Abstract

The mechanism of hydroxyl (·OH) radical generation from O − 2 and H 2O 2 by vanadate [V(V)] and the role of NADH in this reaction have been investigated using electron spin resonance (ESH) and spin trapping techniques. The results show that the reaction of V(V) with O − 2 (generated via xanthine/xanthine oxidase) does not generate any ESR detectable V(IV) ion or ·OH radical and the addition of H 2O 2 has little effect on the radical yield. In the presence of NADH, however, the xanthine/xanthine oxidase/V(V) system generates ·OH as well as V(IV), the formation of both of which could be suppressed by superoxide dismutase. Catalase inhibits the ·OH formation but enhances V(IV) generation. Reaction of V(V) with NADH alone in the presence of phosphate buffer also causes ·OH radical generation albeit at a much reduced rate, and superoxide dismutase reduces the ·OH yield. These observations indicate, in contrast to earlier reports, that O − 2 does not reduce V(V) to V(IV) in the absence of NADH. It is concluded that vanadate generates the ·OH radical via not a Haber-Weiss but a Fenton-like reaction [V(IV) + H 2O 2 → V(V) + ·OH + OH −], the V(IV) and H 2O 2 being generated by V(V)-stimulated, O − 2-dependent NADH oxidation.

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