PYQ DEC 2014 - VedPrep

CSIR NET BIOLOGY (DEC - 2014)
Previous Year Question Paper with Solution.

21. Reaction products inhibit catalysis in enzymes by

(a) covalently binding to the enzyme

(b) altering the enzyme structure

(d) form a complex with the substrate.

Ans. (c)

Sol. For whatever fraction of time a competitive inhibitor molecule is occupying the active site, the enzyme is unavailable for catalysis. The overall effect is as if the enzyme cannot bind substrate as well when the inhibitor is present. Thus, we except that the enzyme would act as if the K_m where increased by the prsence of the inhibitor.

22. Chirality of DNA is due to

(a) the bases

(b) base stacking

(d) deoxyribose

Ans. (d)

Sol. Deoxyribose or more precisely

2-deoxyribose is a monosaccharide with idealized formula

H – (C = 0) – (CH₂) – (CHOH)₃ – H

Deoxyribozymes or DNA enzymes or catalytic DNA or DNAzymes are DNA molecules that have the ability to perform a chemical reaction, such as catalyticaction.

Chirality is a property that a DNAzyme can exploit. DNA occurs in nature as a right-handed double helix and in asymmetric synthesis a chiral molecules from an achiral source.

23. In proteins, hydrogen bonds form as follows : Donor (D) –H– – – Acceptor (A). Hydrogen bond is more favorable if the angle between D – H and A is

(a) < 90°

(b) 180°

(d) 120°

Ans. (b)

Sol. Hydrogen bond : A very weak bond formed by the electrostatic attraction between hydrogen which is covalently bonded to an electro-negative atom, such as oxygen or nitrogen to other electro-negative atom. Hydrogen bond is most favourable and strong when all the three bonded atoms are in a straight line (180°). When hydrogen bonds are structurally constrained, the geometry will losse and all the bonds become weaker.

24. Proton motive force during oxidative phosphorylation is generated in mitochondria by

(a) exchanging products for sodium ions

(b) pumping protons out into interembrane space

(d) hydrolysis of ATP

Ans. (b)

Sol. The ETC creates the pmf by pumping protons from the mitochondrial matrix to the intermembrane space (IMS), a process driven by the oxidation of reducing equivalents from metabolic substrates. The pmf across the mitochondrial inner membrane (IM) manifests as a charge gradient and a chemical gradient .

25. Acetyl-(Ala)₁₈–CONH₂ exists in -helical conformation in solution. Most of the backbone dihedral angles will be

(a) – 60°, –30°

(b) 60°, 30°

(d) –80°, 120°

Ans. (a)

Sol.

Check the angle of -helix. They fall on negative scale (–, –).

26. Enzyme parameters of four isozymes is given below:

Isozmye K_m micromolar V_max

A 0.1 15

B 1.5 45

C 4.0 100

D 0.01 10

These isozymes are localized in different tissues. In liver the substrate concentration is 0.2 micromolar. The liver isozyme is likely to be

(a) A

(b) B

(d) D

Ans. (a)

Sol. Experimentally, the K_m for an enzyme tends to be similar to the cellular concentration of its substrate. An enzyme that acts on a substrate present at a very low concentration in the cell usually has a lower K_m than an enzyme that acts on a substrate that is more abundant. K_m of Isozyme A (0.1) almost similar to liver substrate concentration (0.2).

27. DNA is not hydrolyzed by alkali whereas RNA is readily hydrolyzed. This is due to

(a) the double helical structre of DNA

(b) features observed in RNA such as stem-loop structures

(d) the presence of 2'–OH group in RNA

Ans. (d)

Sol. RNA hydrolysis is a reaction in which a phosphodiester bond in the sugar-phosphate backbone of RNA is broken, cleaving the RNA, molecule. RNA is susceptible to this base-catalyzed hydrolysis because the ribose sugar in RNA has a hydroxyl group at the 2' position. This feature makes RNA chemically unstable compared to DNA, which does not have this 2' OH group and thus is not susceptible to base-catalyzed hydrolysis.

28. The lifetime of a peptide bond in proteins is very large (~1000 years). Which statement below is incorrect with respect to stability of the peptide bond?

(a) The free energy of hydrolysis in negative

(b) The free energy of hydrolysis is positive and large

(d) The peptide bond can be hydrolyzed by 6 N HCl at 100°C.

Ans. (b)

Sol. Peptide bond hydrolysis is favourable and means free energy alwasy negative. Peptide bond easily hydrolysed in presence of 6N HCl.

29. Which of the following statements regarding membrane transport is false?

(a) Polar and charged solutes will not cross cell membranes effectively without specific proteing carriers

(b) Each protein carrier will only bind and transport one (or a few very similar) type of solute

(d) Ions are typically transported by special proteins that form membrane channels.

Ans. (c)

Sol. Due to the non-polar lipid components, polar and charged molecules will not transport directly through the plasma membrane. They require specific transporters which may be operated actively or passively.

For the transporatation of glucose across plasma membrane, cell may use facilitated diffusion or active transport depending upon the specific needs and cell type. For example sugars are transported through active transport from gut to the intestinal epithelial cells where as in the case of RBC glucose is transported by facilitated diffusion. Whether active or facilitated diffusion, the glucose molecule cannot be passed directly through the membrane. They require specific transporters called Glucose Transporters (GLUTs). GLUTs are a class of membrane proteins that facilitate the transport of glucose across plasma membrane.

30. What will happen if histones are depleted from a metaphase chromosome and viewed under a transmission electron microscope?

(a) 30 nm chromatin fibres will be observed

(b) 10 nm chromatin fibres will be observed

(d) A huge number of loops of DNA fibres without scaffold will be observed.

Ans. (c)

Sol. Maximum condensation of chromosome in a cell can be visualized in the metaphase phase of cell division. The metaphase chromosome, actually consists of two sister chromatids connected by a complex called kinetochore at the centromere, will be having an average thickness of 1400 nm (700 nm to each sister chromatids). The formation of chromosome during cell division is by the condensation of genetic material (DNA) with a variety of proteins generally classified as histones and non-histones. The individual DNA double helix will be having a diameter of 2 nm. The 2 nm DNA helix coils around a histone octamer (2 each of H2A, H2B, H3 and H4) to form a "bead or string" like structure having 11 nm thickness. The "beads" are actually the nucleosome particles. Each histone octamer will be surrounded by 1.65 turns of DNA with 146 to 147 nucleotides. The 11 nm "nucleosome beads on string" further coil around with the help of another histone called H1 (linker histone) to form a 30 nm solenoid model. In order to have higher level of coiling, the 30 nm fibre loops around a cellular protein scaffold. The protein scaffold consists mainly of topoisomerase II and other non-histone proteins. Looping 30 nm fibre around the scaffold create 300 nm to 700 nm metaphase sister chromosomes. Chromosome binds to scaffold proteins at specific regions called Scaffold Attachment Region (SARs). In the question only histone proteins are depleted which means the scaffold will not be affected. Thus, the option 'c' i.e., a scaffold and large numebr of loops DNA fibres will be observed under transmission electron microscope.

31. Which of the following statements about meiosis is not true?

(a) Kinetochores of sister chromatids attach to opposite poles in Meiosis I.

(b) Kinetochores of sister chromatids attach to opposite poles in Meiosis II.

(d) Homologous chromosomes are segregated in Meiosis I.

Ans. (a)

Sol. The homologous chromosomes pairs (prophase I) and they separate to two opposite poles after the completion of Meiosis I. Splitting of kinetochore will not occurs during meiosis I (kinetochore : a protien compelx which holds the two sister chromatids together and to which holds the two sister chromatids together and to which the spindle fibres attach during cell division). Splitting of kinetochore and separation of sister chromatids takes place only during meiosis II.

32. In order to prove that liposome can serve as a model membrane (mimicking cellular plasma membrane) and can be used as a target for complement-mediated immunolysis, an experiment as below is designed. To initiate such experiment, hapten-conjugated liposomes are made and loaded with umbelliferyl phosphate (UMP; hydrolysed product of UMP is umbelliferone and is fluorescent). Such loaded, hapten-conjugated liposomes in 10 mM Tris buffered saline, pH 7.4 were mixed with anti-hapten antibodies and fresh guinea pig serum (as a source of complement) to induce immmunolysis of liposomal membrane. To quantify only the membrane lysis component which of the assay sequences below is most appropriate?

(a) Mixture is ultracentrifuged and the supernatant reacted with alkaline phosphatase and fluorescence measured.

(b) Mixture is sequentially reacted with phospholipase and alkaline phosphatase followed by fluorescence measurements.

(d) Mixture is treated with Triton X-100 and fluorescence measured.

Ans. (a)

Sol. If membrane already lysed, UMP fluorescence will directly measured in supernant.

33. If a proteasome inhibitor is added to synchronously cycling human cells in G2 phase, which one of the following events is likely to happen?

(a) Induce re-replication of DNA

(b) Arrest cells in G2 phase

(d) Block chromatin condensation

Ans. (c)

Sol. The addition of proteasome inhibitor ot G₂ cells causes the cells arrest at anaphase. For metaphase to anaphase progression requires the activation of Anaphase Promoting Factor (APF), which causes ubiquitination of two target proteins, one is Securin and other is cyclin-B. Both Securin and cyclin-B are degraded by proteasome; hence cell enters in to anaphase. So, by inhibiting proteasome causes the cell arrest at anaphase.

34. A null mutation is created in a gene which is responsible for specific phosphorylation at 6^th carbon position of mannose on acid hydrolases occurring in cis-Golgi. The following statements are given towards explaining the effect of this mutation:

P. The lysosomes will be devoid of lysosomal enzymes

Q. Lysosomal enzymes will be secreated out

R. Lysosomal enzymes will get localized in cytoplasm.

Which statement or combination of statements will explain the effect of mutation if the acid hydrolases in the mutant do not get degraded?

(a) P and R

(b) Q and R

(d) P and Q

Ans. (d)

Sol. In the cis-Golgi a GlcNAc phosphotranferase adds a GlcNAc-1-phosphate residue onto the 6-hydroxl group of a specific mannose residue within the oligosachharide. This forms a phosphodiester : Man phosphate GlcNAc. Once the phosphodiester has been formed the lysosomal enzyme will be translocated through the Golgi apparatus to the trans-Golgi.

35. A newly identified sequence was experimentally tested by in vitro transport assay using a radiolabelled protein containing the sequence to test import into mitochondria. Trnsport assay was done for a short time with or without membrane potential and after the assay, the mitochondria were either treated or not treated with proteinase K. At the end of the assay the mitochondria were pelleted and total protein of the pellet was isolated and separated on SDS-PAGE and autoradiographed. A represntative autoradiogram is shown below. Based on this experimental data, which of the following statements is not correct?

(a) The protein goes into the matrix

(b) Not all of the added protein was imported

(d) The protein in associated with the outer mitochondrial

Ans. (d)

Sol. Protein import into mitochondria required membrane potential. Presence of Proteinase K will degrade protein outside of mitochondria.

36. Leader sequence in some of the protozoan parasites is transcribed elsewhere in the parasite genome and gets joined with several transcripts to make the functional RNA. The joining of the two transcripts occur by the process of

(a) alternate splicing

(b) trans splicing

(d) RNA editing

Ans. (b)

Sol. Trans splicing : A type of RNA splicing found in some eukaryotes where exons of two separate primary transcripts were joined end to end (ligated) to form a single functional mRNA. The usual splicing where the introns are removed and the remaining exons are joined with in a primary transcript is called cis-splicing. Trans splicing is responsible for the oncogenic properties of some transcripts (fusion transcripts). The process of trans-splicing was first discovered in Trypnosoma and later shown to occur in Caenorhabditis elegans and other nematodes.

