Question

The sodium potassium uses energy in the form of ATP hydrolysis to move 3 sodium ions out of the cell and 2 potassium ions into the cell. The concentration of sodium inside is 14mM and outside is 143mM. The concentration of potassium inside is 157mM and outside is 4mM. Membrane potential is -50mV and temperature is 310K. Demonstrate that hydrolysis of ATP to ADP provides enough energy to move 3 sodium ions out and 2 potassium ions in to the cell. (Energy released from ATP hydrolysis is –50kJ/mol).

Note: I calculated the free energy for Na to be 10.9 kJ/mol (so 32.7 kJ/mol for 3 Na ions) and K to be 14.4 (so 28.8 kJ/mol). The sum of these two energies is 61.5 kJ/mol. I'm not sure where to go from here.

The following formula was used to generate the answeres above:

$\Delta G=RTln(\frac{c2}{c1})+ZF\Delta \psi$

Answer 1

The mechanism of action of the Na+/K+ ATPase, outlined in Figure, it has two conformations, E1 and E2.

In its E1 conformation, the Na+/K+ATPase has three high-affinity Na+-binding sites and two low-affinity K+-binding sites on the cytosolic-facing surface of the protein.

The Km for binding of Na+ to these cytosolic sites is 0.6 mM, a value considerably lower than the intracellular Na+ concentration of ≈12 mM; as a result, Na+ ions normally will fill these sites.

Conversely, the affinity of the cytosolic K+-binding sites is low enough that K+ ions, transported inward through the protein, dissociate from E1 into the cytosol despite the high intracellular K+ concentration.

During the E1 → E2 transition, the three bound Na+ ions move outward through the protein. Transition to the E2 conformation also generates two high-affinity K+sites and three low-affinity Na+ sites on the exoplasmic face.

Because the Km for K+ binding to these sites (0.2 mM) is considerably lower than the extracellular K+ concentration (4 mM), these sites will fill quickly with K+ ions.

In contrast, the three Na+ ions, transported outward through the protein, will dissociate into the extracellular medium from the low-affinity Na+ sites on the exoplasmic surface despite the high extracellular Na+ concentration.

Similarly, during the E2 → E1 transition, the two bound K+ ions are transported inward.

In order for the pump to turn one cycle (exporting three Na+ ions and importing two K+ ions), one molecule of ATP must be hydrolyzed. When ATP is hydrolyzed, its gamma phosphate doesn’t simply float away, but is actually transferred onto the pump protein. This process of a phosphate group binding to a molecule is called phosphorylation.

As with most cases of ATP hydrolysis, a phosphate from ATP is transferred onto another molecule. In a phosphorylated state, the Na+/K+ pump has more free energy and is triggered to undergo a conformational change.

This change allows it to release Na+ to the outside of the cell. It then binds extracellular K+, which, through another conformational change, causes the phosphate to detach from the pump. This release of phosphate triggers the K+ to be released to the inside of the cell.

Essentially, the energy released from the hydrolysis of ATP is coupled with the energy required to power the pump and transport Na+ and K+ ions.

The hydrolysis of one ATP molecule releases 7.3 kcal/mol of energy (∆G = −7.3 kcal/mol of energy).

It takes 2.1 kcal/mol of energy to move one Na+ across the membrane (∆G = +2.1 kcal/mol of energy)

Three sodium ions could be moved by the hydrolysis of one ATP molecule. The ∆G of the coupled reaction must be negative. Movement of three sodium ions across the membrane will take 6.3 kcal of energy (2.1 kcal × 3 Na+ ions = 6.3 kcal). Hydrolysis of ATP provides 7.3 kcal of energy, more than enough to power this reaction. Movement of four sodium ions across the membrane, however, would require 8.4 kcal of energy, more than one ATP molecule can provide.