# Determine vmax and enzyme concentration (Michaelis–Menten)

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

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

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?

• In your text book you should have that $K_m=(k_{-1}+k_2)/k_1$ so this gives you the units. You could try a double reciprocal (Lineweaver-Burk) plot $1/[S],\;vs\;1/V$ assuming $V$ is the initial rate. May 7, 2021 at 10:41

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:

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

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 $$V_\mathrm{max} = \pu{2.10 mM s-1}$$.

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).
• Thank you very much for your answer! It truly helped! May 9, 2021 at 12:08

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 $$k_b = V_{\max}/E_0$$.

• Thank you, I didn't realize that K$_M$ would remain constant which really made it clearer! But I want to calculate [E]$_0$ as well, how can I do that if I don't have k$_b$? May 7, 2021 at 12:25
• You can calculate $k_b$ for the first set of data by $k_b = V_\{max}/[E_0]$ since $[E_0]$ is given (initial enzyme concentration). For the second case, calculate $[E_0]$ by $V_{\max}/k_b$ since $k_b$ is constant. May 7, 2021 at 21:57