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The enzyme glycogen synthase kinase \(3 \beta(\operatorname{GSK}-3 \beta)\) plays a central role in Alzheimer's disease. The onset of Alzheimer's disease is accompanied by the production of highly phosphorylated forms of a protein referred to as " \(\tau . " \mathrm{GSK}-3 \beta\) contributes to the hyperphosphorylation of \(\tau\) such that inhibiting the activity of this enzyme represents a pathway for the development of an Alzheimer's drug. A compound known as Ro \(31-8220\) is a competitive inhibitor of GSK-3 \(\beta\). The following data were obtained for the rate of GSK-3 \(\beta\) activity in the presence and absence of Ro \(31-8220[\text { A. Martinez et al., } J .\) Medicinal Chemistry \(45(2002): 1292]:\) $$\begin{array}{ccc} & \mathbf{R}_{0}\left(\boldsymbol{\mu} \mathbf{M} \mathbf{~ s}^{-1} \mathbf{)}\right. \\ {[S](\boldsymbol{\mu} \mathbf{M})} & \mathbf{R}_{0}\left(\boldsymbol{\mu} \mathbf{M} \mathbf{~} \mathbf{s}^{-1}\right),[\boldsymbol{I}]=\mathbf{0} & {[\mathbf{I}]=\mathbf{2} \mathbf{0} \mathbf{0} \boldsymbol{\mu} \mathbf{M}} \\\ \hline 66.7 & 4.17 \times 10^{-8} & 3.33 \times 10^{-8} \\ 40.0 & 3.97 \times 10^{-8} & 2.98 \times 10^{-8} \\ 20.0 & 3.62 \times 10^{-8} & 2.38 \times 10^{-8} \\ 13.3 & 3.27 \times 10^{-8} & 1.81 \times 10^{-8} \\ 10.0 & 2.98 \times 10^{-8} & 1.39 \times 10^{-8} \\ 6.67 & 2.31 \times 10^{-8} & 1.04 \times 10^{-8} \end{array}$$ Determine \(K_{m}\) and \(R_{\max }\) for GSK-3 \(\beta\) and, using the data with the inhibitor, determine \(K_{m}^{*}\) and \(K_{I}\)

Short Answer

Expert verified
Determine the enzyme kinetics by utilizing the Lineweaver-Burk plot, in the absence and presence of the competitive inhibitor Ro 31-8220. The values of \(K_m\), \(R_{\max}\), \(K_m^*\), and \(K_I\) for GSK-3 β in both scenarios can be found by following a series of steps involving the Michaelis-Menten equation and its modified version for competitive inhibition.

Step by step solution

01

Calculate the initial velocity equation without the inhibitor

Using the Michaelis-Menten equation, the initial velocity (\(v\)) is given by: \[v = \frac{R_{\max} [S]}{K_m + [S]}\] Where: - \(R_{\max}\) is the maximum reaction rate (reaction rate when the enzyme is saturated) - \(K_m\) is the Michaelis constant - \([S]\) is the substrate concentration We will use this equation and the given data to calculate initial velocities without the inhibitor and move on to Step 2.
02

Lineweaver-Burk plot without the inhibitor

To determine the \(K_m\) and \(R_{max}\) for GSK-3 beta without the inhibitor, plot the Lineweaver-Burk equation (reciprocal form of the Michaelis-Menten equation): \[\frac{1}{v} = \frac{1}{R_{\max}} + \frac{K_m}{R_{\max} [S]}\] Where: - \(v\) is the initial velocity - \([S]\) is the substrate concentration The slope of this plot will give the value of \(\frac{K_m}{R_{\max}}\) and the intercept on the y-axis will give the value of \(\frac{1}{R_{\max}}\). Using these values, you can calculate \(K_m\) and \(R_{\max}\).
03

Calculate the initial velocity equation with the competitive inhibitor

For the case with competitive inhibition, the modified Michaelis-Menten equation is given by: \[v^* = \frac{R_{\max} [S]}{K_m(1+\frac{[I]}{K_I}) + [S]}\] Where: - \(v^*\) is the initial velocity with the competitive inhibitor - \([I]\) is the inhibitor concentration - \(K_I\) is the inhibitor constant We will use this equation and the given data to calculate initial velocities with the inhibitor and move on to Step 4.
04

Lineweaver-Burk plot with the competitive inhibitor

To determine the \(K_m^*\) and \(K_I\) for GSK-3 beta with the competitive inhibitor, plot the Lineweaver-Burk equation for competitive inhibition: \[\frac{1}{v^*} = \frac{1}{R_{\max}} + \frac{K_m^*(1+\frac{[I]}{K_I})}{R_{\max} [S]}\] Where: - \(v^*\) is the initial velocity with the competitive inhibitor - \([S]\) is the substrate concentration - \([I]\) is the inhibitor concentration The slope of this plot will now give the value of \(\frac{K_m^*(1+\frac{[I]}{K_I})}{R_{\max}}\) while the intercept on the y-axis will still give the value of \(\frac{1}{R_{\max}}\). Using these values and the previous \(R_{\max}\) result, you can calculate \(K_m^*\) and \(K_I\). By following the stated steps, you will determine the values of \(K_m\), \(R_{\max}\), \(K_m^*\), and \(K_I\) for GSK-3 beta in both the presence and the absence of the competitive inhibitor, Ro 31-8220.

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Key Concepts

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

Michaelis-Menten Equation
The Michaelis-Menten equation is a fundamental concept in enzyme kinetics, describing how the rate of enzyme-catalyzed reactions depends on the concentration of substrate available. It's given by:\[v = \frac{R_{\max} [S]}{K_m + [S]}\]where:
  • \(v\) is the initial reaction velocity.
  • \(R_{\max}\) represents the maximum reaction rate when the enzyme is fully saturated with substrate.
  • \(K_m\) is the Michaelis constant, indicative of the substrate concentration at which the reaction rate is half of \(R_{\max}\).
This equation assumes that the enzyme-substrate complex formation is at equilibrium. It simplifies the understanding of how enzymes work under varying substrate concentrations.
In the context of GSK-3 β, the Michaelis-Menten equation helps in determining the kinetic parameters \(K_m\) and \(R_{\max}\) from experimental data without inhibition.
Competitive Inhibition
Competitive inhibition occurs when a molecule similar in shape to the substrate competes for the active site of the enzyme. This form of inhibition is reversible and directly affects the apparent affinity of the enzyme for the substrate, which is expressed by the Michaelis constant \(K_m\).
The presence of a competitive inhibitor increases the \(K_m\) value because the substrate has to compete with the inhibitor, requiring a higher substrate concentration to reach \(R_{\max}/2\). However, \(R_{\max}\) stays unchanged as increasing substrate concentrations can still achieve maximum reaction velocities.
The modified Michaelis-Menten equation for competitive inhibition is:\[v^* = \frac{R_{\max} [S]}{K_m(1+\frac{[I]}{K_I}) + [S]}\]where:
  • \(v^*\) is the initial velocity in the presence of inhibitor.
  • \([I]\) is the inhibitor concentration.
  • \(K_I\) is the inhibitor constant, representing the inhibitor's affinity for the enzyme.
In the case of GSK-3 β, Ro 31-8220 acts as a competitive inhibitor, allowing us to study enzyme inhibition mechanisms and explore potential therapeutic angles for diseases such as Alzheimer's.
Lineweaver-Burk Plot
The Lineweaver-Burk plot, also known as the double reciprocal plot, is a graphical representation of enzyme kinetics based on the Michaelis-Menten equation. It transforms the hyperbolic curve seen in direct plots of \(v\) versus \([S]\) into a linear form. This linearization aids in directly determining kinetic parameters:\[\frac{1}{v} = \frac{1}{R_{\max}} + \frac{K_m}{R_{\max} [S]}\]where:
  • The y-intercept (\(\frac{1}{R_{\max}}\)) directly provides the inverse of the maximum reaction rate.
  • The slope (\(\frac{K_m}{R_{\max}}\)) gives a ratio involving the Michaelis constant and \(R_{\max}\).
  • More resolved lines in this plot help reduce perturbation by errors in single data points.
This plot becomes particularly insightful when analyzing competitive inhibition, as it distinctly shows the change in \(K_m\) without altering \(R_{\max}\). In studying GSK-3 β enzyme kinetics, the Lineweaver-Burk plot offers a clear visualization to distinguish between reactions with and without the inhibitor Ro 31-8220.

