Abstract
The complete stoichiometry of the metabolism of the cytochrome b5 (cyt b5)-requiring substrate, methoxyflurane, by purified cytochrome P-450 2B4 was compared to that of another substrate, benzphetamine, which does not require cyt b5 for its metabolism. Cyt b5 invariably improved the efficiency of product formation. That is, in the presence of cyt b5 a greater percentage of the reducing equivalents from NADPH were utilized to generate substrate metabolites, primarily at the expense of the side product, superoxide. With methoxyflurane, cyt b5 addition always resulted in an increased rate of product formation, while with benzphetamine the rate of product formation remained unchanged, increased or decreased. The apparently contradictory observations of increased reaction efficiency but decrease in total product formation for benzphetamine can be explained by a second effect of cyt b5. Under some experimental conditions cyt b5 inhibits total NADPH consumption. Whether stimulation, inhibition, or no change in product formation is observed in the presence of cyt b5 depends on the net effect of the stimulatory and inhibitory effects of cyt b5. When total NADPH consumption is inhibited by cyt b5, the rapidly metabolized, highly coupled (approximately equal to 50%) substrate, benzphetamine, undergoes a net decrease in metabolism not counterbalanced by the increase in the efficiency (2-20%) of the reaction. In contrast, in the presence of the slowly metabolized, poorly coupled (approximately equal to 0.5-3%) substrate, methoxyflurane, inhibition of total NADPH consumption by cyt b5 was never sufficient to overcome the stimulation of product formation due to an increase in efficiency of the reaction.
Highlights
The complete stoichiometry of the metabolism of the cytochrome b5-requiring substrate, methoxyflurane, by purified cytochrome P-450 2B4 was compared to that of another substrate, benzphetamine, which does not require cyt b5 for its metabolism
When total NADPH consumption is inhibited by cyt b5, the rapidly metabolized, highly coupled (Х50%) substrate, benzphetamine, undergoes a net decrease in metabolism not counterbalanced by the increase in the efficiency
The reactivity and overall stoichiometry of the reactions catalyzed by the cytochrome P-450 reconstituted system varied with the experimental conditions, the effect of cytochrome b5 was similar in all cases; addition of cytochrome b5 to the reconstituted system in the presence of both methoxyflurane and benzphetamine resulted in improved efficiency of product formation at the expense of the side products, superoxide, hydrogen peroxide, and water
Summary
The complete stoichiometry of the metabolism of the cytochrome b5 (cyt b5)-requiring substrate, methoxyflurane, by purified cytochrome P-450 2B4 was compared to that of another substrate, benzphetamine, which does not require cyt b5 for its metabolism. An intriguing long standing question has been why is the metabolism of methoxyflurane and a minority of other substrates, such as nifedipine, p-nitroanisole, prostaglandin, lauric acid, n-methylcarbazole, chlorobenzene, p-nitrophenetole, 7-ethoxycoumarin, benzo(a)pyrene, and lidocaine, and testosterone -hydroxylation so stimulated by the presence of cytochrome b5 while the metabolism of most other substrates, including benzphetamine, is minimally stimulated, not effected, or slightly inhibited in the presence of cytochrome b5 (Aoyama et al, 1990; Canova-Davis and Waskell, 1984; Hoffman et al, 1989; Okita et al, 1981; Peyronneau et al, 1992; Vatsis et al, 1982) The answer to this question depends on understanding the individual steps of the catalytic reaction cycle of cytochrome P-450: 1) binding of substrate; 2) one electron reduction of the ferric substrate-enzyme complex by cytochrome P-450 reductase; 3) the binding of oxygen to the ferrous enzyme. In uncoupled reactions oxygen can dissociate from the oxyferrous enzyme to regenerate the ferric cytochrome and superoxide anion which in turn can dismutate to hydrogen peroxide according to Equation 1: HO21⁄7 ϩ O2. ϩ Hϩ 3 O2 ϩ H2O2
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