PYQ JUNE 2021 SIFT 2

CSIR NET BIOLOGY (JUNE - 2021 Sift 2)
Previous Year Question Paper with Solution.

21. Phosphofructokinase catalyses one of the regulatory steps in glycolysis. Which one of the following metabolic changes leads to the activation of phosphofructokinase?>

(a) Increased ATP concentration>

(b) Decreased AMP concentration>

(d) Increased fructose 2,6, bisphosphate concentration>

Ans. (d)

Sol. Phosphofructokinase (PFK1) is an allosteric enzyme. It performs both positive as well as negative regulations. Fructose-2, 6-bisphosphate and AMP are positive modulators whereas, citrate and ATP are negative modulators of PKF1.>

22. What is the net charge of the peptide Tyr-Val-Arg at pH 5.0? The pKas of alpha amino and carboxyl groups are 9.6 and 2.3, respectively. The pKas of Tyr and Arg side chains are 10.46 and 12.48, respectively.>

(a) 1.0>

(b) 5>

(d) 11>

Ans. (a)

Sol. >

23. Protein X is an all helical protein with 100 amino acids including 2 cysteines and a pI of 7.0. Which one of the following graphs best describes the solubility of this protein under different ammonium sulphate (salt) concentrations?>

(a) >

(b) >

(d) >

Ans. (b)

Sol. Protein solubility depends on pH, ionic strength (salt concentration), solubility and temperature.>

24. A. 5'- AGTAGTATCAACTATCATGA-3'>

3'- TCATCATAGTTGATAGTACT-5'>

B. 5'- GACGTGCCAGGTGCGAGGTC-3'>

3'- CTGCACGGTCCACGCTCCAG-5'>

C. 5'- TACGATGCACATGCTTGGAC-3'>

3'- ATGCTACGTGTACGAACCTG-5'>

D. 5'- GAACGCTACGTTGCGATCCG-3'>

3'- CTTGCGATGCAACGCTAGGC-5'>

Arrange the DNA fragments (A to D) in the order of decreasing melting temperature.>

(a) B>D>C>A>

(b) C>A>B>D>

(d) A=B>C>D>

Ans. (a)

Sol. The helical structure of dsDNA is stabilized by non-covalent interactions. These interactions include stacking interactions (major) between adjacent base pairs and hydrogen bonding (minor) between complementary strands. Stacking interactions include hydrophobic interactions and van der Waals interactions between base pairs. Higher the value of GC content, higher will be the hydrogen bonding as well as stacking interactions.>

25. Heating of some nucleic acids shows an increase in the absorbance at 260 nm (A260) typified by the plot shown above. The sharp transition midpoint is defined as melting temperature (Tm). Which one of the following nucleic acid samples is NOT expected to generate such a typical profile upon heating of its solution?>

(a) Double stranded DNA>

(b) Double stranded RNA>

(d) Single stranded DNA having imperfect secondary structures>

Ans.>

Sol. *Heating of some nucleic acids shows an increase in the absorbance at 260 nm (A260) typified by the plot shown above. The sharp transition midpoint is defined as melting temperature (Tm). All of the options are available to provide profile upon heating.>

26. The Ramachandran plot graphically shows which combination of torsional angles phi and psi of amino acid residues contained in a peptide are possible. Examination of the plot below shows that only certain regions of the conformational space are permissible.>

Why are all the theoretical combinations of f and ? not possible?>

(a) Two atoms cannot occupy the same space>

(b) The geometry of the peptide bond that links two amino acid residues restricts and angles>

(d) The tertiary fold of polypeptides restricts conformational space>

Ans. (a)

Sol. The phi and psi can have any value between +180° and –180° (i.e., 360° of rotation for each). But not all combinations are possible due to physical clashes of atoms in 3-dimensional space. Atoms take up space and two atoms cannot occupy the same space at the same time. These physical clashes are called steric interference.>

27. Consider the following statements:>

A. Coenzyme B12 is an organometallic compound>

B. Pyridoxal phosphate is a co-factor used by many amino- transferases.>

C. The affinity of biotin for avidin is one of the highest binding affinities known in biochemistry>

D. Enzymes catalyse biochemical reactions by lowering the energy of the transition state>

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

(a) A, B, C and D>

(b) A, B, C and D>

(d) B, C and D only>

Ans. (a)

Sol. Aminotransferase enzyme catalyses the trasamination reaction. Here, pyridoxal phosphate works as cofactor. Avidin, Streptavidin and Neutravidin have high affinity for biotin. Enzyme lowers the activation energy of the substrate thus lowering the energy for transition state.>

28. The pKa of the ionizable groups in the tripeptide shown below are indicated in the structure.>

The isoelectric point (pI) of this peptide is>

(a) 10.15>

(b) 6>

(d) 7.5>

Ans. (a)

Sol. pI = (9.8 + 10.5)/2 = 20.3/2 = 10.15.>

29. What is the fold difference between v at [S] =K_m and v at [S] = 1000K_m, where v is the initial velocity of an enzyme catalyzed reaction, [S] is substrate concentration and K_m is the Michaelis constant?>

(a) 1.998>

(b) 1000>

(d) 3.998>

Ans. (a)

Sol. >

30. Purine and pyrimidine nucleotides serve as monomeric units of the nucleic acid polymers DNA and RNA. Mentioned below are some of the statements with respect to the de novo synthesis of nucleotides. Which one of the following statements is INCORRECT?>

(a) Biosynthesis of both purine and pyrimidine nucleotides begin with ribose-5-phosphate and purine or pyrimidine rings are built on it.>

(b) The first purine nucleotide biosynthesized by de novo pathway is inosinic acid or inosine-monophosphate.>

(d) Thymidylate or TMP is synthesized as deoxy-TMP from deoxy-UMP by thymidylate synthetase.>

Ans. (a)

Sol. In de novo pathways, the framework for a pyrimidine base is assembled first and then attached to ribose. In contrast, the framework for a purine base is synthesized piece by piece directly onto a ribose ring. The purine ring is assembled from a velocity of precursors : glutamine (N3 and N9), glycine (C4, C5 and N7), aspartate (N1), N¹⁰-formyltetrahydrofolate (C2 to C8) and CO₂ (C6).>

31. Cyanide, a chemical warfare agent, is toxic because it:>

A. binds to the heme a3 in mitochondrial cytochrome c oxidase (in complex IV)>

B. inhibits electron transport and thus oxidative phosphorylation>

C. directly blocks mitochondrial DNA replication>

D. blocks the protein trafficking inside the mitochondria by affecting TIM and TOM channels>

Choose the combination with all correct statements.>

(a) A and C>

(b) A and B>

(d) A, B, and D>

Ans. (b)

Sol. Cyanide, azide and carbon monoxide bind with complex IV and inhibit the terminal transfer of electrons to oxygen. Cyanide and azide bind with the Fe³⁺ form (i.e., oxidized form) of the cytochrome a₃ and carbon monoxide binds with the Fe²⁺ form (i.e., reduced form) of the cytochrome a₃.>

32. Which one of the following statements about Short Interspersed Nuclear Elements (SINEs) is true?>

(a) SINEs represent a class of retrotransposons.>

(b) SINEs can transpose independently>

(d) SINEs are normally transcribed by RNA polymerase I.>

Ans. (a)

Sol. There are two classes of transposable elements – Class I (Retrotransposons) and Class II (Transposons). Class I transposons elements are of two types – with LTR (eg. Ty element in Drosophila, Copia in yeast) and without LTR retrotransposon. Non LTR retrotransposons are of two types – LINES and SINES. LINES are autonomous and retrotranspositionally active elements. SINES and 100-400 bp long non-autonomous sequences and thus require the enzymatic machinery of LINES for retro-transposition.>

33. Membrane-enclosed organelles often have a characteristic position in the cytosol. In animal cells, for example, the Golgi apparatus is located close to the nucleus. Which component is directly involved in ensuring correct Golgi localization in animal cells?>

(a) Actin cytoskeleton>

(b) Microtubules>

(d) Peroxisomes>

Ans. (b)

Sol. Microtubules present in all eukaryotic cells and perform diverse functions. They determine the positions of cell organelles within the cell, direct interacellular transport of transport vesicles and organelles, segregation of chromosomes during cell divison by forming mitotic spindle. They form the structural framework of flagella and cilia, which functions as motile structures. Non-motile cilia function as sensory and signaling devices on the surface of the cell. In plant cells, organized arrays of microtubules help to direct the pattern of cell wall synthesis and also determine the plane of formation of the cell wall during cytokinesis.>

34. Signal sequences direct proteins to the correct intracellular locations. Which one of the following sequences is typically used to import proteins into the nucleus?>

(a) -Pro-Pro-Lys-Lys-Lys-Arg-Lys-Val->

(b) -Leu-Ala-Leu-Lys-Leu-Ala-Gly-Leu-Asp-Ile->

(d) -Lys-Asp-Glu-Leu-COO->

Ans. (a)

Sol. A nuclear-localization signal (NLS) directs nuclear proteins from cytosol to the nucleus (import) selectively. In many nuclear proteins, the NLSs consist of one or two short non-cleavable sequnces that are rich in the positively charged basic amino acids lysine and arginine, with the precise sequence varying for different proteins. It was first identified.>

35. Which one of the following statements about stem cells is correct?>

(a) Stem cells cannot be maintained in culture since they require a distinct in vivo microenvironment known as niches.>

(b) During asymmetric stem cell division, only one of the daughter cells is retained as a stem cell.>

(d) Hematopoietic stem cells (HSCs) are totipotent stems cells.>

Ans. (b)

Sol. Stem cells are unspecialized (undifferentiated) cells that can differentiate into other cells and self-regenerate. Cell division in stem cells may be symmetric or asymmetric. In symmetric division, both daughter cells are identical as parent stem cells and both of the daughters are stem cells whereas in asymmetric divison one of two cells is identical to parent stem cell and the second differentiate into a more specialized cell.>

36. Following statements were made about stress response in prokaryotes:>

A. PerR functions as a major peroxide sensor in many Gram- positive bacteria.>

B. Extreme acidic pH induces RecA-mediated DNA damage, which in turn may induce virulence gene expression in some pathogenic bacteria.>

C. Induced expression of heat shock proteins neither protects the cells from heat nor plays a role in bacterial virulence.>

D. In Gram-negative bacteria, heat shock σ factor regulates the transcription of the major heat shock proteins.>

E. GroES is an ATP-dependent chaperonin but GroEL may function in ATP-independent manner.>

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

(a) A, C, D>

(b) A, B, D>

(d) B, D, E>

Ans. (b)

Sol. PerR functions as a peroxide responsive repressor and is a member of the Fur family of small, dimeric, metal-responsive transcriptional regulators (1). PerR is a global regulator that responds primarily to H2O2. It substitutes for OxyR in many Gram-positive bacteria. Heat-shock proteins are involved in several processes in bacterial cells, including assisting the folding of newly synthesized proteins, preventing aggregation of proteins under stress conditions and recovering proteins that have been partially or completely unfolded by stresses such as a sudden temperature increase. Statement A, B and D is true.>

37. Following statements were made about transposons:>

A. Transposons have inverted terminal repeats and their integration generates inverted repeats at the flanks of the target site in the host genome.>

B. A composite transposon can transpose as a unit.>

C. The transposition event may cause deletions or inversions or move a host sequence to a new location>

D. The transposition event may cause deletions or inversions but cannot move a host sequence to a new location.>

E. Replicative transposition proceeds through cointegration.>

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

(a) A, B, D>

(b) B, C, E>

(d) C and E only>

Ans. (b)

