Homework Exercises for Mathematics 6770 - Fall 2014
Remark: Solutions may include maple files or matlab files.
Assignment 1: (due Sept. 23, 2014)
1. 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
= 2k−1 b − 2k1 a2 + k−2 c − k2 ab,
dt
db
= k1 a2 − k−1 b + k−2 c − k2 ab,
dt
dc
= k2 ab − k−2 c,
dt
(1)
(2)
(3)
where a = [A], b = [B], and c = [C].
(b) QSS Approach:
If k−1 is large compared to other rate constants, then b equilibrates rapidly
(as an exponential) to its quasi-steady state,
b=
k1 a2 + k−2 c
.
k−1 + k2 a
(4)
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
(5)
If k−1 ≫ k2 a, this further reduces to
dc
k2 k1 3
=
a − k−2 c.
dt
k−1
1
(6)
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.
Fast-Equilibrium Approach:
k1
k−1 large implies A + A −←→− B is almost in equilibrium. Thus, the fast equilibrium approximation is
k−1
b=
k1 a 2
.
k−1
(7)
The new slow variable equation is
d
(a + 2b) = −3k2 ab + 3k−2 c,
dt
(8)
which makes the resulting system
k1 3
−3 kk2−1
a + 3k−2 c
da
=
1
dt
a
1 + 4 kk−1
(9)
dc
k1 k2 3
=
a − k−2 c.
dt
k−1
(10)
2. The length of microtubules changes by a process called treadmilling, in which
monomer is added to one end of the microtubule and taken off at the other end.
To model this process, suppose that monomer A1 is self-polymerizing in that it
can form dimer A2 via
k+
A1 + A1 −→A2 .
(11)
Furthermore, suppose A1 can polymerize an n-polymer An at one end making an
n + 1-polymer An+1
k+
A1 + An −→An+1 .
(12)
Finally, degradation can occur one monomer at a time from the opposite end at
rate k− . Find the steady state distribution of polymer lengths after an initial
amount of monomer A0 has fully polymerized.
Solution
The differential equations describing the dynamics are
∑
∑
da1
= −2k+ a21 + 2k− a2 + k−
a j − k+ a 1
aj ,
dt
j=3
j=2
∞
∞
(13)
and
dan
= k+ an−1 a1 − k+ an a1 − k− an + k− an+1 ,
dt
∑
for n ≥ 2. The quantity A0 = ∞
j=1 jaj is conserved by these equations.
2
(14)
Table 1: Data for Problem 3.
Substrate
Reaction
Concentration (mM)
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
In steady state, the system for n ≥ 2 is a linear system of equations with solution
an = αλn ,
(15)
with λ = kk+−a1 . Of course, it must be that λ < 1 for this solution to be physically
meaningful. To determine α, we substitute this solution into Equation 14 with
(
)
∞
∑
1
k−
n
n = 2 and find that α = k+ . Since
αλ = α
− 1 − λ , it follows that
1
−
λ
j=2
the total amount of monomer is
2
k−
a1
,
(k+ a1 − k− )2
(16)
√
A0
(1
+
2κ
−
1 + 4κ),
2κ2
(17)
A0 =
which in turn implies that
a1 =
where κ =
k+ A 0
.
k−
3. 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
1.
(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 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
(c) Rewrite the equation (??) as
Vmax −
3
V
Km = V.
s
(18)
V vs V/s: Regression
0.45
0.4
0.35
0.3
V
0.25
0.2
0.15
0.1
0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
V/s
Figure 1:
For each experiment, there is one such equation, so with seven data points
we have the linear system of equations




1.0 −0.40
0.04
 1.0 −0.40 
 0.08 




(
)  1.0 −0.34  (
)  0.17 

 Vmax


Vmax
1.0 −0.24 
0.24 
A
=
=
(19)



