Electrical consequences of ionic gradients

Objectives

1.
Recognize that the movement of ions can generate an electrical potential difference across a membrane.
2.
Define the concept of the equilibrium potential and apply the Nernst equation to calculate it.
3.
Describe how the resting membrane potential is generated in a cell and use the Goldman-Hodgkin-Katz (GHK) equation to calculate membrane potential.
4.
Explain the relationship between the GHK equation and the Nernst equation.
5.
Explain how changes in membrane permeability to permeant ions can change the membrane potential.
6.
Describe the Donnan effect and its consequences for living cells.

Ions are typically present at different concentrations on opposite sides of a biomembrane

For any membrane in a living cell, biologically important ions are distributed asymmetrically on opposite sides of the membrane. In this chapter we focus on the plasma membrane (PM) of a cell, which separates the intracellular and extracellular environments. Across the PM, concentrations of the three common monovalent ions, Na ⁺ , K ⁺ , and Cl ⁻ , are different. Ionic distributions for a “typical” mammalian cell are shown in Table 4.1 . It is clear that the asymmetrical distributions of Na ⁺ , K ⁺ , and Cl ⁻ ions give rise to concentration gradients of these ions across the PM. Such ion concentration gradients can drive the diffusional movement of the ions across the PM, which is selectively permeable to these ions. However, because ions carry electrical charge, their diffusional movement across the PM gives rise to electrical effects, which we now examine.

TABLE 4.1

Intracellular and Extracellular Concentrations of the Common Monovalent Ions for a Typical Mammalian Cell

Ion	Intracellular (mM)	Extracellular (mM)
K ⁺	140	5
Na ⁺	10	145
Cl ⁻	6	106

Selective ionic permeability through membranes has electrical consequences: The nernst equation

Consider a cell with the ion distributions shown in Table 4.1 . Because the K ⁺ concentration is higher inside the cell than outside, if the PM is selectively permeable only to K ⁺ , we expect that K ⁺ ions will move down their concentration gradient, out of the cell ( Fig. 4.1 ).

Fig. 4.1, When the plasma membrane is selectively permeable only to K + ions, the K + concentration gradient (high concentration inside and low concentration outside) drives net movement of K + ions out of the cell.

When the positive K ⁺ ions leave the cell, however, they introduce positive charges to the exterior of the PM while leaving behind an equal number of negative charges on the intracellular side. This means that an electrical potential difference develops as K ⁺ ions diffuse out of the cell. As K ⁺ ions exit the cell, the interior of the cell becomes progressively more negative while the exterior becomes correspondingly more positive. The effect of the developing electric field is to oppose further movement of K ⁺ ions (i.e., the negative interior tends to attract positive K ⁺ , whereas the positive exterior tends to repel K ⁺ ). This analysis suggests that the “leakage” of K ⁺ ions cannot continue indefinitely because eventually a strong enough electric field will build up to balance exactly the tendency of K ⁺ to move out of the cell, down its concentration gradient. When the electrical forces exactly balance the driving force of the concentration gradient, we say that electrochemical equilibrium is reached. At electrochemical equilibrium, there can be no further net movement of K ⁺ ions into or out of the cell. We can also conclude that when K ⁺ ions are in electrochemical equilibrium across the PM, an electrical potential difference exists across the PM, with the inside being more negative than the outside. This electrical potential difference at which no net movement of K ⁺ occurs is the equilibrium potential for K ⁺ and is given the symbol E _K . For any ion, X ^z , whose extracellular and intracellular concentrations are [X ^z ] _o and [X ^z ] _i , respectively, the equilibrium potential can be calculated using the Nernst equation ^a

a Named after Hermann Walther Nernst, who first derived the equation in 1889. Nernst received the Nobel Prize in Chemistry in 1920. Because the equilibrium potential for an ion is defined by the Nernst equation, it is also known as the Nernst potential .

