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CHEM 210 BIOCHEMISTRY MODULE 4 EXAM ACTUAL
2026/2027 | Metabolism: Glycolysis, Gluconeogenesis,
Regulation & Pyruvate Dehydrogenase | Geneva College |
Complete Questions & Verified Answers | Pass Guaranteed -
A+ Graded
Section 1: Glycolysis - Reactions & Energetics (Q1-15)
Q1. A first-year nursing student is reviewing the glycolytic pathway. She correctly
notes that the entire sequence occurs in the cytosol and begins with glucose. What is
the complete stoichiometric summary of glycolysis under aerobic conditions?
A. Glucose + 2 NAD⁺ + 2 ADP + 2 Pi → 2 lactate + 2 ATP + 2 H₂O
B. Glucose + 2 NAD⁺ + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 2 ATP + 2 H⁺ + 2 H₂O
[CORRECT]
C. Glucose + 2 NAD⁺ + 4 ADP + 4 Pi → 2 pyruvate + 2 NADH + 4 ATP + 2 H₂O
D. Glucose + 4 NAD⁺ + 2 ADP + 2 Pi → 2 pyruvate + 4 NADH + 2 ATP + 4 H⁺
Rationale: Glycolysis consumes one glucose, two NAD⁺, two ADP, and two Pi to
produce two pyruvate, two NADH, two net ATP, two H⁺, and two H₂O. The gross ATP
production is four, but two are invested in the energy investment phase, yielding a
net of two ATP. Option A describes anaerobic fermentation to lactate; options C and
D have incorrect stoichiometry.
Correct Answer: B
Q2. During a sprint, a muscle cell's flux through glycolysis increases dramatically.
Which step is considered the committed step of glycolysis, meaning once this
reaction occurs, the molecule is obligated to proceed through the rest of the
pathway?
A. Phosphoglucose isomerase (step 2)
B. Phosphofructokinase-1 (PFK-1) (step 3) [CORRECT]
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C. Aldolase (step 4)
D. Pyruvate kinase (step 10)
Rationale: The phosphorylation of fructose-6-phosphate to fructose-1,6-
bisphosphate by PFK-1 is the committed step of glycolysis because fructose-1,6-
bisphosphate is utilized only in glycolysis, whereas glucose-6-phosphate can enter
glycogen synthesis or the pentose phosphate pathway. Hexokinase (step 1) is also
regulated but is not the committed step for the entire pathway.
Correct Answer: B
Q3. A biochemistry exam asks for the net ATP yield from glycolysis per molecule of
glucose. A student answers "4 ATP." What is the correct net yield, and why is the
student's answer incorrect?
A. 2 ATP; the student reported gross ATP without subtracting the 2 ATP invested in
steps 1 and 3 [CORRECT]
B. 2 ATP; the student confused NADH yield with ATP yield
C. 4 ATP; the student is actually correct because net and gross are identical
D. 6 ATP; the student forgot to include substrate-level phosphorylation from the TCA
cycle
Rationale: Glycolysis generates 4 ATP via substrate-level phosphorylation (steps 7
and 10, each occurring twice per glucose) but consumes 2 ATP in the energy
investment phase (steps 1 and 3), yielding a net of 2 ATP. The student reported gross
ATP. NADH is produced but not counted as ATP in this direct calculation.
Correct Answer: A
Q4. A patient with a rare mutation has a hexokinase variant that cannot bind glucose.
Which metabolic consequence would most directly result from the inability to
phosphorylate glucose at the entry point of glycolysis?
,3
A. Accumulation of fructose-1,6-bisphosphate and activation of gluconeogenesis
B. Trapping of glucose outside the cell and inability to proceed past the first step of
glycolysis [CORRECT]
C. Diversion of all glucose into the pentose phosphate pathway via glucose-6-
phosphate dehydrogenase
D. Spontaneous conversion of glucose to pyruvate via non-enzymatic glycolysis
Rationale: Hexokinase phosphorylates glucose to glucose-6-phosphate, trapping it
intracellularly because the phosphate prevents diffusion back across the membrane.
Without this step, glucose cannot enter glycolysis, glycogen synthesis, or the pentose
phosphate pathway. Fructose-1,6-bisphosphate is downstream and would decrease,
not accumulate.
Correct Answer: B
Q5. In the second reaction of glycolysis, phosphoglucose isomerase catalyzes the
reversible conversion of glucose-6-phosphate to fructose-6-phosphate. What is the
biochemical significance of this isomerization?
