Here are four problems (5 pts each) involving the calculation of AG\" for metabo

Question

Here are four problems (5 pts each) involving the calculation of AG" for metabolic reactions we have discussed, based on experimentally determined redox potentials. Calculate the AG" for each reaction using the equation AG nFAEo and the values for Eo given in Table 1. Show your work and circle your answer. Review the example below, used in lecture 29, the reduction of pyruvate by NADH, catalyzed by lactate dehydrogenase Table 1. Reduction potentials for reduction half reactions: Eo a +0.82 V 112 O2 2H 2e H20 Eo +0.03 V fumarate 2H 2e succinate Eo" -0.17 V oxaloacetate 2H 2e malate -0.19 V pyruvate 2H 2e lactate -0.38 V a-ketoglutarate CO2 2H' 2e isocitrate Eo -0.22 V FAD 2H 2e FADH2 Eo -0.32 V NAD 2H 2e NADH H Eo" +0.06 V CoQ 2H 2e CoQH2

Explanation / Answer

Fumarate+2H+2e---> Succinate Eo =0.03V (1)

FAD +2H+ 2e- ------>FADH2 , EO= -0.22V   (2)

Reversing reaction 1, Succinate ----->Fumurate +2H+2e-   Eo=-0.03V (1A)

Addition of 1A and 2 gives succinate +FAD--->Fumatrate + FADH2, Eo= -0.03-0.22= -0.25V

2.

deltaG= -nFEO= -2*96500*(-0.25) =48250 Joules ( n is number of electrons transferred)

Reaction 3 when reversed, Malate -----<oxaloacetate+2H+ 2e-   EO= 0.17V (3)

NAD+ 2H+ 2e- ----------à NADH +H+ Eo= 0.32V (4)

Additino of Eq.3 and Eq.4 gives

Malate+NAD--àoxaloacetate +H+NADH, Eo= 0.32+0.17=0.49V

deltaGo= -2*96500*0.49 =-94750 joules

3.

FADH2---->FADH+2H+2e- ------> EO= 0.22V (4)

COQ+2H+2e- ----->COQH2 EO= 0.06V (5)

Additino of Eq.4 and 5 gives

FADH2+COQ--------> FADH +COQH2 EO= 0.28V

deltaG=-2*96500*0.28=-54040 Joules

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