BCH 4053 Exam 2 V1 | BCH 4053 Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 2) | University of Central Florida
1. Which of the following best describes the Michaelis constant (Km)?
A. The velocity of the reaction when the enzyme is fully saturated with substrate.
B. The substrate concentration at which the reaction rate is exactly half of Vmax.
C. The equilibrium constant for the dissociation of the enzyme-substrate complex.
D. The total concentration of enzyme active sites available for catalysis.
Answer: B
Explanation: The Michaelis constant, Km, represents the specific substrate concentration
required for an enzyme to reach 50% of its maximum velocity. It is a fundamental
parameter in Michaelis-Menten kinetics that reflects the affinity of an enzyme for its
substrate under specific conditions. A low Km value indicates high affinity, meaning the
enzyme reaches half-saturation at a lower substrate concentration.
2. In a Lineweaver-Burk plot, how is the value of Vmax determined?
A. It is the x-intercept of the plotted line.
B. It is the reciprocal of the y-intercept.
C. It is the slope of the line divided by Km.
D. It is the negative reciprocal of the x-intercept.
Answer: B
Explanation: The Lineweaver-Burk plot is a double-reciprocal representation of the
Michaelis-Menten equation where 1/v is plotted against 1/[S]. The y-intercept of this linear
plot is defined as 1/Vmax, meaning Vmax can be calculated by taking the inverse of the
intercept. This linearization allows for more accurate determination of kinetic constants
compared to a hyperbolic curve.
3. How does a competitive inhibitor affect the kinetic parameters of an enzyme?
A. It decreases Vmax and increases Km.
B. It decreases Vmax while Km remains unchanged.
C. Km increases while Vmax remains unchanged.
D. It increases Vmax and decreases Km.
Answer: C
,Explanation: Competitive inhibitors bind to the active site of the enzyme, directly
competing with the substrate for binding. This results in an increase in the apparent Km
because a higher concentration of substrate is needed to achieve half-maximal velocity.
However, because high substrate concentrations can eventually displace the inhibitor, the
maximum velocity (Vmax) of the reaction remains unaffected.
4. Which amino acid residue in the catalytic triad of chymotrypsin acts as the nucleophile that
attacks the substrate carbonyl group?
A. Glycine
B. Histidine
C. Aspartate
D. Serine
Answer: D
Explanation: In the serine protease mechanism, Serine 195 acts as the primary
nucleophile that performs the initial attack on the peptide bond. The nucleophilicity of this
serine is enhanced by the nearby Histidine 57, which acts as a general base. This
interaction is further stabilized by Aspartate 102, completing the catalytic triad necessary
for proteolysis.
5. What is the specific role of the oxyanion hole in the chymotrypsin mechanism?
A. It facilitates the binding of the aromatic side chain of the substrate.
B. It acts as a base to deprotonate the active site serine.
C. It excludes water from the active site during the acylation phase.
D. It stabilizes the tetrahedral intermediate through hydrogen bonding.
Answer: D
Explanation: The oxyanion hole is a specialized region in the active site of chymotrypsin
formed by the backbone amide hydrogens of Serine 195 and Glycine 193. It specifically
stabilizes the negatively charged oxygen atom of the tetrahedral intermediate formed
during catalysis. By lowering the activation energy of this high-energy transition state, the
oxyanion hole significantly increases the rate of the reaction.
6. Regarding the Bohr effect, what happens to hemoglobin’s affinity for oxygen as the pH
decreases?
A. Affinity increases, shifting the curve to the left.
B. The Hill coefficient increases to compensate for lower pH.
C. Affinity remains constant but the Vmax of binding decreases.
D. Affinity decreases, shifting the curve to the right.
, Answer: D
Explanation: The Bohr effect describes the inverse relationship between acidity (or CO2
concentration) and hemoglobin’s oxygen affinity. As pH decreases, specific amino acid
residues in hemoglobin become protonated, favoring the formation of salt bridges that
stabilize the T-state (tense state). This leads to a decreased affinity for oxygen, facilitating
its release in metabolically active tissues where CO2 and H+ levels are high.
7. Which of the following molecules acts as a heterotropic allosteric effector that stabilizes
the T-state of hemoglobin?
A. Oxygen
B. Nitric Oxide
C. Carbon Monoxide
D. 2,3-Bisphosphoglycerate (2,3-BPG)
Answer: D
Explanation: 2,3-Bisphosphoglycerate (2,3-BPG) is a highly anionic molecule that binds to
the central cavity of the hemoglobin tetramer. This binding occurs specifically in the T-
state, stabilizing the deoxygenated form and reducing oxygen affinity. Without 2,3-BPG,
hemoglobin would bind oxygen too tightly to effectively release it to the tissues.
8. What does a Hill coefficient (n) greater than 1 indicate about a protein’s binding
properties?
A. There is positive cooperativity in binding.
B. The binding is non-cooperative.
C. There is negative cooperativity in binding.
D. The protein has only a single binding site.
Answer: A
Explanation: The Hill coefficient is a measure of the degree of cooperativity in a ligand-
binding process. A value of n > 1 indicates positive cooperativity, where the binding of one
ligand molecule increases the affinity of the remaining sites for subsequent ligands.
Hemoglobin is a classic example of this behavior, exhibiting a Hill coefficient of
approximately 2.8.
