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
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To Aurice B, Max, Elisa, and Eric
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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
[email protected]. 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 ˚ A. 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. i˜i1 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. i˜i2 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 i˜i3 . i˜i4 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 i˜i1
Chloride Calcium Sodium Potassium Magnesium None of the above
√
Channel
Ii1
Chloride Calcium Sodium Potassium Magnesium None of the above
i˜i2
i˜i3
i˜i4 √
i˜i5
√ √ √
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. i˜i5 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. ¯Na m3 . Therefore, since 120(0.73)3 ≈ 47, b. Since h is assumed to be 1, GNa = G ( 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 s˜(t)
S1 S0
0 ˜i(t) (pA)
Figure 6.18: State occupancy s˜(t), single-channel current i˜(t), sin˜g (t), and gle-channel gating charge q single-channel gating current i˜g (t) (Problem 6.20).
−3 q˜g (t)
Q1
Q2 = Q0
˜ig (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
132
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 (i˜(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
134
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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