= Solution
On times $\tau\gg\epsilon$, the <quasi-steady-state approximation> sets
$$
g_1=g_2=0.
$$
The second equation gives
$$
v_2=\frac{\beta}{\gamma}uv_1.
$$
In the first equation the terms $-\beta uv_1+\gamma v_2$ then cancel, leaving
$$
u(1-v_1-v_2)=\alpha v_1.
$$
Solving these two equations gives
$$
v_1=\frac{u}{\alpha+u+(\beta/\gamma)u^2},
\qquad
v_2=\frac{(\beta/\gamma)u^2}
{\alpha+u+(\beta/\gamma)u^2}.
$$
Since $\dot p=k_3c_2=k_3e_0v_2$,
$$
\boxed{
\frac{dp}{dt}
=A\frac{u^2}{\alpha+u+(\beta/\gamma)u^2},
\qquad
A=\frac{k_3e_0\beta}{\gamma}=k_2e_0s_0
}.
$$
At small substrate concentration this allosteric rate is quadratic, $dp/dt\sim(A/\alpha)u^2$, so its graph leaves the origin with zero slope. At large $u$ it saturates at $A\gamma/\beta=k_3e_0$. A <Michaelis-Menten equation> also saturates, but its rate $V_{\max}u/(K_M+u)$ is linear near the origin. The allosteric curve is therefore more sigmoidal:
$$
\begin{array}{c|cc}
&u\to0&u\to\infty\\ \hline
\text{allosteric}&O(u^2)&k_3e_0\\
\text{Michaelis--Menten}&O(u)&V_{\max}
\end{array}
$$
This is the <quasi-steady rate law for a two-substrate allosteric enzyme>.
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