NP11 – Efficiency of fat synthesis
Assignment calculations - Synthesizing glycerol and 3 fatty acids (C18:0)
from solely glucose
Source of acetyl-CoA for fatty acid synthesis
Acetyl-CoA from the mitochondrion is shuttled outside to the cytoplasm with citrate
Cost: 2 ATP + NADHc (for the transport of acetyl-CoA) and you obtain 1 NADPH = - 3.5 ATP for every shuttle
Fatty acid synthesis (in cytoplasm)
The base acetyl-ACP: having it doesn’t cost energy
For each elongation of 2 C: You use 1 ATP (for every 2nd, 3rd, 4th... acetyl-CoA) + 2 NADPH (from the 2nd fatty
acid) (one is produced in the A. step, but the 2nd one come from the PPP)
Esterification
You need to have an activated fatty acid à add a CoA group to each fatty acid (costs energy)
à for each fatty acid you need to invest 2 ATP equivalent
Then you can add the activated fatty acyl-CoA to the glycerol-phosphate (also activated by 1 NADH)
Points to take into account
• NADPH is not equal to NADH!!
• Focus only on C-atoms, not on energy needed (ATP, NADPH)
• NADPH is neglected and if it is taken into account, the end products of PPP are not
• For C18 FA, number of acetyl-CoA needed (9, so 8 times an extension = spiral)
C6 à 2 acetyl-CoA (2 x C2) so you lose CO2
Conclusion of the assignment
Answer = 16 mole of glucose
1. Synthesis of fatty acid (C18:0)
Per acetyl-CoA: 3.5 ATP needed (1 NADHc + 2 ATP) and 1 NADPH given
For the base (so first part of the fatty acid): need 2 NADPH
For every elongation (from 2nd acetyl-CoA)à costs 1 extra ATP/ elongation
Usage for FA elongation: 1 ATP + 2 NADPH à done 8 times
Per elongation: you need 1 acetyl-CoA, 4.5 ATP and 1 NADPH (since you have already 1 NADPH from shuttling
Ac CoA to the cytoplasm)
In total, needed for 1 fatty acid stearate
- For the base: 1 acetyl-CoA + 2 ATP + 1 NADHc – 1 NADPH
Assignment calculations - Synthesizing glycerol and 3 fatty acids (C18:0)
from solely glucose
Source of acetyl-CoA for fatty acid synthesis
Acetyl-CoA from the mitochondrion is shuttled outside to the cytoplasm with citrate
Cost: 2 ATP + NADHc (for the transport of acetyl-CoA) and you obtain 1 NADPH = - 3.5 ATP for every shuttle
Fatty acid synthesis (in cytoplasm)
The base acetyl-ACP: having it doesn’t cost energy
For each elongation of 2 C: You use 1 ATP (for every 2nd, 3rd, 4th... acetyl-CoA) + 2 NADPH (from the 2nd fatty
acid) (one is produced in the A. step, but the 2nd one come from the PPP)
Esterification
You need to have an activated fatty acid à add a CoA group to each fatty acid (costs energy)
à for each fatty acid you need to invest 2 ATP equivalent
Then you can add the activated fatty acyl-CoA to the glycerol-phosphate (also activated by 1 NADH)
Points to take into account
• NADPH is not equal to NADH!!
• Focus only on C-atoms, not on energy needed (ATP, NADPH)
• NADPH is neglected and if it is taken into account, the end products of PPP are not
• For C18 FA, number of acetyl-CoA needed (9, so 8 times an extension = spiral)
C6 à 2 acetyl-CoA (2 x C2) so you lose CO2
Conclusion of the assignment
Answer = 16 mole of glucose
1. Synthesis of fatty acid (C18:0)
Per acetyl-CoA: 3.5 ATP needed (1 NADHc + 2 ATP) and 1 NADPH given
For the base (so first part of the fatty acid): need 2 NADPH
For every elongation (from 2nd acetyl-CoA)à costs 1 extra ATP/ elongation
Usage for FA elongation: 1 ATP + 2 NADPH à done 8 times
Per elongation: you need 1 acetyl-CoA, 4.5 ATP and 1 NADPH (since you have already 1 NADPH from shuttling
Ac CoA to the cytoplasm)
In total, needed for 1 fatty acid stearate
- For the base: 1 acetyl-CoA + 2 ATP + 1 NADHc – 1 NADPH
