Step 1: Understanding the Question:
This question asks us to identify the graph in which the Michaelis constant ($K_m$) of the enzyme-catalyzed reaction is exactly equal to 20.
Step 2: Key Formula or Approach:
The Michaelis constant ($K_m$) is defined as the substrate concentration $[S]$ at which the initial reaction velocity ($V$) is exactly half of the maximum velocity ($V_{max}$).
\[ V = \frac{V_{max}}{2} \quad \text{when} \quad [S] = K_m \]
We can solve this graphically by:
1. Identifying $V_{max}$ from the plateau of the curve.
2. Calculating $V_{max}/2$.
3. Finding the substrate concentration $[S]$ on the x-axis corresponding to this half-maximal velocity on the y-axis.
Step 3: Detailed Explanation:
Let's analyze the graphs provided:
• In all four plots, the dashed horizontal line representing $V_{max}$ is at $V = 16$.
- Therefore, $V_{max} = 16$ units.
- Half of the maximum velocity is:
\[ \frac{V_{max}}{2} = \frac{16}{2} = 8 \text{ units} \]
• We are looking for the plot where $K_m = 20$. This means that at a substrate concentration $[S] = 20$, the velocity $V$ must be exactly 8.
• Let's check each graph at $[S] = 20$:
- Plot (a): At $[S] = 20$ on the x-axis, follow the grid line vertically to the curve. The corresponding value on the y-axis (velocity) is exactly 8. This fits our condition perfectly. Thus, $K_m = 20$.
- Plot (b): At $[S] = 20$, the velocity is around 4. The velocity reaches 8 at $[S] = 40$. Thus, $K_m = 40$.
- Plot (c): The velocity increases very steeply. At $[S] = 20$, the velocity is already near $V_{max}$ (around 16). The velocity is 8 at a much lower concentration, around $[S] = 5$. Thus, $K_m \approx 5$.
- Plot (d): At $[S] = 20$, the velocity is around 4. The curve is sigmoidal and doesn't hit 8 until a much higher concentration.
Step 4: Final Answer:
Only Plot (a) shows a half-maximal velocity of 8 at $[S] = 20$, so the correct option is (A).