PYQ DEC 2013 - VedPrep

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

21. Which one of the following bases has the largest hydrogen bonding possibility?

(a) Adenine

(b) Guanine

(d) Uracil

Ans. (b)

Sol. Guanine has the highest ratio of observed to expected hydrogen bonds; 183 hydrogen bonds observed, with only 69.4 expected.

22. Glucose residues in amylose are linked by

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Guanine has two tautomeric forms, the major keto form (see figures) and rare ennol form. It binds to cytosine through three hydrogen bonds. In cytosine, the amino group act as the hydrogen bond donor and the C-2 carbonyl and the N-3 amine as the hydrogen bond acceptors. Guanine has the C-6 carbonyl group that acts as the hydrogen bond acceptor, while a group at N-1 and the amino group at C-2 act as the hydrogen bond donors.

Guanine can be hydrolyzed with strong acid to glycine, ammonia, carbon dioxide and carbon monoxide. Guanine is first deaminated to xanthine. Its high melting point of 350°C reflects the intermolecular hydrogen bonding between the oxo and amino groups in the molecules in the crystal. Because of this intermolecular bonding, guanine is relatively insoluble in water, but it is soluble in dilute acid and bases.

23. The interaction energy between atom A and B is ~400 kJ mol^–1. The type of interaction between them is

(a) pi – pi

(b) covalent

(d) hydrogen bond

Ans. (b)

Sol. Covalent Bonds : Organic molecules, including biomacromolecules are held together with covalent bonds. Such bonds are usually strong, the stranded free energy of fromation being in the order of –400 kj/mol.

24. Enzymes help to lower the activation energies of reactions by

(a) covalent interaction with substrates

(b) binding only with the solvent molecules

(d) forming weak interactions with substrates

Ans. (d)

Sol. The difference between the energy levels of the ground state and transition state is called as activation energy. Raising the temperature, thereby increasing the number of molecules with sufficient energy to overcome this energy barrier, can increase reaction rates. Alternatively adding a catalyst can lower activation energy. Enzymes-biocatalyst lowers the activation energy. Enzyme-biocatalyst lowers the activation energy. Some weak interactions are formed in ES complex, but fall complement of possible weak interactions between substrate and enzyme are formed only when the substrate reaches the transition state. The weak bonding between the enzyme and substrate provide the major driving force for enzymatic catalysis. The free energy (binding energy) relearned by formation of these interactions partly offsets the energy required to get to top of energy hill and results in lower net activation energy.

25. Tryptic digest of a heptapeptide {built from 3 lysine (K), 2 alanine (A), 1 tyrosine (Y) and 1 phenylalanine (F)} yield tri and tetrapeptide. Which of the following is the correct sequence of the heptapeptide?

(a) KAYAKFK

(b) YKAAFKK

(d) KYAAKFK

Ans. (c)

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

26. 100 ml of 0.1 M sodium acetate solution has a pH of 8.90. To this solution 1000 µl of 1M acetic acid (pK_a = 4.76) of pH 2.80 is added. The pH of this mixture will be

(a) 8.90

(b) 4.76

(d) 5.76

Ans. (d)

Sol. pH = pK_a + = 4.76 + = 4.76 + = 4.76 + log10 = 4.76 + 1 = 5.76

27. What are A, B and C in the following reactions?

(a) Pyruvate, ribose 5-phosphate, glycogen

(b) Ribose 5-phosphate, glycogen, pyruvate

(d) Glycogen, citrate, ribose 5-phosphate.

Ans. (c)

Sol. Ribose 5-phosphate is a product produced in the non-oxidative phase of the pentose phosphate pathway. Glucose 1-phosphate and uridine triphosphate work together to activate UDP-glucose which acts as a precursor of glycogen. Pyruvate is produced from glucose 6-phosphate through glycolysis.

28. Michaelis-Menten enzyme kinetics for a simple reaction involving an enzyme (E) and substrate (S) is given by the scheme:

The description of k_m, k_cat and their relationship is provided in the following statements.

P. K_m represents association constant of the ES complex

Q. K_m represents the dissociation constant of the ES complex

R. k_cat is he rate constant for the chemical convesion of the ES complex to substrate bound enzyme and product.

S. k_cat/K_m is a rate constant that refers to the properties and reactions of the free enzyme and free substrate.

Which of the combinations of above statements is true?

(a) P and R

(b) Q and S

(d) Q and R

Ans. (b)

Sol. Km is the concentration of substrate which permits the enzyme to achieve half Vmax. kcat/KM is the second-order rate constant that determines the reaction rate when the enzyme is mostly free at a very low concentration of the substrate. Both parameters provide critical information on how the enzyme lowers the energy barriers along the reaction pathway for catalysis.

29. A 26-residue peptide composed of alanine and leucine shows a circular dichroism (CD) spectrum characteristic of -helix at 50°C in 5 mM phosphate buffer at pH 7.4. Deconvolution of the spectrum indicates 60% -helical and 40% random conformation. When the peptide solution is cooled gradually to 25°C, and the CD spectra are recorded at different temperatures, the most likely observation will be that

(a) the % helical content will decrease and % random conformation will increase.

(b) the % helical content will increase and % random conformation will decrease

(d) there will be transition from -helix to -hairpin.

Ans. (b)

Sol. CD measurements detect these differences by passing left and right circularly polarized light through an optically active sample. Option (b) is correct.

30. It has been observed that for the DNA double helix melting, the value of (enthalpy change of denaturation) are 80 and 90 kcl/mole at 70° and 80°C, respectively. Assuming that ((constant-pressure heat capacity change) is independent of temperature, estimate associated with the denaturation of DNA at 37°C. This estimated value of (kcl/mole) is

(a) 27

(b) 37

(d) 57

Ans. (c)

Sol. From Kierchhoff's law the change in enthalpy and change in heat capacity are related as :

= ; given that, = 90 – 80 = 10 kcal/mol and = 80 – 70 = 10°C

Hence, = 10/10 =1

We can calculate the at 37°C and we can use the value of temperature either 80°C or 70°C. For example, if we use the value of temperature 80°C, then

For example, if we use the value of temperature 70°C, then

31. Out of the list given below, which is the correct order of increasing lipid bilayer permeability?

(a) N₂ > ethanol > H₂O > glucose > Ca⁺² > RNA

(b) H₂O > glucose > ethanol > N₂ > Ca⁺² > RNA

(d) ethanol > RNA > ca⁺² > H₂O > glucose > N₂

Ans. (a)

Sol. Permeability of phospholipid bilayers. Small uncharged molecules can diffuse freely through a phospholipid bilayer. However, the bilayer is impermeable to larger polar molecules (such as glucose and amino acids) and to ions.

32. Which of the following is not a post-translational modification in a mammalian system?

(a) Palmitoylation

(b) Glycosylation

(d) Phosphorylation

Ans. (c)

Sol. Posttranslational modifications (PTMs) are covalent processing events that change the properties of a protein by proteolytic cleavage and adding a modifying group, such as acetyl, phosphoryl, glycosyl , palmitoyl and methyl, to one or more amino acids.

33. One highly pathogenic DNA virus enters into the host cells by endocytosis replicates in the nucleus followed by cell lysis. You have drugs at your disposal that block

P. acidification of vesicles

Q. mitoochondrial transport

R. nuclear export

S. exocytosis

Identify the right combination to prevent its infection.

(a) P and Q

(b) Q and S

(d) P and S

Ans. (c)

Sol. Option P and R is correct.

34. When cells enter mitosis. their existing array of cytoplasmic microtubules has to be rapidly broken down and replaced with the mitotic spindle, which pulls the chromosomes into the daughter cells. The enzyme katanin is activated during the onset of mitosis and chops microtubules into short pieces. The possible fate of the microtubule fragments created by katanin will be

(a) depolymerization

(b) aggregation

(d) translocation

Ans. (a)

Sol. It has been shown that katanin is responsible for severing microtubules during M-phase in Xenopus laevis. The disassembly of microtubules from their interphase structures is necessary to prepare the cell and the mitotic spindle for cell division.

35. Cell cycle is regulated by varius cyclins and cyclin dependent kinases (CDK). On receiving mitotic stimuli. cyclin D, the first cyclin expreseed, binds with existing CDK4 to form the active cyclin D-CDK4 complex. This is trun phosphorylates retinoblastoma protein (Rb) which activates E2F to further activate the transcription of various downstream cyclins. In a particular cell type there is a mutation in Rb such that it cannot be phosphorylated. What will be the correct expression pattern oc cyclin E is these cell after mitotic stimulation?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Phosphorylated Rb activates E2F which further transcribe downstream cyclins. So if Rb is mutated further cyclins will not be formed. Graph c is correct.

36. A bacterial culture was in log phase in the following figure. At time X, an antibacterial compound was added to the culture.

Which of the following lines in the growth curve represents the antibacterial activities of the compound?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Antibacterial agents are a group of materials that fight against pathogenic bacteria. Thus, by killing or reducing the metabolic activity of bacteria, their pathogenic effect in the biological environments will be minimized.

37. E. coli proliferates faster on glucose than it does on lactose because lactose is

(a) taken up more slowly than glucose

(b) not hydrolyzed by E. coli

(d) toxic to the cells

Ans. (a)

Sol. E. coli grows faster on glucose than on any other carbon source. For example, if E. coli is placed on an agar plate containing only glucose and lactose, the bacteria will use glucose and lactose, the bacteria will use glucose first and lactose second. When glucose is available in the environment, the synthesis of -galactosidase is under repression due to the effect of catabolite repression caused by glucose. The catabolite repression caused by glucose. The catabolite repression in this case is achieved through the utilization of phosphotransferase system.

38. In the lysogenic state of phage

(a) both cI and cro are on

(b) both cI and cro and off

(d) cI is off while cro is on

Ans. (c)

Sol. In the lysogenic state of phage cI is on while cro is off.

39. In bacteria chromosomal DNA replication starts at

(a) one specific locus

(b) several specific loci

(d) from several loci, randomly

Ans. (a)

Sol. In bacteria chromosomal DNA replication starts at one specific locus i.e., Ori site.

