Download Ionic Equilibrium

BENG 171 - Bioelectricity
Ionic Equilibrium
(more or less inspired by Plonsey and Barr, Bioelectricity, Kluwer Academic/Plenum Publishers, 1988)
In order to understand the flow of ions across the cell membrane in a nerve, we will first look at a
simpler problem. We will analyze the flow of ions across a semi-permeable membrane in a
concentration cell. A concentration cell is an electrochemical cell that has two equivalent half-cells of
the same material differing only in concentrations (from Wikipedia). In our case, consider a
concentration cell with a membrane separating the two half cells that is permeable to only one
particular type of ion and not to others. In the concentration cell below we dissolve 8.7 g of NaCl
(molecular weight 58 g/mol) into 1.0 L of deionized water on the right half cell and 0.87 g of NaCl into
1.0 L of deionized water in the left half cell. The fluid on the right side has a sodium concentration of
[Na+] = 150 mM and the fluid on the left side has {Na+] = 15 mM. The membrane is permeable to Na +
but not to Cl-. Here, the left side of the cell represents the inside of a cell and the left side represents
the interstitial space outside of the cell.
intracellular
+
+ - +
-
extracellular
+
- + + +
+ + - + - -
[Na+] = 15 mM
[Cl-] = 15 mM
[Na+] = 150 mM
[Cl-] = 150 mM
Figure 1 – Initial configuration of a concentration cell with sodium chloride solutions
and a membrane permeable only to sodium ions. We will treat the left side as the
inside of a cell and the right as the interstitial space outside the cell.
Since the membrane is semi-permeable, sodium ions will begin to diffuse from the region of high
concentration to the region of low concentration. In our example, this means that they will move from
right to left in the x direction. In the discussion that follows, we will restrict our analysis of
movement to the x direction. However, the theory can be generalized easily to three dimensions with
a more thorough use of vector calculus.
Diffusion of a single ion
In general ionic equilibrium involves the movement of ions due to two factors, a concentration gradient
and an electric field. Movement of ions in the x direction due to a concentration gradient is described
by a one dimensional version of Fick's Law 1:
j D , x=−D
dC
dx
where ion flux due to diffusion is j D, x (mol/sec-cm2), C (mol/cm3) is the ionic concentration
expressed as a scalar field, and D (cm2/sec) is the diffusion constant which describes how easily this ion
can move in solution. This type of ionic diffusion is due to Brownian motion of molecules in solution
and, therefore, the diffusion constant, D , is related to temperature.
1 Adolf Fick (1828-1901) not only developed this law of diffusion, he also devised the first technique to measure cardiac
output and his nephew invented the contact lens!
BENG 171 - Bioelectricity
Ohm's Law describes ion flux due to an electric field:
j E , x=
− Z C d 
∣Z∣ d x
where ion flux due to an electric field is j E, x (mol/sec-cm2), C (mol/cm3) is the ionic concentration at
the point of interest, Z is the valence of the ion in question (ie, +1 for Na +), and  (cm2/V-sec) is the
mobility constant which describes how easily this ion can move in solution.
Mobility and diffusivity sound like they should be very similar quantities. The terms arose from
scientists working in different fields of specialization. In a 1905 paper, Einstein related the variables
that had been used to describe the two types of ion movements:
D=
 RT
∣Z∣F
or,
=
D ∣Z∣F
RT
where T is the temperature (in Kelvin degrees), R is the universal gas constant (8.314 J/K ˚ mole at
27 ˚ C), F is Faraday's constant (96487 coulombs/mol). Recall that the diffusion constant is related to
Brownian motion of molecules in solution. Einstein's equation quantified this relationship between
diffusivity and temperature. Many measurements are made at room temperature (27 ˚ C or 300 K)
where
RT 8.314×300
=
=25.8 mV at 27 ˚ C 
F
96487
Substituting the expression relating  and D into the equation for j E, x results in the following:

j E , x=−
j E , x=

D∣Z∣F Z d 
C
R T ∣Z∣ d x
−D Z F C d 
RT
dx
Now, we can sum the two ionic fluxes to find the total flux, j x :
j x =−D
dC DZFC d
−
dx
RT d x
It is helpful to note here that ionic flux is in units of mol/sec-cm 2 whereas current density is expressed
in units of coulombs/sec-cm2. The number of coulombs in a mole of ion is a product of the charge on a
mole of protons (Faraday's constant, F) and the valence of the particular ion of interest (Z). Thus, we
can write the current density in the x direction, J x (A/cm2), in terms of the ionic flux:
J x =F Z j x=−D Z F

d C ZC F d

dx
RT d x

This is the Nernst-Planck equation and it applies to each ion individually.
BENG 171 - Bioelectricity
Equilibrium state of a single ion
Equilibrium is defined as the state where the ionic flux across the membrane due to the concentration
gradient is equal to the ionic flux due to the electric field. In other words, for a single ion to be in
equilibrium, the net ionic flux for that ion must be zero, j x=0 . Thus,
d C ZC F d

