CHEM 210 Module 6 Exam ()
(PDF) | Biochemistry | Geneva College
CHEM 210 Module 6 Exam ()
Biochemistry | Geneva College
Q1. Which of the following best describes the primary structural difference between a nucleoside
and a nucleotide?
A. Nucleosides contain a phosphate group; nucleotides do not.
B. Nucleotides contain a nitrogenous base; nucleosides do not.
C. Nucleosides consist of a base linked to a sugar; nucleotides additionally contain one or more
phosphate groups.
D. Nucleotides are found only in RNA; nucleosides are found only in DNA.
Correct Answer: C. Nucleosides consist of a base linked to a sugar; nucleotides additionally contain
one or more phosphate groups.
Rationale: A nucleoside is formed by the covalent attachment of a nitrogenous base (purine or
pyrimidine) to a pentose sugar (ribose or deoxyribose). A nucleotide is a nucleoside that has been
phosphorylated at the 5′ position of the sugar. Options A and B are incorrect because they misstate
the defining components. Option D is incorrect because both nucleosides and nucleotides are found
in both RNA and DNA.
Q2. In the context of protein structure, which level of organization is defined by the three-
dimensional arrangement of all atoms in a single polypeptide chain?
A. Primary structure
B. Secondary structure
C. Tertiary structure
D. Quaternary structure
Correct Answer: C. Tertiary structure
Rationale: Tertiary structure refers to the complete three-dimensional fold of a single polypeptide,
including all side-chain interactions and disulfide bonds. Primary structure is the linear amino acid
sequence. Secondary structure refers to local repetitive patterns such as α-helices and β-sheets.
Quaternary structure describes the arrangement of multiple polypeptide subunits.
Q3. Which of the following amino acids is most likely to be found buried in the hydrophobic core of a
globular protein?
A. Lysine
B. Glutamate
pg. 1
,C. Leucine
D. Serine
Correct Answer: C. Leucine
Rationale: Leucine has a nonpolar, aliphatic side chain that is hydrophobic and therefore
thermodynamically favors burial away from water. Lysine and glutamate are charged and
hydrophilic. Serine is polar and often found on the protein surface or in active sites where its
hydroxyl group can participate in hydrogen bonding.
Q4. An enzyme follows Michaelis–Menten kinetics. If the substrate concentration is equal to the Km
value, what is the initial reaction velocity?
A. Zero
B. One-quarter of Vmax
C. One-half of Vmax
D. Equal to Vmax
Correct Answer: C. One-half of Vmax
Rationale: By definition, Km is the substrate concentration at which the initial velocity is one-half of
the maximum velocity (Vmax). This relationship is a direct consequence of the Michaelis–Menten
equation: v = (Vmax[S])/(Km + [S]). When [S] = Km, v = Vmax/2.
Q5. A competitive inhibitor is added to an enzyme-catalyzed reaction. Which of the following kinetic
changes is expected?
A. Vmax decreases; Km remains unchanged
B. Vmax remains unchanged; apparent Km increases
C. Both Vmax and Km decrease
D. Both Vmax and Km increase
Correct Answer: B. Vmax remains unchanged; apparent Km increases
Rationale: A competitive inhibitor binds reversibly to the free enzyme at the active site, preventing
substrate binding. This can be overcome by increasing substrate concentration, so Vmax is
unchanged. However, a higher substrate concentration is required to reach half-maximal velocity, so
the apparent Km increases.
Q6. Which of the following statements about allosteric enzymes is correct?
A. They always exhibit Michaelis–Menten kinetics.
B. They typically display a sigmoidal velocity-versus-substrate curve.
C. They are unaffected by effector molecules.
D. They consist of a single polypeptide chain.
Correct Answer: B. They typically display a sigmoidal velocity-versus-substrate curve.
pg. 2
,Rationale: Allosteric enzymes are typically multisubunit proteins that exhibit cooperative substrate
binding, producing a sigmoidal (S-shaped) curve rather than the hyperbolic curve of Michaelis–
Menten enzymes. They are regulated by allosteric effectors that bind at sites distinct from the active
site.
Q7. In glycolysis, which enzyme catalyzes the committed step that irreversibly converts fructose-6-
phosphate to fructose-1,6-bisphosphate?
