BCH 4053 Exam 2 V2 | BCH 4053 Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 2) | University of Central Florida
1. For an enzyme that follows Michaelis-Menten kinetics, what does the Michaelis constant
(Km) represent?
A. The enzyme concentration at which the reaction velocity is maximal.
B. The dissociation constant for the enzyme-product complex.
C. The substrate concentration at which the reaction velocity is half of Vmax.
D. The total number of substrate molecules converted to product per unit time.
Answer: C
Explanation: The Km value is defined as the substrate concentration required to reach half
the maximum velocity of an enzymatic reaction. It provides a measure of the affinity
between the enzyme and the substrate, where a lower Km suggests higher affinity. This
parameter is a fundamental constant in enzyme kinetics that characterizes the efficiency of
the enzyme under specific conditions.
2. Which type of inhibition can be overcome by increasing the substrate concentration to a
very high level?
A. Noncompetitive inhibition
B. Uncompetitive inhibition
C. Competitive inhibition
D. Irreversible inhibition
Answer: C
Explanation: Competitive inhibitors bind to the active site of the enzyme, directly
competing with the substrate for binding. Because the inhibitor and substrate compete for
the same site, increasing the substrate concentration shifts the equilibrium toward
substrate binding, effectively displacing the inhibitor. Consequently, the Vmax remains
unchanged in competitive inhibition, although the apparent Km increases.
3. In the catalytic triad of chymotrypsin, which amino acid residue acts as the general base to
activate the nucleophile?
A. Aspartate 102
B. Serine 195
C. Histidine 57
D. Glycine 193
,Answer: C
Explanation: Histidine 57 plays a critical role in the catalytic mechanism by accepting a
proton from the Serine 195 hydroxyl group. This deprotonation increases the
nucleophilicity of the Serine, allowing it to attack the carbonyl carbon of the peptide bond.
The Aspartate 102 residue helps to stabilize the resulting positive charge on the Histidine
during this transition.
4. What is the primary effect of 2,3-bisphosphoglycerate (2,3-BPG) on hemoglobin oxygen
binding?
A. It stabilizes the T-state of hemoglobin, decreasing oxygen affinity.
B. It increases the affinity of hemoglobin for oxygen in the lungs.
C. It promotes the transition from the T-state to the R-state.
D. It binds directly to the iron atom in the heme group.
Answer: A
Explanation: 2,3-BPG is an allosteric effector that binds to the central cavity of the
hemoglobin tetramer specifically in the T-state. By stabilizing the T-state (the
deoxygenated form), it reduces the affinity for oxygen, which facilitates the release of
oxygen in peripheral tissues. Without 2,3-BPG, hemoglobin would bind oxygen too tightly
to effectively deliver it to the body’s cells.
5. On a Lineweaver-Burk plot, what does the y-intercept represent?
A. 1/Km
B. -1/Km
C. 1/Vmax
D. Km/Vmax
Answer: C
Explanation: The Lineweaver-Burk plot is a double-reciprocal transformation of the
Michaelis-Menten equation, resulting in a linear graph. The y-intercept occurs where 1/[S]
is zero, which corresponds to the reciprocal of the maximum velocity, or 1/Vmax. This
graphical representation allows for the easy determination of kinetic constants and the
identification of inhibition patterns.
6. Which of the following describes an uncompetitive inhibitor?
A. It binds only to the enzyme-substrate (ES) complex, decreasing both Vmax and Km.
B. It binds only to the free enzyme, increasing Km.
C. It binds to both the free enzyme and the ES complex with equal affinity.
D. It covalently modifies the active site, permanently inactivating the enzyme.
, Answer: A
Explanation: Uncompetitive inhibitors do not bind to the free enzyme but instead
recognize the enzyme-substrate complex after the substrate has bound. This binding locks
the substrate in the active site, effectively lowering the concentration of productive ES
complexes and thus decreasing Vmax. The apparent Km also decreases because the
equilibrium of substrate binding is pulled toward the formation of the ESI complex.
7. What is the function of the oxyanion hole in serine proteases?
A. To shuttle protons away from the active site to the solvent.
B. To provide a hydrophobic pocket for large side chains like Phenylalanine.
C. To act as a metal ion cofactor site for Zinc or Magnesium.
D. To stabilize the tetrahedral intermediate by forming hydrogen bonds with the negatively
charged oxygen.
Answer: D
Explanation: The oxyanion hole is a specialized region in the active site of enzymes like
chymotrypsin that stabilizes the transition state. During catalysis, the carbonyl oxygen of
the substrate becomes a negatively charged oxyanion as it forms a tetrahedral
intermediate. Hydrogen bonds from the backbone amides of Glycine 193 and Serine 195
stabilize this charge, lowering the activation energy for the reaction.
8. Which statement best describes the Bohr effect?
A. High oxygen concentrations promote the release of CO2 from hemoglobin.
B. Lowering the pH increases the affinity of hemoglobin for oxygen.
C. The binding of oxygen to one subunit increases the affinity of other subunits.
D. Decreasing pH and increasing CO2 concentration promote the release of oxygen from
hemoglobin.
Answer: D
Explanation: The Bohr effect describes the physiological regulation of hemoglobin’s
oxygen binding affinity by hydrogen ions and carbon dioxide. As tissues metabolize, they
produce CO2 and protons, which lower the local pH and stabilize the T-state of hemoglobin.
This reduced affinity ensures that oxygen is released specifically in areas where it is most
needed for cellular respiration.
9. How does an increase in the number of double bonds in a fatty acid affect its melting
point?
