PYQ DEC 2018 - VedPrep

CSIR NET BIOLOGY (DEC - 2018)
Previous Year Question Paper with Solution.

21. Which one of the following statements is not correct ?

(a) Together with proteins, rRNA provides a site for polypeptide synthesis.

(b) All DNA molecules are unbranched polymers of nucleotides.

(d) A tRNA anticodon may pair with more than one codon.

Ans. (c)

Sol. Whenever nucleic acid molecules associate by complementary base pairing, the two strands are antiparallel. Just as the two paired strands of DNA are antiparallel, the template strand of the DNA and the complementary RNA strand are also antiparallel. Therefore, when transcription takes place, as RNA synthesized in its 5' 3' direction, the DNA template is read in its 3' 5' direction.

22. Which one of the following statements is true ?

(a) The specific rotation of enantiomers as only time but no concentration units.

(b) The rate constant of a fist order reaction has only time btuy no concentation units.

(d) The bond dissociation energy (kJ/mol) of –C–C– will be gerater than –CC–.

Ans. (b, c)

Sol. Unit of rate constant for first order reaction (k) = min^–1 or s^–1

The rate constant of a first order reaction has only time unit. It has no concentration unit.

This means the numerical value of k for a first order reaction is independent of the unit in which concentration is expressed.

If concentration unit is changed the numerical value of k for a first order reaction will not change.

A reaction is said to be first order if its rate is determined by the change of one concentration term only.

One can say that a first order reaction is one whose rate varies as 1st power of the concentration of the reactant i.e., the rate increases as number of times as the number of times the concentration of reactant is increased.

23. Which one of the following statements on protein conformation is not true ?

(a) Dihedral angles of side-chains in amino acids are depicted in the Ramachandran plot.

(b) Infrared spectroscopy can be used to deduce hydrogen bonding in peptides.

(c) Three-dimensional structures of protein composed of ~100 amino acids can be obtained by nuclear magnetic resonance spectroscopy.

(d) Globular proteins have -helical and -sheet components.

Ans. (a)

Sol. A Ramachandran plot, originally developed in 1963 by G. N. Ramachandran, C. Ramakrishnan, and V. Sasisekharan, is a way to visualize energetically allowed region for backbone dihedral angles against of amino acid residues in protein structure. In the figure, the definition of the and backbone dihedral angles (called and ' by Ramachandran). The angle at the peptide bond is normally 180°, since the partial double bond character keeps the peptide planar.

Dihedral angle values are circular and 0° is the same as 360°, the edges of the Ramachandran plot "wrap" right-to-left and bottom-to-top. For instance, the small strip of allowed values along the lower-lift edge of the plot are a continuation of the large, extended chain region at upper left.

Figure : Conversion of L-amino Acids. 3 PG : 3-phospho-D-glycerate, Pyr : pyruvate, OAA : oxaloacetate, 2 KG : 2-oxoglutarate.

From above conversion, we can see that the amino acids Met, Thr, Lys, Ile, Val and Leu are biosynthesized from oxaloacetate and pyruvate in most bacteria.

24. Choose the correct answer from the following statements on biosynthesis.

(a) In the biosynthesis of palmitate, all the carbon atoms are derived from activated malonate.

(b) The amino acids Met, Thr, Lys, Ile, Val and Leu are biosynthesized from oxaloacetate and pyruvate in most bacteria.

(d) Tryptophan is converted to L-DOPA in the biosynthesis of epinephrine.

Ans. (b)

Sol. 3C and 4C carbohydrates are precursor of amino acids biosynthesis."From Pyruvate(3C): Alanine, Isolucine, Leucine and Valine are synthesized and we called it as the pyruvate family biosynthesis. First of all oxaloacetate(4C) is converted into aspartate and then from aspartate asparagine, methionine,Threonine and Lysine are synthesized and these are called as Aspartate Family.

25. Which one of the following statements on nucleic acids is not true ?

(a) The conformation of ribose in DNA is -2'-deoxy-D-ribofuranose.

(b) Hydrolysis of RNA takes place under alkaline condition unlike DNA, as the 2'-hydroxyl in RNA acts as a nucleophile in an intamolecular displacement.

(d) In DNA, deamination of cytosine to uracil can occur in a non-enzymatic manner.

Ans. (a)

Sol. In Nucleic Acid :

1. The covalent backbone of DNA and RNA is subject to slow, nonenzymatic hydrolysis of the phosphodiester bonds. In the test tube, RNA is hydrolyzed rapidly under alkaline conditions, but DNA is not; the 2'-hydroxyl groups in RNA (absent in DNA) are directly involved in the process. Cyclic 2', 3'-monophosphates are the first products of the action of alkali on RNA, and are rapidly hydrolyzed further to yield a mixture of 2'- and 3'-nucleoside monophosphates.

2. Three major forms of DNA are double stranded and connected by interactions between complementary base pairs. These are terms A-form, B-form and Z-form DNA.

A, B and Z forms of the DNA molecule in the deoxyribonucleic acid (DNA) molecule different combinations of monomeric compounds nucleotides linked together in a long chain are used to encode the information about the structure of proteins. Two chains of deoxyribonucleotides interact with each other following a certain rule the principle of complementarity (adenine forms hydrogen bonds and pairs with thymine and guanine with cytosine) to form a double helix. The resulting molecule can be extremely long. At the human cells nuclei 2 meters of DNA is stored in a volume of 40 cubic micrometers.

3. In DNA, deamination of cytosinie to uracil can occur in a non-enzymatic manner.

Once replication is complete, the most common kind of damage to nucleic acids is one in which the normal A, C, G and T bases are changed into chemically modified based are changed into chemically modified based that usually differ significantly from their natural counterparts. The only exceptions are the deamination of cytosine to uracil and the deaminatio of 5-methylcytosine to thymine. In these cases the product is a G:U or G:T mismatch. Specific enzymes called DNA glycosylases can recognize uracil in DNA or the thymine in a G:T mismatch and can selectively remove the base by cleaving the bond between the base and the deoxyribose sugar. Many of these enzymes are specific for the different chemically modified bases that may be present in DNA.

26. Following are statements on -turns :

P. All the 20 coded amino acids have equal propensity to form β-turns.

Q. Pro cannot occur in -turns.

R. Pro-Gly sequences strongly favours -turns.

S. In Asn-Glu -turns, can have positive values.

Choose the combination with all correct statements.

(a) Q, S

(b) P, R

(d) R, S

Ans. (d)

Sol. A -turn is a region of the protein involving four consecutive residues where the polypeptide chain folds back on itself by nearly 180°. It is these chain reversals which give a protein its globularity rather than linearity. The -turn was originally identified, in model building studies, by Venkatachalam. He proposed three distinct conformations based on phi, psi values (designated I, II and III) along with their related turns (mirror images) which have the phi, psi signs reversed (I', II' and III'), each of which could form a hydrogen bond between the main chain C = O(i) and the N-H(i + 3).

The Pro-D-Tyr sequence, which strongly favours a type II -turn, was used to encourage the adoption of a -turn by the remaining four residues, Gly-Ile-Leu-Gln. The two β-turn conformation is a likely one for cyclic hexapeptides.

Type III -turns (which are excluded from our -turn definition) correspond to a 310-helix. In the simulations, the interconversion of -turns and 310-helices was frequently observed, including -turns with angles that match those of the 31-helix (especially in the higher temperature replicas) although they were short turns only comprised of a few residues and not continuous extended helices.

27. DNA melting temperature (T_m) was found to be 47°C and entalpy measured at T_m was 0.032 kJ. The entropy changes would be

(a) 1 × 10^–3 kJ

(b) 1 × 10^–4 kJ

(d) 6 × 10^–2 kJ

Ans. (b)

Sol. Entropy = Enthalpy/Tm

Enthalpy = 0.032 kJ = 32 J

Tm = 47 C or 320 K

Entropy = 32/320 = 0.1 J

= 0.0001 kJ

28. Match the following bonds with their approximate energies :

P. Hydrogen bond 1. 0.5 kcal

Q. van der Waals forces 2. 40 kcal

R. Covalent bond 3. 80 kcal

S. Ionic bond 4. 3 kcal

(a) P–4, Q–3, R–2, S–1

(b) P–2, Q–1, R–3, S–4

(d) P–4, Q–1, R–3, S–2

Ans. (d)

Sol. Hydrogen Bonds : Hydrogen bonds are relatively weak interactions, which nonetheless are crucial for biological macromolecules such as DNA and proteins. These interactions are also responsible for many of the properties of water that make it such a special solvent. Hydrogen bonds are much weaker than covalent bonds. They have energies of 1-3 kcal mol^–1 (4-13 kJ mol^–1).

Van Der Waals Forces : The basic of a van der Waals interaction is that the distribution of electronic charge around on atom changes with time. At any instant, the charge distribution is not perfectly symmetric. This transient asymmetry in the electronic charge aroun an atms acts through electrostatic interactions to induce a complementary asymmetry in the electron distribution around its neighbouring atoms. Energies associated with van der Waals interactions are quite small; typical interactions contribute from 0.5 to 1.0 kcal mol^–1 (from 2 to 4 kJ mol^–1) per atom pair.

Covalent Bonds : The strongest bonds that are present in biochemicals are covalent bonds, such as the bonds that hold the atoms together within the individual bases. A covalent bond is formed by the sharing of a pair of electrons between adjacent atoms. A typical carbon-carbon (C-C) covalent bond has a bond length of 1.54 Å bond energy of 85 kcal mol^–1 (356 kJ mol^–1).

29. For a evesible non-competitive inhibition of an enzyme, choose the plot that you would use to determine K_m :

—— No inhibitor ------ + inhibitor

(a)

(b)

(c)

(d)

Ans. (c)

Sol. K_m does into change in case of non-competitive inhibition.

30. If one of the two fatty acyl chains is removed from the phosphoglyceride by hydolysis in solution, such phospholipid will form

(a) liposomes

(b) micelles

(d) symmetric phospholipid bilaye

Ans. (b)

Sol. Phospholipid is the most common group of lipids. In fact, cell membranes as well as organized cellular compartments are all made up of these phospholipids. They can form structures called micelles, in which when phospholipids congregate, the hydrophobic fatty acid tails join together in the center of the sphere away from the aqueous environment and the polar heads are exposed to the outside. Structures such as liposomes can also be artificially formed from these lipids : using high frequency sound waves to sonicate the sample containing phospholipids and molecules of interest to create phospholipid vesicles that contain the molecules of interest. Phospholipids and glycolipids have amphipathic characteristics which enzbles them to form a micelle or a "lipid bilayer". Due to the hydrophobic hydrocarbon tail and the hydrophilic polar head groups, the lipids arrange in a form where the polar groups face water while the tail is away from water. One formation is the micelle where the lipids arrange themselves in a circle with the head groups making the circumference while the tails are inside. A more favourable formation is the lipid bilayer or the bimolecular sheet. This arrangement has the lipids form a barrier where the polar head groups face the aqueous media and the hydrophobic tail face inside away from water.

31. In metazoan cell cycle, metaphase to anaphase transition is regulated by the activity of

(a) Cdk1/cyclin B

(b) APC/C

(d) Wee1

Ans. (a, b)

Sol. The catastrophic destruction of cyclin — CDK activity at the end of mitosis is a sine qua non of cell cycle reguation. Mitotic cyclins are eliminated by the ubiquitin system via highly conserved, cell cycle-regulated E3 activity termed the anaphase-promoting complex (APC), also known as the cyclosome or APC/C. The APC is turned on at the metaphase to anaphase transition and turned off at the G1 to S phase transition. Genetic screens for budding yeast mutants that stabilize Clb2p turned up three known CDC genes required for the metaphase to anaphase transition : CDC16, CDD23 and CDC27. At the same time, purification of the Xenopus APC particle yielded vertebrate homologues of Cdc16p, Cdc23p and Cdc27p. Further purification revealed a total of 12 subunits in the yeast APC and a least eight corresponding subunits in the Xenopus APC particle. Finally, parallel cytological screens for fission yeast mutants with altered nuclear structure (nuc) or with a spetum through the undivided nucleus identified Nuc2 and Cut9 as homologues of Cdc27p and Cdc16p respectively.

32. Which one of the following statements is true about human chromosomes ?

(a) The chromosomes that have highest gene density generally localize towards the centre of the nucleus.

(b) The chromosomes that have highest gene density generalyl localize near the nuclea periphery to facilitate rapid transport of the nascent transcripts.

(d) Chromosomal positioning in the nucles is absolutely random.

Ans. (a)

Sol. The genome is highly organized within interphase nucleic. Each chromosome and gene occupies a preferred nuclear position. The prototypical example of this is human chromosomes (HSA) 18 and 19 is proliferating cells. The gene-rich HSA19 is located in the center of the nucleus, while the gene-poor HSA18 locates to the nuclear periphery. Indeed, in proliferating human cells the radial position (position relative to the edge and center of the nucleus) is highly correlated with gene density, whereby CTs generally follow the pattern of HSA18 and 19, with the gene-rich chromosomes in the center of the nucleus and gene-poor chromosome towards the periphery. Chromosome positioning patterns can also be correlated with chromosome size. in this case, the small chromosomes tend to position to the nuclear interior and the larger chromosomes prefer the periphery.

33. Which one of the following activities is not involved in protein folding in the endoplasmic reticulum ?

(a) Peptidyl-prolyl isomerase

(b) Protein disulphide isomerase

(d) Protein ubiqitination

Ans. (d)

Sol. In yeast and animal cells, secretory proteins are synthesized on ribosomes bound to the endoplasmic reticulum (ER), from where they are subsequently translocated into the lumen of the ER by s signal sequence/receptor-mediated mechanism.

The ER membranes harbour (among other proteins) the enzymes responsible for dolichol-dependent protein glycosylation and the lumen of the ER contains enzymes involved in protein folding such as protein disulphide isomerase (PDI), peptidyl prolyl isomerase (PI) or chaperones such as BIP or their yeast equivalent (Kar2 proteins).

34. A single protofilament of microtubule grows at the speed of 2 µm/min. Considering that there is no catastrophe in the microtubule nucleation and the size of the tubulin unit is of the order of 5 nm, how many tubulin units are added to the growing microtubule per minute ?

(a) 400

(b) 1600

(d) 5200

Ans. (d)

Sol. Each microtubule is made up of 13 protofilament. Each protofilament is made up of and tubulins. The rate of eleongation of a single protofilament is 2 µm/min or 2000 nm/min.

Number of tubulin units added per min to elongate a single protofilament

Total number of tubulin units required to be added per min = 400 × 13 = 5200.

35. mRNA of a gene was depleted in human cells using siRNA that arrest cells in the G2 phase of the cell cycle. In order to test whether the G2 arrest is due to an off-target or an on-target effect of siRNA mediated mRNA depletion, an investigato can

P. re-introduce an ectopic copy of the gene coding for the wild-type mRNA and protein.

Q. re-introduce an ectopic copy of the gene that is different in its mRNA sequence at the siRNA target site but encodes for the same protein.

R. re-introduce an ectopic copy of the gene that codes for different mRNA and protein.

S. utilize few more siRNAs targeting different regions of the mRNA in question.

Choose the combination with correct statements.

