# Enzyme Kinetics - Given Km find substrate concentration at a certain velocity

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 $$K_\text{m}$$:

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

• 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. – Curt F. Apr 29 '19 at 21:04
• 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. – Curt F. Apr 29 '19 at 21:05
• My typo, now corrected. – 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: 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 ($$v_{\circ}$$).