*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com1Model Question Paper-01General Science (A) & Life Sciences (B & C)Time: 3:00 Hrs. Max. Marks: 200This Test Booklet will contain 145 (20 Part A +50 Part B + 75 Part C) Multiple Choice Questions (MCQs). Candidates will be required to answer 15 in part A, 35 in Part B and 25 questions in Part C respectively (No. of questions to attempt may vary from exam to exam). In case any candidate answers more than 15, 35 and 25 questions in Part A, B and C respectively only first 15, 35 and 25 questions in Parts A, B and C respectively will be evaluated. Questions in Parts A and B carry two marks each and Part B questions carry four marks each. There will be negative marking @25% for each wrong answer. Below each question, four alternatives or responses are given. Only one of these alternatives is the CORRECT answer to the question.PART-A (General Science)1. A shooting star is1. A small star moving away from the earth at a very high speed2. A fast moving satellite that shines by sunlight3. A heavenly object that shines because it is heated by the friction of the earths atmosphere as it falls at a great speed4. A star of an extremely high density2. By what process is heat transmitted from the filament of an evacuated electric bulb to the glass?1. Conduction2. Convection3. Radiation4. Heat cannot be transmitted through a vacuum3. If the length of a heater coil is reduced by 10 percent of its original length, then the power consumed by the heater will1. Increase over 10%2. Decreased by 10%3. Increase by 0.5%4. Decreased by 0.5%*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com24. The total output of photosynthesis including the organic matter used up in respiration during the period of measurement is called:1. Net primary productivity2. Gross primary productivity3. Net community productivity4. Secondary productivity5. Haemoglobin is 1. The coloring matter of leaves of plants2. The coloring matter of blood3. A compound present in milk4. A compound that transmits signals to the brain6. In solution of HF and acetic acid,1. CH3COOH behaves as a base2. CH3COOH remains unionized3. HF remains unionized4. HF behaves as a base7. When a pencil is partly immersed in water in a beaker and held in a slanting position, the immersed portion appears1. Bent towards the bottom2. Bent towards the water surface3. Bent in a zigzag manner4. As if it was not immersed8. A codon:1. Is a sequence of three bases in a row2. Signals that three particular amino acids be incorporated into a growing peptide3. Helps position in the new amino acids correctly by hydrogen bonding with an anticodon of tRNA4. May have more than one meaning depending upon its location in the mRNA polymer*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com39. Which of the following are false statements?I) Ozone is an allotropic form of oxygen.II) Phosphorus is present in DNA and RNAIII) Halogens occur in free state in nature.IV) The compounds of chlorine are used as cooling agents1. I, IV 2. II, IV3. III, IV 4. I, II10. The incorrect gas is a mixture of1. Silk Polyamide 2. Butane and propane3. Methane and ethylene 4. Carbon dioxide and oxygen11. RuBisCO is made up of:1. 14 sub units 2. 15 sub unit3. 16 sub units 3. 18 sub unit 12. A veena player compares with a tuning fork the fundamental frequency generated by one of the strings of the veena and hears 4 beats per sec. He then tightens the string a bit and hears only 3 beats per second. Then1. The string has a higher frequency than the tuning fork and must be tightened more2. The string has a lower frequency than the tuning fork and must be tightened more to make the frequencies equal3. The string has a lower frequency and its tension has to be relaxed to make the frequencies equal4. The string has a higher frequency than the fork and has to be loosened for equality with fork13. A. It is advantageous to use microbial enzymes in organic synthesis than mammalian enzymes.R. Microorganisms can be made available. If needed, in much higher amounts than the animals.1. Both A and R are correct and R is the correct explanation of A 2. Both A and R are incorrect3. A is correct and R is incorrect4. Both A and R are correct and R is not the explanation of A*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com414. TATA box is found in:1. Promoter region of gene rich in A, T bases2. Promoter region of gene rich in G.C. bases3. Operator region of gene rich in A, T, bases4. Operator region of gene rich in G.C. bases15. The species that will not give positive test for CN ions is 1. NaCN 2. Ca(CN)23. CH3CN 4. [Fe(H2O)4 Cl2]CN16. Metamorphic rocks originate from1. Igneous rocks2. Sedimentary rocks3. Both igneous and sedimentary rocks4. None of these17. What would be the output of following program1. 2 2. 3 3. 4 4. 618. On a PC, how much memory is available to application software?1. 1024 Kilobytes 2. 760 Kilobytes3. 640 Kilobytes 3. 560 Kilobytes*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com519. A cube of side 1 cm is painted by putting a lacquer of thickness , negligible compared to the side of the cube. The volume of the painted cube is approximately 1. 1 + cm3 2. 1 + 3 cm3 3. 1 + 33 cm3 4. 1 + 3 cm3 20. Which of the following straight lines passes through the point (1,1)? 1. y = 2x + 3 2. 2y = x6 3. x = 1 4. x = y + 1PART-B (Life Sciences)21. Which of the following is true for non-competitive inhibition?1. E+1=E1 2. ES + 1 = ESI 3. E+I=ET,ES+I=ESI 4. None of these22. Which of the following hormones is not a secretion product of human placenta? 1. Human chorionic gonadotropin2. Prolactin3. Estrogen4. Progesterone23. DNA repair is carried out by which of the polymerases in eukaryotes?1. and 2. and 3. and 4. and .24. The world conservation was prepared by1. UNEP 2. UNEP and WWF3. IUCN, UNEP and WWF 4. IUCN and WWF*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com625. DNA fingerprinting refers to1. Molecular analysis of profiles of DNA samples2. Analysis of DNA samples using imprinting devices3. Techniques used for molecular analysis of different specimens of DNA4. Techniques used for identification of fingerprints of individual26. Which one of the following genes is defective in patients suffering from severe combined immunodeficiency syndrome (SCID)?1. Cystic fibrosis transmembrane conductor (CFTR)2. Adenosinedeaminase 3. Ribonucleotide reductase4. a 2-microglobulm27. Mitochondria and chloroplast carry out oxidative phosphorylation and photophosphorylation, respectively, by means of 1. Conformational coupling 2. Chemiosmotic coupling3. High energy intermediate coupling4. Sliding filaments28. Two populations of land snails have been effectively isolated from each other for a long period. According to the biological species concept, which of the following would demonstrate that the two populations have become separate species?1. The two populations behave differently when subjected to same dose of pesticides2. Sterile hybrids are produced when member of the two populations are experimentally mated3. DNA nucleotide sequence are different between two populations4. The two populations have different electrophoretic pattern of proteins29. Hydrolysis of starch is more efficiently catalysed by1. Sulphuric acid 2. Diastase3. Both 4. None of these30. Duodenum has characteristic Brunners gland which secrete two hormones called1. Kinase, estrogen*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com72. Secretin, cholecystokinin3. Prolactin, parathormone4. Estradiol, progesterone 31. A pure or nearly pure water contains a BOD of approximately1. 30 mg/L 2. 20 30 mg/L3. 0 3 mg/L 4. 10 12 mg/L32. Which of the following is an advantage of confocal microscopy over conventional fluorescence microscopy?1. The interaction of a laser beam with the cell surface allows the imaging of individual macromolecules.2. The use of electrons instead of light to image the specimen results in greatly increased resolving power.3. Optical sections can be taken at different depths in a specimen.4. Only scattered light enters the microscope lens, making the object appear illuminated against a dark background.33. Which of the following properties is common to all cytoskeletal motor proteins (such as kinesins, dyneins, and myosis)?1. An actin-binding domain 2. Two globular-head domains3. The ability to bind to biological membranes 4. ATPase activity34. Two varieties of maize averaging 48 and 72 inches in height, respectively, are crossed. The F1 progeny is quite uniform averaging 60 inches in height. Of the 500 F2 plants, the shortest 2 are 48 inches and the tallest 2 are 72 inches. What is the probable number of polygenes involved in this trait?1. Four.2. Eight.3. Sixteen.4. Thirty two.