PYQ NOV 2020 SIFT 1

CSIR NET BIOLOGY (November - 2020 Sift 1)
Previous Year Question Paper with Solution.

21. A stochiometric mixture of and a anomers of D-glucose in water exhibits

(a) net optical rotation proportional to the sum of the optical activities of each anomer

(b) no optical activity as the signs of optical rotation are opposite and they cancel each other

(d) no optical activity as they form a racemic mixture

Ans. (a)

Sol. Glucose is a mixture of and -anomers, primarily the -anomer. The optical rotation of the -anomer is +112.2° (c = 10% in water, 20°C) and the -anomer is +18.7° (c = 10% in water, 20°C). When D-glucose is dissolved in water, the optical rotation gradually changes (mutarotates) with time and approaches a final equilibrium value of +52.7° (c = 110%, 20°C) due to the formation of an equilibrium mixture consisting of approximately one third and two thirds -D-glucose.

22. Which one of the following options represents a series of the amino acids with the decreasing pK_a values of their side chains?

(a) Arg Lys Cys His

(b) Lys Arg Cys His

(d) Arg Cys Lys His

Ans. (a)

Sol. Proline is an amino acid and considered as a stress hormone. Pyruvate is the end product of glycolysis. Kaurenoic acid is a primary metabolite, whereas the closely related compound abietic acid is considered secondary metabolite.

23. The rate constant for conversion of a substrate into the product is 10^–4 s^–1 while the reverse rate constant is 10^–6 s^–1. An enzyme enhances the rate of this reaction by 100-fold. The equilibrium constant for this enzyme catalyzed reaction is

(a) 100

(b) 10000

(d) 1000

Ans. (a)

Sol. Suppose that in the absence of the enzymes the forward rate constant (kf) is 10^–4 s^–1 and the reverse rate constant (kr) is 10^–6 s^–1. The equilibrium constant (K_eq) is given by the ratio of the two rate constants.

K_eq = [B]/[A] = kf/kr = 10^–4/10^–6 = 100

The equilibrium concentration of B is 100 times that of A whether or not an enzyme is present. However, in the absence of an enzyme the reaction could take more than an hour to approach this equilibrium, whereas in the presence of an enzyme, equilibrium could be attained within a second, hence the 100-fold increase in rate (this value is not required to calculate K_eq).

24. The following statements were made of suggest the existence of an enzyme-substance complex.

A. At constant concentration of enzyme, the reaction rate increase with increasing substrate concentration

B. An enzyme-catalyzed reaction has a maximal velocity

C. At constnat concentration of the enzyme and substrate, an increase in the reaction rate is observed

D. An enzyme catalyzed reaction is not influenced by high substrate concentration.

Which of the above statements suggest the existance of an enzyme-substrate complex?

(a) A and B

(b) B and C

(d) D only

Ans. (a)

Sol. For a given enzyme concentration, the rate of reaction increases with increasing substrate concentration (A is correct) up to a point, above which any further increase in substrate concentration produces no significant change in reaction rate (C is wrong). This is the point where an enzyme catalyzed reaction reaches maximum velocity (B is correct). This is because the active sites of the enzyme molecules at any given moment are virtually saturated with substrate. The enzyme/substrate complex has to dissociate before the active sites are to accommodate more substrate. Statement D is faulty.

25. Some metabolic aspects of the Red Blood Cell (RBC) are proposed in the following statements:

A. Synthesis of fatty acids does not occur in the RBC

B. The pentose phosphate pathway is operative in the RBC

C. RBC cannot synthesize reduced glutathione (GSH)

D. RBC does not contain enzymes like adenosine deaminase and pyrimidine nucleotidase

E. NADH-dependent methemoglobin reductase system is present in RBC

Which one of the following combinations is NOT correct?

(a) A and B

(b) B and C

(d) D and E

Ans. (c)

Sol. The pentose phosphate pathway is also important in the red blood cell, where NADPH is required to maintain an adequate pool of reduced glutathione, which is used to remove hydrogen peroxide. (Option c not true, since its mentioning that RBC cannot produce reduced glutathione).

In the red blood cell, adenine genertaed from purine degradation can be deaminated to Inosine by adenosine deaminase (ADA), phosphorylated to AMP by adenosine kinase, or transported into extracellular fluid.

26. An enzyme catalyzed reaction comparing proton and deuteron transfer yielded the following kinetic date

K_cat [H]. s^–1 70.7

K_M [H], mM 1.03

K_cat [H], s^–1 10.3

K_M [H], mM 1.06

Which of the following represents the turn over number (M^–1 – S^–1) for proton transfer, deutreron transfer and the knetic isotope effect, respectively?

(a) 67640, 9717, 14

(b) 6864, 972, 7

(d) 68640, 9717, 7

Ans. (d)

Sol. Catalytic efficiency = K_cat/K_m

For proton transfer, K_H = 70.7/(1.03 × 10^–3) = 68640

For deuteron transfer, K_D = 10.3/(1.06 × 10^–3) = 9717

Kinetic isotope effect = K_H/K_D = 68640/9717 = 7.06

27. The 15 base-paired nucleic acid molecule shown below is disolved in an squeous buffer of pH 7.3

5' – A G U U C G G A U A U C G U G – 3'

3' – U C A U G C C U A U A G C A C – 5'

Which one of the following statement is INCORRECT?

(a) It can be a double-stranded DNA molecule in a B-form helix

(b) It can be a double-stranded DNA molecule in a A-form helix

(d) It can be a double-stranded RNA molecule in a B-form helix

Ans. (d)

Sol. Double-stranded RNA cannot adopt B-form due to the presence of 2'OH group in their ribose sugars.

28. Which one of the following is NOT involved in the formation of lipid rafts?

(a) Sphingomyelin

(b) Chloesterol

(d) Phosphatiduylserine

Ans. (d)

Sol. The plasma membranes of cells contain combinations of glycosphinoglipids, cholesterol and protein receptors organised in glycolipoprotein lipid microdomains termed lipid rafts. It has been proposed that they are specialized membrane microdomains which ompartmentalize cellular processes by serving as organising centers for the assembly of signaling molecules. Also, lipid rafts are enriched in sphingolipids such as sphingomyelin, which is typically elevated by 50% compared to the plasma membrane. To offset the elevated sphingolipid levels, phosphatidylcholine levels are decreased which results in similar choline-containing lipid levels between the rafts and the surrounding plasma membrane.

29. Myosin molecules that assemble into bipolar filaments in the muscle are:

(a) Myosin I

(b) Myosin II

(d) Myosin V

Ans. (b)

Sol. Myosin-II molecules contain two motor domains and a coiled-coil tail domain that assembles the protein into bipolar filaments. Historically, the term 'conventional myosin' was applied to the filament forming myosin-II family because the muscle isoforms were the originally described proteins.

30. The total number of cellsa in 1 ml of a bacterial cuture was esrtimated to be 2.7 × 10⁶. The culture was diluted 27-fold and 100 ul seeded per well of 96-well plate. What is the final cell number per well?

(a) 1 × 10⁵

(b) 2.7 ×b 10⁴

(d) 1 × 10⁴

Ans. (d)

Sol. CFU/ml = 2.7 × 10⁶; DF = 27;

Volume planted in 1 well = 100 microL = 0.1 ml;

Let cell numer in 1 well = x

CFU/ml = (x× DF)/Volume planted in well in ml;

2.7 × 10⁶ = (x × 27)/0.1'

x = 1 × 10⁴.

31. Following statements are made about KDEL receptor

A. KDEL receptor is found in COPI vesciles

B. KDEL receptor affinity for KDEL sequence is high at higher pH

C. KDEL receptor is synthesized on the ER

D. KDEL receptor is present in the ER and Golgi

Identify the option that contains all correct statements about the KDEL RECEPTOR.

(a) A and B only

(b) A, B, C

(d) C and D only

Ans. (c)

Sol. Proteins synthesized on ribosomes of Endoplasmic Reticulum (ER) include, secretory protins, integral membrane proteins and soluble proteins that residue within compartmetns of endomembrane system such as ER, golgi complex, lysosomes, endosomes, vesicles and plant vacuoles. The KDEL receptor is a Golgi/intermediate compartment located integral membrane protein that carries out the retrieval of escaped ER proteins bearin a C-terminal KDEL sequence. KDEL receptor shuttles between the cis Golgi and the ER compartments. Resident proteins of the ER contain amino acid sequences that lead to their retrieval from the Golgi complex if they are accidentally incorporated into a Golgi bound transport vesicle. Soluble ER proteins bear the retrieval signal KDEL.

Retrieval is accomplished as soluble ER proteins bind to KDEL receptor residing in the membranous wall of cis Golgi compartmetns. The KDEL receptors, in turn, bind to proteins of the COPI coat, which allows the entire complex to be recycled back to the ER. Recognition of ER proteins by the KDEL receptor is pH dependent, which binding occurring under acidic conditions in the Golgi and release under conditions of higher pH in the ER.

32. Following statements were made about the characteristics of cyclin proteins:

A. Synthesis of M-cyclin is dependent on the cyclin mRNA that is newly transcribed after every cycle

B. Destruction of M-cyclin toward the end of mitosis is driven by ubiquitin independent proteolytic system

C. G1 cyclins can be activated by mitogenic factors

D. Retinoblastoma (Rb) is a key target of the activated cyclin D-Cdk 4/6 complex

E. While cyclin A1 expression is ubiquitous, cyclin A2 expression is restricted to the germ cell lineages.

Which one of the following options contains a combinations of all correct statements?

(a) A, B, D

(b) B, C, E

(d) A, C, D

Ans. (d)

Sol. The G1 cylcins are composed of the D-type cyclins that include cyclins D1, D2 and D3. Along with their partners, CDK4 and CDK6, G1 cyclins act early in the G1 phase of the cell cycle. The levels of G1 cyclin are low in G0 phase and increase progressively upon addition of growth factors or mitogens to the cells. The mechanisms by which mitogens or growth factors activate cyclin D1 are complex and occur at both transcriptional and post-transcriptional levels. Thus statement (c) is correct.

In most cell types, however, M-cyclins synthesis increases during G2 and M, owing primarily to an increase in M-cyclin gene transcription. This increase in M-cyclin protein leads to a gradual accumulation of M-Cdk (the complex of Cdk1 and M-cyclin) as the cell approaches mitosis. Stimuation of G0 cells with nitrogens induces expression of CDK4, CDK6, D-type cyclins and the E2F transcription factors, all encoded by delayed response genes. Rb protein initially inhibits E2F activity. When signaling from mitogens is sustained, the resulting cyclin D-CDK4/6 complexes begin phosphorylating Rb, releasing some E2F, which stimulates transcription of the genes encoding cyclin E, CDK2 and E2F (autostimulation). The cyclin E-CDK4 complexes further phosphorylate Rb, resulting in positive feedbck loops. Thus statement (d) is correct.

33. A protein found in the golgi compartment was found to be degraded rapidly at mitosis. Golgi extrats (isolated from G₁ and M phase cells treated with deubiqutination inhibitors) were treated with glycosidase and western blots done using antibodies against the protein. The following pattern was obtained.

Identify the statement the CANNOT be made bassed on the data

(a) The protein is glycosylated in G₁ and M-phase

(b) The protein is glycosylated and ubiquitinated in M-phase

(d) The protein is not ubiquitinated in G₁

Ans. (c)

Sol. The protein is glycosylated in M phase otherwise we wouldn't have ssen the main band wouldn't have appeared at the same place as that in G₁ phase.

34. Which one of the following RNAs possesses the peptidyltransferase activity?

(a) tRNA

(b) 5S rRNA

(d) 23S rRNA

Ans. (d)

Sol. Peptidyl ransferae activity is carried out by the ribosome. Peptidyl transferase activity is not mediated by an ribosomal proteins but by ribosomal RNA (rRNA), a ribozyme. Ribozymes are the only enzymes which are not made up of proteins, but ribonucleatides. All other enzymes are made up of proteins. This RNA relic is the most significant piece of evidence supporting the RNA World hypothesis.

In Prokaryotes, the 50S (23S component) ribosome subunit contains the peptidyl transferase component and acts as a ribozyme. The peptidyl transferase center on the 50S subunit lies at the lower tips (acceptor ends) of the A and O-site tRNAs.

In Eukaryotes, the 60S (28S component) ribosome subunit contains the peptidyl transferase component and acts as the ribozyme.

35. UV-induced DNA damage causes advancing replication forks to stall. To avoid a collapse of these stalled replication forks the cell uses:

(a) non-homologous end joining

(b) lesion bypass

(d) base excision repair

Ans. (b)

Sol. Nucleotide excision repair and translesion DNA synthesis are two processes that operate at arrested replication forks to reduce the frequency recombination and promote cell survival following UV induced DNA damage. In the abence of repair or when the repair capacity of the cell has been exceeded, translesion synthesis (lesion bypass) by polymerase V (Pol V) allows DNA synthesis to resume and is required to protect the arrested replication fork from degradation.

