SOLUTIONS
In the following sections, solutions are given for some of the exercises.
14.1. SOLUTIONS, CHAPTER 1: VECTOR ANALYSIS
Answer 1a. 11.22
Answer 2. 17
Answer 3. 0.686
Answer 5a. A · B = 8 − 10 + 2 = 0
Answer 5b. A · B = 0, so one knows the vectors are perpendicular. That’s because A · B =
|A||B|cos(θ). With neither |A| and |B| zero, the result can be zero only if cos(θ) = 0,
and that will occur only if the vectors are perpendicular (angle of 90 degrees).
Answer 7a. 4
Answer 7b. 0
Answer 9. 42.
Answer 11. The gradient points in the −r direction, i.e., in the “uphill” direction of ψ = 1/r.
Answer 13. The result is the same as the law of cosines.
Answer 15a. No; instead ∇(1/R) = −∇ (1/R). Note the negative sign.
Answer 15b. Yes.
1
2
SOLUTIONS FOR CHAPTER 1: VECTOR ANALYSIS
Answer 17d. Yes.
Answer 19a. With φ = 1, a constant, both the gradient and divergence of φ are zero. Therefore,
the first term on the left side of Eq. (1) and the first term on the right side become zero.
Answer 19b. The result is called Gauss’s theorem.
Answer 19c. With ψ = 1 a constant, Gauss’s theorem again results, this time expressed in terms
of variable φ rather then ψ.
Answer 20. Yes, it does satisfy Laplace’s equation. See also the text by Stratton where this
equation is used.
Answer 21. The key to these questions is taking advantage of the fact that surface S is a closed
surface. In (a), K · dS is the portion of the surface oriented in the direction of K. In (b),
r · K is a constant, so the result is a multiple of the integral of dS over the whole surface.
One may be able to see the general answers by picturing or computing the results for a
particular example, such as a cube.
Answer 22. 2.31
Answer 23. 1.4554
Answer 24. 0.01
Answer 25. 0.001924
Answer 26. –0.2354
BIOELECTRICITY: A QUANTITATIVE APPROACH
3
14.2. SOLUTIONS, CHAPTER 2: SOURCES AND FIELDS
Answer 1a. Because conductivity is the reciprocal of resistivity, the corresponding units will
be 1/(Ωcm) or Siemens/cm.
Answer 1b.
σ = 0.01 S/cm
Answer 3. It is given that
φ=
Io
4πσr
Consequently,
J = −σ∇φ =
1
−Io
∇( )
4π
r
so that evaluating the gradient gives
J=
Io
= 1r
4πr2
Thus, J has magnitude 1 mA/cm2 and is directed radially.
Answer 5. Take the derivative with respect to x twice to find the divergence. Then find where
the divergence is zero, which is x = 0.33 cm.
Answer 7.
φ=
Answer 9a. The magnitude is
Io ρ
4π50
√
=
= 7.1uV (microVolts)
4πr
4π 52 + 52
√
22 + 32 + 42 = 5.38 µA.
Answer 9b. The vector area is A = U × V /2. The division by 2 occurs because the cross
product’s magnitude is that of the parallelogram formed by the vectors, and the triangle
is half that. Performing the cross product, A = 0.5az .
Answer 9c. Taking the dot product J · A = 4 · 0.5 = 2 µA.
Answer 11. Begin with the solution to Exercise 10. Twice differentiate with respect to x (not
r). The solution then is
A(x) =
.
Answer 13a. ∇Φ = k sech2 (x)
Answer 13b. J = −σ∇Φ = −kσ sech2 (x)
Io 3(x − e)2
4πσ
r5
4
SOLUTIONS FOR CHAPTER 2: SOURCES AND FIELDS
Answer 13c. ∇ · ∇Φ = k sech2 (x) tanh(x)
Answer 13d. The answer is the same as that for 13c, since the same question is asked, using
different words.
Answer 15a. The peak of the gradient is at x = 0.
Answer 15b. The maximum slope occurs at the peak of the gradient, so again at x = 0.
Answer 15c. The numerical value of the maximum slope is sech2 (x) with x = 0, i.e. 1.
Answer 17a. In equation form, the potential with resistivity 1 from source 1 can be written as
φ11 = (ρ1 /4π)(1/r1 ) and φ12 = (ρ2 /4π)(1/r1 ). The ratio φ12 /φ11 = ρ2 /ρ1 = 2. The
same argument will apply to potentials from the other source. Therefore, the ratio asked
equals 2.
Answer 17b. By definition J = −σ∇φ. With φ11 = (ρ1 /4π)(1/r1 ), the equation for J includes
no σ (or ρ) term. Therefore, the result is not a function of the resistivity, and the requested
ratio is 1.
Answer 17c. Again inspect the equation φ11 = (ρ1 /4π)(1/r1 ). Note that ∇φ has a direction
determined by r, not by ρ, since ρ is a constant insofar as the gradient operation, i.e., ρ
is not a function of x, y, z. Thus the dot product is 1, since the directions of J does not
change.
Answer 19a. Suitable units for k are µA/cm4 .
Answer 19b. At (1, 1, 1) the magnitude of the divergence is 3 µA/cm3 .
Answer 19c. At the origin, the divergence is greater than zero. This result is found by using the
definition of divergence, Eq. (1.23).
Answer 21. Hint: Ignore any interaction between the electrodes. Find the potential on an
imaginary electrode of the same size and location as a real arising from a point current
source at an electrode’s center. Make use of linearity to find results for two electrodes as
the sum of results for each one separately.
Answer 23a. k can be mV/cm3
Answer 23b. Electric field E = −∇Φ = −3kx2 .
Answer 23c. Sources Iv = ∇ · J = −∇ · σE = 6kx.
Answer 23d. There are sources, proportional to x. There is no source density at x = 0. Source
densities are negative when x < 0. Loosely speaking, these sources are associated with
currents from regions with more positive x to those with less positive x, and analogously
for negative x, but no current across the plane x = 0.
Answer 25. Hint: Since any potential field can be considered the superposition of point-source
fields, choose an arbitrarily located point source and demonstrate the validity of the
statement for this case.
BIOELECTRICITY: A QUANTITATIVE APPROACH
5
Answer 27. Hint: Think of the two sides of the tank as plates of a parallel plate capacitor, since
each one is a conducting region separated by a dielectric. Then use Gauss’s law with a
volume enclosing a square centimeter of surface at one interface, or use reference results
for capacitance with parallel plates.
Answer 29a. a = 10−3 cm. Thus ri = Ri /a2 = 102 /10−6 = 108 Ohms/cm.
Answer 29b. a = 10−3 cm and b = 2 × 10−3 cm. Thus re = Re /(b2 − a2 ), so re =
600/(3 × 10−6 = 2 × 108 Ohms/cm.
Answer 29c. a = 10−3 cm, so rm = Rm /circumference = Rm /4a = 10000/4 × 10−3 =
2500 × 103 = 2.5 × 106 Ωcm.
Answer 29d. a = 10−3 cm, so cm = Cm × circumference = Cm × 4a = 1.2 × 4× 10−3 =
4.8 × 10−3 microfarads/cm.
Answer 31. The resistance of the total path is the resistance of the top membrane Rtop , the resistance of the material inside the box, Rinside , and the resistance of the bottom membrane,
Rbottom .
Rtop = Rm /Atop . Atop , the area of the top surface, is Atop = a×L = 10×100×10−8 =
10−5 cm2 . Thus Rtop = 10000/10−5 = 109 Ohms. Rbottom = Rtop .
The resistance of the inside of the box is relatively insignificant, being Rinside = Ri ×
a/(L × a) = Ri /L = 100/(100 × 10−4 ) = 104 Ohms.
Thus the total resistance of the path Rtotal ≈ 2 × 109 , and the steady-state current is only
0.1/2 × 109 = 0.05 × 10−9 Amperes, or 0.05 nanoamperes.
Answer 33. Hint: use Gauss’s theorem. Draw an imaginary sphere just outside the inner sphere.
Answer 34. See Smythe, pp. 7–9, on “Lines of Force."
Answer 35. Hint: recall the derivation of the potential from a dipole, as the difference in potential
of two monopoles of opposite sign. Here, think of the quadrupole as two dipoles of
opposite sign with the second moved away from the first by a small amount d. Then
proceed in an analogous fashion.
6
SOLUTIONS FOR CHAPTER 3: BIOELECTRIC POTENTIALS
14.3. SOLUTIONS, CHAPTER 3: BIOELECTRIC POTENTIALS
Answer 1. 148,440 Coulombs
Answer 2. 6.674E11 lbs
Answer 3. 5.092E-8 (m/sec)/(V/m). Use Einstein’s equation.
Answer 4. 24.14 millivolts. The difference arises from the difference in temperature.
Answer 5. 1,074 miles per hour. Balance kinetic and thermal energy. Use
3 RT
mv 2
=
2
2 No
(1)
σ = (Di F Zi )Zi Ci (F/RT )
(2)
Answer 8.
because E = −∇Φ and J = σE
Answer 9A. 9.456E-5 Amperes
Answer 9B. -8.78E-5 Amperes
Answer 10. 4E-7 moles/cm3
Answer 11. 1E-12 moles
Answer 12. 7.448E-6 moles/(cm2 /sec)
Answer 13. 2.98E-9 moles/sec
Answer 14. 29.56 cm/sec
Answer 15. 5.37E-5 seconds
Answer 16. 53.1 millivolts
Answer 17. -83.6 millivolts
Answer 18. -60 millivolts
Answer 19. Three ion species are not in equilibrium. (At this transmembrane potential, no
individual ion is in equilibrium.)
Answer 20. Two ion species are not in equilibrium. Cl− is in equilibrium.
Answer 21. 143.6 millivolts
Answer 22A. 24.1 Angstroms
Answer 22B. 1.15 microfarads per cm2
Answer 23. 4,000 square micrometers
Answer 24. 4.85E-5 microfarads
Answer 25. 558,036,000 Ohms
BIOELECTRICITY: A QUANTITATIVE APPROACH
7
Answer 26. 0.515 picocoulombs. Use Q = CV .
