A fluorescent probe, dansylhydrazine, was coupled to the purple membrane with a water-soluble carbodiimide. Fluorescence spectroscopy before and after proteolysis with papain indicated that the label was attached to the carboxyl-terminal tail of bacteriorhodopsin (residues 232–248). Reaction with [3H]dansylhydrazine showed incorporation of 0.7 mol/mol of bacteriorhodopsin in this region. A minor site (0.3 mol/mol) was also labeled in the reaction, but its fluorescence was almost completely quenched by energy transfer to retinal. The conformation and flexibility of the C-terminal tail was studied by proteolysis and fluorescence polarization. Papain removes the C-terminal tail from the purple membrane in two steps, suggesting a segmented structure. At pH 8, 25°C, the first-order rate constants were k1 = 0.23 min−1 and k2 = 0.011 min−1 for unmodified membrane. The cleavage mechanism was found to be sequential removal of the two tail segments. The labeled membrane had a similar mechanism but the first step was much slower: k1 = 0.026 min−1 and k2 = 0.009 min−1. The steady-state polarization of the dansyl fluorescence on the tail was 0.24 at 25°C, indicating a rigid environment. During proteolysis, the polarization decreased to 0.10 after 4 h. The time course of polarization decrease during proteolysis closely matched the rate of release of the inner tail segment. A model was derived for the polarization change during papain cleavage, on the basis of the assumption of only two tail states: covalently bound to bacteriorhodopsin and free in solution. This model gave a good fit to the proteolysis kinetics. Thus, the release of the outer segment does not affect the polarization, and hence the flexibility, of the inner segment. Time-resolved fluorescence anisotropy indicated no tail motion over two excited-state lifetimes. We conclude that the labeled region of the C-terminal tail of bacteriorhodopsin is rigidly held at the membrane surface.
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