21. The first step in glycogen breakdown releases glucose units as

(a) glucose 6-phosphate

(b) glucose 1-phosphate

(c) glucose

(d) glucose and glucose 6-phosphate

Ans. (b)

Sol. The breakdown of glycogen back to glucose is called glycogenolysis. Although major amounts of glycogen are stored in both muscle tissue and the liver, glycogenolysis can occur in the liver (and kidney and intestinal cells) but not in muscle tissue because one essential enzyme (glucose-6-phosphatase) is lacking. The first step is glycogen breakdown is the cleavage of linkages, catalyzed by glycogen chain as glucose-1-phosphate :

The second step in glycogen breakdown is the cleavage of the linkages by hydrolysis, which is carried out by the debranching enzyme. When the branch points are eliminated, the phosphorylase can then continue to act on the rest of the chain.

In the debranching reaction, phosphoglucomutase catalyzes the isomerization of glucose-1-phosphate to glucose-6-phosphate :

glucose-1-phosphate glucose-6-phosphate

22. Which one of the statements on protein conformation, detailed below is incorrect ?

(a) L-amino acids can occur in Type I -turns where are both positive.

(b) A peptide rich in proline is unlikely to adopt -helical structure.

(c) Proline residues have high propensity to occur in -turns.

(d) The dihedral angles of amino acids in unfolded proteins are exclusively positive.

Ans. (d)

Sol. Option (A) is correct because Type I' beta turns usually have 0, +30, +60 and +90 dihedral angles and whichever Ramachandran plots we consider are usually for L-Type amino acids only.

Option (B) is correct because prolines are helix breakers as these do not support H-bonding. It also causes a bend in the structure.

Optcion (C) is correct because peptide bonds involving the imino nitrogen of proline readily assume the cis configuration, a form that is particularly amenable to a tight turn. So, proline residues readily occur in beta turns.

Option (D) should be wrong because we cannot randomly predict if the dihedral angle has to be positive or negative as the structure is unfolded.

23. Choose the incorrect statement from the following statements made for an enzyme-catalyzed reaction.

(a) The kinetic properties of allosteric enzymes do not diverge from Michaelis-Menten behaviour.

(b) In feedback inhibition, the product of a pathway inhibits an enzyme of the pathway.

(c) An antibody that binds tightly to the analog of the transition state intermediate of the reaction S P, would promote formation of P when the analog is added to the reaction.

(d) An enzyme with K_cat = 1.4 × 10⁴ s^–1 and K_m = 9 × 10^–5 M has activity close to the diffusion controlled limit.

Ans. (a)

Sol. Option (A) is incorrect because Allosteric enzymes show relationships between V₀ and [S] that differ from Michaelis-Menten kinetics. They do exhibit saturation with the substrate when [S] is sufficiently high, but for some allosteric enzymes, plots of V0 versus [S] produce a sigmoid saturation curve, rather than the hyperbolic curve typical of non-regulatory enzymes. On the sigmoid saturation curve we can find a value of [S] at which V₀ is half-maximal, but we cannot refer to it with the designation Km, because the enzyme does not follow the hyperbolic Michaelis-Menten relationship. Instead, the symbol [S]05 or K05 is often used to represent the substrate concentration giving half-maximal velocity of the reaction catalyzed by an allosteric enzyme.

Option (B) is correct because in feedback inhibition products inhibit either the first enzyme of a straight chain pathway or the first enzyme of the branch reaction.

Option (C) is also correct because in enzymes, if an antibody is developed to bind to a molecule that's structurally and electronically similar to the transition state of a given chemical reaction, the developed antibody will bind to and stabilize the transition state just like a natural enzyme, lowering the activation enery of the reaction and thus catalyzing the reaction. By raising an antibody to bind to a stable transition-state analog, a new and unique type of enzyme is produced.

Option (D) is correct because diffusion controlled limits range between 10⁸ to 10⁹/M/s and calculating k_cat/K_m values also give us value in this range.

24. On seqences analysis of a doble stranded DNA, the results showed the content of cytosine, C was 20%. What is the amount of A and T put together ?

(a) 20%

(b) 30%

(c) 50%

(d) 60%

Ans. (d)

Sol. According to Chargaff's rules states that DNA from any cell to all organisms should have a 1 : 1 ratio. Base Pair i.e. A is equals to T and G is equals to C or Purine = pyrimidine.

G (purine) = C (pyrimidine) and

A (purine ) = T (pyrimidine)

Total G + C in Double stranded DNA = 20 + 20 = 40%

Total A + T in double stranded DNA will be = 100 – 40 = 60%.

25. Sugar puckering in double stranded nucleic acids is exclusively

(a) C-2' endo in douuble starnded DNA.

(b) C-3' endo in douuble starnded DNA.

(c) C-2' endo in douuble starnded RNA.

(d) C-2' endo in hybrid duplex with one strand as DNA and the other as RNA.

Ans. (d)

Sol. The hybrid is found in a standard A-form conformation with all the sugars in the C3' endo puckering. The 5'-terminal base dC of the DNA strand was clearly visible in the electron density map of the present structure, in contrast to the previously reported structure d(TTCTTBr⁵CTTC).r(gaagaagaa) where the 5'-terminal base dT was not visible, leaving the terminal rA unpaired.

26. Thermodynamics of proteins folding is depicted as a free enegy funnel below :

Given below are regions in the diagram (Column X) and their representation (Column Y)

X Y

P. 1. Native structure

Q. 2. Structure with highest entropy

R. 3. Molten globule

S. 4. Discrete folding intermediates

Choose the option that shows all the correct matches.

(a) P–2, Q–3, R–4, S–1

(b) P–1, Q–2, R–3, S–4

(c) P–3, Q–4, R–2, S–1

(d) P–4, Q–1, R–2, S–3

Ans. (a)

Sol.

The figure is 'molten globule' concept showing different intermediates observed in the folding of globular proteins. It is very easy to identify 'D' here, it denotes the native structure. If you know only this match, you can easily omit options B and D. Concentrating on options A and C now. Nothing is common in them except D-(i) match. As it can be seen in the figure that energy is increasing from bottom to top and matched to (ii). Hence, it you through the options A and C. Option A is correct as A-ii, D-i.

27. Equal volumes of pH 4.0 and 10.0 solutions are mixed. What will be the approximate pH of the final solution ?

(a) 7.0

(b) 5.0

(c) 6.0

(d) 4.0

Ans. (d)

Sol. This question can have more than one answer correct depending on context. As the nature of two solutions is not known we may assume different conditions as follows.

Condition 1 : If we consider that solution 1 is a strong acid (like HCl) and solution 2 is a strong base (Like NaOH).

For HCl, pH = –(log H⁺)

4 = –(log H⁺)

H⁺ = 10^–4 mol L^–1

For NaOH, pH 10

pH = –(log H⁺)

10 = –(log H⁺)

H⁺ = 10^–10 mol L^–1

OH^– = 10^–4 mol L^–1

Hence, both will be neutralized and net H⁺ ions in the solution would be contributed by water, i.e. 10^–7 and hence pH of the resulting solution would be –log 10^–7 = 7.

Condition 2 : This may be true particularly when base is not strong and does not actively contribute enough OH ions to neutralize the acid.

Since pH of the first solution = 4. Hence [H⁺] = 10^–4 mol L^–1

pH of the second solution = 10. Hence [H⁺] = 10^–10 mol L^–1

On mixing 1L of each solution, molar concentration of total H⁺ is halved.

[H⁺] =

Small values like 10^–10 could be ignored.

[H⁺] = 0.5 × 10^–4 mol L^–1 = 5 × 10^–5 mol L^–1

pH = –log (5 × 10^–5) = –log10^–5 = 5 – 0.69 = 4.31

In addition to this if both the solution together constitue a buffer, then solution may acquire any pH depending on strength of each component.

28. The inborn error of amino acid metabolism, alkaptonuria, is due to the lack of one of the following enzymes :

(a) Fumaryl acetoacetate hydrolase

(b) -ketoacid decarboxylase

(c) Homogentisate oxidase

(d) p-hydroxyphenylpyvate dehdoxylase

Ans. (c)

Sol. The concept of inborn errors or metabolism was formulated by Sir Archibald Garrod in 1908 as a result of his studies on a condition called alkaptonuria, a rare inherited deficiency of homogentisic acid oxidase.

Inherited errors of metabolism are important consideration in differential diagnosis of illness presenting in infancy.

Alkaptonuria : This rate autosomal recessive deficiency of homogentisic acid oxidase is an example of an inborn metabolic error that does not produce serious effect until adult life. Classically the patient's urine darkens on standing and the sweat may also be black! Homogentistic acid accumulates in connective tissues, principally cartilage, where the darkening is called ochronosis. This accumulation casuses joint damage. The underlying condition cannot be cured; treatment is symptomatic only.

29. Some coenzymes taht seve as transient carriers of specific chemical grops are shown below :

Coenzyme Chemical group transferred

P. Coenzyme A 1. Electrons

Q. Flain adenine dinucleotide 2. Acyl groups

R. Pyridoxal phosphate 3. Hydride ions

S. Nicotinamide adenine dinucleotide 4. Amino groups

Choose the combination with all correct matches.

(a) P–1, Q–2, R–3, S–4

(b) P–2, Q–1, R–4, S–3

(c) P–3, Q–4, R–2, S–1

(d) P–4, Q–3, R–1, S–2

Ans. (b)

Sol. Coenzyme A serve as transient carriers of acyl group; FAD is a second electron carrier used by a cell during cellular respiration; Pyridoxal phosphate functions as a carrier of amino groups and as an electron sink by facilitating dissociation of the -hydrogen of the amino acid : NAD helps in the transfer of hydride ions.

30. The Na⁺/K⁺ ATPase pump is found on the plasma membrane of most animal cells. A mutation in the intrinsic phosphorylation site of the pump is most likely to affect

(a) the outward movement of Na⁺ only.

(b) inward movement of K⁺ only.

(c) both the inward and outward movement of K⁺ and Na⁺.

(d) has no effect on pump activity but affects its stability.

Ans. (c)

Sol. The Na⁺/K⁺-ATPase is an example of a P-type ion pump. The "P" stands for phosphorylation, indicating that, during the pumping cycle, the hydrolysis of ATP leads to the transfer of the released phosphate group to an aspartic acid residue of the transport protein, which in turn causes an essential conformational change within the protein. Conformational changes are necessary to change the affinity of the protein for the two cations that are transported.

Consider the activity of the protein. It must pick up sodium or potassium ions from a region of low concentrations, which means that the protein must have a relatively high affinity for the ions. Then the protein must release the ions on the other side of the membrane into a much greater concentration of each ion. To do this, the affinity of the protein for that ion must decrease. Thus, the affinity for each ion on the two sides of the membrane must be different. This is achieved by phosphorylation, which changes the shape of the protein also serves to expose the ion binding sites to different sides of the membrane, so if any change in intrinsic phosphorylation site it will affect both inward and outward movement of Na and K ions.

31. The site of the division plane during cytokinesis of animal cells is determined

(a) by position of nucleus

(b) by the central spindle

(c) by the pre-prophase band

(d) randomly

Ans. (b)

Sol. The mechanism by which the spindle determines the division plane in animal cells is perhaps the most enduring mystery in cyrokinesis. Cleavage signals are sent to the cell cortex by microtubules of the spindle, but the source and molecular basis of these signals are not clear and probably vary in different cell types.

The central spindle contains numerous regulatory proteins and may have multiple functions in cytokinesis : it not only helps to determine the site of cleavage but it may also help to govern the inward movement of the cleavage furrow and the membrane deposition that occurs there.

As the cleavage furrow deepens, the contractile ring eventually meets the central spindle, which is then compacted into a structure called the midbody, which contains bundles of antiparallel microtubules and a dense protein matrix at its midline.

32. Which one of the following statements is not true about nucelosomal organization of core particle?

(a) The typical structure of DNA is altered in the middle of the core particle.

(b) In core particle, DNA is organized as flat super helix with 1.65 turns around the histone octamer.

(c) While forming 30 nm fibers, generally 6 nucleosomes per turn organize into a two-start helix.

(d) The N-terminal histone tails in a core particle are strictly ordered and exit from the nucleosomes between turns of the DNA.

Ans. (d)

Sol. Nucleosomes have a common structure :

Nucleosomal DNA is divided into the core DNA and linker DNA depending on its susceptibility to micrococcal nuclease.

There are 1.65 turns of DNA wound around the histone octamer.

The core DNA is the length of 146 bp that is found on the core particles produced by prolonged digestion with micrococcal nuclease.

Linker DNA is the region of 8-114 bp that is susceptible to early cleavage by the enzyme.

Changes in the length of linker DNA account for the variation in total length of nucleosomal DNA.

Linker histone is associated with linker DNA and may lie at the point where DNA enters or leaves the nucleosome.

The structure of the DNa is altered so it has an increased number of base pairs per turn in the middle but a decreased number at the ends.

33. Table below shows the list of organelles (Column A); and the signals (Column B) that proteins to the organelle.

Column A Column B

P. Lysosome 1. Stretch of amino acid sequence rich in Lys and Arg residues.

Q. Mitochondria 2. C-terminal tripeptide.

R. Nucleus 3. N-terminal amphipathic helix rich in Lys and Arg.

S. Peroxisome 4. Mannose-6-phosphate.

Choose the option that shows all correct matches.

(a) P–2, Q–3, R–4, S–1

(b) P–2, Q–4, R–3, S–1

(c) P–4, Q–3, R–1, S–2

(d) P–4, Q–3, R–2, S–1

Ans. (c)

Sol. Lysosomes : Lysosomes are composed of soluble and transmembrane proteins that are targeted to lysosomes in a signal-dependent manner. The majority of soluble acid hydrolyases and modified with mannose 6-phosphate (M6P) residues, allowing their recognition by M6P receptors in the Golgi complex and ensuring transport to the endosomal/lysosomal system.

Mitochondria : Most mitochondrial proteins are encoded by the nuclear genome and translocated into the mitochondria. Like the proteins destined for other subcellular organelles, the mitochondrially targeted proteins possess targeting signals within their primary or secondary structure that direct them to the organelle with the assistance of elaborate protein translocating and folding machines. Mitochondrial preproteins that carry the cleavable N-terminal mitochondrial targeting sequence (MTS). This signal is classically characterized as an N-terminal motif predicted to form an amphipathic helix that is 15-70 residues in length and enriched in postively charged basic residues.

Nucleus : Nuclear localization signals have since been identified in many other. Most of these sequences, like that of T antigen, are short stretches, like that of T antigen, are short stretches rich in basic amino acid residues (lysine and arginine). Both the Lys-Arg and Lys-Lys-Lys-Lys sequences are required for nuclear targeting, but the ten amino acids between these sequences can be mutated without affecting nuclear localization.

Peroxisomes : Peroxisomes are ubiquitous membrane bound organelles involved in fatty acid oxidation and respiration. Proteins destined for peroxisomes are synthesized exclusively on free polysomes and are post-translationally imported into the organelle wihout any detectable modifications. Recent works has shown that a number of mammalian peroxisomal proteins and one insect peroxisomal protein, firefly luciferase are targeted to peroxisomes of CV-1 cells by a tripeptide sequence at the carboxyl termini of these proteins. This tripeptide has the generalized structure Ser/Ala/Cys-Lys/Arg/His-Leu and has been termed the peroxisomal targeting signal or PTS.

34. Following statements are made about chromatin remodelling in human cells :

P. Local chromatin conformation may play more important role than the local DNA sequence of the promoter.

Q. Histones in nucleosome can undergo many different covalent modifications, which in turn, alter the chromatin architecture locally.

R. Chromatin remodelling is a developmentally regualted passive process which does not require ATP.

S. Several histone vaiants exist, which replace the standard histones in specific types of chromatin.

Select the option that has the combination of all correct answers.

(a) P, R, S

(b) P, Q, R

(c) P, Q, S

(d) Q, R, S

Ans. (c)

Sol. Chromatin remodeling is the rearrangement of chromatin from a condensed state to a transcriptionally accessible state, allowing transcription factors or other DNA binding proteins to access DNA and control gene expression. Statement P, Q and S is true.

