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For the squid giant axon, the membrane potential computed by the Goldman equation is -60mV. And the Nernst potentials are (the differences between the K+ and the Na+'s Nernst potential and the membrane potential let the mechanism work):

``K+ = -74mV Na+ = 55mV Cl- = -60mV``

I am wondering if the membrane potential is 0mV, whether the K+ and Na+ Nernst potentials still cause the mechanism to work, so can 0mV be a reasonable value? (That is, I think the value of potential is not important at all as long as the mechanism works well, is this true?)

The value of the membrane potential does not affect the location of the resting potential. The membrane potential affects only the fraction of channels that are open, thus allowing net current to change, which allows for the depolarization and repolarization of the membrane (the action potential). During the action potential, the resting potential does not change.

The value of the resting potential is governed only by the distribution of ions (how many are on one side of the membrane vs. how many are on the other). This is relatively constant even as ions cross the membrane for the action potential; it is really not a very large fraction of ions moving at all compared to the reservoir they are in, so the reversal potentials are preserved over time.

## Membrane Potential A voltmeter could then mearuse the potental difference across the cell membrane.

outside the cell is considered OmV

work must be performed to separate opposite charges.

So almost all the extracellular and intracellular ions are paired + with -, with just a small number of excess neg charges just inside the membrane paired with a small excess of positive charges just outside the membrane.

ion channels through the membrane make the cell membrane a leaky capacitor.

If a pos ion move out of a cell the cell will become more ______.

Reciprocoal of resistance

Term used in the Equivalent Circuit Model of the electrical properties of the cell membrane.

conductance depnds on poreptiess of both the membrane and concnetration of ions in solution.

resting potential is largely neg bc intracellular con of K is much higher than extracellular conc and the cell membrane is slightly permebale to K

when K flows down its con gradient, leaving the cell, the cell membrane would become more neg, which this electrical force would tend to oppose the movement of K out of the cell.

if the membrane potnetial became less negative for any reason the electrical force holding K in the cell would decrease, so additonal k would leave the cell making the cell more neg again

As EC con of K is raised, the cheomical force driving K out of the cell decreases in size. Ek is the chemical force so it decreases in size.

## Calculate the price of your order Fill in the order form and provide all details of your assignment. Choose the payment system that suits you most. No need to work on your paper at night. Sleep tight, we will cover your back. We offer all kinds of writing services.  ## Physiology of Membrane Potential

### Understanding Membrane Potential

The following points should help you to understand how membrane potential works

• The difference between the electrical and chemical gradient is important.
• Represents the difference in electrical charge across the membrane
• Represents the difference in the concentration of a specific ion across the membrane.
• Channels that are always open
• Permit unregulated flow of ions down an electrochemical gradient.
• Actively transports Na + out of the cell and K + into the cell.
• Helps to maintain the concentration gradient and to counteract the leak channels.

### Membrane Potential and Physiology of Human Nerve Cells

Human nerve cells work mainly on the concept of membrane potentials. They transmit chemicals known as serotonin or dopamine through gradients. The brain receives these neurotransmitters and uses it to perform functions.

• Na + has a much higher concentration outside of the cell and the cell membrane is very impermeable to Na +
• K + has a high concentration inside the cell due to the fact that the cell membrane is very permeable to K +
• A - is used to refer to large ions that are found completely inside of the cell and cannot penetrate the cell membrane.

Concentration (in Millimoles/ Liter) and permeability of Ions Responsible for Membrane Potential in a Resting Nerve Cell

ION Extracellular Intracellular Relative Permeability
Na + 150 15 1
K + 5 150 25-30
A - 0 65 0

Check out this YouTube video if you want to know more about how the Na + /K + pump and how the membrane potential works. www.youtube.com/watch?v=iA-Gdkje6pg

### How To Calculate A Membrane Potential

The calculation for the charge of an ion across a membrane, The Nernst Potential, is relatively easy to calculate. The equation is as follows: (RT/zF) log([X]out/[X]in). RT/F is approximately 61, therefore the equation can be written as

• R is the universal gas constant (8.314 J.K -1 .mol -1 ).
• T is the temperature in Kelvin (°K = °C + 273.15).
• z is the ionic charge for an ion. For example, z is +1 for K + , +2 for Mg 2 + , -1 for F - , -1 for Cl - , etc. Remember, z does not have a unit.
• F is the Faraday's constant (96485 C.mol -1 ).
• [X]out is the concentration of the ion outside of the species. For example the molarity outside of a neuron.
• [X]in is the concentration of the ion inside of the species. For example, the molarity inside of a neuron.

