PYQ DEC 2011 - VedPrep

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

21. In which form of DNA, the number of base pairs per helical turn is 10.5 ?

(a) A

(b) B

(d) Z

Ans. (b)

Sol.

22. On the molar scale which of the following interactions in a nonpolar environment provides highest contribution to the biomolecule ?

(a) van der Waals interaction

(b) Hydrogen bonding

(d) Hydrophobic interaction

Ans. (c)

Sol. Hydrophobic interactions are relatively stronger than other weak intermolecular forces (i.e., Van der Waals interactions or Hydrogen bonds)

23. Michaelis and Menten derived their equation using which of the following assumption ?

(a) Rate limiting step in the reaction is the breakdown of ES complex to product and free enzyme.

(b) Rate limiting step in the reaction is the formation of ES complex

(d) Non-enzymatic degradation of the substrate is the major step.

Ans. (a)

Sol. Michaelis and Menten derived their equation using rate limiting step in the reaction is the breakdown of ES complex to product and free enzyme.

24. Equilibrium constant (K) of noncovalent interaction between the two non-bonded atoms of two different groups was measured at 27°C. It was observed that K = 100 M^–1. The strength of this noncovalent interaction in terms of Gibbs free energy change is :

(a) 2746 kcal/mole

(b) –2746 kcal/mole

(d) –247 kcal/mole

Ans. (b)

Sol. = + 2.303 RT log Q; where, Q = and R = 1.987 cal/K-mol

When a reaction is at equilibrium, substitution of = 0, then

= –2.303 RT log K_eq; where, Q = K_eq =

= –2.303 × 1.987 cal/K-mol × 298 × log 100 = –4.576061 × 298 × 2 = –2727 kcal/mole.

25. Biosynthesis of tyrosine is detailed below :

(a) ATP, phosphoenolpyruvic acid, 3-enolpyruyl shikimic acid-5-phosphate, p-hydroxyphebylpyruvic acid.

(b) GTP, pyridoxal phosphate, 3-enolpyruvyl shikimic acid-5-phosphate, phenylpyruvic acid.

(d) ATP, 3-phosphohydroxypyruvic acid, 3-enolpyruvyl shikimic acid-5-phosphate, pyridoxylphosphate.

Ans. (a)

Sol. A : ATP, B : phosphoenolpyruvic acid, C : 3-enolpyruyl shikimic acid-5-phosphate, D : p-hydroxy-phebylpyruvic acid.

26. Denaturation profiles of DNA are shown below. The differences in the profiles arise because

(a) the DNA is single stranded but of different sizes.

(b) A + T content of A > B > C and the DNA aer fro complex genomes.

(d) G + C content is identical but A + T content in A > B > C in DNA of comparable sizes isolated from simple genomes.

Ans. (c)

Sol. G + C content of C > B > A in DNA of comparable sizes isolated from simple genomes.

27. If van der Waals interaction is described by the following relation,

Where is the free energy of the van der Waals interaction, A and B are constants, r is the distance between two non-bonded atoms 1 and 2, and q₁ and q₂ are partial charges on the dipoles 1 and 2. In this relation, the paarmeter A describes

(a) electron shell attraction

(b) electron shell repulsion

(d) dipole-dipole repulsion

Ans. (d)

Sol. Dipole-dipole repulsion. Option (d) is correct.

28. The pH of blood of a healthy person is maintained at 7.40 + 0.05. Assuming that this pH is maintained entirely by the bicarbonate buffer (pKa₁ and pKa₂ of carbonic acid are 6.1 and 10.3, respectively), the molar ratio of [bicarbnoate]/[carbonic acid] in the blood is

(a) 0.05

(b) 1

(d) 20

Ans. (d)

Sol. Henderson-Hasselbalch equation is, PH = pK_a +

29. The hydrolysis of pyrophosphate to orthophosphate is important for several biosynthetic reactions. In E. coli, the molecular mass of the enzyme pyrophosphatase is 120 kDa, and it consists of six identical subunits. The enzyme activity is defined as the amount of enzyme that hydrolyzes 10 µmol of pyrophhosphate in 15 mintes at 37°C under standard assay condition. The purified enzyme has a V_max of 2800 units per milligram of the enzyme.

How many moles of the substrate are hydrolysed per second per milligram of the enzyme when the substrate concentration is much greater than K_m?

(a) 0.05 µmol

(b) 62 µmol

(d) 1 µmol

Ans. (c)

Sol. Given 1U = 10 µmol PP_i/15 min (for this enzyme)

V_max = 2800 U/mg enzyme

30. In contast with plant cells, the most distinctive feature of cell division in animal cells is

(a) control of cell cycle transitions by protein kinases

(b) enzymes responsible for DNA replication

(d) pattern of chromosome movement.

Ans. (d)

Sol. The most distinctive feature of cell division in animal cells is pattern of chromosome movement.

31. The membrane lipid molecules assemble spontaneously into bilayers when placed in water and form a closed spherical structure known as

(a) lysosome

(b) peroxisome

(d) endosome

Ans. (c)

Sol. The membrane lipid molecules assemble spontaneously into bilayers when placed in water and form a closed spherical structure known as liposome

32. Most common type of phospholipid in the cell membrane of nerve cells is

(a) phospatidylcholine

(b) phosphatidylinositol

(d) sphingomyelin

Ans. (d)

Sol. Most common type of phospholipid in the cell membrane of nerve cells is sphingomyelin.

33. The erythocyte membrane cytoskeleton consists of a meshwork of proteins underylying the membrane. The principle component spectrin has -subunits which assemble to form tetramers. The cytoskeleton is anchored to the membrane through linkages with the transmembrane proteins band 3 and glycophorin C. The cytosolic domain of band 3 also serves as the binding site of glycolytic enzymes such as glyceraldehyde 3-phosphate dehydrogenase. Analysis of the blood sample of a patient with haemolytic anemia shows spherical red blood cells. The patient carries

(a) a mutation in glycophorin C

(b) a mutant spectrin with increased tetramerization propensity.

(d) mutant glyceraldehyde 3-phosphate dehydrogenase.

Ans. (c) The patient carries mutant spectrin defective in dimerization ability.

Sol.

34. Maturation-promoting factor (MPF) controls the initiation of mitosis in eukaryotic cells. MPF kinase activity requires cyclin B. Cyclin B is required for chromosome condensation and breakdown of the nuclear envelop into vesicles. Cyclin B degradation is followed by chromosome decondensation, nuclear envelop reformation and exit from mitosis. This requires ubiquitination of a cyclin destruction box motif in cyclin B. RNase-treated Xenopus egg extracts and sperm chromatin were mixed. MPF activity increased with chromosome condensation and nuclear envelope breakdown. However, this was not followed by chromosome decondensation and nuclear envelope reformation because

(a) RNase contamination persisted in the system

(b) cyclin B was missing from the system

(d) cyclin B lacing the cycling destruction box had been overexpressed.

Ans. (d)

Sol. This was not followed by chromosome decondensation and nuclear envelope reformation because cyclin B lacing the cycling destruction box had been overexpressed.

35. The bacterial flagellar motor is a multi-protein complex. Rotation of the flagellum requires movement of protons across the membrane facilitated by a multi-protein complex. The flagellar motor proteins combine to create a proton channel that drives mechanical rotation. In a screeen for mutants, some non-motile ones were selected. These could have

(a) mutations in tubulin and actin proteins

(b) mutations in kinesin proteins

(d) mutations in the charged residues lining the ridge of the FliG subunit.

Ans. (d)

Sol. Option (d) is correct.

36. A single-strand nick in the parental DNA helix just ahead of a replication fok causes the replication fork to break. Recovery from this calamity requires

(a) DNA ligase

(b) DNA primase

(d) homologous recombination

Ans. (d)

Sol. A single-strand nick in the parental DNA helix just ahead of a replication fok causes the replication fork to break. Recovery from this calamity requires homologous recombination.

37. -amanitin inhibits

(a) only RNA polymerase I

(b) only RNA polymerase II

(d) all RNA polymerases

Ans. (b)

Sol. A major practical distinction between the eukaryotic RNA Pol is drawn from their response to the bicyclic octapeptide -amanitin. In all eukaryotic cells the activity of RNA polymerase II is rapidly inhibited by low concentrations of -amanitin. RNA polymerase I is not inhibited. The response of RNA polymerase III to -amanitin is less well conserved; in animal cells it is inhibited by high levels, but in yeast and insects it is not inhibited.

38. Reverse tarnscriptase has both ribonuclease and polymerase acitivity. Ribonuclease activity is required for

(a) the synthesis of new RNA strand

(b) the degradation of RNA strand

(d) the degradation of DNA strand

Ans. (b)

Sol. Reverse tarnscriptase has both ribonuclease and polymerase acitivity. Ribonuclease activity is required for the degradation of RNA strand.

39. In gene regulation, open reading frame (ORF) implies

(a) intervening nucleotide sequence in between two genes

(b) a series of triplet codons not interrupted by a stop codon

(d) the exonic sequence of a gene that corresponds to the 5'UTR of the mRNA and thus does not code for te protein

Ans. (b)

Sol. In gene regulation, open reading frame (ORF) implies a series of triplet codons not interrupted by a stop codon.

40. While replicating DNA, the rate of misincorporation by DNA polymerase is 1 in 10⁵ nucleotides. However, the actual error arte in the replicated DNA is 1 in 10⁹ nucleotides. This is achieved mainly due to

(a) spontaneous excision of misincorporated nucleotides

(b) 3' 5' proofreading activity of DNA polymerase

(d) 5' 3' proofreading activity.

Ans. (b)

Sol. The actual error arte in the replicated DNA is 1 in 10⁹ nucleotides. This is achieved mainly due to 3' 5' proofreading activity of DNA polymerase.

41. Amino acid selenocysteine (Sec) is incorporated into polypeptide chain during translation by

(a) charging of Sec into tRNA^ser followed by incorporation through serine codon

(b) charging of serine into tRNA^ser followed by modification of serine into selenocysteine and then incorporation through serine codon

(d) charging of serine into tRNA^ser, modification of serine into selenocysteine and then incorporation through a specially placed stop codon

Ans. (d)

Sol. Amino acid selenocysteine (Sec) is incorporated into polypeptide chain during translation by charging of serine into tRNA^ser, modification of serine into selenocysteine and then incorporation through a specially placed stop codon.

42. Synthesis of normal hemoglobin requires coordinated synthesis of globin and -globin. Thalassemias are genetic defects perturbed in this coordinated synthesis. Patients suffering from deficiency of -globin chains (-thalassemia) could also be due to mutations affecting the biosynthesis of -globin mRNA.

The following statements describe the genesis of non-functional β-globin leading to β-thalassemia.