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com835. The term zygotic induction refers to1. Embryogenesis of the fertilized egg2. Process of fertilization3. Prophage induction in a F ~ (F minus) recipient bacteria after Hfr strain mediated conjugation4. Prophage entering the lytic cycle after UV irradiation of a lysogen36. What would be the effect of addition of 2,4 D on the production of berberine by cell culture of Thalictrum minus.1. To stimulate growth and thereby urease secondary metabolite production2. Stimulate dedifferentiation and thereby decrease secondary metabolite production3. Stimulate proliferation and reduce secondary metabolite production4. None of the above37. Which of the following techniques is NOT ideal for immobilizing cell free enzymes?1. Physical entrapment by encapsulation2. Covalent chemical bonding to surface carriers3. Physical bonding by flocculation4. Covalent chemical bonding by cross linking the precipitate38. The full length coding sequence of an eukaryotic gene was expressed in bacteria and the protein was purified. However, in the functional assay, no activity was detected for the purified protein. The reason could be:1. The host bacteria produced an enzyme that inhibited the activity of the expressed eukaryotic protein2. The purified protein was contaminated with bacteria3. The host bacteria did not produce the essential co factors4. No post translational modification on the protein expressed in bacteria39. Which of the following fluorescent probes is used to monitor the progress of amplification in Real time PCR?1. SYBR green 2. Rhodamine3. FITC 4. Cyan blue*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com940. The linked characters will always inherit together till they are1. Delinked due to segregation 2. Masked by dominance 3. Mutated 4. Separated due to crossingover.41. Nullisomy is the term used for the condition when an organism has1. An additional chromosome 2. One chromosome less than normal3. A complete set of chromosomes except one homologous pair4. None of the above42. Which part of the human body is most affected by chronic lead toxicity?1. Muscles and bones 2. Nervous system3. Reproductive system 4. Vascular system43. Pyrology is the study of1. Distribution of animals in relation to temperature2. Role of fire in ecology3. Distribution of plants in forests in relation to sunlight4. Variation in species distribution in relation to ambient temperature44. When nonsense mutations occur in the reading frame of mRNA, protein synthesis gets terminated at the nonsense mutation to deliver a truncated polypeptide. However, in certain bacterial strains, this does not happen; these bacterial cells are able to synthesize full-length polypeptide. This phenomenon is due to:1. Compensatory frame shift mutation that occurs elsewhere in the mRNA2. Involvement of suppressor RNAs3. Polypeptide splicing at the broken point4. Post-transcriptional editing of the nonsense mutation45. The net ecosystem production (NEP) is a function of1. Gross primary production (GPP) and is independent of autotrophic respiration (Ra)*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com102. GPP, Ra, heterotrophic respiration (Rh) and can be either positive or negative3. GPP, Ra, heterotrophic respiration (Rh) and can be positive4. Net primary production (NPP)46. Which of the following accounts for the extent of proliferation observed in the retrovirus-infected cells not treated with growth factors?1. The v erbB product has a function similar to that of activated EGF receptor.2. EGF binds to both the EGF receptor and the v erbB product.3. The v erbB product activates both the LCGF and the EGF receptors.4. The v erbB product antagonizes the action of the EGF receptor.47. Which of the following takes place during anaphase of mitosis in an animal cell? 1. Kinetochore microtubules elongate to push chromosomes towards the Metaphase plate.2. The chromosomes align on the metaphase plate.3. Sister chromatids remain attached to each other at the centromere and move toward the pole as a unit.4. Polar microtubules elongate and slide to push the spindle poles apart48. Larger islands may have greater species diversity than smaller islands because larger islands1. Are in the tropics2. Are farther from continents than smaller islands are3. Have more habitats than smaller islands do4. Have greater genetic drift than smaller islands do49. Two of the premises that form the basis of Darwins concept of natural selection are1. Ecotype and race2. heritability and fitness3. uniformitarianism and catastrophism4. Geographic and reproductive is isolation 50. Blood fibrinogen is converted into fibrin during1. CO2 transport 2. Oxygen transport3. An immune repose 4. Clot formation*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com1151. In calves that consume large quantities of milk, the curdling of milk takes place due to 1. A large variety of useful bacteria2. The high acid content of gastric juice3. The action of rennin4. The action of pepsin52. Honey is formed by1. Worker bees from juicy plants2. The action of digestive juices of worker bees on nectar collected from flowers3. The selective absorption of fructose by worker bees from juicy plants and its processing in their guts4. Worker bees in the nectarines of flowers53. Acetyl CoA, a common key compound in intermediary metabolism, is produced in the body byA. Breakdown of glucose under aerobic conditionsB. Anaerobic breakdown of glucoseC. oxidation of fatty acidsD. Breakdown of anti ketogenic amino acids1. A, C 2. A, B, D3. B, D 4. B, C, D54. A normal couple has five children, two of whom suffer from a somewhat uncommon genetic disorder that has, however, appeared occasionally in this familial line. What kind of gene is involved in this case?1. Codominant 2. completely dominant 3. Completely recessive 4. Incompletely dominant55. Given below are two statements, one labeled as Assertion (A) and the other as Reason (R).Assertion (A): Sonora 64 is a dwarf variety of wheat that played a significant role in the green revolution of India. It is called so because:Reason (R): A scientist, P. Sonora, developed it in 1964, through mutation breeding.In view of the above two statements, which one of the following is correct?*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com121. Both A and R are true and R is the correct explanation of A2. Both A and R are true but R is not the correct explanation of A3. A is true but R is false4. A is false but R is true56. Ribosomes similar to those of bacteria are found in1. Pancreatic endoplasmic reticulum 2. Liver mitochondria 3. Skeletal muscle cytoplasm 4. Daisy chloroplasts.57. Bateson used the terms coupling and repulsion for linkage and crossing over. Name the correct parental or coupling type along with its cross over or repulsion?1. Coupling AABB, aabb; Repulsion AABB, AABB2. Coupling AABB, aaBB; Repulsion AABB, aaBB3. Coupling aaBB, aabb; Repulsion AABB, aabb4. Coupling AABB, aabb; Repulsion aaBB, aaBB58. The biggest drawback of using cost benefit analysis in environmental impact assessment is 1. Its theoretical approach2. Conversion of intangible aspects into monetary units3. Choice of discount rate4. Rudimentary nature of the process59. What is the disadvantage in using electrostatic precipitators for air pollution control1. Small particles cant be removed2. High pressure drop3. Problem in handling hot gases4. High initial cost60. In addition to the requirement of macronutrients, plants also need micronutrients.1. Fe, Mn, Cu, Zn, Mo, B and Cl2. Cs, Sr, I, Mn, Zn, Ba and Cl*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com133. C, H, O, S, P Ca and K4. C, Na, Mn, Mg, Cu, Mo and Cl61. Isozymes can be characterized by1. The different chemical reactions that they catalyze2. The differences in their elution profile from a size-exclusion column3. Differences in their amino acid sequences4. All of the above62. Sodium Dodecyl Sulphate (SDS) is used while separating proteins by polyacrylamide gel electrophoresis because1. It helps in solubilization of proteins thereby making it easier to separate 2. It binds to proteins and confers uniform negative charge density thereby making them move during electrophoresis3. Decreases the surface tension of the buffer used for electrophoresis4. Stabilizes the proteins63. When prospective neuroectoderm from an early amphibian gastrula is transplanted in the prospective epidermal region of a recipient (early gastrula) embryo, the donor tissue will give rise to:1. neural tube.2. epidermis.3. neural tube and notochord.4. neural tube and epidermis.64. A group of six cells called 'equivalence group cells' divide to form the vulval structure in Caenorhabditis elegans. They are called so because:1. they have similar fates during development of vulva.2. all the six cells are competent to form vulva and can replace each other under various experimental conditions.3. they are all under the influence of the anchor cell, signals from which initiate vulval development.4. they interact with each other to form the vulval structure.