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:
Vm=FRTln(∑PiCinM++PiCoutN−∑PiCoutM++PiCinN−)
Then we consider that frog muscle cell membrane is permeable to Na and K only and this makes the equation become:
Vm=FRTln(PK+[K+]in+PNa+[Na+]inPK+[K+]out+PNa+[Na+]out)
Then, we can work on the equation to obtain PNa+/PK+ as:
eRTVmF=[K+]in+PK+PNa+[Na+]in[K+]out+PK+PNa+[Na+]out
⟹PK+PNa+=[Na+]out−eRTVmF[Na+]ineRTVmF[K+]in−[K+]out
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:
PK+PNa+=[109Lmmol]−e(8.314molKJ)(291.15K)(−0.085V)(96485molC)[4Lmmol]e(8.314molKJ)(291.15K)(−0.085V)(96485molC)[124Lmmol]−[2.2Lmmol]
⟹PK+PNa+=0.01826
In conclusion, the rate between the permeabilities PNa+/PK+ is equal to 0.01826 under these conditions at the frog muscle cell membrane.
Reference:
- Bergethon, P. R., & Simons, E. R. (2012). Biophysical chemistry: molecules to membranes. Springer Science & Business Media.
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