BCH 4053 Exam 3V3 | BCH 4053 Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 3) | University of Central Florida
1. The Michaelis constant (Km) is defined as the substrate concentration at which the
reaction rate is half of the maximum velocity. Which statement regarding Km is correct?
A. A higher Km value indicates a higher affinity for the substrate.
B. A lower Km value indicates a higher affinity for the substrate.
C. Km is dependent on the total enzyme concentration used in the assay.
D. Km is equal to Vmax divided by two.
Answer: B
Explanation: The Km value provides an inverse measure of the affinity of an enzyme for its
substrate. A lower numerical value for Km signifies that the enzyme reaches half-saturation
at a lower substrate concentration, implying stronger binding. In contrast, Vmax depends
on enzyme concentration, whereas Km is an intrinsic property of the enzyme-substrate
pair.
2. In a Lineweaver-Burk plot, what does the x-intercept represent?
A. The ratio of Vmax to Km.
B. The reciprocal of the maximal velocity (1/Vmax).
C. The turnover number (kcat) of the enzyme.
D. The negative reciprocal of the Michaelis constant (-1/Km).
Answer: D
Explanation: The Lineweaver-Burk plot is a double-reciprocal transformation of the
Michaelis-Menten equation. The x-axis intercept is found by setting 1/v to zero, which
solves to -1/Km. This graphical representation allows for easier determination of kinetic
constants compared to a hyperbolic curve.
3. Which type of inhibition can be overcome by significantly increasing the substrate
concentration?
A. Noncompetitive inhibition
B. Competitive inhibition
C. Uncompetitive inhibition
D. Mixed inhibition
,Answer: B
Explanation: Competitive inhibitors bind to the same active site as the substrate, creating
a competition for the enzyme. By increasing the concentration of the substrate, the
probability of the substrate binding instead of the inhibitor increases. Consequently, the
Vmax remains unchanged while the apparent Km increases.
4. What characterizes uncompetitive inhibition in terms of kinetic parameters?
A. Vmax stays the same and Km increases.
B. Both Vmax and Km decrease proportionally.
C. Vmax decreases and Km stays the same.
D. Vmax increases and Km decreases.
Answer: B
Explanation: Uncompetitive inhibitors bind only to the enzyme-substrate (ES) complex,
not the free enzyme. This binding removes ES from the system, shifting the equilibrium to
favor more ES formation, which decreases the apparent Km. Simultaneously, the effective
concentration of active enzyme complexes decreases, leading to a proportional decrease in
Vmax.
5. A molecule that binds to a site other than the active site and reduces the overall Vmax
without changing Km is known as a:
A. Suicide substrate.
B. Competitive inhibitor.
C. Pure noncompetitive inhibitor.
D. Transition state analog.
Answer: C
Explanation: Noncompetitive inhibitors bind to an allosteric site regardless of whether the
substrate is bound. Because they do not interfere with substrate binding, the Km remains
unchanged. However, they decrease the catalytic efficiency, which results in a lower Vmax.
6. The ‘turnover number’ of an enzyme is mathematically defined as:
A. Km / Vmax
B. Vmax * Km
C. [E]total / Vmax
D. Vmax / [E]total
Answer: D
, Explanation: The turnover number, or kcat, represents the number of substrate molecules
converted to product per unit time when the enzyme is fully saturated. It is calculated by
dividing the maximum velocity (Vmax) by the total concentration of enzyme active sites
([E]total). This constant is a measure of the intrinsic catalytic power of an enzyme.
7. Which of the following describes the Bohr effect in hemoglobin?
A. Binding of oxygen increases as the pH decreases.
B. Fetal hemoglobin binds oxygen less tightly than adult hemoglobin.
C. The binding of 2,3-BPG stabilizes the R-state of hemoglobin.
D. Hemoglobin’s affinity for oxygen decreases as pH decreases and CO2 increases.
Answer: D
Explanation: The Bohr effect describes the regulation of hemoglobin’s oxygen binding by
hydrogen ions and carbon dioxide. As tissues metabolize, they release CO2 and lower the
local pH, which shifts hemoglobin into the T-state (tense state). This physiological
mechanism promotes the release of oxygen where it is needed most.
8. Which molecule is a heterotropic negative effector of hemoglobin oxygen binding?
A. Oxygen
B. Nitric oxide
C. Carbon monoxide
D. 2,3-Bisphosphoglycerate (2,3-BPG)
Answer: D
Explanation: 2,3-BPG is a small molecule that binds in the central cavity of the hemoglobin
tetramer. It specifically stabilizes the T-state, thereby reducing the affinity for oxygen and
facilitating its release. This is crucial for adaptation to high altitudes and during pregnancy
for oxygen transfer to the fetus.
9. Glucose and Galactose are examples of:
A. Enantiomers
B. Epimers
C. Anomers
D. Constitutional isomers
Answer: B
Explanation: Epimers are a type of diastereomer that differ in configuration at only one
chiral center. Glucose and Galactose differ specifically at the C-4 position. Because they
differ at only one of several chiral centers, they are categorized as epimers.