Sol. Transposons are found in both prokaryotes and eukaryotes. Transposons can move by the conservative or replicative process. A conservative (or cut and paste) transposition involves the excision of the sequence from its original position followed by its reinsertion elsewhere. This requires two elements : the transposase and the inverted repeats at the ends of the transposon. Some transposons are capable of replicative transposition (or copy and paste), during which the transposon creates a second copy of itself. During transposition, transposable elements generally produce short direct repeats at the site of insertion, which flanks the transposable element. Composite transposon consists of two IS elements surrounding a central block of protein-coding gene(s). A composite transposon can transpose as a unit.>

38. Following statements were made about cell cycle regulation:>

A. De novo synthesis and destruction of Cyclin B are essential for cell cycle progression in yeast.>

B. De novo synthesis and destruction of Cyclin B and the related Cyclin dependent Kinase (CDK) are essential for cell cycle progression.>

C. CDK activity is regulated by both activating and inhibitory phosphorylation.>

D. Retinoblastoma (Rb) functions as an inhibitor of G2 to M transition.>

E. Inactivation of Sic 1 is essential for transition into S phase.>

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

(a) A, C, D>

(b) B, C, D>

(d) B, C, E>

Ans. (c)

Sol. Cyclin B is not present in case of yeast. Hence, the first statement is not appropriate. During different phases of cell cycle, concentration of cyclins fluctuates but CDK concentration remains constant. Cyclin and CDK from the active Cyclin-CDK complex. Retinoblastoma (Rb) inhibits G1 to S phase transition. Sic 1 binds CDK1-Cib1 and prevents premature S-phase entry. Inhibition of Sic 1 will result into entry of cell into S-phase.>

39. Following statements were made about protein trafficking in cells:>

A. Cargo selection occurs when coat proteins bind to the sorting signals either directly or indirectly via adaptor complexes.>

B. Protein export from the ER is exclusively mediated through the COPII-coated vesicles.>

C. Identical coat protein is used in the exocytic pathway and/or endocytic pathway.>

D. Tethering of the vesicles involves small guanosine triphosphatases (GTPases) of the Rab family.>

E. Clathrin-coated vesicles transport proteins from the plasma membrane to the trans-Golgi network to late endosomes.>

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

(a) A, B, D>

(b) A, B, E>

(d) C, D, E>

Ans. (a)

Sol. >

The membrane tethering processes before membrane docking and fusion work in association with the Rab family of small monomeric GTPases. Cells contain many types of Rab proteins, each associated with the membrane of a particular organelle and transport vesicles.>

40. The following statements are made with reference to the structure of the nucleosome:>

A. The histone tetramer in the core of the nucleosome comprises H2A, H2B, H3 and H4.>

B. The N-terminal tails of the core histones are believed to stabilize the 30 nm fiber of nucleosomal DNA by their interactions with adjacent nucleosomes.>

C. The post-translational modifications of the N-terminal tails as well as globular domains of the core histones modulate transcriptional events>

D. According to the zigzag model of the 30 nm fiber the linker DNA circles around the central axis of the fiber as the DNA moves from one nucleosome to the next.>

Which of the following combinations represents all correct statements?>

(a) B and C only>

(b) A, B and C>

(d) B and D only>

Ans. (a)

Sol. The nucleosome core particles contains a histone octamer that consists of two copies each of H2A, H2B, H3 and H4 ([H3-H4]2 tetramer flanked by two H2A-H2B dimers). The N-terminal tails of the histones are frequently modified. The post-translational modifications of the histones N-terminal tails alters the function of chromatin. The second level of organization is the coiling of the 11 nm fiber of nucleosomes into a fiber of ~30 nm diameter. The structure of this fiber requires the histone tails and is stabilized by linker histones.>

41. A mutation in which one of the following sigma factors may be a possible cause for E. coli failing to adapt in response to thermal stress?>

(a) >

(b) >

(d) >

Ans. (b)

Sol. The heat-shock response (HSR), a universal cellular response to heat, is crucial for cellular adaptation. In Escherichia coli, the HSR is mediated by the alternative sigma factor, sigma32.>

42. Which one of the following statements about Cre-mediated site-specific recombination at loxP sites is incorrect?>

(a) When loxP sites flanking a test sequence are oriented in same direction, Cre mediates the excision of the intervening sequence>

(b) LoxP sites in inverted orientation around an intervening sequence lead to inversion upon action by Cre recombinase>

(d) LoxP-Cre system cannot be used to generate translocation between chromosomes.>

Ans. (d)

Sol. Cre is a 38 kDa tyrosine recombinase protein (belongs to 'integrase' family) from bacteriophage P1 which mediates site specific recombination between loxP sites. A loxP site (34 bp long) consists of two 13 bp inverted repeats separated by an 8 bp spacer or core region where recombination takes place. The natural function of the Cre recombinase is to mediate recombination between loxP sites.>

43. Given below are a set of enzymes in Column A and enzyme activities in Column B.>

Choose the option that matches the contents of column A with that of column B.>

(a) A: iv; B: iii; C: ii; D: i,>

(b) A: iii; B: iv; C: ii; D: i,>

(d) A: iii; B: iii, iv; C: ii; D: i>

Ans. (a)

Sol. A. DNA topoisomerase I : single strand nicking>

B. DNA topoisomerase II : double strand break and ligation>

C. Polymerase : synthesis of Okazaki fragments>

D. Polymerase : leading strand synthesis>

44. Sequencing of the human genome has identified 67 putative gene encoding K⁺ channels. To characterize the function of one of these genes, a researcher would typically carry out the following experiments:>

A. in vitro transcription of the cloned cDNA for this gene in a cell-free system to produce the corresponding mRNA>

B. Injecting above mentioned mRNA into frog oocytes>

C. Measuring channel-protein activity using a patch-clamping technique>

D. Knocking out the gene encoding the homolog of this channel in frog oocytes and measuring channel activity using a patch-clamping technique>

Choose the correct combination of experiments that would help in the characterization of the K⁺ channel.>

(a) A, B, and C>

(b) B, C, and D>

(d) D, A and B>

Ans. (a)

Sol. The shotgun phase of the Human Genome Project itself consisted of three steps:>

Obtaining a DNA clone to sequence>

Sequencing the DNA clone>

Assembling sequence data from multiple clones to determine overlap and establish a contiguous sequence.>

Patch-clamp is used to evaluate current or voltage in the membrane associated with ion channel activity via direct measurement in real time using ultra-sensitive amplifiers, high-quality data acquisition systems, and powerful software to evaluate the results.>

45. The following observations are being made in the context of the regulation of iron homeostasis in mammalian cells.>

A. The levels of transferrin receptor mRNA increase 30-fold in the absence of iron>

B. Certain mutations in the 3'UTR of transferrin receptor mRNA fail to support increase in mRNA levels under low iron concentrations>

C. The level of transferrin receptor mRNA continues to be high in low iron concentration even when the cells are treated with a-amanitin>

D. The transferrin receptor mRNA levels decrease rapidly when iron is added to the cells.>

Based on these observations, which one of the following statements represents the most likely scenario?>

(a) Transferrin receptor is only transcriptionally regulated, and the 3' end of the gene is the regulatory site>

(b) Transferrin receptor is post-transcriptionally regulated, and the 3'UTR is the regulatory site>

(c) Iron induces expression of transferrin receptor, and the transferrin receptor protein is degraded in the presence of iron>

(d) Regulation of transferrin receptor is not sensitive to iron>

Ans. (b)

Sol. Synthesis of protein transferrin receptor responsible for the transport of protein transferrin is regulated post-transcriptionally by iron regulatory protein binding to an iron-responsive element present at the 3'-UTR of the mRNA.>

46. A heterozygote of E. coli was produced with the following combination of mutations:>

trpR⁺trpO^–trpE⁺/trp^R+trp^O+trpE^–>

where R is the repressor, O is the operator and trpE encodes the first enzyme in the biosynthetic cascade for tryptophan. Assume all other enzymes required are wild type. Which one of the following is the most likely phenotype of this E. coli ?>

(a) Synthesizes tryptophan irrespective of tryptophan status in the medium>

(b) Synthesizes tryptophan only when tryptophan is absent>

(d) Cannot synthesize tryptophan under any condition>

Ans. (a)

Sol. Trp operon is a repressible operon. Tryptophan acts as the negative regulator of the trp operon. In the absence of Trp, an aporepressor called trpR does not bind to operator sequence. RNA pol binds to the operator and forms full length polycistronic mRNA. In the presence of Trp, trp is not synthesized by E. coli. Aporepressor binds the operator and stops the formation of transcript.>

47. Precise recognition of tRNAs by their cognate aminoacyl-tRNA synthetases is crucial for the fidelity of protein synthesis. In the context of the aminoacylation of tRNAAla with its cognate aminoacyl-tRNA synthetase (AlaRS) and based on the studies on the molecules of Escherichia coli origin, following statements are made. Which one of the statements is INCORRECT?>

(a) Anticodon of tRNAAla makes important contribution to the specificity of its aminoacylation by AlaRS AlaRS>

(b) Mutational analyses have shown that for aminoacylation of the tRNAAla by AlaRS, the presence of a wobble pair in the acceptor stem (G3:U70) is the most crucial element.>

(d) A microhelix lacking a clover leaf structure and harboring only the acceptor stem sequence of the tRNAAla is specifically aminoacylated by AlaRS.>

Ans. (a)

Sol. In general, aminoacyl-tRNA synthetases recognize the anticodon loops and acceptor arms of tRNA molecules. Because some amino acids are decoded by as many as six codons, some tRNAs that encode for the same amino acid may not share any nucleotide of the anticodon, which renders recognition difficult for the synthetase. For example, in the case of serine-specific aminoacyl-tRNA synthetase, the long variable arm of tRNA^Ser (not anticodon loop) functions as an important discriminatory element. Similarly, in the case of tRNA^Ala, recognition is exclusively based on a G3-U70 base pair.>

48. In the figure above, replication of DNA beginning from the origin of replication of the chromosome of a newly identified bacterium having a double stranded circular DNA genome is shown. Characterization of DNA polymerase responsible for genome replication showed that DNA synthesis occurred in 5' to 3' direction and it depends on the presence of a primer (as is the case in Escherichia coli). Polarities of DNA (5' or 3') are as shown. Replication begins at a point marked 'o' on the left of the bubble, and both the parent strands were replicated concurrently. The longer arrow inside the bubble shows the leading strand, whereas the shorter arrows (marked a, b, c) show the Okazaki fragments. The model depicts a:>

(a) bidirectional mode of replication wherein synthesis of the Okazaki fragment marked 'c' occurs prior to those marked 'a' and 'b'>

(b) bidirectional mode of replication wherein synthesis of the Okazaki fragment marked'a' occursc prior to those marked 'b' and 'c'>

(c) unidirectional mode of replication wherein synthesis of the Okazaki fragment marked 'c' occurs prior to those marked 'a' and 'b'.>

(d) unidirectional mode of replication wherein synthesis of the Okazaki fragment marked 'a' occurs prior to those marked 'b' and 'c'.>

Ans. (c)

Sol. As compared to the leading strand, lagging strand is synthesized discontinuously from multiple primers (semi-discontinuous replication). Short pieces of DNA synthesized repeatedly on the lagging-strand template are termed as Okazaki fragments.>

49. Which one of the following statements is NOT a correct feature of Escherichia coli RNA polymerase?>

(a) Presence of the subunit along with core RNA polymerase is required for its promoter-specific binding.>

