Net water produced from complete oxidation of an 8 carbon saturated fatty acid

[jk50]

A Berkeley Review passage asks how much water would be produced from the complete oxidation of 1 mole of caprylic acid (8 carbon saturated fatty acid). The answer states that 8 net H20 would be produced, however using the method that Lehningers Biochem teaches on p. 675 I got 11 H20. Can anyone confirm if I did it correctly and if TBR is wrong?:

Step 1: 3 rounds of B-ox produces 4 acetyl CoA, 3 NADH, 3 FADH2, and wastes 3H2O; according to Lehninger's p.675, 1 H2O is produced per 2 electron carrier; there are 6 two electron carriers (3 FADH2 + 3 NADH) produced, so 6 H2O are produced, minus the 3 H20 wasted in the B-ox process is a net 3 H2O.

Step 2: feed the 4 acetyl CoA into the Krebs Cycle: the Krebs wastes a net 2 H2O per acetyl Coa, so 8 H2O used up, 12 NADH made, 4 FADH2 made, and 4 GTP, that becomes ATP, made. Once again, according to lehninger, there would be 16 H2O made (because 16 two electron carriers) minus the 8 H20 wasted in the krebs = 8 net H20.

Step 3: add step 1 and step 2 together = net total of 11 H2O produced from the full oxidation of caprylic acid. TBR says that 8 h2O are made due to a balancing of C8H1602 + 11O2 ---> 8CO2 + 8H20. Can anyone confirm if my way is right or TBR?? On the mcat should i just make a balanced equation like they did or do my way? Also, lehningers teaches 2.5 ATP produced per NADH and 1.5/FADH2, but TBR teaches 3 ATP/ NADH, and 2ATP/FADH2; on the mcat which should we use?

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[kadler538]

Edit: You're right, didn't take into account the other electron transport molecules made from oxidation....

Ah, so I believe that the NADH and FADH2 produced by oxidation turn out to be insignificant in terms of H2O production, as "The water produced from ATP synthesis is negated because it would be used up in regenerating ADP and Pi"....This was all I could find after looking into it....

 

[jk50]

That makes sense, but doesn't your method not take into account the 3 NADH and 3 FADH2 produced from the 3 B-ox cycles? Factoring those in would produce an additional 6 H2O, but subtracting for the 3 H2O used in the B-ox process, that would equal 3 net H20, added to the 8 you calculated equals 11 H2O net, which is what i got. Idk if for the mcat we aren't supposed to factor the h2o produced from oxidizing the FADH2 and NADH made in b oxidation. Lehningers does in chapter 17, but maybe for the mcat we don't need to?