CHEM 210 BIOCHEMISTRY MODULE 6 EXAM ACTUAL
2026/2027 | Carbohydrate Metabolism, Glycolysis,
Gluconeogenesis & Regulation | Geneva College | Complete
Questions & Verified Answers | Pass Guaranteed - A+ Graded
[Section 1: Carbohydrates & Glycobiology (Q1-12)]
Q1. In a Fischer projection of an aldose, the hydroxyl group on the lowest-numbered
chiral carbon (C-5 in aldohexoses) is oriented to the right. This stereochemical
designation is:
A. L-configuration
B. D-configuration
C. Racemic mixture
D. Meso compound
Correct Answer: B. D-configuration [CORRECT]
Rationale: Naturally occurring monosaccharides are typically D-configuration, defined by
the hydroxyl on the right at the highest-numbered chiral carbon. A describes the mirror
image, C describes equal enantiomer mixtures, and D describes an achiral molecule
with internal symmetry.
Correct Answer: B
Q2. A biochemistry student examines the Haworth projection of D-glucopyranose and
observes the hydroxyl group at the anomeric carbon (C-1) is trans to the CH₂OH group
at C-5. This anomeric form is:
A. The β-anomer
B. The open-chain aldehyde form
,C. The α-anomer
D. The L-enantiomer
Correct Answer: C. The α-anomer [CORRECT]
Rationale: In D-glucopyranose, the α-anomer has the C-1 hydroxyl trans (downward) to
the C-6 CH₂OH group, while the β-anomer has it cis (upward). A describes the opposite
anomer, B describes the linear form, and D incorrectly assigns L-configuration to a
D-sugar.
Correct Answer: C
Q3. A food chemist isolates a disaccharide from sugar cane that is non-reducing and
composed of α-D-glucose and β-D-fructose linked through their anomeric carbons. This
disaccharide is:
A. Lactose
B. Maltose
C. Sucrose
D. Cellobiose
Correct Answer: C. Sucrose [CORRECT]
Rationale: Sucrose is uniquely linked via an α(1→2)β glycosidic bond between glucose
and fructose, making both anomeric carbons involved and the molecule non-reducing. A
is a reducing β(1→4) galactose-glucose disaccharide, B is a reducing α(1→4)
glucose-glucose disaccharide, and D is a reducing β(1→4) glucose-glucose
disaccharide.
Correct Answer: C
Q4. A clinical laboratory identifies a reducing disaccharide in a milk sample that
contains β-galactose linked (1→4) to glucose. This disaccharide is:
,A. Sucrose
B. Maltose
C. Lactose
D. Trehalose
Correct Answer: C. Lactose [CORRECT]
Rationale: Lactose consists of β-galactose linked (1→4) to glucose with a free anomeric
carbon on the glucose moiety, making it a reducing sugar. A is non-reducing, B contains
α(1→4) glucose-glucose linkages, and D is a non-reducing α(1→1) glucose-glucose
disaccharide.
Correct Answer: C
Q5. A carbohydrate researcher treats two disaccharides with maltase. Only one is
hydrolyzed. The hydrolyzed disaccharide contains an α(1→4) glycosidic bond, while the
resistant disaccharide contains a β(1→4) linkage. These disaccharides are respectively:
A. Cellobiose and maltose
B. Maltose and cellobiose
C. Sucrose and lactose
D. Lactose and maltose
Correct Answer: B. Maltose and cellobiose [CORRECT]
Rationale: Maltase specifically hydrolyzes α-glycosidic bonds in maltose (α(1→4)),
whereas cellobiose contains β(1→4) linkages resistant to maltase. A reverses the order,
and C and D pair disaccharides with incorrect linkages.
Correct Answer: B
Q6. A glycogen storage disease specialist notes that glycogen has branch points every
8-12 glucose residues, whereas amylopectin has branches every 24-30 residues. This
structural difference means glycogen:
, A. Is indigestible by human enzymes
B. Provides more rapid glucose mobilization due to increased non-reducing ends
C. Contains only linear α(1→4) chains
D. Is composed of β-glucose units
Correct Answer: B. Provides more rapid glucose mobilization due to increased
non-reducing ends [CORRECT]
Rationale: The high branching frequency of glycogen creates numerous non-reducing
ends accessible to glycogen phosphorylase, enabling rapid glucose release. A describes
cellulose, C describes amylose, and D incorrectly assigns β-linkages to glycogen.
Correct Answer: B
Q7. A nutritionist explains to a patient that humans cannot digest dietary cellulose
because human digestive enzymes lack the ability to hydrolyze:
A. α(1→4) glycosidic bonds between glucose units
B. β(1→4) glycosidic bonds between glucose units
C. α(1→6) glycosidic branch points
D. Peptide bonds in associated proteins
Correct Answer: B. β(1→4) glycosidic bonds between glucose units [CORRECT]
Rationale: Human amylases hydrolyze α(1→4) linkages in starch but cannot cleave the
β(1→4) linkages that characterize cellulose, making it indigestible fiber. A describes
starch linkages, C describes branch points in glycogen, and D is irrelevant to
polysaccharide digestion.