Answer 27. 1/10000 Siemens/cm2 . Recall that conductivity is the reciprocal of resistivity.
Answer 28. 20 milliseconds
Answer 29. 66 milliseconds. Found from the number of time constants required.
Answer 30A. (Coulombs/sec)/m2
Answer 30B. (moles/sec)/m2
Answer 30C. Coulombs/Volt
Answer 30D. m2 /s
Answer 30E. Volts
Answer 30F. (m/sec)/(V/m) = m2 /(Vsec)
Answer 31. JK = gK ∗ (Vm − EK )
Answer 32. JNa = gNa ∗ (Vm − ENa )
Answer 33. JL = gL ∗ (Vm − EL )
Answer 34. –67.99 millivolts. Note the negative sign.
Answer 35. –86.14 millivolts. Note the negative sign.
Answer 36. 56.04 millivolts. Note the positive sign.
Answer 37. –58.22 millivolts. Note the negative sign.
Answer 38. 17.3 microamperes per square centimeter (uA/cm2 ). Note that the current is outward.
Answer 39. –0.960 microamperes per square centimeter (uA/cm2 ). Note that the current is
inward.
Answer 40. 10.38 microamperes per square centimeter (uA/cm2 ).
Answer 41. –22.27 millivolts per millisecond. Take the time derivative of V = Q/C, and
substitute Jion to replace dQ/dt.
Answer 42. –7.49 microamperes per square centimeter (uA/cm2 ). Note that the answer is
sensitive to precision in the individual current components, because that is followed by
an addition of terms that usually have different signs.
Answer 43. –23.69 millivolts
Answer 44. –37.88 millivolts. By definition, a condition that must be met to have a resting
(unchanging) potential is Jion = 0.
Answer 45. –5.30 millivolts per millisecond
Answer 50. F , Faraday’s constant
Answer 51. Einstein’s equation shows how the diffusion coefficient can be converted into the
conductivity, which is then used in the “electrical” term.
8
SOLUTIONS FOR CHAPTER 4: CHANNELS
14.4. SOLUTIONS, CHAPTER 4: CHANNELS
Answer 1. 741 Ohms
Answer 2. 37,620,000 channels open
Answer 3. The fraction of potassium channels open is 0.0198, or about 2%.
Answer 4. The number of open potassium channels is 4,021.
Answer 5. 5.62 nanoamperes (nA)
Answer 6. N p
Answer 7. N q
Answer 8. 1
√
Answer 9.
N pq
Answer 10. M = 10000 q/p
Answer 11. A = 10000 q/Dp
Answer 12. a = 50 q/(πDp)
Answer 13a. 8.69 µm
Answer 14a. In phase 1, the probability p1 that a K channel is open is 0.147
14A Solution in more detail:
1 The clamp voltage Vm1 exceeds the resting voltage Vr by vm + 1 = (−40 −
(−60)) = 20 mV.
1
2 Using vm
in the Hodgkin–Huxley equation for αn gives αn = 0.1582 msec−1 .
3 Similarly, βn = 0.09735 msec−1 .
4 Thus n, the probability of an individual particle in the channel being in the open
position, is αn /(αn + βn ) = 0.619.
5 Thereby, the probability that a K + channel is open is n4 , or 0.147.
Answer 14b. Multiplying cell surface area (600 µm2 ) by the channel density (40 ch/µm), one
finds the number of K + channels to be 24,000. This number times p1 gives 3,525 channels
open.
Answer 14c. The fluctuation is 4 × 54.8 channels
Answer 14d. 1.41E-9 Amperes, where –9 means ×10−9 .
Answer 15a. Probability p2 = 0.642.
BIOELECTRICITY: A QUANTITATIVE APPROACH
9
Answer 15b. 15,400 channels
Answer 15c. 12.32E-9 Amperes
Answer 17. 4.37
Answer 22. 8.74
Answer 19. 2.82E-9 Amperes.
Answer 20. The ratio is different because phase 2 has a number of open channels that changes
with time, even though the phase 2 voltage is constant. That is, immediately after the
transition (Ex. 19), the number of open channels is the same as the steady state in phase
1, because there has been no time for the expected numbers to change. (Even then, the
current is not the same as the current in phase 1 at steady state, because the membrane
voltage is immediately different.) The number of open channels, and thus the current,
evolves as the membrane moves to the steady state for phase 2.
Answer 21a. Probability p1 is 0.025.
Answer 21b. 592 open channels
Answer 21c. The expected fluctuation is 4 × 24 channels.
Answer 21d. 1.48E-10 Amperes
Answer 22a. Probability p2 is 0.894.
Answer 22b. The expected number of open channels is 21,467.
Answer 22c. 2.898E-8 Amperes
Answer 23. Yes, at steady state.
Answer 24. The ratio is 36.3, i.e., more channels in steady state in phase 2.
Answer 25. The ratio is 196, so a much higher current flows at the steady state of phase 2. It is
interesting to note that this ratio is higher than the ratio of the numbers of open channels.
Answer 26. The ratio is 2.86, much lower than that of the preceding question.
28 Exercise [28] restatement from text: In the model cell, examine N a+ and K + currents
with Vm1 of -57 mV and Vm2 of -20 mV.
1
a. At t1 , what is IK
?
10
SOLUTIONS FOR CHAPTER 4: CHANNELS
1
b. At t1 , what is INa
?
2
c. At t2 , what is IK ?
2
?
d. At t2 , what is INa
Answer 28a. 9.74E-11 Amperes
Detail of 28a:
1 The current desired is the product of the K + current through an open channel
times the expected number of open K + channels.
2 (Vm − EK ) = 23 mV, so the current through an open channel is Ichan =
γ × (Vm − EK ) = 2.3E − 13 Amperes.
3 Because vm + 1 = 3 mV, αn is 0.069 and βn is 0.1204, making n = αn /(αn +
βn ) = 0.364.
4 The expected number of open channels is Nopen = Ntotal × n4 = 24000 ×
0.01765 = 424 channels.
5 Thus the current from potassium ions is 424 × 2.3E − 13 = 97.4 picoamperes.
Answer 28b. -5.77E-12 Amperes
Detail of 28b:
1 The current desired is the product of the N a+ current through one channel times
the expected number of open N a+ channels.
2 Vm − ENa = −117 mV, so the current through an open channel is Ichan =
γ × (Vm − ENa ) = −1.17E − 12 Amperes.
1
= 3 mV, αm is 0.274, βm is 3.386, so m is 0.075. Similarly, αh is
3 Because vm
0.0603, and βh is 0.063, so h is 0.4889.
4 The expected number of open channels is Nopen = Ntotal × m3 h = 5.
5 Thus the current due to sodium ion movement is 5 × −1.17E − 12 or –5.8
picoamperes.
Answer 28c. 6.09E-9 Amperes. (Similar to 28A, but different Vm .)
Detail of 28c:
1 The current desired is the product of the K + current through an open channel
times the expected number of open K + channels.
2 (Vm − EK ) = 40 mV, so the current through an open channel is Ichan =
γ × (Vm − EK ) = 6.0E − 13 Amperes.
1
= 40 mV, αn is 0.3157 and βn is 0.0782, making n = αn /(αn +
3 Because vm
βn ) = 0.8063.
4 The expected number of open channels is Nopen = Ntotal × n4 = 24000 ×
0.4228 = 10147 channels.
5 Thus the current from potassium ions is 10147×6.0E −13 = 6088 picoamperes.
Answer 28d. –1.34E-10 Amperes. (Similar to 28B, but different Vm .)
Detail of 28d:
1 The current desired is the product of the N a+ current through one channel times
the expected number of open N a+ channels.
BIOELECTRICITY: A QUANTITATIVE APPROACH
11
2 Vm − ENa = 40 mV, so the current through an open channel is Ichan = γ ×
(Vm − ENa ) = −0.8 picoamperes.
1
= 40 mV, αm is 1.9308, βm is 0.4335, so m is 0.8167. Similarly,
3 Because vm
αh is 0.00947, and βh is 0.7310, so h is 0.01279.
4 The expected number of open channels is Nopen = Ntotal × m3 h = 167.
5 Thus the current due to sodium ion movement is 167 × −0.8 or –133.8 picoamperes.
Answer 29a. 7.81E-11 Amperes
Answer 29b. -4.43E-12 Amperes
Answer 29c. 3.39E-9 Amperes
Answer 29d. -1.61E-10 Amperes
Answer 30. Yes. (Note that these exercises allow comparison of currents at steady-state only.)
32 Question 32, from text. In the model cell, evaluate the requested currents if Vm1 is
–55.5 mV, and Vm2 is –24.5 mV.
1
?
a. At t1 , what is IK
1
b. At t1 , what is INa ?
b
?
c. At tb , what is IK
b
d. At tb , what is INa ?
2
?
e. At t2 , what is IK
2
f. At t2 , what is INa ?
Answer 32a. 1.34E-10 Amperes
Answer 32b. -8.40E-12 Amperes
Answer 32c. 5.54E-10 Amperes
Detail of c:
1 The current desired is the product of the K + current through an open channel
times the expected number of open K + channels.
2 (Vm − EK ) = 55.5 mV, so the current through an open channel is Ichan =
γ × (Vm − EK ) = 0.555 picoamperes.
1
= 35.5 mV, αn is 0.2766 and βn is 0.0802, making n = 0.4515 at
3 Because vm
t = ta (using the result of exercise 31).
4 The expected number of open channels is Nopen = Ntotal × n4 = 24000 ×
0.0416 = 997 channels.
5 Thus the current from potassium ions is 998 × 0.555 = 553 picoamperes.
Answer 32d. –9.27E-10 Amperes
Detail of 32d:
1 The current desired is the product of the N a+ current through one channel times
the expected number of open N a+ channels.
