More from Purushotam Singh Thakur

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DBT-BET 2016 SOLVED PAPER DBT 2016, , LECTURE BY: PURUSHOTAM SINGH THAKUR, We have a Telegram group and channel, YOU can join that for more Quiz and Discussion., I am Admin there and post Quizzes Daily., Telegram Group: @baayo2, Telegram channel: @baayo_channel, , Join our Telegram Channel for CSIR/DBT/GATE/ICMR and various other entrance exams.

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BET-2016, Section- A, All 50 questions are compulsory, 1. Two sides of an isosceles triangle measure 3 cm and 7 cm. Which one of the following is the, measure (cm) of the third side?, (A) 9, (B) 7, (C) 5, (D) 3, ANS: B, ISOSCALES TRIANGLE has 2 same sides., Property = Sum of any 2 sides of a triangle > 3rd side, 2. At 9 AM, the shadow of a 5 (feet) tall boy is 8. At the same time, shadow of a flagpole beside is, 28 feet. What is height of the flagpole?, (A) 8.5, (B) 16, (C) 17.5, (D) 20, ANS: C, 5 / 8 = x / 28, x = (5 / 8) (28) = 17.5, 3. What number should appear next in this series: 8, 12, 10, 14, 12, 16….?, (A) 10, (B) 14, (C) 18, (D) 12, ANS: B, 4. Two numbers are more than a third number by 20 % and 50 %, respectively. The ratio of the, first two numbers is, (A) 2:4, (B) 3:5, (C) 4:5, (D) 5:7, ANS: C, Let 3rd number = x, 1st number = x + 0.2x = 1.2x, 2nd number = x + 0.5x = 1.5x, Ratio = 1.2x / 1.5x = 12 / 15 = 4/ 5

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9. The sides of a rectangle are in the ratio of 4:3 and its area is 108 𝒎^𝟐. The perimeter of the, rectangle in cm is, (A) 22, (B) 32, (C) 42, (D) 52, ANS: C, Let length be 3x and breadth be 4x., Our ratio= 3x /4x = 3/4, Ratio of l/b = 3/4 (given), So, ratio is same., Area of rectangle = l× b, Given Area = 108 cm^2, So, l×b = 108 cm^2, (3x)(4x) = 108, 12x^2 = 108, x^2 = 108/12= 9, x= √9 = 3 cm, Length = 3x = 3×3 = 9 cm, Breadth = 4x = 4×3 = 12 cm, Perimeter of rectangle= 2(l +b), = 2(9+12) = 2×21 = 42 cm., 10. If each side of a cube is increased by 1%, the percentage change in the volume would be, approximately, (A) 1, (B) 2, (C) 3, (D) 4, ANS: C, V = L^3, dV/V =3dL/ L, In percentage:, = dV/V ×100 =3dL/ L ×100, dL/L×100 = (1% given), dV/V ×100 =3dL/ L ×100, = 3 × 1% = 3%

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11. A brick weighs ¾ of itself and ¾ of a Kg. The weight of the brick in Kg is, (A) 2, (B) 3, (C) 4, (D) 5, ANS: B, Let the weight of Brick= x, According to question:, 3/4 weight of Brick + 3/4 of kg = weight of Brick, (3/4) (x) + (3/4) (1000 gram) = x, 0.75x +0.75×1000 = x, 750= x - 0.75x, 750= 0.25x, x = 750/0.25 = 3000 gram = 3 kg., So, option B is the correct answer., 12. In an enzyme assay, the corrected absorbance reading obtained on addition of 100 μL of the, enzyme extract is 0.025. Given that one unit of the enzyme is the amount of enzyme required to, increase the absorbance by 0.001 units under assay conditions, the enzyme activity (units/ml) of, the extract is:, (A) 2.5, (B) 2500, (C) 25, (D) 250, ANS: D, 0.01 absorbance = 1 unit, 1 absorbance = 1 / 0.001 unit, 0.025 absorbance = (0.025/0.001) = 25 unit, Enzyme activity = Units / mL = 25 / (100 x 10^-3 mg) = 25 / 0.1 = 250, 13. The cell-free extract prepared from E. coli cells over-expressing enzyme β-glucosidase showed, the activity of 1.5 units per ml (protein concentration 2 mg per ml). The Ni-NTA purified, preparation showed the activity of 75 units per ml (protein concentration 100 μg per ml). Calculate, the fold purification of the enzyme achieved?, (A) 0.001, (B) 0.02, (C) 50, (D) 1000

