Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 6C (Mathematical Biology)

The first reaction is reversible, with reverse rate $k_{-1}{c}_1$. Applying the law of mass action to each reaction givesEach term has the sign dictated by whether the corresponding species is consumed or produced.

Adding the equations for the free enzyme and its two complexes givesThusThis conservation law says that the enzyme is neither created nor destroyed: each enzyme molecule is either free, bound in $C_1$, or bound in $C_2$.

Use enzyme conservation to writeSince $\tau =k_1{e}_{0}t$, $u=s/s_0$, $v_i=c_i/e_0$, and $ϵ=e_0/s_0$, substitution into the mass-action equations givesandwhereThe factors of $ϵ$ multiplying $d{v}_i/d\tau$ identify the complex concentrations as fast variables in a singular perturbation.

On times $\tau \gg ϵ$, the quasi-steady-state approximation setsThe second equation givesIn the first equation the terms $-\beta u{v}_1+\gamma v_2$ then cancel, leavingSolving these two equations givesSince $\dot{p}=k_3{c}_2=k_3{e}_0{v}_2$,

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 $A\gamma /\beta =k_3{e}_0$. 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:This is the quasi-steady rate law for a two-substrate allosteric enzyme.