Multiple Choice
For a Michaelis-Menten enzyme, what is the value of Vmax if at 1/10 Km, the V0 = 1 μmol/min.
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6. Enzymes and Enzyme Kinetics
Calculating Vmax
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?
A
1050 µM/s.
B
100 µM/s.
C
150 µM/s.
D
105 µM/s.
Verified step by step guidance1
First, convert the enzyme concentration from mg/L to mol/L. Use the molecular weight (MW) of the enzyme to do this conversion. The formula is: \( \text{Concentration (mol/L)} = \frac{\text{Concentration (mg/L)}}{\text{MW (g/mole)}} \times \frac{1}{1000} \).
Substitute the given values into the formula: \( \text{Concentration (mol/L)} = \frac{0.05 \text{ mg/L}}{5000 \text{ g/mole}} \times \frac{1}{1000} \).
Calculate the enzyme concentration in mol/L using the above formula.
Next, use the Michaelis-Menten equation to find the maximum reaction velocity \( V_{max} \). The equation is: \( V_{max} = k_{cat} \times [E] \), where \( [E] \) is the enzyme concentration in mol/L and \( k_{cat} \) is the turnover number.
Substitute the values of \( k_{cat} = 1 \times 10^4 \text{ s}^{-1} \) and the calculated enzyme concentration into the equation to find \( V_{max} \) in units of \( \mu M/s \).
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