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ANS: D, Initial Specific activity = Initial Activity / Protein in mg = 1.5/2 = 0.75, Final Specific activity = Final Activity / Protein in mg = 75 / 0.1 = 750, Purification fold = Final Specific activity / Initial Specific activity = 750 / 0.75 = 1000, , 14. Enzyme X showed its activity on substrate A (375 units per ml), substrate B (185 units per ml), and substrate C (75 units per ml). With respect to substrate A, the percent activities on substrate B, and C are --------, respectively., (A) 0.49 & 0.2, (B) 2.02 & 5, (C) 49 & 20, (D) 202 & 500, ANS: C, Percentage activity of B w.r.t. A = B/A = (185/375) × 100= 49 %, Percentage activity of C w.r.t. A= C/A = (75/375) × 100= 20%, 15. Protein ‘A’ from Pseudomonas putida contains 135 amino acids. The number of nucleotides, present in the gene encoding the protein will be -----., (A) 405, (B) 408, (C) 411, (D) 421, ANS: B, No. of Nucleotides: 3 x 105 + 3 = 405 + 3 = 408, 16. The molarity of a 15 % of NaCl solution in water is ----., (A) 2.56, (B) 0.256, (C) 25.6, (D) 0.025, ANS: A, 15% NaCl means in 15 g in 100 mL, So, given mass= 15 g, Vol= 100 mL= 0.1 L, Molecular mass of NaCl= 23+ 35.5= 58.5 g, Molarity = (mass/Molecular mass)/Vol. in Litre, Molarity= (15/58.5)/ 0.1, =0.256410256/0.1, =2.56410256= 2.56, SOLVED BY: PURUSHOTAM SINGH THAKUR

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17. The reaction velocity, (V) vs substrate concentration [S] profile was performed for Enzyme A, using 1 μg enzyme per assay. Similar experiment was carried out under identical conditions except, that the concentration of enzyme used was 2 μg per assay. Under these conditions, the kinetic, constants, (A) 𝐾𝑚 and Vmax will remain unchanged, (B) 𝐾𝑚 will change while Vmax will remain same, (C) 𝐾𝑚 will remain same but Vmax will increase, (D) 𝐾𝑚 and Vmax will increase, ANS: C, (Turn over number=Kcat. =K2 ), Kcat. = k2=Vmax. /Eo, Vmax. = K2 x [Eo], , 18. Which one of the following bacterial cell suspensions at 𝑨𝟓𝟒𝟎𝒏𝒎 = 0.2 will have the maximum, number of cells per ml?, (A) Mycoplasma, (B) Pseudomonas, (C) Streptococcus, (D) Bacillus, ANS: A, (Mycoplasma has small size so max. cells per mL.), 19. Two solutions of a substance (non-electrolyte) are mixed in the following manner: 480 ml of, 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity (M) of the final, mixture?, (A) 1.250, (B) 1.344, (C) 1.433, (D) 1.4795, ANS: B, M1V1 + M2V2 = M3V3, Here V3= V1 +V2, M1V1 + M2V2 = M3(V1+V2), 480×1.5 + 520×1.2= M3(480+520), 720 +624 = M3× 1000, M3 = 1344/1000, M3= 1.344 M

