PYQ JUNE 2018 - VedPrep

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

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

(a) Allosteric enzymes do not obey Michaelis-Menten kinetics.

(b) The free-energy change provides information about the spontaneity but not rate of a reaction.

(d) A k_cat/Km(M^–1 S^–1) of ~ 2 × 10⁸ for an enzyme indicates that the value is close to diffusion-controlled rate of encounter.

Ans. (c)

Sol. Out of all the given options, the option C is incorrect because competitive and non-competitive inhibition are very much kinetically distinguishable. In competitive inhibition, an inhibitor prevents binding of the substrate to the enzyme active site and it can be overcoem by high concentration of substrate i.e., it is reversible but in non-competitive inhibition, the inhibitor reduces the activity of the enzyme and binds equally well to the enzyme alone as well as enzyme substrate complex and it cannot be reversed by increasing substrate concentration.

22. Which one of the following peptides can conexist in both cis-and trans-conformation?

(a) Ala-Ala-CONH₂

(b) Pro-Gly-CONH₂

(d) Val-Pro-CONH₂

Ans. (d)

Sol. Proline is the only amino acid which can exist in cis-confirmation. If the proline is present in the middle then in that case the peptide will exist in cis and trans confirmation.

The peptide bond nearly always has the trans configuration since it is more favourable than cis, which is sometimes found to occur with proline residues. For proline residues, the cyclic nature of the side chain means the both cis and trans configuration have more equivalent energies. Thus proline is found in the cis configuration more frequently than other amino acids. The omega torsion angle of proline will be close to zero degrees or the cis configuration or most often, 180° for the trans configuration.

23. Which one of he following pair of amino acids are glucogentic and ketogeneic in nature?

(a) Alanine and Lysine

(b) Lysine and Leucine

(d) Aspartate and Lysine

Ans. (c)

Sol. A glucogenic amino acid is an amino acid that can be converted into glucose through gluconeogenesis. This is in contrast to the ketogenic amino acids, which are converted into ketone bodies.

The production of glucoe from glucogenic amino acids involves these amino acids being converted to alpha keto acids and then to glucose, with both processes occuring in the liver. The mechanisms predominates during catabolysis, rising as fasting and starvation increase in severity.

Amino acids that are both the glucogenic and ketogenic (mnemonic "PITTT") :

Phenylalanine

Isoleucine

Threonine

Tryptophan

Tyrosine

Only leucine and lysine are not glucogenic (they are only ketogenic).

24. The [OH^–] of 0.1 N HCl solution is

(a) 10^–14 M

(b) 10^–13M

(d) 10^–7 M

Ans. (b)

Sol. HCl is a strong acid and that will dissociate in water completely.

1 normal HCl would have 1 gm-equivalent of H⁺ ions in one liter (equivalent to 1 M of HCl).

Explanation : 0.1 N HCl solution means

[H⁺] = 0.1 = 1 × 10^–1

Thus, pH = – log (1 × 10^–1) or pH = 1

Since, pH + pOH = 14

So, pOH = 13

Also, pOH = –log[OH^–]

So, [OH^–] = –antilog (13)

Solving for antilog,

we get, [OH^–] = 10^–13 M

25. Following are structures of stereoismers of aldohexoes which differ in the stereochemistry:

Based on above structures, following information was given below:

P. D-glucose and D-mannose are empires becaure they differ in the stereochemistry at C-2 position.

Q. D-glucose and D-galacotse are empires because they differ in the sterochemistry at C-4 position.

R. D-mannose and D-glucose are empires because they differ in thesterochemistry at C-3 position.

S. D-galactose and D-glucose are empires because they differ in the stereochemistry at C-5 potion.

Choose one of the correct combinations of above statements:

(a) P nad Q

(b) R amd S

(d) P and S

Ans. (a)

Sol. D-glucose and D-mannose, which differ only in the stereochemistry at C-2 are epimers, as are D-glucose and D-galactose (which differ at C-4). Some sugars occur naturally in their L form; examples are L-arabinose and the L-isomers of some sugar derivatives the are common components of glyco-conjugates. Monosaccharides also exist in cyclic structure.

Figure : Cyclic structures of monosaccharides.

26. Following statements are made related to protein structure:

P. The hydrogen bonding patterns between the CO and NH groups are :

Q. In a turn, there are 10 atoms between the hydrogen bond donor and acceptor

R. In a turn, there are 6 atoms between the hdrogen bond donor and acceptor.

S. Parallel sheets have evenly spaced hydrogen bonds, which bridge the strands at an angle.

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

(a) P and Q

(b) P and Q

(d) Q and S

Ans. (d)

Sol. Beta turn is formed by four residues in which the hydrogen bond is present between the first and third amino acid. So the atoms between first and third residues will be 10. Parallel sheets have evenly spaced hydrogen bonds which are present in the titled form. It means they bridge the strands at an algle i.e., in tilted form.

27. The following statements are made on nucleic and struture:

P. In the B-form of DNA, the sugar pucker is C2' exo.

Q. In RAN, the sugar pucker is C3' exo.

R. The wobble bas pair is formed between G and A in RNA.

S. A change in the sugar pucker from C2' endo in the B form of DNA to C3' endo alters the width and depth of the major groove.

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

(a) P and R

(b) Q and S

(d) Q and R

Ans. (c)

Sol. The G-U wobble base pair is a fundamental unit of RNA secondary structure that is present in nearly every class of RNA from organisms of all three phylogenetic domains. It has comparable thermodynamic stability to Watson-Crick base pairs and is nearly isomorphic to them. Therefore, it often substitutes for G.C or A.U base pairs. Hence statement C is incorrect.

B-DNA, the form described by Watson and Crick, adopts 2'-endo (example at right), whereas RNA and the A form of DNA adopt the 3'-endo twist. Hence statement A is correct, whereas statement 2 is incorrect as it says 3' exo.

Simple modification of the sugar puckering of 2' deooxyribose leads to a reversible change between two stable forms of DNA which resemble very closely the canonical A and B duplex forms. A and B forms differ in the width and depth of the major groove. Hence statement D is correct.

28. The V_max and K_m form a Lineweaver-Burk plot of an enzyme reaction where = 40 µM^–1 min at = 0 and = 1.5 × 10² nM^–1 at = 0 are

(a) 0.025 µM^–1 min^–1 and 0.67 × 10^–2 µM

(b) 0.025 µM^–1 min and 0.67 × 10^–2 µM^–1

(d) 0.038 µM min^–1 and 0.67 × 10^–2 µM

Ans. (a)

Sol. Explanation :

1/V_max = 40 μM^–1 min

So, V_max = 0.025 μM min^–1

Also, 1/[S] = –1.5*10² mM^–1 at 1/V = 0

From the graph, when

1/V = 0, 1/[S] = –1/K_m

Thus, –1/K_m = –1.5*10² mM^–1

So K_m = 1/1.5*10² mM^–1

= 0.067*10^–2 mM

29. From the following statements :

P. Biosynthesis of proteins and nucleic acids from precurosors results in producton of chemcial energy in the form of ATP, NADH, NADPH and FADD₂.

Q. The spontaneity of a reaction in cells dones not whether for the reaction is positive or negative.

R. Both oxidative phosphorylation and photophosphorylation involve oxidation of H₂O to O₂.

S. Only chemical potential energy contributes to proton motive force in mitochondria.

Which one of the following combinations represents all incorrect statements?

(a) P, Q, R

(b) Q, R, S

(d) P, R, S

Ans. (d)

Sol. Biosynthetic reaction never results in the formation of ATP, NADH, NADPH or FADH₂. Oxidative phosphorylation results in the formation of water from oxygen. Proton motive force is generated due to the oxidation of NADH and FADH₂. So the A, C and D are wrong statements.

30. A plasmid with a linking number (Lk) of 200, opological windng (Tw) numberof 200 and a writhing number (Wr) of 0 was transformed into E. coli. The plasmid was re-isolated from the culture of the transformant. re-isolated plasmid was found to possess the same molecular weight as the original plasmid, but it possessed a writhing number of –5. Following statement are made about this observation.

P. Lk of the re-isolated plasmid would be 195

Q. Lk of the re-isolated plasmid would remain 200

R. Tw of the re-isolated plasmid would remain 200

S. Tw of the re-isolated plasmid would be 195

Which one of the following combinatons of the above statements is the correct representation of the facts?

(a) P only

(b) P and R

(d) S only

Ans. (b)

Sol. In vivo, Topo-II (Gyrase) introduces negative supercoiling. During Topo-II action, DNA is broken and reunited and thus Lk number changes.

Lk = Tw + Wr thus 200 + (–5 ) = 195.

31. Following statements are being made about the orientation of the N-glycosidic bond between the base and the sugar in the following DNA duplexes.

P. 'anti' for B form DNA duplexes

Q. 'syn' for B form DNA duplexes

R. 'anti' for A form DNA duplexes

S. 'syn' for A form DNA duplexes

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

(a) P and R

(b) Q and R

(d) Q and S

Ans. (a)

Sol. The bases can exist in two distinct orientations about the N-glycosidic bond. These conformations are identified as, syn and anti. It is the anti conformation that predominates in naturally occurring nucleotides. i.e., in both B and A forms of DNA duplexes. There is only one known case where a purine adopts a syn conformation. The unusual form is to find in an even more unusual structure namely left handed Z-DNA.

32. The cell maintains a high concentrations of protons inside the lysosome because of

(a) antiporter in the lysosomal membrane

(b) ATP-powered proton pump in the lysosomal membrane

(d) facilitated diffusion proton uniporter in the lysosomal membrane

Ans. (b)

Sol. Lysosomes contain acid hydrolases which require an acidic medium to function. To maintain their acidic environment, H⁺ ions (protons) are concentrated by a proton pump in the lysosomal membrane, which actively transports protons into it from the cytosol. This pumping requires expenditure of energy in the form of ATP hydrolysis as it maintains approximately a hundredfold higher H⁺ concentration inside the lysosome.

33. It is known that there is a large difference in the DNA content between two organisms with similar developmental complexity. This is due to large differences in he number of

(a) transposable elements, repetitive DNA and coding sequences

(b) transposable elements and repetitive DNA

(d) introns and tansposable elements

Ans. (b)

Sol. The large difference in DNA of closely related species lies in non-coding DNA and most of its part belongs to repetitive DNA including transposabel elements.

34. Ability of a membrane protein to span the lipid bilayer stricity depends on the presence of

(a) Zinc finger domain

(b) -helices

(d) antiparallel -sheet

Ans.

Sol. Many integral membane proteins (called transmembrane proteins) span the lipid bilayer, with portions exposed on both sides of the membrane. The membrane spanning portions of these proteins are usually -helical regions of 20 to 25 nonpolar amino acids. The hydrophobic side chains of these amino acids interact with the fatty acid chains of membrane lipids and the formation of an α helix neutralizes the polar character of the peptide bonds.

35. Which one of the listed is a P-type ion transporter?

(a) Mg⁺² and Fe⁺²

(b) Mg⁺² and Fe⁺³

(d) Na+ –K⁺, Ca⁺² and H⁺

Ans. (d)

Sol. P-type ion transporter forms a phosphorylated (P) intermediate state during their ion transport cycle. Members of this family generate and maintain crucial (electro) chemical gradients across cellular membranes, by translocating cations, heavy metals and lipids. The ions bind to the transmembrane domain, coordinated by negatively charged and polar residues in a region between transmembrane helices M4, M5, M6 and M8. The number of transported ions varies : the fungal H⁺-ATPase transports one H⁺, SERCA transports two Ca²⁺ out versus two to three H⁺ in and the NKA transports three Na⁺ out and two K⁺ in.

36. An intron was cloned within a transposable element. Absence of the intron following transposition of th element, will indicate that it

(a) follows conservative mode of transposition

(b) follows replicative mode of transposition

(d) is an insertion element

Ans. (c)

Sol. The conservative mode of transposition is a 'cut and paste' mechanism where the whole transposable element is transposed to some other part of the genome or another locus in the same chromosome. The Replicative mode of transposition is a "copy paste" mechanism of transposition. An insertion element codes for a protein. In this case, the intron that was cloned in the transposabel element was absent after transposition of the element. This indicates that the element is a retrotransposon. Because, they are first transcibed into RNA, then converted back into identical DNA sequences using reverse transcription and these sequences are then inserted into the genome at target sites. Introns cannot be transcribed into mRNA. Hence, are lost.

37. An integral membrane protein (P) has been identified as a cell surface protein of hepatocytes and assigned to bind to hepatitis B virus (HBV) and promote its entry into cytosal and in turn, helps in the entry of the HBV particles. P wa successfully cloned and expressed in animal cells in culture wherein its N-terminal is exposed on the surface while the C-terminal resides in the cytosol. The recombinant protein P so expressed retains its complete structure and function. From the list of experiments given below, which one of the experiments will you perform to show that C-terminal of the protein P via interacing with F-actin helps in HBV entry?

(a) Incubating radiolabelled HBV with hepatocytes in cultrue and follow up its association with F-actin by immunoprecipitation analysis using anti-F-actin antibody

(b) Incubating radiolabelled HBV with hepatocyres over-expressing the N-terminal mutant of P and repeat the rest of the experiment as in '1'.

(c) Incubating radiolabelled HBV with hepatocytes over-expressing the N-terminal mutant of P and repeat the rest of the experiment as '1'.

(d) Using wild-type P as well as C-terminal mutant of P and their individual overexpression in a heterologous cell line (completely devoid of endogenous P protein) and then repeat experiment as in '1'

Ans. (d)

Sol. To understand the function of protein P, one would use normal wild-type P and C-termini mutant P protein. In the mutant condition virus entry will not occur which will prove that C termini is crucial of protein function.

38. The cell membrane of neuron maintains intracellular conditions that differ from those of the extracellular environment. Such difference in intra-and extracellular conditions are critical to he function of the nerve cell as the nerve cell membrane resembles a charged capacitor. Assuming the electric field (E) across a parallel-plate capacitor is uniform and if membrane thickness is 7 nm and potential difference across the membreane is –60 mV, calculate E of the membrane

(a) 6 × 10⁵ V m^–1

(b) 7 × 10⁵ V m^–1

(d) 6.6 × 10⁶ V m^–1

Ans. (c)

Sol. The Electric Field strength (E) in a parallel plate capacitor is obtained as Voltage (V) applied to plates divided by Distance (d) between the plate

Here, it is given :

V = 60 mV

d = 7 nm or 7*10^–6 mm

Substituting the values in equation

E = 60 mV/7*10^–6 mm

= 8.57*10⁶ V/m

= 8.6*10⁶ V/m.

39. The liposome preparations ('X' and 'Y') are made using basic lipid composition as phosphaidylcholine (PC) and cholesterol (Chol). In 'Y' ganglioside (asialo-GM₁) is incroporated during the preparation besides PC and Chol. In an attempt to find out the localization of asialo-GM₁ in the membrane bilayer of 'Y' (taking 'X' as a negative control) and considering liposome as a true depiction of lipid bilayer structure of cellular membrane, following reagents and provided as probes:

P. Phospholipase A

Q. Galactose binding lectin

R. Exoglycosidase

S. Cyclodextrin

(a) Only P

(b) Both R and S

(d) Both P and S

Ans. (c)

Sol. Option (b) is correct because galactose binding proteins are capable of binding strongly to the oligosaccharide moiety of ganglioside GM₁.

Option (c) is correct because GM₁ gangliosidosis is an autosomal recessive lysosomal storage diseases associated with a neuro-degenerative disorder or dwarfism and skeletal abnormalities respectively.

These diseases are caused by the deficiencies in the lysosomal enzyme beta-D-galactosidase (beta-Gal), which lead to accumulations of the beta-Gal substrates, GM₁ ganglioside and keratan sulphate. Beta-Gal is an exoglycosidase that catalyzes the hydrolysis of terminal beta-linked galactose residues. So, both galactose binding lectins and exoglycosidase have affinity to GM₁ and can be utilized for localizing the gangliosides located on membrane surface.

40. A group of scientists performed an experiment where they artificially fused mouse cells with monkey cells. The resulting fused cells were labelled fluorescently tagged antibodies against mouse and monkey surface receptor proteins, X and Y respectively. At the time 0 minute just after fusion events, two receptors were confined to their own half in the heterokaryon. However, such surface receptors (X and Y) intermixed on the cell surface after 60 minutes. Which one of the given statements correctly reflects the outcome of the experiment?

(a) The proteins in cytoplasm are in a dynamic state

(b) The proteins on the membrane surface are in a dynamic state

(d) Membrane surface proteins are in a static phase.

Ans. (b)

Sol. The fluorescent tags are attached to surface receptors X and Y on mouse and monkey cells respectively. After fusion both the tags remaini on the surface and are not lost with time. Only with time they change their position.

This shows that membrane proteins are not static but dynamic. Since the tags are not lost with time so, they surface proteins are not exchanged with the cytoplasmic protein. Since the tags are specific so for surface proteins so no conclusion about the cytoplasmic proteins can be drawn for this experiment.

41. Eg5 is a well-studied protein in Xenopus. To understand the functon of Eg5 in mammalian cells, a team of researcher treated mammalian cells during late G2 phase with Eg5-inhibitor. The following diagrams represent images of mitotic cells. Based on the below information, what function might be attributed to Eg5 during mitosis?

(a) Eg5 inhibits actin dynamics

(b) Eg5 can activate GPCR signaling

(d) Eg5 can impac mitochondrial dynamics

Ans. (c)

Sol. In the control part of the question, it is shown that chromosomes are no equator because of the movement of spindle fibre but in experiment part there is no proper alignment of chromosomes in the equator. This is because of the failure of the spindle fibre movement. As stated in the experiment, inhibitor of Eg5 was used this indicate that Eg5 is working as motor protein with microtubule spindle fibres.

42. The following ntracellular event occurs in a cell that is subjected to conditions of starvation. Which one of the following statements correctly represents the event shown below?

(a) The cell is undergoing apoptotic cell death with the help of lysosomes (A)

(b) The cell is undergoing autophagy by formation of autophagolysosomes (C)

(d) The cell is undergoing autophagy and fusion occurs between lysosome (B) and autophagolysosome (A).

Ans. (b)

Sol. During autophagy a double isolation membrane engulfs cytoplasmic material including proteins, protein aggregates and organelles such as mitochondria, ribosomes and fragments of the ER. The membrane forms a vesicle termed autophagosome, which eventually fuses with the lysosome to form autophagolysosome. After fusion the inner membrane is lysed and atuphagic cargo is digested by the lysosomal proteases. This frees up amino acids, lipids and nucleic acids which can be recycled as building blocks for new synthesis or used to provide energy to support cellular function.

43. Match the enzymes in column A with their respective biological function in Column B

Column A Column B

P. Lipases 1. Catalysis of ATP-dependent ranslocaton of the

aminophospholipids, phosphatidylethanolamine and

phosphatidylserine from the exracellular to the cyosolic leaflet

of the plasma membrane

Q. Fippases 2. Catalysis of ATP-dependent translocaion of plasma membrane

phospholpids from the cytosolic to the extracellular leaflet.

R. Floppases 3. Catalyze hydrolysis of triacylglycerols

S. Scramblases 4. Catalyze the movement of any membrane phospholipid across

the bilayer down its concentration gradient.

Choose the correct combinations of answers from the options given below:

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

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

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

Ans. (a)

Sol. Lipases hydrolyse the ester bond of lipids and form glycerol and free acids. Flippases trigger movement of vesicle from extra cellular site to cytosic leaflet in ATP dependent manner. Floppases trigger movement of lipids from cytoslic leaflet to exoplasmic leaflet in ATP dependent manner. Scramblases are responsible for the passive transport of lipids in both leaflet in ATP independent manner.

44. Phosphorylaion of eIF2 α-subunits (at Ser51) leads to

(a) inactivation of Met-tRNA binding activity of eIF2B.

(b) seuestration of eIF2B because of tight binding between eIF2 and eIF2B

(d) enhanced guanine exchange activity of eIF2B

Ans. (b)

Sol. eIF2B is ablt ot exchange GDP for GTP only if eIF2 is in its unphosphorylated state. The phosphorylated eIF2, however, due to its stronger binding, acts as an inhibitor due to its stronger binding, acts as inhibitor of its own GEF (eIF2B). Since the cellular concentration of eIF2B is much lower than that of eIF2, even a small amount of phosphorylated eIF2 can completely abolish eIF2B activity by sequestration.

45. Transcriptional regulation of trp operon by tryptophan involves binding of trypophan to

(a) the repressor protein and inhibition of transcription by its interaction with the operator region.

(b) RNA polymerse and inhibition of transcription

(d) the reressor protein leading ot its interaction with the sigma subunit and inhibition of transcription.

Ans. (a)

Sol. When tryptophan is present, these tryptophan repressor dimmers bind to tryptophan causing a change in the repressor conformation, allowing the repressor to bind to the operator. This prevents RNA polymerase from binding to and transcribing the operon, so tryptophan is not produced from its precursor. When tryptophan is not present, the repressor is in its inactive conformation and cannot bind the operator region, so transcription is not inhibited by the repressor.

46. E. coli takes 40 minutes to duplicate its genone using a bdirectional made of replcation. If E. col were to use unidirectiional mode of replication to synthesize a full copy of DNA complementary to just one of the strnads of the genome, it would take

(a) 40 minutes

(b) 80 minutes

(d) 60 minutes

Ans. (b)

Sol. Unidirectional replication requires double time as compared to bidirectional replication. Therefore, 80 minutes would be required to replicate its genome by unidirectional replication as compared to 40 min in bidirectional mode of replication.

47. Genes translocated to the heterochromatic regions of chromosomes are silenced. In S. pombe, a translocaion event was detected wherein a gene of inerest was translocated to the centromere region and is silinced. Mutagenesis leading to loss-of-function of the following taget genes was done to allow expression of the gene of interest from its new locus.

P. Mutation in histone deacetylase (Clr3)

Q. Mutation in histone acetyltransferase (HAT-8)

R. Mutation in histone H3 lysine 9 methyl transferase (Clr4).

S. Loss of Dicer, an RNA processing enzyme

Which of the above events could allow the expression of this gene from thecentromeric region?

(a) P, Q and R

(b) P, R and S

(d) P and R only

Ans. (b)

Sol. Genes translocated to centromeric region can be expressed by Mutation in histone deacetylase (Clr3), Mutation in histone acetyltransferase (HAT-8) and Mutation in histone H3 lysine 9 methyl transferase (Clr4).

48. Antinomycin D inhibits the process of transcription in both prokaryotic and eukaryotic organisms. The following staements are made about actinomycin D-mediated inhibiton of transcription:

P. Actinomycin D inhibits transcription from a double standed DNA template by either E.coli. or yeast RNA polymerases

Q. Acinomycin D inhibits transcription from a single standed RNA template by eukaryotic viral RNA polymerases

R. Actinomycin D inhibits transcription from single stranded × 174 DNA template by E. coli RNA polymerase immediately after viral DNA entry.

S. Actinomycin D inhibits transcription from double stranded RNA template by eukaryotic RNA polymerase II.

Which of the combinations of the above statements is a true representation of the mechanism of actinomycin D mediated inhibition?

(a) P only

(b) P, Q and R

(d) P and S only

Ans. (a)

Sol. The planar portion of the antibiotic actinomycin D intercalates into the double helical DNA between successive G = C base pairs, deforming the DNA. The two cyclic peptide structures of actinomycin D bind to the minor groove of the double helix. This prevents movement of the polymerase along the template. The drug interacts with single and double stranded DNA and RNA-DNA hybrids, but not with RNA. Furthermore, ActD can inhibit the annealing of DNA to complementary DNA or RNA. It was recently demonstrated that ActD also inhibits elongation of DNA-dependent DNA synthesis by the HIV-1 RT enzyme.

49. During maturation process of some RNA molecules, formaton of 2'-5' phosphodiester bond takes place. Following statements are made about this phenomenon.

P. Spliceosome mediated removal of intronic sequences occurs through the formation of 2'-5' phosphodiester bond.

Q. Removal of group Ii introns occurs through the formation of 2'-5' phosphodiester bond

R. Enzymatic remvoal of introns from the yeast tRNA precursors involves 2'-5' phosphodiester bond.

S. RNaseP mediated 5'-end maturation of tRNA precursors involves formation of 2'-5' phosphodiester bond.

Which one of the following combinations of the statements is a true representation?

(a) P only

(b) P and S

(d) R and S

Ans. (c)

Sol. In group II introns the reaction pattern is similar except for the nucleophile in the first step, which in this case is the 2'-hydroxyl group of an A residue within the intron. A branched lariat 2'-5' structure is formed as an intermediate.

Within the spliceosome, a series of RNA-RNA, RNA-protein and protein-protein interactions is needed to idenify and remove intronic regions and join exons, producing a maturee transcript. The mature spliceosome carries out splicing through two transesterification reactions. First, the 2'-OH of a branch point nucleotide performs a nucelolytic attack on the first nucleotide of the intron, forming a lariat intermediate. Second, the 3'-OH from the free exon performs a nucleolytic attack on the last nucleotide of the intron, thereby joining exons and releasing the lariat intron. Intron identification relies on certain sequences, including the 5' splice site, branch point (and downstream polypyrimidine tract) and 3' splice site.

50. During translation in prokaryotes, when ribosomes reach the termination codon, the termination codon is recognized by the class 1 release factor (RF1 or RF2) leading to the release of the polypeptide. A second class II release factor (RF3) facilitates the termination process. Which of the following statements regarding the mechanism of action of the release factors is incorrect?

(a) Class I release factors decode the stop codons while the RF3 is a GTPase that simulates recycling of the class I release factors.

(b) Free RF3 has a higher affinity for GTP and GDP

(d) RF1 and RF2, individually possess another stretch of trpeptide sequences which are involved in the recognition of the terminatioin codons.

Ans. (b)

Sol. Free RF3 binds GDP very strongly and the spontaneous dissociation of the guanine nucleotide is slow, implying that cytoplasmic RF3 is in the GDP, rather than the GTP, conformation. When RF3.GDP associates with a ribosome in complex with a class 1 RF, the GDP on RF3 can rapidly dissociates and the exchanged for GTP. RF3 in the GTP conformation has a high affinity for the ribosome in the absence of a class 1 RF and its binding destabilizes that of RF1 or RF2. The formation of RF3.GTP therefore leads to rapid dissociation of RF1 or RF2 from the ribosome. Subsequent hydrolysis of GTP on RF3 leads to its fast dissociation in the GDOP conformation, the normal free state of the factor.

51. E. coli DNA ligase catalyses formation of a phosphodiester bond between the adjoinng 3' hydroxyl. and the 5' phosphoryl ends in DNA duplexes. The energetic need for this reaction is mey by the hydrolysis of NAD⁺ to NMN⁺ and AMP in a three-step reaction. Following statements are being made about the mechanism of this reaction.

P. AMP is linked to the 5' phosphorly end of the nicked DNA

Q. Adenylyl group of NAD⁺ is transferred to the -amino group of Lys in DNA lgase to form a phosphoamide adduct.

R. DNA ligase catalyses the formation of a phosphodiester bond by the nucleophilic attack of the 3' hydroxyl group onto the phosphate and releases AMP.

Based on the statements made above, identify he correct sequence of the reaction steps.

(a) P-Q-R

(b) P-R-Q

(d) R-P-Q

Ans. (c)

Sol. All DNA ligases catalyse the synthesis of phosphodiester bonds in a very similar manner, by esterification of a 5'-phosphoryl to a 3' hydroxyl group. The reaction mechanism can be split into three distinct catalytic events.

The first involves activation of the ligase through the formation of a covalent protein-AMP intermediate. The nucleotide has been shown to be linked to the enzyme through a phosphoramidate bond to the -amino group of a conserved active site lysine. In the second step of the reaction, the AMP moiety is transferred from the ligase to the 5'-phosphate group at the single strand break site.

Finally, DNA ligase-catalyse the DNA ligation step with loss of free AMP. In spite of these similarities between the two classes of enzymes the manner by which the eubacterial and 'eukaryotic' proteins become activated is rather different. For eukaryotic ligases, the enzyme AMP complex is formed after reaction of the enzyme and ATP with the release of free pyrophosphate. The bacterial ligases become adenylated in an unusual reaction which involves cleavage of NAD⁺ and release of nicotinamide mononucleotide.

52. If a disease caused by an intracellular pathogen is associated with host anti-inflammatory response, which one of the following may lead to an effective therapeutic approach?

