 Solution

Use the logic approximation for gene regulation, zero basal production and initially vanishing X, Y and Z. Let $K=K_{xy}=K_{xz}=K_{yz}=K_{xx}/2$, $Y_{\infty }=Z_{\infty }={\beta }_y/\alpha$, and assume the activated Q level exceeds the X activation threshold. The circuit combines a coherent feed-forward loop with negative autoregulation. While X is below its self-repression threshold it obeys $\dot{X}={\beta }_x-\alpha X$, hence X first crosses $K$ atY production begins then. While X remains above $K$, $Y=Y_{\infty }[1-e^{-\alpha (t-t_X)}]$. The AND gate for Z first becomes active when Y reaches $K$, soThe approximation drops the short X build-up time because ${\beta }_x\gg {\beta }_y$. Strictly the input must last longer than this onset time to produce a nonzero amount of Z. This requires ${\beta }_y/\alpha >K$: at equality the threshold is approached only asymptotically, and below it Z never switches on. No numerical delay follows without the threshold-to-production ratio.

In the strong negative autoregulation limit, X is held close to $K_{xx}=2K$; a smooth repression law regularizes the rapid switching of the ideal Boolean rule. After the input disappears at $t_0$, production of X stops and $X\simeq 2K{e}^{-\alpha (t-t_0)}$. It takes $(\ln 2)/\alpha$ for X to fall below $K$. Until then Y remains at $Y_{\infty }>K$ and the Z AND gate stays on, so Z stays at $Z_{\infty }$. After that Z decays by dilution, taking another $(\ln 2)/\alpha$ to halve. ThereforeMore generally replace the first $\ln 2$ by $\ln [X(t_0)/K]$. This distinguishes disappearance of the external ligand from disappearance of the persistent X transcription factor.