solutions to problems in cellular biophysics …phstudy.technion.ac.il/~wn116029/exam...

SOLUTIONS TO PROBLEMSIN

CELLULAR BIOPHYSICS

VOLUME 2: ELECTRICALPROPERTIES

Thomas Fischer Weiss

Department of Electrical Engineering and Computer ScienceMassachusetts Institute of Technology

Spring 1997

Date of last modification: October 25, 1997

iii

To Aurice B, Max, Elisa, and Eric

v

PREFACE

The textbooks Cellular Biophysics, Volume 1: Transport (Weiss, 1996a) and CellularBiophysics, Volume 2: Electrical Properties (Weiss, 1996b) contain a collection of exer-cises and problems that have been developed over many years. These problems andexercises allow students to test their comprehension of the material and they also ex-tend the material contained in the textbooks. For learning the material, there is nosubstitute for attempting to solve problems — challenging problems. Solving problemscan reveal which aspects of the material are understood and which are not yet graspedand require further study. This solution book contains solutions to all the exercisesand problems in Volume 2; a companion solution book (Weiss, 1997) contains solutionsto all the exercises and problems in Volume 1. The purpose of making the solutionsavailable is to allow students to check their work. Properly used, these solutions canbe helpful for learning the material. By comparing their solutions with those in the so-lution book, students can obtain an objective evaluation of their comprehension of thesubject material. However, the solutions contained here are more extensive than wouldbe expected for a typical student. Hopefully, these more extensive solutions further ex-plicate the material. There is one caveat concerning a possible misuse of this solutionbook. If a problem is assigned in a subject that uses these textbooks, a student mightbe 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 strugglingwith the problem. This is where learning takes place and the subject material is assim-ilated. My advice to students is: resist the temptation to use the solution texts in acounterproductive manner. Remember as you struggle to solve a problem, “no pain, nogain.”

The practice of the faculty has been to assign problems for homework one week andthen to issue solutions to students the following week when the work was due. Thesesolutions started out as handwritten and progressed over the years to more elaboratetypeset solutions. Therefore, when I started the project of producing these solutionsbooks, I gathered together all previous solutions — multiple solutions were typicallyavailable for the same problem assigned in different years. These solutions were writtenby various faculty who have taught the subject as well as by graduate student teachingassistants. In reviewing these solutions, I found with some chagrin that several solu-tions were incorrect despite the fact that they had been issued many times (by facultyincluding primarily myself), checked by several teaching assistants, and presumablyread by many hundreds of students. I apologize to students who were led astray byerrant solutions, but these same students should feel guilty for not having caught theerrors. The solutions I have found in error, I have tried to fix. However, my experiencewith compiling past solutions has left me a bit pessimistic about eliminating all errorsfrom these solutions. I invite the reader to communicate with me to point out any re-maining errors. I can be reached via email at [email protected]. I will post errors in thetexts and in the problem solution texts on my homepage on the world wide web whosecurrent address is http://umech.mit.edu:80/weiss/home.html. My homepage canbe reached through the MIT home page which links to the Department of ElectricalEngineering and Computer Science homepage.

As with the textbooks, these solution books were typeset in TEX with LATEX macroson 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 encapsu-lated postscript files that were included electronically in the text. Mathematical anno-tations were obtained by typesetting the mathematical expressions with Textures andsaving the typeset version as a file that was read by Adobe Illustrator. The subject istaught with the use of software (Weiss et al., 1992) designed to complement other ped-agogic 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 Gerdand Tom Perkins. My secretaries (Susan Ross and Janice Balzer) were helpful in compil-ing the solutions from past years. Faculty who have taught the material also developedproblems and solutions over the years. In particular, I wish to acknowledge the contri-butions 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 asthe writing has progressed, has continued to support me despite my obsession withwriting texts.

Chapter 1

INTRODUCTION TO ELECTRICALPROPERTIES OF CELLS

Exercises

Exercise 1.1 Graded and action potentials can be produced by passing electric currentsthrough cellular membranes. The distinction between these two potentials lies in thecurrent-voltage relation of the membrane. The relation between current and voltage isdiscontinuous for generating action potentials and continuous for generating gradedpotentials. Action potential have a sharp threshold for brief current pulses. Currentamplitudes below this threshold elicit no action potential; current amplitudes above thisthreshold produce a full action potential. In general, the amplitude of graded potentialsis a monotonic function of the current amplitude.

Exercise 1.2 The current enters the micropipet and leaves at the tip which is located inthe cytoplasm of the cell. The current flows outward through the membrane and back tothe reference electrode as shown in Figure 1.1. If the current flows outward through themembrane and the membrane can be represented by a battery in series with a resistancethen 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 thresholdfor eliciting an action potential are graded potentials.

I e(t

)

I e(t

)

+ −Vm(t)

+

−

+

−

Rm

V om

Vm(t)

Figure 1.1: Flow of cur-rent resulting from anintracellular electrode andthe resulting change inmembrane potential —schematic diagram andelectric network (Exer-cise 1.2).

1

2 CHAPTER 1. INTRODUCTION TO ELECTRICAL PROPERTIES OF CELLS

b. True. Decrement-free conduction is a property of the conduction of action poten-tials.

c. False. Electrically excitable cells are defined as cells that can produce action po-tentials.

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 simpleexample, graded potentials exhibit spatial and temporal summation to produceaction potentials.

Exercise 1.4 About 100 mV.

Exercise 1.5 No action potential is produced for a stimulus that is below a thresholdvalue. Stimuli above this threshold value produce an action potential of the same wave-form independent of the stimulus amplitude.

Exercise 1.6 Refractoriness refers to the increase in the threshold for eliciting an actionpotential just after an action potential has occurred. Refractoriness can be quantifiedby determining the minimum amplitude of the current stimulus that is required to elicitan 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 betweenthe pre- and post-synaptic cells, i.e., current flows between these two cells. In chemicaltransmission there is virtually no direct electrical communication between the pre- andpost-synaptic cells, i.e., no appreciable current flows between these two cells. In chem-ical transmission there is an electrically elicited chemical secretion in the pre-synapticcell. The secreted chemical neurotransmitter acts on the post-synaptic cell to producean electrical effect.

Exercise 1.8 The largest myelinated nerve fibers in vertebrates have diameters of about20 µm and the largest unmyelinated fibers in invertebrates have diameters of about1 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 agiant axon than in a myelinated fiber. This makes collection of cytoplasm for chemicalanalysis much simpler in a giant axon.

Chapter 2

LUMPED-PARAMETER ANDDISTRIBUTED-PARAMETER MODELSOF CELLS

Exercises

Exercise 2.1 For simplicity, consider the relations among the resistances for the samecylindrical conductor through which current flows in the longitudinal direction as shownin Figure 2.1. The resistivity and conductivity are material properties that relate theelectric field intensity to the current density at each point in a conductor via Ohm’s law

J = σeE or E= ρJ.The resistivity has units ofΩ·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 = ρLA,

which for the cylindrical conductor isR = ρL/(πa2) and has units of Ω. The resistanceper unit length is simply

r = RL= ρA,

which for the cylindrical conductor is r = ρ/(πa2) and has units of Ω/cm. The lengthin the definition of resistance per unit length is in the direction of current flow. Theresistance per unit length is an appropriate measure of specific resistance for a one-dimensional resistance, i.e., one with a constant cross-sectional area. For example, it

Ja

L

E

Figure 2.1: Geometry used to define various types of resis-tances (Exercise 2.1).

3

4 CHAPTER 2. LUMPED- AND DISTRIBUTED-PARAMETER MODELS

is used to specify resistances of wires. To get a certain resistance of some wire whoseresistance 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 directionof current flow. This is an appropriate measure of the specific resistance of a two-dimensional resistance where current flow is orthogonal to the surface area, i.e., forexample a sheet of material of constant thickness. Thus, to obtain a given resistancewith a sheet of material of known resistance of a unit area, one need only specify thesurface area required.

Exercise 2.2 In an electrically small cell, the potential difference across the membranedoes 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 re-sistance ri ∝ 1/D2. Hence, an increase in axon diameter results in a decrease in thelongitudinal resistance. Therefore, the longitudinal current will spread further alongthe axon and the conduction velocity will increase. This effect is made clear by theexperiments with space clamps in which a wire placed in the axon greatly increases theconduction velocity. The membrane resistance rm ∝ 1/D. Thus, an increase in axondiameter results in a decrease in the membrane resistance. A decrease in the membraneresistance favors current flow through the membrane rather than longitudinally downthe axon. This effect will reduce the conduction velocity. To summarize, an increasein axon diameter has two opposing effects on the conduction velocity — a decreasein longitudinal resistance leads to an increase in conduction velocity and a decreasein membrane resistance results in a decrease in conduction velocity. However, the de-crease in longitudinal resistance is greater than the decrease in membrane resistancebecause ri ∝ 1/D2 and rm ∝ 1/D. Therefore, an increase in the axon diameter resultsin an increase in conduction velocity.

Exercise 2.4 The equation states that the rate of increase in internal longitudinal cur-rent with distance equals minus the current per unit length that leaves the inner con-ductor, i.e., it equals the current per unit length that enters the inner conductor. Thus, ifthe internal longitudinal current increases with distance then there must be net currentflowing through the membrane into the internal conductor.

Exercise 2.5 The theory predicts that the conduction velocity is proportional to theaxon diameter of unmyelinated fibers of identical membrane properties. The difficultymay be due to the fact that nerve fibers from species that are phylogenetically distantmay 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 ac-tion potential is estimated to be 2.9 ms in seawater and 3.8 ms in oil. The distancebetween stimulation and recording electrodes is 13 mm. Hence, the conductionvelocity is 4.5 mm/ms = 4.5 m/s in seawater.

EXERCISES 5

Axon in:

seawater

oil

seawater

oil

Figure 2.2: Method used to estimate the conduction veloc-ity from the measurements (Exercise 2.6).

Axon in:

seawater

oil

0 7.6 9 20Distance (mm)

Figure 2.3: Action potentials plotted versus distance foran axon in seawater and in oil (Exercise 2.6).

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 inseawater 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 wave-forms versus z requires sketching Vm(z, to) = f(to − z/4.5) for seawater andVm(z, t) = f(to − z/3.4) for oil. Thus, the waveforms are the same as the timewaveforms 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 and17 mm at a conduction velocity of 3.4 mm/ms. The two waveforms must be scaledappropriately to plot them on the same abscissa scale (Figure 2.3). Note that inseawater the onset of the action potential occurs at t = 2.9 ms at location z = 13mm. Hence, Vm(13,2.9) = f(2.9 − 13/4.5) = f(0). The point in space at whichthis 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 ≈ 7.6mm. Thus, in seawater the spatial extent of the action potential is broader and itsleading edge travels farther than in oil.

Exercise 2.7 In the absence of the platinum wire, the time between peaks of the actionpotentials is about 0.74 ms. Since the conduction distance is 1.6 cm, the conductionvelocity is about 16/0.74 = 22 mm/ms = 22 m/s. One could easily discern a time dif-ference of 5 µs in the text figure (Weiss, 1996b), and the time difference of the twoaction potentials is less than that duration. Hence, the conduction velocity is greater


z (mm)0 100 200

0

1

z (mm)−200 −100 00

1

f(z,2

)

f(z,2

)

Figure 2.4: Wave propagation in the+and− z direction (Exercise 2.10). Thearrows show the direction of propa-gation.

than 3200 m/s. Insertion of the platinum wire increased the conduction velocity morethan 145 times.

Exercise 2.8 No. The equation was derived under the assumption that the extracellularpotential is independent of radial distance from the axon. This may be a reasonableassumption in an extracellular volume of limited spatial extent, but will not be true ina 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 istaken into account in Equation 2.38. Both equations are correct, but Equation 2.38 makesthe dependence of the conduction velocity of an unmyelinated fiber on fiber radiusapparent. The derivation that leads to Equation 2.38 also assumes that ri ro which isusually a good assumption except when axons are subjected to insulating extracellularmedia 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 mustsatisfy the relation t1 − z2/ν = t2 − z2/ν . Figure 2.20 (Weiss, 1996b) shows a plot forz = 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 betransformed according to z2 = 100(2 − t1). The result is shown in the left panelof Figure 2.4.

b. To find f(z,2) for a conduction velocity of ν = −100 mm/ms, distance must betransformed according to z2 = 100(t1 − 2). The result is shown in the right panelof Figure 2.4.

Exercise 2.11 For a wave propagating in the +z-direction with propagation velocity 10mm/ms, f(z, t)must have the form g(t−z/ν) = g(t−z/10), where t is in ms and z isin mm. Therefore, f(20,3) = g(3 − 20/10) = g(1). But, from the figure it is apparentthat at z = 0, f(0, t) = g(t) and g(1) = 0.5. Hence, f(20,3) = 0.5. The problem canalso 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

−10

10

0

20

−0.6 −0.4 −0.2

−0.6 −0.4 −0.2

−20

−40

−60

−80

−100

Ii(z)

Io(z)

Km(z)

µA

µA

/cm

Figure 2.5: Plots of the longitudinal andmembrane current per unit length in a coreconductor (Exercise 2.12). In the lowestpanel, the length of the arrow is propor-tional to current for the longitudinal cur-rents and to the current per unit length forthe membrane current per unit length.

c. The membrane current per unit length is

Km(z) = −∂Ii(z)∂z= −5e−5z for z < 0.

Km(z) is plotted in Figure 2.5.

Note that the decrease in the external longitudinal current with distance is associatedwith an inward current per unit length through the membrane in order to satisfy Kirch-hoff’s current law.

Exercise 2.13

a. The longitudinal current density Ji(z) in the cytoplasm is equal to the internallongitudinal current Ii(z) divided by the cross sectional area of the cell.

Ji(z) = Ii(z)πa2 =−1 µA

π(0.01 cm)2e5z = −3183 e5z µA

cm2 .

b. The membrane current per unit lengthKm(z) can be found using the core-conductorrelation,

Km(z) = −dIi(z)dz= − d

dz(−e5z) = 5 e5z µA

cm.

c. The membrane current density Jm(z) is the current through a unit area of mem-brane. This current can be found from Km(z),

Jm(z) = Km(z)2πa= 5 µA/cm

2π0.01 cme5z = 79.58 e5z µA

cm2 .


Ii(−1) Ii(0)Km(z)

− 1 0z (cm)

Figure 2.6: Volume element showing the cur-rent variables associated with the inner conduc-tor (Exercise 2.14).

d. The total current Im flowing through the part of the membrane from z = −1 cmto z = 0 can be found by integrating Km(z) over this region.

Im =∫ 0

−1Km(z)dz =

∫ 0

−15 e5z dz = e5z

∣∣∣0

−1= (1− e−5) µA ≈ 1 µA

Exercise 2.14

a. As illustrated in Figure 2.6, the longitudinal current flowing into the intracellularvolume element at z = −1 is Ii(−1) = −e−5, and the longitudinal current flowingout of the volume element at z = 0 is Ii(0) = −1. Therefore, the net longitudinalcurrent 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 ∫ 0

−1Km(z)dz =

∫ 0

−1−5e5z dz = e5z

∣∣∣0

−1= 1− e−5.

c. The net longitudinal current flowing into the volume element equals the currentthat leaves the volume element through the membrane.

Problems

Problem 2.1

a. If ro = ri, thenρoAo= ρiAi,

where Ao and Ai are the cross-sectional areas. Therefore,

ρoπb2 −πa2 = ρi

πa2 ,

(ρo + ρi)a2 = ρib2

b = a√ρo + ρiρi

.

b. For both ro = ri and ρo = ρi, b = a√

2. Therefore, for a = 500 µm, b = 707 µmand for a = 1 µm, b = 1.4 µm

PROBLEMS 9

0 2 4 6 8 10

+50

0

−50

−100

Time (ms)

Millivol

ts

Vm(3,t) Vm(5,t)

0 2 4 6 8 10

+50

0

−50

−100

Time (ms)

Millivol

ts

Vm(5,t) Vm(3,t)

Propagation in +z-direction Propagation in −z-direction

−2 2 6 10 14 18

+50

0

−50

−100

Distance (cm)

Millivol

ts

Vm(z,5)

−4 0 4 8 12 16

+50

0

−50

−100

Distance (cm)

Millivol

ts

Vm(z,5)

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 inanimals. For these cells, the radius of the extracellular space for which the internaland external resistances per unit length are equal is 41% larger than the radiusof the fiber. This dimension is generally small compared to the typical volumeof extracellular space in situ. Thus, for any fiber that is in a large volume ofextracellular space, such as is the case in situ, the external resistance per unitlength 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 velocityof 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 theplot versus t reversed in time and with the axis scaled. It is helpful to locatethe peak value of the action potential in order to sketch it. The peak valueoccurs 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.


b. With the action potential propagating in the −z-direction at a conduction velocityof 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 likethe 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 cm5 ms

= 2 cm/ms = 20 m/s.

b) The platinum wire provides another path for longitudinal current flow that is inparallel with the cytoplasm. The resistance per unit length of the platinum wireis 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 thesetwo resistance is ri = rplrcy/(rpl + rcy) ≈ rpl. If the extracellular resistance isnegligible, the conduction velocities of the axon are

ν1 =√Km

2πar1and ν2 =

√Km

2πar2,

for different values of the intracellular resistance per unit length with all otherfactors unchanged. Thus,the ratio of conduction velocities is

ν1

ν2=√r2

r1.

Therefore, ν2 = 20√

16000/160 = 200 m/s.

c) With the platinum wire in place, the new conduction velocity is 20000 cm/s. Sincethe action potential travels 10 cm, the peak is delayed by 0.5 ms as is shown inFigure 2.8.

d) Each of the two voltages must have the form of a travelling wave. Therefore,

V1(t) = f(t − z

ν1

)and V2(t) = f

(t − z

v2

),

which implies that

V1

(t + z

ν1

)= f(t) and V2

(t + z

v2

)= f(t).

Equating these two expressions yields

V1

(t + z

ν1

)= V2

(t + z

ν2

)

PROBLEMS 11

0 5 10 15 20 25

+50

0

−50

−100

Time (ms)

Millivol

ts

V1(t)V2(t)

Figure 2.8: Action potential with (V2(t))and without (V1(t)) the platinum wire (Prob-lem 2.3).

which implies that

V2(t) = V1

(t + z

ν1− zν2

),

so that

V2(t) = V1

(t + 10

2000− 10

20000

)= V1

(t + 4.5× 10−3

)t in s.

Problem 2.4

a. From the core conductor model (with no externally applied currents)

∂2Vm(z, t)∂z2 = (ri + ro)Km(z, t).

For an action potential propagating in the ±z direction, Vm(z, t) = f(t ∓ z/ν) sothat the membrane potential must satisfy the wave equation

∂2Vm(z, t)∂z2 = 1

ν2

∂2Vm(z, t)∂t2

.

Therefore,

Km(z, t) = 1(ri + ro)ν2

∂2Vm(z, t)∂t2

.

The membrane potential and its first and second partial derivatives are shown inFigure 2.9.

b. For the linear model of the membrane, the membrane current per unit length is

Km = cm∂Vm(zo, t)∂t+ gm(Vm(zo, t)− Vom).

For the time interval 0 < t < tm both ∂Vm(zo, t)/∂t and Vm(zo, t) − Vom are pos-itive quantities. Therefore, the linear cable model predicts that Km > 0 in thisinterval. However, the core conductor model for a propagated action potentialshows that Km is positive in the interval 0 < t < ti but is negative in the intervalti < t < tm. Therefore, the linear resistance/capacitance model of a membrane isnot consistent with the membrane current during a propagated action potential.


0

0

0

t

t

t

Vm(zo, t)

V om

dVm(zo, t)

dt

d2Vm(zo, t)

dt2

tm

ti

Figure 2.9: Membrane potential and its firsttwo time derivatives plotted versus time(Problem 2.4). The time of occurrence of themaximum value of the action potential is tm,and the point of inflection at the onset of theaction potential is ti.

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 elec-trodes. The onset of the extracellular potential should occur earlier in the record-ing 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 follow-ing argument shows. The extracellular potential is related to the external longitu-dinal current as follows

∂Vo(z, t)∂z

= −roIo(z, t).Integration of this equation over z for electrodes separated by the distance δ yields

Vo(z + δ, t)− Vo(z, t) = ∆Vo(z, t) = −roδIo(z, t).But the partial derivative of Vm(z, t) is related to the longitudinal current by

∂Vm(z, t)∂z

= (ri + ro)Io(z, t).

Therefore,

∆Vo(z, t) = − roδri + ro

∂Vm(z, t)∂z

.

If an action potential is travelling in the ±z direction then Vm(z, t) = f(t ∓ z/ν)so that

∆Vo(z, t) = ± roδν(ri + ro)

∂Vm(z, t)∂t

.

Thus, Vo(z, t) shows an initial increase in potential when the action potentialpasses the electrode in the +z-direction (left to right) and an initial decrease inpotential when the action potential passes the electrode in the −z-direction (rightto 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 shouldcome earlier at this electrode. Since the action potential passes both electrodesfrom left to right, the extracellular potential should be initially positive. This fitswith record 2.

B In this case, the stimulating electrode is closer to V1(t) so that the onset shouldcome earlier at this electrode. Since the action potential passes V1(t) from rightto left, the initial polarity of this potential should be negative. Since the actionpotential passes V2(t) from left to right, the initial polarity of this potential shouldbe positive. This fits with record 7.

C In this case, the stimulating electrode is closer to V2(t) so that the onset shouldcome earlier at this electrode. Since the action potential passes V1(t) from rightto left, the initial polarity of this potential should be negative. Since the actionpotential passes V2(t) from left to right, the initial polarity of this potential shouldbe positive. This fits with record 8.

D In this case, the stimulating electrode is closer to V2(t) so that the onset shouldcome earlier at this electrode. Since the action potential passes both V1(t) andV2(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)∆z

= Kei(z, t)−Km(z, t).

Taking lim∆z→0 of this equation yields

∂Ii(z, t)∂z

= Kei(z, t)−Km(z, t).

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)∆z

= Km(z, t)−Keo(z, t).


Taking lim∆z→0 of this equation yields

∂Io(z, t)∂z

= Km(z, t)−Keo(z, t).

The voltage across ri∆z is

Vi(z, t)− Vi(z +∆z, t) = ri∆zIi(z +∆z, t),which yields

Vi(z +∆z, t)− Vi(z, t)∆z

= −riIi(z +∆z, t).Taking lim∆z→0 of this equation yields

∂Vi(z, t)∂z

= −riIi(z, t).

Similarly, the voltage across ro∆z is

Vo(z, t)− Vo(z +∆z, t) = ro∆zIo(z +∆z, t),which yields

Vo(z +∆z, t)− Vo(z, t)∆z

= −roIo(z +∆z, t).Taking lim∆z→0 of this equation yields

∂Vo(z, t)∂z

= −roIo(z, t).

By inspection of the network, Vm(z, t) = Vi(z, t)− Vi(z, t). Therefore, the partialderivative of this equation with respect to z yields

∂Vm(z, t)∂z

= ∂Vi(z, t)∂z

− ∂Vo(z, t)∂z

= −riIi(z, t)+ roIo(z, t).

b. The partial derivative of the last equation with respect to z yields

∂2Vm(z, t)∂z2 = −ri ∂Ii(z, t)∂z

+ ro ∂Io(z, t)∂z,

= ri(Km(z, t)−Kei(z, t))+ ro(Km(z, t)−Keo(z, t)),= (ro + ri)Km(z, t)− roKeo(z, t)− riKei(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

∆Vo(z, t) = − rori + ro Vm(z, t).

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 ac-tion potential is travelling in the −z-direction. The figure is not consistent withVm(z, t) = f(t − z/ν) for which the action potential is travelling in the +z-direction.