P. Mutation in the promoter region of the -globin gene.

Q. Mutation in the splice junction of the -globin gene.

R. Mutation in the intron I of the -globin gene.

S. Mutations towards the 3' end of the -globin gene that codes for polyadenylation site.

Which of the following combinations is correct ?

(a) P, Q and S

(b) P, Q and R

(d) R, S and P

Ans. (c)

Sol. The genesis of non-functional -globin leading to -thalassemia by Mutation in the splice junction of the -globin gene, Mutation in the intron I of the -globin gene and Mutations towards the 3' end of the -globin gene that codes for polyadenylation site.

43. Bacteriophage has two modes in its life cycle, lytic and lysogenic. In the lysogenic mode, the expression of all the phage genes are repressed while the expression of repressor gene switches between on and off position depending on the concentration of repressor. The following statements are made :

P. Repressor may act both as a positive regulator and negative regulator.

Q. Expression of repressor gene, cI is independent of the expression of cII and cIII genes.

R. Mutation of cI gene will cause it to form clear plaques on both wild type E. coli and E. coli (⁺).

S. Mutation at operators, O_L and O_R will allow the phage to act as a virulent phage.

The correct statements are

(a) P and Q

(b) Q and R

(d) S and P

Ans. (d)

Sol. As per the condition mentioned in question, repressor may act both as a positive regulator and negative regulator and Mutation at operators, O_L and O_R will allow the phage to act as a virulent phage.

44. Pre-mRNAs are rapidly bound by snRNPs which carry out dual steps of RNA splicing, that removes the intron and joints the upstream and downstream exons. The following statements describe some facts related to this event :

P. Almost all introns begin with GU and end with AG sequences and hence all the GU and AG sequences are spliced out of RNA.

Q. U2 RNA recognizes important sequences at the 3' acceptro end of the intron.

R. The splicesome uses ATP to carry out accurate removal of intron.

S. An unsual linkage with 2' OH gropu of guanosine within the intron from a 'lariat' structure.

Which of the following combinations is correct ?

(a) P and Q

(b) Q and R

(d) S and P

Ans. (a)

Sol. Correct option should be P and Q. During RNA splicing, Almost all introns begin with GU and end with AG sequences and hence all the GU and AG sequences are spliced out of RNA and U2 RNA recognizes important sequences at the 3' acceptor end of the intron.

45. In human, protein coding genes are mainly organized as exons and introns. There are intergenic regions that transcribe into various types of non-coding RNA (non translating into protein). Some introns may harbor also transcriptions units, which are

(a) always other protein coding genes

(b) protein coding gene and RNA coding genes

(d) pseudo genes

Ans. (b)

Sol. Homing introns encode an endonuclease that is responsible for their movement. The endonuclease cleaves a recognition sequence within the target gene and generates short overhanging ends. This double-stranded break triggers a gene conversion event in which the intact version of the gene is copied and used to repair the break. Group II homing introns are retro-elements that use an RNA intermediate. The proteins they encode have both endonuclease and reverse transcriptase activity.

46. Insulin and other growth factors stimulate a pathway involving a protein kinase mTOR, which in its turn augments protein synthesis. mTOR essentially modifies protein(s) which in their unmodified form act as inhibitors of protein synthesis. The following proteins are possible candidates :

P. eEF-1

Q. eIF-4E-BP1

R. eIF-4E

S. PHAS-1

Which of the following sets is correct ?

(a) P and Q

(b) Q and S

(d) Q and R

Ans. (d)

Sol. During this translation regulation eIF-4E-BP1 and PHAS-1 bind at eIF-4E and inhibit initiation of translation.

47. For continuation of protein synthesis in bacteria, ribosomes need to be released from the mRNA as well as to dissociate into subunits. These processes do not occur spontaneously. They need the following possible conditions :

P. RRF and EF-G aid in this process.

Q. An intrinsic activity of ribosomes and an unchanged tRNA are required.

R. IF-1 promotes dissociation of ribosomes.

S. IF-3 and IF-1 promote dissociation of ribosomes.

Which of the following sets is correct ?

(a) P and S

(b) P and Q

(d) Q and S

Ans. (a)

Sol. To release ribosome from mRNA after translation both subunits are dissociated with the help of RRF and EF-G while initiation factors IF-1 and IF-3 promote dissociation of ribosomes during initiation.

48. Graft rejection does not involve

(a) erythrocytes

(b) T-cells

(d) polymorphonuclear leukocytes

Ans. (a)

Sol. Graft rejection does not involve erythrocytes.

49. Toxic shock is caused by

(a) toxins produced by some bacteria

(b) excessive stimulation of a large proportion of T-cells by bacterial superantigens

(d) excessive production of immunoglobulins

Ans. (b)

Sol. Toxic shock is caused by excessive stimulation of a large proportion of T-cells by bacterial superantigens.

50. Indirect immunofluroscence involves fluroescently labelled

(a) immunoglobulin-specific antibodies

(b) antigen-specific antibodies

(d) carrier-specific antibodies

Ans. (a)

Sol. Indirect immunofluroscence involves fluroescently labelled immunoglobulin-specific antibodies.

51. A fixed smear of a bacterial culture is subjected to the following solutions in the order listed below and appeared red.

P. Carbofuchsin (heated)

Q. Acid-alcohol

R. Methylene blue

Bacteria stained by this method can be identified as

(a) non-acid fast E. coli

(b) acid fast Mycobacterium sp.

(d) Gram-negative Mycobacterium sp.

Ans. (b)

Sol. Acid alcohol has the ability to completely decolorize all non-acid-fast organisms, thus only leaving behind red-colored acid-fast organisms, like M. tuberculosis. The slides are then stained a second time with methylene blue that serves as a counterstain.

52. Survival of intracellular pathogens depends on the levels of pro-inflammatory and anti-inflammatory cytokines in macrophages. In an experimental condition, Mycobacteria infected macrophages were treated with IL-6 or IL-2 for 4 hours at 37°C. Untreated cells were used as control. Cells were lysed and number of bacteria in each experimental set was counted by measuring colony forming unit (CFU). Which of the following observations is true ?

(a) IL-12 treated cells contain more intracellular bacteria than control.

(b) IL-12 treated cells contain less intracellular bacteria than control.

(d) IL-6 treated cell contain less intracellular bacteria than control.

Ans. (c)

Sol. IL-6 treated cells contain more intracellular bacteria than control, because IL-6 involve in activation of B-cells that release antibody, no role on intracellular pathogen.

53. Many cancers carry mutant p53 genes, while some cancers have normal p53 genes. p53 activates p21 activates p21 (Waf-1) which inhibits G1/S-Cdks and phsophorylation of the retinoblastoma protein (Rb). Cancers with normal p53 genes could

(a) express non-phosphorylatable form of Rb.

(b) express high levels of p53-deubiquitinases.

(d) express inactive forms of G1/S cyclinss.

Ans.

Sol.

54. A nerve impulse or actin potential is generated from transient changes in the permeability of the axon membrane to Na⁺ and K⁺ ions. The depolarizatio of the membrane beyond the threshold level leads to Na⁺ flowing into the cell and a change in membrane potential to a positive value. The K⁺ channel then opens allowing K⁺ to flow outwards ultimately restoring membrane potential to the resting value. The Na⁺ and K⁺ channels operate in oppostie directions becuase

(a) there is an electrochemical gradient growth generated by proton transport.

(b) there is a difference in Na⁺ and K⁺ concentrations on either side of the membrane.

(d) Na⁺ is dependent on ATP whereas K⁺ is not.

Ans. (b)

Sol. The sodium-potassium pump system moves sodium and potassium ions against large concentration gradients. It moves two potassium ions into the cell where potassium levels are high, and pumps three sodium ions out of the cell and into the extracellular fluid. So, there is a difference in Na+ and K+ concentrations on either side of the membrane.

55. A bacterial response regulator turns on gene A in its phosphorylated form. Teh amount of "A" shows a sharp and steep rise at a threshold concentration of the signal sensed by the congnate sensor. This is most likely due to

(a) increased phosphatase activity of the sensor at the threshold concentration.

(b) decreased phosphorylation of the response regulator by the sensor.

(d) negatie feedback in gene A expression.

Ans. (c)

Sol. Response regulators are proteins that undergo transient phosphorylation, connecting specific signals to adaptive responses. Remarkably, the molecular mechanism of response regulator activation remains elusive, largely because of the scarcity of structural data on multidomain response regulators and histidine kinase/response regulator complexes.

56. You are given a group of four mice. Each mouse is immunized with keyhole limpet hemocyanin or azobenzene arsonate or lipopolysaccharide or dextran. Four weeks later, sera were collected from these mice and antigenspecific IgG1 and IgG2a ELISA were performed. Only one of the mice showed positive response. It was

(a) keyhole limpet hemocyanin primed mouse

(b) azobenzene arsonate primed mouse

(d) dextran primed mouse

Ans. (a)

Sol. Keyhole limpet haemocyanin (KLH) is a very large, copper-containing protein molecule derived from the haemolymph of the inedible mollusc, Megathura crenulata. KLH is a highly immunogenic T-cell dependent antigen that is used increasingly in immunotoxicological studies, particularly in those involving animals.

57. T-cell proliferation in vivo is to be analyzed. The clels are labelled with CFSE (a fluorescent probe) and injected in CD86-deficient mice and BALB/c mice along with the required antigens. Three days later, the cells are recovered and analyzed by flow cytometry. Which one of the following is logically correct ?

—— BALB/c ------ CD86-deficient

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Along with CD80, CD86 provides costimulatory signals necessary for T cell activation and survival. Depending on the ligand bound, CD86 can signal for self-regulation and cell-cell association, or for attenuation of regulation and cell-cell disassociation. BALB c mouse is inbred, and genetic monitoring results confirm it to be a BALB/c nude. The animal lacks a thymus, is unable to produce T-cells, and is therefore immunodeficient.

58. Intracellular transport and cytoskeletal organization of a cell is regulated by nucleotide exchange of different small molecular weight GTPases of Ras super family. Overexpression of which of the following GTPase modulates the actin-cytoskeleton of HeLa cells ?

(a) Ran is GDP bound form

(b) Ran in GTP bound

(d) Rho in GDP bound form

Ans. (c)

Sol. Rho GTPases regulate cytoskeletal dynamics, cell polarity, cell migration, cell-cycle progression, vesicle trafficking and cytokinesis. Most studies into the function of Rho GTPases in mammalian systems have used cultured cells that express constitutively active and/or dominant-negative mutants.

59. Tumor cells were isolataed from a breast cancer patient. These cells were injected into nude mice and they were divided into four groups. Group 1 received EGF receptor-conjugated with methotrexate; Group 2 received transferrin rreceptor-conjugated with methotrexate; Group 3 received mannose receptor-conjugated with methotrexate; Group 4 received same amount of the free drug. In which of the following cases tumorigenic index would be minimum ?