36. Which one of the snRNAs given below base pairs with 5' splice site of pre mRNA?

(a) U₁

(b) U₂

(d) U₅

Ans. (a)

Sol. U1 snRNP initially recognizes the 5'ss by base pairing between U₁ snRNA and the 5'ss exon-nitron junction (at positions –3 to +6); these are highly complementary sequences. Non-snRNP splicing factors interact with the 3'ss, resulting in the 5'ss being brought to the proximity of the 3'ss. The U₁/5'ss base pairing is then weakened in an ATP-dependent step, allowing U₂ snRNP to base pair with the branch site.

37. Which one ofthe following statements about human LINEs (long interspersed nuclear elements) is FALSE?

(a) LINEs are located primarily in euchromatic regions

(b) LINEs cannot encode all the products needed for their retrotransposition, and are dependent on SINEs for some components.

(d) Active LINE-1 elements possess an internal promoter located within the 5' untranslated region.

Ans. (b)

Sol. Retrotransposons (also called Class I transposable elements or transposons via RNA intermediates) are a type of genetic component that copy and paste themselves into different genomic locations (transposon) by converting RNA back into DNA through the process reverse transcription using an RNA transposition intermediate. LINE-1 (L1) retrotransposons make up a significant portion of the human genome, with an estimated 500,000 copies per genome. Genes encoding for human LINE1 usually have their transcription inhibited by methyl groups binding to its DNA carried out by PIWI proteins and enzymes DNA methyltransferases. L1 retrotransposition can disrupt the nature of genes transcribed by passing themselves inside or near genes which could in turn lead to human disease. LINE1s can only retrotranspose in some cases to form different chromosome structures contributing to differences in genetics between individuals.

38. Which one of the following statements about Gal gene expression is FALSE?

(a) Gal14 is a positive regulator of Gal genes

(b) The UAS region that regulates the Gal gene expression harbons short, phased AT repeats every 10 base pairs

(d) Gal80 is negatively regulated by Gal13

Ans. (c)

Sol. If a mutant Gal80 results in constitutive Gal1 expression, the simplest model is that Gal80 negatively regulates the Gal genes. Since Gal4 positively regulates and Gal80 negatively regulated Gal1 expression, we have to figure out how these two gene products work together to achieve such regulation. Assuming that Gal4 and Gal80 act in series there are two formal possibilities :

Model 1 is that Gal4 positively regulates Gal1 and that Gal80 negatively regulates Gal4; the presence of galactose somehow inhibits Gal80 function thus releasing Gal4 to positively activate Gal1 expression.

Model 2 is that Gal80 negatively regulates Gal1 and Gal4 negatively regulates Gal80; here the presence of galactose positively activates Gal4 which in turn negatively regulates Gal80, thus relieving inhibition of Gal1 expression.

39. Following statements were made about regulation of eukaryotic gene expression.

A. It is usually regulated at the level of initation of transcription by altering the chromating architecture

B. The eukaryotic genome is divided into domains by insulator elements

C. A chromatin remodeling complex binds to the promoter of a gene in sequence specific manner

D. Architecutural proteins regulate gene expression by promoting DNA bending

E. The chromatin remodeling complex can alter nucleosomal architecture, but cannot displace them.

The option with all the correct statements is

(a) A, B, D

(b) A, C, E

(d) B, C, D

Ans. (a)

Sol. Remodelling complex binds to chromatin to the acetylated regions not sequence and it alters the nucleosome which is visible. DNA remains in a compact form, which renders the genes to the be inactive, to make the gene available to the TF those regions need to be released from compaction. Methylation and deacetylation caues compaction and acetylation causes compaction and acetylation causes decondensation, before the initiation causes decondensation, before the initiation of transcription, certain amino acid residues of the histone gets acetylated and the remodelling complex, recognizing these spots removes the histone thereby decondense the chromosome at that place, now transcription can be initiated. Insulators ahve 2 functions they are present inbetween heterochromatin and euchromatin junction point thereby preventing the extension of heterochromatin and they are placed between 2 genes when they are regulated by the same enhancer to regulate the activation of the genes.

40. Following statements were made about catalytic introns:

A. Group I introns may undergo self-splicing by transesterification

B. Group II introns do not require any factor/protein for autosplicing either in vivo or in vitro

C. Certain introns of both the group I and II classes may contain open reading frames which are translated into protein

D. Generally, group I introns migrate by DNA-mediated mechanisms, wherease group II introns migrate by RNA-mediated mechanisms

E. Ribonuclease P (RNase P) is essential for bacteria but not eukaryotes

Which one of the following combinations represents statements which are all correct?

(a) A, B, D

(b) B, D, E

(d) C, D, E

Ans. (c)

Sol. Group I and group II self-splicing introns. Intron itself folds into a specific conformation within the precursor RNA and catalyzes the chemistry of its own release.

Some group II introns self-splice in vitro, but the reaction is generally slow (k_obs = 0.2-1.0 × 10^–2/min) and requires nonphysiological conditions e.g., high concentrations of monovalent salt and/or Mg++ (reflecting that proteins are needed to help fold group II intron RNAs into the catalytically actinve structure for efficient splicing (indicating statement 2nd to be incorrect). Group I introns encode sequence specific double strand (ds) endoDNases, which recognize the cleave intronless genes to insert a copy of the intron by a ds-break repair mechanism.

Group II introns are mobile genetic elements that are found in bacterial and organellar genomes and are thought to be ancestors of spliceosomal introns and retrotransposons in eukaryotes.

41. Following statements were made about human mitochondrial genome.

A. The replication of both the H and L strands is unidirectional and begins at specific origins

B. Majority of the mitochondrial genome encode for protein products

C. Though the mitochondrial genome is extremely compact, the genes never show any sequences overlap

D. The CR/D-loop region of mitochondrial genome exhibits triple stranded structure

E. Transription of mtDNA starts bi-directionally from a common promoter region in the CR/D-loop region and continues round the circle

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

(a) A, B, D

(b) A, D, E

(d) B, C, D

Ans. (b)

Sol. Mammalian mtDNA molecules replicated unidirectionally from two spatially and temporally distinct, strand specific origins [Thus statement (a) is correct]. A round of replication begins at OH with the synthesis of a daugher H-strand and continues along the parental L-strand to produce a full H-strand circle. Only after the replication fork has passed the scond replication fork has passed the second replication origin. OL, is synthesis of the L-strand initiated which proceeds in a direction opposite to that of H-strand replication. Both major transcription promoters in human mitochondria can function bidirectionally in vitro as well as in vivo [Thus statement (e) is corret]. Mammalian mtDNA displays exceptional economy of organisation. The genes lack introns and except for one regulatory region, intergenetic sequences are absent or limited to a few bases. Both rRNA and tRNA molecules are unusually small. Some of the protein genes are overlapping and in many cases part of the termination codons are not encoded but are generated post-transcriptionally by polyadenylation of the mRNAs. In vertebrate cells that are metabolically active, a large proportion of the mtDNA duplexes contain a short tree-stranded structure, called the displacement loop or D-Loop, in which a short nucleic acid strand complementary to the L-strand, displaces the H-strand. The D-loop region is bounded by the genes for tRNAPhe and tRNAPro and has evolved as the major control site for mtDNA expression, containing the leading strand origin of replication and the major promoters for transcription [Thus statement (d) is correct].

42. The following figure represent the growth curve of wild type E. Coli grown in a medium containing both glucose and lactose.

In E. Coli, the cataboplite repression of lactose operon by glucose has been explained by the different levels of cAMP in presence and absence of glucose. This model was challenged and experiments showed that catabolite repression occured because the activity of lactose permease was inhibited in the presence of glucose. Considering the second model, which one of the following plots correctly represents the growth pattern of a mutant E. Coli with a loss of function mutation in the lacl gene growing in a medium containing glucose and lactose?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Lacl-mutants are constitutive, meaning that they always express -galactosidase at high levels regardless of whether ther is an inducer present or not so as long as lactose is present in the cell it will be broken down but since glucose is also present. Lac Y is not able to uptake any glucose, so when lactose gets exusted bacterial growth stops.

43. Which one of the following statments about eukaryotic RNA processing is INCORRECT?

(a) Termination of transcription occurs co-transcriptionally at the polyadenylation site

(b) Phosphorylation of the Ser 2 of the RNA polymerase II CTD is required for the recruitment of polyadenylation factors.

(d) Xrn 2 is the nuclease that degrades the cleaved RNA to release RNA polymerase II from the template

Ans. (a)

Sol. Contranscriptional degradation of nascent RNA has not been directly demonstrated, however, two exonucleases, Rat1 and Xrn1, both contribute to contranscriptional degradation of nascent RNA, but this degradation is not sufficient to cause polymerase release.

44. RNA silencing is an important strategy to control viral infection in plants. The following statements were made regarding RNA silencing.

A. It is given by small interfering RNA (SiRNA) derived from double-stranded form of viral RNA.

B. siRNA reuires RISC for it function

C. Many plant viruses encode proteins that act as suppressors of RNA silencing

D. Viral P19 protein activates RNA induced silencing complex

Which one of the following combination of statements is correct?

(a) A, B and C

(b) B, C and D

(d) A, B and D

Ans. (a)

Sol. P19 is a viral suppressor of RNA silencing, which functions by sequenstering viral-derived sRNAs and preventing their incorporation into the RISC complex. P19 also functions to suppress the pathway by preventing the accumulation of AGO1, a component of anti-viral RISC, by inducing the accumulation of a host miRNA, miR168, which then downregulates AGO1 expression. Thus option D is incorrect.

45. The following statements are made with reference to the fact that tRNAs are known to possess T in their sequence.

A. RNA polymerase III utilizes TTP as one of the substrates

B. Like any other RNA polymerase, RNA polymerase III also reaches a looped structure it binds to S-adenosylmethionine to methylate C-5 position of the incorporated U to result in a thymine in a co-transcriptional manner.

C. During transcription of tRNA genes at the designated positions, DNA polymerase replaces RNA polymerase III to incorporate T in the tRNA trnascript.

D. A specific methyltransferase utilizes a methyl group donor to post-transcriptionally modify the specific U residues into T residues

E. Uracil to thymine conversion occurs in a large number of tRNAs in the TVC loop.The option with all the correct statements is

(a) A and B only

(b) B and C only

(d) D and E only

Ans. (d)

Sol. tRNA undergoes post transcriptional processing to get converted into mature tRNA for final function. The processing includes the base modifications at several places in tRNA sequence. The clover leaf secondary structure of tRNA contains a T arm which is a 4 to 5 bp stem containing the sequence TC, where is pseudouridine. S-adenosyl-l-methionine dependent methyltransferase TrmA catalyzes the site-specific formation of 5-methyluridin (m5U) in the T-P-loop (position 54) or tRNA.

46. Contents in Column I and II with respect to bacterial transcriptional regulation

Which one of the options below correctly matches contents in column I and Column II?

(a) (i)-B, (ii)-D, (iii)-A, (iv)-C

(b) (i)-C, (ii)-B, (iii)-A, (iv)-D

(d) (i)-D, (ii)-C, (iii)-B, (iv)-A

Ans. (a)

Sol. In absence of glucose cAMP helps polymerases to bind to the promoter of Lac and initiate transcription of the structurla genes. Proper transcription initiation takes place only when there is abortive initiation. Abortive initiation means small pieces of RNA less than 10 nucleotided long is produced which are later degraded. In Ara operon, when arabinose concentration is low and glucose levels are high, AraC protein binds to both Aral and AraO₂ to bring them together in a DNA loop. The operon is repressed in this state. FMN synthesis is regulated by FMN riboswitch.

47. Which one of the following plant proteins is targeted by HC-toxin produced by the maize fungal pathogen Cochliobolus carbonum?

(a) H⁺-ATPase

(b) Histone deacetylase

(d) MTA nucleosidase

Ans. (b)

Sol. HC-toxin is an inhibitor of higstone deacetylases (HDACs) in many organisms, including plants, insects and mammals.

48. Which one of the following is NOT an extracellular matrix protein?

(a) Keratin

(b) Laminin

(d) Vitronectin

Ans. (a)

Sol. Tissue are nto just tightly packed with cells; most of the volume contains extracellular space and is filled with complex meshwork of proteins called the extracellular matix (ECM). The components of ECM in most tissues are secreted by fibroblasts and are categorized into proteoglycans and fibrous proteins (Collagen, elastin, fibronectin and laminin). These components give structural support and facilitate cellular communication.

49. If you require to generate anergy in T cells, which of the following conditions will you use?

(a) Stimulate naive T cells with antibodies that bind both T cell receptor and CD28

(b) Stimulate naive T cells with antibodies that bind only the T cell receptor

(d) Permit naive T cell interactions with activated macrophages in presence of IL-4.

Ans. (b)

Sol. TH cell recognition of an antigenic peptide-MHC complex sometimes results in a state of nonersponsiveness called clonal anergy, marked by the inability of cells to proliferate in response to a peptide MHC complex. T-cell anergy can also be induced in vitro by incubating T cells with normal. CDC80/86-expressing APCs in the presence of the Fab portion of anti-CD28, which blocks the interaction of CD28 with CD80/86.

50. Which one of the following signaling molecules is NOT a protein or a peptide?

(a) Transforming growth factor (TGF-)

(b) Erythropoietin

(d) Epinephrine

Ans. (d)

Sol. Here all the given options are proteins except epinephrine. Chemically, epinephrine is a catecholamine hormone, a sympathomimetic monoamnie derived from the amino acids phenylalanine and tyrosine. It is neither a protein nor a peptide.

