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The Effects of Reversible Inhibitors The MichaelisMenten rate equation for reversible mixed inhibition is written as $$ V_{0}=\frac{V_{\max }[\mathrm{S}]}{\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]} $$ Apparent, or observed, \(K_{\mathrm{m}}\) is equivalent to the [S] at which $$ V_{0}=\frac{V_{\max }}{2 \alpha^{\prime}} $$ Derive an expression for the effect of a reversible inhibitor on apparent \(K_{\mathrm{m}}\) from the previous equation.

Short Answer

Expert verified
\( K_{\mathrm{m,app}} = \frac{\alpha K_{\mathrm{m}}}{\alpha^{\prime}} \) is the expression for the effect of a reversible inhibitor on apparent \( K_{\mathrm{m}} \).

Step by step solution

01

Understand the Given Equations

The problem provides us with two main equations involving reversible mixed inhibition. They are:1. The rate equation: \( V_{0} = \frac{V_{\max }[\mathrm{S}]}{\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]} \) 2. The equation for apparent Km at half Vmax: \( V_{0} = \frac{V_{\max }}{2 \alpha^{\prime}} \). We need to derive an expression for apparent Km using these equations.
02

Substitute Apparent Km Condition

In the given condition, apparent Km is defined as \([\mathrm{S}]\) at which \( V_{0} = \frac{V_{\max }}{2 \alpha^{\prime}} \).Substitute this condition into the first equation:\[ \frac{V_{\max }}{2 \alpha^{\prime}} = \frac{V_{\max }[\mathrm{S}]}{\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]} \]
03

Simplify the Equation

Next, simplify the equation by multiplying both sides by the denominator \( (\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]) \) to get:\[ V_{\max }\alpha^{\prime}[\mathrm{S}] = \frac{V_{\max }}{2 \alpha^{\prime}} (\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]) \] Cancelling \( V_{\max } \) from both sides gives us:\[ 2\alpha^{\prime}[\mathrm{S}] = \alpha K_{\mathrm{m}} + \alpha^{\prime}[\mathrm{S}] \]
04

Isolate [S]

Solve the resulting equation to isolate \( [\mathrm{S}] \):\[ \alpha^{\prime}[\mathrm{S}] = \alpha K_{\mathrm{m}} \]\[ [\mathrm{S}] = \frac{\alpha K_{\mathrm{m}}}{\alpha^{\prime}} \]
05

Write Final Expression

The expression for the apparent \(K_{\mathrm{m}}\) in terms of the reversible inhibitor is:\[ K_{\mathrm{m,app}} = \frac{\alpha K_{\mathrm{m}}}{\alpha^{\prime}} \]This expression shows how the apparent \( K_{\mathrm{m}} \) is affected by the reversible inhibitor.

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Key Concepts

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

Reversible Inhibition
Reversible inhibition refers to a process where an inhibitor binds to an enzyme and can be easily removed or displaced. This means the enzyme's activity can be restored. This type of inhibition is not permanent. - Inhibitors can attach to different sites on the enzyme, including the active site or other places that affect enzyme function. - Depending on where the inhibitor binds, it may change the enzyme's structure in a way that changes how the enzyme functions. In reversible inhibition, inhibitors are classified into three main types:
  • Competitive inhibitors: They bind to the active site, blocking substrate access.
  • Non-competitive inhibitors: They bind to a different part of the enzyme, reducing its activity without competing with the substrate.
  • Mixed inhibitors: They can influence the binding and processing of the substrate, contributing to the equation modifications observed in enzyme kinetics.
The effects of reversible inhibitors are crucial in understanding how enzymes can be controlled in biological systems or drug development.
Michaelis-Menten Equation
The Michaelis-Menten equation is a key formula in enzyme kinetics that describes how the reaction rate is related to the concentration of substrate available to the enzyme. This equation is represented as:\[ V_{0} = \frac{V_{\max }[S]}{K_{m} + [S]} \]where:
  • \(V_{0}\): the initial reaction velocity, which is the rate of reaction when the reaction begins.
  • \(V_{\max}\): the maximum rate of reaction when the enzyme is saturated with substrate.
  • \([S]\): the concentration of the substrate.
  • \(K_{m}\): the Michaelis constant, the substrate concentration at which the reaction velocity is half of \(V_{\max}\).
This equation forms the basis for understanding enzyme activity and helps in analyzing how changes in substrate concentration impact enzyme-catalyzed reactions. It is also essential for evaluating the effects of inhibitors on enzymes.
Apparent Km
The apparent \(K_{m}\) is a calculated version of the Michaelis constant that is observed when the enzyme is in the presence of inhibitors. It provides insight into how inhibitors influence enzyme activity.
In reversible inhibition, inhibitors alter the enzyme's affinity for the substrate, which is reflected by changes in the observed \(K_{m}\). The apparent \(K_{m}\) is particularly relevant in reversible mixed inhibition, where the formula is:\[ K_{m,app} = \frac{\alpha K_{m}}{\alpha^{'}} \]This equation demonstrates how the presence of inhibitors modifies the enzyme's effective \(K_{m}\). Here's what the parameters mean:
  • \(\alpha\) and \(\alpha'\): These parameters represent how the inhibitor impacts the binding of substrate and enzyme.
  • \(K_{m}\): The original Michaelis constant without inhibitors.
Understanding apparent \(K_{m}\) helps in determining the concentration of substrate needed under inhibited conditions.
Mixed Inhibition
Mixed inhibition is a type of reversible inhibition where the inhibitor can bind to both the enzyme alone and the enzyme-substrate complex. It leads to a decrease in enzyme activity by affecting both substrate binding and catalysis. Mixed inhibitors impact the enzyme in a couple of ways:- They alter the maximal reaction velocity (\(V_{max}\)).- They modify the apparent \(K_{m}\), showing changes in substrate affinity and processing capability.Mixed inhibition is particularly interesting because it combines elements of both competitive and non-competitive inhibition. This gives it the ability to bind to multiple sites and affect enzyme activity diversely.
In the Michaelis-Menten context, mixed inhibition introduces additional factors \(\alpha\) and \(\alpha'\) in the equation, which adjust the observed \(K_{m}\) and thus explain changes in reaction rates. This makes mixed inhibition crucial to drug design and therapeutic applications, where understanding enzyme modulation is key.

