PYQ JUNE 2011 - VedPrep

CSIR NET BIOLOGY (JUNE - 2011)
Previous Year Question Paper with Solution.

21. The area of allowed regions in the Ramachandran map will be least for

(a) Gly

(b) L-Ala

(d) -methyl L-valine

Ans. (c)

Sol. The proline Ramachandran plot is restricted by pyrrolidine ring, where the flexibility in pyrrolidine ring couples to back bone.

22. Small RNAs with internally complementary sequences that form hairpin-like structure, synthesized as precursor RNAs and cleaved by endonucleases to form short duplexes are called

(a) snRNA

(b) mRNA

(d) miRNA

Ans. (d)

Sol. miRNA is derived from a single RNA that pairs with itself through complementary sequence. It's loop region is cut out by Dicer to produce double stranded miRNA precursor.

23. The free energy of a dissolved solute

(a) increases with solute concentration

(b) decreases with solute concentration

(d) depends only on temperature

Ans. (b)

Sol. Gibbs free energy depend on concentration of reactant and product. It decreases with solute concentration.

24. A protein in 100 mM KCl solution was heated and the observed T_m (midpoint of unfolding) was 60°C. When the same protein solution in 500 mM KCl was heated, the observed T_m was 65°C. What is the most probable reason for this increase in T_m?

(a) Hydrogen-bonding is increased.

(b) Hydrophobic interaction is decreased and electrostatic repulsion is increased.

(d) van der Walls interaction is increased.

Ans. (b)

Sol. Hydrophobic interaction is decrease as electrostatic repulsion is increased.

25. An amino acid contains no ionizable group in its side chain (R). It is titrated from pH 0 to 14. Which of the following ionizable state is not observed during the entire titration in the pH range 0-14 ?

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

26. An -helix in a peptide or protein is characterized by hydrogen bonds and characteristic dihedral angles. Choose the right combination.

(a) Hydrogen bonding between the amide CO of residue i and amide NH of residue i + 4. Dihedral angles in the region ~ –50°, ~ –60°.

(b) Hydrogen bonding between the amide NH of residue i and amide CO of residue i + 4. Dihedral angles in the region of ~ –50°, ~ –60°.

(c) Hydrogen bonding between the amide CO of residue i and amide NH of residue i + 4. Dihedral angles in the region of ~ –50°, ~ +60°.

(d) Hydrogen bonding between the amide CO of residue i and amide NH of residue i + 3. Dihedral angles in the region of ~ –50°, ~ –60°.

Ans. (a)

Sol. The alpha helix is stabilized by hydrogen bonds (shown as dashed lines) from the carbonyl oxygen of one amino acid to the amino group of a second amino acid. Because the amino acids connected by each hydrogen bond are four apart in the primary sequence, these main chain hydrogen bonds are called "n to n+4".

27. Precursors of the atoms in the purine skeleton are

(a) N1, Asp; C2 and C8, formate; N3 and N9, guanidine of Arg; C4, C5 and N7, Gly; C6, CO₂.

(b) N1, Asp; C2 and C8, citrate; N3 and N9, amide nitrogen of Gln; C4, C5 and N7, Gly; C6, CO₂.

(d) N1, Glu; C and C8, acetate; N3 and N9, amide nitrogen of Asn; C4, C5 and N7, Gly; C6, CO₂.

Ans. (c)

Sol.

28. Values of T_m (midpoint of denaturation), ΔH_m (enthalpy change at T_m) and (constant-pressure heat capacity change) of a protein are measured in a differential scanning calorimeter. , the Gibbs free energy change at any temperature T(K) can be estimated using the following form of the Gibbs-Helmhotz equation with the values obtained from these measurements :

The stability curve for the protein simulated using the observed thermodynamic values is given below. The shape of the stability curve is due to

(a) hydrogen-bonding and electrostatic interactions only

(b) van der Waals and electriostatic interactions only

(d) only hydrophobic interaction

Ans. (d)

Sol. Hydrophobic interactions and hydrogen bonds both make large contributions to protein stability.

29. Transport of water across aquaporins is regulated by the presence of which of the following sequence of three highly conserved amino acids?

(a) Ala-Asn-Pro

(b) Pro-Asn-Ala

(d) Pro-Ala-Asn

Ans. (c)

Sol. The asparagine-proline-alanine sequences (NPA motifs) are highly conserved in aquaporin water channel family. Crystallographic studies of AQP1 structure demonstrated that the two NPA motifs are in the narrow central constriction of the channel, serving to bind water molecules for selective and efficient water passage.

30. Polar bears maintain their body temperature because they have more of

(a) transducin protein

(b) uncoupling protein

(d) F₀F₁ ATPase

Ans. (b)

Sol. Uncoupling proteins play a role in normal physiology, as in cold exposure or hibernation, because the energy is used to generate heat (see thermogenesis) instead of producing ATP. Some plants species use the heat generated by uncoupling proteins for special purposes.

31. Which of the cyclins have/has essential functions in S-phase of cell cycle?

(a) A-type

(b) B-type

(d) Both B-type and D-type

Ans. (a)

Sol. The cyclin A/CDK2 complex terminates the S phase by phosphorylating CDC6 and E2F1, and drives the cell-cycle transition from S phase to G2 phase, and subsequently activates CDK1 by cyclin A leading the cells to enter the transition to M phase. Upon mitosis, CDK1 activity is maintained by the complex cyclin B/CDK1

32. Na⁺-K⁺ ATPase is a tetramer of 2 and 2 subunits. On which of the following subunits are the Na⁺ and K⁺ binding sites present.

(a) Both on

(b) Both on

(d) Na⁺ on and K⁺ on

Ans. (a)

Sol. Na,K-ATPase is an oligomeric protein composed of alpha subunits, beta subunits and FXYD proteins. The catalytic alpha subunit hydrolyzes ATP and transports the cations.

33. Both halophytes and glycophytes compartmentalize cytotoxic ions into thhe intracellular compartment or actively pump them out of the cell to the apoplasts with the help of membrane transport proteins. Among these, the Na⁺ - H⁺ antiporter, NHX1, is localized in the

(a) plasma membrane

(b) chloroplast (inner envelope)

(d) tonoplast

Ans. (d)

Sol. The sodium/proton exchanger (NHX) mediates Na? and H? countertransport in plants and plays an important role in regulating intracellular pH and maintaining ion and osmotic balance. it is located on tonoplast.

34. Budding yeast cells that are deficient for Mad2, a component of the spindle-attachment-check point, are killed by treatment with benomyl, which causes microtubules to depolymerise. In the absence of benomyl, however, the cells are perfectly viable. Which explanation out of the following is able to justify this observation ?

(a) In the absence of benomyl, the majority of spindles form normally and the spindle-attachment checkpoint (Mad2) plays no role.

(b) In the presence of benomyl, the majority of spindles form normally and Mad2 plays critical role in cell survival.

(c) Other than the role in cell survival, microtubule depolymerization affects oxidative phosphory-lation in the absence of Mad2.

(d) Benomyl also affects protein synthesis in the absence of Mad2.

Ans. (a)

Sol. In the absence of benomyl, the majority of spindles form normally and the spindle-attachment checkpoint (Mad2) plays no role. Option (a) is correct.

35. The overall length of the cell can be measured from the doubling time of a population of exponentially proliferating cells. The doubling time of a population of mouse L cells was determined by counting the numer of cells in samples of culture at various times. What is the overall length of the cell cycle in mouse L cells ?

(a) 30h

(b) 20h

(d) 40h

Ans. (b)

Sol. 20 hr is correct answer.

36. During receptor-mediated endocytosis, apolipo-protein B on the surface of a LDL particle binds to the LDL-recepto present in coated pits containing clathrin. The receptor-LDL complex is interalized by endocytosis, trafficked to lysosomes and the LDL-receptor is finally recycled. A patient reports with familial hypercholesterolemia. This could be due to

(a) mutation in the LDL molecules.

(b) defect in LDL-receptor recycling.

(d) defect in cholesterol binding with its receptor.

Ans. (c)

Sol. Mutations in the APOB, LDLR, LDLRAP1, or PCSK9 gene cause familial hypercholesterolemia. Changes in the LDLR gene are the most common cause of this condition. The LDLR gene provides instructions for making a protein called a low-density lipoprotein receptor.

37. Eukaryotic genomes are organized into chromosomes and can be visualized at mitosis by staining with specific dyes. Heat denaturation followed by staining withh Glesma produced alternate dark and light bands. The dark bands obtained by the process are mainly

(a) AT-rich and gene rich regions.

(b) AT-rich and gene desert regions.

(d) GC-rich and gene desert regions.

Ans. (c)

Sol. R-banding is a cytogenetics technique that produces the reverse of the G-band stain on chromosomes. R-banding is obtained by incubating the slides in hot phosphate buffer, then a subsequent treatment of giemsa dye. Resulting chromosome patterns shows darkly stained R bands, the complement to G-bands.

38. In eukaryotic chromatin, 30 nm fiber (solenoid) can open up to give rise to two kinds of chromatin. In one type (A), the promoter of a gene within the open chromatin is occupied by a nucleosome whereas in the other (B), the promoter is occupied by histone H1.

The following possibilities are suggested.

P. The gene in (A) is repressed

Q. The gene in (B) is repressed

R. The gene (A) is active

S. The gene (B) is active

Which of the following sets is correct ?

(a) P and S

(b) P and Q

(d) R and S

Ans. (a)

Sol. The genes whose promoter is wrapped on nucleosomes, is repressed whereas nucleosomes are removed from the promoters of genes which are to be expressed. Hence, gene A is repressed and gene B is active.