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com1465. In higher plants, the red/far-red sensory photoreceptor, phytochrome, is a light-regulated kinase. Which of the following classes of kinases does it represent?1. Two-component sensor regulator (histidine kinase).2. Two-component sensor regulator (serine/threonine kinase).3. Leucine rich repeat (LRR) receptor kinase.4. Calcium-dependent protein kinase.66. Absorption of UV radiation by proteins and nucleic acids is due to transition of Electrons between the1. Vibrational energy levels 2. Rotational energy levels3. Nuclear energy levels4. Electronic energy levels67. A myasthenia gravis patient develops muscle paralysis because1. the nerve terminal at the neuromuscular junction fails to release acetylcholine.2. although enough acetylcholine is released at the neuromuscular junction, it is destroyed by acetylcholinesterase.3. the patient develops immunity against his own acetylcholine receptor.4. the patient develops antibody against his own acetylcholine.68. During the course of prolonged starvation and fasting, glucose or glycogen is synthesized from non -carbohydrate precursors by the process of1. Glycogenesis 2. Glycolysis3. Gluconeogenesis 4. Glycogenolysis69. Deletion of the leader sequence of trp operon of E. coli would result in1. decreased transcription of trp operon.2. increased transcription of trp operon.3. no effect on transcription.4. decreased transcription of trp operon in the presence of tryptophan.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com1570. In the endodermis of higher plants, the role of Casperian strip is to control the water movement so that it flows1. between the cells.2. through the plasma membrane.3. through the cell wall.4. through the transfusion tissue.*****PART-C71. To approximate the actual concentration of enzyme in a bacterial cell, assume that the cell contains equal concentration of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100000. Assume also that the bacterial cell is cylinder having a diameter of 1m and height 2.0 m, that the cytosol (specific gravity 1.20) is 20% soluble protein by weight, and that the soluble protein consists entirely of enzyme. Calculate the average molecular concentration of each enzyme in this hypothetical cell.1) 2.4 x 10-5 M2) 1.4 x 10-6 M3) 1.4 x 10-4 M4) 2.4 x 10-6 M72. The following is the pathway for the biosynthesis of molecule D catalyzed by the enzyme E1, E2, and E3. A B C DE1 E2 E3The reaction catalyzed by E1 in the absence of the other enzymes. You find that the rate of the reaction decreases as you add increasing concentration of D. What does this tell about the mechanism of regulation of enzyme E11) Enzyme E3 is probably regulated by feed-back inhibition2) Enzyme E1 is probably regulated by feed-back inhibition3) Enzyme E1 is probably regulated by competitive inhibition4) Enzyme E3 is probably regulated by competitive inhibition*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com1673. The stem of bamboo, a tropical grass, can grow at phenomenal rate of 0.3m/day under optimal condition. Given that the stems are composed almost entirely of cellulose fibers oriented at the direction of growth, calculate the number of sugar residues per second that must be added enzymatically to growing cellulose chain to account for the growth rate. Each D-glucose unit in the cellulose molecule is about 0.45nm long.1) 5,800 residues2) 7,700 residues3) 10,700 residues4) 20,800 residues74. When a small amount of sodium dodecyl sulphate (Na+CH3 (CH2)11OSO-3) is dissolved in water, the detergent ions enter the solution as a monomeric species. As more detergent is added, a concentration is reached (the critical micelle concentration) at which the monomers associated to form micelles. The critical micelle concentration of SDS is 8.2mM. The micelle has an average particle weight (the sum of molecular weight of the constituent monomer) of 18,000. Calculate the number of detergent molecules in the averaged micelle.1) 23 SDS molecules/micelle 2) 43 SDS molecules/micelle3) 63 SDS molecules/micelle4) 83 SDS molecules/micelle75. The sequence of monosaccharides including position and configuration of glycosidic bonds in a glycoprotein is to be determined. Which one of the following methods can be employed?1. Glycoprotein removal of oligosaccharides by alkaline hydrolysis nuclear magnetic resonance analysis of cleaved mixture of oligosaccharides2. Two dimensional nuclear magnetic resonance spectroscopic analysis of the glycoprotein3. Glycoprotein release of oligosaccharides with endoglycosidases followed by purification to separate oligosaccharides enzymatic hydrolysis of purified oligosaccharides with specific glycosidases mass spectroscopic analysis of smaller oligosaccharides4. Glycoprotein treat with trypsin followed by MALDI analyses of tryptic peptides76. In multipass transmembrane proteins the polypeptide chain passes back and forth repeatedly across the lipid bilayer. It is thought that an internal signal peptide serves as a start-transfer signal in these proteins to initiate translocation, which continues until a stop-transfer peptide is reached. In double-pass transmembrane proteins, for example, the polypeptide is released into the bilayer at this point. By recognizing the first appropriate hydrophobic segment to emerge from the ribosome, the SRP sets the *MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com17reading frame, if translocation is initiated, the next appropriate hydrophobic segment will be recognized as a stop-transfer peptide, causing the region of the polypeptide chain in between to be threaded across the membrane. A similar scanning process continues until all of the hydrophobic regions in the protein have been inserted into the membrane. Because membrane proteins are always inserted from the cytosolic side of the ER in this programmed manner, all copies of the same polypeptide chain will have the same orientation in the lipid bilayer. This generates:-1) a molecular scaffold which help itself to pass through various barriers2) an asymmetrical ER membrane in which the protein domains exposed on one side are different from those domains exposed on the other.3) a symmetrical ER membrane helps the protein domains exposed only one side of the membrane4) Nothing happens77. The lipid-linked precursor oligosaccharide is linked to the dolichol by a high energy pyrophosphate bond, which provides the activation energy that drives the glycosylation reaction. The entire oligosaccharide is built up sugar by sugar on this membrane-bound lipid molecule prior to its transfer to a protein. The sugars are first activated in the cytosol by the formation of nucleotide-sugar intermediates, which then donate their sugar (directly or indirectly) to the lipid in an orderly sequence. All of the diversity of the linked oligosaccharide structures on mature glycoproteins results from later modification of the original precursor structure. This oligosaccharide trimming or processing continues in the:1) Endoplasmic reticulum2) Golgi complex3) Cytosol4) Vesicles78. The proteins which take basic stain are known as the basic proteins. The most important basic proteins of the nucleus are nucleoprotamines and the nucleohistones. The nucleoprotamines are simple and basic proteins having very low molecular weight (about 4000 daltons). The most abundant amino acid of these proteins has the pH 10 to 11. The protamines usually remain bounded with the DNA molecules by the salt linkage. The protamines occur in the spermatozoa of the certain fishes. The nucleohistones have high molecular weight, i.e., 10,000 to 1,00,000 daltons. The histone proteins remain associated with the DNA by the ionic bonds and they occur in the nuclei of most organisms. According to the composition of the amino acids following types of histone proteins have been recognized, as:1. Tyrosine, Threonine and Histidine2. Lysine, Alanine and Serine3. Arginine, Lysine and Histidine.4. Proline, Serine and Threonine79. Drosophila larvae are small, white, wormlike creatures. When they first emerge from their eggs they are about 1-2 mm in length, but rapidly grow to about 7-8 mm before pupating. During the rapid larval growth stage, the larva's cells need large quantities of proteins. Polytene chromosomes contain many copies of *MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com18DNA molecules that have replicated several