In total, for the two-electron reduction of Q (ca. 800 meV of energy release), two substrate protons (yellow) are consumed (at fairly low energy cost, as it occurs at the membrane-water interface), and two pumped protons (green) are taken up to E channel (hydrophilic central axis), roughly half membrane dielectric length costing ca. 200 meV (assuming a proton motive force of 200 mV). The pumped protons are eventually released to the P side of the membrane, which would require an additional ca. 200 meV of energy. Thus, the two other pumped protons (for a stoichiometry of four protons) must be recruited from the N side via antiporter-like subunits ND2/4/5 and released to the P side costing additional ~400 meV. In total, energy released upon Q reduction can be completely utilized to pump protons at maximal proton motive force of 200 mV. Concludingly, the above mechanism, which is thermodynamically feasible, does not include the SQ species, but instead relies on anionic quinol, which has been suggested to be the key species based on experiments, theory and simulations (section 5). The conserved tyrosine of NDUFS2 subunit, which assists in trapping Q with a hydrogen bond, may also be the first proton donor (X) upon two-electron reduction of Q. The exergonic protonation of QH -and X - by uptake of protons from the N side drives the uphill proton pumping in complex I.