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乳过氧化物酶和甲状腺过氧化物酶的光谱研究:天然酶、化合物II和化合物III之间的相互转化。

Spectral studies with lactoperoxidase and thyroid peroxidase: interconversions between native enzyme, compound II, and compound III.

作者信息

Kohler H, Taurog A, Dunford H B

机构信息

Department of Pharmacology, University of Texas Health Science Center, Dallas 75235.

出版信息

Arch Biochem Biophys. 1988 Aug 1;264(2):438-49. doi: 10.1016/0003-9861(88)90309-8.

Abstract

Spectral scans in both the visible (650-450 nm) and the Soret (450-380 nm) regions were recorded for the native enzyme, Compound II, and Compound III of lactoperoxidase and thyroid peroxidase. Compound II for each enzyme (1.7 microM) was prepared by adding a slight excess of H2O2 (6 microM), whereas Compound III was prepared by adding a large excess of H2O2 (200 microM). After these compounds had been formed it was observed that they were slowly reconverted to the native enzyme in the absence of exogenous donors. The pathway of Compound III back to the native enzyme involved Compound II as an intermediate. Reconversion of Compound III to native enzyme was accompanied by the disappearance of H2O2 and generation of O2, with approximately 1 mol of O2 formed for each 2 mol of H2O2 that disappeared. A scheme is proposed to explain these observations, involving intermediate formation of the ferrous enzyme. According to the scheme, Compound III participates in a reaction cycle that effectively converts H2O2 to O2. Iodide markedly affected the interconversions between native enzyme, Compound II, and Compound III for lactoperoxidase and thyroid peroxidase. A low concentration of iodide (4 microM) completely blocked the formation of Compound II when lactoperoxidase or thyroid peroxidase was treated with 6 microM H2O2. When the enzymes were treated with 200 microM H2O2, the same low concentration of iodide completely blocked the formation of Compound III and largely prevented the enzyme degradation that otherwise occurred in the absence of iodide. These effects of iodide are readily explained by (i) the two-electron oxidation of iodide to hypoiodite by Compound I, which bypasses Compound II as an intermediate, and (ii) the rapid oxidation of H2O2 to O2 by the hypoiodite formed in the reaction between Compound I and iodide.

摘要

对天然态的乳过氧化物酶和甲状腺过氧化物酶、化合物II和化合物III进行了可见光区域(650 - 450纳米)和索雷特区域(450 - 380纳米)的光谱扫描。每种酶的化合物II(1.7微摩尔)通过加入略微过量的过氧化氢(6微摩尔)制备,而化合物III通过加入大量过量的过氧化氢(200微摩尔)制备。在这些化合物形成后,观察到在没有外源供体的情况下它们会缓慢地重新转化为天然态酶。化合物III回到天然态酶的途径涉及化合物II作为中间体。化合物III重新转化为天然态酶伴随着过氧化氢的消失和氧气的生成,每消失2摩尔过氧化氢大约生成1摩尔氧气。提出了一个方案来解释这些观察结果,该方案涉及亚铁酶中间体的形成。根据该方案,化合物III参与一个反应循环,有效地将过氧化氢转化为氧气。碘化物对乳过氧化物酶和甲状腺过氧化物酶的天然态酶、化合物II和化合物III之间的相互转化有显著影响。当用6微摩尔过氧化氢处理乳过氧化物酶或甲状腺过氧化物酶时,低浓度的碘化物(4微摩尔)完全阻断了化合物II的形成。当用200微摩尔过氧化氢处理这些酶时,相同低浓度的碘化物完全阻断了化合物III的形成,并在很大程度上防止了在没有碘化物时发生的酶降解。碘化物的这些作用很容易通过以下两点来解释:(i)化合物I将碘化物双电子氧化为次碘酸盐,绕过化合物II作为中间体;(ii)化合物I与碘化物反应形成的次碘酸盐将过氧化氢快速氧化为氧气。

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