35. Irrespective of the chromosomal configuration, a single X-chromosome remains active in all diploid human somatic cell lines. Which one of the following mechanisms best accounts for the above phenomenon ?

(a) A maternally inherited X-chromosome is developmentally programmed to remain active by avoiding DNA methylation.

(b) Chromosome specific expession and binding of rox1 to one of the X-chromosomes protects it from Xist mediated silencing.

(c) The T-six gene produces just enough of the Xist antisense RNA to block one Xic locus.

(d) A cell produces just enough of the blocking factor to block one Xic locus.

Ans. (d)

Sol. A single X-chromosome remains active in all diploid somatic cells, regardless of sex chromosome constitution. The cell appears capable of counting the number of Xic elements present and hence, the number of X chromosomes. A popular mechanism to account for this phenomenon was proposed by Rastan (1983) and is known as the blocking factor model. The model states that a diploid cell produces just enough of this factor to block one Xic at random, during initiation of inactivation.

36. In eukaryotic cells, replication initiation from a replication origin occuurs only once per cell cycle and S-phase CDKs play a vital ole in the regulation DNA replication. In budding yeast a protein complex known as the origin recognition complex (ORC) is associated with DNA replication origin during G₁; however, origins fire only once at the beginning of S-phase. DNA replication does not start in G₁ because

P. MCM helicases are inactive in G₁.

Q. Spindle checkpoint is active in G₁.

R. DNA polymerase is not recruited in G₁.

S. ORC and initiation factors Cdc6 and Cdt1 do not recruit MCM helicases to the site of replication initiati in G₁.

Which one of the above statements are correct ?

(a) P and Q

(b) P and R

(c) Q and R

(d) Q and S

Ans. (b)

Sol. Initiation of eukaryotic DNA replication is divided into two steps. First step is formation of pre-initiation complex which occurs in G1 phase. This involves the binding of protein ORC to the origin along with othe proteins like cdc6 and cdt1, which finally recruit the helicase MCM. Second step is activation of pre-replicative complex by the S cyclin-CDK via phosphorylation. This activates MCM which starts the DNA unwinding by helicase activity. Activation also recruits DNA polymerases DNa pol alpha, delta and epsilon to initiate DNA replication. So, ideally the synthesis of DNA starts in S phase.

37. Measurements of the rate of actin treadmilling in vivo show that it can be seveal times higher than can be achieved with pre actin in vitro. The treadmilling in vitro can be enhanced by providing

P. profilin that binds G-actin on the site opposite the nucleotide binding cleft.

Q. cofilin binds specifically to the ADP containing F-actin and destabilizes the acting filament.

R. buffer with ATP and low levels of cations.

S. buffer with ADP and low levels of cations.

Which of the above statements are correct ?

(a) P and S

(b) P and R

(c) R and S

(d) P and Q

Ans. (d)

Sol. ATP hydrolysis on actin is the key reaction that allows filament treadmilling. It regulates barbed-end dynamics and length fluctuations at steady state and specifies the functional interaction of actin with essential regulatory proteins such as profilin and ADF/cofilin. Profilin is a major regulator of actin assembly that binds ATP–G-actin with high affinity. Profilin–actin exclusively feeds barbed-end growth, thus pointed ends only disassemble in the presence of profilin. As a result, profilin enhances treadmilling.

38. Following are the list of some cellular eceptors (Column X) and with possible functional characteistics (Column Y) :

Which one of the following is the correct match ?

(a) P–1, Q–2, R–3

(b) P–2, Q–3, R–1

(c) P–3, Q–2, R–1

(d) P–1, Q–3, R–2

Ans. (c)

Sol. The major role of ASGPR is the binding, internalization, and subsequent clearance from the circulation of glycoproteins that contain terminal galactose or N-acetylgalactosamine residues. More specifically, in most mammals, the asialoglycoprotein receptor removes glycoproteins that have had some of their sugars, particularly a terminal sialic acid, removed from the end of the protein. The transferrin receptor is a membrane glycoprotein whose only clearly defined function is to mediate cellular uptake of iron from a plasma glycoprotein, transferrin. Steroid receptors are a class of molecules that function as both signal transducers and transcription factors. From cloned sequences it is apparent that steroid receptors and other transcription factors belong to a superfamily of proteins that appear to function by similar mechanisms.

39. During replication, RNaseH removes all of the RNA primer except the ribonucleotide diectly linked to the DNA end. This is because

(a) it can degrade RNA and DNA from their 5' end

(b) it can only cleave bonds between two ribonucleotides

(c) it can degrade RNA and DNA from their 3' end

(d) activity of RNaseH is inhibited by the presence of duplex containing both strands as DNA.

Ans. (b)

Sol. RNase H removes all of the RNa primer except the ribonucleotide directly linked to the DNA end. This is because RNase H can only cleave bonds between two ribonucleotides. The final ribonucleotide is removed by a 5' exonuclease that degrades RNA or DNA from their 5' ends.

40. Eukaryotic mRNAs are modified to possess a 5' cap structure. Which one of the following is an incorrect statement about the function of the 5' cap structure?

(a) It protects the mRNA from 5' 3' exoribonuclease attack.

(b) It facilitates splicing of the nascent transcripts.

(c) It protects the transcript from degradation by RNAse III family enzymes.

(d) It facilitates attachment to 40S subunit of ribosome.

Ans. (c)

Sol. The cap structure is characteristic of all RNA polymerase transcripts and consists of an inverted 7-methyl guanosine linked via a 5'-5' triphosphate bridge to the first transcribed residue, m7G(5')ppp(5')X. The cap structure has been implicated in many aspects of RNA metabolism. It provides resistance to 5'-3' exonucleases and contributes to a variety of cellular processes including pre-mRNA splicing, polyadenylation, RNA nuclear export and mRNA translation. Capping occurs co-transcriptionally and is one of the earlies RNa modifications. Decapping enzymessuch as DCP2, with additional cofactors, hydrolyze the 5' cap, exposing the mRNA to decay that is carried out by XRN1, a processive exoribonuclease that completely hydrolyzes decapped (5' monophosp horylated) RNA in the 5' 3' direction Bacterial RNase III plays important roles in the processing and degradation of RNA transcripts.

41. In a human cell line, a large fractio of double-strand DNA breaks are repaired by non-homologous end joining (NHEJ). An inhibitor of endonulcease will affect

(a) recruitment of DNA-dependent kinase.

(b) gap trimming.

(c) DNA unwinding.

(d) Pairing of micro-homology regions.

Ans. (b)

Sol. Inhibitor of Flap endonuclease will affect gap trimming as FEN is responsible for removing the flap. It is a structure-specific nuclease which possesses double-strand specific 5' to 3' exonuclease activity. Mechanism of DSB microhomology-mediated alternative NHEJ involves recognition of microhomology by MRN complex and PARP1, followed by recruitment of FEN1 (Flap endonuclease), which can remove the flap. Further, recruitement of XRCC1-LIGASE III at the site helps in ligating the DNA ends leading to an intact DNA.

42. In an experiment, the student has infected mammalian host cell with cytoplasmic RNA virus. The virus growth was monitored by measuring the intracellular viral RNA at different time intervals. It was observed that viral RNA titre progressively went down with time, particularly 12 hours post infection. Following are few possibilities which can explain this observation.

P. The virus infection triggered unregulation of miRNAs that might have downregulated the host factor critical for viral RNA replication.

Q. The virus might encode miRNAs that regulate (inhibits) its won replication.

R. One of the viral proteins inhibits replication of the viral RNA to restrict rapid proliferation.

S. Viral RNA goes to nucleaus with time and thus not detectable in the cytoplasmic 12 hours post infection.

Which one of the following options has all correct statements ?

(a) P, Q and R

(b) P, R and S

(c) P, Q and S

(d) Q, R and S

Ans. (a)

Sol. Statement C is correct as seen incase where Nuclear Factor NF45 Interacts with Viral Proteins of Infectious Bursal Disease Virus and Inhibits Viral Replication, so there is a possibility that viral proteins can inhibit viral replication when associated with some cellular protein. Studies of virus-encoded miRNAs, predominantly from the double straded DNA Herperviridae family, have shown that virus encoded miRNAs target viral and host transcripts, facilitate host immune response evasion and share targets with endogenous host miRNAs.

Statement D is incorrect as Cytoplasmic RNA viruses replicate/transcribe their genomes resulting in high levels of cytoplasmic viral RNA and small regulatory RNAs.

Statement A and B seems correct as there is evidence that most times host miRNAs play a role in viral life-cycles and promote infection through complex regulatory pathways. miRNAs can also be encoded by a viral genome and be expressed in the host cell. Viral miRNAs can share common sequences with host miRNAs or have totally different sequences.

They can regulate a variety of biological processes involved in viral infection, including apoptosis, evasion of the immune response or modulation of viral life-cycle phases.

43. The different arms in the tRNA structure are shown in column A. The specific signatures associated with the different arms are shown in column B.

Column A Column B

P. Acceptor arm 1. Dihydrouriding

Q. Anticodon arm 2. 7 bp stem and CCA sequence

R. arm 3. 5 bp stem

S. D-arm 4. Pseudouridine

Choose the correct matches from the following.

(a) P–2, Q–4, R–1, S–3

(b) P–1, Q–3, R–4, S–2

(c) P–4, Q–1, R–2, S–3

(d) P–2, Q–3, R–4, S–1

Ans. (d)

Sol. (1) The 3' end terminates into 5'CCA3' sequence that is always unpaired. (2) The terminal A residue is the site at which the amino acid is bound covalently. (3) Then come the first loop containing 7 unpaired bases. This loop is designated as "T ΨC loop" because it always contains a sequence 5' ribothymidine pseudouridine-cytidine 3'. This loop is involved in binding to ribosomes. (4) After the "5'-T Ψ C-3' loop", in the 5' direction there occurs a loop of variable size, called the extra loop or the "lump". The lump may contain 3 to 21 bases. (5) The third loop contains 7 unpaired bases and and it has the "anticodon". Anticodon consists of 3 bases. At the 3'-end of the anticodon, there is a purine (A or G) while at the 5'-end, there is always uracil (U). At the time of protein synthesis, anticodon pairs with its complementary "codon" on mRNA. (6) The fourth loop is larger than others and contains 8-12 unpaired bases. It is designated as "D-loop" because it is rich in dihydrouridine (UH2). The enzyme aminoacyl synthetase binds to this loop.

44. In a strain of E. coli, a fusion between the Iac and trp operon took place and the new locus structure is shown below. The strain lacks the wild-type trp operon.

Given below are some of the potential scenarios :

P. Tryptophan will be synthesized in a medium containing lactose and tryptophan.

Q. Tryptophan synthesis will be repressed in a medium containing glucose.

R. Tryptophan synthesis will take place only in the absence of sufficient tryptophan in the medium.

Choose the option that correctly describes the behaviours of the fusion operon.

(a) P and Q

(b) P and R

(c) R only

(d) Q and R

Ans. (a)

Sol. The operon created is fused operon with the regulatory genes of the lac operon and structural genes of trp operon. So, expression of trp structural genes will be under the regulation of lac regulatory genes. It is no longer under the regulation of tryptophan presence or absence. Expression of the lac operon occurs when lactose in present and glucose is absent. Presence of glucose suppresses the lac operon by catabolite repression.

45. Following statements have been about removal of supercoiling produced by DNA unwinding at the replication fork :

P. Accumulation of supercoils is the result of DNA helicase activity during unwinding DNA.

Q. Problem of DNA supercoiling is valid only for circular chromosomes of bacteria and not for the linear chromosomes.

R. Supercoiling of DNA is removed by topoisomerases by breaking either one or both strands of DNA on the unreplicated DNA in front of replication fork.

S. Both topoisomerase I and topoisomerase II can remove positive supercoiling during replication.

Which one of the following options has all correct statements ?

(a) P, Q, R

(b) P, Q, S

(c) P, R, S

(d) Q, R, S

Ans. (c)

Sol. The supercoils introduced by the action of the DNA helicase are removed by topoisomerases that act on the unreplicated double-stranded DNA in front of the replication fork. These enzymes do this by breaking either one or both strands of the DNA without letting go of the DNA and passing the same number of DNA strands through the break. This action relieves the accumulation of supercoils. In this way, topoisomerases act as a "swivelase" that rapidly dissipates the accumulation of super coils induced by DNA unwinding.

Type I topoisomerases clave one strand of DNA and Type II topoisomerases cleave both the strands to resolve the supercoils. Both Type I and II can resolve positive supercoils.

46. Phosphorylation of the -subunit of eIF2 at Ser 51 position in Saccharomyces cerevisiae leads to sequestration of eIF2B, a guanosine exchange factor. This phenomenon is

(a) known to activate translation of the capped mRNAs in the cytosol.

(b) known to activate translation of many key mRNAs possessing short ORFs (uORFs) in the mRNA sequence that preceds the main ORF.

(c) an essential requirement for translation of IRES containing mRNAs.

(d) an essential requirement for the transport of mature mRNAs out of the nucleus.

Ans. (b)

Sol. Phosphorylation of the α-subunit of eIF2 at Ser 51 position in Saccharomyces cerevisiae leads to sequestration of eIF2B, a guanosine exchange factor. This phenomenon is known to activate translation of many key mRNAs possessing short ORFs (uORFs) in the mRNA sequence that preceds the main ORF.

47. E. coli mutants isolated from a genetic screen showed following classes of mutations.

P. Point mutations in IacI.

Q. Deletions immediately downstream of the transcription start site of the IacZYA mRNA.

R. Duplications of part of whole of IacY.

S. Duplication of part of whole of IacA.

Choose the option which is likely to result in constitutive expression of Iac operon ?

(a) Both P and Q

(b) Both Q and R

(c) Both R and S

(d) Only P

Ans. (a)

Sol. Constiutive expression of lac operon can be achieved by mutation in lac repressor (LacI) and any mutation in operator region which lie immegiately down stream to transcription start site of LacZYA mRNA.

48. For Escherichia coli chromosomal DNA replication, which one of the following statements is true ?

(a) DNA polymerase I is the main polymerase required for DNA replication.

(b) DNA polymerase I thought identified orignially by Kornberg as the one responsible for replication, is not important for the DNA replication process.

(c) Requirement of DNA polymerase I is in the context of removal of RNA primer needed for DNA synthesis and then fill in the same with DNA equivalent.

(d) DNA polymerase I is the primary enzyme for the error prone DNA synthesis in response to SOS.

Ans. (c)

Sol. For Escherichia coli chromosomal DNA replication, requirement of DNA polymerase I is in the context of removal of RNA primer needed for DNA synthesis and then fill in the same with DNA equivalent.

49. Following observations were made about variations among genomes of eukaryotic organisms :

P. Single nucleotide polymorphism are the numerically most abundant type of genetic variants.

Q. Both, interspersed and tandern repeated sequences can show polymorphic variation.

R. Mitotic recombination between mispaired repeats change in copy numbre and generates minisatellite diversity in population.

S. Smaller variable segments in the genome can be identified by paired end mapping technique.

Select the option with all correct statements.

(a) P, Q, R

(b) P, R, S

(c) Q, R, S

(d) P, Q, S

Ans. (d)

Sol. SNP are the most common type of genetic variation among people. They occur almost once in every 1000 nucleotides on average which means there are roughly 4-5 million SNPs in a person genome (1). Both, in tersepersed and tandem repeated sequences can show polymorphic variation (2). Smaller variable segments in the genome can be identified by paired end mapping technique (3). Here in given option A, B and D is correct.

50. In an experiment it was observed that a protein was unregulated in the cancer tissues (compared to control tissues) that showed correlation with disease progression. Following are a few possibilities which can explain the above observation.

P. A mutation could be located in the 3'UTR of the corresponding mRNA at a miRNA binding site.

Q. A mutation changes the conformation of the protein, resulting in its better stability.

R. A mutation in the corresponding mRNA promotes ribosome read-through of the terminator codon resulting in increased synthesis of the protein.