Rationale (BCH4053 Exam 3) | University of Central Florida
1. The Michaelis constant (Km) is defined as the substrate concentration at which the
reaction rate is half of the maximum velocity. Which statement regarding Km is correct?
A. A higher Km value indicates a higher affinity for the substrate.
B. A lower Km value indicates a higher affinity for the substrate.
C. Km is dependent on the total enzyme concentration used in the assay.
D. Km is equal to Vmax divided by two.
Answer: B
Explanation: The Km value provides an inverse measure of the affinity of an enzyme for its
substrate. A lower numerical value for Km signifies that the enzyme reaches half-saturation
at a lower substrate concentration, implying stronger binding. In contrast, Vmax depends
on enzyme concentration, whereas Km is an intrinsic property of the enzyme-substrate
pair.
2. In a Lineweaver-Burk plot, what does the x-intercept represent?
A. The ratio of Vmax to Km.
B. The reciprocal of the maximal velocity (1/Vmax).
C. The turnover number (kcat) of the enzyme.
D. The negative reciprocal of the Michaelis constant (-1/Km).
Answer: D
Explanation: The Lineweaver-Burk plot is a double-reciprocal transformation of the
Michaelis-Menten equation. The x-axis intercept is found by setting 1/v to zero, which
solves to -1/Km. This graphical representation allows for easier determination of kinetic
constants compared to a hyperbolic curve.
3. Which type of inhibition can be overcome by significantly increasing the substrate
concentration?
A. Noncompetitive inhibition
B. Competitive inhibition
C. Uncompetitive inhibition
D. Mixed inhibition
,Answer: B
Explanation: Competitive inhibitors bind to the same active site as the substrate, creating
a competition for the enzyme. By increasing the concentration of the substrate, the
probability of the substrate binding instead of the inhibitor increases. Consequently, the
Vmax remains unchanged while the apparent Km increases.
4. What characterizes uncompetitive inhibition in terms of kinetic parameters?
A. Vmax stays the same and Km increases.
B. Both Vmax and Km decrease proportionally.
C. Vmax decreases and Km stays the same.
D. Vmax increases and Km decreases.
Answer: B
Explanation: Uncompetitive inhibitors bind only to the enzyme-substrate (ES) complex,
not the free enzyme. This binding removes ES from the system, shifting the equilibrium to
favor more ES formation, which decreases the apparent Km. Simultaneously, the effective
concentration of active enzyme complexes decreases, leading to a proportional decrease in
Vmax.
5. A molecule that binds to a site other than the active site and reduces the overall Vmax
without changing Km is known as a:
A. Suicide substrate.
B. Competitive inhibitor.
C. Pure noncompetitive inhibitor.
D. Transition state analog.
Answer: C
Explanation: Noncompetitive inhibitors bind to an allosteric site regardless of whether the
substrate is bound. Because they do not interfere with substrate binding, the Km remains
unchanged. However, they decrease the catalytic efficiency, which results in a lower Vmax.
6. The ‘turnover number’ of an enzyme is mathematically defined as:
A. Km / Vmax
B. Vmax * Km
C. [E]total / Vmax
D. Vmax / [E]total
Answer: D
, Explanation: The turnover number, or kcat, represents the number of substrate molecules
converted to product per unit time when the enzyme is fully saturated. It is calculated by
dividing the maximum velocity (Vmax) by the total concentration of enzyme active sites
([E]total). This constant is a measure of the intrinsic catalytic power of an enzyme.
7. Which of the following describes the Bohr effect in hemoglobin?
A. Binding of oxygen increases as the pH decreases.
B. Fetal hemoglobin binds oxygen less tightly than adult hemoglobin.
C. The binding of 2,3-BPG stabilizes the R-state of hemoglobin.
D. Hemoglobin’s affinity for oxygen decreases as pH decreases and CO2 increases.
Answer: D
Explanation: The Bohr effect describes the regulation of hemoglobin’s oxygen binding by
hydrogen ions and carbon dioxide. As tissues metabolize, they release CO2 and lower the
local pH, which shifts hemoglobin into the T-state (tense state). This physiological
mechanism promotes the release of oxygen where it is needed most.
8. Which molecule is a heterotropic negative effector of hemoglobin oxygen binding?
A. Oxygen
B. Nitric oxide
C. Carbon monoxide
D. 2,3-Bisphosphoglycerate (2,3-BPG)
Answer: D
Explanation: 2,3-BPG is a small molecule that binds in the central cavity of the hemoglobin
tetramer. It specifically stabilizes the T-state, thereby reducing the affinity for oxygen and
facilitating its release. This is crucial for adaptation to high altitudes and during pregnancy
for oxygen transfer to the fetus.
9. Glucose and Galactose are examples of:
A. Enantiomers
B. Epimers
C. Anomers
D. Constitutional isomers
Answer: B
Explanation: Epimers are a type of diastereomer that differ in configuration at only one
chiral center. Glucose and Galactose differ specifically at the C-4 position. Because they
differ at only one of several chiral centers, they are categorized as epimers.