A. Hexokinase
B. Phosphofructokinase-1
C. Aldolase
D. Pyruvate kinase
Correct Answer: B. Phosphofructokinase-1
Rationale: Phosphofructokinase-1 (PFK-1) catalyzes the phosphorylation of fructose-6-phosphate to
fructose-1,6-bisphosphate. This is the committed and rate-limiting step of glycolysis, subject to
extensive allosteric regulation. Hexokinase catalyzes the first step (glucose to glucose-6-phosphate).
Aldolase cleaves fructose-1,6-bisphosphate. Pyruvate kinase catalyzes the final step.
Q8. Which of the following is a net product of glycolysis per molecule of glucose?
A. 2 ATP, 2 NADH, 2 pyruvate
B. 4 ATP, 2 NADH, 2 pyruvate
C. 2 ATP, 2 FADH2, 2 pyruvate
D. 4 ATP, 4 NADH, 1 pyruvate
Correct Answer: A. 2 ATP, 2 NADH, 2 pyruvate
Rationale: Glycolysis produces 4 ATP by substrate-level phosphorylation but consumes 2 ATP in the
preparatory phase, yielding a net gain of 2 ATP. Two molecules of NAD+ are reduced to NADH, and
one glucose molecule yields two molecules of pyruvate. FADH2 is not produced in glycolysis.
Q9. In gluconeogenesis, which of the following enzymes is used to bypass the irreversible pyruvate
kinase step of glycolysis?
A. Pyruvate carboxylase and phosphoenolpyruvate carboxykinase
B. Hexokinase and glucose-6-phosphatase
C. Phosphofructokinase-1 and fructose-1,6-bisphosphatase
D. Aldolase and triose phosphate isomerase
Correct Answer: A. Pyruvate carboxylase and phosphoenolpyruvate carboxykinase
Rationale: The conversion of pyruvate to phosphoenolpyruvate requires two enzymes: pyruvate
carboxylase (which converts pyruvate to oxaloacetate in the mitochondria) and
phosphoenolpyruvate carboxykinase (which converts oxaloacetate to PEP). The other irreversible
steps of glycolysis are bypassed by glucose-6-phosphatase and fructose-1,6-bisphosphatase.
pg. 3
, Q10. Which of the following best describes the role of the citric acid cycle in cellular metabolism?
A. It produces ATP exclusively through substrate-level phosphorylation.
B. It oxidizes acetyl-CoA to CO2 and generates reduced coenzymes for oxidative phosphorylation.
C. It is the primary site of fatty acid synthesis.
D. It converts glucose directly to lactate under anaerobic conditions.
Correct Answer: B. It oxidizes acetyl-CoA to CO2 and generates reduced coenzymes for oxidative
phosphorylation.
Rationale: The citric acid cycle (Krebs cycle) oxidizes acetyl-CoA to two molecules of CO2 per acetyl
group. The energy released is captured as NADH and FADH2, which donate electrons to the electron
transport chain. One GTP (or ATP) is also produced per turn via substrate-level phosphorylation. It is
not the primary site of fatty acid synthesis, nor does it directly produce lactate.
Q11. Which of the following molecules is the primary electron carrier that delivers electrons from
Complex I and Complex II to Complex III in the electron transport chain?
A. NADH
B. FADH2
C. Coenzyme Q (ubiquinone)
D. Cytochrome c
Correct Answer: C. Coenzyme Q (ubiquinone)
Rationale: Coenzyme Q (ubiquinone) is a lipid-soluble mobile electron carrier that accepts electrons
from both Complex I (NADH dehydrogenase) and Complex II (succinate dehydrogenase) and delivers
them to Complex III. Cytochrome c is a water-soluble carrier that shuttles electrons from Complex III
to Complex IV.
Q12. The chemiosmotic theory of oxidative phosphorylation proposes that ATP synthesis is driven
by:
A. Direct chemical coupling between electron transport and ATP synthase
B. A proton electrochemical gradient across the inner mitochondrial membrane
C. Substrate-level phosphorylation in the mitochondrial matrix
D. The direct transfer of phosphate from NADH to ADP
Correct Answer: B. A proton electrochemical gradient across the inner mitochondrial membrane
Rationale: The chemiosmotic theory, proposed by Peter Mitchell, states that electron transport
creates a proton gradient (proton-motive force) across the inner mitochondrial membrane. The flow
of protons back through ATP synthase drives the synthesis of ATP from ADP and Pi.
Q13. Which of the following is the primary function of the electron transport chain?