39. The 5' cap of RNA is required for the

(a) stability of RNA only

(b) stability and transport of RNA

(d) methylation of RNA

Ans. (b)

Sol. The 5' 7 methyl guanosine cap require for both stability and transport of RNA. 5' cap protect mRNA from 5'-3' exonuclease and Mtr-1 protein bind at 5' cap to facilitate mRNA transport out of the nucleus into cytoplasm.

40. A culture of an E. coli strain that is lysogenic for phage lambda is grown at 32°C. Induction of the prophage from the host chromosome will occur when the culture is exposed to

(a) 40°C

(b) ultraviolent radiation

(d) wild type E. coli culture

Ans. (b)

Sol. At lysogenic phase, phage DNA integrates with host chromosome to form prophage. It would distinguish from host chromosome when bacteria or bacterial DNA get damaged. UV radiation has ability to damage bacterial DNA.

41. The fidelity of replicative base selection can be reduced by a factor of 10² when the repair of DNA synthesis involves

(a) AP endonuclease

(b) ABC excinuclease

(d) TLS DNA polymerase

Ans. (d)

Sol. TLS DNA polymerase (Translesion polymerase) is DNA polymerase V. It lack 3'-5' exonuclease activity i.e. proof reading activity. So, DNA polymerised from DNA polymerase V have reduced fidelity by 10². DNA polymerase IV also lack proof reading activity.

42. What is the minimum number of NTPs required for the formation of one peptide bond during protein synthesis?

(a) One

(b) Two

(d) Six

Ans. (c)

Sol. Total 4 NTPs are required to form one peptide bond 2 ATPs required for charging of two amino acids on their cognate tRNA.

43. Lac repressor inhibits expression of genes in Iac-operon whereas purine biosynthesis is repressed by the Pur repressor. The two proteins have 31% identical sequences and have similar three-dimensional structures. The gene regulatory properties of these proteins differ in relation to

P. binding of small molecules to the repressor.

Q. presence of recognition sites on the genome.

R. oligometric nature of the repressor.

S. DNA binding property.

The correct statements are

(a) P and Q

(b) P, Q and R

(d) Q, R and S

Ans. (b)

Sol. Genes taking part in purine and pyrimidine biosynthesis are repressed by the pur repressor.

31% identical in sequence (protein) with the lac operon with a similar three-dimensional structure.

Opposite behavior between pur repressor and the lac repressor. Lac repressor is released from DNA by binding to a small molecule the pur repressor binds DNA specifically, blocking transcription, only when bound to a small molecule (co-repressor).

For the pur repressor, the corepressor can be either guanine or hypoxanthine.

44. Two E. coli cultures A and B are taken. Culture A was earlier grown in the presence of optimum concentration of gratuitous inducer IPTG. Both the cultures are now used to inoculate fresh medium containing sub-optimal concentration of gratuitous inducer. It was observed that culture B was unable to utilizee lactose, whereas culture A did so efficiently. The reason behind this is

(a) pretreatment with IPTG has resulted in a mutation as a result of which Iac operon is constitutively expressed.

(b) IPTG has made the cell membrane more porous to small molecules and so lactose is taken up more efficiently by A as compared to B.

(c) in culture A, lactose permease was induced to high level, during pretreatment with IPTG, which allowed the prefeential uptake a lactose.

(d) in culture A, IPTG activated a receptor which bound lactose more efficiently, thereby triggering a signal.

Ans. (c)

Sol. The reason behind this is in culture A, lactose permease was induced to high level, during pretreatment with IPTG, which allowed the prefeential uptake a lactose.

45. It has been observed that in 5-10% of the eukaryotic mRNAs with multiple AUGs, the first AUGs, the first AUG is not the initiation site. In such cases, the ribosome skips over or more AUGs before encountering the favourable one and initiating translation. This is postulated to be due to the presence of the following consensus sequence (s) :

P. CCA CC AUG G

Q. CCG CC AUG G

R. CCG CC AUG C

S. AAC GG AUG A

Which of the following sequences sets related to the above postulations is correct ?

(a) P and Q

(b) P and R

(d) Q and S

Ans. (a)

Sol. The Kozak consensus sequence plays a major role in the initiation of the translation process. The Kozak consensus sequence for initiation of translation in vertebrates is (GCC) GCCRCCATGG, where R is a purine (A or G) (Kozak, 2002).

46. Presence of circular mRNAs for a specific protein in a eukaryotic cell reflects a rapid rate of synthesis of that protein. Following mechanisms are suggested :

P. eIF-4G and PABP promote this process through 5'-3' interaction of mRNA.

Q. ribosomes are less active in recognizing circular mRNA.

R. PABP and eIF-4A promote this process.

S. ribosomes can reinitiate translation without being disassembled.

Which of the following is correct ?

(a) P and S

(b) Q and S

(d) Q and R

Ans. (c)

Sol. eIF 4E bind on 5'cap region and PABP bind on poly A tail. eIF 4G interact with both of these to cause cicularization of mRNA.

47. siRNAs and miRNAs are used for achieving gene silencing. Although, major steps are similar there are distinct differences in the key players of the two processing pathways. Following statements are related to some characteristic features of gene silencing.

P. Both siRNAs and miRNAs are processed by cytoplasmic endonuclease Dicer.

Q. Drosha is needed for processing miRNAs and precursor siRNAs.

R. Both siRNAs and miRNAs show association with Argonaute protein.

S. Both the processing pathways involve RISC complex.

Which of the following combinations is not correct ?

(a) P and R

(b) R and S

(d) S and P

Ans. (c)

Sol. Mammalian Dicer is found in nucleus. Along with this dicer is also present in cytoplasm in other organisms. Drosha process only miRNA not siRNA.

48. Genetic studies demonstrated that TBP mutant cell extracts are deficient in transcription of genes from all three promoters viz. class I, II and III. Following statements describe characteristic features of TBP.

P. TBP is considered as an universal basal transcription factor.

Q. TBP is not required for transcription of archaeal genes.

R. TBP is involved in recognizing TATA box.

S. TBP operates at all promoters regardless of their TATA content.

Which of the following combination is not correct ?

(a) P and S

(b) R and S

(d) P and R

Ans. (c)

Sol. TBP is transcription factor in both eukarya and archea domain. It is universal transcription factor. It recognises TATA box hence depends on TATA content.

49. During generation of an action potential, depolarization is due to

(a) K⁺ efflux

(b) Na⁺ efflux

(d) K⁺ influx

Ans. (c)

Sol. Depolarization is caused by a rapid rise in membrane potential opening of sodium channels in the cellular membrane, resulting in a large influx of sodium ions. Membrane Repolarization results from rapid sodium channel inactivation as well as a large efflux of potassium ions resulting from activated potassium channels.

50. G protein-linked receptors are transmembrane protein of

(a) single-pass

(b) three-pass

(d) seven-pass

Ans. (d)

Sol. G protein-coupled receptors (GPCRs), also known as seven-(pass)-transmembrane domain receptors, 7TM receptors, heptahelical receptors, serpentine receptors, and G protein-linked receptors (GPLR), form a large group of evolutionarily-related proteins that are cell surface receptors that detect molecules outside the cell and activate cellular responses.

51. T_H2 response is generated and maintained mainly by which of the following pair of cytokines?

(a) IL-4 and IL-10

(b) IL-12 and IFN-

(d) IL-2 and IL-12

Ans. (a)

Sol. Once Th2 cells are fully differentiated, the production of Th2 cytokines including IL-4, IL-5 and IL-13 requires re-stimulation by antigens through TCR, by cytokines such as IL-33, or possibly by other inflammatory molecules such as cysteinyl leukotrienes."Response is maintained by IL-10

52. Which of the following molecules is involved in Ca²⁺ dependent cell-cell adhesion?

(a) Calmodulin

(b) Cadherin

(d) Calpain

Ans. (b)

Sol. Cadherins are transmembrane proteins that mediate Calcium dependent cell–cell adhesion in animals. By regulating contact formation and stability, cadherins play a crucial role in tissue morphogenesis and homeostasis.

53. With which protein of Yersinia would integrin proteins of mammalian cells interact for internalization?

(a) Pilin

(b) Fimbrin

(d) Adherin

Ans. (c)

Sol. Yersinia pseudotuberculosis is a Gram-negative bacterium and zoonotic pathogen responsible for a wide range of diseases, ranging from mild diarrhea, enterocolitis, lymphatic adenitis to persistent local inflammation. The Y. pseudotuberculosis invasin D (InvD) molecule belongs to the invasin (InvA)-type autotransporter proteins, but its structure and function remain unknown.

54. Mouse bone marrow cells were fractionated to derive stem cell antigen-1⁺ (Sca-1⁺) cells. These cells were cultured with interleukin-3, or granulocyte-macrophage colony stimulating factor, or macrophage-colony stimulating factor, or granulocyte colony stimualting factor. Most numerous and varied colonies were obtained in the culture stimulated with

(a) interleuin-3.

(b) granulocyte-macrophage colony stimulating factor.

(d) granulocyte-colony stimulating factor.

Ans. (a)

Sol. The IL3 regulates the growth and differentiation of hematopoietic progenitor cells and functionally activates mature neutrophils or macrophages. IL-3 induces MHC class II and B7. 2 expression on eosinophils and renders them capable of supporting T cell proliferation to superantigen and antigen-derived peptides.

55. Cancer causing genes can be fuctionally classified into mainly three types :

1. genes that induce cellualr proliferation

2. tumor suppressor genes

3. genes that regulate apoptotic pathway

Epstein-Barr virus that causes cancer by modulating apoptotic pathway, contians a gene having sequence homology with which of the following genes?

