lecture 6: dendrites and axons
DESCRIPTION
Lecture 6: Dendrites and Axons. Cable equation Morphoelectronic transform Multi-compartment models Action potential propagation. Refs: Dayan & Abbott, Ch 6, Gerstner & Kistler, sects 2.5-6; C Koch, Biophysics of Computation , Chs 2,6. Longitudinal resistance and resistivity. - PowerPoint PPT PresentationTRANSCRIPT
Lecture 6: Dendrites and Axons
• Cable equation
• Morphoelectronic transform
• Multi-compartment models
• Action potential propagation
Refs: Dayan & Abbott, Ch 6, Gerstner & Kistler, sects 2.5-6; C Koch, Biophysics of Computation, Chs 2,6
Longitudinal resistance and resistivity
Longitudinal resistance and resistivity
2aLr
R LL Longitudinal resistance
Longitudinal resistance and resistivity
2aLr
R LL Longitudinal resistance
Longitudinal resistivity rL ~ 1-3 kmm2
Longitudinal resistance and resistivity
2aLr
R LL Longitudinal resistance
Longitudinal resistivity rL ~ 1-3 kmm2
x
V
r
aL
x
V
RV
RI
LLLL
211
Cable equation
Cable equation
rightLleftLemm x
Vra
xV
ra
iixatV
cxa
22
)(22
current balance:
Cable equation
rightLleftLemm x
Vra
xV
ra
iixatV
cxa
22
)(22
xxV
ra
xxV
ra
xV
ra
LrightLleftL
222
current balance:
on rhs:
Cable equation
rightLleftLemm x
Vra
xV
ra
iixatV
cxa
22
)(22
xxV
ra
xxV
ra
xV
ra
LrightLleftL
222
emL
m iixV
axart
Vc
2
21
current balance:
on rhs:
Cable equation:
Linear cable theory
m
restm r
VVi
Ohmic current:
Linear cable theory
m
restm r
VVi
rVVu
Ohmic current:
Measure V relative to rest:
Linear cable theory
m
restm r
VVi
rVVu
emL
m iru
xu
ra
tu
c
2
2
2
Ohmic current:
Measure V relative to rest:
Cable equation becomes
Linear cable theory
m
restm r
VVi
rVVu
emL
m iru
xu
ra
tu
c
2
2
2
mmmL
m crr
ar 2
Ohmic current:
Measure V relative to rest:
Cable equation becomes
Now define electrotonic length and membrane time constant:
Linear cable theory
m
restm r
VVi
rVVu
emL
m iru
xu
ra
tu
c
2
2
2
mmmL
m crr
ar 2 emm iru
xu
tu
2
22
Ohmic current:
Measure V relative to rest:
Cable equation becomes
Now define electrotonic length and membrane time constant:
Linear cable theory
m
restm r
VVi
rVVu
emL
m iru
xu
ra
tu
c
2
2
2
mmmL
m crr
ar 2 emm iru
xu
tu
2
22
Ohmic current:
Measure V relative to rest:
Cable equation becomes
Now define electrotonic length and membrane time constant:
Note: cable segment of length has longitudinal resistance = transverse resistance:
Linear cable theory
m
restm r
VVi
rVVu
emL
m iru
xu
ra
tu
c
2
2
2
mmmL
m crr
ar 2 emm iru
xu
tu
2
22
22 ar
ar
R Lm
Ohmic current:
Measure V relative to rest:
Cable equation becomes
Now define electrotonic length and membrane time constant:
Note: cable segment of length has longitudinal resistance = transverse resistance:
dimensionless units:
/ˆ/ˆ ttxx
extTextLT iriirri ˆˆ
dimensionless units:
/ˆ/ˆ ttxx
extTextLT iriirri ˆˆ
Removes , m from equation.
dimensionless units:
),(),(),(),(
2
2
txirtxux
txut
txuem
/ˆ/ˆ ttxx
extTextLT iriirri ˆˆ
Removes , m from equation.
Now remove the hats:
dimensionless units:
),(),(),(),(
2
2
txirtxux
txut
txuem
/ˆ/ˆ ttxx
extTextLT iriirri ˆˆ
Removes , m from equation.