A. To synthesize ATP directly
B. To generate a proton gradient by transferring electrons from reduced coenzymes to oxygen
pg. 4
(PDF) | Biochemistry | Geneva College
CHEM 210 Module 6 Exam ()
Biochemistry | Geneva College
Q1. Which of the following best describes the primary structural difference between a nucleoside
and a nucleotide?
A. Nucleosides contain a phosphate group; nucleotides do not.
B. Nucleotides contain a nitrogenous base; nucleosides do not.
C. Nucleosides consist of a base linked to a sugar; nucleotides additionally contain one or more
phosphate groups.
D. Nucleotides are found only in RNA; nucleosides are found only in DNA.
Correct Answer: C. Nucleosides consist of a base linked to a sugar; nucleotides additionally contain
one or more phosphate groups.
Rationale: A nucleoside is formed by the covalent attachment of a nitrogenous base (purine or
pyrimidine) to a pentose sugar (ribose or deoxyribose). A nucleotide is a nucleoside that has been
phosphorylated at the 5′ position of the sugar. Options A and B are incorrect because they misstate
the defining components. Option D is incorrect because both nucleosides and nucleotides are found
in both RNA and DNA.
Q2. In the context of protein structure, which level of organization is defined by the three-
dimensional arrangement of all atoms in a single polypeptide chain?
A. Primary structure
B. Secondary structure
C. Tertiary structure
D. Quaternary structure
Correct Answer: C. Tertiary structure
Rationale: Tertiary structure refers to the complete three-dimensional fold of a single polypeptide,
including all side-chain interactions and disulfide bonds. Primary structure is the linear amino acid
sequence. Secondary structure refers to local repetitive patterns such as α-helices and β-sheets.
Quaternary structure describes the arrangement of multiple polypeptide subunits.
Q3. Which of the following amino acids is most likely to be found buried in the hydrophobic core of a
globular protein?
A. Lysine
B. Glutamate
pg. 1
,C. Leucine
D. Serine
Correct Answer: C. Leucine
Rationale: Leucine has a nonpolar, aliphatic side chain that is hydrophobic and therefore
thermodynamically favors burial away from water. Lysine and glutamate are charged and
hydrophilic. Serine is polar and often found on the protein surface or in active sites where its
hydroxyl group can participate in hydrogen bonding.
Q4. An enzyme follows Michaelis–Menten kinetics. If the substrate concentration is equal to the Km
value, what is the initial reaction velocity?
A. Zero
B. One-quarter of Vmax
C. One-half of Vmax
D. Equal to Vmax
Correct Answer: C. One-half of Vmax
Rationale: By definition, Km is the substrate concentration at which the initial velocity is one-half of
the maximum velocity (Vmax). This relationship is a direct consequence of the Michaelis–Menten
equation: v = (Vmax[S])/(Km + [S]). When [S] = Km, v = Vmax/2.
Q5. A competitive inhibitor is added to an enzyme-catalyzed reaction. Which of the following kinetic
changes is expected?
A. Vmax decreases; Km remains unchanged
B. Vmax remains unchanged; apparent Km increases
C. Both Vmax and Km decrease
D. Both Vmax and Km increase
Correct Answer: B. Vmax remains unchanged; apparent Km increases
Rationale: A competitive inhibitor binds reversibly to the free enzyme at the active site, preventing
substrate binding. This can be overcome by increasing substrate concentration, so Vmax is
unchanged. However, a higher substrate concentration is required to reach half-maximal velocity, so
the apparent Km increases.
Q6. Which of the following statements about allosteric enzymes is correct?
A. They always exhibit Michaelis–Menten kinetics.
B. They typically display a sigmoidal velocity-versus-substrate curve.
C. They are unaffected by effector molecules.
D. They consist of a single polypeptide chain.
Correct Answer: B. They typically display a sigmoidal velocity-versus-substrate curve.
pg. 2
,Rationale: Allosteric enzymes are typically multisubunit proteins that exhibit cooperative substrate
binding, producing a sigmoidal (S-shaped) curve rather than the hyperbolic curve of Michaelis–
Menten enzymes. They are regulated by allosteric effectors that bind at sites distinct from the active
site.
Q7. In glycolysis, which enzyme catalyzes the committed step that irreversibly converts fructose-6-
phosphate to fructose-1,6-bisphosphate?
