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More Info on CSIR JRF Exam December 2011 CSIR used three (3) different Booklets for June 2011 exam. All of them have same questions except their order (sequence). The below question paper is Booklet A. Instructions: This booklet contains 145 (20 Part A + 50 Pat B + 75 part C) Multiple Choice Questions (MCQs). You are required to Answer a maximum of 15, 35 and 25 questions from Part A, B and C, respectively. If more than required number of questions are answered, only first 15, 35 and 25 questions in Part A, B and C respectively, will be taken for evaluation. Total marks: 200 and Time: 3 hours.

Answer Key with Explanation

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CSIR JRF/NET Life Sciences December 2011

Part A 1. What is the angle θ in the quadrant of a circle shown below?

1. 135° 2. 90° 3. 120° 4. May have any value between 90° and 120°

2. In ∆ABC, angle A is larger than angle C and smaller than angle B by the same amount. If angle B is 67°, angle C is 1. 67° 2. 53° 3. 60° 4. 57° 3. See the following mathematical manipulations. (i) Let x = 5 (ii) then x2-25~x-5 (iii) (x - 5) (x + 5) = x - 5 (iv) x + 5 = 1 [cancelling (x - 5) from both sides] (v) 10 = 1 [Putting x = 5] Which of the above is the wrong step?

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(1) (i) to (ii) (3) (iii) to (iv) (2) (ii) to (ii i) (4) (iv) to (v) 4. Inner planets of the solar system are rocky, whereas outer planets are gaseous. One of the reasons for this is that 1. solar heat drove away the gases to the outer region of the solar system 2. gravitational pull of the sun pulled all rocky material to the inner solar system 3. outer planets are larger than the inner planets 4. comets delivered the gaseous materials to the outer planets 5. The number of craters observed due to meteoritic impacts during the early stages of formation, is less on the Earth than that on the Moon because 1. formation of craters on the Earth was difficult due to the presence of hard rocks 2. impacting bodies on the Earth were smaller in size 3. craters on the Earth are now covered by ocean water 4. earlier craters are not preserved due to continuous modification of Earth's surface by geological processes 6. During a total solar eclipse occurring at noon, it becomes dark enough for a few minutes for stars to become visible. The stars that are seen are those which will be seen from the same location 1. on the following night only 2. on the night one month later 3. on the night three months later 4. on the night six months later 7. The variation of solubilities of two compounds X and Y in water with temperature is depicted below. Which of the following statements are true ?

1. Solubility of Y is less than that of X 2. Solubility of X varies with temperature 3. Solubilities of X and Y are the same at 75 °C 4.Solubilities of X and Y are independent of temperature. 8. Living beings get energy from food through the process of aerobic respiration. One of the reactants are 1. CO2 2. Water vapor 3. O2 4. Phosphorous 9. Restriction endonuclease cleaves DNA molecules at specific recognition sites. One such enzyme has four recognition sites

on circular DNA molecule. After complete digestion, how many fragments would be produced upon reaction with this enzyme. 1. 4 2. 5 3. 3 4. 6 10. Which of the statements about the concentration of Co2 in the Earth’s atmosphere is true? 1. It was the highest in the very early atmosphere of the Earth 2, It has steadily decreased since the formation of the Earth's atmosphere 3. It has steadily increased since the formation of the Earth’s atmosphere 4. Its levels today are the highest in the Earth’s history 11. Magnesium powder, placed in an air-tight glass container at 1.0 bar, is burnt by focusing sunlight. Part of the magnesium burns off and some is left behind. The pressure of the air in the container after it has returned to room temperature is approximately 1. 1 bar 2. 0.2 bar 3. 1.2 bar 4. 0.8 bar 12. When a magnet is made to fall free in air, it falls with an acceleration of 9.8 m s-2. But when it is made to fall through a long aluminium cylinder, its acceleration decreases. Because 1. a part of the gravitational potential energy is lost in heating the magnet 2. a part of the gravitational potential energy is lost in heating the cylinder 3. the said experiment was done in the magnetic northern hemisphere 4. the cylinder shields the gravitational force 13. A solid cube of side L floats on water with 20% of its volume under water. Cubes identical to it are piled one-by one on it. Assume that the cubes do not slip or tupple, and the contact between their surfaces is perfect. How many cubes are required to submerge one cube completely? 1. 4 2. 5 3. 6 4. Infinite 14. An overweight person runs 4 km everyday as an exercise. After losing 20% of his body weight, if he has to run the same distance in the - same time, the energy expenditure would be 1. 20% more 2. the same as earlier 3. 20% less 4. 40% less 15. On exposure to desiccation, which of the following bacteria are least likely to experience rapid water loss 1. Isolated rods 2. Rods in chain 3. Cocci in chain 4. Cocci in ciusters

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16. A cupboard is filled with a large number of balls of 6 different colors. You already have one ball of each color. If you are blind folded, how many balls do you need to draw to be sure of having 3 color matched pairs of balls? 1. 3 2. 4 3. 5 4. 6 17. The conductance of a potassium chloride solution is measured using the arrangement depicted below. The specific conductivity of the solution in Sm-1, when there is no deflection in the galvanometer, is

1. 1 2. 0.5 3. 2 4. 1.5 18. What is the half-life of the radio isotope whose activity profile is shown below?

1. 1 day 2. 3 days 3. 2 days 4. 4 days 19. A bell is rung before giving food to a dog. After doing this continuously for 10 days, which of the following is most likely to happen 1. The dog learns to ignore the bell 2. The dog salivates on hearing the bell 3. The dog ignores food and runs towards the bell 4. The dog will not eat food without hearing the bell 20. For an elastic material, strain is proportional to stress. A constant stress is applied at t1. Which of the following plots characterizes the strain in the material?

Part B 21. On the molar scale which of the following interactions in a nonpolar environment provides highest contribution to the bio-molecule? 1. van der Waals interaction 2. Hydrogen bonding 3. Salt bridge 4. Hydrophobic interaction 22. Michaelis and Menten derived their equation using which of the following assumption? 1. Rate limiting step in the reaction is the breakdown of ES complex to product and free enzyme 2. Rate limiting step in the reaction is the formation of ES complex 3. Concentration of the substrate can be ignored 4. Non-enzymatic degradation of the substrate is the major step

23. In which form of DNA, the number of base pairs per helical turn is 10.5? 1. A 2. B 3. X 4. Z 24. In contrast with plant cells, the most distinctive feature of cell division in animal cells is 1. control of cell cycle transitions by protein kinases 2. enzymes responsible for DNA replication 3. ubiquitin-dependent pathway for protein degradation 4. pattern of chromosome movement 25. Most common type of phospholipids in the cell membrane of nerve cells is 1. phosphatidylcholine

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2. phosphatidylinositol 3. phosphatidylserine 4. sphingomyelin 26. Reverse transcriptase has both ribonuclease and polymerase activities. Ribonuclease activity is required for 1. the synthesis of new RNA strand 2. the degradation of RNA strand 3. the synthesis of new DNA strand 4. the degradation of DNA strand 27. The membrane lipid molecules assemble spontaneously into bilayers when placed in water and form a closed spherical structure known as 1. lysosome 2. peroxisome 3. liposome 4. endosome 28. In gene regulation. Open Reading Frame (ORF) implies 1. intervening nucleotide sequence in between two genes 2. a series of triplet codons not interrupted by a stop codon 3. a series of triplet codons that begins with a start codon and ends with a stop codon 4. the exonic sequence of a gene that corresponds to the 5 'UTR of the mRNA and thus does not code for the protein 29. Amino acid selenocysteine (Sec) is incorporated into polypeptide chain during translation by 1. charging of Sec into tRNAser followed by incorporation through serine codon 2. charging of serine into tRNAser followed by modification of serine into selenocysteine and then incorporation through serine codon 3. charging of Sec into tRNAser and then incorporation through selenocysteine codon 4. charging of serine into tRNAser, modification of serine into selenocysteine and then incorporation through a specially placed stop codon 30. α-amanitin inhibits 1. only RNA polymerase I 2. only RNA polymerase II 3. only RNA polymerase III 4. all RNA polymerases 31. While replicating DNA, the rate of misincorporation by DNA polymerase is 1 in 105 nucleotides. However, the actual error rate in the replicated DNA is 1 in 109 nucleotides incorporated. This is achieved mainly due to 1. Spontaneous excision of misincorporated nucleotides 2. 3'-5' proof reading activity of DNA polymerase 3. Termination of DNA polymerase at misincorporated sites 4. 5'-3' proofreading activity 32. Toxic shock is caused by 1. toxins produced by some bacteria 2. excessive stimulation of a large proportion of T cells by bacterial superantigens 3. abnormal cytokine production by B cells