The only difference in the Goldman-Hodgkin-Katz equation is that is adds together the concentrations of all permeable ions as follows Figure 3. (Clockwise From Upper Left) 1) The charges are equal on both sides therefore the membrane has no potential. 2)There is an unbalance of charges, giving the membrane a potential. 3) The charges line up on opposite sides of the membrane to give the membrane its potential. 4) A hypothetical neuron in the human body a large concentration of potassium on the inside and sodium on the outside.

## B3. Thermodynamics of Membrane Potential

• • Contributed by Henry Jakubowski
• Professor (Chemistry) at College of St. Benedict/St. John's University

How can this equilibrium potential be calculated? The total thermodynamic driving force, (&DeltaG_), for an ion (X) moving across the membrane (from inside to outside) of transmembrane potential &Delta&Psi is the sum of the &DeltaGs arising from concentration gradients and from the electrical potential:

The &DeltaGelect pot can be determine from the simple general chemistry equation:

where F, the Faraday constant (the electric charge on one mol of electric charge = 23,000 cal/V.mol), n (or z) is the number of moles of charge transferred per mol of ions moving (+ 1 for Na and K ions) and &Delta&Psi ( or E) is the transmembrane electrical potential in volts. Hence for the movement of an ion from inside to outside the cell,

This shows that the total driving force for ion movement consists of a chemical potential (first term) plus an electrical potential (second term). The equation can be expanded and rearranged as follows, where ( chi _) and ( chi _) are the concentrations of ion (X) inside and outside the cell, respectively:

[&DeltaG_x = RT ln chi_ - RTln chi_ - zFDY]

Consider (ce>), where (chi_/chi_) is about 1/20 = 0.05. (The table above suggests that it is more like 1/35, but some texts use the value of 20.) What would the transmembrane potential have to be so that no driving force would exist for (ce>) ions to move across the membrane (i.e. under equilibrium conditions)? Under these conditions, &Delta G x = 0, so

[&Delta&Psieq = (0.0615/z) log(Xout/Xin) at 310K or 37oC) label<5>]

[&Delta&Psieq = (0.0591/z) log(Xout/Xin ) at 298K or 25oC label<6>]

Equations ef<3>- ef <6>are versions of the Nernst equation. For K+ ions, &Delta&Psi is about -0.075 V or -75 mV (assuming 37oC = 310K, R = 1.98 cal/(K.mol), and z = +1). That is, since the inside is more negative than the outside, then K ions would tend to stay inside even if the chemical potential of K ions is greater on the inside. A similar calculation for Na ions, with a Xout/Xin of about 12, gives an equilibrium Na ion potential of about +55 mV. Why is the actual transmembrane potential close to the equilibrium potential of K ions and not that of Na ions? The reason has to do with the fact that the permeability of K+ ions is 100 fold greater than of Na ions. Hence it is the K ions which mostly determine the cell resting potential. (Remember glial cells have only a nongated K channel.)

In general, the &Delta&Psieq is determined by more than one ion (usually Na and K), so the actual &Delta&Psieq is between the &Delta&Psieq for each ion alone. A more general equation can be derived which shows that the &Delta&Psieq is determined not only by the concentrations but also the permeability of these ions. (In a liposome made with encapsulated KCl at a higher concentration than the outside concentration, and with both sides being electrically neutral, if no ions could flow across the membrane, no membrane potential would arise.) The equation that considers both concentration AND permeability effects is the Goldman equation and is given by:

where PX is the permeability of ion X. This treatment recognizes the importance of thermodynamics (chemical and electrical potentials) and kinetics (permeability coefficients). Remember this potential would not be possible if the ion gradient was not maintained by Na/K ATPase. To see the effects on the transmembrane voltage use the Goldman Equation Calculators below and plug in the following generally realistic values for concentrations and permeabilities of ions in neurons:

ion inside (mM) outside (mM) permeability coeff (cm/s)
K+ 140 5 (1 imes 10^<-8>)
Na+ 12 145 (1 imes 10^<-10>)
Cl- 3 145 (1 imes 10^<-10>)

Then increase the permeability of Na ions and watch the effect on the transmembrane voltage. Reset the Na ion permeabilty and change that for chloride ion.