12
SOLUTIONS FOR CHAPTER 4: CHANNELS
2 Vm − ENa = −84.5 mV, so the current through an open channel is Ichan =
γ × (Vm − ENa ) = −0.845 picoamperes.
2
3 Because vm
= 40 mV, αm is 1.615, βm is 0.5565. Thus at t = ta the value of
m is 0.7437. Similarly, αh is 0.011864, and βh is 0.6341, so at t = ta h is
0.18365
4 The expected number of open channels is Nopen = Ntotal × m3 h = 1097.
5 Thus the current due to sodium ion movement is 1097 × −0.845 or –927.2
picoamperes.
Answer 32e. 4.81E-9 Amperes
Answer 32f. –1.53E-10 Amperes
Answer 33a. 6.21E-11 Amperes
Answer 33b. -3.38E-12 Amperes
Answer 33c. 5.537E-10 Amperes
Answer 33d. –1.92E-9 Amperes
Answer 33e. 7.59E-9 Amperes
Answer 33f. –1.08E-10 Amperes
Answer 34. No, as judged by these exercises, never the same sign. INa and IK have opposite
signs most of the time because (Vm − EK ) has a different sign from (Vm − ENa ). For
the sign to be the same, Vm would have to be outside the range EK to ENa .
Answer 35. It is not always true, because |INa | is larger at time tb .
BIOELECTRICITY: A QUANTITATIVE APPROACH
14.5. SOLUTIONS, CHAPTER 5: ACTION POTENTIALS
Answer 1a. gK = ḡK n4
Answer 1b. gNa = ḡNa m3 h
Answer 1c. IK = gK (Vm − EK )
Answer 1d. INa = gNa (Vm − ENa )
Answer 1e. IL = gL (Vm − EL )
Answer 1f. Iion = IK + INa + IL
Answer 1g. Vdot = Is − Iion /Cm
Answer 2a. vm = Vm + 60
Answer 2b. αn = .01(10 − vm )/(exp((10 − vm )/10) − 1)
Answer 2c. βn = 0.125 exp(−vm /80)
Answer 2d. αm = 0.1(25 − vm )/(exp((25 − vm )/10) − 1)
Answer 2e. βm = 4 exp(−vm /18)
Answer 2f. αh = 0.07 exp(−vm /20)
Answer 2g. βh = 1/(exp((30 − vm )/10) + 1)
Answer 2h. dn/dt = αn (1 − n) − n βn
Answer 2i. dm/dt = αm (1 − m) − m βm
Answer 2j. dh/dt = αh (1 − h) − h βh
Answer 3a. ∆Vm = Vdot ∆t
Answer 3b. ∆n = dn/dt ∆t
Answer 3c. ∆m = dm/dt ∆t
Answer 3d. ∆h = dh/dt ∆t
Answer 3e. Vm (t0 + dt) = Vm + ∆Vm
Answer 3f. n(t0 + ∆t) = n + ∆n
Answer 3g. m(t0 + ∆t) = m + ∆m
Answer 3h. h(t0 + ∆t) = h + ∆h
13
14
SOLUTIONS FOR CHAPTER 5: ACTION POTENTIALS
Answer 4. If
Im = IC + Iion
then
IC = Im − Iion
For IC to be zero, one must have
IC = Im − Iion = 0
Thus
Im = Iion
In an isolated patch,
Im = Is
so the condition for IC to be zero, and thus no Vm change, is
Is = Iion
Answer 5a. gK 0.73497 mS/cm2
Answer 5b. gNa 4.15057 mS/cm2
Answer 5c. VK [−11.5 − (−72.1)] = +60.6 mV
Answer 5d. VNa [−11.5 − (+52.4]) = −63.9 mV
Answer 5e. IK 44.54 µA/cm2
Answer 5f. INa –265.2 µA/cm2
Answer 5g. IL 11.3 µA/cm2
Answer 5h. Iion –209.4 µA/cm2
Answer 5i. V̇m = 209 volts/sec
Answer 6a. vm 48.5 mV
Answer 6b. αn 0.393371 msec−1
Answer 6c. βn 0.068174 msec−1
Answer 6d. αm 2.597745 msec−1
Answer 6e. βm 0.27032 msec−1
Answer 6f. αh 0.006193 msec−1
Answer 6g. βh 0.864127 msec−1
BIOELECTRICITY: A QUANTITATIVE APPROACH
Answer 6h. dn/dt 0.21891 msec−1
Answer 6i. dm/dt 1.40176 msec−1
Answer 6j. dh/dt –0.40895 msec−1
Answer 7a. ∆Vm 10.47 mV
Answer 7b. ∆n 0.010945 no units
Answer 7c. ∆m 0.070088
Answer 7d. ∆h –0.020447
Answer 7e. Vm (t0 + dt) –1.03 mV
Answer 7f. n(t0 + dt) 0.388945 no units
Answer 7g. m(t0 + dt) 0.487088
Answer 7h. h(t0 + dt) 0.456553
Answer 8. Is = −Iion = 209.4 µA/cm2
Answer 9a. gK 11.95 mS/cm2
Answer 9b. gNa 10.87 mS/cm2
Answer 9c. VK [−11.5 − (−72.1)] = +60.6 mV
Answer 9d. VNa [−11.5 − (+52.4)] = −63.9 mV
Answer 9e. IK 724.01 µA/cm2
Answer 9f. INa –694.59 µA/cm2
Answer 9g. IL 11.31 µA/cm2
Answer 9h. Iion 40.731 µA/cm2
Answer 9i. V̇m –40.73 mV/msec
Answer 10a. vm 48.5 mV
Answer 10b. αn 0.393371 msec−1
Answer 10c. βn 0.068174 msec−1
Answer 10d. αm 2.597745 msec−1
Answer 10e. βm 0.27032 msec−1
15
16
SOLUTIONS FOR CHAPTER 5: ACTION POTENTIALS
Answer 10f. αh 0.006193 msec−1
Answer 10g. βh 0.864127 msec−1
Answer 10h. dn/dt 0.043058 msec−1
Answer 10i. dm/dt –0.141257 msec−1
Answer 10j. dh/dt –0.08432 msec−1
Answer 11a. ∆Vm –2.037 mV
Answer 11b. ∆n 0.002153
Answer 11c. ∆m –0.007063
Answer 11d. ∆h –0.004216
Answer 11e. Vm (t0 + dt) –13.54 mV
Answer 11f. n(t0 + dt) 0.76115
Answer 11g. m(t0 + dt) 0.94794
Answer 11h. h(t0 + dt) 0.09978
Answer 12. Is = −Iion = −40.73 µA/cm2
Answer 13. Both set A and set B have the same value for Vm and hence IL . With the state
variables of Set A, INa is greater and IK is smaller than with the state variables of Set B.
Thus set A leads to a negative total ionic current (and thus a positive ∆Vm . In contrast, set
B leads to a positive total ionic current (since potassium dominates) and thus a negative
∆Vm .
Answer 14a. Set u = 0.1(10 − vm ). Then eu = 1 + u + u2 /2 near u = 0, i.e., near vm = 10.
Substituting the expansion into the expression gives αn = 0.1/(1 + u/2) near u = 0.
Answer 14b. At vm = 10 the result of Ex. 14A reduces to αn = 0.1.
Answer 15a. 50 mV/msec
Answer 15b. –50 mV/msec
Answer 15c. 259.4 mV/msec
Answer 15d. 9.27 mV/msec
BIOELECTRICITY: A QUANTITATIVE APPROACH
17
Answer 16a. A computer procedure that computes every α and β for a given vm , including the
code needed for the special cases is:
// alphab sets values for an, bn, am, bm, ah, bh
// from the value of vm given as the function
parameter
// with consideration of special cases
// of alpha_n and alpha_m.
// Sept 5, 2005
int alphab(double vm){
double u, au;
//n
u = (10-vm)/10;
if (u > 0){au = u;} else{au = -u;}
if (au > 1.E-7){an=.01*(10.-vm)/(exp((10-vm)/10)-1);
}
else{ an = 0.1; }
bn=0.125*exp(-vm/80);
//m
u = (25-vm)/10;
if (u > 0){au = u;} else{au = -u;}
if (au > 1.E-7){am=0.1*(25-vm)/(exp((25-vm)/10)-1);
}
else {am = 1.;}
bm=4*exp(-vm/18);
//h
ah=0.07*exp(-vm/20);
bh=1/(exp((30-vm)/10)+1);
return 0;
}
Answer 16b. Table of α and β values:
i
0
1
2
3
4
5
6
7
8
9
10
vm
0.000
5.000
10.000
15.000
20.000
25.000
30.000
35.000
40.000
45.000
50.000
an
0.06
0.08
0.10
0.13
0.16
0.19
0.23
0.27
0.32
0.36
0.41
bn
0.13
0.12
0.11
0.10
0.10
0.09
0.09
0.08
0.08
0.07
0.07
am
0.22
0.31
0.43
0.58
0.77
1.00
1.27
1.58
1.93
2.31
2.72
bm
4.00
3.03
2.30
1.74
1.32
1.00
0.76
0.57
0.43
0.33
0.25
The α and β values have units of msec−1 .
ah
0.07
0.05
0.04
0.03
0.03
0.02
0.02
0.01
0.01
0.01
0.01
bh
0.05
0.08
0.12
0.18
0.27
0.38
0.50
0.62
0.73
0.82
0.88
18
SOLUTIONS FOR CHAPTER 5: ACTION POTENTIALS
Answer 18: Here are some suggestions to consider for writing code that will be usable for a
range of questions:
(1) Keep track of the iteration number (here called the “loop count”) as an integer. (Most
people do this instinctively.)
(2) Keep track of elapsed time as an integer. (Most program writers do not do this
instinctively.) Integer microseconds work well.
(3) Enter and record ∆t as an integer value (µsec). Whenever it is needed for computation, convert it then to a floating point form.
(4) Convert explicitly between the iteration number and the elapsed time, using ∆t.