(a) Treatment with TGF-

(b) Treatment with macrophage activanig agent

(d) Treatment with IL-4 and IL-10

Ans. (b)

Sol. Since disease is caused due to host anti-inflammatory response. For therapy host pro-inflammatory response should be activated, which in this question can be achieved by activating macrophages. Rest all other options are anti-inflammatory effects.

53. Which one of the following is not true for alternative pathway of complemen activation?

(a) Alternative pathway uses the same membrane-attack complex as the classical pathway

(b) Alternative pathway does not require antigen-antibody niteractions

(d) Certain microbial surfaces have physico-chemical propeties that may result in activation of alternative pathway.

Ans. (c)

Sol. The alternative pathway provides an evolutionarily old and quite primitive innate immune defense mechanisms. Activation begins with the C₃ molecules.

Native C₃ contains an integral thiol ester in its chain that, under normal physiologic conditions, undergoes contnuous low-grade hydrolysis to create a "C₃b like" molecule, C₃(H₂O).

54. Which one of the following permits the rapid diffusion of small, water-soluble molecules between the cytoplasm of adjacent cells?

(a) Tight junctions

(b) Anchoring junctions

(d) Gap junctions

Ans. (d)

Sol. With he exception of a few terminally differentiated cells such as skeletal muscle cells and blood cells, most celsl in animal tissues are in communication with their neighbours via gap junctions. Each gap junction appears asa patch where the membranes of two adjacent cells are separated by a uniform narrow gap of about 2-4 nm.

The gap is spanned by channel-forming proteins (connexins). The channels they form (connexons) allow inorganic ions and other small water soluble molecules to pass directly from the cytoplasm of one cell to the cytoplasm of the other, thereby coupling the cells both electrically and metabolically.

55. Cervical cancer-causing papilloma virus produces two oncoproteins E6 and E7 which are responsible for interfering with cell cycle regulation by

(a) inactvating pRb and p53, respectively

(b) modulating p53 and pRb, respectively

(d) activating expression of p21.

Ans. (b)

Sol. The E6 protein is thought to promote cell proliferation by stimulating degradation of the tumor suppressor p53 protein via the formation of a trimeric complex comprising E6, p53 and the cellular ubiquitination enzyme E6-AP. E6-stimulated degradation interferes with such biological functions of p53 : thus perturbin the control of cell cycle progression, leading finally to incrased tumor cell growth.

56. The extracellular matrix (ECM) is complex combination of secreted proteins that is involved in holding cells and tissues together. The components of ECM form a network by binding to each other and communicate with cells by binding to adhesion receptors on the cell surface. ECM comprises manily two classes of macromolecules, proteglycans and very high molecular weight large proteins. Which one of the following statements regarding ECM constituents is incorrect?

(a) Proteoglycans are a subset of secreted or cell surface-attached glycoproteins containing covalently linked specialized polysaccharide chains called glycosaminoglycans (GAGs).

(b) GAGs are long branched polymers of specific repeating disaccharides of sialic acid and glucose or galactose

(c) Major types of GAGs present n ECM are haparan sulphate, chondroitin sulphate, dermatan sulphate, keratan sulphate and hyaluronan

(d) Major types of large proteins present n ECM are collagen, laminin, elastin and fibronectin

Ans. (b)

Sol. Glycosaminoglycans (GAGs) are large linear (unbranched) (hence statement B is incorrect) polysaccharides constructed of repeating disaccharide units with the primary configurations containing an amino sugar (either GlcNAc ro GalNAc) and an uronic acid (either glucuronic acid and/or iduronic acid). There are five identified glycosaminoglycan chains : Hyaluronan, Chondroitin, Dermatan, Heparin/heparan, Keratan (Hence statement C is correct). Proteoglycans consist of a core protein and one or more covalently attached glycosaminoglycan chains. Virtually all mammalian cells produce proteoglycans and secrete them into the ECM, insert them into the plasma membrane or store them in secretory granules. Hence statement A is correct. The major components of the EMC are fibrillar proteins that provide tensile strength and elasticity (e.g., various collagens and elastins), adhesive glycoproteins (e.g., fibronectin, laminins and tenascins) and proteoglycans. Hence statement D is also correct.

57. Present-day cancer treatment uses many approaches. Beyond surgery and radition treatment, which are most often employed in cases of larger, more discrete tumors, drug therapies can be used to target residual tumor cells and to attack dispersed cancers. Chemotherapes by anti-cancer drugs are mostly aimed at blocking DNA snythesis and cell division.

A list of anti-cancer drugs is given in column A, their chemical nature in column B and their mechanism of action n column C.

Which one of the following is the most appropriate match?

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

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

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

Ans. (d)

Sol. Methotrexate : Methotrexate inhibits dihydrofolic acid reductase, the enzyme that reduces dihydrofolate to tetrahydrofolate, which is utilized in purine nuceleotide synthesis.

Etoposide : Etoposide derives from podophyllotoxin, etoposide targets DNA topoisomerase II activities thus leading to the production of DNA breaks and elicting a response that affects several aspects of cell metabolisms.

Paclitaxel : Lead alkaloids isolated from the trees are taxol and camptothecins that currently have annual sales in billion dollars. Paclitaxel is also commonly known as taxol. 5-flurouracil : 5-FU acts in several ways, but principally as a thymidylate synthase (TS) inhibitor. Interrupting the action of this enzyme blocks synthesis of the pyrimidine thymidine, which is a nucleoside required for DNA replication.

58. Given below are the type or vaccination (column A), the diseases or conditions against which these vaccination types are used (column B) and the advantages or disadvantages for using these vaccination types (column C).

Which one of the following combinations is the most appropriate match?

(a) P-a-3, Q-b-2, R-c-1

(b) P-b-1, Q-a-2, R-c-3

(d) P-b-3, Q-c-1, R-a-2

Ans. (b)

Sol. Live attenuated vaccines — Because of their capacity for transient growth, such vaccines provide prolonged immune-system exposure to the individual epitopes on the attenuated organisms, resulting in increased immunogenicity and production of memory cells. As a consequence, these vaccines often require only a single immunization, eliminatin the need for repeated booster.

Diphtheria and tetanus vaccines, for example, can be made by purifying the bacterial exotoxin and then inactivating the toxin with foraldehdy to form a toxoid—Not stable and recurrent booster is required since its a protein and inactivated vaccines are stable.

59. Activation-induced cytidine deaminase (AID) is the key mediator of somatic hypermutation, gene conversion and class-switch recombination. In order to ascertain the role of AID in class-switch recombination, immune response against a target antigen was compared between AID knock-out mice (AID–/–) with that of mice retaining a functional copy of the AID gene (AID +/–). Development of IgM antibodies against the target antigen was then measured following successive immunization and plotted graphically.

Which one of the following is the most appropriate representation of the experiment?

–––––––– (AID + / –) ------------- (AID – / –)

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Primary response always shows higher production of IgM whereas IgG expression occurs in secondary response, which is due to class switching mechanism. In the given graphs of IgM and IgG production, AID containing mouse hsould be able to perform class switching and as a result in secondary response will produce more amount of IgG compared to IgM. In case of AID deficient mouse there should not be any class switching as a result secondary response will not produce IgG. But it can still produce IgM in both primary and secondary response.

60. G protein-coupled receptors (GPCRs) are used to detect and respond to many differenr types of signals, including neurotransmitters, hormones involved in glycogen and fat metabolism and even photons of light. which one of the following statements regarding GPCR s incorrect?

(a) G protein are a large family with a common structure of seven membrane spanning -helices.

(b) GPCRs are coupled to trimeric G proteins comprising three subunits and .

(c) The subunit is a GTPase switch protein that alternates between an active ('on') state with bound GTP and an inactive ('off') state with GDP.

(d) The 'on' from gets bound to and subunis and activates a membrane-bound effector like adenylyl cyclase, phospholipase C or ion channel.

Ans. (d)

Sol. A remarkable property of GPCRs which make it distinguishable from other classes of receptors is the presence of seven TM (7TM) -helical region. Members of the GPCR superfamily share the same basic architecture i.e., T73 -helices, an extracellular amino-terminal segment and an intracellular carboxy-terminal tail. Hence statement A is correct.

Heterotrimeric G proteins (G, G/G subunits) constitute one of the most important components of cell signaling cascade. G Protein Coupled Receptors (GPCRs) perceive many extracellular signals and tarnsduce them to hetertrimeric G proteins which further transduce these signals intracellular to appropriate downstream effectors and thereby play an important role in various signaling pathways. Hence statement B is correct.

61. Fertilization is sea urchin egg involves Ca²⁺ release from the endoplasmic reticulum for cortical granule reactivation. The major molecule responsible for releasing Ca²⁺ from nitracellular stores is

(a) zona pellucida glycoproteins

(b) protamines

(d) N-acetylglucosaminidase

Ans. (c)

Sol. In echinoderms eggs as in the many other cell types the ER is the main organelle involved in calcium storage and calcium release events. Calcium accumulation in the ER is drived by a SERCA pump (Smooth endoplasmic reticulum calcium ATPase pump) and elementary release events from the ER either by IP3 or RyR receptors transmembrane calcium that open upon binding their specific enabling calcium to diffuse out along a concentration gradient.

62. What is the observed phenotype when the ultrabithorax gene is deleted in Drosophilla?

(a) The third thoracic segment is transformed into another second thoracic segment resulting in a fly with four wings

(b) Since it specifies the second thoracic segment, insted of antenna leg grows out of the head socket.

(d) Since this gene falils to be expressed in the second thoracic segment, he antennae sprout in the leg position.

Ans. (a)

Sol. A particular class of selector genes is formed by the Hox genes, which specify different structures along the anterior posterior axis of metazoans.

In Drosophila, mutations in these genes frequently transform one structure into another keeping the coordinates provided by underlying postitional information, established in part by the activity of signaling pathways. Mutations in the Ultrabithorax (Ubx) Hox gene illustrates this assertion.

Wings and halteres are homologous structures located in the second and third thoracic segments, respectively. These appendages greatly differ in size and pattern and derive from imaginal discs, the wing and haltere discs, which also differ in size but bear a similar morphology.

Ubx, which is expressed in the haltere disc but not in the wing disc, determines the difference between these two structures : mutations or depletion of Ubx transform halterees into wings whereas Ubx ectopic expression changes wings into halters.

63. Which one of the following statements with respect to amphibian development is correct?

(a) The organizer is itself induced by the Nieuwkoop Centre located in the dorsal-most mesodermal cells.

(b) The organizer functions by secreting proteins like Noggin, Chordin an Follistatin that blocks BMP signal that would otherwise dorsalize the mesoderm.

(d) Wnt signalling causes a gradient of -catenin long the anterior-posterior axis of he neural tube that appears to sepecify the regionalization of he neural tube.

Ans. (d)

Sol. Wnt/-catenin signalling occurs in a direct and long-range fashion within the ectoderm and there is an endogenous AP gradient of Wnt/-catenin signalling in the presumptive neural plate of the Xenopus gastrula.

An activity gradient of Wnt/-catenin signalling acts as transforming morphogento pattern the Xenopus central nervous system. Wnt signalling alsos affects patterning within the hypothalamus, suggesting that this pathway is invovled in both the initial anteroposterior subdivision of ventral CNS midline fates and in the subsequent regionalisation of the hypothalamus.

64. Following statements were mde regarding vulval development in Caenorhabditis elegans:

1. The six vulval precusor cells (VPCs) are influenced by the anchor cell to form an equivalence group.

2. In the loss-of-function lin-12 mutants, both cells become uterine wheareas in gain-of-function mutants, both become anchor cell.

3. If the anchor cell s destroyed early in development, all the six VPCs divide once and contribute towards he formation of hypodermal cells.

4. The anchor cell/ventral uterine precursor decision is due to Notch-Delta mediated mechnism of restricting adjacent cell fates.

5. The paracrine factor secreted by the anchor cell directly activates the North-delta pathway.

Which one of the following options represents a combination of correct statements?

(a) 1, 3 and 4

(b) 1, 2 and 4

(d) 2, 4 and 5

Ans. (a)

Sol. 6 vulval precursor cells form an equivalence group and are influenced by the anchor cell. If gonadal anchor cell is destroyed by an way then all fate and differentiate into °VPC adopt 3 hypodermis. The P5 and P4 cells receive the signal through the L-2 N-12 (Notch) protein on their cell membrane.

65. Which one of the following statements regarding limb regeneration in Salmander s correct?

(a) A normal limb is regenerated after amputation, irrespective of whether the cut was made below the elbow or through the humerus.

(b) It occurs by compensatory regeneration and does not include formation of an apical entodermal cap.

(c) Regeneration occurs through formation of a blastema, which essentially consists of unspecified multipotential progenitor cells.

(d) Prolifertion of the blastem cells does not require nerves or factors secreted by the nerves.

Ans. (a)

Sol. When an adult salamender 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 not a new elbow. Experiments done with regeneration of a salamander forelimb showed that in the amputation made blow the elbow and in the amputation throught the humerus, the correct positional information was respecified in both cases. Hence statement A is correct.

Salamanders accomplish this feat by dedifferentiation and respecification. Upon limb amputation, a plasma clot forms and within 6 to 12 hours, epidermal cells from the remaining sumps migrate to cover the wound surface, forming the wound epidermis. This single-layered structure is required for the regeneration of the limb and it proliferates to form the apical ectodermal cap. Hence statement B is incorrect.

The formerly well-structured limb region at the cut edge of the stump forms a proliferating mass of indistinguishable, dedifferentiated cells (Hence not progenitor cells, which makes statement C also incorrect) just beneath the apical ectodermal cap. This dedifferentiation cell mass icalled the regeneration blastema. Thise cells will continue to proliferate and will eventually redifferentiate to form the new structures of the limb. The proliferation of the salamander limb regeneration blastema is dependent on the presence of nerves. Hence statement D is also incorrect.

66. Change in leaf morphology is observed during transition from vegetative to reproductive phase in serveral plants.

The following statements are proposed to explain the above observation:

P. Alteration in the gene content of leaves of reproductive phase from those of vegetative phase.

Q. Differential methylation pattern influencing leaf development and morphology.

R. Mutation in transcription factor that prevents its association with promoter elements of genes regulating leaf development.

S. Small RNA mediated inhibition of gene expression of a homeotic gene.

Which one of the following options represents a correct combination of statements that could explain the observed changes?

