Question

For a typical mammalian cell, the intracellular and extracellular concentrations of the chloride ion, Cl", are 4 mM and 150 mM, respectively. For the calcium ion, Ca2+, the intracellular and extracellular concentrations are 0.2 p.M and 1.8 mm, respectively. What are the equilibrium potentials in millivolts, mV, for each ion at 37 °C? . Veg for CI= V for Ca2+ SS

Answer 1

The equilibrium potential of an ion based on its valency and its concentration gradient across the membrane is calculated by Nernst equation.

The Nernst equation is represented as:

V_eq = RT/zF ln([X]_out /[ X]_in)

where

V_eq = equilibrium potential for the ion

R = universal gas constant = 8.314 J.K^-1.mol^-1

T = Temperature in Kelvin = (273 + 37^oC) = 310K

z = valency of the ion

F = Faraday's constant = 96485 C.mol^-1

[X]_out = extracellular concentration of ion

[X]_in =intracellular concentration of ion

As [X]_out for Cl^- is 150mM and [X]_in for Cl^- is 4mM

Therefore the equilibrium potential for Cl^- in millivolt is

V_eq for Cl^- = ([8.314][310])/[(-1)(96485)] ln (150/4) = -96.861mV

As [X]_out for Ca²⁺is 1.8mM and [X]_in for Ca²⁺ is 0.2 $\mu$ M = 0.0002mM

Therefore the equilibrium potential for Ca²⁺ in millivolt is

V_eq for Ca²⁺ = ([8.314][310])/[(+2)(96485)] ln (1.8/0.00002) = 121.66mV