(5) Keep track of the stimulus start/stop time as integer values.
(6) The goal of items 1 to 5 is to avoid issues of floating point round-off in keeping time,
identifying the stimulus starting time, etc.
(7) Keep track of the interval between times that output lines will be created with a
distinct integer variable. In general, do not write output for every iteration.
(8) If there is a stimulus, start the first stimulus at t = 0. Begin calculation, however,
before that time, so that a baseline can be established.
Generally the program is easiest to manage if elapsed time is converted explicitly from
the iteration number, because, in this way, the iteration number increases from zero as
expected, while elapsed time can begin with a negative value. Additionally, program
decisions about whether it is time to start or stop the stimulus, or to print output, can
be parametrized on the basis of elapsed time, rather than iteration number. Importantly,
interval ∆t can be varied, which often is desirable for testing, independently of the
specification of when the stimulus should occur or output lines should be displayed.
If there is a discrepancy between the output table as printed and your results, consider the
following aspects of how the table above is computed and verify their correspondence.
Here, line k + 1 is found from line k by:
(1) Using vm for line k to compute α and β values for line k, and ∆n, etc, for the time
of line k. Note that vm for line k = 1 is not used.
(2) Using n, m, h from line k (not line k + 1) to get currents and vdot at the time of line
k, and from these estimate vm at time k + 1. Note that newer values of n, m, h for
time k + 1 are not used.
(3) Being sure the stimulus does not extend into the period 150 to 200. That is, at time
150 the stimulus must stop.
Answer 19. The extended output table for ∆t = 50µsec is:
loopcnt time Is
0
-200
0.0
1
-150
0.0
2
-100
0.0
3
-50
0.0
4
0 200.0
vm
-60.000
-60.000
-60.000
-60.000
-60.000
vdot
-0.0
-0.0
-0.0
-0.0
200.0
n
0.31768
0.31768
0.31768
0.31768
0.31768
m
0.05293
0.05293
0.05293
0.05293
0.05293
h
0.59612
0.59612
0.59612
0.59612
0.59612
BIOELECTRICITY: A QUANTITATIVE APPROACH
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
50 200.0
100 200.0
150
0.0
200
0.0
250
0.0
300
0.0
350
0.0
400
0.0
450
0.0
500
0.0
550
0.0
600
0.0
650
0.0
700
0.0
750
0.0
800
0.0
850
0.0
900
0.0
950
0.0
1000
0.0
-50.000
-40.339
-30.965
-31.769
-31.928
-31.077
-28.950
-25.320
-19.905
-12.300
-2.017
11.185
26.277
39.355
45.063
44.797
44.475
43.148
41.972
40.115
193.2
187.5
-16.1
-3.2
17.0
42.5
72.6
108.3
152.1
205.7
264.0
301.8
261.6
114.1
-5.3
-6.4
-26.5
-23.5
-37.1
-33.6
19
0.31768
0.31934
0.32308
0.32925
0.33510
0.34081
0.34667
0.35303
0.36032
0.36908
0.37999
0.39383
0.41141
0.43299
0.45755
0.48256
0.50628
0.52874
0.54978
0.56953
0.05293
0.06726
0.09804
0.14894
0.19253
0.23132
0.26851
0.30769
0.35276
0.40786
0.47705
0.56279
0.66261
0.76511
0.85187
0.91069
0.94580
0.96677
0.97913
0.98649
0.59612
0.59343
0.58617
0.57257
0.55988
0.54761
0.53503
0.52129
0.50556
0.48727
0.46662
0.44473
0.42290
0.40186
0.38180
0.36273
0.34462
0.32741
0.31106
0.29554
After time 200 µsec, voltage vm does not continue to fall. Rather, it rises to a substantially
higher value in the course of generating an action potential.
Answer 20. ∆t = 2µsec
dTime:
StimAmplitude:
StimDuration:
Hh:
Hh:
Hh:
Hh:
Hh:
Hh:
2 microseconds
200
150
EK
-72.100 ENa
52.400
gbarK
36.0 gbarNa 120.0
n
0.31768 m
0.05293
Cm
1.00 Temper
6.30
gK
0.367 gNa
0.011
Vm
-60.000 Vr
-60.000
loopcnt
0
25
50
75
100
125
150
time
Is
-200
0.0
-150
0.0
-100
0.0
-50
0.0
0 200.0
50 200.0
100 200.0
Vm
-60.000
-60.000
-60.000
-60.000
-60.000
-50.155
-40.599
EL -49.187
gL
0.3
h 0.59612
gL
vdot
-0.0
-0.0
-0.0
-0.0
200.0
193.7
188.7
0.300
n
0.31768
0.31768
0.31768
0.31768
0.31768
0.31842
0.32099
m
0.05293
0.05293
0.05293
0.05293
0.05293
0.05921
0.07966
h
0.59612
0.59612
0.59612
0.59612
0.59612
0.59497
0.59039
20
SOLUTIONS FOR CHAPTER 5: ACTION POTENTIALS
175
200
225
250
275
300
325
350
375
400
425
450
475
500
525
550
575
600
150
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
-31.218
-31.529
-30.967
-29.207
-25.894
-20.557
-12.559
-1.283
12.984
27.507
37.908
42.692
43.894
43.580
42.646
41.374
39.856
38.137
-12.5
1.8
22.6
50.0
85.5
132.1
192.6
260.8
302.2
260.3
146.4
49.7
3.6
-14.3
-22.7
-28.2
-32.6
-36.3
0.32577
0.33172
0.33761
0.34371
0.35040
0.35820
0.36781
0.38013
0.39608
0.41601
0.43895
0.46304
0.48679
0.50950
0.53097
0.55117
0.57010
0.58782
0.11756
0.16300
0.20431
0.24456
0.28738
0.33714
0.39886
0.47712
0.57220
0.67487
0.76850
0.84137
0.89273
0.92743
0.95056
0.96592
0.97612
0.98288
0.58045
0.56747
0.55475
0.54155
0.52703
0.51029
0.49086
0.46934
0.44718
0.42554
0.40482
0.38509
0.36631
0.34845
0.33147
0.31531
0.29995
0.28533
Answer 20a. vm 49.692 − 39.355 = 3.337 mV
Answer 20b. n 0.46304 − 0.43299 = 0.03005
Answer 20c. m 0.84137 − 0.76511 = 0.07626
Answer 20d. h 0.38509 − 0.40186 = −0.01677
Answer 21a. Yes, linear for t ≤ 100 µsec.
Answer 21b. No, linear for t > 200 µsec.
Answer 22a. For duration 150, just-above-threshold stimulus 45 µA/cm2 .
Answer 22b. For duration 150, just-above-threshold stimulus at end of stimulus vm 6.46 mV.
Answer 22c. For duration 300, stimulus 23 µA/cm2 .
Answer 22d. For duration 300, vm at end of stimulus 6.38 mV.
The above values depend critically on the fact that the evaluation is for a membrane patch,
so that no current is moving longitudinally along intracellular or extracellular path.
Answer 23a. For Is =50, the time to peak is 3100 µsec.
Answer 23b. For Is =200, the time to peak is 750 µsec.
Answer 23c. For Is =500, the time to peak is 450 µsec.
BIOELECTRICITY: A QUANTITATIVE APPROACH
21
Answer 24. The time duration until stable return within the envelope of the initial conditions is
24100 µsec.
Answer 25a. The EL (that corresponds to the new gL ) is 264.4 mV.
Answer 25b. Some answers are more fun if you see for yourself. Inspect the return to baseline
carefully, perhaps by plotting vm (t).
Answer 26a. Is =50, 2nd AP at 16800 microseconds from start of first stimulus.
Answer 26b. Is =200, 2nd AP at 8150.
Answer 26c. Is =500, 2nd AP at 5150.
22
SOLUTIONS FOR CHAPTER 6: IMPULSE PROPAGATION
14.6. SOLUTIONS, CHAPTER 6: IMPULSE PROPAGATION
Answer 1a. µF/cm
Answer 1b. Ωcm
Answer 1c. Ω/cm
Answer 1d. Ωcm
Answer 2a. Recall that the HH model has a linear response to transmembrane voltage changes
near the resting potential, so the membrane resistance can be found from the conductances
in the HH model. Use Table 13.5, which gives resting values of HH membrane, to find
the conductances (g values) of HH membrane at rest. Using those values,
Rm = 1/(gK + gN a + gL ) = 1, 475 ohm − cm2
(3)
Answer 2b. The resistance R in ohms of a segment of membrane is
R = Rm /As = Ri /(2πaL) = rm /L
(4)
Comparing the last two parts of the equation shows that rm = Rm /(2πa). Using the
result for Rm found in the answer to the preceding question, one has rm = Rm /(2π0.003)
or rm = 78, 248 Ωcm.
Answer 2c. The intracellular resistance per unit length characterizes the volume inside the
membrane in a way that takes into account the medium’s resistivity and the fiber’s radius.
After ri has been found, one need only make a later choice of a length L (distance along
the axis) to find resistance R of the segment. Thus resistance R will be
R = ri L = Ri L/Ai = Ri L/(πa2 )
(5)
To make the last two terms equal, one makes ri = Ri /(πa2 ). In this fiber, the result is
ri = 5, 305, 169 ohm/cm.
Answer 2d. The connection of the intracellular resistance per length with the region inside the
membrane (in the preceding part) seems obvious, but the statement that “Extracellular
currents flow to twice the membrane radius” seems much less so. Indeed, in the absence
of a physical boundary of some kind, asserting that currents flow between radius a and
radius 2a is at best arbitrary. Even so, the statement recognizes that extracellular current
flows most intensely near the membrane, and it provides a specific basis for computing
re , even if it is only an estimate. Note that the cross-sectional area inside a is one-third
the cross-section between a and 2a, i.e., assuming 2a as a limiting radius for extracellular
current assumes that extracellular current flows through a greater cross-section, but still
within a radius of the fiber membrane. With this assumption, then using similar reasoning
to that of the previous part, one has
R = Re L/Ae = Re L/[π((2a)2 − a2 )] = re L
so that re = Re /[π((2a)2 − a2 )]. In this fiber, the result is re = 589, 463 ohm/cm.