RNA editing : Process of changing the specific nucleotide sequence of RNA after it has been produced by transcription. The post-transcriptional modification should not be confucsed with RNA editing. The post-transcriptional modification like splicing 5' capping 3' polyadenylation are not considered as RNA editing.

Alternatie splicing : A method of splicing by which single gene can produce multiple proteins with different functions. In alternate splicing the differential inclusion or exclusion of mRNA region create different mRNA sequences. Alternate splicing increases the diversity of protein coded by a single gene. Alternate splicing is a universal phenomenon and in human about 40-60% genes show alternate splicing.

37. Small nucleolar RNAs used to process and chemically modify rRNAs are called

(a) scaRNAs

(b) siRNAs

(d) snRNAs

Ans. (c)

Sol. SnoRNA : Small nucleolar RNA, they are the components of SNoRNP (small nucleolar ribonucleoproteins). SnoRNP consists of small nucleolar RNA (snoRNA) packed with specific proteins. SnoRNA helps in the processing and modification of pre-rRNA.

Sca RNA : Small Cajal body specific RNA, a class of small nuclear RNA (snRNA) localized to Cajal bodies (CBs).

Cajal body : They are non-membrane bound special small sub-organelles present in the nucleus of actively dividing cells and they mainly consist of proteins and scaRNA. First described as nucleolar accessory bodies by Santiago Ramon Cajal. Functions of Cajal body (CBs) may be (1) snRNPs biogenesis, (2) maturation and recycling of histon mRNA and (3) telomere maintenance. Addition of nucleotide to telomere with the hep of RNA requires the helps of Cajal bodies.

38. During replication, the RNA primer to degraded by the 5'–3' exonuclease activity of

(a) RNase H1 (ribonuclease H1)

(b) FEN-1 (flap endonuclease 1)

(d) DNA polymerase

Ans. (b)

Sol. In prokaryotes, replacement of RNA with DNA is accomplished by the combined action of a ribonuclease (RNase) and a polymerase. RNase H reomves all but the last ribonucleotide from the 5' end of the Okazaki fragment. DNA polymerase I both extends thte 3' end of the previous Okazaki fragment and removes the final 5' ribonucleotide from the Okazaki fragment through a 5'-3' exonuclease activity. The nicks that remain are sealed by DNA ligase. The process is similar in eukaryotes, except that the single ribonucleotide left by RNase H is removed by the 'flap endonuclease' FEN-1. The gap is filled in by one of the major replicative polymerase, most likely pol . pol is also able to displace the 5' end of an Okazaki fragment, including the DNA synthesized by pol . Any single-stranded DNA displaced in this way is able processed by FEN-1. Finally, DNA ligase seals the nicks to form a complete daughter strand.

39. Which one of the following statements about eukaryotic translation is not true? In eukaryotic translation,

(a) ribosome binding site on mRNA is called Kozak consensus sequences

(b) initiator tRNA is tRNA_i^f–met

(d) translocation factor is eEF2.

Ans. (b)

Sol. In eukaryotic translation, initiator tRNA does not have formylated methionine.

40. A gene producing red pigment was placed near centromeres of fission yeast and thus subjected to position effect varigation and produced white colonies. A screen for mutants that increased the red pigment production was undertaken. Which of the following genes, when mutated, is likely to produce this genotype?

(a) Histone deacetylase

(b) Histone acetylase

(d) TATA binding factor

Ans. (a)

Sol. By the mutation of deacetylase gene causes to increased production of product. Normally, deacetylase represses the gene expression but mutant deacetylase can't repress the gene expression.

41. Puromycin is an antibiotic used to inhibit protein synthesis. Given below are few statements about the antibiotic.

1. It enters the E-site of the ribosome where it prevents the release of dea cetylated tRNA after the action of peptidyl tansferase.

2. It blocks the trnaslocation process by binding to the translocation factor Ef-G.

3. Puromycin resembles the initiator tRNA, tRNA_i^f–met and binds exclusively to the P-site.

4. It resembles the aminoacyl tRNA and binds to the A-site of the ribosome.

5. Puromycin inhibits only prokaryotic and eukaryotic protein synthesis.

Which of the above statments(s) is/are true?

(a) 1 and 5

(b) 2 only

(d) 3 and 5

Ans. (c)

Sol. tRNA and it can bind to the ribosomal A site and participate in peptidebond formation. The product of this reaction, instead of being translocated to the P site, dissociates from the ribosome, causing prematurechain termination.

42. Total RNA was isolated separately from cytosol and nuclei of human cells growing a cell culture. Each sample was incubated under renaturating condition. Given below are the statements made about the outcome of the experiment.

P. RNA isolated from nuclei will from RNA-DNA duplexes because of the presence of introns in the primary RNA.

Q. Cytosolic RNAs usually will not form RNA-DNA duplexes.

R. Both cytosolic and nuclear RNA will not form RNA-DNA duplexes as transcription and splicing occur simultaneously.

S. Cytosolic RNA will form RNA-DNA duplexes because unspliced cytosolic RNAs are exceptionally stable.

(a) Only R

(b) P and S

(d) Only S

Ans. (c)

Sol. In the experiment mentioned in question, it can be interpreted that the genome of multicellular animals contain many potential origins of replication and during early development, when embryos are undergoing rapid cell divisons, origin sites are uniformly activated.

43. Following are certain statements related to eukaryotic DNA replication.

P. The genome of multicellular animals contain many potential origins of replication.

Q. During early development, when embryos are undergoing rapid cell divisons, origin sites are uniformly activated.

R. "Pulse-chase" technique is used to label sites of DNA replication.

S. The rate of elongation of different DNA chains during genome replication varies drastically.

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

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (a)

Sol. Eukaryotes often have multiple origins of replication on each linear chromosome that initiate at different times (replication timing), with up to 100,000 present in a single human cell. Having many origins of replication helps to speed the duplication of their (usually) much larger store of genetic material. The segment of DNA that is copied starting from each unique replication origiin is called a replicon.

During the embroynic S-M cycles, entry into S phase must be controlled post-transcriptionally and must be regualted in the presence of constitutive cyclin E kinase activity. The addition of a GI phase during embrogeneis requires that extrinsic developmental cues influence the onset of S phase. In Drosophilia, this is mediated at least in part through cyclin E. During the S-G cycle that produces polytene cells, entry into S phase can no longer have the completion of mitosis as a prerequisite. In addition of differential control of the onset of S phase, the actual parameters of S phse are altered during development. By parameters of S phase, we mean the intrinsic properties of DNA replication.

The parameters of DNA replication changed in modifed cell cycles include replication origin usage and activation, the rate of replication fork movement and the block to re-replication.

By pulse-labelling cells with nucleotides prior to this treatment, it is possible to directly examine sites of DNA synthesis. Originally used with radioactively labelled nucleotides to accurately measure the rate of replication fork progression more recent studies have used approach to compare the efficiency of origin firing between early and late replicating regions.

44. A promoter delection study was done in order to determine the binding sites for a transcription factor on the promoter, which is activated on treatment with the drug 'X'. The following were made-

Luciferase assay reveled the following results:

The following statments can be made.

P. Region between – 1800 and – 1210 contains a binding site for the activator.

Q. Region between – 868 and – 1210 contains a binding site for a repressor.

R. Region between – 1210 and –868 contains a binding site for the activator.

Which of the above is/are true?

(a) P and R

(b) Q and R

(d) Q only

Ans. (a)

Sol. By observing Luciferase activity we can say that Region between – 1800 and – 1210 contains a binding site for the activator and region between – 1210 and –868 contains a binding site for the activator.

45. The expression of a hypothetical gene was analyzed by Northern and Western blot hybridizations under control and induced conditions. The results are summarized below. Expression of genes can be regulated by

P. Control at transcription initiation

Q. Alternative splicing

R. Control of translation initiation

S. Protein stability

Which of the above regulatory mechanisms can explain the observations shown in the figures?

(a) Only Q

(b) Both, P and Q

(d) P,Q,R and S

Ans. (d)

Sol. Norhtern blot for RNA and Western blot for protein. Absence of bands in control for both means regulatory mechanism operating at transcription and translation also. Alternative splicing is a regulated processs during gene expression that results in a single gene encoding for multiple proteins.

46. The following graph represents the expression of tryptophan synthetas3e (TS) in E. coli cells in absence or presence of tryptophan.

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Option (b) is correctly represents the both scenario of absence and presence of tryptophan in E. coli having altered leader sequence.

47. A patient with ER⁺/PR⁺ breast cancer was cured with a drug 'T', whereas a second patient did not respond to 'T'. Which one of the following is the best therapy that you should suggest for the second patient?

(a) Surgery, followed by HER-2/neu targeted drugs.

(b) A drug that targets triple negative (Er^–/PR^–/HER–2^–) breast cancer.

(d) Surgery, followed by radiation only.

Ans. (b)

Sol. Triple-negative breast cancer (TNBC) is a term that has historically been applied to cancers that lack expression of the estrogen receptor (ER), progesterone receptor (PR), and human epidermal growth factor receptor 2 (HER2). TNBC tends to behave more aggressively than other types of breast cancer. Option b is correct.

48. If you run a pentavalent IgM through SDS-polyacrylamide gel electrophoresis, how many bands you are supposed to get by Western blotting using alkaline phosphatase conjugated secondary antibody?

(a) Five

(b) Four

(d) One

Ans. (d)

Sol. One band will be observed.

49. Glycosaminoglycans are usually linked to proteins to form proteoglycans. Which of the following is not a proteoglycan?

(a) Hyaluronan

(b) Aggrecan

(d) Syndecan–1

Ans. (a)

Sol. Hyaluronan is a uniformly repetitive, linear glycosaminoglycan (GAG) composed of disaccharides of glucuronic acid and N-acetylglucosamin : [–(1, 4)-GlcUA-(1, 3)-GlcNAc-]_n. Depending on the tissue source, the polymer usually consists of 2,000-25,000 disaccharides, giving rise to molecular weights ranging from 10⁶ to 10⁷ Da and extended lengths of 2-25 µm.

Unlike other GAGs, however, hyaluronan contains no sulfated groups or epimerized uronic acid residues; it is synthesized at the plasma membrane ratherr than in the Golgi; it is most likely elongated at the reducing rather than non-reducing terminus; it is huge-not only in molecular weight but most notaby in the space in molecular weight but most notaby in the space it occupies in solution; it exhibits exceptional physical properties; it is uniquely important in regulating cell behaviour. Finally, hyaluronan is not covalently linked to protein or synthesized on a protein backbone; thus it is not a proteoglycan.

50. Some T-lymphocytes respond to antigenic stimulation by synthesizing a growth factor that causes T-cell proliferation thereby increasing the responsive T-lymphocytes resulting in amplification of the immune response. This is an example of

(a) endocrine signaling

(b) paracrine signaling

(d) cyclic signaling

Ans. (c)

Sol. Autocrine signals are produced by signalling cells that can also bind to the ligand that is released, which means the signalling cell and the target can be the same or a similar cell. In autocrine signalling, a cell responds to stimulants it produces. An example of this occurs during the immune responses. The T cells of the immune system help destroy harmful invaders and upon detecting their presence they produce and secrete growth factors to which they themselves respond. The result is an increase in their numbers and an ensuing increase in the magnitude of the defensive responses. Whether a cell responds to a signal and how it responds is determined by the set of receptors it has in place when it receives the signal. Much of the development is based on using these differences in receptors and pathways.

