BMB 462 Lecture 9 Outline of Last Lecture I. Neuronal SignalingII. Regulation of Transcription by Steroid Hormone ReceptorsIII. Energy Catabolism in the 4 basic biochemical building blocksIV. Energy storage in lipidsV. Lipid digestion, absorption, and transport in mammalsVI. Mobilization of Triacylglycerol from adipose tissueVII. Glycerol MetabolismVIII. Fatty Acid CatabolismOutline of Current Lecture I. Continuation of Fatty Acid Catabolisma. Beta-oxidation of fully saturated fatty acidsb. ATP yieldc. Beta-oxidation of mono- and polyunsaturated fatty acidsd. Odd chain fatty acidse. Peroxisomal beta-oxidationf. Other types and uses of beta-oxidationII. Ketone Bodiesa. Synthesisb. CatabolismCurrent LectureConcepts to remembers from previous courses/lectures:- NAD+/NADH+H and FAD/FADH2 structures- Lipoprotein Lipases: lipases hydrolyze lipids and remove the fatty acids. So Lipoprotein lipases specifically break down triacylglycerol in lipoproteins (i.e. the chylomicron)- Shuttle systems (i.e. the malate-aspartate shuttle or the glycerol-I3-dehydrogenase shuttle system)I. Continuation of Fatty Acid Catabolisma. Beta-oxidation of fully saturated fatty acidsi. Oxidizing the beta-carbon of a fatty acid is a series of repeated oxidationsThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.ii. Beta-oxidation removes 2 carbons as acetyl CoA1. Acetyl-CoA then moves into the Citric Acid Cycle2. The Citric Acid Cycle reduces the coenzymes NAD+ and FAD+ to get NADH+H and FADH2. These move into the Electron Chain Transport (ETC) and oxidative phosphorylation3. Both NADH and FADH2 come from Beta-oxidation and the Citric acid cyclea. e- go through the ETC to oxygen (as an electron acceptor) where they form H2O.b. At the same time, a proton gradient is generated which turns ATPase in order to make ATP.iii. The 4 steps:1. Dehydrogenation (oxidation) – Done by acyl-CoA dehydrogenase. Remove H (and therefor e-) and oxidize the Beta Carbon, creating a double bond between the alpha and beta carbonsa. When you oxidize something, you have to have a complementary reduction reaction. Here, 2 H and 2 e- are removed from the fatty acid and transported to FAD, creating FADH2b. The CoA acts as an e- sink, making the fatty acid easy to oxidize. This is why the fatty acid must be activated before beta-oxidation occurs2. Hydration – enoyl-CoA hydratase (the ‘ene’ in enoyl indicates a double bond in the substrate) adds H2O and “moves” all of the oxidation from the first step to the beta carbona. This creates a hydroxyl group on the beta carbon3. Dehydrogenation (oxidation) – Now that all the oxidation is on thebeta Carbon, the cell can do a second oxidation. Beta-hydroxyacyl CoA dehydrogenase removes 2 e- & 2 H+ and moves them to NAD+ to get NADH+H4. Thiolysis – thiolase (aka the acyl-CoA acetyltransferase) cause a group transfer reaction. It moves the Beta-Carbon double bonded to oxygen to CoA, and removes 2 carbons in the form of acetyl CoA. It also adds a CoA to the new end of the fatty acid. 5. Beta-oxidation generates 1 FADH2, 1 NADH, 1 acetyl CoA, and 1 acyl-CoA that is now 2 carbons shorter than the initial fatty acid during the rounda. FADH2 and NADH move to the ETC to be oxidized so they can return to accept more e- and to power the ATP synthaseb. Acetyl CoA goes to the Citric Acid Cycle to be oxidized furtherc. The Acyl CoA enters the next round of Beta-oxidationiv. Compared with TCA Cycle (The Citric Acid Cycle)1. TCA starts with succinate – it has 2 fully reduced Carbons in the middlea. The carbons are oxidized with succinate dehydrogenase, which creates FADH2 and fumarate (a molecule with a double bond)2. Fumarase adds water to move the oxidation to the beta-carbon, then there’s a 2nd oxidation which moves e- to NAD, reducing it to NADH+H3. Moral: When nature finds a functional process, it will reuse it.b. ATP yieldi. At the end of oxidation, after repeating the process, the cell is left with 4 carbons. This is broken into 2 acetyl-CoA molecules and the catabolism via beta-oxidation is completeii. Acetyl-CoA yielded = N/2, where N is the number of Carbons in the starting fatty acid1. Acetyl-CoA is equivalent to 1 FADH2, 3 NADH, and 1 GTP (an ATP equivalent)iii. The amount of FADH2 and NADH generated = (N/2) – 1iv. FADH2 = 1.5 moles of ATP; NADH = 2.5 moles ATPc. Beta-oxidation of mono- and polyunsaturated fatty acidsi. Monounsaturated fatty acids1. The double bond is typically at the 9th Carbon, so the fatty acid cannot be dehydrogenated to put a double bond between Carbon 2 and Carbon 3 since there is already a double bond between Carbon 3 and Carbon 42. It is impossible to have back to back double bonds3. The cell needs to use a new enzyme to continue beta-oxidation: enoyl-CoA isomerasea. Enoyl-CoA isomerase flips the double bond from position 3to position 2, and makes it a trans bond if it was cisi. Called cis-trans isomerizationb. You now have the substrate needed for enoyl hydratase and beta oxidation can continue as normal4. Because the dehydrogenase couldn’t function, 1 less FADH2 is produced in the round of beta-oxidation that encountered the double bondii. Polyunsaturated fatty acids1. A double bond on an odd carbon is addressed in the same way as a monounsaturated fatty acid2. When beta-oxidation reaches the round with a double bond on aneven carbon (carbon 4 at this point), the round starts normally with acyl-CoA dehydrogenase adding a double bond to Carbon 2.a. But now there is a 2nd double bond (the fatty acid is now a 2, 4 diene) so the hydratase can’t add waterb. Instead, 2, 4-dienoyl-CoA reductase reduces the carbons in the fatty acidi. NADPH is used to provide e- which removes 1 of the double bonds (the 2nd double bond is left at position 3)c. The enoyl-CoA isomerase is again used to flip the double bond to position 2, and beta-oxidation can now proceed normally with the hydratase3. In eliminating the even double bond, the cell uses NADPH and generates NADH and FADH2a. NADPH and NADH are energetically equivalent, so the 2 cancel outiii. Short cut to calculating ATP generation for unsaturated fatty acids:1. Any time there’s a double bond at an odd position, you generate 1.5 ATP less (because one less FADH2 is produced)2. Any time there’s a double bond at an even carbon, you generate 2.5 ATP less (because one less net NADH is produced)d. Odd chain
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