Question

Consider the reversible product-formation reaction in an enzyme-catalyzed bioreaction: $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons}(\mathrm{ES}) \stackrel{k_{2}}{\stackrel{k_{3}}{\sum}} \mathrm{E}+\mathrm{P} $$ Develop a rate expression for product-formation using the quasi-steady-state approximation and show that $$ v=\frac{d[\mathrm{P}]}{d t}=\frac{\left(v_{s} / K_{m}\right)[\mathrm{S}]-\left(v_{p} / K_{p}\right)[\mathrm{P}]}{1+\frac{[\mathrm{S}]}{K_{m}}+\frac{[\mathrm{P}]}{K_{p}}} $$ $$ \text { where } K_{m}=\frac{k_{-1}+k_{2}}{k_{1}} \text { and } K_{p}=\frac{k_{-1}+k_{2}}{k_{-2}} \text { and } V_{x}=k_{2}\left[\mathrm{E}_{0}\right], V_{p}=k_{-1}\left[\mathrm{E}_{0}\right] \text {. } $$

Short answer

The rate expression for product formation is: \[ v = \frac{(V_{s}/K_{m})[\mathrm{S}] - (V_{p}/K_{p})[\mathrm{P}]}{1 + (\mathrm{S}/K_{m}) + (\mathrm{P}/K_{p})} \]

Step-by-step solution

- Write Down the Reaction

Consider the reversible product-formation reaction in an enzyme-catalyzed bioreaction: \[ \mathrm{E} + \mathrm{S} \stackrel{k_{1}}{\rightleftharpoons} (\mathrm{ES}) \stackrel{k_{2}}{\stackrel{k_{3}}{\sum}} \mathrm{E} + \mathrm{P} \]

- Apply the Quasi-Steady-State Approximation (QSSA)

Assume that the concentration of the intermediate complex (ES) is at steady state, so that \[ \frac{d[\mathrm{ES}]}{dt} = 0 \]This leads to \[ k_{1}[\mathrm{E}][\mathrm{S}] = (k_{-1} + k_{2})[\mathrm{ES}] \]

- Solve for [ES]

Using the steady-state assumption, rearrange for \([\mathrm{ES}]\):\[ [\mathrm{ES}] = \frac{k_{1}[\mathrm{E}][\mathrm{S}]}{k_{-1} + k_{2}} \]This expression describes the concentration of the enzyme-substrate complex.

- Express [E] in Terms of Total Enzyme Concentration

Use the conservation of total enzyme \([\mathrm{E}_{0}]\):\[ [\mathrm{E}_{0}] = [\mathrm{E}] + [\mathrm{ES}] \]and substituting for \([\mathrm{ES}]\) gives:\[ [\mathrm{E}] = [\mathrm{E}_{0}] - [\mathrm{ES}] \]

- Substitute [E] Back into [ES]

Substitute for \([\mathrm{E}]\) in the expression for \([\mathrm{ES}]\):\[ [\mathrm{ES}] = \frac{k_{1}[\mathrm{E}_{0}][\mathrm{S}]}{k_{-1} + k_{2} + k_{1}[\mathrm{S}]} \]

- Write the Rate Expression for Product Formation

The rate of product formation, \(v\) is given by:\[ v = k_{2}[\mathrm{ES}] \]Substitute for \([\mathrm{ES}]\):\[ v = k_{2} \cdot \frac{k_{1}[\mathrm{E}_{0}][\mathrm{S}]}{k_{-1} + k_{2} + k_{1}[\mathrm{S}]} \]

- Introduce Michaelis Constants and Maximum Velocities

Introduce the Michaelis constants:\[ K_{m} = \frac{k_{-1} + k_{2}}{k_{1}} \text{ and } K_{p} = \frac{k_{-1} + k_{2}}{k_{-2}} \]Define the maximum velocities:\[ V_{s} = k_{2}[\mathrm{E}_{0}] \text{ and } V_{p} = k_{-1}[\mathrm{E}_{0}] \]

- Rewrite the Rate Expression in Terms of the Constants

Rewrite the rate expression using the constants defined earlier:\[ v = \frac{(V_{s}/K_{m})[\mathrm{S}]}{1 + (\mathrm{S}/K_{m}) + (\mathrm{P}/K_{p})} - \frac{(V_{p}/K_{p})[\mathrm{P}]}{1 + (\mathrm{S}/K_{m}) + (\mathrm{P}/K_{p})} \]

- Simplify the Final Rate Expression

Combine the terms to give the final rate expression:\[ v = \frac{(V_{s}/K_{m})[\mathrm{S}] - (V_{p}/K_{p})[\mathrm{P}]}{1 + (\mathrm{S}/K_{m}) + (\mathrm{P}/K_{p})} \]

Key concepts

quasi-steady-state approximation

The quasi-steady-state approximation (QSSA) simplifies the complex equations of enzyme kinetics. It assumes the concentration of the enzyme-substrate complex (ES) remains constant over time. This is valid because the complex forms and breaks down quickly compared to the product formation. Mathematically, we can write this as:
\[\frac{d[\text{ES}]}{dt} \approx 0 \/\].
Using this assumption, the rate of formation and breakdown of the complex can be equated to yield: \[\text{k}_{1}[\text{E}][\text{S}] = (\text{k}_{-1} + \text{k}_{2})[\text{ES}]\]
This simplification allows us to solve for \[\text{[ES]}\] and subsequently understand how the enzyme and substrate concentrations affect the reaction.

Michaelis-Menten kinetics

Michaelis-Menten kinetics describe the rate of enzymatic reactions by incorporating the enzyme, substrate, and the enzyme-substrate complex into a single equation. The key idea is that the formation of the enzyme-substrate complex is in a state of dynamic equilibrium. Using the QSSA, we derived \[\text{[ES]} = \frac{k_1[\text{E}][\text{S}]}{k_{-1} + k_2}\]
With this, the rate of product formation is \[v = k_2[\text{ES}]\].
Introducing the Michaelis constant \[K_m = \frac{k_{-1} + k_2}{k_1}\] simplifies the expression. This results in
\[v = \frac{V_{max}[\text{S}]}{K_m + [\text{S}]}\]
where \[V_{max}\] is the maximum rate, given by \[V_{max} = k_{\text{cat}}[\text{E}_0]\], with \[k_{\text{cat}} = k_2\].

rate expression for product formation

By combining QSSA and Michaelis-Menten kinetics, we derive the rate expression for product formation. Introducing \[K_p = \frac{k_{-1} + k_{2}}{k_{-2}}\] and the velocities \[V_s = k_2[\text{E}_0]\] and \[V_p = k_{-1}[\text{E}_0]\], we obtain:
\[v = \frac{(V_s / K_m)[\text{S}] - (V_p / K_p)[\text{P}]}{1 + (\text{S} / K_m) + (\text{P} / K_p)}\]
This equation combines the substrate concentration, enzyme kinetics, and product formation into a single, elegant expression. It tells us how the rate of product formation is influenced by substrate and product concentrations, the enzyme's efficiency, and the reaction's reversibility.