PYQ JUNE 2012 - VedPrep

CSIR NET BIOLOGY (JUNE - 2012)
Previous Year Question Paper with Solution.

21. Which nitrogen of adenosine gets protoanted if the pH of the nucleoside is lowered from 7 to 3 ?

(a) N1

(b) N3

(d) N9

Ans.

Sol. N1 will get protonated if the pH is lowered from 7 to 3 as N1 is more basic site second protonation will be N3.

22. The oligopeptide F-A-R-P-M-T-S-R-P-G-F, is treated with trypsin, chymotryipsin and caboxypeptidase-B. Apart from the origianl peptide, the number of fragments obtained will be

(a) 4

(b) 3

(d) 0

Ans. (d)

Sol. The endopeptidases cleave proteins in the middle of their chains with specificity. Trypsin, cleaves the peptide bonds in which basic amino acids (lysine and arginine) contribute the carboxyl group. Carboxypeptidase B, on the other hand, has an optimal pH range of 7.0–9.0 and cleaves successively at the carboxyl termini of the basic residues arginine and lysine. Chymotrypsin prefers cleaving after large hydrophobic amino acids such as phenylalanine, tyrosine and tryptophan, and also leucine and methionine.

23. Which one of the following interaction plays a major role in stabilizing B-DNA ?

(a) Hydrogne bond

(b) Hydrophobic interaction

(d) Ionic interaction

Ans. (b)

Sol. B-DNA is most common DNA found in living organisms. This contain Purines (A and G) and Pyrimidines (C and T), which bind itself by double and triple H-bonds.

24. Consider a 51-residue long protein containing only 100 bonds about which rotation can occur. Assume that 3 orientations per bond are possible. Based on these assumptions, how many conformations will be possible for this protein ?

(a) 3¹⁰⁰

(b) 100³

(d) 51 × 100 × 3

Ans. (a)

Sol. Conformation for protein is possible = no. of orientation no. of bonds.

Therefore answer is (a)

25. A plot of V/[S] versus V is generated for an enzyme catalyzed reaction, and a straight line is obtained. Indicate the information that can be obtained from the plot.

(a) V_max and turnover number K_m can be obtained only from a plot of 1/V versus 1/[S].

(b) K_m/V_max from the slope.

(d) Only K_m and turnover number.

Ans. (c)

Sol. The turnover number is Vmax divided by KM. The turnover number is Vmax divided by the substrate-bound enzyme concentration.

26. Phosphoglucomtuase is added to 0.1 M glucose-1-phosphate (G-1-P). The standard free energy change of the reaction, G-6-P G-1-P is 1.8 kcal/mole at 25°C. The equilibrium concentrations of G-6-P and G-1-P, respectively, are

(a) 96 mM, 45 mM

(b) 100 mM, 0 mM

(d) 0 mM, 100 mM

Ans. (a)

Sol. The phosphoglucomutase is an enzyme that transfers a phosphate group on an α-D-glucose monomer from the 1' to 6' position in the forward direction or the 6' to the 1' position in the reverse direction.

27. Differential scallning calorimetric study of calf thymus DNA was carried out to measure midpoint of thermal denauturation (T_m), (enthalpy change at T_m) and (constant-pressure heat capacity change). It has been observed that = 0, T_m = 75.5°C and = 50.4 kcal/mole. The Gibbs free energy change at 37°C is

(a) 25.5 kcal/mole

(b) 2.6 kcal/mole

(d) 5.6 kcal/mole

Ans. (d)

Sol. Several formulas are used to calculate T_m values. Some formulas are more accurate in predicting melting temperatures of DNA duplexes.

One problem in nuclei acid thermodynamics is to determine the thermodynamic parameters for forming double-stranded nucleic acid AB from single-stranded nucleic acids A and B.

AB = A + B

The equilibrium constant for this reaction is K = . According to thermodynamics, the relation between free energy, and K is = –RTln K, where R is the ideal gas law constant and T is the kelvin temperature of the reaction. This gives, for the nucleic acid system, = –RTln.

The melting temperature, T_m, occurs when half of the double-stranded nucleic acid has dissociated. If no additional nucleic acids are present [A], [B] and [AB] will be equal and equaal to half the initial concentration of double-stranded nucleic acid, [AB]_initial. This gives an expression for the melting point of a nucleic acid duplex of

T_m =

Because , T_m is also given by

T_m = .

28. Phosphatidyl serine, an important component of biological membrane, is located in

(a) the outer leaflet but flip flops to inner leaflet under specific conditions.

(b) both the leaflets.

(d) the inner leaflet but flip flops to outer leaflet under specific conditions.

Ans. (d)

Sol. Phosphatidyl serine a phospholipids is present preferentially in inner leaflet (cytosoli side). Negatively charged phospholipids are kept on inner leaflet by flippase. It is flipped to outer leaflet upon cell damage. Scrambalase is the enzyme that will transport negatively charged phospholipid from inner leaflet to outer leaflet.

29. Major disadvantage of using liposome as a targeted drug delivery vehicle is that

(a) it gets internalized by phagocytosis inside lysosomes.

(b) it is very unstable and has low shelf-life.

(d) its drug entrapment efficienty is very low.

Ans. (a)

Sol. Following such binding (irrespective of whether it is specific or nonspecific), the liposome can be internalized into the cell by the mechanism of endocytosis (or phagocytosis) inside lysosomes.

30. Major stimulus for spore formation in bacteria is

(a) nutrition limitation

(b) heat stress

(d) pH stress

Ans. (c)

Sol. Endospore formation is usually triggered by a lack of nutrients, and usually occurs in gram-positive bacteria. In endospore formation, the bacterium divides within its cell wall, and one side then engulfs the other. Endospores enable bacteria to lie dormant for extended periods, even centuries.

31. ATP-binding cassette (ABC) transporters

(a) are all P-glycoproteins.

(b) are found only in eukaryotes.

(d) affect translocation by forming channels.

Ans. (c)

Sol. ABC transporter consist of 2 distinct domains TMD (Transmembrane domain) and NBD (Nucleoside binding domain). ATP-binding cassette transporters (ABC-transporter) are members of a protein superfamily that is one of the largest and most ancient families with representatives in all extant phyla from prokaryotes to humans.

32. All cytosolic proteins have nuclear export signals that allow them to be removed from the nucleus when it reassembles after

(a) meiosis

(b) mitosis

(d) DNA replication

Ans. (b)

Sol. RNA molecules, which are made in the nucleus, and ribosomal subunits, which are assembled there, are exported to the cytosol after end of mitosis.

33. Cystic fibrosis (CF) transmembrane conductance regulator (CFTR) protein is known to be a cAMP-dependent Cl^– channel. CF patients (with mutant CFTR proteins) show reduced Cl^– permeability and as a result exhibit elevated Cl^– level in sweat. To prove this, CFTR proteins (both wild type and mutant) are inserted in a model membrane (liposome) and Cl^– transport is followed with radioactive Cl^–. It is known that topology of CFTR in membrane is very important for its function. Despite no proteolytic degradation or denaturation of CFRT proteins, wild-type CFTR failed to transport Cl^– in liposome.

Which of the following is the correct explanation of this ?

(a) CFTR protein gets mutated during insertion in liposome.

(b) CFTR protein loses affinity with Cl^– ions.

(d) CFTR protein loses channel forming property in liposomes.

Ans. (c)

Sol. Cystic fibrosis is an autosomal recessive genetic disorder affecting most critically the lungs and also pancreas, liver and intestine. It is characterized by abnormal transport of chloride and sodium across anepithelium, leading to thick, viscous secretions.

34. Assuming that histone octamer forms a cylinder 9 nm in diameter and 5 nm in height and that the human genome forms 32 million nucleoomes, what fraction (approximately) of the volume of nucleus (6 µm diameter) is occupied by histone octamers ?

(a) 1/21

(b) 1/11

(d) 10/11

Ans. (b)

Sol. Volume of the nucleus (spherical), v × 3.14 × (3 × 10³ nm)³ = 1.13 × 10¹¹ nm³

Volume of the histones octamers (cylindrical), = 3.14 × (4.5 nm)² × 5 nm × 32 × 10⁶ = 1.02 × 10¹⁰ nm³

35. Hoechst 33342 is a membrane-pemenat dye that fluorescse when it binds to DNA through an intercalating process. If a population of cells is incubated briefly with Hoechst dye and sorted in a flow cytometer, the cells display various levels of fluorescence in different phases of cell cycles as shown in figure below (marked as X, Y and Z). Which of the following is correct ?

(a) X is G₁, Y is G₂ + M and Z is S

(b) X is G₁, Y is S and Z is G₂ + M

(d) X is S, Y is G₁ and Z is G₂ + M

Ans. (b)

Sol. Flow cytometry is a laser based, biophysical technology employed in cell counting, sorting, biomarker detection and protein engineering, by suspending cells in a stream of fluid and passing them by an electronic detection apparatus. Flow cytometry is routinely used in the diagnosis of health diorders, especially blood cancers but has many other applications in basic research, clinicla practice and clinical trials.

From the graph it is revealed that X is in G₁, Y is in S and Z is in G₂ + M.

36. The scatter plot of growth rate and growth yield for 100 random environmental isolates of bacteria is shown below. Which of the following can be inferred from the data ?

(a) The two parameters are not related.

(b) Growth rate is inversely proportional to growth yield.

(d) High growth rate cannot be accompanied by high growth yield.

