Biomedical Engineering Reference
In-Depth Information
(Williams, 2003),
Nk 0 exp Fx β
Nk B T
k N N 1
(
F
)=
(3.82)
Because all N bonds are uncoupled, their failure is independent from one another.
Hence, we describe the complete failure as a Markov process, such that the total
time required for all bonds to fail is simply the sum of lifetimes of each step in the
unbinding pathway.
N
(
N
1
) (
N
2
) → ··· →
1
0
τ N
τ N 1
τ N 2
τ 2 τ 1
where the lifetime of each step is given by the inverse of Equation 3.82,
1
τ N
(
F
)=
(3.83)
k N
(
F
)
N
1
Thus, the rate of unbinding all N bonds is given by the inverse of the sum of each
individual lifetime in the sequence
1
N
n = 1
1
n = 1 τ n =
1
nk 0 exp Fx β
nk B T
k N
(
F
)=
(3.84)
0
We will first consider when the N bonds are loaded quickly, such that the force on
the system drives the bonds apart irreversibly. In general, however, the scheme above
illustrating the steps in the unbinding pathway will also have arrows pointing to the
left to account for rebinding. When unbinding is completely irreversible, we describe
the process by first-passage statistics. As the force increases rapidly on the cluster of
bonds, the failure of the first bond will result in a sudden increase in load on the
remaining bonds. Therefore, it is reasonable to assume that for fast loading rates, the
failure of the remaining bonds will occur soon after, if not immediately after, the rup-
ture of the first bond. Under this assumption, we can treat the dynamics in the same
manner as we would treat a single bond (Evans & Williams, 2002; Williams, 2003).
That is, the failure of all N bonds will occur with a rate given by Equation 3.84,
and most frequently at a force F . As explained in Equation 3.55, Evans & Williams
(2002) showed that for first-passage processes, the most probable rupture force F
follows from a relationship between loading rate and unbinding rate:
F
F =
r f dln k
(
F
)
k
(
F
)
(3.85)
d F
Operating the relationship in Equation 3.85, on the transition rate in Equation 3.84, a
transcendental equation for the most probable force of failure F is found (Williams,
2003):
n 2 exp
F x β
nk B T
N
n = 1
1
r f =
x β
k 0 k B T
1
(3.86)
Search WWH ::




Custom Search