The Km is the substrate (oxygen, in this case) concentration at which the activity of the enzyme is one-half its maximal activity when it is fully saturated with substrate. The activity of the enzyme varies in a continuous way with the substrate concentration, according to the Michaelis-Menten equation:
V = Vmax[S]/{Km + [S]}
Water in equilibrium with air (about 21% oxygen) has an oxygen concentration of 270 µM according to this document, which does not specify temperature or pressure, so I assume it refers to standard temperature and pressure: http://butane.chem.uiuc.edu/pshapley/GenChem1/L23/web-L23.pdf
3-12% oxygen therefore corresponds to about 38 to 162 µM. If oxygen were the only substrate to consider, you could use these numbers to calculate the activity of the enzyme relative to Vmax using the M-M equation. However, I suspect that the enzyme has another substrate that takes part in the reaction. The rate equations for 2-substrate enzymes (there are several, one for each kinetic mechanism) are more complicated, and takes into consideration the Kms and concentrations of both substrates.
Other considerations are whether there is any allosteric regulation of the enzyme activity in vivo, and whether the enzyme is expressed, and to what level.
The concentration of a gas in solution depends on its partial pressure above the solution according to Henry's law. The constants Hcp have been collected in a review by Sander in doi: 10.5194/acp-15-4399-2015.
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