During the L-->M reaction of the bacteriorhodopsin photocycle the proton of the retinal Schiff base is transferred to the anionic D85. This step, together with the subsequent reprotonation of the Schiff base from D96 in the M-->N reaction, results in the translocation of a proton across the membrane. The first of these critical proton transfers occurs in an extended hydrogen-bonded complex containing two negatively charged residues (D85 and D212), two positively charged groups (the Schiff base and R82), and coordinated water. We simplified this region by replacing D212 and R82 with neutral residues, leaving only the proton donor and acceptor as charged groups. The D212N/R82Q mutant shows essentially normal proton transport, but in the photocycle neither of this protein nor of the D212N/R82Q/D96N triple mutant does a deprotonated Schiff base (the M intermediate) accumulate. Instead, the photocycle contains only the K, L, and N intermediates. Infrared difference spectra of D212N/R82Q and D212N/R82Q/D96N demonstrate that although D96 becomes deprotonated in N, D85 remains unprotonated. On the other hand, M is produced at pH > 8, where according to independent evidence the L<==>M equilibrium should shift toward M. Likewise, M is restored in the photocycle when the retinal is replaced with the 14-fluoro analogue that lowers the pKa of the protonated Schiff base, and now D85 becomes protonated as in the wild type.(ABSTRACT TRUNCATED AT 250 WORDS)