Ans. (d)

Sol. Evolutionary biologists predicted a rate-yield trade-off subject to differential selective pressures in different environments. Such a trade-off can explain the wide variation in growth rates and growth yields across the microbial world. When there is high growth rate there is not so high growth yield.

37. During cell cycle regulation in eukaryotes, there are post-translational modifications of protein factors, which act as switch for different phases of cell cycle. A cell population of yeast was transfected with gene for Wee1 kinase (modifies Cdc2 protein). Assuming that the transfection efficiency was 50% only. Which of the following graphical representation of the result is most appropriate ?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Wee1 is a nuclear kinase belonging to the Ser/Thr family of protein kinases in the fission yeast Schizosaccharomyces pombe (S. pombe). Wee1 has a molecular mass of 96 kDa and is a key regulator of cell cycle progression. It influences cell size by inhibiting the entry into mitosis, through inhibiting Cdk1.

38. Site-specific recombination results in precise DNA rearrangement, which is limited to specific sequences. The enzymes that are important to carry out the process are

(a) DNA polymerase and DNA gyrase

(b) nuclease and ligase

(d) restriction endonuclease and DNA polymerase

Ans. (b)

Sol. DNA ligase, an enzyme commonly used in molecular biology laboratories to join together DNA fragments. A DNA polymerase is an enzyme that helps catalyze the polymerization of DNA bases (deoxytribonucleotides) into a DNA strand.

39. Which of the following statements is not true about small interfering RNA (siRNA) ?

(a) siRNA has a 21-25 nucleotide sequence with 2 nucleotides overhanging at the 3' end.

(b) siRNA is processed by the RNA-protein complex RISC.

(d) siRNA does not generally act at the level of transcription.

Ans. (b)

Sol. siRNA is member of RNA interference. siRNA is processed by the RNA-protein complex RISC.

40. Presence of an internal ribosome entry site (IRES) in mRNA

(a) inhibits its translation

(b) promotes its post-transcription processing

(d) promotes its translation under adverse conditions

Ans. (d)

Sol. Presence of an internal ribosome entry site (IRES) in mRNA promotes its translation under adverse conditions.

41. Regulatory elements for expresssion of ribosomal RNA genes reside in the

(a) transcribed spacer region

(b) non-transcribed spacer region

(d) 5' flanking region of individual ribosomal RNA genes

Ans. (b)

Sol. Regulatory elements for expresssion of ribosomal RNA genes reside in the non-transcribed spacer region.

42. Double standed DNA replicates in a semi-conservative manner. In an in vitro DNA synthesis reaction, dideoxy CTP and dideoxy CMP were individually added in excess (in separate reaction tubes) in addition to dNTPs and other necessary reagents. Rate of DNA synthesis in two separate reaction tubes. Which of the following graphs represents the expected data ?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. In vitro DNA synthesis is a process of synthesis of DNA in laboratory condition i.e. in test tube by polymerase chain reaction process. The polymerase chain reaction (PCR) is a biochemical technology in molecular biology to amplify a single or a few copies of a peice of DNA across several orders of magnitude, generating thousands to millions of copies of a particular DNA sequence. In the process the concentration of ddCTP will decrease, whereas ddCMP will increase which is visible in choice no 1.

43. A synthetically prepared mRNA contains repetitive AU sequences. The mRNA was incubated with mammalian cell extract which contains ribosomes, tRNAs and all the factors required for protein synthesis. Assuming no initiation codon is required for protein synthesis, which of the following peptides will most likely be synthhesized ?

(a) A single peptide composed of the same amino acid sequence.

(b) A single peptide with alternating sequence of two amino acids.

(d) Three different peptides each sequence composed of a single amino acid.

Ans. (b)

Sol. If synthetic RNA polymer of alternating AU sequences is used for in-vivo translation, then peptide sequence have repetitive amino acid sequence of two different amino acids.

44. In semi-conservative mode of DNA replication, two parental strands unwind and are used for synthesis of new strands following the rule of complementary base pairing. Synthesis of complementary strands require that DNA synthesis proceeds in opposite direction, while the double helix is progressively unwinding and replicating in only one direction. One of the DNA strands is continuously synthesized in the same direction as the advancing replication fork and is called leading strand whereas the other strand is synthesized discontinuously in segments and is referred to as lagging strand. These short fragments made discontinuously are labeled as Okazaki fragments. These Okazaki fragments need to be matured into continuous DNA strand by which one of the following combination of enzymes ?

(a) DNA Pol III and DNA ligase

(b) DNA Pol I and DNA ligase

(d) DNA gyrase and DNA ligase

Ans. (b)

Sol. During replication, removal of Okazaki fragments (Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging template strand during DNA replication) and gap filling is done by DNA polymerase I and DNA ligase.

45. The Iac operon in E. coli, is controlled by both the Iac represssor and the catabolite activation protein CAP. In an in vivo experiment with Iac operon, the following observations were made :

P. cAMP levels are high.

Q. Repressor is bound with allolactose.

R. CAP is interacting with RNA polymerase.

Which one of the following conclusions is most appropriate based on the above observations ?

(a) Glucose and lactose are present

(b) Glucose is present and lactose is absent

(d) Glucose is absent and lactose is present

Ans. (d)

Sol. In an experiment medium when only lactose is present the lac operon

Allow lactose to bind to the repressor.

cAMP concentration is high in the cell.

CRP/CAP was interacting with RNA polymerase.

46. Which of the following statements is incorrect in relation to treatment of pre-B cells with phorbol esters?

(a) Phorbol esters acivate NF_kB for translocation into the nucleus.

(b) Phorbol esters activate protein kinase C

(d) Phorbol estes remove the inhibitor from inactivate complex in the cytoplasm

Ans. (c)

Sol. Phorbol esters are the tetracyclic diterpenoids generally known for their tumor promoting activity. The phorbol esters mimic the action of diacyl glycerol (DAG), activator of protein kinase C, which regulates different signal transduction pathways and other cellular metabolic activities.

47. Mycobacterium tunerculosis is an intra-cellular bacterium. It prefers to infect

(a) macrophages

(b) B-cells

(d) neutrophils

Ans. (a)

Sol. M. tuberculosis is an intracellular pathogen that infects phagocytic APCs (antigen presenting cells) in lungs, including alveolar macrophages, lung macrophages and dendritic cells.

48. Integrin molecule link extracellular matrix (ECM) to the actin cytoskeleton of cell. Integrin binds to which of the following ECM macromolecules ?

(a) Laminin

(b) Collagen

(d) Vitonectin

Ans. (c)

Sol. Integrin binds to the Fibronectin ECM molecules. Fibronectin is a high-molecular weight (~44 kDa) glycoprotein of the extracellular matrix that binds to membrane-spanning receptor proteins called integrins.

49. CD19 is a marker for

(a) B-cells

(b) T-cells

(d) NK cells

Ans. (a)

Sol. CD19 is a marker for B-lymphocytes. B-lymphocytes antigen CD19 also known as CD19 (Cluster of Differentiation 19), is a protein that in humans is encoded by the CD19 gene. It is found on the surface of B-cells a type of white blood cell.

50. Which one of the following matces of oncogene-protein is not correct ?

(a) erbA-thyroid hormone receptor

(b) erbB-epidermal growth factor receptor

(d) fos-platelet-derived growth factor receptor

Ans. (d)

Sol. c-Fos is a cellular proto-oncogene belonging to the immediate early gene family of transcription factors. c-Fos has aleucine-zipper DNA binding domain and a transactivation domain at the C-terminus. Transcription of c-Fos is upregulated in response to many extracellular signals, e.g., growth factors.

51. Upon ligand binding, cell surface receptors move laterally to be capped and internalized. Leishmania, a protozoan parasite, can use several receptors on macrophages to get internalized. One of them is Toll-like receptor 2 (TLR2) that binds lipophosphoglycan on Leishmania. Once internalized, the parasite is destroyed in the phagolysosom. Which of the following treatments of Leishmania-infected macrophages will result in lowerst parasite number in macrophages ?

(a) Membrane cholesterol depleing drug, -methyl cyclodextrin (-MCD)

(b) Ammonium chloride that increases lysosomal pH

(d) Anti-TLR2 antibody

Ans. (c)

Sol. Leishmania species are obligate intracellular parasites of cells of the macrophage-dendritic cell lineage. In the mammalian host, Leishmania reside mainly in long-lived resident macrophages where they differentiate from promastigotes to amastigotes and then survive in the hostile, acidic and higher temperature environment of a phago-lysosome-like organelle. Cholesterol depletion from macrophage plasma membranes using methyl-beta-cyclodextrin (MbetaCD) results in a significant reduction in the extent of leishmanial infection. The weak base ammonium chloride has been previously reported to inhibit lysosomal movements and phagosome-lysosome (Ph-L) fusion in cultured mouse macrophages (Me), thus reducing delivery, to an intraphagosomal infection, of endocytosed solutes that have concentrated in secondary lysosomes.

52. In a tissue, cells are bound together by physical attachement between cell to cell or between cell to extracellular matrix. Following are some of the characteristics of cell junctions :

1. Adherenes junctions are cell-cell anchoring junctions connecting actin filament is one cell with that in the next cell.