( Box 4.1 ):

BOX 4.1

Movement of Ions Driven By an Electrical Potential Gradient and the Origin of the Nernst Equation

In Chapter 2 , diffusion, or the movement of molecules resulting from the presence of concentration gradients, was discussed. In an analogous way, we now examine the movement of electrically charged molecules (ions). Because an ion is charged, when placed in an electric field, it will experience a force, causing it to move. It is reasonable that the speed at which an ion moves in solution should depend on the strength of the electric field (the stronger the electric field, the faster the ion will move) and the charge on the ion (the higher the electrical charge on an ion, the faster it will move in an electric field). In an electric field, the electrical potential, E , changes with distance, x , that is, Δ E /Δ x , or dE/dx in derivative notation. Analogous to the case of diffusion, where dC/dx is a concentration gradient, the change in electrical potential with distance, dE/dx, is an electrical potential gradient . If the speed of an ion is s , the relationship between ion speed, the electrical potential gradient, and the ionic charge is:

s = u z \frac{d E}{d x}

$s = u z \frac{d E}{d x}$

where z is the ionic charge (e.g., +1 for Na ⁺ , +2 for Ca ²⁺ , −2 for SO ₄ ²⁻ ) and u is a proportionality constant known as the “ionic mobility.” Because dE / d x has units of volts per centimeter (V/cm) and s must have units of centimeters per second (cm/sec) and z is just the number of charges on an ion and therefore is dimensionless, u must have units of (cm ² /sec)/V for all the units to work out in Equation B4.1 .

Knowing the speed of ion movement, we can easily figure out the flux of ions moving under the influence of an electric field. Imagine a cylindrical volume of solution containing the ions of interest ( Fig. 4.2 ). In the figure the electrical potential is more negative at the right, so positive ions (cations) would naturally move toward the right. Because flux is the quantity of ions passing through unit area per unit time, to derive an expression for the flux, J, we need only to find out the quantity of ions flowing through area, A, in a given period of time, Δ t. Because the ions are moving at speed s toward the right, within the period Δ t, any cation within a distance of s × Δ t to the left of the area A would pass through A. The cylindrical volume containing these ions that would pass through A is s × Δ t × A. The number of moles of ions in this volume is thus C × s × Δ t × A, where C is the concentration of the ion. Taking the number of moles of ions that would pass through A and dividing by the area, A, and by the time interval, Δ t, gives the flux of ions driven by the electrical field:

Fig. 4.2, Flux of Positive Ions (cations) Being Driven by an Electric Field, dE/dx .

J_{electr} =- [(C × s ×Δ t × A) /A] / Δ t =- C × s

$J_{electr} =- [(C × s ×Δ t × A) /A] / Δ t =- C × s$

Substituting for s ( Equation B4.1 ) gives:

J_{electr} = - z u C \frac{d E}{d x}

$J_{electr} = - z u C \frac{d E}{d x}$

The minus sign takes into account the fact that for cations ( z being a positive number), ion drift is toward the negative direction of the electric field, whereas for anions ( z being a negative number), ion drift is toward the positive direction of the electric field.

Equation B4.3 bears a strong resemblance to Equation 2.3 from Chapter 2 that describes diffusion flux. Whereas a concentration gradient drives the diffusive flux of molecules or ions, an electrical potential gradient (an electric field) drives the electrical flux of ions.

In view of the foregoing, the total (net) flux of an ion is the sum of the flux caused by diffusion and the flux driven by an electric field:

J_{total} = J_{diffusion} + J_{electr} =- D \frac{d C}{d x} - z u C \frac{d E}{d x}

$J_{total} = J_{diffusion} + J_{electr} =- D \frac{d C}{d x} - z u C \frac{d E}{d x}$

This is the Nernst-Planck equation quantifying the ionic flux driven by a concentration gradient and an electrical potential gradient.

At electrochemical equilibrium the flux driven by the concentration gradient is exactly balanced by the flux driven in the opposite direction by the electrical potential gradient, so the net flux must be zero. Therefore:

J_{total} = J_{diffusion} + J_{electr} = - D \frac{d C}{d x} - z u C \frac{d E}{d x} = 0

$J_{total} = J_{diffusion} + J_{electr} = - D \frac{d C}{d x} - z u C \frac{d E}{d x} = 0$

which means that:

z u C \frac{d E}{d x} =- D \frac{d C}{d x}

$z u C \frac{d E}{d x} =- D \frac{d C}{d x}$

Einstein derived the relationship between the diffusion coefficient ( D ) and the mobility ( u ) of an ion:

