Assume that frog muscle cell membrane is permeable to Na and K only. The membrane permeabilities
for Na and K are 𝑃𝐾 and 𝑃𝑁𝑎, respectively. At 180C the resting membrane potential is Δ𝑉 = -85 mV. What
is ratio of permeability 𝑃 / 𝑃 ? Suppose that 𝐶𝐾 = 124 mmole/liter, 𝐶𝐾 = 2.2 mmole/liter and 𝐾𝑁𝑎 𝑖𝑛 𝑜𝑢𝑡
𝐶𝑁𝑎 = 4 mmole/liter, 𝐶𝑁𝑎 = 109 mmole/liter.
We will assume we're under the conditions that also satisfy the Goldman-Hodgkin-Katz constant field equation: this expression gives, at equilibrium state, the transmembrane potential in terms of the specific membrane permeabilities for each ion and their intra- and extracellular concentrations:
"V_m=\\frac{RT}{F}\\ln(\\large{\\frac{\\sum P_iC_{out}^{M+}+P_iC_{in}^{N-}}{\\sum P_iC_{in}^{M+}+P_iC_{out}^{N-}}})"
Then we consider that frog muscle cell membrane is permeable to Na and K only and this makes the equation become:
"V_m=\\frac{RT}{F}\\ln(\\large{\\frac{P_{K^+}[K^+]_{out}+P_{{Na}^+}[{Na}^+]_{out}}{P_{K^+}[K^+]_{in}+P_{{Na}^+}[{Na}^+]_{in}}})"
Then, we can work on the equation to obtain "P_{{Na}^+}\/P_{K^+}" as:
"e^{\\frac{V_mF}{RT}}={\\frac{[K^+]_{out}+ \\large{\\frac{P_{{Na}^+}}{P_{K^+}} } [{Na}^+]_{out}}{[K^+]_{in}+\\large{\\frac{P_{{Na}^+}}{P_{K^+}}}[{Na}^+]_{in}}}"
"\\implies \\frac{P_{{Na}^+}}{P_{K^+}}= \\large{ \\frac{e^{\\frac{V_mF}{RT}}[K^+]_{in}-[K^+]_{out} } { [{Na}^+]_{out} - e^{\\frac{V_mF}{RT}}[{Na}^+]_{in}} }"
Then we substitute all the values (T=18 °C; F = 96485 C/mol; Vm=-0.085 V; [K+]in=124 mmol/L; [K+]out=2.2 mmol/L; [Na+]in=4 mmol/L; [Na+]out=109 mmol/L) and find:
"\\frac{P_{{Na}^+}}{P_{K^+}}= \\frac{e^{\\frac{(-0.085\\,V)(96485\\frac{C}{mol})}{(8.314\\frac{J}{molK})(291.15\\,K)}}[124\\frac{mmol}{L}]-[2.2\\frac{mmol}{L}] } { [109\\frac{mmol}{L}] - e^{\\frac{(-0.085\\,V)(96485\\frac{C}{mol})}{(8.314\\frac{J}{molK})(291.15\\,K)}}[4\\frac{mmol}{L}]}"
"\\implies \\frac{P_{{Na}^+}}{P_{K^+}}= 0.01826"
In conclusion, the rate between the permeabilities "P_{{Na}^+}\/P_{K^+}" is equal to 0.01826 under these conditions at the frog muscle cell membrane.
Reference:
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