/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 The enzyme glycogen synthase kin... [FREE SOLUTION] | 91影视

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Chapter 36: Problem 21

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 尾 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 尾, 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 尾, 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 尾 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

The chlorination of vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}+\) \(\mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3},\) is believed to proceed by the following mechanism: \\[ \begin{array}{l} \mathrm{Cl}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{Cl} \\ \mathrm{Cl} \cdot+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \\ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \cdot+\mathrm{Cl}_{2} \stackrel{k_{3}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}+\mathrm{Cl} \\ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \cdot+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \cdot \stackrel{k_{4}}{\longrightarrow} \text { stable species } \end{array} \\] Derive the rate law expression for the chlorination of vinyl chloride based on this mechanism.

Determine the predicted rate law expression for the following radical-chain reaction: \\[ \begin{array}{l} \mathrm{A}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{A} \\ \mathrm{A} \cdot \stackrel{k_{2}}{\longrightarrow} \mathrm{B} \cdot+\mathrm{C} \end{array} \\] $$\begin{array}{l} \mathrm{A} \cdot+\mathrm{B} \cdot \stackrel{k_{3}}{\longrightarrow} \mathrm{P} \\\ \mathrm{A} \cdot+\mathrm{P} \stackrel{k_{4}}{\rightarrow} \mathrm{B} \end{array}$$

The pyrene/coumarin FRET pair \(\left(r_{0}=39 \AA\right)\) is used to study the fluctuations in enzyme structure during the course of a reaction. Computational studies suggest that the pair will be separated by \(35 \AA\) in one conformation and \(46 \AA\) in a second configuration. What is the expected difference in FRET efficiency between these two conformational states?

(Challenging) Cubic autocatalytic steps are important in a reaction mechanism referred to as the "brusselator" (named in honor of the research group in Brussels that initially discovered this mechanism): $$\begin{array}{l} \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{X} \\ 2 \mathrm{X}+\mathrm{Y} \stackrel{k_{2}}{\longrightarrow} 3 \mathrm{X} \\ \mathrm{B}+\mathrm{X} \stackrel{k_{3}}{\longrightarrow} \mathrm{Y}+\mathrm{C} \\\ \mathrm{X} \stackrel{k_{4}}{\longrightarrow} \mathrm{D} \end{array}$$ If \([\mathrm{A}]\) and \([\mathrm{B}]\) are held constant, this mechanism demonstrates interesting oscillatory behavior that we will explore in this problem. a. Identify the autocatalytic species in this mechanism. b. Write down the differential rate expressions for \([\mathrm{X}]\) and \([\mathrm{Y}]\) c. Using these differential rate expressions, employ Euler's method (Section 35.6 ) to calculate \([\mathrm{X}]\) and \([\mathrm{Y}]\) versus time under the conditions \(\mathrm{k}_{1}=1.2 \mathrm{s}^{-1}, \mathrm{k}_{2}=0.5 \mathrm{M}^{-2} \mathrm{s}^{-1}, \mathrm{k}_{3}=\) \(3.0 \mathrm{M}^{-1} \mathrm{s}^{-1}, \mathrm{k}_{4}=1.2 \mathrm{s}^{-1},\) and \([\mathrm{A}]_{0}=[\mathrm{B}]_{0}=1 \mathrm{M} .\) Begin with \([\mathrm{X}]_{0}=0.5 \mathrm{M}\) and \([\mathrm{Y}]_{0}=0.1 \mathrm{M} .\) A plot of \([\mathrm{Y}]\) versus \([\mathrm{X}]\) should look like the top panel in the following figure. d. Perform a second calculation identical to that in part (c), but with \([\mathrm{X}]_{0}=3.0 \mathrm{M}\) and \([\mathrm{Y}]_{0}=3.0 \mathrm{M} .\) A plot of \([\mathrm{Y}]\) versus \([\mathrm{X}]\) should look like the bottom panel in the following figure. e. Compare the left and bottom panels in the following figure. Notice that the starting conditions for the reaction are different (indicated by the black spot). What DO the figures indicate regarding the oscillatory state the system evolves to?

In this problem you will investigate the parameters involved in a single- molecule fluorescence experiment. Specifically, the incident photon power needed to see a single molecule with a reasonable signal-to-noise ratio will be determined. a. Rhodamine dye molecules are typically employed in such experiments because their fluorescence quantum yields are large. What is the fluorescence quantum yield for Rhodamine B (a specific rhodamine dye) where \(k_{r}=1 \times 10^{9} \mathrm{s}^{-1}\) and \(k_{i c}=1 \times 10^{8} \mathrm{s}^{-1} ?\) You can ignore intersystem crossing and quenching in deriving this answer. b. If care is taken in selecting the collection optics and detector for the experiment, a detection efficiency of \(10 \%\) can be readily achieved. Furthermore, detector dark noise usually limits these experiments, and dark noise on the order of 10 counts \(s^{-1}\) is typical. If we require a signal- tonoise ratio of \(10: 1,\) then we will need to detect 100 counts \(\mathrm{s}^{-1} .\) Given the detection efficiency, a total emission rate of 1000 fluorescence photons \(s^{-1}\) is required. Using the fluorescence quantum yield and a molar extinction coefficient for Rhodamine \(\mathrm{B}\) of \(\sim 40,000 \mathrm{M}^{-1} \mathrm{cm}^{-1},\) what is the intensity of light needed in this experiment in terms of photons \(\mathrm{cm}^{-2} \mathrm{s}^{-1} ?\) c. The smallest diameter focused spot one can obtain in a microscope using conventional refractive optics is approximately one-half the wavelength of incident light. Studies of Rhodamine B generally employ 532 nm light such that the focused-spot diameter is \(\sim 270 \mathrm{nm}\). Using this diameter, what incident power in watts is required for this experiment? Do not be surprised if this value is relatively modest.
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