Shi X, Dalal N S
Department of Chemistry, West Virginia University, Morgantown 26506.
Arch Biochem Biophys. 1993 Dec;307(2):336-41. doi: 10.1006/abbi.1993.1597.
The mechanism of hydroxyl (.OH) radical generation from O2- and H2O2 by vanadate [V(V)] and the role of NADH in this reaction have been investigated using electron spin resonance (ESR) and spin trapping techniques. The results show that the reaction of V(V) with O2- (generated via xanthine/xanthine oxidase) does not generate any ESR detectable V(IV) ion or .OH radical and the addition of H2O2 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 O2- 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) + H2O2-->V(V) + .OH+OH-], the V(IV) and H2O2 being generated by V(V)-stimulated, O(2-)-dependent NADH oxidation.
利用电子自旋共振(ESR)和自旋捕获技术,研究了钒酸盐[V(V)]由超氧阴离子(O2-)和过氧化氢(H2O2)生成羟基(·OH)自由基的机制以及烟酰胺腺嘌呤二核苷酸(NADH)在该反应中的作用。结果表明,V(V)与O2-(通过黄嘌呤/黄嘌呤氧化酶生成)的反应不会产生任何ESR可检测到的V(IV)离子或·OH自由基,添加H2O2对自由基产率影响很小。然而,在NADH存在的情况下,黄嘌呤/黄嘌呤氧化酶/V(V)体系会生成·OH以及V(IV),超氧化物歧化酶可以抑制这两者的形成。过氧化氢酶抑制·OH的形成,但会增强V(IV)的生成。在磷酸盐缓冲液存在下,V(V)单独与NADH反应也会产生·OH自由基,尽管速率大大降低,超氧化物歧化酶会降低·OH的产率。与早期报道相反,这些观察结果表明,在没有NADH的情况下,O2-不会将V(V)还原为V(IV)。得出的结论是,钒酸盐通过类似芬顿反应而非哈伯-维伊斯反应生成·OH自由基[V(IV)+H2O2→V(V)+·OH+OH-],V(IV)和H2O2是由V(V)刺激的、O(2-)依赖的NADH氧化产生的。