Proton transport by a bacteriorhodopsin mutant, aspartic acid-85-->asparagine, initiated in the unprotonated Schiff base state.

At alkaline pH the bacteriorhodopsin mutant D85N, with aspartic acid-85 replaced by asparagine, is in a yellow form (lambda max approximately 405 nm) with a deprotonated Schiff base. This state resembles the M intermediate of the wild-type photocycle. We used time-resolved methods to show that this yellow form of D85N, which has an initially unprotonated Schiff base and which lacks the proton acceptor Asp-85, transports protons in the same direction as wild type when excited by 400-nm flashes. Photoexcitation leads in several milliseconds to the formation of blue (630 nm) and purple (580 nm) intermediates with a protonated Schiff base, which decay in tens of seconds to the initial state (400 nm). Experiments with pH indicator dyes show that at pH 7, 8, and 9, proton uptake occurs in about 5-10 ms and precedes the slow release (seconds). Photovoltage measurements reveal that the direction of proton movement is from the cytoplasmic to the extracellular side with major components on the millisecond and second time scales. The slowest electrical component could be observed in the presence of azide, which accelerates the return of the blue intermediate to the initial yellow state. Transport thus occurs in two steps. In the first step (milliseconds), the Schiff base is protonated by proton uptake from the cytoplasmic side, thereby forming the blue state. From the pH dependence of the amplitudes of the electrical and photocycle signals, we conclude that this reaction proceeds in a similar way as in wild type--i.e., via the internal proton donor Asp-96. In the second step (seconds) the Schiff base deprotonates, releasing the proton to the extracellular side.