(a) Bax

(b) Bcl-2

(d) Caspase-3

Ans. (b)

Sol. EBV BHRF1 is a viral homolog of the human cellular Bcl-2 protein both in structure and function (as noted above), indicating that BHRF1 might interact indirectly with Bcl-2 via BIK and prevent apoptosis during EBV replication, prolong the lifespan of EBV-infected cells, and potentiate viral persistence and spread.

56. Toll-like receptor 4 is associated with responsiveness to LPS, an endotoxin that causes lethal endotoxic shock. The mice deficient in Toll-like receptor 4 and BALB/c mice were injected with Escherichia coli. In addition, some BALB/c mice were also injected with the same bacteria alone or with anti-interleukin-10 (IL-10) antibody. The mice resistant to the lethal effect of the bacteria were

(a) BALB/b mice receiving the bacteria.

(b) BALB/b mice receiving the bacteria and the anti-IL-10 antibody.

(d) BALB/c mice receiving the bacteria.

Ans. (b)

Sol. BALB/b mice receiving the bacteria and the anti-IL-10 antibody. Option (b) is correct.

57. Glucose is mobilized in muscle when epinephrine activates GS. In an experiment in which muscle cells were stimulated with epinephrine, glucose mobilization was observed even after withdrawal of epinephrine. This could be

(a) due to the presence of a cAMP phosphodiesterase inhibitor.

(b) very low rates of cyclic AMP formation.

(d) due to the absence of protein kinase A.

Ans. (a)

Sol.

58. Intracellular pathogens like Mycobacteria, Salmonella, Leishmania and Listeria survive in macrophages by modulating host cellular machinery. In order to study the fate of these intracellular pathogens in macrop-hages, cells were labelled with lysotracker Red and infected with GFP-labelled organisms. After 2 hours at 37°C, cells were fixed, stained with anti-transferrin receptor antibody and probed with secondary antibody conjugated-blue dyes. Cells were viewed under confocal microscope.

Observation : GFP-labelled Mycobacteria, Salmonella and Listeria were localized in the same compartement labelled with blue dyes, whereas GFP-Leishmania colocalized with red labelled compartment. Which of the following statement is true based on these observations ?

(a) Mycobacteria, Salmonella and Listeria reside in the lysosomes.

(b) Leishmania residue in lysosome like compartment.

(d) Mycobacteria, Salmonella and Listeria lyse the phagosomal membrane and reside in cytosol.

Ans. (b)

Sol. Two major APCs, macrophages and dendritic cells (DCs), play critical roles in mediating resistance and susceptibility during Leishmania infection. Macrophages are the primary resident cell for Leishmania : they phagocytose and permit parasite proliferation."Leishmania reside in lysosome like compartment.

59. Macrophages were controlled from BALB/c mice, CD40-deficient mice, CD86-deficient mice and ICAM-1-deficient mice. These macrophages were co-cultured with LCMV peptide-specific T cells in presence of the LCMV peptide for three days. The cells were recovered and co-cultured with BALB/c-derived marophages in presence of the peptide. During the last twelve hour of the co-culture, ³H-thymidine was added to the cultures. The cells were harvested and ³H-thymidine incorporation was assessed. The highest incorporation was observed in

(a) BALB/c macrophage-T cell co-culture.

(b) CD40-deficient macrophage-T cell co-culture.

(d) ICAM-1-deficient macrophage-T cell co-culture.

Ans. (a)

Sol. The highest incorporation was observed in BALB/c macrophage-T cell co-culture.

60. A patient undergoes liver transplantation and during the course of post-operative treatment, becomes susceptible to infection. The patient can be treated in two different modes and can have alternative outcomes. Which of the following statement is correct ?

(a) Treatment with immunostimulatory drugs reducing the infection but rejecting the transplant.

(b) Treatment with immunostimulatory durgs reducing the infection and retaining the transplatnt.

(d) Treatment with antibiotics reducing the infection but rejecting the transplant.

Ans. (c)

Sol. Treatment with antibiotics reducing the infection but retaining the trasplant. The correct option is (c).

61. Based on the structural regions of a nuclear receptor shown in the diagram, the following predictions were made.

P. Region F is responsibe for binding to ligands and contains two zinc-finger-like binding motifs.

Q. Receptors with A/B domains generally associate with chaperones and do not bind to DNA.

R. Region E indicate that receptors associate with chaperones which protect the nuclear hormone receptors.

S. Region C contains the P-box and the D-box required for dimerization of the receptor and creates contact with DNA phosphate backbone.

Which one of the following is true ?

(a) P and Q

(b) Q and R

(d) R and S

Ans. (c)

Sol.

62. Monoclonal antibodies (mAb) can be potentially used as therapeutic agents. The major advantage is that they can specifically target aberrant cells. However, there is a practical difficulty. Monoclonals are raised in mouse and therefore it is expected that an immune reaction will develop if these are injected into humans. It is therefore necessary to humanize monoclonal antibody by

(a) expressing the genes for the mAb in cultured human cells and isolating the mAb from these cells.

(b) replacing the Fv region of a mAb with one derived from a human IgG.

(d) taking a human IgG and replacing the CDRs by those derived from the mouse mAb.

Ans. (d)

Sol. A type of antibody made in the laboratory by combining a human antibody with a small part of a mouse or rat monoclonal antibody. The mouse or rat part of the antibody binds to the target antigen, and the human part makes it less likely to be destroyed by the body's immune system.

63. Cytopalsmic determinants coding for anterior structure of Drosophila embryo, if injected elsewhre in the recipient embryo, would lead to

(a) normal development

(b) formation of additional ectopic head

(d) a phenotype with two heads and two tails

Ans. (b)

Sol. Recent evidence suggests that Bicoid acts in two ways to specify the anterior of the Drosophila embryo. First, it acts as a repressor of posterior formation. It does this by binding to and suppressing the translation of caudal RNA, which is found throughout the egg and early embryo.

64. In amphibian oocyte, the germpasm which gets segregated during cleavage to give rise to primordial germ cell (PGCs) is normally

(a) distributed evenly throughout the oocyte

(b) localized at animal pole

(d) aggregated in central part of oocyte

Ans. (c)

Sol. The germ plasm of anuran amphibians (frogs and toads) collects around the vegetal pole in the zygote. During cleavage, this material is brought upward through the yolky cytoplasm, and eventually becomes associated with the endodermal cells lining the floor of the blastocoel.

65. In mature Arabidopsis embryo, root apical meristem consists of cell derived from

(a) embryo and apical suspensor cell

(b) embryo only

(d) hypophysis only

Ans. (a)

Sol.

66. cAMP signaling plays a very important role in the development and differentiation of Dictyostelium discodideum. This morphogen is synthesized by different adenyl cyclases expressed at different stages of its life cycle. The following statments (P-S) refer to the effect of mutations in different adenyl cyclase genes :

P. aca deficient cells can be allowed to aggregate by exposing them to pulses of cAMP.

Q. acb deficient cells would normal fruiting bodies and the spores can germinate when exposed to favourabel conditions.

R. acg deficient cells develop normally and the spores germinate in the spore head itself.

S. spores formed from the acg deificient cells will germinate irrespective of the osmotic conditions.

Which of the above statement are correct ?

(a) P and S

(b) P only

(d) R and S

Ans. (a)

Sol.

67. During early cleavage of Caenorhabditis elegans embryos, each asymmetrical division produces one founder cells which produces differentiated descendants and one stem cell. The very first cell division produces one anterior founder cell, namely AB and one posterior stem cell, namely P1. When these blastomees are experimentally separated and allowed to proceed further with development, one could get the following possible outcomes :

(a) P1 cell would develop autonomously while the AB would show conditional development.

(b) P1 cell would show conditional development while AB would show autonomous development.

(d) both would show conditional specification and result in regulative development.

Ans. (a)

Sol. P1 cell would develop autonomously while the AB would show conditional development. Option (a) is correct.

68. Injection of noggin mRNA into a 1-cell, UV-irradiated embryos of frog completely rescues dorsal development and allows the formation of a complete embryo. Some of the following statements (P-S) could possibly explain this observation.

P. Noggin is a secreted protein which induces dorsal ectoderm to form neural tissue and it dorsalizes the mesoderm cells which would otherwise contribute to ventral mesoderm.

Q. Noggin binds directly to BMP4 and BMP2 thus preventing complex formation with their receptors.

R. Noggin along with other molecules prevent BMP from binding to and inducing ecotoderm and mesoderm cells near the organizer.

S. Noggin is a secreted protein which induces the dorsal ectoderm to form the epidermis and it ventralizes the mesoderm cells which would otherwise contribute to dorsal mesoderm.

Which of the above statements are correct ?

(a) P, Q and R

(b) P and Q

(d) P and S

Ans. (b)

Sol. This protein is involved in the development of many body tissues, including nerve tissue, muscles, and bones. Noggin's role in bone development makes it important for proper joint formation. Noggin interacts with members of a group of proteins called bone morphogenetic proteins (BMPs).

P. Noggin is a secreted protein which induces dorsal ectoderm to form neural tissue and it dorsalizes the mesoderm cells which would otherwise contribute to ventral mesoderm.

Q. Noggin binds directly to BMP4 and BMP2 thus preventing complex formation with their receptors.

69. In case of sea urchin, which of the following is the correct sequence of events taking place during thei nteraction of sperm and egg?

(a) Chemoattraction of sperm to the egg by soluble molecules secreted by the egg exocytosis of the sperm acrosomal vesicle to release its enzymes binding of the sperm to the extracellular matrix of the egg passage of sperm through this extracellular matrix fusion of egg and sperm cell membranes.

