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8885dc19690-7503/1/0411:32 AM Page701mac76mac76:385 19.1 Electron-Transfer Reactions in Mitochondria dation of Oxidation of Intermembrane Cyt Q QH2. trix (N side) H。 Q+ 2Hp cyt c,(reduced) QH2+ 2Hp+Q+ cyt ci(reduced Net equation: QH2+ 2 cyt cn(oxidized )+ 2HN -)Q+2 cyt ci (red FIGURE 19-12 The Q cycle. The path of electrons through Complex QH2 donates one electron(via the Rieske Fe-S center)to cytochrome Ill is shown by blue arrows. On the P side of the membrane, two mol- C1, and one electron(via cytochrome b) to a molecule of Q near the ecules of QH2 are oxidized to Q near the P side, releasing two pro- N side, reducing it in two steps to QH2. This reduction also uses two tons per Q(four protons in all) into the intermembrane space. Eac Electron transfer through Complex Iv is from cyto- bound to the complex until completely converted to chrome c to the Cua center, to heme a, to the heme water. a3-CuB center, and finally to O2(Fig. 19-14. For every ur electrons passing through this complex, the enzyme The Energy of Electron Transfer Is Efficiently consumes four"substrate"H from the matrix( side) Conserved in a Proton Gradient in converting O2 to 2H,O. It also uses the energy of this redox reaction to pump one proton outward into the in- The transfer of two electrons from NADH through the termembrane space (P side) for each electron that respiratory chain to molecular oxygen can be written as passes through, adding to the electrochemical potential NADH+H++O2→→NAD++H2O roduced by redox-driven proton transport through Complexes I and III. The overall reaction catalyzed by This net reaction is highly exergonic edox pair NAD NADH, E is-0.320 V, and for the pair O2/H2O E°is0.816VThe△E"° for this reaction is therefo 4 Cyt c(reduced)+8H+O2—→ 1. 14 V, and the standard free-energy change(see eqn 4 cyt c(oxidized)+ 4Hp+ 2H20 (19-4) 13-6, p 510)is This four-electron reduction of o involves redox cen- △G°=-n△E° (196 ters that carry only one electron at a time, and it must occur without the release of incompletely reduced =-2965kJN·mol)(1.14V) intermediates such as hydrogen peroxide or hydroxyl -220 k/mol (of NADH) free radicals--very reactive species that would damage This standard free-energy change is based on the as- cellular components. The intermediates remain tightly sumption of equal concentrations (1 M) of NADH andElectron transfer through Complex IV is from cytochrome c to the CuA center, to heme a, to the heme a3–CuB center, and finally to O2 (Fig. 19–14). For every four electrons passing through this complex, the enzyme consumes four “substrate” H from the matrix (N side) in converting O2 to 2H2O. It also uses the energy of this redox reaction to pump one proton outward into the intermembrane space (P side) for each electron that passes through, adding to the electrochemical potential produced by redox-driven proton transport through Complexes I and III. The overall reaction catalyzed by Complex IV is 4 Cyt c (reduced) 8H N O2 On 4 cyt c (oxidized) 4H P 2H2O (19–4) This four-electron reduction of O2 involves redox centers that carry only one electron at a time, and it must occur without the release of incompletely reduced intermediates such as hydrogen peroxide or hydroxyl free radicals—very reactive species that would damage cellular components. The intermediates remain tightly bound to the complex until completely converted to water. The Energy of Electron Transfer Is Efficiently Conserved in a Proton Gradient The transfer of two electrons from NADH through the respiratory chain to molecular oxygen can be written as NADH H 1 2 O2 On NAD H2O (19–5) This net reaction is highly exergonic. For the redox pair NAD/NADH, E is 0.320 V, and for the pair O2/H2O, E is 0.816 V. The E for this reaction is therefore 1.14 V, and the standard free-energy change (see Eqn 13–6, p. 510) is G n E (19–6) 2(96.5 kJ/V mol)(1.14 V) 220 kJ/mol (of NADH) This standard free-energy change is based on the assumption of equal concentrations (1 M) of NADH and 19.1 Electron-Transfer Reactions in Mitochondria 701 bH bL Cyt c1 Cyt c Oxidation of first QH2 Fe-S Q 2H+ •Q – QH2 Q bH bL Cyt c1 Cyt c Fe-S 2H+ Q Oxidation of second QH2 Matrix (N side) Intermembrane space (P side) QH2 2H+ •Q – QH2 QH2 cyt c1 (oxidized) QH2 2 cyt c1 (oxidized) 2HP cyt c1 (reduced) Q 2 cyt c1 (reduced) •Q •Q 4HP 2HP 2HN 2HN QH2 QH2 cyt c1 (oxidized) Q cyt c1 (reduced) Net equation: FIGURE 19–12 The Q cycle. The path of electrons through Complex III is shown by blue arrows. On the P side of the membrane, two molecules of QH2 are oxidized to Q near the P side, releasing two protons per Q (four protons in all) into the intermembrane space. Each QH2 donates one electron (via the Rieske Fe-S center) to cytochrome c1, and one electron (via cytochrome b) to a molecule of Q near the N side, reducing it in two steps to QH2. This reduction also uses two protons per Q, which are taken up from the matrix. 8885d_c19_690-750 3/1/04 11:32 AM Page 701 mac76 mac76:385_reb: