 Solution

Both motifs have a direct path $X\to Z$ and an indirect path $X\to Y\to Z$. C1 coherent feed-forward loop uses three activating edges. I1 incoherent feed-forward loop activates $Y$ and $Z$ from $X$, but lets $Y$ repress $Z$. The AND condition means that the direct input must be active and the intermediate input must permit output: present for an activator, absent for a repressor. ^[ref]

For a simple threshold model letThe coherent output obeys $\dot{Z}={\beta }_{Z}H(X-K_{XZ})H(Y-K_{YZ})-{\alpha }_{Z}Z$. Starting from $Y=0$ under a sustained high input, $Y=Y_{\infty }(1-e^{-{\alpha }_{Y}t})$. Provided $0<K_{YZ}<Y_{\infty }$, the gene expression delay isShorter input pulses do not turn on output gene expression. This is a persistence detector: it rejects brief positive inputs. Removing $X$ immediately removes the direct permission for gene expression, even if $Y$ remains high. The concentration of $Z$ subsequently decays with its own lifetime rather than vanishing instantly.

The incoherent output instead obeys $\dot{Z}={\beta }_{Z}H(X-K_{XZ})H(K_{YZ}-Y)-{\alpha }_{Z}Z$. Gene expression begins immediately, then stops when accumulated repressor crosses threshold at $t_d$. With zero initial output,A maintained stimulus therefore generates a pulse in this ideal Boolean model. Delayed inhibition can accelerate an early response and limit its duration. The ideal final zero output is conditional on crossing the repression threshold; graded response laws or basal gene expression can give a nonzero final level and need not adapt perfectly. If $Y_{\infty }\le K_{YZ}$, the coherent circuit does not activate, while the incoherent circuit fails to terminate its gene expression. These boundary cases show why regulatory signs alone do not fix function.