(b) Chemoattraction of sperm to the egg by soluble molecules secreted by the egg binding of the sperm to the extracellular matrix of the egg exocytosis of the sperm acrosomal vesicle to release its enzymes passage of sperm through this extracellular matrix fusion of egg and sperm cell membranes.

(c) Chemoattraction of sperm to the egg by soluble molecules secreted by the egg binding of the sperm to the extracellular matrix of the egg passage of sperm through this extracellular matrix exocytosis of the sperm acrosomal vesicle to release its enzymes fusion of egg and sperm cell membranes.

(d) Chemoattraction of sperm to the egg by soluble molecules secreted by the egg passage of sperm through this extracellular matrix binding of the sperm to the extracellular matrix of the egg exocytosis of the sperm acrosomal vesicle to release its enzymes fusion of egg and sperm cell membranes.

Ans. (a)

Sol. Option A is correct.

Chemoattraction of sperm to the egg by soluble molecules secreted by the egg exocytosis of the sperm acrosomal vesicle to release its enzymes binding of the sperm to the extracellular matrix of the egg passage of sperm through this extracellular matrix fusion of egg and sperm cell membranes.

70. When a wrist blastema from a recently cut Axolotl forelimb is placed on a host hind limb cut at the mid thigh level, it will generate only a wrist. The host (whose own hind limb was removed) will fill the gap and regenerate the limb up to the wrist. However, if the donor blastema is treated with retinoic acid on grafting, the wrist blastema will regenerate a complete limb and will not allow the host to fill the gap. This hahppends because retinoic acid

(a) helps in the proximalization of the blastema and activates the Hox genes differentially across the blastema.

(b) helps in the distalization of the blastema and activates the Hox genes differentially across the blastema.

(d) helps vigorous proliferation of the cells at the cut surface.

Ans. (a)

Sol. Retinoic Acid help in proximisation of blastema and activate Hox gene.

71. The transition to flowering in plants requires

(a) growth of plants under long-day conditions.

(b) growth of plants under short-day conditions.

(d) synthesis of the flowering hormone florigen.

Ans. (c)

Sol. At the time of flowering, the SAM switches fate and starts producing flowers instead of leaves. Correct timing of flowering in part determines reproductive success, and is therefore under environmental and endogenous control.

72. The dwarf pea mutant (Ie) used by Mendel was defective in which of the following enzyme involved in gibberellin biosynthesis ?

(a) ent-Kaurene synthase (b) GA 3-hydroxylase

Ans. (b)

Sol. GA 3 beta-hydroxylase catalyzes conversion of GA 20 (inactive) to GA1 by hydroxylation of C-3. Lack of enzyme GA 3 beta-hydroxylase cause dwarfism in plants.

73. Which of the following statements with respect to alternate oxidase activity in cyanide-resistant respiration in plants, is not correct ?

(a) Alternate oxidase accepts electrons directly from cytochrome C.

(b) Some plants exhibit thermogenesis during inflorescence development.

(d) When electrons pass to alternate oxidase, two sites of protein pumping are bypassed.

Ans. (a)

Sol. This enzyme was first identified as a distinct oxidase pathway from cytochrome c oxidase as the alternative oxidase is resistant to inhibition by the poison cyanide. Present in plants like Pea, spinach.

74. Fill in the blanks (a, b, c and d) in the following statements with a proper combination of m, n, o and p.

Wherein m represents – longer

n represents – shorter

o represents – prevents

p represents – induces

Short day (SD) plants flower when night lengths are a than a critical dark period. Interruption of the dark period by a brief light treatment b flowering in SD plants. Long day (LD) plants flower when night length is c than a critical period. Shortening of the night with a brief light treatment d flowering in LD plants.

(a) a-m, b-o, c-n, d-p

(b) a-n, b-p, c-m, d-o

(d) a-m, b-p, c-n, d-o

Ans. (a)

Sol. Short day plants flower when night lengths are longer than a critical dark period. Interruption of the dark period by brief light treatment prevents flowering in SD plants. Long day plants flower when night length is shorter than a critical period. Shortening of night with a brief light treatment induces flowering in LD plants

75. Mutants in Constans (CO) of Arabidopis taliana results in the flowering phenotype. Transcript levels of CO were determined in long day and shot day seedlings. Which of the following would likely represent the transcript profile of CO ?

Long day seedlings Short day seedings

(a) — —

(b) — —

(d) — —

Ans. (a)

Sol. The CONSTANS (CO) gene is a key regulatory component of the photoperiodic flowering time pathway. In long day flowering plants, such as Arabidopsis and barley, CONSTANS activates FT expression and flowering in long days. In rice, a short day flowering plant, Hd1, the ortholog of CONSTANS, activates flowering in short days and represses flowering in long days.

76. Following are some statements for synthesis of secondary metabolites in plants :

P. terpenes are synthesized by shikimic acid pathway and mevalonic acid pathway.

Q. alkaloids are nitrogen containing compounds and are synthesized by shikimic acid pathway.

R. phenolic compounds are synthesized by shikimic acid pathway and mevalonic acid pathway.

S. both alkaloids and terpenese are synthesized by mevalonic acid pathway and MEP pathway.

Which one of the following combinations of the above statement is true ?

(a) P and S

(b) P and R

(d) Q and S

Ans. (c)

Sol. Correction in point R. ( Malonic instead of mevalonic)

77. The dependence of the rate of sucrose uptake with respect to sucrose concentration in plant cell was studied and data is shown in the following graph.

From the above data it can be inferred that

(a) the sucrose upatake is energy independent and no special carrier is involved.

(b) the sucrose uptake is energy depenedent and a specia carrier is involved.

(d) at higher concentration of sucrose the uptake is energy dependent and carrier mediated.

Ans. (c)

Sol. Sucrose uptake shows a biphasic dependence on sucrose concentration. At lower concentration, uptake can be described as a carrier-mediated process.

78. During episodes of anoxia in plants, pyruvate produced in glycolysis is initially fermented to lactate. During later stage, there is an increase in the fermentation to ethanol and decrease in the fermentation to lactate, a phenomena which helps plants survive anoxia. Which of the following statements is correct about this change of fermentation flux from locate towards ethanol?

(a) The cytosolic pH increases, thus activating both lactate dehydrogenase and pyruvate decarboxylase activity.

(b) The cytosolic pH increases, thus inhibiting lactate dehydrogenase activity and activating pyruvate decarboxylase activity.

(d) The cytosolic pH decreases, thus inhibitng lactate dehydrogenase and activating pyruvate decarboxylase activity.

Ans. (d)

Sol. Acidification of the cytoplasm is a commonly observed response to oxygen deprivation in plant tissues that are intolerant of anoxia.Cells under anoxia experience an energy crisis; an early response thereof is a rapid cytoplasmic acidification of roughly half a pH unit. Depending on the degree of anoxia tolerance, this pH remains relatively stable for some time, but then drops further due to an energy shortage.

79. Atmospheric CO₂ contains the naturally occuring stable carbon isotopes ¹²C and ¹³C in the proportion of 98.9% and 1.1%, respectively. Following are some of the statements regarding CO₂ assimilation :

P. Both C₃ and C₄ plants assimilate less ¹³CO₂ than ¹²CO₂.

Q. Both C₃ and C₄ plants assimilate less ¹²CO₂ than ¹³CO₂.

R. C₃ plants assimilater lesser ¹³CO₂ than ¹²CO₂ as compared to C₄ plants.

S. C₄ plants assimilater lesser ¹³CO₂ than ¹²CO₂ as compared to C₃ plants.

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

(a) P and Q

(b) P and R

(d) P and S

Ans. (b)

Sol. The biochemical pathways have differing affinities for the lighter and heavier isotopes. This is due to the fact that 12C can "slide" through the C3 pathway more easily than 13C can. Same is true with the C4 pathway, just not as much.

80. During urine formation, the filtration of blood at the glomerular is

(a) an active process

(b) an osmotic process

(d) a non energy-mediated transport process

Ans. (c)

Sol. Glomerular filtration is a pressure dependent physical process. Glomerular filtrate is filter out in Bowman's capsule under net glomerular filtration pressure. It is calculated as

NFP = GBHP – (CHP + BCOP)

where, NFP is net filtration pressure

GBHP is Glomerular blood hydrostatic pressure

CHP is Capsular hydrostatic pressure

BCOP Blood colloid osmotic pressure

81. If the core body temperature of a human rises above normal, which of the following processes would be initiated sequentially for thermo-regulation?

(a) Peripheral vasodilation, increased rate of respiration, tacyacrdia.

(b) Peripheral vasoconstriction, increased rate of respiration, bradycardia.

(d) Peripheral vasodilation, decreased rate of respiration, bradycardia.

Ans. (a)

Sol. During increase in body temperature, thermo-regulation occurs by peripheral vasodilation and increase in respiration rate. It further leads to tachycardia.

82. Graves disease is associated with

(a) insufficiency of throid hormones

(b) excess of thyroid hormones

(d) excess of growth hormones

Ans. (b)

Sol. Graves diseases cause due to hyperthyroidism means excess secretion of thyroid hormones.

83. The time taken for atrial systole and diastole in a normal heart are t_as and t_ad seconds, respectively. If ventricular systole takes t_vs seconds, calculate the ventricular diastolic time (seconds).

(a) (t_as + t_ad) – t_vs

(b) (t_as + t_ad) + t_vs

(d) (t_as + t_ad) × t_vs

Ans. (a)

Sol. Ventricular diastolic time can be calculated as (t_as + t_ad) – t_vs.

84. Spinal cord of an animal was transected at C1/C2 level. The respiration of the animal stopped and it needed artificial respiration. However, the heart continued to beat although at a slower rate. Some of the explanations given were :

P. respiration regulatory centre is located in the medulla.

Q. respiration regulatory centre is located above the C1/C2 cut.

R. heart regulatory centre is above the C1/C2 cut.

S. heart has autoregulation.

Which of the following is most appropriate ?