(a) P, Q, R only

(b) R and S only

(d) Q, R, S only

Ans. (c)

Sol. To check that it is an off target arrest or on target arrest are have to reintroduced a different m-RNA which codes for same protein or introduce different Si RNA's, So that by changing m-RNA can tell us that if the base is missing it its a case of off target or it is by differ siRNA. So statement Q and S is correct.

36. An investigator expresses a GFP-fused protein that localizes to the outer membrane of Golgi apparatus. Upon visualising GFP-signal in the fluorescence microscope, it was noted that GFP is pericentrosomal in its localization (figure A). Treatment of such GFP expressing cells with a newly identified drug disrupted the Golgi into small vesicles (figure B).

Following is a list of potential targets of the drug :

P. Dynein complex

Q. Myosin

R. Microtubules

S. Dicer

Choose the combination with all correct targets.

(a) P, Q and S only

(b) Q and S only

(d) P only

Ans. (c)

Sol. Disruption of normal Microtubule cytoskeleton dynamics with antibodies or agents such as colchicine and nocodazole that cause Microtubule depolymerization or paclitaxel, which stabilizes Microtubules, result in dispersal of the Gogi into smaller units or "mini stacks" spread throughout the cell, cells lacking dynein are unable to concentrate organelles such as Golgi and lysosomes in the perinuclear region.

37. Following observations was made when a mammalian cell in one phase of cell cycle was fused with a cell in another phase of cell cycle :

P. Fusion of a cell in G1-phase with S-phase caused the G1-nucleus to enter S-phase.

Q. Upon fusion of a G-cell with an S-phase cell, G2-cell does not enter S-phase.

R. Upon fusion of a G1-cell with G2-cell, G1 nucleus enters G2 phase.

S. Fusion of an S-phase cell wit a M-phase causes the S-phase cell to immediately enter mitosis.

Choose the combination with all correct statements.

(a) P, Q, R

(b) P, R, S

(d) P, Q, S

Ans. (d)

Sol. Rao and Johnston : mixing together in the same cytoplasm (heterokaryon) to determine whether they could influence one another.

S + G1 : induces the G1 nuclei to start S phase. Suggests that the S phase nucleus contains a diffusible factor that will induce replication.

S + G2 phase : the G2 nucleus does not do S phase. Something about G2 phase is refractory to the diffusible factor from S phase.

G1 + G2 phase : no S or M phase.

M phase + interphase : induces inappropriate mitosis, G1 and G2 do not influence each other.

38. Which one of the following proteins is most likely to be found in the inter-membrane space of mitochondria ? A protein containing

(a) an N-terminal matrix targeting sequence followed by hydrophobic stop-transfer anchor sequence.

(b) an N-terminal matrix targeting sequence followed by a cleavable hydrophobic sequence that blocks complete translocation.

(d) a protein with an outer membrane localization sequence followed by a matrix targeting signal.

Ans. (b)

Sol. The mitochondrial matrix targeting sequence at the N terminus is recognized by the import receptor, which targets the protein to the matrix. Part of the way through translocation, the hydrophobic sequence will stop to go through the translocon. This sequence would diffuse out of the translocon and imbed in the inner mitochondrial membrane. The rest of the protein would remain in the intermembrane space. This could not be an outer mitochondrial membrane proteins, because those proteins require a special outer membrane localization sequence so that the protein does not cross through the translocon to the matrix.

39. What would be the tripeptide produced by traslation of the transcript produced by the following DNA sequence?

3'-AAGTACTCT-5'

(a) Arg-Phe-Trp

(b) Arg-Leu-Gly

(d) Phe-Met-Arg

Ans. (d)

Sol. Transcript sequence of the template mentioned in question would be 5'-UUCATGAGA which will give rise to Phe-Met-Arg.

40. Which one of the following statements is generally true about RNA polymerase II ?

(a) It is dedicated to transcribing RNA from a single transcription unit, generally a large transcript which is then processed to yield three types of ribosomal RNA.

(b) It transcribes varieties of small non-coding RNAs which are expressed in all cell types.

(d) It is exclusively involved in synthesis of rRNA and tRNA.

Ans. (c)

Sol. Transcription by RNA polymerase II (Pol II) is required to produce mRNAs and some noncoding RNAs (ncRNAs) within mammalian cells. This coordinated process is precisely regulated by multiple factors, including many recently discovered ncRNAs.

Table : Major classes of RNA transcribed by the three eukaryotic RNA polymerases and their functions.

41. Presence of selenocysteine in proteins in E. coli is a consequence of

(a) post-translational modification of cysteine present in special structural regions of the proteins by SelB and SelC.

(b) post-translational modification of serine present in special structural regions of the proteins by SelB and SelC.

(c) aminoacylation of a special tRNA (tRNA^SeCys) by serine tRNA synthetase with serine followed by further modification of the attached serine to selenocysteine followed by its transport to the ribosome by SelB.

(d) aminoacylation of a special tRNA (tRNA^SeCys) by serine tRNA synthetase with selenocysteine followed by transport to ribosome by SelB.

Ans. (c)

Sol. Computational and experimental approaches have identified 25 selenoproteins in the human proteome. These selenoproteins can be classified according to their biologcal function into 6 different groups :

1. Peroxidase and reductase activities

2. Hormone metabolism

3. Protein folding

4. Redox signaling

5. Sec synthesis and

6. Selenium transport

The existence of a selenium-laden tRNA was first observed in experiments performed in rat liver where radiolabeled [Se]-selenite was associated with a tRNA-bound selenocysteine residue. Presence of selenocysteine in proteins in E. coli is a sonsequence of aminoacylation of a special tRNA (tRNA Cys) by serine tRNA synthetase with serine followed by further modification of the attached serine to selenocysteine folowed by its transport to the ribosome by SelB.

42. In an experiment, intact chromatin was isolated and digested with micrococcal nuclease in independent tubes for 30 min, 1h, 2h and 4h. Further, the DNA was purified from each tube, separated on agarose gel and Southern hybridization was performed with rRNA gene probe and a centromeric DNA probe. Which one of the following patterns of signal intensity from both of the probes is likely to be obtained following Southern hybridization ?

(a) With increasing time, compared to centromeric probe, a rapid increase in signal intensity of rRNA gene probe was observed.

(b) With increasing time, compared to centromeric probe, a rapid decrease in signal intensity of rRNA gene probe was observed.

(d) Treatment with micrococcal nuclease would instantly degrade the DNA, hence, no hybridization signal would be obtained in any of the samples.

Ans. (b)

Sol. Micrococcal nuclease eleaves initially in the linker region as an endonuclease and in a second stage reduce the size of the chromatin fragment by an exonucleolytic digestion proceeding bi-directionally from the initial cleavage point. By decompacting chromatin, HMGN proteins facilitate access to the linker DNA and increase the rate of the initial endonucleolytic attack. By stabilizing the structure of the nucleosome core, they prevent octamer sliding, protect several bases at the ends of the chromatin particle and decrease the rate of the exonucleolytic digestion that is why with time the nucleosome at the centromere will be protected so hybridization, but with more time the DNA will be degraded, so the hybridization will be very low.

43. A DNA segment was cloned into the active site region of IacZ gene and the recombinant plasmid introduced into IacZ^– strain of E. coli and plated on a medium containing X-gal. The colonies showed blue colour. Which one of the following statements is correct ?

(a) The nature of the cloned DNA segment need not be special as cloning of any DNA in IacZ will result in disruption of its reading frame and production of blue colour on X-gal plates.

(b) The cloned DNA segment need not be a Group I intron whose removal from the precursor IacZ tarnscript in E. coli results in production of mature IacZ mRNA which can then produce active LacZ protein.

(d) The cloned sequence is likely to be an anti-terminator sequence which allows full length transcription of IacZ.

Ans. (b)

Sol. Blue colour of colonies is due to beta galactosidase activity which shows lacZ gene exptession. It was observed that the insertion of DNA segment did not impact the activity of lacZ gene expression and shows blue colour colonies. Possible reason for this observation is that inserted DNA segment could be a group I intron which possess self spliced property so it get removed during processing and it will not be a part of mature DNA.

44. An intron in a yeast reporter gene carries a mutation in the splice site branch point (UACUAAC to UACA*AAC). To suppress the mutation, a libranry of point mutants of snRNAs was introduced into the mutant strain. The suppressor is most likely to have a point mutation in

(a) U1 snRNA

(b) U2 snRNA

(d) U6 snRNA

Ans. (b)

Sol. The suppressor pint mutation should be in U2sn RNA as the branch point is recognized by its snRNA. This is a kind of intergenic suppressor mutation as gene for the original mutation and the suppressor mutation is different.

45. A researcher wanted to identify the enhancer sequence of a newly discovered gene. Shown below are the relevant regions of some of the reporter constructs the researcher designed to identify the enhancer. Which of the below constructs can be used to identify the enhancer ?

(a) P only

(b) Q only

(d) R only

Ans. (a)

Sol. Enhancer can only be identified where transcription can take place. Only option P has promotor. So, transcription can take place at option P and enhancer can only be identified at option P.

46. In a genetic assay, randomly generated fragments of yeast DNA were cloned into a bacterial plasmid containing gene 'X' essential for yeast viability on minimal media. The recombinant plasmid was used to transform a yeast strain deficient in recombination and lacking 'X' gene. Transformants, which survive on minimal media and form colonies should essentially have

(a) yeast centromeric sequence which ensures integrity of the plasmid after transformation.

(b) enhancers for the essential gene missing in the transformed strain.

(d) yeast autonomous replicating sequence.

Ans. (d)

Sol. Bacterial plasmid transformed in yeast cell require autonomous replicating sequence for replication. Because bacterial plasmid will use yeast replicating machinery to divide. It do not required bacterial origin for replication in yeast. Bacterial plasmid do not have centromere unlike yeast.

47. Many cytotoxic T lymphocytes initiate killing of target cells via delivery of molecles that could induce target-cell damage directly. Which one of the following is the most appropriate ?

(a) Interferon

(b) Peroxynitite

(d) Granzyme

Ans. (d)

Sol. Granzymes are serine proteases released by cytoplasmic granules within cytotoxic T cells and natural killer (NK) cells. They induce programmed cell death (apoptosis) in the target cell, thus eliminating cells that have become cancerous or are infected wiht viruses or bacteria. Granzymes also kill bacteria and inhibit viral replication. In NK cells and T cells, granzymes are packaged in cytotoxic granules with peroforin. Granzymes can also be detected in the rough endomplasmic reticulum, golgi complex and the trans-golgi reticulum.

48. Viruses adopt different strategies to suppress immune response of the host. Which one of the following statements is not correct ?

(a) Human Immunodeficiency Virus (HIV) destroys CD4⁺ T-cells.

(b) Epstein-Bar Virs (EBV) produces a homolog of human IL-10.

(d) Human Cytomegalo Virus (CMV) establishes latent infection in bone marrow stem cells.

Ans. (c)

Sol. Epithelial cells that are infected with influenza virus produce inflammatory cytokines acting as chemoattractants for homin macrophages and dendritic cells (DC). DCs take up influenza viral particles to trigger their maturation and pusuant migration to the lymph, where they initiate antigen-specific T cell maturation. These influenza-specific effector T cells thene enter the respiratory tract to counteract viral titres through cytokine expression and the direct lysis of infected cells, with activated CD8⁺ effector cytotoxic T cells (CTLs) representing the main constituents of this response by their release of perforins and granzymes and the engagement of tumor necrosis factor (TNF) receptors. Influenza-specific CD4⁺ T helper cells can act directly and indirectly in viral clearance, primarily by producing cytokines that induce the functions of B cells and CD8⁺ T cells and which have also been reported to directly eliminate infected cells themselves.

49. Which one of the following inactivates the serine/threonin protein kinase, mTOR, related to cell growth in mammalian system ?

(a) Rifamycin

(b) Rapamycin

(d) Chloramphenicol

Ans. (b)

Sol. Protein kinases are key regulators of a cell's communication system, passing messages along the signaling cascade. They are enzymes that attach phosphate groups to serine/threonine proteins or tyrosine proteins in the cell, which then effectively passes the message like a bucket brigade.

The tyrosine protein kinases that do not have the outside receptor portion or the transmembrane portion are called non-receptor tyrosine kinases. They work to relay the message in the cytoplasm of the cell to the nucleus (downstream signaling). There are 10 families, such as the Jak, abl and src. The other type of protein kinase is called the serine/threonine protein kinase. One example of a serine/threonine protein kinase is the mammalian target of rapamycin (mTOR), which plays a central role in cell proliferation (turning on the cell cycle) and cell metabolism and in regulating cell growth and angiogenesis. mTOR was named as such because Rapamycin, an immunosuppressant, was able to inhibit this kinase in organ transplant patients.

50. Match the following cell origin with their nomenclature.

Tumor cell origin Nomenclature

P. Muscle cell 1. Carcinoma

Q. Germ cell 2. Sarcoma

R. Epithelial cell 3. Leukemia

S. White blood cell 4. Teratocarcinoma

(a) P–1, Q–3, R–4, S–2

(b) P–4, Q–1, R–2, S–3

(d) P–2, Q–4, R–1, S–3

Ans. (d)

Sol.

51. Susceptible individuals were infected with pathogen A and pathogen B separately. Pathogen A has a very short incubation period and disease symptoms are already underway by the time memory cells are activated. Pathogen B on the other hand has a long incubation period which allows the memory cells to be activated and respond. Which one of the following will be the most appropriate vaccination strategy against both pathogens A and B ?

(a) Repeated vaccination against both A and B for maintaining high levels of neutrallizing antibodies.

(b) Repeated vaccination against A and a single injection of pathogen B vaccine for maintaining high levels of neutalizing antibodies.

(c) Single injection of pathogen A vaccine and repeated vaccination against pathogen B for maintaining high levels of neutralizing antibodies.

(d) Single injection of both pathogen A and B vaccine so that memory cells can respond by producing high levels of serum antibodies.

Ans. (b)

Sol. Memory cell production plays an important role in immunity and partly it is dependent on incubation period of the pathogen. For example, influenza virus has a very short incubation period, i.e., less than 3 days. Individuals infected with influenza readily show disease symptoms by the time immune memory cells are activated. Therefore, maintaining high levels of neutralizing antibodies in free circulation by repeated immunization is necessary for effective protection against influenza. For pathogens with a longer incubation period, for example, poliovirus does not need higher levels for circulating neutralizing antibodies at the time of infection because poliovirus requires more than 3 days to start infection of central nervous system. Due to this longer incubation period, sufficient time is provided for memory B cells to respond with production of high levels of serum antibodies. Thus, the polio vaccine is also designed to induce production of high levels of immune memory cells. If an immunized individual is later exposed to the poliovrius, these memory cells will respond immediately and within 2 weeks of time produce high levels of serum antibody, which protects the individual from infection.

52. Bacteria adopt different strategies to evade host defense mechanisms. From the lists of various different mechanisms and bacterial strategies against host defense given below, select the option representing all correct pairing.

Host defense mechanisms Bacterial strategies against host defense

P. Phagocytosis 1. Changes of bacterial surface charge,

making it more positive.

Q. Release antibodies, like IgG 2. Capsular polysaccharides, such as that of

Klebsiella pneumoniae

R. Antibody-mediated agglutination 3. Release of soluble proteins like protein

A of Staphylococcus aureus.

S. Antimicrobial peptides 4. Secretion of elastase to inactivate C3a

and C5a.

