Transcript ppt - Neurodynamics Lab

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