(6)
BIOELECTRICITY: A QUANTITATIVE APPROACH
23
Answer 3a. rm = 63, 700 Ωcm
Answer 3b. cm = 0.0377 µF/cm
Answer 3c. ri = 1, 273, 000 ohm/cm
Answer 3d. re = 169, 000 ohm/cm.
Answers 4A–C are taken from material in the text of Chapter 6.
Answer 4a. Intracellular current is
Ii =
∂Vm
−1
[
− Ire ]
(ri + re ) ∂x
(7)
When I = 0 because there is no stimulus, the 2nd term drops out.
Answer 4b. Because I = Ii + Ie , the extracellular current Ie = I − Ii , and thus,
Ie =
∂Vm
1
[
+ Ire ]
(ri + re ) ∂x
(8)
When I = 0 (no stimulus) the second term again drops out.
Answer 4c. The transmembrane current
Im =
∂ 2 Vm
1
(
− re ip )
(2πa)(ri + re ) ∂x2
(9)
where ip = 0 because there is no stimulus. As the question asks for Im , i.e., current per
cm2 , the Im equation requires that dimensions (e.g., of a) are in cm.
Answer 5a. Vm (x) = av tanh(x) has negative Potentials for x < 0, positive for x > 0. The
action potential is moving left, toward the more negative region.
Answer 5b. Ii (x) = −ai sech2 (x). Intracellular current is flowing left.
Answer 5c. Ie (x) = +ae sech2 (x). Extracellular current is flowing right.
Answer 5d. Im (x) = −2am sech2 (x) tanh(x). Membrane current is outward left, inward right.
Answers 5a–d. The expressions given depend on undefined constants av , ai , ae , and am , which
are proportionality constants that include such factors as the radius and the axial resistances. The proportionality constants are not defined in detail here, so that this exercise
can focus on the curve shapes rather than magnitudes.
Answer 6a. (v0 − 2v1 + v2 )/(R + r) milliamperes
Answer 6b. (v1 − 2v2 + v3 )/(R + r) milliamperes
Answer 6c. (v2 − 2v3 + v4 )/(R + r) milliamperes
24
SOLUTIONS FOR CHAPTER 6: IMPULSE PROPAGATION
Answer 7. The difference in Question 7 (relative to Exercise 6) is that in Ex. 7 it is asked that
membrane current be found “per unit area.” Dividing the earlier result by the geometric
factors of one segment, one has
3
Im
=
(v2 − 2v3 + v4 )
1
2
mA/cm
(2πa∆x)
(R + r)
(10)
Answer 8. The transmembrane current at crossing 0, at the left end, is the same as the intracellular axial current between crossings 0 and 1. Thus I 0 = (v1 − v0 + rS)/(R + r)
milliamperes.
Answer 9. The expression must take into account the different manner for computing the derivative, and the different area of the segments at the fiber’s ends. Thus
0
Im
=
(v1 − v0 + rS)
2
(πa∆x)
(R + r)
(11)
1
Answer 10a. Im
= +1.085 mA/cm2
Answers 10A-b. Note that the computation can be simplified by substituting λ algebraically
into the equation for Im .
0
Answer 11a. Im
= −6.508 mA/cm2
1
Answer 11b. Im
= +6.508 mA/cm2
Answers 11a–b. These answers can be obtained by finding the solution again from scratch, or
as a variation on the answers to exercise 10, looking at the relationships of the given
voltages.
Answer 12. From Chapter 6, the rate of change of Vm with time t is
∂Vm (xo , t)
[Im (xo , t) − Iion (xo , t)]
=
∂t
Cm
Answer 13a. 6.18 milliseconds
Answer 13b. 240.5 mV/msec
Answer 13c. 10.3 millivolts
Answer 13d. –7.84 microamperes/cm2
Answer 13e. –248.4 microamperes/cm2
(12)
BIOELECTRICITY: A QUANTITATIVE APPROACH
25
Answer 14a. 2.5 msec
Answer 14b. 280.2 mV/msec
Answer 14c. 12.4 millivolts
Answer 14d. –3.57 µA/cm2
Answer 14e. –284 µA/cm2
Answer 15a. The mesh ratio is 3.01.
Answer 15b. The mesh ratio indicates instability. The bad outlook is tempered by the HH
membrane having fairly low Rm , and the resistance present along the extracellular as
well as intracellular axial direction, which is not taken into account in the mesh ratio
calculation.
Answer 15c. The manifestation of instability will be that, at some time step, values of Vm will
begin to oscillate with increasing amplitude, and after a few time steps will be out of the
computable range.
Answer 15d. Stability can be markedly improved by reducing ∆t to a lower value. For example,
making ∆t = 1 µsec would reduce the mesh ratio to 0.30, a change in mesh ratio that
likely would correspond to a stable calculation.
Answer 15e. Try it and see.
26
SOLUTIONS FOR CHAPTER 7: ELECTRICAL STIMULATION OF EXCITABLE TISSUE
14.7. SOLUTIONS, CHAPTER 7: ELECTRICAL STIMULATION OF EXCITABLE
TISSUE
Answer 1. rheobase
Answer 2. threshold
Answer 3. chronaxie
Answer 4a. seconds
Answer 4b. Volts
Answer 4c. Volts
Answer 5. a/ 1 − exp[−500/(RC)]
Answer 6. W
Answer 7. 1000U/R
Answer 8. 1000W/a
Answer 9. W
Answer 10. Begin by rearranging the Weiss–Lapicque equation so that e−T /τ is the only term
on the left. Rewrite the rearranged equation twice, once for each condition. Divide the
first equation by the second, and combine terms so that there is only one exponential term
on the left. Take the natural log, and rearrange the result to solve for τ . The result is:
τ = (D − d)/ log
where log is the natural log.
Answer 11. 20 millivolts
Answer 12. 2,000 Ωcm2
Answer 13. 125.07 µA/cm2
Answer 14. 2.0 msec.
Answer 17. 0.115214 cm
Answer 18. λ = 0.212 cm
Answer 18. λ = 0.212 cm
Answer 18. λ = 0.212 cm
I (a − i)
i (a − I)
BIOELECTRICITY: A QUANTITATIVE APPROACH
27
Answer 19. τ = 1.8 msec
Answer 21a. P (x) = S/
(h2 + (x − e)2 )
Answer 21b. A(x) = S(−1/r3 + 3(x − e)2 /r5 ). Note the marked difference between the
answer 21A (which naively seems to be the effect of the stimulus on the fiber), and
Answer 21B. Although A(x) is not an equation for Vm produced by the stimulus, it does
at least provide the initial direction of change for Vm .
Answer 21c. A(x) = −S(1/r3 − 3e2 /r5 )
Answer 22a. The distance where A(x) > 0 is 8.6 cm.
Answer 22b. The distance where A(x) > 0 is 1.4 cm.
Answer 22c. No. In stimulation, extent of change is important, but amplitude is also important,
and often more so. A significant point of interest is, however, that both source and sink
stimuli produce regions of positive A(x), as this result indicates that either polarity might
serve as a stimulus, but would stimulate to different regions of the fiber.
Answer 23a. A(x) = 3S/(4h2 )
Answer 23b. A(x) = S(1/h2 − 1/(h2 + d2 )( 3/2) + d2 /(h2 + d2 )( 5/2))
Answer 24. A(x) > 0 is a region where, at least initially, Vm is growing more positive, and
thus is more likely to reach threshold and initiate an action potential.
Answer 25. 3 extrema
Answer 26. 4
Answer 27. The ratio L2 /L1 = 2. The answer shows that the portion of the fiber being depolarized by the stimulus is affected markedly by the location of the electrode relative to
the fiber.
Answer 28. The ratio L2 /L1 = 2. Note that a ratio other than 1 occurs because of a change
in the separation of the electrodes, not a change in orientation of the electrode pair. The
changed ratio indicates that more of the fiber is being affected, i.e., L2 > L1 .
Answer 29. A(x = 0.3) = 28.5 mV/cm2
Answer 30a. 8,100 Joules
Answer 30b. 2,160 Joules
Answer 31a. About 3.2E9 stimuli
Answer 31b. About 5.7E9 stimuli
Answer 32a. 175 mV/msec
Answer 32b. 150 mV/msec
28
SOLUTIONS FOR CHAPTER 7: ELECTRICAL STIMULATION OF EXCITABLE TISSUE
Answer 33a. 1.33 Volts
Answer 33b. 0.995 Volts
Answer 34a. 1.19 milliamperes
Answer 34b. 2.50 milliamperes
Answer 35. The design meets both requirements.
Answer 36. Design fails requirement 1 but meets requirement 2.
Answer 37. Design meets requirement 1 but fails requirement 2.
Answer 38. Design meets both requirements.
Answer 39. The authors know the answer to this question, but have chosen not to include it
here, so as not to take away the reader’s enjoyment of finding the solution.
BIOELECTRICITY: A QUANTITATIVE APPROACH
29
14.8. SOLUTIONS, CHAPTER 8: EXTRACELLULAR FIELDS
Answer 1. At t = 10 milliseconds. On the graph, one sees the midpoint of the upstroke at about
32 mm. Dividing by the velocity (4 mm/msec), one finds that this action potential occurs
about 8 msec after the one at x = 0. Additionally, parameter t1 adds a 2 msec time delay
to the upstroke.
r . Matlab is the registered
Answer 2. See Figure 14.1. The graphs here were made with Matlab
trademark of The Mathworks Inc.
Figure 1. Vm (t) at position x = 24 mm.
Answer 3. In an action potential the activation phase always comes first. Therefore, on the time
waveform, the activation phase (fast upstroke) is on the left (lower time). On an action
potential waveform traveling toward higher x, the activation phase is on the right (higher
x), since activation is moving to the right. Otherwise the action potentials appear largely
the same, though real action potentials may show more subtle differences.