(a) Q and R

(b) P and S

(d) P and R

Ans. (c)

Sol. Changes in leaf morphology are observed during transition from vegetative to reproductive phase in several plants. Due to the epigenetic modification, differential methylation pattern of gene influences leaf development and morphology. Also small RNA mediated inhibitor of gene expression of a homeotic gene occurs. There is no alteration of gene content during the transition and no mutation in any transcription factor.

67. cAMP signalling plays a very important role in the development of Dictyostelium discoideum. Below are few statements related to it.

1. Every amoeba at the time of aggregation has the potential to make, receive and relay cAMP.

2. acb^– mutants develop normally but the spores formed appear glassy and are unable to germinate.

3. The spores formed by the acg^– mutants germinate in the sorus itself.

4. cAMP is continously secreted in nanomolar amounts during aggregation.

Which combination of the above statements is correct?

(a) 1 and 4

(b) 1 and 2

(d) 2 and 4

Ans. (b)

Sol. In Dictyostelium, extracellular cAMP is sensed by cell surface receptors (Klein et al., 1988). Four different cAMP receptors, cAR1, cAR2, cAR3 and cAR4, are expressed sequentially during development. These receptors are homologous to 7-transmembrane G protein-coupled receptors found in plants and mammals. After aggregation of social amoebas, extracellular cAMP binding to surface receptors and intracellular cAMP binding to cAMP dependent protein kinase (PKA) act together to induce prespore differentiation. Adenylyl cyclase B (ACB) produces cAMP for maturation. These combined data show that loss of ACG is most deleterious for prespore differentiation, while loss of ACB has the strongest effect on spore maturation. However, the two enzymes show considerable functional redundancy and the most severe phenotypes on both prespore and spore differentiation are evident when they are both lost.

68. Torpedo is known to serve as a receptor for Gurken. Deficiencies of the torpedo gene in Drosophila cause ventralization of the embryo. In an experiment, the germ cell precursors from a wild-type embryo were transplanted into embryos whose mother carried the torpedo mutation. Also, the reverse experiment, i.e., transplantaion of germ cell precursors from torpedo mutants into wild-type embryos was done. The torpedo deficient germs cells developed in a wild-type female showed normal dorsovenral axis, while the wild-type germ cells developed in a torpedo deficient female showed ventralized egg. Some of the following statements are drawn from the above experiments and some from known facts to understand the functioning of Torpedo.

1. Zygotic contribution of Torpedo is essential for the development of ddorsoventral axis.

2. Maternal contribution of Torpedo is essential for the development of dorsoventral axis.

3. Since Torpedo is a raceptor for Guken and follicle cells surround the part of the oocyte where Gurrken ins expressed, it is likely that Torpedo is expressed in fillicle cells.

4. Gurken signalling initially dorsalizes the follicle cells, which in turn send signal o organize the dorso-ventral polarity in oocyte.

5. Gurken signalling initially dorsalizes the nurse cells which hdlep in generation of dosrso-ventral polarity in oocyte.

Which one of the following combination of statements is most appropriate?

(a) 2, 3 and 4

(b) 1, 3 and 4

(d) 1, 4 and 5

Ans. (a)

Sol. The oocyte nucleus is originally located at the posterior end of the oocyte, away from the nurse cells. It then moves to an anterior dorsal position and signals the overlying follicle cells to become the more columnar dorsal follicle cells. Hence statement D is correct and statement E incorrect.

The dorsalizing signal from the oocyte nucleus appears to be the product of the Gurken gene. The Gurken signal is received by the follicle cells through a receptor encoded by the torpedo gene. Hence statement C is correct. Maternal deficiency of torpedo causes the ventralization of the embryo. Moreover, the torpedo gene is active in the ovarian follicle cells, not in the embryo. Hence statement B is also correct.

69. Given below are few statements regarding the role of Dsheveled (Dsh) and -catenin (-cat) in the development of se urchin.

1. Dsh is localized in the vegetal cortex of thee oocyte before fertilization and in the region of the 16-cell embryo about to become the micromeres.

2. Dsh is localized in he cytosal of the oocyte during oogenesis and in the micromere forming blastomeres of a 16-cell embryo.

3. -cat accumulates predominantly in the micromeres and somewhat in the veg₂ tier cells.

4. Treatment of embryos with lithium chloride does not allow the accumulation of -cat in the nuclei of all blastula cells, and the animal cells thus become specified as endoderm and mesoderm.

5. When -cat is prevented from entering the nucleus, the embryo develops as a cillated ectodermal ball.

Which one of the following options represents a combination of correct statements?

(a) 2, 3 and 5

(b) 1, 3 and 4

(d) 2, 4 and 5

Ans. (c)

Sol. In the early sea urchin embryo, Dsh is concentrated in punctate structures within the cytoplasm of vegetal blastomeres. In these cells, Dsh stabilizes -catenin and causes it to accumulate in nuclei, resulting in the activation of transcriptional gene regulatory networks that drive mesoderm and endoderm formation.

Before oogenesis, -catenin and DSH present in crystal, evenly distributed. However during oogenesis DHS is localized at vegetal cortex to prevent degradation of -catenin.

After 4th cleavage -catenin get incorporate in large micromere, therefore autonomous behaviour is primarily depend upon maternal factor.

70. Given below are certain statements regarding plant-pathogen interactions:

P. The pattern recognition receptor (PRR), upon perceiving pathogen or microbe-associated pattern (PAPMs/MAMPs), activates plant defenses resulting in pattern triggered immunity (PTI).

Q. AvrPto is a resistance gene in tomato that acs against pathogenic attack by the bacterium Pseudomonas syrinage pv. tomato.

R. The effector molecules produced by pathogen is recognized by resistance (R) gene present in plants resulting into a defense strategy known as effector triggered immunity (ETI).

S. Defense mechanisms triggered in plants during PTI are usuaally stronger thn those during ETI.

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

(a) P and Q

(b) R and S

(d) Q and S

Ans. (c)

Sol. The first interaction between the plants and microbes take place in apoplast and is mediated by the recognition of microbial elicitors by the receptor proteins of the plants. These microbial elicitors, also known as pathogen-associated molecular patterns (PAMPs) are recognized by the membrane localized pattern recognized by the membrane-localized receptors (PRRs) of plants.

The bacterial flagellin and elongation factor (EF)-Tu peptide surrogates, flg22 and elf18 and chitin, are common examples of PAMPs, which are recognized by the plant PRRs that include the three receptor-like kinases, flagellin-sensitive22, EF-Tu receptor (EFR) and chitin elicitor receptor kinase1 (CERK1). The successful recognition of microbial derived PAMPs by PRRs of the plants activates a first line of defense which is known as PAMP-triggered immunity (PT1). So A is correct.

However, resistance (R) proteins of plants recognize these effector proteins of pathogens and can induce a second line of defense which is known as the effector-triggered immunity (ETI).

ETI is quantitatively stronger and faster than PTI and can result in a localized cell death (hypersensitive response) to kill both pathogen and pathogen infected plant cells. C is correct and D is wrong. The avrPto gene of Psudomonas syringae pv tomato triggers race-specific resistance in tomato plant carrying Pto, a resistance gene encoding a protein kinase. When introduced into P.s. tabaci, avrPto triggers resistance in tobacco W38 plants that carry the corresponding R gene. The AvrPto protein is believed to be secreted into host cells through the bacterial type III secretion pathway, where it activates disease resistance in tomato by interacting with Pto. So the AvrPto gene is from the bacteria not from the plant. So B is also wrong. So correct combination should be A and C.

71. To characterize the mechanism/s by which heat-stress is perceived in Arabidopsis, a team of researchers fused a Heat Shock promoter with lucferase gene. Transgenic plants having promoter:lucferase fusion were raised. Such plants revealed strong luciferase expression upon heat-stress but the showed no expression under unstresed control condition. Subsequently, these transgenic plans were mutagenized by EMS and seeds from F₂ generation were obtained. To analyze the downstream positive regulators of heat-stress, the researcher should analyze seedling that are

(a) expressing luciferase in the presence of heat-stress

(b) not expressing luciferase in presence of heat-stress

(d) not expressing luciferase in the absence of heat-stress.

Ans. (b)

Sol. The question is about the perceiving of heat stress by Arabidopsis. So obviously under heat stress conditions, the heat stress promoter should be recognized and that leads to strong expression of luciferase (which is the same as option 1, so these seedlings cannot be selected to analyze any downstream expression regulation). And control, does not express luciferase as the promoter has not been recognized as stress conditions are not there (again same condition as Option 2)

So now, the EMS mutagenesis, might mutate the heat stress promoter which renders it not capable of perceiving any heat stress conditions. In this case, if still luciferase is expressed means the expression is clearly controlled by some downstream positive regulators triggered by stress conditions justifying option 3 to be correct.

On the other hand, EMS mutating the luciferase enzyme, will not express luciferase in any conditions, so any kind of expressional regulation cannot be analyzed in seedlings that are not expressing luciferase even in stress conditions. So option 2 cannot be correct.

72. Two near inbred parental lnes P1 and P2 of an angiosperm species are crossed to produce F₁ sees in which, the ploidy of the endosperm in 6N. If plants generate from these F₁ seeds are backcrossed with P1, what will be the ploidy of the somatic cell ni the next generation?

(a) 2N

(b) 4N

(d) 6N

Ans. (b)

Sol. Endosperm will have two polar nuclei and one male gamete [3N = 2N + N]. It is given in the statement that the ploidy of endosperm is 6N. Hence, 6N = 4N + 2N. All gametes will be 2N and the ploidy of somatic cells in the next generation will be 4N [2N + 2N].

73. Dark gown Arabidopsis seedlnig show 'triple response' when exposed to ethylene hormone. Which one of the following options is characteristic of 'triple response'?

(a) Reduced shoot elongation, increased shoot thickness and tightening of apical hook

(b) Reduced shoot elongation, reduced shoo thickness and loosening of apical hook

(d) increased shoot elongation, reduced shoot thickness and tightening of apical hook.

Ans. (a)

Sol. Reduced shoot elongation, increased shoot thickness and tightening of apical hook is special feature of triple response, if Arabidopsis seedlings are exposed to ethylene hormone.

74. Brassinosteroids are a group of steroi hormones that function in a variety of cellular and development contexts in plants. Which one of he following acts as an inhibitor of the brassinosteroid receptor?

(a) BRI 1

(b) BKI 1

(d) BSK 1

Ans. (b)

Sol. BK1 is the only reported inhibitor of receptor kinases in Arabidopsis, which negatively regulates BRI 1 in the brassinosteroid pathway. BKI 1 can also interact with another important LRR-RLK, ERECTA (ER). Phenotypic analysis showed that BKI 1 and ER together regulate plant architecture, including pedicel orientation, which is a newly reported phenotype in the Brand ER-mediated developmental processes. Gene expression analysis revealed that BKI 1 regulates a subset of ER responsive genes.

75. Which one of the following metabolites moves from mitochondria to peroxisome during the operaton of the C₂ oxidative photosyntheric cycle?

(a) Glycerate

(b) Glycolate

(d) Serine

Ans. (d)

Sol. Glycine leaves the peroxisome and enters the mitochondrion, where two molecules of glycine are converted to serine and CO₂. The newly formed serine diffuses from the mitochondrion back to the peroxisome, where it is converted to glycerate.

76. Which one of the following statements related to components/features of senescence in plants is incorrect?

(a) Programmed cell death in plants my generate functional cells or tissues.

(b) Senescenece can be induced by application of cytokinins and delayed by overexpression of salicylic acid.

(d) Leaf senescence is regulated by NAC and WRKY genes families.

Ans. (b)

Sol. Cytokinin production slows down the process of senescence. Many studies have shown that exogenous abscisic acid (ABA) promotes leaf abscission and senescence.

77. Which one of the following secondary metabolities is characterized by the presence of a central carbon atom that is bound by a sulphur to a glycone group, and by a nitrogen to sulfonated oxime group?

(a) Alkaloids

(b) Terpenes

(d) Glucosinolates

Ans. (d)

Sol. Glucosinolates are sulphur-rich, anionic -thioglycosides. They contain a central carbon atom that is bound by a sulphur to a glycone group and by a nitrogen to a sulphonated oxime group.

In addition, the central carbon is bound to a side group. Different glucosinolates have different side groups as defined by the amino acids from which they are derived. The amino acid precursors include alanine, valine, isoleucine, leucine, methionine, phenylalanine, tyrosine and tryptophan and chain elongated forms of methionine and phenylalanine.

78. The table given below represents the types of inercellular transport in "Column I" in land plants and their transport pathways in "Column II".

Column I Column II

P. Apoplastic 1. Via interconnecting plasmodesmata

Q. Symplastic 2. Via the water filled spaces of the cell wall matrices and lumen of xylem tracheary elements.

R. Tanscellular 3. Via the vacuole across the tonoplast followed by exist across the plasma membrane before regaining entry to the adjacent cell through cell through the plasma membrane.

Which one of the following combinations matches column I correctly with column II.

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

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

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

Ans. (b)

Sol. The apoplastic (non-living) pathway provides a route toward the vascular stele through free spaces and cell walls of the epidermis and cortex. The symplastic (living) route to the vascular stele involves cell to cell transport by plasmodesmata. Plasmodesmata are channels of cytoplasm lined by plasma membrane that transverse cell walls. Transcellular transport involves passage through two plasma membranes and can occur by secretion and subsequent endocytosis, diffusion or transporter activity.

79. Stomata from etached epidermis of common dayflower (Commelina communis) were treated with saturating photon fluxes of red light. In a parllel treatment. stomata treated with red light were aslo illuminated with blue light (indicated by arrow). From the graphs shown below, selevt the correct patern of stomata opening (solid lines and dotted lines represent stomatal aperture under red and blue lights, respectively).

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Both red and blue light stimulate stomatal opening. Because chlorophyll also absorbs these wavelenths, sensitivity to red and blue light is consistene with a role of guard cells in opening stomata under conditions conducive for photosynthesis. Indeed, in most species guard cells are the only epidermal cells that contain chloroplasts and guard cell chlorophll is implicated as a photreceptor in the light responses of stomata. However, the greater quantum efficiency of blue light over red light in stimulating stomatal opening suggests that guard cells also possess a specific blue-light photoreceptor. So most appropriate response in seen in graph 1.

80. Following are certain statements regarding Rubisco, the predominant protein in plant leaves that catalyzes the initial reaction of the Calvin-Benson cycle.

P. During the oxygenase activity of Rubisco, O₂ is used as substrate to produce three-carbon molecule, 3-phosphoglycerate and two-carbon molecule, 2-phosphoglycolate.