51. The following statements are made with reference to characeristics of glycosaminoglycans and proteoglycans, which are major constituents of extracellular matrix. Which one of them is INCORRECT?

(a) Glycosaminoglycans are very long polysaccharide chains composed of repeating disaccharide units of an amino sugar and a uronic acid

(b) Except for hyaluronic acid, all glycosaminoglycans are covalently attached to protein as proteoglycans

(c) Glycoproteins contain less carbohydrate usually in the form of relatively short, branched oligosaccharide chains whereas proteoglycans contain more carbohydrate in the form of long unbranched glycosaminoglycan chains.

(d) Like O-linked and N-linked glycoproteins, in proteoglycan also glycosaminoglycans are linked to serine or threonine and asparagine residues.

Ans. (d)

Sol. The basic proteoglycan unit consists of a "core protein" with one or more covalently attached glycosaminoglycan (GAG) chain(s). The point of attachment is a serine (Ser) residue to which the glycosaminoglycan is joined through a tetrasaccharide bridge (e.g. chondroitin sulfate-Glc-A-Gal-Gal-Xyl-PROTEIN). The Ser residue is generally in the sequence -Ser-Gly-X-Gly- (where X can be any amino acid residue but proline) germ cells and symmetrical division of the microscope fails to generate germ cells.

52. Fibrinogen, Factor 1, is a soluble glycoprotein present abundantly in plasma. Following biochemical characteristics are given below about this glycoprotein :

A. It is a dimer of the three polypeptide chains

B. It is a dimer of two polypeptide chains

C. The chains are covalently linked by 29 inter-and intra-chain disulfide bonds

D. The chains are covalently linked by 19 inter-and intra-chain disulfide bonds

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

(a) A and C

(b) B and C

(d) B and D

Ans. (a)

Sol. Fibrinogen is a dimer with each half molecule composed of three diffferent chains (Alpha, Beta, Gamma). Fibrinogen contains 29 disulphide bonds.

53. Receptor tyrosine kinases (RTKS) formn a large and important class of cell surface receptors whose ligands are soluble or membrane bound protein or peptide hormones. Which of the following statement is INCORRECT?

(a) All RTKs are transmembrane proteins with extracellular ligand-binding site and a cytosolic domain

(b) Binding of ligand causes most RTKs to dimerize

(d) Adapter proteins and enzymes involved in signalling pathways bind to different phosphotyrosine residues via a conserved polypeptide domain called SH2 domain

Ans. (c)

Sol. All RTKs comprise an extracellular domain containing a ligand-binding site, a single hydrophobic transmembrane helix and a cytosolic domain that includes a region with protein tyrosine kinase activity. Binding of ligand causes most RTKs to dimerize; the protein kinase of each receptor (since in statement C it mentions cytosolic tyrosine kinase, it is wrong) monomer then phosphorylastes a distinct set of tyrosine residues in the cytosolic domain of its dimer partner, a process phosphotyrosines serve as docking sites for other proteins containing SH2 domain involved in RTK mediated signal transduction.

54. A transgenic mouse was generated for both heavy and light chain specific for MHC molecule H-2K^m so that all the B cells in this mouse therefore had BCRS specific only for H-2K^m and made only anti- H2K^M specific antibiodies. By appropriate brdeeding, the H-2K^m specific immunoglobuling transgene was introduced into mice bearing different MHC genotypes H-2Kⁿ and H-2K^m/n. With respect to selection of B-cells within the bone marrow of H-2Kⁿ and H-2K^m/n mice.

Which of the following statements is correct?

(a) In H-2Kⁿ mice, the transgenic cantibodies will not be detected on the surface of any of the B cells

(b) In H-2K^m/n mice, the transgenic antibodies will not be detected on the surface of any of the B cells

(c) In H-2Kⁿ mice, the transgenic antibodies will be deteched on the surface of some of the B-cells but not released in the serum.

(d) In H-2K^m/n mice, the transgenic antibodies will be detected on the surface of some of the B cells.

Ans. (d)

Sol. Mice with H-2K^m/n expresses both m and n MHC proteins so some B cells expressing n proteins can bind to the antibodies generated in H-2K^m, so statement D is correct.

55. Following diagram displays the MHC haplotype of a heterrozygous H-2k/d mouse.

How many types of MHC molecules will be expressed on activated macrophage and a normal fibroblast (NF)?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. 14 MHC molecules will be expressed on activated macrophages and 18 MHC molecules will be expressed on normal fibroblast cells.

56. Prostate cancer cells were treated with Drug A and Drug B in order to check the efficancy of the drugs in arresting the growth of cells. The following results were obtained

Which one of the following statements is not correct

(a) Drug A targets the Wnt signaling pathway but does not lead to death of cells

(b) Drug B targets the Wnt signaling pathway and leads to G₁ cell cycle arrest.

(d) Drug A kills cells via the mitochondrial-independent pathway

Ans. (a)

Sol. When the Wnt pathway is not active, the destruction complex of -catenin, formed by axin, APC, CK1 and GSK3, phosphorylates -catenin should be less, but since it is showing high, statement A is incorrect. And since it is seen high in case of drug B, hence statement B is correct.

The extrinsic apoptosis pathway is activated through the binding of a ligand to a death receptor, which in turn leads, with the help of the adapter proteins (fadd/tradd), to recruitment, dimerization and activation of caspase-8. Expression level of caspase-8 is high in drug A suggesting it is killing cells via extrinsic pathway.

The intrinsic or mitochondrial apoptosis pathway can be activated through various cellular stresses that lead to cytochrome c release from the mitochondria and the formation of the apoptosome, comprised of APAF1, cytochrome c, ATP and caspase-9 resulting in the activation of caspase-9. Expression level of caspase-9 is high in drug A suggesting it is killing cells via intrinsic pathway. Also, concentration of caspase-9 in drug A is less, hence it is mitochondria independent. Statement D is correct.

57. Plants perceive effector proteins from pathogen to mount a strong defense response. The following statements were made regarding signal transduction events upon perception of a pathogen.

A. Influx of Ca²⁺ and H⁺ ions into the cell

B. Influx of K⁺ and Cl^– ions into the cell

C. Efflux of Ca²⁺ and H⁺ ions out of the cell

D. Efflux of K⁺ and Cl^– ions out of the cell

Which one ofthe following combination of statements is correct?

(a) A and B only

(b) A and D only

(d) C and D only

Ans. (b)

Sol. Recognition of the elicitor by its receptor a 91-kDa plasma membrane protein, rapidly stimulates large, transient influxes of Ca²⁺ and H⁺ and effluxes of K⁺ and Cl^–. Pharmacological studies revealed that this pattern of ion fluxes and is particular extracellular Ca²⁺, is necessary for the production of reactive oxygen species (oxidative burst), defense-related gene activation and phytoalexin production.

58. Leishmanai major, the causative protozoan paratiste of cutaneous leishmaniasis, resides and multiplies within the phagosomes of macrophages. Resistance to the infection correlated very well with the development of a Th1 response. In order to prove this hypothesis, three groups of BALB/c mice were taken: control group, group. All the groups were infected with L. Major and then parasite load was measured in each group after 4 weeks, which one of the following result will justify the experimental outcome confirming the hypothesis?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. has been widely characterized in the modulation of the immune response against protozoan infections as mentioned resistance to infection correlates to development of Th1 response hence if not present then highest parasite load will be observed.

59. If an early cleavage stage wild type Drosophila embryo is injected with bicoid mRNA at the posterior pole then

(a) head structures develop at posterior pole, while the same is inhibited at anterior pole

(b) head structures form at both poles

(d) duplication of usual posterior structures occurs

Ans. (b)

Sol. The Bicoid (Bcd) protein was shown to be expressed in a concentration gradient with the highest Bcd concentration at the anterior end of the embryo and decreasing towards the posterior. Thus when injected at the posterior end will lead to development of Head portion.

60. If the blastomeres of a 4-called sea ruchin embryo are isolated, each blastomere can form a pluteus larvae. Thisis an example of

(a) autonomous specification

(b) conditional specification

(d) mosaic development

Ans. (b)

Sol. A second mode of commitment involves interactions with neighbouring cells. In this type of specification, each cell originally has the ability to become many different cell types. However, the interactions of the cell with other cells restricts the fate of one or both tof the participants. This mode of commitment is sometimes called conditional specification upon the conditions in which the cell finds itself.

Sea urchin blastomeres separate from each other by vigorous shaking (or later, bu placing them in calcium free seawater). To Driesch's surprise, each of the blastomeres from a 2-cell embryo developed into a complete larva. Similarly, when Driesch separated the blastomeres from 4 and 8 cell embryos, some of the isolated cells produced entire pluteus larvae.

61. Which one of the following plant homeotic genes does NOT encode MADS-domain transcription factor involved in floral organ specification?

(a) AP2

(b) AP1

(d) AG

Ans. (a)

Sol. The MADS domain homeotic proteins Apetala1 (AP1), APETALA3 (AP3), pistillata (PI) and AGAMOUS (AG) combinatorially specify the identity of Arabidopsis floral organs. Among all the options AP2 is not involved.

62. The development of anthers and male gametophytes is highly conserved among angiosperms. Following are some of the events associated with pollen development, in random order.

A. Microsporogenesis in pollen sac to produce a tetrad

B. Asymmetric division formaing immature pollen grain

C. Archespore division

D. Division of generative cell to form two sperm cells

E. Callase digestion to form free microsperes.

Which of the following option represent the correct series of events during pollen development?

(a) A, B, C, D, E

(b) C, A, E, B, D

(d) C, E, A, B, D

Ans. (b)

Sol. Anther develoment in Arabidopsis has been divided into 15 stages, which commence from division of a single archesporial cell; defined cell types and adaxial abaxial polarity are established, resulting in the formation of the mature microsporangia. Pollen development comprises three major stages : (i) microsporogenesis (differentiation of the sporogenosu cells and meiosis); (ii) post-meiotic development of microspores; and (iii) microspore mitosis. Sporogenous cells (pollen mother cells or meiocytes) are encased in a tapetum-derived callose wall and undergo meiosis to form tetrads of haploid microspores. The free microspores then go through one/two rounds of mitotic divisions and pollen wall formation continuous. The final mature pollen may be tricellular or as seen in some species it can be bicellular with the final mitotic division occurring the pollen tube. During pollen mitosis I, the microspore divides asymmetrically to produce a vegetative and a generative cell. Following the first mitotic division, a second mitosis produces two sperm cell enabling double fertilization to produce the embryo and endosperm. The asymmetric division is essential for the establishment of male germ cells and symmetrical division of the microsproe fails to generate germ cells.

63. The hedgehog pathway is extremely important in vertebrate limb development, netural differentiation and facial morpho-genesis. In accordance with the above statement, what would happen in mice that are homozygous for a mutant allele of sonic hedgehog (shh)?

(a) Limbs would form normally but the mice would have facial abnormalities

(b) midline of the face would be reduced and a single eye would form in the center of the forehead

(d) Mutations of shh may activate tumor formation if Patched protein can inhibit the Smoothened protein

Ans. (b)

Sol. Injection of noggin mRNA into 1-cell UV-irradiated Xenopus embryos completely rescued the dorsal axis and allowed the formation of a complete embryo. (Option B is correct).

Noggin could accomplish two of the major functions of the Spemann-Mangold organizer : it induced dorsal ectoderm to form neural tissue, and it dorsalized mesoderm cells that would otherwise contribute to the ventral mesoderm. Newly transcribed (as opposed to maternal) noggin mRNA is first localized in the dorsal blastopore lip region and then becomes expressed in the notochord (option C is not true). Noggin binds to BMP4 and BMP2 and inhibits their binding to receptors and hence prevents ventralization. (Option D is not true). Since lithium chloride treatment already dorsalizes the entire gastrula, noggin will not be able to ventralize the ventral part of the embryo and cannot rescue the embryo. (option A is not true).

64. In xenopus, the Noggin protein, accomplishes two major functions of the organizer: it induces dorsal ectoderm to form neural tissues, and it dorsalizes medoderm cells. Which one of the following observations is correct with respect to Noggin?

A. If a plasmid clone expressing Noggin protein is microinjected into a lithium chloride treated Xenopus gastrula, it should rescue the abnormalities induced by lithium chloride treatment

B. In the Antennapedia mutants, the Antennapeida gene is expressed in head region only and thus promotes leg-like strtucture in head socket

C. In addition to promoting thoracic structures, Antennapeida protein binds to and represses the enhancer of homothorax and some other genes responsible for antenna specification

D. Antennapedia has no role in thoracic region specification and thus recessive Antennapedia mutants show no phenotypic effect

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

(a) A only

(b) A and C

(d) B only

Ans. (b)

Sol. The Antennapedia (Antp) gene in necessary for the proper development of thoracic segments in both larval and adult stages of Drosophila.

Loss-of-function mutations in the regulatory region of this gene result in the development of the second leg pair into ectopic antennae (D is wrong).

Antp also functions in the imaginal discs (the developing adult cells) to repress head development and to promote thoracic development.