Now remove the hats:
(t really means t/m, x really means x/)
Stationary solutions0
tuNo time dependence:
Stationary solutions0
tu
)(ˆ)()()(
2
2
xixirxux
xueem
No time dependence:
Static cable equation:
Stationary solutions0
tu
)(ˆ)()()(
2
2
xixirxux
xueem
No time dependence:
Static cable equation:
General solution where ie = 0:
Stationary solutions0
tu
)(ˆ)()()(
2
2
xixirxux
xueem
)exp()exp()( 21 xcxcxu
No time dependence:
Static cable equation:
General solution where ie = 0:
Stationary solutions0
tu
)(ˆ)()()(
2
2
xixirxux
xueem
)exp()exp()( 21 xcxcxu
)()(ˆ 0 xixie
No time dependence:
Static cable equation:
General solution where ie = 0:
Point injection:
Stationary solutions0
tu
)(ˆ)()()(
2
2
xixirxux
xueem
)exp()exp()( 21 xcxcxu
)()(ˆ 0 xixie
xixu e)( 021
No time dependence:
Static cable equation:
General solution where ie = 0:
Point injection:
Solution:
Stationary solutions0
tu
)(ˆ)()()(
2
2
xixirxux
xueem
)exp()exp()( 21 xcxcxu
)()(ˆ 0 xixie
xixu e)( 021
No time dependence:
Static cable equation:
General solution where ie = 0:
Point injection:
Solution:
Stationary solutions0
tu
)(ˆ)()()(
2
2
xixirxux
xueem
)exp()exp()( 21 xcxcxu
)()(ˆ 0 xixie
xixu e)( 021
)(ˆe)( 21 xixdxu e
xx
No time dependence:
Static cable equation:
General solution where ie = 0:
Point injection:
Solution:
Solution for general ie:
Boundary conditions at junctions
Boundary conditions at junctions
V continuous
Boundary conditions at junctions
V continuous
Sum of inward currents must be zero at junctionxV
a 2
Boundary conditions at junctions
0
xV
V continuous
Sum of inward currents must be zero at junction
closed end:
xV
a 2
Boundary conditions at junctions
0
xV
V continuous
Sum of inward currents must be zero at junction
closed end:
xV
a 2
open end: V = 0
Green’s functionResponse to delta-function current source (in space and time)
Green’s function
)()(),(),(),(
2
2
txtxux
txut
txu
Response to delta-function current source (in space and time)
Green’s function
)()(),(),(),(
2
2
txtxux
txut
txu
Response to delta-function current source (in space and time)
Spatial Fourier transform:
Green’s function
)()(),(),(),(
2
2
txtxux
txut
txu
)(),(),(),( 2 ttkutkuk
ttku
Response to delta-function current source (in space and time)
Spatial Fourier transform:
Green’s function
)()(),(),(),(
2
2
txtxux
txut
txu
)(),(),(),( 2 ttkutkuk
ttku
)(])1(exp[),( 2 ttktku
Response to delta-function current source (in space and time)
Spatial Fourier transform
Easy to solve:
Green’s function
)()(),(),(),(
2
2
txtxux
txut
txu
)(),(),(),( 2 ttkutkuk
ttku
)(])1(exp[),( 2 ttktku
Response to delta-function current source (in space and time)
Spatial Fourier transform
Easy to solve:
Invert the Fourier transform:
Green’s function
)()(),(),(),(
2
2
txtxux
txut
txu
)(),(),(),( 2 ttkutkuk
ttku
)(])1(exp[),( 2 ttktku
),(4
exp4
)(),(
2
txGt
xt
t
ttxu
Response to delta-function current source (in space and time)
Spatial Fourier transform
Easy to solve:
Invert the Fourier transform:
Green’s function
)()(),(),(),(
2
2
txtxux
txut
txu
)(),(),(),( 2 ttkutkuk
ttku
)(])1(exp[),( 2 ttktku
),(4
exp4
)(),(
2
txGt
xt
t
ttxu
),(ˆ),(),( txittxxGxdtdtxu e
t
Response to delta-function current source (in space and time)
Spatial Fourier transform
Easy to solve:
Invert the Fourier transform:
Solution for general ie(x,t) :
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
u vs t at various x: x vs tmax:
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
u vs t at various x: x vs tmax:
At what t does u peak?
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
0log4 2
12
t
tx
tdtd
u vs t at various x: x vs tmax:
At what t does u peak?
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
0log4 2
12
t
tx
tdtd
021
41 2
2
tt
x
u vs t at various x: x vs tmax:
At what t does u peak?
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
0log4 2
12
t
tx
tdtd
021
41 2
2
tt
x02)2( 22 xtt
u vs t at various x: x vs tmax:
At what t does u peak?
Pulse injection at x=0,t=0:
tx
tt
txu4
exp4
1),(
2
0log4 2
12
t
tx
tdtd
021
41 2
2
tt
x02)2( 22 xtt
2411
22x
t
u vs t at various x: x vs tmax:
At what t does u peak?