Fluorophores that insert into membranes and whose flourescence changes with transmembrane potential have been developed. An example is di-4-ANEPPS, a AminoNaphthylEthenylPyridinium dye, whose structure is shown below.

When bound to membranes, this probe has an excitation and emission maximum of 475 and 617 nm, respectively. When the membrane potential is made more negative (hyperpolarized), the fluorescence intensity (excitation 440 nm) decreases while the emission increases when excited at 530 nm.

In the 1940's, ways were developed to measure the actual transmembrane potential of cells. Varying the outside sodium and potassium concentrations would change the experimental transmembrane potential, as indicated by the Nernst equation. The experimental resting potentials of glial cells always matched the theoretical potassium equilibrium potentials, supporting the view that the transmembrane potential was associated only with open, nongated potasisum channels. This was not observed with neurons, suggesting that channels other than for potassium were open. Using radioactive tracers and potential measurements, it became clear that nerve cells were permeable not only to potassium, but also to sodium and chloride. How do these work in establishing the resting potential? Consider the simplest case when just potassium channels are present, along with an unequal distribution of other ions. Now add some sodium channels. Two forces act to drive sodium into the cell - the chemical potential since sodium is higher on the outside, and the electrical potential since the inside of the cell is negative. The equilibrium potential of a cell if it were only permeable to sodium is +55 mv, so there is a great electrical drive for sodium to enter through the nongated, open sodium pores we just added. As sodium enters, the cell starts to "depolarize" and have a more positive voltage. However, since in our example, there are many more open potassium channels, the resting potential deviates only a small amount from the potassium potential, since as the potential becomes more positive, more potassium flows out down the concentration gradient. Eventually the enhanced potassium efflux equals the sodium influx, and a new resting membrane potential of -60 mV is established, which is typical of neurons.

Now we can answer question 3 above: How is the resting electrochemical potential and the transmembrane ion distribution maintained? For a resting electrical potential to be maintained once established, the rate of sodium influx must equal the rate of potassium efflux through the open, non-gated channels in order to maintain separation of charge (and the electrical potential) across the membrane. But, if influx and efflux were allowed to continue, the actual ion gradients (chemical potential) across the membranes would collapse. This problem is solved by the Na/K ATPase. In the resting cells, the passive fluxes of sodium and potassium ions are exactly balanced by the active fluxes of these ions mediated by the Na/K ATPase. The cell is not really at equilibrium but at a steady state.

## Equilibrium potentials and different permeabilities.

that was a long stream of consciousness. Not completely sure about the question, but it seems like you may be neglecting the forces caused by the concentration gradient. Just because Na and K are in electric equilibrium does not mean they are in equilibrium.

In a real cell, there's also ion pumps, Cl-, and other antions belonging to protein constituents.

If I didn't answer your question it may help to restate it a single sentence.

Thanks for the response Py.
Sorry about that. I'm not sure if it's possible to communicate what I mean in a single sentence.
I'll try to make it succinct.
I'm confused about the Equilibrium potential of a cell. The Goldman equation, from my understanding, identifies the potential difference across the membrane for given ion concentrations and permeabilities at which the net flux of ions is zero: so there's no change in the system.
This doesn't make sense to me. Let's say we have a cell with only two ions, K+ and Na+ and that their charges at exactly countered by the presence of some negative charge (i.e. to begin with, the potential difference across the membrane is zero) with the K+ concentration higher inside than outside and the opposite is true for Na+. At the equilibrium potential given by the Goldman equation, the net flux of cations (the only ions in the scenario that can move) is equal in both directions. However, the concentration of K+ is different on the two sides of the membrane, and so K+ is more likely to move out, and the opposite is true for Na+. K+ might move 'back and forth' more frequently than Na+ if it has a higher permeability, but over time Na+ will nonetheless creep across, and K+ will also move out. The only point I can see an equilibrium being a 'true equilibrium' is when the proportion of K+ and Na+ is the same on both sides, and the net flux of cations is equal in both directions.
I found this:
"It is important to keep in mind that neither sodium ions nor potassium ions are at equilibrium at that steady value of potential: sodium ions are continually leaking into the cell and potassium ions are continually leaking out. If this were allowed to continue, the concentration gradients for sodium and potassium would eventually run down and the membrane potential would decline to zero as the ion gradients collapsed."

