Transcript Document

Synaptic integration – cable theory
Isopotential sphere
Current injected into a spherical cell will distribute
uniformly across the surface of the sphere. The current
flowing across a unit area of the membrane:
Im  Cm
dV m

dt
Vm
Rm
For a finite step of current:
V m  I m R m (1  e
 t / m
)
0tT
I m  I 0 / 4 a
where
 m R m Cm
2
The input resistance:
- time constant
 t / m
tT
For long impulse Im (t -> inf)
Vm ( )  I m Rm
def
R input  R N 
After the current step:
Vm  I m Rm e
For a sphere, a relationship between Im and I0 is:
- stan ustalony
Vm ( )
I0
The input resistance of a sphere:
RN 
Rm
4 a
2
Nonisopotential cell (cylinder)
Assumptions:
1.Uniform membrane. Membrane parameters are constant and they are
not dependent on the membrane potential.
2.The current flows along dimension x, radial current is zero.
3. Extracellular resistance, r0, is 0.
Vm is function of time and distance from the site of injection. The
decrease in Vm with distance is given by Ohm’s law:
Vm ( x, t )
x
  ri ii
x
t

Vm
rm
We obtain the cable equation
1  Vm
Vm
2
 cm
2
t

Vm
rm
In another form:
 Vm
2
  im
Vm
im  iC  iionic  c m
ri  x
The decrease in ii with distance is equal to the current that
flows across the membrane:
 ii
Recalling

2
x
2
m
Vm
t
 Vm
We obtain:
 Vm ( x , t )
2
x
2
  ri
 ii
x
 
 ri im
rm
ri
- space or length constant
The cable equation
Let:
X  x/
T  t / m
The cable equation becomes:
 Vm
2
X
2

Vm
T
 Vm  0
We will be using solutions to this equation to derrive equations that describe a number of specific situations like
inifite cable, finite cable, and finite cable with lumped soma.
Solutions of the cable equations – infinite cable
We will consider solution to the cable equation for a situation where a step current is injected into an infinitely long
cable
A general solution to the cable equation is
Vm ( X , T ) 
ri I 0    X
X
e
erfc
(


4 
2 T
X
T )  e erfc (
X
2 T
erfc(x) – complementary error function
erfc(x) =1- erf (x) =1-
x
2
p
òe
0
- y2
dy =
¥
2
p
òe
x
erf (0) = 0, erf (¥) =1, erf (-x) = -erf (x)
- y2
dy


T )

Solutions of the cable equations – infinite cable
Steady – state solution
Vm ( X , T ) 
ri I 0    X
X
e
erfc
(


4 
2 T
ri I 0 
e
X
ri I 0 
e
X
2 T


T )

Interpretaton of : 
reflects steady – state
properties of the cable. It
is the distance at which
potential has decayed to
1/e of the value at x = 0.
2
or
Vm ( X , T ) 
X
(T   )
We are seeking steady – state solution
Vm ( X , T ) 
T )  e erfc (
x/
2
The input resistance - infinite cable
RN 

Vm ( x  0, t   )
I0
1
Rm
2
2 a
*

ri 

ri
2
Ri
a
2

2
R m Ri / 2
2 a
rm / ri
3/2

rm ri
The input resistance – semi-infinite cable
2
R N ( semi )  2 * R N 
R m Ri / 2
a
3/2
Solutions of the cable equations – infinite cable
Transient solution
Vm ( X , T ) 
ri I 0    X
X
e
erfc
(


4 
2 T
Transient solution at x, X = 0
erfc (  T )  1  erf ( T )
erfc ( T )  1  erf ( T )
Infinite cable
Vm (0, T ) 
ri I 0 
erf ( T )
2
Semi-infinite cable
V m ( 0 , T ) semi  2V m ( 0 , T ) inf
T )  e erfc (
X
X
2 T


T )

Transient solution and estimation of the membrane time constant
Transient solution at x, X = 0
Vm (0, T ) 
ri I 0 
2
erf ( T )
Transient solution for isopotential sphere
Vm = I m Rm (1- e-t/t m )
Membrane time constant
 m R m Cm
The membrane potential of the infinite cable changes faster toward its steady-state value than that of
the ispotential sphere. This was extremly important result of Rall’s work. It demonstrated why early
estimates of membrane time constants from neurons were too low. The response to current step
(charging curve) was assumed to be single exponential end the membrane time constant was
estimated from the time to reach 63% of the steady-state value. Using point 0.63 on the error function
gives estimate 0.4, instead of (correct value) 1.
Solutions of the cable equations – infinite cable
Full solution
Vm ( X , T ) 
ri I 0    X
X
e
erfc
(