40. The 'Uvr ABC' repair rechanism is involved in repairing

(a) missing bases

(b) strand break

(d) DNA damage caused by 'bulky' chemical adducts

Ans. (d)

Sol. The UvrABC system repairs damged DNA. Initially, a complex comprising two molecules of UvrA and one molecule of Uvrb forms and then binds to DNA. Both the formation and binding of this complex to DNA requires ATP. It seems likely that the UvrA-UvrB complex initially binds to an undamaged segment and translocates along the DNA helix until a distortion caused by an adduct is recognized; this translocation along the helix also requires ATP. An ATP-dependent conformational change in the damaged DNA region bound to the UvrA-UvrB complex then produces a bend or kink, in the DNA backbone. After the UvrA dimer was dissociated the UvrC protein, which has endonuclease activity, binds to the damaged site. The interaction of UvrC and the bend DNA is thought to open up space within the DNA, allowing the catalytic residues of the enzyme to access their target. The precise position of the cleavage sites is determined by the nature of the DNA damage. In the case of thymine dimers, UvrC cleaves two phosphodiester bonds : one is located eight nucleotides 5' to the lesin and one is located four or five nucleotides 3' to the lesion. After UvrC has cleaved the damaged strand at two points, the fragment with the adduct is removed by a helicase and degraded; the gap left in the strand then is repaired by the combined actions of DNA polymerase I and DNA ligase.

41. Aminoacyl tRNA synthetases face two importan challenges:

(i) they must recognize the correct set of tRANs for a particular amino acid

(ii) they must charge all of these isoaccepting tRNAs with the correct amino acid

Both of these processes are carried out with high fidelity by the following possible mechanisms

P. The discrimination ability resides predominantly at the acceptor stem of the tRNAs.

Q. The specificity is contributed by the anticodon loop in tRNAs.

R. The specificity is embedded in the amino acyl synthetase at the 'N' terminus

S. The specificity is contributed by the variable loop of the tRNA.

Which of the following is correct?

(a) P and Q

(b) P and R

(d) P and S

Ans. (a)

Sol. Aminoacyl tRNA synthesis charges cognate tRNA by the discrimination ability resides predominantly at the acceptor stem of the tRNAs and the specificity is contributed by the anticodon loop in tRNAs.

42. Bacteriophage is a temperate bacteriophage and has two modes in its life cycle, lysogenic and lytic. Several genes are involved in these two processes like N, cI, cII, cIII, Q, int, xis, etc.

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Option (a) correctly mentions lysogeny and lytic phase in phase.

43. The complex responses to different types of DNA damage in both prokaryotes and eukaryotes fall into three main categories:

1. damage bypass

2. damage reversal

3. damage removal

Many repair proteins are isolated like

P. DNA methyltransferase

Q. DNA glycosylase

R. DNA plymerase IV

Which one of the following represents the correct combination?

(a) 1–P, 2–Q, 3R

(b) 1–Q, 2–R, 3P

(d) 1–R, 2–Q, 3P

Ans. (c)

Sol. Damage bypass occurs by DNA polymerase IV and damage reversal is done by DNA methyl-transferase while damage removal is done by DNA glycosylase.

44. During heat shock, mammalian cells shut down global protein synthesis while inducing heat shock proteins (HSPs). The possible molecular regulation(s) that could explain the phenomenon are:

P. mRNA of all proteins, expect those of HSPs, undergoes degradation during heat shock.

Q. Cap-dependent transiation of most mRNAs is affected during heat shock due to denaturation of cap binding protein, eIF-4E.

R. Translation initiation of HSP mRNAs takes place through their internal ribosome entry sites (IRES).

S. HSP mRNAs are abundant during heat shock and thus they complete other mRNAs for ribosome binding and translation

Which of the following sets is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (b)

Sol. Mammalian cells shut down global protein synthesis while inducing heat shock proteins (HSPs) because Cap-dependent transiation of most mRNAs is affected during heat shock due to denaturation of cap binding protein, eIF-4E and Translation initiation of HSP mRNAs takes place through their internal ribosome entry sites (IRES).

45. In recent years, genome-wide transcription study using high throughput sequence analysis has revealed some novel results that include:

1. Presence of RNA plymerase in both intra-and intergenic regions of the genome

2. Existence of non-coding RNAs generated from mRNA coding genes.

3. Existence of sense and antisense transcripts generated from the promoter and untraslated region of many annotated genes.

Possible interpretations of the above results are:

P. RNA polymerase can loosely bind to an part of the genome but its affinit becomes strong only ehen its reaches the promoter.

Q. Binding of RNA polymerase to non-promoter regions of the genome leads to the generation of various non-coding regulatory RNAs.

S. Sense and antisense transcripts are generated from the promoter and untranslated regions of protein coding genes by a novel mechanism of bidirectional transcription.

Identify the correct combination of the above interpretations

(a) P and Q

(b) Q and S

(d) Q and R

Ans. (b)

Sol. From the statements mentioned in question we can interprete that binding of RNA polymerase to non-promoter regions of the genome leads to the generation of various non-coding regulatory RNAs and sense and antisense transcripts are generated from the promoter and untranslated regions of protein coding genes by a novel mechanism of bidirectional transcription.

46. In a cell free extract containing DNA polymerase I, mg²⁺, dATP, dGTP, dCTP and dTTP (³H), the following DNA molecules were added:

P. Single stranded closed circular DNA molecule containing 824 nucleotides.

Q. Single stranded closed circular DNA molecule having 1578 nucleotides bas paired with a linear single stranded DNA molecular of 824 nucleotides having a free-3'–OH group.

R. Double standed linear DNA molecule containing 1578 nucleotides having free-3'-OH group at both ends.

S. Double stranded closed circular DNA molecular having 824 nucleotides.

The rate of DNA synthesis was measured by incorporation of ³H thymidine in the DNA molecule and expressed as the percentage of DNA synthesis relative to total DNA input.

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Graph in option (c) is correct.

47. A bacterial strain can use carbohydrates and hydrocarbons as growth substrates. The strain uses glucose following a minimal lag period after innoculation, regardless of the other carbohydrates and hydrocarbons in the growth medium. The following observations were also made.

P. In the absence of glucose, lactose is used after a lag period of about three times as long as the lag period for glucose utilization.

Q. The presence of hydrocarbons does not affect the lag period for the utilization of lactose.

R. The utilization pattern for the hydrocarbons in similar to that of lactose.

S. Branched hydrocarbons are not immediately utilized if straight chain hydrocarbons are initially present.

Which one of the following specific regulatory mechanisms is consistent with the above observations related to carbohydrate hydrocarbon utilization?

(a) Diauxie

(b) End point repression

(d) Transcription attenuation

Ans. (c)

Sol. Statements mentioned in question is related to catabolite repression where Lactose consumption is suppressed in presence of glucose.

48. The cylindrical channels in gap junctions are made of

(a) Fibronectin

(b) Vitronectin

(d) Cyclin

Ans. (d)

Sol. Each gap junction channel is formed by two hemichannels (connexons), representing hexamers of connexin (Cx) subunits, contributed by each of the coupled cells.

49. The cylindrical channels in gap junctions are made of

(a) connexin

(b) collagen

(d) N-CAM

Ans. (a)

Sol. Gap junctions are specialized sites composed of large, tightly packed intercellular channels, which connect cytoplasm of adjacent cells. Each cylindrical channel, 10-12 nm long and 2.8-3.0 nm in diameter, consists of a pair of half-channels, termed connexons, which are embedded in the cell membrance.

50. A tumour suppressor protein

(a) is one whose function brings about regression of a tumour.

(b) is one where mutations are shown to cause or are associated with tumours

(d) inhibits the progression of the cell cycle buy phosphorylating cyclins.

Ans. (b)

Sol. The p53 gene like the Rb gene, is a tumor suppressor gene, i.e., its activity stops the formation of tumors. If a person inherits only one functional copy of the p53 gene from their parents, they are predisposed to cancer and usually develop several independent tumors in a variety of tissues in early adulthood.

51. Immediate hypersensitivity reactions are associated with

(a) IgG

(b) IgE

(d) IgA

Ans. (b)

Sol. IgE antibodies on mast cells play a central role in an immedite hypersensitivity (anaphylactic) reaction. Patients may have developed IgE-antibodies against a component of the local anaesthetic. When this agent is administered again in these patients, it will bind to the IgE-molecules on the surface of the mast cells. This immediately activates the mast cells. This immediately activates the mast cells, which release histamine into the surrounding tissues, which causes vasodilation, increased vascular permeability and smooth muscle contraction.

52. Which of the following phenomena is observed in compatible plant-pathogen interactions?

(a) Virulence in pathogen

(b) Hypersensitive response in host

(d) Avirulence in pathogen

Ans. (a)

Sol. The term virulence has a conflicting history among plant pathologists. Here we define virulence as the degree of damage caused to a host by parasite infection, assumed to be negatively correlated with host fitness, and pathogenicity the qualitative capacity of a parasite to infect and cause disease on a host.

53. When a membrane is depolarized to a voltage value more positive than the treshold voltage, it leads to the generation of

(a) Donnan potential

(b) action potential

(d) electrochemical potential

Ans. (b)

Sol. If, however, depolarization reaches a critical threshold level, the potential across the membrane does not continue to increase linearly with current. Rather, a completely new type of signal called an action potential (often also reffered to as an impulse or spike) is generated. The action potential is characterized by a rapid increase in voltage, crossing the zero mark, and becoming transiently positive (overshooting potential). The membrane potential then rapidly returns toward the resting value, often accompanied by a brief undershooting.

54. Vascular endothelial (VE)-cadherin is an important cell adhesion molecule for endothelial cells. Endothelial cells that are unable to express VE-cadherin still can adhere to one another via N-cadherin (neural cadherin), but these cells do not survive. Which of the following is the most appropriate reason for this?

(a) N-cadherin uses VE-cadherin as coreceptor for adhesion

(b) Ve-cadherin acts as coreceptor for VEGF (Vascular endothelial growth factor)–mediated signal transduction in endothelial cells.

(d) Loss of VE-cadherin impairs Ca²⁺ –homeostasis of vascular endothelial cells leading to their death.

Ans. (d)

Sol. VE-cadherin regulates various cellular processes such as cell proliferation and apoptosis and modulates vascular endothelial growth factor receptor functions. Calcium is an essential second messenger in endothelial cells and plays a pivotal role in regulating a number of physiologic processes, including cell migration, angiogenesis, barrier function, and inflammation.

55. Clearance of phagocytosed intracellular parasite like Leishmania requires the involvement fo reactive oxygen species (ROS) and reactive nitrogen species (RNS). Administration of IFN- to macrophages harbouring an intracellular pathogen leads to the production of ROS and RNS by JAK/STAT pathway. A macrophage cell line J774 infected with Leishmania is given the following treatments.

A. IFN-

B. IFN- + AMT, a potent iNOS inhibitor.

C. IFN- + apocynin, a NADPH oxidase inhibitor

D. IFN- + NMMA (N-monomethyl L-arginine), an arginine analogue

What will be the most appropriate graph showing the survival of parasites after these treatments?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Graph (b) is correct.