2. Desmosomes are cell-matrix anchoring junctions connecting actin filament in one cell to extracellular matrix.

3. Gap junctions are channel forming junctions allowing passage of small water soluble molecules from cell to cell.

4. Tight junctions are occluding junctions, which seal gap between two cells.

5. Hemidesmosomes are cell-matrix anchoring junctions connecting intermediate filament in one cell to extracellular matrix.

Which of the following combination of statements is not correct ?

(a) 1 and 2

(b) 1 and 3

(d) 4 and 5

Ans. (a)

Sol. A desmosome is a cell structure specialized for cell to cell adhesion. A type of junctional complex, they are localized spot-like adhesions randomly arranged on the lateral sides of plasma membranes. So, Desmosomes cannot join cell matrix and connect actin to cell-matrix.

53. Oncogenes are tumor suppressor genes are termed as cancer-critical genes. Increassingly powerful tools are now available for systematically searching the DNA or mRNAs of cancer cells for either significant mutations or altered expresssion. To identify independently an oncogene or a tumor suppressor gene, which of the following would be the most convincing tests to use ?

(a) Transgenic mice that overexpresss the candidate oncogene and knockout mice that lack candidate tumor suppressor gene.

(b) Transgenic mice that overexpress the candidate tumor suppressor gene and knockout mice that lack candidate oncogene.

(d) Knockout mice that lack the candidate oncogene and tumor suppressor gene.

Ans. (a)

Sol. In an experimental condition, use of transgenic mouse over expressing oncogene and loss of function in tumor suppressor gene will help in assessing the role of oncogene and rumor suppressor gene in cancer.

54. A large protein of a pathogenic bacterium has been enzymatically digested to generate a mixture of peptides ranging in size frorm 3 to 8 amino acids in length. Peptide mixtures were then administerd in experimental animals to generate peptide-specific antibodies. In order to develop diagnostics for the bacteria, the anitsera were used for western blotting to detect bacterial antigen. Western blooting failed despite the use of a wide range of antisera concentrations. What is the most probable cause of the problem ?

(a) Peptide-specific antibody mixture is unstable

(b) Peptide specific antibodies were not generated as adjuvant was not administered

(d) Peptide-specific antibodies could not regognize the bacterial antigen

Ans. (c)

Sol. The most probable cause may be that, Peptide specific antibodies were not raised as peptide were not linked to carrier.

55. A mouse was primed with trinitrophenyl-lipopolysaccharide (TNP-LPS) whereas another mouse was primed with TNP-Keyhole limpet hemocyanin (TNP-KLH). Afterr three weeks, tese mice were sacrificed and splenic cells were fractionated to B cells and T cells. B cells from TNP-LPS primed mice were co-cultured with T cells from TNP-LPS or TNP-KLH-primed mice. Similarly, B cells from TNP-KLHH primed mice were co-culutred with the T cells from TNP-LPS or TNP-KLH primed mice. So, we have four co-cultures :

P. B^{TNP – LPS} × T^{TNP – LPS}

Q. B^{TNP – LPS} × T^{TNP – KLH}

R. B^{TNP – KLH} × T^{TNP – LPS}

S. B^{TNP – KLH} × T^{TNP – KLH}

Among these co-cultures, where do you expect the highest IgG production ?

(a) P

(b) Q

(d) S

Ans. (d)

Sol. The aim of the present study was to improve the detection of intracellular antibodies to such a degree that specific antibodies of the IgM class could be demonstrated in addition to specific antibodies of the lgG class. By using trinitrophenyl (TNP)-alkaline phosphatase (AP) conjugates for detection of anti-TNP antibody-forming cells during the immune response against TNP-conjugated keyhole limpet hemocyanin (TNP-KLH), the detection of intracellular antibodies may be improved for two reasons. First, when TNP-AP conjugates, Second, TNP-AP conjugates are small, when compared to the relatively large antigen-enzyme conjugates used in our earlier studies. Smaller conjugates may penetrate better in the tissues and may be more difficult to remove once they are bound to specific antibodies in the cell. It consist of one AP molecule with several TNP groups, are used for detection of intracellular anti-TNP.

56. Given below are fate maps of two organisms and the patterns by which embryos undergo cleavage. Which of the following is/are the right combination(s) ?

(a) Q only

(b) Q and P

(d) Q and S

Ans. (a)

Sol. Rotational cleavage involves a normal first division along the meridional axis, giving rise to two daughter cells. The way in which this cleavage differs is that one of the daughter cells divides meridionally, whilst the other divides equatorially.

57. Spermatogonial stem clel undergoes extensive metamorphosis to become a spermatozoan. Meiosis lead to the formation of spermatid containing 22 autosomes and one sex chromosome. A male mouse was found in a colony which always produced only female pups upon mating. Which one of the following is a possible reason ?

(a) Spermiogenesis was defective

(b) All spermatogonial stem cells contained only X and no Y chromosome

(d) Activation of X chromosome linked postmeiotic death related gene may lead to such a situration

Ans. (c)

Sol. A male mouse always produce female puff on mating. The reason is activation of lethal gene associated with Y-chromosome.

58. In case of Xenopus laevis, which cells make up the Nieuwkoop centre and Spemann's organizer ?

(a) Endothermal and mesodemal respectively

(b) Mesodermal and endodermal respectively

(d) Ectodermal and endodermal respectively

Ans. (a)

Sol. Signals from the Nieuwkoop center induce the Spemann-Mangold organizer, thus the Nieuwkoop Center is known as the organizer of the organizer. Even with the BCNE center removed from the blastula, the Nieuwkoop Center is able to induce formation of the Spemann-Mangold organizer. The organizer is a unique region in the gastrulating embryo that induces and patterns the body axis. This organizer affects embryonic development by self-differentiation, regulation of morphogenesis and secretion of inducing signals.

59. The ced-9 gene appears to be a binary switch that regulates cellualr survival and apoptosis in nematodes. Considereing that CED-9 protein can bind to and inactive CED-4, which of the following would lead to apoptosis ?

(a) Activation of ced-9 gene

(b) Loss of function of CED-3

(d) Loss of function of CED-4

Ans. (c)

Sol. D1 and D2 proteins of PSII binds to P680 and primary acceptor pheophytin.

60. The functionality of the pax6 gene in the formation of optic and nasal structures may be attributed to the following :

P. Pax6 makes the optic vesicle competent and allows lens formation.

Q. The optic vesicle can induce any part of the head ecoderm to form the nasal and optic structures, due to the presence of Pax6.

R. Pax6 renders the head ecotderm competent to receive signals from the optic vesicle.

S. Apart from the optic vesicle, the head ectoderm may also be induced by BMP4 and FGF8, so Pax6 is not exclusive for lens formation.

Which of the above attributions are true ?

(a) P and S

(b) R and S

(d) R only

Ans. (b)

Sol. Pax6 is a transcription factor that plays a major role in ocular morphogenesis. The expression of Pax6 in adult mammals is restricted to the eye, brain, and pancreas. Pax-6 is essential for the formation of lens placodes from surface ectoderm.

61. In an experiment, sperm removed from epididymis of a male mouse was added in a dish containing appropirate media and oocyte. No fertilization was seen. However, when sperm from epidydmis were directly placed in uterus of an ovulated female, she became pregnant. These observations suggest that :

(a) the sperm need to travel some distance to attain fertilizing ability

(b) the oocyte secretes some biochemicals or factors which help sperm to fertilize

(d) the contents of female reproductive tract interact with sperm and activate it for fertilization

Ans. (d)

Sol. Under experimental conditions, when sperm and egg are kept together in test tube, they fail to fertilize the egg. But when the similar sperms are directly placed in uterus of model organism fertilization is observed. The reason for this contrast result in that the content of female reproductive tract interacts with sperm and activates fertilizing activity.

62. The following statements have been proposed for plant vegetative development :

P. Lateral roots develop from epidermal cells.

Q. Shoot axillary meristem develops from shoot apical meristem during differentiation of leaf primordia.

R. Root cap is made up of dead cells.

S. Lateral meristems and cylindrical meristems found in roots and shoots ressult in secondary growth.

Which of the above statements are true ?

(a) P and Q

(b) Q and S

(d) R and S

Ans. (b)

Sol. Shoot apical meristem gives shoot axillary meristmom and lateral meristem and radial meristem are responsible for secondary growth in roots.

63. The pattern of embryonic cleavage specific to a species is determined by two major parameters

P. the amount and distribution of yolk protein within the cytoplasm.

Q. the factors in the egg cytoplasm that influence the angle of mitotic spindles and the timing of its formation.

Which of the following statemens is true ?

(a) Species having telolecithal egg follow a holoblastic cleavage

(b) Species having isolecithal egg follow a holoblastic cleavage

(d) Species having isolecithal egg follow a meroblastic cleavage

Ans. (b)

Sol. In the absence of a large concentration of yolk, four major cleavage types can be observed in isolecithal cells (cells iwth a small even distribution of yolk) or in mesolecithal cells (moderate amount of yolk in a gradient) bilateral holoblastic, radial holoblastic, rotational holoblastic and spiral holoblastic, clevage.

64. The fate of a cell or a tissue is specified when it is capable of differentiating autonomously on being placed in a neutral environment with respect to the developmental pathway. An embryo will show a developmental pattern based on its type of specification :

Based on the above facts it can be said that the potency of a cell is

P. equal to its normal fate in regulative development.

Q. greather than its normal fate in regulative development.

R. equal to its normal fate in mosaic development.

S. greather than its normal fate in mosaic development.

Which of the above statements are true ?

