The G protein coupled receptor rhodopsin contains a pocket within its seven-transmembrane helix (TM) structure, which bears the inactivating 11-cis-retinal bound by a protonated Schiff-base to Lys296 in TM7. Light-induced 11-cis-/all-trans-isomerization leads to the Schiff-base deprotonated active Meta II intermediate. With Meta II decay, the Schiff-base bond is hydrolyzed, all-trans-retinal is released from the pocket, and the apoprotein opsin reloaded with new 11-cis-retinal. The crystal structure of opsin in its active Ops* conformation provides the basis for computational modeling of retinal release and uptake. The ligand-free 7TM bundle of opsin opens into the hydrophobic membrane layer through openings A (between TM1 and 7), and B (between TM5 and 6), respectively. Using skeleton search and molecular docking, we find a continuous channel through the protein that connects these two openings and comprises in its central part the retinal binding pocket. The channel traverses the receptor over a distance of ca. 70 A and is between 11.6 and 3.2 A wide. Both openings are lined with aromatic residues, while the central part is highly polar. Four constrictions within the channel are so narrow that they must stretch to allow passage of the retinal beta-ionone-ring. Constrictions are at openings A and B, respectively, and at Trp265 and Lys296 within the retinal pocket. The lysine enforces a 90 degrees elbow-like kink in the channel which limits retinal passage. With a favorable Lys side chain conformation, 11-cis-retinal can take the turn, whereas passage of the all-trans isomer would require more global conformational changes. We discuss possible scenarios for the uptake of 11-cis- and release of all-trans-retinal. If the uptake gate of 11-cis-retinal is assigned to opening B, all-trans is likely to leave through the same gate. The unidirectional passage proposed previously requires uptake of 11-cis-retinal through A and release of photolyzed all-trans-retinal through B.