DBT BET JRF Exam Question Paper 2016 and 2017
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About This Presentation
DBT BET JRF Exam Question Paper with Answer Key
2016 and 2017
https://www.biologyexams4u.com/
Slide Content
1
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
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
3. What number should appear next in this series: 8, 12, 10, 14, 12, 16….?
(A) 10
(B) 14
(C) 18
(D) 12
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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
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
3. What number should appear next in this series: 8, 12, 10, 14, 12, 16….?
(A) 10
(B) 14
(C) 18
(D) 12
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
2
5. Identify from the options given below the unwritten number (?) in the series: 360, ?,
180, 60, 15, 3
(A) 180
(B) 300
(C) 360
(D) 320
6. For the chemical reaction C2H5OH(ℓ) + 3O2 (g) → 2CO2 + 3H2O (ℓ), if change in
internal energy at 298K is 1368 KJ/mole, the change in enthalpy will be
(A) greater than 1368
(B) less than 1368
(C) equal to zero
(D) negative
7. If a drop of sweat evaporates from body, the work done is
(A) w = ∆U
(B) w = ∆H
(C) w = ∆G
(D) w = P(V2-V1)
8. Dissociation of an aqueous acid (HA) gives HA + H2O H3O
+
+ A
-
, then pKa is
(A) ∆G°R
(B) ∆ H°R
(C) ∆G°/2.303RT
(D) Equilibrium constant Ka
9. The sides of a rectangle are in the ratio of 4:3 and its area is 108 cm
2
. The
perimeter of the rectangle in cm is
(A) 22
(B) 32
(C) 42
(D) 52
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
5. Identify from the options given below the unwritten number (?) in the series: 360, ?,
180, 60, 15, 3
(A) 180
(B) 300
(C) 360
(D) 320
6. For the chemical reaction C2H5OH(ℓ) + 3O2 (g) → 2CO2 + 3H2O (ℓ), if change in
internal energy at 298K is 1368 KJ/mole, the change in enthalpy will be
(A) greater than 1368
(B) less than 1368
(C) equal to zero
(D) negative
7. If a drop of sweat evaporates from body, the work done is
(A) w = ∆U
(B) w = ∆H
(C) w = ∆G
(D) w = P(V2-V1)
8. Dissociation of an aqueous acid (HA) gives HA + H2O H3O
+
+ A
-
, then pKa is
(A) ∆G°R
(B) ∆ H°R
(C) ∆G°/2.303RT
(D) Equilibrium constant Ka
9. The sides of a rectangle are in the ratio of 4:3 and its area is 108 cm
2
. The
perimeter of the rectangle in cm is
(A) 22
(B) 32
(C) 42
(D) 52
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
3
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
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
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
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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
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
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
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
4
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
16. The molarity of a 15 % of NaCl solution in water is ----.
(A) 2.56
(B) 0.256
(C) 25.6
(D) 0.025
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
18. Which one of the following bacterial cell suspension at A540nm = 0.2 will have the
maximum number of cells per ml?
(A) Mycoplasma
(B) Pseudomonas
(C) Streptococcus
(D) Bacillus
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) 1.479
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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
16. The molarity of a 15 % of NaCl solution in water is ----.
(A) 2.56
(B) 0.256
(C) 25.6
(D) 0.025
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
18. Which one of the following bacterial cell suspension at A540nm = 0.2 will have the
maximum number of cells per ml?
(A) Mycoplasma
(B) Pseudomonas
(C) Streptococcus
(D) Bacillus
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) 1.479
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
5
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
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
22. Zymomonas mobilis cells are grown in a chemostat 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
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
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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
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
22. Zymomonas mobilis cells are grown in a chemostat 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
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
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
6
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
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
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.
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
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?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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
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
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.
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
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?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
7
(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.
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)
31. Which one of the following side chains of an amino acid is responsible for
fluorescence in proteins?
(A) Indole ring
(B) Guanidino group
(C) Phenolic group
(D) Imidazole group
32. DNA molecules labeled with
15
N and
14
N can be separated by
(A) Pulse field gel electrophoresis
(B) Density gradient ultracentrifugation
(C) Capillary electrophoresis
(D) Differential centrifugation
33. Match the chromatographic technique from Group A with the appropriate
elution conditions given in Group B
Group A Group B
(P) Chromatofocusing (i) Decreasing [(NH4)2SO4] gradient
(Q) DEAE-Sephadex (ii) pH gradient
(R) G-150 Sephadex (iii) Isocratic gradient
(S) Phenyl Speharose (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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(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.
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)
31. Which one of the following side chains of an amino acid is responsible for
fluorescence in proteins?
(A) Indole ring
(B) Guanidino group
(C) Phenolic group
(D) Imidazole group
32. DNA molecules labeled with
15
N and
14
N can be separated by
(A) Pulse field gel electrophoresis
(B) Density gradient ultracentrifugation
(C) Capillary electrophoresis
(D) Differential centrifugation
33. Match the chromatographic technique from Group A with the appropriate
elution conditions given in Group B
Group A Group B
(P) Chromatofocusing (i) Decreasing [(NH4)2SO4] gradient
(Q) DEAE-Sephadex (ii) pH gradient
(R) G-150 Sephadex (iii) Isocratic gradient
(S) Phenyl Speharose (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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
8
34. Cytoskeletal organization of a cell is regulated by
(A) Ras GTPase
(B) Rab GTPase
(C) Rho GTPase
(D) Ran GTPase
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) sphinogolipid
(B) saturated fatty acid
(C) unsaturated fatty acid
(D) cholesterol
36. In glycolysis, the conversion of 1 mol of fructose 1,6-biphosphate to 2 mol of
pyruvate results in the formation of
(A) 1 mol NAD+ and 2 mol of ATP
(B) 2 mol of NAD+ and 4 mol of ATP
(C) 2 mol of NADH and 2 mol of ATP
(D) 2 mol of NADH and 4 mol of ATP
37. In diabetic ketoacidosis, increase in which of the following would cause
elevated production of ketone bodies?
(A) Proteolysis
(B) Urea production
(C) Insulin release
(D) Lipolysis
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
34. Cytoskeletal organization of a cell is regulated by
(A) Ras GTPase
(B) Rab GTPase
(C) Rho GTPase
(D) Ran GTPase
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) sphinogolipid
(B) saturated fatty acid
(C) unsaturated fatty acid
(D) cholesterol
36. In glycolysis, the conversion of 1 mol of fructose 1,6-biphosphate to 2 mol of
pyruvate results in the formation of
(A) 1 mol NAD+ and 2 mol of ATP
(B) 2 mol of NAD+ and 4 mol of ATP
(C) 2 mol of NADH and 2 mol of ATP
(D) 2 mol of NADH and 4 mol of ATP
37. In diabetic ketoacidosis, increase in which of the following would cause
elevated production of ketone bodies?
(A) Proteolysis
(B) Urea production
(C) Insulin release
(D) Lipolysis
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
9
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.
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)
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%
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.
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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.
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)
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%
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.
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
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
10
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 labeled
lysosomes.
(B) GFP-Mycobacterium will be separated from lysotracker Red labeled
lysosomes.
(C) GFP-Mycobacterium will not be detected as they are degraded in the
cell.
(D) Lysotracker Red labeled lysosomes will be degraded in GFP -
Mycobacterium infected cells
45. A linear DNA fragment which has 3 restriction sites for BamH1, is labeled
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 labeled
and unlabeled fragments will be produced?
(A) 3 labeled and 4 unlabeled
(B) 3 labeled and 5 unlabeled
(C) 4 labeled and 5 unlabeled
(D) 4 labeled and 6 unlabeled
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
47. Approximately how many molecules of CO2 are produced daily by oxidative
metabolism in adult human? Avagadro’s number is 6.02 X 10
23
.
(A) 1.2 X 10
24
(B) 1.2 X 10
23
(C) 1.2 X 10
26
(D) 1.2 X 10
25
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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 labeled
lysosomes.
(B) GFP-Mycobacterium will be separated from lysotracker Red labeled
lysosomes.
(C) GFP-Mycobacterium will not be detected as they are degraded in the
cell.
(D) Lysotracker Red labeled lysosomes will be degraded in GFP -
Mycobacterium infected cells
45. A linear DNA fragment which has 3 restriction sites for BamH1, is labeled
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 labeled
and unlabeled fragments will be produced?
(A) 3 labeled and 4 unlabeled
(B) 3 labeled and 5 unlabeled
(C) 4 labeled and 5 unlabeled
(D) 4 labeled and 6 unlabeled
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
47. Approximately how many molecules of CO2 are produced daily by oxidative
metabolism in adult human? Avagadro’s number is 6.02 X 10
23
.
(A) 1.2 X 10
24
(B) 1.2 X 10
23
(C) 1.2 X 10
26
(D) 1.2 X 10
25
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
11
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
49. Which one of the following is not a deficiency disorder?
(A) Beriberi
(B) Night Blindness
(C) Poliomyelitis
(D) Pernicious Anemia
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
Section-B
(Answer any 50 out of 150 questions)
51. Which one of the following virus is the odd one out?
(A) Epstein-Barr Virus
(B) Human Immunodeficiency Virus.
(C) Human Papilloma Virus
(D) Human T-cell leukemia Virus
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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
49. Which one of the following is not a deficiency disorder?
(A) Beriberi
(B) Night Blindness
(C) Poliomyelitis
(D) Pernicious Anemia
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
Section-B
(Answer any 50 out of 150 questions)
51. Which one of the following virus is the odd one out?
(A) Epstein-Barr Virus
(B) Human Immunodeficiency Virus.
(C) Human Papilloma Virus
(D) Human T-cell leukemia Virus
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
12
52. DAPI 4',6-diamidino-2-phenylindole is a fluorescent stain used for staining of:
(A) DNA only
(B) DNA and RNA
(C) DNA and proteins
(D) protein Only
53. During antigen presentation by antigen presenting cells, viral antigens are
processed and presented to T cells in the context of MHC antigens. Viral
antigens are processed
(A) By proteasomes and are presented along with MHC class I antigen
(B) In endosomes and are presented along with MHC class I antigen
(C) In endosomes and are presented along with MHC class II antigen
(D) By proteasomes and are presented along with MHC class II antigen
54. Which of the following is not specifically protected by intellectual property
legislation?
(A) Industrial designs.
(B) Patents.
(C) Trademarks.
(D) Trade secrets.
55. Teichoic acid possibly plays a role in growth of bacterial cell by regulating the
activity of
(A) lysin
(B) autolysin
(C) peptidase
(D) protease
56. The fungus that causes athlete’s foot is
(A) Aspergillus
(B) Trichophyton
(C) Neurospora
(D) Trichoderma
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
52. DAPI 4',6-diamidino-2-phenylindole is a fluorescent stain used for staining of:
(A) DNA only
(B) DNA and RNA
(C) DNA and proteins
(D) protein Only
53. During antigen presentation by antigen presenting cells, viral antigens are
processed and presented to T cells in the context of MHC antigens. Viral
antigens are processed
(A) By proteasomes and are presented along with MHC class I antigen
(B) In endosomes and are presented along with MHC class I antigen
(C) In endosomes and are presented along with MHC class II antigen
(D) By proteasomes and are presented along with MHC class II antigen
54. Which of the following is not specifically protected by intellectual property
legislation?
(A) Industrial designs.
(B) Patents.
(C) Trademarks.
(D) Trade secrets.
55. Teichoic acid possibly plays a role in growth of bacterial cell by regulating the
activity of
(A) lysin
(B) autolysin
(C) peptidase
(D) protease
56. The fungus that causes athlete’s foot is
(A) Aspergillus
(B) Trichophyton
(C) Neurospora
(D) Trichoderma
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
13
57. The following amino acid is
(A) hydrophobic
(B) nucleophilic
(C) aromatic
(D) basic
58. A DNA fragment was cloned at the EcoRI restriction site of a plasmid
vector that contains antibiotic resistance genes, kan
R
and spec
R
. It was
observed that all the positive clones (containing the DNA fragment of interest)
grew on medium with kanamycin but not on media containing spectinomycin.
In the absence of any other confounding factors, which one of the following
statements would explain this observation?
(A) The spec
R
gene contains a mutation, which generates a stop codon
within the gene.
(B) The EcoRI site was located within the spec
R
gene.
(C) The cloned fragment of interest produces a protein, which confers
resistance to spectinomycin.
(D) The cloned DNA fragment was lethal to the cell.
59. A cDNA encoding an eukaryotic gene was ligated to an expression vector
which was then introduced into E. coli for expression of the protein. The
experiment resulted in poor expression of the protein. However, expression
improved significantly in a Rosetta strain of E. coli which demonstrate that the
initial poor level expression was due to
(A) absence of capping at 5’ end of the transcript
(B) absence of polyadenylation at 3’ end of the transcript
(C) codon bias
(D) lack of splicing machinery in E. coli.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
57. The following amino acid is
(A) hydrophobic
(B) nucleophilic
(C) aromatic
(D) basic
58. A DNA fragment was cloned at the EcoRI restriction site of a plasmid
vector that contains antibiotic resistance genes, kan
R
and spec
R
. It was
observed that all the positive clones (containing the DNA fragment of interest)
grew on medium with kanamycin but not on media containing spectinomycin.
In the absence of any other confounding factors, which one of the following
statements would explain this observation?
(A) The spec
R
gene contains a mutation, which generates a stop codon
within the gene.
(B) The EcoRI site was located within the spec
R
gene.
(C) The cloned fragment of interest produces a protein, which confers
resistance to spectinomycin.
(D) The cloned DNA fragment was lethal to the cell.
59. A cDNA encoding an eukaryotic gene was ligated to an expression vector
which was then introduced into E. coli for expression of the protein. The
experiment resulted in poor expression of the protein. However, expression
improved significantly in a Rosetta strain of E. coli which demonstrate that the
initial poor level expression was due to
(A) absence of capping at 5’ end of the transcript
(B) absence of polyadenylation at 3’ end of the transcript
(C) codon bias
(D) lack of splicing machinery in E. coli.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
14
60. In Drosophila embryogenesis, the signal received from Gurken proteins
by follicle cells results in posteriorization of these cells. Knocking out the gene
for Gurken in Drosophila will NOT result in failure of
(A) Accumulation of maternal mRNAs
(B) rearrangement of maternal mRNA at the two ends of the embryo
(C) establishment of gradients of Gurken
(D) establishment of anterior-posterior axis
61. A patient suffering from an infectious disease had high levels of TNF
alpha and IL-6. Assuming there are no other interactions, which T cell
population is likely to dominate in his blood sample?
(A) Th2
(B) Th17
(C) Treg
(D) Naive T cells
62. The niche of stem cells in human skin is the
(A) dermal papillae region of hair follicle
(B) bulge region of the hair follicle
(C) sebaceous gland
(D) malphigian layer
63. In persons suffering from sickle cell anemia, the 6
th
amino acid of beta
globin protein is valine instead of glutamic acid. This has occurred due to
substitution mutation leading to a change in glutamic acid codon to valine
codon in the gene. The mutation is a
(A) transition where a purine base changes to another purine base
(B) transition where pyrimidine base changes to another pyrimidine base
(C) transversion where pyrimidine base changes to a purine base
(D) transition where purine base changes to a pyrimidine base
64. A covalently closed circular DNA was in relaxed state in water at 30
0
C.
What will happen to the supercoiling state of the molecule, if water temp
increases to 60
0
C or decreases to 10
0
C?
(A) The DNA will remain in relaxed state also at 60
0
C as well as at 10
0
C
(B) The DNA will undergo positive supercoiling at 60
0
C as well as at 10
0
C
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
60. In Drosophila embryogenesis, the signal received from Gurken proteins
by follicle cells results in posteriorization of these cells. Knocking out the gene
for Gurken in Drosophila will NOT result in failure of
(A) Accumulation of maternal mRNAs
(B) rearrangement of maternal mRNA at the two ends of the embryo
(C) establishment of gradients of Gurken
(D) establishment of anterior-posterior axis
61. A patient suffering from an infectious disease had high levels of TNF
alpha and IL-6. Assuming there are no other interactions, which T cell
population is likely to dominate in his blood sample?
(A) Th2
(B) Th17
(C) Treg
(D) Naive T cells
62. The niche of stem cells in human skin is the
(A) dermal papillae region of hair follicle
(B) bulge region of the hair follicle
(C) sebaceous gland
(D) malphigian layer
63. In persons suffering from sickle cell anemia, the 6
th
amino acid of beta
globin protein is valine instead of glutamic acid. This has occurred due to
substitution mutation leading to a change in glutamic acid codon to valine
codon in the gene. The mutation is a
(A) transition where a purine base changes to another purine base
(B) transition where pyrimidine base changes to another pyrimidine base
(C) transversion where pyrimidine base changes to a purine base
(D) transition where purine base changes to a pyrimidine base
64. A covalently closed circular DNA was in relaxed state in water at 30
0
C.
What will happen to the supercoiling state of the molecule, if water temp
increases to 60
0
C or decreases to 10
0
C?
(A) The DNA will remain in relaxed state also at 60
0
C as well as at 10
0
C
(B) The DNA will undergo positive supercoiling at 60
0
C as well as at 10
0
C
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
15
(C ) The DNA will undergo positive supercoiling at 60
0
C and will undergo
negative supercoiling at 10
0
C
(D) The DNA will undergo negative supercoiling at 60
0
C and will undergo
positive supercoiling at 10
0
C
65. 1000 color blind males underwent random mating with 1000 normal
females (neither carrier nor color blind). After how many generations, the
allele frequency in both the sexes in the population will reach equilibrium?
(A) One
(B) Two
(C ) Three
(D ) More than three
66. Acclimatization to high altitude in humans induces differentiation of
(A) basophil
(B) platelets
(C) erythrocytes
(D) neutrophils
67. The protection against smallpox afforded by prior infection with cowpox
presents
(A) specificity.
(B) antigenic cross-reactivity.
(C) innate immunity.
(D) passive protection.
68. What is glycosuria?
(A) Low amount of sugar in urine
(B) Low amount of fat in urine
(C) Average amount of carbohydrate in urine
(D) High amount of sugar in urine
69. The mechanism that permits immunoglobulins to be synthesized in either
a membrane bound or secreted form is
(A) allelic exclusion
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(C ) The DNA will undergo positive supercoiling at 60
0
C and will undergo
negative supercoiling at 10
0
C
(D) The DNA will undergo negative supercoiling at 60
0
C and will undergo
positive supercoiling at 10
0
C
65. 1000 color blind males underwent random mating with 1000 normal
females (neither carrier nor color blind). After how many generations, the
allele frequency in both the sexes in the population will reach equilibrium?
(A) One
(B) Two
(C ) Three
(D ) More than three
66. Acclimatization to high altitude in humans induces differentiation of
(A) basophil
(B) platelets
(C) erythrocytes
(D) neutrophils
67. The protection against smallpox afforded by prior infection with cowpox
presents
(A) specificity.
(B) antigenic cross-reactivity.
(C) innate immunity.
(D) passive protection.
68. What is glycosuria?
(A) Low amount of sugar in urine
(B) Low amount of fat in urine
(C) Average amount of carbohydrate in urine
(D) High amount of sugar in urine
69. The mechanism that permits immunoglobulins to be synthesized in either
a membrane bound or secreted form is
(A) allelic exclusion
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
16
(B) codominant expression
(C) differential RNA processing
(D) class switching
70. Which of the following is NOT involved in antigen- antibody complexes?
(A) Van der Waals forces
(B) Hydrogen bonds
(C) Covalent bonds
(D) Electrostatic interactions
71. Iron uptake by the cells from extracellular environment is mediated
through
(A) LDL receptor
(B) Mannose receptor
(C) Transferrin receptor
(D) Mannose 6-phosphate receptor
72. Which of following viruses causes liver cancer?
(A) Papilloma virus.
(B) Herpes simplex virus.
(C) Hepatitis B virus.
(D) Hepatitis C virus.
73. Which one of the following enzymes is an established intracellular anti-
oxidant?
(A) Lactate dehydrogenase
(B) Phenylalanine hydroxylase
(C) Superoxide dismutase
(D) γ-Secretase
74. A given cytokine has different biological effects on different target cells.
This is known as
(A) pleiotropy
(B) redundancy
(C) synergy
(D) totipotency
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(B) codominant expression
(C) differential RNA processing
(D) class switching
70. Which of the following is NOT involved in antigen- antibody complexes?
(A) Van der Waals forces
(B) Hydrogen bonds
(C) Covalent bonds
(D) Electrostatic interactions
71. Iron uptake by the cells from extracellular environment is mediated
through
(A) LDL receptor
(B) Mannose receptor
(C) Transferrin receptor
(D) Mannose 6-phosphate receptor
72. Which of following viruses causes liver cancer?
(A) Papilloma virus.
(B) Herpes simplex virus.
(C) Hepatitis B virus.
(D) Hepatitis C virus.
73. Which one of the following enzymes is an established intracellular anti-
oxidant?
(A) Lactate dehydrogenase
(B) Phenylalanine hydroxylase
(C) Superoxide dismutase
(D) γ-Secretase
74. A given cytokine has different biological effects on different target cells.
This is known as
(A) pleiotropy
(B) redundancy
(C) synergy
(D) totipotency
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
17
75. Influenza virus contains 11 proteins, of which the two major glycoproteins
are H and N. What do H and N stand for in influenza strains nomenclature?
(A) H: Hemagglutinin; N: Nucleoprotein
(B) H: Hemagglutinin; N: Neuraminidase
(C) H: Hyaluronidase; N: Nucleoprotein
(D) H: Hyaluronidase; N: Neuraminidase
76. Inosine in the tRNA anticodon will base pair with all except which one of
the following bases in the codon of mRNA?
(A) adenine
(B) uracil
(C) cytosine
(D) guanine
77. In Electrophoresis if the buffer pH is above the isoelectric point of the
protein, the protein will
(A) migrate towards the anode
(B) migrate towards the cathode
(C) not migrate at all
(D) migrate partly to anode and partly to cathode
78. In the ABO blood group system in humans, alleles I
A
and I
B
are
codominant and both are dominant to the allele i. If a type B with genotype (I
B
i
) woman marries a type A with genotype (I
A
i ). The probable children to the
couple would be of the type
(A) A, B, O and AB
(B) A and B only
(C) A and AB only
(D) O and AB only
79. Circulating testosterone is mostly present in bound form, which of the
following proteins binds weakly to the major amount of circulating
testosterone?
(A) Transferrin
(B) Prothrombin
(C) Fibrinogen
(D) Albumin
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
75. Influenza virus contains 11 proteins, of which the two major glycoproteins
are H and N. What do H and N stand for in influenza strains nomenclature?
(A) H: Hemagglutinin; N: Nucleoprotein
(B) H: Hemagglutinin; N: Neuraminidase
(C) H: Hyaluronidase; N: Nucleoprotein
(D) H: Hyaluronidase; N: Neuraminidase
76. Inosine in the tRNA anticodon will base pair with all except which one of
the following bases in the codon of mRNA?
(A) adenine
(B) uracil
(C) cytosine
(D) guanine
77. In Electrophoresis if the buffer pH is above the isoelectric point of the
protein, the protein will
(A) migrate towards the anode
(B) migrate towards the cathode
(C) not migrate at all
(D) migrate partly to anode and partly to cathode
78. In the ABO blood group system in humans, alleles I
A
and I
B
are
codominant and both are dominant to the allele i. If a type B with genotype (I
B
i
) woman marries a type A with genotype (I
A
i ). The probable children to the
couple would be of the type
(A) A, B, O and AB
(B) A and B only
(C) A and AB only
(D) O and AB only
79. Circulating testosterone is mostly present in bound form, which of the
following proteins binds weakly to the major amount of circulating
testosterone?
(A) Transferrin
(B) Prothrombin
(C) Fibrinogen
(D) Albumin
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
18
80. T. brucei complex transmitted to man by byte of several species of tsetse fly
can cause
(A) Thymoma
(B) Sleeping sickness
(C) Toxoplasmosis
(D) Leishmaniasis
81. In immune responses at mucosal surfaces, particularly in the context of gut
immunology, which of the following is NOT true?
(A) Mucosal surfaces have continuous and close contact with numerous
and diverse commensal microorganisms.
(B) M cells transport microbes and antigens from the gut lumen to gut
associated lymphoid tissue.
(C) B cells activated in mucosal tissues give rise to plasma cells
preferentially secreting IgG.
(D) Inflammation is prevented at mucosal surfaces, generally by
recruitment of T regulatory cells.
82. You have developed a H-2 k/d mouse model with grafted thymus for studying
T cell maturation. The phenotype of the grafted thymus is H-2d/d. You have
immunized the mouse with antigens derived from Listeria monocytogenes and
would like to assess the activation of cytotoxic T cells. Which among the
following would be your choice of target cell?
(A) Cells expressing MHC class I molecules taken from H-2k/k mouse
(B) Cells expressing MHC class II molecules taken from H-2k/k mouse
(C) Cells expressing MHC class I molecules taken from H-2d/k mouse
(D) Cells expressing MHC class II molecules taken from H-2d/k mouse
83. While analyzing protein sequences of rpoB gene among different bacteria,
different changes in amino acid sequences were observed among the
homologs due to base substitution. If transition is more frequent than
transversion, single mutation is more frequent than double mutation, which
one of the following changes will occur most frequently than others?
(A) Ala Gly
(B) Ala Ser
(C) Ala Leu
(D) Ala Val
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
80. T. brucei complex transmitted to man by byte of several species of tsetse fly
can cause
(A) Thymoma
(B) Sleeping sickness
(C) Toxoplasmosis
(D) Leishmaniasis
81. In immune responses at mucosal surfaces, particularly in the context of gut
immunology, which of the following is NOT true?
(A) Mucosal surfaces have continuous and close contact with numerous
and diverse commensal microorganisms.
(B) M cells transport microbes and antigens from the gut lumen to gut
associated lymphoid tissue.
(C) B cells activated in mucosal tissues give rise to plasma cells
preferentially secreting IgG.
(D) Inflammation is prevented at mucosal surfaces, generally by
recruitment of T regulatory cells.
82. You have developed a H-2 k/d mouse model with grafted thymus for studying
T cell maturation. The phenotype of the grafted thymus is H-2d/d. You have
immunized the mouse with antigens derived from Listeria monocytogenes and
would like to assess the activation of cytotoxic T cells. Which among the
following would be your choice of target cell?
(A) Cells expressing MHC class I molecules taken from H-2k/k mouse
(B) Cells expressing MHC class II molecules taken from H-2k/k mouse
(C) Cells expressing MHC class I molecules taken from H-2d/k mouse
(D) Cells expressing MHC class II molecules taken from H-2d/k mouse
83. While analyzing protein sequences of rpoB gene among different bacteria,
different changes in amino acid sequences were observed among the
homologs due to base substitution. If transition is more frequent than
transversion, single mutation is more frequent than double mutation, which
one of the following changes will occur most frequently than others?