times side by side to provide additional copies of DNA as templates for transcription. The extra copies of DNA cause the chromosomes to expand to an unusually large size, so they can easily be visualized under a microscope, to confirm the site specific expression of gene (transcription) which method would you prefer?I) In situ hybridization using 3H containing probesII) In vitro isolation of cromatin from polytene chromosomeIII) cDNA analysis by cloning in bacterial cellIV) Southern Hybridization method1) I only2) I and III only3) I and IV only4) II and III only80. A man who had purple ears came to the attention of a human geneticist. The human geneticist did a pedigree analysis and made the following observations:In this family, purple ears proved to be an inherited trait due to a single genetic locus. The man's mother and one sister also had purple ears, but his father, his brother, and two other sisters had normal ears. The man and his normal-eared wife had seven children, including four boys and three girls. Two girls and two boys had purple ears. The purple-ear trait is most probably: 1. autosomal, dominant 2. autosomal, recessive 3. sex-linked, dominant 4. sex-linked, recessive81. This antigen is not found in man, however a great variety of other cells and tissues could sensitize man to this antigen. Following an attack of infectious mononucleosis, an infection by the Epstein-Barr virus, antibodies are produced which react not only to the virus but also to a completely unrelated antigen the sheep red cell. This is known as the heterophile antibody response and forms the basis of the Paul-Bunnel test for infectious mononucleosis. Name the antigen:1) MHC antigens2) Monoclonal antibodies3) Unspecified antigens4) Forssman antigen 82. During many important cell processes, many proteins need to undergo degradation to culminate a part of the process. For example, during cell cycle, cycling proteins need to be degraded to allow the cells to exit *MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com19mitosis. This is achieved by selective ubiquitination of cyclin followed by its degradation by proteasomes. The specific protein factor that is involved in this process is called Anaphase Promoting Complex (APC). APC is possibly a protein which is known as1. E1 enzyme.2. E2 enzyme.3. E3 enzyme.4. Protease.83. Digestion of 4kb DNA molecule with EcoRI yields two fragments of 1 kb and 3 kb each. Digestion of same molecule with HindIII yields fragments of 1.5kb and 2.5kb. Finally digestion with EcoRI and HindIII in combination yields fragments of 0.5 kb. 1 kb, and 2.5 kb. Select from the following the correct restriction map indicating the position of the EcoRI and HindIII cleavage sites.EcoRIHindIII1.0 kb 2.5kb2.5 kb4)EcoRIHindIII1.0 kb 2.5kb2.5 kb3)EcoRIHindIII1.5 kb 1.0kb2.5 kb2)EcoRIHindIII1.5 kb 1.0kb2.5 kb1)84. Upon studying a considerable number of different crosses in Drosophila, Morgan reached the conclusion that all genes of this fly were clustered into four linked groups corresponding to the four pairs of chromosomes. Further studies revealed that linkage is not absolute and it is broken frequently. It is broken in prophase by a process called1. Recombination.2. Jumping of genes.3. Integration.4. Mutation.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com2085. DNA repair, synthesis and recombination are intimately connected and inter dependent. An apparent commonality between processes of DNA replication and repair in the enzymatically catalyzed synthesis of DNA polynucleotide segments, which can be assembled with preexisting polynucleotides, leading to repair or replication. Synthesis of these polynucleotide segments is catalyzed by a group of enzymes DNA-dependant DNA polymerases. In the case of E.coli, DNA polymerase has been isolated in three distinctforms whereas five main types of polymerase have been isolated from mammalian cells. All the polymerases synthesize polynucleotides only in the 5' 3' direction. If polynucleotide chains could be elongated in 3' 5' direction, the hypothetical growing 5' terminus, rather than the incoming nucleotide, would carry a triposphate that is unsuitable for further elongation. The 3' 5 exonuclease activity is not associated with all the polymerases and only present in:(A) All E. coli DNA polymerases but not all mammalian polymerases.(B) Pol I, Pol II, Pol III, Pol o, Pol |.(C) Pol I, Pol II, Pol o, Polc, Pol (D) Pol I, Pol II, Polo, Pol c.The correct statements are1. (A) and (B).2. (A), (B) and (C).3. (A) and (C).4. (A), (C) and (D).86. Bacteriophage genetic circuit may be represented as follows:The control of gene expression occurred during the phage infection may be described as follows:(A) N and Q protein act as antiterminator(B) CI acts only as repressor(C) CII act as a retroregulator(D) CI and CII both act as positive and negative regulatorWhich one of the statements are correct?1. (A), (B) and (C).2. (B), (C) and (D).3. (A) and (D) only.4. (A), (C) and (D).*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com2187. A type of gangrene develops in necrotic (dead) tissue that has lost its blood supply. As a consequence, spores of certain obligate anaerobes, can germinate and vegetative cells can proliferate there. As they do, they secrete hydrolytic enzymes and cytotoxins that kill and digest surrounding host cells, expanding the necrotic area in which the pathogen cells grow. Select the correct pathogen of this cause:1) Streptococcal pharyngitis2) Staphylococcus aureus3) Clostridium perfringens4) Bacillus anthracis88. Adrenergic hormones bind to a family of G protein-linked receptors known as adrenergic receptors. The individual members of this family differ mainly in their preference for epinephrine or norepinephrine and in which G protein is linked to the receptor. They can be broadly classified into - and -adrenergic receptors. The -adrenergic receptors bind both epinephrine and norepinephrine. The -adrenergic receptors bind epinephrine much better than norepinephrine. A) - receptors are located on the smooth muscles that regulate blood flow to visceral organs. B) - receptors are found on smooth muscles associated with arterioles that feed the heart smooth muscles of the bronchioles in the lungs, and skeletal muscles.C) - receptors are located near the medulla of KidneyD) - receptors are found on pyramids of KidneyFind the appropriate locations of these receptors in effect to its action:1) A Only2) B Only3) Both A and B4) A, B, C and D89. A young dicot seedling (e.g soyabean) is subjected to gravity stimulus by laying it horizontally on a surface the shoot bends upwards and root bends downward. Indicate the reason.1. Redistribution of auxin throughout the seedlings is responsible for stimulatory unequal growth in shoots and roots.2. Redistribution of auxin in shoots while cytokinine in roots is responsible for stimulatory unequal growth.3. Redistribution of auxin in roots while cytokinine in shoots is responsible for stimulatory unequal growth.4. Redistribution of cytokinine throughout the seedlings is responsible for stimulatory unequal growth in shoots and roots.90. Quorum sensing is a type of decision-making process used by decentralized groups to coordinate behavior. Many species of bacteria use quorum sensing to coordinate their gene expression according to the local density of their population. Similarly, some social insects use quorum sensing to make collective decisions about where to nest. In addition to its function in biological systems, quorum sensing has several useful applications for computing and robotics.Quorum sensing can function as a decision-making process by:*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com221) assessing the number of other components they interact with and a standard response once a threshold number of components is detected.2) gene transcription to be activated, the cell must encounter signaling molecules secreted by other cells in its environment.3) regulating a host of different processes, essentially serving as a simple communication network4) All of these91. Plants grown in greenhouse at 25C when exposed first to 35C for 6 hours and subsequently to 42C for 12 hours adapt better to the high temperature (42C) in comparison to those directly transferred to 42C for the same duration. What is the phenomenon called and what is its main physiological basis?1. Acquired thermo-tolerance because of the induction of mutagensresulting into improved stability of all the proteins.2. Induced thermo-tolerance because of the induction of heat shock proteins.3. Induced thermo-tolerance because changes in RNA polymerase II resulting in efficient and improved transcription.4. Acquired thermo-tolerance because of efficient post