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

Put $M=[E_m]$, $r=V_R[R]$ and $b=V_B[B]$. Saturated CheR supplies a constant receptor methylation flux, while CheB removes it with Michaelis-Menten kinetics:At a finite steady state, $r(K_B+M_0)=b{M}_0$, givingIt requires $0<r<b$ and $M_0$ below the total available receptor pool. The steady state is stable because $d\dot{M}/dM=-b{K}_B/(K_B+M{)}^2<0$. If $r\ge b$, the ideal equation has no finite positive steady state; actual methylatable sites would become exhausted.

Attractant binding reduces receptor signalling, reducing CheA-dependent phosphorylation of CheY and initially reducing tumbling frequency, so smooth runs become longer. A spatially uniform stimulus provides no preferred direction. Perfect adaptation means that after this transient the tumbling frequency returns exactly to its pre-step value despite continued attractant presence.

The equation above alone does not specify attractant occupancy or its influence on activity. If binding simply changes activity from $M$ to $fM$ with $0<f<1$, without changing either modifying flux, $M$ remains $M_0$ and the output remains $f{M}_0$: there is no adaptation. Thus the usual fall followed by recovery requires the additional feedback mechanism developed in c or d, rather than following from the stated Michaelis-Menten kinetics alone.