336
CHAPTER 11. THE RETINA AND VISION
Chapter 12
Biochemical Reactions:
Solutions
Ex. 1: Consider the simple chemical reaction in which two monomers of A combine
to form a dimer B, according to
k+
A + A −→
←− B.
(12.1)
k−
(a) Use the law of mass action to find the differential equations governing
the rates of production of A and B.
(b) What quantity is conserved? Use this conserved quantity to find an
equation governing the rate of production of A that depends only on the
concentration of A.
(c) Non-dimensionalize this equation and show that these dynamics depend
on only one non-dimensional parameter.
Solution
(a) Using the law of mass action, we find the two differential equations
da
dt
db
dt
=
2k− b − 2k+ a2 ,
= k+ a2 − k− b,
(12.2)
(12.3)
where a = [A], b = [B] and where the factor of 2 in the first equation
denotes that 2 molecules of species A are used up or produced by each
of the reactions.
337
338
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
db
(b) Since da
dt + 2 dt = 0, the conserved quantity is a + 2b = B0 . Using this
conserved quantity, the equation for a becomes
da
= k− (B0 − a) − 2k+ a2 .
dt
(c) Introduce dimensionless variables u =
a
B0 ,
and t =
(12.4)
τ
k−
and find
du
2k+ B0 2
u .
=1−u−
dτ
k−
The sole dimensionless parameter is
(12.5)
2k+ B0
k− .
Ex. 2: In the real world trimolecular reactions are rare, although trimerizations are
not. Consider the following trimerization reaction in which three monomers
of A combine to form the trimer C,
k1
A + A −→
←− B,
k−1
k2
A + B −→
←− C.
k−2
(a) Use the law of mass action to find the rate of production of the trimer
C.
(b) Suppose k−1 >> k−2 , k2 A. Use the appropriate quasi-steady state approximation to find the rates of production of A and C, and show that
the rate of production of C is proportional to [A]3 . Explain in words why
this is so.
Solution
(a) Using the law of mass action, we find
da
dt
db
dt
dc
dt
=
2k−1 b − 2k1 a2 + k−2 c − k2 ab,
(12.6)
= k1 a2 − k−1 b + k−2 c − k2 ab,
(12.7)
= k2 ab − k−2 c,
(12.8)
(b) where a = [A], b = [B], and c = [C]. If k−1 is large compared to
other rate constants, then b equilibrates rapidly (as an exponential) to
its quasi-steady state,
k1 a2 + k−2 c
b=
.
(12.9)
k−1 + k2 a
339
If we substitute this quasi-steady state approximation into the equations
governing c and a, we find
dc
k2 k1 a3 − k−2 k−1 c
=
,
dt
k−1 + k2 a
da
dc
= −3 .
dt
dt
(12.10)
If k−1 >> k2 a, this reduces to
dc
k2 k1 3
a − k−2 c.
=
dt
k−1
(12.11)
Here, the rate at which the reaction takes place is represented by a cubic
term because, since k−1 is large, the dimer B comes apart quickly unless
it is stabilized by a collision with a third monomer of A.
Ex. 3: Consider an enzymatic reaction in which an enzyme can be activated or inactivated by the same chemical substance, as follows:
k1
E + X −→
←− E1 ,
(12.12)
k−1
k2
E1 + X −→
←− E2 ,
(12.13)
k−2
k
3
E1 + S −→
P + Q + E.
(12.14)
Suppose further that X is supplied at a constant rate and removed at a rate
proportional to its concentration. Use quasi-steady-state analysis to find the
nondimensional equation describing the degradation of X,
dx
βxy
.
=γ−x−
dt
1 + x + y + αx2
(12.15)
Identify all the parameters and variables and the conditions under which the
quasi-steady state approximation is valid.
Solution
We begin by defining X to be the concentration of X, e1 to be the concentration of E1 , and similarly for s, e and e2 . We then use the law of mass action
to write the differential equations for the reaction system
dX
dt
de1
dt
de2
dt
= k−1 e1 + k−2 e2 − k1 eX − k2 e1 X + a − bX,
(12.16)
= k−2 e2 − k2 e1 X − k3 e1 s − k−1 e1 + k1 eX,
(12.17)
= k2 e1 X − k−2 e2 .
(12.18)
340
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
The fact that X is added at a constant rate and removed at a rate proportional
to its concentration is represented by the terms a − bX in the equation (12.16)
for X. We also have the conservation law
e + e1 + e2 = et ,
(12.19)
where et is the (constant) total amount of enzyme. Notice that, because of
this conservation law, we do not need an additional equation for the dynamics
of e.
Next, we non-dimensionalize the equations. This is often the most difficult
part of the problem as there is no unique way to do this. However, for a
system of ordinary differential equations, as this is, we must rescale each of
the dependent variables and the one independent variable t.
A reasonable scaling for e, e1 , and e2 is et , so we set e = et z, e1 = et z, and
e2 = et z. We do not yet know how to scale X, so for the present we leave it
unscaled. The removal rate of X provides a typical time constant, so we set
t = bτ . With these new variables, the equations become
dX
k−2 et
k1 et
a
k−1 et
k2 et
=
z1 +
z2 −
zX −
z1 X + − X, (12.20)
dτ
b
b
b
b
b
dz1
k2
k3
k−1
k1
k−2
=
z 2 − z1 X − z1 s −
z1 + (1 − z1 − z2 )X,(12.21)
dτ
b
b
b
b
b
dz2
k−2
k2
(12.22)
=
z1 X −
z2 .
dτ
b
b
1 = z + z1 + z2 .
(12.23)
Notice that the equations for z1 and z2 describe exponential processes with
k−2
exponential rates of decay at least as big as k−1
b and b . Thus, if these two
numbers are large compared to one it is legitimate to approximate the solution
of these equations by their steady state solutions,
0
=
0
=
k−2
k2
k3
k−1
k1
z2 − z 1 X − z 1 s −
z1 + (1 − z1 − z2 )X, (12.24)
b
b
b
b
b
k2
k−2
z1 X −
z2 ,
(12.25)
b
b
from which we find that
z1
=
z2
=
k1 k−2 X
,
k−2 k3 s + k−2 k−1 + k−2 k1 X + k1 k2 X 2
k2
z1 X.
k−2
(12.26)
(12.27)
We substitute these into the equation for X, finding
dX
a
et
k−2 k1 k3 sX
= −X −
,
dτ
b
b k−2 k−1 + k−2 k3 s + k−2 k1 X + k2 k1 X 2
(12.28)
341
Now, to make the equation look “pretty”, we set s = Sy, X = X̄x, where
S = kk−1
, X̄ = kk−1
and find
3
1
dx
yx
et k1 k3 S
a
−x−
=
dτ
b k−1 1 + y + x +
X̄b
which is
with γ =
k2 k−1 2
k−2 k1 x
dx
βxy
=γ−x−
dτ
1 + y + x + αx2
k1 a
k−1 b ,
β=
k 1 et
b ,
α=
,
(12.29)
(12.30)
k2 k−1
k−2 k1 .
Ex. 4: There is an alternate derivation of the Michaelis-Menten rate equation that
does not require that = es00 be small. Instead, if one or both of k−1 and
k2 are much larger than k1 e0 , then the formation of complex c is a rapid
exponential process, and can be taken to be in quasi-steady state. Make this
argument systematic by introducing appropriate non-dimensional variables
and then find the resulting quasi-steady state dynamics.
Solution
We introduce dimensionless variables
s
c
σ= , x= ,
s0
e0
τ = k1 e0 t,
(12.31)
in terms of which equations (1.9) and (1.11) become
dσ
dt
= αx − σ(1 − x),
dx
+ Kx =
dt
s0
σ(1 − x),
e0
(12.32)
(12.33)
k2 +k−1
where α = kk1−1
>> 1. Because K is large compared to
s0 , and K =
k 1 e0
one, the dynamics of x are such that x equilibrates rapidly and exponentially,
regardless of the amplitude of es00 . Thus, in quasi-steady state
Kx =
s0
σ(1 − x),
e0
(12.34)
which in dimensioned variables is exactly (1.23) found before. Thus, this approximation gives the same answer as the quasi-steady state analysis described
in the text.
Ex. 5: An enzyme-substrate system is believed to proceed at a Michaelis- Menten
rate. Data for the (initial) rate of reaction at different concentrations is shown
in Table 12.1.
342
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
Table 12.1: Data for Problem 5.
Substrate Concentration (mM) Reaction Velocity (mM/s)
0.1
0.04
0.2
0.08
0.5
0.17
1.0
0.24
2.0
0.32
3.5
0.39
5.0
0.42
(a) Plot the data V vs. s. Is there evidence that this is a Michaelis-Menten
type reaction?
(b) Plot V vs. V /s. Is this data well approximated by a straight line?
(c) Use linear regression and (1.25) to estimate Km and Vmax . Compare
the data to the Michaelis-Menten rate function using these parameters.
Does this provide a reasonable fit to the data?
Solution
(b) Rewrite the equation (1.25) as
Vmax −
V
Km = V.
s
For each experiment, there is one such equation,
points we have the linear system of equations



1.0 −0.40
 1.0 −0.40 




 1.0 −0.32  


 Vmax
Vmax

A
=
=
 1.0 −0.26 

Km
K
m
 1.0 −0.16 




 1.0 −0.11 

1.0 −0.08
(12.35)
so with seven data
0.04
0.08
0.16
0.26
0.32
0.39
0.42





 = b. (12.36)




To find the linear regression solution (also called the least-squares solution) of this system, solve the normal equations
Vmax
T
A A
= AT b,
(12.37)
Km
which in this case is the 2 × 2 system
7.0000 −1.7354
Vmax
1.6700
=
,
−1.7354 0.5351
−0.2967
Km
(12.38)
343
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.4
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.05
0
Figure 12.1:
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Figure 12.2: Michaelis-Menten growth curve for data of Problem 5.
344
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
having the solution Vmax = 0.5160, Km = 1.1189. A plot of the line
y = Vmax − Km x is shown with data points in Fig. 12, and a plot of the
reaction velocity with data points is shown in Fig. 12. This solution can
be found by running the Matlab code ex Burk.m.
Ex. 6: Using the quasi-steady-state approximation, show that the velocity of the
reaction for an enzyme with an allosteric inhibitor (Section 1.5.2) is given by
k2 e0 K3
s(κ3 i + κ2 + s)
V =
.
(12.39)
K3 + i (κ3 i(s + K1 ) + (s + κ2 )(s + κ1 )
Identify all parameters. Under what conditions on the rate constants is this
a valid approximation? Show that this reduces to (1.38) in the case K1 = κ1 .
Solution
Applying the law of mass action to the reaction in Fig.1.1, we find that
de
dt
dy
dt
dz
dt
ds
dt
= (k−1 + k2 )x + k−3 y − k1 se − k3 ie,
(12.40)
= k−1 z + k3 ie − k1 sy − k−3 y,
(12.41)
= k3 ix + k1 sy − (k−3 + k−1 )z.
(12.42)
= −k1 (e + y)s + k−1 z + k−1 x
(12.43)
where e = [E], y = [EI], and z = [EIS]. We do not need a differential equation
for x = [ES] because of the conservation law
e0 = e + x + y + z.
(12.44)
If k−3 and either of k−1 or k2 are large compared to k1 e0 , then reactions of e, y,
and z are fast compared to s, regardless of the amount of s and i. In this case,
dy
dz
the quasi-steady state approximation is valid and we set de
dt = dt = dt = 0.
This gives the three equations
(k−1 + k2 )(e0 − e − y − z) + k−3 y − k1 se − k3 ie = 0,
k−1 z + k3 ie − k1 sy − k−3 y = 0,
k3 i(e0 − e − y − z) + k1 sy − k−3 z − k−1 z
=
0.
(12.45)
(12.46)
(12.47)
To find the velocity of the reaction, i.e., the rate at which product is produced,
we need to solve (12.45)–(12.47) for x. Since this is a linear system this can
be done by hand, although it is tedious. It is much more convenient to use a
symbolic manipulation package to find the solution. A simple Maple code (in
file allosteric.mp) for solving these equations is
345
readlib(unassign):
unassign(’x’,’y’,’z’):
fun1 := (k1*s + k3*i)*(e0-x-y-z) - km1*x - km3*y - k2*x:
fun2 := (km3+k1*s)*y - km1*z - k3*i*(e0-x-y-z) :
fun3 := (km3 + km1)*z - k3*i*x - k1*s*y :
sols := solve({fun1=0,fun2=0,fun3=0},{x,y,z}):
assign(sols):
simplify(x);
This gives the solution
x=
(k3 i + k−3 + k−1 + k1 s)k1 se0 k−3
(k−3 + k3 i)(k3 i(k1 s + k−1 ) + ((k2 + k−1 + k1 s)(k−1 + k−3 + k1 s))
(12.48)
It remains only to write the velocity V in the form
V
dp
= k2 x
dt
k2 e0 K3
s(κ3 i + κ2 + s)
=
K3 + i κ3 i(s + K1 ) + (s + κ2 )(s + κ1 )
=
(12.49)
2
−1
where Vmax = k2 e0 4, κ3 = kk31 , K1 = kk−1
, κ1 = k−1k+k
, κ2 = k−3k+k
which
1
1
1
is the required answer. Notice that the maximum velocity occurs when all the
enzyme is in the ES form (so there is no) inhibition), and so the maximum
velocity is k2 e0 . In the special case that K1 = κ1 , V reduces to
k2 e0 K3
s
V =
(12.50)
K3 + i (s + K1 )
which is the same as (1.38).
Ex. 7: In the case of noncompetitive inhibition, the inhibitor combines with the
enzyme-substrate complex to give an inactive enzyme-substrate-inhibitor complex which cannot undergo further reaction, but the inhibitor does not combine directly with free enzyme or affect its reaction with substrate. Use the
quasi-steady-state approximation to determine that the velocity of this reaction is
s
V = Vmax
.
(12.51)
Km + s + Ki i s
Identify all parameters. Compare this velocity with the velocity for other
types of inhibition discussed in the text.
Solution
346
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
For this case the reaction scheme is
S+E
k1
−→
←−
k−1
k
2
C1 −→E
+ P,
C1 + I
k3
−→
←−
k−3
C2 .
and from the law of mass action the governing equations are
ds
dt
de
dt
dc1
dt
dc2
dt
dp
dt
= −k1 se + k−1 c1 ,
(12.52)
= −k1 se + (k−1 + k2 c1 ,
(12.53)
= k1 se − (k−1 + k2 + k3 i)c1 + k−3 c2 ,
(12.54)
= k3 c1 i − k−3 c2 ,
(12.55)
= k2 c1 ,
(12.56)
where s = [S], i = [I], c1 = [C1 ], and c2 = [C2 ]. We know that e+c1 +c2 = e0 .
Observe that if k−1 + k2 + k3 i and k−3 are large compared to k1 e0 , then
the reactions of c1 and c2 are fast compared to s and the quasi-steady state
dc2
1
approximation is valid. We set dc
dt = dt = 0, so that
0
= k1 s(e0 − c1 − c2 ) − (k−1 + k2 + k3 i)c1 + k−3 c2 ,
(12.57)
0
= k3 c1 i − k−3 c2 .
(12.58)
We need to know c1 since this determines the velocity of the reaction. We
easily find that
k−3 c2 = k3 c1 i,
(12.59)
so that
c1 =
k1 se0
k−1 + k2 + k1 s +
k3
k−3 is
,
(12.60)
yielding the velocity of reaction
V = K2 c1 = Vmax
where Km =
k−1 +k2
,
k1
Ki =
k−3 k1
k3 ,
s
Km + s +
i
Ki s
,
(12.61)
and Vmax = k2 e0 .
Ex. 8: Suppose the maximum velocity of a chemical reaction is known to be 1 mM/s,
and the measured velocity V of the reaction at different concentrations s is
shown in Table 12.
(a) Plot the data V vs. s. Is there evidence that this is a Hill type reaction?
347
Table 12.2: Data for Problem 8.
Substrate Concentration (mM) Reaction Velocity (mM/s)
0.2
0.01
0.5
0.06
1.0
0.27
1.5
0.50
2.0
0.67
2.5
0.78
3.5
0.89
4.0
0.92
4.5
0.94
5.0
0.95
V
(b) Plot ln( Vmax
−V ) vs. ln(s). Is this approximately a straight line, and if
so, what is its slope?
(c) Use linear regression and (1.50) to estimate Km and the Hill exponent n.
Compare the data to the Hill rate function with these parameters. Does
this provide a reasonable fit to the data?
Solution
(c) Rewrite the equation (1.50) as
−n ln Km + n ln s = ln
V
Vmax − V
,
(12.62)
and view n and −n ln Kn as unknown variables. For each experiment,
there is one such equation, so with ten data points we have the linear
system of equations