4. excessive production of immunoglobulins 33. Ethylene binding to its receptor, does NOT lead to 1. dimerization of the receptor 2. phosphorylation of the receptor 3. activation of CTR Raf kinase 4. endocytosis of ethylene-receptor complex 34. Graft rejection does not involve 1. Erythrocytes 2. T cells 3. macrophages 4. polymorphonuclear leukocytes 35. The blastopore region of amphibian embryo that secretes BMP inhibitors and dorsalizes the surrounding tissue is known as 1. Brachet's cleft 2. Nieuwkoop center 3. Spemann's organizer 4. Hensen's node 36. During development of embryos in plants, PIN proteins are involved in 1. establishment of auxin gradients 2. regulation of gene expression 3. induction of programmed cell death 4. induction of cell division 37. Which of the following maternal effect gene products regulate production of anterior structures in Drosophila embryo? 1. Bicoid and Nanos 2. Bicoid and Hunchback 3. Bicoid and Caudal 4. Nanos and Caudal 38. Which of the floral whorls is affected in agamous (ag) mutants? I. Sepals and petals 2. Petals and stamens 3. Stamens and carpels 4. Sepals and carpels 39. Which of the following set of cell organelles are involved in the biosynthesis of jasmonic acid through octadecanoid signalling pathway? 1. Chloroplast and peroxisomes 2. Chloroplast and mitochondria 3. -Mitochondria and peroxisomes 4. Golgi bodies and mitochondria 40. Which of the following is NOT a prosthetic group of nitrate reductase 1. FAD 2. Heme 3. Mo 4. Pterin 41. Chloroplast distribution in a photosynthesizing cell is governed by blue light sensing phototropin 2

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(PHOTO2). When the cells are irradiated with high intensity blue light, the chloplasts 1. move to the side walls 2. aggregate in the middle of the cell 3. are sparsely distributed 4 aggregate in small clusters 42. Which of the following acts as a branch point for the biosynthesis of sesquiterpenes and triterpenes? 1. Farnesyl pyrophosphate 2. Geranyl pyrophosphate 3. Isopentyl pyrophosphate 4. Hydroxymcthylglutaryl-CoA 43. Which of the following waves is likely to be absent in a non11al frog ECG? 1. P 2. Q 3. T 4. R 44. The atmosphere in a sealed space craft contains 1. pure oxvgen 2. a mix of oxygen and nitrogen 3. a mix of oxygen and carbon dioxide 4. pressurized atmospheric air available normally on earth 45. in a normal human eye, for sharp image formation on the retina maximum dioptric power is provided by the 1. retina 2. cornea 3. anterior surface of the lens 4. posterior surface of the lens 46. In this flow diagram name the chemicals A, B, C and D in proper sequence.

1. Renin, Angiotensin II, Angiotensin I, Angiotensinogen 2. Angiotensin I, Angiotensinogen, Angiotensin II, Renin 3. Renin, Angiotensin I, Angiotensin II, Angiotensinogen 4. Renin, Angiotensinogen, Angiotensin I. Angiotensin II

47. 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? 1. 1/2 2. 1/4 3. 1/8 4. 1/16 48. The base analog 2-aminopurine pairs with thymine, and can occasionally pair with cytosine. The type of mutation induced by 2-aminopurine is 1. transversion 2. transition 3. deletion 4. nonsense 49. What kind of aneuploid gametes will be generated if meiotic non-disjunction occurs at first division? (‘n' represents the haploid number of chromosomes) 1. only n + 1 and n 2. only n - 1 and n 3. both n + 1 and n - 1 4. either n + 1 or n – 1 50. A single-strand nick In the parental DNA helix just ahead of a replication fork causes the replication fork to break. Recovery from this calamity requires 1. DNA ligase 2. DNA primase 3. site-specific recombination 4. homologous recombination 51. In transverse sections of a young stem, if vascular canals and carinal canals are present, then the plant belongs to 1. Lycopodiales 2. Isoetales 3. Selaginellales 4. Equisetales 52. Horse shoe crabs belong to the group 1. Onychophora 2. Chelicerata 3. Uniramia 4. Crustacea 53. Which of the following groups of species are typical of grassland habitats in India 1. Black buck, wolf, great Indian bustard, lesser florican 2. Spotted deer, dhole, peacock, finch-lark 3. Sambar, tiger, paradise fly catcher 4. Otter, cormorant, darter, pelican 54. Batrachochytrium dendrobatidis, a fungus, has been implicated in the decline of populations of 1. fish 2. frogs 3. pelicans 4. bats 55. The Hutchinsonian concept of ecological niche is based on

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1. microhabitat occupied 2. multidimensional hypervolume 3. role played in the ecosystem 4. a combination of role played and microhabitat occupied 56. Which of the following is NOT a physiological characteristic of early successional plants? 1. High respiration rate 2. Inhibition by far-red light 3. High transpiration rate 4. Low photosynthetic rate 57. 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 1. L - I 2. I - D 3. I - L 4. L - D 58. Wetlands are conserved internationally through an effort called as 1. Basel Convention 2. Rio Convention 3. Montreal Convention 4. Ramsar Convention 59. The first living beings on earth were anaerobic because 1 there was no oxygen in air 2. oxygen damages proteins 3. oxygen interferes with the action of ribozymes 4. they evolved in deep sea 60. Which of the following processes interferes in sequence-based phylogeny? 1. Horizontal gene transfer 2. Adaptive mutations 3. DNA repair 4. Reverse transcription 61. The peacock's tail is an example of 1. natural selection 2. diversifying selection 3. sexual selection 4. group selection 62. A specialist species has a 1. wider niche and high efficiency of niche utilization 2. narrower niche and high efficiency of niche utilization 3. wider niche and low efficiency of niche utilization 4. narrower niche and low efficiency of niche utilization 63. To keep them in a totipotent state, embryonic stem cells need to be maintained in a medium supplemented with

1. growth hormone 2. leukemia inhibiting factor 3. nestin 4: insulin 64. Which of the following features is NOT shown by glyphosate, a broad spectrum herbicide? 1. Little residual soil activity 2. Ready translocation in phloem 3. Inhibition of a chloroplast enzyme catalyzing the synthesis of aromatic amino acids 4. Inhibition of early steps in the biosynthesis of branched chain amino acids 65. The rattans and canes that we use in furniture belong to 1. bamboos 2. palms 3. arborescent lilies 4. legumes 66. The presence of Salmonella in tap water is indicative of contamination with 1. industrial effluents 2: human excreta 3. agriculture waste 4. kitchen waste 67. Indirect immunofluorescence involves fluorescently labeled 1. immunoglobulin-specific antibodies 2. antigen-specific antibodies 3. hapten-specific antibodies 4. carrier-specific antibodies 68. 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? 1. 13 2. 5 3. 2 4. 75 69. Measurement and mapping with spatial resolution the membrane potential of a cell, which is too small for microelectrode impalement, is done using 1. radioisotope 2. voltage-sensitive dye 3. pH sensitive chemical 4. vital dyes 70. One of the methods for finding common regulatory motifs present in a set of coregulated genes is 1. Prosite 2. MEME 3. Mat Inspector 4. PSSM

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Part C 71. Equilibrium constant (K) of noncovalent interaction between 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: 1. 2746 kcal/mole 2. -2746 kcal/mole 3. 247 kcal/mole. 4. -247 kcal/mole 72. If van der Waals interaction is described by the following relation,

where ∆Gvan is the free energy of the van der Waals interaction, A and B are constants, r is the distance between two nonbonded atoms 1 and 2, and q1 and q2 are partial charges on the dipoles 1 and 2. In this relation, the parameter A describes (1) electron shell attraction (2) electron shell repulsion (3) dipole-dipole attraction (4) dipole-dipole repulsion 73. 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 [bicarbonate]/[carbonic acid] in the blood is (1) 0.05 (2) 1 (3) 10 (4) 20 74. The hydrolysis of pyrophosphate to orthophosphate is important for several biosynthetic reactions. In E. coli, the molecular mass of the enzyme pyrophosphatase is 120 kD, and it consists of six identical subunits. The enzyme activity is defined as the amount of enzyme that hydrolyzes 10 µmol of pyrophosphate in 15 minutes at 37 °C under standard assay condition. The purified enzyme has a Vmax 0f 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 Krn? (1) 0.05 µmol (2) 62 µmol (3) 31.1 µmol (4) 1 flmol 75. Denaturation profiles of DNA are shown below.