translational modification of proteins.92. Several atoms are assembled and held together to form thousands of molecules which participate in the building structural and functional biological systems. A bond is any force which holds two atoms together. The formation of bond between two atoms is due to some redistribution or regrouping of electrons to form a more stable configuration. The regrouping of electrons in the combining atoms may take place in which of the following way(s):1) By a transfer of one or more electrons from one atom to another2) By a sharing of one or more pairs of electrons between the combining atoms3) By a combination of the two processes of transfer and haring4) All of these93. In the 1920, while working with Streptococcus pneumonia, the agent that causes pneumonia, Griffith injected mice with different types of bacteria. For each of the following bacteria types injected, indicate whether the mice lived or died:A) Type IIRB) Type IIIRC) Heat Killed IIISD) Type IIR + Heat Killed IIIS*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com23Results:A- LivedB- DiedC- LivedD- DiedFind the appropriate event(s) from the above result1) Both A and B2) Both A and B and C3) C only4) All A; B; C and D94. Random copolymers were used in some of the experiments that revealed the characteristics of the genetic code. For each of the following ribonucleotide mixtures, give the expected codons and their frequencies, and give the expected proportions of the aminoacids that would be found in a polypeptide directed by the copolymer in a cell-free Protein synthesizing system.1) 2 A : 6 C2) 4 G : 1 C3) 1 A : 3 U : 1 C4) 1 A : 1 U : 1 G : 1 C95. During development and differentiation, there is a dynamic program of differential expression of sets of genes. In bacteria, phage infections are among the simplest examples of developmental process. Typically, only a subset of the phage genome, offer referred to as immediate early genes, are expressed in the host immediately after phage infection. As time passes, early genes start to be expressed, and the immediate early genes and bacterial genes are turned off. In the final stage of phage infection, the early genes give way to late genes. One of the simplest way it is achieved is through:(A) expression of cascade of o factors(B) expression of new RNA polymerases(C) expression of different holoenzymes(D) expression of different transcription factorsThe correct reasons are1. (A), (D)2. (A), (C), (D)3. (A), (B), (D)4. (A), (B), (C)*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com2496. During receptor-mediated endocytosis, ligand first binds with cell surface receptor, then traffic through Rab5 positive early endosomal compartment. Finally, it moves to the Lamp1 positive lysosomes via Rab7 positive late compartment. In order to understand the trafficking of ligand A in epithelia cells, cells were allowed to internalize ligand A for various period of times at 37C. Finally, cells were stained with anti-ligand antibody and probed with secondary antibody labelled with Alexa-Red fluorescence dye. Same cellswere also co-stained with anti-Rab5, anti-Rab 7 or anti-Lamp1 antibody and probed with appropriate secondary antibody labelled with Alexa-green fluorescence dyes. Cells were viewed in confocal microscope and observations are (I) 5 min internalize ligand (Red) in cells are colocalize with anti Rab5 antibody but not with anti-Lamp1 antibody and (II) 90 min internalized ligand are colocalized with anti-Lamp1 antibody but not with anti-Rab5 antibody. The following conclusions could be arrived at from the above bservations.(A) Ligand A travels to early endosomal compartment by 5 min.(B) Ligand A travels to early endosomal compartment by about 90 min.(C) Ligand A travels to lysozome by about 5 min.(D) Ligand A travels to early endosome by about 90 min.Identify the correct inferences.1. (A) and (B)2. (B) and (D)3. (C) and (D)4. (D) and (C)97. Fur color in the babbit, a furry little animal and popular pet, is determined by a pair of allele, B and b. BB and Bb babbits are black, and bb babbits are white. A farmer wants to breed babbits for sale. True-breeding white (bb) female babbits breed poorly. The farmer purchase a pair of black babbits, and these mate and produce six black and two white offspring. The farmer immediately sells his white babbits, and then he comes to consult you for a breeding strategy to produce more white babbits. If he performed crosses between pairs of F1 black babbits, what proportion of the F2 progeny would be white?1) 1/9 2) 2/93) 3/9 4) 6/998. Genes a and b are sex-linked and are located 7 mu apart of the X chromosome of Drosophilia. A female of genotype a+b/ab+ is mated with a wild type (a+b+/Y). What is the probability that one of her sons will wither a+ b+ or a b+ in phenotype?1) 0.25 %2) 0.50 %3) 0.75 %4) 1.00 %*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com2599. Two protein kinases, K1 and K2 function sequentially in regulating intracellular pathway in response to extracellular signal. The following observations are made:(i) Response is observed even in the absence of extracellular signal when a mutation permanently activates K1.(ii) Response is observed even in the absence of extracellular signal when K1 contains an activating mutation and K2 with inactivating mutation.(iii) No response in the cells is detected even in the presence of extracellular signal when both kinases are inactivated by mutation.Which one of the following is correct?1. K1 activates K22. K2 activates K13. K1 inhibits K24. K2 inhibits K1100. It is claimed that the mean () arsenic concentration in the ground water of village is 20g/L with = 3. In a random sample of 16 measurements what values of arsenic concentration should lead to rejection of the claim with 95% confidence?1. values lower then 18.53 and values higher that 21.472. values higher than 21.473. values lower than 18.534. values lower than 18.07 and values higher than 21.93101. Pattern baldness is more frequent in males than in females. This appreciable difference in frequency is assumed to result from-Y-linkage of this trait1) X-linked recessive with sex-limited inheritance2) Sex-influenced autosomal inheritance3) Excessive beer-drinking in males, consumption of gin being approximately equal between the sexes.102. The structure of a protein is known from X-ray diffraction studies which gave 30% o- helix, 50% |-sheet and 20% random coil. Circular dichroism (CD) measurements gave 50% o-helix, 40% |-sheet and 10% random coil. What could not be a possible explanation for these observations.1. Protein structure in the crystal is different from that in the solution.2. CD analysis for structural components is not appropriate for this protein.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com263. Contributions from other chromophores also contribute to the CD spectrum of the protein4. Protein contains high content of disulphide bonds.103. Which of the following redox reaction would be expected to proceed as written? (Assume standard conditions and the presence of appropriate enzymes; E0values are shown below.)Half-reaction E0` (V)Fumarate + 2H+ + 2e- succinate 0.031Oxaloacetate + 2H+ + 2e- malate -0.166Pyruvate + 2H+ + 2e- lactate -0.185Acetaldehyde + 2H+ +2e- ethanol -0.197NAD+ + H+ + 2e- NADH -0.320Acetoacetat + 2H+ +2e--hydroxybutyrate -0.3461. Malate + NAD+ oxalaceteate + NADH + H+2. Acetoacetate + NADH + H+ -hydroxybutyrate + NAD+3. Pyruvate + -hydroxybutyrate lactate + acetoacetate4. Malate + pyruvate oxaloacetate + lactate104. Which of the following sequences of events occurs when E. coli are released from catabolite repression by transfer to low-glucose medium?1. cAMP levels rise, cAMP binds to CAP, cAMP-CAP complex binds to a site on DNA and activates transcription.2. cAMP levels rise, cAMP binds to CAP, cAMP-CAP complex binds to a site on DNA and represses transcription.3. cAMP levels rise, cAMP binds to CAP, cAMP-CAP complex is removed from a site on DNA and activates transcription.4. cAMP levels fall, cAMP is removed from CAP, CAP binds to a site on DNA and activates transcription.105. Opsonisation of a bacterium is a process by which specific antibody binds with the surface molecule of the bacteria. In an experimental condition, macrophage were infected with either WT: Mycobacteria or with opsonised: Mycobacteria for 2 hrs at 37C. Subsequently, cells were washed and further incubated for 24 *MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com27hrs at 37C. Finally, bacterial load in macrophages were determined by colony forming unit (CFU). Which of the following observation is true?1. WT: Mycobacteria inhibits its transport to the lysosomes and survive in macrophages.2. Opsonised: Mycobacteria inhibits its transport to the lysosomes and survive in macrophages.3. WT: Mycobacteria are targeted to the lysosomes and killed in macrophages.4. Opsonised: Mycobateria are targeted to the lysosomes and survive in macrophages.106. The molar absorption coefficient (extinction coefficient) of NADH at 340 nanometers is 6,220 liters per mole per centimeter, whereas that of NAD at 340 nanometers is 0. What absorbance will be observed when light at 340 nanometers passes through a 1- centimeter cuvette containing 10- micromolar NADH and 10- micromolar NAD?1. 