Pulse injection at x=0,t=0:
2
1)/(414
2max
m
x
m xxt
tx
tt
txu4
exp4
1),(
2
0log4 2
12
t
tx
tdtd
021
41 2
2
tt
x02)2( 22 xtt
2411
22x
t
u vs t at various x: x vs tmax:
At what t does u peak?
Restoring, m:
Compare with no-leak case:
tx
ttxu
4exp
4
1),(
2
Compare with no-leak case:
tx
ttxu
4exp
4
1),(
2
Just diffusion, no decay
Compare with no-leak case:
tx
ttxu
4exp
4
1),(
2
Just diffusion, no decay
Now when does it peak for a given x?
Compare with no-leak case:
tx
ttxu
4exp
4
1),(
2
0log4 2
12
t
tx
dtd
Just diffusion, no decay
Now when does it peak for a given x?
Compare with no-leak case:
tx
ttxu
4exp
4
1),(
2
0log4 2
12
t
tx
dtd
021
4 2
2
tt
x
Just diffusion, no decay
Now when does it peak for a given x?
Compare with no-leak case:
tx
ttxu
4exp
4
1),(
2
0log4 2
12
t
tx
dtd
021
4 2
2
tt
x 22 xt
Just diffusion, no decay
Now when does it peak for a given x?
Compare with no-leak case:
tx
ttxu
4exp
4
1),(
2
0log4 2
12
t
tx
dtd
021
4 2
2
tt
x 22 xt
2
2
max 2 x
t m
Just diffusion, no decay
Now when does it peak for a given x?
Restoring, m:
Finite cableMethod of images:
Finite cable
)],2(),2([),;,( 000000,0 ttxnLxGttxnLxGtxtxGn
L
Method of images:
Finite cable
)],2(),2([),;,( 000000,0 ttxnLxGttxnLxGtxtxGn
L
),(ˆ),;,(),( 0000,0
0
00 txitxtxGdxdttxu eL
t L
Method of images:
General solution:
Finite cable
)],2(),2([),;,( 000000,0 ttxnLxGttxnLxGtxtxGn
L
),(ˆ),;,(),( 0000,0
0
00 txitxtxGdxdttxu eL
t L
Method of images:
General solution:
Morphoelectronic transform
Frequency-dependent morphoelectronic transforms
Multi-compartment models
Discrete cable equations
)()( 11,11,
VVgVVg
A
Ii
dt
dVc e
mm
Discrete cable equations
)()( 11,11,
VVgVVg
A
Ii
dt
dVc e
mm
)/( 2aLrL Resistance between compartments:
Discrete cable equations
)()( 11,11,
VVgVVg
A
Ii
dt
dVc e
mm
)/( 2aLrL Resistance between compartments:
Current between compartments: )/()( 12 LrVVa L
Discrete cable equations
)()( 11,11,
VVgVVg
A
Ii
dt
dVc e
mm
)/( 2aLrL
)2)(/()( 12 aLLrVVa L
Resistance between compartments:
Current between compartments:
Current per unit area:
)/()( 12 LrVVa L
Discrete cable equations
)()( 11,11,
VVgVVg
A
Ii
dt
dVc e
mm
)/( 2aLrL
)2)(/()( 12 aLLrVVa L
Resistance between compartments:
Current between compartments:
Current per unit area:
)/()( 12 LrVVa L
21, 2 Lra
gL
Action potential propagation
extL
NaNaKKLLm IxV
ra
VVhmgVVngVVgtV
c
2
234
2)()()(
Action potential propagation
extL
NaNaKKLLm IxV
ra
VVhmgVVngVVgtV
c
2
234
2)()()(
hVhdtdh
VmVmdtdm
VnVndtdn
V hmn )()()()()()(
Action potential propagation
extL
NaNaKKLLm IxV
ra
VVhmgVVngVVgtV
c
2
234
2)()()(
hVhdtdh
VmVmdtdm
VnVndtdn
V hmn )()()()()()(
a “reaction-diffusion equation”
Action potential propagation
extL
NaNaKKLLm IxV
ra
VVhmgVVngVVgtV
c
2
234
2)()()(
hVhdtdh
VmVmdtdm
VnVndtdn
V hmn )()()()()()(
)(),( stxVtxV
a “reaction-diffusion equation”
Moving solutions: look for solution of the form
Action potential propagation
extL
NaNaKKLLm IxV
ra
VVhmgVVngVVgtV
c
2
234
2)()()(
hVhdtdh
VmVmdtdm
VnVndtdn
V hmn )()()()()()(
)(),( stxVtxV
a “reaction-diffusion equation”
Moving solutions: look for solution of the form
Ordinary DE
Action potential propagation
extL
NaNaKKLLm IxV
ra
VVhmgVVngVVgtV
c
2
234
2)()()(
hVhdtdh
VmVmdtdm
VnVndtdn