(A) Ala Gly
(B) Ala Ser
(C) Ala Leu
(D) Ala Val
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
19
84. If G2/M check point and DNA repair proteins are working optimally in a cell
and there are some DNA damages during G2 phase , which one of the
following would be the response of the cells to damage?
(A) Cell cycle will continue and cell will undergo mitosis.
(B) Cell will undergo cytokinesis.
(C) Cells will enter into mitosis and arrest at metaphase.
(D) Cells will arrest at G2 until the damages are repaired.
85. Transferrin binds with its receptor and recruits clathrin via which one of the
following proteins?
(A) Caveolin
(B) AP1 adaptor
(C) AP2 adaptor
(D) Dynamin
86. Which one of the following is the major product of fatty acid synthase?
(A) Acetyl-CoA
(B) Palmitoyl-CoA
(C) Acetoacetate
(D) Palmitate
87. Intracellular transport in mammalian cells through vesicular fusion is
regulated by which among the following GTPases?
(A) Rho
(B) Ran
(C) Rab
(D) Ras
88. Which one of the following processes is NOT regulated by the kidney?
(A) Body temperature
(B) Blood pressure
(C) Coordinated muscle movement
(D) Ionic balance of the blood
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
84. If G2/M check point and DNA repair proteins are working optimally in a cell
and there are some DNA damages during G2 phase , which one of the
following would be the response of the cells to damage?
(A) Cell cycle will continue and cell will undergo mitosis.
(B) Cell will undergo cytokinesis.
(C) Cells will enter into mitosis and arrest at metaphase.
(D) Cells will arrest at G2 until the damages are repaired.
85. Transferrin binds with its receptor and recruits clathrin via which one of the
following proteins?
(A) Caveolin
(B) AP1 adaptor
(C) AP2 adaptor
(D) Dynamin
86. Which one of the following is the major product of fatty acid synthase?
(A) Acetyl-CoA
(B) Palmitoyl-CoA
(C) Acetoacetate
(D) Palmitate
87. Intracellular transport in mammalian cells through vesicular fusion is
regulated by which among the following GTPases?
(A) Rho
(B) Ran
(C) Rab
(D) Ras
88. Which one of the following processes is NOT regulated by the kidney?
(A) Body temperature
(B) Blood pressure
(C) Coordinated muscle movement
(D) Ionic balance of the blood
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
20
89. Rapid removal of large quantities of blood will cause an animal to go into
hypovolemic shock and may even cause death. If the weight of a mouse is
20 gms, what is the amount of maximum blood one can withdraw from a
single bleed to prevent the mouse from going into hypovolemic shock?
(A) 2.0 ml
(B) 3.0 ml
(C) 0.2 ml
(D) 1.0 ml
90. In order to have a desired shRNA cassette integrated in target cells, which of
the following gene transfer vectors is preferable?
(A) Baculovirus vector
(B) Herpes virus vector
(C) Adenoviral vector
(D) Lentiviral vector
91. Which one of the following statements regarding naturally occurring
Agrobacterium strains is true?
(A) The T-DNA region of Agrobacterium does not contain functional
genes.
(B) All the virulence genes of Agrobacterium are constitutively
expressed.
(C) Agrobacterium-induced galls require exogenous application of
phytohormones for their growth.
(D) Agrobacterium-induced galls in nature do not require bacterial
persistence for their growth.
92. The integration of T-DNA in the plant nuclear genome is most likely due to
(A) homologous recombination
(B) non-homologous recombination
(C) non-homologous end joining
(D) single-stranded recombination during transcription
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89. Rapid removal of large quantities of blood will cause an animal to go into
hypovolemic shock and may even cause death. If the weight of a mouse is
20 gms, what is the amount of maximum blood one can withdraw from a
single bleed to prevent the mouse from going into hypovolemic shock?
(A) 2.0 ml
(B) 3.0 ml
(C) 0.2 ml
(D) 1.0 ml
90. In order to have a desired shRNA cassette integrated in target cells, which of
the following gene transfer vectors is preferable?
(A) Baculovirus vector
(B) Herpes virus vector
(C) Adenoviral vector
(D) Lentiviral vector
91. Which one of the following statements regarding naturally occurring
Agrobacterium strains is true?
(A) The T-DNA region of Agrobacterium does not contain functional
genes.
(B) All the virulence genes of Agrobacterium are constitutively
expressed.
(C) Agrobacterium-induced galls require exogenous application of
phytohormones for their growth.
(D) Agrobacterium-induced galls in nature do not require bacterial
persistence for their growth.
92. The integration of T-DNA in the plant nuclear genome is most likely due to
(A) homologous recombination
(B) non-homologous recombination
(C) non-homologous end joining
(D) single-stranded recombination during transcription
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
21
93. Variation in transgene expression levels among independent transgenic
events generated using the same transgene construct could be due to
(i) position effects
(ii) strength of the promoter used to express the transgene
(iii) variation in copy number of the transgene
(iv) presence of restriction enzyme sites in the transgene sequence
(A) (i) and (iv)
(B) (ii) and (iii)
(C) (i) and (iii)
(D) (iii) and (iv)
94. A typical flowering plant has a combination of sporophytic (diploid) and
gametophytic (haploid) tissues. Which of the following types of plant
tissues do NOT contribute to the formation of a mature fruit?
(A) sporophytic tissue from the previous generation
(B) gametophytic tissue from the previous generation
(C) sporophytic tissue from the next generation
(D) gametophytic tissue from the next generation
95. The submerged leaves of an aquatic plant are feathery (to avoid damage
due to flowing underwater currents) while leaves on the surface are
padded (to help in floatation). The correct statement related to the above
phenotype is
(A) The gene content of the submerged leaves is different from that of leaves
on the surface.
(B) The submerged and floating leaves have differential expression of
structural and/or regulatory genes.
(C) The observed phenotypic variation in leaves is not influenced by
diverse growth conditions.
(D) These phenotypic variations are due to transient somatic mutations.
96. Which one of the following statements related to the development of
insect-resistant transgenic plants is correct?
(A) The targeted insects cannot develop resistance against transgenic
plants generated using a single candidate gene.
(B) A transgenic plant developed using multiple genes might be more
effective in inducing and maintaining resistance.
(C) The level of transgene expression does not influence the efficacy of
the transgenic plant.
(D) It is easier to propagate and maintain multi-copy integration events
than single copy events of insect resistant plants.
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93. Variation in transgene expression levels among independent transgenic
events generated using the same transgene construct could be due to
(i) position effects
(ii) strength of the promoter used to express the transgene
(iii) variation in copy number of the transgene
(iv) presence of restriction enzyme sites in the transgene sequence
(A) (i) and (iv)
(B) (ii) and (iii)
(C) (i) and (iii)
(D) (iii) and (iv)
94. A typical flowering plant has a combination of sporophytic (diploid) and
gametophytic (haploid) tissues. Which of the following types of plant
tissues do NOT contribute to the formation of a mature fruit?
(A) sporophytic tissue from the previous generation
(B) gametophytic tissue from the previous generation
(C) sporophytic tissue from the next generation
(D) gametophytic tissue from the next generation
95. The submerged leaves of an aquatic plant are feathery (to avoid damage
due to flowing underwater currents) while leaves on the surface are
padded (to help in floatation). The correct statement related to the above
phenotype is
(A) The gene content of the submerged leaves is different from that of leaves
on the surface.
(B) The submerged and floating leaves have differential expression of
structural and/or regulatory genes.
(C) The observed phenotypic variation in leaves is not influenced by
diverse growth conditions.
(D) These phenotypic variations are due to transient somatic mutations.
96. Which one of the following statements related to the development of
insect-resistant transgenic plants is correct?
(A) The targeted insects cannot develop resistance against transgenic
plants generated using a single candidate gene.
(B) A transgenic plant developed using multiple genes might be more
effective in inducing and maintaining resistance.
(C) The level of transgene expression does not influence the efficacy of
the transgenic plant.
(D) It is easier to propagate and maintain multi-copy integration events
than single copy events of insect resistant plants.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
22
97. When the gene and the promoter used for modification of a plant using
transformation technology are derived from sexually compatible species,
the modified plant thus generated is known as a
(A) cisgenic plant
(B) selfgenic plant
(C) intragenic plant
(D) hemilogous plant
98. A transgenic plant segregates in a 3:1 ratio for the transgenic: non-
transgenic phenotype in T1 progeny derived by self-pollination. This
indicates that
(A) the transgenic plant cannot contain more than one copy of the
transgene.
(B) the transgenic plant may contain two unlinked copies of the
transgene.
(C) the transgenic plant contains more than two unlinked copies of the
transgene.
(D) the transgenic plant may contain two tightly linked copies of the
transgene.
99. Which one of the following reporter genes can be used for real-time
visualization of living cells/tissues in transgenic plants?
(A) gus
(B) gfp
(C) cat
(D) beta-galactosidase
100. Which of the following statements represent correct features of sexual and
asexual modes of reproduction in flowering plants?
(i) In sexual reproduction, progeny are genetically different from each
other.
(ii) In asexual mode of reproduction, progeny are genetically identical
to each other but different from the parent.
(iii) Development of greater adaptive ability is possible only in case of
progeny derived by sexual reproduction.
(iv) A minor change in the habitat may adversely affect all offspring
derived by asexual reproduction.
(A) i, iii and iv
(B) i, ii and iii
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
97. When the gene and the promoter used for modification of a plant using
transformation technology are derived from sexually compatible species,
the modified plant thus generated is known as a
(A) cisgenic plant
(B) selfgenic plant
(C) intragenic plant
(D) hemilogous plant
98. A transgenic plant segregates in a 3:1 ratio for the transgenic: non-
transgenic phenotype in T1 progeny derived by self-pollination. This
indicates that
(A) the transgenic plant cannot contain more than one copy of the
transgene.
(B) the transgenic plant may contain two unlinked copies of the
transgene.
(C) the transgenic plant contains more than two unlinked copies of the
transgene.
(D) the transgenic plant may contain two tightly linked copies of the
transgene.
99. Which one of the following reporter genes can be used for real-time
visualization of living cells/tissues in transgenic plants?
(A) gus
(B) gfp
(C) cat
(D) beta-galactosidase
100. Which of the following statements represent correct features of sexual and
asexual modes of reproduction in flowering plants?
(i) In sexual reproduction, progeny are genetically different from each
other.
(ii) In asexual mode of reproduction, progeny are genetically identical
to each other but different from the parent.
(iii) Development of greater adaptive ability is possible only in case of
progeny derived by sexual reproduction.
(iv) A minor change in the habitat may adversely affect all offspring
derived by asexual reproduction.
(A) i, iii and iv
(B) i, ii and iii
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
23
(C) ii and iv
(D) ii, iii and iv
101. Which one of the options given below represents the correct combination
of plant defense responses listed in Column A and Column B?
Column A Column B
(i) Anatomical defense (1) salicylic acid
(ii) Elicitors (2) chitinases
(iii) Systemic response (3) waxes
(iv) PR proteins (4) beta-glucans
(A) i – 2, ii – 3, iii – 1, iv - 4
(B) i - 3, ii – 4, iii – 1, iv – 2
(C) i – 4, ii – 1, iii – 2, iv - 3
(D) ii – 3, iii – 4, i – 2, iv - 1
102. Non-conditional, negative selection marker genes are usually expressed
under a
(A) developmentally regulated promoter
(B) substrate-induced promoter
(C) constitutive promoter
(D) minimal promoter
103. A method in which a strong enhancer is randomly inserted in a plant
genome by transformation, resulting in mutant plants with dominant
phenotypes, is known as
(A) enhancer trapping
(B) TILLING
(C) activation tagging
(D) gene trapping
104. Which one of the following organisms is used by plant biologists to study
translocation in phloem?
(A) aphids
(B) nematodes
(C) grasshoppers
(D) butterflies
105. In a genetic transformation experiment, a researcher failed to add the
antibiotic selection agent in the shoot regeneration medium for selection of
transgenic plants. In the absence of any other confounding factors, which
one of the following statements is expected to be correct?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(C) ii and iv
(D) ii, iii and iv
101. Which one of the options given below represents the correct combination
of plant defense responses listed in Column A and Column B?
Column A Column B
(i) Anatomical defense (1) salicylic acid
(ii) Elicitors (2) chitinases
(iii) Systemic response (3) waxes
(iv) PR proteins (4) beta-glucans
(A) i – 2, ii – 3, iii – 1, iv - 4
(B) i - 3, ii – 4, iii – 1, iv – 2
(C) i – 4, ii – 1, iii – 2, iv - 3
(D) ii – 3, iii – 4, i – 2, iv - 1
102. Non-conditional, negative selection marker genes are usually expressed
under a
(A) developmentally regulated promoter
(B) substrate-induced promoter
(C) constitutive promoter
(D) minimal promoter
103. A method in which a strong enhancer is randomly inserted in a plant
genome by transformation, resulting in mutant plants with dominant
phenotypes, is known as
(A) enhancer trapping
(B) TILLING
(C) activation tagging
(D) gene trapping
104. Which one of the following organisms is used by plant biologists to study
translocation in phloem?
(A) aphids
(B) nematodes
(C) grasshoppers
(D) butterflies
105. In a genetic transformation experiment, a researcher failed to add the
antibiotic selection agent in the shoot regeneration medium for selection of
transgenic plants. In the absence of any other confounding factors, which
one of the following statements is expected to be correct?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
24
(A) The number of regenerating shoots would be reduced in this
experiment as compared to those in which the antibiotic is added.
(B) The number of regenerating shoots is comparable to results
obtained in the “negative control” of the experiment.
(C) Transgenic plants cannot be generated from the above experiment.
(D) The regenerating shoots would consist of a mixture of transgenic
and non-transgenic plants.
106. Orientation of a cloned passenger gene cassette in a binary plasmid
vector, containing the expression cassette of a selectable marker gene,
can be checked by PCR using
(A) passenger gene-specific primers
(B) selectable marker gene-specific primers
(C) a combination of passenger gene-specific and vector-specific
primers
(D) vector-specific primers
107. Somatic hybrid plants with full or nearly full complement of the two parental
species are called
(A) asymmetric hybrids
(B) symmetric hybrids
(C) complete hybrids
(D) chimeric hybrids
108. In CAGE,
(A) only 5′-end information of the transcript is analysed
(B) only 3′-end information of the transcript is analysed
(C) both 5′-end and 3'-end information of the transcript are analysed
(D) complete exonic sequence of the transcript is analysed.
109. Which one of the following statements about associative nitrogen fixing
bacteria is NOT true?
(A) Rhodobacter and Rhodospirillum are autotrophic photosynthetic N-
fixing bacteria
(B) Clostridium, Desulfotomaculum and Desulfovibrio are heterotrophic
anaerobic N-fixing bacteria
(C) Azospirillum is an associative autotrophic microaerophilic nitrogen
fixing bacteria
(D) Beijerinckia is an heterotrophic aerobic associative nitrogen fixing
bacteria
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(A) The number of regenerating shoots would be reduced in this
experiment as compared to those in which the antibiotic is added.
(B) The number of regenerating shoots is comparable to results
obtained in the “negative control” of the experiment.
(C) Transgenic plants cannot be generated from the above experiment.
(D) The regenerating shoots would consist of a mixture of transgenic
and non-transgenic plants.
106. Orientation of a cloned passenger gene cassette in a binary plasmid
vector, containing the expression cassette of a selectable marker gene,
can be checked by PCR using
(A) passenger gene-specific primers
(B) selectable marker gene-specific primers
(C) a combination of passenger gene-specific and vector-specific
primers
(D) vector-specific primers
107. Somatic hybrid plants with full or nearly full complement of the two parental
species are called
(A) asymmetric hybrids
(B) symmetric hybrids
(C) complete hybrids
(D) chimeric hybrids
108. In CAGE,
(A) only 5′-end information of the transcript is analysed
(B) only 3′-end information of the transcript is analysed
(C) both 5′-end and 3'-end information of the transcript are analysed
(D) complete exonic sequence of the transcript is analysed.
109. Which one of the following statements about associative nitrogen fixing
bacteria is NOT true?
(A) Rhodobacter and Rhodospirillum are autotrophic photosynthetic N-
fixing bacteria
(B) Clostridium, Desulfotomaculum and Desulfovibrio are heterotrophic
anaerobic N-fixing bacteria
(C) Azospirillum is an associative autotrophic microaerophilic nitrogen
fixing bacteria
(D) Beijerinckia is an heterotrophic aerobic associative nitrogen fixing
bacteria
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25
110. According to the Seeds Act of 1966, which one of the following functions is
NOT applicable to a Seed Certification Agency?
(A) It should be an autonomous body.
(B) It should involve itself in the production and marketing of seeds.
(C) It should have close linkages with technical and other related
institutions.
(D) Its long-term objective should be to operate on no-profit no-loss
basis.
111. Which one of the combinations given below represents populations with a
similar genetic composition?
(A) F1 and F2
(B) F2 and BC1F2
(C) RIL and DH
(D) DH and F2
112. Which one of the following statements represents a correct definition of
Gene Pyramiding?
(A) Introducing different genes for resistance to a specific pest in a
single genotype
(B) Introducing different genes for resistance to a specific pest in
different genotypes
(C) Introducing a single gene for resistance to multiple pests in a single
genotype
(D) Introducing multiple genes conferring resistance to multiple pests in
different genotypes.
113. A marker closely linked to the gene of interest for an agronomic trait will be
used for
(A) foreground selection
(B) background selection
(C) both foreground and background selection
(D) selection of unlinked genes
114. The vitrification of cultured explants is generally caused by
A. low light irradiance, high temperature and intensive sterilization
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
110. According to the Seeds Act of 1966, which one of the following functions is
NOT applicable to a Seed Certification Agency?
(A) It should be an autonomous body.
(B) It should involve itself in the production and marketing of seeds.
(C) It should have close linkages with technical and other related
institutions.
(D) Its long-term objective should be to operate on no-profit no-loss
basis.
111. Which one of the combinations given below represents populations with a
similar genetic composition?
(A) F1 and F2
(B) F2 and BC1F2
(C) RIL and DH
(D) DH and F2
112. Which one of the following statements represents a correct definition of
Gene Pyramiding?
(A) Introducing different genes for resistance to a specific pest in a
single genotype
(B) Introducing different genes for resistance to a specific pest in
different genotypes
(C) Introducing a single gene for resistance to multiple pests in a single
genotype
(D) Introducing multiple genes conferring resistance to multiple pests in
different genotypes.
113. A marker closely linked to the gene of interest for an agronomic trait will be
used for
(A) foreground selection
(B) background selection
(C) both foreground and background selection
(D) selection of unlinked genes
114. The vitrification of cultured explants is generally caused by
A. low light irradiance, high temperature and intensive sterilization
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
26
B. high auxin, low temperature and high light irradiance
C. high agar, high nutrients and high pH
D. low pH, low temperature and low micronutrient concentration
115. The Rht mutations in wheat that were pivotal for 'Green Revolution' reduce
plant height due to impairment in
(A) gibberellic acid biosynthesis pathway
(B) gibberellic acid signaling pathway
(C) auxin biosynthetic pathway
(D) auxin response pathway
116. Steady state for a closed system implies
(A) Input = Output
(B) Accumulation = 0
(C) Input – Output = Accumulation
(D) System is in equilibrium
117. In a lab scale reactor, the size of the air bubbles in the reactor is primarily
determined by
(A) agitator speed
(B) viscosity of the broth
(C) density of the broth
(D) size of the holes in the sparger
118. Generally fungal fermentations are pseudoplastic. Hence in large reactors
the cells
(A) are better aerated near walls of the reactor
(B) are better aerated in the centre of the reactor
(C) are evenly aerated
(D) get better aerated at the top rather than bottom
119. Mixing time increases with the volume of the reactor because of increase
in the
(A) circulation time
(B) shear
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B. high auxin, low temperature and high light irradiance
C. high agar, high nutrients and high pH
D. low pH, low temperature and low micronutrient concentration
115. The Rht mutations in wheat that were pivotal for 'Green Revolution' reduce
plant height due to impairment in
(A) gibberellic acid biosynthesis pathway
(B) gibberellic acid signaling pathway
(C) auxin biosynthetic pathway
(D) auxin response pathway
116. Steady state for a closed system implies
(A) Input = Output
(B) Accumulation = 0
(C) Input – Output = Accumulation
(D) System is in equilibrium
117. In a lab scale reactor, the size of the air bubbles in the reactor is primarily
determined by
(A) agitator speed
(B) viscosity of the broth
(C) density of the broth
(D) size of the holes in the sparger
118. Generally fungal fermentations are pseudoplastic. Hence in large reactors
the cells
(A) are better aerated near walls of the reactor
(B) are better aerated in the centre of the reactor
(C) are evenly aerated
(D) get better aerated at the top rather than bottom
119. Mixing time increases with the volume of the reactor because of increase
in the
(A) circulation time
(B) shear
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27
(C) turbulence
(D) flow rate
120. The Reynolds number is a ratio of
(A) external forces/inertial forces
(B) inertial forces/axial dispersion
(C) inertial forces/viscous forces
(D) external forces/ viscous forces
121. The equation τ = μ du/dy is also termed as
(A) pressure flux equation
(B) force flux equation
(C) mass flux equation
(D) momentum flux equation
122. Match the Transport process with appropriate time constants given below.
Transport process Time constants
1. Oxygen transfer p. V/Q
2. Heat transfer q. 4V/(1.5ND
3
)
3. Flow r. VρCp/UA
4. Mixing s. 1/KLa
Where V is volume; Q is pumping rate; N is agitation rate; D is diameter of the
impeller; A is area and ρ is the density of the liquid
(A) 1-s, 2-r, 3-p, 4-q
(B) 1-s, 2-r, 3-q, 4-p
(C) 1-s, 2-q, 3-p, 4-r
(D) 1-q, 2-r, 3-p, 4-s
123. When a liquid of density 1200 kg/m
3
and viscosity 0.01 kg m
-1
s
-1
flows
through a 1 cm ID pipe at 0.2 m/s, the flow is referred as a
(A) Piston flow
(B) Turbulent flow
(C) Laminar flow
(D) Transition flow
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(C) turbulence
(D) flow rate
120. The Reynolds number is a ratio of
(A) external forces/inertial forces
(B) inertial forces/axial dispersion
(C) inertial forces/viscous forces
(D) external forces/ viscous forces
121. The equation τ = μ du/dy is also termed as
(A) pressure flux equation
(B) force flux equation
(C) mass flux equation
(D) momentum flux equation
122. Match the Transport process with appropriate time constants given below.
Transport process Time constants
1. Oxygen transfer p. V/Q
2. Heat transfer q. 4V/(1.5ND
3
)
3. Flow r. VρCp/UA
4. Mixing s. 1/KLa
Where V is volume; Q is pumping rate; N is agitation rate; D is diameter of the
impeller; A is area and ρ is the density of the liquid
(A) 1-s, 2-r, 3-p, 4-q
(B) 1-s, 2-r, 3-q, 4-p
(C) 1-s, 2-q, 3-p, 4-r
(D) 1-q, 2-r, 3-p, 4-s
123. When a liquid of density 1200 kg/m
3
and viscosity 0.01 kg m
-1
s
-1
flows
through a 1 cm ID pipe at 0.2 m/s, the flow is referred as a
(A) Piston flow
(B) Turbulent flow
(C) Laminar flow
(D) Transition flow
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
28
124. In a dynamic gassing out technique for determination of KLa in a bioreactor,
upon restart of the aeration the increase in DO slows down with time because
(A) the cultures specific OUR declines to reach a steady state
(B) KLa changes when the cells get sufficient oxygen supply
(C) C*-CL increases
(D) C*-CL decreases
125. Under unaerated conditions in a fermenter, the power consumed by a single
impeller is 10 KW and upon changing the agitation rate from 200 to 600 rpm,
the new power consumption (KW) would be -----. (Assuming that the power
number remains constant)
(A) 270
(B) 90
(C) 30
(D) 10
126. Fill in the blanks with appropriate option: Vinegar is a liquid consisting of
acetic acid produced by the fermentation of _______________ by
_____________________.
(A) glucose, S. cerevisiae
(B) ethanol, A. aceti
(C) lactose, L. acidophilus
(D) starch, B. amyloliquefaciens
127. A food stored in a refrigerator has caused an outbreak of food borne illness.
Which one of the following is most likely to be the pathogen responsible for this?
(A) S. typhimurium
(B) E. coli
(C) L. monocytogenes
(D) B. cereus
128. To reduce the tartness in wine due to malic acid, a secondary fermentation
is carried out using
(A) Saccharomyces cerevisiae
(B) Oenococcus oeni
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
124. In a dynamic gassing out technique for determination of KLa in a bioreactor,
upon restart of the aeration the increase in DO slows down with time because
(A) the cultures specific OUR declines to reach a steady state
(B) KLa changes when the cells get sufficient oxygen supply
(C) C*-CL increases
(D) C*-CL decreases
125. Under unaerated conditions in a fermenter, the power consumed by a single
impeller is 10 KW and upon changing the agitation rate from 200 to 600 rpm,
the new power consumption (KW) would be -----. (Assuming that the power
number remains constant)
(A) 270
(B) 90
(C) 30
(D) 10
126. Fill in the blanks with appropriate option: Vinegar is a liquid consisting of
acetic acid produced by the fermentation of _______________ by
_____________________.
(A) glucose, S. cerevisiae
(B) ethanol, A. aceti
(C) lactose, L. acidophilus
(D) starch, B. amyloliquefaciens
127. A food stored in a refrigerator has caused an outbreak of food borne illness.
Which one of the following is most likely to be the pathogen responsible for this?
(A) S. typhimurium
(B) E. coli
(C) L. monocytogenes
(D) B. cereus
128. To reduce the tartness in wine due to malic acid, a secondary fermentation
is carried out using
(A) Saccharomyces cerevisiae
(B) Oenococcus oeni
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29
(C) Aspergillus niger
(D) Pichia pastoris
129. Which one of the following statements is INCORRECT with respect to food
rheology?
(A) Yoghurt demonstrates thixotropy
(B) Ketchup is an example of shear thinning, pseudoplastic liquid
(C) Wheat dough is viscoelastic
(D) Molten chocolate is a perfect Newtonian fluid
130. Which one of the following enzyme(s) would result in high glucose yield
from starch hydrolysis?
(A) α- amylase
(B) β- amylase
(C) α- amylase and amyloglucosidase
(D) α- amylase and β- amylase
131. The major metabolite formed and the succession of microorganisms
responsible for Sauerkraut fermentation are
(A) citric acid, yeast- Coliform- Leuconostoc
(B) acetic acid, Aspergillus- Lactobacilli- Leuconostoc
(C) lactic acid, Coliform- Leuconostoc- Lactobacilli
(D) malic acid, Leuconostoc- Lactobacilli- Coliform
132. Omega 3 fats are good for health and play an important role in prevention of
heart disease. If you have to formulate a food product rich in omega 3 fats which
one of the following options would you choose?