 = b.
Km
Km
 1.0 −0.16 
 0.32 




 1.0 −0.11 
 0.39 
1.0 −0.08
0.42
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,
(20)
Km
which in this case is the 2 × 2 system
(
)(
) (
)
7.0000 −1.7354
Vmax
1.6600
=
,
−1.7354 0.5383
Km
−0.2933
(21)
having the solution Vmax = 0.5084, Km = 1.0942. A plot of the line y =
Vmax − Km x is shown with data points in Fig. 1, and a plot of the reaction
velocity with data points is shown in Fig. 2. This solution can be found by
running the Matlab code ex Burk.m.
4. 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 2.
(a) Plot the data V vs. s. Is there evidence that this is a Hill type reaction?
4
0.45
0.4
Reaction Velocity V
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
Substrate Concentration S
Figure 2: Michaelis-Menten growth curve for data of Problem 3.
Table 2: Data for Problem 4.
Substrate
Reaction
Concentration (mM)
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
5
V
(b) Plot ln( Vmax
) vs. ln(s). Is this approximately a straight line, and if so,
−V
what is its slope?
(c) Use linear regression 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 (??) as
(
−n ln Km + n ln s = ln
V
Vmax − V
)
,
(22)
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




 1.0 0.4055 


0
(

(
) 
) 
 1.0 0.6931  −n ln Km
 0.7082 
−n ln Km


A
=
=
 1.0 0.9163 
 1.2657  = b.
n
n




 1.0 1.2528 
 2.0907 




 1.0 1.3863 
 2.4423 




 1.0 1.5041 
 2.7515 
1.0 1.6094
2.9444
(23)
To find the linear regression solution of this system, solve the normal equations
)
(
−n ln Km
T
= AT b,
(24)
A A
n
which in this case is the 2 × 2 system
) (
)
)(
(
3.8616
−n ln Km
10.0000 5.4649
,
=
n
25.8358
5.4649 12.8990
(25)
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. 3, and a plot of the reaction velocity with
data points is shown in Fig. 4. This solution can be found by running the
Matlab code ex Hill.m.
5. 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).
(a) Write the reaction scheme and differential equations, and nondimensionalize,
to get the system of equations
dσ
= −σ + α1 (µ1 + σ)x + α2 (µ2 + σ)y,
dt
6
(26)
3
2
1
ln(V/(V
max
-V)
0
-1
-2
-3
-4
-5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
ln(S)
Figure 3:
1
0.9
0.8
Velocity V
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Substrate Concentration S
Figure 4: Hill curve for data of Problem 4.
dx
1
=
σ(1 − x) − x,
dt
λ1
dy
1
ϵ2
=
σ(1 − y) − y.
dt
λ2
ϵ1
(27)
(28)
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 quasi-steady 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 + ϵ1 λ1 α1 x + ϵ2 λ2 α2 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
k
2
S + E1 −→
←− C1 −→E1 + P,
k−1
7
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
= k−1 c1 + k−3 c2 − k1 se1 − k3 se2 ,
dt
dc1
= k1 se1 − (k−1 + k2 )c1 ,
dt
dc2
= k3 se2 − (k−3 + k4 )c2 .
dt
(29)
(30)
(31)
Since the total amounts of the enzymes are conserved,
e1 + c1 = et ,
e2 + c2 = eT ,
(32)
(33)
and we can eliminate e1 and e2 to find
ds
= k−1 c1 + k−3 c2 − k1 s(et − c1 ) − k3 s(eT − c2 ),
dt
dc1
= k1 s(et − c1 ) − (k−1 + k2 )c1 ,
dt
dc2
= k3 s(eT − c2 ) − (k−3 + k4 )c2 .
dt
