Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2007 / B / 5 / iii

Table of contents
- Solution iii

Solution

The conserved total is

e_{0} = [E] (0)

under the chosen initial conditions. Substitute

E = e_{0} - X - Y - E_{i}

into the equations from the previous part. This closes the subsystem for

S, X, Y, E_{i}

; product can be recovered afterward by integrating

\dot{P} = k_{3} Y

.

Let

k_{Σ} = k_{- 1} + k_{2}

and

K_{m} = k_{Σ} / k_{1}

. The required nondimensionalization is completed by

σ = \frac{s _{0}}{K _{m}}, ρ = \frac{k _{- 1}}{k _{2}}, ε = \frac{e _{0}}{s _{0} + K _{m}}, ψ = \frac{k _{3} + k _{4}}{k _{Σ}}, ϕ = \frac{k _{4}}{k _{Σ}}, T = ε k_{Σ} t .

Indeed

X = e_{0} σ x / (1 + σ)

and the free-enzyme fraction is

1 - σ x / (1 + σ) - y - e_{i}

. Substitution and division by the indicated time scale give

s_{T} ε x_{T} ε y_{T} ε (e_{i})_{T} = - s [(1 + σ) - σ x - (1 + σ) y - (1 + σ) e_{i}] + \frac{ρ}{1 + ρ} x, = s [(1 + σ) - σ x - (1 + σ) y - (1 + σ) e_{i}] - x, = \frac{σ}{( 1 + σ ) ( 1 + ρ )} x - ψ y, = ϕ y .

For example

k_{2} / k_{Σ} = 1/ (1 + ρ)

explains the coefficient in the third equation. The small multiplier on the intermediate-species derivatives identifies their potentially fast adjustment relative to substrate depletion.

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