Foundations I Fall, 2016 Ionic Bases of Cellular Potentials
Ionic Bases of Cellular Potentials All living cells have membrane potentials The resting membrane potential comes about from an unequal distribution of permeant ions inside and outside the cell The unequal distribution is made possible by the existence of a cell membrane
I. Physical Bases of The Resting Membrane Potential Early observations suggest that cells possess a surface barrier Cell loses contents when damaged Intracellular injections of dyes spreads throughout cell but not outside Some types of molecules enter cell from outside but others won't
Ernest Overton (1910) Highly lipophillic substances entered relatively easily But some hydrophillic substances also entered, but slowly ergo, hypothesis The Porous Lipid Skin Model pores ~ 6Å the penetration of uncharged molecules (through the membrane) is governed by their size and lipid solubility the penetration of charged substances (through the pores) is governed by their size and charge.
Gorter and Grendel (1925) surface area of monomolecular layer of extracted lipids ~ 2X that of cells they came from "chromocytes [red blood cells] are covered by a layer of fatty substances that is two molecules thick" (Gorter and Grendel 1925).
So the membrane must be a bilayer
Electron microscopic evidence of a bilayer
Singer-Nicholson Fluid Mosaic Lipid Bilayer (1972) The bilayer is not symmetric polar heads hydrophobic tails Most of the mass is not lipid about 1/2 protein, 1/3 lipid and the rest carbohydrate At body temperature the lipid bilayer has the consistency of olive oil
Intrinsic Proteins Surface Proteins glycoproteins - short sugars - Cell surface receptors, ion channels, etc. antigens Proteoglycans long polysaccharides - cell-cell recognition Cytoplasmic Proteins cytoskeleton actin ankyrin spectrin
Biophysical and Thermodynamic Origins of the Resting Membrane Potential
The Principle of Electrical Neutrality (PEN) In a chemical compound or in a solution, the sum of cation charges must equal the sum of anion charges
In 1911 Donnan studied the conditions under which equilibrium is established between two electrolytic solutions separated by a semipermeable membrane-- that is, by a membrane through which the solvent and some, but not all, of the dissolved ions can pass. In the absence of such a membrane, the solvent and every species of dissolved ion will diffuse freely from each solution into the other, until the composition of the two solutions becomes the same. Frederick G. Donnan The semipermeable membrane, however, prevents the transfer of at least one ionic species, and the preservation of electrical neutrality limits the diffusion of that species' oppositely charged partner. Nevertheless, some movement of mobile ions does occur, and the compositions of the solutions change; as a result, the final distribution of the ionic species is unequal, and there is a measurable difference in the electric potential of the solutions on each side of the membrane. This was predicted some years earlier by J.W. Gibbs on thermodynamic grounds and thus the equilibrium is known as the Gibbs- Donnan equilibrium.
Calculation of the Gibbs-Donnan Equilibrium Cl − ] K + [ ] K + ] = [ out in Cl − [ [ ] in out or K + ]*[ Cl − ] K + Cl − [ ]*[ ] = [ out out in in
But... What about the water? At Gibbs-Donnan Equilibrium, the system is in electrical and chemical equilibrium, but not in osmotic equilibrium...
Control of cellular volume X 1. Make cell impermeable to water X 2. Limit volume with a rigid cell wall 3. Balance the osmotic pressure add an impermeant cation (Na+) outside
II. Ionic Basis of the Resting Membrane Potential The plasma membrane is semi-permeable (high permeability to K+ and Cl-, low permeability to most everything else) so ions distribute themselves across membrane in order to satisfy the PEN and the Gibbs- Donnan Equilibrium, at which point the diffusional forces are exactly balanced by the electrical gradient across the membrane. How is the equilibrium point described?