A. It cleaves the 6-carbon sugar into two 3-carbon fragments
B. It converts an aldose to a ketose, preparing the molecule for aldol cleavage
[CORRECT]
C. It phosphorylates the sugar at the C-1 position
D. It oxidizes the sugar, generating the first NADH of glycolysis
Rationale: Phosphoglucose isomerase converts the aldose glucose-6-phosphate to
the ketose fructose-6-phosphate; the open-chain ketose form is necessary for
subsequent aldol cleavage by aldolase. Phosphorylation at C-1 is performed by PFK-
1; cleavage is performed by aldolase; oxidation occurs at GAPDH.
Correct Answer: B
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Q6. A medical student is asked to identify the products of the aldolase reaction in
glycolysis. Which pair of triose phosphates is generated from one molecule of
fructose-1,6-bisphosphate?
A. Two molecules of glyceraldehyde-3-phosphate
B. One glyceraldehyde-3-phosphate and one dihydroxyacetone phosphate
[CORRECT]
C. One glucose-6-phosphate and one fructose-6-phosphate
D. Two molecules of dihydroxyacetone phosphate
Rationale: Aldolase cleaves the 6-carbon fructose-1,6-bisphosphate between C-3 and
C-4, yielding one glyceraldehyde-3-phosphate (an aldotriose) and one
dihydroxyacetone phosphate (a ketotriose). Only glyceraldehyde-3-phosphate
proceeds directly through the remainder of glycolysis; dihydroxyacetone phosphate
must be isomerized by triose phosphate isomerase.
Correct Answer: B
Q7. After aldol cleavage, only one of the two triose phosphates can directly continue
through the payoff phase of glycolysis. Which enzyme ensures that both halves of
the original glucose molecule are ultimately metabolized through the same
downstream sequence?
A. Phosphoglycerate kinase
B. Triose phosphate isomerase [CORRECT]
C. Phosphoglycerate mutase
D. Enolase
Rationale: Triose phosphate isomerase rapidly interconverts dihydroxyacetone
phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P), ensuring that both
trioses proceed through G3P. Because the equilibrium favors DHAP but G3P is
continuously consumed, the reaction is driven forward. The other enzymes act on 3-
phosphoglycerate or downstream intermediates.
Correct Answer: B
CHEM 210 BIOCHEMISTRY MODULE 4 EXAM ACTUAL
2026/2027 | Metabolism: Glycolysis, Gluconeogenesis,
Regulation & Pyruvate Dehydrogenase | Geneva College |
Complete Questions & Verified Answers | Pass Guaranteed -
A+ Graded
Section 1: Glycolysis - Reactions & Energetics (Q1-15)
Q1. A first-year nursing student is reviewing the glycolytic pathway. She correctly
notes that the entire sequence occurs in the cytosol and begins with glucose. What is
the complete stoichiometric summary of glycolysis under aerobic conditions?
A. Glucose + 2 NAD⁺ + 2 ADP + 2 Pi → 2 lactate + 2 ATP + 2 H₂O
B. Glucose + 2 NAD⁺ + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 2 ATP + 2 H⁺ + 2 H₂O
[CORRECT]
C. Glucose + 2 NAD⁺ + 4 ADP + 4 Pi → 2 pyruvate + 2 NADH + 4 ATP + 2 H₂O
D. Glucose + 4 NAD⁺ + 2 ADP + 2 Pi → 2 pyruvate + 4 NADH + 2 ATP + 4 H⁺
Rationale: Glycolysis consumes one glucose, two NAD⁺, two ADP, and two Pi to
produce two pyruvate, two NADH, two net ATP, two H⁺, and two H₂O. The gross ATP
production is four, but two are invested in the energy investment phase, yielding a
net of two ATP. Option A describes anaerobic fermentation to lactate; options C and
D have incorrect stoichiometry.
Correct Answer: B
Q2. During a sprint, a muscle cell's flux through glycolysis increases dramatically.
Which step is considered the committed step of glycolysis, meaning once this
reaction occurs, the molecule is obligated to proceed through the rest of the
pathway?
A. Phosphoglucose isomerase (step 2)
B. Phosphofructokinase-1 (PFK-1) (step 3) [CORRECT]
,2
C. Aldolase (step 4)
D. Pyruvate kinase (step 10)
Rationale: The phosphorylation of fructose-6-phosphate to fructose-1,6-
bisphosphate by PFK-1 is the committed step of glycolysis because fructose-1,6-
bisphosphate is utilized only in glycolysis, whereas glucose-6-phosphate can enter
glycogen synthesis or the pentose phosphate pathway. Hexokinase (step 1) is also
regulated but is not the committed step for the entire pathway.
