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10.37 Exam 2
25 April, 2007
100 points
Problem 1: 35 points
A protein and ligand bind reversibly with
K
d
=
10 nM . The association rate constant
k
=
2
x
10
4
M
−
1
s
1
. The two species are mixed at an initial protein concentration of 3 nM
on
and an initial ligand concentration of 0.2 nM.
a) At equilibrium, what fraction of the ligand will be complexed with protein? (15 points)
b) At what time will the fraction of ligand in complex reach 95% of the equilibrium
value? (20 points)
Justify any assumptions you make to simplify equations.
P
+
L
←→
C
k
off
[
P
]
eq
[
L
]
eq
K
d
=
=
k
on
[
C
]
eq
Using a batch reactor mole balance and looking at the reaction stoichiometry, it is easy to
see that given the initial conditions any unit of complex formed takes away a unit of
protein and ligand:
[
P
]
+
[
C
]
=
[
P
]
0
[
L
]
+
[
C
]
=
[
L
]
0
Using these in the equilibrium equation we can get a quadratic equation in [C]
eq
.
(
[
P
]
0
−
[
C
]
eq
)(
[
L
]
0
−
[
C
]
eq
)
K
d
=
[
C
]
eq
0
=
[
C
]
eq
2
−(
[
L
]
0
+
[
P
]
0
+
K
d
)
[
C
]
eq
+
[
P
]
0
[
L
]
0
=
0
[
C
]
=
(
[
L
]
0
+
[
P
]
0
+
K
d
)±
(
[
L
]
0
+
[
P
]
0
+
K
d
)
2
−
4[
P
]
0
[
L
]
0
=
0
eq
2
Reject the positive root, it is too large (larger than the initial amount of ligand and
protein).
[
C
]
eq
=
0.0456
nM
[
C
]
eq
=
0.228
=
23%
[
L
]
0
At this point, it is interesting to look at different approximations to the expression.
Good Approximation: [P]
0
>> [C]
This leads to
(
[
P
]
0
)(
[
L
]
0
−
[
C
]
eq
)
K
d
≈
[
C
]
eq
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].
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C
]
=
[
L
]
0
[
P
]
0
0.462
nM
eq
K
d
+
[
P
]
0
[
C
]
eq
=
0.231
=
23%
[
L
]
0
Bad Approximation: [L]
0
>> [C]
(
[
P
]
0
−
[
C
]
eq
)(
[
L
]
0
)
K
d
≈
[
C
]
eq
[
C
]
eq
=
[
L
]
0
[
P
]
0
=
0.0588
K
d
+
[
L
]
0
[
C
]
eq
=
0.294
=
29%
[
L
]
0
The error in [C]
eq
of the bad approximation is about 30% of the true answer, whereas the
good approximation is only off by about 1%.
By noticing that the “good” approximation is a
good
approximation, the dynamic
equation becomes easier to solve.
(As an aside, an even better approximation would be
just to neglect the second order term that is O([C]
eq
2
).)
Start with the full dynamic equation:
d
[
C
]
=
k
on
[
L
][
P
]
−
k
off
[
C
]
=
k
on
(
[
L
]
0
−
[
C
]
)
(
[
P
]
0
−
[
C
]
)
−
k
off
[
C
]
dt
Make an appropriate approximation:
(
[
P
]
0
−
[
C
]
)
≈
[
P
]
0
d
[
C
]
≈
k
on
[
L
]
0
[
P
]
0
−
k
on
[
P
]
0
[
C
]
−
k
off
[
C
]
dt
Rearrange and solve using the integrating factor:
d
[
C
]
+
[
C
]
(
k
on
[
P
]
0
+
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