Past exam of the biological physics course of the University of Oxford / 2021 / 2 / c / Solution

Let attractant reduce signalling activity per methylated receptor by a factor $f\in (0,1)$, so the output is $A=fM$. Before stimulation take $f=1$ and $M=M_0=K_{B}r/(b-r)$. Perfect adaptation requires the final methylated receptor concentration $M^{\prime }=M_0/f$. With $b^{\prime }=V_B^{\prime }[B]$, the new steady state condition isEquivalently $b^{\prime }=fb+(1-f)r$. Since $f<1$, this reduces demethylation relative to the initial condition. Immediately after binding, activity and tumbling fall. The excess CheR flux then increases receptor methylation, compensating the activity loss until $f{M}^{\prime }=M_0$. The methylated population must have sufficient remaining capacity to reach $M^{\prime }$.

Increasing CheR concentration raises $r$ and the baseline activity; increasing CheB concentration raises $b$ and lowers it. If the attractant-dependent ratio $c=V_B^{\prime }/V_B$ remains fixed after such a change, the new pre- and post-step outputs areTheir equality requires $c=f+(1-f)r/b$. Thus a ratio tuned for one $[R]/[B]$ generally loses perfect adaptation when either enzyme level changes independently. For a concrete example, initially $r/b=1/2$ and $f=1/2$ require $c=3/4$; changing $r/b$ to $1/4$ requires $c=5/8$, so keeping $c=3/4$ fails. Proportionally scaling both enzyme levels preserves the ratio and the adaptation condition, while changing response speed. This is fine tuning, rather than robust perfect adaptation.