Question

In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102. In the living E. coli cells, [ATP] = 7.9 mM; [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

Answer 1

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

$q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}$

$q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}$

so,

$q$ ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable  until q = Keq