Tag Archives: anaerobic condition

Energy yield of glycogen

Glycogen, together with lipids, is a reserve of energy for use when needed. Although quantitatively lower than lipids, glycogen has some metabolic advantages.

  • It is mobilized faster than lipids.
  • It is a storage of glucose used to maintain the blood glucose level during fasting, physical activity, and between meals.
  • It is an energy source used both in aerobic and anaerobic conditions.

By considering the last point, the energy yield of glucose released from glycogen is different under aerobic and anaerobic conditions. The metabolic bases of these differences are analyzed below.

CONTENTS

Energy yield of glycogen stores under anaerobic conditions

Under anaerobic conditions, the oxidation of glucose to lactate via anaerobic glycolysis yields two molecules of ATP.
Below, the yield of ATP from anaerobic oxidation of glucose released from glycogen by the action of glycogen phosphorylase (EC 2.4.1.1), and debranching enzyme (EC 3.2.1.33) is considered.
Note: glycogen phosphorylase releases about 90% of stored glucose in the form of glucose-1-phoshate (G-1-P), and debranching enzyme the remaining 10% in the form of glucose.


Glycogen phosphorylase and the oxidation of G-1-P under anaerobic conditions

Glycogen synthesis from glucose requires 2 ATP for each molecule of glucose.
The release of glucose-1-phosphate by the action of glycogen phosphorylase allows the recovery of one of the 2 molecules of ATP used in the preparatory phase of glycolysis. The anaerobic oxidation of glucose-6-phosphate, produced from glucose-1-phosphate (G-1-P) by the action of phosphoglucomutase (EC 5.4.2.2), yields therefore three molecules of ATP and not two, because:

  • one molecule of ATP, instead of two, is used in the preparatory phase of glycolysis, because hexokinase reaction (EC 2.7.1.1) is bypassed;
  • four molecules of ATP are produced in the payoff phase of glycolysis.

The cost-gain rate is 1/3, namely, there is an energy yield of about 66,7%.
The overall reaction is:

Glycogen(n glucose residues) + 3 ADP + 3 Pi → Glycogen(n-1 glucose residues) + 2 Lactate + 3 ATP

By considering the two molecules of ATP used in the synthesis of glycogen and the anaerobic oxidation of glucose-1-phosphate to lactate, there is a yield of one molecule of ATP for each molecule of glucose stored. The overall reaction is:

Glucose + ADP + Pi → 2 Lactate + ATP

Debranching enzyme and the oxidation of glucose under anaerobic conditions

By considering the glucose released by the action of debranching enzyme, the yield of ATP is zero because:

If we now consider the oxidation to lactate of all glucose released from glycogen, there is an energy yield equal to:

1-{[(1/3)*0,9]+[(2/2)*0,1]}=0,60

Then, under anaerobic conditions, there is an energy yield of 60%, hence, glycogen is a good storage form of energy.

Energy yield of glycogen stores under aerobic conditions

Under aerobic conditions, the oxidation of glucose to CO2 and H2O via glycolysis, pyruvate dehydrogenase complex, Krebs cycle, mitochondrial electron transport chain, and oxidative phosphorylation yields about 30 molecules of ATP.
Below, the yield of ATP from aerobic oxidation of glucose released from glycogen by the action of glycogen phosphorylase and debranching enzyme is considered.

Glycogen phosphorylase and the oxidation of G-1-P under aerobic condition

The oxidation of glucose-6-phosphate, produced from glucose-1-phosphate by the action of phosphoglucomutase, to CO2 and H2O yields 31 molecules of ATP, and not 30, because only one molecule of ATP is used in the preparatory phase of glycolysis. The cost-gain rate is 1/31, namely, there is an energy yield 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

By considering the two molecules of ATP used in the synthesis of glycogen and the aerobic oxidation of glucose-1-phosphate to CO2 and H2O, there is a yield of 29 molecules of ATP for each molecule of glucose stored.
The overall reaction is:

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

Debranching enzyme and the oxidation of glucose under aerobic condition

By considering the glucose released by the action of debranching enzyme, there is a yield of 30 molecules of ATP, because two molecules of ATP are used in the preparatory phase of glycolysis. The cost-gain rate is 2/30, namely, there is an energy yield of about 93,3%.

If we now consider the oxidation to CO2 and H2O of all glucose released from glycogen, there is an energy yield equal to:

1-{[(1/31)*0,9]+[(2/30)*0,1]}=0,96

Energy yield of glycogen aerobic anaerobic conditionsThen, under aerobic conditions, there is an energy yield of 96%, hence, glycogen is a extremely efficient storage form of energy, with a gain of 36% compared to anaerobic conditions.

References

Nelson D.L., Cox M.M. Lehninger. Principles of biochemistry. 6th Edition. W.H. Freeman and Company, 2012

Stipanuk M.H., Caudill M.A. Biochemical, physiological, and molecular aspects of human nutrition. 3rd Edition. Elsevier health sciences, 2013 [Google eBooks]

Voet D. and Voet J.D. Biochemistry. 4th Edition. John Wiley J. & Sons, Inc. 2011