SOLUTIONS TO PROBLEMS
IN
CELLULAR BIOPHYSICS
VOLUME 2: ELECTRICAL
PROPERTIES
Thomas Fischer Weiss
Department of Electrical Engineering and Computer Science
Massachusetts Institute of Technology
Spring 1997
Date of last modification: October 25, 1997
ii
iii
To Aurice B, Max, Elisa, and Eric
iv
v
PREFACE
The textbooks Cellular Biophysics, Volume 1: Transport (Weiss, 1996a) and Cellular
Biophysics, Volume 2: Electrical Properties (Weiss, 1996b) contain a collection of exercises and problems that have been developed over many years. These problems and
exercises allow students to test their comprehension of the material and they also extend the material contained in the textbooks. For learning the material, there is no
substitute for attempting to solve problems — challenging problems. Solving problems
can reveal which aspects of the material are understood and which are not yet grasped
and require further study. This solution book contains solutions to all the exercises
and problems in Volume 2; a companion solution book (Weiss, 1997) contains solutions
to all the exercises and problems in Volume 1. The purpose of making the solutions
available is to allow students to check their work. Properly used, these solutions can
be helpful for learning the material. By comparing their solutions with those in the solution book, students can obtain an objective evaluation of their comprehension of the
subject material. However, the solutions contained here are more extensive than would
be expected for a typical student. Hopefully, these more extensive solutions further explicate the material. There is one caveat concerning a possible misuse of this solution
book. If a problem is assigned in a subject that uses these textbooks, a student might
be tempted to just reproduce the solution after only a cursory reading of the problem.
This certainly saves time. However, it short-circuits the process of actively struggling
with the problem. This is where learning takes place and the subject material is assimilated. My advice to students is: resist the temptation to use the solution texts in a
counterproductive manner. Remember as you struggle to solve a problem, “no pain, no
gain.”
The practice of the faculty has been to assign problems for homework one week and
then to issue solutions to students the following week when the work was due. These
solutions started out as handwritten and progressed over the years to more elaborate
typeset solutions. Therefore, when I started the project of producing these solutions
books, I gathered together all previous solutions — multiple solutions were typically
available for the same problem assigned in different years. These solutions were written
by various faculty who have taught the subject as well as by graduate student teaching
assistants. In reviewing these solutions, I found with some chagrin that several solutions were incorrect despite the fact that they had been issued many times (by faculty
including primarily myself), checked by several teaching assistants, and presumably
read by many hundreds of students. I apologize to students who were led astray by
errant solutions, but these same students should feel guilty for not having caught the
errors. The solutions I have found in error, I have tried to fix. However, my experience
with compiling past solutions has left me a bit pessimistic about eliminating all errors
from these solutions. I invite the reader to communicate with me to point out any remaining errors. I can be reached via email at tfweiss@mit.edu. I will post errors in the
texts and in the problem solution texts on my homepage on the world wide web whose
current address is http://umech.mit.edu:80/weiss/home.html. My homepage can
be reached through the MIT home page which links to the Department of Electrical
Engineering and Computer Science homepage.
As with the textbooks, these solution books were typeset in TEX with LATEX macros
on a Macintosh computer using Textures. Spelling was checked with the LATEX spell
vi
checker Excalibur. Theoretical calculations were done with Mathematica and MATLAB.
Graphic files were imported to Adobe Illustrator for annotation and saved as encapsulated postscript files that were included electronically in the text. Mathematical annotations were obtained by typesetting the mathematical expressions with Textures and
saving the typeset version as a file that was read by Adobe Illustrator. The subject is
taught with the use of software (Weiss et al., 1992) designed to complement other pedagogic materials we have used. Some of the problems reflect access to this software,
but the software is not required to solve any of the problems.
I wish to acknowledge support from a faculty professorship donated to MIT by Gerd
and Tom Perkins. My secretaries (Susan Ross and Janice Balzer) were helpful in compiling the solutions from past years. Faculty who have taught the material also developed
problems and solutions over the years. In particular, I wish to acknowledge the contributions of my colleagues Denny Freeman and Bill Peake. Finally, my immediate family
(Aurice, Max, Elisa, Eric, Kelly, Nico, Sarah, Madison, and Phoebe), which has grown as
the writing has progressed, has continued to support me despite my obsession with
writing texts.
Chapter 1
INTRODUCTION TO ELECTRICAL
PROPERTIES OF CELLS
Exercises
Exercise 1.1 Graded and action potentials can be produced by passing electric currents
through cellular membranes. The distinction between these two potentials lies in the
current-voltage relation of the membrane. The relation between current and voltage is
discontinuous for generating action potentials and continuous for generating graded
potentials. Action potential have a sharp threshold for brief current pulses. Current
amplitudes below this threshold elicit no action potential; current amplitudes above this
threshold produce a full action potential. In general, the amplitude of graded potentials
is a monotonic function of the current amplitude.
Exercise 1.2 The current enters the micropipet and leaves at the tip which is located in
the cytoplasm of the cell. The current flows outward through the membrane and back to
the reference electrode as shown in Figure 1.1. If the current flows outward through the
membrane and the membrane can be represented by a battery in series with a resistance
then the membrane potential Vm (t) increases, i.e., the membrane depolarizes.
Exercise 1.3
a. False. In electrically excitable cells, the potentials generated below the threshold
for eliciting an action potential are graded potentials.
Ie(t)
+
Ie(t)
Rm
Vm (t)
Vm(t)
Vmo
+−
1
+
−
−
Figure 1.1: Flow of current resulting from an
intracellular electrode and
the resulting change in
membrane potential —
schematic diagram and
electric
network
(Exercise 1.2).
2
CHAPTER 1. INTRODUCTION TO ELECTRICAL PROPERTIES OF CELLS
b. True. Decrement-free conduction is a property of the conduction of action potentials.
c. False. Electrically excitable cells are defined as cells that can produce action potentials.
d. False. The mechanisms appear to be similar but not identical in all cells.
e. False. Graded potentials are involved in signal generation in neurons. As a simple
example, graded potentials exhibit spatial and temporal summation to produce
action potentials.
Exercise 1.4 About 100 mV.
Exercise 1.5 No action potential is produced for a stimulus that is below a threshold
value. Stimuli above this threshold value produce an action potential of the same waveform independent of the stimulus amplitude.
Exercise 1.6 Refractoriness refers to the increase in the threshold for eliciting an action
potential just after an action potential has occurred. Refractoriness can be quantified
by determining the minimum amplitude of the current stimulus that is required to elicit
an action potential as a function of time after an initial action potential has occurred.
This threshold of current is elevated immediately after an action potential has occurred.
Exercise 1.7 In electrical transmission there is direct electrical communication between
the pre- and post-synaptic cells, i.e., current flows between these two cells. In chemical
transmission there is virtually no direct electrical communication between the pre- and
post-synaptic cells, i.e., no appreciable current flows between these two cells. In chemical transmission there is an electrically elicited chemical secretion in the pre-synaptic
cell. The secreted chemical neurotransmitter acts on the post-synaptic cell to produce
an electrical effect.
Exercise 1.8 The largest myelinated nerve fibers in vertebrates have diameters of about
20 µm and the largest unmyelinated fibers in invertebrates have diameters of about
1 mm. Thus, the cross sectional areas are 3.14 × 10−4 and 0.785 mm2 , respectively.
Thus, there is about 2500 times more cytoplasm available per unit length of fiber in a
giant axon than in a myelinated fiber. This makes collection of cytoplasm for chemical
analysis much simpler in a giant axon.
Chapter 2
LUMPED-PARAMETER AND
DISTRIBUTED-PARAMETER MODELS
OF CELLS
Exercises
Exercise 2.1 For simplicity, consider the relations among the resistances for the same
cylindrical conductor through which current flows in the longitudinal direction as shown
in Figure 2.1. The resistivity and conductivity are material properties that relate the
electric field intensity to the current density at each point in a conductor via Ohm’s law
J = σe E or E = ρJ.
The resistivity has units of Ω·cm and is independent of the dimensions of the conductor.
The total resistance of a conductor of cross-sectional area A and length L is
R=
ρL
,
A
which for the cylindrical conductor is R = ρL/(π a2 ) and has units of Ω. The resistance
per unit length is simply
R
ρ
r =
= ,
L
A
which for the cylindrical conductor is r = ρ/(π a2 ) and has units of Ω/cm. The length
in the definition of resistance per unit length is in the direction of current flow. The
resistance per unit length is an appropriate measure of specific resistance for a onedimensional resistance, i.e., one with a constant cross-sectional area. For example, it
a
E
J
Figure 2.1: Geometry used to define various types of resistances (Exercise 2.1).
L
3
4
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
is used to specify resistances of wires. To get a certain resistance of some wire whose
resistance per unit length is known, one need only specify the length of wire needed.
Finally, the resistance of a unit area is
R = RA,
which for the cylindrical conductor is R = ρL and has units of Ω · cm2 not Ω/cm2 .
The area in the definition of resistance of a unit area is orthogonal to the direction
of current flow. This is an appropriate measure of the specific resistance of a twodimensional resistance where current flow is orthogonal to the surface area, i.e., for
example a sheet of material of constant thickness. Thus, to obtain a given resistance
with a sheet of material of known resistance of a unit area, one need only specify the
surface area required.
Exercise 2.2 In an electrically small cell, the potential difference across the membrane
does not vary over the surface of the cell. In an electrically large cell it does.
Exercise 2.3 Consider a unit length of an axon of diameter D. The longitudinal resistance ri ∝ 1/D 2 . Hence, an increase in axon diameter results in a decrease in the
longitudinal resistance. Therefore, the longitudinal current will spread further along
the axon and the conduction velocity will increase. This effect is made clear by the
experiments with space clamps in which a wire placed in the axon greatly increases the
conduction velocity. The membrane resistance rm ∝ 1/D. Thus, an increase in axon
diameter results in a decrease in the membrane resistance. A decrease in the membrane
resistance favors current flow through the membrane rather than longitudinally down
the axon. This effect will reduce the conduction velocity. To summarize, an increase
in axon diameter has two opposing effects on the conduction velocity — a decrease
in longitudinal resistance leads to an increase in conduction velocity and a decrease
in membrane resistance results in a decrease in conduction velocity. However, the decrease in longitudinal resistance is greater than the decrease in membrane resistance
because ri ∝ 1/D 2 and rm ∝ 1/D. Therefore, an increase in the axon diameter results
in an increase in conduction velocity.
Exercise 2.4 The equation states that the rate of increase in internal longitudinal current with distance equals minus the current per unit length that leaves the inner conductor, i.e., it equals the current per unit length that enters the inner conductor. Thus, if
the internal longitudinal current increases with distance then there must be net current
flowing through the membrane into the internal conductor.
Exercise 2.5 The theory predicts that the conduction velocity is proportional to the
axon diameter of unmyelinated fibers of identical membrane properties. The difficulty
may be due to the fact that nerve fibers from species that are phylogenetically distant
may have membranes whose electrical properties differ appreciably.
Exercise 2.6
a. As shown in Figure 2.2, the time from the stimulus artefact to the onset of the action potential is estimated to be 2.9 ms in seawater and 3.8 ms in oil. The distance
between stimulation and recording electrodes is 13 mm. Hence, the conduction
velocity is 4.5 mm/ms = 4.5 m/s in seawater.
EXERCISES
5
Axon in:
seawater
oil
Figure 2.2: Method used to estimate the conduction velocity from the measurements (Exercise 2.6).
seawater
oil
Axon in:
seawater
Figure 2.3: Action potentials plotted versus distance for
an axon in seawater and in oil (Exercise 2.6).
oil
0
7.6 9
Distance (mm)
20
b. By a similar argument, the conduction velocity is 3.4 m/s in oil.
c. Assume that the time dependence of the action potential is the same in oil and in
seawater except for a shift in conduction time. Therefore, in seawater Vm (13, t) =
f (t − 13/4.5), and in oil Vm (13, t) = f (t − 13/3.4). A sketch of the two waveforms versus z requires sketching Vm (z, to ) = f (to − z/4.5) for seawater and
Vm (z, t) = f (to − z/3.4) for oil. Thus, the waveforms are the same as the time
waveforms except for a time reversal and a change in the abscissa scale. Note,
that in 5 ms the wave travels 22.5 mm at a conduction velocity of 4.5 mm/ms and
17 mm at a conduction velocity of 3.4 mm/ms. The two waveforms must be scaled
appropriately to plot them on the same abscissa scale (Figure 2.3). Note that in
seawater the onset of the action potential occurs at t = 2.9 ms at location z = 13
mm. Hence, Vm (13, 2.9) = f (2.9 − 13/4.5) = f (0). The point in space at which
this onset occurs at t = 2 ms must be determined. Therefore, f (0) = f (2 − z/4.5)
so that z = 4.5 · 2 ≈ 9 mm. Similarly, in oil this onset occurs at z = 3.8 · 2 ≈ 7.6
mm. Thus, in seawater the spatial extent of the action potential is broader and its
leading edge travels farther than in oil.
Exercise 2.7 In the absence of the platinum wire, the time between peaks of the action
potentials is about 0.74 ms. Since the conduction distance is 1.6 cm, the conduction
velocity is about 16/0.74 = 22 mm/ms = 22 m/s. One could easily discern a time difference of 5 µs in the text figure (Weiss, 1996b), and the time difference of the two
action potentials is less than that duration. Hence, the conduction velocity is greater
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
1
f (z, 2)
f (z, 2)
6
0
0
100
z (mm)
200
1
0
−200
−100
z (mm)
0
Figure 2.4: Wave propagation in the +
and − z direction (Exercise 2.10). The
arrows show the direction of propagation.
than 3200 m/s. Insertion of the platinum wire increased the conduction velocity more
than 145 times.
Exercise 2.8 No. The equation was derived under the assumption that the extracellular
potential is independent of radial distance from the axon. This may be a reasonable
assumption in an extracellular volume of limited spatial extent, but will not be true in
a large extracellular volume where the potential will decrease with radial distance.
Exercise 2.9 In Equation 2.35 (Weiss, 1996b), ri depends upon a. This dependence is
taken into account in Equation 2.38. Both equations are correct, but Equation 2.38 makes
the dependence of the conduction velocity of an unmyelinated fiber on fiber radius
apparent. The derivation that leads to Equation 2.38 also assumes that ri ro which is
usually a good assumption except when axons are subjected to insulating extracellular
media such as oil.
Exercise 2.10 Since the wave propagates at constant velocity, it has the form f (z, t) =
g(t − z/ν). Thus, for two different combinations of t and z, the argument of g must
satisfy the relation t1 − z2 /ν = t2 − z2 /ν. Figure 2.20 (Weiss, 1996b) shows a plot for
z = 0 so that t1 = t2 − z2 /ν. Therefore, z2 = ν(t2 − t1 ), where z is in mm, t is in ms,
and ν is in mm/ms.
a. To find f (z, 2) for a conduction velocity of ν = 100 mm/ms, distance must be
transformed according to z2 = 100(2 − t1 ). The result is shown in the left panel
of Figure 2.4.
b. To find f (z, 2) for a conduction velocity of ν = −100 mm/ms, distance must be
transformed according to z2 = 100(t1 − 2). The result is shown in the right panel
of Figure 2.4.
Exercise 2.11 For a wave propagating in the +z-direction with propagation velocity 10
mm/ms, f (z, t) must have the form g(t − z/ν) = g(t − z/10), where t is in ms and z is
in mm. Therefore, f (20, 3) = g(3 − 20/10) = g(1). But, from the figure it is apparent
that at z = 0, f (0, t) = g(t) and g(1) = 0.5. Hence, f (20, 3) = 0.5. The problem can
also be solved graphically by sketching f (20, t) which is f (0, t) delayed 20/10 = 2 ms.
This waveform has value 0.5 at t = 3 ms.
Exercise 2.12
a. The plot of Ii (z) = −e−5z for z < 0 is shown in Figure 2.5.
b. Since there are no external currents, Ii (z) + Io (z) = 0. Therefore Io (z) = −Ii (z) =
e−5z for z < 0. The plot is shown in Figure 2.5.
EXERCISES
7
20
Io (z)
−0.4
0
−0.2
−10
Ii (z)
−0.6
−0.4
µA
−0.6
10
−20
−0.2
−40
Km (z)
−60
−80
µA/cm
−20
Figure 2.5: Plots of the longitudinal and
membrane current per unit length in a core
conductor (Exercise 2.12). In the lowest
panel, the length of the arrow is proportional to current for the longitudinal currents and to the current per unit length for
the membrane current per unit length.
−100
c. The membrane current per unit length is
Km (z) = −
∂Ii (z)
= −5e−5z for z < 0.
∂z
Km (z) is plotted in Figure 2.5.
Note that the decrease in the external longitudinal current with distance is associated
with an inward current per unit length through the membrane in order to satisfy Kirchhoff’s current law.
Exercise 2.13
a. The longitudinal current density Ji (z) in the cytoplasm is equal to the internal
longitudinal current Ii (z) divided by the cross sectional area of the cell.
Ji (z) =
µA
Ii (z)
−1 µA
=
e5z = −3183 e5z
.
2
2
πa
π (0.01 cm)
cm2
b. The membrane current per unit length Km (z) can be found using the core-conductor
relation,
d
dIi (z)
µA
Km (z) = −
=−
(−e5z ) = 5 e5z
.
dz
dz
cm
c. The membrane current density Jm (z) is the current through a unit area of membrane. This current can be found from Km (z),
Jm (z) =
µA
Km (z)
5 µA/cm 5z
.
=
e = 79.58 e5z
2π a
2π 0.01 cm
cm2
8
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
Km (z)
Ii (−1)
Ii (0)
Figure 2.6: Volume element showing the current variables associated with the inner conductor (Exercise 2.14).
−1
z (cm)
0
d. The total current Im flowing through the part of the membrane from z = −1 cm
to z = 0 can be found by integrating Km (z) over this region.
Im =
Z0
−1
Km (z) dz =
Z0
−1
0
5 e5z dz = e5z = (1 − e−5 ) µA ≈ 1 µA
−1
Exercise 2.14
a. As illustrated in Figure 2.6, the longitudinal current flowing into the intracellular
volume element at z = −1 is Ii (−1) = −e−5 , and the longitudinal current flowing
out of the volume element at z = 0 is Ii (0) = −1. Therefore, the net longitudinal
current flowing into the element is Ii (−1) − Ii (0) = −e−5 − (−1) = 1 − e−5 .
b. The membrane current flowing out of the intracellular volume element between
−1 to 0 is
Z0
Z0
0
Km (z) dz =
−5e5z dz = e5z = 1 − e−5 .
−1
−1
−1
c. The net longitudinal current flowing into the volume element equals the current
that leaves the volume element through the membrane.
Problems
Problem 2.1
a. If ro = ri , then
ρo
ρi
=
,
Ao
Ai
where Ao and Ai are the cross-sectional areas. Therefore,
ρi
,
π a2
= ρi b2
s
ρo + ρi
b = a
.
ρi
ρo
π b 2 − π a2
(ρo + ρi )a2
=
√
b. For both ro = ri and ρo = ρi , b = a 2. Therefore, for a = 500 µm, b = 707 µm
and for a = 1 µm, b = 1.4 µm
PROBLEMS
9
Propagation in −z-direction
Propagation in +z-direction
+50
Vm(3,t)
Vm(5,t)
Millivolts
Millivolts
+50
0
−50
−100
2
4
6
Time (ms)
8
−50
10
0
+50
2
4
6
Time (ms)
8
10
0
4
8
12
Distance (cm)
16
+50
Vm(z,5)
Millivolts
Millivolts
Vm(3,t)
−100
0
0
−50
−100
Vm(5,t)
0
−2
Vm(z,5)
0
−50
−100
2
6
10 14
Distance (cm)
18
−4
Figure 2.7: Graphs of the action potential plotted versus t (upper panels) and z (lower panels)
(Problem 2.2). The left panels show the results for an action potential propagating in the
+z-direction and the right panels show the results for an action potential propagating in the
−z-direction. In the graphs versus t, the action potential recorded at z = 3 cm is shown
(dashed) for reference.
c. The range of a of 1-500 µm spans most of the range of unmyelinated fibers in
animals. For these cells, the radius of the extracellular space for which the internal
and external resistances per unit length are equal is 41% larger than the radius
of the fiber. This dimension is generally small compared to the typical volume
of extracellular space in situ. Thus, for any fiber that is in a large volume of
extracellular space, such as is the case in situ, the external resistance per unit
length is much less than the internal resistance per unit length.
Problem 2.2
a. With the action potential propagating in the +z-direction at a conduction velocity
of 2 cm/ms, Vm (z, t) = f (t − z/2) where z is in cm and t in ms.
i. The position z = 5 cm is 2 cm in the +z-direction from the plotted response.
Hence the response at z = 5 is delayed by 2/2 = 1 ms as shown in Figure 2.7.
ii. Since Vm (z, t) = f (t−z/2), Vm (z, t) plotted versus z at a fixed t looks like the
plot versus t reversed in time and with the axis scaled. It is helpful to locate
the peak value of the action potential in order to sketch it. The peak value
occurs at Vm (3, 2) = f (2 − (3/2)) = f (0.5). Thus, Vm (z, 5) = f (5 − z/2) =
f (0.5) for z = 9 cm. The graph is shown in Figure 2.7.
10
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
b. With the action potential propagating in the −z-direction at a conduction velocity
of 2 cm/ms, Vm (z, t) = f (t + z/2) where z is in cm and t in ms.
i. The position z = 5 cm is 2 cm in the +z-direction from the plotted response.
Hence the response at z = 5 is advanced by 2/2 = 1 ms as shown in Figure 2.7.
ii. Since Vm (z, t) = f (t + z/2), Vm (z, t) plotted versus z at a fixed t looks like
the plot versus t with the axis scaled. The peak value occurs at Vm (3, 2) =
f (2 + (3/2)) = f (3.5). Thus, Vm (z, 5) = f (5 + z/2) = f (3.5) for z = −3 cm.
The graph is shown in Figure 2.7.
Problem 2.3
a) The conduction velocity of the peak of the action potential is
ν=
10 cm
= 2 cm/ms = 20 m/s.
5 ms
b) The platinum wire provides another path for longitudinal current flow that is in
parallel with the cytoplasm. The resistance per unit length of the platinum wire
is rpl = 160 Ω/cm. The resistance per unit length of the cytoplasm is rcy =
10π /(π )(0.025)2 = 1.6 × 104 Ω/cm. Hence, the parallel combination of these
two resistance is ri = rpl rcy /(rpl + rcy ) ≈ rpl . If the extracellular resistance is
negligible, the conduction velocities of the axon are
s
s
Km
Km
and ν2 =
,
ν1 =
2π ar1
2π ar2
for different values of the intracellular resistance per unit length with all other
factors unchanged. Thus,the ratio of conduction velocities is
s
r2
ν1
=
.
ν2
r1
p
Therefore, ν2 = 20 16000/160 = 200 m/s.
c) With the platinum wire in place, the new conduction velocity is 20000 cm/s. Since
the action potential travels 10 cm, the peak is delayed by 0.5 ms as is shown in
Figure 2.8.
d) Each of the two voltages must have the form of a travelling wave. Therefore,
z
z
V1 (t) = f t −
and V2 (t) = f t −
,
ν1
v2
which implies that
z
z
V1 t +
= f (t) and V2 t +
= f (t).
ν1
v2
Equating these two expressions yields
z
z
V1 t +
= V2 t +
ν1
ν2
PROBLEMS
11
+50
Millivolts
V2(t)
V1(t)
0
Figure 2.8: Action potential with (V2 (t))
and without (V1 (t)) the platinum wire (Problem 2.3).
−50
−100
0
5
10 15
Time (ms)
20
25
which implies that
V2 (t) = V1
z
z
t+
,
−
ν1
ν2
so that
V2 (t) = V1
10
10
−
t+
2000 20000
= V1 t + 4.5 × 10−3 t in s.
Problem 2.4
a. From the core conductor model (with no externally applied currents)
∂ 2 Vm (z, t)
= (ri + ro )Km (z, t).
∂z2
For an action potential propagating in the ±z direction, Vm (z, t) = f (t ∓ z/ν) so
that the membrane potential must satisfy the wave equation
∂ 2 Vm (z, t)
1 ∂ 2 Vm (z, t)
=
.
∂z2
ν2
∂t 2
Therefore,
Km (z, t) =
1
∂ 2 Vm (z, t)
.
(ri + ro )ν 2
∂t 2
The membrane potential and its first and second partial derivatives are shown in
Figure 2.9.
b. For the linear model of the membrane, the membrane current per unit length is
Km = cm
∂Vm (zo , t)
o
+ gm (Vm (zo , t) − Vm
).
∂t
o are posFor the time interval 0 < t < tm both ∂Vm (zo , t)/∂t and Vm (zo , t) − Vm
itive quantities. Therefore, the linear cable model predicts that Km > 0 in this
interval. However, the core conductor model for a propagated action potential
shows that Km is positive in the interval 0 < t < ti but is negative in the interval
ti < t < tm . Therefore, the linear resistance/capacitance model of a membrane is
not consistent with the membrane current during a propagated action potential.
12
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
Vm (zo , t)
0
t
tm
Vmo
ti
dVm (zo , t)
dt
0
t
d2 Vm (zo , t)
dt2
0
Figure 2.9: Membrane potential and its first
two time derivatives plotted versus time
(Problem 2.4). The time of occurrence of the
maximum value of the action potential is tm ,
and the point of inflection at the onset of the
action potential is ti .
t
Problem 2.5 There are two factors to consider in sorting the records.
• The distance between the stimulating electrode and the two sets of recording electrodes. The onset of the extracellular potential should occur earlier in the recording electrode that is closer to the stimulating electrode.
• The direction the action potential is travelling as it passes the recording electrode.
This direction determines the polarity of the extracellular potential as the following argument shows. The extracellular potential is related to the external longitudinal current as follows
∂Vo (z, t)
= −ro Io (z, t).
∂z
Integration of this equation over z for electrodes separated by the distance δ yields
Vo (z + δ, t) − Vo (z, t) = ∆Vo (z, t) = −ro δIo (z, t).
But the partial derivative of Vm (z, t) is related to the longitudinal current by
∂Vm (z, t)
= (ri + ro )Io (z, t).
∂z
Therefore,
∆Vo (z, t) = −
ro δ ∂Vm (z, t)
.
ri + ro
∂z
If an action potential is travelling in the ±z direction then Vm (z, t) = f (t ∓ z/ν)
so that
ro δ
∂Vm (z, t)
∆Vo (z, t) = ±
.
ν(ri + ro )
∂t
Thus, Vo (z, t) shows an initial increase in potential when the action potential
passes the electrode in the +z-direction (left to right) and an initial decrease in
potential when the action potential passes the electrode in the −z-direction (right
to left).
PROBLEMS
13
Using these results V1 (t) and V2 (t) can be predicted for each stimulating configuration.
A In this case, the stimulating electrode is closer to V1 (t) so that the onset should
come earlier at this electrode. Since the action potential passes both electrodes
from left to right, the extracellular potential should be initially positive. This fits
with record 2.
B In this case, the stimulating electrode is closer to V1 (t) so that the onset should
come earlier at this electrode. Since the action potential passes V1 (t) from right
to left, the initial polarity of this potential should be negative. Since the action
potential passes V2 (t) from left to right, the initial polarity of this potential should
be positive. This fits with record 7.
C In this case, the stimulating electrode is closer to V2 (t) so that the onset should
come earlier at this electrode. Since the action potential passes V1 (t) from right
to left, the initial polarity of this potential should be negative. Since the action
potential passes V2 (t) from left to right, the initial polarity of this potential should
be positive. This fits with record 8.
D In this case, the stimulating electrode is closer to V2 (t) so that the onset should
come earlier at this electrode. Since the action potential passes both V1 (t) and
V2 (t) from right to left, the initial polarity of these potentials should be negative.
This fits with record 4.
All the other records do not fit with recording arrangements A-D. Thus, records for 1,
3, 5, and 6 correspond to E.
Problem 2.6
a. Kirchhoff’s current law at node a yields
Ii (z, t) + Kei (z, t)∆z − Km (z, t)∆z − Ii (z + ∆z, t) = 0,
which can be rearranged to yield
Ii (z + ∆z, t) − Ii (z, t)
= Kei (z, t) − Km (z, t).
∆z
Taking lim∆z→0 of this equation yields
∂Ii (z, t)
= Kei (z, t) − Km (z, t).
∂z
Similarly, Kirchhoff’s current law at node d yields
Io (z, t) + Km (z, t)∆z − Keo (z, t)∆z − Io (z + ∆z, t) = 0,
which can be rearranged as
Io (z + ∆z, t) − Io (z, t)
= Km (z, t) − Keo (z, t).
∆z
14
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
Taking lim∆z→0 of this equation yields
∂Io (z, t)
= Km (z, t) − Keo (z, t).
∂z
The voltage across ri ∆z is
Vi (z, t) − Vi (z + ∆z, t) = ri ∆zIi (z + ∆z, t),
which yields
Taking lim∆z→0
Vi (z + ∆z, t) − Vi (z, t)
= −ri Ii (z + ∆z, t).
∆z
of this equation yields
∂Vi (z, t)
= −ri Ii (z, t).
∂z
Similarly, the voltage across ro ∆z is
Vo (z, t) − Vo (z + ∆z, t) = ro ∆zIo (z + ∆z, t),
which yields
Taking lim∆z→0
Vo (z + ∆z, t) − Vo (z, t)
= −ro Io (z + ∆z, t).
∆z
of this equation yields
∂Vo (z, t)
= −ro Io (z, t).
∂z
By inspection of the network, Vm (z, t) = Vi (z, t) − Vi (z, t). Therefore, the partial
derivative of this equation with respect to z yields
∂Vi (z, t) ∂Vo (z, t)
∂Vm (z, t)
=
−
= −ri Ii (z, t) + ro Io (z, t).
∂z
∂z
∂z
b. The partial derivative of the last equation with respect to z yields
∂ 2 Vm (z, t)
∂z2
∂Ii (z, t)
∂Io (z, t)
+ ro
,
∂z
∂z
= ri (Km (z, t) − Kei (z, t)) + ro (Km (z, t) − Keo (z, t)),
= −ri
= (ro + ri )Km (z, t) − ro Keo (z, t) − ri Kei (z, t).
Problem 2.7
a. The action potential must be traveling in the −z-direction. The reason is as follows.
The core conductor model shows that
ro
Vm (z, t).
∆Vo (z, t) = −
ri + ro
Figure 2.32 (Weiss, 1996b) demonstrates that on an appropriate scale Vm (zo , t) =
Vm (z, to ). This would be the case if Vm (z, t) = f (t + z/ν) for which the action potential is travelling in the −z-direction. The figure is not consistent with
Vm (z, t) = f (t − z/ν) for which the action potential is travelling in the +zdirection.
PROBLEMS
15
b. Note that the difference in scale factor for plotting Vm (z, t) versus t and versus z
is simply the conduction velocity. Hence, ν = 1 cm/ms = 10 m/s.
c. The attenuation of the action potential recorded extracellularly is simply the voltage divider ratio shown above. Therefore,
−9 = −
ro
100,
ri + ro
so that ri /ro = 10.11.
Problem 2.8 Both the magnitude of the extracellular potential and the conduction velocity depend upon the intracellular and extracellular resistance per unit length. Designate the extracellular resistance per unit length as ro1 and ro2 , for Experiments 1 and
2, respectively. Assume that the intracellular resistance per unit length is the same for
both experiments. The core conductor model implies that
2π a(ro + ri )ν 2 = Km ,
where Km is a property of the membrane and not of the dimensions of the cell. In
Experiment 1 the axon is in a large volume of sea water so that ro1 ri . Therefore, the
conduction velocity for Experiment 1 implies that
2π a(ri )(36)2 = Km .
From Experiment 2, the ratio of the external to the transmembrane potential is 75/100 =
3/4. Therefore,
ro2
3
= ,
ro2 + ri
4
which can be solved to give ro2 = 3ri . The conduction velocity of the axon in Experiment
2 can be found from the relation
2π a(ro2 + ri )ν22 = Km ,
which can be written as
2π a(4ri )ν22 = Km .
Therefore, a combination of expressions for the conduction velocity in the two experiments yields 4ν22 = 362 or ν2 = 18 m/s.
Problem 2.9 The change in extracellular potential is
∆Vo = −
ro
∆Vm ,
ro + ri
where the resistances per unit length are
ro =
ρo
ρi
and ri =
,
Ao
Ai
16
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
the resistivities of seawater and axoplasm are ρo and ρi , respectively, and the crosssectional areas of seawater and axoplasm are Ao and Ai , respectively. Substitution
yields
!
ρo /Ao
∆Vm
∆Vo = −
ρo /Ao + ρi /Ai
Since ρo = ρi ,
∆Vo = −
Ai
∆Vm .
Ai + A o
But Ai + Ao is the cross-sectional area of the seawater and axon, i.e., the cross-sectional
area of the trough. Therefore,
!
π (250)2
∆Vm = 0.58∆Vm = 58 mV.
∆Vo = −
(600)2
Problem 2.10 The extracellular potential difference ∆Vo (t) is the difference in potential
between the two voltage electrodes which are spaced 5 mm apart. At each electrode,
the extracellular potential is
Vo (z, t) − Vo (−∞) = −
ro
(Vm (z, t) − Vm (∞)) .
ro + ri
Therefore, since the potential at a location 5 mm further along the cell is delayed by
5/ν, if the conduction velocity ν is expressed in mm/ms,
∆Vo (t) = Vo (l, t) − Vo (l + 5, t) = −
Since ri = 4ro ,
∆Vo (t) −
ro
(Vm (l, t) − Vm (l, t − 5/ν)) .
ro + ri
1
(Vm (l, t) − Vm (l, t − 5/ν)) .
5
a. For ν = 1 mm/ms, the delay at the two electrodes is 5 ms. Thus, the two action
potentials are separated in time. The recorded potential is approximately two
monophasic action potentials of opposite sign separated by 5 ms as shown in
Figure 2.10.
b. For ν = 10 mm/ms, the delay at the two electrodes is 0.5 ms. Thus, the two action
potentials overlap in time. The recorded potential is a diphasic action potential as
shown in Figure 2.10.
Problem 2.11 The extracellular action potential and the conduction velocity depend
upon the longitudinal resistances per unit length.
a. By using the scales provided, the extracellularly recorded action potential can be
estimated and the results are shown in Table 2.1. The extracellularly recorded action potential is proportional to the transmembrane action potential through the
relation ∆Vo = −(ro /(ro + ri ))∆Vm . Note however, that the extracellular action
potential is diphasic because it is recorded both by the active and the reference
PROBLEMS
Millivolts
40
ν = 1 mm/ms
0
2
−40
40
Millivolts
17
4
6
8
t (ms)
Figure 2.10: The extracellular potential at two conduction velocities (Problem 2.10). The potentials recorded
at each recording electrode are shown as dashed lines,
the difference of potential is shown as a solid line.
ν = 10 mm/ms
0
2
4
6
8
t (ms)
−40
Extracellular potential (mV p-p)
ro /ri
ri /ro
Calculated ν/νintact
Estimated conduction delay (ms)
Conduction distance (mm)
Estimated ν (m/s)
Estimated ratio ν/νintact
Intact
Extruded
Inflated
83
0.71
1.4
1
2
40
20
1
3.1
0.016
63
0.19
7.3
28
3.8
0.19
117
1.4
0.71
1.2
0.58
25
43
2.2
Table 2.1: Summary of calculations of parameters for the intact, extruded, and inflated axons
(Problem 2.11).
18
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
electrodes whereas the transmembrane action potential is monophasic. Let us
make the simple assumption that when the resistive divider ratio is 1, the peak-topeak amplitude of the extracellular potential is twice the amplitude of the transmembrane action potential. For example, the intact axon has a peak-to-peak value
of 83 mV and a transmembrane potential of 100 mV. Therefore, for this case
ro /(ro + ri ) = 83/(2 · 100) = 0.42, and the solution is ro /ri = 0.71. Similar
calculation are given in the table for all 3 conditions of the axon.
b. For the intact axon
2
2π a(ri−intact + ro )νintact
= Km ,
where ri−intact is ri for the intact axon. From Table 2.1, ri−intact = 1.4ro so that
2
2π a(2.4ro )νintact
= Km .
For the extruded axon
2
2π a(ri−extr uded + ro )νextr
uded = Km ,
where ri−extr uded is ri for the extruded axon. The quantity 2π a is the perimeter
of a circular axon. It has been assumed that extrusion reduces the cross-sectional
area of the axon without changing its perimeter. This makes sense if extrusion
changes the cross-section of the axon from circular to elliptic. Therefore, it is
assumed that the perimeter of the extruded axon is also 2π a, where a is the
radius of the intact axon. Because ri−extr uded = 63ro ,
2
2π a(64ro )νextr
uded = Km .
2
2
)/(64νextr
Therefore, (2.4νintact
uded ) = 1, so that νextr uded /νintact = 0.19. This
result is given in Table 2.1 in the row labeled “Calculated ν/νintact ”. For the
inflated axon
2
2π a(ri−inf lated + ro )νinf
lated = Km ,
where ri−inf lated is ri for the inflated axon. Since ri−inf lated = 0.71ro ,
2
2π a(1.7ro )νinf
lated = Km .
Therefore, the ratio of the conduction velocity of the inflated axon to that of the
intact axon can be computed and is given in Table 2.1. Thus, extruding the axon
increases the intracellular resistance per unit length and decreases the conduction
velocity. Reinflation reverses these trends.
c. To estimate the conduction velocity, first estimate the time it takes for the initiation of the action potential. This conduction delay can be estimated from the
stimulus artefact — discernible as a small excursion on the baseline preceding the
action potential in each trace — to the onset of the action potential. These estimated conduction delays are indicated in Table 2.1. By dividing these times into
the conduction distance for each case, the conduction velocity can be estimated.
Ratios of conduction velocities of the extruded and inflated axons to that of the
intact axons are estimated and these can be compared with those values calculated
PROBLEMS
19
from the amplitude of the action potential. The ratios for the extruded axon are
0.19 for both the estimated and the calculated conduction velocity. The ratios for
the inflated axon are 2.2 for the estimated and 1.2 for the calculated conduction
velocity. So these two methods of computing the ratio of conduction velocities —
one based on the amplitude of the action potential and the other based directly on
measurements of the conduction velocity — agree qualitatively. However, there
are quantitative differences for the inflated axon. There are a number of possible
sources of error in these estimates. For example, the conduction delay has been
defined in a somewhat arbitrary manner. Another method of estimating the conduction delay, and hence the conduction velocity, would give somewhat different
results. In addition, the estimate of the peak value of the monophasic extracellular action potential from the measured peak-to-peak value of the diphasic action
potential is crude. Overlap of the waveforms recorded by the two electrodes results in an underestimate of the monophasic action potential, and thus leads to
an underestimate of ro /ri .
Problem 2.12 The conduction velocity, ν, of an unmyelinated axon in an extensive extracellular solution is
s
Km a
,
ν=
2ρi
where a is the radius, ρi is the resistivity of cytoplasm, and Km is a constant. Since
all the fibers are identical except for length and diameter, the ratios of velocities of two
axons equals the square root of their radii,
s
a1
ν1
=
.
ν2
a2
a. Based on the above relation, the conduction velocities are computed for all axons
and shown in Table 2.2. The conduction times are computed by dividing the length
of the axon by the conduction velocity. This time is shown in column a.
b. If all the axons had the same diameters as the smallest diameter axon, they would
have the same conduction velocity as the smallest axon. The conduction times
based on this velocity are shown in column b.
c. If all the axons had the same diameter as the giant axon (largest diameter), they
would have the same conduction velocity as the giant axon. The conduction times
based on this velocity are shown in column c.
d. The maximum difference in conduction times of the different fibers is: 5.01 ms
in part a; 13.38 ms in part b; and 6 ms in part c. Therefore, the differences in
diameter and length do give significant synchronization over that which would
occur if all the fibers were of small diameter (5.01 versus 13.38 ms). However, if
all the fibers where as large as the giant fiber then synchronization would decrease
by only 1 ms. However, that would require that all nine fibers were giants. So the
hypothesis is generally true although somewhat simplistic.
20
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
Axon #
Length
mm
Diameter
µm
Conduction
Velocity
(m/sec)
1
2
3
4
5
6
7
8
9
23
22
22
24
43
48
76
91
142
131
90
86
148
227
291
291
317
447
10.83
8.97
8.77
11.51
14.25
16.14
16.14
16.84
20.00
Conduction Time
(msec)
a
b
c
2.12
2.45
2.51
2.09
3.02
2.97
4.71
5.4
7.1
2.57
2.45
2.51
2.68
4.79
5.35
8.47
10.14
15.83
1.15
1.10
1.10
1.20
2.15
2.40
3.80
4.55
7.1
Table 2.2: Conduction velocity and conduction times for axons in the stellate ganglion (Problem 2.12).
Problem 2.13
a. The second partial derivative of the relation Vm (z, t) = Vi (z, t) − Vi (z, t) yields
∂ 2 Vi (z, t) ∂ 2 Vo (z, t)
∂ 2 Vm (z, t)
=
−
.
∂z2
∂z2
∂z2
The core conductor equations yield
∂Vi (z, t)
∂Ii (z, t)
= −ri Ii (z, t) and
= −Km (z, t).
∂z
∂z
A combination of these three equations yields
∂ 2 Vm (z, t)
∂ 2 Vo (z, t)
=
r
K
(z,
t)
−
.
i
m
∂z2
∂z2
b. Examination of Equation 2.25 (Weiss, 1996b) shows that
−ro Ke (z, t) = −
so that
Ke (z, t) =
∂ 2 Vo (z, t)
,
∂z2
1 ∂ 2 Vo (z, t)
.
ro
∂z2
c. Note that because the extracellular potential is independent of z the effective external current, which depends on the second partial derivative of the potential
with respect to z, is zero. This stimulus is ineffective in stimulating the cell.
d. A potential field whose second partial derivative with respect to z is not zero will
stimulate the cell. For example, suppose
Vo (z, t) =
z2
sin 2π f t.
2
PROBLEMS
21
The equivalent external current for this stimulus is
Ke (z, t) =
1 ∂ 2 Vo (z, t)
1
=
sin 2π f t.
ro
∂z2
ro
22
CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS
Chapter 3
LINEAR ELECTRICAL PROPERTIES OF
CELLS
Exercises
Exercise 3.1 The space and time constants are measures of the spatial and temporal
scales over which the membrane potential of the cell changes. They are defined as
1
cm
and τM =
,
λC = p
g
gm (ri + ro )
m
where gm and cm are the conductance and capacitance per unit length of the cell, respectively, and ri and ro are the resistances per unit length inside and outside the cell,
respectively.
Exercise 3.2 The time constant of a cylindrical cell is
τM =
cm
2π aCm
Cm
= τM =
=
,
gm
2π aGm
Gm
so that the time constant depends only on the specific capacitance and conductance of
the membrane and not on the dimensions of the cell. The space constant of a cylindrical
cell is
1
1
=q
,
λC = p
gm (ri + ro )
2π aGm ((ρi /(π a2 )) + ro )
so that the space constant depends upon the dimensions of the cell.
Exercise 3.3 An increase in the membrane conductance allows more of the current from
one section of a cell to cross the membrane, and hence, less of the current flows longitudinally down the cell. Therefore, the spatial spread of longitudinal current is decreased.
As an extreme example, imagine if the conductance were infinite (i.e., a short circuit)
then no current would flow longitudinally down the cell and the space constant would
be zero.
Exercise 3.4 An infinitesimal electrode is one that is so small that making it smaller
(e.g., by a factor of 10) would not appreciably change the electrical responses to stimulation through this electrode. To be infinitesimal, an electrode must have dimensions
that are small compared to the space constant λC of the cell.
23
24
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
Exercise 3.5
a. True. An electrically small cell has an equivalent electric network that is a lumpedparameter model consisting of a resistance and a capacitance. The membrane
potential is governed by a first-order, ordinary differential equation with constant
coefficients whose solution is an exponential function of time.
b. True. This follows from the definition of an electrically small cell.
c. False. An electrically large cell has an equivalent network that is a distributedparameter network with a series resistance and a shunt element that contains a
parallel combination of a resistance and a capacitance. The membrane potential
is governed by a partial differential equation with constant coefficients whose
solution for a step of current is given in terms of error functions.
d. True. For an electrically large cell, the steady state value of the membrane potential
satisfies a second-order ordinary differential equation with exponential solutions.
e. False. See d).
Exercise 3.6 Gm is the conductance per unit area of membrane, Gm is the total conductance of the membrane of a cell. Gm = AGm where A is the surface area of the
cell. gm is the membrane conductance per unit length of a cylindrical cell. Therefore,
gm = 2π aGm , where a is the radius of the cell.
Exercise 3.7 The characteristic conductance is the conductance between the inside and
outside of an infinite length cable. Alternatively, if the cable has a finite length and
is terminated in a conductance that equals the characteristic conductance, then the
conductance between the inside and outside equals the characteristic conductance.
Exercise 3.8 If the dimensions of a cell are small compared to the space constant then
the membrane potential does not change appreciably along the surface of the cell. This
is what is meant by an electrically small cell.
Exercise 3.9 The normalized extracellular potential is
vo (z)
= e−z/λC ,
vo (0)
so that
log10
Therefore, for z = λC
log10
vo (z)
vo (0)
vo (λC )
vo (0)
=
−z
log10 e.
λC
= − log10 e.
Therefore, the logarithm of the normalized potential is reduced by log10 e when z = λC .
Exercise 3.10
PROBLEMS
25
a. The external current produces a discontinuity in the external longitudinal current
at z = 0. Since there is no external current applied intracellularly in this case, the
internal longitudinal current is continuous.
b. As ri /ro approaches zero the internal and external longitudinal currents in the
extrapolar region remain equal in magnitude and opposite in sign. However, because the internal resistance is small, the internal voltage is small. In the limit as
bi (λ) → 0 and v
bo (λ) → −v
bm (λ).
ri /ro → 0, v
bi (λ) ≈ v
bo (λ). In addition, in this limit the
bm (λ) → 0. Hence, v
c. For λ → ∞, v
longitudinal currents become independent of λ. Therefore, in this limit the two
longitudinal conductors are in parallel and the voltage across them is
ri ro
vo (z) = vi (z) = −
zIe .
ri + ro
bo = vo /(ro λC Ie /2) and v
bi = vi /(ro λC Ie /2)
The normalized variables are defined as v
so that
2α
2ri
bo (λ) = v
bi (λ) = −
v
λ,
λ=−
ri + ro
α+1
bi (λ) = v
bo (λ)
where α = ri /ro . Therefore, as α → 0, the magnitude of the slope of v
versus λ decreases.
Exercise 3.11 No. The step response of the cable is attenuated as a function of position. Thus, the step response does not have the form of a wave propagated at constant
velocity, i.e., vm (z, t) 6= f (t − z/ν).
Exercise 3.12 As shown in Figure 3.37 the step response of an electrically large cell is
more rapid than the step response of an electrically small cell with the same membrane
time constant.
Exercise 3.13 Equation 3.5 is the step response of a small cell driven by a current applied across the membrane. Equation 3.52 is the step response of a large cell driven
by a current applied extracellularly; the form of the solution for an intracellularly applied current is similar. Thus, the difference in the equations predominantly reflects
the difference in the step responses of small and large cells.
Problems
Problem 3.1
a. The conductance per unit length is
gm =
1
λ2C ri
where ri =
ρi
.
π a2
Therefore,
gm =
π a2
λ2C ρi
=
π (0.025)2
= 0.182 × 10−3 S/cm.
(0.6)2 · 30
26
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
Rs
Im(t)
Im(t)
+
+
Rm
Rm
Vm(t)
Cm
Source
+
Vmo −
Vm(t)
Cm
Source
+
Vmo −
−
−
Figure 3.1: The left panel shows the circuit model appropriate for parts a and b of Problem 3.2,
and the right panel is for part c.
b. The conductance per unit area is
Gm =
a
gm
0.025
=
= 1.16 × 10−3 S/cm2 .
=
2
2 · 30
2π a
2
·
(0.6)
2λC ρi
c. The cell space constant is related to the cable parameters as
λ2C =
1
gm ri
=
1
a
.
=
2
(2π aGm )(ρi /(π a ))
2Gm ρi
Therefore,
√
√
√
1
1
1
=p
a= p
a = 3.8 a.
λC = √
gm ri
2Gm ρi
2 · 1.16 × 10−3 · 30
Therefore, the space constant is: 8.5 mm for a diameter of 1 mm, 2.7 mm for
a diameter of 0.1 mm, 0.85 mm for a diameter of 0.01 mm, and 0.27 mm for a
diameter of 0.001 mm.
d. The time constant of the cell is
τM =
cm
2π aCm
Cm
1 × 10−6
=
=
=
= 0.86 ms.
gm
2π aGm
Gm
1.16 × 10−3
Problem 3.2 The specification of the problem suggests the cell/source models shown
in Figure 3.1.
a. If the source is a current source where Im (t) = Iu(t), where u(t) is the unit step
function, then Kirchhoff’s current law yields
Iu(t) = Cm
o
dVm (t) Vm (t) − Vm
.
+
dt
Rm
PROBLEMS
27
t
CmVδ(t)
τ
Vm(t)
V/Rmu(t)
IRm
V/Rm
Vmo
t
Figure 3.2: The voltage response to a step of current (left panel) and the current response to a
step of voltage (right panel) for a space-clamped axon (Problem 3.2).
This is a first-order, ordinary, linear differential equation in Vm (t) with constant
o to
coefficients. Therefore, Vm (t) charges exponentially from its initial value Vm
o
its final value Vm + IRm with a time constant τM = Rm Cm . Therefore,
o
+ IRm (1 − e−t/τM ),
Vm (t) = Vm
which is shown in Figure 3.2.
o + V u(t) then Kirchb. If the source is an ideal voltage source such that Vm (t) = Vm
hoff’s current law yields
Im (t) = Cm
o
dVm (t) Vm (t) − Vm
+
,
dt
Rm
which gives
Im (t) = Cm V
du(t)
V
u(t)
= Cm V δ(t) +
u(t),
+V
dt
Rm
Rm
where δ(t) is the unit impulse function. Im (t) is shown in Figure 3.2.
c. If the source is an ideal voltage source in series with a resistance Rs , Kirchhoff’s
current law yields
o
dVm (t) Vm (t) − Vm
Im (t) = Cm
,
+
dt
Rm
and
Im (t) =
V u(t) − Vm (t)
.
Rs
Combining these two equations to eliminate Im (t) yields, after some rearrangement of terms
Vo
V
dVm (t) Vm (t)
Cm
+
=
u(t) + m ,
dt
Rp
Rs
Rm
where Rp = Rm Rs /(Rm + Rs ). Hence, Vm (t) has an initial value
Rs
o
,
Vm
Vm (0) =
Rm + R s
and a final value
Vm (∞) =
Rs
Rm + Rs
o
Vm
Rm
+
Rm + Rs
V,
28
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
r=0.01
1
r=0.1
v(t0)
0.8
0.6
r=1
Figure 3.3: The effect of the series resistance on
the potential across the membrane with normalized
series resistance as the parameter (Problem 3.2).
0.4
0.2
r=10
r=100
0
1
2
t0
3
4
5
and a time constant of τ = Rp Cm , so that
Vm (t) =
Rs
Rm + Rs
o
Vm
+
Rm
Rm + Rs
V 1 − e−t/τ .
To sketch this response for different values of Rs /Rm , it is helpful to express the
response in normalized coordinates. Define
1
Rs
o
v(t) =
Vm (t) −
Vm
,
V
Rm + Rs
where r = Rs /Rm , and t 0 = t/Rm Cm , then
v(t 0 ) =
1 0
1 − e−t (1+r )/r .
1+r
Therefore, v(t 0 ) represents the increment in voltage caused by the source divided
by the magnitude of the source voltage. The sketch shown in Figure 3.3 shows
v(t 0 ) for t 0 ≥ 0 (provided Rs 6= 0 which was assumed implicitly in the derivation).
Note that as the series resistance is decreased, the response approaches that for
an ideal (zero series resistance) voltage source. The membrane potential follows
the voltage step more faithfully in time and approaches a final value of 1. For
example, if the series resistance is 1% of the membrane resistance, the final value
of the voltage change across the membrane is 99% of the voltage step, and the
time constant for the change in voltage is 1% of the membrane time constant. As
the series resistance increases, it takes more time to charge the capacitance. For
Rs Rm , Rp → Rm and τ → Rm Cm . In addition, as Rs increases, the final value
of Vm is less than 1 because of the voltage division that occurs between Rs and
Rm . This problem illustrates the importance of minimizing the series resistance
when the potential across the membrane of a cell is to be controlled electronically,
i.e., with a voltage-clamp circuit.
Problem 3.3
a. Kirchhoff’s current law for the node in the cell yields
im (t) = u(t) = C
dvm (t)
+ Gvm (t).
dt
PROBLEMS
29
The solution to this first-order differential equation with constant coefficients is
1
1 − e−t/τM u(t),
vm (t) =
G
where τM = C/G. The series resistance has no effect on the potential across the
membrane when the membrane is driven by a current source. The source current
equals the membrane current.
b. Kirchhoff’s current law for the node in the cell yields
dvm (t)
vm (t) − u(t)
+ Gvm (t) = 0,
+C
Rs
dt
which can be rewritten as follows
1 + Rs G
Rs G
C dvm (t)
+
G
dt
The solution is
vm (t) =
!
vm (t) =
1
u(t).
Rs G
1
1 − e−t/τM u(t),
1 + Rs G
where τM = Rs C/(1 + Rs G). Thus, because of the presence of the series resistance, the potential across the membrane does not equal the source potential. The
final value of the potential is less than the source potential and the membrane potential changes with a time constant that depends on the series resistance. As the
series resistance approaches zero, the membrane potential approaches the source
potential. For a further analysis of the role of the series resistance see Problem 3.2.
Problem 3.4 This problem is solved in the Section 3.4.2 which shows that
vm1 (z) =
vm2 (z) =
ro λC1
I1 e−|z|/λC1 ,
2
ro λC2
I2 e−|z|/λC2 .
2
The space constant is
1
1
1
=
≈√
=q
λC = p
ri gm
(ri + ro )gm
(ρi /π a2 )(2π aGm )
s
a
.
2ρi Gm
Therefore,
s
λC1
=
s
λC2
=
10−2
= 0.1 cm = 1 mm,
· 5 × 10−3
2 · 102
2 · 102
10−3
= 0.2 cm = 2 mm.
· (1/8) × 10−3
a. From the results obtained above,
A=
vm1 (−0.1)
vm1 (0)e−0.1/λC1
e−1
=
=
= e−0.5 = 0.61.
vm2 (−0.1)
vm2 (0)e−0.1/λC2
e−0.5
30
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
Ie
−
Ve
+
Sea water
Axon
Figure 3.4: Coordinate system
for solving cable problem (Problem 3.5) and a schematic diagram
of the solution as a superposition
of solutions obtained at one electrode.
Sea water
−l
2
vm (z)
0
l
2
z
z
b. Since vm1 (0) = vm1 (0)
ro λC1
ro λC2
I1 =
I2 ,
2
2
which implies that
B=
λC2
I1
=
= 2.
I2
λC1
Problem 3.5 The coordinate system shown in Figure 3.4 is used to solve this problem.
Section 3.4.2 (Weiss, 1996b) gives the time-independent solution for a constant current
applied at z = 0 which can be represented as a current per unit length of the form
ke (z) = Ie δ(z). The solution is
vm (z) =
ro λC
Ie e−|z|/λC .
2
This solution is used to solve the present problem.
a. The external current per unit length is
ke (z) = Ie δ(z + l/2) − Ie δ(z − l/2).
Because the cable equation is linear and space invariant, solutions can be superimposed as shown in Figure 3.4. Therefore,
vm (z) =
ro λC
ro λC
Ie e−|z+l/2|/λC −
Ie e−|z−l/2|/λC .
2
2
This relation is evaluated in three separate regions