The differences in the profiles arise because 1. The DNA is single stranded but of different sizes 2. A + T content of A > B > C and the DNA are from complex genomes 3. G + C content of C > B > A in DNA of comparable sizes isolated from simple genomes 4. G + C content is identical but A + T content in A>B>C in DNA of comparable sizes isolated from simple genomes

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76. Biosynthesis of tyrosine is detailed below:

Shikimic acid shikimic acid-5-phosphate C → Chorismic acid → prephenic acid D tyrosine. Identify A, B, C and D. 1. ATP, phosphoenolpymvic acid, 3-enolpyruyl shikimic acid-5-phosphate, phydroxyphenylpyruvic acid 2. GTP, pyridoxal phosphate, 3-enolpyruvyl shikimic acid-5-phosphate, phenylpyruvic acid, 3. NADP, 3-phosphohydroxypyruvic acid, 3-enolpyruvyl shikimic acid-5-phosphate, p-hydroxyphenylpyruvic acid 4. ATP, 3-phosphohydroxyp yruvic acid, 3-enolpyruvyl shikimic acid-5-phosphate, pyridoxylphosphate 77. A nerve impulse or action potential is generated from transient changes in the permeability of the axon membrane to Na+ and K+ ions. The depolarization 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 opposite directions because 1. there is an electrochemical gradient growth generated by proton transport 2. there is a difference in Na+ and K+ concentrations on either side of the membrane 3. Na+ is a voltage-gated channel, whereas K+ is ligand-gated 4. Na+ is dependent on ATP whereas K+ is not 78. The erythrocyte membrane cytoskeleton consists of a meshwork of proteins underlying the membrane. The principal component spectrin has a, p 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 mutation in glycophorin C 1. a mutation in glycophorin C 2. a mutant spectrin with increased tetramerization propensity 3. mutant spectrin defective in dimerization ability 4. mutant glyceraldehyde 3-phosp'hate dehydrogenase 79. 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 (not translating into protein). Some introns may harbor also transcription units, which are 1. always other protein coding genes 2. protein coding gene and RNA coding genes 3. always RNA coding genes 4. pseudo genes 80. 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 envelope into vesicles. Cyclin B degradation is followed by chromosome decondensation, nuclear envelope 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 1. RNase contamination persisted in the system 2. cyclin B was missing from the system 3. ubiquitin ligase had been overexpressed 4. cyclin B lacking the cyclin destruction box had been overexpressed 81. Many cancers carry mutant p53 genes, while some cancers have normal p53 genes, p53 activates (Waf1) which inhibits G1/S-Cdks and phosphorylation the retinoblastoma protein (Rb). Cancers with normal p53 genes could I. express non-phosphorylatable f0rm of Rb 2. express high levels ofp53-deubiquitinases 3. express inactive forms of G1/S-Cdks 4. express inactive forms of G1/S cyclins 82. A fixed smear of a bacterial culture is subjected to the following solutions in the order listed below and appeared red. (a) Carbolfuchsin (heated) (b) Acid-alcohol (c) Methylene blue Bacteria stained by this method can be identified as (1) Non-acid fast E. coli (2) Acid-fast Mycobacterium sp. (3) Gram-positive E. coli

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(4) Gram-negative Mycobacterium sp. 83. In an in vitro experiment using radio labeled 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 A. There is no difference in the mobility of labeled DNA fragments between the Set 1 and Set 2 B. There is distinct difference in the mobility of the newly synthesized labeled DNA fragments between Set 1 and Set 2 C. The mobility of the newly synthesized labeled DNA fragments in case of Set 1 is faster as compared to the Set 2 D. The mobility of the newly synthesized labeled DNA fragments in case of Set 1 is slower as compared to the Set 2 Which of the following combinations represent correct observations? (1) A and B (2) B and C (3) A and D (4) B and D 84. Synthesis of normal hemoglobin requires coordinated synthesis of α globin and β globin. Thalassemias are genetic defects perturbed in this coordinated synthesis. Patients suffering from deficiency of p globin chains (β-thalassemia) could also be due to mutations affecting the biosynthesis of β globin mRNA. The following statements describe the genesis of non-functional p globin leading to pthalassemia. A. Mutation in the promoter region of the β globin gene. B. Mutation in the splice junction of the β globin gene. C. Mutation in the intron I of the β globin gene. D. Mutations towards the 3' end of the β globin gene that codes for polyadenylation site. Which of the following combinations is correct? (1) A, B and D (2) A, B and C (3) B, C and D (4) C, D and A 85. Pre-mRNAs are rapidly bound by snRNPs which carry out dual steps of RNA that removes the introns and joins the upstream and downstream exons. The following statements describe some facts related to this event. A. Almost all introns begin with GU and end with AG sequences and hence all the GU or AG sequences are spliced out of RNA. B. U2 RNA recognizes important sequences at the 3'acceptor end of the intron. C. The spliceosorhe uses A to carry out accurate removal of intron. D. An unusual linkage with 2' OH group of guanosine within the intron forms a 'Lariat' structure. Which of the following combinations is correct? (1) A and B (2) B and C (3) C and D (4) D and A 86. 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: A. RRF and EF-G aid in this process B. intrinsic activity of ribosomes and an uncharged tRNA are required C. IF-I promotes dissociation of ribosomes D. IF-3 and IF-1 promote dissociation of ribosomes Which of the following sets is correct? (1) A and D (2) A and B (3) A and C (4) B and D

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87. 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: A. eEF-I B. eIF-4E-BP1 C. eIF-4E D. PHAS-1 Which of the following sets is correct '? (1) A and B (2) B and D (3) A and C (4) Band C 88. Bacteriophage λ, 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: A. Repressor may act both as a positive regulator and a negative regulator B. Expression of repressor gene, cI is independent of the expression of cll and cIII genes. C. Mutation of cI gene will cause it to form clear plaques on both wild type E.coli and E.coli (λ+) D. Mutation at operators, OL and OR will allow the phage to act as a virulent phage. The Correct statements are (1) A and B (2) B and C (3) C and D (4) D and A 89. Survival of intracellular pathogens depends all the levels of pro-inflammatory and anti-inflammatory cytokines in macrophages. In an experimental condition, Mycobacreria infected macrophages were treated with lL-6 or IL-12 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? 1. IL-12 treated cells contain more intracellular bacteria than control 2. IL-12 treated cells contain less intracellular bacteria than control 3. IL-6 treated cells contain more intracellular bacteria than control 4. IL-6 treated cells contain less intracellular bacteria than control 90. 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 screen for mutants, some non-motile ones were selected. These could have 1. mutations in tubulin and actin proteins 2. mutations in kinesin proteins 3. mutated H+-ATPase 4. mutations in the charged residues lining the ridge of the FliG subunit 91. A bacterial response regulator turns on gene A in its phosphorylated form. The amount of "A" shows a sharp and steep rise at a threshold concentration of the signal sensed by the cognate sensor. This is most likely due to 1. increased phosphatase activity of the sensor at the threshold concentration 2. decreased phosphorylation of the response regulator by the sensor 3. co-operativity in binding of the response regulator to the target gene A 4. negative feedback in gene A expression 92. Intracellular transport and cytoskeletal organization of a cell is regulated by nucleotide exchange of different small molecular weight GTPases OfRas super family. Over expression of which of the following GTPase modulates the actin-cytoskeleton of HeLa cells? (1) Ran in GDP bound form (2) Ran in GTP bound form (3) Rho in GTP bound form (4) Rho in GDP bound form

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93. 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 arid antigen-specific IgG1 and IgG2a ELISA were perforn1ed. Only one of the mice showed positive response. It was 1. keyhole limpet hemocyanin-primed mouse 2. azobenze arsonate-primed mouse 3. lipopolysaccharide-primed mouse 4, dextran-primed mouse 94. Tumor cells were isolated 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 receptor-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? 1. Free drug 2. EGF receptor-conjugated drug 3. Transferrin receptor-conjugated drug 4. Mannose receptor-conjugated drug 95. When the prospective neurons from an early gastrula of a frog were transplanted into the prospective 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. A. The early gastrula shows conditional development whereas the late gastrula shows autonomous development. B. The early gastrula shows autonomous development whereas the late gastrula shows conditional development. C. The prospective neurons from the early gastrula are only specified whereas those from the late gastrula are determined. D. 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? (1) A and B (2) A and C (3) A and D (4) B and C 96. AP1 (APETLA1) is one of the floral meristem identifying genes. In wild type Arabidopsis thaliana plants transformed with AP1: :GUS, β-glucuronidase (GUS) activity is seen in floral meristem, only after the commitment to flowering. Ectopic expression of AP1::GUS in the EMBRYONIC FLOWER (emf) mutant background results in GUS activity throughout the shoots in four day old seedlings. These observations suggest that API is: I. not involved in flowering 2. involved in repression of flowering 3. involved in promoting flowering 4. stimulation of flowering in the emf background 97. In case of morphallactic regeneration: 1. there is repatterning of the existing tissues with little new growth 2. there is repatterning of the existing tissues after the stem cell division has taken place 3. there is cell division of the differentiated cells which maintain their differentiated state to finally form a complete organism. 4. there is dedifferentiation of the cells at the cut surface which become undifferentiated. These undifferentiated cells then divide to rediffere11tiate to form the complete structure 98. 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 diagrarnrnatic representation of the different cells formed during development from the morula with the help of different molecules. Identify the molecules 1-4, sequentially.