56. A ligand recognize two different cell surface receptors, A and B, on the same cell type. Receptor A, after binding with the ligand is internalized along with the ligand whereas receptor B, after binding with the ligand, initiates tyrosine kinase activity of the intracellular domain. One particular diseasse is associated with the loss of receptor mediated signal transduction of the ligand. Different observes inferred that the disease may be resulted due to

P. loss of binding affinity of receptor A due to mutation in the extracellular domain.

Q. loss of binding affinity of receptor B due to mutation in the extracellular domain

R. mutation inhe tyrosine kinase domain rendering it inactive.

S. mutation in the intracellular domain rendering it incapable of endocytosis.

Which combination of the above inferences do you think is appropriate for the cause of the disease?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (b)

Sol. Statement Q and R is correct. Mutation in the binding affinity of receptor and mutation in tyrosine kinase domain can cause loss of receptor mediated signal transduction.

57. An important role of Fas and Fas ligand is to mediate elimination of tumor cells by killer lymphocyes. In a study of 35 primary lung and colon tumors, half the tumors were found to have amplified and overexpressed a gene for a "secreted protein" that binds to Fas ligand. The main reason for survival of these tumor cells by this "secreted Fas-ligand binding protein" may be attributed to its

(a) decoy recetor activity

(b) anti-proliferative activity

(d) anti-contact inhibition activity

Ans. (a)

Sol. Decoy receptors recognize certain inflammatory cytokines with high affinity and specificity, but are structurally incapable of signaling or presenting the agonist to signaling receptor complexes. They act as a molecular trap for the agonist and for signaling receptor components.

58. A BALB/c mouse was thymectomized on the first day after birth (mouse 1) whereas another was thymectomized on day 7 after birth (mouse 2). A third mouse underwent the amd operation on day 21 after birth. After 56 days, sera were prepared from these mice and also from control mice, which had sham operation. The sera were checked for anti-DNA antibodies. Which one of the following observations is them most plausible?

(a) Both mouse 1 and mouse 2 had anti-DNA antibodies but mouse 3 did not have anti-DNA antibodies.

(b) Only mouse 1 had anti-DNA antibodies

(d) Only the control mice had anti-DNA antibodies.

Ans. (b)

Sol. Option (b) is correct.

59. A potentially valuable therapeutic approach for killing tumor cells without affecting normal cells is the use of immunotoxins. Immunotoxins consist of particular cell-specific monoclonal antibodies coupled to lethal toxins. Which of the following molecular approaches is not appropriate for killing tumor-cells?

(a) Cell surface receptor binding polypeptide chain of toxin molecules should be replaced by monoclonal antibodies which are specific for a particular tumor cell.

(b) Constant region Fc domain of tumor cell-specific monocional antibody should be replaced by toxin molecules.

(d) Inhibitor polypeptide chain of toxin molecules should be conjugated to the F(ab) domain to tumor-specific monoclonal antibody.

Ans. (c)

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

60. An action potential was generated on a nerve fibre by a threshold electrical stimulus. When a second stimulus was applied, no matter how strong it was, during the absolute refractory period of the action potential, the nerve fibre was unable to generate second action potential. This observation was explained in the following statements:

P. A large fraction of potassium channels was voltage inactivated

Q. The critical number of sodium channels required to produce on action potential could not be recruited.

R. A large fraction of sodium channels required to produce an action potential could not be recruited.

S. The critical number of potassium channels required to produce an action potential could not be recruited.

Which one of the following is true?

(a) Only P

(b) P and Q

(d) R and S

Ans. (c)

Sol. The period immediately following the firing of a nerve fiber when it cannot be stimulated no matter how great a stimulus is applied. It is caused by the voltage gated sodium channels shutting and not opening for a short period of time.

61. Match the following :

Disease Pathogen/Causative factor

P. Creutzfeldt-jakob 1. Fungi

Q. Pneumocystis 2. Virus

R. Legionnaires diseases 3. Prion

S. Rabies 4. Bacteria

(a) P–4, Q–3, R–2, S–1

(b) P–3, Q–1, R–4, S–2

(d) P–2, Q–4, R–1, S–3

Ans. (b)

Sol. Prions : During the past two decades, evidence has linked some degenerative disorders of the central nervous system to infectious particles that consist only of protein. These "proteinaceous infectious particles" have been named prions (preeons). The known prion diseases include Creutzfeldt-Jakob disease (in humans), scrapie (in sheep) and bovine spongiform encephalopathy ("mad cow disease" in cattle); all known prion diseases frequently result in brain tissue that is riddled with holes. The organisms causing early infections are Moraxella catarrhalls, Haemophilus influenzae and S.pneumoniae; whereas Gram-negative bacteria or S.aureus are associated with the late HCAP. Viruses may cause early and late infections while yeasts, fungi, Legionella and Pneumocystis carinii are typically late pathogens. Legionella is a pathogenic gram negative bacterium, including species that cause legionellosis or Legionnaires disease, most notably L. pneumophila. It amy be readily visualized with a silver stain. Legionella is common in many environments, with at least 50 species and 70 serogroups identified.

62. The change in the state of specification of imaginal disc of Drosophilla to that of a different disc type is known as

(a) transdetermination

(b) transdifferentiation

(d) transduction

Ans. (a)

Sol. Drosophila imaginal disc cells have the ability to undergo transdetermination, a process whereby determined disc cells change fate to that of another disc identity. For example, leg disc cells can transdetermine to develop as wing cells.

63. Over-expression of a dominant negative FGF receptor during amphibian development would prevent formation of

(a) trunk and tail

(b) head and trunk

(d) head and forelimbs

Ans. (a)

Sol. Option (a) is correct.

64. The cell death pathway in C. elegans can be schematically represented as:

Based on the above, which one of the following statements is true?

(a) A loss-of-function allele of ced9 would lead of survival of cells that normally die

(b) A loss-of-function allele of ced9 would lead to excessive cell death

(d) Neither loss or gain-of-function of ced9 would make any change to the cell death pathway.

Ans. (b)

Sol. Conversely, mutations that inactivate ced-9 cause cells that normally live to undergo programmed cell death; these mutations result in embryonic lethality, indicating that ced-9 function is essential for development.

65. During double fertilization in plants, one sperm fuses with the egg cell and the other sperm fuses with

(a) synergid cell

(b) central cell

(d) nucellar cell

Ans. (b)

Sol. In double fertilization occurs. One sperm nucleus unites with the egg nucleus, forming a 2n zygote and the other sperm nucleus migrates and unites with the polar nuclei of the central cell, forming a 3n endosperm nucleus. Most of the plants have two polar nuclei. Consequently, the second fertilization involves the fusion of three nuclei and the phenomenon is called triple fusion.

66. A set of experiments that were carred out of demonstrate the effect of Apical Ectodermal Ridge (AER) of the chick limb bud on the underlying mesenchyme are enlisted below, with their expected outcomes:

1. Removal of the AER of forelimb leads of cessation of limb development.

2. If and extra AER is placed in the foremlimb bud, duplication of the distal region of the wing takes place.

3. If an extra AER is placed in the formelimb bud, a leg develops instead of a wing.

4. If AER forelimb bud is replaced with beads soaked in FGF2, a normal wing develops.

5. If a nonlimb mesencyme is placed below an AER, the AER directs the mesencyme to form a normal wing.

Which of the above statements are correct?

(a) 1, 3 and 5

(b) 3, 4 and 5

(d) 1, 2 and 4

Ans. (d)

Sol. The apical ectodermal ridge (AER) is a structure that forms from the ectodermal cells at the distal end of each limb bud and acts as a major signaling center to ensure proper development of a limb. Statement 1, 2 and 4 is correct.

67. Hensen's node is estabilished as the avian equivalent of the amphibian dorsal blastopore lip. The following observations are preseumed to be in support of the same.

P. It is the refgion whose cells acan both induce and pattern a sceond embryonic axis when transplanted into other locations of the gastrula.

Q. It is equivalent in terms oftissue structure

R. It expresses the same marker genes as the Spemann's organizer in amphibians

S. The same micro RNA can interfere with the formation of pre-chordal plate in both Hensen's node and spemann's organizer

Choose the correct set among the following:

(a) P and S

(b) P and R

(d) P and Q

Ans. (b)

Sol. Hensen's node, also called the chordoneural hinge in the tail bud, is a group of cells that constitutes the organizer of the avian embryo and that expresses the gene HNF-3. During gastrulation and neurulation, it undergoes a rostral-to-caudal movement as the embryo elongates.

68. In amphibians, when due to some injury, the eye lens is damaged, the fully differentiated iris cells can regenerate the lens. It is achieved through the following possible processes:

P. Iris cells through some signaling undergo dedifferentiation and transdifferentiation into lens cells to regenerate the lens

Q. Iris cells transform into lens cells spontaneously

R. Iris cells induce in a stepwise manner, specific genes responsible for their dedifferentiation and then conversion to lens cells.

S. Stem cells present in iris tissue differentiate into lens cells.

Which of the following is correct?

(a) P and Q

(b) P and R

(d) Q and R

Ans. (b)

Sol. In frogs the lens is regenerated by transdifferentiation of the cornea and is limited only to a time before metamorphosis. On the other hand, regeneration in newts is mediated by transdifferentiation of the pigment epithelial cells of the dorsal iris and is possible in adult animals as well.

69. The control of flowering is a complex process involving several key regulatory genes. Some statements on flower development are given below:

P. Two major types of genes regulate floral development: meristem identity genes and floral organ identify genes.

Q. The important genes is Arabidopsis that play key regulatory roles in meristem identify are: APETALA1, LEAFY and SUPPERSSOR OF CONSTANS1.

R. The genes that determine floral organ identity were discovered as floral homeotic mutants.

S. Most plant homeotic genes belong to a class of related sequences known as FAD box genes.

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

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (a)

Sol. Floral organ identity genes encode proteins that regulate the expression of other genes that have products involved with the formation or function of floral organs. The Arabidopsis floral meristem-identity genes APETALA1 (AP1) and LEAFY (LFY) confer floral identity on developing floral primordia, whereas TERMINAL FLOWER (TFL) is required to repress their expression within shoot and inflorescence meristems. Mutations in the PISTILLATA (PI) gene of Arabidopsis thaliana cause homeotic conversion of petals to sepals and of stamens to carpels. It is thus classed as a B function floral homeotic gene and acts together with the product of the other known B function gene, APETALA3 (AP3).

