Calculate the DeltaG°’ for the oxidation of NADH by O 2 . Assuming that the phos

Question

Calculate the DeltaG°’ for the oxidation of NADH by O2. Assuming that the phosphorylation of ADP to ATP requires 30.5 kJ/mol, how many moles of ATP can be produced per mole of NADH? How many pairs of electrons (reduced NAD+ molecules) are generated per glucose molecule fully oxidized to CO2? Taking this into consideration, what is the overall energy (DeltaG°’) yield of NADH oxidation? How many moles of ATP can this amount of energy produce? How does this number compare with the number usually stated for complete aerobic oxidation of glucose? What factors might lead to this discrepancy?:

Explanation / Answer

delta G o' for the oxidation of NADH by oxygen will be equal to -nFdelta Eo'

where delta Eo' is 1.136V , F is 96.5 kJ/mol V and n is the number of electrons involved

deltaGo' = -2*96.5*1.136= - 219 kJ/mol

If the standard free energy of formation of ATP is 30.5 kJ/mol ,then the number of moles of ATP produced will be 219/30.05 =7.1. It means that 7 ATP will be produced per NADH if the electron transport chain is 100% efficient.But in reality 3 ATP is produced per NADH.

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