 Solution

Let the common production and removal constants be $\beta$ and $\alpha$, so an unrepressed protein approaches $S=\beta /\alpha$. A simple constructive motif isChoose thresholds $K_{XY}=K_{X{Z}_1}=0.2S$, $K_{X{Z}_2}=0.4S$, and $K_{Y{Z}_3}=0.5S$. The logic approximation for gene regulation gives $\dot{Y}=\beta H(X-0.2S)-\alpha Y$, ${\dot{Z}}_1=\beta H(X-0.2S)-\alpha Z_1$, ${\dot{Z}}_2=\beta H(X-0.4S)-\alpha Z_2$, and ${\dot{Z}}_3=\beta H(Y-0.5S)-\alpha Z_3$. These unequal thresholds are permitted in this part; the production and degradation rates remain equal.

After input onset, $X=S[1-e^{-\alpha (t-T_1)}]$. Production of $Z_1$ and $Y$ begins at $T_1+0.2231/\alpha$, $Z_2$ at $T_1+0.5108/\alpha$, and $Z_3$ at $T_1+(0.2231+\ln 2)/\alpha =T_1+0.9163/\alpha$. Thus the required on-order is obtained. Take an input long enough that all these events occur, for example $T_2-T_1=6/\alpha$.

At input termination, $X=X_2{e}^{-\alpha (t-T_2)}$, where $X_2=S[1-e^{-\alpha (T_2-T_1)}]$. The higher direct threshold is crossed first: $Z_2$ production stops at $t_2=T_2+{\alpha }^{-1}\ln [X_2/(0.4S)]$, and $Z_1$ and $Y$ production stop at $t_1=T_2+{\alpha }^{-1}\ln [X_2/(0.2S)]$. At that instant $Y_1=S\{1-e^{-\alpha [t_1-T_1-0.2231/\alpha ]}\}$, so $Z_3$ stops later at $t_3=t_1+{\alpha }^{-1}\ln [Y_1/(0.5S)]$. For the example, the delays after $T_2$ are approximately $0.914$, $1.607$ and $2.300$ in units of ${\alpha }^{-1}$:Protein concentrations rise or decay continuously after these switches; “off” denotes cessation of production, not instantaneous disappearance. The first panel below marks the thresholds and the resulting concentrations.