4 
2 T
T )  e erfc (
X
X
2 T


T )

Observations:
1.At large t, the distribution of
potential along the cable is the
steady – state solution.
2.At intermediate values of t, the
decay of potential with distance
along the cable is greater than at
the steady-state.
3.At x = 0 the charging curve is
described by erfc(T1/2).
Solution of the cable equation as a function of time and distance
for a step of current injected at x = 0 into a semi-finite cable.
4.At values of x increasing from 0,
the charging curves show slower
rising phases and reach a smaller
steady-state value.
Solutions of the cable equations – finite cable
Steady – state solution
I0
The cable equation:
 Vm
2
X
2

Vm
T
 Vm  0
x=0
Under steady – state conditions
Vm
T
- open end
2
X
2
We will consider two boundary conditions for x = l
- sealed end
0
The cable equation can be reduced to:
 Vm
x=l
 Vm
A general solution:
V m (  , X )  A1e
We introduce new variables:
X  x/
- electrotonic distance
L  l/
- electrotonic length
Hiperbolic cosine and sine:
X
 A2 e
X
cosh( X ) 
e
X
e
X
2
sinh( X ) 
e
X
e
2
X
Solutions of the cable equations – finite cable
Steady – state solution
The general solution of the cable equation can be rewritten using the hyperbolic functions as:
V m (  , X )  B1 cosh( X )  B 2 sinh( X )
or
V m (  , X )  C1 cosh( L  X )  C 2 sinh( L  X )
at X = L,
V m (  , L )  V L  C1 cosh( 0 )  C 2 sinh( 0 )  C1
Let BL = C2/VL and substitute it back into the equation.
V m (  , X )  V L [cosh( L  X )  B L sinh( L  X )]
BL is the boundary condition for different end terminations
At X = 0:
V m (  , 0 )  V 0  V L [cosh( L )  B L sinh( L )]
or
V L  V 0 /[cosh( L )  B L sinh( L )]
Substitute this back into the equation, we obtain
Solutions of the cable equations – finite cable
Steady – state solution, boundary conditions
the steady state solution of the cable equation for a finite-length cable.
Vm ( , X )  V0
cosh( L  X )  B L sinh( L  X )
cosh( L )  B L sinh( L )
Effects of the boundary conditions
BL  G L / G
1. If B L  1
G L - conductance of the terminal membrane
that is G L  G 
Vm ( , X )  V0e
X
than
the same as for a semi-infinite cable
2. If B L  0 that is G L  G 
Vm (  , X )  V0
(sealed end)
cosh( L  X )
finite cable with sealed end
finite cable with open end
semi-infinite cabl.
cosh( L )
3. If B   that is G L  G 
L
Vm (  , X )  V0
G  - conductance of a semi-infinite cable
(open end)
sinh( L  X )
sinh( L )
Comparison of voltage decays along finite cables of
different electrotonic lenghts and with different
terminations. Current injected at x = 0.
Solutions of the cable equations – finite cable
Transient solution
The complete solution of the cable equation can be written as:
V m (t , X )  C 0 ( X ) e
 t / 0
 C1 ( X )e
 t / 1
 ...  C n ( X ) e
 t / n
where
 n  n / L,
1   n   0 / n
2
The first term of the equation:
C 0 ( X )e
 t / 0
Corresponds to membrane time constant
0  m
if the cable has uniform membrane properties
Charging curves for finite cables of different
electrotonic lenghts. A step curent is injected and the
voltage is measured at x = 0.
Solutions of the cable equations – finite cable
– AC current
The steady-state decay of potential along a cable is much greater if the applied current changes in
time.
The AC length constant depends on DC length constant and frequency as follows:
 AC   DC
2
1
1  ( 2 f  m )
2
f – frequency (in Hz)
Voltage attenuation along a finite cable (L = 1) for current
injections DC to 100 Hz at X = 0 (soma)
Rall model for neurons
Now, we can use the decriptions for the semi-infinite and finite-length cables to derrive Rall
model of a neuron. This model provides a theoretical support for unerstanding the spread of
current and voltage in complicated dendritic trees.
Assumptions:
1.
Uniform membrane properties Ri, Rm, Cm
2.
R0 = 0
3. Soma represented as an isopotential sphere
4.
All dendrited terminated at the same electrotonic length.
Rall model for neurons
Schematic diagram of a neuron with a branched dendritic tree. X1, X2, X3 are three representative
branch points, d – is the diameter of the respective branches. Final branches are assumed to extend
to infinity and thus are semi-infinite cables.
Let’s recall the input resistance of semi-infinite cable
R N ( semi )  2 * R N 
R m Ri / 2
a
3/2
Rall model for neurons
Input resistance of semi-infinite cable
R m Ri / 2
RN 
a
3/2
Input conductance of the semi-infinite cable
GN 
d
3/2
,
d  2a
2 R m Ri
3/2
,
G N ( X 3 )  G N ( d 3111 )  G N ( d 3112 ) 
 K [( d 3111 )
This can be simplified
G N  Kd
Total conductance is sum of all parallel conductances,
hence a total input conductance at X3 is

