CHEM 210 Biochemistry Module 1 to 8 Exams &
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1. A researcher is studying an enzyme that catalyzes the conversion of substrate A to product B.
The reaction follows Michaelis-Menten kinetics. At a substrate concentration of 2 mM, the initial
velocity is 25% of Vmax. What is the Km of this enzyme?
A. 0.5 mM
B. 2 mM
C. 6 mM
D. 8 mM
Answer: C
Rationale: Using the Michaelis-Menten equation: v = Vmax[S]/(Km+[S]). Given v = 0.25 Vmax and
[S]=2 mM, we have 0.25 = 2/(Km+2). Solving: Km+2 = 8, so Km = 6 mM. Option C is correct. Other
options do not satisfy the equation.
2. In the context of protein structure, which of the following best describes the thermodynamic
driving force for the folding of globular proteins in an aqueous environment?
A. Maximization of hydrogen bonding between backbone amide groups
B. Minimization of solvent-exposed hydrophobic side chains
C. Formation of extensive disulfide bonds stabilizing the native state
D. Optimization of electrostatic interactions among charged residues
Answer: B
Rationale: The hydrophobic effect, which minimizes exposure of nonpolar side chains to water, is the
primary driving force for protein folding. While hydrogen bonding and electrostatic interactions
contribute to stability, they are not the main driving force. Disulfide bonds are not present in all proteins
and are not the primary driving force.
3. Which of the following statements about the regulation of glycogen phosphorylase is correct?
A. Phosphorylation by phosphorylase kinase activates the enzyme and is reversed by protein phosphatase 1.
B. AMP allosterically inhibits the enzyme, while ATP and glucose-6-phosphate activate it.
C. The enzyme exists in an active R state and an inactive T state; phosphorylation shifts the equilibrium toward
the T state.
D. Insulin stimulates glycogenolysis by activating glycogen phosphorylase via a cAMP-dependent cascade.
Answer: A
Rationale: Glycogen phosphorylase is activated by phosphorylation (by phosphorylase kinase) and
inactivated by dephosphorylation (by protein phosphatase 1). Option A is correct. Option B is wrong
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,because AMP activates, ATP and G6P inhibit. Option C is wrong because phosphorylation shifts to the
active R state. Option D is wrong because insulin promotes glycogen synthesis, not breakdown.
4. A patient presents with elevated levels of orotic acid in the urine. Deficiency of which enzyme is
most likely responsible?
A. Carbamoyl phosphate synthetase I
B. Ornithine transcarbamoylase
C. UMP synthase
D. Dihydroorotate dehydrogenase
Answer: B
Rationale: Orotic aciduria is classically associated with deficiency of ornithine transcarbamoylase
(OTC), an enzyme of the urea cycle. OTC deficiency leads to accumulation of carbamoyl phosphate,
which spills over into pyrimidine synthesis, causing orotic acid accumulation. Option B is correct. CPS I
deficiency causes hyperammonemia without orotic aciduria. UMP synthase deficiency causes orotic
aciduria but is less common and typically presents with megaloblastic anemia. Dihydroorotate
dehydrogenase deficiency is rare and does not typically cause orotic aciduria.
5. In the electron transport chain, the complex that directly reduces oxygen to water is:
A. Complex I (NADH dehydrogenase)
B. Complex II (succinate dehydrogenase)
C. Complex III (cytochrome bc1 complex)
D. Complex IV (cytochrome c oxidase)
Answer: D
Rationale: Complex IV (cytochrome c oxidase) catalyzes the reduction of O2 to H2O using electrons from
cytochrome c. Complex I and II transfer electrons to ubiquinone. Complex III transfers electrons from
ubiquinol to cytochrome c. Thus, D is correct.
6. A laboratory technician is performing a Western blot to detect a protein of interest. After
transferring the proteins to a membrane, the technician incubates the membrane with a primary
antibody, then with a secondary antibody conjugated to horseradish peroxidase. What step is
essential to prevent non-specific binding of the antibodies?
