BCH 4053 Exam 3 V1 | BCH 4053
Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 3) | University
of Central Florida
1. Which of the following parameters represents the substrate concentration at which the
reaction velocity is half of the maximum velocity?
A. Vmax
B. kcat
C. Km
D. kcat/Km
Answer: C
Rationale: The Michaelis constant, Km, is defined as the substrate concentration at half-
maximal velocity. It serves as a measure of the affinity between the enzyme and its
substrate under specific conditions. A smaller Km value indicates higher affinity, as less
substrate is needed to saturate half of the enzyme active sites.
2. In a Lineweaver-Burk plot, what does the y-intercept represent?
A. -1/Km
B. Vmax/Km
C. 1/Km
,D. 1/Vmax
Answer: D
Rationale: The Lineweaver-Burk plot is a double-reciprocal representation of the
Michaelis-Menten equation. The y-intercept is determined by setting the reciprocal of the
substrate concentration to zero, which simplifies the equation to 1/Vmax. This graphical
method allows for a more accurate determination of Vmax compared to a standard
hyperbolic curve.
3. Which type of inhibition can be overcome by increasing the substrate concentration?
A. Competitive inhibition
B. Noncompetitive inhibition
C. Uncompetitive inhibition
D. Irreversible inhibition
Answer: A
Rationale: Competitive inhibitors bind to the same active site as the substrate, directly
competing for enzyme access. By significantly increasing the substrate concentration, the
substrate effectively outcompetes the inhibitor for the active site. Consequently, the Vmax
remains unchanged in competitive inhibition, although the apparent Km increases.
4. How does an uncompetitive inhibitor affect the Lineweaver-Burk plot?
A. The lines intersect at the y-axis.
, B. The lines are parallel to each other.
C. The lines intersect at the x-axis.
D. The lines intersect in the second quadrant.
Answer: B
Rationale: Uncompetitive inhibitors bind only to the enzyme-substrate (ES) complex
rather than the free enzyme. This results in a proportional decrease in both Vmax and Km,
which maintains a constant slope (Km/Vmax). On a double-reciprocal plot, this manifests
as parallel lines for the inhibited and uninhibited reactions.
5. Which amino acid residue in the catalytic triad of chymotrypsin acts as the general base to
activate the nucleophile?
A. Serine 195
B. Aspartate 102
C. Glycine 193
D. Histidine 57
Answer: D
Rationale: Histidine 57 plays a critical role in the chymotrypsin mechanism by accepting a
proton from Serine 195. This deprotonation increases the nucleophilicity of the serine
oxygen, allowing it to attack the carbonyl carbon of the peptide bond. Without this general
base catalysis, the reaction would proceed at a physiologically irrelevant rate.
Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 3) | University
of Central Florida
1. Which of the following parameters represents the substrate concentration at which the
reaction velocity is half of the maximum velocity?
A. Vmax
B. kcat
C. Km
D. kcat/Km
Answer: C
Rationale: The Michaelis constant, Km, is defined as the substrate concentration at half-
maximal velocity. It serves as a measure of the affinity between the enzyme and its
substrate under specific conditions. A smaller Km value indicates higher affinity, as less
substrate is needed to saturate half of the enzyme active sites.
2. In a Lineweaver-Burk plot, what does the y-intercept represent?
A. -1/Km
B. Vmax/Km
C. 1/Km
,D. 1/Vmax
Answer: D
Rationale: The Lineweaver-Burk plot is a double-reciprocal representation of the
Michaelis-Menten equation. The y-intercept is determined by setting the reciprocal of the
substrate concentration to zero, which simplifies the equation to 1/Vmax. This graphical
method allows for a more accurate determination of Vmax compared to a standard
hyperbolic curve.
3. Which type of inhibition can be overcome by increasing the substrate concentration?
A. Competitive inhibition
B. Noncompetitive inhibition
C. Uncompetitive inhibition
D. Irreversible inhibition
Answer: A
Rationale: Competitive inhibitors bind to the same active site as the substrate, directly
competing for enzyme access. By significantly increasing the substrate concentration, the
substrate effectively outcompetes the inhibitor for the active site. Consequently, the Vmax
remains unchanged in competitive inhibition, although the apparent Km increases.
4. How does an uncompetitive inhibitor affect the Lineweaver-Burk plot?
A. The lines intersect at the y-axis.
, B. The lines are parallel to each other.
C. The lines intersect at the x-axis.
D. The lines intersect in the second quadrant.
Answer: B
Rationale: Uncompetitive inhibitors bind only to the enzyme-substrate (ES) complex
rather than the free enzyme. This results in a proportional decrease in both Vmax and Km,
which maintains a constant slope (Km/Vmax). On a double-reciprocal plot, this manifests
as parallel lines for the inhibited and uninhibited reactions.
5. Which amino acid residue in the catalytic triad of chymotrypsin acts as the general base to
activate the nucleophile?
A. Serine 195
B. Aspartate 102
C. Glycine 193
D. Histidine 57
Answer: D
Rationale: Histidine 57 plays a critical role in the chymotrypsin mechanism by accepting a
proton from Serine 195. This deprotonation increases the nucleophilicity of the serine
oxygen, allowing it to attack the carbonyl carbon of the peptide bond. Without this general
base catalysis, the reaction would proceed at a physiologically irrelevant rate.
