BCH 4053 Exam 2 V3 | BCH 4053 Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 2) | University of Central Florida
1. Which of the following describes the Michaelis constant (Km)?
A. The maximum velocity of an enzyme-catalyzed reaction
B. The turnover number of the enzyme
C. The rate constant for the formation of the enzyme-substrate complex
D. The substrate concentration at which the reaction velocity is half-maximal
Answer: D
Explanation: The Km is defined as the substrate concentration required to reach 1/2
Vmax. It provides an estimate of the affinity of the enzyme for its substrate, with a lower
Km indicating higher affinity. This parameter is independent of enzyme concentration but
specific to a particular enzyme-substrate pair under specific conditions.
2. In competitive inhibition, how are Vmax and Km affected?
A. Vmax decreases, Km increases
B. Vmax stays the same, Km decreases
C. Vmax stays the same, Km increases
D. Vmax decreases, Km stays the same
Answer: C
Explanation: Competitive inhibitors bind to the active site, effectively competing with the
substrate. Since the inhibitor can be outcompeted by high concentrations of substrate,
Vmax remains unchanged. However, more substrate is required to reach the half-maximal
velocity, resulting in an apparent increase in Km.
3. What role does Histidine 57 play in the catalytic triad of chymotrypsin?
A. It acts as a general base to activate Serine 195
B. It acts as a nucleophile to attack the carbonyl carbon
C. It stabilizes the tetrahedral intermediate in the oxyanion hole
D. It provides a hydrophobic pocket for bulky amino acid side chains
Answer: A
Explanation: Histidine 57 functions as a general base by abstracting a proton from the
hydroxyl group of Serine 195. This increases the nucleophilicity of Serine 195, allowing it
,to attack the carbonyl carbon of the substrate. This step is crucial for the formation of the
covalent acyl-enzyme intermediate during the catalytic cycle.
4. On a Lineweaver-Burk plot, the y-intercept represents which of the following?
A. -1/Km
B. 1/Km
C. Km/Vmax
D. 1/Vmax
Answer: D
Explanation: The Lineweaver-Burk plot is a double-reciprocal transformation of the
Michaelis-Menten equation. The equation follows the form y = mx + b, where the y-
intercept is equal to 1/Vmax. This graphical representation allows for the precise
determination of kinetic parameters and the identification of inhibition types.
5. Which statement is true regarding the melting points of fatty acids?
A. Melting point decreases with increasing chain length
B. Saturated fatty acids have lower melting points than unsaturated ones
C. Trans-fatty acids have significantly lower melting points than cis-fatty acids
D. Melting point decreases with an increase in the number of double bonds
Answer: D
Explanation: The introduction of cis-double bonds creates a ‘kink’ in the hydrocarbon
chain, preventing tight packing of the molecules. This disruption of van der Waals
interactions leads to a lower melting point compared to saturated counterparts of the same
length. Consequently, highly unsaturated fats are typically liquid at room temperature.
6. How does cholesterol affect membrane fluidity at high temperatures?
A. It increases fluidity by disrupting phospholipid packing
B. It promotes the formation of micelle structures within the bilayer
C. It has no effect on membrane fluidity or stability
D. It decreases fluidity by restricting the movement of phospholipid fatty acid chains
Answer: D
Explanation: Cholesterol acts as a bidirectional regulator of membrane fluidity. At high
temperatures, the bulky steroid rings interfere with the motion of the fatty acid tails,
thereby stabilizing the membrane and decreasing fluidity. At low temperatures, it prevents
the membrane from freezing by hindering the close packing of phospholipids.
, 7. Which type of glycosidic bond is found in cellulose?
A. Alpha(1->4)
B. Alpha(1->6)
C. Beta(1->6)
D. Beta(1->4)
Answer: D
Explanation: Cellulose is a structural polysaccharide composed of linear chains of glucose
units linked by beta(1->4) glycosidic bonds. This configuration allows for the formation of
rigid, insoluble microfibrils stabilized by extensive hydrogen bonding. Unlike starch or
glycogen, most animals lack the enzymes necessary to hydrolyze these beta-linkages.
8. Enzymes increase the rate of reaction by:
A. Increasing the change in free energy (delta G)
B. Decreasing the entropy of the surroundings
C. Shifting the equilibrium constant toward products
D. Lowering the activation energy of the transition state
Answer: D
Explanation: Enzymes function as biological catalysts that lower the activation energy
barrier required for a reaction to proceed. They stabilize the transition state relative to the
ground state of the reactants through various mechanisms like acid-base or covalent
catalysis. Importantly, enzymes do not alter the overall equilibrium position or the delta G
of the reaction.
9. In uncompetitive inhibition, the inhibitor binds to:
A. The free enzyme only
B. The substrate before it enters the active site
C. Both the free enzyme and the enzyme-substrate complex
D. The enzyme-substrate complex only
Answer: D
Explanation: Uncompetitive inhibitors bind specifically to the enzyme-substrate (ES)
complex rather than the free enzyme. This binding effectively removes ES from the
reaction, resulting in a decrease in both the apparent Vmax and the apparent Km. This type
of inhibition is characteristically represented by parallel lines on a Lineweaver-Burk plot.
