Assuming 2 ATP per NADH and 1 ATP per FADH, how many moles of ATP can be generated in E. coli cell when one mole of myristic acid (14:0) is completely oxidized during beta-oxidation? Assume only FADH is produced during beta-oxidation.
a. 67 ATP/mole 14:0
b. 68 ATP/mole 14:0
c. 69 ATP/mole 14:0
d. 70 ATP/mole 14:0
Hence for 14- carbon containing Myristate, from C14 to C2 Acetyl Co-A, 6 more Acetyl Co-A will be produced, by removal of 2 carbons in each step:
C14 -----> C12 ----> C10 --- > C8 ----- > C6 ----- > C4 ------ > Acetyl Co-A
Acetyl Co-A Acetyl Co-A Acetyl Co-A Acetyl Co-A Acetyl Co-A Acetyl Co
During each round of beta-oxidation of a fatty acid,
1 NADH+ H + 1 FADH2+ 1molecule of Acetyl CO-A is yielded.
Considering:
Thus, for Myristate.
For 10 Acetyl CO-A= 7 x 10 = 70 ATP
For .6 NADH+ H = 2 x 6 = 12 ATP
For 6 FADH2 = 1 x 6 = 6 ATP
Total = 88
Activation of fatty acid, requires attachment of Acetyl Co-A-SH, this utilizes 2 ATP.
Net ATP yield = 88-2 =86 ATP
If only FADH2 are produced and in E.coli
7 x 10 = 70 ATP
Assuming 2 ATP per NADH and 1 ATP per FADH, how many moles of ATP can...
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