What is the concentration of enzyme- substrate complex by applying steady state approximation ?

Question

Accepted Answer

To determine the concentration of the enzyme-substrate complex $[ES]$ using the steady-state approximation, we need to analyze the given reaction mechanism:

1. $E + S \xrightleftharpoons[k_{a}']{k_{a}} ES$
2. $ES \xrightarrow{k_{b}} P$

The steady-state approximation assumes that the concentration of the enzyme-substrate complex $[ES]$ remains constant over time. This implies that the rate of formation of $[ES]$ is equal to the rate of its breakdown.

**Step-by-step Explanation:**

1. **Formation of $[ES]$:**
   The rate of formation of $[ES]$ is given by the forward reaction:
   $$
   \text{Rate of formation} = k_{a}[E][S]
   $$

2. **Breakdown of $[ES]$:**
   The breakdown of $[ES]$ occurs via two pathways:
   - The reverse reaction to form $E + S$: $k_{a}'[ES]$
   - The conversion to product $P$: $k_{b}[ES]$

Therefore, the total rate of breakdown of $[ES]$ is:
   $$
   \text{Rate of breakdown} = k_{a}'[ES] + k_{b}[ES]
   $$

3. **Steady-State Approximation:**
   According to the steady-state approximation, the rate of formation equals the rate of breakdown:
   $$
   k_{a}[E][S] = k_{a}'[ES] + k_{b}[ES]
   $$

Simplifying, we get:
   $$
   k_{a}[E][S] = (k_{a}' + k_{b})[ES]
   $$

Solving for $[ES]$, we find:
   $$
   [ES] = \frac{k_{a}[E][S]}{k_{a}' + k_{b}}
   $$

This derivation matches Option 2: $(\frac{k_{a}}{k_{a}' + k_{b}})[E][S]$.

**Explanation of Incorrect Options:**

1. **Option 1: $(\frac{k_{b}}{k_{a}' + k_{a}})[E][S]$**
   - This option incorrectly uses $k_{b}$ in the numerator, which represents the rate constant for the conversion of $[ES]$ to product $P$. The numerator should represent the formation of $[ES]$, which is governed by $k_{a}$.

2. **Option 3: $(\frac{k_{a}}{k_{a}'k_{b}})[E][S]$**
   - This option incorrectly uses the product $k_{a}'k_{b}$ in the denominator. The denominator should be the sum $k_{a}' + k_{b}$, representing the total breakdown rate of $[ES]$.

3. **Option 4: $(\frac{k_{a}'}{k_{a} + k_{b}})[E][S]$**
   - This option incorrectly places $k_{a}'$ in the numerator, which represents the reverse reaction rate constant. The numerator should be $k_{a}$, representing the forward reaction rate constant for the formation of $[ES]$.

In conclusion, Option 2 is the correct answer as it accurately represents the concentration of the enzyme-substrate complex $[ES]$ under the steady-state approximation.