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(a) From the following mechanism, derive Eq. 19a, which Michaelis and Menten proposed to represent the rate of formation of products in an enzyme-catalyzed reaction. (b) Show that the rate is independent of substrate concentration at high concentrations of substrate. $$ \begin{aligned} &\mathrm{E}+\mathrm{S} \rightleftarrows \mathrm{ES} \quad k_{1}, k_{1}^{\prime} \\ &\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad k_{2} \end{aligned} $$ where \(E\) is the free enzyme, \(S\) is the substrate, ES is the enzyme-substrate complex, and \(P\) is the product. Note that the steady-state concentration of free enzyme will be equal to the initial concentration of the enzyme less the amount of enzyme that is present in the enzyme-substrate complex: \([\mathrm{E}]=[\mathrm{E}]_{0}-[\mathrm{ES}]\)

Short Answer

Expert verified
The Michaelis-Menten equation, derived from the proposed mechanism, is \(v = \frac{V_{max}[S]}{K_m + [S]}\). At high substrate concentrations, the rate of product formation becomes independent of substrate concentration and approaches the maximum rate \(V_{max}\).

Step by step solution

01

Understanding the Mechanism

The given mechanism is a simplified model of enzyme kinetics, where the enzyme (E) binds to the substrate (S) to form an enzyme-substrate complex (ES). This complex then converts to the product (P) and releases the enzyme back. The rate constants for formation and breakdown of the complex are denoted by \(k_1\) (forward) and \(k_1'\) (reverse), and the rate constant for the formation of product from the enzyme-substrate complex is denoted by \(k_2\).
02

Applying the Steady-State Approximation

To derive the Michaelis-Menten equation, we apply the steady-state approximation, which assumes that the formation rate of the ES complex is equal to its breakdown rate. Mathematically, this can be expressed as \(k_1[E][S] = k_1'[ES] + k_2[ES]\). The concentration of free enzyme [E] can be expressed in terms of the initial enzyme concentration [E]_0 and the concentration of the ES complex [ES].
03

Deriving the Rate Equation

To find the rate of product formation, we focus on the conversion of ES to product. This rate is given by \(v = k_2[ES]\). Using the steady-state approximation, we can express [ES] in terms of [E]_0 and [S], and substitute it into the rate equation.
04

Solving for [ES]

To solve for [ES], we rearrange the steady-state equation to \([ES] = \frac{k_1[E][S]}{k_1' + k_2}\). Substituting [E] with \([E]_0 - [ES]\), we get \([ES] = \frac{k_1([E]_0 - [ES])[S]}{k_1' + k_2}\). Solving for [ES] provides us with the expression to substitute back into the rate equation.
05

Obtaining Michaelis-Menten Equation

Substituting [ES] in the rate equation and simplifying, we obtain the Michaelis-Menten equation: \(v = \frac{V_{max}[S]}{K_m + [S]}\), where \(V_{max} = k_2[E]_0\) is the maximum rate, and \(K_m = \frac{k_1' + k_2}{k_1}\) is the Michaelis constant.
06

Analyzing High Substrate Concentration Limit

At high substrate concentrations ([S] >> K_m), the \([S]\) term in the denominator becomes much larger than \(K_m\). This simplifies the Michaelis-Menten equation to \(v \approx \frac{V_{max}[S]}{[S]} = V_{max}\). This indicates that the rate is independent of substrate concentration and the reaction is limited by the turnover number of the enzyme.

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Key Concepts

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

Enzyme Kinetics
Enzyme kinetics is the study of the rates at which enzyme-catalyzed reactions proceed and the factors affecting them. By understanding kinetics, we can infer details about the mechanism of the enzyme action. Enzymes speed up reactions by lowering the activation energy, enabling the transformation of substrates into products more efficiently. The Michaelis-Menten equation is a cornerstone in enzyme kinetics, describing how the rate of product formation varies with substrate concentration under steady-state conditions.

Core to enzyme kinetics is the concept of enzyme-substrate (ES) complex formation and breakdown. The enzyme binds with its substrate, forming an ES complex, which then transforms into the product, releasing the enzyme for further catalysis. These steps are associated with specific rate constants that quantify the speed of each reaction. Understanding these principles is essential to grasping the broader topic of enzyme kinetics.
Steady-State Approximation
The steady-state approximation is a simplification used in enzyme kinetics to assume that the concentration of the enzyme-substrate complex (ES) remains constant over time. It implies that the rate of formation of the ES complex is equal to its rate of conversion to product and enzyme.

Applying this assumption allows us to derive the Michaelis-Menten equation by focusing on the critical steps of enzyme action without analyzing every intermediate state. Although it is an approximation, this concept is crucial for understanding enzyme kinetics in a manageable and practical way, especially in cases where detailed mechanisms of the enzyme's action are not known.
Rate of Reaction
The rate of reaction in enzyme kinetics refers to the speed at which product forms from the substrate in the presence of an enzyme. It's usually expressed as the concentration of product formed per unit time. In the Michaelis-Menten equation, the rate (\(v\)) of enzyme-catalyzed reaction is closely allied with the concentration of the substrate (\(S\)) and enzyme-substrate complex (\(ES\)).