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Most popular questions from this chapter

Protection of an Enzyme against Denaturation by Heat When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at \(45{ }^{\circ} \mathrm{C}\) lost \(50 \%\) of its activity in \(12 \mathrm{~min}\), but when incubated at \(45^{\circ} \mathrm{C}\) in the presence of a very large concentration of one of its substrates, it lost only \(3 \%\) of its activity in \(12 \mathrm{~min}\). Suggest why thermal denaturation of hexokinase was retarded in the presence of one of its substrates.

The Turnover Number of Carbonic Anhydrase Carbonic anhydrase of erythrocytes \(\left(M_{\mathrm{r}} 30,000\right)\) has one of the highest turnover numbers known. It catalyzes the reversible hydration of \(\mathrm{CO}_{2}\) : $$ \mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3} $$ This is an important process in the transport of \(\mathrm{CO}_{2}\) from the tissues to the lungs. If \(10.0 \mu \mathrm{g}\) of pure carbonic anhydrase catalyzes the hydration of \(0.30 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in \(1 \mathrm{~min}\) at \(37^{\circ} \mathrm{C}\) at \(V_{\max }\), what is the turnover number \(\left(k_{\text {cat }}\right)\) of carbonic anhydrase (in units of \(\min ^{-1}\) )?

Applying the Michaelis-Menten Equation IV Researchers discover an enzyme that catalyzes the reaction \(\mathrm{X} \rightleftharpoons \mathrm{Y}\). They find that the \(K_{\mathrm{m}}\) for the substrate \(\mathrm{X}\) is \(4 \mu \mathrm{M}\), and the \(k_{\text {cat }}\) is \(20 \mathrm{~min}^{-1}\). a. In an experiment, \([\mathrm{X}]=6 \mathrm{mM}\), and \(V_{0}=480 \mathrm{nM} \mathrm{min}^{-1}\). What was the \(\left[\mathrm{E}_{\mathrm{t}}\right]\) used in the experiment? b. In another experiment, \(\left[\mathrm{E}_{\mathrm{t}}\right]=0.5 \mu \mathrm{M}\), and the measured \(V_{0}=5 \mu \mathrm{M} \mathrm{min}^{-1}\). What was the \([\mathrm{X}]\) used in the experiment? c. The researchers discover that compound \(Z\) is a very strong competitive inhibitor of the enzyme. In an experiment with the same \(\left[E_{t}\right]\) as in (a), but a different \([\mathrm{X}]\), they add an amount of \(\mathrm{Z}\) that produces an \(a\) of 10 and reduces \(V_{0}\) to \(240 \mathrm{nM} \mathrm{min}^{-1}\). What is the \([\mathrm{X}]\) in this experiment? d. Based on the kinetic parameters given, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.

Kinetic Inhibition Patterns Indicate how the observed \(K_{\mathrm{m}}\) of an enzyme would change in the presence of inhibitors having the given effect on \(a\) and \(\alpha^{\prime}\) : a. \(\alpha>\alpha^{\prime} ; \alpha^{\prime}=1.0\) b. \(\alpha^{\prime}>\alpha\) c. \(\alpha=\alpha^{\prime} ; \alpha^{\prime}>1.0\) d. \(\alpha=\alpha^{\prime} ; \alpha^{\prime}=1.0\)

Perturbed \(\mathbf{p} \boldsymbol{K}_{\mathrm{a}}\) Values in Enzyme Active Sites Alanine racemase is a bacterial enzyme that converts \(\mathrm{L}\)-alanine to \(\mathrm{D}\) alanine, which is needed in small amounts to synthesize the bacterial cell wall. The active site of alanine racemase includes a Tyr residue with a p \(K_{\mathrm{a}}\) value of \(7.2\). The \(\mathrm{p} K_{\mathrm{a}}\) of free tyrosine is 10 . The altered \(\mathrm{p} K_{\mathrm{a}}\) of this residue is due largely to the presence of a nearby charged amino acid residue. Which amino acid(s) could lower the \(\mathrm{p} K_{\mathrm{a}}\) of the neighboring Tyr residue? Explain your reasoning.

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