The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because?.

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08-08-2022
Chemistry

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The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because?.

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Tutorconsortium365 Tutorconsortium365 · Answer 1 · 2022-08-12T08:31:22+02:00

The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because. it is the rate-limiting step

What is phosphorylation?

The first step in the metabolism of carbohydrates is frequently their phosphorylation. Because the phosphate group stops the molecules from migrating back across the transporter, phosphorylation enables cells to store carbohydrates. Glucose phosphorylation is a crucial step in the metabolism of sugar. In the first phase of glycolysis, D-glucose is converted to D-glucose-6-phosphate using the chemical equation D-glucose + ATP D-glucose-6-phosphate + ADP G° = 16.7 kJ/mol (° signifies measurement under standard conditions).

The rate-limiting stage in the liver's metabolism of glucose is the initial rate of phosphorylation of glucose (ATP-D-glucose 6-phosphotransferase) and non-specific hexokinase. Hepatic cells are freely permeable to glucose (ATP-D-hexose 6-phosphotransferase).

encouraging certain glucose transporters to translocate to the cell membrane.

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