Question

,0+2015MADHE 5. (8 points) Consider the oxidation of a mole of NADH: 083 V NAD +20 +H - NADH E' =-0.32 V What is the AG for the oxidation of a mole of NADH by oxygen? How many moles of protons are pumped as a result of the oxidation of a mole of NADH? If all of the protons could be used for ATP production by an ATP synthase with 10 c subunits, how many moles of ATP could be produced? If the AG for ATP synthesis is about 50 kJ/mol, what percentage of the energy from NADH oxidation is captured in ATP formation?

Answer 1

Free energy change AG = nF4E. where n is the number of moles F is the Faraday's constant and AE, is the change in standard half pottential. in this question to of /₂O2 + 2e + 2H+ H₂O is given as 0.83v and E° of NADP+ 2e + H7-> NADIH is given as -0.3au then AE = 0.83 - (-032) = 1.150 Number of moles (n) is given as 1 Then, AG=DF AEO F-96,485c/mol We know that IV = 15/16 Then Aa=t-Imote ) (964854 ) Mote / = -1 X 96485 X 1.15) - - 110,957.75 J

Free energy =-110957.75 J = -110.95 KJ

The number of protons pumped from a mole of NADH in electron transport chain is 10

3.33 protons yield one ATP

Then the number of atp yielded is

10:3.33 = 3.003

3 ATPs are yielded