Question

Hexokinase catalyzes the phosphorylation of glucose and fructose by ATP. K_M for glucose is 0.15 mmol/L, whereas that for fructose is 1.5 mmol/L. Assume Vmax is the same for both glucose and fructose and the enzyme displays hyperbolic kinetics. For which substrates does hexokinase have the great affinity? Provide an explanation. Also, the comparison of the two sugars for hexokinase indicates which sugar is preferred as a nutrient?

Answer 1

An empirical method to determine the affinity of the enzyme to the substrate is 1/Km

for glucose is 6.667

for fructose is 0.667

The enzyme has higher affinity, also take a look at the graph, the red line (glucose) has a steep slope and reaches the Vmax at a very low substrate concentration, what does not happen with the fructose (green line) it needs a very high concentration of fructose to reach the Vmax.

This is the main reason why glucose is the preferred nutrient in the phosphorylation.