A. Blocking the membrane with a protein solution such as BSA or non-fat dry milk
B. Using a high-salt wash buffer to remove unbound antibodies
C. Adding a reducing agent like -mercaptoethanol to the sample buffer
D. Performing electrophoresis at low voltage to ensure complete transfer
Answer: A
Rationale: Blocking with a protein solution (e.g., BSA, milk) saturates non-specific binding sites on the
membrane, reducing background. Option A is essential. High-salt washes help remove unbound
antibodies but are not the primary step to prevent non-specific binding. Reducing agents are used in
sample preparation for denaturing gels, not for blocking. Electrophoresis conditions affect transfer
efficiency, not antibody specificity.
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,7. Which of the following best explains why the melting temperature (Tm) of a DNA duplex
increases with increasing salt concentration?
A. Salt ions stabilize the phosphodiester bonds between nucleotides.
B. Salt ions neutralize the negative charges on the phosphate backbone, reducing electrostatic repulsion.
C. Salt ions promote the formation of Hoogsteen base pairs.
D. Salt ions increase the hydrophobic effect that drives base stacking.
Answer: B
Rationale: DNA strands are negatively charged and repel each other. High salt concentrations provide
cations (e.g., Na+) that shield the negative charges, reducing repulsion and stabilizing the duplex. This
raises the Tm. Option B is correct. Salt does not directly stabilize phosphodiester bonds or promote
Hoogsteen pairing. The hydrophobic effect is not directly enhanced by salt in this context.
8. In the biosynthesis of palmitate, the acetyl-CoA carboxylase reaction produces malonyl-CoA.
Which of the following is a direct regulatory mechanism for this enzyme in mammals?
A. Allosteric activation by palmitoyl-CoA
B. Phosphorylation by AMP-activated protein kinase (AMPK) leading to inactivation
C. Allosteric inhibition by citrate
D. Proteolytic cleavage by caspase-3
Answer: B
Rationale: Acetyl-CoA carboxylase is inactivated by phosphorylation via AMPK, which is activated by
high AMP (low energy). This prevents fatty acid synthesis when energy is scarce. Option B is correct.
Palmitoyl-CoA is an allosteric inhibitor, not activator. Citrate is an allosteric activator, not inhibitor.
Proteolytic cleavage is not a normal regulatory mechanism for this enzyme.
9. A researcher is analyzing a metabolic pathway and observes that the committed step is catalyzed
by an enzyme that is inhibited by the final product of the pathway. This type of regulation is
known as:
A. Feed-forward activation
B. Feedback inhibition
C. Covalent modification
D. Competitive inhibition
Answer: B
Rationale: Feedback inhibition occurs when the end product of a pathway inhibits an early enzyme,
typically the committed step, to regulate flux. This is a common regulatory mechanism in metabolic
pathways. Option B is correct. Feed-forward activation involves a metabolite early in the pathway
activating a later enzyme. Covalent modification involves reversible modification like phosphorylation.
Competitive inhibition is a type of enzyme inhibition but not specific to pathway regulation.
10. Which of the following statements about the lac operon is correct?
A. The lac repressor binds to the operator in the presence of allolactose.
B. Catabolite repression by glucose is mediated by low levels of cAMP, which reduces CAP binding.
C. The lac operon is induced when glucose is present and lactose is absent.
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, D. CAP-cAMP complex binds to the operator to enhance transcription.
Answer: B
Rationale: When glucose is present, cAMP levels are low, so CAP cannot bind to the CAP site, and
transcription is low. This is catabolite repression. Option B is correct. The lac repressor binds in the
absence of allolactose. The operon is induced when lactose is present and glucose is absent. CAP-cAMP
binds to the CAP site (upstream of promoter), not the operator, to enhance transcription.
