Question

Here are four problems (5 pts each) involving the calculation of DG°' for metabolic reactions we have discussed, based on experimentally determined redox potentials. Calculate the DG°' for each reaction using the equation DG°' = -nFDE₀' and the values for E₀' given in Table 1. Show your work and circle your answer.

Table 1. Reduction potentials for reduction half reactions:

1/2 O₂ + 2H⁺ + 2e^- ® H₂O                                              E₀' = +0.82 V

fumarate + 2H⁺ + 2e^- ® succinate                                  E₀' = +0.03 V

oxaloacetate + 2H⁺ + 2e^- ® malate                                 E₀' = -0.17 V

pyruvate + 2H⁺ + 2e^- ® lactate                                       E₀' = -0.19 V

a-ketoglutarate + CO₂ + 2H⁺ + 2e^- ® isocitrate               E₀' = -0.38 V

FAD + 2H⁺ + 2e^- ® FADH₂                                             E₀' = -0.22 V

NAD⁺ + 2H⁺ + 2e^- ® NADH + H⁺                                    E₀' = -0.32 V

CoQ + 2H⁺ + 2e^- ® CoQH₂                                            E₀' = +0.06 V

Problem 3. malate + NAD⁺ ® oxaloacetate + NADH + H⁺

Problem 4. FADH₂ + CoQ ® FAD + CoQH₂

Answer 1

malate + NAD⁺ -------------------> oxaloacetate + NADH + H⁺

malate -----------------> oxaloacetate + 2H⁺ + 2e^-           E0 = 0.17V

NAD⁺ + 2H⁺ + 2e^- ---------> NADH + H⁺                                    E₀' = -0.32 V

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malate + NAD⁺ -------------------> oxaloacetate + NADH + H⁺     E0cell= -0.15V

DG°' = -nFDE₀'

        = -2*96500*-0.15   = 28950J

FADH₂ + CoQ ------------> FAD + CoQH₂

FADH₂   ------------> FAD + 2H⁺ + 2e^-      E0 = 0.22V

CoQ + 2H⁺ + 2e^- -----------> CoQH₂        E₀' = +0.06 V

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FADH₂ + CoQ ------------> FAD + CoQH₂   E0cell =0.28V

DG°' = -nFDE₀'

        = -2*96500*0.28 = -54040J