(b) Presence of the subunit along with core RNA polymerase is not necessary for the core RNA polymerase to bind to the DNA template.>

(d) Mutations in subunit in its rifampicin resistance defining region (RRDR) confer rifampicin resistance phenotype>

Ans. (d)

Sol. In eubacteria, single type of RNA polymerase appears to be responsible for the synthesis of all different types of RNA such as mRNA, rRNA and tRNA. RNA polymerase in E. coli is a multisubunit enzyme made up of five different polypeptides – The holoenzyme can be separated into two components, the core enzyme and the sigma factor (the -factor). Rifampicin is a semisynthetic derivative of rifamycins. Rifampicin binds in a pocket of the -subunit and blocks the path of the elongating RNA.>

50. If TLR2 is knocked out from human monocyte-derived macrophages, PAMP recognition by which one of the following TLRs will most probably get affected?>

(a) TLR9>

(b) TLR3>

(d) TLR5>

Ans. (c)

Sol. TLR2 forms heterodimers with TLR1 and TLR6, which is the initial step in a cascade of events leading to significant innate immune responses, development of adaptive immunity to pathogens and protection from immune sequelae related to infection with these pathogens.>

51. >

With reference to the signaling pathway shown above, which one of the following options correctly identifies the intracellular components?>

(a) a=Grb2; b=SOS; c=Ras-GTP; d=ERK,>

(b) a=Grb2; b=Ras-GDP; c=MEK; d=ERK,>

(d) a=RTK; b=SOS; c=Ras-GDP; d=ERK,>

Ans. (a)

Sol. >

52. Nitric oxide (NO) acts as intracellular second messenger by stimulating>

(a) Phosphodiesterase>

(b) Nitric oxide synthase>

(d) Guanylyl cyclase>

Ans. (d)

Sol. The neurotransmitter acetylcholine stimulates NO synthesis by activating a GPCR on the membranes of the endothelial cells that line the interior of the vessel. The activated receptor triggers IP3 synthesis and Ca²⁺ release, leading to stimulation of enzyme nitric oxide synthase that synthesizes NO. NO then diffuses to neighboring smooth muscle cells in the walls of blood vessels where it interacts with iron bound to the active site of the enzyme guanylyl cyclase. This increases enzymatic activity, resulting in the synthesis of the second messenger cGMP, which in turn activates cGMP-dependent PKG and/or other effector proteins. PKG (protein kinase G) is a serine/threonine-specific protein kinase. In this signaling process, guanylyl cyclase acts both as an intracellular receptor for NO and as an intracellular signaling protein.>

53. Which one of the following pathogens does not have the ability to survive within macrophages?>

(a) Schistosoma mansoni>

(b) Mycobacterium tuberculosis>

(d) Leishmania donovani>

Ans. (a)

Sol. While calpains are invariably reported to be exclusively intracellular (except in diseased or injured tissues), our data show that schistosomes display unique, constitutive, functional extracellular calpain activity. However, the adult schistosomes are capable of avoiding the immune recognition system by coating their outer tegument with antigens from the hosts. Several studies have shown that the adult Schistosoma parasites were covered with immunoglobulins, ß2 microglobulin, complement components, and other host antigens.>

54. Bone marrow cells were sorted using 4 cell-surface markers, CD45R (B220), CD43 (leukosialin), membrane-associated Immunoglobulin M (mIgM) and membrane-associated Immunoglobulin D (mIgD). The sorted cells were then analyzed for immunoglobulin (Ig) gene rearrangements for the heavy chain. With this information, the sorted cells were categorized into different stages of B-cell development. The observations and the inferences drawn are given below.>

Which of the above inferences are correct?>

(a) B and D only>

(b) A, B, C and E only>

(d) B, D, and E only>

Ans. (c)

Sol. A, C and E inferences are correct.>

55. Virus infected cells are cleared by cytolysis by natural killer (NK) cells and CD8+ Cytotoxic T lymphocytes (CTLs). Which of the following graphs correctly represents the changes in the cell numbers of the two cell types during the course of a viral infection, considering that the virus is cleared by 14th day?>

(a) >

(b) >

(d) >

Ans. (a)

Sol. When a virus infects cells, innate immunity is first activated. Natural killer cells will respond against viruses. It will decrease viral infection by cytolysis of virus infected cells. This response is transient. After some days acquired immunity will come in action and virus specific CD8⁺ and CTL cells will be produced. These cells will lower down further infection of viruses.>

56. The gradient of oxygen available to cells at inner regions of a tumour tissue environment is typically low that creates a hypoxic micro-environment. If enough oxygen is supplied to the cancer cells residing in hypoxic micro-environment, which one of the following processes may NOT occur?>

(a) HIF- stable in the cells under hypoxic conditions, may undergo oxygen-dependent hydroxylation, targeting it for ubiquitination and proteolysis by tumour-suppressor protein VHL>

(b) Warburg effect will be reversed and conversion of glucose to lactate will not take place as sufficient oxygen will be available for oxidative phosphorylation>

(d) Lactate generation in the tumour microenvironment, which contributed to M2 polarization of tissue-associated macrophages, will continue.>

Ans. (b)

Sol. Most cancer cells are characterized by increased glucose uptake and aerobic glycolysis. A very high rae of glycolysis followed by lactic acid fermentation, even in the presence of oxygen and fully functioning mitochondria is known as the aerobic glycolysis (or 'Warburg effect'). Increased aerobic glycolysis is uniquely observed in cancer cells. In low oxygen concentrations, cells adjust their oxygen consumption and switch to anaerobic glycolysis to prevent oxygen depletion. Glycolysis is a less efficient process of generating ATP but is less oxygen demanding and thus protects cells from total anoxia and cell death. HIF- and HIF-. Both the HIF- and HIF- subunits are expressed constitutively, but only HIF- is affected by oxygen levels. Under normoxia, HIF- is rapidly degraded.>

57. Column "X" represents a list of different viruses and column "Y" represents the mechanisms generally adopted by viruses to evade host defense.>

Identify the correct match.>

(a) A – i; B – ii; C – iii; D – iv>

(b) A – ii; B – iii; C – iv>

(d) A – iv; B – i; C – ii; D – iii>

Ans. (d)

Sol. A. Hepatitis C : By evading the action of the major antiviral cytokine>

B. Herpes simplex virus (HSV) : By inhibiting antigen delivery to class 1 MHC receptors on virus-infected cells plus preventing presentation of viral antigens to CD⁸⁺ T cells.>

C. Vaccinia virus : By evading antibody-mediated destruction through complement activation>

D. Human immunodeficiency virus (HIV) : By constantly changing their antigens called antigenic variations>

58. Although a majority of G protein-coupled receptors (GPCRs) act through adenylyl cyclase, many of them (GPCR) exert their effects by activating the plasma membrane-bound enzyme phospholipase C- (PLC-). These receptors activate the so-called inositol phospholipid signaling pathway mainly via a G protein called G_q, which activates PLC- in much the same way that Gs activates adenylyl cyclase. Mentioned below are some of the steps with functional characteristics of inositol phospholipid signaling pathway, one of which is not correct. Select the INCORRECT option.>

(a) PLC- acts on phosphatidyl inositol 4,5-biphosphate (PIP₂), which is present in the inner half of the plasma membrane lipid bilayer. PLC->

(b) Activated PLC- cleaves the PIP₂ to generate two products: inositol 1,4,5-triphosphate (IP₃) and diacylglycerol (DAG)>

(c) IP3 is a small lipid-soluble molecule that binds to IP₃ receptor on plasma membrane thereby increasing Ca²⁺ concentration in the cytosol>

(d) DAG gets further cleaved to release arachidonic acid, which is used in the synthesis of eicosanoids including prostaglandins. DAG also activates Ca²⁺ dependent protein kinase C or PKC>

Ans. (c)

Sol. Inositol 1,4,5-trisphosphate (IP3), a second messenger formed in response to stimulation of cell surface receptors, binds to and activates IP3 receptors (IP3Rs, internal calcium release channels) to liberate calcium from intracellular stores. Both DAG and IP3 act as important second messengers. DAG remains in the membrane where it recruits and activates protein kinase C. IP3 stimulates the opening of IP3-mediated Ca²⁺ channels on intracellular organelles that store Ca²⁺ such as the endoplasmic reticulum.>

59. The embryonic stem cells in mammals are derived from:>

(a) Blastocoel>

(b) Inner cell mass>

(d) Trophoendoderm>

Ans. (b)

Sol. Embryonic stem cells (ESCs) are found in the inner cell mass of the human blastocyst, an early stage of the developing embryo lasting from the 4th to 7th day after fertilization.>

60. Which one of the following transcription factors is important for delimiting the meristematic and elongation zones of roots?>

(a) SCARECROW (SCR)>

(b) SHORT ROOT (SHR)>

(d) SPEECHLESS (SPCH)>

Ans. (c)

Sol. PLETHORA (PLT) transcription factor gradients are unique in their ability to guide the progression of cell differentiation at different positions in the growing Arabidopsis thaliana root, which contrasts with well-described transcription factor gradients in animals specifying distinct cell fates within an essentially static context.>

61. Movement of epithelial sheet spreading as a unit to enclose deeper layers of the embryo is termed as>

(a) Epiboly>

(b) Emboly>

(d) Ingression>

Ans. (a)

Sol. Epiboly describes one of the five major types of cell movements that occur in the gastrulation stage of embryonic development of some organisms. Epiboly is the spreading and thinning of the ectoderm while the endoderm and mesoderm layers move to the inside of the embryo.>

62. What would be the effect of retinoic acid (RA) treatment on the 'positional information' of blastema cells present on the amputated newt limb?>

(a) RA will have no effect>

(b) The cells will become respecified to more proximal position>

(d) Cells will lose their positional information and remain as dedifferentiated cells>

Ans. (b)

Sol. Retinoic acid administration changes the PI in a proximal direction so that a complete limb can be regenerated from a hand. Rather than identifying all the genes altered by RA treatment of the limb, we have eliminated many off-target effects by using retinoic acid receptor selective agonists.>

63. In an experiment, activin-secreting beads were placed on unspecified cells from an early Xenopus embryo. The activin then diffused from the beads. If the beads contained 1nM of activin, it elicited expression of Xbra gene in cells near to the beads. If the beads contained 4 nM activin, the expression ofXbra was elicited in cells, but only at a distance of several cell diameters away from the beads. In the latter case, expression of goosecoid gene was observed near the source bead. Beads with no activin did not elicit the expression of the two genes.>

Following statements were made regarding the observations and the role of activin in determining cell fate.>

A. High concentration of activin activates goosecoid, whereas lower concentrations activate Xbra.>

B. Lower concentrations of activin help specify the dorsal-most structures of the frog's embryo>

C. Concentrations of activin that do not lead to expression of the two genes specifies the cell to become blood vessels and heart>

Which of the above statement(s) are correct?>

(a) A only>

(b) C only>

(d) B and C>

Ans. (c)

Sol. Activin acts as a morphogen. cell receiving highest concentration of activin will express gossecoid gene. Cells receiving moderate concentration will express Xbra gene and the cell receiving lowest concentration will express no gene (Default pathway).>

64. The specification of sea urchin micromeres involves the activation of a repressor protein Pmar1, which represses the expression of hesC, which also encodes a repressor protein. One of the genes controlled by HesC is Delta, whose expression is used as a marker for micromere lineage. The image below represents a sea urchin embryo on which whole mount in situ hybridization (WMISH) was performed using delta probe, indicated by the area 'A'. The rest of the embryo is labeled 'B'.>