(a) Q and R

(b) P and S

(d) Q and S

Ans. (a)

Sol. Based on the fact in question, the potency of a cell is equal to normal fate in mosaic development an greater in regulative development.

Mosaic and regulative development : The molecules that underlie these concepts are becoming more defined and understood. To oversimplify : mosaic development depends on agents, such as transcription factors, being placed locally in the egg by the mother. Regulative development depends in part on long-range gradients of positional information, such as the provided by the Hedgehog protein, that can pattern many cells at once. Regulative development can also be driven by short-range singals that trigger changes in cell identity in nearby neighbours.

65. In the context of the proximal-distal growth and differentiation of a tetrapod limb following experiments were visualized :

P. If the apical ectodermal ridge (AER) is removed at any time during the limb development, further development of distal limb skeletal elements ceases.

Q. If leg mesenchyme is placed directly beneath the wing AER, distal hindlimb structurse develop at the end of the limb.

R. If an extra AER is grafted onto an existing limb bud, supernumerary structures are formed usually at the distal end of the limb.

S. If leg mesenchyme is placed directly beneath the wing AER, proximal hindlimb structuers develp at the end of the limb.

Which of the above experiments would show the possible interactions between the AER and the limb bud mesenchyme directly beneath it during limb development.

(a) P and Q only

(b) Q and R only

(d) P, Q and R

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. AER (apical ectodermal ridge) maintains outgrowth of the limb bud, keeping underlying mesenchymal cells in the PZ (progress zone) in an undifferentiated state. If leg mesenchyme is placed directly beneath the wing AER, distal hindlimb structures (toes) develop at the end of the limb. (However, if this mesenchyme is placed farther from the AER, the hindlimb mesenchyme becomes integrated into wing structures). When the AER is removed, distal limb structures fail to form, with the degree of severity in limb truncation correlating with the timing of AER removal. These early studies established the AER as the major signaling center determining PD outgrowth. Furthermore, limb outgrowth defects following removal of the AER can be rescued by application of FGFs, suggesting that the function of the AER is defined by secreted FGFs.

66. Photosystem II functions as a light-dependent water-plastoquinoe oxidoreductase. What are the names of two reaction center proteins that bind electron transfer prosthetic groups, such as P₆₈₀, pheophytin and plastoquinone ?

(a) CP43 and CP47

(b) D1 and D2

(d) 33 kDa and 23 kDa

Ans. (b)

Sol. The D1 and D2 proteins occur as a heterodimer that form the reaction core of PSII, a multisubunit protein-pigment complex containing over forty different cofactors, which are anchored in the cell membrane in cyanobacteria, and in the thylakoid membrane in algae and plants.

67. Plants have evolved with multiple photoreceptors, which can perceive specific wavelength of light. Which one of the following statements is correct about the photoreceptors ?

(a) Pytochrome A can perceive far red and blue light.

(b) Phytochrome C anc perceive far red light.

(d) Phytochrome B can predominantly perceive far red light.

Ans. (a)

Sol. Plants are coated with a network of light-sensing photoreceptors that detect different wavelengths of light. Light signals perceived by the phytochrome family of red (R) and far-red (FR) light-absorbing photoreceptors direct plant growth and development throughout their lifecycle.

68. Which one of the following statement describes the process of phloem loading ?

(a) Triose phosphate is transported from the chloroplast to cytosol.

(b) Sugars are transported into the sieve elements and companion cells.

(d) Solutes are transported from roots to the shoots.

Ans. (b)

Sol. Sugars produced in sources, such as leaves, need to be delivered to growing parts of the plant via the phloem in a process called translocation, or movement of sugar. The points of sugar delivery, such as roots, young shoots, and developing seeds, are called sinks. Phloem loading refers to the transfer of sugar from mesophyll cells (source) to sieve tube elements, and phloem unloading refers to the transfer of sugar from sieve tube elements to roots or other storage cells.

69. Which one of the following combinations of secondary metabolite biosynthetic pathways result in the biosynthesis or terpenes ?

(a) Mevalonic acid and MEP pathways

(b) Malonic acid and MEP pathways

(d) Shikimic acid and mevalonic acid pathways

Ans. (a)

Sol. The major biosynthetic pathways for terpenes synthesis are : Mevalonic acid and MEP pathways.

70. The following reactions are part of the citric acid cycle. The numbers in parenthesis indicate the number of carbon atoms in each molecule.

Isocitrate (6) -ketoglutarate (5) Succinyl-CoA (4) Syccubate (4) Fumarate (4)

Which one of the following sequences of reaction systems A D is correct ?

(a)

(b)

(c)

(d)

Ans.

Sol. The Krebs Cycle is the central metabolic pathway in all aerobic organisms. The cycle is a series of eight reactions that occur in teh mitochondrion. These reactions take a two carbon molecule (acetate) and completely oxidize it to carbon dioxide. The cycle is summarized in the following chemical equation :

acetyl CoA + 3 NAD + FAD + ADP + 2 CO₂ + CoA + 3 NADH⁺ + FADH⁺ + ATP

71. The respiratory chain is relatively inaccessible to experimental manipulation in intact mitochondria. Upon distrupting mitochondria with ultrasound, however, it is possible to isolate functional submitochondrial particles, which consist of broken cristae that have resealed inside out into small closed vesicles. In these vesicles the components that originally faced the matrix are now exposed to the surrounding medium. This arrangement help in studying of electon transport and ATP synthesis because

(a) it is difficult to manipulate the concentrations of small molecules (NADH, ATP, ADP, P_i) in the matrix of intact mitochondria

(b) in broken cristae, the enzymes and other molecules responsible for electron transport are more active.

(d) purification of intact mitochondria is not possible

Ans. (a)

Sol. The microsome obtained from inner membrane of mitochondria is preferred to investigate because it is easy to maintain the different concentration of NAD, FAD and ADP with isolated microsome.

72. Following are some facts regarding localization of photosynthetic supramoleuclar complexes on plastic lamellae :

P. PSII is preferentially located on granal lamellae.

Q. ATP synthase and PSI are preferentially located on stromal lamellae.

R. PSI and PSII are located adjacent to each other on stromal lamellae.

S. Cyt b₆/f complex is not a membrane-bound complex.

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

(a) P and Q

(b) R and S

(d) Q and R

Ans. (a)

Sol. Photosystem I (PS I) and II (PS II) are found in the thylakoid membranes inside the chloroplast. The photosystems are responsible for carrying out photochemical phase or light reaction of photosynthesis. PS2 is located on granal lamellae ( inner side of thylakoid membrane) but the stroma lamellae lacks PS2 and has only PS1.

73. Upon absorption of a photon, a chlorophyll molecule gets converted to its excited state when the energy of the photon is

(a) more than of the ground state of the pigment molecule

(b) equal to that of the pigment molecule's excited state

(d) equal to the energy gap between ground state energy and the excited state energy

Ans. (d)

Sol. The first and foremost step of photosynthesis is the excitation of chlorophyll molecule which occurs with the absorption of light. This reaction is photochemical reaction since involves light and results in chemical change by transfer of electrons. The pigments present in the photosystems absorb photon of light. When the energy of photon is equal to the free energy gap between ground state energy and excited state energy.

74. Following are certain facts about the effect of abscisic acid (ABA) on the development and physiology effect of plants

P. ABA promotes leaf senescene independent of ethylene

Q. ABA promotes shoot growth and inhibits root growth at low water potentials.

R. ABA inhibits gibberllin induced enzyme production

S. Seed dormanyc is controlled by the ratio of ABA and gibberllin

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

(a) P, Q and R

(b) Q, R and S

(d) P, R and S

Ans. (d)

Sol. ABA is a key hormone that regulates water status and stomatal movement. Under drought conditions, plants produce and accumulate increased amounts of ABA in the guard cells, and this induces stomatal closure to conserve water. It was first detected as a growth inhibitor in diffusates from the young fruit. ABA regulates leaf senescence through the signal transduction of Ca²⁺ secondary messenger action. Embryonic ABA plays a central role in induction and maintenance of seed dormancy and also inhibits the transition from embryonic to germination growth.

75. Red and far-red lights are preceived by plants through various photoreceptors including phytochromes. The activation of phytochrome is caused by

(a) conversion of Pr to Pfr from through the effect of red light

(b) repression of Pr from through the effect of far-red light

(d) presence of red and far-red light at different fluence rates

Ans. (a)

Sol. Pr absorbs red light (~667 nm) and is immediately converted to Pfr. Pfr absorbs far-red light (~730 nm) and is quickly converted back to Pr. Absorption of red or far-red light causes a massive change to the shape of the chromophore, altering the conformation and activity of the phytochrome protein to which it is bound.

76. While studying the primary effects of different abiotic stresses on plants, a researcher observed water potential reduction and cellular dehydrations. Which of the following combination of abiotic stresses may cause the observed effect ?

(a) Water deficit, salinity and chilling

(b) Salinity, high-temperature and flooding

(d) Freezing, chilling and flooding

Ans. (c)

Sol. Plants experience water stress either when the water supply to their roots becomes limiting, or when the transpiration rate becomes intense. Water stress is primarily caused by a water deficit, such as a drought or high soil salinity. For every one degree C below freezing, the water potential decreases by 1.2 MPa.