(a) Free drug

(b) EGF receptor-conjugated drug

(d) Mannose receptor conjugated drug

Ans. (b)

Sol. Tumorigenicity refers to the ability of cultured cells to develop viable tumors in immune-deficient animals. The goal of this protocol is to illustrate tumorigenicity assay by subcutaneous tumor-cell-transplantation in nude mice. Target cells are transplanted to 6-week-old nude mice subcutaneously and the tumor growth is monitored over a period of observation or treatment. When tumor grows to a pre-determined size or by the end of the limited period, the nude mice will be euthanatized and the xenograft will be removed for further examination."Mutations in EGF R have since been associated with a number of different cancers including non-small cell lung cancer, breast cancer, metastatic colorectal cancer and glioblastoma. Activation of EGF R leads to increased cell proliferation and cell growth and decreased programmed cell death (apoptosis).

60. The blastopore region of amphibian embryo that secretes BMP inhibitors and dorsalizesthe surrounding tissue is known as

(a) Brachet's cleft

(b) Nieuwkoop center

(d) Hensen's node

Ans. (c)

Sol. The blastopore region of amphibian embryo that secretes BMP inhibitors and dorsalizesthe surrounding tissue is known as Spermann's organizer.

61. Which of the floral whorls is affected in agamous (ag) mutants ?

(a) Sepals and petals

(b) Petals and stamens

(d) Sepals and carpels

Ans. (c)

Sol. Floral whorls is affected in agamous (ag) mutants are Stamens and carpels.

62. Which of the following maternal effect gene products regulate production of anterior structures in Drosophila embryo ?

(a) Bicoid and Nanos

(b) Bicoid and Hunchback

(d) Nanos and Caudal

Ans. (b)

Sol. Bicoid protein enters the nuclei of early-cleavage embryos, where it activates the hunchback gene. The transcription of hunchback is seen only in the anterior half of the embryo—the region where Bicoid protein is found. Mutants deficient in maternal and zygotic Hunchback protein lack mouthparts and thorax structures.

63. AP1 (APETALA 1) is one of th floral mristem identifying genes. In wild type Arabidopsis thaliana platns transformed wih API :: GUS, glucuronidase (GUS) activity is seen in floral meristem, only after the commitment to flowering. Ectopic expression of API :: GUS in the Embryonic flower (emf) mutant background result in GUS activity throughout the shoots in the four day old seedlings. These observations suggest that API is

(a) not involved in flowering

(b) involved in repression of flowering

(d) stimulation of flowering in the emf background

Ans. (d)

Sol. Floral organ identity genes encode proteins that regulate the expression of other genes that have products involved with the formation or function of floral organs. The Arabidopsis floral meristem-identity genes APETALA1 (AP1) and LEAFY (LFY) confer floral identity on developing floral primordia, whereas TERMINAL FLOWER (TFL) is required to repress their expression within shoot and inflorescence meristems. LFY executes its meristem identity role in part by activating AP1 expression directly.

64. When the prospective neurons from an early gastrula of a frog were tarnsplanted into the prespective epidermis region, the donor cells differentiated into epidermis. However, when a similar experiment was done with the late gastrula of frog, the prospective neurons developed into neurons only. These observations could possibly be explained by the following phenomena.

P. The early gastrula show conditional development whereas the last gastrula shows autonomous development.

Q. The early gastrula show autonomous development whereas the late gastrula show conditional development.

R. The prospective neurons from the early gastrula are only specified whereas those from the late gastrula are determined.

S. The prospective neurons from the early gastrula are determined whereas those from the late gastrula are specified.

Which of the conclusions drawn above are correct ?

(a) P and Q

(b) P and R

(d) Q and R

Ans. (b)

Sol. One of the most widely used herbicides is glyphosate, the active ingredient in Roundup. Glyphosate is a competitive inhibitor of the enzyme EPSP synthase which is required to convert shikimate 3-phosphate to EPSP.

65. The figure below represents a late zebrafish gastrula. The following concepts may be proposed during further development of the embryo.

P. The concentration of FGF decreases from the yolk towards the epidermis, along with the increase of BMP activity from the dorsal to the ventral axis.

Q. Increase in FGF activity in the epidermis with concomitant decrease in BMP activity towards the ventral axis.

R. Neural induction in zebrafish is independent of the organizer and depends on activation of BMP signalling.

S. In comparison, both Xenopus and chick embryos require activation FGF for neural induction to occur in addition to BMP inhibitors.

Which of the above statements are true ?

(a) P and R

(b) Q and R

(d) R and S

Ans. (c)

Sol. We show that in vegetal (trunk and tail) regions of the zebrafish gastrula, neural specification is initiated at all DV positions of the ectoderm in a manner that is unaffected by levels of Bmp activity and independent of organiser-derived signals. Instead, we find that Fgf activity is required to induce vegetal prospective neural markers and can do so without suppressing Bmp activity.

66. In case of morphallactic regeneration :

(a) there is repatterning of the existing tissues with little new growth.

(b) there is repatterning of the existing tissues after the stem cell divison has taken place.

(c) there is cell division of the differentiated cells which maintain their diferentiated state to finally form a complete organism

(d) there is dedifferentiation of the cells at the cut surface which become undifferentiated. These undifferentiated cells then divide to redifferentiate to form the complete structure.

Ans. (a)

Sol. In morphollactic regeneration, regeneration occurs through the repatterning of existing tissues, and there is little new growth. Such regeneration is seen in hydras.

67. With respect to the extraembroynic structures formed in the mammals, the possible functional attributes have been designated :

P. Allantoin stores urinary waste and helps mediate gas exchange. It is derived from splanchnopleure at the caudal end of the primitive streak.

Q. Amnion is a water sac and protects the embryo and its surrounding amniotic fluid. The epithelium is derived from somatopleure.

R. Chorion is essential for gas exchange in amniote embryos. It is generated from the splanchnopleure.

S. Yolk sac is the last embronic membrane to form and is derived from somatopleure.

Which of the above statements are correct ?

(a) P and Q

(b) P and R

(d) P and S

Ans. (a)

Sol. The C₄ plants show saturation at about 360 µ IL-1 while C₃ responds to increased CO₂ concentration and saturation is seen only beyound 450 µ IL-1. Thus current availability of CO₂ levels is limiting to the C₃ plants. In C₄ plants, the CO₂ compensation point is zero or nearly zero, reflecting their very low levels of photorespiration. In C₄ plants, photosynthetic rates saturate internal concentration values of about 15 pascals, reflecting the effective CO₂ concentrating mechanisms operating in these plants.

68. The decision to become either a trophoblast or inner cell mass blastomere is one of the first decisions taken by any mammalian embryo. Below is a diagrammatic representation of the different cells formed during development from the morula with the help of different molecules. Identify the molecules 1-4 sequentially.

(a) cdx 2, Oct 4, Nanog, Stat 3

(b) cdx 2, Nanog, Stat 3, Oct 4

(d) cdx 2, Oct 4, Stat 3, Nanog

Ans. (d)

Sol. Essential to the transient pluripotency of ICM is expression of transcription factors oct4, nanog and sox2. These three regulatory transcription factors are necessary to maintain uncommitted stem-cell like state and functional pluripotency of ICM, enabling ICM cells to give rise to epiblast and all associated derived cell types. Only trophoblast cells synthesize transcription factor cdx2 , which downregulates oct4 and nanog.

69. Which of the following is not a prosthetic group of nitrate reductase ?

(a) FAD

(b) Heme

(d) Pterin

Ans. (d)

Sol. Nitrate reductase is a homodimeric enzyme with three electron-transferring prosthetic groups per subunit: flavin (FAD), heme and Mo cofactor (Moco). Moco consists of Mo covalently bound to two S atoms in the tricyclic molecule pterin.

70. Which of the following acts as a branch point for the biosynthesis of sesquiterpenes and triterpenes ?

(a) Farnesyl pyrophosphate

(b) Geranyl pyrophosphate

(d) Hydroxymethylglutaryl-CoA

Ans. (a)

Sol. In the MVA pathway, FPP is the branching point of sesquiterpene and triterpene biosynthesis.

71. Which of the following set of cell organelles are involved in the biosynthesis of jasmonic acid through octadecanoid signalling pathway ?

(a) Chloroplast and peroxisomes

(b) Chloroplast and mitochondria

(d) Golgi bodies and mitochondria

Ans. (a)

Sol. Jasmonic acid (JA) is synthesized through the octadecanoid pathway that involves the translocation of lipid intermediates from the chloroplast membranes to the cytoplasm and later on into peroxisomes.

72. During development of embroys in plants, PIN proteins are involved in

(a) establishment of auxin gradients

(b) regulation of gene expression

(d) induction of cell division

Ans. (a)

Sol. In plants, the PIN proteins are integral membrane proteins that transport the anionic form of the phytohormone auxin across membranes. Most of the PIN proteins (e.g. PIN1/2/3/4/7 in the model plant Arabidopsis thaliana) localize at the plasma membrane (PM) where they serve as secondary active transporters involved in the efflux of auxin.

73. Chloroplast distribution in a photosynthesizing cell is governed by blue light sensing phototropin 2 (PHOT2). When the cells are irradiated with high intensity blue light, the chloroplasts

(a) move to the side walls

(b) aggregate in the middle of the cell

(d) aggregate in small clusters

Ans. (a)

Sol. A transition mutation is a base-pair substitution point mutation that substitutes a purine for the other purine or a pyrimidine for the other pyrimidine, so that the purine/pyrimidine axis of the DNA molecule is maintained. The four positive transitions are :

G A A G Purine/purine

C T T C Pyrimidine/pyrimidine

The term transition was first defined by Ernst Freese in 1959.

By studying mutations induced by the mutagenic base along 5-bromouracil or 2-aminopurine, he showed that there are two types of mutagenesis, those that undergo reversion by the base analoges, called transitions and those that do not, called transversions. Transitions result from base mispairing due to the ambiguous base-pairing properties of base analogues. 2-Aminopurine mispaired with a thymine results in the insertion of an adenine in the next round of DNA replication, creating a G to A transition. Because these mutations were due to mispairing by the base analog, it was reasoned that the mutations would be revertible with another base analogue.

Transitions are the errors that occur most frequently during DNA replication. DNA polymerases are far more likely to misinsert the wrong purine or pyrimidine than to insert a purine instead of a pyrimidine or vice versa. Transitions are also induced by a number of exogenous and endogenous DNA-damaging agents the behave like base analog in vivo. Such mutagens include alkylating and some oxidative agents.