(a) P–1, Q–2, R–3, S–4

(b) P–2, Q–3, R–4, S–1

(d) P–3, Q–1, R–4, S–2

Ans. (b)

Sol. The polysaccharide capsules are effective physical barriers that protect the bacteria from being killed. The fact that bacteria capsules are commonly hydrophilic and negatively charged diminishes their removal through phagocytosis, so A will match with X. Cell-bound or soluble Protein A iw produced by Staphylococcus aurcus. Protein A attaches to the Fe region of IgG and blocks the cytophilic (cell binding) domain of the Ab so it is again a bacteria strategy to defend of the Ab so it is again a bacteria strategy to defend against host. Thus, the ability of IgG to act as an opsonic factor is inhibited and opsonin-mediated ingestion of the bacteria is blocked, so b will match with Y. In Gram-positive bacteria, the major bactorial lipid component, phosphatidylglycerol can be chemically modified by bacterial enzymes to convert the lipid from anionic to caticnic or zwitterionic form. This process leads to increased levels of resistance of the bacteria against polycationic antimicrobial agents so d will match with W. C3a and C5a are anaphylatoxins that get inhibited so do not induce anti body mediated agglutination.

53. Bacterial chemotaxis response is mediated by histidine-kinase-associated receptors that activate a two-component signalling pathway which enables chemotaxis receptors to control the flagellar motors. When bacteria move towards attractant, they produce smooth swimming by rotating flagella counter-clockwise, whereas when bacteria move away from repellent, they produce increased tumbling by rotating flagella clockwise. Which of the following characteristics regarding chemotaxis receptor is not true ?

(a) The receptors are dimeric transmembrane proteins that bind specific attractants and repellents on the outside of the plasma membrane.

(b) The cytosolic tail of the receptor is stabl associated with a histidine kinase CheA via an adapter protein CheW.

(d) The binding of an attractant increases the activity of the receptor whereas binding of a repellent decreases the activity.

Ans. (d)

Sol. In bacterial chemotaxis, transmembrane chemoreceptors, the CheA histidine kinase and the CheW coupling protein assemble into signaling complexes tha allow bacteria to modulate their swimming behaviour in response to environment stimuli. In addition to membrane spanning helices, most chemoreceptors consist of periplasmic, HAMP and cytoplasmic domains.

Receptors bind attractants and repellents via the periplasmic domain. The HAMPL domain acts as a signal conversion module and transfers the input signal from the periplasmic domain to the cytoplasmic domain where CheA and CheW interact with the receptor. CheW serves as an adaptor that is essential for the formation of receptor CheW-CheA complex and CheA activation. Hence statement (A) and (B) both are correct.

Clustering primarily to the poles of E. coli, diameric chemoreceptors interact to form trimers of dimers that are approximately hexagonally arranged in the plane of the inner membrane. Hence statement (C) is correct.

Upon negative stimuli (repellent increases or attractant decreases) one stream of the phosphoryl groups is transferred to the response regulator, CheY. The phosphorylated CheY intercts directly with the flagellar motor to promote clockwise flagellar rotation that switches the default swimming behaviour of the bacterium to immediate tumbling. Since the receptor is activated among repellent binding, statement (D) is false.

54. Following are a list of extracellular matrix proteins (column A) aloing with their functional characteristics (column B) :

Column A Column B

P. Connexin 1. The chief endothelial cell proteins that are recognized by the white blood cells integrins and member of immunoglobulin (Ig) superfamily.

Q. Plasmodesmata 2. Cell surface carbohydrate binding proteins that mediate a variety of transient cell-cell adhesion interactions in the bloodstream.

R. ICAM 3. Four-pass transmembane proteins which is the major constituent of gap junctions in foming a continuous aqueous channel.

S. Selectin 4. It is the only class of intercellular junctions in plants that directly connect the cytoplasm of adjacent cells.

Which one of the following is the correct match ?

(a) P–1, Q–4, R–3, S–2

(b) P–2, Q–3, R–4, S–1

(d) P–4, Q–1, R–2, S–3

Ans. (c)

Sol. Plant cells have on class of intercellular junctions, plasmodesmata. Like gap junctions, they directly connect the cytoplasms of adjacent cells. Some members of the immunoglobulin family are calcium-independent adhesive transmmembrane glycoproteins (IgCAM), whose macro-molecules contain an immuno-globulin domain (Ig).

Selectins are cell-surface carbohydrate-binding proteins (lectins) that mediate a variety of transient cell-cell adhesion interactions in the bloodstream. Their main role, in vertebrates at least, is in governing the traffic of white blood cells into normal lymphoid organs and any inflammed tissues.

55. A western bolt analysis after treating cancer cells with a prospective anti-cancer drug is shown below.

The following assmptions were made :

P. The drug may have arrested the growth of cells at the G1 phase.

Q. The drug targeted the JAK-STAT signalling pathway.

R. The drug led to apoptosis of the cells.

S. Drug-induced apoptosis was through the extrinsic or mitochondrial-independent pathway.

Which one of the following combination is correct ?

(a) Only Q and S

(b) P, Q and R

(d) Q, R and S

Ans. (b)

Sol. Due to expression of CDK6 and CyclinD1 after treatment with drug shows that drug arrested growth of cells at G1 stage. Drug targeted JAK-STAT signalling pathway due to low expression of phosphorylated STAT3 and cleaved PARP high concentration shows drug led to apoptosis of cells.

56. Which one of the following statements regarding clonal selection hypothesis is not correct ?

(a) Mature B lympocytes bear Ig receptors on their cell surface and all receptors on a single B-cell have variable specificity for antigen.

(b) On antigen stimulation, B-cell matuers, migrates to lymphoid organs and replicates. Its clonal descendents bear the same receptor as parental B-cell and secrete antibodies with identical specificity.

(c) After immne response, more B-cells bearing receptors will remain in the host and act as memory cells of mounting enchanced secondary response.

(d) B-cells with receptors for self-antigens are deleted during embryonic development.

Ans. (a)

Sol. In the process of clonal selection, an antigen binds to a particular B cell and stimulates it to divide repeatedly into a clone of cells with the same antigenic specificity as the original parent cell. Each individual cell has only one specific type of receptor. Specificity is shown because only lymphocytes whose receptors are specific for a given epitope on an antigen will be clonally expanded and thus mobilized for an immune response. Self non-self discrimination is accomplished by the elimination during development, of lymphocytes bearing self-reactive receptors. Immunologic memory cells is a consequence of clonal selection.

57. Membrane-bound, Golgi-derived structures containing proteolytic enzymes in sperms of sea urchin are called

(a) cortical granules

(b) micromers

(d) macromeres

Ans. (c)

Sol. Cortex contains the cortical granules are homologous to the acrosomal vesicle of sperm; both the Gogi-derived organelles containing proteolytic enzymes. However, whereas each sperm contains about 15,000 cortical granules. Moreover, in addition to digestive enzymes the cortical granules contain mucopolysaccharides, adhesive glycoproteins and hyaline protein. the enzymes and mucopolysaccharides are active in preventing other sperm from entering the egg after the first sperm has entered and the hyaline and adhesive glycoproteins surround the early embryo and provide support for the cleavage stage cells, called blastomeres.

58. In case of Hydra, the major head indcer of the hypostome organizer is a set of Wnt proteins acting through the canocial -catenin pathway. What would be the result, if a transgenic Hdyra is made to globally mis-express the downstream Wnt effector -catenin?

(a) Ectopic buds will be formed all along the body axis and even on the top of the newly formed buds.

(b) Ectopic tentacles form at all levels.

(d) There would be no change observed.

Ans. (a)

Sol. In Hydra, Wnt3 as a growth factor fulfilling the criteria of the head activator, expressed locally at the tip of the head in intact Hydra, rapidly re-expressed in head-regenerating tips after amputation, and able to trigger an autocatalytic feedback loop. So, if hydra miss-express Wnt protein then ectopic buds will be formed all along body axis and even on top of newly formed buds.

59. Both TGF- and Sonic hedgehog signals play important roles in both neurulation and cell-fate patterning of the neural tube. Which of the following statements is true ?

(a) High levels of BMP specify the cells to become epidermis.

(b) Very low levels of BMP specify the cells to become epidermis.

(d) Intermediate levels of BMP do not effect the formation of neural crest cells.

Ans. (a)

Sol. The bone morphogenetic proteins (BMPs) are a group of powerful morphogens that are critical for development of the nervous system. The effects of BMP signaling on neural stem cells are myriad and dynamic, changing with each stage of development. During early development inhibition of BMP signaling differentiates neuroectoderm from ectoderm and BMP signaling helps to specify neural crest. Thus modulation of BMP signaling underlies formation of both the central and peripheral nervous systems. BMPs secreted from dorsal structures then form a gradient which helps pattern the dorsal-ventral axis of the developing spinal cord and brain.

60. In which state of Arabidopsis embryogenesis is hypophysis first observed ?

(a) Octant

(b) Dermatogen

(d) Transition

Ans. (b, c)

Sol. As a Crucifer, Arabidopsis follows the onagrad type of embryogenesis which is characterized by a high degree of reproducibility.

Stage of Arabidopsis embryogenesis.

A. Early embryo, wih a single cell in the embryo proper.

B. Early embryo wiht 2 cells in the embryo proper.

C. Octant stage : Four of eight cells in two tiers are visible. Cells of the upper and lower tier (u.t. and l.t.) of the octant will give rise to specific parts of the seedling. Together with descendants of the uppermost suspensor cell (hypophyseal cell) the eight 'octant' cells will form all the sturcutres of the seedling.

D. Dermatogen stage : A tangenital division of each of the eight 'octant' cells produces inner cells and epidermis (protoderm) cells.

E. Early Globular Stage : Divisions of the inner cells immediately after the dermatogen stage are oriented in the apical-basal dimension, endowing the embryo with a morphologically recognizable axis.

F. Triangular Stage : Now a polarized pattern of major elements is recognizable (see text) : u.t. cells have generated two symmetrically positioned cotyledon primordia and l.t. cells a radially patterned cylinder.

G. Heart Stage : Cotyledon outgrowth. Subsequently, cells between the outgrowing cotyledons initiate the primary shoot meristem.

H. Mid-torpedo Stage : Enlargement of cotyledons and hypocotyl and further eleaboration of the radial pattern.

61. Human sperms are allowed to fertilize ova having non-functional ovastacin. The following possibilites may be of significance in the fusion of these gametes :

1. The sperms will not fertilite ova.

2. The sperms will bind and penetrate the zona pellucida but will not be able to fuse with ovum membrane.

3. ZP2 will not be clipped by cortical granule protease.

4. CD9 protein of egg membrane microvilli will not be able to interact with sperm membrane proteins in the absence of ovastacin.

5. Polyspermy may occur frequently.

Which combination of statements represent the outcome of the above event ?

(a) 1 and 2

(b) 3 and 5

(d) 2 and 3

Ans. (b)

Sol. The mouse zona pellucida is composed of three glycoproteins (ZP1, ZP2 and ZP3) of which ZP2 is proteolytically cleaved after gamete fusion to prevent polyspermy. This cleavage is associated with exoctyosis and corteical granules that are peripherally located subcellular organelles unique to ovulated eggs. Ovastacin is an oocyte-specific member of the astacin family of metalloendoproteases. Using specific antiserum, ovastacin is detected in cortical granules before but no after, fertilization. Recombinant ovastacin cleaves ZP2 in native zonae pellucidae, documenting that ZP2 is a direct substrate of this metalloendoprotease. Female mice lacking ovastacin does not claeave ZP2 after fertilization and mouse sperm binds equally well to ovastacin lacking two cell embryos as they did to normal eggs.

Since non-functions ovastacin containing ova were selected, ZP2 will not be cleaved and so there will be constant chances of polyspermy. The sperm egg binding will not be affected, as the cortical reaction takes place after the membrane fusion. Once a sperm has penetrated the egg, evastacin from corticoil vesicles initiates zona hardening to prevent polyspermy.

62. Temporal expression of N-cadherin is extremely important during early development of the mammalian embryos. Accordingly, which one of the following statements about N-cadherin is true ?

(a) Injection of N-cadherin antibodies just prior to condensation of mesenchymal cells will aid cartiage formation.

(b) Presence of N-cadherin just prior to condensation wil facilitate nodule formation and development of the limb skeleton.

(c) The border between the nervous system and skin will form properly only if epidermal cells are experimentally made to express N-cadherin.

(d) Expression of N-cadherin is redundant during separation of neural and epidermal precursor cells.

Ans. (b)

Sol. As cells leave the influence of the AER, they undergo condensation and subsequent chondrogenic differentiation to form a central cartilaginous core, first forming small aggregates that then condense to form precartilaginous condensation and finally cartilage nodules. This central cartilaginous core is then shaped to become the cartilaginous anlage of the future bony skeleton.

One of the best studied models of chrondrogenesis (the processes of mesenchymal condensation and chondrogenic differentiation) is the developing limb bud, where chondrocyte differentiation is initiated by the migration and subsequent condensation of mesenchymal cells to form tightly packed aggregates. TGF-β signalling is important in thse initial stages, its key role being to up-regulate the expression of the extracellular matrix (ECM) molecule, fibronectin, the cell adhesion moelcule, N-cadherin, and the transcription factor Sox9. Fibronectin is critical for the migration of the limb bud mesenchymal cells and together with N-cadherin and Sox9, plays an important role in the aggregation process. Hence statement 1 is false, as injecting N-cadehrin antibodies will hinder with the cartilage formation. Therefore statement (B) is correct.

63. The zygote of C. elegans exhibits rotational cleavage. When the first two blastomeres formed (P1 and AB) ae experimentally separated, the following outcomes may be possible :

P. The P1 cell in isolation generates all the cells it would normally make, showing autonomous specification.

Q. The P1 cell in isolation generates all the cells it would normally make, showing conditional specification.

R. The AB cell in isolation generates a small fraction of cell types it would normally make, showing conditional specification.

S. The AB cell in isolation generates a small fraction of cell types it would normally make, showing autonomous specification.

Which one of the above combination of statements is true ?

(a) P and R

(b) Q and R

(d) P and S

Ans. (a)

Sol. C. elegans demonstrates both the conditional and autonomous modes of cell specification. Both modes can be seen if the first two blastomeres are experimentally separated. The P1 cell develops autonomously withou the presence of AB. It makes all the cells that it would normally make and the result is the posterior half of an embryo. However, the AB cell, in isolation, makes only a fraction of the cell types that it would normally make. For instance, the resulting ABa blastomere fails to make the anterior pharyngeal muscle that it would hve made in an inact embryo. Therefore is conditional and it needs the descendants of the P1 cell to interact with it.

64. Given below are some of the statements regarding regeneration :

P. The type of regeneration characteristic of mammalian liver is considered as compensatory regeneration.

Q. Regrowth of hair shaft from follicular cells exemplifies stem cells mediated regeneration.

R. Regeneration occuring through the repatterning of existign tissues with little new growth is known as morphallaxis.

S. Adult structures undergoing dedifferentiation forming a blastema, that then redifferentiates to form the lost structure, is called epimorphosis.

Choose the most appropriate combination of correct statements.

