CHEM 301 Biochemistry Midterm ACTUAL
EXAM 2026/2027 | 80 Q&A | Verified Q&A |
Pass Guaranteed - A+ Graded
SECTION A: MULTIPLE CHOICE (Questions 1–50)
Q1: Which amino acid has a side chain that forms a covalent disulfide bond in proteins?
A. Methionine
B. Cysteine [CORRECT]
C. Serine
D. Lysine
Correct Answer: B
Rationale: Cysteine contains a thiol (-SH) group that oxidizes to form disulfide bonds (-S-S-), stabilizing
tertiary and quaternary protein structure. Why A is wrong: Methionine has a thioether group (sulfur in
an ether linkage) that cannot form disulfide bonds. Why C is wrong: Serine has a hydroxyl (-OH) group
capable of hydrogen bonding but not disulfide formation. Why D is wrong: Lysine has an amino group
(positive charge at physiological pH) with no sulfur atom.
Verified Source: Lehninger, 8th ed., Ch. 3 – Amino Acids
Topic: Protein Structure – Disulfide Bonds
Q2: A Lineweaver-Burk plot shows lines intersecting on the 1/[S] axis (x-axis). What type of inhibition is
this?
A. Competitive [CORRECT]
B. Noncompetitive
C. Uncompetitive
D. Mixed
Correct Answer: A
Rationale: Competitive inhibition: the inhibitor binds the active site; Vmax remains unchanged (same y-
intercept = 1/Vmax), but Km increases (steeper slope). Lines intersect on the x-axis because -1/Km
changes while 1/Vmax stays constant. Why B is wrong: Noncompetitive inhibition would show lines
,intersecting on the y-axis (1/Vmax changes, Km unchanged). Why C is wrong: Uncompetitive inhibition
yields parallel lines (both Vmax and Km decrease proportionally). Why D is wrong: Mixed inhibition
shows lines intersecting somewhere between the axes.
Verified Source: Lehninger, Ch. 6 – Enzyme Kinetics, Fig. 6.15
Topic: Enzyme Inhibition – Graph Interpretation
Q3: Which of the following amino acids is classified as aromatic and contains a phenolic hydroxyl group?
A. Phenylalanine
B. Tyrosine [CORRECT]
C. Tryptophan
D. Histidine
Correct Answer: B
Rationale: Tyrosine has a benzene ring with a hydroxyl group (phenol), making it both aromatic and
capable of phosphorylation. Why A is wrong: Phenylalanine is aromatic but lacks the phenolic -OH (just
a benzyl group). Why C is wrong: Tryptophan is aromatic (indole ring) but has no phenolic hydroxyl.
Why D is wrong: Histidine is basic and heteroaromatic (imidazole ring) but not phenolic.
Verified Source: Lehninger, Ch. 3; Voet & Voet, Ch. 4
Topic: Amino Acid Classification
Q4: The α-helix is stabilized primarily by:
A. Disulfide bonds between cysteine residues
B. Hydrogen bonds between carbonyl oxygen (i) and amide hydrogen (i+4) [CORRECT]
C. Ionic interactions between acidic and basic side chains
D. Hydrophobic interactions between nonpolar side chains
Correct Answer: B
Rationale: The α-helix features intrachain hydrogen bonds between the carbonyl oxygen of residue i and
the amide hydrogen of residue i+4, creating 3.6 residues per turn. Why A is wrong: Disulfide bonds
stabilize tertiary/quaternary structure, not the repeating helical pattern. Why C is wrong: Salt bridges
contribute to tertiary stability but are not the primary helix-stabilizing force. Why D is wrong:
Hydrophobic interactions drive tertiary folding and membrane protein insertion but do not define the α-
helix backbone pattern.
Verified Source: Lehninger, Ch. 4 – Protein Structure
Topic: Secondary Structure – α-Helix
,Q5: A patient presents with chronic fatigue, muscle pain, and dark urine after exercise. A muscle biopsy
shows absent phosphorylase activity. Which metabolic pathway is primarily affected?
