Ultimate ACS Biochemistry Exam Study Guide: 200 Practice Questions on Enzyme Kinetics, Metabolism, DNA Replication & Lab Techniques (With Answers & Explanations)

Ultimate ACS Biochemistry Exam Study
Guide
200 Practice Questions on Enzyme Kinetics,
Metabolism, DNA Replication & Lab
Techniques (With Answers & Explanations)

Section 1: Enzyme Kinetics and Mechanisms (Questions 1-40)
1. The Michaelis-Menten constant (Km) is defined as:
A) The maximum velocity of the reaction.
B) The substrate concentration at which velocity is half of Vmax.
C) The turnover number of the enzyme.
D) The dissociation constant of the ES complex when kcat >> k-1.
Answer: B
Rationale: By definition, Km is the substrate concentration ([S]) required for an
enzyme to reach one-half of its maximum velocity (Vmax). While it approximates
Kd under certain conditions, the strict definition is the substrate concentration at
1/2 Vmax.
2. If an enzyme has a Vmax of 100 µmol/min and a Km of 5 mM, what is the
initial velocity (Vo) when [S] = 5 mM?
A) 100 µmol/min
B) 50 µmol/min
C) 25 µmol/min
D) 10 µmol/min
Answer: B
Rationale: By definition, at [S] = Km, Vo = 1/2 Vmax. Therefore, Vo = 100
µmol/min / 2 = 50 µmol/min. This is derived from the Michaelis-Menten equation:
Vo = (Vmax[S])/(Km+[S]).

,3. A Lineweaver-Burk plot is constructed by plotting:
A) 1/[S] on the y-axis versus 1/Vo on the x-axis.
B) 1/Vo on the y-axis versus 1/[S] on the x-axis.
C) Vo on the y-axis versus [S] on the x-axis.
D) [S] on the y-axis versus Vo on the x-axis.
Answer: B
Rationale: The Lineweaver-Burk plot is the double reciprocal of the Michaelis-
Menten equation. The y-intercept is 1/Vmax, the x-intercept is -1/Km, and the
slope is Km/Vmax.
4. In a Lineweaver-Burk plot, the y-intercept represents:
A) -1/Km
B) 1/Km
C) -1/Vmax
D) 1/Vmax
Answer: D
Rationale: The equation of the line is 1/Vo = (Km/Vmax)(1/[S]) + 1/Vmax. When
x (1/[S]) is 0, y equals 1/Vmax.
5. The catalytic efficiency of an enzyme is expressed as:
A) Vmax/Km
B) Km/Vmax
C) kcat/Km
D) Vmax/[S]
Answer: C
Rationale: kcat (the turnover number) divided by Km is the rate constant for an
enzyme to convert substrate to product under substrate-limited conditions ([S] <<
Km). It is a measure of how efficiently an enzyme processes a substrate.
6. A competitive inhibitor:
A) Binds to the allosteric site of the enzyme.
B) Binds reversibly to the ES complex.
C) Binds to the active site, preventing substrate binding.
D) Decreases Vmax.
Answer: C
Rationale: Competitive inhibitors structurally resemble the substrate and compete

,for binding to the active site. This increases the apparent Km (requires more
substrate to reach half Vmax) but does not affect Vmax.
7. The effect of a competitive inhibitor on kinetic parameters is:
A) Increased Vmax, no change in Km.
B) Decreased Vmax, no change in Km.
C) Increased Km, no change in Vmax.
D) Decreased Km, no change in Vmax.
Answer: C
Rationale: Since the inhibitor competes with the substrate for the active site, a
higher substrate concentration is needed to achieve the same velocity, thus
increasing Km. Because the inhibitor can be overcome with high substrate, Vmax
remains unchanged.
8. A non-competitive inhibitor:
A) Binds to the active site.
B) Binds to an allosteric site, which distorts the active site.
C) Competes with the substrate for the active site.
D) Has a structure similar to the substrate.
Answer: B
Rationale: Non-competitive inhibitors bind to a site other than the active site
(allosteric site). This binding changes the conformation of the enzyme, reducing its
catalytic activity (decreasing Vmax). It cannot be overcome by adding more
substrate.
9. The effect of a non-competitive inhibitor on kinetic parameters is:
A) Increased Vmax, no change in Km.
B) Decreased Vmax, no change in Km.
C) Increased Km, no change in Vmax.
D) Decreased Km, no change in Vmax.
Answer: B
Rationale: The inhibitor reduces the concentration of functional enzyme, thus
lowering Vmax. Since it doesn't affect substrate binding, Km remains unchanged.
10. An uncompetitive inhibitor binds:
A) Exclusively to the free enzyme (E).
B) Exclusively to the enzyme-substrate complex (ES).

, C) Equally to E and ES. K. K. K. K.




D) To the substrate itself.
K. K. K.




Answer: B K.




Rationale: Uncompetitive inhibitors bind to a site that is only formed when the
K. K. K. K. K. K. K. K. K. K. K. K.




substrate is bound to the enzyme (the ES complex). This effectively "traps" the ES
K. K. K. K. K. K. K. K. K. K. K. K. K. K.




complex.
K.




11. The effect of an uncompetitive inhibitor on kinetic parameters is:
K. K. K. K. K. K. K. K. K.




A) Increased Vmax, decreased Km. K. K. K.




B) Decreased Vmax, decreased Km. K. K. K.




C) Increased Vmax, increased Km. K. K. K.




D) Decreased Vmax, no change in Km. K. K. K. K. K.




Answer: B K.




Rationale: Binding to the ES complex appears to make the enzyme more "affine"
K. K. K. K. K. K. K. K. K. K. K. K.




(lower Km) but also inactivates the complex (lower Vmax). Both Vmax and Km
K. K. K. K. K. K. K. K. K. K. K. K. K.




are decreased by the same factor.
K. K. K. K. K. K.




12. The turnover number (kcat) of an enzyme is:
K. K. K. K. K. K. K.




A) The rate at which substrate binds to the enzyme.
K. K. K. K. K. K. K. K.




B) The maximum number of substrate molecules converted to product per enzyme
K. K. K. K. K. K. K. K. K. K.




molecule per second.
K. K. K.




C) The number of enzyme molecules in a given reaction.
K. K. K. K. K. K. K. K.




D) The time it takes for an enzyme to turn over.
K. K. K. K. K. K. K. K. K.




Answer: B K.




Rationale: kcat is a measure of catalytic rate. It is calculated as Vmax / [E]t, where
K. K. K. K. K. K. K. K. K. K. K. K. K. K. K.




[E]t is the total enzyme concentration. It represents how fast an enzyme can "turn
K. K. K. K. K. K. K. K. K. K. K. K. K. K.




over" one substrate molecule into product.
K. K. K. K. K. K.




13. The initial velocity (Vo) of an enzyme-catalyzed reaction is measured
K. K. K. K. K. K. K. K. K.




under conditions where:
K. K. K.




A) Substrate concentration is in large excess. K. K. K. K. K.




B) The reaction has reached equilibrium.
K. K. K. K.




C) Product concentration is high. K. K. K.




D) The enzyme concentration is changing.
K. K. K. K.




Answer: A K.




Rationale: Initial velocity is measured at the very beginning of the reaction (t=0)
K. K. K. K. K. K. K. K. K. K. K. K.