(Image from Revilla-i-Domingo et al (2007), PNAS 104: 12383-12386)>

The table below summarizes a set of experiments (column A) and the area in which hybridization is observed (column B)>

Which one of the following options is a correct match between columns A and B?>

(a) A - ii; B – iii; C – iii>

(b) A – ii; B – iii; C –iv>

(d) A – i; B – ii; C - iii>

Ans. (a)

Sol. Correct option is (a)

65. Induction is an extrinsic process that depends on the position of a cell in the embryo. It is a process whereby one cell or group of cells can influence the developmental fate of another, and is a common strategy to control differentiation and pattern formation in development. The following statements were made regarding induction in a developing embryo.>

A. The inductive signal can be a protein secreted from the inducing cells that binds to receptors of a responding cell.>

B. Response to inductive signals depends on competence of the inducing cell.>

C. Instructive induction occurs when the responding cell is already committed to a certain fate.>

D. Lateral inhibition is an induction that results in differentiation of individual cells in a regularly spaced pattern.>

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

(a) A and C>

(b) B and D>

(d) B and C>

Ans. (c)

Sol. The cell-cell interactions (or communication) between two or more groups of cells of different history and properties is called induction. This is a process by which a cell or group of cells instructs neighboring cells to adopt a particular fate. There are at least two components to every induction. The first component is the inducer : the tissue that produces a signal (or signals) that changes the cellular behavior of the other tissues. The second component, the tissue being induced, is the responder. Not all tissues can respond to the signal being produced by the inducer. The response to inductive signals depends on the ability of the cell to receive the signal and react in an appropriate manner. This ability is called competence.>

There are two major modes of induction – instructive and permissive. In instructive induction, the responding cell has a choice of fates and will follow one developmental pathway following induction and an alternative pathway in absence of the induction. A signal from the inducing cell is necessary for the responding cell. Without the inducing cell, the responding cell would not be capable of differentiating in that particular way. In permissive induction, the responding tissue has already been specified and needs only an environment that allows the expression of these traits. Here, the responding cells is already committed to a certain fate.>

66. Following are certain statements related to seed maturation:>

A. Seed maturation involves mainly the accumulation of storage products, such as starch, lipids and proteins.>

B. A large number of chaperones including the family of LATE EMBRYO ABUNDANT (LEA) proteins, play a crucial role in the dessication process of seeds>

C. Moisture content gets reduced with the maturation of seed>

D. ABA and gibberellin both promote seed dormancy>

Which one of the following options has all correct statements?>

(a) A and C only>

(b) B and D only>

(d) A, B and C only>

Ans. (d)

Sol. The ABA/GA ratio determines whether the seed germinates or remains quiescent. These hormones act in an antagonistic manner : ABA is required for dormancy maintenance while GA triggers dormancy release. Although ABA and GA play the main role, auxin, cytokinins and jasmonate have been shown to partly contribute to seed germination.>

67. Given below are statements related to expression of some of the pattern-forming genes in vertebrate limb bud:>

A. Lmx1b gene is expressed in dorsal mesenchyme>

B. Shh is expressed in the posterior region.>

C. Wnt7a gene is expressed in dorsal ectoderm.>

D. Hoxa13 and Hoxd13 are expressed in the distal region.>

E. Tbx5 and FGF10 are expressed in the lateral plate mesoderm involved in formation of limb bud.>

The Nail-Patella syndrome in human and the syndrome in mouse exhibiting footpads on both dorsal and ventral surfaces of limb are associated with which of the above mentioned gene functions?>

(a) B, D and E>

(b) B and D only>

(d) A and C only>

Ans. (d)

Sol. Lmx1b was diffusely expressed in the undifferentiated dorsal mesoderm of the emerging limb bud. Wnt7a is first expressed in the dorsal flank ectoderm between stages 14 and 16 and, more intensely, in the dorsal limb bud ectoderm commencing at the same stages. Lmxl expression begins between stages 14 and 16 in the dorsal limb bud mesenchyme.>

68. The early embryonic development in amphibians and aves serve as two different model plans of development. In the former the germ layer formation is initiated from a fluid-filled ball like blastula, while in the latter the germ layer formation is initiated on a flat blastodisc. Given below are some of the terms for amphibian embryo in column I and from avian embryo in column II:>

Which of the following is the all correct match of the terms in column I with that of column II?>

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

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

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

Ans. (b)

Sol. P. Blastocoel : Between epiblast and hypoblast>

Q. Blastopore lip : Primitive streak>

R. Dorsal lip of blastopore : Hensen's node>

S. Blastopore : Primitive groove>

T. Nieuwkoop center : Posterior Marginal Zone (PMZ)>

69. -thioglucosidases, also known as myrosinases, are the enzymes that are known to hydrolyse which one of the following plant natural products?>

(a) Glucosinolates>

(b) Terpenoids>

(d) Phenolics>

Ans. (a)

Sol. Glucosinolates are hydrolyzed by thioglucosidases called myrosinases to isothiocyanates, thiocyanates, nitriles, epithionitriles etc., which are bioactive compounds .>

70. The high affinity ammonium uptake system in plant's roots involves transporters in the AMT/Rh family. Which of the following AMT genes is expressed in cortex and endodermis?>

(a) AMT1.1>

(b) AMT1.2>

(d) AMT1.5>

Ans. (b)

Sol. AMT1.2 encodes an ammonium transporter protein believed to act as a high affinity transporter. It is expressed in the root, primarily in endodermal and cortical cells, and contributes to ammonium uptake in the root.>

71. Which one of the following are the correct encoding sites of large and small subunits of Rubisco enzyme in red and brown algae?>

(a) Large subunit in chloroplast and small subunit in nucleus.>

(b) Large subunit in nucleus and small subunit in chloroplast.>

(d) Both large and small subunits in chloroplast.>

Ans. (d)

Sol. Plant RuBisCo is a multisubunit enzyme composed of eight large subunits and eight small subunits. Large subunits are encoded by the chloroplast genome whereas smaller subunits are encoded by nuclear genome. However, in non-green algae i.e, red and brown algae both subunits of RuBisCo come from chloroplast.>

72. Which one of the following correctly states the action of sucrose phosphate synthase enzyme?>

(a) UDP-glucose and Fructose-6-phosphate are used as substrates.>

(b) UDP-glucose and Fructose-6-phosphate are the products.>

(d) Sucrose-6-phosphate and UDP-glucose are the products>

Ans. (a)

Sol. The synthesis of sucrose, a disaccharide of glucose and fructose plac in the cytosol of the plant cells. Sucrose is synthesized in the cytosol from triose phosphates. As in starch synthesis, triose phosphate is converted into glucose 1-phosphate. Glucose-1-phosphate is activated to UDP-glucose. This reaction is catalyzed by the enzyme UDP-glucose pyrophosphorylase. Enzyme sucrose phosphate synthase catalyzes the transfer of the glucose residue from UDP-glucose to fructose 6-phosphate forming sucrose 6-phosphate. Finally, the sucrose 6-phosphate cleaves the phosphate from sucrose 6-phosphate, yielding sucrose.>

73. The following statements refer to photosystem structure and function involved in light-dependent reaction of photosynthesis:>

A. The antenna or light harvesting complex absorbs light energy and transfers it to the reaction centre.>

B. The first electron is released from P680 and transferred to QA to produce a semiquinone >

C. D1, a protein subunit of the plant core complex is encoded by gene psbD>

Which one of the following combinations of above statements is INCORRECT?>

(a) A and B>

(b) B and C>

(d) Only C>

Ans. (b)

Sol. In light dependent reaction of photosystem during photosynthesis, The first electron is released from P680 and transferred to QA to produce a semiquinone Q_B^– and D1, a protein subunit of the plant PSπ core complex is encoded by gene psbD.>

74. Following are certain statements regarding photorespiration pathway in plants:>

A. The first two-carbon (2C) compound synthesized by the action of Rubisco in the chloroplast is glycolate.>

B. Glycolate exits the chloroplast and enters peroxisomes.>

C. Glycolate that is synthesized during C₂ cycle enters the chloroplast from mitochondria.>

D. Glycine is transported from peroxisomes to mitochondria.>

Which one of the following combinations is INCORRECT?>

(a) A and B only>

(b) A and C only>

(d) A, C and D>

Ans. (b)

Sol. >

Statements A and C is false.>

75. Plant pathogens produce effector molecules that aid in colonization of their host cells. Column X denotes the name of effector molecules and Column Y denotes the potential functions:>

Which one of the following is the correct match:>

(a) A – ii, B – iii, C – iv, D – i>

(b) A – iii, B – iv, C – ii, D – i>

(d) A – iv, B – ii, C – i, D – iii>

Ans. (b)

Sol. A. HC-toxin : Inhibits Histone deacetylases>

B. Fusicoccin : Activates H⁺-ATPase>

C. GA₃ : Accelerates growth>

D. TAL : Activates specific host gene expression>

76. Plants are known to synthesize more than 30,000 terpenoids, involving four stages of biosynthesis. Following are the list of biosynthetic steps (Column X) and the key class of enzymes involved (Column Y):>

Which of the following is the correct match?>

(a) A – i, B – iii, C – iv, D – ii>

(b) A – ii, B – iii, C – i, D – iv>

(d) A – iii, B – iv, C – i, D – ii>

Ans. (d)

Sol. A. Biosynthesis of two basic five-carbon unit : HMG-CoA synthase>

B. Repetitive additions of C₅ units : Prenyltransferases>

C. Formation of the basic terpenoid skeletons : Terpene synthases>

D. Modification of terpenoid skeletons : P450 monoxygenases>

77. Following statements are made about nitrate transporters in plant cells:>

A. Nitrate uptake displays two saturable phases, with Km in the micromolar (µM) range for the high-affinity system and in the millimolar (mM) range for the low-affinity system.>

B. The Arabidopsis AtNRT1·1 is a dual affinity nitrate transporter.>

C. AtNRT 1·2 participates in high-affinity uptake.>

D. AtNRT 2·1 and AtNRT 2·2 are involved in low-affinity uptake.>

Select the option that has the combination of all correct statements.>

(a) A and D only>

(b) B and D only>

(d) A, C and D>

Ans. (c)

Sol. The AtNRT1. 1 (CHL1) transporter provides a primary mechanism for nitrate uptake in Arabidopsis and is expected to localize to the epidermis and cortex of the mature root, where the bulk of nitrate uptake occurs. Statement A and B is true.>

78. Glycophytes are salt-sensitive plants while halophytes are salt-tolerant plants. The following statements were made to explain the difference between the glycophytes and halophytes.>

A. Glycophytes enhance the uptake of ions.>

B. Glycophytes actively pump ions back into the soil.>

C. Halophytes have ability to resist net ion uptake in the shoot.>

D. Halophytes have greater capacity for vacuolar sequestration of ions.>

Select the option that has the combination of all correct statements.>

(a) A, B and C A>

(b) A and D only>

(d) B, C and D>

Ans. (d)

Sol. Halophytes employ effectively salt-tolerance mechanisms to avoid salt damage, whereas glycophytes are noneffective under salt stress conditions due to limited salt-tolerance mechanisms.>

79. The time taken by synaptic vesicles to travel from the soma of a motor neuron in the spinal cord to its neuromuscular junction in a person's foot by fast axon transport is about->

(a) 5-10 seconds>

(b) 10-15 minutes>

(d) 2-3 days>

Ans. (d)

Sol. Distance from the soma of a motor neuron in the spinal cord to its neuromuscular junction in a person's foot = 1 m = 100 cm. The fast transport of vesicles reaches speeds of between 5 and 40 centimeters per day, the slower movement of soluble proteins averages only about 0.8 cm per day.>

