PYQ DEC 2015 - VedPrep

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

21. A cell line deficient in salvage pathway for nucleotide biosynthesis was fed with medium containing ¹⁵N labelled amino acids. Purines were then extracted. Treatment with which one of the following amino acids is likely to produce ¹⁵N labelled purines?

(a) Aspartic acid

(b) Glycine

(d) Aspartamine

Ans. (a, b, c)

Sol. Aspartame is an artificial non-saccharide sweetener 200 times sweeter than sucrose and is commonly used as a sugar substitute in foods and beverages.

22. The ionic strength of a 0.2 M Na₂HPO₄, solution will be

(a) 0.2 M

(b) 0.4 M

(d) 0.8 M

Ans. (c)

Sol. For .02 M Na₂HPO₄ the ionic strength is :

23. Enzymes accelerate a reaction by which one of the following strategies?

(a) Decreasing energy required to form the transition state.

(b) Increasing kinetic energy of the substrate.

(d) Increasing the turn over number of enzymes.

Ans. (a)

Sol. Enzymes increase reaction rates by decreasing the amount of energy required to form a complex of reactants that is competent to produe reaction products. This complex is known as the activated state of transition state complex for the reaction. Enzymes and other catalysts accelerate reactions by lowering the energy of the transition state. The free energy required to form an activated complex is much lower in the catalyzed reaction. The amount of energy required to achieve the transition state is lowered; consequently, at any instant a greater proportion of the molecules in the population can achieve the transition state. The result is that the reaction rate is increased.

24. The genome of a bacterium is composed of a single DNA molecule which 10⁹ bp long. How many moles of genomic DNA is present in the bacterium? [Consider Avogadro no. = 6 × 10²³]

(a)

(b)

(d) 6 × 10²³

Ans. (a)

Sol. 1 mol = Avogardo number = 6 × 10²³ molecules of DNA

1 DNA molecule = moles

25. Which one of the following statement is correct?

(a) In all L-amino acids, only the carbon atom is chiral.

(b) Deoxyriboseis opticaly inactive.

(d) Phosphatidylcholine isolated from biological membranes is optically active.

Ans. (d)

Sol. The biologically active species discussed here include oxidized phosphatidylcholines with biologic acitivity similar to platelet-activating factor (PAF), oxidized phosphatidylcholines that stimulate responses through ways other than the PAF receptor and the lysophosphatidylcholines that result from the enzymatic metabolism of these modified phospholipids.

26. Four single amino acid mutants (a to d) of a protein in the epitope-region of a monoclonal antibody X were made and expressed in E. coli. The lysates from the four E. coli cultures expressing these four proteins were run or an SDS-PAGE gel and subsequently transferred to nitrocellulose membrane and Western blotted using a monoclonal antibody X aised against the wild-type protein. The results are presented in the figure below

The four single mutation, upon sequencing, were found to be Valine (V) to Alanine (A); Glycine (G) to Proline (P); Alanine (A) to Aspartic acid (D) and isoleucine (I) to eucine (L).

(a) b is due to V A and c is due to G P.

(b) b is due to G P and d is due to V A.

(d) c is due to V A and a is due to I L.

Ans. (a)

Sol. b is due to Valine to Alanine and c is due to Glycine to Proline.

27. The exact backbone dihedral angles in a folded protein can be obtained by

(a) deconvolution of its circular dichroism spectra obtained at different pH and temperature.

(b) estimating the number of protons that exchange with deuterium on treating the protein with D₂O.

(d) analysis of the crystal structure of the protein obtained by X-ray diffraction at high resolution.

Ans. (d)

Sol. XRD finds the geometry or shape of a molecule using X-rays. XRD techniques are based on the elastic scattering of X-rays from structures that have long range order. The X-rays get diffracted by a crystal because the wavelength of X-rays is similar to the inter-atomic spacing in the crystals.

28. It takes 40 minutes for a typical E. coli cell to completely replicate its chromosome. Simultaneous to the ongoing replication, 20 minutes of a fresh round of replication is completed before the cell divides. What would be the generation time of E. coli growing at 37°C in complex medium?

(a) 20 minutes

(b) 40 minutes

(d) 30 minutes

Ans. (a)

Sol. Generation time : The rate of exponential growth of a bacterial culture in optimum condition. In simple term, it is the time required to double the population. Generation time (G) is defined as the time (t) per generation (n = number of generations) [G = t/n].

29. Glycophorin having one highly hydrophobic domain is able to span a phospholipid bilayer membrane only

(a) once

(b) twice

(d) thrice

(d) four times

Ans. (a)

Sol. Glycophorins are the membrane proteins of RBC, they facilitate the transport of sugar molecules through the membrane. They are highly glycosylated with many sialic acid residues.

30. Given below are events in the cell cycle:

P. Phosphorylation of lamin A, B, C.

Q. Phosphorylation of Rb (Retinoblastoma protein)

R. Polyubiquitination of securin.

S. Association of inner nuclear membrane proteins and nuclear pore complex proteins with chromosomes.

Which one of the following reflects the correct sequence of events in the mammalian cell cycle?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Dephosphorylated Rb a tumor suppressor gene inhibits cell cycle. When Rb is phosphorylated, it becomes inactive and cells proceed to cell division. Phosphorylation of lamin activates the nuclear membrane disorganization. Securin activate separase, a protease which degrade the cohesion proteins that links the sister chromatids during cell division. Once the cohesion proteins are completely degraded and the siter chromatids are separated, the securing should be inactivated. In the phosphorylated state of securing, the APC (anaphase promoting complex) cannot initiate the onset of anaphase. Thus, in order to initiate anaphase, the APC complex uniquitinates sequrin and target it for degradation.

31. Membrane proteins are synthesized on endoplasmic reticulum and transported to various oraganelle membranes. One hypothesis for membrane protein sorting is hydrophobicity matching i.e. the proteins with a shorter transmembrane portion would partition into thinner membranes. You are given the following three observations:

P. It was found that transmembrane portions of proteins in Golgi membranes.

Q. Presence of cholesterol increases the thickness of the bilayer.

R. The phospholipid composition of Golgi and plasma membranes are same.

Which of the following statement is correct?

(a) Proteins in plasma membrane have longer transmembrane portion than proteins in Golgi membranes.

(b) Proteins in Golgi membranes have longer transmembrane portion than proteins in plasma membranes.

(d) Cholesterol is more in Golgi membrane than in plasma membrane.

Ans. (a)

Sol. A protein consists of a polypeptide backbone with attached side chains. Each type of protein differs in its sequence and number of amino acids; therefore, it is the sequence of the chemically different side chains.

32. A culture medium contains two carbon sources, one is preferred carbon (glucose) and the second is a non-perferred source (lactose). Which one below is correct regarding the nature of growth curve of E. coli cultured in this medium.

(a) Growth curve will be same as when grown in presence of only glucose.

(b) Growth curve will be same as when grown in presence of only lactose.

(d) Two lag phase will be observed between the two exponential phases.

Ans. (c)

Sol. Lag time is defined as the initial period in the life of a bacterial population when cells are adjusting to a new environment before starting exponential growth.

33. Which one of the following statements correctly applies to proteins which are translated on the rough endoplasmic reticulum?

(a) Cytoplasmic proteins which are targeted to the nucleus in response to hormone stimuli.

(b) Proteins targeted to lysosomes, plasma membrane and cell exterior.

(c) Proteins which are targeted to the nucleus through endoplasmic reticulum lumen as the lumen is in direct connection with the intermembrane space of the nucleus.

(d) All proteins which get targeted to peroxisomes.

Ans. (b)

Sol. Eukaryotic cells posses distinct membrane bound organelles which are absent in prokaryotic cells. The membrane bound organelles have different functions and these organelles provide discrete compartments in which specific cellular activities take place. The complex internal organization of eukrayotic cells generate hardship for transport of proteins to their destinations.

Many protiens destined for the endoplasmic reticulum, the golgi apparatus, lysosomes, the plasma membrane and secretion from the cell are synthesized on ribosomes that are bound to the membrane of endoplasmic reticulum.

34. Lipid rafts are rich in both sphingolipids and cholesteroll plays a central role in raft formation since lipid rafts apparently do not formin its absence. Why do you think cholesterol is essential for the formation of lipid rafts?

(a) Cholesterol decreases the mobility of sphingolipids in the lipid bilayer.

(b) Large head groups of sphingolipids repel each other in presence of cholesterol.

(d) The planar cholesterol molecules are postulated to fill the voids that form undeneath the large head groups of the sphingolipids.

Ans. (d)

Sol. Cholesterol is thought to serve as a spacer between the hydrocarbon chains of the sphingolipids and to function as a dynamic glue that keeps the raft assembly together. Cholesterol partitions between the raft and the nonraft phase, having higher affinity to raft sphingolipids than to unsaturated phospholipids.

35. Following is the domain organization of three proteins that are targetted to the mitochondria.

Based on the domain organization in the above figure and assuming the left box to be having the mitochondrial sorting signal, predict the most likely sub-compartment of the mitochondria in which the protein will be found.

(a) A in matrix; B in inner membrane; C in inter-membrane space.

(b) A in inner membrane ; B in inter-membrane space; C in outer membrane.

(d) A in matrix; B and C are in inter-membrane space.

Ans. (a)

Sol. Signal sequences target proteins to different parts of cell. Once the signal sequence is cleaved by proteases protein targeting stops there only.

36. The frequency of cells in a population that are undergoing mitosis (the mitotic index) is a convenient way to estimate the length of the cell cycle. In order to measure the cell cycle in the liver of the adult mouse by measuring the mitotic index, liver slices are prepared and stained to easily identify cells undergoing mitosis. It was observed that only 3 out of 25,000 cells are found to be undergoing mitosis. Assuming that M-phase lasts 30 minutes, calculate the approximate length of the cell cycle in the liver of an adult mouse?

(a) 76 hours

(b) 50 hours

(d) 21 hours

Ans.

Sol. Options are not correct.

Mitotic index =

If mitosis (M-phase) is 30 minutes (0.5 hours) long and the frequency of cells in mitosis is 0.00012, then 0.5 hours is 0.00012 of the length of the cell cycle. Thus, the cell cycle is 0.5/0.00012 = 4167 hours in length.

37. Which one of the following chemicals is a DNA intercalator?

(a) 5-Bromouracil

(b) Ethyl methanesulfonate

(d) UV

Ans. (c)

Sol. Intercalation is a reversible inclusion of molecules to another molecule which have a layered structural organization. 5-bromouracil, ethyl methane sulphonate, acridine orange and UV are DNA damaging agents. Among these, the acridine orange intercalate to double stranded DNA. Apart from its mutation potential, acridine orange is a nucleic acid specific fluorescent dye which can be used for the detection and labelling of nucleic acid samples in molecular biology experiments. Ethidium bromide (EtBr) is another DNA intercalating agent which is also used as fluorescent tag for DNA and RNA.

38. An antibiotic that resembles the 3' end of a charged tRNA molecule is

(a) Streptomycin

(b) sparsomycin

(d) Tetracycline

Ans. (c)

Sol. Puromycin is an antibiotic obtained from Streptomyces alboniger. Puromycin cause premature chain termination in protein synthesis. Part of the molecule resembles the 3' end of the aminoacylated tRNA. Puromycin enters the A site and transfers to the growing chain, causing the formation of a puromycylated nascent chain. This results in premature of release of polypeptide from the ribosome.

39. -amanitin is a fungal toxin which inhibits eukaryotic RNA polymerases. The three eukaryotic RNA polymerases show differential sensitivity to this toxin. Which one of the following order (higher to lower) is correct in respect of sensitivity towards -amanitin?