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20. A 29 % 𝑯𝟐𝑺𝑶𝟒 solution having a molarity of 3.60, would have a density (𝒈. 𝒎𝒍–𝟏) of --. (MW of, 𝑯𝟐𝑺𝑶𝟒 is 98), (A) 1.45, (B) 1.64, (C) 1.88, (D) 1.22, ANS: D, Molarity= (mass %) ×10 × density/Molecular mass, 3.60= 29 ×10 × density /98, (3.60×98) / (29×10) = density, density = 352.8/ 290 = 1.21655172, density = 1.22 g /mL., 21. Calculate the turnover number (sec-1) for an enzyme, if Km = 0.001 μM; Vmax = 10 μM/sec; and, E0= 0.001 μM., (A) 1x10^4, (B) 1x10^5, (C) 1x10^6, (D) 1x10^7, ANS: A, (Turn over number=Kcat. =K2 ), Kcat. = k2=Vmax. /Eo, Kcat. = 10/0.001=1×10^4., 22. Zymomonas mobiles cells are grown in a chemo state culture in a 60 litre (l) fermenter. The Ks, and μmax for the organism is 0.2 g.l-1 and 0.2 h-l, respectively. What flow rate (l.h-1) is required for a, steady-state substrate concentration of 0.2 g.l-1?, (A) 6, (B) 8, (C) 10, (D) 12, ANS: A, In chemostat D= u, So, according to Monad's equation:, D= u = (umax. × S) / (Ks +S), D= (0.2×0.2) / (0.2+0.2), D= 0.04/0.4= 0.1, Also, we know D= F/V, So, F= D×V, F = 0.1×60 = 6 L/h.

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23. Consider three independently, assorting gene pairs, A/a, B/b, and C/c., The probability of obtaining an offspring, of AABbCc from parents that are AaBbCC, and AABbCc is, (A) 4/9, (B) 3/4, (C) 1/8, (D) 1/9, ANS: C, AABbCc = (1/2) (2/4) (1/2) = 1/8, , SOLVED BY: PURUSHOTAM SINGH THAKUR

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24. Given that husband and wife are both heterozygous for a recessive allele for albinism. If they, have dizygotic twins, the probability that both the twins will have the same phenotype for, pigmentation will be, (A) 5/8, (B) 1/4, (C) 3/4, (D) 1/16, ANS: A, For Heterozygous parents,, Dizygotic twins that will be affected by albinism = (1/4) x (1/4), Dizygotic twins that will not be affected = (3/4) x (3/4), Dizygotic twins having either albinism or not affected by albinism, = (1/4) x (1/4) + (3/4) x (3/4), = (1/16) + (9/16), = 10/16, = 5/8, 25. In a family with three children, what is the probability that two are boys and one is a girl?, (A) 2/3, (B) 1/2, (C) 3/8, (D) 1/3, ANS: C, In a family of 3 children Probability of 1 girl and two boys is=, (3 POSSIBILITIES ARE THERE), =BBG+BGB+GBB, = (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2) +(1/2)(1/2)(1/2), =1/8 +1/8 +1/8 =3/8=0.375, OR, You can use this formula:, nCr(p)^r ×(q)^n-r, n=3(no. Of children), r=1(one girl), p=1/2(prob. of girl), q=1/2(prob. of boy), 3C1×(1/2)^1×(1/2)^3-1, {3C1=3!/[(3-1)!×1!], =3×2!/(2!×1!)=3}, 3×(1/2)^1×(1/2)^2, = 3×(1/2)×(1/4), =3/8=0.375

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26. LDL binds with cell surface receptor and gets internalized via clathrin mediated endocytosis., This process helps in maintaining the cholesterol-LDL level in the plasma. However, in a disease, known as familial hypercholesterolemia (FH), very high levels of plasma cholesterol is found. This, could be due to mutation in which one of the following genes in FH patients?, (A) Clathrin, (B) LDL, (C) LDL receptor, (D) Adaptor, ANS: C, 27. Digestion of a 5Kb linear DNA fragment with EcoRI generates two fragments of 2 Kb and 3 Kb,, while digestion of the same molecule with HindIII yields three fragments of 0.7 Kb, 3.5 Kb and 0.8, Kb. When the same DNA is digested with both the enzymes, it yields fragments of 0.7 Kb, 1.3 Kb,, 2.2 Kb and 0.8 Kb. The right sequence of restriction sites in the DNA fragment is, (A) One EcoRI site in between two HindIII sites, (B) One HindIII site in between two EcoRI sites, (C) Two HindIII sites followed by only one EcoRI site, (D) One EcoRI site followed by two HindIII sites., ANS: A, 28. GFP, when overexpressed in a cell, remains mostly in the cytosol. A GFP construct is modified, such that the resultant GFP protein will have a conjugated peptide Pro-Lys-Lys-Lys-Arg-Lys-Val at, its N-terminus. If such a GFP construct is expressed in a cell, the modified GFP protein will be, localized in the, (A) lysosome, (B) Golgi bodies, (C) nucleus, (D) endoplasmic reticulum, ANS: C, 29. Succinate dehydrogenase converts succinate to fumarate. Which one of the following is TRUE, when the competitive inhibitor malonate is added in the reaction mixture?, (A) Both Km and Vmax increase., (B) Both Km and Vmax decrease., (C) Km increases and Vmax remains the same., (D) Km increases and Vmax decreases., ANS: C, 30. In a mammalian cell, protein synthesis is regulated at the level of initiation by various kinases., During viral infection, which one of the following kinases is involved in regulating the step of, formation of eIF2.GTP.Met tRNAi ternary complex in the host?, (A) Heme-regulated inhibitor kinase (HRI), (B) Protein kinase RNA dependent (PKR), (C) GCN2-like kinase, (D) PKR-like endoplasmic reticulum kinase (PERK), ANS: B

