Equation 13.9 presents the simple Michaelis-Menten situation where the reaction is considered to be irreversible ([P] is negligible). Many enzymatic reactions are reversible, and \(P\) does accumulate. a. Derive an equation for \(v\), the rate of the enzyme-catalyzed reaction \(\mathrm{S} \rightarrow \mathrm{P}\) in terms of a modified Michaelis-Menten model that incorporates the reverse reaction that will occur in the presence of product, \(\mathbf{P}\) b. Solve this modified Michaelis-Menten equation for the special situation when \(v=0\) (that is, \(S \rightleftharpoons P\) is at equilibrium, or in other words, \(\left.K_{\mathrm{eq}}=[\mathrm{P}] /[\mathrm{S}]\right)\) (J. B. S. Haldane first described this reversible Michaelis-Menten modification, and his expression for \(K_{\mathrm{eq}}\) in terms of the modified M-M equation is known as the Haldane relationship.)

Short Answer

Expert verified
The modified Michaelis-Menten equation incorporating the reverse product-forming reaction is \(v = \frac{V_{max}[S]}{K_m + [S]} - \frac{V_{reverse}[P]}{K_m + [P]}\). When \(v=0\), the system is at equilibrium, and the Haldane relationship gives that \(K_{eq} = \frac{[P]}{[S]}\).

Step by step solution

01

Understand the basics

The Michaelis-Menten equation in a simple irreversible situation is usually represented as \(v = \frac{V_{max}[S]}{K_m + [S]}\), where \(v\) is the rate of the reaction, \(V_{max}\) is the maximum rate achieved by the system, \(K_m\) is the Michaelis-Menten constant, and \([S]\) is the concentration of the substrate.
02

Incorporating the reverse reaction

When considering a reversible scenario, product \(P\) can also revert to substrate \(S\). Hence, the rate \(v\) should be the difference between the forward (Product-forming) and reverse (Substrate-forming) reaction rates. This can be represented by \(v = \frac{V_{max}[S]}{K_m + [S]} - \frac{V_{reverse}[P]}{K_m + [P]}\), where \([P]\) is the product concentration and \(V_{reverse}\) is the maximum rate of the reverse reaction.
03

Setting the rate of reaction to zero

When \(v=0\), the system is at equilibrium. Using the modified Michaelis-Menten equation, \(0 = \frac{V_{max}[S]}{K_m + [S]} - \frac{V_{reverse}[P]}{K_m + [P]}\). This can be rearranged to \(\frac{V_{max}[S]}{V_{reverse}[P]} = \frac{K_m + [P]}{K_m + [S]}\). If we set \(\frac{V_{max}}{V_{reverse}}\) to be equal to \(K_{eq}\), which is the equilibrium constant, our equation becomes \(K_{eq} = \frac{[P]}{[S]}\). This relationship between \(K_{eq}\), \([S]\), and \([P]\) is known as the Haldane relationship.

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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 chemical reactions that are catalyzed by enzymes. In biochemistry, understanding how enzymes work and the rates at which they convert substrates into products is fundamental. The rate of an enzyme-catalyzed reaction is affected by factors such as enzyme and substrate concentrations, temperature, and pH.

The most well-known model of enzyme kinetics is the Michaelis-Menten equation, which describes the rate of enzymatic reactions by relating the reaction rate to the concentration of the substrate. This equation assumes that the formation of the enzyme-substrate complex (ES) is a quick process and that the product formation is the rate-limiting step, which is true for many enzymatic reactions.

However, real-world reactions can be more complex, with inhibitors, different substrates, and reversible reactions, among other factors, affecting the rate. The foundation of enzyme kinetics helps us understand these more complex situations by starting with the simple principle that enzyme activity can be expressed quantitatively and modeled mathematically.
Reversible Enzymatic Reactions
Unlike the simple model of enzymatic reactions that are considered irreversible, many biological processes involve reversible enzymatic reactions where the products can convert back into the substrates. This is an important concept because in living cells, the balance between the forward and reverse reactions is essential for homeostasis.

In this case, the modified Michaelis-Menten equation must account for the product's ability to revert back to the substrate. Consequently, we add a term to the equation to account for this reverse process. This modified equation more accurately reflects the dynamic nature of biological systems, where the accumulation of product can shift the reaction in the opposite direction. Knowledge of the reversible reactions is crucial for understanding metabolic pathways and for designing drugs that can either inhibit or enhance specific enzymatic reactions.
Haldane Relationship
The Haldane relationship is an important concept in enzyme kinetics for reversible reactions. It was first described by J.B.S. Haldane and relates the equilibrium constant (\(K_{eq}\)) for the conversion of substrate (S) to product (P) with the enzyme's kinetic parameters.

This relationship provides a link between the enzyme kinetics and the thermodynamics of the reaction. Essentially, it expresses the equilibrium constant in terms of the maximum rates of the forward and reverse reactions (\(V_{max}\) and (\(V_{reverse}\)) respectively), and the Michaelis constants for the substrate and product.

