C785 Biochemistry - Enzyme Inhibition & Kinetics Practice Pack 2026
|WGU
1. In Michaelis-Menten kinetics, what does the Michaelis constant (Km)
represent?
A. The equilibrium constant for the dissociation of the product
B. The maximum velocity of the enzyme-catalyzed reaction
C. The substrate concentration at which the reaction rate is half of Vmax
D. The total concentration of the enzyme used in the assay
Answer: C
Rationale: Km is defined as the substrate concentration at which the reaction velocity is
exactly half of the maximum velocity (Vmax).
2. How does a competitive inhibitor affect the kinetic parameters of an enzyme?
A. It decreases Vmax and increases Km
B. It decreases both Vmax and Km
C. It increases Km while Vmax remains unchanged
D. It decreases Km while Vmax remains unchanged
Answer: C
Rationale: Competitive inhibitors bind to the active site, increasing the apparent Km
(lowering affinity) but can be outcompeted at high substrate concentrations, leaving Vmax
unchanged.
,3. In a Lineweaver-Burk plot, what does the y-intercept represent?
A. -1/Km
B. Km/Vmax
C. 1/[S]
D. 1/Vmax
Answer: D
Rationale: The Lineweaver-Burk equation is 1/v = (Km/Vmax)(1/[S]) + 1/Vmax. The y-
intercept occurs when 1/[S] is zero, which is 1/Vmax.
4. Which type of inhibition involves the inhibitor binding only to the enzyme-
substrate (ES) complex?
A. Competitive inhibition
B. Non-competitive inhibition
C. Mixed inhibition
D. Uncompetitive inhibition
Answer: D
Rationale: Uncompetitive inhibitors bind specifically to the ES complex and not the free
enzyme, leading to a decrease in both Vmax and Km.
5. What is the primary effect of a non-competitive inhibitor on an enzyme’s
kinetics?
A. Vmax increases and Km decreases
B. Vmax remains the same and Km increases
C. Both Vmax and Km increase
D. Vmax decreases and Km remains the same
Answer: D
Rationale: Non-competitive inhibitors bind to a site other than the active site, reducing the
turnover number (Vmax) without affecting the enzyme’s affinity for the substrate (Km).
, 6. Enzymes speed up reactions by doing which of the following?
A. Increasing the Gibbs free energy of the reaction
B. Changing the equilibrium constant (Keq)
C. Lowering the activation energy (Ea)
D. Raising the temperature of the cellular environment
Answer: C
Rationale: Enzymes stabilize the transition state, thereby lowering the activation energy
barrier required for the reaction to proceed.
7. An enzyme with a low Km value has which of the following characteristics?
A. Low affinity for its substrate
B. High sensitivity to temperature changes
C. A very slow reaction rate
D. High affinity for its substrate
Answer: D
Rationale: A low Km indicates that the enzyme reaches half-saturation at a low substrate
concentration, meaning it has a high affinity for the substrate.
8. What does the x-intercept on a Lineweaver-Burk plot represent?
A. -1/Km
B. Km/Vmax
C. 1/Vmax
D. Vmax/Km
Answer: A
Rationale: The x-intercept occurs when 1/v = 0 in the Lineweaver-Burk plot, which
mathematically corresponds to -1/Km.
|WGU
1. In Michaelis-Menten kinetics, what does the Michaelis constant (Km)
represent?
A. The equilibrium constant for the dissociation of the product
B. The maximum velocity of the enzyme-catalyzed reaction
C. The substrate concentration at which the reaction rate is half of Vmax
D. The total concentration of the enzyme used in the assay
Answer: C
Rationale: Km is defined as the substrate concentration at which the reaction velocity is
exactly half of the maximum velocity (Vmax).
2. How does a competitive inhibitor affect the kinetic parameters of an enzyme?
A. It decreases Vmax and increases Km
B. It decreases both Vmax and Km
C. It increases Km while Vmax remains unchanged
D. It decreases Km while Vmax remains unchanged
Answer: C
Rationale: Competitive inhibitors bind to the active site, increasing the apparent Km
(lowering affinity) but can be outcompeted at high substrate concentrations, leaving Vmax
unchanged.
,3. In a Lineweaver-Burk plot, what does the y-intercept represent?
A. -1/Km
B. Km/Vmax
C. 1/[S]
D. 1/Vmax
Answer: D
Rationale: The Lineweaver-Burk equation is 1/v = (Km/Vmax)(1/[S]) + 1/Vmax. The y-
intercept occurs when 1/[S] is zero, which is 1/Vmax.
4. Which type of inhibition involves the inhibitor binding only to the enzyme-
substrate (ES) complex?
A. Competitive inhibition
B. Non-competitive inhibition
C. Mixed inhibition
D. Uncompetitive inhibition
Answer: D
Rationale: Uncompetitive inhibitors bind specifically to the ES complex and not the free
enzyme, leading to a decrease in both Vmax and Km.
5. What is the primary effect of a non-competitive inhibitor on an enzyme’s
kinetics?
A. Vmax increases and Km decreases
B. Vmax remains the same and Km increases
C. Both Vmax and Km increase
D. Vmax decreases and Km remains the same
Answer: D
Rationale: Non-competitive inhibitors bind to a site other than the active site, reducing the
turnover number (Vmax) without affecting the enzyme’s affinity for the substrate (Km).
, 6. Enzymes speed up reactions by doing which of the following?
A. Increasing the Gibbs free energy of the reaction
B. Changing the equilibrium constant (Keq)
C. Lowering the activation energy (Ea)
D. Raising the temperature of the cellular environment
Answer: C
Rationale: Enzymes stabilize the transition state, thereby lowering the activation energy
barrier required for the reaction to proceed.
7. An enzyme with a low Km value has which of the following characteristics?
A. Low affinity for its substrate
B. High sensitivity to temperature changes
C. A very slow reaction rate
D. High affinity for its substrate
Answer: D
Rationale: A low Km indicates that the enzyme reaches half-saturation at a low substrate
concentration, meaning it has a high affinity for the substrate.
8. What does the x-intercept on a Lineweaver-Burk plot represent?
A. -1/Km
B. Km/Vmax
C. 1/Vmax
D. Vmax/Km
Answer: A
Rationale: The x-intercept occurs when 1/v = 0 in the Lineweaver-Burk plot, which
mathematically corresponds to -1/Km.
