Past exam of the biological physics course of the University of Oxford / 2015 / 1 / d / Solution

Let $S=\beta /\alpha$ be the unrepressed steady state, $x=X/S,y=Y/S,z=Z/S$, $h=K/S$, and $\tau =\alpha t$. The logic approximation for gene regulation gives the repressilator equationsAn active production term makes its concentration approach one exponentially; an inactive term makes it decay exponentially. Oscillations require an accessible threshold, $0<h<1$. Equal thresholds and rate constants alone do not specify a numerical period.

Initially $(x,y,z)=(0,1,0)$, so $x=1-e^{-\tau }$, $y=1$, $z=0$ until $x$ reaches $h$ at ${\tau }_1=-\log (1-h)$. Then $Y$ production stops and $y=e^{-(\tau -{\tau }_1)}$. It reaches $h$ after another $\log (1/h)$, allowing $Z$ production. When $z$ rises through $h$, $X$ production stops. As $x$ falls through $h$, $Y$ production resumes; its next upward crossing represses $Z$. The downward crossing of $z$ then restarts $X$. This ordered sequence repeats, giving displaced pulses and rounded exponential rises and falls.

Every event can be constructed without a numerical differential-equation solver: a rising variable starting at $v_0<h$ takes $\log [(1-v_0)/(1-h)]$ to reach $h$, while a decaying variable starting at $v_0>h$ takes $\log (v_0/h)$. Advance all three variables with their current production terms to the earliest such event, switch the affected production term, and continue. The original plot uses the explicitly illustrative choice $h=1/2$. If $h\ge 1$, the unrepressed levels cannot cross the threshold and this oscillatory sequence does not follow.