If we take maximum fast transport, then time take by synaptic vesicles will be 100/40 = 2.5 days.>

80. The release of which neurotransmitter from the rods of retina is reduced when light strikes its outer segment?>

(a) Glutamate>

(b) Acetylcholine>

(d) Glycine>

Ans. (a)

Sol. Glutamate is the neurotransmitter of the neurons of the vertical pathways through the retina. All photoreceptor types, rods and cones, use the excitatory amino acid glutamate to transmit signals to the next order neuron in the chain.>

81. The principal product of fat digestion by pancreatic lipase is the free fatty acids (FFAs) and which one of the following?>

(a) 3- monoacylglycerols>

(b) 2- monoacylglycerols>

(d) 1, 3- diacylglycerols>

Ans. (b)

Sol. Pancreatic lipase exhibits optimal activity under alkaline conditions and hydrolyzes triglycerides to fatty acids and glycerol, but mono- and diglycerides are also end products.>

82. Which one of the following tissues normally DOES NOT produce ghrelin that stimulates food intake?>

(a) Stomach>

(b) Pancreas>

(d) Adrenal>

Ans. (d)

Sol. Secreted predominantly from the stomach, ghrelin is the natural ligand for the growth hormone secretagogue receptor in the pituitary gland, thus fulfilling criteria of a brain-gut peptide. Ghrelin cells are found mainly in the stomach and duodenum, but also in the jejunum, lungs, pancreatic islets, gonads, adrenal cortex, placenta, and kidney. It has also been shown that ghrelin is produced locally in the brain.>

83. Human chorionic gonadotropin (hCG) is a placental gonadotropin that controls hormonal secretions from corpus luteum during initial stage of pregnancy. Following statements are made about hCG:>

A. It is a glycoprotein that contains galactose and hexosamine.>

B. It is a heterodimer with a larger alpha subunit and smaller beta subunit.>

C. It is a heterodimer with a smaller alpha subunit and larger beta subunit.>

D. hCG is identical to beta subunit of LH and FSH.>

E. It appears as early as 6 days after conception in blood and 14 days after conception in urine.>

Which one of the following has all correct combination of statements?>

(a) A, B and D>

(b) A, C and E>

(d) A, C and D>

Ans. (b)

Sol. A glycoprotein composed of two subunits, the alpha (small) and beta (large) subunits. HCG is primarily catabolized by the liver, although about 20% is excreted in the urine. The beta subunit is degraded in the kidney. HCG maintains the corpus luteum for continued secretion of progesteron so as to maintain the pregnancy.>

84. The amount of hemoglobin in blood is one of the important health markers. Following statements are made regarding hemoglobin degradation when older red blood cells (RBCs) are destroyed by tissue macrophages.>

A. The globin protein of the hemoglobin is split off and heme is converted first to bilirubin by the action of heme oxygenase.>

B. The globin protein of the hemoglobin is split off and heme is converted first to biliverdin by the action of heme oxygenase.>

C. Carbon monoxide (CO) is formed in the process.>

D. Nitric oxide (NO) is formed in the process.>

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

(a) A and C>

(b) B and C>

(d) B and D>

Ans. (b)

Sol. Heme Biliverdin Unconjugated bilirubin Conjugated bilirubin Urobilinogen Stercobilin>

85. Catecholamines (i.e. dopamine, norepinephrine and epinephrine) are important adrenal medullary hormones which play a role in the response for emergency situations. The following statements are made with reference to this:>

A. Catecholamines can easily cross the blood-brain barrier.>

B. Dopa decarboxylase (DD) is a soluble enzyme that converts L- dopa to dopamine.>

C. Dopamine -hydroxylase (DBH) is a particulate enzyme carrying copper in its active site and it converts dopamine to epinephrine.>

D. Phenylethanolamine-N-methyltransferase (PNMT) is a soluble enzyme that is induced by glucocorticoids.>

Which one of the following has the correct combination of statements?>

(a) A and B>

(b) B and D>

(d) A and C>

Ans. (b)

Sol. Catecholamines (i.e. dopamine, norepinephrine and epinephrine) are important adrenal medullary hormones which play a role in the response for emergency situations. Dopa decarboxylase (DD) is a soluble enzyme that converts L- dopa to dopamine and phenylethanolamine-N-methyltransferase (PNMT) is a soluble enzyme that is induced by glucocorticoids.>

86. The following statements were proposed by a researcher on the characteristic features of stretch receptors in atria and the effect of these receptors' activity on blood pressure regulation:>

A. The activity of type A receptors are increased by burst of impulses during atrial systole>

B. The activity of type B receptors are increased by burst of impulses at the time of peak atrial filling during late diastole of atria>

C. The discharge of type B atrial receptors is increased when venous return is increased>

D. The activity of type B atrial receptors is increased by positive pressure breathing>

E. The increased activity of most of the atrial receptors initiates reflex circulatory adjustment by increasing blood pressure>

F. The heart rate is decreased reflexly by the increased activity of atrial receptors>

Choose all CORRECT statements from the following options:>

(a) A, B and C>

(b) B, C and D>

(d) D, E and F>

Ans. (c)

Sol. Stretch receptor in Atria directly affect blood pressure. The discharge of type B atrial receptors is increased when venous return is increased, the activity of type B atrial receptors is increased by positive pressure breathing and the increased activity of most of the atrial receptors initiates reflex circulatory adjustment by increasing blood pressure>

87. The explanations for increased conduction velocity of action potentials in a myelinated nerve fibre (MNF) as compared to that of a non-myelinated fibre (NNF) of same diameter are suggested below:>

A. Much higher number of ions traverse a unit length of an MNF membrane compared to that of an NNF during conduction of action potential as the ionic currents are restricted to the membrane at the Nodes of Ranvier>

B. The energy expenditure required to maintain ionic gradient after conduction of action potential in an MNF is higher than that of an NNF>

C. The action potentials of MNF do not have a hyperpolarizing effect like that of an NNF>

D. The relative refractory period is not extended in an MNF like that of an NNF>

E. The axolemma at the nodes of Ranvier lacks K⁺ ion channels>

F. MNFs are metabolically less efficient than NNFs>

Choose all correct statements from the following options:>

(a) A, B and C>

(b) B, C and D>

(d) D, E and F>

Ans. (c)

Sol. A special type of nerve impulse propagation called saltatory conduction occurs along myelinated axons. It occurs because of the uneven distribution of voltage-gated channels. Few voltage-gated channels are present in regions where a myelin sheath covers the axolemma (plasma membrane). By contrast, at the nodes of Ranvier (where ther is no myelin sheath), the axolemma has many voltage-gated channels. Hence, flow of Na⁺ and K⁺ ions across the membrane mainly occurs at the nodes. Thus, action potentials literally jump from one node to the next as they propagate along a myelinated axon and for this reason such propagation is called saltatory conduction. Propagation via saltatory conduction occurs faster than continuous conduction in nerve fibers of the same axon diameter.>

88. The mechanisms of regulation of H⁺ secretion by kidneys in acidosis have been suggested in the following statements:>

A. Acidosis inhibits the secretion of cortisol by adrenal cortex>

B. The transcription of Na⁺- H⁺ antiporter gene is decreased by cortisol>

C. The translation of mRNA of 1Na⁺- 3HCO₃- symporter gene is decreased by cortisol>

D. The secretion of endothelin-1(ET-1) from the proximal tubule is enhanced in acidosis>

E. ET-1 stimulates the phosphorylation and subsequent insertion of the Na⁺- H⁺ antiporter into the apical membrane of proximal tubular cells>

F. The insertion of 1Na⁺- 3HCO₃- symporter into the basolateral membrane of proximal tubular cells is also increased by ET-1>

Choose all CORRECT statements from the following options:>

(a) A, B and C>

(b) B, C and D>

(d) D, E and F D>

Ans. (d)

Sol. H ion secretion is regulated by kidney during acidosis as>the secretion of endothelin-1(ET-1) from the proximal tubule is enhanced in acidosis, ET-1 stimulates the phosphorylation and subsequent insertion of the Na⁺- H⁺ antiporter into the apical membrane of proximal tubular cells and the insertion of 1Na⁺- 3HCO₃- symporter into the basolateral membrane of proximal tubular cells is also increased by ET-1.>

89. Consider the following statements about thermoregulation in animals.>

A. It uses external environment to regulate internal body temperature>

B. It does not use external environment to regulate internal body temperature>

C. It can vary internal temperature considerably>

D. It can maintain thermal homeostasis in a narrow range of temperatures>

Which one of the following options correctly describes a poikilothermic ectotherm?>

(a) A and D>

(b) B and C>

(d) B and D>

Ans. (c)

Sol. Poikilothermic animals like reptiles, fishes and amphibians are not able to maintain their body temperature at a constant level. In thse organisms, the body temperature fluctuates with changes in the environmental temperature Ectothermic animals obtain much of their heat from the environment.>

90. A newborn baby got mixed up with other babies in a hospital. If the mother is of O blood group and is Rh +ve and the father is of AB blood group and is Rh -ve, which one of the following can be their baby?>

(a) AB and Rh +ve>

(b) O and Rh –ve>

(d) B and Rh –ve>

Ans. (c)

Sol. A and Rh +ve can be their baby.>

91. Two genes a and b are located at a distance of 10cM. Individuals of the genotype AaBb are sib-mated. The two genes are linked in trans. What percentage of the progeny is expected to have the genotype aabb?>

(a) 0.25>

(b) 0.01>

(d) 25>

Ans. (a)

Sol. The genotype of F1 progeny will be AaBb. Genotype AaBb can form a maximum of four different types of gametes : AB, ab, Ab and aB. If genes are 10 map units apart, genotype AaBb will form four different types of gametes (AB 5%, ab 5%, Ab 45% and aB 45%). Hence, self-cross progeny with genotype aabb will be 0.5 × 0.5 = 0.25.>

92. Which one of the following statements is true regarding heritability of a quantitative character?>

(a) The estimate obtained from a given population and in one set of environment can be extrapolated to other population and sets of environment>

(b) The estimate is a population as well as an individual parameter.>

(d) Heritability indicates the degree to which a trait is genetic.>

Ans. (c)

Sol. Heritability is a concept that describes how much of total phenotypic variation in a population is due to variation in genetic factor. In this context, a hgih heritability indicates that much of the phenotypic variation can be attributed to genetic factors, with the environment having less impact on the expression of the trait. With a low heritability, environmental factors are likely to have a greater impact on phenotypic variation within the population.>

93. Chromosomal inversions are balanced rearrangements and thus do not change the overall amount of genetic material. While inversions can exist in homozygous condition, some only exist as heterozygotes. In the latter condition, the breakpoint disrupts:>

(a) pairing of the homologous chromosomes>

(b) a non-coding region of the genome>

(d) a gene with an essential function>

Ans. (d)

Sol. A inversion heterozygote needs to form a loop with one chromosome so that the chromosome pairs can align in meiosis. Recognize that here heterozygote is in reference to chromosome and not genes. An inversion heterozygote can still be homozygote for all its genes.>

94. A Drosophila male carrying an X-linked temperature sensitive recessive mutation that is lethal at 29°C but viable at 18°C is mated to:>

A. a normal female>

B. a female containing attached X-chromosome>

If the eggs laid in both the cases are reared at 29°C, what will be malefemale ratio in the given progeny?>

(a) .A-1:2, B- 1:1>

(b) A- 1:1, B- only females>

(d) A- 1:0, B- 1:2>

Ans. (b)

