and covers a layer of epidermal cells.

102. Which of these programs is used to conserve a species facing extinction ?

(a) Captive breeding

(b) Natural resources

(d) Edge effects

Ans. (a)

Sol. This process happens before species are reintroduced into an area where they once lived.

103. A grasshopper popultion is being assessed by capture-mark-release-recapture method. On the first day, 100 grasshopper were captured from a given area in 1 hour time, marked and released. On the next day during recapture, 10 marked and 90 unmarked grasshoppers could be found in the same time period from same area. What will be the estimated population size in the given area ?

(a) 80

(b) 100

(d) 10,000

Ans. (c)

Sol. Total population size =

Where, N = Total population size.

M = Marked initially.

R = Marked recaptures.

T = Total in second sample.

N = = 1000.

104. Free living nitrogen fixers can survive in different ecological niches. Identify the incorrect combination from the following list :

(a) Azotobacter-acidic soil

(b) Deraxia-alkaline soil

(d) Frankai-neutral soil

Ans. (a)

Sol. Azotobacter species are generally found in slightly acidic to alkaline soils, which often governs the occurrence of certain species. So, option a is incorrect combination.

105. A plot of soil contaminated with diesel oil was inoculated with oyster mushrooms. After 4 weeks, more than 95% of the polycyclic aromatic hydrocarbons had been reduced to non-toxic compounds. This process is called

(a) phytoremediation

(b) chemoremediation

(d) zooremediation

Ans. (c)

Sol. Bioremediation is the process of cleaning up of degrading environmental contaminants using biological organisms especially microbes. Here mushrooms are used. Mushrooms belong to fungi. Therefore myco (fungus) remediation.

106. The following table gives vascular tissue characteristics of four divisions of Tracheophyta.

Identify the correct combinations :

(a) P–1, Q–2, R–3, S–4

(b) P–2, Q–1, R–4, S–3

(d) P–3, Q–4, R–1, S–2

Ans. (b)

Sol. Psilophyta have tracheids.

Lycopodiphyta have well -developed tracheid and pits in lateral wall."Sphenophyta- primitive tracheids and pits in lateral wall"Pteridophyta- Pteridophytes evolved a system of xylem and phloem to transport fluids and thus achieved greater heights than was possible for their avascular ancestors, having tracheids, vessels and well-developed phloem.

107. Which of the following is not an advantage to speed based reproduction ?

(a) Reserve food material is provided for the developing embryo.

(b) Seed coat protects the embryo and allows it to remain dormant until favourable environmental conditions are available.

(d) The female gametophyte remains on the sporophyte which provides and noruishment.

Ans. (c)

Sol. After fertilisation, a tiny plant called an embryo is formed inside a seed. The seed protects the embryo and stores food for it. The parent plant disperses or releases the seed. If the seed lands where the conditions are right, the embryo germinates and grows into a new plant. The sporophyte produces spores by meiosis. The spores grow into gametophytes. The gametophytes produce gametes (eggs and sperm). System c is not an advantage of seed based reproduction.

108. As per the International Code of Botanical Nomenclature, 2006 (Vienna Code), which of the following is a Nothospecies ?

(a) Polypodium vulgare subsp. prionodes (Asch.) Rothm.

(b) Polypogon monspeliensis (L.) Desf.

(d) Agrostis stolonifera L. × Polypogon monspeliensis (L.) Desf.

Ans. (d)

Sol. Nothospecies : A hybrid which is formed by direct hybridization of two species, not other hybrids. Option d is a nothospecies.

109. Which of the following groups have only two wings ?

(a) Honey bee, beetle, ant.

(b) Butterfly, housefly, fruitfly.

(d) Housefly, fruitfly, mosquito.

Ans. (d)

Sol. Flies and mosquitoes are classified as order Diptera, which mean two wings. The insects in this order have only one pair of membranous flying wings. The second pair of wings are reduced to small knobs, called halteres for the purpose of balancing. Their body is relatively soft and hairy.

