Chapter 6 – Forces 257
Answers
a
The Kratsky–Porod model states that the end-to-end separation is:
R
l R
l
R
l
WLC
p
max
p
max
p
2
2
2
2
1
(
) =
−
−
−(
)
(
)
exp
/
The local pointing direction of a filamentous “floppy” molecule changes often
over distances that are relatively short compared to its maximum end-to-end
length Rmax, also known as the molecule’s contour length, indicating that its per
sistence length lp«Rmax where, and so:
R
l R
l
l R
WLC
p
max
p
p
max
2
2
2
2
2
1 1 1
2
2
(
) =
−
−
+
+
+…
(
)
(
) ≈
/
α
α
!
if Rmax/lp = α and α » 1. In the Gaussian chain approximation (Equation 8.42),
the mean squared distance of one end of a polymer to the other for n random
“polymer walk steps” of size b, if we take other end as the origin, is:
R
bx
p x
b n
GC
2
2
2
(
) =
(
)
=
−∞
+∞
∫
d
where p
n
x
n
=
−
1
2
2
2
2
π
exp
(from Equation 8.42). The length b is the Kuhn length,
which is roughly 2lp for a floppy polymer (see Equation 8.47) thus:
R
R
WLC
GC
2
2
(
) (
)
≈
In the case a stiff molecule, lp»Rmax so:
R
l R
l
R
WLC
p
max
p
max
2
2
2
2
2
2
1
1
2
(
) =
−
−
+ −
(
)+ −
(
)
+…
≈
α
α
!
In other words, the end-to-end distance is the same as the contour length of the
molecule, with is the definition of the rodlike limit.
b
Here, Rmax is 15,000 × 0.34 nm (see section 2.3.5) or ~5,100 nm or 5.1 µm. The bead
is a vertical distance 1 µm from the point of attachment of the DNA on the cover
slip, so the distance between the center of the bead at this maximum deflection
of 38% of its own radius (i.e., 0.688 µm) and its point of attachment to the sur
face of the coverslip with some simple trigonometry is √(1.02 + 0.6882) = 1.214
µm (note the point of attachment of the DNA on the bead at maximum deflec
tion will not be vertically directly below the center of the bead since the direc
tion of force from the tether will act through the bead center by conservation of
angular momentum—that is, the bead must rotate as it is laterally deflected). So,
the mean end-to-end distance of the DNA tether is 1.214 minus the bead radius,
that is, 1.214 – 0.5 or 0.714 µm. So, assume the floppy molecule result of part (a),
this implies:
lp ≈ 0.714/(2 × 5.1) = 0.050 µm or 50 nm.