Biomedical Engineering Reference
In-Depth Information
It may be questioned why the addition of these currents is necessary. After all, the Hodgkin-
Huxely model successfully reproduced the properties of the action potential. There are at least three
reasons why the extra currents, and complexity, are important. First, the giant squid axon does not display
all of the behavior of mammalian neurons. Second, is often the case that a genetic mutation, drug, or
hormone will only effect one type of ion channel. For example, Nifedipine specifically blocks a certain
type of Calcium ion channel, while allowing other Calcium channels to remain functional. Third, when
a single action potential fires,
Na
+
,
K
+
, and other ions cross the membrane. Surprisingly, it does not
require many ions to cross the membrane to achieve the
100
mV
change in
V
m
during the action
potential. Therefore, the intracellular and extracellular concentrations do not change significantly during
a single action potential. If many action potentials fire in rapid succession, however, the concentrations
will eventually begin to change. Cells therefore have a variety of
pumps
and
exchangers
that slowly act
in the background, transporting ions back across the membrane. As these pumps are typically working
against a concentration gradient, they require ATP. Similar to other currents, they may be added to
I
ion
.
We will consider how to incorporate changing concentrations in Sec. 3.4.5.
≈
3.4.2 TheTraubModel of the Pyramidal Neuron
A model with additional currents is the Traub model. We present the model here as an example of how
extra currents lead to more realistic behavior and because it will be encountered again in Ch. 7.
I
ion
=
I
kCa
I
Na
=
g
na
m
3
h
[
V
m
−
E
Na
]
I
K
=
I
Na
+
I
K
+
I
Ca
+
g
k
n
4
y
E
K
]
I
Ca
=
g
Ca
s
5
r
[
V
m
−
E
Ca
]
I
kCa
=
[
V
m
−
g
kCa
q
[
V
m
−
V
k
]
where the gates are defined in Table 3.1.
Table 3.1:
gate
α
β
−
V
m
)
exp
60
−
V
m
10
0
.
04
(
60
0
.
005
(V
m
−
45
)
s
exp
V
m
−
45
10
−
1
−
1
0
.
32
(
12
−
V
m
)
exp
12
−
V
m
4
0
.
28
(V
m
−
40
)
exp
V
m
−
40
5
m
−
1
−
1
0
.
128 exp
17
−
V
m
18
4
exp
20
−
V
m
5
h
+
1
0
.
5exp
10
−
V
m
40
0
.
032
(
15
−
V
m
)
n
exp
15
−
V
m
5
−
1
0
.
028 exp
15
−
V
m
15
2
0
.
4
y
+
exp
85
−
V
m
10
exp
40
−
V
m
10
+
1
+
1
0
.
025
(
200
−
x)
exp
200
−
x
20
r
0
.
005
−
1
0
.
005
(
200
−
x)
exp
V
20
exp
200
−
x
20
q
0
.
002
−
1