(A) rice bran oil because it contains C 18:2 ∆9c, 12c
(B) flax seed oil because it contains C 18:3 ∆9c, 12c, 15c
(C) olive oil because it contains C 18:1 ∆9c
(D) black- currant seed oil because it contains C 18:3 ∆ 6c, 9c, 12c
133. For a well-established enzyme assay method where the reaction kinetics
should remain zero order, what percentage of substrate is converted to product
during the incubation period?
(A) 80 – 90%
(B) 50 – 60%
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(C) Aspergillus niger
(D) Pichia pastoris
129. Which one of the following statements is INCORRECT with respect to food
rheology?
(A) Yoghurt demonstrates thixotropy
(B) Ketchup is an example of shear thinning, pseudoplastic liquid
(C) Wheat dough is viscoelastic
(D) Molten chocolate is a perfect Newtonian fluid
130. Which one of the following enzyme(s) would result in high glucose yield
from starch hydrolysis?
(A) α- amylase
(B) β- amylase
(C) α- amylase and amyloglucosidase
(D) α- amylase and β- amylase
131. The major metabolite formed and the succession of microorganisms
responsible for Sauerkraut fermentation are
(A) citric acid, yeast- Coliform- Leuconostoc
(B) acetic acid, Aspergillus- Lactobacilli- Leuconostoc
(C) lactic acid, Coliform- Leuconostoc- Lactobacilli
(D) malic acid, Leuconostoc- Lactobacilli- Coliform
132. Omega 3 fats are good for health and play an important role in prevention of
heart disease. If you have to formulate a food product rich in omega 3 fats which
one of the following options would you choose?
(A) rice bran oil because it contains C 18:2 ∆9c, 12c
(B) flax seed oil because it contains C 18:3 ∆9c, 12c, 15c
(C) olive oil because it contains C 18:1 ∆9c
(D) black- currant seed oil because it contains C 18:3 ∆ 6c, 9c, 12c
133. For a well-established enzyme assay method where the reaction kinetics
should remain zero order, what percentage of substrate is converted to product
during the incubation period?
(A) 80 – 90%
(B) 50 – 60%
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30
(C) 30 – 20%
(D) 6 – 10%
134. As the viscosity increases, the method used for mixing is changed in the
following order
(A) Paddle, Turbine, Propeller, Helical ribbon and Screw
(B) Propeller, Turbine, Paddle, Helical ribbon and Screw
(C) Propeller, Paddle, Turbine, Helical ribbon and Screw
(D) Propeller, Turbine, Paddle, Screw and Helical ribbon
135. On scale up, the effective area available for heat transfer in a bioreactor
(A) Increases
(B) Decreases
(C) Depends on viscosity of the medium
(D) Remains constant
136. A fermenter produces 100 kg lysine per day. If the volumetric productivity is
0.8 gl
-1
h
-1
, what is the volume of the fermenter in liter?
(A) 2200
(B) 3200
(C) 4200
(D) 5200
137. Chemostat cultivation is carried out at steady state with a dilution rate of 0.1
h
-1
. Given μm = 1.1 h
-1
and Ks=0.1 g/L. What will be steady state substrate
concentration?
(A) 0.01 g/L
(B) 0.1 g/L
(C) 1.0 g/L
(D) 10.0 g/L
138. For degradation of crude oil in ocean by microorganisms, the two limiting
nutrients are
(A) nitrates and ammonia
(B) nitrates and phosphates
(C) sulphates and iron
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(C) 30 – 20%
(D) 6 – 10%
134. As the viscosity increases, the method used for mixing is changed in the
following order
(A) Paddle, Turbine, Propeller, Helical ribbon and Screw
(B) Propeller, Turbine, Paddle, Helical ribbon and Screw
(C) Propeller, Paddle, Turbine, Helical ribbon and Screw
(D) Propeller, Turbine, Paddle, Screw and Helical ribbon
135. On scale up, the effective area available for heat transfer in a bioreactor
(A) Increases
(B) Decreases
(C) Depends on viscosity of the medium
(D) Remains constant
136. A fermenter produces 100 kg lysine per day. If the volumetric productivity is
0.8 gl
-1
h
-1
, what is the volume of the fermenter in liter?
(A) 2200
(B) 3200
(C) 4200
(D) 5200
137. Chemostat cultivation is carried out at steady state with a dilution rate of 0.1
h
-1
. Given μm = 1.1 h
-1
and Ks=0.1 g/L. What will be steady state substrate
concentration?
(A) 0.01 g/L
(B) 0.1 g/L
(C) 1.0 g/L
(D) 10.0 g/L
138. For degradation of crude oil in ocean by microorganisms, the two limiting
nutrients are
(A) nitrates and ammonia
(B) nitrates and phosphates
(C) sulphates and iron
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31
(D) phosphates and nitrites
139. The first circumpolar expedition in the history of Marine Sciences was
(A) Challenger
(B) Beagle
(C) International Indian Ocean Expedition
(D) Voyager
140. Paralytic shellfish poisoning (PSP) is caused by the consumption of
molluscan Shellfish contaminated with
(A) Brevetoxins
(B) Domoic Acid
(C) Saxitoxins
(D) Okadaic acid
141. Biofilm/capsular material produced by the soil bacteria is detected by
(A) Saffranin
(B) Malachite green
(C) Basic fuchsin
(D) Congo red
142. A compound secreted by a soil bacterium capable of reducing the surface
tension at air-water interface is called as
(A) Osmolyte
(B) Detergent
(C) Primary metabolite
(D) Biosurfactant
143. Match the organism in Group A with the appropriate habitat from Group B.
Group A Group B
P) Pseudomonas i) Hot springs
Q) Thermus ii) Rhizosphere
R) Photobacterium iii) Soil
S) Thiobacillus iv) Marine
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(D) phosphates and nitrites
139. The first circumpolar expedition in the history of Marine Sciences was
(A) Challenger
(B) Beagle
(C) International Indian Ocean Expedition
(D) Voyager
140. Paralytic shellfish poisoning (PSP) is caused by the consumption of
molluscan Shellfish contaminated with
(A) Brevetoxins
(B) Domoic Acid
(C) Saxitoxins
(D) Okadaic acid
141. Biofilm/capsular material produced by the soil bacteria is detected by
(A) Saffranin
(B) Malachite green
(C) Basic fuchsin
(D) Congo red
142. A compound secreted by a soil bacterium capable of reducing the surface
tension at air-water interface is called as
(A) Osmolyte
(B) Detergent
(C) Primary metabolite
(D) Biosurfactant
143. Match the organism in Group A with the appropriate habitat from Group B.
Group A Group B
P) Pseudomonas i) Hot springs
Q) Thermus ii) Rhizosphere
R) Photobacterium iii) Soil
S) Thiobacillus iv) Marine
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32
(A) P-ii, Q-i R-iv S-iii
(B) P-i, Q-iii, R-ii, S-iv
(C) P-iv, Q-iii, R-ii, S-i
(D) P-ii, Q-iv, R-ii S-i
144. Match the type of Organism in Group A with appropriate phenotype/property
from Group B.
Group A Group B
P) Pseudomoans aeruginosa i) Sulfur oxidizing organism
Q) Staphylococcus aureus ii) Elemental N2 fixation
R) Thiobacillus ferroxidance iii) Biofilm formation
S) Bradyrhizobium iv) Quorum sensing
(A) P-ii, Q-iii, R-iv, S-i
(B) P-iii, Q-iv, R-i, S-ii
(C) P-iii, Q-iv, R-ii, S-i
(D) P-i, Q-iii, R-ii, S-iv
145. Which one of the following reactions is performed by Cytochrome-P450 to
detoxify the xenobiotics?
(A) Hydroxylation
(B) Ligation
(C) Hydrolysis
(D) Group transfer
146. Which one of the following is NOT true regarding Zika virus?
(A) It was first isolated in Uganda
(B) It is transmitted by Anopheles mosquito
(C) It is an RNA virus
(D) It may result in microcephaly
147. The vaccine currently used in India’s Pulse polio campaign is
(A) heat killed
(B) chemically attenuated
(C) live attenuated
(D) vaccine with adjuvant
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(A) P-ii, Q-i R-iv S-iii
(B) P-i, Q-iii, R-ii, S-iv
(C) P-iv, Q-iii, R-ii, S-i
(D) P-ii, Q-iv, R-ii S-i
144. Match the type of Organism in Group A with appropriate phenotype/property
from Group B.
Group A Group B
P) Pseudomoans aeruginosa i) Sulfur oxidizing organism
Q) Staphylococcus aureus ii) Elemental N2 fixation
R) Thiobacillus ferroxidance iii) Biofilm formation
S) Bradyrhizobium iv) Quorum sensing
(A) P-ii, Q-iii, R-iv, S-i
(B) P-iii, Q-iv, R-i, S-ii
(C) P-iii, Q-iv, R-ii, S-i
(D) P-i, Q-iii, R-ii, S-iv
145. Which one of the following reactions is performed by Cytochrome-P450 to
detoxify the xenobiotics?
(A) Hydroxylation
(B) Ligation
(C) Hydrolysis
(D) Group transfer
146. Which one of the following is NOT true regarding Zika virus?
(A) It was first isolated in Uganda
(B) It is transmitted by Anopheles mosquito
(C) It is an RNA virus
(D) It may result in microcephaly
147. The vaccine currently used in India’s Pulse polio campaign is
(A) heat killed
(B) chemically attenuated
(C) live attenuated
(D) vaccine with adjuvant
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33
148. Which one of the following enzymes helps in the survival of Helicobacter
pylori in the stomach?
(A) carbonic anhydrase
(B) β-lactamase
(C) urease
(D) transpeptidase
149. Which one of the following is INCORRECT about Streptokinase?
(A) It was originally obtained from β hemolytic streptococci
(B) It is highly antigenic
(C) It is used as a therapeutic anti-thrombogenic agent
(D) It is used as a therapeutic fibrinolytic agent
150. The utility of positron emission tomography to diagnose tumors is based on
(A) uptake of 2-fluoro-2 deoxyfructose
(B) uptake of 2-fluoro-2-deoxyglucose
(C) activation of hypoxia-inducible transcription factor HIF-1
(D) detection of hypoxia by pimonidazole
151. Which one of the following imaging techniques is non-invasive?
(A) Ultrasound imaging
(B) Contrast CT imaging
(C) Nuclear imaging
(D) PET imaging
152. Amphotericin B is clinically used against which one of the following
pathogens?
(A) Herpes simplex virus I
(B) M. tuberculosis
(C) Candida spp.
(D) P. vivax
153. The therapeutic index (TI) indicates the relative margin of safety of a drug
and it is deduced from the values of mean lethal dose (LD50) and median
effective dose (ED50). Which of the following represents TI of a drug?
(A) TI= LD50 x ED50
(B) TI= LD50 + ED50
(C) TI= LD50 / ED50
(D) TI= LD50 – ED50
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148. Which one of the following enzymes helps in the survival of Helicobacter
pylori in the stomach?
(A) carbonic anhydrase
(B) β-lactamase
(C) urease
(D) transpeptidase
149. Which one of the following is INCORRECT about Streptokinase?
(A) It was originally obtained from β hemolytic streptococci
(B) It is highly antigenic
(C) It is used as a therapeutic anti-thrombogenic agent
(D) It is used as a therapeutic fibrinolytic agent
150. The utility of positron emission tomography to diagnose tumors is based on
(A) uptake of 2-fluoro-2 deoxyfructose
(B) uptake of 2-fluoro-2-deoxyglucose
(C) activation of hypoxia-inducible transcription factor HIF-1
(D) detection of hypoxia by pimonidazole
151. Which one of the following imaging techniques is non-invasive?
(A) Ultrasound imaging
(B) Contrast CT imaging
(C) Nuclear imaging
(D) PET imaging
152. Amphotericin B is clinically used against which one of the following
pathogens?
(A) Herpes simplex virus I
(B) M. tuberculosis
(C) Candida spp.
(D) P. vivax
153. The therapeutic index (TI) indicates the relative margin of safety of a drug
and it is deduced from the values of mean lethal dose (LD50) and median
effective dose (ED50). Which of the following represents TI of a drug?
(A) TI= LD50 x ED50
(B) TI= LD50 + ED50
(C) TI= LD50 / ED50
(D) TI= LD50 – ED50
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
34
154. Which one of the following is a frequently used radio-opaque dye?
(A) Barium chloride
(B) Barium sulphate
(C) Barium nitrite
(D) Barium iodide
155. The name of the blood vessel marked [?] in the figure below is
(A) pulmonary vein
(B) pulmonary artery
(C) cardiac vein
(D) cardiac artery
(A) pulmonary vein
(B) pulmonary artery
(C) cardiac vein
(D) cardiac artery
156. Which one of the following is a reason for administering general anesthesia
to experimental animals?
(A) It reduces the toxicity of a test drug
(B) It provides reasonable degree of muscle relaxation
(C) It maintains the normal level of haemoglobin
(D) It maintains the normal body temperature
157. During the pathogenesis of rabies, the replication of virus first starts in the
(A) Spinal ganglia
(B) Muscle fibers and connective tissues
(C) Lungs
(D) Central nervous system
?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
154. Which one of the following is a frequently used radio-opaque dye?
(A) Barium chloride
(B) Barium sulphate
(C) Barium nitrite
(D) Barium iodide
155. The name of the blood vessel marked [?] in the figure below is
(A) pulmonary vein
(B) pulmonary artery
(C) cardiac vein
(D) cardiac artery
(A) pulmonary vein
(B) pulmonary artery
(C) cardiac vein
(D) cardiac artery
156. Which one of the following is a reason for administering general anesthesia
to experimental animals?
(A) It reduces the toxicity of a test drug
(B) It provides reasonable degree of muscle relaxation
(C) It maintains the normal level of haemoglobin
(D) It maintains the normal body temperature
157. During the pathogenesis of rabies, the replication of virus first starts in the
(A) Spinal ganglia
(B) Muscle fibers and connective tissues
(C) Lungs
(D) Central nervous system
?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
35
158. Route of entry for canine distemper virus is
(A) Skin dermal fibroblasts
(B) Respiratory epithelial cells
(C) Lymphocytes
(D) Mesenchymal stem cells
159. Callipyge gene mutation in sheep leads to increased muscle growth. The
gene presents a typical case of Polar over-dominance. Which of the four
genotypes given below shows increased muscle growth? (N= wild type allele,
C= mutant allele for callipyge)
(A) N
mat
C
pat
(B) N
pat
C
mat
(C) C
mat
C
pat
(D) N
Pat
N
pat
160. The technique used in animal biotechnology for the rapid multiplication and
production of animals with a desirable genotype is:
(A) protoplast fusion and embryo transfer
(B) hybrid selection and embryo transfer
(C) in vitro fertilization and embryo transfer
(D) artificial insemination
161. A primary culture of neurons isolated from the mouse brain responded
positively when exposed to a chemical for a few seconds, but the neurons were
less responsive and eventually stopped responding when exposed to the same
chemical for prolonged duration. However, if the culture was left for some time
and washed, the neurons started responding to the same chemical at the same
concentration as before. The reduced response was due to
(A) increased apoptosis of the neurons.
(B) necrosis of the neurons.
(C) change in the pH of the medium.
(D) desensitization/down-regulation of the receptors.
162. If a neuron is tyrosine hydroxylase positive, it could be either:
(A) Noradrenergic or Histaminergic
(B) Dopaminergic or Serotonergic
(C) Noradrenergic or Dopaminergic
(D) Serotonergic or Noradrenergic
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158. Route of entry for canine distemper virus is
(A) Skin dermal fibroblasts
(B) Respiratory epithelial cells
(C) Lymphocytes
(D) Mesenchymal stem cells
159. Callipyge gene mutation in sheep leads to increased muscle growth. The
gene presents a typical case of Polar over-dominance. Which of the four
genotypes given below shows increased muscle growth? (N= wild type allele,
C= mutant allele for callipyge)
(A) N
mat
C
pat
(B) N
pat
C
mat
(C) C
mat
C
pat
(D) N
Pat
N
pat
160. The technique used in animal biotechnology for the rapid multiplication and
production of animals with a desirable genotype is:
(A) protoplast fusion and embryo transfer
(B) hybrid selection and embryo transfer
(C) in vitro fertilization and embryo transfer
(D) artificial insemination
161. A primary culture of neurons isolated from the mouse brain responded
positively when exposed to a chemical for a few seconds, but the neurons were
less responsive and eventually stopped responding when exposed to the same
chemical for prolonged duration. However, if the culture was left for some time
and washed, the neurons started responding to the same chemical at the same
concentration as before. The reduced response was due to
(A) increased apoptosis of the neurons.
(B) necrosis of the neurons.
(C) change in the pH of the medium.
(D) desensitization/down-regulation of the receptors.
162. If a neuron is tyrosine hydroxylase positive, it could be either:
(A) Noradrenergic or Histaminergic
(B) Dopaminergic or Serotonergic
(C) Noradrenergic or Dopaminergic
(D) Serotonergic or Noradrenergic
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
36
163. Pain sensation is a subjective and conscious feeling. However, although the
autonomic organs viz. brain, heart etc. do not get represented in the cerebral
cortex, one feels pain in those parts as well. The reason is
(A) these parts receive less blood supply.
(B) increased pH in those parts.
(C) it is a referred pain.
(D) these organs are metabolically compromised.
164. Neurons are formed from precursor proneural cells due to
(A) relatively low level notch activity.
(B) relatively high level notch activity.
(C) inactivation of notch.
(D) activation of delta.
165. A neuron at resting state when treated with “X” showed transmembrane
potential -50mV, while when treated with “Y” it showed -90mV. Given such
a condition, which of the following statements would be most appropriate?
(A) “X” induced depolarization, while “Y” induced hyperpolarization.
(B) The threshold for inducing a response by the neuron was higher for
“X” than that for “Y”.
(C) Both the treatments induced depolarization of the neuron.
(D) Both the treatments induced hyperpolarization of the neuron.
166. In a population which is in Hardy-Weinberg equilibrium, the frequency of a
recessive allele for a certain genetic trait is 0.40. What percentage of individuals
would be expected to show the dominant trait in the next generation?
(A) 16%
(B) 32%
(C) 84%
(D) 96%
167. Color blindness in human is a sex-linked recessive trait. If two individuals
with normal color vision have a color blind son, what will be the genotypes of the
parents?
(A) XcXc and XcY
(B) XCXc and XCY
(C) XcXc and XCY
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
163. Pain sensation is a subjective and conscious feeling. However, although the
autonomic organs viz. brain, heart etc. do not get represented in the cerebral
cortex, one feels pain in those parts as well. The reason is
(A) these parts receive less blood supply.
(B) increased pH in those parts.
(C) it is a referred pain.
(D) these organs are metabolically compromised.
164. Neurons are formed from precursor proneural cells due to
(A) relatively low level notch activity.
(B) relatively high level notch activity.
(C) inactivation of notch.
(D) activation of delta.
165. A neuron at resting state when treated with “X” showed transmembrane
potential -50mV, while when treated with “Y” it showed -90mV. Given such
a condition, which of the following statements would be most appropriate?
(A) “X” induced depolarization, while “Y” induced hyperpolarization.
(B) The threshold for inducing a response by the neuron was higher for
“X” than that for “Y”.
(C) Both the treatments induced depolarization of the neuron.
(D) Both the treatments induced hyperpolarization of the neuron.
166. In a population which is in Hardy-Weinberg equilibrium, the frequency of a
recessive allele for a certain genetic trait is 0.40. What percentage of individuals
would be expected to show the dominant trait in the next generation?
(A) 16%
(B) 32%
(C) 84%
(D) 96%
167. Color blindness in human is a sex-linked recessive trait. If two individuals
with normal color vision have a color blind son, what will be the genotypes of the
parents?
(A) XcXc and XcY
(B) XCXc and XCY
(C) XcXc and XCY
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
37
(D) XCXC and XcY
168. A human is phenotypically female, but her interphase somatic nuclei do not
display the existence of Barr bodies. Which of the following conditions could
explain the above feature?
(A) Klinefelter syndrome
(B) 2n+XXX
(C) Turner syndrome.
(D) 2n+YY
169. For the pedigree depicted below, which mode(s) of inheritance CANNOT be
possible?
(A) X-linked dominant
(B) Autosomal dominant
(C) X-linked recessive
(D) Autosomal recessive
170. C banding of human chromosomes specifically reveals
(A) polymorphism of constitutive heterochromatin of chromosomes 1, 9, 16
and Y
(B) polymorphism of constitutive heterochromatin of chromosomes 3, 7, 12
and X
(C) polymorphism of facultative heterochromatin of chromosome X
(D) satellite sequences
171. Which one of the following expressions is likely to retrieve more matches in
a database search?
(A) D-A-V-I-D
(B) [DE]-A-V-I-[DE]
(C) [DE]-[AVILM]-X-E
(D) D-A-V-E
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(D) XCXC and XcY
168. A human is phenotypically female, but her interphase somatic nuclei do not
display the existence of Barr bodies. Which of the following conditions could
explain the above feature?
(A) Klinefelter syndrome
(B) 2n+XXX
(C) Turner syndrome.
(D) 2n+YY
169. For the pedigree depicted below, which mode(s) of inheritance CANNOT be
possible?
(A) X-linked dominant
(B) Autosomal dominant
(C) X-linked recessive
(D) Autosomal recessive
170. C banding of human chromosomes specifically reveals
(A) polymorphism of constitutive heterochromatin of chromosomes 1, 9, 16
and Y
(B) polymorphism of constitutive heterochromatin of chromosomes 3, 7, 12
and X
(C) polymorphism of facultative heterochromatin of chromosome X
(D) satellite sequences
171. Which one of the following expressions is likely to retrieve more matches in
a database search?
(A) D-A-V-I-D
(B) [DE]-A-V-I-[DE]
(C) [DE]-[AVILM]-X-E
(D) D-A-V-E
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
38
172. Which substitution matrices should you prefer to find distantly related
orthologs through BLAST search?
(A) BLOSUM 40 and PAM 250
(B) BLOSUM 82 and PAM 250
(C) BLOSUM 40 and PAM 120
(D) BLOSUM 60 and PAM 120
173. Match the items in Group 1 with an appropriate description in Group 2
Group I Group 2
P. UPGMA 1. Protein sequence database
Q. CLUSTALW 2. Phylogenetic Analysis
R. SWISS-PROT 3. 3-D structure visualization
S. RasMol 4. Multiple sequence alignment
(A) P-4, Q-1, R-2, S-3
(B) P-2, Q-4, R-1, S-3
(C) P-2, Q-3, R-1, S-4
(D) P-2, Q-1, R-4, S-3
174. The rate of substitutions in a certain region of DNA of length 1000 bases is
estimated as 10
-9
per base per year. If two species diverged approximately 10
million years ago, the fraction of sites that differ between them should be
approximately
(A) 0.2%
(B) 2%
(C) 20%
(D) 25%
175. When searching the Blocks and PRINTS databases, a match is judged
significant if
(A) a single motif is matched
(B) two motifs are matched
(C) the E-value is above e-4
(D) a combined E-value above a given threshold is reported for a
multiple -motif match
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
172. Which substitution matrices should you prefer to find distantly related
orthologs through BLAST search?
(A) BLOSUM 40 and PAM 250
(B) BLOSUM 82 and PAM 250
(C) BLOSUM 40 and PAM 120
(D) BLOSUM 60 and PAM 120
173. Match the items in Group 1 with an appropriate description in Group 2
Group I Group 2
P. UPGMA 1. Protein sequence database
Q. CLUSTALW 2. Phylogenetic Analysis
R. SWISS-PROT 3. 3-D structure visualization
S. RasMol 4. Multiple sequence alignment
(A) P-4, Q-1, R-2, S-3
(B) P-2, Q-4, R-1, S-3
(C) P-2, Q-3, R-1, S-4
(D) P-2, Q-1, R-4, S-3
174. The rate of substitutions in a certain region of DNA of length 1000 bases is
estimated as 10
-9
per base per year. If two species diverged approximately 10
million years ago, the fraction of sites that differ between them should be
approximately
(A) 0.2%
(B) 2%
(C) 20%
(D) 25%
175. When searching the Blocks and PRINTS databases, a match is judged
significant if
(A) a single motif is matched
(B) two motifs are matched
(C) the E-value is above e-4
(D) a combined E-value above a given threshold is reported for a
multiple -motif match
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
39
176. Two species are found to share a cluster of 8 genes, but the genes are in
different orders in the two species. The orders are represented by signed
permutations as given below:
Species X 1,2,3,4,5,6,7,8
Species Y 1, 2,-5,-4,-3,8,6,7
The transformation between the two gene orders
(A) cannot be achieved by inversions alone.
(B) can be achieved by one translocation and one inversion.
(C) can be achieved by three inversions.
(D) requires six separate genome rearrangement events.
177. As the E value of a BLAST search becomes smaller
(A) the value K also becomes smaller
(B) the score tends to be larger
(C) the probability p tends to be larger
(D) the extreme value distribution becomes less skewed
178. In a genome with average GC-content 40%, the expected frequency of
occurrence of the tetranucleotide AACG is
(A) 0.24 %
(B) 0.36 %
(C) 0.42 %
(D) 0
179. Consider that you have collected X-ray diffraction data on three different
protein crystals, referred to as Protein A, B and C. Data set for crystal A has the
best diffraction spot at 8A, crystal B at 5A and crystal C has the best diffraction at
3A. Which data set is likely to yield a higher resolution crystal structure?
(A) Protein A
(B) Protein B
(C) Crystal C
(D) All will have same resolution structure
180. In a protein crystallization experiment, you have identified an optimal protein
concentration and precipitant concentration which yield crystals. Now if you are
asked to grow crystals again with a protein solution that is four times more
concentrated, how will you alter the precipitant concentration?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
176. Two species are found to share a cluster of 8 genes, but the genes are in
different orders in the two species. The orders are represented by signed
permutations as given below:
Species X 1,2,3,4,5,6,7,8
Species Y 1, 2,-5,-4,-3,8,6,7
The transformation between the two gene orders
(A) cannot be achieved by inversions alone.
(B) can be achieved by one translocation and one inversion.
(C) can be achieved by three inversions.
(D) requires six separate genome rearrangement events.
177. As the E value of a BLAST search becomes smaller
(A) the value K also becomes smaller
(B) the score tends to be larger
(C) the probability p tends to be larger
(D) the extreme value distribution becomes less skewed
178. In a genome with average GC-content 40%, the expected frequency of
occurrence of the tetranucleotide AACG is
(A) 0.24 %
(B) 0.36 %
(C) 0.42 %
(D) 0
179. Consider that you have collected X-ray diffraction data on three different
protein crystals, referred to as Protein A, B and C. Data set for crystal A has the
best diffraction spot at 8A, crystal B at 5A and crystal C has the best diffraction at
3A. Which data set is likely to yield a higher resolution crystal structure?