ro λC (z+l/2)/λC

(z−l/2)/λC

−
e
I
e
for z < −l/2,

e



 2
r o λC
vm (z) =
Ie e−(z+l/2)/λC − e(z−l/2)/λC
for −l/2 < z < l/2,


2




 ro λC Ie e−(z+l/2)/λC − e−(z−l/2)/λC
for z > l/2.
2
PROBLEMS
31
These equations can be written alternatively as


l


λ
I
sinh
for z < −l/2,
ez/λC
r

o
C
e


2λ
C


z
−ro λC Ie e−l/(2λC ) sinh
for −l/2 < z < l/2,
vm (z) =

λC




l


e−z/λC for z > l/2.
 −ro λC Ie sinh
2λC
The solution depends upon a number of parameters. In order to gain some insight into this solution, it is helpful to use normalized variables. Define v =
vm /(ro λC Ie ), λ = z/λC , and L = l/λC . Therefore,

λ

 sinh(L/2)e
−e−L/2 sinh λ
v(λ) =


− sinh(L/2)e−λ
for λ < −L/2,
for −L/2 < λ < L/2,
for λ > L/2.
The normalized membrane potential (Figure 3.5) depends on only one parameter
L which is the inter-electrode distance divided by the cell space constant. When
the inter-electrode distance is much larger than the space constant (i.e., L 1),
the solutions under the two electrodes are separated in space. When the interelectrode distance is comparable in length to the space constant (i.e., L ≈ 1), the
responses at the two electrodes overlap. When this distance is very small (i.e.,
L 1), the membrane potential has a small magnitude and a spatial distribution
that resembles the spatial derivative of the solution at one electrode. This results
because the stimulus approaches a current dipole at λ = 0.
b. The relation between the membrane potential and the longitudinal currents is
given by
dvm (z)
= −ri ii (z) + ro io (z).
dz
The longitudinal current must also satisfy Kirchhoff’s current law which implies
that
(
−Ie for |z| < l/2,
ii (z) + io (z) =
0
for |z| > l/2.
Combining these equations yields
 ri
dvm (z)
1


−
Ie

dz
ri + ro
ri + ro io (z) =
dvm (z)
1



ri + ro
dz
for |z| < l/2,
for |z| > l/2.
Derivatives with respect to z yield
 l
ro


Ie sinh
ez/λC



r
+
r
2λ
C

 i r o ri
z
o
−l/(2λC )
I
−
Ie
e
cosh
−
io (z) =
e

ri + ro
λC
ri + ro




l
ro


Ie sinh
e−z/λC
ri + r o
2λC
for z < −l/2,
for −l/2 < z < l/2,
for z > l/2.
32
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
v(λ)
i(λ)
0.5
0.4
−6
0.2
L = 10
−6
−4
−2
−0.2
2
4
−4
λ
6
−2
2
4
6
λ
2
4
6
λ
2
4
6
λ
−0.5
−1
−0.4
−1.5
0.3
0.2
-6
-4
-2
0.1
L=1
−6
−4
−2
−0.1
-0.5
2
4
6
λ
-1
−0.2
-1.5
−0.3
−6
0.04
−6
−4
−2
−0.02
−0.04
−2
−0.5
0.02
L = 0.1
−4
2
4
6
λ
−1
−1.5
−2
Figure 3.5: The spatial distribution of membrane potential (left column) and the external longitudinal current (right column) in normalized coordinates (Problem 3.5). The parameters are
L = 10, L = 1, L = 0.1 and α = 1.
PROBLEMS
33
The same normalized variables used for vm (z) are used for io (z). In addition,
i = io (ri + ro )/(ro Ie ) and α = ri /ro so that