A. Glycolysis
B. Glycogenolysis [CORRECT]
C. Gluconeogenesis
D. β-Oxidation
Correct Answer: B
Rationale: This describes McArdle disease (Type V glycogen storage disease) caused by muscle glycogen
phosphorylase deficiency, blocking glycogen breakdown (glycogenolysis). Why A is wrong: Glycolysis
proceeds normally if glucose is available; the defect is upstream glucose release from glycogen. Why C is
wrong: Gluconeogenesis occurs primarily in liver; muscle lacks glucose-6-phosphatase and cannot
perform gluconeogenesis. Why D is wrong: β-Oxidation of fatty acids is unaffected; symptoms worsen
during exercise precisely because fatty acid oxidation cannot compensate fast enough for the missing
glycogenolytic glucose.
Verified Source: Lehninger, Ch. 15 – Glycogen Metabolism
Topic: Clinical Correlation – Metabolic Disease
Q6: In the Michaelis-Menten equation, Km is defined as:
A. The maximum velocity of the reaction
B. The substrate concentration at which V0 = ½ Vmax [CORRECT]
C. The rate constant for product formation
D. The enzyme concentration required for catalysis
Correct Answer: B
Rationale: Km (Michaelis constant) is operationally defined as the substrate concentration yielding half-
maximal velocity. It approximates the dissociation constant (Kd) when kcat << k-1. Why A is wrong:
Vmax is the maximum velocity, not Km. Why C is wrong: kcat (turnover number) represents the rate
constant for product formation from the ES complex. Why D is wrong: Enzyme concentration affects
Vmax but is not Km; Km is an intrinsic property of the enzyme-substrate pair.
Verified Source: Lehninger, Ch. 6 – Enzyme Kinetics
Topic: Enzyme Kinetics – Definitions
Q7: Which of the following is a nonpolar, aliphatic amino acid?
A. Asparagine
B. Valine [CORRECT]
C. Cysteine
D. Glutamic acid
, Correct Answer: B
Rationale: Valine has an isopropyl side chain (pure hydrocarbon, no functional groups), making it
nonpolar and hydrophobic. Why A is wrong: Asparagine has an amide group (polar, uncharged). Why C
is wrong: Cysteine has a thiol group (polar, reactive, capable of hydrogen bonding). Why D is wrong:
Glutamic acid has a carboxylate group (negatively charged, highly polar).
Verified Source: Lehninger, Ch. 3 – Amino Acid Properties
Topic: Amino Acid Classification – Hydrophobicity
Q8: The quaternary structure of hemoglobin consists of:
A. Four identical α-subunits
B. Two α and two β subunits [CORRECT]
C. A single polypeptide chain with four domains
D. Two identical β-subunits only
Correct Answer: B
Rationale: Adult hemoglobin (HbA) is a tetramer: two α chains (141 aa) and two β chains (146 aa), each
with a heme prosthetic group. Why A is wrong: Four identical subunits describe some enzymes (e.g.,
lactate dehydrogenase in some isoforms) but not hemoglobin. Why C is wrong: That describes tertiary
structure with domains, not quaternary assembly of separate chains. Why D is wrong: Hemoglobin
requires both α and β chains for cooperative oxygen binding.
Verified Source: Lehninger, Ch. 5 – Protein Function; Voet & Voet, Ch. 9
Topic: Quaternary Structure – Hemoglobin
Q9: Which type of enzyme inhibition can be overcome by increasing substrate concentration?
A. Noncompetitive
B. Uncompetitive
C. Competitive [CORRECT]
D. Mixed
Correct Answer: C
Rationale: Competitive inhibitors bind the active site; sufficiently high [S] can outcompete the inhibitor,
restoring Vmax (though more substrate is needed). Why A is wrong: Noncompetitive inhibitors bind
elsewhere; increasing [S] cannot overcome binding to a non-active site. Why B is wrong: Uncompetitive
inhibitors bind only the ES complex; increasing [S] actually worsens inhibition by generating more ES.
Why D is wrong: Mixed inhibition affects both Vmax and Km; substrate increase only partially
overcomes inhibition.