77. Phenylalanine ammonia-lyase (PAL) and chalcone synthase (CHS) are involved in biosynthesis of phenolic compounds in plants. Following are some statements regarding the actions of PAL and CHS :

P. Substrates for PAL and CHS are phenylalanine and chalcone, respectively.

Q. PAL converts phenylalanine to trans-cinnamic acid.

R. PAL converts phenylalanine to p-coumaric acid.

S. p-coumaroyl-CoA is converted to chalcones by CHS.

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

(a) P and Q

(b) P and R

(d) Q and S

Ans. (d)

Sol. Phenylalanine ammonia lyase converts phenylalanine to trans-cinnamic acid and free ammonia. Chalcone synthase (CHS) catalyzes condensation of three molecules of malonyl-CoA to p-coumaroyl-CoA and Claisen condensation to yield naringenin chalcone.

78. Which of the following statemetns regarding aquaporins or water channel is not correct ?

(a) Aquaporins are found in both plant and animal membranes

(b) Aquaporins cannot transport unchanged molecules like NH₃

(d) Activity of aquaporin is regulated by pH and reactive oxygen species.

Ans. (b)

Sol. Aquaporins can transport unchanged molecules like NH₃ (aquaglyceroporins).

79. Which one of the following changes will occur in the cell membrane of nodal tissue of heart, which results in an increased heart rate due to stimulation of sympathetic nerves ?

(a) Opening of sodium channels is facilitated

(b) Potassium conductance is increased

(d) 'h' channels are inhibited

Ans. (c)

Sol. Sympathetic system increase heart rate by regulating T type calcium channel.

80. A person takes 1.0 ml of insulin injection daily at 8.00 AM. His son gave him 1.5 ml insulin at 8.00 AM considering the father will go to party and eat more during lunch. The father also avoided breakfast, as he planned to eat more during lunch. Which one of the following events will occur ?

(a) Father will be normoglycemic

(b) Father will be in hypoglycemic condition before lunch

(d) Blood glucose of father will be low after taking lunch

Ans. (b)

Sol. Because he skipped breakfast too, which is essential for sustainable maintenance of glucose level.

81. Level of follicle stimulating hormone (FSH) during infancy and adulthood is the same but spermatogenesis is seen only during adulthood. mRNA levels coding for FSH receptor are also found to be the same in testis of both age groups. Which of the following investigations will clarify this paradox a little more ?

(a) Culture testicular cells and add LH to see testosterone production

(b) Culture testicular celsl and add testosterone to see comparative rise in FSH and mRNA from both age groups

(d) Add both LH and FSH to testicular cells and evalutate cAMP production

Ans. (c)

Sol. The level of FSH during infacy and adult conditions are almost same, but sperm production is observed only in adults. It is due to the fact that culture leydig cells and add FSH bring comparative rise in cAMP in both age group.

82. The intestinal absorption of glucose is impaired by the use of ouabain, an inhibitor of Na⁺/K⁺ ATPase. Indicate the correct explanation.

(a) The inhibitor has blocked the transport of Na⁺ from intestinal lumen to epithelial cells

(b) The inhibitor has blocked the transport of Na⁺ from epithelial cells to the intestinal lumen

(d) The inibitor has blocked Na⁺ transport from the interstitial space to epithelial cells

Ans. (c)

Sol. Ouabain inhibt sodium potassium ATPase by blocking Na⁺ transport from epithelial cells to the interstitial space.

83. The stereocilia of auditory hair cells are arranged in rows but the heights of stereocilia are not the same in all the rows. Though the height of sterocilia is the same within a particular row, the heights incresase in subsequenct rows. When the sterecilia of shorter rows are mechanically pushed towards the taller rows, the hair cells are depolarized but a push on opposite direction hyperpolarizes them. The significane of this graded height of sterecilia is :

P. Each row of stereocilia may be displaced independent of other rows in physiological conditions.

Q. The tip of the taller stereocilia will show greater displacement as compared to shorter ones when all the rows arem oving in the same axis.

R. The hair cells will be depolarized or hyperpolarized in different grades when the axis of displacement is changed.

S. The taller stereocilia are involved with depolarization and shorter ones are responsible for hyperpolarization.

Which one of the following is correct ?

(a) P only

(b) Q only

(d) P and S

Ans. (c)

Sol. The tip of the taller stereocilia will show greater displacement as compared to shorter ones when all the rows arem oving in the same axis and the hair cells will be depolarized or hyperpolarized in different grades when the axis of displacement is changed.

84. GnRH is secreted during infancy (0-6 months) and puberty onwards (4 years and above) in mokeys. However, i.v injection of GnRH during pre-pubertal period (about 2 years of age) led to elevated LH and FSH in blood compared to untreated 2 years old monkey. This suggests that :

P. Hypothalamus is active during pre-pubertal period.

Q. GnRH action on pituitary is age dependent.

R. Pituitary matures during adulthood.

S. Pitutary is active in all the stages of developemnt in monkey

Which one of the following is true ?

(a) P and Q

(b) Q and R

(d) S only

Ans. (d)

Sol. Condition mentioning in question occurs because Pitutary is active in all the stages of developemnt in monkey.

85. A person has been suffering from night blindness. On consultation, the doctor advised the person to eat carrots and/or cod fish oil. After some time having seen no imporvement, the doctor gave the person vitamin A injection. Still no marked improvement was seen. The doctor mooted several suggestions indicating lack of the following enzymes for the failure of treatment :

P. Retinol dehydrogenase

Q. Retinal reductase

R. Retinal isomerase

S. Retinal synthase

Which one of the following is correct ?

(a) P only

(b) Q only

(d) R and S

Ans. (a)

Sol. Night blindness may also be due to the result of Zn deficiency which reduces the activity of an enzyme that helps vitamin A. This enzyme is retinol dehydrogenase.

86. A person suffering from thyrotoxicosis has extremely high level of thyroid hormone in blood. There is a failure of feedback regulation in hypothalamic-pituitary-thyroid axis. The detailed blood investigation exhibit high level of the following :

P. Thyroid stimulating hormone (TSH)

Q. Thyroid stimulating immunoglobulin (TSI)

R. Thyrotrophin releasin hormone (TRH)

S. Parathyroid hormone (PTH)

In your opinion, which one of the following is the reasone for such thyrotoxicosis ?

(a) P only

(b) Q only

(d) R and S

Ans. (b)

Sol. Thyrotoxicosis is caused due to high level secretion of thyroid stimulating in immunoglobular.

87. How many genetically different gametes can be made by an individual of genotype AaBbccDDEe, assuming they are independently assorting ?

(a) 3

(b) 5

(d) 22

Ans. (c)

Sol. Three genes can have 2 different types of alleles. The genes that show only 1 type of allele need not to be considered. Hence, 2*2*2 = 8.

88. Mutation at two different loci of the same gene X results in altered functions. These two mutated versions of the gene X are called

(a) alleles

(b) complementation groups

(d) linkage groups

Ans. (a)

Sol. A change on one locus to another locus of the same gene results in mutation. The two copies of mutated same gene termed as allele.

89. A gene encoding tRNA undergoes a mutational event in its anticodon region that enables it to recognize a mutatnt nonsense codon and permit completion of translation. Such a mutation is konwn as

(a) silent mutation

(b) neutral mutation

(d) nonsense suppressor

Ans. (d)

Sol. A tRNA gene is mutated at anticodon region to recognize stop codon an allow translation beyond pre mature stop codon in mRNA transcribed from mutated protein coding gene, such that functional protein is produced. Such a phenomenon is termed as Nonsense suppressive mutation.

90. Two puer lines of corn have mean cob length of 9 and 3 inches, respectively. The polygenes involved in this trait exhibit additive gene action. Crossing these two lines is expected to produce a progeny population with mean cob length (in inches) of

(a) 12.0

(b) 7.5

(d) 2.75

Ans. (c)

Sol. Cob length of dominant pure lines (CC) = 9 inches

Contribution of each allele = = 4.5 inches

Cob length of recessive pure lines (cc) = 3 inches

Contribution of each allele = = 1.5 inches

On crossing of both pure lines, we get heterozygous progeny (Cc).

Cob length in heterozygous progeny = 4.5 + 1.5 = 6 inches.

91. When two independent pure lines of pea with white flowers are crossed, the F1 progeny has purple flowers. The F2 progeny obtained from selfing shows both purple and white flower in a ratio of 9 : 7. The following conclusions were made :

P. Two different genes are involved, mutations in which lead to formatino of white flowers.

Q. These two genes show independent assortment.

R. This is an example of complementary gene action.

S. This is an example of duplicate genes.

Which of the abvoe conclusions are correct ?

(a) P and R only

(b) P and S only

(d) P, Q and R

Ans. (d)

Sol. F2 progeny phenotypic ratio 9:7 represent duplicate recessive epistasis. So, it is clear that two different genes are involved, mutations in which lead to formatino of white flowers, these two genes show independent assortment, this is an example of complementary gene action.

92. A cell undergoing meiosis produces four daughter cells, two of which are aneuploids, while two are haploid. This can occur due to :

(a) Non-disjunction during first meiotic division only

(b) Non-disjunctions during second meiotic division only

(d) Non-disjunction during both first and second meiotic divisions

Ans. (b)

Sol. Nondisjuction during meiosis 2 will produce aneuploids as n + 1 and n – 1 and haploids as n and n. Thus half gametes are affected by non disjunction during meiosis 2 while all gametes are affected during meiosis I.

