Figure 5

Balance between forward and back reactions leads to a high Hill coefficient when the reactions are saturated. The equilibrium output signal, A^∗, is obtained by solving $\dot{A}=0$ in Eq. (6) as a function of the input signal, S₀. The signal is given as $Ŝ=\varphi S_0/\sigma B_0$. The total amount of reactant is N=R+A. The half-maximal concentrations are set to m=μ=1. The three curves illustrate the solutions for N=1,10,100, with increasing Hill coefficients for higher N values and greater reaction saturation levels.