K 
2 R m Ri
Input conductance of the cable d3111
G N ( d 3111 )  K ( d 3111 )
3/2
A similar equation exists for cable d3112
3/2
 ( d 3112 )
3/2
]
If instead of a branch point at X3 , cable d211 was
extended to infinity and detached at the same spot, its
input conductane would be:
G N ( d 211 )  K ( d 211 )
3/2
Rall recognized that if
( d 211 )
3/2
 ( d 3111 )
3/2
 ( d 3112 )
3/2
than having branch point X3 with branches d3111 and d3112
is exactly equivalent to extending branch d211 to infinity!
Rall model for neurons
If we had done the same operation at the branch
point along d212, then at X2 there would be two
semi-infinite cables d211 i d212 attached to branch
d11. As before, if
( d 11 )
3/ 2
 ( d 211 )
3/ 2
 ( d 212 )
3/ 2
Applying 3/2 power rule:
then
G N ( X 2 )  G N ( d 211 )  G N ( d 212 )  K ( d 11 )
which is equivalent to extending cable d11 to infinity.
3/2
dP
3/2


dD
3/ 2
one can reduce the entire dendritic tree to an
equivalent semi-infinite cylinder.
A number of studies has suggested that the
branching in dendrites o spinal motor neurons,
cortical neurons and hippocampal neurons closely
approximates the 3/2 power rule.
Rall model for neurons
Equivalent finite cylinder
Dendrites are usually better represented by finite-length cables,
because usually l < 2. The equation for the input conductance
of finite cable is:
d
GN 
3/2
tanh( L )
2 R m Ri
and is also proportional to d3/2. L – electrotonic
length, assumed to be the same for all dendrites.
Using the relations:
 
