{'fraction_non_alphanumeric': 0.056283357593401265, 'fraction_numerical': 0.024260067928190198, 'mean_word_length': 4.557951482479784, 'pattern_counts': {'":': 0, '<': 0, '<?xml version=': 0, '>': 0, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Reaction of phthalate dioxygenase reductase with NADH and NAD: kinetic and spectral characterization of intermediates.

Phthalate dioxygenase reductase (PDR) is an electron transferase that contains FMN, which accepts a hydride from NADH, and a [2Fe-2S] center, which transfers electrons to phthalate dioxygenase. The reduction of PDR by NADH has been studied by stopped-flow spectroscopy. Data from studies using both portio- and deuterio-NADH were analyzed by nonlinear curve fitting and numerical simulation techniques. The results of these analyses indicate that the reductive half-reaction of PDR consists of five distinct kinetic phases: (a) NADH binds to form a primary Michaelis complex (MC-1) (Kd = 50 microM). (b) The enzyme undergoes a structural change (116 +/- 5 s-1) resulting in a charge-transfer complex (CT-1). (c) The next phase in the reaction shows a deuterium isotope effect of 7.0 when (4R)-[2H]NADH (NADD) is substituted for NADH, identifying this step as the one involving hydride transfer. The rate of hydride transfer from NADH to FMN is 70 s-1, and this process results in a charge-transfer intermediate between the flavin hydroquinone anion and NAD (CT). (d) Internal electron transfer from the flavin to the iron-sulfur center, which is only 35 +/- 4 s-1, then results in an intermediate consisting of a reduced [2Fe-2S] center and a neutral flavin semiquinone (SQ). It is surprising that this rate is so slow, since the shortest interatomic distance between these centers is only 4.7 A [Correll, C. C., et al. (1992) Science 258, 1604-1610]. The 2-electron-reduced form of PDR (SQ in Figure 1) binds weakly to the reaction product, NAD (Kd = 3.7 mM), but forms a tight complex with NADH (Kd = 10 microM). (e) Two molecules of the reduced iron-sulfur flavin semiquinone (SQ) form of PDR then undergo a relatively slow second-order disproportionation reaction, resulting in one molecule of 3-electron-reduced PDR and one molecule of 1-electron-reduced PDR. The latter reacts rapidly with excess NADH to form a 3-electron-reduced PDR.