Q. In red and brown algae, the large subunit of Rubsco is localilzed in the chlorplast while small subunit is localized in the nucleus.

R. The bound sugar phosphates in Rubisco are specifically removed by an ATP depenent enzyme, Rubisco activase.

S. The active form or Rubisco catalyzes carboxylation or oxygenation reactions in five steps.

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

(a) P, Q and R

(b) P, Q and S

(d) P, R and S

Ans. (d)

Sol. Rubisco is one of the slowest and largest enzymes, with a molecular mas of 560 kDa. In land plants and green algae, the chloroplast rbcL gene encodes the 55 kDa large subunit whereas a family of rbcS nuclear genes encodes nearly identical 15 kDa small subunits; in nongreen algae both the rbcL and rbcS genes are chloroplast encoded. So statement B is wrong. Correct combination is A, C and D.

81. Sympathetic post-ganglionc neurons that are cholinergic, innervate

(a) sweat glands

(b) parotid glands

(d) pancreas

Ans. (a)

Sol. The postganglionic sympathetic neurons that innervate the sweat glands are, before innervating the sweat glands, adrenergic neurons. Following the innervation of the sweat glands of the palm are innervated by adrenergic neurons. The cells of the adrenal medulla are actually postganglionic cells; they are innervated by preganglionic cholinergic sympathetic neurons.

82. Melanopsin is found in which cell of the retina?

(a) Cones

(b) Rods

(d) Bipolar cells

Ans. (c)

Sol. In humans, melanopsin is found in intrinsically photosensitive retinal ganglion cells (ipRGCs). It is also found in the iris of mice and primates. Melanopsin is also found in rats, amphioxus, and other chordates. ipRGCs are photoreceptor cells which are particularly sensitive to the absorption of the short-wavelength (blue) visible light and communicate information directly to the area of the brain called the suprachiasmatic nucleus (SCN), also known as the central "body clock", in mammals. Melanopsin plays an important non-image-forming role in the setting of circadian rhythms as well as other functions.

83. Prestin, a membrane protein, is found ni whch one of the following cells of the organ of Corti?

(a) Inner hair cells

(b) Inner phalangeal cells

(d) Outer phalangeal cells

Ans. (c)

Sol. Prestin is a protein, evolved from a family of anion transporters, that is only present in the outer hair cells of mammals. In response to changes in membrane potential, prestin changes conformation within the plasma membrane between long and short forms. Because prestin serves, by itself, as a motor, electromotility is fast ordered of magnitude faster than skeletal muscle contraction.

This means that the outer hair cell shortens and lengthens in response to sinusoidal deflection of the hair bundle of physiological frequencies.

The shortening and lengthening of the outer hair cell pulls the basilar membrane up and down, thereby amplifying the excursion of the basilar membrane within the organ of Corti. Thus, outer hair cells boost the stimulus that reaches the inner hair cells.

84. Two individuals A and B, each of 75 kg body weight, have similar volume of body water. Both of them had high salt snack. However, individual A also had a glass of alcoholic drink. Based on this information, which one of the following statements is true?

(a) A will have lower circulating level of antidiuretic hormone (ADH) than B.

(b) B will have lower circulating level of ADH and A.

(d) The reabsorption of water in kidney will be more in A and B.

Ans. (c)

Sol. Antidiuretic hormone helps to control blood pressure by acting on the kidneys and the blood vessels. Its most important role is to conserve the fluid volume of your body by reducing the amount of water passed out in the urine. It does this by allowing water in the urine to be taken back into the body in a specific area of the kidney. Dietary sodium intake increases vasopressin secretion in man and the increased urine flow results from alcohol's acute inhibition of the release of antidiuretic hormone (ADH).

When A consumed a slated diet and alcohol there will be increase in ADH release and subsequently inhibition in the release, but the inhibition of ADH release is not possible in Individual B, so option (A).

85. The different segments of renal tubule (column A) aand the mechanism of Na⁺ transport iin the apical membrane of tubular cells (column B) are tabulated below:

Column A Column B

P. Proximal tubule 1. Na⁺ – Cl^– symporter

Q. Thick ascending loop of Henle 2. Diffusion through Na⁺ – selective

Channel (ENaC)

R. Early distal tubule 3. Na⁺ – glucose symporter

S. Late distal tubule and collecting duct. 4. 1 Na⁺ – 1 K⁺ – 2 Cl^– Symporter

Select the option with the correctmatches:

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

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

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

Ans. (a)

Sol. In proximal tubule, sodium/proton exchanger enables reabsorption of bicarbonate. Glucose is transported along with Na⁺ via Na/glucose symporters.

In thick ascending loop of Henle, Three-ion contransporter (sodium/potassium/chloride) and the sodium/potassium ATPase, which as before maintains the sodium concentration gradient. Sodium is actively pumped out, while potassium and chloride diffuse down their electrochemical gradients through channels in the tubule wall and into the bloodstream. Apporixmately 5% of sodium filtered load is reabsorbed in the early distal tubule via sodium-chloride cotransporter sodium reabsorption in the laste distal tubule takes place through special channels ENaCs.

86. The peaks of the compound action potential (i.e., A, B and C) recorded from a mmmalian mixed nerve were affected after application of increasing pressure on the nerve. Some probable changes of compound action are stated below:

1. 'A' peak was inhibited by lower intensity of pressure

2. 'C' peak was inhibited by higher inensity of pressure

3. 'B' peak was inhibited by lower intensity of pressure

4. 'C' peak was inhibited by lower intensity of pressure

5. 'A' peak was inhibited by higher intensity of pressure

Select the option with the combination of correct staements.

(a) 1 and 2

(b) 2 and 3

(d) 4 and 5

Ans. (a)

Sol. On application of pressure on the nerve, action potential are affected. Here, peaks of compound action potential (i.e., A, B, C) are inhibited by increase in pressure. As a is first peaks, it is inhibited by lower intensity of pressure whereas C peaks is inhibited by higher intensity of pressure.

87. The figure below represents normal sex determination, differentiation and development in humans. Identify A, B, C and D.

(a) A = WT1 (Wilm's Tumor 1), B = MIS (Mullerian Inhibitory Substance), C = SRY, D = Testosterone.

(b) A = GnRH, B = FSH, C = Testosterone, D = Reductase

(d) A = WT1, B = LH, C = ABP (Androgen Binding Protein), D = Inhibin

Ans. (c)

Sol. In figure A is SRY gene, B is MIS, C is testosterone and D is DHT.

88. The Cl^– content of red blood cells (RBCs) in the venous blood was found to be higher than that in arterial blood in a human subject. Followig proposal were mae to explain these observations:

P. The high pCO₂ in venous plasma leads to increase diffusion of CO₂ into RBC and the formation of H₂CO₃.

Q. content in the RBC of venous blood becomes much greater than that in plasma.

R. The excess HCO₃^– leaves the RBC of venous blood along with Na⁺ to plasma by a Na⁺ – symporter.

S. The increased Na⁺ in the venous plasma is transproted to the RBC along with Cl^–.

Select the combination with incorrect statements from the following options:

(a) P and S

(b) Q and R

(d) R and S

Ans. (d)

Sol. The high pCO₂ in venous blood results in diffusion of CO₂ into RBCs. It combines with water to form H₂CO₃, which quickly dissociates into hydrogen ions and bicarbonate ions in presence of carbonic anyhdrase. In order to maintain neutrality bicarbonate ions move out of the cell and chloride ions moves in, this is done by an antiporter chloride bicarbonate exchanger (Band 3 protein).

89. The different waves of normal electro cardiogram (ECG) of a human subject are shown below. The relationship of the events of caardiac cycle to these ECG waves are proposed in the following statements:

1. The P wave occurs due to the depolarization of atria

2. The atrial repolarizton is responsible for the T wave.

3. The QRS complex occurs during ventricular depolarization

4. Q – T intrval indicates plateau protion of auricular action potential.

Select the combination with incorrect statements from the following options:

(a) 1 and 2

(b) 2 and 3

(d) 2 and 4

Ans. (b)

Sol. The normal ECG consists of a P wave, a QRS complex and a T wave. A wave representing repolarization of the atria cannot be seen because it occurs during the QRS complex. The time between the beginning of the P wave and the beginning of the QRS complex is the PQ or PR interval, during which the atria contract and begin to relax. The ventricles begin to depolarize at the end of the PR interval. The QT interval extends from the beginning of the QRS complex to the end of the T wave represents the approximate length of time required for the ventricles to contract and begin to relax.

90. The excitation of auditory hair cells by the displacement of stereocilla has been explained in the following proposed statements:

P. The gradual increased height of stereocillia is required for the transduction process.

Q. The changes of membrane potential of auditory hair cells are proportional to he direction and magnitude of the displacement of stereocillia.

R. The higher concentration of K⁺ in endolymph and higher concentration of Na⁺ in perilymph are not required for the excitation of hair cells.

S. The mechanically sensitive cation channels on the top of stereocillia are not adapted to maintain displacement of stereocilia.

Select the combination with incorrect statemens from the following options:

(a) P and Q

(b) Q and R

(d) P and R

Ans. (c)

Sol. Endolymph should have higher K⁺ concentration as compared to Na⁺ and Na⁺ concentration should be higher in parilymph for the excitation of hair cells. Also the displacement of stero cilia is maintained by mechanically sensitive cation channels.

91. In an organism, allele for red eye colour is dominant over the allele for white eye colour. A cross is made between a white eyed male and a red eyed female. In the progeny all males are red eyes while the females are white eyed. The reciprocal cross leads to all red eyed progeny. Based on the above information which one of the following conclusions is correct?

(a) This is a a sex-limied trait, and the male is the homomoporphic sex

(b) This is a sex-linked trait, with male being the homomorphic sex

(d) This is a case of autosomal inheritance, with incomplete penetrance.

Ans. (b)

Sol. The trait given here is sex linked which is explained by taking X and Y. Here males are XX (homomorphic) and females are heteromorphic (XY).

The first cross is between white eyed male and red eyed females.

X^rX^r × X^RY X^rX^R (male) and X^RY (female) red eyed male and female white eyed.

The second cross is between red eyed male with white eyed female.

X^RX^R × X^rY X^rX^R (male) and X^RY (female) all red eyed.

The given example shows that the trait is sex linked and male is homomorphic.

92. A male snail homozygous for dextral alleles is crossed with a female homozygous for sinistral alleles. All the F1 individuals showed sinistral phenotype. When F1 progeny snails were self-fertilized all individuals of F2 progeny had dextral coiling. This experiment demonstrated

(a) dominant epistasis as extral allele is dominant over sinistral allele.

(b) Recessive epistasis as in F2 dextral allele appeared in homozygous condition

(d) maternal inheritance as the mitochondrial genes of the F1 mother has governed the phenotype of the F2 individuals.

Ans. (c)

Sol. Shell coiling in the case of snail is an example of maternal effect where nuclear gene in the ooplasm of mother determines the phenotype of the offsprin irrespective of its genotype. We can say mother's genotype will be offspring phenotye.

93. A virgin Drosophila female was crossed with a wild-type male. The F₁ progeny obtained had four types of males as shown below:

Assuming that white eye and corssveinless mutations are X-linked and recessive, the following statements were made:

1. F₁ females were also of four types as that of males.

2. The white eyed crossveinless male flies appeared due to independent assortument.

3. The map distance between the genes for white eye and crosseveinless is estimated to be 12 cM.

4. The map distance between white eye and corssveinless is estimated to be 6 cM.

5. All F₁ females are expected to be wild-type.

6. The F₁ wild type males appeared due to crossing over.

The combination with correct statements is:

(a) 3, 5, 6

(b) 1, 2, 4

(d) 2, 4, 5

Ans.

Sol. Statement(e) and (f) are true because it states that (e). All F₁ females are expected to be wild type — since the male parent is wildtype and the mutations are X-linked recessive (given) all the female progeny must be wildtype. F.F₁ wildtype males appeared due to crossing over. This is possible when there occurs a crossing over within the XX female chromosome. Option (A) seems to be the best option because only in this option both statements (e) and (f) fall together along.

94. The locations of five overlapping deletions have been mapped to a Drrosophila chromosome as shown below:

(Horizontal lines in the above figure indicate the deleted regions)

Recessive mutations a, b, c, d and e are known to be located within this region, but the order of mutations on the chromosome is not known. When the flies homozygous for the recessive mutations are corssed with flies homozygous for the deletions, the following results are obtained (letter 'm') represeents mutant phenotype an '+' represents the wild-type). On he basis of the above data, the relative order of the five mutant genes on the chromosome is

(a) b c d e a

(b) a b c d e

(d) c d b e a

Ans. (a)

Sol. Del-1 b c d

Del-2 c d

Del-3 d e

Del-4 e a

Del-5 e a

95. The pedigree given below follows the inheritance pattern of a late-onset (after age of 30 years) genetic disease that is 100% penetrant. Affected individuals are indicated by a solid circle (woman) or solid square (males). RFLP analysis of DNA from each individual is shown below in the pedigree.

Which grandchildren (IIIb and IIId) will be affected by the disease after attaining the age of 30 years?

(a) Only IIIb

(b) Both IIb and IIIc

(d) Both IIIb and IIId

Ans. (d)

Sol. RFLP, is a tehcnique that exploits variations in homologous DNA sequences. In the given problem, the individuals Ia, Ic and IIc are affected that means they must carry the affected allele and must show respective band in RFLP. The analysis of the pattern in which only last 2 bands are present indicates that the disease is occurring. So these are responsible for disease, hence IIIb and IIId will be affected.

96. A chemist synthesizes three new chemical compounds in the laboratory and names them as X, Y and Z. After analysing mutagenic potential of mutations induced by these compounds to be reversed by other known mutagens and obtained the following results.

Assuming that X, Y and Z caused any of the three types of mutations, transition, transversion or single base deletion, what conclusions can you make about the naure of mutations produced by these compounds?

(a) X causes trnasversion; Y causes transition; Z causes single base deletion

(b) X causes transition; Y causes transversion; Z causes single base deletion

(d) X causes transversion; y causes single base deletion; Z causes transition

Ans. (b)

Sol. Treatment of cytosine with nitrous acid produces uracil which pairs with adenine to produce a CG-to-TA transition mutation during replication. Likewise, nitrous acid modifies adenine to produce hypoxanthine, a base that pairs with cytosine rather than thymine, which results in an AT-to-CG transition mutation. So transition mutations can be reverted by Nitrous acid. Mutations induced by hydroxylamine can only be CG-to-TA transitions.

