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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.

Use the source's explicit graded production rules. X is already present at a constant concentration; the input at zero activates it rather than starting accumulation of X. Assume its active level exceeds both relevant X thresholds, put ${\alpha }_x={\alpha }_y={\alpha }_z=\alpha$, let $Y_s={\beta }_y/\alpha$, and take $0<h=K_{yz}/Y_s<1$. Before the input Z is not produced, so use the initial steady value $Z(0)=0$.

For I1 incoherent feed-forward loop, activation of X starts Y production; for I4 incoherent feed-forward loop it stops Y production. ConsequentlyThe two delays are the times for Y to cross its output-regulation threshold. In I1 incoherent feed-forward loop, initially absent Y permits the high output rate, then accumulated Y represses it. In I4 incoherent feed-forward loop, initially abundant Y activates the high output rate, then Y's decay removes that activation. Thus both have production ${\beta }_{\mathrm{high}}$ before their respective delay and ${\beta }_{\mathrm{low}}$ afterwards.

With $Z_s={\beta }_{\mathrm{low}}/\alpha$ and $F={\beta }_{\mathrm{high}}/{\beta }_{\mathrm{low}}$, their common piecewise response, with $d=d_1$ or $d_4$, isThe output is continuous at the production switch. There is an overshoot precisely when $F(1-e^{-\alpha d})>1$; otherwise it rises monotonically to $Z_s$. For a threshold not extremely close to its limiting values and the supplied $F\gg 1$, a rapid pulse followed by relaxation to a lower nonzero level is expected. This can provide fast initial response or a transient strong signal, but it is not perfect adaptation to the original zero output.

The original plots below use the illustrative values $h=1/4$ and $F=10$. At $h=1/2$ the two Y curves are complementary and their output delays and Z curves are identical. These threshold values are examples, not extra source data. “AND” here must be read with the stipulated graded rates: replacing them by an ordinary all-or-nothing AND rule would give a different response.

For the final low steady value $Z_s={\beta }_{\mathrm{low}}/\alpha$, the first crossing of $Z_s/2$ occurs during high production if $F(1-e^{-\alpha d})\ge 1/2$. Solving $F(1-e^{-\alpha t_{1/2}})=1/2$ then gives, for either motif,This half-final response of a graded incoherent feed-forward loop is a first-crossing measure. A simple output produced at the final low rate from zero would instead take $(\ln 2)/\alpha$. This comparison explains the acceleration of the first half-final response, not the time needed to settle after a potentially large overshoot.

The condition matters because the source does not fix $K_{yz}/Y_s$. Define $\delta =1-e^{-\alpha d}$, equal to $h$ for I1 incoherent feed-forward loop and $1-h$ for I4 incoherent feed-forward loop. If the high phase ends before half-final output is reached, solve the second branch of part (a) instead:The expressions agree at the boundary. The $F$-only answer is therefore valid for both motifs when $F\min (h,1-h)\ge 1/2$, as holds for large F and a fixed interior threshold. Merely saying $F\gg 1$ does not cover a threshold that approaches zero or one even faster. If the Y threshold is never crossed, the assumed final low state itself changes and one must reconsider the steady state before using this response-time definition.

At the final steady state, the ligands control activity as well as the amount of each factor. Assuming active X exceeds its thresholds and the high Y level exceeds $K_{yz}$, the normalized output production rates areFor I1 incoherent feed-forward loop, active X makes Y abundant, but Y can repress Z only if its own activating ligand is present. Without $S_y$, Y is inactive and the output remains high. For I4 incoherent feed-forward loop, active X represses Y to below its threshold, so the output is low regardless of $S_y$. If X is inactive, the direct permissive arm is absent and both outputs are zero even when Y is active.

Thus I1 incoherent feed-forward loop can retain two-input combinatorial control at steady state as well as generate the transient in part (a). I4 incoherent feed-forward loop loses steady-state sensitivity to $S_y$ once X has suppressed Y, under these particular assumptions. That gives a functional reason the first motif may be more useful in many regulatory contexts. It is not a proof of an evolutionary frequency: promoter implementation, thresholds, noise, costs and other functions also influence how common a motif is. The argument follows the supplied rate table, including its nonzero ${\beta }_{\mathrm{low}}$, rather than a generic Boolean AND truth table.