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BCH 4053 Exam 2 V1 | BCH 4053 Biochemistry I | Actual Q&A with Rationale (BCH4053 Exam 2) | University of Central Florida

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BCH 4053 Exam 2 V1 | BCH 4053 Biochemistry I | Actual Q&A with Rationale (BCH4053 Exam 2) | University of Central Florida

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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 describes the effect of 2,3-bisphosphoglycerate (2,3-BPG) on

hemoglobin?

A. It increases the affinity of hemoglobin for oxygen by stabilizing the R-state.


B. It decreases oxygen affinity by stabilizing the T-state, facilitating oxygen release in

tissues.


C. It binds to the iron atom in the heme group to prevent carbon monoxide poisoning.


D. It acts as a competitive inhibitor for oxygen at the heme binding site.


Answer: B


Rationale: 2,3-BPG is a heterotropic allosteric effector that binds to the central cavity of

the hemoglobin tetramer. By binding to the T-state (deoxy-hemoglobin), it stabilizes this

low-affinity conformation and shifts the equilibrium away from the R-state. This

mechanism is essential for the efficient delivery of oxygen to peripheral tissues where

oxygen tension is low.


2. In the Michaelis-Menten model, what does the constant Km represent?

A. The maximum velocity of the reaction when the enzyme is saturated.

,B. The substrate concentration at which the reaction velocity is half of Vmax.


C. The number of substrate molecules converted to product per unit time.


D. The equilibrium constant for the dissociation of the product from the enzyme.


Answer: B


Rationale: Km, the Michaelis constant, is defined as the substrate concentration required

for an enzyme to reach one-half of its maximum velocity. It provides an inverse measure of

the enzyme’s affinity for its substrate, with a low Km indicating high affinity. This value is

characteristic for a specific enzyme-substrate pair under defined conditions of temperature

and pH.


3. Which type of inhibition can be overcome by increasing the substrate concentration?

A. Competitive inhibition


B. Uncompetitive inhibition


C. Non-competitive inhibition


D. Mixed inhibition


Answer: A


Rationale: Competitive inhibitors resemble the substrate and compete for binding at the

active site of the enzyme. Because the inhibitor and substrate compete for the same site,

adding excess substrate eventually outcompetes the inhibitor. Consequently, the Vmax of

the reaction remains unchanged, even though the apparent Km increases.

, 4. What is the characteristic appearance of a Lineweaver-Burk plot for uncompetitive

inhibition compared to no inhibitor?

A. The lines intersect at the y-axis.


B. The lines intersect at the x-axis.


C. The lines intersect in the second quadrant.


D. The lines are parallel to each other.


Answer: D


Rationale: Uncompetitive inhibitors bind only to the enzyme-substrate (ES) complex, not

the free enzyme. This binding reduces both the apparent Vmax and the apparent Km by the

same factor, resulting in a shift of both intercepts on a double-reciprocal plot. Because the

ratio of the slopes remains constant, the resulting lines for inhibited and uninhibited

reactions are parallel.


5. Which amino acids constitute the catalytic triad found in serine proteases like

chymotrypsin?

A. Aspartate, Histidine, and Serine


B. Serine, Histidine, and Lysine


C. Cysteine, Histidine, and Aspartate


D. Serine, Glutamate, and Histidine


Answer: A

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