To find the **Km** (Michaelis constant) from the given enzyme kinetics data, recall that **Km** is the substrate concentration \([S]\) at which the reaction rate (\(V_0\)) is **half of Vmax**. From the table: - At high \([S]\), \(V_0\) plateaus at 110 μM/min, so **Vmax ≈ 110 μM/min**. - Half of Vmax is **110 / 2 = 55 μM/min**. Looking at the data: - When \([S] = 20\) μM, \(V_0 = 55\) μM/min. So, **Km ≈ 20 μM**. **Correct answer:** - 20 μM

To determine the Km (Michaelis constant) from the given enzyme kinetics data, we need to understand that **Km** is defined as the substrate concentration \([S]\) at which the initial reaction velocity \(V_0\) is **half of the maximum velocity (Vmax)**. **Step-by-step explanation:** 1. **Identify Vmax:** - From the data, as \([S]\) increases, the initial velocity \(V_0\) approaches a plateau. - The highest observed \(V_0\) is **110 μM/min**, so we take this as an approximation of **Vmax**. 2. **Calculate half of Vmax:** - Half of Vmax = \(110 / 2 = 55\, \text{μM/min}\). 3. **Find the substrate concentration at which \(V_0 \approx 55\, \text{μM/min}\):** - Looking at the data: - At \([S] = 20\, \text{μM}\), \(V_0 = 55\, \text{μM/min}\). 4. **Interpretation:** - Since at \([S] = 20\, \text{μM}\), \(V_0\) is exactly half of Vmax, this indicates: - **Km ≈ 20 μM**. **Conclusion:** The Michaelis constant **Km** for this reaction is approximately **20 μM** because at this substrate concentration, the initial velocity is about half of the maximum velocity observed.