Quasi-Steady-State Rate Law for a Two-Site Enzyme Mechanism: 20-Bio-A5 Systems Analysis & Control, December 2017, Q1

Question 1: Quasi-Steady-State Rate Law for a Two-Site Enzyme Mechanism (10 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.

Given. A two-site sequential mechanism: a first substrate binds reversibly to free enzyme ($k_1,k_{-1}$), a second substrate binds reversibly to $ES$ to form $ES_2$ ($k_2,k_{-2}$), and $ES_2$ decomposes irreversibly to release product and regenerate free enzyme ($k_{cat}$).

Find. The closed set of governing equations (mass balance + quasi-steady-state balances + rate expression) that determine $dP/dt$ as a function of $S$, $E_0$, and the five rate constants — not the eliminated closed-form rate law itself.

Approach. Apply the quasi-steady-state assumption (QSSA) to both enzyme-bound intermediates ($ES$ and $ES_2$), close the system with the total-enzyme mass balance, and state the product-formation rate in terms of $[ES_2]$ — four equations in total, left unsolved as instructed.

Total-enzyme mass balance. The enzyme is distributed among three forms at every instant: $$E_0=[E]+[ES]+[ES_2].$$
QSSA on the central complex $ES$. $ES$ is formed from $E+S$ and consumed by dissociation back to $E+S$ and by the forward second-binding step to $ES_2$ (with the reverse of that step feeding back in): $$\frac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES][S]+k_{-2}[ES_2]=0.$$
QSSA on the doubly-bound complex $ES_2$. $ES_2$ is formed from $ES+S$ and is consumed both by dissociation back to $ES+S$ and by the irreversible catalytic step: $$\frac{d[ES_2]}{dt}=k_2[ES][S]-k_{-2}[ES_2]-k_{cat}[ES_2]=0.$$
Rate of product formation. $P$ is formed only by the irreversible catalytic step, which simultaneously regenerates free enzyme: $$\frac{dP}{dt}=k_{cat}[ES_2].$$

Equations 1–4 form a closed algebraic system: eliminating $[E]$ between equations 1–3 gives $[ES]$ and $[ES_2]$ explicitly in terms of $S$, $E_0$, and the five rate constants, which substituted into equation 4 yields $dP/dt(S,E_0,k_1,k_{-1},k_2,k_{-2},k_{cat})$ — the algebra is not required by the question.

Concept & Theory

Topic. Quasi-steady-state derivation of a multi-intermediate (two-substrate-binding-site) enzyme rate law.

Key relations. QSSA sets $d[\text{complex}]/dt=0$ for every enzyme-bound species; the total-enzyme mass balance $E_0=\sum[\text{enzyme forms}]$ closes the system; the observable rate is always the flux through the irreversible product-forming step.

Why this works. Enzyme concentrations are typically orders of magnitude below substrate concentrations, so the bound intermediates equilibrate on a timescale much faster than $S$ or $P$ change — QSSA replaces their differential equations with algebraic ones, which is what makes a multi-step mechanism tractable without integrating every species simultaneously.

Common pitfall. Forgetting that $ES_2$ feeds back into the $ES$ balance through $k_{-2}[ES_2]$ — treating the second binding step as an irreversible sink from $ES$ silently drops a reaction and gives a rate law with the wrong functional dependence on $S$ at high substrate concentration.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 3 (enzyme kinetics, QSSA, multi-intermediate mechanisms).

Question 1 of 6: Quasi-Steady-State Rate Law for a Two-Site Enzyme Mechanism

Question 1: Quasi-Steady-State Rate Law for a Two-Site Enzyme Mechanism (10 marks)