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

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:  

$g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}}$

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:

$E_{ion} = \frac{RT}{zF}*Ln(\frac{[ion]_{out}}{[ion]_{ins}})$

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.    

$E_{ion} = \frac{(8.314 J*K^{-1}*mol^{-1})((15 + 273)K)}{(+1)(96500 C*mol^{-1})}*Ln(\frac{[500mM]}{[70mM]}) = 48.78 mV$  

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

$g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}} = \frac{-318*10^{-9} A}{20*10^{-3} V - 48.78*10^{-3} V} = 1.11*10^{-5} \Omega$

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

I hope it helps you!