93. The auxotrophic strains of E. coli A (met^– bio^– thr⁺ leu⁺ thi⁺) and B (met⁺ bio⁺ thr^– leu^– thi^–) were incubated together for 18 hours in a liquid complete medium and then ~10⁸ cells were plated on a minimal medium. Prototrophs were observed at a frequency of 1 × 10^–7 cells. This may have happened by a process of genetic recombination between the two strains or by mutation of the strains. Which of the following control experiments would help rule out the possibility of mutation ?

(a) Planting starins A and B directly on minimal medium

(b) Growing the mixture of strains A and B for 18 hours and then plating on complete medium

(c) Growing strains A and B individually in a liquid complete medium for 18 hours and then plating them on a minimal medium

(d) Growing the obtained prototrophs in a liquid complete medium for 18 hours and then plating them on a minimal medium

Ans. (c)

Sol. By inhibiting recombination between both auxotrophic strains of E.coli, possibility of mutation to produce prototroph can be rule out by growing them individually in separate media and plating them on minimal medium.

94. Following are four modes of inheritance

P. X-linked recessive

Q. X-linked dominant

R. Autosomal recessive

S. Autosomal dominant

Which of the above modes of inheritance can explain the pedigree shown below ?

(a) P and R

(b) Q and R

(d) S only

Ans. (c)

Sol. By observing pedigree chart the trait may be autosomal recessive or autosomal dominant.

95. Three E. coli mutants are isolated which require compound 'A' for their growth. The compounds B, C and D are known to be involved in the biosynthetic pathway to A. In order to determine the pathway, the mutants were grown in a medium supplement with one of the compounds, A to D. The results obtained are summarized below :

'+' indicates growth; '0' indicate lack of growth. Which of the following equations represents the biosynthetic pathway of A ?

(a) B C D A

(b) C D B A

(d) A C D B

Ans. (a)

Sol. The correct sequence of compounds involved in biosynthetic pathway is B C D A.

96. Four different mutant lines showing similar phenotype were identified from a genetic screen. When genetic crosses among these mutants were carried out, the first mutant was found to complement the second, third and fourth mutant lines. However, no other complementation was observed. How many complementation groups do the four mutant lines belong to ?

(a) one

(b) two

(d) four

Ans. (b)

Sol. As first mutant complement with second, third and fourth mutant lines mutant 1 is in separate complementation group then mutant 2, 3 and 4. Hence, there are two complentation groups.7

97. Why lysogenic cycle is more benefical to a virus than lytic cycle under certain circumstances ?

(a) The lysogenic cycle prevents local extinction of the host while still retaining its infection potential

(b) By integrating with the bacterial chromosomes, the genetic instructions for the virus become refreshed after one or more replication events during binary fission.

(d) Lysogeny causes more mutations to occur in the virus, creating more variants upon which natural selection can operate.

Ans. (a)

Sol. Lysogenic cycle is more benefical to a virus than lytic cycle because the lysogenic cycle prevents local extinction of the host while still retaining its infection potential.

98. The rates of mutation in E. coli from Iac^– to Iac⁺ are determined using a medium containing lactose, as the only source of energy. The principle of spontaneity of mutations can be said to e violated if :

(a) the rate of mutation increases during starvation

(b) in the presence of lactose, the rate of mutation from Iac^– and Iac⁺ is increased but overall rate of mutation is not

(d) the rate of mutation in Iac gene is always less than that in other genes

Ans. (b)

Sol. The principle of spontaneity of mutations can be said to e violated if in the presence of lactose, the rate of mutation from Lac^– and Lac⁺ is increased but overall rate of mutation is not.

99. Which of the following organism is widely used as biocontrol agent in organic farming ?

(a) Rhizobium tropicii

(b) Trichoderma viridis

(d) Nostroc muscorum

Ans. (b)

Sol. Trichoderma viridis, is used as a biocontrol agent in organic farming.

100. A paraphyletic group

(a) contains unrealted organisms

(b) includes the most recent common ancestor but not all of its descendents

(d) contains all the representatives of a clade and the most recent common ancestor

Ans. (b)

Sol. A paraphyletic group contains the most recent common ancestor but not all of its descendants, e.g., reptilia contains the last common ancestor of descendants of that ancestor but not birds and mammals.

101. Which of the following is not an adaptive modification in a xerophytic plant ?

(a) Strongly developed sclerenchyma

(b) Sunken stomata

(d) Presence of lacunar tissues

Ans. (d)

Sol. Xerophytic plants are those which grown in and condition, where evaporation is more than precipitation and hence plant has sunken stomata.

102. If milk is left open, lactose is fermented first to produce acid. This is followed by proteolytic bacterial acitivity which increases the pH. Ultimately milk fats are degraded to produce rancidity. This is an example of

(a) ecological succession

(b) microbial antagonism

(d) microevolution

Ans. (a)

Sol. Ecological succession is the phenomenon or process by which an ecological community undergoes more or less orderly and predictable changes following disturbance or initial colonization of new habitat.

103. Symbiotic biological nitrogen fixation takes place with the association between a plant and a nitrogen fixing prokaryote as shown in the following table :

The correct combination is

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

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

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

Ans. (b)

Sol. Carbon dioxide equivalent is highest for CFCs. (amongst given option. It is 140 ~ 11,700)

104. Secondary sewage treatment involves

(a) physical removal of solids from polluted water by filtration and sedimentation

(b) removal of chemical remains by precipation

(d) removal of microbial pathogens by chlorination or ozonization

Ans. (c)

Sol. Secondary treatment is the removal of biodegradable organic matter (in solution or suspension) from sewage or similar kinds of wastewater. The aim is to achieve a certain degree of effluent quality in a sewage treatment plant suitable for the intended disposal or reuse option.

105. Based on per molecule, which of the following gas has the most powerful greenhouse effect ?

(a) CO₂

(b) CH₄

(d) CFCs

Ans. (d)

Sol. The infrared bands of chlorofluorocarbons and chlorocarbons enhance the atmospheric greenhouse effect. A single molecule of CFC-12 can hold nearly 11,000 times the heat of carbon dioxide, making it an extraordinarily potent greenhouse gas.

106. A species has the following population characteristics :

P. Reduction in population size ≥ 90% over the last 10 years or 32 generations.

Q. Geographic range : Extent of occurence : < 100 km² and Area of occupancy : < 10 km².

R. Population size less than 50 matured individuals.

S. Probability of extinction in the wild is at least 50% within the next 10 years or 3 generations.

To which of the following categories the species will be assigned according to IUCN categorization of threatened species (version 3.1) ?

(a) Endangered

(b) Vulnearble

(d) Extinct in the wild

Ans. (c)

Sol. A species is critically endangered when it is facing an extremely high risk of extinction in the wild in the immediate future. To be defined as critically endangered, a species must meet any of the following criteria.

Population reduction : ≥ 80-90% population decline.

Geographic range

Extent of occurrence : <100 km²

Area of occupancy : <10 km²

Population size : <250 mature individuals

Extinction probability (in the wild) : at least 50% within 10 years or 3 generations.

107. Which of the follwoing hypothesis best explains the occurence of Himalayan floral elements in Western Ghats of India ?

(a) Continental drift theory

(b) Deccan trap theory

(d) Coromandel coast hypothesis

Ans. (c)

Sol. Some scientists believe that the Himalayas are a major factor in the current ice age, because these mountains have increased Earth's total rainfall and therefore the rate at which carbon dioxide is washed out of the atmosphere, decreasing the greenhouse effect.

108. Identify a, b and c in the figure :

(a) a = mitochondria; b = multicelluarity; c = chloroplast

(b) a = mitochondria; b = chloroplast; c = multicelluarity

(d) a = chloroplast; b = nucleus; c = mutlicelluarity

Ans. (b)

Sol. Among the most informative mitochondrial genomes have been those of protists (primarily unicellular eukaryotes), some of which harbor the most gene-rich and most eubacteria-like mitochondrial DNAs (mtDNAs) known. Chloroplasts were originally established in eukaryotes by the endosymbiosis of a cyanobacterium; they then spread through diversification of the eukaryotic hosts and subsequent engulfment of eukaryotic algae by previously nonphotosynthetic eukaryotes. Brown algae are exclusively multicellular and found in marine habitats, most typically in the intertidal zone.

109. In which of the following classes of vertebrates there are groups of animals without limbs?

(a) Fish, reptiles and mammals

(b) Reptiles only

(d) Amphibianas only

Ans. (c)

Sol. Reptiles and mammals are classes of vertebrates without limbs.

110. The schematic section given below of animal indicates that the animal is

(a) triploblastic, coelomic, invertebrate

(b) triploblastic, acoelomic invertebrate

(d) triploblastic, coelomic, vertebrate

Ans. (a)

Sol. AS VENTRAL nerve cord is there and coelom is also there along with mesoderm so it is triploblastic, coelomate, invertebrate.

111. At a given time, the age class distribution of a population was as shown in the figure. Which of the following can be infered from the figure ?

(a) Age class 2 has maximum fecundity

(b) Age class 2 has maximum survival

(d) Age class distribution is not at equilibrium

Ans. (d)

Sol. The fecundity is defined as the maximum reproductive output potential of an individual under ideal environmental conditions. Age class distribution is not at equilibrium.

112. While studying the diversity of 4 communities, 5 species and 50 individuals were recorded from each community. The number of individuals under each species was listed as mentioned in the following table. In which of the following communities Pielou's Evenness Index (e) will be 1 ?

