Question

Consider a hypothetical cell whose plasma membrane is impermeant to everything. The membrane is studded with channel proteins for Ca2+, but these channels are initially closed. The concentration of Ca2+ outside the cell is 100 mM and the concentration of Ca2+ inside the cell is 1 mM. Initally, the membrane potential is 0 mV.

After a while, the net flow of Ca2+ ceases because dynamic equilibrium has been reached. What is the membrane potential of the cell when it reaches dynamic equilibrium? Show how you arrived at this answer.

Answer 1

When thermodynamic equilibrium has been achieved (there is no more net flux of ions), the resting membrane potential becomes equal to Nernst potential, which can be calculated using the Nernst equation formula

where E is the membrane potential, R is universal gas constant, T is temperature in Kelvin, "z" is the charge on the ion, F is the Faraday's constant and C_o and C_i mean the concentration of the particular ion (Ca²⁺) in the ECF (outside the cell) and ICF (inside the cell) respectively.

Since T is not given, it is assumed to be the standard temperature, 298.15 K (25 °C). F = 96485 C mol^-1 and R = 8.314 J K^-1 mol^-1. It is also given that C_o = 100 mM and C_i = 1 mM. Substituting these values, we get the membrane potential E = +59.15626 mV. But, here, +59.16 mV is considered to be 0 mV (just changing the baseline value), possibly because other ions in the cell counter this +59.16 mV.

When Ca²⁺ concentration is in equilibrium the concentrations outside and inside become equal i.e., 50 mM each. At that state, E (because of Ca²⁺) should be zero. But since +59.16 mV is the zero, then zero will be −59.16 mV (because the other ions that counter the Ca²⁺ do not change). In other words, under the given conditions, the membrane potential at dynamic equilibrium will be −59.16 mV.