A typical glycogen particle has 55,000 glucose residues and 2,000 non-reducing e

Question

A typical glycogen particle has 55,000 glucose residues and 2,000 non-reducing ends. The efficiency of glucose storage in the form of glycogen can be calculated from the number of ATP's required for the synthesis of the glycogen particle, and the number of ATP's that are gained from its complete consumption. Remember that different energy yields are obtained through glycolysis for residues removed by glycogen phosphorylase and glycogen debranching enzyme. For the purpose of this calculation, use a value of 2.5 ATP per each NADH formed in glycolysis. 25 ATP's per glucose are generated from reactions that occur downstream of glycolysis. Calculate the efficiency as net ATP's gained from glycogen particle/total ATP's gained from glycogen particle.

Explanation / Answer

Answer

Glucose 1-Phosphate Glucose 6-Phosphate

In the muscle, and in most of the other organs and tissues, glucose from glycogenolysis enters the glycolytic pathway as glucose 6-phosphate catalyzed by hexokinase. Therefore, glycogen phosphorylase, releasing an already “activated” glucose molecule, saves an ATP. An ATP molecule is required to synthesize another glycolytic intermediate, the fructose 1,6-bisphosphate.
In this way, some of the activation energy required for glycogen synthesis is conserved: the net yield of ATP per glucose molecule by glycolysis to lactate is 3 rather than 2, an advantage for the working muscle. The overall equation is:

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

In the liver, glucose 6-phosphate from glycogen is dephosphorylated by glucose 6-phosphatase and then released into the bloodstream. These are the steps in the removal of glucose units, as phosphorylated glucose, by hepatic glycogenolysis:

Glycogen(n glucose residues) + Pi Glucose 1-Phosphate + Glycogen(n-1 glucose residues)

glucose-1-phosphate glucose-6-phosphate

glucose-6-phosphate + H2O glucose + Pi

The overall equation is:

Glycogen(n glucose residues) + H2O Glycogen(n-1 glucose residues) +Gglucose

Since a typical glycogen has 55000 glucose residues so the efficiency of glucose storage in form of glycogen in form of ATP will be: 3*55000= 165000 .

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