S. A mutation in the corresponding mRNA increased the stability of the RNA due to change in secondary structure.

Which one of the following combinations represents the most likely explanations?

(a) P, Q and R

(b) Q, R and S

(c) R, S and P

(d) P, Q and S

Ans. (d)

Sol. In the termination codon is read through by the ribosome because of the mutation, it will result in a different protein, not increased synthesis. Hence this cannot be an explanation for upregulated synthesis.

Evidence for statemetn A to be a possible explanation : The let-7 family includes 12 miRNA members, which are thought to play important roles as tumor suppressors. Let-7 binds to the 3'-UTRs of key oncogenes including RAS and MYC and inhibits their expression. Hence if this 3'-UTR site is mutated in the corresponding mRNA, then miRNA cannot bind and oncogenes expression will be upregulated.

Evidence for statement B and D to be possible explanation : The cytoplasmic level of a messenger RNA and hence protein, depends not only upon its rates of synthesis, processing and transport, but its decay rate as well. Sequences elements within an mRNA, together with the protein and/or small noncoding RNA factors that bind these elements, dictate its decay rate. Genetic alternations in mRNA stability can lead to various diseases, including cancer, heart disease and immune disorders. Hence, a mutation in the corresponding mRNA may make it more stable for increased translation.

51. To prepare individual tissue cells from a primary culture, the cell-cells and cell-matrix interaction must be broken. To achieve this, one would not use

(a) EDTA

(b) Trypsin

(c) Collagenase

(d) Sepaase

Ans. (d)

Sol. Cell culture is the process by which cells are grown under controlled conditions, generaly outside their natural environment. After the cells of interest have been isolated from living tissue, they can subsequently be maintained under carefully controlled conditions.

The first step in isolating cells of a uniform type form a tissue that contains a mixture of cell types is to disrupt the extracellular matrix that holds the cells together. The best yields of viable dissociated cells are usually obtained from fetal or neonatal tissues. The tissue sample is typically treated with proteolytic enzymes (such as trypsin and collagenase) to digest proteins in the extracellular matrix and with agents (such as ethylenediaminetetraacetic acid or EDTA) that bind or chelate, the Ca²⁺ on which cell-cell adhesion depends. The tissue can then be teased apart into single living cell by gentle agitation.

52. Which one of the following does not belong to human antimicrobial proteins and peptides at epithelial surfaces forming part of innate immunity ?

(a) Lactoferrin

(b) Defensin

(c) Calprotectin

(d) Vimentin

Ans. (d)

Sol. Lactoferin is found primarily in mucosal secretions, sythesized by epithelial cells. Lactoferrin is considered a first-line deense protein involved in protection against a multitude of microbial infections. -defensins are secreted by most leucocytes and epithelial cells. -Defensins are produced by the respiratory epithelium and the alveolar macrophage and secreted into the airway surface fluid. Calprotectin is also an epithelial cell-derived antimicrobial peptide Vimentin : Vimentin is a type III cytosol but is also expresse on the host cell surface.

53. Fruit bats are known to harbour and spread several viruses that can infect othe animals and humans. Which one of the following viruses is not reported to spread by fruit bats ?

(a) Ebola

(b) Nipah

(c) SARS

(d) HIV

Ans. (d)

Sol. In recetn years, bats have gained notoriety after being implicated in numerous (Emerging infectious diseases) EIDs. Bat-borne viruses that can affect humans and have caused EIDs in human fall into different families : paramyxoviruses including Hendra viruses and Nipah viruses; Ebola hemorrhagic fever filoviruses; Marburg hemorrhagic fever filoviruses and sudden acute respiratory syndrome-like coronoaviruses (sars-coV).

54. In a type I hypersensitivity-mediated asthmatic response, which one of the following is thought to contribute significantly to the brochospasm and build-up of mucuous seen in asthmatics ?

(a) Thromboxane

(b) Leukotriene

(c)

(d) Chondoitin

Ans. (b)

Sol. Leukotrienes play a key role in asthma in three ways : causing inflammation, broncho-construction and mucus production. The cysteinyl leukotrienes (LTC 4, LTD 4 and LTE 4) have been shown to be the most potent bronchoconstrictor in humans and are believed to play a crucial role in asthmatic airway obstruction. Leukotrienes may attract white blood cells to the lungs, increasing swelling of the lung lining. Leukotrienes also increase mucus production and make it easier for fluids to accumulate (an important part of inflammation).

55. Which one of the following influenza A virus subtypes caused severe avian flu and was responsible for disease outbreak in the year 1997 in Hong Kong ?

(a) H1N1

(b) H7N7

(c) H3N2

(d) H5N1

Ans. (d)

Sol. Bird flu, also called avian influenza, a viral respiratory disease mainly of poultry and certain other bird species, including migratory waterbirds, some imported pet birds and ostriches that can be transmitted directly to humans. The first known cases in humans were reported in 1997, when an outbreak of avian influenza in 1997, when an outbreak of avian influenza A virus subtype H5N1 in poultry in Hong Kong led to severe illness in 18 people one-third of whom died.

56. In a laboratory experiment it was observed that both 'Virus A' and 'Virus B' could infect a mammalian host cell, when infected individually. Interestingly, if the cells were first infected with 'Virus A' (with large MOI), Virus B failed to infect the same cell. If the Virus B (with large MOI) is added first followed by Virus A, both the virus can infect the cells. Howeer, infection with 'Virus A' was found to be in lesser extent. Considering X and Y are the receptors/co-receptors which may be involved for the virus entry, following are few possibilities that can explain the observation.

P. 'Virus A' uses 'X' as receptor and Y as coreceptor.

Q. 'Virus B' uses exclusively 'Y' as receptor for entry.

R. Both 'Virus A' and 'Virus B' need X as receptor.

Choose the option with all correct statements.

(a) P, Q and R

(b) P and Q

(c) Q and R

(d) P and R

Ans. (b)

Sol. Statement P and Q is correct.

57. While testing the effect of several potent anti-cancer compounds on cycling human oral cancer cells, a student observed that a major percentage of cells showed dose-dependent cell death after 12 hours of drug treatment. However, the remaining cells populated the culture dish once the compounds were removed and the cells were cultured in complete medium. The student made the following assmptions :

P. Not all cells were equally affected by the compounds on as they were not synchronized before treatment.

Q. The compound selectively killed cells which were in G₀ phase.

R. The cancer stem cells were impervious to the effects of the compounds and therefore repopulated the culture.

S. The cancer cells differentiated into a mesenchymal phenotype and grew in fresh culture medium containing inhibitors of epithelial-to-mesenchymal transition (EMT).

Which one of the following combination of assumptions would best justify the results ?

(a) Q and R

(b) P and R

(c) Q and S

(d) P and Q

Ans. (b)

Sol. Depending on their mechanism of action, the efficacy of chemotherapoy drugs may be influenced markedly by the time of exposure (phase specific or time-dependent drugs) or by the dose that can be administered (phase nonspecific or dose-dependent drugs). The efficacy of phase-specific anti cancer drugs is time-dependent, as only a fraction of tumor cells are in appropriate cell cycle phase for chemotherapy-mediated killing at any given time. Since the question mentions a dose-dependent cell death, there is no question that the drug will be specific to any phase. Hence statement B cannot be the correct assumption.

Also, statement C is correct. Despite the recent advances in treatment modalities, non-small cell lung cancer remains to be one of the leading causes of mortality with a 5-year survival rate of 15%. This is owed to the increasingly emergent concept of cancer stem cells (CSCs) a subpopulation of the tumor mass, alleged to be impervious to radiation and chemotherapeutic treatments, making them responsible for recurrence of the disease in cancer patients.

58. There are number of specific T-cell surface molecules involed in various functions of adaptie immune response. Column X represents a list of T-cell surface molecules and Column Y with the possible functional characteristics :

Column X Column Y

P. T-cell receptor 1. binds to CD40 on B-cells and APCs and triggers activation of APCs and activation/differentiation of B-cells.

Q. CD28 2. binds to MHC class I molecules and restricts T cytotoxic cells to recognizing only peptide presented on MHC class I.

R. CD8 3. binds to B7-1, 2 or CD80/86 on B-cells and APCs, which triggers T-cell activation.

S. CD154 4. consists of two polypeptide chains and and some consist of polypeptide chains and .

Which of the following option has all correct matches ?

(a) P–1, Q–2, R–3, S–4

(b) P–4, Q–1, R–2, S–3

(c) P–3, Q–4, R–1, S–2

(d) P–4, Q–3, R–2, S–1

Ans. (d)

Sol. T cell receptor consists of two polypeptides alpha and beta and in some cases gamma and delta.

CD28 binds to B7 and acts costimulation for activatino of T cell.

CD8 expressed on Tc cells binds to MHC class I and makes its activation self MHC restricted.

CD154 or CD40 L is responsible for binding to CD40 which helps in B cell activation and differentiation.

59. The extracellular matrix contains a number of non-collagen proteins that typically have multiple domains, each with specific binding sites for other matrix molecules and cell surface receptors. These proteins therefore contribut to both organizing the matrix and helping cells attach to it. The most well characterized matrix protein of this kind is fibronectin. Which one of the following characteristics is not true for fibronectin ?

(a) It is a large glycoprotein found in all vertebrates and important for many cell-matrix interactions.

(b) It is composed of three polypeptides that are disulfide bonded into a crosslink structure.

(c) In human genome, there is only one firbonectin gene containing about 50 exons, but transcripts can be spliced in different ways to produce many different fibronectin isofoms.

(d) Fibronectin binds to integrin through an RGD motif. Even very short peptide containing RGD sequence can inhibit attachment of cells to fibronectin matrix.

Ans. (b)

Sol. Firbonectin is the principal adhesion protein of connective tissues. Fibronectin is a dimeric glycoprotein consisting of two polypeptide chains, each containing nearly 2500 amino acids. Fibronectin are proteins that connect cells with collagen fibers in the ECM, allowing cells to move through the ECM. So the given option B is incorrect.

60. Which one of the following best describes death-upon-detachment ?

(a) Necroptosis

(b) Anoikis

(c) Extravasation

(d) Metastasis

Ans. (b)

Sol. Anoikis is a form of programmed cell death that occurs in anchorage-dependent cells when they detach from the surrounding extracellular matrix (ECM). Usually cells stay close to the tissue to which they belong since the communication between proximal cells as well as between cells and ECM provide essential signals for growth or survival. When cells are detached from the ECM, there is a loss of normal cell-matrix interactions and they may undergo anoikis. However, metastatic tumor celsl may escape from anoikis and invade other organs.

61. Homeobox transcription factors (Hox proteins), play important roles in specifying whether a particular mesenchymal cell will become stylopod, zeugopod or autopod. Based on the expression pattens of these genes, a model was proposed wherein these Hox genes specify the identity of a limb region. What would be the observed phenotype for human homozygous for a HOXD13 mutation ?

(a) No zeugopod formation.

(b) Abnormalities of the hands and feet wherein the digits fuse.

(c) Deformities in stylopods.

(d) No femur or patella formation.

Ans. (b)

Sol. Hox genes specify the identity of the limb region (as per question), so whenever there is a mutation in HOXD13 some or the other abnormalities will be observed in the hands and feet wherein the digits fuse. Disease caused is called as Synpolydactyly 1 (SPD1) : Limb malformation that shows a chracteristic manifestation in both hands and feet.

62. In Xenopus embyos, -catenin plays an important role in the Dorsal/Ventral axis is development. What would you expect if the endogenous glycogen synthase kinase 3 (GSK3) is knocked out b a dominant-negative form of GSK3 in the ventral cells of the early embryo ?

(a) Blocking of GSK3 on the ventral side has no effect. A normal embryo will form.

(b) The resulting embryo will only have vental sides.

(c) A second axis will form.

(d) The dorsal fate is suppressed.

Ans. (c)

Sol. -catenin is necessary for formin the dorsal axis, since experimental depletion of -catenin transcripts with antisense oligonucleotides results in the lack of dorsal structures. Moreover, the injection of exogenous -catenin into the ventral side of the embryo produces a secondary axis. -catenin is part of the Wnt signal transduction pathway and is negatively regulated by the glycogen synthase kinase 3. GSK-3 also plays a critical role in axis formation by suppressing dorsal fates. Activated GSK-3 blocks axis formation when added to the egg. If endogenous GSK-3 is knocked out by a dominant negative protein in the ventral cells of the early embryo, a second axis forms.

63. When 8-cell embryo of tunicates is separated into 4 blastomere pairs and allowed to grow independently in culture medium, then each blastomere pair can form most of the cell types; however, cells for nervous system are not developed. The following statements are formed from the above obsevations :

P. Nervous system development demostrated autonomous specification.

Q. The other tisse types are formed due to conditional specification.

R. All the tissue types, except nervous tissues that developed, demonstrated autonomous specification.

S. Nervous system development demonstrated conditional specification.

The correct combination of statements that explains the above result is

(a) P and Q

(b) Q and R

(c) R and S

(d) P and S

Ans. (c)

Sol. Early tunicate cells are specified autonomously, each cell acquiring a specific type of cytoplasm that will determine its fate. When the 8-cell embryo is separated into its four doubles (the right and left sides being equivalent), mosaic determination is the rule. The animal posterior pair of blastomeres gives rise to the ectoderm and the vegetal posterior pair produces endoderm, mesenchyme and muscle tissue, just as expected from the fate map. Conditional specification however, also plays an important role in tunicate development. One example of this process involves the development of neural cells. The nerve producing cells are generated from both the animal and vegetal anterior cells, yet neither the anterior or posterior cell of each half can produce them alone.

64. Given below are statements related to different aspects of plant growth and development.

P. Leaf longevity is increased in ethylene-insensitive mutants etr1 and ein2 of ARabidopsis.

Q. Programmed cell death (PCD) is responsible for the formation of prickles, thorns and spines in plants.

R. Senescence and PCD occur only in the development of vegetative tissues and does not occur in reproductive tissues.

S. Redifferentiation of organelles is an integral component duing initial stages of senescence in plants.

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

(a) P, R and S

(b) Q and R

(c) P, Q and S

(d) R and P

Ans. (c)

Sol. Ethylene signaling in Arabidopsis begins at a family of five ethylene receptors that regulate activity of a downstream mitogen-activated protein kinase kinase kinase, CTR1. Triple and quadruple loss-of-function ethylene receptor mutants display a constitutive ethylene response phenotype, indicating they function as negative regulators in this pathway. No ethylene-related phenotype has been described for single loss-of-function receptor mutants, although it was reported that etr1 loss-of-function mutants display a growth defect limiting plant size. In actuality, this apparent growth defect results from enhanced responsiveness to ethylene; a phenotype manifested in all tissues tested. "Programmed cell death (PCD) in plants is a crucial component of development and defence mechanisms. In animals, different types of cell death (apoptosis, autophagy, and necrosis) have been distinguished morphologically and discussed in these morphological terms. PCD is largely used to describe the processes of apoptosis and autophagy (although some use PCD and apoptosis interchangeably) while necrosis is generally described as a chaotic and uncontrolled mode of death. In plants, the term PCD is widely used to describe most instances of death observed. "Leaf senescence is manifested by color change from green to yellow (due to chlorophyll degradation) or to red (due to de novo synthesis of anthocyanins coupled with chlorophyll degradation) and frequently culminates in programmed death of leaves. However, the breakdown of chlorophyll and macromolecules such as proteins and RNAs that occurs during leaf senescence does not necessarily represent a one-way road to death but rather a reversible process whereby senescing leaves can, under certain conditions, re-green and regain their photosynthetic capacity. This phenomenon essentially distinguishes senescence from programmed cell death, leading researchers to hypothesize that changes occurring during senescence might represent a process of trans-differentiation, that is the conversion of one cell type to another. In this review, we highlight attributes common to senescence and dedifferentiation including chromatin structure and activation of transposable elements and provide further support to the notion that senescence is not merely a deterioration process leading to death but rather a unique developmental state resembling dedifferentiation.

65. The following demonstrates proposed functions of different genes which determine the decision to become either trophoblast or inner cell mass (ICM) blastomere during early mammalian development :

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

(a) The interplay between Cdxx2 and Oct4 can influence the formation of ICM.

