Question

Consider the following reaction sequence: $$ \mathrm{S}+\mathrm{E} \underset{k_{2}}{\rightleftharpoons}(\mathrm{ES})_{1} \rightleftharpoons_{k_{4}}^{k_{3}}(\mathrm{ES})_{2} \stackrel{k_{g}}{\longrightarrow} \mathrm{P}+\mathrm{E} $$ Develop a suitable rate expression for production formation \(\left[v=k_{3}(\mathrm{ES})_{2}\right]\) by using (a) the equilibrium approach, and (b) the quasi-steady-state approach.

Short answer

Equilibrium: \( v = k_3 K_1 K_2 [S][E] \) Quasi-steady-state: \( v = \frac{k_3^2 k_1 [S][E]}{k_4 (k_2 + k_3)} \)

Step-by-step solution

- Define equilibrium constants for sequestration reactions (S and E form complexes)

For the reaction sequence, establish equilibrium constants: For the first equilibrium: \[ K_1 = \frac{[ES]_1}{[S][E]} \] For the second equilibrium: \[ K_2 = \frac{[ES]_2}{[ES]_1} \]

- Express complexes in terms of reactants and intermediates under equilibrium approach

Using equilibrium constants, rewrite concentrations of intermediate complexes as: \[ [ES]_1 = K_1 [S][E] \] \[ [ES]_2 = K_2 [ES]_1 = K_2 K_1 [S][E] \]

- Develop rate expression using equilibrium concentrations

Under equilibrium, the rate of product formation is: \[ v = k_3 [ES]_2 = k_3 (K_2 K_1 [S][E]) \] Thus, the rate expression is: \[ v = k_3 K_1 K_2 [S][E] \]

- Define rate equations under quasi-steady-state approach

Under the quasi-steady-state approach, assume intermediate complexes change slowly: \[ \frac{d[ES]_1}{dt} = k_1[S][E] - k_2[ES]_1 - k_3[ES]_1 = 0 \] \[ \frac{d[ES]_2}{dt} = k_3[ES]_1 - k_4[ES]_2 = 0 \]

- Solve quasi-steady-state equations

From \(\frac{d[ES]_1}{dt} = 0\), we get: \[ [ES]_1 = \frac{k_1[S][E]}{k_2 + k_3} \] From \(\frac{d[ES]_2}{dt} = 0\), we get: \[ [ES]_2 = \frac{k_3[ES]_1}{k_4} = \frac{k_3 k_1 [S][E]}{k_4(k_2 + k_3)} \]

- Develop rate expression using quasi-steady-state concentrations

Under quasi-steady-state, the rate of product formation is: \[ v = k_3 [ES]_2 = \frac{k_3^2 k_1 [S][E]}{k_4 (k_2 + k_3)} \] Thus, the rate expression is: \[ v = \frac{k_3^2 k_1 [S][E]}{k_4 (k_2 + k_3)} \]

Key concepts

Equilibrium Approach

The equilibrium approach is a method to simplify the rate expression for enzyme-catalyzed reactions. In this method, all intermediate steps are considered to be in equilibrium.
This implies that the forward and reverse reactions occur at the same rate.
This provides a steady concentration of intermediate complexes.
For the given reaction sequence: $$\text{S}+\text{E} \rightleftharpoons (\text{ES})_1 \rightleftharpoons (\text{ES})_2 \rightarrow \text{P}+\text{E}$$ we define equilibrium constants for each step:

• For the first equilibrium: \( K_1 = \frac{[\text{ES}]_1}{[\text{S}][\text{E}]} \)


• For the second equilibrium: \( K_2 = \frac{[\text{ES}]_2}{[\text{ES}]_1} \)

Using these, we can express the concentrations of intermediate complexes in terms of reactants and other intermediates.
Thus, we get: \( [\text{ES}]_1 = K_1 [\text{S}][\text{E}] \)
and \( [\text{ES}]_2 = K_2 K_1 [\text{S}][\text{E}] \).
By substituting these into the rate expression for product formation: \( v = k_3 [\text{ES}]_2 = k_3 K_1 K_2 [\text{S}][\text{E}] \).
This provides the rate expression under the equilibrium approach.

Quasi-Steady-State Approach

The quasi-steady-state approach is another method often used in enzyme kinetics.
This approach assumes that the concentrations of intermediate complexes change very slowly over time. As a result, their rate of formation is roughly equal to their rate of breakdown. For the same reaction sequence:

• \text{d[\text{ES}]_1}/\text{dt} = k_1[\text{S}][\text{E}] - k_2[\text{ES}]_1 - k_3[\text{ES}]_1 = 0


• \text{d[\text{ES}]_2}/\text{dt} = k_3[\text{ES}]_1 - k_4[\text{ES}]_2 = 0

From these, we first solve for [\text{ES}]_1:
\( [\text{ES}]_1 = \frac{k_1[\text{S}][\text{E}]}{k_2 + k_3} \).
Then, solving for [\text{ES}]_2, we get:
\( [\text{ES}]_2 = \frac{k_3 [\text{ES}]_1}{k_4} = \frac{k_3 k_1 [\text{S}][\text{E}]}{k_4 (k_2 + k_3)} \).
Inserting these into the rate of product formation, we find:
\( v = k_3 [\text{ES}]_2 = \frac{k_3^2 k_1 [\text{S}][\text{E}]}{k_4 (k_2 + k_3)} \).
This provides the rate expression under the quasi-steady-state approach.

Rate Expression

A rate expression provides a mathematical way to describe the speed of a reaction.
For enzyme-catalyzed reactions, this typically involves product formation.
It connects the concentration of substrate and enzyme to the rate at which the product forms.
In our case, for reaction sequence: $$\text{S}+\text{E} \rightleftharpoons (\text{ES})_1 \rightleftharpoons (\text{ES})_2 \rightarrow \text{P}+\text{E}$$ we found two different rate expressions under different assumptions.
Using the equilibrium approach, the rate expression is:
\( v = k_3 K_1 K_2 [\text{S}][\text{E}] \).
Under the quasi-steady-state approach, it's slightly different:
\( v = \frac{k_3^2 k_1 [\text{S}][\text{E}]}{k_4 (k_2 + k_3)} \).
Both expressions highlight how enzyme and substrate concentrations influence the reaction rate uniquely.

Intermediate Complexes

Intermediate complexes are temporary structures formed during a chemical reaction.
In enzyme kinetics, these complexes form between the enzyme and substrate.
They play a critical role in determining the reaction rate.
For our reaction sequence: $$\text{S}+\text{E} \rightleftharpoons (\text{ES})_1 \rightleftharpoons (\text{ES})_2 \rightarrow \text{P}+\text{E}$$, we have two main intermediates: (\text{ES})_1 and (\text{ES})_2.

• (\text{ES})_1 is the initial complex formed by the enzyme and substrate.
• (\text{ES})_2 is a secondary complex that eventually breaks down to produce the product.

Understanding these complexes is key to comprehending the overall reaction.
Each step involving these intermediates can be analyzed to derive rate expressions using different kinetic approaches, like the equilibrium or quasi-steady-state methods.
The transitions between these intermediate states determine the reaction's efficiency and rate.