ECE 795:
Quantitative
Electrophysiology
Notes for Lecture #4
Wednesday, October 4, 2006
7. CHEMICAL SYNAPSES AND GAP
JUNCTIONS
We will look at:
¾ Chemical synapses in the nervous
system
¾ Gap junctions in cardiac cells and
nervous tissue
2
Chemical synapses:
The specialized contact zones between
neurons are called synapses.
In the nervous system, chemical synapses
are much more common than electrical
synapses (gap junctions).
Most chemical synapses are unidirectional
— the presynaptic neuron releases
neurotransmitter across the synaptic cleft to
the postsynaptic terminal, which leads to
activation of a neurotransmitter-gated ion
channel. 3
Chemical synapses (cont.):
In the electron micrograph below, a presynaptic
body in the inner hair cell is seen to hold a cluster
of neurotransmitter vesicles. A thickening of the
cell membranes is observed between the pre- and
post-synaptic terminals, and a very narrow
synaptic cleft exists.
4
(from Francis et al., Brain Res. 2004)
Chemical synapses (cont.):
(from Koch)
5
Chemical synapses (cont.):
Equivalent electric circuit of a fast chemical
synapse:
(from Koch)
6
Chemical synapses (cont.):
The postsynaptic current (PSC) has the
same form as a voltage-gated ion channel:
but the conductance gsyn(t) is controlled by
the reception of neurotransmitter (rather
than the transmembrane potential), which
has a waveform that is often approximated
by a so-called alpha function:
7
Chemical synapses (cont.):
The direction of the postsynaptic current
depends on the value of Esyn:
¾ if Esyn > Vrest, then Isyn will be a negative
(i.e., inward) current, which will
depolarize the cell.
Consequently, this current is referred to
as an excitatory postsynaptic current
(EPSC), and the resulting membrane
depolarization is referred to as an
excitatory postsynaptic potential (EPSP).
8
Chemical synapses (cont.):
¾ if Esyn < Vrest, then Isyn will be a positive
(i.e., outward) current, which will
hyperpolarize the cell. Because
hyperpolarization takes the membrane
potential further away from the threshold
potential, this is a form of inhibition.
Consequently, this current is referred to
as an inhibitory postsynaptic current
(IPSC), and the resulting membrane
hyperpolarization is referred to as an
inhibitory postsynaptic potential (IPSP).
9
Chemical synapses (cont.):
¾ if Esyn ≈ Vrest, then Isyn will be a negligible
when the membrane is at rest.
However, if current is injected into the
membrane by a propagating EPSP or
action potential or an applied current
source, the increased conductance of
gsyn(t) will tend to “shunt” this injected
current, such that the membrane is
locked at Vrest. Because this prevents
action potential generation, it is referred
to as shunting inhibition.
10
Chemical synapses (cont.):
(from Koch)
11
Chemical synapses (cont.):
Shunting inhibition is most effective if placed
on the path between an excitatory synapse
and the soma.
(from Koch)
12
Gap junctions in cardiac cells:
13
Gap junctions in cardiac cells (cont.):
14
Gap junctions in cardiac cells (cont.):
Cable analysis of Purkinje fibers gives
λ ≈ 1 mm and τ = 18 ms.
15
Gap junctions in cardiac cells (cont.):
Experimental estimation of gap junction resistance
in dissociated cardiac cells.
16
Gap junctions in cardiac cells (cont.):
17
Gap junctions in cardiac cells (cont.):
Estimation of gap
junction resistance
in chick embryo
cell pairs.
18
Gap junctions in nervous tissue:
Gap junctions are found between :-
– some neurons, mainly during
development,
– glial cells, and
– glial cells & neurons.
For more details, see:
Brain Research Reviews 32(1), 2000.
19
8. DENDRITIC TREES
We will look at:
¾ Properties of infinite, semi-infinite &
finite cables
¾ Branching in passive dendritic trees
¾ Equivalent cylinder of a dendritic tree
¾ Compartmental modeling
20
Dendritic tree morphology:
(from Koch)
21
Steady-state response of a finite cable:
The steady-state response of a cable of
finite length l in absolute units (or L = l/λ in
electrotonic units) can be described by one
of the equivalent forms:
where:
22
Input impedance of a semi-infinite cable:
Consider a current that is injected into the
intracellular space at the origin of a semi-
infinite cable (i.e., starts at x = 0 and heads
off only in one direction to x = +∞), with the
return electrode in the extracellular space at
the origin.
