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20-Bio-A5 Systems Analysis & Control · May 2017

Question 1 of 6: Quasi-Steady-State Equations for a Substrate-Inhibited Mechanism

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

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National Exams / EGBC — May 2017 — 04-Bio-A5 Enzyme and Microbial Kinetics. Three-hour open-book examination; any non-communicating calculator is permitted. Six questions constitute a complete paper (10 + 25 + 15 + 20 + 15 + 15 = 100 marks). Content spans quasi-steady-state derivation of a substrate-inhibited enzyme mechanism, diffusion-reaction in an immobilized-enzyme bead (Thiele modulus / effectiveness factor), the rapid-equilibrium integrated batch equation for substrate inhibition, a growth-associated-product batch fermentation, Contois-kinetics chemostat design, and fed-batch quasi-steady-state theory.

Reference texts: Bailey & Ollis, Biochemical Engineering Fundamentals (2nd ed.) — enzyme kinetics, substrate inhibition, immobilized-enzyme diffusion-reaction and effectiveness factors, Monod/Contois growth models, chemostat and fed-batch theory. All quantities are used exactly as printed on the exam (SI/CGS mixed units as given).

Question 1: Quasi-Steady-State Equations for a Substrate-Inhibited 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. The three-step mechanism above: productive binding ($k_1,k_{-1}$), catalysis/product release ($k_2,k_{-2}$, written reversibly), and non-productive substrate-inhibition binding of a second S onto ES to form $ES_2$ ($k_3,k_{-3}$).

Find. The closed set of governing equations (mass balance + quasi-steady-state balances + rate expression) that determine $dP/dt$ as a function of $S$, $E_0$ and the six rate constants — not the eliminated closed-form rate law itself.

E + S ES E + P k1 k-1 k2 k-2 k3 k-3 (+S) ES2 (inhibited)
Fig. 1 — substrate-inhibition mechanism: E and S bind reversibly to ES, which either releases product (E + P) or binds a second S to form the dead-end complex $ES_2$.

Approach. Apply the quasi-steady-state assumption (QSSA) to both enzyme-bound intermediates ($ES$ and $ES_2$), close the system with the total-enzyme mass balance, and state the product-formation rate in terms of $[ES]$ — four equations in total, left unsolved as instructed.

  1. Total-enzyme mass balance. The enzyme is distributed among three forms at every instant: $$E_0=[E]+[ES]+[ES_2].$$
  2. QSSA on the inhibited complex $ES_2$. $ES_2$ is formed only from $ES+S$ and decays only back to $ES+S$, so at quasi-steady state its net rate of formation is zero: $$\frac{d[ES_2]}{dt}=k_3[ES][S]-k_{-3}[ES_2]=0 \quad\Rightarrow\quad [ES_2]=\frac{k_3}{k_{-3}}[ES][S].$$
  3. QSSA on the central complex $ES$. $ES$ is formed from $E+S$ and from $E+P$ (reverse of the product step), and is consumed by dissociation back to $E+S$, by forward catalysis to $E+P$, and by the side reaction to $ES_2$ (with the reverse of that side reaction feeding back in): $$\frac{d[ES]}{dt}=k_1[E][S]+k_{-2}[E][P]+k_{-3}[ES_2]-\big(k_{-1}+k_2+k_3[S]\big)[ES]=0.$$
  4. Rate of product formation. $P$ is formed only by the (reversible) catalytic step: $$\frac{dP}{dt}=k_2[ES]-k_{-2}[E][P].$$

Equations 1–4, together with the initial condition $P(0)=0$, form a closed algebraic-differential system: eliminating $[E]$ and $[ES_2]$ between equations 1–3 gives $[ES]$ explicitly in terms of $S$, $P$, $E_0$ and the six rate constants, which substituted into equation 4 yields $dP/dt(S,P,E_0,k_1,k_{-1},k_2,k_{-2},k_3,k_{-3})$ — the algebra is not required by the question.

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