74. Ethylene binding to its receptor, does not lead to

(a) dimerization of the receptor

(b) phosphorylation of the receptor

(d) aggregate in small clusters

Ans. (b)

Sol. In the current view, ethylene is sensed by a family of receptors that show similarity to the bacterial two-component histidine kinases and function as negative regulators of the pathway. Binding of the ethylene gas turns off the receptors, resulting in the inactivation of another negative regulator of ethylene signaling, CTR1, a Raf-like protein kinase that directly interacts with the receptors. EIN2, a protein of unknown biochemical activity that functions as a positive regulator of the pathway, acts downstream of CTR. Activation of EIN2 by ethylene leads to the activation of EIN3 and EIN3-like transcription facctor. In the absence of ethylene, the levels of EIN3 protein and extremely low because of the function of two F-box-containing proteins, EBF1 and EBF2 that targets EIN3 for proteasome-mediated degradation. In the presence of ethylene, the EIN3 protein accumulates in the nucleus and initiates a transcriptional cascade, resulting in the activation and repression of hundreds of genes.

75. Which of the following features is not shown by glyphosate; a broad spectrum herbicide ?

(a) Little residual soil activity.

(b) Ready translocation in phloem.

(d) Inhibiton of early steps in the biosynthesis of branched chain amino acids.

Ans. (d)

Sol. In plants, glyphosate disrupts the shikimic acid pathway through inhibition of the enzyme 5-enolpyruvylshikimate-3-phosphate (EPSP) synthase. The resulting deficiency in EPSP production leads to reductions in aromatic amino acids that are vital for protein synthesis and plant growth.

76. Following are some statements regarding plant growth hormones.

1. Ethylene regulates sabscission.

2. Gibberllins do not play any role in flowering.

3. Auxin and cytokinin promote cell division.

4. Over experssion of cytokinin oxidase would promote root growth.

5. ABA inhibits root growth and promotes shoot growth at low water potential.

6. ABA promotes leaf senescence independent of ethylene.

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

(a) 1, 3 and 6

(b) 2, 3 and 4

(d) 2, 4 and 5

Ans. (a)

Sol. Ethylene causes abscission in vivo by inhibiting auxin synthesis and transport or enhancing auxin destruction, thus lowering the diffusible auxin level.

In the root meristem, auxin induces the meristematic cell division, whereas cytokinin promotes the cell to switch from the meristematic to differentiated state through inhibiting auxin signaling.

ABA also induces senescence through core ABA signaling in an ethylene-independent way.

77. Following are some statements about low temperature stress in plants.

P. Fatty acid composition of mitochondria isolated from chilling resistant and chilling sensitive plants differs significantly.

Q. Ratio of unsaturated fatty acids to saturated fatty acids is lower in chilling resistant species.

R. The cellular water does not freeze even at –40°C, because of the presence of solutes and toher antifreeze proteins.

S. Heat shock proteins do not play any role during low temperature stress.

Which one of the following combination of above statement is correct ?

(a) P and Q

(b) P and R

(d) Q and S

Ans. (b)

Sol. The AFPs are a diverse group of proteins that are present in cells at low concentrations. They depress the freezing point and inhibit ice recrystallization of cells.

An increased chilling tolerance of elymus, seems to be caused by a high content of unsaturated fatty acid residues in the total membrane lipids; linoleic (41.9%) and -linolenic (23.4%) acids predominated among these fatty acids.

78. Following are some of the statements regarding the effect of CO₂ concentration on photosynthesis in plants.

P. With elevated CO₂ levels, C₃ plants are much more responsive than C₄ plants under well watered conditions.

Q. In C₃ plants, increasing intracellular CO₂ partial pressure can stimulate photosynthesis ony over a narrow range.

R. In C₄ plants, CO₂ compensation point is nearly zero.

Which one of the following combination of above statements is/are correct ?

(a) P and Q

(b) Q and R

(d) R only

Ans. (c)

Sol. An increase in CO₂ concentration increases the rate of photosynthesis. But the higher concentration of CO₂ can inhibit the photosynthesis. In the atmosphere, there is 0.03-0.04% concentration of CO₂. Increase in concentration up to 0.05% can increase the rate of photosynthesis.

At high light intensities, both C₃ and C₄ plants show increase in the rates of photosynthesis. The fact that C₃ plants respond to higher CO₂ concentration by showing increased rates of photosynthesis leading to higher productivity has been used for some greenhouse crops such as tomatoes and bell pepper.

79. The quantum yield of photosynthetic carbon fixation in a C₃ plant and C₄ plant is studied as a function of leaf temperature. Following are some statements based on this study :

P. At lower temperature the quantum yield of C₃ plant is lower than C₄ plant.

Q. In C₄ plant quantum yield does not show a temperature dependence.

R. Since the photorespiration is low in C₄ plants because of CO₂ concentating mechanism, quantum yield is not affected.

S. At higher temperature the quantum yield of C₃ plant is lower than C₄ plant.

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

(a) P, Q and S

(b) Q, R and S

(d) P, R and S

Ans. (b)

Sol. Although under optimal conditions it is expected that C₄ plants should have a lower quantum yield than C₃ plants because of the additional energy expense of the C4 cycle, under current atmospheric conditions the quantum yield of C₃ plants is significantly reduced because of photorespiration.

However C₄ plants have a higher temperature optimum than C₃ plants, Photosynthesis is usually inhibited when leaf temperatures exceed about 38°C.

80. Following are some statements for synthesis of jasmonic acid in plants :

1. 12-oxo-phytodienoic acid is produced in chloroplast and transported peroxisome.

2. Action of lipoxygenase, allene oxide synthase and allene oxide cyclase takes place in peroxisome.

3. 12-oxo-phytodicnoic acid is a first reduced and then converted to jasmonic acid by β-oxidation.

4. Final production of jasmonic acid takes place in chloroplast.

5. Action of allene oxide synthase and allene oxide cyclase takes place in chloroplast.

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

(a) 1, 2 and 3

(b) 2, 4 and 4

(d) 1, 3 and 5

Ans. (d)

Sol. In Arabidopsis, there are three pathways for the synthesis of JAs, including the octadecane pathway starting from a-linolenic acid and the hexadecane pathway starting from hexadecatrienoic acid. All three pathways require three reaction sites: the chloroplast, peroxisome, and cytoplasm. 12-oxo-Phytodienoic acid (OPDA) is a primary precursor of (-)-jasmonic acid (JA), able to trigger autonomous signaling pathways that regulate a unique subset of jasmonate-responsive genes, activating and fine-tuning defense responses, as well as growth processes in plants.

81. In a normal human eye, for sharp image formation on the retina, maximum dioptic power is provided by the

(a) retina

(b) cornea

(d) posterior surface of the lens

Ans. (b)

Sol. In humans, the total optical power of the relaxed eye is approximately 60 dioptres. The cornea accounts for approximately two thirds of this refractive power (about 40 dioptres) and the crystalline lens contributes the remaining one third (about 20 dioptres).

82. Which of the following waves is likely to be absent in a normal frog EGG ?

(a) P

(b) Q

(d) R

Ans. (b)

Sol. In frog, ECG only P, R and T waves exist. Q wave is absent. In human ECG, if Q wave is absent it represents coronary artery disease and strongly associated with fibrosis of the septum with or without infarction.

83. In tis flow diagram name the chemicals A, B, C and D in proper sequence.

(a) Renin, Angiotensin II, Angiotensin I, Angioten-sinogen.

(b) Angiotensin I, Angiotensinogen, Angiotensi II, Renin.

(d) Renin, Angiotensinogen, Angiotensin I, Angiotensin II.

Ans. (d)

Sol. Decrease artrial pressure is sensed by juxtaglomerular cells near DCT which release renin. Renin converts angiotensinogen into angiotensin 1 and angiotensin converting enzyme from lungs convert angiotensin 1 to angiontensin 2 which results in renal retention of salt and water and vasoconstruction [Renin Angiotensin System (RAS)].

84. An isolated cartoid sinus was prepared so that the pressure may be regulated by a pump and the resultig discharge in single carotid sinus nerve fibre could be recoded. The following are the possible observations.

P. No discharge when carotid sinus perfusion pressure was below 30 mm Hg.

Q. Linear increase in discharge frequency when cartoid sinus perfusion pressure was gradually increased from 70 to 110 Hg.

R. Increase in discharge frequency was more prominent in greater pulsatile changes of cartoid sinus pressure keeping the mean pressure identical in all cases.

S. Increase in discharge was more prominent in the falling phase of pulsatile change of cartoid sinus pressure than in the rising phase.

Which one of the following is correct ?

(a) P, Q and R

(b) P and R

(d) S only

Ans. (a)

Sol. There is no discharge when carotid sinus perfusion pressure was below 30 mm Hg, linear increase in discharge frequency when cartoid sinus perfusion pressure was gradually increased from 70 to 110 Hg and increase in discharge frequency was more prominent in greater pulsatile changes of cartoid sinus pressure keeping the mean pressure identical in all cases.

85. A 1 meter tall object was placed 10 meter in front of a normal eye. The size of the image on the retina will be (consider distance between lens and retina = 1.7 cm)

(a) 0.17 mm

(b) 1.7 mm

(d) 170 µm

Ans. (b)

Sol. From Lens equation, we have ; where, d₀ = object distance, d_i = image distance and f = focal length.

= 58.82 – 0.10 = 58.72

So, d_i = 0.017 m

For image height, we have ; where, h_o = height of the object, h_i = height of the image.

h_i = –0.0017 m = –1.7 mm

86. For a normal heart, the time taken for atrial systole and diastole are A_s and A_d seconds, respectively, while the same for ventricular systole and diastole are V_s and V_d. Which one of the following equations is correct ?

(a) A_s + A_d = V_s + V_d

(b) A_s + A_d < V_s + V_d

(d) A_s + A_d > V_s + V_d

Ans. (a)

Sol. Time taken for atrial systole and diastole is equal to time taken for ventricular systole and diastole.

87. A monkey undergoes cerebellectomy. After the post-operative recovery, the monkey was given a task to press a bar. The possible observations are :

P. Its hand would overshoot the target while reaching the bar.

Q. It would be unable to move forelimbs.

R. It would shoe intention tremor while trying to press the bar.

S. It would press the bar with mouth instead of hand.

Which one of the following is correct ?

(a) P and R

(b) Q only

(d) Q and S

Ans. (a)

Sol. After cerebellectomy in among monkey, its hand would overshoot the target while reaching the bar and it would shoe intention tremor while trying to press the bar.

88. During the Spanish conquest of the Inca Empire at the high altitude in Peru, many soldiers fell sick. It was found that the sickness was due to the low partial pressure of O₂ in the atmosphere at that altitude. To determine the reason, blood was collected from those patients. The circulating erythropoietin (EPO) level were estimated and the O₂- dissociation curve of haemoglobin were drawn and compared with the same in native people as depicted below.

Which one of the following combinations is logically correct ?

(a) P and R

(b) P and S

(d) Q and S

Ans. (a)

Sol. Hb saturation with oxygen is correctly represented in option (P) and erythropoietin is high in native of Inca empire than patient. Hence, option (R) is also correct.

