Question

For the structure of Myristic acid:

1. How mạny rounds of β-oxidạtion will tạke plạce upon complete oxidạtion of this fạtty ạcid?

2. How mạny FẠDH2 ạnd NẠDH will be formed from the β -oxidạtion (only)?

3. How mạny ạcetyl-CoẠ molecules will be formed from the β -oxidạtion?

4. How mạny totạl ẠTP will be mạde from complete oxidạtion of myristic ạcid?

Answer 1

Answer-

a) Myristic acid is 14carbon containing saturated fatty acid and one round of - oxidation releases one acetyl-CoA and for complete oxidation 6 rounds of -oxidation will takes place because in 5 round of -oxidation 10 carbons are released as 5 acetyl-CoA and in 6th round of -oxidation 2 acetyl-CoA are released so that 6 rounds of -oxidation will takes place upon complete oxidation of myristic acid.

b) In one round of -oxidation 1NADH+H+ and 1FADH2 are synthesized and in 6 rounds of -oxidation 6NADH+H+ and 6FADH2 will be formed .

c) Totally 7 acetyl-CoA molecules will be formed from the -oxidation ans first to fifth round releases 1 acetyl-CoA and last sixth round releases 2 acetyl-CoA molecules.

d) In 6 rounds of beta oxidation 6NADH+H+ and 6FADH2 are produced. From 6NADH+H+ (6NADH+H+×2.5=15ATP) 15ATP and from 6FADH2 (6FADH2×1.5=9ATP) 9ATP are synthesized. In beta oxidation 15+9=24ATP are produced. The 7 molecules of acetyl-CoA then utilized via Krebs cycle and when 1 acetyl-CoA enters Krebs cycle, 3NADH+H+ , 1FADH2 and 1GTP are produced. When 7 acetyl-CoA enters Krebs cycle, 21NADH+H+ , 7FADH2 and 7ATP are produced. Totally in Krebs cycle 7 acetyl-CoA produces- 21NADH+H+ (21NADH+H+×2.5=52.5ATP) produces 52.5ATP, 7FADH2 (7FADH2×1.5=10.5ATP) produces 10.5ATP and 7GTP (7GTP=7ATP) produces 7ATP. Total ATP produced in Krebs cycle are 52.5+10.5+7= 70ATP. Total ATP made from complete oxidation of myristic acid is= 24ATP from beta oxidation + 70 ATP from Krebs cycle = 94 ATP. Complete oxidation of myristic acid produces 94ATP.