Question

6. Draw the structure and calculate the yield, in ATP equivalents, upon complete aerobic oxidation of melissic acid (30:0). 7. Draw the structure and calculate the yield, in ATP equivalents, upon complete aerobic oxidation of lignoceric acid (24:0). 8. Draw the structure and calculate the yield, in ATP equivalents, upon complete aerobic oxidation of montanic acid (28:0). 9. Draw the structure and calculate the yield, in ATP equivalents, upon complete aerobic oxidation of lacceroic acid (32:0).

Answer 1

- 6. Melissic Acid (30:0) (Even Chain saturated) - It is a saturated fatty acid and follows ß- oxidation that occurs in the mitochondria. . - Involves sequential removal of two carbons from the fatty acid chain until the last too. carbons are left B-oxidation (it occurs under aerobic conditions) involves four steps: Oxidation Hydrolysis Oxidation Theolysis Since the oxidation occurs at the ß- carbon followed by double bond formation between & & B carbons, the oxidation is called ß-oxidation The double bond is removed by further cocidation & remoral of the 2 carbons form the chain including a carbon. - last two carbons The of process repeate the chain are until the obtained. - The two carbon chain breaking down after each cycle is in form of Acetyl Est. for oxidation, the fatty aced is first converted to its acul form (activation of the hilo the fatty fatty acid) & then transported across acid) membeane- s

ATP AMP APP i Fatty acid Patty acyl-CoA Structure of Melissic Acid (30:0) - 30 stands for number of carbon atoms in the chain o sländs for number of double bends (no double bonds) 29 27 25 23 21 19 17 15 15 11 9 7 5 3 e-S-GA Meliscyl CoA B 0 e-on Melissic acid Carbon next to the e carbon is a. ß- oxidation of Melissic Acidi) (begins with Melissye CoA) for simlicity the carbons Chain has been indicated as R after ß carlson Oxidation R-CH₂-CH₂-C-S-CA (300 B2 T Melissyl CoA -PAD (Acyl Cost dehydrogenase R _ c = c - C-slot Trans A²-enoje COMO | Enoyl-CoA hydratase seydrolysis

04 1 oxidation OS) -s-coA 3 Hydroxyache CoA -NADT L-B-Hydroxylacyl CA dehydrogenase NADH the Thislysis R - C Ca - C-s-CoA B-koto acyl Cot COA-SHA 1 Thiolase R-c-S-LOA & Cuz-c-S-CoA Acyl COA Acetyl CoA (280) 13 cycles of B-oxidation (20) Chz-C-s-cot & crz - C-s-con Acetyl Cost Acetyl Con (Total 14 cycles) Total number of Acetyl con released - 15 Total number of FADH2 released - 14 Total number of NADH released - 14 Oxidation of l Acetyl Cot gives 1o ATP IFADH giver 1.5 ATP INADH gives 2.5 ATP i. Total ATP yield = 15x10 & 14x 1.5 + 14x2.5 - 150 +21+ 35 = 206 ATPS)

Similar to the will occure of the sixth question , The ß osudation. Lignoceric acid (24:0). structure : Lignocaric acid Lignoceryl CoA (246) K u indergoes Il eylles op buidation 2 Acetyl COA Total Actyl CoA FADHL _Nhok - 12 - 11 - || Total ATP yield - = 124 10 + 11X15 + 11x 25 120 t 16.5 +27.5 164 ATPs

8. Montanic Acid (28:0) ДАЛИЛЛИЛЛои 28 Montanic acid structure Montanyl CoA (280) Undergoes 3 cycles of B-oxidation 2 Acetyl con Released Total ATP yield Acelyl CoA - 14 14 x 10 = 140 FADH2 BX 15 = 19.5 NADH 13 X 2.5 = 32.5 = 1192 ATPS 13 - 9. Lacceroic Acid (32:0) Lacceroic acid structure . Laccersyt coA (320) by Undergoes 15 cycles of ß- oxidation JAcety CoA Released Total ATP yield Acetylcoh - 16 16810 160 FAOL - 15 15x15 - 22.5 Мои - 15 15*2.5 37.5 1220 APPS