1. Fundamentals of Biochemistry Third Edition Fundamentals of Biochemistry Third Edition Chapter 18 Electron Transfer and Oxidative Phosphorylation Chapter 18 Electron Transfer and Oxidative Phosphorylation Copyright © 2008 by John Wiley & Sons, Inc. Donald Voet Judith G. Voet Charlotte W. Pratt

2. Figure 18-2 part 2 Parts of the Mitochondrion Text – Figure 18-2

3. Figure 18-5 The reducing equivalents produced in the cytosol have to be shuttled into the mitochondrion if they are to be used for oxidative phosphorylation Text – Figure 18-5

4. Figure 18-6 ATP produced in the mitochondrion by oxidative phosphorylation is transported back into the cytosol Text – Figure 18-6

5. Figure 18-20 The overall scheme by which a proton gradient is generated by the transfer of electrons from NADH to Oxygen Text – Figure 18-20

6. Figure 18-7 Electrons originating from NADH jump through a series of transmembrane protein complexes, from cofactor to cofactor, of increasing redox potentials, ultimately to O 2 note that ATP is not formed directly!

7. Figure 18-8 Migration of the electrons through the protein complexes causes protons to be pumped from the matrix and into the intermembrane space Text – Figure 18-8

8. Page 605 Structures of Fe-S clusters that are used for electron transfer in Complex I (and elsewhere) Text – Figures, pg. 605

9. Figure 18-10 Structures of flavin and quinone cofactors used for electron transfer in Complex I (and elsewhere). note that these organic cofactors can do single electron transfer, whereas NADH only does 2 electron transfer

10. Figure 18-11 Structure of the peripheral (i.e. non-membrane part of) complex I, including a series of electro transfer cofactors Text – Figure 18-11

11. Figure 18-13 Structure of the Complex II, by which redox equivalents can enter oxidative phosphorylation via succinate instead of NADH Text – Figure 18-13

12. Box 18-1c Examples of heme cofactors

13. Figure 18-14 Structure of the Complex III, which receives electrons from quinone and delivers them to cytochrome Text – Figure 18-14

14. Figure 18-15 The ‘Q cycle’ ‘Q cycle’: the complex mechanism by which redox equivalents (i.e. electrons) are delivered from Complex 1 to quinones, which go on to deliver their electrons to complex IV. very complex, but some basic points can be made note that the quinone gets reduced on the right (matrix) side and oxidized on the left (inter-membrane side) note that the reduction of an organic cofactor by a metal center generally involves an uptake of protons, whereas reduction of a metal center by an organic factor generally involves a release of protons since those two processes occur on different sides, proton pumping results the Q cycle represents just one way that protons get pumped. Other protons are believed to be pumped by protein conformational changes driven by electron transfer (through complexes I and IV); the protein bacteriorhodopsin serves as a model system for studying the coupling between protein conformation changes and proton transfer

15. Figure 18-17 Structure of the Complex IV, which receives electrons from cytochrome and reduces O 2 to H 2 O Text – Figure 18-17

16. Figure 18-18 Cofactors in Complex IV involved in the 4-electron reduction of O 2 to H 2 O Text – Figure 18-18

17. Figure 18-19 Proposed mechanism of 4-electron reduction of O 2 to H 2 O by the Fe-Cu binuclear cluster Text – Figure 18-19

18. Figure 18-20 Electron transfer and ATP formation are coupled via a proton gradient (i.e. a proton gradient is generated by electron transfer, and then used to drive ATP synthesis) the idea that ATP synthesis occurrs by way of a proton gradient was part of the chemiosmotic hypothesis proposed by Peter Mitchell. The mystery of how this occurs was not worked out until decades later. Paul Boyer proposed a controversial binding- change rotating mechanism, which was eventually shown to be astonishingly accurate by crystal structures of the F 1 -ATPase (by Andrew Leslie and John Walker).

