Question 1 of 5: Enzymes — Definition, Classification & First-Principles Derivation of the Michaelis–Menten Equation

Question 1: Enzymes — Definition, Classification & 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.

What is an enzyme? An enzyme is a biological catalyst — almost always a protein (though a small number of catalytic RNAs, "ribozymes," also exist) — that accelerates the rate of a specific biochemical reaction by many orders of magnitude without itself being consumed. It works by binding its substrate(s) at a precisely shaped active site and stabilizing the reaction's transition state, thereby lowering the activation energy of the reaction pathway; the enzyme is regenerated unchanged at the end of each catalytic cycle and can turn over many substrate molecules per second.

Classes of enzymes. The International Union of Biochemistry and Molecular Biology (IUBMB) groups every enzyme into one of six major classes by the type of chemical reaction it catalyzes (a seventh class, translocases, was split out of the hydrolases in a 2018 IUBMB update for enzymes that move ions/molecules across membranes, but the classical six-class scheme below is what a 2019 exam expects):

Approach to the derivation. The classical (Michaelis–Menten/Briggs–Haldane) derivation starts from the elementary two-step reaction mechanism for a single-substrate enzyme reaction, writes mass balances on the free and bound enzyme, and closes the system with the steady-state approximation on the enzyme–substrate complex.

1. Elementary mechanism. Free enzyme E binds substrate S reversibly to form the enzyme–substrate complex ES, which then 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 appears only in the breakdown step, so the reaction velocity is $$v=\frac{d[P]}{dt}=k_2[ES]$$ which is unknown until [ES] is expressed in terms of measurable quantities ([S] and total enzyme).
3. Enzyme mass balance. Enzyme is neither created nor destroyed — it is only ever free (E) or bound (ES) — so the total enzyme concentration is conserved at every instant: $$[E]_0=[E]+[ES]\;\Rightarrow\;[E]=[E]_0-[ES]$$
4. Steady-state approximation on [ES]. After a brief initial transient, [ES] is formed and consumed at essentially equal rates, so its net rate of change is taken as zero: $$\frac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES]\approx 0$$ Substituting [E]=[E]₀−[ES] from Step 3, $$k_1([E]_0-[ES])[S]=(k_{-1}+k_2)[ES]$$
5. Solve for [ES]. Expanding and collecting [ES] terms, $$k_1[E]_0[S]=[ES]\big(k_1[S]+k_{-1}+k_2\big)\;\Rightarrow\;[ES]=\frac{[E]_0[S]}{[S]+\dfrac{k_{-1}+k_2}{k_1}}$$ Defining the Michaelis constant $K_m\equiv(k_{-1}+k_2)/k_1$ (a lumped rate-constant ratio, units of concentration) collapses this to $$[ES]=\frac{[E]_0[S]}{K_m+[S]}$$
6. Substitute back into Step 2. With [ES] known, and defining the maximum velocity $V_{max}\equiv k_2[E]_0$ (the rate at total saturation, [S]→∞, when essentially all enzyme is in the ES form), $$\boxed{v=\frac{V_{max}[S]}{K_m+[S]}}$$ This is the Michaelis–Menten equation: at low [S] («K_m) the rate is first-order in substrate (v≈(V_max/K_m)[S]); at high [S] (»K_m) the rate saturates at V_max, zero-order in substrate.

Assumptions. (1) The mechanism is exactly the two-step E+S⇌ES→E+P scheme — no other enzyme forms (e.g. no significant substrate or product inhibition complex) participate. (2) The product-formation step is irreversible ([P] does not react back to ES) — valid for initial-rate measurements before product has accumulated. (3) [S]≫[E]₀ so that the substrate bound in ES is a negligible fraction of total substrate, keeping [S] essentially constant over the measurement window (initial rate conditions). (4) The steady-state approximation d[ES]/dt≈0 holds — true after a short pre-steady-state transient (typically milliseconds) has decayed. (5) The system is well mixed and isothermal, and each elementary step obeys simple mass-action (first- or second-order elementary) kinetics.