They have a pair of large compound eyes, a pair of very short antennae and sucking mouth. Example of such flies are Housefly, fruitfly, mosquito.

110. In a study of sexual isolation in a species of salamander, scientists brought together males and females from different populations and from the same population. They observed the frequency of mating and calculated a sexual isolation index. One graph shows the relationship between mating frequency and genetic distance, and the other shows the relationship between sexual isolation index and geographic isolation.

Choose the appropriate for X₁, Y₁, X₂ and Y₂ in the figures, above.

(a) X₁ = Geographic distance; Y₁ = Sexual isolation index; X₂ = Genetic distance; Y₂ = Mating frequency.

(b) X₁ = Geographic distance; Y₁ = Mating frequency; X₂ = Genetic distance; Y₂ = Sexual isolation index.

(d) X₁ = Genetic distance; Y₁ = Geographic distance; X₂ = Sexual isolation index; Y₂ = Mating frequency.

Ans. (a)

Sol. Islands epitomize allopatric speciation, where geographic isolation causes individuals of an original species to accumulate sufficient genetic differences to prevent them breeding with each other when they are reunited. Increase in geographic distance will increase sexual isolation index. Genetic distance is a measure of the genetic divergence between species or between populations within a species, whether the distance measures time from common ancestor or degree of differentiation. Populations with many similar alleles have small genetic distances. With increase in genetic distance, mating frequency is reduced.

111. India has currently 17 biosphere reserves representing different ecosystems. These conservation areas significantly differ from the conventional protected areas of the country. Identify the correct combination of attributes (1 to 7) that best explains the concept of biosphere reserve.

1. Conservation,

2. Education,

3. Human habitation allowed,

4. Human habitation not allowed,

5. Strong legal back-up,

6. No supporting act,

7. Research.

(a) 1, 2, 3, 6, 7

(b) 1, 2, 4, 6, 7

(d) 1, 4, 5, 7

Ans. (a)

Sol. Biosphere reserve integrate three main "functions": Conservation of biodiversity and cultural diversity. Economic development that is socio-culturally and environmentally sustainable. Logistic support, underpinning development through research, monitoring, education and training.

112. Followings are the niche characteristics of the constituent species and resoure partitioning pattern in different ecosystems. Which of these would lead to competitive exclusion of species ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Competitive interactions among the populations of two species will lead to the exclusion of one of the species when the realized niche of the superior competitor encompasses the fundamental niche of the inferior competitor. In graph b, niche overlap is maximum and resource partitioning is minimum so, competitive exclusion will take place.

113. Environmental conditions can influece accumulation of species in successional communities. Curves representing changes in forest species over time are given in the figure below. Which of the following keys is correct for the curves ?

(a) A = xeric, B = mesic, C = intermediate

(b) A = intermediate, B = xeric, C = mesic.

(d) A = mesic, B = intermediate, C = Xeric

Ans. (d)

Sol. Many factors affect small-scale species richness, including geographic (e.g. species pool, dispersal), biotic (e.g. competition, predation, facilitation) and abiotic (e.g. resource availability, environmental heterogeneity, disturbance frequency and intensity). The factors related to these patterns of small- scale species richness include (1) geographic factors such as scale of observation, available species pool and dispersal patterns, (2) biotic factors such as competition or predation and (3) abiotic environmental factors such as site resource availability, disturbance. Mesic species will survive in mesic biome is one that enjoys a fairly constant supply of moisture and will decline after that . Xeric species requires time to increase their number.

114. A plant with blue-coloured flowers was observed to attract a large number of pollinators. However, these flowers were not producing any nectar. Which of the following can be a logical explanation to the observation ?

(a) There could be another species in the vicinity that has blue flowers and is rich in nectar.

(b) There is no other species with blue flowers in the vicinity so pollinators are compelled to visit this species.

