1
Homework 4 Answer
(Due 02/11/2009 Wednesday)
P2.31, P2.34
P3.2, P3.3, P3.4, P3.7, P3.10, P3.15, P3.16, P3.19, P3.21, P3.22, P3.23, P3.25
P2.31)
DNA can be modeled as an elastic rod that can be twisted or bent. Suppose a DNA molecule of
length
L
is bent such that it lies on the arc of a circle of radius
R
c
.
The reversible work involved in
bending DNA without twisting is
w
bend
=
BL
2
R
2
c
where
B
is the bending force constant. The DNA
in a nucleosome particle is about 680 Å in length. Nucleosomal DNA is bent around a protein complex
called the histone octamer into a circle of radius 55 Å. Calculate the reversible work involved in
bending the DNA around the histone octamer if the force constant
B
= 2.00
×
10
–28
J m.
The bending work is given by:
(
)
(
)
(
)
J
10
2.25
m
10
55
2
m
10
680
m
J
10
2.00
R
2
L
B
w
19

2
10

10
28
2
c
bend
×
=
×
×
×
×
×
=
=
P2.34)
The formalism of Young’s modulus is sometimes used to calculate the reversible work
involved in extending or compressing an elastic material. Assume a force
F
is applied to an elastic rod
of crosssectional area
A
0
and length
L
0
. As a result of this force, the rod changes in length by
Δ
L.
Young’s modulus
E
is defined as
E
=
tensile stress
tensile strain
=
F
/
A
0
Δ
L
/
L
0
=
FL
0
A
0
Δ
L
a.
Derive Hooke’s law from the Young’s modulus expression just given.
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2
b.
Using your result from part (a), show that the reversible work involved in changing by
Δ
L
the
length
L
0
of an elastic cylinder of crosssectional area
A
0
is
w
=
1
2
Δ
L
L
0
⎛
⎝
⎜
⎞
⎠
⎟
2
EA
0
L
0
.
a) Hooke’s law says that the force with which a mass on a spring opposes displacement is directly
proportional to the displacement:
d
k
F
=
Young’s modulus shows the same dependency of the force:
L
A
L
F
E
0
0
Δ
=
L
k
L
A
L
E
F
0
0
Δ
′
=
Δ
=
,
where k’ is a constant given by:
L
A
E
k
0
0
=
′
b) The work is given by the integral:
2
0
0
0
2
0
0
2
L
0
L
0
twist
L
L
L
A
E
2
1
L
L
A
E
2
1
L
k
2
1
ds
L
k
ds
F
w
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ Δ
=
Δ
=
Δ
′
=
Δ
′
=
=
∫
∫
Δ
Δ
P3.2)
The function
f
(
x,y
) is given by
f
(
x,y
) =
xy
sin 5
x
+
x
2
y
ln
y
+
3
e
−
2
x
2
cos
y
. Determine
∂
f
∂
x
⎛
⎝
⎜
⎞
⎠
⎟
y
,
∂
f
∂
y
⎛
⎝
⎜
⎞
⎠
⎟
x
,
∂
2
f
∂
x
2
⎛
⎝
⎜
⎞
⎠
⎟
y
,
∂
2
f
∂
y
2
⎛
⎝
⎜
⎞
⎠
⎟
x
,
∂
f
∂
y
∂
f
∂
x
⎛
⎝
⎜
⎞
⎠
⎟
y
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
x