University of Washington
Department of Chemistry
Chemistry 453
Winter Quarter 2012
Homework Assignment 8; Due Friday 3/09/12
Text Problems: 24.3, 24.5, 24.7
24.3)
24.5) Assume one dimensional diffusion. The solute is in the bottom of the cylinder so
we integrate from zero to infinity. But before doing that we have to normalize the
probabioity for motion in only the +x direction:
∞
4
1
1 = c ∫ e − x /4 Dt =
4π Dt = π Dt
2
0
∞
x =
4
Dt
1
1 4 Dt
xe − x /4 Dt =
=2
∫
π
π Dt 0
π Dt 2
∞
4
1
1 4 Dt
x =
x 2 e − x / 4 Dt =
4π Dt = 2 Dt
∫
π Dt 0
π Dt 4
Then just substitute D for sucrose (see 24.3) and t into these expressions.
2
24.7)
1) Consider a one dimensional random walk composed of N jumps where for each
jump the probability of a net excursion by +m (i.e. to the right) is p and the
probability of an excursion by –m (i.e. to the left) is q. If p≠q this is called a
biased random walk and the probability of observing a displacement of m jumps
is
⎡ ( m − N ( p − q ) )2 ⎤
1
⎥
Ppq ( m, N ) =
exp ⎢ −
8 Npq
⎢
⎥
8π Npq
⎣
⎦
a) Show that if p=q=1/2 Ppq ( m, N ) becomes the unbiased random walk distribution.
P ( m, N ) =
m2
Solution:
⎡ m2 ⎤
1
exp ⎢ −
⎥ For the unbiased distribution calculate m and
2π N
⎣ 2N ⎦
⎡ ( m − N ( 1 − 1 ) )2 ⎤
⎡ ( m − N ( p − q ) )2 ⎤
1
1
2
2
⎥
⎥=
Ppq ( m, N ) =
exp ⎢ −
exp ⎢ −
8 Npq
8N / 4
⎢
⎥
⎢
⎥
8π Npq
8π N / 4
⎣
⎦
⎣
⎦
⎡ m2 ⎤
1
exp ⎢ −
⎥
2π N
⎣ 2N ⎦
=
+∞
∫
m =
1
2π N
mP ( m )dm =
−∞
+∞
∫
m2 =
m 2 P ( m )dm =
−∞
+∞
⎡ m2 ⎤
m
exp
⎢−
⎥ dm = 0
∫−∞
⎣ 2N ⎦
+∞
⎡ m2 ⎤
2
m
exp
⎢ − 2 N ⎥ dm
∫
⎣
⎦
−∞
1
2π N
1 2
( 2 N ) 2π N = N
2π N 4
=
+∞
b) Assuming a biased random walk, calculate m =
∫ mP ( m, N ) dm and calculate
pq
−∞
+∞
also m
∫ m P ( m, N ) dm . Show that these results equal the unbiased results
=
2
2
pq
−∞
when p=q.
+∞
1
m = ∫ mPpq ( m )dm =
8π Npq
−∞
y = m − N ( p − q)
∴ m =
1
8π Npq
⎡ ( m − N ( p − q ) )2 ⎤
⎥dm
∫ m exp ⎢⎢−
8 Npq
⎥
−∞
⎣
⎦
+∞
+∞
⎡
2
⎤
y
∫ ( y + N ( p − q ) ) exp ⎢− 8 Npq ⎥dy =
⎣
−∞
N ( p − q ) +∞
⎦
⎡
y2 ⎤
exp
−
∫ ⎢⎣ 8 Npq ⎥⎦dy = N ( p − q )
8π Npq −∞
Note when p=q… m = 0
+∞
m
2
1
= ∫ m Ppq ( m )dm =
8π Npq
−∞
2
y = m − N ( p − q)
∴ m2 =
=
=
1
8π Npq
+∞
⎡ ( m − N ( p − q ) )2 ⎤
⎥dm
∫ m exp ⎢⎢−
8 Npq
⎥
−∞
⎣
⎦
+∞
∫ ( y + N ( p − q ))
−∞
2
2
⎡
y2 ⎤
exp ⎢ −
⎥dy
⎣ 8 Npq ⎦
1
8π Npq
+∞
1
8π Npq
2
⎡2
⎤
2
⎢⎣ 4 ( 8 Npq ) 8π Npq + N ( p − q ) 8π Npq ⎥⎦
∫(
−∞
)
⎡
y2 ⎤
2
y 2 + N 2 ( p − q ) exp ⎢ −
⎥dy
⎣ 8 Npq ⎦
= 4 Npq + N 2 ( p − q )
2
When p=q=1/2 m 2 = N
2) A macromolecule has a diffusion coefficient D = 6.9 ×10−11 m 2 s −1 at T=298K and
for a solvent viscosity of η = 0.891×10−3 kg m −1s −1 .
