Multiple Choice
Suppose an enzyme (MW = 5,000 g/mole) has a concentration of 0.05 mg/L. If the kcat is 1 x 104 s-1, what is the theoretical maximum reaction velocity for the enzyme?
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6. Enzymes and Enzyme Kinetics
Calculating Vmax
Multiple Choice
Carbonic anhydrase catalyzes the hydration of CO2. The Km of carbonic anhydrase for CO2 is 12 mM. The initial velocity (V0) of the enzyme-catalyzed reaction was 4.5 μmole*mL-1*sec-1 when [CO2] = 36 mM. Calculate the Vmax of carbonic anhydrase.
A
8.1 x 102 M s-1.
B
6 x 10-3 M s-1.
C
2.5 x 10-4 M s-1.
D
7.3 x 10-5 M s-1.
Verified step by step guidance1
Understand that the problem involves enzyme kinetics, specifically using the Michaelis-Menten equation: V0 = (Vmax * [S]) / (Km + [S]), where V0 is the initial velocity, Vmax is the maximum velocity, [S] is the substrate concentration, and Km is the Michaelis constant.
Identify the given values: Km = 12 mM, V0 = 4.5 μmole*mL^-1*sec^-1, and [CO2] = 36 mM. Note that V0 is given in different units, so it may need conversion to match the units of Vmax.
Rearrange the Michaelis-Menten equation to solve for Vmax: Vmax = V0 * (Km + [S]) / [S].
Substitute the known values into the rearranged equation: Vmax = 4.5 μmole*mL^-1*sec^-1 * (12 mM + 36 mM) / 36 mM.
Convert the units of V0 from μmole*mL^-1*sec^-1 to M s^-1 if necessary, to ensure consistency with the units of Vmax, and then perform the calculation to find Vmax.
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