Problem 1.
The binding of protein P with ligand L to form complex C is reversible, as told
k on
off
P + L ⎯⎯→
C and C ⎯k⎯
→P + L
We are given a table with various initial concentrations of L in order to estimate kon and koff and
also Kd for the reaction.
dCC
= k on C P C L − k off CC
dt
Also from material balances and stoichiometry, we have CP+CC=CP0 and CL+CC=CL0, therefore
dCC
= k on (C P0 - C C )(CL0 − CC ) − k off CC
dt
In this problem, we can safely assume that CL0-CC≈CL0 since CL0>> CP0 in all three cases of
different CL0.
Thus, the integrated analytic expression for CC becomes
CC =
k on C P0 C L0
C C
{1 - exp[-(k on C L0 + k off )t]} = P0 L0 {1 − exp[-(k on C L0 + k off )t]}
k on C L0 + k off
C L0 + K d
where K d ≡
k off
k on
Therefore, if we plot CC w.r.t time for each cases of CL0, we can fit according to an exponential
C P0 C L0
. Values for a and b are shown in the
C L0 + K d
y=a[1-exp(-bt)], where b is konCL0+koff, and a is
following table.
L0(uM)
a
b
1
0.903
1.1113
5
1.0436
4.7047
15
0.9932
15.1079
One important observation in this table is that parameter a does not change much when initial
ligand concentration is changed, indicating CL0=1μM is already above the saturating value.
Therefore, value for koff can not be obtained accurately from this design of experiments. We can
only conclusively obtain the value for kon.
So we fit a linear express of b vs. CL0 to get kon.
kon=0.0010 nM-1 sec-1
An estimate on koff would be
0<=koff<<konCL0,min =0.0010 nM-1 sec-1*1uM=1 sec-1
Similarly an estimate on Kd is
0<= Kd<< CL0,min=1μM.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
CL0=1uM
0.4
0.35
Concentration of C(nM)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.05
0.1
0.15
0.2
0.25
time(t)
0.3
0.35
0.4
0.45
0.5
0.3
0.35
0.4
0.45
0.5
CL0=5uM
1
0.9
Concentration of C(nM)
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.05
0.1
0.15
0.2
0.25
time(t)
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
CL0=15uM
1
0.9
Concentration of C(nM)
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.05
0.1
0.15
0.2
0.25
time(t)
0.3
0.35
0.4
0.45
0.5
16
14
12
10
b
8
6
4
2
0
-2
0
5
10
15
CL0(uM)
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Problem 2.
For a steady state chemostat, the material balance on cell mass yields
D≡
F
=μ
V
the volumetric productivity is
F(X- X0)=μV(X- X0)
For a batch reactor, the material balance on cell mass yields
dX
= μX
dt
where the initial condition is X(t=0)=X0.
Therefore, we have X = X 0 exp(μt )
The volumetric productivity is
V(X - X 0 )
V(X - X 0 )
=
1
X
t + t turn
ln
+ t turn
X0
μ
Therefore, the ratio of the two
μV(X - X0)
V(X - X 0 )
1
X
ln
+t
μ X 0 turn
= ln
X
+ μt turn
X0
In practice, for chemostat, in order to maximize the productivity of biomass (DX), the operating
condition for μ is close μmax. Therefore the ratio above is approximately
ln
X
+ μ max t turn
X0
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Problem 3.
The expression that would be suitable to describe the change of protein expression is:
CP =
k Pk r
{1 − exp[−(γ P + μ )t ]}
γ r (γ P + μ )
where the meaning of each symbol is in accord with what we did in class.
The time required to change from an “off state” to an “on state” (95% of the steady-state value) is
0.95 = 1 − exp[−(γ P + μ )t ] or t = −
ln 0.05
γP + μ
a) if cells are rapidly growing with a doubling time 30 min and stable protein with a degradation
half-time one day, i.e.
ln2
μ
= 30min and
ln2
γP
= 1 day
So the half-time for switching is the time need to reach
1
= 1 − exp[−(γ P + μ )t ]
2
1
1
- ln(1 - 0.95 × ) - ln(1 - 0.95 × )
2 =
2 ≈ 27.9 min
=
ln 2
ln 2
(γ P + μ )
+
30 min 1day
0.95 ×
t switching
b) if cells are not growing at all and the protein with a degradation half-time one hour, i.e.
ln2
γP
= 1 hr
So the half-time for switching is
t switching
1
1
- ln(1 - 0.95 × ) - ln(1 - 0.95 × )
2 =
2 = 0.93hr
=
ln
2
(γ P + μ )
+0
1hr
So in both cases, the switching times are much much longer than that of the current electronic
circuits, which is on the order of ~ns-μs. Thus, it would not be promising in realizing a computer
for practical uses.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].