Problem Set 3 Solutions
1. We have a formula for the thermally-averaged dipole-dipole
interaction potential energy, for two identical dipoles of moment p:
| < U > = - |
2
3 |
|
k2 p4
kB T r6 |
|
|
We need the dipole moment for a water molecule, which must be on the order
of an electron charge times about 0.5 A or about 0.8 ×10-29
C·m. This estimate is a bit too large because the electron charges
are rather spread out in the H2O; the real value is about 0.6
×10-29 C·m, which is
2 Debye units (recall 1 Debye = 3×10-30
C·m).
We just plug in p = 6×10-30
C·m, k = 9×109 N·m2/C2,
kB T = 4×10-21 J,
and r = 3×10-10 m. The resulting
potential is
| < U > = - |
2
3 |
|
(9×109)2 (6×10-30)4
(4×10-21)
(3×10-10)6 |
|
|
= 2.4 ×10-20 J = 6 kB
T
This estimate neglects the dielectric screening caused by the water
molecules in the 3 A between our two interacting water molecules. At a
range of 3 A, a value of dielectric constant of e
= 5 is reasonable, which reduces k to k/e. Thus
our estimated interaction potential ends up reduced by a factor of 25,
to < U > = 9.6 ×10-22 J
= 0.24 kB T.
By comparison, a hydrogen bond has a cohesive energy somewhat larger
than this, about 5 kB T.
2. We will consider the electric field due to the point charge
a distance r from the charge, and the dipole whose orientation is subject
to thermal fluctuation.
The interaction energy between a point dipole p and a static
electric field E is U = -p·E
= -p E cosq, where
q is the angle between the electric field and
the dipole direction. The probability of each orientation is proportional
to the Boltzmann distribution, exp[pE cosq/(kB
T)]. We will call the combination p E/(kB T) = a.
Integrating over all directions of the dipole using the Boltzmann distribution,
including the normalization, gives the averaged potential
| < U > = - p E |
|
ó
õ |
p
0 |
dqsinqcosqexp[a
cosq] |
|
ó
õ |
p
0 |
dqsinqexp[a
cosq] |
|
|
|
These are just the integrals encountered in the theory of the random flight
polymer. Here their meaning is different, but the calculation is just a
repeat of the random-flight polymer elasticity calculation that you can
find in the notes...
| = - p E |
é
ê
ë |
cotha - |
1
a |
ù
ú
û |
|
|
For U << kB T, we have a << 1, so that we can keep
just the linear term in the power series expansion of the above,
| < U > = - |
p E a
3 |
= - |
p2 E2
3 kB T |
|
|
Finally we should plug in the value of E = k q / er2
for the E-field a distance r from the charge, giving
| < U > = - |
p k2 q2
3 kB T e2
r4 |
|
|
3. The charge q simply induces a dipole moment proportional
to the field at the molecule, p = aE = ak
/ (er2).
Note that this induced dipole moment is always in the direction
of the electric field, no matter what the orientation is of the molecule.
Therefore the interaction potential energy is simply
(b) Averaging over the orientations of the molecule does not change this,
since the induced dipole moment is always in the same direction:
4. (a) This problem involves a `two-state system', where one
state the random (open) coil, and where the other is the (closed) hairpin.
The Boltzmann distribution tells us
|
Prandom
Phairpin |
= |
exp[-Grandom/(kB
T)]
exp[-Ghairpin/(kB
T) |
= exp[DG/(kB
T)] |
|
Since Prandom + Phairpin = 1 we have then
| exp[DG/(kB
T)] = |
1 - Phairpin
Phairpin |
|
|
or
|
DG
kB T |
= ln |
æ
ç
è |
|
1 - Phairpin
Phairpin |
ö
÷
ø |
|
|
Reading off the hairpin probabilies at 40, 50 and 60 C (see graph), Phairpin
= 0.9, 0.5 and 0.1, and plugging in gives us DG
/ (kB T) = -2.2 , 0, and +2.2. The
change in kB T over the 20 degree range of these measurements
is only a few percent, so we just take kB T = 4×10-21
J, which means
DG = -8 ×10-21
J, 0 and +8×10-21 J at 40,
50 and 60 C.
(b) The dependence of DG on temperature is
nearly linear,
DG = (8 ×10-21
) (T - 323 )/10 J, where T is now in Kelvins
(recall 50 C = 273+50 K = 323 K). Rewriting this
| DG = -2.6
×10-19 J + (0.8×10-21
J/K) T = DE - T DS |
|
The simple identification of this linear dependence with energy and entropy
differences as indicated above gives us
DE = -2.6 ×10-19
J = 65 kB T.
The energy (or enthalpy depending on your taste) difference is about
65 kB T, or about 8 kB T per base pair.
DS = -8 ×10-22
J/K = -58 kB.
This entropy difference can be interpreted in terms of the ratio of
the number of states of the random coil vs the hairpin:
|
Wrandom
Whairpin |
= exp[ (Srandom -
Shairpin)/kB ] = exp[ -DS
] = exp[58] = 1.5 ×1025 |
|
Rather mind-bogglingly, the random coil has about 1025 times
as many states as the hairpin. Each base in the random coil therefore has
something like 103 times the number of states as when it is
a hairpin.
5. (a) We suppose G1 -
G0 = DG. The relative probability
of these states is given by the Boltzmann distribution as
|
P1
P0 |
= exp[-DG / (kB
T)] |
|
Using P1 + P0 = 1 tell us the individual probabilities,
| P1 = |
1
1 + exp[DG / (kB T)] |
|
|
and
| P0 = |
1
1 + exp[-DG / (kB T)] |
|
|
(b) The next step is to follow the directions, and to suppose that the
energies of the two states are E0 = 0 and E1 = -Ebinding,
and the entropies are S0 = Scst -
kB logc and S1 = 0. Therefore the free energies are
| G0 = E0 -
T S0 = - T Scst + kB
T logc |
|
| G1 = E1 -
T S1 = -Ebinding |
|
and the relative free energy is
| G1 - G0
= -Ebinding + T Scst -
kB T logc |
|
Finally we find the probability for the protein being bound,
| P1 = |
1
1 + exp[-Ebinding/(kB
T) + Scst/kB] / c |
|
|
(c) The entropy goes as -logc since 1/c is proportional
to the volume available per free protein, which is in turn proportional
to the total number of states. Entropy is then given by kB times
the log of the number of states.
(d) If the DNA site is occupied with probability P1 = 0.95
at concentration c = 10-6 M, we will
need (106 M) exp[-Ebinding/kB
T + Scst/kB] = 0.05. This determines exp[-Ebinding/kB
T + Scst/kB] = 5×10-8
M, and therefore
The plot of P1 vs. concentration on log scale is below:
(e) Kd = 5 ×10-8
M = 50 nM, a typical dissociation constant for a DNA-binding protein.
Note: John apologizes for the bad wording, and messy notation of problem
5 in the assignment.
File translated from TEX by TTH,
version 2.53.
On 5 Mar 2001, 23:05.