Glutamic acid at position 113 in bovine rhodopsin ionizes to form the counterion to the protonated Schiff base (PSB), which links the 11-cis-retinylidene chromophore to opsin. Photoactivation of rhodopsin requires both Schiff base deprotonation and neutralization of Glu-113. To better understand the role of electrostatic interactions in receptor photoactivation, absorbance difference spectra were collected at time delays from 30 ns to 690 ms after photolysis of rhodopsin mutant E113Q solubilized in dodecyl maltoside at different pH values at 20 degrees C. The PSB form (pH 5. 5, lambda(max) = 496 nm) and the unprotonated Schiff base form (pH 8. 2, lambda(max) = 384 nm) of E113Q rhodopsin were excited using 477 nm or 355 nm light, respectively. Early photointermediates of both forms of E113Q were qualitatively similar to those of wild-type rhodopsin. In particular, early photoproducts with spectral shifts to longer wavelengths analogous to wild-type bathorhodopsin were seen. In the case of the basic form of E113Q, the absorption maximum of this intermediate was at 408 nm. These results suggest that steric interaction between the retinylidene chromophore and opsin, rather than charge separation, plays the dominant role in energy storage in bathorhodopsin. After lumirhodopsin, instead of deprotonating to form metarhodopsin I(380) on the submillisecond time scale as is the case for wild type, the acidic form of E113Q produced metarhodopsin I(480), which decayed very slowly (exponential lifetime = 12 ms). These results show that Glu-113 must be present for efficient deprotonation of the Schiff base and rapid visual transduction in vertebrate visual pigments.