Determine vmax and enzyme concentration (Michaelis–Menten) I am given that the enzyme concentration is $\pu{15 nM}$ and the following data:
$$
\begin{array}{rl}
\hline 
S/\pu{mM} &amp; V/\pu{mM s^{-1}} \\
\hline
1 &amp; 0.202 \\
2 &amp; 0.368 \\ 
5 &amp; 0.729 \\ 
10 &amp; 1.08 \\ 
20 &amp; 1.43 \\ 
50 &amp; 1.77 \\
100 &amp; 1.92 \\
\hline
\end{array}
$$
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?
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 $k_b = V_{\max}/E_0$.