- The reductive half-reaction of xanthine oxidase. Reaction with aldehyde substrates and identification of the catalytically labile oxygen.
The reductive half-reaction of xanthine oxidase. Reaction with aldehyde substrates and identification of the catalytically labile oxygen.
The kinetics of xanthine oxidase has been investigated with the aim of addressing several outstanding questions concerning the reaction mechanism of the enzyme. Steady-state and rapid kinetic studies with the substrate 2,5-dihydroxybenzaldehyde demonstrated that (kcat/Km)app and kred/Kd exhibit comparable bell-shaped pH dependence with pKa values of 6.4 +/- 0.2 and 8.4 +/- 0.2, with the lower pKa assigned to an active-site residue of xanthine oxidase (possibly Glu-1261, by analogy to Glu-869 in the crystallographically known aldehyde oxidase from Desulfovibrio gigas) and the higher pKa to substrate. Early steps in the catalytic sequence have been investigated by following the reaction of the oxidized enzyme with a second aldehyde substrate, 2-aminopteridine-6-aldehyde. The absence of a well defined acid limb in this pH profile and other data indicate that this complex represents an Eox.S rather than Ered.P complex (i.e. no chemistry requiring the active-site base has taken place in forming the long wavelength-absorbing complex seen with this substrate). It appears that xanthine oxidase (and by inference, the closely related aldehyde oxidases) hydroxylates both aromatic heterocycles and aldehydes by a mechanism involving base-assisted catalysis. Single-turnover experiments following incorporation of 17O into the molybdenum center of the enzyme demonstrated that a single oxygen atom is incorporated at a site that gives rise to strong hyperfine coupling to the unpaired electron spin of the metal in the MoV oxidation state. By analogy to the hyperfine interactions seen in a homologous series of molybdenum model compounds, we conclude that this strongly coupled, catalytically labile site represents a metal-coordinated hydroxide rather than the Mo=O group and that this Mo-OH represents the oxygen that is incorporated into product in the course of catalysis.