I'm not sure if this is referring to the same thing as me, it's not that clear. I think it is saying the same thing as me concerning the concentrations changing because they would move with greater probability in one direction compared to the other. I don't know why the potential would go to zero, though.
Hopefully this is a little clearer.
Thanks again,
Nobahar.

Partly, there's not a single question as such, it's more how I imagine the process to work, but for some reason it may not be consistent with other sources. I was under the impression that the Goldman equation identifies the potential difference at which the net flux of ions is zero.

Say there was only K+ that could move across the membrane, and the concentration was higher inside than outside. The positive charges all have a corresponding negative charge that is impermeable, so despite the difference in concentration, the potential difference across the membrane is zero.

Since the concentration of K+ is higher inside the cell, the K+ ions would move down their concentration gradient due to random motion but as this happens, a potential difference across the membrane would form. This would oppose their movement out of the cell and when the negative charge inside the cell relative to outside the cell has a certain value, the charge will draw the ions into the cell with a strength such that the net flux is zero. Yet the concentrations will not be equal inside and outside and it is an equilibrium. As far as I am aware, a stable one.

The difficult part is extending this to more than one ion. If the Goldman equation represents the potential difference at which the net flow of ions is zero, I don't see how it represents a stable equilibrium. The Goldman equation takes into account permeabilities, and, from my understanding at least, puts the potential difference for equilibrium at a position between the equilibrium potentials for the individual ions under 'conditions' in which they were only permeable. Its weighted by the permeabilities but why? I don't see why the permeabilities are important. Overtime, the ions will continue to move down their concentration gradients. As I said, the only 'conditions' I can see that give rise to a stable equilibrium is when the net fluc of ions is zero (i.e. they head in both directions at the same rate), and the concentrations have equal proportions on both sides (i.e. the probability that a given ion moving across the membrane is K+ is the same in both directions) - if they didn't the concentrations would continue to change.

Something in my reasoning may be incorrect. So, if there is a single question, its where is the reasoning wrong? If it is wrong. Which it must be somewhere, since I don't see why the potential difference across the membrane would have to be zero. You could have a potential difference across the membrane if the proportions of each specific ions where equal both sides but the concentrations were different, by my understanding.

Thanks for the quick response, by the way. I'll need to do some more reading on this, but the books I find skip an explanation.

## THE PHARMACOLOGY OF THE INSECT NERVOUS SYSTEM

### B DIVALENT CATIONS

The resting potential of the cockroach giant axon is little affected by a tenfold decrease or a 30-fold increase in the external calcium concentration. Similarly, addition of 52 mM Sr + or Mg 2+ to the normal 1.8 mM Ca 2+ has no affect on the resting potential ( Narahashi, 1966a ). Complete removal of Ca 2+ causes a progressive depolarization ( Narahashi, 1966a ) as well as the addition of 52 mM Ba 2+ to the solution ( Narahashi, 1961 ).

Calcium (Ca 2+ ) ( Narahashi and Yamasaki, 1960b ) and Ba 2+ ions ( Narahashi, 1961 ) both markedly affect the time course of the action potential (1) by slowing down the rising and the falling phase of the spike and (2) by increasing the amplitude and duration of the negative afterpotential. The sodium-inactivation curve is shifted along the potential axis toward lower membrane potentials by an increase in external divalent ion concentration ( Narahashi, 1966a ). This shift, which also occurs in high K + solutions, at least partly accounts for the so-called K + − Ca 2+ antagonism ( Narahashi, 1966a Pichon and Boistel, 1967b ).

These experiments show that divalent cations stabilize the membrane and suggest that they contribute very little to the inward current during the spike ( Narahashi, 1966a ). This has been confirmed by voltage-clamp experiments on isolated giant axons by Pichon (1968) .