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(1) cdx2, Oct 4, Nanog, Stat 3 (3) cdx 2, Nanog, Oct 4, Stat 3 (2) cdx 2, Nanog, Stat 3, Oct 4 (4) cdx 2, Oct 4, Stat 3, Nanog 99. With respect to the extra embryonic structures formed in the mammals, the possible functional attributes have been designated: A. Allantoin stores urinary waste and helps mediate gas exchange. It is derived from splanchnopleure at the caudal end of the primitive streak. B. Amnion is awater sac and protects the embryo and its surrounding amniotic fluid. This epithelium is derived from somatopleure. C. Chorion is essential for gas exchange in amniote embryos. It is generated from the splanchnopleure. D. Yolk sac is the last embryonic membrane to form and is derived from somatopleure. Which of the above statements are correct? (1) A and B (2) A and C (3) Band C (4) A and D 100.

The figure above represents a late zebrafish gastrula. The following concepts may be proposed during further development of the embryo. A. 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. B. Increase in FGF activity in the epidermis with concomitant decrease in BMP activity towards the ventral axis C. Neural induction in zebrafish is independent of the organizer and depends on activation of BMP signaling. D. In comparison, both Xenopus and chick embryos require activation of FGF for neural induction to occur in addition to BMP inhibition. Which of the above statements are true? (1) A and C (2) B and C (3) A and D (4) C and D 101. Following are some of the statements regarding the effect of CO2 concentration on photosynthesis in plants. A. With elevated CO2 levels, C3 plants are much more responsive than C4 plants under well watered conditions. B. In C3 plants, increasing intracellular CO2 partial pressure can stimulate photosynthesis only over a narrow range. C. In C4 plants, C02 compensation point is nearly zero. Which one of the following combination of above statements is correct? (1) A and B (2) B and C (3) A and C (4) Only C 102. 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. A. At lower temperature the quantum yield of C3 plant is lower than C4 plant. B. In C4 plant quantum yield does not show a temperature dependence. C. Since the photorespiration is low in C4 plants because of C02 concentrating mechanism, quantum yield is not affected.

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D. 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? (1) A, Band D (2) B, C and D (3) A, Band C (4) A, C and D 103. Following are some statements regarding plant growth hormones. A. Ethylene regulates abscission. B. Gibberlins do not play any role in flowering. C. Auxin and cytokinin promote cell division. D. Over expression of cytokinin oxidase would promote root growth. E. ABA inhibits root growth and promotes shoot growth at low water potential. F. ABA promotes leaf senescence independent of ethylene. Which one of the following combination of above statements is correct? (1) A, C and F (2) B, C and D (3) D, E and F (4) B, D and E 104. Following are some statements for synthesis of jasmonic acid in plants A. 12-oxo-phytodienoic acid is produced in chloroplast and transported to peroxisome. B. Action of lipoxygenase, allene oxide synthase and allene oxide cyclase takes place in peroxisome. C. 12-oxo-phytodicnoic acid is first reduced and then converted to jasmonic acid by β oxidation. D. Final production of jasmonic acid takes place in chloroplast. E. Action of allene oxide synthase and allene oxide cyclase takes place in chloroplast. Which one of the following combination of above statements is correct? (1) A, Band C (2) B, D and E (3) C, D and E (4) A, C and E 105. Following are some statements about low temperature stress in plants. A. Fatty acid composition of mitochondria isolated from chilling resistant and chilling sensitive plants differs significantly. B. Ratio of unsaturated fatty acids to saturated fatty acids is lower in chilling resistant species. C. The cellular water does not freeze even at -40°C, because of the presence of solutes and other antifreeze proteins. D. Heat shock proteins do not play any role during low temperature stress. Which one of the following combination of above statement is correct? (1) A and B (2) A and C (3) B and C (4) B and D 106. An isolated carotid sinus was prepared so that the pressure may be regulated by a pump and the resulting discharge in single carotid sinus nerve fibre could be recorded. The following are the possible observations. A. No discharge when carotid sinus perfusion pressure was below 30 mm Hg. B. Linear increase in discharge frequency when carotid sinus perfusion pressure was gradually increased from 70 to 110 mm Hg. C. Increase in discharge frequency was more prominent in greater pulsatile changes of carotid sinus pressure keeping the mean pressure identical in all cases. D. Increase in discharge was more prominent in the falling phase of pulsatile change of carotid sinus pressure than in the rising phase. Which one of the following is correct? (1) A, B and C (2) A and C (3) Band D

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(4) D only 107. For a normal heart, the time taken for atrial systole and diastole are As and Ad seconds, respectively, while the same for ventricular systole and diastole are Vs and Vd. Which one of the following equations is correct? (1) As + Ad = Vs + Vd (2) As + Ad < Vs + Vd (3) As + Ad = Vs + Vd = 0 (4) As + Ad > Vs + Vd 108. 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 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 hemoglobin were drawn and compared with the same in native people as depicted below.

Which one of the following combinations is logically correct? (1) A and C (2) A and D (3) B and C (4) B and D 109. A monkey undergoes cerebellectomy. After the post-operative recovery, the monkey was given a task to press a bar. The possible observations are: A. Its hand would overshoot the target while reaching the bar. B. It would be unable to move forelimbs. C. It would show intention tremor while trying to press the bar. D. It would press the bar with mouth instead of hand. Which one of the following is correct? (1) A and C (2) B only (3) D only (4) B and D 110. 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 cms) (1) 0.17 mm (2) 1.7 mm (3) 3.4 mm (4) 170 µm

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

1. (A) is FSH and (B) is estrogen 2. (A) is FSH and (B) is LH 3. (A) is estrogen and (B) isLH 4. (A) is LH and (B) is FSH 112. 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. A. More number of recombination events occur between A and B as compared to Band C. B. Lesser number of recombination events occur between C and D as compared to B and C. C. 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. D. The physical distance between A and B is less than that between C and D, and thus the region between A and B is less recombinogenic. E. Although the physical distance between C and D is more than that between Band C, the region between C and D is'less recombinogenic. F. Although the physical distance between C and D is more than Band C, the region between C and D is more recombinogenic. Which statements are correct? (1) A and B (2) C and E (3) D and F (4) A, C and E 113. Consider the following crosses involving grey (wild-type) and yellow body colour true breeding Drosophila:

Assuming 200 F2 offsprings are produced in cross 2, which one of the following outcome is expected? 1. 97 grey males, 54 yellow females, 49 grey males

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2. 102 yellow males, 46 yellow females, 52 grey females 3. 52 grey males, 49 yellow males, 48 yellow females, 51 grey females 4. 98 grey males, 94 yellow females, 2 yellow males, 6 grey females 114. The ABO blood type in human is 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 0 blood group. What is the probability that their next female child has normal vision and 0 blood group? (1) ¼ (2) ¾ (3) 1/8 (4) 1 115. In E. coli four Hfr strains donate the following genetic markers in the order shown below:

Which of the following depicts the correct 0rder 6tthe markers and the site of integration of the F-factor in the four Hfr strains?

116. The following is a schematic representation of region (showing six bands) of the polytene chromosome of Drosophila, along with the extent of five deletions (Dell to Del5):

Recessive 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 fable 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? (1) a-3; b-1; c-2; d-4; e-5; f-6. (2) a-2; b-1; c-3; d-4; e-5; f-6. (3) a-4; b-1; c-2; d-3; e-5; f-6. (4) a-6; b- 2; c-3; d-4; e-1; f-5. 117. Assuming a 1: 1 sex ratio, what is the probability that three children from the same parents will consist of two daughters and one son?