(A) Protein A
(B) Protein B
(C) Crystal C
(D) All will have same resolution structure
180. In a protein crystallization experiment, you have identified an optimal protein
concentration and precipitant concentration which yield crystals. Now if you are
asked to grow crystals again with a protein solution that is four times more
concentrated, how will you alter the precipitant concentration?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
40
(A) Increase precipitant concentration
(B) Decrease precipitant concentration
(C) Keep the precipitant concertation the same
(D) Remove the precipitant altogether
181. Which of the following statements is true for two different tripeptides
consisting of either glycine or proline?
(A) Glycine tripeptide will have relatively larger allowed area on the
Ramachandran plot.
(B) Proline tripeptide will have relatively larger allowed area on the
Ramachandran plot.
(C) Both the tripeptides will fall primarily in the disallowed regions of the
Ramachandran plot.
(D) Both the tripeptides will fall primarily in the overlapping allowed regions of the
Ramachandran plot.
182. In a Fluorescence Resonance Energy Transfer (FRET) experiment, the
donor is Cyan Fluorescent Protein (CFP) which has excitation and emission
maxima at 435 and 485 nm, respectively. The acceptor in this experiment is
Yellow Fluorescent Protein (YFP) that has excitation and emission maxima at
513 and 545 nm, respectively . Under the conditions, where a significant FRET is
observed, what will happen to the FRET signal if the emission maxima of CFP is
made to shift to 475nm.
(A) Increase
(B) Decrease
(C) Remains unchanged
(D) May increase or decrease
183. Consider that the following polypeptides are being analyzed by Mass-
spectrometry (MS). These peptides are generated by proteolytic cleavage of a
protein that is expressed in mammalian cells and it has undergone glycosylation,
phosphorylation and acetylation. What will the expected positioning of the
spectra corresponding to these peptides (smaller to larger)?
(A) Asn-Lys-Ser/Asn-Lys-Thr/ Ser-Lys-Asn/ Thr-Arg-Asn
(B) Thr-Arg-Asn/ Ser-Lys-Asn/ Asn-Lys-Thr/ Asn-Lys-Ser
(C) Asn-Lys-Thr/ Asn-Lys-Ser/ Thr-Arg-Asn/ Ser-Lys-Asn
(D) Ser-Lys-Asn/ Asn-Lys-Thr/ Asn-Lys-Ser/ Thr-Arg-Asn
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(A) Increase precipitant concentration
(B) Decrease precipitant concentration
(C) Keep the precipitant concertation the same
(D) Remove the precipitant altogether
181. Which of the following statements is true for two different tripeptides
consisting of either glycine or proline?
(A) Glycine tripeptide will have relatively larger allowed area on the
Ramachandran plot.
(B) Proline tripeptide will have relatively larger allowed area on the
Ramachandran plot.
(C) Both the tripeptides will fall primarily in the disallowed regions of the
Ramachandran plot.
(D) Both the tripeptides will fall primarily in the overlapping allowed regions of the
Ramachandran plot.
182. In a Fluorescence Resonance Energy Transfer (FRET) experiment, the
donor is Cyan Fluorescent Protein (CFP) which has excitation and emission
maxima at 435 and 485 nm, respectively. The acceptor in this experiment is
Yellow Fluorescent Protein (YFP) that has excitation and emission maxima at
513 and 545 nm, respectively . Under the conditions, where a significant FRET is
observed, what will happen to the FRET signal if the emission maxima of CFP is
made to shift to 475nm.
(A) Increase
(B) Decrease
(C) Remains unchanged
(D) May increase or decrease
183. Consider that the following polypeptides are being analyzed by Mass-
spectrometry (MS). These peptides are generated by proteolytic cleavage of a
protein that is expressed in mammalian cells and it has undergone glycosylation,
phosphorylation and acetylation. What will the expected positioning of the
spectra corresponding to these peptides (smaller to larger)?
(A) Asn-Lys-Ser/Asn-Lys-Thr/ Ser-Lys-Asn/ Thr-Arg-Asn
(B) Thr-Arg-Asn/ Ser-Lys-Asn/ Asn-Lys-Thr/ Asn-Lys-Ser
(C) Asn-Lys-Thr/ Asn-Lys-Ser/ Thr-Arg-Asn/ Ser-Lys-Asn
(D) Ser-Lys-Asn/ Asn-Lys-Thr/ Asn-Lys-Ser/ Thr-Arg-Asn
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
41
184. Assuming that the sequence of CDRs of an antibody are heavily enriched
with Tyrosine and Serine, what is likely to be the driving force stabilizing its
interaction with the antigen?
(A) Hydrophobic interaction
(B) Hydrogen bonding
(C) Van-der Waals interaction
(D) Covalent interactions
185. Hydrophobic amino acids are prevalent in transmembrane regions of
membrane embedded proteins because
(A) Hydrophobic amino acids destabilize the membrane bilayer and increase
membrane fluidity
(B) Phospholipid tails are hydrophobic and therefore membrane embedded
regions can be stabilized through hydrophobic interactions
(C) The side chains of hydrophobic amino acids interact covalently with
phospholipids
(D) It is just a random occurrence originated from heavy bias of hydrophobic
amino acids in protein sequences
186. Bryostatin is an anticancer agent obtained from
(A) Tethya crypta
(B) Salinospora tropica
(C) Bugula neritina
(D)Trididemnum solidum
187. Synthesis of White spot syndrome virus envelop takes place in
(A) endoplasmic reticulum
(B) nucleus
(C) mitochondria
(D) cytoplasm
188. Primary etiology of Epizootic Ulcerative Syndrome is a Fungus
Aphanomyces invadans. Its zoospores have the property to
(A) colonize gill epithelium
(B) cause ulcerations on skin
(C) travel long distances and penetrate deep in to the tissue
(D) cause multiple infections with protozoans
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
184. Assuming that the sequence of CDRs of an antibody are heavily enriched
with Tyrosine and Serine, what is likely to be the driving force stabilizing its
interaction with the antigen?
(A) Hydrophobic interaction
(B) Hydrogen bonding
(C) Van-der Waals interaction
(D) Covalent interactions
185. Hydrophobic amino acids are prevalent in transmembrane regions of
membrane embedded proteins because
(A) Hydrophobic amino acids destabilize the membrane bilayer and increase
membrane fluidity
(B) Phospholipid tails are hydrophobic and therefore membrane embedded
regions can be stabilized through hydrophobic interactions
(C) The side chains of hydrophobic amino acids interact covalently with
phospholipids
(D) It is just a random occurrence originated from heavy bias of hydrophobic
amino acids in protein sequences
186. Bryostatin is an anticancer agent obtained from
(A) Tethya crypta
(B) Salinospora tropica
(C) Bugula neritina
(D)Trididemnum solidum
187. Synthesis of White spot syndrome virus envelop takes place in
(A) endoplasmic reticulum
(B) nucleus
(C) mitochondria
(D) cytoplasm
188. Primary etiology of Epizootic Ulcerative Syndrome is a Fungus
Aphanomyces invadans. Its zoospores have the property to
(A) colonize gill epithelium
(B) cause ulcerations on skin
(C) travel long distances and penetrate deep in to the tissue
(D) cause multiple infections with protozoans
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
42
189. Gonad Inhibiting Hormone (GIH) in crustaceans is synthesized in
(A) Thoracic ganglion
(B) Sinus gland
(C) X-organ
(D) Y-organ
190. The synthetic equivalent of neuro-pharmacologically active peptides
obtained from the marine snail Conus magus is
(A) Zinconotide
(B) Discodermin
(C) Didemnins
(D) Dolastatins
191. Viral encephalopathy and retinopathy in fishes are caused by
(A) Betanodavirus
(B) Rhabdovirus
(C) Baculovirus
(D) Rotavirus
192. A pigment isolated from marine red algae that finds application in flow
cytometry is
(A) Xanthophyll
(B) Phycoerythrin
(C) Chlorophyll
(D) Fluorescamine
193. Which of the following electron acceptor that is used by the bacteria is
mainly responsible for marine corrosion?
A) O2
B) NO3
-
C) SO4
-2
D) CO2
194. Which one of the following bacterial species is responsible for the
Scombroid fish poisoning?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
189. Gonad Inhibiting Hormone (GIH) in crustaceans is synthesized in
(A) Thoracic ganglion
(B) Sinus gland
(C) X-organ
(D) Y-organ
190. The synthetic equivalent of neuro-pharmacologically active peptides
obtained from the marine snail Conus magus is
(A) Zinconotide
(B) Discodermin
(C) Didemnins
(D) Dolastatins
191. Viral encephalopathy and retinopathy in fishes are caused by
(A) Betanodavirus
(B) Rhabdovirus
(C) Baculovirus
(D) Rotavirus
192. A pigment isolated from marine red algae that finds application in flow
cytometry is
(A) Xanthophyll
(B) Phycoerythrin
(C) Chlorophyll
(D) Fluorescamine
193. Which of the following electron acceptor that is used by the bacteria is
mainly responsible for marine corrosion?
A) O2
B) NO3
-
C) SO4
-2
D) CO2
194. Which one of the following bacterial species is responsible for the
Scombroid fish poisoning?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
43
(A) Aermonas hydrophila
(B) Photobacterium phosphoreum
(C) Vibrio parahaemolyticus
(D) E. coli
195. The larval stage of Macrobrachium rosenbergii is referred to as
(A) Nauplii
(B) Zoea
(C) Mysis
(D) Megalopa
196. The group of marine algae which have urea cycle is
(A) Green algae
(B) Cyanobacteria
(C) Diatoms
(D) Red algae
197. Which of the following algae is responsible for the red color of RED SEA?
(A) Trichodesmium erythraeum
(B) Noctiluca scintillans
(C) Karenia brevis
(D) Chaetoceros sp.
198. An oligotrophic lake has
(A) High level of nutrients in water
(B) High aquatic productivity
(C) Algal blooms
(D) Low nutrients and low productivity
199. After sodium chloride, which one of the following compounds has the
maximum concentration in sea water?
(A) Magnesium sulphate
(B) Calcium sulphate
(C) Magnesium chloride
(D) Potassium sulphate
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
(A) Aermonas hydrophila
(B) Photobacterium phosphoreum
(C) Vibrio parahaemolyticus
(D) E. coli
195. The larval stage of Macrobrachium rosenbergii is referred to as
(A) Nauplii
(B) Zoea
(C) Mysis
(D) Megalopa
196. The group of marine algae which have urea cycle is
(A) Green algae
(B) Cyanobacteria
(C) Diatoms
(D) Red algae
197. Which of the following algae is responsible for the red color of RED SEA?
(A) Trichodesmium erythraeum
(B) Noctiluca scintillans
(C) Karenia brevis
(D) Chaetoceros sp.
198. An oligotrophic lake has
(A) High level of nutrients in water
(B) High aquatic productivity
(C) Algal blooms
(D) Low nutrients and low productivity
199. After sodium chloride, which one of the following compounds has the
maximum concentration in sea water?
(A) Magnesium sulphate
(B) Calcium sulphate
(C) Magnesium chloride
(D) Potassium sulphate
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
44
200. Which of the following is an example of an intellectual property right?
(A) A book that you own
(B) The copyright on a book
(C) The deed to a plot of land
(D) An airline ticket
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
200. Which of the following is an example of an intellectual property right?
(A) A book that you own
(B) The copyright on a book
(C) The deed to a plot of land
(D) An airline ticket
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
Q. No. Key
41 3✓
42 1✓
43 2
44 1***
45 4
46 1
47 4
48 4✓
49 3
50 4
51 2✓
52 1 or 2**
53 1
54 4✓
55 2
56 2
57 1✓
58 2
59 3
60 1
61 2✓
62 2
63 *
64 3✓
65 4
66 3
67 2
68 4✓
69 3
70 3
71 3
72 3 or 4**
73 3
74 1
75 2
76 4
77 1
78 1
79 4
80 2
Q. No. Key
81 3
82 3
83 4
84 4
85 3
86 4
87 3
88 3✓
89 3
90 4
91 4✓
92 3✓
93 3✓
94 4
95 2
96 2
97 3
98 4✓
99 2✓
100 1
101 2
102 1
103 3
104 1
105 4
106 3
107 2
108 1
109 2***
110 2
111 3
112 1✓
113 1✓
114 1
115 2
116 4✓
117 1
118 2
119 1
120 3
Q. No. Key
1 2
2 3
3 2
4 3
5 3
6 2
7 4
8 3
9 3
10 3
11 2 ✓
12 4
13 4
14 3
15 2✓
16 1
17 3
18 1
19 2
20 4
21 1
22 1
23 3
24 1
25 3
26 3✓
27 1
28 3✓
29 3
30 2
31 1✓
32 2
33 1
34 3
35 3✓
36 4
37 4
38 1
39 3✓
40 2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
41 3✓
42 1✓
43 2
44 1***
45 4
46 1
47 4
48 4✓
49 3
50 4
51 2✓
52 1 or 2**
53 1
54 4✓
55 2
56 2
57 1✓
58 2
59 3
60 1
61 2✓
62 2
63 *
64 3✓
65 4
66 3
67 2
68 4✓
69 3
70 3
71 3
72 3 or 4**
73 3
74 1
75 2
76 4
77 1
78 1
79 4
80 2
Q. No. Key
81 3
82 3
83 4
84 4
85 3
86 4
87 3
88 3✓
89 3
90 4
91 4✓
92 3✓
93 3✓
94 4
95 2
96 2
97 3
98 4✓
99 2✓
100 1
101 2
102 1
103 3
104 1
105 4
106 3
107 2
108 1
109 2***
110 2
111 3
112 1✓
113 1✓
114 1
115 2
116 4✓
117 1
118 2
119 1
120 3
Q. No. Key
1 2
2 3
3 2
4 3
5 3
6 2
7 4
8 3
9 3
10 3
11 2 ✓
12 4
13 4
14 3
15 2✓
16 1
17 3
18 1
19 2
20 4
21 1
22 1
23 3
24 1
25 3
26 3✓
27 1
28 3✓
29 3
30 2
31 1✓
32 2
33 1
34 3
35 3✓
36 4
37 4
38 1
39 3✓
40 2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
Q. No. Key
161 4
162 3
163 3
164 1
165 1
166 3
167 2✓
168 3
169 3✓
170 1
171 3
172 1
173 2
174 2
175 4
176 1 or 2**
177 2
178 2
179 3
180 2
181 1✓
182 2
183 2
184 2
185 2
186 3
187 2
188 3
189 3
190 1
191 1
192 2
193 3
194 *
195 2 ***
196 3
197 1
198 4
199 1
200 2
* Benefit of marks to all those candidates who attempted **Both options are correct
***Option changed ✓ No Change
Q. No. Key
121 4
122 1
123 3
124 4
125 1
126 2
127 3✓
128 2
129 4
130 3
131 3
132 2
133 4
134 2✓
135 2
136 4
137 1***
138 2
139 *
140 3
141 4
142 4
143 1
144 2
145 1
146 2✓
147 3
148 1 or 3**
149 3✓
150 2
151 1
152 3
153 3
154 2
155 2
156 2
157 2
158 2
159 1
160 3
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
161 4
162 3
163 3
164 1
165 1
166 3
167 2✓
168 3
169 3✓
170 1
171 3
172 1
173 2
174 2
175 4
176 1 or 2**
177 2
178 2
179 3
180 2
181 1✓
182 2
183 2
184 2
185 2
186 3
187 2
188 3
189 3
190 1
191 1
192 2
193 3
194 *
195 2 ***
196 3
197 1
198 4
199 1
200 2
* Benefit of marks to all those candidates who attempted **Both options are correct
***Option changed ✓ No Change
Q. No. Key
121 4
122 1
123 3
124 4
125 1
126 2
127 3✓
128 2
129 4
130 3
131 3
132 2
133 4
134 2✓
135 2
136 4
137 1***
138 2
139 *
140 3
141 4
142 4
143 1
144 2
145 1
146 2✓
147 3
148 1 or 3**
149 3✓
150 2
151 1
152 3
153 3
154 2
155 2
156 2
157 2
158 2
159 1
160 3
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
1
BET 2017
Question Paper
Part A
1. For a single substrate reaction, doubling the substrate concentration increases the rate by 10
fold. The order of the reaction is
a. 2
b. 3
c. 4
d. Between 3 and 4
2. An enzyme reaction follows Michaelis Menten kinetics. What will be the reaction velocity at a
substrate concentration = Km/3?
a. Vmax
b. Vmax/3
c. Vmax/2
d. Vmax/4
3. What will be the required volumes of 1N HCl and 4N NaOH to prepare one litre solution of pH
7?
a. 500 ml; 500 ml
b. 800 ml; 200 ml
c. 600 ml; 400 ml
d. 200 ml; 800 ml
4. During a batch culture experiment, the following data was obtained
Time (h) Substrate (g/l) Cell (g/l) Product (g/l)
2 40 1 2
8 20 5 7
What will be the product yield (g) per gram of substrate?
a. 10
b. 0.25
c. 0.5
d. 2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
BET 2017
Question Paper
Part A
1. For a single substrate reaction, doubling the substrate concentration increases the rate by 10
fold. The order of the reaction is
a. 2
b. 3
c. 4
d. Between 3 and 4
2. An enzyme reaction follows Michaelis Menten kinetics. What will be the reaction velocity at a
substrate concentration = Km/3?
a. Vmax
b. Vmax/3
c. Vmax/2
d. Vmax/4
3. What will be the required volumes of 1N HCl and 4N NaOH to prepare one litre solution of pH
7?
a. 500 ml; 500 ml
b. 800 ml; 200 ml
c. 600 ml; 400 ml
d. 200 ml; 800 ml
4. During a batch culture experiment, the following data was obtained
Time (h) Substrate (g/l) Cell (g/l) Product (g/l)
2 40 1 2
8 20 5 7
What will be the product yield (g) per gram of substrate?
a. 10
b. 0.25
c. 0.5
d. 2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
2
5. In a bacterial growth experiment, the concentration of cells increased from 10,000 cells/ml to
30,000 cells/ml in 3 h during the exponential growth phase. The doubling time (h) of the
bacteria is:
a. 2.5
b. 3.0
c. 1.0
d. 1.9
6. A 20mer DNA contains A, G and C only. In how many ways can this DNA sequence be
constructed?
a. 4
20
b. 20
4
c. 3
20
d. 20
3
7. Match the Matrices in Group I with appropriate gradient elution condition from Group II
Group I Group II
1 DEAE-Sephacel P Isocratic gradient
2 Phenyl-Sepharose Q Increasing concentrations of sodium
chloride
3 Chromatofocusing R Decreasing concentrations of ammonium
sulphate
4 Sephadex-G100 S pH gradient
a. 1-Q; 2-R; 3-S; 4-P
b. 1-Q; 2-R; 3-P; 4-S
c. 1-R; 2-Q; 3-P; 4-S
d. 1-S; 2-P; 3-Q; 4-R
8. The activity of Enzyme X (total volume 5.3 ml) is 2.34 micromoles of product formed per min.
The total protein content of this solution is 0.8 mg. What is the specific activity?
a. 2.93
b. 1.87
c. 18.2
d. 15.5
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
5. In a bacterial growth experiment, the concentration of cells increased from 10,000 cells/ml to
30,000 cells/ml in 3 h during the exponential growth phase. The doubling time (h) of the
bacteria is:
a. 2.5
b. 3.0
c. 1.0
d. 1.9
6. A 20mer DNA contains A, G and C only. In how many ways can this DNA sequence be
constructed?
a. 4
20
b. 20
4
c. 3
20
d. 20
3
7. Match the Matrices in Group I with appropriate gradient elution condition from Group II
Group I Group II
1 DEAE-Sephacel P Isocratic gradient
2 Phenyl-Sepharose Q Increasing concentrations of sodium
chloride
3 Chromatofocusing R Decreasing concentrations of ammonium
sulphate
4 Sephadex-G100 S pH gradient
a. 1-Q; 2-R; 3-S; 4-P
b. 1-Q; 2-R; 3-P; 4-S
c. 1-R; 2-Q; 3-P; 4-S
d. 1-S; 2-P; 3-Q; 4-R
8. The activity of Enzyme X (total volume 5.3 ml) is 2.34 micromoles of product formed per min.
The total protein content of this solution is 0.8 mg. What is the specific activity?
a. 2.93
b. 1.87
c. 18.2
d. 15.5
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
3
9. Match the techniques in Group I with applications given in Group II
Group I Group II
1 Salting out P pI determination
2 Ultracentrifugation Q Protein precipitation
3 Dialysis R Sedimentation coefficient
4 Isoelectric focusing S Removal of low molecular
weight impurities
a. 1-Q; 2-R; 3-S; 4-P
b. 1-P; 2-R; 3-Q; 4-S
c. 1-S; 2-P; 3R-; 4-Q
d. 1-Q; 2-S; 3-R; 4-P
10. What would be the number of protein molecules present in 1.0 mg of protein having a
molecular weight of 25 kDa?
a. 2.4×10
15
b. 2.4×10
16
c. 2.4×10
17
d. 2.4×10
18
11. Calculate the ionic strength (M) of 50 ml of 0.75 % (w/v) NaCl solution?
a. 0.128
b. 0.256
c. 7.8
d. 0.064
12. A cell suspension (1.5×10
5
cells per ml)
was treated with 1mM HgCl2 for 30 min. After
treatment, the cell suspension was diluted 10 fold and 100 microliter was plated which gave 5
colonies. Calculate the percentage of cells that survived?
a. 0.00033
b. 3.3
c. 0.033
d. 0.33
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
9. Match the techniques in Group I with applications given in Group II
Group I Group II
1 Salting out P pI determination
2 Ultracentrifugation Q Protein precipitation
3 Dialysis R Sedimentation coefficient
4 Isoelectric focusing S Removal of low molecular
weight impurities
a. 1-Q; 2-R; 3-S; 4-P
b. 1-P; 2-R; 3-Q; 4-S
c. 1-S; 2-P; 3R-; 4-Q
d. 1-Q; 2-S; 3-R; 4-P
10. What would be the number of protein molecules present in 1.0 mg of protein having a
molecular weight of 25 kDa?
a. 2.4×10
15
b. 2.4×10
16
c. 2.4×10
17
d. 2.4×10
18
11. Calculate the ionic strength (M) of 50 ml of 0.75 % (w/v) NaCl solution?
a. 0.128
b. 0.256
c. 7.8
d. 0.064
12. A cell suspension (1.5×10
5
cells per ml)
was treated with 1mM HgCl2 for 30 min. After
treatment, the cell suspension was diluted 10 fold and 100 microliter was plated which gave 5
colonies. Calculate the percentage of cells that survived?
a. 0.00033
b. 3.3
c. 0.033
d. 0.33
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
4
13. Purification data for an enzyme is given below:
Steps Purification step Volume
(ml)
Total protein
(mg)
Total activity
(micromoles per min)
I Cell-free extract 20 100 150
II Ni-NTA chromatography 4 10 120
What is the fold-purification?
a. 8
b. 13.5
c. 10.5
d. 18
14. Competent cells prepared in your lab have a transformation efficiency of 10
8
cfu/µg of
plasmid DNA. These competent cells (100 µl) were transformed with 10 ng of plasmid DNA
using heat shock method followed by addition of 900 µl of LB medium. 100 µl of transformed
cells were plated on antibiotic containing plates. The number of colonies expected to be on
the plate would be:
a. 10
5
b. 10
4
c. 10
3
d. 10
2
15. A disease has a prevalence of 1 in 1000 in the general population. A diagnostic kit for the
disease has 10% false positives and no false negative. In a general population, if a person
tested with the kit gives a positive result, the probability that he DOES NOT have the disease
is approximately:
a. 9%
b. 90%
c. 99%
d. 99.9%
16. From the start point on the runaway to final take off, an aircraft takes 50 seconds. During this
period it covers a distance of 1 km. What is the acceleration (m/s
2
)
during take-off?
a. 0.6
b. 0.8
c. 1.0
d. 1.2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
13. Purification data for an enzyme is given below:
Steps Purification step Volume
(ml)
Total protein
(mg)
Total activity
(micromoles per min)
I Cell-free extract 20 100 150
II Ni-NTA chromatography 4 10 120
What is the fold-purification?
a. 8
b. 13.5
c. 10.5
d. 18
14. Competent cells prepared in your lab have a transformation efficiency of 10
8
cfu/µg of
plasmid DNA. These competent cells (100 µl) were transformed with 10 ng of plasmid DNA
using heat shock method followed by addition of 900 µl of LB medium. 100 µl of transformed
cells were plated on antibiotic containing plates. The number of colonies expected to be on
the plate would be:
a. 10
5
b. 10
4
c. 10
3
d. 10
2
15. A disease has a prevalence of 1 in 1000 in the general population. A diagnostic kit for the
disease has 10% false positives and no false negative. In a general population, if a person
tested with the kit gives a positive result, the probability that he DOES NOT have the disease
is approximately:
a. 9%
b. 90%
c. 99%
d. 99.9%
16. From the start point on the runaway to final take off, an aircraft takes 50 seconds. During this
period it covers a distance of 1 km. What is the acceleration (m/s
2
)
during take-off?
a. 0.6
b. 0.8
c. 1.0
d. 1.2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
5
17. A ladder of 5 m length is standing against a wall. The distance between the wall and the
base of the ladder is 4 m. If the top of ladder slips down by 0.5 m, the foot of the ladder will
shift by:
a. 0.25 m
b. 0.33 m
c. 0.5 m
d. 0.6 m
18. A person jogs from his home to the playground at 6 km/h. He walks back from the
playground to his home at 4 km/h. What is his average speed (km/h)?
a. 4.25
b. 4.8
c. 5
d. 5.25
19. Which one of the following numbers is equal to three times the sum of its digits?
a. 15
b. 12
c. 24
d. 27
20. If 5 ml of 20% ethanol is mixed with 25 ml of 80% ethanol, the resulting solution will
approximately be:
a. 50% ethanol
b. 70% ethanol
c. 100% ethanol
d. 80% ethanol
21. When DNA is extracted from bacterial cells and analyzed for base composition, it is found
that 38% of bases are Cytosine (C). What percentage of bases is Adenine (A)?
a. 12
b. 24
c. 38
d. 62
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17. A ladder of 5 m length is standing against a wall. The distance between the wall and the
base of the ladder is 4 m. If the top of ladder slips down by 0.5 m, the foot of the ladder will
shift by:
a. 0.25 m
b. 0.33 m
c. 0.5 m
d. 0.6 m
18. A person jogs from his home to the playground at 6 km/h. He walks back from the
playground to his home at 4 km/h. What is his average speed (km/h)?
a. 4.25
b. 4.8
c. 5
d. 5.25
19. Which one of the following numbers is equal to three times the sum of its digits?
a. 15
b. 12
c. 24
d. 27
20. If 5 ml of 20% ethanol is mixed with 25 ml of 80% ethanol, the resulting solution will
approximately be:
a. 50% ethanol
b. 70% ethanol
c. 100% ethanol
d. 80% ethanol
21. When DNA is extracted from bacterial cells and analyzed for base composition, it is found
that 38% of bases are Cytosine (C). What percentage of bases is Adenine (A)?
a. 12
b. 24
c. 38
d. 62
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6
22. Equal volumes of cell suspensions of Escherichia coli, Saccharomyces cerevisiae,
Streptococcus lactis and Mycoplasma pneumoniae, all have the same OD600nm = 0.50. Which
cell suspension would have the minimum and maximum number of cells, respectively?
a. Saccharomyces cerevisiae and Mycoplasma pneumoniae
b. Saccharomyces cerevisiae and Streptococcus lactis
c. Escherichia coli and Mycoplasma pneumoniae
d. Mycoplasma pneumoniae and Streptococcus lactis
23. A bag contains 4 red, 5 green and 7 yellow balls. If 2 balls are picked simultaneously in a
random manner from the bag, the probability of both being green is:
a. 1/16
b. 1/8
c. 1/12
d. 5/16
24. The spectroscopic method for detection of functional groups is:
a. CD spectroscopy
b. FTIR spectroscopy
c. ESR spectroscopy
d. UV-VISIBLE spectroscopy
25. Match the native microbial sources in Group I with the products in Group II
Group I Group II
M. Leuconostoc mesenteroides 1. Lysine
N. Lactococcus lactis 2. Cheese
O. Brevibacterium brevis 3. Dextran
P. Penicillium roqueforti 4. Nisin
a. M-2. N-1. O-4. P-3.
b. M-1. N-2. O-3. P-4.
c. M-3. N-4. O-1. P- 2.
d. M-4. N-3. O-2. P-1.
