Question

15. Given the following concentrations of ions in the intracellular and extracellular fluids of a cell; inside the cell: Na+ 12mM. K+ 150mM, CI-9mm, Ca2+0.0001mM and outside the cell: Na+ 145mM, K+ 5mM, CI- 125mM, and Ca2+ 2.5mM. Calculate the equilibrium potential for K+ (Ek) if its extracellular concentrations rise from 5 mM to 10 mM. Use the Nernst equation and the ion concentration to perform the following calculation. Comparing this to the normal Ek, is the change a depolarization or hyperpolarization.

Answer 1

15.

The Nernst equation to calculate the equilibrium potential for K+ is:

Veq. = RT/zF * ln ([X]out/[X]in)

where R is the universal gas constant (8.314 Joules per Kelvin per mole); T is the temperature in Kelvin (in the living cell the temperature is 36.5°C = 273.15 + 36.5°C); z is the valence of the ion (+1 for K+); F is the Faraday's constant (96485 coulombs per mole); [X]out is the concentration of K+ in the extracellular fluid; [X]in is the concentration of K+ intracellular.

Substituting the values we obtain:

Veq. =  8.314*308.65/1*96485 * ln([5mM]/[150mM])

Veq. = -90.45 mV for K+

When extracellular concentration raises from 5 mm to 10 mm, we get:

Veq. =  8.314*308.65/1*96485 * ln([10mM]/[150mM])

Veq. = -72.02 mV for K+

Because the equilibrium potential is more positive when [k+] is raised, then, a depolarization is stimulated.

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