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氧化和还原自由基探针与硫辛酰胺脱氢酶的反应。

Reactions of oxidizing and reducing radical probes with lipoamide dehydrogenase.

作者信息

Elliot A J, Munk P L, Stevenson K J, Armstrong D A

出版信息

Biochemistry. 1980 Oct 28;19(22):4945-50. doi: 10.1021/bi00563a001.

Abstract

One-electron redox radicals generated by radiation--chemical methods have been reacted with the oxidized (E) form of pig heart lipoamide dehydrogenase. The reducing radicals eaq- and CO2-. and O2-. do not measurably inactivate the enzyme, whereas the oxidizing species .OH and Br2-. do. The CO2-. anion forms the semiquinone radical .EH on the millisecond time scale, whereas at longer times only EH2 is observed. Evidence suggests that Br2-. oxidizes adjacent sulfhydryl groups to form a disulfide in a manner similar to the reaction of Cu2+ ions. With .OH, destruction of the flavin adenine dinucleotide (FAD) moiety is responsible for at least 50% of the enzyme inactivation. This destruction appears to be a result of secondary reactions which transfer damage from remote initial sites of attack to the flavin. Pathways for migration of eaq- damage also appear to exist.

摘要

通过辐射化学方法产生的单电子氧化还原自由基已与猪心硫辛酰胺脱氢酶的氧化态(E)形式发生反应。还原性自由基水合电子(eaq-)、二氧化碳阴离子自由基(CO2-)和超氧阴离子自由基(O2-)不会使该酶显著失活,而氧化性物种羟基自由基(·OH)和溴自由基阴离子(Br2-)则会使其失活。CO2-阴离子在毫秒时间尺度上形成半醌自由基(·EH),而在更长时间内仅观察到二氢硫辛酰胺脱氢酶(EH2)。有证据表明,Br2-以类似于Cu2+离子反应的方式氧化相邻的巯基以形成二硫键。对于·OH,黄素腺嘌呤二核苷酸(FAD)部分的破坏至少导致50%的酶失活。这种破坏似乎是二级反应的结果,这些二级反应将损伤从远程初始攻击位点转移到黄素上。水合电子损伤的迁移途径似乎也存在。

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