Question

In the text, the steady-state rate law was derived with the assumption that the reaction is irreversible and/or only the initial velocity was determined. Derive the steady-state rate law for the reversible enzyme reaction: $$ \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{X} \rightleftharpoons \mathrm{E}+\mathrm{P} $$ Show that the rate law can be put into the form $$ v=\frac{\left.V_{\mathrm{S}} / K_{\mathrm{S}} \mid \mathrm{S}\right]-V_{\mathrm{p}} / K_{\mathrm{p}}[\mathrm{P}]}{1+[\mathrm{S}] / K_{\mathrm{S}}+[\mathrm{P}] / K_{\mathrm{p}}} $$ where \(V_{\mathrm{S}}\) and \(V_{\mathrm{P}}\) are the maximum velocities for the forward and reverse reactions, and \(K_{\mathrm{s}}\) and \(K_{\mathrm{p}}\) are the Michaelis constants for the forward and reverse reactions. When equilibrium is reached, \(v=0\). Calculate the ratio of the equilibrium concentrations of \(S\) and \(P,[P] /[S]\), in terms of the four steady-state parameters. This relationship is called the Haldane relationship and is a method for determining the equilibrium constant of the overall reaction.

Short answer

The steady-state rate law is derived to be \[ v=\frac{\left.V_{\mathrm{S}} / K_{\mathrm{S}} \cdot [S]\right]-V_{\mathrm{p}} / K_{\mathrm{p}}\cdot[P]}{1+[\mathrm{S}] / K_{\mathrm{S}}+[\mathrm{P}] / K_{\mathrm{p}}} \]. At equilibrium, \([P]/[S] = \frac{V_SK_P}{V_PK_S}\).

Step-by-step solution

Write the reaction scheme

The reaction involves two reversible steps: the enzyme-substrate complex formation and breakdown: \( E + S \rightleftharpoons X \rightleftharpoons E + P \). Each link in this chain has a forward and backward rate constant associated with it.

Define the rate expressions for each step

The formation and dissociation of the enzyme-substrate complex \(X\) are governed by \( k_1[E][S] - k_{-1}[X] \) and \( k_2[X] - k_{-2}[E][P] \), respectively.

Apply the steady-state approximation

Assume the concentration of the intermediate \(X\) remains constant over time, leading to the equation \( k_1[E][S] - k_{-1}[X] = k_2[X] - k_{-2}[E][P] \).

Solve for the concentration of the intermediate

Rearrange the steady-state equation to isolate \([X]\), yielding \([X] = \frac{k_1[E][S] + k_{-2}[E][P]}{k_{-1} + k_2} \).

Derive the net rate of reaction

Consider the net formation of \(P\), given by \(v = k_2[X] - k_{-2}[E][P]\). Substituting \([X]\) from Step 4 gives \( v = \frac{k_2k_1[E][S] - k_{-1}k_{-2}[E][P]}{k_{-1} + k_2} \).

Simplify using Michaelis-Menten constants

Recognize \(V_S = k_2[E_{total}], V_P = k_{-1}[E_{total}], K_S = \frac{k_{-1} + k_2}{k_1}, K_P = \frac{k_{-1} + k_2}{k_{-2}} \). Substitute in the rate expression so \(v = \frac{\frac{V_S[S]}{K_S} - \frac{V_P[P]}{K_P}}{1 + \frac{[S]}{K_S} + \frac{[P]}{K_P}} \).

Analyze at equilibrium

At equilibrium, the net reaction rate \(v=0\). This gives \(\frac{V_S[S]}{K_S} = \frac{V_P[P]}{K_P}\). Rearranging provides the ratio \([P]/[S] = \frac{V_SK_P}{V_PK_S}\), indicating equilibrium concentrations.

Key concepts

Reversible Enzyme Reactions

In biochemical systems, many enzyme-catalyzed reactions are reversible. This means the enzyme can catalyze both the forward and reverse processes. The general form for a reversible enzyme reaction can be described as \( E + S \rightleftharpoons X \rightleftharpoons E + P \), where \(E\) is the enzyme, \(S\) is the substrate, \(X\) is the enzyme-substrate complex, and \(P\) is the product.

Reversibility in reactions is significant as it allows cells to dynamically control metabolic pathways. The direction a reaction proceeds depends on the concentrations of substrates and products, which can shift under varying cellular conditions.

This reversibility is essential in maintaining homeostasis and energy efficiency within a cell. By facilitating reactions in either direction, enzymes help balance metabolic processes in response to changes in the cell's environment. Understanding reversible enzyme reactions is crucial for grasping the complex regulation of biochemical pathways.

Enzyme-Substrate Complex

The enzyme-substrate complex \(X\) is a transitory structure formed when an enzyme binds to its substrate. This binding is a key step in catalyzing a reaction, as the complex brings the enzyme and substrate close together, enabling the conversion to product.

The formation and breakdown of this complex is described by rate constants: \( k_1[E][S] - k_{-1}[X] \) for the formation and \( k_2[X] - k_{-2}[E][P] \) for breakdown.

These reactions depict how the enzyme temporarily interacts with the substrate to lower the activation energy and speed up the conversion to product.

• The balance of formation and breakdown rates is crucial for regulating reaction speeds.
• It's described by the steady-state approximation, assuming \(X\) remains constant over time.

In summary, the enzyme-substrate complex is essential for catalytic activity, allowing enzymes to efficiently turn substrates into products by forming specific and transient complexes.

Michaelis-Menten Constants

Michaelis-Menten constants are vital parameters that describe the kinetics of enzyme-catalyzed reactions. They are used to express how enzymes interact with substrates and influence reaction rates. Two constants, \(K_S\) and \(K_P\), relate to the reversible reactions under study.

\(K_S\) and \(K_P\) are derived from the rate constants involved in forming and breaking the enzyme-substrate complex: \(K_S = \frac{k_{-1} + k_2}{k_1}\) and \(K_P = \frac{k_{-1} + k_2}{k_{-2}}\).

These constants provide:

• Insight into the affinity between enzyme and substrate or product, indicating the concentration needed to reach half the maximum reaction velocity.
• Knowledge about the efficiency of the enzyme acting on different substrates or products.

In practice, smaller \(K_S\) or \(K_P\) values indicate higher affinity, meaning less substrate or product is needed for effective enzyme activity. These constants are essential for understanding how enzymes behave under various conditions.

Haldane Relationship

The Haldane relationship provides a link between the equilibrium concentrations of substrates and products in reversible enzyme reactions. It relates to the steady-state and maximum reaction velocities of enzyme catalysis, offering a means to determine the equilibrium constant for the transformation of substrates to products.

When the reaction reaches equilibrium (i.e., net rate \(v = 0\)), the expression \(\frac{V_S[S]}{K_S} = \frac{V_P[P]}{K_P}\) holds.

By rearranging this equation, we derive:

• The equilibrium concentrations ratio: \([P]/[S] = \frac{V_SK_P}{V_PK_S}\).
• This ratio directly links the kinetic parameters with equilibrium behavior, integrating dynamic and static aspects of reaction chemistry.

The Haldane relationship is fundamental for biochemists as it offers a deeper understanding of reaction dynamics and equilibrium, crucial for interpreting cellular metabolism and enzyme function.