E2341: DNA molecule
Submitted by: Eyal Gavish
The problem:
The DNA molecule forms a double stranded helix with hydrogen bonds stabilizing the double helix.
Under certain conditions the two strands get separated resulting in a sharp ”phase transition” (in
the thermodynamic limit). We model the described above as a zipper consisting of N parallel links
that can be opened from one side of the zipper. If links 1, 2, 3, ..., p are opened, then the energy to
open link p + 1 is ε and the energy to open link p + 2 is infinity. The last link p = N cannot be
opened. Each open link has g orientations corresponding to the rotational freedom about the bond.
Assume a large number of links N .
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Define x = ge−ε/T and find the canonical partition function Z(β, x).
Find the average number of open links < p > as a function of x.
Find the linear approximation for < p >.
Approximate <p>
N for large x.
Describe the dependence of <p>
N on x.
Find expressions for the entropy S(x) and the heat capacity CV (x) at x = 1.
What is the order of the phase transition?
The solution:
(1) The partition function of one open link is e−βε + e−βε + ... + e−βε = ge−βε = x. For p open links
we multiply the p partition functions and get g p e−βεp = xp . So the partition function of the system
would be:
Z(β, x) =
N
−1
X
p=0
xp =
xN − 1
x−1
(2) The average number of open links is:
< p >=
N
−1
X
p=0
P rob(p) · p =
N
−1
X
p=0
N −1
xp
1 ∂ X p
∂
N xN
x
·p= x
x =x
ln(Z) = N
−
Z
Z ∂x
∂x
x −1 x−1
p=0
(3)
< p >=
N xN +1 − N xN − xN +1 + x
xN +1 − xN − x + 1
Now we expand the term for < p > to a series at x = 1. We expand the numerator and the
denominator separately to a third order and then divide them. We start with the numerator:
1
1
N xN +1 − N xN − xN +1 + x ≈ (x − 1)2 [N 2 − N ] + (x − 1)3 [2N 3 − 3N 2 + N ]
2
6
The series expansion for the denominator:
1
1
xN +1 − xN − x + 1 ≈ (x − 1)2 N + (x − 1)3 [3N 2 − 3N ]
2
6
So we get:
< p >≈
1
2 (x
1
1
2
2
− 1)2 [N 2 − N ] + 61 (x − 1)3 [2N 3 − 3N 2 + N ]
2 (N − N ) + 6 (x − 1)(2N − 3N + 1)
=
1
1
2
3
2
1 + 61 (x − 1)(3N − 3)
2 (x − 1) N + 6 (x − 1) [3N − 3N ]
1
Now we expand the term
1+
1
6 (x
1
1+ 16 (x−1)(3N −3)
to a series at x = 1:
1
1
≈ 1 − (3N − 3)(x − 1)
6
− 1)(3N − 3)
So we have:
1
1
1
< p >≈ [ (N − N 2 ) + (x − 1)(2N 2 − 3N + 1)] · [1 − (3N − 3)(x − 1)]
2
6
6
And by taking < p > to first order and taking N to be large we get:
1
1
< p >≈ N [1 + N (x − 1)]
2
6
(4)
1
<p>
xN
1
x
1
1
−
= N
−
·
=
·
1
N
x −1 N x−1
N 1−
1 − xN
1
x
And for large x and large N we get:
<p>
≈1
N
(5) We can see that for x = 0 we get
x = 1.
<p>
N
≈ 0. The linear approximation gives us
<p>
N
=
1
2
for
The dependence of <p>
N on x converges to a Heaviside step function as N becomes larger, so for
N → ∞ the dependence is:

0 for x < 1
<p>  1
for x = 1 ≡ H(x − 1)
=
 2
N
1 for x > 1
(6) We first find the entropy:
S(x) = −
∂F (β, x)
∂ 1
1 ∂x ∂
∂
= −β 2 ( ln(Z)) = ln(Z)+
ln(Z) = ln(Z)+βεx
ln(Z) = ln(Z)+βε < p >
∂T
∂β β
β ∂β ∂x
∂x
By using L’Hopital’s rule we calculate the limit of ln(Z) at x = 1 and get ln(Z) ≈ ln(N ). By
assuming that βε is not zero we get ln(N ) << βε· < p >= βε N2 . So the entropy at x = 1 is:
S(x = 1) ≈ βε < p >
And in the neighborhood of the point x = 1 and for N → ∞ we get:
S(x)
<p>
≈
= H(x − 1)
βεN
N
We Now calculate the heat capacity:
CV (x) = T
∂S(x)
∂S(x)
CV (x)
∂ S(x)
= βεx
⇒
=x (
)
∂T
∂x
βεN
∂x N
In the limit N → ∞ and in the neighborhood of x = 1 the entropy is a step function. So the heat
capacity in that limit is:
CV (x)
CV (x = 1)
= δ(x − 1) ⇒
=∞
βεN
βεN
(7) The entropy which is the first derivation of the Helmholtz free energy is a step function, so the
phase transition is of the first order.
2