Correct Answer: B
Q8. A clinical chemistry student is asked to identify which of the following sugars would
test positive as a reducing sugar in Benedict's assay:
2026/2027 | Carbohydrate Metabolism, Glycolysis,
Gluconeogenesis & Regulation | Geneva College | Complete
Questions & Verified Answers | Pass Guaranteed - A+ Graded
[Section 1: Carbohydrates & Glycobiology (Q1-12)]
Q1. In a Fischer projection of an aldose, the hydroxyl group on the lowest-numbered
chiral carbon (C-5 in aldohexoses) is oriented to the right. This stereochemical
designation is:
A. L-configuration
B. D-configuration
C. Racemic mixture
D. Meso compound
Correct Answer: B. D-configuration [CORRECT]
Rationale: Naturally occurring monosaccharides are typically D-configuration, defined by
the hydroxyl on the right at the highest-numbered chiral carbon. A describes the mirror
image, C describes equal enantiomer mixtures, and D describes an achiral molecule
with internal symmetry.
Correct Answer: B
Q2. A biochemistry student examines the Haworth projection of D-glucopyranose and
observes the hydroxyl group at the anomeric carbon (C-1) is trans to the CH₂OH group
at C-5. This anomeric form is:
A. The β-anomer
B. The open-chain aldehyde form
,C. The α-anomer
D. The L-enantiomer
Correct Answer: C. The α-anomer [CORRECT]
Rationale: In D-glucopyranose, the α-anomer has the C-1 hydroxyl trans (downward) to
the C-6 CH₂OH group, while the β-anomer has it cis (upward). A describes the opposite
anomer, B describes the linear form, and D incorrectly assigns L-configuration to a
D-sugar.
Correct Answer: C
Q3. A food chemist isolates a disaccharide from sugar cane that is non-reducing and
composed of α-D-glucose and β-D-fructose linked through their anomeric carbons. This
disaccharide is:
A. Lactose
B. Maltose
C. Sucrose
D. Cellobiose
Correct Answer: C. Sucrose [CORRECT]
Rationale: Sucrose is uniquely linked via an α(1→2)β glycosidic bond between glucose
and fructose, making both anomeric carbons involved and the molecule non-reducing. A
is a reducing β(1→4) galactose-glucose disaccharide, B is a reducing α(1→4)
glucose-glucose disaccharide, and D is a reducing β(1→4) glucose-glucose
disaccharide.
Correct Answer: C
Q4. A clinical laboratory identifies a reducing disaccharide in a milk sample that
contains β-galactose linked (1→4) to glucose. This disaccharide is:
,A. Sucrose
B. Maltose
C. Lactose
D. Trehalose
Correct Answer: C. Lactose [CORRECT]
Rationale: Lactose consists of β-galactose linked (1→4) to glucose with a free anomeric
carbon on the glucose moiety, making it a reducing sugar. A is non-reducing, B contains
α(1→4) glucose-glucose linkages, and D is a non-reducing α(1→1) glucose-glucose
disaccharide.
Correct Answer: C
Q5. A carbohydrate researcher treats two disaccharides with maltase. Only one is
hydrolyzed. The hydrolyzed disaccharide contains an α(1→4) glycosidic bond, while the
resistant disaccharide contains a β(1→4) linkage. These disaccharides are respectively:
A. Cellobiose and maltose
B. Maltose and cellobiose
C. Sucrose and lactose
D. Lactose and maltose
Correct Answer: B. Maltose and cellobiose [CORRECT]
Rationale: Maltase specifically hydrolyzes α-glycosidic bonds in maltose (α(1→4)),
whereas cellobiose contains β(1→4) linkages resistant to maltase. A reverses the order,
and C and D pair disaccharides with incorrect linkages.
Correct Answer: B
Q6. A glycogen storage disease specialist notes that glycogen has branch points every
8-12 glucose residues, whereas amylopectin has branches every 24-30 residues. This
structural difference means glycogen:
, A. Is indigestible by human enzymes
B. Provides more rapid glucose mobilization due to increased non-reducing ends
C. Contains only linear α(1→4) chains
D. Is composed of β-glucose units
Correct Answer: B. Provides more rapid glucose mobilization due to increased
non-reducing ends [CORRECT]
Rationale: The high branching frequency of glycogen creates numerous non-reducing
ends accessible to glycogen phosphorylase, enabling rapid glucose release. A describes
cellulose, C describes amylose, and D incorrectly assigns β-linkages to glycogen.
Correct Answer: B
Q7. A nutritionist explains to a patient that humans cannot digest dietary cellulose
because human digestive enzymes lack the ability to hydrolyze:
A. α(1→4) glycosidic bonds between glucose units
B. β(1→4) glycosidic bonds between glucose units
C. α(1→6) glycosidic branch points
D. Peptide bonds in associated proteins
Correct Answer: B. β(1→4) glycosidic bonds between glucose units [CORRECT]
Rationale: Human amylases hydrolyze α(1→4) linkages in starch but cannot cleave the
β(1→4) linkages that characterize cellulose, making it indigestible fiber. A describes
starch linkages, C describes branch points in glycogen, and D is irrelevant to
polysaccharide digestion.
Correct Answer: B
Q8. A clinical chemistry student is asked to identify which of the following sugars would
test positive as a reducing sugar in Benedict's assay:
