Question

The phosphorylation of glucose in the liver can be catalyzed by hexokinase and glucokinase. The enzymes have the following kinetic parameters: glucokinase Km = 10 mM, Vmax = 100 mmoles/min; hexokinase Km = 0.1 mM, Vmax = 10 mmoles/min. Of the total glucose-6- phosphate is simultaneously produced by both enzymes, what percentage of the glucose-6- phosphate formed will be made by glucokinase when the initial concentration of glucose is 10.0 mM.

Answer 1

According to the Michaelis-Menten mechanism of enzyme-catalyzed reactions, we can write as shown below.

d[P]/dt = k_b[E]₀[S]₀/(Km + [S]₀)

Here, k_b[E]₀ = Vmax and [S]₀ = 10.0 mM

For glucokinase, Km = 10 mM, Vmax = 100 mmol/min

Therefore, d[P]/dt = 100*10/(10 + 10), i.e. 50 mmol/min

For hexokinase, Km = 0.1 mM, Vmax = 10 mmol/min

Therefore, d[P]/dt = 10*10/(0.1 + 10), i.e. 9.901 mmol/min

Hence, the percentage of the glucose-6- phosphate formed that is made by glucokinase can be calculated as shown below.

50*100/(50+9.901) ~ 83.5%