89. In an animal experiment :

P. Electrical stimulation of an area in the brain (A) increased a function (F) which was prevented by systemic injection of adrenergic antagonistic, prazosin.

Q. Injection of carbachol (cholinergic against) into A also increased function F which was, however, not prevented by systemic injection of adrenergic antagonistic, prazosin.

The results are likely to be due to the stimulation of

(a) nonadrenergic and cholinoceptive neurons.

(b) cholinergic and non-adrenoceptive neurons.

(d) both neurons and fibres passing through 'A'.

Ans. (b)

Sol. P. Electrical stimulation of an area in the brain (A) increased a function (F) which was prevented by systemic injection of adrenergic antagonistic, prazosin.

Q. Injection of carbachol (cholinergic against) into A also increased function F which was, however, not prevented by systemic injection of adrenergic antagonistic, prazosin

These observations are due to cholinergic and non-adrenoceptive neurons.

90. The graph reperesents relative plasma concnetration of hormones (A and B) during reproductive cycle in a normal female. Which one of the following combinations is correct ?

(a) (A) is FSH and (B) is estrogen

(b) (A) is estrogen and (B) is LH

(d) (A) is LH and (B) is FSH

Ans. (c)

Sol. Option (c) is correct where FSH cause maturation of ovum and LH is responsible for ovulation.

91. A plant of the genotype AaBb is selfed. The two genes are linked and are 50 map units apart. What proportion of the progeny will have the genotype aabb ?

(a) ½

(b) ¼

(d) 1/16

Ans. (d)

Sol. Types of gametes from AaBb :

Proportion of the progeny of genotype, aabb =

92. The basic analog 2-aminopurine pairs with thymine, and can occasionally pair with cytoskine. The type of mutation induced by 2-aminopurine is

(a) transversion

(b) transition

(d) nonsense

Ans. (b)

Sol. Base analog 2-aminopurine will cause transition mutation which is a type of substitution mutation.

93. What kind of aneuploid gametes will be generated if meiotic non-disjunction occurs at first divison ? ('n' represents the haploid number of chromosomes)

(a) only n + 1 and n

(b) only n – 1 and n

(d) either n + 1 or n – 1

Ans. (c)

Sol. Aneuploidy is caused due to non-disjunction during meiosis. Non-disjunction during meiosis 1 leads to production of only two type of gametes one having one extra chromosome (n+1) and another with one less chromosome (n–1).

94. Assuming a 1 : 1 sex ration, what is the probability that three children from the same parents will consist of two daughters and one son ?

(a) 0.375

(b) 0.125

(d) 0.75

Ans. (a)

Sol. Each possible birth order = 1/8

95. Consider the following crosses involving grey (wild-type) and yellow body colour trud breeding Drosphila :

Assuming 200 F2 offsprings are produced in cross 2, which one of the following outcome is expected ?

(a) 97 grey males; 54 yellow females, 49 grey males.

(b) 102 yellow males, 46 yellow females, 52 grey females.

(d) 98 grey males, 94 yellow females, 2 yellow males, 6 grey females.

Ans. (c)

Sol.

Here, ratio of all type of progeny is 1:1:1:1.

Option (c) approximately matched to this ratio.

96. The ABO blood type in human in under the control of autosomal multiple alleles. Colour blindness is recessive X-linked trait. A male with a blood type A and normal vision marries a female who also has blood type A and normal vision. The couple's first child is a male who is colour blind and has O blood group. What is the probability that their next female child has normal vision and O blood group ?

(a) ¼

(b) ¾

(d) 1

Ans. (c)

Sol. There is a ½ chance that one parent will give the recessive O-chromosome.

Probability of recessive alleles given by parents to child =

Probability that a girl will have normal vision = 1

Probability that a child will be a girl = 1/2

Probability that their child will be a girl and will have O blood group =

97. The following is a schematic representation of region (showing six bands) of the polytene chromosome of Drosophila, along wth the exten of five deletions (Del1 to Del5) :

Recessie alleles a, b, c, d, e and f are known to correspond to each of the bands (1 to 6), but their order is not known. When the recessive alleles are placed against each of these deletions, the following results are obtained. The plus (+) in the table indicates wild type phenotype of the corresponding allele, while a minus (–) indicates the phenotype governed by the corresponding mutant allele.

Which one of the following indicates the correct location of the recessive alleles on the bands of the polytene chromosomes ?

(a) a-3, b-1, c-2, d-4, e-5, f-6

(b) a-2, b-1, c-3, d-4, e-5, f-6

(d) a-6, b-2, c-3, d-4, e-1, f-5

Ans. (c)

Sol. Aquatic ecosystem is the most diverse ecosystem in the world. The first life originated in the water and first organisms were also aquatic where water was the principal external as well as internal medium for organisms. Thus water is the most vital factor for the existence of all living organisms. Water covers about 71% of the earth of which more than 95% exists in gigantic oceans. A very less amount of water is contained in the rivers (0.00015%) and lakes (0.01%), which comprise the most valuable fresh water resources. Global aquatic ecosystems fall under two broad classes defined by salinity – freshwater ecosystem and the saltwater ecosystem. Freshwater ecosystems are inland water that have low concentrations of salts (< 500 mg/L). The salt-water ecosystem has high concentration of salt content (averagin about 3.5%). An aquatic ecosystem (habitats and organisms) includes rivers and streams, ponds and lakes, oceans and bays, and swamps and marshes and their associated animals. These species have evolved and adapted to watery habitats over millions of years. Aquatic habitats provide the food, water, shelter and space essential for the survival of aquatic animals and plants. Aquatic biodiversity is the rich and harbours variety of plants and animals — from primary producers algae to tertiary consumers large fishes, intermittently occupied by zooplankton, small fishes, aquatic insects and amphibians. Many of these animals and plants species live in water; some like fish spend all their lives underwater, whereas others, like toads and frogs, may use surface waters only during the breeding season or as juveniles.

98. The following figure depicts the relationship between a genetic map for four genes (A, B, C and D) and their corresponding physical map :

The following statements are made to explain this relationship :

1. More number of recombination events occur between A and B as compared to B and C.

2. Lesser number of recombination events occur between C and D as compared to B and C.

3. Although the physical distance between A and B is less than that between C and D, the region between A and B is more recombinogenic.

4. The physical distance between A and B is less than between C and D, and thus the region between A and B is less recombinogenic.

5. Although the physical distance between C and D is more than that between B and C, the region between C and D is less recombinogenic.

6. Although the physical distance between C and D is more than B and C, the region between C and D is more recombinogenic.

Which statements are correct ?

(a) 1 and 2

(b) 3 and 5

(d) 1, 3 and 5

Ans. (b)

Sol. Although the physical distance between A and B is less than that between C and D, the region between A and B is more recombinogenic and although the physical distance between C and D is more than that between B and C, the region between C and D is less recombinogenic.

99. In E. coli four Hfr stains donate the following genetic markers in the order shown below :

Which of the following depicts the correct order of the markers and the site of integration of the F-factor in the four Hfr strains ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. By observing all Hfr strains the correct order of markers are shown in option (b).

100. Aquatic primary production was measured using Light-and-Dark Bottle technique. If the initial oxygen concentration was I and the final oxygen concentration in the light bottle was L and that in the dark bottle D, the gross productivity (in terms of oxygen released) is given as

(a) L-I

(b) I-D

(d) L-D

Ans. (d)

Sol. The difference in dissolved oxygen over time between the bottles stored in the light and in the dark is a measure of the total amount of oxygen produced by photosynthesis. The total amount of oxygen produced is called the gross productivity. Only respiration can occur in the bottle stored in the dark. The decrease in dissolved oxygen (DO) in the dark bottle over time is a measure of the rate of respiration. Divide your production figure by the amount of energy you put into it ( L-D).

101. Which of the following processes interferes in sequence-basd phylogeny ?

(a) Horizontal gene transfer

(b) Adaptive mutations

(d) Reverse transcription

Ans. (a)

Sol. Toxic shock syndrome (TSS) is a series but uncommon infection caused by either Staphy-lococcus aureus bacteria or bystreptococcus bacteria.

Originally linked to the use of tampons, especially high-absorbency ones and those that are not changed frequently, it's now also known to be associated with the contraceptive sponge and diaphragm birth control methods. TSS also can arise from wounds secondary to minor trauma or surgery incisions where bacteria have been able to enter the body and cause the infection.

TSS also can affect anyone who has any type of staph infection, including pneumonia, abscess, skin or wound infection, the blood infection septicemia or the bond infection osteomyelitis.

Most often, streptococcal TSS appears after bacteria have invaded areas of injured skin, such as cuts and scrapes, surgical wounds and even chickenpox blisters.

Symptoms of TSS can include sudden high fever, a faint feeling, diarrhoea, headache, a rash and muscle aches.

102. Which of the following groups of species are typical of grassland habitats in India ?

(a) Black buck, wolf, great Indian bustard, lesser florican

(b) Spotted deer, dhole, peacock, finch-lark

(d) Otter, cormorant, darter, pelican

Ans. (a)

Sol. Many rare species such as The Bengal Florican, One-horned Rhinoceros, Pygmy Hog, Hispid Hare, Wild Buffalo, Hog Deer, Swamp Deer in Terai grassland, the Great Indian Bustard in dry, short grasslands, the Lesser Florican in monsoonal grasslands of western India are some examples of animals and birds that thrive in grasslands.

103. The atmosphere in a sealed space craft contains

(a) pure oxygen

(b) a mix of oxygen and nitrogen

(d) pressurised atmospheric air available normally on earth

Ans. (b)

Sol. Batrachochytrium dendrobatidis is a chytrid fungus that causes the disease chytridiomycosis. In the decade after it was first discovered in amphibians in 1998, the disease devastated amphibian populations around the world, in a global decline towards multiple extinctions, part of the Holocene extinction. Some amphibian species appear to have an innate capacity to withstand chytridiomycosis infection. Even within species that generally succumb, some populations survive, possibly demostrating that these traits or alleles of species are being subjected to evolutionary selection.

104. The Hutchinsonian concept of ecological niche is based on

(a) microhabitat occupied

(b) multidimensional hypervolume

(d) a combination of role played and microhabitat occupied

Ans. (b)

Sol. The Hutchinsonian niche is an "n-dimensional hypervolume", where the dimensions are environmental conditions and resources, that define the requirements of an individual or a species to practice its way of life, more particularly, for its population to persist. G. Evelyn Hutchinson more than a half century ago proposed that one could characterize the ecological niche of a species as an abstract mapping of population dynamics onto an environmental space, the axes of which are abiotic and biotic factors that influence birth and death rates.