Sol. In case where male drosophila having X-linked temperatures sensitive recessive mutation is crossed with a normal female, the laid eggs at 29°C will produce equal ratio of male and female i.e. 1 : 1. But if the same male drosophila is crossed with a female containing attached X chromosome the male progeny would not be able to survive. Hence, only female progeny will be produced.>

95. Sampling of 200 persons for their ABO blood group was done from an urban area. The types of blood group observed in the given population are as follows:>

A = 60, B = 32, AB = 10 and O = 98>

Which of the following gives the correct frequency of blood group determining alleles I^A, I^B and I^O in the given population?>

(a) I^A = 0.19, I^B = 0.11, I^O = 0.7>

(b) I^A = 0.27, I^B = 0.63, I^O = 0.09>

(d) I^A = 0.38, I^B = 0.22, I^O = 0.7>

Ans. (a)

Sol. As we know that, p +q + r = 1; since I^A = p, I^B = q and I^O = r>

r² = I^OI^O = 98/200 = 0.49>

r = I^O = = 0.7>

Genotype of blood type A = I^AI^A and I^AI^O>

If a population is in Hardy-Weinberg equilibrium these together should be>

(p + r)² = p² + 2pr + r² = 0.30 + 0.49 = 0.79>

Since, p² + 2pr = 60/200 = 0.30.>

p + r = = 0.88>

p + 0.7 = 0.88>

p = 0.88 – 0.7 = 0.18>

q = 1 – (p + r) = 1 – 0.88 = 0.12.>

96. The following figure represents a physical map and a genetic map for 5 different genes (a to e).>

Which one of the following statements based on the above is correct?>

(a) The region between b and c is more recombinogenic than the other loci>

(b) In comparision to the region between a and b, the region between d and e is more recombinogenic>

1 Kb 1 cM>

(d) If more markers were mapped between d and e, the genetic distance between d and e is likely to decrease>

Ans. (b)

Sol. Genetic maps having limited accuracy because it is based on the assumption that crossing overs occur randomly along chromosomes and its frequency between any two genes is generally proportional to the distance between those genes. This assumption is only partly correct because crossovers occur with greater frequency at some points rather than at others. Some regions of chromosomes, called recombination hotspots, are more likely to be involvd in crossing overs than others. Regions with a higher than average probability of crossing over are 'hot spots', whereas regions that participate less commonly than an average segment are 'cold'.>

97. A cross is made between Drosophila stocks, each with an independent mutant allele, resulting in white eye color. The mutant alleles (named w¹ and w²) are recessive, X-linked and caused by a deletion in the w⁺ allele. The wild type phenotype is red eye color.>

The F₁ females are then crossed with wild type males. In the progeny, all females have red eye color, 1 out of 10,000 males was observed to have red eye color, while the remaining had white eyes.>

Which one of the following could possibly explain the occurrence of red eyed males in the progeny?>

(a) One of the mutant alleles has a high rate of spontaneous reversion>

(b) There is an intragenic recombination between the w¹ and w² alleles during meiosis of>

(d) The w¹ and w² alleles show intragenic complementation in red eyed males though it is a rare event.>

Ans. (b)

Sol. A cross is made between Drosophila stocks, each with an independent mutant allele, resulting in white eye color. The mutant alleles (named w¹ and w²) are recessive, X-linked and caused by a deletion in the w⁺ allele. The wild type phenotype is red eye color. The F₁ females are then crossed with wild type males. In the progeny, all females have red eye color, 1 out of 10,000 males was observed to have red eye color, while the remaining had white eyes. There is an intragenic recombination between the w¹ and w² alleles during meiosis of F₁ females.>

98. Doubled haploids (DH) are plants derived from single immature pollen and doubled artificially to form diploids. A DH population was created from F1 progeny derived from a cross between two parents (P1 and P2), one resistant (R) and the other sensitive (S) to white rust. The parents, F1 and DH population were profiled with a set of co-dominant markers, which is represented below.>

The following table summarizes the proposed percentage of the 4 different types (1 to 4) of DH progeny, assuming that the DNA marker is (i) unlinked to the gene governing the trait and (ii) linked at a distance of 10cM.>

Which one of the following options correctly represents the expected ratio for unlinked and linked, respectively?>

(a) A, ii>

(b) A, i>

(d) A, iii>

Ans. (b)

Sol. If genes are unlinked, then there will be independent assortment and all progeny will have equal chance i.e., 25%. If two genes are linked and the distance between them is 10 cM, then 10% (5% and 5%) recombinant will appear and progeny with parental combination will be 90% (45% and 45%).>

99. Recently, the Bhaironghati area of Uttarkashi has been declared a conservation centre for a species referred to as the 'Ghost of the Mountains'. Which one of the following animals does this refer to?>

(a) Marbled cat>

(b) Himalayan ibex>

(d) Snow leopard>

Ans. (d)

Sol. Often referred to as the "Ghost of the Mountains," the snow leopard is an elusive big cat that reigns among the peaks of the Himalayas. Snow leopards act as an ambassador of the planet's highest places and are considered sacred by the people who live there.>

100. The Covid-19 pandemic had a major impact on CO₂ emissions due to the disruption of industrial activities caused by it. Which of the following conuntries/regions had the smallest fall of CO₂ emissions in % terms during the year 2020?>

(a) India>

(b) USA>

(d) European Union>

Ans. (c)

Sol. Carbon dioxide (CO₂)—a greenhouse gas—has become a major concern as climate change becomes a bigger issue. The top five CO₂-producing nations in 2020 were China, the United States, India, Russia, and Japan.>

101. Select the correct combination of factors which drive the extinction vortex>

(a) reduced population size, loss of genetic diversity, inbreeding>

(b) emigration, fragmented habitat, inbreeding>

(d) catastrophic events, reduced population size, outbreeding>

Ans. (a)

Sol. Demographic factors that are involved in extinction vortices include reduced fecundity, changes in dispersal patterns, and decreased population density. The main modern causes of extinction are the loss and degradation of habitat (mainly deforestation), over exploitation (hunting, overfishing), invasive species, climate change, and nitrogen pollution.>

102. The fall armyworm is a recent, fast invading, destructive herbivore in agricultural systems in north India. This pest species belongs to which one of the following Orders?>

(a) Hemiptera>

(b) Homoptera>

(d) Lepidoptera>

Ans. (d)

Sol. The fall armyworm is a species in the order Lepidoptera and one of the species of the fall armyworm moths distinguished by their larval life stage. The term "armyworm" can refer to several species, often describing the large-scale invasive behavior of the species' larval stage.>

103. Which one of the following ecosystems is unique to India?>

(a) Mangroves>

(b) Cold deserts>

(d) Riparian forest>

Ans. (c)

Sol. Myristica swamps, which were once widespread across the Konkan coast, are now a fast-shrinking, fragmented, and endangered ecosystem restricted to small patches found in southern Kerala, Uttara Kannada district of Karnataka, Goa, and recently discovered in the northern Western Ghats of Maharashtra.>

104. Which of the following constitute the largest reservoir of carbon in the global carbon cycle?>

(a) The atmosphere>

(b) The plant biomass on land>

(d) The ocean>

Ans. (d)

Sol. The largest reservoir of the Earth's carbon is located in the deep-ocean, with 37,000 billion tons of carbon stored, whereas approximately 65,500 billion tons are found in the globe.>

105. Select the correct statement from the options given below to complete the following.In the 1960s, experiments were conducted to test the theory of island biogeography. The main findings of these studies indicate that over a long period of time .........................>

(a) the rate of extinction and colonization are not equal to each other.>

(b) the colonization rates gradually exceed extinction rates>

(d) the overall rate of colonization will be balanced by the rate of extinction>

Ans. (c)

Sol. The theory of island biogeography proposes that the number of species on any island is determined by a balance between the rate at which new species colonize it (i.e., the rate of immigration of species from the mainland to the island) and the rate at which populations of established species become extinct (i.e., the rate of extinction of species on the island).>

106. Select the correct statement that best describes animal territories>

(a) Are always inherited from the parent>

(b) Are always non-overlapping with neighbours>

(d) Are always guarded and defended>

Ans. (d)

Sol. Territory is defined as that area of the habitat defendend by individuals of a particular species agaisnt other members of the same or different species. Territories are usually held because they contain a food source or because they provide some site needed in breeding. In contrast to territory, home ranges are not defended or exclusively held and may overlap among individuals. Territories are defendend and usually do not overlap.>

107. Match the organisms in column A with their status in Column B.>

(a) A, C are X; B, D are Y>

(b) A, D are X; B>

(d) A, B are X; C, D are Y>

Ans. (b)

Sol. P. Kallimodon : Fossil>

Q. Tuatara : Fossil>

R. Xiphosura : Living fossil>

S. ankylosphenodon : Living fossil>

108. Select the correct combination of species (column X) and the known cause of their population declines.>

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

(b) A - iv; B - iii; C - i; D – ii>

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

Ans. (b)

Sol. P. Honey bee : Neonicotinoids>

Q. Gyps vulutre : Diclofenac>

R. Shellfish : Methylmercury>

S. Minnow : Neonicotinoids>

109. According to the APG-IV (Angiosperm phylogeny group-IV) which of the following groups of angiosperms first diverged from the common ancestor of the angiosperms?>

(a) Nymphaeales>

(b) Monocots>

(d) Ranunculales>

Ans. (a)

Sol. According to the APG-IV (Angiosperm phylogeny group-IV), Nymphaeales groups of angiosperms first diverged from the common ancestor of the angiosperms.>

110. The table below lists the name of organisms and different aspects of nervous system.>

Which one of the following options represents a correct match between the two columns?>

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

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

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

Ans. (b)

Sol. P. Protozoan : Stimulus response coordination>

Q. Jellyfish : Diffuse nevous system with small ganglial centres>

R. Flatworms : Two cephalic ganglia joined by a commisure>

S. Bony fish : Central and peripheral nervous system>

111. The following statements were made describing the properties of a UPGMA tree (Unweighted Pair Group Method with Arithmetic Mean):>

A. It describes species relationships and is therefore the best method to describe a new species.>

B. It is a method of hierarchical clustering.>

C. The raw data is a similarity matrix and the initial tree is rooted.>

D. It permits lineages with largely different branch lengths and corrections for multiple substitutions.>

Which one of the following options represents the correct properties?>

(a) A and B>

(b) B and C>

(d) C and D>

Ans. (b)

Sol. UPGMA is one of the agglomerative clustering methods. This algorithm constructs a rooted tree that reflects the structure present in a pairwise similarity matrix (or a dissimilarity matrix).>

112. The microclimate at the ground level is very important to plant life. Each curve (A-D) in the diagram below shows the temperature profiles collected above and below the bare ground (non-vegetated) shown with respect to the distance from the ground level (at different times over a 24-hour period).>

Consider the following four different time points within a diel period (i-iv). i. Immediately after sunrise ii. Noon iii. Immediately before sunset iv. Midnight. Match the temperature profiles with the correct diel period.>

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

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

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

Ans. (b)

Sol. The correct option is (b).>

113. The largest reservoir of nitrogen in the global nitrogen cycle is the atmosphere. Options A-D below represent important pathways in the removal of nitrogen from the atmosphere at different rates.>

A. Biological fixation in oceans>

B. Fixation by lightning>

C. Biological fixation in natural terrestrial systems>

D. Industrial nitrogen fixation>

Arrange the above pathways from the lowest to the highest rate.>

(a) D < B < A < C>

(b) B < D < C < A>

(d) A < B < D < C>

Ans. (c)