Verified Source: Lehninger, Ch. 6 – Enzyme Inhibition
Topic: Inhibition Mechanisms
EXAM 2026/2027 | 80 Q&A | Verified Q&A |
Pass Guaranteed - A+ Graded
SECTION A: MULTIPLE CHOICE (Questions 1–50)
Q1: Which amino acid has a side chain that forms a covalent disulfide bond in proteins?
A. Methionine
B. Cysteine [CORRECT]
C. Serine
D. Lysine
Correct Answer: B
Rationale: Cysteine contains a thiol (-SH) group that oxidizes to form disulfide bonds (-S-S-), stabilizing
tertiary and quaternary protein structure. Why A is wrong: Methionine has a thioether group (sulfur in
an ether linkage) that cannot form disulfide bonds. Why C is wrong: Serine has a hydroxyl (-OH) group
capable of hydrogen bonding but not disulfide formation. Why D is wrong: Lysine has an amino group
(positive charge at physiological pH) with no sulfur atom.
Verified Source: Lehninger, 8th ed., Ch. 3 – Amino Acids
Topic: Protein Structure – Disulfide Bonds
Q2: A Lineweaver-Burk plot shows lines intersecting on the 1/[S] axis (x-axis). What type of inhibition is
this?
A. Competitive [CORRECT]
B. Noncompetitive
C. Uncompetitive
D. Mixed
Correct Answer: A
Rationale: Competitive inhibition: the inhibitor binds the active site; Vmax remains unchanged (same y-
intercept = 1/Vmax), but Km increases (steeper slope). Lines intersect on the x-axis because -1/Km
changes while 1/Vmax stays constant. Why B is wrong: Noncompetitive inhibition would show lines
,intersecting on the y-axis (1/Vmax changes, Km unchanged). Why C is wrong: Uncompetitive inhibition
yields parallel lines (both Vmax and Km decrease proportionally). Why D is wrong: Mixed inhibition
shows lines intersecting somewhere between the axes.
Verified Source: Lehninger, Ch. 6 – Enzyme Kinetics, Fig. 6.15
Topic: Enzyme Inhibition – Graph Interpretation
Q3: Which of the following amino acids is classified as aromatic and contains a phenolic hydroxyl group?
A. Phenylalanine
B. Tyrosine [CORRECT]
C. Tryptophan
D. Histidine
Correct Answer: B
Rationale: Tyrosine has a benzene ring with a hydroxyl group (phenol), making it both aromatic and
capable of phosphorylation. Why A is wrong: Phenylalanine is aromatic but lacks the phenolic -OH (just
a benzyl group). Why C is wrong: Tryptophan is aromatic (indole ring) but has no phenolic hydroxyl.
Why D is wrong: Histidine is basic and heteroaromatic (imidazole ring) but not phenolic.
Verified Source: Lehninger, Ch. 3; Voet & Voet, Ch. 4
Topic: Amino Acid Classification
Q4: The α-helix is stabilized primarily by:
A. Disulfide bonds between cysteine residues
B. Hydrogen bonds between carbonyl oxygen (i) and amide hydrogen (i+4) [CORRECT]
C. Ionic interactions between acidic and basic side chains
D. Hydrophobic interactions between nonpolar side chains
Correct Answer: B
Rationale: The α-helix features intrachain hydrogen bonds between the carbonyl oxygen of residue i and
the amide hydrogen of residue i+4, creating 3.6 residues per turn. Why A is wrong: Disulfide bonds
stabilize tertiary/quaternary structure, not the repeating helical pattern. Why C is wrong: Salt bridges
contribute to tertiary stability but are not the primary helix-stabilizing force. Why D is wrong:
Hydrophobic interactions drive tertiary folding and membrane protein insertion but do not define the α-
helix backbone pattern.
Verified Source: Lehninger, Ch. 4 – Protein Structure
Topic: Secondary Structure – α-Helix
,Q5: A patient presents with chronic fatigue, muscle pain, and dark urine after exercise. A muscle biopsy
shows absent phosphorylase activity. Which metabolic pathway is primarily affected?