105. A specialist species has a

(a) wider niche and high efficiency of niche utilization

(b) narrower niche and high efficiency of nich utilization

(d) narrower niche and low efficiency of niche utilization

Ans. (b)

Sol. Specialist species are animals that require very unique resources. Often, these species have a very limited diet, or need a specific habitat condition to survive. A specialist species, has a narrow niche and is not adaptable to change in their environment. Tiger salamanders are an example of specialists.

106. The rattans and canes that we use in furniture belong to

(a) bamboos

(b) palms

(d) legumes

Ans. (b)

Sol. Around 20% of rattan palm species are economically important and are traditionally used in Southeast Asia in producing wickerwork furniture, baskets, canes, woven mats, cordage, and other handicrafts. Rattan canes are one of the world's most valuable non-timber forest products.

107. Horse shoe crabs belong to the group

(a) Onychophora

(b) Chelicerata

(d) Crustacea

Ans. (b)

Sol. Horseshoe crab, (order Xiphosura), common name of four species of marine arthropods (class Merostomata, subphylum Chelicerata) found on the east coasts of Asia and of North America.

108. The presence of Salmonella in tap water is indicative of contamination with

(a) industrial effluents

(b) human excreta

(d) kitchen waste

Ans. (b)

Sol. It is generally accepted that Salmonella present in water can be traced back to its animal origins. This pathogen may directly be transported from feces or exudates of wild animals by rain water runoff to rivers or ponds used for irrigation.

109. Which of the following is not a physiological characteristic of early successional plants ?

(a) High respiration rate

(b) Inhibition by far-red light

(d) Low photosynthetic rate

Ans. (d)

Sol. Studies suggest that photosynthetic flexibility to be high for early successional annuals, intermediate for midsuccessional species, and low for late successional species.

110. In transverse sections of a young stem, if vallecular canals and carinal canals are present, then the belongs to

(a) lycopodiales

(b) isoetales

(d) equisteales

Ans. (d)

Sol. A valecular canal is a large air canal in the cortex below each sulcus . Then comes the endodermis covering the vascular bundles and pith . Carinal canals are present in vascular bundles. It is present on the inner side of vascular bundles. The function of the carinal canal is a water-conducting channel. The anatomy of Equisetum stems is characterized by the occurrence of vallecular and carinal canals.

111. Batrachochytrium dendrobatidis, a fungus, has been implicated in the decline of populations of

(a) fish

(b) frogs

(d) bats

Ans. (b)

Sol. Nitrogen is one of the primary nutrients critical for the survival of all living organisms. It is a necessary component of many biomolecules, including proteins, DNA and chlorophyll. Although nitrogen is very abundant in the atmosphere as dinitrogen gas (N₂), it is largely inaccessible in this form to most organisms, making nitrogen a scarce resource and often limiting primary productivity in many ecosystems. Only when nitrogen is converted from dinitrogen gas into ammonia (NH₃) does it become available to primary producers, such as plants.

Fig. : Major transformations in the nitrogen cycle.

In addition to N₂ and NH₃ nitrogen exists in many different forms, including both inorganic (e.g., ammonia, nitrate) and organic (e.g., amino and nucleic acids) forms. Thus, nitrogen undergoes many different transformations in the ecosystem, changing from one form to another as organisms use it for growth and in some cases energy. The major transformations of nitrogen into its many oxidation states is key to productivity in the biosphere and is highly dependent on the activities of a diverse assemblage of microorgansims, such as bacteria, archaea and fungi.

112. Wetlands are conserved internationally through an effort called as

(a) Basel convention

(b) Rio convention

(d) Ramsar convention

Ans. (d)

Sol. The Ramsar Convention is the intergovernmental treaty that provides the framework for the conservation and wise use of wetlands and their resources. The Convention was adopted in the Iranian city of Ramsar in 1971 and came into force in 1975. Wetland conservation is aimed at protecting and preserving areas where water exists at or near the Earth's surface, such as swamps, marshes and bogs. Wetlands cover at least six per cent of the Earth and have become a focal issue for conservation due to the ecosystem services they provide.

113. Chlorophyll pigment composition and carbohydrate food reserves of some algal groups are given below :

Pigments : (i) Chlorophyll a and b (ii) Chlorophyll a and c

Carbohydrate food reserve; (a) Paramylon; (b) Starch; (c) Laminarin; (d) Leucosin.

Identify the correct combination of the characters for the given groups.

(a) Euglenophyta-(i and a); Bacillariophyta-(ii and d); Phaeophyta-(ii and c); Chlorophyta-(i and b).

(b) Euglenophyta-(ii and a); Bacillariophyta-(ii and d); Phaeophyta-(i and c); Chlorophyta-(i and b).

(d) Euglenophyta-(i and d); Bacillariophyta-(ii and a); Phaeophyta-(ii and c); Chlorophyta-(i and b).

Ans. (a)

Sol. The chloroplasts found in Euglena contain chlorophyll a and b which aids in the synthesis of carbohydrates to be stored as starch granules and paramylon. Paramylon is made in the pyrenoids of Euglena. Chrysolaminarin is a storage polysaccharide typically found in photosynthetic heterokonts. It is used as a carbohydrate food reserve by phytoplankton such as Bacillariophyta (similar to the use of laminarin by brown algae). The photosynthetic pigments in phaeophyta include chlorophyll-a and chlorophyll-c. Phaeophyta stores food in the form of complex polysaccharides, alcohol and sugar. The chief source of carbohydrate is a substance known as laminarine.In Chlorophyceae, the stored food material is starch and the major pigments are chlorophyll a and b.

114. Identify the synapomorphies in the following cladogram :

(a) (a) seeds with long terminal wing; (b) ovules 1-20 per scale; (c) resin canals; (d) 1 ovule per scale.

(b) (a) resin canals; (b) seeds with long terminal wing; (c) 1 ovule per scale; (d) ovules 1-20 per scale.

(d) (a) seeds with long terminal wing; (b) ovules 1-20 per scale; (c) 1 ovule per scale; (d) resin canals.

Ans. (b)

Sol. Although resins occur in a variety of flowering resinous plants, they mostly occur in Gymnosperms (a large group of cone-producing plants that include conifers, cycads, Ginkgo, etc.), especially of the Pinaceae family (such as pine, fir, and cedar trees). Plants classified as gymnosperms with naked seeds arising from woody cones rather than flowers, the Pine Family (Pinaceae) contains many genera with winged seeds, including Pinus (Pine). Trees in the pine family have cone scales with a pair of winged ovules (seeds) on the upper surface; members of the araucaria family have only one ovule per scale. There are generally two to nine erect or inverted ovules for each fertile scale. The seeds of Cupressaceae have compressed, narrow wings on both sides.

115. During a field study, three insects with the following characteristics were observed :

P. elongate, membranous wings with net-like venation, long and slender abdomen, large compound eyes.

Q. small bodied, sucking mouth parts, narrow wings frigned with setae.

R. sclerotized forewings, membranous hindwings, chewing mouth parts.

They can be identified to their respective orders as

(a) P-Orthoptera; Q-Hemiptera; R-Coeloptera

(b) P-Odonata; Q-Coeloptera; R-Hemiptera

(d) P-Odonata; Q-Thysanoptera; R-Coleoptera

Ans. (a)

Sol. Inbreeding coefficient : An inbreeding coefficient (F) is used for quantitative calculation of the effects of inbreeding. F reflects the extent to which heterozygosity is reduced relative to the Hardy Weinberg expection derived from allelic frequencies.

F = 1 – (H/2pq)

where H is the actual observed frequency of heterozygous individuals in the population. If there are no heterozygotes, F = 1.0, whereas if the number of heterozygotes equals the Hardy-Weinberg predicted value of 2pq, then there is no effect of inbreeding and F = 0.

Two different ways to calculate F : Two very different methods can be used to calculate the value of the inbreeding coefficient F. They both arrive at the same number, but the methods that are used are so different that careful examination is needed to realize that they aer wimply two different ways of achieving the same goal. Our current textbook uses a procedure based on summation of pathway diagrams to arrive at a final value for F. Although this method is effective, it involves a procedure that is not intuitively self-explanatory and that must be basically memorized. These notes will therefore begin with a more direct calculation, and then relate the calculation to the use of pathway diagrams.

Significance of inbreeding coefficient : The inbreeding coefficient (F) is the probability that the two alleles at a genetic locus in an individual are identical by descent (derived from one particular allel in their ancestry). This is referred to as autozygosity and is genrally understood to be limited to recent aucestry, since broad populations also share common alleles (referred to as allozygosity). Inbreeding causes a population to become completely homozygous over a number of generations, whereas random mating maintains heterozygosisty at a level of 2pq. The balance between the two depends on the F value, with increased homozygosity as F becomes larger and increased heterozygosity as F becomes smaller.

116. From among the five animals listed below, match the two attributes-amniotic egg, and endothermy, with the correct animal(s) :

1. Fish

2. Frog

3. Crocodile

4. Pigeon

5. Zebra

(a) amniotic egg : 2, 3, 4; endothermy : 4, 5

(b) amniotic egg : 3, 4, 5; endothermy : 4, 5

(d) amniotic egg : 2, 3, 4; endothermy : 3, 4, 5

Ans. (b)

Sol. Michaelis and Menten postulated that the enzyme first combines reversibly with its substrate to form an enzyme substrate complex in a relatively fast reversible step :

E + S ES ...(1)

The ES complex then breaks down in a slower second step to yield the free enzyme and the reaction product P.

ES E + P ...(2)

In this model the second reaction (2) is slower and therefore limits the rate of the overall reaction. It follows that the overall rate of the enzyme-catalyzed reaction must be proportional to the concentration of the species that reacts in the second step, that is ES.

117. Which of the following is not true for a critically endangered species ?

(a) Reduction of population breeding ability due to increased relatedness through the action of incompatibility mechanism in plants or behavioural difficulties in animals.

(b) The individuals of the species which have declined to low numbers aer still a genetically open system.

(c) Loss of some alleles from the species causing loss of genetic divesity with consequenct inability to respond rapidly to selection.

(d) Expression of deleterious alleles and increased homozygosity increases mortality of young and inbreeding depression leads to reduced offspring fitness.

Ans. (b)

Sol. A species is classified as critically endangered when its population has declined at least 90 percent and the cause of the decline is known. Genetically open system simply means that genes can flow between organisms via breeding, for example in the case of species, there is flow of genes between members of same species via breeding, but no flow between members of different species, hence species is a genetically closed system.

118. In the global nitrogen cycle, the following microbial organisms are involved in three important process-denitrification, nitrification and nitrogen fixation.

1. Rhizobium

2. Nitrosomonas

3. Nitrobacter

4. Pseudomonas

5. Azotobacter

Which of the following is the correctly matched pair of process and its causative species ?