(b) The ICM would form even if expression of Oct4 was inhibited.

(c) YAP and TEAD4 aer upstream components of Cdx2 and can be inhibited by Nanog.

(d) The expression of Stat3 is optional for maintaining pluripotency of the ICM.

Ans. (a)

Sol. Oct4 forms a repressive complex with and inhibits the transcription of Cdx2, a transcription factor essential for trophectoderm specification [63]. In the absence of Oct4, the embryo develops to the blastocyst stage and appears to be morphologically normal. Expression of CDX2 is diffusely evident in all cells of an 8–16 cell embryo, but expression becomes progressively more intense in nuclei of outer cells during the blastocyst stage.

66. During wing development in chick, if Apical Ectodermal Ridge (AER) is removed, the limb development ceases, on the other hand placing leg mesenchyme directly beneath the wing AER, distal hindlimb structures develop at the end of the wing, and if limb mesenchyme is replaced by non-limb mesenchyme beneath AER, the AER regresses. This may demonstrate that :

P. the limb mesenchyme cells induce and sustain AER.

Q. the mesenchyme cells specify the type : wing or limb.

R. the AER is responsible for specifying the type : wing or limb.

S. the AER is responsible for sustained outgrowth and development of the limb.

T. the AER does not specify the type : wing of limb.

Which combinatiosn of above statements is demonstated by the experiment ?

(a) P, Q, R and S only

(b) P, Q, S and T only

(c) R, S and T only

(d) P and T only

Ans. (b)

Sol. The proximal distal growth and differentiation of the limb bud is made possible by a series of interaction between the limb bud mesenchyme and the AER. Although the mesenchyme cells induces and sustain the AER and determine the eype of limb to be formed, the AER is responsible for the sustained outgrowth and development of the limb. The AER keeps the mesenchyme cells directly beneath it in a state of mitotic proliferation and prevents them from forming cartilage. Statement A, B and D and E are correct according to the explanation. Hence, option B is correct here.

67. The following statements regarding the generation of dorsal/ventral axis in Drosophila was made :

P. Gurken protein moves along with the oocyte nucleus and signals follicle cells to adopt the ventral fate.

Q. Maternal deficiencies of either the gurken or topedo gene cause ventraization of the embryo.

R. Gurken is active only in the oocyte and Torpedo is active only in the somatic follicle cells.

S. The Pipe proteins is made in the dorsal follicle cells.

T. The highest concentration of Dorsal is in the dorsal cell nuclei, which becomes the mesoderm.

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

(a) P and T

(b) R and S

(c) Q and R

(d) Q and T

Ans. (c)

Sol. Maternal effect mutations of the torpedo and gurken genes produce very similar phenotypes where the follicle cells and the resulting embryos are ventralized/dorsalized.

Gurken is active only in oocyte, germiline cells and torpedo is active only in follicle cells, somatic cells.

Before gurken and torpedo where cloned and identified ad homologs of germline cells and somatic cells respectively, a key experiment was performed that predicted how Gurken and Torpedo might be functioning.

68. Which one of the following describes the functions of silicon is plants ?

(a) Constitutent of amino acids.

(b) Contributes to cell wall rigidity and elasticity.

(c) Constituent of the photosynthesis reaction centre.

(d) Maintenance of cell turgor and electroneutrality.

Ans. (b)

Sol. Silicon is deposited primarily in the endoplasmic reticulum, cell walls and intercellular spaces as hydrated, amorphous silica (SiO₂.nH₂O). Recent evidences show that silicon contributes to cell wall rigidity and strengthening besides increasing cell wall elasticity during extension of growth. In the primary cell walls, silicon interacts with cell wall constituents such as crosslinks obviously increase cell wall elasticity during extension growth.

Plants deficient in silicon are more susceptible to loadging (falling over) and fungal infection. A sufficient supply of silicon permits rice plants to keep the leavse erect, uncurved and thus improves the assimilatory system, counteracts the negative effects of a large nitrogen supply on light interception.

69. Most of the plant disease resistance (R) gene products conatin :

(a) G-Box domains.

(b) Transcription repression domains.

(c) Leucine-rich repeats.

(d) Enzymatic activities.

Ans. (c)

Sol. Resistance genes (R-Genes) are genes in plant genomes that convey plants disease resistance against pathogens by producing R proteins. The main class of R-genes consist of a nucleotide binding domain (NB) and a leucine rich repeat (LRR) domain(s) and are often referred to as (NB-LRR) R-genes. Generally, the NB domain binds either ATP/ADP or GTP/GDP. The LRR domain is often involved in protein-protein interactions as well as ligand binding. NB-LRR R-genes can be further subdivided into toll interleukin 1 receptor (TIR-NB-LRR) and coiled coil (CC-NB-LRR).

70. Out of several gibberellins identified in plants, which one of the following is not bioactive ?

(a) GA₁

(b) GA₃

(c) GA₄

(d) GA₅

Ans. (d)

Sol. Gibberellins are now known to be regulators of many phases of higher plant development, including seed germination, stem growth, induction of flowering, pollen development and fruit growth. THe concentration of bioactive GAs in plants is in range 10^–11–10^–9 g/g fresh weight, depending on the tissue and species and is closely regulated. This chapter describes the biosynthesis and catabolism of GAs and examines the regulatory processes that optimize the levels of bioactive GAs within plant tissues.

Only a few of the 136 known GAs have intrinsic biological activity. Not surprisingly, many of the GAs that were identified in the earliest years of GA research are the ones which possess the highest biological activity and are candidates for the role of an active hormone. These include GA₁, GA₃, GA₄, GA₅, GA₆ and GA₇.

71. Nitrogenase, a complex metal containing enzyme is involved in convesion of N₂ to NH₃. Which one of the following metals is not involved in the activity of nitrogenase ?

(a) Molybdenum (Mo)

(b) Iron (Fe)

(c) Vanadium (V)

(d) Cobalt (Co)

Ans. (d)

Sol. All nitrogenases are two-component systems made up of Component I (also known as dinitrogenase) and Component II (also known as dinitrogenase reductase). Component I is a MoFe protein in molybdenum nitrogenase and a Fe protein in iron-only nitrogenase. Component II is a Fe protein that contains the Fe-S cluster, which transfers electrons to Component I. Component I contains 2 key metal clusters : the P-cluster and the FeMo-cofactor (FeMo-co). Mo is replaced by V or Fe in vanadium nitrogenase and iron-only nitrogenase respectively. During catalysis, electrons flow from a pair of ATP molecules within Component II to the Fe-S cluster, to the P-cluster and finally to the Fe-Mo-co, where reduction of N₂ to NH₃ takes place.

72. In certain plants, the mechanism where timing of anther dehiscence and stigma receptivity do not coincide to avoid self-pollination is called

(a) dichogamy

(b) herkogamy

(c) monoecy

(d) dioecy

Ans. (a)

Sol. Dichogamy : In sexual flowers, when two sexes mature at different intervals and thus avoid self pollination is known as dichogamy. When stamens mature earlier than the stigma, it is known as protandry and the flowers are called protandous e.g., Coriander, Jasmine, Sunflower, Lady's finger, etc. When, stigma matures earlier than the stamens, it is known as protogyny and the flowers are called protogynous. e.g., Rose, Tobacco, Crucifers, etc.

In this mechanism, the stigma and anther mature at different times. Depending on who matures first, dichogamy can be further divided into two :

Protandry : In this type, the androecium matures earlier than the gynoecium. Ex : maize palnt.

Protogyny : In this type, the gynoecium matures earlier than the androecium.

73. Jasmonate is known to inhibit root growth while auxin facilitates root growth. Upon infection with pathogenic bacteria that produce coronatine, we may expect the following in plants :

P. Upregulation of COI-1 gene and inhibition of root growth.

Q. Upregulation of AuxI-1 gene and inhibition of root growth.

R. Inhibition of AuxI-1 gene and promotion of root growth.

S. Inhibition of COI-1 gene and promotion of root growth.

Which one of the following is correct ?

(a) P, Q and R

(b) Only P

(c) Only Q

(d) Only R

Ans. (b)

Sol. The expression of the JA receptor coronatine insensitive1 (coi1) and the key JA signaling regulator MYC2 is upregulated in response to AI stress in the root tips. hence, upregulation of COI-1 gene will occur over here which will further inhibit root growth statement A is correct.

Either of the options A or B is correct. This process together with COI1-mediated AI-induced root growth inhibition under AI stress was controlled by ethylene but not auxin. Option A includes both the regulatory mechanisms, upregulation as well as inhibition of Aux-I which is not possible. Hence, we can say that only A statement is correct or option B is correct.

74. Following are certain statements with regard to plant respiration :

P. Metabolism of glucose into pyruvate through glycolysis generatse NADH and not NADPH.

Q. Metaboism of glucose through oxidative pentose phosphate cycle does not produce NADPH.

R. Cyanide forms a complex with haem iron of cytochrome oxidase leading to prevention of change in valency, which in turn stops electron transport in the respiratory chains.

S. Alternative oxidase is insensitive to cyanide and has higher K_m than that of cytochrome oxidase.

Which one of the following combinations is correct ?

(a) P, Q and R

(b) Q, R and S

(c) Q and S

(d) P, R and S

Ans. (d)

Sol. One of the major differences between glycolysis and the pentose phosphate pathway is that glycolysis generates NADH, whereas the pentose phosphate pathway generates NADPH, NADH is used in oxidative phosphorylation to provide ATP and NADPH is used to synthesize fatty acids and steroids. Because the pentose phosphate pathway generates NADPH, tissues that are highly active in the synthesis of fatty acids and steroids actively employ the pentose phosphate pathway.

The hypothesis that under hypoxic conditions the ratio of cyanide-sensitive to cyanide-insensitive respiration may correlate with waterlogging sensitivity has not been supported by results from several studies. The affinity of cytochrome oxidase for O₂ (Km < 0.1 µM) is actually greater than tha of the alternative oxidase (Km = 1 to µ2M). Cyanide acts an inhibito in ETC, it forms a complex with Fe of cytochrome oxidase which stops Electron transport.

75. Blue light receptor cry1 binds to COP1 and SPA1 complex by interacting with C-terminal region of cry1 (CCT) in a light dependent manner and regulates photomorphogenesis via transcription factor HY5. Read the following statements :

P. cry1 binds to COP1 and SPA1 complex leading to degradation of HY5.

Q. cry1 binds to COP1 and SPA1 complex and prevents degradation of HY5.

R. CCT is overexpressed and the plants are kept in dark.

S. CCT is overexpressed and the plants are kept in light.

Which of the following combination of above statements will result in photomorphogenesis ?

(a) Only P

(b) Only Q

(c) P, Q and R

(d) Q, R and S

Ans. (d)

Sol. Arabidopsis COP1 is a constitutive repressor of photomorphogenesis that interacts with photomorphogenesis-promoting factors such as HY5 to promote their proteasome-mediated degradation. SPA1 is a repressor of phytochrome A-mediated responses to far-red light. Here we report that COP1 acts as part of a large protein complex and interacts with SPA1 in a light-dependent manner. We further demonstrate the E3 ubiquitin ligase activity of COP1 on HY5 in vitro and the alteration of that activity by SPA1. Thus, the COP1-SPA1 interaction defines a critical step in coordinating COP1-mediated ubiquitination and subsequent degradation of HY5 with PHYA signaling.

Dark-grown transgenic Arabidopsis seedlings expressing the C-terminal domains (CCT) of the cryptochrome (CRY) blue light photoreceptors exhibit features that are normally associated only with light-grown seedlings, indicating that the signaling mechanism of Arabidopsis CRY is mediated through CCT. The phenotypic properties mediated by CCT are remarkably similar to those of the constitutive photomorphogenic1 (cop1) mutants. Here we show that Arabidopsis cryptochrome 1 (CRY1) and its C-terminal domain (CCT1) interacted strongly with the COP1 protein. Coimmunoprecipitation studies showed that CRY1 was bound to COP1 in extracts from both dark- and light-grown Arabidopsis. An interaction also was observed between the C-terminal domain of Arabidopsis phytochrome B and COP1, suggesting that phytochrome signaling also proceeds, at least in part, through direct interaction with COP1. These findings give new insight into the initial step in light signaling in Arabidopsis, providing a molecular link between the blue light receptor, CRY1, and COP1, a negative regulator of photomorphogenesis.

76. After absorbign light, chlorophyll molecules in green plants exist in singlet and triplet states. Following are cetains statements on singlet and triplet states of chloophyll molecules ;

P. Singlet state is short lived compared to triplet state.

Q. Singlet state is short lived compared to triplet state.

R. Singlet state contains electrons with anti-parallel spins while triplet state has electrons with parallel spins.

S. Singlet state contains electrons with paralle spins white triplet stae has electnos with anti-parallel spins.

Which one of the following combinations is correct ?

(a) P and Q

(b) Q and R

(c) P and R

(d) Q and S

Ans. (c)

Sol. Excited electronic states are due to the promotion of an electron from the ground state distribution, changing the electron density configuration to one of higher energy. The excited singlet state is generally short-lived (10-8-10-9 sec) with the valence electrons having opposite spins. An excited triplet state is longer lvied (milliseconds or more) with the two electrons having parallel spms.

77. Only membes of the plant kingdom and many bacteria have capability of biological nitrogen reduction. In this regard, following statements are given :

P. Nitrogen is normally taken by the plant in their fully oxidized form but needs to be reduced before incorporation in organic molecles.

Q. Conersion of oxidized nitrogen into reduced nitrogen energy in the form of NAD(P)⁺.

R. The metal associated with the enzyme nitrate reductase is magnesium.

S. Nitrate reduction takes place in the cytoplasm, whereas nitrite reduction takes place in chlooplast matrix.

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

(a) P and R

(b) P, Q and R

(c) Q and S

(d) P and S

Ans. (d)

Sol. Statement A is correct as Nitrogen is taken in the form of Nitric oxide and Nitrous oxide which is further reduced by the process of Nitrogen metabolism and assimilation. According to this option A is incorrect. Nitrate reductase are molybdoenzymes that reduce nitrate (NO₃^–) to nitrite (NO₂^–), it does not involve Mg rather includes FAD, Mo and Heme as cofactors. Statement 3 is also incorrrect. As you can see in the options only option D is having 1 without 3. Hence, option D is correct, 4 statement is correct as Nitrite reduction takes place in chloroplast matrix and nitrate reduction in cytoplasm.

78. During interaction with host, phytopathogens are known to deliver effector proteins directly into the host cells. The following statements were made regarding the role of these effector proteins.

P. May promote pathgen virulence.

Q. May elicit aviulence response.

R. May sppress defense response.

S. May promote plant growth.

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

(a) P, Q and S

(b) P, R and S

(c) P, Q and R

(d) Q, R and S

Ans. (c)

Sol. Pathogens attempting to invade plants have to overcome the first level of immunity in the prospective host and do so by developing effector proteins. The first purpose of effector proteins is to overcome this innate basic resistance of plants Statement C is correct.

Effector proteins and PAMPs, do not always use the terms pathogenicity, virulence and avirulence appropriately and at the same time they do not trouble to give a clear new definition of these phenomena. Currently, the substances (effectors) that some phytopathogen secretes into the host plant cells are taken to be either virulence or avirulence factors or sometimes also pathogenicity factors. Effector molecules are likewise primarily termed avirulence factors, but they are also factors that contribute to pathogen virulence. Statement A is correct.

Avirulence factors elicity avirulent responses in the host plant, e.g. a response consisting in host cell necrosis at the site of inoculation (HR). Statement B is correct.

79. Which one of the following agents cause relaxation of mesangial cells ?

(a) Histamine

(b) Thrombaxane A₂

(c) Norepinephrine

(d) Dopamine

Ans. (d)

Sol. Mesangial cells are specialized pericytes with functional properties similar to that of smooth muscle cells. In addition to providing structural support for the glomerular capillary loop, its contractile properties help regulate glomerular filtration.