PROBLEMS 15

b. Note that the difference in scale factor for plotting Vm(z, t) versus t and versus zis simply the conduction velocity. Hence, ν = 1 cm/ms = 10 m/s.

c. The attenuation of the action potential recorded extracellularly is simply the volt-age divider ratio shown above. Therefore,

−9 = − rori + ro100,

so that ri/ro = 10.11.

Problem 2.8 Both the magnitude of the extracellular potential and the conduction ve-locity depend upon the intracellular and extracellular resistance per unit length. Desig-nate the extracellular resistance per unit length as ro1 and ro2, for Experiments 1 and2, respectively. Assume that the intracellular resistance per unit length is the same forboth experiments. The core conductor model implies that

2πa(ro + ri)ν2 = Km,

where Km is a property of the membrane and not of the dimensions of the cell. InExperiment 1 the axon is in a large volume of sea water so that ro1 ri. Therefore, theconduction velocity for Experiment 1 implies that

2πa(ri)(36)2 = Km.

From Experiment 2, the ratio of the external to the transmembrane potential is 75/100 =3/4. Therefore,

ro2

ro2 + ri =34,

which can be solved to give ro2 = 3ri. The conduction velocity of the axon in Experiment2 can be found from the relation

2πa(ro2 + ri)ν22 = Km,

which can be written as2πa(4ri)ν2

2 = Km.Therefore, a combination of expressions for the conduction velocity in the two experi-ments yields 4ν2

2 = 362 or ν2 = 18 m/s.

Problem 2.9 The change in extracellular potential is

∆Vo = −(

roro + ri

)∆Vm,

where the resistances per unit length are

ro = ρoAo and ri = ρiAi ,


the resistivities of seawater and axoplasm are ρo and ρi, respectively, and the cross-sectional areas of seawater and axoplasm are Ao and Ai, respectively. Substitutionyields

∆Vo = −(

ρo/Aoρo/Ao + ρi/Ai

)∆Vm

Since ρo = ρi,∆Vo = −

(Ai

Ai +Ao)∆Vm.

But Ai+Ao is the cross-sectional area of the seawater and axon, i.e., the cross-sectionalarea of the trough. Therefore,

∆Vo = −(π(250)2

(600)2

)∆Vm = 0.58∆Vm = 58 mV.

Problem 2.10 The extracellular potential difference∆Vo(t) is the difference in potentialbetween the two voltage electrodes which are spaced 5 mm apart. At each electrode,the extracellular potential is

Vo(z, t)− Vo(−∞) = − roro + ri (Vm(z, t)− Vm(∞)) .

Therefore, since the potential at a location 5 mm further along the cell is delayed by5/ν , if the conduction velocity ν is expressed in mm/ms,

∆Vo(t) = Vo(l, t)− Vo(l+ 5, t) = − roro + ri (Vm(l, t)− Vm(l, t − 5/ν)) .

Since ri = 4ro,

∆Vo(t)− 15(Vm(l, t)− Vm(l, t − 5/ν)) .

a. For ν = 1 mm/ms, the delay at the two electrodes is 5 ms. Thus, the two actionpotentials are separated in time. The recorded potential is approximately twomonophasic action potentials of opposite sign separated by 5 ms as shown inFigure 2.10.

b. For ν = 10 mm/ms, the delay at the two electrodes is 0.5 ms. Thus, the two actionpotentials overlap in time. The recorded potential is a diphasic action potential asshown in Figure 2.10.

Problem 2.11 The extracellular action potential and the conduction velocity dependupon the longitudinal resistances per unit length.

a. By using the scales provided, the extracellularly recorded action potential can beestimated and the results are shown in Table 2.1. The extracellularly recorded ac-tion potential is proportional to the transmembrane action potential through therelation ∆Vo = −(ro/(ro + ri))∆Vm. Note however, that the extracellular actionpotential is diphasic because it is recorded both by the active and the reference

PROBLEMS 17

0

40

2 4 6 8

−40

t (ms)Millivol

ts

0

40

2 4 6 8

−40

t (ms)Millivol

ts

ν = 1 mm/ms

ν = 10 mm/ms

Figure 2.10: The extracellular potential at two conduc-tion velocities (Problem 2.10). The potentials recordedat each recording electrode are shown as dashed lines,the difference of potential is shown as a solid line.

Intact Extruded Inflated

Extracellular potential (mV p-p) 83 3.1 117ro/ri 0.71 0.016 1.4ri/ro 1.4 63 0.71Calculated ν/νintact 1 0.19 1.2Estimated conduction delay (ms) 2 7.3 0.58Conduction distance (mm) 40 28 25Estimated ν (m/s) 20 3.8 43Estimated ratio ν/νintact 1 0.19 2.2

Table 2.1: Summary of calculations of parameters for the intact, extruded, and inflated axons(Problem 2.11).


electrodes whereas the transmembrane action potential is monophasic. Let usmake the simple assumption that when the resistive divider ratio is 1, the peak-to-peak amplitude of the extracellular potential is twice the amplitude of the trans-membrane action potential. For example, the intact axon has a peak-to-peak valueof 83 mV and a transmembrane potential of 100 mV. Therefore, for this casero/(ro + ri) = 83/(2 · 100) = 0.42, and the solution is ro/ri = 0.71. Similarcalculation are given in the table for all 3 conditions of the axon.

b. For the intact axon2πa(ri−intact + ro)ν2

intact = Km,where ri−intact is ri for the intact axon. From Table 2.1, ri−intact = 1.4ro so that

2πa(2.4ro)ν2intact = Km.

For the extruded axon

2πa(ri−extruded + ro)ν2extruded = Km,

where ri−extruded is ri for the extruded axon. The quantity 2πa is the perimeterof a circular axon. It has been assumed that extrusion reduces the cross-sectionalarea of the axon without changing its perimeter. This makes sense if extrusionchanges the cross-section of the axon from circular to elliptic. Therefore, it isassumed that the perimeter of the extruded axon is also 2πa, where a is theradius of the intact axon. Because ri−extruded = 63ro,

2πa(64ro)ν2extruded = Km.

Therefore, (2.4ν2intact)/(64ν2

extruded) = 1, so that νextruded/νintact = 0.19. Thisresult is given in Table 2.1 in the row labeled “Calculated ν/νintact”. For theinflated axon

2πa(ri−inf lated + ro)ν2inf lated = Km,

where ri−inf lated is ri for the inflated axon. Since ri−inf lated = 0.71ro,

2πa(1.7ro)ν2inf lated = Km.

Therefore, the ratio of the conduction velocity of the inflated axon to that of theintact axon can be computed and is given in Table 2.1. Thus, extruding the axonincreases the intracellular resistance per unit length and decreases the conductionvelocity. Reinflation reverses these trends.

c. To estimate the conduction velocity, first estimate the time it takes for the initi-ation of the action potential. This conduction delay can be estimated from thestimulus artefact — discernible as a small excursion on the baseline preceding theaction potential in each trace — to the onset of the action potential. These esti-mated conduction delays are indicated in Table 2.1. By dividing these times intothe conduction distance for each case, the conduction velocity can be estimated.Ratios of conduction velocities of the extruded and inflated axons to that of theintact 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 are0.19 for both the estimated and the calculated conduction velocity. The ratios forthe inflated axon are 2.2 for the estimated and 1.2 for the calculated conductionvelocity. 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 onmeasurements of the conduction velocity — agree qualitatively. However, thereare quantitative differences for the inflated axon. There are a number of possiblesources of error in these estimates. For example, the conduction delay has beendefined in a somewhat arbitrary manner. Another method of estimating the con-duction delay, and hence the conduction velocity, would give somewhat differentresults. In addition, the estimate of the peak value of the monophasic extracellu-lar action potential from the measured peak-to-peak value of the diphasic actionpotential is crude. Overlap of the waveforms recorded by the two electrodes re-sults in an underestimate of the monophasic action potential, and thus leads toan underestimate of ro/ri.

Problem 2.12 The conduction velocity, ν , of an unmyelinated axon in an extensive ex-tracellular solution is

ν =√Kma2ρi

,

where a is the radius, ρi is the resistivity of cytoplasm, and Km is a constant. Sinceall the fibers are identical except for length and diameter, the ratios of velocities of twoaxons equals the square root of their radii,

ν1

ν2=√a1

a2.

a. Based on the above relation, the conduction velocities are computed for all axonsand shown in Table 2.2. The conduction times are computed by dividing the lengthof 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 wouldhave the same conduction velocity as the smallest axon. The conduction timesbased on this velocity are shown in column b.

c. If all the axons had the same diameter as the giant axon (largest diameter), theywould have the same conduction velocity as the giant axon. The conduction timesbased on this velocity are shown in column c.

d. The maximum difference in conduction times of the different fibers is: 5.01 msin part a; 13.38 ms in part b; and 6 ms in part c. Therefore, the differences indiameter and length do give significant synchronization over that which wouldoccur if all the fibers were of small diameter (5.01 versus 13.38 ms). However, ifall the fibers where as large as the giant fiber then synchronization would decreaseby only 1 ms. However, that would require that all nine fibers were giants. So thehypothesis is generally true although somewhat simplistic.


Axon # Length Diameter Conduction Conduction Timemm µm Velocity (msec)

(m/sec) a b c

1 23 131 10.83 2.12 2.57 1.152 22 90 8.97 2.45 2.45 1.103 22 86 8.77 2.51 2.51 1.104 24 148 11.51 2.09 2.68 1.205 43 227 14.25 3.02 4.79 2.156 48 291 16.14 2.97 5.35 2.407 76 291 16.14 4.71 8.47 3.808 91 317 16.84 5.4 10.14 4.559 142 447 20.00 7.1 15.83 7.1

Table 2.2: Conduction velocity and conduction times for axons in the stellate ganglion (Prob-lem 2.12).

Problem 2.13

a. The second partial derivative of the relation Vm(z, t) = Vi(z, t)− Vi(z, t) yields

∂2Vm(z, t)∂z2 = ∂

2Vi(z, t)∂z2 − ∂

2Vo(z, t)∂z2 .

The core conductor equations yield

∂Vi(z, t)∂z

= −riIi(z, t) and∂Ii(z, t)∂z

= −Km(z, t).

A combination of these three equations yields

∂2Vm(z, t)∂z2 = riKm(z, t)− ∂

2Vo(z, t)∂z2 .

b. Examination of Equation 2.25 (Weiss, 1996b) shows that

−roKe(z, t) = −∂2Vo(z, t)∂z2 ,

so that

Ke(z, t) = 1ro∂2Vo(z, t)∂z2 .

c. Note that because the extracellular potential is independent of z the effective ex-ternal current, which depends on the second partial derivative of the potentialwith 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 willstimulate the cell. For example, suppose

Vo(z, t) = z2

2sin 2πft.

PROBLEMS 21

The equivalent external current for this stimulus is

Ke(z, t) = 1ro∂2Vo(z, t)∂z2 = 1

rosin 2πft.

Chapter 3

LINEAR ELECTRICAL PROPERTIES OFCELLS

Exercises

Exercise 3.1 The space and time constants are measures of the spatial and temporalscales over which the membrane potential of the cell changes. They are defined as

λC = 1√gm(ri + ro) and τM = cmgm ,

where gm and cm are the conductance and capacitance per unit length of the cell, re-spectively, 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 = cmgm = τM =2πaCm2πaGm

= CmGm

,

so that the time constant depends only on the specific capacitance and conductance ofthe membrane and not on the dimensions of the cell. The space constant of a cylindricalcell is

λC = 1√gm(ri + ro) =

1√2πaGm((ρi/(πa2))+ 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 fromone section of a cell to cross the membrane, and hence, less of the current flows longitu-dinally 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 wouldbe 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 stim-ulation through this electrode. To be infinitesimal, an electrode must have dimensionsthat 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 lumped-parameter model consisting of a resistance and a capacitance. The membranepotential is governed by a first-order, ordinary differential equation with constantcoefficients 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 distributed-parameter network with a series resistance and a shunt element that contains aparallel combination of a resistance and a capacitance. The membrane potentialis governed by a partial differential equation with constant coefficients whosesolution 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 potentialsatisfies 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 con-ductance of the membrane of a cell. Gm = AGm where A is the surface area of thecell. gm is the membrane conductance per unit length of a cylindrical cell. Therefore,gm = 2πaGm, where a is the radius of the cell.

Exercise 3.7 The characteristic conductance is the conductance between the inside andoutside of an infinite length cable. Alternatively, if the cable has a finite length andis terminated in a conductance that equals the characteristic conductance, then theconductance 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 thenthe membrane potential does not change appreciably along the surface of the cell. Thisis what is meant by an electrically small cell.

Exercise 3.9 The normalized extracellular potential is

vo(z)vo(0)

= e−z/λC ,

so that

log10

(vo(z)vo(0)

)= −zλC

log10 e.

Therefore, for z = λClog10

(vo(λC)vo(0)

)= − 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 currentat z = 0. Since there is no external current applied intracellularly in this case, theinternal longitudinal current is continuous.

b. As ri/ro approaches zero the internal and external longitudinal currents in theextrapolar region remain equal in magnitude and opposite in sign. However, be-cause the internal resistance is small, the internal voltage is small. In the limit asri/ro → 0, vi(λ)→ 0 and vo(λ)→ −vm(λ).

c. For λ → ∞, vm(λ) → 0. Hence, vi(λ) ≈ vo(λ). In addition, in this limit thelongitudinal currents become independent of λ. Therefore, in this limit the twolongitudinal conductors are in parallel and the voltage across them is

vo(z) = vi(z) = −(rirori + ro

)zIe.

The normalized variables are defined as vo = vo/(roλCIe/2) and vi = vi/(roλCIe/2)so that

vo(λ) = vi(λ) = −(

2riri + ro

)λ = −

(2αα+ 1

)λ,

where α = ri/ro. Therefore, as α → 0, the magnitude of the slope of vi(λ) = vo(λ)versus λ decreases.

Exercise 3.11 No. The step response of the cable is attenuated as a function of posi-tion. Thus, the step response does not have the form of a wave propagated at constantvelocity, 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 ismore rapid than the step response of an electrically small cell with the same membranetime constant.

Exercise 3.13 Equation 3.5 is the step response of a small cell driven by a current ap-plied across the membrane. Equation 3.52 is the step response of a large cell drivenby a current applied extracellularly; the form of the solution for an intracellularly ap-plied current is similar. Thus, the difference in the equations predominantly reflectsthe difference in the step responses of small and large cells.

Problems

Problem 3.1

a. The conductance per unit length is

gm = 1

λ2Cri

where ri = ρiπa2 .

Therefore,

gm = πa2

λ2Cρi

= π(0.025)2

(0.6)2 · 30= 0.182× 10−3 S/cm.


Vm(t)

+

−

CmRm

Im(t)

+

−

Source Vm(t)

+

−

CmRm

Im(t)

+

−

Source

Rs

V om V o

m

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 = gm2πa

= a2λ2Cρi

= 0.0252 · (0.6)2 · 30

= 1.16× 10−3 S/cm2.

c. The cell space constant is related to the cable parameters as

λ2C =

1gmri

= 1(2πaGm)(ρi/(πa2))

= a2Gmρi

.

Therefore,

λC = 1√gmri =1√

2Gmρi

√a = 1√

2 · 1.16× 10−3 · 30

√a = 3.8

√a.

Therefore, the space constant is: 8.5 mm for a diameter of 1 mm, 2.7 mm fora diameter of 0.1 mm, 0.85 mm for a diameter of 0.01 mm, and 0.27 mm for adiameter of 0.001 mm.

d. The time constant of the cell is

τM = cmgm =2πaCm2πaGm

= CmGm

= 1× 10−6

1.16× 10−3 = 0.86 ms.

Problem 3.2 The specification of the problem suggests the cell/source models shownin Figure 3.1.

a. If the source is a current source where Im(t) = Iu(t), where u(t) is the unit stepfunction, then Kirchhoff’s current law yields

Iu(t) = CmdVm(t)dt+ Vm(t)− V

om

Rm .

PROBLEMS 27

Vm(t)

t

Vmo

IRm

τ

t

V/Rm

CmVδ(t)V/Rmu(t)

Figure 3.2: The voltage response to a step of current (left panel) and the current response to astep 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 constantcoefficients. Therefore, Vm(t) charges exponentially from its initial value Vom toits final value Vom + IRm with a time constant τM = RmCm. Therefore,

Vm(t) = Vom + IRm(1− e−t/τM ),which is shown in Figure 3.2.

b. If the source is an ideal voltage source such that Vm(t) = Vom +Vu(t) then Kirch-hoff’s current law yields

Im(t) = CmdVm(t)dt+ Vm(t)− V

om

Rm ,

which gives

Im(t) = CmV du(t)dt+ V u(t)Rm = CmVδ(t)+ V

Rmu(t),

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’scurrent law yields

Im(t) = CmdVm(t)dt+ Vm(t)− V

om

Rm ,

and

Im(t) = Vu(t)− Vm(t)Rs .

Combining these two equations to eliminate Im(t) yields, after some rearrange-ment of terms

CmdVm(t)dt+ Vm(t)Rp = V

Rs u(t)+VomRm ,

where Rp = RmRs/(Rm +Rs). Hence, Vm(t) has an initial value

Vm(0) =( RsRm +Rs

)Vom,

and a final value

Vm(∞) =( RsRm +Rs

)Vom +

( RmRm +Rs

)V,


0 1 2 3 4 5

0.2

0.4

0.6

0.8

1

v(t′ )

t′

r=0.01

r=0.1

r=1

r=10r=100

Figure 3.3: The effect of the series resistance onthe potential across the membrane with normalizedseries resistance as the parameter (Problem 3.2).

and a time constant of τ = RpCm, so that

Vm(t) =( RsRm +Rs

)Vom +

( RmRm +Rs

)V(1− e−t/τ

).

To sketch this response for different values ofRs/Rm, it is helpful to express theresponse in normalized coordinates. Define

v(t) = 1V

(Vm(t)−

( RsRm +Rs

)Vom

),

where r = Rs/Rm, and t′ = t/RmCm, then

v(t′) = 11+ r

(1− e−t′(1+r)/r

).

Therefore, v(t′) represents the increment in voltage caused by the source dividedby the magnitude of the source voltage. The sketch shown in Figure 3.3 showsv(t′) for t′ ≥ 0 (providedRs 6= 0 which was assumed implicitly in the derivation).Note that as the series resistance is decreased, the response approaches that foran ideal (zero series resistance) voltage source. The membrane potential followsthe voltage step more faithfully in time and approaches a final value of 1. Forexample, if the series resistance is 1% of the membrane resistance, the final valueof the voltage change across the membrane is 99% of the voltage step, and thetime constant for the change in voltage is 1% of the membrane time constant. Asthe series resistance increases, it takes more time to charge the capacitance. ForRs Rm,Rp →Rm and τ →RmCm. In addition, asRs increases, the final valueof Vm is less than 1 because of the voltage division that occurs between Rs andRm. This problem illustrates the importance of minimizing the series resistancewhen 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) = Cdvm(t)dt+Gvm(t).

PROBLEMS 29

The solution to this first-order differential equation with constant coefficients is

vm(t) = 1G(1− e−t/τM

)u(t),

where τM = C/G. The series resistance has no effect on the potential across themembrane when the membrane is driven by a current source. The source currentequals the membrane current.

b. Kirchhoff’s current law for the node in the cell yields

vm(t)−u(t)Rs +Cdvm(t)

dt+Gvm(t) = 0,

which can be rewritten as follows

CGdvm(t)dt

+(

1+RsGRsG

)vm(t) = 1

RsGu(t).

The solution is

vm(t) = 11+RsG

(1− e−t/τM

)u(t),

where τM = RsC/(1 + RsG). Thus, because of the presence of the series resis-tance, the potential across the membrane does not equal the source potential. Thefinal value of the potential is less than the source potential and the membrane po-tential changes with a time constant that depends on the series resistance. As theseries resistance approaches zero, the membrane potential approaches the sourcepotential. 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) = roλC1

2I1e−|z|/λC1 ,

vm2(z) = roλC2

2I2e−|z|/λC2 .

The space constant is

λC = 1√(ri + ro)gm ≈

1√rigm =1√

(ρi/πa2)(2πaGm)=√

a2ρiGm

.

Therefore,

λC1 =√

10−2

2 · 102 · 5× 10−3 = 0.1 cm = 1 mm,

λC2 =√

10−3

2 · 102 · (1/8)× 10−3 = 0.2 cm = 2 mm.

a. From the results obtained above,

A = vm1(−0.1)vm2(−0.1)

= vm1(0)e−0.1/λC1

vm2(0)e−0.1/λC2= e−1

e−0.5 = e−0.5 = 0.61.


Sea water

Sea water

Axon

+−

Ie

Ve

−l

2

l

2z0

z

vm(z)

Figure 3.4: Coordinate systemfor solving cable problem (Prob-lem 3.5) and a schematic diagramof the solution as a superpositionof solutions obtained at one elec-trode.

b. Since vm1(0) = vm1(0)roλC1

2I1 = roλC2

2I2,

which implies that

B = I1I2= λC2

λC1= 2.

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 currentapplied at z = 0 which can be represented as a current per unit length of the formke(z) = Ieδ(z). The solution is

vm(z) = roλC2Iee−|z|/λC .

This solution is used to solve the present problem.

a. The external current per unit length is

ke(z) = Ieδ(z + l/2)− Ieδ(z − l/2).Because the cable equation is linear and space invariant, solutions can be super-imposed as shown in Figure 3.4. Therefore,

vm(z) = roλC2Iee−|z+l/2|/λC − roλC2

Iee−|z−l/2|/λC .

This relation is evaluated in three separate regions

vm(z) =

roλC2Ie(e(z+l/2)/λC − e(z−l/2)/λC

)for z < −l/2,

roλC2Ie(e−(z+l/2)/λC − e(z−l/2)/λC

)for −l/2 < z < l/2,

roλC2Ie(e−(z+l/2)/λC − e−(z−l/2)/λC

)for z > l/2.

PROBLEMS 31

These equations can be written alternatively as

vm(z) =

roλCIe sinh(l

2λC

)ez/λC for z < −l/2,

−roλCIee−l/(2λC) sinh(zλC

)for −l/2 < z < l/2,

−roλCIe sinh(l

2λC

)e−z/λC for z > l/2.

The solution depends upon a number of parameters. In order to gain some in-sight into this solution, it is helpful to use normalized variables. Define v =vm/(roλCIe), λ = z/λC , and L = l/λC . Therefore,

v(λ) =

sinh(L/2)eλ for λ < −L/2,−e−L/2 sinhλ for −L/2 < λ < L/2,− sinh(L/2)e−λ for λ > L/2.

The normalized membrane potential (Figure 3.5) depends on only one parameterL which is the inter-electrode distance divided by the cell space constant. Whenthe 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 inter-electrode distance is comparable in length to the space constant (i.e., L ≈ 1), theresponses 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 distributionthat resembles the spatial derivative of the solution at one electrode. This resultsbecause the stimulus approaches a current dipole at λ = 0.

b. The relation between the membrane potential and the longitudinal currents isgiven by

dvm(z)dz

= −riii(z)+ roio(z).The longitudinal current must also satisfy Kirchhoff’s current law which impliesthat

ii(z)+ io(z) =−Ie for |z| < l/2,0 for |z| > l/2.

Combining these equations yields

io(z) =

(

1ri + ro

)dvm(z)dz

−(

riri + ro

)Ie for |z| < l/2,(

1ri + ro

)dvm(z)dz

for |z| > l/2.

Derivatives with respect to z yield

io(z) =

(ro

ri + ro)Ie sinh

(l

2λC

)ez/λC for z < −l/2,

−(

rori + ro

)Iee−l/(2λC) cosh

(zλC

)−(

riri + ro

)Ie for −l/2 < z < l/2,(

rori + ro

)Ie sinh

(l

2λC

)e−z/λC for z > l/2.