The most natural nondimensionalization of the concentrations is
c1
,
x =
et
c2
y =
,
eT
s
σ =
,
s0
(34)
(35)
(36)
(37)
(38)
(39)
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
µet
µeT
dσ
= k−1
x + k−3
y − µk1 σet (1 − x) − µk3 σeT (1 − y), (40)
dt
s0
s0
1 dx
= k1 s0 σ(1 − x) − (k−1 + k2 )x,
(41)
µ dt
1 dy
= k3 s0 σ(1 − y) − (k−3 + k4 )y.
(42)
µ dt
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
8
and y are “fast” variables, compared to σ. Notice that all the linear terms
proportional to σ on the right hand side of (40) are multiplied by either et or
eT , suggesting that this variable is slow. This also suggests that we should
1
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
which is
dσ
dt
dx
ϵ1
dt
dy
ϵ2
dt
= −σ +
et (k−1 + k1 s0 σ)
eT (k−3 + k3 σ)
x+
y (43)
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
=
(44)
(45)
= −σ + α1 (µ1 + σ)x + α2 (µ2 + σ)y,
(46)
1
σ(1 − x) − x,
λ1
1
=
σ(1 − y) − y.
λ2
(47)
=
+k2
+k4
k1 et +k3 eT
t +k3 eT
, ϵ2 = k1ke−3
, λ1 = k−1
, λ2 = k−3
,
k−1 +k2
+k4
k1 s 0
k3 s 0
k3 e T
k1 e t
α1 = k1 et +k3 eT , and α2 = k1 et +k3 eT , with α1 + α2 = 1.
where ϵ1 =
(48)
µ1 =
k−1
,
k1 s0
µ2 = kk3−3
,
s0
(b) Under the assumption that both ϵ1 and ϵ2 are small, we set ϵ1 = ϵ2 = 0 and
find the quasi-steady state solution
σ
x =
,
(49)
λ1 + σ
σ
y =
.
(50)
λ2 + σ
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
.
(51)
dt
λ1 + σ
λ2 + σ
(c) The equation (51) can be solved by separating variables. First, we rewrite it
as
(λ1 + σ)(λ2 + σ)
dσ = −dt.
(52)
σ(θ1 σ + θ2 )
where
θ1 = λ1 + λ2 − α1 (µ1 + λ2 ) − α2 (µ2 + λ1 )
θ2 = λ1 λ2 − α1 µ1 λ2 − α2 µ2 λ1 .
(53)
(54)
(It is important to notice that α1 + α2 = 1.) Notice that in original dimensioned parameters these are
k1 e t k2 + e T k4 k1
(55)
θ 1 = k3
(k1 et + k3 eT )k1 k3 s0
1
k1 et k2 (k−3 + k4 ) + k3 eT k4 (k−1 + k2 ), (56)
θ2 =
(k1 et + k3 eT )k1 s0 k3 s0
9
1
0.9
σ
0.8
0.7
0.6
y
0.5
x
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
time (dimensionless)
Figure 5: Plots of σ, x and y for the parameters µ1 = 0.5, µ2 = 0.6, λ1 = 0.9, λ2 = 0.8,
α1 = 0.3, α2 = 0.7.
both positive.
We integrate both sides of (52) to find
−
σ
λ1 λ2 ln(σ)
λ1 λ2 λ1 + λ2 θ2
−
+(
−
+ 2 ) ln(θ1 σ + θ2 ) = t + K,
θ1
θ2
θ2
θ1
θ1
(57)
for some constant K. To determine K, we apply the initial condition σ(0) = 1,
from which it follows that
K=−
1
λ1 λ2 λ1 + λ2 θ2
+(
−
+ 2 ) ln(θ1 + θ2 ).
θ1
θ2
θ1
θ1
(58)
Thus,
1 − σ λ1 λ2 ln(σ)
λ1 λ2 λ1 + λ2 θ 2
−
+(
−
+ 2 ) ln
θ1
θ2
θ2
θ1
θ1
(
θ1 σ + θ2
θ1 + θ2
)
= t.
(59)
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.5.
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 quasisteady state approximation we have constructed a solution that is incorrect
close to t = 0. To correct this requires a knowledge of how 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
10
discussed briefly in the Appendix of this Chapter, 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 (26)-(28) 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 + ϵ1 λ1 α1 x + ϵ2 λ2 α2 y) = α1 (µ1 − λ1 )x + α2 (µ2 − λ2 )y
dt
(60)
(since α1 + α2 = 1).
Now observe that
µ1 − λ1 = −
k2
< 0,
k1 s0