So only AT-GC transitions can be reverted by hydroxylamine. Intercalating agents such as proflavin, acridine and ethidium bromide can cause either additions or deletions, frameshift mutations induced by intercalating agents can be reverted by a second treatment with those same agents. So single base deletion induced by Z can be reverted by acridine orange.

97. An individual is having an inversion in heterozygous condition. The regions on normal chromosome are marked A, B, C, D, E, F, G while the chromosome havnig inversion has the regions as a, b, c, d, e, f, g. The diagram given below shows pairing of these two homologous chromosomes durnig meiosis and the site of a crossing over is indicated:

The following statements are given to describe the inversion and the consequence of crossing over shown in the above diagram:

1. This is a pericentric inversion

2. This wikll generate a dicentric and an acentric chromosome following separaion of chromosomes after crossing over.

3. This will generate two monocentric recombinant chromosomes following separation of chromosomes after crossing over.

4. All the gametes thus formed will have delection and/or duplication and will be non-viable.

5. 50% of the gametes having recombinant chromatid will be non-viable, while 50% gametes having non-recombinant chromatid will survive.

6. This is a paracentric inversion

Which combination of the above statements describe the inversion and meiotic consequences correctly?

(a) 1, 2 and 3

(b) 1, 3 and 5

(d) 3, 4 and 6

Ans. (c)

Sol. The figure represents a paracentric inversion; centromere not involved in inversion. The characteristic dicentric (a chromatid with two centromeres) and acentric chromatid (having no centromere) is formed. During Anaphase when chromatids separate they result in deletion in recombinant gametes hence non viable whereas other two non recombinant gametes will be viable.

98. Two Hfr strains, Hfr-1 (arg⁺ leu⁺ gal⁺ str^s) and Hfr-² (arg⁺ his⁺ gal⁺ pur^r str^s) were mated with a F strain (arg^– leu^– gal^– his^– pur^s str^r). The results of the interrrupted mating experiment are shown as plots 'P' and 'Q' respectively.

Based on these results, identify which of the options accurately reflects the order of loci?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. The two step approach : (1) determine the underlying principle and (2) draw a diagram. Here the principle is clearly that each Hfr strain donates genetic markers from a fixed point on the circular chromosome and that the earliest markers are donated with the highest frequency. By visualizing the time graph for both Hfr strains the order can be deduced. The marker which has less time enters early and vice versa.

99. Bipinnaria and brachiolaria are the larval forms of

(a) Crustacea

(b) Arthropoda and Mollusca, respectively

(d) Asteroidea

Ans. (d)

Sol.

Asteroidea (sea stars) transmit through bipinnaria and brachiolaria larval forms.

100. Scientists discovered two new plant species, 'A' and 'B' that look similar excep that, species 'A' bears flowers and leaves that are twice the size of those in species 'B'. Which method should the scientists use to approprately investgate if species 'A' is a result of gene duplication ni species 'B'?

(a) Sequence similarity, gene structure and gene size.

(b) Plant size, physical proximity of gene and genome size

(d) Sequence length, gene structure and chromosome count.

Ans. (c)

Sol. Gene duplications can also be identified through the use of next-generation sequencing platforms. The simplest means to identify duplications in genomic resequencing data is through the use of paired-end sequencing reads. Sequence analysis allows for the interrogation of all bases within a gene and is used routinely by clinical laboratories to detect point mutations and small insertions and deletions.

101. The animals belonging to phylum Onychophora

(a) have arthropodan characteristics and thus also considered as a class of Arthropoda

(b) have both annelidan and arthropodan characteristics

(d) serve as a conncting link between Annelida and Mollusca.

Ans. (b)

Sol. The Onycophora are a small phylum of about 70 species of quite fascinating animals often referred to by the common name of Velvet Worms or Walking Worms. Onycophorans have a mixture of morphological traits resent in Annelid worms and the Arthropods. They do not represent a mission link between the worms an the arthropods. Instead, they are looked on as a separate line of evolution that arose independently from some long forgotten ancestor.

102. Which one of the following parameters is not used in phenetic classification of bacteria?

(a) Trophism

(b) Susceptibility of a bacteria to a particular bacteriophage

(d) 16S rRNA sequence

Ans. (d)

Sol. In numerical taxonomy (also called computer or phenetic taxonomy) many (50 to 200) biochemical, morphological and cultural characteristics, as well as susceptibilities to antibiotics and inorganic compounds, are used to determine the degree of similarity between organisms. The sequencing of ribosomal RNA (rRNA) genes, which have been highly conserved through evolution, allows phylogenetic comparisons to be made between species whose total DNAs are essentially unrelated. It also allows phylogenetic classification at the genus, family and higher taxonomic levels. The rRNA sequence data are usually not used to designated genera or families unless supported by similarities in phenotypic tests.

103. Which of the following groups represents SAR clade of protists?

(a) Euglenozoans, Red algae, Parabasilids

(b) Brown algae, Forams, Radiolarians

(d) Charophyes, Choanoflagellates, Tubulinids

Ans. (b)

Sol. Sar or Harosa (informally the SAR supergroup) is a clade that includes stramenopiles (heterokonts), alveolatesand Rhizaria. The first letter of each group provides the "SAR" in the name.

The SAR include taxa that are very diverse including some of the most important photosynthetic organisms on Earth. This supergroup can be divided into three clades : the Stramenopiles, the Alveolates and the Rhizaria. Brown Alga is a Stramenopile, while Foram and Radiolarians belong to rhizaria.

104. Following table presents bryophyte phyla with their selected characteristics. In the below table, phyla A, B and C represnets.

(a) A - Marchantiophyta, B - Bryophyta, C - Anthocerotophyta

(b) A - Bryophyta, B - Marchantiophyta, C- Antthocerotophyta

(d) A - Bryophyta, B - Anthocerotophyta, C - Marchantiophyta.

Ans. (b)

Sol. In Bryophyta, reddish-brown, multicellular, branched rhizoids are found and also numerous chloroplasts are present per cell. Stomata are only present in sporophyte capsule. In marchantiophyta rhizoids are one-celled and unbranched but cells contain numerous chloroplast. Stomata is absent in both generations. In anthocerotophyta, one-celled/unicellular rhizoids are present. The gametophyte cells contain one or few chloroplasts. The stomata are present on the capsule wall and widely separated from each other. They arise from median longitudinal division of sporophytic epidermal cells. Thus A-Bryophyta, B-Marchantiophyta and C-anthocerotophyta.

105. In order to survive in a non-aquatic environment, plants acquired several daptatons with specialized functions. Given below is a list of features/characteristics (Column a) and their potenial role (Column B)

Which one of the following options represents a correct match between the adaptatons and their functions?

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

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

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

Ans. (b)

Sol. Waxy cuticle prevents water loss and maintains tissue hydration. Thickened/lignified cell wall provides mechanical support. The pigmentation protects against excess light and prevents ROS formation. The Homoiohydry maintains the vascular system that plays an important role in transportation.

106. Following table shows a list of clades and plants:

Which one of the following is a correct match for the above?

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

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

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

Ans. (d)

Sol. The basal angiosperms are the flowering plants which diverged from the lineage leadint to ... Nymphaeales (water lilies, together with some other aquatic plants) and Austrobaileyales (woody aromatic plants including star anise). Eudicots includes — The rose family (Rosaceae) provides fruits such as strawberries, raspberries, apples, cherries, peaches and plums and others. Orchids are the largest family of monocots among the angiosperms (flowering plants), with between twenty-five thousand and thirty thousand species and new species are continually being described.

Piperales, order of flowering plants comprising 3 families, 17 genera and 4,170 species. Along with the orders Laurales, Magnoliales and Canellales, Piperales forms the Magnoliid clade, which is an early evolutionary branch in the angiosperm tree; the clade corresponds to part of the subclass Magnoliidae under the old Cronquist botanical classification system.

107. The following table shows names of bones (Column A) and specifiic features (Column B)

Which one of the following options gives the correct match of the bones with their specific fetures?

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

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

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

Ans. (a)

Sol. The odontoid process (also dens or odontoid peg) is a protuberance (process or projection) of the Axis (second cervical vertebra). It exhibits a slight constriction or neck, where it joins the main body of the vertebra. The deltoid tuberosity is located on the lateral surface of the humerus a little more than a third of the way along its length and is the attachment site for the deltoid muscle.

Acromian process is a bony process of the scapula which forms a part of the pectoral girdle. The articular surface of the ulna is called sigmoid cavity/notch.

108. The figure below shows the nervous system of Mollusca with ganglia and the connecting nerves. The connecting nerves are labelled as A, B, C and D. Which one of the following options has correct labelling of A, B, C and D?

(a) A – Cerebral commissure, B – Left Cerebropedal connective

C – Pedal commissure, D – Left Pedal-visceral connective

(b) A – Cerebral connective, B – Left Cerebropedal commissure

C – Pedal connective, D – Left Pedal-visceral commissure

C – Pedal commissure, D – Left Pedal-visceral connective

(d) A – Cerebral commissure, B – Left Cerebropedal commissure

C – Pedal commissure, D – Pedal-caudal connective

Ans. (a)

Sol. The nervous system of all molluscs, is bascially different from vertebrates nervous system. Molluscs, except, highly developed cephalopods, have no brain in the strict sense. The mollusc nervous system is referred to as a tetraneural nervous system, because there are four main neural strands : A, B, C and D. In the given diagram, A is Cerebral commissure; B is Left Cerebropedal connective; C is Pedal commissure and D is Left Pedal visceral connective.

109. Given below are some properties related to botanical and zoological nomenclature

P. Absence of tautonyms

Q. Presence of genus and species ranks only

R. Absence of principle of coordination

S. Presence of only holotype and neotype

Select the correct combination tha distinguishes botanical nomenclature from zoological nomenclature system.

(a) P, Q and S

(b) P, Q and S

(d) P, R and S

Ans. (c)

Sol. Correct combination that distinguishes botanical nomenclature from zoological nomenclature system is as follows : In plant nomenclature (ICBN), tautonyms are not valid i.e., generic name and specific name should not be same in plants, e.g., Magnifera magnifera. But tautonyms are valid for animal nomenclature (ICZN–International Code of Zoological Nomenclature) e.g., Naja naja (Indian cobra), Raitus rattus (Rat). Hence statement A is correct.

article 4.1 of the ICBN states that A plant may be assigned to taxa of the following ranks (in descending sequence) regnum, subregnum, divisco, subdivisio, classis, subclassis, ordo, subordo, familia, subfamilia tribus, subtribus, genus, subgenus, sectio, sebsectio, series, subseries, speices, subspecies, varietas, subvarietas, forma, subforma.

article 24.1 : The name of an infraspecific taxon is a combination of the name of a species and an infraspecific epithet connected by a term denoting its rank. An item often overlooked in this respect is that the infraspecific epithet should be preceded by a term denoting its rank. This applies not only when introducing a new taxon, but also when quoting one, since the article clearly states that this term is part of the name. Therefore Discoaster tanii nodifer, which is commonly used, should correctly by cited as Discoaster tanii subsp. nodifer. Although the term is abbreviated as 'subsp.' in the ICBN, the abbreviated 'ssp.' is also very common and can be considered correct. Hence statement B is wrong as there are other ranks conidered too.

Statement of the Principle of Coordination — A name established for a taxon at any rank in the family group, the phytoflagellates excepted, is deemed to be established with the same author and date for taxa based upon the same name bearing type (type genus) at other ranks in the family group, with appropriate mandatory change of nomenclature in accordance with the Principle of Coordination is not considered validly published under the botanical Code unless it appears in print and is applied to an accepted taxon. Hence statement C is right. Statement D is wrong as there are other type specimen terms like isotype, lectotype etc. Therefore the correct combination is A and C.

110. Given below are biodiversity hotsports in decreasing order of endemic plant species recorded in them. Select he correct order.

(a) Western Ghats and Sri Lanka > Indo-Burma > Sundaland > Philippines

(b) Philippines > Sundaland > Indo-Burma > Western Ghats and Sri Lanka

(d) Western Ghats and Sri Lanka > Sundaland > Philippines > Indo-Burma

Ans. (c)

Sol. Biodiversity is closely related to sustainable development and is essential for ecosystem stability. It directly or indirectly benefits human life.

It is estimated that there are four thousand species of flowering plants known from the Western Ghats and 1,500 (nearly 38%) of these are endemic.

Indo Burma — A conservative estimate of total plant diversity in the hotspot reveals about 13,500 vascular plant species, of which about 7,000 (52%) are endemic.

Sundaland : In terms of vascular plant diversity and endemism, no truly reliable estimates exist. However, by extrapolating from what is known of the principal islands and countries that comprise the hotspot, total vascular plant diversity is estimated at roughly 25,000 species and the number of endemics at 15,000.

At the very least one-third of the more than 6000 vascular plant species native to the Philippines are endemic.

111. Which of the following options lists ecosystems in increasing order of plant productivity per day per unit leaf area?

(a) Tropical forests, hot deserts, temperate forests

(b) Hot deserts, temperate forests, tropical forests

(d) Tropical forests, temperate grasslands, hot doserts.

Ans. (d)

Sol. Tropical forests are most dense with multiple layers of trees. As a result most leaves are overshadowed by each other and many of the understory plants to receive diffuse sunlight or no light at all. This imposes a constraint for light to get to all chloroplasts and gaseous diffusion limitations impose a constraint for CO₂ to get all internal cells. Thus the plant productivity per day unit area is least for tropical forests.

The leaves of temperate forest trees are relatively better exposed to sunlight and therefore their productivity per day per unit leaf area is more than tropical forests. The leaves of plants in hot deserts are fully exposed to sunlight hence their productivity per day per unit leaf area will be maximum.

112. A general increase in the average body mass of animal population within a species with latitude is known as

(a) Allen's rule

(b) Bergmann's rule

(d) Hamilton's rule

Ans. (b)

Sol. Bergmann's rule is an ecogeographical rule that states that within a broadly distributed taxonomic clade, populations and species of larger size and found in colder environments and species of smaller size are found in warmer regions. According to Bergmann's rule, animals have large body size in cold climates as compared to warm climates. Large bodies have a smaller surface area to volume ratios so lesser heat is dissipated in the environment. In warm climates the animals are required to expel heat, so bodies are smaller and more linear.

