Glycogen: an efficient storage form of energy in aerobic conditions

What is the net energy yield for the oxidation of a glucose unit from glycogen in aerobic conditions?

Aerobic Conditions: Glycogen Structure
Fig. 1 – Glycogen Structure

In aerobic conditions, the oxidation of a free glucose to CO2 and H2O (glycolysis, Krebs cycle and oxidative phosphorylation) leads to the net production of about 30 molecules of ATP.

Glucose from the action of glycogen phosphorylase: glucose-1-phosphate release (about 90% of the removed units).

Glycogen synthesis from free glucose costs two ATP units for each molecule; a glucose-1-phosphate is released by the action of glycogen phosphorylase with recovering/saving one of the two previous ATP molecules.
Therefore in aerobic condition, the oxidation of glucose starting from glucose-6-phosphate and not from free glucose yields 31 ATP molecules and not 30 (one ATP instead of two is expended in the activation phase, 30 ATP are produced during Krebs cycle and oxidative phosphorylation: 31 ATP gained).
The net rate between cost and yield is 1/31 (an energy conservation of about 97%).
The overall reaction is:

glycogen(n glucose residues) + 31 ADP + 31 Pi → glycogen(n-1 glucose residues) + 31 ATP + 6 CO2 + 6 H2O

If we combine glycogen synthesis, glycogen breakdown and finally the oxidation of glucose to CO2 and H2O we obtain 30 molecules of ATP per stored glucose unit, that is the overall reaction is:

glucose + 29 ADP + 30 Pi → 29 ATP + 6 CO2 + 6 H2O

Glucose from the action of debranching enzyme: free glucose release (about 10% of the removed units).

The net yield in ATP between glycogen synthesis and breakdown is two ATP molecules expended because of free glucose is released.
In this case the oxidation of glucose starts from the not-prephosphorylated molecule so we obtain 30 ATP molecules.
The net rate between cost and yield is 2/30 (a energy conservation of about 93,3%).
Considering the oxidation of the glucose units from glycogen to CO2 and H2O we have an energy conservation of:

1-(((1/31)*0,9)+((2/30)*0,1))=0,9643

Conclusion

In aerobic conditions, there is the conservation of about 97% of energy into the glycogen molecule, an extremely efficient storage form of energy.

References

Arienti G. “Le basi molecolari della nutrizione”. Seconda edizione. Piccin, 2003

Cozzani I. and Dainese E. “Biochimica degli alimenti e della nutrizione”. Piccin Editore, 2006

Giampietro M. “L’alimentazione per l’esercizio fisico e lo sport”. Il Pensiero Scientifico Editore, 2005

Mahan LK, Escott-Stump S.: “Krause’s foods, nutrition, and diet therapy” 10th ed. 2000

Mariani Costantini A., Cannella C., Tomassi G. “Fondamenti di nutrizione umana”. 1th ed. Il Pensiero Scientifico Editore, 1999

Nelson D.L., M. M. Cox M.M. Lehninger. Principles of biochemistry. 4th Edition. W.H. Freeman and Company, 2004

Stipanuk M.H.. “Biochemical and physiological aspects of human nutrition” W.B. Saunders Company-An imprint of Elsevier Science, 2000

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