(a) RNA Pol III > RNA Pol II > RNA Pol I

(b) RNA Pol II > RNA Pol III > RNA Pol I

(d) RNA Pol II > RNA Pol I > RNA Pol III

Ans. (b)

Sol. -Amanitin is a cyclic peptide of 8 amino acid produced by a mushroom Amanita. It inhibits the transcription process of protein synthesis by inhibiting RNA polymerase II which synthesize the mRNA. Amanitin can also inhibit other RNA polymerase such as RNA polymerase I and RNA polymerases III which synthesis rRNA and tRNA respectively. The sensitivity of -amanitin towards RNA Polymerase enzyme shows differential sensitivity as follows RNA POL II > RNA POL III > RNA POL I.

RNA POL I is least inhibited by -amanitin.

40. In eukaryotic replication, helicase loading occurs at all replicators during

(a) G₀ phase

(b) G₁ phase

(d) G₂ phase

Ans. (b)

Sol. Helicase loading is carefully regulated to control the lcoation and freqeuncy of DNA replication initiation. In eukaryotic cells helicase loading is tightly restricted to the G₁ phase of the cell cycle. This is to ensure that no origin of replication is initiated more than once per cell cycle.

41. You have labelled DNA in a bacterium by growing cells in medium containing either ¹⁴N nitrogen or the heavier isotope, ¹⁵N. Furthermore, you have isolated pure DNA from these organisms, and subjected it to CsCl density gradient centrifugation leading to their separration of light (¹⁴N) and heavy (¹⁵N) forms of DNA to different locations in the centifuge tube. In the next experiment, bacteria were regrown first in medium containing ¹⁵N, so that all the DNA made by cells will be in heavy form. Then these cells were transferred to medium containing only ¹⁴N and allowed the cells to divide for one generation. DNAs were extracted and centrifuged as above in the CsCl gradient. A hybrid DNA band was observed at a position located between and equidistant from the ¹⁵N and ¹⁴N DNA bands. Based on the above observation, which one of the follloiwng conclusions is correct?

(a) Replication of DNA is conservation.

(b) Replication of DNA is semi-conservation.

(d) Replication by rolling circle mode.

Ans. (b)

Sol. Shikimic acid pathway is a metabolic seven step pathway in bacteria, fungi and plants for the synthesis of aromatic amino acids such as Phenylalanine, tyrosine and tryptophan. Shikimic acid pathway is completely absent in animals and hence all the above amino acids are essential amino acids to animals Mevalonic acid pathway is for the production of isoprenoids.

42. Although ribonucleoside triphosphates (rNTPs) are present at approximately 10-fold higher concentration than deoxyribonucleoside triphosphate (dNTPs) in the cell, but they are incorporated into DNA at a rate that is more than 1000-fold lower than dNTPs. This is because

(a) DNA polymerase cannot discriminate between dNTPs and rNTPs. But as soon as rNTPs are incorporated in the DNA chain, they are hydrolyzed due to the presence of 2-OH group.

(b) DNA polymerase cannot discriminate between dNTPs and rNTPs. But as soon as rNTPs are incorporated in the DNA chain, they are excised by the proofreading activity of DNA polymerase.

(c) DNA polymerase efficiently discriminate between rNTPs and dNTPs, because its nucleotide binding pocket cannot accommodate a 2-OH on the incoming nucleotide.

(d) DNA polymerase cannot discriminate between rNTPs and dNTPs. Since the rate of transcription in cell is 10⁶ times faster than replication, it cannot compete with RNA polymerase for rNTPs.

Ans. (c)

Sol. Although ribonucleoside triphosphates (rNTPs) are present at approximately 10-fold higher concentration than deoxyribonucleoside triphosphate (dNTPs) in the cell, but they are incorporated into DNA at a rate that is more than 1000-fold lower than dNTPs. This is because DNA polymerase efficiently discriminate between rNTPs and dNTPs, because its nucleotide binding pocket cannot accommodate a 2'-OH on the incoming nucleotide.

43. The mismatch repair activity of E. coli repairs misincorporated bases which is not removed by the proofreading activity of DNA polymerase. However, while doing so, it has to decide which strand of the DNA is newly synthesized and which one is parental. Mismatch repair system does it by which one of the following ways?

(a) It recognizes nearby GATC sequence

(b) It recognizes any nearby palindromic sequence

(d) It recognises the hemi-methylated GATC sequence nearby

Ans. (d)

Sol. The mismatch repair activity of E. coli repairs misincorporated bases which is not removed by the proofreading activity of DNA polymerase. However, while doing so, it has to decide which strand of the DNA is newly synthesized and which one is parental. Mismatch repair system recognises the hemi-methylated GATC sequence nearby

44. Enlisted below the different types of RNAs produced in the cell (Column A) and their functions (Column B), but not in the same order.

Column A Column B

P. SnRNAs 1. turn off gene expression by directing degradation of selective

mRNAs.

Q. siRNAs 2. regulate gene expression by blocking translation of selective

mRNAs.

R. miRNAs 3. function in a variety of processes including splicing of pre-

mRNA.

S. SnoRNAs 4. used to process and chemically modify rRNAs.

Choose the correct combination.

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

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

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

Ans. (b)

Sol. Table : Principle Types of RNAs Produced in Cells

45. In prokaryotes, the initiator tRNA is first charged with a methionine, followed by the addition of a formy group to the methionine by the enzyme Met-tRNA transformylase. Given below are several statements in the context.

1. All prokaryotic proteins have formylmethionine at their amino terminal end.

2. Deformylase removes the formyl group from the amino termina methionine.

3. All prokaryotic proteins have methionine at their amino terminal methionine.

4. Aminopeptidases often remove the amino terminal methionine.

5. Aminopeptidases remove amino terminal formylmethionine.

Which of the above statement(s) are most likely to be true?

(a) 1 only

(b) 2 and 3

(d) 2 and 4

Ans. (d)

Sol. N-formyl methionine is the first amino aicd to be incorporated into a polypeptide chain. An enzyme known as a deformylase removes the formyl group from the amino terminus during or after the synthesis of the polypeptide chain. In fact, many prokaryotic protein chain. In fact, many prokaryotic proteins do not even start with a methionine; aminopeptidases often removes the amino terminal methionine as well as one or two additional amino acids.

46. A hypothetical operon involved in the synthesis of an amino acid 'X' is 'ON' (transcribing) in the presence of low levels of 'X' and 'OFF' (not transcribing) in presence of high level of 'X'. The symbols a, b and c (in the table below) represents a structural gene for the synthesis of X (X-synthase), the operator region and gene encoding the repressor – but not necessarily in that order. From the following data, in which superscripts denote wild type or defective genotype, identity which are the genes for X-synthase, operator region and the repressor.

The respective genes for 'X'-synthase, the operator region and repressor are

(a) a, b, c

(b) c, a, b

(d) b, a, c

Ans. (d)

Sol. By observing 'X' synthase activity in presence of low level of X and high level of X with their genotype we can say that 'b' gene is X synthase, 'a' is operator region and 'c' is repressor.

47. A protein has 4 equally spaced trypsin sensitive sites which results in peptide fragments A₁, A₂, A₃, A₄ and A₅ upon digestion with trypsin. The peptides A₂ and A₅, represent N-terminal and C-terminal fragments respectively. Now you are asked to synthesize this protein. At time t = 0 you added all the 20 amino acids labeled with ¹⁴C and initiated the synthesis. At time t = 4, full length protein is synthesized. If you stop the synthesis of the protein in time t = 1 and digest the protein with trypsin, which peptide will have maximum ¹⁴C label than others?

(a) A₃

(b) A₁

(d) A₂

Ans. (d)

Sol. The protein sequence looks like as : N-terminal A₂-X-Y-Z-A₅ C-terminal, where, the order of the rest 3 peptides is not known. The N-terminal of the protein is first to be synthesized and there will be maximum incorporation of ¹⁴C labelled amino acids. Therefore, at time t = 1, the peptide fragment having maximum ¹⁴C label will be A₂.

48. Cytotoxic T-cells express

(a) CD8 marker and are class II MHC resticted.

(b) CD4 marker and are class I MHC resticted.

(d) CD8 marker and are class I MHC resticted.

Ans. (d)

Sol. Cytotoxic T-lymphocytes express the surface receptor CD8, which recognizes class I MHC molecules. When foreign antigen, e.g., from virally infected or tumour cells, is expressed together with MHC class I molecule, this is recognized by the T-cell receptor/CD3 complex on the cytotoxic T-cell. This then releases various proteases that puncture the cell membrane, destroying the abnormal cell.

The T-lymphocytes with suppressor function form an ill-defined group that also expresses the CD8 marker. There is no doubt that suppressor activity exists, but the existence of specific T-suppressor cells remains unconfirmed.

49. The mutation in an oncogene falls under which of the following classes?

(a) Loss of-function mutation. (b) Frameshift mutation

Ans. (c)

Sol. Loss of function mutation : Gene lost its function due to mutation.

Frame shift mutation : The readiing frame of mRNA changes due to point mutation.

Gain of function mutation : Give enhanced or new activities to a protein.

Dominant negative mutation : Produce an altered gene product which acts antagonistically to the gene product of tis wild allele. Usually, these mutations will be dominant.

Oncogenes are tumour producing genes. Usually these genes are very essential genes for the normal functioning of the cells and we call them as proto-oncogenes. Protooncogenes code for proteins that help to regulate cell growth and differentiation. Mutations of proto-oncogenes cause the formation of oncogenes.

50. Which of the following is not a cell adhesion protein?

(a) Cadherin

(b) Selectin

(d) Laminin

Ans. Laminins are fibrous proteins; they are present in the nucleus as nuclear lamins. Nuclear lamins interact with membrane-associated proteins to form the nuclear lamina on the interior of the nuclear envelope.

Cadherins : they are calcium dependent transmembrane adhesion proteins forming adherence junctions.

Selection : are a type of lectins that functions as adhesion proteins.

Sol.

51. Which of the following is not a second messenger?

(a) Cyclic GMP

(b) Diacylglycerol

(d) Phosphatidylinositol

Ans. (d)

Sol. Cyclin guanosine monophosphate (cGMP) is a cyclic nucleotide derived from guanosine triphosphate (GTP). cGMP acts as a second messenger much like cyclic AMP.

Inositol triphosphate (IP3) and diacylglycerol (DAG) are important second messengers. Their formation begins with the binding of an extracellular regulatory molecule to a membrane receptor that activates a trimeric G protein.

An important intracellular second-messenger signaling system, the phosphatidylinositol system, employs two second-messenger lipids, both of which are derived from phosphatidylinositol. One is diacylglycerol (diglyceride), the other is triphosphoinositol. In this system a membrane receptor acts upon an enzyme, phospholipase C, located on the inner surface of the cell membrane.

52. Which of the following statements about the nuclear receptor superfamily in not-true?

(a) The receptors are always cytosolic, where they remain associated with heat-shock proteins and have variable ligand binding domains in the N-terminal region.

(b) The receptors have characteristics repeat of the C₄ zinc-finger motif.

(c) The receptors are either homodimeic or heterodimeric, and in the absence of their hormone ligand, the heterodimeric receptors repress transcription, when bound to their response elements.

(d) The receptors have a unique N-terminal rergion of variable length and may contain a nuclear localization signal between the DNA and ligand-binding domains.

