0
$\begingroup$

From studying $\beta$-oxidation of free fatty acids, I've learned how to compute for the ATP equivalence of even-chain and odd-chain free fatty acids.

What really bothers me now is how to compute for the ATP equivalence of an unsaturated fatty acid. Since one of the major difference of the process is in unsaturated fatty acids, normal $\beta$-oxidation rounds will occur until the unsaturation is encountered. The cis-isomer will be converted to trans-isomer completely skipping the step where the fatty acyl-CoA molecule is oxidized to the trans-form with the release of FADH2.

Is the computation same and I just need to subtract a molecule of FADH2 per unsaturation (double bond)?

$\endgroup$
1
$\begingroup$

Is the computation same and I just need to subtract a molecule of FADH2 per unsaturation (double bond)?

Definitely subtract a molecule of $\ce{FADH2}$. After all, you can skip the first of the the four steps in each cycle of remove two carbons (formation of double bond, adding water across the double bond, oxidizing the double bond to a carbonyl group, oxidative decarboxylation). You would also have to check whether any of the steps are coupled to ATP hydrolysis (e.g. the formation of the link to the acyl carrier protein, or transport steps) if you want to get an accurate number of ATP molecules synthesized per fatty acid turned into carbon dioxide.

The cis-isomer will be converted to trans-isomer completely skipping the step where the fatty acyl-CoA molecule is oxidized to the trans-form with the release of $\ce{FADH2}$.

Most unsaturated fatty acids in unprocessed food are in the cis-form. To turn them into the requisite $\ce{\Delta^2}$ trans-double bond, you need one or two additional enzymes, see https://www.ncbi.nlm.nih.gov/books/NBK22387/. For turning cis- into trans- and, if necessary, moving the double bond into the correct position, an isomerase is used. If there are two double bonds, one is reduced first by a reductase.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.