Chemical reaction kinetics: Kinetics of multi-step reactions
The Hill equation
Now suppose that in an enzymatic reaction the E enzyme must bind on \(n\) sites to form complex C. The reaction scheme is now as follows: \[\text{E}+n\,\text{S }{\mathop{\rightleftharpoons}\limits_{k_{2}}^{k_1}} \text{ C}{\mathop{\longrightarrow} \limits_{}^{k_3}} \text{ E} + \text{P}\] For the concentration \(c\) of the complex C, we can then write: \[\frac{\dd c}{\dd t}=k_1s^ne - (k_2+k_{3})c\] The pseudo steady state approximation implies \(\frac{dc}{dt}=0\) and thus: \[k_1s^ne - (k_2+k_3)c=0\] Substitution of \(c+e=e_0\) and isolation of \(c\) then gives \[c=\frac{e_0s^n}{K_m+s^n}\] with the Michaelis-Menten constant \[K_m=\frac{k_2+k_3}{k_1}\tiny.\] The reaction rate \(r\) is then given by \[r=\frac{V_{\max}\cdot s^n}{K_m+s^n}\] where \(V_{\max}=n\,k_3e_0\). This is called the Hill equation.