a) Assuming the molecular is approximately spherical in solution, calculate
the translational frictional coefficient ζtr and the radius.
kT
kT
D= B = B
6πη R
f
1.38 × 10−23 JK −1 ) ( 298K )
(
k BT
∴R =
=
= 3.55 ×10−9 m
6πη D ( 6π ) ( 0.891× 10−3 kg m −1s −1 )( 6.9 × 10−11 m 2 s −1 )
∴ f = 6πη R = ( 6π ) ( 0.891× 10−3 kg m −1s −1 )( 3.55 × 10−9 m ) = 5.96 × 10−11 kgs −1
b) Calculate the root-mean-squared displacement of the molecule after 1
millisecond.
2
∆r = 6 D∆τ = 6 × 6.9 × 10−11 m 2 s −1 × 0.001s = 4.14 × 10−13 m 2
∴ rrms =
∆r 2 = ( 41.4 × 10−14 m 2 )
1/ 2
= 6.43 × 10−7 m
c) For the purpose of calculating the translational friction f, many rod-like
polymers can be approximated in solution as a chain of beads, each bead
of diameter d. For such a polymer the frictional coefficient is
3πη Nd
ftr =
ln N
where N is the number of beads in the polymer chain and η is the solvent
viscosity. . Suppose the macromolecule in part a aggregates as a linear, rod-like
hexamer. Calculate the coefficient of translational friction and the coefficient of
translational diffusion. Calculate the rms displacement after 1 millisecond.
Assume η = 0.891×10−3 kg m −1s −1
−3
−1 −1
−9
3πη Nd ( 3π ) ( 0.891×10 kg m s ) ( 6 ) ( 2 × 3.55 × 10 m )
=
ftr =
ln N
ln ( 6 )
3.58 × 10−10
=
kg s −1 = 2.00 ×10−10 kg s −1
1.79
−23
−1
k BT (1.38 × 10 JK ) ( 298K )
D=
=
= 2.06 × 10−11 m 2 s −1
f tr
2.00 ×10−10 kg s −1
∆r 2 = 6 D∆τ = 6 × 2.06 × 10−11 m 2 s −1 × 0.001s = 1.24 × 10−13 m 2
∴ rrms =
∆r 2 = (12.4 × 10−14 m 2 )
1/2
= 3.52 × 10−7 m
d) Suppose the molecule in part a forms a hexamer in solution, but the
aggregate is roughly spherical, with a volume equal to about six times the
volume of the spherical monomer. Calculate the coefficient of frictional
and the coefficient of translational diffusion. Calculate the rms
displacement after 1 millisecond.
1/3
4π R13
⎛ 3V ⎞
V1 =
⇒⎜ 1⎟
3
⎝ 4π ⎠
= R1
1/3
1/3
4π R63
⎛ 18V1 ⎞
1/3 ⎛ 3V1 ⎞
V6 = 6V1 =
⇒ R6 = ⎜
⎟ =6 ⎜
⎟
3
⎝ 4π ⎠
⎝ 4π ⎠
∴ R6 = 61/3 × 3.55 ×10−9 m = 6.45 × 10−9 m
= 61/3 R1
f 6 = 61/3 × ζ 1 = 61/3 × 5.96 × 10−11 kgs −1 = 1.08 × 10−10 kgs −1
D1 6.9 × 10−11 m 2 s −1
= 3.81×10−11 m 2 s −1
D6 = 1/3 =
6
1.81
e) Based on your answers in part a-d, can translational diffusion coefficients
be used to detect aggregation? How sensitive are diffusion coefficients to
aggregate geometry?
Looks promising. Translational friction of the linear aggregate is twice that of the
spherical aggregate geometry. In fact, dynamic light scattering…which measures
D…is used to monitor for aggregation in macromolecular solutions.
