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I am a bit confuse with the oxidation of fatty acids in the mitochondria and the perixosome. Anyone can list out the steps and explain to me? Is there any difference between the oxidaion of fatty acids in mitochondria and oxidation of fatty acids in the perixosome?
Thanks in Advance!
yes there's a difference, since the peroxisome use different enzymes (peroxidases and so on). I'm more into the fat oxidation as in the metabolism. Here are the steps from the B-oxidation in the mitochondria ( thus: the acyl-CoA has been transported into the mitochomndrium uusing the acyl-carnitine shuttle):
The acyl-CoA is oxidized to trans - delta 2- enoyl - coA by acyl CoA-dehydrogenase. Because a double bonding is formed FADH2 is formed and is used in the phosphorylation.
Next the eboyl is hydratized by the enzyme hydratase. Now L - 3 - hydroxyacyl- CoA is formed.
This molecule is again oxidized But now a double bonding at O is formed using the enzyme L - 3 - hydroxyacyl- CoA dehydrogenase. Hence, NADH and 1 proton is formed and used in the phosphotylation.
This molecule contains 2 double bons for O and now the molecule is ready to split off 1 acetyl-CoA. Therefore the thio-ester must be hydrolized. The enzyme 3 - oxoacylCoA thiolase is responsible for this.
--> The nett result from 1 cycle B-oxidation in the mitochondrium is thus a fatty acid that is 2 carbons shorter and 1 acetyl-CoA that is used in the Krebs-cycle. Products formed in the B-oxidation are FADH2 and NADH + H(+) per cycle. FADH2 will eventually give 2 ATP and the NADH gives 2,5 ATP per molecule.
(all from the gray matter above. Had to learn it for molecular biology.....)
2.5 ATP? Wasn't it 3?
I need to revise these things...
"As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife" - J. Craig Venter
nope because NADH will release 10 protons in the oxidative forsforylation ( 4 in NADH-Q-reductase, 2 in cytochrome - c-reductase and 4 in cytochrome-c-oxidase), Since production of 1 mnolecule needs 4 protons ( 3 for ATP-synthase and 1 for the ATP/ADP translocase). Therefore, from 1 NADH 10/4 = 2,5 ATP will be formed.
(2 NADH produces 5 ATP, 2 FADH2 produces 3 ATP. For FAD, the elctrons are directly given to ubichinone, there fore releasing a protoninflux in the intermembranous space of the mitochondrium of 6 protons. thus 1 FADH2 gives us 6/4 = 1,5 ATP)
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