1.0 −1.6094
−4.5951
 1.0 −0.6931 
 −2.7515 




 1.0

 −0.9946 
0








0

 1.0 0.4055  
 1.0 0.6931  −n ln Km
 0.7082 
−n ln Km



 = b.
A
=
=


n
n
 1.0 0.9163 
 1.2657 
 1.0 1.2528 
 2.0907 




 1.0 1.3863 
 2.4423 




 1.0 1.5041 
 2.7515 
1.0 1.6094
2.9444
(12.63)
To find the linear regression solution of this system, solve the normal
equations
−n ln Km
T
A A
= AT b,
(12.64)
n
348
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
1
0.9
0.8
0.7
V
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
s
3
3.5
4
4.5
5
3.5
4
4.5
5
Figure 12.3:
1
0.9
0.8
0.7
V
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
s
3
Figure 12.4: Hill curve for data of Problem 8.
which in this case is the 2 × 2 system
10.0000 5.4649
−n ln Km
3.8616
=
,
5.4649 12.8990
n
25.8358
(12.65)
having the solution n = 2.4, Km = 1.47. A plot of the line y = n ln s −
n ln Km is shown with data points in Fig. 12, and a plot of the reaction
velocity with data points is shown in Fig. 12. This solution can be found
by running the Matlab code ex Hill.m.
Ex. 9: (a) Find the velocity of reaction for an enzyme with three active sites. Under
what conditions does the velocity reduce to a Hill function with exponent
three? Identify all parameters.
(b) What is the relationship between rate constants when the three sites are
independent? What is the velocity when the three sites are independent?
Solution
349
(a) For an enzyme with three active sites, the reaction mechanism is represented by
k1
k
2
S + E −→
←− C1 −→E + P,
(12.66)
k−1
k3
S + C1
−→
←−
k
4
C2 −→C
1+P
(12.67)
k−3
k5
S + C2
−→
←−
k
6
C3 −→C
2 + P.
(12.68)
k−5
Using the law of mass action, one can write down the rate equations
for the 6 concentrations [S], [E], [C1 ], [C2 ], [C3 ], and [P]. However,
because the amount of product [P] can be determined by quadrature,
and because the total amount of enzyme molecule is conserved, we only
need four equations for the four quantities [S], [C1 ], [C2 ], and [C3 ]. These
are
ds
dt
dc1
dt
dc2
dt
dc3
dt
= −k1 se + k−1 c1 − k3 sc1 + k−3 c2 − k5 sc2 + k−5 c3 ,(12.69)
= k1 se − (k−1 + k2 )c1 − k3 sc1 + (k4 + k−3 )c2 ,
(12.70)
= k3 sc1 − (k4 + k−3 )c2 − k5 sc2 + (k6 + k−5 )c3 ,
(12.71)
= k5 sc2 − (k6 + k−5 )c3 ,
(12.72)
where s = [S], c1 = [C1 ], c2 = [C2 ], c3 = [C3 ], and e + c1 + c2 + c3 = e0 .
Proceeding as before, we invoke the quasi-steady-state assumption that
dc1
dc2
dc3
dt = dt = dt = 0 (Under what conditions is this a valid assumption?),
and solve for c1 , c2 , and c3 to get (this is easily done using the maple file
three sites.mp)
c1
=
c2
=
c3
=
K2 K3 se0
,
+ K3 + K3 K2 s + K3 K2 K1
K3 s2 e0
,
s3 + K3 s2 + K3 K2 s + K3 K2 K1
s3 e0
,
s3 + K3 s2 + K3 K2 s + K3 K2 K1
s3
s2
(12.73)
(12.74)
(12.75)
where
K1 =
k−1 + k2
,
k1
K2 =
k4 + k−3
,
k3
K3 =
k6 + k−5
.
k5
(12.76)
It follows that the velocity of the reaction is
V
= k2 c1 + k4 c2 + k6 c3
k2 K2 K3 s + k4 K3 s2 + k6 s3
= e0 3
.
s + K3 s2 + K3 K2 s + K3 K2 K1
(12.77)
(12.78)
350
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
We recover a Hill function with exponent three in the limit that K2 → 0,
K3 → 0, while K1 K2 K3 remains bounded and nonzero.
(b) With three independent binding sites, k4 = 2k2 , k6 = 3k2 , k1 = 3k5 ,
k3 = 2k5 , k−5 = 3k−1 , k−3 = 2k−1 , so that K3 = 2K2 , and K1 = 13 K2 .
It follows that
k2 s
V =3
,
(12.79)
s + K2
where K2 =
k2 +k−1
.
k5
Ex. 10: (a) Derive the expression (1.55) for the fraction of occupied sites in a Monod–
Wyman–Changeux model with n binding sites.
(b) The principle of detailed balance says that for a system of reactions to
be in thermal equilibrium (as opposed to merely at a steady state), each
individual reaction must be at equilibrium. Thus, for example, for the
cycle of reactions
k1
k2
k3
k−1
k−2
k−3
−→
−→
A −→
←− B ←− C ←− A
(12.80)
to be in thermal equilibrium, we must have [A] = K1 [B], [B] = K2 [C],
and [C] = K3 [A], where Ki = k−i /ki , i = 1, 2, 3. Hence, for this reaction
cycle, detailed balance implies the relation k1 k2 k3 = k−1 k−2 k−3 .
Modify the Monod–Wyman–Changeux model shown in Fig. 1.3 to allow
transitions between states R1 and T1 , and between states R2 and T2 .
Use the principle of detailed balance to derive an expression for the equilibrium constant of each of these transitions. Do these transitions change
the expression for Y , the fraction of occupied sites?
Solution
(a) It is usually the case that to prove a result for general n, it is often
instructive to work through the proofs for several special cases. The case
of n = 2 is done in the text, so here we begin with the case n = 3.
The binding diagram for n = 3 is shown in Fig.0a. Assuming that each
reaction is at equilibrium we get
=
3sK1−1 r0 ,
(12.81)
r3
=
=
3s K1−2 r0 ,
s3 K1−3 r0 ,
(12.82)
(12.83)
t1
=
3sK3−1 t0 ,
(12.84)
t2
= 3s2 K3−2 t0 ,
= s3 K3−3 t0 .
r1
r2
2
and similarly,
t3
(12.85)
(12.86)
351
2sk1
3sk1
R0
k2
R1
k-1
2k-1
3sk3
2sk3
sk1
R2
3k-1
R3
k-2
T0
T1
k-3
2k-3
T2
sk 3
3k-3
T3
Figure 12.5: Diagram of the states of the protein and the possible transitions in an
eight-state Monod-Wyman-Changeux model (i.e., the n = 3 case.
By definition, the saturation function is
r1 + 2r2 + 3r3 + t1 + 2t2 + 3t3
Y =
.
3(r0 + r1 + r2 + r3 + t0 + t1 + t2 + t3 )
(12.87)
Using the above relations, we quickly see that
r1 + 2r2 + 3r3
=
=
3sK1−1 r0 + 6(sK1−1 )2 r0 + 3(sK1−1 )3 r0
3r0 sK1−1 (1 + K1−1 )2 ,
(12.88)
t1 + 2t2 + 3t3
=
3sK3−1 t0 + 6(sK3−1 )2 t0 + 3(sK3−1 )3 t0
=
3t0 sK3−1 (1 + K3−1 )2 .
and
(12.89)
Similarly,
3(r0 + r1 + r2 + r3 )
= 3r0 (1 + 3sK1−1 + 3(sK1−1 )2 + (sK1−1 )3 )
(12.90)
= 3r0 (1 + sK1−1 )3 .
and
3(t0 + t1 + t2 + t3 )
Finally, we use that
= 3t0 (1 + 3sK3−1 + 3(sK3−1 )2 + (sK3−1 )3 )
= 3t0 (1 + sK3−1 )3 .
(12.91)
t0 = K2−1 r0 ,
(12.92)
so that
Y
=
r1 + 2r2 + 3r3 + t1 + 2t2 + 3t3
3(r0 + r1 + r2 + r3 + t0 + t1 + t2 + t3 )
=
3r0 sK1−1 (1 + K1−1 )2 + 3t0 sK3−1 (1 + K3−1 )2
3r0 (1 + sK1−1 )3 + 3t0 (1 + sK3−1 )3
=
3sK1−1 (1 + K1−1 )2 + 3K2−1 sK3−1 (1 + K3−1 )2
,
3(1 + sK1−1 )3 + 3K2−1 (1 + sK3−1 )3
(12.93)
352
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
as expected.
The calculation for the general case (depicted in Fig. 0a) should now be
clear. We first observe that
n
n
iri + 1 iti
1
n .
Y = n
(12.94)
n( 0 ri + 0 ti )
From the assumption that all reactions are at chemical equilibrium we
find
r1
r2
r3
ri
= nsK1−1 r0 ,
n−1
n−1
=
sK1−1 r1 =
n(sK1−1 )2 r0 ,
2
2
n−2
n−1
=
n(sK1−1 )3 r0 ,
3
2
..
.
n!
=
(sK1−1 )i r0 .
(n − i)!i!
(12.95)
(12.96)
(12.97)
(12.98)
Thus,
n
iri
= r0
i=1
n
i=1
i
n!
(sK1−1 )i
(n − i)!i!
= r0 n(sK1−1 )
n
i=1
= r0 n(sK1−1 )
n−1
i=0
(n − 1)!
(sK1−1 )i−1
(n − i)!(i − 1)!
(n − 1)!
(sK1−1 )i
(n − 1 − i)!i!
= r0 n(sK1−1 )(1 + sK1−1 )n−1 .
(because of the binomial expansion formula (1 + x)k =
Similarly,
n
n
i=0
rn
= nr0
n
0
(12.99)
k
k!
j
j=0 (k−j)!j! x ).
n!
(sK1−1 )i
(n − i)!i!
= nr0 (1 + sK1−1 )n
(12.100)
The same argument works for the T states (using, of course, K3 instead
of K1 ). We substitute these expressions into (12.94), and use that t0 =
K2−1 r0 to find
Y =
sK1−1 (1 + sK1−1 )n−1 + K2−1 [sK3−1 (1 + sK3−1 )n−1 ]
.
(1 + sK1−1 )n + K2−1 (1 + sK3−1 )n
as required.
(12.101)
353
nsk1
(n-1)sk1
R0
R1
2k-1
k-1
k2
R2 . . .
sk1
...
sk 3
nk-1
Rn
k-2
nsk3
T0
(n-1)sk3
T1
k-3
T2
2k-3
nk-3
Tn
Figure 12.6: Diagram of the states of the protein and the possible transitions in the
general case.
sk1
2sk1
R0
R1
k-1
k2
k4
k-2
k-4
2sk3
T0
k-3
R2
2k-1
k5
k-5
sk3
T1
2k-3
T2
Figure 12.7: Diagram of the states of the protein and the possible transitions when
R1 -T1 and R2 -T2 are allowed.
354
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
(b) The new binding diagram is shown in Fig.0b. Applying detailed balance
around the reaction loop R0 -T0 -T1 -R1 gives
2sk2 k3 k−4 k−1 = 2sk1 k4 k−3 k−2 ,
(12.102)
K1 K4 = K2 K3 ,
(12.103)
so that
where, as usual, Ki = k−i /ki . Thus, K4 is determined by K1 , K2 and
K3 . Similarly, from the R1 -T1 -T2 -R2 reaction loop
K1 K5 = K3 K4 .
(12.104)
Do these transitions make any difference to the calculation of Y ? No,
they do not, because the additional equilibrium constraints
r1
r2
= K 4 t1 ,
= K 5 t2 ,
(12.105)
(12.106)
are automatically satisfied. To see this, notice that the equilibrium conditions
r1 = 2sK1−1 r0 ,
t1 = 2sK3−1 t0 ,
(12.107)
used before imply that
r1
=
2sK1−1 r0
=
2sK1−1 K2 t0
t1 K 3
2sK1−1 K2
2s
=
= K1−1 K2 K3 t1
= K 4 t1 .
(12.108)
It follows similarly that r2 = K5 t2 is automatically satisfied. Thus,
because of the constraints on the reaction rates, the additional transitions are automatically in equilibrium and so the expression for Y is
unchanged.
Ex. 11: Suppose that a substrate can be broken down by two different enzymes with
different kinetics. (This happens, for example, in the case of cAMP or cGMP,
which can be hydrolyzed by two different forms of phosphodiesterase—see
Chapter 11).
(a) Write down the reaction scheme and differential equations, and nondimensionalize, to get the system of equations
dσ
dt
dx
1
dt
dy
2
dt
= −σ + α1 (µ1 + σ)x + α2 (µ2 + σ)y
=
=
1
σ(1 − x) − x, soleq : ex8.2
λ1
1
σ(1 − y) − y.
λ2
(12.109)
(12.110)
(12.111)
355
where x and y are the nondimensional concentrations of the two complexes. Identify all parameters.
(b) Apply the quasi-steady-state approximation to find the equation governing the dynamics of substrate σ. Under what conditions is the quasisteady state approximation valid?
(c) Solve the differential equation governing σ.
(d) For this system of equations, show that the solution can never leave the
positive octant σ, x, y ≥ 0. By showing that σ + 1 x + 2 y is decreasing
everywhere in the positive octant, show that the solution approaches the
origin for large time.
Solution
(a) The chemical equations for the two reactions are
k1
−→
←−
S + E1
k
2
C1 −→E
1 + P,
k−1
k3
S + E2
−→
←−
k
4
C2 −→E
2 + P.
k−3
As usual we let lower letters denote concentrations. Applying the law of
mass action gives
ds
dt
dc1
dt
dc2
dt
= k−1 c1 + k−3 c2 − k1 se1 − k3 se2 ,
(12.112)
= k1 se1 − (k−1 + k2 )c1 ,
(12.113)
= k3 se2 − (k−3 + k4 )c2 .
(12.114)
Since the total amounts of the enzymes are conserved,
e1 + c1
e2 + c2
= et ,
= eT ,
(12.115)
(12.116)
and we can eliminate e1 and e2 to find
ds
dt
dc1
dt
dc2
dt
= k−1 c1 + k−3 c2 − k1 s(et − c1 ) − k3 s(eT − c2 ), (12.117)
= k1 s(et − c1 ) − (k−1 + k2 )c1 ,
(12.118)
= k3 s(eT − c2 ) − (k−3 + k4 )c2 .
(12.119)
The most natural nondimensionalization of the concentrations is
c1
x =
,
(12.120)
et
356
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
y
=
σ
=
c2
,
eT
s
,
s0
(12.121)
(12.122)
where s0 is the initial amount of substrate. Since it is not immediately
obvious how to scale time, we set t = µτ , and wait until later to determine
µ. This change of variables gives
dσ
dt
1 dx
µ dt
1 dy
µ dt
= k−1
µet
µeT
x + k−3
y − µk1 σet (1 − x) − µk3 σeT (1 (12.123)
− y),
s0
s0
= k1 s0 σ(1 − x) − (k−1 + k2 )x,
(12.124)
= k3 s0 σ(1 − y) − (k−3 + k4 )y.
(12.125)
Now we want to exploit the fact that there is far more substrate than
either enzyme, and pick µ in a way that emphasizes that under these
conditions, x and y are “fast” variables, compared to σ. Notice that all
the linear terms proportional to σ on the right hand side of (12.123) are
multiplied by either et or eT , suggesting that this variable is slow. This
1
also suggests that we should pick µ = k1 et +k
in which case we find
3 eT
dσ
dt
k1 et + k3 eT dx
k−1 + k2 dt
k1 et + k3 eT dy
k−3 + k4 dt
= −σ +
=
=
et (k−1 + k1 s0 σ)
eT (k−3 + k3 σ)
x+
(12.126)
y
s0 (k1 et + k3 s0 eT )
s0 (k1 et + k3 eT )
k1 s0
σ(1 − x) − x,
k−1 + k2
k3 s0
σ(1 − y) − y.
k−3 + k4
(12.127)
(12.128)
which is
dσ
dt
dx
1
dt
dy
2
dt
where 1 =
k−1
k 1 s0 ,
µ2 =
= −σ + α1 (µ1 + σ)x + α2 (µ2 + σ)y,
=
=
1
σ(1 − x) − x,
λ1
1
σ(1 − y) − y.
λ2
k1 et +k3 eT
k1 et +k3 eT
k−1 +k2 , 2 = k−3 +k4
k−3
k 1 et
k3 s0 , α1 = k1 et +k3 eT , and
(12.129)
(12.130)
(12.131)
k−1 +k2
k−3 +k4
k1 s0 , λ2 = k3 s0 , µ1
k 3 eT
k1 et +k3 eT , with α1 + α2 =
, λ1 =
=
α2 =
1.
(b) Under the assumption that both 1 and 2 are small, we set 1 = 2 = 0
and find the quasi-steady state solution
x =
y
=
σ
,
λ1 + σ
σ
.
λ2 + σ
(12.132)
(12.133)
357
We substitute these expressions into the differential equation for σ to get
the equation for σ,
dσ
µ1 + σ
µ2 + σ
= −σ + α1 σ
+ α2 σ
.
dt
λ1 + σ
λ2 + σ
(12.134)
(c) The equation (12.134) can be solved by separating variables. First, we
rewrite it as
(λ1 + σ)(λ2 + σ)
dσ = −dt.
(12.135)
σ(θ1 σ + θ2 )
where
θ1
= λ1 + λ2 − α1 (µ1 + λ2 ) − α2 (µ2 + λ1 )
(12.136)
θ2
= λ1 λ2 − α1 µ1 λ2 α2 µ2 λ1 .
(12.137)
(It is important to notice that α1 + α2 = 1.) Notice that in original
dimensioned parameters these are
k1 et k2 + eT k4 k1
(12.138)
(k1 et + k3 eT )k1 k3 s0
1
k1 et k2 (k−3 + k4 ) + k3 eT k4 (k−1 (12.139)
+ k2 ),
(k1 et + k3 eT )k1 s0 k3 s0
θ1
= k3
θ2
=
both positive.
We integrate both sides of (12.135) to find
−
σ λ1 λ2 ln(σ)
λ 1 λ2 λ1 + λ2 θ 2
−
+(
−
+ 2 ) ln(θ1 σ+θ2 ) = t+K, (12.140)
θ1
θ2
θ2
θ1
θ1
for some constant K. To determine K, we apply the initial condition
σ(0) = 1, from which it follows that
K=−
1
λ1 λ 2
λ1 + λ2
θ2
+(
−
+ 2 ) ln(θ1 + θ2 ).
θ1
θ2
θ1
θ1
(12.141)
Thus,
1 − σ λ1 λ2 ln(σ) λ1 λ2 λ1 + λ2 θ2
−
+(
−
+ 2 ) ln
θ1
θ2
θ2
θ1
θ1
θ1 σ + θ2
θ1 + θ2
= t. (12.142)
We can now plot the solutions. (This is most easily done by considering
t to be a function of σ, and then reversing the axes.) The results, for one
particular choice of the parameters, are shown in Fig. 1. This calculation
is done in maple using the file two enzymes.mp.
It is important to realize that the graphs for x and y are incorrect at
small times, since at t = 0, x and y are nonzero. This is not physically
possible, since all the enzymes start in the unbound form. Hence, using
the quasi-steady state approximation we have constructed a solution that
is incorrect close to t = 0. To correct this requires a knowledge of how
358
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
1.0
0.8
σ
0.6
y
0.4
x
0.2
0
1
2
time (dimensionless)
3
4
Figure 12.8: Plots of σ, x and y for the parameters µ1 = 0.5, µ2 = 0.6, λ1 =
0.9, λ2 = 0.8, β = 2 This figure needs to be redone using newly defined
parameters.
to construct asymptotic series solutions to equations of this type (i.e., to
singular perturbation problems), a subject which is beyond the scope of
this text. This is discussed briefly in Keener and Sneyd (1998), as well
as in Murray (1989) and Lin and Segel (1988).
However, for larger times this solution is reasonably accurate.
(d) To show that solutions of equations (12.109)-(12.111) cannot leave the
positive octant, it suffices to show that they cannot cross the boundary.
The boundary consists of the three planes, x = 0, y = 0, and σ = 0.
When x = 0, dx/dt = λσ1 which is strictly positive everywhere in the
first octant (and hence on the boundary), except at the origin (which is
a critical point of the system). Similarly, when y = 0, dy/dt = λσ2 > 0
except at the origin, and when σ = 0, dσ/dt = α1 µ1 x+α2 µ2 y > 0 except
at the origin. It follows that no solution can leave the first octant.
To show that the solution approaches the origin for large time, we add
the three equations together, weighted by 1, λ1 α1 , and λ2 α2 respectively
to find
d
(σ + 1 λ1 α1 x + 2 λ2 α2 y) = α1 (µ1 − λ1 )x + α2 (µ2 − λ2 )y (12.143)
dt
(since α1 + α2 = 1).
Now observe that
µ1 − λ1 = −
k2
< 0,
k1 s0
µ2 − λ2 = −
k4
< 0,
k3 s0
(12.144)
359
so that σ + 1 λ1 α1 x + 2 λ2 α2 y decreases whenever x and y are nonnegative. However, we already know that x and y cannot be zero unless
σ = 0 as well. Thus σ + 1 λ1 α1 x + 2 λ2 α2 y is always decreasing, and so
approaches zero for large time.
360
CHAPTER 12. BIOCHEMICAL REACTIONS: SOLUTIONS
Chapter 13
Cellular Homeostasis:
Solutions
Ex. 1: A rule of thumb (derived by Einstein) is that the diffusion coefficient for a
globular molecule satisfies D ∼ M −1/3 where M is the molecular weight.
Determine how well this relationship holds for the substances listed in Table
2.2 by plotting D and M on a log-log plot.
Ex. 2: A fluorescent dye with a diffusion coefficient of D = 10−7 cm2 /s and binding
equilibrium of Keq = 30mM is used to track the spread of hydrogen (Dh =
4.4 × 10−5 cm2 /s). Under these conditions the measured diffusion coefficient
is 8 × 10−6 cm2 /s. How much dye is present? (Assume the dye is a fast buffer
of hydrogen.)
Solution
Use that
Def f =
0
Dh + DB Kweq
so that
w0 = Keq (
1+
w0
Keq
Dh − Def f
).
Def f − DB
With these numbers we find that w0 = 0.14M.
361
(13.1)
(13.2)
362
CHAPTER 13. CELLULAR HOMEOSTASIS: SOLUTIONS
Ex. 3: Segel, Chet and Henis [?] used (13.3) to estimate the diffusion coefficient for
bacteria. With the external concentration C0 at 7 × 107 ml−1 , at times t = 2,
5, 10, 12.5, 15, and 20 minutes, they counted N of 1,800, 3,700, 4,800, 5,500,
6,700, and 8,000 bacteria, respectively, in a capillary of length 32 mm with 1
µl total capacity. In addition, with concentrations external concentrations C0
of 2.5, 4.6, 5.0, and 12.0 ×107 bacteria per milliliter, counts of 1,350, 2,300,
3,400, and 6,200 were found at t = 10 minutes. Estimate D.
Solution
Segel, Chet and Henis [?] used (13.3) to estimate the diffusion coefficient for
bacteria. With the external concentration C0 at 7 × 107 ml−1 , at times t = 2,
5, 10, 12.5, 15, and 20 minutes, they counted N of 1,800, 3,700, 4,800, 5,500,
6,700, and 8,000 bacteria, respectively, in a capillary of length 32 mm with 1
µl total capacity. In addition, with concentrations external concentrations C0
of 2.5, 4.6, 5.0, and 12.0 ×107 bacteria per milliliter, counts of 1,350, 2,300,
3,400, and 6,200 were found at t = 10 minutes. Estimate D.
According to the formula 13.3)
D=
πN 2
.
4C02 A2 T
(13.3)
2
the ratio Nt should be a constant. In fact, this number varies between 1.6×106
and 3.2 × 106 , giving an estimate of D.
Ex. 4: Calculate the effective diffusion coefficient of oxygen in a solution containing
3
1.2 × 10−5 M/cm myoglobin. Assume that the rate constants for the uptake
of oxygen by myoglobin are k+ = 1.4 × 1010 cm3 M−1 s−1 , and k− = 11 s−1 .
Solution
Use that (2.29)
Def f =
0
D + DB Kweq
1+
w0
Keq
(13.4)
to calculate that Keq = 5.11 × 10−5 cm2 s−1 , which is essentially indistiguishable from that of myoglobin.
Ex. 5: Find the maximal enhancement for diffusive transport of carbon dioxide via
binding with myoglobin using Ds = 9 × 10−4 cm2 /s, k+ = 2 × 108 cm3 /M · s,
k− = 1.7 × 10−2 σ. Compare the amount of facilitation of carbon dioxide
transport with that of oxygen at similar concentration levels.
Solution
363
Calculate ρ = DDc skk+−e0 , with Dc = 5.1 × 10−7 cm2 /s for myoglobin, Ds =
?? × 10−4 cm2 /s for carbon dioxide, k+ = 2 × 108 cm3 M−1 s−1 , and k− =
1.7 × 10−2 s−1 , and find ρ = 5.75 × 106 e0 cm3 M−1 . Compare this with oxygen
using Ds = 2.1 × 10−5 cm2 /s, k+ = 1.4 × 1010 cm3 M−1 s−1 , and k− = 11s−1 ,
and find ρ = 4.66 × 107 e0 cm3 M−1 , which is larger by a factor of about ?.
Ex. 6: Devise a model to determine the rate of production of product for a “onedimensional” enzyme capsule of length L in a bath of substrate at concentration S0 . Assume that the enzyme is confined to the domain 0 ≤ x ≤ L
and there is no flux through the boundary at x = 0. Assume that the enzyme cannot diffuse within the capsule but that the substrate and product
can freely diffuse into, within, and out of the capsule. Show that the steady
state production per unit volume of enzyme is less than the production rate of
a reactor of the same size in which substrate is homogeneously mixed (infinite
diffusion).
Solution
We assume that the reaction is a standard Michaelis-Menten type reaction,
with
k
1
k
2
S + E −→
←− C−→P + E.
k−1
Then the concentrations of substrate, enzyme, and product in the enzymatic
region are governed by the system of equations
∂s
∂t
∂c
∂t
∂p
∂t
= Ds
∂2s
+ k−1 c − k1 se,
∂x2
= k1 se − k−1 c − k2 c,
(13.5)
(13.6)
2
= Dp
∂ p
+ k2 c,
∂x2
(13.7)
where, because enzyme does not diffuse, c + e = e0 . Using the quasi-steady
state approximation
e0 s
c=
,
(13.8)
(s + Km )
with Km =
k−1 +k2
.
k1
It follows that at steady state,
0
0
e0 s
(s + Km )
= Dp pxx + Ds sxx
= Ds sxx − k2
(13.9)
(13.10)
It follows that the total rate of production is
R = −Dp px (L) = Ds sx (L).
(13.11)
364
CHAPTER 13. CELLULAR HOMEOSTASIS: SOLUTIONS
We introduce nondimensional variables σ =
0 = σyy − α
s
Km ,
σ
,
(σ + 1)
y=
x
L,
and find
(13.12)
2
where α = LDks2 Ke0m , with σ (0) = 0 and σ(0) = Ks0m and the rate of production
is
Ds Km
Ds Km
k2 Le0
R=
σy (1) =
(13.13)
σy (1) = k2 Le0
σy (1),
L
k2 L2 e0
α
so that the effectiveness of the reaction, i.e., the rate of production per total
amount of enzyme is
k2
E = σy (1).
(13.14)
α
The easiest way to solve this problem numerically is using “shooting”, i.e.,
solve an initial value problem with σ (0) = 0, σ(1) = σ1 to determine both σ0
and the rate of production, and then to plot rate of production as a function
of σ0 .
Run the maple code ex2.m to see this solution.
Ex. 7: Develop and analyze a model for the diffusion and consumption of oxygen
in a cylindrical muscle. Assume that the consumption of oxygen by muscle
at rest is 5 × 10−8 mol/cm3 s, and the concentration of myoglobin is about
2.8 × 10−7 mol/cm3 . Suppose that a muscle fiber is a long circular cylinder
(radius a = 2.5 × 10−3 cm) and that diffusion takes place only in the radial
direction. We suppose that the oxygen concentration at the boundary of
the fiber is a fixed constant and that the distribution of chemical species
is radially symmetric. Use parameter values Ds = 10−5 cm2 /s, Dc = 5 ×
10−7 cm2 /s, k+ = 2.4 × 1010 cm3 /mol · s, and k− = 65/s (Wyman, 1966).
Use (and justify) the quasi-steady state approximation to determine the oxygen concentration needed outside the muscle to prevent the oxygen concentration in the muscle from falling below zero (oxygen debt).
Solution
Under the assumptions stated in the problem, the steady-state equations governing the diffusion of oxygen and oxymyoglobin are
1 d
ds
Ds
r
− f − g = 0,
(13.15)
r dr
dr
1 d
dc
Dc
r
+ f = 0,
(13.16)
r dr
dr
where, as before, s = [O2 ], c = [MbO2 ], and f = −k− c+k+ se. The coordinate
r is in the radial direction of the muscle fiber. The new term in these equations
is the constant g, corresponding to the constant consumption of oxygen. The
365
boundary conditions are s = sa , dc/dr = 0 at r = a, and ds/dr = dc/dr = 0
3
at r = 0. For muscle, sa is typically 3.5 × 10−8 mol/cm (corresponding to
the partial pressure 20 mm Hg).
+
Introducing nondimensional variables σ = kk−
s, u = c/e0 , and r = ay, we
obtain the differential equations
1 d
1 d
dσ
du
1
y
− γ = σ(1 − u) − u = −2
y
,
(13.17)
y dy
dy
y dy
dy
c
where 1 = e0 kD+sa2 , 2 = kD
2 , γ = g/k− . Using the parameters appropriate
−a
for muscle, we estimate that 1 = 2.3 × 10−4 , 2 = 1.2 × 10−3 , γ = 3.3 × 10−3 .
While these numbers are not as small as for the experimental slab described
earlier, they are still small enough to warrant the approximation that the
quasi-steady state (2.37) holds in the interior of the muscle fiber.
It also follows from (13.17) that
1 d
1 d
dσ
du
1
y
+ 2
y
= γ.
y dy
dy
y dy
dy
(13.18)
We integrate (13.18) twice with respect to y to find
1 σ + 2 u = A ln y + B +
γ 2
y .
4
(13.19)
The constants A and B are determined by boundary conditions. Since we
want the solution to be bounded at the origin, A = 0, and B is related to the
concentration at the origin.
Now suppose that there is just enough oxygen at the boundary to prevent
oxygen debt. In this model, oxygen debt occurs if σ falls to zero. Marginal
oxygen debt occurs if σ = u = 0 at y = 0. For this boundary condition, we
take A = B = 0. Then the concentration at the boundary must be at least as
large as σ0 , where, using the quasi-steady state σ(1 − u) = u,
σ0 + ρ
γ
σ0
=
,
σ0 + 1
41
(13.20)
and where ρ = 2 /1 . Otherwise, the center of the muscle is in oxygen debt.
Note also that σ0 is a decreasing function of ρ, indicating a reduced need for
external oxygen because of facilitated diffusion.
A plot of this critical concentration σ0 as a function of the scaled consumption
γ
41 is shown in Fig. 13.1. For this plot ρ = 5, which is a reasonable estimate
for muscle. The dashed curve is the critical concentration when there is no
facilitated diffusion (ρ = 0). The easy lesson from this plot is that facilitated
diffusion decreases the likelihood of oxygen debt, since the external oxygen
concentration necessary to prevent oxygen debt is smaller in the presence of
myoglobin than without.
A similar lesson comes from Fig. 13.2, where the internal free oxygen content
σ is shown, plotted as a function of radius y. The solid curves show the
366
CHAPTER 13. CELLULAR HOMEOSTASIS: SOLUTIONS
Figure 13.1: Critical concentration σ0 plotted as a function of oxygen consumption
γ
41 . The dashed curve is the critical concentration with no facilitated diffusion.
Figure 13.2: Free oxygen σ as a function of radius y. Solid curves show oxygen
concentration in the presence of myoglobin (ρ = 5), the lower of the two having
the critical external oxygen concentration. The dashed curve shows the oxygen
concentration without facilitation at the critical external concentration level.
internal free oxygen with facilitated diffusion, and the dashed curve is without.
The smaller of the two solid curves and the dashed curve have exactly the
critical external oxygen concentration, showing clearly that in the presence of
myoglobin, oxygen debt is less likely at a given external oxygen concentration.
The larger of the two solid curves has the same external oxygen concentration
as the dashed curve, showing again the contribution of facilitation toward
preventing oxygen debt. For this figure, ρ = 5, γ/1 = 14.
Ex. 8: Suppose a membrane that contains water-filled pores separates two solutions.
(a) Suppose that the solution on either side of the membrane contains an impermeant solute. Show that the hydrostatic pressure of the water within
the pores must be less than the hydrostatic pressure of the solutions on
either side of the membrane.
(b) Show that if the solute can permeate the pore freely, no such drop in
hydrostatic pressure exists.
(c) Show that if the solution on one side of the membrane contains both
permeant and impermeant solutes, while the solution on the other side
contains only the permeant solute, it is possible for water to flow against
its chemical potential gradient (at least temporarily). This problem of
wrong-way water flow has been observed experimentally, and it is discussed in detail by Dawson (1992).
Solution
(a) Suppose the hydrostatic pressure is P1 , Pp , and P2 for side one, the pore,
and side two, respectively. Similarly, the osmotic (or chemical) potential
is π1 , πp , π2 .
Ex. 9: Red blood cells have a passive exchanger that exchanges a single Cl− ion for
a bicarbonate (HCO−
3 ) ion. Develop a model for this exchanger and find the
flux.
367
Solution
This is exactly the exchanger model (2.55) with m = n = 1.
Ex. 10: Modify (2.62) to take into account the fact that the concentration of ATP
affects the rate of reaction (since pumping should stop if there is no ATP).
Solution
P]
Replace kp by kp KAT[AT
.
P +[AT P ]
Ex. 11: Generalize (2.63) to account for the fact that with each turn of the sodium–
potassium pump three sodium ions are exchanged for two potassium ions.
Solution
Replace [Na+ ] by [Na+ ]3 , and [K+ ] [K+ ]2 .
Ex. 12: Almost immediately upon entering a cell, glucose is phosphorylated in the
first reaction step of glycolysis. How does this rapid and nearly unidirectional
reaction affect the transmembrane flux of glucose as represented by (2.51)?
How is this reaction affected by the concentration of ATP?
Solution
To solve this problem one must first modify the differential equation (2.45),
replacing −J by the removal term −kr si , where kr is the ATP dependent rate
of phosphorylation. Now the system of equations is nonlinear and the formula
for J is the same as before (2.51) where si is the solution of the quadratic
equation
C2 si + C1 si − se kk− k+ c0 = 0,
(13.21)
where C2 = 2kp kr (k + k− + k+ se ), and C1 = kk− k+ c0 + kr (2k+ kse + 4kk− +
2
2se k+ k− + 2k−
).
Ex. 13: How does the concentration of ATP affect the rate of the sodium–potassium
pump?
368
CHAPTER 13. CELLULAR HOMEOSTASIS: SOLUTIONS
Ex. 14: The process by which calcium is taken up into the sarcoplasmic reticulum
(SR) in muscle and cardiac cells is similar to the sodium–potassium ATPase,
but simpler. Two intracellular calcium ions bind with a carrier protein with
high affinity for calcium. ATP is dephosphorylated, with the phosphate bound
to the carrier. There is a conformational change of the carrier protein that
exposes the calcium to the interior of the SR and reduces the affinity of the
binding sites, thereby releasing the two ions of calcium. The phosphate is
released and the conformation changed so that the calcium binding sites are
once again exposed to the intracellular space.
Formalize this reaction and find the rate of calcium uptake by this pump.
Solution
The system of differential equations is (where sc = u, and scp = v, ci p = w)
dse
dt
dce
dt
dv
dt
dsi
dt
dw
dt
= k−e u − ke se ce + J
(13.22)
= ke se ce − k−e sc + k−a scp − ka sc
(13.23)
= ka sc − k−a scp + k−i si ci p − ki v
(13.24)
= ki v − k−i si ci p − J
(13.25)
= ki v − k−i si w − kp w + k−p ce
(13.26)
with the conservation law ce + w + v + u = c0. Solving these in steady state
we find that
Kp se − K−p si
J = c0
(13.27)
se ke + K0 + si Ki + se ∗ si Kei
where Kp = kp ki ke ka , K−p = k−i k−a k−p k−e , Ke = ke (ki kp + k−a kp + ki ka +
ka kp ), Ki = k−i (k−e k−a + k−p ka + k−p k−e + k−p k−a ), Kei = ke k−i (k−a + ka ),
Ke = k−i (k−a + k−a kp + ke ka k−i ), K0 = ki k−p k−e + ki k−e kp + kik−p ka +
ki ka kp + k−a k−p k−e + k−e k−a kp
Ex. 15: A 1.5 oz bag of potato chips (a typical single serving) contains about 200 mg
of sodium. When eaten and absorbed into the body, how many osmoles does
this bag of potato chips represent?
Solution
The gram molecular weight of sodium is 23, while the gram molecular weight
of chloride is 35.5. Therefore 200 grams of sodium is 8.7 moles of sodium.
Since salt has equal amounts of sodium and chloride, 200 grams of sodium
gives 17.4 osmoles.
369
Ex. 16: Generalize formula (2.96) to take into account that the two fluids have different
densities and to allow the columns to have different cross-sectional areas.
Solution
If the initial heights are h1 and h2 with cross-sectional areas A1 and A2 , then
the total volume is V = A1 h1 + A2 h2 . Suppose that fluid is drawn out of
vessel 2 into vessel 1, so that the new height in vessel 1 is h1 + δh1 and the
new height in vessel 2 is h2 − δh2 . The pressure balance equation is
ρ1 gh1 + ρ2 gδh1 = ρ2 g(h2 − δh2 ) −
nkT
.
(h1 + δh1 )A1
(13.28)
Because total volume is conserved, A1 δh1 = A2 δh2 . This leads to the quadratic
equation for δh1 ,
ρ2 (A2 −A1 )δh21 +A2 (ρ1 h1 +ρ2 h1 −ρ2 h2 −ρ2
A1
nkT
h1 )δh1 +A2 (ρ1 h21 −ρ2 h2 h1 +
)=0
A2
gA1
(13.29)
Ex. 17: Two columns with cross-sectional area 1 cm2 are initially filled to a height of
one meter with water at T = 300K. Suppose 1 gm of sugar is dissolved in one
of the two columns. How high will the sugary column be when equilibrium is
reached? Hint: The weight of a sugar molecule is 3 × 10−22 gm, and the force
of gravity on 1cm3 of water is 980 dynes.
Solution
Using that a dyne is 10−5 N, we calculate that the density of water is ρ =
103 kg/m3 . A gram of sugar contains n = 13 × 1022 molecules. Using (2.86)
we find h1 − h2 = 0.143m so that the height of the sugar-containing column
is h1 = 1.07m.
Ex. 18: Suppose an otherwise normal cell is placed in a bath of high extracellular
potassium. What happens to the cell volume and resting potentials?
Ex. 19: Suppose 90% of the sodium in the bath of a squid axon is replaced by inert choline, preserving electroneutrality. What happens to the equilibrium
potentials and membrane potentials?
Ex. 20: Determine the effect of temperature (through the Nernst equation) on cell
volume and membrane potential.
370
CHAPTER 13. CELLULAR HOMEOSTASIS: SOLUTIONS
Ex. 21: Many animal cells swell and burst when treated with the drug ouabain. Why?
Hint: Ouabain competes with K+ for external potassium binding sites of the
Na+ –K+ ATPase. How would you include this effect in a model of cell volume
control?
Ex. 22: Modify the pump–leak model to include a calcium current and the 3-for-1
sodium–calcium exchanger. What effect does this modification have on the
relationship between pump rate and membrane potential? Must include a
hint here as to the form of the voltage-dependence. J.S.
Ex. 23: Because there is a net current, the sodium–potassium pump current must be
voltage dependent. Determine this dependence by including voltage dependence in the rates of conformational change in expression (2.63). How does
voltage dependence affect the pump–leak model of Chapter 2?
Ex. 24: Intestinal epithelial cells have a glucose–sodium symport that transports one
sodium ion and one glucose molecule from the intestine into the cell. Model
this transport process. Is the transport of glucose aided or hindered by the
cell’s negative membrane potential?
Chapter 14
Membrane Ion Channels:
Solutions
Ex. 1: Derive the extended independence principle. Assume there are more than one
species of ion present, all with the same valence, and assume that the reversal
potential is given by the GHK potential. Show that
Pj [Sj ]e exp(−V F/RT )
I
j Pj [Sj ]i −
j
=
,
(14.1)
I
j Pj [Sj ]i −
j Pj [Sj ]e exp(−V F/RT )
where the sum over j is over all the ionic species. Subscripts i and e denote
internal and external concentrations respectively.
Solution
This is a rather tricky question, although it looks simple. The most important
thing to realise is that the method for the single ion case described in the
book doesn’t work in this more general case. The reason for this is that
when there are two or more ions present, one cannot derive an independence
principle without assuming something about the I −V curve. In the single-ion
case one can, because then the reversal potential is determined by underlying
thermodynamic principles, and is not dependent on the properties of the I −V
curve. But, as we discuss in the text, this is not true when multiple ion types
are present.
Note that in the discussion at the beginning of section 3.2 there is no mention
of the I − V curve for the ions, and thus the independence principle depends
only on the expression for the Nernst potential. However, if we try to repeat
this argument for the multiple ion case, we fail. For instance, if we have ion
types S1 and S2 , with concentrations c1 and c2 , with the usual convention for
the i and e subscripts, we get
Jin
k1,e c1,e + k2,e c2,e
=
.
Jout
k1,i c1,i + k2,i c2,i
371
(14.2)
372
CHAPTER 14. MEMBRANE ION CHANNELS: SOLUTIONS
There is now no way to cancel out the rate constants so as to leave only the
concentration terms; the plus signs get in the way.
However, the desired result is obtained very easily if we assume that the
current of S1 through the channel obeys the GHK current relation
F2
c1,i − c1,e exp(−V F/RT )
I1 = P1
V
,
(14.3)
RT
1 − exp(−V F/RT )
and similarly for the S2 current. Then a simple addition and cancellation of
the common terms gives
I
I
=
=
I1 + I2
I1 + I2
P1 c1,i + P2 c2,i − (P1 c1,e + P2 c2,e ) exp(−V F/RT )
.
P1 c1,i + P2 c2,i − (P1 c1,e + P2 c2,e ) exp(−V F/RT )
(14.4)
(14.5)
It should now be clear how this generalises to the desired result.
******************************
Jim, can you check this carefully. I don’t like to disagree with Hille as it makes
me think I’ve got something wrong.
Although this extended independence principle is given in Hille’s book (equation ??), its derivation is not. In fact, Hille seems to claim implicitly that one
can obtain (14.5) from (14.3) without invoking the form of the I − V curve.
I just don’t see how this can be done. Perhaps I’m just missing something
important.
******************************
Ex. 2: Show that the GHK equation (3.2) satisfies both the independence principle
and the Ussing flux ratio, but that the linear I −V curve (3.1) satisfies neither.
Solution
GHK equation
Consider the GHK equation first. In order to see whether it satisfies the
Ussing flux ratio we need first to find some expression for the one way fluxes.
In this case this is most simply done by letting either the external or the
internal concentrations be zero. For instance, the outward flux is obtained by
setting the external concentration of the ion to be zero. Thus, setting ce = 0
in (2.77) we get
zF V
ci
,
|Jout | = PS
(14.6)
F
RT 1 − exp −zV
RT
373
and similarly, by setting ci = 0 in (2.77) we get
F
zF V ce exp −zV
RT
.
|Jin | = PS
F
RT 1 − exp −zV
RT
(14.7)
Note that we take the absolute value of the fluxes as the Ussing flux ratio
assumes that all fluxes are counted as positive.
Taking the ratio now gives
Jin ce
−zV F
=
exp
,
Jout ci
RT
(14.8)
as required. Hence the GHK current equation obeys the Ussing flux ratio.
It also satisfies the independence principle. This follows immediately from
the fact that Jin and Jout are linear functions of the concentrations. Thus, a
doubling of ci , say, results in a doubling of Jout , which is just the independence
principle.
It’s important to note that the Ussing flux ratio and independence are not
equivalent. There are simple models for ionic flow that satisfy the Ussing flux
ratio but not independence; they are discussed in Keener and Sneyd (1998).
The linear equation
First we note that, since the current does not depend linearly on either ci or
ce , the linear I − V curve cannot satisfy the independence principle.
However, it is now not possible to use the above procedure to find the inward
and outward fluxes, as the linear I − V curve is not valid in the limit where
either the internal or external concentration tends to zero. (As a side note,
it is important to note how this makes the GHK formulation more useful for
many situations. Although in reality there is never just a single ion type
present and so this singular limit never causes any problems, it is nice not to
have to deal with the associated mathematical difficulties that can arise when
the linear I − V curve is used.)
Instead of calculating the inward and outward fluxes separately, we now just
use the formulation of the Ussing flux ratio in (3.14). In other words, if we
assume that ci = ci , and thus the outward fluxes are the same, then the Ussing
flux ratio can be reformulated as
I
I
=
=
=
− Jin
Jout
Jout − Jin
J
Jout
1 − J in
out
in
Jout 1 − JJout
1−
1−
Jin
Jout
Jin
Jout
(14.9)
(14.10)
(14.11)
374
CHAPTER 14. MEMBRANE ION CHANNELS: SOLUTIONS
=
=
−V F ce
ci exp
RT
−V F ce
1 − ci exp RT
F
ci − ce exp −V
RTF .
ci − ce exp −V
RT
1−
(14.12)
(14.13)
Now we need only note that the linear I − V curve does not satisfy this
equation, and thus does not satisfy the Ussing flux ratio.
The astute reader may be wondering why we bothered to rederive (14.13)
when it is already derived in the book. The answer is that the derivation in
the book relies on the use of the independence principle, and thus is no good
for our purposes. After all, if the linear I − V curve doesn’t satisfy the independence principle, it thus will automatically not satisfy any other equation
that depends on the independence principle for its derivation. However, the
derivation above does not use the independence principle, which shows that
we may indeed use (14.13) to see whether the linear I − V curve obeys the
Ussing flux ratio.
Ex. 3: In section 3.3.1 we used the PNP equations to derive I − V curves when two
ions with opposite valence are allowed to move through a channel. Extend this
analysis by assuming that two types of ions with positive valence, and one type
of ion with negative valence are allowed to move through the channel. Show
that in the high concentration limit, although the negative ion still obeys a
linear I − V curve, the two positive ions do not. Further details can be found
in Chen, Barcilon and Eisenberg (1992).
Solution
The equations
First we write down the differential equations and boundary conditions that
apply to this three-ion case. Let c1 and c2 denote the concentrations of the
positive ions S1 and S2 , and c3 the concentration of the negative ion S3 . Then
we just extend (3.26)–(3.28) in the book by adding a second positive ion to
get
dφ
dc1
+ c1 ,
dx
dx
dφ
dc2
+ c2 ,
dx
dx
dφ
dc3
− c3 ,
dx
dx
−J¯1
=
−J¯2
=
−J¯3
=
d2 φ
dx2
= −λ2 (c1 + c2 − c3 ),
(14.14)
(14.15)
(14.16)
(14.17)
375
where the terms are as defined in the book. The first three equations govern
the flux of the ions through the channel, while the last equation is Poisson’s
equation.
When formulating the boundary conditions we must remember to keep the
solutions on either side of the membrane electrically neutral. This is most
easily done with the introduction of two new parameters, l and r, that govern
the relative proportions of c1 and c2 on either side of the membrane. Thus,
c1 (0) = lci ,
c2 (0) = (1 − l)ci ,
c3 (0) = ci ,
φ(0) = v,
c1 (1) = rce ,
c2 (1) = (1 − r)ce ,
c3 (1) = ce ,
φ(1) = 0.
(14.18)
Now that we have the equations, we want to construct solutions for the longchannel limit (λ 1) and the short channel limit (λ 1).
The long-channel limit
Just as in the book, letting λ → ∞ gives
c1 + c2 = c3 ,
(14.19)
which just happens to satisfy both boundary conditions (and thus we can
avoid the difficulties of a full matched asymptotic expansion procedure).
If we now add the three flux differential equations we get
d
¯
(c1 + c2 + c3 ) = −J,
dx
(14.20)
where J¯ = J¯1 + J¯2 + J¯3 is an unknown constant to be determined from the
boundary conditions. Thus
¯ + K̄,
c1 + c2 + c3 = −Jx
(14.21)
for some other constant, K̄. Applying the boundary conditions now gives
K̄ = 2ci and −J¯ = 2ce + 2ci . Thus
c1 + c2 + c3 = 2[(ce − ci )x + ci ],
(14.22)
whence it follows (using (14.19)) that
c1 + c2 = c3 = (ce − ci )x + ci .
(14.23)
Hence, the concentration profile of c3 in the channel is linear, but those of c1
and c2 not necessarily so. The profiles of c1 and c2 sum to a linear profile,
but this does not mean that the individual profiles must be linear also.
We now solve for φ(x). Forming (14.14) + (14.15) - (14.16) gives
(c1 + c2 + c3 )
dφ
˜
= −J,
dx
(14.24)
376
CHAPTER 14. MEMBRANE ION CHANNELS: SOLUTIONS
where J˜ = −J¯1 − J¯2 + J¯3 . Thus
[(ce − ci )x + ci ]
dφ
˜
= −J,
dx
(14.25)
which can now be solved for φ by separation of variables. One integration
gives
J˜
φ(x) = −
(14.26)
ln[2(ce − ci )x + 2ci ] + K̃,
2(ce − ci )
for some other constant K̃. We now determine J˜ and K̃ from the boundary
conditions φ(0) = v, φ(1) = 0. Plugging in these conditions and solving then
gives
−
J˜
2(ce − ci )
K̃
v
,
(14.27)
ln(2ce )v
.
ln ccei
(14.28)
=
ln
= −
ci
ce
Substituting into the expression for φ then gives
φ(x)
=
=
v
ln(2c )v
ln[2(ce − ci )x + 2ci ] − e ci
ln ce
ln ccei
v
(ce − ci )x + ci
ln
.
ce
ln ccei
(14.29)
(14.30)
Since we are going to be using the ratio ci /ce often in the following pages, we
define
c = ci /ce
(14.31)
just to make things look a little simpler.
Finally, we want to find each flux. This is most easily done by solving for c1 ,
c2 and c3 . Of course, we already have an expression for c3 (x), namely (14.23
above. Hence, it follows immediately that
−J¯3
=
=
dφ
dc3
− c3
dx
dx
ce − ci
[ln(c) − v].
ln(c)
(14.32)
(14.33)
Since ln(c) is the Nernst potential of S3 , we see immediately that S3 obeys a
linear I − V curve, as claimed.
Calculating the fluxes of c1 and c2 takes somewhat longer, but is no more
difficult in concept. We shall illustrate the procedure for c1 only, since the
calculations for c2 are practically identical.
377
Start with the differential equation for c1 :
dc1
dφ
+ c1
= −J¯1 .
dx
dx
(14.34)
Since φ(x) is known, this differential equation can be solved by using an
integrating factor. Multiply through by eφ to get
d
(c1 eφ ) = −J¯1 eφ .
dx
(14.35)
eφ = [(1 − c)x + c]v/ ln(c) .
(14.36)
Now note that
Thus, integrating (14.35), we get
φ
c1 e =
ln(c)
v + ln(c)
−J¯1 [(1 − c)x + c]
1−c
v+ln(c)
ln(c)
+ K,
(14.37)
for some constant K.
J¯1 and K can now be determined from the boundary conditions. At x = 0 we
get
¯ v+ln(c) ln(c)
−J1 c ln(c)
v
lci e =
+K
(14.38)
v + ln(c)
1−c
¯ v
ln(c)
−J1 ce
=
+ K,
(14.39)
v + ln(c)
1−c
while at x = 1
rce =
ln(c)
v + ln(c)
−J¯1
1−c
+ K.
Solving these two equations for −J¯1 and K gives
v + ln(c) (lci ev − rce )(1 − c)
¯
−J1 =
,
ln(c)
cev − 1
lci ev − rce
K = rce −
,
cev − 1
(14.40)
(14.41)
(14.42)
and thus, finally,
1
c1 = v
(lci ev −rce )[(1−c)x+c]+ci ev (r−l)[(1−c)x+c]v/ ln(c) . (14.43)
ce − 1
Note that in passing we have obtained the expression for the flux of S1 , namely
(14.41). It’s interesting to note a few things about this flux expression. Firstly,
it has the correct reversal potential (fortunately); although it looks like it has
multiple zeros it actually doesn’t, as when v = − ln(c) the zeros on the top and
bottom lines cancel out. Secondly, it looks very like the GHK flux equation,
378
CHAPTER 14. MEMBRANE ION CHANNELS: SOLUTIONS
20
0
J1
-20
-40
-40
-20
0
v
20
40
Figure 14.1: Plot of J¯1 against v. The flux is mostly a fairly straight function
of v, except around the reversal potential where it looks more like a GHK flux
relationship. Parameters were chosen arbitrarily as ci = 1, ce = 2, r = 0.7, l = 0.4.
with the exception that is multiplied by the linear v term on the top of the
fraction. Thus, in this case the flux of S1 is a combination of both the GHK
and the linear flux expressions.
The calculations for J¯2 and c2 are essentially identical, so we won’t reproduce
them.
Short-channel limit
In the limit of a short channel, or small ionic concentrations on either side
of the membrane, the calculations are exactly the same as those in the book
(equations (3.30)–(3.33)). Thus,
d2 φ
= 0,
dx2
(14.44)
from which it follows that dφ/dx = −v and that the flux of each ion satisfies
the GHK current equation (equation (3.34) of the book).
Ex. 4: Suppose that a channel consists of k identical, independent subunits, each of
which can be open or closed, and that a current can pass through the channel
only if all units are open.
379
(a) Let Sj denote the state in which j subunits are open. Show that the
conversions between states are governed by the reaction scheme
kα
α
−→
S0 −→
←− S1 . . . Sk−1 ←− Sk .
β
(14.45)
kβ
(b) Derive the differential equation for xj , the proportion of channels in state
j.
j
k
k!
(c) Show that xj =
nj (1 − n)k−j , where
= j!(k−j)!
is the binomial
k
j
coefficient, is a stable invariant manifold for the system of differential
equations, provided that
dn
= α(1 − n) − βn.
dt
(14.46)
Solution
(a) First, let α be the rate at which a single subunit opens, and let β be
the rate at which a single subunit closes. The channel can move from
state S0 to state S1 by having any one of its subunits open. Since there
are k subunits, this can occur in any of k different ways. With the rate
of opening of an individual subunit being α, it follows that the rate at
which S0 converts to S1 is given by kα. However, the reverse conversion
from S1 to S0 can occur only at the base rate β, since there is only a
single subunit that can close.
In general, the transition from state Sj to state Sj+1 can be written as
(k−j)α
Sj
−→
←−
Sj+1 .
(14.47)
(j+1)β
(b) The solution to part (a) above gives immediately the solution to part (b)
just by applying the law of mass action:
dxj
= (k − j + 1)αxj−1 + (j + 1)βxj+1 − [jβ + (k − j)α]xj .
dt
(14.48)
(c) Strictly speaking, since we are given the answer we need only check that
it works. However, before doing that it is useful to gain some insight into
where this answer comes from.
Note that, from conservation we must have x0 + x1 + · · · + xk = 1. Thus,
we have k + 1 bits that change over time but always must add to give 1.
The trick is in using the identity
1 = [n + (1 − n)]k ,
(14.49)
and recognizing that the right hand side has the desired number of terms
(i.e., k + 1 of them). Hence, if we ensure that x0 corresponds to the first
380
CHAPTER 14. MEMBRANE ION CHANNELS: SOLUTIONS
term of [n + (1 − n)]k , x1 to the second, and so forth, we automatically
ensure that the xj s always sum to one. That is where the expression for
xj comes from; it is the j + 1st term in the expansion of [n + (1 − n)]k .
However, it remains to be seen whether we can find a differential equation
for n that
is consistent with this. Firstly, take d/dt of the equation
j
xj =
nj (1 − n)k−j to get
k
dxj
dn k
=
jnj−1 (1 − n)k−j − (k − j)nj (1 − n)k−j−1 . (14.50)
dt
dt j
Secondly, take the expressions for xj , xj+1 and xj−1 , substitute them
into the right hand side of (14.48), and equate to (14.50). This gives
dn k
j−1
k−j
j
k−j−1
=
− (k − j)n (1 − n)
jn (1 − n)
dt j
k
− [jβ + (k − j)α]
nj (1 − n)k−j
j
k
+ (k − j + 1)α
nj−1 (1 − n)k−j+1
j−1
k
+ (j + 1)β
ni+1 (1 − n)k −(14.51)
j + 1.
j+1
k
Now just divide through by
nj−1 (1 − n)k−j−1 to get
j
(j − kn)
dn
= jα(1 − n)2 + (k − j)βn2 − [iβ + (k − j)α]n(1 − n), (14.52)
dt
and thus, factoring out α and β on the right hand side, and cancelling
out the common factor, we get
dn
= α(1 − n) − βn,
dt
(14.53)
as required.
Note that this argument works for all 0 < j < k. The arguments for
j = 0 and j = k are very similar, just simpler.
Ex. 5: Consider the model of the Na+ channel shown in Fig. 3.7. Show that, if α
and β are large compared to γ and δ, x21 is given (approximately) by
α
α+β
2
x21
=
dh
dt
= γ(1 − h) − δh,
h,
(14.54)
(14.55)
381
while, conversely, if γ and δ are large compared to α and β, then (approximately)
γ
2
x21 = m
,
(14.56)
γ+δ
dm
= α(1 − m) − βm.
(14.57)
dt
Solution
Case I: α and β large
In problems of this sort, where there are some fast rates and some slow ones,
the general procedure is to divide all the variables into groups such that every
variable within a group is at equilibrium with all the other variables in that
group. When α and β are large we see that there is an obvious grouping to use,
by letting group I consist of S00 , S10 and S20 , while letting group II consist of
S01 , S11 and S21 . Note that, since α and β are large all the variables in group
I are in equilibrium with each other, but not necessarily with the variables in
group II.
Within group I we see that
2αx00
αx10
= βx10
(14.58)
=
(14.59)
2βx20 .
We also let
x = x00 + x10 + x20 .
(14.60)
Similarly, within group II we have
2αx01
αx11
= βx11
(14.61)
=
(14.62)
2βx21 .
and we also let
h = x01 + x11 + x21 .
(14.63)
Note that h + x = 1.
Using (14.61) and (14.62) to express x01 and x11 in terms of x21 , and substituting into (14.63) then gives
2
α
x21 =
h.
(14.64)
α+β
Finally, the differential equation for h is
dh
dt
as required.
= γ(x00 + x10 + x20 ) − δ(x01 + x11 + x21 )
= γx − δh
= γ(1 − h) − δh,
(14.65)
382
CHAPTER 14. MEMBRANE ION CHANNELS: SOLUTIONS
Case II: γ and δ large
This is slightly more complicated, due to the presence of three variable groupings rather than two, as in the previous case.
We let group I consist of x00 and x01 . We have γx00 = δx01 , and we let
x = x00 + x01 .
Similarly, we let group II consist of x10 and x11 . We have γx10 = δx11 , and
we let y = x10 + x11 .
Finally, we let group III consist of x20 and x21 . We have γx20 = δx21 , and we
let h = x20 + x21 .
It now follows that h = γδ x21 + x21 and thus
x21 =
γ
γ+δ
h.
(14.66)
Next we derive the differential equations for x, y and h. Inspection of the
binding diagram quickly gives
dx
dt
dy
dt
dh
dt
= βy − 2αx,
=
2αx + 2βh − (β + α)y,
= αy − 2βh.
(14.67)
(14.68)
(14.69)
We now use the same trick that we used in Question 4. Since x + y + h = 1,
we can write x + y + h = (n + (1 − n))2 , to get h = n2 , y = 2n(1 − n), and
x = (1 − n)2 . Substituting these expressions into the differential equation for
h (or x or y of course) then gives
dn
= α(1 − n) − βn,
dt
(14.70)
as required.
Ex. 6: Show that (3.64) has two negative real roots. Show that when β = 0 and
1
a ≤ λ−λ
(3.65)–(3.67) have two possible solutions, one with α + δ = −λ1 ,
1 −λ2
γ = −λ2 , the other with α + δ = −λ2 , γ = −λ1 . In the first solution
inactivation is faster than activation, while the reverse is true for the second
solution.
Solution
First we have to show that the equation
f (λ) ≡ λ2 + (α + β + γ + δ)λ + (α + δ)(β + γ) − αβ = 0
(14.71)
383
has two real roots. This is most easily done by noticing that f (λ) is a quadratic
in λ, with f (0) > 0 and f (0) > 0. Hence, just from inspection of the graph
of f it follows immediately that the roots of f have negative real part. Furthermore,
(α+β +γ +δ)2 −4[(α+δ)(β +γ)−αβ] = [(a+δ)−(β +γ)]2 +4αβ > 0, (14.72)
from which it follows that the roots must be real and negative.
Now let β = 0. We then have to solve the equations
α+γ+δ
= −λ1 − λ2 ,
γ(α + δ) = λ1 λ2 ,
α = a(λ1 − λ2 ).
(14.73)
(14.74)
(14.75)
Letting z = α + δ, we get from the first two equations
z 2 + (λ1 + λ2 )z + λ1 λ2 = 0,
(14.76)
2z = −λ1 − λ2 ± (λ1 − λ2 ).
(14.77)
and thus
Hence
α + δ = −λ2 ,
or α + δ = −λ1 ,
(14.78)
which are the two desired solutions.
There is a further constraint on the solutions however. Since all the parameters
must be non-negative (from the physiological constraints), these solutions are
only acceptable if we can solve for α and δ to get both positive. This can only
occur if
α < −λ1 ,
(14.79)
1
that is, if a ≤ λ−λ
, as required. Note that λ1 and λ2 are both negative,
1 −λ2
with λ2 < λ1 < 0, and thus if α < −λ1 we must also have α < −λ2 .
Ex. 7: Write a computer program to simulate the response of a stochastic three-state
Na+ channel (Fig. 3.8A) to a voltage step. Take the ensemble average of many
runs to reproduce the macroscopic behavior of Fig. 3.6. Using the data from
simulations, reconstruct the open-time distribution, the latency distribution,
and the distribution of N , the number of times the channel opens. From these
distributions calculate the rate constants of the simulation.
Solution
Ex. 8: Find the differential equations describing the interaction of a two-state channel
with a use-dependent blocker.
Solution
384
CHAPTER 14. MEMBRANE ION CHANNELS: SOLUTIONS
Chapter 15
Excitability: Solutions
Ex. 1: Show that, if k > 1, then (1 − e−x )k has an inflection point, but (e−x )k does
not.
Solution
It is easy to see that (e−x )k has no inflection point. Differentiating twice gives
k(k − 1)(e−x )k−2 which is never zero as long as k > 1.
On the other hand if we differentiate (1−e−x )k twice we get (1−e−x )k−2 e−2x (k 2 −
kex ) which is equal to zero when x = 0 or x = ln k.
Ex. 2: Explain why replacing the extracellular sodium with choline has little effect on
the resting potential of an axon. Calculate the new resting potential with 90%
of the extracellular sodium removed. Why is the same not true if potassium
is replaced?
Solution
Because the sodium conductance is so low at steady-state the sodium Nernst
potential has almost no effect on the resting potential. Thus, removal of the
extracellular sodium, although greatly affecting the sodium Nernst potential,
has little effect on the resting potential.
For the required calculation we use the linear I − V curve with the conductances gNa = 0.01 mS/cm2 , gK = 0.367 mS/cm2 (which are reasonable
estimates for the resting values in squid giant axon, see Chapter 2, Section
2.8.1) and the concentrations given in Table 2.1. This
Vr
g N a V N a + gK V K
gN a + gK
0.01 × 56 − 0.367 × 77
=
mV
0.367 + 0.01
= −73 mV,
=
385
(15.1)
(15.2)
(15.3)
386
CHAPTER 15. EXCITABILITY: SOLUTIONS
which is reasonably close to the value of -65 mV given in Table 2.1.
If 90% of the extracellular sodium is replaced by choline then [Na+ ]e = 43.7
mM and thus
43.7
RT
VN a =
ln
= −3.5 mV.
(15.4)
F
50
Then
Vr =
−0.01 × 3.5 − 0.367 × 77
mV = −75 mV.
0.367 + 0.01
(15.5)
Because the potassium conductance is high at rest, the potassium Nernst
potential has a large effect on the resting potential. Hence, removal of 90% of
the external potassium would have a large effect on the resting potential.
Ex. 3: Plot the nullclines of the Hodgkin–Huxley fast subsystem. Show that vr and
ve in the Hodgkin–Huxley fast subsystem are stable steady states, while vs is
a saddle point. Compute the stable manifold of the saddle point and compute
sample trajectories in the fast phase-plane, demonstrating the threshold effect.
Solution
This exercise is just to reproduce Fig. 4.10. This is most easily done using
XPPAUT, the file for which is given below.
XPPAUT code for Exercise 3.
# set the variables. v is the deviation from rest.
v(0) = 0
m(0) = 0
#
# set the pseudo-steady state parameters
n0=0.3176
h0=0.596
#
# set the remaining parameters
param gna=120
gk=36
gl=0.3 cm=1.0
param vna=115
vk=-12 vl=10.6
#
# set the functions
alpha_m(v)=0.1*(25-v)/(exp((25-v)/10)-1)
beta_m(v)=4*exp(-v/18)
#
v’=(1/cm)*(-gk*(n0^4)*(v-vk) - gna*(m^3)*h0*(v-vna) - gl*(v-vl))
m’=alpha_m(v)*(1-m) - beta_m(v)*m
#
@ total=200, xp=v, yp=m, dt=0.001, xlo=0, xhi=120, ylo=0, yhi=1.2,
@ meth=gear, bound=1000
done
387
Ex. 4: Show how the Hodgkin–Huxley fast subsystem depends on the slow variables;
i.e., show how the v nullcline moves as n and h are changed, and demonstrate
the saddle-node bifurcation in which ve and vs disappear.
Solution
The task here is just to reproduce Fig. 4.11, which may be done using the
same XPPAUT file as given in the previous exercise.
Ex. 5: Plot the nullclines of the fast–slow Hodgkin–Huxley phase-plane and compute
a complete action potential.
Solution
Here one just has to reproduce Fig. 4.13, which can be done using the XPPAUT code shown below.
XPPAUT code for Exercise 5
# initialise the variables. v is the deviation from rest.
v(0)=-0.19571
n(0)=0.31468
#
# set the remaining parameters
param Iapp=0
param gna=120
gk=36
gl=0.3 cm=1.0
param vna=115
vk=-12 vl=10.6
#
# define the functions
alpha_m(v)=0.1*(25-v)/(exp((25-v)/10)-1)
beta_m(v)=4*exp(-v/18)
alpha_n(v)=0.01*(10-v)/(exp((10-v)/10)-1)
beta_n(v)=0.125*exp(-v/80)
minf(v)=alpha_m(v)/(alpha_m(v)+beta_m(v))
#
# define the ODEs
v’=(1/cm)*(-gk*(n^4)*(v-vk) - gna*(minf(v)^3)*(0.8-n)*(v-vna) - gl*(v-vl) + Iapp)
n’=alpha_n(v)*(1-n) - beta_n(v)*n
#
@ total=200, xp=v, yp=n, dt=0.01, xlo=-20, xhi=120, ylo=0, yhi=1.2,
@ meth=gear, bound=1000
@ autovar=Iapp,ntst=50,nmax=2000,npr=2000,dsmin=0.001,dsmax=1,ds=0.1
@ parmin=0,parmax=500,autoxmin=0,autoxmax=500,autoymin=0,autoymax=130
done
Ex. 6: How does the phase plane of the fast-slow Hodgkin-Huxley equations change
with applied current? How much applied current in the fast-slow HodgkinHuxley equations is needed to generate oscillations? Plot a typical oscillation
388
CHAPTER 15. EXCITABILITY: SOLUTIONS
0.9
0.8
n
0.7
0.6
0.5
0.4
0
20
40
60
80
100
v
Figure 15.1: The phase plane and a typical oscillatory trajectory in the fast-slow
Hodgkin-Huxley model when Iapp = 100 µA/cm2 (Exercise 6).
in the phase plane. Plot the maximum of the oscillation against the applied
current to construct a bifurcation diagram.
Solution
The XPPAUT file from the previous exercise can be used to calculate the
nullclines, a typical oscillation, and plot the bifurcation diagram. When Iapp =
100 µA/cm2 we get the phase plane shown in Fig. 15.1. There is only one
minor complication, that XPPAUT does not plot nullclines very well. These
are often better done separately in a program such as Maple, which allows for
more detailed control over the plotting. Below, we give the Maple code that
we used to plot the nullclines in Fig. 15.1.
The Maple code for plotting the nullclines of Fig. 15.1. Note how we step v along in small
increments, and for each value of v we calculate the corresponding value of n by solving the
dv/dt = 0 equation numerically.
Iapp:=100: gna:=120:
vna:=115:
vk:=-12:
gk:=36:
vl:=10.6:
gl:=0.3:
hhdata:=array(1..241,1..3):
count:=0:
for v from -5.1 by 0.5 to 115.1 do
count:=count+1:
# The counter for the index of hhdata
389
oscmax
100
v (mV)
80
60
HB2
40
20
HB1
steady state
0
2
1
3
4 5 6
2
3
10
4 5 6
2
2
3
4 5
100
Iapp (µA/cm )
Figure 15.2: Bifurcation diagram showing the steady state value of v, and the
maximum of v over an oscillation, plotted against Iapp . Calculated from the fastslow Hodgkin-Huxley model (Exercise 6). Stable branches are shown as solid lines,
unstable branches as dotted lines. The two Hopf bifurcations are labeled HB1 and
HB2.
alpha_m:=0.1*(25-v)/(exp((25-v)/10)-1):
beta_m:=4*exp(-v/18):
alpha_n:=0.01*(10-v)/(exp((10-v)/10)-1):
beta_n:=0.125*exp(-v/80):
ninf:=alpha_n/(alpha_n+beta_n):
minf:=alpha_m/(alpha_m+beta_m):
#
vfun:=-gk*(n^4)*(v-vk) - gna*(minf^3)*(0.8-n)*(v-vna) - gl*(v-vl) + Iapp:
hhdata[count,1]:=v:
hhdata[count,2]:=fsolve(vfun=0,n,0..1): # Do the numerical solve
hhdata[count,3]:=ninf: # Might as well plot out ninf while we’re at it.
od:
writedata(hh_fastslow,hhdata);
# Print the data to a file for plotting with IGOR
Following the procedure outlined in the hints gives the bifurcation diagram
shown in Fig. 15.2. The two Hopf bifurcations are labeled HB1 and HB2.
Ex. 7: Suppose that in the Hodgkin–Huxley fast–slow phase-plane, v is slowly decreased to v ∗ < v0 (where v0 is the steady state), held there for a considerable
time, and then released. Describe what happens in qualitative terms, i.e.,
without actually computing the solution. This is called anode break excitation (Hodgkin and Huxley, 1952d. Also see Peskin, 1991). What happens if
v is instantaneously decreased to v ∗ and then released immediately? Why do
these two solutions differ?
390
CHAPTER 15. EXCITABILITY: SOLUTIONS
0.7
0.6
n
0.5
0.4
a
0.3
b
0.2
c
-20
0
20
40
60
v (mV)
80
100
Figure 15.3: Schematic diagram of anode break excitation.
Solution
Anode break excitation is sketched in Fig. 15.3. Suppose the cell is at rest at
the steady state (where the two nullclines intersect. The nullclines are the two
solid lines.) If the cell is hyperpolarised, v is decreased and thus the voltage
moves along the dotted line, along the line labeled a. If v is then held at this
negative value, n will slowly equilibrate to the steady value of n given by that
steady v, i.e., n will decrease until it reaches the dn/dt = 0 nullcline, along
the line labeled b. Note that this will only happen if v is held at this low
value for long enough, as n does not change quickly (n is the slow variable).
If v is now released from the voltage clamp, the system is free to return to the
steady state. However, the trajectory starts from below the N-shaped curve,
and thus v quickly increases until it hits the dv/dt = 0 nullcline (the solution
follows the dotted line labeled c). The trajectory then follows the usual path
back to the steady state. This corresponds to a large action potential, called
anode break excitation.
If v is decreased and then immediately released there is no time for n to
equilibrate to its steady state (i.e., no time to move along path b). Thus, upon
release, the solution just returns to the steady state along path a, moving in
the opposite direction to the arrow.
The crucial fact that causes the difference between these two solutions is that
n is slow. An action potential follows upon release of v only if n is given time
to reach its pseudo-steady state (along path b).
Ex. 8: In the text, the Hodgkin-Huxley equations are written in terms of v = V −Veq .
391
Show that in terms of V the equations become
Cm
dV
dt
= −ḡK n4 (V − VK ) − ḡNa m3 h(V − VNa )
−ḡL (V − VL ) + Iapp ,
dm
dt
dn
dt
dh
dt
(15.6)
= αm (1 − m) − βm m,
(15.7)
= αn (1 − n) − βn n,
(15.8)
= αh (1 − h) − βh h,
(15.9)
where (in units of (ms)−1 ),
αm
=
βm
=
αh
=
βh
=
αn
=
βn
=
−40 − V
,
exp −40−V
−1
10
−V − 65
,
4 exp
18
−V − 65
,
0.07 exp
20
1
,
exp( −35−V
)+1
10
−55 − V
,
0.01
exp( −55−V
)−1
10
−V − 65
0.125 exp
,
80
0.1
(15.10)
(15.11)
(15.12)
(15.13)
(15.14)
(15.15)
and
ḡNa = 120,
VN a = 55,
ḡK = 36,
VK = −77,
ḡL = 0.3,
VL = −54.4,
Veq
(15.16)
= −65.(15.17)
Solution
The only nontrivial task here is to show that the equilibrium voltage is indeed
-65 mV. We use the XPPAUT file below. This is very similar to the XPPAUT
file of Exercise 9, just written in terms of V instead of v. By running this
file we see quickly that Veq = −65. Once Veq is determined, the equations
are easily transformed by setting v = V + 65, which gives (4.20)–(4.29), as
required.
The XPPAUT file for the full Hodgkin-Huxley model, expressed in terms of V (the actual
voltage) instead of v (the deviation of the voltage from the rest state).
# initialise the variables. V is the voltage.
V(0)=-65.0
392
CHAPTER 15. EXCITABILITY: SOLUTIONS
n(0)=0.31768
m(0)=0.052934
h(0)=0.59611
#
# set the parameters
param Iapp=0
param gna=120
gk=36
gl=0.3 cm=1.0
param Vna=50
Vk=-77 Vl=-54.4
#
# define the functions
alpha_m(V)=0.1*(-40-V)/(exp((-40-V)/10)-1)
beta_m(V)=4*exp((-V-65)/18)
alpha_n(V)=0.01*(-55-V)/(exp((-55-V)/10)-1)
beta_n(V)=0.125*exp((-V-65)/80)
alpha_h(V)=0.07*exp((-V-65)/20)
beta_h(V)=1/(exp((-35-V)/10)+1)
#
# define the ODEs
V’=(1/cm)*(-gk*(n^4)*(V-Vk) - gna*(m^3)*h*(V-Vna) - gl*(V-Vl) + Iapp)
n’=alpha_n(V)*(1-n) - beta_n(V)*n
m’=alpha_m(V)*(1-m) - beta_m(V)*m
h’=alpha_h(V)*(1-h) - beta_h(V)*h
#
@ total=200, xp=t, yp=v, dt=0.1, xlo=0, xhi=200, ylo=-80, yhi=30,
@ meth=gear, bound=1000
done
Ex. 9: Solve the full Hodgkin–Huxley system numerically with a variety of constant
current inputs. For what range of inputs are there self-sustained oscillations?
Construct the bifurcation diagram as in Exercise 6.
Solution
We use the XPPAUT file given below to solve the Hodgkin-Huxley model. A
typical oscillation (for Iapp = 100 µA/cm2 ) is shown in Fig. 15.4, while the
bifurcation diagram, calculated the same way as in Exercise 6, is shown in
Fig. 15.5.
XPPAUT file for the Hodgkin-Huxley model, expressed in terms of v, the deviation of the
voltage from rest.
# initialise the variables. v is the deviation from rest.
v(0)=0.0
n(0)=0.31768
m(0)=0.052934
h(0)=0.59611
#
# set the remaining parameters
param Iapp=0
param gna=120
gk=36
gl=0.3 cm=1.0
param vna=115
vk=-12 vl=10.6
#
393
100
0.7
n
80
v
0.6
60
n
0.5
40
v
20
0.4
0
0
10
20
30
40
50
time
Figure 15.4: Oscillation in the Hodgkin-Huxley model when Iapp = 100 µA/cm2 . n
is plotted against the right-hand axis, and v against the left.
maxosc
80
60
v
40
HB2
20
HB1
steady state
0
2
1
3
4 5 6
2
10
3
4 5 6
2
3
4 5
100
Iapp
Figure 15.5: Bifurcation diagram of the Hodgkin-Huxley model, showing the maximum of the oscillation as a function of the applied current.
394
CHAPTER 15. EXCITABILITY: SOLUTIONS
# define the functions
alpha_m(v)=0.1*(25-v)/(exp((25-v)/10)-1)
beta_m(v)=4*exp(-v/18)
alpha_n(v)=0.01*(10-v)/(exp((10-v)/10)-1)
beta_n(v)=0.125*exp(-v/80)
alpha_h(v)=0.07*exp(-v/20)
beta_h(v)=1/(exp((30-v)/10)+1)
#
# define the ODEs
v’=(1/cm)*(-gk*(n^4)*(v-vk) - gna*(m^3)*h*(v-vna) - gl*(v-vl) + Iapp)
n’=alpha_n(v)*(1-n) - beta_n(v)*n
m’=alpha_m(v)*(1-m) - beta_m(v)*m
h’=alpha_h(v)*(1-h) - beta_h(v)*h
#
@ total=200, xp=t, yp=v, dt=0.1, xlo=0, xhi=200, ylo=-10, yhi=100,
@ meth=gear, bound=1000
@ autovar=Iapp,ntst=50,nmax=2000,npr=2000,dsmin=0.001,dsmax=1,ds=0.1
@ parmin=0,parmax=500,autoxmin=0,autoxmax=500,autoymin=0,autoymax=130
done
Ex. 10: The Hodgkin–Huxley equations are for the squid axon at 6.3◦ C. Using that
the absolute temperature enters the equations through the Nernst equation,
determine how changes in temperature affect the behavior of the equations.
In particular, simulate the equations at 0◦ C and 30◦ C to determine whether
the equations become more or less excitable with an increase in temperature.
Solution
It is easiest to use the Hodgkin-Huxley equations written in terms of V , the
voltage, rather than v, the deviation from rest. This is because a change in
temperature will change the Nernst potentials of Na+ and K+ , and will thus
change Veq . We use the same XPPAUT file as used in Exercise 8, changing
the Nernst potentials for each temperature.
Since 6.3◦ C is equal to 279.45 K, the Nernst potentials at 0◦ C can be found
◦
by multiplying VN a , VK and VL by 273.15
279.15 . Hence, at 0 C
VN a =
273.15
× 50 = 48.87,
279.45
VK = −75.26,
VL = −53.17,
(15.18)
VL = −59.
(15.19)
while at 30◦ C,
VN a =
303.15
× 50 = 54.24,
279.45
VK = −83.5,
Solutions of the Hodgkin-Huxley equations, with these values of the Nernst
potentials, are shown in Fig. 15.6. At both temperatures the threshold value of
V (0) is between −59 and −58 mV, but the responses at the higher temperature
are larger and faster. Whether or not you call this more excitable is a matter
of definition.
395
40
20
T=0 degrees C
V
0
-20
-40
-60
-80
0
5
10
40
15
t
20
25
30
20
T=30 degrees C
V
0
-20
-40
-60
-80
0
5
10
15
t
20
25
30
Figure 15.6: Solutions of the Hodgkin-Huxley model at two different temperatures.
Each curve was calculated with a different initial value of V . In the top panel,
V (0) = −62, −60, −59, and −58 (from the lowest curve to the uppermost curve).
In the lower panel V (0) = −65, −63, −61, −59, and −58 (from the lowest curve to
the uppermost curve).
396
CHAPTER 15. EXCITABILITY: SOLUTIONS
Ex. 11: Show that a Hopf bifurcation occurs in the generalized FitzHugh–Nagumo
model only when fv (v ∗ , w∗ ) = −gw (v ∗ , w∗ ), assuming that
fv (v ∗ , w∗ )gw (v ∗ , w∗ ) − gv (v ∗ , w∗ )fw (v ∗ , w∗ ) > 0.
(15.20)
Solution
The generalized FitzHugh-Nagumo system is
dv
dt
dw
dt
= f (v, w)
(15.21)
= g(v, w).
(15.22)
The Jacobian matrix at the steady state is
fv / fw /
gv
gw
(15.23)
where all the partial derivatives are evaluated at the steady state, (v ∗ , w∗ ).
Thus, the eigenvalues of the Jacobian are the roots of
1
λ2 − λ(fv / + gw ) + (fv gw − fw gv ) = 0.
(15.24)
A necessary condition for there to be a Hopf bifurcation is that the roots are
complex conjugate with zero real part, which happens when
fv
+ gw
fv gw − fw gv
=
0
(15.25)
> 0,
(15.26)
as required. (Although this is not a sufficient condition, in practice it nearly
always suffices.)
Ex. 12: Morris and Lecar (1981) proposed the following two-variable model of membrane potential for a barnacle muscle fiber:
dV
+ Iion (V, W ) = Iapp ,
dT
(15.27)
dW
= φΛ(V )[W∞ (V ) − W ],
dT
(15.28)
Cm
where V = membrane potential, W = fraction of open K+ channels, T =
time, Cm = membrane capacitance, Iapp = externally applied current, φ =
397
Cm = 20 µF/cm
2
gCa = 4.4 mS/cm
gL = 2 mS/cm
2
Iapp = 0.06 mA/cm
2
gK = 8 mS/cm
V2 = 18 mV
V3 = 2
V4 = 30 mV
= 120 mV
2
φ = 0.04 (ms)−1
V1 = −1.2 mV
0
VCa
2
VK0 = −84 mV
VL = −60 mV
Table 15.1: Typical parameter values for the Morris–Lecar model.
maximum rate for closing K+ channels, and
0
0
= gCa M∞ (V )(V − VCa
) + gK W (V − VK0 ) + gL (V − V(15.29)
L ),
1
V − V1
M∞ (V ) =
,
(15.30)
1 + tanh
2
V2
1
V − V3
W∞ (V ) =
,
(15.31)
1 + tanh
2
V4
V − V3
Λ(V ) = cosh
.
(15.32)
2V4
Iion (V, W )
Typical rate constants in these equations are shown in Table 15.
(a) Compute the phase portrait of the Morris–Lecar equations. Plot the
nullclines and show some typical trajectories, demonstrating how the
model is excitable.
(b) Does the Morris–Lecar model exhibit anode break excitation (see Exercise 7)? If not, why not?
Solution
(a) We plot the nullclines using the simple Maple routine shown below (XPPAUT is not very good at plotting nullclines and saving them to a file),
and compute the trajectories using the XPPAUT file shown below. The
phase plane, with two typical trajectories is shown in Fig. 15.7.
(b) Consider the trajectory labeled B in Fig. 15.7. If we had hyperpolarised
the cell and held it fixed at this lower voltage we would have ended up
right on the dw/dt = 0 nullcline, just where trajectory B starts, for
example. However, since the dw/dt = 0 nullcline does not dip very low,
when we release the voltage clamp the solution is “sucked” back into the
steady state, without first giving an action potential, as it does for the
Hodgkin-Huxley model.
So the Morris–Lecar model does not exhibit anode break excitation.
398
CHAPTER 15. EXCITABILITY: SOLUTIONS
0.6
0.5
0.4
0.3
w
0.2
A
0.1
B
0.0
-0.1
-60
-40
-20
V (mV)
0
20
40
Figure 15.7: Phase plane of the Morris–Lecar model, with the nullclines shown as
dotted lines, and two typical trajectories (labeled A and B) shown as solid lines.
The Maple code used to plot the nullclines of the Morris–Lecar model.
Iapp:=60: gca:=4.4:
gk:=8:
gl:=2:
vca:=120:
vk:=-84: vl:=-60:
vee1:=-1.2: vee2:=18: vee3:=2:
vee4:=30: phi:=0.04:
mlcdata:=array(1..221,1..3):
count:=0:
for V from -70.1 by 0.5 to 40.1 do
count:=count+1:
# The counter for the index of mlcdata
minf:=0.5*(1+tanh((V-vee1)/vee2)):
winf:=0.5*(1+tanh((V-vee3)/vee4)):
lambda:=cosh((V-vee3)/(2*vee4)):
Iion:=gca*minf*(V-vca)+gk*w*(V-vk)+gl*(V-vl):
vfun:=-Iion + Iapp:
mlcdata[count,1]:=V:
mlcdata[count,2]:=fsolve(vfun=0,w,0..1): # Do the numerical solve
mlcdata[count,3]:=winf: # Might as well plot out winf while we’re at it.
od:
writedata(mlc_fastslow,mlcdata);
# Print the data to a file for plotting with IGOR
The XPPAUT code used to compute the trajectories of the Morris–Lecar model.
399
# Declare the variables
V(0)=-36.755
w(0)=0.070198
#
# Declare the parameters
#
param
Iapp=60.0
cm=20.0
gca=4.4
gk=8
param
gl=2
phi=0.04 vee1=-1.2
param
vee2=18 vee3=2 vee4=30
param
vca=120
vk=-84
vl=-60
#
minf(V)=0.5*(1+tanh((V-vee1)/vee2))
winf(V)=0.5*(1+tanh((V-vee3)/vee4))
lambda(V)=cosh((V-vee3)/(2*vee4))
Iion(V,w)=gca*minf(V)*(V-vca)+gk*w*(V-vk)+gl*(V-vl)
#
# Define right hand sides of the pdes
#
V’ = (1/cm)*( -Iion(V,w) + Iapp )
w’ = phi*lambda(V)*(winf(V)-w)
#
#
@ total=300,dt=0.1,xp=v,yp=w,xlo=-70, xhi=40,
@ ylo=0, yhi=1, meth=gear
done
Ex. 13: The Pushchino model is a piecewise linear model of FitzHugh–Nagumo type
proposed as a model for the ventricular action potential. The model has
f (v, w) = F (v) − w,
1
g(v, w) =
(v − w),
τ (v)
where

