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Chapter 13: Problem 16

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?

Short Answer

Expert verified
In relation to enzyme A, substrate \(\mathrm{S}\) is the preferred one because it has a lower \(K_m\) value, indicating a greater affinity. However, substrate \(\mathrm{T}\) has a greater catalytic efficiency because of its larger \(k_2 / K_m\) ratio.

Step by step solution

01

Interpret \(K_m\) Values

Part a. According to Michaelis-Menten kinetics, an enzyme has a higher affinity for the substrate with the lower \(K_m\) value. Here, \(\mathrm{S}\) has a \(K_m\) of 1mM, whereas \(\mathrm{T}\) has a \(K_m\) of 10mM. Therefore, enzyme A prefers substrate \(\mathrm{S}\) because it has a lower \(K_m\) value (\(K_m^{\mathrm{S}}< K_m^{\mathrm{T}}\)). The lower the \(K_m\), the less substrate is needed to reach half the maximum speed of the reaction.
02

Calculate Catalytic Efficiency

Part b. The catalytic efficiency of an enzyme is determined by the equation \(k_2 / K_m\). This indicates how efficiently an enzyme converts a substrate into a product. Here, the catalytic efficiency for substrate \(\mathrm{S}\) is \(2 \times 10^{4} / 1 = 2 \times 10^{4}\) and for \(\mathrm{T}\) it is \(4 \times 10^{5} / 10 = 4 \times 10^{4}\). Therefore, enzyme A has a greater catalytic efficiency for substrate \(\mathrm{T}\) because the ratio of \(k_2 / K_m\) is larger for \(\mathrm{T}\) as compared to \(\mathrm{S}\) (\(k_2 / K_m\) for \(\mathrm{T}> k_2 / K_m\) for \(\mathrm{S}\)).

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Most popular questions from this chapter

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"?

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 \(\mathbf{A}\)

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