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

The reaction catalyzed by hexokinase in E. coli is not at equilibrium and it favors the reactant side.

Step-by-step explanation:

The reaction catalyzed by hexokinase in E. coli is Glucose + ATP → Glucose 6-phosphate + ADP. To determine if the reaction is at equilibrium, we need to compare the concentrations of the reactants and products. In living E. coli cells, the concentrations are as follows: [ATP] = 7.9 mM, [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Since the concentration of ATP is higher than the concentration of ADP, it indicates that the reaction is not at equilibrium and it favors the reactant side.

Answer 2

Step-by-step 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=((1mm)* (1.04 mm))/((7.9mm)* (2mm)) \\\\=6.582* 10^(-2)`

so,

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