Question

Show all work please, do not just give a number!

Determine the number of ATP derived from each component and name the additional enzymes necessary to recognize the double bonds and account for the cost in ATP or NADH.

a) palmitic acid (even number of carbons)

b) oleic acid (even number of carbons, one double bond)

c) linoleic acid (even number of carbons, two double bonds)

Answer 1

The oxidation of fatty acids takes place by the process known as β-oxidation. In β-oxidation, the long chain fatty acids are degraded in two carbon units (Acetyl-CoA) by their oxidation at the C3 or β-carbon.

a) Palmitic Acid: is a saturated fatty acid having 16 carbons (C16). During β-oxidation, this fatty acid gets degraded into 8 molecules of Acetyl-CoA in 7 cycles.

During 1 cycle, we get 1 NADH+H, 1 FADH2 molcules + 1 Acetyl-CoA molecule

therefore during 7 cycles we get 7 NADH+H, 7 FADH2 molcules + 8 Acetyl-CoA molecules (in the last cycle we get 2 acetyl-CoA molecules.

1 molecule of acetyl-CoA in mitochondria yields 3 NADH+H, 1 FADH2 and 1 GTP (equivalent to ATP).

therefore 8 acetyl-CoA molecules yield= 24 NADH+H, 8 FADH2 and 8 ATP

Total number of NADH+H molecules = 31 NADH+H + 15 FADH2 + 8ATP

= 31 x 3= 93 ATP + 15 x 2 = 30 ATP + 8 ATP

= 131 ATP molecules.

No. of ATP molecules consumed during Activation of fatty acid= 2 ATP

therfore Net ATP produced= 131-2 = 129 ATP molecules.

b) Oliec acid: is a monounstaurated C18 fatty acid.

It means that it will produce 9 molecules ofv Acetyl CoA during β-oxidation.

but there is one double bond and to overcome this problem, one additional enzyme enoyl-CoA isomerase is required at one point due to which 1 FADH2 molecule will not be produced.

therefore we can get 8 NADH+H + 7 FADH2 + 9 Acetyl-CoA

1 molecule of acetyl-CoA in mitochondria yields 3 NADH+H, 1 FADH2 and 1 GTP (equivalent to ATP).

therefore 9 acetyl-CoA molecules yield= 27 NADH+H, 9 FADH2 and 9 ATP

Total number of NADH+H molecules = 35 NADH+H + 16 FADH2 + 9 ATP

= 35 x 3=105 ATP + 16 x 2 = 32 ATP + 9 ATP

= 146 ATP molecules.

No. of ATP molecules consumed during Activation of fatty acid= 2 ATP

therfore Net ATP produced= 146 -2 = 144 ATP molecules.

C Linoleic acidis a unstaurated C18 fatty acid with 2 double bonds.

It means that it will produce 9 molecules ofv Acetyl CoA during β-oxidation.

but there are two double bonds and overcome this problem, two additional enzyme enoyl-CoA isomerase and 2,4-dienoyl CoA reductase is required which results in 2 FADH2 being less produced

therefore we can get 8 NADH+H + 6 FADH2 + 9 Acetyl-CoA

1 molecule of acetyl-CoA in mitochondria yields 3 NADH+H, 1 FADH2 and 1 GTP (equivalent to ATP).

therefore 9 acetyl-CoA molecules yield= 27 NADH+H, 9 FADH2 and 9 ATP

Total number of NADH+H molecules = 35 NADH+H + 15 FADH2 + 9 ATP

= 35 x 3=105 ATP + 16 x 2 = 30  ATP + 9 ATP

= 144  ATP molecules.

No. of ATP molecules consumed during Activation of fatty acid= 2 ATP

therfore Net ATP produced= 144- 2 = 142 ATP molecules

Here i have used the formulae that 1 NADH+H yields 3 ATP molecules and 1 FADH2 yields 2 ATP molecules.

However if u use 2.5 ATP for NADH+H and 1.5 ATP for FADH2, the number of ATP will change accordingly. However the method of calculation will be same.

Hit like if u understand. Thank u