CHEM 210 MODULE 6 EXAM 2026/2027 | Nucleic Acid
Structure & Function | Geneva College / Portage Learning |
Verified Q&A | Pass Guaranteed - A+ Graded
Time Allowed: 75 minutes
Total Questions: 25
Point Value: 2 points each (50 points total)
Q1: Which functional group distinguishes an aldose from a ketose?
A. Hydroxyl group (-OH)
B. Carbonyl group position (aldehyde vs. ketone). [CORRECT]
C. Carboxylic acid group (-COOH)
D. Amino group (-NH₂)
Correct Answer: B
Rationale: Aldoses contain an aldehyde carbonyl group at C-1 (terminal carbon), while
ketoses contain a ketone carbonyl group at C-2 (internal carbon). This structural
difference affects their chemical reactivity and metabolic pathways. A is incorrect
because both aldoses and ketoses possess multiple hydroxyl groups (polyhydroxy
compounds). C is incorrect because carboxylic acids characterize aldonic acids or
uronic acids (oxidized sugars), not the fundamental aldose/ketose classification. D is
incorrect because amino groups characterize amino sugars (glucosamine,
galactosamine), not the aldose/ketose distinction.
,Q2: In the Fischer projection of D-glyceraldehyde, which configuration correctly
describes the chiral center?
A. The hydroxyl group on the chiral carbon points to the left
B. The hydroxyl group on the chiral carbon points to the right. [CORRECT]
C. The hydroxyl group points upward toward the aldehyde
D. There is no chiral center in glyceraldehyde
Correct Answer: B
Rationale: The D/L system in carbohydrate chemistry is defined by the configuration of
the highest-numbered chiral carbon (farthest from the carbonyl). In D-glyceraldehyde,
the hydroxyl group on C-2 (the only chiral center) points to the right in Fischer
projection. L-glyceraldehyde has the hydroxyl pointing left. This convention extends to
all monosaccharides—those with the bottommost hydroxyl on the right are D-sugars. C
is incorrect because vertical bonds in Fischer projections point away from the viewer
(into the page), not toward functional groups. D is incorrect because glyceraldehyde has
one chiral center (C-2).
Q3: Which monosaccharide is a C-4 epimer of glucose?
A. Galactose. [CORRECT]
B. Fructose
C. Mannose
D. Idose
Correct Answer: A
, Rationale: Epimers are diastereomers that differ in configuration at exactly one chiral
center. Galactose differs from glucose only at C-4 (hydroxyl left vs. right in Fischer
projection), making it the C-4 epimer. Mannose (C) is the C-2 epimer of glucose.
Fructose (B) is a ketose (C-2 carbonyl), not an epimer of glucose (aldohexose). Idose
(D) is an aldohexose but differs from glucose at multiple centers (C-2, C-3, C-4), making
it an epimer of galactose, not glucose.
Q4: Mutarotation refers to which phenomenon?
A. Enzymatic conversion of glucose to fructose
B. The equilibration between α and β anomers in solution via the open-chain form.
[CORRECT]
C. The oxidation of reducing sugars to aldonic acids
D. The polymerization of monosaccharides into polysaccharides
Correct Answer: B
Rationale: Mutarotation is the change in optical rotation observed when a pure anomer
(α or β) is dissolved in water and equilibrates to a mixture of both anomers via the
transient open-chain aldehyde form. For glucose, pure α-D-glucose ([α]D = +112°) or
β-D-glucose ([α]D = +18.7°) mutarotates to the equilibrium mixture ([α]D = +52.7°, ~36%
α, 64% β). A describes isomerization (phosphoglucose isomerase reaction). C describes
oxidation ( Tollens' or Benedict's test). D describes glycoside bond
formation/polymerization.
Q5: Which structural feature makes cellulose resistant to human digestive enzymes?
A. β-1,4-glycosidic linkages between glucose units. [CORRECT]
Structure & Function | Geneva College / Portage Learning |
Verified Q&A | Pass Guaranteed - A+ Graded
Time Allowed: 75 minutes
Total Questions: 25
Point Value: 2 points each (50 points total)
Q1: Which functional group distinguishes an aldose from a ketose?
A. Hydroxyl group (-OH)
B. Carbonyl group position (aldehyde vs. ketone). [CORRECT]
C. Carboxylic acid group (-COOH)
D. Amino group (-NH₂)
Correct Answer: B
Rationale: Aldoses contain an aldehyde carbonyl group at C-1 (terminal carbon), while
ketoses contain a ketone carbonyl group at C-2 (internal carbon). This structural
difference affects their chemical reactivity and metabolic pathways. A is incorrect
because both aldoses and ketoses possess multiple hydroxyl groups (polyhydroxy
compounds). C is incorrect because carboxylic acids characterize aldonic acids or
uronic acids (oxidized sugars), not the fundamental aldose/ketose classification. D is
incorrect because amino groups characterize amino sugars (glucosamine,
galactosamine), not the aldose/ketose distinction.
,Q2: In the Fischer projection of D-glyceraldehyde, which configuration correctly
describes the chiral center?
A. The hydroxyl group on the chiral carbon points to the left
B. The hydroxyl group on the chiral carbon points to the right. [CORRECT]
C. The hydroxyl group points upward toward the aldehyde
D. There is no chiral center in glyceraldehyde
Correct Answer: B
Rationale: The D/L system in carbohydrate chemistry is defined by the configuration of
the highest-numbered chiral carbon (farthest from the carbonyl). In D-glyceraldehyde,
the hydroxyl group on C-2 (the only chiral center) points to the right in Fischer
projection. L-glyceraldehyde has the hydroxyl pointing left. This convention extends to
all monosaccharides—those with the bottommost hydroxyl on the right are D-sugars. C
is incorrect because vertical bonds in Fischer projections point away from the viewer
(into the page), not toward functional groups. D is incorrect because glyceraldehyde has
one chiral center (C-2).
Q3: Which monosaccharide is a C-4 epimer of glucose?
A. Galactose. [CORRECT]
B. Fructose
C. Mannose
D. Idose
Correct Answer: A
, Rationale: Epimers are diastereomers that differ in configuration at exactly one chiral
center. Galactose differs from glucose only at C-4 (hydroxyl left vs. right in Fischer
projection), making it the C-4 epimer. Mannose (C) is the C-2 epimer of glucose.
Fructose (B) is a ketose (C-2 carbonyl), not an epimer of glucose (aldohexose). Idose
(D) is an aldohexose but differs from glucose at multiple centers (C-2, C-3, C-4), making
it an epimer of galactose, not glucose.
Q4: Mutarotation refers to which phenomenon?
A. Enzymatic conversion of glucose to fructose
B. The equilibration between α and β anomers in solution via the open-chain form.
[CORRECT]
C. The oxidation of reducing sugars to aldonic acids
D. The polymerization of monosaccharides into polysaccharides
Correct Answer: B
Rationale: Mutarotation is the change in optical rotation observed when a pure anomer
(α or β) is dissolved in water and equilibrates to a mixture of both anomers via the
transient open-chain aldehyde form. For glucose, pure α-D-glucose ([α]D = +112°) or
β-D-glucose ([α]D = +18.7°) mutarotates to the equilibrium mixture ([α]D = +52.7°, ~36%
α, 64% β). A describes isomerization (phosphoglucose isomerase reaction). C describes
oxidation ( Tollens' or Benedict's test). D describes glycoside bond
formation/polymerization.
Q5: Which structural feature makes cellulose resistant to human digestive enzymes?
A. β-1,4-glycosidic linkages between glucose units. [CORRECT]