Ans.

Sol. As each species has 10 individuals so it is more even.

113. Average annual precipitatian and temperature are two important determinats of world's major biomass. Which of the following combination is correct ?

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

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

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

Ans. (a)

Sol. The average temperature in tropical rainforests ranges from 21 to 30°C. The environment is pretty wet in tropical rainforests, maintaining a high humidity of 77% to 88% year-round. The yearly rainfall ranges from 200 to 1000 cm,and it can rain hard. Annual precipitation over 140 cm Mean annual temperature is between 4 and 12°C. Average dewpoints in summer range from 67.8 to 71.6°F (20 to 22°C). Winters in Savannah are mild and sunny. The average daily high temperatures close to 60°F (16°C). Temperatures ranging between –30°C-30°C (–22°F-86°F) drop below freezing on an annual basis, resulting in defined growing seasons during the spring, summer, and early fall. Precipitation is relatively constant throughout the year, ranging between 75 cm and 150 cm (29.5-59 in).

114. According to fossil records, the earliest fossils of liverworts are found in the late Devonian, of mosses in the early Cretaceous and vascular plants in the late Silurian/early Devonian. Anthoceros (hronworts) fossils have not been discovered. Reading fossil records we would say that vascular plants appeared first, then mosses and the liverworts.

However, pohylogenetic relationships (shown in the figure) suggest otherwise. It my be that

1. evolutionary history can be read directly from the fossil record.

2. the moss lineage goes back to at least early Silurian/early Devonian.

3. fossils can only set a maximum age for a lineage.

4. fossils can only set a minimum age for a lineage.

5. the divergence between liverworts and the rest of tland plants goes back to at least the early Ordovician

Which of the following statements is correct ?

(a) 1, 2, 3, 5

(b) 2, 4, 5

(d) 2, 3, 5

Ans. (b)

Sol. Figure shows that the moss lineage goes back to at least early silurian. Fossils can only set a minimum age for a lineage. The divergence between liverworts and rest of land plants goes back to at least early Ordovician.

115. In Lotak and Voltera's two species competition model :

Where N represents population size, r growth rate and K maximum carrying capacity for species 1 and 2. The interspecific competition coefficient <1 will mean

(a) individuals of species 2 have less inhibiting effect on individuals of species 1 than individuals of species 1 on others of their own species

(b) individuals of species 2 have greater inhibiting effect on individuals of species 1 than individuals of species 1 on others of their own species

(c) individuals of species 1 have less inhibiting effect on individuals of species 1 than individuals of species 1 on others of their own species

(d) individuals of species 1 have greater inhibiting effect on individuals of species 1 than individuals of species 1 on others of their own species

Ans. (a)

Sol. When < 1, the effects of species 1 (interspecific competition) is less than the effect of species 1 on its own members (intraspecific competition). Coversely, when > 1, the effect of species 2 on species 1 is greater than the effect of species 1 on its own members.

116. Sexual selection resulsts in variation in the reproductive success of males, often due to female choice with particular phenotypes. This type of sexual selection occurs because

(a) males cannot complete with other males

(b) cost of breeding is higher for females as compared to males

(d) males are a limiting resource for females

Ans. (b)

Sol. Cost of breeding is higher for females as compared to males.

117. Among the follwong events in the history of life

1. Prokaryotic cell

2. Eukaryotic cell

3. Natural selection

4. Organic molecules

5. Self-replicating molecules

Which is the correct chronological order ?

(a) 4, 5, 3, 1, 2

(b) 4, 5, 1, 2, 3

(d) 4, 5, 1, 3, 2

Ans. (a)

Sol. In the history of life, the first organic molecules formed about 4 billion years ago. This may have happened when lightning sparked chemical reactions in Earth's early atmosphere. RNA may have been the first organic molecule to form a s well as the basis of early life. The first self-replicating molecules were nucleotides. After natural selection, prokaryotic cells formed and at last eukaryotic cells were formed.

118. The Hardy-Weinberg principle comes from considering what happends when Mendelian genes act in a population. The model predicts that there will be no change in allele frequencies when

(a) migration into the population occurs at a steady rate

(b) the population suffers a bottleneck

(d) no evolutionary process is at work

Ans. (d)

Sol. Hardy-Weinberg law states that both alleles and genotypic frequencies in a population remains constant from generation to generation until some specific disturbing influences are introduced like non random mating, mutation, selection, random genetic drift, gene flow and meiotic drive.

119. Which of the following is responsible for initiation of maternal behaviour in the first-time pregnant rats after parturition ?

(a) Higher prolactin levels in blood

(b) Stimulation of secondary receptors during delivery

(d) Presence of male rats

Ans. (a)

Sol. Prolactin is inducer for lactation and maternal behaviour.

120. Northern elephant seals had been reduced to about 20 individuals in the 1800s. Biologists studied variation in proteins in the species. They found no genetic differences in the protein among different individuals. This lack of variation is due to :

(a) the fact that elephant seals live in a constant environment where there is no need for genetic variation

(b) population bottleneck and genetic drift

(d) a very low rate of mutation

Ans. (b)

Sol. They were killed for their blubber from which oil is obtained. Due to their mass killing once their overall population let was 20 to 100. All the current northern elephant seals are the descendants of the same population i.e. they are brothers and sisters.

They faced bottleneck at that time and were near to extinct. Thus this lack of variation is due to population bottleneck and genetic drift bottleneck is an evolutionary event in which a significant percentage of population or species is killed or otherwise prevented from reproducing.

121. There are 'n' number of alleles for a given locus in a diploid population. The proportion of all homozygotes in the population (i) if all alleles are equally abundant and (ii) if all alleles are not equally abundant, will be

(a) (i) 1/n (ii) < 1/n

(b) (i) 1/n (ii) > 1/n

(d) (i) 1/n² (ii) < 1/n²

Ans. (b)

Sol. Correct option is (b)

122. Which of the following is a prediction of the neutral theory of molecular evolution that is supported by data ?

(a) Humans and chimps differ more in DNA sequences of pseudogenes than in coding regions of functional genes.

(b) Humans and chimps differ more in DNA sequences of coding regions of functional genes than of pseudogenes.

(d) The more advanced species have more number of functional genes.

Ans. (a)

Sol. Correct option is (a)

123. In hymenopteran insects, males are haploid and females are diploid. All fertilized eggs give rise to femalesa and unfertilised eggs give rise to males. As a result, if a female mates with a single male, the females in the progeny are related to each other by 75%. But if the mother had mated with many males, the mean genetic relatedness of female progeny is correctly represented by :

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Correct option is (a)

124. To replace animal use in testing hepatic toxicity of a drug on trial, which one of the following would be used in vitro to be closest to the in vivo scenario ?

(a) Liver cells

(b) Hepatic cell lines

(d) Co-culture of liver parenchymal cells and kupffer cells

Ans. (c)

Sol. To replace animal use in testing hepatic toxicity of a drug on trial, liver slices are used in vitro to be closest to the in vivo.

125. Which of the following does not represent a strategy of phytoremediation ?

(a) Phytodegradation

(b) Phytomining

(d) Chelate mediated extraction of pollutants

Ans. (b)

Sol. Phytomining is use of plants to extract metal compound of high economic value.

126. The word fermentation is used in biochemistry and microbial technology to denote different phenomena. If the former is called C and latter is called T, which of the following statements is true ?

(a) All C is T but all T is not C

(b) All T is C but all C is not T

(d) C is always anaerobic process, while T can be aerobic or anaerobic

Ans. (d)

Sol. As per question, C is always anaerobic process, while T can be aerobic or anaerobic.

127. Which of the following statements is not true during the infection of plant cells with Agrobacterium ?

(a) The protein products of virulence genes virA and VirG perceive acetosyringone

(b) The VirB proteins forms a connection between Agrobacterium and the plant cell and facillitates T-DNA transfer into the plant

(d) The T-DNA, after becoming coated with VirF binds to phophorylated VIPI, which, allows the complex to enter the plant's nucleus

Ans. (d)

Sol. Agrobacterium is a genus of Gram-negative bacteria established by H.J. Conn that uses horizontal gene transfer to cause tumors in plants.

128. Which is the best method for checking mycoplasma contamination in a mammalian cell line ?

(a) Southern hybridization

(b) ELISA

(d) Western hybridization

Ans. (c)

Sol. PCR is the best method for checking mycoplasma contamination in a mammalian cell line.

129. Among existing technologies, which of the following vector systems would you prefer to use for generating a libranry for 140 kb eukaryotic genomic DNA fragments, while giving due consideration to size as well as stability of the insert ?

(a) Phage

(b) Cosmid

(d) Yeast artificial chromosome (YAC)

Ans. (c)

Sol. COSMID-37 to 52 kb

BAC-150-350 kb

YAC-100-3000 kb

Lamda-below 100 kb

130. If r denotes the correlation coefficient and m denotes the slope of regression line, interchanging X and Y axes would

(a) change m but not r

(b) change r but not m

(d) not change r or m

Ans. (a)

Sol. If r denotes the correlation coefficient and m denotes the slope of regression line, interchanging X and Y axes would change m but not r.

131. The use of biotinylated secondary antibody in ELISA

(a) increases the sensitivity of the assay but compromises the specificity

(b) increases the sensitivity of the assay without compromising the specificity

(d) decreases both sensitivity and specificity

Ans. (b)

Sol. Biotin labeled Antibodies (Abs) increase sensitivity. They reduce cross reactivity and steric hindrance. It results in higher sensitivity and other specificity.