D = \frac{u R T}{F}

$D = \frac{u R T}{F}$

where R is the universal gas constant, T is the absolute temperature in Kelvins (Celsius temperature plus 273.15), and F is Faraday’s constant (96,485 coulombs/mole). Using Equation B4.7 , we can rewrite Equation B4.6 :

z u C \frac{d E}{d x} = - \frac{u R T}{F} \frac{d C}{d x}

$z u C \frac{d E}{d x} = - \frac{u R T}{F} \frac{d C}{d x}$

which rearranges to:

\frac{d E}{d x} = \frac{- R T}{z F} \frac{1}{C} \frac{d C}{d x}

$\frac{d E}{d x} = \frac{- R T}{z F} \frac{1}{C} \frac{d C}{d x}$

Equation B4.9 can be integrated across the thickness of the membrane:

\int_{x_{1}}^{x_{2}} \frac{d E}{d x} d x = \frac{- R T}{z F} \int_{x_{1}}^{x_{2}} \frac{1}{C} \frac{d C}{d x} d x

$\int_{x_{1}}^{x_{2}} \frac{d E}{d x} d x = \frac{- R T}{z F} \int_{x_{1}}^{x_{2}} \frac{1}{C} \frac{d C}{d x} d x$

The result of integration is:

\begin{matrix} E_{2} - E_{1} = \frac{- R T}{F} (\ln C_{2} - In C_{1}) \\ = \frac{- R T}{z F} \ln \frac{C_{2}}{C_{1}} = \frac{R T}{z F} \ln \frac{C_{1}}{C_{2}} \end{matrix}

$\begin{matrix} E_{2} - E_{1} = \frac{- R T}{F} (\ln C_{2} - In C_{1}) \\ = \frac{- R T}{z F} \ln \frac{C_{2}}{C_{1}} = \frac{R T}{z F} \ln \frac{C_{1}}{C_{2}} \end{matrix}$

In other words, if a membrane is selectively permeable to a particular ion, and the ion is in electrochemical equilibrium across the membrane, we can calculate the membrane potential, ( E ₂ – E ₁ ), that would be established just by knowing the concentration of the ion on the two sides of the membrane ( C ₁ and C ₂ ).

The membrane potential of a cell is defined to be the potential of the inside relative to the outside (i.e., E _in − E _out ; subscripts 2 and 1 taken to be in and out, respectively). The membrane potential that is established when an ion, X ^z , is in electrochemical equilibrium across the membrane is referred to as the equilibrium potential for that ion and is given the symbol E _X (e.g., E _K , E _Cl , and E _Na are the equilibrium potentials for K ⁺ , Cl ⁻ , and Na ⁺ , respectively). By representing the extracellular and intracellular concentrations of X ^z as [X ^z ] _o and [X ^z ] _i , respectively, we can rewrite Equation B4.11 in a form that is one of the most important equations in cellular physiology—the Nernst equation:

E_{X} = \frac{R T}{z F} ln \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}

$E_{X} = \frac{R T}{z F} ln \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}$

E_{X} = \frac{R T}{z F} ln \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}

$E_{X} = \frac{R T}{z F} ln \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}$

where R is the universal gas constant, T is the absolute temperature in Kelvins (Celsius temperature plus 273.15), z is the electrical charge on the ion (e.g.,+1 for K ⁺ ,+ 2 for Ca ²⁺ , –1 for Cl ⁻ ), and F is Faraday’s constant (96,485 coulombs/mol). Some alternative forms of the Nernst equation that may be more convenient for use in computation are shown in Box 4.2 . The membrane potential ( V _m ) for a cell is defined as the electrical potential inside the cell measured relative to the electrical potential outside . Because the extracellular electrical potential is a reference level against which the intracellular potential is measured, we can define the extracellular electrical potential to be zero (0).

BOX 4.2

Alternative Forms of the Nernst Equation That May Be More Convenient for Calculations

In the Nernst equation for an ion, X ^z , the natural logarithm, ln, can be converted to base-10 logarithm: ln([X ^z ] _o /[X ^z ] _i ) = 2.303·log([X ^z ] _o /[X ^z ] _i ), to give the equivalent expression:

E_{X} = \frac{2.303 R T}{z F} log \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}

$E_{X} = \frac{2.303 R T}{z F} log \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}$

At 37°C, the group of constants RT/F = 26.7 mV. For computation at 37°C, either of the following two forms of the Nernst equation can be used (to give E in units of mV):

E_{X} = \frac{26 .7}{z} ln \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}

$E_{X} = \frac{26 .7}{z} ln \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}$

