Question 2 of 5: First-Principles Derivation of the Michaelis–Menten Equation

Question 2: First-Principles Derivation of the Michaelis–Menten Equation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Approach. Start from the elementary two-step enzyme mechanism, write a mass balance on total enzyme, and close the system with the pseudo-steady-state approximation on the enzyme–substrate complex (the Briggs–Haldane derivation).

1. Elementary mechanism. Free enzyme E binds substrate S reversibly to form the enzyme–substrate complex ES, which breaks down irreversibly to release product P and regenerate free enzyme: $$E+S\ \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}\ ES\ \xrightarrow{k_2}\ E+P$$
2. Rate of product formation. Product is generated only in the breakdown step, so the reaction velocity is $$v=\frac{d[P]}{dt}=k_2[ES]$$ — not yet useful, since [ES] is not directly measurable.
3. Total-enzyme mass balance. Enzyme is neither created nor destroyed; at every instant it is either free or bound: $$[E]_0=[E]+[ES]\quad\Rightarrow\quad[E]=[E]_0-[ES]$$
4. Pseudo-steady-state approximation on [ES]. Because $k_1,k_{-1},k_2\gg$ the rate at which [S] itself changes, [ES] is formed and consumed at essentially equal rates after a brief initial transient: $$\frac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES]\approx0$$ Substituting [E] from Step 3, $$k_1\big([E]_0-[ES]\big)[S]=(k_{-1}+k_2)[ES]$$
5. Solve for [ES]. Expanding and collecting terms in [ES], $$k_1[E]_0[S]=[ES]\Big(k_1[S]+k_{-1}+k_2\Big)\quad\Rightarrow\quad[ES]=\frac{[E]_0[S]}{[S]+\dfrac{k_{-1}+k_2}{k_1}}$$ Defining the Michaelis constant $K_m\equiv(k_{-1}+k_2)/k_1$ (a lumped ratio of rate constants, units of concentration), $$[ES]=\frac{[E]_0[S]}{K_m+[S]}$$
6. Substitute back into the rate expression. Defining the maximum velocity $V_m\equiv k_2[E]_0$ (the rate at total saturation, $[S]\to\infty$, when essentially all enzyme is bound as ES), $$\boxed{v=\frac{V_mS}{K_m+S}}$$

Assumptions. (1) Only the elementary scheme $E+S\rightleftharpoons ES\to E+P$ operates — no substrate/product inhibition complexes. (2) The product step is irreversible (valid for initial-rate measurements, before product accumulates enough to drive a reverse reaction). (3) $[S]\gg[E]_0$, so the substrate consumed in forming ES is a negligible fraction of total substrate, keeping [S] essentially constant over the measurement window. (4) $d[ES]/dt\approx0$ (pseudo-steady state), valid once the sub-second pre-steady-state transient has decayed. (5) The system is well mixed and isothermal, and each step follows simple mass-action kinetics.