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(1) 0.375 (2) 0.125 (3) 0.675 (4) 0.75 118. 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. 1. Euglenophyta - (i and a); Bacillariophyta - (ii and d); Phaeophyta - (ii and c); Chlorophyta - (i and b) 2. Euglenophyta - (ii and a); Bacillariophyta - (ii and d); Phaeophyta -- (i and c); Chlorophyta - (i and b) 3. Euglenophyta - (i and a); Bacillariophyta - (ii and b); Phaeophyta - (i and c); Chlorophyta - (ii and d) 4. Ettglenophyta - (i and d); Bacillariophyta - (ii and a); Phaeophyta - (ii and c); Chlorophyta - (i and b) 119. Identify the synapomorphies in the following cladogram:

1. (a) seeds with long terminal wing; (b) ovules 1-20 per scale; (c) resin canals; (d) 1 ovule per scale 2. (a) resin canals; (b) seeds with long terminal wing; (c) 1 ovule per scale; (d) ovules 1-20 per scale 3. (a) resin canals; (b) ovules 1-20 per scale; (c) seeds with long terminal wing; (d) 1 ovule per scale 4. (a) seeds with long terminal wing; (b) ovules 1-20 per scale; (c) 1 ovule per scale; (d) resin canals 120. From among the five animals listed below, match the two attributes - amniotic egg and endothermy, with the correct animal(s): (a) fish (b) frog (c) crocodile (d) pigeon (e) zebra 1. Amniotic egg:b, c, d; Endothermy: d, e 2. Amniotic egg: c, d, e; Endothermy: d, e 3. Amniotic egg: a, b, c, d; Endothermy: c, d, e 4. Amniotic egg: b, c, d; Endothermy: c, d, e 121. During a field study, three insects with the following characteristics were observed: A: elongate, membranous wings with netlike venation, long and slender abdomen, large, compound eyes B: small bodied, sucking mouth parts, narrow wings fringed with setae C: sclerotized forewings, membranous hindwings, chewing mouth parts They can be identified to their respective orders as 1. A - Orthoptera; B-Hemiptera; C-Coleoptera 2. A - Odonata; B-Coleoptera; C-Hemiptera 3. A - Orthoptera; B - Odonata; C-Coleoptera 4. A - Odonata;.B-Thysanoptera; C-Coleoptera 122. Several distinct time periods and different routes might explain the entrance of marsupials into Australia. (i) Late Jurassic - early therians arrived in Antarctica - Australia where the marsupials subsequently evolved. (ii) Early to middle Cretaceous - early marsupials arrived in Australia from northern regions and then radiated in isolation. (iii) Paleocene - marsupials entered Australia from South-East Asia. (iv) Eocene- chance dispersal of marsupials into Australia- Which of the following is the correct combination? (1) (i) (ii) (iii) (2) (i) (iii) (iv) (3) (ii) (iii) (iv) (4) (i) (ii) (iv)

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123. Which of the following is NOT true for a critically endangered species? 1. Reduction of population breeding ability due to increased relatedness through the action of incompatibility mechanisms in plants or behavioral difficulties in animals. 2. The individuals of the species which have declined to low numbers are still a genetically open system. 3. Loss of some alleles from the species causing loss of genetic diversity with consequent inability to respond rapidly to selection. 4. Expression of deleterious alleles and increased homozygosity increases mortality of young, and inbreeding depression leads to reduced offspring fitness. 124. Ecological compression differs from character displacement in that it operates on a 1. shorter timescale and does not involve heritable change. 2. longer timescale and does not involve heritable change. 3. shorter timescale and involves heritable change. 4. longer timescale and involves heritable change. 125. 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 1. Productivity/Biomass ratios are higher and energy transfer rates to higher trophic levels are faster. 2. Productivity/Biomass ratios are lower and the energy transfer rates to higher trophic levels are slower. 3. Productivity/Biomass ratios are lower and the energy transfer rate to higher trophic levels are faster. 4. Productivity/Biomass ratios are higher and the energy transfer rate to higher trophic levels are slower. 126. Which of the following graphs illustrates the current consensus on the role of disturbance on the species richness of a community?

127. In the global nitrogen cycle, the following microbial organisms are involved III three important process - denitrification, nitrification and nitrogen fixation. (a) Rhizobium (b) Nitrosomonas (c) Nitrobacter (d) Pseudomonas (e) Azotobacter. Which of the following is the correCtly matched pair of process and its causative species? 1. Denitrification - (b); nitrogen fixation - (c) and (e); nitrification - (d) 2. Denitrification - (d); nitrogen fixation - (a) and (e); nitrification - (c) 3. Denitrification - (e); nitrogen fixation - (a) and (d); nitrification - (d) 4. Denitrification - (b); nitrogen fixation - (a) and (d); nitrification - (c) 128. Suppose you discovered a new species about which you know only two facts: it is small-sized (<10 cm) and short-lived (<20d). Which of the following reproductive strategies is most likely to be true for this species? 1. Breeds early and more than once in life and produces large number of small-sized offspring. 2. Breeds late and only once in life and produces large number of small-sized offspring. 3. Breeds early and only once in life and produces large number of small-sized offspring. 4. Breeds early and only once in life and produces a small number of large-sized offspring. 129. 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 (1) equal to the cost of care.

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(2) more than the cost of care by 25%. (3) double the cost of care. (4) four times the cost of care. 130. 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 (1) 0.42 (2) 0.28 (3) 0.33 (4) 0.67 131. Which of the following behavioral changes are expected in a rat when its nucleus accumbens is experimentally ablated? (1) Aggressive behavior increases (3) Nest-building activity increases (2) Exploratory behavior decreases (4) Level of parental care drops 132. Number of trials required for rats to learn a task when they were exposed to various conditions were as follows: - Experimental Conditions Observations A Light: light dark cycle - 12h: 12h N - trials B Bright light - 24 h Significantly more trials than 'N' C Bright light - 24 h + continuous physical disturbance Significantly more trials than 'N' D Dark light - 24h + continuous physical disturbance Significantly more trials than 'N' Which of the following inferences is most appropriate? (1) Continuous light enhanced learning (2) Continuous darkness inhibited learning (3) Physical activity inhibited learning (4) Learning was reduced by sleep loss 133. 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 behavior will species X show? 1. It will sing the song of species Y that it had heard in the critical period. 2. It will sing the song of its own species X. 3. It will not sing at all. 4. It will sing a song not sung by either X or Y. 134. Enzymes are nowadays used extensively in bioprocessing industries. Enzyme 1 is used for treatment of hides to provide a finer texture, in leather processing and manufacture of glue. Enzyme 2 is used for clarification of fruit juices. Identify Enzymes 1 and 2 Enzyme 1 Enzyme 2 1. Amylase Pectinase 2. Protease Amylase 3. Protease Pectinase 4. Pectinase Amylase 135. 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 he given a. immediately after birth b. after the onset of the condition. In case of B, the immunization has to be done

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c. before pregnancy. d. late in the pregnancy. The correct combination is (1) A/a (2) A/b (3) B/c (4) B/d 136. 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 labeled fragment X

The pattern obtained in Southern hybridization is shown below:

Based on the above, which of the following interpretations is correct: 1. All the plants (P1, P2 and P3) contain two copies of the transgene 2. P1, P3 contain one and P2 contains two copies of the transgene 3. P1 contains two, whereas P2 and P3 contain one copy of transgene each 4. P1 and P2 contains two and P3 contains one copy of the transgene 137. The following are statements about molecular markers in the context of plant breeding. A. Molecular markers can be used for elimination of undesirable traits. B. Molecular markers cannot be used for estimation of the genetic contribution of each individual parent in a segregating population. C. Molecular markers are used for mapping of QTLs, which is' also possible by conventional techniques. D. 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? (1) A and B (2) A and C (3) A and D (4) B and C 138. In 'TaqMan' assay for detection of base substitutions (DNA variant), probes (oligonucleotides) with fluorescent dyes at the 5' -end and a quencher at 3' -end are used. While the probe is intact, the proximity of the. quencher reduces the fluorescence 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 primers 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 reporter 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 fluorescence for the dye attached to the wild type probe, sample for individual II shows maximum fluorescence 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? 1. Individual I is homozygous for the variant allele. 2. Individual II is homozygous for variant allele. 3. Individual II is homozygous for wild type allele.

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4. Individual III is homozygous for wild type allele. 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 fracture repair. Few methods to achieve the above are given below: A. Develop a collagen gel containing plasmids carrying the human parathyroid hormone gene and place in the gap between the ends of the broken leg. B. Develop a gel matrix disc containing genetically modified stem cells to excrete BMP4 and VEGF-A and implant it at the site of the wound. C. Make scaffolds of material that resemble normal extracellular matrix that could be molded to form the shape of a bone needed and seed them with bone marrow stem cell. D. 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 (1) A and B (2) A, Band C (3) A and C (4) C and D 140. Cre/loxP system is used by phage PI to remove terminally 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 ?

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

What type of mutation has occurred in the gene? (1) Nonsense (2) Missense (3) Frameshift (4) Transversion

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142. T cell proliferation in vivo is to be analyzed. The cells are labeled 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?