Answer 4. See Figure 14.2. Here V̇mmax = 95.5 mV/ms. For comparison, measured transmembrane potentials from healthy tissue usually have V̇mmax between 100 and 200 mV/ms, as
abnormalities inevitably leads to a lower value. This (unnatural) AP also is on the slow
side.
Answer 5a. Amax = 23.8708, xa = 32.0070, va = −14.9099, i = 640. Here xa is the x
coordinate of the point of maximum slope, va is the value of Vm at that point, and Amax
is the magnitude of the slope at that point.
Answer 5b. Vm = −Amax (x − xa) + va
Answer 5c. Bmax = 6.2494, xr = 19.9950, vr = −10.0319, i = 400
Answer 5d. Vm = Bmax (x − xr) + vr
30
SOLUTIONS FOR CHAPTER 8: EXTRACELLULAR FIELDS
max = 95.5 mV/ms.
Figure 2. V̇m (t) at position x = 24 mm. V̇m
Answer 5e. xcross = 29.35, Vmcross = 48.4
Answer 5f. Activation crossing at x = 33.9 mm. Recovery crossing at x = 12.0 mm.
Answer 6b. With the velocity doubled, the AP waveform of Vm (x) occupies twice the distance
along x. Also, the leading edge is about twice the distance from the origin.
Answer 6d. With velocity doubled, the AP wave shape of Vm (t) remains the same, except that
the time of onset is shorter.
Answer 7. The axial current is plotted in Figure 14.3.
Figure 3. Ii (x) at time t = 10 msec. The vertical scale extends to 2.5E-4 mA.
BIOELECTRICITY: A QUANTITATIVE APPROACH
31
Answer 8. The transmembrane current is plotted in Figure 14.4.
Figure 4. Transmembrane current im (x)) at time t = 10 msec.
Answer 17. The transmembrane current is plotted in Figure 14.5. In the figure, each panel
shows the potential that arises from the lumped monopolar sources (the solid line) and the
potential that arises from the continuous sources (the dashed line). Panels A, B, and C are
for distances h of 0.1, 1.0, and 10.0 millimeters, as marked. All plots are for time t = 10
msec. When h is small, as in panel A, the approximation of the continuous by lumped
sources is poor, so lumpiness is striking. Conversely, as h increases to 1 millimeter, the
potentials as generated by the lumped sources are smoother. At 10 millimeters, the lumped
solution becomes similar to those from the continuous sources, even though distance h
still is less than the distance between source M1 and source M3 .
Answer 27. V = 7.162 × 10−5 V. This value was found using Eq. (8.71). The potentials are to
be found at points far enough from the cell that a simplification is possible, because the
solid angle is virtually a constant and so can be factored from the equation.
32
SOLUTIONS FOR CHAPTER 8: EXTRACELLULAR FIELDS
Figure 5. Extracellular potential Φe (x) from Lumped Monopolar and Continuous sources.
BIOELECTRICITY: A QUANTITATIVE APPROACH
33
14.9. SOLUTIONS, CHAPTER 9: CARDIAC ELECTROPHYSIOLOGY
Answer 1. Cardiac electrophysiology, compared to that of nerve. Correct statements are numbers 1, 3, and 5.
Answer 2. Cardiac structure. Correct choices are 1, 2, 3, and 4.
Answer 3. AV node propagation: choice 2
Answer 4. Correct choices are 1, 2, 3, 4, 5, 9
Answer 5. 10
Answer 6. 4.1E9 ohms
Answer 7. Multiply milliseconds by 0.001 to convert the values to seconds.
Answer 8. 22 ± 3 ms
Answer 9. 44 ± 4 ms
Answer 10. Purkinje strand. The correct responses are 2 and 3.
Answer 11. 300E6 ohms
Answer 12. 100 ± 20 mV
Answer 13. 0.3 ± 0.05 m/s
Answer 14. 0.0015 ± 6E-4 m
Answer 15. 0.036 ± 0.005 V. Some variability depends on exactly how the edges are defined.
Answer 16. 40 ± 15 mV Making an enlarged figure is helpful. Note the calibration bar.
Answer 17. 15 ± 5 mV Making an enlarged figure is helpful. Note the calibration bar.
Answer 18. Constants in the differential equation are a = 1/(cqR) and b = u/(cq).
34
SOLUTIONS FOR CHAPTER 9: CARDIAC ELECTROPHYSIOLOGY
Answer 19. In the solution for v, a = uR and b = cqR.
Answer 20. About Purkinje fibers. Choices are 3 and 5.
Answer 21. Connexon selectivity. Choices are 1, 2, and 3.
Answer 22. Choices are all the statements.
Answer 23. 80% of 8E10 cells.
Answer 24. 4.2E-4 s
Answer 25. vm rises 0.04 V
Answer 26. 2.451 megaohms
Answer 27. Cell 2 has transmembrane voltage of 0.019 V relative to rest.
Answer 28. 0.021 V
Answer 29. Time required to reach threshold is 2.3541E-4 s.
Answer 30. Nominal velocity 0.288 m/s
Answer 31a. 0.143 m/s
Answer 31b. 0.282 m/s
Answer 31c. 0.144 m/s
Answer 31d. 0.659 m/s
Answer 32. Increasing the velocity. Choose 1 and 4. Choices other than 4 are evaluated in the
preceding question, while 4 is evaluated in an earlier section.
Answer 33. Select choices 1, 2, 6, 7, and 8.
Answer 34. 1, 2, 4, and 5
BIOELECTRICITY: A QUANTITATIVE APPROACH
35
Answer 35. 0.002618 S/cm
Answer 36. 0.523599
Answer 37. 923.99784 1/cm−1
Answer 38. 209 Ωcm
Answer 39. 0.083 cm
Answer 40. 0.1617 cm
Answer 41. 0.0265 V
Answer 42. 0.0707 V
Answer 43. –0.0900 V
Answer 44. –0.0242 V
Answer 45. 0.0604 V
Answer 46. 0.0465 V
Answer 47. –0.097 V
Answer 48. Select 2, 4, 5, 6, 7
Answer 49. Select 2 and 3. The design fails because requirement 1 is not satisfied.
Answer 50. Select 1 and 3. The design fails because requirement 2 is not satisfied.
Answer 51. 0.8
Answer 52. –0.12 V. Note that ∆vm has a negative sign. It is the value across the excitation
wave, i.e., the value in the tissue ahead of the excitation wave minus the value in the
region the excitation wave has passed.
Answer 53. –0.6389
36
SOLUTIONS FOR CHAPTER 9: CARDIAC ELECTROPHYSIOLOGY
Answer 54. 250 Ωcm
Answer 55. 300 Ωcm
Answer 56. 0.0042E-4 V
Answer 57. φLL = 5.04E − 5 V
Answer 58. VII = 9.62E − 5 V
Answer 59. VII = −1.07E − 4 V
Answer 75a.
v
−σφ∇2
1
dV
r
σM ∇φH · dS Hin
1
· dS Hin
σM φH ∇
−
r
r
H
H
1
1
· dS Lin −
· dS L
−
σM φL ∇
σL φL ∇
r
r
L
L
1
· dsB
−
σM φB ∇
(13)
r
B
=
In Eq. (??), vector surface element dS Hin points into the heart region H, vector surface
element dS B points out of the body surface, vector surface element dS Lin points into
the lung region L, and vector surface element dS L points out of the lung region L. The
lung conductivity is designated σL , while the remaining torso conductivity is described
by σM .
In getting Eq. (??) from Eq. (13.18), several particular aspects of current flow in the torso
volume conductor were used. First, ∇2 φ is zero at every point within the volume, since
the volume has been drawn in such a way as to exclude the active tissue in the heart and to
exclude regions where σ makes a transition (i.e., the site of secondary sources). Second,
current flow across the lung boundary is continuous. Finally, the normal component of
electric field, ∇φ · n, approaches zero as a point approaches the body surface, since the
exterior of the body surface is a nonconductor (hence no current can leave the volume
conductor).
Answer 75b.
V
−σφ∇2
1
dV = 4πδ(r)σB φb
r
(14)
where φb and σB are the potential and conductivity at the point b, where r = 0. This
point is chosen arbitrarily and located just inside the body surface. This is the point for
which potentials will be determined.
BIOELECTRICITY: A QUANTITATIVE APPROACH
37
Answer 75c.
∇φH · dS Hin
1
· dS Hin
φH ∇
4πσB φb = σM
− σM
r
r
H
H
1
1
· dS L − σM
· dS B
−(σL − σM ) φL ∇
ΦB ∇
r
r
L
B
(15)
Answer 75d. As a first step we get:
σM
∇φH cdotdS H
ar · nH
σM
−
φH H 2
dSH
φb = −
4πσB H
r
4πσB H
rH
ar · nL
ar · nB
(σL − σM )
σM
+
φL L 2 dSL +
φB B 2
dSB
4πσB
r
4πσ
rB
B B
L
L
(16)
where unit normal vector nH points out of heart region H, nL points out of lung region
L, and nB points out of the body region B. Vectors arH , arL , and arB are all of unit
magnitude and their directions are associated with the radius vectors (rH , rL , rB ) from
b to the heart, lung, and body surface, respectively.
Using the equation for the solid angle in the equation above (??) gives
φb = −
+
∇φH · dS H
σM
σM
φH dΩH −
4πσB H
4πσB H
r
σM
(σL − σM )
φL dΩL +
φB dΩB
4πσB
4πσB B
L
(17)
This expression gives the potential at one point on the body surface in terms of several integrals. The integrals thereby can be interpreted as identifying the major factors affecting
the potential at a body surface point.
The first two integrals take into account the potentials and the gradients of potential (i.e.,
the currents) that exist on the cardiac surface. Although shown here as two separate
integrals, it is useful to note that the potentials and gradients on the cardiac surface are
not independent of each other.