for v < v1 ,
 −30v,
γv − 0.12,
for v1 < v < v2 ,
F (v) =

−30(v − 1), for v > v2 ,
2,
for v < v1 ,
τ (v) =
16.6, for v > v1 ,
(15.33)
(15.34)
(15.35)
(15.36)
with v1 = 0.12/(30 + γ) and v2 = 30.12/(30 + γ).
Simulate the action potential for this model. What is the effect on the action
potential of changing τ (v)?
Solution
The differential equations are
dv
dt
dw
dt
= F (v, w)
(15.37)
= g(v, w).
(15.38)
400
CHAPTER 15. EXCITABILITY: SOLUTIONS
1.0
0.8
0.6
w 0.4
0.2
0.0
-0.2
0.0
0.4
v
0.8
1.0
0.8
0.6
v
0.4
0.2
0.0
0
40
80
t
120
Figure 15.8: Typical action potentials in the Pushchino model, shown in the phase
plane (upper panel) and as a function of time (lower panel). In the upper panel the
dashed lines are the nullclines while the solid line and the dotted line (both labeled
with arrows) are the action potentials. The dotted line corresponds to the action
potential where τ = 2.
Note that 1/τ (v) is playing the role of , making w a slow variable compared
to v.
As usual, we use XPPAUT to simulate the model, using the file given below.
After some playing around, it turns out that γ = 1 is a reasonable value to
use. A typical action potential is shown in Fig. 15.8.
Varying τ has the same effect as varying in the FitzHugh-Nagumo model.
As τ is decreased, w becomes less of a slow variable (i.e., it becomes faster
with respect to v) and the trajectories follow the nullclines less closely. The
dotted line in Fig. 15.8 is the action potential corresponding to τ = 2 (for all
v).
The XPPAUT code for the Pushchino model (Exercise 13).
v(0)=-36.755
401
w(0)=0.070198
#
param
gam=1
!v1=0.12/(30+gam)
!v2=30.12/(30+gam)
#
# Do the piecewise definitions using Heaviside functions
bigF(v)=-30*v*heav(v1-v) + (gam*v-0.12)*heav(v-v1)*heav(v2-v) -30*(v-1)*heav(v-v2)
tau(v)=2*heav(v1-v) + 16.6*heav(v-v1)
#
v’ = bigF(v)-w
w’ = (1/tau(v))*(v-w)
#
@ total=300,dt=0.1,xp=v,yp=w,xlo=-1, xhi=1.2,
@ ylo=-0.5, yhi=1, dt=0.01, meth=gear
done
Ex. 14: Perhaps the most important example of a nonphysiological excitable system
is the Belousov–Zhabotinsky reaction. This reaction denotes the oxidation
of malonic acid by bromate in acidic solution in the presence of a transition
metal ion catalyst. Kinetic equations describing this reaction are (Tyson and
Fife, 1980)
du
dt
dv
dt
= −f v
u−q
+ u − u2 ,
u+q
= u − v,
(15.39)
(15.40)
where u denotes the concentration of bromous acid and v denotes the concentration of the oxidized catalyst metal. Typical values for parameters are
≈ 0.01, f = 1, q ≈ 10−4 . Describe the phase portrait for this system of
equations.
Solution
This is an example of a problem where numerical methods don’t help a great
deal, and one has to rely on understanding the general nature of the problem
in order to sketch the phase plane. (The fact that and q are so small hints
at this problem.)
The first task is to sketch the nullclines. Clearly, one is the straight line u = v,
while the other is the line
v=
1
u+q
u(1 − u)
.
f
u−q
(15.41)
This second curve has an asymptote at u = q (very close to 0), and, for all
u away from q, is approximately equal to u(1 − u). We can thus sketch the
nullclines, as shown in Fig. 15.9.
402
CHAPTER 15. EXCITABILITY: SOLUTIONS
du/dt=0
dv/dt=0
v
(u0 ,v0)
q
1
u
Figure 15.9: Sketch of the phase plane of the Belousov–Zhabotinsky reaction. (The
sketch is not to scale, since q = 0.0001). The vertical dotted line is the asymptote
of the dv/dt = 0 nullcline. A typical oscillatory trajectory is shown as a dashed
line, with arrows to show the direction of movement. Double arrow heads denote a
fast portion of the trajectory. The steady state is u0 = v0 = 0.014089.
The steady state, u0 , is given by the root of the equation
u2 − u(1 − q − f ) − q(1 + f ) = 0,
(15.42)
√
√
which, when f = 1 has the approximate solution u ≈ 2q = 2 × 10−2 . In
this case numerical solution shows that u0 = v0 = 0.014089.
Because u is very fast with respect to v we know that the solutions will move
almost horizontally, except on the du/dt = 0 nullcline. Thus, by determining
the direction of the trajectories on the nullclines we can sketch in a typical
solution, as shown by the dashed curve. The horizontal parts of the trajectory
are very fast. Hence, a typical oscillatory solution consists of fast jumps
between an upper and a lower value of u.
Chapter 16
Intercellular Communication:
Solutions
Ex. 0a: (a) Verify the
last two steps of (5.2). Hint: Use Stirling’s formula n! ≈
2π
x n
n+1 −n
n
e
n , and show that limn→∞ ln[(1 − n ) ] = −x.
(b) Verify that the Poisson
∞ distribution P (k) =
verifying that k = k=0 kP (k) = m.
e−m mk
k!
has mean m, by
(c) Verify that the sum of k identical Gaussian distributions with mean µ
and variance σ 2 is a Gaussian distribution with mean kµ and variance
kσ 2 .
Solution
(a) Firstly we show that limn→∞ (1 − x/n)n = e−x .
x n
x
lim ln 1 −
= lim n ln 1 −
n→∞
n→∞
n
n
ln 1 − nx
= lim
n→∞
1/n
=
lim
n→∞
1
1−x/n
x
n2
−1/n2
= −x.
Now take the exponential of both sides to get the result.
Secondly, we use Stirling’s formula to get
2π
n+1 −n
n
e
n!
n
lim
= lim
n→∞ nk (n − k)!
n→∞ k
n (n − k)n−k+1 e−(n−k)
403
(16.1)
2π
n−k
404
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
= e−k lim
n→∞ nk (n
−k
= e
lim
n→∞
= e−k lim
n→∞
nn+1
− k)n−k+1
n−k+1
n
n−k
1+
k
n−k
n−k
·
n
n−k
= e−k ek
=
1.
(16.2)
The last two steps of (5.2) follow from these two results.
(b)
k =
=
=
∞
k=0
∞
k=0
∞
k=1
kP (k)
ke−m mk
k!
e−m mk
(k − 1)!
= me−m
∞
mk−1
(k − 1)!
k=1
−m m
= me
since em =
∞
mk
k=0 k!
e
= m,
(16.3)
(its Taylor series).
(c)
Ex. 2: Motivated by the smallness of ci with respect to ce , simplify the Llinás model
of presynaptic Ca2+ channels by setting ci ≈ 0 (Peskin, 1991). Show that this
change makes no difference to the behavior of the model as shown in Figs. 5.5
and 5.6.
Solution
One needs only to replot the solutions (using your favorite plotting package)
to see this is so. This is a simple exercise, but useful to ensure that students
understand how to plot these curves. The major concern is to get the units
correct.
If we express the concentrations in units of mM (i.e., millimoles per litre) then
this is equivalent to moles/m3 . Similarly, if we let PCa = 10 µm s−1 this is
equivalent to 10−5 m s−1 . Then, since all the length units are in metres, the
405
current density is calculated in units of amperes per square meter, which is
equivalent to pA(µm)−2 .
Ex. 3: (a) In response to voltage jumps, the model of the presynaptic Ca2+ current
discussed in the text cannot give the curves shown in Fig. 5.7. Why not?
(b) Calculate and plot the solution to (5.8) if V = Vs + V1 (H(t) − H(t − 4)),
where H is the Heaviside function.
Solution
(a) Because, in the model, an instantaneous jump in voltage gives an instantaneous jump in ICa , while the curves in Fig. 5.7 give smooth changes in
ICa .
(b) Suppose that o(0) = ô(Vs ) (the steady-state value of o). If V is jumped
up to V1 at t = 0, and then jumped back down to Vs at t = 4, we get
o1 (t),
0 < t ≤ 4,
o(t) =
(16.4)
4 < t,
o2 (t),
where
o1 (t)
=
o2 (t)
=
k1 (V1 )s0
(1 − e−(k1 (V1 )+k2 )t ) + ô(Vs )e−(k1 (V1 )+k2 )t(16.5)
],
k1 (V1 ) + k2
k1 (Vs )s0
(1 − e−(k1 (Vs )+k2 )(t−4) ) +
k1 (Vs ) + k2
o1 (4)e−(k1 (Vs )+k2 )(t−4) .
Finally
ICa