E_{X} = \frac{61.5}{z} log \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}

$E_{X} = \frac{61.5}{z} log \frac{{[X^{z}]}_{o}}{{[X^{z}]}_{i}}$

Using the concentrations given in Table 4.1 , we calculate the equilibrium potential for K ⁺ at 37°C to be:

E_{K} = \frac{61.5}{+ 1} \log \frac{5}{140} = 61 .5 (- 1 .45) = - 89 .1 mV

$E_{K} = \frac{61.5}{+ 1} \log \frac{5}{140} = 61 .5 (- 1 .45) = - 89 .1 mV$

This is the potential inside the cell relative to the outside, and it is negative, as we deduced earlier.

The Nernst equation can be used to calculate the equilibrium potential for any permeant ion, as long as the inside and outside concentrations for that ion are known. For example, for the ionic distributions shown in Table 4.1 , if the PM were permeable only to Na ⁺ ions, the sodium equilibrium potential, E _Na , for our cell would be:

E_{Na} = \frac{61.5}{+ 1} \log \frac{145}{10} = 61 .5 (+ 1 .16) =+ 71 .5 mV

$E_{Na} = \frac{61.5}{+ 1} \log \frac{145}{10} = 61 .5 (+ 1 .16) =+ 71 .5 mV$

at 37°C. The sign for E _Na is positive, because, as positively charged Na ⁺ ions leak into the cell, down their concentration gradient, they make the inside of the cell more positive while leaving behind a corresponding excess of negative charges on the outside. That is, the inside of the cell becomes more positive relative to the outside; hence E _Na is positive. It is equally straightforward to verify that for chloride ions, E _Cl equals –76.8 mV at 37°C for our cell.

Some ion movement is required to establish physiological membrane potentials. Therefore we are justified in asking whether such movements significantly alter the ion concentrations inside the cell. After all, if ions enter or leave the cell, the intracellular ion concentration must change. In turn, we may ask whether the concentrations used in the Nernst equation should be corrected for the effect of such ion movements. The calculation in Box 4.3 shows that the number of ions that move into or out of the cell to establish a V _m is so small that the cellular ion concentrations are essentially undisturbed.

BOX 4.3

The Ion Movement Needed to Establish a Physiological Membrane Potential Does Not Significantly Change Ion Concentrations Inside the Cell

Realizing that ions must move across the plasma membrane to establish a membrane potential ( V _m ), we may ask whether such ion movements (e.g., leakage of K ⁺ ) will significantly alter the intracellular concentration of the ion of interest. To answer this question, we need to know one important property of biological membranes: the membrane capacitance, C. The capacitance is a measure of the amount of charge, q, that can be separated by the membrane at a given V _m :

C = \frac{q}{v_{m}}

$C = \frac{q}{v_{m}}$

The amount of charge is then just q = C · V _m . The relevant units are the coulomb for electrical charge, the volt for electrical potential, and the farad (symbol F) for capacitance; 1 farad equals 1 coulomb per volt. The capacitance of biological membranes is typically 1 μF/cm ² of membrane area (1 × 10 ⁻⁶ F/cm ² ). A spherical cell with a radius of 10 μm has a membrane surface area of:

A_{cell} = 4 π r^{2} = 1257 μ m^{2} = 1.257 × 10^{- 5} {cm}^{2}

$A_{cell} = 4 π r^{2} = 1257 μ m^{2} = 1.257 × 10^{- 5} {cm}^{2}$

The capacitance for such a cell is:

\begin{matrix} C = 1 × 10^{- 6} {F/cm}^{2} × (1.257 × 10^{- 5} {cm}^{2}) \\ = 1.257 × 10^{- 11} F \end{matrix}

$\begin{matrix} C = 1 × 10^{- 6} {F/cm}^{2} × (1.257 × 10^{- 5} {cm}^{2}) \\ = 1.257 × 10^{- 11} F \end{matrix}$

If the V _m of this cell is equal to the potassium equilibrium potential, E _K = −89.1 mV (i.e., −0.0891 V; see main text), the amount of charge separated by the cell membrane is:

\begin{matrix} q = C \cdot E_{K} = (1.257 \times 10^{- 11} F) × (0.0891 V) \\ = 1.120 × 10^{- 12} coulombs \end{matrix}