9. Which of the following is a structural characteristic of 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 (E).
Rationale (BCH4053 Exam 2) | University of Central Florida
1. Which of the following best describes the Michaelis constant (Km)?
A. The velocity of the reaction when the enzyme is fully saturated with substrate.
B. The substrate concentration at which the reaction rate is exactly half of Vmax.
C. The equilibrium constant for the dissociation of the enzyme-substrate complex.
D. The total concentration of enzyme active sites available for catalysis.
Answer: B
Explanation: The Michaelis constant, Km, represents the specific substrate concentration
required for an enzyme to reach 50% of its maximum velocity. It is a fundamental
parameter in Michaelis-Menten kinetics that reflects the affinity of an enzyme for its
substrate under specific conditions. A low Km value indicates high affinity, meaning the
enzyme reaches half-saturation at a lower substrate concentration.
2. In a Lineweaver-Burk plot, how is the value of Vmax determined?
A. It is the x-intercept of the plotted line.
B. It is the reciprocal of the y-intercept.
C. It is the slope of the line divided by Km.
D. It is the negative reciprocal of the x-intercept.
Answer: B
Explanation: The Lineweaver-Burk plot is a double-reciprocal representation of the
Michaelis-Menten equation where 1/v is plotted against 1/[S]. The y-intercept of this linear
plot is defined as 1/Vmax, meaning Vmax can be calculated by taking the inverse of the
intercept. This linearization allows for more accurate determination of kinetic constants
compared to a hyperbolic curve.
3. How does a competitive inhibitor affect the kinetic parameters of an enzyme?
A. It decreases Vmax and increases Km.
B. It decreases Vmax while Km remains unchanged.
C. Km increases while Vmax remains unchanged.
D. It increases Vmax and decreases Km.
Answer: C
,Explanation: Competitive inhibitors bind to the active site of the enzyme, directly
competing with the substrate for binding. This results in an increase in the apparent Km
because a higher concentration of substrate is needed to achieve half-maximal velocity.
However, because high substrate concentrations can eventually displace the inhibitor, the
maximum velocity (Vmax) of the reaction remains unaffected.
4. Which amino acid residue in the catalytic triad of chymotrypsin acts as the nucleophile that
attacks the substrate carbonyl group?
A. Glycine
B. Histidine
C. Aspartate
D. Serine
Answer: D
Explanation: In the serine protease mechanism, Serine 195 acts as the primary
nucleophile that performs the initial attack on the peptide bond. The nucleophilicity of this
serine is enhanced by the nearby Histidine 57, which acts as a general base. This
interaction is further stabilized by Aspartate 102, completing the catalytic triad necessary
for proteolysis.
5. What is the specific role of the oxyanion hole in the chymotrypsin mechanism?
A. It facilitates the binding of the aromatic side chain of the substrate.
B. It acts as a base to deprotonate the active site serine.
C. It excludes water from the active site during the acylation phase.
D. It stabilizes the tetrahedral intermediate through hydrogen bonding.
Answer: D
Explanation: The oxyanion hole is a specialized region in the active site of chymotrypsin
formed by the backbone amide hydrogens of Serine 195 and Glycine 193. It specifically
stabilizes the negatively charged oxygen atom of the tetrahedral intermediate formed
during catalysis. By lowering the activation energy of this high-energy transition state, the
oxyanion hole significantly increases the rate of the reaction.
6. Regarding the Bohr effect, what happens to hemoglobin’s affinity for oxygen as the pH
decreases?
A. Affinity increases, shifting the curve to the left.
B. The Hill coefficient increases to compensate for lower pH.
C. Affinity remains constant but the Vmax of binding decreases.
D. Affinity decreases, shifting the curve to the right.
, Answer: D
Explanation: The Bohr effect describes the inverse relationship between acidity (or CO2
concentration) and hemoglobin’s oxygen affinity. As pH decreases, specific amino acid
residues in hemoglobin become protonated, favoring the formation of salt bridges that
stabilize the T-state (tense state). This leads to a decreased affinity for oxygen, facilitating
its release in metabolically active tissues where CO2 and H+ levels are high.
7. Which of the following molecules acts as a heterotropic allosteric effector that stabilizes
the T-state of hemoglobin?
A. Oxygen
B. Nitric Oxide
C. Carbon Monoxide
D. 2,3-Bisphosphoglycerate (2,3-BPG)
Answer: D
Explanation: 2,3-Bisphosphoglycerate (2,3-BPG) is a highly anionic molecule that binds to
the central cavity of the hemoglobin tetramer. This binding occurs specifically in the T-
state, stabilizing the deoxygenated form and reducing oxygen affinity. Without 2,3-BPG,
hemoglobin would bind oxygen too tightly to effectively release it to the tissues.
8. What does a Hill coefficient (n) greater than 1 indicate about a protein’s binding
properties?
A. There is positive cooperativity in binding.
B. The binding is non-cooperative.
C. There is negative cooperativity in binding.
D. The protein has only a single binding site.
Answer: A
Explanation: The Hill coefficient is a measure of the degree of cooperativity in a ligand-
binding process. A value of n > 1 indicates positive cooperativity, where the binding of one
ligand molecule increases the affinity of the remaining sites for subsequent ligands.
Hemoglobin is a classic example of this behavior, exhibiting a Hill coefficient of
approximately 2.8.
9. Which of the following is a structural characteristic of 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 (E).
