Chapter 13

According to the Michaelis-Menten equation, what is the \(v / V_{\max }\) ratio when \([\mathrm{S}]=4 K_{\text {w? }} ?\)

The following kinetic data were obtained for an enzyme in the absence of any inhibitor \((1),\) and in the presence of two different inhibitors (2) and (3) at \(5 \mathrm{m} M\) concentration. Assume \(\left[\mathrm{E}_{T}\right]\) is the same in each experiment. $$\begin{array}{cccc} & (1) & (2) & (3) \\ {[\mathrm{S}]} & v(\mu \mathrm{mol} / & v(\mu \mathrm{mol} /) & v(\mu \mathrm{mol} /) \\ (\mathrm{m} M) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec} \\ 1 & 12 & 4.3 & 5.5 \\ 2 & 20 & 8 & 9 \\ 4 & 29 & 14 & 13 \\ 8 & 35 & 21 & 16 \\ 12 & 40 & 26 & 18 \end{array}$$ Graph these data as Lineweaver-Burk plots and use your graph to find answers to a. and b. a. Determine \(V_{\max }\) and \(K_{m}\) for the enzyme. b. Determine the type of inhibition and the \(K_{1}\) for each inhibitor.

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 \(\mathbf{A}\)

Liver alcohol dehydrogenase (ADH) is relatively nonspecific and will oxidize ethanol or other alcohols, including methanol. Methanol oxidation yields formaldehyde, which is quite toxic, causing, among other things, blindness. Mistaking it for the cheap wine he usually prefers, my dog Clancy ingested about \(50 \mathrm{mL}\) of windshield washer fluid (a solution \(50 \%\) in methanol). Knowing that methanol would be excreted eventually by Clancy's kidneys if its oxidation could be blocked, and realizing that, in terms of methanol oxidation by ADH, ethanol would act as a competitive inhibitor, I decided to offer Clancy some wine. How much of Clancy's favorite vintage \((12 \% \text { ethanol })\) must he consume in order to lower the activity of his ADH on methanol to \(5 \%\) of its normal value if the \(K_{m}\) values of canine ADH for ethanol and methanol are 1 millimolar and 10 millimolar, respectively? (The \(K_{1}\) for ethanol in its role as competitive inhibitor of methanol oxidation by ADH is the same as its \(K_{m \cdot}\) ) Both the methanol and ethanol will quickly distribute throughout Clancy's body fluids, which amount to about 15 L. Assume the densities of \(50 \%\) methanol and the wine are both \(0.9 \mathrm{g} / \mathrm{mL}\).

Measurement of the rate constants for a simple enzymatic reaction obeying Michaelis-Menten kinetics gave the following results: \(k_{1}=2 \times 10^{8} M^{-1} \sec ^{-1}\) \(k_{-1}=1 \times 10^{3} \sec ^{-1}\) \(k_{2}=5 \times 10^{3} \mathrm{sec}^{-1}\) a. What is \(K_{\mathrm{S}},\) the dissociation constant for the enzyme- substrate complex? b. What is \(K_{m},\) the Michaelis constant for this enzyme? c. What is \(k_{\text {cat }}\) (the turnover number) for this enzyme? d. What is the catalytic efficiency \(\left(k_{\mathrm{cat}} / K_{m}\right)\) for this enzyme? e. Does this enzyme approach "kinetic perfection"? (That is, does \(k_{\mathrm{cat}} / K_{m}\) approach the diffusion-controlled rate of enzyme association with substrate? f. If a kinetic measurement was made using 2 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? g. Again, using 2 nanomoles of enzyme per mL of reaction mixture, what concentration of substrate would give \(v=0.75 V_{\max } ?\) h. If a kinetic measurement was made using 4 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? What would \(K_{m}\) equal under these conditions?

Triose phosphate isomerase catalyzes the conversion of glyceraldehyde-3-phosphate to dihydroxyacetone phosphate. Glyceraldehyde-3-P \(\rightleftarrows\) dihydroxyacetone-P The \(K_{m}\) of this enzyme for its substrate glyceraldehyde- -phosphate is \(1.8 \times 10^{-5} M .\) When [glyceraldehydes-3-phosphate ] \(=30 \mu M\), the rate of the reaction, \(v,\) was \(82.5 \mu \mathrm{mol} \mathrm{mL}^{-1} \mathrm{sec}^{-1}\) a. What is \(V_{\max }\) for this enzyme? b. Assuming 3 nanomoles per mL of enzyme was used in this experiment \((\left[E_{\text {total }}\right]=3 \text { nanomol/mL), what is } k_{\text {cat }}\) for this enzyme? c. What is the catalytic efficiency \(\left(k_{\mathrm{ca}} / K_{m}\right)\) for triose phosphate isomerase? d. Does the value of \(k_{\mathrm{ca}} / K_{m}\) reveal whether triose phosphate isomerase approaches "catalytic perfection"? e. What determines the ultimate speed limit of an enzyme-catalyzed reaction? That is, what is it that imposes the physical limit on kinetic 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"?

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"?

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.)

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?