Oxidation of myristic acid in a skeletal muscle cell

[Biochem_Student]

Hello Folks,

 

I'm a Biochemistry student trying to do some extra work over the summer and to get ahead in my next year at university. I've been doing some problems/questions based on what is available for download at my university intranet site/black board. One of the practice questions has me stumped and I hoped to ask for some help if at all possible?

 

The whole point of the exercise is to calculate the maximum realizable ATP yield per mole of myristic acid being fully oxidized to CO2 and H2O in a skeletal muscle cell under aerobic conditions. I'm trying to understand the step-by-step calculation so as to compare to other substrates such as glucose if that makes sense? I've never really done anything like this before and Biochemistry isn't my field (I'm more animals, plants and ecology).

So far what I know is that Myristoyl CoA is the product of the activation of myristic acid (C14:0) in the cytoplasm of a cell with a low energy charge, prior to transport to the mitochondria for beta-oxidation. The overall reaction of beta-oxidation of myristoyl CoA in the mitochondrial matrix of an animal cell is:

6 FAD + 6 NAD+ + 6 CoASH + 6 H2O + H(CH2CH2)6CH2CO-SCoA --> 7 CH3CO-SCoA + 6 FADH2 + 6 NADH + 6 H+

Edited July 10, 2015 by Biochem_Student

[CharonY]

To provide some pointers, these calculations generally involve the summation of ATP generated via substrate-level phosphorylation as well as the reducing equivalents. The latter (NADH) then indirectly can result in ATP production via the respiratory chain. The precise ratio can vary a bit and tends to be a semi-empirical estimate.

Edited July 12, 2015 by CharonY

[jimmydasaint]

A clue may be to work out how many carbons enter the Krebs cycle per glucose molecule and then calculate the number of ATP s generated in total from the Krebs cycle and the election transport chain. Then, using these figures, work out number of ATP s for myristic acid which has more carbon atoms than glucose. Come back to us for more clues...