1. Membrane potentials Thermodynamic treatment: Nernst-Planck flux equation Nernst equationConstant field equation Equivalent electrical circuit: Parallel conductance model Active transport: pumps, carriers

2. Electrochemical potential u i is the electrochemical potential (contribution of one mole of the i th constituent to free energy and is the drive for flux across the membrane) Solute transport occurs from region of higher to lower electrochemical potential For membrane transport: u i = u i o + RTlnC i + z i FV + … z i FV is the contribution of the electrical potential ie the work required to bring a mole of ions with charge z from zero potential to potential V (F = 9.65 x 10 4 coulombs/mole) u i o is the standard electrochemical potential at C= 1M, V=0

3. Equilibrium At equilibrium the electrochemical potential is the same on each side of membrane; u i o + RTlnC 1 + z i FV 1 = u i o + RTlnC 2 + z i FV 2 V 2 -V 1 = E = RT ln C 1 z i F C 2 Nernst potential Z = valence R = gas constant: 8.314 (volts x coulombs/K O x mol) T = absolute temperature F = Faraday constant: 96,500 coulombs/mol

4. Nernst Equation: how it works IN = 100 mM KClOUT = 10 mM KCl = K + = Cl - - + - + - - - + `` `  (-)  (+) A membrane separates solutions of 100 mM and 10 mM KCl: The membrane is permeable to K +, but not to Cl - K + ions diffuse across the membrane from left to right The movement of K + makes the inside negative with respect to outside - this membrane potential prevents further net movement of K + Equilibrium is quickly established; the magnitude of  is computed from the Nernst equation

5. For a real cell Area of cell membrane = 1000 µm 2 (   ) Radius = 8.92 µm Volume = 2974 x 10 -15 liters (4/3  r 3 ) 140 mM K + 5 mM Na + 145 mM Cl - 5 mM K + 140 mM Na + 145 mM Cl - K+K+ V m = 58.17 log [5]/[140] = -87 mV Questions: 1) How many K + ions did the cell lose? 2) Did the K + concentration inside the cell change?

6. How many K + ions did the cell lose? Use Q= CV Know V m = -87 mV Calculate C A = 1000 µm 2 = 10 -5 cm 2 (1 µm = 10 -4 cm) C m = 1 µF/cm 2 x A (cm 2 ) = 10 -6 F/cm 2 x 10 -5 cm 2 = 10 -11 F = 10 pF Q = 0.087 V x 10 -11 F = 8.7 x 10 -13 coulombs Using Faraday’s constant: 8.7 x 10 -13 C / 96,500 C/mol = 8.7 x 10 -18 moles Using Avogadro’s #: 8.7 x 10 -18 moles x 6 x 10 23 molecules/mole = 5 x 10 6 K + ions A single K channel ~2 pA current passes 10 7 ions/sec

7. Did the K + concentration inside the cell change? 140 mM K + = 0.14 moles/liter 0.14 moles/l x 6 x 10 23 molecules/mole = 0.84 x 10 23 molecules/liter Cell volume is 2974 x 10 -15 liters 0.84 x 10 23 molecules/liters x 2974 x 10 -15 liters = 2.3 x 10 11 K + ions in the cell So losing 5 x 10 6 ions is not much K i does not change (in some cases K o can!)

8. Ion equilibrium potentials in some cells

9. Ions are not in equilibrium across membranes

10. Membrane potentials with more than one ion InOut 100 mM KCl10 mM KCl 10 mM NaCl 100 mM NaCl If membrane is permeable only to K + - what is  ? -------- ++++++++ -------- ++++++++ If membrane is permeable only to Na + ? What if the membrane is 10 times more permeable to K + than to Na + ?  V m in this case is somewhat more positive than E k -------- ++++++++  can’t use Nernst equation in this situation – ions are not in equilibrium.

11. Ions are not in equilibrium Need to calculate ion fluxes at steady-state Nernst-Planck Flux Equation J = L dµ/dx Flux (J) proportional to gradient of electrochemical potential J = L d [RTlnC + zFV] dx = L [RT dC + zF dV] C dx dx For a neutral diffusing substance (z=0), flux given by Fick eqn J = -D dC/dx = L RT dC C dx so L = -DC/RT Substituting and rearranging: J = -D dC + C zF dV dx RT dx Nernst-Planck Equation

12. V d µoµo V(x) Constant field approximation: dV = V dx d Integrate Nernst-Planck eqn from 0 to d across membrane: J = -zF D V RT d C (d) e zFV/RT - C (0) e zFV/RT - 1 The concentration of an ion just inside the membrane is related to Its concentration in the bulk solution by a partition coefficient: C (membrane) = k C (bulk solution) Permeability coefficient P= Dk/d J = -zFV P RT C (1) e zFV/RT - C (2) e zFV/RT - 1 J -mol/cm 2 s D= cm 2 /s P= cm/s

