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Chapter 13: Problem 6

The general rate equation for an ordered, single-displacement reaction where \(A\) is the leading substrate is $$v=\frac{V_{\max }[\mathrm{A}][\mathrm{B}]}{\left(K_{\mathrm{S}}^{\mathrm{A}} K_{m}^{\mathrm{B}}+K_{\mathrm{m}}^{\mathrm{A}}[\mathrm{B}]+K_{\mathrm{m}}^{\mathrm{B}}[\mathrm{A}]+[\mathrm{A}][\mathrm{B}]\right)}$$ Write the Lineweaver-Burk (double-reciprocal) equivalent of this equation and from it calculate algebraic expressions for the following: a. The slope b. The \(y\) -intercepts c. The horizontal and vertical coordinates of the point of intersection when \(1 / v\) is plotted versus \(1 /[\mathrm{B}]\) at various fixed concentrations of \(\mathbf{A}\)

Short Answer

Expert verified
The Lineweaver-Burk Plot of the given reaction will be a straight line with the slope \(\frac{K_m}{V_{\text{max}}}\), y-intercept \(\frac{1}{V_{\text{max}}}\) and a point of intersection at \(\frac{1}{[B]} = \frac{K_m}{V_{\text{max}}}\) and \(\frac{1}{v} = \frac{K_m}{V_{\text{max}}}\).

Step by step solution

01

Write the double-reciprocal form

Rewrite the rate equation in Lineweaver-Burk's form, ie. \(1/v\) form. This can be done by taking reciprocal of the given rate equation: \[ \frac{1}{v} = \frac{K_{\mathrm{S}}^{\mathrm{A}} K_{m}^{\mathrm{B}}+K_{\mathrm{m}}^{\mathrm{A}}[\mathrm{B}]+K_{\mathrm{m}}^{\mathrm{B}}[\mathrm{A}]+[\mathrm{A}][\mathrm{B}]} {V_{\max}[\mathrm{A}][\mathrm{B}]} \]
02

Simplify the equation further

Divide the numerator and the denominator of the right side by \([B]\): \[ \frac{1}{v} = \frac{K_{\mathrm{S}}^{\mathrm{A}} K_{m}^{\mathrm{B}}+[A]K_{m}^{\mathrm{B}}+[\mathrm{A}][\mathrm{B}]} {V_{\max}[\mathrm{A}]} \]
03

Express it for different concentrations of A

The above equation can be written generally as: \[ \frac{1}{v} = \frac{1}{V_{\text{max}}} + \frac{K_m}{V_{\text{max}}}.\frac{1}{[B]} \] where \(K_m = K_{\mathrm{S}}^{\mathrm{A}} K_{m}^{\mathrm{B}}+[A]K_{m}^{\mathrm{B}}\) this shows that the plot of \( \frac{1}{v}\) vs \( \frac{1}{[B]}\) is a straight line of slope \(\frac{K_m}{V_{\text{max}}}\) and y intercept \(\frac{1}{V_{\text{max}}}\)
04

Calculate the intercepts and coordinates

From the equation in step 3 we can find: a. The slope of this line is \( \frac{K_m}{V_{\text{max}}}\) , b. The \(y\) -intercept of the line is \( \frac{1}{V_{\text{max}}}\), c. The horizontal and vertical coordinates of the point of intersection would be \(\frac{1}{[B]} = \frac{K_m}{V_{\text{max}}}\) and \(\frac{1}{v} = \frac{K_m}{V_{\text{max}}}\), since at the point of intersection \( \frac{1}{v} = \frac{1}{[B]}.\) (At constant [A], as [B] approaches infinity, 1/[B] approaches 0, so the plot intersects x-axis at 1/[B]=0)

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Most popular questions from this chapter

Acetylcholinesterase catalyzes the hydrolysis of the neurotransmitter acetylcholine: Acetylcholine \(+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) acetate \(+\) choline The \(K_{m}\) of acetylcholinesterase for its substrate acetylcholine is \(9 \times 10^{-5} M .\) In a reaction mixture containing 5 nanomoles/mL of acetylcholinesterase and \(150 \mu M\) acetylcholine, a velocity \(v_{\mathrm{o}}=\) \(40 \mu \mathrm{mol} / \mathrm{mL} \cdot\) sec was observed for the acetylcholinesterase reaction. a. Calculate \(V_{\max }\) for this amount of enzyme. b. Calculate \(k_{\text {cat }}\) for acetylcholinesterase. c. Calculate the catalytic efficiency \(\left(k_{\mathrm{cat}} / K_{m}\right)\) for acetylcholinesterase. d. Does acetylcholinesterase approach "catalytic perfection"?

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"?

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 \([\mathrm{S}]=4 K_{\text {w? }} ?\)

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.)
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