Question

Membrane Conductance You are doing a measurement to determine the conductance of an artificial membrane. The membrane is selectively permeable to Na . The temperature is 15 degrees C The external concentrate of Na is 500 mM The internal concentration of Na is 70 mM Using a voltage clamp apparatus you clamp the membrane voltage (Vm) at 20 mV At this clamped voltage you measure a current of -318 nA What is the membrane's conductance

Answer 1

g = 1.11x10⁻⁵Ω.

Step-by-step explanation:

The membrane conductance (g) can be calculated by dividing membrane current (I) through the driving force (Vm - E) as follows:

`g_(ion) = (I_(ion))/(V_(m) - E_(ion))`

where Vm: is the membrane potential and
`E_(ion)`: is the equilibrium potential for the ion or reversal potential.

The equilibrium potential for the ion can be calculated using the Nernst equation:

`E_(ion) = (RT)/(zF)*Ln(([ion]_(out))/([ion]_(ins)))`

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.

`E_(ion) = ((8.314 J*K^(-1)*mol^(-1))((15 + 273)K))/((+1)(96500 C*mol^(-1)))*Ln(([500mM])/([70mM])) = 48.78 mV`

Now, we can calculate the membrane conductance (g) using equation (1):

`g_(ion) = (I_(ion))/(V_(m) - E_(ion)) = (-318*10^(-9) A)/(20*10^(-3) V - 48.78*10^(-3) V) = 1.11*10^(-5) \Omega`

Therefore, the membrane conductance is 1.11x10⁻⁵Ω.

I hope it helps you!