BCH 4053 Exam 2 V3 | BCH 4053
Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 2) | University
of Central Florida
1. In Michaelis-Menten kinetics, what does the KM value represent?
A. The maximum velocity of the reaction.
B. The total enzyme concentration.
C. The substrate concentration at which the reaction rate is half of Vmax.
D. The dissociation constant for the product-enzyme complex.
Answer: C
Rationale: The Michaelis constant (KM) is a specific substrate concentration where the
initial reaction velocity equals half of the maximal velocity (Vmax/2). It serves as an
indicator of the affinity between the enzyme and its substrate, with a low KM indicating
high affinity. This parameter is crucial for understanding how enzymes function under
physiological substrate levels.
2. Which type of inhibition is characterized by the inhibitor binding only to the enzyme-
substrate (ES) complex?
A. Competitive inhibition
B. Mixed inhibition
,C. Uncompetitive inhibition
D. Noncompetitive inhibition
Answer: C
Rationale: Uncompetitive inhibitors bind exclusively to the ES complex rather than the
free enzyme, effectively preventing the complex from proceeding to product. This
mechanism results in a proportional decrease in both Vmax and KM values on a
Lineweaver-Burk plot. Because the inhibitor removes ES from the equilibrium, it appears
as though the enzyme has a higher affinity for the substrate.
3. During a competitive inhibition, how are the Vmax and KM affected?
A. Vmax decreases, KM stays the same.
B. Vmax increases, KM increases.
C. Vmax stays the same, KM decreases.
D. Vmax stays the same, KM increases.
Answer: D
Rationale: Competitive inhibitors compete with the substrate for the active site of the free
enzyme. This competition can be overcome by increasing the substrate concentration,
which is why the Vmax remains unchanged. However, because more substrate is required
to reach Vmax/2, the apparent KM value increases.
, 4. What is the primary role of the catalytic triad in the mechanism of chymotrypsin?
A. To bind the peptide backbone of the substrate.
B. To stabilize the oxyanion hole.
C. To prevent the hydrolysis of water molecules.
D. To enhance the nucleophilicity of the active site Serine.
Answer: D
Rationale: The catalytic triad of chymotrypsin consists of Asp102, His57, and Ser195.
Through a charge-relay system, Asp102 stabilizes His57, which in turn acts as a base to
deprotonate Ser195. This process turns the Serine hydroxyl group into a highly reactive
alkoxide ion capable of nucleophilic attack on the substrate carbonyl carbon.
5. Which of the following describes an epimer?
A. Sugars that are mirror images of each other.
B. Sugars that differ in configuration around only one specific carbon atom other than the
anomeric carbon.
C. Sugars that differ only in the configuration of the anomeric carbon.
D. Sugars that have the same molecular formula but different functional groups.
Answer: B
Rationale: Epimers are a subtype of diastereomers that differ in the stereochemistry at
only one chiral center. For example, D-glucose and D-galactose are C-4 epimers because
Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 2) | University
of Central Florida
1. In Michaelis-Menten kinetics, what does the KM value represent?
A. The maximum velocity of the reaction.
B. The total enzyme concentration.
C. The substrate concentration at which the reaction rate is half of Vmax.
D. The dissociation constant for the product-enzyme complex.
Answer: C
Rationale: The Michaelis constant (KM) is a specific substrate concentration where the
initial reaction velocity equals half of the maximal velocity (Vmax/2). It serves as an
indicator of the affinity between the enzyme and its substrate, with a low KM indicating
high affinity. This parameter is crucial for understanding how enzymes function under
physiological substrate levels.
2. Which type of inhibition is characterized by the inhibitor binding only to the enzyme-
substrate (ES) complex?
A. Competitive inhibition
B. Mixed inhibition
,C. Uncompetitive inhibition
D. Noncompetitive inhibition
Answer: C
Rationale: Uncompetitive inhibitors bind exclusively to the ES complex rather than the
free enzyme, effectively preventing the complex from proceeding to product. This
mechanism results in a proportional decrease in both Vmax and KM values on a
Lineweaver-Burk plot. Because the inhibitor removes ES from the equilibrium, it appears
as though the enzyme has a higher affinity for the substrate.
3. During a competitive inhibition, how are the Vmax and KM affected?
A. Vmax decreases, KM stays the same.
B. Vmax increases, KM increases.
C. Vmax stays the same, KM decreases.
D. Vmax stays the same, KM increases.
Answer: D
Rationale: Competitive inhibitors compete with the substrate for the active site of the free
enzyme. This competition can be overcome by increasing the substrate concentration,
which is why the Vmax remains unchanged. However, because more substrate is required
to reach Vmax/2, the apparent KM value increases.
, 4. What is the primary role of the catalytic triad in the mechanism of chymotrypsin?
A. To bind the peptide backbone of the substrate.
B. To stabilize the oxyanion hole.
C. To prevent the hydrolysis of water molecules.
D. To enhance the nucleophilicity of the active site Serine.
Answer: D
Rationale: The catalytic triad of chymotrypsin consists of Asp102, His57, and Ser195.
Through a charge-relay system, Asp102 stabilizes His57, which in turn acts as a base to
deprotonate Ser195. This process turns the Serine hydroxyl group into a highly reactive
alkoxide ion capable of nucleophilic attack on the substrate carbonyl carbon.
5. Which of the following describes an epimer?
A. Sugars that are mirror images of each other.
B. Sugars that differ in configuration around only one specific carbon atom other than the
anomeric carbon.
C. Sugars that differ only in the configuration of the anomeric carbon.
D. Sugars that have the same molecular formula but different functional groups.
Answer: B
Rationale: Epimers are a subtype of diastereomers that differ in the stereochemistry at
only one chiral center. For example, D-glucose and D-galactose are C-4 epimers because
