Rapid-Equilibrium Rate Law for a Mixed-Inhibition Mechanism: 20-Bio-A5 Systems Analysis & Control, Undated paper, Q1

Question 1: Rapid-Equilibrium Rate Law for a Mixed-Inhibition 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 three-branch mechanism: productive binding/catalysis $E+S\rightleftharpoons ES\rightarrow E+P$; inhibitor $I_1$ binds only the $ES$ complex (uncompetitive branch, dissociation constant $K_{i,1}=k_{-i,1}/k_{i,1}$); inhibitor $I_2$ binds only the free enzyme $E$ (competitive branch, $K_{i,2}=k_{-i,2}/k_{i,2}$).

Find. The closed set of rapid-equilibrium and mass-balance equations that determine $v=dP/dt$ as a function of $S$, $E_0$, $[I_1]$, $[I_2]$ and the rate constants — not necessarily reduced to one final formula.

Fig. 1 — mixed-inhibition mechanism: $I_1$ binds the $ES$ complex (uncompetitive branch, top), $I_2$ binds free $E$ (competitive branch, bottom); catalysis proceeds only through the central $ES$ pathway.

Approach. Apply rapid equilibrium (all binding steps at equilibrium, $k_2\ll$ the binding/dissociation rates) to write each enzyme-bound species as a dissociation-constant expression in $[E]$, close the system with the total-enzyme balance, and state the rate through the single catalytic step.

Dissociation constants (rapid equilibrium). $$K_s=\frac{k_{-1}}{k_1}=\frac{[E][S]}{[ES]},\qquad K_{i,1}=\frac{k_{-i,1}}{k_{i,1}}=\frac{[ES][I_1]}{[ESI_1]},\qquad K_{i,2}=\frac{k_{-i,2}}{k_{i,2}}=\frac{[E][I_2]}{[EI_2]}.$$
Express every bound species in terms of $[E]$. $$[ES]=\frac{[E][S]}{K_s},\qquad [ESI_1]=\frac{[ES][I_1]}{K_{i,1}}=\frac{[E][S][I_1]}{K_sK_{i,1}},\qquad [EI_2]=\frac{[E][I_2]}{K_{i,2}}.$$
Total-enzyme mass balance. $$E_0=[E]+[ES]+[ESI_1]+[EI_2]=[E]\left(1+\frac{S}{K_s}+\frac{S\,I_1}{K_sK_{i,1}}+\frac{I_2}{K_{i,2}}\right).$$
Rate of product formation. Only $ES$ turns over: $$v=\frac{dP}{dt}=k_2[ES]=k_2\frac{[E]S}{K_s}.$$ Substituting $[E]$ from Step 3 gives the (unsolved, as instructed) system that determines $v(S,E_0,I_1,I_2)$; eliminating $[E]$ explicitly yields $$v=\frac{k_2E_0S}{K_s\left(1+\dfrac{I_2}{K_{i,2}}\right)+S\left(1+\dfrac{I_1}{K_{i,1}}\right)}.$$

Concept & Theory

Topic. Rapid-equilibrium derivation of a two-inhibitor mixed-inhibition rate law.

Key relations. Rapid equilibrium replaces every fast binding step with its dissociation-constant algebraic relation ($K=k_{\text{off}}/k_{\text{on}}$); the total-enzyme balance sums every enzyme-containing species; only the catalytically competent complex ($ES$) contributes to $v$.

Why this works. When $k_2$ is much smaller than the binding/dissociation rate constants, every binding equilibrium re-establishes essentially instantaneously relative to the slow catalytic step, so each complex concentration can be written algebraically instead of via its own differential equation.

Common pitfall. Binding $I_1$ to $ES$ (uncompetitive) is not interchangeable with binding $I_2$ to $E$ (competitive) — the former's term scales with $S$ (it only matters once $ES$ has formed), the latter's does not; swapping which inhibitor multiplies which side of the denominator gives the wrong inhibition type entirely.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 3 (enzyme inhibition mechanisms, rapid-equilibrium and QSSA rate laws).

Question 1 of 6: Rapid-Equilibrium Rate Law for a Mixed-Inhibition Mechanism

Question 1: Rapid-Equilibrium Rate Law for a Mixed-Inhibition Mechanism (10 marks)