Batch Substrate-Inhibited Kinetics — Time to Half-Consume Substrate: 20-Bio-A5 Systems Analysis & Control, May 2017, Q3

Question 3: Batch Substrate-Inhibited Kinetics — Time to Half-Consume Substrate (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. Rapid-equilibrium substrate-inhibition rate law $v=\dfrac{V_{\max}S}{K_m'+S+S^2/K_s}$; $S_0=10$ mol/L, $K_m'=5$ mol/L, $K_s=1$ mol/L; $V_{\max}$ (i.e. $k_2E_0$) is not given numerically.

Find. The time $t_{1/2}$ for $S$ to fall to $S_0/2$, expressed in terms of the unknown $V_{\max}$; and the behaviour of this result as $S_0\to0$ compared with the non-inhibited (simple Michaelis–Menten) batch case.

Approach. Write the batch mass balance $-dS/dt=v$, separate variables, and integrate the rapid-equilibrium rate law in closed form — the reactor volume $V_L$ cancels (it appears on both sides of a concentration-based balance) so it never enters the answer.

Batch balance. $$-\frac{dS}{dt}=\frac{V_{\max}S}{K_m'+S+S^2/K_s}.$$
Separate and integrate. $$-\left(\frac{K_m'}{S}+1+\frac{S}{K_s}\right)dS=V_{\max}\,dt \ \Rightarrow\ K_m'\ln\frac{S_0}{S}+(S_0-S)+\frac{S_0^2-S^2}{2K_s}=V_{\max}t.$$
Evaluate at $S=S_0/2=5$ mol/L. $$5\ln 2+(10-5)+\frac{10^2-5^2}{2(1)}=3.466+5.000+37.500=45.97,$$ so $$\boxed{t_{1/2}=\frac{45.97}{V_{\max}}}$$ (time units follow from those of $V_{\max}$, e.g. minutes if $V_{\max}$ is in mol/(L·min)).
Limit $S_0\to0$. As $S_0\to0$ (with $S=S_0/2\to0$ too), the quadratic term $(S_0^2-S^2)/(2K_s)$ vanishes as $S_0^2$, strictly faster than the logarithmic and linear terms, so the integrated equation collapses to $$K_m'\ln\frac{S_0}{S}+(S_0-S)=V_{\max}t,$$ which is exactly the classical integrated batch equation for a non-substrate-inhibited (simple Michaelis–Menten) reaction with Michaelis constant $K_m'$. In other words, substrate inhibition is a high-concentration effect; at vanishingly small $S_0$ the inhibition term contributes nothing and $t_{1/2}$ approaches the plain-MM value using the same $K_m'$.

Result	Value
$t_{1/2}\cdot V_{\max}$ coefficient	45.97 (mol/L)
$t_{1/2}$	$45.97/V_{\max}$
$S_0\to0$ limiting equation	$K_m'\ln(S_0/S)+(S_0-S)=V_{\max}t$ (plain MM)

Concept & Theory

Topic. Rapid-equilibrium substrate-inhibition kinetics, integrated batch-reactor form.

Key relations. $v=V_{\max}S/(K_m'+S+S^2/K_s)$; integrated batch form $K_m'\ln(S_0/S)+(S_0-S)+(S_0^2-S^2)/(2K_s)=V_{\max}t$; reduces to plain MM ($K_m'\ln(S_0/S)+(S_0-S)=V_{\max}t$) when $S\ll K_s$.

Why this works. The extra $S^2/K_s$ denominator term represents the non-productive $ES_2$ complex of Question 1 siphoning enzyme away from the productive pathway at high $S$; integrating $1/v$ term-by-term (each of $K_m'/S$, 1, and $S/K_s$ has an elementary antiderivative) is what makes the closed form possible without numerically solving the ODE.

Common pitfall. Forgetting that $V_{\max}$ is not given here — the correct answer is a formula for $t_{1/2}$ in terms of $V_{\max}$, not a numeric time; students who assume a $V_{\max}$ value or drop it from the answer are not doing what the question actually supports.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 3 (rapid-equilibrium enzyme kinetics, substrate inhibition, integrated batch reactor design equations).

Question 3 of 6: Batch Substrate-Inhibited Kinetics — Time to Half-Consume Substrate

Question 3: Batch Substrate-Inhibited Kinetics — Time to Half-Consume Substrate (15 marks)