Ans. (a)

Sol. Nuclear receptors superfamily act as ligand-inducible transcription factors by directly interacting as monomers, homodimers or heterodimers with the retinoid X receptor with DNA response elements of target genes, as well as by "cross-talking" to other signalling pathways. The effects of nuclear receptors on transcription are mediated through recruitment of coregulators. A subset of receptors binds corepressor factors and actively represses target gene expression in the absence of ligand. Corepressors are found within multicomponent complexes that contain histone deacetylase activity. Deacetylation leads to chromatin compactation and transcriptional repression. The receptors have characteristics repeat of the C₄ zinc-finger motif. This has a unique N-terminal region of variable length and may contain a nuclear localization signal between the DNA nd ligand-binding domains.

53. Physical attachment between cells is very important in imparting strength in tissues. Various physical cell junctions in vetebrate epithelial tissues are classified according to their primary functions. Enlisted below in column A is the major function of a particular junction and column B enlists cell junctions, but not in the same order.

Column A Column B

P. Seals gap between epithelial cells i. Desmosomes

Q. Connects actin filament bundle in one ii. Hemidesmosomes

cell with that in the next cell

R. Connects intermediate filaments in one iii. Tight junction

cell to those in the next cell

S. Anchors intermediate filaments in a cell iv. Adherens junction

to extracellular matrix

Choose the correct combination :

(a) P-i, Q-ii, R-iii, S-iv

(b) P-ii, Q-iii, R-iv, S-i

(d) P-iv, Q-i, R-ii, S-iii

Ans. (c)

Sol.

54. G-protein coupled receptors (GPCRs) consist of three protein subunits and . In stimulated state, -subunit is GDP bound and GPCR is inactive. When GPCR gets activated, it acts like guanine nucleotide exchange (GEF) factor and induces -subunit to release its bound GDP allowing GTP to bind in its place. In order to regulate G-protein activity by regulating GDP/GTP concentration, -subunit acts as

(a) GTPase

(b) GDP kinase

(d) cAMP-specific phosphodiesterase

Ans. (a)

Sol. Signal deactivation is achieved by G alpha-mediated GTP hydrolysis (GTPase activity) which is enhanced by the GTPase-accelerating protein (GAP) activity of "regulator of G-protein signaling" (RGS) proteins.

55. Cellular level of tumour suppressor protein p53 is maintained by the ubiquitin ligase protein, Mdm2. Over expressiion of Mdm2 was found to convert a normal cell into cancer cells by destabilizing p53. Another protein p19^ARF inhibits the activity of Mdm2 thus stabilizing p53. Loss of p19^ARF function also converts normal cells into cancer cells. Based on the above information, which one of the following statement is correct?

(a) Both Mdm2 and p19 ^ARF are oncogenes.

(b) Both Mdm2 and p19 ^ARF are tumous suppressor genes.

(d) p19^ARF is an oncogene but Mdm2 is a tumor suppressor gene.

Ans. (c)

Sol. MDM2 is an oncogene with both p53-dependent and p53-independent oncogenic activities, and often has increased expression levels in a variety of human cancers. MDM2 is highly regulated; the levels and function of MDM2 are regulated at the transcriptional, translational and post-translational levels. p14/p19ARF (ARF) is a tumor suppressor gene that is frequently mutated in human cancer. ARF has multiple tumor suppressor functions, some of which are mediated by signalling to p53.

56. The relation cellular immune response generated against hepatitis C virus is the critical determinant of the outcome of infection. Given below are the representative figures of cellular immune response in column I and various outcome of infection in column II.

Column I Column II

P. i. Acute

Q. ii. Resolution

R. iii. Chronic

Choose the best possible combination.

(a) P-ii, Q-iii, R-i

(b) P-i, Q-iii, R-ii

(d) P-i, Q-ii, R-iii

Ans. (b)

Sol. Hepatitis C virus (HCV) causes both acute and chronic infection. Acute HCV infections are usually asymptomatic and most do not lead to a life-threatening disease. Around 30% (15–45%) of infected persons spontaneously clear the virus within 6 months of infection without any treatment. Option b is correct.

57. There are various subclasses of antibodies found in body fluids and body secretion. Many different functions may be attributed to these subclasses. Given below in column I is major functions of different subclasses and column II consists of the name of subclass.

Column I Column II

P. Binds to macrophages by F_C i. IgA

Q. Binds to mast cells and basophills ii. IgD

R. First B cell receptor iii. IgE

S. No major specific function known iv. IgG

other than antigen binding

T. Protector of mucous membrane v. IgM

Select the correct combination:

(a) P-i, Q-ii, R-iii, S-iv, T-v

(b) P-ii, Q-iii, R-iv, S-v, T-i

(d) P-iv, Q-iii, R-v, S-ii, T-i

Ans. (d)

Sol. IgG : IgG is the most versatile immunoglobulin becasue it is capable of carrying out all of the functions of immunoglobulin molecules. Binding to cells—Macrophages, monocytes, PMNs and some lymphocytes have F_c receptors for the F_c region of IgG.

IgM : IgM is the first Ig to be made by the foetus and the first Ig to be made by a virgin B cells when it is stimulated by antigen. Cell surface IgM functions as a receptor for antigen on B cells. Surface IgM is non-covalently associated with two additional proteins in the membrane of the B cell called Ig-alpha and Ig beta.

IgA : IgA is he 2nd most common serum Ig. IgA is the major class of Ig in secretions tears, saliva, colostrum, mucus. The secretory piece helps IgA to be transported across mucosa and also protects it from degradation in the secretions.

IgD : IgD is primarily found on B cell surfaces where it functions as a receptor for antigen. IgD on the surface of B cells has extra amino acids at C-terminal end for anchoring to the membrane.

IgE : IgE is the least common serum Ig since it binds very tightly to F_c receptors on basophils and mast cells even before interacting with antigen.

58. In chick, development of wing feather, thigh feather and claws depends on epithelia specificity conferred by induction from mesenchymal components from different sources of the dermis. This may be attributed to

(a) autocrine interaction

(b) regional specificity of induction

(d) inactivation of genetic interactions

Ans. (b)

Sol. In chick, development of wing feather, thigh feather and claws depends on epithelial specificity conferred by induction from mesenchymal components from different sources of the dermins.

59. Alveolar cells of the lung arise from which one of the following layer(s)?

(a) Mesoderm

(b) Endoderm

(d) Both Ectoderm and Endoderm

Ans. (b)

Sol. The epithelial lining of the alveoli, bronchi, larynx, and glands of the trachea develop from the endoderm layer.

60. Migration of individual cells from the surface into the embryo's interior is termed as

(a) ingression

(b) involution

(d) delamination

Ans. (a)

Sol. Invagination is the infolding of one part within another part of a structure, a folding that creates a pocket. The term, originally used in embryology, has been adopted in other disciplines as well.

The inner membrane of a mitochondrion invaginates to form cristae, thus providing a much greater surface area to accommodate the protein complexes and other participants that produce adenosine triphosphate (ATP).

61. Floral organ development is contoled by overlapping expression of'A' class, 'B' class and 'C' class genes in different whorls. In anArabidopsis mutant, the flowers had sepals, sepals, carpels and carpels in the four whorls. Mutation in which one of the following is the cause for the mutant phenotype?

(a) 'A' class gene alone

(b) 'B' class gene alone

(d) 'C' class gene alone

Ans. (b)

Sol. Mutations in type B genes – These mutations affect the corolla and the stamen, which are the intermediate verticils. Two mutations have been found in A. thaliana, APETALA3 and PISTILLATA, which cause development of sepals instead of petals and carpels in the place of stamen.

62. Instructive and permissive interactions are two major modes of inductive interaction during development. The following compares some properties ofcell lines and cord blood stem cells. Cell lines, which are stoerd in liquid nitrogen, can be retrieved for experiments, where they behave as per their original self. Cord blood can also be retrieved from liquid nitrogen for procuring stem cells. Unlike cell lines, the stem cells can be additionally induced to undergo differentiation into desired lineages, which are very different from their original self. The behaviour of cell lines and stem cells is analogous to which of the interactions?

(a) Both cell lines and stem cells show instructive interactions?

(b) Cell lines show instructive interaction whereas stem cells show permissive interaction.

(d) Both types of cells show permissive instruction.

Ans. (c)

Sol. INSTRUCTIVE AND PERMISSIVE INTERACTIONS INSTRUCTIVE - a signal from the inducing cell is necessary for initiating new gene expression in the responding cell. PERMISSIVE - the responding tissue has already been specified; needs only an environment that allows the expression of those traits.

63. Following are certain statements regarding morphogen gradients and cell specification.

P. Morphogens are always transcription factors.

Q. Morphogens can be paracrine factors that are produced in one group of cells and travel to another population of cells.

R. When the concentration of a morphogen drops below a certain threshold cells stop differentiating and never get determined to another fate.

S. Morphogen gradients are involved in conditional specification.

Which combination of the above statement is true?

(a) P and Q

(b) Q and S

(d) P and R

Ans. (b)

Sol. Morphogens are signaling molecules produced in a restricted region of a tissue; they provide positional information by diffusing from their source to form long-range concentration gradients. Morphogens are substances that establish a graded distribution and elicit distinct cellular responses in a dose dependent manner. They function to provide individual cells within a field with positional information, which is interpreted to give rise to spatial patterns.

64. Successful fertilization in sea urchin demands specific interaction between proteins and receptors of sperms and eggs. In view of the above, which one of the following combinations is correct?

(a) Bindin in acrosomes and bindin receptors on egg vitelline membrane.

(b) Bindin in egg membrane and bindin receptors in acrosomes.

(d) Proteasomes on egg membranes and complex sugars on sperm membranes.

Ans. (a)

Sol. Bindin is a 30,000-mol-wt protein of sea urchin sperm that is responsible for the specific adhesion of the sperm acrosomal process to the vitelline layer covering the egg plasma membrane during fertilization. Bindin isolated from the sperm acrosome results in insoluble particles that cause homospecific eggs to aggregate, whereas no aggregation occurs with heterospecific eggs. Therefore, Bindin is concluded to play a critical role in fertilization, yet its function has never been tested.

65. Following statements are made in relation to the five widely recognized stages of Arabidopsis embryogenesis:

P. The fusion of haploid egg and sperm takes place in globular stage.

Q. Rapid cell division in two regions on either side of the future shoot apex forms heart stage.

R. The cell elongation throughout the embryo axis and further development result in Torpedo stage.

S. The embryo loses water and becomes metabolically inactive in the zygotic stage.

Which combination of the above statement is correct?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (b)

Sol. The embryo arises from the zygote.

Major transitions have been determined for Arabidopsis. While in plants (unlike animals), embryogenesis does not directly generate adult tissues and organs, patterns are formed that persist :

— Apical baseal axial development.

— Radial tissue patterns in stems and roots.

— Establishment of primary meristems.

Arabidopsis embryos pass through four distinct developmental stages :

— Gobular stage : Afterfirst zygotic division, a series of highly ordered divisions follow, producing an 8-cell globular embryo. Radial patterning is apparent.

— Heart stage : Forms following rapid cell divisions on either side of the future shoot apex. Outgrowths that will give rise to cotyledons produce bilateral symmetry.

— Torpede stage : Results from cell elongation through the axis and further development of cotyledons.

— Maturation : Embryo and seed lose water and become metabolically quiescent as they enter dormancy.

66. The following are statements regarding the development and maintenance of anterior and posterior compartments in each segment of Drosophila:

1. Expression of wingless and engrailed is activated by pair-rule genes.

2. Continued expression of wingless and engrailed is maintained by interaction between the cells expressing engrailed and wingless proteins.

