Biomedical Engineering Reference
In-Depth Information
k
0
k
B
T
≥
/
than unity. That is, the mode is only defined when
r
f
x
β
. Caution should
be taken when applying the first-passage solution for the mean or mode to force
spectra in general because, in principle, the first-passage model is correct only when
(1) the bond is driven at very large loading rates, which negate rebinding effects
and/or (2) the linkage to the bond is very soft, such as a long polymer, which imparts
a large degree of entropic freedom to the unbound state. Otherwise, as the loading
rate is decreased, the force spectrum can enter into a linear-response regime whereby
the rebinding rate is effective at small forces, and the simple first-passage model is
no longer valid. This near-equilibrium scenario is detailed below.
3.4.2 R
EVERSIBLE
T
WO-
S
TATE
A
PPROXIMATION
As explained at the beginning of this section, the complete description of bond rup-
ture includes the dynamics of both the bound state (on) and the unbound state (off):
d
d
F
p
on
r
f
(
F
)=
−
k
off
(
F
)
p
on
(
F
)+
k
on
(
F
)
p
off
(
F
)
(3.57)
where
k
off
(
F
)
and
k
on
(
F
)
are the rates of leaving the bound and unbound states,
respectively, and
p
on
are, respectively, the probabilities of finding the
system bound or unbound at a force
F
. These probabilities have initial conditions
(
F
)
or
p
off
(
F
)
(
)=
,
(
)=
p
on
0
1
p
off
0
0
(3.58)
and
p
on
(
F
)+
p
off
(
F
)=
1. At equilibrium d
F
/
d
t
=
r
f
=
0, Equation 3.57 yields the
principle of detailed balance
k
off
(
F
)
p
on
(
F
)=
k
on
(
F
)
p
off
(
F
)
(3.59)
which states that the number of transitions per unit time between the pair of states
are equal. Rewriting this equality in terms of the relative population of the two states,
we have
p
off
(
F
)
k
off
(
F
)
)
=
)
=
exp
[
−
Δ
G
(
F
)
/
k
B
T
]
(3.60)
p
on
(
F
k
on
(
F
where Δ
G
is the free energy difference between the bound and unbound states
when held at an external load
F
. It turns out that, under the simple analytic transition
rates derived above, a straightforward expression for Δ
G
(
F
)
can be found. Starting
from the rebinding rate in Equation 3.26, we can expand the term in parentheses to
find
(
F
)
k
on
exp
2
k
cant
x
2
β
k
B
T
F
2
2
k
cant
+
1
k
on
(
F
)=
−
Fx
β
−
/
(3.61)
exp
Δ
G
0
k
B
T
F
2
2
k
cant
=
k
off
(
F
)
−
/
(3.62)
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