51. Binding of two ligands to their binding proteins were investigated. Following binding isotherms were obtained.

Which of the following statements is correct?

(a) A is obtained with an oligomeric protein and B is obtained with a monomeric protein

(b) B is obtained with protein with positive cooperatively

(d) The profile B is not possible.

Ans. (b)

Sol. B ligand shows more positive cooperativity with protein.

52. A pharmacy student designed a drug to specifically target the receptors for retinoic acid in order to prevet stem cell differentiation. After oin vitro trial, the investigator found that the cells underwent differentiation and the drug seemed to be ineffective. The following reasons were given by the student:

P. The size of the drug exceeded the size of molecules that could cross the membrane

Q. The drug was small in size but hydrophobic in nature

R. The drug did not bind to its receptors

Which of the above could be the probable reason for drug ineffectiveness?

(a) Only R

(b) P and R

(d) Only Q

Ans. (a, b)

Sol. A drug's ability to affect a given receptor is related to the drug's affinity (probability of the drug occupying a receptor at any given instant) and intrinsic efficacy (intrinsic activity—degree to which a ligand activates receptors and leads to cellular response). A drug's affinity and activity are determined by its chemical structure.

53. A researcher wanted to immunize individuals of a particular area with viral infections. The researcher dveloped two different vaccine types (A and B) with the following properties:

(i) When vaccine type A specific for a viral strain is administered to individuals, they fail to develop circulating antibody response with very poor immunological memory. Hence it has to be administered in repetitive doses.

(ii) When vaccine type B specific for a viral strain is administed to individuals, they fail to develop circulating antibody response at the time of infection but they develop strong immunological memory.

If two viral strains V1 (incubation period-2 days) and V2 (incubation period-15 days) are likely to infect the area, which of the following vaccine combination would provide maximum immunization?

(a) V1 specific type A and V1 specific type B

(b) V1 specific type A and V2 specific type B

(d) V2 specific type A and V2 specific type B

Ans. (b)

Sol. Memory cells confer immediate protection and generate secondary responses which are more rapid and of higher magnitude as compared to primary responses. In the B cell system immediate protection is mediated by long-lived plasma cells that are present in the bone marrow and secrete antibodies in an antigen-independent fashion thus maintaining constant levels in serum.

54. Which one of the following statements about cell-cell interactions is not true?

P. Cadherins are transmemrane linker proteins which carry out Ca²⁺-mediated adhesion between adjacent cells

Q. Integrins are transmembrane adhesion proteins that mediate homophilic adhesion through actin and intermediate filaments.

R. Selectins are cell surface lectins that mediate a variety of transient, cell-cell adhesion interactions in the bloodstream.

S. ICAMs (intracellular cell adhesion molecules) and VCAMs (Vascular cell adhesion molecules) are members of immunoglobulin (Ig) superfamily.

(a) Only P

(b) Only Q

(d) Both P and S

Ans. (b)

Sol. Integrins function as transmembrane linkers (or "integrators"), mediating the interactions between the cytoskeleton and the extracellular matrix that are required for cells to grip the matrix. Most integrins are connected to bundles of actin filaments.

55. A technician wanted to make rabbit antiserum specific for mouse IgG. The technician injected rabbit with purified mouse IgG but obtained antiserum which reacted strongly with each of the other mouse isotypes. Which of the procedures mentioned below will allow him to make antiserum specific for IgG only?

(a) Injecting rabbit with purified F(ab)₂ 'region of the IgG antibody

(b) Injecting rabbit with purified heavy chain of the IgG antibody

(d) Injecting rabbit with purified F(ab)' region of the IgG antibody

Ans. (b)

Sol. Immunoglobulin isotypes are selectively distributed in the body. IgG and IgM predominate in plasma, whereas IgG and monomeric IgA are the major isotypes in extracellular fluid within the body.

56. In order to precipitate a particular protein by its specific antiserum, it was found that the protei formed cross-linked lattice with specific polyclonal antiserum but failed to precipitate with specific monoclonal antiserum. Which of the following would accurately justify the reason for this behavior?

(a) The protein has multiple copies of the same epitope specific for the monoclonal antibody.

(b) The protein has multiple distinct epitopes but each has a single copy.

(d) The protein has multiple copies of different epitoopes.

Ans. (b)

Sol. Polyclonal antibodies are made using several different immune cells. They will have the affinity for the same antigen but different epitopes, while monoclonal antibodies are made using identical immune cells that are all clones of a specific parent cell. Polyclonal antisera have been used to determine the seroprevalence of antibodies to a number of infectious organisms and the degree of antibody production.

57. Activation of the Wnt singnal transduction pathway is extremely important during early development. Of the various pathways, which one of the following is most likely to induce cytoskeletal changes, like cell shape and movement?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The Wnt signaling pathway is an ancient and evolutionarily conserved pathway that regulates crucial aspects of cell fate determination, cell migration, cell polarity, neural patterning and organogenesis during embryonic development.

58. The spliting or migration of one sheet of cells into two sheets as seen during hypoblast formation in bird embrygenesis is termed as

(a) delamination

(b) ingression

(d) invagination

Ans. (a)

Sol. Gastrulation is the process by which cellular movements cause the embryonic germ layers to become rearranged into their final positions. Several types of characteristic cell movements may be employed to affect gastrulation, depending on the spatial orientation of the germ layers and the type of embryo one is considering. The movements of gastrulation are accompanied by tissue interactions that pattern the body.

Types of movements associated with gastrulation, depending on the animal.

Invagination : Infolding or "pocket formation" within a sheet of cells.

Delamination : Splitting of a single sheet into two parallel sheets that separate from each other.

Involution : Turning inward by an expanding outer layer, turning the corner.

Ingression : Cells migrate invidividually from an outer layer to the interior of the embryo.

59. Lens formation requires sequential events whereby the anterior neural plate signals the anterior ectoderm to promote secretion of Pax 6, which renders the anterior ectoderm more receptive to secretions from the optic vesicle. The above can be best explained by which of the following phenomenon?

(a) Instructive interactions only

(b) Epithelial-Mesenchymal interactions

(d) Induction and competence

Ans. (d)

Sol. Induction refers to the change in fate of a group of cells in response to signals from other cells. The signal-receiving cells must be capable of responding, a property termed competence.

60. The group of cells of amphibian blastula capable of inducing the organizer is called as

(a) Hensen's node

(b) Nieuwkoop centre

(d) Hypoblast

Ans. (b)

Sol. The organizer is induced from a dorsal vegetal region called the Nieuwkoop center. There are many different developmental potentials throughout the blastula gene embros. The vegetal cap can give rise to only endodermal cell types while the animal cap can give rise to only epidermal cell types. The marginal zone, however, can give rise to most structures in the embryo including mesoderm.

61. For successful fertilization in sea urchin, interaction between the surface of the egg and acrosomal proteins, specifically a 30.5 kDa protein called binding, is necessary. The following factors could affect this interaction and prevent fertilization:

P. Removal of egg jelly polysaccharides

Q. Removal of bindin receptors on the egg vitelline membrane

R. Removal of bindin receptors from the egg jelly.

S. Removal of bindin receptors from a single cluster on the vitelline membrane.

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

(a) P and S

(b) Only Q

(d) Only R

Ans. (c)

Sol. Bindin is an insoluble protein coating the sperm acrosome process and mediating the adhesion of sperm to sea urchin eggs. Bindin isolated from the sperm acrosome results in insoluble particles that cause homospecific eggs to aggregate, whereas no aggregation occurs with heterospecific eggs. Therefore, Bindin is concluded to play a critical role in fertilization, yet its function has never been tested. Each of these sulfated polysaccharides induces the acrosome reaction in conspecific but not in heterospecific spermatozoa. These results demonstrate that species specificity of fertilization in sea urchins depends in part on the fine structure of egg jelly sulfated polysaccharide.

62. A mutant embryo of Drosophila in which one of the major sex determining gene, sex lethal, can only undergo default splicing, was allowed to develop. The following statements are towards explaining the determination of sex of the embryo:

P. The embryo will develop into a male fly

Q. The embryo will develop into a female fly

R. Sex lethal gene product directly regulates sex specific alternate splicing of double sex RNA

S. Sex lethal gene product regulates sex specific splicing of transformer RNA which in turn regulates splicing of double sex RNA.

The correct combination of above statements to explain sex determination of the given embryo is

(a) P and R

(b) P and S

(d) Q and R

Ans. (b)

Sol. In Drosophila, dosage compensation takes place in males by hypertranscription of the single X chromosome and is mediated essentially by a group of genes known as male-specific lethal genes (msl1, msl2, msl3, and maleless [mle]). Three additional genes are also involved in dosage compensation: mof, roX1, and roX2.

63. A two-celled embryo is made of blastomeres A and B. If the two blastomeres are experimentally separated, the 'A' blastomere generates all the cells if would normally make. However, the 'B' blastomere in isolation makes only a small fraction of cells it would normally make. Based on the above observations only, which one of the following conclusions is correct?

(a) 'A' blastomere is autonomously specified while 'B' blastomere is conditionally specified.

(b) 'A' blastomere is conditionally specified, while 'B' blastomere is autonomously specified

(d) Descendents of 'B' blastomere can either be autonomously specified or conditionally specified.

Ans. (a)

Sol. In autonomous specification, if a particular blastomere is removed from an embryo early in its development, that isolated blastomere will produce the same cells that it would have made if it were still part of the embryo. In contrast to the autonomous specification, this type of specification is a cell-extrinsic process that relies on cues and interactions between cells or from concentration-gradients of morphogens. Inductive interactions between neighboring cells is the most common mode of tissue patterning.

64. A mutant was experimentally generated which had wings reduced to haltere like structure. The following statements are put forward regarding this phenotype:

P. ultrabithorax gene ectopically expressed in second thoracic segment

Q. antennapedia gene ectopically expressed in second thoracic segment

R. A homeotic mutation

S. A mutation in gap gene.

The following combination of statements will be most appropriate explaining the molecular basis of mutant phenotype:

(a) P and Q

(b) Q and R

(d) P and R

Ans. (d)

Sol. These mutations lead to transformations of particular body regions in the fly, for example, a segment is transformed so that it resembles a neighboring segment. This will also cause changes in the appendages of segments, such as haltere to wing transformations. Hox gene Ultrabithorax regulates distinct sets of target genes at successive stages of Drosophila haltere morphogenesis.

65. Following are certain statements regarding the activities of homeotic genes of classes A, B and C involved in floral organ identity:

P. Activity of A alone specifies sepals

Q. Activity of B alone specifies petals

R. Activities of B and C from stamens

S. Activity of C alone specifies carpels

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

(a) P, Q and R

(b) P, Q and S

(d) P, R and S

Ans. (d)

Sol. The ABC model posited that floral organ identity is controlled by three gene functions – A, B and C – that act in combination; A-function alone specifies sepal identity, A- and B-functions together control petal identity; B- and C-functions together control stamen identity; C-function alone specifies carpel identity.

66. In chloroplast, the site of coupled oxidation-reduction reactions is the

(a) outer membrane

(b) inner membrane

(d) stromal space

Ans. (c)

Sol. Chloroplasts are the organelles within plant and algal cells where photosynthesis occurs. The oxidation-reduction reactions involved in oxidation of water, reduction of NADP⁺ and synthesis of ATP via a chemiosmotic mechanism are located in the thylakoid membranes of the chloroplast.