Consistent with this thorax-determining role, dominant mutations of Antp cause striking homeotic transformations of head structures into thoracic structures (typically, antennae are transformed into legs). Because the Antp gene is not needed for normal head development, the dominant transformation appears to result from the expression of the thorax-specifying functions of Antp in the wrong tissue (A is correct).

65. Maternal effects genes are extremely important in establishing the anterior-posterior polarity of the Drosophila embryo. Mutant phenotypes arise when genes of this family are mutated. The following table enlists genes and phenotypes observed on mutation of these genes but not correctly matched.

Which one of the following combinations is correctly matched?

(a) A-ii, B-i, C-ii, D-iv

(b) A-i, B-iii, C-iv, D-ii

(d) A-iii, B-iv, C-ii, D-i

Ans. (c)

Sol. Option A is wrong as mutation (in shh) would certainly interfere in limb development as well.

SHH (Sonic Hedgeohog Gene Regulation) is involved in the separation of the single eye field into two bacterial fields. Although not proven, it is thought that SHH emitted from the prechordal plate suppresses Pax6 which causes the eye filed to divide into two. If the SHH gene is mutated, the result is cyclopia, a single eye in the centre of the face. (Hence option B is correct and option C is wrong).

The SHH ligand inactivates the Patched receptor allowing Smoothened (SMO) to become active. The SMO activate the GLI proteins, a family of trancription factors that turn on the expression of different target genes, giving raise to cell proliferation and tumorigenesis. Patched protein's inhibition on SMO must be removed in order to activate tumor formation (Option D is wrong).

66. In C. elegans the SKN-1 protein controls the fate of the EMS blastomere which generates the posterior pharynx. With reference ot the above which one of the following statements is INCORRECT?

(a) The MS blastomere is able to generate phyrangeal tissue even in isolation

(b) Embroys from skn-1 (skin excess) deficient mothers lack both pharynegeal mesoderm and endoderm derivatives of EMS

(d) SKN-1 activates MED transcription factors whose level of activity controls the fate of EMS lineage

Ans. (c)

Sol. By the 4-cell stage of C. elegans embryogenesis, a ventral blastomere, called EMS is already committted to producing pharyngeal and intestinal cell types. Maternally expressed SKN-1 specifies the fate of a single cell, the EMS blastomere (A is correct). Recessive maternal effect mutation in the gene SKN-1 prevent intestinal cell (B is correct). SKN mutants undergo embryonic lethality and must be maintained with genetic balancers that require careful monitoring and labor intensive strategies to obtain large populations. (Thus C is incorrect).

67. Antennapedia protein expresses in the thoracic segment of the fly and not in the head region. However, a dominant mutation of Antennapedia replaces antenna with leg-like structure. The following statements were made with reference to Antennapedia :

P. In a dominant Antennapedia mutant, the Antennapedia gene is expressed in the head as well as in thorax.

Q. In the Antennapedia mutants, the Antennapedia gene is expressed in head region only and thus promotes leg-like structure in head socket.

R. In addition to promoting thoracic structures, Antennapedia protein binds to and represses the enhancer of homothorax and some other genes responsible for antenna specification.

S. Antennapedia has no role in thoracic region specification and thus recessive Antennapedia mutants show no phenotypic effect.

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

(a) P only

(b) P and R

(d) Q only

Ans. (b)

Sol. The homeotic selector genes in the Antennapedia complex (contains the homeotic genes labial, Antennapedia, sex combs reduced, deformed and proboscipedia) control the fate of the anterior development of the fruit fly (head and anterior thorax). The labial and deformed genes specify the head segments, while sex combs reduced and Antennapedia contributes to giving the thoracic segments their identities. The proboscipedia gene appears to act only in adults, but in its absence, the labial palps of the mouth are transformed into legs.

68. Which of the following is a plant secondary metabolite?

(a) Kaurenoic acid

(b) Abietic acid

(d) Pyruvate

Ans. (b)

Sol. Secondary metabolites (SM) are compounds that are not necessary for a cell (organism) to live, but play a role in the interaction of the cell (organism) with the environment. These compounds are often involved in plants protection against biotic or abiotic stresses. Secondary metabolites are from different metabolites families that can be highly inducible in response to stresses. Primary metabolites perform essential metabolic roles by participating in nutrition and reproduction. Abietic acid and dehydroabietic acid are the main components of pine resin obtain from Pinus sp.

69. Which one of the following represnets the largest outflux of nitrogen from the atmospheric reservoir?

(a) Biological nitrogen fixation

(b) Nitrogen fixation due to lightning

(d) Fixation by Haber's process for fertilizer production

Ans. (a)

Sol. A relatively small amount of ammonia is produced by lightning. Some ammonia also is produced industrially by the Haber-Bosch process, using an iron based catalyst, very high pressures and fairly high temperature. But the major conversion of N₂ into ammonia and thence into proteins, is achieved by microorganisms in the process called nitrogen fixation (or dinitrogen fixation).

70. Which one of the following metabolites formed during Calvin-Benson cycle in chloroplast is involved in starch biosynthesis and can also be transported to cytosol?

(a) Triose phosphate

(b) Glyceraldehyde-3 phosphate

(d) Ribulose 1,5-bisphosphate

Ans. (a)

Sol. After reduction of phosphoglyceric acid to triose phosphate, fixed carbon can follow several cources. First, triose phosphate may move to the cytoplasm in exchange for phosphate by means of a translocator located in the chloroplast envelope membranes. In the cytoplasm triose phosphate is converted into sucrose, much of which is exported from the cell. Within the chloroplast the triose phosphate also may enter a complex sequence of interconversions by which RuBP, consumed as CO₂ as fixed by Rubisco, is regenerated. Finally, within the cycle fructose-6-phosphate may be diverted to starch syntehsis in the chloroplast.

71. Which one of the following amino acids is characteristic of and is conserved in response regulator of plant two-component systems?

(a) Aspartic acid

(b) Glutamic acid

(d) Histidine

Ans. (a)

Sol. In the field of molecular biology, a two component regulatory system serves as a basic stimulus response coupling mechanism to allow organisms to sense and respond to changes in many different environmental conditions. Two component systems typically consist of a membrane bound histidine kinase that senses a specific environmental stimulus and a corresponding response regulator that mediates the cellular response, mostly through differential expression of target genes. Aspartic acid characteristic of and is conserved in response regulator of plant two component systems.

72. Which one of the following essential plant mineral nutrients complexes with mannitol, mannan, polymannuronic acid, and other constituents of cell wall?

(a) Silicon

(b) Chlorine

(d) Boron

Ans. (d)

Sol. Boron complexes with mannitol, mannan, polymannuronic acid and other constituents of cell walls. Involved in cell elongation and nucleci acid metabolism. Silicon is deposited as amorphous silica in cell walls. It contributes to rigidity and elasticity. Chlorine is required for the photosynthetic reactions involved in Oxygen evolution. Mn is required for activitiy of some dehydrogenses, decarboxylases kinases, oxidases and peroxidases. Involved with other cation-activated enzymes and photosynthetic O₂ evolution.

73. The gene encoding for mevalonate-3-kinase is disrupted in a certain plant. Which one of the following statements concering the above plant is correct?

(a) Mevalonic acid 3, 5-biphosphate will be synthesized, while there will no synthesis of Mevalonic acid 5-phosphate

(b) Mevalonic acid 3, 5-biphosphate and isopentenyl diphosphate will be synthesized

(d) Acetoacetyl-coA and Mevalonic acid will not be synthesized.

Ans. (c)

Sol. Mevalonate is a key intermediate, and mevalonate kinase a key early enzyme, in isoprenoid and sterol synthesis. As the second enzyme in the Mevalonate pathway, it catalyzes the phosphorylation of Mevalonic acid to produce Mevalonate-5-phosphate. A 5-10% reduction in mevalonate kinase activity is associated with the mevalonate kinase deficiency (MVD) resulting in accumulation of intermediate mevalonic acid.

74. Following are certain statements regarding C₂, C₃, C₄ and CAM carbon metabolism in plants.

A. In C₃ cycle, one molecule of 3-phosphoglycerate formed during carboxylation phase is utilized for the biosynthesis is sugars, fatty acid and amino acids

B. During C₂ cycle glycine is transported from peroxisome to mitochondria and glycerate is transported from peroxisome to chloroplast

C. The concentration of CO₂ in bundle sheath of C₄ plants is several fold lower than the external atmosphere

D. The stomata is CAM plants open at night

Which one of the following combination of statements is correct?

(a) A and B

(b) B and C

(d) B and D

Ans. (d)

Sol. During C₂ cycle, glycine is transported from peroxisome to mitochondria and glycerate is transported from peroxisome to chloroplast. For CAM, at night, when the air is cooler and moister, the stomata open to allow entry of CO₂, which is then fixed into oxaloacetate by PEP carboxylase. The oxaloacetate is reduced to maltate and stored in the vacuoles, to protect cytosolic and plastid enzymes from the low pH produced by malic acid dissociation.

75. Followings are certain statements regarding phytochrome-mediated signal transduction in Arabidopsis:

A. phyA nd phyD are photo-stable, while phy, phyc and phyE are photo-labile

B. Prolonged light exposure and the conversion of PRA to PERA cause phyA concentration to drop.

C. Exposure of light and concersion of PR to PFR cause movement of phytochromes from the cytosol into the nucleus.

D. phyB lacks nuclear localization sequence and depends on transporter protein for nuclear import

Which one of the combinations of above statements is correct?

(a) A and B

(b) A and C

(d) B and C

Ans. (b)

Sol. phyA is an exceptional phytochrome among different types of phytochrome available. It is heat liable and unstable in the presence of light that means the active form of phyA is Pr (phyB-E are heat stable).

Phytochromes are synthesized in the cytosol as the Pr form are converted into Pfr form upon absorbing R light. This photoactivated Pfr translocates from the cytosol to the nucelus where regulation of light responsive genes occurs through several TFs. Prolonged light exposure and the conversion of PrA to PfrA cause phyA concentration to drop approximately 100-fold as PfrA is rapidly degraded.

76. Plant growth hormone receptors, namely TIR1, GID1, 2011 and PYL1 were knocked out independently in Arabidopsis plant. The resultant plants are named as tir1, gid1, coit and pyl1. Which one of the following statements regarding the above Arabidopsis plants is correct?

(a) tiris defective in ABA signaling while coil in gibberlline signalling

(b) gidi is defective in auxin signaling while pyll in JA signalling

(d) tir7 is defective in gibberellin signaling while gidi in auxin signaling

Ans. (c)

Sol. The auxin receptor TIR1 and the JA receptor COI1 (option c) form SCFTIR1 and SCFCOI1 complexes with CUL1. The GA receptor GID1 and SL receptor D14 facilitate the recruitment of their target proteins DELLA and D53 to the respective SCF complexes.

The core regulatory components of the ABA signaling pathway are the PYR/PYL1 receptor (option 3), the negative regulator PP2C and the positive regulator SnRK2.

77. Following is the list of cellular response of root hair to bacterial Nod factor and timing of their induction, which may or may not be matched

Which one of the following is the correct match?

(a) A-i, B-ii, C-iii

(b) A-ii, B-i, C-iii

(d) A-i, B-iii, C-ii

Ans. (a)

Sol. Electrophysiological and cell biologist approaches as well as pharmacological studies with specific blockers have provided insight into the signal transduction cascades that are activated in the root epidermis by Nod factors. Oneo f the first responses is plasma membrane depolarization, which is observed –1 min after Nod factor addition. This depolarization is caused by a very rapid calcium influx, induced within seconds of Nod factor addition, followed by a chloride efflux. Upon depolarization of the membrane, an efllux of potassium ions is induced, by which the membrane can repolarize. Several minutes after Nod after application, calcium oscillations occur in the perinuclear region of the root hairs and propagate radially through the cell.

78. Which one is not the function of P-cells in the collecting ducts?

(a) Na⁺ reabsroption

(b) K⁺ secretion

(d) H⁺ secretion

Ans. (d)

Sol. The principal cells are responsible for sodium reabsorption via the amiloride-sensitive sodium channel ENaC. Principal cells also secrete potassium.

79. Which one of the following is NOT produced by -amylase digestion of ingested amylopectins?

(a) Glucose

(b) Maltose

(d) -limit dextrins

Ans. (a)

Sol. Pancreatic -amylase hydrolyses amylose into maltose and maltotriose and amylopectin into maltose, maltotriose and -limit dextrins and these oligosaccharides are further degraded to glucose in unstirred water layer by maltase and sucrase-isomaltase at the surface of the micro-villus.

80. Which one of the following causes constriction of blood vessels?

(a) Carbon monoxide

(b) Nitric oxide

(d) Atrial natriuretic peptide

Ans. (c)

Sol. Vasoconstriction is the narrowing (constriction) of blood veseels by small muscles in their walls.

Carbon monoxide (CO) is an autocrine and paracrine vasodiltor in cerebral and systemic circuulations. Nitric oxide also dilates blood vessles, raising blood supply and lowering blood pressure. It is a mediator of vasodilation in blood vessels.