(a) P only

(b) Q and R only

(d) Q, R and S only

Ans. (c)

Sol. Due to transection of spinal cord at C1/C2 region stops respiration because respiratory center lie in medulla oblongata and this transection cut out the connection between respiratory center and respiratory muscles. Cardiovascular center in medulla oblongata is also unable to maintain heart rate properly but it still continues to pump at slow rate because of autoregulation.

85. Which of the following graphs represents a normal sexual cycle in a normal human female ?

—— estrogen ------- progesteone

(a)

(b)

(c)

(d)

Ans. (b)

Sol. As per the facts that ovulation occurs at about 14 days of menstrual cycle, progesterone is released from corpus luteum after ovulation. High level of estrogen prior to ovulation is due to granulosa cells and menstruation occurs in initial 3-5days of menstrual cycle, graph in option b is correct.

86. An organism having heart for circulation, excretes through green glands. It has several ganglia and tactile organs on its body and its larval form is very different than its adult form. The organism is most likely to respire by :

P. exchanging oxygen and carbon dioxide through an extensive tracheal system.

Q. gaseous exchange over thinner areas of cuticle or by gills.

R. an efficient tracheal system that delivers oxygen directly to the tissues.

S. a double transport system, where the circulating fluid contains a dissolved respiratory pigment.

Chosse the correct option.

(a) P and R

(b) Only S

(d) Q and S

Ans. (c)

Sol. As per the characters given in question, the organism is a crustacean which comes under class arthropoda. They perform gaseous exchangement either through body surface (cuticle) or gills.

87. In a stressful condition, ACTH secretion was increased and as a result glucocorticoid concentration was elevated in blood. One or a combination of the following changes most likely taking place in this condition :

P. decreased circulating eosinophils and basophils.

Q. reduced IL2 release.

R. potentiated inflammatory response to tissue injury.

S. increased mitotic activity of lymphocytes in lymph nodes.

The correct answer is

(a) Q and R

(b) P and Q

(d) R and S

Ans. (b)

Sol. During stress conditions, immune response decrease due to low eosinophil, basophil and interleukin.

88. The following is the biochemical pathway for purple pigement production in flowers of sweet pea :

Colourless precursor 1 Colourless precursor 2 Purple pigment

Recessive mutation of either gene A or B leads to the formation of white flowers. A cross is made between two parents with the genotype : AaBb × aabb. Considering that the two genes are not linked, the phenotypes of the expected progenies are

(a) 9 purple : 7 white

(b) 3 white : 1 purple

(d) 9 purple : 6 light purple : 1 white

Ans. (b)

Sol. Parent : AaBb × aabb

If genes are not linked, their would be four different gametes i.e., AB, Ab, aB, ab. Individual A and Individual B cannot produce purple phenotype for this alleles of both genes must be dominant.

89. Aneuploid females with only one X-chromosome is a characteristic of individuals with

(a) cri du chat syndrome

(b) Klinefelter syndrome

(d) Turner syndrome

Ans. (d)

Sol. Turner syndrome results when one normal X chromosome is present in a female's cells and the other sex chromosome is missing or structurally altered. The missing genetic material affects development before and after birth.

90. A mechanism that can cause a gene to move from one linkage group to another is

(a) crossing over

(b) inversion

(d) duplication

Ans. (c)

Sol. A mechanism that can cause a gene to move from one linkage group to another is translocation.

91. A mother of blood group O has a group O child. The father could be of blood type

(a) A or B or O

(b) O only

(d) AB only

Ans. (a)

Sol. If a mother of blood group O has a group O child. The father could be of blood type A (I^AI⁰), blood group B (I^BI⁰) or blood group O.

92. The total variance in a phenotypic character can be split into two components-genetic (V_G) and environmental (V_E). The heritability of a phenotypic trait can be expressed quantitatively as heritability coefficient (h²) which is calculated as h² =

(a) V_G – V_E

(b) V_E/V_G

(c)

(d)

Ans. (c)

Sol. The heritability of a phenotypic trait can be expressed quantitatively as heritability coefficient (h²) which is calculated as h² = V_G/V_P. Here V_P = V_G + V_E.

93. In Neurospora a cross between the genotypes 'A' and 'a' results in an ascus with ascospores of genotypes as shown below :

Statements P and S are events that could have occurred during meiosis

P. Crossing over between the centromere and the gene.

Q. Segregation of alleles 'A' and 'a' in meiosis I.

R. Segregation of alleles 'A' and 'a' in meiosis II.

S. Assortment of alleles 'A' and 'a'.

Which of the above events could correctly explain the observation shown in the figure ?

(a) P followed by R

(b) P followed by Q

(d) S alone

Ans. (a)

Sol. This octad ratio i.e., 1:2:1 arises when alleles are segregated in II meiotic division which occurs when crossinover takes place between centromere and gene A.

94. When F1 female Drosophila of te genotype a⁺ a b⁺ b c⁺ c is test crossed, the following progenies were obtained :

*The progeny has been as classes derived from the female gamete.

Statements 1 to 6 as given below are conclusions derived from the above result

1. genes a and b are linked in cis.

2. genes a and b are linked in trans.

3. genes a and b are linked in cis while b and c are linked in trans.

4. the genotype of the parents are a⁺ a⁺ b⁺ b⁺ and aabb.

5. the genotype of the parents are a⁺ a⁺ bb and aab⁺ b⁺.

6. genes a and b are 10 cM apart.

Which of the above statements are correct ?

(a) 3 alone

(b) 1, 5 and 6

(d) 1, 4 and 6

Ans. (c)

Sol. Genes a and b are linked in trans, the genotype of the parents are a⁺ a⁺ bb and aab⁺ b⁺ and genes a and b are 10 cM apart.

Distance between gene A and B

95. The following is a hypothetical pathway for the development of wild type (red) eye color in an intersect :

Enzymes A and B are encoded by the genes a⁺ and b⁺, respectively.

The following statements are made regarding inheritance of the genes involved in the development of eye color :

P. when two heterozygous individuals of the genotype a⁺ ab⁺ b are mated, progenies with red, organge, brown and white eye color will be observed irrespective of whether the genes are independently assorting or showing incomplete linkage.

Q. when two heterozygous individuals of the gentoype a⁺ ab⁺ b are mated, progenies with red, orange, brown and white eye color will be observed in a ratio of 9 : 3 : 3 : 1, when the genes are independently assorting.

R. when a heterozygous individuals of the genotype a⁺ b/a b⁺ is test crossed, progenies with red and white eye color will be more in number.

S. when a heteozygous individual of the genotype a⁺ b/a b⁺ is test crossed, progenies with orange and brown eye colour will be more in number.

Which of the above statements is true ?

(a) P and R

(b) Q and R

(d) P, Q and S

Ans. (d)

Sol. As per the pathway mentioned in question following statements are true.

S. when a heteozygous individual of the genotype a⁺ b/a b⁺ is test crossed, progenies with orange and brown eye colour will be more in number.

96. Mendel crossed tall pea plants with dwarf ones. The F1 plants were all tall. When these F1 plants were selfed to produce F2 generation, he got a 3 : 1 tall to dwarf ratio in the offspring. What is the probability that out of three plants (of F2 generation) picked up at random two would be dwarf and one would be tall ?

(a) 3/4

(b) 3/8

(d) 9/32

Ans. (c)

Sol. By using binomial expansion,

where, P = Probability, n = Total number of events, x = number of desired events, p = individual probability of desired events and q = individual probability of other events.

97. Affected individuals from the pedigree given below are suffering from albinism, an autosomal recessive disease. Identify the confirmed carrier individuals in this pedigree assuming that the members coming from outside the family are homozygous for the dominant allele.

(a) III-2, III-3, III-5, III-6, II-1, II-3 and II-6.

(b) III-2, III-3, III-5, III-6, II-2, II-4, II-5 and I-2.

(d) III-1, III-4, III-7, II-2, II-4 and II-5.

Ans. (c)

Sol. Individuals mentioned below would be carrier for autosomal recessive albinism disease III-2, III-3, III-5, III-6, II-2, II-4 and II-5.

98. Imperfect fungi is a group represented bu fungal species which have

(a) simple mycelia

(b) unknown mechanism of sexual reproduction

(d) lost its survival mechanism against harsh environment

Ans. (b)

Sol. The fungi imperfecti, are fungi which do not fit into the commonly established taxonomic classifications of fungi that are based on biological species concept because their sexual form of reproduction has never been observed. They are known as Imperfect Fungi because only their asexual and vegetative phases are known.

99. Which of the following is not a characteristic of phylum chordata ?

(a) Pharyngeal slits

(b) Amniotic egg

(d) Notochord

Ans. (b)

Sol. Chordates have five basic traits: a notochord, a dorsal hollow (tubular) nerve cord, pharyngeal gill arches or slits, and a post-anal tail. Amphibians, falling under phylum chordata don't have amniotic eggs.

100. The most commonly used method of estimating primary productivity of a pond involvs measurement of the amount of

(a) CO₂ utilized

(b) autotroph biomass

(d) organic carbon

Ans. (c)

Sol. Primary productivity is a term that is used to describe the production levels of primary producers (plants) in an ecosystem. Primary production measures the new growth and biomass that is added to the ecosystem through primary producers.

101. A much greater proportion of energy fixed by autotrophs is transferred to the herbivore level in the open ocean ecostem than in a forest ecosystem because

(a) aquatic autotrophs are small.

(b) aquatic herbivores are more efficient feeders.