70. A few statements on early developmental stages in plants are given below:

P. The cells of flower are diploid in nature

Q. Only some specialized cells in the reproductive organs undergo meiosis to produce haploid cells.

R. The haploid cells produced in Q above, undergo a few normal mitotic cell divisions.

S. All the progeny cells produced in R above, differentiate either into haploid egg cells or into haploid sperm cells.

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

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (a)

Sol. A flowering plant has diploid somatic cells, haploid gametophytes and a triploid endosperm tissue (formed due to triple fusion between the central cell having 2 haploid polar nuclei and the haploid male gamete). Ovary, anther, and zygote are diploid as the initial two are places where gametes are formed and the zygote is formed after the fusion of haploid gametes. Pollen and egg are haploids formed after meiosis and are male and female gametophyte.

71. Cells from an early frog blastula were removed from the animal pole and used to replace cells from the vegetal pole of the blastula. The following events may be expected.

P. Transplanted cells would develop normally as part of the cells of the vegetal pole

Q. Tansplanted cells would develop as cells of the animal pole of the adult on the vegetal pole.

R. Region of the animal pole from where the cells eere removed would be missing in the adult.

S. Remaining cells in the animal pole would compensate for the cells that were removed.

Which of the following statements are true?

(a) Q, R and S

(b) P, Q and S

(d) P, R and S

Ans. (b)

Sol. Sexual reproduction requires genetic material (DNA) from two parents. The parent plants have male and female sex cells, called gametes. The genetic material from the male and female gametes combines to produce offspring. This process is called fertilization. Statement P, Q and S is correct.

72. Capacitation of mammalian sperms allows them to be activated within the uterus and facilitate fertilization. The following statements were made regarding events occuring during capacitation:

P. Removal of chloesterol from sperm head

Q. Removal of non-covalently bound glycoproteins

R. Increased expression of fibronectin

S. Decreased permeability of calcium ions.

Identify the correct statements:

(a) Q, R and S

(b) P, Q and S

(d) P, R and S

Ans. (c)

Sol. During capacitation, several changes in the sperm membrane have been described: increase in membrane fluidity, lateral movement of cholesterol to the apical region of the sperm head, and cholesterol efflux from the sperm plasma membrane to the extracellular environment.

73. Forelimb of human and fippers of whale are embryologically homologous structures. What does the study of homologous structures tell us about evolution?

P. This is the example of adaptive radiation, occurred due to similar group of organisms inhabiting different environments

Q. This is the example of divergent evolution, occurred due to similar group of organisms inhabiting different environments

R. Similar group of organisms with mutations and variations getting naturally selected in different environments.

S. This is the example of convergent evolution, occurred due to similar group of organisms inhabiting different environments.

Which of the following is the correct combination?

(a) P, Q and R

(b) P and S

(d) Only S

Ans. (a)

Sol. Homology is the relationship between structures or DNA derived from the most recent common ancestor. A common example of homologous structures in evolutionary biology are the wings of bats and the arms of primates. Homologies are the result of divergent evolution. Divergent evolution is the process in which organisms from the same common ancestor evolve and accumulate differences, often resulting in a new species. This may occur due to pressures such as changes in abiotic or biotic factors within the environment.

74. Which of the following mechanisms is not involved in providing photoprotection to plants?

(a) Degradation of D1 protein

(b) Zeaxanthin formation

(d) Thermal dissipation

Ans. (c)

Sol. Photoprotection is the biochemical process that helps organisms cope with molecular damage caused by sunlight. Plants and other oxygenic phototrophs have developed a suite of photoprotective mechanisms to prevent photoinhibition and oxidative stress caused by excess or fluctuating light conditions. Photolysis of water don't provide photoprotection.

75. Under which condiions do members of the family Graminease synthesize and release phytosiderophores?

(a) Iron defiiciency

(b) Phosphorus deficiency

(d) Availability of phsphorus complexes in rhizosphere

Ans. (a)

Sol. Phytosiderophore (PS) release, which occurs mainly under iron deficiencies, has been speculated to be a general adaptive response to enhance the acquisition of micronutrient metals.

76. The membranes of chilling-sensitive plants are characterized by

(a) higher proportion of saturated fatty acids

(b) lower transition temperature

(d) lower transition temperature and higher proportion of unsaturated fatty acids.

Ans. (a)

Sol. It has been found that chilling-sensitive plants have higher levels of saturated fatty acids in their membranes, so their membranes tend to solidify at higher temperatures than those of tolerant plants.

77. Which of the following plant hormones can mimic the det1 mutation, causing de-etiiolation and chloroplast development in dark?

(a) Cytokinin

(b) Gibberellin

(d) Ethylene

Ans. (a)

Sol. When grown in the absence of light, Arabidopsis thaliana deetiolated (det) mutants develop many of the characteristics of light-grown plants, including the development of leaves and chloroplasts, the inhibition of hypocotyl growth elongation, and elevated expression levels of light-regulated genes. We show here that dark-grown wild-type seedlings exhibit similar phenotypic traits if any one of a variety of cytokinins are present in the growth medium. We further show that the striking phenotype of det mutants is unlikely to be caused by different levels of cytokinins in these mutants. The three major Arabidopsis cytokinins, zeatin, zeatin riboside, and isopentenyladenosine, accumulate to similar levels in wild-type seedlings grown in either the light or the dark. There is no consistently different pattern for the levels of these cytokinins in wild-type versus det1 or det2 mutants. However, det1 and det2 have an altered response to cytokinin in a detached leaf senescence assay and in tissue culture experiments.

78. Asada-halliwell pathway protects plants against oxidative stress during unfavorable environmental growth regimes. The following are some statements related to the stress-tolerance mechanism through this pathway in plants.

P. Oxygen accepts electrons as an alternative electron acceptor.

Q. Hdrogen peroxide is reduced by catalase to form water

R. Ascorbate is oxidized and regenerated.

S. Glutathione is oxidized and reduced.

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

(a) Q, R and S

(b) P, Q and R

(d) P, R and S

Ans. (d)

Sol. Statement P, R and S is correct.

79. The following are certain facts regarding biological nitrogen fixation in plants:

P. Oxygen irreversibly inactivates nitrogenase enzyme inovolved in nitrogen fixation.

Q. The nod genes that code for nodulation proteins are activated by NodD.

R. The two components of nitrogenease enzyme complex, the Fe protein and MoFe protein, can show catalytic activity independently.

S. During the reaction catalyzed by nitrogenase enzyme, the Fe protein reduces the MoFe ptotein while the MoFe protein reduces N₂.

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

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (d)

Sol. The nod genes are the key bacterial determinants of the signal exchange between the two symbiotic partners. The product of the nodD gene is a transcriptional activator protein that functions as receptor for a flavonoid plant compound. Nitrogenase is a complex, bacterial enzyme that catalyzes the ATP-dependent reduction of dinitrogen (N2) to ammonia (NH3). In its most prevalent form, it consists of two proteins, the catalytic molybdenum-iron protein (MoFeP) and its specific reductase, the iron protein (FeP).

80. Secondary metabolites are diverse array of organic compounds in plants. The following are certain statements about secondary metabolites:

P. They product plants against being eaten by herbibores and gainst being infected by microbial pathogens.

Q. Terpenes, the largest class of secondary metabolites are synthesized by methylerythritol phosphate (MEP) pathway and shikimic acid pathway

R. The most abundant classes of phenolic compounds in platns are derived from phenylalanine.

S. Alkaloids are nitrogen containing secondary metabolities in plants.

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

(a) P,Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (c)

Sol. Statement P, R and S is correct.

81. The following are some statements about long distance translocation of photoassimilates in higher plants:

P. Sugars are translocated in the holem by mass transfer along a hydrostatic pressure

Q. Gibberellic acid stimulates the unloading of sugars from phloem tissue into apoplasts

R. Munch pressure-flow hypothesis is crucial to drive translocation in the phloem.

S. Allocation and partition of carbon within a source leaf determine the phloem loading phenomenon.

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

(a) P, Q and S

(b) P, Q and R

(d) Q, R and S

Ans. (c)

Sol. The mechanism by which sugars are transported through the phloem, from sources to sinks, is called pressure flow. At the sources (usually the leaves), sugar molecules are moved into the sieve elements (phloem cells) through active transport. The pressure flow hypothesis introduced by Ernst Münch in 1930 describes a mechanism of osmotically generated pressure differentials that are supposed to drive the movement of sugars and other solutes in the phloem, but this hypothesis has long faced major challenges. Statement p,r and s is correct.

82. Directional growth of plants induced by light is called phototropism. Some statements on phototropism are given below:

P. Phototropism is a photomorphogenetic response.

Q. PHOT1 and PHOT2 genes mediate phototropism.

R. CRY1 and CRY2 genes although help to perceive blue light are not involved in phototropism.

S. Preception of the blue light by phyA photoreceptor initiates phototropism

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

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (a)

Sol. Phototropism, or the differential cell elongation exhibited by a plant organ in response to directional blue light, provides the plant with a means to optimize photosynthetic light capture in the aerial portion and water and nutrient acquisition in the roots. Phot1 primarily functions in chloroplast accumulation process, whereas phot2 mediates both chloroplast avoidance and accumulation responses.CRY1 and CRY2 act primarily in the nucleus, whereas CRY3 probably functions in chloroplasts and mitochondria. Plants depend on cryptochromes and other photoreceptors to sense environmental cues, such as irradiance, day-night transition, photoperiods, and light quality for optimal growth and development.

83. Plants make several hormones that are important for growth and development. Some statements on plant hormones are given below:

P. Auxin is produced primarily in the root apices

Q. Cytokinins are a smaller group of related compounds

R. Gibberellins are a large group of related compounds defined not by their biological functions but by their structures.

S. Brassinosteroids are an important class of plant hormones, which control a broad spectrum of development response including pollen tube growth.

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

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (b)

Sol. Statement Q, R and S is correct.

84. The only bone marrow cell that never appears in peripheral blood is

(a) myeloblast

(b) myelocyte

(d) megaloblast

Ans. (d)

Sol. The only bone marrow cell that never appears in peripheral blood is myeloblast. The myeloblast is a unipotent stem cell that differentiates into granulocyte cells. Myeloblasts reside extravascularly in the bone marrow. Hematopoiesis takes place in the extravascular cavities between the sinuses of the marrow. The wall of the sinuses is composed of two different types of cells, endothelial cells and advenititial reticular cells. The hemopoietic cells are aligned in cords or wedges between these sinuses, with myeloblasts ando ther granular progenitors concentration in the subcortical regions of these hemopoietic cords.

85. Various types of excitable tissues when stimulated showed response as shown in the figures below. Which one of them is an example of fast adapting tissue?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Figure (b) is example of fast adapting tissue because it shows response at similar stimulus intensity for only initial time after that there is no response.