=0
dx
RT dx
From this equation, we can derive a relationship between concentration and the electrical potential
difference between the left and right sides of our concentration cell when the ion is in equilibrium.
d C −Z C F d 
=
dx
RT d x
d C −Z F
=
d
C
RT
right
∫
left
dC −ZF
=
C
RT
%right
ln C∣%left =
right
∫ d
left
−ZF ∣%right
Φ %left
RT
ln C right −ln C left =
−ZF
 right − left 
RT
Recall, the left side of our concentration cell represents the intracellular space (denoted by a subscript i)
and the right represents the extracellular space (denoted by a subscript e).
ln C e−ln C i =
ln
 
Ce
Ci
=
−ZF
 e−i 
RT
−ZF
e−i 
RT
 
C
RT
ln e =i − e
ZF
Ci
Finally, we note that the difference in potential across the cell membrane is referred to as the
transmembrane potential, V m =i − e . When the permeable ion is in equibrium, the transmembrane
potential difference will be:
V m=
 
C
RT
ln e
ZF
Ci
This equation describes the Nernst equilibrium state for a single ion. As we continue our study of cell
membrane dynamics we will encounter several ions of interest. To differentiate the Nernst equilibrium
potential of each ion, we will adopt the following notation. The Nernst equilibrium potential for some
ion A is expressed as:
BENG 171 - Bioelectricity
E A=
 
[ A]e
RT
ln
Z AF
[ A]i
For our example, we can calculate the sodium Nernst potential at a temperature of 27 ˚ C :
 

+

[ Na ]e 25.8 mV
1 RT
150 mM
E Na =
ln
=
ln
=59.4 mV
+
Z Na F
1
15 mM
[Na ]i
The sodium Nernst potential is the transmembrane potential at which there is no net flux of sodium
ions across the membrane.
+ Vm
intracellular
Φi
[Na+] = 15 mM
[Cl-] = 15 mM
+
+
+
+
+
+
-
extracellular
- Φ
e
- [Na+] = 150 mM
[Cl-] = 150 mM
Figure 2 – Measuring the voltage across the semi-permeable membrane. If the
membrane is permeable to sodium ions only, the transmembrane potential
measured after ion flux is allowed to reach equilibrium is the sodium Nernst
potential, ENa. At equilibrium, the intracellular space has a net positive charge
and the extracellular space has a net negative charge.
BENG 171 - Bioelectricity
Equilibrium for multiple ionic species
Suppose a concentration cell contains multiple ionic species (e.g., Na +, K+, Cl-) and the membrane is
permeable to each in varying degrees. Permeability is a function of the number and state (ie, open or
closed) of a number of types of ionic channels and ion pumps. The units of permeability are cm/sec.
You might think of permeability as a quantity similar to the velocity that ions will move through the
pores in the membrane. The higher the permeability, the faster the ions will flow. The faster the ions
flow through the pores, the larger the ionic current through the membrane will be. Permeability is
measured independently of the driving forces (electric field and concentration gradient) which actually
cause ions to flow through the pores in the membrane. Ionic current is actually a function of both the
permeability and the driving force.
Our imaginary concentration cell has three ionic species within the two chambers and each ionic
species has its own Nernst potential:
 
 
 
[ Na+]e
RT
E Na=
ln
F
[ Na+ ]i
[K + ]e
RT
EK=
ln
F
[ K + ]i
E Cl=
[Cl - ]i
RT
ln
F
[Cl - ]e
Note, the chloride ion has a slightly different form due to its negative valence number. When the
membreane is permeable to all three ions, the equilibrium condition was described by Goldman,
Hodgkin, and Katz:

PNa [Na+ ]e PK [ K + ]e PCl [Cl - ]i
RT
V m=
ln
F
P Na [ Na+ ]i P K [ K + ]i PCl [Cl -]e