V hmn )()()()()()(
)(),( stxVtxV
a “reaction-diffusion equation”
Moving solutions: look for solution of the form
extL
NaNaKKLLm Idt
Vdsr
aVVhmgVVngVVg
dtdV
c 2
2
234
2)()()(
Ordinary DE
Action potential propagation
extL
NaNaKKLLm IxV
ra
VVhmgVVngVVgtV
c
2
234
2)()()(
hVhdtdh
VmVmdtdm
VnVndtdn
V hmn )()()()()()(
)(),( stxVtxV
a “reaction-diffusion equation”
Moving solutions: look for solution of the form
extL
NaNaKKLLm Idt
Vdsr
aVVhmgVVngVVg
dtdV
c 2
2
234
2)()()(
Ordinary DE
HH solved iteratively for s (big success of their model)
Propagation speed
a/s2 must be independent of a
Propagation speed
a/s2 must be independent of a
as
Propagation speed
a/s2 must be independent of a
as
This is probably why the squid axon is so thick.
Multi-compartment model
Multicompartment calculation
Myelinated axonsNodes of Ranvier: active Na channels
Myelinated axonsNodes of Ranvier: active Na channels
Myelinated axons
Treat as multilayer capacitor each layer of thickness a:
Nodes of Ranvier: active Na channels
Myelinated axons
a
daLcC m
mm 2Treat as multilayer capacitor each layer of thickness a:
Nodes of Ranvier: active Na channels
Myelinated axons
a
daLcC m
mm 2Treat as multilayer capacitor each layer of thickness a:
Integrate up from a1 to a2 (inverse capacitances add)
Nodes of Ranvier: active Na channels
Myelinated axons
a
daLcC m
mm 2
Ldcaa
ada
LdcC mm
a
ammm 2)/log(
211 12
2
1
Treat as multilayer capacitor each layer of thickness a:
Integrate up from a1 to a2 (inverse capacitances add)
Nodes of Ranvier: active Na channels
Myelinated axons
a
daLcC m
mm 2
Ldcaa
ada
LdcC mm
a
ammm 2)/log(
211 12
2
1
Lx
V
r
a
t
VC
Lm
2
221
Treat as multilayer capacitor each layer of thickness a:
Integrate up from a1 to a2 (inverse capacitances add)
Negligible leakage between nodes: cable equation becomes
Nodes of Ranvier: active Na channels
Myelinated axons
a
daLcC m
mm 2
Ldcaa
ada
LdcC mm
a
ammm 2)/log(
211 12
2
1
Lx
V
r
a
t
VC
Lm
2
221
mLmLm drcaaa
rCLa
D2
)/log( 1221
21
Treat as multilayer capacitor each layer of thickness a:
Integrate up from a1 to a2 (inverse capacitances add)
Negligible leakage between nodes: cable equation becomes
Diffusion constant:
Nodes of Ranvier: active Na channels
How much myelinization? optimal a1/a2
Find the value of y = a1/a2 that maximizes D
How much myelinization? optimal a1/a2
21112 log0log20log yyyy
dyd
yy
Find the value of y = a1/a2 that maximizes D
How much myelinization? optimal a1/a2
21112 log0log20log yyyy
dyd
yy
Find the value of y = a1/a2 that maximizes D
221 6.0e/ aaa
How much myelinization? optimal a1/a2
21112 log0log20log yyyy
dyd
yy
Find the value of y = a1/a2 that maximizes D
221 6.0e/ aaa
Agrees with experiment
Speed of propagation
mLm dreca
D4
22
2
2
xV
DtV
Diffusion equation
with diffusion constant
Speed of propagation
mLm dreca
D4
22
2
2
xV
DtV
22
2
/,const axyyV
tV
Diffusion equation
with diffusion constant
Speed of propagation
mLm dreca
D4
22
2
2
xV
DtV
22
2
/,const axyyV
tV
Diffusion equation
with diffusion constant
Speed of propagation proportional to a2
Speed of propagation
mLm dreca
D4
22
2
2
xV
DtV
22
2
/,const axyyV
tV
Diffusion equation
with diffusion constant
Speed of propagation proportional to a2
(cf a1/2 for unmyelinated axon)