<blockquote>
  <p>Apparently the resting potential of -65mV is reached when the two forces, diffusion and electrical gradient are in equilibrium.</p>
</blockquote>

<p>That is not true. You are confusing a neuron's <strong>resting potential</strong> with what is actually a type of ion's <strong>equilibrium potential</strong>. </p>

<p>An <strong>equilibrium potential</strong> for a particular type of ion--such as Na+ or K+--is the exact voltage across the membrane that exactly opposes the diffusional force for that ion, and thereby prevents a net flow of that ion across the membrane. So, for example, in the case of Na+ (sodium ion), in mammalian neurons, a typical value given in textbooks is something like +55 mV. In other words, the interior of the neuron needs to be 55 mV (relative to the outside of the neuron) in order to prevent all the Na+ ions from rushing into the neuron. There are equilibrium potentials for all the ions that are used in neurons to influence the voltage across the membrane: Na+, K+, Cl-, Ca++. If a fictional neuron were to be at the equilibrium potential for Na+, and there were no other ions at play, <em>there would be no need for pumps to maintain any concentration gradient of Na+.</em>, because, by definition, there would be no net flow of charge, and so the situation would be stable.</p>

<p>However, the <strong>resting potential</strong> refers to the voltage of the neuron when it is at rest--not having an action potential--but that voltage is going to be some "weighted average" of the equilibrium potentials of all of the types of ions: Na+, K+, Cl-, and Ca++. Of these, K+ (potassium ion) dominates, and so we find that typical values for the resting potential of neurons is around -65mV, which is closer to the equilibrium potential for K+ (-90mV), than for Na+ (+55mV) or Ca++ (+155mV!). </p>

<p>But since the typical resting potential of the neuron actually <em>is not equal to any of the equilibrium potentials for any type of ion</em>, that means by definition all of the ions <em>are</em> going to have net movement across the membrane. If ion channels are left open and ions are therefore allowed to flow across the membrane, Na+ is going to flow in, K+ is going to flow out, Ca++ is going to flow in, etc. (I'm leaving Cl- out for simplicity).</p>

<p>Since these ions are flowing with a net direction (either into or out of the neuron), that means concentrations are changing; e.g., K+'s large concentration gradient is getting smaller as it flows out of the neuron. Over time, these ion flows move the concentrations of ions towards equality on both sides of the membrane, and therefore move the neuron's resting membrane potential toward 0 mV. In some sense, the "battery is running down". This would render the neuron incapable of having an action potential, and thus acting as a signaling machine, and therefore must be prevented.</p>

<p>In order to prevent that rundown of concentration gradients, ion pumps <strong>actively maintain</strong> the concentration gradients of the ions. One of the relevant ones here is the <strong>Na+/K+ exchanger</strong>, which uses ATP to power its action of pumping <em>in</em> two Na+ ions for every three K+ ions it pumps <em>out</em>. Using ATP requires producing ATP, which requires glucose, and so just maintaining "ready to go neurons" is metabolically demanding.</p>

CC BY-SA 3.0
2015-10-06T16:07:20.713Z
2015-10-10T00:04:37.300Z
2015-10-10T00:04:37.300Z
Chelonian
<p>You're on the right track: it is a dynamic equilibrium that must be maintained actively, not a chemically automatic equilibrium that occurs passively (without biological action). Excerpts from <a href="https://en.wikipedia.org/wiki/Membrane_potential#Resting_potential" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>Values of resting membrane potential in most animal cells usually vary between the potassium reversal potential (usually around -80 mV) and around -40 mV. The resting potential in excitable cells (capable of producing action potentials) is usually near -60 mV—<strong>more depolarized voltages would lead to spontaneous generation of action potentials</strong>. Immature or undifferentiated cells show highly variable values of resting voltage, usually significantly more positive than in differentiated cells.<sup>[23]</sup> In such cells, the resting potential value correlates with the degree of differentiation: undifferentiated cells in some cases <strong>may not show any transmembrane voltage difference at all</strong>.</p>
<p>Maintenance of the resting potential can be metabolically costly for a cell because of its requirement for <strong>active pumping of ions to counteract losses due to leakage channels</strong>. The cost is highest when the cell function requires an especially depolarized value of membrane voltage. For example, the resting potential in daylight-adapted blowfly (Calliphora vicina) photoreceptors can be as high as -30 mV.<sup><a href="http://europepmc.org/articles/PMC2229225/pdf/jg1043593.pdf" rel="nofollow noreferrer">[24]</a></sup> This elevated membrane potential allows the cells to respond very rapidly to visual inputs; the cost is that maintenance of the resting potential <strong>may consume more than 20% of overall cellular ATP</strong>.<sup><a href="http://www.nature.com/neuro/journal/v1/n1/full/nn0598_36.html" rel="nofollow noreferrer">[25]</a></sup></p>
<p>On the other hand, the high resting potential in undifferentiated cells can be a metabolic advantage. This apparent paradox is resolved by examination of the origin of that resting potential. Little-differentiated cells are characterized by extremely high input resistance,<sup>[23]</sup> which implies that few leakage channels are present at this stage of cell life. As an apparent result, potassium permeability becomes similar to that for sodium ions, which places resting potential in-between the reversal potentials for sodium and potassium as discussed above. The <strong>reduced leakage currents also mean there is little need for active pumping in order to compensate, therefore low metabolic cost</strong>. <sub>[Emphasis added.]</sub></p>
</blockquote>
<p>Thus it turns out that resting potential is an interesting, functionally variable quality of different cells, and sometimes costs quite a bit of energy to maintain!</p>
<p><strong>References</strong></p>
<p><sup>[23]</sup> Magnuson, D. S., Morassutti, D. J., Staines, W. A., McBurney, M. W., &amp; Marshall, K. C. (1995). In vivo electrophysiological maturation of neurons derived from a multipotent precursor (embryonal carcinoma) cell line. <em>Developmental Brain Research, 84</em>(1), 130–141.</p>
<p><sup>[24]</sup> Juusola, M., Kouvalainen, E., Järvilehto, M., &amp; Weckström, M. (1994). Contrast gain, signal-to-noise ratio, and linearity in light-adapted blowfly photoreceptors. <em>The Journal of General Physiology, 104</em>(3), 593–621. Retrieved from <a href="http://europepmc.org/articles/PMC2229225/pdf/jg1043593.pdf" rel="nofollow noreferrer">http://europepmc.org/articles/PMC2229225/pdf/jg1043593.pdf</a>.</p>
<p><sup>[25]</sup> Laughlin, S. B., van Steveninck, R. R. D. R., &amp; Anderson, J. C. (1998). The metabolic cost of neural information. <em>Nature Neuroscience, 1</em>(1), 36–41. Retrieved from <a href="http://www.nature.com/neuro/journal/v1/n1/full/nn0598_36.html" rel="nofollow noreferrer">http://www.nature.com/neuro/journal/v1/n1/full/nn0598_36.html</a>.</p>