Correct Answer: B
Q3. A biochemistry exam asks for the net ATP yield from glycolysis per molecule of
glucose. A student answers "4 ATP." What is the correct net yield, and why is the
student's answer incorrect?
A. 2 ATP; the student reported gross ATP without subtracting the 2 ATP invested in
steps 1 and 3 [CORRECT]
B. 2 ATP; the student confused NADH yield with ATP yield
C. 4 ATP; the student is actually correct because net and gross are identical
D. 6 ATP; the student forgot to include substrate-level phosphorylation from the TCA
cycle
Rationale: Glycolysis generates 4 ATP via substrate-level phosphorylation (steps 7
and 10, each occurring twice per glucose) but consumes 2 ATP in the energy
investment phase (steps 1 and 3), yielding a net of 2 ATP. The student reported gross
ATP. NADH is produced but not counted as ATP in this direct calculation.
Correct Answer: A
Q4. A patient with a rare mutation has a hexokinase variant that cannot bind glucose.
Which metabolic consequence would most directly result from the inability to
phosphorylate glucose at the entry point of glycolysis?
,3
A. Accumulation of fructose-1,6-bisphosphate and activation of gluconeogenesis
B. Trapping of glucose outside the cell and inability to proceed past the first step of
glycolysis [CORRECT]
C. Diversion of all glucose into the pentose phosphate pathway via glucose-6-
phosphate dehydrogenase
D. Spontaneous conversion of glucose to pyruvate via non-enzymatic glycolysis
Rationale: Hexokinase phosphorylates glucose to glucose-6-phosphate, trapping it
intracellularly because the phosphate prevents diffusion back across the membrane.
Without this step, glucose cannot enter glycolysis, glycogen synthesis, or the pentose
phosphate pathway. Fructose-1,6-bisphosphate is downstream and would decrease,
not accumulate.
Correct Answer: B
Q5. In the second reaction of glycolysis, phosphoglucose isomerase catalyzes the
reversible conversion of glucose-6-phosphate to fructose-6-phosphate. What is the
biochemical significance of this isomerization?
A. It cleaves the 6-carbon sugar into two 3-carbon fragments
B. It converts an aldose to a ketose, preparing the molecule for aldol cleavage
[CORRECT]
C. It phosphorylates the sugar at the C-1 position
D. It oxidizes the sugar, generating the first NADH of glycolysis
Rationale: Phosphoglucose isomerase converts the aldose glucose-6-phosphate to
the ketose fructose-6-phosphate; the open-chain ketose form is necessary for
subsequent aldol cleavage by aldolase. Phosphorylation at C-1 is performed by PFK-
1; cleavage is performed by aldolase; oxidation occurs at GAPDH.
Correct Answer: B
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Q6. A medical student is asked to identify the products of the aldolase reaction in
glycolysis. Which pair of triose phosphates is generated from one molecule of
fructose-1,6-bisphosphate?
A. Two molecules of glyceraldehyde-3-phosphate
B. One glyceraldehyde-3-phosphate and one dihydroxyacetone phosphate
[CORRECT]
C. One glucose-6-phosphate and one fructose-6-phosphate
D. Two molecules of dihydroxyacetone phosphate
Rationale: Aldolase cleaves the 6-carbon fructose-1,6-bisphosphate between C-3 and
C-4, yielding one glyceraldehyde-3-phosphate (an aldotriose) and one
dihydroxyacetone phosphate (a ketotriose). Only glyceraldehyde-3-phosphate
proceeds directly through the remainder of glycolysis; dihydroxyacetone phosphate
must be isomerized by triose phosphate isomerase.
Correct Answer: B
Q7. After aldol cleavage, only one of the two triose phosphates can directly continue
through the payoff phase of glycolysis. Which enzyme ensures that both halves of
the original glucose molecule are ultimately metabolized through the same
downstream sequence?
A. Phosphoglycerate kinase
B. Triose phosphate isomerase [CORRECT]
C. Phosphoglycerate mutase
D. Enolase
Rationale: Triose phosphate isomerase rapidly interconverts dihydroxyacetone
phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P), ensuring that both
trioses proceed through G3P. Because the equilibrium favors DHAP but G3P is
continuously consumed, the reaction is driven forward. The other enzymes act on 3-
phosphoglycerate or downstream intermediates.
Correct Answer: B