86. Thyroxin releasing hormone (TRH) receptor belongs to

(a) nuclear receptor family

(b) receptor tyrosine kinase family

(d) guanylate cyclase receptor family

Ans. (c)

Sol. Thyrotropin releasing hormone initiates its biological actions by binding to a membrane anchored receptor that, like the GHRH, somatostatin and CRH receptors, belongs to a family of G protein-coupled receptors containing seven highly conserved transmembrane domains.

87. Myoglobin (Mb) in muscles, Hemoglobin A (HbA) in adult RBC, Hemoglobin C (HbC) in patients with thinner RBC adn Hemoglobin S (HbS) in sickle cell disease are four differnet hemoproteins. Oxygen saturation kinetics of these four proteins is different. Which of the following profile is most plausible?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Oxygen saturation kinetics of Mb, HbA, HbC and HbS are correctly represented in figure (a).

88. Water and electrolytes like Na⁺ and Cl^– are lost from the body in diarrhoea. Oral administration of NaCl solution in this condition does not improve the situation. When glucose is admisnistered with normal NaCl solution through oral route, the absorption of electroyes along with water occurs and the patient recovers.

P. Glucose enhancess ATP production in the mucosal cells of small intestine and thus facilitates sodium absorption.

Q. Glucose inhibits the diarrheal toxin induced cAMP production in the mucosal cells of small intestine.

R. Na⁺ is co-transported with glucose on the apical surface of the mucosal cells of small intentine.

S. The epithelial sodium channels (ENaC) are activated by glucose in colon.

Which one of the following is true?

(a) Only P

(b) P and Q

(d) R and S

Ans. (c)

Sol. Water and electrolytes like Na⁺ and Cl^– are lost from the body in diarrhoea. Oral administration of NaCl solution in this condition does not improve the situation. When glucose is admisnistered with normal NaCl solution through oral route, the absorption of electroyes along with water occurs and the patient recovers. Here, Na⁺ is co-transported with glucose on the apical surface of the mucosal cells of small intentine.

89. A patient has episodes of painful spontaneous mucles contraction, fo9llowed by periods of paralysis of the affected muscles. It was identified as primary hyperkalemic paralysis, and inherited disorder. The possible causes of the paralysis are

P. The elevation of extracellular K⁺ causes hyperpolarization of skeletal muscle cells.

Q. The hyperpolarization of the muscle cell membrane inactivates sodium channels.

R. The elevation of extracellular K⁺ causes depolarization of skeletal muscle cells.

S. The sodium channels are voltange inactivated in depolarized state.

Which one of the following is true

(a) Only P

(b) P and Q

(d) R and S

Ans. (d)

Sol. A patient has episodes of painful spontaneous mucles contraction, followed by periods of paralysis of the affected muscles. It was identified as primary hyperkalemic paralysis, and inherited disorder. The possible causes of the paralysis may be the elevation of extracellular K⁺ causes depolarization of skeletal muscle cells or the sodium channels are voltange inactivated in depolarized state.

90. Maintaining the salt concentration and volume of plasma are two key parameters for physiological processes achieved by kidney. Which one of the following structural and functional combinations is the most efficient renal regulatory system is mammals?

Ans. (a)

Sol. Most efficient renal regulatory system which maintain salt concentration and plasma volume should have Large glomerulus, long proximal and distal tubules, long Henle's loop with transepithelial potential in proximal tubule, countercurrent multiplier, ADH responsiveness of distal tubule.

91. Which one of the following graphs best represents the homone profile in a rat right after mating?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Graph (c) correctly represent the hormone profile in rat right after mating because rat follow estrus cycle.

92. A cross is made between a pure breeding plant having red coloured flowers with a pure breeding plant having white coloured flowers. Such a cross is called as

(a) test cross

(b) monohybrid cross

(d) back cross

Ans. (b)

Sol. A cross is made between a pure breeding plant having red coloured flowers with a pure breeding plant having white coloured flowers. Such a cross is called as monohybrid cross.

93. A cis-trans complementation test is carried out to identify

(a) If two mutations are allelic in nature

(b) If two genes interact with one another

(d) to understand the dominance/recessive relationship between alleles.

Ans. (a)

Sol. Complementation Test : Complementation test is carried out to differentiate between allelic and non allelic mutations, even when the latter affect the same phenotype.

This is based on a simple relationship : while non-allelic mutations always complement, allelic mutations do not, +az. In cis arrangement all combinations will give wild-type phenotype and thus are non-informative. In trans, however, non-alleles will give a mutant phenotype. Thus, complementation test is always carried out in trans configuration. Although simple, it is a very strong test and often resorted to by bacterial geneticists when they have a collection of mutant alleles affecting the same phenotype. In fact, the complementation test with a collection of mutants belonging to the same gene has even helped in the identification of functional subdivisions within a gene, appropriately referred to as complementation groups.

94. The following is the inheritance pattern of a trait under obervation:

P. The trait often skips a generation

Q. The number of affected males and females is almost equal

R. The trait is often found in pedigress with consanguineous marriages.

The trait is likely to be

(a) autosomal recessive

(b) autosomal dominant

(d) sex-linked dominant

Ans. (a)

Sol. As per the inheritance pattern of trait is mentioned in question the trait is likely to be autosomal recessive.

95. Which of the following representations of chromosomal arrangement in meiotic metaphase I best explains the law of Independent Assortment?

% of cells with arrangement

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Option (a) correctly represents chromosomal arrangement in meiotic metaphase 1 as per law of independent assortment.

96. Which of the following statements are true for Robertsonian translocations?

1. The size of the homologous chromosome involved in translocation will differ.

2. Genes on the chromosome involved in translocation will show linkage with genes with which it normally independently assorts.

3. There will be change in the physical map, but no change in the genetic map.

4. It can be identified by G-bandling of chromosomes

5. It can be identified by C-banding of chromosomes.

6. It can lead to Down's syndrome

Which one of the following combinations is correct?

(a) 1, 3 and 4

(b) 1, 4 and 6

(d) 1, 3 , 5 and 6

Ans. (c)

Sol. In humans, when a Robertsonian translocation joins the long arm of chromosome 21 with the long arm of chromosome 14 (or 15), the heterozygous carrier is phenotypically normal because there are two copies of all major chromosome arms and hence two copies of all essential genes. However, the progeny of this carrier my inherit an unbalanced trisomy 21, causing Down Syndrome. About one in a thousand newborns has a Robertsonian translocation. The most frequent forms of Robertsonian translocations are between chromsomes 13 and 14, 14 and 21 and 14 and 15 and occur when the long arms of acrocentric chromosomes fuse at the centromere and two short arms are lost.

A Robertsonian translocation in balaned form results in no excess or deficit of genetic material and causes no health difficulties. In unbalanced forms, Robertsonian translocations cause chromosomal deletions or addition and result in syndromes of multiple malformations, including trisomy 13 (Patau syndrome) and trisomy 21 (Down syndrome).

A Robertsonian translocation result when the long arms of two acrocentric chromosomes fuse at the centromere and the two short arms are lost. If, for example, the long arms of chromosomes 13 and 14 fuse, no significant genetic material is los-and the person is completely normal in spite of the translocation. Common Robertsonian translocations are confined to the acrocentric chromosomes 13, 14, 15, 21 and 22, because the short arms of these chromosomes encode for rRNA which is present in multiple copies.

Most people with Robertsonian translocations have only 45 chromosomes in each of their cells, yet all essential genetic material is present and they appear normal. Their children, however, may either be normal and carry the fusion chromosome (depending which chromosome is represented in the gamete) or they may inherit a missiing or extra long arm of an acrocentric chromosome. Genetic counseling and genetic testing is offered to families that may be carriers of chromosomal translocations.

Rarely, the same translocation may be present homozygously if heterozygous parents with the same Robertsonian translocation have children. The result may be viable offspring with 44 chromosomes.

97. Ames test is used to evaluate mutagens in the environment. Which of the following statements about Ames test are true?

1. The mutagenic effect of a compound is tested using an auxotrophic strain of Salmonella typhimurium.

2. The mutagenic effect of a compound is tested using His^– strain of Escherichia coli.

3. Using appropriate strains, compound causing base substitutions and frameshift mutations can be distinguished.

4. Liver enzymes are important as they are activated by test compound to evaluate its mutagenicaity potential.

5. Many compounds may have to be converted to bioactive metabolites, which is carried out by the enzymes from the liver.

(a) 1, 3 and 4

(b) 1, 2 and 4

(d) 1 and 5 only

Ans. (c)

Sol. Ames test is used to evaluate mutagens in the environment. In Ames test the size of the homologous chromosome involved in translocation will differ, there will be change in the physical map, but no change in the genetic map and it can be identified by C-banding of chromosomes.

98. The genetic map of three genes in Drosophila melanogaster is given below:

A crossm, as given below is made between individuals of the genotype:

The female F₁ progent are test-crossed and 1000 progent are obtained. Assuming that there has been no double crossover, what is the expected number of progeny with the genotype?

Select the set which shows the correct number of expected progeny:

(a) P-100; Q-50; R-850

(b) P-50; Q-25; R-425

(d) P-0; Q-425; R-75

Ans. (b)

Sol.

Expected number of genotype of progeny of and

Expected number of genotype of progeny of

Total parental genotype of = Total progeny – (number of recombinants produced as a result of single crossover between a and b gene + number of recombinants produced as a result of single crossover between b and c gene)

= 1000 – [10% of 1000 + 5% of 1000] = 1000 – 150 = 850

Expected number of genotype of progeny of

99. The following pedigree shows the inheritance pattern of a rare recessive disorder with complete penetrance.

A child from marriage between indiviuduals II-2 and II-3 will show the disorder only if the parents carry the recessive allele. What is the probability that the child will show the disorder?

(a) 1/9, and the probability of the parents of carry the recessive allele is 2/3.

(b) 1/4, and the probability of the parents to carry the recessive allele is 3/4.

(d) 1/64, and the probability of the parents to carry the recessive allele is 3/4.

Ans. (a)

Sol. We can get II-1 and II-4 carrying disorder only if I-1, I-2, I-3 and I-4 all are heterozygous (carrier) for the trait. So, II-2 and II-3 can be carrier with probability 2/3 while progeny from marriage between II-2 and II-3 would be probable to show the disorder = .