In Manduca sexta, Ca 2+ and Mg 2+ ions cannot replace Na + in action-potential production ( Pichon et al., 1972 ), whereas high Mg 2+ solutions have very little effect on resting and action potentials in intact thoracic connectives of Carausius morosus ( Treherne and Maddrell, 1967 ). This lack of effect is most probably due to the fact that the nerve sheath acts as a diffusion barrier against Mg 2+ ions.

## Response to a disturbance of equilibrium

Fig. 2. Ion movement followed by a change in concentration of solute in the OUT compartment.

Suppose the concentration of solute is suddenly changed so that the concentration in the outer compartment goes from 5 mM to 50 mM. The disturbance of equilibrium weakens the concentration gradient so that it can no longer hold as much positive charge outside the membrane. A transient net movement of cations toward the inner compartment takes place. The driving force for the movement (which is against the concentration gradient) is the electrical potential difference, which is now too great to be maintained by the weaker concentration gradient.

During the equilibration period, which is of very short duration, ion flux still continues in both directions (not shown, for clarity).

Figure 3 represents the new steady state. It looks a great deal like figure 1, but the concentration in the outer compartment is 50 mM instead of 5 mM, fewer charges are displaced across the membrane, and the potential difference has been reduced in magnitude.

During all of these adjustments, the concentration of the inner compartment remained exactly the same, for practical purposes. The concentration of the outer compartment was changed only when we deliberately changed the concentration. As long as there is selective permeability ion movements during equilibration do not have a measurable effect on chemical concentrations in either compartment.

## Facilitated Diffusion (Uniporter)

The easiest way to transport a molecule is down it's concentration gradient. These uniporters transport molecules that are thermodynamically favoured to enter the cell but can't because they are not able to diffuse across the lipid bilayer. These molecules include amino acids, nucleosides, sugars etc.
For this course we will discuss one uniporter, the glucose uniporter.

### GLUT1 (mammalian glucose transporter)

Used by most mammalian cells to get glucose across the membrane.
Know a lot about its function and kinetics through studies that place the protein in liposomes (see figure above).
This transporter (and all uniporters) use the concentration gradient of the glucose to drive transport.

The transporter can work in reverse thus if the concentration of glucose is higher on the inside can transport glucose out of the cell.
From Figure 15-5 you can see that the kinetics of transport can be thought of in terms of the Michaelis-Menten equation where v = Vmax ([glucose]/[glucose] + Km).

From Figure 15-5 you can also see that without the transporter to facilitate diffusion then the rate of entry of glucose into the cell almost zero.
There is a maximal rate of transport and the rate of transport is dependent on the glucose concentration. Thus the Km can be calculated from the half maximal rate. The Km for glucose is 1.5 mM. This reflects the affinity that the transporter has for glucose. The lower the Km the greater the affinity. As we will see with the Ca+2 ATPase it's internal Ca+2 binding sites have a Km = 0.0001 mM.

It is also clear that there is a favourable free energy for the transport of glucose.
The concentration of glucose in the blood around 3.6 mM - 5.0 mM After glucose is transported into the cell it is phosphorylated to form glucose 6-phosphate, which cannot leave the cell. Because this reaction is the first step in the metabolism of glucose which is rapidly used the favourable free energy of glucose transport is maintained.

### Nernst potential

All cells have a membrane potential (an electrical potential) that exists across the cell membrane. Researchers use microelectrodes to measure the voltage difference between the outside and inside of the cell. You can measure the membrane potential of a cell = the voltage difference between the inside and the outside of the cell.

### Nernst equation:

Used to calculate the exact electrical potential at equilibrium that is generated for a known concentration difference in a specific ion, separated by a membrane permeable to that ion.
Walther Nernst (1888) derived this equation, based purely on theoretical considerations.

The free energy associated with the transport of an ion (X) across the membrane from the outside to the inside can be written out as:
D G = RTln([Xi]/[Xo]) + zFEm

This is because there are no bonds broken or generated and no heat generated so D G 01 is zero.
As well because the ion is charged there is both a chemical component RTln([Xi]/[Xo]) and an electrical zFEm component.
At equilibrium then D G is zero and so:
zFE = - RTln([Xi]/[Xo])

Thus the equilibrium potential for ion X is:
Ex = - RT ln [X]i
. zF . [X]o
OR
Ex = RT ln [X]o
. zF . [X]i
R = universal gas constant, T = absolute temperature, z =valence of ion (i.e. Cl- = -1), F = Faraday's constant
Note: the valence of the ion is very important to remember!!

