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:
Vmβ=FRTβln(βPiβCinM+β+PiβCoutNβββPiβCoutM+β+PiβCinNβββ)
Then we consider that frog muscle cell membrane is permeable to Na and K only and this makes the equation become:
Vmβ=FRTβln(PK+β[K+]inβ+PNa+β[Na+]inβPK+β[K+]outβ+PNa+β[Na+]outββ)
Then, we can work on the equation to obtain PNa+β/PK+β as:
eRTVmβFβ=[K+]inβ+PK+βPNa+ββ[Na+]inβ[K+]outβ+PK+βPNa+ββ[Na+]outββ
βΉPK+βPNa+ββ=[Na+]outββeRTVmβFβ[Na+]inβeRTVmβFβ[K+]inββ[K+]outββ
Then we substitute all the values (T=18 Β°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+βPNa+ββ=[109Lmmolβ]βe(8.314molKJβ)(291.15K)(β0.085V)(96485molCβ)β[4Lmmolβ]e(8.314molKJβ)(291.15K)(β0.085V)(96485molCβ)β[124Lmmolβ]β[2.2Lmmolβ]β
βΉPK+βPNa+ββ=0.01826
In conclusion, the rate between the permeabilities PNa+β/PK+β 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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