CC BY-SA 3.0
2014-04-24T17:12:33.700Z
@Vadim: Welcome :) Anything else you wanted to know? (I ask because you haven't accepted an answer; no pressure though.)
Nick Stauner
CC BY-SA 3.0
2015-10-06T16:08:45.627Z
I don't see how your citations actually answers the question, and doesn't correct the confusion between equilibrium potential and resting membrane potential, which is a fundamental point. See my answer.
Chelonian
CC BY-SA 3.0
2014-03-08T11:30:48.543Z
2014-03-08T11:30:48.543Z
2020-06-17T09:55:31.633Z
Nick Stauner
<p>I am confused by the following. Apparently the resting potential of -65mV is reached when the two forces, diffusion and electrical gradient are in equilibrium. So why does the book say </p>

<p>"The electrical potential difference across the membrane [..] charge is maintained by the work of the ion pumps."</p>

<p>Why should it be "maintained by the pumps" if it is an equilibrium anyway?</p>

<p>One possible answer I can imagine is: the equilibrium is too unstable and so the pumps are stabilizing it. But I have no idea if this is correct. Thanks.</p>

<neuroscience>
The function of pumps in forming the resting potential
Vadim
# The function of pumps in forming the resting potential

(Asked by: username_0 on 2014-03-08)

I am confused by the following. Apparently the resting potential of -65mV is reached when the two forces, diffusion and electrical gradient are in equilibrium. So why does the book say

"The electrical potential difference across the membrane [..] charge is maintained by the work of the ion pumps."

Why should it be "maintained by the pumps" if it is an equilibrium anyway?

One possible answer I can imagine is: the equilibrium is too unstable and so the pumps are stabilizing it. But I have no idea if this is correct. Thanks.

**Tags:** neuroscience


---

## 2 answers

### Answer by username_1 on 2015-10-06. Score: 3

> Apparently the resting potential of -65mV is reached when the two forces, diffusion and electrical gradient are in equilibrium.

That is not true. You are confusing a neuron's **resting potential** with what is actually a type of ion's **equilibrium potential**.

An **equilibrium potential** for a particular type of ion--such as Na+ or K+--is the exact voltage across the membrane that exactly opposes the diffusional force for that ion, and thereby prevents a net flow of that ion across the membrane. So, for example, in the case of Na+ (sodium ion), in mammalian neurons, a typical value given in textbooks is something like +55 mV. In other words, the interior of the neuron needs to be 55 mV (relative to the outside of the neuron) in order to prevent all the Na+ ions from rushing into the neuron. There are equilibrium potentials for all the ions that are used in neurons to influence the voltage across the membrane: Na+, K+, Cl-, Ca++. If a fictional neuron were to be at the equilibrium potential for Na+, and there were no other ions at play, *there would be no need for pumps to maintain any concentration gradient of Na+.*, because, by definition, there would be no net flow of charge, and so the situation would be stable.

However, the **resting potential** refers to the voltage of the neuron when it is at rest--not having an action potential--but that voltage is going to be some "weighted average" of the equilibrium potentials of all of the types of ions: Na+, K+, Cl-, and Ca++. Of these, K+ (potassium ion) dominates, and so we find that typical values for the resting potential of neurons is around -65mV, which is closer to the equilibrium potential for K+ (-90mV), than for Na+ (+55mV) or Ca++ (+155mV!).

But since the typical resting potential of the neuron actually *is not equal to any of the equilibrium potentials for any type of ion*, that means by definition all of the ions *are* going to have net movement across the membrane. If ion channels are left open and ions are therefore allowed to flow across the membrane, Na+ is going to flow in, K+ is going to flow out, Ca++ is going to flow in, etc. (I'm leaving Cl- out for simplicity).

Since these ions are flowing with a net direction (either into or out of the neuron), that means concentrations are changing; e.g., K+'s large concentration gradient is getting smaller as it flows out of the neuron. Over time, these ion flows move the concentrations of ions towards equality on both sides of the membrane, and therefore move the neuron's resting membrane potential toward 0 mV. In some sense, the "battery is running down". This would render the neuron incapable of having an action potential, and thus acting as a signaling machine, and therefore must be prevented.

In order to prevent that rundown of concentration gradients, ion pumps **actively maintain** the concentration gradients of the ions. One of the relevant ones here is the **Na+/K+ exchanger**, which uses ATP to power its action of pumping *in* two Na+ ions for every three K+ ions it pumps *out*. Using ATP requires producing ATP, which requires glucose, and so just maintaining "ready to go neurons" is metabolically demanding.


---

### Answer by username_2 on 2014-03-08. Score: 2

You're on the right track: it is a dynamic equilibrium that must be maintained actively, not a chemically automatic equilibrium that occurs passively (without biological action). Excerpts from [Wikipedia](https://en.wikipedia.org/wiki/Membrane_potential#Resting_potential):

> Values of resting membrane potential in most animal cells usually vary between the potassium reversal potential (usually around -80 mV) and around -40 mV. The resting potential in excitable cells (capable of producing action potentials) is usually near -60 mV—**more depolarized voltages would lead to spontaneous generation of action potentials**. Immature or undifferentiated cells show highly variable values of resting voltage, usually significantly more positive than in differentiated cells.[23] In such cells, the resting potential value correlates with the degree of differentiation: undifferentiated cells in some cases **may not show any transmembrane voltage difference at all**.
>
> Maintenance of the resting potential can be metabolically costly for a cell because of its requirement for **active pumping of ions to counteract losses due to leakage channels**. The cost is highest when the cell function requires an especially depolarized value of membrane voltage. For example, the resting potential in daylight-adapted blowfly (Calliphora vicina) photoreceptors can be as high as -30 mV.[[24]](http://europepmc.org/articles/PMC2229225/pdf/jg1043593.pdf) This elevated membrane potential allows the cells to respond very rapidly to visual inputs; the cost is that maintenance of the resting potential **may consume more than 20% of overall cellular ATP**.[[25]](http://www.nature.com/neuro/journal/v1/n1/full/nn0598_36.html)
>
> On the other hand, the high resting potential in undifferentiated cells can be a metabolic advantage. This apparent paradox is resolved by examination of the origin of that resting potential. Little-differentiated cells are characterized by extremely high input resistance,[23] which implies that few leakage channels are present at this stage of cell life. As an apparent result, potassium permeability becomes similar to that for sodium ions, which places resting potential in-between the reversal potentials for sodium and potassium as discussed above. The **reduced leakage currents also mean there is little need for active pumping in order to compensate, therefore low metabolic cost**. [Emphasis added.]

Thus it turns out that resting potential is an interesting, functionally variable quality of different cells, and sometimes costs quite a bit of energy to maintain!

**References**

[23] Magnuson, D. S., Morassutti, D. J., Staines, W. A., McBurney, M. W., & Marshall, K. C. (1995). In vivo electrophysiological maturation of neurons derived from a multipotent precursor (embryonal carcinoma) cell line. *Developmental Brain Research, 84*(1), 130–141.

[24] Juusola, M., Kouvalainen, E., Järvilehto, M., & Weckström, M. (1994). Contrast gain, signal-to-noise ratio, and linearity in light-adapted blowfly photoreceptors. *The Journal of General Physiology, 104*(3), 593–621. Retrieved from <http://europepmc.org/articles/PMC2229225/pdf/jg1043593.pdf>.

[25] Laughlin, S. B., van Steveninck, R. R. D. R., & Anderson, J. C. (1998). The metabolic cost of neural information. *Nature Neuroscience, 1*(1), 36–41. Retrieved from <http://www.nature.com/neuro/journal/v1/n1/full/nn0598_36.html>.

> **Comments:**
> **username_2**: @username_0: Welcome :) Anything else you wanted to know? (I ask because you haven't accepted an answer; no pressure though.)  
> **username_1**: I don't see how your citations actually answers the question, and doesn't correct the confusion between equilibrium potential and resting membrane potential, which is a fundamental point. See my answer.  

---