The third integral takes into account the lack of uniformity in the volume conductor.
Specifically, in this derivation the integral takes into account the different conductivity
assigned to the lung volume. In problems with other inhomogeneities, additional integrals
would be present to take into account their effects. For example, the skeletal muscle is
another inhomogeneity (which also is anisotropic) that is thought to play a significant
role in determining the body surface potentials.
Finally, the fourth integral shows that the potential at a point on the body surface also
depends on the potentials at other points on the body surface. Why is this so? The body
surface defines the interface between the conducting torso and the nonconducting air,
locating an obvious discontinuity in conductivity. Accordingly, secondary sources arise
at this surface and the potential of any surface point will depend on the field generated
by all such secondary sources. In fact, the secondary source distribution will be such
that the field it generates when added to the field from all other sources (i.e., applied
sources) must satisfy the aforementioned boundary condition, namely, ∇φ · an = 0 on
38
SOLUTIONS FOR CHAPTER 9: CARDIAC ELECTROPHYSIOLOGY
B. In general, the body surface sources cause the potentials measured there to be higher
than they otherwise would be, by approximately a factor of 2 or 3.
While the integrals in (14.17) arise from field theoretic considerations, note that the
contribution from the lung and body surface contains the secondary source double layers
described by (8.64). In the sense of (14.17) the first two terms could be thought to
represent primary sources, namely, a double layer (with strength σM φH , as shown in
the first integral) and a single layer (with strength σM ∇φH · an , as shown in the second
integral). However, such sources are actually equivalent sources, and (14.17) can only be
used to calculate fields outside the heart.
The integrals in the expression above make unequal contributions to the body surface
potentials. A first approximation to the resulting body surface potential’s variation from
point to point can often be obtained after ignoring the last two integrals altogether. Further, because of their different r dependence, the contributions from the first and second
integrals may be quite unequal.
Answer 75e. The equation is the same except that the lung integral drops out.
BIOELECTRICITY: A QUANTITATIVE APPROACH
39
14.10. SOLUTIONS, CHAPTER 12: FUNCTIONAL ELECTRICAL STIMULATION
Answer 24. The exercises ask for portions of computer code that handle the Frankenhauser–
Huxley (FH) membrane model. The responses below are taken from a report by Chad
R. Johnson, which includes a complete program listing. (Johnson CR. 2005. A mathematical model of peripheral nerve stimulation in regional anesthesia, pp. 359–367. PhD
dissertation, Department of Biomedical Engineering, Duke University.)
Answer 24A. Code to initialize the necessary FH variables.
function [] = iFH(vm)
% IFH Initialize FH membrane model.
%
iFH initializes the parameters and gating
%
variables for the FH membrane model.
%
%
INPUT
vm: Transmembrane potential vector.
% -- PREAMBLE -----------------------------------------global
global
global
global
pFH;
CONST;
DATA;
UDATA;
% -- pFH STRUCTURE ------------------------------------% Set FH parameters
pFH = struct( ...
’Vrest’, -70.0,
’Nai’,
13.74,
’Nao’,
114.5,
in structure.
...
...
...
% mV
% mM
% mM
’Ki’,
120.0, ...
% mM
’Ko’,
2.5, ...
% mM
’El’,
0.026, ...
% mV
’PNa’,
8.0e-3, ... % cm/sec
’PK’,
1.2e-3, ... % cm/sec
’PP’,
0.54e-3, ... % cm/sec
’gbarl’, 30.3, ...
% mS/cmˆ2
’gmyel’,
DATA.gmyel, ... % mS/cmˆ2
’R’,
8314.4, ... % mJ/(K mole)
’T’,
UDATA.temp, ...
% K
’Q10’,
0, ...
’F’,
96485.0, ... % coulombs/mole
’D’,
0, ...
% Constant, defined below.
’Rv’,
0, ...
% Indices for active nodes.
’M’,
0, ...
% Indices for myelination.
’malpha’,
0, ...
% Current rate constants...
40
SOLUTIONS FOR CHAPTER 12: FUNCTIONAL ELECTRICAL STIMULATION
’mbeta’,
’halpha’,
’hbeta’, 0, ...
’nalpha’,
’nbeta’, 0, ...
’palpha’,
’pbeta’, 0, ...
’gates’, 0, ...
’I’,
0);
0, ...
0, ...
0, ...
0, ...
% -- INITIALIZE FH MODEL
% Gating variables.
% Ionic currents.
------------------------------
% Find indices for nodes and non-nodes.
pFH.Rv = find(DATA.ldx);
pFH.M = find(1-DATA.ldx);
% Define constant D (RT/F).
pFH.D = pFH.R.*pFH.T./pFH.F;
% Recalculate Nernst potentials for given temperature.
pFH.El = (pFH.T./293.15).*pFH.El;
% Define Q10’s (from Rattay review, 1993).
pFH.Q10.malpha = 1.8.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.mbeta = 1.7.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.halpha = 2.8.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.hbeta = 2.9.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.nalpha = 3.2.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.nbeta = 2.8.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.palpha = 3.0ˆ((pFH.T-293.15)./10.0);
pFH.Q10.pbeta = 3.0.ˆ((pFH.T-293.15)./10.0);
% Q10’s for PNa and PK
% from Frankenhaeuser and Moore:
% Q10.PNa = 1.3; Q10.PK = 1.2.
pFH.Q10.PNa = 1.3.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.PK = 1.2.ˆ((pFH.T-293.15)./10.0);
% Q10 for leakage current: added by CRJ.
pFH.Q10.gbarl = 1.2.ˆ((pFH.T-293.15)./10.0);
% Initialize rate constant vectors, msec-1.
pFH.halpha = pFH.Q10.halpha.*0.1.* ...
(-10.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)+10.0)./6.0));
pFH.hbeta = pFH.Q10.hbeta.*4.5./ ...
(1.0+exp((45.0-vm(pFH.Rv))./10.0));
pFH.malpha = pFH.Q10.malpha.*0.36.* ...
BIOELECTRICITY: A QUANTITATIVE APPROACH
(vm(pFH.Rv)-22.0)./ ...
(1.0-exp((22.0-vm(pFH.Rv))./3.0));
pFH.mbeta = pFH.Q10.mbeta.*0.4.* ...
(13.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)-13.0)./20.0));
pFH.nalpha = pFH.Q10.nalpha.*0.02.* ...
(vm(pFH.Rv)-35.0)./ ...
(1.0-exp((35.0-vm(pFH.Rv))./10.0));
pFH.nbeta = pFH.Q10.nbeta.*0.05.* ...
(10.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)-10.0)./10.0));
pFH.palpha = pFH.Q10.palpha.*0.006.* ...
(vm(pFH.Rv)-40.0)./ ...
(1.0-exp((40.0-vm(pFH.Rv))./10.0));
pFH.pbeta = pFH.Q10.pbeta.*0.09.* ...
(-25.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)+25.0)./20.0));
% Initialize gating variable vectors.
pFH.gates.m = pFH.malpha./(pFH.malpha+pFH.mbeta);
pFH.gates.h = pFH.halpha./(pFH.halpha+pFH.hbeta);
pFH.gates.n = pFH.nalpha./(pFH.nalpha+pFH.nbeta);
pFH.gates.p = pFH.palpha./(pFH.palpha+pFH.pbeta);
% Adjust pFH.Vrest to retain equilibrium.
INa = pFH.Q10.PNa.*pFH.PNa.*(pFH.gates.m(1).ˆ2.0).* ...
pFH.gates.h(1).*(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
IK = pFH.Q10.PK.*pFH.PK.*(pFH.gates.n(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Ko-pFH.Ki.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
IP = pFH.Q10.PNa.*pFH.PP.*(pFH.gates.p(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
Il = pFH.Q10.gbarl.*pFH.gbarl.*(-pFH.El);
Iion = INa+IK+IP+Il;
while (abs(Iion)>CONST.myzero)
INa = pFH.Q10.PNa.*pFH.PNa.*(pFH.gates.m(1).ˆ2.0).* ...
pFH.gates.h(1).*(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
41
42
SOLUTIONS FOR CHAPTER 12: FUNCTIONAL ELECTRICAL STIMULATION
(1.0-exp(pFH.Vrest./pFH.D)));
IK = pFH.Q10.PK.*pFH.PK.*(pFH.gates.n(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Ko-pFH.Ki.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
IP = pFH.Q10.PNa.*pFH.PP.*(pFH.gates.p(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
Il = pFH.Q10.gbarl.*pFH.gbarl.*(-pFH.El);
Iion = INa+IK+IP+Il;
pFH.Vrest = pFH.Vrest - ...
((UDATA.dt.*1.0e-3)./(mean(DATA.Ctot))).*Iion;
end
% Adjust El to get Iion=0 at rest.
pFH.El = (INa+IK+IP)./( pFH.Q10.gbarl.*pFH.gbarl);
% Initialize current vectors: include all nodes.
pFH.I.Na = zeros(size(vm));
pFH.I.K = zeros(size(vm));
pFH.I.P = zeros(size(vm));
pFH.I.l = zeros(size(vm));
pFH.I.myel = zeros(size(vm));
Answer 24B. Code to compute the membrane currents for the current membrane state:
function [] = iFH(vm)
% IFH Initialize FH membrane model.
%
iFH initializes the parameters and gating
%
variables for the FH membrane model.
%
%
INPUT
vm: Transmembrane potential vector.
% -- PREAMBLE ------------------------------------global
global
global
global
pFH;
CONST;
DATA;
UDATA;
% -- pFH STRUCTURE --------------------------------
BIOELECTRICITY: A QUANTITATIVE APPROACH
% Set FH parameters in structure.
pFH = struct( ...
’Vrest’, -70.0, ...
% mV
’Nai’,
13.74, ...
% mM
’Nao’,
114.5, ...
% mM
’Ki’,
120.0, ...
% mM
’Ko’,
2.5, ...
% mM
’El’,
0.026, ...