(d) Pollinators may be able to see only blue colour.

Ans. (a)

Sol. Statement a is logical explanation. There could be another species in vicinity that has blue flowers and rich in nectar, that's why blue flowers attract pollinator in search of nectar.

115. Three islands have identical habitat characteristics. On first island rodent species A is present at a density 325/km². Second islands has only speices B at a density of 179/km². On the third islands, both A and B co-exist with densities 297/km² and 150/km² respectively. Which of the following can be inferred from this ?

(a) The two species do not compete with each other.

(b) The intra-species competition is more intense than inter-species competition.

(d) The inter and intra species competition are of the same intensity.

Ans. (b)

Sol. Intraspecific competition occurs between individuals of the same species. Intraspecific competition is usually more intense than interspecific competition because the individuals have the same niche so are competing for exactly the same resources.

116. A few males and females of a species were introduced to a new island. Their population was monitored overr several generations and followed a pattern shown in the figure. Which of the characteristics of the species does not explain the pattern ?

(a) Skewed sex ratio (more females than males).

(b) Large litter size.

(d) Effects of intra-uterine development on fecundity.

Ans. (a)

Sol. Skewed sex ratio means less number of females per equal number of males. Less number of females means less number of total live births provided the fertility ratio being constant or declining. Thus, the population growth rates gets affected negatively with increasing skewed sex ratio.

117. In pre-industrial period in England, peppered moths had light coloration which effectively camouflaged them against light coloured trees and lichens. During industrial revolution, many lichens died out and trees and lichens. During industrial revolution, many lichens died out and trees became blackened by soot from factories and interestingly, dark coloured moths were predominantly seen. This happened due to

(a) natural selection of dark coloured moths which initially present in fewer numbers.

(b) new mutation which arose due to environmental pollution.

(d) natural selection of the camouflaging mechanism of the moths.

Ans. (a)

Sol. The evolution of the peppered moth is an evolutionary instance of directional colour change in the moth population as a consequence of air pollution during the Industrial Revolution. The frequency of dark-coloured moths increased at that time, an example of industrial melanism.

118. The speciation in which a population splits into two geographically isolated populations experience dissimilar selective pressure and genetic drift is known as

(a) sympatric speciation

(b) parapatric speciation

(d) allopatric speciation

Ans. (d)

Sol. In allopatric speciation, a population splits into two geographically isolated populations due to formation of a barrier between portions of a population, for example, because of mountain building as depicted in the animal to the left.

The isolated populations then experience differentiating genotypic and phenotypic divergence as a result of different selective pressures in their differing environments. Additionally, genetic drift will occur differently, eventually differentiating the population's genotypes and phenotypes.

119. Evolution of multi-gene family occurs by

(a) only gene duplication

(b) only unequal crossing-over

(d) both duplication and unequal crossing over

Ans. (d)

Sol. Evolution of multigene families by gene duplication and subsequent diversification is analyzed assuming a haploid model without interchromosomal crossing over. Chromosomes with more different genes are assumed to have higher fitness. Advantageous and deleterious mutation and duplication/deletion also affect the evolution as in previous studies. In addition, negative selection on the total number of genes (copy number selection) is incorporated in the model.

120. In an experiment has continued for more than 50 years, corn has been propagated by bredding only from plants with the highest amount of oil in the seeds. The average oil content is now much greater than any of the plants in the original population. The following hypotheses were proposed as explanations for this observation.

P. Mutations occurred that increased the oil content in seeds.

Q. Plants with high oil content were stimulated to produce offspring with more oil in the their seeds.

R. The breeding led to increased frequency of alleles at multiple loci, so that new combinations of genes for even higher oil content were formed.

Which of the following represents a combination of correct statements ?

(a) P and Q only

(b) P and R only

(d) P, Q and R

Ans. (b)

Sol. Mutational effects can be beneficial, harmful, or neutral, depending on their context or location. Breeding leads to increased frequency of alleles at multiple loci, leading to new combinations.