0.031 2. 0.0623. 0.124 4. 0.31107. During vertebrate limb development, a specialised ectodermal structure, called Apical Ectodermal Ridge (AER), forms at the dorso-ventral ectodermal boundary at the distal tip of the developing limb bud.The following experimental facts about the AER is available:(A) FGF 2, 4, and 8 are expressed in the AER(B) Removal of the AER causes cessation of limb growth(C) Removal of AER along with implantation of beads soaked in FGF 8 or(D) FGF 4 or FGF 2 protein rescues the AER removal phenotype and gives rise to normal limbWhich of the following statements cannot be made based on the above facts?1. FGF 2, 4, and 8 are secreted proteins.2. FGF 2, 4, and 8 are necessary and sufficient for AER function3. FGF 2, 4, and 8 are sufficient for AER function4. FGF 2, 4, and 8 have largely redundant functions108. A solution contains DNA polymerase I, Mg2+ salts of dATP, dGTP, dCTP, and dTTP, and an appropriate buffer. Which of the following DNA molecules would serve as a template for DNA synthesis when added to this solution?1. A single-stranded closed circle2. A single-stranded closed circle base-paired to a shorter linear strand with a 3`-terminal hydroxyl.3. A single-stranded closed circle base-paired to a shorter linear strand with a 3`-terminal phosphate.4. A double-stranded closed circle*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com28109. When bacteriophage lambda infects a sensitive bacterium, one of the first messenger RNA species synthesized is very short, beginning at a site PL and extending just through an adjacent gene N. After the appearance of the gene N protein, messages become much longer, still beginning at PL but extending far beyond gene N. The N gene encodesa) An antiterminator activator for a promoter beyond gene Nb) A new sigma factor acting on a promoter beyond gene Nc) An activator for a promoter beyond agene Nd) An antirepressor that removes a protein repressor bund at gene N110. In many different contexts of cell differentiation, two distinct cell populations emerge from a uniform cell population. This process is referred to as lateral inhibition. Which one of the following must not be true about lateral inhibition?(A) lateral inhibition results from morphogen action(B) lateral inhibition requires direct cell cell contacts(C) lateral inhibition requires reciprocal signalling between twoneighbouring cells(D) lateral inhibition is preceded by stochastic changes in gene expression in two neighbouring cells1. (D)2. (A) and (D)3. (B) and (C)4. (A)111. In agamous mutant (flower within flower phenotype) which of the following statements is valid?1. Class A genes are expressed in the first two whorls, Class B genes are expressed in the second and third whorls and Class C genes are expressed in the third and fourth whorls.2. Class A genes are not expressed. Class B and C genes are expressed in all the whorls.3. Class A genes are not expressed. Class B genes are expressed in the second and the third whorlsand Class C genes are expressed in all the whorls.4. Class A genes are expressed in all the whorls. Class B genes are expressed in the second and the third whorls.112. During wing development in the chicken embryo, the digit pattern (2-3-4) is thought to be controlled by a morphogen concentration gradient that originates in the posterior of the young wing bud as indicated in the diagrams above. An agar plug soaked in retinoic acid (RA) can mimic the action of the morphogen. Which of the following digit patterns would be expected to result if an agar plug soaked in retinoic acid were paced in the anterior of a developing wing bud?*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com291. 2 3 4 only 4. 4 3 2 only3. 2 3 4 4 3 2 5. 4 3 2 2 3 4113. Pyruvate kinase transfers a phosphate group from phosphenolpyruvate to ADP, forming pyruvate and ATP. The reaction catalyzed by this enzyme is essentially irreversible. Which of the following is the best explanation for the irreversible nature of this reaction?1. The binding of pyruvate to the active site is very weak relative to the binding of phosphoenol pyruvate.2. The reaction is coupled to the pyruvate dehydrogenase reaction.3. The hydrolysis of ATP is highly favorable.4. The change in free energy (G) for the overall reaction is large and negative.114. Aspartate kinase is a key enzyme in the lysine amino-acid biosynthesis in plants. With an objective of increasing the lysine content in maize seeds, maize plants were transformed with E.coli aspartate kinase with a strong seed specific plant promoter. Resulting transgenic plants were found to express the transgene; however, the content of lysine did not increase. Which of the following option best explain the possible reason?1. Bacterial proteins are not stable in plants.2. Bacterial proteins are not properly folded in plants.3. Proper post-translational modification did not take place in plants.4. Lysine causes feed back inhibition of aspartate kinase*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com30115. A response was observed when a specific site in a rat brain was stimulated by passing electrical pulses through indwelling electrode implanted surgically. In another experiment in another rat, a cannula was surgically implanted instead of the electrode and stimulated the area by injecting excitatory neurotransmitter. However, the result of the two experiments did not match. The possibilities of variations in the results could be due to:1. animal variations only.2. stimulation of cell bodies or nerve fibers only.3. difference in anatomical brain areas only.4. variations in all the reasons mentioned in 1, 2 and 3.116. Match the following1. Reflection a) Light passing from one medium to another is bent at an angle2. Transmission b) Light bounces back off surface of an object3. Refraction c) Light is captured inside of object4. Absorption d) Light passes directly through object5. Fluorescence e) Light changes to a different wavelength1. 1-b; 2-d; 3-a; 4-c; 5-e 2. 1-d; 2-b; 3-c; 4-a; 5-e3. 1-e; 2-c; 3-a; 4-d; 5-a 4. None of these117. Identify the position of each of the following on the accompanying graph:1. Organisms divide at their most rapid rate2. New cells are produced at same rate as old cells die3. Lag phase*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com314. Log phase5. Many cells undergo involution and deathChoose correct one of these:1. 1-c; 2-b; 3-a; 4-b; 5-d 2. 1-b; 2-a; 3-b; 4-c; 5-d3. 1-b; 2-c; 3-a; 4-b; 5-d 4. None of these118. In order for chemical reaction to proceed inside living cells, the energy of activation must be supplied. Usually most reactions require more energy than the cell has. How then do these reactions proceed?1. Cells produce copious amounts of energy in the form of ATP2. Cellular enzymes lower the activation energy of reactions3. Cells borrow energy from adjacent cells4. The ribosomes actively synthesize proteins for energy conservation119. In study of the biosynthesis of a particular secretory glycoprotein, the first step was to fractionate a crude RNA extract using an oligo-dT column. The RNA bound to the column was eluted and translated in vitro in the presence of [3H] leucine and [14C] mannose. An antibody specific for the secretory glycoprotein was added and the resulting immunoprecipitate was analyzed by sodium dodecyl sulfate (SDS) polyacrylamide gel electrophoresis. Distribution of radioactivity as a function of position in the gel was then analyzed as shown in Figure 1. The in vitro translation experiment was repeated in the presence of rough microsomes, which were then solubilized. Immunoprescipitation and electrophoresis were performed as in the precvious experiment. The results are shown in figure.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com32Why is the oligo dT coumn used in this experiment?1) Only hnRNA will bind to ligo-dT2) Poly RNA is purified by the procedure3) Ribosomal RNA is bound to the column.4) Intact, rather than partially hydrolyzed, RNA is retained on the column.120. In a haploid organism, the loci A/a and D/d are 8 map units apart. In a cross Ad X aD, what will be the proportion of each of the following progeny classes: (a) Ad (b) Recombinants.1. 