67. Which one of the following statements regading seed germination of a wild-type plant is not correct?

(a) Low ABA and high bioactive GA can break seed dormancy.

(b) Light accompained with high temperature can break seed dormancy

(d) Degradation of carbohydrates and storage proteins provide nourishment and energy to support seeddling growth.

Ans. (b)

Sol. Scarification, hot water, dry heat, fire, acid and other chemicals, mulch, and light are the methods used for breaking seed coat dormancy. Physiological conditions causing internal dormancy arise from the presence of germination inhibitors inside the seed. Since light stimulates the synthesis of GAs as well as the responsiveness to GAs, temperature-induced changes in dormancy may indirectly change the capacities to synthesize GAs and to respond to GAs. GA sensitivity is also directly controlled by temperature. High temperatures generally reinforce dormancy or may even induce it. Low temperatures may also induce dormancy in some circumstances, but in many species they are stimulatory.

68. Light is the dominant environmental signal that controls stomatal movement in leaves of well-watered plants grown in natural environment. Which one of the following wavelengths of light is responsible for such regulation?

(a) Red light

(b) Blue light

(d) Fr-red light

Ans. (b)

Sol. Blue-light signals are utilized by the plant in many responses, allowing the plant to sense the presence of light and its direction.

Stomatal response to the blue light : Stomata have a major regulatory role in gas exchange in leaves and they can often affect yields of agricultural crops. Several characteristics of blue-light dependent stomatal movements make guard cells a valuable experimental system for the study of blue-light responses:

The stomatal response to blue light is rapid and reversible and it is localized in a single cell type, the guard cell.

The stomatal response to blue light regulates stomatal movements throughout the life of the plant. This is unlike phototropism or hypocotyl elongation, which are functionally important at early stages of development.

The signal transduction cascade that links the perception of blue light with the opening of stomata is understood in considerable detail.

69. Which one of the following is not the main factor that contributes to water potential during plant growth under normal conditions?

(a) Solute potential

(b) Hydrostatic pressure

(d) Temperature

Ans. (d)

Sol. Water potential is influenced by three factors : Solute concentration pressure and gravity. By taking each component, water potential () of any solution can be represented as follows :

Here, represents solute potential (i.e., concentration of solute), denotes pressure potential and represents gravity potential.

1. Solute potential () : The solute potential is also called osmotic potential. The amount by which the water potential is reduced as a result of the presence of solute is called solute potential. It is equivalnet to osmotic pressure, expressed in bars but has a negative value.

2. Pressure potential () : The positive pressure (such as wall pressure or turgor pressure) operating in the plant cell is termed as pressure potential. It is usually positive and increases the water potential in the system.

3. Gravity pressure () : It denotes the effect of gravity on water potential. It depends on the height of water above the reference state of water, the density of water and the acceleration due to gravity. If the vertical distance are small (less than 5 metres) the is negligible and hence ignored.

70. The plant hormone indole-3-acetic acid (IAA) is present in most plants. The structure of this hormone is related to which one of the following amino acids?

(a) Glutamic acid

(b) Asparatic acid

(d) Tryptophan

Ans. (d)

Sol.

Indole-3-acetic acid is probably synthesized from tryptophan. Two mechanisms of synthesis exist, both of which involved the decarboxylation and removal of the amino group from the side chain of tryptophan (13-15). Regulation of IAA activity in plants is controlled not only by synthesis of the hormone but also by conjugation of IAA through the carboxy group with sugars, amino acids and myo-inositol (16-18) and possibly degradation (9).

71. Which one of the following is not a characteristic property of carotenoids?

(a) They possess complex porphyrin ring

(b) They are integral constituent of thylakoid membrane

(d) They product plants from damages caused by light.

Ans. (a)

Sol. Carotenoids are tetraterpene pigments, which exhibit yellow, orange, red and purple colors. Carotenoids are the most widely distributed pigments in nature and are present in photosynthetic bacteria, some species of archaea and fungi, algae, plants, and animals. Carotenoids contain a porphyrin ring, contain a long carbon chain and two small six-carbon rings.

72. Following are certain statements regarding C₃, C₄ and CAM plants

P. The ratio of water loss to CO₂ uptake is higher in CAM plants than it is either C₃ and C₄ plants.

Q. The rate of photosynthesis attains maximum rate at lower intracellular CO₂ partial pressure in C₄ plants than in C₃ plants.

R. Plants with C₄ metabolism need less rubisco than C₃ plants to achieve a given rate of photosynthesis.

S. Plant with C₄ metabolism need less rubisco than C₃ plants to achieve a given rate of photosynthesis.

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

(a) P and Q

(b) P and R

(d) Q and S

Ans. (d)

Sol. Statement Q and S is correct.

73. Following are certain statements regarding secondary metabolites found in plants

P. All terpenes are derived from a six-carbon element.

Q. Alkaloids are nitrogen containing compounds.

R. Pyrethroids, a monoterpene ester found inthe leaves and flower of Chrysanthemum species, show insecticidal activity

S. Limonoids are groups of alkaloids and have antiherbivoral activity.

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

(a) P and Q

(b) P and S

(d) R and S

Ans. (c)

Sol. An alkaloid is a class of naturally occurring organic nitrogen-containing compounds that are frequently found in the plant kingdom. Many alkaloids are valuable medicinal agents that can be utilized to treat various diseases including malaria, diabetics, cancer, cardiac dysfunction etc. Pyrethroids are found in many commercial products used to control insects, including household insecticides, pet sprays and shampoos.

74. Light is crucial for plant growth and development. Following are certain statements related to photoreceptors in model plant Arabidopsis thaliana.

P. Among the five phytochrome genes, representing a gene family, PHYB plays a predominant role in red-light perception.

Q. Cryptochromes are involved in the regulation of flowering time and hypocotyl length.

R. PhyA photoreceptor is predominantly involved in far-red light perception.

S. The LOV domain of phytochrome C (PHYC) is an important domain for signal transmission.

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

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (a)

Sol. Phytochromes are a class of photoreceptor in plants, bacteria and fungi used to detect light. They are sensitive to light in the red and far-red region of the visible spectrum and can be classed as either Type I, which are activated by far-red light, or Type II that are activated by red light. 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.

75. One of the important functions of program cell death (PCD) in plants is protection aganist pathogenes. PCD also appears to occur during the differentiation of xylem tracheary elements that leads to nuclei and chromatin degradation. These changes result from the activation of certain genes. Following are certain genes encoding

P. Topoisomerase

Q. Nuclease

R. RNA polymerase

S. Protease

Which one of the following combinations of the above is involved in differentiation of xylem tracheary elements?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (d)

Sol. TEs undergo a very well-defined process of differentiation that involves specification, enlargement, patterned cell wall deposition, programmed cell death and cell wall removal. This process is coordinated such that adjacent TEs are joined together to form a continuous network. Protease and Topoisomerase are involved in differentiation of xylem tracheary elements.

76. Following are some statements related to osmotic stress in plants.

P. The accumulation of ions during osmotic adjustment is predmoinatly restricted to the vacuoles.

Q. In order to maintain the water potential equilibrium within the cell, other solutes called compatible solutes or compatible osmolytes accumulate in the cytoplasm.

R. Galactose is one of the compatible osmolytes involved in osmotic stress in plants.

S. There are mainly four groups of molecules that frequently serve as compatible solutions.

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

(a) P, Q, and R

(b) Q, R and S

(d) P, R and S

Ans. (c)

Sol. Significant changes in water potentials in the environment can impose osmotic stress to plants, which disrupts normal cellular activities, or even causes plant death. Under natural conditions, high salinity and drought are the major causes of osmotic stress to plants. The cellular responses to this type of stress deal with the activity of water channels (aquaporins) and electrolyte transporters, and the accumulation of osmolytes.

77. Which one of the following cells in the renal corpuscle can influence glomerular filtration by its contraction?

(a) Podocytes

(b) Endothelial cells of glomerular capillaries

(d) Mesangial cells

Ans. (d)

Sol. Mesangial cells have properties of both smooth muscle cells and tissue macrophages. Similarly to smooth muscle cells, mesangial cell contraction and relaxation potentially modify glomerular hemodynamics by regulating the surface area available for filtration and the glomerular ultrafiltration coefficient. Like macrophages, mesangial cells synthesized and release a number of mediators of inflammation, including prostaglandins (PG), superoxide, interleukin-1, platelet-derived growth factor, platelet activating factor and eicosanoids. Both alterations in glomerular hemodynamis and increased accumulation of macromolecules in the mesangium have been implicated in the pathogenesis of glomerular injury. The turnover of mesangial celsl in the glomeruli of intact, healthy animals is tightly regulated and quite slow. Under disease conditions, however, the turnover of mesangial cells is increased.

78. Production of excessive amount of corticotropin (ACTH) occurs in which one of the following?

(a) Graves' disease

(b) Cushing's syndrome

(d) Alport's syndrrome

Ans. (b)

Sol. Hyperadrenocorticalism (Cushing's syndrome)

Types : Cushing's syndrome results from the effects of chronically increased cortisol levels. Different mechanisms cause two types of Caushing's syndrome.

(a) ACTH dependent

1. Pituitary Cushing's syndrome or Cushing's disease accounts for approximately 70% of the cases of Cushing's syndrome and is more common in middle aged women.

(a) It results from an overperoduction of ACTH by the pituitary, which results in bilateral adrenal hyperplasia.

(b) The source of the excess ACTH has been debated.

(i) Pituitary tumors, either chromorphobic or basophilic adenomas, probably accound for the majority of cases. Some autopsy series have shown pituitary tumors in atleast 90% of patients who have Cushing's disease.

(ii) However, in the remaining patients no tumor was found. This rasises the possibility of an abnormality in the hypothalmic-pituitary axis, resulting in increased ACTH secretion.

2. Ectopic Cushing's syndrome also represents approximately 15% of the cases and is more common in older men.

(a) In this form, ACTH is produced by an extra-adrenal, extrapituitary neoplasm. The result is a hyperplasis of the adrenocortical tissue with consequent hypercortisolism.

(b) The cause is most commonly a small cell carcinoma of the lung, but the syndrome can also occur with bronchial carcinoids, thymomas and tumors of the pancreas and liver.

(b) ACTH independent : Adrenal Cushing's syndrome accounts for approximately 15% of the cases.

1. It is caused by an excess of cortisol that is produced autonomously by the adrenla cortex. This can be due to an adenoma, a carcinoma or bilateral nodular dysplasis and ectopic cortisol-producing tumors.

2. The remaining adrenocorticol tissue atrophies and ACTH levels are low because of suppression by the excess cortisol.

79. The type-I glomus cells present in the carotid bodies contain granules which release some substances during hypoxia. Which one of the following is released in hypoxia?

(a) Serotonin

(b) GABA

(d) IL-B

Ans. (c)

Sol. Glomus cells contain biogenic amines (acetylcholine, norepinephrine, 5-HT and dopamine) and neuropeptides (substance P, exkephalins, endothelins, etc.). It is now well diocumented that dopamine release under hypoxia generates action potential via D₂ receptor, albeit acetylcholine is also released from glomus cells in response to hypoxia.

80. Which one of the following functions is not served by the plasma proteins?

(a) Blood clotting

(b) O₂ trnasport

(d) Buffering capacity of blood

Ans. (b)

Sol. Many functions of plasma are carried out by plasma proteins.

The plasma proteins are the one group of plasma constitutents not present just for the ride.