3) Consider the following data for the proteins myoglobin and hemoglobin
Specific
Frictional
Protein
Diffusion
Molecular
Volume V2
ratiof/f0
Coefficient
Weight (kg) Dx1011m2s-1 (mLg-1)
Myoglobin
16.900
11.3
0.74
1.11
Hemoglobin 64.500
6.9
0.75
1.16
a) Calculate the radii of myoglobin and hemoglobin assuming they are unhydrated
spheres.
Solution:
4π R03 MV2
=
3
NA
For Mb:
1/3
⎛ 3 MV2 ⎞
R0 = ⎜
⎟
⎝ 4π N A ⎠
⎛ 3 (16900 gmol −1 )( 0.74cm3 g −1 ) ⎞
⎟
=⎜
⎜ 4π
⎟
6.023 ×1023 mol −1
⎝
⎠
1/3
= 1.18 × 10−7 cm
For Hb:
⎛ 3 ( 64500 gmol −1 )( 0.75cm3 g −1 ) ⎞
⎟ = 2.68 ×10−7 cm
R0 = ⎜
⎜ 4π
⎟
6.023 × 1023 mol −1
⎝
⎠
b) Calculate the radii and volumes of myoglobin and hemoglobin that would account
for the frictional ratios, assuming they are hydrated spheres.
Solution:
1/3
For Mb : R = R0
f
= (1.18 ×10−9 m ) (1.11) = 1.31× 10−9 m
f0
3
4π 3 4π
R =
1.31×10−9 m ) = 9.41× 10−27 m3
(
3
3
f
= ( 2.68 ×10−9 m ) (1.16 ) = 3.11× 10−9 m
For Hb : R = R0
f0
VMb =
3
4π 3 4π
R =
2.68 × 10−9 m ) = 8.06 × 10−26 m3
(
3
3
c) For each protein, calculate δ1, the mass of water bound per mass of protein.
Assume the density of water is 1g mL-1.
VHb =
There are several ways to do this...the easiest is to use the friction ratio provided.
For Mb:
3
⎛ f ⎞ V2 + δ1V1
⎜ ⎟ =
V2
⎝ f0 ⎠
⎡⎛
⎢⎜
⎢⎣⎝
For Hb:
V ⎡⎛
δ1 = 2 ⎢ ⎜
V1 ⎢⎝
⎣
V
δ1 = 2
V1
3
⎤ 0.74
f ⎞
⎡(1.11)3 − 1⎤ = 0.27
⎥=
−
1
⎟
⎦
f0 ⎠
1 ⎣
⎥⎦
3
⎤ 0.75
f ⎞
⎡(1.16 )3 − 1⎤ = 0.42
⎟ − 1⎥ =
⎦
f0 ⎠
1 ⎣
⎥⎦
Note: You can also determine the number of water molecules that hydrate a molecule
of protein from this ratio.
M Mb
16900
= ( 0.27 )
≈ 254
In the case of Mb... # water molecules = δ1
18
M water
M Hb
64500
In the case of Hb:... # water molecules = δ1
= ( 0.42 )
≈ 1505
M water
18
d) Myoglobin is a oxygen storage protein found in the body tissues. Hemoglobin is
an oxygen transport protein. Hemoglobin is composed of n sub-units each roughly
the size of myoglobin. Based on your answers in part b, how many myoglobinlike sub-units does hemoglobin contain?
Four
4) Collagen is a structural protein with a molar mass of 350 kg, a specific volume of
V2=0.70 mLg-1, and a diffusion coefficient of D2=0.71x10-11 m2s-1. The frictional ratio
2/3
f ( 2 / 3) P
(i.e. the Perrin factor) for a long cylinder is
=
where P is the length to
f 0 ln P + 0.312
diameter ratio: P=L/d.
1/3
a) Calculate f0 for collagen assuming η=0.891x10-3 kgm-1s-1.
Solution: As the hint says...assume the protein is unhydrated. Then...