A. Glycolysis
B. Glycogenolysis [CORRECT]
C. Gluconeogenesis
D. β-Oxidation
Correct Answer: B
Rationale: This describes McArdle disease (Type V glycogen storage disease) caused by muscle glycogen
phosphorylase deficiency, blocking glycogen breakdown (glycogenolysis). Why A is wrong: Glycolysis
proceeds normally if glucose is available; the defect is upstream glucose release from glycogen. Why C is
wrong: Gluconeogenesis occurs primarily in liver; muscle lacks glucose-6-phosphatase and cannot
perform gluconeogenesis. Why D is wrong: β-Oxidation of fatty acids is unaffected; symptoms worsen
during exercise precisely because fatty acid oxidation cannot compensate fast enough for the missing
glycogenolytic glucose.
Verified Source: Lehninger, Ch. 15 – Glycogen Metabolism
Topic: Clinical Correlation – Metabolic Disease
Q6: In the Michaelis-Menten equation, Km is defined as:
A. The maximum velocity of the reaction
B. The substrate concentration at which V0 = ½ Vmax [CORRECT]
C. The rate constant for product formation
D. The enzyme concentration required for catalysis
Correct Answer: B
Rationale: Km (Michaelis constant) is operationally defined as the substrate concentration yielding half-
maximal velocity. It approximates the dissociation constant (Kd) when kcat << k-1. Why A is wrong:
Vmax is the maximum velocity, not Km. Why C is wrong: kcat (turnover number) represents the rate
constant for product formation from the ES complex. Why D is wrong: Enzyme concentration affects
Vmax but is not Km; Km is an intrinsic property of the enzyme-substrate pair.
Verified Source: Lehninger, Ch. 6 – Enzyme Kinetics
Topic: Enzyme Kinetics – Definitions
Q7: Which of the following is a nonpolar, aliphatic amino acid?
A. Asparagine
B. Valine [CORRECT]
C. Cysteine
D. Glutamic acid
, Correct Answer: B
Rationale: Valine has an isopropyl side chain (pure hydrocarbon, no functional groups), making it
nonpolar and hydrophobic. Why A is wrong: Asparagine has an amide group (polar, uncharged). Why C
is wrong: Cysteine has a thiol group (polar, reactive, capable of hydrogen bonding). Why D is wrong:
Glutamic acid has a carboxylate group (negatively charged, highly polar).
Verified Source: Lehninger, Ch. 3 – Amino Acid Properties
Topic: Amino Acid Classification – Hydrophobicity
Q8: The quaternary structure of hemoglobin consists of:
A. Four identical α-subunits
B. Two α and two β subunits [CORRECT]
C. A single polypeptide chain with four domains
D. Two identical β-subunits only
Correct Answer: B
Rationale: Adult hemoglobin (HbA) is a tetramer: two α chains (141 aa) and two β chains (146 aa), each
with a heme prosthetic group. Why A is wrong: Four identical subunits describe some enzymes (e.g.,
lactate dehydrogenase in some isoforms) but not hemoglobin. Why C is wrong: That describes tertiary
structure with domains, not quaternary assembly of separate chains. Why D is wrong: Hemoglobin
requires both α and β chains for cooperative oxygen binding.
Verified Source: Lehninger, Ch. 5 – Protein Function; Voet & Voet, Ch. 9
Topic: Quaternary Structure – Hemoglobin
Q9: Which type of enzyme inhibition can be overcome by increasing substrate concentration?
A. Noncompetitive
B. Uncompetitive
C. Competitive [CORRECT]
D. Mixed
Correct Answer: C
Rationale: Competitive inhibitors bind the active site; sufficiently high [S] can outcompete the inhibitor,
restoring Vmax (though more substrate is needed). Why A is wrong: Noncompetitive inhibitors bind
elsewhere; increasing [S] cannot overcome binding to a non-active site. Why B is wrong: Uncompetitive
inhibitors bind only the ES complex; increasing [S] actually worsens inhibition by generating more ES.
Why D is wrong: Mixed inhibition affects both Vmax and Km; substrate increase only partially
overcomes inhibition.
Verified Source: Lehninger, Ch. 6 – Enzyme Inhibition
Topic: Inhibition Mechanisms
