Answer to Question #210411 in Electricity and Magnetism for Jay

Question #210411

Assume that frog muscle cell membrane is permeable to Na and K only. The membrane permeabilities

for Na and K are 𝑃𝐾 and π‘ƒπ‘π‘Ž, respectively. At 180C the resting membrane potential is Δ𝑉 = -85 mV. What

is ratio of permeability 𝑃 / 𝑃 ? Suppose that 𝐢𝐾 = 124 mmole/liter, 𝐢𝐾 = 2.2 mmole/liter and πΎπ‘π‘Ž 𝑖𝑛 π‘œπ‘’π‘‘

πΆπ‘π‘Ž = 4 mmole/liter, πΆπ‘π‘Ž = 109 mmole/liter.


1
Expert's answer
2021-06-25T10:44:14-0400

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=RTFln⁑(βˆ‘PiCoutM++PiCinNβˆ’βˆ‘PiCinM++PiCoutNβˆ’)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:


Vm=RTFln⁑(PK+[K+]out+PNa+[Na+]outPK+[K+]in+PNa+[Na+]in)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 PNa+/PK+P_{{Na}^+}/P_{K^+} as:


eVmFRT=[K+]out+PNa+PK+[Na+]out[K+]in+PNa+PK+[Na+]ine^{\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}}}


β€…β€ŠβŸΉβ€…β€ŠPNa+PK+=eVmFRT[K+]inβˆ’[K+]out[Na+]outβˆ’eVmFRT[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 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:


PNa+PK+=e(βˆ’0.085 V)(96485Cmol)(8.314JmolK)(291.15 K)[124mmolL]βˆ’[2.2mmolL][109mmolL]βˆ’e(βˆ’0.085 V)(96485Cmol)(8.314JmolK)(291.15 K)[4mmolL]\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}]}


β€…β€ŠβŸΉβ€…β€ŠPNa+PK+=0.01826\implies \frac{P_{{Na}^+}}{P_{K^+}}= 0.01826


In conclusion, the rate between the permeabilities PNa+/PK+P_{{Na}^+}/P_{K^+} 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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment