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Chapter 36: Problem 20

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.

Short Answer

Expert verified
The short answer for this problem involves calculating the Michaelis-Menten constant (K_m) for invertase using the given data. First, linearize the Michaelis-Menten equation using the Lineweaver-Burk plot. Calculate the reciprocal values for initial reaction rate (1/R鈧�) and initial sucrose concentration (1/[S鈧�]). Plot these values, and find the slope and y-intercept of the resulting linear regression line. Finally, calculate K_m using the formula \(K_m = slope \times \frac{1}{y-intercept}\).

Step by step solution

01

Understand the Michaelis-Menten equation

The Michaelis-Menten equation is given by: \(R_0 = \frac{V_{max} \times [S_{0}]}{K_m + [S_{0}]}\) Here, R鈧� is the initial reaction rate, [S鈧�] is the initial concentration of the substrate (sucrose), V_max is the maximum rate of the reaction, and K_m is the Michaelis constant. We need to fit the given data to this equation to determine K_m.
02

Linearize the Michaelis-Menten equation using the Lineweaver-Burk plot

A Lineweaver-Burk plot is a double reciprocal plot of the Michaelis-Menten equation, which allows us to linearize the equation and easily determine K_m and V_max. The linear form of the equation is given by: \(\frac{1}{R_0} = \frac{1}{V_{max}} + \frac{K_m}{V_{max} \times [S_{0}]}\) Here, \(\frac{1}{R_0}\) is the y-axis, and \(\frac{1}{[S_{0}]}\) is the x-axis. The slope of the line will be \(\frac{K_m}{V_{max}}\), and the y-intercept will be \(\frac{1}{V_{max}}\).
03

Calculate the reciprocal values for initial reaction rate and sucrose concentration

Use the given data to calculate the reciprocal values for the initial reaction rate (1/R鈧�) and the initial sucrose concentration (1/[S鈧�]). $$\begin{array}{ccc} \text { [S鈧�]^{-1} (M^{-1}) } & \text { R鈧�^{-1} (M^{-1}s) } \\\ \hline 1/0.029 & 1/0.182 \\\ 1/0.059 & 1/0.266 \\\ 1/0.088 & 1/0.310 \\\ 1/0.117 & 1/0.330 \\\ 1/0.175 & 1/0.362 \\\ 1/0.234 & 1/0.361 \end{array}$$
04

Plot the Lineweaver-Burk plot and find the slope and y-intercept

Plot the calculated reciprocal values of initial reaction rate (1/R鈧�) against the reciprocal values of initial sucrose concentration (1/[S鈧�]). The slope of the resulting linear regression line will be \(\frac{K_m}{V_{max}}\) and the y-intercept will be \(\frac{1}{V_{max}}\).
05

Calculate the Michaelis constant (K_m) and maximum rate (V_max) from the Lineweaver-Burk plot

Use the slope and y-intercept obtained in step 4 to calculate the Michaelis constant (K_m) and the maximum rate (V_max). \(\frac{K_m}{V_{max}} = slope\) \(K_m = slope \times V_{max}\) \(\frac{1}{V_{max}} = y-intercept\) \(V_{max} = \frac{1}{y-intercept}\) Now, plug the value of V_max back into the equation for K_m: \(K_m = slope \times \frac{1}{y-intercept}\) The resulting value for K_m is the Michaelis constant for invertase.

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Key Concepts

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

Understanding Enzyme Kinetics
When we talk about enzyme kinetics, we are discussing the study of how enzymes bind to substrates and turn them into products.

Enzymes are biological catalysts that increase the rate of chemical reactions without being consumed or altered in the process. Enzyme kinetics is crucial in understanding biological processes and can inform us about the mechanisms of enzyme activity, their efficiency, and how they're affected by different conditions and inhibitors.

The classic model used to describe enzyme kinetics is the Michaelis-Menten equation. This model assumes that the initial reaction speed is most rapid and then slows down as the substrate concentration decreases or as the product accumulates. It is essential to grasp that in a Michaelis-Menten scenario, one enzyme molecule can only bind to one substrate molecule at a time, leading to a saturation point known as Vmax, where increasing substrate concentration does not increase the reaction rate.
Explaining Reaction Rate
The reaction rate is a measure of how quickly a reactant is converted into a product by an enzyme. In our example, we consider the initial reaction rate, denoted as R0, which tells us how fast the reaction is proceeding right when it starts.

