# Why is the resting potential of a neuron so close to the equilibrium potential of K⁺?

I know this has something to do with the K+ leak channel. I just don't understand how.

I know that 3 Na+ are pumped out for every 2 K+ pumped in. This makes the cell interior net negative.

I know that K+ is allowed to leak out slowly via K+ leak channels.

How does this place the resting potential of the cell so much closer to the equilibrium potential of K+ than Na+?

## 1 Answer

Lets just go with the basic Hodgkin-Huxley equation:

$$C_M \frac{\text{d}V}{\text{d}t}=-g_{Na}(V-E_{Na}) -g_K(V-E_K) -g_L(V-E_L)$$

At rest ${\large\frac{\text{d}V}{\text{d}t}}=0$ and therefore $V$ is dependent on the conductances ($g_X$) of different ions.
Since $g_K \sim 30\times g_{Na}$, the resting potential is closer to $E_K$ (Nernst equilibrium potential of K+).

What it basically means is that since potassium has higher membrane conductance, it diffuses faster thereby attaining equilibrium faster than sodium (in isolated cases assuming only one ion at a time). When both ions are present the resting potential is because of both the ions but since K+ diffuses faster, it has a major contribution to the resting potential compared to Na+. Leak current that is described in the above equation is a general leak (non-specific). Other channels like rectifiers and voltage gated channels are important when talking about action potential and all these channels contribute to the conductance; however you can group all that into a single metric called membrane conductance. Voltage gated channels can be considered separately when studying the dynamics (in this case the conductances themselves become a function of voltage).

• Well, thank you. However, could I ask for a slightly less technical answer? Basically the potassium ions are 30x more conductive, so they have greater determining power over what the resting potential is? So K+ leak channels don't really have anything to do with resting potential. – kevin Feb 5 '15 at 6:53