$\begin{matrix} q = C \cdot E_{K} = (1.257 \times 10^{- 11} F) × (0.0891 V) \\ = 1.120 × 10^{- 12} coulombs \end{matrix}$

To convert the amount of electrical charge into the quantity of K ⁺ ions that had to move to establish E _K , we make use of Faraday’s constant ( F = 96,485 coulombs/mole; note the distinction between Faraday’s constant, F, and the farad, F):

\begin{matrix} {Amount of K}^{+} moved = q / F \\ = (1.120 × 10^{- 12} coulombs) / \\ (96, 485 coulombs/mol) \\ = 1.161 × 10^{- 17} mol \end{matrix}

$\begin{matrix} {Amount of K}^{+} moved = q / F \\ = (1.120 × 10^{- 12} coulombs) / \\ (96, 485 coulombs/mol) \\ = 1.161 × 10^{- 17} mol \end{matrix}$

To assess whether the leakage of 1.161 × 10 ⁻¹⁷ mol of K ⁺ ions out of the cell significantly lowers the K ⁺ content of the cell, we need to calculate the moles of K ⁺ originally present in the cell. The number of moles of K ⁺ present inside the cell is just [K ⁺ ] _in × Vol _cell . The volume of the cell is:

V o l_{cell} = 4 π r^{3} /3 = 4189 μ m^{3} = 4.189 × 10^{- 12} L

$V o l_{cell} = 4 π r^{3} /3 = 4189 μ m^{3} = 4.189 × 10^{- 12} L$

(1 liter = 10 ¹⁵ μm ³ ). The total amount of K ⁺ initially present in the cell must have been:

\begin{matrix} K^{+} content = (0 .140 mol/L) × (4.189 × 10^{- 12} L) \\ = 5.864 × 10^{- 13} mol \end{matrix}

$\begin{matrix} K^{+} content = (0 .140 mol/L) × (4.189 × 10^{- 12} L) \\ = 5.864 × 10^{- 13} mol \end{matrix}$

The fraction of the K ⁺ content that had to move out of the cell to establish E _K is simply:

\begin{matrix} \frac{{Amount of K}^{+} moved}{K^{+} content in cell} = \frac{1.160 × 10^{- 17} mol}{5.864 × 10^{- 13} mol} \\ = 1.979 × 10^{- 5} \approx \frac{20}{1, 000, 000} = 0.002 % \end{matrix}

$\begin{matrix} \frac{{Amount of K}^{+} moved}{K^{+} content in cell} = \frac{1.160 × 10^{- 17} mol}{5.864 × 10^{- 13} mol} \\ = 1.979 × 10^{- 5} \approx \frac{20}{1, 000, 000} = 0.002 % \end{matrix}$

Thus for every 1 million K ⁺ ions in the cell, roughly 20 must leak out to establish E _K = −89.1 mV. This K ⁺ leakage would diminish the intracellular K ⁺ content by only approximately 0.002%—a negligibly small fraction. Therefore it is clear that the ion movement needed to establish a V _m , although electrically significant, does not change ion concentrations much.

At first sight the magnitudes of the electrical potentials calculated earlier seem somewhat small. However, it must be remembered that these potentials exist across the PM, which is only approximately 5 nanometers thick (1 nanometer = 1 × 10 ⁻⁹ meter). Box 4.4 gives some observations about the V _m and the strength of electrostatic forces acting on oppositely charged ions separated by the PM.

BOX 4.4

The Strength of Electrical Forces and the Principle of Electroneutrality

Typical physiological membrane potentials ( V _m ) are on the order of many tens of millivolts (mV) across a membrane that is approximately 5 nanometers in thickness (1 nm = 10−9 m). Because the electric field, ℰ, is defined as the electrical potential per unit distance, the electric field across the PM when V _m = 80 mV is:

ℰ = \frac{80 × 10^{- 3} V}{50 × 10^{- 10} m} = 16,000,000 \frac{V}{m}

$ℰ = \frac{80 × 10^{- 3} V}{50 × 10^{- 10} m} = 16,000,000 \frac{V}{m}$

This is about a thousand times stronger than typical field strengths used in electrophoresis.

