 Solution

The sequential-enzyme Hill limit uses the intended distinguished limit for $n\ge 2$ holds the other rates fixed while $k_1\to 0$, $k_n\to \infty$, and $k_1{k}_n$ tends to a positive finite constant. Then for every $1\le j<n$, ${\varphi }_j={\prod }_{i=1}^j{k}_i/(k_{-i}+k_i^{+})=O(k_1)\to 0$, whereas ${\varphi }_n$ remains finite and nonzero. Hence intermediate complexes have negligible occupancy relative to the unbound and fully bound states.

In the quasi-steady-state approximation rate, all intermediate numerator and denominator terms disappear. Multiplying by $1/{\varphi }_n={\prod }_{i=1}^n{K}_i$ givesThis is the Hill equation with maximal rate $E_0{k}_n^{+}$ and effective half-saturation concentration $(\prod K_i{)}^{1/n}$. Here its power $n$ follows from the sequential binding mechanism and the suppression of intermediate states.

The phrase “$k_1{k}_n$ finite” needs this scaling interpretation for a nontrivial limiting rate. Merely bounding that product, while permitting unrelated rates to vary, does not ensure that all intermediate weights vanish relative to the terminal one. The derivation explicitly supplies the usual fixed-other-rates, positive-product assumptions. For $n=1$, the rate in the box is already the exact one-state quasi-steady-state approximation rate, without this two-endpoint limiting argument.