µ2 − λ2 = −
k4
< 0,
k3 s0
(61)
so that σ + ϵ1 λ1 α1 x + ϵ2 λ2 α2 y decreases whenever x and y are non-negative.
However, we already know that x and y cannot be zero unless σ = 0 as well.
Thus σ + ϵ1 λ1 α1 x + ϵ2 λ2 α2 y is always decreasing, and so approaches zero for
large time.
6. ATP is known to inhibit its own dephosphorylation. One possible way for this to
occur is if ATP binds with the enzyme, holding it in an inactive state, via
k4
S1 + E −→
←− S1 E.
k−4
(a) Add this reaction to the Sel’kov model for glycolysis and derive the corresponding equations governing glycolysis of the form
dσ1
= ν − f (σ1 , σ2 ),
dτ
dσ2
= αf (σ1 , σ2 ) − ησ2 .
dτ
(62)
(63)
Explain from the model why this additional reaction is inhibitory.
(b) Give an analysis of these equations using xpp. In particular, modify the
file selkov.ode and find phase portraits of periodic solutions as well as the
bifurcation diagram, similar to Fig. 1.9 in the text.
Solution
11
The addended Sel’kov model with s1 = [S1 ], s2 = [S2 ], e = [E], c1 = [ES2γ ], c2 =
[S1 ES2γ ],and c3 = [S1 E] is
ds1
=
dt
ds2
=
dt
dc1
=
dt
dc2
=
dt
dc3
=
dt
with conservation equation
−k1 s1 c1 + k−1 c2 − k4 s1 e + k−4 c3 + v1
(64)
−γk3 sγ2 e + k−3 γc1 + k2 c2 − v2 s2
(65)
−k1 s1 c1 + k−1 c2 + k2 c2 + k3 esγ2 − k−3 c1
(66)
k1 s1 c1 − k−1 c2 − k2 c2
(67)
k4 s1 e − k−4 c3
(68)
e = e 0 − c1 − c2 − c3 .
(69)
( )1/γ
3
1 s1
Nondimensionalizing this using σ1 = k2k+k
, σ2 = kk−3
s2 , x = c1 /e0 , y = c2 /e0 ,
−1
z = c3 /e0 , and t =
dσ1
dτ
dσ2
dτ
dx
ϵ
dτ
dy
ϵ
dτ
dz
ϵ
dτ
k−1 +k2
τ
e 0 k1 k2
gives
k−1 + k2
k−1
k4 (k−1 + k2 )
k−4
σ1 x +
y−
σ1 (1 − x − y − z) +
z + ν (70)
k2
k2
k1 k2
k2
[
]
γk−3
γk−3 γ
= α y−
σ (1 − x − y − z) +
x − ησ2
(71)
k2 2
k2
k−3
= −σ1 x + y +
(σ γ (1 − x − y − z) − x)
(72)
k−1 + k2 2
= −
= σ1 x − y
(73)
k4
k−4
σ1 (1 − x − y − z) −
z,
k1
k−1 + k2
( )1/γ
k−1 +k2
v1
k3
1 k2
where ϵ = (ke−10 k+k
,
ν
=
,
α
=
, and η =
2
k2 e 0
k1
k−3
2)
state
=
σ2γ
x =
1 + σ2γ + βσ1 + σ1 σ2γ
σ1 σ2γ
y =
1 + σ2γ + βσ1 + σ1 σ2γ
βσ1
z =
,
γ
1 + σ2 + βσ1 + σ1 σ2γ
where β =
(k−1 +k2 )k4
.
k1 k−4
(74)
v2 (k−1 +k2 )
.
e 0 k1 k2
In quasi-steady
(75)
(76)
(77)
Thus,
dσ1
= ν − f (σ1 , σ2 )
dτ
dσ2
= αf (σ1 , σ2 ) − ησ2 ,
dτ
12
(78)
(79)
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
Figure 6: Phase portrait of periodic solutions with ν = 0.0285, η = 0.1, α = 1, γ = 2, and
β = 0.0001.
where
f (σ1 , σ2 ) =
σ1 σ2γ
.
1 + σ2γ + βσ1 + σ1 σ2γ
(80)
A typical phase portrait is shown in Fig. 6.
Assignment 2: (due Oct. 21, 2014)
1. Suppose a semi-infinite tube with cross-sectional area A initially has only water,
and that the concentration of a chemical species (with diffusion coefficient D), in
a large bath at the end of the tube has fixed concentration C0 . Find the total
number of molecules in the tube at time t.
Solution
Derivation of (2.16) follows by using the Sine-Fourier transform,
∫ ∞
ĉ =
c(x, t) sin ωxdx
(81)
0
which, applied to the heat equation yields
ĉt + ω 2 Dĉ = ωC0
13
(82)
The solution of this is
)
C0 (
2
ĉ =
1 − exp(−ω Dt)
ωD
∫
) sin ωx
2 C0 ∞ (
c(x, t) =
1 − exp(−ω 2 Dt)
dω
πD 0
ω
so that
(83)
(84)