One can appreciate the strength of the electrostatic (or “coulombic”) interaction by calculating the attractive force between oppositely charged ions separated on the two sides of the PM. The magnitude of the electrostatic force, F _{electrostatic} , depends on q, the amount of charge separated by the membrane, and ℰ, the electric field across the membrane:

F_{electrostatic} = \frac{q × ℰ}{2}

$F_{electrostatic} = \frac{q × ℰ}{2}$

In Box 4.3 , it was shown that to establish E _K = −89.1 mV (or −0.0891 V) across the PM of a 20-μm spherical cell, 1.12 × 10 ⁻¹² C of charges are separated on the two sides of the membrane. Using Equation B4.16 , we can estimate the magnitude of the attractive force (in newtons) exerted by the separated charges on each other across the 5 nm thickness of the PM:

\begin{matrix} F_{electrostatic} = \frac{1}{2} \cdot (1.12 \times 10^{- 12} C) \times \frac{0.0891 V}{5 \times 10^{- 9} m} \\ = 9.98 \times 10^{- 6} N \end{matrix}

$\begin{matrix} F_{electrostatic} = \frac{1}{2} \cdot (1.12 \times 10^{- 12} C) \times \frac{0.0891 V}{5 \times 10^{- 9} m} \\ = 9.98 \times 10^{- 6} N \end{matrix}$

We recall from Box 4.3 that the membrane area of the 20-μm spherical cell is 1.257 × 10 ⁻⁹ m ² . Therefore the force per unit area of membrane is 7940 N·m ⁻² . Because each newton (N) is equal to 0.225 pounds of force, this means that for a square meter of membrane area, the attractive force between the charges would be approximately 1800 pounds, or nine-tenths of a ton! Incidentally, this remarkable strength of the electrical forces underlies the Principle of Electroneutrality, which states that, in a solution, the number of positive ionic charges is balanced by an equal number of negative ionic charges, so that overall, the solution carries no net electrical charge.

The most important observation from the preceding discussion is that selective permeability of the PM to ions can have profound electrical consequences. For a typical cell (with ionic concentrations similar to those in Table 4.1 ), if the PM is selectively permeable to K ⁺ , the resulting K ⁺ efflux will tend to drive the cell’s V _m to more negative values. Alternatively, if the PM is selectively permeable to Na ⁺ , the resulting Na ⁺ influx will tend to drive the cell’s V _m to more positive values. The linkage between the V _m and the selective ionic permeabilities of a cell underlies the mechanism by which living cells regulate their electrical properties. This linkage is more fully explored in Chapter 7 .

The stable resting membrane potential in a living cell is established by balancing multiple ionic fluxes

Cell membranes are permeable to multiple ions

The concept of the equilibrium, or Nernst, potential for a particular ion (e.g., Na ⁺ , K ⁺ , or Cl ⁻ ) was developed by examining the fluxes of that ion in an idealized cell whose membrane is permeable only to that ion. The PM of a real cell, however, has moderate to significant permeability to all three common monovalent ions. During the earlier discussion on the Nernst potential, we noted that it is the selective permeability of the PM to various ions that allows the cell to regulate its V _m . Just how does this regulation take place? How do the permeabilities of K ⁺ , Na ⁺ , and Cl ⁻ contribute to the cell’s resting membrane potential ?

If the cell is permeable to all three ionic species, fluxes of all three ions will occur across the PM. As we have seen, ion fluxes into and out of the cell have electrical consequences; namely, they change the V _m of the cell. For example, efflux of K ⁺ tends to drive V _m toward more negative values, whereas influx of Na ⁺ tends to drive V _m toward more positive values. Similarly, influx of Cl ⁻ brings negative charges into the cell and would make V _m more negative. With all these fluxes occurring simultaneously (and “fighting” with each other), eventually a steady state will be established—a steady state in which the cell will have a stable, nonvarying V _m . What is the implication of a stable, nonvarying V _m ? We know that whenever net movement of electrically charged ions into or out of the cell occurs, V _m will change. We can therefore conclude that a stable, nonvarying V _m implies that no net charge movement occurs into or out of the cell in the steady state. In other words, all fluxes tending to make the cell more negative are exactly balanced by fluxes that tend to make the cell more positive.

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Objectives

Ions are typically present at different concentrations on opposite sides of a biomembrane

Selective ionic permeability through membranes has electrical consequences: The nernst equation

The stable resting membrane potential in a living cell is established by balancing multiple ionic fluxes

Cell membranes are permeable to multiple ions

You're Reading a Preview

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