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Given a very unusual cell with an intracellular fluid containing 1 mM calcium phosphate and an extracellular fluid containing 100 mM calcium phosphate. If the cell membrane contains open ion channels that are permeable only to calcium ions, what is the membrane potential (Vm) when the system attains equilibrium

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Answer:

M.P = 140.46 Vm

Explanation:

Membrane potential can calculated using the following equation:

(61/z) ln ([X]out/[X]in)

where z is the charge of Ca2+ ion and X represent the concentrations By applying necessary calculations…

Membrane potential = (61 / 2) ln (100 x 10^-3 / 1 x 10^-3)

                                  = 30.5 x 4.60 = 140.46 Vm

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