−6 −4 −2 2 4 6

−0.4

−0.2

0.2

0.4

−6 −4 −2 2 4 6

−0.3

−0.2

−0.1

0.1

0.2

0.3

−6 −4 −2 2 4 6

−0.04

−0.02

0.02

0.04

−6 −4 −2 2 4 6

−1.5

−1

−0.5

0.5

-6 -4 -2 2 4 6

-1.5

-1

-0.5

−6 −4 −2 2 4 6

−2

−1.5

−1

−0.5

v(λ)

λ

λ

λ λ

λ

λ

i(λ)

L = 10

L = 1

L = 0.1

Figure 3.5: The spatial distribution of membrane potential (left column) and the external lon-gitudinal current (right column) in normalized coordinates (Problem 3.5). The parameters areL = 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)/(roIe) and α = ri/ro so that

i(λ) =

sinh(L/2)eλ for λ < −L/2,−e−L/2 coshλ−α for −L/2 < λ < L/2,sinh(L/2)e−λ for λ > L/2.

The external longitudinal current is plotted in normalized coordinates in Fig-ure 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. Atλ = 0, the external longitudinal current approaches the negative of the magnitudeof 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 isvery small (i.e., L 1), the external longitudinal current has a small magnitudein the extrapolar region and contains a sharp spike of current centered on λ = 0which is the location of the two electrodes. Thus, very little of the external cur-rent enters the cell and appreciable external current occurs only in the interpolarregion where the current equals the negative of the external current.

c. The voltage Ve is computed from io(z) obtained in part b,

Ve = −∫ l/2−l/2

roio(z)dz,

= −∫ l/2−l/2

((ro

ri + ro)dvm(z)dz

−(rirori + ro

)Ie)dz,

= −(

rori + ro

)(vm(l/2)− vm(−l/2))+

(rirori + ro

)lIe,

=(r2o λC

ri + ro

)Ie(1− e−l/λC

)+(rirolri + ro

)Ie.

Therefore,

Re = VeIe =r2o λC

ri + ro(1− e−l/λC

)+ rirolri + ro .

This resistance is examined in two limits. For l/λC 1

Re ≈ r2o λC

ri + ro (1− (1− l/λC))+rirolri + ro ,

Re ≈ r2o l

ri + ro +rirolri + ro ,

Re ≈ rolri + ro (ro + ri),

Re ≈ rol.

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


+ −

Ie

Ve

z0−d

2

d

2

Figure 3.6: The arrangement for measuring the membraneresistance of an axon (Problem 3.6).

resistance per unit length times the distance between the electrodes through whichthe external current flows. For l/λC 1

Re ≈ r2o λC

ri + ro +rirolri + ro ,

Re ≈ rirolri + ro .

Thus, if the distance between the electrodes is much greater than a space con-stant, the equivalent resistance is a resistance per unit length that is the parallelcombination of ri and ro times the inter-electrode distance.

Problem 3.6 To analyze this problem, define the coordinates shown in Figure 3.6. Themethod is to determine the membrane potential versus position and the external longi-tudinal current and then to match boundary conditions.

a. In general the time-independent solution of the cable equation can be written as

vm(z) = A1e−z/λC +A2ez/λC or vm(z) = A1 sinh(z/λC)+A2 cosh(z/λC).

Because of the symmetry of the stimulation and recording arrangement, the ex-pression with hyperbolic functions is more convenient for this problem. It is ap-parent that vm(z) is an odd function of z since current flows into the membranefor z < 0 and out of the membrane for z > 0. Therefore, the form of the solutionis

vm(z) = A sinh(z/λC).

The membrane potential is related to the longitudinal currents by the relation

dvm(z)dz

= −riii(z)+ roio(z).

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) = 1ri + ro

dvm(z)dz

+ riri + ro Ie,

= A(ri + ro)λC cosh(z/λC)+ ri

ri + ro Ie.

PROBLEMS 35

The longitudinal current io(±d/2) = Ie, i.e., the external current equals the exter-nal longitudinal current at the edges of the oil gap. Therefore,

io(±d/2) = Ie = A(ri + ro)λC cosh(±d/2/λC)+ ri

ri + ro Ie,

which can be solved for AA = roλCIe

cosh(d/(2λC)).

Therefore, the solutions are

vm(z) = roλCIesinh(z/λC)

cosh(d/(2λC)),

io(z) = rori + ro Ie

cosh(z/λC)cosh(d/(2λC))

+ riri + ro Ie

The external potential difference between the two electrodes is

Ve =∫ d/2−d/2

roio(z)dz,

= 2r2o λC

ri + ro Iesinh(z/λC)

cosh(d/(2λC))+ rirori + ro Iez

∣∣∣∣∣d/2

−d/2,

= 2r2o λC

ri + ro Ie tanh(d/(2λC))+ rirodri + ro Ie.

Therefore,

R = VeIe= 2r2

o λCri + ro tanh(d/(2λC))+ rirod

ri + ro .

b. The value of the resistance for small values of d is

R ≈ 2r2o λC

ri + rod

2λC+ rirodri + ro ,

≈ r2o d

ri + ro +rirodri + ro ,

≈ rod.

The value of the resistance for large values of d is

R ≈ 2r2o λC

ri + ro +rirodri + 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. Fromthe measurements shown in Figure 3.7,

initial slope = 22× 103

0.41= 5.4× 104 Ω/cm,

final slope = 15.5× 103

1.2= 1.3× 104 Ω/cm,

final intercept = 7× 103 Ω.


JJJ

JJ

J

X0

10

20

0 5 10

R (

kΩ

)

d (mm)

J

J

J

JJ

J

J

J

J

J

J

J

Figure 3.7: Estimation of parameters from mea-surements (Problem 3.6). The initial slope is indi-cated by a line with short dashes, the final slopewith a line with long dashes. The line showing thefinal slope is also shown translated to the origin.

Therefore, from the initial slope ro = 5.4× 104 Ω/cm and from the final slope

rirori + ro = 1.3× 104 Ω/cm,

from which ri = 1.7× 104 Ω/cm. From the intercept

2r2o λC

ri + ro = 7× 103 Ω,

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(0.085)2(1.7× 104 + 5.4× 104)

= 1.9× 10−3 S/cm.

The specific resistance of the membrane can be obtained from the conductanceper 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 withspace constant λC = 1/

√(gm(ri + ro)). The general solution is of the form

vm(z) = A1e−z/λC +A2ez/λC for z > 0.

On physical grounds, the solution will converge for z →∞. Since vm(0) = V1, thesolution is

vm(z) = V1e−z/λC for z > 0,

which is plotted in Figure 3.8.

b. The core conductor equation yields

ii(z) = − 1ri + ro

dvm(z)dz

= − 1ri + ro

−V1

λCe−z/λC =

√gmri + ro V1e−z/λC .

PROBLEMS 37

vm(z)

V1

λC z

Figure 3.8: Membrane potential versus position ina dendrite (Problem 3.7).

c. Combining the results of parts a and b yields

ii(z) =√

gmri + ro vm(z),

so thatvm(0)ii(0)

= 1√gm/(ri + ro) .

d. If ri ro then

Gc =√

gmri + ro ≈

√gmri=√

2πaGmρi/(πa2)

=√

2π2a3Gmρi

.

e. At the bifurcation, the internal current divides between the two branches, ii =2ib = 2Gbvm, where ib is the internal longitudinal current in each branch of radiusb, and Gb is the characteristic conductance of these branches. The current in theunbranched dendrite is ii = Gavm, where Ga is the characteristic conductance ofthe dendrite of radius a. Therefore, if Ga = 2Gb then the electrical load on thecell is the same, i.e., √

2π2a3Gmρi

= 2

√2π2b3Gm

ρi,

which shows that b = a/41/3.

Problem 3.8 This problem investigates the stimulation of a cell by an external currentand the role of the space constant of the cell in determining the effectiveness of thecurrent stimulus.

a. The symmetry of the stimulation arrangement determines which solution is cor-rect. We expect the membrane current to enter the fiber for z < 0 and to leave thefiber for z > 0. Hence, the membrane current per unit length is an odd functionof z. Because for the steady-state cable equations km(z) = gmvm, the membranepotential is also an odd function of z. This conclusion also follows from the coreconductor relation for which the membrane current per unit length is proportionalto the second derivative of the membrane potential. Both arguments show thatthe membrane potential is an odd function of z. Therefore, the solution has theform vm(z) = A sinh(z/λC).


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)dz

= −riii(z)+ roio(z).

Hence,AλC

cosh(z/λC) = −riii(z)+ roio(z).

A boundary condition is that ii(±L) = 0. Since ii(z)+ io(z) = I, this implies thatio(±L) = I. Therefore,

AλC

cosh(L/λC) = roI,

and

A = roλCIcosh(L/λC)

.

Therefore,

vm(z) = roλCIcosh(L/λC)

sinh(z/λC).

This equation is expressed as

vm(z)roλCI

= sinh(z/λC)cosh(L/λC)

,

which is plotted in Figure 3.9. The membrane potential difference is small foran 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 coreconductor equation to yield

dvm(z)dz

= −ri(I − io(z))+ roio(z) = (ri + ro)io(z)− riI,

which can be solved for io(z) to yield

io(z) = 1ri + ro

dvm(z)dz

+ riri + ro I,

= Iri + ro

(ri + ro cosh(z/λC)

cosh(L/λC)

),

which can be written as

io(z)I= 1

1+ (ro/ri)(

1+ (ro/ri)cosh(z/λC)cosh(L/λC)

).

The internal longitudinal current is ii(z) = I − io(z) which is

ii(z)I= (ro/ri)

1+ (ro/ri)(

1− cosh(z/λC)cosh(L/λC)

).

The longitudinal currents are plotted in Figure 3.9.

PROBLEMS 39

−0.1 −0.05 0.05 0.1

−1

−0.5

0.5

1

−1 −0.5 0.5 1

−1

−0.5

0.5

1

−10 −5 5 10

−1

−0.5

0.5

1

−0.1 −0.05 0.05 0.1

0.5

1

−1 −0.5 0.5 1

0.5

1

−10 −5 5 10

0.5

1

vm(z)

roλCI

io(z)

I

ii(z)

I

z/λC

z/λC

io(z)/I

ii(z)/I

L

λC= 0.1

L

λC= 1

L

λC= 10

0.1

0.1

0.1

0.1

1

1

1

1

10

10

10

10

Figure 3.9: The membrane potential vm(z) (left panels) and the external (solid lines) and inter-nal (dashed lines) longitudinal currents (right panels) are plotted versus z/λC for the differentvalues of L/λC shown at the left (Problem 3.8). The longitudinal currents are shown for valuesof ro/ri of 0.1, 1, 10.


I

V+ −

I

V+ −

−L L0z

L/λC ¿ 1

L/λC À 1

Figure 3.10: Schematic diagram of currentflow in and around a muscle fiber (Prob-lem 3.8). The length of each arrow indi-cates the magnitude of the longitudinal cur-rent and the current per unit length for themembrane current per unit length. For anelectrically small cell (L/λC 1), no ap-preciable current enters the cell and mostof the current flows extracellularly betweenthe electrodes. For an electrically large cell(L/λC 1), appreciable current enters thecell.

These results are examined for electrically small cells first and then for large elec-trically large cells. For an electrically small muscle cell, L λC , which impliesthat sinh(z/λC) ≈ z/λC , cosh(L/λC) ≈ 1, and cosh(z/λC) ≈ 1. Therefore,

vm(z) ≈ rozI, km(z) ≈ gmrozI, io(z) ≈ I, and ii(z) ≈ 0.

Thus, as shown in Figure 3.9 for electrically small muscle cells (L/λC = 0.1), theexternal 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 themembrane potential is small. Thus, for electrically small cells very little currententers 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 lon-gitudinal current increases as current enters the cell (Figure 3.10). The externallongitudinal current is a minimum and the internal longitudinal current is a maxi-mum at z = 0. These extrema depend upon the ratio ro/ri. As this ratio increasesthe minimum external longitudinal current decreases and the maximum internallongitudinal current increases indicating that more current enters the muscle cellas the external longitudinal resistance is increased.

d. The voltage V = vo(−L)− vo(L), which can be found from vo(z), is

dvo(z)dz

= −roio(z).

PROBLEMS 41

0 1 2 3 4 5

0.2

0.4

0.6

0.8

1

R2roL

rori

= 10

rori

= 1

rori

= 0.1

L/λC

Figure 3.11: The resistance R is plot-ted versus L/λC for different values ofro/ri (Problem 3.8).

Hence,

V = −∫ L−Ldvo(z) =

∫ L−Lroio(z)dz.

Substitution of io(z) into the integral yields

V = I rori + ro

∫ L−L

(ri + ro cosh(z/λC)

cosh(L/λC)

)dz,

which can be integrated to yield

V = Iro

ri + ro(

2riL+ 2λCrosinh(L/λC)cosh(L/λC)

),

= I2Lrirori + ro

(1+ ro

ritanh(L/λC)L/λC

),

Therefore,

R = VI= 2Lrirori + ro

(1+ ro

ritanh(L/λC)L/λC

),


R2Lro

= 11+ (ro/ri)

(1+ (ro/ri)tanh(L/λC)

L/λC

).

The resistance is plotted in Figure 3.11. For an electrically small muscle cell, LλC and tanh(L/λC) ≈ L/λC . Therefore, R ≈ 2Lro. Thus, for an electrically smallcell, since no appreciable current enters the cell, the external longitudinal currentequals the external current (Figure 3.9). Therefore, the resistance between theelectrodes is the external resistance per unit length, ro, times the length of theexternal conductor which is the length of the muscle cell, 2L, (Figure 3.11). Theexternal potential is simply the external current flowing through a resistance perunit length ro. Therefore, the membrane potential difference reflects the changein 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 ≈ 2Lrirori + ro .

Therefore, the equivalent resistance per unit length is just the parallel combinationof ri and ro and the total resistance between electrodes is this equivalent resistanceper unit length times the length of the muscle, 2L.


e. The fact thatR changed at all suggests that the cell is neither an electrically smallcell 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 largecell which implies that the length of the cell is roughly comparable to its spaceconstant.

Problem 3.9 The solution of the cable equation for an impulse of charge is given byEquation 3.45 (Weiss, 1996b) which is

vm(z, t) = roλCQe/τM√4πt/τM

e−(z/λC)2/(4t/τM)e−t/τMu(t),

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 lengthwhich is cmvm(z, t) to yield

Q(t) =(∫∞−∞roλCQecm/τM√

4πt/τMe−(z/λC)

2/(4t/τM)e−t/τM dz)u(t),

= roλCQecm/τM√4πt/τM

(∫∞−∞e−(z/λC)

2/(4t/τM) dz)e−t/τMu(t).

The integral equals√

4πλ2Ct/τM so that

Q(t) = roλ2CQecmτM

e−t/τMu(t).

This solution needs to be related to the problem at hand. The total charge on the cableat t = 0 is Qo so that

Qo = roλ2CQecmτM

= roQecmgm(ri + ro)cm/gm =

rori + roQe,

which equates the solution for an external impulse of charge delivered at t = 0 at z = 0to the solution for an initial voltage distribution in this problem. For the problem athand, the total charge on the cable can be expressed simply as

Q(t) = Qoe−t/τMu(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 inspace and time to a cable. The solution to this problem in normalized variables isgiven in Equation 3.42 (Weiss, 1996b) and is

vm(λ, τ) = A√4πτ

e−λ2/4τe−τ, for τ > 0,

PROBLEMS 43

where λ = z/λC , τ = t/τM . At the peak of the response ∂vm(λ, τ)/∂τ = 0

∂vm(λ, τ)∂τ

= A(

1√4πτ

(λ2

4τ2 − 1

)e−λ

2/4τe−τ −(

12√

4πτ3/2

)e−λ

2/4τe−τ),

= A√4πτ

e−λ2/4τe−τ

(λ2

4τ2 − 1− 12τ

).

The derivative is zero for the value τ = τp that satisfies the equation

λ2 = 4τ2p + 2τp,

which in unnormalized form is(zλC

)2

= 4( tpτM

)2

+ 2( tpτM

).

b. 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 thefact that the end-plate currents, that give rise to the measured end-plate po-tentials, deviate from impulses. The measurements for small values of z, andt 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 givendistance. To find τp for λ = 1 the following quadratic equation must besolved

4τ2p + 2τp − 1 = 0.

The roots are

τp = −2±√4+ 168

= −14± 1

4

√5.

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 representedby a parallel combination of a resistance and a capacitance. Therefore, the membranepotential change must satisfy the cable equation

λ2C∂2vm(z, t)∂z2 = vm(z, t)+ τM ∂vm(z, t)∂t

,

but since the foot of the action potential travels at a constant conduction velocity ν in thepositive z-direction, the membrane potential must have the form vm(z, t) = f(t−z/ν)which satisfies the wave equation.


a. The proposed form of the solution does satisfy the wave equation. It only remainsto determine if that form satisfies the linear cable equation. Therefore, try thesolution

vm(z, t) = Aeα(t−z/ν),whose partial derivatives are

∂vm(z, t)∂t

= Aαeα(t−z/ν),

∂2vm(z, t)∂z2 = A

(αν

)2

eα(t−z/ν).

Substitution of these partial derivatives into the cable equation yields

λ2CA

(αν

)2

eα(t−z/ν) = Aeα(t−z/ν) + τMAαeα(t−z/ν),

λ2C

(αν

)2

= 1+ τMα.

This is a polynomial in α which can be written as

α2 − τM(νλC

)2

α−(νλC

)2

= 0,

which has two roots

α = τM2

(νλC

)2

±√√√√(τM

2

(νλC

)2)2

+(νλC

)2

.

Therefore, the proposed solution satisfies both the linear cable equation and thewave equation for any value of A and for two values of α,

b. The solution should converge as t → −∞. Therefore, α > 0 and

α = τM2

(νλC

)21+

√√√√1+(

2λCτMν

)2 .

Problem 3.12

a. The steady-state solution for this current distribution is a superposition of solu-tions given in Section 3.4.2 (Weiss, 1996b) to yield

vm(z,∞) = −roλC2Ie(e−|z|/λC − e−|z−L|/λC

).

For Ie = 1 mA, Vo = 0.2 mV. Therefore, vm(0,∞) = −100Vo = −20 mV. Therefore,

vm(0,∞) = −20× 10−3 = −roλC2

1× 10−3,

PROBLEMS 45

z0L

vo(z,∞) (mV)

0.2

−0.2

Figure 3.12: Extracellular potential (Prob-lem 3.12). The space constant is 2 mm.

and roλC = 40. If the term in the interpolar region is ignored as stated in theproblem, then from the core conductor model

vo(0,∞)vm(0,∞) = −

rori + ro =

Vo−100Vo

.

Therefore,ro

ri + ro =1

100,

which implies that ri = 99ro. The inside resistance per unit length is

ri = ρiπa2 =

25π(0.02)2

≈ 2× 104 Ω/cm.

Therefore, ro ≈ ri/100 ≈ 2× 102 Ω/cm. A combination of these results yields

λC = 40ro≈ 40

2× 102 = 0.2 cm.

The extracellular potential is shown plotted in Figure 3.12.

b. Note that for Ie = −2 mA, the membrane resistance of the 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 (roλC/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 distri-bution 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 establishosmotic 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 tendto decrease the conduction velocity. Second, the concentration of intracellular so-lutes will increase. Since most of the intracellular solutes are ionized, the increasein ion concentration will increase the conductivity of the cytoplasm which will tendto increase the conduction velocity of the action potential. A more quantitativeanalysis is required to determine which if any of these effects predominates.

b. A quantitative analysis includes a determination of the change in radius of thecell, the change in conductivity of the cytoplasm, and the change in conductionvelocity.


• Dependence of intracellular concentration of ions on extracellular concentra-tion. At osmotic equilibrium,

CiΣ = CoΣ ,where CiΣ and CoΣ are the total concentrations of solute osmolarities insideand outside the cell. If the outside concentration is doubled then the insideconcentration will double by the efflux of water to halve the water volumein the cell. Let us assume that the cell water volume is approximately equalto the cell volume. The volume of a cylinder of radius a and of length L isV = πa2L. Hence, halving the volume of a cylindrical cell whose length isconstant results in a change in radius of 1/

√2.

• Dependence of conductivity of cytoplasm on ion concentration. The conduc-tivity (σi) of the cytoplasm, assuming it acts as an electrolyte, is (Weiss, 1996a,Chapter 7)

σi =∑nunz2

nF2cn,

whereun is the molar mechanical mobility of ionn, zn is its valence, F is Fara-day’s constant, and cn is its concentration. Thus, doubling the concentrationof all the ions should double the conductivity.

• Dependence of conduction velocity on cytoplasmic conductivity. If it is as-sumed that ri ro, then the conduction velocity is

ν =√Km

2πari=√

Km2πaρi/(πa2)

=√Kma2ρi

=√Kmaσi

2.

c. Thus, the conduction velocities are

ν1 =√Kmaσi

2,

ν2 =√Km(a/

√2)(2σi)

2= (2)1/4

√Kmaσi

2= (2)1/4ν1 = 1.2ν1,

where it is assumed thatKm, which is a property of the membrane alone, does notchange as the cell shrinks. Thus, the effect of the change in conductivity exceedsthe 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 ap-propriate to model this cell as an electrically small cell. Hence, a lumped-parametermodel of the cell, as shown in Figure 3.13, is appropriate. Since the specific capac-itance 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 = 6× 10−12

2× 10−3 = 3 nS.

In addition, Vom = −50 mV.

PROBLEMS 47

Vm(t) i(t)C

+

+

−−GV om

Figure 3.13: Equivalent electric circuit of a smallcell (Problem 3.14).

0 2 4 6 8

−50

−60

Vm(t)

t

Figure 3.14: Step response of a small cell (Prob-lem 3.14).

b. Before the onset of the current, Vm(t) = Vom = −50 mV. The current is in the direc-tion to hyperpolarize the membrane. Hence, the membrane potential decreasesexponentially 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

λC ≈ 1√gmri =1√

10−4 · 104= 1 cm.

The time constant is

τM = cmgm =150× 10−9

10−4 = 1.5 ms.

b. The cell is 4 space space constants long. Hence, it is an electrically large cell. Thestep 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 thiscell. A more involved treatment is required to get a more accurate answer for acell that is 4 space constants long. The approximate solution is

vm(0, t) ≈ λCri2Ie erf(t/τM)u(t) = 5× 103Ie erf(t/1.5)u(t),

where t is in ms.

c. Note the axial electrode has a resistance per unit length that is 2000 times smallerthan 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


0 1 2 3 4 5

0.2

0.4

0.6

0.8

1

clamped

unclamped

Time (ms)

Figure 3.15: The unclamped cable response isa plot of vm(0, t)/(5 × 103Ie) = erf(t/1.5)u(t);the clamped cable response is a plot ofvm(0, t)/(5 × 103Ie) = (1 − e−t/1.5)u(t) (Prob-lem 3.15).

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 twoends of the cylindrical cell, is

R = 1gmL

= 110−4 · 4

= 2.5× 103 Ω.

Therefore,vm(0, t) = 2.5× 103Ie

(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, incrementalportions of the membrane can be represented by a parallel network consisting of aresistance 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 willalso be assumed that the potential change is negligibly small at a recording electrodelocated 5 or more space constants away from a stimulating electrode.

a. The intracellular electrode is one space constant away from an electrode thatpasses an outward current through the membrane which depolarizes the mem-brane at that location transiently. Thus, the potential change across the membraneat this site is a positive, pulsatile waveform that starts at zero. The potential atthe external remote reference electrode is zero. Thus, the potential difference be-tween 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 recordingelectrode, and therefore, produces a positive potential at the active extracellularelectrode. 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 differ-ence 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 toan 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, thesolution is that of the voltage response of a parallel resistance and capacitance toa brief pulse of current. The answer is v8(t).