(a) S only

(b) R and S only

(d) P, Q, R and S

Ans. (d)

Sol. Regeneration is the ability of the fully developed organism to replace tissues, organs and appendages. Some amphibians like salamanders how a remarkable capacity for regeneration, being able to regenerate complete new tails and limbs as well as some internal tissues. Some insects and other arthropods can regenerate lost appendages, such as legs. Another striking case of regeneration in vertebrates is the zebrafish, which can regenerate the heart after removal of part of the ventricle.

Compensatory regeneration is regeneration in which cells divide, but maintain their differentiated function; they produce cells similar to themselves and do not form a mass of undifferentiated tissue; characteristic of mammalian liver regeneration.

Regrowth of hair shaft from follicular cells exemplifies stem cell mediated regeneration. The stem cells that regenerate a frog tail and a salamender limb have very different properties from a planarian stem cell. These multipotent tissue-specific stem cells are probably very similar to the stem cells in our own bodies that renew or repair tissues such as our skin or muscle.

Regeneration could lead to progress in the development of medical ways of repairing tissues such as the mammalin heart and the spinal cord. A distinction has been drawn between two types of regeneration. In epimorphosis, regeneration involves growth o f a new, correctly patterned structure as a limb. In morphallaxis, there is little new cell division and growth and regeneration of structure occurs mainly by the repatterning of existing tissue; regeneration of the head in Hydra is a good example.

The cellular events during regeneration may be considered under two headings, regeneration events in the blastema starting from a wound surface, commonly called epimorphosis and regeneration events in the old tissues, commonly called morphallaxis.

The mechanism involves the defifferentiation of adult structures to form an undifferentiated mass of cells that then becomes respecified. This type of regeneration is called epimorphosis and is characteristic of regenerating limbs. When an adult salamander limb is amputated, the remaining cells are able to reconstruct a complete limb, with all its differentiated cells arranged in the proper order. In other words, the new cells construct only the missing structures and no more. For example, when a wrist is amputated, the salamander forms a new wrist and no a new elbow.

65. Following are certain statements regarding apomixis in plants :

P. Apomixis cannot be used to maintain hybrid vigor over many generations in plants.

Q. In sporophytic apomixin maternal genotype is maintained.

R. There is an event of meiosis during gametophytic apomoxis and is also referred as apomeiosis.

S. In diplospory, meiosis of the megaspore mother cells is aborted, resulting in two unreduced spores, out of which one forms the female gametophyte.

Which one of the following combinations is correct ?

(a) P and Q

(b) P and R

(d) Q and S

Ans. (d)

Sol. Apomictic plants can also form embryos directly from a chromosomally unreduced female gametophyte (apomeiosis) in which the egg cell develops autonomously into an embryo by parthenogenesis (gametophytic apomixis), be means of two types, apospory and diplospory. In apospory, the embryo and endosperm develop in unreduced embryo sac in the ovule. In this case, the megaspore mother cell in the sexual ovule starts to develop but stops at some stage and one ore more somatic cells in the ovule and their nuclei start to develop, resembling megaspore mother cells. Before the mature embryo sac formation, the megaspore or young embryo sac formation, the megaspore or young embryo is aborted and replaced by developing aposporous sacs. Apospory is by far the most common mechanism in higher plants and has been reported in Beta, Brachiaria, Cenchrus, Chloris, Compositae, Erichloa, Heteropogon, Hieracium, Hyparrhenia, Hypericum, Panicum, Paspalum, Pennisetum, Poaceae, Ranunculus, Sorghum, Themeda and Urochloa.

Apomixis occurs in 500-600 species that belong to abuot 40 plant families and so is thought to have arisen multiple times during evolution. Consequently, there are also many different ways a clonal seed can be produced. In some species, such aas citrus varieties and mango (Magnifera indica), an embryo is formed directly from a sporophytic cell of tghe integuments and invades the embryo sac, competing with the sexually produced embryo. In such sporophytic apomixis, a diploid cell gives rise to the next generation, thereby maintaining the maternal genotype. Because this an occur along side sexual reproduction, sporophytic apomicts often produce a mix of clonal and sexual progeny.

66.

The following assumptions were derived from the above experiment :

P. Medium A contained bFGF and PDGF.

Q. Medium B contained retinoic acid.

R. Cells cultured in Medium B were determined to become functional neurons prior to addition of the medium.

Which one of the following combinations represents correct statements ?

(a) P and Q only

(b) P, Q and R

(d) P and R only

Ans. (a)

Sol. PDGF has been shown to mediate interactions among glial cells in vitro. More recent evidence has indicated that PDGF may also be involved in controlling communication between neurons and glial cells and among neurons. Retinoic acid is an important molecular taking part in the development and homeostasis of nervous system. Neural stem cells (NSCs) are pluripotent cells that can differentiate into three main neural cells including neuron, astrocyte and oligodendrocyte.

67. Which one of the following components is expected to be most abundant in the phloem sap of a plant ?

(a) Proteins

(b) Organic acids

(d) Phosphates

Ans. (c)

Sol. Plant phloem sap is rich in nutrients; it contains high qualitities of sugars, amino acids, organic acids, vitamins and inorganic ions. Because phloem sap is rich in nutrients and free of feeding deterrents and toxins, it is exclusively consumed by many phloem sap feeding insects which facilitates transmission of several vector-borne plant pathogens.

Sucrose is the main sugar in phloem sap and its concentration varies between species. Other sugars such as hexoses (e.g., glucose and fructose), raffinos-oligosaccharides and polyols (mannito and sorbitol) are also found in phloem sap.

68. Which one of the following mineral deficiency will first be visible in younger leaves ?

(a) Calcium

(b) Nitrogen

(d) Molybdenum

Ans. (a)

Sol. Calcium : Deficiency of calcium is first manifested as a scorching of the tip or margin of leaf. Since the element is immobile is young trees, deficiency symptoms is generally seen on shade leaves in mature trees. Calcium also plays an immense role in stability and flow of latex.

69. The CO₂ compensation point for C₃ plants is greater than C₄ plants because in C₃ plants

(a) dark respiration is higher

(b) dark respiration is lower

(d) photorespiration is absent.

Ans. (c)

Sol. C₄ cycle is more efficient than C₃ cycle. C₄ plants shows a higher rate of photosynthesis and greater increase in dry weight than the C₃ plants. These plants can carry on photosynthesis in much lower concentration while C₃ plants are better suited to high CO₂ concentration while C₃ plants and the rate of photosynthesis at optimum light intensities may be twice that of C₃ plants.

In C₃ plants, 30% of CO₂ initially fixed into carbohydrate is released via photorespiration, while in C₄ plants, photorespiration is negligible. Hence, entire carbohydrate is accumulated.

70. Which one of the following best describes the function of Caspaian bands during the translocation of nutrients and water across the root ?

(a) Block apoplastic nutrient transport

(b) Block symplastic nutrient transport

(d) Help in creating passage cells

Ans. (a)

Sol. Inside a plant, the apoplast is the space outside the plasma membrane within which material can diffuse freely. It is interrupted by the Casparian strip in roots, by air spaces between plant cells and by the plant cuticle. The Casparian strip is a band of cell wall material deposited in the radial and transverse walls of the endodermis and is chemically different from the rest of the cell wall the cell wall being made of lignin and without suberin whereas the Casparian strip is made of suberin and sometimes lignin.

71. The following are some statements regarding glycolysis :

P. Glycolysis is not regulated by pyruvate kinase.

Q. Lactate can be an end product of glycolysis.

R. Glycolysis cannot function anaerobically.

S. In eythocytes, the second site in glycolysis for ATP generation can be bypassed.

From the above, choose the combination with both incorrect statements.

(a) P and Q

(b) Q and S

(d) P and R

Ans. Glycolysis is regualted by three common enzymes phosphofructokinase-I, hexokinase and pyruvate kinase. Glycolysis can operate both aerobically and anaerobically. Lactate is by-product of anaerobic glycolysis.

Sol.

72. Following statements were made with respect to symbiotic association of rhizobia with legumes :

P. nodD is a regulatoy gene.

Q. Nod factors aer lipochitin oligosaccharides.

R. Nod factors predominantly have 4 linked N-acetyl-D-glucosamine backbone.

S. Receptors for Nod factors are protein kinases with extracellular sugar-binding Lys M domain.

Which one of the following combinations represents all correct statements ?

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (d)

Sol. Nodulation (Nod) factors are key signal moelcules that play a pivotal role during initiatio of nodule development and bacterial invasion. Nod factors consist of an oligomeric backbone of -1, 4-linked N-acetyl-D-glucosaminyl residues, N-acylated at the nonreading terminal residue and thus are lipo-chitooligosaccharides.

73. Following observations were recorded while studying physiological parameters of sorghum and wheat under similar conditions :

1. Sorghm RUBISCO exhibits relatively higher affinity for CO₂ compared to that of wheat.

2. Light saturation of net photosynthetic flux is relatively lower for sorghum compared to that of wheat.

3. Warburg effect is difficult to recod fo sorghm and could be said as "not measrable" whereas it coud be easily recorded for wheat.

4. Temperature optimum for net photosynthesis is lower for sorghum compared to that of wheat.

5. ¹³C/¹²C ratio of assimilate is relatively higher for sorghum compared to that of wheat.

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

(a) Only 1, 2 and 3

(b) Only 2, 3 and 5

(d) Only 1, 3 and 5

Ans. (d)

Sol. More light generally equates to higher levels of photosynthesis. However, as the light intensity increases, the photosynthetic rate eventually reaches a maximum point. This point where the light intensity does not increase the photosynthesis rate is called the light saturation point. C₄ plants show saturation at about 360 µl/L while C₃ responds to increased CO₂ concentration and saturation is seen only beyond 450 µl/L. Warburg effect states that if the concentration of oxygen increases the RuBisCO gets affinity for it and a process which is a waste of energy carried out known as photorespiration. This is observed in C₃ plants like wheat in which glycolate acts as a substrate and reduce the rate of photosynthesis. C₄ plants like sorghum are very efficient due to their adaptation to warmer, drier environments. They metabolize almost all the 13C from the CO₂ they take up and, therefore, C₄ plants retain more of the heavy isotope of carbon, 13C, in their tissues.

74. Following are certain statements regarding respiatory metabolism in plants :

P. Respiratoy quotient during partial breakdown of carbohydrate (alcoholic fementation) will be infinity.

Q. Respiratory quotient indirectly provides information about (i) nature of the substrate used fo respiration and (ii) the relative rate of competing respiratory processes.

R. Breakdown of organic acids in mature fruit will exhibit a respiratory quotient value of more than one since organic acids are relatively oxygen-rich compared to other common substrates.

S. Anabolic metabolism can influence respiratory quotient by removing reduction equivalents for respiration leading to decrease in oxygen uptake.

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

(a) Only P

(b) Only Q and R

(d) P, Q, R and S

Ans. (d)

Sol. The Respiratory Quotient : The ratio between CO₂ release and O₂ uptake, measured at the same time, is called the respiratory quotient (RQ). This quotient is defined as follows:

The RQ can directly provide information on the nature of the substrate used for respiration and on the relative rate of competing respiratory processes. This can be demonstrated by the balanced equations of various CO₂ producing and O₂ consuming metabolic pathways.

The value of RQ depends upon the nature of the respiratory substrate and its oxidation

1. When carbohydrates such as hexose sugars are oxidised in respiration, the value of RQ is 1 or unity because vol. of CO₂ evolved equals to the vol. of O₂ absorbed as is shown by the following equation :

C₆H₁₂O₆ + 6O₂ (Glucose) 6CO₂ + 6H₂O

RQ = Vol. of CO₂/Vol. of O₂ = 6/6 = 1 or unity.

2. When fats are the respiratory substrates, the value o RQ becomes less than one because the fats are porrer in oxygen in comparison to carbon and they required more O₂ for their oxidation which is obvious from the following equation :

2C₅₁H₉₈O₆ (Tripalmitin) + 145O₂ 102CO₂ + 98H₂O

RQ = Vol. of CO₂/Vol. of O₂ = 102/145 = 0.7 (less than one)

(Fats are oxidised in respiration usually during the germination of fatty seeds).

3. When organic acids are oxidised in respiration the value or RQ becomes more than one. It is because organic acids are rich in O₂ and require less O₂ for their oxidation e.g.,

C₄H₆O₅ (Malic acid) + 3O₂ 4CO₂ + 3H₂O

RQ = Vol. of CO₂/Vol. of O₂ = 4/3 = 1.3 (more than one)

4. Partial oxidation of carbohydrates : In some succulent plants like Opuntia, Bryophyllum etc. carbohydrates are incompletely oxidised to organic acids in dark without the evolution of CO₂ hence the value of RQ remains O.

2C₆H₁₂O₆ (Glucose) + 3O₂ 3C₄H₆O₅ (malic acid) + 3H₂O

RQ = Vol. of CO₂/Vol. of O₂ = 0/3 = 0

5. During anaerobic respiration due to the absence of O₂ the value of RQ is always very high rather infinite.

C₆H₁₂O₆ (Glucose) 2CO₂ + 2C₂H₅OH (Alcohol)

RQ = Vol. of CO₂/Vol. of O₂ = 2/0 = (infinite)

75. Sieve elements of phloem conduct sugars and other organic materials throughout the plant. The following statements were made about characteristics of sieve elements in seed plants :

P. Angiosperms contain sieve plate pores.

Q. There are no sieve plates in gymnosperms.

R. P-protein is present in all eudicots and many monocots.

S. There is no P-protein in angiosperms.

Which of the following combination is correct ?

(a) Q, R and S

(b) P, Q and R

(d) P, R and S

Ans. (d)

Sol. The incorrect statement is (d). Early stages of sieve element differentiation are characterized by the appearance of structurally distinct cytoplasmic proteins, collectively called P-protein. P-protein as a structureal entity, has been observed in sieve elements of all dicotyledons examined and in the majority of monocotyledons, although conspicuously absent in families such as the Poaccae.

76. The plant hormones, auxins and cytokinins, and their interactions play an important role in regulating apical dominance. The following figure represents an experiment related to the study of gene interactiosn that influence axillary bud outgrowth or dormancy. Q, Z and M represent genes involved in phytohormone patway.

Based on the above figure, the following statements were made :

1. 'X' is an auxin that maintains expression of 'Q' and 'Z' and represses 'M'.

2. 'Y' is a cytokinin that promotes axillary bud growth and is induced by 'M'.

3. Decapitation (removal of apex) activates 'X'.

4. 'X' is a cytokinin that represses 'M'.

Which one of the following options represents correct statement(s) ?

(a) 1 and 3 only

(b) 2 and 4 only

(d) 3 only

Ans. (c)

Sol. In many plants the growing shoot apex inhibits the outgrowth of axillary buds, a phenomenon termed 'apical dominance'. Removal of the shoot apex leads to the release of dormant axillary buds below it to form branches. Apical dominance allows plants to focus resources into the main axis of growth, while activation of dormant buds allows for recovery after damage or loss of the main shoot. It has been known for a long time that auxihn is a major signal for apcial dominance stump at the shoot apex, it can replace the growing apex in restoring branch inhibition. Auxin is mainly synthesized inthe shoot apex in young leaves and subsequently transported basically (downwards from the tip to the base) in the polar auxin transport stream (PATS), Cytokinins represent a possible second messenger for auxin signalling in regualting bud activity. Endogenous CKs can be transported acropetally (towards the shoot apex) in the xylem sap, enter axillary buds and promote their outgrowth.

