Given the following data for an enzyme-catalyzed reaction:| [S] (mM) | Initial velocity (\(v_0\)) (μmol/min) ||----------|-----------------------------------|| 0.5 | 2.0 || 1.0 | 3.3 || 2.0 | 4.4 || 5.0 | 5.0 |What is the approximate \(V_{max}\) for this enzyme under these conditions?

• A. 10.0 μmol/min
• B. 3.3 μmol/min
• C. 2.0 μmol/min
• D. 5.0 μmol/min

Answer: (D)

Verified step by step guidance

1. Step 1: Understand the concept of \(V_{max}\). In enzyme kinetics, \(V_{max}\) represents the maximum velocity of the reaction when the enzyme is saturated with substrate. It is the plateau of the Michaelis-Menten curve where increasing substrate concentration no longer increases the reaction rate.
2. Step 2: Analyze the data provided in the table. Look at the relationship between substrate concentration ([S]) and initial velocity (\(v_0\)). Notice that as [S] increases, \(v_0\) approaches a maximum value.
3. Step 3: Identify the highest initial velocity (\(v_0\)) in the table. This value corresponds to the approximate \(V_{max}\) because it represents the point where further increases in [S] do not significantly increase the reaction rate.
4. Step 4: Compare the given answer choices to the highest \(v_0\) value in the table. The correct \(V_{max}\) should match the highest observed velocity.
5. Step 5: Conclude that under these experimental conditions, the approximate \(V_{max}\) is the highest \(v_0\) value observed in the table, which is 5.0 μmol/min.