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22. Equal volumes of cell suspensions of Escherichia coli, Saccharomyces cerevisiae,
Streptococcus lactis and Mycoplasma pneumoniae, all have the same OD600nm = 0.50. Which
cell suspension would have the minimum and maximum number of cells, respectively?
a. Saccharomyces cerevisiae and Mycoplasma pneumoniae
b. Saccharomyces cerevisiae and Streptococcus lactis
c. Escherichia coli and Mycoplasma pneumoniae
d. Mycoplasma pneumoniae and Streptococcus lactis
23. A bag contains 4 red, 5 green and 7 yellow balls. If 2 balls are picked simultaneously in a
random manner from the bag, the probability of both being green is:
a. 1/16
b. 1/8
c. 1/12
d. 5/16
24. The spectroscopic method for detection of functional groups is:
a. CD spectroscopy
b. FTIR spectroscopy
c. ESR spectroscopy
d. UV-VISIBLE spectroscopy
25. Match the native microbial sources in Group I with the products in Group II
Group I Group II
M. Leuconostoc mesenteroides 1. Lysine
N. Lactococcus lactis 2. Cheese
O. Brevibacterium brevis 3. Dextran
P. Penicillium roqueforti 4. Nisin
a. M-2. N-1. O-4. P-3.
b. M-1. N-2. O-3. P-4.
c. M-3. N-4. O-1. P- 2.
d. M-4. N-3. O-2. P-1.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
7
26. Match the enzyme in Group I with the application/function in Group II
Group I Group II
M. Pectinase 1. Lactose free milk products
N. Papain 2. High Fructose Corn Syrup
O. Glucose isomerase 3. Juice Clarification
P. β- Galactosidase 4. Meat tenderisation
a. M-4. N-3. O-2. P-1.
b. M-3. N-4. O-2. P-1.
c. M-1. N-2. O-4. P- 3.
d. M-2. N-1. O-3. P-4.
27. Histone code implies which of the following?
a. Combination of different histone proteins to form a nucleosome.
b. Modification of histones.
c. Different histone proteins present in different eukaryotic organisms.
d. Different order of histone proteins in a given nucleosome.
28. The A260 of a plasmid solution after 100-fold dilution is 0.2. Given that A260 of 1.0
represents 50 µg/ml of DNA and the total volume of isolated plasmid solution is 50 µl,
what will be the concentration and amount respectively of the isolated plasmid?
a. 1.0 µg/µl and 50 µg
b. 1.0 mg/µl and 50 mg
c. 10 µg/µl and 50 µg
d. 10 mg/µl and 50 mg
29. A gene was cloned into a unique HindIII restriction site present in the ampicillin resistance
gene of a vector that contains both ampicillin and kanamycin resistance genes. To select
for only recombinant clones, the transformation mixture should be plated on which of the
following plates?
a. Ampicillin containing plate
b. Ampicillin plus Kanamycin containing plate
c. Ampicillin containing plate followed by replica-plating on kanamycin containing plate
d. Kanamycin containing plate followed by replica-plating on ampicillin containing plate
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
26. Match the enzyme in Group I with the application/function in Group II
Group I Group II
M. Pectinase 1. Lactose free milk products
N. Papain 2. High Fructose Corn Syrup
O. Glucose isomerase 3. Juice Clarification
P. β- Galactosidase 4. Meat tenderisation
a. M-4. N-3. O-2. P-1.
b. M-3. N-4. O-2. P-1.
c. M-1. N-2. O-4. P- 3.
d. M-2. N-1. O-3. P-4.
27. Histone code implies which of the following?
a. Combination of different histone proteins to form a nucleosome.
b. Modification of histones.
c. Different histone proteins present in different eukaryotic organisms.
d. Different order of histone proteins in a given nucleosome.
28. The A260 of a plasmid solution after 100-fold dilution is 0.2. Given that A260 of 1.0
represents 50 µg/ml of DNA and the total volume of isolated plasmid solution is 50 µl,
what will be the concentration and amount respectively of the isolated plasmid?
a. 1.0 µg/µl and 50 µg
b. 1.0 mg/µl and 50 mg
c. 10 µg/µl and 50 µg
d. 10 mg/µl and 50 mg
29. A gene was cloned into a unique HindIII restriction site present in the ampicillin resistance
gene of a vector that contains both ampicillin and kanamycin resistance genes. To select
for only recombinant clones, the transformation mixture should be plated on which of the
following plates?
a. Ampicillin containing plate
b. Ampicillin plus Kanamycin containing plate
c. Ampicillin containing plate followed by replica-plating on kanamycin containing plate
d. Kanamycin containing plate followed by replica-plating on ampicillin containing plate
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
8
30. A student sequenced a DNA using Sanger‟s method and obtained the following
autoradiogram.
The sequence of DNA is:
a. 5‟ CTTAG 3‟
b. 5‟ GAATC 3‟
c. 5‟ CTAAG 3‟
d. 5‟ AATTG 3‟
31. From among the options given below, RNA polymerase II transcribes which one of the
following?
a. rRNA
b. tRNA
c. MicroRNA
d. None of the given options
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
30. A student sequenced a DNA using Sanger‟s method and obtained the following
autoradiogram.
The sequence of DNA is:
a. 5‟ CTTAG 3‟
b. 5‟ GAATC 3‟
c. 5‟ CTAAG 3‟
d. 5‟ AATTG 3‟
31. From among the options given below, RNA polymerase II transcribes which one of the
following?
a. rRNA
b. tRNA
c. MicroRNA
d. None of the given options
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
9
32. The melting curve of a DNA solution is represented below.
It suggests that:
a. The given DNA has one stretch with a biased base composition
b. The given DNA has two stretches with a biased base composition
c. The given DNA has three stretches with a biased base composition
d. The given DNA has an unbiased base composition
33. The DNA binding domain of a transcription factor that is specifically required for the
regulation of gene A is exchanged with the DNA binding domain of another transcription
factor that is required specifically for the regulation of gene B. This chimeric transcription
factor will regulate:
a. Gene A only
b. Gene B only
c. Both Gene A and B
d. Neither Gene A nor Gene B
34. Protein synthesis in eukaryotic cells terminates at the stop codon because:
a. mRNA synthesis stops at the stop codon.
b. The tRNA corresponding to the stop codon cannot bind to an amino acid.
c. There is no naturally occurring tRNA with an anticodon corresponding to the stop
codon.
d. The conformation around stop codons prevents binding of aminoacyl tRNA.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
32. The melting curve of a DNA solution is represented below.
It suggests that:
a. The given DNA has one stretch with a biased base composition
b. The given DNA has two stretches with a biased base composition
c. The given DNA has three stretches with a biased base composition
d. The given DNA has an unbiased base composition
33. The DNA binding domain of a transcription factor that is specifically required for the
regulation of gene A is exchanged with the DNA binding domain of another transcription
factor that is required specifically for the regulation of gene B. This chimeric transcription
factor will regulate:
a. Gene A only
b. Gene B only
c. Both Gene A and B
d. Neither Gene A nor Gene B
34. Protein synthesis in eukaryotic cells terminates at the stop codon because:
a. mRNA synthesis stops at the stop codon.
b. The tRNA corresponding to the stop codon cannot bind to an amino acid.
c. There is no naturally occurring tRNA with an anticodon corresponding to the stop
codon.
d. The conformation around stop codons prevents binding of aminoacyl tRNA.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
10
35. A DNA fragment was digested with a restriction enzyme X with the recognition sequence
5‟ GGATCC 3‟
3‟ CCTAGG 5‟
that would cleave at the position of the arrow as shown. You want to clone this fragment
in a vector which does not contain the restriction site for X. Which one of the following
enzymes will you use to digest the vector?
The sequence of the restriction site and position of cleavage for each enzyme is shown
below:
a. Restriction enzyme A: 5‟ GAATTC 3‟
3‟ CTTAAG 5‟
b.
Restriction enzyme B: 5‟ CTCGAG 3‟
3‟ GAGCTC 5‟
c.
Restriction enzyme C: 5‟ AGATCT 3‟
3‟ TCTAGA 5‟
d.
Restriction enzyme D: 5‟ GTCGAG 3‟
3‟ CAGCTC 5‟
36. Which one of the following is NOT important for regulation of the tryptophan operon by
attenuation?
a. Presence of two adjacent codons for tryptophan in the leader peptide sequence
b. Coupled transcription-translation
c. Concentration of tRNA charged with tryptophan
d. The operator sequence of tryptophan operon
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
35. A DNA fragment was digested with a restriction enzyme X with the recognition sequence
5‟ GGATCC 3‟
3‟ CCTAGG 5‟
that would cleave at the position of the arrow as shown. You want to clone this fragment
in a vector which does not contain the restriction site for X. Which one of the following
enzymes will you use to digest the vector?
The sequence of the restriction site and position of cleavage for each enzyme is shown
below:
a. Restriction enzyme A: 5‟ GAATTC 3‟
3‟ CTTAAG 5‟
b.
Restriction enzyme B: 5‟ CTCGAG 3‟
3‟ GAGCTC 5‟
c.
Restriction enzyme C: 5‟ AGATCT 3‟
3‟ TCTAGA 5‟
d.
Restriction enzyme D: 5‟ GTCGAG 3‟
3‟ CAGCTC 5‟
36. Which one of the following is NOT important for regulation of the tryptophan operon by
attenuation?
a. Presence of two adjacent codons for tryptophan in the leader peptide sequence
b. Coupled transcription-translation
c. Concentration of tRNA charged with tryptophan
d. The operator sequence of tryptophan operon
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11
37. A plasmid DNA when digested with EcoRI gave a single band of 16 Kb. When the same
plasmid was digested with BamHI it gave two bands of 6Kb and 4 Kb. The plasmid has:
a. Single site of EcoRI and 2 sites of BamHI
b. Single site of EcoRI and 3 sites of BamHI
c. Single site of EcoRI and 2 sites of BamHI
d. 2 sites of EcoRI and 2 sites of BamHI
38. Myeloma cells fused with spleen cells in hybridoma technology are:
a. Immortal and antibody producing cells
b. Mortal and antibody producing cells
c. Hypoxanthine guanine phosphoribosyl transferase lacking cells
d. Thymidine kinase lacking cells
39. Enzyme inactivation by suicide inhibitors should be:
a. allosteric-site directed
b. active-site directed
c. regulatory-site directed
d. both allosteric and regulatory-site directed
40. Resolution in adsorption chromatography is achieved at:
a. Elution stage only
b. Binding stage only
c. Equilibration stage only
d. Both binding and elution stage
41. Protein kinases phosphorylate proteins at hydroxyl groups on amino acid side chains.
Which one of the following groups of amino acids contain side chain hydroxyl groups?
a. Aspartate, glutamate and serine
b. Serine, threonine and tyrosine
c. Lysine, arginine and proline
d. Threonine, phenylalanine and arginine
42. In a segment of a transcribed gene, the non-template strand of DNA has the following
sequence 5‟..AGCTCACTG..3‟. What will be the corresponding 5‟ to 3‟ sequence in the
RNA produced from this segment of the gene?
a. CAGUGAGCU
b. AGCUCACUG
c. CAGTGAGCT
d. UCGAUGAC
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37. A plasmid DNA when digested with EcoRI gave a single band of 16 Kb. When the same
plasmid was digested with BamHI it gave two bands of 6Kb and 4 Kb. The plasmid has:
a. Single site of EcoRI and 2 sites of BamHI
b. Single site of EcoRI and 3 sites of BamHI
c. Single site of EcoRI and 2 sites of BamHI
d. 2 sites of EcoRI and 2 sites of BamHI
38. Myeloma cells fused with spleen cells in hybridoma technology are:
a. Immortal and antibody producing cells
b. Mortal and antibody producing cells
c. Hypoxanthine guanine phosphoribosyl transferase lacking cells
d. Thymidine kinase lacking cells
39. Enzyme inactivation by suicide inhibitors should be:
a. allosteric-site directed
b. active-site directed
c. regulatory-site directed
d. both allosteric and regulatory-site directed
40. Resolution in adsorption chromatography is achieved at:
a. Elution stage only
b. Binding stage only
c. Equilibration stage only
d. Both binding and elution stage
41. Protein kinases phosphorylate proteins at hydroxyl groups on amino acid side chains.
Which one of the following groups of amino acids contain side chain hydroxyl groups?
a. Aspartate, glutamate and serine
b. Serine, threonine and tyrosine
c. Lysine, arginine and proline
d. Threonine, phenylalanine and arginine
42. In a segment of a transcribed gene, the non-template strand of DNA has the following
sequence 5‟..AGCTCACTG..3‟. What will be the corresponding 5‟ to 3‟ sequence in the
RNA produced from this segment of the gene?
a. CAGUGAGCU
b. AGCUCACUG
c. CAGTGAGCT
d. UCGAUGAC
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
12
43. If the DNA content of a diploid cell in the G1 phase of the cell cycle is X, then the DNA
content of the same cell at metaphase of meiosis I would be:
a. 2X
b. 4X
c. 0.5X
d. X
44. The α-helix in a protein is primarily due to:
a. Intramolecular hydrogen bond
b. Intermolecular hydrogen bond
c. van der Waals interaction between amino acids
d. covalent interactions
45. Amino acids with asymmetric Cβ atoms are:
a. Pro, Met
b. Lys, Ile, Val
c. Thr, Ile
d. Cys, Ser, Met
46. The strength of the hydrogen bond represented by D-H...A (where D is the donor atom, H
is the hydrogen atom and A is the acceptor atom) depends on
a. the D-H bond length and the nature of the D and A atoms.
b. the nature of the D and A atoms and the D...A distance.
c. the D...A distance and linearity of the angle DHA.
d. the H...A distance, linearity of the angle DHA and the nature of the atoms D & A.
47. The van der Waals energy of a single water molecule is:
a. 0
b. -0.2 Kcal/mol
c. -0.5 Kcal/mol
d. 0.2 Kcal/mol
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
43. If the DNA content of a diploid cell in the G1 phase of the cell cycle is X, then the DNA
content of the same cell at metaphase of meiosis I would be:
a. 2X
b. 4X
c. 0.5X
d. X
44. The α-helix in a protein is primarily due to:
a. Intramolecular hydrogen bond
b. Intermolecular hydrogen bond
c. van der Waals interaction between amino acids
d. covalent interactions
45. Amino acids with asymmetric Cβ atoms are:
a. Pro, Met
b. Lys, Ile, Val
c. Thr, Ile
d. Cys, Ser, Met
46. The strength of the hydrogen bond represented by D-H...A (where D is the donor atom, H
is the hydrogen atom and A is the acceptor atom) depends on
a. the D-H bond length and the nature of the D and A atoms.
b. the nature of the D and A atoms and the D...A distance.
c. the D...A distance and linearity of the angle DHA.
d. the H...A distance, linearity of the angle DHA and the nature of the atoms D & A.
47. The van der Waals energy of a single water molecule is:
a. 0
b. -0.2 Kcal/mol
c. -0.5 Kcal/mol
d. 0.2 Kcal/mol
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13
48. What is the minimum number of edges that meet at every branch node in a phylogenetic
tree?
a. 1
b. 2
c. 3
d. 4
49. According to the Induced-fit theory, an agonist is defined as:
a) A compound which produces an exothermic effect
b) A compound which occupies the receptor for a longer period
c) A compound that induces a specific conformational change in the macromolecule
d) A compound which hits the receptor center more often.
50. A recombinant protein is found to be expressed very poorly in E.coli. It is hypothesized
that the expression is blocked at the translational step. The first experimental technique to
test this is:
a. PCR followed by sequencing
b. Quantitative RT PCR
c. Western Blot
d. EMSA
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
48. What is the minimum number of edges that meet at every branch node in a phylogenetic
tree?
a. 1
b. 2
c. 3
d. 4
49. According to the Induced-fit theory, an agonist is defined as:
a) A compound which produces an exothermic effect
b) A compound which occupies the receptor for a longer period
c) A compound that induces a specific conformational change in the macromolecule
d) A compound which hits the receptor center more often.
50. A recombinant protein is found to be expressed very poorly in E.coli. It is hypothesized
that the expression is blocked at the translational step. The first experimental technique to
test this is:
a. PCR followed by sequencing
b. Quantitative RT PCR
c. Western Blot
d. EMSA
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
14
Part B
51. There are 3 genes A, B and C that are functionally related. There is a point
mutation in gene A due to which gene B is not expressed resulting in a non-
functional gene C product. What is the possible relationship between these 3
genes?
a. A is an enzyme, B and C are substrates of A
b. A is transcription factor for B and B is needed for C to be functional.
c. B is transcription factor for C and A
d. C is enzyme that requires A and B as its substrate
52. Detectable serum antibody against a T-independent pathogen is a good indication
that:
a. A functional B-cell system exists
b. A functional T-cell system exists
c. The patient has immune suppression
d. Both T and B cell systems are not functional
53. If a 1000 kb fragment of DNA has 10 evenly spaced and symmetric replication
origins and DNA polymerase moves at 1 kb per minute, how many minutes will it
take to produce two daughter molecules ignoring the potential problem at the end of
the linear piece of DNA? Assume that the 10 origins are evenly spaced from each
other, none starting from the ends of the chromosome.
a. 20
b. 30
c. 50
d. 100
54. In an experiment, 4 different N-terminal blocked purified proteins were treated with
glutaraldehyde, individually. One of the proteins did not get modified. What may be
the reason?
a. The protein lacks histidine
b. The protein lacks phenylalanine
c. The protein lacks lysine
d. The protein lacks arginine
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
Part B
51. There are 3 genes A, B and C that are functionally related. There is a point
mutation in gene A due to which gene B is not expressed resulting in a non-
functional gene C product. What is the possible relationship between these 3
genes?
a. A is an enzyme, B and C are substrates of A
b. A is transcription factor for B and B is needed for C to be functional.
c. B is transcription factor for C and A
d. C is enzyme that requires A and B as its substrate
52. Detectable serum antibody against a T-independent pathogen is a good indication
that:
a. A functional B-cell system exists
b. A functional T-cell system exists
c. The patient has immune suppression
d. Both T and B cell systems are not functional
53. If a 1000 kb fragment of DNA has 10 evenly spaced and symmetric replication
origins and DNA polymerase moves at 1 kb per minute, how many minutes will it
take to produce two daughter molecules ignoring the potential problem at the end of
the linear piece of DNA? Assume that the 10 origins are evenly spaced from each
other, none starting from the ends of the chromosome.
a. 20
b. 30
c. 50
d. 100
54. In an experiment, 4 different N-terminal blocked purified proteins were treated with
glutaraldehyde, individually. One of the proteins did not get modified. What may be
the reason?
a. The protein lacks histidine
b. The protein lacks phenylalanine
c. The protein lacks lysine
d. The protein lacks arginine
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
15
55. Which of the following signatures (the stretch of amino acids) in a protein will target
it to the nucleus?
a. Arg-Glu-Glu-Trp-Glu-Cys
b. Arg-Lys-Lys-Lys-Arg-Lys
c. Trp-Phe-Phe-Phe-Phe-Gly
d. Phe-Pro-Pro-Arg-Tyr-Tyr
56. Which one of the following rearrangements is NOT permitted during somatic
recombination in the heavy chain and light chain immunoglobulin loci?
a. DH:JH
b. VL:JL
c. VH:JH
d. VH:DH
57. Junctional diversity in CDR3 during gene rearrangement results from the addition
of:
a. Switch region nucleotides
b. P and N nucleotides
c. V, D and J nucleotides
d. Recombination signal sequences
58. The function of negative selection of thymocytes in the thymus is to eliminate:
a. Single-positive thymocytes
b. Double-positive thymocytes
c. Alloreactive thymocytes
d. Autoreactive thymocytes
59. Antigen recognition by T cells in the absence of co-stimulation results in:
a. Upregulation of B7.1
b. T-cell apoptosis
c. T-cell anergy
d. Upregulation of B7.2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
55. Which of the following signatures (the stretch of amino acids) in a protein will target
it to the nucleus?
a. Arg-Glu-Glu-Trp-Glu-Cys
b. Arg-Lys-Lys-Lys-Arg-Lys
c. Trp-Phe-Phe-Phe-Phe-Gly
d. Phe-Pro-Pro-Arg-Tyr-Tyr
56. Which one of the following rearrangements is NOT permitted during somatic
recombination in the heavy chain and light chain immunoglobulin loci?
a. DH:JH
b. VL:JL
c. VH:JH
d. VH:DH
57. Junctional diversity in CDR3 during gene rearrangement results from the addition
of:
a. Switch region nucleotides
b. P and N nucleotides
c. V, D and J nucleotides
d. Recombination signal sequences
58. The function of negative selection of thymocytes in the thymus is to eliminate:
a. Single-positive thymocytes
b. Double-positive thymocytes
c. Alloreactive thymocytes
d. Autoreactive thymocytes
59. Antigen recognition by T cells in the absence of co-stimulation results in:
a. Upregulation of B7.1
b. T-cell apoptosis
c. T-cell anergy
d. Upregulation of B7.2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
16
60. Mutation of homeotic genes often result in which one of the following developmental
defects in Drosophila?
a. Absence of a group of contiguous segments
b. Transformation of one segment to another
c. Tumor formation in imaginal discs
d. Absence of every other segment along the antero-posterior axis
61. Which one of the following graphs represent the kinetics of ion transport through a
membrane channel?
a.
b.
.
a. . b.
.
c. d.
62. Transition type of gene mutation is caused when
a. GC is replaced by TA
b. CG is replaced by GC
c. AT is replaced by CG
d. AT is replaced by GC
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
60. Mutation of homeotic genes often result in which one of the following developmental
defects in Drosophila?
a. Absence of a group of contiguous segments
b. Transformation of one segment to another
c. Tumor formation in imaginal discs
d. Absence of every other segment along the antero-posterior axis
61. Which one of the following graphs represent the kinetics of ion transport through a
membrane channel?
a.
b.
.
a. . b.
.
c. d.