n
 j(V1 ) s0 o1 (t) ,
n
s0
n
=
 j(Vs ) s0 o2 (t) ,
n
s0
0 < t ≤ 4,
(16.6)
(16.7)
4 < t.
Note that, although o(t) is continuous, ICa is not. Plots for three values
of V1 are shown in Fig. 16.1.
Ex. 4: Calculate the analytic solution to (5.8) when V is a given function of t.
Solution
If V is a given function of t, then k1 and k2 are also. Thus, we need to solve
do
+ (k1 (t) + k2 (t))o = s0 k1 (t).
dt
(16.8)
406
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
0
2
ICa ( pA/(µm) )
-200
-400
V1 = -40 mV
V1 = -20 mV
V1 = -10 mV
-600
-800
-1000
0
2
4
time (ms)
6
8
Figure 16.1: Synaptic suppression in the model of presynaptic Ca2+ channels
(Exercise 3). Plots of ICa , calculated from (16.7), for Vs = −70 mV and
V1 = −40, −20, −10 mV.
t
Multiply by the integrating factor e (k1 (s)+k2 (s))ds to get
t
t
d
o(t)e (k1 (s)+k2 (s))ds = s0 k1 (t)e (k1 (s)+k2 (s))ds .
dt
(16.9)
Integrate both sides and solve for o(t) to get
t
t
t
τ
o(t) = s0 e− (k1 (s)+k2 (s))ds
k1 (τ )e (k1 (s)+k2 (s))ds dτ +Ce− (k1 (s)+k2 (s))ds ,
(16.10)
for some constant C. Use the initial condition to solve for C, to get
t
t
t
τ
−
(k1 (s)+k2 (s))ds
−
(k (s)+k2 (s))ds
(k1 (s)+k2 (s))ds
0
o(t) = s0 e
k1 (τ )e
dτ +o(0)e 0 1
.
0
(16.11)
Ex. 5: (a) In Fig. 5.11, why are some of the curves positive and some negative?
(b) Construct a simple function F (t) with the same qualitative shape as the
function W (t) used in the Magleby and Stevens model (Fig. 5.10).
(c) Calculate the analytic solution to (5.44) for that F . Compare to the
curves shown in Fig. 5.11.
Solution
(a) Note that the current is plotted in Fig. 5.11, not the conductance, and so
the curves change sign when V is on different sides of the Nernst potential
of the ACh-sensitive channels (around 15 mV). Thus, when V < 15 mV,
the current is negative, and when V > 15 mV, the current is positive.
407
current (arbitrary units)
0.05
0.00
-0.05
-0.10
0
1
2
3
4
5
time (ms)
Figure 16.2: Plots of IACh calculated from (16.15) of Exercise 5. The corresponding
values for V are, from top to bottom, 32, 20, −30, −56, −82, −106 and −161 mV,
the same values as in Fig. 5.11. The only parameter not specified in the text is b
(in β(V )) which we set arbitrarily at 1.
(b) A bit of fiddling shows that W (t) = t2 e−10t is a reasonable approximation
of the curve in Fig. 5.10.
(c) We need to solve
dx
+ αx = βW (t).
dt
Multiplying by the integrating factor eαt and integrating gives
t
−αt
ατ
x(t) = e
x(0) + β
e W (τ ) dτ .
(16.12)
(16.13)
0
Substituting in W (t) = t2 e−10t and assuming that x(0) = 0 then gives
t
x(t) = βe−αt
τ 2 e(α−10)τ dτ
0
2
[t (α − 10)2 + 2t(10 − α) + 2]e(α−10)t − 2
= βe−αt
(16.14)
.
(α − 10)3
The current, IACh , is then given by
IACh = x(t)(V − VACh ),
(16.15)
where VACh = −15 mV. Plots of IACh for various values of V are given
in Fig. 16.2.
408
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
Ex. 6: In the Magleby and Stevens model, a simple choice for the release function
f (t) results in end-plate conductances with considerable qualitative similarity
with those in Fig. 5.11. Suppose there is a sudden release of ACh into the
synaptic cleft at time t = 0. We take f (t) = γδ(t), where δ is the Dirac delta
function. Show that the resulting differential equation is
dc
= −ke c,
dt
c(0) = γ,
(16.16)
for t ≥ 0. Solve for c, and, assuming that γ is small, substitute this expression
for c into the differential equation for x, (5.40). Look for a solution for x of
order γ, and show that
x(t) =
−ke t
γβN
− e−αt ,
e
K(α − ke )
(16.17)
which is always positive. Sketch the solution.
Solution
With f (t) = γδ(t), (5.37) becomes
dc
= γδ(t) − ke c,
dt
c(0) = 0.
(16.18)
As usual, we get rid of the delta function by integrating across the singularity,
i.e., by integrating from − to and letting → 0. Thus
dc
γδ(t) dt −
ke c dt = γ −
ke c dt.
(16.19)
dt =
− dt
−
−
−
Letting → 0, and assuming that c is bounded, now gives
c(0+ ) − c(0− ) = γ,
(16.20)
and thus, since c(0− ) = 0 (from the original initial condition) we have
dc
= −ke c,
dt
c(0) = γ.
t > 0,
(16.21)
(16.22)
The equation for x now becomes
dx
γβN e−ke t
= −αx +
dt
K + γe−ke t
(16.23)
which, if γ is assumed to be small, can be approximated by
dx
γβN −ke t
.
= −αx +
e
dt
K
(16.24)
409
0.25
0.20
0.15
x
0.10
0.05
0.00
0
1
2
3
4
5
t
Figure 16.3: Plot of x(t) calculated as in Exercise 6, using the arbitrarily chosen
parameters γ = 0.1, K = 10, α = 1, β = 1, N = 100 and ke = 2.
Let x = γx0 . Then
dx0
βN −ke t
= −αx0 +
e
dt
K
(16.25)
which has solution (using the integrating factor eαt )
x0 =
βN
(e−ke t − e−αt ).
K(α − ke )
(16.26)
Note that we have used the initial condition x0 (0) = 0. Hence,
x=
γβN
(e−ke t − e−αt ),
K(α − ke )
(16.27)
as required. A plot of x is given in Fig. 16.3.
Ex. 7: Peskin (1991) presented a more complex version of the Magleby and Stevens
model. His model is based on the reaction scheme
−→ ACh,
ACh + R
ACh + E
rate rT per unit volume,
β
k1
−→
←−
k2
ACh · R
k3
−→
←−
k4
ACh · E
−→
←−
(16.28)
ACh · R∗ ,
(16.29)
E,
(16.30)
α
γ
−→
where E is some enzyme that degrades ACh in the synaptic cleft. (The assumption of enzymatic degradation of ACh is one of the ways in which the
Peskin model differs from the Magleby and Stevens model. The other difference is that the Peskin model does not assume that the amount of ACh bound
to its receptor is negligible.)
410
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
(a) Write down the equations for the 6 dependent variables. Use conservation
laws to eliminate two of the equations.
(b) Assume that the reactions involving ACh with R, and ACh with E (with
reaction rates ki , i = 1, 4), are fast to obtain expressions for [R] and [E]
in terms of the other variables. Substitute these expressions into the
differential equations for [ACh · R∗ ] and [ACh] − [R] − [E] to end up with
two differential equations in [ACh · R∗ ] and [ACh].
Answer: if we let A denote [ACh], and R∗ denote [ACh · R∗ ] the differential equations are
d
dt
dR∗
dt
A−
C̄1 − R∗
C̄2
−
KR A + 1 KE A + 1
βKR A
∗
(16.31)
,
(C̄1 − R∗ ) − αR
KR A + 1
KE A
= rT − γ C̄2
,
(16.32)
KE A + 1
=
where C̄1 , C̄2 , KR and KE are constants.
(c) Solve these equations when the stimulus is a small sudden release of
ACh (i.e., assume that rT = δ(t) and look for solutions of O()). Show
that the solution has the same form as (5.62) but that the exponential
coefficients are given by the roots of a quadratic polynomial. Choose
arbitrary values for the parameters and plot the solution.
Hint: look for solutions of the form R∗ = x, A = y and derive the
differential equations
dx
dt
dy
dx
+ a2
dt
dt
= a1 y − αx,
(16.33)
= δ(t) − a3 y,
(16.34)
for some constants a1 , a2 and a3 . Then integrate over the singularity at
t = 0 to show that y(0) = 1/a2 . Solve by finding the eigenvalues and
eigenvectors of the resultant homogeneous linear system.
Solution
(a) Let A = [ACh], R = [R], E = [E], R∗ = [ACh · R∗ ], C1 = [ACh · R], and
C2 = [ACh · E]. Then
dA
dt
dR
dt
dE
dt
dR∗
dt
= rT − k1 AR + k2 C1 − k3 AE + k4 C2 ,
(16.35)
= −k1 AR + k2 C1 ,
(16.36)
= −k3 AE + k4 C2 + γC2 ,
(16.37)
= βC1 − αR∗ ,
(16.38)
411
dC1
dt
dC2
dt
= k1 AR − k2 C1 − βC1 + αR∗ ,
(16.39)
= k3 AE − k4 C2 − γC2 .
(16.40)
From the conservation of E and R we have
E + C2
∗
R + R + C1
= constant = C̄2 ,
(16.41)
= constant = C̄1 .
(16.42)
Thus we have the four equations
dA
dt
dR
dt
dE
dt
dR∗
dt
(16.43)
= rT − k1 AR + k2 (C̄1 − R − R∗ ) − k3 AE + k4 (C̄2 − E),
= −k1 AR + k2 (C̄1 − R − R∗ ),
(16.44)
= −k3 AE + k4 (C̄2 − E) + γ(C̄2 − E),
(16.45)
= β(C̄1 − R − R∗ ) − αR∗ .
(16.46)
(b) Assuming that the two enzymatic reactions are fast we have the pseudoequilibrium
k1 AR
= k2 C1 ,
(16.47)
k3 AE
= k4 C2 ,
(16.48)
and thus
R
=
E
=
C̄1 − R∗
,
KR A + 1
C̄2
,
KE A + 1
where KR = k1 /k2 and KE = k3 /k4 .
d
(A − R − E) = rT − γ(C̄2 − E), it follows that
Since dt
d
KE A
C̄1 − R∗
C̄2
A−
−
= rT − γ C̄2
.
dt
KR A + 1 KE A + 1
KE A + 1
(16.49)
(16.50)
(16.51)
Furthermore, from (16.46) we have
dR∗
dt
= β(C̄1 − R − R∗ ) − αR∗
=
βKR A
(C̄1 − R∗ ) − αR∗ .
KR A + 1
(16.52)
(c) Let A = y, R∗ = x. Then
dx
βKR y
=
(C̄1 − x) − αx,
dt
KR y + 1
(16.53)
412
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
and thus, to lowest order in ,
dx
= βKR C̄1 y − αx.
dt
(16.54)
Similarly,
d
KE y
C̄1 − x
C̄2
y −
−
= δ(t) − γ C̄2
.
dt
KR y + 1 KE y + 1
KE y + 1
(16.55)
Since
(C̄1 − x)(1 + KR y)−1 = C̄1 − (x + C̄1 KR y) + O(2 ),
(16.56)
it follows that, to lowest order in ,
d
[y(1 + C̄1 KR + C̄2 KE ) + x] = δ(t) − γ C̄2 KE y.
dt
(16.57)
Then, let a1 = βKR C̄1 , a2 = 1 + C̄1 KR + C̄2 KE , and a3 = γ C̄2 KE to
get
dx
dt
dx
dy
+ a2
dt
dt
= a1 y − αx,
(16.58)
= δ(t) − a3 y.
(16.59)
To cope with the delta function we integrate both equations across the
singularity, i.e., from − to and then let → 0. Assuming bounded
solutions, this gives x(0+ ) − x(0− ) = 0 and y(0+ ) − y(0− ) = 1/a2 . Thus,
(16.58) and (16.59) become
dx
dt
dx
dy
+ a2
dt
dt
= a1 y − αx,
(16.60)
= −a3 y
(16.61)
on t > 0, with the initial conditions x(0) = 0, y(0) = 1/a2 .
A slight rearrangement now gives
d x
x
=M
.
y
dt y
where
M=
−α
α
a2
a1
−(a1 +a3 )
a2
(16.62)
.
The eigenvalues of M are the roots of
αa3
a1 + a3
2
λ +λ α+
+
= 0.
a2
a2
(16.63)
(16.64)
413
Since
2
2
2
a1 + a3
a3
a1
αa3
a3
a1
α+
α+
+
−4
= α−
+2
> 0,
a2
a2
a2
a2
a2
a2
(16.65)
and since
a1 + a3
α+
>0
(16.66)
a2
it follows that the roots of (16.64) are both real and negative. Call these
roots λ1 and λ
to
2 , where λ2 < λ1 < 0. The eigenvector corresponding
a1
a1
λ1 is
, and the eigenvector corresponding to λ2 is
.
λ1 + α
λ2 + α
Thus, the general solution is
x
a1
a1
λ1 t
= κ1
(16.67)
e + κ2
eλ2 t ,
y
λ1 + α
λ2 + α
where the constants κ1 and κ2 are determined from the initial conditions,
y(0) = 1/a2 , x(0) = 0. Hence
κ1
=
κ2
=
1
,
a2 (λ1 − λ2 )
1
,
a2 (λ2 − λ1 )
(16.68)
(16.69)
from which we get, finally,
1
1
x
a1
a1
=
eλ1 t +
eλ2 t .
y
a2 (λ1 − λ2 ) λ1 + α
a2 (λ2 − λ1 ) λ2 + α
(16.70)
In Fig. 16.4 we plot x(t) and y(t) for arbitrary choices of the parameters.
Ex. 8: Solve the above exercise (and obtain the same solution!) by nondimensionalizing, finding a small parameter, and then solving in terms of an asymptotic
expansion.
Hints:
(a) Introduce the dimensionless variables τ = γt, r = R/C̄1 , r∗ = R∗ /C̄1 ,
e = E/C̄2 , a = A/C̄1 and let γ/k2 = be the small parameter. What
are the initial conditions?
(b) Look for solutions of the form a = a1 , e = e0 + e1 , r = r0 + r1 and
r∗ = r0∗ + r1∗ . At order 0 obtain algebraic equations for r0 and e0 and
solve the differential equation for r0∗ .
(c) At order obtain algebraic equations for r1 and e1 , and a differential
equation for r1∗ .
414
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
2.0
x
y
1.5
1.0
0.5
0.0
0
1
2
3
4
5
time
Figure 16.4: Plot of x(t) and y(t) from (16.70) (Exercise 7), using the arbitrarily
chosen parameters a1 = 0.1, a2 = 0.5, a3 = 1, α = 0.5.
(d) Finally, consider terms of order 2 . Add the differential equations in the
appropriate manner (as in the previous question) so that unwanted terms
cancel. Thus derive the second differential equation.
(e) Check that you have derived the correct differential equations by converting to the original variables.
Solution
(a) With that change of variables the differential equations become
da
dτ
dr
dτ
de
dτ
dr∗
dτ
= 2 µ1 (t) − µ2 ar + (1 − r − r∗ ) − µ3 ae + µ4 (1 − e),(16.71)
= −µ2 ar + (1 − r − r∗ ),
(16.72)
= −µ3 µ5 ae + µ4 µ5 (1 − e) + (1 − e),
(16.73)
= µ6 (1 − r − r∗ ) − µ7 r∗ ,
(16.74)
δ(t)
2 k4
1
where µ1 (t) = C̄
, µ2 = C̄k12k1 , µ3 = C̄k22k3 , µ4 = C̄
, µ5 = C̄
, µ6 =
C̄1 k2
C̄2
1γ
2
β/γ and µ7 = α/γ. Note that µ1 has a coefficient of . One of those
epsilons comes from the nondimensionalisation, the other comes from the
assumption that rT = δ(t).
The initial conditions are a(0) = 0, r(0) = 1, e(0) = 1 and r∗ (0) = 0.
(b) Substitute in a = a1 , e = e0 + e1 , r = r0 + r1 and r∗ = r0∗ + r1∗ and
equate coefficients of 0 (i.e., the constant terms). This gives
1 − r0 − r0∗
=
0,
(16.75)
415
1 − e0
dr0∗
dt
=
0,
(16.76)
= µ6 (1 − r0 − r0∗ ) − µ7 r0∗ .
(16.77)
Thus, e0 = 1. Furthermore,
dr0∗
= −µ7 r0∗ ,
dt
r0∗ (0) = 0,
(16.78)
which has the solution r0∗ = 0. It follows that r0 = 1. Note that r0 and
e0 both satisfy the initial conditions also.
(c) Now go to order . Equating the coefficients of gives
0
= −µ2 a1 r0 − r1 − r1∗ ,
0 = −µ3 µ5 a1 e0 − µ4 µ5 e1 ,
dr1∗
= µ6 (−r1 − r1∗ ) − µ7 r1∗ ,
dt
(16.79)
(16.80)
(16.81)
and thus
r1
e1
dr1∗
dt
= −µ2 a1 − r1∗ ,
−µ3
=
a1 ,
µ4
= µ6 µ2 a1 − µ7 r1∗ .
(16.82)
(16.83)
(16.84)
(d) Finally, go to order 2 , where all kinds of unwanted terms (a2 , for
instance) start appearing. The trick is to recognise that by forming
da1
dr1
1 de1
dt − dt − µ5 dt all the unwanted terms cancel out, leaving only the
term in e1 , which is already known from the previous step.
By equating coefficients of powers of 2 , and adding the equations in the
indicated manner, we get
da1
1
dr1
1 de1
e1 ,
−
−
= µ1 (t) +
dτ
dτ
µ5 dt
µ5
and thus, using (16.82) and (16.83),
d
µ3
µ3
+ r1∗ = µ1 (t) −
a1 .
a1 1 + µ2 +
dt
µ4 µ5
µ4 µ5
(16.85)
(16.86)
Equations (16.84) and (16.86) correspond to (16.58) and (16.59) of Exercise 7.
(e) To check that we have obtained the same model as in Exercise 7 we just
need to rewrite (16.84) and (16.86) in the original variables, a straightforward exercise of which no further details are given.
416
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
Ex. 9: Solve (5.50) and plot the solution (Peskin, 1991). Compare to the plot of x(t)
from (5.49).
Solution
Let
φ(t) =
−ke t
βN
− e−αt
e
K(α − ke )
(16.87)
so that gs (t)/γ = φ(t). Then we can write (5.50) as
dV1
Vs − Vr
gr
φ(t).
V1 =
+
dt
Cm
Cm
(16.88)
Multiplying both sides by the integrating factor egr t/Cm gives
d gr t/Cm V s − Vr
φ(t)egr t/Cm ,
=
V1 e
dt
Cm
(16.89)
and hence, integrating both sides and dividing by egr t/Cm , we get
(Vs − Vr )βN
e−ke t
e−αt
V1 =
+ Ke−gr t/Cm ,
−
K(α − ke )
αCm − gr
ke Cm − gr
(16.90)
for some constant K. From the initial condition V1 (0) = 0 (which is a result
of the initial condition V (0) = Vr ) it follows that
(Vs − Vr )βN
1
1
K=
,
(16.91)
−
K(α − ke )
ke Cm − gr
αCm − gr
and thus
V1 =
(Vs − Vr )βN
K(α − ke )
e−ke t − e−gr t/Cm
e−αt − e−gr t/Cm
−
αCm − gr
ke Cm − gr
.
(16.92)
In Fig. 16.5 we plot x(t) and V1 (t) on the same graph, with V1 plotted against
the right-hand axis in order to facilitate comparison. As can be seen, V1
responds more slowly than x, due to the time delay inherent in the membrane
conductance.
Ex. 10: By linking the output of the Llinás model to the input of the single-domain/boundcalcium model, and then linking this to the input of the Magleby and Stevens
model (the rate of production of ACh) construct a unified model for the synaptic cleft that connects the presynaptic action potential to the postsynaptic
voltage via the concentration of ACh in the synaptic cleft. Solve the model
numerically and compare to the simpler model presented briefly in Fig. 5.8.
417
40
0.14
x
V1
0.12
30
0.10
0.08
20 mV
0.06
0.04
10
0.02
0.00
0
0
1
2
3
4
5
time
Figure 16.5: Comparison of x(t) (Exercise 6) and V1 (t) (Exercise 9), using the same
parameters as in Fig. 16.3. Additional parameters are Vs = −15, Vr = −70, Cm = 1,
gr = 1. (Cm and gr are chosen arbitrarily.) V1 is plotted against the right-hand
axis, x against the left.
Solution
Briefly, the main problem is getting the units correct throughout. Some parameters just have to be guessed. Time in ms, voltage in mV, concentration
in µM. This question will probably be expanded in presentation, with a lot
more hints thrown in. The solution will be greatly expanded with a lot more
graphs. This is just the quick and dirty version.
Code for the full model connecting the presynaptic voltage to the postsynaptic voltage.
v1(0)=-50
w(0)=0
o(0)=0
c(0)=0
o1(0)=0
o2(0)=0
o3(0)=0
o4(0)=0
a(0)=0
x(0)=0
y(0)=0
v2(0)=-70
###########################################################
# FHN params
Vr=-70
# Llinas params
param k10=2, k20=1, z1=1, n=5,ci=0,ce=40,PCa=0.00001,s0=100
# Voltage-type params
param FRT=.0386 F=96490.0
418
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
# Calcium parameters. radius and area are in um, vol in litres.
param kout=1 radius=10
!area=4*pi*radius^2
!vol=(4/3)*pi*(radius^3)*1e-15
# Calcium-stimulated secretion params
param kp1=3.75e-3, km1=4e-4
param kp2=2.5e-3, km2=1e-3
param kp3=5e-4, km3=0.1
param kp4=7.5e-3, km4=10
# Magleby parameters
param bigA=0.008 bigB=1.43 bigN=10
param smalla=0.00315 smallb=1
param k1m=1000, k2m=500, ke=10
# Post-synaptic voltage params
param Vs=-15, gr=1, Cm=1
###########################################################
# Llinas functions
k1(x)=k10*exp(z1*x*FRT)
fac(x)=2*x*FRT
j(x)=PCa*2*F*fac(x)*(ci-ce*exp(-fac(x)))/(1-exp(-fac(x)))
ICa(x,y)=j(x)*(s0/n)*((y/s0)^n)
# calcium-stimulated secretion functions
tau1(x)=1/(kp1*x+km1)
tau2(x)=1/(kp2*x+km2)
tau3(x)=1/(kp3*x+km3)
tau4(x)=1/(kp4*x+km4)
# Magleby functions
alpha(x)=bigB*exp(bigA*x)
beta(x)=smallb*exp(smalla*x)
###########################################################
aux ICaplot=j(v1)*(s0/n)*((o/s0)^n)
aux R=o1*o2*o3*o4*3e6
###########################################################
# FHN odes for the presynaptic voltage
v1’=100*(0.0001*(v1-Vr)*(70-(v1-Vr))*((v1-Vr)-7)-w)
w’=0.25*(v1-Vr-5*w)
# Llinas odes
o’=k1(v1)*(s0-o) - k20*o
# Calcium odes. Conversion factors are included explicitly, for clarity
c’= -(ICa(v1,o)*1e-12*1e3)*area/(2*vol*F) - kout*c
# Calcium-stimulated secretion odes
o1’=kp1*c - o1/tau1(c)
o2’=kp2*c - o2/tau2(c)
o3’=kp3*c - o3/tau3(c)
o4’=kp4*c - o4/tau4(c)
# Magleby odes
419
voltage (mV)
-20
presynaptic
postsynaptic
-40
-60
-80
0
2
4
6
8
time (ms)
10
12
14
Figure 16.6: Plots of the presynaptic and postsynaptic voltages calculated from the
model of Exercise 10.
x’=-alpha(v2)*x + beta(v2)*y
y’=alpha(v2)*x + k1m*a*(bigN-x-y) - (beta(v2)+k2m)*y
a’=3e6*o1*o2*o3*o4 - ke*a - k1m*a*(bigN-x-y) + k2m*y
# Post-synaptic voltage ode
v2’= (1/Cm)*(-gr*(v2-Vr) - x*(v2-Vs))
###########################################################
@ total=15, dt=0.01, xlo=0, xhi=15,
@ meth=gear
done
ylo=-85, yhi=0, yp1=v1, yp2=v2, bounds=10000, nplot=2
420
CHAPTER 16. INTERCELLULAR COMMUNICATION: SOLUTIONS
Chapter 17
Nonlinear Wave
Propagation: Solutions
Ex. 1: Show that the (U, W ) phase-plane of the bistable equation, (7.6) and (7.7),
has three steady states, two of which are saddle points. What is the nature of
the third steady state? Show that the slope of the unstable manifold at the
origin is given by the positive root of λ2 − cλ + f (0) = 0 and is always larger
than c. What is the slope of the stable manifold at (U, W ) = (1, 0)? Show
that the slopes of both these manifolds are increasing functions of c.
Solution
As described in the text, the phase plane of the bistable equation is described
by the two equations
U
= W,
(17.1)
W
= cW − f (U ),
(17.2)
where f has a generic “cubic” shape, with three roots, at U = 0, α and 1.
Thus, the steady states are obtained by setting the two right hand sides equal
to zero, from which we get W = 0 and f (U ) = 0. It follows immediately
(from the constraints on f (U )) that there are three steady states in the phase
plane, (U, W ) = (0, 0), (α, 0) and (1, 0).
To determine the linear stability of each steady states we construct the linearized system around that steady state. At the steady state U = U0 , the
linearized system is governed by the matrix
0
1
A=
.
(17.3)
−f (U0 ) c
As usual, the eigenvalues and eigenvectors of A determine the linear stability
of U0 . (If this procedure is unfamiliar, we suggest very strongly that you look
421
422
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
through Strogatz (1994), for an introduction to linear stability analysis). The
eigenvalues of A are the roots of the equation
λ2 − cλ + f (U0 ) = 0
(17.4)
and thus the eigenvalues, λ±, are given by
2λ± = c ± c2 − 4f (U0 ).
(17.5)
Each eigenvalue has an associated eigenvector, ξ± , obtained by solving the
linear system
(A − λ± I)ξ± = 0.
(17.6)
This immediately gives
ξ± =
1
λ±
.
(17.7)
We are now in a position to answer all the various bits of this question. First
note that if f (U0 ) < 0 then (17.4) must necessarily have one real positive root,
and one real negative root, in which case (U0 , 0) will be a saddle point. Since,
by the way we have constructed f we know that f (0) < 0 and f (1) < 0,
we then conclude that (0, 0) and (1, 0) are saddle points. Furthermore, since
f (α) > 0, it follows that, at the steady state (α, 0), the eigenvalues can be
either real or complex, but in either case must have positive real part, since
c > 0. Thus (α, 0) is linearly unstable.
Next we consider the slope of the unstable manifold at the origin. From (17.7)
we see that the unstable manifold at the origin is in the direction of the vector
(1, λ+ ), where
2λ+ = c + c2 − 4f (0) > c.
(17.8)
Since this vector has slope λ+ , it follows that the slope of the unstable manifold
at the origin is λ+ . From (17.8) it is clear that λ+ is an increasing function
of c.
Similarly, the stable manifold at (1, 0) is in the direction of the eigenvector
(1, λ− ), where
(17.9)
2λ− = c − c2 − 4f (1) < 0.
It remains only to show that (17.9) defines an increasing function of c. However, this is easily seen by differentiating with respect to c:
2λ− (c) = 1 − c
c2
− 4f (1)
>0
(17.10)
where
the last inequality results from the fact that f (1) < 0 and thus
2
c − 4f (1) > c.
423
Ex. 2: Use cable theory to find the speed of propagation for each of the axons listed
in Table 6.1, assuming that the ionic currents are identical and the speed of
propagation for the squid giant axon is 21 mm/ms.
Solution
According to (7.21)
s=
cλm
,
τm
(17.11)
where c depends only on the ionic conductances. Here, we are assuming that
c is the same for all the cell types in Table 6.1.
For the squid giant axon
cλm
c × 6.5 mm
= 21 mm/ms,
=
τm
1 ms
(17.12)
from which it follows that c = 21/6.5 = 3.23. We now just use that value for
c to calculate the wave speeds for the other axons.
lobster giant axon s = 3.23 ×
2.5 mm
2 ms = 4 mm/ms
mm
crab giant axon s = 3.23 × 2.4
7 ms = 1.1 mm/ms
4 mm
earthworm giant axon s = 3.23 × 3.6
ms = 3.6 mm/ms
5.4 mm
marine worm giant axon s = 3.23 × 0.9 ms = 19.4 mm/ms
mm
mammalian cardiac cell s = 3.23 × 1.5
8.4 ms = 0.58 mm/ms
mm
barnacle muscle fiber s = 3.23 × 2.8
4.6 ms = 2 mm/ms
Ex. 3: Construct a traveling wave solution to the piecewise linear bistable equation,
(7.1) and (7.3). Show that the wave travels with speed √1−2α
. (Construct
α−α2
the solution by using the techniques presented in Section 7.3.1.)
Solution
We want to solve
Vt
= Vxx + f (V )
f (V ) = −V + H(V − α).
(17.13)
(17.14)
Change variables to ξ = x − ct to get V + cV − V + f (V ) = 0. We look
for a solution of the form shown in Fig. 17.1. Note that the ξ axis is divided
into two regions; region I, ξ < 0, where V > α, and region II, ξ > 0, where
V < α. We solve the differential equation on each region separately and then
stick the pieces together by ensuring the solution has a continuous derivative
at ξ = 0.
424
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
V
α
II
I
ξ
Figure 17.1: Schematic diagram of the traveling wave for question 3.
Region I On region I we have
VI + cVI − VI = −1,
VI (0) = α,
(17.15)
which has solution
√
VI = 1 + (α − 1)eλ+ ξ ,
(17.16)
where 2λ+ = −c + c2 + 4 > 0. Here, we have used the fact that the
solution must be bounded as ξ → −∞ in order to eliminate one of the
possible exponential solutions.
Region II On region II we have
VII
+ cVII
− VII = 0,
VII (0) = α,
(17.17)
which has solution
VII = αeλ− ξ ,
√
(17.18)
where 2λ− = −c − c2 + 4 > 0. Again we have used the fact that VII
must be bounded as ξ → ∞.
Now force the derivative to be continuous at ξ = 0, i.e., VI (0) = VII
(0). This
gives
λ+ (α − 1) = αλ−
(17.19)
from which it follows that
(1 − 2α)2
α − α2
(17.20)
vt = vxx + v(0.1 − v)(v − 1)
(17.21)
c2 =
as required.
Ex. 4: (a) Solve the bistable equation
numerically and plot a traveling wave.
425
(b) Solve the FitzHugh-Nagumo equations
vt
= vxx + v(0.1 − v)(v − 1) − w
wt
=
0.1(v − w)
(17.22)
(17.23)
numerically and plot a traveling wave.
Solution
(a) There are a number of ways of solving this nonlinear diffusion equation in
one spatial dimension, and they can be programmed relatively simply in
almost any language. However, probably the easiest way to solve PDEs
like this is to use the capabilities of XPPAUT. This will not be nearly
as fast as a compiled program such as Fortran, but is much easier to
program.
If we discretise in space, but not in time, we get a large system of (stiff)
ODEs, one for each spatial grid point, which we can then solve with
XPPAUT. In fact, XPPAUT is designed to make it easy to solve such
systems. Let xi , 0 ≤ i ≤ n, denote the ith spatial grid point, let vi denote
the approximate solution at xi , and use centred differences to discretise
vxx . Then
dvi
vi+1 − 2vi + vi−1
=
+ vi (0.1 − vi )(vi − 1)
dt
(∆x)2
(17.24)
for 1 ≤ i < n. At the boundaries we specify no flux by introducing a
phantom grid point x−1 , such that v−1 = v1 , and similarly at the other
end. Thus
dv0
dt
dvn
dt
=
=
2v1 − 2v0
+ v0 (0.1 − v0 )(v0 − 1)
(∆x)2
−2vn + 2vn−1
+ vn (0.1 − vn )(vn − 1)
(∆x)2
(17.25)
(17.26)
For initial conditions, we put the solution at the leftmost grid points at
0.9. This initiates a waves which travels from left to right across the
domain.
This large system of ODEs can be programmed in XPPAUT as follows:
# Declare the variables
# Initial conditions
v[0..5](0)=0.9
# Initialise the rest of the domain
v[6..100](0)=0
#
param delx=1.0 diff=1.0
#
!lam=diff/(delx*delx)
#
# Define right hand sides of the pdes
426
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
1.0
0.8
t=10
0.6
v
t=50
t=80
0.4
0.2
0.0
0
20
40
60
80
100
x
Figure 17.2: Traveling wave in the bistable equation
bistab(v)=v*(1-v)*(v-0.1)
#
v[1..99]’ = lam*(v[j-1]-2.0*v[j]+v[j+1]) + bistab(v[j])
# No-flux boundary conditions
v0’ = lam*(-2.0*v0+2.0*v1) + bistab(v0)
v100’ = lam*(2.0*v99-2.0*v100) + bistab(v100)
#
#
@ total=200, dt=0.1, xlo=0, xhi=200, ylo=0, yhi=1, meth=gear, yp2=v4,yp3=v50,yp4=v80,nplot=4
done
All that remains is to save the output into a file, read it into IGOR
(or the plotting program of your choice) and plot the waves. A typical
solution is shown in Fig. 17.2.
(b) The procedure for solving the FitzHugh-Nagumo model is similar. Again,
it is easiest to use XPPAUT. The code is
# Declare the variables
v[0..5](0)=1.0
v[6..100](0)=0
w[0..100](0)=0
#
param delx=4.0 diff=5.0
#
!lam=diff/(delx*delx)
#
# Define right hand sides of the pdes
bistab(v)=v*(1-v)*(v-0.1)
#
v0’ = lam*(-2.0*v0+2.0*v1) + bistab(v0) -w0
v[1..99]’ = lam*(v[j-1]-2.0*v[j]+v[j+1]) + bistab(v[j]) - w[j]
427
0.6
0.4
v
t=300
t=200
0.2
0.0
-0.2
0
100
200
x
300
400
Figure 17.3: Traveling wave in the FitzHugh-Nagumo equations
v100’ = lam*(2.0*v99-2.0*v100) + bistab(v100) -w100
#
# The diff eqn for w
w[0..100]’ = 0.01*(v[j]-w[j])
# No-flux boundary conditions
#
#
@ total=500, dt=0.1, xlo=0, xhi=500, ylo=-0.3, yhi=1,
done
meth=gear, yp2=v4,yp3=v50,yp4=v80,nplot=4
A typical solution is shown in Fig. 17.3
Ex. 5: Find the speed of traveling fronts for barnacle muscle fiber using the Morris–
Lecar model (Chapter 4, (15.27)–(15.28)).
Answer: About 6 cm/s.
Solution
Information must be combined from two other chapters. The details of the
model are in Chapter 4, while the details of the diffusion are in Chapter 6.
From (6.14) we have
τm
∂V
∂2V
+ Rm (Iion − Iapp ) = λ2m 2 ,
∂t
∂x
(17.27)
∂V
λ2 ∂ 2 V
.
+ Iion − Iapp = m
∂t
Rm ∂x2
(17.28)
and thus
Cm
428
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
20
x = 5 cm
x = 9 cm
V (mV)
0
-20
-40
0
500
1000
1500
time (ms)
2000
2500
3000
Figure 17.4: Traveling waves in the Morris-Lecar model of question 5, plotted as
functions of time at two fixed points in space.
The values of λm and Rm are given in Table 6.1. All the other model parameters are given in Table 15. There is only one tricky one, the value for Iapp ,
where one must be careful to use the correct units. Since Cm is in units of
2
µF/cm and V and T are in units of mV and ms respectively, we see that Iapp
2
2
must be given in units of µA/cm . Hence, Iapp = 60 µA/cm .
As in question 4 we use XPPAUT and the method of lines, i.e., we discretize
the equations in space and then use XPPAUT to solve the resulting stiff
system of ODEs. Then we saved the results to a file and used IGOR to plot
the waves and calculate the wave speed. The XPPAUT code is given below.
From the two sample waves in Fig. 17.4 we see that the wave travelled 4 cm in
about 0.66 seconds, giving a wave speed of close to 6 cm/sec. As can be seen
from the code, these waves were calculated with no-flux boundary conditions
on a domain of length 10 cm. Initially V was set at 0 at the left-most 6 grid
points.
XPPAUT code for Exercise 5.
# Declare the variables
V[0..5](0)=0.0
V[6..100](0)=-36.755
w[0..100](0)=0.070198
#
# Declare the parameters
#
param
Iapp=60.0
cm=20.0
param
gl=2
phi=0.04
param
vca=120
vk=-84
param
taum=4.6
Rm=0.23
#
param delx=0.1
gca=4.4
vee1=-1.2
vl=-60
lamm=0.28
gk=8
vee2=18
vee3=2
vee4=30
429
!diff=lamm*lamm/Rm
#
!lam=diff/(delx*delx)
#
minf(V)=0.5*(1+tanh((V-vee1)/vee2))
winf(V)=0.5*(1+tanh((V-vee3)/vee4))
lambda(V)=cosh((V-vee3)/(2*vee4))
Iion(V,w)=gca*minf(V)*(V-vca)+gk*w*(V-vk)+gl*(V-vl)
#
# Define right hand sides of the pdes
#
V[1..99]’ = (1/cm)*( lam*(V[j-1]-2.0*V[j]+V[j+1]) - Iion(V[j],w[j]) + Iapp )
# No-flux boundary conditions
V0’ = (1/cm)*( lam*(-2.0*V0+2.0*V1) - Iion(V0,w0) + Iapp )
V100’ = (1/cm)*( lam*(2.0*V99-2.0*V100)
- Iion(V100,w100) + Iapp )
w[0..100]’ = phi*lambda(V[j])*(winf(V[j])-w[j])
#
#
@ total=300, dt=0.1, xlo=0, xhi=300, ylo=-70, yhi=40, meth=gear, yp2=V4,yp3=V50,yp4=V80,nplot=4
done
Ex. 6: Show that 1/p (λ1 ) + 1/p (λ2 ) + 1/p (λ3 ) = 0, where p is defined by (7.34).
Hence, show that h(1) = 0, where h is defined by (7.40). Show also that
λ1 /p (λ1 ) + λ2 /p (λ2 ) + λ3 /p (λ3 ) = 0 and thus h (1) = 0. Finally, show that
h (1) = p (λ1 )/λ21 − 2.
Solution
Use the fact that λ1 , λ2 and λ3 are all roots of p(λ). Thus, p(λ) = (λ−λ1 )(λ−
λ2 )(λ−λ3 ) and so p (λ1 ) = (λ1 −λ2 )(λ1 −λ3 ), p (λ2 ) = (λ2 −λ1 )(λ2 −λ3 ) and
p (λ3 ) = (λ3 − λ2 )(λ3 − λ1 ). From these identities it follows immediately that
1/p (λ1 ) + 1/p (λ2 ) + 1/p (λ3 ) = 0 and λ1 /p (λ1 ) + λ2 /p (λ2 ) + λ3 /p (λ3 ) = 0.
p (λ1 )
1
1
1
1)
It now follows that h(1) = 1+ pp (λ
(λ2 ) + p (λ3 ) = p (λ1 )( p (λ1 ) + p (λ2 ) + p (λ3 ) ) = 0
as required. Furthermore,
h (s) = −1 −
λ2 p (λ1 )
λ3 p (λ1 )
exp(−λ
exp(−λ3 ln(s)/λ1 ),
ln(s)/λ
)
−
2
1
λ1 s p (λ2 )
λ1 s p (λ3 )
(17.29)
and thus
h (1)
as required.
λ2 p (λ1 ) λ3 p (λ1 )
−
λ1 p (λ2 ) λ1 p (λ3 )
λ1
λ2
λ3
p (λ1 )
+
+
= −
λ1
p (λ1 ) p (λ2 ) p (λ3 )
= 0
= −1 −
(17.30)
(17.31)
(17.32)
430
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
Finally, a straightforward calculation shows that
λ2
p (λ1 )
λ3
λ22
λ23
h (1) =
+
+
+
λ1
p (λ2 ) p (λ3 ) λ1 p (λ2 ) λ1 p (λ3 )
λ1
p (λ1 )
λ22
λ23
− =
+
+
λ1
p (λ1 ) λ1 p (λ2 ) λ1 p (λ3 )
p (λ1 )
λ22
λ23
= −1 +
+
λ21
λ1 p (λ2 ) λ1 p (λ3 )
p (λ1 )
λ21
= −1 +
1
−
λ21
p (λ1 )
p (λ1 )
= −2 +
λ21
(17.33)
(17.34)
(17.35)
(17.36)
(17.37)
as required, where we have used the fact that
λ21
λ22
λ23
+
+
= 1.
p (λ1 ) p (λ2 ) p (λ3 )
(17.38)
Ex. 7: The results of Exercise 6 can be generalized. Use contour integration in the
complex plane to show that for an nth order polynomial p(z) = z n + · · · with
simple roots zk , k = 1, . . . , n,
n
k=1
zkj
p (zk )
= 0,
(17.39)
provided that n > j + 1. In addition, show that
n
zkn−1
= 1.
p (zk )
(17.40)
k=1
Solution
Firstly, note that 1/p(z) has simple poles at z = zi . Secondly,
residue at zi = lim
z→zi
1
z − zi
= .
p(z)
p (zi )
(17.41)
Now take as our contour, C, a circle of radius R, centred at the origin. If R
is large enough, C will contain all the poles of 1/p(z) and we know from the
residue theorem that the integral of 1/p(z) around C is equal to 2πi times the
sum of the residues at the poles, zi .
However, in the limit as R → ∞ we know that the integral of 1/p(z) around
C tends to zero. (This is most easily seen by letting z = Reiθ and converting
the integral around C to an integral over θ. Then
2π
2π
1
iReiθ
f (R) dθ
(17.42)
dz =
dθ =
p(Reiθ )
0
0
C p(z)
431
for some function f (R) = O(R1−n ). As R → ∞, |f | → 0 and so the integral
vanishes.)
Thus, letting R → ∞, it follows immediately from the residue theorem that
0=
n
1
.
p (zk )
k=1
The exact same argument can be applied to every function
and thus
n
zkj
0=
,
0 ≤ j < n − 1.
p (zk )
(17.43)
zj
p(z) ,
for j < n − 1,
(17.44)
k=1
When j = n−1, the integral around C now becomes of order Rn /Rn (because
of the additional factor of R introduced by the change of integration variable
to dθ). In this case
lim
R→∞
C
z n−1
dz
p(z)
=
2π
lim
R→∞
0
Rn−1 ei(n−1)θ iReiθ
dθ
p(Reiθ )
(17.45)
2π
=
idθ
(17.46)
0
= 2πi.
(17.47)
It remains only to apply the residue theorem again. Equating the integral
over C to the sum of the residues gives
n
zkn−1
2πi = 2πi
,
p (zk )
(17.48)
k=1
as required.
Ex. 8: Show that the duration of the absolute refractory period of the traveling pulse
for the generalized FitzHugh–Nagumo model is (approximately)
w0
dw
Tar =
,
(17.49)
w− G− (w)
where w0 is that value of w for which c(w) = 0.
Solution
From Fig. 7.8, and from the definition of w+ and w− , we see that the refractory
phase starts at w = w− , as that is where the back of the wave jumps down. For
all values of w such that c(w) < 0 it is not possible to initiate another wave,
as no wave front from V− (w) to V+ (w) exists (the only possible transitions
432
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
from V− (w) to V+ (w) are moving backwards, and thus cannot be wave fronts).
This is exactly what is meant by the absolute refractory period; the period
when no stimulus, no matter how large, can initiate another wave front. So
the absolute refractory period lasts as long as w is between w− and w0 .
Since
dw
dt
= G− (w) during the refractory period it follows that
w0
Tar =
w−
dw
,
G− (w)
(17.50)
as required.
Of course, once w < w0 a stimulus of sufficient strength can initiate another
wave, and so the system, although still refractory, is not absolutely refractory.
Ex. 9: Show that in the limit as the period approaches infinity, (7.61) and (7.62)
reduce to the equations for a solitary pulse (7.38) and (7.39).
Solution
Since the period of the wave is ξ2 , letting the period tend to infinity corresponds to taking the limit P → ∞. Note that we don’t let ξ1 → ∞ also,
because we want the solitary pulse to have a finite width. Since λ1 < 0 and
the real parts of λ2 and λ3 are positive, it follows that the terms involving eλ1 P
will disappear, while the terms involving eλ2 P and eλ3 P will become large.
Taking the limit as P → ∞ of (7.61) thus gives
−1
e−ξ1 λ2
e−ξ1 λ3
+ + + 2 α = 0,
−p (λ1 )
p (λ2 )
p (λ3 )
(17.51)
which is the same as (7.39).
Similarly, taking the limit as P → ∞ of (7.62) gives
−eξ1 λ1
1
1
+
+
+ 2 α = 0,
−p (λ1 ) p (λ2 ) p (λ3 )
from which, using the identity
1
p (λ1 )
+
1
p (λ2 )
+
1
p (λ3 )
eξ1 λ1
1
− + 2 α = 0,
p (λ1 ) p (λ1 )
and hence (7.38).
(17.52)
= 0 it follows that
(17.53)
433
Ex. 10: (Fisher’s equation.) Suppose f (0) = f (1) = 0, f (0) > 0, f (1) < 0, and