λ

for λ < −L/2,
 sinh(L/2)e
−L/2
−e
cosh λ − α for −L/2 < λ < L/2,
i(λ) =


sinh(L/2)e−λ
for λ > L/2.
The external longitudinal current is plotted in normalized coordinates in Figure 3.5. When the inter-electrode distance is much larger than the space constant
(i.e., L 1), the solutions under the two electrodes are separated in space. At
λ = 0, the external longitudinal current approaches the negative of the magnitude
of the external current source multiplied by a current divider ratio (ro /(ro + ri )).
When the inter-electrode distance is comparable in length to the space constant
(i.e., L ≈ 1), the responses at the two electrodes overlap. When this distance is
very small (i.e., L 1), the external longitudinal current has a small magnitude
in the extrapolar region and contains a sharp spike of current centered on λ = 0
which is the location of the two electrodes. Thus, very little of the external current enters the cell and appreciable external current occurs only in the interpolar
region where the current equals the negative of the external current.
c. The voltage Ve is computed from io (z) obtained in part b,
Ve
= −
Z l/2
ro io (z) dz,
−l/2
Z l/2 ro
ri ro
dvm (z)
−
Ie dz,
ri + r o
dz
ri + ro
−l/2
ri ro
ro
(vm (l/2) − vm (−l/2)) +
lIe ,
= −
ri + ro
ri + ro
!
rr l ro2 λC
i o
−l/λC
=
+
Ie .
Ie 1 − e
ri + ro
ri + ro
= −
Therefore,
Re =
r 2 λC Ve
ri ro l
= o
.
1 − e−l/λC +
Ie
ri + ro
ri + ro
This resistance is examined in two limits. For l/λC 1
Re
Re
Re
Re
ro2 λC
ri ro l
,
(1 − (1 − l/λC )) +
ri + ro
ri + ro
ro2 l
ri ro l
≈
+
,
ri + ro
ri + ro
ro l
≈
(ro + ri ),
ri + ro
≈ ro l.
≈
Thus, if the distance between the electrodes is much less than the space constant,
relatively little current enters the cell. The equivalent resistance is the external
34
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
+
−
Ve
Ie
Figure 3.6: The arrangement for measuring the membrane
resistance of an axon (Problem 3.6).
−d
0
2
d
z
2
resistance per unit length times the distance between the electrodes through which
the external current flows. For l/λC 1
Re
≈
Re
≈
ro2 λC
ri ro l
+
,
ri + ro
ri + ro
r i ro l
.
ri + ro
Thus, if the distance between the electrodes is much greater than a space constant, the equivalent resistance is a resistance per unit length that is the parallel
combination of ri and ro times the inter-electrode distance.
Problem 3.6 To analyze this problem, define the coordinates shown in Figure 3.6. The
method is to determine the membrane potential versus position and the external longitudinal current and then to match boundary conditions.
a. In general the time-independent solution of the cable equation can be written as
vm (z) = A1 e−z/λC + A2 ez/λC or vm (z) = A1 sinh(z/λC ) + A2 cosh(z/λC ).
Because of the symmetry of the stimulation and recording arrangement, the expression with hyperbolic functions is more convenient for this problem. It is apparent that vm (z) is an odd function of z since current flows into the membrane
for z < 0 and out of the membrane for z > 0. Therefore, the form of the solution
is
vm (z) = A sinh(z/λC ).
The membrane potential is related to the longitudinal currents by the relation
dvm (z)
= −ri ii (z) + ro io (z).
dz
The longitudinal currents have the property
ii (z) + io (z) − Ie = 0 for |z| < d/2,
so that a combination of these equations yields
io (z) =
=
ri
dvm (z)
1
+
Ie ,
ri + ro
dz
ri + ro
ri
A
cosh(z/λC ) +
Ie .
(ri + ro )λC
ri + ro
PROBLEMS
35
The longitudinal current io (±d/2) = Ie , i.e., the external current equals the external longitudinal current at the edges of the oil gap. Therefore,
io (±d/2) = Ie =
A
ri
cosh(±d/2/λC ) +
Ie ,
(ri + ro )λC
ri + ro
which can be solved for A
A=
ro λC Ie
.
cosh(d/(2λC ))
Therefore, the solutions are
sinh(z/λC )
,
cosh(d/(2λC ))
ri
ro
cosh(z/λC )
+
Ie
Ie
ri + ro cosh(d/(2λC )) ri + ro
vm (z) = ro λC Ie
io (z) =
The external potential difference between the two electrodes is
Z d/2
ro io (z) dz,
Ve =
−d/2
=
d/2
2ro2 λC
ri ro
sinh(z/λC )
Ie
Ie z
,
+
ri + ro cosh(d/(2λC )) ri + ro
−d/2
=
2ro2 λC
ri ro d
Ie tanh(d/(2λC )) +
Ie .
ri + ro
ri + ro
Therefore,
R=
2ro2 λC
ri ro d
Ve
=
tanh(d/(2λC )) +
.
Ie
ri + ro
ri + ro
b. The value of the resistance for small values of d is
2ro2 λC d
ri ro d
+
,
ri + ro 2λC
ri + ro
ro2 d
ri ro d
+
,
≈
ri + ro
ri + ro
≈ ro d.
R ≈
The value of the resistance for large values of d is
R≈
2ro2 λC
ri ro d
+
.
ri + ro
ri + ro
Therefore, the measurements are used to estimate the slope for small values of d,
and the straight line (slope and intercept) fit to the data at large values of d. From
the measurements shown in Figure 3.7,
22 × 103
= 5.4 × 104 Ω/cm,
0.41
15.5 × 103
final slope =
= 1.3 × 104 Ω/cm,
1.2
final intercept = 7 × 103 Ω.
initial slope =
36
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
J
J
R (kΩ)
20
JJ
J
JJ
JJ
J
JJ
0X
10
J
0
J
J
J
J
Figure 3.7: Estimation of parameters from measurements (Problem 3.6). The initial slope is indicated by a line with short dashes, the final slope
with a line with long dashes. The line showing the
final slope is also shown translated to the origin.
5
d (mm)
10
Therefore, from the initial slope ro = 5.4 × 104 Ω/cm and from the final slope
ri ro
= 1.3 × 104 Ω/cm,
ri + ro
from which ri = 1.7 × 104 Ω/cm. From the intercept
2ro2 λC
= 7 × 103 Ω,
ri + ro
p
which is solved to yield λC = 0.085 cm. The space constant is λC = 1/ gm (ri + ro )
so that
gm =
1
λ2C (ri
+ ro )
=
1
= 1.9 × 10−3 S/cm.
(0.085)2 (1.7 × 104 + 5.4 × 104 )
The specific resistance of the membrane can be obtained from the conductance
per unit length if the radius of the axon is known.
Problem 3.7
a. The time-independent solution
of the cable equation is exponential in space with
p
space constant λC = 1/ (gm (ri + ro )). The general solution is of the form
vm (z) = A1 e−z/λC + A2 ez/λC for z > 0.
On physical grounds, the solution will converge for z → ∞. Since vm (0) = V1 , the
solution is
vm (z) = V1 e−z/λC for z > 0,
which is plotted in Figure 3.8.
b. The core conductor equation yields
1
1
dvm (z)
−V1 −z/λC
=−
e
=
ii (z) = −
ri + r o
dz
ri + ro λC
s
gm
V1 e−z/λC .
ri + ro
PROBLEMS
37
V1
vm (z)
Figure 3.8: Membrane potential versus position in
a dendrite (Problem 3.7).
z
λC
c. Combining the results of parts a and b yields
s
ii (z) =
gm
vm (z),
ri + ro
so that
vm (0)
1
.
=p
ii (0)
gm /(ri + ro )
d. If ri ro then
s
Gc =
gm
≈
ri + r o
s
gm
=
ri
s
2π aGm
=
ρi /(π a2 )
s
2π 2 a3 Gm
.
ρi
e. At the bifurcation, the internal current divides between the two branches, ii =
2ib = 2Gb vm , where ib is the internal longitudinal current in each branch of radius
b, and Gb is the characteristic conductance of these branches. The current in the
unbranched dendrite is ii = Ga vm , where Ga is the characteristic conductance of
the dendrite of radius a. Therefore, if Ga = 2Gb then the electrical load on the
cell is the same, i.e.,
s
s
2π 2 a3 Gm
2π 2 b3 Gm
=2
,
ρi
ρi
which shows that b = a/41/3 .
Problem 3.8 This problem investigates the stimulation of a cell by an external current
and the role of the space constant of the cell in determining the effectiveness of the
current stimulus.
a. The symmetry of the stimulation arrangement determines which solution is correct. We expect the membrane current to enter the fiber for z < 0 and to leave the
fiber for z > 0. Hence, the membrane current per unit length is an odd function
of z. Because for the steady-state cable equations km (z) = gm vm , the membrane
potential is also an odd function of z. This conclusion also follows from the core
conductor relation for which the membrane current per unit length is proportional
to the second derivative of the membrane potential. Both arguments show that
the membrane potential is an odd function of z. Therefore, the solution has the
form vm (z) = A sinh(z/λC ).
38
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
b. A core conductor model for incremental quantities relates the longitudinal current,
for which boundary conditions are given, to the membrane potential as follows:
dvm (z)
= −ri ii (z) + ro io (z).
dz
Hence,
A
cosh(z/λC ) = −ri ii (z) + ro io (z).
λC
A boundary condition is that ii (±L) = 0. Since ii (z) + io (z) = I, this implies that
io (±L) = I. Therefore,
A
cosh(L/λC ) = ro I,
λC
and
A=
ro λC I
.
cosh(L/λC )
Therefore,
vm (z) =
ro λC I
sinh(z/λC ).
cosh(L/λC )
This equation is expressed as
sinh(z/λC )
vm (z)
=
,
ro λC I
cosh(L/λC )
which is plotted in Figure 3.9. The membrane potential difference is small for
an electrically small cells (L/λC = 0.1) and large for a large electrically large cell
(L/λC = 10).
c. The external longitudinal current is obtained by eliminating ii (z) from the core
conductor equation to yield
dvm (z)
= −ri (I − io (z)) + ro io (z) = (ri + ro )io (z) − ri I,
dz
which can be solved for io (z) to yield
io (z) =
=
ri
1
dvm (z)
+
I,
ri + r o
dz
ri + ro
cosh(z/λC )
I
,
ri + ro
ri + r o
cosh(L/λC )
which can be written as
1
cosh(z/λC )
io (z)
=
1 + (ro /ri )
.
I
1 + (ro /ri )
cosh(L/λC )
The internal longitudinal current is ii (z) = I − io (z) which is
ii (z)
(ro /ri )
cosh(z/λC )
=
1−
.
I
1 + (ro /ri )
cosh(L/λC )
The longitudinal currents are plotted in Figure 3.9.
PROBLEMS
39
vm (z)
ro λ C I
1
io (z)
I
ii (z)
I
1
0.5
L
= 0.1
−0.1
λC
−0.05
0.05
0.5
0.1
−0.5
−1
−0.1 −0.05
1
0.05
1
0.5
L
=1
λC
−1
−0.5
0.5
−1
−0.5
1
−1
0.5
1
0.1
10
5
−0.5
ii (z)/I
1
0.5
−5
io (z)/I
10
1
0.1
−1
−10
0.1
1
10
0.5
1
−0.5
L
= 10
λC
0.1
1
0.5
10
1
z/λC
10
−10
0.1
−5
z/λC
5
10
Figure 3.9: The membrane potential vm (z) (left panels) and the external (solid lines) and internal (dashed lines) longitudinal currents (right panels) are plotted versus z/λC for the different
values of L/λC shown at the left (Problem 3.8). The longitudinal currents are shown for values
of ro /ri of 0.1, 1, 10.
40
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
I
+
V
−
Figure 3.10: Schematic diagram of current
flow in and around a muscle fiber (Problem 3.8). The length of each arrow indicates the magnitude of the longitudinal current and the current per unit length for the
membrane current per unit length. For an
electrically small cell (L/λC 1), no appreciable current enters the cell and most
of the current flows extracellularly between
the electrodes. For an electrically large cell
(L/λC 1), appreciable current enters the
cell.
L/λC ¿ 1
I
+
V
−
L/λC À 1
−L
0
z
L
These results are examined for electrically small cells first and then for large electrically large cells. For an electrically small muscle cell, L λC , which implies
that sinh(z/λC ) ≈ z/λC , cosh(L/λC ) ≈ 1, and cosh(z/λC ) ≈ 1. Therefore,
vm (z) ≈ ro zI,
km (z) ≈ gm ro zI,
io (z) ≈ I, and ii (z) ≈ 0.
Thus, as shown in Figure 3.9 for electrically small muscle cells (L/λC = 0.1), the
external longitudinal current io (z) is approximately equal to the external current I,
and the internal longitudinal current io (z) is approximately equal to zero. Also the
membrane potential is small. Thus, for electrically small cells very little current
enters the cell. The external current flows between electrodes outside the cell
(Figure 3.10).
As L/λC increases, the external longitudinal current decreases and the internal longitudinal current increases as current enters the cell (Figure 3.10). The external
longitudinal current is a minimum and the internal longitudinal current is a maximum at z = 0. These extrema depend upon the ratio ro /ri . As this ratio increases
the minimum external longitudinal current decreases and the maximum internal
longitudinal current increases indicating that more current enters the muscle cell
as the external longitudinal resistance is increased.
d. The voltage V = vo (−L) − vo (L), which can be found from vo (z), is
dvo (z)
= −ro io (z).
dz
PROBLEMS
41
1
ro
= 0.1
ri
0.8
ro
=1
ri
R 0.6
2ro L 0.4
0.2
0
Figure 3.11: The resistance R is plotted versus L/λC for different values of
ro /ri (Problem 3.8).
ro
= 10
ri
1
2
3
L/λC
4
Hence,
V =−
5
ZL
−L
dvo (z) =
ZL
−L
ro io (z) dz.
Substitution of io (z) into the integral yields
ZL ro
cosh(z/λC )
V =I
dz,
ri + ro
ri + ro −L
cosh(L/λC )
which can be integrated to yield
V
sinh(L/λC )
ro
,
2ri L + 2λC ro
= I
ri + ro
cosh(L/λC )
2Lri ro
ro tanh(L/λC )
= I
1+
,
ri + ro
ri
L/λC
Therefore,
R=
V
2Lri ro
ro tanh(L/λC )
=
1+
,
I
ri + ro
ri
L/λC
which can be written as
1
tanh(L/λC )
R
=
1 + (ro /ri )
.
2Lro
1 + (ro /ri )
L/λC
The resistance is plotted in Figure 3.11. For an electrically small muscle cell, L λC and tanh(L/λC ) ≈ L/λC . Therefore, R ≈ 2Lro . Thus, for an electrically small
cell, since no appreciable current enters the cell, the external longitudinal current
equals the external current (Figure 3.9). Therefore, the resistance between the
electrodes is the external resistance per unit length, ro , times the length of the
external conductor which is the length of the muscle cell, 2L, (Figure 3.11). The
external potential is simply the external current flowing through a resistance per
unit length ro . Therefore, the membrane potential difference reflects the change
in external potential caused by the external current (Figure 3.9).
For a large electrically large muscle cell, L λC , tanh(L/λC ) ≈ 1 so that
R≈
2Lri ro
.
ri + ro
Therefore, the equivalent resistance per unit length is just the parallel combination
of ri and ro and the total resistance between electrodes is this equivalent resistance
per unit length times the length of the muscle, 2L.
42
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
e. The fact that R changed at all suggests that the cell is neither an electrically small
cell nor a large electrically large cell. Recall that for both L λC and for L λC ,
R is independent of gm . However, the effect of SDS has a large effect on R.
Therefore, the cell is an electrically large cell, but not a large electrically large
cell which implies that the length of the cell is roughly comparable to its space
constant.
Problem 3.9 The solution of the cable equation for an impulse of charge is given by
Equation 3.45 (Weiss, 1996b) which is
ro λC Qe /τM −(z/λC )2 /(4t/τM ) −t/τM
vm (z, t) = p
e
e
u(t),
4π t/τM
where Qe is the external charge delivered by an impulse of current in space and time.
To find the total charge on the cable for this case, integrate the charge per unit length
which is cm vm (z, t) to yield
!
Z∞
ro λC Qe cm /τM −(z/λC )2 /(4t/τM ) −t/τM
p
e
e
dz u(t),
Q(t) =
4π t/τM
−∞
Z ∞
ro λC Qe cm /τM
−(z/λC )2 /(4t/τM )
p
e
dz e−t/τM u(t).
=
4π t/τM
−∞
q
The integral equals 4π λ2C t/τM so that
Q(t) =
ro λ2C Qe cm −t/τM
e
u(t).
τM
This solution needs to be related to the problem at hand. The total charge on the cable
at t = 0 is Qo so that
Qo =
ro λ2C Qe cm
ro Qe cm
ro
=
=
Qe ,
τM
gm (ri + ro )cm /gm
ri + r o
which equates the solution for an external impulse of charge delivered at t = 0 at z = 0
to the solution for an initial voltage distribution in this problem. For the problem at
hand, the total charge on the cable can be expressed simply as
Q(t) = Qo e−t/τM u(t),
which states that the total charge on the cable is an exponential function of time.
Problem 3.10
a. This physical situation corresponds to the application of an impulse of current in
space and time to a cable. The solution to this problem in normalized variables is
given in Equation 3.42 (Weiss, 1996b) and is
bm (λ, τ) = √
v
A
2
e−λ /4τ e−τ , for τ > 0,
4π τ
PROBLEMS
43
bm (λ, τ)/∂τ = 0
where λ = z/λC , τ = t/τM . At the peak of the response ∂ v
!
!
bm (λ, τ)
∂v
1
1
λ2
2 /4τ −τ
2 /4τ −τ
−λ
−λ
√
−1 e
e −
e
= A √
e
,
∂τ
4π τ 4τ 2
2 4π τ 3/2
!
1
λ2
2 /4τ −τ
A
−λ
.
e
−1−
e
= √
4τ 2
2τ
4π τ
The derivative is zero for the value τ = τp that satisfies the equation
λ2 = 4τp2 + 2τp ,
which in unnormalized form is
b.
z
λC
2
=4
tp
τM
2
+2
tp
τM
.
i. The theory predicts that a plot of 4(tp /τM )2 + 2(tp /τM ) plotted versus z2
should be a straight line with a slope of 1/λ2C that goes through the origin.
The data fall on a straight line, but the line does not go through the origin.
This disagreement between measurement and theory may well be due to the
fact that the end-plate currents, that give rise to the measured end-plate potentials, deviate from impulses. The measurements for small values of z, and
t are particularly sensitive to these deviations.
ii. The slope of the line is (2.6 − 0.2)/12 = 0.2. Therefore, λ2C = 5 and λC = 2.2
mm.
iii. The equation given in part a gives the time it takes the peak to travel a given
distance. To find τp for λ = 1 the following quadratic equation must be
solved
4τp2 + 2τp − 1 = 0.
The roots are
τp =
−2 ±
√
1 1p
4 + 16
=− ±
5.
8
4 4
√
Only the positive root is physically plausible. Hence, τp = ( 5 − 1)/4 = 0.31.
Hence, tp = 0.31τM .
Problem 3.11 During the foot of the action potential, the membrane can be represented
by a parallel combination of a resistance and a capacitance. Therefore, the membrane
potential change must satisfy the cable equation
λ2C
∂ 2 vm (z, t)
∂vm (z, t)
= vm (z, t) + τM
,
∂z2
∂t
but since the foot of the action potential travels at a constant conduction velocity ν in the
positive z-direction, the membrane potential must have the form vm (z, t) = f (t − z/ν)
which satisfies the wave equation.
44
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
a. The proposed form of the solution does satisfy the wave equation. It only remains
to determine if that form satisfies the linear cable equation. Therefore, try the
solution
vm (z, t) = Aeα(t−z/ν) ,
whose partial derivatives are
∂vm (z, t)
∂t
2
∂ vm (z, t)
∂z2
= Aαeα(t−z/ν) ,
2
α
= A
eα(t−z/ν) .
ν
Substitution of these partial derivatives into the cable equation yields
λ2C A
α
ν
2
eα(t−z/ν)
2
2 α
λC
ν
= Aeα(t−z/ν) + τM Aαeα(t−z/ν) ,
= 1 + τM α.
This is a polynomial in α which can be written as
2
α − τM
ν
λC
2
ν
α−
λC
2
= 0,
which has two roots
τM
α=
2
ν
λC
2
v
!
u
u τ ν 2 2 ν 2
t M
±
+
.
2 λC
λC
Therefore, the proposed solution satisfies both the linear cable equation and the
wave equation for any value of A and for two values of α,
b. The solution should converge as t → −∞. Therefore, α > 0 and


v
u
2
2
u
τM ν
2λC


t
α=
1 + 1 +
.
2 λC
τM ν
Problem 3.12
a. The steady-state solution for this current distribution is a superposition of solutions given in Section 3.4.2 (Weiss, 1996b) to yield
vm (z, ∞) = −
ro λC −|z|/λC
− e−|z−L|/λC .
Ie e
2
For Ie = 1 mA, Vo = 0.2 mV. Therefore, vm (0, ∞) = −100Vo = −20 mV. Therefore,
vm (0, ∞) = −20 × 10−3 = −
ro λC
1 × 10−3 ,
2
PROBLEMS
45
vo (z, ∞) (mV)
0.2
L
0
z
Figure 3.12:
Extracellular potential
lem 3.12). The space constant is 2 mm.
(Prob-
−0.2
and ro λC = 40. If the term in the interpolar region is ignored as stated in the
problem, then from the core conductor model
ro
Vo
vo (0, ∞)
=−
=
.
vm (0, ∞)
ri + ro
−100Vo
Therefore,
1
ro
,
=
ri + ro
100
which implies that ri = 99ro . The inside resistance per unit length is
ri =
ρi
25
=
≈ 2 × 104 Ω/cm.
2
πa
π (0.02)2
Therefore, ro ≈ ri /100 ≈ 2 × 102 Ω/cm. A combination of these results yields
λC =
40
40
≈
= 0.2 cm.
ro
2 × 102
The extracellular potential is shown plotted in Figure 3.12.
b. Note that for Ie = −2 mA, the membrane resistance ofpthe cell is reduced, i.e.,
the membrane conductance is increased. Since λC = 1/ (ri + ro )gm for a linear
cable, λC decreases if gm increases. Since vo (z, ∞) is proportional to (ro λC /2)Ie ,
the magnitude of vo (z, ∞) decreases. Hence the correct solution is (9). Of course,
since the voltage current relation of the membrane is nonlinear, the spatial distribution of the membrane and extracellular potential will not be exponential.
Problem 3.13
a. Doubling the extracellular osmolarity will drive water out of the cell to establish
osmotic equilibrium (Weiss, 1996a, Chapter 4). This will have two effects. First,
the radius of the cell will decrease so as to halve the water volume. This will tend
to decrease the conduction velocity. Second, the concentration of intracellular solutes will increase. Since most of the intracellular solutes are ionized, the increase
in ion concentration will increase the conductivity of the cytoplasm which will tend
to increase the conduction velocity of the action potential. A more quantitative
analysis is required to determine which if any of these effects predominates.
b. A quantitative analysis includes a determination of the change in radius of the
cell, the change in conductivity of the cytoplasm, and the change in conduction
velocity.
46
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
• Dependence of intracellular concentration of ions on extracellular concentration. At osmotic equilibrium,
CΣi = CΣo ,
where CΣi and CΣo are the total concentrations of solute osmolarities inside
and outside the cell. If the outside concentration is doubled then the inside
concentration will double by the efflux of water to halve the water volume
in the cell. Let us assume that the cell water volume is approximately equal
to the cell volume. The volume of a cylinder of radius a and of length L is
V = π a2 L. Hence, halving the volume of √
a cylindrical cell whose length is
constant results in a change in radius of 1/ 2.
• Dependence of conductivity of cytoplasm on ion concentration. The conductivity (σi ) of the cytoplasm, assuming it acts as an electrolyte, is (Weiss, 1996a,
Chapter 7)
X
2 2
σi =
un zn
F cn ,
n
where un is the molar mechanical mobility of ion n, zn is its valence, F is Faraday’s constant, and cn is its concentration. Thus, doubling the concentration
of all the ions should double the conductivity.
• Dependence of conduction velocity on cytoplasmic conductivity. If it is assumed that ri ro , then the conduction velocity is
s
s
s
s
Km
Km
Km a
Km aσi
=
.
=
=
ν=
2π ari
2π aρi /(π a2 )
2ρi
2
c. Thus, the conduction velocities are
s
Km aσi
,
ν1 =
2
s
s
√
Km (a/ 2)(2σi )
Km aσi
1/4
ν2 =
= (2)
= (2)1/4 ν1 = 1.2ν1 ,
2
2
where it is assumed that Km , which is a property of the membrane alone, does not
change as the cell shrinks. Thus, the effect of the change in conductivity exceeds
the effect of the change in radius, and the conduction velocity increases.
Problem 3.14
a. Since the dimensions of this cell are much smaller than a space constant, it is appropriate to model this cell as an electrically small cell. Hence, a lumped-parameter
model of the cell, as shown in Figure 3.13, is appropriate. Since the specific capacitance and the surface area of the cell are given,
C = 10−6 · 6 × 10−6 = 6 pF.
Since the time constant, τM = C/G,
G=
o = −50 mV.
In addition, Vm
6 × 10−12
= 3 nS.
2 × 10−3
PROBLEMS
47
+
C
G
+ o
−Vm
i(t)
Vm (t)
−
Figure 3.13: Equivalent electric circuit of a small
cell (Problem 3.14).
−50
Vm (t)
−60
0
2
Figure 3.14: Step response of a small cell (Problem 3.14).
4
t
6
8
o = −50 mV. The current is in the direcb. Before the onset of the current, Vm (t) = Vm
tion to hyperpolarize the membrane. Hence, the membrane potential decreases
exponentially with a time constant of 2 ms to a final value that is −50×10−3 −30×
10−12 /(3×10−9 ) = −60×10−3 V or −60 mV. The response is shown in Figure 3.14.
Problem 3.15
a. The cell space constant is
1
1
=p
= 1 cm.
λC ≈ √
−4
gm ri
10 · 104
The time constant is
τM =
cm
150 × 10−9
=
= 1.5 ms.
gm
10−4
b. The cell is 4 space space constants long. Hence, it is an electrically large cell. The
step response of an electrically large cell of infinite length is given in Equation 3.55
(Weiss, 1996b). This relation gives an approximation to the step response of this
cell. A more involved treatment is required to get a more accurate answer for a
cell that is 4 space constants long. The approximate solution is
vm (0, t) ≈
λC ri
Ie erf(t/τM )u(t) = 5 × 103 Ie erf(t/1.5)u(t),
2
where t is in ms.
c. Note the axial electrode has a resistance per unit length that is 2000 times smaller
than
that of cytoplasm. Therefore, the space constant has increased by a factor of
√
2000 ≈ 45, to about 45 cm. Therefore, the use of the axial electrodes has made
48
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
1
unclamped
0.8
0.6
Figure 3.15: The unclamped cable response is
a plot of vm (0, t)/(5 × 103 Ie ) = erf(t/1.5)u(t);
the clamped cable response is a plot of
vm (0, t)/(5 × 103 Ie ) = (1 − e−t/1.5 )u(t) (Problem 3.15).
clamped
0.4
0.2
0
1
2
3
Time (ms)
4
5
this cell an electrically small cell whose step response is given in Equation 3.54
(Weiss, 1996b),
vm (0, t) = RIe 1 − e−t/τM u(t).
The total resistance of the membrane of the cell, ignoring the resistance of the two
ends of the cylindrical cell, is
R=
Therefore,
1
gm L
=
1
10−4
·4
= 2.5 × 103 Ω.
vm (0, t) = 2.5 × 103 Ie 1 − e−t/1.5 u(t).
where t is in ms.
The two results are compared in Figure 3.15.
Problem 3.16 Since the cell’s voltage response remains in its linear range, incremental
portions of the membrane can be represented by a parallel network consisting of a
resistance and capacitance. Thus, solutions for a-d are governed by the cable model,
and the solution for e is governed by that of a lumped parameter network model. It will
also be assumed that the potential change is negligibly small at a recording electrode
located 5 or more space constants away from a stimulating electrode.
a. The intracellular electrode is one space constant away from an electrode that
passes an outward current through the membrane which depolarizes the membrane at that location transiently. Thus, the potential change across the membrane
at this site is a positive, pulsatile waveform that starts at zero. The potential at
the external remote reference electrode is zero. Thus, the potential difference between the remote electrode and the intracellular electrode is negative. Therefore,
the answer is v7 (t).
b. The stimulating electrode hyperpolarizes the membrane at the active recording
electrode, and therefore, produces a positive potential at the active extracellular
electrode. Thus, v(t) is a positive pulse and equals v3 (t).
c. The stimulus current depolarizes the membrane as in part a. There is no difference of potential across the membrane at the remote voltage measuring electrode.
Therefore, the response is the same as in part a which is v7 (t).
PROBLEMS
49
d. This current hyperpolarizes the membrane. However, the response of a cable to
an impulse of current delivered at the recording electrode is unbounded at t = 0.
Therefore, the answer is v13 (t).
e. This is a space-clamped axon driven by a hyperpolarizing current pulse. Thus, the
solution is that of the voltage response of a parallel resistance and capacitance to
a brief pulse of current. The answer is v8 (t).
Problem 3.17 Reduction of the extracellular osmolarity results in the flow of water into
the cell so that the cell will swell to achieve osmotic equilibrium (Weiss, 1996a, Chapter 4). If the intracellular ions are assumed to be relatively impermeant compared with
i
i
water, then the volumes in two solutions satisfy the equation NΣi = CΣ1
V1 = CΣ2
V2 ,
i
i
where NΣ is the total quantity of intracellular solute, CΣ is total concentration of intracellular solute, and V is the water volume which equals the cell volume in this case. In
addition, at osmotic equilibrium CΣi = CΣo . The volume of a cylindrical cell is V = π a2 L,
where a is the radius and L is the length of the cylinder. Since the length of the cell is
o
o
assumed not to change as the cell swells, a21 CΣ1
= a22 CΣ2
which implies that
v
u o
uC
a2
= t Σ1
o .
a1
CΣ2
Dilution of the intracellular solution also reduces the cytoplasmic conductivity and
raises the cytoplasmic resistivity. The conductivity of a dilute solution is proportional
to the concentration of ions (see solution to Problem 3.13). Therefore,
o
Ci
CΣ1
ρ2
= Σ1
=
o .
i
ρ1
CΣ2
CΣ2
a. The conduction velocity of an unmyelinated axon is
s
Km
,
ν=
2π a(ro + ri )
which for ro ri , and for ri = ρi /(π a2 ) equals
s
Km a
,
ν=
2ρi
so that for two different solutions
s
a2 ρ1
ν2
=
=
ν1
a1 ρ2
o
CΣ2
o
CΣ1
!1/4
,
o
o
/CΣ1
=
under the assumption that Km does not change as the cell swells. For CΣ2
1/4
0.78, the new conduction velocity is ν(0.78)
= 0.94ν.
b. The space constant of the axon is
1
,
λC = p
(ro + ri )gm )
50
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
which for ro ri , ri = ρi /(π a2 ), and gm = 2π aGm equals
s
λC =
so that
λC2
=
λC1
a
,
2ρi Gm
s
a2 ρ1
,
a1 ρ2
o
o
which is the same ratio as the ratio of velocities. Therefore, For CΣ2
/CΣ1
= 0.78,
1/4
the new space constant is λC (0.78)
= 0.94λC .
c. The membrane time constant is τM = cm /gm = Cm /Gm which is independent of
the resistivity of cytoplasm and the dimensions of the cell so that the new time
constant equals τM .
Problem 3.18
a. The core conductor equations for incremental quantities include the relation vm (z, t) =
vi (z, t) − vi (z, t). The second partial derivative of this relation with respect to z
yields
∂ 2 vm (z, t)
∂ 2 vi (z, t) ∂ 2 vo (z, t)
=
−
.
∂z2
∂z2
∂z2
But
∂ii (z, t)
∂vi (z, t)
= −ri ii (z, t) and
= −km (z, t).
∂z
∂z
A combination of these three equations yields
∂ 2 vm (z, t)
∂ 2 vo (z, t)
=
r
k
(z,
t)
−
.
i
m
∂z2
∂z2
But, in the cable model the relation between membrane current per unit length
and membrane potential is given by
km (z, t) = gm vm (z, t) + cm
∂vm (z, t)
.
∂t
Combining these last two equations yields
∂ 2 vm (z, t)
∂vm (z, t) ∂ 2 vo (z, t)
= ri gm vm (z, t) + ri cm
.
−
2
∂z
∂t
∂z2
If the equation is divided by ri gm then
λ2C
2
∂ 2 vm (z, t)
∂vm (z, t)
2 ∂ vo (z, t)
−
λ
=
v
(z,
t)
+
τ
,
m
M
C
∂z2
∂t
∂z2
where λ2C = 1/(ri gm ) and τM = cm /gm .
PROBLEMS
51
b. The time independent cable equation for this case is obtained by setting the time
derivatives to zero,
λ2C
2
d2 vm (z, t)
2 d vo (z)
−
v
(z,
t)
=
−λ
.
m
C
dz2
dz2
The rightmost term for vo (z) = (1/2)z2 u(z) is evaluated as follows
λ2C
d2 vo (z, t)
= λ2C u(z).
dz2
Therefore, we seek a solution to the equation
λ2C
d2 vm (z, t)
− vm (z, t) = −λ2C u(z).
dz2
To obtain the solution to this problem, we take advantage of the linearity and
space invariance of the cable equation. The solution to the equation
λ2C
d2 vm (z, t)
− vm (z, t) = −λ2C ro Ie δ(z),
dz2
is (Weiss, 1996b, Section 3.4.2)
vm (z) =
ro λC
Ie e−|z|/λC .
2
Thus, except for differences in scale factors, the solution for a spatial impulse can
be used to determine the solution for a spatial step. This, is achieved by integrating
the solution for the spatial impulse, suitably scaled,
Z
λC z −|u|/λC
vm (z) =
e
du.
2 −∞
The integral is expressed in parts,

Z

λC z u/λC


e
du
for z < 0,

2 Z−∞
Z
vm (z) =

λC 0 u/λC
λC z −u/λC


e
du +
e
du for z > 0.

