AMATH/BIOL 382 Problem Set 3
Solutions
1. (i) Running the file
# Assignment 3 Problem 2i
s1’=Vmax1Vmax2*S1/(Km2+s1)
s2’=Vmax2*S1/(Km2+s1)Vmax3*s2/(Km3+s2)
# parameters
par Vmax1=1,Vmax2=2,Vmax3=4,Km2=4,Km3=3
done
yields steady state (
s
1
, s
2
) = (4
,
1).
(ii) Perturbing
V
max
2
to
V
max
2
= 2
.
01 results in a new steady state of (
s
1
, s
2
) = (3
.
96
,
1). This
corresponds to a sensitivity of zero in
s
2
and of
3
.
96

4
2
.
01

2
=

4
Checking at
V
max
2
= 1
.
99 gives a similar result
4
.
04

4
1
.
99

2
=

4
The absolute sensitivity of
s
1
with respect to
V
max
2
is thus 4.
The relative senstivity is
2
4
(

4) =

2. The relative sensitivity in
s
2
is zero.
(iii) With this new nominal parameter set, we find steady state (
s
1
, s
2
) = (
1
3
,
1). Making
a slight change to
V
max
2
= 1
.
01 results in a new steady state of (
s
1
, s
2
) = (0
.
329
,
1). This
corresponds to a sensitivity of zero in
s
2
and of
0
.
329

0
.
333
1
.
01

1
=

0
.
43
The absolute sensitivity of
s
1
with respect to
V
max
is thus 0.43. The relative senstivity is
1
1
/
3
(

0
.
43) =

1
.
3. The relative sensitivity in
s
2
is zero.
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 Spring '10
 Leroy
 Homeostasis, Steady State, #, #metabolic

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