Problem Set 5 - Solutions

1. (a) For this two-state equilibrium, we have [K]/[U] = k/k¢.

To remember where this comes from, consider the rate equation for the unknots:

d[U]
dt		 = -k [U] + k¢[K]
 
In equilibrium, d[U]/dt = 0, which tells us [K]/[U] = k/k¢

(b) The relative concentrations of K and U in equilibrium just gives the free energy difference, via the Boltzmann distribution:

[K]
[U] 		 =  e-GK/(kB T)
e-GU/(kB T) 		 = e-DG/ (kB T) 
 
Therefore we have k/k¢ = [K]/[U] = exp[ -DG /(kB T), or
DG = kB T ln(k¢/k) = kB T ln([U]/[K])
 
(c) From the above, if [K]/[U] = 0.03, we have

DG = 3.5 kB T = 1.4 ×10-20   J

Note that as usual, T = 300 K.

2. (a) In general this cannot be a thermal equilibrium reaction. One way to see this is to note that there is no pathway from C to B (i.e. a zero reaction rate), which means that if this system were in thermal equilibrium, the free energy of state B would have to be infinitely greater than that of C. However, you would not come to this conclusion by looking at the (finite-rate) transitions between B and C passing through state A. So the reaction shown cannot reach an equilibrium which is described using free energies.

(b) Rate equations:

d A
dt		 = -(k1 + k3¢) A + k1¢B + k3 C
 
 
d B
dt		 = -(k2 + k1¢) B + k1 A 
 
 
d C
dt		 = -k3 C + k3¢A + k2 B
 
(c) In the steady state, we have no change of concentrations, i.e.
0 = -(k1 + k3¢) A + k1¢B + k3 C
 
 
0 = -(k2 + k1¢) B + k1 A 
 
 
0 = -k3 C + k3¢A + k2 B
 
By conservation of particles we know that A + B + C = N, some fixed number of molecules. This can be verified by adding the three equations together which shows d(A+B+C)/dt = 0.

Therefore we can just eliminate C = N-A-B. The final two equations then immediately tell us

A = [(k2 + k1¢)/k1] B 
 
 
k3 N = (k3 + k3¢)A + (k2 + k3) B
 
Or plugging together,
k3 N =  é
ê
ë 		 (k3 + k3¢)(k2 +k1¢)
k1		 + k2 + k3  ù
ú
û 		 B 
 
which tells us
B
N		 =  k1 k3
(k3+k3¢)(k2 + k1¢) + k1(k2+k3) 
 
Finally, the rate of transitions from B to C per molecule is just
k2 B
N		 =  k1 k2 k3
(k3 + k3¢)(k2 +k1¢) + k1 (k2+k3)
 
(d) There is a constant flux of particles going from B to C in the steady state, with a rate per particle as given above. This means that the molecules must be undergoing cycles from A to B to C and back to A. Therefore this reaction could describe a simple cycling system, for example a rotary motor. The lack of thermal equilibrium indicates that this reaction must be driven by some external source of energy.

Note that the transitions from B to C stop if either k1, k2 or k3 are zero, since zeroing out these rates `breaks' the cycle.

Note also that there are finite rates in the steady state for particles to go the `wrong way' from A to C or from B to A. This is typical for small machines (e.g. protein motors) where thermal fluctuations can sometimes cause them to run `backwards'.

3. (a) For a solid cylinder, the bending modulus
B = (p/4) Y r4
where r is the cross-sectional radius and Y is the Young modulus. Inverting this gives
Y = 4 B /(pr4) = 4×1.5 ×10-23 / (p×[12 ×10-9]4) Pa = 9.2 ×108 Pa = 0.92 GPa

(b) I'll assume that the displacement takes the form of a smooth bend. For a small bend, and for a rod of length L, the displacement x and the radius of curvature of the bend R are related by

x =  L2
2R
 
so that R = L2 / 2x, and therefore 1/R2 = 4x2 /L4

Since the bending energy is E = BL /(2 R2) we have

E =  2 B x2
L3
 
This is the work done to displace the end by an amount x.

(c) The energy has the form of a harmonic oscillator energy,

E =  k
2		 x2
 
where the force constant is k = 4 B / L3 The fluctuation amplitude is just
< x2 > =  kB T
k
 
which we computed on a previous problem set and in class. Plugging in gives
< x2 > =  kB T L3
4 B
 
If we put in L = 100 microns, B = 1.5×10-23 J·m, and kB T = 4.1 ×10-21 J we obtain

< x2 > = 70 ×10-12 m
i.e. a deflection of about ±8 microns, easily observable in the microscope.

4. Below I have plotted the data given in the problem set, along with 10 curves, the interpolation formula given for persistence length A = 10, 20, 30, 40, 50, 60, 70, 80, 90 and 100 nm. Note that kB T/(1 nm) = 4.1 pN.

Just by eye we can see that A = 40 to 50 is giving the best fit. More careful fitting (note that the length of the molecule actually appears to be a bit longer than 16.3 microns) can give closer agreement, and still indicates A » 50 nm.

A slightly fancier way to extract the persistence length is to plot 1/Öf versus extension z, as shown below:

Since we have the asymptotic behavior for large forces

x = L  æ
ç
ç
ç
è 		 1 -    æ
 ú
Ö
kB T
4 A f
 		 ö
÷
÷
÷
ø 
 
we can extract the molecule length from the x-intercept, giving about 17.3 microns, indeed apparently a little longer than the 16.3 microns expected

As I recall these experiments used some linker DNA segments at their ends which slightly increased the DNA length.

The persistence length can be extracted from the extrapolation of the high-force tangent onto the 1/Öf axis (shown below as the straight line). This extrapolated intercept on the 1/Öf axis is equal to Ö{[4A/(kB T)]} = 6.5 (pN)-1/2.

This indicates A = (42.25 / 4 ) (kB T / pN) and since 1 kB T = 4.1 pN·nm we have
A = 42.25 ×4.1 / 4 nm = 43.3 nm.
 

Here is a final plot using these two parameters (L = 17.3 microns, A = 43.3 nm):

 Now the fit is pretty good at the larger forces.  It is not so good at the lower forces, but it turns out that this is due to a number of factors, not the least of which is a slight disagreement between the exact solution of the persistent-chain model, and our interpolation formula.

File translated from TEX by TTH, version 2.53.
On 3 May 2001, 16:46.