f (u) > 0 for 0 < u < 1. For example, just take f (u) = u(1 − u). We are going
to look for traveling waves in the equation
ut = uxx + u(1 − u).
(17.54)
This equation, sometimes known as Fisher’s equation, is commonly used in
mathematical ecology and population biology to model traveling waves. In
this context u typically corresponds to a population density or a gene density.
(a) Convert to the traveling wave coordinate ξ = x + ct, where c ≥ 0 is the
wave speed, and show that the PDE converts to the system of ODEs
u
w
= w
(17.55)
= cw − u(1 − u).
(17.56)
(b) What are the steady states? Determine the stability of the steady states
and sketch the phase plane, including the directions of the eigenvectors
at the saddle point.
(c) Show that a biologically realistic traveling wave cannot exist when c < 2.
(d) Starting at the saddle point (1, 0), trace the stable manifold backwards
in time. Show that as we move backwards in time, this stable manifold
cannot leave the first quadrant across the u axis.
(e) Show that if c ≥ 2 there exists a line, u = mw, such that the stable
manifold of the saddle point cannot be above this line.
(f) What can we conclude about the existence of traveling wave solutions to
Fisher’s equation?
Solution
(a) In the traveling wave coordinate ξ = x + ct we get
cu = u + u(1 − u),
(17.57)
where a prime denotes differentiation with respect to ξ. Introducing the
subsidiary variable w = u then gives the required system.
(b) The steady states are clearly (0, 0) and (1, 0). The Jacobian matrix
around a steady state (u0 , w0 ) is
0
1
J=
.
(17.58)
−f (u) c
At (0, 0) the eigenvalues are
2λ± = c ±
c2 − 4.
(17.59)
434
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
w
(0,0)
(1,0)
u
Figure 17.5: Sketch of the phase plane of Fisher’s equation (Question 10).
Thus, when 0 < c < 2 the origin is an unstable spiral. When c > 2 the
origin becomes an unstable node.
At (1, 0) the eigenvalues are
2λ± = c ±
c2 + 4,
(17.60)
and thus (1, 0) is a saddle point.
The directions of the eigenvectors at the saddle point are determined by
solving the system Jη = λ± η. For example,
suppose that the eigenvalue
η1
λ− has corresponding eigenvector η =
. Then, since Jη = λ− η it
η2
follows from the first component of the matrix equation that η2 = λ− η1 .
Since λ− < 0 we thus know that η must point into quadrants II and IV.
This gives us enough information to sketch the phase plane, as shown in
Fig. 17.5.
(c) First note that a traveling wave corresponds to a heteroclinic orbit, starting at (0, 0) and ending at (1, 0). If the origin is an unstable spiral then
any trajectory leaving the origin will pass through all four quadrants as
it spirals out, and thus u will become negative along the trajectory. In
general, u corresponds to a population density or a gene density and thus
cannot be negative. Thus no acceptable traveling wave solution exists
when c < 2.
(d) From Fig. 17.5 we see that the only possible heteroclinic orbit from (0, 0)
to (1, 0) is the stable manifold of (1, 0). If we can trace the stable manifold
backwards in time and show that it begins at the origin we will thus have
constructed a traveling wave solution.
435
w=mu
w
heteroclinic
orbit
(0,0)
(1,0)
u
Figure 17.6: Sketch of the phase plane of Fisher’s equation (Question 10), showing
the line L and a sketch of the heteroclinic orbit.
Now, on the u axis, w = −f (u) < 0 when 0 < u < 1, and thus for
0 < u < 1 all the trajectories on the u axis point down (illustrated in
Fig. 17.5). It follows immediately that if we trace the stable manifold
backwards in time it cannot cross the u axis between 0 < u < 1.
(e) Consider the line L, given by w = mu. If we can show that no trajectories
cross L from above to below then we will have the desired result. This
holds if w /u ≥ m which is true only if
m2 − cm + (1 − u) ≤ 0.
(17.61)
Thus, if we can choose an m such that
m2 − cm + 1 ≤ 0.
(17.62)
we will have the desired result.
Consider the quadratic
m2 − cm + 1 = 0.
(17.63)
√
This has roots c/2 ± c2 − 4/4. If c ≥ 2 there are two positive roots of
(17.63) and thus a range of m for which (17.62) holds. Hence the result.
The phase plane and the line L are sketched in Fig. 17.6.
(f) For any c ≥ 2, follow the stable manifold of (1, 0) backwards in time. It
can’t leave the first quadrant across the u axis, and neither can it cross
436
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
the line L. It is thus forced to head to the origin as ξ → −∞, giving a
heteroclinic orbit and thus a traveling wave solution.
It thus follows that a traveling wave solution exists for all c ≥ 2.
Ex. 11: Here we look at Fisher’s equation (question 10) again, but from a slightly
different perspective. Suppose f (0) = f (1) = 0, f (0) > 0, f (1) < 0, and
f (u) > 0 for 0 < u < 1.
(a) Show that when c = 0 there are no trajectories with u ≥ 0 connecting
the two rest points u = 0 and u = 1.
(b) Show that if there is a value of c for which there is no heteroclinic connection with u ≥ 0, then there is no such connecting trajectory for any
smaller value of c.
(c) Show that if there is a value of c for which there is a connecting trajectory
with u ≥ 0, then there is a similar connecting trajectory for every larger
value of c.
(d) Let µ be the smallest positive number for which f (u) ≤ µu for 0 ≤ u ≤ 1.
√
Show that a heteroclinic connection exists for all c ≥ 2 µ.
Solution
Recall from question 10 that we need to consider the ODE system
u
w
= w
(17.64)
= cw − f (u).
(17.65)
As in question 10 we have a saddle point at (1, 0), an unstable steady
state at
the origin, and the stable manifold at (1, 0) has slope λ− = 12 (c− c2 + 4f (0)).
The phase plane is just the same as is sketched in Fig. 17.5.
(a) When c = 0 we have w = −f (u), which is always negative when 0 <
u < 1. Consider any solution leaving the origin. It cannot head into the
first quadrant as then it would have u > 0 and w > 0 which contradicts
w = −f (u). Similarly, it cannot head into the fourth quadrant as then
it would have u > 0 with w < 0 which contradicts u = w. Hence the
result.
(b) Suppose that a heteroclinic connection does not exist for c = c0 . We
know that, since the stable manifold of (1, 0) cannot enter quadrant I
across the u axis, it must enter quadrant I across the w axis (Fig. 17.7).
Consider any c1 < c0 . Since λ− is an increasing function of c, it follows
that l− (c1 ) < l− (c0 ), i.e., the stable manifold at (1, 0) has rotated in a
clockwise direction (Fig. 17.7).
Suppose a heteroclinic connection exists for c1 . Then we would have
the situation shown in Fig. 17.7 in which the heteroclinic connection for
437
w
c=c0
c=c1< c0
heteroclinic
orbit
(0,0)
(1,0)
u
Figure 17.7: Sketch of the situation when a heteroclinic connection exists for c1 < c0 ,
with no such connection existing for c0 . As discussed in question 11 this leads to a
contradiction at the intersection point of the two solutions.
c1 intersects the stable manifold for c0 , and has a greater slope at the
point of intersection. However, from the differential equations we see
that w /u is an increasing function of c, i.e., for a fixed u and w, as c
decreases, the slope of the trajectory there must also decrease. Hence, the
situation illustrated in Fig. 17.7 is not possible, and no such heteroclinic
connection can exist for c1 .
(c) Suppose that a heteroclinic connection exists for c = c0 and let c = c1 >
c0 . Now we have the situation sketched in Fig. 17.8. The stable manifold
for c = c1 initially lies below the stable manifold for c = c0 (since λ−
is an increasing function of c). The two solutions cannot cross (or we
would get the same contradiction as we got in the previous part of this
question), and the stable manifold cannot enter the first quadrant across
the u axis. The only possibility is that it connects to the origin, thus
forming a heteroclinic connection, as required.
(d) The argument is just the same as in question 10 so we won’t show all the
details. Just as before we construct a line w = mu and choose m so that
solutions cannot cross this line from above to below. In terms of f (u)
the required condition is
m2 − cm +
f (u)
≤ 0,
u
(17.66)
for all 0 < u < 1. By the definition of µ we see that f (u)/u ≤ µ for all
0 < u < 1 and thus it suffices to have
m2 − cm + µ ≤ 0.
(17.67)
438
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
w
c=c0
c = c1 > c0
(0,0)
(1,0)
u
Figure 17.8: Sketch of the situation when a heteroclinic connection exists for c0 .
√
This has a range of possible m when c2 ≥ 4µ, i.e., c ≥ 2 µ, as required.
Ex. 12: Construct a traveling wave solution for the equation vt = vxx + f (v) with
v(−∞, t) = 0 and v(+∞, t) = 1 where f (v) = 0 for 0 < v < q, f (v) = 1 − v
for q < v < 1. What is the speed of propagation? For what values of q does
the wave exist? Plot the traveling wave on the same graph as the traveling
wave for the piecewise linear bistable equation, making sure that q and α are
chosen so that the two waves are traveling with the same speed.
Solution
We use the same techniques as in question 3. In the traveling wave coordinate
ξ = x + ct we have v − cv + f (v) = 0. Just as for the piecewise linear bistable
equation (question 3) we divide the ξ axis into two regions: Region I (ξ < 0)
and Region II (ξ > 0). We also set v(0) = q, which specifies the position of
the wave along the ξ axis.
In Region I denote the solution by vI . We have
vI − cvI = 0
and thus
(17.68)
A cξ
(17.69)
e ,
c
for some constant A. Note that we have eliminated the arbitrary constant by
applying the boundary condition at −∞.
vI =
439
In Region II denote the solution by vII . We have
vII
− cvII
− vII = −1,
(17.70)
vII = 1 + Beλ− ξ ,
(17.71)
and thus
for some constant B, and where
2λ− = c −
c2 + 4.
(17.72)
Note that we only use the negative root in order to satisfy the boundary
condition at ∞.
Now apply the constraints vI (0) = vII (0) = q and vI (0) = vII
(0), which give
A/c = 1 + B = q and A = λ− B. Thus, B = q − 1, A = cq and
c=
λ− (q − 1)
.
q
(17.73)
Solving this last equation for c then gives
1−q
c= √ .
q
(17.74)
Clearly, the wave exists for all q such that 0 < q < 1.
We now need to choose q and a so that
1−q
1 − 2α
.
√ =√
q
α − α2
(17.75)
In Fig. 17.9 we plot the wave speed as a function of q and of α for the two
different equations. Clearly, if we choose α = 0.4 and q = 0.7 the waves travel
with similar speeds. Plots of the two waves are given in Fig. 17.10. As one
would expect, the traveling wave of the bistable equation rises more slowly,
due to the fact that f (u) < 0 when u is close to zero, and thus pulls the wave
front down.
Ex. 13: Show that the equation vt = vxx +f (v), where f (0) = f (1) = 0, has a traveling
1
wave solution connecting v = 0 to v = 1 only if 0 f (v) dv > 0. Interpret the
results of questions 3 and 12 in light of this result. (Hint: convert to traveling
wave coordinates, multiply by v and integrate.)
Solution
Let ξ = x + ct. Then the PDE becomes cv = v + f (v). Multiply by v to get
c(v )2 = v v + f (v)v .
(17.76)
440
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
14
12
c
10
8
6
4
2
0
0.0
0.2
0.4
0.6
0.8
1.0
q, α
Figure 17.9: Wave speed as a function of q (dotted line) and of α (solid line).
The dotted line refers to question 12 while the solid line is for the piecewise linear
bistable equation (question 3).
1.0
0.8
v
0.6
0.4
0.2
0.0
-10
-5
0
ξ
5
10
Figure 17.10: The traveling wave of question 12 (dotted line) and of the piecewise
linear bistable equation (solid line).
441
Let
F (v) =
v
f (s)ds.
(17.77)
0
d
dξ F (v)
Then
= f (v)v . Hence, from (17.76) we see that
c(v )2 =
1 d 2
d
(v ) + F (v).
2 dξ
dξ
(17.78)
Now integrate from ξ = −∞ (i.e., v = 0) to ξ = ∞ (i.e., v = 1) to get
ξ=∞
1
(v )2 ξ=∞
c
(v ) dξ =
+ F (v)|ξ=−∞ = 0 +
f (v) dv.
2 ξ=−∞
−∞
0
Hence
∞
2
1
f (v) dv
0
c = ∞
.
(v )2 dξ
−∞
(17.79)
(17.80)
It follows that c > 0 (and thus a traveling wave solution exists) only when
1
f (v) dv > 0, as required.
0
1
For the bistable equation, 0 f (v) dv > 0 only when α < 1/2, while for the
choice of f in question 12 the integral is always positive when 0 < q < 1.
This explains why the wave exists for all q, 0 < q < 1, but only exists for
0 < α < 1/2.
442
CHAPTER 17. NONLINEAR WAVE PROPAGATION: SOLUTIONS
Chapter 18
The Circulatory System:
Solutions
Ex. 1: Equation (8.7) was derived assuming that the radius of the vessel and pressure
drop along the vessel were constant, but then it was used in (8.14) as if the
radius was variable. Under what conditions is this a valid approximation?
Solution
In the derivation of (8.7) we had to assume that the vessel radius remains
constant, so that there was never any axial flow. This let us conclude that Px
was independent of r and x, which allowed an easy integration to get (8.7).
However, by writing down A(P ) in (8.14) we are implicitly assuming that the
vessel radius will change with P , i.e., as P decreases down the length of the
vessel (dP/dx < 0), the vessel radius gradually deceases also.
Clearly, there is a fundamental inconsistency in the way we are using these two
equations. However, if we assume that the vessel has a very low compliance,
and thus the cross-sectional area A does not vary greatly with P , then it will
be a good approximation to derive the equation for a flux in a compliant vessel
(8.14) from that for the flux in a non-compliant vessel (8.7).
Of course, the situation is quite different if the vessel is highly compliant. We
are all probably quite familiar with the situation of applying pressure to a
highly compliant tube (for instance, blowing up a balloon). As the pressure
increases, the flow doesn’t increase all that much, the radius of the vessel just
expands.
Ex. 2: Show that
lim
x→y
x4 − y 4
4
= y,
3
3
x −y
3
443
(18.1)
444
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
so that (8.21) follows.
Solution
lim
x→y
x4 − y 4
x3 − y 3
1 − (y/x)4
x→y
1 − (y/x)3
1 − α4
= y lim
α→1 1 − α3
−4α3
= y lim
by L Hopital s Rule
α→1 −3α2
4y
=
.
3
lim x ·
=
Ex. 3: Simplify the six-compartment model of the circulation by assuming that there
are no pressure drops over the arterial and venous systems (either systemic or
pulmonary), and thus, for example, Psa = Ps1 . Assume also that the systolic
compliances are negligible (why?). Solve the resultant equations and compare
with the behavior of the three-compartment model presented in the text. How
do the parameter values change? Are the sensitivities altered? (Calculate the
sensitivities using a symbolic manipulation program.)
Solution
Simplifying the model in this way is equivalent to making the four assumptions
Rpa
Rpv
Rsv
Rsa
=
=
0,
0,
(i.e., Pp1 = Ppa ),
(i.e., Pp2 = Ppv ),
(18.2)
(18.3)
=
=
0,
0,
(i.e., Ps2 = Psv ),
(i.e., Ps1 = Psa ).
(18.4)
(18.5)
So the model equations (8.63), (8.66), (8.69) and (8.72) are no longer used in
the model, which leaves us with the equations
Systemic arteries:
Q = F (Cld Ppv − Clσ Psa ),
Vsa
=
V0s
+ Csa Psa .
(18.6)
(18.7)
Systemic veins:
Psa − Psv
,
Rs
= Csv Psv .
Q =
Vsv
(18.8)
(18.9)
445
Pulmonary arteries:
Q = F (Crd Psv − Crσ Ppa ),
Vpa
=
V0p
+ Cpa Ppa .
(18.10)
(18.11)
Pulmonary veins:
Ppa − Ppv
,
Rp
= Cpv Ppv .
Q =
Vpv
(18.12)
(18.13)
Volume conservation
Vsa + Vsv + Vpa + Vpv = Vt .
(18.14)
We thus have nine equations to solve for the nine unknowns (4 pressures, 4
volumes, and Q). However, the model is much easier to solve and understand
if we assume that the systolic compliances are zero. This is a reasonable
approximation, for during systole the ventricular muscle is highly contracted,
with only a very low compliance. We then get the following equations:
Q = F Cld Ppv ,
Ppa − Ppv
Q =
,
Rp
Psa − Psv
Q =
,
Rs
Q = F Crd Psv ,
(18.18)
Vsa
= V0s + Csa Psa ,
(18.19)
Vsv
= Csv Psv ,
(18.20)
Vpa
V0p
=
+ Cpa Ppa ,
= Cpv Ppv ,
(18.21)
(18.22)
= Vsa + Vsv + Vpa + Vpv .
(18.23)
(18.15)
(18.16)
(18.17)
and
Vpv
Vt
Note that the 9 equations fall into two natural groups. The first group involves
only Q and the pressures, while the second group is used to solve for the
volumes.
It’s now relatively simple to solve for the unknowns (using Maple, or even
solving by hand) to get
1
Psa = Q Rs +
,
(18.24)
F Crd
Q
Psv =
,
(18.25)
F Crd
1
Ppa = Q Rp +
,
(18.26)
F Cld
Q
Ppv =
,
(18.27)
F Cld
446
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
and
Vt − V0s − V0p
Q=
Cra Rs + Cpa Rp +
Csa +Crv
F Crd
+
Cpv +Cpa
F Cld
.
(18.28)
Clearly, the solution is very similar to that of the three compartment model
presented in the text, with only slight differences caused by the different model
structures.
The following code solves the nine differential equations for the unknowns. In order to find,
for instance, Psa in the form given above, it is easiest to just solve the first four equations
for the pressures as functions of Q.
readlib(unassign):
unassign(’QQ’,’Psa’,’Psv’,’Ppa’,’Ppv’,’Vsa’,’Vsv’,’Vpa’,’Vpv’,’sols’):
fun1
fun2
fun3
fun4
:=
:=
:=
:=
QQ
QQ
QQ
QQ
-
F*Cld*Ppv:
(Psa-Psv)/Rs:
F*Crd*Psv:
(Ppa-Ppv)/Rp:
fun5
fun6
fun7
fun8
fun9
:=
:=
:=
:=
:=
Vsa - V0s - Csa*Psa:
Vsv - Csv*Psv:
Vpa - V0p - Cpa*Ppa:
Vpv - Cpv*Ppv:
Vt - Vsa - Vsv - Vpa - Vpv:
eqns := {fun1=0,fun2=0,fun3=0,fun4=0,fun5=0,fun6=0,fun7=0,fun8=0,fun9=0}:
vars := {Psa,Psv,Ppa,Ppv,Vsa,Vsv,Vpa,Vpv,QQ}:
##
## To find the pressures as functions of Q, just uncomment out the following two lines
#eqns := {fun1=0,fun2=0,fun3=0,fun4=0}:
#vars := {Psa,Psv,Ppa,Ppv}:
##
sols := solve(eqns,vars):
assign(sols):
simplify(QQ);
We now have to determine values for the parameters. Determining the parameters from the values for the solution may seem a rather odd way of
proceeding, but it has its value in this situation. The form of the solution
helps us understand how the various pressures depend on the blood flow and
the compliances, but it is only the pressures and the volumes that can be
measured.
First note that finding the compliances, given the pressures and volumes, is
very much easier than the other way around (as done in the first part of the
question). For we can just use (18.15)-(18.23) directly, to get
Cld = Q/(F Ppv ),
Psa − Psv
Rs =
,
Q
Vsa − V0s
Csa =
,
Psa
Crd = Q/(F Psv ),
Ppa − Ppv
Rp =
,
Q
Vsv
Csv =
Psv
(18.29)
(18.30)
(18.31)
447
Psa
Psv
Ppa
Ppv
Vsa
Vsv
Vpa
Vpv
Q
Normal
100
2
15
5
1.0
3.5
0.4
0.1
5.6
Csa
0.0018
−5
−5
−5
−5
17
−5
−0.3
−5
−5
Csv
1.75
−92
−92
−92
−92
−17
8
−5
−92
−92
Cpa
0.0013
−0.5
−0.5
−0.5
−0.5
−0.1
−0.5
5
−0.5
−0.5
Cpv
0.02
−3
−3
−3
−3
−0.5
−3
−0.1
97
−3
Rs
17.5
93
−5
−5
−5
17
−5
−0.3
−5
−5
Rp
1.78
−0.3
−0.3
66
−0.3
−0.06
−0.3
3
−0.3
−0.3
Cld
0.014
3
3
−31
−97
0.8
3
−2
−97
2
Crd
0.035
91
−8
93
92
16
−8
4
93
93
F
80
93
−5
62
−5
17
−5
3
−5
95
Table 18.1: Sensitivities for the circulatory system model of Question 3 (expressed
as percentages).
, Cpa =
Vpa − V0s
,
Ppa
Cpv =
Vpv
.
Ppv
(18.32)
We now use the values for the pressures and volumes given in the text, i.e.,
Vsa = 1, Vsv = 3.5, Vpa + Vpv = 0.5 (liters), and Psa = 100, Psv = 2, Ppa = 15,
Ppv = 5 (mm Hg). From these volumes we know that Vt = 5 liters. We also
use that V0s + V0p = 1.2 liters. Finaly, total cardiac output is about 5.6
liters/min (Q = 5.6), with a heart rate of 80 beats per minute (F = 80) and
a stroke volume of 0.07 liters.
These values give give Cld = 0.014, Crd = 0.035, Rs = 17.5, Rp = 1.78, and
Csv = 1.75.
It remains only to calculate the values for Csa , Cpa and Cpv . There are slightly
more difficult as they need values for V0s , V0p and Vpv , which we don’t have
explicitly. All we have are the relationships Vpa +Vpv = 0.5 and V0s +V0p = 1.2.
However, all we need do is follow the calculation in the text, and calculate Csa
from the stroke volume. As shown in the text, we get Csa = 0.0018 liters/mm
Hg, and we can thus find V0s from the equation V0s = Vsa − Psa Csa = 0.82
liters. Thus, V0p = 0.38 liters.
We now (arbitrarily) choose Vpa = 0.4 > V0p which then gives Cpa =
0.0013 liters/mm Hg, and Cpv =
Vpv
Ppv
Vpa −V0s
Ppa
=
= 0.1/5 = 0.02 liters/mm Hg.
Sensitivities
Now that we have values for the parameters, we can calculate the sensitivities.
∂y
This requires calculation of xy ∂x
, for y = Psa , Psv , Ppa , Ppv , Vsa , Vsv , Vpa , Vpv , Q
and x = Csa , Csv , Cpa , Cpv , Rp , Rs , Cld , Crd , F . Clearly, this is best done using
a symbolic calculation package such as Maple. The results from the calculation are shown in Table 18.1, and the Maple code used to generate the table
is shown below.
448
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
This is the Maple code that we used to find the table of sensitivities. As always, elegance
is not guaranteed.
readlib(unassign):
unassign(’QQ’,’Psa’,’Psv’,’Ppa’,’Ppv’,’Vsa’,’Vsv’,’Vpa’,’Vpv’,’sols’,’F’,
’Cld’,’Crd’,’Csa’,’Csv’,’Cpa’,’Cpv’,’Rs’,’Rp’,’Vt’,’V0p’,’V0s’):
unassign(’uu’,’pp’):
#
#
#
#
We do the same solve as the previous Maple code, just storing all the pressures,
etc, in arrays. This is so the sensitivities can all be calculated in a loop later.
uu (for unknowns) stores the pressures, volumes and Q.
pp (for parameters) stores the compliances, resistances and F.
uu:=array(1..9):
pp:=array(1..9):
sens:=array(1..9,1..9):
# We store all the sensitivies in this array.
uu[1]:=Psa:
uu[5]:=Vsa:
pp[1]:=Csa:
pp[5]:=Rs:
uu[2]:=Psv:
uu[6]:=Vsv:
pp[2]:=Csv:
pp[6]:=Rp:
uu[3]:=Ppa:
uu[7]:=Vpa:
pp[3]:=Cpa:
pp[7]:=Cld:
uu[4]:=Ppv:
uu[8]:=Vpv:
pp[4]:=Cpv:
pp[8]:=Crd:
fun1
fun2
fun3
fun4
:=
:=
:=
:=
uu[9]
uu[9]
uu[9]
uu[9]
-
fun5
fun6
fun7
fun8
fun9
:=
:=
:=
:=
:=
uu[5] - V0s - pp[1]*uu[1]:
uu[6] - pp[2]*uu[2]:
uu[7] - V0p - pp[3]*uu[3]:
uu[8] - pp[4]*uu[4]:
Vt - uu[5] - uu[6] - uu[7] - uu[8]:
uu[9]:=QQ:
pp[9]:=F:
pp[9]*pp[7]*uu[4]:
(uu[1]-uu[2])/pp[5]:
pp[9]*pp[8]*uu[2]:
(uu[3]-uu[4])/pp[6]:
eqns := {fun1=0,fun2=0,fun3=0,fun4=0,fun5=0,fun6=0,fun7=0,fun8=0,fun9=0}:
vars := {Psa,Psv,Ppa,Ppv,Vsa,Vsv,Vpa,Vpv,QQ}:
sols := solve(eqns,vars):
assign(sols):
values:={Psa=100,Psv=2,Ppa=15,Ppv=5,Vsa=1,Vsv=3.5,Vpa=0.4,Vpv=0.1,QQ=5.6,
Csa=0.0018,Csv=1.75,Cpa=0.0013,Cpv=0.02,Rs=17.5,Rp=1.78,Cld=0.014,
Crd=0.035,F=80,Vt=5,V0s=0.82,V0p=0.38}:
Digits:=3:
# We only want three significant figures.
# Now calculate all the sensitivities in a double loop, and substitute in the values
# above (in the values array).
#
for i from 1 to 9 do
for j from 1 to 9 do
sens[i,j]:= 100*subs(values,(pp[j]/uu[i])*diff(uu[i],pp[j]))
od:
od:
print(sens);
449
Some features of the model remain the same as the three-compartment model
discussed in the text. For instance, the steady state is very sensitve to Csv , but
not to Csa . Also, a decrease in Cld again gives a large increase in pulmonary
blood volume, and pulmonary edema. It is a fun exercise to look through
the table of sensitivities for expected, or not so expected, relationships. For
instance, an increase in Rs gives an increase in Psa and a decrease in Psv ,
as expected, and similarly for Rp . However, Q depends most sensitively on
Csv , Crd and F . The dependence on F is easily understandable, but the high
sensitivity to only these two compliances is less expected. A small stiffening
of the systemic veins (a small decrease in Csv ) gives a large increase in Q, as
a result of the decreased storage capacity of the veins, while a slight stiffening
of the right heart ventricle gives a greatly decreased blood flow, due to its
decreased pumping effectiveness. Overall, the system is more sensitive to the
right heart compliance than the left heart compliance.
Ex. 4: Show that (8.76)–(8.84) can be derived from (8.63)–(8.75) by letting Rs and
Rp approach zero and by letting Cpa = Cpv and Rpa = Rpv .
Solution
First note that, if Rs and Rp are zero it follows from (8.67) and (8.73) that
Ps1 = Ps2 and Pp1 = Pp2 . If we now just write Ps in place of Ps1 we see
immediately that the first six equations of the full model are just the same
as the first five equations of the three-compartment model, i.e., (8.63)–(8.68)
are the same as (8.76)–(8.80).
It remains only to derive the final four equations of the three-compartment
model. First, from (8.69) and (8.72) it follows that
Q=
Ppa − Pp1
Pp1 − Ppv
=
,
Rpa
Rpv
(18.33)
from which, if we assume that Rpa = Rpv , we get
Ppa − Pp1 = Pp1 − Ppv ,
and thus
(18.34)
Ppa + Ppv
.
2
(18.35)
2Q =
Ppa − Ppv
,
Rpa
(18.36)
Q=
Ppa − Ppv
,
Rp
(18.37)
Pp1 =
Now add (8.69) and (8.72) to get
and define Rp = 2Rpa to get
450
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
which is (8.81) as required.
Next, add (8.71) and (8.74) to get
Vpa + Vpv
Cpa
(Ppa + Ppv + 2Pp1 )
2
= Cpa (Ppa + Ppv ).
=
(18.38)
(18.39)
Defining Vp = Vpa + Vpv and Cp = 2Cpa then gives
Vp =
Cp
(Ppa + Ppv )
2
(18.40)
which is (8.83) as required.
The final equation, (8.84), follows immediately from (8.75) and the definition
Vp = Vpa + Vpv .
Ex. 5: In the three-compartment circulatory model the base volume of the pulmonary
circulation V0p was calculated using the constraint V0s + V0p = 1.2 liters. Using
the fact that the systolic pressure in the pulmonary artery is about 22 mm
Hg and the diastolic pressure is about 7 mm Hg, calculate a new value for
the volume of the pulmonary circulation. How does this change to the model
affect the results?
Solution
To solve this problem we use the same approach as was used in the text to
determine Csa . First, note that, since we now have
Vp = V0p +
Cp
(Ppa + Ppv ),
2
(18.41)
and V0p is unknown, we can no longer calculate Cp directly once Vp is known.
Thus we must calculate Cp first. We do this using the equation
∆V = Cp ∆P.
(18.42)
The stroke volume is the same as given in the text (∆V = 0.07 liters) and
∆P = 22 − 7 = 15 mm Hg, as given in the question. Thus, Cp = 0.0047.
Once Cp is known, we substitute the values Vp = 0.5 liters, Ppa = 15 mm Hg
and Ppv = 5 mm Hg into (18.41) and solve for V0p to get V0p = 0.45 liters. This
is close to the value of V0p = 0.38 liters calculated for a slightly different model
in Question 3. In this version of the model, the base volume of the pulmonary
circulation is almost the entire pulmonary volume (Vp = 0.5). This means
that, to get the same blood flow, the pulmonary circulation can have a much
smaller compliance as it doesn’t have to expand greatly to accomodate the
pulse from each heartbeat. This relationship between V0p and Cp is clear from
(18.41); given a constant Vp , Cp must decrease as V0p increases, and conversely.
451
With these new values of V0p and Cp we can recalculate the table of sensitivities
(using a Maple program very similar to that given above, just with some minor
modifications mandated by the slightly different model). The values come out
similar to those of Table 8.2.
Ex. 6: Find the pressures as a function of cardiac output assuming Vl = 0, Vr = 0,
where Vl is the basal volume of the left heart, and similarly for Vr (cf. equation
(8.44)). Show that
Q∞ =
2Cp Vl /Cld + 2(Csv + Csa )Vr /Crd + 2(Vt − V0p − V0s )
.
Cp Rp + Csv Rsv + Csa (Rsa + 2Rsv )
(18.43)
Solution
We need to solve the same equations as those for the three-compartment
model in the text (8.76)–(8.84) with two exceptions. Note first the equations
relating Q to stroke volume, (8.41) and (8.42). If the resting volumes of the
right and left heart are not zero, they must be taken into account when Vstroke
is calculated. If we again make the simplifying assumption that the systolic
compliances are zero, this gives
Q = F (Vl + Cld Ppv ),
for the left heart,
(18.44)
Q = F (Vr + Crd Psv ),
for the right heart.
(18.45)
Using these two equations in place of (8.77) and (8.82), and using Maple (code
given below) to solve for the pressures, then gives
Vr
1
Psa = Q
+ Rsa + Rsv −
,
(18.46)
F Crd
Crd
Vr
1
Ps = Q
+ Rsv −
,
(18.47)
F Crd
Crd
Q
Vr
Psv =
−
,
(18.48)
F Crd
Crd
Vl
1
+ Rp −
,
(18.49)
Ppa = Q
F Cld
Cld
Q
Vl
Ppv =
−
.
(18.50)
F Cld
Cld
Finally, using Maple again to solve for Q, and then taking the limit as F → ∞
gives the desired result.
Maple code for Exercise 6.
452
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
readlib(unassign):
unassign(’QQ’,’Psa’,’Psv’,’Ppa’,’Ppv’,’Ps’,’Vsa’,’Vsv’,’Vp’,’sols’,’F’,
’Cld’,’Crd’,’Csa’,’Csv’,’Cpa’,’Cpv’,’Rs’,’Rp’,’Vt’,’V0p’,’V0s’,’Vl’,’Vr’):
fun1
fun2
fun3
fun4
fun5
:=
:=
:=
:=
:=
QQ
QQ
QQ
QQ
QQ
-
F*(Vl+Cld*Ppv):
(Psa-Ps)/Rsa:
F*(Vr+Crd*Psv):
(Ppa-Ppv)/Rp:
(Ps-Psv)/Rsv:
fun6
fun7
fun8
fun9
:=
:=
:=
:=
Vsa - V0s - Csa*(Psa+Ps)/2:
Vsv - Csv*(Psv+Ps)/2:
Vp - V0p - Cp*(Ppa+Ppv)/2:
Vt - Vsa - Vsv - Vp:
eqns := {fun1=0,fun2=0,fun3=0,fun4=0,fun5=0}:
vars := {Psa,Psv,Ppa,Ppv,Ps}:
sols := solve(eqns,vars):
assign(sols):
simplify(Psa); simplify(Ps); simplify(Psv); simplify(Ppa); simplify(Ppv);
unassign(’QQ’,’Psa’,’Psv’,’Ppa’,’Ppv’,’Ps’,’Vsa’,’Vsv’,’Vp’,’sols’,’F’,
’Cld’,’Crd’,’Csa’,’Csv’,’Cpa’,’Cpv’,’Rs’,’Rp’,’Vt’,’V0p’,’V0s’,’Vl’,’Vr’):
eqns := {fun1=0,fun2=0,fun3=0,fun4=0,fun5=0,fun6=0,fun7=0,fun8=0,fun9=0}:
vars := {Psa,Psv,Ppa,Ppv,Ps,Vsa,Vsv,Vp,QQ}:
sols := solve(eqns,vars):
assign(sols):
limit(QQ,F=infinity);
Ex. 7: Explore the behavior of the autoregulation model with R = R0 (1 + A[O2 ]v ) /(1+
B[O2 ]v ).
Solution
First note that we must have A > B so that R is an increasing function of
[O2 ]v . Next, let Ov and Oa denote [O2 ]v and [O2 ]v respectively. Then we
have the following two equations to solve for Q:
(Oa − Ov )Q = M,
Pa − Pv
= QR0
1 + AOv
1 + BOv
(18.51)
.
(18.52)
We treat Oa as a given constant, and Ov as variable. Thus we proceed by
eliminating Ov from these two equations. This gives the quadratic equation
R0 (1 + AOa )Q2 − (Pa BOa + R0 AM + Pa )Q + Pa BM = 0,
which we have to solve for Q.
(18.53)
453
This quadratic has either zero or two positive real roots. Clearly the first
option is not acceptable, so let us consider only the case where there are two
positive real roots. Call these roots Q1 and Q2 , with Q1 < Q2 . When Pa = 0
we have Q1 = 0, Q2 = AM/(1 + AOa ), while as Pa → ∞,
Q1
→
Q2
→
BM/(1 + BOa ),
and
(A − B)M
1 + BOa
+
Pa .
1 + BOa + AOa + ABOa2
R0 (1 + AOa )
(18.54)
(18.55)
(This second limit is calculated by assuming a solution of the form Q2 =
q0 + q1 Pa , as Pa → ∞. One then substitutes this expression into (18.53) and
sets to zero the coefficients of the two highest powers of Pa . This gives two
equations to solve for q0 and q1 . Note that the resultant expression is not an
exact solution of the quadratic, only approximately correct as Pa gets large.)
We now just copy what was done in the text. We want our expression for
Q2 to agree with the straight line approximation of the experimental data, as
shown in Fig. 8.10. So, as in the text, we set
(A − B)M ∗
1 + BOa∗ + AOa∗ + AB(Oa∗ )2
1 + BOa∗
R0 (1 + AOa∗ )
=
=
Q∗
,
3
2Q∗
,
3P ∗
and
(18.56)
(18.57)
where we use the same standard values Q∗ = 5.6 liters/min, P ∗ = 100 mm
Hg, Oa∗ = 104 mm Hg, Ov∗ = 40 mm Hg, and M ∗ = Q∗ (Oa∗ − Ov∗ ).
This system has three unknowns but only two equations, so we choose B to
be a given fraction of A. If we choose B = A/12 we get two possible solutions:
A = 0.024 and R0 = 9.3, or A = 0.046 and R0 = 6.45. (Real solutions don’t
always exist. If, for instance, we chose B = A/2, we would find that A was
complex. A moment’s work with Maple quickly shows us that B = A/12 is
about the largest B that actually gives a real solution.)
In summary, this more complex model can give the same kind of agreement
with the experimental data as does the simpler model, and is thus largely not
worth the additional effort.
Ex. 8: What symptoms in the circulation would you predict from anemia?
Hint: Anemia refers simply to an insufficient quantity of red blood cells, which
results in decreased resistance and oxygen-carrying capacity of the blood.
Solution
We see from Table 8.2 that as Rsv decreases, Q increases greatly (with a sensitivity of -85 %). Thus anemia causes an increased cardiac output, which
serves partially to offset the effects of a lower oxygen-carrying capacity. However, the danger is that this causes an increased workload on the heart. It
454
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
can happen that, during exercise, the heart of an anemic person cannot cope
with the additional workload, often resulting in cardiac failure.
Ex. 9: Devise a simple model for the fetal circulation that treats the systemic flow
and the placental flow as parallel flows. (Suppose that the placental circulation
and the systemic circulation each consist of two compliance vessels). Find the
ratio of the systemic flow to the placental flow. How should the flow be split
between system and placenta to support the highest metabolic rate?
Solution
A diagram of the simple model is given in Fig. 18.1. There are two compliance
vessels for the systemic circulation, and two for the placental circulation.
As before, we write down the governing steady-state equations.
Systemic loop:
Qs
=
Qs
=
Vsa
=
Vsv
=
Qu
=
Qu
=
Vua
=
Vuv
=
Pa − Ps
,
Rsa
Ps − Pv
,
Rsv
Csa
(Ps + Pa ),
2
Csv
(Ps + Pv ).
2
(18.58)
(18.59)
(18.60)
(18.61)
Placental loop:
P a − Pu
,
Rua
Pu − Pv
,
Ruv
Cua
(Pu + Pa ),
2
Cuv
(Pu + Pv ).
2
(18.62)
(18.63)
(18.64)
(18.65)
Conservation equations:
Qs + Qu
= Q,
(18.66)
Vsa + Vsv + Vua + Vuv
= Vt .
(18.67)
We treat these as ten equations for the ten unknowns (two flows, four pressures, and four volumes). From (18.58) and (18.59) we get
Rsa =
Pa − P s
,
Qs
Rsv =
P s − Pv
Qs
(18.68)
455
Heart
Q
Qs
Pa
R sa C sa V sa
R ua C ua V ua
Ps
Qu
Pu
Rsv Csv Vsv
R uv C uv V uv
Placenta
Figure 18.1:
Pv
456
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
and thus
Rsa + Rsv =
P a − Pv
Qs
(18.69)
Similarly, from (18.62) and (18.63) we get
Rua + Ruv =
It thus follows that
Pa − Pv
Qs
Qs
Rua + Ruv
=
.
Qu
Rsa + Rsv
(18.70)
(18.71)
It should come as no surprise to learn that the ratio of the flows is inversely
related to the ratio of the resistances.
For the second part of the question, we first have to realise that the highest
systemic metabolic rate is attained when the systemic oxygen transport is
greatest. This oxygen transport will be a function, not only of the flow Qs ,
but also of the flow Qu , since oxygen must be first picked up in the placenta
before it can be transferred to the foetal systemic circulation. When Qs is
large (Qs ≈ Q, say) then Qu will be small. In this case little oxygen can be
picked up in the placenta, and so little oxygen can be transferred to the foetal
circulation. On the other hand, when Qs is very small, little oxygen can be
transferred to the foetal circulation, because the flow rate is too small. In
other words, at the two extremes Qs = Q or Qs = 0, the possible systemic
metabolic rate is very low. Between these two extremes there is some value
for Qs at which the possible systemic metabolic rate is highest.
To calculate this “best” value of Qs , we assume that the rate at which oxygen
can enter the blood stream in the placenta is ku Qu , for some given constant
ku , and we assume that oxygen can enter the systemic circulation at the rate
ks Qs , for some constant ks . At steady state, these rates must be equal and
thus
ks Qs = ku Qu = ku (Q − Qs ),
(18.72)
from which it follows that
Qs =
ku Q
.
ks + ku
(18.73)
Thus, if we know the transfer rates, ku and ks , we can calculate the value of
Qs so that all the oxygen picked up in the placenta gets transferred to the
systemic circulation.
In actual fact this isn’t a very good model, as it assumes that the oxygen
transport rates can get very large as ku or ks increase. It would be better to
