161
Solution to Problems in Chapter 12, Section 12.7
12.1. From the data in Table 12.2, CD = 4.255 and CM =0.77. From equations (12.3.14a,b), the
drag force and torque are:
FD = τwApCD T= τwApCMhc
12.2. For a spherical cell adherent via a single microvillus, the force balance equations are given
by equations (12.4.10) to (12.4.12). The orientations are shown in Figure S12.2.1.Since the angle
θ is given, the force can be determined from equation (12.4.10)
Fb = 32.054R2τw/cosθ
12.3. To simplify the analysis, let
κ
=k−1
0
k1NR0
,
α
=xaF
kBTNR0
, and θ = NC/NR0. For a single bond,
162
zero at large values of κ, the presence of an applied force leads to a rapid decline in attachment
above a critical value of κ. As a result the curves for α > 0 end indicating that adhesion is not
supported above that value of κ.
12.4. The rate coefficient k-1 equals ln(2)/t1/2. Taking the logarithms of equation (12.2.4).
ln k−1
( )
=ln k−1
0
( )
+xaF
kBT
(S12.4.1)
163
12.5. The force balance is described by Equations (12.4.10) and (12.4.11). Equation (12.4.12)
provides a geometric relation for the angle θ between the microvillus and the surface (Figure
12.15). The bond force is given by Equations (12.4.13) and (12.4.14). Equation (12.4.13)
applies if the bond force is less than the critical force of 45 pN. Otherwise equation (12.4.14)
applies. The angle θ can be eliminated from Equations (12.4.10) to (12.4.12) by dividing
equation (12.4.11) by Equation (12.4.10) and substituting into Equation (12.4.12). The result is:
164
Figure S12.5.1. Change in tether length with time.
165
Since the force per bond is declining due to tether formation, this expression cannot be
integrated analytically. Integration was done using a simple Euler method with the same time
step used for the integration of the force and tether length. Results are shown in Figure S12.5.3
for
kr
0=
1 s-1 and xa = 0.005 nm and 0.001 nm. Since tether formation reduces the force on the
bonds (Figure S12.5.2), the bond lifetimes are longer when tethers form.
Figure S12.5.3. Cell lifetimes for adhesion with and without tether formation.
12.6. (a) On a basis of 2 moles of A being produced, 0.5 moles of A2 react with another 0.5
moles of A2. In a mixture of x0 initial numbers of molecules of A2, the probability that there are
x molecules of A2 at time t + Δt, Px(t + Δt), is equal to the probability that there are x molecules
of A2 at time t, Px(t), plus the rate that x+2 molecules have collided with x+1 molecules to leave
x molecules in time interval Δt, minus the probability that x molecules have reacted in time
interval Δt. Mathematically, we have;
166
dP
6
dt
=−15kP
6
(S12.6.3d)
167
168
Solution to Problems in Chapter 13, Section 13.8
13.1. The following MATLAB code was developed to solve Equation (13.3.2) with constant k’1
and with k’1 given by equation given by equation (13.3.5).
k1=3.5e6;
169
Figure S13.1.1
13.2. The molecular weight of hemoglobin is 64,000. A mass concentration of hemoglobin of
0.1 g ml-1 corresponds to a molar concentration of 1.550 mM. Assuming 97% saturation which
occurs at 100 mm Hg and normal oxygen concentration is blood (1.34 x 10-4 M) and a Hct =
0.45, Equation (13.2.3) becomes
CBT = Cpl + Hct4CHbS = 0.00286 M
13.3. For a thickness of 20 µm, the permeability through the layer of edema is 1.20 x 10-2 cm s-1.
To incorporate this permeability into the model, compute an overall permeability that includes
the permeability of the edema layer (Ped) plus the permeability of the cell and plasma layers is
(Figure S13.3.1)
1
170
Figure S13.3.1
13.4. Using Equation (13.5.7)
RC
2−R0
2+2R0
2ln R0
RC
⎛
⎝
⎜
⎞
⎠
⎟ =
4CRC
DO2
R
O2
Substituting
3 x 10-4 cm
( )
2
+R0
22 ln R0
3 x 10-4 cm
⎛
⎝
⎜⎞
⎠
⎟−1
⎡
⎣
⎢⎤
⎦
⎥=
4 0.14x 10-6 mole cm-3
( )
1.92 x 10-5 cm2 s-1
( )
2 x 10-6 mole cm-3 s-1
( )
Rearranging yields:
5.286 x 10-6 =R0
22 ln R0
3 x 10-4
⎛
⎝
⎜ ⎞
⎠
⎟ −1
⎡
⎣
⎢
⎤
⎦
⎥
Choose and initial guess for R0 and solve iteratively to find that R0 = 15.3 µm.
13.5. For a venular oxygen concentration of 8 mM, solving Equation (13.5.12) for L yields.
2
171
8 x 10-6 =0.55CRC+0.45 1.1194CRC+4 5.1 x 10-6
( )
CRC
3.617 x 10-8
⎛
⎝
⎜
⎞
⎠
⎟
2.7
1+
CRC
3.617 x 10-8
⎛
⎝
⎜
⎞
⎠
⎟
2.7
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
(S13.5.2)
Solving iteratively yields CRC = 7.15 x 10-8 mole cm-3. Using this oxygen concentration,
Equation (13.5.7) becomes.
2.6556 x 10-6 =R0
22 ln R0
3 x 10-4
⎛
⎝
⎜ ⎞
⎠
⎟ −1
⎡
⎣
⎢
⎤
⎦
⎥
(S13.5.3)
Solving iteratively yields R0 = 12.15 µm. The capillary spacing is not this close. As a result,
during heavy exercise, anoxic regions develop. Energy is provided by anaerobic metabolism
resulting in lactic acid production.