Because they are the largeset of the plasma constituents, plasma proteins usually do not exit through the narrow pores in the capillary walls.

Plasma proteins exist in a colloidal dispersion.

Plasma proteins establish an osmotic gradient between blood and intestitial fluid.

Plamsa proteins are partially responsible for the plasma's capacity to buffer changes in pH.

Some plasma proteins are involved in the process of blood clotting.

Inactive precursor the antibodies or immunoglobulins are important for the body's defense.

The plasma proteins are generally synthesized by the liver, with the exception of the antibodies, which are produced by lymphocytes.

81. Different frequencies of sound were presented on the ear and the movement of basilar membrane was experimentally determined. The characteristics of movmement of basilar membrane after presentation of 100 Hz sound are described in the following statements:

P. The base of basilar membrane showed resonance

Q. The apex of basilar membrane showed resonance

R. A wave travelled from the base to apex of basilar membrane but the maximum displacement was noted near the apex.

S. A wave travelled from the base to apex of basilar membrane but the maximum displacement was noted near the base.

Which one of the following is correct?

(a) P and R

(b) Q and S

(d) S only

Ans. (c)

Sol. At 100 Hz sound, A wave travelled from the base to apex of basilar membrane but the maximum displacement was noted near the apex.

82. The changes in left venticular stoke work (LVSW) according to the different left venticular end-diastolic pressures (LVEDP, which indicates the initial myocardial fiber length) in a dog, under control conditions, were recorded, which follows Starling's law of the heart. This LVSW-LVEDP relationship was investigated in the same dog after constant infusion of norepinephrine, and these two data sets were plotted. Which one of the following graphs correctly/represents the result obtained?

Control ———

Norepinephrine ---------

(a)

(b)

(c)

(d)

Ans. (b)

Sol. When the changes in left venticular stoke work (LVSW) according to the different left venticular end-diastolic pressures (LVEDP, which indicates the initial myocardial fiber length) in a dog, under control conditions, were ecorded, which follows Starling's law of the heart. This LVSW-LVEDP relationship was investigated in the same dog after constant infusion of norepinephrine, and these two data sets were plotted, graph presented in option (b) is correct.

83. When a nerve fiber is stimulated with increasing strength of stimulus, the action potential fails to generate even though the threshold level may be passed. The following statements may explain this accommodation of nerve fibre:

P. The critical number of open sodium channels required to trigger the action potential may never be attained due to slow depolarization

Q. Potassium channels open in response to slow depolarization, which makes the nerve fiber refractory to depolarization.

R. The low threshold sodium channels remain open, which increases the threshold of firing of action potential.

S. The efflux of sodium and influx of potassium due to operation of Na⁺, K⁺ –ATPase oppose the depolarization.

Which one of the following is correct?

(a) P only

(b) P and Q

(d) R and S

Ans. (b)

Sol. The critical number of open sodium channels required to trigger the action potential may never be attained due to slow depolarization and potassium channels open in response to slow depolarization, which makes the nerve fiber refractory to depolarization.

84. The heart rate shows variation during respiratory rhythm in most human subjects. Which one of the following statements describing the changes of heart rate during respiratory phases is true?

(a) The heart rate is accelerated during expiration, but no change occurs during inspiration.

(b) The heart rate is accelerated during inspiration and decelerated during expiration.

(d) The heart rate is accelerated during inspiration and no change occurs during expiration.

Ans. (b)

Sol. In human, the heart rate is accelerated during inspiration and decelerated during expiration.

85. At 17 years, a 7 feet tall human was diagnosed with gigantism caused by pituitary tumor. The condition was surgically corrected by removal of the person's pituitary gland. Doctors advised horrmonal theraphy. The possible hormonal therapies that would be required for survival are

1. Thyroid hormone

2. Glucocorticoids

3. Glucagon

4. Growth hormone

5. Insulin

Which one of the following combination can be used?

(a) 1 and 2 only

(b) 2 and 4 only

(d) 1, 3 and 5

Ans. (a)

Sol. In hormonal therapy after removal of pituitary gland, there must be requirement of thyroid hormone and glucocorticoids for survival of patients primarily because secretion of these hormones is directly regulated by pituitary gland.

86. Given below are few statements with reference to blood clot formation which results from triggered chain of reactions:

P. Conversion of fibrinogen to fibrin

Q. Activation of factor XIII, which stabilizes fibrin mesh work

R. Activation of factor XII, which promotes plasmin activation

S. Enhancement of platelet aggregation.

Which one of the following combination of statements is correct with reference to roles of rhrombin in hemostasis?

(a) Q, R and S

(b) P, Q and S

(d) P, Q and S

Ans. (b)

Sol. Thrombin stimulates Conversion of fibrinogen to fibrin, activation of factor XIII, which stabilizes fibrin mesh work and enhancement of platelet aggregation.

87. 5-Bromouracil is a base analog that can cause mutation when incorportated into DNA. Which of the following is the most likely change that 5-bromouracil induces?

(a) T : A to C : G

(b) T : A to A : T

(d) C : G to A : T

Ans. (a)

Sol. 5-Bromouracil is a base analog that can cause mutation when incorportated into DNA. It will induce substitution mutation (transition) which is in T : A to C : G.

88. The following pedigree shows the inheritance of a common phenotype controlled by an autosomal recessive allele. The probability of carries in the population is 1/3. What is the probability that a child from parents II-3 and II-4 will show the phenotype?

(a) 1/16

(b) 1/18

(d) 3/16

Ans. (c)

Sol. Child from parents II-3 and II-4 will show the phenotype with probability = .

89. An interrupted mating experiment was performed between Hfr Str^s a⁺ b⁺ c⁺ and F^– Str^r a^– b^– C^– strains. The genotype of majority of streptomycin resistant (Str^r) exconjugant after 10, 20 and 30 minutes of interrupted mating is given below:

10 min a⁺ b^– c^–

20 min a⁺ b^– c⁺

30 min a⁺ b⁺ c⁺

The most probable gene order would be

(a) a b c

(b) c a b

(d) a c b

Ans. (d)

Sol. Hfr Str^s a⁺ b⁺ c⁺ × F^– Str^r a^– b^– c^–

a⁺ b^– c^– 10 min

a⁺ b^– c⁺ 20 min

a⁺ b⁺ c⁺ 30 min

The correct gene order would be (a c b).

90. Two plants with white flowers are crossed. White flowers arise due to recessive mutation. All F1 progeny have red flowers. When he F1 plants are selfed, both red and white flowered progeny are observed. In what ratio will red-flowered plants and white-flowered plants occur?

(a) 1 : 1

(b) 3 : 1

(d) 15 : 1

Ans. (c)

Sol. Two plants with white flowers are crossed. White flowers arise due to recessive mutation. All F1 progeny have red flowers. When he F1 plants are selfed, both red and white flowered progeny are observed. This information suggests this is an example of duplicate recessive epistasis. Hence, the ratio of red flower plant and white flower plants would be 9:7.

91. Autoteraploids arise by the doubling of 2n complement to 4n. There are three different pairing possibilities at meiosis in tetraploids as given below:

P. Two bivalents

Q. One quadriavlent

R. One univalent + one trivalent

Which of the above pairings can lead to production of diploid gamete?

(a) Only P

(b) Q and R

(d) P and Q

Ans. (d)

Sol. Two bivalents and One quadrivalent can lead to production of diploid gamete.

92. The following is the amino acid sequence of a part of a protein encoded by gene 'X'.

...Phe Leu Val Pro Ser Tyr Cys...

A mutant for gene 'X' is isolated following treatment with a mutagen. The amino acid sequence of the same region encoded by the mutant gene is as follows:

...Phe Leu Phe Arg Arg Ile...

Which of the following mutagens is most likely to have been used?

(a) 5-bromoouracil

(b) 2-aminopurine

(d) Acidine orange

Ans. (d)

Sol. By observing the alteration of amino acids in mutant gene X and comparing it with normal protein encoded by gene X, it appears example of frameshiting after amino acid leucine it can be achieved by an entercalating agent i.e. acridine orange.

93. A co-transduction experiment was peformed to decipher the linear order of 4 genes: a, b, c and d. Three sets of experiments were done where transductions were selected for a (Set-1) or b (Set-2) or c (Set-3) and screened for co-transduction of the other markers.

Based on the frequencies shown above, identify the most likely order in the genome.

(a) a b c d

(b) b c d a

(d) a d b c

Ans. (d)

Sol. By observing co-transduction and their frequencies the order of genes in the genome would be (a d b c).

94. A hypothetical bochemical pathway for the formation of eye color in insect is given below.

Two autosomal recessive mutants 'a' and 'b' are identified which block the pathway as shown above. Considering that the mutants are not linked, what will be the phenotype of the F2 progeny if crosses were made between parents of the genotype aaBB × AAbb, and the F1 progeny are intercrossed?

(a) 9 orange-brown : 3 orange : 3 brown : 1 colorless

(b) 9 orange-brown : 7 colorless

(d) 15 orange-brown : 1 colorless

Ans. (a)

Sol. Parents : aaBB × AAbb

Gametes : aB Ab

F1 progeny : AaBB

Self cross : AaBb × AaBb

Gametes : AB, ab, Ab, aB AB, ab, Ab, aB

F2 self cross progeny :

So, phenotype of F2 progeny = 9 orange-brown : 3 brown : 3 orange : 1 colourless

95. In Neurospora, the mutant stp exhibits stop and-and-start growth. When a female of stp strain is crossed with a normal strain acting as a male, all progeny individuals showed stp mutant phenotype. However, the reciprocal cross resulted in all normal progeny individuals. These results can be explained on the basis of

P. maternal inheritance

Q. sex limited inheritance

R. sex influenced inheritance

S. stp. mutation may be located in mitochondrial DNA

The most appropriate statement or combination of the above statements for explaining the experiments results is

(a) P and R

(b) R only

(d) Q and S

Ans.

Sol.

96. An analysis of four microsatellite markers was carried out in a family showing a genetic discorder. The results are summarized below:

Based on the above, which of the markers shows linkage to the disorder?

(a) M1

(b) M2

(d) M4

Ans. (b)

Sol. Through analysis of all microsatellite markers in all family members M2 marker shows linkage to the disorder.

97. Schizocoelous coelom formation, mouth formation from embryonic blastopore, spiral and determinate cleavage are characteristic of

(a) deuterostomes

(b) pseudocoelomates

(d) protostomes

Ans. (c)

Sol. The fate of the blastopore is used to classify animals that have three germ layers into two large categories, protostomes and deuterostomes. Most adult animals have two external openings, the mouth animals have two external openings, the mouth and the anus, into the digestive tract. During gastrulation, the blastopore is the opening into the primitive gut. During further development, the blastopore becomes the mouth in animals classified as protostomes; the blastopore becomes the mouth in animals classified as protostomes; the blastopore becomes the anus in deutreostomes. Organisms belonging to the phyla Mollusca (clams and snails), Arthropoda (insects and crustaceans) and Annelida (earthworms) are protostomes; members of the phyla Echinodermata (starfish) and Chordata (fish and humans) are deuterostomes. The type of cleavage and the development of the body cavity are other important differences between the protostomes and deuterostomes.

Protostomes exhibit spiral cleavage in which the blastomeres divide at acute angles to one another and aer not aligned over one another. If one of these blastomeres is removed from the embryo, neither the removed blastomere nor the remaining cells develop into an individual. This type of determinate cleavage indicates that the fate of dterminate deavage indicates that the fate of the daughter cells is determined early in development. A final characteristic of protostomesisn that the body cavity or coelom develops as a split within the middle of the mesodermal layer. This type of coelom formation is termed schizocoelous development.