The presence of actin and myosin allows the mesangial cells to contract in the presence of vasoactive agents like angiotensin II, vasopressin, nore pinephrine, leukotrienes, thromboxanes and platelet activating leukotrienes, thromboxanes and platelet activating factors. However, prostag-landins, atrial peptides and dopamine cause mesangial relaxation. The surrounding mesangial matrix consists of glycosamin-oglycans, fibronectins, laminin and other collagens.

80. A patient comes to the hospital complaining of vomiting and diarrhoea. The doctor suggested that the patient take glucose and electrolyte solution orally. Which one of the following membrane prroteins is likely to be involved in rehydrating the patient ?

(a) Cystic fibrosis transmembarne regulator (CFTR).

(b) Sodium glucose transporter proteins 1 (SGLT1).

(c) Insulin receptor protein (IRP).

(d) Sucrase-isomaltase protein (SIP).

Ans. (b)

Sol. Sodium glucose transporter proteins 1 (SGLT1) will be involved in rehydrating the patient.

81. Which one is required for vitamin B₁₂ absorption in small intestine ?

(a) Cobalophilin

(b) Hephaestin

(c) Hepicidin

(d) Na⁺-cotransporter

Ans. (a)

Sol. Cobalophilin is a Vitamin B12 binding protein that is secreted is saliva that binds the vitamin in the stomach when it is released from dietary proteins by the action of gratic acid and pepsin. Going through the other options, Hephaestin maintenacne of Fe and possibly copper. Hepcidin is a key regulatory of the entry of Iron into the circulation in mammals and is central regulator of Fe homeostasis.

82. Which one is the correct sequence of events that takes place dring phototransduction when light falls onto the retina ?

(a) Closure of Na⁺ channels activation of transducin decreased release of glutamate decrease in intracellular cGMP structural changes in rhodopsin.

(b) Decreased release of glutamate structural changes in rhodopsin activation of transducin decrease in intracellula cGMP closure of Na⁺ channels.

(c) Structural changes in rhodospin activation of transducin decreased in intracellular cGMP closre of Na⁺ channels decreased release of glutamate.

(d) Decrease in intracellular cGMP activation of transdcin decreased release of glutamate structural changes in rhodopsin closure of Na⁺ channels.

Ans. (c)

Sol. When the retinal moiety in the rhodopsin molecule absorbs a photon, its configuration changes from the 11-cis isomer to all-trans retinal; this change then triggers a series of alterations in the protein component of the molecule. The changes then triggers a series of alterations in the protein component of the molecule. The changes lead, in turn to the activation of an intracellular messenger called transductin, which activates a phosphodiesterase that hydrolyzes cGMP. All of these events take place within the disk membrane. The hydrolysis by phosphodiesterase at the disk membrane lowers the concentration of cGMP throughout the outer segment, and thus reduces the number of cGMP molecules that are available for binding to the channels in the surface of the outer segment membrane, leading to channel closure.

83. A specialized area of teratogenesis involves the misregulation of the endocrine system. Which one of the following statements regarding endocrine disptors is true ?

(a) They can act as antagonist and inhibit the binding of a hormone to its receptors or block the synthesis of a hormone.

(b) They do not affect the synthesis, elimination or transportation of a hormone in the body.

(c) They do not mimic the effect of natural hormones.

(d) Low dose exposure to endocrine disruptors is not sufficient to prodce significant disabilities later in life.

Ans. (a)

Sol. Mimic or partly mimic naturally occurring hormones in the body like estrogens (the female sex hormone), androgens (the male sex hormone) and thyroid hormone, potentially producing overstimulation.

Bind to a receptor within a cell and block the endogenous hormone from binding. The normal signal then fails to occur and the body fails to respond properly. Examples of chemicals that block or antagonize hormones are anti-estrogens and anti-androgens.

Interferer or block the way natural hormones or their receptors are made of controlled, for example, by altering their metabolism in the liver.

84. During prolonged illumination, hodopsin is desesitized which leads to the termination of visual response. The associated proteins (Column A) and their effects (Column B) are given below :

Column A Column B

P. Phosphoryalted opsin 1. Phosphorylates opsin

Q. Rhodopsin kinase 2. Binds to phosphorylated opsin

R. Arrestin 3. Decreases activation of transducin

S. Phosphatase 4. Reveses the termination process

Which one of the matched combinations is correct ?

(a) P–4, Q–2, R–1, S–3

(b) P–2, Q–3, R–4, S–1

(c) P–3, Q–1, R–2, S–4

(d) P–2, Q–4, R–3, S–1

Ans. (c)

Sol. Photoactivated rhodpsin interacts not only with transducin, but with two more proteins : a protein kinase that specifically phosphorylates R (in contrast to dark-adapted rhodpsin) at multiple sites and an abundant soluble protein of 48 KDal (called 48 K-protein, S-antigen or arrestin) that specifically binds to phosphorylated R. Phosphorylation partially suppresses the ability of R to catalyze transducin-mediated phosphodiesterase activation even in the absence of arrestin. Binding of arrestin to the phosphorylation R potentiates this inhibitory effect, most probably because arrestin competes with transducin for binding on the phosphorylated R. Phosphorylation, in conjunction with arrestin binding, therefore appears to be a mechanism that terminates the active state of the receptor, R.

85. Estrs cycle in rats is controlled by pituitary and gonadal hormones. While treating a set of rats with vitamin D, a student accidentally injected the rats with an inhibitor of 17-hydroxypregnenolone and checked vaginal smear for 10 consecutive days.

Which one of the following observations is corect ?

(a) The smears showed well formed nucleated epithelial cells throughout the period.

(b) The smears initially showed normal estrus stage but eventually entered a prolonged diestus stage.

(c) The smears showed leukocytes and few epithelial cells.

(d) The cells showed metestrs for 3 days and then returned to the poestrus stage.

Ans. (b)

Sol. Consistent with the C57B1/16J mice, WT mice shows 4-5 days cycle with four stages, proestrous, estrous, diestrous and metaestrous.

1. Proestrous : it is the phase at which estrogen is lowest and gradually increases. The vaginal smear reveals a discharge of nucleated epithelial cells.

2. Estrous : This is the phase of highest estrogen level after which estrogen declines. It is a phase after which the epithelium of the vagina sloughs off having reached its maximum thickness, which shows cornified cells.

3. Metaestrous : A phase of declining concentration of estrogen. With declining concentrations epithelial cells along with may leukocytes slough off from the vagina.

4. Diestrous : With low levels of E2, the vaginashows similar profile as metaestrous but with fewer leukocytic cells.

Vitamin D3 deficiency delays puberty and causes prolonged estrous cycles characterized by extended periods of diestrus and reduced frequency of proestrus and estrus. 17-Hydroxy-pregnenolon is important for the synthesis of estrogen. Inhibiting 17-HOP results in estrogen deficiency.

Ovariectomy (lack of estrogen) in adult female mic has been shown to result in extended diestrous phase similar to Vitamin D3 deficiency. Thereby combining the above statements option B holds true.

86. The CO₂ dissociation curves of oxygenated and deoxygeneated blood are gien along with dissolved CO₂ below :

Following are the statements deduced fom the curves above and/or based on the knowledge about CO₂ transpot, which may or may not be corect :

P. The deoxygenated haemoglobin has greater affinity for CO₂ than oxygenated haemoglobin.

Q. The deoxygeneated haemoglobin does not bind with free H⁺ ions released during the fomation of from CO₂.

R. The haemoglobin saturation with O₂ has no effect on CO₂ dissocation curve.

S. O₂ and CO₂ bind to haemoglobin at different sites.

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

(a) P and Q

(b) Q and R

(c) R and S

(d) P and S

Ans. (d)

Sol. As per CO₂ dissocation curve, The deoxygenated haemoglobin has greater affinity for CO₂ than oxygenated haemoglobin. O₂ and CO₂ bind to haemoglobin at different sites.

87. Given below are the different intervals/durations of electrocardiogram of a human subject (column A) and the events in heart during the process (column B).

Column A Column B

P. PR interval 1. Ventricular action potential

Q. QRS duration 2. Atrioventricular conducution

R. QT interval 3. Venricular depolarization

S. ST interval 4. Plateau portion of the ventricular action potential

Which one of the following options is a correct match of entries in columns A and B ?

(a) P–1, Q–4, R–2, S–3

(b) P–2, Q–3, R–1, S–4

(c) P–4, Q–2, R–3, S–1

(d) P–3, Q–1, R–4, S–2

Ans. (b)

Sol. The QRS complex represents rapid ventricular depolarization, as the action potential spreads through ventricular contractile fibers. The S-T segment, which begins at the end of the S wave and ends at the beginning at the end of the S wave and ends at the beginning of the T wave, represents the time when the ventricular contractile fibers are depolarized during the plateau phase of the action potential. PR includes both atria and ventricular condution. The P wave represents atrial depolarization, which spreads from the SA node through contractile fibers in both atria. The second wave, called the QRS complex, begins as a downward deflection, continues as a large, upright, triangular wave and ends as a downward wave. The QRS complex represents rapid ventricular depolarization. The Q-T interal extends from the start of the QRS complex to the end of the T wave. It is the time from the beginning of ventricular depolarization to the end of ventricular repolarization.

88. The pathway of synthesis of aldosteron in zona glomerulosa along with the intracellular locations is shown below :

The enzyme below are required for different steps of synthesis of aldosterone :

1. 21-Hydoxylase

2. P450 side chain cleavage enzyme

3. -Hydroxy steroid dehydogenase

Which one of the following option represents correct matches for P, Q and R ?

(a) P–1, Q–2, R–3

(b) P–3, Q–1, R–2

(c) P–2, Q–3, R–1

(d) P–2, Q–1, R–3

Ans. (c)

Sol.

89. Given below is a figure of proopiomelanocortin (POMC) polypeptide and its cleavage products (marked as P, Q, R, S) which have different hormonal activities. The name of the cleaved products obtained from POMC are shown in the table below the diagram.

1. Adernocorticotropic hormone

2. -lipotropin

3. -melanocyte-stimulating hormone

4. -endotropin

Which one of the following option represents P, Q, R and S correctly ?

(a) P–1, Q–2, R–3, S–4

(b) P–2, Q–3, R–1, S–4

(c) P–1, Q–4, R–3, S–2

(d) P–3, Q–2, R–4, S–1

Ans. (a)

Sol. POMC is cleaved to produce Adernocorticotropic hormone, -lipotropin, -melanocyte-stimulating hormone and -endotropin which are represented in diagram as P, Q, R and S respectively.

90. The blood plasma proteins (albumin and globulins) from a healthy person were separated by electrophoresis as shown below. The diagnosis of acute inflammation can be done based on one of the following observations :

(a) Increase in both 1 and 2; decrease in albumin.

(b) Increase in albumin; decrease in 1, 2 and .

(c) Increase in albumin and decrease in -globulin.

(d) Only decrease in albumin.

Ans. (a)

Sol. Analbuminaemia or analbuminemia is a genetically inherited metabolic defect characterised by an impaired synthesis of serum albumin. Hypoalbuminemia can result from decreased albumin prodution, defective synthesis because of hepatocyte damage, deficient intake of amino acids, increased losses of albumin via GI or renal processes and most commonly, acute or chronic inflammation. Both alpha 1 and alpha 2 globulins increase during acute inflammation.

91. Column X lists two diseases and column Y lists name of proteins whic are commonly used for routine clinical diagnosis of these diseases :

Column X Column Y

P. Myocardial Infarction 1. Amylase

Q. Panceatitis 2. Ceratine kinase

3. Lipase

4. Troponin

Find out the correct combination

(a) P–2, Q–1, P–4, Q–3

(b) P–3, Q–4, P–2, Q–1

(c) P–1, Q–2, P–3, Q–4

(d) P–4, Q–3, P–1, Q–2

Ans. (a)

Sol. Troponin is a protein released from myocytes when irreversible myocardial damage occurs. It is highly specific to cardiac tissue and accurately diagnoses myocardial infarction with a history of ischaemic pain or ECG changes reflecting ischaemia. Creatine kinase (CK) and its isoenzyme CK-MB are important tools for the diagnosis of acute myocardial infarction. The content of CK-MB relative to total CK in myocardial cells is variable; it is low in normal myocardium and increased several-fold in hypoxic myocardium.

Acute pancreatitis is diagnosed when a patient presents with two of three findings, including abdominal pain suggestive of pancreatitis, serum amylase and/or lipase levels at least three times the normal level, and characteristic findings on imaging.

So, P-2, Q-1, P-4, Q-3

92. Assuming that the A, B, C and D genes are not linked, the probability of a progeny being AaBBccDd from a cross between AABbccDd and aaBBccDD parents will be

(a) 4/32

(b) 3/16

(c) 1/4

(d) 3/32

Ans. (c)

Sol. Genotype of Parents – AABbccDd × aaBBccDD

Gametes from parent 1 – A parent 2 – a

B or b B

c c

D or d D

Genotype for individual gene pairs in progeny – Aa

BB/bb

cc

DD/Db

Probability of (AaBBccDd) Genotyp of progeny – 1 × 1/2 × 1 × 1/2 = 1/4

(As genes are not linked individual probability as multiplied by product rule of probability for indepenent events).

93. The new born baby of a mother having blood group AB, h⁺ and father having blood group O, Rh^–, got mixed with other babies in the hospital. The baby with which of the following blood gorups is expected to be of the said couple ?

(a) O, Rh⁺

(b) O, Rh^–

(c) AB, Rh^–

(d) B, Rh⁺

Ans. (d)

Sol. Genes from the mother = IA and IB Genes from the father = i and i Genotype of children = IAi, IBi = A blood group and B blood group Rhesus antiger is represented by D Mother can have genotype DD or Dd Father has genotype dd So children can have either Dd, or dd (the chance of Dd is higher = Rhesus positive). So the most probable child is B+ve.

94. A Lod scoe of 3 represents a Recombination Frequency (RF) that is

(a) 3 times as likely as the hypothesis of no linkage.

(b) 30 times as likely as the hypothesis of no linkage.

(c) 100 times as likely as the hypothesis of no linkage.

(d) 1000 times as likely as the hypothesis of no linkage.

Ans. (d)

Sol. The small numbers of progeny in human families mean that it is impossible to deterine linkage on the basis of single matings. To obtain reliable RF values, large sample sizes are necessary. However, if the results of many identical matings can be combined thena more reliable estmate can be made. The standard way of doing so is to calculate Lod scores. Lod stands for "log of odds". The method simply calculates the probability of obtaining a set of results in a family on the basis of independent assortment and a specific degree of linkage. Then the ratio (odds) of the two probabilities is calculated and the logarithm of this number is calculated which is the Lod.

Note that a Lod score of 3 represents an RF value that is 1000 times (thati s 10³ times) as likely as the hypothesis of no linkage.

95. Centromere positions can be mapped in linea tetrads in some fungi. A cross was made between two strains a b and a⁺ b⁺ and 100 linear tetrads were analysed. The genes a and b are located on two arms of the chromosome. The tetrads were divided into 5 classes as shown below :

Based on the above observation, the following conclusions were drawn :

P. Class 1 is a result of a cross over between 'a' and the centromere.

Q. Class 2 is a result of a double cossover involving 3 strands between 'a' and the centromere.

R. Class 5 is a result of a double crossover between 'a'-centromere and 'b'-centromere, involving three strands.

S. Class 4 is a result of a doble crossover, involving all the 4 strands.

Which one of the following options represents all correct statements ?

(a) P and Q

(b) P and R

(c) Q and S

(d) R and S

Ans.

Sol. If we analyze the arrangement in tetrads in class 1 the a is in 2 : 2 : 2 : 2 ratio and b is in 4 : 4 ratio. Hence the first calss of tetrads is definitely a crossover product between a and the centromere. Hence statement A is absolutely correct. DCO between gene and centromere involving two strands can form 4 : 4 ratio but in statement B since three strands is given it is incorrect. In class 5 both a and b are in 2 : 4 : 2 ratio which is crossover product between a and cetromere and b and cetromere, hence a DCO involving 3 strands. So, statement C is correct. Class 4 is PDS that could result from DCO between two strands not all 4. So, D is incorrect. Hence, correct combination is option B.

