= 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 $M=M_0=K_Br/(b-r)$. <Perfect adaptation> requires the final methylated <receptor> <concentration> $M'=M_0/f$. With $b'=V'_B[B]$, the new <steady state> condition is
$$
r=\frac{b'M'}{K_B+M'},\qquad
\boxed{V'_B=\frac{r}{[B]}\left(1+\frac{fK_B}{M_0}\right)
=V_B\frac{M_0+fK_B}{M_0+K_B}.}
$$
Equivalently $b'=fb+(1-f)r$. 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 $fM'=M_0$. The methylated population must have sufficient remaining capacity to reach $M'$.
Increasing <CheR> <concentration> raises $r$ and the baseline activity; increasing <CheB> <concentration> raises $b$ and lowers it. If the attractant-dependent ratio $c=V'_B/V_B$ remains fixed after such a change, the new pre- and post-step outputs are
$$
A_{\rm before}=\frac{K_Br}{b-r},\qquad
A_{\rm after}=\frac{fK_Br}{cb-r}.
$$
Their 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>.
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