2 −∞
2 0
Evaluation of the integrals yields
 2
z

λ


 C eu/λC −∞
2
vm (z) =
2
2

λ
−λC −u/λC 
z

 C +
e
0
2
2
The solution is
for z < 0,
for z > 0.
 2

λ


 C ez/λC
for z < 0,
2
vm (z) =
2
2

λ
λ 

 C + C 1 − e−z/λC
for z > 0.
2
2
vm (z) is plotted in Figure 3.16. To summarize, the external potential which varies
parabolically in space acts as a discontinuous external current. The cable model
smooths out this discontinuous current to produce a spatially smoothed membrane potential.
52
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
2
2vm (z)
λ2C
1
-4
-2
0
z/λC
Figure 3.16: Membrane potential versus position for an applied extracellular potential (Problem 3.18).
2
4
Problem 3.19 The extracellular potential is (Weiss, 1996b, Section 3.4.2)
ro2 λC
ri
z
−|z|/λC
+2
vo (z) − vo (−∞) = −
Ie e
u(z) .
2(ri + ro )
ro
λC
Note that the first term decreases to zero as z → ∞. Therefore, the only term that
contributes is the second term whose derivative with respect to z is
−
1 dvo (z)
ri ro
=
.
Ie dz
ri + ro
Problem 3.20 The steady-state extracellular potential in response to a step of current
is
vo (z, ∞)
= e−z/λC .
vo (0, ∞)
Therefore, the normalized logarithmic potential is
vo (z, ∞)
= −z/λC log10 e.
log10
vo (0, ∞)
The measurements (Weiss, 1996b, upper panel of Figure 3.59) show that the normalized
logarithmic potential is −0.8 for z = 2 mm. Therefore, (−2/λC ) log10 e = −0.8 so that
λC = 1.09.
The time it takes the step response to reach 1/2 of its maximum value is given
approximately by the relation (Weiss, 1996b, Figure 3.31)
τ1/2 =
λ + 0.5
.
2
In unnormalized coordinates
t1/2
(z/λC ) + 0.5
,
=
τM
2
so that
t1/2 =
τM
2λC
z+
τM
.
4
The measurements (Weiss, 1996b, lower panel of Figure 3.59) show that the intercept
is at 6.7 ms and the slope is about (28.3 − 6.67)/2 = 10.8 ms/mm. From the intercept
τM = 26.7 ms. From the slope λC = τM /(2 · 10.8) = 1.24 mm. Thus, the two estimates
of the space constant from these two sets of measurements agree within 14%.
PROBLEMS
53
∆vo (z, ∞) (mV)
10
8
6
Figure 3.17: Estimation of space constant
from measurements of steady-state value of
the extracellular potential (Problem 3.21).
4
2
0
0
2
4
z (mm)
6
8
Problem 3.21 Measurement 1 gives information about the longitudinal resistances,
∆Vo (z, t) ro
∆V (z, t) = r + r .
m
o
i
Measurement 2 gives information about other cable parameters since
ro
ro
ro λC
Ie e−|z|/λC .
∆vm (z, ∞) =
∆vo (z, ∞) = −
ro + ri
ro + ri
2
a. The spatial dependence of the steady-state extracellular potential is exponential
and the space constant can be obtained from the initial slope of the potential as
shown in Figure 3.17. Therefore, λC = 1.5 mm.
b. From Measurement 1, ro /(ro + ri ) = 40/120 = 1/3 implies that ri /ro = 2. In
addition, Measurement 2 shows that
ro
r o λC
Ie = 10 mV,
∆vo (0, ∞) =
r o + ri
2
which implies that
10
−2
1
=
3
ro · 0.15
5 × 10−7 .
2
The solution for ro = 8 × 105 Ω/cm.
c. The above results are combined to yield ri = 16 × 105 Ω/cm.
d. The space constant is
1
1
=q
,
λC = 0.15 = p
gm (ro + ri )
gm · 24 × 105
which can be solved to yield gm = 1.85 × 10−5 S/cm.
e. The specific membrane conductance is
Gm =
1.85 × 10−5
gm
=
= 1.18 × 10−3 S/cm2 .
2π a
2π 25 × 10−4
f. The longitudinal resistance per unit length of cytoplasm is ri = ρi /(π a2 ) so that
ρi = ri π a2 = (16 × 105 )π (50 × 10−4 )2 = 126 Ω·cm.
54
CHAPTER 3. LINEAR ELECTRICAL PROPERTIES OF CELLS
Chapter 4
THE HODGKIN-HUXLEY MODEL
Exercises
Exercise 4.1
a. Experiment 2 shows the result for changing intracellular sodium concentration.
Note that the resting potential is not changed appreciably, but the peak of the
action potential (which approaches the Nernst equilibrium potential for sodium)
is changed appreciably. The Nernst equilibrium potential for sodium is VNa =
o
i
(RT /F ) ln(cNa
/cNa
). Hence, the largest value of intracellular sodium concentration corresponds to the lowest Nernst equilibrium potential. Therefore, curve 3
must correspond to the highest intracellular sodium concentration.
b. Experiment 1 shows the result for changing extracellular potassium concentration. Note that the resting potential is changed appreciably, the peak of the action
potential is not changed much, and the undershoot of the action potential (which
approaches the Nernst equilibrium potential for potassium) is changed appreciao
i
bly. The Nernst equilibrium potential for potassium is VK = (RT /F ) ln(cK
/cK
).
Thus, the largest value of the Nernst equilibrium potential for potassium occurs
at the highest value of the extracellular potassium concentration. Therefore, curve
1 must correspond to the highest extracellular potassium concentration.
Exercise 4.2
a. True. After the step of membrane potential has occurred, the factors n(t), m(t),
and h(t) are solutions to linear, first-order, ordinary differential equations with
constant coefficients. Hence, the solutions are exponential functions of time.
b. False. After the step of current has occurred, the factors n(t), m(t), and h(t)
are solutions to linear, first-order, ordinary differential equations with voltage
dependent coefficients. Furthermore, the voltage changes with time. Hence, the
solutions are not exponential functions of time.
c. False. In general, if the membrane potential is not constrained to be constant, the
factors n(t), m(t), and h(t) are solutions to linear, first-order, ordinary differential equations with voltage dependent coefficients. Hence, the solutions are not
exponential functions of time.
55
56
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
Exercise 4.3
a. False. Action potentials cannot occur under voltage-clamp. The applied voltage
precludes action potentials.
b. True. According to the core-conductor model, the primary effect of a space clamp
is to increase the speed of a propagating action potential. In fact, in a well-clamped
cell, the speed can be so fast that the action potential occurs simultaneously at all
points along the cell. For that reason, a space-clamped action potential is often
called a membrane action potential.
c. True. “Unclamped” is the normal condition for an axon. Therefore, action potentials occur in unclamped axons.
Exercise 4.4
a. In general, at rest m is small and h is relatively large. For a large depolarization, the
reverse is true — m is large and h is small. In both cases, the sodium conductance
is small compared to the potassium conductance. This question can be answered
quantitatively by computing the relevant values from the Hodgkin-Huxley model
or by using a simulation of the Hodgkin-Huxley model (Weiss et al., 1992). At rest
m∞ ≈ 0.05 and h∞ ≈ 0.6 so that
3
h∞ ≈ 120(0.05)3 0.6 = 0.009 mS/cm2 .
GNa = GNa m∞
In contrast, at rest
GK = GK n4∞ ≈ 36(0.3)4 = 0.3 mS/cm2 .
Thus, at rest GK ≈ 33GNa . Now consider a strong depolarization to 0 mV. At that
potential n∞ = 0.9, m∞ = 0.96, h∞ = 0.003. Therefore,
GNa ≈ 120(0.96)3 0.003 = 0.3 mS/cm2 ,
and
GK ≈ 36(0.9)4 = 24 mS/cm2 .
Thus, for a large, maintained depolarization GK ≈ 80GNa .
b. Although the sodium conductance is small both at rest and after a prolonged, large
depolarization, the response of the sodium conductance to a change in potential is
quite different in the two cases. At rest, the sodium conductance can show a rapid
increase because it is m that must increase to increase GNa appreciably and m
has a fast time course. However, after a prolonged depolarization it is h that must
increase to increase GNa appreciably and h changes much more slowly. Therefore,
after a prolonged depolarization the membrane is not capable of exhibiting a rapid
change in sodium conductance which is necessary for excitation. This is exactly
the situation that occurs during the peak of the action potential — the membrane is
depolarized, h is low, and GNa cannot change rapidly. This phenomenon accounts
for the refractory properties of excitable membranes.
57
2
GN a (mS/cm )
EXERCISES
6
6
5
5
4
4
3
3
2
2
1
1
0
1
2
3
0
t (ms)
2
4
6
8
10
Figure 4.1: The sodium conductance is shown as a function of time for a step in membrane
potential for two different time scales from (Exercise 4.5). The membrane potential is stepped
from −50 to −25 mV (solid line) and from −25 to −50 mV (dashed line).
Exercise 4.5
a. Upon a depolarization, m increases rapidly and h decreases slowly. Therefore,
the maximum conductance can be estimated by assuming that at the peak value
m has reached its final value and h has not changed much. The initial value of the
conductance is
3
GNa (−25, 0) = GNa m∞
(−50)h∞ (−50) = GNa (0.16)3 (0.26) = 0.0011GNa .
The maximum value of the conductance is
3
(−25)h∞ (−50) = GNa (0.73)3 (0.26) = 0.10GNa .
(GNa )max ≈ GNa m∞
The final value of the conductance is
3
(−25)h∞ (−25) = GNa (0.73)3 (0.02) = 0.0078GNa .
GNa (−25, ∞) = GNa m∞
The time dependence of the sodium conductance is shown in Figure 4.1 for the
value GNa = 120 mS/cm2 .
b. Upon repolarization from a depolarized state, m decreases rapidly and h increases
slowly. Therefore, the maximum value of GNa occurs at t = 0.
(GNa )max
3
≈ GNa (−25, 0) = GNa m∞
(−25)h∞ (−25)
≈ GNa (0.73)3 (0.02) = 0.0078GNa ,
whereas the final value of the conductance is
3
(−50)h∞ (−50) = GNa (0.16)3 (0.26) = 0.0011GNa .
GNa (−50, ∞) = GNa m∞
c. The membrane potential is the same both for t < 0 in part a and for t → ∞ in part
b. Hence, the sodium conductance is the same in both cases.
d. The maximum value of the conductance in part a exceeds that in part b because h
changes so much more slowly than m and it is h that controls the maximum value
in the two cases.
58
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
Exercise 4.6
a. True. The leakage conductance is a constant, independent of the membrane potential.
b. False. The sodium conductance is proportional to the product of m3 and h. Each
factor, m and h, is the solution to a first-order differential equation. Hence, neither
of these factors can change instantaneously as the membrane potential changes.
Therefore, the sodium conductance does not change instantaneously.
c. False. The potassium conductance is proportional to n4 where n is the solution to
a first-order differential equation and does not change instantaneously. Therefore,
the potassium conductance does not change instantaneously.
d. False. The leakage conductance is constant. Therefore, the leakage current to a
step of membrane potential will show a step in leakage current.
e. True. The sodium conductance is continuous at t = 0. But, the sodium current
is proportional to the product of the sodium conductance and the difference in
membrane potential from the Nernst equilibrium potential for sodium. Since this
potential difference is discontinuous, the sodium current is discontinuous at t = 0.
The discontinuity is not always obvious. If the initial membrane potential is close
to the rest potential, then GNa is small at the time of the step, and the resulting
step in sodium current is also small.
f. True. The potassium conductance is continuous at t = 0. But, the potassium
current is proportional to the product of the potassium conductance and the difference in membrane potential from the Nernst equilibrium potential for potassium. Since this potential difference is discontinuous, the potassium current is
discontinuous at t = 0. As with the sodium current, the step change in potassium
current will be small if the initial voltage is such that the potassium conductance
is small.
g. False. The factors n(t), m(t), and h(t) are all solutions to first-order differential
equations with voltage dependent parameters. A change in a parameter results in
a change in the time derivative of each factor, but not in the instantaneous value
of each factor.
h. True. The time constants τn , τm , and τh are all instantaneous functions of the
membrane potential. Hence, an instantaneous change in Vm (t) results in an instantaneous change in each time constant.
i. True. The steady-state values n∞ , m∞ , and h∞ are all instantaneous functions of
the membrane potential. Hence, an instantaneous change in Vm (t) results in an
instantaneous change in each steady-state value.
Exercise 4.7 This quote gives an incorrect description of the source in a voltage clamp.
The principle of the voltage clamp is to provide a source that acts as an ideal voltage
source in that the voltage across the membrane will be independent of the current
EXERCISES
59
through the membrane. An ideal voltage source has an equivalent resistance of zero.
This is approximated in practical voltage-clamp systems by a feedback amplifier that
senses the membrane potential and passes a current through the membrane to produce
the command voltage. A practical voltage-clamp system has an equivalent resistance
that is much less than the resistance of the membrane.
Exercise 4.8 This statement is consistent with the analysis in Section 4.4 (Weiss, 1996b)
concerning threshold. In particular, Figures 4.43 and 4.44 (Weiss, 1996b) show that the
threshold in response to a brief current stimulus occurs when the ionic current changes
from outward to inward. The outward current is carried predominantly by potassium
and the inward current predominantly by sodium. Therefore, threshold occurs approximately when the magnitude of the sodium current exceeds that of the potassium current. The statement is approximately true with the caveats that other ions may also
contribute to the balance between inward and outward currents. In addition, the analysis presented in Section 4.4 (Weiss, 1996b) is approximate in that the ionic current versus
membrane potential characteristic was analyzed at one point in time only. A careful examination of threshold requires analysis of all 4 state variables of the Hodgkin-Huxley
model, i.e., Vm , m, n, and h.
Exercise 4.9 Consider just the ionic current branches in the Hodgkin-Huxley model of
the membrane
Jion
= GK (Vm , t)(Vm (t) − VK ) + GNa (Vm , t)(Vm (t) − VNa ) + GL (Vm (t) − VL ),
Jion
= (GK (Vm , t) + GNa (Vm , t) + GL )Vm (t) −
(GK (Vm , t)VK + GNa (Vm , t)VNa + GL VL ),
Jion
= G(Vm , t)(Vm (t) − V (t)),
where
G(Vm , t) = GK (Vm , t) + GNa (Vm , t) + GL ,
GK (Vm , t)VK + GNa (Vm , t)VNa + GL VL
.
V (Vm , t) =
GK (Vm , t) + GNa (Vm , t) + GL
Thus, both the equivalent conductance G(Vm , t) and the equivalent potential V (Vm , t)
depend upon time and the membrane potential.
Exercise 4.10 The article claims that the sodium-potassium pumps repolarize the membrane during an action potential. This is wrong. Repolarization occurs because sodium
channels close and potassium channels open during repolarization. Transport of sodium
and potassium is passive during the action potential — no pumps are involved. [45
words]
Exercise 4.11
a. The answer is v, a capacitance current. As an action potential propagates along an
axon, current flows inward at the peak of the action potential and outward both
ahead and behind the peak. The current that flows outward in the part of the
axon ahead of the action potential tends to depolarize that part of the membrane.
The initial current flows primarily through the membrane capacitance and this is
readily seen in Figure 4.32 (Weiss, 1996b).
60
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
2
Jion (mA/cm )
0.1
5
−0.1
−0.2
−0.3
10
15
20
25
t (ms)
Figure 4.2: The ionic current is shown for
the default parameters and for a change in
potential from −80 to −40 mV (dashed line)
and when the external sodium concentration
is doubled (solid line) (Exercise 4.12).
b. The answer is i, an ionic current carried by sodium ions. The early inward current
results because the sodium conductance increases. Sodium current then flows
down the electrochemical gradient for sodium, i.e., from outside to inside.
c. The initial outward current is readily explained in terms of linear membrane properties. The action potential causes the intracellular potential to increase and this
increase in potential tends to drive currents outward. The early inward current is
not consistent with linear membrane properties. This effect can only be understood by taking into account the fact that the action potential causes the sodium
conductance to change, as described in part b.
Exercise 4.12
a. An increase in the external sodium concentration increases VNa . Since
f
JNa = GNa (Vm − VNa ),
an increase VNa decreases JNa , especially when m3 h is large (i.e. near the negative
peak of JNa ). An increase in external sodium concentration has no affect on JK
or JL . Therefore the net change in Jion = JNa + JK + JL is to make the negative
peak even more negative and to make the steady state value a little more negative.
Thus, the magnitude of Jp increases, and the magnitude of Jss decreases. The exact
solution for these parameters is shown in Figure 4.2. It is reasonable to compute
the affect of doubling the extracellular sodium concentration on the ionic current
predicted by the theory. However, this theoretical computation is difficult to test
experimentally since doubling the sodium concentration approximately doubles
the extracellular osmolarity and causes water efflux from the axon so that the
volume approximately halves. Such large changes in osmolarity are known to
cause irreversible changes in the electrical properties of axons.
b. An increase in the external potassium concentration increases VK . Since
f
JK = GK (Vm − VK ),
an increase in VK decreases JK , especially when n4 is large (i.e. for large time). An
increase in external potassium concentration has no affect on JNa or JL . Therefore
the net change in Jion = JNa + JK + JL is to make the negative peak slightly more
negative and to make the steady state value appreciably more negative. Thus, the
magnitude of Jss decreases appreciably, and the magnitude of Jp increases a small
amount. The exact solution for these parameters is shown in Figure 4.3.
PROBLEMS
61
2
Jion (mA/cm )
0.1
5
10
15
20
25
t (ms)
Figure 4.3: The ionic current is shown for the
default parameters and for a change in potential from −80 to −40 mV (dashed line) and
when the external potassium concentration is
doubled (solid line) (Exercise 4.12).
10
15
20
25
t (ms)
Figure 4.4: The ionic current is shown for the
default parameters and for a change in potential from −80 to −40 mV (dashed line) and
when the final potential is changed from −40
to −30 mV (solid line) (Exercise 4.12).
−0.1
−0.2
0.25
2
Jion (mA/cm )
−0.3
−0.25
5
−0.50
−0.75
−1.00
f
c. An increase in Vm increases n, increases m, and decreases h. This change affects
f
JNa directly (by changing (Vm − VNa )) and indirectly (by changing m and h) and
f
it affects JK directly (by changing (Vm − VK )) and indirectly (by changing n). At
the negative peak, JNa is the largest component of current, and is most affected
by the change in m (m increases from 0.37 to 0.63, h barely has changed because
f
of its long time constant, and |Vm − VNa | diminishes about 10 percent). Since m
f
increases and (Vm − VNa ) is negative, Jp becomes more negative. For large time t,
f
JK is the largest current component. The value of n increases as does (Vm − VK ),
therefore Jss becomes more positive. Therefore, the magnitude of Jp increases and
of Jss decreases. The exact solution for these parameters is shown in Figure 4.4.
i affects the values of m, n, and h at time t = 0; m increases, n
d. An increase in Vm
increases, and h decreases. These changes have no long term effects, and Jss is
unchanged. However, these changes do affect Jp . Although the peak value of m
f
does not change much (it is determined largely by Vm ), the value of h at the time
of the peak is reduced (in proportion to the change in its initial value from 0.97 to
0.86). Therefore, the magnitude of the peak sodium current is smaller, and Jp is
more positive. Therefore, the magnitude of Jp is decreased and the magnitude of
Jss does not change appreciably. The exact solution for these parameters is shown
in Figure 4.5.
Problems
Problem 4.1
a. There are only three ions for which data are available so that a model of the membrane of the cell that allows conduction by all three ions is shown in Figure 4.6.
62
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
2
Jion (mA/cm )
0.1
5
10
15
20
−0.1
25
t (ms)
−0.2
−0.3
Figure 4.5: The ionic current is shown for the
default parameters and for a change in potential from −80 to −40 mV (dashed line) and
when the initial potential is changed from −80
to −70 mV (solid line) (Exercise 4.12).
Jm
JC
JK
JNa
GK
GNa
Intracellular
Vm
Membrane
JCl
GCl(Vm,t)
Cm
+
VK −
+
+
VNa −
+
VCl −
−
Extracellular
Figure 4.6: A network model for a patch of membrane of a plant cell (Problem 4.1).
To determine which ions make appreciable contributions to the resting and action
potential, it is essential to determine the Nernst equilibrium potential for each ion.
In general, the Nernst equilibrium potential is defined as
Vn =
co
co
RT
60
ln n
≈
log10 n
i
i mV.
zn F
zn
cn
cn
The Nernst equilibrium potential for the three ions are evaluated as follows
VK
VNa
VCl
0.06
= −180 mV,
60
0.15
= −156 mV,
≈ 60 log10
60
0.05
≈ −60 log10
= 198 mV.
100
≈ 60 log10
Since the resting potential is near the potassium equilibrium potential, we hypothesize that at rest GK GNa and GK GCl . During the peak of the action potential,
the membrane potential is near the chloride equilibrium potential. Therefore, we
hypothesize that during the peak of the action potential GCl GK and GCl GNa .
However, the results given thus far cannot explain the undershoot of the action
potential. Thus, either the Nernst equilibrium potential for potassium is not accurately known or some other ion is involved whose Nernst equilibrium potential
is less than −180 mV.
PROBLEMS
63
o
b. To test the model proposed in part a experimentally, increase cK
and measure
both the resting potential and the peak of the action potential. The resting potential should increase by 60 mV per decade change in potassium concentration
o
and the action potential peak should remain constant. In addition, increasing cCl
should not affect the resting potential of the cell, but should decrease the peak
of the action potential by −60 mV per decade increase in extracellular chloride
concentration.
Problem 4.2 In assessing the experimental results, assume that it is the change in potassium concentration that is primarily responsible for the change in resting potential. In
Figure 4.2 (Weiss, 1996b) the maximum reduction in extracellular potassium concentration is to replace seawater with 33% seawater. Since the normal potassium concentration of seawater is about 10 mmol/L (Weiss, 1996a, Figure 7.1), the reduction is to
3.33 mmol/L. In this range of extracellular potassium concentration, the resting potential is relatively insensitive to changes in potassium concentration. As indicated in
Figure 7.22 (Weiss, 1996a) in this range of concentration the expected decrease in resting potential is less than 2 mV. This is consistent with the results shown in Figure 4.2
(Weiss, 1996b). In Figure 4.4 (Weiss, 1996b) the maximum reduction in intracellular
potassium concentration is to replace half the potassium by sodium. Since the normal
potassium concentration of axoplasm is about 400 mmol/L (Weiss, 1996a, Figure 7.1),
the reduction is to 200 mmol/L. As indicated in Figure 7.22 (Weiss, 1996a) in this range
of concentration the expected increase in resting potential is less than 5 mV. This is
consistent with the results shown in Figure 4.4 (Weiss, 1996b).
Problem 4.3
a. The impulse of current flows through the capacitance and delivers a charge Q to
the capacitance at t = 0. Therefore, the change in potential is ∆Vm = Q/Cm so
that
Q
10−8 C
= 0.5 µF.
=
Cm =
∆Vm
20 × 10−3 V
p
b. The maximum value of the membrane potential during an action potential Vm
p
must be less than the Nernst equilibrium potential for sodium, i.e., VNa > Vm = 0.
Therefore,
!
o
cNa
RT
VNa =
ln
> 0.
i
F
cNa
o
i
o
i
/cNa
> 1 which shows that cNa
> cNa
= 10 mmol/L.
Thus, cNa
Problem 4.4
a. Since the capacitance current precedes the ionic current onset, the ionic current
can be ignored in the equivalent circuit shown in Figure 4.7. The total series resistance of a unit area of membrane is Rs , the capacitance per unit area of membrane
is Cm , and the potential difference applied by the voltage-clamp circuit is V (t).
64
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
JC (t)
Rs
+
V (t)
−
Cm
Figure 4.7: Equivalent circuit of the membrane for computing
the capacitance current under voltage clamp (Problem 4.4).
For a step of amplitude V , the current through the capacitance implied by this
network is
V −t/τs
JC (t) =
e
,
Rs
where τs = Rs Cm . The measurements (Weiss, 1996b, Figure 4.13) show that the
initial current density is about 4.2 mA/cm2 in response to a depolarization of 40
mV. The time constant is about 10 µs. Therefore, the series resistance is Rs =
40/4.2 = 9.5 Ω · cm2 .
b. The membrane capacitance is Cm = 10 × 10−6 /9.5 = 1.05 µF/cm2 .
c. The series resistance is Rs = 9.5 Ω · cm2 .
d. The step response predicted by the simple model shown in Figure 4.7 implies that
the capacitance current has a discontinuity at t = 0. The measurements show a
rapid rise in capacitance current but not a discontinuous rise. The calculations
shown with the measurements (Weiss, 1996b, Figure 4.13) take a number of factors into account that the simple model shown here does not take into account.
For example, the voltage-clamp circuit produces a voltage that does not rise instantaneously but has a finite rise time. In addition, the step response of the
recording electronics is not an instantaneous step but has some rapid rise time.
These factors reduce the rate of rise of the measured capacitance current.
Problem 4.5
a. The sodium current is JNa = GNa (28 × 10−3 − VNa ) and the sodium equilibrium
o
i
potential is VNa = (RT /F ) ln(cNa
/cNa
). The early component of the ionic current,
which is carried mainly by sodium ions, is inward for the lowest 2 curves, which
implies that VNa > 28 mV for these curves. The early component is zero for the
second curve from the top which implies that VNa = 28 mV for this curve. The
early component is outward for the upper curve which implies that VNa < 28 mV
for this curve. Since the three upper curves differ only a little, they must have been
obtained for the concentrations 140, 150, 160 mmol/L. Putting these observations
together shows that the large inward current must correspond to 450 mmol/L,
and that the curve for 150 mmol/L must correspond to the membrane potential
at the sodium equilibrium potential. The labeled curves are shown in Figure 4.8.
b. Since the sodium equilibrium potential was 28 mV for an extracellular concentration of 150 mmol/L, the intracellular concentration of sodium can be determined
by the relation
150
28 mV = 59 log10 i ,
cNa
PROBLEMS
65
Ionic current
2
2 (mA/cm )
1
140
−1
160
−1
1
2
3
4
Time (ms)
5
Figure 4.8: Ionic current in response to the
same voltage clamp but in solutions of different sodium concentration which is indicated
in mmol/L as a parameter (Problem 4.5). The
dashed curve corresponds to 150 mmol/L.
450
i
from which cNa
≈ 50 mmol/L.
c. The membrane potential profile is the same for all 4 curves. Hence, the potassium
current is the same for all 4 curves. Since the sodium current is zero for the curve
obtained at 150 mmol/L, this curve equals the potassium current. Therefore, at
150 mmol/L, JNa150 = 0 and JK150 = J150 . Since the potassium current is the same
at all concentration, at 450 mmol/L JK450 = J150 and JNa450 = J450 − J150 . Plots of
the ionic, sodium and potassium currents at extracellular sodium concentrations
of 150 and 450 mmol/L are shown in Figure 4.9.
d. In the Hodgkin-Huxley model, the conductances are independent of concentration. Therefore, the sodium conductance can be computed from the results at 450
mmol/L and the sodium current from the results at 50 mmol/L.
GNa =
JNa450
J450 − J150
JNa450
=
=
,
28 − 59 log10 (450/50)
−28
−28
from which the sodium current at 50 mmol/L is
JNa50 =
J450 − J150
(28 − 59 log10 (50/50)) = −(J450 − J150 ) = −JNa450 .
−28
A sketch of the ionic, sodium, and potassium current at an extracellular concentration of 50 mmol/L is shown in Figure 4.9.
Problem 4.6
a. The current densities of sodium and potassium during an action potential are
JNa (t) = GNa (t)(Vm (t) − VNa ) and JK (t) = GK (t)(Vm (t) − VK ).
Thus, the flux of sodium and potassium are
φNa (t) =
GNa (t)
GK (t)
(Vm (t) − VNa ) and φK (t) =
(Vm (t) − VK ).
F
F
The total quantity of sodium and potassium transferred through the membrane
per cm2 of area and for each action potential,
Z∞
Z∞
GNa (t)
GK (t)
(Vm (t) − VNa ) dt and ∆nK =
(Vm (t) − VK ) dt.
∆nNa =
F
F
0
0
66
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
3
coN a = 450 mmol/L
2
1
−1
Jion
JK
1
−1
2
3
4
5
Time (ms)
JNa
3 co = 150 mmol/L
Na
2
JK = Jion
1
2
Jion (mA/cm )
−2
−1
1
2
3
4
5
4
5
−1
Figure 4.9: Ionic (solid lines), sodium (dashed
lines), and potassium (dotted lines) currents at
extracellular sodium concentrations of 450, 150,
and 50 mmol/L (Problem 4.5).
−2
3 co = 50 mmol/L
Na
2
Jion
J
K
1
JNa
−1
1
2
3
−1
−2
VNa
40
Vm
0
30
−20
2
GN a
20
−40
10
−60
0
1
2
Time (ms)
G (mS/cm )
Vm (mV)
20
−40 mV
3
4
Figure 4.10: Method for estimating the transfer of sodium during an action potential (Problem 4.6). The sodium conductance is approximated by a rectangular pulse (shown shaded) and
the membrane potential is approximated as a constant (thick horizontal line).
PROBLEMS
67
b. Using the results shown in Figure 4.34 (Weiss, 1996b), a rough approximation to
GNa (t) and Vm (t) makes the integration simple as shown in Figure 4.10. Approximate
the sodium conductance as a rectangular pulse of amplitude 25 mS/cm2 for an interval of 0.4 ms, and approximate Vm (t) − VNa ≈ −40 mV over the same interval.
Therefore,
(25 × 10−3 S/cm2 ) · (−40 × 10−3 V) · (0.4 × 10−3 s)
,
9.65 × 104 C/mol
∆nNa
≈
∆nNa
≈ −4 pmol/cm2 .
The negative sign indicates that there is a net transfer of sodium into the cell
during each action potential.
c. Each centimeter length of axon contains a total quantity of sodium of niNa =
i
cNa
π a2 . In N action potentials the amount of sodium that enters this length
of axon is ∆nNa 2π aN. Therefore, to change the concentration by about 10%
∆nNa 2π aN =
i
π a2
cNa
.
10
Therefore,
i
acNa
,
20∆nNa
N
=
N
=
N
≈ 1.5 × 107 .
(250 × 10−4 cm) · (50 × 10−3 mol/cm3 )
20(4 × 10−12 mol/cm2 )
,
So that about 15 million action potentials are required to raise the intracellular
sodium concentration 10%. If the cell generated a high rate of action potentials of
100/s, it would take
1.5 × 107
≈ 1.7 days
100 · 60 · 60 · 24
for the concentration to change 10% by this mechanism. This analysis has ignored
the effect of active transport of sodium on the intracellular sodium concentration.
Problem 4.7 The potassium conductance is defined as
GK (Vm , t) = GK n4 (Vm , t),
where
dn(Vm , t)
+ n(Vm , t) = n∞ (Vm ).
dt
From Figure 4.25 (Weiss, 1996b) it is apparent that n∞ (75) ≈ 1, n∞ (−100) ≈ 0.02, and
τn (−100) ≈ 5 ms. Therefore, n starts at a value of 1 and decreases exponentially to
0.02 with a time constant of about 5 ms,
τn (Vm )
n(Vm , t) = n∞ − (n∞ − no )e−t/τn for t ≥ 0,
n(Vm , t) ≈ 0.02 − (0.02 − 1)e−t/5 for t ≥ 0,
n(Vm , t) ≈ 0.02 + 0.98e−t/5 for t ≥ 0,
68
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
GK mS/cm
2
40
30
20
Figure 4.11: The potassium conductance versus
time (Problem 4.7).
10
0
1
2
3
t (ms)
4
5
where t is in ms. Therefore, the potassium conductance is
4
for t ≥ 0,
GK (Vm , t) ≈ 36 0.02 + 0.98e−t/5
where GK is expressed in mS/cm2 and t in ms. Note that this is approximately
GK (Vm , t) ≈ 36e−4t/5 for t ≥ 0,
so that the time constant for the conductance is about 1.25 ms. Thus, the initial value
of the potassium conductance is about 36 mS/cm2 and the final value is about 36 · (2 ×
10−2 )4 ≈ 5.8 × 10−6 mS/cm2 . The exact solution is shown in Figure 4.11.
Problem 4.8
a. The sodium current density is