77. Which one of the following is not secreted by capillary endothelium ?

(a) Prostacyclin

(b) Guanosine

(d) Nitric oxide

Ans. (b)

Sol. Endothelial cells synthesize and secrete prostacyclin, endothelin, clotting factors, nitric oxide, prostaglandins and cytokines. Vascular endothelial cells, however, lack 5-lipoxygenase and are not able to synthesize leukotrienes. Compounds not metabolized by the pulmonary capillary bed include epinephrine, dopamine, histamine, isoproterenol, angiotensin II and substance P.

78. The "Mayer waves" in the blood pressure originate due to

(a) systole and diastole of ventricle.

(b) inspiration and expiration.

(d) Bainbridge reflex.

Ans. (c)

Sol. Mayer waves are cyclic changes or waves in arterial blood pressure brought about by oscillations in baroreceptor and chemoreceptor reflex control systems. The waves are sen both in the ECG and in continuous blood pressure curves and have a frequency about 0.1 Hz (10-second waves).

Mayer waves can be defined as arterial blood pressure (AP) oscillations at frequencies slower than respiratory frequency and which show the strongest, significant coherence with efferent sympathetic nervous activity (SNA). In humans, AP oscillations which meet these properties have a characteristic frequency of aprox. 0.1 Hz; 0.3 Hz in rabbits and 0.4 Hz in rats.

79. The maturation of red blood cells does not depend on

(a) folic acid

(b) vitamin B₁₂

(d) tocopherol

Ans. (d)

Sol. Erythropoiesis is the process which produces red blood cells (erythrocytes). It is stimulated by decreased O₂ in circulations, which is detected by the kidneys, which then secrete the hormone erythropoietin.

Essential for the maturation of red blood cells are Vitamin B₁₂ (cobalamin) and Vitamin B₉ (Folic acid). Lack of either causes maturation failure in the process of erythropoiesis, which mainfests clinically as reticulocytopenia, an abnormally low amount of reticulocytes.

80. Which one of the following is not a function of angiotensin II ?

(a) Facilitates the release of norepinephrine from post-ganglionic sympathetic neurons.

(b) Increases the sensitivity of baroreflex by acting on brain.

(d) Increases the secretion of vasopressin.

Ans. (b)

Sol. Angiotensin II (Ang II) raises blood pressure (BP) by a number of actions, the most important ones being vasoconstriction, sympathetic nervous stimulation, increased aldosterone biosynthesis and renal actions.

Other Ang II actions include induction of growth, cell migration and mitosis of vascular smooth muscle cells, increased synthesis of collagen type I and III in fibroblasts, leading to thickening of the vascular wall and myocardium and fibrosis.

Function of angiotensin II :

(i) Facilitates the release of norepinephrine from post-ganglionic sympathetic neurons.

(ii) Produces anterior contraction.

(iii) Increase the secretion of vasopressin.

81. The conduction velocity of action potential in a myelinated nerve fibre was much greater than that of an nmyelinated fibre of the same diameter. The following statements were proposed to explain this observation :

1. The speed of conduction in a nerve fibre is determined by the plasma membrane resistance and axial resistance of axonal cytoplasm.

2. The electrical porperties of myelinated and unmyelinated nerve fibres are not similar.

3. The myelin sheath decreases the effective membrane resistance.

4. The magnitude of an electronic potential decreases more with distance along the axon in myelinated nerve fibres than that of unmyelinated fibres.

5. The voltage-gated Na⁺ channels are highly concentrated at the nodes of Ranvier.

Choose one of the following combinations with both incorrect statements.

(a) 1 and 2

(b) 2 and 3

(d) 4 and 5

Ans. (c)

Sol. Both membrane and axonal resistance is required to calculate the speed of conductance. Statement (a) is true.

Myelin has both passive and active properties : it increases the axon's membrane resistance and decreases the membrane capacitance. Statement (b) is true and (c) is false.

The movement of sodium ions to depolarize the membrane can only occur at the Node of Ranvier as the sodium voltage gated channels are found only at the nodes of Ranvier. Statement (e) is correct.

82. A four yea old boy was brought to hospital for weak bones in spite of sufficient intake of calcium in his diet. The attending doctor examined the functioning of the following organs :

P. Liver

Q. Kidney

R. Lung

S. Pancreas

Which one of the following options represents a combination of probable malfunctioning ogans ?

(a) P and Q

(b) Q and R

(d) P and S

Ans. (a)

Sol. Malfunctioning of liver and kidney would lead to insufficient uptake of calcium because these both organs are related to activation of vitamin D which is necessary for calcium absorption from intestine.

83. The oxygen-haemoglobin dissocation curve illustrates the elationship between pO₂ i blood and the number of O₂ molecules bound to haemoglobin. The 'S' shape of the curve has been explained in the following proposed statements :

1. The quaternary structue of haemoglobin determines its affinity tio O₂.

2. In deoxyaemoglobin, the globin units are tightly bound in a T-configuration.

3. The interactions between globin subunits ae altered when O₂ binds with deoxyhaemoglobin.

4. The affinity to O₂ in T-configuration of haemoglobin is increased.

5. In the relaxed configuration of haemoglobin, the affinity to O₂ is reduced.

Choose one of the following combinations with the incorect statements.

(a) 1 and 2

(b) 2 and 3

(d) 4 and 5

Ans. (d)

Sol. 'Co-operativity' between the component parts of haemoglobin means that oxyhaemoglobin has a substantailly different quaternary structure to deoxyhaemoglobin. As a molecule of oxygen binds to haem, it pulls the Fe2+ ion closer towards the plane of the protoporphyrin ring, slightly flattening the ring and so changing its shape. This small movement within the centre of the globin chain is transmitted to the surface of the molecule. Ionic interactions holding the four globin chains together are distracted and as they re-form in a different position, the quaternary structure is altered, which increases the oxygen binding affinity of the other globin chains. Statement (a) and (c) are true.

In fully deoxygenated haemoglobin, the molecule's quaternary structure is described as the 'T' or 'tense' form in which the crevices are small, making it difficult for oxygen to gain access to the haem. Statement (b) is true and (d) is false.

Relaxed or R state, exhibits increaed oxygen binding affinity. Statement (e) is false.

84. The changes in left atrial, left venticular and aortic pressure in a cardiac cycle are shown below in the figure :

Given below are the events of cardiac cycle (column A) associated with marked points (A, B, C, D) in the figure (column B).

Column A Column B

P. Aortic valve opens 1. D

Q. Mitral valve closes 2. B

R. Mitrral valve opens 3. A

S. Aortic valve closes 4. C

Choose the option that matches the events with marked points in the figure.

(a) P–2, Q–3, R–1, S–4

(b) P–1, Q–4, R–2, S–3

(d) P–3, Q–2, R–4, S–1

Ans. (a)

Sol. The mitral valve and tricuspid valves are known as the atrioventricular (A-V) valves and connect the atrium to the ventricles. The semilunar valves are located between the aorta and the left ventricle and between the pulmonary artery and the right ventricle.

At the beginning of the cardiac cycle, both the atria and ventricles are relaxed (diastole). Blood is flowering into the right atrium from the superior and inferior venae cavae and the coronary sinus. Blood flows into the left atrium from the fuor pulmonary veins. The two atrioventricular valves, the tricuspid and mitral valves are both open, so blood flowsunimpeded from the atria and into the ventricles. Approximately 70-80% of ventricular filling occurs by this method. The two semilunar valves, the pulmonary and aortic vlaves are closed, preventing backflow of blood into the right and left ventricles from the pulmonary trunk on the right and the aorta on the left.

85. A person recovered from a moderate degree of haemorrhagic shock. The participating physiological mechanisms in this recovery process are proposed in the following statements :

1. The decrease in arterial pressure after haemorrhage causes inhibition of sympathetic vasoconstrictor system.

2. After haemorrhage, the angiotensin II level in blood is increaed which causes increased re-abosorption of Na⁺ in renal tubules.

3. The increased secretion of vasopressin after haemorrhage increases water retention by the kidneys.

4. After haemorrhage, the reduced secretion of epinephrine and norepinephrine from adrenal medulla induces decreased peripheral resistance.

5. In haemorrhage, the central nervous system ischemic response elicits sympathetic inhibition.

Choose one of the following combinations with both the correct statements.

(a) 1 and 2

(b) 2 and 3

(d) 4 and 5

Ans. (b)

Sol. After haemorrhage, the angiotensin II level in blood is increaed which causes increased re-abosorption of Na⁺ in renal tubules. and The increased secretion of vasopressin after haemorrhage increases water retention by the kidneys would help a person to recover a moderate degree of haemorrhagic shock.

86. A healthy individual was immersed in water up to neck in an upright posture for 3 h. The plasma concentration of atrial natriuretic peptide (ANP), renin and aldosterone were measured for 5 h at 1 h intervals including the immersion period. The results are graphically presented below :

The results of this experimental condition (EC) are explained in the following proposed statements which may be correct or incorrect.

1. ANP secretion is proportional to the degree of stretch of atria.

2. The decreased plasma renin concentration in EC is due to increase in sympathetic activity.

3. The decreased aldosterone level in EC is the effect of plasma renin level.

4. The effect of gravity on the circulation is counteracted in EC.

5. The central venous pressure is decreased in EC.

Choose one of the following combinations with all correct statements.

(a) 1, 2, 3

(b) 1, 3, 4

(d) 2, 3, 4

Ans. (b)

Sol. By observing results of experiment, we can say that ANP secretion is proportional to the degree of stretch of atria.The decreased aldosterone level in EC is the effect of plasma renin level. The effect of gravity on the circulation is counteracted in EC.

87. Deamination of bases is a common chemical event that produces spontaneous mutation. Which one of the following bases will be formed by deamination of 5-methylcytosine ?

(a) Uracil

(b) Thymine

(d) Guanine

Ans. (b)

Sol. 5-methylcytosine will be deaminated to produce thymine. Spontaneous deamination of 5-methylcytosine results in thymine and ammonia. This is the most common single nucelotide mutation. In DNA, this reaction, if detected prior to passage of the replication fork, can be corrected by the enzyme thymine-DNA glycosylase, which removes the thymine base in a G/T mismatch. This leaves an abasic site that is repaired by AP endonucleases and polymerase as with uracil-DNA glycosylase.

88. The pedigree below represents the inheritance of an autosomal recessive trait. What is the probability that individual '6' is a heterozygote ?

(a) 1/4

(b) 1/2

(d) 1/3

Ans. (c)

Sol. Spontaneous deamination of 5-methylcytosine results in thymine and ammonia. This is the most common single nucelotide mutation. In DNA, this reaction, if detected prior to passage of the replication fork, can be corrected by the enzyme thymine-DNA glycosylase, which removes the thymine base in a G/T mismatch. This leaves an abasic site that is repaired by AP endonucleases and polymerase as with uracil-DNA glycosylase.

89. The allele l in Drosophila is recessive, sex-linked and lethal when homozygous or hemizygous. If a female of the genotype Ll is crossed with a male, what is the ratio of females : males in the progeny ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. As per the information given in question, the gene l is X-linked recessive gene. Since, it is lethal in homozygous condition in female and hemizygous condition in male, a female of the genotype Ll is crossed with a male to produce female : male as 2 : 1.

90. Following statements have been made about recombination in a diploid organism :

P. Recombination could be identified by genotyping parents and offspring for a pair of loci.

Q. Recombination frequency does not exceed 0.5, and therefore, 50 cM would be the maximum distance between two loci.

R. Recombination is a reciprocal process. However, a non-reciprocal exchange may cause gene conversion.

S. Occasionally non-homologous recombination happens and this functions as a source of chromosomal rearrangement.

Select the combination with all correct statements.

(a) P, Q, R

(b) P, Q, S

(d) P, R, S

Ans. (d)

Sol. Recombination could be identified by genotyping parents and offspring for a pair of loci, recombination is a reciprocal process. However, a non-recipocal exchange may cause gene conversion and occasionally non-homologous recombination happens and this functions as a source of chromosomal rearrangement.

91. A quadatic check of gene combinations and disease reaction types in a host-pathogen system where the gene-for-gene concept operates is represented below :

The following statements were made about the above genotypes :

P. AR genotype had incompatible (resistant) reactions.

Q. Ar genotype had compatible (susceptible) reactions.

R. a genotype had compatible (susceptible) reactions.

S. aR genotype had incompatible (resistant) reactions.

Choose the combination with all correct statements :

(a) P, Q and S

(b) P, Q and R

(d) P, R and S

Ans. (b)

Sol. As pre gene for gene concept, AR genotype had incompatible (resistant) reactions, Ar genotype had compatible (susceptible) reactions and a genotype had compatible (susceptible) reactions.

92. Angelman syndrome (AS) and Prader-Willi Syndrome (PWS) have very distinct symptoms. Factors responsible for the occurrence of these syndromes are given below :

1. Microdeletion of 15q11-13 in paternal chromosome.

2. Uniparental disomy of matenal chromosome 15.

3. Lack of functional maternal copy of ubiquitin ligase E3A.

4. Lack of SNURF-SNRPN transcript, which is produced only from paternal chromosome.

5. Deficiencies of small nucleolar RNAs, which are encoded from the introns of SNURF-SNRPN transcript from paternal chromosome.

Which of the following combination of answers is correct for Angelman and Prader-Willi Syndromes ?

(a) PWS-1, 3, 4; AS-2, 5

(b) PWS-2 only; AS-1, 3, 4, 5

(d) PWS-1, 2; AS-3, 4, 5

Ans. (c)

Sol. Prader-Willi Syndrome is caused when Microdeletion of 15q11-13 in paternal chromosome, Uniparental disomy of matenal chromosome 15, Lack of SNURF-SNRPN transcript, which is produced only from paternal chromosome and Deficiencies of small nucleolar RNAs, which are encoded from the introns of SNURF-SNRPN transcript from paternal chromosome whereas Angeiman syndrome arise due to Lack of functional maternal copy of ubiquitin ligase E3A.

93. Using interrupted mating, four Hfr strains were analysed for the sequence in which they transmitted a number of different genes to a F^– strain. Each Hfr strain was found to transmit its genes in a unique order as smmarized in the table (only the first five genes were scored).

Which one of the following correctly represents the gene sequence in the original strain from which the Hf strains were derived as well as the place of integration and polarity of the F plasmid ?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Only one F factor is integrated into each Hfr strain. Different Hfr strains have the F factor integrated into the chromosome at different locations and in different orientations.

Therefore, Hfr strains differ with respect to where the transfer of donor genes begins and in what order donor genes transfer. So in this question if analysiis is done it clear that hte sequence should be abdfhgjkml (Circular). Site of integration of F plasmid is between A and L (ABDFH-strain 1), between B and D (BALMK-strain 2) betweem M and L (MKJGH-strain 3) and between F and D (FHGJK-strain 4).

94. Several mutants (1-4) are isolated, all of which require compound E for growth. The compounds A to D in the biosynthetic pathway to E are known, but thei order in the pathway is not known. Each compound is tested for its ability to support the growth of each mutant (1-4). In the following table, a plus sign indicates growth and a minus sign indicates no growth. What is the order of the compounds (A to E) in the pathway ?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Growth will be supported by a particular compound if it is later in the pathway than the enzymatic step blocked in the mutant. Restated the more mutants a compound supports, the later in the pathway growth must be. In this example, compound E supports growth of all mutants and can be considered the end product of the pathway. Alternatively, compound A does not support the growth of any mutant and can be considered the starting substrate for the pathway.