11. A novel enzyme catalyzes the conversion of substrate S to product P. The steady-state kinetic
parameters are: Km = 0.5 mM, Vmax = 100 mol/min. In the presence of 2 mM inhibitor I, the
apparent Km increases to 2.0 mM, while Vmax remains unchanged. Which of the following best
describes the inhibition mechanism and the expected effect on the enzyme's specificity constant
(kcat/Km)?
A. Competitive inhibition; kcat/Km decreases by a factor of 4
B. Competitive inhibition; kcat/Km remains unchanged
C. Mixed inhibition; kcat/Km decreases by a factor of 2
D. Noncompetitive inhibition; kcat/Km remains unchanged
Answer: A
Rationale: The increase in Km with no change in Vmax is characteristic of competitive inhibition. The
specificity constant kcat/Km is reduced because Km increases while kcat (Vmax/[E]total) is unchanged.
In this case, Km increases from 0.5 to 2.0 mM (4-fold), so kcat/Km decreases 4-fold.
12. In a patient with a rare glycogen storage disease, liver biopsy reveals normal glycogen
phosphorylase activity but a defect in the debranching enzyme. Which of the following metabolic
consequences is most likely to be observed during a fast?
A. Normal blood glucose levels due to gluconeogenesis
B. Accumulation of glycogen with short outer chains
C. Severe hypoglycemia and accumulation of limit dextrin
D. Increased lactate production from Cori cycle
Answer: C
Rationale: Deficiency of debranching enzyme (Cori disease) prevents complete breakdown of glycogen,
leading to accumulation of limit dextrin (glycogen with short outer branches). This impairs glucose
release, causing severe fasting hypoglycemia. Glycogen phosphorylase is normal but cannot fully
degrade glycogen without debranching enzyme.
13. During a study of fatty acid oxidation, mitochondria from rat liver are incubated with
[1-14C]palmitate (16:0). After one round of -oxidation, where is the radiolabel most likely to be
found?
A. Acetyl-CoA
B. Butyryl-CoA
C. Myristoyl-CoA
D. Propionyl-CoA
Answer: A
Page 4
Final Exam () Portage Learning
Questions and Verified Answers, 100% Guaranteed
Pass ||Already Graded A+
1. A researcher is studying an enzyme that catalyzes the conversion of substrate A to product B.
The reaction follows Michaelis-Menten kinetics. At a substrate concentration of 2 mM, the initial
velocity is 25% of Vmax. What is the Km of this enzyme?
A. 0.5 mM
B. 2 mM
C. 6 mM
D. 8 mM
Answer: C
Rationale: Using the Michaelis-Menten equation: v = Vmax[S]/(Km+[S]). Given v = 0.25 Vmax and
[S]=2 mM, we have 0.25 = 2/(Km+2). Solving: Km+2 = 8, so Km = 6 mM. Option C is correct. Other
options do not satisfy the equation.
2. In the context of protein structure, which of the following best describes the thermodynamic
driving force for the folding of globular proteins in an aqueous environment?
A. Maximization of hydrogen bonding between backbone amide groups
B. Minimization of solvent-exposed hydrophobic side chains
C. Formation of extensive disulfide bonds stabilizing the native state
D. Optimization of electrostatic interactions among charged residues
Answer: B
Rationale: The hydrophobic effect, which minimizes exposure of nonpolar side chains to water, is the
primary driving force for protein folding. While hydrogen bonding and electrostatic interactions
contribute to stability, they are not the main driving force. Disulfide bonds are not present in all proteins
and are not the primary driving force.
3. Which of the following statements about the regulation of glycogen phosphorylase is correct?
A. Phosphorylation by phosphorylase kinase activates the enzyme and is reversed by protein phosphatase 1.
B. AMP allosterically inhibits the enzyme, while ATP and glucose-6-phosphate activate it.
C. The enzyme exists in an active R state and an inactive T state; phosphorylation shifts the equilibrium toward
the T state.