Sol. B < C < D < A is the correct order in the removal of nitrogen from the atmosphere at different rates.>

114. In a high-altitude meadow region, it was observed that over the last five years 20 forb species flowered 2-3 weeks earlier than their long-term average time of flowering. At the same time, their fruit production has fallen. The following statements were proposed as reasonable explanations for why this is happening:>

A. The forbs are responding to a warming climate but pollinators are not available at the same time>

B. Early flowering has increased competition for pollinators>

C. Flowering and fruiting success are unrelated phenomena in forbs>

D. Animals that eat fruits are not available at the right time so fruiting has stopped>

Which one of the following options represents statements with correct reasonable explanations?>

(a) A and C>

(b) C and D>

(d) B and D>

Ans. (c)

Sol. In a high-altitude meadow region, it was observed that over the last five years 20 forb species flowered 2-3 weeks earlier than their long-term average time of flowering. At the same time, their fruit production has fallen. The forbs are responding to a warming climate but pollinators are not available at the same time and early flowering has increased competition for pollinators>

115. Convergent evolution creates:>

(a) Analogous structures>

(b) Homologous structures>

(d) Pleiotropic structures>

Ans. (a)

Sol. Character refers to any heritable trait possessed by an organism. Biologists distinguish two different types of characters : homoplasy (homoplasious character) and homology (homologous character). A homoplasy is a character that species share through convergent evolution. It is a similarity in appearance but not in origin. The wings of birds and bats are an example of a homoplasious character. The homoplasy shared between species that were not present in the common ancestor. These characters are not an indicator of phylogenetic relationship. Homoplasy also refers to analogous structures which has similarity in function but not in origin.>

116. >

Select the correct option that describes the above graph regarding length of bird's bill:>

(a) Neutral selection>

(b) Directional selection>

(d) Mutational selection>

Ans. (b)

Sol. As per the graph mentioned in question, length of bird's bill is directionally selected.>

117. Eusocial societies are NOT characterised by which of the following?>

(a) Altruism>

(b) Kin selection>

(d) Equal reproductive opportunities>

Ans. (d)

Sol. Eusocial societies are not characterised by Kin selection.>

118. The table below enlists name of scientists and different areas of scientific contribution>

Which one of the following options represents a correct match between the scientist and the area of his/her scientific contribution?>

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

(b) A-v; B- i, C-ii; D-iv; E-iii>

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

Ans. (d)

Sol. P. Alfred Wallace : Sociobiology>

Q. Konard Lorenz : Ethology>

R. Joseph Banks : Botany>

S. E.O. Wilson : Sociobiology>

T. Robert MacArthur and E.O. Wilson : Island biogeography>

119. Fragmentation breaks up contiguous tracts of natural habitats into smaller patches. In a fragmented landscape where a previously large forest has become a mosaic of patches of different sizes, the following statements can be made about the fragment size and its species diversity.>

A. Smaller fragments will always have lower species richness than larger fragments>

B. Species richness will depend on fragment size.>

C. Species richness will depend on physical connectivity between fragments>

D. Species richness cannot be compared between large and small fragments>

Select the option where both the statements are correct>

(a) A and B>

(b) B and C>

(d) B and D>

Ans.>

Sol.>

120. Which of the following is NOT the function of dispersal behaviour in which organisms move away from their natal homes?>

(a) Tracking resource availability>

(b) Providing mating opportunities>

(d) Avoiding pathogens>

Ans. (c)

Sol. Preventing species extinction is not the function of dispersal behaviour in which organisms move away from their natal homes.>

121. Colour blindness affects approximately 1 in 12 men (8%). In a population that is in Hardy-Weinberg Equilibrium (HWE) where 8% of men are colour-blind due to a sex-linked recessive gene. What proportion of women are expected to be carriers?>

(a) 0.92>

(b) 0.85>

(d) 0.15>

Ans. (d)

Sol. Since males need only one copy of the recessive allele to have the disease, the frequency of diseased males will have the frequency of the recessive allele i.e., 0.08. The frequency of the dominant allele = 1 – 0.08 = 0.92.>

By the Hardy-Weinberg principle, chance of being carrier is 2pq, which equals (2 × 0.92 × 0.08) >= 0.1472 ~ 0.15.>

122. Which of the following pairs of traits is most likely in a species when maternal investment is very high?>

(a) Multiple reproductive events and high maternal mortality>

(b) Slow developmental rates and low maternal fecundity>

(d) Few reproductive events and high maternal mortality>

Ans. (c)

Sol. Parental investment refers to the energy and time each sex invests in producing and rearing off spring; it is, in effect, an estimate of the energy expended by males and females in each reproductive event. In most species, females invest maximum in producing and rearing offspring.>

123. A researcher studying a cricket species finds that individuals on either side of a large river have different call frequencies on average. The following statements were made:>

A. The different call frequencies may signal incipient speciation>

B. The change in call frequency can lead to allopatric speciation>

C. Individuals of one population transplanted to the other population (across the river) may have lower chance of finding mates than residents>

D. Call frequencies have changed from ultrasound to infrasound across the river>

If the call helps attract mates which of the above statements are correct?>

(a) A, B and C>

(b) A, C and D>

(d) A, B and D>

Ans. (a)

Sol. Allopatric speciation – also referred to as geographic speciation, vicariant speciation, or its earlier name the dumbbell model – is a mode of speciation that occurs when biological populations become geographically isolated from each other to an extent that prevents or interferes with gene flow. Statement A, B and C are correct.>

124. Biased gene conversion (BGC) has been proposed to cause changes in allele frequencies in a population. Select the statement that is NOT correct about BGC.>

(a) BGC is present in bacteria and eukaryotes suggesting it may be present in the Last Universal Common Ancestor (LUCA). BGC>

(b) BGC can favor the fixation of deleterious donor alleles. BGC>

(d) BGC selects against maladaptations resulting in fixation of only advantages mutations>

Ans. (d)

Sol. Classical genetic studies show that gene conversion can favour some alleles over others. Molecular experiments suggest that gene conversion could favour GC over AT base pairs, leading to the concept of biased gene conversion towards GC. The expected consequences of such a process is the GC-enrichment of DNA sequences under gene conversion.>

125. When species express a suite of correlated traits (e.g., behavior, morphology, function), within a given context or across contexts, it is referred to as>

(a) A syndrome>

(b) Trait flexibility>

(d) Character displacement>

Ans. (a)

Sol. A behavioral syndrome is a suite of correlated behaviors expressed either within a given behavioral context (e.g., correlations between foraging behaviors in different habitats) or across different contexts (e.g., correlations among feeding, antipredator, mating, aggressive, and dispersal behaviors).>

126. Which one of the following is the most sensitive immunoassay?>

(a) Immuno-electrophoresis>

(b) Rocket electrophoresis>

(d) Passive agglutination>

Ans. (d)

Sol. In passive particle (indirect) agglutination, soluble antigens are coupled to large particles such as erythrocytes or latex spheres. When specific antibodies are added to coated erythrocytes or latex spheres, antibody bridges are formed between the particles, and agglutination occurs.>

127. What is the nature of India's indigenous COVID-19 vaccine?>

(a) It is an mRNA vaccine (for expression of viral spike protein)>

(b) It is a preparation of inactivated whole virus>

(d) It is a recombinant, replication-deficient chimpanzee adenovirus vector encoding the SARS-CoV-2 Spike (S) glycoprotein>

Ans. (b)

Sol. Inactivated (killed) vaccines are produced by inactivation of pathogenic organisms by heat or chemical treatment so that organisms are unsable to multiply in the host. COVAXIN, India's indigenous COVID-19 inactivated vaccine by Bharat Biotech is developed in collaboration with the ICMR. The vaccine is developed using Whole-Virion Inactivated Vero Cell derived platform technology. Attenuated vaccines are prepared by attenuating pathogenic organisms by growing them in unfavourable conditions which result in gene mutations due to which organism looses pathogenicity but retains their capacity for transient growth.>

128. Which one of the following statements related to genetic transformation of plants is correct?>

(a) Negative selection markers need to be expressed only under strong constitutive promoters for development of transgenic plants>

(b) A transgenic plant containing two linked copies of the transgene in heterozygous condition would segregate in a 3:1(transgenic : nontransgenic) ratio for the transgenic phenotype on self pollination.>

(d) In a binary vector system of Agrobacterium tumefaciens, the oncogenes are located on the Helper plasmid.>

Ans. (b)

Sol. A selectable marker gene encodes a product that allows the artificial selection of transformed or transfected cells. Selectable marker genes can be divided into several categories depending on whether they confer positive or negative selection. Positive selectable marker genes are defined as those that promote the growth of transformed or transfected cells (i.e., conferring resistance to transformed or transfected cells) whereas negative (or counterselectable) selectable marker genes result in the death of the transformed or transfected cells (i.e., conferring sensitivity to transformed or transfected cells).>

129. Which of the following components is typically NOT utilized while capturing images using a confocal laser scanning microscope (CLSM)?>

(a) CCD camera>

(b) Pinhole>

(d) Detectors>

Ans. (a)

Sol. In cameras, CCD enables them to take in visual information and convert it into an image or video. They are, in other words, digital cameras. This allows for the use of cameras in access control systems because images no longer need to be captured on film to be visible. "CLSM is one of the most important advances achieved during recent decades in the field of fluorescence imaging and is considered as an essential tool in biological research.>

130. The electrodes (also called leads) used for human electroencephalography in a 10-20 lead system are labeled with an uppercase letter and a subscript number or a letter. Which one of the following is INCORRECT meaning of the labels of electrodes?>

(a) The uppercase letter is an abbreviation of the location of electrode on the head>

(b) The electrodes on the midline of head are labeled with the subscript letter x along with the uppercase letter>

(d) The electrodes over right hemisphere are labeled with a subscript even number along with the uppercase letter>

Ans. (c)

Sol. An EEG is a test that detects abnormalities in your brain waves, or in the electrical activity of your brain. During the procedure, electrodes consisting of small metal discs with thin wires are pasted onto your scalp. The electrodes detect tiny electrical charges that result from the activity of your brain cells. Statement C is incorrect.>

131. Which one of the following techniques CANNOT be used for separation, detection or visualization of DNA?>

(a) Western Blotting>

(b) Polyacrylamide gel electrophoresis>

(d) Denaturing high performance liquid chromatography>

Ans. (a)

Sol. Western blot is often used in research to separate and identify proteins. In this technique a mixture of proteins is separated based on molecular weight, and thus by type, through gel electrophoresis. These results are then transferred to a membrane producing a band for each protein.>

132. The units of molar extinction coefficient are>

(a) L mol^-1cm^-1>

(b) cm^-1mg ml^-1>

(d) mol cm ml^-1>

Ans. (a)

Sol. The SI units of e are m²/mol, but in practice they are usually taken as M^–1cm^–1.>

133. Two experiments were performed on a peptide sample 'X'. In experiment 1, treatment of 'X' with dithiothreitol (DTT), followed by blocking of free sulphydryl groups, yielded two polypeptides whose amino acid sequences are as shown below:>

I. Ala-Phe-CysA³-Met-Tyr-CysA⁶-Leu-Trp-CysA⁹-Asn>

II. Val-CysB²-Trp-Val-Ile-Phe-Gly-CysB⁸-Lys>

In experiment 2, 'X' was treated with chymotrypsin, a protease that cleaves the carboxy-terminal of aromatic residues. The amino acid composition of five peptides obtained from this experiment are shown below>

