Questions

1. An ideal freely-jointed chain (FJC, often also called a Gaussian chain) consists of N rigid segments of length b, freely hinged where they join. You may neglect possible consequences of interference between different parts of the chain.

(a) (i) Derive an expression for 〈R²〉, the mean square end-to-end distance for a FJC.

     (ii) The corresponding result for a wormlike chain (WLC), which models a polymer as a continuous filament with a non-zero bending modulus, is



where L is the contour length and A the persistence length. Evaluate this expression in the limits L < i>A and L >> A . Show your working and comment on both results.

(b) Write down an expression for the partition function of the FJC in the case that a force ƒẑ is applied between the ends of the chain to separate them. In the low-force limit the chain behaves as a linear spring. Without further calculation, deduce from your expression for the partition function how the spring constant varies with temperature. Explain your reasoning.

(c) Show that the average value of the projection of a dipole p in the direction of an electric field E, under the influence of the randomising effect of temperature, is given by the Langevin function L(α):




where α is a dimensionless parameter that describes the relative strengths of the interaction energy between the dipole and the electric field (pE) and the thermal energy kBT.

α =pE/ kBT

(d) Show that for α << 1 the Langevin function becomes




2 The human genome contains approximately 3 × 109 base-pairs. With the aid of the dimensions shown in Figure 1, estimate the capacity of DNA as a data-storage medium, and compare this capacity to that of other storage media.


Figure 1


Figure 1 Dimensions of DNA

3 Go to www.pdb.org (the Protein Data Bank, PDB) and explore links under ‘PDB-101’ at bottom left, including ‘Understanding PDB Data’ and ‘Molecule of the Month’ (MOM).

To go straight to a database entry for a particular structure click on its PDB ID in MOM or enter the ID in the search box near the top of the page. Use an interactive viewer, e.g. Jmol, to look at structures – see the options given under the thumbnail picture of the molecule on the RHS of the structure page. You may have to install Java on your computer. N.B. Alternative viewers (available on the web, and see menu) include Rasmol, the Swiss-PDB Viewer and Pymol– these make it easier to change the way the data is displayed (Pymol was used to make the images used in the lectures). Suggestions for display settings below refer to Jmol.

Read the entries for the following molecular machines in the MOM archive.

(a) DNA polymerase (MOM March 2000; PDBID 1tau).

Set Display: cartoon; Colour: secondary structure.

Click and drag with the left mouse button to rotate the molecule.

What are the two dominant secondary structure motifs of the polypeptide chain?

What other important macromolecule is present?

Play with display settings and compare the ball-and-stick representation (coloured by element), which shows the position of every atom, with the cartoon, which is a simplified representation of the same structural data. Too much information can be confusing.

(b) DNA (MOM November 2001; PDBID 1bna)

Set Display: ball and stick; Colour: by clement.

What elements are represented by red, orange, blue and grey spheres?

Identify C-G and A-T base pairs, and satisfy yourself that you can identify the locations of three hydrogen bonds between C and G, and two between A and T.

4 Proteins that span biological membranes often contain alpha helices. What type of amino acid would you expect to find in such a helix? Why is an alpha helix particularly suited to exist in the hydrophobic environment of the interior of a membrane?

5. Describe the hierarchical structure of a folded protein, and the forces that are important in protein folding.

Answers

(a)(i) Let ni be a unit vector in the direction of the ith segment of the chain. Mean square end to end distance


Problem Equation 1


Note that ni.ni = 1 (always) and that < b>ni.ni> = 0 for i ≠ j.

(See ‘Modelling DNA and RNA’ Section 2.1.1.)

(ii) In the limit L < i>A


Equation 2


The expected end-to-end distance is equal to the contour length as expected for a polymer whose length is much shorter than the persistence length – i.e. a stiff, straight rod.

In the limit L >> A:

R2〉 = 2AL

which is equivalent to the FJC result with an effective segment length of 2A.

(See Phillips Chapter 10 and Nelson Chapter 9 for discussion of wormlike chain.)

(b) See ‘Modelling DNA and RNA, Section 2.1.2. Here we need to incorporate the integral over all orientations of each segment.


