BME 6938 Neurodynamics Instructor: Dr Sachin S Talathi The cell membrane-equivalent circuit Pore Resistance The bilipid layer: Capacitance The First ODE-For XPPAuto Passive Membrane with time dependent input current dVm CM = gM ( E R - Vm ) + I(t) dt Look up nice tutorial on using xppauto on bards webpage at http://www.math.pitt.edu/~bard/bardware/tut/start.html XPPAuto ODE File Comment Define Parameters The ODE Initial Conditions The Function Aux File End of File # passive membrane with step function #current: passive.ode parameter R_M=10000, C_m=1, I_0=2, E=70 parameter t_on=5, t_off=10,Vm=0 dV/dt = (1000*(Vm-V)/R_M + I_0*f(t))/C_M V(0)=0 # define a pulse function f(t)=heav(t_off-t)*heav(t-t_on) # track the current aux ibar=f(t)*I_0 done The Cable Theory for Passive Cell Use the mathematical frame work of linear cable theory and the elec. circuit representation of neuronal cell membrane to understand how membrane potential is affected in function of neuronal cell geometry. Important to understand concepts like synaptic integration Assumptions: Membrane parameters are linear and independent of mem. potential (passive) ; current entering the cable flows linearly (homogeneous); resistance of extracellular medium is zero (cell immersed in homogeneous isopotential medium, the reference) Kirchoff’s Current law applied to cable dV j V j - E r V j -1 - 2V j + V j +1 j Cm + = + Iext dt Rm Ra Cm: Membrane Capacitance (F) Rm: Membrane Resistance (Ohm) Ra: Axial Resistance (Ohm) Iext: External current (Amp) Ijext The Cable Equation dV (x,t) V (x,t) - E r d 2 ¶ 2V (x,t) pdCM + pd =p + pdiext (x,t) 2 dt RM 4RA ¶x CM: Specific Capacitance (F/cm2) RM: Specific Resistance (Ohm-cm2) RA: Specific Axial Resistance (Ohm-cm) iext: Current density (Amp/cm2) The Cable Equation: Rescaling Variables ~ dV (X,T) ¶ 2V (X,T) + V (X,T) = i ext (X,T) 2 dT ¶X x ® lX t ® tT l= dRM 4RA t = CM RM Recap The Cable Equation IM IL d ~ dV( X,T) ¶ 2V( X,T) + V( X,T) = RM i ext ( X,T ) 2 dT ¶X x ® lX IM: Membrane Current (Amp/cm2) CM: Specific Capacitance (F/cm2) RM: Specific Resistance (Ohm-cm2) RA: Specific Axial Resistance (Ohm-cm) RM rm = pdl cm = pdlCM ra = 4RA l pd 2 t ® tT l= dRM 4RA Space Constant t = CM RM Time Constant The Cable Equation: Steady State ~ ¶ 2V (X) - V (X) = - i ext (X) 2 ¶X Longitudinal Current: Input Resistance V = Ra IL Im V (x) V (x,t) - V (x + Dx,t) = Ra IL (x,t) = RL Dx V (x+Δx) Il (x) V (x + Dx,t) - V (x,t) = -RL IL (x,t) Dx ¶V (x,t) = -RL IL (x,t) ¶x Ra RL: Cytoplasmic Resistance per unit length (Ohm/cm) = l The Cable Equation: Steady State ~ ¶ 2V (X) - V (X) = - i ext (X) 2 ¶X Green’s Function G(X): Solution to Steady State Cable Equation for ~ - i ext (X) = d (X) with boundary conditions: V (X) ® 0 X ®±¥ 1 -X G(X) = e (Infinite cable) 2 ¥ Solution to Steady State V (X) = Cable Equation: ~ ò G(X - X')i(X')dX' -¥ Steady State: Boundary Conditions Cable Type Semi-Infinite Cable Schematic Diagram Boundary Condition V (0) = V0 V (¥) = 0 Finite Cable Sealed End (closed circuit) Finite Cable Open End (open circuit) Finite Cable Clamped End V (0) = 0 I(L) = dV dX V (0) = V0 V (L) = 0 V(0) = V0 V(L) = VL =0 X =L Semi-Infinite Cable: Constant Current I0 V (X) = V0e -X V (X = 0,t ® ¥) RN = IL (0) def lRa 4lRA RN = lRL = = l pd 2 RN = 2 RM RA p d -3 / 2 Finite Cable: Constant Current Length of Cable: l Dimensionless Length: L = l l V (X) = Asinh(L - X) + Bcosh(L - X) General Solution: cosh(L - X) + BL sinh(L - X) V (X) = V0 cosh(L) + BL sinh(L) V(X = 0) = V0 GL Conductance of terminal end BL = G¥ Conductance of semi-infinite cable Finite Cable: Sealed End BL = 0 (G L << G¥ ) I0 V(X = 0) = V0 I(X = L) = 0 V0 cosh(L - X) V (X) = cosh(L) L Finite Cable: Open End BL = ¥ (G L >> G ¥ ) I0 V(X = 0) = V0 V(X = L) = 0 V0 sinh(L - X) V (X) = sinh(L) L Finite Cable: Clamped End BL = Constant I0 V(X = 0) = V0 V(X = L) = VL V0 sinh(L - X) + VL sinh(X) V (X) = sinh(L) L Steady State Solution Cable Type Semi-Infinite Cable Finite Cable Sealed End Finite Cable Open End Solution Boundary Condition V (X) = V0e -X V (0) = 0 V (¥) = 0 V0 cosh(L - X) V (X) = cosh(L) V (0) = 0 V0 sinh(L - X) V (X) = sinh(L) V (0) = 0 V (L) = 0 V sinh(L - X) + VL sinh(X) Finite Cable Clamped End V (X) = 0 sinh(L) I(L) = dV dX V (0) = 0 V (L) = VL =0 X =L Steady State Solution Cable Equation: Transient Solution ~ dV (X,T) ¶ 2V (X,T) + V (X,T) = i ext (X,T) 2 dT ¶X Green’s Function G(X,T) ~for infinite cable: solution of above equation for: i ext = (a /2)d (X).d (T) With initial condition: V(X,0) = (a /2)d (X) and Boundary condition: G(X) = C.Q(T)e æ X2ö -çT + 4T ÷ø è C =a p t General Solution to Cable Equation: t V (X) = ¥ ~ ò dT' ò dX'G(X - X';T - T') i -¥ -¥ Hint: Use the formula: ext (X - X';T - T') Ralls Model-Equivalent cylinder Basic Idea: Impedence matching Assumptions: 1. The membrane properties are identical for soma and dendritic branches. 2. Membrane properties are uniform and voltage independent 3. All dendritic branches terminate at the same electrotonic length (and the tip of dendrite ends are sealed) Class assignment: Please read sections 4.5.1.3 and 4.5.2 on your own. Synaptic Integration Model for current injection into neuron through synapse-alpha function Use XPP AUTO to answer following Questions (Cable.ode) 1. Sketch the potential at the soma for the synaptic input at compartments 0, 5, 10, and 20. 1a.How do the peak amplitudes depend on distance? 1b. How about the time to peak? 1c.Does the peak appear to decay slower or faster for more distant inputs? 1d. How does the potential scale across various compartments For synaptic input at different locations on the cable