A. It increases the melting point by allowing tighter packing.
B. It decreases the melting point by creating kinks that prevent tight packing.
Rationale (BCH4053 Exam 2) | University of Central Florida
1. For an enzyme that follows Michaelis-Menten kinetics, what does the Michaelis constant
(Km) represent?
A. The enzyme concentration at which the reaction velocity is maximal.
B. The dissociation constant for the enzyme-product complex.
C. The substrate concentration at which the reaction velocity is half of Vmax.
D. The total number of substrate molecules converted to product per unit time.
Answer: C
Explanation: The Km value is defined as the substrate concentration required to reach half
the maximum velocity of an enzymatic reaction. It provides a measure of the affinity
between the enzyme and the substrate, where a lower Km suggests higher affinity. This
parameter is a fundamental constant in enzyme kinetics that characterizes the efficiency of
the enzyme under specific conditions.
2. Which type of inhibition can be overcome by increasing the substrate concentration to a
very high level?
A. Noncompetitive inhibition
B. Uncompetitive inhibition
C. Competitive inhibition
D. Irreversible inhibition
Answer: C
Explanation: Competitive inhibitors bind to the active site of the enzyme, directly
competing with the substrate for binding. Because the inhibitor and substrate compete for
the same site, increasing the substrate concentration shifts the equilibrium toward
substrate binding, effectively displacing the inhibitor. Consequently, the Vmax remains
unchanged in competitive inhibition, although the apparent Km increases.
3. In the catalytic triad of chymotrypsin, which amino acid residue acts as the general base to
activate the nucleophile?
A. Aspartate 102
B. Serine 195
C. Histidine 57
D. Glycine 193
,Answer: C
Explanation: Histidine 57 plays a critical role in the catalytic mechanism by accepting a
proton from the Serine 195 hydroxyl group. This deprotonation increases the
nucleophilicity of the Serine, allowing it to attack the carbonyl carbon of the peptide bond.
The Aspartate 102 residue helps to stabilize the resulting positive charge on the Histidine
during this transition.
4. What is the primary effect of 2,3-bisphosphoglycerate (2,3-BPG) on hemoglobin oxygen
binding?
A. It stabilizes the T-state of hemoglobin, decreasing oxygen affinity.
B. It increases the affinity of hemoglobin for oxygen in the lungs.
C. It promotes the transition from the T-state to the R-state.
D. It binds directly to the iron atom in the heme group.
Answer: A
Explanation: 2,3-BPG is an allosteric effector that binds to the central cavity of the
hemoglobin tetramer specifically in the T-state. By stabilizing the T-state (the
deoxygenated form), it reduces the affinity for oxygen, which facilitates the release of
oxygen in peripheral tissues. Without 2,3-BPG, hemoglobin would bind oxygen too tightly
to effectively deliver it to the body’s cells.
5. On a Lineweaver-Burk plot, what does the y-intercept represent?
A. 1/Km
B. -1/Km
C. 1/Vmax
D. Km/Vmax
Answer: C
Explanation: The Lineweaver-Burk plot is a double-reciprocal transformation of the
Michaelis-Menten equation, resulting in a linear graph. The y-intercept occurs where 1/[S]
is zero, which corresponds to the reciprocal of the maximum velocity, or 1/Vmax. This
graphical representation allows for the easy determination of kinetic constants and the
identification of inhibition patterns.
6. Which of the following describes an uncompetitive inhibitor?
A. It binds only to the enzyme-substrate (ES) complex, decreasing both Vmax and Km.
B. It binds only to the free enzyme, increasing Km.
C. It binds to both the free enzyme and the ES complex with equal affinity.
D. It covalently modifies the active site, permanently inactivating the enzyme.
, Answer: A
Explanation: Uncompetitive inhibitors do not bind to the free enzyme but instead
recognize the enzyme-substrate complex after the substrate has bound. This binding locks
the substrate in the active site, effectively lowering the concentration of productive ES
complexes and thus decreasing Vmax. The apparent Km also decreases because the
equilibrium of substrate binding is pulled toward the formation of the ESI complex.
7. What is the function of the oxyanion hole in serine proteases?
A. To shuttle protons away from the active site to the solvent.
B. To provide a hydrophobic pocket for large side chains like Phenylalanine.
C. To act as a metal ion cofactor site for Zinc or Magnesium.
D. To stabilize the tetrahedral intermediate by forming hydrogen bonds with the negatively
charged oxygen.
Answer: D
Explanation: The oxyanion hole is a specialized region in the active site of enzymes like
chymotrypsin that stabilizes the transition state. During catalysis, the carbonyl oxygen of
the substrate becomes a negatively charged oxyanion as it forms a tetrahedral
intermediate. Hydrogen bonds from the backbone amides of Glycine 193 and Serine 195
stabilize this charge, lowering the activation energy for the reaction.
8. Which statement best describes the Bohr effect?
A. High oxygen concentrations promote the release of CO2 from hemoglobin.
B. Lowering the pH increases the affinity of hemoglobin for oxygen.
C. The binding of oxygen to one subunit increases the affinity of other subunits.
D. Decreasing pH and increasing CO2 concentration promote the release of oxygen from
hemoglobin.
Answer: D
Explanation: The Bohr effect describes the physiological regulation of hemoglobin’s
oxygen binding affinity by hydrogen ions and carbon dioxide. As tissues metabolize, they
produce CO2 and protons, which lower the local pH and stabilize the T-state of hemoglobin.
This reduced affinity ensures that oxygen is released specifically in areas where it is most
needed for cellular respiration.
9. How does an increase in the number of double bonds in a fatty acid affect its melting
point?
A. It increases the melting point by allowing tighter packing.
B. It decreases the melting point by creating kinks that prevent tight packing.