3. Hedgehog is expressed wingless expressing cells and forms short range gradient.

4. Hedgehog is transcription factor.

5. Engrailed is a secretory factor and binds with the patched receptor of the wingless expressing cells.

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

(a) 3 and 5

(b) 3, 4 and 5

(d) 1 and 2

Ans. (d)

Sol. After activation by pair-rule genes, both an extracellular signal, wingless, and autoregulation are required for engrailed expression.

67. In C. elegans, an anchor cell and a few hypodermal cells take part in the formation of vulva. The experiment performed to understand the role of these cells in vulva formation and the results obtained are as follows:

• If the anchor cell iskilled by laser beam, hypodermal cells do not participate in vulva formation and no vulva develops.

• If six hypodermal cells closely located with anchor cell (called vulva precursor cells) are killed, no vulva develops.

• If the three central vulval precursors are destroyed, the three outer cels, which normally form hypodermis, take the fate of vulval cells instead.

Following are certain statements regarding vulva formation:

1. Anchor cells acts as an inducer.

2. Six hypodermal cells with the potentiality to form vulva from an equivalence group.

3. Three, out of six, hypodermal cells participate in vulva formation.

4. The central cell functions as the 1° cell and the two cells on both side act as the 2° cells.

5. The 1° cell secretes a short range juxtacrine signal.

Which combinations of the above statements have been derived from the above experimental results?

(a) 1, 2 and 3 (b) 1, 2 and 4 (c) 4 and 5 (d) 2, 4 and 5

Ans. (a)

Sol. The AC is the key organizer of vulval patterning and morphogenesis, and as a consequence, some mutations that affect the development.

68. The following are some important features which are commonly associated with animal development:

P. Position of anus development with respect to blastopore.

Q. Method of cell division.

R. Mechanism of coelom formation.

S. Cleavage pattern during egg development.

Based on the above, which one of the following combinations differentiate the development of deuterostomes from that of protostomes?

(a) P, Q and R

(b) Q, R and S

(d) P and Q

Ans. (c)

Sol. The key difference between protostomes and deuterostomes is the fate of the blastopore during their embryonic development. The blastopore in a protostome develops into a mouth, whereas the blastopore in deuterostomes develops into an anal opening. In protostomes, the coelom forms when the mesoderm splits through the process of schizocoely, while in deuterostomes, the coelom forms when the mesoderm pinches off through the process of enterocoely. In Protostomes undergo spiral cleavage, while deuterostomes undergo radial cleavage. In both deuterostomes and protostomes, a zygote first develops into a hollow ball of cells, called a blastula. In deuterostomes, the early divisions occur parallel or perpendicular to the polar axis. This is called radial cleavage, and also occurs in certain protostomes, such as the lophophorates.

69. Coupling of the reaction centers of oxidative phosphorylation is achieved by which one of the following?

(a) Making a complex of all four reaction centers

(b) Locating all four complexes in the inner membrane.

(d) Pumping of protons.

Ans. (c)

Sol. The cytochrome bc1 complex (ubiquinol: cytochrome c oxidoreductase complex, is an energy-transducing, electron-transfer enzyme located in the inner mitochondrial membrane of oxygen-utilizing eukaryotic cells, where it participates in cell respiration.

70. Phenylalanine, a precursor of most of the phenolics in higher plants is a product of which one of the following pathways?

(a) Shikimic acid pathway

(b) malonic acid pathway

(d) Methylerythritol pathway

Ans. (a)

Sol. Phenolic acids are a diverse class of plant polyphenols and most studied too, and are produced through shikimic acid by phenylpropanoid pathway, during monolignol pathway as by-products, by the breakdown of cell wall polymers such as lignin and some are produced by microbes also.

71. For which one of the following physiological studies ¹²CO₂and ¹³CO₂ are used?

(a) Estimate the rate of photosynthesis

(b) Determine rate of photosynthesis

(d) The ratio of C₃ and C₄ pathways of CO₂ fixation.

Ans. (d)

Sol. C₃ plants include around 95% of the shrubs, trees, and plants. While C₄ plants are defined as the plants that use the C₄ pathway during the dark reaction. The chloroplasts of these plants are dimorphic and unlike C₃ plants the leaves of C₄ plants possess kranz anatomy. C₄ plants include around 5% of plants on earth. Radioactive carbon is used to study ration of C₃ and C₄ pathways of carbon dioxide fixation.

72. Gibberellic acid (GA) controls seed germination by directing breakdown of the stored starch. In which one of the following tissues of the barley seed, -amylase gene is induced in response to GA?

(a) Endosperm

(b) Coleoptile

(d) Embryo

Ans. (c)

Sol. During seed germination, plant embryo produces gibberellin which triggers the aleurone cells to release -amylase for the hydrolysis of starch stored inthe endosperm.

73. The photosynthetic assimilation of atmospheric CO₂ by leaves yield sucrose and starch as end products of two gluconeogenic pathways that are phsically separated. Which one of the following combination of cell organelles are involved in such physical separation of the process?

(a) Sucrose in cytosol and starch in mitochondria.

(b) Sucrose in chloroplasts and starch in ctosol.

(d) Sucrose in cytosol and starch in chloroplasts.

Ans. (d)

Sol. Sucrose is the principal form of carbohydrate translocated throughout the plant by the phloem. Starch is an insoluble stable carbohydrate reserve that is present in almost all plants. Both starch and sucrose are synthesized from the triose phosphate that is generated by the Calvin cycle.

74. The following are the statements about pyruvate kinase (PK):

P. ATP is an allosteric inhibitor of PK.

Q. Fructose 1, 6 bisphosphate is anactivator of PK.

R. ADP is an allosteric inhibitor of PK.

S. Alanine is an allosteric modulator of PK.

Which of the above statement(s) are true?

(a) P, Q, R

(b) P, Q, S

(d) Only P

Ans. (b)

Sol. Pyruvate kinase activity is regulated by :

Its own substrate PEP and fructose 1, 6-bisphosphate, an intermediate in glycolysis, both of which enhance enzymatic activity. Thus, glycolysis is driven to operate faster when more substrate is present.

ATP is a negative allosteric inhibitor. This accounts for parallel regulation with PFK 1.

It is not known whether citrate plays a role in negative allosteric inhibition, however it is believed that acetyl-CoA does.

Alanine, a negative allosteric modulator.

75. A practical class was going on where the students were demonstrating ATP synthesis in vitro using active mitochondria. Some students added one of the following to their tubes:

P. Dinitrophenol (DNP), an uncoupler.

Q. Mild acidification of the medium.

R. Glutathione, that permeabilizes both the membranes.

S. An outer membrane permeabe H⁺ quencher compound, Elila.

In which one of the above, ATP synthesis will be detected?

(a) P

(b) Q

(d) S

Ans. (b)

Sol. The reaction of ATP synthase requires delivery of protons, magnesium, ADP and phosphate, and consumption of formed ATP. In eukaryotes, the ATP synthase complex is located in the inner membrane of mitochondria, with ATP synthesis reaction occurring on the membrane side toward matrix compartment. In vitro ATP synthesis using mitochondria can be done after mild acidification of medium.

76. Following are certain statements regarding terpene class of secondary metabolites in plants:

P. Isopentenyl diphosphate and its isomer combine to form larger terpenes.

Q. Diterpenes are 20 carbon compounds.

R. All terpenes are derived from the union of 4-carbon elements.

S. Pyrethroids are monoterpene esters.

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

(a) P, Q and R

(b) P, Q and S

(d) P, R and S

Ans. (b)

Sol. Diterpenes are, by definition, C20 compounds based on four isoprene (C5H8) units and can be found in plants, fungi, bacteria, and animals in both terrestrial and marine environments. Pyrethroids are monoterpene esters found in flowers and leaves of certain Chrysanthemum species. These are secondary metabolites. These are used commercially for making insecticides. The toxin disturbs the nervous system and ultimately leads to paralysis and death of the insect.

77. The nodulation (nod) genes are classified as common nod genes or host specific nod genes. Some statements related to such classification are given below:

P. nodA is a common nod gene and nodC is a host specific gene.

Q. nodB is a common nod gene and nodP is a host specific gene.

R. nodQ is a common nod gene and nodA is a host specific gene.

S. nodH is a common nod gene and nodQ is a host specific gene.

Choose the correct answer form the above statements:

(a) P and Q

(b) R and S

(d) Q only

Ans. (d)

Sol. Genes for Nodulation : The Rhizobial gens that participate in nodule formation are called nodulation (nod genes). There are two classes of nod genes. Only one of the node gene i.e. the regulatory nod gene D constitutively expressed and its product (Nod D) regulates the transcription of all other nod genes.

78. Following are certain statements regarding CO₂ assimilation in higher plants:

P. The action of aldolase enzme during Calvin-Benson cycle produces fructose 1, 6-bisphosphate.

Q. The conversion of glycineto serine takes place in mitochondria during C₂ oxidative photosynthetic carbon cycle.

R. During C₄ carbon cycle, NAD-malic enzyme releases the CO₂ from the 4-carbon acid, malate yielding a 3-carbon acid, pyruvate.

S. Malic acid during crassulacean acid metabolism (CAM) is stored in mitochondria during dark and released back to cytosol during day.

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

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (a)

Sol. Green plants contain in their chloroplasts unique enzymatic machinery that catalyzes the conversion of CO₂ to simple (reduced) organic compounds a process called CO₂ assimilation.

In CO₂ assimilation :

The action of aldolase enzyme during Calvin-Benson cycle produces fructose 1, 6-bisphosphate.

The conversion of glycine to series takes place in mitochondria during C₂ oxidation photosynthetic carbon cycle.

During C₄ carbon cycle, NAD-malic enzyme releases the CO₂ from the 4-carbon acid, malate yielding a 3-carbon acid, pyruvate.

79. Many factors releated to the role of abscisic acid (ABA) in contributing to drought, cold and salt resistance in plants are listed below:

P. The transcription factors DREB 1 and DREB 2 bind to the cis-acting elements of the promoter of ABA responsive genes in an ABA dependent manner.

Q. ABA induces many genes such as LEA and RD29.

R. ABA-responsive gene contain six nucleotide ABRE elements in the promoter.

S. Nine-nucleotide dehydration-responsive elements (DRE) are present in ABA responsive genes.

Which one of the following combinations of the above statements is correct with respect to ABA?

(a) P, Q and R

(b) P, R and S

(d) P only

Ans. (a, b, c)

Sol. The dehydration-responsive element (DRE), which has a core sequence of A/GCCGAC, is a cis-acting element originally isolated from the promoter of the RD29A gene of Arabidopsis (Arabidopsis thaliana) and is involved in both cold- and dehydration-inducible gene expression via an abscisic acid (ABA)-independent pathway . The RD29 (Responsive to Desiccation) genes RD29A and RD29B are such genes induced by desiccation, cold and high salt stresses. DREBs (dehydration responsive element binding) are important plant transcription factors (TFs) that regulate the expression of many stress-inducible genes mostly in an ABA-independent manner and play a critical role in improving the abiotic stress tolerance of plants by interacting with a DRE/CRT cis-element present in the promoter region of various abiotic stress-responsive genes.

80. Examples of many factors that regulate plant height in response to gibberellic acid (GA) are listed below:

P. Binding of a GA-bound repressor to the promoter of the DELLA domain-containing GRAS protein gene and blocking its expression.

