Ionic Equilibrium [Na + ] = 150 mm [Cl - ] = 150 mm

Transcription

1 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. 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 intracellular extracellular [Na + ] = 15 mm [Cl - ] = 15 mm [Na + ] = 150 mm [Cl - ] = 150 mm 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 d C d x where ion flux due to diffusion is j D, x (mol/sec-cm 2 ), C (mol/cm 3 ) is the ionic concentration expressed as a scalar field, and D (cm 2 /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 ( ) not only developed this law of diffusion, he also devised the first technique to measure cardiac output and his nephew invented the contact lens!

2 Ohm's Law describes ion flux due to an electric field: j E, x = Z C Z where ion flux due to an electric field is j E, x (mol/sec-cm 2 ), C (mol/cm 3 ) is the ionic concentration at the point of interest, Z is the valence of the ion in question (ie, +1 for Na + ), and (cm 2 /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: or, D= RT Z F = 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 d d x RT F = =25.8 mv at 27 C Substituting the expression relating and D into the equation for j E, x results in the following: j E, x = D Z F R T Z Z C d d x j E, x = D Z F C RT d d x Now, we can sum the two ionic fluxes to find the total flux, j x : j x = D d C d x D Z F C R T 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-cm 2. 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/cm 2 ), in terms of the ionic flux: J x =F Z j x = D Z F d C d x Z C F d RT d x d d x This is the Nernst-Planck equation and it applies to each ion individually.

3 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 d x Z C F d R T d x =0 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. right left d C d x = Z C F d R T d x d C C = Z F R T d right dc C = ZF d RT left ln C %right %left = ZF RT Φ %right %left ln C right ln C left = ZF RT right left 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). lnc e ln C i = ZF RT e i ln C e C = ZF i RT e i RT ZF ln C e C = i e i 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 = RT ZF ln C e C i 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:

4 E A = R T Z A F ln [ A] e [ A] i For our example, we can calculate the sodium Nernst potential at a temperature of 27 C : E Na = 1 R T Z Na F ln [ Na + ] e [Na + ] = 25.8mV i 1 ln 150mM 15 mm =59.4 mv The sodium Nernst potential is the transmembrane potential at which there is no net flux of sodium ions across the membrane. + V m - intracellular Φ i [Na+] = 15 mm [Cl-] = 15 mm extracellular 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, E Na. At equilibrium, the intracellular space has a net positive charge and the extracellular space has a net negative charge. Φ e [Na+] = 150 mm [Cl-] = 150 mm

5 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: E Na = R T F ln [ Na+ ] e [ Na + ] i E K = R T F ln [K + ] e [ K + ] i E Cl = R T F ln [Cl- ] i [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: V m = R T F ln P Na [Na + ] e P K [ K + ] e P Cl [Cl - ] i P Na [ Na + ] i P K [ K + ] i P Cl [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 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.

6 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 K Cl 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: P K : P Na : P Cl = 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?

7 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 K Cl 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.

8 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, ϵ 0 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 A = k ϵ 0 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 ϵ 0 d =(3.0)(8.854 x ) 3x10 9 = F /m 2 1.0μ F/ cm 2 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

9 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 K 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

10 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, ) 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 :

11 E Na = 1 RT Z Na F ln ( [ Na+ ] e ) mv =25.8 [ Na + ] i +1 ln ( 150mM 15 mm ) =59.4 mv 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. + V m - 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 V m (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, R m, 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

12 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 dv m dt dv V m +R m C m m E dt Na =0 dv m = 1 (V dt R m C m E Na ) m The solution to this equation with an initial condition of V m (t=0)=0 is: V m (t)=e Na (1 e t/ R mc m ) I Na = E Na R m e t / R m C m 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. Transmembrane potential Time Ionic current Time Figure 8 Voltage and current transients in the concentration cell from Figure 6 with sodium chloride added at t = 0..