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BME 100 Computer Project 2
BME 100 Computer project 2
Nigel Chou
1)
since units of
s
nM
=
dt
dM
i
,
therefore units of
s
1
,
s
nM
=
=
m
o
d
k
j
K
,
j
P
and
j
M
are nM (given)
The terms in
j
K
and
j
P
cancel out, therefore units of
s
nM
=
k
Since units of
s
nM
=
dt
dP
i
,
therefore units of
s
1
=
=
p
p
d
k
These results are summarized in table 1.
Table 1:
Dimensions of parameters
Parameter
Units
K
j
nM
P
j
nM
M
i
nM
k
nM s
1
k
0
nM s
1
d
m
s
1
k
p
s
1
d
p
s
1
m
m
d
d
dt
t
d
1
=
∴
=
τ
dt
dM
K
dt
dm
M
m
K
i
j
i
i
i
j
1
=
∴
=
dt
dM
K
dt
dm
M
m
K
i
j
i
i
i
j
1
=
∴
=
,
0
1
1
)
(
L
LR
C
k
K
k
=

and
LR
R
R
N
N
N

=
0
)
(
,
We can find the rate of change of
LR
n
with
:
(
29
LR
LR
LR
R
LR
R
LR
R
LR
R
LR
R
R
LR
R
LR
R
R
LR
LR
R
L
R
LR
R
LR
n
n
K
N
N
N
N
N
K
N
N
N
N
K
k
N
N
k
N
N
K
k
k
N
k
N
C
k
N
d
dt
dt
dN
N
d
dn


=


=

=

=

=
=





)
1
(
)
(
)
(
)
(
)
(
)
(
1
)
(
)
(
1
)
(
)
(
1
)
(
1
0
0
0
0
0
1
0
1
0
1
1
1
0
1
0
0
1
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Hence the dimensionless form of equation 1 is:
LR
LR
LR
LR
n
n
K
d
dn


=
)
1
(
τ
(1)
Using the additional relations:
0
*
*
)
(
'
P
M
M
N
K
K
=
,
0
)
(
'
P
M
M
N
K
K
=
,
0
2
0
1
2
)
(
'
)
(
R
P
N
k
N
k
k

=
,
'
)
(
)
(
*
2
0
1
*
2
2
k
N
k
N
k
P
T
E

=
and
1
*
=
+
P
P
n
n
We can find the rate of change of
*
P
n
with
:
*
*
*
*
2
*
*
2
*
*
*
*
2
2
0
*
*
0
*
*
2
0
0
0
2
*
*
0
*
*
2
0
0
2
1
*
0
*
*
2
0
1
0
0
2
0
1
0
1
*
*
*
2
2
0
*
0
*
'
'
1
'
)
1
(
'
'
'
'
'
)
(
'
)
(
'
)
(
'
)
(
)
(
'
'
)
(
'
'
)
(
)
(
)
(
'
1
*'
)
(
'
)
(
'
)
(
)
(
)
(
'
)
(
)
(
1
1
*
)
(
)
(
)
(
1
)
(
1
1
1
2
1
p
M
p
p
M
p
LR
p
M
p
p
M
p
LR
p
P
M
p
P
p
P
M
p
P
R
LR
P
M
p
P
P
M
p
P
T
E
R
P
M
p
P
p
P
M
p
P
T
E
R
p
P
P
M
P
T
E
P
M
P
T
E
P
p
p
p
n
K
n
k
n
K
n
n
k
n
K
n
k
n
K
n
n
k
N
N
K
N
N
k
N
N
K
N
N
N
N
k
N
K
N
N
k
N
K
N
N
N
N
k
k
N
K
N
N
k
N
k
N
K
N
N
N
N
k
N
k
N
k
N
K
N
N
k
N
K
N
N
k
N
d
dt
dt
dN
N
d
dn
+


+

=
+

+
=
+

+
=
+

+
=
+

+
=
+

+
=
=




Hence the dimensionless form of equation 2 is:
*
*
*
*
2
*
*
2
*
'
'
1
'
)
1
(
'
p
M
p
p
M
p
LR
p
n
K
n
k
n
K
n
n
k
d
dn
+


+

=
(2)
2)
At t = 0, no ligand has bound to receptor, hence there is no LR complex. Thus:
0
)
(
0
)
(
0
0
0
=
=
=
=
R
R
LR
t
LR
N
N
N
n
and
1
)
(
)
(
)
(
0
0
0
0
=
=
=
=
R
R
R
LR
t
R
N
N
N
N
n
At t = 0, none of protein P has been phosphorylated since there is no E
1
(the activated kinase domain of
the receptor complex) to phosphorylate it. Thus:
0
)
(
0
)
(
0
0
*
0
*
=
=
=
=
R
P
p
t
p
N
N
N
n
and
1
)
(
)
(
)
(
0
0
0
0
=
=
=
=
R
P
P
p
t
p
N
N
N
N
n
3)
At steady state,
0
=
dt
dn
LR
and
0
*
=
dt
dn
P
2
BME 100 Computer Project 2
Since τ is proportional to t,
0
)
1
(
=


=
=
LR
LR
LR
LR
LR
n
n
K
dt
dn
d
dn
τ
Thus
0
=


LR
LR
LR
LR
n
n
K
K
LR
LR
LR
K
K
n
+
=
1
Similarly for
n
P*
,
0
'
'
1
'
)
1
(
'
*
*
2
*
*
2
*
*
=
+


+

=
=
p
M
p
p
M
p
LR
P
P
n
K
n
k
n
K
n
n
k
dt
dn
d
dn
Substituting for
LR
n
in this equation, we get
0
'
'
1
'
)
1
(
1
'
*
*
*
*
2
*
*
2
=
+


+

+
P
M
P
P
M
P
LR
LR
n
K
n
k
n
K
n
K
K
k
0
'
0
.
1
1
'
)
1
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This lab report was uploaded on 04/09/2008 for the course BME 100 taught by Professor Yuan during the Fall '07 term at Duke.
 Fall '07
 YUAN

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