96. Two yellow mice with straight hair were crossed and the following progeny was obtained :

1/2 yellow, straight hair

1/6 yellow, fuzzy hair

1/4 gray, straight hair

1/12 gray, fuzzy hair

In order to provide genetic explanation for the results and assign genotypes to the parents and progeny of this cross, the following statements were given :

P. The 6 : 2 : 3 : 1 ratio obtained here indicates recessive epistasis.

Q. This cross concerns two independent characteistics body colour and type of hair.

R. The deviation of dihybrid ratio from 9 : 3 : 3 : 1 to 6 : 2 : 3 : 1 may be due to one of the genes being a recessive lethal.

S. The lethal allele is associated with straight hair.

The most appropriate combination of statementsto provide genetic explanation for this result is :

(a) Q and R

(b) P only

(c) Q, R and S

(d) P, R and S

Ans. (a)

Sol. This cross concerns two independent characteistics body colour and type of hair and The deviation of dihybrid ratio from 9 : 3 : 3 : 1 to 6 : 2 : 3 : 1 may be due to one of the genes being a recessive lethal.

97. A family was examined for a given trait which is represented in the pedigere shown below. Further, the degree of expression of the trait is highly variable among membes of the family; some are only affected while others developed severe symptoms at an early age.

The following statements are made to explain the pattern of inheritance shown in the pedigree :

P. X-linked dominant mutation

Q. X-linked recessive mutation

R. Mitochondrial inheritance

S. Variable expression can be due to heteroplasty

The best possible explanation for this inheritance is

(a) P and S

(b) R and S

(c) Q only

(d) P only

Ans. (b)

Sol. In the pedigree it is clearly shown that whenever mother is affected 100% of the progenies are affected irrespective of their sexes and also affected father is not inheriting the trait to any progeny. This is definitely because of Mitochondrial inheritance as it is only inherited from the mother through egg cytoplasm. Since the amount of defective mitochondria might be unequally distributed during oogenesis a process called as heteroplasmy, it affects the degree of expression of the trait. So, both statement 3 and 4 are correct.

98. The location of six deletions (shown as solid line underneath the chromosome) has been mapped to the Drosophila chromosome as shown in the diagram given below :

Recessive mutations a, b, c, d, e, f and g are known to be located in the region of deletions, but the order of mutations on the chromosome is not known. When flies homozygous for the recessive mutations are crossed with flies homozygous for the deletion, the following results were obtained, where the letter 'm' represents a mutant phenotype and '+' represents the wild type.

The relative order of the seven mutant genes on chromosome is :

(a) b c e a f g d

(b) b c d f g e a

(c) b c d e a f g

(d) c d e a g f d

Ans. (c)

Sol. The key principle here is that point mutations can recombine with deletions that do not extend past the mutation, but they cannot recombine to yield wild-type phages with deletions that do extend past the mutation. In a simpler way we can find the sequence by observing the length of mutation and which gene functions are getting affected by that and then comparing the overlapping regions we can predict the correct sequence. In this question if we focus on the last gene in the map, it is affected by only deletion 5. In the table function of g gene is affected by only deletion 5. So g must be last in the sequence. Similarly by overlapping the deletions and comparing the loss of function of genes we can predict the sequence which is option C here. The correct order is bcdeafg.

99. In the following pedigree, individulas with shaped circule or shaded square show presence of a recessive autosomal trait.

The calculated risk of occurence of this trait for III-1 is

(a) ½

(b) ¼

(c) 1/8

(d) 1/3

Ans. (d)

Sol. The trait given in autosomal recessive. In this pedigree that man's (II-3) parents must have both been heterozygotes Aa because they produced affected aa child. So among expected progeny of man's (II-3) parents AA(1/4) AA(1/2) aa(1/4) : 2/3 probability that the man (II-3) is a carrier of the trait. The woman II-4 is affected by the trait, so the must have genotype aa. So expected progeny genotype is Aa or aa. So, overall probability of the son to have the diseae is 2/3 × 1/2 = 1/3.

100. Which one of the following is a fungal disease of plants ?

(a) Cucumber mosaic

(b) Fire blight of pear

(c) Crown gall

(d) Apple scab

Ans. (d)

Sol. Apple scab or black spot is caused by the fungus Venturia inaequalis. It infects leaves, shoots, buds, blosooms and fruit. It occurs almost everywhere apples are grown and is the most serious and widespread disease of this crop, especially important in regions wiht high rainfall and relative humidity during the growing season.

The fungus may infect developing flowers, but is more usually seen on reproductive parts after the fruit has set. Infection of fruit stalks usually causes the young fruit to fall.

101. Which one of the following statements is correct ?

(a) Ectomycorrhizal association predominantly reduce phosphorus limitation, and endomycorrhizal assocations reduce both nitrogen and phosphorus limitation.

(b) Ectomycorrhizal association predominantly reduce phosphorus limitation, and ectomycorrhizal assocations reduce both nitrogen and phosphorus limitation.

(c) Ecto and endo-mycorhizal assocations do not reduce nitrogen and phosphorus limitation.

(d) Ecto and endo-mycorhizal assocations are able to only phosphorus limitation.

Ans. (b)

Sol. Plants adjust nutrient absorption to maximize absorption of growth-limiting nutrients and reduce capacity of absorb non-limiting nutrients.

Symbiotic associations also reduce nutrient limitation by specific elements. Endomycorrhizal associations reduce phosphorus limitation and ectomycorrhizal associations reduce both nitrogen and phosphorus limitation. Analogously, symbiotic association with nitrogen-fixing bacteria reduces nitrogen limitation for the host plant and inderectly for other plants in the ecosystem.

102. Which of the following describes the identification features of non-poisonous snakes ?

(a) Cylindrical tail and small belly scales.

(b) Cylindrical tail, broad transverse belly scales and 4^th infralabial scale is the largest.

(c) Flat tail, broad transverse scales and 3^rd supralabial scale touches eye and nose.

(d) Cylindrical tail, broad transverse belly scales and a loreal pit between eye and nostril.

Ans. (a)

Sol. The identification of snakes is very simple and the following key shall provide the purpose for distruction between poisonous snake and non-poisonous snakes.

1. Tail : If the tail of the snake is cylindrical it may be poisonous or non-poisonous. But if the tail is laterally compressed and oarshaped, the snake is poisonous, e.g., Enhydrian or Hydrophis (sea snake). If the tail is provided with rings which produce a sound when the snake moves, the snake is poisonous e.g., Crotallus (rattle snake). In a few non-poisonous snakes such as Typhlop (blind snake) and Eryx johnii (Dumai) the tail is blunt.

2. Head : If the head of snake is oval, it may be poisonous or non-poisonous. If it is pear shaped, the snake is viper and poisonous. If there is a cervical hood the snake is poisonous i.e., Naja naja (cobra) or Naja bungarus (king kobra).

3. Cephalle scales : If the head is covered with large shields it may be poisonous or non-poisonous but if it is covered with small scales, the snake is poisonous e.g., vipers.

Table of identification of Poisonous and Non-Poisonous Snakes.

103. A road is constructed through a wet tropical forest, following which the population of a species of forest butterfly declines. Which of the following is not a possible explanation for the road causing a decline in the butterfly population ?

(a) Road facilitates immigration of gap-loving species which compete with the forest species.

(b) Road facilitates increased movement of individuals of the forest butterfly within the forest, which reduces genetic diversity.

(c) Road internally fragments the habitat and negatively affects important micro-habitat conditions for the forest butterfly.

(d) Road facilitates invasion by non-native plants that displace native host and nectar plants of the forest butterfly.

Ans. (b)

Sol. The effect of roads on butterflies by surveying abundance, species richness and composition, and mortality in ten grassland patches along high-traffic roads (~50–100 vehicles per hour) and ten reference grassland patches next to unpaved roads with very little traffic (<1 vehicle per day) in southern Poland. Five 200-m transects parallel to the road were established in every grassland patch: at a road verge, 25 m from the verge, in the patch interior, and 25 m from the boundary between the grassland and field and at the grassland-arable field boundary. Moreover, one 200-m transect located on a road was established to collect roadkilled butterflies. The butterfly species richness but not abundance was slightly higher in grassland patches adjacent to roads than in reference grassland patches. Butterfly species composition in grasslands adjacent to roads differed from that in the reference patches. Proximity of a road increased variability in butterfly abundances within grassland patches. Grassland patches bordering roads had higher butterfly abundance and variation in species composition in some parts of the grassland patch than in other parts."These negative genetic effects, which include reduced genetic diversity, limit the potential for populations to respond to selective agents such as disease epidemics and global climate change. We provide clear evidence from a forward-time, agent-based model (ABM) that corridors can facilitate genetic resilience in fragmented habitats across a broad range of species dispersal abilities and population sizes. Our results demonstrate that even modest increases in corridor width decreased the genetic differentiation between patches and increased the genetic diversity and effective population size within patches.

Habitat fragmentation is one of the most important processes contributing to population decline, biodiversity loss, and alteration of community structure and ecosystem. Some of the primary adverse impacts of roads and highways on wildlife and wildlife habitats.

104. Tropical regions may have more species diversity because of the following possible reasons, except

(a) tropical regions have had more time to diversify under relatively stable climatic conditions than temperate regions.

(b) tropical regions have high spatial heteogeneity.

(c) greate biological competition in the tropics leads to narrower niches.

(d) lower predation intensity in the tropics allows survival of more prey species.

Ans. (d)

Sol. The competition hypothesis is based on the idea that competition is the most important factor of evolution in the tropics, whereas natural selection at higher latitudes is controlled mainly by physical factors such as drought and cold (Dobzhansky 1950). Such catastrophic mortality factors are said to be rare in the tropics and thus conpetition for resources becomes keener and niches become smaller, resulting in a greater opportunity for new species to evolve. The predation hypothesis contradicts the competition hypothesis. It claims that ther are more predators and/or parasites in the tropics and that these hold down individual prey populations enough to lower the level of competition between and among them. The lowered level of competition then allows the addition and co-existance of new intermediate prey type, which in turn support new predators in the system. However, the predation hypothesis does not explain why there are more predators and/or parasites in the tropics to begin with.

105. The table given below lists types of plant communities and types of growth forms.

Plant communities Growth forms

P. Dry gasslands 1. Chamaephytes

Q. Semi desert 2. Cryptophytes

R. Tropical forests 3. Hemicryptophytes

S. Tundra 4. Phanerophytes

Which of the following is the best match for the plant communities with most dominant growth form generally present in that community ?

(a) P–4, Q–3, R–2, S–1

(b) P–4, Q–2, R–3, S–1

(c) P–1, Q–4, R–2, S–3

(d) P–2, Q–1, R–3, S–4

Ans. (d)

Sol. Generally, the biological spectra are worked out and compared with Raunkiaer's normal spectrum. The percentage of phanerophytes in different floras ranges from zero to over 74% in the tropical rain forest. The higher percentage of phanerophytes is indicative of phanerophytic climate. High percentage of chamephytes (above 50%) indicates an extremely cold climate. High percentage of hemicryptophytes in temperate forest vegetation indicates the conditions favourable for the development of extensive grassland.

106. Given below ae names of the animals in column X and accesory respiratory organs in teleost fishes (column Y)

Column X Column Y

P. Anabas 1. Labyrinthine organ

Q. Clarias 2. Air sacs

R. Amphipnous 3. Suprabranchial cavity

S. Channa 4. Arborescent organ

The correct match of the animals with the acccssory respiratory organs they have are :

(a) P–4, Q–3, R–2, S–1

(b) P–3, Q–4, R–1, S–2

(c) P–1, Q–4, R–2, S–3

(d) P–2, Q–1, R–3, S–4

Ans. (c)

Sol. Generally adult fishes depended chiefly on pharyngeal gills for aquatic respiration. However, other devices also occur to supplement or replace gill respiration. All such additional respiratory organs, other than gills, are known as accessory respiratory organs.

Accessory respiratory organs dissected on the left side in some air-breathing teleost fishes.

107. The terms expressing some of the developmental events or specific body structures are given in column X and the names of animals that are associated with them in column Y :

Column X Column Y

P. Torsion 1. Starfish

Q. Metagenesis 2. Obelia

R. Apolysis 3. Taenia

S. Pedicellaria 4. Apple snail

The correct match of the terms in column X with the name of animals in column Y is :

(a) P–1, Q–2, R–3, S–4

(b) P–4, Q–2, R–3, S–1

(c) P–3, Q–1, R–2, S–4

(d) P–2, Q–4, R–1, S–3

Ans. (b)

Sol. Torsion of the Apple Snail: The twisting of the visceral mass on head- foot axis is termed torsion, a phenomenon peculiar to gastropod molluscs. Metagenesis is the term used for the process of alternation of generation. Obelia alternates between asexual and sexual modes of generation. Polyps asexually produce medusae, while medusae sexually produce polyps. In Taenia solium, gravid proglottides are regularly detached from the posterior end of strobila and are sent out with the faeces of the host. The shedding of gravid proglottides by the tapeworm is called apolysis. Pedicellariae are minute stalked appendages that are found in among the spines of echinoids.

108. The table lists characteristic anatomical features and names of plants.

Anatomical features Plants

P. Prostele, xylem core surrounded by phloem 1. Lycopodim

Q. Siphonostele, centre pith present or medulated protostele 2. Marsilea rhizome

R. Eustele, conjoint vasculature on edges of the pith 3. Selaginella species

4. Equisetum

(a) P–3, Q–2, R–4

(b) P–1, Q–3, R–4

(c) P–3, Q–1, R–2

(d) P–1, Q–2, R–4

Ans. (a)

Sol. The stem of Selaginella is formed of vascular stele which is protostele. Xylem is present in the center and is surrounded by phloem and phloem is surrounded by pericycle. Amphiphloic siphonostele- In the rhizome of Marsilea, where phloem is present as a ring outside the ring of xylem and the pith is encircled by inner and outer endodermis, pericycle, xylem, and phloem. The horsetail, Equisetum, has a stelar pattern that is unique among spore-bearing plants, which resembles a eustele with several air canals.

109. The table given below provides a list of female gametophyte features and plant genera

Female Gametophyte Plant Genera

P. Monosporic, 8 nucleate 1. Allium

Q. Monosporic, 4 nucleate 2. Oenothera

R. Bisporic, 8 nucleate 3. Peperomia

S. Tetrasporic, 16 nucleate 4. Polygonum

(a) P–4, Q–3, R–1, S–2

(b) P–4, Q–2, R–3, S–1

(c) P–1, Q–2, R–4, S–3

(d) P–4, Q–2, R–3, S–1

Ans. (b)

Sol. The megaspore mother cell undergoes meiosis. Depending upon the participation of the number of megaspore nuclei, the development of embryo sac may be monosporic, bisporic or tetrasporic. When only one megaspore nucleus forms the embryo sac, the development is monosporic. The monosporic embryo sac may be of two polygonum type or oenothera type. The polygonum type of embryo sac is eight nucleated and seven called. The organised embryo sac comprises a 3-celled egg apparatus, three antipodal cells and two polar nuclei lying in the central cell. The oenothera type of embryo sac is four nucleate and 4-celled comprising a 3-celled egg apparatus and a single polar nucleus. There are no antipodals.

When two megaspore nuclei aparticipate in the formation of the embryo sac, the development is said to be bisporic. The bisporic embryo sac are of two types : allium type and endymion type. The bisporic embryo sacs are 8-nucleate and 7-celled. While the allium type develops from the chalazal dyad, the endymion type is derived from the micropylar dyad.

110. Following are some generalization related to wood anatomy of higher plants :

P. The axial system of conife woods consist mainy or entirely of tracheids.

Q. The rays of confiers typically contain only parenchyma cells.

R. The rays of angiosperms typically contain both scerenchyma cells and tracheids.

S. Angiosperm wood may be either diffsue-porous or ring-porous.

Which one of the following options represents all correct statements ?

(a) P and Q only

(b) P and S only

(c) Q and R only

(d) R and S only

Ans. (b)

Sol. Vascular tissue of angiosperm is organized into discrete strands called vascular bundles, each containing xylem and phloem. Storage parenchyma and fibres are generally present and sclereids rarely present.

