 Solution

Use the logic approximation for gene regulation: production is either $\beta$ or zero, and every species has linear removal rate $\alpha$. Put $C=\beta /\alpha$ and assume zero basal expression, inactive X before time zero, and all required thresholds strictly between $0$ and $C$. The input is $u(t)=1$ for $0<t<T$ and zero otherwise. The common ordinary differential equation is $\dot{c}=\beta g-\alpha c$, with Boolean production gate $g$. Both networks therefore havewhere the latter uses the stipulated attained steady state. Define the threshold-crossing delayBecause $K_{YY1}>K_{YZ}$, $t_1>t_r$. In both networks, $Y_1$ stays zero until $t_1$ and then follows $Y_1=C[1-e^{-\alpha (t-t_1)}]$ until the OFF response. Write $K_A$ for the activation threshold of $Y_1$ on Z in diagram (a), and $K_I$ for the level of $Y_1$ that blocks Y's repression in diagram (b); these thresholds were not supplied numerically.

Diagram (a). All three inputs X, Y and $Y_1$ must permit Z production. Since Y has already crossed its Z threshold before $Y_1$ starts, the final permission arrives atThus $Z=0$ until $t_a$ and $Z=C[1-e^{-\alpha (t-t_a)}]$ subsequently. Solving $Z=C/2$ givesFor the frequently adopted but extra convention that both required thresholds are $C/2$, this is $3\ln 2/\alpha$. At the OFF step, X's direct AND input disappears, so $Z=C{e}^{-\alpha (t-T)}$ immediately. Y also decays immediately. $Y_1$ remains at $C$ while Y remains above $K_{YY1}$; it starts decaying at $T+{\alpha }^{-1}\ln (C/K_{YY1})$. This is a coherent feed-forward loop persistence response: activation is delayed, while removal of the direct required input switches production off promptly.

Diagram (b). The production gate is $g_Z=u[\neg (Y>K_{YZ})\vee (Y_1>K_I)]$. Z initially rises, Y switches Z production off at $t_r$, and $Y_1$ restores production at $t_b=t_1+d(K_I)$. The continuous solution during the ON phase isThe derivative changes from positive to negative at $t_r$, and from negative to positive at $t_b$. Consequently the first temporary minimum following the initial pulse occurs atIts value is $Z(t_b)=K_{YZ}{e}^{-\alpha (t_b-t_r)}$. At the OFF step Z again decays immediately because X is required for production, with Y and $Y_1$ behaving as above. If a required threshold is unattainable, these late switching events do not occur; equal production and removal rates alone do not fix threshold values.

The plotted thresholds are explicitly illustrative, with $K_{YZ}/C=0.25$, $K_{YY1}/C=0.55$ and $K_A/C=K_I/C=0.45$.