81. The action potential recorded from pace maker tissue (SA/AV node) , of mammalian heart is shown in the following diagram:

On the basis of mechanism of generation of action potential in pace maker tissue (SA/AV node), the following statement were proposed from the above figure

A. 'funny' channels are activated at 'b'

B. Outward flow of K⁺ occurs at 'a'

C. T-Ca⁺⁺ channels are closed at 'c'

D. Fast Na⁺ channels are opened at 'd'

E. The upward phast at 'd' is largely due to inward movement of Na⁺

Which one of the following combinations represent correct statements?

(a) A and B

(b) B and C

(d) D and E

Ans. (a)

Sol. At the end of repolarization, when the membrane potential is very negative (about –60mV), ion channels open that conduct slow, inward (depolarizing) Na⁺ currents. These currents are called "funny" currents. Repolarization occurs (Phase 3) as K⁺ channels open (increased gK⁺) thereby increasing the outward directed, hyperpolarizing K⁺ currents. At the same time, the L-type Ca⁺⁺ channels become inactivated and close, which decreases gCa⁺⁺ and the inward depolarizing Ca⁺⁺ currents.

82. Testosterone is a steroid hormone about which a set of statements are given below :

A. It is secreated by Leydig cells of testes

B. It is secreated by Sertoli cells of testes

C. It inhibits secretion of luteinizing hormone from anterior pituitary

D. It does not inhibit secretion of follicle-stimulating hormone from anterior pituitary

Which one of the following combinations represents correct statements about testosterone?

(a) A and C

(b) B and C

(d) A and D

Ans. (a)

Sol. Testosterone is secreted by cells that lie between the seminiferous tubules, known as the Leydig cells. LH release is stimulated by gonadotropin-releasing hormone (GnRH) and inhibited by estrogen in females and testosteron in males.

83. Insulin is a heterodimer made up of A and B peptide chains joined by the intra and interchain disulfide bridges formed between amino acid of respetive chains as suggested below :

A. A₆ A₁₁

B. A₇ B₇

C. A₂₀ B₁₉

D. A₅ B₁₅

Which one of the following represents the correct disulfide brideges joining A and B chains of insulin hormone?

(a) A, B and D

(b) A, C and D

(d) B, C and D

Ans. (c)

Sol. Two peptides comprising the insulin hormone include : a. 21 amino acid peptide, corresponding to residues 90-110 of the INS gene product, frequently referred to as the A-chain; and b. 30 amino acid peptide, corresponding to residues 25-54 of the INS gene product, referred to as the B-chain. The chains are stabilized by three disulphide bridges (two inter-chain and one intra-chain), which are present at the following location A₆-A₁₁, A₇-B₇ and A₂₀-B₁₉.

84. What kind of inheritance is indicative in the pedigree chart shown below?

(a) Y-linked

(b) X-linked dominant

(d) Autosomal dominant

Ans. (d)

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

85. Which of the following statements about R-bandin of metaphase chromosomes is FALSE?

(a) The heat treatment preferentially denatures the GC rich DNA to produce R-banding pattern.

(b) This is essentially the reverse of the G-banding pattern

(d) The DNA of R-bands generally replicate early during the S-phase of Cell cycle.

Ans. (a)

Sol. Except option 1, rest all options are correct about R-banding. R name is given to this banding pattern as it is reverse of G-banding pattern. The R banding donot shows any bands stained by Quinacrine, hence it is Q negative. The most common R-banding method involves heat denaturing chromosomes in hot acidic saline followed by Giemsa staining. This method is thought to preferentially denature AT-rich DNA and to stain the under-denatured GC-rich regions.

86. Cytoplasmic male sterility (CMS) in plants is caused by mutation inthe mitochondrial genome. CMS can be restored by a nuclear gene, restorer of fertility (Rf), which is a dominant character. If a male sterile pea plant is pollinated by a fertile male pea plant with Rf in heterzygous condition, the progeny obtained will have

(a) all male sterile progeny

(b) all fertile progeny

(d) 75% of the progeny fertile and 25% male sterile

Ans. (c)

Sol. The highest forest area (41.35%) is found between 1000 and 2000 m followed by forest area between 500 m and 1000 m is 16.57% while the lowest forest area (0.08%) is found under >4000 m. Forest area <500 m is 11.56% whereas 7.14% forest is found between 3000 and 4000 m.

The Uttarakhand Himalaya contains rich forest diversity. Forests are distributed along the altitudinal gradients from broad leaf deciduous forests to pine, mixed-oak, coniferous forests and alpine meadows.

87. Mutation is essential for genetic variation. Which one of the following events can lead to variation amongst the gametes produced by males of Drosophila melanogaster?

[Crossing over does not occur in D. melanogaster males]

(a) Segregation

(b) Imprinting

(d) Independent Assortment

Ans. (d)

Sol. Correct answer should be option 2. Imprinting in drosophila involve epigenetic effects on heterochromatin that results from transmission through males or females and produce parent-of-origin effects on the expression of visible markers in or near the heterochromatic regions. The link between imprinting and heterochromatin in Drosophila is supported by the observation that mutations in Su(var) 3-9, a histone H3 methyl transferase and hterochromatin protin 1 (HP1), a major component of hterochromatin, modify both imprinting and heterochromatin formation.

88. Which one of the following is a correct statement about the difference between a F⁺ cell and Hfr cell?

(a) In F⁺ cell F factor is an intergral part of bacterial chromosome, while in Hfr cell F fractor is present as an episome

(b) F⁺ and Hfr are synonymous

(d) In F⁺ cell F factor is in the form of an episome while in Hfr cell, F factor is ingegrated into the bacterial chromosome.

Ans. (d)

Sol. A high-frequency recombination cell (Hfr cell) (also called an Hfr strain) is a bacterium with a conjugative plasmid (for example, the F-factor) integrated into its chromosomal DNA. The integration of the plasmid into the cell's chromosome is through homologous recombination. If F factor plasmid is inserted into host chromosome (Hfr cell), this will result in the transfer of the entire DNA complex.

When an F factor becomes integrated into the chromosome of an F⁺ cell, it makes the cell a high frequency of recombination (Hfr) cell.

89. Gene 'A' in mouse is needed for its normal growth [normal size]. Mouse strains with deletion in gene 'A' (A^del) have been developed.

When the following cross is made :

All progeny (both males and females) are of normal size.

When the F1, progeny with the following genotypes are crossed :

50% of the pogeny obtained are small in size. The above observation can be explained by

(a) Maternal inheritance

(b) Mathernal imprinting

(d) Maternal effect

Ans. (b)

Sol. In the 1st cross all the progeny are normal because they inherited the normal copy of the gene from the father. But in the second cross, the father had one deleted and one normal copy of the gene. Those who inherited the normal gene were of normal size. This shows that the gene is maternally imprinted.

90. Following statements were made about sex determination in Drosophila melangoaster

A. It is achieved by a balance of female determinants on the X-chromosome and male determinants on the autosomes.

B. A Drosophila with 0.66 value of X : A ratio would develop interesex type

C. Due to noninvolvement of Y chromosome in sex determination process, XO Drosophila develop as normal fertile male

D. The high value of X : A ratio facilitates activation of feminizing switch gene Sex-lethal (sxl).

E. Sex specific expression of sxl causes selective activation of dosage compensation genes in female Drosophilka.

Select the option with combination of all correct statements.

(a) A, B, E

(b) A, B, D

(d) B, C, D

Ans. (b)

Sol. In Drosophila, sex determination is achieved by a balance of female determinants on the X chromosome and male determinants on the autosomes [(a) is correct]. Normally, flies have either one or two X chromosomes and two sets of autosomes. If there is but one X chromosomes in a diploid cell (1X:2A), the fly is male. If there are two X chromosomes in a diploid cell (2X:2A), the fly is female. Thus, XO Drosophila are sterile males [(c) is wrong, as it is saying fertile male].

The first phase of Drosophila sex determination involves reading the X:A ratio. High values of the X:A ratio are responsible for activating the feminizing switch gene Sexlethal (Sxl) [(d) is correct].

The msl loci, under the control of the Sxl gene, regulate the dosage compensatory transcription of the male X chromosome [(e) is wrong, as it is saying in female drosophila].

91. In D. Melanogaster, traits 'a', 'b', and 'c' result from recessive alleles that are located on one of the autosomes. The resulting F₁ females wre test corssed and the following progeny (total of 1000) were obtained :

a⁺b⁺c⁺ 350

a⁺bc 25

abc 380

ab⁺c 35

abc⁺ 100

abc⁺ 10

a⁺ b c⁺ 85

a⁺b⁺c 15

The following conclusions were made from the above data:

A. The order of genes is a b c and the distance between a and b is 8.5 cM.

B. The order of genes is a c b and the distance between a and c is 8.5 cM.

C. The order of genes is b c a and the distance between b and c is 21 cM.

D. The order of genes is c b a and the distance between b and c is 8.5 cM

Which one of the following options represent statement(s) that is/are correct?

(a) A only

(b) A and D

(d) D only

Ans. (c)

Sol. Parental combinations are always most common types of a⁺ b⁺ c⁺ (350 progeny) and abc (380 progeny) and parental combinatinos. Double cross overs are least common types so abc⁺ (10) and a⁺ b⁺ c (15) are double cross overs. We can determine the gene order by comparing the alleles present in the double crossoverrs with those present in the nonrecombinants. Compare any double-crossover progeny abc⁺ with any of the nonrecombinant, a⁺ b⁺ c⁺. Middle gene is the one that is switched relative to the other two in doubles vs. parentals therefore here the gene order is acb or bca. Rewrite all the combinations in correct gene order. a + cb (25) and ac⁺ b⁺ (35) and SCOI where cross is between locus a and c. acb⁺ (100) and a⁺ c⁺ b (85) is SCOII showing cross between locus b and c. Next we deterine the recombination frequency and draw a genetic map : Recombination frequency (a-c) : 25 + 35 + 10 + 15/1000 (Total progeny) × 100 = 8.5 cM. Recombination frequency (b-c) : 100 + 85 + 10 + 15/1000 × 100 = 21 cM.

92. The table given below shows the Lod score values of three different pairs of genes studied for assessing if they are linked pairs:

The following conclusions were made from the data given above:

A. For gene pair 1 the probability of the genes being linked is 10 times more likely than them assorting independently.

B. For gene pair 2, the Lod score 2 indicates that the probability of the genes being linked to twice more likely than assorting independently

C. The genes of pair 1 and 2, can both be considered as linked, while the genes of pair 3 exhibits independent assortment.

D. The genes of pair 3 can be considered as linked.

Which one of the following options represents statement(s) that is/are correct?

(a) A and C

(b) B and D

(d) A and B

Ans. (c)

Sol. A Lod (Logarithm of the odds) score of 3 means that odds are 1,000 : 1 that the two genes are linked and therefore inherited together. 1000 to 1 means 10³ times more likely that the genes are linked. So similarly, Lod score 1 means, the genes are 10 times more likely to be linked and assorting independently. Positive Lod score indicates linkage. A Lod score of 3 or more than 3 is a definitive for the genes to be linked.

93. A scientist synthesized four new chemicals which had mutagenic potential and named then as C1, C2, C3 and C4. he tried to analyse the nature of mutations caused by them and obtasined the following results.

Which one of the following answers describes the mutagenic potential of the chemicals?

(a) C1 causes transitions, C2 causes transversions or large deletions, C3 causes transitions and C4 single base insertions or deletions

(b) C1 causes transversions, C2 causes transitions, C3 causes transversions and C4 causes large deletions

(d) C1 and C3 causes transversions while C2 and C4 causes transitions

Ans. (a)

Sol. 2-Amino purine, nitrous oxide and hydroxyl-amine causes transition mutation, acridine orange causes frame shift mutation (either by insertion or deletion), if there is no reversion by any of these then it is transversion mutation.

94. Fertilization of gametes containing chromosome with duplications or deletions often results in children with disabilities. What is the probability of a couple where the male is karyotypically normal and the female has a pericentric inversion in 26% of meiotic divisions?

In this case consider that fertilization with a gamete containing chromosomes with duplications or deletion will result in diabilites.

(a) 26%

(b) 13%

(d) 50%

Ans. (b)

Sol. Crossing over in the region of pericentric inversion result in two normal gametes and two abnormal gametes (having deletion/duplication of a few genes). So, for each crossing over in pericentric region : 50% gametes-normal and 50% gametes abnormal. For crossing over frequency of 26%, 26/2 = 13% gametes will be abnormal resulting in a child with disability.

95. In a mammal, coat colour is governed by gene B, The coat colour is either black or borwn, depending on whether the genotype is BB or Bb. It is not known which of these genotypes lead to the black and brown colours. The genotype bb results in albino coat colour. Further, the genotype ce suppresses the expression of coat colour resulting in albino

(a) BB cc and Bb CC, respectively

(b) Bb cc and Bb CC, respectively

(d) bb Cc and BB CC, respectively

Ans. (a)

Sol. Since the resulting F1 ahs no albino, this means in F1, genotypes bb (w.r.t to b gene) and cc (w.r.t to c gene) is not possible as they both will cause albino. Moreover, in F1 both black brown is present, so BB and Bb (w.r.t b gene) both has to be present to form 2 different colours and C_ (w.r.t c gene) to allow the colour formation. Option A is the only possible cross that will satisfy the above conditions. Option B will result in bb (albino); Option C and Option D can form only Bb (only one colour).

