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Biochemistry Test Bank – Topic 2 In every slide, you will find a question with 2 – 5 choices. The answers of the questions will be in the last slide, so don’t forget to write your answers (To check them ).

Biochemistry Test Bank – Topic 2

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Page 1: Biochemistry Test Bank – Topic 2

Biochemistry Test Bank – Topic 2In every slide, you will find a question with 2 – 5 choices.

The answers of the questions will be in the last slide, so don’t forget to write your answers (To check them 💪).

Page 2: Biochemistry Test Bank – Topic 2

1. KM, the Michaelis constant, is defined as that concentration of substrate at which enzyme is working at maximum velocity:

A. True.

B. False.

Page 3: Biochemistry Test Bank – Topic 2

2. KM, the Michaelis constant is defined as the dissociation constant of the enzyme-substrate complex:

A. True.

B. False.

Page 4: Biochemistry Test Bank – Topic 2

3. When the velocity of enzyme activity is plotted against substrate concentration, which of the following is obtained?

A. Hyperbolic curve.

B. Parabola.

C. Straight line with positive slope.

D. Straight line with negative slope.

Page 5: Biochemistry Test Bank – Topic 2

4. Which of the following is the correct Line weaver-Burk equation?

A.1

𝑉0=

𝐾𝑀

𝑉𝑀𝑎𝑥[𝑆]+

1

𝑉𝑀𝑎𝑥

B.1

𝑉𝑀𝑎𝑥=

𝐾𝑀

𝑉0 𝑆+

1

𝑉0

C. 𝑉0 =𝑉𝑀𝑎𝑥 𝑆

𝐾𝑀+ 𝑆

D. 𝑉𝑀𝑎𝑥 =𝑉0 𝑆

𝐾𝑀 𝑆

Page 6: Biochemistry Test Bank – Topic 2

5. The catalytic efficiency of two distinct enzymes can be compared based on which of the following factor?

A. KM.

B. Product formation.

C. Size of the enzymes.

D. pH of optimum value.

Page 7: Biochemistry Test Bank – Topic 2

6. From the Lineweaver-Burk plot, the X-axis experimental data are expressed as:

A. 1/V.

B. V.

C. 1/S.

D. S.

Page 8: Biochemistry Test Bank – Topic 2

7. In enzyme kinetics KM implies:

A. The substrate concentration that gives one half Vmax.

B. The dissocation constant for the enzyme substrate comples.

C. Concentration of enzyme.

D. Half of the substrate concentration required to achieve Vmax.

Page 9: Biochemistry Test Bank – Topic 2

8. What is enzyme kinetics?

A. Study of [S].

B. Study of rate of reaction.

C. Study of initial velocity.

D. All of the above.

Page 10: Biochemistry Test Bank – Topic 2

9. What is the effect on initial velocity if the substrate concentration is low?

A. Increases rapidly.

B. Increases slowly.

C. Decreases.

D. Remains constant.

Page 11: Biochemistry Test Bank – Topic 2

10. What is the effect on initial velocity if the substrate concentration is high?

A. Increases rapidly.

B. Increases slowly.

C. Decreases.

D. Remains constant.

Page 12: Biochemistry Test Bank – Topic 2

11. In how many forms, enzymes exists in an enzymatic reaction?

A. 1.

B. 2.

C. 3.

D. 4.

Page 13: Biochemistry Test Bank – Topic 2

12. In which form the enzyme exists when the maximum velocity is observed?

A. E form.

B. ES form.

C. Both.

D. None.

Page 14: Biochemistry Test Bank – Topic 2

13. When the reaction reaches its plateau state, which type of kinetics is seen?

A. Zero order kinetics.

B. First order kinetics.

C. Second order kinetics.

D. None of the above.

Page 15: Biochemistry Test Bank – Topic 2

14. Turnover number is also known as?

A. Km.

B. Kcat.

C. K1.

D. K-1.

Page 16: Biochemistry Test Bank – Topic 2

15. In which condition, V0=Vmax?

A. Low [S].

B. High [S].

C. Low V0.

D. High V0.

Page 17: Biochemistry Test Bank – Topic 2