assume that the transport rates are given by ku (Qu ) and ks (Qs ), where ku
and ks are nonlinear, saturating, functions. We could, for instance, use the
equations of the countercurrent or cocurrent mechanisms discussed in section
20.2.1 of Keener and Sneyd (1998). Solution of the nonlinear equation
ku (Qu ) = ks (Qs )
(18.74)
457
would then determine the best flow ratio.
Ex. 11: (a) Choose some parameter values and plot the solution to the filtration
equations, (8.32) and (8.33), in the q,Pc phase plane. What is the relationship between the nullcline q = 0 and the solution?
(b) Modify the model for capillary filtration by allowing the plasma osmotic
pressure to vary along the capillary distance. Show that πc = RT cc Qqi ,
where cc is the concentration of osmolites at x = 0, i.e., in the influx.
(c) For the same parameter values as in the first part of this question, calculate and plot the solution numerically.
(d) If incoming pressure is unchanged from the first model, what is the effect
of osmotic pressure on the filtration rate?
(e) What changes must be made to the incoming pressure and the length of
the capillary to maintain the same filtration rate? Hint: Study the phase
portrait for this system of equations.
Solution
Part (a)
In the text we are given the parameter values Pi = −3, πi = 8, πc = 28. We
choose the other parameters arbitrarily as Qi = 2, β = 1, L = 10, ρ = 1 and
Kf = 2. A plot of the solution is given in Fig. 18.2. Note that q decreases
over the first half of the domain, but increases over the second half.
The nullcline q = 0 is the vertical line
P c = P i + πc − π i .
(18.75)
For the parameter values here this corresponds to Pc = 17 which is the place
where the solution is horizontal. Thus, the q = 0 nullcline corresponds to
the place where the solution (in the phase plane) turns around and starts to
increase.
Part (b)
First note that the osmotic pressure of the incoming fluid is given by πc (x =
0) = RT cc . As the flow decreases along the length of the capillary, the volume
per unit time passing a fixed point must be decreasing. However, since the
osmolites are trapped in the capillary, the osmotic pressure must be increasing
as the flow decreases. In fact, if the volume per unit time passing a fixed point
is halved, this must mean that the concentration of the osmolites in the fluid
458
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
x=L
2.0
x=0
1.5
q 1.0
0.5
0.0
16.0
16.5
17.0
Pc
17.5
18.0
Figure 18.2: Plot of the solution to the filtration equations, (8.32) and (8.33), in
the q,Pc phase plane.
is doubled, and thus the osmotic pressure is doubled. Because of this linear
relationship it follows immediately that
πc (x) = RT cc ·
Qi
.
q(x)
(18.76)
Note that, to obtain the same parameter values as in the previous model we
must choose RT cc equal to the previous value of πc , i.e., RT cc = 28.
Part (c)
With this equation for πc , the model becomes
dPc
= −ρq,
dx
dq
RT cc Qi
= Kc −Pc + Pi +
− πi ,
dx
q
q(0) = Qi ,
q(L) = Qi .
(18.77)
(18.78)
(18.79)
(18.80)
It is not a trivial matter to obtain a numerical solution to these equations.
The difficulty is that the boundary conditions are both expressed in terms of
q; one thus has to fiddle around with the starting value for Pc until one finds
a Pc (0) that gives a solution with the desired boundary conditions. (Note
that we always have q(0) = Qi , and thus we have no flexibility in the initial
condition for q). It is a worthwhile exercise to do just that, to use xppaut to
find the solutions for a range of starting values for Pc . One very quickly learns
how the solution behaves.
459
2.2
q'=0 nullcline
x=0
2.0
q 1.8
x=L
1.6
1.4
16
18
20
Pc
22
24
26
Figure 18.3: Plot of the solution to the filtration equations (18.77)–(18.80), which
include the effects of varying osmolarity along the capillary.
To be quite rigorous one would use the shooting method to obtain the solution
to any desired accuracy. That is, by systematic variation of Pc (0) one could
find a value of Pc (0) that gives q(L) = Qi to any desired accuracy. In Fig.
18.3 we plot the solution calculated using xppaut, the .ode file for which is
#
par Qi=2 beta=1 L=10 Kf=2
par P_i=-3 pii=8 pic=28 dum=17
!rho=beta*beta/Kf
#
q(0)=0
P(0)=25.5
#
P’=-rho*q
q’=Kf*(-P+P_i+pic*Qi/q-pii)
#
@ total=10 yp=q xp=p xlo=0 xhi=50 ylo=0 yhi=3
done
Part (d)
The most important thing to note about this model is the changed shape of
the q = 0 nullcline. In the model in the text, the q = 0 nullcline was vertical,
the solution went down and across until it reached this nullcline, and then
turned around and began to increase. In this revised version of the model
460
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
the same thing happens. The solution goes down and across until the q = 0
nullcline is reached, and then q starts to increase. However, because of the
different shape of the q = 0 nullcline, q can no longer decrease so far before
the nullcline is reached.
Thus, for the same incoming pressure, the decrease in q is much less, and thus
the filtration rate is decreased, when the osmolarity is allowed to vary. This
makes good intuitive sense. As q starts to decrease, the osmotic pressure of the
fluid inside the capillary will increase, thus drawing water into the capillary
and counteracting the decrease in q. From inspection of Fig. 18.3 we see that,
if the incoming pressure is maintained at 18, then there will be almost no
decrease in q along the length of the capillary. In order for q to satisfy the
boundary conditions the incoming pressure must be raised to 25.5.
Part (e)
To maintain the same filtration rate (i.e., to get the same decrease in q) the
incoming pressure must be greatly increased. Then, to maintain the boundary
conditions, the length of the capillary must also be increased. In other words,
to counteract the effects of increasing osmotic pressure one must drive fluid
at high pressure through a long capillary.
Ex. 12: Derive a simplified windkessel model by starting with a single vessel with
volume V (t) = V0 + CP (t). Assume that the flow leaves through a resistance
R and that there is an inflow (from the heart) of Q(t). Derive the differential
equation for P and compare it to (8.154).
Solution
This derivation is very similar to that of (8.154). If we assume a single vessel
with volume V (t) = V0 + CP (t), then the output from the vessel (assuming
output against zero pressure) is P (t)/R. Since the input is just Q(t), as before,
conservation of volume gives
dV
P (t)
= Q(t) −
.
dt
R
(18.81)
Now just use the relation V (t) = V0 + CP (t) to get
dV
dP
P (t)
=C
= Q(t) −
,
dt
dt
R
and thus
Q(t) = C
dP
P (t)
+
.
dt
R
(18.82)
(18.83)
Clearly this result is very similar to (8.154), the only difference being that
the expression θ(P ) is replaced by the constant compliance, C. Remember
461
that c in the expression for θ(P ) is compliance per unit length, while C is
compliance. It is thus intuitively reasonable that in the simplified model an
integral of the compliance per unit length is replaced by the compliance.
Ex. 13: Frank (1899) described a method whereby the flux of blood out of the heart
could be estimated from a knowledge of the pressure pulse, even when the
arterial resistance is unknown. Starting with (8.154), assume that during the
second part of the arterial pulse, Q(t) ≡ 0. Write down equations for the first
and second parts of the pulse, eliminate R, and find an expression for Q. Give
a graphical interpretation of the expression for Q.
Solution
The crucial assumption to make in this question is that the pressure pulse
curve consists of two rather different parts. During the first part, the rising
phase, there is inflow from the heart, while during the second part, the falling
phase, inflow from the heart has stopped, and the vessel is ejecting its contents
(refer to Fig. 18.4 for a schematic diagram of a pressure pulse).
Now assume that, during the second part of the pulse, Q ≡ 0. Thus, during
the falling phase of the pulse we have
θ(P )
dP
P
+
= 0,
dt
R
(18.84)
R=
−P
.
θ(P )dP/dt
(18.85)
and so
Now consider the rising phase of the pressure pulse. In this phase we have
Q = θ(P )
dP
P
+ .
dt
R
(18.86)
In order to eliminate R between (18.85) and (18.86) we now use the simple
(but rather elegant) trick that Otto Frank used. At P = P0 , draw a horizontal
line across the pressure pulse. This will intersect the pressure pulse curve at
two places, both with the same value of P but with different slopes, one
positive and one negative. Call these two slopes P1 and P2 . We now know
that
−P0
(18.87)
R=
θ(P0 )P2
and
Q = θ(P0 )P1 +
whence it follows that
P0
R
Q = θ(P )(P1 − P2 ).
(Since this works for any P we have dropped the subscript 0).
(18.88)
(18.89)
462
CHAPTER 18. THE CIRCULATORY SYSTEM: SOLUTIONS
P
slope = P'2
P0
slope = P'1
time
Figure 18.4: Schematic diagram of a pressure pulse.
This is our desired expression for Q. Given a pressure pulse curve, for every
value of P just calculate the difference between the slopes (noting, of course,
that P2 < 0) and substitute into (18.89) to get Q as a function of P , and thus
as a function of time also.
Chapter 19
Blood: Solutions
Ex. 1: What is the volume (per mole) of an ideal gas at room temperature (27◦ C)
and 1 atm pressure? What is its volume at body temperature (98◦ F)?
Solution
From the ideal gas law P V = nRT (where n is the number of moles) we see
that the volume, V , of a mole of ideal gas is given by
V =
RT
.
P
(19.1)
Since R = 8.315 J mol−1 ·K−1 , T = 300.15 K (corresponding to 27◦ C), and
P = 1 atm = 1.01325 × 105 N · m−2 , it follows that
V
=
8.315 × 300.15 3
m
1.01325 × 105
0.025 m3
=
25 liters.
=
(19.2)
(19.3)
(19.4)
Secondly, to calculate the temperature at 98◦ F we need to convert the temperature to Kelvin. What on earth is this conversion factor? I have no idea.
Ex. 2: Suppose that a carrier (like hemoglobin) of a molecule (like oxygen) has n
independent binding sites, with individual binding and unbinding rates k+
and k− . Let cj denote the concentrations of the state with j molecules bound.
Assume concentrations are in steady state.
n j
n
n!
(a) Show that cj =
x c0 , where
= j!(n−j)!
is the binomial coeffij
j
cient where x = s0 /K, K = k− /k+ , and s0 is the concentration of the
carrier molecule.
Hint: Keep track of the total number of binding sites.
463
464
CHAPTER 19. BLOOD: SOLUTIONS
(b) Find the saturation function in the case that n = 4.
(c) Show that the four equilibrium constants K1 , K2 , K3 , K4 defined in (9.11)
are given by (K1 , K2 , K3 , K4 ) = K( 14 , 23 , 32 , 4).
(d) Estimate K to give a good fit of this model to the hemoglobin saturation
curve. How does this curve compare with the curve (9.13)?
(e) Determine whether the hemoglobin binding sites are independent. How
close are the equilibrium constants here to those found in the text?
Solution
Part (a)
We have the binding scheme
nk+ s0
c0
−→
←−
(n−1)k+ s0
c1
k−
−→
←−
(n−2)k+ s0
−→
←−
c2
2k−
···
(n−k−1)k+ s0
3k−
−→
←−
kk−
(n−k)k+ s0
ck
−→
←−
(k+1)k−
···
k+ s0
−→
←−
cn (19.5)
nk−
State c0 can convert to state c1 in n different ways (since there are n unoccupied binding sites) and each site reacts at speed k+ s0 . Thus we get
nxc0 = c1 ,
(n − 1)xc1 = 2c2 ,
(n − 2)xc2 = 3c3 ,
···,
(19.6)
from which it follows that
c2
=
n−1
n(n − 1) 2
xc1 =
x c0 ,
2
2
..
.
ci
=
(19.7)
(19.8)
n(n − 1) · · · (n − i − 1) i
x c0 ,
1 · 2 · 3···i
(19.9)
which is the desired answer.
Part (b)
The fraction of occupied sites, Y , is given by
Y
=
=
c1 + 2c2 + 3c3 + 4c4
4(c1 + c2 + c3 + c4 )
x + 3x2 + 3x3 + x4
.
1 + 4x + 6x2 + 4x3 + x4
(19.10)
(19.11)
465
percent saturation
80
60
40
20
0
0
20
40
60
80
100
Oxygen partial pressure (mm Hg)
120
Figure 19.1: Plot of the saturation function when n = 4.
Part (c)
From the reaction scheme in (19.5) we see that kj = (n − j + 1)k+ and
k−j = jk− . Hence
k−j
j
=
K.
(19.12)
kj
n−j+1
When n = 4 we thus have K1 = K/4, K2 = 2K/3, K3 = 3K/2 and K4 = 4K,
as required.
Part (d)
It’s not really possible to get a good fit to the hemoglobin saturation curve
by using independent binding sites, as illustrated in Fig. 19.1. The symbols
are the approximate experimental data as given in Table 19.1, and the solid
curve is drawn from (19.11) using K = 12σ mm Hg. Clearly, if we assume
independent and identical binding sites, the saturation function has the incorrect shape and cannot be made to fit the experimentally measured saturation
function well.
Part (e)
We’ve already answered this part of the question. We must conclude from
the result of (d) that the binding sites are not independent. The measured
values of K1 , · · · , K4 are very different from the values obtained if we assume
independent binding sites, with K = 12σ mm Hg.
466
CHAPTER 19. BLOOD: SOLUTIONS
PO2 (mm Hg)
3.08
4.61
6.77
10.15
12.31
15.38
18.77
22.77
25.85
30.15
36.00
45.23
51.69
61.85
75.38
87.08
110.5
percent saturation
2.21
3.59
6.08
10.50
14.09
19.34
28.45
40.33
50.0
60.50
69.89
80.11
83.98
88.95
93.37
95.86
98.07
Table 19.1: Approximate numerical data for the hemoglobin saturation curve.
Ex. 3: Construct a Monod–Wyman–Changeux model (Section 1.6) for oxygen binding to hemoglobin and determine the saturation function. Find parameter
values so that the saturation curve agrees (at least approximately) with the
data in Fig. 19.1. Compare your results to the fit of (9.13).
In the Monod–Wyman–Changeux model, each binding site is independent.
Thus, on the surface it appears that the Monod–Wyman–Changeux model
is quite different in principle from the model discussed in the text, in which
binding to the fourth binding site is enhanced. However, by considering the
solution to Exercise 1.55 of Chapter 1 (and using the parameter values found in
the first part of the question), show that this is not really true. In other words,
show that the Monod–Wyman–Changeux model indirectly incorporates an
enhanced binding to subsequent binding sites, once the initial site is bound.
Solution
Most of the work for this question has already been done in Chapter 1, where
we discussed Monod–Wyman–Changeux models. The MWC model is shown
in Fig. 19.2. Equation (1.55) gives the saturation function for an MWC model
with n binding sites. In our case n = 4, and so we get immediately that
3
3
s
s
1 s
s
1
+
1
+
+
K1
K1
K2 K3
K3
Y =
.
(19.13)
4
4
1 + Ks1 + K12 1 + Ks3
Note also that, as we showed in Exercise ?? of Chapter 1, the incorporation
of transitions from Ri to Ti doesn’t change the saturation function. By ap-
467
3sk1
4sk1
R0
R1
k-1
k2
k4
k-2
4sk3
T0
k-3
2k-1
k-4
k5
2k-3
R3
3k -1
k6
k-5
T2
3k -3
R4
4k-1
k-6
2sk3
3sk 3
T1
sk1
2sk1
R2
k7
k-7
sk3
T3
4k-3
T4
Figure 19.2: Schematic diagram of the Monod–Wyman–Changeux model for oxygen
binding to hemoglobin.
propriate choice of the parameters K1 , K2 and K3 we can make this MWC
saturation function agree closely with the experimental data shown in Table
19.1; typical results are shown in Fig. 19.2.
We did not determine the parameters by a formal fitting procedure to the
data, but merely by trial and error. With only three parameters to twiddle
this is not a difficult procedure. There are many ways of finding the “best-fit”
parameters, in, say, the least-squares sense, but such methods would require
considerably more programming. Interested readers are referred to ?? for
details on such methods.
In a MWC model each binding site is considered to be identical. This appears
different from the model discussed in the text, in which the binding of oxygen
to the fourth binding site proceeds at quite a different rate. However, this is
not really the case. From the principle of detailed balance we know that
K1 K 4 = K 2 K 3 ,
K1 K 5 = K 4 K 3 ,
K1 K 6 = K 5 K 3 ,
K1 K 7 = K 6 K 3 .
(19.14)
From these equations and from the values for K1 , K2 and K3 given in the
caption to Fig. 19.3 we find that K4 = 5.7, K5 = 0.025, K6 = 1.2 × 10−4 and
K7 = 5.3 × 10−7 .
Since K2 1 it follows that the state R0 is greatly preferred over the state
T0 . Hence, upon the addition of oxygen, most R0 gets converted to state
R1 . Of course, since K3 /K1 is so small we know that oxygen will bind much
more easily to hemoglobin when it is in the T conformation, but this does
not happen for the first binding site as hemoglobin is “locked” into the R
conformation. However, K7 < K6 < K5 < K4 and thus, as hemoblobin binds
more oxygen the rate at which R is converted to T increases. In other words,
the binding of oxygen tends to favour the conversion of the R conformation
to the T conformation, and thus facilitates the binding of further oxygen.
Ex. 4: (a) If a 25 mM solution of sodium bicarbonate is equilibrated with carbon
dioxide at 40 mm Hg partial pressure, the pH is found to be 7.4. If the
468
CHAPTER 19. BLOOD: SOLUTIONS
percent saturation
80
60
40
20
0
0
20
40
60
80
100
Oxygen partial pressure (mm Hg)
120
Figure 19.3: Plot of the saturation function of the Monod–Wyman–Changeux model
in the previous figure. The solid curve is the MWC model with the parameters
K1 = 1000 σ mm Hg, K2 = 1250, K3 = 4.54 σ mm Hg. The dashed curve is a plot
of (9.13) using the parameter values given in the text.
partial pressure of carbon dioxide is increased until the pH is 6.0, what
is the bicarbonate concentration? What is the carbon dioxide partial
pressure at this pH? What is the difference if this experiment is carried
out in whole blood instead of sodium bicarbonate solution?
(b) Pick reasonable values for K1 , K2 and K̄1 and find the oxygen saturation
curve as a function of h. (Hint: the parameters must be chosen so that
hemoglobin acts as a hydrogen ion buffer at physiological concentrations.
So, for instance, pick K1 , K2 and K̄1 so that (φ(h))1/4 = 26 σ mm Hg
at pH = 7.4. You also have to ensure that K̄1 > K1 and that K2 is not
too large. Why?)
(c) In this model how much CO2 is transported from the tissues to the
lungs? How much O2 is transported? (Determine this by solving (9.27)
- (9.31) numerically to find h, X, Y , Z and W at a given [O2 ] and
[CO2 ].) Remove the Bohr and Haldane effects by setting K1 = K̄1 . How
does this change the amount of oxygen and carbon dioxide transported?
Typical parameter values are: PCO2 in arterial blood, 39 mm Hg ; PCO2
in venous blood, 46 mm Hg; PO2 in arterial blood, 100 mm Hg; PO2 in
venous blood, 40 mm Hg; R1 R2 = 106.1 M−1 ; THb = 3 mM; n = 10,
[HCO−
3 ] = 25 mM and pH = 7.4 in arterial blood.
Solution
469
Part (a)
We begin by using the information to calculate the equilibrium constant of
the reaction
k
+
−→
H+ + HCO−
3 ←− CO2 + H2 O.
k−
In the body this reaction is catalysed by carbonic anhydrase, and we shall
sometimes refer to it as the carbonic anhydrase reaction. Since the partial
pressure of CO2 is held fixed at 40 mm Hg, and since the solubility of CO2 is
3.3 × 10−5 M/mm Hg, it follows that [CO2 ] = 40 × 3.3 × 10−5 M = 1.32 mM.
Similarly, since the pH is 7.4 it follows that [H+ ] = 10−4.4 mM = 4 × 10−5
mM.
Letting K = k− /k+ we know that, at equilibrium,
K=
[H+ ][HCO−
3]
.
[CO2 ]
(19.15)
Hence, to calculate K it remains only to calculate [HCO−
3 ]. We started with
25 mM of HCO−
3 . For each CO2 molecule that reacts with water we get one
hydrogen ion and one bicarbonate ion. Thus we know that
+
[HCO−
3 ] = 25 + [H ],
where [HCO−
3 ] is in units of mM. Substituting into (19.15) then gives
K=
[H+ ](25 + [H+ ])
.
[CO2 ]
(19.16)
Since [H+ ] is so much less than 25 we can ignore its square. Hence
K=
(4 × 10−5 ) · 25
= 7.5 × 10−4 mM.
1.32
(19.17)
If the pH is increased to 6, [H+ ] is increased to 10−3 mM. Again, since this is
so much smaller than 25 mM there is no appreciable change to the bicarbonate
concentration. However, we must have
[CO2 ] =
[H+ ][HCO−
3]
= 33.3 mM
K
(19.18)
from which it follows that the partial pressure of CO2 is
PCO2 =
0.033
= 103 mm Hg.
3.3 × 10−5
(19.19)
In whole blood hemoglobin acts as a hydrogen ion buffer; as CO2 reacts with
water the H+ that is produced is taken up by hemoglobin, greatly decreasing
any changes in blood pH. Hence, to raise the pH of whole blood by a given
470
CHAPTER 19. BLOOD: SOLUTIONS
amount would require a far greater CO2 partial pressure than would be needed
for a bicarbonate solution.
As a side note, the most important system for controlling the extracellular
acid-base balance is the bicarbonate buffer system. The extracellular fluid
contains large amounts of bicarbonate ions, mostly as Na+ HCO−
3 . Addition
of excess H+ ions merely drives the carbonic anhydrase reaction from left to
right (as written above); the additional CO2 produced can be removed at
the lungs. Conversely, addition of a strong base and the consequent removal
of H+ just results in a lowering of [CO2 ] and the production of additional
H+ . Since the overall levels of bicarbonate and carbon dioxide are controlled,
respectively, by the kidneys and the lungs, this allows precise control over the
pH of the extracellular fluid.
Part (b)
We need to choose K1 , K̄1 and K2 so as to get reasonable behavior of the O2
saturation curve. We know (from the experimental data) that, at blood pH
(approximately 7.4), the half-maximal saturation occurs at about [O2 ] = 26σ
mm Hg. Thus we must have φ(10−7.4 ) = (26σ)4 mm Hg.
We also require that K̄1 > K1 . Thus we choose K̄1 = 10K1 (this is not based
on experimental data; it is just a reasonable choice that we use to illustrate
the model). Then,
K2 + h
φ(h) = 10K1
.
(19.20)
10K2 + h
Clearly, if K2 is very much larger than h, any changes in h will not be noticeable. So, again quite arbitrarily, we choose K2 = 10−7.4 M, the same as the
normal blood pH.
With these constraints on K̄1 and K2 we can now choose K1 to ensure that
φ(10−7.4 ) = (26σ)4 mm Hg. This gives
10K1
2 × 10−7.4
= (26σ)4 ,
11 × 10−7.4
(19.21)
or, since we are just going to plot the saturation curve against the partial
pressure of O2 , not its concentration,
10K1
2 × 10−7.4
= 264 ,
11 × 10−7.4
from which it follows that K1 =
(19.22)
20
4
11 26 .
Plots of the saturation curve with these parameters, and for three different
values of the pH, are given in Fig. 19.4. We see that, as the pH increases (i.e.,
as [H+ ] decreases), the saturation curves shifts to the left, thus increasing
the amount of O2 bound at a given partial pressure. Conversely, as the pH
decreases, the amount of bound O2 decreases.
471
fraction saturation
0.8
0.6
pH 7
pH 6
pH 8
0.4
0.2
0.0
0
20
40
60
O2 partial pressure (mm Hg)
80
Figure 19.4: Oxygen saturation curves at three different pH values, calculated with
4
4
−7.4
M.
the parameter values K1 = 20
11 26 M , K̄1 = 10K1 , K2 = 10
Part (c)
There are many ways of solving the equations numerically, but probably the
easiest way is just to write a short Maple routine. The Maple code we used
is given below. As always, we make no claims that this is the most efficient
possible code, or the shortest, or the cleverest, but it was quick and easy to
write and it works.
The trickiest part of this question is to make sure you get the units right. Since
we are going to be working with concentrations of H+ , the easiest thing to do
4
is just to use Molar units throughout. Thus, instead of using K1 = 20
11 26 , we
4
−6
−1
use K1 = 20
(26σ
)
M,
where
σ
=
1.4
×
10
M
(mm
Hg)
.
O2
O2
11
The next tricky thing is to calculate the value of T0 . We choose the arterial
blood as the reference state. We now just fiddle T0 until [HCO−
3 ] comes out
to be close to 25 mM. (Because we are not trying for exact answers, close is
good enough). Using the Maple code to try a few values of T0 we quickly
find that T0 = −0.022 M works well. With this value of T0 it turns out that
[HCO−
3 ] = h + n(Z + W ) − T0 = 25 mM, while n(Z + W ) = 3 mM. The value
of h in the arterial blood is so small it can be ignored.
Finally, note that in the calculation of the total oxygen or carbon dioxide
content we include the gas in the dissolved state. We include the dissolved
gas in our calculation of total transport, although it makes only a small change
to the answers.
We are now ready to answer the question; the results are summarised in Table
19.2. We see that, with the Bohr effect, CO2 is transported at a concentration
472
CHAPTER 19. BLOOD: SOLUTIONS
Total CO2
Total O2
With Bohr effect
Arterial Venous
26.27
30.96
3.09
1.93
Without Bohr effect
Arterial
Venous
36.03
37.27
3.13
2.6
Table 19.2: Summary of results for part (c) of Question 4. All concentrations are
in mM.
of 30.96 − 26.27 ≈ 4.7 mM, and O2 is transported at a concentration of 3.09 −
1.93 ≈ 1.2 mM. However, if the Bohr effect is removed then CO2 is transported
at 37.27 − 36.03 ≈ 1.3 mM, and O2 is transported at approximately 0.5 mM.
We emphasise that the exact numbers are not meant to be taken too seriously
here, but are just an illustration of the effect. Different choices for K1 , K̄1
and K2 will give different quantitative results.
The Maple code for part (c)
readlib(’unassign’):
unassign(’X’,’Y’,’Z’,’W’,’sols’,’h’,’K1’,’K2’,’K1bar’,
’THb’,’n’,’T0’,’R’,’Ox’,’C’):
sigmaO := 1.4*(10^(-6)): sigmaC := 3.3*(10^(-5)):
PCarterial:=39:
PCvenous:=46:
POarterial:=100: POvenous:=40:
# Solubilities
# Convert the partial pressures to Molar concentrations
# Comment out whichever you don’t need
C:=PCarterial*sigmaC: Ox:=POarterial*sigmaO:
# Arterial concentrations
#C:=PCvenous*sigmaC: Ox:=POvenous*sigmaO:
# Venous concentrations
# Make sure all the concentration units are Molar. That’s why
# we have to include the sigmaO in the formula for K1.
# Comment out whichever you don’t need
K1 := (20/11)*(26*sigmaO)^4: K1bar:=10*K1:
# with Bohr effect
#K1 := (26*sigmaO)^4:
K1bar:=K1:
# without Bohr effect
K2 := 10^(-7.4):
THb := 0.003: n:=10:
T0:=-0.022:
R:=10^(6.1):
fun1 := (Ox^4)*X - K1*Y:
fun2 := h*X - K2*Z:
fun3 := (Ox^4)*Z - K1bar*W:
fun4 := X+Y+Z+W - THb:
sols := solve({fun1=0,fun2=0,fun3=0,fun4=0}, {X,Y,Z,W}):
assign(sols):
# Now find h numerically
hfun := R*h*(h + n*(Z+W) - T0) - C:
h:=fsolve(hfun=0,h,0..0.00001);
# Now calculate the total bicarbonate and oxygen
bicarb := h + n*(Z+W) - T0;
total_co2 := bicarb + C;
heme_oxygen := Y+W;
473
total_oxygen := heme_oxygen + Ox;
Ex. 6: Find an analytic relationship for the critical stability curve (Section 9.2.4),
F (0)
relating dF (0) to X/d as follows: Use that F (N ) = 1+N
7 to solve (9.49) for
N0 as a function of X/d and then determine dF (0) using that F (N0 ) = N0 /X.
Solution
Since
F (N0 ) =
it follows that
F (0)
1 + N0p
(19.23)
F (N0 )
−pN0p−1
.
=
F (N0 )
1 + N0p
(19.24)
For convenience we introduce the notation
α=
2nπ
,
2 + X/d
φ=
1 αx
.
2 sin α
(19.25)
Then, from (9.49) and (19.24) it follows that
pN0p
= φ,
1 + N0p
and hence
N0 =
φ
p−φ
(19.26)
p1
.
(19.27)
Now use the fact that, at steady state
N0
.
X
(19.28)
F (0)
N0
,
=
1 + N0p
X
(19.29)
F (N0 ) =
Hence,
and thus
dF (0)
N0
= d(1 + N0p )
X
p1
d
p
φ
=
,
X p−φ
p−φ
which is the required expression.
(19.30)
(19.31)
474
CHAPTER 19. BLOOD: SOLUTIONS
Ex. 7: A deficiency of vitamin B12 , or folic acid, is known to cause the production of
immature red blood cells with a shortened lifetime of one-half to one-third of
normal. What effect does this deficiency have on the population of red blood
cells?
Solution
A decreased lifetime corresponds to a decrease in X. The result of a decrease
in X depends on the value of dF (0). If dF (0) > 0.4 or so, then a decrease in
X from the normal value of 120 days may result in the steady state becoming
unstable and the appearance of oscillations in the number of red blood cells.
If dF (0) is too small, the steady state never loses stability, for any value of X.
Ex. 8: Suppose X → ∞ in the red blood cell production model. Show that
dN
= F (N (t − d)) − βN.
dt
(19.32)
(a) Find the stability characteristics for the steady-state solution of this
equation.
(b) Show that the period of oscillation T = 2π/ω at a Hopf bifurcation point
is bounded between 2d and 4d.
∞
Hint: Differentiate the equation N (t) = 0 n(x, t)dx with respect to t and
use the partial differential equation nt + nx = −βn and the initial condition
n(x, t) = F (N (t − d)) to eliminate n(x, t).
Solution
∞
If we let X → ∞, then N (t) = 0 n(x, t) dx. Now integrate the conservation
equation (9.34) from x = 0 to x = ∞. This gives
0
∞
∂n
dx +
∂t
0
∞
∂n
dx = −β
∂x
∞
n(x, t) dx,
(19.33)
0
and thus
∂
N (t) − n(0, t) = −βN (t).
∂t
(19.34)
The boundary condition (9.36) now gives
∂N (t)
= F (N (t − d)) − βN (t)
∂t
as required.
(19.35)
475
Part (a)
The steady state, N0 , is given by the equation F (N0 ) = βN0 . To determine
the stability of this steady state we linearise around N0 . If N = N0 + δN ,
then
d
δN (t)
dt
= F (N0 + δN (t − d)) − β(N0 + δN )
(19.36)
= F (N0 ) + F (N0 )δN (t − d) − βN0 − βδN
(19.37)
= F (N0 )δN (t − d) − βδN.
(19.38)
Looking for solutions of the form δN = eλt gives
λ = F (N0 )e−λd − β.
(19.39)
Recall that F (N0 ) < 0 and that β > 0. It thus follows that λ cannot be real
and positive. It can be real and negative, in which case the steady state is
stable. Just as in the example discussed in the text, instability can only arise
at a Hopf bifurcation, where λ = iω. Substituting this into (19.39) gives
β
ω
= F (N0 ) cos(ωd)
= −F (N0 ) sin(ωd).
(19.40)
(19.41)
We can eliminate ω by squaring the equations and adding, to get
ω 2 = (F (N0 ))2 − β 2 .
(19.42)
Note that we must have β < |F (N0 )|. With this expression for ω, (19.40) is
a relationship between N0 and d that must be satisfied in order for a Hopf
bifurcation to exist. Or, since F (N0 ) = βN0 , (19.40) can be interpreted as a
relationship between β and d that ensures a Hopf bifurcation.
Part (b)
From (19.40) we see that, in order for β to be positive we must have cos(ωd) <
0. Similarly, from (19.41) we see that we must have sin(ωd) > 0. Combining
these two constraints gives
π
< ωd < π,
(19.43)
2
from which it follows that
2d <
as required.
2π
< 4d,
ω
(19.44)
476
CHAPTER 19. BLOOD: SOLUTIONS
Ex. 10: Suppose X is finite and β = 0 in the red blood cell production model. Show
that the evolution of N is described by the delay differential equation
dN
= F (N (t − d)) − F (N (t − d − X)).
dt
(19.45)
Solution
Integrate (9.34) from x = 0 to x = X to get
X
0
X ∂n
∂n
dx + inf
dx
0 ∂t
0 ∂x
X
∂
n(x, t) dx + n(X, t) − n(0, t)
∂t 0
dN
+ n(X, t) − n(0, t)
dt
dN
+ n(X, t) − F (N (t − d)).
dt
= inf
(19.46)
=
(19.47)
=
=
(19.48)
(19.49)
However, since the death rate of the cells is zero (β = 0) we know that the
number of cells of age X is just equal to the number of cells that entered the
bloodstream X time units before. Thus,
n(X, t) = F (N (t − d − X)),
(19.50)
dN
= F (N (t − d)) − F (N (t − d − X)),
dt
(19.51)
in which case
as required.
Ex. 11: Sketch the phase portraits for the equations (9.89)–(9.90) in Case III (α >
0, β > 0) as follows:
(a) ξ <
1
U0 , ξ
<
1
β.
(For example,
i. ξ = 1.0, β = 0.5, α = 0.5, β0 = 1.0, γ0 = 0.2, and
ii. ξ = 1.5, β = 0.5, α = 1.3, β0 = 1.0, γ0 = 0.2.)
(b) ξ < U10 , ξ >
0.2.)
1
β
(For example, ξ = 1.8, β = 0.6, α = 0.5, β0 = 1.0, γ0 =
(c) ξ > U10 , ξ <
0.2.)
1
β.
(For example, ξ = 2.2, β = 0.3, α = 0.5, β0 = 1.0, γ0 =
(d) ξ > U10 , ξ >
0.2.)
1
β.
(For example, ξ = 2.2, β = 2.0, α = 0.5, β0 = 1.0, γ0 =
477
10
8
V
6
4
dV/dt = 0
2
dU/dt = 0
0
0.0
0.2
0.4
U
0.6
0.8
Figure 19.5: Phase plane for Question 11, part (a)i. Parameters are ξ = 1, β = 0.5,
α = 0.5, β0 = 1 and γ0 = 0.2
Locate each of these cases in Fig. 9.13.
Solution
The easiest way to plot these phase portraits is to use your favourite differential equations package. Here, we used XPPAUT to calculate the nullclines and
trajectories, wrote the curves to disk, and then plotted them using a scientific
plotting package called Igor Pro (ref). This separation of the calculation stage
from the plotting stage lets us use specialist packages for the different jobs,
giving a higher quality output.
The XPPAUT code is
par xi=2.2 beta=0.3 alpha=0.5 beta0=1.0 gamma0=0.2
u’=gamma0*((beta0+beta*v)*(1-u*fun(alpha*v)*exp(-alpha*v))-(v*fun(v)+1)*u)
v’=v*(1-xi*u*fun(v))
#
fun(v)=alpha*v/(1-exp(-alpha*v))
@yp=v xlo=0 xhi=0.8 yhlo=0 yhi=10
The phase planes for (a)i, (a)ii, (b), (c) and (d) are given in Figs. 19.5–19.9
respectively. The location of each of these cases in Fig. 9.13 is shown in Fig.
19.10.
478
CHAPTER 19. BLOOD: SOLUTIONS
10
8
saddle point
V
6
4
dU/dt = 0
2
dV/dt = 0
0
0.1
0.2
0.3
0.4
0.5
0.6
U
Figure 19.6: Phase plane for Question 11, part (a)ii. Parameters are ξ = 1.5,
β = 0.5, α = 1.3, β0 = 1 and γ0 = 0.2
10
8
dU/dt = 0
6
V
dV/dt = 0
4
2
0
0.2
0.3
0.4
0.5
0.6
0.7
U
Figure 19.7: Phase plane for Question 11, part (b). Parameters are ξ = 1.8, β = 0.6,
α = 0.5, β0 = 1 and γ0 = 0.2
479
10
8
6
V
saddle point
4
stable steady
state
2
dV/dt = 0
dU/dt = 0
0
0.0
0.1
0.2
0.3
U
0.4
0.5
Figure 19.8: Phase plane for Question 11, part (c). Parameters are ξ = 2.2, β = 0.3,
α = 0.5, β0 = 1 and γ0 = 0.2
10
8
dU/dt = 0
6
V
dV/dt = 0
4
2
0
0.0
0.2
0.4
0.6
0.8
1.0
U
Figure 19.9: Phase plane for Question 11, part (d). Parameters are ξ = 2.2, β = 2,
α = 0.5, β0 = 1 and γ0 = 0.2
480
CHAPTER 19. BLOOD: SOLUTIONS
4
3
{0,∞}
{0}
(d)
ξ
(c)
2
(b)
{V }
(a)ii
p
1
(a)i
{∞}
0
0
1
2
3
4
1/β
Figure 19.10: The approximate location in Fig. 9.13 of each of the above phase
planes.
Chapter 20
Muscle: Solutions
Ex. 1: Derive a differential equation for the load in a three-element Hill model with
an elastic element in series with a contractile element (as shown in Fig. 10.8)
in parallel with an additional elastic element of length L.
Remark: The total load in the whole unit is divided between the two parallel
subunits.
Solution
A schematic diagram of a three-element Hill model is shown in Fig. 20.1. From
this diagram we see that
F1 + F2 = p,
(20.1)
and thus
β(L − L0 ) + α(x − x0 ) = p,
(20.2)
from which it follows that
β(x + l − L0 ) + α(x − x0 ) = p.
Thus, following the same procedure as in the book, we have
dp
dp dx
dL
=
= (β + α)
+v .
dt
dx dt
dt
(20.3)
(20.4)
Ex. 2: When a muscle cell dies, its ATP is depleted, with the result that the power
stroke stalls and calcium cannot be withdrawn using the Ca2+ ATPase. How
does this explain rigor mortis?
Solution
481
482
CHAPTER 20. MUSCLE: SOLUTIONS
x
l
F2
F1
p
Figure 20.1: A three-element Hill model.
At rest, every cell in the body is under a high calcium load, since the calcium
concentration outside the cell is much higher than the calcium concentration
inside the cell. The reason for this is that prolonged high calcium concentrations are toxic to cells, but the cell still wishes to use calcium as an intracellular
second messenger. Models of calcium dynamics are discussed extensively in
Keener and Sneyd (1998) and Goldbeter (199?). All cells expend considerable
amounts of energy keeping their intracellular calcium concentrations low.
Thus, when the calcium ATPase stops working at death, calcium can no longer
be removed from the muscle cell, and the intracellular calcium concentration
will begin to rise. This prevents the disassociation of the actin and myosin
filaments which remain locked together, thus causing rigidity in the muscles.
This rigidity lasts until the muscle proteins are broken down, about 15 to 25
hours after death.
Ex. 3: Derive the force–velocity relationship (10.33) for the Huxley model.
Solution
From the expressions given for f and g in the Huxley model ((20.17) and
483
(20.18)), the steady state solution for n is

