Determine vmax and enzyme concentration (Michaelis–Menten)

Question

I am given that the enzyme concentration is $\pu{15 nM}$ and the following data:

$$ \begin{array}{rl} \hline S/\pu{mM} & V/\pu{mM s^{-1}} \\ \hline 1 & 0.202 \\ 2 & 0.368 \\ 5 & 0.729 \\ 10 & 1.08 \\ 20 & 1.43 \\ 50 & 1.77 \\ 100 & 1.92 \\ \hline \end{array} $$

And calculated that the Michaelis–Menten constant was $K_\mathrm{M} = 9.41$ (unsure about the units).

I then want to calculate $v_\mathrm{max}$ and $[\ce{E}]_0$ for the same reaction given that the reaction rate is $\pu{0.101 mM s-1}$ at a substrate concentration of $\pu{24 mM}.$

I know that once I have $v_\mathrm{max}$ I could just use $v_\mathrm{max} = k_\mathrm{b}[\ce{E}]_0$ to calculate the enzyme concentration, but I am unsure as to how I can calculate $v_\mathrm{max}$ from one set of data. Could someone give me a hint of how to begin this problem?

Answer 1

Michaelis-Menten kinetics is given by the equation:

$$V = V_\mathrm{max}\frac{S}{K_M+S} \tag1$$

Where $V_\mathrm{max} = k_\mathrm{cat}\cdot [\ce{E_T}]$ and $K_M = \dfrac{k_\mathrm{on} + k_\mathrm{cat}}{k_\mathrm{off}}$ (using the conventional description of Briggs and Haldane's derivation of the Michaelis-Menten equation; Ref.1):

$$\ce{E + S <=>[$k_\mathrm{on}$][$k_\mathrm{off}$] [ES] ->[$k_\mathrm{cat}$] E + P} \tag2$$

The derivation of the equation has taken a few assumption. Accordingly, as a rule of thumb, the enzyme kinetics follow Michaelis-Menten equation if $K_M \gt 5 \times [\ce{E_{T}}]$. Since calculated $K_M$ is $\pu{0.939 mM}$ (see Vinícius Godim's answer elsewhere) and $[\ce{E_{T}}]$ of the first kinetics is $\pu{15.0 nM}$, this requirement is satisfied.

However, OP's confusion is how to calculate the $[\ce{E_{T}}]$ of the second kinetics using the same enzyme and substrate in different set. OP didn't aware about the following fact of the enzyme kinetics: If you have used same enzyme and substrate under similar conditions (temperature, etc.) the second time, the constants $K_M, k_\mathrm{on}, k_\mathrm{off}$, and $k_\mathrm{cat}$ you found in the first time are going to be the same.

Thus, for second kinetics, $K_M = \pu{0.939 mM}$ and $k_\mathrm{cat} = \dfrac{V_\mathrm{max-1}}{[\ce{E_{T1}}]} = \dfrac{\pu{2.10 mM s-1}}{\pu{15.0 \times 10^{-6} mM}} = \pu{1.40 \times 10^{5} s-1}$.

Therefore, since it has been calculated that $V_\mathrm{max-2} = \pu{0.141 mM s-1}$ (see Vinícius Godim's answer elsewhere), $[\ce{E_{T2}}]$ can be calculated as follows:

$$V_\mathrm{max-2} = k_\mathrm{cat}[\ce{E_{T2}}] \ \Rightarrow \ [\ce{E_{T2}}] = \frac{V_\mathrm{max-2}}{k_\mathrm{cat}} = \frac{\pu{0.141 mM s-1}}{\pu{1.40 \times 10^{5} s-1}} = \pu{1.01 \times 10^{-6} mM}\\ = \pu{1.01 nM}$$

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Late edit:

I think it is beneficial to show how to find $K_M$ and $V_\mathrm{max}$ from the given set of kinetic data. The method is called Lineweaver–Burk plot, which is a plot of $\dfrac{1}{[\ce{S}]}$ versus $\dfrac{1}{V}$:

Since the equation of the plot is:

$$\frac{1}{[V]} = \frac{K_M}{V_\mathrm{max}}\cdot \frac{1}{[\ce{S}]} + \frac{1}{V_\mathrm{max}}$$

which is a straight-line equation ($y = mx + c = 4.4754x + 0.4767$). Thus, positive intercept is equal to $\dfrac{1}{V_\mathrm{max}}$, which is $0.4767$, and slope is equal to $\dfrac{K_M}{V_\mathrm{max}}$, which is $4.4754$. When resolve you would find, $K_M = \pu{9.39 mM}$ and $V_\mathrm{max} = \pu{2.10 mM s-1}$.

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Reference:

1. George Edward Briggs and John Burdon Sanderson Haldane, "A Note on the Kinetics of Enzyme Action," Biochem. J. 1925, 19(2), 338-339 (DOI: 10.1042/bj0190338).

Answer 2

Michaelis-Menten kinetics is given by

$$V = V_{\max} \frac{S}{K_M + S}$$

Taking the inverse of both sides to linearise the equation,

$$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$

Taking $y = 1/V$ and $x = 1/S$ in $y = A + Bx$ leads by linear regression to $A = 0.4767$, $B = 4.4754$ (not including units). Therefore,

$$V_\max = \frac{1}{A} = \pu{2.10 mMs^{-1}}$$

$$K_M = B V_\max = \pu{9.39 mM}$$

For the new situation with different $V$ and $S$ and necessarily different $[E_0]$ , $V_{\max}$ would change because $V_{\max} = k_b [E_0]$ but $K_M$ remains constant, thus

$$V_{\max} = V \frac{K_M + S}{S} = \pu{0.101mM s^{-1}} \frac{(\pu{9.39 mM} + \pu{24 mM})}{\pu{24 mM}} = \pu{0.141 mM s^{-1}}$$

You can then calculate $k_b$ by $k_b = V_{\max}/E_0$.