19. Figure 18-21b Electron transfer and ATP formation are coupled via a proton gradient (i.e. a proton gradient is generated by electron transfer, and then used to drive ATP synthesis)

20. Figure 18-22a One of the first figures prepared of the structure of the F1- ATPase (  ) 3 with  similar to , making a pseudo-symmetric C 6 hexamer only the active sites in the  subunits are active; the  subunits are effectively structural hole down the middle, with the  subunit (blue) acting as an axle for rotation

21. Figure 18-22b In the absence of the central (asymmetric)  subunit, the three  subunits would be identical, but the  subunit breaks the symmetry

22. Figure 18-23 A rough model for the complete F 0 -F 1 ATPase

23. Figure 18-24 Based on the structural data, the ‘binding- change’ mechanism can be understood as: The central  subunit rotates through 3 different positions (driven by protons flowing down their gradient, as described later) In a given orientation of the  subunit, the 3  subunits are driven to 3 different configurations: one that prefers to have an empty active site (O above), one that prefers to bind ATP (T above), and one that prefers to bind ADP (L above). Note that taking an active site that has ADP (and P i ) bound and driving i to a configuration that prefers ATP will effectively synthesize ATP! Text – Figure 18-24

24. Figure 18-25 How proton flow through the F 0 complex drives the rotation of the  subunit is not understood fully, but current models show that a ring of ~10 transmembrane (c) subunits [the number may vary between species] are caused to shift by proton flow between the a and c subunits; a stator complex (subunit b) holds the a and  /  subunits in a fixed orientation so that only the c and  subunits turn. Coupling to proton flow Text – Figure 18-25

25. Figure 18-25 Crystal structures of the ATPsynthase components provided compelling evidence for a rotational motor mechanism (consistent withBoyer’s visionary hypothesis) – essentially a wheel (  /  ) and axle (  ). Morover, direct evidence for rotation was provided later by attaching an actin polymer to the tip of the  subunit, sticking the  /  subunits to a glass slide, and providing ATP (note that in this situation the complex acts in the reverse direction, consuming ATP and causing rotation). The link below (from R. Berry) shows a slightly different experiment in which the  subunit is labeled with a pair of microscopic beads, which can be see easily by light microscopy. http://www.youtube.com/watch?v=QOiwpW1sivA Visualizing the rotational mechanism of F 0 F 1 ATPase

26. Figure 18-27a Text – Figure 18-27

27. Figure 18-25 Some final accounting Stoichiometric accounting 1 NADH  10 H + pumped 10 H + pumped  ~ 1 complete turn of the ATPase 1 complete turn of the ATPase  3 ATP (one from each  subunit) electrons from FADH2 enter at Complex II which leads to only 8 H + pumped [  2.4 ATP] some fraction of the H + (gradient) get used for other transport processes, or are lost by leakage, so the actual (so-called P/O ratios for NADH and FADH2 are closer to 2.5 and 1.5 based on those values, 1 glucose  2ATP+2GTP+10NADH+2FADH2  32 ATP

28. Figure 18-25 Some final accounting Energy accounting (efficiency) Energy available by oxidation of NADH by (½)O 2 (per NADH molecule (2e - )):  G =  G 0’ +RT ln ([product]/[reactants]) = -n F  E 0’ +RT ln((1./P O2 (atm)) 1/2 ) + RT ln([NAD + ]/([NADH]([H+]/10 -7 M)))  E 0’ = 0.815V (for reduction of O 2 ) – (-0.315V) (for reduction of NAD + = 1.13V the ratios in the 2 nd and 3 rd terms in the equation for  G above are close enough to 1 that their values don’t matter so much; the first term in the equation for  G dominates since the voltage difference is so great)  G  - 2 x 96,500C/mol x 1.13V  218kJ/mol ± several kJ for the terms omitted

29. Figure 18-25 Some final accounting Energy accounting (efficiency) Energy recovered (at the H + gradient stage) (per 10H + pumped): Free energy for H+ gradient depends on the concentration gradient and electrostatic potential difference (  ), which is positive the intermembrane space compared to the matrix  G transporting one H+ = RTln [H + outside ]/[H + inside ] + z F  [if the pH gradient is 0.75 units (higher conc. outside vs inside) and the  is +0.168 (outside vs inside)] [z is the charge on the ion transported = +1]  G transporting one H+ = 20.5 kJ/mol  G transporting 10 H+ = 205 kJ/mol So, energy efficiency is about 205/218, which is > 90% efficiency for forming the H + gradient

30. Figure 18-25 Some final accounting Energy accounting (efficiency) Energy recovered (at the final ATP synthesis stage) assuming 2.5 ATP’s formed: The text give  G  45 kJ/mol for ADP+P i  ATP under physiological conditions  G for 2.5 ATP  112 kJ/mol So, total energy efficiency is about 112/218  50% for forming ATP, with most of the loss apparently coming in the second step (i.e. conversion of the H + gradient to ATP)