f1

[1 − exp(−φ/v)] exp(xg2 /v),


 f1 + g1 2
x
f1
φ
n(x) =
1 − exp
−1
,

2

f
+
g
h
v

1
 1
0,
x > h,
x < 0,
0 < x < h,
(20.5)
where φ = (f1 + g1 )h/2.
We assume that each crossbridge acts as a linear spring, with spring constant
k, and thus
∞
p(t) = ρ
kx n(x, t) dx.
(20.6)
−∞
It thus follows that
2
0
h f1 + g1
x
φ
−φ/v
xg2 /v
)
xe
dx+
x 1 − exp
−1
p = (1−e
dx.
2
kf1 ρ
h
v
−∞
0
(20.7)
Evaluating these integrals (using integration by parts for the first one, and a
u-substitution for the second) gives
2
f1 + g1
h2
−v
v
−φ/v
+
)
p = (1 − e−φ/v )
1
−
(1
−
e
kf1 ρ
g22
2
φ
2
2
h2
v
v
h
=
− (1 − e−φ/v )
+
2
g22
2 φ
2
h
v 2φ2
v
=
1 − (1 − e−φ/v ) 1 +
2
φ
φ h2 g22
h2
v
v (f1 + g1 )2
=
(20.8)
1 − (1 − e−φ/v ) 1 +
2
φ
φ
2g22
which is the desired answer.
Ex. 4: Calculate the response of the Huxley model to a step change in length. In
other words, calculate how n changes as a function of t and x, and hence
calculate how the tension changes as a function of time.
Solution
At steady state, the fraction of bound crossbridges is given by
n(x) =
f (x)
.
f (x) + g(x)
(20.9)
After a step decrease in length, the x position of each crossbridge will be
decreased by ∆x, say, but immediately after the length change, the distribution of bound crossbridges will not alter. Hence, immediately after the length
484
CHAPTER 20. MUSCLE: SOLUTIONS
change
n(x, t = 0+ ) =
f (x + ∆x)
.
f (x + ∆x) + g(x + ∆x)
(20.10)
Thus, to find the evolution of n over space and time, we need to solve the
differential equation
∂n(x, t)
∂t
n(x, 0)
= (1 − n(x, t))f (x) − ng(x),
=
f (x + ∆x)
.
f (x + ∆x) + g(x + ∆x)
(20.11)
(20.12)
Since x is just carried along as a parameter, this is just an ordinary differential
equation in t. Note that v does not enter into the equation, since the length
is decreased and then held fixed, i.e., v = 0.
To solve, we just use the integrating factor e (f +g)dt = e−(f (x)+g(x))t (remembering that we are integrating with respect to t, and thus x can be treated as
a constant). Multiplying both sides by the integrating factor gives
∂ −(f (x)+g(x))t = f (x)e−(f (x)+g(x))t ,
ne
∂t
(20.13)
and thus
n(x, t) =
f (x)
+ Ke−(f (x)+g(x))t ,
f (x) + g(x)
(20.14)
for some constant K. Determining K from the initial condition then gives
f (x)
f (x)
f (x + ∆x)
n(x, t) =
+e−(f (x)+g(x))t
−
.
f (x) + g(x)
f (x + ∆x) + g(x + ∆x) f (x) + g(x)
(20.15)
Finally,
∞
p(x, t) = ρk
x n(x, t) dx.
(20.16)
−∞
We now evaluate each of these expressions for the particular choice of f and
g used by Huxley, i.e.,

x < 0,
 0,
f1 x/h,
0 < x < h,
f (x) =
(20.17)

0,
x > h,
g2 ,
x < 0,
(20.18)
g(x) =
x > 0.
g1 x/h,
For this choice of f and g we have
f (x)
= α(H(0) − H(h)),
f (x) + g(x)
(20.19)
485
where α = f1 /(f1 + g1 ) and H is the Heaviside function (H(x) = 0 when
x < 0, and H(x) = 1 when x > 1.) We also have

−∆x < x < 0,
 α,
f (x + ∆x)
f (x)
−α,
h − ∆x < x < h,
−
=
f (x + ∆x) + g(x + ∆x) f (x) + g(x) 
0,
otherwise.
(20.20)
Substitute these expressions into (20.15) and (20.16) to get
h
h
0
−g2 t
−(f1 +g1 )xt/h
p(x, t) =
αx dx + α
xe
dx −
xe
dx
−∆x
0
2
=
−g2 t
ρkαh
αe
−
2
2
(∆x)
−α
2
h−∆x
h
x e−(f1 +g1 )xt/h dx.
(20.21)
h−∆x
At this stage, it’s not really very useful to evaluate the final integral analytically. (It is relatively easily done, particularly with the help of Maple, but just
gives a somewhat long answer that is difficult to interpret). But it is useful
to note that p → ρkαh2 /2 = ps , say, as t → ∞, and that p < ps .
Now, at steady state,
p = ρkα
h
x dx,
(20.22)
0
and thus it follows that ps is the steady state force exerted by the muscle.
Hence we see from (20.21) that immediately after the length decreases the
force decreases also, and then slowly increases back to the resting level. This
is all as expected intuitively and is agrees qualitatively with the sketch given
in Fig. 10.10A of the book.
In Fig. 20.2 we plot the solution for p.
Ex. 5: Show that when
g(x)
fmax e(x−h)/λ , x < h,
0,
x > h,
= fmax 1 − e(x−h)/λ ,
r(x)
= rmax
f (x)
=
e(x−h)/λ − α
,
1−α
(20.23)
(20.24)
(20.25)
the Huxley model reproduces the Hill force–velocity curve exactly. Note that
the behavior of g and r for x > h are irrelevant, as no crossbridges are ever
bound there. Unfortunately, although this choice for f and g gives good
agreement with the force–velocity curve, it gives poor agreement with the
energy flux data.
Solution
486
CHAPTER 20. MUSCLE: SOLUTIONS
0.40
force
0.35
0.30
0.25
0.20
-1.0
-0.5
0.0
0.5
time
1.0
1.5
2.0
Figure 20.2: The solution for p, calculated from (20.21), using the parameter values
ρ = 1, k = 1, f1 + g1 = 1/2, g1 = 3/32, g2 = 2, ∆x = 0.3.
We begin by calculating n(x), the steady-state crossbridge distribution at a
constant velocity. It’s easiest to begin by calculating the solution for x > h.
In this region (since f (x) = 0 here) we have the differential equation
v
dn
= g(x)n.
dx
(20.26)
Fortunately, we don’t actually have to solve this equation for n, except to
note that n ≡ 0 is a solution. Clearly, since f (x) = 0 for x > h, and thus no
crossbridges can ever attach there, the identically zero solution is the one we
want. So, n(x) = 0 when x > h.
Next we solve for the region x < h. Since f (x) + g(x) = fmax when x < h, it
follows that, when x < h, n(x) satisfies the differential equation
−v
dn
= f (x) − fmax n.
dx
(20.27)
To characterize the solution completely we need a boundary condition, and
this is supplied by the solution for x > h. Because we wish n(x) to be
continuous, and because n(x) = 0 for x > h, it follows that we just take
n(h) = 0
(20.28)
as our boundary condition.
This differential equation can be solved using an integrating factor to get
fmax λ
x−h
fmax (x − h)
n(x) =
exp
− exp
.
(20.29)
fmax λ − v
λ
v
487
To find p(v), we now substitute the expression for r(x) into the formula
∞
p = ρ
r(x)n(x) dx
−∞
h
= ρ
r(x)n(x) dx
−∞
=
=
h ρrmax λfmax
e(x−h)/λ − α e(x−h)/λ − efmax (x−h)/v
(1 − α)(fmax λ − v) −∞
1 λrmax ρ 2λfmax α − fmax λ + 2αv
.
(20.30)
2 α−1
fmax λ + v
(To do this last step we used Maple. It is straightforward to do the integral
directly as it involves only integrating exponential functions, but the details
are messy and not particularly enlightening.)
It helps to write (20.30) in a slightly different form. At v = 0 we get
p0 = p(0) =
λrmax ρ 2α − 1
·
.
2
α−1
and thus, substituting into (20.30) we get
2α
fmax λ + ( 2α−1
)v
p(v) = p0
.
fmax λ + v
(20.31)
(20.32)
It remains only to derive constraints on α so that our solution has the correct
form. First, notice that the Hill equation
v=
b(p0 − p)
p+a
(20.33)
implies that
b − (a/p0 )v
.
v+b
Thus, for (20.32) to have the same form as (20.34) we need
p = p0
2α
< 0.
2α − 1
(20.34)
(20.35)
Furthermore, in order for p0 to be greater than zero, we also need
2α − 1
> 0.
α−1
(20.36)
Fortunately, both these constraints can be satisfied by setting 0 < α < 1/2.
In summary therefore, when 0 < α < 1/2 we have shown that this choice of
f and g satisfies Hill’s equation exactly, with
b
= fmax λ
2α
a =
.
1 − 2α
and
(20.37)
(20.38)
488
CHAPTER 20. MUSCLE: SOLUTIONS
Ex. 6: Assuming that the crossbridges act as linear springs, calculate the force–
velocity curve for (10.38) and show that in the limit as → 0 and δ → 0
it does not produce zero force for some positive velocity. Give an intuitive
explanation for this.
Solution
Since the crossbridges act as linear springs, we have
for some constant k. Hence
p(v) = kρ
r(x) = kx,
(20.39)
xn(x, v) dx
(20.40)
∞
−∞
h
= kρ
α(x − h)
dx
x 1 − exp
v
h−
h−
+ kρ
x[1 − exp(−α/v)] dx.
(20.41)
(20.42)
δ
Since we certainly wish to avoid calculating these integrals (although that is
easily done using Maple), we just take the desired limits immediately. In the
limit as → 0 and δ → 0 the first integral disappears and we get
lim
→0,δ→0
p(v)
= ρk
h
x(1 − e−α/v ) dx
(20.43)
0
2
=
ρkh 1 − e−α/v .
2
(20.44)
From this we see that p(v) is nonzero for all v < ∞, i.e., no matter how fast
the muscle is contracting, it always exerts a force. The explanation for this can
be readily seen if we first think about why, in the usual Huxley model, there is
some finite velocity at which the muscle generates zero force. This is because,
as v increases, more crossbridges are carried into the region x < 0 where
they exert a negative force. Once v gets high enough, so many crossbridges
are carried into the x < 0 region that the negative force they exert exactly
balances the positive force exerted by all the crossbridges that are bound with
x > 0.
However, for the choice of f and g used here, in the limit as and δ tend to
zero, all the crossbridges detach at x = 0 and thus never exert a negative force.
Thus there can never be a balance between the positive and negative forces,
and the force exerted by the muscle can never be zero for any contraction
velocity.
489
Ex. 7: This exercise and the next are based on the discrete binding site models of
T.L. Hill (1974,1975). Suppose that each crossbridge is within reach of no
more than two binding sites at one time, and that adjacent crossbridges do
not “see” the same two binding sites. Suppose also that adjacent binding sites
are separated by a distance ∆, and let x denote the distance of the crossbridge
from one of the binding sites, binding site 0, say. Define x such that if the
crossbridge is bound to site 0 and has x = 0, it exerts no force. Also, let
ni (x, t), i = 0, −1, denote the fraction of crossbridges with displacement x
that are bound to binding site i.
(a) Show that the conservation equations are
dn0
= f (x)[1 − n0 (x) − n−1 (x)] − g(x)n0 (x),
(20.45)
dx
dn−1
−v
(20.46)
= f (x − ∆)[1 − n0 (x) − n−1 (x)] − g(x − ∆)n−1 (x),
dx
where as usual, v denotes the steady contraction velocity.
(b) Derive expressions for the isometric distributions of n0 and n−1 . Compute the isometric force. Show that if the Huxley model is modified to
include two binding sites, the isometric force is increased.
(c) Compute the force–velocity curve. Hint: For each v solve the differential
equations numerically, using the boundary conditions n0 (h) = 0, n−1 (h+
∆) = 0, then substitute the result into the expression for the force and
integrate numerically.
(d) Modify the model to include slippage of the crossbridge from one binding
site to another. Show that in the limit as slippage becomes very fast, the
two differential equations (20.45) and (20.46) reduce to a single equation.
−v
Solution
Part (a)
In Fig. 20.3 we give a schematic diagram of the crossbridge and the two binding
sites. From there we see that, if a crossbridge is distance x away from site
0, it must be distance x − ∆ away from site -1. Thus, the rate at which it
binds and unbinds to site -1 must be, respectively, f (x − ∆) and g(x − ∆).
Furthermore, the fraction of crossbridges that are unbound to either site, and
thus available for binding, is 1 − n0 − n−1 . Hence, at steady state (so that
∂n0 /∂t = ∂n−1 /∂t = 0) we have
dn0
(20.47)
= f (x)[1 − n0 (x) − n−1 (x)] − g(x)n0 (x),
dx
dn−1
−v
= f (x − ∆)[1 − n0 (x) − n−1 (x)] − g(x − ∆)n−1 (x),(20.48)
dx
as required. We emphasise that n−1 (x) is the proportion of crossbridges that
are distance x from site 0, but are bound to site -1. These crossbridges are
thus extended only a distance x − ∆.
−v
490
CHAPTER 20. MUSCLE: SOLUTIONS
Figure 20.3: Schematic diagram of the crossbridge model with two actin binding
sites.
Part (b)
The isometric case is when v = 0, in which case we get
f (x)[1 − n0 (x) − n−1 (x)]
f (x − ∆)[1 − n0 (x) − n−1 (x)]
= g(x)n0 (x)
= g(x − ∆)n−1 (x).
(20.49)
(20.50)
Solving for n0 and n−1 gives
n0 (x)
=
n−1 (x)
=
f (x)
f (x−∆)
g(x−∆) g(x)
,
f (x − ∆)
f (x)
g(x) g(x
− ∆) + f (x − ∆) + g(x − ∆)
To calculate the isometric force we need to calculate the integral
∞
∞
p=ρ
r(x)n0 (x) dx + ρ
r(x − ∆)n−1 (x) dx.
−∞
(20.51)
+ f (x) + g(x)
(20.52)
(20.53)
−∞
Note that the second integral on the right hand side using the argument x − ∆
in the force function. That is because a crossbridge that is distance x away
from site 0, but is attached to site −1, is only extended a distance x − ∆.
These isometric distributions and the isometric force can be calculated explicitly for the Huxley model, i.e., using

x < 0,
 0,
f1 x/h,
0 < x < h,
f (x) =
(20.54)

0,
x > h,
g2 ,
x < 0,
g(x) =
(20.55)
g1 x/h,
x > 0.
r(x)
= kx.
(20.56)
Plots of n0 (x) and n−1 (x) are given in Fig. 20.4. We see that n0 and n−1 are
piecewise constant, and that only n−1 is nonzero for any x > h. A moment’s
thought explains why this is so. If x > h then the crossbridge cannot bind to
site 0 as it is too far away, but it can still bind to site −1 as long as x < h + ∆.
Because the isometric distributions are piecewise constant, to evaluate the
force it is easiest to break the integrals up into regions in which n0 and n−1
are piecewise constant. For the first integral we get
∆
∞
h
xf (x)
xf (x)
r(x)n0 (x) dx = k
dx
dx+k
f (x−∆)
f
(x)
+
g(x)
−∞
0
∆ g(x−∆) g(x) + f (x) + g(x)
(20.57)
491
0.5
n0
n-1
0.4
0.3
0.2
0.1
0.0
-0.5
0.0
0.5
x
1.0
1.5
Figure 20.4: Isometric distributions of n0 and n−1 . Parameters are h = 1, ∆ = 0.1,
f1 = 0.25, g1 = 0.25, g2 = 2.
while the second gives
∞
r(x − ∆)n−1 (x) dx
−∞
∆
(x − ∆)f (x − ∆)
h
= k
f (x)
g(x) g(x
h+∆
+ k
h
(20.58)
dx
− ∆) + f (x − ∆) + g(x − ∆)
(x − ∆)f (x − ∆)
dx.
f (x − ∆) + g(x − ∆)
(20.59)
Using Maple to evaluate these integrals and add them (the Maple code is
given below) gives
ptwo
sites
= ρk
f1 h(hg1 + hf1 + f1 ∆)
.
(g1 + f1 )(2f1 + g1 )
(20.60)
This is the desired expression for the isometric force.
Finally, we compare this expression to the expression calculated for the Huxley
model with a single binding site. When there is only a single binding site, we
have
h
f (x)
ρk
f1 h2
pone site = ρk
.
(20.61)
dx =
·
2 f1 + g1
0 f (x) + g(x)
Subtracting these expressions give
ptwo
sites
− pone
site
=
ρk f1 h(hg1 + 2f1 ∆)
·
> 0.
2 (g1 + f1 )(2f1 + g1 )
(20.62)
Hence it follows that the isometric force is greater when there are two binding
sites.
492
CHAPTER 20. MUSCLE: SOLUTIONS
Here is the Maple code we used to do these calculations. As usual, we make no claims that
this is the most efficient possible Maple code, but it does at least work.
# fm and gm denote the shifted functions
# The other notation should be self-explanatory.
f := f1*x/h: fm := f1*(x-Delta)/h:
g:=g1*x/h: gm := g1*(x-Delta)/h:
n0low := f/(g + f):
n0high := f/( fm*g/gm + f + g):
nm1low := fm/( f*gm/g + fm+gm):
nm1high := fm/(fm+gm):
# Calculate the force generated by the $n_0$ crossbridges
n0int := k*int(x*n0low,x=0..Delta) + k*int(x*n0high,x=Delta..h):
simplify(n0int);
# Calculate the force generated by the $n_{-1}$ crossbridges
nm1int := k*int((x-Delta)*nm1low,x=Delta..h) + k*int((x-Delta)*nm1high,x=h..h+Delta):
simplify(nm1int);
# Now add the two integrals together to get the total force.
dum1:=simplify(n0int+nm1int);
# Now calculate the isometric force of the Huxley model with a single binding site
dum2:=k*int(x*n0low,x=0..h);
#Subtract the two forces to see which is greater
dum1-dum2;
Part (c)
As it says in the hint, this part of the question must be done numerically.
For each v we have to solve the differential equations numerically, and then
integrate numerically to find the force.
There are a number of ways such numerical computations can be performed.
Maple has built-in ordinary differential equation solvers, as have Mathematica
and Matlab. However, once this level of numerical complexity is reached, it
is often easier in the long run to write your own code in a language such as
C or Fortran. The problem with programs such as Maple is that they are
rather a black box, and when they go wrong, or give funny answers, it can be
very difficult to find out where exactly the problem lies. If you write the code
yourself, at least you know exactly what it is doing, and why.
Hence we have chosen to write a program in Fortran (the full code is given
below). For ease of programming, the differential equations are solved using
the simple forward Euler method, and stability is ensured by using a small
space step. The forward Euler algorithm has many disadvantages, but for a
493
small problem like this it is quick and easy to program, and quite sufficient
for the task at hand.
We discretize space in increments of dx. We take our initial condition to
n0 (h + ∆) = n−1 (h + ∆) = 0, and then take steps of −dx until we reach
x = −2, which seems (by trial and error) to be far enough.
At the ith space step (so that xi = h + ∆ − idx), we have
n0 (xi+1 )
n−1 (xi+1 )
∆x (20.63)
f (xi )(1 − n0 (xi ) − n−1 (xi )) − g(xi )n0 (x
i) ,
v
∆x f (xi − ∆)(1 − n0 (xi ) − n−1 (xi ))
= n−1 (xi ) +
(20.64)
v
= n0 (xi ) +
− g(xi − ∆)n−1 (xi ) .
(20.65)
Once n0 and n−1 are obtained the force can be calculated from the equation
∞
∞
p(v) = ρk
xn0 (x) dx + ρk
(x − ∆)n−1 (x) dx.
(20.66)
−∞
−∞
To evaluate these integrals numerically we use the trapezoidal rule. If a function θ(x), say, is known at n equally spaced points xi , i = 1, n, then
xn
θ(x1 )
θ(xn )
+ θ(x2 ) + · · · + θ(xn−1 ) +
,
(20.67)
θ(x) dx ≈ dx
2
2
x1
where dx is the spacing between the points.
Results are shown in Figs. 20.5 and 20.6. Fig. 20.5 shows the solutions for n0
and n−1 , while Fig. 20.6 shows the computed force-velocity curve.
Here is the complete Fortran code used to obtain these solutions.
use globalparams
double precision :: force
integer :: i
k=1.0d0; rho=1.0d0; h=1.0d0; delta=0.1d0; f1=0.25d0; g1=0.25d0; g2=2.0d0
delx = 0.001d0
lower = -2.0d0
numsteps=nint((h+delta - lower)/delx)
!
!
!
!
!
! Choose a nice small number for stability
! The lower bound for the integration.
! Solve from x=lower to x=h+delta
The solution for n$_0$ is stored in the array n0.
The solution for n$_{-1}$ is stored in the array nm1.
We keep the solutions in arrays (instead of just printing out each value to a file
as we compute it) because we need to use n0 and nm1 later in the integration
subroutine
allocate(n0(numsteps))
! It’s most efficient to use allocatable arrays
allocate(nm1(numsteps))
open(1,file=’c11q7.out’)
! Put the solution into this file
494
CHAPTER 20. MUSCLE: SOLUTIONS
0.4
v=0.01
0.3
n0
n-1
0.2
0.1
0.0
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
-0.5
0.0
0.5
1.0
-0.5
0.0
0.5
1.0
0.35
0.30
v=0.1
0.25
0.20
0.15
0.10
0.05
0.00
-2.0
-1.5
0.15
-1.0
v=0.5
0.10
0.05
0.00
-2.0
-1.5
-1.0
x
Figure 20.5: Computed solutions of the two binding site model for three different
values of v. Note that as v gets smaller the solutions resemble the piecewise constant
isometric solutions. Parameters are h = 1, ∆ = 0.1, f1 = 0.25, g1 = 0.25, g2 = 2.
495
0.30
0.25
0.20
p
0.15
0.10
0.05
0.00
0.0
0.2
0.4
0.6
0.8
1.0
v
Figure 20.6: The computed force-velocity curve for the two-site model. Parameters
are k = 1, ρ = 1, h = 1, ∆ = 0.1, f1 = 0.25, g1 = 0.25, g2 = 2.
496
CHAPTER 20. MUSCLE: SOLUTIONS
do i=1,100
v = i*0.01d0
call solve_for_n
call integrate(force)
print*,v,force
write(1,*) v,force
end do
close(1)
end
!------------------------! We solve the differential equations using the simplest, quickest method, the
! forward Euler method. It’s not a great method, but good enough for rough work
! like this, and it’s easy to program.
subroutine solve_for_n
use globalparams
integer :: i
double precision :: x
n0(1)=0.0d0; nm1(1)=0.0d0
! Initial conditions
do i=1,numsteps-1
x = h+delta - i*delx
n0(i+1) = n0(i) + (delx/v)*( f(x)*(1.0d0-n0(i)-nm1(i)) - g(x)*n0(i) )
nm1(i+1) = nm1(i) + (delx/v)*( f(x-delta)*(1.0d0-n0(i)-nm1(i)) &
- g(x-delta)*nm1(i) )
end do
end
!------------------------! Integrate r(x)n0(x) and r(x-Delta)nm1(x) using the trapezoidal rule. Note the
! shift in the second integral (i.e., in sum2) where we use x2 in place of x1
subroutine integrate(force)
use globalparams
integer :: i
double precision :: force,x1,x2
sum1 = (h+delta)*n0(1)/2.0d0
sum2 = h*nm1(1)/2.0d0
do i=2,numsteps-1
x1 = h+delta - float(i)*delx
x2 = h - i*delx
! Shifted by delta: we need r(x-delta)*nm1(x)
sum1 = sum1 + x1*n0(i)
sum2 = sum2 + x2*nm1(i)
end do
sum1 = delx*k*(sum1 + lower*n0(numsteps)/2.0d0)
sum2 = delx*k*(sum2 + (lower-delta)*nm1(numsteps)/2.0d0)
force = sum1+sum2
end
! -----------------------function f(x)
use globalparams
double precision :: f,x
if (x<0) f=0
if (x>h) f=0
! Define the function f(x)
497
if (x<=h .and. x>=0) f=f1*x/h
end
!-------------------------function g(x)
use globalparams
double precision :: g,x
! Define the function g(x)
if (x<0) g=g2
if (x>=0) g=g1*x/h
end
!-------------------------module globalparams
implicit none
! Always a good idea to specify no implicit typing
double precision :: f1,g1,g2,h,delta,k,rho,v,delx,lower
double precision,dimension(:),allocatable :: n0,nm1
integer :: numsteps
end module globalparams
!-------------------------
Part (d)
To include a fast slippage of the crossbridge from one binding site to another,
we assume that the transitions
k1
n−1
−→
←−
n0
(20.68)
k−1
occur rapidly. Thus,
k1 n−1 = k−1 n0 ,
and so
n−1 = Kn0 ,
where K =
(20.69)
k−1
.
k1
(20.70)
Next we add the equations (20.47) and (20.48) to get
∂
n(x) = [f (x) + f (x − ∆)](1 − n(x)) − [g(x)n0 (x) + g(x − ∆)n−1 (x)],
∂x
(20.71)
where we have defined n(x) = n0 (x) + n−1 (x). Noting that n = (1 + K)n0 =
(1 + 1/K)n−1 then gives
∂
g(x) + Kg(x − ∆)
−v n(x) = [f (x) + f (x − ∆)](1 − n(x)) −
n(x).
∂x
1+K
(20.72)
Thus, the original two equations in n0 and n−1 condense down to a single
equation in n.
−v
498
CHAPTER 20. MUSCLE: SOLUTIONS
Ex. 8: Consider the general binding site model (10.49) that incorporates the discrete
distribution of binding sites. Why is this equation much harder to integrate
than the models we have discussed previously? The isometric solution is
∞
considerably easier to calculate. Let nu (x) = 1 − −∞ n(x + i∆). Show that
∞
1 − nu (x) =
1+
f (x+i∆)
−∞ g(x+i∆)
∞ f (x+i∆)
−∞ g(x+i∆)
.
(20.73)
Hence calculate the isometric solution n(x).
Solution
The difficulty in integrating the general binding site model arises from how to
cope with the boundary conditions. In the previous question we could specify
the boundary conditions n0 (h + ∆) = n−1 (h + ∆) = 0, and this was enough
to let us compute the solution numerically, i.e., we just started at x = h + ∆,
secure in the knowledge that we knew the solution there, and then decreased
x in jumps of dx, calculating the solution as we go.
But the steady state solution of (10.49) cannot be calculated this way, as
dn/dx depends on the solution along the entire range of x values. The only
realistic boundary conditions are to assume that n → 0 as x → ±∞, and
these give no way of “starting” the solution for n. One possible way around
this problem is to pick a starting distribution for n(x), and then step the
full partial differential equation (10.49) forward in time until a steady state is
reached. The techniques for doing this numerically are beyond the scope of
this book.
To calculate the isometric solution is easier. First, for simplicity let ni =
n(x + i∆), and similarly for fi and gi . Then, the isometric steady solution is
given by
i=∞
f (x) 1 −
(20.74)
ni = g(x)n(x).
i=−∞
But since
i=∞
n(x + i∆) =
i=−∞
i=∞
n(x + (i + j)∆),
(20.75)
i=−∞
for any integer j, substituting x + j∆ for x in (20.74), it follows that
i=∞
f (x + j∆) 1 −
ni
= g(x + j∆)n(x + j∆),
(20.76)
i=−∞
for any integer j, and thus
nj =
fj
(1 −
ni ).
gj
i
(20.77)
499
Now just sum (20.77) from j = −∞ to j = ∞ to get
nj = (1 −
j
and thus (since
j
nj =
ni )
i
i
fj
gj
j
,
(20.78)
ni )
nj =
j
fj
j gj
1+
fj
j gj
,
(20.79)
as required.
To get the isometric solution we substitute (20.79) into (20.74) to get
n(x) =
f (x)
1
·
g(x) 1 + j
fj
gj
.
(20.80)
Ex. 9: A muscle fiber must be able to produce a force even at negative velocities. For
example, if you slowly lower a brick onto a table, your bicep is extending and
simultaneously resisting the freefall of the brick. Investigate the behaviour of
the Huxley model when v < 0.
(a) Plot n(x) for v = −0.01, −0.1 and −1.
(b) Derive an expression for the maximum force as v → −∞.
Solution
For convenience, we let v = −u, where u > 0. Then, to find n(x), we need to
solve the differential equation
u
dn
= f (x)(1 − n) − g(x)n,
dx
where f and g are as given for the Huxley model, i.e.,