4π R03 MV2
=
NA
3
For Collagen:
1/3
⎛ 3 MV2 ⎞
R0 = ⎜
⎟
⎝ 4π N A ⎠
⎛ 3 ( 350000 gmol −1 )( 0.70cm3 g −1 ) ⎞
⎟
=⎜
⎜ 4π
⎟
6.023 × 1023 mol −1
⎝
⎠
1/3
= 4.60 ×10−7 cm
f 0 = 6πη R0 = ( 6π ) ( 0.001kgm −1s −1 )( 4.60 × 10−9 m ) = 8.67 × 10−11 kgs −1
b) Calculate f for collagen assuming T=298K
−23
−1
k T (1.38 ×10 JK ) ( 298 K )
= 5.79 ×10−10 kgs −1
c) f = B =
−11 2 −1
0.71×10 m s
D
d) From the frictional ratio, assuming collagen is a long cylinder, determine the
length-to-diameter ratio. Hint: Do this graphically…vary P until you get the
observed Perrin factor.
f 57.9
=
= 6.68 ...so very non-spherical
f 0 8.67
2/3
f ( 2 / 3) P
To determine P graph
=
vs. P, and read off P for a friction ration of
f 0 ln P + 0.312
1/3
2/3
f ( 2 / 3) P
6.68 =
. From the graph below for a friction ratio of 6.68 P is around
f 0 ln P + 0.312
350.
1/3
Collagen friction ratio vs. P.
8
7
6
f/f0
5
4
3
2
1
0
0
100
200
300
400
500
P=L/d
5) Chromatin is a complex of DNA and proteins (mostly histones) found in all
eukaryotic cells. Chromatin is a composed of a sequence of spherical particles called
nucleosomes. The distribution of DNA and protein within the nucleosome particle
can be determined qualitatively using a random flight polymer model for the
nucleosomal DNA and by viewing the nucleosome as a hydrodynamic sphere.
a) Assuming the nucleosome particle is spherical, calculate its Stokes radius. Assume
the viscosity of the solution is 0.01 gm cm-1 s-1. Assume the diffusion coefficient for
the nucleosome particle is 4.37x10-7 cm2/s at T=293K.
f = 6πη R =
k BT
D
1.38 × 10−23 JK −1 ) ( 293K )
(
k BT
∴R =
=
= 4.91×10−9 m
−1 −1
−11
2
6πη D ( 6π ) ( 0.001kgm s )( 4.37x10 m / s )
b) The DNA within the nucleosome particle is about 200 base pairs in length. Assuming
each base pair in DNA is separated from the adjacent base pairs by about 3.4x10-10 m,
approximately how long is a piece of DNA 200 base pairs in length? Based on this
nucleosome.
b
gc
h
Length of DNA 200 base pairs 3.4 × 10 −8 cm / base pair = 68 × 10 −7 cm
Because the length of the DNA is much greater than the diameter of the nucleosome
particle we assume the DNA must be tightly folded in the nucleosome
c) In addition to the DNA there are eight histone proteins (i.e. the histone octamer) in
the nucleosome particle. How are the proteins and DNA packed in the nucleosome?
To answer this question, assume the eight proteins form a unhydrated, spherical
complex with specific volume 0.74 cm3/g. Assume the molecular weight of the
histone octamer is 2.73x105 gm/mole. Calculate the radius of this hypothetical protein
sphere.
Weight Histone Pr otein Complex = Weight Nucleosome − Weight DNA
= 2.73 × 105 gm / mole − ( 200 base pairs )( 660 gm / mole base pairs ) ≈ 1.38 × 105 gm / mole
1/3
R protein
⎛ 3 MV2 ⎞
=⎜
⎟
⎝ 4π N A ⎠
⎛ 3 (138, 000 g / mole )( 0.74cc / g ) ⎞
⎟ = 3.5 ×10−7 cm = 3.5 × 10−9 m
=⎜
−1
23
⎜ 4π
⎟
( 6.02 ×10 mole )
⎝
⎠
d) Assume the 200 base pair DNA behaves as a random flight polymer. Calculate the
root mean square (rms) end-to-end distance. Comparing the protein sphere radius
with the rms end-to-end distance for the DNA, is most of the DNA packed outside or
inside the protein core of the nucleosome? Explain.
e)
2
RDNA
= A N = ( 3.4 × 10−10 m ) ( 200 )
1/ 2
= 4.81× 10−9 m
2
>> R protein , the DNA occupies a larger volume than the protein and
Note because RDNA
is thus likely wrapped around the exterior of the protein. This is in fact the result that was
found when the crystal structure of the nucleosome was finally reported in the 1990’s.