95. Following is the pictue of an inversion heterozygote undergoing a single crossing over event :

The following statements are given towards explaining the consequences at the end of meiosis :

1. The resultant two chromosomes will have deletions and duplications.

2. A dicentric and an acentric chromosome will be formed.

3. The inversion does not alow crossing over to occur, so even if a crossing over is initiated, it will fail to occur.

4. The crossing over is considered suppressed by inversion as the acentric chromosome will not segregate normally.

5. All the gametes formed with cross-over chromatids at the end of meiosis will be non-viable as they cary large deletion or duplication.

6. The gametes having non-crossover (parental) chromatid will survive.

Which combination of statements is correct ?

(a) 2 and 5

(b) 1 and 3

(d) 1, 5 and 6

Ans. (d)

Sol. As per the picture given in question, The resultant two chromosomes will have deletions and duplications, All the gametes formed with cross-over chromatids at the end of meiosis will be non-viable as they cary large deletion or duplication and The gametes having non-crossover (parental) chromatid will survive.

96. Polymorphic DNA sequence are used for molecular identification. Short tandem repeats (STRs) and Single Nucleotide Polymorpism (SNPs) are used as polymorphic markers. The table below summarizes the status of autosomal SNP, autosomal STR, mitochondrial SNP, Y-linked STR for four individuals related to each other, representing parents and their two children. Based on the below data, identify the individuals representing the two parents.

(a) Individuals A and D

(b) Individuals A and C

(d) Individuals C and D

Ans. (b)

Sol. Since Y linked STR is present in A and B, both are male and one among them is the male parent. If we observe the autosomal STR for D, it must have inherited 13 from A and 14 from C, so definitely C should be female parent. The autosomal SNP for B confirms C as the female parent as it must have inherited C from A and G from C. D has inherited C from A as well as C.

97. Given below are some statements related to lower eumetazoans. Select the incorrect statement.

(a) Ctenophores are diploblastic with radial symmetry.

(b) Placozoans, with weakly differentiated tissue layers aer not diploblasts.

(d) Hydrozoans, a Cnidarian class, often have colonial polyps in their life cycle.

Ans. (a, b)

Sol. The ctenophores are diploblastic animals with a modified radial or biradial symmetry. Their tentacles (see Figure 1) have adhesive structures called colloblasts that do not seem to be homologous to the nematocysts (=cnidae) of the Cnidaria.

98. Which of the following plastid coding region(s) have been recommended as a core barcode by Plant Working Group of the consortium for the Barcode of Life ?

(a) Col and rbcL

(b) rbcL and matK

(d) rbcL only

Ans. (b)

Sol. Based on assessments of recoverability, sequence quality and levels of species discrimination, a core two-locus combination or rbcL and matK as the plant barcode was recommended. This combination was shown to successfully discriminate among 907 samples from 550 species at the species level with a probability of 72%. The group admits that the two-locus barcode is far away from perfection due to the limited identification rate and thus further research for other appropriate candidates is necessary. Additional combinations of noncoding and coding plastid regions have been tested for barcoding purposes.

99. Based on the type of excretion of nitrogenous waste, animals can be categorized as ammonotelic, ureotelic and uricotelic. Given below are combinatiosn of groups of organisms an type of excretion. Select the correct combination.

(a) Poriferans, adult ampibians, cartilaginous fishes are ammonotelic.

(b) Ascaris, cockroaches, prawn are uricotelic.

(d) Humans, sharks and aquatic anurans are uretelic.

Ans.

Sol. Ammonotelism :

1. Chief nitrogenous waste is ammonia.

2. It easily and rapidly dissolves in liver cells by deamination of amino acids.

3. Ammonia is formed in liver cells by deamination of amino acids.

4. Ammonia is very toxic.

Examples : Protozoans (e.g., Amoeba, Paramecium) Sponges (i.g., Sycon) Cnidarians (e.g., Hydra), Liverfluke, Tapeworm, Ascaris, Nereis, Earthworm, Leech, Prawn, Pila, Bony fishes, Salamanders, Amphibian tadpoles (e.g., tadpole of frog) and crocodiles.

100. Basal angiosperms are not represented by the members of

(a) Chloranthales

(b) Nymphaeales

(d) Amborellales

Ans.

Sol. The basal angiosperms consist of three sequential cladogram branches, in what has been termed the ANITA grade of angiosperm evolution. These branches are : first, the Amborellales, second, the Nymphaealse and third, the Austrobaileyales, based upon molecular phylogenies. Although molecular evidence places the Amborellales at the very base of the angiosperm clade, the oldest fossil flowers found thus far belong to the Nymphaeales, on branch up from the base of the clade.

101. Given below is a list of bacteria either functioning as methanogens or methanotophs :

P. Methanobacterium sp

Q. Methanococcus sp

R. Methylomonas sp

S. Methylosinus sp

Which of the following options classifies the above list correctly ?

(a) Methanogen-P; Methanotrophs-Q, R, S

(b) Methanogen-P, Q, R; Methanotrophs-S

(d) Methanogen-P, S; Methanotrophs-Q, R

Ans. (c)

Sol. Microorganisms that create methane as a byproduct of their metabolism are known as methanogens. During sewage treatment, they are typically detected in anaerobic sludge. These bacteria can be found in the rumen as well (a part of the stomach of cattle). Example - Methanobacterium, Methanococcus. Methanotrophs can be found all over the world, wherever there is methane for them to eat. They can live in low oxygen environments and high oxygen environments, and in extremely hot or extremely cold regions too. Because methane is a significant greenhouse gas. Example - Methylomonas, methylosinus.

102. The table given below provides a list of groups of Arthropods (P-S) and some features (1-5).

P. Onychophoans 1. includes insects

Q. Trilobites 2. have cephalothoax and often princer-like appendage

R. Hexapods 3. marine arthoropods that disappeared in the Permian extinction

S. Cheliceates 4. are considered related to arthropods but have unjointed appendages

5. only arthoropods group without antennae

Which one of the following options represents the correct match between the arthopod groups with these features ?

(a) P–4, Q–3, R–1, S–5

(b) P–1, Q–3, R–4, S–2

(d) P–4, Q–3, R–5, S–2

Ans. (a)

Sol. Lobopod, collective name for two phyla of animals : Onychophora and Tardigrada. Phyla Onychophora and Tardigrada have long been considered separate from their close arthropod relatives. Both groups have similar paired locomotory appendages, called lobopodia (or lobopods); a body cavity (hemocoel); a cuticle (skin) secreted by surface cells and shed periodically (molting); a gut that is usually a straight tube; and separate sexes and gonads.

Tribites are a well known fossil group of extinct marine arthropods that form the class Trilobita. Trilobites finally disappeared in the mass extinction at the end of the Permian about 250 million years ago. the trilobites were among the most successful of all early animals, roaming the oceans for over 270 million years.

The subphyllum Hexapoda (from the Greek for six legs) constitutes the larges number of species of arthropods and includes the insects as well as three much smaller groups of wingless arthropods : Collembola, Protura and Diplura (all of these were once considered insects). Hexapods have bodies ranging in length from 0.5 mm to over 300 mmm which are divided into an anterior head, thorax and posterior abdomen.

The subphylum Chelicerata constitutes one of the major subdivisions of the phylum Arthropoda. It contains the horseshoe crabs, sea spiders, arachnids (including scorpions and spiders) and several extinct lineages such as the eurypterids.

103. In the following table, a list of threat categories and animals of India is given in an alphabetical order :

Animals Threat category

P. Bengal florican 1. Critically endangered

Q. Ganga river dolphin 2. Endangered

R. Indian Rhinoceros 3. Vulnerable

S. Indian Vulture

Which of the following options show correct combination of animals and their threat category as per Red Data list of IUCN ?

(a) P–1, Q–1, R–2, S–3

(b) P–1, Q–2, R–3, S–1

(d) P–3, Q–1, R–2, S–2

Ans. (b)

Sol. The bengal florican (Houbaropsis bengalensis), also called Bengal bustard, is a bustard species native to the Indian subcontinent, Cambodia and Vietnam. It is listed as Critically Endangered on the IUCN Red List because fewer than 1,000 individuals were estimated to be alive as of 2017.

The Ganges River Dolphin is one of the world's most endangered freshwater mammals.

The Indian rhinoceros (Rhinoceros unicornis), also alled the greater one-horned rhinoceros and great Indian rhinoceros, is a rhinoceros native to the Indian subcontinent. It is listed as Vulnerable on the IUCN Red List, as populations are fragmented and restricted to less than 20,000 km² (7,700 sq mi). Moreover, the extent and quality of the rhino's most important habitat, alluvial grassland and riverine forest, is considered to be in decline due to human and livestock encroachment. As of 2008, a total of 2,575 mature individuals were estimated to live in the wild.

The Indian vulture in an Old World vulture native to India, Pakistan and Nepal. It has been listed as Critically Endangered on the IUCN Red List since 2002, as the population severely declined. Indian vultures died of renal failure caused by diclofenace poisoning. It breeds mainly on hilly crags in central and peninsular India.

104. Following table shows presence (+) and absence (–) of selected distinguishing characters of different plant taxa :

Base on the above, which of the following shows correct identity of taxa P, Q, R and S ?

(a) P-Hornworts, Q-Oaks, R-Ferns, S-Pines

(b) P-Ferns, Q-Oaks, R-Hornworts, S-Pines

(d) P-Ferns, Q-Pines, R-Hornworts, S-Oaks

Ans. (b)

Sol. Hornworts are a group of non-vascular Embryophytes (land plants) constituting the division Anthocerotophyta. They did not have wood, flower and seeds. Oaks have a ring-porous xylem anatomy, allowing rapid sap movement in large diameter, early-wood vessels when soil water is plentiful. They have all of the characters. Pine is a non-flowering tree i.e. it does not produce flowers. It reproduces from seeds that grow within pine cones instead of flowers, but it have rest all three characters. Ferns are plants that do not have flowers. Ferns generally reproduce by producing spores. Similar to flowering plants, ferns have roots, stems and leaves.

105. Following table shows an alphabetical list of certain domesticated crops and places of origin :

Crop Place of origin

P. Barley 1. China

Q. Maize 2. Fertile Crescent

R. Mung Bean 3. India

S. Rice 4. Southern Mexico

T. Wheat

Based on the above, which one of the following options represent the correct match between crops and their place of origin ?

(a) P–3; Q–4; R–1; S–2; T–2

(b) P–2; Q–4; R–1, 3; S–1; T–2

(d) P–2; Q–4; R–3; S–1, 3; T–2

Ans. (d)

Sol. Barley and Wheat has been continuously cultivated for >8,000 years in southern Central Asia, east of the Fertile Crescent. Mexico has about 1.5 million corn farmers. Green gram or moong bean is one of the main pulse crops of India. The total area under its cultivation in India is 3 million hectares with average production of 1 million tonnes.

106. A list of floral fomulae and plant families are given in the following table. Which of the following options most appropriately matches given plant families with their representative floral formulae ?

(a) P–4, Q–2, R–1, S–3

(b) P–4, Q–3, R–1, S–2

(d) P–1, Q–3, R–2, S–4

Ans. (b)

Sol. Floral formual is the representation of the structure of a flower with the use of numbers, letters and various symbols in a compact form. A floral formula provides information about the number of part of a flower, their arrangement and their relation with one another, here, K represents Calyx, C represents Corola, A represent Androexium, G represents Gynoecium.

Solanaceae : The flower is actinomorphic bisexual, calyx-sepals 5 in number-gamosepalous, corollapetals 5 in number-gamopetalous, androecium stamens 5-monodelphous gynoecium-bicarpellary-syncarpous-superior ovary.

Liliaceae : Actinomorphic, bisexual, calyx-sepals 6-gamosepalous, corolla-petals-6-gamopetalous, androexium-stamens 6-epiphyllous, gynoecium-carpels 3-syncarpous-superior ovary.

Brassicaceae : Actinomorphic, bisexual, calyx-sepals 4-corolla-petals-4-androexium-stamens 6-gynoecium-carpels 2-syncarpous-superior ovary.

Fabaceae : Zygomorphic, calyx-sepals 5-gamosepalous-corolla-petals-5-androecium-stamens-10-gynoecium-carpels 1-superior ovary.

107. Which of the following is the correct increasing order for the daily net primary productivity (NPP) per unit leaf area in different ecosystems ?

(a) Deserts < Temperate forests < Tropical forests

(b) Deserts < Tropical forests < Temperate forests

(d) Tropical forests < Temperate forests < Deserts

Ans. (c)

Sol. Net Primarily Production, NPP, is the net amount of primarily production after the costs of plant respiration are included. Therefore

NPP = GPP – R

Where,

Gross Primary Production, GPP, is the total amount of CO₂ that is fixed by the plant in photosynthesis.

Respiration, R, is the amount of CO₂ that is lost from an organism or system from metabolic activity. Respiration can be further divided into components that reflect the source of the CO₂.

108. The equilibrium model of island biogeography proposed by MacArthur and Wilson assumes that the number of species on an island represents a balance between

(a) resource consumption rate and predation rate

(b) birth rate and death rate

(d) speciation rate and hybridization rate

Ans. (c)

Sol. In MacArthur and Wilson's model, the equilibrium number of species on an island depends on the balance between rate of colonization and the rate of extinction. MacArthur and Wilson argued that the colonization rate should decrease as species number increases. Initially, any species reaching the island would be a new to the island and therefore represent a colonization event. Eventually, the colonization rate should decrease until all the species found on the mainland are also found on the island. If a species goes locally extinct on the island, then that species may again eventually recolonize the island, contributing again to the colonization rate. The extinction rate, however, should increase as species number increases. Initialy, extinction rates are zero because there are no species to go extinct. However, as the number of species on the island increases, the number of species that can potentially go extinct will also increase, thereby increasinig the possible extinction rate.

109. A population grows according to the logistic growth equation, where is the rate of population growth, r is the intrinsic rate of increase, N is population size and K is the carrying capacity of the environment. According to this equation, population growth rate is maximum at

(a) K/4

(b) K/2

(d) 2K

Ans. (b)

Sol. Population growth is maximum, when

110. What is the significance of upwelling zone for marine ecosystems ?

(a) It is responsible for uniformity of temperature in ocean to support the marine life.

(b) It brings nutrients from deeper zones to relatively nutrient poor ocean surface thus increasing marine productivity.

(d) It helps in circulating decomposers from the bottom of ocean to surface for proper decomposition of dead material on the surface.

Ans. (b)

Sol. Winds blowing across the ocean surface push water away. Water then rises up from beneath the surface to replace the water that was pushed away. This process is known as "upwelling". Water that rises to the surface as a result of upwelling is typically colder and is rich in nutrients. These nutrients "fertilize" surface waters, meaning that these surface waters often have high Biological productivity. Therefore, good fishing grounds are typically found where upwelling is common.

111. Given below in column A are schematic representations of three types of pairwise species interactions and the name of some interactions are in column B. Select the best match for interaction between column A and B in each schematic figure.