96. In echinoderms, sperm direction is provided by

(a) Calcineurin

(b) Bindin

(d) EBRI

Ans. (c)

Sol. In many species, sperm are attracted toward eggs of their species by chemotaxis, that is, by following a gradient of a chemical secreted by the egg. One chemotactic molecule, a 14-amino acid peptide called resct, has been isolated from the egg jelly of the sea urchin Arbacia punctulata. Resact diffuses readily in seawater and has a profound effect at very low concentrations when added to a suspension of Arbacia sperm.

97. Which among the following states has the highest forest cover in terms of percentage of geographical area?

(a) Chattisgarh

(b) Uttarakhand

(d) Odisha

Ans. (b)

Sol. Madhya Pradesh has the largest forest cover in terms of area in the country followed by Arunachal Pradesh.

98. Which one of the following approaches is INCORRECT in classifying organisms from kingdom Animalia?

(a) Spicules were used as a primary criterion for the classification of phylum Porifera

(b) Protostomes were classified into two major lineages; Lophotrochzoans and Ecdysozoans

(d) Ambulacrians included Echinoderms and Hemichordates

Ans. (c)

Sol. The siter group relationship between echinoderms and hemichordates (Ambulacraria) is based on morphological and molecular characters. Initially hemichordates were considered a subphylum of the chordates but they are now put as a separate phylum. Evolutinoary history has taken the echinoderms to the point where they are very much unlike any other animal group.

99. Which of the following phytopathogens has predominantly necrotrophic mode of colonization?

(a) Phytophthora infestans

(b) Erwinia spp

(d) Puccinia graminis

Ans. (b)

Sol. Necrotrophic pathogens are bacterial, fungal and oomycete species that hve very destructive pathogenesis strategies resulting in extensive necrosis, tissue maceration and plant rots. Among the given options Erwinia species has necrotrophic mode of colonization. Also, the oomycete Phytophthora infestans, behaves as biotroph during much of its interaction with its potato and tomato hosts, but may shift to necrotrophy late in infection.

100. Column I list names of organisms and column II list stages in life cycle

Which one of the following options represents all correct matches between the two columns?

(a) A-i, B-ii, C-iii, D-v

(b) A-iii, B-ii, C-i, D-v

(d) A-ii, B-v, C-iii, D-iv

Ans. (c)

Sol. Hexacanth is the larval stage of pork tapeworm. Planula, plural planulae, free-swimming of crawling larval type common in many species of the phylum Cnidaria like obelia. Tornaria isa characteristic, free-swimming larga of the some species of the Enteropneusta or acorn worms (Hemichordata).

101. Given below are statements regarding apomixis, i.e., asexual reproduction through seeds

A. Sporophytic apomicts often produce a mix of clonal and sexual progeny

B. In gametophytic apomixis, the unreduced central cell gives rist to apomictic embryo

C. In pseudogamy the endosperm is formed in the absence of fertilization

D. Apomixis can potentially be used to maintain hybrid vigour over many generations.

Which one of the following options represents the combinations of all correct statements?

(a) A and C

(b) B and C

(d) A and D

Ans. (d)

Sol. Pseudogamy and hybridogenesis are sometimes classified together as different forms of 'sperm dependent parthenogenesis'. But in hybridogenesis, the ovum is haploid and the sperm nucleus is actually incorporated into it to form a diploid zygote.

Apagamy or gametophytic apomixis give rise to embryo development with the help of female gamete only, but do not involve any somatic cells like central cells.

Sporophytic apomicts give rise to Embryo development with the help of spermatic cells which is sometimes referred to as recurrent apomixis.

Hence, the correct combination should be D.

102. The figrue below shows reptilian skull types (A to C), class of roganisms (i to iii) and examples of animals (X to Z).

Which one of the following options shows correct matches between the three components?

(a) A-ii, Y; B, iii, Z; C, i, X

(b) B-ii, Y; A, i, X; C, iii, Z

(d) B-iii, Y; A, ii, X; C, i, Z

Ans. (c)

Sol. A. Image is of Anapsis type of skull, e.g., Turtles; B. image is of Euryapsin type of skull, e.g., Ichthyosours; C. image is of Diapsid, e.g., Snake.

103. The following diagrams (A to D) show characteristic arrangement of stamens within a flower and (i) and (v) enlist families of plants

i. Cucurbitaceae

ii. Fabaceae

iii. Malvaceae

iv. Asteraceae

v. Euphoribaceae

which one of the following options has all correct matches between the arrangment of stamens and the families they belon to?

(a) A-v, B-ii, C-iii, D-iv

(b) A-iii, B-i, C-iv, D-ii

(d) A-v, B-i, C-ii, D-iv

Ans. (c)

Sol. The stamens are five to numerous, and connate at least at their bases, but often forming a tube around the pistils. The pistils are composed of two to many connate carpels.

The stamens are arranged in bundles of two in the flowers of fabaceae family. 9+1 is usually the arrangement of the stamen. Due to the presence of the two bundles of stamen, it is termed diadelphous.

There are usually four or five stamens. The filaments are fused to the corolla, while the anthers are generally connate (syngenesious anthers), thus forming a sort of tube around the style (theca). They commonly have basal and/or apical appendages.

Thladiantha has five stamens with monothecous anthers; the filaments of four stamens are coherent at the base into two pairs, whereas the fifth stamen stands apart.

104. The diagram below shows the floral diagram of the family Malvaceae. Parts of the diagram have been labelled from A-D.

Which of the following combinations represent all correct matches between labels and floral characters?

(a) A-iv, B-v, C-iii, and D-i

(b) A-i, B-v, C-iv, and D-ii

(d) A-i, B-ii, C-iii, and D-iv

Ans. (a)

Sol. Floral diagram is a graphic representation of flower structure. It shows the number of floral organs, their arrangement and fusion. Different parts of the flower are represented by their respective symbols. Floral diagrams are useful for flower identification or can help in understanding angiosperm evolution. The floral formula of the family Malvaceae consists of 3-8 epicalyx, and calyx which consists of five gamosepalous sepals. The corolla constitutes five polypetalous petals with numerous stamens and a carpellary and multicarpellary syncarpous superior ovary.

A : Epicalyx

B : Syncarpous ovary

C : Fused sepals

D : Syncarpous ovary.

105. Mycorrhizal fungi are associated with a large variety of plant species. The diagram below shows the cost-benefit curves from individual plants with or without mycorrhizal fungi associated with the roots across a soil nutrient concentration gradient.

Which one of the following options best describes the association between the plant and mycorrhiza when soil nutrient concentrations are high?

(a) Parasitism

(b) Mutualism

(d) Commensalism

Ans. (a)

Sol. Mycorrhizae are especially beneficial for the plant partner in nutrient-poor soils. In fertilized soil, however, mycorrhizal benefits may be reduced because plants can acquire sufficient nutients in the absence of AMF. Mycorrhizal fungi might be considered to the parasitic on plants when net cost of the symbiosis exceeds net benefits.

106. In a population showing exponential growth, per capita growth rate will

(a) decrease as popultion size increases

(b) increase as population size increases

(d) increase initially and then saturate at large population sizes

Ans. (a)

Sol. In exponential growth, a population's per capita (per individual) growth rate stays the same regardless of population size, making the population grow faster and faster as it gets larger.

107. Identify the plot that depicts the change in metabolic rate of an endotherm with respect to change in environmental temperature:

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Metabolic rates in endotherms are responsive to environmental conditions, so that heat production increases as environmental temperatures drop below the thermal neutral zone. So, according to the options, option 3 defines this situation as temperature increases, metabolic rate decreases.

108. The linkage density in a food web is function of the connectance and the number of species in it. It is defined as the average number of feeding links per species. Which one of the following would have the highest linkage density?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Connectance has become a fundamental quantity for nearly all other measures of food web structure and dynamics. The third important quantity is another ratio, linkage density,

L D = L / S ;

L = average number of links

S = species

This value represents the number of links added to the network for every additional species in the ecological system. So, C have maximum linkage density.

109. Given below are growth and survivorship curves

Select the correct combination of growth curves from survivroship curves from figrue B given above, that would best represent r and K strategies, respectively.

(a) r = a and I; K = b and n

(b) r = b and n; K = a and I

(d) r = b and m; K = a and m

Ans. (b)

Sol. The Type II curve, characteristic of small mammals, fishes and invertebrates is the opposite : it describes oranisms with a high death rate (or low survivorship rate) and organisms which are considered R-selected species tend to be small, reach sexual maturity early in life and reproduce a single time and the growth rate is definately low. Hence option B will be correct.

110. Ozone concentration in the atmosphere is measured in Dobson units (Du). An ozone hole is said to have occurred when the concentration of ozone in the region falls below 220 Du. The diagram of the globe below shows the location of three study sites. The table below it shows values of ozone concentrations recorded at the sites for two dates in 2009.

Which one of the following options represents the correct matches between the study sites and the recorded ozone values?

(a) A-i, B-iii and C-iv

(b) A-ii, B-iii and C-i

(d) A-i, B-ii and C-iii

Ans. (c)

Sol. Total ozone varies with latitude, longitude and season, with the largest values at high latitudes and the lowest values in tropical regions. The variations are demonstrated here with two week averages of total ozone in 2009 as measured with a satellite instrument. Total ozone shows little variation in the tropics (20°N-20°S latitudes) over all seasons. Total ozone outside the tropics varies mroe strongly with time on a daily to seasonal basis as ozone rich air is moved from the tropics and accumulates at highe latitudes. The low total ozone values over Antarctica in September constitute the "ozone hole" in 2009. Since the 1980s, the ozone hole in the late winter and early spring represents the lowest values of total ozone that occur all seasons and latitudes.

111. A population of crickets invading a new grassland showed a population growth pattern as shown in the figure

Following is the list of potential interpretations:

A. Environment is damaged due to population overshooting its K

B. The resources did not recover and population dies out

C. Carrying capacity is lowered due to shift in environmental conditions

Which one of the following options/combination of options can correctly explain the cricket growth pattern?

(a) A only

(b) B only

(d) B and C only

Ans. (c)

Sol. This graph represents a condition of over-shooting. Normal condition of carrying capacity represents a S shaped curve but when overshooting take place the line exceeds above the normal one. Overshooting gives us information how the Earth citizens exceed the exploitation of worlds scarce resources and nowadays it is 30%. It means that we are using the earth scare resources 30% faster than they can be regenerated. The ecological footprint of an average person is way bigger than the area he/she lives in.

112. Given below are the spectra of plant forms (on the basis of where the plants bear their buds) in different biomes (A to C).

Which one of the following options correctly identifies the biomes represented in graphs A to C?

(a) A-Mediterrranean; B-Arctic; C-Tropical

(b) A-Tripical; B-Desert; C-Temperate

(d) A-Desert; B-Arctic; C-Temerate

Ans. (b)

Sol. Phaneropytes are large shrubs and trees in which the overwintering (perennating) buds are located high above the ground. The buds are thus at risk of exposure to drought stress or frost and such plants occur mainly in regions where frost and drought are uncommon, such as the tropics.

Hemicryptophyte are plant whose perennating buds are at ground level, the aerial shoots dying down at the onset of unfavourable conditions.

Therophytes are plant that completes their its life cycle rapidly during periods when conidtions are favourable and survives unfavourable conditions (e.g., cold, heat or competiton) as seed; it is thus an annula or ephemeral plant. Therphpytes are very typical of desert.

113. Growth patterns of two species (grown alone or together) are shown in Figures A and B.

Match the growth patterns with the correct type of interaction represented by them

(a) A-mutualism, B-commensalism

(b) A-competition, B-parasitism

(d) A-competition, B-resource partitioning

Ans. (d)

Sol. Resource partitioning is the division of limited resources by species to help avoid competition in an ecological niche. In any environment, organisms compete for limited resources, so organisms and different species must find ways to coexist with one another whic is well evident in graph B and when it is a condition of competition one Species will be affected and the other will be benefitted and such condition is exhibited in Graph A.

In Graph A the species 1 has got a population size near 100 whereas comparing it with Species B the population size is less than 50.

114. The following four climatograms depict observation of mean temperature and rainfall for four locations. The solid lines depict temperature while the dashed line depicts rainfall.

A. B.

C. D.

Given below are five biomes:

i. Desert

ii. Savannah

iii. Tropical rain forest

iv. Mediterranean

v. Conifer ecosystem

Which of the following combinations correctly matches the climtrograms to the biomes?

(a) A-iv, B-ii, C-i, D-v

(b) A-v, B-iv, C-ii, D-iii

(d) A-v, B-iii, C-ii, D-i

Ans. (c)

Sol. The tropical rainforest is a hot, moist biome wher it rains all year long. The Savanna biome is associated with climates having seasonal precipitation accompaned with a seasonal drought. The desert has some extreme climate. The average temperatures rises to a very high point as deserts are usually very hot. The average rainfall deserts get actually go really especially in the middle of the year. Mediterranean forests, woodlands and scrub is a biome generally characterized by dry summers and rainy winters, although in some areas rainfall may be uniform.