100. DNA from a strain of bacteria with genotype a⁺ b⁺ c⁺ d⁺ e⁺ was isolated and used to transfrom a strain of bacteria that was a^– b^– c^– d^– e^–. The transformed cells were tested for the presence of donated genes. The following genes are found to be co-transformed:

P. a⁺ and d⁺

Q. b⁺ and e⁺

R. C⁺ and d⁺

S. c⁺ and e⁺

The order of genes on the bacterial chromosome is

(a) a – b – c – d – e

(b) a – d – c – e – b

(d) a – d – e – b – e – c

Ans. (b)

Sol. By observing co-transformed genes, the order of genes on the bacterial chromose is a – d – c – e – b.

101. Chinase Brake ferm (Pteris vittata) is a hyperaccumulator of

(a) Cadmium

(b) Arsenic

(d) Chromium

Ans. (b)

Sol. Arsenic (As) has been categorized as a toxic and carcinogenic element and contribute to environmental and human health problems worldwide; the highest number of cases has been reported in South-East Asian countries (Mandal and Suzuki 2002). Ma et al. (2001) discovered the first known vascular platn, Pteris vittanta L. commonly known as Chinese brake fern to hyperaccumulate aresenic. The brake fern takes up high concentrations of arsenic (as high as 2.3%) from soil and allocated most of it to the above ground pars for final storage (Tu and Ma 2002).

102. The fungal group presently classified under protists is

(a) Zygomycetes

(b) Oomycetes

(d) Discomycetes

Ans. (b)

Sol. Oomycota or oomycetes (oomycetes) form a distinct phylogenetic lineage of fungus like eukaryotic microorganisms. They are filamentous, microscopic, absorptive organisms that reproduce both sexually and asexually. Oomycetes occupy both saprophytic and pathogenic lifestyles and inlcude some of the most notorious pathogens of plants, causing devastating diseases such as late blight of potato and sudden oak death. This group was originally classified among the fungi (the name "oomycota" means "egg fungus") and later treated as protists, based on general morphology and lifestyle. A cladistic classification based on modern insights supports a relatively close relationship with photosynthetic organisms such as brown algae and diatoms, within the heterokonts.

103. The fungs associated with human oral or vaginal infection is

(a) Fusarium

(b) Aspergillus

(d) Pneumcytis

Ans. (c)

Sol. Candidiasis is an infection caused by a yeast (a type of fungus) called Candida. Candida normally lives on the skin and inside the body, in places such as the mouth, throat, gut, and vagina, without causing any problems.

104. Which one of the following advancements in animal classification is correct?

(a) Protosomes Pseudocoelomates Deuterostomes Eucoelomates

(b) Acoelomates Protostomes Eucoelomates Deuterostomes

(d) Protostomes Deuterostomes Acoelomates Eucoelomates.

Ans. (c)

Sol. Pseudocoelomate metazoans have a fluid-filled body cavity, the pseudocoelom, which, unlike a true coelom, does not have a cellular peritoneal lining. Most of the pseudocoelomates (e.g., the classes Nematoda and Rotifera) are small and none possess an independent vascular system. A (eu)coelomate has a true coelom (fluid filled body cavity) which surrounds the gut and is lined on both sides by mesoderm. Phylum Annelida. Phylum Arthropoda. Based on the embryonic development, metazoans are divided into protostomes and deuterostomes. Protostomes are primitive invertebrates while deuterostomes include chordates and echinoderms. This division is helpful in understanding the relationships between different groups of animals.

105. The following graph is for a logistically growing population, with N_t plotted on the X-axis, What is the parameter plotted on the Y-axis?

(a) dN/dt

(b) N_t+1

(c)

(d) K

Ans. (a)

Sol. Logistic growth model.

This is most commonly shown in differential form as: dN/dt = rN((K – N)/K), where dN/dt (e.g., the rate of change in population size, N, at a given moment in time, t) is a function of the current size of the population (N) and the intrinsic or instantaneous rate of change (r).

106. Annual weeds of arable lands are classified as

(a) phonerophytes

(b) therophytes

(d) geophytes

Ans. (b)

Sol. Therophytes are the plants of deserts (they make up nearly 50% of the flora of Death Valley, USA) sand dunes and repeatedly disturbed habitats. They also include the annula weeds of arabic lands, gardens and urban wastelands.

107. The most common vegetation in the Western Ghats of India trpical moist deciduous forest but that in Deccan plateau is depleted throm forest. The possible reason is

(a) richer soil of Western Ghats compared to Deccan plateau

(b) extensive deforestation in Deccan plateau comapred to Western Ghats.

(d) higher temperature in Deccan plateau compared to Western Ghats.

Ans. (c)

Sol. There is heavy rainfall on the western coast of India but very little in the Deccan because the Deccan plateau is situated in the rain shadow of the Western Ghats. The Deccan Plateau to the east of the Western Ghats receives significantly less rainfall than the coasts.

108. Which one of the following plants group combinations reflects correct increasing order of the number of species it has?

(a) Gymnosperms, Bryophytes, Algae, Angiosperms

(b) Angiosperms, Algae, Gymnosperms, Fungi

(d) Angiosperms, Gymnosperms, Bryophtes, Algae

Ans. (a)

Sol. While in principal species can be counted, it is not clear the species number is an appropriate measure of biodiversity.

Table : Estimate of species number by taxa

109. Coelomates have

1. Fluid filled body cavity

2. a complete lining called peritoneum, derived from mesoderm covering the body cavity.

3. a complete lining called peritoneum, derived from ectoderm covering the body cavity.

4. Round worm as representative of this group

5. Flat worm as representative of this group

Select the correct combination

(a) 1, 3 and 4

(b) 1, 3 and 5

(d) 1 and 2

Ans. (d)

Sol. Coelomate animals or Coelomata (also known as eucoelomates – "true coelom") have a body cavity called a coelom with a complete lining called peritoneum derived from mesoderm (one of the three primary tissue layers).

110. Most biologists agree that seaweeds are protists. Some biologists think that at least some seaweeds should be considered plants, not protists, Which of the following would support the latter one?

(a) Certain seaweeds contain several kinds of specialised cells

(b) Certain seaweeds have multicellular organization

(d) Certain seaweeds undergo sexual and asexual reproduction.

Ans. (b)

Sol. Seaweed, or macroalgae, refers to thousands of species of macroscopic, multicellular, marine algae. Seaweeds are technically not plants but algae. They may be single cellular or multi-cellular, but generally they are non-flowering, contain chlorophyll but lack true stems, roots, leaves, and vascular tissue.

111. In the following diagram, two models of succession are represented. In this diagram A, B, C and D are species and arrows indicate 'is replaced by'

Based on the above, which statements is correct?

(a) Figure 1 represents facilitation model and figure 2 represents tolerance model

(b) Figure 1 represents tolerance model and figure 2 represents facilitation model

(d) Figure 1 represents tolerance model and figure 2 represents inhibition model.

Ans. (c)

Sol. Fig.1 The facilitation model suggests that the presence of an initial species aids and increases the probability of the growth of a second species.

Fig.2. Inhibition model

The only possibility for new growth/colonization in this successional sequence arises when a disturbance leads to dominating species being destroyed, damaged, or removed. This frees up resources and allows for the invasion of other species that were not previously present.

112. Which of following Shows the correct systematic hieracrchy?

(a) Kingdom Phylum Subphylum Superclass Class Subclass Cohort Order Suborder Superfamily Family Subfamily Genus Subgenus Species Subspecies.

(b) Kingdom Phylum Subphylum Superclass Cohort Superclass Class Subclass Superfamily Family Family Subfamily Order Suborder Genus Subgenus Species Subspecies.

(d) Kingdom Phylum Class Cohort Family Order Genus Species.

Ans. (a)

Sol. Taxonomic or systematic hierarchy is a system by which various taxonomic categories are arranged in a proper descending order. Kingdom is the highest rank and species is the lowest or basic rank, i.e. species < genus < family < order < class < phylum/division < Kingdom. Option a is correct systematic hierarchy.

113. Lindeman's efficiency between trophic levels is depicted by the formula : Efficiency = A/B; where, A and B, respectively are:

(a) 'assimilation at trophic level n' and 'assimilation at trophic level n–1'.

(b) 'intake at trophic level n' and assimilation at trophic level n–1'.

(d) 'intake a trophic level n' and 'net productivity at trophic level n–1'

Ans. (a)

Sol. Cological efficiency describes the efficiency with which energy is transferred from one trophic level to the next. It is determined by a combination of efficiencies relating to organismic resource acquisition and assimilation in an ecosystem. Ecological efficiency is a combination of several related efficiences that describes resource utilization and the extent to which resources are converted into biomass.

Efficiency within a Trophic Level

Production Efficiency

Proportion of assimilation energy that goes toward net production

Production Efficiency =

Whole community data are difficult to collect but individual species data are known and production efficiency varies between different groups of animals.

Birds and Mammals (endothermus) 1%-3%

Insects (ectotherms) 10%-41%

Habitat differences do not account for this 10x difference in efficiency. This is a fundamental difference between endothermic and ecothermic organisms, birds and mammals compared to all other organisms.

Efficiency between Trophic Levels

Trophic Level Assimilation Efficiency (Lindeman's Efficiency)

Trophic Level Assimialtion Efficiency

= (100) = % Efficiency

In terrestrial systems, Lindeman's Efficiency is approximately 10%.

In aquatic and marine systems, Lindeman's Efficiency is 15%-20%.

114. The following matrix shows the relationship between probability of death and duration of species association. In the above, A, B, C and D are:

(a) A-parasites, B-Parasitoids, C-Grazers, D-Predators

(b) A-Carnivores, B-Herbivores, C-Parasites, D-Parasitoids

(d) A-Predators, B-Parasitoids, C-Parasites, D-Carnivores

Ans. (a)

Sol. A : Parasitism is a close relationship between species, where one organism, the parasite, lives on or inside another organism, the host, causing it some harm, and is adapted structurally to this way of life.

B : Parasitoids are a group of organisms that live entire life attached to or within a single host. They ultimately kill and sometimes consumes,the host. The majority of parasitoids are wasps (Hymenoptera) or files (Diptera), and they are numerically very abundant and important in nearly all terrestrial ecosystems.

C : A grazer is defined as any species that moves from one victim to another, feeding on part of each victim without actually killing it outright. The "victim" is to the grazer as prey is to the predator.

D : In social predation, a group of predators cooperates to kill prey. This makes it possible to kill creatures larger than those they could overpower singly; for example, hyenas, and wolves collaborate to catch and kill herbivores as large as buffalo, and lions even hunt elephants.

115. An observation was made on a species experiments three factors A, B and C in order to infer a density dependent population regulation by a factor. The following graph shows the relationship graph shows the relationship between the adverse effect of the factors in terms of number and population density.