(d) terrestrial autotrophs have more indigestible tissues.

Ans. (d)

Sol. Due to presence of more morphological and structural complexity, autotrophs have more indigestible tissues.

102. Following figure shows McArthur and Wilson's equilibrium model of biota on a single island. In this figure, terms A, B, C and D in order are

(a) extinction, imigration, equilibrium number of species, size of species pool.

(b) immigration, extinction, equilibrium number of species, size of species pool.

(d) immigration, extinction, size of species pool, equilibrium number of species.

Ans. (b)

Sol.

103. Name the ectothermic animal that can termoregulate by behavioural means rather than by physiological means.

(a) Bumble bee in an orchord

(b) Tuna fish in the ocean

(d) Flatworm in a pond

Ans. (c)

Sol. Lizards regulate their internal body temperature by moving back and forth between 15 degrees and 45 degrees C environments to maintain colonic and brain temperatures.

104. Which of the following is a characteristic of an early seral community ?

(a) Narrow niche specialization

(b) High species diversity

(d) Open mineral cycling

Ans. (d)

Sol. Mineral cycling is the movement and exchange of inorganic and organic matter back into the production of matter.

105. In India, brown antlered deer (sangai) is found only in the floating landmasses of

(a) Wular lake

(b) Sasthamkotta lake

(d) Lok Tak lake

Ans. (d)

Sol. The brow-antlered deer or the dancing deer is found in its natural habitat only at Keibul Lamjao National Park over the floating biomass locally called phumdi in the southeastern part of Loktak Lake.

106. Two new plant species, A and B were described in 1872. Subsequently it was found that the type of species A was never designated and for species B there was one speciment designated as type but missing. As per International Code of Botanical Nomenclature (ICBN), typification should be

(a) neotype for A only

(b) neotype for both A and B

(d) lectotype for both A and B

Ans. (b)

Sol. A lectotype is a specimen served as the type specimen when no holotype was indicated at the time of publication. A neotype is a specimen served as the type specimen when the holotype is lost or destroyed.

107. Assume a new subspecies Ficus callosa subsp. microcarpa has been published by Jacobs. The nomenclature of the resulting entities would be

(a) F. callosa and F. callosa subsp. microcarpa Jacobs.

(b) F. callosa subsp. microcarpa Jacobs and other yet to be named subspecies of F. callosa.

(d) F. callosa subsp. callosa and F. callosa subsp. microcarpa Jacobs.

Ans. (b)

Sol. F.callosa and F.callosa subsp. microspora Jacobs.

108. An organism hasthe following architectural pattern :

1. Multicellular with germ layers.

2. A coelom derived from the mesoderm.

3. Primary bilateral symmetry with secondary radial symmetry.

4. Presence of endoskeletal plates.

Such an organism is most likely to

P. have mesohyl as its connective tissue.

Q. undergo torsion, whereby the mouth and anus are properly oriented.

R. be devoid of a brain but have calcareous spicules.

S. have comb plates to help in locomotion.

Which of the following is/are true ?

(a) P and R

(b) R only

(d) Q and R

Ans. (b)

Sol. Phylum Echinodermata have all the underlying features.

109. Identify the apomorphic characters marked in the cladogram :

(a) a-amniotic egg; b-4-chambered heart; c-anapsidan skull; d-diapsidan skull; e-synapsid skull.

(b) a-amniotic egg; b-synapsidan skull; c-4-chambered heart; d-anapsidan skuull; e-diapsidan skull.

(d) a-amniotic egg; b-synapsidan skull, c-anapsidan skull; d-diapsidan skull; e-4-chambered heart.

Ans. (d)

Sol. Amniotic egg present in all of these, synapsidan skull ( one temporal opening on each side of skull) feature present in mammalians, testudines have anapsidan skull ( lacks one or more skull openings) , lepidosauria, crocodile and aves have diapsidan skull ( has two pairs of temporal openings in skull). Chocodile and aves have 4-chambered heart.

110. According the MacArthur and Wilson's equilibrium theory, which of the following is true ?

(a) Larger islands and islands closer to continent are expected to have more species than smaller and isolated islands.

(b) Smaller islands and islands far form the continent are expected to have more species than larger and isolated islands.

(c) Smaller islands and islands closer to the continent are expected to have more species than far away smaller and isolated islands.

(d) More species are expected on all islands irrespective of their size and distance from the continent.

Ans. (a)

Sol. MacArthur and Wilson thus assume that there will be an equilibrial point where the immigration rate equals the extinction rate. They further hypothesize that an increase in island size will lower extinction curves while a decrease in distance between the island and the source region will raise immigration curves.

111. Primary production in aquatic ecosystem is measured using Light-and-Dak-Bottle tehcnique. In this method, as an indirect measure of photosynthetic production, dissolved oxygen concentration of the pond water enclosed in a BOD bottle is measured initially (I) and after a fixed duration of incubation in a light bottle (L) and a dark bottle (D). Then, the gross and net primary production are estimated as

(a) (L-D) and (L-I) respectively

(b) (L-I) and (L-D) respectively

(d) (L-D) and (I-D) respectively

Ans. (a)

Sol. To calculate NPP, you take the total amount of carbon that the plant fixes (or turns into usable material) and subtract the amount of carbon lost during respiration. The total amount of carbon taken in by the plant is known as the ?gross primary productivity? (GPP).

112. Identify the most appropriate cladogram that can be constructed using the data matrix given below, assuming '0's are plesiomorphic and '1's aer apomorphic characters.

(a)

(b)

(c)

(d)

Ans. (b)

Sol. A primitive or ancestral character state is called plesiomorphy. Apomorphic is unique to a group or species, rather than being derived from an ancestral group or species. A has the maximum number of plesiomorphic characters so is ancestor of all.

113. Gause's Competitive exclusion principle states that two species with identical niches cannot coexist indefinitely. Which of the following statements is the most appropriate regarding the validity of the principle ?

(a) It depends on how one defines niche.

(b) There are in nature many instances of continued coexistence of closely related species.

(d) It does not predict the outcome where both the species are equally strong competitiors.

Ans. (a)

Sol. The competitive exclusion principle says that two species can't coexist if they occupy exactly the same niche (competing for identical resources). Two species whose niches overlap may evolve by natural selection to have more distinct niches, resulting in resource partitioning.

114. The graph below shows the relationships of per capita population growth rate (r), fecundity (b) and age at first reproduction in an animal species.

What is the most important conclusion to be drawn from the graph ?

(a) The later age of first reproduction, the lowerr is the population growth rate achieved.

(b) The population growth rate decreases as first reproduction is postponed to a later state, regardless of the fecundity.

(d) As the age at first reproduction is postponed further, the benefits of increasing fecundity on the population growth rate become progressively negligible.

Ans. (d)

Sol. Fecundity is the reproductive output, usually of an individual or number of offspring produced or the capacity of reproduction . In the given curve, the population growth rate decreases as first reproduction is postponed to a later stage regardless of fecundity.

115. In a lake ecosystem, bottom-up effects (B) refer to control of a lower trophic level by the higher trophic levels and top-down effects (T) refer to the opposite. In a lake with three trophic levels-Phytoplankton (P), Zooplankton (Z) and primary Carnivore (C),

(a) P and C are controlled by B, and Z is controlled by T.

(b) P, Z and C are all controlled by T.

(d) P is controlled by T, Z is controlled by B and C is controlled by T.

Ans. (b)

Sol. Statements and options are not related.

Two models of community organization are common: the bottom-up model and the top-down model. Bottom-up model postulates a unidirectional influence from lower to higher trophic levels. In this case, the presence or absence of mineral nutrients controls plant numbers, which control herbivore numbers, which in turn control predator numbers.

In contrast, the top-down model postulates the opposite. Predation mainly controls community organization because predators limit herbivores, herbivores limit plants, and plants limit nutrient levels through their uptake of nu

116. Recent studies on Archea suggest that life could have originated

(a) extraterrestrially and seeded through meteorite impacts.

(b) in shallow coastal areas.

(d) in hot, terestrial habitats.

Ans. (c)

Sol. Archaea are microorganisms that define the limits of life on Earth. They were originally discovered in extreme environments, such as hydrothermal vents and terrestrial hot springs. They were also found in a diverse range of highly saline, acidic, and anaerobic environments

117. If the ratio of the number of non-synonymous to synonymous substitutions per site in protein coding gene is greater than one, it is an evidence of selection that is

(a) positive

(b) negative

(d) random

Ans. (a)

Sol. Some of the mutations are minor and do not change anything. These DNA mutations are called synonymous mutations. Others can change the gene that is expressed and the phenotype of the individual. Mutations that do change the amino acid, and usually the protein, are called nonsynonymous mutations. If this ratio is nearly 1, it refers to neutral evolution, when dN/dS < 1, negative selection pressure is assumed and when dN/dS > 1, the sequence evolves under positive selection.

118. The frequencies of allels 'A' and 'a' in a population at Hardy-Weinberg equilibrium are 0.7 and 0.3 respectively. In a random sample of 250 individuals taken from the population, how many are expected to be heterozygous ?

(a) 112

(b) 81

(d) 145

Ans. (c)

Sol. Frequency of allele A = p = 0.7 = 70% and frequency of allele a = q = 0.3 = 30%

Total number of individuals (p2 + 2pq + q2) = 250 and heterozygous genotype (2pq) = 2 × 0.7 × 0.3 = 0.42 = 42%

So, the number of heterozygous genotypes 250*42/100= 105

119. In an altruistic act, if a donor sacrifices 'C' offspring which helps the recipient of gain 'B' offspring and the donor is related to the recipient by a coefficient , under which condition would kin selection favour this altruistic trait ?

