From what I understand about the Michaelis Menten Model

  • Km defines the amount of substrate required to reach half-saturation.
  • 1/2 Vmax corresponds to Km on the x axis
  • Generic formula is v = Kcat [ES] [S] / (Km + [S])

Given Km and the desired velocity in terms of reaction efficiency, can one just plug in Km into the equation and get a relational expression with [S] on one side?

Or do you manipulate the 1/2 Vmax expression?


You can solve for [S]. There are two common ways to write the equation, like you have it: $$v_0 = k_\text{cat} [E_0] \frac{[S]}{K_\text{m} + [S]}$$

or after dividing both numerator and denominator by [S]:

$$v_0 = k_\text{cat} [E_0] \frac{1}{\frac{K_\text{m}}{[S]} + 1}$$

In this form, it is easier to see that [S] occurs once and you can solve for it.

Get [S] out of the denominator of the "big" fraction and isolate the "little" fraction:

$$\frac{K_\text{m}}{[S]} + 1 = \frac{k_\text{cat} [E_0]}{v_0}$$

$$\frac{K_\text{m}}{[S]} = \frac{k_\text{cat} [E_0]}{v_0} - 1$$

Finally, take the reciprocal and multiply by $K_\text{m}$:

$${[S]} = \frac{K_\text{m}}{\frac{k_\text{cat} [E_0]}{v_0} - 1}$$

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    $\begingroup$ I'd recommend writing the ES term as $\ce{[E_0][S]}$ to explicitly indicate you are talking about the total enzyme concentration times the substrate concentration. People might get confused by $\ce{[ES]}$, thinking you mean the concentration of the enzyme-substrate complex, as this notation is often used in e.g. derivations of the MM model. $\endgroup$ – Curt F. Apr 29 '19 at 21:04
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    $\begingroup$ Also, when you do that, you will find that there is an $\ce{[S]}$ on both sides of the equation, so your "explicit" solution for $\ce{[S]}$ really isn't. $\endgroup$ – Curt F. Apr 29 '19 at 21:05
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    $\begingroup$ My typo, now corrected. $\endgroup$ – Karsten Theis Apr 29 '19 at 21:43

You can also solve for $\mathrm{[S]} $ using Lineweaver-Burk plot (or double reciprocal plot). One way of writing the Michaelis Menten kinetic equation given in Wikipedia is: $$v_{\circ} = V_\mathrm{Max} \frac {[S]}{K_\mathrm{M} + [S] } = = k_\text{cat} \mathrm{[E]_{\circ}} \frac {[S]}{K_\mathrm{M} + [S]} $$

If you take the reciprocal of both sides of the equation, you get:

$$\frac{1}{v_{\circ}} = \frac{K_\mathrm{M} + \mathrm{[S]}}{V_\mathrm{Max} \mathrm{[S]}} = \left(\frac{K_\mathrm{M}}{V_\mathrm{Max}}\right)\frac{1}{\mathrm{[S]}} + \frac{1}{V_\mathrm{Max}}$$

Thus, a plot of $\frac{1}{\mathrm{[S]}}$ vs $\frac{1}{v_{\circ}}$ is a straight line, which is called Lineweaver-Burk plot:

Lineweaver-Burk plot

Now, if you know the values of $K_\mathrm{M}$ and $V_\mathrm{Max}$ for your enzyme and substrate, you can draw a plot upper-hand ($x$-$\text{intercept} = -\frac{1}{K_\mathrm{M}}$ and $y$-$\text{intercept} = \frac{1}{V_\mathrm{Max}}$) and determine your substrate concentration ($\mathrm{[S]}$) for each desired velocity ($v_{\circ}$).


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