121. Knox genes code transcriptional factors important for the regualtion of indeterminate growth in plant shoots. These genes also regulate patterns of development of plant organs such as leaves and flowers. The figure represents a phylogenetic tree of the multigene family in some land plants. The circles represent genes that act to maintain shoot apical meristem (equivalent to stem cells). Orthologues are genes that duplicate due to speciation and paralogues are genes that duplicate within a species.

From the figure, the following inferences were made.

P. Multiple gene duplication occurred in vascular plants.

Q. Gene duplications may have enabled shoot diversification in vascular plants.

R. Shoot apical meristems are regulated by ortho-logous genes in vascular plants.

S. Shoot apical meristems are regulated by paralogous genes in vacular plants.

Which of the following represents a combination of correct inferences ?

(a) P, Q and S

(b) P, Q and R

(d) Q and S only

Ans. (a)

Sol. Statement P, Q and S is correct.

122. The Galapagos finches were an important clue to Darwin's thinking about the origin of speices. These finches are believed to have descended from a single ancestral species that colonized the Galapagos archipelago, America, over a short period of time. The Galapagos finches differ in their break shape and size. Different species feed on seeds that vary in size and hardness. Which of the following is the most likely explanation for these patterns ?

(a) The finches represent an example of directional trend in break size from small to big.

(b) Beak shapes changed in responses to different seed types and these changes were inherited by subsequent generations.

(c) The ancestral finch already had all the beak variations and different lineages formed that were specialized to eat diffeent seed types.

(d) The finches represent an example of adaptive radiation in which beak variation was generated by mutation followed by selection by different seed types.

Ans. (d)

Sol. Darwin's finches are a good example of adaptive radiation. It is an evolutionary process starting from a point in a geographical area, giving rise to new species depending upon habitat.

123. In order to demonstrate that the long tails of males attracted females in a bird species, experiments captured and cut the tails of 'n' number of males and monitored the number of females mated by each male. They had two types of controls in the experiment.

P. 'n' males that were not captured.

Q 'n' males that were captured, had their tails cut and then stitched back to attain the original size.

The males with cut tails mated with a significantly smaller number of females than both the controls. Which of the following alternative explanations is not ruled out by the experiment ?

(a) The stress of cutting tails affected the performance of males.

(b) The time wasted in the capture reduced mating opportunities of males.

(d) Females chose males randomly.

Ans. (c)

Sol. Males with long tails also carry genes with preference for long tails. They become correlated. Once female preference passes through a population, males with long tails are getting selected for twice: once for fitness, and once because they are preferred by females.

124.

In the phylogenetic tree above, branch-lengths are drawn proportional to the number of changes along a lineage. The following inferences were made from this tree.

P. Bacteria are more closely related to Eukarya than to Archea.

Q. Bacteria and Archea are more similar to each other than either is to Eukarya.

R. Archea and Eukarya diverged from each other after their common ancestor diverged from bacteria.

Which of the following represents a combination of correct inferences ?

(a) P, Q and R

(b) P and Q only

(d) P and R only

Ans. (c)

Sol. Bacteria and archaea were on Earth long before multicellular life appeared. They are ubiquitous and have highly diverse metabolic activities. This diversity allows different species within clades to inhabit every imaginable surface where there is sufficient moisture. Statement Q and R is correct.

125. One aims to find out the role of a gene product in macrophages by using a transgenic mouse expressing the genes under a promotor. Which of the following is the most appropriate promoter ?

(a) Actin promoter

(b) MHC Class II promoter

(d) IL-2 promotor

Ans. (c)

Sol. CD11b is the alpha chain of the Mac-1 integrin and is preferentially expressed in myeloid cells (neutrophils, monocytes, and macrophages). We have previously shown that the CD11b promoter directs cell-type-specific expression in myeloid lines using transient transfection assays.

