How efficient is the sodium-potassium pump ?

Question

I am reading about transportation of ions in a cell. It is necessary to transport sodium back out and potassium back in, against their electrochemical gradient. This task is carried out by sodium-potassium pump, which transports three sodium ions out across the cell membrane for every two potassium ions carried inward. The source of energy for this transport process was hydrolysis of ATP. My question is the order of the efficiency of a sodium-potassium pump.

Answer 1

I couldn’t find a value for this but I have calculated an efficiency of 84 %.

Caution: this seems too high to me, so I show below my calculation, fully dissected, in case anyone can spot an error.

First of all the parameters. Ion concentrations are:

internal Na⁺ = 12 mM; external Na⁺ = 140 mM; internal K⁺ = 140 mM; external K⁺ = 5 mM; and membrane potential = -60 mV (negative inside the cell)

I use these values for constants:

gas constant R = 2 cal mol^-1 K^-1

Faraday constant F= 23,062 cal mol^-1 V^-1

and I’ll assume that the temperature, T, is 37℃ = 310 K

The Na⁺/K⁺ pump uses 1 molecule of ATP to move 3 Na⁺ out of the cell and 2 K⁺ into the cell.

The efficiency of the pump is therefore energy required to move these ions/energy released by ATP hydrolysis.

How much energy is needed to move the ions? In each case there is a contribution from the fact that the ions are moved along a concentration gradient (the chemical component) and from the movement along a gradient of electrical potential (because of the membrane potential). In considering these components we have to think about whether energy needs to be used when the ions move, or whether it is released.

For the chemical component we can use

ΔG = RTln(ratio of concentration, inside and outside), where ln = natural logarithm

The sign of this ΔG will depend upon how the ratio is expressed (in/out or out/in) but the absolute value will be the same. So I calculate an absolute value and then deduce the sign of the free energy change.

For Na⁺ ΔG = modulus(2 x 310 x ln(12/140)) = 1,523 cal mol^-1

Since Na⁺ is being moved up a concentration gradient, work must be done so the free energy change is positive.

Applying the same formula and logic for the movement of K⁺:

ΔG = 2,065 cal mol^-1.

(This value is greater than that for Na⁺ because the concentration gradient is steeper.)

Next I calculate the electrical components using:

ΔG = zFμ, where z is the charge on the ion = 1 in both cases, and μ is the membrane potential (in volts).

So for both Na⁺ and K⁺ the absolute ΔG value is 1,383 cal mol^-1.

However, in this case Na⁺ is moving out, against the membrane potential while K⁺ is moving in with the membrane potential. Thus the value of ΔG is positive for Na⁺ but negative for K⁺.

Now I combine the chemical and electrical components to get the net free energy change per mole of ions moved:

Na⁺ ΔG = 1,523 + 1,383 = 2,906 cal mol^-1

K⁺ ΔG = 2,065 -1,383 = 682 cal mol^-1

Now I consider how to combine these two values. Each cycle of the pump uses 1 ATP to move 3 Na⁺ and 2 K⁺. Therefore each mole of ATP moves 3 moles of Na⁺ and 2 moles of K⁺.

Energy from 1 mole ATP that is used to move ions = (3 x 2,906) + (2 x 682)

= 10,082 cal

= 10.1 kcal

How much energy is released from hydrolysis of 1 mol of ATP? Estimates vary but a commonly accepted value seems to be 12 kcal mol^-1

Thus efficiency of the pump = 100*(10.1/12) = 84%.

I think the free energy of ATP hydrolysis depends on ATP concentration, pH, and the amount of Mg present. For a well-fed brain, this is around 14 kcal/mol (see Journal Of Physical Chemistry B, 114 (49), (2010). 16137-16146). This gives an efficiency of around 70%, still too high but better...