Chemiosmotic Coupling

Chemiosmotic Coupling - Chemiosmotic Coupling Bryant Miles...

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Chemiosmotic Coupling Bryant Miles The four major complexes of the electron transport chain operate independently in the inner mitochondrial membrane. Each of the complexes is an aggregate of proteins that are held firmly together by noncovalent forces. There is no experimental evidence that the four complexes associate with one another in the membrane. Each of these complexes has its own rate of lateral diffusion through the bilipid membrane which shows that the complexes do not move together. Kinetic studies of reconstituted electron transport systems support the theory of four independent complexes. The model for the electron transport system is shown above. Four independent complexes are transferring electrons through the mobile electron carriers of CoQ and cytochrome c . CoQH 2 is produced by both complex I and complex II and delivers the electron to complex III via the Q-cycle. Complex III reduces cytochrome c which is a water soluble electron carrier located in the intermembrane space of the mitochondria. The reduced cytochrome c carries the electrons to complex IV which transfers the electrons to molecular oxygen. The process of electron transfer is coupled to transporting protons from the matrix to the intermembrane space of the mitochondria. This generates a chemical potential and an electrostatic potential. This potential energy is used to drive the synthesis of ATP. Complex I: NADH + 5H + N + Q ± NAD + + QH 2 + 4H + P Complex II: FADH 2 + Q ± FAD + QH 2 Complex III: QH 2 + 2H + N + 2Cyt c (Fe 3+ ) ± 2Cyt c (Fe 2+ ) + Q + 4H + P Complex IV: 4Cyt c (Fe 2+ ) + 8H + N + O 2 ± 4Cyt c (Fe 3+ ) + 4H + P + 2H 2 O Overall reaction beginning with NADH: NADH + 11H + N + ½O 2 ± NAD + + 10H + P + H 2 O (10H + P /2e - ) Overall reaction beginning with FADH 2 :
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FADH 2 + 6H + N + ½O 2 ± FAD + 6H + P + H 2 O (6H + P /2e - ) NADH + H + + ½O 2 ± NAD + + H 2 O NAD + + 2e - + H + ± NADH E o ’ = 0.315 V ½O 2 + 2e - + 2H + ± H 2 O E o ’ = +0.816 V Δ E o ’ = +0.816 V – ( 0.315 V) = 1.136 V Δ G o ’ = nF Δ E o ’ = 219 kJ/mol FADH 2 + ½O 2 ± FAD + H 2 O FAD + 2e - + H + ± FADH 2 E o 0.000 V ½O 2 + 2e - + 2H + ± H 2 O E o ’ = +0.816 V Δ E o ’ = +0.816 V – (0 V) = 0.816 V Δ G o ’ =
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