Fatty Acid Catabolism Flashcards

Question

Step 3 of FA Transport (carnitine shuttle)

Answer 1

Deacylation of Fatty acyl-carnitine catalyzed by carnitine palmitoyltransferase II (CPT 2) (reverse of step 1)

Answer 2

CPT1 is strongly inhibited by malonyl-CoA (regulation): 1) Malonyl-CoA is the first committed intermediate of fatty acid synthesis. 2) Conditions favoring fatty acid synthesis prevent fatty acids from entering the mitochondria where β-oxidation will occurs (catabolism) Summary: If trying to synthesize FA, we don't want to break them down so malonyl-CoA inhibits CPT1 so synthesis occurs and not catalysis

Answer 3

1. Occurs in the mitochondria matrix 2. Modification occurs at the β-carbon 3. Chain is shortened by 2 carbons at a time 4. Cyclic pathway of 4 reactions

Answer 4

1. Oxidation of the β-carbon in a series of steps in which two carbons shorten the fatty acyl chain at a time 2. This four step process will release a two carbons fragment, acetyl-CoA, at each cycle. Example: Palmitic acid (16:0)

Answer 5

Dehydrogenation of acyl-CoA in trans-Δ2-Enoyl-CoA catalyzed by Acyl-CoA Dehydrogenases (ACAD)

Answer 6

1. These enzymes are bound to FAD 2. Upon dehydrogenation, FADH2 will be generated. 3. 2 Electrons will be transferred to the electron-transferring flavoprotein (ETF) which is also bound to an FAD molecule. 4. These electrons will be passed to CoQ via the ETF-Q oxidoreductase.

Answer 7

1)complex I 2) complex II 3) Glycerol 3-phosphate 4) FAD

Answer 8

Hydration of trans- Δ2-Enoyl-CoA to form L-3-hydroxyacyl-CoA catalyzed by enoyl-CoA hydratase

Answer 9

Dehydrogenation of L-3-hydroxyacyl-CoA to form 3-ketoacyl-CoA catalyzed by 3-L-hydroxyacyl-CoA dehydrogenase -Upon dehydrogenation, NADH will be generated.

Answer 10

Dehydrogenation of L-3-hydroxyacyl-CoA to form 3-ketoacyl-CoA catalyzed by 3-L-hydroxyacyl-CoA dehydrogenase -Upon dehydrogenation, NADH will be generated.

Answer 11

Thiolytic cleavage of 3-ketoacyl-CoA to form a Cn-2 acyl-CoA and acetyl-CoA catalyzed by thiolase -A new CoA is added to the end of Cn-2 acyl chain.

Answer 12

1 cycle of β-oxidation produces: 1 FADH2 1 NADH + H+ 1 acetyl-CoA 1 acetyl-CoA (TCA cycle) produces: 3 NADH 1 FADH2 1 GTP (ATP) ** 1 FADH2 = 1.5 ATP ** 1 NADH + H+ = 2.5 ATP

Answer 13

1) Need 7 rounds of B-oxidation (because the last cycle produces 2 acetyl-coa) -8 acetyl-CoA will be produced (C16/2) 2) 7FADH2, 7 NADH, and 8 acetyl-CoA are produced (1.5ATP x 7 + 2.5ATP x 7 + ((24x2.5) + (8 x 1.5) + (1 x 8 atp)) = 108 ATP (B-oxidation)- 2ATP investment (FA activation) = 106 ATP

Answer 14

The last oxidation cycle produces a 3 carbons molecule called propionyl-CoA which is converted to Succinyl-CoA (TCA cycle intermediate). Therefore, propionyl-CoA (succinyl-CoA) will generate: 1) 1 GTP (Need 1 ATP to convert propionyl coa to succinyl coa so production of 1 gtp gets cancelled out) 2) 1 FADH2 3) 1 NADH 4) an extra TCA cycle intermediate.

Answer 15

Things to remember: If the double bond is located at odd carbon number eg. D9, D11 or D13: 1) Use of Enoyl-CoA isomerase 2) Skip reaction 1 of b-oxidation cycle so minus 1 FADH2 in in net yield

Answer 16

Things to remember: IF the double bond is located at odd carbon number eg. D9, D11 or D13: 1. Use of Enoyl-CoA isomerase 2. Skip reaction 1 of b-oxidation cycle so minus 1 FADH2 in in net yield IF the double bond is located at even carbon number eg. D8, D12 or D14: 1. Use of 2,4 Dienoyl-CoA reductase and Enoyl-CoA isomerase 2. Oxidation of one molecule of NADPH (equivalent to 2.5 ATP) go through all four steps of b-oxidation cycle (plus the two extra steps to resolve the double bond in C4)

Answer 17

eg. Δ9, Δ11 or Δ13 1) Use of Enoyl-CoA isomerase 2) Skip reaction 1 of β-oxidation cycle so minus 1 FADH2 in in net yield