(a) B > C

(b) B > C

(d) B – C > 0

Ans. (d)

Sol. Kin selection by definition is a unique type of natural selection where the reproductive success of relatives is favored over an individual's success. In kin selection, an individual will often sacrifice themselves or their chances of reproduction to help ensure that relatives have a chance to reproduce and pass on their traits. Kin selection plays an important role in the evolution of certain species.

120. The photoreceptor commonly involved in light entrainment of the biological clock in flies, moulds and plants is

(a) pytochrome

(b) rhodopsin

(d) cryptochrome

Ans. (b)

Sol. Cryptochromes are photoreceptors that regulate entrainment by light of the circadian clock in plants and animals. They also act as integral parts of the central circadian oscillator in animal brains and as receptors controlling photomorphogenesis in response to blue or ultraviolet (UV-A) light in plants.

121. Using molecular clock, it was estimated that two species A and B must have diverged from their common ancestor about 9 × 10⁶ years ago. If the raete of divergence per base pair is estimated to be 0.0015 per million years, what is the proportion of base pairs that differ between the two species now ?

(a) 0.0270

(b) 0.0135

(d) 0.0035

Ans. (a)

Sol. As we know that, D = 2 × r × t; where, "D = proportion of base pairs that differ between the two sequences.

r = the rate of divergence per base pair per million years.

t = the time in million years since the species common ancestor.

2 = two diverging lineages and D = 2 × 0.0015 × 9 = 0.0270.

122. What will be the approximate effective population size in a panmictic population of 240 with 200 females and 40 polygamous males ?

(a) 160

(b) 133

(d) 67

Ans. (b)

Sol. As we know that, effective population size, Ne= 4NmNf/Nm+Nf—

Where,

Nm = Number of breeding males

Nf = Number of breeding females.

Ne = 4*40*200/40+200 =32000/240 = 133

123. An animal was first maintained in a constant environmental condition for several days until a consistent biological rhythm (B) was established. The animal was then exposed to an experimental physical rhythm (E) which modulates the phase and period of B. However, upon withdrawal of E, the B gradually regained its pattern of pre-exposure condition. From these observations which one or more of the following should be the most logical inference ?

P. E is a Zeitgeber

Q. E is masking agent

R. E causes entertainment of B

S. B is a conditioned to E

The correct answer is

(a) P and R

(b) Q and S

(d) S only

Ans. (a)

Sol. The term zeitgeber (literally, time giver or time cue) refers to environmental variables that are capable of acting as circadian time cues. The light/dark cycle is the most important zeitgeber, but other stimuli such as melatonin can also function as zeitgebers. E is a zeitgeber and acts as masking agent for B.

124. The following geological eras mark the advent of important events in the history of earth-origin of terrestrial plants, origin of mammals, and breakup of the land mass Pangaea;

P. early Cambrian

Q. late Devonian

R. late Cretaceous

S. late Jurassic

Identify the correct match of the events with the geological era.

(a) Origin of terestrial plants (R); origin of mammals-(S); break-up of Pangaea-(P)

(b) Origin of terestrial plants (Q); origin of mammals-(S); break-up of Pangaea-(R)

(d) Origin of terestrial plants (Q); origin of mammals-(R); break-up of Pangaea-(S)

Ans. (b)

Sol. Origin of terrestrial plants- late Devonian"Origin of mammala- late Jurassic"Break-up of Pangaea- late Cretaceous

125. Shown in the graph below are the fitness costs and benefits of four alternative behavioural phenotypes (P, Q, R, S). Give sufficient evolutionary time, which phenotype(s) is likely to evolve as an adaptation ?

(a) Q and R

(b) Q only

(d) S only

Ans. (b)

Sol. Behavioral phenotype, Q is likely to evolve as an adaptation in sufficient evolutionary time due to high benefit and low cost.

126. Which of the following food crops has recently been genetically engineered to obtain edible vaccine to develop immunity against hepatitis B ?

(a) Banana

(b) Maize

(d) Tomato

Ans. (c)

Sol. Edible vaccines are plant parts which contain an antigen introduced using recombinant DNA technology and therefore stimulate antibody formation in the host. First successful attempt for the development of edible vaccine against hepatitis B was made in recombinant potato. Later other plants such as transgenic lupin and lettuce have also been used to make edible vaccines hepatitis B.

127. Release of nutrients, oxidants or electron donors into the environment to stimulate naturally occuring microorganisms to degrade a contaminant, is referred to as

(a) biostimulation

(b) phytoremediation

(d) bioremediation

Ans. (a)

Sol. Biostimulation involves the modification of the environment to stimulate existing bacteria capable of bioremediation. This can be done by addition of various forms of rate limiting nutrients and electron acceptors, such as phosphorus, nitrogen, oxygen, or carbon (e.g. in the form of molasses).

128. ELISA assay uses

(a) an enzyme which can react with secondary antibody.

(b) an enzyme which can react with the antigen.

(d) a radiolabelled secondary antibody.

Ans. (c)

Sol. The enzyme-linked immunosorbent assay (ELISA) is an immunological assay commonly used to measure antibodies, antigens, proteins and glycoproteins in biological samples. Some examples include: diagnosis of HIV infection, pregnancy tests, and measurement of cytokines or soluble receptors in cell supernatant or serum. In ELISA, various antigen-antibody combinations are used, always including an enzyme-labeled antigen or antibody, enzyme activity is measured using a substrate that changes color when modified by the enzyme.

129. Routinely used glucose biosensor estimates blood glucose level by sensing the concentration of

(a) glucose

(b) oxygen

(c)

(d) H₂O₂

Ans. (b)

Sol. Three general strategies are used for the electrochemical sensing of glucose; by measuring oxygen consumption, by measuring the amount of hydrogen peroxide produced by the enzyme reaction or by using a diffusible or immobilized mediator to transfer the electrons from the GOx to the electrode.

130. Which of the following methods of plant transformation can be used to introduce a gene into chloroplast genome ?

(a) Agrobacterium-mediated transformation

(b) Particle delivery system

(d) Electroporation

Ans. (b)

Sol. The plasmid DNA is coated on the surface of the microparticles of either gold or tungsten and then shot on to the abaxial surface of 4- to 6-week-old sterile leaves using a gene gun. The bombarded leaves are incubated for 48 h in the dark, cut into small discs and placed on regeneration medium supplemented with the appropriate antibiotic and hormones. Primary shoots generally arise within 2–3 months.

131. Yeast artificial chromosome (YAC) vectors contain selectable markers. Loss of which marker at the cloning site distinguishes the re-ligated YACs from the orignial vector marker ?

(a) TRP1

(b) SUP4

(d) CEN

Ans. (b)

Sol. The yeast SUP4 gene contains a cloning site used as a colour marker for selection of YACs containing exogenous insert DNA.

132. The µ and of wing lenth (a normally distributed parameter) in a population of fruitflies are 4 and 0.2 mm, respectively. In a random sample of 400 fruitflies, how many individuals are expected to have wing lengths greater than 4.4 mm ?

(a) 20

(b) 64

(d) 336

Ans. (a)

Sol.

For rejection of the hypothesis,

Where, µ = sample mean, = standard deviation and n = size of the sample and z* = 1.96 (at 95% confidence interval)

L = lower estimate of the pouplation and U = upper estimate of the population.

= 4 ± 1.96 × 0.01 = 4 ± 0.01 = 4 ± 0.0196 = 4.0196, 3.9804

It means that out of 100 fruitflies of exactly 4 mm length we are confident that 95 of them are derived from fruitflies of lengths between 4.0196 mm and 3.9804 mm. The other 5 are probably outside these limits.

So, total number of fruitflies whose length is beyond the limit of upper and lower estimate = 0.05 × 400 = 20.

Hence, the number of fruitflies whose length is greater than 4.4 mm = 20/2 = 10.

133. Industrial products in which bacteria are employed for production are shown in the following table :

The correct combination is

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

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

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

Ans. (a)

Sol. 2,3-Butanediol - Bacillus polymyxa

Dextran - Leuconostoc

Glutamic acid- Brevibacterium

Cobalmin- Propionibacterium

134. A transposon carrying a promoterless -galacrosidase (Iac Z) was used to create insertional mutation in the vir region of Ti-plasmid of Agrobacterium tumefaciens. All the mutants in which Iac Z fusion was in frame were divided into the following three gorups :

A. The virulence of the bacteria was completely lost and the Iac Z was induced by acetosyringone.

B. The virulence of the bacteria was reduced and the Iac Z was induced by acetosyringone.

C. The virulence of the bacteria was completely lost and Iac Z was not induced by aceptosyringone.

Which of the following assumptions are valid about these mutants ?

(a) In group A, the insertion could be in vir B, C, D or G; in group B the insertion could be vir C or E; and in group C the insertion could be in vir A or G.

(b) In group A, the insertion could be in vir A, B, C, or D; in group B the insertion could be vir C or D; and in group C the insertion could be in vir G.

(c) In group A, the insertion could be in vir A; in group B the insertion could be vir B; and in group C the insertion could be in vir C.

(d) In group A, the insertion could be in vir G; in group B the insertion could be vir B, D and E; and in group C the insertion could be in vir A.

Ans. (a)

Sol.

Agrobacterium tumefaciens is a plant pathogen with the capacity to deliver a segment of oncogenic DNA carried on a large plasmid called the tumor-inducing or Ti plasmid to susceptible plant cells.