126. Which of the following genes was engineered in the "Flavr Savr" transgenic tomato variety ?

(a) 1-amino cyclopropane-1-carboxylic acid synthase

(b) 1-amino cyclopropane-1-carboxylic acid oxidase

(d) Polygalacturonase

Ans. (d)

Sol. 'Flavr Savr' tomato has increased shelf life by silencing the polygalactouronase gene involved in ripening by anti sense technology.

127. For developing transgenic mice, embryonic stem cells are engineered to express the transgene. These cells are selected by

(a) novabiocin

(b) neomycin

(d) penicillin

Ans. (b)

Sol. PGK–Neo (a hybrid gene consisting of the phosphoglycerate kinase I promoter driving the neomycin phosphotransferase gene) is a widely used cassette employed as a selectable marker for homologous recombination in embryonic stem cells.

128. Microbial leaching involves the process of dissolution of metals from ore breaking rocks using microorganisms. Which one of the following bacteria helps in leaching copper from its ore ?

(a) Acidithiobacillus ferroxidans

(b) Pseudomonas putida

(d) Rhodopseudomonas capsulate

Ans. (a)

Sol. The Acidithiobacillus group of microorganisms is important for dissolution of Cu, Zn, Fe and As, from different ores and waste. They accelerate the dissolution of sulfide minerals, which leads to an increased production acid mine drainage.

129. In the case of monoclonal antibody production by hybridoma technology, myeloma cells used lack the enzyme hypoxanthine-guanine phosphoriboxyl transferase (HGPRT) such that fusec cells can only survive when selected on hypoxanthine-aminopterin-thymidine (HAT). What is the role of aminopterin in this medium ?

(a) To be used as cell cycle inhibitor of myeloma cells.

(b) To block the pathway for nucleotide synthesis.

(d) To facilitate production of antibody producing B-cells.

Ans. (b)

Sol. Aminopterin present in HAT media blocks the power of cells to synthesize nucleotides by the de novo synthesis pathway. Hypoxanthine and deoxythymidine allow cells with functional hypoxanthine-guanine phosphoribosyltransferase (HGPRT) genes to survive through salvage pathways.

130. Molar absorption coefficient of phenylalanine is 200 M^–1 cm^–1 at 257 nm. What concentration (g/L) of this amino acid will give an absorption of 1 in a cell of 0.5 cm path length at 257 nm ?

(a) 3.30

(b) 0.33

(d) 0.17

Ans. (c)

Sol. From Beer-Lambert law, A =

= 1.65 g/L

Where, A = absorbance, c = concentration, = molar absorption coefficient and M = molecular weight of the amino acid.

131. Which of the following atomic nuclei cannot be probed by nuclear magnetic resonance spectroscopy ?

(a) ¹H

(b) ³¹P

(d) ¹⁵N

Ans. (c)

Sol. ¹H : most commonly used.

¹⁵N, ³¹p : used in biochemical studies.

¹⁸O : not used.

132. t_1/2 of an irreversible first order reaction, S P is 1 hour. The time (in hours) required to reach 75% completion is

(a) 1.5

(b) 2.0

(d) 3.0

Ans. (b)

Sol. …(1)

Put the value of equation (1) in above equation, we have

t_75% = 2 × t_50% = 2 × 1 = 2.

Similarly, t_87.5% = 3 × t_50% and t_99.9% = 10 × t_50%.

133. Mouse erythroleukemia (MEL) cells are used an an in vitro cell culture model for understanding erythropoiesis. These cells are arrested at the stage of pro-erythroblast due to transformation. These cells could be induced by heme to differentiate further so as to synthesize hemolobin. The most probable molecular mechanism for this could be that heme may suppress and/or downregulate an endogenous heme-regualted inhibitor (HRI) kinase, an inhibitor of globin synthesis. This downregulation in turn promotes differentiation. To validate this hypothesis which of the following approaches is not appropriate ?