The input impedance for the semi-infinite
cable is:
23
Input impedance of a semi-infinite cable
(cont.):
For a semi-infinite cable the relative
membrane potential is:
Assuming re ≈ 0, the intracellular axial
current is:
24
Input impedance of a semi-infinite cable
(cont.):
Since λ ≈ (rm/ri)1/2 when re ≈ 0:
and the input impedance is:
25
Input impedance of a finite cable:
The input impedance Zin for a finite cable
will depend on the cable’s:
¾ length (l in absolute units, or L = l/λ in
electrotonic units), and
¾ termination impedance (ZL).
Ii
X= 0 X=L ZL
26
Input impedance of a finite cable:
The termination impedance is determined by
the physical configuration of the end of the
fiber.
Some common boundary conditions include:
¾ semi-infinite cable,
¾ sealed-end,
¾ killed-end, and
¾ arbitrary-impedance.
27
Input impedance of a finite cable (cont.):
¾ A semi-infinite cable termination of a
finite cable corresponds to ZL = Z0. In
this case, the finite cable is simply
considered to be the proximal section of
a semi-infinite cable.
28
Input impedance of a finite cable (cont.):
¾ A sealed-end termination corresponds to
having a patch of membrane covering the
end of the fiber, such that ZL ≈ ∞.
In this case, the internal axial current
must be zero at the end of the fiber,
i.e., Ii(X=L) = 0.
¾ Using this boundary condition:
29
Input impedance of a finite cable (cont.):
In the case of a sealed-end termination, the
shorter the finite cable, the greater the effect
of the infinite termination impedance on the
input impedance.
30
Input impedance of a finite cable (cont.):
¾ A killed-end termination corresponds to
having a direct opening to the
extracellular fluid at the end of the fiber,
such that ZL ≈ 0. In this case, the
transmembrane potential must be zero at
the end of the fiber, i.e., Vm(X=L) = 0.
¾ Using this boundary condition:
31
Input impedance of a finite cable (cont.):
(from Koch)
32
Input impedance of a finite cable (cont.):
¾ An arbitrary-impedance termination is
often used to describe the boundary
between parent and daughter branches
in dendritic trees.
¾ Using this boundary condition:
33
Steady-state response of a finite cable:
(from Koch)
34
Time-dependent response of a finite cable:
(from Koch)
35
Branching in passive dendritic trees:
(from Koch)
36
Branching in passive dendritic trees (cont.):
Assuming sealed ends on the daughter
branches 1 and 2, their input impedances
are:
respectively.
Thus, the parallel daughter branches are
equivalent to a termination impedance given
by:
37
Branching in passive dendritic trees (cont.):
Consequently, the input impedances of the
parent branch is:
Multiple branches can be solved recursively
in this manner.
38
Branching in passive dendritic trees (cont.):
Once the input impedance at the site of
current inject is calculated, the voltage at
this site can be determined via:
and given V0, we can compute the voltage at
any point in the tree. This is achieved by
calculating how much current flows into
each of the daughter branches.
39
Branching in passive dendritic trees (cont.):
In daughter branch 1:
Thus:
⇒ the current divides between the daughter
branches according to the input conductances
40
Equivalent cylinder of a dendritic tree:
Assume L1 = L2 and d1 = d2
(from Koch)
41
Equivalent cylinder of a dendritic tree (cont.):
In the case of d03/2 = 2d13/2, all derivatives of
the voltage profile are continuous ⇒ the
voltage decay can be described by a single
expression.
Why?
42
Equivalent cylinder of a dendritic tree (cont.):
For a semi-infinite cable:
If d03/2 = 2d13/2:
⇒ Input impedances are matched
43
Equivalent cylinder of a dendritic tree (cont.):
Consequently:
and:
⇒ A parent branch of length L0 with two identical
impedance-matched daughter branches of
length L1 is equivalent to a single cylinder of
length L0 + L1.
44
Equivalent cylinder of a dendritic tree (cont.):
Rall showed that the entire dendritic tree can
be collapsed into a single equivalent cylinder
if:
1. Rm and Ri are the same in all branches,
2. all terminals end in the same boundary
condition,
3. all terminal branches end at the same
electrotonic distance L = Σi Li, and
4. d03/2 = d13/2 + d23/2 at every branch point.
Assumptions 1 & 2 are reasonable, but 3 & 4 are
only met in a few remarkable neuron types. 45
Compartmental modeling:
(from Koch)
46
