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CHEM120 Biochemsitry - Final Assessment Review - Mod 4 2025

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CHEM120 Biochemsitry - Final Assessment Review - Mod 4 2025CHEM120 Biochemsitry - Final Assessment Review - Mod 4 2025CHEM120 Biochemsitry - Final Assessment Review - Mod 4 2025

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CHEM210 Biochemistry

Final Assessment Review

Module 4 (Questions & Solutions)

2025




©2025

, 1. Case Study – Glycosidic Bond Formation Mechanism:
A researcher studies the enzymatic synthesis of maltose from two
glucose molecules under acid‑catalyzed conditions. Kinetic analysis
shows that the reaction proceeds via the elimination of water and
formation of a new bond between the anomeric carbon of one glucose
and the hydroxyl group at C4 of another.
Question: Which mechanism best describes this glycosidic bond
formation?
A. Base‑catalyzed nucleophilic substitution
B. Acid‑catalyzed dehydration reaction via an oxocarbenium intermediate
C. Radical-mediated polymerization
D. Direct enzymatic oxidation-reduction reaction

ANS: B. Acid‑catalyzed dehydration reaction via an oxocarbenium
intermediate
Rationale: Under acidic conditions, the reaction proceeds by
protonating the carbonyl oxygen to form an oxocarbenium ion;
subsequent nucleophilic attack by a hydroxyl group leads to water loss
and formation of the glycosidic bond.

---

2. Case Study – Anomeric Preferences:
A student analyzes the equilibrium between the open-chain and cyclic
forms of D-glucose in aqueous solution. Using polarimetry, the student
finds that the predominant cyclic form is the β-anomer.
Question: What is the primary reason the β‑anomer of
D‑glucopyranose predominates in solution?
A. The β‑anomer has fewer intramolecular hydrogen bonds.
B. The equatorial orientation of the –OH group at the anomeric carbon
minimizes steric hindrance.
C. The α‑anomer is unstable due to rapid racemization.
D. The β‑anomer is produced enzymatically in the cell.
©2025

, ANS: B. The equatorial orientation of the –OH group at the anomeric
carbon minimizes steric hindrance.
Rationale: In β‑D‑glucopyranose, the anomeric –OH group is equatorial,
which minimizes steric strain relative to the axial position in the
α‑anomer; thus, β‑glucose is thermodynamically favored.

---

3. Case Study – Conformational Preferences of Cyclic Sugars:
During an NMR study, researchers compare the conformations of
D‑glucose in its pyranose form.
Question: Which ring size and conformation is most common for
D‑glucose in solution?
A. Five‑membered furanose ring in a planar conformation
B. Six‑membered pyranose ring in a chair conformation
C. Six‑membered pyranose ring in a boat conformation
D. Five‑membered furanose ring in a skew form

ANS: B. Six‑membered pyranose ring in a chair conformation
Rationale: D‑Glucose predominantly cyclizes to form a six‑membered
pyranose ring; the chair conformation minimizes torsional strain and
steric interactions, making it the most stable form.

---

4. Case Study – Branching in Carbohydrate Polymers:
A biochemist extracts two carbohydrate polymers from plant tissues. One
is found to be linear with minimal branching, and the other exhibits
extensive branching with α(1→6) linkages.
Question: Which polymers are most likely represented by these two
samples?
A. Amylopectin (linear) and Amylose (branched)
B. Amylose (linear) and Amylopectin (branched)
C. Cellulose (linear) and Glycogen (branched)
©2025

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