CHEM 210 Biochemistry Module 4: The Definitive 300+-Question Exam Bank on Carbohydrate Structure, Glycolysis, Gluconeogenesis, and Metabolic Integration – With Answers and Clinical Rationales (2026/2027)

CHEM 210 Biochemistry Module 4:
The Definitive 300+-Question Exam
Bank on Carbohydrate Structure,
Glycolysis, Gluconeogenesis, and
Metabolic Integration – With Answers
and Clinical Rationales (2026/2027)




SECTION 1: CARBOHYDRATE FUNDAMENTALS (Questions 1-40)

1. Which monosaccharide is not a 6-carbon monosaccharide?

 a) Fructose
 b) Ribose
 c) Mannose
 d) Galactose

Answer: b) Ribose

Rationale: Ribose is a pentose (5-carbon sugar), while fructose, mannose,
and galactose are all hexoses (6-carbon sugars). Ribose is a critical
component of RNA and nucleotides .

,2. Which of the following monosaccharides is NOT an aldose?

 a) Erythrose
 b) Dihydroxyacetone
 c) Glucose
 d) Glyceraldehyde
 e) Ribose

Answer: b) Dihydroxyacetone

Rationale: Dihydroxyacetone is a ketose (contains a ketone group), while
erythrose, glucose, glyceraldehyde, and ribose are all aldoses (contain an
aldehyde group). Ketoses have the carbonyl group on an internal carbon .




3. When two monosaccharides are epimers:

 a) They differ only in configuration about the penultimate carbon
 b) One is an aldose, the other a ketose
 c) They differ only in the configuration about one carbon atom
 d) They form O-glycosidic bonds
 e) They are oligosaccharides

Answer: c) They differ only in the configuration about one carbon
atom

Rationale: Epimers are diastereomers that differ in configuration at exactly
one chiral carbon. For example, D-glucose and D-galactose differ only at C-
4, making them C-4 epimers .

,4. D-glucose and D-galactose are best described as:

 a) Anomers
 b) Aldoses
 c) Ketoses
 d) Epimers
 e) Pentoses

Answer: d) Epimers

Rationale: D-glucose and D-galactose differ only in the configuration
around C-4, making them C-4 epimers. Epimers are a specific type of
diastereomer that differ at exactly one chiral center .




5. Which pair is anomeric?

 a) D-glucose and D-fructose
 b) D-glucose and L-fructose
 c) D-glucose and L-glucose
 d) alpha-D-glucose and beta-D-fructose
 e) alpha-D-glucose and beta-D-glucose

Answer: e) alpha-D-glucose and beta-D-glucose

Rationale: Anomers are cyclic sugars that differ only in configuration at the
anomeric carbon (C-1). Alpha and beta forms of the same sugar are
anomers. Epimers differ at other chiral carbons, not the anomeric carbon .

, 6. What is an anomeric carbon?

Answer: The new chiral center formed in ring closure; it was the carbon
containing the carbonyl group in the straight-chain form. It is attached to
an -OR group, -OH group, carbon, and hydrogen .

Rationale: When a linear monosaccharide cyclizes, the carbonyl carbon (C-
1 for aldoses, C-2 for ketoses) becomes a chiral center. This new chiral
center is called the anomeric carbon and can exist in alpha or beta
configurations.




7. Which carbon in a glucose molecule becomes the anomeric carbon
upon ring formation?

Answer: Carbon-1 (C-1), which is at the top of the Fischer structure .

Rationale: For aldoses like glucose, the anomeric carbon is C-1, which was
the aldehyde carbon in the straight-chain form. This carbon reacts with the
C-5 hydroxyl group to form the cyclic hemiacetal.




8. Pyranose rings are formed by 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