= Solution
The <ADP> addition contains
$$
n_{\mathrm{ADP}}=(10\times10^{-6}\,\mathrm L)(20\times10^{-3}\,\mathrm{mol\,L^{-1}})=0.200\,\mathrm{\mu mol}.
$$
During the initial stimulated interval, the plotted <oxygen> concentration falls from about $240$ to $207\,\mathrm{\mu M}$. The chamber volume after this addition is $1.900+0.050+0.050+0.010=2.010\,\mathrm{mL}$, giving
$$
\Delta n_{\mathrm{O_2}}\simeq33\,\mathrm{\mu mol\,L^{-1}}\times0.002010\,\mathrm L=0.0663\,\mathrm{\mu mol}.
$$
The conventional <phosphorylation-to-oxidation ratio> counts <ATP> formed per oxygen atom reduced, so its denominator is twice the number of oxygen molecules:
$$
\boxed{\mathrm{P/O}\simeq\frac{0.200}{2(0.0663)}\simeq1.5.}
$$
This assumes the added <ADP> is phosphorylated and uses the displayed total oxygen loss over the pulse; it is an apparent experimental ratio. A separate correction for background respiration or <proton leak> would change the denominator and must be specified explicitly. Dividing by oxygen molecules without the factor of two would instead give an ATP:$\mathrm O_2$ ratio near $3$.
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