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31. Which one of the following side chains of an amino acid is responsible for fluorescence in, proteins?, (A) Indole ring, (B) Guanidine group, (C) Phenolic group, (D) Imidazole group, ANS: A, 32. DNA molecules labelled with 15N and 14N can be separated by, (A) Pulse field gel electrophoresis, (B) Density gradient ultracentrifugation, (C) Capillary electrophoresis, (D) Differential centrifugation, ANS: B, 33. Match the, chromatographic technique, from Group A with the, appropriate elution, conditions given in Group B., Group A, (P)Chromatofocusing, (Q) DEAE-Sephadex, (R) G-150 Sephadex, (S) Phenyl Seahorse, , Group B, (i) Decreasing [(𝑁𝐻4)2𝑆𝑂4], gradient, (ii) pH gradient, (iii) Isocratic gradient, (iv) Increasing NaCl gradient, , (A) P-ii, Q-iv, R-iii, S-i, (B) P-i, Q-iv, R-iii, S-ii, (C) P-iv, Q-iii, R-ii, S-i, (D) P-iii, Q-i, R-ii, S-iv, ANS: A, 34. Cytoskeletal organization of a cell is regulated by, (A) Ras GTPase, (B) Rab GTPase, (C) Rho GTPase, (D) Ran GTPase, ANS: C, 35. In comparison to animals residing in a warm climate, animals living in cold climate need, thermal insulation. The cell membranes of the latter would have a relatively higher content of ……, (A) sphingolipids, (B) saturated fatty acid, (C) unsaturated fatty acid, (D) cholesterol, ANS: C

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38. Which one of the following modes of inheritance is seen in Cystic Fibrosis?, (A) Autosomal recessive, (B) Autosomal dominant, (C) Sex linked, (D) Spontaneous mutation, ANS: A, 39. Statins are very effective against hypercholesterolemia, a major cause of atherosclerosis., These drugs reduce plasma cholesterol levels by, (A) Preventing absorption of cholesterol from the intestine., (B) Increasing the excretion of cholesterol from the body via conversion to bile acids., (C) Inhibiting the conversion of 3-hydroxy-3-methylglutaryl-CoA to mevalonate in the pathway for, cholesterol biosynthesis., (D) Increasing the rate of degradation of 3-hydroxy-3-methylglutaryl CoA reductase., ANS: C, 40. Measles, Mumps, Rubella-MMR combined vaccine represents which one of following vaccine, categories?, (A) Inactivated/killed, (B) Live, attenuated, (C) Subunit, (D) Toxoid (inactivated toxin), ANS: B, 41. A haemophiliac man marries a normal woman. They have a daughter who does not show, symptoms of haemophilia. If she marries a haemophiliac man, what will be the probability of their, son displaying symptoms of haemophilia?, (A) 0%, (B) 25%, (C) 50%, (D) 100%, ANS: C, MALE= X^hY, FEMALE= X^h^X (carrier) or XX, DAUGHTER= X^h^X, MARRIAGE= (X^h^X) x (X^hY), SON HAEMOPHILIC = X^hY = 1/2 = 50%, 42. The conventional treatment for methanol toxicity is to administer ethanol. Which of the, following explains the basis of this treatment?, (A) Ethanol acts as a competitive inhibitor to methanol, (B) Ethanol acts as a non-competitive inhibitor to methanol, (C) Ethanol destroys the enzymatic activity of alcohol dehydrogenase, (D) Ethanol blocks the entry of methanol within the cells., ANS: A

