BIOCHEM 210 MODULE 3 EXAM 2026/2027 | Hemoglobin,
Myoglobin & Protein Function | Portage Learning | Verified
Q&A | Pass Guaranteed - A+ Graded
Q1: According to the Michaelis-Menten model, which assumption is made about the
enzyme-substrate complex (ES)?
A. ES breaks down only to form product
B. ES formation is the rate-limiting step
C. ES is in rapid equilibrium with free enzyme and substrate, and its concentration
remains constant during the initial phase [CORRECT]
D. ES concentration increases continuously throughout the reaction
Correct Answer: C
Rationale: The steady-state assumption (Briggs-Haldane modification) states that [ES]
remains constant during the initial velocity phase because its rate of formation equals
its rate of breakdown (to E + P or E + S). This allows derivation of the Michaelis-Menten
equation. Option A is incorrect because ES can dissociate back to E + S. Option B is
incorrect because product formation (kcat) is typically rate-limiting, not ES formation.
Option D violates the steady-state condition that defines initial velocity measurements.
Q2: An enzyme has a Km of 5 mM and a Vmax of 100 μmol/min. At what substrate
concentration will the reaction velocity be 25 μmol/min?
A. 1.0 mM
B. 1.67 mM [CORRECT]
C. 5.0 mM
D. 10.0 mM
Correct Answer: B
Rationale: Using the Michaelis-Menten equation: v = (Vmax × [S])/(Km + [S]).
Rearranging: 25 = (100 × [S])/(5 + [S]). Solving: 25(5 + [S]) = 100[S]; 125 + 25[S] = 100[S];
125 = 75[S]; [S] = 1.67 mM. At [S] = Km (5 mM), v would be 50 μmol/min (half Vmax). At
1.0 mM (A), v = 16.7 μmol/min. At 10 mM (D), v = 66.7 μmol/min.
Q3: Which type of inhibition results in an increased apparent Km but unchanged Vmax?
, A. Non-competitive inhibition
B. Uncompetitive inhibition
C. Competitive inhibition [CORRECT]
D. Mixed inhibition
Correct Answer: C
Rationale: Competitive inhibitors bind reversibly to the active site, competing with
substrate. This increases the apparent Km (more substrate needed to reach half Vmax)
but Vmax remains unchanged because sufficient substrate can outcompete the
inhibitor. On Lineweaver-Burk plots, competitive inhibition shows lines intersecting on
the y-axis (1/Vmax unchanged), with different x-intercepts (-1/Km). Non-competitive (A)
decreases Vmax with unchanged Km. Uncompetitive (B) decreases both Vmax and Km.
Mixed (D) affects both parameters with lines intersecting left of the y-axis.
Q4: The catalytic efficiency of an enzyme is best described by which parameter?
A. Vmax alone
B. Km alone
C. kcat/Km [CORRECT]
D. Turnover number (kcat) alone
Correct Answer: C
Rationale: Catalytic efficiency is defined as kcat/Km (the specificity constant), with units
of M⁻¹s⁻¹. This second-order rate constant reflects how efficiently an enzyme converts
substrate to product at low substrate concentrations, combining both binding affinity
(1/Km) and catalytic rate (kcat). Values approaching 10⁸-10⁹ M⁻¹s⁻¹ approach
diffusion-controlled limits. Vmax (A) depends on enzyme concentration. Km (B) alone
doesn't indicate catalytic rate. kcat (D) alone doesn't reflect substrate binding.
Q5: In a Lineweaver-Burk plot, the y-intercept represents:
A. -1/Km
B. 1/Km
C. 1/Vmax [CORRECT]
D. Vmax
Correct Answer: C
Rationale: The Lineweaver-Burk equation (double reciprocal): 1/v = (Km/Vmax)(1/[S]) +
1/Vmax. The y-intercept (when 1/[S] = 0) equals 1/Vmax. The x-intercept (when 1/v = 0)
Myoglobin & Protein Function | Portage Learning | Verified
Q&A | Pass Guaranteed - A+ Graded
Q1: According to the Michaelis-Menten model, which assumption is made about the
enzyme-substrate complex (ES)?
A. ES breaks down only to form product
B. ES formation is the rate-limiting step
C. ES is in rapid equilibrium with free enzyme and substrate, and its concentration
remains constant during the initial phase [CORRECT]
D. ES concentration increases continuously throughout the reaction
Correct Answer: C
Rationale: The steady-state assumption (Briggs-Haldane modification) states that [ES]
remains constant during the initial velocity phase because its rate of formation equals
its rate of breakdown (to E + P or E + S). This allows derivation of the Michaelis-Menten
equation. Option A is incorrect because ES can dissociate back to E + S. Option B is
incorrect because product formation (kcat) is typically rate-limiting, not ES formation.
Option D violates the steady-state condition that defines initial velocity measurements.
Q2: An enzyme has a Km of 5 mM and a Vmax of 100 μmol/min. At what substrate
concentration will the reaction velocity be 25 μmol/min?
A. 1.0 mM
B. 1.67 mM [CORRECT]
C. 5.0 mM
D. 10.0 mM
Correct Answer: B
Rationale: Using the Michaelis-Menten equation: v = (Vmax × [S])/(Km + [S]).
Rearranging: 25 = (100 × [S])/(5 + [S]). Solving: 25(5 + [S]) = 100[S]; 125 + 25[S] = 100[S];
125 = 75[S]; [S] = 1.67 mM. At [S] = Km (5 mM), v would be 50 μmol/min (half Vmax). At
1.0 mM (A), v = 16.7 μmol/min. At 10 mM (D), v = 66.7 μmol/min.
Q3: Which type of inhibition results in an increased apparent Km but unchanged Vmax?
, A. Non-competitive inhibition
B. Uncompetitive inhibition
C. Competitive inhibition [CORRECT]
D. Mixed inhibition
Correct Answer: C
Rationale: Competitive inhibitors bind reversibly to the active site, competing with
substrate. This increases the apparent Km (more substrate needed to reach half Vmax)
but Vmax remains unchanged because sufficient substrate can outcompete the
inhibitor. On Lineweaver-Burk plots, competitive inhibition shows lines intersecting on
the y-axis (1/Vmax unchanged), with different x-intercepts (-1/Km). Non-competitive (A)
decreases Vmax with unchanged Km. Uncompetitive (B) decreases both Vmax and Km.
Mixed (D) affects both parameters with lines intersecting left of the y-axis.
Q4: The catalytic efficiency of an enzyme is best described by which parameter?
A. Vmax alone
B. Km alone
C. kcat/Km [CORRECT]
D. Turnover number (kcat) alone
Correct Answer: C
Rationale: Catalytic efficiency is defined as kcat/Km (the specificity constant), with units
of M⁻¹s⁻¹. This second-order rate constant reflects how efficiently an enzyme converts
substrate to product at low substrate concentrations, combining both binding affinity
(1/Km) and catalytic rate (kcat). Values approaching 10⁸-10⁹ M⁻¹s⁻¹ approach
diffusion-controlled limits. Vmax (A) depends on enzyme concentration. Km (B) alone
doesn't indicate catalytic rate. kcat (D) alone doesn't reflect substrate binding.
Q5: In a Lineweaver-Burk plot, the y-intercept represents:
A. -1/Km
B. 1/Km
C. 1/Vmax [CORRECT]
D. Vmax
Correct Answer: C
Rationale: The Lineweaver-Burk equation (double reciprocal): 1/v = (Km/Vmax)(1/[S]) +
1/Vmax. The y-intercept (when 1/[S] = 0) equals 1/Vmax. The x-intercept (when 1/v = 0)
