Titration of aspartate-85 in bacteriorhodopsin: What it says about chromophore isomerization and proton release

Titration of Asp-85, the proton acceptor and part of the counterion in bacteriorhodopsin, over a wide pH range (2-11) leads us to the following conclusions: 1) Asp-85 has a complex titration curve with two values of pK(a); in addition to a main transition with pK(a) = 2.6 it shows a second inflection point at high pH (pK(a) = 9.7 in 150-mM KCI). This complex titration behavior of Asp-85 is explained by interaction of Asp-85 with an ionizable residue X'. As follows from the fit of the titration curve of Asp-85, deprotonation of X' increases the proton affinity of Asp-85 by shifting its pK(a) from 2.6 to 7.5. Conversely, protonation of Asp-85 decreases the pK(a) of X' by 4.9 units, from 9.7 to 4.8. The interaction between Asp-85 and X' has important implications for the mechanism of proton transfer. In the photocycle after the formation of M intermediate (and protonation of Asp-85) the group X' should release a proton. This deprotonated state of X' would stabilize the protonated state of Asp-85. 2) Thermal isomerization of the chromophore (dark adaptation) occurs on transient protonation of Asp-85 and formation of the blue membrane. The latter conclusion is based on the observation that the rate constant of dark adaptation is directly proportional to the fraction of blue membrane (in which Asp-85 is protonated) between pH 2 and 11. The rate constant of isomerization is at least 10(4) times faster in the blue membrane than in the purple membrane. The protonated state of Asp-85 probably is important for the catalysis not only of all-trans double left right arrow 13-cis thermal isomerization during dark adaptation but also of the reisomerization of the chromophore from 13-cis to all-trans configuration during N --> O --> bR transition in the photocycle. This would explain why Asp-85 stays protonated in the N and O intermediates.