(a) Transfect MEL cells with HRI kinase gene.

(b) Knock down HRI kinase gene in MEL cells.

(d) Measure HRI kinase activity as a function of differentiation.

Ans. (c)

Sol. Determining the rate of protein synthesis in situ as a function of differentiation can't validate the hypothesis provide in question but by observing transfect MEL cells with HRI kinase gene, knock down HRI kinase gene in MEL cells and measure HRI kinase activity as a function of differentiation, hypothesis can be validated.

134. While attempting to create a disease model of poliomyelitis in mice, it was found that mice cannot be infected with the said virus. Since human beings are susceptible to this viral infection, which kind of transgenic mice should be generated to have a transgenic mouse model that can be infected with polio virus ? Select the right approach from below :

(a) A mouse expressing surface protein of polio virus.

(b) A mouse expressing human receptor gene which makes cell surface protein for docking and internalization of polio virus.

(d) A mouse expressing human receptor gene which makes cell surface protein for docking and internalization of polio virus along with a gene designed to express surface protein of this virus at puberty.

Ans. (b)

Sol. A mouse expressing human receptor gene which makes cell surface protein for docking and internalization of polio virus.

135. Protoplast fusion is used in plant tissue culture for various applications. In protoplast fusion :

P. naked plant cells are used.

Q. transfer of organelles is not possible.

R. partial genome transfer is invovled.

S. cells from two different plants can be mixed together and forced to fuse.

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

(a) P, Q and R

(b) P, R and S

(d) Q, R and S

Ans. (b)

Sol. Protoplast fusion is a physical phenomenon,during fusion two or more protoplasts come in contact and adhere with one another either spontaneously or in presence of fusion inducing agents. Statement P, R and S is correct.

136. Which of the following is a mismatch between the plant drug and its source ?

(a) Codeine – Papaver somniferum

(b) Vinblastine – Catharanthus roseus

(d) Digitaline – Artemisia annua

Ans. (d)

Sol. Digitalis, drug obtained from the dried leaves of the common foxglove (Digitalis purpurea) and used in medicine to strengthen contractions of the heart muscle.

137. Which of the following curves correctly represents the process of ethanol production by yeast ?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The production process of bioethanol depends on the feedstock; it is based on technologies that go from the simple conversion of sugars by fermentation to the multistage conversion of lignocellulosic biomass into ethanol. Curve (d) is correct.

138. Inbreeding for 5 generations led to production of homozygous transgenic mice. However, these homozygous males or females were infertile. Which of the following approach is most preferable and economical to obtain heterozygous transgenic animals continuously ?

(a) More transgenic founder (1^st animal) should be generated.

(b) Crossing (breeding) of transgenic mice with wild type mice in earlier generations should be done for continued production of transgenic heterozygous offspring.

(d) Homozygous transgenic mice should be mated with wild type mice for continued production of transgenic heterozygous offspring.

Ans. (b)

Sol. Crossbreeding, sometimes called "designer crossbreeding", is the process of breeding such an organism, While crossbreeding is used to maintain health and viability of organism.

139. Following are few statements for regeneration of plants from explants/tissues.

P. Cytokinin is required for shoot development.

Q. Auxin is required for shoot development.

R. Auxin to cytokinin ratio is very important.

S. Jasmonic acid is required for both root and shoot development.

Which of the following combinations of above statements is true ?

(a) P and R

(b) Q and S

(d) Q and R

Ans. (a)

Sol. The plant hormones auxin and cytokinin are critical for plant regeneration in tissue culture, with cytokinin playing an instrumental role in shoot organogenesis. It is well established that the auxin-cytokinin ratio used in culture media determines the degree of shoot and/or root formation in tissue culture.