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Most popular questions from this chapter

For phenanthrene, the measured lifetime of the triplet state \(\tau_{p}\) is \(3.3 \mathrm{s}\), the fluorescence quantum yield is \(0.12,\) and the phosphorescence quantum yield is 0.13 in an alcohol-ether glass at \(77 \mathrm{K}\). Assume that no quenching and no internal conversion from the singlet state occurs. Determine \(k_{p}, k_{i s c}^{T},\) and \(k_{i s c}^{S} / k_{f}\)

The rate of reaction can be determined by measuring the change in optical rotation of the sample as a function of time if a reactant or product is chiral. This technique is especially useful for kinetic studies of enzyme catalysis involving sugars. For example, the enzyme invertase catalyzes the hydrolysis of sucrose, an optically active sugar. The initial reaction rates as a function of sucrose concentration are as follows: $$\begin{array}{cc} \text { [Sucrose] }_{\mathbf{0}}(\mathbf{M}) & \mathbf{R}_{\mathbf{0}}\left(\mathbf{M} \mathbf{~ s}^{-\mathbf{1}}\right) \\ \hline 0.029 & 0.182 \\ 0.059 & 0.266 \\ 0.088 & 0.310 \\ 0.117 & 0.330 \\ 0.175 & 0.362 \\ 0.234 & 0.361 \end{array}$$ Use these data to determine the Michaelis constant for invertase.

A proposed mechanism for the formation of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) from \(\mathrm{NO}_{2}(g)\) and \(\mathrm{O}_{3}(g)\) is \\[ \begin{array}{c} \mathrm{NO}_{2}(g)+\mathrm{O}_{3}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{3}(g)+\mathrm{O}_{2}(g) \\ \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)+\mathrm{M}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{M}(g) \end{array} \\] Determine the rate law expression for the production of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) given this mechanism.

The Kermack-McKendrick model was developed to explain the rapid rise and fall in the number of infected people during epidemics. This model involves the interaction of susceptible (S), infected (I), and recovered (R) people through the following mechanism: \\[ \begin{array}{l} \mathrm{S}+\mathrm{I} \stackrel{k_{1}}{\longrightarrow} \mathrm{I}+\mathrm{I} \\\ \mathrm{I} \stackrel{k_{2}}{\longrightarrow} \mathrm{R} \end{array} \\] a. Write down the differential rate expressions for \(S,\) I, and \(R\) b. The key quantity in this mechanism is called the epidemiological threshold defined as the ratio of \([\mathrm{S}] k_{1} / k_{2}\). When this ratio is greater than 1 the epidemic will spread; however, when the threshold is less than 1 the epidemic will die out. Based on the mechanism, explain why this behavior is observed.

a. For the hydrogen-bromine reaction presented in Problem P36.7 imagine initiating the reaction with only \(\mathrm{Br}_{2}\) and \(\mathrm{H}_{2}\) present. Demonstrate that the rate law expression at \(t=0\) reduces to \\[ \left(\frac{d[\mathrm{HBr}]}{d t}\right)_{t=0}=2 k_{2}\left(\frac{k_{1}}{k_{-1}}\right)^{1 / 2}\left[\mathrm{H}_{2}\right]_{0}\left[\mathrm{Br}_{2}\right]^{1 / 2} \\] b. The activation energies for the rate constants are as follows: $$\begin{array}{cc} \text { Rate Constant } & \Delta \boldsymbol{E}_{a}(\mathbf{k J} / \mathbf{m o l}) \\ \hline k_{1} & 192 \\ k_{2} & 0 \\ k_{-1} & 74 \end{array}$$ c. How much will the rate of the reaction change if the temperature is increased to \(400 .\) K from \(298 \mathrm{K} ?\)

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