JNa (Vm (t), t) = GNa (Vm (t), t)(Vm (t) − VNa ),
where
GNa (Vm (t), t) = GNa m3 (Vm (t), t)h(Vm (t), t).
Because, Vm (t) is constant in the three intervals t < 0, 0 < t < 0.1, and t >
0.1, m(Vm (t), t) and h(Vm (t), t) are exponential functions of time in all three
intervals. The parameters need to be determined for these three intervals. From
Figure 4.25 (Weiss, 1996b), m∞ (−100) ≈ 0, h∞ (−100) ≈ 1, τm (−100) ≈ 0.03 ms,
τh (−100) ≈ 2 ms, m∞ (60) ≈ 1, h∞ (60) ≈ 0, τm (60) ≈ 0.1 ms, and τh (60) ≈ 1
ms. Therefore, for t < 0, GNa (−100, t) ≈ 0 and JNa (−100, t) ≈ 0. For 0 < t < 0.1
m(60, t) ≈ 1 − e−t/0.1 and h(60, t) ≈ e−t/1 ,
where t is in ms. Therefore,
3
GNa (60, t) ≈ 120 1 − e−t/0.1 e−t/1 mS/cm2 .
During the interval 0 < t < 0.1, Vm (t) = VNa . Therefore, JNa (60, t) = 0. At
the end of this interval, m(60, 0.1) ≈ 0.63, h(60, 0.1) ≈ 0.9, and GNa (60, 0.1) ≈
120 · (0.63)3 · 0.9 = 27 mS/cm2 For t > 0.1
m(−100, t) ≈ 0.63e−(t−0.1)/0.03 and h(60, t) ≈ 0.9 + 0.1 1 − e−(t−0.1)/2 .
PROBLEMS
69
Therefore,
3 GNa (−100, t) ≈ 120 0.63e−(t−0.1)/0.03
0.9 + 0.1 1 − e−(t−0.1)/2
mS/cm2 .
Thus, h(60, t) ≈ 1 so that
GNa (−100, t) ≈ 30e−(t−0.1)/0.01 mS/cm2 ,
so that the conductance is approximately exponential with a time constant of 0.01
ms. The current is
JNa (−100, t) ≈ 30e−(t−0.1)/0.01 (−100 − 60) × 10−3 = −4.8e−(t−0.1)/0.01 mA/cm2 .
The exact values of the variables are shown in Figure 4.12
b. A change in VNa changes only JNa and none of the other variables. VNa is defined
approximately as
!
!
o1
o2
cNa
cNa
1
2
VNa = 60 log10
and VNa = 60 log10
,
i
i
cNa
cNa
1
2
and VNa
are the old and new values of VNa . Note
where VNa
!
o1
cNa
/10
2
1
= VNa
VNa = 60 log10
− 60 log10 10 = 60 − 60 = 0.
i
cNa
For 0 < t < 0.1
3
JNa (60, t) ≈ 120 1 − e−t/0.1 e−t/1 (60 − 0) × 10−3 mA/cm2 ,
3
JNa (60, t) ≈ 7.2 1 − e−t/0.1 e−t/1 mA/cm2 .
For t > 0.1
JNa (−100, t) ≈ 30e−(t−0.1)/0.01 (−100 − 0) × 10−3 mA/cm2 ,
JNa (−100, t) ≈ 3e−(t−0.1)/0.01 mA/cm2 .
The exact values of the current is shown in Figure 4.12
Problem 4.9
a. This experiment is designed to measure the sodium inactivation factor h(Vm , t)
for Vm = −60 mV.
b. The membrane potential profile starts at Vm = −120 mV which sets h∞ (−120) ≈ 1.
The step to Vm = −60 mV starts a change in h. The step to Vm = −40 mV now
results in a transient change GNa which is due to a change in m. The assumption
of the method is that the change in m is so rapid that m reaches its final value
before h has changed appreciably. The current density is
3
Jp (T ) ≈ GNa m∞
(−40)h(−60, T )(−40 − VNa ) × 10−3 ,
70
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
50
Vm (t) (mV)
30
2
−0.1
0.1
0.2
GN a (t) (mS/cm )
20
0.3
t (ms)
−50
10
−100
−0.1
0
0.1
0.2
4
0.6
m(t)
2
0.3
t (ms)
2
JN a (t) (mA/cm )
part a
0.4
−0.1
0.2
−0.1
0
−0.1
0.1
h(t)
0.2
0.2
0.3
t (ms)
−4
0.3
t (ms)
2
0.3
t (ms)
2
JN a (t) (mA/cm )
part b
1
−0.1
0.1
−1
0.9
0.2
−2
0.1
0.95
0.1
−2
Figure 4.12: The sodium variables versus time (Problem 4.8).
0.2
0.3
t (ms)
PROBLEMS
71
1
0.8
y(T)
0.6
Figure 4.13: Method to estimate sodium inactivation (Problem 4.9).
0.4
0.2
10
20
T
30
2
(mA/cm )
5
40
10
t (ms)
15
t (ms)
5
−0.1
−0.1
−0.2
−0.2
−0.3
10
15
−0.3
Sodium only
Sodium and potassium
Figure 4.14: A sequence of two-step membrane potential profiles of sodium current (left panel)
and both sodium and potassium current (right panel) (Problem 4.9). The delay T ranges from
0 to 12 ms in 1 ms increments.
and
3
(−40)h(−120, 0)(−40 − VNa ) × 10−3 .
Jp (0) ≈ GNa m∞
Therefore,
y(T ) =
Jp (T )
h(−60, T )
=
,
Jp (0)
h(−120, 0)
but h(−120, 0) ≈ 1. Therefore, y(T ) = h(−60, T ). From Figure 4.25 (Weiss,
1996b) it can be found that h∞ (−60) ≈ 0.6, the initial value of h is h∞ (−120) ≈ 1,
and τh (−60) ≈ 8 ms. Therefore,
y(T ) = 0.6 − (0.6 − 1)e−t/8 = 0.6 + 0.4e−t/8 ,
where t is in milliseconds as shown in Figure 4.13.
The accuracy of the method can be assessed by comparing the method of estimation of
h with the value of h. The above analysis was based on the sodium current alone. In
the further analysis presented below, the method will be assessed both when only the
sodium current is included and when the sodium and potassium currents are included.
The method is shown for both of these cases in Figure 4.14. The results show that both
JNa and JNa + JK exhibit a negative peak in inward current whose magnitude decreases
as T increases. To assess the method outlined above, the minima of the inward current
were computed numerically and these minima were normalized by the minimum when
T = 0. These results are compared to h in Figure 4.15. The results show that at least
for the voltage levels investigated, there are small but systematic errors in estimating
72
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0
Sodium only
5
10
15
0.2
Sodium and potassium
0
20
Time (ms)
5
10
15
20
Figure 4.15: A comparison of the minimum inward current versus duration of the pulse T (solid
line) to h(T ) for Vm = −60 mV (dashed line) (Problem 4.9). The value of Jp (T )/Jp (0) is plotted
versus T .
h with this method. The maximum errors are larger when the potassium current is
included. A more careful analysis would explore these results for a range of membrane
potentials.
Problem 4.10 The membrane ionic current during the interval (0, tp ) is assumed to be
carried predominantly by sodium. Hence,
Jm (z, t) ≈ Cm
∂Vm (z, t)
+ JNa (z, t).
∂t
From the core conductor model
Jm (z, t) =
1
∂ 2 Vm (z, t)
.
2
2π a(ri + ro )ν
∂t 2
Therefore,
∂ 2 Vm (z, t)
∂Vm (z, t)
1
+ JNa (z, t).
≈ Cm
2π a(ri + ro )ν 2
∂t 2
∂t
Integrate this equation over the interval (0, tp ). Three terms need to be evaluated. The
integral of the left-hand side is
!
1
∂Vm (z, t)
∂Vm (z, t)
= 0,
−
2π a(ri + ro )ν 2
∂t
∂t
tp
0
because the derivatives are zero at both 0 and tp . The integral of the first term on the
left-hand side is
Z tp
∂Vm (z, t)
Cm
dt = Cm (Vm (z, tp ) − Vm (z, 0)).
∂t
0
If these results are combined they yield
0 ≈ Cm (Vm (z, tp ) − Vm (z, 0)) +
Z tp
0
0 ≈ Cm (Vm (z, tp ) − Vm (z, 0)) + F
JNa (z, t) dt,
Z tp
0
φNa (z, t) dt,
PROBLEMS
73
R tp
where φNa (z, t) is the outward flux of sodium. The term 0 φNa (z, t) dt represents
the total efflux of sodium during the rising phase of the action potential. The equation
can be solved for Cm as follows
Cm
≈
R tp
−F 0 φNa (z, t) dt
,
Vm (z, tp ) − Vm (z, 0)
≈
(9.65 × 104 ) · (1.5 × 10−12 )
≈ 1.2 µF.
(50 − (−70)) × 10−3
Problem 4.11 The capacitance current density is defined as
JC (t) = Cm
∂Vm (t)
.
∂t
Therefore, during an action potential JC (t) must be both positive and negative. Furthermore, JC (t) = 0 at the peak of the action potential. Therefore, the capacitance current
must be J2 (t).
The sodium current density is defined as
JNa (t) = GNa (t)(Vm (t) − VNa ).
Since GNa (t) > 0 and Vm (t) − VNa < 0 during an action potential, therefore, JNa (t) < 0
during an action potential. Therefore, the sodium current density must be J3 (t).
The potassium current density is defined as
JK (t) = GK (t)(Vm (t) − VK ).
Since GK (t) > 0 and Vm (t) − VK > 0 during an action potential, therefore, JK (t) > 0
during an action potential. Therefore, the potassium current density must be J1 (t).
Problem 4.12 From the core conductor model
Jm (z, t) =
∂ 2 Vm (z, t)
1
.
2π a(ri + ro )
∂z2
Since the action potential is propagating at constant velocity ν,
Jm (z, t) =
1
∂ 2 Vm (z, t)
.
2π a(ri + ro )ν 2
∂t 2
Therefore, Jm (z, t) = 0 when ∂ 2 Vm (z, t)/∂t 2 = 0. Also the capacitance current is
JC (z, t) = Cm
∂Vm (z, t)
.
∂t
Therefore, JC (z, t) = 0 when ∂Vm (z, t)/∂t = 0. Taking these results into account leads
to the answers shown in Table 4.1.
Problem 4.13
74
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
t0
a
b
c
d
e
f
g
h
i
j
k
None
t1
t2
t3
t4
√
√
√
√
√
Table 4.1: Properties at different points in a
propagated action potential (Problem 4.12).
√
√
√
a. The membrane potential at which the early, transient current reverses polarity
is approximately equal to the Nernst equilibrium potential for sodium. These
f
f
potentials are Vm = 54 mV in solution A and Vm = 36 mV in solution B. Therefore,
i
i
A
B
60 log(cNa
/cNa
) ≈ 54 and 60 log(cNa
/cNa
) ≈ 36. Subtracting the second equation
A
B
A
B
from the first, yields 60 log(cNa /cNa ) ≈ 18 which implies that cNa
/cNa
≈ 2.
b. The late, maintained current is carried by potassium. However, at low membrane
potential, the potassium conductance is small and the reversal of the potassium
f
current cannot be discerned. However, at each value of Vm , JKA > JKB which imf
f
plies that GK (Vm − VKA ) > GK (Vm − VKB ) which shows that VKA < VKB . Expressing
oA
i
the Nernst potential in terms of concentrations shows that (RT /F ) ln(cK
/cK
)<
oB
i
oA
oB
(RT /F ) ln(cK /cK ) which implies that cK < cK .
Problem 4.14
a. Leakage current (vi). The leakage current is JL = GL (Vm − VL ) where GL and
VL are constants. Hence, the leakage current waveform has the same shape as
the membrane potential waveform. In addition, since GL is relatively small, the
leakage current during an action potential is relatively small compared with the
sodium and potassium currents.
b. Potassium current (iv). The potassium current is JK = GK (Vm , t)(Vm − VK ). Since
GK ≥ 0 and Vm ≥ VK , JK ≥ 0 during an action potential. Also the potassium
current is larger than the leakage current and the onset of the potassium current
occurs later than Vm since n has a relatively slow response to a change in potential.
c. Sodium current (iii). The sodium current is JNa = GNa (Vm , t)(Vm − VNa ). Since
GNa ≥ 0 and Vm ≤ VNa , JNa ≤ 0 during an action potential. Also the sodium current is larger than the leakage current and the sodium current onset occurs rapidly
as Vm changes since m has a relatively fast response to a change in potential.
d. Total membrane current (ix). The core conductor model shows that Jm is proportional to ∂ 2 Vm /∂z2 which, for an action potential traveling at constant velocity,
PROBLEMS
75
is proportional to ∂ 2 Vm /∂t 2 . The waveform looks like the shape of the second
derivative of Vm .
e. External longitudinal current (xii). The core conductor model shows that Io =
(1/(ro + ri ))∂Vm /∂z. For an action potential traveling at constant velocity in the
+z-direction −ν∂Vm /∂z = ∂Vm /∂t. Therefore, Io = −(1/(ro + ri )ν)∂Vm /∂t. The
waveform looks like the negative of the first derivative of Vm . The waveform is
large and negative before the peak in the action potential, zero at the peak of the
action potential, and positive immediately after the peak of the action potential.
However, the external longitudinal current is not a component of the membrane
current.
f. Capacitance current (vii). The capacitance current is JC = Cm dVm /dt. Hence,
the capacitance current is proportional to the first derivative of the membrane
potential and should look like the negative of the external longitudinal current
(see part e).
Problem 4.15
a. The current stimulus delivers a charge that equals the area of the current pulse,
Q = (0.3 × 10−3 ) · (0.1 × 10−3 ) = 30 nC.
This charge causes a depolarization of 30 mV. Therefore, the total capacitance of
the membrane is
Q
30 × 10−9
C=
=
= 1 µF.
∆Vm
30 × 10−3
The specific capacitance of the membrane is 1 µF/cm2 , and the capacitance of the
axon is C = π dLCm where d and L are the area and length of the axon. Therefore,
L=
1 µF
C
=
= 6.4 cm.
π dCm
π · 0.05 cm · 1 µF/cm2
b. Replacing half the sodium in the sea water with impermeant ions, changes the
Nernst equilibrium potential for sodium,
!
!
o
o
/2
cNa
cNa
RT
RT
−
,
ln
ln
∆VNa =
i
i
F
F
cNa
cNa
RT
1
=
ln
F
2
≈ −59 log10 2 = −18 mV.
The resting potential V1 is relatively insensitive to the sodium concentration so it
will change very little. Because of the reduction of the sodium equilibrium potential, V1 will decrease slightly. The difference between V2 and V1 is determined by
the current pulse so that V2 will change by the same amount as V1 , namely very
little. V3 approaches the sodium equilibrium potential. Hence, V3 will decrease by
about −18 mV. V4 approaches the potassium equilibrium potential and will not
change.
76
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
Problem 4.16 In this problem, the Nernst equilibrium potential for ion n is assumed to
be
!
o
cn
59
(mV).
Vn =
log10
i
zn
cn
a. The voltage-clamp results show that after the initial pulse of capacitance current
which occurs near t = 0, there is an early outward current which is the sodium
current. This implies that Vm − VNa > 0 so that VNa < 50 mV. Therefore,
!
400
VNa = 59 log10
< 50 mV,
i
cn
i
> 57 mmol/L. The current-clamp results reveal an action
which implies that cNa
potential with a peak voltage of 30 mV. Therefore, VNa > 30 mV, and
!
400
> 30 mV,
VNa = 59 log10
i
cn
i
< 124 mmol/L. Putting these results together yields 57 <
which implies that cNa
i
cNa < 124 mmol/L.
b. The voltage-clamp results show that the late current component carried by potassium is outward which shows that VK < 50 mV. The current-clamp results are more
revealing. They show that the undershoot of the action potential which approaches
the potassium equilibrium potential reaches −70 mV. Therefore, VK < −70 mV.
Therefore,
!
10
VK = 59 log10
< −70 mV,
i
cK
i
> 154 mmol/L.
which implies that cK
Problem 4.17
a. The resting potential according to the Hodgkin-Huxley model will be
o
=
Vm
GK
GNa
GL
VK +
VNa +
VL ,
Gm
Gm
Gm
where the Nernst equilibrium potential at a potential of 6.3◦ is
Vn =
o
co
cn
RT
≈
55.7
log
log10 n
10
i
i mV,
F log10 e
cn
cn
and at the resting potential Gm = 0.374+0.0109+0.3 = 0.685 mS/cm2 . Therefore,
o
=
Vm
0.374
0.0109
0.3
(−72.3) +
(55.3) +
(−49) = −60.1 mV.
0.685
0.685
0.685
b. Since, at rest the membrane potential is constant, the capacitance current is zero.
The total membrane current is determined by the stimulus which is zero prior to
the occurrence of the pulse of membrane current. Since the ionic current is the
difference between the membrane current and the capacitance current, the ionic
current is also zero.
PROBLEMS
77
c. For t > 0.5 ms, Jm = 0. Hence, in this interval JC +Jion = 0. Therefore, JC = −Jion
and, in particular, (JC )max = −(Jion )min .
d. Note that the maximum value of Vm is 44.2 mV and the minimum value is −71.2
mV. In part a, it was found that the Nernst equilibrium potentials were +55.3 for
sodium and −72.3 for potassium. Therefore, during the action potential VNa >
Vm > VK . JNa = GNa (Vm − VNa ) so that during the action potential JNa < 0.
Similarly, JK = GK (Vm − VK ) so that during the action potential JK > 0. However,
JL = GL (Vm − VL ), and Vm < VL during a portion of the action potential while
Vm > VL during other portions of the action potential. Therefore, JL is both
positive and negative.
e. The space constant is
λC ≈ √
1
,
gm ri
when ro ri . Let a be the radius of the axon, then gm = 2π aGm and ri =
ρi /(π a2 ) so that
s
s
a
0.025
= 0.43 cm.
=
λC =
2Gm ρi
2 · 0.685 × 10−3 · 100
f. The time constant of the membrane is
τM =
Cm
10−6
=
= 1.46 ms.
Gm
0.685 × 10−3
Problem 4.18
a. The propagated action potential starts at the resting value with a slope of 0, i.e., at
the point t1 . During the onset of the action potential, dVm /dt is large and positive.
Therefore, the arrows must be as shown in Figure 4.16.
b. JC = Cm dVm /dt. Therefore, the capacitance current is zero when dVm /dt =
0. This occurs at t1 (at the resting potential) and at t5 (the peak of the action
potential).
c. During a propagated action potential, the core conductor model implies that the
membrane current density is
Jm =
1
∂ 2 Vm
.
2π a(ri + ro )ν 2 ∂t 2
Thus, the Jm = 0 when ∂ 2 Vm /∂t 2 = 0 which is zero when ∂Vm /∂t has a maximum
or a minimum value. These occur at the times t2 , t3 , and t4 .
Problem 4.19 It is helpful to determine the Nernst equilibrium potentials for each ion.
Note that the concentration ratios are all factors of 10 which makes the determination
78
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
300
t3
dVm /dt (mV/ms)
200
Figure 4.16: Phase plane plot showing phase
trajectory during a propagated action potential
(Problem 4.18). Arrows show the direction for
increasing time.
100
0
−100
t1
t5
t4
t2
−80
−40
0
Vm (mV)
40
simple. In addition, approximate the Nernst potential as 60 mV per decade of concentration for a univalent ion. Therefore,
!
o
cK
60
16
VK ≈
= −60 mV,
log
= 60 log
i
zK
160
cK
!
o
cNa
400
60
= 60 log
= +60 mV,
log
VNa ≈
i
zNa
40
cNa
!
o
cCl
100
60
= −60 log
= 0 mV,
log
VCl ≈
i
zCl
100
cCl
!
o
cCa
60
20
= +30 mV.
log
VCa ≈
=
30
log
i
zCa
2
cCa
Since the resting potential of the cell equals the Nernst equilibrium potential for potassium, a model consist with these results is that at rest the membrane is permeable only
to potassium ions. That is, at rest the conductance of the membrane to ions other than
potassium is negligible.
The investigators differ in the results they obtained on the effect of calcium concentration on the peak value of the action potential. Consider the model of investigator A
first. First note that
!
!
o
o
cCa
cCa
60
= 30 log
,
VCa ≈
log
i
i
zCa
cCa
cCa
so that the Nernst equilibrium potential for calcium changes by 30 mV/decade of calcium concentration. The data of investigator A is consistent with Vp = VCa . Hence, investigator A would be likely to conclude that the action potential was due to a transient
increase in the calcium conductance of the membrane. Thus, the model of investigator
A (Figure 4.17) consists of two parallel branches — a potassium branch, that predominates at rest, and a calcium branch, that predominates during the peak of the action
potential.
PROBLEMS
GK
VK
79
+
Intracellular
Vm
Membrane
GCa
+
−
VCa
Figure 4.17: Investigator A’s membrane
model (Problem 4.19). At rest GK GCa .
During the peak of the action potential
GK GCa .
+
−
−
Extracellular
+
GK
GN a
GK
GN a
+
−
VN a
+
−
VK
−
+
−
VN a
+
−
Intracellular
Vm
Membrane
GCl
Vm
VK
+
VCl
+
−
−
Extracellular
Figure 4.18: Investigator B’s membrane models (Problem 4.19). In the model in the left panel —
at rest GK GNa and during the peak of the action potential both GK and GNa are appreciable.
In the model in the right panel — at rest GK GNa and GK GCl and during the peak of the
action potential both GNa and GCl are appreciable.
Since the peak of the action potential is independent of calcium concentration in
investigator B’s data, investigator B would be led to conclude that the calcium conductance was small during the peak of the action potential. Investigator B could propose
several simple models to account for her data. Note that the action potential normally
reaches +30 mV. Only two ions have positive Nernst potentials — calcium, which is not
involved in investigator B’s data, and sodium. Therefore, it is clear that the membrane
conductance to sodium must be appreciable during the peak of the action potential.
The membrane must be permeable to potassium at rest and could be highly permeable
to potassium and/or to chloride during the peak of the action potential. These models
are shown in Figure 4.18.
It is important to keep in mind that investigators A and B can propose models that
can account for their data, either investigator A or B (or perhaps both) must be wrong because of the fundamental disagreement in their measurements. The discrepancy must
be due to differences in experimental techniques (ruling out such possibilities as typographical errors, fraud, etc.).
80
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
Interval
Sign of Vm (zo , t)
Sign of Jm (zo , t)
Sign of Pm (zo , t)
τ0
τ1
τ2
τ3
τ4
−
−
+
+
+
+
−
−
−
+
−
+
−
−
+
Table 4.2: (Problem 4.20).
Problem 4.20
a. The power density flowing into a unit area of membrane located at zo at time t is
Pm (zo , t) = Vm (zo , t)Jm (zo , t).
According to the core conductor model
Jm (zo , t) =
∂ 2 Vm (z, t)
1
.
2π aν 2 (ri + ro )
∂t 2
Therefore,
Pm (zo , t) =
1
∂ 2 Vm (z, t)
V
(z
,
t)
.
m
o
2π aν 2 (ri + ro )
∂t 2
Therefore, power flows into the membrane during any interval of time for which
Pm (zo , t) > 0 as is shown in Table 4.2. During intervals τ0 , τ2 , and τ3 , Pm (zo , t) <
0 and there is a net flow of energy out of the membrane.
b. The net flow of energy density into the potassium branch is
Z tb
EK (zo ) =
PK (zo , t) dt,
ta
where the power flowing into the potassium branch is
PK (zo , t) = Vm (zo , t)JK (zo , t) = Vm (zo , t)GK (zo , t)(Vm (zo , t) − VK ).
Since GK (zo , t) ≥ 0 and Vm (zo , t) − VK ≥ 0, PK (zo , t) ≥ 0 when Vm (zo , t) ≥ 0 and
PK (zo , t) ≤ 0 when Vm (zo , t) ≤ 0. Therefore, during intervals τ0 and τ1 there is a
net flow of energy out of the potassium branch.
c. The net flow of energy density into the sodium branch is
Z tb
ENa (zo ) =
PNa (zo , t) dt,
ta
where the power flowing into the sodium branch is
PNa (zo , t) = Vm (zo , t)JNa (zo , t) = Vm (zo , t)GNa (zo , t)(Vm (zo , t) − VNa ).
Since GNa (zo , t) ≥ 0 and Vm (zo , t) − VNa ≤ 0, PNa (zo , t) ≥ 0 when Vm (zo , t) ≤ 0
and PNa (zo , t) ≤ 0 when Vm (zo , t) ≥ 0. Therefore, during intervals τ2 , τ3 and τ4
there is a net flow of energy out of the sodium branch.
PROBLEMS
d.
81
i. The power into the capacitance is
PC (zo , t) = Vm (zo , t)JC (zo , t) = Vm (zo , t)
∂Vm (zo , t)
.
∂t
During interval τ2 power is being delivered to the capacitance which is storing
electrostatic energy. During this interval the capacitance is not supplying
energy.
ii. Since the ionic conductances are non-zero, they cannot supply energy.
iii. “Metabolic storehouses” are not included in the model. Hence, they cannot
supply energy in this model.
iv. The power into the potassium battery is
VK JK (zo , t) = VK GK (zo , t)(Vm (zo , t) − VK ).
Since VK < 0, GK (zo , t) ≥ 0, and Vm (zo , t) − VK ≥ 0, the power flowing into
VK is always negative. That is, VK always delivers energy. The power into the
sodium battery is
VNa JNa (zo , t) = VNa GNa (zo , t)(Vm (zo , t) − VNa ).
Since VNa > 0, GNa (zo , t) ≥ 0, and Vm (zo , t) − VNa ≤ 0, the power flowing
into VNa is always negative. That is, VNa always delivers energy.
v. Since the batteries represent the chemical potential energy stored in concentration differences across the membrane, the answers are the same as in part
iv.
vi. There are no “external sources” in the model. Hence, they cannot supply
energy in this model.
vii. During interval τ2 no energy is supplied by the capacitance or by the potassium branch. Energy is supplied by the sodium branch. The leakage branch
needs to be examined. The power flowing into the leakage branch is
PL (zo , t) = Vm (zo , t)JL (zo , t) = Vm (zo , t)GL (Vm (zo , t) − VL ).
Since GL ≥ 0, power flows out of the leakage branch if Vm (zo , t) < 0 and
Vm (zo , t) > VL or if Vm (zo , t) > 0 and Vm (zo , t) < VL . Since VL < 0, the second case is not possible. This leaves the only possibility as VL < Vm (zo , t) < 0.
But this condition does not occur during interval τ2 . Hence, energy is supplied only by the sodium branch in this interval.
viii. See part vii.
Therefore, the following statements apply to interval τ2 — iv, v, and vii.
Problem 4.21 Three components of the current can be identified chronologically in
the phase-plane trajectory shown in Figure 4.102 (Weiss, 1996b). Starting at a current
of 10 mA, the initial portion of the trajectory is linear with dIm (t)/dt = −0.2Im (t).
Therefore, this portion of the trajectory satisfies the differential equation
dIm (t)
+ 0.2Im (t) = 0,
dt
82
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
Im (t)
Rs
Cm
+
V (t)
−
Figure 4.19: Model for capacitance current with series resistance (Problem 4.21).
which has a solution of the form Im (t) = Ae−0.2t for t > 0 where t is in µs. Since the
trajectory starts at 10 mA at t = 0, the solution is Im (t) = 10e−0.2t for t > 0, where
Im (t) is expressed in mA. Thus, Im (t) decays exponentially with a time constant of 5
µs and represents the capacitance current at the onset of the transition in membrane
potential. This is followed by a negative current component, presumed to be carried
by sodium ions, which is followed by a positive component, presumed to be carried by
potassium ions.
a. Based on the above discussion, the answers are i. 10 mA, ii. −5 mA, and iii. 6 mA.
b. Since the peak sodium current is negative (inward), the membrane potential must
be less than VNa . Therefore, VNa > 50 mV.
o
i
c. Since VNa > 50 mV and VNa = (RT /F ) ln(cNa
/cNa
) then to reduce VNa to 50 mV
i
o
(while keeping cNa constant) requires decreasing cNa
.
d. The phase-plane trajectory shows that the capacitance current is over before the
ionic current changes, i.e., the linear portion of the trajectory intersects the origin
at which both Im (t) = 0 and dIm (t)/dt = 0. Therefore, a simple model of this
portion of the trajectory is shown in Figure 4.19. The capacitance Cm is the total
capacitance of the membrane of the axon under voltage clamp, and the resistance
Rs is the total resistance in series with the membrane and the external voltage
source. It includes the resistance of the sea water and the electrodes. The current
produced by a step of potential of amplitude V is obtained as follows. If the voltage
across the capacitance does not change instantaneously then the total voltage will
appear across the resistance to produce and initial current V /Rs . This current
will decay exponential to zero with time constant τs = Rs Cm . The phase-plane
trajectory indicates that in response to a step of 100 mV, the initial current is 10
mA. Therefore, Rs = 100/10 = 10 Ω. Earlier it was determined that the time
constant is 5 µs. Hence, the capacitance is Cm = 5/10 = 0.5 µF.
Problem 4.22 The membrane current in steady state is
Jm = GNa (Vm − VNa ) + GK (Vm − VK ),
since other ions are negligible and the capacitance current is zero in steady state. Since
o
i
Vm = 0, and since VNa = 0 because cNa
/cNa
= 1, it follows that 1.5 = −GK VK mA/cm2 .
If the extracellular sodium concentration is increased by a factor of ten, then VNa ≈
60 log10 10 = 60 mV. Since the potential has not changed, the potassium current density
is the same. Therefore,
Jm = 5(0 − 60) × 10−3 + 1.5 = 1.2 mA/cm2 .
PROBLEMS
83
5
2
Jion (t) (mA/cm )
Jp
Jp∗
−50
50
100
Vmf (mV)
Figure 4.20: Current-voltage relations
of the peak sodium current (Problem 4.23). The solid line is a plot of
the peak value of the ionic current Jp
and the dashed line is the instantaneous value Jp∗ both plotted against
f
the membrane potential Vm .
−5
Problem 4.23
a. Note that the initial hyperpolarization of −80 mV is provided to increase the the
early sodium current (by making h∞ ≈ 1 initially) and to reduce the initial potassium current (by making n∞ ≈ 0 initially). This increases the separation of sodium
from potassium currents. Therefore, it is assumed that the early component of
the ionic current is due primarily to sodium ions. More specifically, assume that
f
Jp is the peak sodium current. Therefore, Jp should reverse its sign at Vm = VNa .
f
Figure 4.104 (Weiss, 1996b) shows that Jp reverses its polarity at Vm = 45 mV for
both solutions A and B. Therefore, the sodium concentrations of solutions A and
B must be approximately equal.
b. In the Hodgkin-Huxley model, the conductances do not change instantaneously.
f
Therefore, the locus of Jp∗ versus Vm must be a straight line. Therefore, it is only
necessary to find two points through which the straight line passes. Assume that
f
Jp∗ is due to sodium ions only. Then Jp∗ = 0 when Vm = VNa = 45 mV. In addition,
f
for Vm = −30 mV it must be that Jp∗ = Jp as shown in Figure 4.20. The slope of this
line is approximately equal to the value of the sodium conductance at a membrane
potential of −30 mV at the time of occurrence of the second step which is from
f
−30 mV to Vm .
Problem 4.24
a. The current density is
J(V ) = G(V )(V − V0 ).
Therefore, G(V ) is multiplied by the factor V − V0 which is a straight line when
plotted versus V . The results are shown in Figure 4.21. Note that with a constant
conductance, the J-V characteristic is a straight line whose intercept depends upon
84
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
2
J (mA/cm )
−60 mV
4
0 mV
2
−100
−50
60 mV
50
−2
100
V (mV)
−4
−60 mV
4
0 mV
2
60 mV
−100
−50
50
100
−2
−60 mV
4
0 mV
2
60 mV
−100
−50
50
100
Figure 4.21: Current-voltage relations for
three different conductance-voltage characteristics (Problem 4.24) The parameter shown
to the right of each trace is V0 . The upper
panel shows the results for a constant conductance 30 mS/cm2 . The middle panel shows
the results for conductance that is discontinuous with a value of 15 mS/cm2 for V < 0
and a value of 30 mS/cm2 for V > 0. The
lower panel is for a conductance that is 0
for V < −30 mV, is 30 mS/cm2 for V > 30
mV, and changes linearly with V in the range
−30 < V < 30 mV.
PROBLEMS
85
V0 . With the piecewise constant conductance, the J-V characteristic is a piecewise
straight line. However, with the saturating conductance, the J-V characteristic has
a negative slope conductance region that depends upon the bias voltage V0 .
b. The measurements show that both the potassium and sodium conductances saturate as the membrane potential increases (Weiss, 1996b, Figure 4.27). In addition,
the measurements show that the peak sodium current has a negative slope conductance region (Weiss, 1996b, Figure 4.16). The results of part a show that a
negative slope conductance region can be obtained with a conductance, whose
voltage dependence saturates as a function of membrane potential, in series with
an appropriate bias potential.
Problem 4.25
a. Sodium. The electrical responses of the giant axon of the squid can be accounted
for by four distinct currents: one due to flow of sodium ions through the cell
membrane, one due to flow of potassium ions through the cell membrane, one
due to flow of current through the membrane capacitance, and one due to leakage
currents. Since the capacitance and leakage currents have been subtracted, the
measured membrane currents must consist of sodium and/or potassium. Since
there is no potassium in the solution bathing the outside of the cell, the transient
inward current must be carried by sodium ions.
b. Outward currents would normally result from outward flow of sodium or potassium ions. However, all of the intracellular cations have been replaced by cesium,
to which the membrane is relatively impermeant. Thus there is very little outward
current.
Problem 4.26
a. Assume that steady-state is reached and that no action potentials are produced
by this current. Assume that the capacitance current is zero at steady state and
that if the depolarization produced by the current is sufficient then sodium will
be inactivated. Also ignore the leakage current. Then the total membrane current
is carried by potassium to yield
Jm = GK n4∞ (Vm )(Vm − VK ).
For simplicity, assume that the depolarization is sufficiently large that n∞ (Vm ) ≈
1. Then 4 ≈ 36(Vm + 72) × 10−3 which implies that Vm ≈ 39 mV. Examination
of the dependence of the activation/inactivation factors on Vm (Weiss, 1996b, Figure 4.25) reveals that at this potential h∞ ≈ 0 and n∞ ≈ 0.95. Taking these values
into account, estimate Vm again as 4 ≈ 36 · 0.954 (Vm + 72) × 10−3 which implies
that Vm ≈ 64 mV. This calculation could be iterated further. However, it is clear
that this more accurate estimate of Vm results in a value of n∞ that is closer to 1
and a value of h∞ that is closer to zero. Thus, depolarizing the membrane with an
outward current can reduce h sufficiently to inactivate the sodium conductance.
86
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
b. Depolarizing the membrane to −10 mV will set h∞ to near zero (Weiss, 1996b,
Figure 4.25). A more precise value can be obtained from the equations which yield
h∞ ≈ 0.006. Although the depolarization caused by the current pulse,
∆Vm =
(100 × 10−6 A/cm2 )(0.5 × 10−3 s)
= 0.05 V = 50 mV,
1 × 10−6
is large, the membrane is depolarized and the sodium conductance is inactivated.