113. Ruderal species are those which are found in the environment with

(a) low disturbance, high competition

(b) high disturbance, low competition

(d) high distrubance, high competition

Ans. (b)

Sol. Grime (1979) expanded the r- and K-selected system and recognized three categories of plants, i.e., ruderals (R), competitors (C) and stress tolerants (S). Grime visualizes two groups of environmental factors, the stress factors and the disturbance factors. The stress factors are those which reduce biomass production, e.g. low light, low moisture, etc. The disturbance factors remove part or whole of biomass, e.g., grazing. Four combinations of these groups are possible.

(i) high stress, high disturbance

(ii) low stress, high disturbance

(iii) low stress, low disturbance, and

(iv) high stress, lwo disturbance.

The first combination is too harsh to permit colonisation. The combnation (ii) is suitable for ruderal species, combination (iii) for competitive species and combination (iv) for stress tolerants. The ruderal species are similar to r-selected species, which colonise disturbed patches of habitats. The R species exhibit rapid growth, early maturation and have high reproductive rates. Competitive species are found in habitats that are rich in resources, relatively undisturbed and with dense populations. Competitors tend to allocate carbon to resource gaining structures including roots, leaves and stems and are long-lived. S species have also long life spans and occur in stressed environments such as deserts, arctic tundra, peat bogs and serpentine soils and grow slowly.

114. Following table gives a list of international environmental agreements and areas covered.

Which of the following is correct combination?

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

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

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

Ans. (b)

Sol. The Basal Convention was adopted to Control of Trans boundary Movements of Hazardous Wastes and their Disposal. The Cartagena Protocol is an international agreement which aims to ensure the safe handling, transport and use of living modified organisms (LMOs). Kyoto Protocol is an international agreement linked to the United Nations Framework Convention on Climate Change, which commits its Parties by setting internationally binding greenhouse gas emission reduction targets. The Stockholm Convention is a global treaty to protect human health and the environment from persistent organic pollutants. (POPs).

115. The following figure is a 'risk graph' that illustates the percent risk a sepecies faces towards extinction. The following are ranks assigned according to IUCN's red-list category:

1. Critically endangered

2. Near threatened

3. Vulnerable

4. Least concern

Which one of the following is the most appropriatte match between the percent-risk and their assigned rank?

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

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

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

Ans. (a)

Sol. The Critically Endangered (CR) category comprises of organisms taht are facing an extremely high risk of extinction in the wild. Since in the "risk graph" a represents maximum percent risk so it denotes the CR category. Next in the order in Vulnerable (VU), category which comprises of organisms considered to be facing a high risk of extinction in the wild thus it is represented by b.

The percent risk c represents Near Threatened (NT), category which comprises of organisms close to qualifying for or is likely to quality for a threatened category in the near future. The least percent risk d represents Least Concern (LC) category. The organisms listed to this category are widespread and abundant.

116. The complexity of a food web in a community s quantified using certain parameters which are defined below. Which of the following is an incorrect representation?

(a) Chain length =

(b) Connectance =

(d) Linkage density =

Ans. (c)

Sol. Hierarchically higher species can prey on lower-positio ned species; however, the opposite does not occur. This constraint sets the maximum possible number of potential links in a food web in an ecosystem. The maximum possible number of potential links is given by the formula :

L_max = (all the possible pairs among n species) – (the pairs among B basal species)

Where n are the animal species of which b are the basal species. The basal species do not eat within the food web and all the basal species are at the lowest position, above which non-basal species were arranged in a hierarchy.

117. The below graph illustrates two lines that represent the immigration and extinction rates for an island based on its distance from mainland (solid line) and its size (dotted line) Which of the following is true for this island?

——— Distance of island from mainland

--------- Size of island

(a) It is close to the mainland and is very small

(b) It is far from mainland and is very large

(d) It is far from the mainland and is very small.

Ans.

Sol. According to "equilibrium model of island biogeography" the farther islands have lower probability of immigration as compared to nearer islands. It also states that species numbers tends to increase with increasing area and with greater area the rate of extinction declines. The graph shows a higher immigration rate thereby indicating that the island is closer/nearer to the mainland. In addition, it shows a lower extinction rate thus indicating that it is a large island.

118. Following are certain statements regarding energy efficiencies of ectotherms and endotherms:

P. Ectotherms have high assimilation efficiency but low production efficiency

Q. Ectotherms have low assimilation efficiency but high production efficiency

R. Endotherms have high assimilation efficiency but low production efficiency

S. Endotherms have low assimilation efficiency but high production efficiency

Which one of the following represents the combination of correct statements?

(a) P and Q

(b) Q and R

(d) P and R

Ans. (b)

Sol. The assimilation efficiency is the proportion of ingested food that is assimilated. Since endotherms tend to digest food more completely than ectotherms so they have higher assimilation efficiency than the latter. The production efficiency is the proportion of assimilated food that goes into new consumers biomass. Since endotherms have high respiration rates and are also required to maintain body temperature so they have low production efficiency as compared to ectotherms.

119. In a sample from a population there were 65 indiviuals with BB genotype, 30 individuals with Bb genotype and 15 indivduals with bb genotype. The frequency of the 'b' allele is

(a) 0.59

(b) 0.27

(d) 0.73

Ans. (b)

Sol. Given here genotype BB = 65, Bb = 30 and bb = 15 thus allelic frequency of 'b' will be calculated as f(b) = total number of 'b' allele/total number of all alleles 30 from Bb and 15 * 2 from bb.

Thus, 30 + 30/220 = 0.27.

120. A group of palaeontologists digging in an area discovers a pre-hstoric human burial site. The same group, while exploring a nearby area, discoverd fossils most ccurately?

(a) ¹⁴C dating for human remains and ²³⁵U dating for dinosaur remains.

(b) ⁸⁷Rb dating for both human and dinosaur remains.

(d) ¹²⁹I dating for human remains an ¹²⁹Xe for dinosaur remains

Ans. (a)

Sol. The most widely known form of radiometric dating is carbon-14 dating. But carbon-14 dating does not work on dinosaur bones. The reason being, the half-life of carbon-14 is only 5,730 years but Dinosaur bones are millions of years old, some fossils even billions of years old. To determine the ages of these specimens, scientists need an isotope with a very long half-life. Some of the isotopes used for this purpose are uranium-238, uranium-235 and potassium-40, each of which have a half-life of more than a million years.

121. Given below are statements related to the two competing hypotheses on the origin of modern humans: the Out-of-Africa hypothesis and the multi-regional hypothesis. Which of the following statements in incorrect?

(a) Both the hypotheses support tha Homo erectus originated in Africa and expanded to Eurasia.

(b) Mitochondrial DNA (mtDNA) and Y-chormosome DNA evience support the 'Out-of-Africa' hyothesis.

(c) The principal confict between the two hypotheses is tha multi-regional hypothesis does not support african origin of Homo erectus.

(d) The multi-regional hypothesis states that independent multiple origins occured in the million years since Homo erectus came out of Africa.

Ans. (c)

Sol. Both the competing hypotheses agree that Homo erectus originated in Africa and expanded to Eurasia about one million years ago, but they differ in explaining the origin of modern humans.

The out-of-Africa hypothesis proposes that a second migration out of Africa happened about 100,000 years ago, in which anatomically modern humans of Africa origin conquered the world by completely replacing archaic human populations Homo sapiens. The multiregional hypothesis states the independent multiple origins or shared multiregional evolution with continuous gene flow between continental populations.

122. Which of the following statements is true for positive-frequency dependent selection?

(a) Fitness of a genotype increases as it becomes less common

(b) Fitness of a genotype increases as it becomes more common

(d) Fitness of a genotype decreases as it becomes common and gets fixed.

Ans. (b)

Sol. In positive-frequency dependent selection, the fitness of a phenotype increases as it becomes more common. In "positive-frequency dependent selection" fitness increases with frequency so that rare genotypes are eliminated and local genetic diversity reduced.

123. Inclusive fitness of an animal can be measured s a sum of direct6 fitness and indirect. Imagine you have 10 offsprings. Through dillgent parental care, 5 survive to reproduce. You give your life in a heroic deed to save a total of 5 of your nieces and nephews. What is your inclusive fitness?

(a) 15

(b) 12.5

(d) 3.75

Ans. (d)

Sol. Inclusive Fitness = Direct Fitness + Indirect Fitness

= (Survival of offspring) × (r for parent-offspring) + (Survival of non-descendant kin) × (the proper r for each type of relationship)

For the given problem we have

5*0.5 + 5*0.25 = 2.5 + 1.25 = 3.75.

124. Altruism describes a behaviour perfomed by animals that may be disadvantageous to self while benefitting others. Which one of the following statements is incorrect about altruism?

(a) It is he net gain of direct fitness when sociality is facultative

(b) It is under positive selection via indirect fitness benefits that exceed direct fitness costs

(d) It is favoured when rb – c > 0 where c is fitness cost to altruist, b is finess benefit to recipien, andd r is genetic relatedness.

Ans. (a)

Sol. Hamilton's rule that altruism is favoured when rb – c > 0; where c is the fitness cost to the altruist, b is the fitness benefit to the recipient and r is their genetic relatedness. This predicts that altruism is favoured when r or b are higher and c lower. Hamilton's theory is referred to in many ways. Hamilton called it 'inclusive fitness theory', but it is more often referred to as 'kin selection', a term coined by John Maynard Smith. Jerram Brown pointed out that the inclusive fitness of an individual is divided into two components : 'direct fitness' and 'indirect fitness'. Direct fitness is gained through the production of offspring and indirect fitness through aiding the reproduction of nondescendent relatives. A behaviour is only altruistic if it leads to a decrease in direct fitness, so altruisms can only be favoured when an indirect benefit outweighs thie direct cost, as shown by Hamilton's rule.

125. Given are some statements with reference to the use of genes in plant molecular systematics.

P. mtDNA are not preferred over cpDNA or rDNA because they generally show slow rate of sequence evolution and fast rate of structural evolution.

Q. cpDNA are not preferred because of their haploidy, unipaarenal inheritance, and absence of recombination among cpDNA molecules.

R. rDNA such as ITS are preferred for their higher evolutionary rates as well as shorter sequence length.

S. rDNA and cpDNA cannot be used simulaneously in molecular systematics since they represent conflicting patterns of inheritance.

Which of the above statements are incorrect?

(a) P, R and S

(b) P, Q and R

(d) Q and S only

Ans. (d)

Sol. Chloroplast DNA (cpDNA) is a traditional workhorse for reconstructing evolutionary relationshipe among angiosperms. The frequeny use of cpDNA in such analyses is predicated on the apparent simplicity of its inheritance : uniparental through the maternal line and lacking biparental recombination. Hence statement B is wrong. Since the cpDNA molecule is highly conserved, there is very little intra-specific variation and cpDNA based RFLP studies have therefore mostly been conducted on an inter-specific level. By contrast, oplant mitochondria have never been much used in molecular analyses. The major reason is that while the plant mitochondrial DNA (mtDNA) sequence is usually highly conserved, the size and structure of mtDNA molecules may very widely even within individual plants. Moreover, recent studies indicated that substitution rates of mtDNA genes can vary enormously even among closely related plant species. Hence statement A is correct. The 16S-23S internally transcribed spacer (ITS) regions of the rRNA operon might be under minimal selective pressure during evolution and therefore have more variation in the sequences than that of the coding regions of 16S and 23S rRNAs. While the sequence of the rRNA genes evolves slowly, the internal transcribed spacers (ITS1 and ITS2), the external transcribed spacer (ETS) and the intergenic spacer (IGS) evolve rapidly. Hence statement C is correct.

126. Following are key points about the effect of genetic drift:

P. Genetic drift is significant in small populations

Q. Genetic drift can cause allele frequencies to change in a pre-directed way

R. Genetic drift can lead to a loss of genetic varation within population

S. Genetic drift can causes harmful alleles to become fixed.

Which one of the following combination of the above statements are true?

(a) P and Q only

(b) P and R only

(d) P, R and S

Ans. (d)

Sol. Changes in relative allele frequency due only to random sampling error are known as genetic drift, a stochastic process. Hence, statement B is wrong. The smaller the populations, the smaller the subset of potentially successful gametes and the more likely that genetic drift will occur. Hence statement A is right. Genetic drift can happen in two ways : Founder Effect in which a small sample of breeding individuals from a large population colonize a new area. Bottleneck Effect : most members of a large population are removed (perhaps due to some natural disaster such as a hurricane, volcanic eruption, pathogen invasion or other catastrophe that doesn't favour any particular genotype over another) leaving only a few survivors. Through sampling error, genetic drift can-cause populations to lose genetic variation. Hence, statement C is right. Alleles which are not harmful nor beneficial can become fixed by chance through genetic drift can cause alleles that are slightly harmful to become fixed. When this occurs the population's survival can be threatened. Hence, option D is right.

127. Following table contains some of the generalizations of evolutionary biology:

Which of the following is correctmatch between column I and II?

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

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

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

Ans. (b)

Sol. Cope's rule states that animal population lineages tend to increase in body size over evolutionary time.

Dollo's law of irreversibility, states that "an organism never returns exactly to a former state, even if it finds itself placed in conditions of existence identical to those in which it has previously lived ... it always keeps some trace of the intermediate stages through which it has passed."

Van Valen's law states that the probability of extinction for species and higher taxa (such as families and orders) is constant for each group over time; groups grow neither more resistant nor more vulnerable to extinction, however old their lineage is.

Occam's razor (or Ockham's razor) is a principle from philosophy and science. Suppose their exist two explanation s for an occurrence. In this case the simpler one is usually better — also called the Principle of Parsimony.

128. Three anatomical characteristics (A, B and C) of invertebrate nervous system are used to build a generalized cladogram given below. Presence of the anatomical character is indicated by '+'. Based on the pattern of character distribution, pick the correct combination that are represented by A, B and C.

(a) A – unpaired nerve cord, B – Paired nerve cord, C – Cephalic ganglia.

(b) A – cephalic ganglia, B – unpaired nerve cord, C – paired nerve cord.

(d) A – unpaired nerve cord, B – cephalic ganglia, C – paired nerve cord.

