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

Short Answer

Expert verified
The \(v / V_{\max }\) ratio when \([S] = 4K_m\) is 0.8.

Step by step solution

01

Substitution of values

First, substitute the given substrate concentration \([S] = 4K_m\) in the Michaelis-Menten equation to get \(v = \frac{V_{\max} \cdot 4K_m}{K_m + 4K_m} = \frac{4V_{\max}K_m}{5K_m}\).
02

Solving the equation

Simplify the expression by canceling common factors, resulting in the rate of reaction \(v = \frac{4}{5} V_{\max}\).
03

Ratio Calculation

Now, calculate the asked ratio \(v / V_{\max }\) by substituting \(v\) from Step 2 into this ratio. So the ratio is \(\frac{4V_{\max}}{5V_{\max}}\). Again cancel out common factors which gives us 0.8

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Most popular questions from this chapter

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?

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.)

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?

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 \(\mathbf{A}\)

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"?

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