In: Chemistry
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 1. isocitrate + NAD+ ® a-ketoglutarate + CO2 + NADH
Problem 2. succinate + FAD ® fumarate + FADH2
Problem 1. isocitrate + NAD+---------> a-ketoglutarate + CO2 + NADH
isocitrate ---------------> a-ketoglutarate + CO2 + 2H+ + 2e- E0 = 0.38V
NAD+ + 2H+ + 2e-------------> NADH + H+ E0' = -0.32 V
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isocitrate + NAD+---------> a-ketoglutarate + CO2 + NADH E0 cell = 0.06V
G = -nE0cell *F
= -2*0.06*96500 = -11580J
Problem 2. succinate + FAD --------------->fumarate + FADH2
succinate ------------------> fumarate + 2H+ + 2e- E0 = -0.03V
FAD + 2H+ + 2e- ----------> FADH2 E0' = -0.22 V
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succinate + FAD --------------->fumarate + FADH2 E 0cell =- 0.25 V
G = -nE0cell *F
= -2*-0.25*96500 = 48250J