115. Examples of antibiotic resistance highlight important features of natural selection

Which of the following statements is not true?

(a) Evolution by natural selection is progressive, it makes individuals 'better'.

(b) Natural selection acts on individuals but it is populations that change with time

(d) Natural selection acts on phenotype

Ans. (a)

Sol. Antibiotic resitance is a consequence of evolution via natural selection. The antibiotic action is an environmental pressure; those bacteria which have a mutation allowing them to survive will live on to reproduce.

116. The following statements describe the outcomes of genetic drift

A. Genetic drift can eliminate alleles

B. Genetic drift can be associated with population bottleneck

C. Genetic drift it not observed in populations that increase in size, once thye grow through a bottleneck

D. Genetic drift can be associated with founder effect

Which one of the following combinations represents all correct statements?

(a) A, B and C

(b) B, C and D

(d) A, C and D

Ans. (c)

Sol. Genetic drift describes random fluctuations in the number of gene variants in a population. Genetic drift takes place when the occurrence of variant forms of a gene called alleles, increases and decreases by chance over time. These variations in the presence of alleles are measured as changes in allele frequencies. Genetic drift occurs in small populations, where infrequently occurring alleles face a greater chance of being lost. Thus, Genetic drift can have major effects when a population is sharply reduced in size by a natural disaster (bottleneck effect) or when a small group splits off from the main population to found a colony (founder effect).

117. The common cuckoo, a parasitic bird, lays eggs in the nests of other bird species. Soon after the cuckoo egg hatches, the chick shoves the nest owners' eggs out of the nest. This is an example of:

(a) habituation

(b) imprinting

(d) operant conditionaing

Ans. (c)

Sol. Behaviour that are closely controlled by genes with little or no environmental influence are called inate behaviours. These are behaviours that occur naturally in all members of a species whenever they are exposed to a certain stimulus. Innate behaviours do not have to be learned or practiced.

118. Given below are the few statements on concepts related to genome evolution.

A. Presnece of introns in some chloroplast genes suggests that endosymbiosis (leading to organelle evolution) occurred before loss of introns in prokaryotes and supports the hyopthesis that genes originated as interrrupted structures.

B. Negative selection is associated with increased rate of nonsynonymous substitutions as compared to synonymous substitutions.

C. Nucleotide substitution rates during evolution can be inferred from divergence of the sequences that non-functional or neutral.

D. Positive selection is associated with increased rate of nonsynonymous substitutions as compared to synonymous substitutions.

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

(a) A and C only

(b) B and C only

(d) C and D only

Ans. (c)

Sol. A nucleotide substitution that changes the corresponding amino acid in the protein is called a nonsynonymous substitution (denoted as KA), whereas a nucleotide substitution that does not change the amino acid in the protein is called a synonymous substitution (denoted as KS).

However, if a nonsynonymous substitution confers some selective advantage, then it will be rapidly fixed in the population by positive selection. Mutations in which natural selection does not affect the spread of the mutation in a species are termed neutral mutations. The neutral theory of molecular evolution posits that a majority of evolutionary changes are due to stochastic drift of selectively neutral mutations.

119. You wanted to conduct Miller-Urey experiment and used a simplified apparatus with Tungsten electrodes. You heated the glassware at 500°C for 3 hours to remove any organic contaminants. Gases NH₃, CH₄, CO and H₂ were introduced followed by generating electric spark. Which of the essential ingredients did you forget to add?

(a) O₂

(b) H₂O

(d) CHO

Ans. (b)

Sol. The Urey-Miller experiment mixed methane, ammonia, hydrogen, and water — the chemicals that made up the early atmosphere — and ran sparks of electricity through it to mimic lightning. So, he forget to add water .

120. In a population that is in a Hardy-Weinberg equilibrium, 40% of the plants are recessive homozygotes and produce white flowers (WF). It the total number of individuals in the populations is 14000 plants, the numbers of homozygous dominant red flowered (RF) plants and heterozygrous pink flowered (PF) plants would be:

(a) RF-56000 PF-1891

(b) RF-1891 PF-6508

(d) RF-5145 PF-8855

Ans. (b)

Sol. q² = 0.4

q = 0.632

p = 1 – 0.632 = 0.368

RF = p² × 14000

= (0.368)² × 14000 = 1890

PF = 2pq × 14000

= 6510

Values are not exactly what given in options but close enough to choose it as the correct answer. The differences is because of rounding off of decimal points.

121. Given below are names of scientists and phrases describing their work, which may or may not be matched

Which one of the following options represent correct matches between the scientist and his/her work?

(a) i-A; ii-B; iii-C; iv-E

(b) i-B; ii-C; iii-A; iv-D

(d) i-E; ii-D; iii-A; iv-C

Ans. (b)

Sol. Alfred Russel Wallace, proposed that evolution occurs because of a phenomenon called natural selection. In the theory of natural selection, organisms produce more offspring than are able to survive in their environment.

Lyell argued that the formation of Earth's crust took place through countless small changes occurring over vast periods of time, all according to known natural laws. His "uniformitarian" proposal was that the forces molding the planet today have operated continuously throughout its history.

Lamarck is best known for his Theory of Inheritance of Acquired Characteristics, first presented in 1801 (Darwin's first book dealing with natural selection was published in 1859): If an organism changes during life in order to adapt to its environment, those changes are passed on to its offspring.

French naturalist Georges Cuvier developed his theory of catastrophes. Accordingly, fossils show that animal and plant species are destroyed time and again by deluges and other natural cataclysms, and that new species evolve only after that.

122. Which one of the following statements is TRUE?

(a) Nick translation cannot be used for radio labeling DNA fragments for used in Sourthern hybridization

(b) RACE can be used to obtain sequence information of the ends of a cDNA

(c) In a 2-D gel electrophoresis for separating proteins, analyes are first separated on the basis of theeir size and then on the basis of their pl

(d) AFLP markers are typically used to screen heterozygotes from homozygotes

Ans. (b)

Sol. Nick transltaion is a tagging tehcnique in molecular bioloty in which DNA Polymerase I is used to replace some of the nucleotides of a DNA sequence with their labelled analogues, creating a tagged DNA sequence which can be useda s a probe in Fluorescent in situ Hybridization (FISH) or blotting techniques. It can also be used for radiolabelling.

Hence, option A is incorrect. Rapid Amplification of cDNA ends (RACE) provides an inexpensive and powerful tool to quickly obtain full-length cDNA when the sequence is only partially known, hence option B is true.

123. In the context of plant breeding and genetic engineering, which one of the following statements is correct?

(a) To achieve high expression level of a heterologous gene in a transgenic plant a promoter of bacterial origin is often used

(b) F1 progeny derived by crosses between inbred lines with low genetic diversity is more likely to show heterosis

(d) Qualitatibe traits are typically charaterized bly discontinuous variation while quantitative traits often generate continuous phenotpic variation

Ans. (d)

Sol. Qualitative traits are types of traits that fall into distinct classes or categories without variation within those traits. These types of traits are also referred to as discontinuous traits and discrete traits, since there is no variation outside of the specific, aka discrete, trait classes. Quantitative traits are also called continuous traits, and they stand in contrast to qualitative, or discontinuous, traits that are expressed in the form of distinct phenotypes chosen from a discrete set. Continuous variation in the expression of a trait can be due to both genetic and non-genetic factors.

124. Which one of the following approaches/markers would be typically used for discovering polymorphism between two closely related accessions of a crop plant?

(a) AFLP

(b) GBS

(d) RAPD

Ans. (b)

Sol. Genotyping-By-Sequencing (GBS) is novel application of NGS protocols for discovering and genotyping SNPs (Single Nucleotide Polymorphism) in crop genomes and populations. The low cost of GBS makes it an attracitve approach to saturate the mapping and breeding populations with a high density of SNP markers.

125. Given below are a few statements related to plant breeding and genetics:

A. Non-random mating between individuals closely releated to each other would tend to promote a decrease in homozygosity at all loci

B. Quantitative traits are typically influenced by environmental conditions to a greater extent than qualitative traits

C. The over dominance theory for explanation of heterosis assumes that in a heterozygote, only one of the alleles plays a functional role

D. During domestication selected genes/traits show a decrease in diversity.

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

(a) A and D only

(b) B, C and D

(d) B and D only

Ans. (c)

Sol. According to quantitative genetic theory, heterosis can result from dominance, overdominance or epistasis. Overdominance is an intra-allelic interaction in which the presence of multiple alleles leads to greater performance than homozygosity for either allelic state. If overdominance is the predominant basis of heterosis, then populations and breeding strategies that maximize heterozygosity will result in the best performance. Non-random mating leads to departures from Hardy-Weinberg proportions. For example, inbreding and positive assortative mating (where inidividuals prefer to mate with phenotypically similar individuals) yield and excess of homozygotes. By contrast, negative assortative mating (hwere opposites attract and individuals prefer to mate with phenotypically different individuals) results in excess of heterozygotes.

126. Given below is a schematic representation of the T-DNA region of a transgene construct and the Southern blot analysis (using Pst I digested genomic DNA) of five independent transgenic lines (labelled as A to E) developed using the construct. The probe used for hybridization is shown as a black box below the construct (UT : Untransformed Plant)

Based on the above data, which one of the following options gives the correct list of single insertion transgenic events?

(a) B and C only

(b) A, B and E only

(d) E only

Ans. (b)

Sol. 2 recognition sites of Pst I are present within the transgene, so when digested with Pst I, the transgene will get cleaved into three fragments. Now the probe has a sequence that is complementary to some part of two out of those three cleaved fragments, so in southern blot it will be able to detect 2 different sized fragments for single copy of gransgene. Therefore, for single insertion event of the transgene, there will be 3 bands visible 1 for UT and 2 for transgene. Since A, B and E are the ones having the above mentioned 3 bands, so they are the ones having the single insertions of the transgene.

127. A group of researchers are testing two new agents, M1 and M2 for their efficacy in selecting transgenic plants. When they performed tissue culture experiments using three explants, A, B, and C without Agrobacterium trnasformation, and selected the regenerated plants on M1 and M2, the following regeneraton frequencies were obtained.

Based on the above data, which one of the following conclusions is INCORRECT?

(a) M2 is stronger selection agent than M1 for explant types A and B.

(b) A concentration of 100 mg/L of M1 and 15 mg/L of M2 can be used for selection of transformed cells using explant type A.

(d) Use of M1 with explant type B is the most ideal combination for selection of transformants.

Ans. (d)

Sol. The best combination will be with M1 strain because most of the explants are showing negligible results with M2 and morevover explant B type is the best combination to be used for selecting transformants as cumulatively it less 100% and which should be the ideal concentration level.

128. Which one of the following sets of terms is matches correctly?

(a) Fluorescence microscopy : Use of dyes that absorb and emit light of the same wavelength

(b) Confocal microscopy : Use of pinpoint illumination and detection apertures

(d) Gas Chromatogrphy : Uses for analysis of non-volatilized carbohydrate molecules

Ans. (b)

Sol. Fluorescence microscopy used dyes that absorbs light and emit light of different wavelength, hence option A is incorrect. Gas Chromatography (GC) is applied for gases and mixtures of volatile liquids and solid material. Liquid chromatography is used especially for thermal unstable and non-volatile samples, hence option D is incorrect. The confocal microscope essentially removes the out-of-focus light by inserting a pinhole at the image plane such that out of focus light does not rech the detector. Coherent light emitted by the laster system (excitation source) passes through a pinhole aperture that is situated in a conjugate plane (confocal) with a scanning point on the specimen anda second pinhole aperture (detector pinhole aperture) positioned in front of the dectector (a photomultiplier tube). Therefore answer is option (b).

129. The mechanism or oxygen transport by hemocyanin (containing Cu) is described by

Which one of the following techniques can be used to monitor the change in the oxidation state of copper?

(a) Mass spectrometry

(b) Circular dichroism

(d) Fluorescence spectroscopy

Ans. (c)

Sol. X-ray absorption near edge spectroscopy (XANES), provides information concerning the geometry and valence state of metal ions. XANES can be usefully applied to the study to native and metal substituted hemocyanins and can probe differences in geometry between sites having equivalent corrdination number. Cu X-ray absorption edge spectra for a large series of Cu(I) and Cu(II) complexes, have also been reported.

130. Which one of the following proteins produces a dark blue/purple color in presence of 5-bromo-4-chloro-3indolyl phosphate and nitroblue tetrazolium?

(a) Polynucleotide kinase

(b) Alkaline phosphatase

(d) -galactosidase

Ans. (b)

Sol. 5-Bromo-4-chloro-3-indolyl phosphate disodium salt has been used to detect protein-antibody complexes. It is used to conjuction with NBT (nitro blue tetrazolium) for the colorimetric detection of alkaline phosphatas activity.

131. Given below are a few approaches/techniques used in the analysis of plants:

A. phenotupe linked of the marker gene

B. PCR using primers specific to the host genome

C. Southern hybridization

D. PCR using transgene-specific primers.

Which one of the following combination of approaches/techniques can be used to differentiate between transgenic variety if the site of insertion is unknown?