143. The most important property of any microscope is its power of resolution, which is numerically equivalent to D, the minimum distance between two distinguishable objects. D depends on three parameters namely, the angular aperture, α, the refractive index, N, and wavelength, λ, of the incident light. Below are given few possible options to increase the resolution of the microscope. A. Decrease the value of λ or increase either N or α to improve resolution. B. Moving the objective lens closer to the specimen will decrease sin α and improve the resolution. C. Using a medium with high refraction index between the specimen and the objective lens to improve the resolution. D. Increase the wavelength of the incident light to improve the resolution. Which of the following combination of above statements is correct? (1) A and C (2) Band C (3) A and D (4) C and D 144. In an animal experiment; (i) Electrical stimulation of an area in the brain (A) increased a function (F) which was prevented by systemic injection of adrenergic antagonistic, prazosin. (ii) Injection of carbathol (cholinergic agonist) into A also increased function F which was, however, not prevented by systemic injection of adrenergic antagonistic, prazosm. The results are likely to be due to the stimulation of 1. Nonadrenergic and cholinoceptive neurons 2. Cholinergic and non-adrenoceptive neurons 3. adrenergic terminals in 'A' 4. both neurons and fibres passing through 'A' 145. 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)" 1. Ratio of the mean temperatures of the two time periods tested 2. Probability of the error of rejecting a true null hypothesis 3. Probability of the error of accepting a false null hypothesis 4. Probability of the t-test being effective in detecting significant differences in the mean annual temperatures of the two time periods.

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Answers PART-A 1-1, 2-2, 3-4, 4-4, 5-4, 6-1, 7-3, 8-3, 9-1, 10-4, 11-4, 12-1, 13-1, 14-3, 15-4, 16-4, 17-2, 18-1, 19-2, 20-1. 6. If you were looking into the solar system, it would be daytime and you would see the sun. If you wait six months and look at the night sky you are looking out away from the center of the system in the exact opposite direction from six months ago. The night sky as it will look six months from now or as it did look six months ago is the sky blotted out by the luminous sun in the daytime. All those stars are there, but you can't see them because of the brilliant light of the sun. Only the moon is close enough and large enough in the sky for the reflected sunlight off of its surface to make it visible in the daytime when it is on the same side of the earth as the sun. The one exception to seeing the sky "behind the brilliance of the sun" is during a total solar eclipse. When the moon blocks the sun completely over an area of the earth "the stars come out" and you can see the daytime the rest of the sky, the sky you saw at night six months earlier. 10.

PART-B 21-4, 22-1, 23-2, 24-1, 25-4, 26-2, 27-3, 28-2, 29-4, 30-2, 31-2, 32-1, 33-4, 34-1, 35-2, 36-1, 37-2, 38-3, 39-1, 40-4, 41-1, 42-1, 43-2, 44-2, 45-2, 46-4, 47-4, 48-2, 49-3, 50-4, 51-4, 52-2, 53-1, 54-2, 55-2, 56-4, 57-4, 58-4, 59-1, 60-1, 61-3, 62-2, 63-2, 64-4, 65-2, 66-2, 67-1, 68-, 69-2, 70-2. 24. Read http://www.plantcell.org/content/11/4/509.full 29. Unlike other amino acids present in biological proteins, selenocysteine is not coded for directly in the genetic code. Instead, it is encoded in a special way by a UGA codon, which is normally a stop codon. When cells are grown in the absence of selenium, translation of selenoproteins terminates at the UGA codon, resulting in a truncated, nonfunctional enzyme. The UGA codon is made to encode selenocysteine by the presence of a SECIS element (SElenoCysteine Insertion Sequence) in the mRNA. The SECIS element is defined by characteristic nucleotide sequences and secondary structure base-pairing patterns. In bacteria, the SECIS element is located immediately following the UGA codon within the reading frame for the selenoprotein. In archaea and in eukaryotes, the SECIS element is in the 3' untranslated region (3' UTR) of the mRNA, and can direct multiple UGA codons to encode selenocysteine residues. 28. In a random sequence of nucleotides, 1 in every 20 codons in each reading frame is, on average, a termination codon. In general, a reading frame without a termination codon among 50 or more codons is referred to as an open reading frame (ORF). From nelson and Cox, Chapter 27. 32. Toxic shock syndrome (TSS) is a potentially fatal illness caused by a bacterial toxin. Different bacterial toxins may cause toxic shock syndrome, depending on the situation. The causative bacteria include Staphylococcus aureus and Streptococcus pyogenes. TSS is thought to be a superantigen-mediated disease. Superantigens are a group of proteins (S aureus toxins in the case of TSS) that are able to activate the immune system by bypassing certain steps in the usual antigen-mediated immune response sequence. Superantigens are not processed within the antigen-presenting cell before being presented to T cells. Instead, they bind directly to molecules of the major histocompatibility complex, class II, which requires recognition of only one element of the T-cell receptor (V beta) to trigger a massive T-cell activation (figure 1: not shown). Between 5% and 30% of the entire T-cell population may be activated. Conventional antigens activate only about 0.01% to 0.1% of the T-cell population. Superantigens lead to a massive release of cytokines, especially tumor necrosis factor alpha (TNF-alpha), interleukin-1 (IL-1), and IL-6. These cytokines are responsible for a capillary leak syndrome and account for many of the clinical signs of TSS.



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Option 2 says excessive stimulation large proportion of T cells (here only 5-30%). 33. Ethylene binding to its receptor leads to dimerization and undergo autophosporylation (similar to bacterial two component system). The first downstream component affected by ethylene signaling is a protein Ser/Thr kinase (CTR1). From Nelson and Cox. 36. Auxin gradients in the cells are produced and maintained by auxin efflux carriers PIN proteins which contain transmembrane domains and mainly localized in the plasma membranes. PIN proteins are the rate limiting factors of auxin transport and provide vectorial direction for the auxin flow. 41. Chloroplast movement in Arabidopsis, mediated by receptor PHOT2. When light is weak, chloroplasts gather at the upper and lower surface of the mesophyll cells, maximizing light absorption. Under strong light, chloroplasts move to the side walls that are parallel to the incident light, minimizing light absorption. Increase in cellular levels of PHOT2 have been correlated with an increased rate of chloroplast movement during the avoidance response. 42.

63. Leukemia inhibitory factor (LIF) is an indispensable factor for maintaining mouse embryonic stem (ES) cell pluripotency. Removal of LIF pushes stem cells toward differentiation, but they retain their proliferative potential or pluripotency. 64. Glyphosate's mode of action is to inhibit an enzyme involved in the synthesis of the aromatic amino acids: tyrosine, tryptophan and phenylalanine. It is absorbed through foliage and translocated to growing points. Because of this mode of action, it is only effective on actively growing plants; it is not effective as a pre-emergence herbicide. PART-C 71-2, 72-4, 73-4, 74-3, 75-3, 76-1, 77-2, 78-3, 79-2, 80-4, 81- may be 2, 82-2, 83-2, 84-1, 85-1, 86-1, 87-2, 88-4, 89-2, 90-3, 91-1, 92-3, 93-1, 94-3, 95-2, 96-4, 97-1, 98-1, 99-3, 100-1, 101-3, 102-2, 103-1, 104-4, 105-2, 106-, 107-1, 108-1, 109-1, 110-2, 111-3, 112-3, 113-3, 114-1, 115-2, 116-3, 117-2, 118-1, 119-2, 120-2, 121-1, 122-2, 123-2, 124-1, 125-1, 126-4, 127-2, 128-3, 129-, 130-3, 131-1, 132-4, 133-1, 134-3, 135-3, 136-3, 137-2 or 3, 138-2, 139-1, 140-3, 141-3, 142-may be 2, 143-1, 144-1, 145-. 71.

∆Go = - (8.315 J/mol . K)(300 K)(ln 100) Natural log 100 = 4.60 ∆Go = -11474.7 J/mol Convert in to cal, 1 cal = 4.184 J ∆Go = -2746 cal/mol 2 Option is -2746 Kcal/mol, not -2746 cal/mol. If you attempted you will get 4 marks. 72. Answer is 4.