% mV
’PNa’,
8.0e-3, ... % cm/sec
’PK’,
1.2e-3, ... % cm/sec
’PP’,
0.54e-3, ... % cm/sec
’gbarl’, 30.3, ...
% mS/cmˆ2
’gmyel’,
DATA.gmyel, ... % mS/cmˆ2
’R’,
8314.4, ... % mJ/(K mole)
’T’,
UDATA.temp, ...
% K
’Q10’,
0, ...
’F’,
96485.0, ... % coulombs/mole
’D’,
0, ...
% Constant, defined below.
’Rv’,
0, ...
% Indices for active nodes.
’M’,
0, ...
% Indices for myelination.
’malpha’,
0, ...
% Current rate constants...
’mbeta’,
0, ...
’halpha’,
0, ...
’hbeta’, 0, ...
’nalpha’,
0, ...
’nbeta’, 0, ...
’palpha’,
0, ...
’pbeta’, 0, ...
’gates’, 0, ...
% Gating variables.
’I’,
0);
% Ionic currents.
% -- INITIALIZE FH MODEL
----------------------------
% Find indices for nodes and non-nodes.
pFH.Rv = find(DATA.ldx);
pFH.M = find(1-DATA.ldx);
% Define constant D (RT/F).
pFH.D = pFH.R.*pFH.T./pFH.F;
% Recalculate Nernst potentials for given temperature.
pFH.El = (pFH.T./293.15).*pFH.El;
% Define Q10’s (from Rattay review, 1993).
pFH.Q10.malpha = 1.8.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.mbeta = 1.7.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.halpha = 2.8.ˆ((pFH.T-293.15)./10.0);
43
44
SOLUTIONS FOR CHAPTER 12: FUNCTIONAL ELECTRICAL STIMULATION
pFH.Q10.hbeta = 2.9.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.nalpha = 3.2.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.nbeta = 2.8.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.palpha = 3.0ˆ((pFH.T-293.15)./10.0);
pFH.Q10.pbeta = 3.0.ˆ((pFH.T-293.15)./10.0);
% Q10’s for PNa and PK
% from Frankenhaeuser and Moore:
% Q10.PNa = 1.3; Q10.PK = 1.2.
pFH.Q10.PNa = 1.3.ˆ((pFH.T-293.15)./10.0);
pFH.Q10.PK = 1.2.ˆ((pFH.T-293.15)./10.0);
% Q10 for leakage current: added by CRJ.
pFH.Q10.gbarl = 1.2.ˆ((pFH.T-293.15)./10.0);
% Initialize rate constant vectors, msec-1.
pFH.halpha = pFH.Q10.halpha.*0.1.* ...
(-10.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)+10.0)./6.0));
pFH.hbeta = pFH.Q10.hbeta.*4.5./ ...
(1.0+exp((45.0-vm(pFH.Rv))./10.0));
pFH.malpha = pFH.Q10.malpha.*0.36.* ...
(vm(pFH.Rv)-22.0)./ ...
(1.0-exp((22.0-vm(pFH.Rv))./3.0));
pFH.mbeta = pFH.Q10.mbeta.*0.4.* ...
(13.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)-13.0)./20.0));
pFH.nalpha = pFH.Q10.nalpha.*0.02.* ...
(vm(pFH.Rv)-35.0)./ ...
(1.0-exp((35.0-vm(pFH.Rv))./10.0));
pFH.nbeta = pFH.Q10.nbeta.*0.05.* ...
(10.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)-10.0)./10.0));
pFH.palpha = pFH.Q10.palpha.*0.006.* ...
(vm(pFH.Rv)-40.0)./ ...
(1.0-exp((40.0-vm(pFH.Rv))./10.0));
pFH.pbeta = pFH.Q10.pbeta.*0.09.* ...
(-25.0-vm(pFH.Rv))./ ...
(1.0-exp((vm(pFH.Rv)+25.0)./20.0));
% Initialize gating variable vectors.
pFH.gates.m = pFH.malpha./(pFH.malpha+pFH.mbeta);
pFH.gates.h = pFH.halpha./(pFH.halpha+pFH.hbeta);
pFH.gates.n = pFH.nalpha./(pFH.nalpha+pFH.nbeta);
pFH.gates.p = pFH.palpha./(pFH.palpha+pFH.pbeta);
% Adjust pFH.Vrest to retain equilibrium.
INa = pFH.Q10.PNa.*pFH.PNa.*(pFH.gates.m(1).ˆ2.0).* ...
BIOELECTRICITY: A QUANTITATIVE APPROACH
pFH.gates.h(1).*(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
IK = pFH.Q10.PK.*pFH.PK.*(pFH.gates.n(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Ko-pFH.Ki.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
IP = pFH.Q10.PNa.*pFH.PP.*(pFH.gates.p(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
Il = pFH.Q10.gbarl.*pFH.gbarl.*(-pFH.El);
Iion = INa+IK+IP+Il;
while (abs(Iion)>CONST.myzero)
INa = pFH.Q10.PNa.*pFH.PNa.*(pFH.gates.m(1).ˆ2.0).* ...
pFH.gates.h(1).*(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
IK = pFH.Q10.PK.*pFH.PK.*(pFH.gates.n(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Ko-pFH.Ki.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
IP = pFH.Q10.PNa.*pFH.PP.*(pFH.gates.p(1).ˆ2.0).* ...
(pFH.Vrest.*pFH.F./pFH.D).* ...
((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...
(1.0-exp(pFH.Vrest./pFH.D)));
Il = pFH.Q10.gbarl.*pFH.gbarl.*(-pFH.El);
Iion = INa+IK+IP+Il;
pFH.Vrest = pFH.Vrest - ...
((UDATA.dt.*1.0e-3)./(mean(DATA.Ctot))).*Iion;
end
% Adjust El to get Iion=0 at rest.
pFH.El = (INa+IK+IP)./( pFH.Q10.gbarl.*pFH.gbarl);
% Initialize current vectors: include all nodes.
pFH.I.Na = zeros(size(vm));
pFH.I.K = zeros(size(vm));
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46
SOLUTIONS FOR CHAPTER 12: FUNCTIONAL ELECTRICAL STIMULATION
pFH.I.P = zeros(size(vm));
pFH.I.l = zeros(size(vm));
pFH.I.myel = zeros(size(vm));
Answer 24C. Code to advance the values of the gating variables:
function [] = gatesFH(vmNew,dt);
% GATESFH Update FH rate constants and gating variables.
%
GATESFH(vmNew,dt) updates the membrane channel
rate
%
constants using the vector of updated nerve fiber
%
transmembrane potentials vmNew, then finds the gating
%
variables for the next iteration given the integration
%
time step dt.
% -- PREAMBLE -----------------------------------------global pFH;
% -- RATE CONSTANTS -------------------------% Calculate rate constant vectors, in msec-1.
halphaNew = pFH.Q10.halpha.*0.1.* ...
(-10.0-vmNew(pFH.Rv))./ ...
(1.0-exp((vmNew(pFH.Rv)+10.0)./6.0));
hbetaNew = pFH.Q10.hbeta.*4.5./ ...
(1.0+exp((45.0-vmNew(pFH.Rv))./10.0));
malphaNew = pFH.Q10.malpha.*0.36.* ...
(vmNew(pFH.Rv)-22.0)./ ...
(1.0-exp((22.0-vmNew(pFH.Rv))./3.0));
mbetaNew = pFH.Q10.mbeta.*0.4.* ...
(13.0-vmNew(pFH.Rv))./ ...
(1.0-exp((vmNew(pFH.Rv)-13.0)./20.0));
nalphaNew = pFH.Q10.nalpha.*0.02.* ...
(vmNew(pFH.Rv)-35.0)./ ...
(1.0-exp((35.0-vmNew(pFH.Rv))./10.0));
nbetaNew = pFH.Q10.nbeta.*0.05.*...
(10.0-vmNew(pFH.Rv))./ ...
(1.0-exp((vmNew(pFH.Rv)-10.0)./10.0));
palphaNew = pFH.Q10.palpha.*0.006.* ...
(vmNew(pFH.Rv)-40.0)./ ...
(1.0-exp((40.0-vmNew(pFH.Rv))./10.0));
pbetaNew = pFH.Q10.pbeta.*0.09.* ...
(-25.0-vmNew(pFH.Rv))./ ...
(1.0-exp((vmNew(pFH.Rv)+25.0)./20.0));
% -- GATING VARIABLES -------------------------------
BIOELECTRICITY: A QUANTITATIVE APPROACH
% Update gating variables’ vectors for next iteration.
pFH.gates.m = (pFH.gates.m + (dt./2.0).* ...
(pFH.malpha + malphaNew - pFH.gates.m.*
(pFH.malpha + pFH.mbeta)))./ ...
(1 + (dt./2.0).*(malphaNew + mbetaNew));
pFH.gates.h = (pFH.gates.h + (dt./2.0).* ...
(pFH.halpha + halphaNew - pFH.gates.h.*
(pFH.halpha + pFH.hbeta)))./ ...
(1 + (dt./2.0).*(halphaNew + hbetaNew));
pFH.gates.n = (pFH.gates.n + (dt./2.0).* ...
(pFH.nalpha + nalphaNew - pFH.gates.n.*
(pFH.nalpha + pFH.nbeta)))./ ...
(1 + (dt./2.0).*(nalphaNew + nbetaNew));
pFH.gates.p = (pFH.gates.p + (dt./2.0).* ...
(pFH.palpha + palphaNew - pFH.gates.p.*
(pFH.palpha + pFH.pbeta)))./ ...
(1 + (dt./2.0).*(palphaNew + pbetaNew));
% Store current rates for next iteration.
pFH.malpha = malphaNew;
pFH.mbeta = mbetaNew;
pFH.halpha = halphaNew;
pFH.hbeta = hbetaNew;
pFH.nalpha = nalphaNew;
pFH.nbeta = nbetaNew;
pFH.palpha = palphaNew;
pFH.pbeta = pbetaNew;
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