13.6. Integrating equation (13.5.14) using the following MATLAB program, the saturation drops
to 0.4 after 0.0075 s (Figure S13.6.1).
function dc=unload2(t,y);
13.7. The transport processes are the same in each tissue, radial diffusion and zero order
reaction. The only difference between the two regions is the oxygen consumption rate.
172
Rc<r<R1
DO2
r
d
dr
r
dCO2
dr
⎛
⎝
⎜
⎞
⎠
⎟ =RO2
(S13.7.1a)
The constant B is determined from the boundary condition at r = Rc.
173
B=CRC
−RO2
RC
2
4DO2
+ln RC
( )
R‘O2
R0
2
2DO2
+
RO2
−R‘O2
( )
R1
2
2DO2
⎡
⎢
⎢
⎤
⎥
⎥
(S13.7.7)
13.8. Shown in Figure S13.8.1 are the concentration profiles for the two-layer model (Equations
(S13.7.8) and (S13.7.11)) using D O2 = 2.0 x 10-5 cm2 s-1 and CRC = 0.14 mM. For the two layer
model, the decline in concentration occurs primarily near the capillary surface due to the higher
oxygen consumption rate. Nevertheless, the decrease in concentration is less than the change in
concentration for a single layer model with an oxygen consumption rate equal to (RO2+ R’O2)/2.
For the composite model, the flux at r =RC is:
2
⎛
⎞
174
13.9. For steady one-dimensional radial diffusion with a constant oxygen uptake rate, the
conservation relation for oxygen transport is:
175
−RO2R0
2
2
RC
R0
2−1
RC
⎡
⎣
⎢
⎤
⎦
⎥ =−kNOCpl +P
ec Cpl −RO2R0
2
4DO2
RC
2
R0
2−2 ln RC
( )
⎡
⎣
⎢
⎤
⎦
⎥ −B
⎛
⎜
⎜
⎞
⎟
⎟
(S13.9.7)
capillaries in part of the tumor.
176
After substituting for A2 in equation (S13.103b) and integrating Equations (S13.103a) and
(S13.103b), we have
13.11. Since the initial oxygen concentration is much higher than the concentration of NO,
oxygen can be assumed to constant at its initial value. For a second order reaction in NO, the
solution is given by equation (10.2.16a)
13.12. Assume that the reaction mechanism is
NO +S−OHb k
⎯ → ⎯ SNO −Hb
where OHb represents oxygenation hemoglobin.
The rate expression is:
177
−dCS−OHb
dt
=kCS−OHbCNO
(S13.12.1)
capillary. It can be re-written as
0Ck
dr
dC
r
dr
d
r
D
B
BB =−
⎟
⎠
⎞
⎜
⎝
⎛
(0 < r < RC) (S13.13.1)
178
0Ck
dx
dC
)xln(
1
1
dx
Cd
xx
R
DB
B
2
B
2
2
C
B=−
⎥
⎦
⎤
⎢
⎣
⎡⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛α
[exp(-α) < x < 1] (S13.13.11)
0Ck
dx
dC
)xln(
1
1
dx
Cd
xx
R
D2
TT
T
2
T
2
2
C
T=−
⎥
⎦
⎤
⎢
⎣
⎡⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛α
[0 < x < exp(-α)] (S13.13.12)
0
dx
dCB=
at x = 1 (S13.13.13)
dx
dC
D
dx
dC
D
)exp(
Rq T
T
B
B
C
NO +−=
α−α
•
at x = exp(-α) (S13.13.14)
TB CC =
at x = exp(-α) (S13.13.15)
0CT=
at x = 0 (S13.13.16)
The equations listed above are solved by a finite difference method based on the MATLAB code
attached at the end of this problem. The procedure is as follows. Let
2
1x)x(g =
;
⎟
⎟
⎞
⎜
⎜
⎛+=
)xln(
1
1x)x(g2
;
2
B
B
h⎟
⎠
⎞
⎜
⎝
⎛
α
φ
=
;
2
T
T
h⎟
⎠
⎞
⎜
⎝
⎛
α
φ
=
179
0Ch
x
CC
)x(g
x
CC2C
)x(g i
BB
B
i
B
1i
B
i2
2
B
1i
B
i
B
1i
B
i1 =−
Δ
−
+
Δ
+−+−+
(S13.13.24)
i = k+1, k+2, …, k+n-1
0)C(h
x
CC
)x(g
x
0C2C
)x(g 21
TT
T
1
T
2
T
12
2
T
1
T
2
T
11 =−
Δ
−
+
Δ
+−
aa (S13.13.25)
0Ch
x
CC
)x(g2 nk
BB
2
B
nk
B
1nk
B
nk1 =−
Δ
−+
+−+
+
(S13.13.26)
q
x
2
)x(g)x(g
)C(h
x
CC
)x(g2
s
x
2
)x(g)x(g
Ch
x
CC
)x(g2
T
k1k2
2k
T
2
T
k1k
T
k1
B
k1k2
k
B
2
B
k1k
B
k1
=
Δ
+
−
Δ
−
−
Δ
−
−
Δ
−−+
(S13.13.27)
where
k
B
k
T
kCCC ==
. The resulting nonlinear algebraic equations can be solved by the iteration
method. For the parameter values given in Table 13.1 and kT = 0.05 µM-1 s-1, the simulation
results are shown in the following figure. They are similar to those shown in Figure S13.13.1,
based on the first-order kinetics of NO reaction in the surrounding tissue. The concentration
gradients are slightly smaller than those predicted by the first-order kinetics of NO reaction.
Figure S13.13.1
MATLAB code
clear all;
eps = 0.001;
180
% Model Parameters
qNO = 5.5e–9; % M*cm/sec or 5.5e–12 mol/cm2/sec
DB = 4.5e–5; % cm2/sec