D. Insulin stimulates glycogenolysis by activating glycogen phosphorylase via a cAMP-dependent cascade.
Answer: A
Rationale: Glycogen phosphorylase is activated by phosphorylation (by phosphorylase kinase) and
inactivated by dephosphorylation (by protein phosphatase 1). Option A is correct. Option B is wrong
Page 1
,because AMP activates, ATP and G6P inhibit. Option C is wrong because phosphorylation shifts to the
active R state. Option D is wrong because insulin promotes glycogen synthesis, not breakdown.
4. A patient presents with elevated levels of orotic acid in the urine. Deficiency of which enzyme is
most likely responsible?
A. Carbamoyl phosphate synthetase I
B. Ornithine transcarbamoylase
C. UMP synthase
D. Dihydroorotate dehydrogenase
Answer: B
Rationale: Orotic aciduria is classically associated with deficiency of ornithine transcarbamoylase
(OTC), an enzyme of the urea cycle. OTC deficiency leads to accumulation of carbamoyl phosphate,
which spills over into pyrimidine synthesis, causing orotic acid accumulation. Option B is correct. CPS I
deficiency causes hyperammonemia without orotic aciduria. UMP synthase deficiency causes orotic
aciduria but is less common and typically presents with megaloblastic anemia. Dihydroorotate
dehydrogenase deficiency is rare and does not typically cause orotic aciduria.
5. In the electron transport chain, the complex that directly reduces oxygen to water is:
A. Complex I (NADH dehydrogenase)
B. Complex II (succinate dehydrogenase)
C. Complex III (cytochrome bc1 complex)
D. Complex IV (cytochrome c oxidase)
Answer: D
Rationale: Complex IV (cytochrome c oxidase) catalyzes the reduction of O2 to H2O using electrons from
cytochrome c. Complex I and II transfer electrons to ubiquinone. Complex III transfers electrons from
ubiquinol to cytochrome c. Thus, D is correct.
6. A laboratory technician is performing a Western blot to detect a protein of interest. After
transferring the proteins to a membrane, the technician incubates the membrane with a primary
antibody, then with a secondary antibody conjugated to horseradish peroxidase. What step is
essential to prevent non-specific binding of the antibodies?
A. Blocking the membrane with a protein solution such as BSA or non-fat dry milk
B. Using a high-salt wash buffer to remove unbound antibodies
C. Adding a reducing agent like -mercaptoethanol to the sample buffer
D. Performing electrophoresis at low voltage to ensure complete transfer
Answer: A
Rationale: Blocking with a protein solution (e.g., BSA, milk) saturates non-specific binding sites on the
membrane, reducing background. Option A is essential. High-salt washes help remove unbound
antibodies but are not the primary step to prevent non-specific binding. Reducing agents are used in
sample preparation for denaturing gels, not for blocking. Electrophoresis conditions affect transfer
efficiency, not antibody specificity.
Page 2
,7. Which of the following best explains why the melting temperature (Tm) of a DNA duplex
increases with increasing salt concentration?
A. Salt ions stabilize the phosphodiester bonds between nucleotides.
B. Salt ions neutralize the negative charges on the phosphate backbone, reducing electrostatic repulsion.
C. Salt ions promote the formation of Hoogsteen base pairs.
D. Salt ions increase the hydrophobic effect that drives base stacking.
Answer: B
Rationale: DNA strands are negatively charged and repel each other. High salt concentrations provide
cations (e.g., Na+) that shield the negative charges, reducing repulsion and stabilizing the duplex. This
raises the Tm. Option B is correct. Salt does not directly stabilize phosphodiester bonds or promote
Hoogsteen pairing. The hydrophobic effect is not directly enhanced by salt in this context.
8. In the biosynthesis of palmitate, the acetyl-CoA carboxylase reaction produces malonyl-CoA.
Which of the following is a direct regulatory mechanism for this enzyme in mammals?
A. Allosteric activation by palmitoyl-CoA
B. Phosphorylation by AMP-activated protein kinase (AMPK) leading to inactivation
C. Allosteric inhibition by citrate
D. Proteolytic cleavage by caspase-3
Answer: B
Rationale: Acetyl-CoA carboxylase is inactivated by phosphorylation via AMPK, which is activated by
high AMP (low energy). This prevents fatty acid synthesis when energy is scarce. Option B is correct.