62. Transition type of gene mutation is caused when
a. GC is replaced by TA
b. CG is replaced by GC
c. AT is replaced by CG
d. AT is replaced by GC
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
17
63. Which one of the following is NOT enriched in eukaryotic promoters located in
active chromatin?
a. Acetylated histones
b. DNAse I hypersensitive sites
c. Methylated cytosine
d. Bound TFIID
64. One of the reasons why non-substrate inducers (e.g. IPTG) are preferred over
substrate inducers (e.g. lactose) for induction of an inducible operon is because:
a. They directly interact with the promoter sequences
b. They directly interact with the repressor
c. They directly interact with operator region
d. They interact with the activation sequences and induce enhancers
65. In humans, the enzyme having reverse transcriptase activity is:
a. Ribonuclease P
b. Ribonuclease D
c. Recombinase
d. Telomerase
66. To prepare a DNA probe of high specific activity for detecting a single copy gene in
a Zoo-blot experiment, which one of the following procedures would be preferred?
a. 5‟ end labelling
b. 3‟ end labelling
c. In vitro transcription
d. Random primer labelling
67. The function performed in the smooth ER is:
a. Biosynthesis of secretory proteins
b. Folding of membrane proteins
c. Addition of N-linked sugars
d. Detoxification of drugs
68. In human carcinomas, many proteins including most cytoskeletal proteins undergo
modifications, thereby making it difficult for a clinician to identify the origin of the
cancer cells. In this context, which one of the following can be relied upon for
identification of the origin of cancer cells?
a. microtubules
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
63. Which one of the following is NOT enriched in eukaryotic promoters located in
active chromatin?
a. Acetylated histones
b. DNAse I hypersensitive sites
c. Methylated cytosine
d. Bound TFIID
64. One of the reasons why non-substrate inducers (e.g. IPTG) are preferred over
substrate inducers (e.g. lactose) for induction of an inducible operon is because:
a. They directly interact with the promoter sequences
b. They directly interact with the repressor
c. They directly interact with operator region
d. They interact with the activation sequences and induce enhancers
65. In humans, the enzyme having reverse transcriptase activity is:
a. Ribonuclease P
b. Ribonuclease D
c. Recombinase
d. Telomerase
66. To prepare a DNA probe of high specific activity for detecting a single copy gene in
a Zoo-blot experiment, which one of the following procedures would be preferred?
a. 5‟ end labelling
b. 3‟ end labelling
c. In vitro transcription
d. Random primer labelling
67. The function performed in the smooth ER is:
a. Biosynthesis of secretory proteins
b. Folding of membrane proteins
c. Addition of N-linked sugars
d. Detoxification of drugs
68. In human carcinomas, many proteins including most cytoskeletal proteins undergo
modifications, thereby making it difficult for a clinician to identify the origin of the
cancer cells. In this context, which one of the following can be relied upon for
identification of the origin of cancer cells?
a. microtubules
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
18
b. F-actin
c. G-actin
d. Intermediate filament
69. Most organisms can regulate membrane fluidity by changing the lipid composition.
If cells are transferred from a warm environment to a cold one, it can be expected
that:
a. There will be a decrease in the proportion of 16-carbon fatty acids compared to
18-carbon fatty acids
b. There will be an increase in the proportion of 16-carbon fatty acids compared to
18-carbon fatty acids
c. There will be an increase in the proportion of saturated fatty acids
d. Phospholipids will not be incorporated into the membrane
70. Glycosylation of membrane proteins and lipids is carried out by enzymes present in
the lumen of endoplasmic reticulum and Golgi. The glycosylated part of membrane
proteins and lipids in the plasma membrane is likely to be:
a. Oriented towards the cytosol
b. Exposed to the extracellular environment
c. The glycosylated parts of the proteins are towards the cytoplasm but
glycosylated lipids are exposed to the extracellular environment
d. The glycosylated parts of the lipids are towards the cytoplasm but glycosylated
proteins are exposed to the extracellular environment
71. Lysosomal storage diseases are a group of inherited diseases that are
characterized by the accumulation of specific substances or class of substances
within the lysosomes. All of the following mechanisms can cause lysosomal
storage disease EXCEPT:
a. Defects in the enzyme N-acetylglucosamine phosphotransferase
b. Defective or missing acid hydrolases
c. Defects in the transport of lysosomal enzymes to the cell surface
d. Defects in the transport proteins that transport proteins from Golgi to lysosomes
72. The gene encoding an enzyme A that functions in a metabolic pathway for
conversion of metabolite „x‟ to „y‟ was knocked out but it still resulted in the
formation of metabolite „y‟. From this it can be concluded that:
a. Enzyme A is necessary but not sufficient for formation of „y‟
b. Enzyme A is sufficient but not necessary for formation of „y‟
c. Enzyme A is neither necessary nor sufficient for formation of „y‟
d. Enzyme A is both necessary and sufficient for formation of „y‟
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
b. F-actin
c. G-actin
d. Intermediate filament
69. Most organisms can regulate membrane fluidity by changing the lipid composition.
If cells are transferred from a warm environment to a cold one, it can be expected
that:
a. There will be a decrease in the proportion of 16-carbon fatty acids compared to
18-carbon fatty acids
b. There will be an increase in the proportion of 16-carbon fatty acids compared to
18-carbon fatty acids
c. There will be an increase in the proportion of saturated fatty acids
d. Phospholipids will not be incorporated into the membrane
70. Glycosylation of membrane proteins and lipids is carried out by enzymes present in
the lumen of endoplasmic reticulum and Golgi. The glycosylated part of membrane
proteins and lipids in the plasma membrane is likely to be:
a. Oriented towards the cytosol
b. Exposed to the extracellular environment
c. The glycosylated parts of the proteins are towards the cytoplasm but
glycosylated lipids are exposed to the extracellular environment
d. The glycosylated parts of the lipids are towards the cytoplasm but glycosylated
proteins are exposed to the extracellular environment
71. Lysosomal storage diseases are a group of inherited diseases that are
characterized by the accumulation of specific substances or class of substances
within the lysosomes. All of the following mechanisms can cause lysosomal
storage disease EXCEPT:
a. Defects in the enzyme N-acetylglucosamine phosphotransferase
b. Defective or missing acid hydrolases
c. Defects in the transport of lysosomal enzymes to the cell surface
d. Defects in the transport proteins that transport proteins from Golgi to lysosomes
72. The gene encoding an enzyme A that functions in a metabolic pathway for
conversion of metabolite „x‟ to „y‟ was knocked out but it still resulted in the
formation of metabolite „y‟. From this it can be concluded that:
a. Enzyme A is necessary but not sufficient for formation of „y‟
b. Enzyme A is sufficient but not necessary for formation of „y‟
c. Enzyme A is neither necessary nor sufficient for formation of „y‟
d. Enzyme A is both necessary and sufficient for formation of „y‟
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
19
73. The following figure shows the electrophoretic migration of a secretory protein in SDS-
PAGE from the time of its synthesis (0 min) to its secretion (90 min) from the cells.
Assuming that there was no problem in sample preparation and SDS-PAGE, the
reason for a lower band at 30 min and a higher band at 60 and 90 min compared to
that at 0 min could be due to:
a. Changes in the rate of protein synthesis at various time points
b. The protein was associated with other proteins at different time points resulting
in change in its migration
c. The protein underwent processing and posttranslational modifications as a
function of time
d. Small changes observed in the migration of the protein do not give enough
information to derive a meaningful conclusion
74. Four yeast mutants block membrane and secretory proteins in following
compartments:
Mutant A: Golgi
Mutant B: Endoplasmic reticulum
Mutant C: Cytosol
Mutant D: Secretory vesicles
If two new mutants are made that have combined defects of A+C and B+D,
proteins of the new mutants will accumulate in:
a. Golgi and ER, respectively
b. Golgi and secretory vesicle, respectively
c. Cytosol and ER, respectively
d. Cytosol in both cases
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
73. The following figure shows the electrophoretic migration of a secretory protein in SDS-
PAGE from the time of its synthesis (0 min) to its secretion (90 min) from the cells.
Assuming that there was no problem in sample preparation and SDS-PAGE, the
reason for a lower band at 30 min and a higher band at 60 and 90 min compared to
that at 0 min could be due to:
a. Changes in the rate of protein synthesis at various time points
b. The protein was associated with other proteins at different time points resulting
in change in its migration
c. The protein underwent processing and posttranslational modifications as a
function of time
d. Small changes observed in the migration of the protein do not give enough
information to derive a meaningful conclusion
74. Four yeast mutants block membrane and secretory proteins in following
compartments:
Mutant A: Golgi
Mutant B: Endoplasmic reticulum
Mutant C: Cytosol
Mutant D: Secretory vesicles
If two new mutants are made that have combined defects of A+C and B+D,
proteins of the new mutants will accumulate in:
a. Golgi and ER, respectively
b. Golgi and secretory vesicle, respectively
c. Cytosol and ER, respectively
d. Cytosol in both cases
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
20
75. Asynchronous animal cells were stained with a fluorescent DNA binding dye and
analyzed by flow cytometry. The histogram of cell count versus fluorescence
intensity is shown below with or without treatment.
It can be concluded from the histogram that:
a. The treatment inhibits the cells from entering the S-phase
b. The treatment increases the proportion of G1 phase
c. The treatment increases the proportion of G2/M phase
d. The treatment increases the proportion of S phase
76. Which one of the following eIF2α kinases get activated during unfolded protein
response (UPR) in mammalian cells?
a. PKR
b. GCN2
c. PERK
d. HRI
77. E. coli RNA polymerase has six subunits two α, one β, one βᶦ, one ω and one σ.
Which among these subunits imparts specificity of transcription of a gene and how?
a. ω, by binding to -10 sequence
b. σ, by binding to both -10 and -35 sequences
c. β, by binding to -10 sequence
d. Both σ and ω, by binding to -10 and -35 sequences
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
75. Asynchronous animal cells were stained with a fluorescent DNA binding dye and
analyzed by flow cytometry. The histogram of cell count versus fluorescence
intensity is shown below with or without treatment.
It can be concluded from the histogram that:
a. The treatment inhibits the cells from entering the S-phase
b. The treatment increases the proportion of G1 phase
c. The treatment increases the proportion of G2/M phase
d. The treatment increases the proportion of S phase
76. Which one of the following eIF2α kinases get activated during unfolded protein
response (UPR) in mammalian cells?
a. PKR
b. GCN2
c. PERK
d. HRI
77. E. coli RNA polymerase has six subunits two α, one β, one βᶦ, one ω and one σ.
Which among these subunits imparts specificity of transcription of a gene and how?
a. ω, by binding to -10 sequence
b. σ, by binding to both -10 and -35 sequences
c. β, by binding to -10 sequence
d. Both σ and ω, by binding to -10 and -35 sequences
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
21
78. The use of guide RNA to bind to DNA and target the double strand break at a
specific site is a feature of:
a. Gene knockout by homologous recombination
b. Gene knock down by RNAi
c. Genome editing by CRISPR/Cas system
d. Gene silencing by siRNA
79. Epidermolysis bullosa simplex (EBS) disease is caused by a mutation in which one
of the following genes?
a. β –tubulin
b. Keratin
c. Collagen
d. Tau
80. The genes that have been used for reprograming somatic cells of an adult mammal
to stem cells called induced pluripotent stem cells (IPSC) are KLF4, SOX2, OCT4
and c-MYC. These code for:
a. Transcription factors
b. Both transcription factors and RNA binding proteins
c. Chaperone proteins
d. Growth factors
81. Following fertilization, zygote divides rapidly to form a large number of cells within a
short time through a process called cleavage. In some organisms like sea urchin
the cells of the early embryos skip some stages of cell cycle in order to achieve this.
Which of the following steps may be skipped?
a. G1
b. G1 and G2
c. S
d. G0
82. Cell cycle regulatory genes (cdc) were originally discovered by Paul Nurse in yeasts
using genetic approach involving temperature sensitive mutant screening. Which of
the following was used to identify cdc?
a. Genome sequencing
b. cDNA complementation assay
c. RNAi method
d. Homologous recombination
83. Bindin is a protein that is expressed in the tip of sperm head (acrosome) and is
important for sperm-egg interaction. Its distribution pattern changes during sperm
maturation. Which one of the following methods may be used to monitor the
expression pattern of bindin in a simple light microscope?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
78. The use of guide RNA to bind to DNA and target the double strand break at a
specific site is a feature of:
a. Gene knockout by homologous recombination
b. Gene knock down by RNAi
c. Genome editing by CRISPR/Cas system
d. Gene silencing by siRNA
79. Epidermolysis bullosa simplex (EBS) disease is caused by a mutation in which one
of the following genes?
a. β –tubulin
b. Keratin
c. Collagen
d. Tau
80. The genes that have been used for reprograming somatic cells of an adult mammal
to stem cells called induced pluripotent stem cells (IPSC) are KLF4, SOX2, OCT4
and c-MYC. These code for:
a. Transcription factors
b. Both transcription factors and RNA binding proteins
c. Chaperone proteins
d. Growth factors
81. Following fertilization, zygote divides rapidly to form a large number of cells within a
short time through a process called cleavage. In some organisms like sea urchin
the cells of the early embryos skip some stages of cell cycle in order to achieve this.
Which of the following steps may be skipped?
a. G1
b. G1 and G2
c. S
d. G0
82. Cell cycle regulatory genes (cdc) were originally discovered by Paul Nurse in yeasts
using genetic approach involving temperature sensitive mutant screening. Which of
the following was used to identify cdc?
a. Genome sequencing
b. cDNA complementation assay
c. RNAi method
d. Homologous recombination
83. Bindin is a protein that is expressed in the tip of sperm head (acrosome) and is
important for sperm-egg interaction. Its distribution pattern changes during sperm
maturation. Which one of the following methods may be used to monitor the
expression pattern of bindin in a simple light microscope?
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
22
a. Phase contrast microscopy
b. Normarski-contrast microscopy
c. Immuno cytochemistry
d. Immunofluorescence
84. Which of the following mechanisms converts c-erbB (EGFR) to an oncogene?
a. Point mutation
b. Truncation
c. Recombination
d. Gene amplification
85. Midblastula transition (MBT) is a characteristic process that occurs in early
embryogenesis of organisms (like Amphibians) whose eggs are large. MBT refers
to:
a. Transition of structures from early embryonic to late embryonic
b. Transition of gene expression from maternal to zygotic
c. Regulation of transition of primordial germ cells to Gonadal structure
d. Epigenetic modification of gene regulation
86. Introns which are self-splicing and do not require any cofactor for their splicing, are
present in the primary transcripts of:
a. miRNA
b. Nuclear mRNA
c. tRNA
d. Mitochondrial mRNA
87. The phenomenon of transfer of traits from a man to his grandson through his
daughter is known as:
a. Sex influenced inheritance
b. Criss-cross inheritance
c. Y-linked inheritance
d. Sex-limited inheritance
88. If the doubling time of a prokaryotic cell becomes progressively shorter, then it can
be predicted that the ribosome concentration will:
a. remain constant
b. decrease
c. increase
d. remain constant but their composition will be different
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
a. Phase contrast microscopy
b. Normarski-contrast microscopy
c. Immuno cytochemistry
d. Immunofluorescence
84. Which of the following mechanisms converts c-erbB (EGFR) to an oncogene?
a. Point mutation
b. Truncation
c. Recombination
d. Gene amplification
85. Midblastula transition (MBT) is a characteristic process that occurs in early
embryogenesis of organisms (like Amphibians) whose eggs are large. MBT refers
to:
a. Transition of structures from early embryonic to late embryonic
b. Transition of gene expression from maternal to zygotic
c. Regulation of transition of primordial germ cells to Gonadal structure
d. Epigenetic modification of gene regulation
86. Introns which are self-splicing and do not require any cofactor for their splicing, are
present in the primary transcripts of:
a. miRNA
b. Nuclear mRNA
c. tRNA
d. Mitochondrial mRNA
87. The phenomenon of transfer of traits from a man to his grandson through his
daughter is known as:
a. Sex influenced inheritance
b. Criss-cross inheritance
c. Y-linked inheritance
d. Sex-limited inheritance
88. If the doubling time of a prokaryotic cell becomes progressively shorter, then it can
be predicted that the ribosome concentration will:
a. remain constant
b. decrease
c. increase
d. remain constant but their composition will be different
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
23
89. How many DNA molecules of 6 base pairs length are possible where the first base
is a purine and the last base is a pyrimidine?
a. 32
b. 256
c. 4096
d. 1024
90. A recombinant protein is expressed in E. coli under T7 promoter at 37
o
C. However
no biological activity is obtained in the cell lysate. If the same experiment is carried
out at 25
o
C, the cell lysate shows a reasonable biological activity. The most
probable explanation for this is:
a. lower temperature increases recombinant protein stability
b. lower temperature increases rate of production of recombinant protein
c. IPTG used for induction does not get degraded
d. recombinant protein is properly folded at low temperature
91. IPTG is used as an inducer in the T7 expression system for recombinant protein
expression in E.coli. This is because of:
a. availability of special E.coli cells which have the T7 RNA polymerase
gene integrated into their genome under a regulatable promoter
b. T7 promoter is recognized by E.coli RNA polymerase
c. High copy number of plasmid allows sequestration of E.coli RNA
polymerase by T7 promoter
d. IPTG facilitates binding of E.coli RNA polymerase to the T7 promoter.
92. Pyrosequencing derives its name from the fact that:
a. the bases are detected by pyrolysis
b. it detects pyrophosphate released during base incorporation
c. it uses apyrase to detect the bases
d. it generates a pyrogram as an output
93. Which of the following is a reason that geneticists use mtDNA to study the relatedness of
animal populations?
a. mtDNA mutates at a slower rate than nuclear DNA
b. mtDNA transmitted from mother to child, is free from recombination
c. All mitochondrial proteins are coded by mitochondrial genes
d. There are only a few single nucleotide polymorphisms in the mtDNA
94. Lyophilization is a method used for preservation of microbes because:
a. moisture is removed by sublimation
b. moisture is removed by very slow evaporation
c. ice crystals formed at low temperature improves cell viability
d. removal of air during lyophilization prevents formation of free radicals
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
89. How many DNA molecules of 6 base pairs length are possible where the first base
is a purine and the last base is a pyrimidine?
a. 32
b. 256
c. 4096
d. 1024
90. A recombinant protein is expressed in E. coli under T7 promoter at 37
o
C. However
no biological activity is obtained in the cell lysate. If the same experiment is carried
out at 25
o
C, the cell lysate shows a reasonable biological activity. The most
probable explanation for this is:
a. lower temperature increases recombinant protein stability
b. lower temperature increases rate of production of recombinant protein
c. IPTG used for induction does not get degraded
d. recombinant protein is properly folded at low temperature
91. IPTG is used as an inducer in the T7 expression system for recombinant protein
expression in E.coli. This is because of:
a. availability of special E.coli cells which have the T7 RNA polymerase
gene integrated into their genome under a regulatable promoter
b. T7 promoter is recognized by E.coli RNA polymerase
c. High copy number of plasmid allows sequestration of E.coli RNA
polymerase by T7 promoter
d. IPTG facilitates binding of E.coli RNA polymerase to the T7 promoter.
92. Pyrosequencing derives its name from the fact that:
a. the bases are detected by pyrolysis
b. it detects pyrophosphate released during base incorporation
c. it uses apyrase to detect the bases
d. it generates a pyrogram as an output
93. Which of the following is a reason that geneticists use mtDNA to study the relatedness of
animal populations?
a. mtDNA mutates at a slower rate than nuclear DNA
b. mtDNA transmitted from mother to child, is free from recombination
c. All mitochondrial proteins are coded by mitochondrial genes
d. There are only a few single nucleotide polymorphisms in the mtDNA
94. Lyophilization is a method used for preservation of microbes because:
a. moisture is removed by sublimation
b. moisture is removed by very slow evaporation
c. ice crystals formed at low temperature improves cell viability
d. removal of air during lyophilization prevents formation of free radicals
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
24
95. Stoichiometric matrices in metabolic pathways are used in:
a. Gene regulatory network analysis
b. Measuring robustness of a system
c. Flux Balance Analysis
d. Analysis of phenotypic characters from genome.
96. The main difference between domain and motif in protein structure is:
a. Domain can remain stable, independent of the rest of the protein while motif cannot.
b. Domain cannot remain stable, independent of the rest of the protein while motif can.
c. Domain can be predicted but motif cannot be predicted.
d. Both are synonyms and there is no difference
97. Select the best algorithm to do pairwise alignment when two proteins are very
different in length.
a. Smith-Waterman
b. Needleman-Wunsch
c. dot-matrix
d. ClustalW
98. From literature it is known that the length of an E. coli bacterium is 1 µm with a standard
deviation of 0.1 µm. After treatment with chemical “X” the mean length of 100 cells is 1.1 µm with
a standard deviation of 0.1 µm. What will you conclude from this experiment?
a. Treatment with chemical “X” has increased the length with a statistical confidence of
more than 99%
b. The length has increased with statistical confidence of 67%
c. Length has not increased and the observed difference is due to statistical error
d. Length increased can be claimed with 95% statistical confidence.
99. Which one of the following techniques CANNOT be used to remove salt from a protein
solution
a. Ultrafiltration
b. Ion exchange chromatography
c. Gel filtration chromatography
d. Dialysis
100. The jelly roll in protein structure is:
a. made of eight alpha helices
b. made of four alpha/beta motifs
c. made of seven hydrophobic strands and an amphipathic helix
d. formed from eight beta strands
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
95. Stoichiometric matrices in metabolic pathways are used in:
a. Gene regulatory network analysis
b. Measuring robustness of a system
c. Flux Balance Analysis
d. Analysis of phenotypic characters from genome.
96. The main difference between domain and motif in protein structure is:
a. Domain can remain stable, independent of the rest of the protein while motif cannot.
b. Domain cannot remain stable, independent of the rest of the protein while motif can.
c. Domain can be predicted but motif cannot be predicted.
d. Both are synonyms and there is no difference
97. Select the best algorithm to do pairwise alignment when two proteins are very
different in length.
a. Smith-Waterman
b. Needleman-Wunsch
c. dot-matrix
d. ClustalW
98. From literature it is known that the length of an E. coli bacterium is 1 µm with a standard
deviation of 0.1 µm. After treatment with chemical “X” the mean length of 100 cells is 1.1 µm with
a standard deviation of 0.1 µm. What will you conclude from this experiment?
a. Treatment with chemical “X” has increased the length with a statistical confidence of
more than 99%
b. The length has increased with statistical confidence of 67%
c. Length has not increased and the observed difference is due to statistical error
d. Length increased can be claimed with 95% statistical confidence.
99. Which one of the following techniques CANNOT be used to remove salt from a protein
solution
a. Ultrafiltration
b. Ion exchange chromatography
c. Gel filtration chromatography
d. Dialysis
100. The jelly roll in protein structure is:
a. made of eight alpha helices
b. made of four alpha/beta motifs
c. made of seven hydrophobic strands and an amphipathic helix
d. formed from eight beta strands
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
25
101. Which one of the following is NOT a structure or structurally derived database?
a. PDB
b. PROSITE
c. SCOP
d. CATH
102. Which of the following is TRUE?
In the threading approaches for prediction of protein structure, the optimization is
used for:
a. Sequence to sequence alignment
b. Generation of profile by converting three-dimensional structure to one-
dimensional string
c. Alignment of sequence to structure which minimizes energy of the target
sequence in the template fold
d. Alignment of target sequence to one-dimensional profile of a template fold
103. The statistical significance of the BLAST hit is obtained using:
a. extreme value distribution
b. normal distribution
c. Poisson distribution
d. random distribution
104. What is referred by the term k in the following energy expression E = ½ k (b-bo)
2
where b
and bo refer to the bond length and reference bond length respectively?
a. van der Waals radius
b. stretching constant for bond length variation
c. torsional potential
d. kinetic energy of an atom
105. The estimates of number of false positives from a BLAST search can be made using:
a. Percent positives
b. Percent identity
c. E value
d. Bit score
106. The length of Beta Hairpin motif is usually:
a. 2 to 7 residues
b. 12 to15 residues
c. 22 to 25 residues
d. 6 to 10 residues
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
101. Which one of the following is NOT a structure or structurally derived database?
a. PDB
b. PROSITE
c. SCOP
d. CATH
102. Which of the following is TRUE?
In the threading approaches for prediction of protein structure, the optimization is
used for:
a. Sequence to sequence alignment
b. Generation of profile by converting three-dimensional structure to one-
dimensional string
c. Alignment of sequence to structure which minimizes energy of the target
sequence in the template fold
d. Alignment of target sequence to one-dimensional profile of a template fold
103. The statistical significance of the BLAST hit is obtained using:
a. extreme value distribution
b. normal distribution
c. Poisson distribution
d. random distribution
104. What is referred by the term k in the following energy expression E = ½ k (b-bo)
2
where b
and bo refer to the bond length and reference bond length respectively?
a. van der Waals radius
b. stretching constant for bond length variation
c. torsional potential
d. kinetic energy of an atom
105. The estimates of number of false positives from a BLAST search can be made using:
a. Percent positives
b. Percent identity
c. E value
d. Bit score
106. The length of Beta Hairpin motif is usually:
a. 2 to 7 residues
b. 12 to15 residues
c. 22 to 25 residues
d. 6 to 10 residues
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
26
107. A scoring function is used in which one of the following drug design technologies?
a. QSAR
b. Molecular docking
c. Molecular dynamics
d. Pharmacophore mapping
108. BLOSUM matrices are based on:
a. mutations observed throughout a global alignment
b. highly conserved regions in a series of alignments forbidden to contain gaps
c. explicit evolutionary model
d. alignment of same sequences containing highly mutable regions
109. ProDom is a comprehensive set of protein domain families automatically generated
from:
a. Pfam
b. UniProt Knowledge Database
c. Swiss-prot
d. InterPro
110. Which one of the following methods used to find evolutionary trees is also referred
as “the minimum evolution method”?
a. Distance Method
b. Maximum Parsimony Method
c. Fitch and Margoliash Method
d. UPGMA Method
111. Z-score =
a. (score of alignment– mean)/ (standard deviation)
b. (variance -score of alignment)/standard deviation
c. (mean -score of alignment) /standard deviation
d. (score of alignment – mean)/standard deviation
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
107. A scoring function is used in which one of the following drug design technologies?
a. QSAR
b. Molecular docking
c. Molecular dynamics
d. Pharmacophore mapping
108. BLOSUM matrices are based on:
a. mutations observed throughout a global alignment
b. highly conserved regions in a series of alignments forbidden to contain gaps
c. explicit evolutionary model
d. alignment of same sequences containing highly mutable regions
109. ProDom is a comprehensive set of protein domain families automatically generated
from:
a. Pfam
b. UniProt Knowledge Database
c. Swiss-prot
d. InterPro
110. Which one of the following methods used to find evolutionary trees is also referred
as “the minimum evolution method”?
a. Distance Method
b. Maximum Parsimony Method
c. Fitch and Margoliash Method
d. UPGMA Method
111. Z-score =
a. (score of alignment– mean)/ (standard deviation)
b. (variance -score of alignment)/standard deviation
c. (mean -score of alignment) /standard deviation
d. (score of alignment – mean)/standard deviation
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
27
112. In a sequence logo of the type given below, the sizes of the letters are proportional to
the:
a. number of residues in the sequences
b. information content of the respective residues
c. frequencies of the respective residues in the sequences
d. resolution of the output device(terminal/printer)
113. For a homology search program such as BLAST, which one of the following best
describes the scoring pattern?
a. identical residue = 10 pts, conservative substitution = 5 pts, gap = 0 pts
b. identical residue = 10 pts, conservative substitution = 1 to 9 pts, gap = -3 pts
c. identical residue = 10 pts, conservative substitution = 1 to 9 pts, gap = 0 pts
d. identical residue = 10 pts, conservative substitution = 10 pts, gap = -3 pts
114. Given the results of a 'sequence versus fingerprint' search, which of the following would
be considered the best hit?
a. Motifs:7 of 7 p-value:2.2e-08 & e-value:4.5e-05
b. Motifs:8 of 8 p-value:2.2e-08 & e-value:4.5e-05
c. Motifs:8 of 8 p-Value:2.2e-18 & e-value:4.5e-15
d. Motifs:7 of 8 p-Value:2.2e-18 & e-value:4.5e-15
115. In a Multi-Locus Variable number tandem repeat Analysis (MLVA) for Salmonella
enterica subspecies, three tandem repeats loci have been identified inside yohM gene of
S. typhimurium LT2, S.typhi CT18 and S.typhi Ty2 strains. Motif lengths for the loci are
2,3 and 5 respectively. Motif AT and ATG are found to be repeated by 13 and 5 times
respectively in all the strains. Motif ATGTC is repeated 13 times in S. typhimurium LT2,
15 times S.typhi CT18 and 12 times S.typhi Ty2 strains. Which locus is Variable Number
Tandem Repeat (VNTR)?
a. AT
b. ATG
c. ATGTC
d. GTCA
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
112. In a sequence logo of the type given below, the sizes of the letters are proportional to
the:
a. number of residues in the sequences
b. information content of the respective residues
c. frequencies of the respective residues in the sequences
d. resolution of the output device(terminal/printer)
113. For a homology search program such as BLAST, which one of the following best
describes the scoring pattern?
a. identical residue = 10 pts, conservative substitution = 5 pts, gap = 0 pts
b. identical residue = 10 pts, conservative substitution = 1 to 9 pts, gap = -3 pts
c. identical residue = 10 pts, conservative substitution = 1 to 9 pts, gap = 0 pts
d. identical residue = 10 pts, conservative substitution = 10 pts, gap = -3 pts
114. Given the results of a 'sequence versus fingerprint' search, which of the following would
be considered the best hit?
a. Motifs:7 of 7 p-value:2.2e-08 & e-value:4.5e-05
b. Motifs:8 of 8 p-value:2.2e-08 & e-value:4.5e-05
c. Motifs:8 of 8 p-Value:2.2e-18 & e-value:4.5e-15
d. Motifs:7 of 8 p-Value:2.2e-18 & e-value:4.5e-15
115. In a Multi-Locus Variable number tandem repeat Analysis (MLVA) for Salmonella
enterica subspecies, three tandem repeats loci have been identified inside yohM gene of
S. typhimurium LT2, S.typhi CT18 and S.typhi Ty2 strains. Motif lengths for the loci are
2,3 and 5 respectively. Motif AT and ATG are found to be repeated by 13 and 5 times
respectively in all the strains. Motif ATGTC is repeated 13 times in S. typhimurium LT2,
15 times S.typhi CT18 and 12 times S.typhi Ty2 strains. Which locus is Variable Number
Tandem Repeat (VNTR)?
a. AT
b. ATG
c. ATGTC
d. GTCA
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
28
116. Of a population of cells undergoing meiosis, 1% of the cells undergo recombination
between genes A and B. What is the distance between the two genes?
a. 0.25 cM
b. 0.75 cM
c. 0.50 cM
d. 1.00 cM
117. Which one of the following combinations of marker genes and promoters CANNOT be
used for selection of transgenic plants under in vitro conditions?
a. positive selection marker genes under the Agrobacterium-derived NOS promoter
b. conditional negative selection marker genes under an inducible promoter
c. positive selection marker genes under CaMV 35S promoter
d. non-conditional negative selection marker genes under CaMV 35S promoter
118. Several experiments have shown leaky/deregulated expression of reporter genes viz.,
GUS from plant promoters in Agrobacterium cells. Which one of the following approaches
would be most useful in restricting transgene expression in transformed plant cells and
prevent their expression in Agrobacterium?
a. Use of weak promoters to express the transgene
b. Use of 5‟ and 3‟ UTRs flanking the transgene
c. Avoiding use of the polyA signal
d. Use of intron(s) within the transgene sequence
119. A T0 transgenic plant showing two copies of T-DNA on Southern analysis, segregated
in a 3:1 ratio for the transgenic:non-transgenic phenotype among T1 progeny obtained by
self-pollination. Which one of the following statements best explains this observation?
a. The T0 plant contains a single copy of the transgene.
b. The T0 plant contains two linked copies of the transgene.
c. The T0 plant contains two unlinked copies of the transgene.
d. The T0 plant contains at least three copies of the transgene.
