 Solution

Besides the stated scales, set $P=S_{0}p$, $ϵ=E_0/S_0$, ${\alpha }_i=k_i/k_1$, ${\beta }_i=(k_{-i}+k_i^{+})/(k_1{S}_0)$, and ${\kappa }_i=k_i^{+}/(k_1{S}_0)$. Then $r_i={\alpha }_i{c}_{i-1}s-{\beta }_i{c}_i$, and direct substitution, remembering $d/dt=k_1{E}_{0}d/d\tau$, givesThe exact initial values are $s(0)=1$, $p(0)=0$, $c_0(0)=1$, $c_i(0)=0$ for $1\le i\le n$. The enzyme-state equations carry the small parameter because the enzyme concentration scale is small relative to the initial substrate concentration; the substrate evolves on the chosen slower depletion time scale.