Question

Calculate Δ G for ATP hydrolysis in liver at 18 °C. Use the liver concentrations from the first question.

Answer 1

This question is incomplete and the full question can be seen below:  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

The equation for the ATP hydrolysis is:  

           H₂0  

ATP ------------> ADP + P₁                      ΔG = -30.5 $\frac{KJ}{mol}$  

a) calculate ΔG for ATP hydrolysis to rank the following conditions from most favorable to least favorable. Assume a temperature of 37.0° C, R = 8.315 $\frac{J}{mol.k}$  

muscle: [ATP]= 8.1mM; [ADP]= 0.9mM; [P₁]= 8.1mM  

liver: [ATP]= 3.4mM; [ADP]= 1.3mM; [P₁]= 4.8mM  

brain: [ATP}= 2.6mM; [ADP}= 0.7mM; [P₁]= 2.7mM  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

Answer:  

−45.8 KJ/mol

Explanation:  

Equilibrium constant (k) is defined as a measure of the ratio of the equilibrium concentration of the products of a reaction, to the Equilibrium concentration of the reactants with each concentration raised to the power corresponding to the coefficient in the balanced equation of the reaction.

In the reaction in the question given above;  

$K=\frac{[ADP][P_1]}{ATP}$

For muscle;

ADP= 0.9 × 10⁻³

P₁= 8.1 × 10⁻³

ATP= 8.1 × 10⁻³

∴ K = $\frac{(0.9*10^-3)(8.1*10^-3)}{(8.1*10^-3)}$

  K = 0.9 × 10⁻³

For liver;

ADP= 1.3 × 10⁻³

P₁= 4.8 × 10⁻³

ATP= 3.4 × 10⁻³

∴ K = $\frac{(1.3*10^-3)(4.8*10^-3)}{(3.4*10^-3)}$

  K = 1.8 × 10⁻³

For brain;

ADP= 0.7 × 10⁻³

P₁= 2.7 × 10⁻³

ATP= 2.6 × 10⁻³

∴ K = $\frac{(0.7*10^-3)(2.7*10^-3)}{(2.6*10^-3)}$

  K = 7.3 × 10⁻⁴

b) Since we are concerned about calculating ΔG for ATP hydrolysis in liver at 18° C and we've already obtained the liver concentration from part a; we can therefore calculate ΔG as:

ΔG = ΔG° + RTInK

ΔG° = -30.5

R= 8.315 $\frac{J}{mol.k}$  

T= 18° C = 18 + 273.15k = 291.15k

K= 1.8 × 10⁻³

InK = In(1.8 × 10⁻³ )

      ≅ -6.32

∴ ΔG = -30.5 +  8.315 $\frac{J}{mol.k}$ × 291.15k × (-6.32)

   ΔG = -30.5 +  (−15.30016542)

   ΔG = −45.80016542

   ΔG ≅ −45.8 KJ/mol