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

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.