Question

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

Answer 1

• Myristate or myristic acid (1-Tetradecanoic acid) is a 14-carbon saturated fatty acid.
• During Beta-oxidation of fatty acids, in successive steps there are removal of 2-carbon atoms to form Acetyl Co-A.

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.

• Thus, from 1 molecule of 14- carbon containing Myristate: in total 7 molecules of Acetyl Co-A +6 NADH +H + 6 FADH2 will be generated

Considering:

• 1 molecule of Acetyl Co-A, = 10 ATP

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