BCH 4053 Exam 2 V2 | BCH 4053
Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 2) | University
of Central Florida
1. What does the Michaelis-Menten constant, Km, represent in enzyme kinetics?
A. The substrate concentration at which the reaction rate is half of Vmax
B. The equilibrium constant for the reaction
C. The maximum velocity of the reaction
D. The total concentration of the enzyme
Answer: A
Rationale: The Km value provides a measure of the affinity of an enzyme for its substrate.
It is defined as the specific substrate concentration required to reach half of the maximum
velocity (Vmax). A lower Km value suggests higher affinity, meaning the enzyme reaches
half-saturation at a lower substrate level.
2. In a Lineweaver-Burk plot, what does the y-intercept represent?
A. -1/Km
B. Km/Vmax
C. 1/Vmax
D. 1/S
,Answer: C
Rationale: The Lineweaver-Burk plot is a double-reciprocal representation of the
Michaelis-Menten equation. The equation takes the form 1/v = (Km/Vmax)(1/[S]) +
1/Vmax. Therefore, the y-intercept corresponds to the value of 1/Vmax when the
reciprocal of substrate concentration is zero.
3. How does a competitive inhibitor affect the kinetic parameters of an enzyme?
A. It increases the apparent Km while Vmax remains unchanged
B. It decreases both Vmax and Km
C. It decreases Vmax while Km remains unchanged
D. It increases both Vmax and Km
Answer: A
Rationale: Competitive inhibitors bind directly to the active site of the enzyme, competing
with the substrate. This competition increases the amount of substrate needed to reach
half-maximal velocity, thus increasing the apparent Km. Since high concentrations of
substrate can eventually outcompete the inhibitor, the Vmax remains reachable and
unchanged.
4. Which of the following describes an uncompetitive inhibitor?
A. It binds only to the enzyme-substrate (ES) complex
B. It binds to both the free enzyme and the enzyme-substrate complex
, C. It binds only to the free enzyme
D. It binds at a site distinct from the active site and does not affect Km
Answer: A
Rationale: Uncompetitive inhibition occurs when the inhibitor binds only to the ES
complex, preventing the conversion of substrate to product. This binding effectively pulls
the equilibrium toward the formation of the ES complex, which decreases the apparent Km.
Simultaneously, because the ESI complex is unproductive, the Vmax is also decreased.
5. In the chymotrypsin mechanism, what is the role of the catalytic triad member Asp 102?
A. It acts as the primary nucleophile attacking the peptide bond
B. It stabilizes the positive charge on His 57 through hydrogen bonding
C. It provides the proton needed to release the N-terminal fragment
D. It forms a covalent intermediate with the substrate
Answer: B
Rationale: The catalytic triad of chymotrypsin consists of Ser 195, His 57, and Asp 102. Asp
102 functions by forming a hydrogen bond with His 57, which helps to orient the histidine
residue and stabilize the positive charge that develops during the transition state. This
interaction enhances the basicity of His 57, allowing it to deprotonate Ser 195 more
effectively.
Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 2) | University
of Central Florida
1. What does the Michaelis-Menten constant, Km, represent in enzyme kinetics?
A. The substrate concentration at which the reaction rate is half of Vmax
B. The equilibrium constant for the reaction
C. The maximum velocity of the reaction
D. The total concentration of the enzyme
Answer: A
Rationale: The Km value provides a measure of the affinity of an enzyme for its substrate.
It is defined as the specific substrate concentration required to reach half of the maximum
velocity (Vmax). A lower Km value suggests higher affinity, meaning the enzyme reaches
half-saturation at a lower substrate level.
2. In a Lineweaver-Burk plot, what does the y-intercept represent?
A. -1/Km
B. Km/Vmax
C. 1/Vmax
D. 1/S
,Answer: C
Rationale: The Lineweaver-Burk plot is a double-reciprocal representation of the
Michaelis-Menten equation. The equation takes the form 1/v = (Km/Vmax)(1/[S]) +
1/Vmax. Therefore, the y-intercept corresponds to the value of 1/Vmax when the
reciprocal of substrate concentration is zero.
3. How does a competitive inhibitor affect the kinetic parameters of an enzyme?
A. It increases the apparent Km while Vmax remains unchanged
B. It decreases both Vmax and Km
C. It decreases Vmax while Km remains unchanged
D. It increases both Vmax and Km
Answer: A
Rationale: Competitive inhibitors bind directly to the active site of the enzyme, competing
with the substrate. This competition increases the amount of substrate needed to reach
half-maximal velocity, thus increasing the apparent Km. Since high concentrations of
substrate can eventually outcompete the inhibitor, the Vmax remains reachable and
unchanged.
4. Which of the following describes an uncompetitive inhibitor?
A. It binds only to the enzyme-substrate (ES) complex
B. It binds to both the free enzyme and the enzyme-substrate complex
, C. It binds only to the free enzyme
D. It binds at a site distinct from the active site and does not affect Km
Answer: A
Rationale: Uncompetitive inhibition occurs when the inhibitor binds only to the ES
complex, preventing the conversion of substrate to product. This binding effectively pulls
the equilibrium toward the formation of the ES complex, which decreases the apparent Km.
Simultaneously, because the ESI complex is unproductive, the Vmax is also decreased.
5. In the chymotrypsin mechanism, what is the role of the catalytic triad member Asp 102?
A. It acts as the primary nucleophile attacking the peptide bond
B. It stabilizes the positive charge on His 57 through hydrogen bonding
C. It provides the proton needed to release the N-terminal fragment
D. It forms a covalent intermediate with the substrate
Answer: B
Rationale: The catalytic triad of chymotrypsin consists of Ser 195, His 57, and Asp 102. Asp
102 functions by forming a hydrogen bond with His 57, which helps to orient the histidine
residue and stabilize the positive charge that develops during the transition state. This
interaction enhances the basicity of His 57, allowing it to deprotonate Ser 195 more
effectively.