98. An example of the species interaction called commensalism is

(a) nitrogen-fixing bacteria in association with legume plant roots.

(b) microbes in living human gut.

(d) Orchid plant growing on the trunk of a mango tree.

Ans. (d)

Sol. An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected. At best, the orchids benefit from being closer to sunlight and from nutrients from bird/animals droppings from visitors to the trees and from decaying vegetation that gets trapped against their roots. The trees really don't get any benefit and they aren't harmed in any way by the orchids.

99. Which of the following is a correct match of the animal with its taxonomic group?

(a) Hirudinea-Leech; Chelicerata-Horse shoe crab; Cestoda-Tapeworrm; Echinoidea-Sea urchins; Cephalopoda-Octopus; Oligochaeta-Earthworm.

(b) Hirudinea-Earthworm; Chelicerata-Horse shoe crab; Cestoda-Octopus; Echinoidea-Tapeworm; Cephalopoda-Earthworm; Oligochaeta-Leech.

(c) Hirudinea-Tapeworm; Chelicerata-Leech; Cestoda-Tapeworm; Echinoidea-Horse shoe crab; Cephalopoda-Earthworm; Oligochaeta-Octopus.

(d) Hirdinea-Leech; Chelicerata-Tapeworm; Cestoda-Earthworm; Echinoidea-Sea urchins; Cephalopoda-Octopus; Oligochaeta-Horse shoe crab.

Ans. (a)

Sol. The Hirudinea, or true leeches, are highly specialized clitellates, separated from other annelid groups by the presence of an anterior circumoral sucker and a posterior ventral sucker.

Horseshoe crab, common name of four species of marine arthropods (class Merostomata, subphylum Chelicerata).

Tapeworm, also called cestode, any member of the invertebrate class Cestoda.

Sea urchins are members of the phylum Echinodermata, which also includes sea stars, sea cucumbers, brittle stars, and crinoids.

Octopus are the best known of the cephalopods with their charismatic personalities, intelligence, and diversity.

The term 'Earthworm' is a collective name for a group of organisms within the class Oligochaeta which means 'few bristles'.

100. Individuals with greater mass have a smaller surface area to volume ratio, which helps to conserve heat. This is known as

(a) Leibig's

(b) Cope's rule

(d) Bergmann's rule

Ans. (d)

Sol. Bergmann's rule implies that some energetic advantage may be gained through a decreased surface-area-to-volume ratio. We can further generalize that the amount of heat loss depends on both the animal's surface area and the difference in temperature between its body surface and the surroundings. Large mammals, for example, have less heat loss than small mammals because of their larger body mass relative to surface area.

Consider a cube, an imaginary little heat generator, 1 cm in each dimension. The surface area of the cube is 6 cm² while the volume is of course, 1 cm³. Hence, the surface-area-to-volume ratio is now 3 : 1. Assuming that heat storage is related somehow to volume and that heat loss is proportional to surface area, then this doubling of size would seem to favour heat conservation by reducing surface area relative to volume.

101. The population density of an insect species increases from 40 to 46 in one month. If the birth rate during that period is 0.4, what is the death rate?

(a) 0.25

(b) 0.15

(d) 0.40

Ans. (a)

Sol. As we know that, dN/dt = r × N; where, r is instantaneous rate of increase and N is initial population.

This equation can be written as, dN/dt × 1/N = r ...(1)

Instantaneous rate of increase (r) = per capita birth rate (b) – per capita death rate (d)

From equation (1), 6/40 = 0.15 = r

According to this question,

r = b – d

0.15 = 0.4 – d

Hence, d = 0.4 – 0.15 = 0.25

102. In which ecosystem is the detrital pathway of energy flow most important?

(a) Lakes

(b) Grasslands

(d) Oceans

Ans. (c)

Sol. In tropic rainforest ecosystem, detritus food chains (DFC) a major conduit of energy flow. As an organism does or produces biological matter or waste, it enters DFC. Further steps include recycling of nutrient and flow of energy in DFC.

103. What parameter plotted on Y-axis against generation time, would yield the curve shown in the figure?

(a) Surrvivorship

(b) Body size

(d) Intrinsic rate of increase

Ans. (d)

Sol. The intrinsic rate of natural increase can be used to compare growth rates of populations of a species that have different generation times. Some human populations have higher intrinsic rates of natural increase partially because individuals in those groups begin reproducing earlier than those in other groups.

104. Which of the following statements best describe archaebacteria?

(a) Mostly autotrophic, cell wall contains peptidoglycan, 60S ribosomes, live in extreme enviroment.

(b) Divide by fission, not susceptible to lysozyme, live in extreme enviroments, mostly autotrophic.

(d) Chitinous cell wall, obligate aerobic, circular chromosomes.

Ans. (d)

Sol. Organisms in the domains of Archaea and Bacteria reproduce with binary fission. This form of asexual reproduction and cell division is also used by some organelles within eukaryotic organisms (e.g., mitochondria). NAG and NAT are bonded by a (1,3) sugar linkage instead of a (1,4) linkage. This is significant, because it makes these archaea resistant to the enzyme, lysozyme. Archaea is the main group to thrive in extreme environments. Archaea can be both autotrophic and heterotrophic. Archaea are very metabolically diverse. Some species of archaea are autotrophic.

105. Which of the following statements is not correct?

(a) Stomata are present in mosses and hornworts but absent in liveworts.

(b) Only the lycophytes have microphylls and almost all other vascular plants have megaphylls

(d) Monocots have fibrous root system whereas eudicots have taproot.

Ans. (c)

Sol.

106. In group I are given 4 orders of class Insecta. Match each one with a common name (Group II) and its diagnostic characters (Group III)

Group – I Group – II

A. Dermaptera E. Ant

B. Ephemeroptera F. Mayfly

C. Odonata G. Grasshopper

D. Plecoptea H. Damselfly

I. Stonefly

J. Earwig

Group – III

(i) Elongate, membranous wings with net-like venation, abdomen long and slender, compound eyes occupy most of head, hemimetabolous metamorphosis.

(ii) Elongate, chewing mouthparts, thread-like antennae, abdomen with unsegmented forceps-like cerci, hemimetabolous metamorphosis.

(iii) Forewing long, narrow and leathery, hindwing broad and membranous, chewing mouthparts, hemimetabolous metamorphosis.

(iv) Elongate abdomen with two or three tail filaments, two pairs of membranous wings with many veins, forewings triangular, short, bristle-like antennae, hemimetabolous metamorphosis

(v) Adults with reduced mouthparts, elongate antennae, long cerci, nymphs aquatic with gills, hemimetabolous metamorphosis.

(vi) Wings membranous with few veins, well developed ovipositors, sometimes modified into a sting, mouth parts modified for biting and lapping, holometabolous metamorphosis.

(a) A-J-ii

(b) B-I-vi

(d) D-F-i

Ans. (a)

Sol. Earwigs make up the insect order Dermaptera. Earwigs have characteristic cerci, a pair of forcep-like pincers on their abdomen, and membranous wings folded underneath short, rarely used forewings. Earwigs are hemimetabolous, meaning they undergo incomplete metamorphosis, developing through a series of 4 to 6 molts. The developmental stages between molts are called instars.

107. Which of the following is a correct match of the animal with its attribute?

(a) P–3, Q–2, R–1, S–4

(b) P–2, Q–3, R–1, S–4

(d) P–4, Q–3, R–2, S–1

Ans. (a)

Sol. Rotifers are multicellular animals with body cavities that are partially lined by mesoderm (pseudocoelomate). These organisms have specialized organ systems and a complete digestive tract that includes both a mouth and anus.

Sea anemones, named after a terrestrial flower, have a basic radial symmetry with tentacles that surround a central mouth opening.

The first larval stage for crabs, lobsters, shrimp, barnacles, copepods, and some other crustaceans is called a nauplius. The nauplii of different species all look alike so they are hard to tell apart. These larvae live in the water column as part of the zooplankton.

Sea urchins do not have a closed circulatory system and instead rely on an open water vascular system to help with the uptake of nutrients, the flushing out of wastes, locomotion, and respiration.

108. Which of the following is a correct statement?

(a) Euglenids have a spiral or crystalline rod inside flagella.

(b) Pheophytes have a spiral or crystalline rod inside flagella.

(d) Euglenids and pheophytes both have a spiral or crystalline rod inside flagella.

Ans. (a)

Sol. Euglenids only contain paraxial rods in flagella.

109. Which of the following characteristics make Amborella the most basal living angiosperm?

(a) Carpels fused by tissue connection and absence of vessel elements

(b) Absence of carpels and presence of vessel elements

(d) Presence of carpels and absence of vessel elements.

Ans. (d)

Sol. Amborell's lineage diverged from other angiosperms around 130 million years ago, sometime after the first flowering plant appeared. Ambroella has all of the defining features of a flowering plant, but at the same time it seems to have retained some "gymnospermy" characteristics as well. For instance, it lacks the vessel elements for water conduction present in most other flowering plants. Also, while Amborella has carpels, they are incompletely closed. This is significant because the carpel is thought to have originated from a flat, leaf-like structure with ovules on its margins. This structure eventually rolled inward and became enfolded, creating a hollow, enclosed ovary with one or more ovules. Early angiosperms probably had carpels that were not quite fused shut but were sealed with secretions from the carpel, which is the case with Amborella.

110. Following is a table showing selected characteristics of important fungal groups.

Fungal group Characteristic

A No regularly occurring septa in thallus

B Performed septa

C Forms arbuscular mycorrhizae on plant roots

D Have zoospores with flagella

In the above table, the fungal groups A, B, C and D, are respectively

(a) Chytridiomycetes, Ascomycetes, Glomermoycetes, Zygomycetes

(b) Zygomycetes, Ascomycetes, Glomeromycetes, Chytridiomycetes

(d) Chytridiomycetes, Zygomycetes, Ascomycetes, Glomeromycetes.

Ans. (b)

Sol. Like the oomycetes the mycelium of the zygomycetes is non-septate, except where septa may separate structures such as chlamydospores, sporangia and zygospores.

Filamentous ascomycetes produce hyphae divided by perforated septa, allowing streaming of cytoplasm from one cell to another.

The phylum Glomeromycota also called Glomeromycetes is a newly established phylum which contains species that form mutualistic symbiotic association with roots of many plant species. Members of this group include arbuscular mycorrhizal fungi.

Motile fungal spores called zoospores have a single posterior flagellum that pushes them head first through the water.

111. Complete the following hypothetical life table of a species to calculate the net reprodctive rate R₀:

The calculated R₀ will be

(a) 0.75

(b) 1.00

(d) 1.15

Ans. (a)

Sol.

Net reproductive rate R₀ = = 0 + 0 + 0.25 + 0.30 + 0.20 = 0.75

112. Following four types of species were observed in a community:

P. Species A has a large effect on community because of its abundance.

Q. Species B has a large role in community out of proportion to its abudance.

R. Status of species C provides information on the overall health of an ecosystem

S. Significant conservation resources are allocated to species D which in single, large and instantly recognizable.

According to above description, species A, B, C and D are called resptively,

(a) Dominant, keystone, Indicator and Flagship

(b) Keystone, Flagship, Dominant and Indicator

(d) Flagship, Dominant, Keystone and Indiacator

Ans. (a)

Sol. Dominant species: Species that have high abundance relative to other species in a community, and have proportionate effects on environmental conditions, community diversity and/or ecosystem function. Dominant species can be common (widespread) or restricted in their range (limited).

