BCH 4053 Exam 4 V3 | BCH 4053 Biochemistry I | Actual Q&A with
Rationale (BCH4053 Exam 4) | University of Central Florida
1. Which of the following best describes the Michaelis constant (Km)?
A. The maximum velocity of the reaction.
B. The number of substrate molecules converted to product per unit time.
C. The equilibrium constant for the dissociation of the ES complex.
D. The substrate concentration at which the reaction rate is half of Vmax.
Answer: D
Explanation: Km is a fundamental kinetic parameter that reflects the affinity of an enzyme
for its substrate under specific conditions. A low Km indicates that the enzyme reaches
half-maximal velocity at a low substrate concentration, implying high affinity. This value is
mathematically derived from the rate constants of the Michaelis-Menten model.
2. In competitive inhibition, how are the kinetic parameters Vmax and Km affected?
A. Vmax remains unchanged and Km increases.
B. Vmax remains unchanged and Km decreases.
C. Vmax decreases and Km remains unchanged.
D. Both Vmax and Km decrease.
Answer: A
Explanation: Competitive inhibitors bind to the active site of the enzyme, effectively
competing with the substrate. Because the inhibitor can be outcompeted by high
concentrations of substrate, the Vmax remains reachable and thus unchanged. However,
more substrate is required to reach half-Vmax, resulting in an apparent increase in the Km
value.
3. Which enzyme catalyzes the primary ‘committed step’ of glycolysis?
A. Phosphofructokinase-1 (PFK-1)
B. Hexokinase
C. Pyruvate Kinase
D. Glyceraldehyde-3-phosphate dehydrogenase
Answer: A
,Explanation: PFK-1 catalyzes the phosphorylation of Fructose-6-phosphate to Fructose-
1,6-bisphosphate using ATP. This step is considered the committed step because Fructose-
1,6-bisphosphate is exclusively used in the glycolytic pathway. Regulation of PFK-1 is
complex, involving allosteric effectors like ATP, AMP, and Fructose-2,6-bisphosphate.
4. Which molecule is a potent allosteric activator of PFK-1 and inhibitor of Fructose-1,6-
bisphosphatase?
A. Glucose-6-phosphate
B. Citrate
C. ATP
D. Fructose-2,6-bisphosphate
Answer: D
Explanation: Fructose-2,6-bisphosphate is the most significant regulator of glycolytic flux
in the liver. It signals high blood glucose levels and promotes glycolysis by activating PFK-1
while simultaneously inhibiting the gluconeogenic enzyme FBPase-1. This reciprocal
regulation prevents futile cycling between glycolysis and gluconeogenesis.
5. During gluconeogenesis, which enzyme is required to bypass the irreversible step catalyzed
by Pyruvate Kinase?
A. Glucose-6-phosphatase
B. Fructose-1,6-bisphosphatase
C. Pyruvate Carboxylase and PEP Carboxykinase
D. Malate Dehydrogenase
Answer: C
Explanation: The conversion of pyruvate back to phosphoenolpyruvate (PEP) is
energetically unfavorable and requires two steps. Pyruvate carboxylase first converts
pyruvate to oxaloacetate in the mitochondria, requiring biotin and ATP. PEP carboxykinase
then converts oxaloacetate to PEP, utilizing GTP as a phosphate donor and decarboxylation
to drive the reaction.
6. What is the primary role of the Pentose Phosphate Pathway’s oxidative phase?
A. Production of NADH for the electron transport chain.
B. Degradation of glucose to generate ATP.
C. Production of NADPH and Ribose-5-phosphate.
D. Synthesis of glycogen from glucose-6-phosphate.
Answer: C
, Explanation: The oxidative phase of the Pentose Phosphate Pathway is essential for
generating NADPH, which provides reducing power for biosynthetic reactions and
antioxidant defense. It also produces Ribose-5-phosphate, a critical precursor for
nucleotide synthesis. These products are vital for rapidly dividing cells and tissues involved
in fatty acid or steroid synthesis.
7. Which enzyme is responsible for the breakdown of glycogen by cleaving alpha-1,4-
glycosidic bonds?
A. Glycogen Phosphorylase
B. Branching Enzyme
C. Glycogen Synthase
D. Phosphoglucomutase
Answer: A
Explanation: Glycogen phosphorylase uses inorganic phosphate to cleave the alpha-1,4-
linkages at the non-reducing ends of glycogen. This reaction produces Glucose-1-
phosphate, which can then be isomerized to Glucose-6-phosphate. The enzyme acts
processively until it reaches a point four glucose residues away from a branch point.
8. The catalytic triad of chymotrypsin consists of which three amino acids?
A. Serine, Histidine, Lysine
B. Cysteine, Histidine, Aspartate
C. Aspartate, Histidine, Serine
D. Glutamate, Histidine, Serine
Answer: C
Explanation: The catalytic triad is a classic example of enzyme structure-function
relationship in serine proteases. Aspartate 102 helps orient and polarize Histidine 57,
which in turn acts as a general base to deprotonate Serine 195. The resulting nucleophilic
serine alkoxide ion attacks the carbonyl carbon of the peptide substrate.