A little manipulation, and using properties of the Fourier transform, one gets
(2.16). Then integration over all x gives (2.17).
2. The following data were used by Segel, Chet and Henis (1977) 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 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
According to the formula (2.18)
D=
πN 2
,
4C02 A2 T
(85)
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 =0.4-0.9 ×10−4 cm2 /s. For the second data
set, with T fixed, the estimate is D =0.3-0.6×10−4 cm2 /s.
3. Find the maximal enhancement for diffusive transport of carbon dioxide via binding with myoglobin using Ds = 1.92 × 10−5 cm2 /s, k+ = 2 × 108 cm3 /M · s,
k− = 1.7×10−2 /s. Compare the amount of facilitation of carbon dioxide transport
with that of oxygen at similar concentration levels.
Solution
Calculate ρ = DDc ks k+−e0 , with Dc = 4.4 × 10−7 cm2 /s for myoglobin, Ds = 1.92 ×
10−5 cm2 /s for carbon dioxide, k+ = 2 × 108 cm3 M−1 s−1 , and k− = 1.7 × 10−2 s−1 ,
and find ρ = 2.7 × 108 e0 cm3 M−1 . Compare this with oxygen using Ds = 1.2 ×
10−5 cm2 /s, k+ = 1.4×1010 cm3 M−1 s−1 , and k− = 11s−1 , and find ρ = 4.6 × 107 e0 cm3 M−1 ,
which is smaller by a factor of about 10.
4. A fluorescent dye with a diffusion coefficient of D = 10−7 cm2 /s and binding
equilibrium of Keq = 30 mM 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
Dh + DB Kweq0
,
(86)
Deff =
1 + Kweq0
14
(
so that
w0 = Keq
Dh − Deff
Deff − DB
)
.
(87)
With these numbers we find that w0 = 0.14M.
5. 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 (Find an expression for glucose flux that
incorporates this reaction.) How is this reaction affected by the concentration of
ATP?
Solution
To solve this problem one must first write a differential equation for the substrate
si ,
dsi
= k− pi − k+ si ci − kr si
(88)
dt
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 (??) except
that si is the solution of the quadratic equation
C2 si + C1 si − se kk− k+ c0 = 0,
(89)
where C2 = 2kp kr (k + k− + k+ se ), and C1 = kk− k+ c0 + kr (2k+ kse + 4kk− +
2
2se k+ k− + 2k−
).
6. A 1.5 oz bag of potato chips (a typical single serving) contains about 200 mg of
Na+ . 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 mg of sodium is 8.7 ×10−3 moles of sodium. Since
salt has equal amounts of sodium and chloride, 200 grams of sodium gives 0.0174
osmoles.
7. (a) Consider a vertical tube with a cross-sectional area of 1 cm2 . The bottom of
the tube is closed with a semi-permeable membrane and 1 gram of sugar is
placed in the tube. The membrane-closed end of the tube is then put into
an inexhaustible supply of pure water at T = 300K. What will be the height
of the water in the tube at equilibrium? (The weight of a sugar molecule is
3 × 10−22 gm, and the density of water is 1 gm/cm3 ).
(b) Two columns with cross-sectional area 1 cm2 are initially filled to a height of
one meter with water at T = 300◦ K. Suppose 0.001 gm of sugar is dissolved in
one of the two columns. How high will the sugary column be when equilibrium
is reached?.
(c) Suppose in the previous question 1 gm of sugar is dissolved in one of the two
