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Ion Channels and Electrical Activity

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Ion Channels and Electrical Activity 10/10/2012 ION CHANNELS AND ELECTRICAL ACTIVITY Colin Nichols Department of Cell Biology and Physiology Center for Investigation of Membrane Excitability Diseases Box 8228 9611 BJC-IH 362-6630 [email protected] http://www.nicholslab.wustl.edu/nichols.htm http:www.cimed.wustl.edu Many of the figures in the following notes come from Hille, B. ‘Ion Channels of Excitable Membranes’, Sinauer Associates, Sunderland, Mass. This is highly recommended to anyone interested in following up the lectures, and is essential reading to anyone interested in pursuing research on ion channel structure and function. Further background material can be found in: Lodish et al., Molecular Cell Biology, 4th ed, chapter 15 (p. 633-665) and chapter 21 (p. 925-965) and Alberts et al., Molecular Biology of the Cell, 4th ed, chapter 11 (p. 615-650) and p. 779-780. Overview These notes provide additional background to the lectures. The first section is an extended discussion of ion channel structure and function, followed by the classical description of the action potential and the role of ion channels in it. 1 1. Electrical principles A. Electrical properties of membranes In order to understand the electrical properties of cells we need to review the following principles of electricity: 1) Current, 2) Voltage, 3) Resistance/Conductance, 4) Capacitance All matter is made up of charged particles - protons and electrons. Charge is symbolized Q and is measured in Coulombs. The elementary charge of one electron or proton is e = 1.602 X 10-19 Coulombs. Faraday's constant (F) is the number of Coulombs per mole of particles that bear a single + or - charge. F = 9.648 * 104 Coulombs / mole. Charged particles move. They attract and repel each other. 1) Current (I) measures the rate of movement of the charge: I = Q / t = Coulombs / sec = Amps (A) 2) Voltage (V) is a measure of the difference in potential energy experienced by a charged particle in two locations. It is the work required to move a charge from point A to point B: V = Joules / Coulomb = Volts (V) 3) Resistance/Conductance Ohm's Law states that Current through a piece of homogeneous material is proportional to the Voltage applied across the material. Conductance (G) is the proportionality factor between current and voltage. The unit for Conductance is the Siemen (S). Resistance (R) is the inverse of conductance. It is measured in Ohms (). G = 1 / R Conductance = 1 / Resistance Ohm's Law I = G V or 1 Amp = 1 Siemen * 1 Volt V = I R 1 Volt = 1 Amp * 1 Ohm Resistance and Conductance depend on the size and shape of the object that you are passing current through. Resistivity (units = cm) is an intrinsic property of a homogeneous material that reflects its ability to carry current. For a right circular cylinder: R = Resistivity (Length / Area of a Cross Section) Sample Calculation - Consider a cylindrical pore 10 Angstroms in diameter and 50 Angstroms long that spans a lipid bilayer. The pore contains saline with resistivity of 60 cm (the resistivity of the bilayer is about 1015 cm). What is the resistance of this pore to axial current? 2 (Remember: 1 Angstrom = 10-10 m = 10-8 cm = 0.1 nm) 60 cm 50 10-8 cm 9 R = 3.1416 (5 10-8 cm) 2 = 4 X 10 or 4 Giga For Resistors "In Series" For Resistors "In Parallel" RTot = R1 + R2 + R3 . + Rn 1 / RTot = (1 / R1) + (1 / R2) + (1 / R3) . + (1 / Rn) Separation of + and - charges produces a potential difference or Voltage. 4) Capacitance (C) is a measure of how much charge must be separated to give a particular voltage. C (Farads) = Q (Coulombs) / V (Volts) or Q = C V And I = Q / t = C V / 2 Since Q = C * V, for a 1 cm region of membrane to be charged to 60 mV would require: -8 1.0 µF * 0.060 V = 6 X 10 Coulombs or 375 X 109 ions or 0.622 pico moles. t The capacitance of a physical object depends on its geometry. For a parallel plate capacitor: C = o Area / distance between the plates where is the dielectric constant of the material between the plates and o is the permittivity of free space (8.85 X 10-12 Coulomb / Volt Meter). Capacitance increases with increasing surface area and decreases as the separation between the plates becomes greater. The cell membrane with saline on both sides is very similar to a parallel plate capacitor. The lipid bilayer of most cells has a specific capacitance of 1.0 µFarad / cm2. By separating charge on either side of the membrane you develop a potential difference across the membrane. Only a small number of charges must be separated to result in a significant voltage. A simple model cell: Consider a spherical cell with several