132. In a Radioimmunoassay (RIA) for glycocorticoid hormone, radioactive glucocorticoid (tritiated) is added to the RIA cocktail. When the amount of bound hormone was measured no counts were observed. The following explanation (s) were proposed :

P. the radioactive hormone was insufficient.

Q. the radioactive tag to the hormone completely dissociated during storage.

R. antibody was not added to the cocktail.

S. the specific activity of the tritium was low.

Choose the correct option (s)

(a) P only

(b) R only

(d) P and S

Ans. (c)

Sol. In a Radioimmunoassay (RIA) for glycocorticoid hormone, radioactive glucocorticoid (tritiated) is added to the RIA cocktail. When the amount of bound hormone was measured no counts were observed. In this case, the radioactive tag to the hormone completely dissociated during storage and antibody was not added to the cocktail.

133. Which of the following curves correctly represents the process of antibiotic production by Streptomyces sp. ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Figure in option (b) correctly represents the process of antibiotic production by Streptomyces sp.

134. Ten different mouse strains were primed with whole Keyhole limpet hemocyanin (KLH). KLH was broken into ten peptides for in vitro stimulation. The splenocytes fom ten different primed mouse strains were restimulated with each of these ten peptides and the sresponsiveness to these peptides were measured in vitro. It was found that each of these mouse stains had responded to one of the peptides. When the peptide 3 responser was mated with peptide 4 responder, the splenocytes of F1 offspring responded to bothe the peptides. Which of the following is most appropriate ?

(a) Mouse strains responding to peptide 3 or peptide 4 have different MHC haplotypes

(b) Mouse strains responding to peptide 3 or peptide 4 have eithe or thsee T cell receptor

(d) Mouse strains responding to peptide 3 or peptide 4 did not express MHC class-I molecule

Ans. (a)

Sol. Ten different mouse strains were primed with whole Keyhole limpet hemocyanin (KLH). KLH was broken into ten peptides for in vitro stimulation. The splenocytes fom ten different primed mouse strains were restimulated with each of these ten peptides and the sresponsiveness to these peptides were measured in vitro. It was found that each of these mouse stains had responded to one of the peptides. When the peptide 3 responser was mated with peptide 4 responder, the splenocytes of F1 offspring responded to both the peptides. This conclude that Mouse strains responding to peptide 3 or peptide 4 have different MHC haplotypes.

135. For the generation of transgenic plants in crop improvement, one important regulatory gene X was overexpressed in a crop plant. Out of 30 tarnsgenic rice plants generated, 22 showed high levels of gene X experssion. However, rest 8 lines displayed low levels of experssion. One explanation of such observation may be

(a) Suppression effect of the transgene

(b) Knock down effect of the gene X

(d) Co-suppression effect of the transgene

Ans. (c)

Sol. For the generation of transgenic plants in crop improvement, one important regulatory gene X was overexpressed in a crop plant. Out of 30 tarnsgenic rice plants generated, 22 showed high levels of gene X experssion. This can happen only due to Gene silencing effect.

136. Stem cells are widely used for their degenerative property and capacity to differentiate into different lineages. A person with a damaged liver approaches a stem cell therapist. Which of the following therapeutic strategies would be the safest ?

(a) Procuring adult liver cells from a healthy donor and grafting them into the patient

(b) Transformign skin cells from the patient into iPS cells and using them for further differentiation and grafting in liver

(d) Injecting cord blood cells into the liver directly

Ans. (b)

Sol. Transformign skin cells from the patient into iPS cells and using them for further differentiation and grafting in liver is the safest method for patient.

137. Phosphoryaltion of ADP to ATP occurs through energy metabolism, comprising oxidative phosphorylation or substrate-level phosphorylation or photo-phosphorylation (in plants). ATP can also be formed from ADP through the action of adenyalte kinase. Crystal structure determination of adenylate kinase shows that the C-terminal region has the sequence

-Val-Asp-Asp-Val-Phe-Ser-Gln-Val-Cys-Thr-His-Leu-Asp-Thr-Leu-Lys

What can be a possible conforamtino of the sequence ?

(a) A helix that is not amphipathic

(b) RNase protection assay

(d) Chromatin immunoprecipitation assay

Ans. (b)

Sol. In an amphipathic α helix, one side of the helix contains mainly hydrophilic amino acids and the other side contains mainly hydrophobic amino acids. The amino acid sequence of amphipathic helix alternates between hydrophilic and hydrophobic residues every 3 to 4 residues, since the helix makes a turn for every 3.65 residues.

138. A reporter cell line with stably integrated retroviral promoter-luciferae construct was transfected with an expression vector for a cellular protein. The protein seems to regulate the activation of the retroviral promoter as analyzed by luciferase activity assay. Which one of the following techniques will you use to show in vivo recruitment of the cellular protein on the integrated retroviral promoter ?

(a) Electrophoretic mobility shift assay

(b) RNase protection assay

(d) Chromatin immunoprecipitation assay

Ans. (d)

Sol. Chromatin immunoprecipitation technique can be used for study of DNA (Promoter) Protein interaction (Regulatory Protein) under in-vivo condition. Chromatin Immuno-precipitation (ChIP) is a type of immuno-precipitation experimental technique used to investigate the interaction between proteins and DNA in the cell. It aims to determine whether specific proteins are associated with specific genomic regions, such as transcription factors on promoters or other DNA binding sites and possibly defining cistromes. ChIP also aims to determine the specific location in the genome that various histone modifications are associated with indicating the target of the histone modifiers.

139. For 5' end labelling of DNA, the following reactions are carred out sequentially as indicated.

5'-dephosporylated DNA + dATP + T4 polynucleotide kinase (T4PNK) and incubated for 2 hours Ammonium acetate Tris-EDTA ethanol.

If trace amount of is present in the initial DNA mix, which of the following statements would most likely be true ?

(a) ion activates T4PNK, thereby increasing the labeling efficiency.

(b) ion inhibits T4PNK, therfore should not be present in the DNA mix.

(d) ion dephosphorylatse DNA, thereby increasing the labelling efficency.

Ans. (b)

Sol. Ammonium ions inhibit T4PNK (T4 polynucleotide kinase) activity therefore should not be present in DNA mix. T4PNK transfer gamma phosphate from ATP to 5' OH group of single and double strand DNAs and RNAs, oligonucleotide or nucleoside 3' monophosphates.

140. A researcher collected information from four forest areas using a sensor to assess their green cover, observed average spectral values for each of the forests are given in the table below. The forest green cover in the order of highest to lowest is

(a) A > C > B > D

(b) A > D > C > B

(d) D > A > B > C

Ans. (b)

Sol. NORMALIZED differential vegetation index = NIR – VIS/NIR + VIS

NDVI is highest for A.

141. Biologists randomly sampled about 3000 insects from a newly found island. The distribution of their abundance in the sample was as in the figuer given below :

Which of the following can be correctly inferred from the graph ?

(a) Many species have only one individual each on the island

(b) The bar on the extreme right represents a large number of species with very few individuals

(d) All species from the island may not be represented in the sample

Ans. (d)

Sol. All species from the island may not be represented in the sample because complete graph should be bell-shaped.

142. Four different receptros viz, A, B, C and D bind to the same ligand X. In oder to determine which receptor has the highest ligand binding affinity following experiment was carried out. Cells were transfected with green fluorescence protein (GFP)-tagged receptors (A, B, C and D) individually, then incubated with red fluorescence protein (RFP) tagged-'X' and subjected to FACS analysis. Following are the results

Which receptor has the highest ligand beinding affinity ?

(a) A

(b) B

(d) D

Ans. (c)

Sol. Receptor C has the highest ligand binding affinity.

143. If one wishes to design a microarray chip for whole genome expression analysis of an eukaryotic system, which region of the gene should be preferred for selection of unique target sequences ?

(a) Any region of the coding DNA sequenc (CDS)

(b) 3' region of the CDS 3' untranslated reigon (UTR)

(d) 1^st intron only

Ans. (b)

Sol. If one wishes to design a microarray chip for whole genome expression analysis of an eukaryotic system, 3' region of the CDS 3' untranslated reigon (UTR) should be preferred for selection of unique target sequences.

144. During line transect sampling of two solitary species of ground mammals the following observations were made :

Smaller species Larger species

P. Transect length 100 km 100 km

Q. Number of animals sighted 30 36

R. Mean perpendicular distance from transect line 10 m 40 m

Which of the following can be inferred from the data ?

(a) The smaller species is more abundant but seen less frequently

(b) The smaller species is less abundant but seen less frequently

(d) The larger species is seen more frequently, but its abudance cannot be compared with the smaller species

Ans. (a)

Sol.

Where, n = total number of animals sighted, L = length of the transect and Y = mean perpendicular distance.

Density of smaller species = D_S = = 15

Density of larger species = D_L = = 4.5

As we know that number of individual sighted decreases with increasing perpendicular distance from the line.

145. A protein contains 2 Trp and 4 Tyr residues. the molecular mass of the protein is 17000 Da and that of Trp and Tyr are 204 and 180 Da respectively. Values of the absorption coefficient of 1% (g/v) solution of Trp and Tyr in 1-cm cell at 208 nm, are 269.60 and 83.33, respectively. The absorption of 1 mg/ml protein solution in 1 cm cell at 280 nm will be