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73. Answer 4

7.4 = 6.1 + log Bicarbonate/Carbonic acid We have to take pKa1, because it close to physiological pH (7.4) 1.3 = log Bicarbonate/Carbonic acid Anti log 1.3 = Bicarbonate/Carbonic acid 20 = Bicarbonate/Carbonic acid 74. Answer 3 Here enzyme activity (1 unit) 10 µmol of pyrophosphate in 15 minutes or 0.0111 µmol of pyrophosphate per second Vmax = 2800 units/mg enzyme Vmax = 2800 × 0.0111 µmol /mg enzyme = 31.1 µmol 75. Answer 3 More G+C content, more will be Tm. 78. Answer 3 Spectrin, the principal protein of the membrane skeleton, is composed of alpha and beta subunits. These two filamentous polypeptides are aligned side by side to form spectrin heterodimers, which in turn self-associate to form spectrin tetramers and higher molecular weight oligomers. Abnormalities in the skeleton or its attachments have been discovered in a number of congenital hemolytic anemias. For example, molecular defects described in kindreds with hereditary elliptocytosis (HE) include a quantitative deficiency of protein 4.1, defective ankyrin-band 3 binding (5), and a number of qualitative defects in spectrin. Most of alpha- and beta spectrin defects in HE result in diminished self-association of affected spectrin heterodimers, leading to reduced formation of higher molecular weight forms of spectrin and, thus, to skeletal instability. Several previously described defects in alpha spectrin have been distinguished from one another by differing patterns of spectrin peptides generated by limited tryptic digestion. One beta-spectrin mutant exhibited defective binding to ankyrin, while several others displayed apparent alterations in molecular weight as estimated by SDS polyacrylamide gel electrophoresis (PAGE). 79.

At least a dozen studies have found evidence that introns are either directly or indirectly involved in cancer causation. Examples include evidence that introns are involved in transcriptional regulation of apoprotein B, E, and A-1145 and that introns may be involved in regulating neoplasm developments.






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Cytolytic T lymphocyte clones used to study melanomas found the gene coding for the antigen recognized by the cytolytic T lymphocyte was the same gene which codes for N-acetylglucosaminyl-transferase V. The antigenic peptide recognized by the cytolytic T lymphocyte was found to be encoded by a sequence located inside an intron. The researchers found that the mRNA containing the introns coding for the antigen was not found at significant levels in nominal tissues but was observed to be present in close to half the melanoma tissues studied. The researchers concluded that a promoter located near the end of the relevant intron was activated in melanoma cells, resulting in the production of an mRNA that codes for the antigen.

Defective glutathione S-transferase and N-acetyl-transferase enzymes have been associated with an increased risk of developing both lung and bladder cancer. The research results are inconsistent, though, and several studies have failed to find associations. According to some studies, the lung cancer risk is elevated up to 40-fold in subpopulations that contain both the high-risk cytochrome P-450 type Al and glutathione S-transferase Ml genotypes which are a result of mutations in introns or other silent areas of DNA. One study on the glutathione S-transferase M3 gene found a mutant three-base deletion in intron 6 of the wild type glutathione S-transferase allele. This defect may be related to neoplasm development, but exactly how is unknown.

Megonigal, et al. used panhandle PCR to clone MLL genomic breakpoints in two pediatric treatment-related leukemias. The panhandle PCR identified a fusion of MLL intron 6 with a previously uncharacterized sequence in MLL intron 1 which the researchers concluded was consistent with a partial duplication. The breakpoints in both cases were located in Alu repeats, suggesting that the Alu sequences were an important contributor to the rearrangements they found.

Conclusions about introns and cancer are difficult for several reasons, including the fact that it must be determined if mutations in the introns actually contribute to tumorigenesis progression, or if this damage is a result of collateral or unrelated damage. The associations found between introns and cancer may occur by chance, or it may be due to an unknown role of introns or other putative silent areas of DNA. Numerous other studies indicate a regulatory function for introns:

Introns are becoming more widely recognized as having important gene-regulatory roles, such as containing enhancer or silencer elements. While many intronic polymorphisms may be of trivial consequence ... we believe that variation in the number of intronic 37-bp copies in the Kras will serve as an interesting model system for examining how inherited variations in the copy number of oligonucleotide regions within an intron affect the etiology of diseases.

Among the many other intron genetic variations that have been linked to cancer (in this case lung cancer) is a K-ras intron variation involving tandem repeats in the H-ras 3'-untranslated region. Mutations in introns may influence neoplasm development because mutations that occur in introns (or in the DNA sequences that flank the gene) can affect the various steps required for normal expression of the gene even though the intron codes may not be detectable by studies of the protein product of the gene. If the defective expression of a gene product is a tumor suppressor gene or a proto-oncogene, cancer may result.

Regulatory functions of introns may involve controlling gene activity in different developmental stages or responding to immediate biological needs by controlling local gene expressions. This function of introns could occur if, as one theory indicates, exons code for a domain, a polypeptide unit that has a discreet function such as binding to a membrane, or to the catalytic site of an enzyme or serving as a structural unit of a protein.

Some evidence also exists that introns structurally stabilize the pre-mRNA to protect it against degradation. Marculis and Sagan noted that DNA is packed into a chromosome so effectively that it is 1/8,000 of its former size after packing.They concluded from studies on ciliates that at least some introns, and possibly other noncoding DNA, may be involved in DNA packaging in eukaryotes and perhaps in some prokaryotes.

80. Answer 4

Cyclin destruction box plays a major role in the inactivation of MPF, after metaphase.

APF acts on MPF (after metaphase) by binding to MPF at a specific site (cyclin destruction box), so that it inactivates MPF.






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81. Loss of the p53 tumor suppressor function is one of the most common steps in tumorigenesis. Germ-line mutations of p53 are present in cancer-prone families with Li-Fraumeni syndrome, and somatic mutations are found in more than half of all cancer cases. In clear demonstration of the role of p53 as a key tumor suppressor, mice nullizygous of p53 spontaneously develop a broad spectrum of tumors.

The p53 gene encodes a transcription factor. The majority of the mutations in p53 are missense point mutations clustered in the DNA-binding domain that disrupt DNA binding. The activity of p53 is highly regulated by post-translational mechanisms, including protein-protein interactions, acetylation, neddylation, phosphorylation, sumoylation, and ubiquitination. One of the transcriptional targets of p53, MDM2, can regulate both the transcriptional activity and the half-life of p53 in a negative feedback loop. MDM2 binds to the NH2- terminal transactivation domain of p53 and inhibits its transcriptional activity directly. MDM2 also shuttles p53 out of the nucleus by the virtue of the nuclear exporting signal in MDM2. Finally, MDM2 is a ubiquitin ligase that targets p53 for ubiquitin-mediated proteolysis. Ubiquitin-mediated degradation entails linking of ubiquitin by a thioester linkage to a ubiquitinactivating enzyme (E1) before the ubiquitin is transferred to a ubiquitin carrier (E2). E2 acts alone or in conjunction with a ubiquitin ligase (E3) to conjugate ubiquitin to the ε-amino group of lysine residues in substrate proteins to form a glycyllysine isopeptide bond. Multiple rounds of ubiquitin conjugation form polyubiquitin chains on the substrates, which are then degraded by the 26S proteasome complex.

82. Answer 2

The Ziehl–Neelsen stain, also known as the acid-fast stain. It is a special bacteriological stain used to identify acid-fast organisms, mainly Mycobacteria.

Mycobacterium tuberculosis is the most important of this group, as it is responsible for the disease called tuberculosis (TB) along with some others of this genus. It is helpful in diagnosing Mycobacterium tuberculosis since its lipid rich cell wall makes it resistant to Gram stain. It can also be used to stain few other bacteria like Nocardia.

The reagents used are Ziehl–Neelsen carbolfuchsin, acid alcohol and methylene blue. Acid-fast bacilli will be bright red after staining.

83. Answer 2

The enzyme RNase H is a non-specific endonuclease and catalyzes the cleavage of RNA via a hydrolytic mechanism. Members of the RNase H family can be found in nearly all organisms, from archaea to bacteria and eukaryota.






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RNase H’s ribonuclease activity cleaves the 3’-O-P bond of RNA in a DNA/RNA duplex to produce 3’-hydroxyl and 5‘-phosphate terminated products. In DNA replication, RNase H is responsible for removing the RNA primer, allowing completion of the newly synthesized DNA.

Presence RNase H (set 1), degrades RNA and replaced with DNA after replication, which decreases weight of newly synthesized labeled DNA fragments (OH → H).

So, the migration of Set 1 newly synthesized labeled DNA fragments will more compared to Set 2 DNA fragments.

84.

Book- Human genetics: the molecular revolution By Edwin H. McConkey (Page 144)

85.

Option C. ATP required for Spliceosome assembly

Option D is wrong. The 2’ OH of a specific adenosine in the intron acts as a nucleophile, attacking the 5’ splice site to form.

87. The mTOR pathway involves the regulation of a protein kinase variously known as: FKBP12-rapamycin-associated protein (FRAP); mammalian target of rapamycin (mTOR) or rapamycin and FKBP12 target 1 (RAFT1) and its downstream effectors.

