Membrane Potential - Guiding Problems

1) Na⁺ movement at different membrane potentials when [Naᵢ⁺] = [Naₒ⁺]

When the internal and external Na⁺ concentrations are equal, the movement of Na⁺ depends on the membrane potential:

At –50 mV: The inside of the cell is negative, and Na⁺ (a positive ion) will move into the cell.
At +50 mV: The inside of the cell is positive, and Na⁺ will move out of the cell.
At 0 mV: There is no electrical driving force, so there will be no net movement of Na⁺.

2) Equilibrium potential for Na⁺ (Eₙₐ) if [Naₒ⁺] is 10 times [Naᵢ⁺]

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

E_{Na} = \frac{R T}{z F} \ln (\frac{[{Na}_{o}]}{[{Na}_{i}})

At physiological temperature (37°C or 310 K), it simplifies to:

E_{Na} \approx 61.5 mV \times \log_{10} (\frac{[{Na}_{o}]}{[{Na}_{i}})

If [Naₒ⁺] = 10 × [Naᵢ⁺], then:

E_{Na} = 61.5 mV \times \log_{10} (10) = 61.5 mV

Thus, the equilibrium potential for Na⁺ is approximately +61.5 mV.

3) Equilibrium potential for K⁺ (Eₖ) if [Kᵢ⁺] is 10 times [Kₒ⁺]

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

E_{K} = 61.5 mV \times \log_{10} (\frac{[K_{o}]}{[K_{i}]})

If [Kᵢ⁺] = 10 × [Kₒ⁺]:

E_{K} = 61.5 mV \times \log_{10} (\frac{1}{10}) = - 61.5 mV

Thus, the equilibrium potential for K⁺ is approximately –61.5 mV.

4) Membrane potential (Eₘ) with gK = 100, gNa = 1, gCl = 10

$g_{\text{K}}$ $g_{\text{Na}}$ $g_{\text{Cl}}$ $K^+$ $E_K$ $K^+$

$g_{\text{K}} = 100$ ) dominates, Eₘ will be closest to the equilibrium potential for K⁺, which is approximately –61.5 mV.

5) Membrane potential (Eₘ) with gK = 10 , gNa = 100 , gCl = 1

$g_{\text{Na}} = 100$ ) dominates, so the membrane potential will be closest to the equilibrium potential for Na⁺, which we calculated to be +61.5 mV.

Thus, Eₘ is approximately +61.5 mV.

6) Impact of Liddle syndrome on epithelial cell luminal membrane potential and K⁺ secretion

Liddle syndrome increases the activity of epithelial sodium channels (ENaC), leading to increased Na⁺ reabsorption.

Membrane potential: Since more Na⁺ is reabsorbed, the epithelial cell luminal membrane becomes more positive (depolarized).
K⁺ secretion: A more positive membrane potential decreases the electrochemical gradient for K⁺ secretion, leading to reduced K⁺ secretion, which can cause hypokalemia (low blood K⁺ levels).