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

Use the logic approximation for gene regulation, zero basal production and initially vanishing X, Y and Z. Let $K=K_{xy}=K_{xz}=K_{yz}=K_{xx}/2$, $Y_{\infty }=Z_{\infty }={\beta }_y/\alpha$, and assume the activated Q level exceeds the X activation threshold. The circuit combines a coherent feed-forward loop with negative autoregulation. While X is below its self-repression threshold it obeys $\dot{X}={\beta }_x-\alpha X$, hence X first crosses $K$ atY production begins then. While X remains above $K$, $Y=Y_{\infty }[1-e^{-\alpha (t-t_X)}]$. The AND gate for Z first becomes active when Y reaches $K$, soThe approximation drops the short X build-up time because ${\beta }_x\gg {\beta }_y$. Strictly the input must last longer than this onset time to produce a nonzero amount of Z. This requires ${\beta }_y/\alpha >K$: at equality the threshold is approached only asymptotically, and below it Z never switches on. No numerical delay follows without the threshold-to-production ratio.

In the strong negative autoregulation limit, X is held close to $K_{xx}=2K$; a smooth repression law regularizes the rapid switching of the ideal Boolean rule. After the input disappears at $t_0$, production of X stops and $X\simeq 2K{e}^{-\alpha (t-t_0)}$. It takes $(\ln 2)/\alpha$ for X to fall below $K$. Until then Y remains at $Y_{\infty }>K$ and the Z AND gate stays on, so Z stays at $Z_{\infty }$. After that Z decays by dilution, taking another $(\ln 2)/\alpha$ to halve. ThereforeMore generally replace the first $\ln 2$ by $\ln [X(t_0)/K]$. This distinguishes disappearance of the external ligand from disappearance of the persistent X transcription factor.

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.

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.

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.