Problem 3.17 Reduction of the extracellular osmolarity results in the flow of water intothe cell so that the cell will swell to achieve osmotic equilibrium (Weiss, 1996a, Chap-ter 4). If the intracellular ions are assumed to be relatively impermeant compared withwater, then the volumes in two solutions satisfy the equation NiΣ = CiΣ1V1 = CiΣ2V2,where NiΣ is the total quantity of intracellular solute, CiΣ is total concentration of intra-cellular solute, and V is the water volume which equals the cell volume in this case. Inaddition, at osmotic equilibrium CiΣ = CoΣ . The volume of a cylindrical cell isV = πa2L,where a is the radius and L is the length of the cylinder. Since the length of the cell isassumed not to change as the cell swells, a2

1CoΣ1 = a2

2CoΣ2 which implies that

a2

a1=√√√√CoΣ1

CoΣ2.

Dilution of the intracellular solution also reduces the cytoplasmic conductivity andraises the cytoplasmic resistivity. The conductivity of a dilute solution is proportionalto the concentration of ions (see solution to Problem 3.13). Therefore,

ρ2

ρ1= C

iΣ1

CiΣ2= C

oΣ1

CoΣ2.

a. The conduction velocity of an unmyelinated axon is

ν =√

Km2πa(ro + ri) ,

which for ro ri, and for ri = ρi/(πa2) equals

ν =√Kma2ρi

,

so that for two different solutions

ν2

ν1=√a2ρ1

a1ρ2=(CoΣ2

CoΣ1

)1/4

,

under the assumption thatKm does not change as the cell swells. For CoΣ2/CoΣ1 =

0.78, the new conduction velocity is ν(0.78)1/4 = 0.94ν .

b. The space constant of the axon is

λC = 1√(ro + ri)gm),


which for ro ri, ri = ρi/(πa2), and gm = 2πaGm equals

λC =√

a2ρiGm

,

so thatλC2

λC1=√a2ρ1

a1ρ2,

which is the same ratio as the ratio of velocities. Therefore, For CoΣ2/CoΣ1 = 0.78,

the new space constant is λC(0.78)1/4 = 0.94λC .

c. The membrane time constant is τM = cm/gm = Cm/Gm which is independent ofthe resistivity of cytoplasm and the dimensions of the cell so that the new timeconstant equals τM .

Problem 3.18

a. The core conductor equations for incremental quantities include the relationvm(z, t) =vi(z, t)− vi(z, t). The second partial derivative of this relation with respect to zyields

∂2vm(z, t)∂z2 = ∂

2vi(z, t)∂z2 − ∂

2vo(z, t)∂z2 .

But∂vi(z, t)∂z

= −riii(z, t) and∂ii(z, t)∂z

= −km(z, t).

A combination of these three equations yields

∂2vm(z, t)∂z2 = rikm(z, t)− ∂

2vo(z, t)∂z2 .

But, in the cable model the relation between membrane current per unit lengthand membrane potential is given by

km(z, t) = gmvm(z, t)+ cm∂vm(z, t)∂t.

Combining these last two equations yields

∂2vm(z, t)∂z2 = rigmvm(z, t)+ ricm∂vm(z, t)∂t

− ∂2vo(z, t)∂z2 .

If the equation is divided by rigm then

λ2C∂2vm(z, t)∂z2 = vm(z, t)+ τM ∂vm(z, t)∂t

− λ2C∂2vo(z, t)∂z2 ,

where λ2C = 1/(rigm) and τM = cm/gm.

PROBLEMS 51

b. The time independent cable equation for this case is obtained by setting the timederivatives to zero,

λ2Cd2vm(z, t)dz2 − vm(z, t) = −λ2

Cd2vo(z)dz2 .

The rightmost term for vo(z) = (1/2)z2u(z) is evaluated as follows

λ2Cd2vo(z, t)dz2 = λ2

Cu(z).

Therefore, we seek a solution to the equation


Cu(z).

To obtain the solution to this problem, we take advantage of the linearity andspace invariance of the cable equation. The solution to the equation


CroIeδ(z),

is (Weiss, 1996b, Section 3.4.2)

vm(z) = roλC2Iee−|z|/λC .

Thus, except for differences in scale factors, the solution for a spatial impulse canbe used to determine the solution for a spatial step. This, is achieved by integratingthe solution for the spatial impulse, suitably scaled,

vm(z) = λC2∫ z−∞e−|u|/λC du.

The integral is expressed in parts,

vm(z) =

λC2

∫ z−∞eu/λC du for z < 0,

λC2

∫ 0

−∞eu/λC du+ λC

2

∫ z0e−u/λC du for z > 0.

Evaluation of the integrals yields

vm(z) =

λ2C

2eu/λC

∣∣∣z−∞ for z < 0,

λ2C

2+ −λ

2C

2e−u/λC

∣∣∣z0

for z > 0.

The solution is

vm(z) =

λ2C

2ez/λC for z < 0,

λ2C

2+ λ

2C

2

(1− e−z/λC

)for z > 0.

vm(z) is plotted in Figure 3.16. To summarize, the external potential which variesparabolically in space acts as a discontinuous external current. The cable modelsmooths out this discontinuous current to produce a spatially smoothed mem-brane potential.


-4 -2 0 2 4

1

2

2vm(z)

λ2C

z/λC

Figure 3.16: Membrane potential versus posi-tion for an applied extracellular potential (Prob-lem 3.18).

Problem 3.19 The extracellular potential is (Weiss, 1996b, Section 3.4.2)

vo(z)− vo(−∞) = − r2o λC

2(ri + ro)Ie(e−|z|/λC + 2

(riro

)(zλC

)u(z)

).

Note that the first term decreases to zero as z → ∞. Therefore, the only term thatcontributes is the second term whose derivative with respect to z is

− 1Iedvo(z)dz

= rirori + ro .

Problem 3.20 The steady-state extracellular potential in response to a step of currentis

vo(z,∞)vo(0,∞) = e

−z/λC .

Therefore, the normalized logarithmic potential is

log10

(vo(z,∞)vo(0,∞)

)= −z/λC log10 e.

The measurements (Weiss, 1996b, upper panel of Figure 3.59) show that the normalizedlogarithmic potential is −0.8 for z = 2 mm. Therefore, (−2/λC) log10 e = −0.8 so thatλC = 1.09.

The time it takes the step response to reach 1/2 of its maximum value is givenapproximately by the relation (Weiss, 1996b, Figure 3.31)

τ1/2 = λ+ 0.52

.

In unnormalized coordinates

t1/2τM

= (z/λC)+ 0.52

,

so that

t1/2 =(τM2λC

)z + τM

4.

The measurements (Weiss, 1996b, lower panel of Figure 3.59) show that the interceptis at 6.7 ms and the slope is about (28.3− 6.67)/2 = 10.8 ms/mm. From the interceptτM = 26.7 ms. From the slope λC = τM/(2 · 10.8) = 1.24 mm. Thus, the two estimatesof the space constant from these two sets of measurements agree within 14%.

PROBLEMS 53

200

4 6 8

2

4

6

8

10

z (mm)

∆v o

(z,∞

)(m

V)

Figure 3.17: Estimation of space constantfrom measurements of steady-state value ofthe extracellular potential (Problem 3.21).

Problem 3.21 Measurement 1 gives information about the longitudinal resistances,∣∣∣∣ ∆Vo(z, t)∆Vm(z, t)

∣∣∣∣ = roro + ri .

Measurement 2 gives information about other cable parameters since

∆vo(z,∞) = − roro + ri∆vm(z,∞) =

(ro

ro + ri)(roλC

2

)Iee−|z|/λC .

a. The spatial dependence of the steady-state extracellular potential is exponentialand the space constant can be obtained from the initial slope of the potential asshown 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. Inaddition, Measurement 2 shows that

∆vo(0,∞) =(

roro + ri

)(roλC

2

)Ie = 10 mV,

which implies that

10−2 = 13

(ro · 0.15

2

)5× 10−7.

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

λC = 0.15 = 1√gm(ro + ri) =

1√gm · 24× 105

,

which can be solved to yield gm = 1.85× 10−5 S/cm.

e. The specific membrane conductance is

Gm = gm2πa

= 1.85× 10−5

2π25× 10−4 = 1.18× 10−3 S/cm2.

f. The longitudinal resistance per unit length of cytoplasm is ri = ρi/(πa2) so that

ρi = riπa2 = (16× 105)π(50× 10−4)2 = 126 Ω·cm.

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 theaction potential (which approaches the Nernst equilibrium potential for sodium)is changed appreciably. The Nernst equilibrium potential for sodium is VNa =(RT/F) ln(coNa/c

iNa). Hence, the largest value of intracellular sodium concentra-

tion corresponds to the lowest Nernst equilibrium potential. Therefore, curve 3must correspond to the highest intracellular sodium concentration.

b. Experiment 1 shows the result for changing extracellular potassium concentra-tion. Note that the resting potential is changed appreciably, the peak of the actionpotential is not changed much, and the undershoot of the action potential (whichapproaches the Nernst equilibrium potential for potassium) is changed apprecia-bly. The Nernst equilibrium potential for potassium is VK = (RT/F) ln(coK/c

iK).

Thus, the largest value of the Nernst equilibrium potential for potassium occursat the highest value of the extracellular potassium concentration. Therefore, curve1 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 withconstant 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 voltagedependent coefficients. Furthermore, the voltage changes with time. Hence, thesolutions are not exponential functions of time.

c. False. In general, if the membrane potential is not constrained to be constant, thefactors n(t),m(t), and h(t) are solutions to linear, first-order, ordinary differen-tial equations with voltage dependent coefficients. Hence, the solutions are notexponential 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 voltageprecludes action potentials.

b. True. According to the core-conductor model, the primary effect of a space clampis to increase the speed of a propagating action potential. In fact, in a well-clampedcell, the speed can be so fast that the action potential occurs simultaneously at allpoints along the cell. For that reason, a space-clamped action potential is oftencalled a membrane action potential.

c. True. “Unclamped” is the normal condition for an axon. Therefore, action poten-tials occur in unclamped axons.

Exercise 4.4

a. In general, at restm is small andh is relatively large. For a large depolarization, thereverse is true —m is large and h is small. In both cases, the sodium conductanceis small compared to the potassium conductance. This question can be answeredquantitatively by computing the relevant values from the Hodgkin-Huxley modelor by using a simulation of the Hodgkin-Huxley model (Weiss et al., 1992). At restm∞ ≈ 0.05 and h∞ ≈ 0.6 so that

GNa = GNam3∞h∞ ≈ 120(0.05)30.6 = 0.009 mS/cm2.

In contrast, at rest

GK = GKn4∞ ≈ 36(0.3)4 = 0.3 mS/cm2.

Thus, at rest GK ≈ 33GNa. Now consider a strong depolarization to 0 mV. At thatpotential n∞ = 0.9, m∞ = 0.96, h∞ = 0.003. Therefore,

GNa ≈ 120(0.96)30.003 = 0.3 mS/cm2,

andGK ≈ 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, largedepolarization, the response of the sodium conductance to a change in potential isquite different in the two cases. At rest, the sodium conductance can show a rapidincrease because it is m that must increase to increase GNa appreciably and mhas a fast time course. However, after a prolonged depolarization it is h that mustincrease to increaseGNa appreciably and h changes much more slowly. Therefore,after a prolonged depolarization the membrane is not capable of exhibiting a rapidchange in sodium conductance which is necessary for excitation. This is exactlythe situation that occurs during the peak of the action potential — the membrane isdepolarized, h is low, and GNa cannot change rapidly. This phenomenon accountsfor the refractory properties of excitable membranes.

EXERCISES 57

0 1 2 3

1

2

3

4

5

6

0 2 4 6 8 10

1

2

3

4

5

6

t (ms)

GN

a(m

S/cm

2)

Figure 4.1: The sodium conductance is shown as a function of time for a step in membranepotential for two different time scales from (Exercise 4.5). The membrane potential is steppedfrom −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 valuem has reached its final value and h has not changed much. The initial value of theconductance is

GNa(−25,0) = GNam3∞(−50)h∞(−50) = GNa(0.16)3(0.26) = 0.0011GNa.

The maximum value of the conductance is

(GNa)max ≈ GNam3∞(−25)h∞(−50) = GNa(0.73)3(0.26) = 0.10GNa.

The final value of the conductance is

GNa(−25,∞) = GNam3∞(−25)h∞(−25) = GNa(0.73)3(0.02) = 0.0078GNa.

The time dependence of the sodium conductance is shown in Figure 4.1 for thevalue GNa = 120 mS/cm2.

b. Upon repolarization from a depolarized state,m decreases rapidly andh increasesslowly. Therefore, the maximum value of GNa occurs at t = 0.

(GNa)max ≈ GNa(−25,0) = GNam3∞(−25)h∞(−25)≈ GNa(0.73)3(0.02) = 0.0078GNa,

whereas the final value of the conductance is

GNa(−50,∞) = GNam3∞(−50)h∞(−50) = GNa(0.16)3(0.26) = 0.0011GNa.

c. The membrane potential is the same both for t < 0 in part a and for t →∞ in partb. 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 hchanges so much more slowly thanm and it is h that controls the maximum valuein the two cases.


Exercise 4.6

a. True. The leakage conductance is a constant, independent of the membrane po-tential.

b. False. The sodium conductance is proportional to the product ofm3 and h. Eachfactor,m andh, is the solution to a first-order differential equation. Hence, neitherof 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 toa 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 astep of membrane potential will show a step in leakage current.

e. True. The sodium conductance is continuous at t = 0. But, the sodium currentis proportional to the product of the sodium conductance and the difference inmembrane potential from the Nernst equilibrium potential for sodium. Since thispotential difference is discontinuous, the sodium current is discontinuous at t = 0.The discontinuity is not always obvious. If the initial membrane potential is closeto the rest potential, then GNa is small at the time of the step, and the resultingstep in sodium current is also small.

f. True. The potassium conductance is continuous at t = 0. But, the potassiumcurrent is proportional to the product of the potassium conductance and the dif-ference in membrane potential from the Nernst equilibrium potential for potas-sium. Since this potential difference is discontinuous, the potassium current isdiscontinuous at t = 0. As with the sodium current, the step change in potassiumcurrent will be small if the initial voltage is such that the potassium conductanceis small.

g. False. The factors n(t), m(t), and h(t) are all solutions to first-order differentialequations with voltage dependent parameters. A change in a parameter results ina change in the time derivative of each factor, but not in the instantaneous valueof each factor.

h. True. The time constants τn, τm, and τh are all instantaneous functions of themembrane potential. Hence, an instantaneous change in Vm(t) results in an in-stantaneous change in each time constant.

i. True. The steady-state values n∞, m∞, and h∞ are all instantaneous functions ofthe membrane potential. Hence, an instantaneous change in Vm(t) results in aninstantaneous 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 voltagesource 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 thatsenses the membrane potential and passes a current through the membrane to producethe command voltage. A practical voltage-clamp system has an equivalent resistancethat 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 thethreshold in response to a brief current stimulus occurs when the ionic current changesfrom outward to inward. The outward current is carried predominantly by potassiumand the inward current predominantly by sodium. Therefore, threshold occurs approx-imately when the magnitude of the sodium current exceeds that of the potassium cur-rent. The statement is approximately true with the caveats that other ions may alsocontribute to the balance between inward and outward currents. In addition, the analy-sis presented in Section 4.4 (Weiss, 1996b) is approximate in that the ionic current versusmembrane potential characteristic was analyzed at one point in time only. A careful ex-amination of threshold requires analysis of all 4 state variables of the Hodgkin-Huxleymodel, i.e., Vm, m, n, and h.

Exercise 4.9 Consider just the ionic current branches in the Hodgkin-Huxley model ofthe 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 +GLVL),Jion = G(Vm, t)(Vm(t)− V(t)),

where

G(Vm, t) = GK(Vm, t)+GNa(Vm, t)+GL,V(Vm, t) = GK(Vm, t)VK +GNa(Vm, t)VNa +GLVL

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 mem-brane during an action potential. This is wrong. Repolarization occurs because sodiumchannels close and potassium channels open during repolarization. Transport of sodiumand potassium is passive during the action potential — no pumps are involved. [45words]

Exercise 4.11

a. The answer is v, a capacitance current. As an action potential propagates along anaxon, current flows inward at the peak of the action potential and outward bothahead and behind the peak. The current that flows outward in the part of theaxon ahead of the action potential tends to depolarize that part of the membrane.The initial current flows primarily through the membrane capacitance and this isreadily seen in Figure 4.32 (Weiss, 1996b).


5 10 15 20 25

−0.3

−0.2

−0.1

0.1

t (ms)Jion

(mA

/cm

2)

Figure 4.2: The ionic current is shown forthe default parameters and for a change inpotential from −80 to −40 mV (dashed line)and when the external sodium concentrationis doubled (solid line) (Exercise 4.12).

b. The answer is i, an ionic current carried by sodium ions. The early inward currentresults because the sodium conductance increases. Sodium current then flowsdown the electrochemical gradient for sodium, i.e., from outside to inside.

c. The initial outward current is readily explained in terms of linear membrane prop-erties. The action potential causes the intracellular potential to increase and thisincrease in potential tends to drive currents outward. The early inward current isnot consistent with linear membrane properties. This effect can only be under-stood by taking into account the fact that the action potential causes the sodiumconductance to change, as described in part b.

Exercise 4.12

a. An increase in the external sodium concentration increases VNa. Since

JNa = GNa(Vfm − VNa),an increase VNa decreases JNa, especially whenm3h is large (i.e. near the negativepeak of JNa). An increase in external sodium concentration has no affect on JKor JL. Therefore the net change in Jion = JNa + JK + JL is to make the negativepeak 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 exactsolution for these parameters is shown in Figure 4.2. It is reasonable to computethe affect of doubling the extracellular sodium concentration on the ionic currentpredicted by the theory. However, this theoretical computation is difficult to testexperimentally since doubling the sodium concentration approximately doublesthe extracellular osmolarity and causes water efflux from the axon so that thevolume approximately halves. Such large changes in osmolarity are known tocause irreversible changes in the electrical properties of axons.

b. An increase in the external potassium concentration increases VK . Since

JK = GK(Vfm − VK),an increase in VK decreases JK , especially when n4 is large (i.e. for large time). Anincrease in external potassium concentration has no affect on JNa or JL. Thereforethe net change in Jion = JNa + JK + JL is to make the negative peak slightly morenegative and to make the steady state value appreciably more negative. Thus, themagnitude of Jss decreases appreciably, and the magnitude of Jp increases a smallamount. The exact solution for these parameters is shown in Figure 4.3.

PROBLEMS 61

5 10 15 20 25

−0.3

−0.2

−0.1

0.1

t (ms)

Jion

(mA

/cm

2)

Figure 4.3: The ionic current is shown for thedefault parameters and for a change in poten-tial from −80 to −40 mV (dashed line) andwhen the external potassium concentration isdoubled (solid line) (Exercise 4.12).

5 10 15 20 25

−1.00

−0.75

−0.50

−0.25

0.25

t (ms)

Jion

(mA

/cm

2)

Figure 4.4: The ionic current is shown for thedefault parameters and for a change in poten-tial from −80 to −40 mV (dashed line) andwhen the final potential is changed from −40to −30 mV (solid line) (Exercise 4.12).

c. An increase in Vfm increases n, increasesm, and decreases h. This change affects

JNa directly (by changing (Vfm − VNa)) and indirectly (by changing m and h) and

it affects JK directly (by changing (Vfm − VK)) and indirectly (by changing n). Atthe negative peak, JNa is the largest component of current, and is most affectedby the change inm (m increases from 0.37 to 0.63, h barely has changed because

of its long time constant, and |Vfm − VNa| diminishes about 10 percent). Since mincreases and (Vfm−VNa) is negative, Jp becomes more negative. For large time t,JK is the largest current component. The value of n increases as does (Vfm − VK),therefore Jss becomes more positive. Therefore, the magnitude of Jp increases andof Jss decreases. The exact solution for these parameters is shown in Figure 4.4.

d. An increase in Vim affects the values of m, n, and h at time t = 0; m increases, nincreases, and h decreases. These changes have no long term effects, and Jss isunchanged. However, these changes do affect Jp. Although the peak value of mdoes not change much (it is determined largely by Vfm), the value of h at the timeof the peak is reduced (in proportion to the change in its initial value from 0.97 to0.86). Therefore, the magnitude of the peak sodium current is smaller, and Jp ismore positive. Therefore, the magnitude of Jp is decreased and the magnitude ofJss does not change appreciably. The exact solution for these parameters is shownin Figure 4.5.

Problems

Problem 4.1

a. There are only three ions for which data are available so that a model of the mem-brane of the cell that allows conduction by all three ions is shown in Figure 4.6.


5 10 15 20 25

−0.3

−0.2

−0.1

0.1

t (ms)

Jion

(mA

/cm

2)

Figure 4.5: The ionic current is shown for thedefault parameters and for a change in poten-tial from −80 to −40 mV (dashed line) andwhen the initial potential is changed from −80to −70 mV (solid line) (Exercise 4.12).

GK

JK JCl

VK VCl+−

+−

+−

Jm

JC

Cm Vm

+

−

Membrane

Intracellular

Extracellular

GNa

JNa

VNa

GCl(Vm,t)

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 actionpotential, it is essential to determine the Nernst equilibrium potential for each ion.In general, the Nernst equilibrium potential is defined as

Vn = RTznF

lnconcin≈ 60zn

log10concin

mV.

The Nernst equilibrium potential for the three ions are evaluated as follows

VK ≈ 60 log100.0660

= −180 mV,

VNa ≈ 60 log100.1560

= −156 mV,

VCl ≈ −60 log100.05100

= 198 mV.

Since the resting potential is near the potassium equilibrium potential, we hypoth-esize that at restGK GNa andGK GCl. During the peak of the action potential,the membrane potential is near the chloride equilibrium potential. Therefore, wehypothesize that during the peak of the action potentialGCl GK andGCl GNa.However, the results given thus far cannot explain the undershoot of the actionpotential. Thus, either the Nernst equilibrium potential for potassium is not ac-curately known or some other ion is involved whose Nernst equilibrium potentialis less than −180 mV.

PROBLEMS 63

b. To test the model proposed in part a experimentally, increase coK and measureboth the resting potential and the peak of the action potential. The resting po-tential should increase by 60 mV per decade change in potassium concentrationand the action potential peak should remain constant. In addition, increasing coClshould not affect the resting potential of the cell, but should decrease the peakof the action potential by −60 mV per decade increase in extracellular chlorideconcentration.

Problem 4.2 In assessing the experimental results, assume that it is the change in potas-sium concentration that is primarily responsible for the change in resting potential. InFigure 4.2 (Weiss, 1996b) the maximum reduction in extracellular potassium concen-tration is to replace seawater with 33% seawater. Since the normal potassium concen-tration of seawater is about 10 mmol/L (Weiss, 1996a, Figure 7.1), the reduction is to3.33 mmol/L. In this range of extracellular potassium concentration, the resting po-tential is relatively insensitive to changes in potassium concentration. As indicated inFigure 7.22 (Weiss, 1996a) in this range of concentration the expected decrease in rest-ing 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 intracellularpotassium concentration is to replace half the potassium by sodium. Since the normalpotassium 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 rangeof concentration the expected increase in resting potential is less than 5 mV. This isconsistent 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 tothe capacitance at t = 0. Therefore, the change in potential is ∆Vm = Q/Cm sothat

Cm = Q∆Vm

= 10−8 C20× 10−3 V

= 0.5 µF.

b. The maximum value of the membrane potential during an action potential Vpmmust be less than the Nernst equilibrium potential for sodium, i.e., VNa > V

pm = 0.

Therefore,

VNa = RTF ln

(coNaciNa

)> 0.

Thus, coNa/ciNa > 1 which shows that coNa > c

iNa = 10 mmol/L.

