NACS641-apendix-RA - Some useful equations Current I =...

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Unformatted text preview: Some useful equations Current I = (Coulombs/second) = Amperes (A) Ohm’s law V= IR Capacitance C = Q/V (Coulombs/Volts) = (F) Voltage across capacitor V = Q/C Changing the voltage in a capacitor ΔV = ΔQ/C We change the charge by passing current Ic = ΔQ/Δt The change in V depends on the duration of Ic ΔV = Ic∗Δt/C Friday, September 10, 2010 Capacitor + + _ A= Area Q ++ _ _ ++ _ + d = distance (separation of the plates) Q + Capacitance C = Q/V Coulomb/Volt or Farads (F) C =ε ∗A/d ο Friday, September 10, 2010 εο electrostatic permittivity A= C d= C Capacitor rubber membrane Water pressure Small Capacitance Friday, September 10, 2010 Release of pressure Capacitors in parallel a dd larger Capacitance Ion channels as resistors (conductors) Internal conducting solution (ions) External conducting solution (ions) Ion channels Voltage-gated, NT-gated etc. Friday, September 10, 2010 A = Area l = length Ohm’s law R = V Ohms (Ω) /I l= R R = ρ∗l/A ρ resistivity A= R For the same current, a larger R produces larger V Friday, September 10, 2010 Resistors R1 As the # of Rs in parallel increases RT For ion channels is better to think in terms of conductance R1 = 1/g1 decreases! R1 1/RT = 1/R1 +1/R2 More (open) channels in the membrane more conductance R2 g T = g 1 +g 2 R1 R2 RT = R 1 + R 2 Friday, September 10, 2010 Long, thin parts of a neuron have large resistance! Also remember... Current likes to flow through the path w ith less resistance R = 100 Ω IT And I2 I1 R=1Ω IT = I 1 + I 2 Friday, September 10, 2010 Specific membrane resistance cross section of a cell To compare cell with different sizes The specific membrane resistance (resistance per area) 2 RM = Ω∗cm depends on the # of channels per cm More channels make RM smaller For a spherical cell 2 Rin = RM /4πa a = radius Rin determines how much the cell depolarizes in response to a steady current Friday, September 10, 2010 2 Example same RM = 2000 2 Ω∗cm a a Cell diameter is 5 μm Cell diameter is 50 μm -4 a = 25 μm = 25∗10 cm Rin = 2 RM/4πa 2 2/4π(25∗10 -4 Ω∗cm cm) Rin = 2000 Rin = 25 MΩ Friday, September 10, 2010 your numbers here... Rin = 637 MΩ Rin is larger in a smaller cell Specific membrane capacitance of biological membranes 2 CM = 1 μF/cm CM = Q/V -6 Q = 10 C/V∗ 0.08 V -8 2 Q = 8∗10 C/cm 5 Faraday constant ≈ 10 Coulombs/mole 23 -1 Avogadro’s number = 6.02∗10 mole 11 2 Then this is 4.8∗10 ions/cm Friday, September 10, 2010 Specific membrane capacitance of biological membranes 2 CM = 1 μF/cm For a cell at -80 mV how many ions is this? CM = Q/V -6 Q = 10 C/V∗ 0.08 V -8 2 Q = 8∗10 C/cm 5 Faraday constant ≈ 10 Coulombs/mole 23 -1 Avogadro’s number = 6.02∗10 mole 11 2 Then this is 4.8∗10 ions/cm Friday, September 10, 2010 Specific membrane capacitance of biological membranes 2 CM = 1 μF/cm For a cell at -80 mV how many ions is this? CM = Q/V -6 Q = 10 C/V∗ 0.08 V -8 2 Q = 8∗10 C/cm 5 Faraday constant ≈ 10 Coulombs/mole 23 -1 Avogadro’s number = 6.02∗10 mole 11 2 Then this is 4.8∗10 ions/cm Friday, September 10, 2010 Is this a lot??? Let’s assume the cell is 50 μm in diameter -4 a = 25 μm = 25∗10 cm 50 μm 2 surface of sphere A = 4πa -5 cm2 A = 7.85∗10 11 2 4.8∗10 ions in 1 cm 7 So total is 4∗10 ions -8 3 The volume of this cell is 6.55∗10 cm Then this number of ions is -6 10 M If KCl inside is 120 mM this means that only 1/120,000 ions is in excess! Friday, September 10, 2010 Small cell small capacitance Large cell large capacitance “CM” is the same (same membrane) For a spherical cell, the input capacitance 2 Cin = CM∗4πa a = radius More charge (current) is required to change the voltage across a larger cell Friday, September 10, 2010 Solving for Vm a) INa + IK = 0 b) Vin-V0ut = Ek+ IK/gk c) Vin-V0ut = ENa+ INa/gNa http://heart.med.upatras.gr/Prezentare_adi/electric_memb.jpg Vout +++ I --- INa K Vin Vin-V0ut =Vm - http://www.unm.edu/~toolson/equivcirc.gif + d) Ik = gk ( Vm- EK) - e) INa = gNa ( Vm- ENa) Vm = (ENagNa + EKgK) gNa + gK Friday, September 10, 2010 In summary Rin = RM /4πa 2 Cin = CM∗4πa 2 τ = Rin∗Cin The product of input Capacitance and Resistance (τ) determines the time it takes to change the potential Notice that (τ) is not affected by “a” Friday, September 10, 2010 Current in axons and dendrites Current pulse rm Section of axon or dendrite of determined length (x). In this case 1 cm. ra ra axial resistance (Ω/cm) rm membrane resistance (Ω∗cm) Friday, September 10, 2010 Axial resistance increases with distance (x) Total axial resistance ra1 ra2 ra3 ra4 rX = ra∗x (x = 4) (remember RT = R1 + R2) Near the site of injection, the current flows through rm (less resistance) VO Then VO = Im ∗ rm Friday, September 10, 2010 For 1 cm of cytoplasm (dendrites or axon) ρ (Ω∗cm) 3 resistive property of 1 cm of cytoplasm (dendrites or axon) 2 ra = ρ/(πa ) rm = Rm/(2πa) a = radius 1 cm λ =√ (rm/ra) λ =√ R ∗a m 2ρ If R and ρ are constant m λ =√ K a The length constant is proportional to the square root of the radius of the process For neurons is usually 0.1 to 1 mm Friday, September 10, 2010 Changes in length constant Given a dendrite or an axon with the same diameter (a, b, c) If the length constant increases, the potential decreases less with distance. Friday, September 10, 2010 ...
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