I. ([Ala], [Phe])>

II. ([Asn], 2[Cys], [Met], [Tyr])>

III. ([Cys], [Gly], [Lys])>

IV. (2[Cys], [Leu], 2[Trp], [Val])>

V. ([Ile], [Phe], [Val])>

Based on the above results, which cysteine/s are linked by disulfide bond in peptide 'X'?>

(a) A6-B2 and B8-A3 A6-B2>

(b) A3-A9 and B2-A6 A3-A9>

(d) A3-B8 only>

Ans. (b)

Sol. After sequencing is completed, locating the disulfide bonds requires an additional step. A sample of the protein is again claved with chymotrypsin, this time without breaking the disulfide bonds. When the resulting peptides are separated by electrophoresis and compared with the original set of peptides generated by chymotrypsin, two of the original peptides will be missing and a new, larger peptide will appear. The two missing peptides represent the regions of the intact polypeptide that are linked by a disulfide bond.>

134. In 1990, Bhattacharya et al identified that the wrinkled seed character of pea as described by Mendel is caused by a transposon – like insertion in a gene encoding Starch Branching Enzyme (encoded by the R allele). This leads to an RFLP pattern, when genomic DNA of round and wrinkled seed is digested with EcoRI and probed with the cDNA of the R gene product. The following is a representation of the hybridization pattern.>

Based on the above, which one of the following statements is INCORRECT>

(a) Lane 3 represents genomic DNA from plant with wrinkled seeds>

(b) These DNA markers are co-dominant in nature>

(d) If genomic DNA from F₂ progeny as obtained in Mendel's work was analyzed by RFLP, the ratio of progeny with patterns in lane 1, 2 and 3 will be 2 : 1 :1.>

Ans. (a)

Sol. If there is wrinkled seed in lane 3 then rr genotype will be present. The rr genotyep will make a single band in a well.>

135. Consider predators with a choice between two prey types: a big prey 1 which has energy value E1, handling time h1, and search time S1 and; a small prey 2 with energy value E2, handling time h2, and search time S2. According to the optimal foraging (diet) theory, when will the predator preferentially select prey 2?>

(a) When E2/h2 > E1/(h1+S1)>

(b) When the abundance of prey 1 is very high>

(d) When E2/h2=E1/h1>

Ans. (a)

Sol. Optimal foraging theory (OFT) is a behavioral ecology model that helps predict how an animal behaves when searching for food. Although obtaining food provides the animal with energy, searching for and capturing the food require both energy and time. Optimal foraging theory (OFT) focuses on the effects of habitat quality and diet choice on the spatial distribution and movement of animals. This theory addresses the question of how long a forager should spend exploiting one food source before switching to another.>

136. Select the option that represents the correct combination of non-parametric tests and its equivalent parametric test respectively that can be used to compare two or more groups.>

(a) Wilcoxon Rank Sum Test and Paired t-test>

(b) Wilcoxon Rank Sum Test and Spearman correlation>

(d) Mann-Whitney U test and Pearson correlation>

Ans. (a)

Sol. Examples of parametric and non-parametric tests :>

Parametric tests : T-test, Z-test, F-test, ANOVA.>

Non-parametric tests : Chi-square, Mann-Whitney U-test, Kruskal-Wallis H-test, Wilcoxon Signed-Rank test.>

137. A researcher raised antibodies against sheep red blood cells (SRBCs) and purified the IgG fraction. Some of the IgG antibodies were then subjected to enzymatic digestion to have Fab, Fc and F(ab')₂ fractions. He placed each preparation in a separate tube (1 to 3), labeled the three tubes to indicate their contents, and incubated them on ice. After a while he noticed that the label on two of the tubes (1 and 2) had gotten erased. He did a test for tube 1 and found that the preparation in the tube agglutinated SRBCs but did not lyse them in presence of complement. Which preparation was in tube 1?>

(a) Fab>

(b) Fc>

(d) IgG>

Ans. (c)

Sol. Recovered F(ab')₂ fragments are monomeric (by dynamic light scattering), preserve their secondary structure (by circular dichroism) and are as pure as those obtained via Protein A chromatography (from a mixture of F(ab')₂ and Fc fragments).>

138. In a forward genetic screen to investigate the heat stress response in Arabidopsis, a team of researchers identified an uncharacterized gene 'x' that shows some sequence homology to alpha-subunit of heterotrimeric G-protein. Since a typical alpha-subunit of heterotrimeric G-protein localizes at the membrane in a eukaryotic cell, researchers sought to validate whether the protein coded by gene 'x' localizes to membrane in tobacco protoplasts. To achieve this, they cloned the gene in fusion with GFP at its N-terminus, under the control of the CaMV promoter; however, upon expression of this GFP-gene 'x' fusion construct, they did not observe any membrane localization of GFP-signal in tobacco protoplasts. Based on this, they made a few assumptions:>

A. N-terminally tagging of protein 'X' with GFP may block membrane localization of protein X>

B. Tagging of protein 'X' with GFP may alter the conformation of protein X because of its bigger size>

C. Tobacco protoplasts are a heterologous system for the expression of gene 'x,' and thus, the protein X does not localize to the membrane>

D. Gene 'x' is not getting transcribed because of the wrong promoter choice>

Which one of the following options represents all correct assumptions?>

(a) A and B only>

(b) B and D only>

(d) A, B and C>

Ans. (d)

Sol. Biologists use GFP to study cells in embryos and fetuses during developmental processes. Biologists use GFP as a marker protein. GFP can attach to and mark another protein with fluorescence, enabling scientists to see the presence of the particular protein in an organic structure.>

Statement A, B and C is correct.>

139. Given below is a figure representing expression levels of transgenic protein in ten independent transgenic plants generated using the same transformation vector by Agrobacterium-mediated transformation.>

Given below are a few statements to explain the above data:>

A. Plant nos. 4, 9 and 10 that show high expression levels of the transgene would necessarily contain multiple copies of the transgene.>

B. Plant nos. 2 and 7 contain mutations in the coding sequence of the transgene in the construct.>

C. The transgenic plants may contain varying number of transgene copies inserted at different locations in the host genome.>

D. The host genome has no role in influencing expression levels of the transgene.>

E. The stability of the transgenic mRNA and its translatability would not be different among the independent transgenic plants.>

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

(a) A and D only>

(b) B and C only>

(d) A, D and E only>

Ans. (c)

Sol. Statement C and E are correct.>

140. In recent decades, the use of genetic markers has allowed the rapid introgression and selection of desired breeding stocks in advance generations. In this regard, following statements are given:>

A. AFLP markers can discriminate between homozygote and heterozygote genotypes.>

B. Microsatellites (eg. SSR) are capable of detecting higher level of polymorphism than RFLP.>

C. SNPs are more prevalent in the coding regions of the genome.>

D. SNP markers are the most suitable markers for both foreground and background selection.>

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

(a) A, B and C>

(b) A, B and D>

(d) B and D only>

Ans. (d)

Sol. SNPs (single nucleotide polymorphisms) are positions in a genome wherer some individuals have one bp (e.g., a GC) and others have a different bp (e.g., a AT). It represents a variation at a single position in a DNA sequence among individuals. Most SNPs are found in noncoding DNA, such as within introns and intergenic sequences. AFLP is a PCR-based method for arbitrarily amplifying restriction fragments. Like RAPD, AFLP does not require any DNA sequence information from the organism under study and also a dominant marker. However, it is highly reliable and reproducible. It combines the power of RFLP with the flexibility of PCR-based technology.>

141. To investigate the relationship between microtubules and centrioles in fixed HeLa cells using an epifluorescence microscope, a researcher plans to conduct immunostaining using antibodies against tubulin and centrin (centriolar protein). After the incubation with the primary antibodies and wash, she/he plans to use secondary antibodies that bind to the primary antibodies. Below is a list of secondary antibodies carrying various fluorophores (dyes) available to the researcher.>

A. Alexa 568>

B. FITC>

C. Alexa 488>

D. Alexa 647>

Select the correct combinations of the appropriate dyes that the researcher would typically utilize to observe co-localization in an epifluorescence microscope?>

(a) A and C A and C>

(b) B and C B and C>

(d) C and D C and D>

Ans. (a)

Sol. Probes with high fluorescence quantum yield and high photostability allow detection of low-abundance biological structures with great sensitivity. Alexa Fluor 568 dye molecules can be attached to proteins at high molar ratios without significant self-quenching, enabling brighter conjugates and more sensitive detection.>

We offer Alexa Fluor 568 dye conjugated to a variety of antibodies, peptides, proteins, tracers, and amplification substrates optimized for cellular labeling and detection.>

142. >

Find the linear regression equation for the following data pairs (x, y) given in the above table>

(a) y = 4x + 0>

(b) y = 3x +2>

(d) y = 0.33 + 2>

Ans. (b)

Sol. >

Regression line x on y>

3x – 21 = y – 23>

3x – 21 + 23 = y>

3x + 2 = y>

143. The Green Fluorescent Protein (GFP) from the deep-sea jellyfish Aequorea victoria has excitation peaks at 395 nm and 475 nm. Its emission peak is at 509 nm, which is in the green portion of the visible spectrum. In the deep-sea habitat of this marine organism, what is the source of light for excitation of GFP?>

(a) Blue-light emitted from the oxidation of the cofactor coelentrazine is energy transferred to GFP.>

(b) Blue light emitted by aequorin present in A. victoria is absorbed by GFP A. victoria>

(d) Ocean currents provide electrical energy that is converted to light>

Ans. (a)

Sol. Green fluorescent protein (GFP) was originally derived from the jellyfish Aequorea victoria (Prendergast and Mann, 1978). It has 238 amino acid residues and a green fluorophore, which is comprised of only three amino acids: Ser65-Tyr66-Gly67. The green fluorescent protein (GFP) is a protein that exhibits bright green fluorescence when exposed to light in the blue to ultraviolet range.>

144. For the template sequence given below, which one of the following combination of primers can hypothetically be used to amplify the target region (Ignore Tm & length parameters for the primers)? 5'- A T C G A C T A G NNNNNNNNNNNNNNNNN C C T A A T G C A G - 3'>

(a) Primer 1 Primer 2>

5'-TAGCTG-3' 5'–GACGTA-3'>

(b) Primer 1 Primer 2>

5'-CTGCAT-3' 5'–ATCGAC-3'>

5'-ATCGAC-3' 5'–GACGTA-3'>

(d) Primer 1 Primer 2>

5'-TAGCTG-3' 5'–CTGCAT-3'>

Ans. (b)

Sol. PCR relies on a thermostable DNA polymerase, Taq polymerase, and requires DNA primers designed specifically for the DNA region of interest. In PCR, the reaction is repeatedly cycled through a series of temperature changes, which allow many copies of the target region to be produced.>

145. Which one of the following combinations of enzymes used for cloning a linear insert fragment into a digested plasmid vector would have the least probability of generating self-ligated vectors in a cloning experiment following complete digestion of all vector molecules and no further enzymatic treatment of the vector?>

(a) >

(b) >

(d) >

Ans. (c)

Sol. Hindlll and Xhol enzymes will have least probability of generating self-ligated vectors.>

CSIR NET BIOLOGY (JUNE - 2021 Sift 2)
Previous Year Question Paper with Solution.

Locate Us

Company

Courses

Academies

CSIR NET BIOLOGY (JUNE - 2021 Sift 2) Previous Year Question Paper with Solution.

Locate Us

Company

Courses

Academies

Get in touch

CSIR NET BIOLOGY (JUNE - 2021 Sift 2)
Previous Year Question Paper with Solution.