16. Which of the following statement is NOT true?

A. Turnover number is also known as catalytic constant.

B. The reaction for the formation of ES is irreversible.

C. The reaction is in steady state when rate of synthesis is equal to rate of degradation.

D. Plateau is called Vmax.

Page 18: Biochemistry Test Bank – Topic 2

17. Which of the following is the slope of double-reciprocal plot?

A. Km/V0.

B. Vmax/Km.

C. -1/Km.

D. None of the above.

Page 19: Biochemistry Test Bank – Topic 2

18. Which of the following terms are involved in Michaelis-Menten Equation?

A. Vmax.

B. Km.

C. V0.

D. All of the above.

Page 20: Biochemistry Test Bank – Topic 2

19. The point at which the line intersects the y-axis of double-reciprocal plot is numerically equal to?

A. Km/V0.

B. Vmax/Km.

C. -1/Km.

D. 1/Vmax.

Page 21: Biochemistry Test Bank – Topic 2

20. When [S]=KM, the velocity of an enzyme catalyzed reaction is about:

A. 0.1*VMAX.

B. 0.2*VMAX.

C. 0.3*VMAX.

D. 0.5*VMAX.

E. 0.9*VMAX.

Page 22: Biochemistry Test Bank – Topic 2

21. The substrate KM in an enzyme-catalyzed reaction:

A. is usually less than Kd, the dissociation constant.

B. is never less than Kd.

C. cannot be equal to Kd.

D. is estimated from the Y-intercept of a Lineweaver-Burk plot.

E. cannot be derived from simple rate equations.

Page 23: Biochemistry Test Bank – Topic 2

22. In an enzyme catalyzed reaction, __________ provides information on __________ and __________ provides information on __________.

A. KM, chemical step, VMAX, substrate binding.

B. VMAX, chemical step, kCAT, substrate binding.

C. KD, substrate binding, kCAT, chemical step.

D. KM, substrate binding, VMAX, chemical step.

E. kCAT, substrate binding, VMAX, chemical step.

Page 24: Biochemistry Test Bank – Topic 2

23. The initial velocity, Vo, of an enzyme catalyzed reaction reaches Vmax:

A. At [S] = KM.

B. At [S] = 10*KM.

C. At 1/[S] = 1/KM.

D. Only as 1/[S] = 0.

E. The third and fourth choices are both correct.

Page 25: Biochemistry Test Bank – Topic 2

24. The __________ of a double reciprocal plot equals __________ and the __________ equals __________.

A. slope, VMAX, y-intercept, KM.

B. slope, KM, y-intercept, 1/VMAX.

C. slope, KM/VMAX, y-intercept, 1/VMAX.

D. slope, KM/VMAX, y-intercept, VMAX.

E. slope, KM/VMAX, x-intercept, VMAX.

Page 26: Biochemistry Test Bank – Topic 2

25. When [S] = 0.1*KM, the velocity of an enzyme catalyzed reaction is about:

A. 0.1*Vmax.

B. 0.3*Vmax.

C. 0.5*Vmax.

D. 0.7*Vmax.

E. 0.9*Vmax.

Page 27: Biochemistry Test Bank – Topic 2

26. When [S] = 10*KM, the velocity of an enzyme catalyzed reaction is about:

A. 0.1*Vmax.

B. 0.3*Vmax.

C. 0.5*Vmax.

D. 0.7*Vmax.

E. 0.9*Vmax.

Page 28: Biochemistry Test Bank – Topic 2

27. For an enzyme that displays Michaelis-Menten kinetics, the reaction velocity (as a fraction of Vmax ) observed at [S] = 2 KM will be:

A. 0.09.

B. 0.33.

C. 0.66.

D. 0.91.

Page 29: Biochemistry Test Bank – Topic 2

28. Read the informative box below, and answer the question:

A. 1.5 min.

B. 3 min.

C. 6 min.

D. 13.5 min.

E. 27 min.

An enzyme-catalyzed reaction was carried out with

the substrate concentration initially 1,000 times

greater than the Km for that substrate. After 9

minutes, 1% of the substrate had been converted to

product, and the amount of product formed in the

reaction mixture was 12 µmol. If, in a separate

experiment, one-third as much enzyme and twice as

much substrate had been combined, how long

would it take for the same amount (12 µmol) of

product to be formed?