92%, 8%2. 46%, 8%3. 92%, 4%4. 46%, 4%121. Which of the following illustrations explain the correct pairing preceeding recombination between a chromosome (ABC-DEFG/ABC-DEFG) and its inverted homologue (ABC-DGFE/ABC-DGFE). The dotin genotype represent the centromere.122. The specific activity of the DNA fragments in Figure 2 is defined as 32P disintegrations per minute per microgram of DNA. Which of the following best describes the relative specific activities of the fragments in lane1?1. Fragment A has the highest specific activity, followed by B, C, D and E.2. Fragment E has the highest specific activity, followed by D, C, B and A.3. The specific activity depends on the order in which the fragments were replicated.4. All fragments have the same specific activity.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com33123. In a family, father is homozygous dominant (AA) for a gene A and his wife is homozygous for its ressive allele (aa) showing albino phenotype. It was surprising that their child showed the albino phenotype. Which of the following phenomenon can explain the phenotype?1. Nondisjunction2. Uniparental Disomy3. Gene conversion4. All of the above124. The protein 1-antitrypsin (1AT) inhibits the action of the proteolytic enzyme elastase in lung tissue. A mutation has been described in which the only change is the substitution of an Arg residue for a Met residue in 1AT. The Altered 1AT does not inhibit elastase but has the new property of inhibiting the blood coagulation protein thrombin. The sequence around the active site of 1AT, altered 1AT, and the natural thrombin inhibitor (antithrombin) are given below. 1At Met Ser Ile Pro Pro GluAltered 1AT Arg Ser Ile Pro Pro GluAntithrombin Arg Ser Leu Asn Pro AsnWhich of the following would be the best method to separate 1AT from altered 1AT?1. Size exclusion Chromatography2. Ion exchange chromatography3. Thin layer chromatography4. Sucrose gradient centrifugation125. Complete the following sentence using options given below the sentence as a, b, c, d and e.Species are critically endangered when it is not endangered but is facing _________risk of extinction in the wild in the _____________future.[(a) high; (b) very high; (c) extremely high;(d) near; (e) immediate]1. a, a2. b, c3. c, e4. c, d126. An experimenter generates library of plasmids containing 10-15 kilobase (kb) inserts from the genome of a bacterium by partially digesting the bacterial genomic DNA with EcoRI and cloning the resulting fragments into the EcoRI site of a Plasmid vector. The experimenter must then identify the plasmids containing the pur B gene. To do this 5 of the plasmids from the library were digested with EcoRI and the *MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com34digests were separated by gel electrophoresis. In a second experiment, the same 5 plasmids were analyzed by PCR using primers derived from sequences internal to pur B and electrophoresis was performed on the PCR products. Both gels were stained with ethidium bromide to visualize the DNA.Which of the following methods would NOT be a useful alternative to using PCR to determine which plasmids contain purB?1. Testing for complementation of purB auxotroph2. Sequencing the inserts3. Hybridizing the plasmids with a probe complementary to purB4. Foot-printing with DNase127. Which one of the following trait set characterizes best a r selected species?1. Usually a type III survivorship curve, short life span and density dependent mortality2. Usually a type I survivorship curve, short life span and density dependent mortality3. Usually a type I survivorship curve, long life span and density independent mortality4. Usually a type III survivorship curve, short life span and density independent mortality128. Muscle cells were cells incubated in the presence of O2 and then quickly made anoxic. The concentrations of various metabolites were measured immediately following the removal of O2. The results are shown in the figure below.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com35The change in the glucose 6-phosphate concentration can be explained best by which of the following?1. Increased synthesis of glycogen2. Increased conversion to free glucose3. Increased rate of glycolysis4. Decreased synthesis of glucose 6-phsphate129. Researchers studying the regulation of a hormone-responsive gene isolated 750 base pairs of DNA immediately preceding the start site of transcription (+1). They demonstrated that if these sequences are cloned upstream of the bacterial chloramphenicol acetyltransferase (CAT) gene and the DNA then introduced into mammalian cells. CAT enzyme activity increases in response to hormone treatment. To define the sequences involved in the regulation of this gene, they made a series of deletions containing various lengths of the 5`regulatory sequences. They cloned these truncated DNA fragments upstream of the CAT gene as shown in the figure below, introduced the constructs into mammalian cells, and assayed for CAT enzyme activity in the absence (-) and presence (+) of hormone. The figure below gives the results of a representative experiment.Assuming that there is a single hormones-responsive regulatory element in the gene, that element is located between1. 742 and 638 2. 638 and 424 3. 424 and 315 4. 315 and 116*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com36130. Three important biological parameters generation time, population growth rate (r) and metabolic rate per gram body weight are a function of the organisms body size. Which of the curves (a) or (b) represents the correct relation of each of the parameters to body size?1. Generation time (a); population growth rate (b); metabolic rate/g bw (a)2. Generation time (a); population growth rate (b); metabolic rate/g bw (b)3. Generation time (b); population growth rate (a); metabolic rate/g bw (a)4. Generation time (b); population growth rate (b), metabolic rate rate/g bw (a)131. The female of a species of insects lays about 300 eggs in June-July. Half of them hatch successfully (equal proportion of males and females) by October. Forty percent, of the larvae form pupae by January, and adults emerge from one third of the pupae by March. Mating takes place during May, and 20% of the adult insects manage to mate successfully. Thereafter, all the adults die after the females have laid eggs in June-July. There are no sex specific differences in survival, mortality, successful completion of developmental stages and mating success. If 10 fertilized females are released in a very large enclosure in June-July 9,001 how many eggs are likely to b laid during June-July 2005?1. 48000 2. 60002. 36000 3. 24000132. Eukaryotic cell membranes were analyzed for hormone receptors. A membrane preparation was incubated with radiolabeled hormone (3H-hormone) for five minutes. A similar incubation of membranes and 3H-hormone contained, in addition, a 1,000 fold excess of unlabeled hormone. In both cases, the unbound hormone was removed by washing the preparation and the amount of radioactivity remaining in the membrane preparation was determined. The following results were obtained.cpmonly hormone H000 , 53 cpmHormone Unlabeled Excess plus hormone H500 , 13 Which of the following statements is most likely true concerning the binding of 3H-hormone? 1. The total amount of radioactivity bound in the absence of unlabeled hormone represents the amount bound by the receptors.2. Most of the label is bound nonspecidfically.3. The unlabeled hormone competes with 3H-hormone for binding to the receptors.4. The cpm bound in the absence of unlabeled hormone minus the cpm bound in the presence of unlabeled hormone is a measure of nonspecific binding.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com37133. In the process of nitrification by organisms, the respective bacteria A and B in the following reaction are:NH4 +1/2 O2 A NO2- +2 H+ H2ONO2-+1/2 O2 B NO31. Azotobacter, Nitrobacter2. Nitrobacter, Azotobacter3. Nitrosomonas, Nitrobacter4. Nitrobacter, Nitrosomonas134. Inspite of its two-fold cost, sexual reproduction is the most dominant mode of reproduction among the living organisms. Which of the following reasons might account for this?