conducting pores. The cell contains saline and is bathed in saline. The equivalent circuit is a capacitor and a resistor in parallel. If we inject a square pulse of current into the cell with a microelectrode, some of it will charge the membrane capacitance and some will pass through the resistance of the conducting pores. ITot = IR + IC where IR = Vm / R and IC = C Vm / t Vm / t = (ITot / C) - (Vm / (RC)) The solution of this differential equation is: Vm = ITot R (1 - exp(-t / )) where = R C is the membrane time constant, the membrane potential of the cell will change along an exponential time course that is governed by . At equilibrium, when t >> Vm = ITot R , where R = 1 / Gpores is called the Input Resistance of the cell. 3 Surface Area of a Sphere of Radius r is A = 2 Sample Calculation - Consider a spherical cell, 4 r 100 µm in diameter, that has 200 open channels, each with a conductance of 10 pS. R = 0.5 GOhms -4 2 -4 Area = 4 3.1416 (50 X 10 cm) C = 3.1 X 10 µF -4 2 t = 155 msec = 3.1 X 10 cm Vm and C = 3.1 X 10-4 µF (mV) GTot = 200 10 pS 8 so Rin = 1 / GTot = 5 X 10 and = Rin C = 155 msec I = 5 pA time (msec) Modeling changes in Vm in a cell membrane Physical diagrams and electrical equivalence circuits showing how current injection from an electrode (left) or current entry through channels selective for sodium (right) can change the membrane potential. 4 The figures below, taken from Jack, Noble and Tsien Electric Current Flow in Excitable Cells (1983), illustrate the change in potential across a resistor and capacitor in parallel. The dashed and dotted lines in the figure on the right show the fraction of total current that is flowing through the resistor (IR) and charging the capacitor (IC), respectively. We said above that IR = Vm / Rm so the time course of IR will be identical to the change in membrane potential. IC = ITot - IR shows that initially all of the current goes to charging the capacitor. As charge builds up on the capacitor, the rate of addition of more charges decreases exponentially. Notice that at the end of the current pulse IC has the opposite sign. B. The Resting Potential An undisturbed cell at rest contains a slight excess of anions that produces a steady membrane potential, called the resting potential. The resting potential is usually in the range from -30 mV to -90 mV or so. Why do cells have ion channels and a resting potential? Cells live in an environment of dilute salt water. They contain within their cytoplasm a variety of impermeant solutes, including proteins and nucleic acids, but also smaller metabolites like amino acids, Kreb's cycle intermediates and so on. The sum total charge of these impermeant solutes is negative. Ion channels allow the cell to cope with the osmotic difficulties that result from confining these large charged ions inside the plasma membrane. The resting potential is an unavoidable consequence of the cell's strategy for handling changes in osmolarity. A cell needs to accommodate two physical facts - 1) The osmolarity of the cell's contents and the solution it is bathed in must be the same, otherwise water will flow into or out of the cell causing it to swell or contract. Cells will tolerate a bit of swelling, but not much. Since the environment may change at any moment, the cell needs to be able to adjust its internal osmolarity quickly. 2) There must be bulk neutrality of the two solutions - separation of tiny amounts of charge produce a substantial voltage across the membrane. An imbalance in the millimolar range is not physically sustainable. Let's examine several possible strategies the cell could use: 1) The cell could simply make its membrane impermeable to everything - this will not work because then slight changes in the external or internal osmolarity would exert great pressure on the membrane 5 2) The cell could be equally permeable to all ions - that won't work either because it would lead to osmotic imbalance. Ions will enter, causing the cell to expand and eventually burst. If impermeable and freely permeable will not work then the membrane has to be selectively permeable to a subset of ions. In order to allow for rapid adjustment of osmolarity, while preserving bulk neutrality, the membrane should be permeable to a cation and an anion. Most cell membranes are selectively permeable to potassium and chloride, but nearly impermeable to sodium ions, at rest. Consider the distribution of ions that might be typical for a frog neuron: (Concentration in mM) Out In Na+ 117 30 K+ 3 90 Cl- 120 4 Anions- 0 116 In addition to osmotic balance and bulk neutrality there is an additional thermodynamic restriction: All permeable ions will move toward electrochemical equilibrium.
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