= 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 have
$$
Y(t)=C(1-e^{-\alpha t})\quad(0<t<T),\qquad
Y(t)=Ce^{-\alpha(t-T)}\quad(t>T),
$$
where the latter uses the stipulated attained <steady state>. Define the threshold-crossing delay
$$
d(K)=-\alpha^{-1}\ln(1-K/C),\qquad
t_r=d(K_{YZ}),\quad t_1=d(K_{YY1}).
$$
Because $K_{YY1}>K_{YZ}$, $t_1>t_r$. 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.
\b[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 at
$$
t_a=t_1+d(K_A).
$$
Thus $Z=0$ until $t_a$ and $Z=C[1-e^{-\alpha(t-t_a)}]$ subsequently. Solving $Z=C/2$ gives
$$
\boxed{t_{Z,1/2}=d(K_{YY1})+d(K_A)+\frac{\ln2}{\alpha}}.
$$
For the frequently adopted but extra convention that both required thresholds are $C/2$, this is $3\ln2/\alpha$. At the OFF step, X's direct AND input disappears, so $Z=Ce^{-\alpha(t-T)}$ 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.
\b[Diagram (b).] The production gate is $g_Z=u\,[\neg(Y>K_{YZ})\lor(Y_1>K_I)]$. 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 is
$$
Z(t)=\begin{cases}
C(1-e^{-\alpha t}),&0<t<t_r,\\
C(1-e^{-\alpha t_r})e^{-\alpha(t-t_r)},&t_r<t<t_b,\\
C+[Z(t_b)-C]e^{-\alpha(t-t_b)},&t_b<t<T.
\end{cases}
$$
The <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 at
$$
\boxed{t_{\min}=d(K_{YY1})+d(K_I)}.
$$
Its 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.
\Image[/past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2022-regulatory-responses.png]
{title=<Logic approximation for gene regulation>: delayed activation and a transient repression window}
The plotted thresholds are explicitly illustrative, with $K_{YZ}/C=0.25$, $K_{YY1}/C=0.55$ and $K_A/C=K_I/C=0.45$.
Back to article page
