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Consider the following mechanism for the enzymecatalyzed reaction: $$ \begin{array}{r} \mathrm{E}+\mathrm{S} \underset{k-1}{\stackrel{k_{1}}{\sum_{-1}}} \mathrm{ES} \\\ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \end{array} $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of forward reaction is equal to the rate of the reverse reaction.

Short Answer

Expert verified
The rate law for the given enzymatic reaction, in terms of the concentrations of enzyme (E) and substrate (S), is given by \(d[P]/dt = (k_1 k_2 / k_{-1})[E]T[S]\).

Step by step solution

01

Identify the reaction mechanism

The reaction mechanism is given as follows: Enzyme (E) and substrate (S) combine reversibly to form an enzyme-substrate complex (ES) with forward rate constant \(k_1\) and reverse rate constant \(k_{-1}\). This ES complex then breaks down irreversibly into enzyme (E) and product (P) with rate constant \(k_2\). This second step is the rate determining step of the reaction.
02

Write rate expressions for individual steps

The rate of formation of ES complex can be written as \(-d[E]/dt = k_1[E][S] - k_{-1}[ES]\) and the rate of formation of product P is \(d[P]/dt = k_2[ES]\). Here, [E], [S], [ES] and [P] represent the concentrations of enzyme, substrate, enzyme-substrate complex and product respectively.
03

Establish equilibrium condition for ES formation and breakdown

At equilibrium, the rate of formation and the rate of breakdown of ES complex will be equal, i.e., \(k_1[E][S] = k_{-1}[ES]\), leading to the expression for [ES]: \([ES] = (k_1 / k_{-1})[E][S]\).
04

Substitute [ES] in the rate of product formation

Substitute the expression for [ES] from Step 3 into the rate expression \(d[P]/dt = k_2[ES]\) obtained in Step 2. This gives \(d[P]/dt = k_2(k_1 / k_{-1})[E][S] = (k_1 k_2 / k_{-1})[E][S]\).
05

Simplify the rate law

The above can be simplified assuming that the concentration of the free enzyme [E] is much less than the total enzyme concentration [E]T, as most of the enzyme is in the form of the ES complex. So, we can replace [E] in the rate expression with [E]T, and the rate law for the enzymatic reaction is then given by \(d[P]/dt = (k_1 k_2 / k_{-1})[E]T[S]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law
A rate law is an equation that represents how the rate of a chemical reaction depends on the concentration of reactants. In enzyme kinetics, the rate law helps us understand how quickly an enzyme-catalyzed reaction proceeds. This is expressed in terms of the concentration of substrate (the molecule that interacts with the enzyme) and enzyme.

For example, consider a simple enzymatic reaction scenario where an enzyme (E) binds with a substrate (S) to form an enzyme-substrate complex (ES). This is reversible and is represented by:
  • Forward reaction with rate constant \(k_1\): \( E + S \rightarrow ES \)
  • Reverse reaction with rate constant \(k_{-1}\): \( ES \rightarrow E + S \)
To find the rate law, we focus on the rate-determining step, which is the breaking down of ES into enzyme and product (P) with a rate constant \(k_2\).

The rate of product formation can be expressed as \(d[P]/dt = k_2[ES]\). Substitution of the ES concentration simplifies the expression, resulting in a rate law that describes how changes in the concentrations of the reactants affect reaction speed.
Enzyme-Substrate Complex
The enzyme-substrate complex (ES) is central to understanding enzyme reactions. It forms when an enzyme temporarily binds to its substrate. This step is fundamental as it allows the enzyme to catalyze the conversion of the substrate into the product.

In our reaction mechanism:\[ E + S \rightleftharpoons ES \rightarrow E + P \]
the ES complex is the intermediary step that facilitates the reaction. The creation of ES from E and S can be expressed by the equilibrium equation \(k_1[E][S] = k_{-1}[ES]\).

The breaking down of the ES complex into E and P is often the slow and rate-limiting step in the process, making its study crucial for understanding reaction kinetics.
  • This step determines the overall rate of the reaction.
  • Monitoring the concentration of ES helps predict the reaction's progress.
When writing rate equations, the concentration of ES significantly affects the rate of product formation due to its role in facilitating the catalytic process.
Michaelis-Menten Mechanism
The Michaelis-Menten mechanism is a model that describes the kinetics of many enzyme-catalyzed reactions. This model assumes the formation of an enzyme-substrate complex as a crucial step.

The basic steps are described as follows:
  • Enzyme (E) binds to substrate (S), reversibly, forming enzyme-substrate complex (ES).
  • The ES complex then breaks down to release the product (P) and regenerate the enzyme.
This mechanism simplifies the derivation of a rate law. It leads to the Michaelis-Menten equation: \[v = \frac{{V_{max}[S]}}{{K_M + [S]}}\]where:
  • \(v\) is the reaction rate.
  • \(V_{max}\) is the maximum rate achieved by the system, at maximum substrate saturation.
  • \(K_M\) is the Michaelis constant, representing the substrate concentration at which the reaction rate is half of \(V_{max}\).
This equation provides insights into the effects of substrate concentration on the rate of reaction. As substrate concentration ([S]) increases, the reaction rate approaches \(V_{max}\), illustrating saturation kinetics—a hallmark of many enzyme-catalyzed reactions.

Understanding the Michaelis-Menten mechanism allows biochemists to predict how changes in enzyme concentration, substrate availability, and other conditions affect reaction rates.

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Most popular questions from this chapter

The integrated rate law for the zero-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-k t .\) (a) Sketch the follow- ing plots: (i) rate versus \([\mathrm{A}]_{t}\) and (ii) \([\mathrm{A}]_{t}\) versus \(t\). (b) Derive an expression for the half-life of the reaction. (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when \([\mathrm{A}]_{t}=0\)

When a mixture of methane and bromine is exposed to visible light, the following reaction occurs slowly: $$ \mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g) $$ Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

The following gas-phase reaction was studied at \(290^{\circ} \mathrm{C}\) by observing the change in pressure as a function of time in a constant-volume vessel: $$ \mathrm{ClCO}_{2} \mathrm{CCl}_{3}(g) \longrightarrow 2 \mathrm{COCl}_{2}(g) $$ Determine the order of the reaction and the rate constant based on the following data: $$ \begin{array}{rr} \hline \text { Time (s) } & \boldsymbol{P} \text { (mmHg) } \\ \hline 0 & 15.76 \\ 181 & 18.88 \\ 513 & 22.79 \\ 1164 & 27.08 \\ \hline \end{array} $$ where \(P\) is the total pressure.

Radioactive plutonium-239 \(\left(t_{\frac{1}{2}}=2.44 \times 10^{5} \mathrm{yr}\right)\) is used in nuclear reactors and atomic bombs. If there are \(5.0 \times 10^{2} \mathrm{~g}\) of the isotope in a small atomic bomb, how long will it take for the substance to decay to \(1.0 \times 10^{2} \mathrm{~g},\) too small an amount for an effective bomb?

Determine the molecularity and write the rate law for each of the following elementary steps: (a) \(\mathrm{X} \longrightarrow\) products (b) \(\mathrm{X}+\mathrm{Y} \longrightarrow\) products (c) \(\mathrm{X}+\mathrm{Y}+\mathrm{Z} \longrightarrow\) products (d) \(\mathrm{X}+\mathrm{X} \longrightarrow\) products (e) \(\mathrm{X}+2 \mathrm{Y} \longrightarrow\) products

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