Ans. (b)

Sol. Loriciferans contains a paired ventral nerve cord whereas nematodes and nematomorpha contain unpaired ventral nerve cord. The blueprint of arthropod CNS consists of a dorsal cephalic ganglion, the 'brain' followed by a chain of ventral ganglia, the ventral cord. In the Spiralia (e.g. Gnathifera, Trochozoa) the main neural trancts found in the ventro-lateral body region are paired. Within Spiralia, Gnathifera may represent the deepest branching lineage comprising the jaw worms Gnathostomulida and their sister group Micrognathozoa + Syndermata.

Their intra-epidermal unsegmented nervous systems comprise an anterior brain and three to five ventral and two to four dorsal longitudinal nerves, connected by few transverse commissures. Neurites of the stomatogastric nevous system wee found lining the pahrynx and connecting to a prominent buccal ganglion. The entire body of Onychophora, including the stub feet, is littered with numerous papillae : warty protrusions that carry a mechanoreceptive bristle (responsive to mechanical stimuli) at the tip, each of which is also connected to further sensory nerve cells lying beneath. the mouth papillae, the exists of the slime glands, probably also have a function in sensory perception.

Sensory cell known as "sensills" on the "lips" or labrum respond to chemical stimuli and are known as chemoreceptors.

These are also found on the two antennae, which can be regarded as the velvet worm's most important sensory organs. Except in a few (typically subterranean) species, one simply constructed eye (ocellus) lies laterally, just underneath the head, behind each antenna.

129. Protein conformational dynamics cannot be dertermined by which one of the following techniques?

(a) NMR spectroscopy

(b) Differential scanning calorimetry

(d) Fluorescence microscopy

Ans. (b)

Sol. Mass spectra only determine the mass of the protein. It never gives any information either about structure or dynamic features of the protein.

130. Given below are a few statements on Agrobacterium mediated transfromation of plants. Which one of the following statements is correct?

(a) T-DNA transfer occurs from left border to right border

(b) The gfp reporter gene can never be used for selection of transgenic plants

(d) Host plant genes play an important role in influencing transformation frequencies

Ans. (d)

Sol. Virulence genes present in Agrobacterium are required for mediating the transfer of T-DNA from the bacterium into the plant. Thus, overexpression of virulence genes increase the transformation frequencies.

In addition to virulence genes of Agrobacterium, host plant genes are equally essential for mediating the transfer of T-DNA.

131. A uracl containing plasmid was constructed and was used in transformatioin into the wild-type (ung⁺) and uracil-N-glycosylase mutated (ung^–) E. coli cells and scored for transformants in the presence of appropriate antibiotics. Which one of the following statements correctly describes the experimental outcomee?

(a) ung⁺ cells will have fewer tansformants compared to ung^– cells.

(b) ung^– cells will give fewer transformants compared to ung⁺ cells.

(c) No transformants will be obtained in ung^– cells as uracil excision repair will not occur and the plasmid would not replicate.

(d) Presence of uracil in DNA is unnatural and the plasmid DNA with uracils in it will not produce transformants in either ung⁺ cells.

Ans. (a)

Sol. Bacterial transformation is a process of horizontal gene transfer by which some bacteria take up foreign genetic material (naked DNA) from the environment. ung+ cells will have fewer transformants compared to ung- cells due to lack of sequence in wild-type.

132. Which one of the following assay systems can specifcally detect apoptotic cells?

(a) Tetrazolium dye (MTT) based colorimetric assay

(b) FITC – annexin V based FACS analysis

(d) Trypan blue exclusion assay

Ans. (b)

Sol. Annexion V binds to Phosphatidyl Serine (PS) in a calcium dependent manner. PS is normally only found on the intracellular leaflet of the plasma membrane in healthy cells, but during early apoprtosis, membrane asymmetry is lost and PS translocates to the external leaflet. Fluorochrome-labelled Annexin V can then be used to specifically target and identify apoptotic cells.

133. From the various techniques listed below, which one cannot be used to precisely map the transcripion start-site of a gene?

(a) S1 Mapping

(b) Sequencing the region downstream of promoter

(d) Primer Extension Method

Ans. (b)

Sol. S1 nuclease mapping is a nuclease protection assay using nuclease S1. This technique is used to quantify and map RNA transcripts. It is capable of identifying individual RNAs in a mixture of RNA sample of known sequence. It can particularly map introns and 5' and 3' ends of transcribed gene regions. Primer extension is a method typically used to map the 5' end(s) of an RNA, thus defining the transcription start site and providing initial evidence for where the promoter is located within a cloned gene. RACE results in the production of a cDNA copy of the RNA sequence of interest, produced through reverse transcription, followed by PCR amplification of the cDNA copies (RT-PCR). The amplified cDNA copies are then sequenced and, if long enough, should map to a unique genomic region. RACE with high-throughout sequenceing was first introduced in 2009 as Deep-RACE to perform mapping of Transcription start sites (TSS) of 17 genes in a single cell line. However, sequencing the region downstream of promoter cannot be used to map transcription start site.

134. Following are statements to depict relationship among measures of central tendency in a skewed dataset:

P. In positively skewed datasets, mean > median > mode

Q. In positively skewed datasets, made > median > mean

R. In negatively skewed datasets, mean > median > mode

S. In negatively skewed dataasets, mode > median > mean.

Which of the above statements are true?

(a) P and Q

(b) P and R

(d) P and S

Ans. (d)

Sol. In deciding which measure of central tendency to use, in a normal distribution the mean, mode and median should give similar results. However, when scores are badly skewed in eiter direction and data are on an ordinal scale, Vincent (2005) recommends using the median. The mode may be appropriate when only a rough estimate of the measures of central tendency is needed. When data are presented on a ratio or interval scale, the mean is recommended, especially for making further calculations (i.e., standard deviations or standardized scores).

Figure : Normal and skewed distributions.

135. The MALDI mass spectrum of a pepte gave a single peak with m/z of 2000. The ESI mass spectrum of the same peptide gave multiple peaks. These observations indicate that

(a) degradation has occured while acquiring ESI mass spectrum

(b) multiple charged species of the same compound are observed in the ESI spectrum

(d) ESI induces polymeriztion of the peptide.

Ans. (b)

Sol. Fragmentation of the protonated or deprotonated molecular ions generated from ESI is generally limited, and the mass spectra are relatively simple.

However, multiple-protonation of proteins and peptides occurs in the ESI process and the ESI mass spectra might becomes more complicated.

A single protein will show its characteristic cluster ions containing multiple-charged ions. The number of charges on the protein molecules will depend on the molecular weight of the protein and the number of accessible basic sites.

136. A resercher attempted to clone two genes (X and Y) independently in a plasmid vector for overexpression and purification in E.coli. All attempts to clone gene X were unsuccessful whereas gene 'Y' could be cloned easily. When the researcher attempted to clone gene 'X' in the plasmid clone containing gene 'Y', gene 'X' could be cloned. The following statements were proposed to explain the above results.

P. Protein encoded by gene 'Y' is no lethal to the cell.

Q. Gene 'X' has introns, which prevents its expression in E.coli.

R. Expression of 'X' protein is lethal to the cell

S. The 'Y' gene product inhibits the activity of 'X' protein

Which one of the following options represents a combination of correct statements to explain the observations?

(a) Only P and Q

(b) Q, R and S

(d) P, R and S

Ans. (d)

Sol. As Y protein is easily encoded so is not lethal to the cell. While X protein is lethal to cell and 'Y' gene product inhibits activity of 'X' protein.

137. Sub-cellular fractionation-based assays have been used to identify various organelles in the mammalian cell, which of the following microscopy-based method would be the most accurate?

(a) Use of fluorescent probes specific for organelles

(b) Use of organelle specific fluorescent probes followed by microinjection of fluorescent antibodies against organelle-specific protein

(d) Use of organelle specific fluorescent probes followed by cryo-electron microscopy.

Ans. (b)

Sol. Both of these fluorescent labels have different binding sites e.g. that respond to changes in microenvironments (e.g., pH, viscosity, and polarity) or quantities of biomolecules of interest (e.g., ions, reactive oxygen species, and enzymes). So their use in combination will help locate structure in a better way. Further, unlike antibodies, these fluorescent probes can be used to investigate organelle structure and activity in live cells with minimal disruption of cellular function.

138. In order to detect minor variaions in antigen concentration, the following procedures were suggesed. Which one will likely be the best option?

(a) Antigen coated microtitre well add antibody add enzymes conjugated secondary antibody add substrate and measure colour.

(b) Antibody coated microtitre well add antigen add enzyme conjugated secondary antibody add substrate and measure colour.

(c) Preincubate antigen with fixed amount of antibody add to antigen coated well add enzyme conjugated secondary antibody add substrate and measure colour.

(d) Preincubate antigen with fixed amount of antibody add to antibody coated well add enzyme conjugated secondary antibody add substrate and measure colour.

Ans. (c)

Sol. To detect minor variation in antigen out of the options given competitive ELISA can be performed. In which the primary antibody (unlabelled) is incubated with sample antigen. Antibody-antigen complexes are then added to 96-well plates which are precoated with the same antigen.

Unbound antibody is removed by washing the plate. (The more antigen in the sample, the less antibody will be able to bind to the antigen in the well, hence "competition.")

The secondary antibody that is specific to the primary antibody and conjugated with an enzyme is added. A substrate is added and remaining enzymes elicit a chromogenic of fluorescent signal. The strength of the signal inversely proportional to antigen concentration.

139. From the following statements:

P. Surface plasmon resonnce can be used to determine binding constants only in the range of 10² – 10³ M.

Q. de novo sequencing s not possible by mass spectral methods

R. The position of hydrogen atoms in proteins is not directly determined by X-ray diffraction

S. Circular dichroism an nuclear magnetic resonance spectroscopy do not give the same information on protein structure.

Choose the option with all correct statements:

(a) P, Q and R

(b) P, R and S

(d) R and S

Ans. (d)

Sol. Hydrogen has only one electron hence it cannot diffract enough X-rays so their position cannot be determined. Circular dichrosim gives conformational dynamics while NMR predicts overall structure.

140. Three students (P, Q, R) in a research lab were trying to identify proteins that interact with a transcription factor X. P performed gel filtration experimetns and identified that X was found along with proteins A, B, C and D. Q performed co-immunoprecipitation experiments using antibodies to X and identified A, B and C. R did a yeast-2-hybrid screen and identified only B.

The following are likely conclusions that may explain all the results:

1. A, B, C and d are in a complex with X.

2. X directly interacts with B

3. Only A, B and C are in complex with X

4. D is probably weakly associated with X.

Which of the above conclusions best explains all the results?

(a) 1, 2 and 3

(b) 1, 2 and 4

(d) 2, 3 and 4

Ans. (b)

Sol. Co-immunoprecipitation can provide false positives as it is less stringent and somehow weakly associated interacdtion while it must be yeast-2-hybrid.

141. A 127 bp genomic DNA sequence of a prokaryotic gene was cloned under a strong constitutive promoteer along wth a suitable polyA signal and used for development of transgenic tobacco plants. Molecular analysis revealed the presence of three types/lengths of transgene derived mRNAs: 555 bp, 981 bp an 1257 bp– n the leaves of transgenic plants. The following statements were proposed to explain the above results.

P. The three mRNAs represent alternatively spliced transcripts due to the presence of putative intronic sequence in the gene.

Q. The gene sequence was haracterized by the presence of potential polyadenylation signals that resulted in premature termination of transcription.

R. Expression of full-length transcripts (1257 bases) was lethal to the transformed cells.

S. The transgenic plants were chimeric in nature and comprised of a mix of transformed and untransformed cells.

Which of the following combinations of the above statements would correctly explain the obtained results?

(a) P and R

(b) Q and S

(d) R and S

Ans. (c)

Sol. Only A and B options are correct. C statements is wrong because transgenic plant is expressing full transcript. And D statement is also wrong, because mixed expression will not tell about different mRNA transcripts.

142. Which one of the following set of essential components are required for Sanger method of DNA sequencing in a required buffer containing MgCl₂ and Tris-HCl?

(a) DNA template, a primer, 4 deoxyribonucleoides, 4 labelled dideoxyribonucleotides, DNA polymerase.

(b) DNA template, a primer, 4 labelled dideoxyribonucleotides, DNA polymerase, DNA polymerase.

(d) DNA template, a primer, 4 labelled dideoxyribonucleotides, DNA polymerase, telomerase.

Ans. (a)

Sol. DNA ligase is not required in Sanger sequencing.

143. Given below is a table with information on isotopes, their half life and type of particle(s) they emit.

Choose the correct combinaton from the options given below:

(a) P-3-y; Q-2-x, y; R-1-y

(b) P-3-x; Q-1-x; R-2-x, y

(d) P-1-x; Q-2-x; R-3-x, y

Ans. (b)

Sol. The short half-life of carbon-11 (20.38 minutes) creates special challenges for the synthesis of C-11 labeled tracers. C-11 show Positron emission, beta plus decay, or ß+ decay. The half-life of radiocarbon (14C) is 5700 ± 30 yr, which makes it particularly useful for dating in archaeology. Sodium has two radioactive cosmogenic isotopes (22Na, half-life = 2.605 years; 24Na, half-life = ~ 15 hours) that have been used as tracers in hydrologic studies. 24Na emits both beta and gamma particles.

144. A certain protein has been assumed to play an indispensable role in the survival of an intracellular parasite inside that host cells. Which one of the following techniques will best prove the assumpton to be correct?

(a) Treat the parasite-infected host cells with an inhibitor of the protein and check the number of parasies per host cell under the microscope

(b) Check the expression of the protein ini parasite-infected host cells

(d) Treat the parasite-infected host cells with an activator of the protein and check the number of parasites per host cell under the microscope.

Ans. (a)

Sol. Expression and activity of that protein can be checked in parasitic-infected host cells. It can also be checked by increasing the level of protein using activator to check number of parasites per host cell. So, statement A is incorrect for checking.

145. In order to visualize the intracellular organization of a cell, one can utilize various microscopy-based techniques. These include:

1. Differential interference contrast (DIC) microscopy

2. Phase contrast microscopy

3. Dark field microscopy

4. Epifluorescence microscopy

5. Scanning electron microscopy

6. Transmission electron microscopy

7. Confocal microscopy

Which of the above mentioned microscopes can be used to study the intra-cellular dynamics using live cell imging?

(a) 1, 2, 5, 6, 7

(b) 1, 2, 3, 4, 7

(d) 3, 4, 5, 6, 7

Ans. (b)

Sol. Live cells cannot be visualized using electron microscopy or high energy beam would kill them. Also they need to be fixed viewing them.

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