120. Which one of the following statements is correct?
a. All the virulence genes of Agrobacterium tumefaciens are expressed in a
constitutive manner.
b. Opines are a source of iron for Agrobacterium cells.
c. One Agrobacterium cell can generate only one T-DNA molecule for transfer into
the host cell.
d. Integration of T-DNA in the genome can mutate or modulate endogenous plant
genes.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
116. Of a population of cells undergoing meiosis, 1% of the cells undergo recombination
between genes A and B. What is the distance between the two genes?
a. 0.25 cM
b. 0.75 cM
c. 0.50 cM
d. 1.00 cM
117. Which one of the following combinations of marker genes and promoters CANNOT be
used for selection of transgenic plants under in vitro conditions?
a. positive selection marker genes under the Agrobacterium-derived NOS promoter
b. conditional negative selection marker genes under an inducible promoter
c. positive selection marker genes under CaMV 35S promoter
d. non-conditional negative selection marker genes under CaMV 35S promoter
118. Several experiments have shown leaky/deregulated expression of reporter genes viz.,
GUS from plant promoters in Agrobacterium cells. Which one of the following approaches
would be most useful in restricting transgene expression in transformed plant cells and
prevent their expression in Agrobacterium?
a. Use of weak promoters to express the transgene
b. Use of 5‟ and 3‟ UTRs flanking the transgene
c. Avoiding use of the polyA signal
d. Use of intron(s) within the transgene sequence
119. A T0 transgenic plant showing two copies of T-DNA on Southern analysis, segregated
in a 3:1 ratio for the transgenic:non-transgenic phenotype among T1 progeny obtained by
self-pollination. Which one of the following statements best explains this observation?
a. The T0 plant contains a single copy of the transgene.
b. The T0 plant contains two linked copies of the transgene.
c. The T0 plant contains two unlinked copies of the transgene.
d. The T0 plant contains at least three copies of the transgene.
120. Which one of the following statements is correct?
a. All the virulence genes of Agrobacterium tumefaciens are expressed in a
constitutive manner.
b. Opines are a source of iron for Agrobacterium cells.
c. One Agrobacterium cell can generate only one T-DNA molecule for transfer into
the host cell.
d. Integration of T-DNA in the genome can mutate or modulate endogenous plant
genes.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
29
121. Which one of the following statements related to transgene silencing in plants is
INCORRECT?
a. Transgene silencing is usually accompanied by methylation of cytosine residues in
„CG‟ and/or „CNG‟ sites.
b. Events with multi-copy integrations of the T-DNA are more susceptible to
transgene silencing.
c. Transgene silencing always occurs in T0 individuals and never in subsequent
generations.
d. Transgene silencing may lead to silencing of endogenous plant gene homologs.
122. Two independent transgenic plants, one with single copy of gene A and another with
single copy of gene B were expressed under the same seed-specific promoter. In
transgenic plants, seed formation on self-pollination was similar to that of
untransformed plants. When homozygous plants with gene A (male parent) were
crossed with homozygous plants with gene B (female parent), viable seed
formation did not occur although, pollen production was normal. What could be the
possible reason for this observation?
a. Product of gene A is lethal to the male gametophyte.
b. Interaction between products of genes A and B is lethal to the zygote.
c. Interaction between products of genes A and B is lethal to the male gametophyte.
d. Product of gene B is lethal to the zygote.
123. The range of transformation frequencies obtained in independent transformation
experiments using four different constructs with different combinations of promoters (Pr)
and selection marker genes, is given below:
Construct Transformation frequency
CaMV35S Pr – bar – nos polyA 20% – 35%
Nos Pr – bar – nos polyA 2% – 4%
CaMV35S Pr – NPTII
– nos polyA 46% – 49%
Nos Pr – NPTII – nos polyA 24% – 28%
In the absence of any other factors, which one of the following statements is NOT
acceptable as a logical conclusion based on the above data?
a. Use of stronger promoters for expression of selection marker genes can increase
transformation frequencies.
b. Variations in expression levels between the NPTII and bar genes is not influenced
by the nos poly A signal.
c. Production of herbicide resistant plants at high frequency can be achieved with
lower expression levels of the transgene.
d. Use of kanamycin as a selection marker appears to be more favorable for
production of larger number of transgenic plants.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
121. Which one of the following statements related to transgene silencing in plants is
INCORRECT?
a. Transgene silencing is usually accompanied by methylation of cytosine residues in
„CG‟ and/or „CNG‟ sites.
b. Events with multi-copy integrations of the T-DNA are more susceptible to
transgene silencing.
c. Transgene silencing always occurs in T0 individuals and never in subsequent
generations.
d. Transgene silencing may lead to silencing of endogenous plant gene homologs.
122. Two independent transgenic plants, one with single copy of gene A and another with
single copy of gene B were expressed under the same seed-specific promoter. In
transgenic plants, seed formation on self-pollination was similar to that of
untransformed plants. When homozygous plants with gene A (male parent) were
crossed with homozygous plants with gene B (female parent), viable seed
formation did not occur although, pollen production was normal. What could be the
possible reason for this observation?
a. Product of gene A is lethal to the male gametophyte.
b. Interaction between products of genes A and B is lethal to the zygote.
c. Interaction between products of genes A and B is lethal to the male gametophyte.
d. Product of gene B is lethal to the zygote.
123. The range of transformation frequencies obtained in independent transformation
experiments using four different constructs with different combinations of promoters (Pr)
and selection marker genes, is given below:
Construct Transformation frequency
CaMV35S Pr – bar – nos polyA 20% – 35%
Nos Pr – bar – nos polyA 2% – 4%
CaMV35S Pr – NPTII
– nos polyA 46% – 49%
Nos Pr – NPTII – nos polyA 24% – 28%
In the absence of any other factors, which one of the following statements is NOT
acceptable as a logical conclusion based on the above data?
a. Use of stronger promoters for expression of selection marker genes can increase
transformation frequencies.
b. Variations in expression levels between the NPTII and bar genes is not influenced
by the nos poly A signal.
c. Production of herbicide resistant plants at high frequency can be achieved with
lower expression levels of the transgene.
d. Use of kanamycin as a selection marker appears to be more favorable for
production of larger number of transgenic plants.
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
30
124. Callus-mediated regeneration is NOT preferred for micropropagation because:
a. it takes longer time for regeneration.
b. plant regeneration is problematic due to poor organogenesis
c. it leads to generation of variants
d. hardening of plants is difficult to achieve.
125. In a transgenic plant, the phenomenon of Co-suppression is due to:
a. transgene integration within the endogenous gene
b. transgene integration at a locus very close to the endogenous gene
c. similarity between the transgene and endogenous gene sequences
d. lack of similarity between the transgene and endogenous gene sequences
126. In inbred lines, gene and genotypic frequencies are maintained by growing them in
isolation followed by:
a. self-pollination
b. self-pollination and selection
c. pair-wise crossing
d. random mating without selection
127. Homozygous plants from bi-parental mating for development of mapping populations
can be obtained in a short time by:
a. development of RILs
b. development of doubled haploids
c. random mating in F2 followed by selfing in subsequent generations
d. chromosome elimination technique
128. Which one of the following is NOT required for QTL analysis in association mapping?
a. Phenotypic data from multi-location trials
b. Polymorphic markers
c. Genetic linkage map
d. Diverse germplasm lines
129. Which one of the following hormones promote production of seedless grapes?
a. IAA
b. IBA
c. BAP
d. GA3
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
124. Callus-mediated regeneration is NOT preferred for micropropagation because:
a. it takes longer time for regeneration.
b. plant regeneration is problematic due to poor organogenesis
c. it leads to generation of variants
d. hardening of plants is difficult to achieve.
125. In a transgenic plant, the phenomenon of Co-suppression is due to:
a. transgene integration within the endogenous gene
b. transgene integration at a locus very close to the endogenous gene
c. similarity between the transgene and endogenous gene sequences
d. lack of similarity between the transgene and endogenous gene sequences
126. In inbred lines, gene and genotypic frequencies are maintained by growing them in
isolation followed by:
a. self-pollination
b. self-pollination and selection
c. pair-wise crossing
d. random mating without selection
127. Homozygous plants from bi-parental mating for development of mapping populations
can be obtained in a short time by:
a. development of RILs
b. development of doubled haploids
c. random mating in F2 followed by selfing in subsequent generations
d. chromosome elimination technique
128. Which one of the following is NOT required for QTL analysis in association mapping?
a. Phenotypic data from multi-location trials
b. Polymorphic markers
c. Genetic linkage map
d. Diverse germplasm lines
129. Which one of the following hormones promote production of seedless grapes?
a. IAA
b. IBA
c. BAP
d. GA3
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
31
130. Which one of the following molecular markers is associated with bacterial blight
resistance in rice?
a. PiB
b. Rar
c. Xa21
d. Pi9
131. Match the proteins in Group I with members in Group II
Group I Group II
1 Alkaline phosphatase P Extracellular
2 Fo component of ATP synthase Q Cytoplasm
3 Alcohol dehydrogenase R Inner membrane
4 Pectinase S Periplasm
a. 1-S; 2-R; 3-Q; 4-P
b. 1-P; 2-R; 3-Q; 4-S
c. 1-R; 2-S; 3-P; 4-Q
d. 1-Q; 2-P; 3-S; 4-R
132. Application of the herbicide, phosphoinothricin, results in death of plants due to
accumulation of:
a. CO2
b. NH3
c. CN
d. Cl2
133. Which one of the following is NOT a feature of Agrobacterium VirE protein?
a. Its synthesis is induced by VirG protein
b. It contains a nuclear localization signal
c. It protects T-DNA from being destroyed by plant defense mechanism
d. It is a component of T-pilus
134. Transgenic potato plants with high amylose starch were developed by:
a. suppression of starch branching enzymes
b. over-expression of starch branching enzymes
c. over-expression of ADP-glucose pyrophosphorylase
d. suppression of granule-bound starch synthase
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
130. Which one of the following molecular markers is associated with bacterial blight
resistance in rice?
a. PiB
b. Rar
c. Xa21
d. Pi9
131. Match the proteins in Group I with members in Group II
Group I Group II
1 Alkaline phosphatase P Extracellular
2 Fo component of ATP synthase Q Cytoplasm
3 Alcohol dehydrogenase R Inner membrane
4 Pectinase S Periplasm
a. 1-S; 2-R; 3-Q; 4-P
b. 1-P; 2-R; 3-Q; 4-S
c. 1-R; 2-S; 3-P; 4-Q
d. 1-Q; 2-P; 3-S; 4-R
132. Application of the herbicide, phosphoinothricin, results in death of plants due to
accumulation of:
a. CO2
b. NH3
c. CN
d. Cl2
133. Which one of the following is NOT a feature of Agrobacterium VirE protein?
a. Its synthesis is induced by VirG protein
b. It contains a nuclear localization signal
c. It protects T-DNA from being destroyed by plant defense mechanism
d. It is a component of T-pilus
134. Transgenic potato plants with high amylose starch were developed by:
a. suppression of starch branching enzymes
b. over-expression of starch branching enzymes
c. over-expression of ADP-glucose pyrophosphorylase
d. suppression of granule-bound starch synthase
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
32
135. Developing cisgenic disease resistant apple is advantageous over conventional
breeding because:
a. genes cannot be transferred from sexually compatible species through pollination
due to hybridization barriers.
b. there is no linkage drag.
c. apple is vegetatively propagated and produces sterile seeds.
d. cisgenes will not disrupt endogenous genes when introduced by Agrobacterium
mediated transformation.
136. For development of selection marker-free transgenic plants by co-transformation
using Agrobacterium, the marker genes are:
a. segregated out in T0 generation
b. excised out in T0 generation
c. segregated out in T1 generation
d. excised out in T1 generation
137. Breakdown of Bt- mediated insect resistance in crops can be delayed by:
a. Refugia strategy
b. Cultivation of different Bt cotton varieties with same Bt gene
c. Application of Bt Bio-pesticides
d. application of systemic insecticides
138. Which one of the following is matched INCORRECTLY?
a. Nanopore – DNA sequencing
b. beta-carotene – golden rice
c. bar gene – resistance to viruses
d. Figwort Mosaic Virus – constitutive promoter
139. Which one of the following would produce androgenic haploids in anther culture?
a. Anther wall
b. Tapetal layer
c. Connective tissue
d. Young microspores
140. Male sterility in plants is induced by expression of the TA29-barnase-pA cassette
in:
a. Pollen mother cell
b. Stamen
c. Tapetum
d. Pollen grain
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
135. Developing cisgenic disease resistant apple is advantageous over conventional
breeding because:
a. genes cannot be transferred from sexually compatible species through pollination
due to hybridization barriers.
b. there is no linkage drag.
c. apple is vegetatively propagated and produces sterile seeds.
d. cisgenes will not disrupt endogenous genes when introduced by Agrobacterium
mediated transformation.
136. For development of selection marker-free transgenic plants by co-transformation
using Agrobacterium, the marker genes are:
a. segregated out in T0 generation
b. excised out in T0 generation
c. segregated out in T1 generation
d. excised out in T1 generation
137. Breakdown of Bt- mediated insect resistance in crops can be delayed by:
a. Refugia strategy
b. Cultivation of different Bt cotton varieties with same Bt gene
c. Application of Bt Bio-pesticides
d. application of systemic insecticides
138. Which one of the following is matched INCORRECTLY?
a. Nanopore – DNA sequencing
b. beta-carotene – golden rice
c. bar gene – resistance to viruses
d. Figwort Mosaic Virus – constitutive promoter
139. Which one of the following would produce androgenic haploids in anther culture?
a. Anther wall
b. Tapetal layer
c. Connective tissue
d. Young microspores
140. Male sterility in plants is induced by expression of the TA29-barnase-pA cassette
in:
a. Pollen mother cell
b. Stamen
c. Tapetum
d. Pollen grain
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
33
141. Anchorage dependent CHO cells are grown by aeration using micro carrier beads. The
maximum detrimental effect of shear occurs due to agitation if the size (Kolmogorov
scale) of eddies is:
a. lesser than the size of the beads
b. equal to the size of beads
c. larger than the size of beads
d. independent of the size of beads
142. With progress in the growth of E. coli in a minimal medium (constant aeration and
agitation), the dissolved oxygen (DO) initially declined and then started to increase. If the
DO again starts to decrease on addition of glucose, then the most probable explanation
is:
a. growth is limited by glucose
b. growth is limited by oxygen
c. decrease in solubility of oxygen occurred due to glucose addition
d. decrease in the number of cells takes place due to cell lysis
143. An organism obeys Andrews model for growth inhibition with Ks and Ki values of 0.01 g/l
and 1.0 g/l respectively. If the substrate is present at an initial concentration of 1.0g/l,
the specific growth rate of the culture upon entering log phase would be approximately:
a. 1/5
th
of µmax
b. 1/4
th
of µmax
c. 1/3
rd
of µmax
d. 1/2 of µmax
144. In a stirred tank reactor when the agitation rate is increased, the kL and kLa values will:
a. increase and decrease respectively
b. decrease and increase respectively
c. both increase
d. both decrease
145. Which one of the following is correct regarding cell damage in an agitated and sparged
mammalian cell bioreactor?
a. High shear stress arising during mixing of the cell culture is the major cause of cell
damage.
b. Shear stress arising from the breakup of bubbles at the liquid surface is a major cause
of cell damage.
c. Shear stress between bubbles in the foam at the liquid surface is a major cause of cell
damage
d. Using agitator blades of a radial flow type with mirror finish reduces shear and hence
cell damage
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
141. Anchorage dependent CHO cells are grown by aeration using micro carrier beads. The
maximum detrimental effect of shear occurs due to agitation if the size (Kolmogorov
scale) of eddies is:
a. lesser than the size of the beads
b. equal to the size of beads
c. larger than the size of beads
d. independent of the size of beads
142. With progress in the growth of E. coli in a minimal medium (constant aeration and
agitation), the dissolved oxygen (DO) initially declined and then started to increase. If the
DO again starts to decrease on addition of glucose, then the most probable explanation
is:
a. growth is limited by glucose
b. growth is limited by oxygen
c. decrease in solubility of oxygen occurred due to glucose addition
d. decrease in the number of cells takes place due to cell lysis
143. An organism obeys Andrews model for growth inhibition with Ks and Ki values of 0.01 g/l
and 1.0 g/l respectively. If the substrate is present at an initial concentration of 1.0g/l,
the specific growth rate of the culture upon entering log phase would be approximately:
a. 1/5
th
of µmax
b. 1/4
th
of µmax
c. 1/3
rd
of µmax
d. 1/2 of µmax
144. In a stirred tank reactor when the agitation rate is increased, the kL and kLa values will:
a. increase and decrease respectively
b. decrease and increase respectively
c. both increase
d. both decrease
145. Which one of the following is correct regarding cell damage in an agitated and sparged
mammalian cell bioreactor?
a. High shear stress arising during mixing of the cell culture is the major cause of cell
damage.
b. Shear stress arising from the breakup of bubbles at the liquid surface is a major cause
of cell damage.
c. Shear stress between bubbles in the foam at the liquid surface is a major cause of cell
damage
d. Using agitator blades of a radial flow type with mirror finish reduces shear and hence
cell damage
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
34
146. Which one of the statements given below is NOT true? Equilibrium constant (K) of a
chemical reaction at a specific temperature can be determined if the:
a. ΔG
0
of the reaction is known
b. Equilibrium concentrations of reactants and products are known
c. ΔH and ΔS of the reaction are known
d. ΔH and initial concentrations of the reactants and products are known
147. If a big centrifuge with the bowl diameter of 1m rotates at 60 rpm, at what speed (rpm)
does a smaller centrifuge with a diameter of 0.5 m need to be operated for achieving the
same separation factor?
(a) 85
(b) 120
(c) 30
(d) 42
148. What is the linear flow rate (superficial velocity in cm/h) in a chromatographic column of
1.0 cm inner diameter with a bed porosity of 0.8 when the volumetric flow rate is 1
ml/min?
a) 61.1
b) 76.4
c) 38.2
d) 15.2
149. For a cell growth process, the units for yield, productivity and titre are:
a. gl
-1
, gl
-1
h
-1
and gl
-1
b. gg
-1
, gl
-1
h
-1
and g g
-1
l
-1
h
-1
c. gl
-1
, gl
-1
h
-1
and gg
-1
d. gg
-1
, gl
-1
h
-1
and gl
-1
150. In a fermentor the impeller diameter is increased from 0.5 m to 1 m and the rpm is
increased from 100 rpm to 400 rpm. Reynolds number will increase by?
a. 16 fold
b. 8 fold
c. 2 fold
d. 4 fold
151. Zymomonas mobilis is cultivated in a 60 l chemostat. The µmax and Ks values are 0.2 h
-
1
and 0.5 g l
-1
respectively. The flow rate (l h
-1
) required for a steady state substrate
concentration of 0.5 g l
-1
in the reactor is:
a. 6
b. 9
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
146. Which one of the statements given below is NOT true? Equilibrium constant (K) of a
chemical reaction at a specific temperature can be determined if the:
a. ΔG
0
of the reaction is known
b. Equilibrium concentrations of reactants and products are known
c. ΔH and ΔS of the reaction are known
d. ΔH and initial concentrations of the reactants and products are known
147. If a big centrifuge with the bowl diameter of 1m rotates at 60 rpm, at what speed (rpm)
does a smaller centrifuge with a diameter of 0.5 m need to be operated for achieving the
same separation factor?
(a) 85
(b) 120
(c) 30
(d) 42
148. What is the linear flow rate (superficial velocity in cm/h) in a chromatographic column of
1.0 cm inner diameter with a bed porosity of 0.8 when the volumetric flow rate is 1
ml/min?
a) 61.1
b) 76.4
c) 38.2
d) 15.2
149. For a cell growth process, the units for yield, productivity and titre are:
a. gl
-1
, gl
-1
h
-1
and gl
-1
b. gg
-1
, gl
-1
h
-1
and g g
-1
l
-1
h
-1
c. gl
-1
, gl
-1
h
-1
and gg
-1
d. gg
-1
, gl
-1
h
-1
and gl
-1
150. In a fermentor the impeller diameter is increased from 0.5 m to 1 m and the rpm is
increased from 100 rpm to 400 rpm. Reynolds number will increase by?
a. 16 fold
b. 8 fold
c. 2 fold
d. 4 fold
151. Zymomonas mobilis is cultivated in a 60 l chemostat. The µmax and Ks values are 0.2 h
-
1
and 0.5 g l
-1
respectively. The flow rate (l h
-1
) required for a steady state substrate
concentration of 0.5 g l
-1
in the reactor is:
a. 6
b. 9
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
35
c. 4
d. 12
152. Which one of the following statements is WRONG for conventional batch filtration of a
mycelial fermentation broth?
a. Blockage of the membrane pores by cell debris
b. The specific cake resistance remains constant
c. compressible cake deposition on the membrane
d. medium resistance remain constant
153. The reason for the choice of E. coli for the production of ethanol from lignocellulose
is because it:
a. grows efficiently in various hexoses and pentoses
b. can be grown to high cell density
c. can tolerate very high concentration of ethanol
d. is a GRAS organism
154. Two organisms A and B with the same µmax and Yx/s are cultivated independently in
batch culture. They have Ks values of 1 g.l
-1
and 3 g.l
-1
respectively. Given that the
initial substrate concentration was 5 g/l, which of the following is TRUE after complete
exhaustion of the substrate?
a. Organism A will have lower average specific growth rate than B.
b. Organism A will have higher average specific growth rate than B.
c. Both the organisms would have same average specific growth rate.
d. The final biomass achieved in B would be higher than in A.
155. Pseudomonas with the elemental composition of CH3O0.5N0.5 (MW = 30) is grown in
a bioreactor to a final cell mass of 30 g/l. The minimum concentration (g/l) of ammonia
(NH3) (MW = 17) (as the sole nitrogen source) required is:
a. 17
b. 8.5
c. 14
d. 15
156. To reduce the level of deactivation of media components for culturing Lactobacillus
spp., a high temperature and short time (HTST) regime is used for sterilization,
PRIMARILY because:
a. Deactivation energy of contaminants > Deactivation energy of medium components
b. Deactivation energy of contaminants < Deactivation energy of medium components
c. Deactivation energy of contaminants = Deactivation energy of medium components
d. the time required for sterilization is reduced
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
c. 4
d. 12
152. Which one of the following statements is WRONG for conventional batch filtration of a
mycelial fermentation broth?
a. Blockage of the membrane pores by cell debris
b. The specific cake resistance remains constant
c. compressible cake deposition on the membrane
d. medium resistance remain constant
153. The reason for the choice of E. coli for the production of ethanol from lignocellulose
is because it:
a. grows efficiently in various hexoses and pentoses
b. can be grown to high cell density
c. can tolerate very high concentration of ethanol
d. is a GRAS organism
154. Two organisms A and B with the same µmax and Yx/s are cultivated independently in
batch culture. They have Ks values of 1 g.l
-1
and 3 g.l
-1
respectively. Given that the
initial substrate concentration was 5 g/l, which of the following is TRUE after complete
exhaustion of the substrate?
a. Organism A will have lower average specific growth rate than B.
b. Organism A will have higher average specific growth rate than B.
c. Both the organisms would have same average specific growth rate.
d. The final biomass achieved in B would be higher than in A.
