CHEM 210 — Biochemistry Comprehensive Final Exam 2026/2027
UPDATE
1. Which property of water allows it to act as an efficient solvent for polar
molecules?
A. Its ability to form hydrogen bonds
B. Its high heat of vaporization
C. Its low molecular weight
D. Its tetrahedral geometry
Answer: A
Rationale: Water’s polarity and ability to form hydrogen bonds allow it to interact with
and dissolve other polar substances and ions.
2. Calculate the pH of a solution with a hydrogen ion concentration [H+] of 1.0 x
10^-5 M.
A. 5
B. 7
C. 9
D. 3
Answer: A
Rationale: pH is calculated as -log[H+]. -log(1.0 x 10^-5) equals 5.
,3. Which amino acid contains a sulfur atom in its side chain?
A. Tyrosine
B. Serine
C. Alanine
D. Cysteine
Answer: D
Rationale: Cysteine and Methionine are the two sulfur-containing amino acids commonly
found in proteins.
4. What is the primary force stabilizing the alpha-helix secondary structure of
proteins?
A. Disulfide bridges
B. Hydrogen bonding between backbone atoms
C. Hydrophobic interactions
D. Ionic bonds between R-groups
Answer: B
Rationale: Alpha-helices are stabilized by hydrogen bonds between the carbonyl oxygen of
one amino acid and the amino hydrogen of another four residues away.
5. Which of the following describes a non-competitive inhibitor?
A. It binds only to the active site.
B. It increases the Km of the enzyme.
C. It can be overcome by increasing substrate concentration.
D. It binds to a site other than the active site and decreases Vmax.
Answer: D
Rationale: Non-competitive inhibitors bind to an allosteric site, reducing the enzyme’s
turnover number (Vmax) regardless of substrate concentration.
, 6. Hemoglobin’s affinity for oxygen decreases when which of the following
increases?
A. pH
B. Carbon monoxide concentration
C. 2,3-Bisphosphoglycerate (2,3-BPG)
D. Oxygen partial pressure
Answer: C
Rationale: 2,3-BPG binds to the center of the hemoglobin tetramer, stabilizing the T-state
(deoxy) and reducing oxygen affinity.
7. An enzyme with a low Km value has which of the following?
A. Low catalytic efficiency
B. Low affinity for its substrate
C. High Vmax
D. High affinity for its substrate
Answer: D
Rationale: Km represents the substrate concentration at half-maximal velocity; a lower
value indicates the enzyme reaches half-saturation at lower concentrations, implying
higher affinity.
8. Which carbohydrate is a non-reducing sugar?
A. Glucose
B. Lactose
C. Maltose
D. Sucrose
Answer: D
Rationale: Sucrose consists of glucose and fructose linked by their anomeric carbons,
leaving no free aldehyde or ketone group to act as a reducing agent.
UPDATE
1. Which property of water allows it to act as an efficient solvent for polar
molecules?
A. Its ability to form hydrogen bonds
B. Its high heat of vaporization
C. Its low molecular weight
D. Its tetrahedral geometry
Answer: A
Rationale: Water’s polarity and ability to form hydrogen bonds allow it to interact with
and dissolve other polar substances and ions.
2. Calculate the pH of a solution with a hydrogen ion concentration [H+] of 1.0 x
10^-5 M.
A. 5
B. 7
C. 9
D. 3
Answer: A
Rationale: pH is calculated as -log[H+]. -log(1.0 x 10^-5) equals 5.
,3. Which amino acid contains a sulfur atom in its side chain?
A. Tyrosine
B. Serine
C. Alanine
D. Cysteine
Answer: D
Rationale: Cysteine and Methionine are the two sulfur-containing amino acids commonly
found in proteins.
4. What is the primary force stabilizing the alpha-helix secondary structure of
proteins?
A. Disulfide bridges
B. Hydrogen bonding between backbone atoms
C. Hydrophobic interactions
D. Ionic bonds between R-groups
Answer: B
Rationale: Alpha-helices are stabilized by hydrogen bonds between the carbonyl oxygen of
one amino acid and the amino hydrogen of another four residues away.
5. Which of the following describes a non-competitive inhibitor?
A. It binds only to the active site.
B. It increases the Km of the enzyme.
C. It can be overcome by increasing substrate concentration.
D. It binds to a site other than the active site and decreases Vmax.
Answer: D
Rationale: Non-competitive inhibitors bind to an allosteric site, reducing the enzyme’s
turnover number (Vmax) regardless of substrate concentration.
, 6. Hemoglobin’s affinity for oxygen decreases when which of the following
increases?
A. pH
B. Carbon monoxide concentration
C. 2,3-Bisphosphoglycerate (2,3-BPG)
D. Oxygen partial pressure
Answer: C
Rationale: 2,3-BPG binds to the center of the hemoglobin tetramer, stabilizing the T-state
(deoxy) and reducing oxygen affinity.
7. An enzyme with a low Km value has which of the following?
A. Low catalytic efficiency
B. Low affinity for its substrate
C. High Vmax
D. High affinity for its substrate
Answer: D
Rationale: Km represents the substrate concentration at half-maximal velocity; a lower
value indicates the enzyme reaches half-saturation at lower concentrations, implying
higher affinity.
8. Which carbohydrate is a non-reducing sugar?
A. Glucose
B. Lactose
C. Maltose
D. Sucrose
Answer: D
Rationale: Sucrose consists of glucose and fructose linked by their anomeric carbons,
leaving no free aldehyde or ketone group to act as a reducing agent.
