An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol -min¹ when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme? Vmax = μmol-min-¹

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Q23:

**Enzyme Kinetics Problem**

An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol·min⁻¹ when the substrate concentration is five times greater than the \( K_M \). What is the \( V_{max} \) of this enzyme?

\[ V_{max} = \, \_\_\_\_\_\_ \, \text{µmol·min⁻¹} \]

**Explanation:**

This problem involves calculating the maximum reaction velocity (\( V_{max} \)) of an enzyme using the Michaelis-Menten equation:

\[
v = \frac{V_{max} \cdot [S]}{K_M + [S]}
\]

Where:
- \( v \) is the initial reaction velocity (10 µmol·min⁻¹).
- \( [S] \) is the substrate concentration, which is five times the \( K_M \).

To solve for \( V_{max} \), plug in the given values:

\[
10 = \frac{V_{max} \cdot 5K_M}{K_M + 5K_M}
\]

Simplify the expression:

\[
10 = \frac{5V_{max} \cdot K_M}{6K_M}
\]

\[
10 = \frac{5V_{max}}{6}
\]

Multiply both sides by 6:

\[
60 = 5V_{max}
\]

Divide by 5:

\[
V_{max} = 12 \, \text{µmol·min⁻¹}
\]

So, \( V_{max} \) is 12 µmol·min⁻¹.
Transcribed Image Text:**Enzyme Kinetics Problem** An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol·min⁻¹ when the substrate concentration is five times greater than the \( K_M \). What is the \( V_{max} \) of this enzyme? \[ V_{max} = \, \_\_\_\_\_\_ \, \text{µmol·min⁻¹} \] **Explanation:** This problem involves calculating the maximum reaction velocity (\( V_{max} \)) of an enzyme using the Michaelis-Menten equation: \[ v = \frac{V_{max} \cdot [S]}{K_M + [S]} \] Where: - \( v \) is the initial reaction velocity (10 µmol·min⁻¹). - \( [S] \) is the substrate concentration, which is five times the \( K_M \). To solve for \( V_{max} \), plug in the given values: \[ 10 = \frac{V_{max} \cdot 5K_M}{K_M + 5K_M} \] Simplify the expression: \[ 10 = \frac{5V_{max} \cdot K_M}{6K_M} \] \[ 10 = \frac{5V_{max}}{6} \] Multiply both sides by 6: \[ 60 = 5V_{max} \] Divide by 5: \[ V_{max} = 12 \, \text{µmol·min⁻¹} \] So, \( V_{max} \) is 12 µmol·min⁻¹.
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