What does the equation mean in terms of two different ion concentrations separated by a membrane?
Imagine two chambers separted by a membrane which is only permeable to K+ and not to Cl-. The solutions on either side of the membrane contain KCl.

Using electrodes measure the voltage (potential) difference across the membrane when:

The concentration of KCl is equal on either side (0.01M) and so no there is no potential difference.

The membrane potential is: 0 mV

Now increase the KCl concentration by 10 fold in chamber I

K+ flows down its concentration gradient, chamber II becomes more positively charged than I. The process reaches a point where no more K+ ions flow into II becaused balanced by equal flow of K+ ions out due to electrical repulsion - the system has reached an equilibrium.

i) the chemical gradient which drives K+ into chamber II
ii) the electrical gradient which drives K+ out of chamber II
Therefore at equilibrium if one K+ enters II another K+ ion will be repelled - no net flux occurs.

We can use the Nernst equation to calculate what the membrane potential will be at equilibrium.

Each ion has a different potential given the difference in concentration gradients.

Remember the membrane has to be permeable to the ion. Ions can only cross the membrane through pores or channels. If the membrane does not contain the appropriate ion channel then no ion flow and no potential is created.

### Chemical gradients in animal cells

These differences in Nernst potential reflect the differences in the chemical gradients for each ion.
All animal cells maintain chemical gradients across their plasma membrane and organelle membranes. As we will discuss there is large concentration gradient of Ca+2 in all cells such that the cytosol has a very low Ca+2 concentration while the outside of the cell and in the organelles such as the ER, mitochondria Ca+2 is highly concentrated.
All animal cells also are characterized by a large K+ gradient so that the inside of the cell has a higher K+ concentration than the outside. There is more Na+ on the outside compared to the inside.

From Lodish, Molecular Cell Biology, 4th edition

We will concentrate on the protein pumps that are necessary to maintain these gradients and more importantly why the cell would got to all this trouble to use a large of energy to do this.

### Free energy associated with the Na+ electrochemical gradient.

An example of the advantages of creating an electrical/chemical gradient are outlined for Na+:

From Lodish, Molecular Cell Biology, 4th edition
The forces of the ion and the voltage gradients govern the movement of the ions across the membrane. We can calculate the free-energy change ( D G) that corresponds to the transport of an ion across the membrane.
Because ions are also charged the calculation included both a chemical and electrical component.

For instance the the free-energy change generated by the Na+ ion concentration gradient is:
D Gc = RTln([Na+in]/[Na+out])

In our sample cell this corresponds to -1.45 kCal/mol (the change associated with the transport of 1 mole of Na+ from the outside to the inside of the cell).

The free-engery change generated by the membrane electrical potential is:
D Gm = zFEm

where F = Faraday constant, Em is the membrane potential (-70 mV in most animal cells) and z is the valence of the ion (+1 in this case). This would correspond to -1.6 kCal/mol.

Because Na+ is affected by both the Na+ concentration gradient and the membrane potential both are added to gether to give a total of -3.06 kCal/mol.
Therefore because this is less than 0 this favours thermodynamically the movement of Na+ into the cell. This feature of Na+ we will see in different examples in class can drive a number of cellular processes.

## Effects and implications

While cells expend energy to transport ions and establish a transmembrane potential, they use this potential in turn to transport other ions and metabolites such as sugar. The transmembrane potential of the mitochondria drives the production of ATP, which is the common currency of biological energy.

Cells may draw on the energy they store in the resting potential to drive action potentials or other forms of excitation. These changes in the membrane potential enable communication with other cells (as with action potentials) or initiate changes inside the cell, which happens in an egg when it is fertilized by a sperm.

In neuronal cells, an action potential begins with a rush of sodium ions into the cell through sodium channels, resulting in depolarization, while recovery involves an outward rush of potassium through potassium channels. Both these fluxes occur by passive diffusion.