Maskos Z, Winston G W
Department of Biochemistry, Louisiana State University, Baton Rouge 70803-1800.
J Biol Chem. 1994 Dec 16;269(50):31579-84.
The mechanism of reduction of p-nitrosophenol (pNSP) catalyzed by horse liver alcohol dehydrogenase (HADH) and human pi-alcohol dehydrogenase (pi-ADH) has been compared in transient and steady-state experiments. Our results indicate that pNSP reduction catalyzed by these two ADH proceeds by different mechanisms. In one mechanism, shown by Equation 1, pNSP is reduced to p-aminophenol (pAP) via two enzymatic steps (Steps 1 and 3), which are mediated by the nonenzymatic dehydration of p-N-hydroxyaminophenol (pN-OHAP) to 1,4-benzoquinoneimine (BQI) (Step 2). [formula: see text] Pathway (I) is proposed mainly for pi-ADH but can be catalyzed by HADH. However, Step 3 is catalyzed approximately 2 orders of magnitude more slowly by HADH than by pi-ADH. This conclusion is confirmed by the results, which indicate that pi-ADH very efficiently catalyzes the reduction of BQI and 1,4-benzoquinone (BQ) to the corresponding hydroquinones. The kinetic constants determined at pH 7.4 suggest that pi-ADH is a more efficient quinone reductase and nitroso reductase than it is an ethanol oxidase or acetaldehyde reductase. An alternative mechanism of pNSP reduction, shown by Equation 2, is suggested for HADH. In this mechanism, formation of the p-hydroxybenzylnitrenium ion (pNH+P) occurs at the active-site zinc ion of the enzyme (Step 2) and accelerates further nonenzymatic reduction to pAP or hydrolysis to BQ (Step 3). [formula: see text]
在瞬态和稳态实验中，对马肝醇脱氢酶（HADH）和人π-醇脱氢酶（π-ADH）催化对亚硝基苯酚（pNSP）还原的机制进行了比较。我们的结果表明，这两种ADH催化的pNSP还原通过不同的机制进行。在一种机制中，如方程式1所示，pNSP通过两个酶促步骤（步骤1和步骤3）还原为对氨基酚（pAP），这两个步骤由对-N-羟基氨基酚（pN-OHAP）非酶脱水生成1,4-苯醌亚胺（BQI）介导（步骤2）。[公式：见正文]途径（I）主要是针对π-ADH提出的，但也可由HADH催化。然而，步骤3由HADH催化的速度比对π-ADH催化的速度慢约2个数量级。结果证实了这一结论，结果表明π-ADH非常有效地催化BQI和1,4-苯醌（BQ）还原为相应的对苯二酚。在pH 7.4下测定的动力学常数表明，π-ADH作为醌还原酶和亚硝基还原酶比作为乙醇氧化酶或乙醛还原酶更有效。对于HADH，提出了另一种pNSP还原机制，如方程式2所示。在这种机制中，对羟基苄基氮鎓离子（pNH+P）在酶的活性位点锌离子处形成（步骤2），并加速进一步非酶还原为pAP或水解为BQ（步骤3）。[公式：见正文]