Q. Binding of the GA receptor complex to GRAS.

R. Directing GRAS for ubiquitination and degradation by the 26S proteasome.

S. Micro RNA directed down regulation of the GRAS protein expression.

Which one of the following combinations is correct?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (b)

Sol. GRAS proteins are an important family of plant-specific proteins named after the first three members: GIBBERELLIC-ACID INSENSITIVE (GAI), REPRESSOR of GAI (RGA) and SCARECROW (SCR). Some GRAS proteins contain conserved sequence motifs in the N-terminal region, which are expected to be involved in molecular recognition. For example, DELLA proteins possess a conserved DELLA sequence motif in the N-terminal region and the peptide region containing the DELLA motif is found to be conformationally disordered in the free state, but refolds into a helical structure upon binding to GA-bound GA receptor GID1.

81. Ethylene is an important plant hormone that regulates several aspects of plant growth and development. Some statements are given below in relation to ethylene signalling pathways:

P. Unbound ethylene receptors work as positive regulators of the response pathway.

Q. There are more than two ethylene receptors known to date.

R. The carboxy terminal half of the ethyene receptor, ETR1 (Ethylene response 1), contains a domain homologous to histidine kinase catalytic domain.

S. EIN2 (Ethylene insensitive 2) encodes a transmembrane protein. The ein2 mutation promotes ethylene responses in both seedlings and adult Arabidopsis plant.

Which combination of the above statements is correct?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (b)

Sol. Five different types of ethylene receptors are present in Arabidopsis thaliana: ETR1, ERS1, EIN4, ETR2, and ERS2. The amino-terminal half of ETR1 contains a hydrophobic domain responsible for ethylene binding and membrane localization. The carboxyl-terminal half of the polypeptide contains domains with homology to histidine kinases and response regulators, signaling motifs originally identified in bacteria.

82. A diabetic patient developed metabolic acidosis resulting in deep and rapid breathing which is called

(a) Kussmaul breathing

(b) Cheyne-Stokes respiratory pattern

(d) Periodic breathing

Ans. (a)

Sol. Kussmaul breathing is a type of abnormal respiration, characterized by being laboured and very deep, that can be seen in patients with extreme metabolic acidosis.

When patients go into metabolic acidosis, their blood becomes very acidic. The body uses a numebr of measures to compensate, including respiratory compensation. Patients in the early stages may breathe quickly and shallowly. As the acidosis progresses, Kussmaul breathing can develop. In Kussmaul breathing, patients breathe at a normal or slightly slower rate, but their breaths are much deeper than usual. This is a form of hyperventilation, causing carbon dioxide levels in the blood to drop while oxygen rises.

83. Which one of the following is not involved with the pacemaker potential of heart?

(a) "h"-channel

(b) Transient calcium channel

(d) "f" -channel

Ans. (c)

Sol. Pacemaker potential is due to SA node which is found of self excitable fibres. This cells does not have long lasting calcium channel. Instead, transient calcium channel are present through which calcium ion enters into cell. Long lasting calcium channel are present in cardiac muscles.

84. You are asked to identify the stage of estrus cycle in vaginal smear of a mouse containing large number of leukocytes and very few nucleated epithelial cells. Which one of the following will be the correct stage of estrous cycle?

(a) Early estrus, late proestrus

(b) Late estrus, early metestrus.

(d) late diestrus, early proestrus

Ans. (d)

Sol. Cycling changes are usually divided into four stages — estrus, metestrus, diestrus and proestrus. Estrus lasts approximately 10-15 h in rats and approximately 21 h in mice; it is the only stage in the cycle when the female will copulate with the male. Cornified squamous epithelial cells are the dominant cell type in the vaginal smear and are few in number or absent. The squamous epithelial cells vary from flat (occurring singly during early estrus) to curled. Metestrus lasts from 6 to 14 h in rats and approximatley 22 h in mice. It is characterized by the presence of numerous cornified epithelial cells together with irregularly shaped nucleated epithelial cells. Leukocytes are present in considerable numbers. Diestrus lasts 60-70 h in ratsa nd 22-23 h in mice. It is dominated by leukocytes, although some nucleated epithelial cells may be present. Proestrus has a 12-18 h duration in rats and approximately 21 h in mice and denotes the beginning of the next cycle. The vaginal smear is dominated by numerous nucleated epithelial cells, usually arranged in grape-like clusters.

85. Which one of the following neurotransmitters is secreted by the pre-ganglionic neurons of sympathetic nervous system?

(a) Epinephrine

(b) Acetycholine

(d) Norepinephrine

Ans. (b)

Sol. In the autonomic nervous system, acetylcholine (ACh) is the nurotransmitter in the preganglionic sympathetic and parasympathetic neurons. The red ACh in the ganglion. ACh is also the neurotransmitter at the adrenal medulla and serves as the neurotransmitter at all the parasympathetic innervated organs. ACh is also the neurotransmitter at the sweat glands and at the piloerector muscle of the sympathetic ANS.

86. After hemorrhage, a subject develops hypovolemia and hypotension. Following are some of the statements regarding homeostatic measure taken by the body after hemorrhage.

P. Increased release of vasopressin.

Q. Increased water retention and reduced plasma osmolality.

R. Increased rate of afferent discharge from low pressure receptors of vascular system.

S. Decreased rate of afferent discharge from high pressure receptors of vascular system.

Which one of the following is not correct in this condition?

(a) Only P

(b) P and Q

(d) Q and S

Ans. (c)

Sol. After hemorrhage, hypovolemia and hypotension appears in patient under affect of Increased release of vasopressin, Increased water retention and reduced plasma osmolality and Decreased rate of afferent discharge from high pressure receptors of vascular system.

87. A visitor to a region of hot climate is more distressed by the heat than the regular resident. Within a few weeks, the visitor is more comfortable with the heat and capacity for work is increased. Following are some of the explanations given by a researcher regarding acclimatization to heat.

P. Sweating begins at a lower body temperature.

Q. Blood flow through skin is high for any body temperature.

R. There is rise in resting body temperature.

S. Vasoconstriction starts at a lower body temperature.

Which one of the following is not true?

(a) Only P

(b) P and Q

(d) R and S

Ans. (d)

Sol. A visitor to a region of hot climate is more distressed by the heat than the regular resident. Within a few weeks, the visitor is more comfortable with the heat and capacity for work is increased. In this situation, There is rise in resting body temperature and Vasoconstriction starts at a lower body temperature.

88. The difference in circulation between glomerular capillaries (GC) and true capillaries (TC) are described by a researcher in the following statements:

P. The hydrostatic pressure in GC is higher than that in TC.

Q. The endothelial cells are fenestrated in GC but not in TC.

R. Both filtration and fluid movement into capillary takes place in TC but only filtration occurs in GC.

S. The plasma colloid osmotic pressures in both the ends of GC or TC are similar.

Which one of the following is not correct?

(a) Only P

(b) P and Q

(d) Only S

Ans. (d)

Sol. In comparison between glomerular capillaries and true capillaries. The hydrostatic pressure in GC is higher than that in TC, The endothelial cells are fenestrated in GC but not in TC and Both filtration and fluid movement into capillary takes place in TC but only filtration occurs in GC.

89. When rods of retina kept in dark, were exposed to light, phototransduction occurred. Following are some explanations given by a researcher regarding phototansduction:

P. Activation of transducin.

Q. Inhibiton of cGMP phosphodiesterase.

R. Closure of Na⁺ channels.

S. Hyperpolarization of rods.

Which one did not occur in phototransduction?

(a) Only P

(b) Only Q

(d) R and S

Ans. (b)

Sol. During phototransduction, there is Activation of transducin, Closure of Na⁺ channels and Hyperpolarization of rods.

90. The afferent nerve fibres of a stretch reflex were electrically stimulated and the contraction of the muscle innervated by efferent fibres was recorded. The synaptic delay was calculated from the time points of the nerve stimulation and response of the muscle. Which one of the following time durations will be probable value for the observed synaptic delay?

(a) 0.05 msec

(b) 0.5 msec

(d) 5.0 msec

Ans. (b)

Sol. Observed synaptic delay would be 0.5 mili second.

91. A convenient and reasonably reliable indicator of the time of ovulation is usually a rise in the basal body temperature, possibly because progesterone is thermogenic. Of the four situations given below, which one is ideal for ensuring pregnancy after intercourse?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Diagram in option (a) is correct.

92. In an organism, exprssion of gene 'X' is induced in the presence of a hormone. Genetic analysis showed that the hormonal signal is transduced thorugh two proteins Let 1 and Let 2. The expressionof gene 'X' was studied in lines overexpressing (OE) the active Let proteins, knockout (KO) of the Let proteins or combination of both. Resullts of expression of gene 'X' in presence of the hormone is summarized below:

Based on the above, which one of the following pathways best fits the observation made?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. By observation of reaction pathway and expression of gene X, option (b) correctly represent cascade action.

93. Following is ahypothetical biochemical pathway responsible for pigmentation of leaves. The pathway is controlled by two independently assorting genes 'A' and 'B' encoding enzymes as shown below. Mutant alleles 'a' and 'b' code for non-functional proteins. What is the expected progeny after selfing a plant with the genotype AaBb?

(a) Green (9) : White (4) : Yellow (3)

(b) Green (9) : Yellow (4) : White (3)

(d) Green (9) : White (7)

Ans. (a)

Sol. AaBb × AbBb, then gametes : AB, Ab, aB, ab

According to the Punnet square method,

Genotype Phenotype

1 AABB – Green

2 AABb – Green

1 AAbb – Yellow

2 AaBB – Green

4 AaBb – Green

2 Aabb – Yellow

1 aaBB – White

2 aaBb – White

1 aabb – White

Expected ratio of progeny of Green : White : Yellow = 9 : 4 : 3

94. Mutation in gene 'X' leads to lethality in a haploid organism. Which one of the following is best suited to analyse the function of gene 'X'?

(a) Pleiotropic mutants

(b) Temperature-sensitive mutants

(d) Mutants with low penetrance

Ans. (b)

Sol. If mutation in gene 'X' leads to lethality in a haploid organism the gene acts must be temperature sensitive mutant.

95. The following pedigree chart shows inheritance of a given trait. The trait can be called

(a) autosomal dominant

(b) autosomal recessive

(d) sex limited

Ans. (a)

Sol. Autosomal dominance is a pattern of inheritance characteristic of some genetic disease. "Autosomal" means that the gene in question is located on one of the numbered or non-sex, chromosomes. "Dominant" means that a single copy of the disease-associated mutation is enough to cause the disease.

96. In a heterozygous individual for a given gene, if a crossing over has occurred between thegene locus and the centromerer of the chromosome, the segregation of the two alleles of the given gene will occur during meiosis at

(a) either anaphase I or anaphase II

(b) anaphase I only

(d) both anaphase I and II

Ans. (c)

Sol. The homologous chromosomes are segregated during Anaphase I without an exchange of genetic material between the gene and the centromere. If crossing over does occur between the gene and the centromere the second division (MII/meiosis 2) segregation of the two alleles will occur in anphase II.

When non-sister chromatids do cross over between the centromerer and the locus, the A and a alleles are still together in the nuclei at the end of first division of meiosis i.e., no first-division segregation occurs.

However, the second meiotic division does move the A and a alleles into separate nuclei, giving rise to the second-division segregation.

97. Two siblings who inherit 50% of the genome from the mother and 50% from the father show lot of phenotypic differences. Which one of the following events during gametogenesis of the parents will maimally contribute to this difference?

(a) Mutation

(b) Recombination

(d) Environment

Ans. (c)

Sol. Two siblings who inherit 50% of the genome from the mother and 50% from the father show lot of phenotypic differences. It is possible by independent assortment.

98. Consider the following hypothetical pathway:

An individual with 'A' phenotype when crossed with that of 'B' phenotype has a progeny with 'O' phenotype. Which one of the following crosses can lead to the above observation?