The wood of both gymnosperms and woody angiosperms is produced in annual growth rings, but the rings tend to be much more pronounced in angiosperms because of the greater size difference between summer wood cells and spring wood cells. Gymnosperms wood is considered to be "nonporous" as it lacks the vessel cells that are characteristics of angiosperms. Angiosperms wood is considered to be "porous", as the large vessels are readily evident and appear to form rings of large pores when viewed in cross-section. Angiosperm wood can be further divided into the "ring-porous" wood and "diffuse-porous" wood.

111. Following are some of the generalization regarding energy flow in an ecosystem :

P. Assimilation efficiency of carnivorres is higher than herbivores.

Q. Consumption efficiency of aquatic herbivores is higher than terrestrial herbivore.

R. Vertebrates have higher production efficiencies thann invertebrates.

S. Trophic-level transfer efficienc is higher in teresterial food chain than in marine.

Based on the above, select the correct option.

(a) Only P and R

(b) Only P and Q

(c) P, Q and R

(d) P, R and S

Ans. (b)

Sol. Assimilation efficiency depends on both the quality of the food and the physiology of the consumer. Assimilation efficiency is the proportion of ingested energy thati s digested and assimilated (A_n) into the bloodstream.

E_assim =

Unassimilated material returns to the soil as feces, a component of the detrital input to ecosystems.

Assimilation efficiencies are often hiher (5 to 80%) than consumption efficiencies (0.1 to 50%). Carnivores feeding on vertebrates tend to have higher assimilation efficiencies (about 80%) than to terrestrial herbivores (5 to 20%), because carnivores eat food that has less structural material than is present in terrestrial plants. Among herbivores, species that feed on seeds, which have high concentrations of digestible, energy rich storage reserves, have a higher assimilation efficiency than those feeding on leaves. Leaf feeding herbivores, in turn, have higher assimilation efficiencies than those feeding on wood, which has higher concentrations of cellulose and lignin. Many aquatic herbivores have a particularly high assimilation efficiency (up to 80%) because of the low allocation of structure in many algae and other aquatic plants. Even in aquatic ecosystem, however, herbivores that feed on well-defended species have low assimilation efficiencies. Assimilation efficiencies of herbivores feeding on cyanobacteria, for example can be as low as 20%.

112. In an experiment to show that biogeochemical cycles interact, nitorgen fixing vines (Galactia sp.) were grown in plots under norma levels of CO₂ (Control) and under artificially elevated atmospheric CO₂ (Experimental). Effect of elevated CO₂ levels on nitrogen fixation was measured over a period of 7 yeas (Plot A) and the concentrations of iron and molybdenm ain the leaves of these plants were quantified at the end of the study (Plot B).

Which one of the following inferences cannot be made from the above experiment ?

(a) Decreasing rate of N-fixation corelates with decreased levels of leaf iron and molybdenum, two micronutrients essential for N-fixation.

(b) An initial exposure to elevated CO₂ increased N-fixation by these plants.

(c) There is a continuous decrease in N-fixation due to elevated CO₂ treatment.

(d) Plants exposed to continuous elevated levels of CO₂ had lower levels of iron and molybdenum in their leaves.

Ans. (c)

Sol. Nitrogen fixing plant species may respond more positively to elevated atomspheric carbon dioxide concentrations (CO₂) than other species because of their ability to maintain a high internal nutrient supply. Elevated CO₂ increased nodule size and number, specific nitrogenase activity and plant N content and consequently increased biomass and/or seed yield in legumes. Also iron and molybdenon are essential cofctors required for the enzyme nitrogenase which fixes nitrogen. Hence if we observe plot A and B the correct statements are 1 and 4 as elevated CO₂ enhance nitrogen fixation and hence utilization of these micronutrients increases. Now lets compare statements 2 and 3, statement 2 makes more justice to the graph as initial exposure to high CO₂ is increasing nitogen fixation. While in statement 3 in the 6th and 7th year it is becoming more or less stable hence statement 3 is incorrect.

113. Match the following invasive plants to the likely habitats in which they are expected to occur :

Invasive plant Habitat(s) that they invade

P. Eichhornia crassipes 1. Arid and semi-aid habitats

Q. Lantana camara 2. Dry and moist tropical forests

R. Prosopis juiflora 3. Wetlands

(a) P–2, Q–1, R–3

(b) P–1, Q–3, R–2

(c) P–3, Q–2, R–1

(d) P–3, Q–1, R–2

Ans. (c)

Sol. Lantana camera is found predominantly in tropical and sub-tropical environments, but also capable of growing in warmer temperate and semi-arid regions, whereas Eichhornia crassipes is a free floating aquatic plant that has invaded aquatic areas throughout the eastern and southern portions of the United States. Eichhornia crassipes invades lakes, ponds, rivers, marshes and other types of wetland habitats. Puosopls juliflora is called as mesquite which usually invade Arid and semi-arid habitats.

114. Incorporating additional ecological factors into the Lotka-Voterra predator-prey model can change the predator isocline. Given below are three state-space graphs (P–R) representing modification of predator isocline due to the ecological factors listed beow (1-3)

1. Victim abundance acting as predator carrying capacity.

2. Availability of alternate prey victim population.

3. Predator carrying capacity determined by factors other than victim abundance.

Which one of the following options represents all correct matches of the state-space graphs with their ecological factor ?

(a) P–2, Q–3, R–1

(b) P–2, Q–1, R–3

(c) P–3, Q–2, R–1

(d) P–1, Q–2, R–3

Ans. (c)

Sol. In graph P, predator carrying capacity determined by factors other than victim abundance.

In graph Q, availability of alternate prey victim population.

In graph R, victim abundance acting as predator carrying capacity.

115. Antelopes are proposed to form groups to redce the risk of predation. A researche measured the predation of individuals in groups of different sizes. She found that per capita mortality risk decreased with increasing group size for males (solid line) but remained unchanged for females (dashed line). Furthermore, males in all groups experienced greater per capita mortality risk than females. Identify the graph below that best depicts the above findings :

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Per capita mortality risk decreased with increasing group size of males so the solid line must have a decreasing slope from a higher value per capita mortality risk was constant with increasing group size of females so the dashed line must have constant slope. So graph 2 is rejected.

Now, males in All groups must have greater per capita mortality risk than females so the solid line must never have a y-axis value below the dashed line. So, graph 4 is correct (In graph 3, it is seen that some males are having lower mortality risk than females).

116. The prominent mammal species found in four different protected areas are listed below :

Area A : Tiger, Wild dog, Leopard, Elephant

Area B : Common langur, Barking deer, Wild dog, Elephant

Area C : Tiger, Indian rhinoceros, Pygmy hog, Wild pig

Area D : Blackbuck, Indian gazelle, Hyena, Indian wolf

The area with the greatest phylogenetic diversity is

(a) A

(b) B

(c) C

(d) D

Ans. (b)

Sol. All the four animals belong to 4 different mammalian orders thus exhibiting maximum diversity.

Common langur- Primates, Barking deer-Artiodactyla, Wild dog- Carnivora, Elephant- Proboscidea.

117. A field ecologist gathers following data (abundance values) in order to study diversity of species in four plant commnities.

Based on the above observations, the ecologist draws following conclusions :

P. Plant communities C₁ and C₄ show strong similarity with each other.

Q. Plant communities C₁ and C₄ as well as communities C₂ and C₃ show strong similarity with each other.

R. Plant community C₁ is most diverse.

S. Plant community C₄ is most diverse.

Which of the following statements is correct regarding above conclusions ?

(a) All the conclusions are correct

(b) Only conclusions P and S are correct

(c) Only conclusions P and R are correct

(d) Only conclusions Q and S are correct

Ans. (b)

Sol. An index of community similarity that is based on the relative abundance of species within the communities being compared is called as the percent similarity (PS). Based on percentage similarity C1 and C2 show strongest similarity. The maximum values of the diversity index (Shannon index), occurs when all species are present in equal numbers. So here most even distribution of species is seen in C4. So it is the most diverse community among all.

118. Given below are the steps to assess the population size of grasshoppers in a given area :

P. 'n' individuals are collected randomly from the study area in a defined period of time.

Q. The captured individuals are counted, marked and released at the site of collection. Next day, individuals are captured from the same site for same length of time. Number of marked (n_m) and unmarked (n_u) individuals are separated and counted.

R. This capture-release and recature is continued till one gets 100% maked individuals.

S. The size of the population (N) is estimated as follows :

T. The size of population (N) is estimated as follows :

The most appropriate combination of steps for estimating population size using mark-recapture method is :

(a) P, Q and S

(b) P, Q and T

(c) P, Q, R and S

(d) P, Q, R and T

Ans. (a)

Sol. n = collected randomly first and marked. in the next catch, nm = marked and nu = unmarked hence total = (nm + nu). Total population is N. Hence if nm are number of individual marked from the total population (nm + nu). Therefore, the previous marked i.e., n were from how much total population N. Just by simple proportionality theory the answer comes similar to statement D.

119. Which one of the following statements is correct for the process of speciation ?

(a) Allopatric speciation occs between adjacent populations.

(b) Parapatric speciation may occuu between geographically separated populations.

(c) Sympatric speciation occurs within one continuously distributed population.

(d) Sympatric speciation occurs when continuously distributed populations are fragmented.

Ans. (c)

Sol. In sympatric, reproductive isolation evolves between distinct subgroups that arise within one population. Models of sympatric speciation do not require that the populations be either geographically or environmentally separated as their gene pools diverge.

120. Which one of the following does not contribute to microevolutinary change ?

(a) Mutation

(b) Random mating

(c) Genetic drift

(d) Natural selection

Ans. (b)

Sol. Microevolution is simply a change in gene frequency within a population. Evolution at this scale can be observed over short periods of time — for example, between one generation and the next, the frequency of a gene for pesticide resistance in a population of crop pests increases. Such a change might come about because natural selection favoured the gene because the popullation received new immigrants carrying the gene, because some nonresistants genes mutated to the resistant version or because of random genetic drift from one generation to the next. Mutation, migration, genetic drift and natural selection are all processes that can directly affect gene frequencies in a population.

121. According to Hamilton's rule, altruistic behaviou can evolve when rb > c, where b is the extra benefit gained by the recipient as a result of the altruistic act, c is the cost to the actor arising from performing the altruistic act and is the relatedness between the

(a) individual performing the altiustic act and the offsping of the recipient.

(b) individual performing the altruistic act and the recipient.

(c) recipient and the offspring of the individual performing the altrruistic act.

(d) individual performing the altruistic act and the members of its population.

Ans. (b)

Sol. The probability that the altruist and the recipient share a gene is called the coefficient of relatedness (r). The rule states that a social behaviour will be favoured by natural selection if and only if rb > c, where 'b' is the effect of the behaviour on the reproductive success of others, 'c' is the effect on one's own reproductive success and 'r' is the coefficient of relatedness, which measure the statistical association between the genes of the actor who performs the behaviour and the genes of the recipient affected by it.

122. Analyses of nucleotide sequence of ribosomal RNA (rRNA) are particularly useful for evolutionay studies of living organisms because of the following reasons except

(a) rRNA is evolutionarily ancient.

(b) no free-living organism lacks rRNA.

(c) rRNA, since critical for translation, can undergo lateral transfer amongst distant species.

(d) rRNA has evolved slowly over geological time.

Ans. (c)

Sol. rRNA gene sequences have become the gold standard for microbial identification and the inference of deep evolutionary relationships. rRNA genes occur in all cells and organelles and the rRNA genes are the most conservative large sequences in nature. For instance, the eukaryotic and bacterial SSU rRNA genes typically have about 50% identity over the alignable lengths of their SSU rRNAs. The rRNA genes have not undergone significant lateral transfer and the structural properties of the rRNA provdie for optimization of alignments.

123. Depicted below is a phylogenetic tree of selected taxa

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

(a) Group X is monophyletic and group Y is polyphyletic.

(b) Group X is paraphyletic and group Y is monophyletic.

(c) Both group S and Y are monophyletic.

(d) Group X is monophyletic and group Y is paraphyletic.

Ans. (d)

Sol. A paraphyletic group includes the common ancestor but is missing some of the descendants, a classic example being my own kingdom of choice, Protista, Reptiles are often used to demonstrate that as well, since they gave rise to the birds and mammals that are not classified within the reptiles.

A group taht includes all the descendants of a given ancestor along with their last common ancestor is called a clade. A clade is sometimes referred to as a monophyletic group.

124. During the corse of vertebrate evolution, the jaw bones got modified into three ear ossicles in mammals. Which one of the following is a correct match of ear ossicle and its ancestral jaw bone ?

(a) Stapes-Articular; Incus-Hymandibular; Malleus-Quadarate

(b) Stapes-Quadarate; Incus-Articular; Malleus-Hymandibular

(c) Stapes-Quadarate; Incus-Hymandibular; Malleus-Articular

(d) Stapes-Hymandibular; Incus-Quadarate; Malleus-Articular

Ans. (d)

Sol. From fossil data, comparative anatomy and developmental biology it is clear that the bones in the mammalian middle ear, the malleus and incus are homologous to the quadrate and articular, which form the articulation for the upper and lower jaws is non-mammalian jawed vertebrates. The evolution of the stapes from the hyomandibula was an earlier and distinct event (1, 2).

125. Birds in a population show two foraging phenotypes, A and B. Birds of phenotype A search, attack and capture prey from birds of phenotype A. A and B are maintained in the population through negative frequency-dependent selection. The graph below shows the fitness of a (broken line) and B (solid line) at different relative frequencies of A (Frequency of B = 1 – frequency of A).

Which of the following statements does the graph support ?

(a) A outcomes B; at equilibrium, A goes to fixation.

(b) B outcomes A; at equilibrium, B goes to fixation.

(c) A and B are both maintained in the population; the equilibrium frequencies are A = 0.6, B = 0.4.

(d) A and B are both maintained in the population; the equilibrium frequencies are A = 0.9, B = 0.1.

Ans. (c)

Sol. As you can clearly see from the graph that at equilibrium i.e., at the point when A and B meets at the point frequency of A = 0.6 hence B = 0.4 i.e., (Frequency of B = 1 – Frequency of A) hence 3rd is the right answer. Negative frequency dependent selection is a powerful type of balancing selection that maintains many natural polymorphism, but it is also commonly misinterpreted.

126. The Hardy-Weinberg principle states that allele frequencies in a population will remain constant over generations if certain assumptions are net.

P. Random mating

Q. Mate choice

R. Small Population size

S. Large population size

T. Lack of mutations

U. Directional selection

Which of the above factors will cause changes in allele frequencies over generations ?

(a) P, S and U

(b) Q, S and U

(c) P, R and T

(d) Q, R and U

Ans. (d)

Sol. Five assumptions of a Hardy-Weinberg population :

1. No Selection

2. No Mutation

3. No Gene Flow-No migration between populations to change allele frequencies.

4. Infinite Population is necessary because the larger the population size is, the harder to change the allele frequency.

5. Random Mating which is pretty much what it says : Mating being based on nothing other than pure chance, making it random. Hence any deviation from these assumptions will cause allele frequencies to change over time.

127. Given below are few traits and related functions :

Trait Function

P. Aposematism 1. Acquiring food

Q. Basking 2. Avoiding predation

R. Cooperative hunting 3. Territory defence

S. Song 4. Thermoregulation

Match the above given traits to their most likely functions.

(a) P–3, Q–4, R–2, S–1

(b) P–4, Q–2, R–1, S–3

(c) P–2, Q–4, R–1, S–3

(d) P–3, Q–1, R–4, S–2

Ans. (c)

Sol. The term "aposematism" is commonly used as a synonym for warning colouration (i.e., something that is aposematic is warningly coloured). The word literally means "away signal'. Aposematism is the combintion of a conspicous signal and an unprofitable trait in a prey species (i-B), 'Basking' is the most conspicuous thermoregulatory behaviour in many animals that enable them to enhance pysiological performance (ii-D). The cooperative hunting helps in acquiring food for examples species like Hyena, Lions they hunt in packs so as to be efficient in acquiring food in short time (iii-A). Using the responses of territory owners to playback to infer the territory function of acoustic signals is common practice. BIRD song is generally considered to have one or both of two main function : attracting a mate and proclaiming a territory. The great tit (Parous major) isa typical songbird in that song is primarily. (iv-B).