Problem 4.4

a. Since the capacitance current precedes the ionic current onset, the ionic currentcan be ignored in the equivalent circuit shown in Figure 4.7. The total series resis-tance of a unit area of membrane is Rs , the capacitance per unit area of membraneis Cm, and the potential difference applied by the voltage-clamp circuit is V(t).


−

+V (t)

Rs

Cm

JC(t)

Figure 4.7: Equivalent circuit of the membrane for computingthe capacitance current under voltage clamp (Problem 4.4).

For a step of amplitude V , the current through the capacitance implied by thisnetwork is

JC(t) = VRse−t/τs ,

where τs = RsCm. The measurements (Weiss, 1996b, Figure 4.13) show that theinitial current density is about 4.2 mA/cm2 in response to a depolarization of 40mV. 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 thatthe capacitance current has a discontinuity at t = 0. The measurements show arapid rise in capacitance current but not a discontinuous rise. The calculationsshown with the measurements (Weiss, 1996b, Figure 4.13) take a number of fac-tors 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 in-stantaneously but has a finite rise time. In addition, the step response of therecording 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 equilibriumpotential is VNa = (RT/F) ln(coNa/c

iNa). The early component of the ionic current,

which is carried mainly by sodium ions, is inward for the lowest 2 curves, whichimplies that VNa > 28 mV for these curves. The early component is zero for thesecond curve from the top which implies that VNa = 28 mV for this curve. Theearly component is outward for the upper curve which implies that VNa < 28 mVfor this curve. Since the three upper curves differ only a little, they must have beenobtained for the concentrations 140, 150, 160 mmol/L. Putting these observationstogether 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 potentialat 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 concentra-tion of 150 mmol/L, the intracellular concentration of sodium can be determinedby the relation

28 mV = 59 log10150

ciNa,

PROBLEMS 65

−1 1 2 3 4 5

−1

1

2

Time (ms)

Ionic current(mA/cm2)

450

160

140

Figure 4.8: Ionic current in response to thesame voltage clamp but in solutions of differ-ent sodium concentration which is indicatedin mmol/L as a parameter (Problem 4.5). Thedashed curve corresponds to 150 mmol/L.

from which ciNa ≈ 50 mmol/L.

c. The membrane potential profile is the same for all 4 curves. Hence, the potassiumcurrent is the same for all 4 curves. Since the sodium current is zero for the curveobtained at 150 mmol/L, this curve equals the potassium current. Therefore, at150 mmol/L, JNa150 = 0 and JK150 = J150. Since the potassium current is the sameat all concentration, at 450 mmol/L JK450 = J150 and JNa450 = J450−J150. Plots ofthe ionic, sodium and potassium currents at extracellular sodium concentrationsof 150 and 450 mmol/L are shown in Figure 4.9.

d. In the Hodgkin-Huxley model, the conductances are independent of concentra-tion. Therefore, the sodium conductance can be computed from the results at 450mmol/L and the sodium current from the results at 50 mmol/L.

GNa = JNa450

28− 59 log10(450/50)= JNa450

−28= J450 − J150

−28,

from which the sodium current at 50 mmol/L is

JNa50 = J450 − J150

−28(28− 59 log10(50/50)) = −(J450 − J150) = −JNa450.

A sketch of the ionic, sodium, and potassium current at an extracellular concen-tration 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)F(Vm(t)− VNa) and φK(t) = GK(t)F (Vm(t)− VK).

The total quantity of sodium and potassium transferred through the membraneper cm2 of area and for each action potential,

∆nNa =∫∞

0

GNa(t)F

(Vm(t)− VNa)dt and ∆nK =∫∞

0

GK(t)F

(Vm(t)− VK)dt.


−1 1 2 3 4 5

−2

−1

1

2

3

−1 1 2 3 4 5

−2

−1

1

2

3

−1 1 2 3 4 5

−2

−1

1

2

3

JNa

JK

JNa

JK

Time (ms)

Jion

JK = Jion

Jion

Jion

(mA

/cm

2)

coNa = 450 mmol/L

coNa = 150 mmol/L

coNa = 50 mmol/L

Figure 4.9: Ionic (solid lines), sodium (dashedlines), and potassium (dotted lines) currents atextracellular sodium concentrations of 450, 150,and 50 mmol/L (Problem 4.5).

1 2 3 4

0

20

−20

−40

−40

−60

Time (ms)

0

10

20

30

40

VNa

Vm

(mV

)

mVVm

GNa

G(m

S/c

m2)

Figure 4.10: Method for estimating the transfer of sodium during an action potential (Prob-lem 4.6). The sodium conductance is approximated by a rectangular pulse (shown shaded) andthe 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 toGNa(t) andVm(t)makes the integration simple as shown in Figure 4.10. Approximatethe sodium conductance as a rectangular pulse of amplitude 25 mS/cm2 for an in-terval of 0.4 ms, and approximate Vm(t)− VNa ≈ −40 mV over the same interval.Therefore,

∆nNa ≈ (25× 10−3 S/cm2) · (−40× 10−3 V) · (0.4× 10−3 s)9.65× 104 C/mol

,

∆nNa ≈ −4 pmol/cm2.

The negative sign indicates that there is a net transfer of sodium into the cellduring each action potential.

c. Each centimeter length of axon contains a total quantity of sodium of niNa =ciNaπa2. In N action potentials the amount of sodium that enters this lengthof axon is ∆nNa2πaN. Therefore, to change the concentration by about 10%

∆nNa2πaN = ciNaπa2

10.

Therefore,

N = aciNa20∆nNa

,

N = (250× 10−4 cm) · (50× 10−3 mol/cm3)20(4× 10−12 mol/cm2)

,

N ≈ 1.5× 107.

So that about 15 million action potentials are required to raise the intracellularsodium concentration 10%. If the cell generated a high rate of action potentials of100/s, it would take

1.5× 107

100 · 60 · 60 · 24≈ 1.7 days

for the concentration to change 10% by this mechanism. This analysis has ignoredthe effect of active transport of sodium on the intracellular sodium concentration.

Problem 4.7 The potassium conductance is defined as

GK(Vm, t) = GKn4(Vm, t),

where

τn(Vm)dn(Vm, t)

dt+n(Vm, t) = n∞(Vm).

From Figure 4.25 (Weiss, 1996b) it is apparent that n∞(75) ≈ 1, n∞(−100) ≈ 0.02, andτn(−100) ≈ 5 ms. Therefore, n starts at a value of 1 and decreases exponentially to0.02 with a time constant of about 5 ms,

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,


0 1 2 3 4 5

10

20

30

40

t (ms)

GK

mS/cm

2

Figure 4.11: The potassium conductance versustime (Problem 4.7).

where t is in ms. Therefore, the potassium conductance is

GK(Vm, t) ≈ 36(0.02+ 0.98e−t/5

)4for t ≥ 0,

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 valueof 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),

whereGNa(Vm(t), t) = GNam3(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 threeintervals. The parameters need to be determined for these three intervals. FromFigure 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) ≈ 1ms. 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,

GNa(60, t) ≈ 120(1− e−t/0.1

)3e−t/1 mS/cm2.

During the interval 0 < t < 0.1, Vm(t) = VNa. Therefore, JNa(60, t) = 0. Atthe 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,

GNa(−100, t) ≈ 120(0.63e−(t−0.1)/0.03

)3 (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 0.01ms. 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 definedapproximately as

V1Na = 60 log10

(co1Na

ciNa

)and V2

Na = 60 log10

(co2Na

ciNa

),

where V1Na and V2

Na are the old and new values of VNa. Note

V2Na = 60 log10

(co1Na/10

ciNa

)= V1

Na − 60 log10 10 = 60− 60 = 0.

For 0 < t < 0.1

JNa(60, t) ≈ 120(1− e−t/0.1

)3e−t/1(60− 0)× 10−3 mA/cm2,

JNa(60, t) ≈ 7.2(1− e−t/0.1

)3e−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 atVm = −120 mV which setsh∞(−120) ≈ 1.The step to Vm = −60 mV starts a change in h. The step to Vm = −40 mV nowresults in a transient change GNa which is due to a change inm. The assumptionof the method is that the change in m is so rapid that m reaches its final valuebefore h has changed appreciably. The current density is

Jp(T) ≈ GNam3∞(−40)h(−60, T )(−40− VNa)× 10−3,


−0.1 0.1 0.2 0.3

−100

−50

50

−0.1 0 0.1 0.2 0.3

0.2

0.4

0.6

−0.1 0.1 0.2 0.3

0.9

0.95

−0.1 0 0.1 0.2 0.3

10

20

30

−0.1 0.1 0.2 0.3

−4

−2

2

4

−0.1 0.1 0.2 0.3

−2

−1

1

2

t (ms)

t (ms)

t (ms)

t (ms)

t (ms)

t (ms)

Vm(t) (mV)

m(t)

h(t)

GNa(t) (mS/cm2)

JNa(t) (mA/cm2)

JNa(t) (mA/cm2)

part a

part b

Figure 4.12: The sodium variables versus time (Problem 4.8).

PROBLEMS 71

10 20 30 40

0.2

0.4

0.6

0.8

1

T

y(T)

Figure 4.13: Method to estimate sodium inactiva-tion (Problem 4.9).

5 10 15

−0.3

−0.2

−0.1

t (ms) t (ms)

(mA

/cm

2)

5 10 15

−0.3

−0.2

−0.1

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 from0 to 12 ms in 1 ms increments.

andJp(0) ≈ GNam3∞(−40)h(−120,0)(−40− VNa)× 10−3.

Therefore,

y(T) = Jp(T)Jp(0)

= h(−60, T )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 ofh with the value of h. The above analysis was based on the sodium current alone. Inthe further analysis presented below, the method will be assessed both when only thesodium 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 bothJNa and JNa+JK exhibit a negative peak in inward current whose magnitude decreasesas T increases. To assess the method outlined above, the minima of the inward currentwere computed numerically and these minima were normalized by the minimum whenT = 0. These results are compared to h in Figure 4.15. The results show that at leastfor the voltage levels investigated, there are small but systematic errors in estimating


0 5 10 15 20

0.2

0.4

0.6

0.8

1

0 5 10 15 20

0.2

0.4

0.6

0.8

1

Time (ms)

Sodium only Sodium and potassium

Figure 4.15: A comparison of the minimum inward current versus duration of the pulse T (solidline) to h(T) for Vm = −60 mV (dashed line) (Problem 4.9). The value of Jp(T)/Jp(0) is plottedversus T .

h with this method. The maximum errors are larger when the potassium current isincluded. A more careful analysis would explore these results for a range of membranepotentials.

Problem 4.10 The membrane ionic current during the interval (0, tp) is assumed to becarried predominantly by sodium. Hence,

Jm(z, t) ≈ Cm∂Vm(z, t)∂t+ JNa(z, t).

From the core conductor model

Jm(z, t) = 12πa(ri + ro)ν2

∂2Vm(z, t)∂t2

.

Therefore,1

2πa(ri + ro)ν2

∂2Vm(z, t)∂t2

≈ Cm∂Vm(z, t)∂t+ JNa(z, t).

Integrate this equation over the interval (0, tp). Three terms need to be evaluated. Theintegral of the left-hand side is

12πa(ri + ro)ν2

((∂Vm(z, t)∂t

)tp−(∂Vm(z, t)∂t

)0

)= 0,

because the derivatives are zero at both 0 and tp. The integral of the first term on theleft-hand side is ∫ tp

0Cm∂Vm(z, t)∂t

dt = Cm(Vm(z, tp)− Vm(z,0)).

If these results are combined they yield

0 ≈ Cm(Vm(z, tp)− Vm(z,0))+∫ tp

0JNa(z, t)dt,

0 ≈ Cm(Vm(z, tp)− Vm(z,0))+ F∫ tp

0φNa(z, t)dt,

PROBLEMS 73

where φNa(z, t) is the outward flux of sodium. The term∫ tp0 φNa(z, t)dt represents

the total efflux of sodium during the rising phase of the action potential. The equationcan be solved for Cm as follows

Cm ≈ −F ∫ tp0 φNa(z, t)dtVm(z, tp)− Vm(z,0) ,

≈ (9.65× 104) · (1.5× 10−12)(50− (−70))× 10−3 ≈ 1.2 µF.

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. Further-more, JC(t) = 0 at the peak of the action potential. Therefore, the capacitance currentmust 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) < 0during 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) > 0during an action potential. Therefore, the potassium current density must be J1(t).

Problem 4.12 From the core conductor model

Jm(z, t) = 12πa(ri + ro)

∂2Vm(z, t)∂z2 .

Since the action potential is propagating at constant velocity ν ,

Jm(z, t) = 12πa(ri + ro)ν2

∂2Vm(z, t)∂t2

.

Therefore, Jm(z, t) = 0 when ∂2Vm(z, t)/∂t2 = 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 leadsto the answers shown in Table 4.1.

Problem 4.13


t0 t1 t2 t3 t4abc

√d

√ √ √e

√fghij

√k

√None

√

Table 4.1: Properties at different points in apropagated action potential (Problem 4.12).

a. The membrane potential at which the early, transient current reverses polarityis approximately equal to the Nernst equilibrium potential for sodium. These

potentials are Vfm = 54 mV in solution A and Vfm = 36 mV in solution B. Therefore,60 log(cANa/c

iNa) ≈ 54 and 60 log(cBNa/c

iNa) ≈ 36. Subtracting the second equation

from the first, yields 60 log(cANa/cBNa) ≈ 18 which implies that cANa/c

BNa ≈ 2.

b. The late, maintained current is carried by potassium. However, at low membranepotential, the potassium conductance is small and the reversal of the potassium

current cannot be discerned. However, at each value of Vfm, JAK > JBK which im-

plies that GK(Vfm − VAK ) > GK(Vfm − VBK) which shows that VAK < V

BK . Expressing

the Nernst potential in terms of concentrations shows that (RT/F) ln(coAK /ciK) <

(RT/F) ln(coBK /ciK) which implies that coAK < coBK .

Problem 4.14

a. Leakage current (vi). The leakage current is JL = GL(Vm − VL) where GL andVL are constants. Hence, the leakage current waveform has the same shape asthe membrane potential waveform. In addition, since GL is relatively small, theleakage current during an action potential is relatively small compared with thesodium and potassium currents.

b. Potassium current (iv). The potassium current is JK = GK(Vm, t)(Vm−VK). SinceGK ≥ 0 and Vm ≥ VK , JK ≥ 0 during an action potential. Also the potassiumcurrent is larger than the leakage current and the onset of the potassium currentoccurs later than Vm sincen has a relatively slow response to a change in potential.

c. Sodium current (iii). The sodium current is JNa = GNa(Vm, t)(Vm − VNa). SinceGNa ≥ 0 and Vm ≤ VNa, JNa ≤ 0 during an action potential. Also the sodium cur-rent is larger than the leakage current and the sodium current onset occurs rapidlyas 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 propor-tional to ∂2Vm/∂z2 which, for an action potential traveling at constant velocity,

PROBLEMS 75

is proportional to ∂2Vm/∂t2. The waveform looks like the shape of the secondderivative 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. Thewaveform looks like the negative of the first derivative of Vm. The waveform islarge and negative before the peak in the action potential, zero at the peak of theaction potential, and positive immediately after the peak of the action potential.However, the external longitudinal current is not a component of the membranecurrent.

f. Capacitance current (vii). The capacitance current is JC = CmdVm/dt. Hence,the capacitance current is proportional to the first derivative of the membranepotential 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 ofthe membrane is

C = Q∆Vm

= 30× 10−9

30× 10−3 = 1 µF.

The specific capacitance of the membrane is 1 µF/cm2, and the capacitance of theaxon is C = πdLCm where d and L are the area and length of the axon. Therefore,

L = CπdCm

= 1 µF

π · 0.05 cm · 1 µF/cm2 = 6.4 cm.

b. Replacing half the sodium in the sea water with impermeant ions, changes theNernst equilibrium potential for sodium,

∆VNa = RTF

ln

(coNa/2ciNa

)− RTF

ln

(coNaciNa

),

= RTF

ln(

12

)≈ −59 log10 2 = −18 mV.

The resting potential V1 is relatively insensitive to the sodium concentration so itwill change very little. Because of the reduction of the sodium equilibrium poten-tial, V1 will decrease slightly. The difference between V2 and V1 is determined bythe current pulse so that V2 will change by the same amount as V1, namely verylittle. V3 approaches the sodium equilibrium potential. Hence, V3 will decrease byabout −18 mV. V4 approaches the potassium equilibrium potential and will notchange.


Problem 4.16 In this problem, the Nernst equilibrium potential for ion n is assumed tobe

Vn = 59zn

log10

(concin

)(mV).

a. The voltage-clamp results show that after the initial pulse of capacitance currentwhich occurs near t = 0, there is an early outward current which is the sodiumcurrent. This implies that Vm − VNa > 0 so that VNa < 50 mV. Therefore,

VNa = 59 log10

(400

cin

)< 50 mV,

which implies that ciNa > 57 mmol/L. The current-clamp results reveal an actionpotential with a peak voltage of 30 mV. Therefore, VNa > 30 mV, and

VNa = 59 log10

(400

cin

)> 30 mV,

which implies that ciNa < 124 mmol/L. Putting these results together yields 57 <ciNa < 124 mmol/L.

b. The voltage-clamp results show that the late current component carried by potas-sium is outward which shows that VK < 50 mV. The current-clamp results are morerevealing. They show that the undershoot of the action potential which approachesthe potassium equilibrium potential reaches −70 mV. Therefore, VK < −70 mV.Therefore,

VK = 59 log10

(10

ciK

)< −70 mV,

which implies that ciK > 154 mmol/L.

Problem 4.17

a. The resting potential according to the Hodgkin-Huxley model will be

Vom =GKGm

VK + GNaGm VNa +GLGm

VL,

where the Nernst equilibrium potential at a potential of 6.3 is

Vn = RTF log10 e

log10concin≈ 55.7 log10

concin

mV,

and at the resting potentialGm = 0.374+0.0109+0.3 = 0.685 mS/cm2. Therefore,

Vom =0.3740.685

(−72.3)+ 0.01090.685

(55.3)+ 0.30.685

(−49) = −60.1 mV.

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 tothe occurrence of the pulse of membrane current. Since the ionic current is thedifference between the membrane current and the capacitance current, the ioniccurrent is also zero.

PROBLEMS 77

c. For t > 0.5 ms, Jm = 0. Hence, in this interval JC+Jion = 0. Therefore, JC = −Jionand, in particular, (JC)max = −(Jion)min.

d. Note that the maximum value of Vm is 44.2 mV and the minimum value is −71.2mV. In part a, it was found that the Nernst equilibrium potentials were +55.3 forsodium 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 whileVm > VL during other portions of the action potential. Therefore, JL is bothpositive and negative.

e. The space constant is

λC ≈ 1√gmri ,

when ro ri. Let a be the radius of the axon, then gm = 2πaGm and ri =ρi/(πa2) so that

λC =√

a2Gmρi

=√

0.0252 · 0.685× 10−3 · 100

= 0.43 cm.

f. The time constant of the membrane is

τM = CmGm =10−6

0.685× 10−3 = 1.46 ms.

Problem 4.18

a. The propagated action potential starts at the resting value with a slope of 0, i.e., atthe 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 = CmdVm/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 actionpotential).

c. During a propagated action potential, the core conductor model implies that themembrane current density is

Jm = 12πa(ri + ro)ν2

∂2Vm∂t2

.

Thus, the Jm = 0 when ∂2Vm/∂t2 = 0 which is zero when ∂Vm/∂t has a maximumor 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


0 40−40−80−100

0

100

200

300

Vm (mV)

t1

t2

t3

t4

t5

dVm/d

t(m

V/m

s)

Figure 4.16: Phase plane plot showing phasetrajectory during a propagated action potential(Problem 4.18). Arrows show the direction forincreasing time.

simple. In addition, approximate the Nernst potential as 60 mV per decade of concen-tration for a univalent ion. Therefore,

VK ≈ 60zK

log

(coKciK

)= 60 log

(16

160

)= −60 mV,

VNa ≈ 60zNa

log

(coNaciNa

)= 60 log

(40040

)= +60 mV,

VCl ≈ 60zCl

log

(coClciCl

)= −60 log

(100100

)= 0 mV,

VCa ≈ 60zCa

log

(coCaciCa

)= 30 log

(202

)= +30 mV.

Since the resting potential of the cell equals the Nernst equilibrium potential for potas-sium, a model consist with these results is that at rest the membrane is permeable onlyto potassium ions. That is, at rest the conductance of the membrane to ions other thanpotassium is negligible.

The investigators differ in the results they obtained on the effect of calcium concen-tration on the peak value of the action potential. Consider the model of investigator Afirst. First note that

VCa ≈ 60zCa

log

(coCaciCa

)= 30 log

(coCaciCa

),

so that the Nernst equilibrium potential for calcium changes by 30 mV/decade of cal-cium concentration. The data of investigator A is consistent with Vp = VCa. Hence, in-vestigator A would be likely to conclude that the action potential was due to a transientincrease in the calcium conductance of the membrane. Thus, the model of investigatorA (Figure 4.17) consists of two parallel branches — a potassium branch, that predom-inates at rest, and a calcium branch, that predominates during the peak of the actionpotential.

PROBLEMS 79

+−

+−

+

−

Membrane

Intracellular

Extracellular

Vm

GK GCa

VK VCa

Figure 4.17: Investigator A’s membranemodel (Problem 4.19). At rest GK GCa.During the peak of the action potentialGK GCa.

+−

+−

+−

+

−

Membrane

Intracellular

Extracellular

Vm

+−

+−

+

−

Vm

GK

VK

GNa

VNa

GK

VK

GNa

VNa

GCl

VCl

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 theaction potential both GNa and GCl are appreciable.

Since the peak of the action potential is independent of calcium concentration ininvestigator B’s data, investigator B would be led to conclude that the calcium conduc-tance was small during the peak of the action potential. Investigator B could proposeseveral simple models to account for her data. Note that the action potential normallyreaches +30 mV. Only two ions have positive Nernst potentials — calcium, which is notinvolved in investigator B’s data, and sodium. Therefore, it is clear that the membraneconductance 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 permeableto potassium and/or to chloride during the peak of the action potential. These modelsare shown in Figure 4.18.

It is important to keep in mind that investigators A and B can propose models thatcan account for their data, either investigator A or B (or perhaps both) must be wrong be-cause of the fundamental disagreement in their measurements. The discrepancy mustbe due to differences in experimental techniques (ruling out such possibilities as typo-graphical errors, fraud, etc.).


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) = 12πaν2(ri + ro)

∂2Vm(z, t)∂t2

.

Therefore,

Pm(zo, t) = 12πaν2(ri + ro)Vm(zo, t)

∂2Vm(z, t)∂t2

.

