Question

An important step in the glycolytic path is the phosphorylation of glucose by ATP, catalyzed by the enzyme hexokinase and Mg 2+

: glucose + ATP hexokinase ⟶
Mg 2+
​
glucose-6-P ​
+ ADP In the absence of ATP, glucose-6-P is unstable at pH 7, and in presence of the enzyme glucose6-phosphatase, it hydrolyzes to give glucose: glucose-6-P +H2O G-6-phosphatase ⟶
​
glucose + phosphate (a) Using these data, calculate ΔrG ∘′
at pH7 for the hydrolysis of glucose-6-P at 298 K. Reaction D-Glucose + ATP → D-glucose-6-phosphate + ADP ATP + H2O → ADP + phosphate ΔrG ∘(−16.7−31.0
​
(b) If the phosphorylation of glucose is allowed to proceed to equilibrium in the presence of equal concentrations of ADP and ATP, what is the ratio (glucose-6-P)/(glucose) at equilibrium? Assume a large excess of ATP and ADP: [ATP]=[ADP]≫[( glucose )+( glucose-6-P )] (c) In the absence of ATP (and ADP), calculate the ratio [glucose-6-P]/[glucose] at pH 7 if [phosphate] =10 ^−2M.

Answer 1

Explanation:(a) To calculate ΔrG∘' at pH 7 for the hydrolysis of glucose-6-P, we need to use the Gibbs free energy values for the relevant reactions and apply the equation:

ΔrG∘' = ΣΔrG∘' (products) - ΣΔrG∘' (reactants)

Given the following values:

ΔrG∘' for the reaction D-Glucose + ATP → D-glucose-6-phosphate + ADP = -16.7 kJ/mol

ΔrG∘' for the reaction ATP + H2O → ADP + phosphate = -31.0 kJ/mol

ΔrG∘' for the hydrolysis of glucose-6-P can be calculated as follows:

ΔrG∘' = ΔrG∘' (glucose-6-P) - ΔrG∘' (glucose) - ΔrG∘' (phosphate)

= (-16.7 kJ/mol) - 0 - (-31.0 kJ/mol)

= 14.3 kJ/mol

Therefore, ΔrG∘' at pH 7 for the hydrolysis of glucose-6-P at 298 K is 14.3 kJ/mol.

(b) If the phosphorylation of glucose is allowed to proceed to equilibrium in the presence of equal concentrations of ADP and ATP, the ratio (glucose-6-P)/(glucose) at equilibrium can be determined from the ΔrG∘' values of the reactions involved. Since the ΔrG∘' values are given, we can use the following equation:

ΔrG∘' = -RT ln(K)

Where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298 K), and K is the equilibrium constant.

Since ΔrG∘' for the reaction D-Glucose + ATP → D-glucose-6-phosphate + ADP is -16.7 kJ/mol, we can calculate K as follows:

-16.7 kJ/mol = -RT ln(K)

K = exp(-ΔrG∘' / (RT))

Using the values, we find:

K = exp(-(-16.7 kJ/mol) / (8.314 J/(mol·K) * 298 K))

Similarly, we can calculate K for the reaction ATP + H2O → ADP + phosphate using its ΔrG∘' value (-31.0 kJ/mol).

Once we have the equilibrium constants for both reactions, we can determine the ratio (glucose-6-P)/(glucose) at equilibrium.

(c) In the absence of ATP (and ADP), the hydrolysis of glucose-6-P to glucose and phosphate is driven by the presence of phosphate. The ratio [glucose-6-P]/[glucose] can be calculated using the equation:

K = [glucose]/[glucose-6-P] * [phosphate]

Given that [phosphate] = 10^(-2) M, we can rearrange the equation to solve for [glucose-6-P]/[glucose]:

[glucose-6-P]/[glucose] = K/[phosphate]

To calculate the ratio, we need to know the equilibrium constant K for the hydrolysis of glucose-6-P.