Michaelis–Menten Enzyme Kinetics — Competitive, Non-Competitive & Uncompetitive Inhibition: 23-Chem-B4 Biochemical Engineering, December 2016, Q1

Question 1: Michaelis–Menten Enzyme Kinetics — Competitive, Non-Competitive & Uncompetitive Inhibition (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.

Given.

Quantity	Symbol	Value
Michaelis constant	K_m	4.7×10^-5 M
Maximum rate	v_max	22 µmol·L^-1·min^-1
Substrate concentration	S	2×10^-4 M
Inhibitor concentration	[I]	5×10^-4 M
Inhibition constant (all three cases)	K_i	3×10^-4 M

Find. The reaction rate v with (a) a competitive inhibitor, (b) a non-competitive inhibitor, (c) an uncompetitive inhibitor, and the uninhibited rate v₀.

Approach. Each inhibition mode modifies the basic Michaelis–Menten equation v=v_maxS/(K_m+S) in a distinct, diagnostic way — competitive inhibition inflates the apparent K_m only, uncompetitive inhibition deflates both v_max and K_m by the same factor, and non-competitive inhibition deflates v_max only — so all four cases are evaluated from the one set of given numbers.

Check

"Non-competitive" is taken as the classical case (inhibitor binds E and ES with equal affinity, α=α'), i.e. K_m is unchanged and only v_max is scaled down — the standard textbook definition used when a single K_i is given for the mode, as here.

Uninhibited rate. $$v_0=\frac{v_{max}S}{K_m+S}=\frac{22\times(2\times10^{-4})}{4.7\times10^{-5}+2\times10^{-4}}=\boxed{17.81\ \mu\text{mol}\cdot\text{L}^{-1}\text{min}^{-1}}$$
(a) Competitive inhibition — apparent K_m only. The inhibitor competes for the free-enzyme active site, so it raises the apparent K_m by the factor (1+[I]/K_i) while leaving v_max untouched: $$K_m'=K_m\left(1+\frac{[I]}{K_i}\right)=4.7\times10^{-5}\left(1+\frac{5}{3}\right)=1.253\times10^{-4}\ \text{M}$$ $$v_a=\frac{v_{max}S}{K_m'+S}=\frac{22\times(2\times10^{-4})}{1.253\times10^{-4}+2\times10^{-4}}=\boxed{13.52\ \mu\text{mol}\cdot\text{L}^{-1}\text{min}^{-1}}$$
(b) Non-competitive inhibition — apparent v_max only. Binding away from the active site (at E and ES equally) removes a fixed fraction of enzyme from productive turnover regardless of S, scaling v_max down by (1+[I]/K_i) while K_m is unchanged: $$v_{max}'=\frac{v_{max}}{1+[I]/K_i}=\frac{22}{1+5/3}=8.25\ \mu\text{mol}\cdot\text{L}^{-1}\text{min}^{-1}$$ $$v_b=\frac{v_{max}'S}{K_m+S}=\frac{8.25\times(2\times10^{-4})}{4.7\times10^{-5}+2\times10^{-4}}=\boxed{6.680\ \mu\text{mol}\cdot\text{L}^{-1}\text{min}^{-1}}$$
(c) Uncompetitive inhibition — K_m and v_max scaled together. The inhibitor binds only the ES complex, so both constants are divided by the same factor, which is algebraically equivalent to inflating the effective S term: $$v_c=\frac{v_{max}S}{K_m+S(1+[I]/K_i)}=\frac{22\times(2\times10^{-4})}{4.7\times10^{-5}+(2\times10^{-4})(1+5/3)}=\boxed{7.582\ \mu\text{mol}\cdot\text{L}^{-1}\text{min}^{-1}}$$

Fig. 1 — reaction rate at S=2×10^-4 M, [I]=5×10^-4 M under each inhibition mode. Competitive inhibition is the least severe here because [S]>K_m already out-competes much of the inhibitor at the active site; non-competitive and uncompetitive both suppress the rate by more than half since neither can be relieved by adding more substrate.

Case	Reaction rate, v	% of uninhibited rate
No inhibitor (v₀)	17.81 µmol·L^-1min^-1	100%
(a) Competitive	13.52 µmol·L^-1min^-1	75.9%
(b) Non-competitive	6.680 µmol·L^-1min^-1	37.5%
(c) Uncompetitive	7.582 µmol·L^-1min^-1	42.6%

Topic: Reversible enzyme inhibition modalities in Michaelis–Menten kinetics.

Key relations: Competitive: $v=v_{max}S/[K_m(1+[I]/K_i)+S]$ (K_m apparent ↑, v_max unchanged). Non-competitive (classical): $v=[v_{max}/(1+[I]/K_i)]\,S/(K_m+S)$ (v_max apparent ↓, K_m unchanged). Uncompetitive: $v=v_{max}S/[K_m+S(1+[I]/K_i)]$ (both K_m and v_max apparent ↓ by the same factor).

Why this works: Each inhibitor traps a different species on the reaction pathway — competitive inhibitors bind free E (outcompeted by raising S), uncompetitive inhibitors bind only ES (cannot be outcompeted by raising S; in fact more S makes more ES to trap), and non-competitive inhibitors bind E and ES equally (removes a constant fraction of active enzyme irrespective of S). This is exactly why competitive inhibition is overcome at saturating S while the other two are not.

Common pitfall: Applying the "raise K_m" rule to non-competitive or uncompetitive inhibition, or assuming all inhibitor types reduce v_max. Only non-competitive and uncompetitive inhibition reduce the apparent v_max; competitive inhibition leaves the true v_max reachable at sufficiently high S.

Source: Bailey & Ollis, Biochemical Engineering Fundamentals, 2nd ed., Ch. 2 (enzyme kinetics, reversible inhibition).

Question 1 of 5: Michaelis–Menten Enzyme Kinetics — Competitive, Non-Competitive & Uncompetitive Inhibition

Question 1: Michaelis–Menten Enzyme Kinetics — Competitive, Non-Competitive & Uncompetitive Inhibition (20 marks)