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

In the simplified Barkai-Leibler model, let $M$ be the total methylated receptor concentration and $A=f(S)M$ the active, unoccupied part. Ligand-bound methylated receptors contribute zero activity. Because CheB acts only on active receptors,At a finite steady state,The output $A_{\ast }$ is independent of attractant concentration $S$: attractant changes the necessary receptor methylation level, while the flux balance fixes activity. This proves perfect adaptation for $0<r<b$, $f(S)>0$ and enough methylatable receptors. At fixed $S$, $d\dot{M}/dM=-bf(S)K_B/[K_B+f(S)M{]}^2<0$, so this steady state is locally stable. If every receptor is permanently bound, $f=0$, or if the required $M_{\ast }$ exceeds the available pool, compensation fails.

This feedback couples demethylation directly to the output. Lower activity lowers demethylation; continuing CheR action restores it. The structural dependence on active receptors removes the separate adjustment of $V_B^{\prime }$ required in c.^[ref]

Changing CheR or CheB concentration changes $A_{\ast }$ and hence the baseline tumbling frequency, but the output still returns to that new baseline after an attractant step within the operating range. Raising CheR raises $A_{\ast }$; raising CheB lowers it. Adaptation time also changes: linearization gives $\tau =(K_B+A_{\ast }{)}^2/[bf(S)K_B]$. Thus robustness of the return property does not mean an invariant tumbling frequency or response time across different enzyme abundances.

If CheR instead has Michaelis-Menten kinetics on the unmethylated pool $U=E_T-M$, its flux is $r(M)=V_R[R](E_T-M)/(K_R+E_T-M)$. ThenThe left side depends on ligand through $f$, even at fixed output $A$; consequently the equilibrium output generally depends on attractant and perfect adaptation is lost. It is recovered approximately when $E_T-M\gg K_R$ throughout the response, so the CheR flux is effectively saturated.