Determining the Inhibition Constant and Inhibition Type: 20-Bio-A5 Systems Analysis & Control, December 2017, Q2

Question 2: Determining the Inhibition Constant and Inhibition Type (15 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. Initial-rate data at two inhibitor levels:

$I$ (mmol/mL)	$S$ (mmol/mL)	$V$ (mmol/(mL·min))
0	1.0	16.4
0	0.20	9.00
1.2	1.0	13.30
1.2	0.20	5.70

Find. The inhibition constant(s) $K_i$ and the inhibition mechanism (competitive / uncompetitive / noncompetitive / mixed).

Fig. 1 — Lineweaver–Burk (double-reciprocal) plot of the two data sets. Both lines extrapolate to nearly the same intercept ($1/V_{\max}$) but the inhibited line's slope ($K_{m,\text{app}}/V_{\max,\text{app}}$) is much steeper — $K_m$ roughly doubling while $V_{\max}$ barely changes is the signature of a predominantly competitive inhibitor.

Approach. Fit the uninhibited ($I=0$) pair exactly with the standard Michaelis–Menten form (two points, two unknowns) to get $K_m$, $V_{\max}$; then fit the general (mixed) inhibition rate law $V=V_{\max}S/(\alpha K_m+\alpha' S)$ to the $I=1.2$ pair — with $K_m$, $V_{\max}$ already known this is again exactly two equations in the two remaining unknowns $\alpha,\alpha'$ — and convert to $K_i=I/(\alpha-1)$, $K_i'=I/(\alpha'-1)$.

Uninhibited kinetic constants. Solving $V=V_{\max}S/(K_m+S)$ exactly for the two $I=0$ points: $$K_m=0.259\ \text{mmol/mL}, \qquad V_{\max}=20.6\ \text{mmol/(mL}\cdot\text{min)}.$$
General (mixed) inhibition rate law. $$V=\frac{V_{\max}S}{\alpha K_m+\alpha' S}, \qquad \alpha=1+\frac{I}{K_i}\ \text{(binds free }E\text{)}, \qquad \alpha'=1+\frac{I}{K_i'}\ \text{(binds }ES\text{)}.$$
Solve for $\alpha,\alpha'$ from the $I=1.2$ pair. With $K_m$, $V_{\max}$ from step 1, the two $(S,V)$ points at $I=1.2$ give two linear equations in $\alpha,\alpha'$: $$\alpha K_m+\alpha'(1.0)=\frac{V_{\max}(1.0)}{13.30}, \qquad \alpha K_m+\alpha'(0.20)=\frac{V_{\max}(0.20)}{5.70}.$$ Solving simultaneously: $$\alpha=2.00, \qquad \alpha'=1.035.$$
Convert to inhibition constants. $$K_i=\frac{I}{\alpha-1}=\frac{1.2}{1.00}=\boxed{1.20\ \text{mmol/mL}}, \qquad K_i'=\frac{I}{\alpha'-1}=\frac{1.2}{0.035}\approx 34.5\ \text{mmol/mL}.$$
Classify the inhibition. $K_i'$ is roughly 29× larger than $K_i$ — the inhibitor binds essentially only the free enzyme $E$ and has negligible affinity for the $ES$ complex ($\alpha'\approx 1$, so $V_{\max}$ is barely depressed while $K_m$ roughly doubles). This is the limiting behaviour of competitive inhibition, with $\boxed{K_i\approx 1.20\ \text{mmol/mL}}$ the operative constant.

Result	Value
Uninhibited $K_m$, $V_{\max}$	0.259 mmol/mL, 20.6 mmol/(mL·min)
Apparent (inhibited) $K_{m,\text{app}}$, $V_{\max,\text{app}}$	0.500 mmol/mL, 19.95 mmol/(mL·min)
$\alpha$, $\alpha'$	2.00, 1.035
$K_i$ (competitive, dominant)	1.20 mmol/mL
$K_i'$ (uncompetitive component)	≈34.5 mmol/mL (negligible)
Inhibition type	Competitive

Concept & Theory

Topic. Classifying reversible enzyme inhibition from initial-rate data using the general mixed-inhibition model.

Key relations. $V=V_{\max}S/(\alpha K_m+\alpha' S)$ with $\alpha=1+I/K_i$, $\alpha'=1+I/K_i'$; the classical cases are limits of this one model — competitive ($K_i'\to\infty$, $\alpha'=1$), uncompetitive ($K_i\to\infty$, $\alpha=1$), noncompetitive ($K_i=K_i'$), mixed (both finite and unequal).

Why this works. With exactly four independent $(I,S,V)$ measurements and four unknowns ($K_m$, $V_{\max}$, $K_i$, $K_i'$), the system is exactly determined — fitting the general model first and then inspecting the RELATIVE size of $K_i$ and $K_i'$ is more robust than guessing the inhibition type from a qualitative shift in $K_m$/$V_{\max}$ alone.

Common pitfall. Concluding noncompetitive inhibition just because $V_{\max}$ changed at all — a small $V_{\max}$ shift can be experimental scatter (only two $S$ levels were measured) riding on top of a dominant competitive effect; always compute $K_i'/K_i$ and judge by its order of magnitude, not by whether $V_{\max}$ moved at all.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 3 (reversible enzyme inhibition, Lineweaver–Burk analysis).

Question 2 of 6: Determining the Inhibition Constant and Inhibition Type

Question 2: Determining the Inhibition Constant and Inhibition Type (15 marks)