CHEM 210 MODULE 2 EXAM 2026/2027 | Water, pH, Buffers &
Intermolecular Forces | Geneva College / Portage Learning |
Verified Q&A | Pass Guaranteed - A+ Graded
Q1: According to the IUBMB enzyme classification system, which enzyme class
catalyzes reactions involving the cleavage of C-C, C-O, C-N, and other bonds by means
other than hydrolysis or oxidation?
A. Hydrolases
B. Transferases
C. Lyases [CORRECT]
D. Isomerases
Correct Answer: C
Rationale: Lyases (Class 4) catalyze the cleavage of C-C, C-O, C-N, and other bonds by
elimination, leaving double bonds or rings, or conversely adding groups to double
bonds. This differs from hydrolysis (which uses water) or oxidation-reduction. A is
incorrect because hydrolases cleave bonds using water (hydrolysis). B is incorrect
because transferases move functional groups between molecules. D is incorrect
because isomerases catalyze structural rearrangements within a single molecule
without changing its molecular formula.
Q2: In the catalytic mechanism of chymotrypsin, which amino acid residue in the
catalytic triad acts as a general base to abstract a proton from Ser195 during the
nucleophilic attack on the substrate?
A. Asp102
B. His57 [CORRECT]
,C. Ser195
D. Gly193
Correct Answer: B
Rationale: His57 functions as the general base, abstracting the proton from Ser195 to
generate a more nucleophilic alkoxide ion for attack on the carbonyl carbon of the
peptide bond. The imidazole ring of histidine has a pKa (~7) suitable for proton transfer
at physiological pH. A is incorrect because Asp102 orients His57 and stabilizes its
positive charge but does not directly abstract the proton. C is incorrect because Ser195
is the nucleophile that donates the proton, not accepts it. D is incorrect because Gly193
is part of the oxyanion hole and plays no role in proton abstraction.
Q3: The following kinetic data were obtained for an enzyme-catalyzed reaction:
TableCopy
[S] (mM) v₀ (μmol/min)
0.5 25
1.0 40
2.0 57
4.0 74
8.0 88
, Calculate the approximate Km value for this enzyme.
A. 0.5 mM
B. 1.0 mM [CORRECT]
C. 2.0 mM
D. 4.0 mM
Correct Answer: B
Rationale: At [S] = Km, v₀ = Vmax/2. From the data, Vmax ≈ 100 μmol/min (approaching
asymptotically). Half-Vmax (50 μmol/min) occurs between [S] = 0.5 mM (v₀ = 25) and [S]
= 2.0 mM (v₀ = 57). Using the Michaelis-Menten equation and interpolation, or plotting
1/v vs 1/[S], the x-intercept (-1/Km) indicates Km ≈ 1.0 mM. A is incorrect because at
0.5 mM, v₀ is only 25 (Vmax/4). C is incorrect because at 2.0 mM, v₀ = 57 (> Vmax/2). D
is incorrect because 4.0 mM gives v₀ = 74, approaching saturation but well above Km.
Q4: Which of the following represents the correct Lineweaver-Burk (double reciprocal)
transformation of the Michaelis-Menten equation?
A. 1/v₀ = (Km/Vmax)(1/[S]) + 1/Vmax [CORRECT]
B. 1/v₀ = (Vmax/Km)(1/[S]) + Km
C. v₀ = Vmax[S]/(Km + [S])
D. v₀/Vmax = [S]/(Km + [S])
Correct Answer: A
Rationale: Taking reciprocals of the Michaelis-Menten equation (v₀ = Vmax[S]/(Km +
[S])) yields 1/v₀ = (Km/Vmax)(1/[S]) + 1/Vmax. This linear form has slope = Km/Vmax,
y-intercept = 1/Vmax, and x-intercept = -1/Km. B is incorrect because it inverts the slope
relationship and uses incorrect intercepts. C is incorrect because it is the original
Michaelis-Menten equation, not the double-reciprocal form. D is incorrect because it
represents the fractional saturation form, not the Lineweaver-Burk transformation.
