Answer:
g = 1.11x10⁻⁵Ω.
Explanation:
The membrane conductance (g) can be calculated by dividing membrane current (I) through the driving force (Vm - E) as follows:
[tex] g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}} [/tex]
where Vm: is the membrane potential and [tex]E_{ion}[/tex]: is the equilibrium potential for the ion or reversal potential.
The equilibrium potential for the ion can be calculated using the Nernst equation:
[tex] E_{ion} = \frac{RT}{zF}*Ln(\frac{[ion]_{out}}{[ion]_{ins}}) [/tex]
where R: is the gas constant = 8.314 J/K*mol, F: is the Faraday constant = 96500 C/mol, T: is the temperature (K), z: is the ion's charge, [ion]out and [ion]ins: is the concentration of the ion outside and inside, respectively.
[tex] E_{ion} = \frac{(8.314 J*K^{-1}*mol^{-1})((15 + 273)K)}{(+1)(96500 C*mol^{-1})}*Ln(\frac{[500mM]}{[70mM]}) = 48.78 mV [/tex]
Now, we can calculate the membrane conductance (g) using equation (1):
[tex] g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}} = \frac{-318*10^{-9} A}{20*10^{-3} V - 48.78*10^{-3} V} = 1.11*10^{-5} \Omega [/tex]
Therefore, the membrane conductance is 1.11x10⁻⁵Ω.
I hope it helps you!
