Using the Standard reduction potentials for Q and FAD from Table 2 below, show t
ID: 915000 • Letter: U
Question
Using the Standard reduction potentials for Q and FAD from Table 2 below, show that the oxidation of FADH_2 by Q liberates enough energy to drive the synthesis of ATP from ADP and Pi under cellular conditions where [FADH_2] = 5 mM, [FAD] = 0.2 mM, [Q] = 0.1 mM and [QH_2] = 0.05 nM. Assume that DeltaG for ATP synthesis from ADP and Pi = +30 kJ mol^-1. In the electron transport chain, the oxidation of NADH is coupled with the reduction of O_2. coupling data in Table 2 to calculate the overall DeltaG degree for this process. How many ATP can theoretically be made from this redox coupling (to nearest whole number)? Give 2 uses for the "excess energy".Explanation / Answer
ANSWER
FAD + 2H+ + 2e <----->FADH2 Eo = -0.22V
Reverse the reaction
FADH2 <-----------> FAD + 2H+ + 2e
Q + 2H+ + 2e <-------> QH2
Net reaction
FADH2 + Q <--------> FAD + QH2
Eocell = Eorxn - 0.059/n log ( [FAD][QH2] / [FADH2][Q])
Eorxn = +0.04 - ( -0.22) = + 0.26V
Eocell = 0.26 - 0.059/2 log ( [0.2 X 10-3][0.05 X 10-9] / [5.0 X 10-3][0.1 X 10-3])
Eocell = 0.487V
G = - nFEorxn
G = - 2 X 96500 X 0.487 = - 50180J = - 93.99KJ
Hence more energy is released (93.99KJ) by oxidation of FADH2 han consumed (30KJ) by synthesis of ATP
(a) NAD+ + H+ +2e <----> NADH Eo -0.32V
Reverse the reaction
NADH <----> NAD+ + H+ +2e
multiplying by 2
2NADH <----> 2NAD+ + 4H+ +4e Eo = 2 X 0.32V
O2 + 4H+ + 4e <----> 2H2O
Net reaction
2NADH + O2 <----> 2NAD+ 2H2O
Eorxn = 0.82 - ( - 0.64) = 0.18V
G = - nFEorxn
G = - 2 X 96500 X 0.18 = 34.74KJ
(b) energy released during synthesis of ATP = 94KJ
Energy consumed during synthesis of ATP = 30Kj
Therefore no. of ATP that can be synrhesised = 94 / 30 = 3ATP
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