140. A set of neonatal mice are divided into four groups. Group 1 neomates were not primed with any antigen. Group 2 neonates were primed with KLH. Group 3 neonates were primed with KLH but thymectomized. Group 4 neonates were KLH-primed, thymecotomized, but reconstituted with KLH- specific CD4⁺ T-cells. All these mice, when grown adult, were challenged with KLH and anti-KL IgG antibody was measured in sera. Which of the following is the correct order of magnitude of antibody response ?

[> = greater than, > = greater than or equal to]

(a) Group 1 > Group 2 > Group 3 > Group 4.

(b) Group 2 > Group 1 > Group 3 > Group 4.

(d) Group 4 > Group 1 > Group 2 > Group 3.

Ans. (d)

Sol. Order of magnitude of antibody response would be Group 4 > Group 1 > Group 2 > Group 3.

141. Choose the correct sequence of events in a next generation sequencing technology-based whole genome sequencing project.

(a) DNA extraction shearing library preparation sequencing → assembly finishing annotation submission of Genbank.

(b) DNA extraction library preparation sequencing assembly annotation finishing submission of Genbank.

(c) DNA extraction shearing adapter ligation library preparation sequencing assembly finishing annotation submission of Genbank.

(d) DNA extraction adapter ligation library preparation shearing sequencing finishing assembly annotation submission of Genbank.

Ans. (c)

Sol. The correct sequence of events in a next generation sequencing technology-based whole genome sequencing project is DNA extraction shearing adapter ligation library preparation sequencing assembly finishing annotation submission of Genbank.

142. An investigator discovers a new receptor for a known ligand and wanted to identify the binding partner of the receptor i.e. its co-receptor. The antireceptor antibody is not available but anti GFP-antibdoy is availabel. Which one of the following strategies is most likely to identify the co-receptor ?

(a) The GFP-receptor fusion protein is expressed in a cell line and analyzed by LC-MS/MS.

(b) The GFP-receptor fushion protein is expressed in a cell line and the cells positive for GFP were sorted out, lysed and run on a polyacrylamide gel.

(d) The receptor is cloned as a fusion protein of GFP and expressed in stimulated cells. The immuno-precipitated complex obtained by anti-GFP antibody was analyzed by LC-MS/MS.

Ans. (d)

Sol. The receptor is cloned as a fusion protein of GFP and expressed in stimulated cells. The immuno-precipitated complex obtained by anti-GFP antibody was analyzed by LC-MS/MS.

143. The frequency distribution of tree heights in two forest areas with different annual rainfall are given

Which of the following statistical analysis will you choose to test whether rainfall has an effect on tree heights ?

(a) t-test for comparison of means.

(b) A non-parametric comparison of the two groups.

(d) Regression of tree heights on rainfall.

Ans. (b)

Sol. To test whether rainfall has an effect on tree heights in two forest areas, A non-parametric comparison of the two groups would be done.

144. Two species of plants were sampled in 32 quadrats in a forest. The mean and variance for the occurrence of species 1 were 16.2 and 48 and species 2 were 3.6and 3.2 respectively. Which of the following statements about the distribution of the two species in these quadrats is supported by these findings ?

(a) Both species are distributed randomly.

(b) Species 1 is distributed randomly and species 2 is clustered.

(d) Both species are clustered.

Ans. (c)

Sol. The species will show the random distribution, if the ratio of variance and mean tends to one(1).

The species will show that clustered distribution, if the ratio of variance and mean tends to infinity.

145. Poly-L-lysine exists in pre -helix, -sheet and random coiled conformations depending upon the solvent conditions. The values of mean residue ellipticity at 220 nm are –35,700, –13,800 and +3,900 deg cm²dmol^–1 for -helix, -sheet and random coil conformations of this polypeptide, respectively. The polypeptide exists in -helix conformation at pH 10.8 and 25°C. Addition of urea leads to a two state transition between -helix and random coil conformation. It has been observed that of the polypeptide is –14800 deg cm²dmol^–1 in the presence of 6M urea. The percentage of the polypeptide in -helix conformation is