columns. What is the equilibrium height of the two columns?
Solution
15
(a) If ρgh = 25atm = 2.533 × 106 N m−2 , then h = 256.6 m.
(b) Since a gram of sugar has 31 × 1022 molecules, nkT
= 14.12m2 . The column
ρgA
will have height 3.76m.
√
(c) h1 = h20 + 12 h20 + 2nkT
= 1.007m.
ρgA
(d) There is not enough water to balance the osmotic pressure.
8. Ouabain is known to compete with K+ for external potassium binding sites of
the Na,K-ATPase. Many animal cells swell and burst when treated with the drug
ouabain. Why? (Explain this using a model of cell volume control.)
Solution
If ouabain blocks K+ and prevents the pump from working, Na+ builds up inside
the cell. This causes water to enter the cell to balance the osmotic pressure.
When too much water enters the cell, it bursts. This effect could be included in
the model by reducing the number of functional Na,K-ATPase pumps or including
extracellular competition for binding.
Assignment 3 (due Nov. 13, 2014)
1. Suppose the Na+ Nernst potential of a cell is 56 mV, its resting potential is −70
mV, and the extracellular Ca++ concentration is 1 mM. At what intracellular
Ca++ concentration is the flux of a three-for-one Na+ –Ca++ exchanger zero? (Use
that RT /F = 25.8 mV at 27◦ C.)
Solution
Using Equations (2.131) and (2.75),
[
]
F
++
++
[Ca ]i = [Ca ]e exp
(Veq − 3zVN ernst ) ,
(90)
RT
so that [Ca++ ]i = 9.86 × 10−5 mM.
2. Intestinal epithelial cells have a glucose–Na+ symport that transports one Na+
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?
Solution
The current associated with this symport is exactly the same as for NCX, the
sodium-calcium exchanger. It follows from (2.79) that the flux is
J ∼ (ge ne − exp(
FV
)gi ni ).
RT
(91)
Therefore, a negative membrane potential enhances the inward transport of glucose.
3. Explore the behavior of the reduced Hodgkin-Huxley model (you may use the code
hhred.ode). In particular, for what values of applied current are there oscillatory
16
solutions? Produce phase portraits for several different parameter values showing
the different types of possible behaviors, and use xppaut to produce a bifurcation
diagram.
Solution
The bifurcation diagram is shown on the left panel of Fig.7 and a phase portrait
for I0 = 91.8, showing periodic limit cycle solution is shown on the right panel.
There are two Hopf bifurcation points, at I0 = 8.817 and I0 = 372.6.
Figure 7: Left: The bifurcation diagram for the Reduced HH equations and Right: a phase
portrait for I0 = 91.8, showing periodic limit cycle solution.
4. Morris and Lecar (1981) proposed the following two-variable model of membrane
potential for a barnacle muscle fiber:
dV
+ Iion (V, W ) = Iapp ,
dT
(92)
dW
= ϕΛ(V )[W∞ (V ) − W ],
dT
(93)
Cm
where V = membrane potential, W = fraction of open κ+ channels, T = time,
Cm = membrane capacitance, Iapp = externally applied current, ϕ = maximum
rate for closing κ+ channels, and
0
Iion (V, W ) = gCa M∞ (V )(V − VCa
) + gK W (V − VK0 ) + gL (V − VL0 ),
(
(
))
V − V1
1
1 + tanh
,
M∞ (V ) =
2
V2
(
(
))
1
V − V3
W∞ (V ) =
1 + tanh
,
2
V4
(
)
V − V3
.
Λ(V ) = cosh
2V4
(94)
(95)
(96)
(97)