128. Following is a diagrammatic representation of human evolutionary tree.

In the above diagram A, B, C and D respectively represent :

(a) Denisovan, Homo habilis, Homo erectus, Homo neandethalensis

(b) Homo haibilis, Homo erectus, Homo neanderthalensis, Denisovan

(c) Homo erectus, Homo habilis, Homo neanderthalensis, Denisovan

(d) Homo erectus, Denisovan, Homo neanderthalensis, Homo Habilis

Ans. (b)

Sol. A - Homo habilis is an extinct species. This species known as 'handy man' because stone tools were found near its fossil remains and it is assumed this species had developed the ability to modify stone into tools.

B - The extinct ancient human Homo erectus is a species of firsts. It was the first of our relatives to have human-like body proportions, with shorter arms and longer legs relative to its torso.

C - Neanderthals made and used a diverse set of sophisticated tools, controlled fire, lived in shelters, made and wore clothing, were skilled hunters of large animals and also ate plant foods, and occasionally made symbolic or ornamental objects.

D - Alternatively, on arriving in Denisovan habitat, H. sapiens may have outcompeted or killed their cousins, or brought lethal diseases with them. Climatic events could have been crucial too.

129. Which one of the following DNA markers can be used to distinguish between a homozygote and a heterozygote ?

(a) RAPD

(b) AFLP

(c) RFLP

(d) ISSR

Ans. (c)

Sol. Codominant markers are markers for which both alleles are expressed when co-occurring in an individual. Therefore, with codominant markers, heterozygotes can be distinguished from homozygotes, allowing the determination of genotype and allele frequencies at loci.

130. Which one of the following is the most appropriate definition of 'Gene Pyramiding' in plants ?

(a) Introducing different genes for resistance to a specific pest in different genotypes.

(b) Introducing a single gene for resistance to a particular pest in different genotypes.

(c) Introducing different genes for resistance to a single pest in single genotypes.

(d) Introducing a single gene for resistance to multiple pests in different genotypes.

Ans. (c)

Sol. Gene pyramiding includes combination of genes for resistance to multiple races of the same disease, genes for resistance to different diseases and genes for disease and insdect resistance.

In rice, three blast resistance genes have been pyramided into one cultivar. First, three blast resistance genes were mapped on to rice chromosomes 6, 11 and 12. Gene pyramiding started with three NILs, each carrying one of the genes. After two cycles of crossing and selection, a plant containing the three was obtained and it has been used a source for three genes in palnt breeding.

131. In bioremediation by microorganisms detailed below, choose the incorrect option ?

(a) The organic contaminants provide a souce of carbon.

(b) The bacteria do not get net energy be degrading contaminants.

(c) Bacteria can produce oxidized or reduced species that can cause metals to precipitate.

(d) Bacteria act on contaminants by aerobic and anaerobic respiration.

Ans. (b)

Sol. The key players in bioremediation are bacteria-microscopic organisms that live virtually everywhere. Microorganisms are ideally suited to the task of contaminant destruction because they possess enzymes that allow them to sused environmental contaminants as food and because they are so small that they are able to contact contaminants easily.

Microbial transformation of organic contaminants normally occurs because the organisms can use the contaminants for their own growth and reproduction. Organic contaminants serve two purposes for the organisms : they provide a source of carbon, which is one of the basic building block of new cell constituents and they provide electrons, which the organisms can extract to obtain energy.

132. Which one of the following statements regarding normal distibution is not correct ?

(a) It is symmetric around the mean.

(b) It is symmetric around the median.

(c) It is symmetric around the variance.

(d) It is symmetric around the mode.

Ans. (c)

Sol. Unlike skewed distributions, which are asymmetrical, normal distribution are symmetrical; that is, the right and left halves of the curve are mirror images of each other. Consequently the mean, median and mode in a normal distribution are identical and are located at its center as illustrated in Fig. Suppose you cut out the distribution in Fig. and then folded it on the centerline depicting the mean, median and mode.

The normal distribution (Also known as the bell shaped curve, the bell curve or the normal curve)

You can also seed in Fig. that the normal distribution is curved in the shape of a bell. Thus, it is often referred to as a bell-shaped curve or more simply as a bell curve. It is also often called the normal curve.

133. A multimeric protein when run on an SDS gel showed 2 bands at 20 kDa and 40 kDa. However, when the protein was run on a native gel, it showed a single band at 120 kDa. The native form of the protein would be

(a) homotrimer

(b) heterotetramer

(c) heterodimer

(d) heterotrimer

Ans. (b)

Sol. In SDS PAGE, SDS (denaturing agent) cleaves the 120 kd multimeric protein into four subunits 2 × 20 kd + 2 × 40 kd. However, Native PAGE doesn't separate the subunits of proteins, so it gives 120 kd as total weight. Here the four subunits have different weights. So the native form of protein would be heterotetramer.

134. A solution contains NADH and NAD⁺, both at 0.1 mM concentration. If NADH has a molar extinction coefficient of 6220 and that of NAD⁺ is negligible, the optical density measured in a cuvette of 5 mm path length will be

(a) 0.62

(b) 0.062

(c) 0.31

(d) 0.031

Ans. (c)

Sol. Optical density of NADH

= molar absorbance coefficent of NADH × molar concentration of the solution × path length of the light

= 6220 × 0.1 mM × 5 mm = 6220 × 0.0001 M × 0.5 cm = 0.31

Optical density of NAD⁺ = 0

Total optical density = optical density of NADH + optical density of NAD⁺ = 0.31

135. Orientation of a cloned DNA fragment (gene) in a plasmid vector can be checked by

(a) PCR using two gene-specific primers.

(b) Restriction digestion with an enzyme that has a single restriction site within the cloned gene and none in the vector.

(c) PCR using a combination of one gene-specific primer and one vector-specific primer.

(d) Restriction digestion with an enzyme that has two restriction sites within the vector sequence and none in the cloned gene.

Ans. (c)

Sol. The usual PCR technique for fragment orientation involves one primer specific of the vector (upstream of downstream of the cloning site) and on primer within the cloning sequence. If the primer set is correctly oriented, PCR will give a unique band with a specific size according to the distance between primers. If not properly oriented, there will be no amplification. For digestion, principle is the same : using one restriction enzyme specific of the vector (a unique site close to the cloning site) and one restriction specific of the insert but not lying right in the middle of it (it may be the same restriction enzyme specific of the vector).

136. The emission maximum of tryptophan fluorescence in a protein in ~335 nm. This suggests that tryptophan

(a) is in hydrophobic environment.

(b) occurs in a helical segment.

(c) has proximal cysteine residues.

(d) is oxidized.

Ans. (a)

Sol. Tryptophan in a hydrophobic microenvironment inside the protein globule and tryptophan in a hydrophilic microenvironment on the surface of the globule. Judged by the behaviour of glycyltryptophan and acetyl tryptophan amide, molecules in a weakly polar hydrophobic environment will have higher quantum yields than molecules in a polar aqueous environment.

137. To test the impact of cAMP on protein kinase A conformation in cells, an investigator made FRET biosensor by fusing two fluorescent proteins at the N- and C-terminus of protein kinase A. In the absence of cAMP in the cellular milieu, no FRET signal was detected. However, upon cAMP addition, a strong emission at 530 nm was observed. What could be the best configuration of flurorophores that were used by the investigator ?

(a) Green fluorescent protein (GFP) and Red fluorescent protein (RFP).

(b) Cyan fluorescent protein (CFP) and Yellow fluorescent protein (YFP).

(c) Yellow fluorescent protein (YFP) and Red fluorescent protein (RFP).

(d) Red fluorescent protein (RFP) and Cyan fluorescent protein (CFP).

Ans. (b)

Sol. The choice of two fluorophore in the FRET experiment is based on the minimal or no spectral overlap of emission spectra. The most common choice is therefore cyan and yellow colours contributed by CFP and YFP respectively. Another aspect is quantum yield, which also affect the decision of choosing a pair of fluorophore.

138. Following statements were given regarding facto influencing variation in expression levels of transgene in transgenic plants :

P. Difference in restriction enzyme sites within the T-DNA.

Q. Difference in copy number of the transgene.

R. Variations in site of integration of the T-DNA within the plant genome.

S. Presence of multiple promoters within the T-DNA region.

Which one of the following options represents a combination of statements that would not lead to variations in transgene expression levels in transgenic plants generated using the same T-DNA/binary vector ?

(a) P and R only

(b) Q only

(c) R and S only

(d) P and S only

Ans. (d)

Sol. The routine generation of single-copy transgenic events is therefore a major goal for agricultural events is therefore a major goal for agricultural biotechnology. Integration of transfected DNA into the plant genome usually takes place by a process of non-homologous or illegitimate recombination. This random integration of transgenes, but also variations in copy number and the configuration of the transgene, which may result in gene silencing, have been implicated to explain differences in gene expression levels between individual transformants.

139. Based upon phenotypic observation, it was concluded that an unknown gene responsible for an agronomically important trait is present in a particular plant. In order to identify the gene, a researcher proposes to use the following strategies :

P. PCR amplification of the gene.

Q. Map based cloning of the gene.

R. Subtractive DNA hybridization.

S. Genome sequencing.

T. Develop molecular markers linked to the trait.

Which one of the following options is most suitable for identifying the unknown gene ?

(a) P and R

(b) Q and T

(c) R only

(d) P and S

Ans. (b)

Sol. An unknown gene responsible for a certain character can be identified through map based cloning and molecular markers. For example, with the construction of a linkage map to a genomic region linked to vernalization response in carrot, the region was mapped in a linkage group with 78 molecular markers and flanking markers wer at 0.69 and 0.79 cM. This map could be used to develop molecular markers linked to the trait and also in the future to start physical mapping and sequencing using a carrot BAC library.

140. A T₀ transgenic plant contains two unlinked copies of the T-DNA of wich, one is functional and the other is silenced. Segregation of the transgenic to non-transgenic phenotype would occu in a (i) ratio in progeny obtained by backcrossing and in a (ii) ratio in F₁ progeny obtained by self-pollination.

Fill in the blanks with the correct combination (i) and (ii) fom the options given below :

(a) (i)–3 : 1 and (ii)–15 : 1

(b) (i)–1 : 1 and (ii)–3 : 1

(c) (i)–3 : 1 and (ii)–3 : 1

(d) (i)–1 : 1 and (ii)–15 : 1

Ans. (b)

Sol. There are 2 copies of T-DNA, the which is functional lets take it as D and the non-functional to be d, now since this cell contains 2 copies it is Dd. When Dd is backcrossed with either of the parent cell having either D/d the ratio will come as 1 : 1 when Dd is selfed the ratio comes as 3 : 1 it will never come as 15 : 1 because in that case both the genes are dominant, which is not the case stated here.

141. The structure of a protein with 100 residues was determined by X-ray analiysis at atomic resolution and NMR spectroscopy. The following observations are possible.

P. The dihedral angles determined from the X-ray structure and NMR will be identical.

Q. The dihedral angles determined from the X-ray structure will be more accurate.

R. -turns can be determined only by NM.

S. -sheets can be more accurately determined from the X-ray structure.

Indicate the combination with all correct answers.

(a) P and R

(b) Q and S

(c) Q and R

(d) P and S

Ans. (b)

Sol. X-ray diffraction is considered to be the most reliable and most accurate method for protein structure determination and till date, highest number of protein structures have been determined using X-ray diffractions.

However, with increasing technological advancements it is low possible to achieve an equivalent resolution of structure using parallel methods such as NMR and cryo-EM.

142. The below figure shows the fluorescence emission spectra of three different proteins : Protein (X), Protein (Y), and Protein (Z) excited at 280 nm.

Which one of the following statements gives the correct interpretation ?

(a) Proteins (Y) and (Z) have tryptophan while protein (X) has only phenylalanine.

(b) Protein (X) has only tyrosine and protein (Y) has tryptophan on the surface while protein (Z) has tryptophan buried inside.

(c) Protein (X) has tryptophan buried inside while proteins (Y) and (Z) have tryptophan on the surface.

(d) Protein (X) has only tyrosine and protein (Y) has tryptophan buried and protein (Z) has tryptophan on the surface.

Ans. (d)

Sol. The most common use of Trp fluorescence information is to assign a Trp as buried and in a "non-polar" environment if is < ~330 nm; if is longer than ~330 nm, the Trp is assigned a "polar" environment, which almost always is intended to imply solvent exposure.

143. Specimens for light microscopy are commonly fixed with a solution containing chemicals that crosslink/denautre cellular constituents. Commonly used fixatives such as formaldehyde and methanol could act in various ways as descibed below :

P. Formaldehyde crosslinks amino groups on adjacent molecules and stabilizes protein-protein and protein-nucleic acid interactions.

Q. Methanol acts as a denaturing fixative and acts by reducing the solubility of protein molecules by disrupting hydrophobic interactions.

R. Formaldehyde crosslinks lipid tails in biological members.

S. Methanol acts on nucleic acids, crosslinks nucleic acids with proteins and thus stabilizes protein-nucleic acid interactions.

Which one of the following combinations represents all correct statements ?

(a) P and R

(b) Q and R

(c) Q and S

(d) P and Q

Ans. (d)

Sol. Ethanol (CH₃CH₂OH) and methanol (CH₃OH) are considered to be coagulants that denature proteins. They replace water in the tissue environment disrupting hydrophobic and hydrogen bonding thus exposing the internal hydrophobic groups of proteins and altering their tertiary structure and their solubility in water. Methanol is commonly used as a fixative for blood films and 95% ethanol is used as a fixative for cytology smears but both alcohols are usually combined with other reagents when used as fixatives for tissue specimens. Hence 2 is correct.

Formaldehyde is an aldehyde, which mildly oxidize reduced carbon or nitrogen. It creates bonds between proteins, lipids or between lipids and proteins. Formaldehyde crosslinks protein and DNA. Hence statement 1 is correct.

144. Run off transcription assays were performed to establish the specificity of three novel sigma factors for their promoters. Results of the experiments are shown below :

Following inferences were made from these results :

P. initiates trancription from P₂ and from P₁.

Q. can initiate transcription form both promoters.

R. prevents initiation of transcription from P₂.

S. initiates transcription from P₁.

Choose the option that correctly interprets the results.

(a) P, Q and R only

(b) P and Q only

(c) R and S only

(d) Q, R and S only

Ans. (a)

Sol. A run-off transcription assay is conducted in vitro to identify the position of transcription start site (1 base pair upstream) of a specific promoter along with its accuracy and rate of transcription. If a transcription factor bonds to DNA it leads to a gel retardation or shift. The observed pattern shows that initiates transcription from P₂, initiates transcription from P₁, initiates transcription from P₁ and P₂. Hence, statements P, Q and R are correct.

145. A neurophysiologist was interested in using the patch-camp technique. Following statements are related to this technique :

P. Intracellular movement of ion channels.

Q. Post-transcriptional modification of the ion channel protein.

R. Ligand that controls the opening or closing of ion channels.

S. Change in current flow in a single ion channel.

Which one of the following combinations will be achievable using the patch-clamp technique ?

(a) P and Q

(b) Q and R

(c) R and S

(d) S and P

Ans. (c)

Sol. The patch clamp technique is a laboratory techinque in electrophysiology used to study ionic currents in individual isolated living cells, tissue sections or patches of cell membrne.

Patch clamping can be performed using the voltage clamp technique. In this case, the voltage across the cell membrane is controlled by the experimenter and the resulting currents are recorded. Alternatively, the current clamp technique can be used. In this case the current passing across the membrane is controlled by the epxerimenter and the resulting changes in voltage are recorded, generally in the form of action potentials.

For ligand gated ion channels or channels that are modulated by metabotropic receptors the neurotransmitter or drug being studied is usually included in the pipette solution, where it can interact with what used to be the external surface of the membrane.