Based on the above observation, which of the following is correct?

(a) A-Density independent; B-Density dependent; C-Inversely density dependent.

(b) A-Inversely density dependent; B-Density independent; C-Density dependent.

(d) A-Density dependent; B-Density independent; C-Inversely density dependent.

Ans. (b)

Sol. Inverse density dependent factors are those that decrease as the population density increases. Typically, births and immigration are inverse density dependent factors whereas deaths and emigrants are directly related to the population density - they increase as the density increases. These populations can grow rapidly because the initial number of individuals is small and there is no competition for resources. This is called density-independent growth because the density of individuals does not have any effect on future growth.

116. Which of the following biotic provinces are part of Deccan peninsula biogeographic zone of India?

(a) Malabar Cost, Western Plateau, Eastern Plateau

(b) Western Ghats, Central Plateau, Eastern Plateau

(d) Central Plateau, Malabar Cost, Western Ghats

Ans. (c)

Sol. Biogeography of India and Deccan Peninsula : Rodgers and Panwar (1988 and 1992) biogeographically classified the whole of India into 10 major zones, integrating zoological and botanical parameters. The Deccan peninsula (Zone 6), which is the largest zone, includes five biotic provinces with 11 subdivisions. (Table 1)

Table : Biogeographical Classification of the Deccan Peninsula

117. After gull nestlings hatch, the parents remove the egg-shells from the nest. The behaviour is to

(a) Clean the area

(b) reduce infection

(d) minimise nest detection by predators

Ans. (d)

Sol. To protect their chicks from getting attacked by predators, the Common Black-Headed Gulls learned to dispose the empty eggshells, after their chicks hatch, as far away from their nest as possible.

118. Greater male investment in the care of offspring is most likely to lead to

(a) a lek system

(b) stronger female choice

(d) run-away selection

Ans. (c)

Sol. Reversed sexual dimorphism (females being larger than males) occurs in several bird groups, including hawks and vultures. RSD might evolve as the result of sexual selection for small size in males and constraints on the reduction of size in females because of some factor associated with reproduction.

119. A neuron that fires when an individual is eating by hand, also fires when he sees someone else eating with hand. Such neurons are called

(a) mirror neurons

(b) mimicry neurons

(d) reward neurons

Ans. (a)

Sol. A mirror neurons is a neuron that fires both when an animal acts an when the animal observes the same action performed by another. Thus, the neuron "mirrors" the behaviour of the other, as though the observer were itself acting. Such neurons have been directly observed in primate species. Birds have been shown to have imitative resonance behaviours and neurological evidence suggests the presence of some form of mirroring system. In humans, brain activity consistent with that of mirror neurons has been found in the premotor cortex, the supplementary motor area, the primary somatosensory cortex and the inferior parietal cortex.

120. Given below is an evolutionary tree.

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

(a) A-Protosstome; B-Deuterostome; X-Mollusca; Y-Cnidaria; Z-Protozoa

(b) A-Protostome; B-Deuterostome; X-Echindermata; Y-Mollusa; Z-Cnidaria

(d) A-Deuterostome; B-Protostome; X-Echinodermata; Y-Roundworm; Z-Ctenophora

Ans. (b)

Sol. Protostomes are primitive invertebrates while deuterostomes include chordates and echinoderms. Option (b) is correct.

121. Name the common Indian bird that is generally seen in groups (aggregation)

(a) Bulbul

(b) Warbler

(d) Sun bird

Ans. (c)

Sol. Jungle babblers are gregarious birds that forage in small groups of six to ten birds, a habit that has given them the popular name of "Seven Sisters" in urban Northern India, and Saath bhai (seven brothers) in Bengali, with cognates in other regional languages which also mean "seven brothers".

122. Which of the following transenic crop(s) have been approved for commercial cultivation in India?

(a) Cotton

(b) Brinjal

(d) Cotton, Brassica, brinjal

Ans. (a)

Sol. Among crops that were field teste in India during 1996-2001, the main crops are cotton brinjal, tomato and rice. While Bt-cotton, that was grown without approval in large areas (> 10,000 hectares) in Gujarat in the year 2001, was later fied tested (Mahyco-Monsanto) and approved for commercial cultivation in the year 2002. Several other transgenic crops have been approved for large scale field trials.

123. Which of the following diseases does not leave any paleontological evidence?

(a) Tuberculosis

(b) Arthritis

(d) Cholera

Ans. (d)

Sol. That is cholera, because the other diseases like ARTHERITIS is a disease which appear in the Joints between BONES like KNEES which can be detected from the fossils. Similarly rickets is another bone disease, hence it can be observed from fossils. tuberoculosis forms nodules in the lungs (tubercles) that break down the lung tissues and leave evidence forever. Where as CHOLERA doesn't leaves any palentological evidence.

124. Microevolution is the term used for changes in allele frequencies that occur over time. This occurs

P. Within a population at species level

Q. Within a community at genus level

R. Due to appearance of new genes from infections

S. Due to mutation, natural selection, gene flow and genetic drift.

Which of the following combinations is not appropriate?

(a) P and R

(b) P and S

(d) Q and S

Ans. (b)

Sol. Microevolution is the change in allele frequencies that occurs over time within a population. This change is due to four different processes: mutation, selection (natural and artificial), gene flow and genetic drift.

125. The following genotypes were observed in population

Genotype Number

HH 90

Hh 60

hh 50

Which of the following is the correct frequency of H allele and what will be the expected number of HH in the given population?

(a) 0.60 and 72

(b) 0.80 and 96

(d) 0.30 and 90

Ans. (a)

Sol. Since each individual has two alleles to contribute the size of the gene pool is the number of individual times 2. Each HH individual contributes 2 H's to the pool. Each Hh individual contributes one H and one h to the pool. Each hh individual contributes 2 h's to the pool.

Total number of alleles = 180 + 120 + 100 = 400

Total number of H alleles = 180 + 60 = 240

Probability of H alleles = 240/400 = 0.60

Frequency of HH alleles = p² = (0.60)² = 0.36

Expected number of HH in the given population = 0.36 × 200 = 72

126. The first vertebrate animal appeared in which of the following geological ages?

1. Paleozoic era

2. Mesozoic era

3. Ordovician period

4. Cretaceous period

5. Mississippian epoch

6. Paleocene epoh

Which of the following combinations given the best answer?

(a) 1, 3 and 6

(b) 1 and 6

(d) 1 and 3

Ans. (d)

Sol. A time of tremanedous biologic change began with the appearance of skeletonized animals near the Precambrian-Cambrian boundary. Following this event, marine invertebrates began a period of adaptive radiation and evolution, during which the Paleozoic marine invertebrate community greatly diversified. Plants preceded animals onto land during the Middle to Late Ordovician, evolving a variety of innovations that allowed subsequent radiations and diversification within the terrestrial and diversification within the terrestrial envrionment. One of the striking parallels between plants and animals is that they had to solve the same basic problems in making the transition from water to land. For both groups, the method of reproduction proved to be the major barrier to expansion into the various terrestrial environments. But, with the evolution of the seed in plants and the amniote egg in animals, that limitation was removed, allowing for expansion into all terrestrial habitats.

The end of the Paleozoic Era witnessed the greatest mass extinction in Earth's history. Some scientific estimate that more than 90% of marine invertebrate species and upward of 70% of all terrestrial vertebrate species became extinct. Thus, by the end of Permian, a near collapse of both the marine and terrestrial ecosystems had occurred.

127. Which of the following microbial fermentations are anaerobic?

(a) Ethanol and acetone-butanol

(b) Citric acid and propionic acid

(d) Streptomycin and rifampicin

Ans. (a)

Sol. Anaerobic fermentation is a metabolic process done by bacteria and eukaryotes in the absence of air to convert carbohydrates into the products like gases, alcohol, and acids. This fermentation is used to produce various chemicals in industries, i.e., acetic acid. Some examples of anaerobic respiration include ethanol fermentation, lactic acid fermentation (which can result in yogurt and in sore muscles), and in decomposition of organic matter.

128. Encasing of which of the following plant cells in gelatinous matrix is referred as artificial seed?

(a) Microcalli

(b) Somatic embryos

(d) Shoot tips

Ans. (b)

Sol. Artificial seed production through the use of somatic embryos is an important technique for transgenic plants, where a single gene can be placed in a somatic cell and then this gene will be located in all the plants produced from this cell.

129. Co-licalization of two fluorescently labelled proteins in an organelle in cells is usually visualised by

(a) interference-contrast microscopy

(b) scanning electron microscopy

(d) atomic force microscopy

Ans. (c)

Sol. In confocal microscopy, specimens are recorded as a digital image composed of a multi-dimensional array containing many volume elements termend voxel that represent three-dimensional pixels. The size of a voxel (or detection volume) is determined by the numerical aperture of the objective, the illumination wavelength and the confocal detector pinhole diameter. Thus, the colocalization of two fluorescent probes in a speciman such as Alexa Fluor 488 having green emission and Cy3 with orange-red emission is represented in the image by pixels containing both red and green colour contributions (often producing various shades of orange and yellow).

130. Haemoglobin has characteristic cicular dichroism (CD) peaks C in thefar-UV, near UV and Soret regions. Contribution to near-UV CD comes entirely from

(a) aromatic amino acid residues

(b) heme group

(d) peptide bonds and aromatic amino acid residues.

Ans. (c)

Sol. Near-UV CD spectroscopy can reflect the arrangement information of protein side chain chromophore groups heme, tryptophan, phenylalanine, tyrosine and other residues and changes in disulfide bond microenvironment.

131. A weed is assumed to be dispersed randomly in a meadow. What statistical distribution will describe the dispersion correctly?

(a) Binomial

(b) Negative Binomial

(d) Normal

Ans. (c)

Sol. The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

132. Which of the following are the most appropriate spectral bands for vegetation analysis using remote sensing platforms?

(a) Red, Near Infrared

(b) Infrared, Visible

(d) Visible, Microwave

Ans. (a)

Sol. NDVI is a simple numerical indicator that can be used to analyze remote sensing measurements, typically but not necessarily from a space platform and assess whether the target being observed contains live green vegetation or not. The NDVI is calcualted as a ratio between measured reflectivity in the red and near infrared portions of the electromagnetic spectrum. These two spectral bands are chosen because they are most affected by the absorption of chlorophyll in leafy green vegetation and by the density of green vegetation on the surface. Also, in red and near-infrared bands, the contrast between vegetation and soil is at a maximum. The NDVI transformation is computed as the ratio of the measured intensities in the red (R) and near infrared (NIR) spectral bands using the following formula :

133. Foster pups ware presented to a primpiarous rat at the mid gestation period. Which of the following behaviour will be found in the pregnant rat?

