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Capítulo 5
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1. 2 and 3 are correct. 2B + 2H + + 2e - -> 2BH

2. 1 and 2 are correct. The same ones would be true for submitochondrial particles treated with DNP.

3. flavoprotein + 2H + + 2e - -> flavoprotein:H 2
2cytochrome ox + 2e - -> 2cytochrome red
b) AH 2 + 2B ox -> 2B red + A + 2H +
c) Reduced B and oxidized A would be present at highest concentrations.


4. NADH is the strongest reducing agent; O 2 is the strongest oxidizing agent; and O 2 has the greatest affinity for electrons.

5. It could abolish the voltage component of the proton-motive force.

6. The first major drop in potential would be bypassed and electrons would be fed into coenzyme Q from a donor at roughly the same energy level.

7. Yes, because the negatively charged Pi ion would decrease the electrochemical gradient (matrix positive) that exists across the inner mitochondrial membrane.

8. Malate concentrations must be kept high enough so that the reaction is favorable under conditions that operate in the cell.

9. One by substrate-level phosphorylation. Eleven by oxidative phosphorylation (assuming three from each NADH and two from each FADH 2 ). Two CO 2 molecules released. One molecule of FAD reduced. Five pairs of electrons stripped.

10. They move (1) through the F0 base of the ATP synthase during the phosphorylation of ADP and (2) through the H + /Pi transporter during the uptake of Pi.

11. ATP/ADP levels in the cell are much higher than that at standard conditions, which requires a greater proton-motive force and thus the expenditure of more energy for ATP production than would be true at standard conditions.

12. If a potential difference of 59 mV is equivalent to a ΔG of 1.37 kcal/mol, then a proton-motive force of 220 mV is equivalent to 5.1 kcal/mol. The passage of 3 mol of protons across the membrane at this DG is sufficient to release 15.3 kcal/mol. It was estimated in Chapter 3 that the ΔG for ATP synthesis in the cell was about 11.5 kcal/mol (p. 92), which suggests that three protons release enough energy to generate a molecule of ATP.

13. If a potential difference of 59 mV is equivalent to a ΔG of 1.37 kcal/mol, then a proton-motive force of 220 mV is equivalent to 5.1 kcal/mol. The passage of 3 mol of protons across the membrane at this DG is sufficient to release 15.3 kcal/mol. It was estimated in Chapter 3 that the ΔG for ATP synthesis in the cell was about 11.5 kcal/mol (p. 92), which suggests that three protons release enough energy to generate a molecule of ATP.

14. No. Mitochondria would not be able to utilize glucose because the reactions of glycolysis occur in the cytosol. Pyruvate or fatty acids would be a good source of electrons.

15. 23 kcal/mol

16. a.) The protonophore would collapse the H + gradient. The Δp would approach 0, stopping ATP production. b.) In the absence of O2, the movement of H + out of the matrix would stop. If the ATP synthase continues to function until the gradient is dissipated, the pH of the matrix would decrease due to the coupled movement of H + across the membrane into the matrix. c.) Reduced glucose would mean reduced CO 2 production.

17. The proton-motive force (Δp) = 59 mV • ΔpH - Vm. If the ΔpH = 1 and the Vm = +59 mV, then Δp = 59 mV •1 - 59 mV = 0. There is no proton-motive force and thus no ATP synthesis.

18. They would be identical because it is the asymmetric gamma subunit that imparts any differences that exist among the catalytic sites.

19. The stator consists of the alpha, beta, and delta subunits of F 1 , and the a and b subunits of F 0 . The other subunits are part of the rotor. The b subunit of F 0 connects with the delta subunit of F 1 , which holds the alpha and beta subunits in a fixed position.

20. If the proton-motive force should drop, as can occur if a cell is deprived of O 2 , the inhibitor will block the enzyme from catalyzing the reverse reaction in which ATP is hydrolyzed.







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