Reaction rates can be influenced by various factors, including enzyme concentration, substrate concentration, temperature, and pH. In the context of enzyme kinetics, examining the relationship between substrate concentration and reaction rate is fundamental. As we increase the amount of substrate, the reaction rate also increases until all enzyme molecules are busy catalyzing the reaction, and we reach Vmax, the maximum rate.
Deciphering the Lineweaver-Burk Plot
A Lineweaver-Burk plot is a graphical representation that aims to linearize the relationship expressed by the Michaelis-Menten equation, simplifying the determination of two critical parameters: Vmax and Km (the Michaelis constant).

In our exercise, by plotting the reciprocal of the reaction rate (1/R0) against the reciprocal of substrate concentration (1/[S0]), we create a straight line which makes it easier to interpret these parameters. The plot's slope gives the ratio of Km/Vmax, while the y-intercept is equivalent to 1/Vmax. This linear transformation is a widely used method to extract Km and Vmax from reaction rate data.
Invertase Enzyme Catalysis
The invertase enzyme provides a fascinating example of enzyme catalysis. It specifically catalyzes the hydrolysis of sucrose into glucose and fructose.

Enzyme catalysis involves the enzyme binding to a specific substrate鈥攊n this case, sucrose鈥� and converting it into a product faster than would occur without the enzyme. Invertase works on sucrose by breaking the glycosidic bond between glucose and fructose.

The efficiency of invertase and its characteristics, such as the Michaelis constant (Km), are determined by the reaction's kinetics. Km offers a measure of the affinity of the enzyme for its substrate 鈥� a lower Km signifies a higher affinity, meaning that invertase can function effectively even at low sucrose concentrations. This property is vital when applying invertase in various industries, such as food manufacturing, where it's used to make syrups and other sweeteners.

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Most popular questions from this chapter

A central issue in the design of aircraft is improving the lift of aircraft wings. To assist in the design of more efficient wings, wind-tunnel tests are performed in which the pressures at various parts of the wing are measured generally using only a few localized pressure sensors. Recently, pressure- sensitive paints have been developed to provide a more detailed view of wing pressure. In these paints, a luminescent molecule is dispersed into an oxygen- permeable paint and the aircraft wing is painted. The wing is placed into an airfoil, and luminescence from the paint is measured. The variation in \(\mathrm{O}_{2}\) pressure is measured by monitoring the luminescence intensity, with lower intensity demonstrating areas of higher \(\mathrm{O}_{2}\) pressure due to quenching. a. The use of platinum octaethylporphyrin (PtOEP) as an oxygen sensor in pressure-sensitive paints was described by Gouterman and coworkers [Review of Scientific Instruments \(61(1990): 3340] .\) In this work, the following relationship between luminescence intensity and pressure was derived: \(I_{0} / I=A+B\left(P / P_{0}\right),\) where \(I_{0}\) is the fluorescence intensity at ambient pressure \(P_{0},\) and \(I\) is the fluorescence intensity at an arbitrary pressure \(P .\) Determine coefficients \(A\) and \(B\) in the preceding expression using the Stern-Volmer equation: \(k_{\text {total}}=1 / \tau_{l}=k_{l}+k_{q}[Q] .\) In this equation \(\tau_{l}\) is the luminescence lifetime, \(k_{l}\) is the luminescent rate constant, and \(k_{q}\) is the quenching rate constant. In addition, the luminescent intensity ratio is equal to the ratio of luminescence quantum yields at ambient pressure \(\Phi_{0}\) and an arbitrary pressure \(\Phi:\) \\[ \Phi_{0} / \Phi=I_{0} / I \\] b. Using the following calibration data of the intensity ratio versus pressure observed for PtOEP, determine \(A\) and \(B\) : $$\begin{array}{cccc} I_{0} / I & P / P_{0} & I_{0} / I & P / P_{0} \\ \hline 1.0 & 1.0 & 0.65 & 0.46 \\ 0.9 & 0.86 & 0.61 & 0.40 \\ 0.87 & 0.80 & 0.55 & 0.34 \\ 0.83 & 0.75 & 0.50 & 0.28 \\ 0.77 & 0.65 & 0.46 & 0.20 \\ 0.70 & 0.53 & 0.35 & 0.10 \end{array}$$ c. \(A t\) an ambient pressure of 1 atm, \(I_{0}=50,000\) (arbitrary units \()\) and 40,000 at the front and back of the wing. The wind tunnel is turned on to a speed of Mach \(0.36,\) and the measured luminescence intensity is 65,000 and 45,000 at the respective locations. What is the pressure differential between the front and back of the wing?