Equation 3


By inspection of the partition function we see that the probability that the chain will adopt any particular configuration {ni} depends only on ƒẑ /kBT.

Each configuration has a characteristic z, so the probability distribution over possible configurations is unchanged (and therefore 〈 z〉 is unchanged) if f and T are varied such that f/T is constant.

In the low-force regime the chain acts as a linear spring (given), i.e.

z〉∝ƒ

Combining these two pieces of information, we can deduce that

z 〉∝ƒ/T

so

ƒ ∝ Tz

Compare Hooke’s law F = kz: spring constant is proportional to T.

(c) Probability that dipole p makes angle θ with electric field p depends on its potential energy through the Boltzmann distribution:


Equation 4


Average projection of dipole in direction of field:


Equation 5


(d)


Equation 6


2. NB 1 B (byte) = 8 bits

DNA:

One base pair (AT, CG, GC or TA) corresponds to 2 bits = 0.25 bytes

Volume occupied by 1 base pair = 0.34 nm x π x (1 nm)2 ≃ 1 nm3 = 10-27 m3

Storage density ≈ 0.25 B/10-27 m3 = 2.5 x 1026 B m-3

Typical CD:

Data ≈ 0.7 GB (approximately equivalent to a genome)

Disc has radius r = 60 mm, thickness t ≃ 1 mm

Volume = 1 mm x π x (60 mm)2 ≈ 1.1 x 104 mm3 = 1.1 x 10-5 m3

Storage density ≈ 0.7 x 10 B/1.1 x 10-5 m3 = 6 x 1013 B m-3

Typical DVD:

Stores ~10GB in a disc similar to a CD, so it has about ten times the capacity of genome with a density 1015 B m–3.

The calculation of CD storage density includes the plastic support. DNA also needs protein supports (histones) and machinery to compact, organise and read it. See Phillips Chapter 8, Berg Section 31.3.

3 This question is an introduction to what proteins really look like – there is a great deal of interesting information in the Molecule of the Month, entries which could be picked up in a tutorial.

(a) DNA polymerase (MOM March 2000; PDBID 1tau).

Dominant motifs are alpha helices and beta pleated sheets (there are examples of both parallel and antiparallel beta sheets)

See Berg et al Section 2.3 for secondary structure, Section 4.3 for DNA polymerase.

There is a blunt-ended DNA duplex bound to the polymerase active-site cleft.

(b) DNA (MOM November 2001; PDBID 1bna)

red = oxygen, orange = phosphorus (in sugar-phosphate backbone linkage), blue = nitrogen, grey =carbon.

See Berg et al Section 2.3 for protein secondary structure, Sections 4.1 and 4.2 for DNA structure.

4 Alpha helices buried in the interior of a folded protein generally have hydrophobic side chains. If exposed on one side to aqueous solution (usually on the outside of the protein) they may contain a mixture of hydrophobic and hydrophilic residues organized (phased) such that the hydrophilic residues are positioned on the water-exposed side. For a membrane protein, however, the outside is exposed to the hydrophobic middle of the lipid bilayer so one would expect exclusively hydrophobic residues except at the faces exposed to aqueous solution above and below the membrane.

The backbone linkages of a polypeptide contain strong hydrogen bond donors and acceptors. Any configuration in which these strongly polar units did not form hydrogen bonds would have high energy relative to an unfolded chain in which they H-bond with water. The ordered secondary structure of the alpha helix ensures that the hydrogen bonding potential of the backbone carbonyl and amine groups is satisfied by intramolecular pairing(every backbone N-H group is hydrogen bonded to the backbone C=O group of the amino acid four residues earlier) so even in the hydrophobic membrane interior there are no high-energy unpaired hydrogen donors or acceptors.

Students should be encouraged to discuss the nature of the “hydrophobic force”. An interface between a hydrophobic surface and water increases free energy mainly by restricting the number of ways the hydrogen bond network of water can be formed – thus decreasing the entropy.

See e.g. Berg et al Sections 2.3, 2.4.

5 This question is asking for an essay that summarises the content of Lecture 3 (Protein structures) and Lecture 4 (protein folding), plus any further knowledge gained by wider reading.