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Added by Barbara M.
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Since FADH2 reduces NAD+ to give FAD and NADH, we can write the half-reactions as follows: Show more…
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B. Given the redox potentials for the half reactions below, explain why (in terms of redox potential), the oxidation of succinate is coupled to the reduction of FAD and not to the reduction of NAD to NADH. Fumarate + 2H+ + 2e- -> succinate Eo' = 0.03 V FAD + 2H+ + 2e- -> FADH2 Eo' = 0.05 V NAD+ + H+ + 2e- -> NADH Eo' = -0.32 V
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The reduction of NAD+ to NADH and the reduction of FAD to FADH2 proceed by the following half reactions: NAD+ + H+ + 2e- <--------> NADH E°’ = -0.320 V FAD + 2H+ + 2e- <--------> FADH2 E°’ = -0.219 V Calculate delta E°’ and delta G°’ for the oxidation of NADH by FAD. Will this be a favorable reaction?
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The E'° of the NAD+/NADH half reaction is -0.32 V. The E'° of the acetoacetate/β–hydroxybutyrate half reaction is -0.346 V. For a redox reaction based on two redox couples, the "spontaneous" net reaction is: Hint: Use the reduction potential table. A) acetoacetate + NAD+ → + NADH + H+ + β–hydroxybutyrate B) NAD+ + NADH + H+ → acetoacetate + β–hydroxybutyrate C) β–hydroxybutyrate + NAD+ → acetoacetate + NADH + H+ D) β–hydroxybutyrate + NADH + H+ → acetoacetate + NAD+ E) acetoacetate + NADH + H+ → β–hydroxybutyrate + NAD+"
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