Consider the gas-phase isomerization of cyclopropane: Are the following data of the observed rate constant as a function of pressure consistent with the Lindemann mechanism? $$\begin{array}{cccc} \boldsymbol{P}(\text { Torr }) & \boldsymbol{k}\left(\mathbf{1 0}^{4} \mathbf{s}^{-1}\right) & \boldsymbol{P}(\text { Torr }) & \boldsymbol{k}\left(\mathbf{1 0}^{4} \mathbf{s}^{-1}\right) \\ \hline 84.1 & 2.98 & 1.36 & 1.30 \\ 34.0 & 2.82 & 0.569 & 0.857 \\ 11.0 & 2.23 & 0.170 & 0.486 \\ 6.07 & 2.00 & 0.120 & 0.392 \\ 2.89 & 1.54 & 0.067 & 0.303 \end{array}$$

Consider the collision-induced dissociation of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) via the following mechanism: \\[ \begin{array}{l} \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(g) \frac{k_{1}}{\overline{k_{-1}}} \mathrm{N}_{2} \mathrm{O}_{5}(g)^{*}+\mathrm{N}_{2} \mathrm{O}_{5}(g) \\ \mathrm{N}_{2} \mathrm{O}_{5}(g) * \stackrel{k_{2}}{\longrightarrow} \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g) \end{array} \\] The asterisk in the first reaction indicates that the reactan is activated through collision. Experimentally it is found that the reaction can be either first or second order in \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) depending on the concentration of this species. Derive a rate law expression for this reaction consistent with this observation.

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.

The hydrogen-bromine reaction corresponds to the production of \(\operatorname{HBr}(g)\) from \(\mathrm{H}_{2}(g)\) and \(\mathrm{Br}_{2}(g)\) as follows: \(\mathrm{H}_{2}(g)+\operatorname{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) This reaction is famous for its complex rate law, determined by Bodenstein and Lind in 1906: \\[ \frac{d[\mathrm{HBr}]}{d t}=\frac{k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{1 / 2}}{1+\frac{m[\mathrm{HBr}]}{\left[\mathrm{Br}_{2}\right]}} \\] where \(k\) and \(m\) are constants. It took 13 years for a likely mechanism of this reaction to be proposed, and this feat was accomplished simultaneously by Christiansen, Herzfeld, and Polyani. The mechanism is as follows: \\[ \begin{array}{l} \operatorname{Br}_{2}(g) \stackrel{k_{1}}{\sum_{k_{-1}}} 2 \operatorname{Br} \cdot(g) \\ \text { Br' }(g)+\mathrm{H}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \operatorname{HBr}(g)+\mathrm{H} \cdot(g) \\ \text { H\cdot }(g)+\operatorname{Br}_{2}(g) \stackrel{k_{3}}{\longrightarrow} \operatorname{HBr}(g)+\operatorname{Br} \cdot(g) \\ \operatorname{HBr}(g)+\mathrm{H} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \mathrm{H}_{2}(g)+\operatorname{Br} \cdot(g) \end{array} \\] Construct the rate law expression for the hydrogen-bromine reaction by performing the following steps: a. Write down the differential rate expression for \([\mathrm{HBr}]\) b. Write down the differential rate expressions for \([\mathrm{Br} \cdot]\) and [H']. c. Because \(\mathrm{Br} \cdot(g)\) and \(\mathrm{H} \cdot(g)\) are reaction intermediates, apply the steady-state approximation to the result of part (b). d. Add the two equations from part (c) to determine [Br'] in terms of \(\left[\mathrm{Br}_{2}\right]\) e. Substitute the expression for \([\mathrm{Br} \cdot]\) back into the equation for \([\mathrm{H} \cdot]\) derived in part \((\mathrm{c})\) and solve for \([\mathrm{H} \cdot]\) f. Substitute the expressions for \([\mathrm{Br} \cdot]\) and \([\mathrm{H} \cdot]\) determined in part (e) into the differential rate expression for \([\mathrm{HBr}]\) to derive the rate law expression for the reaction.
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