(a) P–3, Q–2, R–4

(b) P–4, Q–2, R–3

(d) P–3, Q–1, R–2

Ans. (c)

Sol. Five different sorts of indirect interactions involving three or four different species populations are depicted in Figure. Pointed arrows indicate beneficial effects whereas circle headed "arrows" depict detrimental interactions. Solid arrows are direct interactions, dashed arrows represent indirect interactions. Two consumers sharing a common prey may compete indirectly via classical exploitation copetition (resource depression). Two prey species may appear to compete because if either increases, a shared predator also increases, which operates to the detriment of the other prey population - Holt called this apparent competition. Three species populations at three different trophic levels result in what has been termed a food chain mutualism (such vertical interactions have also been called cascading effects or trophic cascades).

112. Grassland plots with varying number of grass species were cultivated for 10 years. At the end of the experiment, total plant cover was measured. Soil nitrogen was also measured to assess its utilization by plants. The relationships are shown in the following plots.

Which one of the following inferences can be drawn from the above experiment ?

(a) Grasses in plots with lower species richness enriched soil nitrogen, thereby increasing the plant cover.

(b) Plots with greater species richness showed greater stability and more efficient soil nitrogen utilization.

(c) Plots with greater species richness utilized nitrogen more efficiently, but would not show increased net primary production.

(d) No correlation can be drawn between species richness, community productivity and nitrogen utilization.

Ans. (b)

Sol. Plots with greater species richness showed greater stability and more efficient soil nitrogen utilization. After that long-term Nitrogen addition leads to Loss of Species Richness due to litter accumulation and soil acidification.

113. Complete the follownig sentences with the most appropriate option.

Global analysis of a large number of plant species traits showed that with increases in life lifespan,

(a) specific leaf area increases whereas leaf nitrogen and net photosynthesis rate decrease.

(b) specific leaf area, leaf nitrogen and net photosynthesis rate increase.

(d) specific leaf area decreases whereas leaf nitrogen and net photosynthesis rate increase.

Ans. (c)

Sol. According to Wilson-Mac Arthur's Theory of Island Biogeography the number of species on an island should increase with its size and area where as the number of species should decrease with increasing distance of the island from the source pool.

So for the given graphical representation (a) indicates the species richness in the fragment areas increases with increase in size of the area. This correlates with the graph (d) which shoes the decrease of specise richness with increase in distance from the main land. So the combination of (a) and (d) correctly represent the prediction model.

114. Forest fragments in an agricultural landscape can be viewed as islands of habitat in an ocean of non-habitat MacArth and Wilson's islands biogeography model can be used to predict patterns of species richness in these forest fragments which are represented in the graph below :

Which one of the following combinations of the graphs correctly represents predictions from the model ?

(a) P and R

(b) P and S

(d) Q and S

Ans. (b)

Sol. Species richness is directly proportional to area and increases exponentially with increase in area. The theory that species richness and individual abundance will decrease with reduced patch size. So graph P and S is correct.

115. In order to estimate population size of a fish species in a lake, a researcher captures 100 fish from the lake and marks them with coloured tags. A week later, the researcher returns to the lake and catches 150 fish of the same species and finds that 25 of them are previously tagged ones. Assuming no immigration or emigration occured, the total population size of the fish species in the lake will be

(a) 17

(b) 38

(d) 860

Ans. (c)

Sol.

n₁ = total marked after catch 1

n₂ = total marked after catch 2

n₃ = total caught in catch 2 but marked in catch 1

116. Given below is a graphical representation of plant life histories based on Grime's model in which stress, disturbance and competition are the important factos. Which of the following options correctly represents P, Q and R, respectively in the figure below ?

(a) Perennial herbs, trees and shrubs, annual plants.

(b) Annual plants, perennial herbs, trees and shrubs.

(d) Trees and shrubs, preennial herbs, annual plants.

Ans. (b)

Sol. Grime identified two factor gradients, broadly categorized as disturbance and stress, which limit plant biomass. Stresses include factors such as the availability of water, nutrients, and light, along with growth-inhibiting influences like temperature and toxins. Conversely, disturbance encompasses herbivory, pathogens, anthropogenic interactions, fire, wind, etc. Emerging from high and low combinations of stress and disturbance are three life strategies commonly used to categorize plants based on environment: (1) C-competitors, (2) S-stress tolerators, and (3) R-ruderals. Annual plants are found in low competition and high disturbance zone. Perennial herbs are found in moderate environment. Trees and shrubs found in medium stress conditions.

117. Following table shows attributes of selected species P, Q, R and S :

Based on the above information, which of the following is most likely to become invasive if climate matches between its site of origin and site of colonization ?

(a) Species P and S

(b) Species P only

(d) Species Q and R

Ans. (b)

Sol. Common characteristics of IAS include rapid reproduction and growth, high dispersal ability, phenotypic plasticity (ability to adapt physiologically to new conditions), and ability to survive on various food types and in a wide range of environmental conditions. They also show high competitive ability and are r-selected.

118. In a population of 2000 individals of a plant species, genetic difference at a single locus leads to different flower colours. The alleles are incompletely dominant. The population has 100 individuals with the genotype rr (white flowers), 800 individuals with the genotypes Rr (pink flower) and the remaining have genotype RR (red flowers). What is the frequency of the allele in the population ?

(a) 0.25

(b) 0.50

(d) 1.00

Ans. (a)

Sol.

119. Competition for mates and variance in fitness is higher among females than among males in which of the following animal mating systems ?

(a) Monogamy

(b) Polygyny

(d) Sequential monogamy

Ans. (c)

Sol. Polyandry : Females have more than one male as a mate during a breeding season. While polyandry is now recognized as being widespread, its origin and maintenance remain enigmatic due to variation between the sexes in the costs and benefits or reproduction.

These patterns will be opposite in sex-role-reversed specie, where males become the limiting sex and there is greater variance in success in competition for mates among females than males. Additionally, whereas sex is costly for both sexes, females appear to bear the sharp end of this expense. Aside from the energy and time expenditure required to engage energy and time expenditure required to engage in copulation, polyandrous females, may for example, experience greater predation, be injured and have reduced longevity as a consequences of the receipt of caustic male seminal fluids.

120. Which one of the following will have the least impact on allele frequencies in small populations ?

(a) Inbreeding

(b) Random mating

(d) Outbreeding

Ans. (b)

Sol. We know that in the essence of Hardy-Weinberg equilibrium, if alleles in the gamete pool exactly mirror those in the parent generation and if they meet up randomly (in an infinitely large numer of events), there is no reason—in fact, no way—for allele and genotype frequencies to change from one generation to the next.

In the absence of other factors, you can imagine this process repeating over and over, generation after generation, keeping allele and genotype frequencies the same. Since evolution is a change in allele frequencies in a population over generations, a population in Hardy-Weinberg equilibrium is, by definition not evolving.

121. Consider a single locus with 2 alleles which are at Hardy-Weinberg equilibrim. If the frequency of one of the homozygous genotypes is 0.64, what is the frequency of heterozygotes in the population ?

(a) 0.16

(b) 0.20

(d) 0.36

Ans. (c)

Sol. Consider the case of a gene with alleles, A and a. The frequency of the A allele is p and the frequency of the a allele is q.

We know that p + q = 1, because only these two alleles are present, so the sum of the fractions must be 1. We then know that q = 1 – p and p = 1 – q, by simple math. A homozygous dominant individual, AA, has two alleles both A. A heterozygous individual has two alleles A and a. If we go through each member of the population and count all of the A alleles and a alleles, then the fraction of A alleles gives p and the fraction of the a alleles given q.

If a population is in Hardy-Weinberg equilibrium, knowing the value of either p or q allows us to determine the fraction of individuals who are homozygous and heterozygous in the population.

For instance, if p = 0.8 then q = 1 – p = 0.2.

Furthermore, because we are at equilibrium, we can view the entire gene pool as randomly mixing each generation.

Therefore, we determine the frequency of the AA (homozygous dominant) genotype in the popualtion by calculating the probability of pulling two would be p² = 0.8 × 0.8 = 0.64 or 64%.

Similarly, the frequency of the aa (homozygous recessive) genotype is q² = 0.2 × 0.2 = 0.04, or 4%. the frequency of heterozygotes is a little more subtle because there are two genotypes for being heterozygote, Aa and aA. That is, we can first pull an A allele and then an a allele out of the gene pool, or we can pull an a allele and then an A allele (in other words, the sperm can be A and the ovum a or the sperm can be a and the ovum A).

Therefore, the frequency of heterozygotes in the population is 2pq = 2 × 0.2 = 0.32 or 32%.

122. An autosomal recessive condition affects 1 newborn in 10,000 in a random mating population without any disruptive acting force. What is the approximate expected frequency of carriers in this population ?

(a) 1 in 1000 newborns

(b) 1 in 500 newborns

(d) 1 in 50 newborns

Ans. (d)

Sol. If we assume that the population's in H-W equilirbium, then the frequency of individuals with the diseased genotype is the square of the frequency of the recessive allele. In other words, freq (aa) = q2. Freq (aa) = 1/10000 = 0.0001 and the square root of that is 0.01, which is q, the frequency of the recessive allele. The frequency of the normal allele is p, equal to 1 – q, so p = 0.99. The frequency of carriers to be 2pq, which is 0.0198. Sio 1.98% or 198 people out of 10000, should be carriers of autosomal diesase, if the population is in H-W or 1 amongst 50.

123. An alphabetical list of tropical rainforest mammals from South America and Africa is given below :

South America Africa

P. Agouti 1. Bosman's Potto

Q. Giant Armadillo 2. Chevrotain

R. Paca 3. Pangolin

S. Three toed sloth 4. Royal Antelope

Pair the species in the list of demonstrate the concept of convergent evolution between the two continents.

(a) P–4, Q–3, R–1, S–2

(b) P–1, Q–3, R–4, S–2

(d) P–4, Q–3, R–2, S–1

Ans. (d)

Sol. Agouti-Royal antelope

Giant armadillo-Pangolin

Paca-Chevrotain

Three toad sloth-Bosman's potto.

124. A study tested the importance of learning mechanisms in the deveopment of anti-predator escape responses in tadpoles of a frog species. Tadpoles hatched from eggs in the lab wee kept individually either with predator chemical cues (Prior exposure) or without chemical cues (naive) for 1 week. These individuals were tested for thier escape responses when exposed to a live predator. They were tested either alone or togethe with 3 older experienced tadpoles. The graph below shows the escape response of the test individuals in the four different treatments.

Some of the inferences drawn are given below :

1. Prior exposure to predator cues is necessary for the development of escape response.

2. Prior exposure to predator cues positively influences the development of escape response.

3. The presence of older experienced individuals is necessary for the development of escape response.

4. The presence of older experience individuals positively influences the development of escape response.

5. An individual with prior exposure and with older and experienced individuals showed the strongest escape response.

Which one of the following combination of statements represents the correct inference from the experiment ?

(a) 1 and 3

(b) 2 and 4

(d) 1, 3 and 5

Ans. (b)

Sol. Prior exposure to predator cues positively influences the development of escape response. The presence of older experience individuals positively influences the development of escape response.

125. Break shapes in birds has evolved in response to their diet. The table listing bird species and food type is given below

Bird species Food type

P. Barn swallow 1. Fruits

Q. Great hornbill 2. Insects

R. House sparrow 3. Nectar

S. Purple sunbird 4. Seeds

Match the bird species shown above to theer main food resource.

(a) P–4, Q–1, R–3, S–2

(b) P–2, Q–4, R–1, S–3

(d) P–2, Q–1, R–4, S–3

Ans. (d)

Sol. (a) Barn swallows are (ii) insectivorse. Flies grasshoppers, crickets, dragonflies, beetles, moths and other flying insects make up 99% of their diet.

(b) Great hornbills are predominantly (i) frugivores that feed on both lipid-rich and sugar-rich fruits. Great hornbills are arboreal and live mainly in wet, tall, evergreen.

(d) Purple sunbird is a small subird. Like other subirds they feed mainly on nectar (iii). They have a fast and direct flight and can take nectar by hovering like a hummingbird but often perch at the base of flowers.

126. Four drugs (P, Q, R, S) were used disrupt a biological rhythm in experimental animals. The changes in the pattern of the biological rhythm as compared to untreated are shown below. The black line represents the biological rhythm of the untreated and gray line represents that of the treated animals. Which of the following interpretations from the below experiment is incorrect ?

(a) Drug P can be used to reduce the period length of the rhythm.

(b) Drug Q can be sed for sustained lowering of amplitude of the rhythm without changing its period.

(d) Drug S can be used to reduce the robustness and dampen out the rhythm.

Ans. (c)

Sol. Drug used lowered the amplitude and period of the rhythm.

127. The phylogentic tree below shows evolutionay relationships among 8 species. Males of these species are either blue (b) or red (r) in colour, the colour being indicated next to each species name. Based on the principle of parsimony, which of the following statements best represents the evolution of male body colour in the set of species ?

(a) The most recent common ancestor of all 8 species was blue; red evolved indepedently 5 times.

(b) The most recent common ancestor of all 8 species was blue; red evolved indepedently 4 times.

(d) The most recent common ancestor of all 8 species was blue; red evolved indepedently 2 times.

Ans. (d)

Sol. According to the principle of maximum parsimony, we should first investigate the simplest explanation that is consistent with the facts. (The parsimony principle is also called "Occam's razor" after William of Occam, a 14th-century English philospher who advocated this minimalist problem-solving approach of "shaving away" unnecessary complications). In the case of trees based on morphology, the most parsimonious tree requires the fewest evolutionary events, as measured by the origin of shared derived morphological characters. For phylogenies based on DNA, the most parsimonious tree requires the fewest base change. Here only two changes in lineage of A and common linearge of B and D for development of blue colour from the ancestral red can explain the tree.

128. Given below is a marker profile for two parental lines (P1 and P2) and their derived F1 progeny. The marker that is represented in the above figure is most likely to be

(a) RFLP or SSR

(b) SSR only

(d) RAPD only

Ans. (d)

Sol. The DNA vaccine in the form of gene construct as above is delivered into human body in one of the following forms : (i) direct DNA delivery by microinjection or particle gun; (ii) vector mediated gene transfer in vivo on in vitro when DNA is delivered in vitro, the genetically engineered cells are introduced into human body, so that immunogenic protein is synthesized there are immunity is invoked.

Advantages and disadvantages of DNA vaccine : Many of the disadvantages often encountered with other types osf vaccines, discussed earlier in this chapter, are largely overcome through the use of DNA vaccines.

They provide life long protection and there is no risk of vaccine causing the disease as is the case with vaccines involving dead organisms or attenuated form of virus. The DNA vaccines are also cheap and easy to prepare fast in action and safter to use. The only disadvantage, however, is that the gene construct that is introduced may get integrated with the genomic DNA of the vaccinated individual (animal or human) and may cause cancer. This factor alone calls for extreme precaution that needs to be exercised while using DNA vaccines for control of infectious deseases.

129. Which one of the follownig statements regading restriction/modifying enzymes used in recombinant DNA technology is correct ?

(a) Endonucleases remove nucleotides, one at a time, from the ends of a sequence.

(b) Type II class of restriction enzymes do not recognise palindromic sequence.

(d) Type II class of restriction enzymes can generate either "sticky" (staggered) or "blunt" ends.

