For the reaction: $E+S \xrightarrow[Ka']{Ka} ES$ and $ES \xrightarrow{Kb} P$. 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. $E + S \xrightarrow[Ka']{Ka} ES$
2. $ES \xrightarrow{Kb} P$

The steady-state approximation assumes that the formation and breakdown of the enzyme-substrate complex $[ES]$ reach a dynamic equilibrium, meaning the rate of formation of $[ES]$ is equal to the rate of its breakdown.

### Explanation for Option 2:

**Correct Answer: $(\frac{Ka}{Ka'+Kb})[E][S]$**

- **Formation of $[ES]$:** The rate of formation of the enzyme-substrate complex is given by the forward reaction rate: $v_{\text{formation}} = Ka [E][S]$.
  
- **Breakdown of $[ES]$:** The breakdown of $[ES]$ occurs via two pathways:
  - The reverse reaction to form $E + S$ with rate $Ka' [ES]$.
  - The conversion to product $P$ with rate $Kb [ES]$.

- **Steady-State Assumption:** At steady state, the rate of formation of $[ES]$ equals the rate of its breakdown:
  $$
  Ka [E][S] = Ka' [ES] + Kb [ES]
  $$

- **Solving for $[ES]$:**
  $$
  Ka [E][S] = (Ka' + Kb) [ES]
  $$
  $$
  [ES] = \frac{Ka [E][S]}{Ka' + Kb}
  $$

This derivation matches Option 2, confirming it as the correct answer.

### Explanation for Why Other Options are Incorrect:

1. **Option 1: $(\frac{Kb}{Ka'+Ka})[E][S]$**
   - This option incorrectly places $Kb$ in the numerator, which represents the rate constant for the conversion of $[ES]$ to product $P$. The numerator should reflect the formation rate constant $Ka$, not the breakdown rate constant $Kb$.

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

3. **Option 4: $(\frac{Ka'}{Ka+Kb})[E][S]$**
   - This option incorrectly places $Ka'$ in the numerator, which represents the reverse reaction rate constant. The numerator should reflect the forward reaction rate constant $Ka$.

In conclusion, Option 2 is the correct answer because it accurately represents the steady-state concentration of the enzyme-substrate complex $[ES]$ using the appropriate rate constants and their roles in the reaction mechanism.