Formation of Meta III during the decay of activated rhodopsin proceeds via Meta I and not via Meta II.

Thermal isomerization of the retinal Schiff base C=N double bond is known to trigger the decay of rhodopsin's Meta I/Meta II photoproduct equilibrium to the inactive Meta III state [Vogel, R., Siebert, F., Mathias, G., Tavan, P., Fan, G., and Sheves, M. (2003) Biochemistry 42, 9863-9874]. Previous studies have indicated that the transition to Meta III does not occur under conditions that strongly favor the active state Meta II but requires a residual amount of Meta I in the initial photoproduct equilibrium. In this study we show that the triggering event, the thermal isomerization of the protonated Schiff base, is independent of the presence of Meta II and occurs even under conditions where the transition to Meta II is completely prevented. We have examined two examples in which the transitions from Lumi to Meta I or from Meta I to Meta II are blocked. This was achieved using dry films of rhodopsin and rhodopsin reconstituted into rather rigid lipid bilayers. In both cases, the resulting fully inactive room temperature photoproducts decay specifically by thermal isomerization of the protonated Schiff base C=N double bond to an all-trans 15-syn chromophore isomer, corresponding to that of Meta III. This thermal isomerization becomes less efficient as the conformation of the respective photoproduct approaches that of Meta II and is fully absent in a pure Meta II state. These results indicate that the decay of the Meta I/Meta II photoproduct equilibrium to Meta III proceeds via Meta I and not via Meta II.