Electron transfer is activated by calmodulin in the flavin domain of human neuronal nitric oxide synthase.

The objective of this study was to clarify the mechanism of electron transfer in the human neuronal nitric oxide synthase (nNOS) flavin domain using the recombinant human nNOS flavin domains, the FAD/NADPH domain (contains FAD- and NADPH-binding sites), and the FAD/FMN domain (the flavin domain including a calmodulin-binding site). The reduction by NADPH of the two domains was studied by rapid-mixing, stopped-flow spectroscopy. For the FAD/NADPH domain, the results indicate that FAD is reduced by NADPH to generate the two-electron-reduced form (FADH(2)) and the reoxidation of the reduced FAD proceeds via a neutral (blue) semiquinone with molecular oxygen or ferricyanide, indicating that the reduced FAD is oxidized in two successive one-electron steps. The neutral (blue) semiquinone form, as an intermediate in the air-oxidation, was unstable in the presence of O(2). The purified FAD/NADPH domain prepared under our experimental conditions was activated by NADP(+) but not NAD(+). These results indicate that this domain exists in two states; an active state and a resting state, and the enzyme in the resting state can be activated by NADP(+). For the FAD/FMN domain, the reduction of the FAD-FMN pair of the oxidized enzyme with NADPH proceeded by both one-electron equivalent and two-electron equivalent mechanisms. The formation of semiquinones from the FAD-FMN pair was greatly increased in the presence of Ca(2+)/CaM. The air-stable semiquinone form, FAD-FMNH(.), was further rapidly reduced by NADPH with an increase at 520 nm, which is a characteristic peak of the FAD semiquinone. Results presented here indicate that intramolecular one-electron transfer from FAD to FMN is activated by the binding of Ca(2+)/CaM.