9. Which of the following describes an uncompetitive inhibitor?
A. It binds only to the free enzyme.
B. It increases both Vmax and Km.
C. It binds with equal affinity to the free enzyme and the ES complex.
D. It binds only to the enzyme-substrate (ES) complex.
Answer: D
Rationale (BCH4053 Exam 4) | University of Central Florida
1. Which of the following best describes the Michaelis constant (Km)?
A. The maximum velocity of the reaction.
B. The number of substrate molecules converted to product per unit time.
C. The equilibrium constant for the dissociation of the ES complex.
D. The substrate concentration at which the reaction rate is half of Vmax.
Answer: D
Explanation: Km is a fundamental kinetic parameter that reflects the affinity of an enzyme
for its substrate under specific conditions. A low Km indicates that the enzyme reaches
half-maximal velocity at a low substrate concentration, implying high affinity. This value is
mathematically derived from the rate constants of the Michaelis-Menten model.
2. In competitive inhibition, how are the kinetic parameters Vmax and Km affected?
A. Vmax remains unchanged and Km increases.
B. Vmax remains unchanged and Km decreases.
C. Vmax decreases and Km remains unchanged.
D. Both Vmax and Km decrease.
Answer: A
Explanation: Competitive inhibitors bind to the active site of the enzyme, effectively
competing with the substrate. Because the inhibitor can be outcompeted by high
concentrations of substrate, the Vmax remains reachable and thus unchanged. However,
more substrate is required to reach half-Vmax, resulting in an apparent increase in the Km
value.
3. Which enzyme catalyzes the primary ‘committed step’ of glycolysis?
A. Phosphofructokinase-1 (PFK-1)
B. Hexokinase
C. Pyruvate Kinase
D. Glyceraldehyde-3-phosphate dehydrogenase
Answer: A
,Explanation: PFK-1 catalyzes the phosphorylation of Fructose-6-phosphate to Fructose-
1,6-bisphosphate using ATP. This step is considered the committed step because Fructose-
1,6-bisphosphate is exclusively used in the glycolytic pathway. Regulation of PFK-1 is
complex, involving allosteric effectors like ATP, AMP, and Fructose-2,6-bisphosphate.
4. Which molecule is a potent allosteric activator of PFK-1 and inhibitor of Fructose-1,6-
bisphosphatase?
A. Glucose-6-phosphate
B. Citrate
C. ATP
D. Fructose-2,6-bisphosphate
Answer: D
Explanation: Fructose-2,6-bisphosphate is the most significant regulator of glycolytic flux
in the liver. It signals high blood glucose levels and promotes glycolysis by activating PFK-1
while simultaneously inhibiting the gluconeogenic enzyme FBPase-1. This reciprocal
regulation prevents futile cycling between glycolysis and gluconeogenesis.
5. During gluconeogenesis, which enzyme is required to bypass the irreversible step catalyzed
by Pyruvate Kinase?
A. Glucose-6-phosphatase
B. Fructose-1,6-bisphosphatase
C. Pyruvate Carboxylase and PEP Carboxykinase
D. Malate Dehydrogenase
Answer: C
Explanation: The conversion of pyruvate back to phosphoenolpyruvate (PEP) is
energetically unfavorable and requires two steps. Pyruvate carboxylase first converts
pyruvate to oxaloacetate in the mitochondria, requiring biotin and ATP. PEP carboxykinase
then converts oxaloacetate to PEP, utilizing GTP as a phosphate donor and decarboxylation
to drive the reaction.
6. What is the primary role of the Pentose Phosphate Pathway’s oxidative phase?
A. Production of NADH for the electron transport chain.
B. Degradation of glucose to generate ATP.
C. Production of NADPH and Ribose-5-phosphate.
D. Synthesis of glycogen from glucose-6-phosphate.
Answer: C
, Explanation: The oxidative phase of the Pentose Phosphate Pathway is essential for
generating NADPH, which provides reducing power for biosynthetic reactions and
antioxidant defense. It also produces Ribose-5-phosphate, a critical precursor for
nucleotide synthesis. These products are vital for rapidly dividing cells and tissues involved
in fatty acid or steroid synthesis.
7. Which enzyme is responsible for the breakdown of glycogen by cleaving alpha-1,4-
glycosidic bonds?
A. Glycogen Phosphorylase
B. Branching Enzyme
C. Glycogen Synthase
D. Phosphoglucomutase
Answer: A
Explanation: Glycogen phosphorylase uses inorganic phosphate to cleave the alpha-1,4-
linkages at the non-reducing ends of glycogen. This reaction produces Glucose-1-
phosphate, which can then be isomerized to Glucose-6-phosphate. The enzyme acts
processively until it reaches a point four glucose residues away from a branch point.
8. The catalytic triad of chymotrypsin consists of which three amino acids?
A. Serine, Histidine, Lysine
B. Cysteine, Histidine, Aspartate
C. Aspartate, Histidine, Serine
D. Glutamate, Histidine, Serine
Answer: C
Explanation: The catalytic triad is a classic example of enzyme structure-function
relationship in serine proteases. Aspartate 102 helps orient and polarize Histidine 57,
which in turn acts as a general base to deprotonate Serine 195. The resulting nucleophilic
serine alkoxide ion attacks the carbonyl carbon of the peptide substrate.
9. Which of the following describes an uncompetitive inhibitor?
A. It binds only to the free enzyme.
B. It increases both Vmax and Km.
C. It binds with equal affinity to the free enzyme and the ES complex.
D. It binds only to the enzyme-substrate (ES) complex.
Answer: D