This is usually called the Goldman equation and the equilibrium potential may be referred to as the
Goldman equilibrium potential 2. This is the membrane potential at which the net flow of ions across
the cell membrane is zero. This does not necessarily mean that the net flow of sodium or potassium
ions is zero; it means only that the sum of all ionic currents is zero.
2 Alan Lloyd Hodgkin and Bernard Katz each won Nobel prizes for their contributions to the field of neurophysiology Hodgkin in 1963 and Katz in 1970. David Goldman merely earned his Ph.D. for deriving this significant equation. At
least we are still reminded of his efforts by the common name of the equilibrium equation.
BENG 171 - Bioelectricity
Membrane biophysics review questions
1. The major external and internal ion concentrations for the Aplysia giant nerve cell are
approximately:
Ion
Inside (mmol/l)
Outside (mmol/l)
Na+
61
485
K+
280
10
Cl-
51
485
a) Determine the Nernst potential for sodium, potassium, and chloride. (Express as V m, potential
inside minus potential outside.)
b) Is there a transmembrane potential for which all ions are in equilibrium?
c) The resting potential for the cell is -49 mV. Which ions are in equilibrium and which are not?
For the latter, in what direction do the ions move? What must be true of the net ion movement?
Why?
2. Using the concentrations above, assume that the permeabilities for the resting membrane are in the
ratio:
PK : PNa : PCl = 1 : 0.12 : 1.44
a) Use Goldman's equation to provide the appropriate mathematical expression for the resting
potential of the cell in terms of the given quantities.
b) Evaluate this expression to determine the numerical value of the resting potential in appropriate
units.
c) Assume the ratio of permeabilities is now 0.01 : 1.00 : 0.01 and calculate the new resting
potential.
3. Transmembrane potential arises from two main forces. Name each and identify the mathematical
equations through which each quantifies ionic flux. What mathematical expression evaluates the
flow due to the sum of these two forces?
BENG 171 - Bioelectricity
4. The concentrations of various ions are measured on the inside and outside of a nerve cell. The
following values are obtained at rest when the potential inside the cell is -72 mV with respect to the
outside.
Ion
Inside (mmol/l)
Outside (mmol/l)
Na+
15
145
K+
150
5
Cl-
9
125
Comment on the ability of each of the ionic species to pass through the cell membrane when the cell
is at rest. Be sure to give numerical evidence to support your answer. Assume T=300K.
5. Show that the units balance in the Goldman equation.
BENG 171 - Bioelectricity
Membrane capacitance
The cell membrane is composed of a lipid bilayer that is approximately 3x10 -9 m in thickness. It is able
to maintain a transmembrane voltage by separating charge on either side of the membrane. Thus, it
functions as a capacitor. The definition of capacitance for a parallel plate capacitor is:
C=
k 0 A
d
where k is the relative permittivity of the material separating the two plates of the capacitor, ϵ0 is the
permittivity of free space (8.854x10 -12 F/m), A is the area of the capacitor plates, and d is the separation
distance of the two plates. Specific membrane capacitance is
c m=
C k ϵ0
=
A
d
If we assume that the relative permittivity of the lipid bilayer is similar to that of oil (k=3.0) than we
can estimate the specific membrane capacitance.
c m=
k ϵ0 (3.0)(8.854 x 10−12 )
=
=0.0089 F /m2 ≃1.0 μ F/ cm2
d
3x10−9
3 nm
Figure 3 – The plasma membrane of a cell is composed of a lipid bilayer. Ion channels
facilitate the transport of ions from one side of the membrane to the other.
Selectivity of ion channels
Ion channels are protein structures which loop back and forth through the plasma membrane to form a
pore. The pore allows ions to pass from one side of the plasma membrane to the other. Most channels
will allow only one ion to pass through the pore by filtering out all molecules that do not have a
particular size and ionic charge. An ion in an aqueous solution attracts either the negative or positively
charged region of the water dipolar molecules which form a sphere of hydration around the ion. For an
ion to pass through an ion channel, either the sphere of hydration must fit through the pore, or the
channel must be able to strip off one or more of the water molecules through an interaction with the
BENG 171 - Bioelectricity
charged regions of the channel protein.
Figure 4 – Water forms a sphere of hydration around an ion through the relatively weak
attraction of one side of the dipolar water molecule to the positive or negatively
charged ion in solution.
Consider the problem of selectively allowing potassium ions to pass through a channel while filtering
out sodium ions. They are both positive, monovalent ions which means that they can't be differentiated
by an electrostatic method. The following table gives some properties of the sphere of hydration
surrounding each ion.
Table 1 – Hydration properties of sodium and potassium ions
Sphere of hydration
Free energy of
Ion
radius (Å)
hydration (kcal/mol)
Na+
0.95
298
K+
1.33
280
Potassium channels have an internal structure which binds to the potassium ion in an intermediate step
before allowing the ion to pass through the channel. The energy inherent in this bond is sufficient to