(a) Denitrification-(2); nitrogen fixation-(3) and (4); nitrification-(4)

(b) Denitrification-(4); nitrogen fixation-(1) and (5); nitrification-(3)

(d) Denitrification-(2); nitrogen fixation-(1) and (4); nitrification-(3)

Ans. (b)

Sol. Denitrifying bacteria are microorganisms whose action results in the conversion of nitrates in the soil to free atmospheric nitrogen. Some examples of denitrifying bacteria are Thiobacillus denitrificans, Micrococcus denitrificans, and some species of Serratia, Pseudomonas, and Achromobacter. Two kinds of nitrogen-fixing microorganisms are recognized: free-living (nonsymbiotic) bacteria, including the cyanobacteria (or blue-green algae) Anabaena and Nostoc and genera such as Azotobacter, Beijerinckia, and Clostridium; and mutualistic (symbiotic) bacteria such as Rhizobium, associated with leguminous plants. The nitrification process requires the mediation of two distinct groups: bacteria that convert ammonia to nitrites (Nitrosomonas, Nitrosospira, Nitrosococcus, and Nitrosolobus) and bacteria that convert nitrites (toxic to plants) to nitrates (Nitrobacter, Nitrospina, and Nitrococcus).

119. Which of the following graphs illustrates the current consensus on the role of disturbance on the species richness of a community ?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Hypothesis proposes that species richness should be maximized under intermediate levels of disturbance because at low levels of disturbance superior competitor species monopolize resources and exclude other species, whereas at high disturbance levels only the most resistant species survive.

120. Suppose you discovered a new species about which you know only two facts : it is small-sized (< 10 cm) and short-lived (< 20 days). Which of the following reproductive strategies is most likely to be true for this species ?

(a) Breeds early and more than once in life and produces large number of small-sized offspring.

(b) Breeds late and only once in life and produces large number of small-sized offspring.

(d) Breeds early and only once in life and produes a small number of large sized offspring.

Ans. (c)

Sol. The production of numerous small offspring followed by exponential population growth is the defining characteristic of r-selected species. They require short gestation periods, mature quickly (and thus require little or no parental care), and possess short life spans.

121. Autotrophs in the aquatic ecosystem, unlike their counterparts in the terrestrial ecosystem, are mostly microscopic and very low in indigestible (to the herbivores) matter. This explains the fact that compared to the terrestrial ecosystem, in the aquatic ecosystem

(a) productivity/Biomass ratios are higher and energy transfer rates to higher trophic levels are faster.

(b) Productivity/Biomass ratios are lower and the energy transfer rates to higher trophic levels are slower.

(d) Productivity/Biomass ratios are higher and the energy transfer rate to higher trophic levels are slower.

Ans. (a)

Sol. To achieve a high rate of primary production, an aquatic plant community must achieve a high rate of collection of light energy and an efficient utilization of this absorbed energy by its photosynthetic system, followed by conversion of photosynthate to new cell material. Ecological efficiency of aquatic trophic level as average 10 percent of the energy transferred from one trophic level to another trophic level. Phytoplankton, the primary producers with 1000 units energy decreased 10 percent to 100 units energy for the herbivorous.

122. Ecological compression differs from character displacement in that it operates on a

(a) shorter timescale and does not invovled heritable change

(b) longer timescale and does not involve heritable change.

(d) longer timescale and involves heritable change.

Ans. (a)

Sol. The ecological compression hypothesis predicts that when two species occur together in narrow sympatry, individuals in the overlap zone will use a smaller range of habitats and a larger or unchanged range of prey than individuals in allopatry. Character displacement is the phenomenon where differences among similar species whose distributions overlap geographically are accentuated in regions where the species co-occur, but are minimized or lost where the species' distributions do not overlap. This pattern results from evolutionary change driven by biological competition among species for a limited resource

123. The first living beings on earth were anaerobic because

(a) there was no oxygen in air

(b) oxygen damages proteins

(d) they evolved in deep sea

Ans. (a)

Sol. The first life forms were anaerobic as the primitive atmosphere was reducing and oxygen was absent. They were chemoheterotrophs that used the organic component as a source of energy and carbon.

124. The peacock's tail is an example of

(a) natural selection

(b) diversifying selection

(d) group selection

Ans. (c)

Sol. In genetics, a molecular marker (identified as genetic marker) is a fragment of DNA that is associated with a certain location within the genome. Molecular markers are used in molecular biology and biotechnology to identify a particular sequence of DNA in a pool of unknown DNA.

125. Number of trails required for rats to learn a task when they were exposed to various conditions were as follows :

Experimnental conditions Observations

P. Light : light dark cycle-12h : 12h N-trials

Q. Bright light-24h Significantly more trials than 'N'

R. Bright light-24h + continuous physcial Significantly more trials than 'N'

disturbance

S. Dark light-24h + continuous physical Significantly more trials than 'N'

disturbance

Which of the following inferences is most appropriate ?

(a) Continuous light enhanced learning

(b) Continuous darkness inhibited learning

(d) Learning was reduced by sleep loss

Ans. (d)

Sol. From experiment, it was concluded that when duration of dark cycle ( sleep cycle) were altered using physical disturbances or when there was no dark cycle, rats took more trials than 'N'. Proper duration of light and dark period was required for learning a task.

126. Assume a male sparrow (species X) is hatched and reared in isolation and allowed a critical imprinting period to hear the song of a male of another sparrow (species Y). Now after the isolation, what kind of behaviour will species X show ?

(a) it will the song of species Y that it had heard in the critical period

(b) it will sing the song of its own species X

(d) it will sing a song not sung by either X or Y

Ans. (a)

Sol. Open Reading Frames (ORF) : Reading frames where successive nucleotide triplets can be read as codons specifying amino acids and where the sequence of these triplets is not interrupted by stop codons.

127. The genetic relatedness (r) of an individual to his nephew is 0.25. The alleles that cause uncles to care for nephews will spread, according to Hamilton's Rule, only if the fitness benefit is

(a) equal to the cost of care

(b) more than the cost of care by 25%

(d) four times the cost of care

Ans. (d)

Sol. AGAMOUS (AG) is a C-function floral organ idenity gene and is responsible for formation of stamens and carpels in the third and fourth whorls of the wild type flower. In addition, AG is required for meristem determinancy such that ag mutant flowers are indeterminate and produce an essentially endless number of floral whorls, repeating the basic pattern : (sepal, petal, petal)_n. AG RNA begins to accumulate in stage 3 flowers in cells that will later give rise to stamens and carples.

128. Several distinct time periods and different routes might explain the entrance of marsupials into Australia.

P. Late Jurassic early therians arrived in Antarctica-Australia where the marsupials subsequently evolved.

Q. Early to middle Cretaceous-early marsupials arrived in Australia for northern regions and then radiated in isolation.

R. Paleocene-marsupials entered Australia from South-East Asia.

S. Eocene-chance dispersal of marsupials into Australia.

Which of the following is the correct combination ?

(a) P, Q, R

(b) P, R, S

(d) P, Q, S

Ans. (b)

Sol. According to the continental drift theory, millions of years ago, South America and Australia were joined. This joined continent was present in the Southern hemisphere. The marsupials lived there and moved freely, they must have migrated between these islands. Marsupials make up more than 50% of the mammal species in some South American Paleocene faunas, and adaptive types include insectivores, omnivores, carnivores and small herbivores. Paleocene mammals are not yet known from Antarctica and Australia.

129. The frequencies of two alleles p and q for a gene locus in a population at Hardy-Weinberg equilibrium are 0.3 and 0.7, respectively. After a few generations of inbreeding, the heterozygote frequency was found to be 0.28. The inbreeding coefficient in this case is

(a) 0.42

(b) 0.25

(d) 0.67

Ans. (c)

Sol. Inbreeding coefficient =

Where, H_obs = frequency of observed heterozygote, frequency of p = 0.3 and frequency of q = 0.7.

130. Which of the following behavioural changes are expected in a rat when its nucleus accumbens is experimentally ablated ?

(a) Aggressive behaviour increases

(b) Exploratory behaviour decreases

(d) Level of parental care drops

Ans. (b)

Sol. The nucleus accumbens is considered as the neural interface between motivation and action, playing a key role on feeding, sexual, reward, stress-related, drug self-administration behaviors, etc. Damage to the nucleus accumbens core (AcbC) produces impulsive choice in rats reducing their ability to choose large, delayed rewards in preference to small, immediate rewards, yet these and other similar lesions do not appear to impair rats' ability to discriminate reward size.

131. To keep them in a totipotent state, embryonic stem cells need to be maintained stem cells need to be maintained in a medium supplemented with

(a) growth hormone

(b) leukemia inhibitng factor

(d) insulin

Ans. (b)

Sol. Leukemia inhibitory factor (LIF), a pleiotropic cytokine, has been used extensively in the maintenance of mouse embryonic stem cell totipotency.

132. One of the methods for finding common regulatory motifs present in a set of coregulated genes is

(a) Prosite

(b) MEME

(d) PSSM

Ans. (b)

Sol. Physiology Characteristics of Early and Late Successional Plants

133. Measurement and mapping with spatial resolution of membrane potential of a cell, which is too small for microelectrode impalement, is done using

(a) radioisotope

(b) voltage sensitive dye

(d) vital dyes

Ans. (a)

Sol. The radioisotope was found to be taken up into monolayer cultures via an ATP-independent, saturable process (Km = 40 mM). The presence of sodium on the opposite side of the membrane gave rise to a transstimulation of the 22Na+ flux. Studies utilizing potassium and valinomycin suggested that the transport system was insensitive to changes in the membrane potential.

134. A sample counted for one minute shows a count rate of 752 cpm. For how many minutes should it be counted to have 1% probable error ?

(a) 13

(b) 5

(d) 75

Ans.

Sol.

135. Genomic DNA of transgenic plants (P1, P2 and P3) obtained by transforming with binary vector A whose map is depicted below, was digested with BamH I and Sal I and hybridized with a labelled fragment X

The pattern obtained in Southern hybridization is shown below :

Based on the above, which of the following interpretation is correct ?

(a) All the plants (P1, P2 and P3) contain two copies of the transgene

(b) P1, P3 contains one and P2 contains two copies of the transgene

(d) P1 and P2 contain two and P3 contains one copy of the transgene.

Ans. (c)

Sol. By the pattern obtained in sudden hybridisation, P1 contains two, whereas P2 and P3 contain one copy of transgene each.

136. Cre/loxP system is used by phage PI to remove termianlly redundant sequences that arise during packaging of the phage DNA. Cre-lox system can be used to create targeted deletions, insertions and inversion in genomes of transgenic animals and plants. Consider a series of genetic markers A to K. How should the Lox P sites be positioned in order that Cre recombinase can create an inversion in the EFG segment relative to ABCD and HIJK ?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Generally, two loxP sites are inserted into the introns flanking an essential exon of the target gene.

137. The following are statements about molecular markers in the context of plant breeding

P. Molecular markers can be used for elimination of undesirable traits.

Q. Molecular markers cannot be used for estimation of the genetic contribution of each individual parnt in a segregating population.