(a) B and C only

(b) A and D only

(d) A, B and C

Ans. (c)

Sol. PCR using primers specific to host genome will amplify genome of transgenic as well as non transgenic plant. And, will nto be able to differentiate between them. All the other three methods mentioned in statements A, C and D can be used to differentiate between transgenic and non transgenic plants.

132. Which one of the following genomes is most appropriate hybrids in plants and animals?

(a) Nuclear

(b) Mitochondrial

(d) Mitochondria and chloroplast

Ans. (a)

Sol. Nuclear Genome represents main chromosomal genetic content in both plants and animals. During gametogenesis, only nuclear genome segregates and while fertilization, nuclear genome from both parents compare genome of Progeny. Hybrids may be generated when nuclear genome of two different species come together in individual.

133. In a cloning experiment, a DNA molecule isolated as a Pst I fragment was inserted into the Pst 1 stie of a cloning vector that is 3 kb in length. Digestion of the CONFIRMED clones with Pst resulted in the apperance of a single band corresponding to 3kb in an agarose gel. Based on this information, what is the probable size of the cloned fragment?

(a) 6 kb

(b) 1.5 kb

(d) 4.5 kb

Ans. (c)

Sol. Agarose gel electrophoresis is a type of electrophoresis used to separate nucleic acids (most commonly, DNA) according to their molecular weights. Agarose is a linear polymer that gelatinizes to form a three-dimensional mesh of channels ranging in size from 50 to = 200 nm. The migration of the nucleic acid sample depends on the size and confirmation of the DNA as well as on experimental parameters. As 3 kb single band is observed in agarose gel , so the size of fragment is 3kb.

134. A tree built using BLAST cannot be used to infer phylognetic relationships. Given this, which of the following statements is NOT true about trees generated by BLAST?

(a) It is based on a distance method, where alighment similarity scores are used to cluster sequences

(b) It is an exhaustive tool where similar sequences are found by locating all mathches between multiple sequences simultaneously

(d) It is built by first performance seeding following local alignments.

Ans. (b)

Sol. A blast results is essentially a unidirectional catalog of how similar a sequence is to a list of otherr sequences, while phylogenetic analysis, which presents results in a tree based format, allows one to compare the degree of relationship between all sequences in the set. The "two-dimensional" view of a phylogenetic analysis cna be helpful in, for example identifying whether there are susbets of sequences in a BLAST output that could be grouped as a family. BLAST alone, for example, cannot identify whether three sequences found to be equally similar to a sequences are (option A) actually all highly realted to each other or are (option B) quite divergent from each other (but still equally related to the query sequences). Such information provided by phylogenetic analysis is critical when inferring function based on the sequence similarity.

135. A 100 residue lone protein has single chromporic residue (tyrosine). The UV absorption of this protein and a homologous protein (also with a single tyrosine residue) was monitored at 280 nm at different pH conditions. A plot of the absorbance as a function of pH is shown below. The locations of the tuyrosine residue in the context of the protein sequences is also shown in the figure.

Which one of the following rationalises the differences in the two pH titrations?

(a) removal of the hydroxyl group of tyrosine above pH 11

(b) location of the tyrosine residue in the protein structure

(d) hydrolysis of the polypeptide as a function of pH

Ans. (b)

Sol. Absorption band shifts towards longer wavelengths when solution is made alkaline. Changes of the absorption spectrum of proteins with pH indicate that Tyr residues in the native protein are nto free to ionize, due to restrictions imposed upon the protein configuration by its tertiary structure. Deeply buried Tyr interct with other side chain groups to maintain a rigid structure. Hence for protein B, at a given pH, the absorbane is less, as the only difference is the change in the location of Tyr, which may have caused it to move it a deeper area in the protein, thus making it difficult to ionize as is done by its homologous protein at a lesser pH.

136. Given below are the restriction profiles obtained upon digestion of a 4.8 kb plasmid with four different enzymes (E₁, E₂, E₃ and E₄).]

E₁ : 4.8kb E₁ + E₂ = 500 bp + 4.3 kb

E₂ : 4.8kb E₂ + E₃ = 400 bp + 4.4 kb

E₃ : 4.8kb E₃ + E₄ = 2 kb + 2.8 kb

E₄ : 4.8kb E₁ + E₄ = 1.9 kb + 2.9 kb

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

(a) E₁ and E₄ have two recognition sites each in the plasmid

(b) E₂ recognition sequence is located between E₁ and E₃.

(d) The sequence of recognition sites after digestion of the plasmid with E₂ is E₁–E₃–E₄.

Ans. (b)

Sol. As the smallest fragment is observed on E2 digestion, so E2 recognition sequence is located between E1 and E3.

137. The ¹H NMR spectrum of ¹³CH₃Cl in CDCl₃ shows two lines (double, 1 : 1 intensity) spaced 140 Hz apart. The carbon (¹³C) NMR spectrum of the same molecule shows four lines (quartet, intensity ratio 1:3:3:1). Both ³H and ¹³C have nuclear spin quantum numbers f = 1/2

Based on the above information which one of the following options is correct?

(a) The multiplicity pattern observed follows the 2n/+1 rule

(b) The ¹H–¹³C coupling constant is different in the ¹H (proton) and ¹³C (Carbon) spectra.

(c) The multiplicity patterns are a consequence of the difference in gyromagnetic rations of the ¹H and ¹³C nuclear spins

(d) The multiplicity patterns are dependent on the Larmor precessional frequencies

Ans. (a)

Sol. Multiplicity. The multiplicity (sometimes referred to as splitting) tells how many hydrogen atoms are immediately next door to the hydrogens producing that peak. The multiplicity is the most important piece of information, since it allows you to connect the pieces together to identify the molecule. As Carbon is attached to 3 hydrogen atoms so it gives four lines and Hydrogen is attached to one carbon atom , it shows 2 lines.

138. One curie (Ci) is defined as the quantity of radioactive substance having a decay rate of 2.22 × 10¹² disintegrations per minute (dpm). How much of counts per minutes (cpm) will be recorded for 100 µCi of ³²P when measured in a scientillation counter working at 50% efficiency?

(a) 1.11 × 10⁸

(b) 1.11 × 10⁶

(d) 2.22 × 10⁶

Ans. (a)

Sol. The ratio between CPM detected by the instrument and actual disintegrations per minute (DPM) of the isotope is termed efficiency. DPM are calculated from the equation shown below, where efficiency is expresses as a decimal percent.

Converting counts per minute to disintegratiosn per minute.

DPM = CPM/Efficiency of detector or

CPM = DPM

Givne 1 Ci = 2.22 × 10¹²

DPM = ?

Therefore,

DPM = 100 µCi

= 100 × 2.22 × 10¹² × 10^–6

= 2.22 × 10⁸

Efficiency = 50% = 0.5

Therefore

CPM = DPM × efficiency

= 2.22 × 10⁸ × .5

= 1.11 × 10⁸

139. Given below is a schematic representation of a few restriction sites in the multiple cloning site of a plasmid:

The recognition sequences of the above enzymes and their digestion patterns are shown below:

Which one of the following combinations of enzymes from the vector can be used for cloning a DNA fragment obtained as a SaII – Hind II fragment.

(a) Sma I – Hind III

(b) Xho I – Pst I

(d) Hind III – Pst I

Ans. (c)

Sol. An MCS contains many unique restriction sites to choose from, so compatible restriction enzymes can be used on both the vector and the insert. Here we get an out put as mentioned like Sal I G*TCAG-cohesive cand a blunt cut obviously from Sma.

140. The following disulfide bond containing peptide was trypsin. How many peptide fragments will be produced by the digestion?

(a) Three

(b) Four

(d) Six

Ans. (a)

Sol. Trypsin cleaves the peptide bond between the carboxyl group of arginine or the carboxyl group of lysine and the amino group of the adjacent amino acid. The rate of cleavage occurs more slowly when the lysine and arginine residues are adjacent to acidic amino acids in the sequence or cystine. So, three peptide fragments will be obtained after digestion.

141. The first step in the biosynthesis of valine begines with enzyme catalyzed condensation of two molecules of pyruvic acid. If an equimolar mixture of ¹³CH₃COOH and ¹²CH₃COCOOH are used as substrates for the reaction, which one of the following would represent the correct isotype incorporation pattern of the pro-R and pro-S diastereotopic methly group in valine?

(a) 50% ¹³CH₃ (pro-R), ¹²CH₃ (pro-S)

50% ¹²CH₃ (pro-R), ¹³CH₃ (pro-S)

(b) 75% ¹³CH₃ (pro-R), ¹²CH₃ (pro-S)

25% ¹²CH₃ (pro-R), ¹³CH₃ (pro-S)

25% ¹³CH₃ (pro-R), ¹³CH₃ (pro-S)

25% ¹²CH₃ (pro-R), ¹²CH₃ (pro-S)

(d) 75% ¹²CH₃ (pro-R), ¹³CH₃ (pro-S)

25% ¹³CH₃ (pro-R), ¹³CH₃ (pro-S)

Ans. (c)

Sol. Biosynthesis of Valine requires two molecules of Pyruvate. They undergo condensation reaction to produce alpha acetolactate catalyzed by enzyme acetolactate synthase. Alpha acetoacetate thus formed undergoes reduction by enzyme acetohydroxy acid isomeroreductase forming alpha, beta isovalerate, which undergoes dehydration releasing alpha keto isovalerate which gains amino group from glutamate catalyzed by enzyme valine aminotransferase forming valine.

The first enzyme involved in the pathway (acetoacetate synthase) is neither isotope specific nor stereospecific, hence it can use any pyruve in biosynthesis of valine. Since the reaction mixture has equimolar concentration of C-12 and C-13 isotopes of Pyruvate. Hence it is highly likely that we will get the result given in option (c).

142. The data from an S1 nuclease mapping experiment for a transcript mfg1 using a 5'-end labelled probe are shown below.

Following interpretations were made:

A. The liver RNA is 500 bp in length

B. The start site of the liver mfg1 transcript is 500 bp downstream of the 5' end of probe

C. The kidney makes two mfg 7 transcripts, and the 3' end of one of these is shorter than the other.

Which one of the following options represents the correct combinations of the interpretations?

(a) A, B and

(b) A and C only

(d) B only

Ans. (d)

Sol. The S1 nuclease is an endonuclease that digests single (but not double-stranded) nucleic acid. S1 is used to recognize and cut mismatches or unannealed regions and the products are analyzed on a denaturing polyacrylamide gel. A probe of end-labelled double-stranded DNA in denatured and hybridized to complementary RNA molecules. The labelled DNA hybridizes to the corresponding mRNA and the introns/unannealed parts are cut by S1 nuclease. So liver RNA will be more than 500 bp in lengt.

Since the labelled DNA is used as a probe to map the 5' end of a messenger RNA, statement C is wrong as it mentions 3' end.

143. Given below are a few statistical terms in Column A and their related features and terms in Column B.

Which one of the options given below correctly matches all items of Column A and B?

(a) A-iv, B-iii, C-ii, D-i

(b) A-ii, B-iv, C-i, D-iii

(d) A-ii, B-i, C-iv, D-iii

Ans. (c)

Sol. Standard Deviation (SD) is the square root ot the variance (both population and sample). Coefficient of Variation (CV) describes the relative variation within a sample or population. Chi-square tests are a family of significance tests that give us ways to test hypotheses about distributions of categorical data and a two sample test t-test, is used to compare the population means to each other and again look at the difference.

144. Following are the typical dot plots of cytometric data obtained from peripherial white blood cells:

R1 in plot 1 shows the selection window of cells which were further analysed based on immunostaining (plot 2). Analyze the given data and choose the correct option reprsentings the type of cell population in plot 1 selected and segregated in various quadrants of plot 2.

(a) Plot 1 lymphocytes; Plot 2 (A) CD4^– CD8^– (B) CD4^– CD8⁺ (C) CD4⁺ CD8^– (D) CD4⁺ CD8⁺

(b) Plot 1 neutrophils; Plot 2 (A) CD4⁺ CD4^– (B) CD4^– CD8⁺ (C) CD4⁺ CD8^– (D) CD8^– CD4^–

(d) Plot 1 lymphocytes; Plot 2 (A) CD4^– CD8⁺ (B) CD4^– CD8⁺ (C) CD4⁺ CD^– (D) CD8⁺ CD4^–.

Ans. (a)

Sol. Plot 1 will be lymphocytes as they are of different types which can be separated in lot 2. Plot 2 and 4 quadrants and each quandrant has certain flourescence intensities. Quadrant A should be double negative and quadrant D should be double positive.

145. Given below are a few terms in column A and column B.

Which one of the following options represents all correct matches between terms of Column A and Column B?

(a) A-iv, B-iii, C-i, D-ii

(b) A-iii, B-i, C-iv, D-ii

(d) A-i, B-iv, C-ii, D-iii

Ans. (a)

Sol. In the cisgenic technology, the cisgene must be an identical copy of host's native gene cassette, including its regulatory sequences integrated in the host.

Association mapping (AM) or linkage disequilibrium (LD) is an approach that accounts for thousands of polymorphisms to evaluate the effects of quantitative trait loci (QTL). ZFRs can be tailored to promote.

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CSIR NET BIOLOGY (November - 2020 Sift 1)
Previous Year Question Paper with Solution.