Ans. (d)

Sol. Restriction Endonuclease (Enzyme) : A class of endonucleases that cleaves DNA after recognizing a specific sequence, e.g., BamHI (5'GGATCC3'), EcoRI (5'GAATTC3') and HindIII (5'AAGCTT3'). There are three types of restriction endonuclease enzymes :

Type I : Cuts non-specifically a distance greater than 1000 bp from its recognition sequence and contains both restriction fand methylation activities.

Type II : Cuts at or near a short and often palindromic (q.v.), recognition sequence. A separate enzyme methylates the same recognition sequence. They may make the cuts in the two DNA strands exactly opposite one another and generate blunt ends or they may make staggered cuts to generate sticky ends. The type II restriction enzymes are the ones commonly exploited in recombinant DNA technology.

Type III : Cuts 24-26 bp downstream from a short, asymmetrical recognition sequence. Requires ATP and contains both restriction and methylation activities.

130. Which one of the following is used as a souce of excitation in a confocal microscope ?

(a) Lasers

(b) Electron beam

(d) Masers

Ans. (a)

Sol. With the exception of bioluminescence methods any spatial imaging of ion dynamics requires the use of a conventional fluorescence microscope or of a laser confocal microscope. Fluorescence microscopes use incident illumination (epi-fluorescence) to deliver ligh of a defined wavelength to the speciman; the excitation source is usually a xenon/mercury lamp and the choice of wavelength is done by attached filters. In laser confocal microscopes, the light source is a laser emitting at defined wavelengths; through variable pinhole apertures, the light detected is confined to the focal plane.

131. The following cassette was designed to create estrogen receptor knock-out mice :

What would ensure that proper recombination has taken palce ?

(a) Cells survive when cultued in presence of only G418.

(b) Cells survive when cultured in presence of G418 followed by ganciclovir.

(d) Cells survive when cultured with G418 and die when cultuerd with ganciclovir.

Ans. (b)

Sol. Cells survive in presence of both G418 and gancicliovir, because tk gene is not expressed so not giving sensitivity to thymidine kinase and G418 resistant gene is expressed.

132. Detergents at low concentration generally do not denature proteins and are thus used for extracting proteins in their folded and active form. For isolation of 'Porins', an E. coli membrane protein, which one of the following purification approaches will be most appropriate ?

(a) Use of low concentration of non-ionic detergent without salt.

(b) Use of low concentration of ionic detergent.

(d) Use of salt solution containing ionic detergent.

Ans. (c)

Sol. Ionic detergent like SDS isolate protein but denature it rather the membrance proteins from miscelle with non-inoic detergent and helps us to seprate protein without denaturating it.

133. DNA vaccines offer several advantages over other existing vaccine approaches. Which one of the following statements related to DNA vaccine is not correct ?

(a) The immune response is directed to the antigen encoded by the DNA and able to induce both humoral and cell mediated immunity.

(b) DNA vaccine can induce prolonged expression of the antigen, enhancing the induction of immunological theory.

(c) DNA vaccine could remain and potent for long time without refrigeration, eliminating the challenges of storage and transportation.

(d) DNA vaccine construct can be engineered to carry several antigens to infect host and replicate in neuronal cells.

Ans. (d)

Sol. As proof of the principle of DNA vaccination, immune responses in animals have been obtained using genes from a variety of infectious agents, including influenza virus, hepatitis B virus, human immunodeficiency virus, rabies virus, lymphocytic chorio-meningitis virus, malarial parasites and mycoplasmas.

134. Choose the correct answer from the statements indicated below :

(a) Chi square test is parametric.

(b) Non-parametric test assumes normal distribution.

(d) Non-parametric test is more powerful as compared to parametic test.

Ans. (c)

Sol. Parametric tests can perform well with skewed and non-normal distributions. Parametric tests can also perform well when the spread of each group is different. While nonparametric tests don't assume that one's data follos a normal distribution, they do have other assumptions that can be hard to meet.

Second, in this case the sample size is very small. If you do not meet the sample size guidelines for the parametric tests and are not confident that the data are normally distributed, it is best to use a nonparametric test.

Third, if you have ordinal data, ranked data, or outliers that can't be removed, then it is best to use a nonparamentric test. Typical parametric test can only assess continuous data and the results can be significantly affected by outliers. Conversely, some nonparametric tests can handle ordinal data, ranked data and not be seriously affected by outliers.

135. Given below are four statements related to Agobacterium-mediated transfer of T-DNA into plant cells :

P. Prodcution of single-stranded T-DNA by VirD1 and VirD2 proteins.

Q. Interaction of VirE2 with VIP1 and VirE3.

R. Use of VirB/VirD4 type IV secretion system.

S. Activation of VirA-VirG complex.

The correct sequence of events (from earliest to latest) is

(a) P, Q, S, R

(b) Q, R, P, S

(d) S, P, R, Q

Ans. (d)

Sol. The first-identified and major plant-produced molecule involved in vir activation is acetosyringone, a phenolic compound often found in plant exudates; AS activates the VirA/VirG two-component system, resulting in the induction of the vir gene expression . Two essential proteins for T-DNA processing are VirD1 and VirD2. VirD2 is an endonuclease, which, in association with the VirD1 DNA topoisomerase, mediates the mobilization of the transferable T-DNA from the Ti plasmid via a strand replacement mechanism. Macromolecules are translocated across the bacterial membranes via a T4SS (type IV secretion system) composed of the 11 proteins encoded by the virB operon and VirD4 by a mechanism closely resembling plasmid transfer during bacterial conjugation. Interactions with bacterial factors are likely required to mediate targeting of the exported substrates—i.e., the VirD2–T-strand complex and the effector proteins VirD5, VirE2, VirE3, and VirF—to the VirB/D4 T4SS.

136. Three proteins Bim 1, Bim 2, Bim 3 shown to be involved in repair of DNA double strand breaks. A chomatin immnoprecipitation experiment was performed for the three proteins. The pattern of results obtained is shown below :

Based on the above figure, choose the option that correctly interprets the data.

(a) Bim 1, Bim 2, Bim 3 bind to DNA berak sites.

(b) Bim 1 binds to the break site; Bim 3 binds to the break site and beyond.

(d) Bim 3 binds to DNA irrespective of the break.

Ans. (b)

Sol. Figure shows that Bim 1 expressed 0kb from break site, it binds to break site. While Bim 3 binds to break site and beyond.

137. Given below is a table comprising various terms associated with plant tissue cultue in column A and column B.

Column A Column B

P. Auxin 1. Embryogenesis

Q. Protoplast culture 2. 6-Furfurylaminopurine

R. Cytokinin 3. Pectinase and Cellulase

S. Microscope culture 4. Indole-3-acetic acid

Which one of the following options represents the most appropriate match between all the terms of column A and column B ?

(a) P–2, Q–1, R–4, S–3

(b) P–4, Q–3, R–2, S–1

(d) P–4, Q–1, R–2, S–3

Ans. (b)

Sol. Indole-3-acetic acid (IAA, 3-IAA) is the most common, naturally occurring, plant hormone of the auxin class. It is the best known of th auxins and has been the subject off extensive studies by plant physiologists. IAA is a derivative of indole, containing a carboxymethyl substituent. It is a colourless solid is soluble in polar organic solvents.

138. In an attempt to increase the yield of a commercially important enzyme from natural isolate, several stategies were adopted as follows :

P. Genome was selectively modified to increase yield.

Q. Reappraisa of culture requirements of te modified organism to increase yield.

R. Induced mutants were screened and selected for organism synthesizing improved levels of the enzyme.

S. Organism was genetically modified, so that it produces a factor that enhances stability of the enzyme.

Which one of the following options represents strategies that are appropriate for the purpose ?

(a) P, Q, R and S

(b) Q and R only

(d) P and Q only

Ans. (a)

Sol. All of the four options seem correct to increase the yield.

139. The most important property of any microscope is its resolution (D). Which one of the following wavelengths (nm) would be used to achieve the best resolution using a light microscope with lenses having numerical aperture (NA) of 1.4 ?

(a) 450

(b) 480

(d) 700

Ans. (a)

Sol. Resolution of microscope and other optical devices is governed by formula :

D = ; where, is the wavelength, µ is the refractive index and is the angular aperture.

Hence at fixed aperture, best resolution (least value of D) would be obtained at less wavelength.

140. Detailed NMR spectra of a 20-residues peptide were recorded using a 600 MHz instrument. If the peptide adopts and -helical conformation, which one of the following statements is correct ?

(a) Prominent NH_i – NH_{i + 1} NOE peaks would be observed along with ³J_{NH – HA} coupling constant ~ 8.5 Hz.

(b) Prominent – NH_{i + 1} NOE peaks would be observed along with ³J_{NH – HA} coupling constant ~ 4.8 Hz.

(d) Prominent NH_i – NH_{i + 1} NOE peaks along with ³J_{NH – HA} coupling constants ~ 4.8 Hz.

Ans. (d)

Sol. NMR spectroscopy is the use of NMR phenomena to study the physical, chemical, and biological properties of matter. Chemists use it to determine molecular identity and structure. Medical practitioners employ magnetic resonance imaging (MRI), a multidimensional NMR imaging technique, for diagnostic purposes. A 1H NMR spectrum is composed of a number of peaks rising from the baseline; each set of peaks is produced by hydrogens in the molecule. Some of the peaks are isolated (a singlet), some are two peaks close together (a doublet), others have three peaks (a triplet), etc. 2D CSI ("two-dimensional cluster analyses of chemical shifts to identify protein secondary structure") analyzes paired, two-dimensional scattering diagrams of six chemical shift data sets; i.e., six different chemical shifts 1HN, 13C', and 15NH) are used to identify the secondary structure of amino-acid residues in proteins.

141. Given below are names of statistical distribution (column I) ad their characteristics features (column II):

Column I Column II

P. Binomial distribution 1. Each observation represents one of two outcomes (success or failure).

Q. Poisson distribution 2. Probability distribution that is symmetric about the mean.

R. Normal distribution 3. Probability of a given number of events happening in a fixed interval of time.

Which one of the following represents a correct match between columns I and II ?

(a) P–2, Q–1, R–3

(b) P–1, Q–2, R–3

(d) P–3, Q–2, R–1

Ans. (c)

Sol. Properties of a normal distribution

1. The mean, mode and median are all equal.

2. The curve is symmetric at the center i.e. around the mean.

3. Exactly half of the values are to the left of center and exactly half the values are to the right.

4. The total area under the curve 1.

Properties of a binomial distribution : A distribution is said to be binomial distribution if the following conditions are met.

1. Each trial has a binary outcome (one of the two outcomes is labeled a 'success').

2. The probability of success is known and constant over all trials.

3. The number of trials is specified.

4. The trials are independent. The outcome from one trial does not affect the outcome of successive trials.

Properties of a poisson distribution

1. It describes random events that occur rarely over a unit of time or space.

2. In Poisson distribution, we count only the average number of success in a given unit of time or space.

142. Highly purified peptides P1, P2 and P3 were subjected to MALDI mass spectral analysis. The following obserations were made :

P1 : Showed a m/z of 16 more than the expected value.

P2 : Showed a m/z of 80 more than the expected value. MS/MS spectra of the peptide resulted in a precursor ion with m/z 98 less than the expected m/z.

P3 : Showed a m/z that was double te expected value.

[Note : z = +1 for all the mass spectra.]

Which one of the options given below comprises all correct interpretations ?

(a) P1 : Cys is oxidized; P2 : has undergone oxidation at multiple Met residues; P3 : is a non-covalent dimer.

(b) P1 : Met is oxidized; P2 : is phosphorylated; P3 : is a covalent dimer.

(d) P1 : Cys is oxidized; P2 : phosphorylated and oxidized at Met; P3 : is a non-covalent dimer.

Ans. (b)

Sol. Specific modification in mass spectrometry can be detected using the relative change in molecular weight caused by the modification such as oxidation increases the mass by 16 Da, phosphorylation causes an increase of 80 Da, while dimer formation double the mass.

143. Construction of knockout mice may be performed using the Cre-LoxP system. Eventually, the Cre recombinase of the bacteriophage P1 mediates site-specific recombination at a 34 bp sequence, loxP. Fom the following statements, choose the incorrect event.

(a) The alteration of the chromosomal copy of the target gene requires a guide RNA.

(b) The loxP containing mice should not express Cre recombinase prio to mating.

(d) Inductio of the promoter results in the expression of Cre, recombination at loxP sites and excision of the sequence in between.

Ans. (a)

Sol. Cre/lox is usually used to make knockout alleles, but it can also be used to activate gene expression. The proper insertion of a loxP-flanked "stop" sequence (transcriptional termination element) between the promoter and transgene coding sequence blocks the expression of the gene. A guide RNA (gRNA) is a piece of RNA that functions as a guide for RNA- or DNA-targeting enzymes, with which it forms complexes. Very often these enzymes will delete, insert or otherwise alter the targeted RNA or DNA. So, A is incorrect statement.

144. Monoclonal antibodies can be modified for better research and therapeutic application. Several such approaches are mentioned below (column A) along wit their possible applications (column B).

Which one of the following options represents a correct combination of terms in column A and column B ?

(a) P–1, Q–2, R–3

(b) P–2, Q–1, R–3

(d) P–2, Q–3, R–1

Ans. (d)

Sol. The concept of transition state binding and stabilisation of the same by catalytic enzymes, is an extension of the new transition state theory developed by Evans and Eyring to explain chemical catalysis. The basic proposition is that the rate of a reaction is related to the difference in Gibbs free energy between the ground state and the transition state of the rection. Catalysis proceeds smoothly either by lowering the energy of the transition state i.e., by transition state stabilization or by elevating the energy of the substrate i.e., by substrate destabilization. The idea of creating catalytic proteins along with the transition state analogues directs the generation such designer protein catalysts popularly known as "Catalytic Antibodies" or "Abzymes".

Immunotoxins are antibody toxin bifunctional molecules they rely on intracellular toxin action to kill target cells. Target specificity is determined via the binding attributes of the chosen antibody. Mostly, but not exclusively, immunotoxins are purpose-built to kill cancer cells as part of novel treatment approaches.

Immunotoxins are MAbs linked to bacterial plant, or synthetic toxins. First-generation immunotoxins were made by chemically coupling toxins such as ricin (a plant toxin) or diphteria toxin to MAbs.

145. Given below are names of techniques (column A) and their characteristic features/applications (column B) :

Column A Column B

P. Hybidoma technology 1. Separation of proteins according to charge

Q. MALDI-TOF 2. Identification of protein complexes in cells

R. Ion-exchange chromatography 3. Production of identica antibodies

S. Co-immuunoprecipitation 4. Determination of molecula weight of proteins and/or peptides.

Which one of the following represents a correct match between column A and column B ?

(a) P–2, Q–3, R–4, S–1

(b) P–3, Q–4, R–1, S–2

(d) P–2, Q–4, R–1, S–3

Ans. (b)

Sol. MALDI-TOF is a mass spectrometric method and it is used for protein identification using m/z (mass/charge) ratios. Hybridoma technology involves fusion of B-cells and myeloma cells fro a large scale antibody production. Coimmunoprecipitation makes use of antibody mediated pulling down of protein and their interacting partners.

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CSIR NET BIOLOGY (DEC - 2018)
Previous Year Question Paper with Solution.