A. Hexokinase
B. Phosphofructokinase-1
C. Aldolase
D. Pyruvate kinase
Correct Answer: B. Phosphofructokinase-1
Rationale: Phosphofructokinase-1 (PFK-1) catalyzes the phosphorylation of fructose-6-phosphate to
fructose-1,6-bisphosphate. This is the committed and rate-limiting step of glycolysis, subject to
extensive allosteric regulation. Hexokinase catalyzes the first step (glucose to glucose-6-phosphate).
Aldolase cleaves fructose-1,6-bisphosphate. Pyruvate kinase catalyzes the final step.
Q8. Which of the following is a net product of glycolysis per molecule of glucose?
A. 2 ATP, 2 NADH, 2 pyruvate
B. 4 ATP, 2 NADH, 2 pyruvate
C. 2 ATP, 2 FADH2, 2 pyruvate
D. 4 ATP, 4 NADH, 1 pyruvate
Correct Answer: A. 2 ATP, 2 NADH, 2 pyruvate
Rationale: Glycolysis produces 4 ATP by substrate-level phosphorylation but consumes 2 ATP in the
preparatory phase, yielding a net gain of 2 ATP. Two molecules of NAD+ are reduced to NADH, and
one glucose molecule yields two molecules of pyruvate. FADH2 is not produced in glycolysis.
Q9. In gluconeogenesis, which of the following enzymes is used to bypass the irreversible pyruvate
kinase step of glycolysis?
A. Pyruvate carboxylase and phosphoenolpyruvate carboxykinase
B. Hexokinase and glucose-6-phosphatase
C. Phosphofructokinase-1 and fructose-1,6-bisphosphatase
D. Aldolase and triose phosphate isomerase
Correct Answer: A. Pyruvate carboxylase and phosphoenolpyruvate carboxykinase
Rationale: The conversion of pyruvate to phosphoenolpyruvate requires two enzymes: pyruvate
carboxylase (which converts pyruvate to oxaloacetate in the mitochondria) and
phosphoenolpyruvate carboxykinase (which converts oxaloacetate to PEP). The other irreversible
steps of glycolysis are bypassed by glucose-6-phosphatase and fructose-1,6-bisphosphatase.
pg. 3
, Q10. Which of the following best describes the role of the citric acid cycle in cellular metabolism?
A. It produces ATP exclusively through substrate-level phosphorylation.
B. It oxidizes acetyl-CoA to CO2 and generates reduced coenzymes for oxidative phosphorylation.
C. It is the primary site of fatty acid synthesis.
D. It converts glucose directly to lactate under anaerobic conditions.
Correct Answer: B. It oxidizes acetyl-CoA to CO2 and generates reduced coenzymes for oxidative
phosphorylation.
Rationale: The citric acid cycle (Krebs cycle) oxidizes acetyl-CoA to two molecules of CO2 per acetyl
group. The energy released is captured as NADH and FADH2, which donate electrons to the electron
transport chain. One GTP (or ATP) is also produced per turn via substrate-level phosphorylation. It is
not the primary site of fatty acid synthesis, nor does it directly produce lactate.
Q11. Which of the following molecules is the primary electron carrier that delivers electrons from
Complex I and Complex II to Complex III in the electron transport chain?
A. NADH
B. FADH2
C. Coenzyme Q (ubiquinone)
D. Cytochrome c
Correct Answer: C. Coenzyme Q (ubiquinone)
Rationale: Coenzyme Q (ubiquinone) is a lipid-soluble mobile electron carrier that accepts electrons
from both Complex I (NADH dehydrogenase) and Complex II (succinate dehydrogenase) and delivers
them to Complex III. Cytochrome c is a water-soluble carrier that shuttles electrons from Complex III
to Complex IV.
Q12. The chemiosmotic theory of oxidative phosphorylation proposes that ATP synthesis is driven
by:
A. Direct chemical coupling between electron transport and ATP synthase
B. A proton electrochemical gradient across the inner mitochondrial membrane
C. Substrate-level phosphorylation in the mitochondrial matrix
D. The direct transfer of phosphate from NADH to ADP
Correct Answer: B. A proton electrochemical gradient across the inner mitochondrial membrane
Rationale: The chemiosmotic theory, proposed by Peter Mitchell, states that electron transport
creates a proton gradient (proton-motive force) across the inner mitochondrial membrane. The flow
of protons back through ATP synthase drives the synthesis of ATP from ADP and Pi.
Q13. Which of the following is the primary function of the electron transport chain?
A. To synthesize ATP directly
B. To generate a proton gradient by transferring electrons from reduced coenzymes to oxygen
pg. 4