10. Which component is NOT found in a glycerophospholipid?
A. Glycerol backbone
Rationale (BCH4053 Exam 2) | University of Central Florida
1. Which of the following describes the Michaelis constant (Km)?
A. The maximum velocity of an enzyme-catalyzed reaction
B. The turnover number of the enzyme
C. The rate constant for the formation of the enzyme-substrate complex
D. The substrate concentration at which the reaction velocity is half-maximal
Answer: D
Explanation: The Km is defined as the substrate concentration required to reach 1/2
Vmax. It provides an estimate of the affinity of the enzyme for its substrate, with a lower
Km indicating higher affinity. This parameter is independent of enzyme concentration but
specific to a particular enzyme-substrate pair under specific conditions.
2. In competitive inhibition, how are Vmax and Km affected?
A. Vmax decreases, Km increases
B. Vmax stays the same, Km decreases
C. Vmax stays the same, Km increases
D. Vmax decreases, Km stays the same
Answer: C
Explanation: Competitive inhibitors bind to the active site, effectively competing with the
substrate. Since the inhibitor can be outcompeted by high concentrations of substrate,
Vmax remains unchanged. However, more substrate is required to reach the half-maximal
velocity, resulting in an apparent increase in Km.
3. What role does Histidine 57 play in the catalytic triad of chymotrypsin?
A. It acts as a general base to activate Serine 195
B. It acts as a nucleophile to attack the carbonyl carbon
C. It stabilizes the tetrahedral intermediate in the oxyanion hole
D. It provides a hydrophobic pocket for bulky amino acid side chains
Answer: A
Explanation: Histidine 57 functions as a general base by abstracting a proton from the
hydroxyl group of Serine 195. This increases the nucleophilicity of Serine 195, allowing it
,to attack the carbonyl carbon of the substrate. This step is crucial for the formation of the
covalent acyl-enzyme intermediate during the catalytic cycle.
4. On a Lineweaver-Burk plot, the y-intercept represents which of the following?
A. -1/Km
B. 1/Km
C. Km/Vmax
D. 1/Vmax
Answer: D
Explanation: The Lineweaver-Burk plot is a double-reciprocal transformation of the
Michaelis-Menten equation. The equation follows the form y = mx + b, where the y-
intercept is equal to 1/Vmax. This graphical representation allows for the precise
determination of kinetic parameters and the identification of inhibition types.
5. Which statement is true regarding the melting points of fatty acids?
A. Melting point decreases with increasing chain length
B. Saturated fatty acids have lower melting points than unsaturated ones
C. Trans-fatty acids have significantly lower melting points than cis-fatty acids
D. Melting point decreases with an increase in the number of double bonds
Answer: D
Explanation: The introduction of cis-double bonds creates a ‘kink’ in the hydrocarbon
chain, preventing tight packing of the molecules. This disruption of van der Waals
interactions leads to a lower melting point compared to saturated counterparts of the same
length. Consequently, highly unsaturated fats are typically liquid at room temperature.
6. How does cholesterol affect membrane fluidity at high temperatures?
A. It increases fluidity by disrupting phospholipid packing
B. It promotes the formation of micelle structures within the bilayer
C. It has no effect on membrane fluidity or stability
D. It decreases fluidity by restricting the movement of phospholipid fatty acid chains
Answer: D
Explanation: Cholesterol acts as a bidirectional regulator of membrane fluidity. At high
temperatures, the bulky steroid rings interfere with the motion of the fatty acid tails,
thereby stabilizing the membrane and decreasing fluidity. At low temperatures, it prevents
the membrane from freezing by hindering the close packing of phospholipids.
, 7. Which type of glycosidic bond is found in cellulose?
A. Alpha(1->4)
B. Alpha(1->6)
C. Beta(1->6)
D. Beta(1->4)
Answer: D
Explanation: Cellulose is a structural polysaccharide composed of linear chains of glucose
units linked by beta(1->4) glycosidic bonds. This configuration allows for the formation of
rigid, insoluble microfibrils stabilized by extensive hydrogen bonding. Unlike starch or
glycogen, most animals lack the enzymes necessary to hydrolyze these beta-linkages.
8. Enzymes increase the rate of reaction by:
A. Increasing the change in free energy (delta G)
B. Decreasing the entropy of the surroundings
C. Shifting the equilibrium constant toward products
D. Lowering the activation energy of the transition state
Answer: D
Explanation: Enzymes function as biological catalysts that lower the activation energy
barrier required for a reaction to proceed. They stabilize the transition state relative to the
ground state of the reactants through various mechanisms like acid-base or covalent
catalysis. Importantly, enzymes do not alter the overall equilibrium position or the delta G
of the reaction.
9. In uncompetitive inhibition, the inhibitor binds to:
A. The free enzyme only
B. The substrate before it enters the active site
C. Both the free enzyme and the enzyme-substrate complex
D. The enzyme-substrate complex only
Answer: D
Explanation: Uncompetitive inhibitors bind specifically to the enzyme-substrate (ES)
complex rather than the free enzyme. This binding effectively removes ES from the
reaction, resulting in a decrease in both the apparent Vmax and the apparent Km. This type
of inhibition is characteristically represented by parallel lines on a Lineweaver-Burk plot.
10. Which component is NOT found in a glycerophospholipid?
A. Glycerol backbone