155. Pseudomonas with the elemental composition of CH3O0.5N0.5 (MW = 30) is grown in
a bioreactor to a final cell mass of 30 g/l. The minimum concentration (g/l) of ammonia
(NH3) (MW = 17) (as the sole nitrogen source) required is:
a. 17
b. 8.5
c. 14
d. 15
156. To reduce the level of deactivation of media components for culturing Lactobacillus
spp., a high temperature and short time (HTST) regime is used for sterilization,
PRIMARILY because:
a. Deactivation energy of contaminants > Deactivation energy of medium components
b. Deactivation energy of contaminants < Deactivation energy of medium components
c. Deactivation energy of contaminants = Deactivation energy of medium components
d. the time required for sterilization is reduced
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
36
157. Acetobacter aceti converts alcohol to acetic acid according to the stoichiometric
relation
C2H5OH + O2 CH3COOH + H2O
In a vigorously agitated and aerated reactor containing 20 g/l ethanol, the organism
produces 16 g/l acetic acid and 2 g/l was the residual ethanol concentration. What are the
theoretical and observed yields of acetic acid expressed in g/g ethanol?
a. 60/46 and 16/18
b. 50/46 and 18/16
c. 46/60 and 18/16
d. 26/60 and 16/18
158. A continuous stirred tank bioreactor produces 48 kg lysine.day
-1
. If the volumetric
productivity is 2 g.l
-1
.h
-1
, the volume of the reactor is:
a. 1 m
3
b. 1.5 m
3
c. 0.1 m
3
d. 2 m
3
159. Aqueous Two Phase System (ATPS) is used to isolate a protease from a fermentation
broth. The partition coefficient (K) is 2.5. For 80 % recovery of protease in a single step,
the volume ratios of upper and the lower phases should be:
i. 1.6:1
ii. 3.2:1
iii. 2.4:1
iv. 1.2:1
160. To have an overall yield of greater than 50% in a three step purification process for a
food additive, the minimum average step yield(%) necessary would be around:
a. 95
b. 80
c. 50
d. 20
161. In a fed batch process for the production of an antibiotic, the dissolved oxygen (DO)
level was found to be falling below 30%. If the DO level of 30% is to be maintained in
the reactor (without altering the composition of oxygen-air mixture, aeration rate and
agitation) then we need to:
a. increase the feed rate
b. decrease the feed rate
c. increase the concentration of feed
d. decrease the volume of the broth by partial withdrawal
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
157. Acetobacter aceti converts alcohol to acetic acid according to the stoichiometric
relation
C2H5OH + O2 CH3COOH + H2O
In a vigorously agitated and aerated reactor containing 20 g/l ethanol, the organism
produces 16 g/l acetic acid and 2 g/l was the residual ethanol concentration. What are the
theoretical and observed yields of acetic acid expressed in g/g ethanol?
a. 60/46 and 16/18
b. 50/46 and 18/16
c. 46/60 and 18/16
d. 26/60 and 16/18
158. A continuous stirred tank bioreactor produces 48 kg lysine.day
-1
. If the volumetric
productivity is 2 g.l
-1
.h
-1
, the volume of the reactor is:
a. 1 m
3
b. 1.5 m
3
c. 0.1 m
3
d. 2 m
3
159. Aqueous Two Phase System (ATPS) is used to isolate a protease from a fermentation
broth. The partition coefficient (K) is 2.5. For 80 % recovery of protease in a single step,
the volume ratios of upper and the lower phases should be:
i. 1.6:1
ii. 3.2:1
iii. 2.4:1
iv. 1.2:1
160. To have an overall yield of greater than 50% in a three step purification process for a
food additive, the minimum average step yield(%) necessary would be around:
a. 95
b. 80
c. 50
d. 20
161. In a fed batch process for the production of an antibiotic, the dissolved oxygen (DO)
level was found to be falling below 30%. If the DO level of 30% is to be maintained in
the reactor (without altering the composition of oxygen-air mixture, aeration rate and
agitation) then we need to:
a. increase the feed rate
b. decrease the feed rate
c. increase the concentration of feed
d. decrease the volume of the broth by partial withdrawal
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
37
162. In a batch sterilization process if ∇overall , ∇heating & ∇cooling are 32.2, 9.8 & 10.1
respectively, THEN the holding time at 121ºC of the process is (given that the specific
death rate (k) of microorganisms at 121ºC is 2.54 min
-1
):
a. 4.84 min
b. 6.75 min
c. 2.96 min
d. 9.25 min
163. To achieve complete separation of two pharmaceutical compounds (shown as two
peaks in the chromatogram), the relative separation (RS) should be:
a. < 1
b. ≥ 1.5
c. < 0.5
d. = 0.8
164. The anticancer drug, Halichondrin – D is isolated from:
a. Corals
b. Gorgonians
c. Sponges
d. Sea anemone
165. Heparin is a:
a. Glycosylated lipid
b. Glycoprotein
c. Lipopolysaccharide
d. Sulphated polysaccharide
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
162. In a batch sterilization process if ∇overall , ∇heating & ∇cooling are 32.2, 9.8 & 10.1
respectively, THEN the holding time at 121ºC of the process is (given that the specific
death rate (k) of microorganisms at 121ºC is 2.54 min
-1
):
a. 4.84 min
b. 6.75 min
c. 2.96 min
d. 9.25 min
163. To achieve complete separation of two pharmaceutical compounds (shown as two
peaks in the chromatogram), the relative separation (RS) should be:
a. < 1
b. ≥ 1.5
c. < 0.5
d. = 0.8
164. The anticancer drug, Halichondrin – D is isolated from:
a. Corals
b. Gorgonians
c. Sponges
d. Sea anemone
165. Heparin is a:
a. Glycosylated lipid
b. Glycoprotein
c. Lipopolysaccharide
d. Sulphated polysaccharide
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
38
166. The only naturally transformable marine cyanobacteria is:
a. Agmenellum sp.
b. Spirulina sp.
c. Oscillatoria sp.
d. Nostoc sp.
167. Marine chemosynthesis is mainly based on:
a. Oxygen
b. Carbon dioxide
c. Sulphate
d. Manganese nodules
168. Which one is a DNA replication blocking agent produced by sponges?
a. Clathesine
b. Spongosides
c. Spongin
d. Scleorin
169. Eutrophication in aquatic ecosystems is due to reduction of:
a. Carbon
b. Oxygen
c. Sulphur
d. Hydrogen
170. Halotolerant and halophilic microbes can be isolated from ------ and -------, respectively.
P. Sea Q. River R. Lake S. Estuary
a. S and P
b. P and S
c. Q and P
d. R and S
171. Heavy metal pollutants like Cd and Hg inactivate enzymes by interacting with:
a. cysteine
b. glutamic acid
c. lysine
d. histidine
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
166. The only naturally transformable marine cyanobacteria is:
a. Agmenellum sp.
b. Spirulina sp.
c. Oscillatoria sp.
d. Nostoc sp.
167. Marine chemosynthesis is mainly based on:
a. Oxygen
b. Carbon dioxide
c. Sulphate
d. Manganese nodules
168. Which one is a DNA replication blocking agent produced by sponges?
a. Clathesine
b. Spongosides
c. Spongin
d. Scleorin
169. Eutrophication in aquatic ecosystems is due to reduction of:
a. Carbon
b. Oxygen
c. Sulphur
d. Hydrogen
170. Halotolerant and halophilic microbes can be isolated from ------ and -------, respectively.
P. Sea Q. River R. Lake S. Estuary
a. S and P
b. P and S
c. Q and P
d. R and S
171. Heavy metal pollutants like Cd and Hg inactivate enzymes by interacting with:
a. cysteine
b. glutamic acid
c. lysine
d. histidine
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
39
172. Which of the following class of enzymes initiates aerobic degradation of aromatic
pollutants (like naphthalene) in bacteria?
a. Oxido-reductases
b. Hydrolases
c. Ligases
d. Lyases
173. Which one of the following organisms is NOT able to perform light harvesting reaction?
a. Azospirillum
b. Chlamydomonas
c. Rhodopseudomonas
d. Halobacterium
174. In the soil environment, which one of the following factors is responsible for the biotic
stress on the microbial community?
a. Nutrients
b. Oxidation-reduction potential
c. Moisture
d. Microflora
175. Match the enzymes in Group I with their appropriate role in the environment from Group
II
Group I Group II
1 Laccase P Xenobiotic detoxification
2 Catalse Q Nitrogen fixation
3 Nitrogenase R Lignin degradation
4 Cytochrome p450 S Neutralization of toxic O2
species
a. 1-R; 2-S; 3-Q; 4-P b. 1-P; 2-R; 3-Q; 4-S
c. 1-S; 2-P; 3-R; 4-Q d. 1-R; 2-Q; 3-S; 4-P
176. Which one of the following types of mutation would usually NOT be detected in a
molecular diagnostic test that is based on sequencing each exon of a gene individually
from a male suspected of having an X-linked disorder?
a. Missense mutation
b. Nonsense mutation
c. Deletion of an exon
d. Inversion of a part of the gene
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
172. Which of the following class of enzymes initiates aerobic degradation of aromatic
pollutants (like naphthalene) in bacteria?
a. Oxido-reductases
b. Hydrolases
c. Ligases
d. Lyases
173. Which one of the following organisms is NOT able to perform light harvesting reaction?
a. Azospirillum
b. Chlamydomonas
c. Rhodopseudomonas
d. Halobacterium
174. In the soil environment, which one of the following factors is responsible for the biotic
stress on the microbial community?
a. Nutrients
b. Oxidation-reduction potential
c. Moisture
d. Microflora
175. Match the enzymes in Group I with their appropriate role in the environment from Group
II
Group I Group II
1 Laccase P Xenobiotic detoxification
2 Catalse Q Nitrogen fixation
3 Nitrogenase R Lignin degradation
4 Cytochrome p450 S Neutralization of toxic O2
species
a. 1-R; 2-S; 3-Q; 4-P b. 1-P; 2-R; 3-Q; 4-S
c. 1-S; 2-P; 3-R; 4-Q d. 1-R; 2-Q; 3-S; 4-P
176. Which one of the following types of mutation would usually NOT be detected in a
molecular diagnostic test that is based on sequencing each exon of a gene individually
from a male suspected of having an X-linked disorder?
a. Missense mutation
b. Nonsense mutation
c. Deletion of an exon
d. Inversion of a part of the gene
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
40
177. A couple came for counselling following three first-trimester miscarriages and underwent
chromosomal analysis. The man was found to have pericentric inversion with the
karyotype 46,XY,inv8(p12q22). Which one of the following would be a correct conclusion?
a. This is a normal variant, likely of no significance
b. This is an abnormal chromosome that would cause congenital anomalies if transmitted
to a child.
c. This rearrangement might lead to chromosomal imbalance in an offspring and could
explain multiple miscarriages.
d. This rearrangement might cause dicentric or acentric chromosomes in an offspring,
which would probably not be compatible with survival.
178. Unequal crossing over between two Alu repeats can lead to an LDL receptor gene with
an internal deletion or duplication. Based on this information, which of the following must
be true?
a. The LDL receptor gene contains one and only one Alu repeat sequence.
b. The LDL receptor gene does not contain Alu repeat sequences.
c. The genome contains only one copy of the Alu repeat sequence.
d. The LDL receptor gene contains at least two Alu repeat sequences
179. Linkage analysis is performed in a large family with an autosomal dominant hemolytic
anemia, using a polymorphic marker within the β-globin locus. The LOD score at q=0 is
negative infinity. The LOD score at q=0.01 is -4.5. You conclude that the disorder in this
family is:
a. not due to a -globin gene mutation
b. due to a -globin gene mutation
c. an acquired disorder, due to a somatic gene mutation
d. due to a mutation in a gene on chromosome 11, 10 cM centromeric of -globin
180. In Li-Fraumeni syndrome most frequent mutation occurs in:
a. p53
b. RB
c. BRCA1
d. PTEN
181. Which one of the following promoters is most efficient for transgene expression in
mammary gland of livestock?
a. Beta-casein
b. Prolactin
c. Uromodulin
d. Immunoglobulin
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
177. A couple came for counselling following three first-trimester miscarriages and underwent
chromosomal analysis. The man was found to have pericentric inversion with the
karyotype 46,XY,inv8(p12q22). Which one of the following would be a correct conclusion?
a. This is a normal variant, likely of no significance
b. This is an abnormal chromosome that would cause congenital anomalies if transmitted
to a child.
c. This rearrangement might lead to chromosomal imbalance in an offspring and could
explain multiple miscarriages.
d. This rearrangement might cause dicentric or acentric chromosomes in an offspring,
which would probably not be compatible with survival.
178. Unequal crossing over between two Alu repeats can lead to an LDL receptor gene with
an internal deletion or duplication. Based on this information, which of the following must
be true?
a. The LDL receptor gene contains one and only one Alu repeat sequence.
b. The LDL receptor gene does not contain Alu repeat sequences.
c. The genome contains only one copy of the Alu repeat sequence.
d. The LDL receptor gene contains at least two Alu repeat sequences
179. Linkage analysis is performed in a large family with an autosomal dominant hemolytic
anemia, using a polymorphic marker within the β-globin locus. The LOD score at q=0 is
negative infinity. The LOD score at q=0.01 is -4.5. You conclude that the disorder in this
family is:
a. not due to a -globin gene mutation
b. due to a -globin gene mutation
c. an acquired disorder, due to a somatic gene mutation
d. due to a mutation in a gene on chromosome 11, 10 cM centromeric of -globin
180. In Li-Fraumeni syndrome most frequent mutation occurs in:
a. p53
b. RB
c. BRCA1
d. PTEN
181. Which one of the following promoters is most efficient for transgene expression in
mammary gland of livestock?
a. Beta-casein
b. Prolactin
c. Uromodulin
d. Immunoglobulin
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
41
182. Epinephrine is used along with local anaesthetic because of its:
a. α-adrenergic receptor agonist properties
b. β2 adrenergic receptor agonist properties
c. β1 adrenergic receptor agonist properties
d. nicotinic receptor agonist properties
183. Founder effects and bottlenecks are:
a. expected only in large populations
b. mechanisms that increase genetic variation in a population
c. two different modes of natural selection
d. forms of genetic drift
184. In commercial farms, embryos are routinely recovered by non-surgical methods. On
which day are the embryos recovered from the donor cow after the onset of estrus?
a. 21
b. 7
c. 14
d. 5
185. Human mesenchymal stem cells:
a. Can differentiate into only one type of cell
b. Can differentiate into few types of cells
c. Can differentiate into all types of cells
d. Do not differentiate at all
186. Vectors are important in transmission of diseases. Match group 1 with group 2.
Group 1 Group 2
1 Louse A West Nile Fever
2 Tick B Scrub Typhus
3 Mite C Lyme disease
4 Mosquito D Epidemic Typhus
a. 1-D; 2-C; 3-B; 4-A
b. 1-A, 2-B; 3-C; 4-D
c. 1-B; 2-C; 3-D; 4-A
d. 1-C, 2-D; 3-A; 4-B
187. Which one of the following is an inhibitory neurotransmitter?
a. GABA
b. Glutamate
c. Acetylcholine
d. Dopamine
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
182. Epinephrine is used along with local anaesthetic because of its:
a. α-adrenergic receptor agonist properties
b. β2 adrenergic receptor agonist properties
c. β1 adrenergic receptor agonist properties
d. nicotinic receptor agonist properties
183. Founder effects and bottlenecks are:
a. expected only in large populations
b. mechanisms that increase genetic variation in a population
c. two different modes of natural selection
d. forms of genetic drift
184. In commercial farms, embryos are routinely recovered by non-surgical methods. On
which day are the embryos recovered from the donor cow after the onset of estrus?
a. 21
b. 7
c. 14
d. 5
185. Human mesenchymal stem cells:
a. Can differentiate into only one type of cell
b. Can differentiate into few types of cells
c. Can differentiate into all types of cells
d. Do not differentiate at all
186. Vectors are important in transmission of diseases. Match group 1 with group 2.
Group 1 Group 2
1 Louse A West Nile Fever
2 Tick B Scrub Typhus
3 Mite C Lyme disease
4 Mosquito D Epidemic Typhus
a. 1-D; 2-C; 3-B; 4-A
b. 1-A, 2-B; 3-C; 4-D
c. 1-B; 2-C; 3-D; 4-A
d. 1-C, 2-D; 3-A; 4-B
187. Which one of the following is an inhibitory neurotransmitter?
a. GABA
b. Glutamate
c. Acetylcholine
d. Dopamine
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
42
188. Huntington‟s disease is clinically characterized by chorea (abnormal involuntary
movements). Which part of the brain is responsible for this phenotype?
a. Basal Ganglia
b. Cerebellum
c. Hippocampus
d. Brain stem
189. α-Amanitin is a fungal toxin which inhibits eukaryotic RNA polymerases. The three
eukaryotic RNA polymerases show differential sensitivity to this toxin. Which one of the
following order (higher to lower) is correct with respect to sensitivity towards α-amanitin?
a. RNA POL III > RNA POL II > RNA POL I
b. RNA POL II > RNA POL III > RNA POL I
c. RNA POL I > RNA POL III > RNA POL II
d. RNA POL II > RNA POL I > RNA POL III
190. A primary cell culture can be transformed into a cell line by all EXCEPT:
a. Simian Virus 40
b. Hepatitis B virus
c. Human Papillomavirus
d. Epstein-Barr Virus
191. In a randomly breeding population, an autosomal recessive condition affects 1
newborn in 10,000. The expected frequency of carriers will be nearly:
a. 1 in 25
b. 1 in 50
c. 1 in 100
d. 1 in 1,000
192. Assessment of the extent of DNA double strand breaks in cultured human cells,
following exposure to ionizing radiation, can be done by quantitation of :
a. acetylated histones H3 and H4
b. methylated histones H3 and H4
c. ubiquitylated histone H2A.X
d. phosphorylated histone H2A.X
193. There are reports of more than 100 mutations at different sites of Factor IX that
manifests Haemophilia B. This is an example of:
a. clinical heterogeneity
b. allelic heterogeneity
c. protein heterogeneity
d. locus heterogeneity
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
188. Huntington‟s disease is clinically characterized by chorea (abnormal involuntary
movements). Which part of the brain is responsible for this phenotype?
a. Basal Ganglia
b. Cerebellum
c. Hippocampus
d. Brain stem
189. α-Amanitin is a fungal toxin which inhibits eukaryotic RNA polymerases. The three
eukaryotic RNA polymerases show differential sensitivity to this toxin. Which one of the
following order (higher to lower) is correct with respect to sensitivity towards α-amanitin?
a. RNA POL III > RNA POL II > RNA POL I
b. RNA POL II > RNA POL III > RNA POL I
c. RNA POL I > RNA POL III > RNA POL II
d. RNA POL II > RNA POL I > RNA POL III
190. A primary cell culture can be transformed into a cell line by all EXCEPT:
a. Simian Virus 40
b. Hepatitis B virus
c. Human Papillomavirus
d. Epstein-Barr Virus
191. In a randomly breeding population, an autosomal recessive condition affects 1
newborn in 10,000. The expected frequency of carriers will be nearly:
a. 1 in 25
b. 1 in 50
c. 1 in 100
d. 1 in 1,000
192. Assessment of the extent of DNA double strand breaks in cultured human cells,
following exposure to ionizing radiation, can be done by quantitation of :
a. acetylated histones H3 and H4
b. methylated histones H3 and H4
c. ubiquitylated histone H2A.X
d. phosphorylated histone H2A.X
193. There are reports of more than 100 mutations at different sites of Factor IX that
manifests Haemophilia B. This is an example of:
a. clinical heterogeneity
b. allelic heterogeneity
c. protein heterogeneity
d. locus heterogeneity
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
43
194. Specificity factor which activates ubiquitin ligase activity of APC/C during
separation of sister chromatids is:
a. CDC10
b. CDC20
c. CDH1
d. CDC45A
195. In a diploid organism, loss of function mutations produce dominant phenotypes
when there is:
a. pleiotropy
b. epistasis
c. multiple allelism
d. haploinsufficiency
196. In Mendel‟s dihybrid cross experiment, if the two selected traits were on the same
chromosome (linked) and assuming that there is no recombination, what would be the
expected genotypic ratio in the F2 generation?
a. 1:2:1
b. 9:3:3:1
c. 1:3
d. 1:1
197. Huntington disease is caused by:
a. Expanded dinucleotide repeats sequence in coding region
b. Expanded trinucleotide repeats in non-coding region
c. Expanded trinucleotide sequence in coding sequence
d. Expanded dinucleotide sequence in non-coding sequence
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
194. Specificity factor which activates ubiquitin ligase activity of APC/C during
separation of sister chromatids is:
a. CDC10
b. CDC20
c. CDH1
d. CDC45A
195. In a diploid organism, loss of function mutations produce dominant phenotypes
when there is:
a. pleiotropy
b. epistasis
c. multiple allelism
d. haploinsufficiency
196. In Mendel‟s dihybrid cross experiment, if the two selected traits were on the same
chromosome (linked) and assuming that there is no recombination, what would be the
expected genotypic ratio in the F2 generation?
a. 1:2:1
b. 9:3:3:1
c. 1:3
d. 1:1
197. Huntington disease is caused by:
a. Expanded dinucleotide repeats sequence in coding region
b. Expanded trinucleotide repeats in non-coding region
c. Expanded trinucleotide sequence in coding sequence
d. Expanded dinucleotide sequence in non-coding sequence
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
44
198. The following pedigree shows the inheritance of a very rare human disease. What
is the most likely mode of inheritance for the disease trait and what is the probability that
the second child of III-2 and III-3 will be a son and will also have the disease?
a. Sex-linked recessive; 0.5
b. Autosomal recessive; 0.25
c. Autosomal dominant: 0.75
d. Autosomal dominant; 0.25
199. Microscopic evaluation of a post-mortem brain exhibits hyperchromatic areas when
stained with glial fibrillary acidic protein (GFAP). This suggests activation of:
a. astrocytes
b. oligodendrocytes
c. Microglia
d. neurons
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
198. The following pedigree shows the inheritance of a very rare human disease. What
is the most likely mode of inheritance for the disease trait and what is the probability that
the second child of III-2 and III-3 will be a son and will also have the disease?
a. Sex-linked recessive; 0.5
b. Autosomal recessive; 0.25
c. Autosomal dominant: 0.75
d. Autosomal dominant; 0.25
199. Microscopic evaluation of a post-mortem brain exhibits hyperchromatic areas when
stained with glial fibrillary acidic protein (GFAP). This suggests activation of:
a. astrocytes
b. oligodendrocytes
c. Microglia
d. neurons
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
45
200. Parasite numbers in the blood of a patient with Trypanosoma brucci (Sleeping
sickness) shows the following pattern:
Parasites isolated from population 1 and 2 were found to be antigenecially non-
cross reactive. The parasite distribution seen may be due to:
a. Change in gene expression
b. Loss of antigenic epitopes
c. Post-translational modification of antigen
d. Antigenic switching
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
200. Parasite numbers in the blood of a patient with Trypanosoma brucci (Sleeping
sickness) shows the following pattern:
Parasites isolated from population 1 and 2 were found to be antigenecially non-
cross reactive. The parasite distribution seen may be due to:
a. Change in gene expression
b. Loss of antigenic epitopes
c. Post-translational modification of antigen
d. Antigenic switching
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
Question Correct Option
41 2
42 2
43 1
44 1
45 3
46 4
47 1
48 3
49 3
50 2
51 2
52 1
53 3
54 3
55 2
56 3
57 2
58 4
59 3
60 2
61 4
62 4
63 3
64 2
65 4
66 4
67 4
68 4
69 2
70 2
71 3
72 2
73 3
74 1,3
75 3
76 3
77 2
78 3
79 2
80 1
Question Correct Option
81 2
82 2
83 3
84 2
85 2
86 1,2,3,4
87 2
88 3
89 4
90 4
91 1
92 2
93 2
94 1
95 3
96 1
97 1
98 1
99 2
100 4
101 2
102 3
103 1
104 2
105 3
106 1
107 2
108 2
109 2
110 2
111 4
112 3
113 2
114 3
115 3
116 3
117 4
118 4
119 2
120 4
BIOTECHNOLOGY ELIGIBILITY TEST- 2017
ANSWER KEY
Question Correct Option
1 4
2 4
3 2
4 2
5 4
6 3
7 1
8 1
9 1
10 2
11 1
12 4
13 1
14 1
15 3
16 2
17 2
18 2
19 4
20 2
21 1
22 1
23 3
24 2
25 3
26 2
27 2
28 1
29 4
30 2
31 3
32 2
33 2
34 3
35 3
36 4
37 2
38 1,3
39 2
40 4
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
41 2
42 2
43 1
44 1
45 3
46 4
47 1
48 3
49 3
50 2
51 2
52 1
53 3
54 3
55 2
56 3
57 2
58 4
59 3
60 2
61 4
62 4
63 3
64 2
65 4
66 4
67 4
68 4
69 2
70 2
71 3
72 2
73 3
74 1,3
75 3
76 3
77 2
78 3
79 2
80 1
Question Correct Option
81 2
82 2
83 3
84 2
85 2
86 1,2,3,4
87 2
88 3
89 4
90 4
91 1
92 2
93 2
94 1
95 3
96 1
97 1
98 1
99 2
100 4
101 2
102 3
103 1
104 2
105 3
106 1
107 2
108 2
109 2
110 2
111 4
112 3
113 2
114 3
115 3
116 3
117 4
118 4
119 2
120 4
BIOTECHNOLOGY ELIGIBILITY TEST- 2017
ANSWER KEY
Question Correct Option
1 4
2 4
3 2
4 2
5 4
6 3
7 1
8 1
9 1
10 2
11 1
12 4
13 1
14 1
15 3
16 2
17 2
18 2
19 4
20 2
21 1
22 1
23 3
24 2
25 3
26 2
27 2
28 1
29 4
30 2
31 3
32 2
33 2
34 3
35 3
36 4
37 2
38 1,3
39 2
40 4
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
Question Correct Option
161 2
162 1
163 2
164 3
165 4
166 1
167 2
168 2
169 2
170 1
171 1
172 1
173 1
174 4
175 1
176 4
177 3
178 4
179 1
180 1
181 1
182 1
183 4
184 2
185 2
186 1
187 1
188 1
189 2
190 2
191 2
192 4
193 2
194 2
195 4
196 1
197 3
198 4
199 1
200 4
Question Correct Option
121 3
122 2
123 3
124 3
125 3
126 4
127 2
128 3
129 4
130 3
131 1
132 2
133 4
134 1
135 2
136 3
137 1
138 3
139 4
140 3
141 1
142 1
143 4
144 3
145 2
146 4
147 1
148 2
149 4
150 1
151 1
152 2
153 1
154 2
155 2
156 1
157 1
158 1
159 1
160 2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
161 2
162 1
163 2
164 3
165 4
166 1
167 2
168 2
169 2
170 1
171 1
172 1
173 1
174 4
175 1
176 4
177 3
178 4
179 1
180 1
181 1
182 1
183 4
184 2
185 2
186 1
187 1
188 1
189 2
190 2
191 2
192 4
193 2
194 2
195 4
196 1
197 3
198 4
199 1
200 4
Question Correct Option
121 3
122 2
123 3
124 3
125 3
126 4
127 2
128 3
129 4
130 3
131 1
132 2
133 4
134 1
135 2
136 3
137 1
138 3
139 4
140 3
141 1
142 1
143 4
144 3
145 2
146 4
147 1
148 2
149 4
150 1
151 1
152 2
153 1
154 2
155 2
156 1
157 1
158 1
159 1
160 2
DBT JRF 2016 & 2017 Question Paper www.biologyexams4u.com
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