= Reversible receptor phosphorylation model
<Protein kinase> activity adds modification, while <protein tyrosine phosphatase> activity can remove it. An original one-site model shows why a <receptor phosphorylation assay> cannot determine either process from a single modified fraction. Let $f(t)\in[0,1]$ be the fraction of a fixed <receptor> population carrying one selected <phosphotyrosine residue>. Stipulate constant effective first-order <rate constants> $k_+,k_->0$, with <ATP> availability absorbed into $k_+$, and no <receptor> synthesis, degradation or trafficking:
$$
\frac{df}{dt}=k_+(1-f)-k_-f,\qquad
f_* = \frac{k_+}{k_++k_-},\qquad
\tau=\frac{1}{k_++k_-}.
$$
Here $f_*$ is the <steady state> fraction and $\tau$ the <time constant>. The <ordinary differential equation> has the <exponential function> solution
$$
f(t)=f_*+(f(0)-f_*)e^{-t/\tau}.
$$
The balance equation and a single relaxation step give this result; it is an illustrative model, rather than a fit or a claim about every <RTK>.
With stipulated <rate constants> $(k_+,k_-)=(2,1)\,\mathrm{min}^{-1}$, $f_*=2/3$ and $\tau=1/3\,\mathrm{min}$. Doubling both <rate constants> leaves $f_*=2/3$ but changes $\tau$ to $1/6\,\mathrm{min}$. The balanced <phosphate> turnover per site,
$$
J_*=k_+(1-f_*)=k_-f_*=\frac{k_+k_-}{k_++k_-},
$$
doubles from $2/3$ to $4/3\,\mathrm{min}^{-1}$. Identical <steady state> modification therefore permits different catalytic fluxes. Removing the addition term instead gives $f(t)=f(0)e^{-k_-t}$; this illustrates why a perturbation that stops <protein kinase> activity can expose removal dynamics, subject to the model's unchanged-removal assumption. A maintained <steady state> with ongoing <ATP> consumption is not <chemical equilibrium>.
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