Question

Consider the Na+/K+-ATPase to be present in a membrane with the following conditions: Na+ concentration inside, 50mM, outside, 250mM; K+ concentration inside, 70mM, outside, 5 mM; at 298K with the voltage being 50mV lower in the interior.

3Na+(in) + 2K+(out) + ATP + H2O ↔ 3Na+(out) + 2K+(in) + ADP + Pi

Calculate the free-energy difference for moving both Na⁺ and K⁺ in the observed stoichiometry.

A -12 KJ/mol

B 12 KJ/mol

C 24 kJ/mol

D 31 kJ/mol

Answer 1

Given reaction 3Na+(in) + 2K+(out) + ATP + H2O ↔ 3Na+(out) + 2K+(in) + ADP + Pi

Na+ concentration inside - 50mM, outside - 250mM; K+ concentration inside - 70mM, outside - 5 mM

Equilibrium constant K = [Na+(out)]³ [K+(in)]² / [Na+(in)]³ [K+(out)]²

                                 = [250 mM]³ [70 mM]²/ [50 mM]³ [5 mM]²

                                 = 24500

                  K = 24500

We know that

free-energy difference ΔG^orxn = -RT ln K where R = 8.314 x 10^-3 kJ/mol/K

                      Given that T = 298 K

Hence,

ΔG^orxn = -RT ln K

           = - (8.314 x 10^-3 kJ/mol/K) (298 K) In (24500)

          = -25 KJ/mol

Therefore,

free-energy difference for moving both Na⁺ and K⁺ = -25 KJ/mol