A keystone species is an organism that helps define an entire ecosystem. By keeping populations of mussels and barnacles in check, this sea star helps ensure healthy populations of seaweeds and the communities that feed on them—sea urchins, sea snails, limpets, and bivalves.

An indicator species is an organism whose presence, absence or abundance reflects a specific environmental condition. Indicator species can signal a change in the biological condition of a particular ecosystem, and thus may be used as a proxy to diagnose the health of an ecosystem.

A flagship species is a species selected to act as an ambassador, icon or symbol for a defined habitat, issue, campaign or environmental cause.

113. Which of the following is the correct decressing order for the rate of decomposition of litter constituents?

(a) Hemicellulose, cellulose, lignin, phenol

(b) Cellulose, hemicellulose, phenol, lignin

(d) Lignin, phenol, hemicellulose, cellulose

Ans. (a)

Sol. Cellulose degrades more quickly than lignin does, so the degradation of cellulose is believed to play a dominant role during the earlier stages of foliage decomposition. Lignin, after cellulose, is the most abundant organic material on Earth; it decomposes slowly. The slow rate of lignin decomposition by fungi, actinomycetes, and bacteria is thought to be due to the complexity of its bonds and cross-linkages, and because it has a relatively low nitrogen content.

114. The possible relationship between level of disturbance and species diversity in a biological community are that species diversity.

P. is unaffected by disturbance

Q. is highest at intermediate levels of distrubance

R. decreases exponetially with increasing levels of disturbance

S. starts decreasing only at higher levels of disturbance

Match each graph with its corresponding statements above

(a) 1–S, 2–R, 3–Q, 4–S

(b) 1–R, 2–S, 3–Q, 4–P

(d) 1–R, 2–P, 3–Q, 4–S

Ans. (d)

Sol. Second graph shows species which is unaffected by disturbance. Third graph shows species which is highest at intermediate level of disturbance. First graph represent species that decrease exponentially with increasing level of disturbance.

115. The diagram represents competition between species 1 and species 2 accroding to Lotka-Volterra model of competition

Given the conditions in the diagram, the predicted outcome of competition is

(a) Unstable coexitence between species 1 and 2 because K₁ > K₂/ and K2 > K₁/.

(b) Unstable coexistence between species 1 and 2 because K₁ < K₂/ K₂ < K₁/.

(d) Stable coexistence between species 1 and 2 because K₁ < K₂/ and K₂ < K₁/.

Ans. (a)

Sol. Here, the carrying capacity of species 1 (K₁) is higher than the carrying capacity of species 2 divided by the competition coefficient (K₂/a₂₁) and the carrying capacity of species 2 (K₂) is higher than the carrying capacity of species 1 dividied by the competition coefficient (K₁/a₁₂). Below both isoclines and above both isoclines the populations incrase or decrase as in the first two scenarios and there is un unstable equilibrium point (closed circle) where the isoclines intersect. For points above the dashed pink line (species 2 isocline) and below the solid yellow line (species 1 isocline), the outcome is the same as in the first scenario : competitive exclusion of species 2 by species 1. On the other hand, for points above the solid yellow line (species 1 isocline) and below the dashe pink line (species 2 isocline), the outcome is the same as in the second scenario : competitive exclusion of species 1 by species 2. The two stable equilibrium points are again represented by open circles. In this scenario, the outcome depends on the initial abundances of the two species.

116. Which species concept ultilizes morphological and molecular characters to distinguish between species?

(a) Evolutionary

(b) Ecological

(d) Phylogenetic

Ans. (d)

Sol. Phylogenetics is the study of evolutionary relationships among biological entities – often species, individuals or genes (which may be referred to as taxa).

117. Worker bees, instead of themselves reproducing, help the queen reproduce. This behaviour is explained as an example of

(a) kin selection

(b) group selection

(d) natural selection

Ans. (a)

Sol. The worker bees themselves are the daughters of the queen. In helping the queen to rear offspring, they are therefore helping their mother to have sons and daughters. These sons and daughters are their sibs. Darwin termed thsi 'family selection'. Nowadays, it is called 'kin selection'. There is one mroe technical term-namely inclusive fitness. The idea is that an individual can, in a sense, reproduce via its relatives, as we saw in our exmaple of a worker honey bee. The term 'inclusive fintess' therefore takes account of the helping behaviour an individual gives its relatives. In other words, it is not an organism's individual reproductive success alone that matters, but also the reproductive success of any of its relatives that have been affected by its helping behaviour.

118. The wings of insects and the wings of bats represent a case of

(a) divergent evolution

(b) convergent evolution

(d) neutral evolution

Ans. (b)

Sol. Wings of insects and wings of flying mammals (bats) represent the example of analogous organs. Because wings of insecdts have adopted aerial life like the flying lizard, birds and bats wings of insects superifically look like the wings of flying vertebrates but are basically different, having no bones. Thus, the insect wings are analogous to vertebrate wings.

119. The degree of genetic relatedness between the offspring and their parents is

(a) higher than that between sister and brother

(b) lower than that between sister and brother

(d) dependent on the number of siblings.

Ans. (c)

Sol. The evolution of haplo-diploidy has a number of interesting effects from an evolutionary genetic persepective. First, as male gametes are produced without meiosis, all these gametes are genetically identical. This means that the degree of genetic relatedness in haplo-diploids is different from that in normal diploids. As all female offspring of a mating between a femals and male obtain the same genes from their father and half the same genes from their moterh, who produces her gametes through normal meiosis, these sisters will hvae not half of their genes in common, but three-quarters. This means that in haplodiploids, full sisters (same mother and same father) are more closely related to each other than they are to either of their parents or indeed than full sisters are to one another or to their parents in a normal diploid species. The genetic contribution of aqueen hymenopteran to a daughter or sone is 50% : they contain half her genes. The unusually high genetic relatedness of hymenopteran full sisters means that a female will benefit more copies of her own genes if she rears a full sister than she would by reproducing herself. The high level of relatedness of full sisters can be contrasted to the degree of relatedness between brother and sister. Calculating the degree of relatedness between brother and sister depends on which may around one looks. From the male point of view, half of a male's genes are carried by his sister. All these have come from the mother. However, only a quarter of the sister's genes are carried by her brother because the gets none of the genes that she received from her father.

120. During which geological period was there an explosive increase in the number of many marine invertebrate phyla?

(a) Ordovician

(b) Devonian

(d) Cambrian

Ans. (d)

Sol. The Carmbrian Period was a time of many evolutionary innovation during which almost all of the major invertebrate phyla evolved. The Ordovician Period witnessed striking changes in the marine community, resulting in a dramatic increase in diversity of the shelly fauna, followed by a mass extinction at the end of the Ordovician.

121. In life history evolution there is generally a trade-off between the size and number offspring produced. Some conditions are listed below:

P. Scarcity of food during the early stages of life

Q. Provision of parenal care

R. High mortality during early stages of life

S. Predator's preference for large sized prey

What are the above two conditions that would favour the production of a small number of large-sized offspring?

(a) Q and R

(b) Q and S

(d) P and R

Ans. (c)

Sol. A trade-off (or tradeoff) is a situational decision that involves diminishing or losing one quality, quantity, or property of a set or design in return for gains in other aspects. In simple terms, a tradeoff is where one thing increases, and another must decrease. The trade-off results in a higher reproductive investment per offspring at low food concentrations, whereas females allocate more of their reproductive resources toward increasing offspring number at high food concentrations.

122. In a random sample of 400 individuals from a population with alleles of a trait in Hardy-Weinberg equilibrium, 36 individuals are homozygous for allele a. How many individuals in the sample are expected to carry at leas one allele A?

(a) 36

(b) 168

(d) 196

Ans. (c)

Sol. According to the Hardy-Winberg equilibrium, frequency of genotype aa = q² = 36/400

So, f(a) = q –

f(A) = p = (1 – q) = (1 – 0.3) = 0.7

Expected number of AA in population = p² × 400 = (0.7)² × 400 = 196

Expected number of Aa in population = 2 pq × 400 = 2 × 0.7 × 0.3 × 400 = 168

Total number of expected at least one allele A = 196 + 168 = 364

123. Which of the following statements is no correct regarding effect of genetic drift?

(a) It alters allele frequency substantially only in small populations

(b) It can cause allele frequencies to change at random

(d) It can cause harmful alleles to become eliminated.

Ans. (d)

Sol. Genetic drift is the change in frequency of an existing gene variant in the population due to random chance. Genetic drift may cause gene variants to disappear completely and thereby reduce genetic variation. It could also cause initially rare alleles to become much more frequent, and even fixed. Unlike natural selection, genetic drift does not depend on an allele's beneficial or harmful effects. Instead, drift changes allele frequencies purely by chance, as random subsets of individuals (and the gametes of those individuals) are sampled to produce the next generation.

124. Which of the following is not a benefit for the female adopting polyandry?

(a) Greater probability of getting all her eggs fertilized

(b) Ability o receive more resources from the males

(d) Improved chances of genetic compatibility with her own DNA.

Ans. (c)

Sol. Polyandry, in which a female mates with several males, has the potential advantage of providing the female with a wider range of genetic diversity for her eggs.

125. Two kinds of natural selection (A and B) acting on a trait are shown in the figure below. In each, the top graph shows he trait frequency before and the bottom graph frequency after the action of natural selection. The kind of natural selection in A and B are

(a) A-Directional, B-Disruptive

(b) A-Neutral, B-Disrupive

(d) A-Disruptive, B-Stabilizing

Ans. (c)

Sol. Disruptive selection favors both extreme phenotypes, different from one extreme in directional selection. Stabilizing selection favors the middle phenotype, causing the decline in variation in a population over time.

126. Individual A performs to another individual a behavioral act which has a fitness consequence. Math the behavioral acts (P to T) with the correct fitness consequence (1 to 4)

(a) P–4, Q–3, R–2, S–2, T–1

(b) P–1, Q–2, R–2, S–3, T–4

(d) P–2, Q–2, R–3, S–1, T–4

Ans. (a)

Sol. Option (a) is correct.

127. Assume that individual A wants to do an altruistic act to individual B and that the benefit and cost of doing this act are, in 'fitness' units, 40 and 12, respectively. Accroding to Hamilton's Rule, A should perform the altruistic act only if B is his

(a) nephew

(b) niece

(d) daughter or son

Ans. (d)

Sol. A social behaviour counts as altruistic if it reduces the fitness of the organism performing the behaviour, but boosts the fitness of others.

According to the Hamilton's rule,

r × b > c

r × 40 > 12

r > 12/40 = 0.3

It is possible in case, when B is daughter or son.

128. You want to purify protein of your interest. You can use affinity chromatography to purify as you have nickel columns available in the laboratory. With what molecule will you tag the protein to purify using those columns?

(a) GST

(b) Hstidine

(d) Proline

Ans. (b)

Sol. In recombinant DNA, histidine tag on the desired protein and Nickel resin are commonly used to purify desired protein via affinity chromatography. That is, histidine has strong affinity towards the nickel resin which does not flow through the column.

129. In an experiment to detect a new protein in fixed cells, no secondary antibody tagge with fluorescence dye is available. What should be the best choice out of the following to detect the protein?