Therefore, power flows into the membrane during any interval of time for whichPm(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

EK(zo) =∫ tbtaPK(zo, t)dt,

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 andPK(zo, t) ≤ 0 when Vm(zo, t) ≤ 0. Therefore, during intervals τ0 and τ1 there is anet flow of energy out of the potassium branch.

c. The net flow of energy density into the sodium branch is

ENa(zo) =∫ tbtaPNa(zo, t)dt,

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) ≤ 0and 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 81

d. 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 storingelectrostatic energy. During this interval the capacitance is not supplyingenergy.

ii. Since the ionic conductances are non-zero, they cannot supply energy.

iii. “Metabolic storehouses” are not included in the model. Hence, they cannotsupply energy in this model.

iv. The power into the potassium battery is

VKJK(zo, t) = VKGK(zo, t)(Vm(zo, t)− VK).Since VK < 0, GK(zo, t) ≥ 0, and Vm(zo, t) − VK ≥ 0, the power flowing intoVK is always negative. That is, VK always delivers energy. The power into thesodium battery is

VNaJNa(zo, t) = VNaGNa(zo, t)(Vm(zo, t)− VNa).Since VNa > 0, GNa(zo, t) ≥ 0, and Vm(zo, t) − VNa ≤ 0, the power flowinginto VNa is always negative. That is, VNa always delivers energy.

v. Since the batteries represent the chemical potential energy stored in concen-tration differences across the membrane, the answers are the same as in partiv.

vi. There are no “external sources” in the model. Hence, they cannot supplyenergy in this model.

vii. During interval τ2 no energy is supplied by the capacitance or by the potas-sium branch. Energy is supplied by the sodium branch. The leakage branchneeds 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 andVm(zo, t) > VL or if Vm(zo, t) > 0 and Vm(zo, t) < VL. Since VL < 0, the sec-ond case is not possible. This leaves the only possibility asVL < Vm(zo, t) < 0.But this condition does not occur during interval τ2. Hence, energy is sup-plied 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 inthe phase-plane trajectory shown in Figure 4.102 (Weiss, 1996b). Starting at a currentof 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)dt

+ 0.2Im(t) = 0,


−

+

Im(t)

Rs

Cm V (t)Figure 4.19: Model for capacitance current with series resis-tance (Problem 4.21).

which has a solution of the form Im(t) = Ae−0.2t for t > 0 where t is in µs. Since thetrajectory starts at 10 mA at t = 0, the solution is Im(t) = 10e−0.2t for t > 0, whereIm(t) is expressed in mA. Thus, Im(t) decays exponentially with a time constant of 5µs and represents the capacitance current at the onset of the transition in membranepotential. This is followed by a negative current component, presumed to be carriedby sodium ions, which is followed by a positive component, presumed to be carried bypotassium 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 mustbe less than VNa. Therefore, VNa > 50 mV.

c. Since VNa > 50 mV and VNa = (RT/F) ln(coNa/ciNa) then to reduce VNa to 50 mV

(while keeping ciNa constant) requires decreasing coNa.

d. The phase-plane trajectory shows that the capacitance current is over before theionic current changes, i.e., the linear portion of the trajectory intersects the originat which both Im(t) = 0 and dIm(t)/dt = 0. Therefore, a simple model of thisportion of the trajectory is shown in Figure 4.19. The capacitance Cm is the totalcapacitance of the membrane of the axon under voltage clamp, and the resistanceRs is the total resistance in series with the membrane and the external voltagesource. It includes the resistance of the sea water and the electrodes. The currentproduced by a step of potential of amplitude V is obtained as follows. If the voltageacross the capacitance does not change instantaneously then the total voltage willappear across the resistance to produce and initial current V/Rs . This currentwill decay exponential to zero with time constant τs = RsCm. The phase-planetrajectory indicates that in response to a step of 100 mV, the initial current is 10mA. Therefore, Rs = 100/10 = 10 Ω. Earlier it was determined that the timeconstant 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. SinceVm = 0, and since VNa = 0 because coNa/c

iNa = 1, it follows that 1.5 = −GKVK 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 densityis the same. Therefore,

Jm = 5(0− 60)× 10−3 + 1.5 = 1.2 mA/cm2.

PROBLEMS 83

5

−5

(mV)

50 100

−50

V fm

Jion(t) (mA/cm2)

JpJ∗p

Figure 4.20: Current-voltage relationsof the peak sodium current (Prob-lem 4.23). The solid line is a plot ofthe peak value of the ionic current Jpand the dashed line is the instanta-neous value J∗p both plotted against

the membrane potential Vfm.

Problem 4.23

a. Note that the initial hyperpolarization of −80 mV is provided to increase the theearly sodium current (by making h∞ ≈ 1 initially) and to reduce the initial potas-sium current (by making n∞ ≈ 0 initially). This increases the separation of sodiumfrom potassium currents. Therefore, it is assumed that the early component ofthe ionic current is due primarily to sodium ions. More specifically, assume that

Jp is the peak sodium current. Therefore, Jp should reverse its sign at Vfm = VNa.

Figure 4.104 (Weiss, 1996b) shows that Jp reverses its polarity at Vfm = 45 mV forboth solutions A and B. Therefore, the sodium concentrations of solutions A andB must be approximately equal.

b. In the Hodgkin-Huxley model, the conductances do not change instantaneously.

Therefore, the locus of J∗p versus Vfm must be a straight line. Therefore, it is onlynecessary to find two points through which the straight line passes. Assume that

J∗p is due to sodium ions only. Then J∗p = 0 when Vfm = VNa = 45 mV. In addition,

for Vfm = −30 mV it must be that J∗p = Jp as shown in Figure 4.20. The slope of thisline is approximately equal to the value of the sodium conductance at a membranepotential of −30 mV at the time of occurrence of the second step which is from

−30 mV to Vfm.

Problem 4.24

a. The current density isJ(V) = G(V)(V − V0).

Therefore, G(V) is multiplied by the factor V − V0 which is a straight line whenplotted versus V . The results are shown in Figure 4.21. Note that with a constantconductance, the J-V characteristic is a straight line whose intercept depends upon


−100 −50 50 100

−4

−2

2

4

−100 −50 50 100

−2

2

4

−100 −50 50 100

2

4

J (mA/cm2)

V (mV)

−60 mV

0 mV

60 mV

−60 mV

0 mV

60 mV

−60 mV

0 mV

60 mV

Figure 4.21: Current-voltage relations forthree different conductance-voltage charac-teristics (Problem 4.24) The parameter shownto the right of each trace is V0. The upperpanel shows the results for a constant conduc-tance 30 mS/cm2. The middle panel showsthe results for conductance that is discontin-uous with a value of 15 mS/cm2 for V < 0and a value of 30 mS/cm2 for V > 0. Thelower panel is for a conductance that is 0for V < −30 mV, is 30 mS/cm2 for V > 30mV, 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 piecewisestraight line. However, with the saturating conductance, the J-V characteristic hasa negative slope conductance region that depends upon the bias voltage V0.

b. The measurements show that both the potassium and sodium conductances satu-rate as the membrane potential increases (Weiss, 1996b, Figure 4.27). In addition,the measurements show that the peak sodium current has a negative slope con-ductance region (Weiss, 1996b, Figure 4.16). The results of part a show that anegative slope conductance region can be obtained with a conductance, whosevoltage dependence saturates as a function of membrane potential, in series withan appropriate bias potential.

Problem 4.25

a. Sodium. The electrical responses of the giant axon of the squid can be accountedfor by four distinct currents: one due to flow of sodium ions through the cellmembrane, one due to flow of potassium ions through the cell membrane, onedue to flow of current through the membrane capacitance, and one due to leakagecurrents. Since the capacitance and leakage currents have been subtracted, themeasured membrane currents must consist of sodium and/or potassium. Sincethere is no potassium in the solution bathing the outside of the cell, the transientinward current must be carried by sodium ions.

b. Outward currents would normally result from outward flow of sodium or potas-sium ions. However, all of the intracellular cations have been replaced by cesium,to which the membrane is relatively impermeant. Thus there is very little outwardcurrent.

Problem 4.26

a. Assume that steady-state is reached and that no action potentials are producedby this current. Assume that the capacitance current is zero at steady state andthat if the depolarization produced by the current is sufficient then sodium willbe inactivated. Also ignore the leakage current. Then the total membrane currentis carried by potassium to yield

Jm = GKn4∞(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. Examinationof the dependence of the activation/inactivation factors on Vm (Weiss, 1996b, Fig-ure 4.25) reveals that at this potential h∞ ≈ 0 and n∞ ≈ 0.95. Taking these valuesinto account, estimate Vm again as 4 ≈ 36 · 0.954(Vm + 72)× 10−3 which impliesthat Vm ≈ 64 mV. This calculation could be iterated further. However, it is clearthat this more accurate estimate of Vm results in a value of n∞ that is closer to 1and a value of h∞ that is closer to zero. Thus, depolarizing the membrane with anoutward current can reduce h sufficiently to inactivate the sodium conductance.


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 yieldh∞ ≈ 0.006. Although the depolarization caused by the current pulse,

∆Vm = (100× 10−6 A/cm2)(0.5× 10−3 s)1× 10−6 = 0.05 V = 50 mV,

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 pulsem rises exponentiallyfrom 0 to 1 with a time constant of about 0.23 ms. At the offset of the pulse mfalls 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 0to 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 from0.03 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, andτn(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 S-shaped onset to a value of about 16 mS/cm2 with a time constant of the underlyingfactor n of 1.7 ms. Thus, the transient response will not be completed at the offsetof the 4 ms pulse. After the offset, the conductance declines exponentially to ≈ 0with 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, andτh(10) ≈ 1 ms. Therefore, at the onset of the pulse h falls exponentially from1 to 0 with a time constant of about 1 ms. At the offset of the pulse h risesexponentially from 0 to 1 with a time constant of 2 ms. Therefore, the sodiumconductance has an S-shaped onset to a peak value of roughly 120(1)3e−0.8/1 =54 mS/cm2. The conductance then declines exponentially to approximately 0. Atthe 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 twoterms. During the pulse Vm − VNa ≈ 10− 55 = −45 mV. Thus, at the peak of theconductance, 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 offsetthe direction of current flow will reverse.

PROBLEMS 87

10 2 3

2

0

4

6

t (ms)

JN

a(V

m,t

)(m

A/cm

2)

Figure 4.22: A plot of the sodium current density(Problem 4.29).

Problem 4.28

a. Ans. is (3). The action potential shows a slower depolarization at the onset anda slower repolarization. This is consistent with an increase in the membrane ca-pacitance.

b. Ans. is (4). The peak action potential goes to a higher potential which is consistentwith an increase in the Nernst equilibrium potential for sodium.

c. Ans. is (1). The resting potential is decreased and the undershoot of the actionpotential has been reduced implying that the resting potential is closer to thepotassium equilibrium potential. This is consistent with a reduction in the leakageconductance.

d. Ans. is (2). The action potential shows a more rapid depolarization at the onsetand 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) = GNam3(Vm, t)(Vm − VNa),

wherem(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 currentthat does not inactivate. This is distinctly different from the sodium current fornormal sodium concentration which is an inward current and inactivates.

Chapter 5

SALTATORY CONDUCTION INMYELINATED NERVE FIBERS

Exercises

Exercise 5.1 In the cable model, the current per unit length and the membrane potentialare related by

Km(z, t) = cm∂Vm(z, t)∂t+ gm(Vm(z, t)− Vom).

During any time interval for which ∂Vm(z, t)/∂t > 0 and Vm(z, t) > Vom, this relationimplies that Km(z, t) > 0. Examination of Figure 5.23 (Weiss, 1996b) reveals that at thenode of Ranvier, the membrane current is positive then negative during the onset ofthe membrane potential, an interval in which the cable model predicts a positive (out-ward) current only. Thus, the relation of membrane current to membrane potential atthe node of Ranvier is inconsistent with the cable model. In contrast, at the internodethe membrane current is positive (outward) during the onset of the membrane poten-tial as predicted by the cable model. These results are consistent with the model thataction potentials are initiated at the nodes of Ranvier and propagate in the internodesaccording to the cable model.

Exercise 5.2 An inward current occurs through the membrane at a node of Ranvier dur-ing the initiation of an action potential. This inward current flows outward through theinternodal 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 ofthe internode and the other when an action potential is initiated at the other side of theinternode.

Exercise 5.3 It is simplest to estimate the conduction velocity from measurements ofthe delay of the longitudinal current versus time (Weiss, 1996b, Figure 5.20). Sincethe measurements near the end of the nerve fiber show some nonuniformities in thelongitudinal currents, it is best to estimate the conduction velocity from the centralportion of the fiber. The delay between 1 and 7 mm is 0.83− 0.6 = 0.23 ms so that theconduction 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 fiberdiameter whereas the conduction velocity of an unmyelinated nerve fiber is pro-portional to the square root of fiber diameter. Thus, the relation between the twoconduction velocities for the same fiber diameter is complex. However, for largediameter fibers it is true that for the same diameter fiber, the conduction velocityof myelinated nerve fibers exceeds that of unmyelinated nerve fibers.

b. False. Measurements show that the action potential at a node of Ranvier doesnot differ appreciably from that at a neighboring internode. Hence, the actionpotential cannot be said to hop from node to node.

c. True. The action current at the node of Ranvier differs radically from that at aninternode. The action current is large and has an inward component at the nodeand is smaller and has only an outward component at the internode. Thus, theinward 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 produc-ing saltatory conduction. However, the fact that the nodes of Ranvier contain amuch higher density of voltage-gated sodium channels is also critical. Thus, theinsulatimg 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 inex-citable nodes. The number of nodes past which an action potential can propagate iscalled the safety factor.

Exercise 5.6 Figures 5.9 and 5.10 shows vertebrate myelinated axons with diametersbetween 4 and 17 µm. Generally, vertebrate myelinated axons have diameters from 1 to20 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 positionoccur at the nodes of Ranvier. These cusps can be interpreted by noting that the coreconductor model yields the equation

∂2Vm(z, t)∂z2 = (ri + ro)Km(z, t).

Hence, integrating this equation over distance from a to b, a region that includes a node,gives

∂Vm(z, t)∂z

∣∣∣∣z=b− ∂Vm(z, t)

∂z

∣∣∣∣z=a

= (ri + ro)∫ baKm(z, t)dz.

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 anappreciable value of the integral of the membrane current per unit length even thoughthe width of the node (b − a) is small.

PROBLEMS 91

Exercise 5.8 The large effect on conduction of the action potential of narcotics appliedat the nodes of Ranvier, which occupy only 0.02% of the length of a nerve fiber, occursbecause the action potential is initiated at these nodes. If the action potential is blockedat a few nodes, propagation may still occur beyond this block, however, the conductionvelocity is reduced.

Exercise 5.9 Measurements (Weiss, 1996b, Figure 5.9) show that d/D = 0.74 so thatthe 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 ofsodium channels in a node is Nn

Nn = π11.1 · 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 = π11.1 · 2000 · 25 = 1.7× 106 channels.

c. While the number of sodium channels in the internode is larger than at the node ofRanvier, the density of sodium channels is 40 times larger in the node of Ranvierthan in the internode. It is the density of sodium channels that must reach athreshold 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 thatat t = 0.37 ms dVm(t)/dt = 0, and hence that KC = cmidVm(t)/dt = 0 wherecmi is the membrane capacitance per unit length of the internode. Therefore, atthe maximum of the membrane potential, the capacitance current is zero and thetotal membrane current is through the conductance of the membrane. Therefore,Km(0.37) = Kg(0.37) = gmi(Vm(0.37) − Vom) where Kg is the current per unitlength through the membrane conductance and gmi is the membrane conductanceper unit length of the internode. From these relations, solve for gmi as follows

gmi = Km(0.37)Vm(0.37)− Vom =

8× 10−9 A/cm(50− (−50))× 10−3 V

= 80 nS/cm.

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)dt

≈ 100 mV/0.12 ms = 833 V/s,

Vm(t) ≈ 0, and Km(t) ≈ 16 nA/cm. Therefore,

Kc(0.24) = Km(0.24)− gmi(Vm(0.24)− Vom) = cmidVmdt

∣∣∣∣t=0.24

,


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 = ρiπ(d/2)2

= 110 Ω· cmπ · (5× 10−4 cm)2

= 140 MΩ/cm,

where d/2 is the radius of the axoplasm in the axon. The space constant is

λCi = 1√riigmi =1√

140× 106 Ω/cm · 80× 10−9 S/cm= 0.3 cm = 3 mm.

c. The thickness of the membrane of the internode of a myelinated nerve fiber isappreciably larger than the thickness of the membrane of an unmyelinated nervefiber. Therefore, to determine the specific membrane conductance and capacitanceof the internodal membrane, we use the average radius of the membrane of theinternode in the following, i.e., a = (d/2 +D/2)/2. With this approximation, thespecific conductances and capacitances are

Gmi = gmi2πa

= 80× 10−9 S/cm2π · 6× 10−4 cm

= 21× 10−6 S/cm2,

Cmi = cmi2πa

= 14.4× 10−12 F/cm2π · 6× 10−4 cm

= 3.8× 10−9 F/cm2.

d. For each Schwann cell membrane, the change in membrane potential from its rest-ing value can be represented by an equivalent electric network of a square cen-timeter of membrane that consists of a specific conductance Gm in parallel witha specific capacitance Cm. Each lamella consists of two Schwann cell membraneswith little intervening cytoplasm. Thus, if the internode myelin has n lamellae,then there are 2n layers of Schwann cell membrane in each internode. As indi-cated in Figure 5.1, the membranes in myelin are stacked in series because thesame membrane current flowing from inside the fiber to the extracellular spaceflows through all the membranes. It is assumed that there is no appreciable cur-rent flowing in the longitudinal paths between lamellae because these paths musthave a very high resistance. If it is assumed that the contribution of the axonalmembrane, which is in series with the myelin, is negligible then the capacitanceper 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

Cmi Gmi

Cm Gm

Cm Gm

2n 2n= =

2n

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 istypically seen for cell membranes and a value of the conductance that is about afactor of 10 less than that seen for most cells (Weiss, 1996b, Table 3.1). This resultmay well result because the membranes of Schwann cells contain a low density ofion channels.

e. The time constant is

τM = CmGm =1.15× 10−6 F/cm2

6.37× 10−3 S/cm2 = 0.18 ms,

which is the same as τMi found in part b. The space constant is

λC = 1√gmrii =1√

2π(d/2) ·Gm · ρi/(π(d/2)2)

=√

a2Gmρi

=√√√ 5× 10−4 cm

2 · 6.3× 10−3 S/cm2 · 110 Ω·cm= 0.19 mm.

Thus, an unmyelinated axon with this membrane material and dimensions wouldhave a space constant that is more than a factor of 15 smaller than that of themyelinated axon of the same dimensions.

Since the time constants are equal, and the space constant is larger for the myeli-nated axon, its potential changes will spread over a larger distance — in the in-ternodal regions which make up most of the length of the fiber — than occurs inthe unmyelinated axon. This analysis applies when the cable model is valid. Thecable model applies directly in the internode of a myelinated axon and applies tothe unmyelinated axon at the onset of the action potential.

Problem 5.2


0.8

−0.8

0

5.0

−5.0

0

0 1 2Time (ms)

Toad nodeof Ranvier

Squid giantaxon

JNa

JNa

J(m

A/c

m2)

Figure 5.2: Triangular-wave approximations toionic currents.(Problem 5.2).

a. Action potentials propagate along a squid axon with a constant velocity. Therefore,the transmembrane sodium current density JNa also propagates with a constantvelocity. Hence, the transmembrane sodium current densities at two differentlocations are identical functions of time except for a time shift. This propertyimplies that the number of moles of sodium that enter a squid axon during asingle propagated action potential is the same in each increment of the axon’slength. Therefore, the number of moles of sodium per unit area of membranethat enter the axon multiplied by the circumference of the axon yields the totalnumber 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 sodiumcurrent by Faraday’s constant F . Thus,

N = πdF

∫∞0JNa(t)dt,

whered is the diameter of the fiber. The integral of the current density is estimatedby approximating the current density waveform as a triangle that is 1 ms wide and0.8 mA/cm2 high (Figure 5.2). Then,

N ≈ 500π µm96,500 C/mol

× 12

(0.8

mAcm2

)(1 ms) ≈ 0.6

pmolcm

.

To estimate the the number of moles transported in myelinated nerve fibers, as-sume that sodium currents are only important at the nodes of Ranvier. If l repre-sents the length of a node and L represents the length of the internode, only thenodal fraction l/L of the length is important, and

N = lLπdF

∫∞0JNa(t)dt.

The integral of the current density is estimated by approximating the current den-sity waveform as a triangle that is 1.2 ms wide and 7 mA/cm2 high (Figure 5.2).Then,

N ≈ 0.7 µm2 mm

× 10π µm96,500 C/mol

× 12

(7

mAcm2

)(1.2 ms) ≈ 0.005

fmolcm

.

PROBLEMS 95

b. The ratio of ATP used to pump sodium out of the unmyelinated to that used topump sodium out of the myelinated nerve fiber is equal to the ratio of sodiumions for the two cases,

ratio = 0.6 pmol/cm0.005 fmol/cm

≈ 1.2× 105.

c. Measurements show that a small myelinated nerve fiber (10 µm diameter) canconduct action potentials as quickly as a large (500 µm diameter) unmyelinatednerve fiber. However, the energy required in the myelinated nerve fiber is fiveorders 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 actionpotential is at node 4 at 0.6 ms and at node 12 at 1.2 ms. Since nodes are spaced1 mm apart, the conduction velocity is

ν = 12− 41.2− 0.6

≈ 13 mm/ms.

Nodes 0 and 14 were avoided in estimating the conduction velocity because thecomputations 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, toppanel 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 thehorizontal axis. The cusps in Vm(z, t) plotted versus z that occur at the nodesof Ranvier, do not occur in Vm(z, t) plotted versus t. This demonstrates thatVm(z, t) 6= f(t − z/ν).

c. A myelinated nerve fiber is inhomogeneous; the nodal and internodal myelinatedmembrane do not have the same electrical characteristics. In particular, the cuspsare caused by the large inward membrane current that occurs only at the nodes ofRanvier. This causes a discontinuity in the longitudinal current at the nodes whichis proportional to a discontinuity in ∂Vm(z, t)/∂z according to the core conductormodel. A more formal argument follows from the core conductor model whichyields

∂2Vm(z, t)∂z2 = (ri + ro)Km(z, t).

Integration of this expression over a node of Ranvier yields

∂Vm(z, t)∂z

∣∣∣∣z=b− ∂Vm(z, t)

∂z

∣∣∣∣z=a

= (ri + ro)∫ baKm(z, t)dz.

A large membrane current at the nodes makes the right-hand term appreciableeven when a and b are close together. Therefore, the difference in slope of thepotential is appreciable and gives rise to the cusps.


+

−

+

−

RnCn

va(t)

vb(t)

riL

roL

Figure 5.3: Equivalent electric network of twonodes with the intervening internode (Prob-lem 5.4).

d. The lower panel of Figure 5.31 (Weiss, 1996b) shows that the onset of the calcu-lated action potential contains a point of inflection between−20 and−30 mV whenrecorded at the point of stimulation at node 0. Therefore, the threshold for elicit-ing 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 issimultaneously above −30 mV for about 10 nodes. Therefore, when an action po-tential 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 anabsence of a large inward current, the potential plotted versus position will notcontain cusps but will otherwise change very little. Therefore, nodes beyond theinexcitable nodes can be stimulated to produce action potentials. Assume that theinexcitable 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). Considerthe spatial waveform at the time 0.45 ms. The maximum membrane potential isat node 6. Clear cusps are seen at nodes 5 and 7 indicating that there is a largeinward current at these nodes. However, at nodes 8, 9, and 10, no cusps are seenindicating that the inward current at these nodes is small. Thus, these nodes arenot yet excitable. If they remained inexcitable, then even in the absence of aninward current at these nodes, the membrane potential would be appreciable atsubsequent nodes as the membrane potential profile moves to the right. Thus, theaction 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 ofinternode has an inside resistance of riL and an outside resistance of roL as shownin Figure 5.3.

b. If va(t) = Vau(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 modelwhen vb(t) = Vth which occurs at time t = T as shown in Figure 5.4. The value ofT can be computed as follows

Vth = Va(1− e−T/τ),

from which

T = −τ ln(

1− VthVa

)≈ −Cn(Rn + Lri) ln

(1− Vth

Va

).

Note that T is a positive quantity since 0 ≤ Vth/Va ≤ 1, and therefore, ln(1 −Vth/Va) ≤ 0.

PROBLEMS 97

vb(t)

Va

T t

Vth

Figure 5.4: Step response of simple model of prop-agation in a myelinated nerve fiber (Problem 5.4).