The initial rate of reaction is of particular interest because it reflects the enzyme activity before any significant changes in substrate concentration occur, allowing us to make comparisons under various experimental conditions. Analyzing the rate at which reactions occur reveals the efficiency and capacity of the enzyme, which is essential for many applications in biochemistry and medicine.
Enzyme-Substrate Complex
The enzyme-substrate (ES) complex is an intermediate formed when an enzyme binds to its substrate molecule(s). The formation of the ES complex is reversible and is the first step in the catalytic action of the enzyme. The ES complex is critical because the catalytic event that leads to product formation occurs while the substrate is bound within the complex.

Under the Michaelis-Menten framework, the concentration of the ES complex is a pivot point, connecting the substrate concentration with the rate of reaction. The ability to measure or infer the concentration of this complex lends itself to understanding the catalytic efficiency and affinity of the enzyme for its substrate.
Maximum Rate (Vmax)
The maximum rate (\(V_{max}\)) is the rate achieved by the enzyme-catalyzed reaction when the enzyme is fully saturated with substrate. At this point, all active sites of the enzyme molecules are occupied, and the rate of reaction is at its peak, limited only by the turnover rate of the enzyme.

In the context of the Michaelis-Menten equation, knowing the value of the maximum rate is important because it reflects the intrinsic catalytic power of the enzyme, independent of substrate concentration. It helps in determining enzyme efficiency and is essential for characterizing enzyme behavior in a biochemical reaction.
Michaelis Constant (Km)
The Michaelis constant (\(K_m\)) is a key parameter in enzyme kinetics representing the substrate concentration at which the reaction rate is half of the maximum rate (\(V_{max}\)). It's indicative of the affinity of the enzyme for its substrate; a low \(K_m\) value means high affinity, as the enzyme reaches half of its maximum rate at a low substrate concentration.

\(K_m\) is derived mathematically under the steady-state conditions of enzyme kinetics and is equivalent to the substrate concentration when the rate of formation of the ES complex equals the rate of its breakdown to enzyme and product. This parameter is fundamental in comparing the efficiency of different enzymes or the behavior of an enzyme towards different substrates.

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Most popular questions from this chapter

Write the overall reaction for the mechanism proposed below and identify any reaction intermediates. $$ \begin{aligned} &\text { Step } 1 \mathrm{AC}+\mathrm{B} \longrightarrow \mathrm{AB}+\mathrm{C} \\ &\text { Step } 2 \mathrm{AC}+\mathrm{AB} \longrightarrow \mathrm{A}_{2} \mathrm{~B}+\mathrm{C} \end{aligned} $$

The half-life for the second-order reaction of a substance A is \(50.5 \mathrm{~s}\) when \([\mathrm{A}]_{0}=0.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) one- sixteenth; (b) onefourth; (c) one-fifth of its original value.

The decomposition of A has the rate law Rate \(=k[\mathrm{~A}]^{a}\). Show that for this reaction the ratio \(t_{1 / 2} / t_{3 / 4}\), where \(t_{1 / 2}\) is the halflife and \(t_{3 / 4}\) is the time for the concentration of A to decrease to \(\frac{3}{4}\) of its initial concentration, can be written as a function of \(a\) alone and can therefore be used to make a quick assessment of the order of the reaction in A.

(a) Using a graphing calculator or standard graphing software, such as that on the Web site for this book, make an appropriate Arrhenius plot of the data shown here for the decomposition of iodoethane into ethene and hydrogen iodide, \(\mathrm{C}_{2} \mathrm{H} \mathrm{H}_{5} \mathrm{I}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{HI}(\mathrm{g})\), and determine the activation energy for the reaction. (b) What is the value of the rate constant at \(400^{\circ} \mathrm{C}\) ? $$ \begin{array}{lcccc} T(\mathbf{K}) & 660 & 680 & 720 & 760 \\ k\left(\mathbf{s}^{-1}\right) & 7.2 \times 10^{-4} & 2.2 \times 10^{-3} & 1.7 \times 10^{-2} & 0.11 \end{array} $$

A reaction was believed to occur by the following mechanism. Step \(1 \mathrm{~A}_{2} \longrightarrow \mathrm{A}+\mathrm{A}\) Step \(2 \mathrm{~A}+\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{A}_{2} \mathrm{~B}\) Step \(3 \mathrm{~A}_{2} \mathrm{~B}+\mathrm{C} \longrightarrow \mathrm{A}_{2}+\mathrm{BC}\) (a) Write the overall reaction. (b) Write the rate law for each step and indicate its molecularity. (c) What are the reaction intermediates? (d) A catalyst is a substance that accelerates the rate of a reaction and is regenerated in the process. What is the catalyst in the reaction?

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