(a) The rat shows maternal behaviour after a few days of presentation of pups.

(b) The rat attacks and kills the pups every time they are presented

(d) The rat shows fear response for a few days.

Ans. (a)

Sol. On conduction of the test in the dark phase of the light/dark cycle under dim red light. Animals were habitutated to the test room for at least 24 hours prior to testing. Tests began with the placement of 4 stimulus pups (all the same age between 2-7 days old) in the areas of the cage farthest from the female's nest. A 15-minute retrieval test was conducted during which of the following behaviours were scored : latencies to retrieve each pup to the nest, sniff/lick all pups in th nest and crouch over all pups in the nest. 45 minutes after the presentation of pups each female was observed continuously for 15 minutes. Behaviour was scored once every 15 seconds for a total of 60 observations, during this time hovering, sniffing and licking or crouching over the pups were recorded. Hovering was defined as an upright posture over pups (so that pups had access to the female's ventral surface), including actively sniffing/licking the pups or engaging in self-grooming. In contrast, crouching was recorded when females were in a quiescent, immobile posture with all four limbs supporting a slightly arched or highly arched posture over the pups.

134. Electrons are transferred from reduced co-enzymes like NADH, to pyruvic acid or its derivatives during fermentation. Those final electron acceptors are reduced to the end-products for example lactic acid. propionic acid, etc. The end products depend on the particular microorganism and the substrate.

(a) P–1, Q–2, R–3, S–4

(b) P–4, Q–3, R–1, S–3

(d) P–3, Q–4, R–1, S–2

Ans. (c)

Sol. Option (c) is correct.

135. While designing an experiment for Agrobacterium mediated plant transformation, a student noted down the following points:

P. Ti and Ri plasmids induce crown gall and hairy root disease, respectively

Q. Enzymes octopine synthase and nopalline synthase involved in the synthesis of cotopine and nopaline, respectively are encoded by T-DNA.

R. All the six vir genes, virA, virB, virC, virD, virE and virG are absolutely required for virulence.

S. Almost perfect 25 bp direct repeat sequences flanking all Ti and Ri plassmids in the T-DNA region is essential for T-DNA transfer.

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

(a) P, Q and R

(b) Q, R and S

(d) P, Q and S

Ans. (d)

Sol. Statement P, Q and S is correct.

136. The following are certain facts regarding bioremediation:

P. Biodegradable pastics are made using polyhydroxyalkanoates (PHAs) such as polyhydroxybutyrate (PHB).

Q. Pseudomonas putida F1 bacterial strain in involved in degradation of aromatic hydrocarbons.

R. The bacterium Deinnococcus radiodurans consume and digest toluence and ionic mercury from highly radioactive nuclear waste.

S. Bioaugmentation is a process of imporving the microorrganisms already existing in the system for degradation of xenobiotic compound.

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

(a) P, Q and R

(b) P, Q and S

(d) Q, R and S

Ans. (a)

Sol. Bioremediation is a branch of biotechnology that employs the use of living organisms, like microbes and bacteria to decontaminate affected areas. It is used in the removal of contaminants, pollutants, and toxins from soil, water, and other environments. Statement P, Q and R is correct.

137. A transgenic plant is developed with the following T-DNA construct

In order to analyze the nature of integration, genomic DNA digested with EcoRI was used for Southern hybridization using either probe A or B. The result obtained is as

The following conclusions were made :

1. There are two copies of the T-DNA cassette integrated at one loci and a third copy at another loci.

2. There are two copies of the T-DNA cassette integrated at one loci.

3. Complete T-DNA cassette has been integrated in all cases

4. In one T-DNA cassette there is a truncation towards the RB.

5. In one T-DNA cassette there is a truncation towards the LB.

6. The arrangement of the T-DNA cassettes integrated at the same loci is

7. The arrangement of the T-DNA cassette intergrated at the same loci is

Which of the above are correct?

(a) 2, 5 and 7

(b) 2, 4 and 6

(d) 2, 3 and 5

Ans. (b)

Sol. By observing probe A and probe B and there position in gel, there are two copies of the T-DNA cassette integrated at one loci, in one T-DNA cassette there is a truncation towards the RB and the arrangement of the T-DNA cassettes integrated at the same loci is

138. An unknown peptide was isolated from the leaf of a medicinal plant and purified to homogeneity. The peptide did not yield any sequence when subjected to Edman degradation. However, tryptic digest of the peptide produced a unique sequence. The mass of the intact peptide was found to be 18 Da less than that obtained from the trypsin treated sample. The possible interpretation of the above experimental results could be that

P. the N-terminus of the peptide was blocked by acetylation or methylation.

Q. the peptide was cyclic and contained at least one internal arginine or lysine residue.

R. the peptide was cyclic and contained a lysine or arginine residue at the C-terminus

S. the peptide was cyclized by peptide bond formation between -amino group and -carboxyl group.

Which of the above statements is true?

(a) P and S

(b) P and Q

(d) Q and S

Ans. (d)

Sol. An unknown peptide was isolated from the leaf of a medicinal plant and purified to homogeneity. The peptide did not yield any sequence when sujected to Edman degradation. However, tryptic digest of the peptide produced a unique sequene. The mass of the intact peptide was found to be 18 Da less than that obtained from the trypsin treated sample. This may be correct only if the peptide was cyclic and contained at least one internal arginine or lysine residue or the peptide was cyclized by peptide bond formation between α-amino group and -carboxyl group.

139. A gene is regulated by a noval transcription factor. The following techniques may be used to identify the cis-regulatory element in the 1kb promoter sequence fo the gene where the novel transcription factor binds:

1. Bioinformatics analysis

2. Cell based reporter assay

3. S1 nuclease assay

4. Electrophoretic mobility shift away

5. DNase-1 foot-printing analysis

Which one of the following can help to identify the cis element?

(a) 1 and 2

(b) 3 and 5

(d) 5 only

Ans. (d)

Sol. DNase-1 foot-printing analysis may be used to identify the cis-regulatory element in the 1kb promoter sequence fo the gene where the novel transcription factor binds.

140. A protein D is encoded by a gene, which is 5 kb long and has three HindIII restriction enzyme sites. The first one is 0.5 kb from he transcription start site, the second one is 2.5 kb from the first siter and the third one is 0.5 kb internal to the stop codon. The second site is plymorphic. In order to find out whether fetal cells contain the normal or the mutated gene, total genomic DNA from fetal cells was isolated, completely digested with HindIII, separated in an agarose gel, transferred to membrane and detected by a probe against the region between the second and third restriction site. Which one of the following band patterns will be obtained if the fetal cell is heterozygous?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. A polymorphic site means that the sequence has changed, therefore the enzyme will not be able to act in that position.

141. Given below are the experimental protocols to find out the exact location of repetitive DNA sequence in mitotic chromosome by FISH (fluorescence in situ hybridization). Which one of the protocols will give the correct result?

(a) Mitotic chromosomes were fixed on glass slide incubated with biotinylated telomeric DNA denatured incubated with fluorescently labbled avidin localization observed under fluorescence microscope.

(b) Mitotic chromosomes were fixed on glass slide denatured incubated with FITC labeled unrelated non-reptitive DNA sequence counterstained with propidium iodide localization observed under fluorescence microscope.

(c) Mitotic chromosomes were fixed on glass slide incubated with biotinylated satellite DNA incubated with fluorescently labeled avidin localization observed under fluorescence microscope.

(d) Mitotic chromosomes were fixed on glass slide incubated with repetitive DNA sequence binding protein denatured FITC labeled antibody against the protein localization observed under fluorescence microscope.

Ans. (c)

Sol. Mitotic chromosomes were fixed on glass slide incubated with biotinylated satellite DNA incubated with fluorescently labeled avidin localization observed under fluorescence microscope.

142. At 25°C values of the mean residue ellipiciy at 222 nm, are –33000 and –3000 deg cm² dmol^–1 for a polypeptide existing in -helical () and -structure (), respectively. If this polypeptide undergoes a two-state heat-induced transition, and a value of = –18000 deg cm² dmol^–1 is observed at 60°C, then this observation leads to the conclusion that the -helix conversion to -structure is

(a) 40%

(b) 50%

(d) 60%

Ans. (b)

Sol.

143. An EEG was recorded and its power spectrum analyses were done in rats with implanted electrode for a long time. The power of the EEG waves decressed two months after electrode implantation.

This observation may be due to the following :

P. Glial cells accumulate surrounding the exposed tips of electrodes.

Q. Degeneration of neurons occur surrounding the electrode tips due to metal ion deposition.

R. Coating of electrodes are destroyed with time.

S. The microsocket becomes loose with time.

Which one of the following is true?

(a) only P

(b) P and Q

(d) R and S

Ans. (b)

Sol. Glial cells accumulate surrounding the exposed tips of electrodes and degeneration of neurons occur surrounding the electrode tips due to metal ion deposition.

144. Using FRAP (Fluorescence Recovery After Photo-bleaching) techniques, diffusion coefficient of three integral membrans proteins M1, M2 and M3 in a kideny cell is calcualted as 1μm/s and 0.005 μm/s, respectively. Considering fluid-mosaic nature of biological membrane and relationship of structural organization of integral membrane protein with diffusion coefficient, which protein (s) will have highest number of integral membrane domain?

(a) M2 and M3

(b) M2 only

(d) M1 and M3

Ans. (c)

Sol. The higher the diffusion coefficient, the faster will be the speed (or mobility), which means higher FRAP. Transmembrane protein with multipass condition will have less mobility.

145. In resting cells, proteins X and Y are localized in the cytosol. Upon stimulation with lipopolysaccharide (LPS), both of them are phosphorylated and translocate to the nucleus. You have used antibodies against phosphorylated forms of proteins X and Y which are conjugated to either red, or green or blue dye. Keeping optical aberration of light in mind, which one of the following will be the best for visualizing X and Y in the nucleus by fluorescence microscopy?

(a) Anti green X and anti red Y

(b) Anti red X and anti green Y

(d) Anti blue X and anti green Y

Ans. (c)

Sol. In resting cells, proteins X and Y are localized in the cytosol. Upon stimulation with lipopolysaccharide (LPS), both of them are phosphorylated and translocate to the nucleus. You have used antibodies against phosphorylated forms of proteins X and Y which are conjugated to either red, or green or blue dye. Keeping optical aberration of light in mind, Anti red X and anti blue Y will be the best for visualizing X and Y in the nucleus by fluorescence microscopy.

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Previous Year Question Paper with Solution.