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43. What will be the angular velocity of a rotor in a centrifuge operating at 6000 revolution per, minute?, (A) 62.8 radians per second, (B) 628 radians per second, (C) 6.28 radians per second, (D) 6280 radians per second, ANS: 1 rpm= 2π/ 60 radians per second, Or, 1 rpm = 0.1047 radians per second, So, 6000 rpm= 6000×0.1047, = 628.2 radians per second, So, 628 radians per second is correct answer., 44. Lysosomes of a cell were labelled with lysotracker Red. Subsequently, these cells were infected, with GFP-transfected Mycobacterium and observed under a fluorescence microscope. What will, you observe?, (A) GFP-Mycobacterium will be colocalized with lysotracker Red labelled lysosomes., (B) GFP-Mycobacterium will be separated from lysotracker Red labelled lysosomes., (C) GFP-Mycobacterium will not be detected as they are degraded in the cell., (D) Lysotracker Red labelled lysosomes will be degraded in GFP Mycobacterium infected cells, ANS: B, 45. A linear DNA fragment which has 3 restriction sites for BamH1, is labelled only at the 5/ end., This DNA is partially digested with BamH1 in such a way that all kinds of fragments are generated., Under these conditions, how many labelled and unlabelled fragments will be produced?, (A) 3labelled and 4 unlabelled, (B) 3 labelled and 5 unlabelled, (C) 4 labelled and 5 unlabelled, (D) 4 labelled and 6 unlabelled, ANS: D

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46. In the preparation of humanized antibody, part of the antibody molecule is taken from mouse, and the remaining is taken from that of human, through genetic engineering technique. Which one, of the following statements is true for humanized antibody?, (A) CDRs of mouse IgG is fused with framework regions of human IgG, (B) CDRs of human IgG is fused with framework regions of mouse IgG, (C)CDRs of mouse IgG is fused with CDRs of human IgG, (D) framework regions of mouse IgG is fused with framework regions of human IgG, ANS: A, 47. Approximately how many molecules of CO2 are produced daily by oxidative metabolism in, adult human? Avogadro’s number is 𝟔. 𝟎𝟐2 × 𝟏𝟎^𝟐𝟑 ., (A) 1.2 × 10^24, (B) 1.2 × 10^23, (C) 1.2 × 10^26, (D) 1.2 × 10^25, ANS: A, Moles = mass / Molecular mass = Vol. at STP / 22.4 L = No. of particles / Na, , C + O2 = CO2, Moles = No. of particles / Na, MOLECULES= 2 x 6.022 × 𝟏𝟎^𝟐𝟑, , = 12 x, , 𝟏𝟎^𝟐𝟑, , = 1.2 x 𝟏𝟎^𝟐4, , 48. A 25-year-old man undertakes a prolonged fast for religious reasons. Which one of the, following metabolites will be elevated in his blood plasma after 24 hours?, (A) Lactic acid, (B) Glycogen, (C) Ketone bodies, (D) Non-esterified fatty acids, ANS: D, , 49. Which one of the following is not a deficiency disorder?, (A) Beriberi, (B) Night Blindness, (C) Poliomyelitis, (D) Pernicious Anaemia, ANS: C

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50. Electrophoresis of a purified protein in SDS-PAGE in the presence of 2- mercaptoethanol yields, two bands of 35 kDa and 45 kDa. However, in a gel filtration chromatography, the same protein, elutes as 80 kDa. What conclusion will you draw from the results?, (A) The protein is not purified to homogeneity., (B) Two bands generated in SDS-PAGE due to degradation., (C) The protein is a homodimer, (D)The protein is a heterodimer, , ANS: D, Heterodimer = 35 + 45 = 80 kDa., , SOLVED BY: PURUSHOTAM SINGH THAKUR, , THANK YOU