0 1 2 3

νmax

LriRn

νmax

2

ν

Figure 5.5: Dependence of conduction velocity oninternodal length for a simple model of propaga-tion in a myelinated nerve fiber (Problem 5.4).

c. The conduction velocity is

ν = LT= νmax

(L(ri/Rn)

L(ri/Rn)+ 1

),

where

νmax = −1

CnLri ln(1− Vth

Va

) .The conduction velocity is plotted versus the internodal length in Figure 5.5. WhenL(ri/Rn) 1 then ν = νmaxL(ri/Rn), i.e., the conduction velocity is propor-tional to the internodal length. Under these condition the nodal resistance Rn ismuch larger than the longitudinal resistance of the internode riL. Therefore, thetime constant τ ≈ RnCn is independent of L, and the time delay for triggering anaction potential is constant independent of the internodal length. Under these con-ditions, the time delay for the action potential to travel a distance L is constant sothat the conduction velocity is proportional to L. Conversely, when L(ri/Rn) 1then ν = νmax , i.e., the conduction velocity is independent of internodal length.Under these conditions the longitudinal resistance of the internode riL is muchlarger than the nodal resistance Rn. Therefore, the time constant τ ≈ LriCn isproportional to L. Therefore, the delay for travelling a distance L is proportionalto 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 ismuch smaller than the space constant of the internode λC , the internode can berepresented with a lumped-parameter network. However, when L is much largerthan the space constant of the internode, then the current through the myelin


sheath cannot be ignored and the internode needs to be treated as a cable. Underthese conditions there is appreciable attenuation of the membrane potential in aninternode. 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 membranepotential at one point in space is

Km(t) = gmV(t)+ cm∂V(t)∂t(in the internode),

Km(t) = 1(ri + ro)ν2

∂2V(t)∂t2

(in the node).

Therefore, the subsequent parts of the problem can be solved by evaluating the deriva-tives of V(t) which are shown in Table 5.1 and plotted in Figure 5.6.

Time interval V(t)∂V(t)∂t

∂2V(t)∂t2

0 ≤ t < 0.2 1250t2 2500t 25000.2 ≤ t < 0.4 50+ 500(t − 0.2)− 1250(t − 0.2)2 500− 2500(t − 0.2) −25000.4 ≤ t < 2.0 100− 25(t − 0.4)2 −50(t − 0.4) −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 oftwo terms one equal to gmV and the other to cm∂V/∂t. The individual currentsthrough the membrane conductance and capacitance and the total membrane cur-rent are shown in Figure 5.7. The magnitude of the capacitance current exceedsthat of the current through the conductance. Note that during the onset of theaction 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 = 1riν2

∂2V(t)∂t2

Note that during the onset of the action potential the current is first outward andthen inward. This inward current component is the signature for the membranecurrent during the onset of an action potential.

c. The answer depends upon the space constant defined as λC = 1/√(ri + ro)gm

which for ri ro is

λC ≈ 1√rigm =1√

(2.5× 106 Ω/cm)(10−7 S/cm)= 2 cm.

Thus, the space constant is twice the internodal length. Therefore, in the firstfew millimeters from the active node, the internodal membrane potential will bevery similar to that at the node. Further away from the active node, the internodalpotential will have a smaller magnitude and a slower time course.

PROBLEMS 99

100

0

−100

500

0.5 1 1.5 2

−2000

2000

t (ms)

V(t

)(m

V)

∂V

(t)

∂t

(mV

/ms)

∂2V

(t)

∂t2

(mV

/ms2

)

Figure 5.6: Membrane potential and itsfirst two time derivatives (Problem 5.5).

0 0.5 1 1.5 2

2

4

6

8

10

t (ms)

Km

(t)

(nA

/cm

)

0.5 1 1.5 2

20

40

0.5 1 1.5 2

20

40

60

cm∂V

∂t

gmV

gmV + cm∂V

∂t

Figure 5.7: The membrane current components inan internode (Problem 5.5).


0.5 1 1.5 2

−2000

2000

t (ms)

Km

(t)

(nA

/cm

)Figure 5.8: The membrane current at a node ofRanvier (Problem 5.5).

0 2 4 6 8z/ν (ms)

50

0

-50

Mem

bra

ne

pot

enti

al

(mV

)

Figure 5.9: The action potential of an un-myelinated nerve fiber plotted versus po-sition (Problem 5.6).

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 theform Vm(z, t) = f(t − z/ν). Figure 5.46 (Weiss, 1996b) shows a plot ofVm(z, t) versus t. Therefore, a plot of Vm(z, t) versus z looks similar ex-cept for a reflection of the waveform about the vertical axis and a change inthe horizontal scale as shown in Figure 5.9.

ii. The duration for which the action potential has a positive potential is esti-mated to be about 0.5 ms (Weiss, 1996b, Figure 5.46). Hence, since the actionpotential is travelling at 2 m/s or 2 mm/ms, the spatial extent of the actionpotential 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 isabout 0.3 ms. Each internode is 1.38 mm long. Hence, the total distancebetween these nodes is 4× 1.38 = 5.52 mm. Hence, the conduction velocityν = 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). Theshape of Vm(z, t) as a function of z contains cusps at the nodes while theshape 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 vari-ables will be periodic in z. That is, the membrane variables will be the sameexcept for a delay at corresponding points along the fiber, i.e.,

Vm(z, t) = f(t − z/ν) for z = nL,

PROBLEMS 101

D

d

rr + dr

Figure 5.10: Geometry for computing the resistance and ca-pacitance per unit length (Problem 5.8).

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 nodeswhich 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 be-cause action potentials at a node produce inward current at the node that flows outwardthrough the internode. Action potentials at the nodes on either side of an internode eachproduce 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 tothe right node. Therefore, the outward current caused by an action potential atthe 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 occurslater than the action potential at the right node. Therefore, the action potential ispropagating in the −z-direction.

b. If the action potential were propagating in the +z direction it should arrive at theleft node before it arrives at the right node. The outward current peak should belargest at the nearest node. At the center of the internode, the outward currentfrom 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 betweenthe peaks is about 0.1 ms in trace 3. Since the nodes are separated by 1 mm, theconduction velocity is about 10 mm/ms.

Problem 5.8 In general, a resistance is related to the resistivity by the relation R = ρL/Awhere ρ is the resistivity, L is the length over which the current flows, andA is the cross-sectional area. The geometry for calculating the resistance per unit length of membraneis shown in Figure 5.10. For the cylinder, the current flows in the radial direction andthe cross-sectional area depends upon the radius. Hence, the resistance per unit lengthof an incremental cylindrical shell of radius r is

drm = ρmdr2πr.


This is integrated from d/2 to D/2 to yield

rm =∫D/2d/2

ρm2πr

dr = ρm2π

ln(D/d).

The capacitance of a parallel plate capacitance has the form C = εA/d where ε isthe permittivity, A is the surface area of the plates of the capacitance, and d is theseparation of the plates of the capacitance. Therefore, the same incremental cylindricalshell shown in Figure 5.10 has a capacitance

dcm = ε2πrdr.

Since reciprocal capacitances in series add, we integrate the reciprocal capacitance asfollows

1cm=∫D/2d/2

12πεr

dr = 12πε

ln(D/d),

so that

cm = 2πεln(D/d)

.

Note that the membrane time constant of the myelin in an internode is

τMi = rmcm = ρm2πln(D/d)

2πεln(D/d)

= ρmε,

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

λC = 1√gmri =√rmri,

where rm = 1/gm. In Problem 5.8 it was shown that

rm = ρm2πln(D/d).

Therefore,

λC =√(ρm/2π) ln(D/d)ρi/(π(d/2)2)

=√ρm8ρid2 ln(D/d). =

√√√√ρmD2

8ρi

(dD

)2

ln(D/d),

which yields

λC = K√y2 ln(1/y),

where

K =√ρmD2

8ρi,

and y = d/D. λC/K has a maximum value at d/D ≈ 0.6 (Figure 5.11). The

PROBLEMS 103

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

√ y2ln

(1/y)

y

Figure 5.11: The dependence of the space constant ony = d/D (Problem 5.11).

d/D ¿ 1 d/D ≈ 1

Figure 5.12: Schematic diagrams of two fibers of the sameouter diameter D with different myelin thickness (Prob-lem 5.11).

maximum value of λC is found by taking its derivative with respect to y . Notethat the square root is a monotonic function of its argument so only the argumentneeds to be maximized as follows

ddy(y2 ln(1/y)) = 0.

The derivative yields

y + 2y lny = 0 which implies that y(1+ 2 lny) = 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 diameterD in two limits, d/D 1and d/D ≈ 1 as shown in Figure 5.12. The space constant depends on the ratioof 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 forlongitudinal current is small. The dependence of ri on d is ri ∝ d−2. The ratio ofthese resistances will approach zero as d→ 0.

When d/D ≈ 1, rm ≈ 0 because the myelin thickness approaches zero (assumingthatD is fixed). When d/D ≈ 1, ri decreases to a fixed value which is the longitudi-nal resistance of a cylinder of diameterD. Therefore, the ratio of these resistanceswill approach zero as d/D ≈ 1.

These arguments show that the space constant is zero both when d = 0 and whend = D. Since the space constant is always positive, the dependence of the spaceconstant on d/Dmust have a maximum value between the two limits d/D = 0 andd/D = 1.

c. The conduction velocity depends upon the time it takes to charge a node. Increas-ing the space constant will result in more rapid charging of a node. This argumentsuggests that increasing the space constant will result in increasing the conduc-tion velocity. Although this is not a proof, maximizing the space constant appearsto maximize the conduction velocity.


4 5 6 7

110

90

100

Node Number

Vm

(z,t

o)−

Vo m

(mV

)to = 0.45

Figure 5.13: Method for estimating the discontinuity ofthe slope of the membrane potential at a cusp (Prob-lem 5.10).

d. Figure 5.9 (Weiss, 1996b) shows that measurements of d/D for myelinated nervefibers have an average value of 0.74. In part b it was found that the value of d/Dthat maximizes the space constant in the internode is 0.61. Thus, the value of d/Dthat is found in myelinated fibers is within 18% of the value the theoretical estimatepredicts will maximize the space constant. Furthermore, as shown in Figure 5.11the maximum is fairly broad and the value of the space constant for d/D of 0.74differs by only 5% from the maximum obtained for d/D = 0.61. Since a maximumin the space constant also leads to a maximum in the conduction velocity, thevalue of d/D that is found experimentally corresponds to a value that leads to amaximum in the conduction velocity.

Problem 5.10

a. The core-conductor equations can be used to determine the relation between mem-brane potential Vm(z, t) and extracellular longitudinal current Io(z, t). For a prop-agating action potential, there are no internal or external electrodes and the core-conductor equations reduce to

∂Vm(z, t)∂z

= ∂Vi(z, t)∂z

− ∂Vo(z, t)∂z

= −riIi(z, t)+ roIo(z, t) = (ri + ro)Io(z, t).

Therefore, the external longitudinal current is proportional to the derivative ofmembrane potential with respect to space. The derivatives of membrane potentialcan be estimated just to the left and just to the right of node 6 (Figure 5.13 as

∂Vm(z, t)∂z

∣∣∣∣node 6−≈ 109 mV− 96 mV

2 internodes× 1.38 mm/internode≈ 4.7 V/m,

and

∂Vm(z, t)∂z

∣∣∣∣node 6+≈ 99 mV− 109 mV

1 internode× 1.38 mm/internode≈ −7.2 V/m.

These are proportional to the external longitudinal current

Io(z, t)|node 6− ≈1

ri + ro∂Vm(z, t)∂z

≈ 4.7 V/m140 MΩ/cm

≈ 0.3 nA and

Io(z, t)|node 6+ ≈1

ri + ro∂Vm(z, t)∂z

≈ −7.2 V/m140 MΩ/cm

≈ −0.5 nA,

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 relationbetween membrane current and membrane potential

∂2Vm(z, t)∂z2 = (ro + ri)Km(z, t).

Therefore, Km(z, t) is proportional to the second derivative of membrane poten-tial with respect to distance. From the data in Figure 5.13, the relation betweenmembrane potential and distance is concave up: i.e., the second derivative of mem-brane potential is positive. It follows that the sign of Km at the internode betweennodes 5 and 6 is positive — and the current flow is outward.

Chapter 6

VOLTAGE-GATED ION CHANNELS

Exercises

Exercise 6.1 Conduction current results from charge carriers (e.g., electrons in a metalor ions in an electrolyte) that are relatively free to move. Thus, the application of anelectric field causes a flow of these charge carriers that constitutes a conduction cur-rent. In the presence of polarizable matter, displacement current results from chargecarriers (e.g., protons and electrons in an electric insulator or fixed charged groups ina macromolecule) that are not free to move in a medium. The application of an electricfield causes microscopic separations of charge or alignment of dipoles. The time rateof 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 con-duction, charge groups associated with these channels move. The time rate of changeof 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 ionchannels caused by a change in membrane potential.

Exercise 6.3 The single-channel ionic current random variable is the current through asingle channel as a function of time. This current is a random rectangular wave that hastwo values — 0 when the channel is closed (non-conducting) and I when the channel isopen (conducting). Transitions between these two states occur randomly with rates oftransitions determined by rate constants that may depend upon the membrane poten-tial. The single-open-channel current is the current through a single channel when it isopen. It is designated by I . The average single-channel current is the ensemble averagecurrent of a population of identical, statistically independent channels. In response toa step voltage clamp, this current shows exponential transitions between its initial andfinal 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 anyion that can pass through the channel including potassium

b. True.

107

108 CHAPTER 6. VOLTAGE-GATED ION CHANNELS

V

+

−

+

−

+

−

Vmn

Vmp

Figure 6.1: Equivalent circuit of a patch of mem-brane attached to a cell (Exercise 6.7).

c. False. See part b.

d. False. Gating currents give information about charge movements in the membranebetween any states — conducting and non-conducting states — whereas ionic cur-rent give information about the conducting states only.

Exercise 6.5 Trace 1 shows a single-channel current with two states of conduction: onecurrent is zero and the other is negative. The negative current represents ion flow whenthe channel is open. The magnitude of that current is not changed by the step changein membrane potential. This is inconsistent with the assumption that the open-channelvoltage-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 changein membrane potential. This is expected if the open-channel voltage-current relationis linear. From this short segment of data, one cannot conclude that the probabilitythat the channel is open has or has not changed during the step change in membranepotential. Therefore, the current in trace 2 could be from a voltage-gated ion channelor 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 mem-brane in the outward direction and negative charges in the inward direction. Both ofthese result in an outward current.

Exercise 6.7 An equivalent network of the arrangement for recording from a patch ofmembrane 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 goesfrom −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 activationof the potassium conductance of squid giant axon is not exponential — it has an S-shaped onset.

Exercise 6.9 The model of a sodium channel that consists of four, independent two-state 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 themeasured time constant is close to τm(Vm).

Exercise 6.10 Water channels (Weiss, 1996a) have the following properties: (1) theytransport water, (2) water transport is driven by the difference of hydraulic and osmoticpressure across the membrane, (3) they are ungated and are always open, (4) they areblocked by mercury compounds. Ion channels have the following properties: (1) theytransport ions, (2) ion transport is driven by differences in electrochemical potentialdifference 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 statesof 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 thatthe gate opens when the membrane is depolarized.

b. Since x∞ = 0.5 at Vm = V1/2 −−40 mV. The effective valence of the gating chargecan be obtained from the slope of the logarithm of x∞ plotted versus Vm. Notethat

x∞ = 11+ e−z(F/RT)(Vm−V1/2)

= ez(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 = RTF

slopelog10 e

≈ 602/60

log10 e= 4.6.


−120 −80 −40 0 40

100

−120 −80 −40 0 40

10−1

10−2

10−3

100

10−1

10−2

10−3

Membrane potential,

x∞

(

ms)

τ x

Vm (mV)

Figure 6.2: Estimation of parameters of a two-state gate from the voltage dependence of thesteady-state state occupancy and the time constant (Problem 6.1).

The asymmetry factor ζ is estimated from the dependence of τx on Vm. Theasymptotic values of log10 τx are

limVm→−∞

log10 τx = − log10A− (zF/RT)(ζ − 0.5)(Vm − VO) log10 e,

limVm→∞

log10 τx = − log10A− (zF/RT)(ζ + 0.5)(Vm − VC) log10 e.

Thus, the slopes estimated from Figure 6.2 yields the ratio of the slopes

1/80−2/100

= −58= ζ − 0.5ζ + 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(∞) = zeNx(∞) = zeNx∞.

The two state gate model shows that

x∞ = 11+ e−z(F/RT)(Vm−V1/2)

,

so that

Q∞ = zeN1+ e−z(F/RT)(Vm−V1/2)

.

Since limVm→∞ x∞ = 1, limVm→∞Q∞ = Qmax = zeN . Therefore,

Q∞ = Qmax1+ e−z(F/RT)(Vm−V1/2)

,

PROBLEMS 111

and

Q∞Qmax −Q∞ =

Qmax1+e−z(F/RT)(Vm−V1/2)

Qmax − Qmax1+e−z(F/RT)(Vm−V1/2)

,

= ez(F/RT)(Vm−V1/2)

Note that F/(RT) = e/(kT) where e is the magnitude of the electronic charge andk is Boltzmann’s constant. If this substitution is made and a logarithm is taken ofboth sides of the equation this yields

kT ln

(Q∞

Qmax −Q∞

)= ze(Vm − V1/2).

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. Dividingthis by e yields the energy per unit electronic charge or the energy in electronvolts. 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 meV131 mV

= 1.3e.

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 equalsthe thickness of the membrane) is

1γ= 1

20× 10−12 S= ρdπa2 =

(102 Ω·cm) · (75× 10−8 cm)πa2 .

Therefore, the radius of such a pore is

a =√

102 · 75× 10−8 · 20× 10−12

π= 2.2× 10−8 cm = 2.2 A.

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 theseequations is satisfied if Vn = 50 mV and γ = 20 pS. Alternatively, the singleopen-channel current can be plotted versus the membrane potential as shown inFigure 6.3. The voltage-current characteristic is linear.


J

J

J

J

J

J

−1

400−40−80

−2

−3

Vm (mV)

I(p

A)

Figure 6.3: Single open-channel current versus themembrane potential (Problem 6.4).

b. The conductance can be computed from the slope of the line relating the singleopen-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 relat-ing the single open-channel current to the membrane potential (Figure 6.3) whichyields Vn = 50 mV.

d. If these records were from a sodium channel, then the reversal potential shouldbe equal to the sodium equilibrium potential and the probability that the chan-nel is open should transiently increase following a step depolarization and thendecrease. The reversal potential of 50 mV is close to the sodium equilibrium poten-tial. Furthermore, from the single channel records, it appears that the probabilitythat the channel is open is smaller when the final voltage is −100 mV than whenthe final voltage is larger. However, there is no evidence that the probability isdecreasing with time. Thus, there is no evidence for inactivation, and it appearsunlikely 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 thatx∞(−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

0 0.5 1 1.5

10

20

30

40

t (ms)

g(t

)(p

S)

Figure 6.4: Average conductance (Problem 6.5).

0 0.5 1 1.5

10

20

30

40

t (ms)

g(t

)(p

S)

Figure 6.5: Average conductance (Problem 6.5).

d. This is an activation gate because the channel opens to increase current flow whenthe 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 thatthe channel is in S1 needs to be determined. Since the channel must be in one ofits 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,


g. This is an inactivation gate because the channel closes to reduce current flow whenthe 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.


+20

−60

Mem

bra

ne

pot

enti

al(m

V)

0 1 2 3 4 5 6 7 8 9 10Time (ms)

0

−10

Curr

ent

den

sity

(mA

/cm

2 )

J

1

0

0.5

Sta

teoc

cupan

cypro

bab

ilit

yx

1−x

−20

−30

Figure 6.6: Solution to Prob-lem 6.6.

b. The single open-channel current is

I = γ(Vm − Ve),

whereγ is the single open-channel conductance, andVe is the equilibrium potentialof the channel. I is given for two values of Vm which yields the simultaneousequations

−2× 10−12 = γ(−60− Ve)× 10−3,−0.4× 10−12 = 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 inFigure 6.6. But,

x∞(20) = 0.7 = α(20)α(20)+ β(20)

and τx(20) = 0.001 = 1α(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 themacroscopic current density from a population of identical channels. From suchmeasurementsN = J(t)/(Ix(t)).

PROBLEMS 115

f. The macroscopic ionic current density due to this population of channels is

J(Vm, t) = Nγx(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) = −20mA/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∞ = αα+ β =

11+ e(EO−EC)/(kT) ,

τx = 1α+ β =

1A(e(EC−EB)/(kT) + 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) = 11+ 0.75

= 0.57.

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(∞) = 11+ 1.5

= 0.4.

c. The information given shows that A = 3000/s. Therefore,

τx(−40) = 13000

(e−1.1 + e−0.693

) = 13000(0.33+ 0.5)

= 0.4 ms.

d. The results determined in parts a-c yield

x(t) = 0.4−(0.4− 0.57e−t/0.4

)for t ≥ 0,


e. This gate closes for a depolarization. Therefore, this direction is consistent withthe idea that x represents the sodium inactivation factor h. However, the timecourse of x(t) is about an order of magnitude too fast to represent h(t).

Problem 6.8


0 0.5 1 1.5 2

0.2

0.4

0.6

x(t

)

t (ms)

Figure 6.7: The probability that the channel isopen (Problem 6.7).

a. To find the density of channels, note that the maximum macroscopic conductanceper unit area when all channels are open equals the density of channels times thesingle open-channel conductance, i.e.,

limVfm→∞

G(Vfm,∞) =Nγ.

The maximum macroscopic conductance is 80 mS/cm2 and the single channel con-

ductance can be gotten from the slope of the I-Vfm characteristic and is 12 pA/120 mV =100 pS. Therefore, the density of channels is 80×10−3/10−10 channels/cm2 whichis 8 channels/µm2.

b. The single open-channel current reverses sign at Vfm = 20 mV so that this is theNernst equilibrium potential for the channel. Therefore,

VX = 20 = 59 log10

(150

ciX

),

which yields ciX = 69 mmol/L.

c. The voltage dependence of the macroscopic conductance can be used to determinethe gating charge valence. Note that

G(Vfm,∞) = 80

1+ e−3(Vfm−20)/26,

so that the gating charge valence is z = 3.

d. The gating current is

Jg(Vm, t) = zeN dx(Vm, t)dt

.

Hence, find x(Vm, t) first. At Vfm = −80, x∞(−80) ≈ 0. At Vfm = +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)dt

= 0.5e−tu(t) (1/ms).

PROBLEMS 117

0 1 2 3 4

0.05

0.1

0.15

0.2

t (ms)

Jg

(µA

/cm

2)

Figure 6.8: Gating current (Problem 6.8).

Therefore, the gating current is

Jg(Vm, t) = 3 · 1.6× 10−19 C · 8× 108 channels/cm2 · 0.5e−tu(t) (1/ms),= 0.19e−tu(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

VNa ≈ 60 log10

(20020

)= 60 mV.

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

INazNaF

= −2.4× 10−12

9.65× 104 . ≈ −2.5× 10−17 mol/s

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 trans-ported inward.

Problem 6.10 It is helpful to determine all the Nernst equilibrium potentials and driv-ing voltages at the outset (Table 6.1). Table 6.1 shows that the current through openchannels is: outward for the potassium channels, inward for sodium and calcium chan-nels, 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. Inaddition, the kinetics are first-order for chloride and magnesium and higher-order forthe other channels.


Ion Vn Vm − Vn (mV) Current direction Kinetic(mV) Vm = −60 Vm = 0 Vm = −60 Vm = 0 order

K+ −90 +30 +90 outward outward higherNa+ +60 −120 −60 inward inward higherCl− −30 −30 +30 inward outward firstCa++ +100 −160 −100 inward inward higherMg++ 0 −60 0 inward — first

Table 6.1: Nernst equilibrium potentials, differences of membrane potentials from Nernst equi-librium potentials for permeant ions, current direction, and order of kinetics (Problem 6.10).

