1. Using the reduction half-reactions and standard reductions potentials, determ

Question

1. Using the reduction half-reactions and standard reductions potentials, determine the free energy change for a pair of electrons from NADH to the FMN complex I. Is there enough free energy to make a molecule of ATP from ADP and Pi?

Reduction half-reaction Acetyl CoA + Co, + H +26 Ferredoxin (spinach). Fe+ - 2H + 2eP H2 (at pH 7.0) -Ketoglutarate + CO2 + 2 He + 2eeIsocitrate Lipoyl dehydrogenase (FAD) + 2 H + 2E9 Lipoyl dehydrogenase (FADH2) E (V) -0.48 -0.43 -0.42 +CoA -0.34 -0.32 -0.29 -0.28 -0.23 -0.22 -0.22 0.20 -0.18 0.17 0.02 0.03 0.04 0.08 0.22 0.23 Lipoic acid + 2H + 2eP Dihydrolipoic acid Thioredoxin (oxidized) + 2H + 2e Thioredoxin (reduced) Glutathione (oxidized) + 2 +2eP 2 Glutathione (reduced) Acetaldehyde + 2 H + 2E9 Ethanol Pyruvate + 2 HO + 2E9 Lactate Oxaloacetate + 2 H + 2yG Malate Cytochrome bs (microsomal). Fe Fe Fumarate + 2 H + 2e Succinate ubiquinone (Q) + 2n® + 2yGOH2 Cytochrome b Cytochrome c, Fe- Cytochrome c F Cytochrome a, Fg + -R Cytochrome fr" + plastocyanin, Cu2+ + ee Cu+ No? + 2H +aPNo@ + H2O 0.37 0.42 0.43 0.77 0.82 Photosystem II (P680)

Explanation / Answer

For the given biological process

The relavant half-reactions are,

FMN gets reduced to FMNH2

FMN + 2H+ + 2e- ---> FMNH2      Eo = -0.22 V

NADH gets oxidized to NAD+

NADH --> NAD+ + H+ + 2e-         Eo = -0.32 V

So,

Eo = Eo(cathode) - Eo(anode)

      = -0.22 - (-0.32) = 0.10 V

Free energy change dGo,

dGo = -nFEo

with,

n = 2

F = Faraday's constant

So,

dGo = -2 x 96500 x 0.1 = -19.30 kJ

The free energy change for ATP hydrolysis is -51.8 kJ which is much higher than the free energy change calculated above. Therefore, the energy is not sufficient for the conversion of ATP to ADP.

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