(A) Sexual reproduction generates genetic heterogeneity through recombination(B) Sexual reproduction helps in purging deleterious mutations(C) Sexual reproduction evolved to stay evolutionarily ahead of fast evolving internal parasites.1. (A) only2. (A) and (B)3. (C) only4. (A), (B) and (C)135. The filamentous alga Cladophora is illuminated with light dispersed by a prism. As shown in the diagram above, aerobic bacteria Pseudomonas included in the medium congregate where the alga is illuminated with light at 656 nanometers and at 486 nanometers. Pseudomonas do not congregate in this manner if the Cladophora is removed from the medium. Which of the following is the most likely explanation fro the bacterial movement?1. Bacteria have a phototropic response 2. Starch made by photosynthesis is Secreted from the alga in regions illuminated by light at 656 nm and 486 nm.3. O2 evolved by photosynthesis in the regions illuminated by light at 656 nm and at 486 nm attracts bacteria.4. Alga and bacteria have a symbiotic relationship; the bacteria need O2 suppolied by the alga and the alga requires CO2 supplied by the bacteria.*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com38136. It is found that people with the genetic disease called sickle cell anaemia are resistant to malaria. Which of the following best describes the underlying mechanism?1. Frequency-dependent selection2. Superiority of heterozygotes3. Transient polymorphism4. Balanced polymorphism137. The autoradiograms above (after electrophoresis and Southern blotting) show human DNA digested with a specific restriction enzyme and probed with labeled rRNA. In the autoradiogram on the lert, the probe was 28S rRNA; at the right, the probe was 18S rRNA. If the arrows in the following maps locate the recognition sites the restriction enzyme, which map best explains the results shown above?1)2)3)4)138. A severe winter storm kills many chickadees. An investigation comparing the body size of dead birds with that of survivors reveals that the dead birds included mainly the largest and smallest members of the population. This winter storm exemplifies1. Kin selection 2. Stabilizing selection3. Directional selection 4. Balancing selection139. If a given gene in a randomly mating population has three alleles a, b and c in the ratio of 0.5, 0.2 and 0.3 respectively, what is the expected frequency of genotypes ab and bc in the population at equilibrium?1. 0.1 and 0.062. 0.2 and 0.153. 0.2 and 0.124. 0.04 and 0.09*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com39140. One summer the moose population on Isle Royale was unusually high, and park naturalists noticed signs of malnutrition among the adults. The wolf population was fairly low, near 20. That winter, for the first time in many years, a substantial number of seemingly healthy adult moose as well as calves and crippled animals were killed and eaten by wolves. This description is part of a general situation in which the wolf and moose populations1. Are maintained in a stable equilibrium, from year to year2. Are simultaneously becoming extinct 3. Fluctuate out of phase with each other4. Fluctuate independently of each other141. Animal cell cultures are frequently used for production of therapeutic proteins. NIH3T3 (a fibroblast cell line) and CHO (Chinese hamster ovarian cell line) are some of the popular cell lines used. Choose the best combination of cell line (for transfection) and starting material for purification of human growth hormone, a secretary protein1. NIH3T3 cell pellet2. CHO and cell pellet3. NIH3T3 and culture medium4. CHO and culture medium142. It is now thought that the atmosphere of the primitive Earth was composed largely of carbon dioxide, nitrogen, and water vapor. The composition of certain iron-containing minerals suggests that the carbondioxide began to be replaced by oxygen about 2 billion years ago. Which of the following is the best explanation for the change in atmospheric composition?1. Ozone produced in the upper atmosphere by ultraviolet light broke down to oxygen.2. Photosynthesis was established in primitive bacteria.3. Oxygen was present in volcanic gases and slowly accumulated with time.4. Water was broken down into oxygen and hydrogen by lightning discharges.143. Some individuals suffer form an autosomal recessive disorder known as alpha1-antitrypsinn deficiency. The recessive homozygote for this disorder lacks the enzyme that ordinarily degrades trypsin. The genotype of the recessive disorder may be designated as aa. Normal individuals are either either homozygous dominant (AA) or heterozygous carriers (Aa). Enzymatic tests and colorimetric analysis reveal that a given individual possesses one of three distinct levels of alpha1-antitrypsin enzyme activity. The data below show the levels of trypsin inhibited per milliliter of serum (which contains the antitrypsin enzyme) in three different groups.Groups I. The general populationGroups II. Several families in which some members have alpha1-antitrypin deficiency*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com40Groups III. Patients with alpha1-antitrypsin deficiencyWhen the serum of an adult woman was tested colorimetrically, a level of 0.5 milligram of trypsin inhibited per milliliter of serum was found. This can be interpreted as indicating which of the following?1. She is homozygous dominant2. She is an asymptomatic heterozygous carrier.3. She will exhibit alpha1-antitrypsin deficiency.4. She will always transmit the abnormal recessive gene to her offspring.144. In replicated experiments, plots are planted with flax seeds in densities of 60, 1,440, and 3,600 per square meter, respectively. The factors of soil, water, and light are similar for each plot. Dry weights of the mature plants are obtained.[DISTRIBUTION OF DRY WEIGHTS OF INDIVIDUALS IN POPULATIONS OF FLAXPLANTS SOWN AT DIFFERENT DENSITIES]The experiments are most likely performed to test for 1. Interspecific competition 2. Intraspecific competition3. Competitive exclusion 4. Founder effect*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com41145. Following are some statements about Agrobacterium plant interactions(A) Agrobacterium transfers a part of its chromosome into plant cell.(B) Agrobacterium transfers a part of one of its plasmid DNA into plant cell.(C) All the virulence genes of Agrobacterium are inducible.(D) All the virulence genes of Agrobacterium are functional only inside the bacterial cells.(E) Some of the virulence genes of Agrobacterium are inducible.(F) Some of the virulence genes of Agrobacterium are functional both in bacterial and plant cells.Which of the following combination of statements is true?1. (A), (C) and (D)2. (B), (E) and (F)3. (C), (D) and (E)4. (B), (E) and (F)*MUDRA* Model Question Paper for NET-LS & JRF of UGC-CSIR in Life Sciences (Model Question Paper No. 1) www.mudralifesciences.com42ANSWER SHEET FOR MODEL QUESTION PAPER No.1(Must be printed on 100 gsm or Card Paper)Name of the Candidate:______________________________________________________Address:____________________________________________________________________________________________________________________________________Roll No._______ (Provided at the time of Purchase of Study Material from Mudra Life Science)MUDRALIFE SCIENCESEmail:___________________________________________ Mob:____________________001. 2 1 3 4001. 2 1 3 4002. 2 1 3 4003. 2 1 3 4003. 2 1 3 4005. 2 1 3 4006. 2 1 3 4007. 2 1 3 4008. 2 1 3 400. 2 1 3 4009. 2 1 3 4010. 2 1 3 4011. 2 1 3 4012. 2 1 3 4013. 2 1 3 4014. 2 1 3 4015. 2 1 3 40 2 1 3 4016. 2 1 3 4017. 2 1 3 4018. 2 1 3 4019. 2 1 3 4020. 2 1 3 4021. 2 1 3 4022. 2 1 3 4023. 2 1 3 4024. 2 1 3 4025. 2 1 3 4026. 2 1 3 4027. 2 1 3 4 2 1 3 4028. 2 1 3 4029. 2 1 3 4030. 2 1 3 4031. 2 1 3 4 2 1 3 4032. 2 1 3 4033. 2 1 3 4034. 2 1 3 4035. 2 1 3 4036. 2 1 3 4037. 2 1 3 4 2 1 3 4038. 2 1 3 4039. 2 1 3 4040. 2 1 3 4041. 2 1 3 4042. 2 1 3 4043. 2 1 3 4044. 2 1 3 4 2 1 3 4045. 2 1 3 4046. 2 1 3 4047. 2 1 3 4048. 2 1 3 4049. 2 1 3 4050. 2 1 3 4051. 2 1 3 4 2 1 3 4052. 2 1 3 4053. 2 1 3 4054. 2 1 3 4055. 2 1 3 4056. 2 1 3 4057. 2 1 3 4058. 2 1 3 40. 2 1 3 4059. 2 1 3 4060. 2 1 3 4061. 2 1 3 4062. 2 1 3 4063. 2 1 3 4064. 2 1 3 4065. 2 1 3 4066. 2 1 3 40. 2 1 3 4067. 2 1 3 4068. 2 1 3 4069. 2 1 3 4070. 2 1 3 40. 2 1 3 4071. 2 1 3 4072. 2 1 3 4 2 1 3 4073. 2 1 3 4074. 2 1 3 4075. 2 1 3 4076. 2 1 3 4077. 2 1 3 4078. 2 1 3 4079. 2 1 3 40. 2 1 3 4080. 2 1 3 4081. 2 1 3 4082. 2 1 3 4083. 2 1 3 4084. 2 1 3 4085. 2 1 3 4086. 2 1 3 40. 2 1 3 4087. 2 1 3 4088. 2 1 3 4089. 2 1 3 4090. 2 1 3 4091. 2 1 3 4092. 2 1 3 4093. 2 1 3 40. 2 1 3 4094. 2 1 3 4095. 2 1 3 4096. 2 1 3 4097. 2 1 3 4098. 2 1 3 4099. 2 1 3 4100. 2 1 3 4101. 2 1 3 4 2 1 3 4102. 2 1 3 4103. 2 1 3 4104. 2 1 3 4105. 2 1 3 4 2 1 3 4106. 2 1 3 4107. 2 1 3 4108. 2 1 3 4109. 2 1 3 42 1 3 4110. 2 1 3 4111. 2 1 3 4112. 2 1 3 4113. 2 1 3 4114. 2 1 3 4115. 2 1 3 4116. 2 1 3 4117. 2 1 3 4118. 2 1 3 4119. 2 1 3 4120. 2 1 3 4121. 2 1 3 4123. 2 1 3 4124. 2 1 3 4125. 2 1 3 4 2 1 3 4126. 2 1 3 4127. 2 1 3 4128. 2 1 3 4129. 2 1 3 4130. 2 1 3 4131. 2 1 3 42 1 3 4132. 2 1 3 4133. 2 1 3 4134. 2 1 3 4135. 2 1 3 4136. 2 1 3 4137. 2 1 3 4138. 2 1 3 4139. 2 1 3 4140. 2 1 3 4 2 1 3 4141. 2 1 3 4142. 2 1 3 4143. 2 1 3 4144. 2 1 3 4 2 1 3 4145. 2 1 3 4PART-BPART-CPART-A