Palmitoyl-CoA is an allosteric inhibitor, not activator. Citrate is an allosteric activator, not inhibitor.
Proteolytic cleavage is not a normal regulatory mechanism for this enzyme.
9. A researcher is analyzing a metabolic pathway and observes that the committed step is catalyzed
by an enzyme that is inhibited by the final product of the pathway. This type of regulation is
known as:
A. Feed-forward activation
B. Feedback inhibition
C. Covalent modification
D. Competitive inhibition
Answer: B
Rationale: Feedback inhibition occurs when the end product of a pathway inhibits an early enzyme,
typically the committed step, to regulate flux. This is a common regulatory mechanism in metabolic
pathways. Option B is correct. Feed-forward activation involves a metabolite early in the pathway
activating a later enzyme. Covalent modification involves reversible modification like phosphorylation.
Competitive inhibition is a type of enzyme inhibition but not specific to pathway regulation.
10. Which of the following statements about the lac operon is correct?
A. The lac repressor binds to the operator in the presence of allolactose.
B. Catabolite repression by glucose is mediated by low levels of cAMP, which reduces CAP binding.
C. The lac operon is induced when glucose is present and lactose is absent.
Page 3
, D. CAP-cAMP complex binds to the operator to enhance transcription.
Answer: B
Rationale: When glucose is present, cAMP levels are low, so CAP cannot bind to the CAP site, and
transcription is low. This is catabolite repression. Option B is correct. The lac repressor binds in the
absence of allolactose. The operon is induced when lactose is present and glucose is absent. CAP-cAMP
binds to the CAP site (upstream of promoter), not the operator, to enhance transcription.
11. A novel enzyme catalyzes the conversion of substrate S to product P. The steady-state kinetic
parameters are: Km = 0.5 mM, Vmax = 100 mol/min. In the presence of 2 mM inhibitor I, the
apparent Km increases to 2.0 mM, while Vmax remains unchanged. Which of the following best
describes the inhibition mechanism and the expected effect on the enzyme's specificity constant
(kcat/Km)?
A. Competitive inhibition; kcat/Km decreases by a factor of 4
B. Competitive inhibition; kcat/Km remains unchanged
C. Mixed inhibition; kcat/Km decreases by a factor of 2
D. Noncompetitive inhibition; kcat/Km remains unchanged
Answer: A
Rationale: The increase in Km with no change in Vmax is characteristic of competitive inhibition. The
specificity constant kcat/Km is reduced because Km increases while kcat (Vmax/[E]total) is unchanged.
In this case, Km increases from 0.5 to 2.0 mM (4-fold), so kcat/Km decreases 4-fold.
12. In a patient with a rare glycogen storage disease, liver biopsy reveals normal glycogen
phosphorylase activity but a defect in the debranching enzyme. Which of the following metabolic
consequences is most likely to be observed during a fast?
A. Normal blood glucose levels due to gluconeogenesis
B. Accumulation of glycogen with short outer chains
C. Severe hypoglycemia and accumulation of limit dextrin
D. Increased lactate production from Cori cycle
Answer: C
Rationale: Deficiency of debranching enzyme (Cori disease) prevents complete breakdown of glycogen,
leading to accumulation of limit dextrin (glycogen with short outer branches). This impairs glucose
release, causing severe fasting hypoglycemia. Glycogen phosphorylase is normal but cannot fully
degrade glycogen without debranching enzyme.
13. During a study of fatty acid oxidation, mitochondria from rat liver are incubated with
[1-14C]palmitate (16:0). After one round of -oxidation, where is the radiolabel most likely to be
found?
A. Acetyl-CoA
B. Butyryl-CoA
C. Myristoyl-CoA
D. Propionyl-CoA
Answer: A
Page 4