Take a squid giant axon in out mM mM K+ 400 20 Na+ 50 440 Cl- 52 560 proteins- 385 0
Let’s deal with one ion, K+ flux is net rate of movement and is proportional to concentration flux = r [ K + ] out in in where r is the rate constant flux r [ K + ] in = out out where ρ is the net flux, or r [ K + ] out − r netflux = [ K + ] in = ρ K permeability in out r r = if the membrane were not charged, then in out and since in a squid, [ K + ] in = 20 x [ K + ] out then the flux should be 20 x the flux in out
But since the membrane is negatively charged, it is harder for K+ to move out than it is to move in. Thus the two rate constants differ depending on valence, membrane potential and temperature. zV ' − zV ' ( e (1 − e r = ρ zV ' r − 1) = ρ zV ' ) in out where z = valence and V’ = thermodynamic potential, the membrane potential divided by (RT/F), e.g., V’ =Vm/(RT/F) where F= the faraday constant and R is the gas constant V’=-75/25 = -3.02 At room temperature RT/F =25 so for a cell at-75 mV, e -3.02 using equation above for r , = 0.0488 in r in = 3.02 ρ /(0.0488-1) = 3.175 ρ i.e., the inward flow of a positively charged ion is more than 3 times greater than it would be if there were no membrane potential
What conditions obtain at equilibrium? take ln zV ' = ln[ K + ] out [ K + ] in zV ' zV ' netflux = ρ [ K + ] out − ρ [ K + ] in zV ' − zV ' (1 e − ) e ( − 1) substitute back for V’ zv ' multiply by to get denominators the same e zVmF = ln[ K + ] out zV ' e zV ' ρ [ K + ] out − zV ' RT [ K + ] in netflux = ρ [ K + ] in zV ' zV ' ( e − 1) e ( − 1) solve for Vm at equilibrium, net flux=0 so zv ' e [ K + ] out = [ K + ] in Vm = RT zF ln[ K + ] out rearrange [ K + ] in [ K + ] out zV ' e = [ K + ] Nernst Equation
Squid E Eq in out mM mM mV K+ 400 20 -75 Na+ 50 440 +55 Cl- 52 560 -60 Ca++ 0.0001 10 +125 Typical Mammalian Cell E Eq in out mM mM mV K+ 140 5 -90 Na+ 15 145 +61 Cl- 4 110 -89 Ca++ 0.0001 10 +136
The existence of a membrane potential indicates that The PEN is false! By how much? The electrostatic force is so much stronger than the diffusional 18 force (10 times) that for a cell with a resting membrane potential of -80 mV, for every 100,000 intracellular cations there are ~100,001 anions In a small axon the ratio is even less - perhaps 0.000000007 2 A net difference of 600 charges/ µ m gives a membrane potential of 10 mV The violation of the PEN only occurs within +/- 1 µ m of the cell membrane, held in place by the membrane capacitance. The rest of the outside and inside of the cell has no charge. The PEN is true!
Julius Bernstein 1902 “~The resting membrane potential is due to the selective permeability of the membrane to potassium. It should obey the Nernst equation.”
But there is a difference between the calculated membrane potential based on the Nernst equation and what is observed in real neurons at physiological + slope = 58 mV/log[K ] concentrations of extracellular K+ Text Text What does that difference mean? And what can one conclude from the nature of the deviation?
The neuron at rest is not in equilibrium. It is in steady- state and requires energy input to stay there.
Na-K ATPase
Na-K ATPase
Na+ -K+ ATPase In some cells the the transport ratio is 1:1 But in most cells, the ratio is Na+>K+. The ratio is often given as 3:2 Na+:K+, but in nerve and muscle estimates range from 4:3 to 5:1. In other words, the pump generates a current - it is said to be electrogenic The contribution of the pump to the resting membrane potential is given by: Vm = ( RT zF )( r [ K + ] out + b [ Na + ] out + b [ Na + ] in ) r [ K + ] in where r is the pump ratio and b is the ratio of ρ Na+: ρ K+
Other pumps and transporters Na+ -H+ transporter Ca++ - Mg++ - ATPase K+ - Cl- cotransporter (KCC2) - :Cl - HCO 3 (NDAE) Slower than movement through channels but still very fast
So how does one predict the resting membrane potential? Nernst Equation? Vm = RT zF ln[ K + ] out [ K + ] in The Nernst equation only handles one ion at a time
Constant Field Equation start with net flux for an ion, same as for Nernst zV ' e netflux = ρ K zV ' [ K + ] out − [ K + ] in e zV ' − 1 turn flux (moles/cm2/sec) into current by multiplying by Faraday constant - do this for each ion zV ' e = F ρ Na zV ' [ Na + ] out − [ Na + ] in zV ' e i = F ρ K zV '[ K + ] out − [ K + ] in i zV ' Na e − 1 K e zV ' − 1 zV ' e = F ρ ClzV '[ Cl − ] out − [ Cl − ] in i zV ' Cl e − 1 at equilibrium the membrane potential is not changing (constant field) so the sum of all the currents must equal 0 zV ' zV ' zV ' e e e ρ K ([ K + ] out − [ K + ] in ) + ρ Na ([ Na + ] out − [ Na + ] in ) + ρ Cl ([ Cl − ] in − [ Cl − ] out ) = 0 rearranging gives e zV ' ( ρ K ([ K + ] in + ρ Na ([ Na + ] in + ρ Cl ([ Cl − ] out ρ K ([ K + ] out + ρ Na ([ Na + ] out + ρ Cl ([ Cl − ] in =
Recommend
More recommend