13. What is the permeability constant? Comes from flux equation: J = -DA dc/dx J = flux (moles/second) A= area dc/dx = concentration gradient xx CoCo CiCi  big  small dc/dx ~ k (C o -C i )/  x So J = -DA k (C o -C i )/  x or J = -Dk/  x A (C o -C i ) = -P A (C o -C i ) Permeability constant (can be measured) k Is partition coefficient

14. For membrane permeable to Na +, K +, and Cl - : I Na+ = FzJ Na+ I K+ = FzJ K+ I cl- = FzJ Cl- I total = I Na+ + I K+ + I cl- J = mol/cm 2 s F = 9.65 coul/mol I = current density (coul/cm 2 x sec -1 ) When I total = 0 V o = RT F P Na Na o + P K K o + P Cl Cl i P Na Na i + P K K i + P Cl Cl o V o = zero current potential Goldman-Hodgkin-Katz equation

15. J = -zFV P RT C (1) e zFV/RT - C (2) e zFV/RT - 1 Rectification when C(1) C(2) J=0 When C(1)e zFV/RT = C(2) or V=RTlnC(2)/C(1) zF V  ∞, J  zP C (1) FV RT V  -∞, J  - z P C (2) FV RT Limiting cases: Fluxes depend on ion concentrations

16. What is Rectification? Constant field: concentration asymmetry Permeation mechanism: asymmetric barrier to ion entry or block Gating -100-50 0.5 100 0 mV 1.0 G / G max 50 p -100-5050100 -15 -10 -5 5 10 15 nA I = N p i K Na

17. Resting Potential using the Goldman-Hodgkin-Katz eqn [Na] o = 145 mM [Na] i = ~12 mM [K] o = 4 [K] i = 140 mM K+K+ Na + In real cells there are many more open K + channels than Na + channels at rest V m is somewhat more positive than E K E K = -92.6 mv E Na = +64.9 mv V m = -82 mv The Goldman-Hodgkin-Katz equation (constant field equation) Where P K and P Na are the membrane permeability to K and Na, respectively -------- -------- - -

18. The GHK Equation - continued an equivalent formulation of the GHK equation is the following: where The GHK equation is often expanded to include terms for Cl - Note that internal [Cl] is in the numerator, external in the denominator

19. K + -Dependence of Resting Potential membrane potentials measured using microelectrodes depolarization – V m becomes more positive (less negative) than resting potential hyperpolarization – V m becomes more negative than resting potential increasing the external [K + ] depolarizes the membrane

20. Dependence of the resting membrane potential on [K+] o and on the P Na /P K ratio. The blue line shows the case with no Na+ permeability (i.e., P Na /P K = 0). Orange curves describe the Vm predicted by the GHK equation for  > 0. The deviation of these curves from linearity is greater at low [K+]o, where the Na+ contribution is relatively larger.

22. Application to resting and active membrane in squid axon (Hodgkin & Katz, 1949)

23. P Na PKPK Na+ K+K+ P Na PKPK Na+ K+K+ Neurons can have different P Na /P K EKEK E Na EKEK

24. K + o increases K permeability K o = 4 mM P Na PKPK K o = 10 mM P Na /P K = 1P Na /P K = 0.5 Na+ K+K+ V m is halfway between V na and V K V m closer to V K P Na PKPK External K + can modify K conductance of resting membrane K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ K+K+

25. P Na PKPK Na+K+K+ Hyperkalemia in a neuron with a large resting K conductance P Na PKPK K+K+ EKEK E Na E K* hyperkalemia

26. P Na PKPK Na+ K+K+ EKEK E Na Hyperkalemia in a pacemaker cell P Na PKPK Na+ K+K+ hyperkalemia

27. Mg 2+ K o = 4 mM K+K+ K+K+ K+K+ K+K+ Mg 2+ K o = 10 mM K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ Removal of Mg 2+ block accounts for effects of K o on P K Inward K flux clears Mg from channel K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ K+K+ K+K+

28. Resting potential

29. Active ion transport in nerve 24 Na (Hodgkin & Keynes, 1955)

31. A case of too much K + A 35 year old male with history of bipolar disorder, taking no medications, presented to the emergency department complaining of nausea, vomiting, lethargy, and abdominal pain 5 hours after ingesting an unknown number of digoxin tablets He was found partly conscious by a friend who called for emergency medical help. Paramedics found him to be agitated with a pulse of 30. Upon arrival to the ED, vital signs were blood pressure 162/87, pulse of 30 beats per minute, respiratory rate of 22 per minute and temperature of 97 o F. An ECG showed second degree AV block. The patient was intubated and digoxin antibody administered. The next day his digoxin level was reduced from 2.49 ng/ml to 0.69 ng/ml and he was transferred out of the ICU. How does digoxin work? Can the GHK eqn explain the electrical effects of digoxin?