Understanding the Haldane relationship allows biochemists to predict how changes in enzyme activity can affect the equilibrium position of reversible reactions. This insight is vital for the study of metabolic control and the design of biochemical experiments.
Equilibrium Constant
The equilibrium constant (\(K_{eq}\)) is a number that expresses the ratio of the concentration of the products to the concentration of the reactants for a reversible reaction at equilibrium at a given temperature. It is a measure of the tendency of a reversible reaction to proceed to completion and is a key concept in chemical thermodynamics and kinetics.

For enzyme-catalyzed reactions, the equilibrium constant is derived from the rate constants of the forward and reverse reactions. When the reaction has reached equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, so no net change occurs in the concentration of substrates and products.

This is the special situation referred to in the exercise, where the reaction rate (\(v\) is zero, and the ratio of the products to the reactants corresponds to the equilibrium constant. This is crucial to understanding how enzymes work under different conditions and can impact the direction and rate of the reaction in a biological system.

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Most popular questions from this chapter

Enzyme A follows simple Michaelis-Menten kinetics. a. The \(K_{m}\) of enzyme A for its substrate \(\mathrm{S}\) is \(K_{m}^{\mathrm{S}}=1 \mathrm{m} M .\) Enzyme \(\mathrm{A}\) also acts on substrate \(\mathrm{T}\) and its \(K_{m}^{\mathrm{T}}=10 \mathrm{m} M .\) Is \(\mathrm{S}\) or \(\mathrm{T}\) the preferred substrate for enzyme A? b. The rate constant \(k_{2}\) with substrate \(\mathrm{S}\) is \(2 \times 10^{4} \mathrm{sec}^{-1}\); with substrate \(\mathrm{T}, k_{2}=4 \times 10^{5} \mathrm{sec}^{-1} .\) Does enzyme A use substrate S or substrate T with greater catalytic efficiency?

According to the Michaelis-Menten equation, what is the \(v / V_{\max }\) ratio when \([\mathrm{S}]=4 K_{\text {w? }} ?\)

The following kinetic data were obtained for an enzyme in the absence of any inhibitor \((1),\) and in the presence of two different inhibitors (2) and (3) at \(5 \mathrm{m} M\) concentration. Assume \(\left[\mathrm{E}_{T}\right]\) is the same in each experiment. $$\begin{array}{cccc} & (1) & (2) & (3) \\ {[\mathrm{S}]} & v(\mu \mathrm{mol} / & v(\mu \mathrm{mol} /) & v(\mu \mathrm{mol} /) \\ (\mathrm{m} M) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec} \\ 1 & 12 & 4.3 & 5.5 \\ 2 & 20 & 8 & 9 \\ 4 & 29 & 14 & 13 \\ 8 & 35 & 21 & 16 \\ 12 & 40 & 26 & 18 \end{array}$$ Graph these data as Lineweaver-Burk plots and use your graph to find answers to a. and b. a. Determine \(V_{\max }\) and \(K_{m}\) for the enzyme. b. Determine the type of inhibition and the \(K_{1}\) for each inhibitor.

The citric acid cycle enzyme fumarase catalyzes the conversion of fumarate to form malate. $$\text { Fumarate }+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { malate }$$ The turnover number, \(k_{\mathrm{cat}},\) for fumarase is \(800 / \mathrm{sec} .\) The \(K_{m}\) of fumarase for its substrate fumarate is \(5 \mu M\) a. In an experiment using 2 nanomole/L of fumarase, what is \(V_{\max } ?\) b. The cellular concentration of fumarate is \(47.5 \mu M .\) What is \(v\) when [fumarate] \(=47.5 \mu M ?\) c. What is the catalytic efficiency of fumarase? d. Does fumarase approach "catalytic perfection"?

Liver alcohol dehydrogenase (ADH) is relatively nonspecific and will oxidize ethanol or other alcohols, including methanol. Methanol oxidation yields formaldehyde, which is quite toxic, causing, among other things, blindness. Mistaking it for the cheap wine he usually prefers, my dog Clancy ingested about \(50 \mathrm{mL}\) of windshield washer fluid (a solution \(50 \%\) in methanol). Knowing that methanol would be excreted eventually by Clancy's kidneys if its oxidation could be blocked, and realizing that, in terms of methanol oxidation by ADH, ethanol would act as a competitive inhibitor, I decided to offer Clancy some wine. How much of Clancy's favorite vintage \((12 \% \text { ethanol })\) must he consume in order to lower the activity of his ADH on methanol to \(5 \%\) of its normal value if the \(K_{m}\) values of canine ADH for ethanol and methanol are 1 millimolar and 10 millimolar, respectively? (The \(K_{1}\) for ethanol in its role as competitive inhibitor of methanol oxidation by ADH is the same as its \(K_{m \cdot}\) ) Both the methanol and ethanol will quickly distribute throughout Clancy's body fluids, which amount to about 15 L. Assume the densities of \(50 \%\) methanol and the wine are both \(0.9 \mathrm{g} / \mathrm{mL}\).

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