Past exam of the biological physics course of the University of Oxford / 2018 / 4 / b / Solution

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.