***Note: Use this link to know how to solve this question.

Page 30: Biochemistry Test Bank – Topic 2

29. How can the Michaelis-Menten constant, Km, be derived from this Lineweaver-Burk plot?

A. Km = (-1) * (x-intercept).

B. Km = (slope).

C. Km = (slope) * (1/y-intercept).

D. Km = 1 / (x-intercept).

Page 31: Biochemistry Test Bank – Topic 2

30. What relative values of Km and kcat would describe an enzyme with a high catalytic efficiency?

A. Low Km, low Kcat.

B. High Km, high Kcat.

C. Low Km, high Kcat.

D. High Km, low Kcat.

Page 32: Biochemistry Test Bank – Topic 2

31. Read the informative box below, and answer the question:

A. 25 μM.

B. 50 μM.

C. 100 μM.

D. 250 μM.

E. 475 μM.

Assume that the reaction catalyzed by an enzyme

follows Michaelis-Menten kinetics. The substrate

concentration (Km, Michaelis constant) needed to

reach 50% of the maximum reaction velocity

(Vmax) is 25 μM. What substrate concentration is

required to obtain at least 95% of the maximum

reaction velocity?

***Note: Use this link to know how to solve this question.

Page 33: Biochemistry Test Bank – Topic 2

32. Read the informative box below, and answer the question:

A. 2 nM.

B. 5 nM.

C. 10 nM.

D. 50 nM.

E. 100 nM.

Assume that the reaction catalyzed by an enzyme

follows Michaelis-Menten kinetics. If at a substrate

concentration of 100 nM, the reaction proceeds at

98% of the maximum reaction velocity (Vmax),

what is the Michaelis constant (Km) for this

substrate. Km is the substrate concentration

needed to reach 50% of Vmax.

Page 34: Biochemistry Test Bank – Topic 2

33. The following ranking of substrates is in terms of affinity. Which one of these substrates has the highest Km?

A. D-Glucose.

B. ATP.

C. D-Fructose.

D. Can’t be determined.

D-Glucose > ATP > D-Fructose.

Page 35: Biochemistry Test Bank – Topic 2

34. Read the informative box below, The KM of the enzyme is higher for:

A. A.

B. B.

C. Neither.

D. Can’t tell from this information.

The enzyme trypsin hydrolyzes peptides at the carboxyl side of Arg and Lys

residues. However, the amino acid residue that follows the Arg or Lys can

influence either the KM or the kcat of the reaction. Suppose that the

adjacent curves were generated by comparing two substrates for trypsin:

o Substrate A = Ser-Val-Arg-Pro.

o Substrate B = Ser-Val-Arg-Phe.

Page 36: Biochemistry Test Bank – Topic 2

35. Read the informative box below, The enzyme binds:

A. More tightly to A.

B. More tightly to B.

C. Same to both.

D. Can’t tell.

The enzyme trypsin hydrolyzes peptides at the carboxyl side of Arg and Lys

residues. However, the amino acid residue that follows the Arg or Lys can

influence either the KM or the kcat of the reaction. Suppose that the

adjacent curves were generated by comparing two substrates for trypsin:

o Substrate A = Ser-Val-Arg-Pro.

o Substrate B = Ser-Val-Arg-Phe.

Page 37: Biochemistry Test Bank – Topic 2

36. Read the informative box below, the approximate KM for Substrate A is:

A. 0.0001 M.

B. 0.0005 M.

C. 0.001 M.

D. >0.002 M.

The enzyme trypsin hydrolyzes peptides at the carboxyl side of Arg and Lys

residues. However, the amino acid residue that follows the Arg or Lys can

influence either the KM or the kcat of the reaction. Suppose that the

adjacent curves were generated by comparing two substrates for trypsin:

o Substrate A = Ser-Val-Arg-Pro.

o Substrate B = Ser-Val-Arg-Phe.