Intermolecular Forces | Geneva College / Portage Learning |
Verified Q&A | Pass Guaranteed - A+ Graded
Q1: According to the IUBMB enzyme classification system, which enzyme class
catalyzes reactions involving the cleavage of C-C, C-O, C-N, and other bonds by means
other than hydrolysis or oxidation?
A. Hydrolases
B. Transferases
C. Lyases [CORRECT]
D. Isomerases
Correct Answer: C
Rationale: Lyases (Class 4) catalyze the cleavage of C-C, C-O, C-N, and other bonds by
elimination, leaving double bonds or rings, or conversely adding groups to double
bonds. This differs from hydrolysis (which uses water) or oxidation-reduction. A is
incorrect because hydrolases cleave bonds using water (hydrolysis). B is incorrect
because transferases move functional groups between molecules. D is incorrect
because isomerases catalyze structural rearrangements within a single molecule
without changing its molecular formula.
Q2: In the catalytic mechanism of chymotrypsin, which amino acid residue in the
catalytic triad acts as a general base to abstract a proton from Ser195 during the
nucleophilic attack on the substrate?
A. Asp102
B. His57 [CORRECT]
,C. Ser195
D. Gly193
Correct Answer: B
Rationale: His57 functions as the general base, abstracting the proton from Ser195 to
generate a more nucleophilic alkoxide ion for attack on the carbonyl carbon of the
peptide bond. The imidazole ring of histidine has a pKa (~7) suitable for proton transfer
at physiological pH. A is incorrect because Asp102 orients His57 and stabilizes its
positive charge but does not directly abstract the proton. C is incorrect because Ser195
is the nucleophile that donates the proton, not accepts it. D is incorrect because Gly193
is part of the oxyanion hole and plays no role in proton abstraction.
Q3: The following kinetic data were obtained for an enzyme-catalyzed reaction:
TableCopy
[S] (mM) v₀ (μmol/min)
0.5 25
1.0 40
2.0 57
4.0 74
8.0 88
, Calculate the approximate Km value for this enzyme.
A. 0.5 mM
B. 1.0 mM [CORRECT]
C. 2.0 mM
D. 4.0 mM
Correct Answer: B
Rationale: At [S] = Km, v₀ = Vmax/2. From the data, Vmax ≈ 100 μmol/min (approaching
asymptotically). Half-Vmax (50 μmol/min) occurs between [S] = 0.5 mM (v₀ = 25) and [S]
= 2.0 mM (v₀ = 57). Using the Michaelis-Menten equation and interpolation, or plotting
1/v vs 1/[S], the x-intercept (-1/Km) indicates Km ≈ 1.0 mM. A is incorrect because at
0.5 mM, v₀ is only 25 (Vmax/4). C is incorrect because at 2.0 mM, v₀ = 57 (> Vmax/2). D
is incorrect because 4.0 mM gives v₀ = 74, approaching saturation but well above Km.
Q4: Which of the following represents the correct Lineweaver-Burk (double reciprocal)
transformation of the Michaelis-Menten equation?
A. 1/v₀ = (Km/Vmax)(1/[S]) + 1/Vmax [CORRECT]
B. 1/v₀ = (Vmax/Km)(1/[S]) + Km
C. v₀ = Vmax[S]/(Km + [S])
D. v₀/Vmax = [S]/(Km + [S])
Correct Answer: A
Rationale: Taking reciprocals of the Michaelis-Menten equation (v₀ = Vmax[S]/(Km +
[S])) yields 1/v₀ = (Km/Vmax)(1/[S]) + 1/Vmax. This linear form has slope = Km/Vmax,
y-intercept = 1/Vmax, and x-intercept = -1/Km. B is incorrect because it inverts the slope
relationship and uses incorrect intercepts. C is incorrect because it is the original
Michaelis-Menten equation, not the double-reciprocal form. D is incorrect because it
represents the fractional saturation form, not the Lineweaver-Burk transformation.
