Biomedical Engineering Reference
In-Depth Information
productive collisions. 1 Typical values are between 10 8 M 21 s 21 for 'normal'
molecules to 10 10 M 21 s 21 for molecules of opposite charge and for those
molecules where the entire collisional surface area is effective in capturing the
ligand, allowing surface-diffusion of the ligand to the final binding site. 2
In the case where the free ligand concentration is known, the concentrations
of the other species at equilibrium can be obtained from a partition function
approach:
½ PL
P tot ~
K A ½ L
Z L
½ P
P tot ~
1
Z L
Z L ~1zK A ½ L
ð 10 : 2 Þ
½ PL
L tot ~ K A ½ P
L tot ~ 1
½ L
Z P ~1zK A ½ P
Z P
Z P
In these equations, Z L and Z P are the partition functions, or sums of states,
for the ligand (L) and protein (P) respectively. K A is the association constant.
[P] and P tot are the free protein and total protein concentration, respectively.
[L] and L tot are the free ligand and total ligand concentration, respectively.
[PL] is the concentration of the complex.
This approach is readily extended to include many different binding sites.
The free ligand concentration is often known in enzymatic reactions, since
the quantity of free ligand is much larger than the quantity of enzyme. In that
case, one may make the approximation [L] eq 5 [L] total . In other cases, such as
in acid-base titrations, the ligand (H 3 O + ) concentrations are known from a pH
meter electrode. However, in NMR one often deals with systems in which only
total ligand and/or total protein concentration are known, and which are
typically of the same order of magnitude. An exact, cubic equation is available
for the case of one binding site:
K A ½ PL 2 {K A P tot zL tot z1 = K A
ð
Þ½ PL zK A P tot L tot ~0
q
P tot zL tot z1 = K A
ð 10 : 3 Þ
Þ 2 {4P tot L tot
ð
P tot zL tot z1 = K A
Þ +
ð
P ½
P tot ~
2P tot
Beyond this simple case, numerical methods must be used to obtain the
equilibrium concentrations of the species. We prefer to numerically integrate
kinetic equations until equilibrium is reached. For the simple case shown above
we obtain for the time dependencies of the individual concentrations:
d ½ P t
dt ~{k f ½ P t ½ L t zk b ½ PL t
d ½ L t
ð 10 : 4 Þ
dt ~{k f ½ P t ½ L t zk b ½ PL t
d ½ PL t
dt
~zk f ½ P t ½ L t {k b ½ PL t
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