Channel Ji1 Ji2 Ji3 Ji4 Ji5Chloride

√Calcium

√Sodium

√Potassium

√Magnesium

√None of the above

√

Table 6.2: Macroscopic mem-brane ionic current density (Prob-lem 6.10).

a. The results are summarized in Table 6.2. Since there are no inward currents withfirst-order kinetics, Ji1 is “none of the above.” Ji2 is inward at −60 mV and is zeroat 0 mV which implies that this is magnesium. Ji3 is outward and shows first-order kinetics which means it must be chloride. Ji4 is outward and shows higher-order kinetics which implies it is potassium. Ji5 is inward and shows higher-orderkinetics which implies it could be either calcium or sodium.

b. The results are summarized in Table 6.3. Gating currents are outward for a depo-larization 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 ki-netics. Hence, this could be calcium, sodium, or potassium.

c. The results are summarized in Table 6.4. ii1 is outward and the single open-channel current at Vm = 0 is three times its magnitude at Vm = −60 mV whichimplies that this is potassium. ii2 is inward which implies this is either calciumor sodium. There is no channel for which the current switches from outward toinward as does ii3. ii4 switches from inward to outward which implies that it is

Channel Jg1 Jg2 Jg3 Jg4 Jg5

ChlorideCalcium

√Sodium

√Potassium

√MagnesiumNone of the above

√ √ √ √

Table 6.3: Macroscopic mem-brane gating current density(Problem 6.10).

PROBLEMS 119

Channel ii1 ii2 ii3 ii4 ii5Chloride

√Calcium

√Sodium

√Potassium

√Magnesium


√

Table 6.4: Single-channel ionic cur-rents (Problem 6.10).

Channel Ii1 Ii2 Ii3 Ii4 Ii5ChlorideCalciumSodium

√Potassium

√Magnesium


√ √

Table 6.5: Single-open-channelmembrane ionic currents (Prob-lem 6.10).

the chloride current. ii5 switches from inward to zero which implies that it is themagnesium current.

d. The results are summarized in Table 6.5. The I-Vm characteristics cross the Vmaxis 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 andstate occupancy probability.

i. To determine if the channel is voltage gated determine if the probability thatthe channel is open changes with a change in membrane potential. For chan-nel A at Vm = −60 mV

−1× 10−12 = 25× 10−12(−60− 20)× 10−3x(−60,∞),

so that x(−60,∞) = 0.5. For Vm = −20 mV,

−0.5× 10−12 = 25× 10−12(−20− 20)× 10−3x(−20,∞),

so that x(−20,∞) = 0.5. Therefore, there has been no change in the steady-state value of the probability that the channel is open. Hence, channel Ashows no signs of being voltage-gated.


For channel B at Vm = −60 mV

−0.5× 10−12 = 25× 10−12(−60− 20)× 10−3x(−60,∞),so that x(−60,∞) = 0.25. For Vm = −20 mV,

−0.75× 10−12 = 25× 10−12(−20− 20)× 10−3x(−20,∞),so that x(−20,∞) = 0.75. Therefore, this channel is voltage-gated and theprobability that it is open increases with depolarization.

For channel C at Vm = 10 mV

−0.2× 10−12 = 25× 10−12(10− 20)× 10−3x(10,∞),so that x(10,∞) = 0.8. For Vm = 80 mV,

0.3× 10−12 = 25× 10−12(80− 20)× 10−3x(80,∞),so that x(80,∞) = 0.2. Therefore, this channel is voltage-gated and the prob-ability that it is open decreases with depolarization.

ii. Channel A is not voltage gated, channel B is voltage-gated and is activatedby the depolarization, channel C is voltage-gated and is inactivated by thedepolarization.

b. We express all single-channel currents in units of picoamperes. Channel A is notvoltage gated and the probability that it is open is 0.5. The single-open chan-nel current I changes instantaneously from −2 to −1 as the membrane potentialchanges. Therefore, the average single-channel current i also changes instanta-neously as follows

i(t) =−2× 0.5 for t < 0−1× 0.5 for t ≥ 0,


Channel B is voltage gated and the probability that the channel is open increasesexponentially from 0.25 to 0.75. The single open-channel current I changes in-stantaneously from −2 to −1 pA as the membrane potential changes. Hence,

i(t) =−2× 0.25 for t < 0−1× (0.25+ 0.5

(1− e−t/τ)) for t ≥ 0,


Channel C is voltage gated and the probability that the channel is open decreasesexponentially from 0.8 to 0.2. The single open-channel current I changes instan-taneously from −0.25 to 1.5 pA as the membrane potential changes. Hence,

i(t) =−0.25× 0.8 for t < 01.5× (0.8− 0.6

(1− e−t/τ)) for t ≥ 0,


PROBLEMS 121

0 1 2 3 4

0.2

0.4

0.6

0.8

1

0 1 2 3 4

0.2

0.4

0.6

0.8

1

0 1 2 3 4

0.2

0.4

0.6

0.8

1

1 2 3 4

−2

−1

1

2

1 2 3 4

−2

−1

1

2

1 2 3 4

−2

−1

1

2

1 2 3 4

−1

−0.5

0.5

1

1.5

1 2 3 4

−1

−0.5

0.5

1

1.5

1 2 3 4

−1

−0.5

0.5

1

1.5

Channel A Channel B Channel C

x(V

m,t

)I

pA

i(Vm,t

)pA

t/τ t/τ t/τ

t/τ t/τ t/τ

t/τ t/τ t/τ

Figure 6.9: Probability that channel is open, the single open-channel current, and the averagesingle channel current (Problem 6.11).


0.1 0.2 0.3 0.4 0.5t (ms)

0.1

0.2

0.3

0.4

Jg(t

)(m

A/cm

2)

Figure 6.10: Method for estimating the time constant ofthe gating current (Problem 6.12).

Problem 6.12

a. The maximum value of GNa,peak is about 95 mS/cm2 at 150 mV. If all the sodiumchannels are open at this potential then the single open-channel conductance canbe used to estimate the number of sodium channels. The single open-channel con-ductance can be computed from the slope of the relation of I to Vm and is γ =0.2 pA/(50 mV) = 4 pS. Therefore, ifGNa,peak =Nγ orN = (95 mS/cm2)/(4 pS) =238 channels/µm2.

b. The total gating charge displaced is the integral of the gating current, so that

(Qg)max =∫∞

0Jg(t)dt.

Figure 6.10 shows that the gating current density decays exponentially with a timeconstant 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,

(Qg)max =∫∞

00.4e−t/τ dt,

= 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, thedensity of gates isN = 4× 10−16/9.6× 10−19 = 417 channels/µm2.

c. The plot of bound STX shows that

1n= 0.0025+ 0.0025

0.051c,

which can be rewritten in the form

n = 400cc + 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 ofbound channels is 400 channels/µm2.

PROBLEMS 123

0 0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

t (ms)

m3 m3h

h m3h

m3h

−60 −100

Figure 6.11: The quantitym3h 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 potentialof +150 mV.

d. Therefore, we have the following estimates ofN in channels/µm2: 238 from parta; 417 from part b; and 400 from part c. Thus, the methods in parts b and c givesimilar estimates, but the method in part a gives a lower estimate of the channeldensity. However, the method in part a would be expected to underestimate thechannel density. Note, that because of inactivation, GNa,peak < GNa and it is GNathat represents the conductance of the sodium channels when they are all open.Recall thatGNa = GNam3h. Hence, the quantitym3h can be examined in responseto a voltage clamp pulse to estimate the error (Figure 6.11). The error occursbecause the peak value of m3h is less than 1 so that GNa,peak < GNa. The peakvalue of m3h depends upon the holding potential. When the holding potential isnear the resting potential (−60 mV), the initial value of h ≈ 0.6 and the peak valueofm3h ≈ 0.5. However, when the holding potential is hyperpolarized (−100 mV),the initial of h ≈ 1 and the peak value of m3h ≈ 0.75. Thus, the minimum errorincurred by ignoring inactivation occurs when a hyperpolarizing holding potentialis used. Thus, a better estimate of N ≈ 238/0.75 = 317 channels/µm2 which iscloser to the other two estimates of channel density. If the experiment is donewith a holding potential at the resting potential, then the the estimate is N ≈238/0.5 = 476 channels/µm2. Intermediate holding potentials yield intermediateestimates of channel density.

Problem 6.13

a. The average macroscopic ionic current density is

J(Vm, t) =N i(Vm, t) =Nγ(Vm − VNa)m3(Vm, t),

where N is the density of channels in the membrane, i is the average single-channel current, γ is the single open-channel conductance, andm is the probabil-ity that a gate is in the open conformation. Since m satisfies first-order kinetics,

dm(Vm, t)dt

= α(Vm)(1−m(Vm, t))− β(Vm)m(Vm, t),



dm(Vm, t)dt

+ (α(Vm)+ β(Vm))m(Vm, t) = α(Vm).

Therefore, m is an exponential function of time that depends on

m∞ = α(Vm)α(Vm)+ β(Vm) and τm = 1

α(Vm)+ β(Vm).

If the rate constants are evaluated as a function of membrane potential, it is ap-parent that as the membrane potential changes from −80 to +80 and back to −80mV, α(Vm) changes approximately from 0 to 11 to 0 /ms, and β(Vm) changesapproximately from 18 to 0 to 18 /ms. Therefore, for this same range of poten-tial, 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−104)/56)u(t − 104),

where t is in µs. Because the transients essentially come to completion at eachtransition (Figure 6.12), the solution can also be written as

m(Vm, t) ≈

1− e−t/91 for 0 ≤ t ≤ 104 µs,e−(t−104)/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

J(Vm, t) ≈ 12

(1− e−t/91

)3for 0 ≤ t ≤ 104 µs,

−52(e−(t−104)/56

)3for t ≥ 104 µs.

Note that for t ≥ 104 µs the ionic current can be written as

J(Vm, t) ≈ −52e−3(t−104)/56 for t ≥ 104 µs,

where the time constant at turn off is 56/3 = 18.7 µs. The ionic current is shownin 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 gatingcharge 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 peropen gate, andm is the probability that the gate is open. Substitution of numericalvalues 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

2 4 6 8 10

2 4 6 8 10

2 4 6 8 10

0.2

0.4

0.6

0.8

1

−20

−10

10

−0.3

−0.2

−0.1

0.1

0.2

t (ms)

t (ms)

t (ms)

J(V

m,t

)(m

A/c

m2)

Jg(V

m,t

)(m

A/c

m2)

m(V

m,t

)

Figure 6.12: Time course ofm(Vm, t), J(Vm, t) andJg(Vm, t) in response to a pulse of membrane po-tential (Problem 6.13).

0 200 400 600 800

5

10200 400 600 800

−50−40−30−20−10

10

200 400 600 800

−0.3

−0.2

−0.1

0 200 400 600 800

0.1

0.2

t (µs)

t (µs)

t (µs)

t (µs)

J(V

m,t

)(m

A/c

m2)

Jg(V

m,t

)(m

A/c

m2)

J(V

m,t

)(m

A/c

m2)

Jg(V

m,t

)(m

A/c

m2)

Onset Offset

Figure 6.13: Time course of J(Vm, t) and Jg(Vm, t) in response to a pulse of membrane potentialshowing response at the onset and offset of the membrane potential (Problem 6.13). The arrowsmark the time of offset of the membrane potential in the right panels and time is referred tothis offset in these panels.


βn

4αn 3αn 2αn

2βn 3βn

αn

4βn

C4 C3 C2 C1 O

Figure 6.14: Kinetic diagram for the Hodgkin-Huxley model of the potassium channel in termsof 4 independent, two-state gates (Problem 6.14).

The average macroscopic gating current density is the derivative of the averagemacroscopic gating charge density

Jg(Vm, t) = 19.2dm(Vm, t)

dtnA/cm2,

so that

Jg(Vm, t) ≈

19.2

ddt(1− e−t/91) nA/cm2 for 0 ≤ t ≤ 104 µs,

19.2ddt(e−(t−104)/56) nA/cm2 for t ≥ 104 µs,

where t is in µs and yields

Jg(Vm, t) ≈

0.21e−t/91 mA/cm2 for 0 ≤ t ≤ 104 µs,−0.34e−(t−104)/56 mA/cm2 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 anunderlying time constant of m of 91 µs and a final value of 12 mA/cm2. Duringthis interval the average macroscopic gating current density Jg is a decaying ex-ponential with a time constant of 91 µs. At the offset of the potential both J andJg are exponential but their time constants differ by a factor of 3.

Problem 6.14 If each two-state gate has the kinetic diagram

CgαnzβnOg,

and all four gates are independent then the states can be enumerate as shown in Fig-ure 6.14. State Cn is defined as having n gates closed, state O has all 4 gates open. Startwith state C4 with all 4 gates closed. Since the rate of opening of a gate is αn then therate at which state C3 is reached is 4αn since any one of the 4 independent gates canopen to cause a transition to state C3. Similarly, the rate of transition from state C3 toC4 is the rate at which the one open gate closes which is βn. The state diagram shownin Figure 6.14 is constructed by extending this analysis to all the states.

PROBLEMS 127

0.8

0.6

0.4

0.2

2 4 6 8

10

20

30

40

50

−2

−4

−6

Time (ms)

Vm(t)

−75

−25

m(t)

GNa(t)

JNa(t)

τm= 0.46τm= 0.1

τI= 0.1/3

Figure 6.15: Plots of m(t), GNa(t), and JNa(t)(Problem 6.15).

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 con-stant (0.46 ms) for the increase in m. Thus, m essentially reaches its final valueof 0.73. Therefore,

m(t) =

0.73(1− et/0.46) for 0 ≤ t ≤ 5,0.73e(t−5)/0.1 for t ≥ 5,

where t is in ms. This solution is shown plotted in Figure 6.15.

b. Since h is assumed to be 1, GNa = GNam3. Therefore, since 120(0.73)3 ≈ 47,

GNa(t) =

47(1− et/0.45)3 for 0 ≤ t ≤ 5,47e3(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 =


−25− 55 = −80 mV and for t > 5 Vm − VNa = −75− 55 = −130 mV, so that

JNa(t) = −3.8(1− et/0.45)3 for 0 ≤ t < 5,−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 or33 µs.

e. Measurements of the offset time constant (Weiss, 1996b, Figure 6.59) suggest thatτI ≈ τm whereas the Hodgkin-Huxley model predicts that τI = τm/3. Thus, theHodgkin-Huxley model does not correctly predict the offset time constant of thetail current. This failure of the model implies that the kinetic scheme shown inFigure 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 likew3(t).

b. Only S3 in channel c is conducting. Therefore, gc(t) = 10x3(t) which looks likew2(t).

c. Only S1 in channel a has a non-zero gating charge. Therefore, qa(t) = −x1(t) andiga = −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) andiga = dx1(t)/dt which looks like w5(t).

e. Channel b shows activation followed by inactivation. As indicated in part a, theconductance increases as the channel state progresses from S1 to S2 and decreasesas the channel state progresses from S2 to S3.

f. Channel c exhibits no inactivation. As shown in part b, the channel shows anS-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, thestate occupancy probabilities satisfy the equilibrium equations

dx1(Vm, t)dt

= 0 = −x1(Vm,∞)+ x2(Vm,∞),1 = x1(Vm,∞)+ x2(Vm,∞).

Therefore, x1(Vm,∞) = x2(Vm,∞) = 1/2. When all the rate constants are the same, thechannel is equally likely to occupy any of its states at equilibrium. Therefore, for 3 statesx1(Vm,∞) = x2(Vm,∞) = x3(Vm,∞) = 1/3 and for 4 states x1(Vm,∞) = x2(Vm,∞) =x3(Vm,∞) = x4(Vm,∞) = 1/4.

PROBLEMS 129

2αm

βm

m

m

h

m

h h

m

m

m

m m

mm

m

h h hαm

2βm

2αm

βm

αm

2βm

αh βh αh βh αh βh

m

Figure 6.16: Kinetic diagramfor a channel whose kinet-ics has the form m2h (Prob-lem 6.18).

Problem 6.18 Let the rate constants for the m-gate be αm and βm and those for theh-gate beαh and βh. Then for the channel to be open requires 2m-gates and 1 h-gate tobe open. With 3 gates there are 6 distinguishable states. These are shown in Figure 6.16along with a scheme that has the appropriate kinetics. In the top three states, the h-gateis 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, bothm-gates are open.

Problem 6.19

a. Ans. (16). Externally applied TTX blocks sodium currents so that the ionic currentis predominantly the potassium current JK = GK(Vm−VK). Thus, the shape of thecurrent is the product of the two terms. During the pulse Vm−VK ≈ 10+70 = 80mV. 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 willreverse.

b. Ans. (12). Internally applied TEA blocks the potassium current so that the ioniccurrent is predominantly sodium current JNa = GNa(Vm − VNa). Thus, the shapeof 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 bea 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 timeconstant of m of 0.23 ms to a final value of JNa(10,4) ≈ 120(1)3(−45) ≈ −5.4mA/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 whichdoes not affect the ionic currents particularly if the concentrations of all solutions


−2

−50 +50−150 +150

−1

−3

V fm (mV)

I(p

A)

Figure 6.17: The relation between the singleopen-channel current and membrane potential(Problem 6.20).

are maintained constant. Hence, the ionic current is simply the normal ionic cur-rent which is predominantly the sodium current (12) plus the potassium current(16).

Problem 6.20

a. Linear. A plot of the single open-channel currents (−2, −2.5, −3 pA) against Vfm(−50, −100, −150 mV) is a straight line (Figure 6.17). Therefore the relation islinear.

b. Yes. The channel spends more if its time conducting current at Vm = −150 mVthan at either of the other two potentials. Therefore, the probability that the chan-nel 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 = ∆I∆Vm

= 1 pA100 mV

= 10 pS.

d. Since the relation between I and Vm is linear, it is only necessary to find the valueof Vm for which I = 0. This can be done graphically (Figure 6.17) or analyticallyas follows

I = γ(Vm − Vn),so that

Vn = Vm − Iγ = −50 mV− −2 pA10 pS

= 150 mV.

e. From the fraction of time that the channel is open in each record, it appears that theprobability 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 currentswhen the gate opens and when the gate closes, but at no other times. The gatingcurrent records show gating currents not only at those times but also when thegate is open.

PROBLEMS 131

0

−3

0 8Time (ms)

(pA)

ig(t)

i(t)

s(t)

S0

S1

S2

qg(t)Q1

Q2 = Q0

Figure 6.18: State occupancy s(t),single-channel current i(t), sin-gle-channel gating charge qg(t), andsingle-channel gating current ig(t)(Problem 6.20).

g. When the channel is open, gating currents occur without a change in the ionic cur-rent. Since there are three states and one is a closed state, there must be two openstates. However, there is only one non-zero value of current. Therefore, the twoopen states must have the same conductance. Assume that state 0 correspondsto the closed state (i.e. γ0 = 0) and that states 1 and 2 are open. In part c itwas 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 changein the single-channel ionic current but do generate gating currents. The combi-nation of gating and ionic currents allows a complete reconstruction of the stateoccupancies as shown in Figure 6.18 for the traces at −150 mV. Note that channeltransitions are allowed between states 0 and 1 and between states 1 and 2 andnot between states 0 and 2. Hence, because the gating charge of state 1 differsfrom 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 transitionsoccur wherever there is a gating current. Those, gating currents that are coinci-dent with changes in ionic current must result from transitions between state 0and state 1. Therefore, transitions from state 0 to state 1 open the channel andgenerate positive gating currents (since Q1 > Q0). Transitions from state 1 to state2 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 positivegating currents (since Q1 > Q2). Transitions from state 1 to state 0 close the chan-nel and generate negative gating currents (since Q0 < Q1). All of these transitionsare seen in the traces (Weiss, 1996b, Figure 6.97) (which consist of single-channelionic 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 av-erage single-channel currents are summarized as follows. The average single channel


Average current (i(Vm, t)) Random-variable current (i(Vm, t))Steady-state value Time constant Open current (I) Prob. open

t < 20 ms t > 20 ms t < 20 ms t > 20 ms t < 20 ms t > 20 ms

A ⇑ ⇑ ⇑ NC NC ? ⇓B NC ⇑ ⇓ NC ⇑ ? ⇑C ⇓ ⇓ NC ⇓ ⇓ ? ?D NC NC ⇑ NC NC ? ⇑E ⇑ ⇑ NC ⇑ ⇑ ? ?

Table 6.6: The table indicates the change in the indicated quantity from its value for the de-fault parameters — “NC” means there was no change, “⇑” means there was an increase in thealgebraic value, “⇓” means there was a decrease in the algebraic value, and “?” indicates thatit is difficult to determine whether or not there was a change (Problem 6.21). Changes in theaverage single-channel current and in the single open-channel current amplitude are relativelyeasy to judge from such brief records. Estimates of a change in the probability that the channelis open are not accurately judged with such short records. Hence, those results will not beweighed as heavily in our assessment.

current isi(Vm, t) = Ix(Vm, t),

whereI = γ(Vm − VNa).

The probability that the channel is openx(Vm, t) is governed by the differential equation

dx(Vm, t)dt

= α(Vm)(1− x(Vm, t)

)− β(Vm)x(Vm, t).

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 thatinterval is β(Vm)∆t.

In order to decide which records go with which change in parameter, the results arefirst summarized in Table 6.6.

a. An increase in γ, decreases the steady-state value of the average current in bothtime intervals, but does not affect the time constant. The single-open channelcurrent also decreases in both time intervals, but the probability that the channelis open does not change. Therefore, the answer is C.

b. A increase in Vfm will not affect the average or single-channel currents for t < 20ms. However for t > 20 ms, this increase will increase the probability that thechannel is open, and will increase the single channel current. Hence for t > 20ms, the average current will increase. Therefore, the answer is B.

c. A decrease in the extracellular concentration of sodium, will decrease the Nernstequilibrium potential for sodium. This decrease will not affect the probability thatthe channel is open nor any of the rate constants. However, the driving voltage onthe channel will increase. Hence, the single-channel current will increase for bothtime intervals. The steady-state value of the average current will also increase.Therefore, the answer is E.

PROBLEMS 133

−200 −100 0 100 200

Membrane potential (mV)

0

1

2

3

4

5

Aver

age

ionic

curr

ent

(pA

)

Figure 6.19: Average open channel cur-rent for Problem 6.22.

d. A decrease in both α(Vm) and β(Vm) without changing the ratio, will affect nei-ther 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. How-ever, the time constant will increase. The rates of transitions between states willalso decrease. Therefore, the answer is D.

e. If α(Vm) is decreased, the time constant will increase and the probability thatthe channel is open will decrease. However, the values of the single-open channelcurrents will not be affected. Therefore, steady-state values of the average currentwill 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-channelopen-state conductance and Vn is the Nernst equilibrium potential. The problemstatement indicates (Weiss, 1996b, Problem 6.100) that I=2 pA when Vm = −50mV, and that I=3 pA when Vm = +50 mV. These two conditions can be used toobtain a solution,

γ = 10 pS and Vn = −250 mV.

b. The first and second conducting segments are preceded by negative gating cur-rents. Negative gating currents represent inward motion of positive charge, start-ing from the closed state. Therefore, the first two conducting segments are instate 1. By similar reasoning, the last conducting state is state 3.

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 character-istics,

iss =∑ixi∞I = (x1∞ + x3∞)I = (x1∞ + x3∞)γ(Vm − Vn).

x1∞ is nearly 1 for very negative values of Vm and x3∞ is nearly 1 for very positivevalues ofVm. The sumx1∞+x3∞ is therefore small only nearVm = 0. The solid linein Figure 6.19 shows the relation between iss and Vm. The dashed line shows theopen-channel current γ(Vm−Vn) for the same range of membrane potential. Theimportant point is that the average current approaches the open-channel current


for very negative and for very positive values of Vm. However, the average currentis smaller than the open-channel current for Vm near zero because the gate is openless 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.

135

solutions to problems in cellular biophysics …phstudy.technion.ac.il/~wn116029/exam...

Documents