Inhibitor Constant and Inhibition Type from Initial-Rate Data: 20-Bio-A5 Systems Analysis & Control, Undated paper, Q2

Question 2: Inhibitor Constant and Inhibition Type from Initial-Rate Data (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. Four initial-rate measurements: two without inhibitor ($I=0$) and two at $I=0.6$ mmol/mL, each pair at $S=0.1$ and $S=0.02$ mmol/mL.

$I$ (mmol/mL)	$S$ (mmol/mL)	$V$ (mmol/(mL·min))
0	0.1	1.64
0	0.02	0.9
0.6	0.1	1.33
0.6	0.02	0.57

Find. $K_i$ for the inhibitor, and whether it is competitive, uncompetitive or noncompetitive.

Fig. 2 — Lineweaver–Burk plot: the two lines meet close to the same point on the 1/V axis (Vmax essentially unchanged) while the slope (Km/Vmax) increases with inhibitor — the classic competitive-inhibition signature.

Approach. Fit the Michaelis–Menten (double-reciprocal) line through each pair of points, at $I=0$ and at $I=0.6$, then compare how $V_{\max}$ and $K_m$ shift with $I$ to identify the inhibition type; finally use the appropriate apparent-$K_m$ relation to back out $K_i$.

No-inhibitor fit ($I=0$). Using $1/V=(K_m/V_{\max})(1/S)+1/V_{\max}$ through $(1/S,1/V)=(10,0.6098)$ and $(50,1.1111)$: slope $=0.012533$, intercept $=0.4845$, so $$V_{\max}=2.064\ \text{mmol/(mL}\cdot\text{min)},\qquad K_m=0.02587\ \text{mmol/mL}.$$ (Check: $V=2.064(0.1)/(0.02587+0.1)=1.640$; $V=2.064(0.02)/(0.02587+0.02)=0.900$ — both match the data exactly.)
With-inhibitor fit ($I=0.6$). Through $(10,0.7519)$ and $(50,1.7544)$: slope $=0.025063$, intercept $=0.50126$, so $$V_{\max}'=1.995\ \text{mmol/(mL}\cdot\text{min)},\qquad K_m'=0.05000\ \text{mmol/mL}.$$ (Check: $V=1.995(0.1)/(0.05+0.1)=1.330$; $V=1.995(0.02)/(0.05+0.02)=0.570$ — both match.)
Identify the inhibition type. $V_{\max}'/V_{\max}=1.995/2.064=0.9664$ (a 3.4% shift, within the precision of two-significant-figure rate data — i.e. $V_{\max}$ is unchanged), while $K_m'/K_m=0.05000/0.02587=1.933$ (Km nearly doubles). Unchanged $V_{\max}$ with increased $K_m$ is the defining signature of $$\boxed{\text{competitive inhibition}}$$ (the inhibitor binds only the free enzyme, so it can still be fully out-competed by enough substrate, leaving $V_{\max}$ unaffected).
Solve for $K_i$. For competitive inhibition, $K_m'=K_m(1+I/K_i)$: $$\frac{K_m'}{K_m}=1+\frac{I}{K_i}\ \Rightarrow\ K_i=\frac{I}{K_m'/K_m-1}=\frac{0.6}{1.933-1}=\boxed{0.643\ \text{mmol/mL}}.$$

Result	Value
$V_{\max}$ (no inhibitor)	2.064 mmol/(mL·min)
$K_m$ (no inhibitor)	0.02587 mmol/mL
$V_{\max}'$ (I = 0.6)	1.995 mmol/(mL·min)
$K_m'$ (I = 0.6)	0.05000 mmol/mL
Inhibition type	Competitive
$K_i$	0.643 mmol/mL

it is essential (the only other zero-inhibitor point) and is used above. This inhibitor's behaviour (Km increases, Vmax unchanged) matches the $I_2$ branch of Question 1's mechanism (binds free $E$ only), which is the physical picture of competitive inhibition.

Concept & Theory

Topic. Lineweaver–Burk (double-reciprocal) determination of inhibition type and $K_i$.

Key relations. $1/V=(K_m/V_{\max})(1/S)+1/V_{\max}$; competitive: $K_m^{app}=K_m(1+I/K_i)$, $V_{\max}$ unchanged; uncompetitive: both $K_m^{app}$ and $V_{\max}^{app}$ scale down by the same factor $(1+I/K_i)$; noncompetitive: $V_{\max}^{app}$ scales down, $K_m$ unchanged.

Why this works. The double-reciprocal transform linearises the rate law so the inhibition pattern is read directly off how slope and intercept move: a common y-intercept across inhibitor concentrations (unchanged $1/V_{\max}$) is the geometric signature of purely competitive binding.

Common pitfall. Comparing raw $V$ values at unequal $S$ instead of extrapolated $V_{\max}$/$K_m$ can suggest the wrong inhibition type — always fit the full line (or use two points to solve slope+intercept exactly, as here) rather than eyeballing the raw rate drop.

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

Question 2 of 6: Inhibitor Constant and Inhibition Type from Initial-Rate Data

Question 2: Inhibitor Constant and Inhibition Type from Initial-Rate Data (15 marks)