x < 0,
 0,
f1 x/h,
0 < x < h,
f (x) =

0,
x > h,
g2 ,
x < 0,
g(x) =
g1 x/h,
x > 0.
r(x)
= kx.
(20.81)
(20.82)
(20.83)
(20.84)
500
CHAPTER 20. MUSCLE: SOLUTIONS
Part (a)
Since the muscle is now stretching we know that n(x) = 0 for all x < 0. We
thus have to solve for n only on the two regions 0 ≤ x ≤ h and h ≤ x. Let n1
denote the solution on 0 ≤ x ≤ h, and let n2 denote the solution on h ≤ x.
When 0 ≤ x ≤ h we have
u
dn1
f1 x
=
−
dx
h
(f1 + g1 )x
h
n1 ,
n1 (0) = 0,
(20.85)
which has solution
f1
f1 + g1
n1 (x) =
1−e
−x2 (f1 +g1 )
2uh
.
(20.86)
On the region h ≤ x we have
u
dn2
g1 x
=−
n2 ,
dx
h
(20.87)
which has solution
n2 (x) = Ce
−g1 x2
2hu
,
(20.88)
where C is a constant to be determined from the continuity condition
n1 (h) = n2 (h).
(20.89)
Applying this condition gives
C=
and hence
n2 (x) =
−hf1
f1 hg1
e 2u − e 2u ,
f1 + g1
(20.90)
−g1 x2
−hf1
f1 hg1
e 2u − e 2u e 2hu .
f1 + g1
(20.91)
Plots of n(x), constructed by joining together n1 and n2 , are given in Fig.
20.7 for three different values of v. Clearly, when v is close to 0 the solution is
close to the isometric solution, but as v becomes more negative the crossbridge
distribution is skewed to higher x values.
Part (b)
The force is given by
p = kρ
∞
xn(x) dx = kρ
0
h
∞
x n1 (x) dx + kρ
0
x n2 (x) dx
h
(20.92)
501
0.5
v = -0.01
v = -0.1
v = -1
0.4
0.3
n
0.2
0.1
0.0
0.0
0.5
1.0
x
1.5
2.0
Figure 20.7: Plots of n(x) for three negative values of velocity, i.e., as the muscle
lengthens.
1.0
0.5
0.0
v
-0.5
-1.0
-1.5
-2.0
0.0
0.1
0.2
0.3
0.4
p
Figure 20.8: The force-velocity curve for positive and negative velocities.
502
CHAPTER 20. MUSCLE: SOLUTIONS
An integral like this is most easily done with a symbolic manipulator such as
Maple, which gives the answer
p
=
−h(f1 +g1 )
hkf1 ρ
+ g1 f1 h + hg12 + 2uf1
2g1 ue 2u
2g1 (f1 + g1 )2
−(f1 +g1 )
−(f1 +g1 ) − 2uf1 e 2hu − 2ug1 e 2hu
(20.93)
(20.94)
Since
lim 2g1 u e
u→∞
and
−h(f1 +g1 )
2u
−e
−(f1 +g1 )
2hu
= g1 (f1 + g1 )
1
−h ,
h
f (f + g )
−(f1 +g1 )
1 1
1
lim 2f1 u 1 − e 2hu
.
=
u→∞
h
(20.95)
(20.96)
it follows that
lim p
u→∞
=
=
hkf1 ρ
1
f1 (f1 + g1 )
2
g1 (f1 + g1 )
(20.97)
hg1
−h +
+ g1 hf1 +
2g1 (f1 + g1 )2
h
h
1 kf1 ρ
.
(20.98)
2 g1
Chapter 21
Vision: Solutions
Ex. 1: In this question we study some of the solutions of the model of light adaptation
in turtle cones (Section 11.2.2).
(a) Before we can calculate solutions of the model the expression for g(y)
must be derived. By forcing the steady state of the model to be
V0 = −s1 log(1 + s2 I0 ),
show that
E(1−y)/V ∗
g(y) = ye
1/3 (γ − δ)η(e−E(1−y)/s1 − 1)
δ+
s2 k1 + η(e−E(1−y)/s1 − 1)
(21.1)
.
(21.2)
Plot g(y) (using the parameters given in Table 21.1) and compare to the
functional form found experimentally (11.44), i.e.,
A(y) = 4 +
84
.
1 + (y/0.34)4
(21.3)
(b) Calculate the impulse responses of the model for a range of background
light levels. For each background light level of I0 , let the magnitude of
the impulse be I0 also. Do this for I0 = 0.5, 0.05, 0.005, 0.0005 and
0.00005. How linear are the responses? What can be said about the
contrast sensitivity?
(c) Scale each of the impulse responses calculated above so that each has a
maximum value of 1, and plot them all on the same graph. How do the
impulse responses change as the light level increases?
(d) Replace g(y) by A(y) and recalculate the steady state V0 and the impulse
responses (at the same background light levels as in part (b) above). Are
there any significant differences?
Solution
503
504
CHAPTER 21. VISION: SOLUTIONS
Part (a)
Let a subscript 0 denote the steady state of the model. Then, by setting the
right hand sides of (11.38)–(11.42) to zero, we get
ηI0
,
k1 + ηI0
p0
=
y0
= x30 e−V0 /V ,
g(y0 )
z0
V −E
(21.4)
∗
(21.5)
= x0 (δ + (γ − δ)p0 ),
(21.6)
∗
1−κ
2κ
x30 e−V0 /V +
y0 ,
=
(21.7)
1+κ
1+κ
∗
−6E
1+κ
3Eκ
=
x30 e−V0 /V + 2
Ez0 −
y0 .(21.8)
4+κ
4+κ
4+κ
Substituting (21.5) into (21.7) gives
z0 = y0 ,
(21.9)
V0 = E(1 − y0 ).
(21.10)
and thus, from (21.8), that
Furthermore, from (21.5) it follows that
x0
=
=
y0 eV0 /V
∗
13
y0 eE(1−y0 )/V
∗
13
.
(21.11)
We now have x0 , z0 and V0 expressed in terms of y0 . It remains only to express
p0 in terms of y0 and we are done.
Since V0 = −s1 log(1 + s2 I0 ) it follows that
I0 =
and thus
p0
e−V0 /s1 − 1
,
s2
=
η e−V0 /s1 − 1
,
k1 s2 + η e−V0 /s1 − 1
=
η(e−E(1−y0 )/s1 − 1)
.
s2 k1 + η(e−E(1−y0 )/s1 − 1)
(21.12)
(21.13)
Substituting (21.11) and (21.13) into (21.6) then gives the required result.
In Fig. 21.1 we show plots of g(y) and A(y). Clearly the two functions agree
very well over the plotted range. (For smaller values of y the two functions
differ more, but such low values of y are unphysiological and thus we pay them
little attention).
505
80
g(y)
A(y)
60
40
20
0
0.2
0.4
0.6
0.8
1.0
y
Figure 21.1: Plots of g(y) and A(y), using the parameters given in Table 21.1.
Part (b)
This question involves the numerical solution of the model equations (11.38)
– (11.42). One could do this a number of ways; with Maple for instance, or by
writing a program in Fortran or C. Here, the differential equations contain no
surprises so we choose to use the program xppaut (see Appendix ??) because
of its convenience and ease of use.
In Fig. 21.2 we see responses of the model to flashes of fixed contrast superimposed on various background light levels. All the responses were calculated by
numerical solution of (11.38)–(11.42), using the .ode file given in full below.
To facilitate comparison of the curves, each curve was shifted vertically until
its steady state voltage coincided with 0. In other words, these curves are
really just plotting deviations from the steady-state voltage. To simulate a 5
ms light flash of strength I0 superimposed on a background light level of I0
we let
I(t) = I0 + I0 δ(t)/200.
(21.14)
Note that the factor of 200 is needed because if we approximate an impulse
function by using a square pulse of width 0.005 seconds, then that square
pulse has to be 200 units high. Thus, dividing by 200 gives us a brief flash of
light, of width 5 ms, and of unit height.
Each light flash had the same contrast, and thus if the photoreceptor had
constant contrast sensitivity the amplitude of all the responses would be the
same. However, we see from the figure that this is not exactly true. Instead,
the contrast sensitivity first increases with increasing background light level,
and then decreases again. For a background light level between 0.5 and 0.05,
the contrast sensitivity is approximately constant.
506
CHAPTER 21. VISION: SOLUTIONS
0.1
0.0
V (mV)
4
3
-0.1
1
-0.2
2
0.0
0.1
0.2
0.3
0.4
0.5
t (s)
Figure 21.2: Responses of the model of light adaptation in turtle cones, using the
parameters given in Table 21.1. For each curve a light flash of length 5 ms and
magnitude I0 was superimposed on a background of I0 (so that each stimulus had
the same contrast). The values for I0 were (1) 0.5, (2) 0.05, (3) 0.005 and (4) 0.0005.
So that we can plot the curves together, each curve is shifted vertically so that the
steady state voltage is 0.
507
0.05
0.00
V (mV)
-0.05
-0.10
-0.15
-0.20
0.0
0.1
0.2
0.3
0.4
0.5
t (s)
Figure 21.3: Testing the linearity of the flash responses. The solid curve is the
same as curve 1 of Fig. 21.2, and is the response to a brief flash of light on top of
a background light level of 0.5. The dotted curve is the result of doubling the flash
strength and dividing the response by two.
To see if the responses are linear we double the stimulus strength (i.e., just
put contrast=2 in the .ode file below) and divide the response by two. If
the response is linear the result will exactly match the contrast=1 case. For
the lowest three background light levels this is the case, as the curves are
essentially identical to the naked eye. Fig. 21.3 shows the solutions for I0 =
0.5; the dotted curve was the one obtained by doubling the stimulus strength
and then dividing the response by two. As can be seen from the figure,
although the responses are almost exactly linear, there is a slight decrease of
the amplitude.
Part (c)
In Fig. 21.4 we show the same flash responses as in Fig. 21.2 normalised to a
maximum amplitude of 1. What this figure illustrates is the speed-up of the
kinetics of the flash responses. As the background light level increases, the
flash responses reach their peak sooner, and develop oscillating tails, i.e., become biphasic. This (by the way) is in excellent agreement with experimental
data.
Part (d)
When g(y) is replaced by A(y) the flash responses are essentially the same
(Fig.21.5). There are, however, minor differences. For instance, we no longer
508
CHAPTER 21. VISION: SOLUTIONS
0.4
2
1
0.2
V/Vmax
0.0
-0.2
3
4
-0.4
-0.6
-0.8
-1.0
0.0
0.1
0.2
0.3
0.4
0.5
t (s)
Figure 21.4: The same flash responses as in Fig. 21.2, normalised to a maximum
amplitude of 1.
have V0 = −s1 ln(1 + s2 I0 ) and thus the variable Vshift in the .ode file below
is no longer of any use. Nevertheless, the steady state is easily calculated
and then subtracted from the responses to shift them to the origin. We did
this, for example, in the plotting program we used to display the results (Igor
Pro; see Appendix ??). It would be a simple matter to automate all this by
writing a Fortran program, but the time saved is probably not worth the time
invested.
It is a slightly more involved matter to calcuate V0 as a function of I0 , as no
analytical expression is readily obtainable. One option is to write a program
that uses a nonlinear solver such as Newton’s method to solve for V0 given I0 .
However, xppaut has the ability to do this quickly and easily as it provides a
friendly interface to AUTO, a program designed to track steady states and
bifurcations (see Appendix ??). The .ode file given below shows how to set
the options for the AUTO interface, and the options that we used for this
question. The result is show in Fig. 21.6. The steady state curves for the two
different cases (using g(y) or A(y)) cannot be told apart in this graph.
The solutions above were calculated using the following XPPAUT file.
#
# contrast=0 gives steady-states.
# contrast=1 gives a stimulus of size I0 on a background of I0.
#
par I0=0.05 contrast=1
par s1=1.59 s2=1130 E=-13 Vstar=35.7
par tauy=0.07 k1=35.4 gam=303 delta=5 kappa=0.1
509
0.1
V (mV)
0.0
4
3
-0.1
1
-0.2
0.0
2
0.1
0.2
0.3
0.4
0.5
t (s)
Figure 21.5: Flash responses of the model of light adaptation in turtle cones, using
A(y) instead of g(y). Everything else was done exactly the same as in Fig. 21.2.
0
-2
V0 (mV)
-4
-6
-8
-10
0.0
0.2
0.4
I0
0.6
0.8
1.0
Figure 21.6: Steady state curves V0 (I0 ) of the model of light adaptation in turtle
cones: solid curve, using g(y); dotted curve, using A(y). The curves are indistinguishable at this resolution.
510
CHAPTER 21. VISION: SOLUTIONS
par eta=52.5 tau1=0.012 taum=0.016 tauz=0.04
#
!V0=-s1*ln(1+s2*I0)
#
kernel(t)=(eta/tau1/6)*((t/tau1)^3)*exp(-t/tau1)
s(t)=eta*I0 + contrast*I0*kernel(t)/200
p’=s(t)*(1-p)-k1*p
x’=gfun(y)-(gam-delta)*x*p-delta*x
y’=((x^3)*exp(-V/Vstar)-y)/tauy
z’=(((1-kappa)/(1+kappa))*(x^3)*exp(-V/Vstar)\
+(2*kappa/(1+kappa))*y-z)/tauz
V’=((-6*E/(4+kappa))*(x^3)*exp(-V/Vstar)+\
(2*(1+kappa)/(4+kappa))*E*z-(V-E)-\
(3*E*kappa/(4+kappa))*y)/taum
#
gfun(y)=(y*exp(E*(1-y)/Vstar))^(1/3)*\
(delta+(gam-delta)*eta*\
(exp(-E*(1-y)/s1)-1)/s2/\
(k1+eta*(exp(-E*(1-y)/s1)-1)/s2))
#
# Here’s A(y) (still called gfun just to make it easy).
# For part (d) use this gfun instead of the gfun above.
# gfun(y)=4+84/(1+(y/0.34)^4)
#
# Can’t put in the actual steady states as functions of I0
# so we just put in the dark steady state.
p(0)=0
x(0)=1
y(0)=1
z(0)=1
V(0)=0
#
# We choose to plot out the shifted voltages
# (which is why we bother to calculate V0 at all)
aux Vshift=V-V0
@ yp=Vshift total=2.5 xlo=0 xhi=0.5 ylo=-0.5 yhi=0.1 dt=0.001
@ ntst=50 nmax=2000 npr=2000 dsmin=0.0001 dsmax=0.01 ds=0.001 parmin=0.00001
@ parmax=1 autoxmin=0.00001 autoxmax=1 autoymin=-12 autoymax=0 autovar=v
done
One thing that we should note is that XPPAUT does not handle the initial
conditions very elegantly, as they can’t be specified as functions of I0 . Thus,
every time I0 is changed during calculation of the curves the steady state
must be recomputed (because it is needed to serve as the initial condition).
This is most easily done by running up to t = 5, say, and then using the last
computed point as the initial condition. The AUTO options are set on the
last two lines and are used only for the calculation of V0 (I0 ) when A(y) is
used in place of g(y).
Ex. 2: In 1989, Forti et al. published a model of phototransduction in newt rods.
In many respects their model is very similar to the model discussed here.
However, one major difference is in how they chose to model the initial stages
511
T0
α1
τ1
β1
=
=
=
=
1000 µM
20 s−1
0.1 s−1 µM−1
10.6 s−1
P0
τ2
=
=
=
100 µM
0.5 s−1 µM−1
10 s−1
Table 21.1: Parameter values for the model of the initial stages of the light response
in newt rods (Forti et al., 1989). P0 is the total amount of PDE, and T0 is the total
amount of transducin.
of the light response. Instead of just using a linear filter (which is, to be
sure, a bit of a cheat) they modelled the equations in more detail. Forti et al.
assumed that the initial stages of the light response could be modeled by the
reactions
α1
R
hν
−→
R∗ −→
(21.15)
R∗ + T
−→
R ∗ + T∗
(21.16)
∗
β
−→
T∗ + P
τ1
−→
←−
τ2
T
(21.17)
T∗ + P∗
(21.18)
Here, R denotes rhodopsin, T denotes transducin and P denotes PDE. The
* superscript denotes the activated form, and hν denotes the action of light.
The parameter values of the model are given in Table 21.1. (Actually, these
are not the exact same reactions assumed by Forti et al. as we have omitted
the inactivated rhodopsin state. However, this makes almost no difference to
the results presented here.)
(a) Assuming that rhodopsin in present in unlimited quantities, write down
the differential equations for this system of reactions, and solve numerically. Plot P ∗ (t) for a variety of stimuli, each 10 ms in duration, but
with magnitudes 0.2, 0.5, 1, 2, 5, 10, 20, 50, 100 and 500. What can
you conclude about the linearity of the response when the magnitude of
the stimulus is small? What can you conclude about the linearity of the
response as the magnitude of the stimulus increases?
(b) Compare the model in part (a) with the detailed reaction scheme in
Fig. 11.9. Where do they disagree? Modify the model so that it agrees
better with the more detailed reaction scheme. (Hint: the conservation
equation for transducin must be considered more carefully, as must the
differential equation for T∗ .) Solve the new model numerically for the
same stimuli as part (a), and compare with the solutions of the Forti et
al. model. Explain the differences.
(c) Compare the model of part (a) to the linear cascade model in the book.
Where are they similar? Where are they not similar? Why?
Solution
512
CHAPTER 21. VISION: SOLUTIONS
Part (a)
Denote the light stimulus by I(t), and let R∗ denote the concentration of R∗
with the same convention for P , P ∗ , T and T ∗ . Then we have the conservation
laws
P + P∗
T +T
∗
= P0 ,
(21.19)
= T0 .
(21.20)
Hence,
dR∗
dt
dT ∗
dt
dP ∗
dt
= I(t) − α1 R∗ ,
(21.21)
= R∗ (T0 − T ∗ ) − β1 T ∗ + τ2 P ∗ ,
(21.22)
= τ1 T ∗ (P0 − P ∗ ) − t2 P ∗ .
(21.23)
We solve these equation numerically using xppaut. For each curve the light
stimulus had a duration of 10 ms, and had the magnitude as described in the
caption. Fig. 21.7 shows P ∗ as a function of t for stimuli of seven different
magnitudes. As expected, the responses get larger as the magnitude of the
stimulus increases.
It is more enlightening to look at the scaled responses, rather than the actual
responses, as this shows us the degree of linearity of the response. Fig. 21.8
shows such scaled responses, each scaled to a stimulus strength of 1. In other
words, the solution corresponding to a stimulus strength of 0.2 was scaled
by a factor of 5, the solution corresponding to a stimulus strength of 2 was
scaled by a factor of 1/2 and so on. If the responses were exactly linear, the
scaled responses would be identical. From Fig. 21.8 we see this is true for the
first three (stimulus strengths of 0.2, 1 and 2). However, at the remaining
stimulus strengths (10, 20 100 and 500) the magnitude of the response is
diminished, and at the highest stimulus strengths the kinetics of the response
are considerably faster also. Hence, linearity is lost at these high stimulus
strengths.
To solve the equations using XPPAUT we used the Runge-Kutta option with a step size of
0.001.
# Declare the variables
r(0)=0
g(0)=0
p(0)=0
# Declare the parameters
par str=0.2 alpha1=20 eps=0.5 T0=1000 P0=100 beta1=10.6
par tau1=0.1 tau2=10
#
# Define right hand sides
r’=str*(heav(t)-heav(t-0.01))-alpha1*r
513
30
25
20
P*(t)
500
15
10
100
5
20
0
0
1
2
t
3
4
Figure 21.7: Numerical solutions of (21.21)–(21.23). The light stimulus had a duration of 10 ms, and the magnitude was (from the bottom curve to the top curve) 0.2,
1, 2, 10, 20, 100 and 500 (in units of R∗ s−1 ). So as not to clutter the graph, the
stimulus strength is shown on the three uppermost curves but not on the others.
0.10
P* (scaled)
0.08
0.06
20
100
0.04
500
0.02
0.00
0
1
2
t
3
4
Figure 21.8: Numerical solutions of (21.21)–(21.23), scaled to a stimulus strength
of 1. The light stimulus had a duration of 10 ms, and the magnitude was (from the
top curve to the bottom curve) 0.2, 1, 2, 10, 20, 100 and 500. So as not to clutter
the graph, the stimulus strength is shown on the three lowest curves but not on the
others.
514
CHAPTER 21. VISION: SOLUTIONS
g’=eps*r*(T0-g)-beta1*g+tau2*p
p’=tau1*g*(P0-p)-tau2*p
#
@ total=4, dt=0.001
done
Part (b)
In the more detailed scheme of Fig. 11.9 we see that the Tα subunit of transducin (i.e., the activated form of transducin) is bound to the activated form
of the PDE, and thus (21.23) should really be
τ1
∗
T∗ + P −→
←− P ,
(21.24)
τ2
and the conservation law for transducin should therefore be
T0 = T ∗ + T + P ∗ .
(21.25)
Furthermore, the inactivation of T∗ requires a Tβγ subunit, and so one should
keep track of the Tβγ subunits in order to model the inactivation of transducin.
With these modifications the reaction scheme becomes
α1
R
hν
−→
R∗ −→
(21.26)
R∗ + T
−→
R∗ + T∗ + Tβγ
(21.27)
T∗ + Tβγ
β
−→
T
(21.28)
∗
τ1
−→
←−
τ2
T +P
∗
P ,
(21.29)
with the corresponding differential equations
dR∗
dt
dT ∗
dt
dP ∗
dt
= I(t) − α1 R∗ ,
(21.30)
= R∗ T − β1 T ∗ Tβγ − τ1 T ∗ P + τ2 P ∗ ,
(21.31)
= τ1 T ∗ P − t 2 P ∗ .
(21.32)
The three conservation laws are
T0
Tβγ
P0
= T ∗ + T + P ∗,
(21.33)
= T ∗ + P ∗,
= P + P ∗.
(21.34)
(21.35)
515
4
3
P*(t)
500
2
100
1
20
0
0.0
0.2
0.4
0.6
0.8
1.0
t
Figure 21.9: Numerical solutions of (21.36)–(21.38). The light stimulus had a duration of 10 ms, and the magnitude was (from the bottom curve to the top curve) 0.2,
1, 2, 10, 20, 100 and 500 (in units of R∗ s−1 ). So as not to clutter the graph, the
stimulus strength is shown on the three uppermost curves but not on the others.
Hence the model is
dR∗
dt
dT ∗
dt
dP ∗
dt
= I(t) − α1 R∗ ,
(21.36)
∗
= R∗ (T0 − T ∗ − P ∗ ) − β1 T ∗ (T ∗ + P ∗ ) − τ1 T ∗ (P0 − P ∗ ) + τ(21.37)
2P ,
= τ1 T ∗ (P0 − P ∗ ) − t2 P ∗ .
(21.38)
In Fig. 21.9 we plot numerical solutions of the model equations, calculated
for the same stimuli as in Fig. 21.7. The responses of this second model are
similar to those of the model of Forti et al., except that they are smaller, and
decay more quickly (note the different vertical and horizontal scales).
In Fig. 21.10 we plot the scaled solutions, just as in Fig. 21.8. We see that none
of the responses are linear, even at the lowest stimulus strength, and that the
responses to the stronger stimuli are greatly attenuated by the nonlinearities
in the model.
These results are unsurprising, as the second model contains “stronger” nonlinear terms than does the Forti model. In particular, in the Forti model the
decay of T∗ is always linearly proportional to T ∗ , while in the second model
the decay of T∗ is proportional rather to (T ∗ )2 , as well as to T ∗ P ∗ . Thus,
when T ∗ gets large, it decays much faster in the second model than in the
Forti model. This causes the smaller responses and the faster decay rates.
To solve the equations using XPPAUT we used the Runge-Kutta option with a step size of
0.001.
516
CHAPTER 21. VISION: SOLUTIONS
0.10
0.2
P* (scaled)
0.08
1
0.06
2
0.04
10
20
0.02
0.00
0
1
2
t
3
4
Figure 21.10: Numerical solutions of (21.36)–(21.38), scaled to a stimulus strength
of 1. The light stimulus had a duration of 10 ms, and the magnitude was (from the
top curve to the bottom curve) 0.2, 1, 2, 10, 20, 100 and 500.
# Declare the variables
r(0)=0
g(0)=0
p(0)=0
# Declare the parameters
par str=0.2 alpha1=20 eps=0.5 T0=1000 P0=100 beta1=10.6
par tau1=0.1 tau2=10
#
#
# Define right hand sides
r’=str*(heav(t)-heav(t-0.01))-alpha1*r
g’=eps*r*(T0-g-p)-beta1*g*(g+p)+tau2*p-tau1*g*(P0-p)
p’=tau1*g*(P0-p)-tau2*p
#
@ total=4, dt=0.001, xlo=0, xhi=4
done
Part (c)
Finally, we compare the models in (a) and (b) to the linear cascade model
described in the text, i.e.,
dP ∗
= s(t)(P0 − P ∗ ) − k1 P ∗ ,
dt
where
(21.39)
t
s(t) =
−∞
I(τ )K(t − τ ) dτ.
(21.40)
517
100
P* (scaled)
0.08
500
0.06
0.04
0.02
0.00
0.00
0.05
0.10
t
0.15
0.20
Figure 21.11: Numerical solutions of (21.39), scaled to a stimulus strength of 1.
The light stimulus had a duration of 10 ms, and the magnitude was (from the top
curve to the bottom curve) 0.2, 1, 2, 10, 20, 100 and 500.
We solve (21.39) for the same light inputs as in parts (a) and (b). However,
care is needed, as the function I(t) in the Forti model (having units of R∗ s−1 )
does not have exactly the same meaning as I(t) in the linear cascade model
(which has units of simply R∗ ).
In the linear cascade model the actual light input is l1 I(t), and we do not know
the exact value of l1 . However, we shall just choose l1 so that the response at
the smallest light stimulus has the same magnitude as that of the Forti model,
and then use that value of l1 for all the other responses.
For a light stimulus that is 10 ms in duration with a magnitude of 0.2 R∗ s−1 ,
we choose l1 = 200 s−1 , to get the responses shown in Fig. 21.11. From this
graph we see that the responses are linear at all but the highest two stimulus
strengths, and of very similar magnitude to the responses of the Forti model.
The major difference is in the time scale; the responses of the linear model
are approximately 10 times faster than the responses of the Forti model. This
is as expected from the physiology; the parameters in the Forti model were
chosen to fit data from rods, while the parameters of the linear cascade model
were chosen to fit data from cones. Since the light responses of cones are much
faster than the responses of rods it stands to reason that this difference will
be reflected in the responses of the two models.
In addition to the differences in the speeds, the shapes of the responses differ
as well, with the responses of the Forti model weighted more towards the early
part of the response. The exact reasons for these detailed differences in shape
are not clear, but are obviously a result of the nonlinearities in the initial
stages of the Forti model.
518
CHAPTER 21. VISION: SOLUTIONS
Ex. 3: Show that the current generated by the electrogenic exchange pumps should
not be ignored in the modeling. (Hint: By writing down the balance equations
for Na+ and Ca2+ show that the outward current generated by the electrogenic Na+ –K+ exchanger in the inner segment must be approximately a third
of the total inward light-sensitive current.) How significant is the current
generated by the Na+ –Ca2+ , K+ exchanger in the outer segment? Should
the electrogenic pumps be included in the model if the model is compared to
photocurrent measurements and not to voltage measurements?
Solution
From Fig. 11.7 we see that there are two pumps that need to be considered
in addition to the light-sensitive flux of Na+ and Ca2+ ; the Na+ –Ca2+ , K+
exchanger in the outer segment and the Na+ –K+ exchanger in the inner segment. (As it turns out, we need not consider the flux of K+ in the balance
equations as none of the light-sensitive current is carried by K+ .)
We begin by defining some notation. Let
=
=
flux of Na+ through the light − sensitive channel,
flux of Na+ through the Na+ : Ca2+ , K+ exchanger,
=
flux of Na+ through the Na+ : K+ exchanger,
c1
=
flux of Ca2+ through the light − sensitive channel,
c2
=
flux of Ca2+ through the Na+ : Ca2+ , K+ exchanger.
n1
n2
n3
All these fluxes are counted as positive, and have units of molecules per time.
At steady state we must have no net ionic fluxes and thus
n1 + n 2
c1
= n3 ,
= c2 .
(21.41)
(21.42)
Furthermore, because the exchangers pump ions in certain ratios we must also
have
n2 = 4c2 .
(21.43)
If we now let Jls denote the total light-sensitive current, and assume that
Ca2+ carries a fraction α of the total light-sensitive current we have
c1
=
n1
=
αJls
2q
Jls (1 − α)
,
q
(21.44)
(21.45)
where q is the unit positive charge.
We can now solve for the current carried by the Na+ –K+ exchanger in the
inner segment, which is given by qn3 /2. Since
n3
= n 1 + n2
519
=
=
=
=
Jls (1 − α)
+ 4y2
q
Jls (1 − α)
+ 4y1
q
Jls (1 − α) 2αT
+
q
q
Jls (1 + α)
,
q
it follows that
qn3
=
3
1+α
3
Jls .
(21.46)
Thus, since α ≈ 0.1 we see that the Na+ –K+ exchanger in the inner segment
generates an outward positive current that is approximately one third as big
as the total inward light-sensitive current.
Because Cm dV /dt = I, where I includes all the trans-membrane currents,
we know that the currents generated by the exchange pumps will affect the
voltage responses of the photoreceptor and thus should be included in the
model; the omission of these currents is one of the most severe deficiencies of
the model for light adaptation presented here.
The current generated by the Na+ –Ca2+ , K+ exchanger in the outer segment
is given by n2 q/4 = c2 q = c1 q = αJls /2. Hence, the Na+ –Ca2+ , K+ exchanger generates a much smaller current than the Na+ –K+ exchanger and
can probably be safely ignored in the modelling.
This problem is only partially avoided by comparing the model to photocurrent measurements, rather than to voltage measurements. The photocurrent
is dependent mainly on the number of open channels, which in turn depends
on the cGMP concentration. So a good approximation of the photocurrent
is obtained by assuming that it is proportional to x3 (or to x3 /(K 3 + x3 ),
which is more accurate at higher light levels). With this approximation the
membrane voltage of the photoreceptor never appears in the model and thus
the exchange pump currents can be ignored. This is the approach taken by
Forti et al. (1989), for instance.
However, the photocurrent does depend on the membrane potential to a certain extent, as shown in (11.27), and so a fully accurate model would include
the effects of the transmembrane potential on the photocurrent. It’s not clear
what difference this would make.
Ex. 4: Calculate the response of the lateral inhibition model (Section 11.3) when the
light stimulus is a strip of width 2a, modulated sinusoidally with frequency ω.
Show that as a increases (i.e., as the stimulus goes from a spot to a uniform
field), the gain at x = 0 becomes more band-pass in nature.
520
CHAPTER 21. VISION: SOLUTIONS
Hint: Show that the solution for Iˆ is of the form
!
√
λL̂
[1 + A cosh[x 1 + λ + iω], |x| < a,
(iωτ +1)(1+λ+iω)
√
Iˆ =
B exp(−|x| 1 + λ + iω),
|x| > a,
(21.47)
and then require that Iˆ and its derivative be continuous at |x| = a. Define
the gain as the response divided by the stimulus, and calculate the amplitude
of the gain at x = 0 as a function of ω and a. (The details are given in Peskin
(1976)).
Solution
First consider the response of the lateral inhibition model to a stimulus of the
form
L(x, y, t) = L1 (x)eiωt .
(21.48)
We look for solutions of the form L = L1 (x)eiωt , I = I1 (x)eiωt and E =
E1 (x)eiωt . Substituting these into (11.45)–(11.47) and cancelling out the eiωt
gives
E1
∇2 I1
R1
L1
,
1 + iωτ
= α2 I1 − λE1 ,
= E1 − I1 ,
=
(21.49)
(21.50)
(21.51)
where
α2 = 1 + λ + iω.
For a strip of light of width 2a, modulated sinusoidally, we have
L̄,
|x| < a,
L1 (x) =
0,
otherwise,
(21.52)
(21.53)
where L̄ is a constant. Since there is no y dependence ∇2 I1 = d2 I1 /dx2 .
Hence we need to solve
2
λL̄
d2 I1
,
|x| < a,
α I1 − 1+iωτ
=
(21.54)
2
dx2
α I1 ,
otherwise.
We consider the three regions x < −a, |x| < a and x > a. When x > a we
have
d2 I1
= α2 I1 ,
(21.55)
dx2
which has solution
I1region 1 = Ae−αx .
(21.56)
Note that the other exponential solution is discarded because we require the
solution to be bounded at infinity.
521
Similarly, when |x| < a we have
d2 I1
λL̄
,
= α2 I1 −
dx2
1 + iωτ
(21.57)
which has solution
I1region
2
= Beαx + Ceαx +
λL̄
.
α2 (1 + iωτ )
(21.58)
Because of the inherent symmetry of the problem we know that the solution
must be symmetrical around the origin, and thus B = C. Hence
I1region
λL̄
α2 (1 + iωτ )
λL̄
= D cosh(αx) + 2
,
α (1 + iωτ )
2
= B(eαx + eαx ) +
(21.59)
(21.60)
where D = 2B.
Finally, on the region x < −a we know from symmetry that
I1region
Thus, in summary,
I1 =
!
Ae−α|x| ,
D cosh(αx) +
3
= Aeαx .
λL̄
α2 (1+iωτ )
,
(21.61)
|x| > a,
|x| < a.
(21.62)
We now determine the constants A and D by requiring that
I1region 1 (a)
= I1region 2 (a),
dI1region 1 (a)
=
dx
(21.63)
dI1region 2 (a)
dx
.
(21.64)
This gives the two conditions
D cosh(αa) +
λL̄
= Ae−αa ,
+ iωτ )
αD sinh(αa) = −αAe−αa ,
α2 (1
(21.65)
(21.66)
which we can solve to get
A
=
D
=
λL̄ sinh(αa)
,
α2 (1 + iωτ )
−λL̄
e−αa .
α2 (1 + iωτ )
(21.67)
(21.68)
Thus,
!
I1 =
λL̄ sinh(αa) −α|x|
,
α2 (1+iωτ ) e
λL̄
−αa
α2 (1+iωτ ) (1 − e
cosh(αx)),
|x| > a,
|x| < a.
(21.69)
522
CHAPTER 21. VISION: SOLUTIONS
| G (0,ω) |
1
0.1
a=0
a=
0.01
0.1
1
10
100
ω
Figure 21.12: Plots of |G(0, ω)|2 a=∞ and |G(0, ω)|2 a=0 using the arbitrary parameter values τ = 0.2, λ = 1. We plot them as log-log plots as this is a common way
of plotting gain against frequency (see Fig. 11.4).
The gain, G, at x = 0 is given by
G(0, ω)
=
=
=
R1 (0, ω)
L̄
E1 (0, ω) − I1 (0, ω)
L̄
1
λ
−αa
) .
1 − 2 (1 − e
1 + iωτ
α
(21.70)
(21.71)
(21.72)
To study the behaviour of |G(0, ω)| as a function of a we consider two extreme
cases; a = 0 and a = ∞. When a = 0, G(0, ω) = 1/(1 + iωτ ), which implies
that
1
|G(0, ω)|
=√
.
(21.73)
1 + ω2 τ 2
a=0
This is a monotonic decreasing function of ω.
However,
and thus
G(0, ω)
|G(0, ω)| =
a=∞
2
=
a=∞
1 + iω
1
·
1 + iωτ 1 + λ + iωτ
1 + ω2
1
.
·
2
2
1 + ω τ (1 + λ) + ω 2 τ 2
(21.74)
(21.75)
When τ is small enough this function first increases as ω increases, and then
decreases, i.e., it demonstrates band-pass characteristics, with greater amplitude at a band of intermediate
frequencies. This is illustrated in Fig. 21.12
where we plot |G(0, ω)|2 a=∞ and |G(0, ω)|2 a=0 as functions of ω.
523
Ex. 5: Calculate the response of the lateral inhibition model (Section 11.3) to a moving step of light. Show that as the speed tends to zero, the response approaches
the response to a steady step, while as the speed goes to infinity, the response
behaves like the response to a step of light presented simultaneously to the
entire retina. (This question is taken from Peskin (1976)).
Hint: to exhibit the space-like limit write R as a function of x + ct, where c
is the speed of the moving step. To exhibit the time-like limit, write R as a
function of t + x/c.
Solution
Consider first the space-like limit, i.e., the limit as the speed of the moving
step tends to zero. Define a new travelling wave variable ξ = x + ct, where
c is the speed of the moving step. In this new variable, L(ξ) = H(ξ), where
H is the Heaviside function. We look for solutions of the form E(ξ), I(ξ) and
R(ξ). Since
∂E(x, t)
∂t
∂ 2 I(x, t)
∂x2
dE(ξ)
,
dt
d2 I(ξ)
,
dx2
= c
=
and
(21.76)
(21.77)
it follows that the response is found by solving
cτ E + E
I − cI − I
= L,
= −λ(E − I),
(21.78)
(21.79)
where a prime denotes differentiation with respect to ξ.
We now need to take the limit of these equations as c → 0. However, this
is not such a trivial matter as it might appear at first. Firstly, note that we
must take the limit c → 0 for a fixed t and thus the limit will hold only for
any finite time. (In other words, we shall show that for any finite time, in the
limit as c → 0 the response to the moving step looks like the response to a
fixed step).
Secondly, if we just let c → 0 directly, we get E = L and
I = (1 + λ)I − λL,
(21.80)
which is just the same as (11.52), which gives, as we showed before, the
response to a steady step. Although this gives the correct answer, one should
be a little more careful about taking the limit as c → 0 in (21.78) as in the
limit the highest derivative of E is lost.
A better approach is to write the equation for E as
0,
ξ < 0,
cτ E = −E +
1,
ξ > 0.
(21.81)
524
CHAPTER 21. VISION: SOLUTIONS
On the region ξ < 0, boundedness of the solution requires that E ≡ 0 there,
while on the region ξ > 0 we use the boundary condition E(0) = 0 to get
0,
ξ < 0,
−ξ
E(ξ) =
(21.82)
1 − e cτ ,
ξ > 0.
Clearly, as c → 0 (for fixed t) E just tends toward the Heaviside function, i.e.,
E → L.
It is much easier to take the limit of (21.79) as c → 0, as the highest derivative
is not now lost in the limit; one can safely just set c = 0.
Next we consider the time-like limit. Define the traveling wave variable
η =t+
x
,
c
(21.83)
and let a prime now denote differentiation with respect to η. This gives
τ E + E
1 I − I − (1 + λ)I
c2
= L,
(21.84)
= −λE.
(21.85)
For any fixed x we can let c → ∞ to get η → x and
τ E + E
I + (1 + λ)I
= L,
(21.86)
= λE,
(21.87)
which is the same as (11.48) and (11.49), i.e., we just get the space-independent
behavior as required.
However, it is best to check that taking the limit as c → ∞ in (21.85) gives no
nasty surprises as the highest derivative is lost. What happens is the following.
For any value of c we can solve (21.86) and (21.87) on the two regions ξ < 0
and ξ > 0. Since each region will give two unknown constants in the solution
that gives a total of four unknown constants. These constants may then be
determined from the two boundary conditions at ±∞ and the requirement
that the solution and its derivative be continuous at ξ = 0. However, as
c → ∞ we find that we are no longer able to force the derivative to be
continuous at ξ = 0, and thus, in the limit, we get a solution with a “kink” at
ξ = 0 (i.e., at t = 0). This is, of course, precisely the solution plotted in Fig.
11.12A. Note that the loss of the highest derivative in the differential equation
results merely in the loss of a continuous derivative at ξ = 0, but the solution
itself remains continuous.