rm / ri , Ri   ( d / 2 ) ri , R m   dr m
2
We obtain
GN 
1
 ri
tanh( L )
Applying the equations for input conductance, the
3/2 power rule and assumption of the same L for
all dendrites one can reduce any dendritic tree to a
single equivalent cylinder of diameter d0
electrotonic length L. Furthermore recalling that, L
= l/, for such a tree:
Ltotal =
l0
l0
+
l11
+
l211
l11 l211
+... +
Rall model for neurons – application to synaptic inputs
The main reason for studying the electrotonic properties of neurons is to help to understand the function of
dendritic synaptic inputs. Using his theory, Rall demonstrated that distal synaptic inputs produce measurable
signal at the soma. He estimated that the electrotonic length o spinal motor neurons was about 1.5. He
calculated the changes in membrane potential measured in the soma in response to brief current steps to the
soma, middle region of the dendrites and distal dendrites.
Results:
- the amplitude of synaptic potential measured at soma is attenuted with input distance from the soma
- the rise time and peak amplitude of the inputs is slowed and delayed for more distant inputs
- the decay time of all inputs is the same (this is because the potential due to synaptic input cannot decay
slower than the membrane time constant).
Functional properties of the synapses
Synapses don’t inject step current, they produce brief conductance change often approximated by an alpha
function gs = Gs(t/α)e-αt.
Synaptic conductance gs and
excitatory postsynaptic potential
EPSP. If the time course of gs is
brief compared to m, EPSP will
decay with the membrane time
constant.
Synapse A
Rising time constant Cm/(GsA + Gr)
Decay time constant Cm/ Gr
Synapse A + B
Rising time constant Cm/(GsA + GsA + Gr)
Decay time constant Cm/ Gr
Parralel conductance model representing the
separate synaptic inputs, A and B.
Summary: temporal and spatial summation of synaptic inputs
Dendritic computations – an example
Summation of dendritic inputs in a model neuron.
When four synaptic inputs (A–D) arrive at four separate locations
of a neuron with brief intervals, spatial summation can be
significant only when they synapses are activated in a preffered
order i.e., D to A, but not A to D. From Arbib, M. A., 1989, The
Metaphorical Brain 2: Neural Networks and Beyond, New York:
Wiley-Interscience, p. 60.
Livingstone MS. Mechanisms of direction selectivity in
macaque V1. Neuron. 1998, 20(3):509-26.
Cable and compartmental models of dendritic trees
Various dendritic trees (a, b, c) and their computer models
(d, e).
Dendrites are modeled either as a set of cylindrical membrane cables (B) or as a set of discrete isopotential
RC compartments (C).
B. In the cable representation, the voltage can be computed at any point in the tree by using the continuous cable
equation and the appropriate boundary conditions imposed by the tree. An analytical solution can be obtained for any
current input in passive trees of arbitrary complexity with known dimensions and known specific membrane
resistance and capacitance (RM,CM) and specific cytoplasm (axial) resistance (RA).
C. In the compartmental representation, the tree is discretized into set of interconnected RC compartments. Each is a
lumped representation of the membrane properties of a sufficiently small dendritic segment. Compartments are
connected via axial cytoplasmic resistances. In this approach, the voltage can be computed at each compartment for
any(nonlinear) input and for voltage and time-dependent membrane properties (not only passive membranes).
Dendirtic computations - summary
From Idan Segev and Michael London
Dendritic Processing. In M. Arbib (editor).
The Handbook of Brain Theory and Neural
Networks. THE MIT PRESS Cambridge,
Massachusetts London, England, 2002
Dendritic computations - logical operations
Własności dendrytów umożliwają wykonywanie operacji logicznych. Z Idan Segev and Michael London Dendritic
Processing. Rozdział w M. Arbib (edytor). The Handbook of Brain Theory and Neural Networks. THE MIT PRESS
Cambridge, Massachusetts London, England, 2002
Dendritic computations - coincidence detection
In birds, a special type of neuron is responsible for
computing the time difference between sounds
arriving to the two ears. Coincident inputs from
both ears arriving to the two dendrites are summed
up at the soma and cause the neuron to emit action
potentials. However, when coincident spikes arrive
from the same ear, they arrive at the same dendrite
and thus their summation is sublinear, resulting in a
subthreshold response.
In layer 5 pyramidal neurons excitatory distal synaptic input
(EPSP) that coincides with backpropagation of the action potential
(bAP) results in a large dendritic Ca 2+ spike, which drives a burst
of spikes in the axon. Otherwise, a single AP is generated.
From: Michael London and Michael Hausser. Dendritic Computation. Annu. Rev. Neurosci.2005. 28:503–32
Dendritic computations - feature extraction
Mapping of synaptic inputs onto dendritic branches may play a role in feature extraction responsible for face recognition.
In neurons with active dendrites, clusters of inputs active synchronously on the same branch can evoke a local dendritic spike, which
leads to significant amplification of the input. Specific configuration of inputs may trigger a spike while activation of synapses at
different branches will not generate a response. The features extracted by dendrites may be fed into a soma or features extracted by
multiple neurons are fed to a next cortical cell.
From: Michael London and Michael Hausser. Dendritic Computation. Annu. Rev. Neurosci.2005. 28:503–32
Encoding/decoding information in a neuron
Exploring dendritic input-output relation using information theory. (A) Reduced model of a neuron consisting of a passive dendritic cylinder, a
soma and an excitable axon. The dendritic cylinder is bombarded by spontaneous background synaptic activity (400 excitatory synapses, each
activated 10 times/s; 100 inhibitory synapses, each activated 65 times/s). (B) Soma EPSPs for a proximal (red line) and a distal (blue line)
synapse. (C) Two sample traces of the output spike train measured in the modeled axon. Identical background activity was used in both cases;
the location of only one excitatory synapse was displaced from proximal to distal. This displacement noticeably changes the output spike train.
(D) The mutual information (MI), which measures how much could be known about the input (the presynaptic spike train) by observing the
axonal output, is plotted as a function of the maximal synaptic conductance. The distal synapse transmits significantly less information
compared to the proximal synapse. For strong proximal synapses, the MI is saturated because, for large conductance values, each input spike
generates a time-locked output spike and no additional information is gained by further potentiating this synapse. From: Idan Segev and
Michael London. Untangling Dendrites with Quantitative Models. Science 290, 2000