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六价铬与抗坏血酸盐和过氧化氢的反应会产生羟基自由基并导致DNA损伤:一种四价铬介导的类芬顿反应的作用。

Reaction of Cr(VI) with ascorbate and hydrogen peroxide generates hydroxyl radicals and causes DNA damage: role of a Cr(IV)-mediated Fenton-like reaction.

作者信息

Shi X, Mao Y, Knapton A D, Ding M, Rojanasakul Y, Gannett P M, Dalal N, Liu K

机构信息

Laboratory of Experimental Pathology, National Cancer Institute, Bethesda, MD 20892.

出版信息

Carcinogenesis. 1994 Nov;15(11):2475-8. doi: 10.1093/carcin/15.11.2475.

DOI:10.1093/carcin/15.11.2475
PMID:7955094
Abstract

Incubation of Cr(VI) with ascorbate generated Cr(V), Cr(IV) and ascorbate-derived carbon-centered alkyl radicals, as well as formyl radicals. H2O2 caused generation of hydroxyl radicals (OH) and much higher levels of Cr(V), showing that .OH can be generated via a Cr(IV)-mediated Fenton-like reaction (Cr(IV) + H2O2-->Cr(V) + .OH + OH-). 1,10-Phenanthroline and deferoxamine inhibited the formation of both .OH and Cr(V) from the reaction of Cr(VI) with ascorbate in the presence of H2O2. Electrophoretic assays showed that ascorbate-derived free radicals caused DNA double-strand breaks. .OH radicals generated by Cr(V)- and Cr(IV)-mediated Fenton-like reactions also caused DNA double-strand breaks. HPLC measurements showed that .OH radicals generated by Cr(IV) and Cr(V) from H2O2 caused 2'-deoxyguanine hydroxylation to form 8-hydroxy-2'-deoxyguanine.

摘要

六价铬(Cr(VI))与抗坏血酸盐孵育会生成五价铬(Cr(V))、四价铬(Cr(IV))、抗坏血酸盐衍生的碳中心烷基自由基以及甲酰基自由基。过氧化氢(H2O2)会导致羟基自由基(OH)的生成以及更高水平的五价铬(Cr(V)),这表明羟基自由基(OH)可以通过四价铬(Cr(IV))介导的类芬顿反应(Cr(IV) + H2O2→Cr(V) + OH + OH-)生成。1,10 - 菲咯啉和去铁胺在过氧化氢(H2O2)存在的情况下,抑制了六价铬(Cr(VI))与抗坏血酸盐反应中羟基自由基(OH)和五价铬(Cr(V))的形成。电泳分析表明,抗坏血酸盐衍生的自由基会导致DNA双链断裂。由五价铬(Cr(V))和四价铬(Cr(IV))介导的类芬顿反应产生的羟基自由基(OH)也会导致DNA双链断裂。高效液相色谱测量表明,四价铬(Cr(IV))和五价铬(Cr(V))与过氧化氢(H2O2)反应产生的羟基自由基(OH)会导致2'-脱氧鸟嘌呤羟基化形成8 - 羟基 - 2'-脱氧鸟嘌呤。

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