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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.