No action potential is generated. This phenomenon is called depolarization block.
Problem 4.27
a. Ans. (8). From the dependence of the parameters on the membrane potential
(Weiss, 1996b, Figure 4.25), m∞ (−100) ≈ 0, m∞ (10) ≈ 1, τm (−100) ≈ 0.03 ms,
and τm (10) ≈ 0.23 ms. Therefore, at the onset of the pulse m rises exponentially
from 0 to 1 with a time constant of about 0.23 ms. At the offset of the pulse m
falls exponentially from 1 to 0 with a time constant of 0.03 ms.
b. Ans. (1). From the analysis in part a, m∞ is a rectangular pulse that goes from 0
to 1 at the onset and from 1 to 0 at the offset.
c. Ans. (3). From the analysis in part a, τm is a rectangular pulse that goes from
0.03 to 0.23 ms at the onset and from 0.23 to 0.03 ms at the offset.
d. Ans. (9). From the dependence of the parameters on the membrane potential
(Weiss, 1996b, Figure 4.25), n∞ (−100) ≈ 0, n∞ (10) ≈ 0.9, τn (−100) ≈ 5 ms, and
τn (10) ≈ 1.7 ms. Therefore, in 4 ms n(4) ≈ 0.9(1 − e−4/1.7 ) = 0.81, and GK (4) ≈
36(0.81)4 ≈ 16 mS/cm2 . Therefore, the potassium conductance rises with an Sshaped onset to a value of about 16 mS/cm2 with a time constant of the underlying
factor n of 1.7 ms. Thus, the transient response will not be completed at the offset
of the 4 ms pulse. After the offset, the conductance declines exponentially to ≈ 0
with a time constant that is about 5/4 = 1.25 ms.
e. Ans. (10). From the dependence of the parameters on the membrane potential
(Weiss, 1996b, Figure 4.25), h∞ (−100) ≈ 1, h∞ (10) ≈ 0, τh (−100) ≈ 2 ms, and
τh (10) ≈ 1 ms. Therefore, at the onset of the pulse h falls exponentially from
1 to 0 with a time constant of about 1 ms. At the offset of the pulse h rises
exponentially from 0 to 1 with a time constant of 2 ms. Therefore, the sodium
conductance has an S-shaped onset to a peak value of roughly 120(1)3 e−0.8/1 =
54 mS/cm2 . The conductance then declines exponentially to approximately 0. At
the offset, the conductance goes to zero with a time constant of 0.03/3 = 0.01 ms.
f. Ans. (12). JNa = GNa (Vm − VNa ), i.e., the current is the product of the two
terms. During the pulse Vm − VNa ≈ 10 − 55 = −45 mV. Thus, at the peak of the
conductance, the current is 50 × −45 = −2250 µA/cm2 which is −2.25 mA/cm2 .
At the voltage offset there will be a discontinuity in the current.
g. Ans. (16). JK = GK (Vm − VK ), i.e., the current is the product of the two terms.
During the pulse Vm −VK ≈ 10+70 = 80 mV. Thus, at the peak of the conductance,
the current is 18 × 80 = 1440 µA/cm2 which is 1.44 mA/cm2 . At the voltage offset
the direction of current flow will reverse.
JN a (Vm , t) (mA/cm2 )
PROBLEMS
87
6
4
Figure 4.22: A plot of the sodium current density
(Problem 4.29).
2
0
0
1
2
3
t (ms)
Problem 4.28
a. Ans. is (3). The action potential shows a slower depolarization at the onset and
a slower repolarization. This is consistent with an increase in the membrane capacitance.
b. Ans. is (4). The peak action potential goes to a higher potential which is consistent
with an increase in the Nernst equilibrium potential for sodium.
c. Ans. is (1). The resting potential is decreased and the undershoot of the action
potential has been reduced implying that the resting potential is closer to the
potassium equilibrium potential. This is consistent with a reduction in the leakage
conductance.
d. Ans. is (2). The action potential shows a more rapid depolarization at the onset
and a faster repolarization. This is consistent with an increase in the temperature.
Problem 4.29
a. Since h = 1, the sodium current density is
JNa (Vm , t) = GNa m3 (Vm , t)(Vm − VNa ),
where
m(Vm , t) = m∞ (Vm ) − (m∞ (Vm ) − m(0))e−t/τm (Vm ) .
m(0) = m∞ (−100) ≈ 0, m∞ (0) ≈ 0.96, τm (0) ≈ 0.27 ms, VNa = 59 log 40/400 =
−59. Hence,
JNa (Vm , t) = 120 · (0.96)3 (1 − e−t/0.27 )3 (0 + 59) µA/cm2
= 6(1 − e−t/0.27 )3 mA/cm2 .
A plot of the sodium current density is shown in Figure 4.22.
b. The sodium current under the conditions in this problem is an outward current
that does not inactivate. This is distinctly different from the sodium current for
normal sodium concentration which is an inward current and inactivates.
88
CHAPTER 4. THE HODGKIN-HUXLEY MODEL
Chapter 5
SALTATORY CONDUCTION IN
MYELINATED NERVE FIBERS
Exercises
Exercise 5.1 In the cable model, the current per unit length and the membrane potential
are related by
∂Vm (z, t)
o
+ gm (Vm (z, t) − Vm
).
Km (z, t) = cm
∂t
o , this relation
During any time interval for which ∂Vm (z, t)/∂t > 0 and Vm (z, t) > Vm
implies that Km (z, t) > 0. Examination of Figure 5.23 (Weiss, 1996b) reveals that at the
node of Ranvier, the membrane current is positive then negative during the onset of
the membrane potential, an interval in which the cable model predicts a positive (outward) current only. Thus, the relation of membrane current to membrane potential at
the node of Ranvier is inconsistent with the cable model. In contrast, at the internode
the membrane current is positive (outward) during the onset of the membrane potential as predicted by the cable model. These results are consistent with the model that
action potentials are initiated at the nodes of Ranvier and propagate in the internodes
according to the cable model.
Exercise 5.2 An inward current occurs through the membrane at a node of Ranvier during the initiation of an action potential. This inward current flows outward through the
internodal membrane. Each internode is terminated by nodes of Ranvier at both ends.
Thus, there are two peaks in the outward current through the internodal membrane —
one results when an action potential is initiated at the node of Ranvier on one side of
the internode and the other when an action potential is initiated at the other side of the
internode.
Exercise 5.3 It is simplest to estimate the conduction velocity from measurements of
the delay of the longitudinal current versus time (Weiss, 1996b, Figure 5.20). Since
the measurements near the end of the nerve fiber show some nonuniformities in the
longitudinal currents, it is best to estimate the conduction velocity from the central
portion of the fiber. The delay between 1 and 7 mm is 0.83 − 0.6 = 0.23 ms so that the
conduction velocity is approximately 26 mm/ms which equals 26 m/s.
89
90
CHAPTER 5. SALTATORY CONDUCTION
Exercise 5.4
a. False. The conduction velocity of a myelinated nerve fiber is proportional to fiber
diameter whereas the conduction velocity of an unmyelinated nerve fiber is proportional to the square root of fiber diameter. Thus, the relation between the two
conduction velocities for the same fiber diameter is complex. However, for large
diameter fibers it is true that for the same diameter fiber, the conduction velocity
of myelinated nerve fibers exceeds that of unmyelinated nerve fibers.
b. False. Measurements show that the action potential at a node of Ranvier does
not differ appreciably from that at a neighboring internode. Hence, the action
potential cannot be said to hop from node to node.
c. True. The action current at the node of Ranvier differs radically from that at an
internode. The action current is large and has an inward component at the node
and is smaller and has only an outward component at the internode. Thus, the
inward component of the action current does hop from node to node.
d. False. The insulating effect of the myelin on the internode is important in producing saltatory conduction. However, the fact that the nodes of Ranvier contain a
much higher density of voltage-gated sodium channels is also critical. Thus, the
insulatimg effect of the myelin alone does not account for saltatory conduction.
Exercise 5.5 The action potential in a myelinated nerve fiber can propagate past inexcitable nodes. The number of nodes past which an action potential can propagate is
called the safety factor.
Exercise 5.6 Figures 5.9 and 5.10 shows vertebrate myelinated axons with diameters
between 4 and 17 µm. Generally, vertebrate myelinated axons have diameters from 1 to
20 µm. Diameters of unmyelinated nerve fibers in vertebrates are typically 0.1 to 1 µm,
i.e., much smaller than those of myelinated nerve fibers.
Exercise 5.7 Note that the cusps in the membrane potential plotted versus position
occur at the nodes of Ranvier. These cusps can be interpreted by noting that the core
conductor model yields the equation
∂ 2 Vm (z, t)
= (ri + ro )Km (z, t).
∂z2
Hence, integrating this equation over distance from a to b, a region that includes a node,
gives
Zb
∂Vm (z, t) ∂Vm (z, t) −
= (ri + ro )
Km (z, t) dz.
∂z
∂z
a
z=b
z=a
The cusp is due to an apparent discontinuity in ∂Vm (z, t)/∂z at a node of Ranvier.
This discontinuity is due to the large current flowing at the nodes which results in an
appreciable value of the integral of the membrane current per unit length even though
the width of the node (b − a) is small.
PROBLEMS
91
Exercise 5.8 The large effect on conduction of the action potential of narcotics applied
at the nodes of Ranvier, which occupy only 0.02% of the length of a nerve fiber, occurs
because the action potential is initiated at these nodes. If the action potential is blocked
at a few nodes, propagation may still occur beyond this block, however, the conduction
velocity is reduced.
Exercise 5.9 Measurements (Weiss, 1996b, Figure 5.9) show that d/D = 0.74 so that
the axon diameter d = 11.1 µm for a 15 µm diameter fiber.
a. If the length of a node of Ranvier is assumed to be l = 1 µm, then the number of
sodium channels in a node is Nn
Nn = π 11.1 · 1 · 1000 = 3.5 × 104 channels.
b. Measurements (Weiss, 1996b, Figure 5.10) show that for D = 15 µm L ≈ 2 mm.
Therefore, the number of sodium channels in an internode is Ni
Ni = π 11.1 · 2000 · 25 = 1.7 × 106 channels.
c. While the number of sodium channels in the internode is larger than at the node of
Ranvier, the density of sodium channels is 40 times larger in the node of Ranvier
than in the internode. It is the density of sodium channels that must reach a
threshold value to allow the initiation of an action potential.
Problems
Problem 5.1
a. First find the conductance per unit length of the internode. The figure shows that
at t = 0.37 ms dVm (t)/dt = 0, and hence that KC = cmi dVm (t)/dt = 0 where
cmi is the membrane capacitance per unit length of the internode. Therefore, at
the maximum of the membrane potential, the capacitance current is zero and the
total membrane current is through the conductance of the membrane. Therefore,
o ) where K is the current per unit
Km (0.37) = Kg (0.37) = gmi (Vm (0.37) − Vm
g
length through the membrane conductance and gmi is the membrane conductance
per unit length of the internode. From these relations, solve for gmi as follows
gmi =
Km (0.37)
8 × 10−9 A/cm
= 80 nS/cm.
o =
Vm (0.37) − Vm
(50 − (−50)) × 10−3 V
Next find the capacitance per unit length of the internode. At t = 0.24 ms,
dVm (t)/dt is near its maximum value. At this time,
dVm (t)
≈ 100 mV/0.12 ms = 833 V/s,
dt
Vm (t) ≈ 0, and Km (t) ≈ 16 nA/cm. Therefore,
Kc (0.24) =
o
Km (0.24) − gmi (Vm (0.24) − Vm
)
= cmi
dVm ,
dt t=0.24
92
CHAPTER 5. SALTATORY CONDUCTION
from which it follows that
cmi × 833 = 16 × 10−9 − 80 × 10−9 (0 + 50) × 10−3 = (16 − 4) × 10−9 = 12 nA/cm.
Therefore, cmi = (12 × 10−9 )/833 = 14.4 pF/cm.
b. The time constant of the internode is
τMi = cmi /gmi = (14.4 × 10−12 F/cm)/(80 × 10−9 S/cm) = 0.18 ms.
The resistance per unit length in the internode is
rii =
ρi
110 Ω· cm
=
= 140 MΩ/cm,
2
π (d/2)
π · (5 × 10−4 cm)2
where d/2 is the radius of the axoplasm in the axon. The space constant is
1
1
=p
= 0.3 cm = 3 mm.
λCi = √
6
rii gmi
140 × 10 Ω/cm · 80 × 10−9 S/cm
c. The thickness of the membrane of the internode of a myelinated nerve fiber is
appreciably larger than the thickness of the membrane of an unmyelinated nerve
fiber. Therefore, to determine the specific membrane conductance and capacitance
of the internodal membrane, we use the average radius of the membrane of the
internode in the following, i.e., a = (d/2 + D/2)/2. With this approximation, the
specific conductances and capacitances are
Gmi =
Cmi =
80 × 10−9 S/cm
gmi
=
= 21 × 10−6 S/cm2 ,
2π a
2π · 6 × 10−4 cm
cmi
14.4 × 10−12 F/cm
=
= 3.8 × 10−9 F/cm2 .
2π a
2π · 6 × 10−4 cm
d. For each Schwann cell membrane, the change in membrane potential from its resting value can be represented by an equivalent electric network of a square centimeter of membrane that consists of a specific conductance Gm in parallel with
a specific capacitance Cm . Each lamella consists of two Schwann cell membranes
with little intervening cytoplasm. Thus, if the internode myelin has n lamellae,
then there are 2n layers of Schwann cell membrane in each internode. As indicated in Figure 5.1, the membranes in myelin are stacked in series because the
same membrane current flowing from inside the fiber to the extracellular space
flows through all the membranes. It is assumed that there is no appreciable current flowing in the longitudinal paths between lamellae because these paths must
have a very high resistance. If it is assumed that the contribution of the axonal
membrane, which is in series with the myelin, is negligible then the capacitance
per unit area of Schwann cell membrane Cm is
Cm = 2nCmi = 300 · 3.8 × 10−9 F/cm2 = 1.15 µm/cm2 ,
and the conductance per unit area of Schwann cell membrane Gm is
Gm = 2nGmi = 300 · 21 × 10−6 S/cm2 = 6.37 mS/cm2 .
PROBLEMS
93
Cm
Gm
Cm
Gm
2n
= Cm
2n
Gm = C
mi
2n
Gmi
Gm
Cm
Figure 5.1: Equivalent circuit of myelin lamellae (Problem 5.1).
These rough calculations give a value of the capacitance that is in the range that is
typically seen for cell membranes and a value of the conductance that is about a
factor of 10 less than that seen for most cells (Weiss, 1996b, Table 3.1). This result
may well result because the membranes of Schwann cells contain a low density of
ion channels.
e. The time constant is
τM =
Cm
1.15 × 10−6 F/cm2
=
= 0.18 ms,
Gm
6.37 × 10−3 S/cm2
which is the same as τMi found in part b. The space constant is
λC
1
1
=q
√
gm rii
2π (d/2) · Gm · ρi /(π (d/2)2 )
v
s
u
a
5 × 10−4 cm
t
=
=
= 0.19 mm.
2Gm ρi
2 · 6.3 × 10−3 S/cm2 · 110 Ω·cm
=
Thus, an unmyelinated axon with this membrane material and dimensions would
have a space constant that is more than a factor of 15 smaller than that of the
myelinated axon of the same dimensions.
Since the time constants are equal, and the space constant is larger for the myelinated axon, its potential changes will spread over a larger distance — in the internodal regions which make up most of the length of the fiber — than occurs in
the unmyelinated axon. This analysis applies when the cable model is valid. The
cable model applies directly in the internode of a myelinated axon and applies to
the unmyelinated axon at the onset of the action potential.
Problem 5.2
94
CHAPTER 5. SALTATORY CONDUCTION
0.8
Squid giant
axon
JN a
2
J (mA/cm )
0
−0.8
5.0
Figure 5.2: Triangular-wave approximations to
ionic currents.(Problem 5.2).
Toad node
of Ranvier
0
JN a
−5.0
0
1
Time (ms)
2
a. Action potentials propagate along a squid axon with a constant velocity. Therefore,
the transmembrane sodium current density JNa also propagates with a constant
velocity. Hence, the transmembrane sodium current densities at two different
locations are identical functions of time except for a time shift. This property
implies that the number of moles of sodium that enter a squid axon during a
single propagated action potential is the same in each increment of the axon’s
length. Therefore, the number of moles of sodium per unit area of membrane
that enter the axon multiplied by the circumference of the axon yields the total
number of moles that enter a unit length of axon, which is denoted as N. Finally,
the number of moles of sodium can be obtained by dividing the integral of sodium
current by Faraday’s constant F . Thus,
Z
πd ∞
N=
JNa (t)dt,
F 0
where d is the diameter of the fiber. The integral of the current density is estimated
by approximating the current density waveform as a triangle that is 1 ms wide and
0.8 mA/cm2 high (Figure 5.2). Then,
1
mA
pmol
500π µm
×
0.8
.
N≈
(1 ms) ≈ 0.6
2
96, 500 C/mol 2
cm
cm
To estimate the the number of moles transported in myelinated nerve fibers, assume that sodium currents are only important at the nodes of Ranvier. If l represents the length of a node and L represents the length of the internode, only the
nodal fraction l/L of the length is important, and
Z
l πd ∞
JNa (t)dt.
N=
L F 0
The integral of the current density is estimated by approximating the current density waveform as a triangle that is 1.2 ms wide and 7 mA/cm2 high (Figure 5.2).
Then,
10π µm
1
mA
fmol
0.7 µm
×
×
7
.
N≈
(1.2 ms) ≈ 0.005
2
2 mm
96, 500 C/mol 2
cm
cm
PROBLEMS
95
b. The ratio of ATP used to pump sodium out of the unmyelinated to that used to
pump sodium out of the myelinated nerve fiber is equal to the ratio of sodium
ions for the two cases,
ratio =
0.6 pmol/cm
≈ 1.2 × 105 .
0.005 fmol/cm
c. Measurements show that a small myelinated nerve fiber (10 µm diameter) can
conduct action potentials as quickly as a large (500 µm diameter) unmyelinated
nerve fiber. However, the energy required in the myelinated nerve fiber is five
orders of magnitude smaller than that required for the unmyelinated nerve fiber.
Problem 5.3
a. The lower panel in Figure 5.31 (Weiss, 1996b) shows that the peak of the action
potential is at node 4 at 0.6 ms and at node 12 at 1.2 ms. Since nodes are spaced
1 mm apart, the conduction velocity is
ν=
12 − 4
≈ 13 mm/ms.
1.2 − 0.6
Nodes 0 and 14 were avoided in estimating the conduction velocity because the
computations are clearly affected by the boundary conditions at the ends.
b. No! If Vm (z, t) = f (t − z/ν) then Vm (z, t) plotted versus z (Weiss, 1996b, top
panel of Figure 5.31) would be identical to Vm (z, t) plotted versus t (Weiss, 1996b,
bottom panel of Figure 5.31) except for both a reversal and a scale change of the
horizontal axis. The cusps in Vm (z, t) plotted versus z that occur at the nodes
of Ranvier, do not occur in Vm (z, t) plotted versus t. This demonstrates that
Vm (z, t) 6= f (t − z/ν).
c. A myelinated nerve fiber is inhomogeneous; the nodal and internodal myelinated
membrane do not have the same electrical characteristics. In particular, the cusps
are caused by the large inward membrane current that occurs only at the nodes of
Ranvier. This causes a discontinuity in the longitudinal current at the nodes which
is proportional to a discontinuity in ∂Vm (z, t)/∂z according to the core conductor
model. A more formal argument follows from the core conductor model which
yields
∂ 2 Vm (z, t)
= (ri + ro )Km (z, t).
∂z2
Integration of this expression over a node of Ranvier yields
Zb
∂Vm (z, t) ∂Vm (z, t) −
=
(r
+
r
)
Km (z, t) dz.
i
o
∂z
∂z
a
z=b
z=a
A large membrane current at the nodes makes the right-hand term appreciable
even when a and b are close together. Therefore, the difference in slope of the
potential is appreciable and gives rise to the cusps.
96
CHAPTER 5. SALTATORY CONDUCTION
ri L
Rn
+
va (t)
−
+
Figure 5.3: Equivalent electric network of two
nodes with the intervening internode (Problem 5.4).
Cn vb (t)
ro L
−
d. The lower panel of Figure 5.31 (Weiss, 1996b) shows that the onset of the calculated action potential contains a point of inflection between −20 and −30 mV when
recorded at the point of stimulation at node 0. Therefore, the threshold for eliciting an action potential at a node is in this range. The upper panel of Figure 5.31
(Weiss, 1996b) shows that during an action potential the membrane potential is
simultaneously above −30 mV for about 10 nodes. Therefore, when an action potential occurs about 10 nodes are simultaneously at a potential above threshold.
This makes it clear that if a given node or two are inexcitable, as indicated by an
absence of a large inward current, the potential plotted versus position will not
contain cusps but will otherwise change very little. Therefore, nodes beyond the
inexcitable nodes can be stimulated to produce action potentials. Assume that the
inexcitable nodes are not short circuits but maintain their resting conductances.
This can be seen clearly in the upper panel of Figure 5.31 (Weiss, 1996b). Consider
the spatial waveform at the time 0.45 ms. The maximum membrane potential is
at node 6. Clear cusps are seen at nodes 5 and 7 indicating that there is a large
inward current at these nodes. However, at nodes 8, 9, and 10, no cusps are seen
indicating that the inward current at these nodes is small. Thus, these nodes are
not yet excitable. If they remained inexcitable, then even in the absence of an
inward current at these nodes, the membrane potential would be appreciable at
subsequent nodes as the membrane potential profile moves to the right. Thus, the
action potential can propagate beyond the inactive nodes.
Problem 5.4
a. Since there is no current through the membrane in the internode, a length L of
internode has an inside resistance of ri L and an outside resistance of ro L as shown
in Figure 5.3.
b. If va (t) = Va u(t) then vb (t) = Va (1 − e−t/τ )u(t) where τ = Cn (Rn + L(ri + ro )).
Since ri ro , τ ≈ Cn (Rn + Lri ). An action potential occurs in this simple model
when vb (t) = Vth which occurs at time t = T as shown in Figure 5.4. The value of
T can be computed as follows
Vth = Va (1 − e−T /τ ),
from which
Vth
T = −τ ln 1 −
Va
Vth
≈ −Cn (Rn + Lri ) ln 1 −
.
Va
Note that T is a positive quantity since 0 ≤ Vth /Va ≤ 1, and therefore, ln(1 −
Vth /Va ) ≤ 0.
PROBLEMS
97
Va
vb (t)
Vth
Figure 5.4: Step response of simple model of propagation in a myelinated nerve fiber (Problem 5.4).
T
t
νmax
ν
νmax
2
Figure 5.5: Dependence of conduction velocity on
internodal length for a simple model of propagation in a myelinated nerve fiber (Problem 5.4).
0
1
L
ri
Rn
2
3
c. The conduction velocity is
ν=
L
= νmax
T
where
νmax =
L(ri /Rn )
,
L(ri /Rn ) + 1
−1
Cn Lri ln 1 −
Vth
Va
.
The conduction velocity is plotted versus the internodal length in Figure 5.5. When
L(ri /Rn ) 1 then ν = νmax L(ri /Rn ), i.e., the conduction velocity is proportional to the internodal length. Under these condition the nodal resistance Rn is
much larger than the longitudinal resistance of the internode ri L. Therefore, the
time constant τ ≈ Rn Cn is independent of L, and the time delay for triggering an
action potential is constant independent of the internodal length. Under these conditions, the time delay for the action potential to travel a distance L is constant so
that the conduction velocity is proportional to L. Conversely, when L(ri /Rn ) 1
then ν = νmax , i.e., the conduction velocity is independent of internodal length.
Under these conditions the longitudinal resistance of the internode ri L is much
larger than the nodal resistance Rn . Therefore, the time constant τ ≈ Lri Cn is
proportional to L. Therefore, the delay for travelling a distance L is proportional
to L so that the conduction velocity is constant independent of L.
d. All of the assumptions are independent of L except assumption (1). When L is
much smaller than the space constant of the internode λC , the internode can be
represented with a lumped-parameter network. However, when L is much larger
than the space constant of the internode, then the current through the myelin
98
CHAPTER 5. SALTATORY CONDUCTION
sheath cannot be ignored and the internode needs to be treated as a cable. Under
these conditions there is appreciable attenuation of the membrane potential in an
internode. The larger L, the larger this attenuation. If L is made large enough,
conduction would be blocked.
Problem 5.5 The relation of the membrane current per unit length to the membrane
potential at one point in space is
∂V (t)
(in the internode),
∂t
1
∂ 2 V (t)
(in the node).
(ri + ro )ν 2 ∂t 2
Km (t) = gm V (t) + cm
Km (t) =
Therefore, the subsequent parts of the problem can be solved by evaluating the derivatives of V (t) which are shown in Table 5.1 and plotted in Figure 5.6.
Time interval
V (t)
0 ≤ t < 0.2
0.2 ≤ t < 0.4
0.4 ≤ t < 2.0
1250t 2
50 + 500(t − 0.2) − 1250(t − 0.2)2
100 − 25(t − 0.4)2
∂V (t)
∂t
2500t
500 − 2500(t − 0.2)
−50(t − 0.4)
∂ 2 V (t)
∂t 2
2500
−2500
−50
Table 5.1: Membrane potential and its time derivatives (Problem 5.5).
a. In the internode, the cable model applies and the membrane current is a sum of
two terms one equal to gm V and the other to cm ∂V /∂t. The individual currents
through the membrane conductance and capacitance and the total membrane current are shown in Figure 5.7. The magnitude of the capacitance current exceeds
that of the current through the conductance. Note that during the onset of the
action potential Km > 0, i.e., the membrane current is outward in the internode.
b. Since ri ro , the membrane current in the node is
Km =
1 ∂ 2 V (t)
ri ν 2 ∂t 2
Note that during the onset of the action potential the current is first outward and
then inward. This inward current component is the signature for the membrane
current during the onset of an action potential.
p
c. The answer depends upon the space constant defined as λC = 1/ (ri + ro )gm
which for ri ro is
λC ≈ √
1
1
=p
= 2 cm.
6
ri gm
(2.5 × 10 Ω/cm)(10−7 S/cm)
Thus, the space constant is twice the internodal length. Therefore, in the first
few millimeters from the active node, the internodal membrane potential will be
very similar to that at the node. Further away from the active node, the internodal
potential will have a smaller magnitude and a slower time course.
PROBLEMS
99
V (t) (mV)
100
∂V (t)
(mV/ms)
∂t
0
500
Figure 5.6: Membrane potential and its
first two time derivatives (Problem 5.5).
∂ 2 V (t)
2
(mV/ms )
∂t2
−100
2000
0.5
1
1.5
2
t (ms)
−2000
10
gm V
8
6
4
2
Km (t) (nA/cm)
0
0.5
1
1.5
40
cm
2
∂V
∂t
20
Figure 5.7: The membrane current components in
an internode (Problem 5.5).
0.5
1
1.5
2
60
40
gm V + cm
∂V
∂t
20
0.5
1
t (ms)
1.5
2
Km (t) (nA/cm)
100
CHAPTER 5. SALTATORY CONDUCTION
2000
0.5
1
Figure 5.8: The membrane current at a node of
Ranvier (Problem 5.5).
1.5
2
t (ms)
Membrane potential (mV)
−2000
50
0
Figure 5.9: The action potential of an unmyelinated nerve fiber plotted versus position (Problem 5.6).
-50
0
2
4
z/ν (ms)
6
8
Problem 5.6
a. This part deals with the unmyelinated nerve fiber.
i. Since the action potential is travelling in the +z-direction, it must have the
form Vm (z, t) = f (t − z/ν). Figure 5.46 (Weiss, 1996b) shows a plot of
Vm (z, t) versus t. Therefore, a plot of Vm (z, t) versus z looks similar except for a reflection of the waveform about the vertical axis and a change in
the horizontal scale as shown in Figure 5.9.
ii. The duration for which the action potential has a positive potential is estimated to be about 0.5 ms (Weiss, 1996b, Figure 5.46). Hence, since the action
potential is travelling at 2 m/s or 2 mm/ms, the spatial extent of the action
potential is about 0.5 × 2 = 1 mm.
b. This part deals with the myelinated nerve fiber.
i. The time it takes the action potential to travel from node 6 to node 10 is
about 0.3 ms. Each internode is 1.38 mm long. Hence, the total distance
between these nodes is 4 × 1.38 = 5.52 mm. Hence, the conduction velocity
ν = 5.52/0.3 = 18.4 mm/ms.
ii. No, since Vm (z, t) does not have the property Vm (z, t) = f (t − z/ν). The
shape of Vm (z, t) as a function of z contains cusps at the nodes while the
shape of Vm (z, t) as a function of t contains no cusps.
iii. If the structure of myelinated fiber is periodic in z then all membrane variables will be periodic in z. That is, the membrane variables will be the same
except for a delay at corresponding points along the fiber, i.e.,
Vm (z, t) = f (t − z/ν) for z = nL,
PROBLEMS
101
r
r + dr
d
Figure 5.10: Geometry for computing the resistance and capacitance per unit length (Problem 5.8).
D
where n is an integer and L is the internodal length.
iv. From Figure 5.47 (Weiss, 1996b), the potential is positive for about 8.2 nodes
which span a distance of 8.2 × 1.38 = 11.3 mm.
Problem 5.7 The two peaks in the outward current measured in an internode occur because action potentials at a node produce inward current at the node that flows outward
through the internode. Action potentials at the nodes on either side of an internode each
produce an outward component of membrane current in the intervening internode.
a. Suppose trace 1 was recorded at location A which is closer to the left than to
the right node. Therefore, the outward current caused by an action potential at
the left node should exceed that caused by an action potential at the right node.
Since the larger peak is later in time, the action potential at the left node occurs
later than the action potential at the right node. Therefore, the action potential is
propagating in the −z-direction.
b. If the action potential were propagating in the +z direction it should arrive at the
left node before it arrives at the right node. The outward current peak should be
largest at the nearest node. At the center of the internode, the outward current
from the action potential at each node should have about the same peak value.
Therefore, trace 2 was recorded at A, trace 3 at B, and trace 1 at C.
c. It is simplest to estimate the conduction velocity from trace 3. The time between
the peaks is about 0.1 ms in trace 3. Since the nodes are separated by 1 mm, the
conduction velocity is about 10 mm/ms.
Problem 5.8 In general, a resistance is related to the resistivity by the relation R = ρL/A
where ρ is the resistivity, L is the length over which the current flows, and A is the crosssectional area. The geometry for calculating the resistance per unit length of membrane
is shown in Figure 5.10. For the cylinder, the current flows in the radial direction and
the cross-sectional area depends upon the radius. Hence, the resistance per unit length
of an incremental cylindrical shell of radius r is
drm =
ρm dr
.
2π r
102
CHAPTER 5. SALTATORY CONDUCTION
This is integrated from d/2 to D/2 to yield
rm =
Z D/2
d/2
ρm
ρm
dr =
ln(D/d).
2π r
2π
The capacitance of a parallel plate capacitance has the form C = A/d where is
the permittivity, A is the surface area of the plates of the capacitance, and d is the
separation of the plates of the capacitance. Therefore, the same incremental cylindrical
shell shown in Figure 5.10 has a capacitance
dcm =
2π r
.
dr
Since reciprocal capacitances in series add, we integrate the reciprocal capacitance as
follows
Z D/2
1
1
1
dr =
ln(D/d),
=
cm
2π
r
2π
d/2
so that
cm =
2π .
ln(D/d)
Note that the membrane time constant of the myelin in an internode is
τMi = rm cm =
2π ρm
ln(D/d)
= ρm ,
2π
ln(D/d)
so that the time constant of the myelin is independent of the fiber diameter.
Problem 5.9
a. The space constant of the internode, under the assumption that ri ro , is
s
1
rm
=
,
λC = √
gm ri
ri
where rm = 1/gm . In Problem 5.8 it was shown that
rm =
ρm
ln(D/d).
2π
Therefore,
s
λC =
(ρm /2π ) ln(D/d)
=
ρi /(π (d/2)2 )
which yields
s
v
u
u ρm D 2 d 2
ρm 2
d ln(D/d). = t
ln(D/d),
8ρi
8ρi
D
q
λC = K y 2 ln(1/y),
s
where
K=
ρm D 2
,
8ρi
and y = d/D. λC /K has a maximum value at d/D ≈ 0.6 (Figure 5.11). The
p
y 2 ln(1/y)
PROBLEMS
103
0.4
0.3
0.2
0.1
Figure 5.11: The dependence of the space constant on
y = d/D (Problem 5.11).
0.2 0.4 0.6 0.8
y
d/D ¿ 1
1
d/D ≈ 1
Figure 5.12: Schematic diagrams of two fibers of the same
outer diameter D with different myelin thickness (Problem 5.11).
maximum value of λC is found by taking its derivative with respect to y. Note
that the square root is a monotonic function of its argument so only the argument
needs to be maximized as follows
d
(y 2 ln(1/y)) = 0.
dy
The derivative yields
y + 2y ln y = 0 which implies that y(1 + 2 ln y) = 0.
Hence, y = e−1/2 = 0.61. Therefore, the space constant is maximized for d/D =
0.61.
b. Consider the space constant for fibers of a fixed diameter D in two limits, d/D 1
and d/D ≈ 1 as shown in Figure 5.12. The space constant depends on the ratio