Page 38: Biochemistry Test Bank – Topic 2

37. Read the informative box below, The kcat of the enzyme is higher for:

A. A.

B. B.

C. Neither.

D. Can’t tell from this information.

The enzyme trypsin hydrolyzes peptides at the carboxyl side of Arg and Lys

residues. However, the amino acid residue that follows the Arg or Lys can

influence either the KM or the kcat of the reaction. Suppose that the

adjacent curves were generated by comparing two substrates for trypsin:

o Substrate A = Ser-Val-Arg-Pro.

o Substrate B = Ser-Val-Arg-Phe.

Page 39: Biochemistry Test Bank – Topic 2

38. What is the correct calculation for the rate of an enzyme reaction?

A. Amount of substrate used ÷ time taken.

B. Amount of substrate used + time taken.

C. Amount of substrate used × time taken.

Page 40: Biochemistry Test Bank – Topic 2

39. Given the reaction of an enzyme that follows Michaelis-Menten kinetics, What is the initial reaction velocity at a substrate concentration of 30 mM?

A. 10 μM min-1.

B. 15 μM min-1 .

C. 30 μM min-1 .

D. 60 μM min-1 .

Page 41: Biochemistry Test Bank – Topic 2

40. Read the informative box below, Vmax is equal to:

A. 0.1 (µM/min)-1.

B. 10 (µM/min)-1.

C. 0.1 µM/min.

D. 10 µM/min.

You measure the kinetics of an enzyme E as a function of substrate

concentration first without any inhibitor (see Table) and plot the data

using the double-reciprocal (Lineweaver-Burk) plot (Figure below). The

enzyme concentration is maintained constant at a level of 1 µM (=10-6 M).

Page 42: Biochemistry Test Bank – Topic 2

41. Read the informative box below, Km is equal to:

A. 0.05 µM.

B. 0.5 µM.

C. 5 µM.

D. 50 µM.

E. 500 µM.

You measure the kinetics of an enzyme E as a function of substrate

concentration first without any inhibitor (see Table) and plot the data

using the double-reciprocal (Lineweaver-Burk) plot (Figure below). The

enzyme concentration is maintained constant at a level of 1 µM (=10-6 M).

Page 43: Biochemistry Test Bank – Topic 2

42. Read the informative box below, Kcat is equal to:

A. 5 min-1.

B. 10 min-1.

C. 20 min-1 .

D. Can’t be determined.

You measure the kinetics of an enzyme E as a function of substrate

concentration first without any inhibitor (see Table) and plot the data

using the double-reciprocal (Lineweaver-Burk) plot (Figure below). The

enzyme concentration is maintained constant at a level of 1 µM (=10-6 M).

Page 44: Biochemistry Test Bank – Topic 2

43. Which letter is drawn next to the arrow representing the difference in activation energy between the catalyzed and noncatalyzed reaction?

A. A.

B. B.

C. C.

D. D.

Page 45: Biochemistry Test Bank – Topic 2

44. The turnover number of an enzyme is calculated by dividing the maximal velocity by the enzyme concentration.

A. True.

B. False.

Page 46: Biochemistry Test Bank – Topic 2

45. In a Lineweaver-burk plot, the Y-intercept is -1/KM.

A. True.

B. False.

Page 47: Biochemistry Test Bank – Topic 2

The key answers

Question No. Answers Question No. Answers Question No. Answers Question No. Answers Question No. Answers

1. B 10. B 19. D 28. E 37. A

2. B 11. B 20. D 29. C 38. A

3. A 12. B 21. B 30. C 39. C

4. A 13. A 22. D 31. E 40. D

5. A 14. B 23. D 32. A 41. C

6. C 15. B 24. C 33. C 42. B

7. A 16. B 25. A 34. B 43. C

8. B 17. D 26. E 35. A 44. A

9. A 18. D 27. C 36. A 45. B