CHEM 210 Biochemistry Module 4 Exam: Carbohydrates, Glycolysis,
and Gluconeogenesis 2026/2027 UPDATE
1. Which of the following is an example of a ketotriose?
A. Glyceraldehyde
B. Glucose
C. Ribose
D. Dihydroxyacetone
Answer: D
Rationale: Dihydroxyacetone is the simplest ketotriose, having three carbons and a ketone
functional group.
2. Which carbon atom determines the D or L configuration in an aldohexose like
glucose?
A. C-5
B. C-2
C. C-1
D. C-6
Answer: A
Rationale: In an aldohexose, the D or L configuration is determined by the orientation of
the hydroxyl group on the highest-numbered chiral center, which is C-5.
,3. What is the relationship between D-glucose and D-galactose?
A. Enantiomers
B. Epimers
C. Anomers
D. Constitutional isomers
Answer: B
Rationale: D-glucose and D-galactose differ only in the configuration around C-4, making
them C-4 epimers.
4. The cyclization of an aldose to form a pyranose ring involves the reaction
between:
A. C-1 aldehyde and C-4 hydroxyl
B. C-1 aldehyde and C-5 hydroxyl
C. C-2 ketone and C-5 hydroxyl
D. C-1 aldehyde and C-6 hydroxyl
Answer: B
Rationale: Pyranose rings are six-membered rings formed by the reaction of the C-1
aldehyde group and the C-5 hydroxyl group.
5. In the alpha-anomer of a D-glucose pyranose ring, the hydroxyl group at C-1
is:
A. Always involved in a peptide bond
B. On the same side of the ring as the C-6 hydroxymethyl group
C. In the equatorial position only
D. On the opposite side of the ring from the C-6 hydroxymethyl group
Answer: D
Rationale: In D-sugars, the alpha-anomer has the anomeric hydroxyl group trans (opposite
side) to the CH2OH group at C-6.
, 6. Which enzyme catalyzes the phosphorylation of glucose to glucose-6-
phosphate in most tissues?
A. Hexokinase
B. Phosphofructokinase-1
C. Glucose-6-phosphatase
D. Pyruvate kinase
Answer: A
Rationale: Hexokinase is the enzyme responsible for the first step of glycolysis,
phosphorylating glucose to trap it inside the cell.
7. Which of the following molecules is the ‘committed’ step of glycolysis?
A. Glucose-6-phosphate
B. Fructose-6-phosphate
C. Phosphoenolpyruvate
D. Fructose-1,6-bisphosphate
Answer: D
Rationale: The production of fructose-1,6-bisphosphate by PFK-1 is the primary
committed and regulated step of glycolysis.
8. What is the net gain of ATP molecules per molecule of glucose processed
through glycolysis?
A. 1
B. 32
C. 4
D. 2
Answer: D
Rationale: Glycolysis uses 2 ATP in the preparatory phase and produces 4 ATP in the
payoff phase, resulting in a net gain of 2 ATP.
and Gluconeogenesis 2026/2027 UPDATE
1. Which of the following is an example of a ketotriose?
A. Glyceraldehyde
B. Glucose
C. Ribose
D. Dihydroxyacetone
Answer: D
Rationale: Dihydroxyacetone is the simplest ketotriose, having three carbons and a ketone
functional group.
2. Which carbon atom determines the D or L configuration in an aldohexose like
glucose?
A. C-5
B. C-2
C. C-1
D. C-6
Answer: A
Rationale: In an aldohexose, the D or L configuration is determined by the orientation of
the hydroxyl group on the highest-numbered chiral center, which is C-5.
,3. What is the relationship between D-glucose and D-galactose?
A. Enantiomers
B. Epimers
C. Anomers
D. Constitutional isomers
Answer: B
Rationale: D-glucose and D-galactose differ only in the configuration around C-4, making
them C-4 epimers.
4. The cyclization of an aldose to form a pyranose ring involves the reaction
between:
A. C-1 aldehyde and C-4 hydroxyl
B. C-1 aldehyde and C-5 hydroxyl
C. C-2 ketone and C-5 hydroxyl
D. C-1 aldehyde and C-6 hydroxyl
Answer: B
Rationale: Pyranose rings are six-membered rings formed by the reaction of the C-1
aldehyde group and the C-5 hydroxyl group.
5. In the alpha-anomer of a D-glucose pyranose ring, the hydroxyl group at C-1
is:
A. Always involved in a peptide bond
B. On the same side of the ring as the C-6 hydroxymethyl group
C. In the equatorial position only
D. On the opposite side of the ring from the C-6 hydroxymethyl group
Answer: D
Rationale: In D-sugars, the alpha-anomer has the anomeric hydroxyl group trans (opposite
side) to the CH2OH group at C-6.
, 6. Which enzyme catalyzes the phosphorylation of glucose to glucose-6-
phosphate in most tissues?
A. Hexokinase
B. Phosphofructokinase-1
C. Glucose-6-phosphatase
D. Pyruvate kinase
Answer: A
Rationale: Hexokinase is the enzyme responsible for the first step of glycolysis,
phosphorylating glucose to trap it inside the cell.
7. Which of the following molecules is the ‘committed’ step of glycolysis?
A. Glucose-6-phosphate
B. Fructose-6-phosphate
C. Phosphoenolpyruvate
D. Fructose-1,6-bisphosphate
Answer: D
Rationale: The production of fructose-1,6-bisphosphate by PFK-1 is the primary
committed and regulated step of glycolysis.
8. What is the net gain of ATP molecules per molecule of glucose processed
through glycolysis?
A. 1
B. 32
C. 4
D. 2
Answer: D
Rationale: Glycolysis uses 2 ATP in the preparatory phase and produces 4 ATP in the
payoff phase, resulting in a net gain of 2 ATP.