of rm to ri . When d/D is small, rm is large because the myelin thickness is large
(assuming that D is fixed). But, the dependence of rm on d is logarithmic, i.e.,
rm ∝ ln(D/d). When d/D is small, ri is large because the cross-sectional area for
longitudinal current is small. The dependence of ri on d is ri ∝ d−2 . The ratio of
these resistances will approach zero as d → 0.
When d/D ≈ 1, rm ≈ 0 because the myelin thickness approaches zero (assuming
that D is fixed). When d/D ≈ 1, ri decreases to a fixed value which is the longitudinal resistance of a cylinder of diameter D. Therefore, the ratio of these resistances
will approach zero as d/D ≈ 1.
These arguments show that the space constant is zero both when d = 0 and when
d = D. Since the space constant is always positive, the dependence of the space
constant on d/D must have a maximum value between the two limits d/D = 0 and
d/D = 1.
c. The conduction velocity depends upon the time it takes to charge a node. Increasing the space constant will result in more rapid charging of a node. This argument
suggests that increasing the space constant will result in increasing the conduction velocity. Although this is not a proof, maximizing the space constant appears
to maximize the conduction velocity.
Vm (z, to )−Vmo (mV)
104
CHAPTER 5. SALTATORY CONDUCTION
110
to = 0.45
100
Figure 5.13: Method for estimating the discontinuity of
the slope of the membrane potential at a cusp (Problem 5.10).
90
4
5
6
7
Node Number
d. Figure 5.9 (Weiss, 1996b) shows that measurements of d/D for myelinated nerve
fibers have an average value of 0.74. In part b it was found that the value of d/D
that maximizes the space constant in the internode is 0.61. Thus, the value of d/D
that is found in myelinated fibers is within 18% of the value the theoretical estimate
predicts will maximize the space constant. Furthermore, as shown in Figure 5.11
the maximum is fairly broad and the value of the space constant for d/D of 0.74
differs by only 5% from the maximum obtained for d/D = 0.61. Since a maximum
in the space constant also leads to a maximum in the conduction velocity, the
value of d/D that is found experimentally corresponds to a value that leads to a
maximum in the conduction velocity.
Problem 5.10
a. The core-conductor equations can be used to determine the relation between membrane potential Vm (z, t) and extracellular longitudinal current Io (z, t). For a propagating action potential, there are no internal or external electrodes and the coreconductor equations reduce to
∂Vm (z, t)
∂Vi (z, t) ∂Vo (z, t)
=
−
= −ri Ii (z, t) + ro Io (z, t) = (ri + ro )Io (z, t).
∂z
∂z
∂z
Therefore, the external longitudinal current is proportional to the derivative of
membrane potential with respect to space. The derivatives of membrane potential
can be estimated just to the left and just to the right of node 6 (Figure 5.13 as
∂Vm (z, t) 109 mV − 96 mV
≈ 4.7 V/m,
− ≈
∂z
2
internodes
× 1.38 mm/internode
node 6
and
∂Vm (z, t) 99 mV − 109 mV
≈ −7.2 V/m.
+ ≈
∂z
1 internode × 1.38 mm/internode
node 6
These are proportional to the external longitudinal current
4.7 V/m
1
∂Vm (z, t)
Io (z, t)|node 6− ≈
≈
≈ 0.3 nA and
ri + ro
∂z
140 MΩ/cm
−7.2 V/m
∂Vm (z, t)
1
Io (z, t)|node 6+ ≈
≈
≈ −0.5 nA,
ri + r o
∂z
140 MΩ/cm
where ri + ro has been approximated by ri taken from Table 5.1 (Weiss, 1996b).
The current out of node 6 is the difference between these longitudinal currents,
Im ≈ (−0.5 − 0.3) nA = −0.8 nA.
PROBLEMS
105
b. For a propagating action potential, the core-conductor equations predict a relation
between membrane current and membrane potential
∂ 2 Vm (z, t)
= (ro + ri )Km (z, t).
∂z2
Therefore, Km (z, t) is proportional to the second derivative of membrane potential with respect to distance. From the data in Figure 5.13, the relation between
membrane potential and distance is concave up: i.e., the second derivative of membrane potential is positive. It follows that the sign of Km at the internode between
nodes 5 and 6 is positive — and the current flow is outward.
106
CHAPTER 5. SALTATORY CONDUCTION
Chapter 6
VOLTAGE-GATED ION CHANNELS
Exercises
Exercise 6.1 Conduction current results from charge carriers (e.g., electrons in a metal
or ions in an electrolyte) that are relatively free to move. Thus, the application of an
electric field causes a flow of these charge carriers that constitutes a conduction current. In the presence of polarizable matter, displacement current results from charge
carriers (e.g., protons and electrons in an electric insulator or fixed charged groups in
a macromolecule) that are not free to move in a medium. The application of an electric
field causes microscopic separations of charge or alignment of dipoles. The time rate
of change of this charge movement constitutes a displacement current.
Exercise 6.2 When ion channels open or close, or in general change their state of conduction, charge groups associated with these channels move. The time rate of change
of redistribution of this charge constitutes a gating current. In voltage-gated channels,
this change in state of conduction results from electric forces on charges in the ion
channels caused by a change in membrane potential.
Exercise 6.3 The single-channel ionic current random variable is the current through a
single channel as a function of time. This current is a random rectangular wave that has
two values — 0 when the channel is closed (non-conducting) and I when the channel is
open (conducting). Transitions between these two states occur randomly with rates of
transitions determined by rate constants that may depend upon the membrane potential. The single-open-channel current is the current through a single channel when it is
open. It is designated by I. The average single-channel current is the ensemble average
current of a population of identical, statistically independent channels. In response to
a step voltage clamp, this current shows exponential transitions between its initial and
final values with a time course determined by the rate constants.
Exercise 6.4
a. True. Tetrodotoxin blocks the sodium channel. Hence, it blocks the flow of any
ion that can pass through the channel including potassium
b. True.
107
108
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
−
+
V mp
+
V
−
Figure 6.1: Equivalent circuit of a patch of membrane attached to a cell (Exercise 6.7).
+
V mn
−
c. False. See part b.
d. False. Gating currents give information about charge movements in the membrane
between any states — conducting and non-conducting states — whereas ionic current give information about the conducting states only.
Exercise 6.5 Trace 1 shows a single-channel current with two states of conduction: one
current is zero and the other is negative. The negative current represents ion flow when
the channel is open. The magnitude of that current is not changed by the step change
in membrane potential. This is inconsistent with the assumption that the open-channel
voltage-current relation is linear. Therefore trace 1 cannot result from an ion channel:
neither from a voltage-gated ion channel nor any other ion channel.
The open-channel currents in trace 2 are different before and after the step change
in membrane potential. This is expected if the open-channel voltage-current relation
is linear. From this short segment of data, one cannot conclude that the probability
that the channel is open has or has not changed during the step change in membrane
potential. Therefore, the current in trace 2 could be from a voltage-gated ion channel
or any other ion channel.
a. Trace 2 only.
b. Trace 2 only.
Exercise 6.6 A depolarization of the membrane will move positive charges in the membrane in the outward direction and negative charges in the inward direction. Both of
these result in an outward current.
Exercise 6.7 An equivalent network of the arrangement for recording from a patch of
membrane when that patch remains attached to the rest of the cell is shown in Figure 6.1.
Therefore, V = Vmn − Vmp . With Vmn = −70 mV, Vmp = −70 − V mV. Thus, as V goes
from −70 to −190 mV, the potential across the patch Vmp goes from 0 to +120 mV.
PROBLEMS
109
Exercise 6.8 Activation of the two-state gate model is exponential, whereas activation
of the potassium conductance of squid giant axon is not exponential — it has an Sshaped onset.
Exercise 6.9 The model of a sodium channel that consists of four, independent twostate gates (which is equivalent to the Hodgkin-Huxley model) predicts (Weiss, 1996b,
Section 6.6) that the time constants for tail currents should be τm (Vm )/3 whereas the
measured time constant is close to τm (Vm ).
Exercise 6.10 Water channels (Weiss, 1996a) have the following properties: (1) they
transport water, (2) water transport is driven by the difference of hydraulic and osmotic
pressure across the membrane, (3) they are ungated and are always open, (4) they are
blocked by mercury compounds. Ion channels have the following properties: (1) they
transport ions, (2) ion transport is driven by differences in electrochemical potential
difference across the membrane, (3) they are gated by some physico-chemical variable,
(4) they are blocked by specific pharmacological agents but not by mercury compounds.
Exercise 6.11 Currents from single voltage-gated ion channels exhibit: (1) discrete states
of conduction, (2) rapid transitions between states, (3) probabilistic times of transition,
(4) a probability of state occupancy that depends upon the membrane potential.
Problems
Problem 6.1
a. This is an activation gate because x∞ increases as Vm increases which means that
the gate opens when the membrane is depolarized.
b. Since x∞ = 0.5 at Vm = V1/2 − −40 mV. The effective valence of the gating charge
can be obtained from the slope of the logarithm of x∞ plotted versus Vm . Note
that
1
ez(F /RT )(Vm −V1/2 )
x∞ =
=
.
1 + e−z(F /RT )(Vm −V1/2 )
1 + ez(F /RT )(Vm −V1/2 )
In the limit as Vm becomes arbitrarily negative, the denominator approaches 1 and
x∞ ≈ ez(F /RT )(Vm −V1/2 ) .
Therefore,
log10 x∞ ≈ z(F /RT )(Vm − V1/2 ) log10 e,
so that the slope of this characteristic is
slope = z(F /RT ) log10 e.
Estimation of the slope (Figure 6.2) yields about 2/(60 mV) so that
z=
RT slope
2/60
≈ 60
= 4.6.
F log10 e
log10 e
110
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
100
10−1
10−1
x∞
τx (ms)
100
10−2
10−3
10−2
−120
−80
−40
0
10−3
−120 −80
40
Membrane potential, Vm (mV)
−40
0
40
Figure 6.2: Estimation of parameters of a two-state gate from the voltage dependence of the
steady-state state occupancy and the time constant (Problem 6.1).
The asymmetry factor ζ is estimated from the dependence of τx on Vm . The
asymptotic values of log10 τx are
lim log10 τx
= − log10 A − (zF /RT )(ζ − 0.5)(Vm − VO ) log10 e,
lim log10 τx
= − log10 A − (zF /RT )(ζ + 0.5)(Vm − VC ) log10 e.
Vm →−∞
Vm →∞
Thus, the slopes estimated from Figure 6.2 yields the ratio of the slopes
5
ζ − 0.5
1/80
=− =
,
−2/100
8
ζ + 0.5
whose solution is ζ = 1.5/13 = 0.12.
Problem 6.2
a. The steady-state value of the gating charge is
Q∞ = Qg (∞) = zeN x(∞) = zeN x∞ .
The two state gate model shows that
x∞ =
1
1 + e−z(F /RT )(Vm −V1/2 )
,
so that
Q∞ =
zeN
1 + e−z(F /RT )(Vm −V1/2 )
.
Since limVm →∞ x∞ = 1, limVm →∞ Q∞ = Qmax = zeN . Therefore,
Q∞ =
Qmax
,
−z(F
/RT )(Vm −V1/2 )
1+e
PROBLEMS
111
and
Q∞
Qmax − Q∞
=
Qmax
1+e−z(F /RT )(Vm −V1/2 )
Qmax
Qmax −
1+e−z(F /RT )(Vm −V1/2 )
z(F /RT )(Vm −V1/2 )
,
= e
Note that F /(RT ) = e/(kT ) where e is the magnitude of the electronic charge and
k is Boltzmann’s constant. If this substitution is made and a logarithm is taken of
both sides of the equation this yields
!
Q∞
= ze(Vm − V1/2 ).
kT ln
Qmax − Q∞
Thus, the logarithmic ratio is a linear function of Vm .
b. The right-hand side of the relation found in part a has units of energy. Dividing
this by e yields the energy per unit electronic charge or the energy in electron
volts. Therefore, the slope of the relation plotted in these units is the valence z.
The slope estimated from Figure 6.74 (Weiss, 1996b) is
slope =
168 meV
= 1.3e.
131 mV
Hence, the equivalent valence of the gating charge is 1.3.
Problem 6.3 The resistance of a cylindrical pore of radius a and length d (which equals
the thickness of the membrane) is
1
1
ρd
(102 Ω·cm) · (75 × 10−8 cm)
=
=
=
.
γ
20 × 10−12 S
π a2
π a2
Therefore, the radius of such a pore is
s
102 · 75 × 10−8 · 20 × 10−12
a=
= 2.2 × 10−8 cm = 2.2 Å.
π
Problem 6.4
a. If the open-channel voltage-current characteristic is linear then
I = γ(Vm − Vn ).
Each of the six records provides one constraint: −1 pA = γ(0 mV− Vn ), −1.4 pA =
γ(−20 mV − Vn ), −1.8 pA = γ(−40 mV − Vn ), −2.2 pA = γ(−60 mV − Vn ),
−2.6 pA = γ(−80 mV − Vn ), snf −3 pA = γ(−100 mV − Vn ). Each of these
equations is satisfied if Vn = 50 mV and γ = 20 pS. Alternatively, the single
open-channel current can be plotted versus the membrane potential as shown in
Figure 6.3. The voltage-current characteristic is linear.
112
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
Vm (mV)
−40
0
40
J −1
J
J
J
I (pA)
−80
Figure 6.3: Single open-channel current versus the
membrane potential (Problem 6.4).
−2
J
J
−3
b. The conductance can be computed from the slope of the line relating the single
open-channel current to the membrane potential (Figure 6.3) which yields γ =
20 pS.
c. The equilibrium potential can be computed from the intercept of the line relating the single open-channel current to the membrane potential (Figure 6.3) which
yields Vn = 50 mV.
d. If these records were from a sodium channel, then the reversal potential should
be equal to the sodium equilibrium potential and the probability that the channel is open should transiently increase following a step depolarization and then
decrease. The reversal potential of 50 mV is close to the sodium equilibrium potential. Furthermore, from the single channel records, it appears that the probability
that the channel is open is smaller when the final voltage is −100 mV than when
the final voltage is larger. However, there is no evidence that the probability is
decreasing with time. Thus, there is no evidence for inactivation, and it appears
unlikely that these records resulted from a sodium channel.
Problem 6.5
a. The probability that the gate is in state S2 is x.
b. Figure 6.73 (Weiss, 1996b) shows that x∞ (−80) ≈ 0.05, x∞ (−20) ≈ 0.82, τx (−80) ≈
0.22 ms, and τx (−20) ≈ 0.27 ms. Therefore, the step response from −80 mV to
−20 mV is
x(t) = 0.82 − (0.82 − 0.05)e−t/0.27 for t > 0,
= 0.82 − 0.77e−t/0.27 for t > 0,
where t is in ms.
c. The average single-channel conductance is
g(t) = γx(t) = 40 0.82 − 0.77e−t/0.27 for t > 0,
where g is in pS. This conductance is plotted in Figure 6.4.
PROBLEMS
113
g(t) (pS)
40
30
20
Figure 6.4: Average conductance (Problem 6.5).
10
0
0.5
1
t (ms)
1.5
g(t) (pS)
40
30
20
Figure 6.5: Average conductance (Problem 6.5).
10
0
0.5
1
t (ms)
1.5
d. This is an activation gate because the channel opens to increase current flow when
the membrane is depolarized.
e. The only change is the channel conductance when the gate is open and closed.
Hence, x(t) does not change and is
x(t) = 0.82 − 0.77e−t/0.27 for t > 0,
where t is in ms.
f. Now the channel conducts when the gate is in S1 . Therefore, the probability that
the channel is in S1 needs to be determined. Since the channel must be in one of
its two states at every time, the probability that the channel is in state S2 is
1 − x(t) = 0.18 + 0.77e−t/0.27 for t > 0.
The conductance is
g(t) = 40 0.18 + 0.77e−t/0.27 for t > 0,
which is plotted in Figure 6.5.
g. This is an inactivation gate because the channel closes to reduce current flow when
the membrane is depolarized.
Problem 6.6
a. For state S0 , the current is zero at both values of the membrane potential. Hence,
this state is non-conducting.
114
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
State
occupancy
probability
1
1−x
x
0.5
Membrane
potential
(mV)
Current
density
(mA/cm2)
0
0
−10
Figure 6.6:
lem 6.6.
J
Solution to Prob-
−20
−30
+20
−60
0
1
2
3
4 5 6
Time (ms)
7
8
9
10
b. The single open-channel current is
I = γ(Vm − Ve ),
where γ is the single open-channel conductance, and Ve is the equilibrium potential
of the channel. I is given for two values of Vm which yields the simultaneous
equations
−2 × 10−12
−0.4 × 10−12
= γ(−60 − Ve ) × 10−3 ,
= γ(20 − Ve ) × 10−3 ,
which can be solved to yield γ = 20 pS, and Ve = 40 mV.
c. From the probability of occupancy of state S1 , x(Vm , t), x(20, ∞) = x∞ (20) =
0.7, and the time constant τx (20) = 1 ms which can be estimated as shown in
Figure 6.6. But,
x∞ (20) = 0.7 =
α(20)
1
and τx (20) = 0.001 =
,
α(20) + β(20)
α(20) + β(20)
whose solution is α(20) = 700 s−1 and β(20) = 300 s−1 .
d. Since there are only two states, the probability of occupying state S0 is 1−x(Vm , t)
which is sketched in Figure 6.6.
e. A direct method would be to measure both the single channel current and the
macroscopic current density from a population of identical channels. From such
measurements N = J(t)/(Ix(t)).
PROBLEMS
115
f. The macroscopic ionic current density due to this population of channels is
J(Vm , t) = N γx(Vm , t)(Vm − Ve ),
= 1011 · 20 × 10−12 · x(Vm , t)(Vm − 40) mA/cm2 ,
= 2x(Vm , t)(Vm − 40) mA/cm2 ,
which is shown plotted in Figure 6.6. For t < 0, J(Vm , t) = 2·0.1·(−60−40) = −20
mA/cm2 . For t = 0+, J(Vm , t) = 2 · 0.1 · (20 − 40) = −4 mA/cm2 . For t → ∞,
J(Vm , t) = 2 · 0.7 · (20 − 40) = −28 mA/cm2 .
Problem 6.7 In general, for the two-state gate model
x∞
=
τx
=
1
α
=
,
(E
α+β
1 + e O −EC )/(kT )
1
1
.
=
(E
−E
)/(kT
)
α+β
A e C B
+ e(EO −EB )/(kT )
a. For Vm = −70 mV, (EO − EC )/(kT ) = −1.386 + 1.1 = −0.286. Therefore,
e(EO −EC )/(kT ) = e−0.286 ≈ 0.75,
and
x(0) =
1
= 0.57.
1 + 0.75
b. For Vm = −40 mV, (EO − EC )/(kT ) = −0.693 + 1.1 = 0.407. Therefore,
e(EO −EC )/(kT ) = e0.407 ≈ 1.5,
and
x(∞) =
1
= 0.4.
1 + 1.5
c. The information given shows that A = 3000/s. Therefore,
τx (−40) =
1
3000
e−1.1
+ e−0.693
=
1
= 0.4 ms.
3000(0.33 + 0.5)
d. The results determined in parts a-c yield
x(t) = 0.4 − 0.4 − 0.57e−t/0.4 for t ≥ 0,
which is plotted in Figure 6.7.
e. This gate closes for a depolarization. Therefore, this direction is consistent with
the idea that x represents the sodium inactivation factor h. However, the time
course of x(t) is about an order of magnitude too fast to represent h(t).
Problem 6.8
116
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
0.6
x(t)
0.4
Figure 6.7: The probability that the channel is
open (Problem 6.7).
0.2
0
0.5
1
1.5
t (ms)
2
a. To find the density of channels, note that the maximum macroscopic conductance
per unit area when all channels are open equals the density of channels times the
single open-channel conductance, i.e.,
f
lim G(Vm , ∞) = N γ.
f
Vm →∞
The maximum macroscopic conductance is 80 mS/cm2 and the single channel conf
ductance can be gotten from the slope of the I-Vm characteristic and is 12 pA/120 mV =
100 pS. Therefore, the density of channels is 80×10−3 /10−10 channels/cm2 which
is 8 channels/µm2 .
f
b. The single open-channel current reverses sign at Vm = 20 mV so that this is the
Nernst equilibrium potential for the channel. Therefore,
!
150
,
VX = 20 = 59 log10
i
cX
i
= 69 mmol/L.
which yields cX
c. The voltage dependence of the macroscopic conductance can be used to determine
the gating charge valence. Note that
80
f
G(Vm , ∞) =
f
1 + e−3(Vm −20)/26
,
so that the gating charge valence is z = 3.
d. The gating current is
Jg (Vm , t) = zeN
dx(Vm , t)
.
dt
f
f
Hence, find x(Vm , t) first. At Vm = −80, x∞ (−80) ≈ 0. At Vm = +20, x∞ (20) =
0.5 and τ(20) = 1 ms. Therefore,
x(Vm , t) = 0.5 1 − e−t u(t),
where u(t) is the unit step function. Differentiation of x with respect to t yields
dx(Vm , t)
= 0.5e−t u(t) (1/ms).
dt
PROBLEMS
117
Jg (µA/cm2 )
0.2
0.15
0.1
Figure 6.8: Gating current (Problem 6.8).
0.05
0
1
2
t (ms)
3
4
Therefore, the gating current is
Jg (Vm , t) = 3 · 1.6 × 10−19 C · 8 × 108 channels/cm2 · 0.5e−t u(t) (1/ms),
= 0.19e−t u(t) (µA/cm2 ).
The gating current is plotted in Figure 6.8.
Problem 6.9
a. The single open-channel current is
INa = γNa (Vm − VNa ),
and the sodium equilibrium potential is
200
= 60 mV.
VNa ≈ 60 log10
20
Therefore,
INa = 30 × 10−12 (−20 − 60) × 10−3 = −2.4 × 10−12 = −2.4 pA.
The rate of transport of moles of sodium is
2.4 × 10−12
INa
=−
. ≈ −2.5 × 10−17 mol/s
zNa F
9.65 × 104
Therefore, the rate of transport of sodium ions is
−2.5 × 10−17 · 6 × 1023 = −1.5 × 105 molecules/s.
b. The sodium ion current is negative, i.e., inward. Therefore, sodium ions are transported inward.
Problem 6.10 It is helpful to determine all the Nernst equilibrium potentials and driving voltages at the outset (Table 6.1). Table 6.1 shows that the current through open
channels is: outward for the potassium channels, inward for sodium and calcium channels, is inward for Vm = −60 mV and outward for Vm = 0 mV for chloride channels,
and is inward for Vm = −60 mV and zero for Vm = 0 mV for magnesium channels. In
addition, the kinetics are first-order for chloride and magnesium and higher-order for
the other channels.
118
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
Ion
Vn
(mV)
K+
Na+
Cl−
Ca++
Mg++
−90
+60
−30
+100
0
Vm − Vn (mV)
Vm = −60 Vm = 0
+30
−120
−30
−160
−60
Current direction
Vm = −60 Vm = 0
+90
−60
+30
−100
0
outward
inward
inward
inward
inward
outward
inward
outward
inward
—
Kinetic
order
higher
higher
first
higher
first
Table 6.1: Nernst equilibrium potentials, differences of membrane potentials from Nernst equilibrium potentials for permeant ions, current direction, and order of kinetics (Problem 6.10).
Channel
Chloride
Calcium
Sodium
Potassium
Magnesium
None of the above
Ji1
Ji2
Ji3
√
Ji4
Ji5
√
√
Table 6.2: Macroscopic membrane ionic current density (Problem 6.10).
√
√
√
a. The results are summarized in Table 6.2. Since there are no inward currents with
first-order kinetics, Ji1 is “none of the above.” Ji2 is inward at −60 mV and is zero
at 0 mV which implies that this is magnesium. Ji3 is outward and shows firstorder kinetics which means it must be chloride. Ji4 is outward and shows higherorder kinetics which implies it is potassium. Ji5 is inward and shows higher-order
kinetics which implies it could be either calcium or sodium.
b. The results are summarized in Table 6.3. Gating currents are outward for a depolarization and are capacitative currents and their final values are zero. Therefore,
only Jg4 has the characteristics of a gating current and it shows higher-order kinetics. Hence, this could be calcium, sodium, or potassium.
c. The results are summarized in Table 6.4. ĩi1 is outward and the single openchannel current at Vm = 0 is three times its magnitude at Vm = −60 mV which
implies that this is potassium. ĩi2 is inward which implies this is either calcium
or sodium. There is no channel for which the current switches from outward to
inward as does ĩi3 . ĩi4 switches from inward to outward which implies that it is
Channel
Chloride
Calcium
Sodium
Potassium
Magnesium
None of the above
Jg1
Jg2
Jg3
Jg4
Jg5
√
√
√
√
√
√
Table 6.3: Macroscopic membrane gating current density
(Problem 6.10).
√
PROBLEMS
Channel
119
ĩi1
Chloride
Calcium
Sodium
Potassium
Magnesium
None of the above
√
Channel
Ii1
Chloride
Calcium
Sodium
Potassium
Magnesium
None of the above
ĩi2
ĩi3
ĩi4
√
ĩi5
√
√
√
Table 6.4: Single-channel ionic currents (Problem 6.10).
√
Ii2
Ii3
Ii4
Ii5
√
√
√
√
Table 6.5:
Single-open-channel
membrane ionic currents (Problem 6.10).
√
the chloride current. ĩi5 switches from inward to zero which implies that it is the
magnesium current.
d. The results are summarized in Table 6.5. The I-Vm characteristics cross the Vm
axis at the Nernst equilibrium potential.
Problem 6.11 For each channel, the average single channel current is given by
i(t) = Ix(Vm , t),
where the single open-channel current is
I = γ(Vm − Ve ) = 25 × 10−12 (Vm − 20) × 10−3 .
a. This part of the problem deals only with the steady-state values of the current and
state occupancy probability.
i. To determine if the channel is voltage gated determine if the probability that
the channel is open changes with a change in membrane potential. For channel A at Vm = −60 mV
−1 × 10−12 = 25 × 10−12 (−60 − 20) × 10−3 x(−60, ∞),
so that x(−60, ∞) = 0.5. For Vm = −20 mV,
−0.5 × 10−12 = 25 × 10−12 (−20 − 20) × 10−3 x(−20, ∞),
so that x(−20, ∞) = 0.5. Therefore, there has been no change in the steadystate value of the probability that the channel is open. Hence, channel A
shows no signs of being voltage-gated.
120
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
For channel B at Vm = −60 mV
−0.5 × 10−12 = 25 × 10−12 (−60 − 20) × 10−3 x(−60, ∞),
so that x(−60, ∞) = 0.25. For Vm = −20 mV,
−0.75 × 10−12 = 25 × 10−12 (−20 − 20) × 10−3 x(−20, ∞),
so that x(−20, ∞) = 0.75. Therefore, this channel is voltage-gated and the
probability that it is open increases with depolarization.
For channel C at Vm = 10 mV
−0.2 × 10−12 = 25 × 10−12 (10 − 20) × 10−3 x(10, ∞),
so that x(10, ∞) = 0.8. For Vm = 80 mV,
0.3 × 10−12 = 25 × 10−12 (80 − 20) × 10−3 x(80, ∞),
so that x(80, ∞) = 0.2. Therefore, this channel is voltage-gated and the probability that it is open decreases with depolarization.
ii. Channel A is not voltage gated, channel B is voltage-gated and is activated
by the depolarization, channel C is voltage-gated and is inactivated by the
depolarization.
b. We express all single-channel currents in units of picoamperes. Channel A is not
voltage gated and the probability that it is open is 0.5. The single-open channel current I changes instantaneously from −2 to −1 as the membrane potential
changes. Therefore, the average single-channel current i also changes instantaneously as follows
(
−2 × 0.5 for t < 0
i(t) =
−1 × 0.5 for t ≥ 0,
which is plotted in Figure 6.9.
Channel B is voltage gated and the probability that the channel is open increases
exponentially from 0.25 to 0.75. The single open-channel current I changes instantaneously from −2 to −1 pA as the membrane potential changes. Hence,
(
−2 × 0.25
for t < 0
i(t) =
−t/τ
−1 × 0.25 + 0.5 1 − e
for t ≥ 0,
which is plotted in Figure 6.9.
Channel C is voltage gated and the probability that the channel is open decreases
exponentially from 0.8 to 0.2. The single open-channel current I changes instantaneously from −0.25 to 1.5 pA as the membrane potential changes. Hence,
(
−0.25 × 0.8
for t < 0
i(t) =
1.5 × 0.8 − 0.6 1 − e−t/τ
for t ≥ 0,
which is plotted in Figure 6.9.
PROBLEMS
121
x(Vm , t)
Channel A
Channel B
1
1
0.8
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.4
0.2
0.2
0.2
0
1
2
3
4
t/τ
0
1
2
3
0
4
t/τ
2
2
1
1
1
I pA
2
1
2
3
−1
i(Vm , t) pA
Channel C
1
t/τ
4
1
2
3
−1
t/τ
4
−2
−2
1.5
1.5
1.5
1
1
1
0.5
0.5
0.5
−1
1
2
3
t/τ
4
−0.5
−1
1
2
3
t/τ
2
3
4
t/τ
1
2
3
1
2
3
−1
−2
−0.5
1
4
−0.5
−1
Figure 6.9: Probability that channel is open, the single open-channel current, and the average
single channel current (Problem 6.11).
t/τ
t/τ
4
4
2
Jg (t) (mA/cm )
122
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
0.4
0.3
0.2
Figure 6.10: Method for estimating the time constant of
the gating current (Problem 6.12).
0.1
0.1 0.2 0.3 0.4 0.5
t (ms)
Problem 6.12
a. The maximum value of GNa,peak is about 95 mS/cm2 at 150 mV. If all the sodium
channels are open at this potential then the single open-channel conductance can
be used to estimate the number of sodium channels. The single open-channel conductance can be computed from the slope of the relation of I to Vm and is γ =
0.2 pA/(50 mV) = 4 pS. Therefore, if GNa,peak = N γ or N = (95 mS/cm2 )/(4 pS) =
238 channels/µm2 .
b. The total gating charge displaced is the integral of the gating current, so that
Z∞
Jg (t) dt.
(Qg )max =
0
Figure 6.10 shows that the gating current density decays exponentially with a time
constant of 0.1 ms. Therefore, the gating current density is
Jg (t) = 0.4e−10t mA/cm2 for t > 0,
where t is in ms. Therefore,
Z∞
0.4e−t/τ dt,
(Qg )max =
0
= 0.4 × 10−3 × 10−4 = 4 × 10−8 C/cm2 = 4 × 10−16 C/µm2 .
Each gate has a charge Q = ze = 6(1.6 × 10−19 ) = 9.6 × 10−19 C. Therefore, the
density of gates is N = 4 × 10−16 /9.6 × 10−19 = 417 channels/µm2 .
c. The plot of bound STX shows that
0.0025 1
1
= 0.0025 +
,
n
0.05 c
which can be rewritten in the form
n=
400c
.
c + 20
This equation describes the binding reaction
STX + Channel z STX · Channel.
At high STX concentration all the channels are bound to STX and the density of
bound channels is 400 channels/µm2 .
PROBLEMS
123
1
0.8
0.6
0.4
0.2
0
1
m3
0.6
h
m3
m3 h
0.4
m3 h
0.2
h
0.8
0.4
−60
0.6 0.8
0.2
1
0
t (ms)
0.2
0.4
−100
0.6 0.8
1
Figure 6.11: The quantity m3 h computed for the Hodgkin-Huxley model for the sodium channel
(Problem 6.12). The responses are shown for a voltage step at t = 0 from a holding potential of
−60 mV (left panel) and from a holding potential of −100 mV (right panel) to a final potential
of +150 mV.
d. Therefore, we have the following estimates of N in channels/µm2 : 238 from part
a; 417 from part b; and 400 from part c. Thus, the methods in parts b and c give
similar estimates, but the method in part a gives a lower estimate of the channel
density. However, the method in part a would be expected to underestimate the
channel density. Note, that because of inactivation, GNa,peak < GNa and it is GNa
that represents the conductance of the sodium channels when they are all open.
Recall that GNa = GNa m3 h. Hence, the quantity m3 h can be examined in response
to a voltage clamp pulse to estimate the error (Figure 6.11). The error occurs
because the peak value of m3 h is less than 1 so that GNa,peak < GNa . The peak
value of m3 h depends upon the holding potential. When the holding potential is
near the resting potential (−60 mV), the initial value of h ≈ 0.6 and the peak value
of m3 h ≈ 0.5. However, when the holding potential is hyperpolarized (−100 mV),
the initial of h ≈ 1 and the peak value of m3 h ≈ 0.75. Thus, the minimum error
incurred by ignoring inactivation occurs when a hyperpolarizing holding potential
is used. Thus, a better estimate of N ≈ 238/0.75 = 317 channels/µm2 which is
closer to the other two estimates of channel density. If the experiment is done
with a holding potential at the resting potential, then the the estimate is N ≈
238/0.5 = 476 channels/µm2 . Intermediate holding potentials yield intermediate
estimates of channel density.
Problem 6.13
a. The average macroscopic ionic current density is
J(Vm , t) = N i(Vm , t) = N γ(Vm − VNa )m3 (Vm , t),
where N is the density of channels in the membrane, i is the average singlechannel current, γ is the single open-channel conductance, and m is the probability that a gate is in the open conformation. Since m satisfies first-order kinetics,
dm(Vm , t)
= α(Vm )(1 − m(Vm , t)) − β(Vm )m(Vm , t),
dt
124
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
which can be written as
dm(Vm , t)
+ (α(Vm ) + β(Vm ))m(Vm , t) = α(Vm ).
dt
Therefore, m is an exponential function of time that depends on
m∞ =
α(Vm )
1
and τm =
.
α(Vm ) + β(Vm )
α(Vm ) + β(Vm )
If the rate constants are evaluated as a function of membrane potential, it is apparent that as the membrane potential changes from −80 to +80 and back to −80
mV, α(Vm ) changes approximately from 0 to 11 to 0 /ms, and β(Vm ) changes
approximately from 18 to 0 to 18 /ms. Therefore, for this same range of potential, m∞ (Vm ) goes from 0 to 1 to 0, and τm (Vm ) goes from 56 to 91 to 56 µs.
Therefore,
m(Vm , t) = (1 − e−t/91 )u(t) − (1 − e−(t−10
4 )/56
)u(t − 104 ),
where t is in µs. Because the transients essentially come to completion at each
transition (Figure 6.12), the solution can also be written as
(
for 0 ≤ t ≤ 104 µs,
1 − e−t/91
m(Vm , t) ≈
4
e−(t−10 )/56 for t ≥ 104 µs.
Therefore, the average macroscopic ionic current is
J(Vm , t) = (200 × 108 channels/cm2 ) · (20 × 10−12 S) ·
((Vm − 50) × 10−3 V) · m3 (Vm , t),
J(Vm , t) =
0.4(Vm − 50)m3 (Vm , t) mA/cm2 ,
where Vm is in mV. This expression can be written as