strip away a part of the sphere of hydration in potassium, but not the more tightly bound water
molecules around a sodium ion. Thus, only potassium ions will pass through the potassium channel.
Similarly, the pore in the sodium channel is large enough to pass a sodium ion through (with its sphere
of hydration) but not the larger sphere of hydration surrounding the potassium ion.
Voltage-gated of ion channels
A number of the ion channels that we focus on in this course have voltage-controlled gates which open
to allow ions to pass through the channel or close to block ion flows. Below is an image from a review
article which shows the shape of the membrane protein and illustrates how the sensor portion of the
channel gate operates. The positively charged sensor region of the protein chain is positioned towards
the intracellular side of the membrane at rest. The negative charge which polarizes the membrane at
rest maintains this orientation. When the membrane is depolarized, sodium ions rush into the cell
causing a net positive charge to lie on the intracellular side of the membrane. This positive charge
BENG 171 - Bioelectricity
repels the sensor region of the channel protein which in turn opens the potassium channels. The open
channels allow potassium ions to leave the cell, thus repolarizing the membrane.
Figure 5 – Structure of a voltage-gated potassium channel (PDB 2A79). (A) Side view
showing two of the pore forming domains (cyan and purple) with the associated
ion binding sites in the filter (green spheres) and two of the voltage sensing
domains (orange and green). The four arginine residues responsible for voltage
sensing are shown in ball and stick representation. The approximate location of
the membrane is shown by black lines. Voltage sensor and pore forming domains
are shown from different subunits so no connection between the two is illustrated.
(B) Top view of the complete transmembrane protein structure with the four
subunits indicated in different colours. The voltage sensor from one subunit sits
adjacent to the pore-forming domain of the next subunit. (C) Schematic depiction
of voltage gating. In the closed conformation the intracellular end of the pore is
occluded. Upon membrane depolarisation the voltage sensor moves slightly or
undergoes a conformational change such that the intracellular end of the pore is
opened. (From Corry, “Understanding ion channel selectivity and gating and
their role in cellular signalling, Mol. BioSyst., 2006, 2, 527–535)
Transient analysis in a concentration cell
In the previous analysis of a concentration cell, we used the Nernst equation to calculate the steady
state transmembrane potential for a membrane permeable to a single ion. We calculated the sodium
Nernst potential at a temperature of 27 ̊ C :
BENG 171 - Bioelectricity
( )
[ Na+ ]e 25.8 mV
1 RT
150 mM
E Na=
ln
=
ln
=59.4 mV
+
Z Na F
+1
15 mM
[ Na ]i
(
)
When the solutions are first introduced into the cell, there is no net charge on either side of the
membrane and the transmembrane potential is zero: V m(t=0) = 0.0 mV.
+ Vm
intracellular
-
extracellular
[Na+] = 15 mM
[Cl-] = 15 mM
[Na+] = 150 mM
[Cl-] = 150 mM
Figure 6 – Concentration cell in which the membrane is permeable to sodium ions only.
The initial value of the transmembrane potential when the solutions are first
introduced is Vm(t=0) = 0. Some time later the system will equilibrate and the
steady state transmembrane potential is equal to the Nernst potential.
We have already discussed the capacitive properties of the cell membrane. Now we will develop a
circuit representation of the ionic channels which includes a Nernst-like driving force to create an ionic
current through the channel. While the channels allow ions to pass through the membrane, they are a
restrictor of ionic flow; the analogous circuit element for an ionic channel would be a resistor. The
Nernst-like driving force is represented in a circuit as a battery which is used in conjuntion with the
membrane resistance, Rm, to charge the capacitive membrane.
Figure 7 – The equivalent circuit representation of a cell membrane with an embedded
ionic channel..
To understand the transient behavior of this equivalent circuit, we can apply some basic circuit analysis
techniques. First, we write KVL and KCL equations:
V m – I i R m −E r =0
BENG 171 - Bioelectricity
I i +I c =0
In these equations, I i is I Na , the potassium ionic current and E r is E Na the potassium Nernst
potential. Using the explicit definition of the potassium current, we can derive the following:
I Na=−C m
V m+ Rm Cm
dV m
dt
dV m
−E Na =0
dt
dV m
−1
=
(V −E Na )
dt
Rm C m m
The solution to this equation with an initial condition of V m (t=0)=0 is:
−t / R m C m
V m (t)=E Na (1− e
I Na=
−E Na −t / R
e
Rm
m
)
Cm
60
0.1
50
Ionic current
Transmembrane potential
With E Na = +59 mV, when the sodium chloride solution is added to the concentration cell, sodium
current begins to flow from the extracellular space to the intracellular space, a negative current flow.
The results of this thought experiment are shown in the figure below.
40
30
20
10
0
0
-0.1
-0.2
-0.3
0
Time
0
Time
Figure 8 – Voltage and current transients in the concentration cell from Figure 6 with
sodium chloride added at t = 0..