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At the peak of the graph, is the concentration of Na+ outside the cell more than inside? That must be to overcome the electrical force in the opposite direction.

When is the concentration of Na+ greatest inside the cell? At the peak assuming the Na+ channel close immediately at it then it is also greatest at 3. If there is still passage of Na+ then 3 is the maximum. Is this true or false?

One last question: is the inside of membrane actually negative, or only negative relative to the outside? In a galvanic cell for example, a given rod isn't actually negative, but it is negative relative to another.


2 Answers 2


Very good question. I use it routinely when I teach basic neurophysiology. The key to the answer is to understand that the change in the membrane potential is not caused by transport of large amounts of ions over the membrane. It is simply the change in permeability of the membrane to these ions that change the potential. The membrane potential can be calculated by the Goldman equation from the permeabilities of the individual ions ($\ p_x$ indicates the permeability of ion X).

$\ E_m = \frac{RT}{F} \ln{ \left( \frac{ P_{\text{Na}}[\text{Na}^{+}]_\mathrm{out} + P_{\text{K}}[\text{K}^{+}]_\mathrm{out} + P_{\text{Cl}}[\text{Cl}^{-}]_\mathrm{in} }{ P_{\text{Na}}[\text{Na}^{+}]_\mathrm{in} + P_{\text{K}}[\text{K}^{+}]_{\mathrm{in}} + P_{\text{Cl}}[\text{Cl}^{-}]_\mathrm{out} } \right) }$

It is therefore not the transfer of charges as such that drive the change in membrane potential, but a small amount of ions must be moved to counteract the capacitive properties of the cell membrane. The cell membrane acts as a capactor, which means that the voltage across the cell membrane cannot change instantly, but changes as charges are redistributed between the two sides. These charges are moved by Na and K ions moving through the channels in the membrane.

Because we can model the membrane as a capacitor in an electric equivalent circuit, we can estimate the amount of ions moved during the AP. The charge (Q) across a capacitor (cell membrane) is measured in Coulomb (C) and calculated as the product of the capacitance and the voltage across the capacitor: $\ Q=C_m V_m$. If we assume a spherical cell with diameter 10µm and a membrane capacitance of 1$\mu F/cm^2 $ (= $\ 0.01F/m^2 $) and for simplicity assume that the membrane potential changes by 0.1V during an action potential (from -70 mV to +30 mV), we can calculate the number of charges moved.

The total capacity of the cell membrane is $\ 4\times\pi \times (5\times10^{-6}m)^2 \times 0.01F/m^2 = 3.14\times10^{-10}F $. As we change the voltage by 0.1V, this means that we must move a charge of $\ 3.14\times10^{-10}F\times0.1V=3.14\times10^{-11}C $ across the membrane. Given that the charge carried by one monovalent ion (the elementary charge) is $\ 1.6\times10^{-19}C$ we must move $\ 2\times10^6$ ions.

The volume of the cell is $\ \frac43\pi\times(5\times10^{-6}m)^3 =5.24\times10^{-16}m^3=5.24\times10^{-13}l$. If we assume intracellular concentrations at the start of the action potential to be [Na+] = 0.01M and [K+] = 0.15M we can calculate the number of Na+ and K+ ions in the cell by multiplying with Avogadro's number ($\ A_N = 6.02\times10^{23}$). Thus: number of Na+ ions = $\ 0.01M\times5.24\times10^{-13}l\times6.02\times10^{23}=3\times10^9$ and number of K+ ions = $\ 4.7\times10^{10}$.

Given that $\ 2\times10^6$ Na+ ions were moved into the cell during an AP and the same number of K+ ions moved out, the percent change in concentration can be calculated. The change in $\ [Na^+]$ due to one AP is therefore $\ \frac{2\times10^6}{3\times10^9}\times100\ percent=0.07\ percent $

and the change in $\ [K^+]$ is $\ \frac{4.7\times10^{10}}{3\times10^9}\times100\ percent=0.004\ percent$

In conclusion, the changes in intracellular concentrations of Na and K during a single action potential is insignficant, and can easily be counterbalanced by the continuous action of the Na/K-ATPase. In most cases, even if the nerve cell fires repeatedly at high rates, there is no measurable change in concentrations.


Below is a schoolbook example of the membrane potential combined with the individual conductivities of Na+ and K+ during an action potential (AP) (Fig. 1);

enter image description here Fig. 1. Membrane potential and conductivities of the two most important ions in AP generation. source: Physiology Web

And in Fig. 2 is an example of the currents of Na+ and K+ during an action potential (AP) (Fig. 1). Note that K+ stays high due to amanipulation of ion concetrations.

enter image description here Fig. 2. Sodium and potassium currents during AP generation.

However, you ask about the concentrations of Na+ in the cell. As far as I know, this has not been investigated directly. Intracellular Ca2+ concentrations, for example, can be imaged with various imaging techniques, but absolute Na+ and K+ concentrations have to be measured with intracellular electrodes. So while the conductivities and the currents are depicted quite clearly in the above figures, they do not show the absolute concentrations of Na+ or K+ and I doubt it can be found anywhere in the literature. It can of course be mathematically deduced, by calculating the number of ions entering the cell using the Nernst equation and correcting that for the background effect of the sodium-potassium pump and other factors that alter Na+ concentration in the cell (protein transporters and what not). It will at least give an approximation of [Na+]i.

  • 1
    $\begingroup$ Thank for the information about the current and how absolute concentration weren't measured. Can you just solve this question for me? At the peak of the graph, Is the concentration of Na outside the cell is more than inside? because that must be to overcome the electrical force in the opposite direction Because our book shows that at the peak the concentration of Na outside is lower than inside which is not reasonable to me. $\endgroup$ Commented Nov 25, 2016 at 11:57
  • $\begingroup$ @user3733086 the idea I have is that, even though the peak Na+ concentration inside is still greater than outside, the amount of Na+ that exits the cell is sufficient to revert the the electric potential. Remember that there are other ions intervening, and that the potential differences are measured in mV $\endgroup$ Commented Mar 25, 2017 at 11:53
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    $\begingroup$ See my answer to another question: the emphasis there is on resting potential but it is relevant to this one too. tl;dr: Very few ions actually move across the membrane (relative to the total number of ions present) during an action potential. The intracellular concentration of sodium does not change much during a spike, far less than 1%. $\endgroup$
    – Bryan Krause
    Commented Mar 25, 2017 at 18:45

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