135. To detect mutation (GAG GTG) allele specific hybridization method is used. For members of an affected family are investigated. DNA isolated from blood samples of paretns and two offsprings are spotted on a membrane after appropriate processing and probed with their TGACTCCTGAGGAGAAGTC (first probe) or TGACTCCTGTGGAGAAGTC (second probe) after labelling. While probed with first oligonucleotide, signals are obtained for the positions where DNA are spotted from parents and offspring II. When probed with second oligonucleotide, signals are obtained at position where DNA from the paretns and offspring I spotted. Results are shown below:

On the basis fof the result, which of the following statements is correct ?

(a) Parents are affected

(b) Offspring I is affected

(d) Offspring II is affected

Ans. (b)

Sol. By observing result, it is visualized that first probe is absent in offsping I and due to absence of probe I, offsprin I is affected.

136. ³²P-labelled genomic DNA from Bacillus subtilis (A) and cold genomic DNA from Clostridium jejeuni (B) mixed in various properties was used to transform Bacillus subtilis. The result obtained are tabulated below :

The following interpretation (P to S) could be made :

P. The transformation is dependent on recombination between homologous sequences.

Q. Cells did not distinguish between homologous or heterologous DNA sequences for uptake of DNA

R. DNA uptake is based on specific receptors.

S. DNA degradation dictates transformation efficiency.

Which of the following interpretations are correct ?

(a) P and Q

(b) P and R

(d) P and S

Ans. (a)

Sol. By observing the result in table, we can say that the transformation is dependent on recombination between homologous sequences and cells did not distinguish between homologous or heterologous DNA sequences for uptake of DNA.

137. You are studying the binding of proteins to the cytoplasmic face of cultured liver cells and have found a method that gives a good yield of inside-out vesicles from the plasma membrane. Unfortunately, your preparations are contaminated with variable amounts of right-side-out vesicles. Nothing you have tried avoids this contamination. Somebody suggests that you pass the vesicles over an affinity column made of lectin coupled to Sepharose beads. What is the rational of this suggestion ?

(a) Right-side-out-vesicles will be lysed by lectin coupled to Sepharose beads.

(b) Right-side-out-vesicles will simply bind to the lectin coupled Sepharose beads.

(d) Lecting will bind to only glycoproteins and glycolipids present on the inside the vesicles.

Ans. (b)

Sol. Glycoproteins and polysaccharides react reversibly, via specific sugar residues, with a group of proteins known as lectins. As ligands for purification media, lectins are used to isolate and separate glycoproteins, glycolipids, polysaccharides, subcellular particles and cells, and to purify detergent-solubilized cell membrane components. Substances bound to the lectin are resolved by using a gradient of ionic strength or of a competitive binding substance.

138. A researcher has isolated a restriction endonucleases that cleaves at only one specific 10 base pair site.

P. Would this enzyme be useful in protecting cells from viral infections, given that a typical viral genome is 5 × 10⁴ base pairs long ?

Q. Restriction endonucleases are slow enzymes with turnover number is 1s^–1. Suppose the isolated endonuclease was faster with turnover numbers similar to those for carbonic anhydrase (10⁶ s^–1), would this increased rate be beneficial to host cells, assuming that the fast enzymes have similar levels of specificity ?

The correct combination of answer is

(a) P : No Q : Yes

(b) P : No Q : No

(d) P : Yes Q : Yes

Ans. (a)

Sol. The restrcition endonucleases that cleaves at only one specific 10 base pair site can't cleave viral genome of 50,000 base pairs. Because there is very less chance of presence of restrcition sites in 50,000 base pair long which is 10 base pairs sequence. If turnover of restriction endonucleases is faster than carbonic anhydrase, it would increase rate in host cell.

139. Bacteria often acquire genes by the process of lateral or horizontal transfer. Such foreign genes, if acquire in recent past, may be identified by their atypical GC content, as compared to native genes. Suppose the genomic GC content of a bacterium is 40%. Gene A of this organism contains 1000 bases with 225 G and 215 C. Another gene B of length 800 bases contains 160 G and 140 C. Which one of the following would be the most acceptable hypothesis (given that = 3.841 at 0.05 significance level) ?

(a) A : native, B : foreign

(b) A : foreign, B : native

(d) A : native, B : native

Ans. (b)

Sol. Gene A of this organism contains 1000 bases with 225 G and 215 C. Hence, 225 G = 225 C and 215 C = 215 G

Observed value of total GC content = 440 bp and expected value of total GC content = 400 bp

As we know that, ; where, O = observed value and E = expected value.

Gene B contains 800 bases with 160 G and 140 C. Hence, 160 G = 160 C and 140 C = 140 G

Observed value of total GC content = 300 bp and expected value of total GC content = 320 bp

Given that, = 3.841 at 0.05 significance level i.e., 95% conficence interval. The gene B lies in the significance level means that gene B is a native.

140. The MALDI spectrum of a peptide shows a peak at m/z corresponding to 3600. When the ESI spectrum is recorded, peaks at m/z corresponding to 721,904 and 1801 were obtained. When the MALDI MS/MS spectrum was recorded, lage number of peaks with m/z less than 3600 were observed. The spectral data indicate that the peptide is

(a) highly impure

(b) pure with molecular mass of 3600 and patial sequence of the peptide can be determined.

(d) degraded under condition employed for recording ESI spectrum.

Ans. (b)

Sol. The MALDI spectrum of a peptide shows a peak at m/z corresponding to 3600. When the ESI spectrum is recorded, peaks at m/z corresponding to 721,904 and 1801 were obtained. When the MALDI MS/MS spectrum was recorded, lage number of peaks with m/z less than 3600 were observed. The spectral data indicate that the peptide is pure with molecularr mass of 3600 and patial sequence of the peptide can be determined.

141. In one study, a group of 5 day rat pups were fed for 3 weeks a diet A and the pups gained weight by 300%. In a second stuyd, when the same diet fed for 3 weeks to rats of 350 gms, they did not gain weight significantly. In a third study, a diet B was fed to 250-350 gms rats and it was observed that they delivered normal pups after five weeks. Based on these observations which of the following statements is correct ?

(a) Diet A facilitates weight gain than diet B.

(b) Diet B facilitates pregnancy and child-bearing.

(d) Diet A is more energy containing than diet 'B'. Hence, its quantity should be reduced.

Ans. (c)

Sol. Based on the observations provided in question, More control experiments are to be conducted for definitive conclusion.

142. Proteins in cells can be visualized by the following methods :

P. Express the gene (coding for the said protein) as a fusion with the green fluoroescence protein (GFP) and directly visualize under a fluorescence microscope.

Q. Express the gene (coding for the said protein) as a fusion withe the β-galactosidase gene (Iac Z) and directly visualize under a phase contrast bright field microscope.

R. A flurosecene tagged antibody raised against the said protein could be used for visualization in a fluoroescence microscope.

S. Overexpress the protein and directly visualize it under a scanning electron microscope.

Which of the following methods you would choose to visualize a protein in a living cell ?

(a) P only

(b) P and Q only

(d) S only

Ans. (a)

Sol. Common to all techniques, the critical first step to visualizing protein interactions in living cells is fluorescent tagging of proteins with fluorophores that have suitable spectral properties and minimal impact on the biological functions.

143. Insertion lambda vectors are used to create cDNA libraries. In some insertion vectors, the site of insertion is within the cI gene. The recombinants which are cI can then be screeened by plating on E. coli hfl. The gene hfl encodes a protease that controls lytic-lysogeny decision through which mechanism ?

(a) Hfl protease degrades cI, thereby promoting lytic pathway.

(b) Hfl protease degrades cIII and so cIII cannot interact, with cII. The lysogenic pathwya is thus is preferred.

(d) Hfl protease degrades INT, the protein involved in phage integartion and the lytic pathway is initiated.

Ans. (c)

Sol. Insertion lambda vectors are used to create cDNA libraries. In some insertion vectors, the site of insertion is within the cI gene. The recombinants which are cI can then be screeened by plating on E. coli hfl. Here, Hfl protease degrades cII, and therefore cI synthesis cannot be established, thus lytic pathway is preferred.

144. A 30-residue peptide was treated with trypsin and the tryptic peptides were separated by HPLC. Four peaks A, B, C and D wre obtained. Peptides corresponding to A, B, C and D were reduced and alkylated selectively at cysteine residues. The sequences obtained from A, B, C and D after reduction and alkylation were : A, AEK; B, C(S-alkyl)EPGYR and WC(S-alkyl)SPPK; C, C(S-alkyl) GGK; D, C (S-alkyl) EAFC(S-alkyl)L. The sequence of the 30-residue peptide is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Sequence of amino acid in option (c) is correct.

145. Molecular beacons (MB) and Taqman (TQ) are used as probes in Real time PCR experiments. Both these probes are based on the principle FRET and employ and flurophore (F) and a quencher (Q). However the mechanism by which they function are differnet as illustrated below :

At what stage of the PCR we would be able to detect fluorescence ?

(a) Annealing step for both

(b) Extension step for both

(d) Extension for A and annealing for B

Ans. (c)

Sol. The molecular beacons are the ideal hybridization-based probe for short oligonucleotide detection, and, thus, it is a suitable probe for real-time PCR. In real-time PCR, MB hybridizes with template DNA at the annealing step and produces the fluorescent signal directly. When the reporter dye molecule is released from the probe and is no longer in close proximity to the quencher dye, it can begin to fluoresce. Increase in the fluorescent signal results if the target sequence is complementary to the TaqMan probe.

CSIR NET BIOLOGY (JUNE - 2011)
Previous Year Question Paper with Solution.

Locate Us

Company

Courses

Academies

CSIR NET BIOLOGY (JUNE - 2011) Previous Year Question Paper with Solution.

Locate Us

Company

Courses

Academies

Get in touch

CSIR NET BIOLOGY (JUNE - 2011)
Previous Year Question Paper with Solution.