(a) Protein A-FITC

(b) Protein A-Sepharose

(d) Aviin-FITC

Ans. (a)

130. Two 18-residue helical peptides A and B are enantiomers. They can be distinguished by

(a) recording their MALDI mass spectrum

(b) hydrolysis followed by amino acid analysis

(d) examining their circular dichroism spectra

Ans. (d)

Sol. Circular dichroism (CD) is a form of spectroscopy that uses circularly polarized light and is highly sensitive to the conformations of molecules. Because of this conformational sensitivity, CD is one of the most widely used techniques for characterizing the conformations of proteins, nucleic acids and carbohydrates and for monitoring conformational tarnsitions induced by temperature, solvent composition, lignad binding etc. Two 18-residue helical peptides A and B are enatiomers. They can be distinguished by examining their circular dichroism (CD) spectra.

131. Lower limits of detection by sensors is important. Which method of detecion is more senstive than glass electrode used for pH measurement?

(a) Absorption spectroscopy

(b) Refractive index

(d) Fluorescence spectroscopy

Ans. (d)

Sol. Fluorescence excitation brightness values increased as pH increased, whilst peak fluorescence decreased with increasing culture age at equivalent pH.

132. If a researcher intends to identify a specific brain area activity linked to a cognitive function in human subjects, which one of the following techniques should be used?

(a) CAT

(b) MRI

(d) Patch-clamp

Ans. (c)

Sol. Functionaly magnetic resonance imaging or functional MRI (fMRI) is a functional neuroimaging procedure using MRI technology that measures brain activity by detecting associated changes in blood flow. This technique relies on the fact that cerebral blood flow and neuronal activation are coupled. When an area of the brain is in use, blood flow to that region also increases.

133. Which of the following statements is incorrect for fluorescence in situ hybridization (FISH) technique?

(a) A fluorescence or confocal microscope is used for detection of signal.

(b) A labeled sequence of nucleotides is used

(d) A stringent washing step is essential to remove appearance of non specific signal.

Ans. (c)

Sol. Fluorescence in situ hybridization is a molecular cytogenetic technique that uses fluorescent probes that bind to only particular parts of a nucleic acid sequence with a high degree of sequence complementarity."The basic principle involved is hybridization of nuclear DNA of either interphase cells or of metaphase chromosomes affixed to a microscopic slide, with a nucleic acid probe. The probes are either labeled indirectly with a hapten or directly through incorporation of a fluorophore. FISH is applied to detect genetic abnormalities that include different characteristic gene fusions or the presence of an abnormal number of chromosomes in a cell or loss of a chromosomal region or a whole chromosome.

134. You have isolated a short chain plypeptide from a bacteria which upon administering to animal cells in culture is demonstraed to get transported to mitochondria and to cause mitochondrial membrane disruption. This peptide is being considered for its therapeutic effects on cancer and is to be tested for its anticancer properties in the following models of the disease.

P. Cancer with Apaf-1 mutaion

Q. Cancer where Bax genes aare inactivated

R. Cancer where Bak genes are inacivated

Which of the following is likely to provide best effect (i.e., induce apoptosis)?

(a) Only P

(b) Q and R

(d) P and R

Ans. (b)

Sol. Bax and Bak are members of the Bcl-2 family and core regulators of the intrinsic pathway of apoptosis. Upon apoptotic stimuli, they are activated and oligomerize at the mitochondrial outer membrane (MOM) to mediate its permeabilization, which is considered a key step in apoptosis.

135. A transgenic lettuce plant was generated by over-expressing isopentenyl transferase (IPT) gene under the control of the promoter of senescence activator gene (SAG12).

Following are some statements regarding this transgenic plant. The transgenic plants

P. exhibit delayed senescence

Q. exhibit fas senescence

R. have higher amount of cytokinin during senescence

S. have higher amoun of gibberellins during senescence.

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

(a) P and Q

(b) P and R

(d) R and S

Ans. (b)

Sol. IPT gene required for Cytokinin biosynthesis. Cytokinin delays senescence.

136. During the production of alcohol by fermentation using budding yeast, oxygen supply is kept limited. Why?

(a) Budding yeasts are obligate anaerobes and cannot tolerate oxygen.

(b) Budding yeasts lose mitochondria in the absence of oxygen

(d) Alcohol is oxidized further in the presence of oxygen.

Ans. (c)

Sol. Yeast species either require oxygen for aerobic cellular respiration (obligate aerobes) or are anaerobic, but also have aerobic methods of energy production (facultative anaerobes). Unlike bacteria, no known yeast species grow only anaerobically (obligate anaerobes).

137. A student was asked to design a knockout cassette for specifically deleting the p53 gene from the prostate gland of mice. Which one of the following pairs of cassettes will ensure deletion of the gene?

TSP = Tissue-specific promoter

Cre = Cre recombinase

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Cre/lox is usually used to make knockout alleles, but it can also be used to activate gene expression. The proper insertion of a loxP-flanked "stop" sequence (transcriptional termination element) between the promoter and transgene coding sequence blocks the expression of the gene.

138. Two homologous proteins wee isolated from a psyshrophile (P) and a thermophile (T). The purified proteins were subjected to denaturation, protease digestion and circular dichroism (CD). Following observations were made:

P. The CD spectra of P and T proteins are identical

Q. Their amino acid composition in 95% identical

R. T and P are equally susceptible to proteolysis in the presence or absence of reducing agent.

S. T has higher midpoints of thermal denauration than P.

The reason for enhanced stability in T is due to

(a) altered secondary structure

(b) increased number of disulfides in T

(d) increase in number of salt bridges

Ans. (d)

Sol. Presence of disulfide bonds makes protein more stable at high temperatures.

139. In an attempt to detect protein expression profile in a cell, Western blot technique is employed. Expression of two new proteins is to be followed by probing with respective high affinity antibodies (raised in rabbit). Unfortunately, the two proteins were found to co-migrate in SDS-PAGE profile. Under this situation, using one dimensional SDS-PAGE and by Western blot, which one of the following is the best way to demonstrate the presence of both the proteins?

(a) Develop Western blots with their antibodies in the same gel.

(b) Prior to doing SDS-PAGE /Western blot, one protein vould be removed by immunoprecipitating in the cell extracts.

(d) Subjecting the technique of stripping/reprobing of the gel after trnasferring to nitrocellulose membrane while doing Western blotting.

Ans. (d)

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.

140. Nichrome coated stainless steel electrodes were implanted in a rat brain for chronically recording the electrical activity of deep brain structures. During a study of 3 months the intensity of electrical signals gradually decreased. The following statements may explain the cause of this observation.

P. The deposition of metallic iron from the electrode tips caused degeneration of some neurons

Q. The gradual accumulation of microglia at the electrode tips increased the resistance of electrodes.

R. The neurons at the electrode tips were hyperpolarized gradually

S. he threshold for firing action potential in the neurons at the electrode tips was increased due to prolonged presence of electrodes.

Which one of the following is correct?

(a) P only

(b) P and Q

(d) R and S

Ans. (b)

Sol. The brain requires iron for mitochondrial respiration and synthesis of myelin, neurotransmitters, and monoamine oxidases. Iron accumulates in distinct parts of the brain in patients with neurodegenerative diseases, and some have proposed that neurons die because they contain too much iron. Statement P and Q is correct.

141. The number of seeds in the frui of a plant species, H₀ : µ = 30. A random sample of 9 fruits gives the mean number fo seeds as 24 with a standard deviation of 6.12. (a) What are the confidence limits for the sample mean? (b) Would you reject or accept the null hypothesis a 95% confidence level?

(a) (a) 18 and 30, (b) reject the hypothesis

(b) (a) 20 and 28, (b) reject the hypothesis

(d) (a) 18 and 30, (b) accept he hypothesis

Ans. (b)

Sol. A confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample any by using the specific confidence level of the estimate.

The central limit theorem states that when the sample size is large, approximately 95% of the sample means taken from a population and same sample size will fall within + 1.96 standard errors of the population mean, that is, µ + 1.96.

Now, if a specific sample mean is selected, say , there is a 95% probability that the interval µ + 1.96 contains . Likewise, there is a 95% probability that the interval specified by .

For rejection of the hypothesis, = sample mean

The number of seeds in the fruit of a plant species, H₀ = µ : 30.

= sample mean, s = standard deviation and n = size of the sample.

The value used for the 95% confidence interval of z* is 1.96

Where, L = lower estimate of the population and U = upper estimate of the population.

= 20, 28

Hence, one can say with 95% confidence that the interval between 20 and 28 seeds does not contain the population mean, based on a sample of 30 seeds.

142. A chromatin immunoprecipitation (ChIP) assay was performed to determine specific transcription factor binding sites on the promoter of a gene. Pull down was done using either IgG or anibodies c-myc. A DNA containing c-myc binding regions was used as a control for PCR amplification (input). Which one of the following PCR representations of DNA is correct?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Chromatin immunoprecipitation is a type of immunoprecipitation experimental technique used to investigate the interaction between proteins and DNA in the cell. Chromatin immunoprecipitation (ChIP) assays identify links between the genome and the proteome by monitoring transcription regulation through histone modification (epigenetics) or transcription factor–DNA binding interactions.

143. Following are certain statements regarding somatic hybridizaton, a technique used for plant improvement.

P. Protoplasts of only sexually compatible plant species can be fused

Q. Hybrids are produced with variable and asymmetric amounts of genetic material of parental species.

R. Protoplast fusion permits transfer of gene block or chromosomes

S. Genes to be transferred need to be identified and isolated.

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

(a) P and R

(b) Q and R

(d) Q and S

Ans. (b)

Sol. Somatic hybridization and cybridization have great potential in plant improvement. Somatic hybridization through protoplast fusion provides the ability to combine parent genes in higher plants to overcome sexual incompatibility among plant species or genera. The protoplast fusion gives a unique nuclear-cytoplasmic combination in the hybrid cell. It yields innovative plants with desirable characters.

144. A researcher was repeating a FACS experiment but somehow got confused with the labeling of the tubes. There are four tubes, one control, C (with no fluorescent label), one standard 1 S1 (with PE label) and the last one test, T (which should be FITC positive). Given below is the resul of the FACS experiment.

What should be the correct labeling?

(a) P–S2; Q–S1; R–T; S–C

(b) P–S1; Q–T; R–C; S–S2

(d) P–S1/T; Q–S2; R–S1/T; S–C

Ans. (d)

Sol. FITC reacts with a primary amine on the protein to form a covalent amide bond. FITC-labelled protein substrates/peptides, antibodies, peptide hormones are used as specific probes in enzyme kinetics, immunocytochemistry as well as flow cytometry and identification of receptors on target cells, respectively. PE or APC acts as the reporter which translates the detected event into a quantifiable signal. Graph P and R is S1/T. Q shows S2 and S shows C.

145. Small molecular weight compounds affect the activity of luciferase differently. The oil/water solubility of various compounds is one property important for its effect on luciferase. The straight line in the graph was obtained by plotting the activity data with 50 different compounds. The luciferase activity of two new derivatives of benzene (A and B) are shown below? Which of the following statements is correct?

(a) A is phosphate and B is amine

(b) A is methyl and B is amine

(d) A is amine and B is phosphate

Ans. (b)

Sol. A luciferase assay is used to determine if a protein can activate or repress the expression of a target gene. A is methyl and B is amine.

CSIR NET BIOLOGY (DEC - 2014)
Previous Year Question Paper with Solution.

Locate Us

Company

Courses

Academies

CSIR NET BIOLOGY (DEC - 2014) Previous Year Question Paper with Solution.

Locate Us

Company

Courses

Academies

Get in touch

CSIR NET BIOLOGY (DEC - 2014)
Previous Year Question Paper with Solution.