3

 12 1 − e−t/91
for 0 ≤ t ≤ 104 µs,
J(Vm , t) ≈
3

 −52 e−(t−104 )/56
for t ≥ 104 µs.
Note that for t ≥ 104 µs the ionic current can be written as
J(Vm , t) ≈ −52e−3(t−10
4 )/56
for t ≥ 104 µs,
where the time constant at turn off is 56/3 = 18.7 µs. The ionic current is shown
in Figure 6.12, and a detail at the onset and offset is shown in Figure 6.13
b. Each channel contains 3 independent gates, so the average macroscopic gating
charge density is
Qg (Vm , t) = 3N · ze · m(Vm , t),
where 3N is the number of gates per unit area of membrane, ze is the charge per
open gate, and m is the probability that the gate is open. Substitution of numerical
values yields
Qg (Vm , t) = 3(200 × 108 channels/cm2 ) · 2 · (1.6 × 10−19 C) · m(Vm , t),
Qg (Vm , t) = 19.2m(Vm , t) nC/cm2 .
PROBLEMS
125
m(Vm , t)
1
0.8
0.6
0.4
2
Jg (Vm , t) (mA/cm )
2
J(Vm , t) (mA/cm )
0.2
2
4
6
t (ms)
8
10
2
4
6
t (ms)
8
10
2
4
6
t (ms)
8
10
10
−10
Figure 6.12: Time course of m(Vm , t), J(Vm , t) and
Jg (Vm , t) in response to a pulse of membrane potential (Problem 6.13).
−20
0.2
0.1
−0.1
−0.2
−0.3
J(Vm , t) (mA/cm )
800
2
800
Jg (Vm , t) (mA/cm )
Offset
10
5
200
400 600
t (µs)
10
−10
−20
−30
−40
−50
200
400 600
t (µs)
800
200
400 600
t (µs)
800
2
0
2
Jg (Vm , t) (mA/cm )
2
J(Vm , t) (mA/cm )
Onset
0.2
0.1
0
200
400 600
t (µs)
−0.1
−0.2
−0.3
Figure 6.13: Time course of J(Vm , t) and Jg (Vm , t) in response to a pulse of membrane potential
showing response at the onset and offset of the membrane potential (Problem 6.13). The arrows
mark the time of offset of the membrane potential in the right panels and time is referred to
this offset in these panels.
126
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
C4
4αn
C3
βn
3αn
2βn
C2
2αn
3βn
C1
αn
4βn
O
Figure 6.14: Kinetic diagram for the Hodgkin-Huxley model of the potassium channel in terms
of 4 independent, two-state gates (Problem 6.14).
The average macroscopic gating current density is the derivative of the average
macroscopic gating charge density
Jg (Vm , t) = 19.2
dm(Vm , t)
nA/cm2 ,
dt
so that

d


 19.2 (1 − e−t/91 ) nA/cm2
dt
Jg (Vm , t) ≈
d
4


 19.2 (e−(t−10 )/56 ) nA/cm2
dt
for 0 ≤ t ≤ 104 µs,
for t ≥ 104 µs,
where t is in µs and yields
(
Jg (Vm , t) ≈
0.21e−t/91 mA/cm2
4
−0.34e−(t−10 )/56 mA/cm2
for 0 ≤ t ≤ 104 µs,
for t ≥ 104 µs.
The gating currents are shown in Figures 6.12 and 6.13.
c. The average macroscopic ionic current density J has an S-shaped onset with an
underlying time constant of m of 91 µs and a final value of 12 mA/cm2 . During
this interval the average macroscopic gating current density Jg is a decaying exponential with a time constant of 91 µs. At the offset of the potential both J and
Jg are exponential but their time constants differ by a factor of 3.
Problem 6.14 If each two-state gate has the kinetic diagram
αn
Cg z Og ,
βn
and all four gates are independent then the states can be enumerate as shown in Figure 6.14. State Cn is defined as having n gates closed, state O has all 4 gates open. Start
with state C4 with all 4 gates closed. Since the rate of opening of a gate is αn then the
rate at which state C3 is reached is 4αn since any one of the 4 independent gates can
open to cause a transition to state C3 . Similarly, the rate of transition from state C3 to
C4 is the rate at which the one open gate closes which is βn . The state diagram shown
in Figure 6.14 is constructed by extending this analysis to all the states.
PROBLEMS
127
2
Time (ms)
4
6
8
−25
Vm(t)
−75
0.8
0.6
m(t) 0.4
τm= 0.46
τm= 0.1
0.2
50
Figure 6.15: Plots of m(t), GNa (t), and JNa (t)
(Problem 6.15).
40
GNa(t)
30
20
10
−2
JNa(t)
τI= 0.1/3
−4
−6
Problem 6.15
a. The voltage dependence of sodium activation parameters (Weiss, 1996b, Figure 4.25)
show that m∞ (−75) ≈ 0, m∞ (−25) ≈ 0.73, τm (−75) ≈ 0.1, τm (−25) ≈ 0.46.
Therefore, the duration of the pulse (5 ms) is 11 times longer than the time constant (0.46 ms) for the increase in m. Thus, m essentially reaches its final value
of 0.73. Therefore,
(
0.73(1 − et/0.46 ) for 0 ≤ t ≤ 5,
m(t) =
0.73e(t−5)/0.1
for t ≥ 5,
where t is in ms. This solution is shown plotted in Figure 6.15.
b. Since h is assumed to be 1, GNa = ḠNa m3 . Therefore, since 120(0.73)3 ≈ 47,
(
47(1 − et/0.45 )3 for 0 ≤ t ≤ 5,
GNa (t) =
47e3(t−5)/0.1
for t ≥ 5,
where GNa is in mS/cm2 , and t is in ms.
c. The sodium current density is JNa = GNa (Vm − VNa ). For 0 ≤ t < 5, Vm − VNa =
128
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
−25 − 55 = −80 mV and for t > 5 Vm − VNa = −75 − 55 = −130 mV, so that
(
−3.8(1 − et/0.45 )3 for 0 ≤ t < 5,
JNa (t) =
−6.1e3(t−5)/0.1
for t > 5,
where JNa is in mA/cm2 , and t is in ms.
d. The time constant of tail current at the offset of the potential is τI = 0.1/3 ms or
33 µs.
e. Measurements of the offset time constant (Weiss, 1996b, Figure 6.59) suggest that
τI ≈ τm whereas the Hodgkin-Huxley model predicts that τI = τm /3. Thus, the
Hodgkin-Huxley model does not correctly predict the offset time constant of the
tail current. This failure of the model implies that the kinetic scheme shown in
Figure 6.54 (Weiss, 1996b) is not exactly correct.
Problem 6.16
a. Only S2 in channel b is conducting. Therefore, gb (t) = 10x2 (t) which looks like
w3 (t).
b. Only S3 in channel c is conducting. Therefore, gc (t) = 10x3 (t) which looks like
w2 (t).
c. Only S1 in channel a has a non-zero gating charge. Therefore, qa (t) = −x1 (t) and
iga = −dx1 (t)/dt which looks like w1 (t).
d. Only S1 in channel c has a non-zero gating charge. Therefore, qc (t) = x1 (t) and
iga = dx1 (t)/dt which looks like w5 (t).
e. Channel b shows activation followed by inactivation. As indicated in part a, the
conductance increases as the channel state progresses from S1 to S2 and decreases
as the channel state progresses from S2 to S3 .
f. Channel c exhibits no inactivation. As shown in part b, the channel shows an
S-shaped activation and no inactivation.
g. Channel a is open for t < 0 and then the channel state progresses to S2 and to S3
which are both non-conducting states. Thus, channel a closes on depolarization.
Problem 6.17 Consider the two-state model first. With all rate constants equal to 1, the
state occupancy probabilities satisfy the equilibrium equations
dx1 (Vm , t)
= 0 = −x1 (Vm , ∞) + x2 (Vm , ∞),
dt
1 = x1 (Vm , ∞) + x2 (Vm , ∞).
Therefore, x1 (Vm , ∞) = x2 (Vm , ∞) = 1/2. When all the rate constants are the same, the
channel is equally likely to occupy any of its states at equilibrium. Therefore, for 3 states
x1 (Vm , ∞) = x2 (Vm , ∞) = x3 (Vm , ∞) = 1/3 and for 4 states x1 (Vm , ∞) = x2 (Vm , ∞) =
x3 (Vm , ∞) = x4 (Vm , ∞) = 1/4.
PROBLEMS
129
m
m
h
βm
m
m
2αm
αh βh
h
αm
m
m
αh βh
αh βh
m
m
h
m
2αm
βm
Figure 6.16: Kinetic diagram
for a channel whose kinetics has the form m2 h (Problem 6.18).
m
h
m
h
2βm
αm
h
2βm
m
Problem 6.18 Let the rate constants for the m-gate be αm and βm and those for the
h-gate be αh and βh . Then for the channel to be open requires 2 m-gates and 1 h-gate to
be open. With 3 gates there are 6 distinguishable states. These are shown in Figure 6.16
along with a scheme that has the appropriate kinetics. In the top three states, the h-gate
is open, in the bottom three it is closed. In the left column, both m-gates are closed.
In the center column, one m-gate is open and one is closed. In the right column, both
m-gates are open.
Problem 6.19
a. Ans. (16). Externally applied TTX blocks sodium currents so that the ionic current
is predominantly the potassium current JK = GK (Vm −VK ). Thus, the shape of the
current is the product of the two terms. During the pulse Vm − VK ≈ 10 + 70 = 80
mV. Thus, at the peak of the conductance, the current is 18 × 80 = 1440 µA/cm2
which is 1.44 mA/cm2 . At the voltage offset the direction of current flow will
reverse.
b. Ans. (12). Internally applied TEA blocks the potassium current so that the ionic
current is predominantly sodium current JNa = GNa (Vm − VNa ). Thus, the shape
of the current is the product of the two terms. During the pulse Vm − VNa ≈
10 − 55 = −45 mV. Thus, at the peak of the conductance, the current is 50 × −45 =
−2250 µA/cm2 which is about −2.25 mA/cm2 . At the voltage offset there will be
a discontinuity in the current.
c. Ans. (14). Internally applied pronase blocks sodium inactivation. Therefore,
the sodium current during the pulse should have an S-shaped onset with a time
constant of m of 0.23 ms to a final value of JNa (10, 4) ≈ 120(1)3 (−45) ≈ −5.4
mA/cm2 . At the offset, the current goes instantaneously to 120(1)3 (−100 − 55) ≈
−18.6 mA/cm2 and then decays exponentially to 0 with a time constant 0.03/3 =
0.01 ms.
d. Ans. (17). Externally applied ouabain blocks the sodium/potassium pump which
does not affect the ionic currents particularly if the concentrations of all solutions
130
−150
CHAPTER 6. VOLTAGE-GATED ION CHANNELS
Vmf (mV)
−50
+50
+150
−1
I (pA)
−2
Figure 6.17:
The relation between the single
open-channel current and membrane potential
(Problem 6.20).
−3
are maintained constant. Hence, the ionic current is simply the normal ionic current which is predominantly the sodium current (12) plus the potassium current
(16).
Problem 6.20
f
a. Linear. A plot of the single open-channel currents (−2, −2.5, −3 pA) against Vm
(−50, −100, −150 mV) is a straight line (Figure 6.17). Therefore the relation is
linear.
b. Yes. The channel spends more if its time conducting current at Vm = −150 mV
than at either of the other two potentials. Therefore, the probability that the channel is open is larger at Vm = −150 mV than at either of the other two potentials.
c. The conductance is computed directly from the data as follows:
γ=
1 pA
∆I
= 10 pS.
=
∆Vm
100 mV
d. Since the relation between I and Vm is linear, it is only necessary to find the value
of Vm for which I = 0. This can be done graphically (Figure 6.17) or analytically
as follows
I = γ(Vm − Vn ),
so that
Vn = Vm −
−2 pA
I
= −50 mV −
= 150 mV.
γ
10 pS
e. From the fraction of time that the channel is open in each record, it appears that the
probability that the channel is open is about 0.5, 0.5, and 0.85 for Vm = −50, −100,
and −150 mV, respectively.
f. No. There are too many transitions. The two state model predicts gating currents
when the gate opens and when the gate closes, but at no other times. The gating
current records show gating currents not only at those times but also when the
gate is open.
PROBLEMS
131
S2
s̃(t)
S1
S0
0
ĩ(t)
(pA)
Figure 6.18: State occupancy s̃(t),
single-channel current ĩ(t),
single-channel gating charge q̃g (t), and
single-channel gating current ĩg (t)
(Problem 6.20).
−3
q̃g (t)
Q1
Q2 = Q0
ĩg (t)
0
Time (ms)
8
g. When the channel is open, gating currents occur without a change in the ionic current. Since there are three states and one is a closed state, there must be two open
states. However, there is only one non-zero value of current. Therefore, the two
open states must have the same conductance. Assume that state 0 corresponds
to the closed state (i.e. γ0 = 0) and that states 1 and 2 are open. In part c it
was shown that the conductance is 10 pS when the channel is open. Therefore,
γ1 = γ2 = 10 pS. Thus, transitions between states 1 and 2 generate no change
in the single-channel ionic current but do generate gating currents. The combination of gating and ionic currents allows a complete reconstruction of the state
occupancies as shown in Figure 6.18 for the traces at −150 mV. Note that channel
transitions are allowed between states 0 and 1 and between states 1 and 2 and
not between states 0 and 2. Hence, because the gating charge of state 1 differs
from both state 0 and state 2, the gating charge changes for each state transition.
Therefore, each state transition gives rise to a gating current. State transitions
occur wherever there is a gating current. Those, gating currents that are coincident with changes in ionic current must result from transitions between state 0
and state 1. Therefore, transitions from state 0 to state 1 open the channel and
generate positive gating currents (since Q1 > Q0 ). Transitions from state 1 to state
2 leave the channel open and generate negative gating currents (since Q2 < Q1 ).
Transitions from state 2 to state 1 leave the channel open and generate positive
gating currents (since Q1 > Q2 ). Transitions from state 1 to state 0 close the channel and generate negative gating currents (since Q0 < Q1 ). All of these transitions
are seen in the traces (Weiss, 1996b, Figure 6.97) (which consist of single-channel
ionic currents and gating currents), and no other kinds of transitions are seen.
Therefore, this model fits with the data.
Problem 6.21 The relations between single-channel random variable currents and average single-channel currents are summarized as follows. The average single channel
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CHAPTER 6. VOLTAGE-GATED ION CHANNELS
Average current (i(Vm , t))
Steady-state value
Time constant
t < 20 ms t > 20 ms
A
B
C
D
E
⇑
NC
⇓
NC
⇑
⇑
⇑
⇓
NC
⇑
⇑
⇓
NC
⇑
NC
Random-variable current (ĩ(Vm , t))
Open current (I)
Prob. open
t < 20 ms t > 20 ms t < 20 ms t > 20 ms
NC
NC
⇓
NC
⇑
NC
⇑
⇓
NC
⇑
?
?
?
?
?
⇓
⇑
?
⇑
?
Table 6.6: The table indicates the change in the indicated quantity from its value for the default parameters — “NC” means there was no change, “⇑” means there was an increase in the
algebraic value, “⇓” means there was a decrease in the algebraic value, and “?” indicates that
it is difficult to determine whether or not there was a change (Problem 6.21). Changes in the
average single-channel current and in the single open-channel current amplitude are relatively
easy to judge from such brief records. Estimates of a change in the probability that the channel
is open are not accurately judged with such short records. Hence, those results will not be
weighed as heavily in our assessment.
current is
i(Vm , t) = Ix(Vm , t),
where
I = γ(Vm − VNa ).
The probability that the channel is open x(Vm , t) is governed by the differential equation
dx(Vm , t)
= α(Vm ) 1 − x(Vm , t) − β(Vm )x(Vm , t).
dt
The probability of a transition from the closed to the open state in the interval (t, t +∆t)
is α(Vm )∆t, and the probability of a transition from the open to the closed state in that
interval is β(Vm )∆t.
In order to decide which records go with which change in parameter, the results are
first summarized in Table 6.6.
a. An increase in γ, decreases the steady-state value of the average current in both
time intervals, but does not affect the time constant. The single-open channel
current also decreases in both time intervals, but the probability that the channel
is open does not change. Therefore, the answer is C.
f
b. A increase in Vm will not affect the average or single-channel currents for t < 20
ms. However for t > 20 ms, this increase will increase the probability that the
channel is open, and will increase the single channel current. Hence for t > 20
ms, the average current will increase. Therefore, the answer is B.
c. A decrease in the extracellular concentration of sodium, will decrease the Nernst
equilibrium potential for sodium. This decrease will not affect the probability that
the channel is open nor any of the rate constants. However, the driving voltage on
the channel will increase. Hence, the single-channel current will increase for both
time intervals. The steady-state value of the average current will also increase.
Therefore, the answer is E.
Average ionic current (pA)
PROBLEMS
133
5
4
3
2
Figure 6.19: Average open channel current for Problem 6.22.
1
0
−200 −100
0
100
200
Membrane potential (mV)
d. A decrease in both α(Vm ) and β(Vm ) without changing the ratio, will affect neither the probability that the channel is open nor the single-open channel current.
Therefore, the steady-state values of the average current will be the same. However, the time constant will increase. The rates of transitions between states will
also decrease. Therefore, the answer is D.
e. If α(Vm ) is decreased, the time constant will increase and the probability that
the channel is open will decrease. However, the values of the single-open channel
currents will not be affected. Therefore, steady-state values of the average current
will increase. Therefore, the answer is A.
Problem 6.22
a. The open-state single-channel current I= γ(Vm −Vn ), where γ is the single-channel
open-state conductance and Vn is the Nernst equilibrium potential. The problem
statement indicates (Weiss, 1996b, Problem 6.100) that I=2 pA when Vm = −50
mV, and that I=3 pA when Vm = +50 mV. These two conditions can be used to
obtain a solution,
γ = 10 pS and Vn = −250 mV.
b. The first and second conducting segments are preceded by negative gating currents. Negative gating currents represent inward motion of positive charge, starting from the closed state. Therefore, the first two conducting segments are in
state 1. By similar reasoning, the last conducting state is state 3.
P
c. The steady-state value of the average ionic current is iss = i xi∞ I. Since only
states 1 and 3 are conducting and since they have identical permeation characteristics,
X
iss =
xi∞ I = (x1∞ + x3∞ )I = (x1∞ + x3∞ )γ(Vm − Vn ).
i
x1∞ is nearly 1 for very negative values of Vm and x3∞ is nearly 1 for very positive
values of Vm . The sum x1∞ +x3∞ is therefore small only near Vm = 0. The solid line
in Figure 6.19 shows the relation between iss and Vm . The dashed line shows the
open-channel current γ(Vm − Vn ) for the same range of membrane potential. The
important point is that the average current approaches the open-channel current
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CHAPTER 6. VOLTAGE-GATED ION CHANNELS
for very negative and for very positive values of Vm . However, the average current
is smaller than the open-channel current for Vm near zero because the gate is open
less frequently.
Bibliography
Weiss, T. F. (1996a). Cellular Biophysics. Volume 1: Transport. MIT Press, Cambridge,
MA.
Weiss, T. F. (1996b). Cellular Biophysics. Volume 2: Electrical Properties. MIT Press,
Cambridge, MA.
Weiss, T. F. (1997). Solutions to Problems in Cellular Biophysics. Volume 1: Transport.
MIT Press, Cambridge, MA.
Weiss, T. F., Trevisan, G., Doering, E. B., Shah, D. M., Huang, D., and Berkenblit, S. I.
(1992). Software for teaching physiology and biophysics. J. Sci. Ed. Tech., 1:259–274.
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