The extracellular concentrations of Na+ and K+ are 135mM and 6mM, respectively.
The intracellular concentrations of Na+ and K+ are 10mM and 105mM, respectively.
The membrane potential is 75mV. For Na+/K+ ATPase, 3Na+ are exported and 2K+ are imported. Calculate the total free energy in KJ/mole required to transport the ions.
To calculate the total free energy required to transport the ions, we need to consider the concentration gradients and the membrane potential.
First, we need to calculate the free energy change for transporting each ion against their respective concentration gradients.
For Na+:
ΔG_Na = RT ln (Concentration extracellular / Concentration intracellular)
= (8.314 J/(mol*K)) * (298K) * ln (135 mM / 10 mM)
= 6032 J/mol
For K+:
ΔG_K = RT ln (Concentration intracellular / Concentration extracellular)
= (8.314 J/(mol*K)) * (298K) * ln (105 mM / 6 mM)
= 3668 J/mol
Next, we need to consider the electrical potential across the membrane. The electrical potential can be calculated using the Nernst equation:
ΔΨ = (RT / F) * ln (Concentration intracellular / Concentration extracellular)
= (8.314 J/(mol*K)) * (298K) / (96,485 C/mol) * ln (105 mM / 6 mM)
= 61.54 mV
Now, we calculate the total free energy required to transport the ions by multiplying the electrical potential by the charge of each ion and by taking into account the stoichiometry of the Na+/K+ ATPase:
Total free energy = (3 * ΔG_Na) + (2 * ΔG_K) + (2 * ΔΨ)
= (3 * 6032 J/mol) + (2 * 3668 J/mol) + (2 * 61.54 mV * 96,485 C/mol)
= 18,096 J/mol + 7,336 J/mol + 11,827 J/mol
= 37,259 J/mol
To convert this value to kilojoules per mole (kJ/mol):
Total free energy = 37,259 J/mol * (1 kJ / 1000 J)
= 37.259 kJ/mol
Therefore, the total free energy in kJ/mol required to transport the ions is approximately 37.259 kJ/mol.