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Home ‣ Biochemical Engineering ‣ Enzymes and Kinetics Comments
- Question
Enzymes are basically
Options- A. proteins
- B. vitamins
- C. fat
- D. carbohydrates
- Correct Answer
- proteins
- 1. The active site of an enzyme differs from an antibody-antigen binding site in that the enzyme active site
Options- A. contains modified amino acids
- B. catalyzes a chemical reaction
- C. is complementary to a specific ligand
- D. contains amino acids without side chains Discuss
- 2. The initial velocity, V0, of an enzyme catalyzed reaction reaches Vmax as
Options- A. [S] = KM
- B. [S] = 10 * KM
- C. 1/[S] = 1/KM
- D. 1/[S] ? 0 Discuss
- 3. The rate equation in non-competitive inhibition based on Michaelis Menten equation is given by
Options- A. rmaxS/(Km + S)(1+I/Ki)
- B. rmaxE/(Km (1+I/Ki)+S))
- C. VmaxS/(Km + S)(1+I/Ki)
- D. rmaxS/Km Discuss
- 4. A classical noncompetitive inhibitor has
Options- A. no effect on substrate binding
- B. no effect on substrate binding and vice versa
- C. significant effect on substrate binding
- D. significant effect on substrate binding and vice versa Discuss
- 5. When substrate [S] = KM (Michaelis-Menten constant), the velocity of an enzyme catalyzed reaction is about
Options- A. 0.1 * Vmax
- B. 0.2 * Vmax
- C. 0.5 * Vmax
- D. 0.9 * Vmax Discuss
- 6. Classical noncompetitive inhibition is obtained only under
Options- A. slow equilibrium conditions
- B. moderate equilibrium conditions
- C. rapid equilibrium conditions
- D. non-equilibrium conditions Discuss
- 7. Lock and key theory is based on the compatibility of
Options- A. enzyme and substrate
- B. enzyme and product
- C. enzyme and enzyme substrate complex
- D. enzyme substrate complex and product Discuss
- 8. Michaelis Menten equation can also be written as
Options- A. (-Cs)/r = (Cs/rmax)+(Km/rmax)
- B. 1/r = (1/rmax)+(Km/(rmax.Cs))
- C. r = rmax-(Km.r/Cs)
- D. All of these Discuss
- 9. The rate equation in competitive inhibition based on Michaelis Menten equation is given by
Options- A. rmaxS/(Km (1+I/Ki)+S))
- B. rmaxE/(Km (1+I/Ki)+S))
- C. rmaxI/(Km (1+I/Ki)+S))
- D. rmaxS/(Km (1+I/Ki)) Discuss
- 10. Lag phase occurs mainly due to
Options- A. acclimatization with new environment and substrate
- B. higher growth temperature
- C. substrate disliking
- D. variation in substrate concentration Discuss
Enzymes and Kinetics problems
Search Results
Correct Answer: catalyzes a chemical reaction
Correct Answer: 1/[S] ? 0
Correct Answer: rmaxS/(Km + S)(1+I/Ki)
Correct Answer: no effect on substrate binding and vice versa
Correct Answer: 0.5 * Vmax
Correct Answer: rapid equilibrium conditions
Correct Answer: enzyme and substrate
Correct Answer: All of these
Correct Answer: rmaxS/(Km (1+I/Ki)+S))
Correct Answer: acclimatization with new environment and substrate
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