Water is required for proton transfer from aspartate-96 to the bacteriorhodopsin Schiff base

During the M in equilibrium with N----BR reaction sequence in the bacteriorhodopsin photocycle, proton is exchanged between D96 and the Schiff base, and D96 is reprotonated from the cytoplasmic surface. We probed these and the other photocycle reactions with osmotically active solutes and perturbants and found that the M in equilibrium with N reaction is specifically inhibited by withdrawing water from the protein. The N----BR reaction in the wild-type protein and the direct reprotonation of the Schiff base from the cytoplasmic surface in the site-specific mutant D96N are much less affected. Thus, it appears that water is required inside the protein for reactions where a proton is separated from a buried electronegative group, but not for those where the rate-limiting step is the capture of a proton at the protein surface. In the wild type, the largest part of the barrier to Schiff base reprotonation is the enthalpy of separating the proton from D96, which amounts to about 40 kJ/mol. We suggest that in spite of this D96 confers an overall kinetic advantage because when this residue becomes anionic in the N state its electric field near the cytoplasmic surface lowers the free energy barrier of the capture of a proton in the next step. In the D96N protein, the barrier to the M----BR reaction is 20 kJ/mol higher than what would be expected from the rates of the M----N and N----BR partial reactions in the wild type, presumably because this mechanism is not available.