(a) Aahh × BbHH

(b) AaHh × BBHh

(d) AAHH × BbHh

Ans. (b)

Sol. If an individual with 'A' phenotype when crossed with that of 'B' phenotype has a progeny with 'O' phenotype, it would be possible only by cross of AaHh × BBHh.

99. Three somatic hybrid cell lines, designated as X, Y and Z, have been scored for the presence or absence of chromosomes 1 through 8, as well as for their ability to produce the hypothetical gene product A, B, C and D as shown in the following table:

Which of the following option has most appropriately assigned chromosomes for each of the given genes?

(a) Gene A on chromosome 5, Gene B on chromosome 3, Gene C on chromosome 8 and Gene D on chromosome 1.

(b) Gene A on chromosome 5, Gene B on chromosome 3 only

(d) Gene A on chromosome 5, Gene B on chromosome 3 and Gene D on chromosome 1.

Ans. (d)

Sol. As per the information shown in table, Gene A on chromosome 5, Gene B on chromosome 3 and Gene D on chromosome 1.

100. Of the following, which one of the individuals will not necessary carry the allele responsible for the mentioned trait?

(a) A woman in a family where an autosomal dominant trait is segregating and her mother and son are affected.

(b) A daughter of a man who is affected by an X-linked dominant trait.

(d) A father of a boy affected with X-linked recessive trait.

Ans. (d)

Sol. A father of a boy affected with X-linked recessive trait will not necessary carry the allele responsible for the mentioned trait.

101. The following diagram shows meiotic pairing in an inversion heterozygote and a point where single crossing over has occurred

The resulting gametes produced may have

A. the chromosome having normal gene sequence.

B. the chromosome having inverted gene sequence.

C. a dicentric chromosome with duplication and deletion.

D. an acentric chromosome having duplication and deletion.

E. the chromosome having duplication and deletion.

Which of the following combination will be most appropriate for the diagram shown?

(a) A, B, C and D

(b) A, B and E

(d) A, C, D and E

Ans. (b)

Sol. As per the diagram mentioned in question, the resulting gametes produce may have the chromosome having normal gene sequence, the chromosome having inverted gene sequence and the chromosome having duplication and deletion.

102. Most members of bryophyte phylum Anthocerophyta are characterized by

(a) gametophyte with single chloroplast per cell and multicellular rhizoids; sporophyte without stomata.

(b) gametophyte with single chloroplast per cell and unicellular rhizoids; sporophyte without stomata.

(d) gametophyte with single chloroplast per cell and multicellular rhizoids; sporophyte without stomata.

Ans. (b)

Sol. Bryophytes are gametophyte dominant, meaning that the more prominent, longer-lived plant is the haploid gametophyte. The diploid sporophytes appear only occassionally and remain attached to and nutritionally dependent on the gametophyte. In bryophytes, the sporophytes are always unbranched and produce a single sporangium (spore producing capsule).

103. Identify the correct match between the animal (flatworm, earthworm, roundworm) and its body cavity type (acelomate, coelomate, pseudocoelomate):

(a) Roundworm – pseudocoelomate; Earthworm – acoelomate; Flatworm – coelomate

(b) Roundworm – acoelomate; Earthworm – coelomate; Flatworm – acoelomate

(d) Roundworm – coelomate; Earthworm – pseudocoelomate; Flatworm – acoelomate

Ans. (c)

Sol. Acoelomate : Animal like sponges, coelenterates and flatworms are without a coelom or any other internal cavity except the digestive tract. Such animals are called acoelomate. In them either there is no space between the gut and body wall (sponges and coelenterates) or the space is filled with mesenchyme (flatworms).

Psudocoelomate : Round worms have a body cavity derived directly from the blastocoel of the embryo. It is called pseudocoel because like true coelom, it is not lined by peritoneum but is bounded with ectoderm on the outer side and endoderm on the inner side. The organs float in the fluid that fills the pseudocoel.

Eucoelomate or coelomate : Animals with a tube-within-a-tube body plan have a fluid filled body cavity between the body wall and the digestive tract as like in earthworms. It is derived from embryonic mesoderm and is lined on both sides with a special mesodermal lining called peritoneum or coelomic epithelium. Such a body cavity is called coelom or true body cavity and the animals with coelom are called coelomates or eucoelomates. The internal organs in coelomates hang down into the coelom but do not float free. These are slung in pouches of peritoneum. The vertical portion os peritoneum are called mesenteries.

104. Which one of the following gymnosperm phyla produces motile sperms, bears ovulate and microsporangiate cones on separate plants and has fleshy, coated seeds?

(a) Coniferophyta

(b) Cycadophyta

(d) Gnetophyta

Ans. (b, c)

Sol. In ginkgos and cycads, the microgametophyte produce swimming sperms; in the conifers and angiosperms, the sperms are non-motile.

Evolution of seeds was preceded by the evolution of the vascular cambium. The cells of this new cambium could undergo radial longitudinal divisions, thus allowing the cambium to grow in circumference as wood accumulated.

105. According to 2014 IUCN Red List, which of the following vertebrate classes has the largest percentage of threatened species?

(a) Mammals

(b) Birds

(d) Amphibians

Ans. (d)

Sol. Of the 6,260 amphibian species assessed, nearly one-third of species (32.4%) are globally threatened or extinct, representing 2,030 species. Thirty-eight of the 2,030 species are considered to be Extinct (EX) and one Exitnct in the Wild (EW). Another 2,697 species are not considered to be threatened at present, being classified in the IUCN Categories of Near Threatened (NT) or Least Concern (LC), while sufficient information was not available to assess the status os an additional 1,533 species [IUCN Category Data Deficient (DD)]. It is predicted that a significant proportion of these Data Deficient species are likely to be globally threatened.

106. In the following equations

Q. N_t = N₀e^rt

exponential population growth is described by

(a) P and Q

(b) P only

(d) Q and S

Ans. (b)

Sol. Population is always comprises of exponential growth.

Rate of increase in population is linear proportional to the population in a given time

where r is proportional constant.

Integrating both sides, we have

107. Which gas does not contribute to global warming through its greenhouse effect?

(a) Nitrous oxide

(b) Methane

(d) Nitric oxide

Ans. (d)

Sol. Gases that contribute to the greenhouse effect include :

Water vapor : The most abundant greenhouse gas, but importantly, it acts as a feedback to the climate. Water vapour increases as the Earth's atomsphere warms, but so does the possibility of clouds and precipitation, making these some of the most important feedback mechanisms to the greenhouse effect.

Carbon dioxide (CO₂) : A minor but very important component of the atmosphere, carbon dioxide is released through natural processes such as respiration and volcano eruptions such s respiration and volcano eruptions and through human activities such as deforestation, land use changes and burning fossil fuels. Humans have increased atomspheric CO₂ concentration by a third since the Industrial Revolution began. This is the most important long-lived "forcing" of climate change.

Methane : A hydrocarbon gas produced both through natural sources and human activities, including the decomposition of wastes in landfills, agriculture and especially rice cultivation, as well as ruminant digestion and manure mangement associated with domestic livestock. On a molecule for molecule basis, methane is a far more active greenhouse gas than carbon dioxide, but also one which is much less abundant in the atmosphere.

Nitrous oxide : A powerful greenhouse gas produced by soil cultivation practices, especially the use of commercial and organic fertilizers, fossil fuel combustion, nitric acid production and biomass burning.

Chlorofluorocarbons (CFCs) : Synthetic compounds entirely of industrial origin used in a number of applications, but now largely regulated in production and release to the atmosphere by international agreement for their ability to contribute to destruction of the ozone layer. They are also greenhouse gases.

108. A red coloured tubular flower without any odour is most likely to be pollinated by

(a) Beetles

(b) bees

(d) birds

Ans. (d)

Sol. Pollination syndromes are suites of flower traits that have evolved in response to natural selection imposed by different pollen vectors, which can be abiotic (wind and water) or biotic, such as birds, bees, flies and so forth. These traits include flower shape, size, colour, odour, reward type and amount, nectar composition, timing of flowering, etc. For exmple, tubular red flowers with copious nectar often attract birds; foul smelling flowers attract carriom flies or beetles, etc.

109. Which one of the following would contribute to intrinsic fluorerscence to a protein?

(a) Aromatic amino acids

(b) Disulfide bonds

(d) Branched chain amino acids

Ans. (a)

Sol. The aromatic amino acids tryptophan, tyrosine and phenylalanine are capable of contributing to the intrinsic fluorescences of proteins. When all three residues are present in a protein (termed as class B protein), pure emission from tryptophan can be obtained only by photoselective excitation at wavelengths above 295 nm.

110. Which of the following is the correct match of the algal group with its food reserve?

Algal group Carbohydrate reserve

P. Bacillariiophyceae i. Oil

Q. Xanthophyceae ii. Floridean starch

R. Phaeophyceae iii. Laminarin

S. Rhodophyceae iv. Chrysolaminarin

v. Starch

(a) P-iv, Q-i, R-iii, S-ii,

(b) P-ii, Q-i, R-iii, S-iv

(d) P-i, Q-v, R-iii, S-ii

Ans. (a)

Sol. Reserve Food : The food is reserved in different forms in various algal groups, some of which are summarised here under :

Chlorophyceae : Starch and Oil

Xanthophyceae : Chrysolaminarin (carbohydrate) and Oils

Bacillariophyceae : Chrysolaminarin and Oils

Phaeophyceae : Laminarin (carbohydrate), Mannitol (alcohol) and Oils

Rhodophyceae : Floridean Starch and Galactin-SO₄ polymers

Cyanophyceae : Cyanophycean Starch (glycogen) and Cyanophycin (protein).

111. A researcher conducts a standard test to identify enteric bacteria (A, B, C) on the basis of their biochemical properties. The result is given in the following table:

Based on the above, the identified bacteria A, B and C are most probably

(a) Enterobacter, Salmonela, Escherrichia

(b) Escherichia, Salmonella, Enterobacter

(d) Escherichia, Enterobacter, Salmonella

Ans. (b)

Sol. E.coli organisms are first identified as lactose-fermenting Gram-negative rods on routine culture medium. Other biochemical characteristics include indole production, lack of citrate fermentation, positive methyl red test. Salmonellae yield negative Voges-Proskauer and positive methyl red tests and do not produce cytochrome oxide. The methyl red and Voges-Proskauer tests are part of a battery of biochemical tests known as IMViC used in the clinical lab. Enterobacter aerogenes and Klebsiella pneumoniae are MR– and VP+.

112. Following is a cladogram of the major taxonomic groups of the angiosperms. Groups A-E represent respectively

(a) Astrobaileyales, Nymphaedales, Amborellales, Chloranthaceae, Magnoliids.

(b) Amborellales, Astrobaileyales, Nymphaedales, Magnoliids, Chloranthaceae.

(d) Amborellales, Nymphaedales, Chloranthaceae, Magnoliids, Astrobaileyales.

Ans. (c)

Sol. E. Most magnoliids have features of a relatively archaic nature.

D. Chloranthaceae are soft-wooded aromatic shrubs and trees with opposite saw-toothed leaves and swollen nodes. The flowers are very small, often unisexual, and usually lack petals or tepals, although they are sometimes subtended by leaflike bracts.

C. Astrobaileyales family includes evergreen trees and shrubs with simple leaves and small, bisexual flowers composed of numerous tepals.

B. The Nymphaeaceae are aquatic, rhizomatous herbs. The family is further characterized by scattered vascular bundles in the stems, and frequent presence of latex, usually with distinct, stellate-branched sclereids projecting into the air canals. Hairs are simple, usually producing mucilage (slime).

