Solution

On times

τ ≫ ϵ

g_{1} = g_{2} = 0.

The second equation gives

v_{2} = \frac{β}{γ} u v_{1} .

In the first equation the terms

- β u v_{1} + γ v_{2}

then cancel, leaving

u (1 - v_{1} - v_{2}) = α v_{1} .

Solving these two equations gives

v_{1} = \frac{u}{α + u + ( β / γ ) u ^{2}}, v_{2} = \frac{( β / γ ) u ^{2}}{α + u + ( β / γ ) u ^{2}} .

Since

\overset{p}{˙} = k_{3} c_{2} = k_{3} e_{0} v_{2}

\frac{d p}{d t} = A \frac{u ^{2}}{α + u + ( β / γ ) u ^{2}}, A = \frac{k _{3} e _{0} β}{γ} = k_{2} e_{0} s_{0} .

At small substrate concentration this allosteric rate is quadratic,

d p / d t \sim (A / α) u^{2}

, so its graph leaves the origin with zero slope. At large

u

it saturates at

A γ / β = k_{3} e_{0}

. A Michaelis-Menten equation also saturates, but its rate

V_{m a x} u / (K_{M} + u)

is linear near the origin. The allosteric curve is therefore more sigmoidal:

allosteric Michaelis-Menten u \to 0 O (u^{2}) O (u) u \to \infty k_{3} e_{0} V_{m a x}

Solved by gpt-5.6-sol high.

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