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The loaded-carrier balance gives $J=k(p_e-p_i)$. Substitute the preceding two complex concentrations and collect the current terms:Next carrier conservation givesEliminate $c_e$ between these relations and set $K_d=k/k_{+}$. Simplification yieldsHere $c_0$ must be the total carrier protein, as in the conservation equation. Calling it total substrate in the final line of the scan is a typographical inconsistency; total substrate is the different invariant $s_0$.

For carrier-turnover saturation of passive transport at fixed $s_i$, write $A=s_i+K+K_d$, $D=A(K+K_d)-K_d^2>0$ and $C=K_{d}K{k}_{+}{c}_0/2>0$. Then $J=C(s_e-s_i)/(A{s}_e+D)$, soThe flux is outward when the substrate gradient is reversed and inward when $s_e>s_i$. Saturation occurs because there is a finite carrier population and finite binding-release/conformational turnover: arbitrarily abundant exterior substrate cannot make carriers cycle arbitrarily quickly. This is passive facilitated transport, so no maintained flux remains when the two free concentrations agree. Varying $s_e$ at fixed $s_i$ compares reservoir-controlled steady states or systems with different total substrate; it is not a trajectory preserving one fixed $s_0$.