A. The Amborellaceae are distinctive in being vessel-less, evergreen shrubs with unisexual flowers having an undifferentiated, spiral perianth, numerous, laminar stamens, and an apocarpous, apically-open gynoecium, with 1-ovuled carpels.

113. The phylogenetic tree of amnoote vertebrates is given in diagram. The groups labelled A, B, C, D are

(a) A-Snakes, B-Turtles, C-Birds, D-Mammals

(b) A-Snakes, B-Turtles, C-Mammals, D-Birds

(d) A-Birds, B-Turtles, C-Snakes, D-Mammals

Ans. (c)

Sol. C will be the correct statement for phylogenetic tree of amniote vertebrates.

114. The following are matches made between adult animals and their larval forms:

1. Copepods - Nauplius

2. Sea cucumber - Zoea

3. Sea urchin – Echinopleuteus

4. Crabs - Auricularia

5. Star fish - Bipinnaria

6. Brittle star - Ophiopleuteus

Which one of the combinations below reflects incorrect matches?

(a) 1, 3, 5

(b) 2 and 4

(d) 6 only

Ans. (b)

Sol. The first stage of larval development of sea cucumber is known as an auricularia, and is only around 1 mm (39 mils) in length. This larva swims by means of a long band of cilia wrapped around its body, and somewhat resembles the bipinnaria larva of starfish."Most crabs develop through a few stages (usually 2 to as many as 8 or 9, depending on the group) of planktonic zoea larvae. The final zoea larva then moults into a still-swimming larva called a megalopa, which is morphologically and ecologically a transitional phase between the planktonic zoeae and the benthic adult.

115. Which of the following statements about the birth rates (b₁, b₂) and death rates (d₁, d₂) of species 1 and 2 indicated in the figure is not true?

(a) Birth rates of species 1 are density-independent.

(b) Death rates of both species are density-dependent.

(d) Density-dependent effects on death rates are similar for both the species.

Ans. (d)

Sol. Birth and death rates are more likely a function of population density or abundance. births are a decreasing function of density b(N) and deaths are an increasing function of density d(N). Hence population growth will be zero at some population size. As death rate of both species is increasing with increasing population density, so death rates are similar for both the species.

116. Important chemical reactions involved in nutrient cycling in ecosystems are given below:

The organisms associated with these chemical reactions are

(a) P - Nitrosomonas, Q - Pseudomonas, R - Nostoc, S - Nitrobacter.

(b) P - Pseudomonas, Q - Nitrobacter, R - Nostoc, S - Nitrosomonas.

(d) P - Nostoc, Q -Nitrobacter, R - Nitrobacterx, S - Pseudomonas.

Ans. (c)

Sol.

Fig. : Biological Conversions of Ammonia to Nitrate by Bacteria.

Table : Inorganic Compounds and Ions as Final Electron Acceptors for Bacteria

117. Following table shows the presence (+) or absence (–) of five species in three communities (A, B, C):

Based on the above, which of the following is the correct order of similarity between two pairs of communities?

(a) A and B > B and C > A and C

(b) A and B > A and C > B and C

(d) A and C > A and B > B and C

Ans. (a)

Sol. Due to presence of 4 similarities between A and B (maximum), then B and C ( 2 similarities) and at last A and C (one similarity).

118. Which of the following X-Y relationship does not follow the pattern shown in the graph?

(a) Number of prey killed (Y) in relation to prey density (X).

(b) Photosynthetic rate (Y) in relation to light intensity (X).

(d) Tree species richness (Y) in relation to actual evapotranspiration.

Ans. (d)

Sol. D statement does not follow the x-y relationship in the graph.

119. Following table shows the number of individuals of five tree species in a community:

Tree species Number of individuals

A 50

B 20

C 20

D 05

E 05

Based on the above, the Simpson's diversity (DS) index of the community will be

(a) 0.552

(b) 0.335

(d) 0.345

Ans. (b)

Sol.

120. In a field eperiment, autotrophs are provided a ¹⁴C-labeled carbon compound for photosynthesis. Radioactivity (¹⁴C) levels were then monitored at regular intervals in all the trophic levels. In which ecosystem is the radioactivity likely to be detected fastest at the primary carnivore level?

(a) Open ocean

(b) Desert

(d) Grassland

Ans. (a)

Sol. Carnivores have higher consumption efficiency than herbivores, since more of their food source is consumed than enters into the detrital food chain.

121. Which one of the following conditions is not likely to favour male monogamy?

(a) When the male has to guard his mate against mating by another male.

(b) When the male wants to spend more time for foraging.

(d) When the female guards her mate against seeking other females to mate.

Ans. (b)

Sol. The term 'monogamy' can maen a number of things. Genetic monogamy is unlikely unless it raises (male) reproductive success enough to compensate for the loss of reproductive success (RS) that would have come from additional mating efforts. We except to see maleparental polygyny under these conditions : (i) whenever a female can raise offspring successfully alone and (ii) when male care that sufficiently enhances offspring survivorship and competitive success can be 'generalizable'—no more expensive for several offspring than for one. Consider red-winged black birds, in which males watch and warn at the approach of potential predators; a male can do this effectively for several nests, at the same cost as for one.

122. The origin and diversification of Angiosperms was during which geological period?

(a) Permian

(b) Triassic

(d) Creataceous

Ans. (d)

Sol. Angiosperms, which evolved in the Cretaceous period, are a diverse group of plants which protect their seeds within an ovary called a fruit.

Angiosperms evolved during the late Cretaceous Period, about 125-100 million years ago.

Angiosperms have developed flowers and fruit as ways to attract pollinators and protect their seeds, respectively.

Flowers have a wide array of colors, shapes and smells, all of which are for the purpose of attracting pollinators.

Once the egg is fertilized, it grows into a seed that is protected by a fleshy fruit.

As angiosperms evolved in the Cretaceous period, many modern groups of insects also appeared, including pollinating insects that drove the evolution of angiosperms; in many instances, flowers and their pollinators have coevolved.

Angiosperms did not evolve from gymnosperms, but instead evolved in parallel with the gymnosperms, however, it is unclear as to what type of plant actually gave rise to angiosperms.

123. Which of the following statements about evolution is not true?

(a) Evolution is the product of natural selection.

(b) Evolution is goal-oriented.

(d) Evolution need not always lead to a better phenotype.

Ans. (b)

Sol. Evolution by natural selection is one of the best substaniated theories in the history of science, supported by evidence from a wide variety of scientific disciplines, including paleontology, geology, genetics and developmental biology.

Eukaryotic cells are generally bigger than prokaryotic cells. This allows eukaryotes to have organelles have slower turnover rates of macromolecules and requires the presence of mechanisms to move things around the cell.

Evolution os not goal oriented. Evolution simply depends on the environment which the organisms live and try to survive. The environment is fit for strongest survive and reproduce. Evolution uses the theory of natural selection where their is variation of "survival of the fittest". You have a variation of traits, heredity and different reproduction as a result of natural selection.

124. A population is growing logistically with a growth rate (r) of 0.15 per week, in an environmental with a carrying capacity of 400. What is the maimum growth rate (number of individuals/week) that this population can achieve?

(a) 15

(b) 30

(d) 60

Ans. (a)

Sol. Maximum population growth rate occurs when N = K/2

We have K = 400, R = 0.15/week and N = 400/2 = 200

Logistic growth equation is,

125. In several populations, each of size N = 20, if genetic drift results in a change in the relative frequencies of alleles.

A. What is the rate of increase per generation in the proportion of populations in which the allele is lost or fixed?

B. What is the rate of decrease per generation in each allele frequency class between 0 and 1?

The correct answer for A and B is

(a) A — 0.25, B — 0.125

(b) A — 0.025, B — 0.0125

(d) A — 0.125, B — 0.25

Ans. (c)

Sol. Rate of increase per generation in the proportion of populations in which the alleles is lost or fixed =

Rate of decrease per generation in each allele frequency class between 0 and 1 = 0.025

126. Individual A can derive 'fitness' benefit of 160 units by helping individual B, but incurs a 'fitness' cost of 50 units in doing so. Following Hamilton's rule, A should help B only if B in his

(a) brother or sister

(b) first cousin only

(d) nephew or niece

Ans. (a)

Sol. According to the Hamilton's rule, r × b > c, or r × 160 > 50 or r > 50/160.

So, r > 0.3125

In this case, coefficient of relatedness is more than 0.3125 which is applicable in case of brother or sister.

127. In a population of effective population size N_e, with rate of neutral mutation μ₀, the frequency of heterozygotes per nucleotide per site at equilibrium between mutation and genetic drift is calculated as

(a)

(b)

(c)

(d)

Ans. (b)

Sol. For an effective population size of N and a mutation when an equilibrium is reached between mutation and drift is approximately

The quantity 4Nµ₀ determines the relative influence of mutation and drift. When 4Nµ₀ is very small (close to 0), then the expected heterozygosity at equilibrium will be close to 0, typical of a vary small population experiencing heavy genetic drift. When 4Nµ₀ is very large (>5 or so), the effect of genetic drift si minimal and the expected heterozygosisty is high (>0.83). Under the infinite alleles model, the maximum heterozygosity is H = 1.0. Intermediate values at or rear 4Nµ₀ = 1 show an expected balance between mutation and drift. The balance between mutation and drift at equilibrium forms an important part of the neutral theory of evolution. The neutral theory sees much of allele frequency change as genetic drift operating on neutral mutations.

128. The "Red Queen Hypothesis" is related to

(a) the mating order in the harem of a polygamous male.

(b) the elimination by deleterious mutations by sexual reproduction.

(d) the evolutionary arms race between the host and the parasite.

Ans. (d)

Sol. The two main hypotheses proposed to explain the success of sex despite its two fold disadvantage in the short therm are (i) the "red queen" hypothesis, i.e. an evolutionary arms race between species, especially between hosts and pathogens and (ii) DNA repair. In both cases it is claimed that, if an asexual lineage appeared, it would rapidly suffer from a large disadvantage because of the pressure by pathogens or because of a rapid degenerescence of their genome and that the two-fold advantage of asexual individuals would rapidly be counterbalnaced by one of these phenomena before the complete invasion of the species by parthogenetic females.

129. Application of gene therapy in clinical trials did not succeed due to

(a) poor integration of a gene in the host genome.

(b) lack of expression of integrated gene in cells.

(d) activation of oncogenes consequent to integration of the gene.

Ans. (d)

Sol. It's possible that once introduced into the body, the viruses may recover their original ability to cause disease. Possibility of causing a tumor. If the new genes get inserted in the wrong spot in your DNA, there is a chance that the insertion might lead to tumor formation.

130. A gene expressing a 50 kDa protein from an eukaryote was cloned in a E. coli plasmid under the lac promoter and operator. Upon addition of IPTG, the 50 kDa protein was not detected. Which one of the following explains the above observation?

(a) The cloned sequence lacked the Kozak sequence.

(b) E. coli does not make proteins larger than 40 kDa.

(d) 50 kDa protein contains a nuclear localization signal.

Ans. (c)

Sol. Codon usage bias refers to differences in the frequency of occurrence of synonymous codons in coding DNA. Although translation initiation is the key step in protein synthesis, it is generally accepted that codon bias contributes to translation efficiency by tuning the elongation rate of the process.

131. Neomycin phosphotransferase gene, frequently used as a selection marker during plant transformation, inactivates which one of the following antibiotics?