R. Molecular markers are used for mapping of QTLs, which is also possible by conventinoal techniques.

S. Molecular markers can be used for selection of individuals from a population that are homozygous for the recurrent parent genotype at loci flanking the target locus.

Which of the above statements are true ?

(a) P and Q

(b) P and R

(d) Q and R

Ans. (c)

Sol. Molecular marker technology enables plant breeders to select individual plants based on their marker pattern (genotype) rather than their observable traits (phenotype). This process is called marker assisted breeding (MAB) or marker assisted selection (MAS). These chromosome landmarks can be used in the differentiation of normal and mutated chromosomes. Such markers can also be used in the identification of linkage groups and in physical mapping. Molecular markers and marker-assisted breeding in plants.

138. Enzymes are nowadays used extensively in bio-processing industries.

Enzyme 1 is used for treatment of hides to provide a finer texture, in leather processing and manufacturing of glue.

Enzyme 2 is used for clarification of fruit juices.

Identify enzymes 1 and 2

Enayme Enzyme 2

(a) Amylase Pectinase

(b) Protease Amylase

(d) Pectinase Amylase

Ans. (c)

Sol. Proteases are used in dehairing and dewooling of leather, and improve its quality (cleaner and stronger surface, softer leather, less spots). Pectinases are used for the clarification of the juice by breaking the polysaccharide pectin structure present in the cell wall of plants into galacturonic acid monomers. Pectin structure breakage facilitates the filtration process and it increases the total yield of juice.

139. Stem cell therapies are being used in regenerative medicine like forming new adult bone which usually does not regrow to bridge wide gaps. Successful attempts have now been made in this area because the same paracrine and endocrine factors were found to be involved in both endochondral ossification and fractuire repair. Few methods to achieve the above are given below :

P. Develop a collagen gel containing plasmids carrying te hhuman parathyroid hormone gene and place in the gap between the ends of the broken leg.

Q. Develop a gel matrix disc containing genetically modified stem cells to secrete BMP4 and VEGF-A and implant it at the site of the wound.

R. Make ascaffolds of material taht resembles normal extracellular matrix that could be molded to form the shape of a bodn needed and seed them with bone marrow stem cell.

S. Develop a collagen gel containing plasmids carrying the human bone marrow cells and place them between the ends of the bones.

Which of the above methods would you employ to develop a new functional bone in patients with severely fractured bones ?

(a) P and Q

(b) P, Q and R

(d) R and S

Ans. (a)

Sol. Previous studies have demonstrated that bone morphogenetic protein-4 (BMP4) could participate in vivo endochondral ossification and is one of the main local contributing factors in the early stage of fracture healing. VEGF stimulates wound healing via multiple mechanisms including collagen deposition, angiogenesis and epithelization. PTH enhances the number and the activation of osteoblast through 4 pathways: increasing osteoblast proliferation and differentiation, decreasing osteoblast apoptosis and reducing the negative effects of peroxisome proliferator activator (PPAR)? receptor on osteoblast differentiation.

140. In order to prevent tetanus in neonates, one of the following treatments can be adopted.

A. Treatment of the infant with anti-toxin and the toxoid.

B. Immunize the mother with the toxoid.

In case of A, the treatment can be given

a. immediately after birth.

b. after the onset on the condition.

In case of B, the immunization has to be done

c. before pregnancy.

d. late in the pregnancy.

The correct combination is

(a) A/a

(b) A/b

(d) B/d

Ans. (d)

Sol. Immunisation is the only effective prevention of tetanus. Tetanus toxoid is an effective, safe, stable and inexpensive vaccine that can be given to all ages, to pregnant women and to immunocompromised individuals. To prevent neonatal tetanus, maternal immunization is recommended with two doses of tetanus toxoid 4 weeks apart during pregnancy for women who have never been vaccinated or incompletely vaccinated.

141. In the following statement taken from a research paper, what does p in the parenthesis stand for ?

"The mean temperature of this region now is significantly higher than the one 50 years ago (p < 0.05, t-test)"

(a) ratio of the mean temperatures of the time periods tested.

(b) probability of the error of rejecting a true null hypothesis.

(d) probability of the t-test effective in detecting significant differences in te mean annual temperatures on the two time periods.

Ans. (d)

Sol. The probability of having a female child is 1/2.

The probability that she has normal vision is 1 (because the father's X is normal).

Probability of 'O' type blood is ¼.

Hence normal visioned, type O blood typed girl is = ½ × ¼ = 1/8.

142. In an in vitro experiment using radiolabelled nucleotides, a researcher is trying to analyze the possible products of DNA replication by resolving the products using urea-polyacrylamide gel electrophoresis. In one experimental set up RNase H was added (Set 1), while in another set no RNase H was added (Set 2).

The possible observations of this experiment could be

P. There is no difference in the mobility of labelled DNA fragmetns between the Set 1 and Set 2.

Q. There is distinct differences in the mobility of the newly synthesized labelled DNA fragments between Set 1 and Set 2.

R. The mobility of the newly synthesized labelled DNA fragments in case of Set 1 is faster as compared to the Set 2.

S. The mobility of the newly synthesized labelled DNA fragments in case of Set 1 is slower as compared to the Set 2.

Which of the following combinations represent correct observations ?

(a) P and Q

(b) Q and R

(d) Q and S

Ans. (b)

Sol. Character displacement refers to the phenomenon where differences among similar species whose distribution overlap geographically are accentuated in regions where the species co-occur but are minimized or lost wher the species distributions do not overlap. This pattern results from evolutionary change driven by competition among species for a limited resource (e.g., food). The rationale for character displacement stems from the competitive exclusion principle, also called Gause's Law, which contends that to coexist in a stable environment two competing species must differ in their respective ecological niche; without differentiation, one species will eliminate or exclude the other through competition.

Character displacement was dirst explicity explained by William L. Brown and E.O. Wilson (1956) : "Two closely related species have overlapping ranges. In the parts of the ranges where one species occurs alone, the populations of that species are similar to the othe species and may even by very difficult to distinguish from it. In the area of overlap, where the two species occur together, the populatiosn are more divergent and easily distinguished, i.e., they 'displace' one another in one or more characters. The characters involved can be morphological, ecological, behavioural or physiological they are assumed to be genetically based."

Brown and Wilson (1956) used the term character displacement to refer to instances of both reproductive character displacement or reinforcement of reproductive barriers and ecological character displacement driven by competition. As the term character displacement is commonly used, it generally refers to morphological differences due to competition. Brown and Wilson viewed character displacement as phenomenon involved in speciation stating, "we believe that it is a common aspect of geographical speciation, arising most often as a product of the genetic and ecological interaction of two (or more) newly evolved, cognate species [derived from the same immediate parental species] during their period of first contact" (1956). While character displacement is important in various scenarios of speciation, including adaptive radiations like the cichlid fish faunas in the rift lakes of East Africa (Meyer 1993), it also plays an important role in structuring communities. The results of numerous studies contribute evidence that character displacement often influences the evolution of resource acquisition among members of anecological guild (Dayan and Simberloff 2005).

Competitive relase (Grant 1972), defined as the expansion of an ecological niche in the absence of a competitor, is essentially the mirror image of character displacement. It too was described by Brown and Wilson (1956) : "Two closely related species are distinct where they occur together, but where oen member of the pair occurs alone it converges toward the second, even to the extent of being nearly identical with it in some characters."

143. In 'TaqMan' assay for detection of base substitutions (DNA variant), probes (oligonucleotides) with fluorescent dyes at the 5'-end an a quencher at 3'-end are used. While the probe is intact, the proximity of the quenchor reduces the fluorescene emitted by reporter dye. If the target sequences (wild type or the variant) are present, the probe anneals to the target sequence, downstream to one of the primes used for amplifying the DNA sequence flanking the position of the variants. For an assay two flanking PCR primers, two probes corresponding to the wild type and variant allele and labelled with two different reporters dyes and quencher were used. During extension the probe may be cleaved by the Taq-polymerase separating the reporter dye and the quencher. Three individuals were genotyped using this assay. Sample for individual I shows maximum fluorescene for the dye attached to the wild type probe, sample for individual II shows maximum fluorescene for the dye attached to variant probe and sample for individual III exhibits equal fluorescence for both the dyes. Which of the following statement is correct ?

(a) Individual I is homozgous for the variant allele.

(b) Individual II is homozgous for variant allele.

(d) Individual III is homozgous for wild type allele.

Ans. (b)

Sol. By analysing the observation provide in question, it is revealed that Individual II is homozgous for variant allele.

144. Figures A and B respective represent the dideoxy sequencing gels obtained for patial sequences from 5'-ends of a bacterial gene and its mutant (with a point mutation)

What type of mutation has occured in the gene ?

(a) Nonsense

(b) Missense

(d) Transversion

Ans. (c)

Sol. Sequence on Sanger's gele is read from bottom to top (5' 3'). So, if we read sequence of the strains we observe that,

Wild-type 5' TAC CGT GGA CTT GA 3'

Mutant 5' TAC CGT GAC TTG AG 3'

If we align these sequence we find that a single change caused frameshift (insertion or deletion) which resulted in change of sequence after TACCGTG (shown in bold). If it would have been missense mutation rest of the sequence after one base should not change. In case of nonsense mutation stop codon should be introduced.

145. The most important property of any microscope is its power of resolution, which is numerically equivalent to D, the minimum distance between two distinguishabel objects. D depends on three parameters namely, the angular aperture, , the refractive index, N, and wavelength, , of the incident light. Below are given few possible options to increase the resolution of the microscope.

P. Decrease the value of or increase either N or to improve resolution.

Q. Moving the objective lens closer to the specimen will decrease sin a and improve the resolution.

R. Using a medium with high refraction index between the speciment and the objective lens to improve the resolution.

S. Increase the wavelength of the incident light to improve the resolution.

Which of the following combination of above statements is correct ?

(a) P and R

(b) Q and R

(d) R and S

Ans. (a)

Sol. The most important property of an microscope is not its magnification but its resolving power, or resolution — its ability to distinguish between two very closely positioned objects. Merely enlarging the image of a specimen accomplishes nothing if the image is blurry. The resolution of a microscope lens is numerically equivalent to D, the minimum distance between two distinguishable objects; the smaller the value of D, the better the resolution. D depends on three parameters, all of which must be considered in order to achieve the best possible resolution : the angular aperture, or half-angle of the cone of light entering the objective lens from the specimen; the refractive index, N of the air or fluid medium between the specimen and the objective lens and the wavelength of incident light : D = (0.61) ÷ (N × sin ). Decreasing the value of or increasing either N or α will decrease the value of D and thus improve the resolution.

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

Locate Us

Company

Courses

Academies

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

Locate Us

Company

Courses

Academies

Get in touch

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