Activated mTOR (TORC1) phosphorylates protein phosphatase 2A (PP2A), p70S6K (S6K1) and eIF4E-BP (PHAS-1). Phosphorylation of PP2A prevents the dephosphorylation of p70S6K and eIF4E-BP. The phosphorylation of eIF4E-BP releases eIF4E which can then participate in the formation of the protein translation complex involving capped mRNAs. Phosphorylated p70S6K can phosphorylate the ribosomal S6 subunit which enhances the translation of 5'-terminal oligopyrimidine (TOP)-mRNAs. Enhanced translation of TOP-mRNAs leads to increased synthesis of proteins required for protein synthetic machinery. The combined effect of increased translation initiation and increased ribosomal and other translational proteins leads to enhanced protein synthesis and cell growth.

88. Answer 1

Normal λ infections produce turbid plaques, accounted for by lysogenic bacterial growth within the plaques.






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The cI gene, with the aid of the cII-gene product, is transcribed from a promoter known as pRE, the RE standing for repression establishment (fig. 14.23). Once cI is transcribed, it is translated into a protein called the λ repressor, which interacts at the left and right operators, OL and OR, of the left and right operons. When these operators are bound by cI protein, transcription of the left and right operons (and therefore also the late operon) ceases. Removal of OL and OR, leads to lytic cycle (virulent phage). Book- Principles of Genetics by Tamarin. Page 422 89. The host’s immune response to M. avium infection has been shown to be complex and to involve the participation of a number of cells, such as CD4 T cells, natural killer (NK) cells, mononuclear phagocytes, and perhaps γδT cells. Recent studies in vitro and in vivo have demonstrated that the cytokine interleukin-12 (IL-12) plays an important role in the orchestration of the immune defense of the host against M. avium, and deficiency of the cytokine is associated with accelerated progression of the disease. Furthermore, individuals with a genetic deficiency of functional IL-12 receptors have an increased predisposition for mycobacterial infections. IL-12 participates in the host defense against Mycobacterium tuberculosis as well. For example, IL-12 is secreted by macrophages following infection with M. tuberculosis in vitro and work carried out in mice supports a key role for IL-12 in the mechanisms of immunity to M. tuberculosis. 90. Oligomerization and activation of the FliI ATPase central to bacterial flagellum assembly. 92. The Rho family of GTPases is a family of small (~21 kDa) signaling G protein (more specific, a GTPase), and is a subfamily of the Ras superfamily. The members of the Rho GTPase family have been shown to regulate many aspects of intracellular actin dynamics, and are found in all eukaryotic organisms including yeasts and some plants. Three members of the family have been studied a great deal: Cdc42, Rac1, and RhoA. Rho proteins have been described as "molecular switches" and play a role in cell proliferation, apoptosis, gene expression, and multiple other common cellular functions.

94. The transferrin receptors (TfR) are overexpressed on tumor cells, The drug is selectively up-taken by the tumor cells via Tf-TfR interaction and internalized by endocytosis, which leads to release of Methotrexate in to cells.

Methotrexate is thought to affect cancer and rheumatoid arthritis by two different pathways. For cancer, methotrexate allosterically inhibits dihydrofolate reductase (DHFR), an enzyme that participates in the tetrahydrofolate synthesis. The affinity of methotrexate for DHFR is about one thousand-fold that of folate. DHFR catalyses the conversion of dihydrofolate to the active tetrahydrofolate. Folic acid is needed for the de novo synthesis of the nucleoside thymidine, required for DNA synthesis. Also, folate is needed for purine base synthesis, so all purine synthesis will be inhibited. Methotrexate, therefore, inhibits the synthesis of DNA, RNA, thymidylates, and proteins.

Methotrexate acts specifically during DNA and RNA synthesis, and thus it is cytotoxic during the S-phase of the cell cycle. Logically, it therefore has a greater toxic effect on rapidly dividing cells (such as malignant and myeloid cells, and gastrointestinal and oral mucosa), which replicate their DNA more frequently, and thus inhibits the growth and proliferation of these noncancerous cells, as well as causing the side effects listed below. Facing a scarcity of dTMP, rapidly dividing cancerous cells undergo cell death via thymineless death.

For the treatment of rheumatoid arthritis, inhibition of DHFR is not thought to be the main mechanism, but rather the inhibition of enzymes involved in purine metabolism, leading to accumulation of adenosine, or the inhibition of T cell activation and suppression of intercellular adhesion molecule expression by T cells.In these cases, patients should supplement their diets with folate.






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98. from Development Biology by Gilbert

99.

Chorion

101.






31

102.






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104.

12-oxo-phytodienoic acid (OPDA)

allenoxidsynthase (AOS)

allenoxidcyclase (AOC)

105. Membrane lipids from chilling-resistant plants often have a greater proportion of unsaturated fatty acids than those from chilling-sensitive plants and during acclimation to cool temperatures the activity of desaturase enzymes increases and the proportion of unsaturated lipids rises. This modification lowers the temperature at which the membrane lipids begin a gradual phase change from fluid to semicrystalline and allows membranes to remain fluid at lower temperatures. Thus, desaturation of fatty acids provides some protection against damage from chilling. More info: http://agritech.tnau.ac.in/agriculture/agri_temperature_low.html

108. Erythropoietin (EPO) is a glycoprotein hormone that controls erythropoiesis, or red blood cell production. It is a cytokine for erythrocyte (red blood cell) precursors in the bone marrow.

The Native Population (Inca Empire at high altitude in Peru) will have high concentration of EPO compared to soldiers (Patient).

Hemoglobin saturation is less at low Oxygen tension (pressure).






33

111.

114.

117.

(1/2)2 × (1/2)1

(1/2) × (1/2) × (1/2)

1/8 = 0.125

124.

Book-Ecology: Principles and Applications by J. L. Chapman and M. J. Reiss



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126. Disturbance, defined here as any process that removes biomass from the community, as long been recognized as influencing species coexistence and the maintenance of biodiversity. As a result, there is a considerable body of research examining the effects of disturbance on species diversity, much of which focuses on testing the intermediate disturbance hypothesis (IDH).

The IDH, originally presented as an explanation for species diversity in plants and sessile animals (e.g., corals and trees, predicts that intermediate frequency or intensity of disturbance will maximize diversity. At high disturbance frequency, species diversity is predicted to be low, because only "weedy” species that quickly colonize and reach maturity are able to survive. In contrast, at low disturbance frequency, species diversity is expected to be low, because competitively dominant species exclude competitively inferior species (Figure). Only at intermediate levels of disturbance do a mix of colonizers and competitors co-exist.

(a) Species diversity is low at low disturbance frequency because of competitive exclusion.

(b) Species diversity is higher at intermediate disturbance frequency due to a mix of good colonizer and good competitor species.

(c) Species diversity is low at high disturbance frequency because only good colonizers or highly tolerant species can persist.

127. Denitrification is a microbially facilitated process of nitrate reduction that may ultimately produce molecular nitrogen (N2) through a series of intermediate gaseous nitrogen oxide products.

The process is performed primarily by heterotrophic bacteria (such as Paracoccus denitrificans and various pseudomonads), although autotrophic denitrifiers have also been identified (e.g., Thiobacillus denitrificans).

128. r-selection

In unstable or unpredictable environments, r-selection predominates as the ability to reproduce quickly is crucial. There is little advantage in adaptations that permit successful competition with other organisms, because the environment is likely to change again. Traits that are thought to be characteristic of r-selection include: high fecundity, small body size, early maturity onset, short generation time, and the ability to disperse offspring widely.

Organisms whose life history is subject to r-selection are often referred to as r-strategists or r-selected. Organisms who exhibit r-selected traits can range from bacteria and diatoms, to insects and weeds, to various semelparous cephalopods and mammals, particularly small rodents.

130. The inbreeding coefficient, F (see also F-statistics), is one minus the observed frequency of heterozygotes over that expected from Hardy–Weinberg equilibrium.

where the expected value from Hardy–Weinberg equilibrium is given by

E = 2 × 0.7 × 0.3

= 0.42

For example, for Ford's data above;






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F = 1- 0.28/0.42 = 0.33

131. The nucleus accumbens (NAcc), also known as the accumbens nucleus is a collection of neurons and forms the main part of the ventral striatum. It is thought to play an important role in reward, pleasure, laughter, addiction, aggression, fear, and the placebo effect.

135. The Tetanus Toxoid (TT) vaccine is given during pregnancy to prevent tetanus to mother. Antibodies formed in mother, after the vaccination, are passed on to baby and protect her for a few months after birth. It also helps prevent premature delivery.

137. Marker assisted selection

Sometimes many different genes can influence a desirable trait in plant breeding. The use of tools such as molecular markers or DNA fingerprinting can map thousands of genes. This allows plant breeders to screen large populations of plants for those that possess the trait of interest. The screening is based on the presence or absence of a certain gene as determined by laboratory procedures, rather than on the visual identification of the expressed trait in the plant.

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