(a) Hygromycin

(b) Ampicillin

(d) Kanamycin

Ans. (d)

Sol. The most classical of the selection marker systems is antibiotic resistance. Among the most commonly used resistance genes are the ones encoding neomycin phosphotransferase II (nptII) and the hygromycin phosphotransferase (hpt), both from E. coli. Through phosphorylation, the NPTII protein inactivates a number of aminoglycoside antibiotics, like kanamycin, neomycin geniticin or paromomycin. While geniticin if often used for the selection of transformed mammalian cells, the three others—with different efficacies for different species—are almost exclusively use in plant transformation systems.

132. Which among the following is the simplest method to estimate the concentration of glycerol in an aqueous solution of glycerol?

(a) UV absorption spectroscopy

(b) Gas chromatography

(d) Viscosity measurement

Ans. (d)

Sol. Measuring viscosity via a capillary tube is one of the oldest methods of determining the kinematic viscosity of a sample, requiring prior knowledge of the density and volume of the sample of interest. This fluid is passed through a vertical U-tube of known dimensions and a very small diameter.

133. Which one of the following techniques will you use to identify more than 1000 differentially expressed genes in normal and tumor tissues in one experiment?

(a) RAPD

(b) Genome sequencing

(d) Transcriptome analysis

Ans. (d)

Sol. An RNA sequence mirrors the sequence of the DNA from which it was transcribed. Consequently, by analyzing the entire collection of RNA sequences in a cell (the transciptome) researchers can determine when and where each gene is turned on or off in the cells and tissues of an organism. By this technique we can identify more than 1000 differentially expressed genes in normal and tumour tissues in one single experiment.

134. For identification of three proteins moving together (as a single band) upon loading in a single lane of SDS-PAGE gel, the best method is

(a) one step Western blot

(b) NMR spectroscopy

(d) ESR spectroscopy

Ans. (c)

Sol. Stripping involves washing of nitrocellulose membrane to remove previously bound antibodies, so that new protein can be probed and detected with another set of antibodies.

135. Which isotope below is best suited for metaboic labelling of glyceraldehyde-3 phospho-dehydrogenase?

(a) ¹⁴C

(b) ¹²⁵I

(d) ¹³¹I

Ans. (a)

Sol. Glyceraldehyde-3-phosphate-dehydrogenase (abbreviated as GAPDH or less commonly as G3PDH) is an enzyme of ~37kDa that catalyzes the sixth step of glycolysis and thus serves to break down glucose for energy and carbon molecules. In addition to this long established metabolic function, GAPDH has recently been implicated in several non-metabolic processes, including transcription activation, initiation of apoptosis. C-14 is best suited for metabolic labelling of glyceraldehyde-3-phospho-dehydrogenase.

136. A student noted the following points regarding Agrobacterium tumefaciens:

P. A. tumefaciens is a gram-negative soil bacterium.

Q. Opine catabolism genes are present in T-DNA region of T_i-plasmid

R. Opines are synthesized by condensation of amino acids and α-ketoacids or amino acids and sugars.

S. A callus culture of crown gal tissue caused by A. tumefaciens in plants can be multiplied without adding phytohormones.

Which one of the combinations of above statement is correct?

(a) P, Q and R

(b) P, Q and S

(d) P, R and S

Ans.

Sol. Agrobacterium radiobacter is the causal agent of crown gall disease in over 140 species of eudicots. It is a rod-shaped, Gram-negative soil bacterium. Opines are low molecular weight compounds found in plant crown gall tumors or hairy root tumors produced by pathogenic bacteria of the genus Agrobacterium. Agrobacterium tumefaciens causes crown gall disease on a wide range of host species by transferring and integrating a part of its own DNA, the T-DNA, into the plant genome. Crown gall tumors were initiated in a variety of plant species by infection with Agrobacterium tumefaciens strain B6 and the concomitant changes in the tissue levels of phytohormones, mainly indole-3-acetic acid (IAA) and cytokinins, were analyzed.

137. The following statements are related to plant tissue culture:

A. Friable callus provides the inoculum to form cell-suspension cutures.

B. The process known 'habitutation' refers to the property of callus loosing the requirement of auxin and/or cytokinin during long term culture.

C. Cellulase and pectinase enzymes are usually used for generating protoplast cultures.

D. During somatic embryo development, torpedo stage embryo is formed before heart stage embryo.

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

(a) A, B, and C

(b) A, B and D

(d) B, C and D

Ans. (a)

Sol. Several different culture types most commonly used in plant transformation studies will now be examined in more detail :

Callus

Cell-suspension Cultures

Protoplasts

Root Cultures

Shoot Tip and Meristem Culture

Embryo Culture

Microscope Culture

Cell Suspension Cultures : Callus cultures, broadly speaking, fall into one of two categories : compact or friable. In compact callus the cells are densely aggregated, whereas in friable callus the cells are only loosely associated with each other and the callus become soft and breaks apart easily. Friable callus provides the inoculum to form cell-suspension cultures.

Protoplasts : Protoplasts are plant cells with the cell wall removed. Protoplasts are most commonly isolated from either leaf mesophyll cells or cell suspensions, although other sources can be used to advantage. Two general approaches to removing the cell wall (a difficult task without damaging the protoplast) can be taken-mechanical or enzymatic isolation.

Cellulose and pectinase enzymes are usually used for generating protoplast cultures.

The process known as 'habituation' refers to property of callus loosing the requirement of auxin and/or cytokinin during long-term culture.

138. You have transiently expressed a new protein (for which no antibody is available) in a cell line to establish structure function relationship. Which one of the following strategies is the most straight forward way to examine the expression profile of this new protein?

(a) By metabolic labeling using ³⁵S labeled amino acids.

(b) Making a GFP fusion protein with this new protein.

(d) Running SDS-PAGE and identify the protein.

Ans. (b)

Sol. The green fluorescent protein (GFP) of Aequorea victoria is frequently fused to other proteins to serve as a reporter for gene expression or the localization of proteins in vivo.

139. As cancer progersses, several genome rearrangements including translocation, deletion, duplications etc. occur. If these arrangements are to be identified, which of the following techniques would be most suitable?

(a) RAPD

(b) Micrroarray

(d) Flow cytometry

Ans. (c)

Sol. Multicolor FISH (mFISH) is a method to facilitate analysis of each single chromosome or chromosome part of a metaphase. Thus, marker chromosomes, complex chromosomal rearrangements, and all numerical aberrations can be visualized simultaneously in a single hybridization experiment.

140. A mixture of two proteins was subjected to following three chromatographic columns:

A. Cation exchange,

B. Size exclusion (Sephadex 100) and

C. Reverse phase.

Following elution profiles were obtained.

Which of the following statements is correct?

(a) A is larger and more hydrophobic than B.

(b) B is more anionic and more hydrophobic than A

(b) A is more cationic and smaller than B.

Ans. (c)

Sol. Size exclusion chromatography (SEC) separates molecules based on their size by filtration through a gel. The gel consists of spherical beads containing pores of a specific size distribution. Smaller-sized molecules have more pores that are accessible to them and therefore spend more time inside the pores relative to larger-sized molecules. Therefore, smaller molecules elute last and larger molecules elute first in Size Exclusion Chromatography. So, A is smaller than B. Reversed-phase chromatography is a technique using alkyl chains covalently bonded to the stationary phase particles in order to create a hydrophobic stationary phase, which has a stronger affinity for hydrophobic or less polar compounds. A is also more hydrophobic than B.

141. Glucose in the blood is detected by four different methods (P, Q, R and S). The sensitivity and range of detection of glucose by these four methods is shown below. Clinically relevant concentration of glucose in blood in between 80 - 125 mg/dL. Which of the following method is most appropriate?

(a) P

(b) Q

(d) S

Ans. (b)

Sol. Q is most appropriate method to detect glucose level in blood between 80-125mg/dL.

142. Which one of the following statement is correct?

(a) Electrospray ionization mass spectrum of a compound can be obtained only if it has a net positive charge at pH 7.4.

(b) Helical content of a tryptophan containing peptide can be obtained by examining the fluorescence spectrum of tryptophan.

(d) The chemical shift spread for a compound is more in its ¹H NMR spectrum as compared to its ¹³C NMR spectrum.

Ans. (c)

Sol. Circular Dichroism, an absorption spectroscopy, uses circularly polarized light to investigate structural aspects of optically active chiral media. It is mostly used to study biological molecules, their structure, and interactions with metals and other molecules. Circular dichroism (CD) is an excellent method for rapidly evaluating the secondary structure, folding and binding properties of proteins.

143. A researcher is studying the subcellular localization of a particuar protein 'X' in an animal cell. The researcher performs successive centrifugation at increasing rotor speed. The researcher starts spinning the cellular homogenate the supernatant at 100,000g for 1 hour, collects both the pellet and the final supernatant. On subjecting various pellets and the final supernatant to Western blotting with anti-protein-X antibody, the protein X is observed to be maximally expressed in pellet after centrifugation at 10,000g. Based on the above observation, what will be the most likely localization of protein X.

(a) Nucleus

(b) Ribosomes

(d) Microsomes

Ans. (c)

Sol. Circular dichroism (CD) spectroscopy was carried out to determine whether out mini-proteins were well folded and stable despite the presence of eight surface mutations on the adjacent -strands. The secondary structure of proteins and polypeptides are studied by CD spectroscopy which is based on differences in absorption of left and right circulatory polarized light as a result of structural asymmetry. There are three distinct CD signatures for specific secondary structural conformations in proteins, the -helix, -sheet and a random coil. On the basis of these specific CD profiles, secondary structural details of our proteins can be inferred.

144. Fluorescence recovery after photobleaching in live cell is used to determine

(a) co-localization of proteins

(b) distance between two organelles.

(d) nucleic acid compactness

Ans. (c)

Sol. Fluorescence recovery after photobleaching (FRAP) is an optical technique used to measure the temporal dynamics of fluorescently tagged molecules. It is used to determine diffusion of proteins. This renaissance is driven largely by the advent of confocal microscopy and the introduction of green microscopy and the introduction of green fluorescent protein (GFP) as an endogenous protein marker. After GFP is covalently attached to the protein of interest, its cellular distribution can be visualized at low light intensities that do not damage cellular processes. Photobleaching is the irreversible destruction of fluorescence in a region within the sample by brief exposure to high light intensities. After bleaching a region, the recovery of fluorescence over time there can be recorded to measure the rate of redistribution of fluorescent molecules. This rate of fluorescence redistribution provides information about the processes involved in the movement of the molecule. The movement reflects diffusion, which can be retarded if the fluorescently tagged molecule bind to other molecules. In the latter case, the rate of redistribution of bleached and unbleached molecules contain information about the strength of the binding interaction. Thus, FRAP is a valuable technique to study molecular dynamics in live cells.

145. During an experiment, a student found increased activity of a protein, for which there were three possible explanations, viz., increased expression of the protein, increased phosphorylation, or increased interaction with other effector proteins. After conducting several experiments, the students concluded that increased activity was due to increased phosphorylation. Which one of the following experiments will not support/provide the correct explanation drawn by the student?

(a) Western blot analysis

(b) Analysis of transcription rate

(d) Phosphoamino acid analysis

Ans. (b)

Sol. Genome-wide transcription analysis is a powerful tool for determining the complete set of mRNAs and their relative expression levels as a function of growth conditions. Phosphorylation is a Post-translational modification. Post-translational modification of proteins can be experimentally detected by a variety of techniques, including mass spectrometry, Eastern blotting, and Western blotting. Phosphoamino acid analysis, or PAA, is an experimental technique used in molecular biology to determine which amino acid or acids are phosphorylated in a protein. So, analysis of transcription rate is the only technique which can't provide correct explanation.

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

Locate Us

Company

Courses

Academies

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

Locate Us

Company

Courses

Academies

Get in touch

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