Typical parameter values for these equations are shown in Table 3.
(a) Make a phase portrait for the Morris–Lecar equations. Plot the nullclines and
show some typical trajectories, demonstrating that the model is excitable.
(b) For what range of applied current is this system oscillatory? Use XPP to find
the bifurcation diagram and determine the types of bifurcations. (You can
use the file ML.ode to get started.)
Solution
(a) A phase portrait for Iapp = 137 showing a periodic limit cycle solution is
shown in Fig.8.
(b) There are Hopf bifurcations at Iapp = 94 and Iapp = 212, shown in the
bifurcation diagram in Fig.9.
17
Cm = 20 µF/cm2
Iapp = 0.06 mA/cm2
gCa = 4.4 mS/cm2
gK = 8 mS/cm2
gL = 2 mS/cm2
ϕ = 0.04 ms−1
V1 = −1.2 mV
V2 = 18 mV
V3 = 2
V4 = 30 mV
VK0 = −84 mV
0
VCa
= 120 mV
VL = −60 mV
Table 3: Typical parameter values for the Morris–Lecar model.
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
-80
-60
-40
-20
0
20
40
60
80
100
Figure 8: A phase portrait for Iapp = 137, showing periodic limit cycle solution.
5. The Hodgkin-Huxley equations are
dv
dt
dm
dt
dn
dt
dh
dt
Cm
= −ḡK n4 (v − vK ) − ḡNa m3 h(v − vNa ) − ḡL (v − vL ) + Iapp ,
(98)
= αm (1 − m) − βm m,
(99)
= αn (1 − n) − βn n,
(100)
= αh (1 − h) − βh h,
(101)
with the functions αj and βj , in units of (ms)−1 ,
αm = 0.1
βm
αh
25 − v
( 25−v )
exp 10 − 1
( )
−v
= 4 exp
,
18
( )
−v
= 0.07 exp
,
20
18
,
(102)
(103)
(104)
V
40
20
0
-20
-40
-60
0
50
100
150
iapp
200
250
300
Figure 9: The bifurcation diagram for the Morris Lecar equations.
1
,
+1
10 − v
= 0.01
,
exp( 10−v
)−1
10
( )
−v
.
= 0.125 exp
80
βh =
αn
βn
(105)
exp( 30−v
)
10
(106)
(107)
For these expressions, the potential v is the deviation from rest, measured in
units of mV, current density is in units of µA/cm2 , conductances are in units of
mS/cm2 , and capacitance is in units of µF/cm2 . The remaining parameters are
ḡNa = 120,
ḡK = 36,
ḡL = 0.3,
Cm = 1,
(108)
and with shifted equilibrium potentials
vNa = 115,
vK = −12,
vL = 10.6.
(109)
(a) Simulate these equations with current input Iapp = 0. What are the steady
state values for all variables?
(b) Starting with all variables at steady state, apply a piecewise constant current
that is nonzero for 0.5 ms. Plot the voltage response for several different
values of Iapp . Identify the threshold value of Iapp .
(c) For what range of constant applied current is this system oscillatory? Use
XPP to find the bifurcation diagram and determine the types of bifurcations.
Solution
(a) The steady solution is V = 0, n = 0.3177, m = 0.053, h = 0.596.
(b) With Iapp < 13.3 the response is subthreshold while with Iapp > 13.4 the
response is superthreshold, as depicted in Fig.10.
(c) There are Hopf bifurcations at Iapp = 9.78 and Iapp = 154.4. The bifurcation
diagram is shown in Fig.11.
19
100
80
V (mV)
60
40
20
0
−20
0
2
4
6
8
10
12
14
16
18
20
t (ms)
Figure 10: V as a function of t for the HH equations following a stimulus of length 0.5 ms
with amplitude Iapp = 13.3 (dashed) and Iapp = 13.4 (solid).
V
120
100
80
60
40
20
0
-20
0
50
100
150
200
250
Iapp
Figure 11: The bifurcation diagram for the HH equations.
20