7. DNA stiffness and ability to be bent What role does this play for binding of proteins? Example: 434 repressor; this is a protein from the 434 virus. When the protein binds to the DNA, it stops the RNA from being synthesized (RNA polymerase). This protein binds to a “specific” site with a binding preference over other sites of about 10000. Let’s look at a bit more detail. The “native” binding site is: ACAAGAAAGTTTGT. The middle four bases are marked (6,7,8,9). Replacing the central sequence by ACGT or AGCT reduced the binding affinity by a factor of 50. Replacing the central sequence with AAAT increased the affinity by a factor of 3.

8. By putting a “low-roll” sequence of four bases (AATT) in positions 6,7,8,9 the 434 repressor binds 300 times better than if we put a “high-roll” sequence there (GGCC). This is not enough to explain the 10000 times preference for the binding to the specific site, compared to any random sequence (all 14 base pairs) on DNA. However, it is clear that the adjacent AT base pairs are “bent” easier than the adjacent GC base pairs. The difference must be due to the specific contacts on either side of the center four bases of the binding site (1-5 and 10-14). The minor groove in position 6-9 is compressed to 8.5 Å (normal minor groove is 11.5 Å in B-DNA), and the minor groove at the ends of the binding sequence (1-5 and 10-14) is widened to 11.5 Å.

10. The 434 protein binds to a 14 bp DNA sequence, and forms specific contacts with the bps 1-5 and 10-14; but the middle of the sequence (base pairs 5-9) do not form strong contacts with the protein. The sequence of the center base pairs of the 14 bp binding sequence must have low roll for the specific contacts with the 1-5 and 10-14 base pair regions of the 434 binding sequence, and if this center sequence is AATT (low roll) the repressor binds 300 times stronger than if the center sequence is GCGC (high roll). In addition, the AATT sequence is easier to deform, than the GGCC sequence. These effects can have large influences, but the 434 repressor binds a total of 10,000 times better to its sequence than to other random sequences on the protein. The biological activity is just due to the binding (blocking to RNA polymerase) of the repressor to the specific site, but the total binding specificity is still not explained, but the flexibility and the extent of roll, certainly play a role.

11. The 434-repressor is from the 434 bacteriophage, and binds, as does the λ repressor, to several operator sites on the 434 chromosome. It binds as a dimer, to a total of 14 base pairs. But there are many dissimilarities to the λ repressor. The sequences are very different, and the DNA is bent when the 434 repressor binds to the 434 operator. This is a consequence of the specifics of the H-bonding contacts, which require a bending of the DNA. The width of the grooves are also effected (look at the computer structure). Replacing some of the AAs (to ACGT or AGCT) in the center of the binding sequence (ACAATATATATTGT) changes the affinity of binding, although there are no specific H-binding contacts. This has to do with the compression of the minor groove (probably).

12. Interaction of the 434 repressor with the left half of the 434 operator. The orientation of helix 3 of the helix-turn-helix motif (helix 2 and helix 3) within the major groove of the left half of the 434 operator is shown. Helix 3 lies in the major groove and makes hydrogen bonding contacts with functional groups on base pairs 2 and 4. Helix 2 makes several hydrogen bond contacts, as well.

13. It appears that there may be two major effects taking place in order to explain the very specific binding, a “docking” and “probing”. On a larger scale, the protein must fit together with the DNA: this is the docking. This is done on a more global scale with the intention of bringing the “right” surfaces of the protein and the DNA into close proximity. If this “fitting” is good, then more detailed “testing (with hydrogen bonding, and van der Waals contacts) can take place with the very local specific groups of the protein and the DNA. In this case, the “bending” at the center sequence plays a central role in the docking. If this center sequence allows the DNA to “bend” then the “probing” of very local interactions on a smaller scale can take place in the flanking regions. In this way we see a synergistic action between the different parts of the DNA sequence. The overall action of “docking” and “probing” explains the very large (10000 times) specificity.

14. Docking and Probing

15. How do we analyze the contribution of the distortion of the helix to the binding? M.E. Hogen & R.H. Austin Nature 329, 263-266 (1987).

18. Mol fraction of GC base pairs in the helix Young’s moduli Shear moduli

24. Checking the model: Critique of the model In order to get a better idea of the “rod-bending” model that we have just applied, and to see what might be weak points of the model, we will look at where the equations come from, and what assumptions were made. This is because for these types of models, we have to be careful, and this is a good example for this. We cannot go into the derivations of all the formulas, but we do want to point out the meaning of several of the terms, and under which circumstances we would expect the equations to hold. Just a few definitions to all start at the same place: Young’s modulus and Shear modulus

25. Young’s modulus (E) is a parameter that relates stress to strain:

26. The model used for the bending of DNA is that of a thin rod. It is assumed to bend in such a way that there is no strain at the center axis of the rod (pure bending). We pick a coordinate system with “z” in the direction of the straight rod (at the local point) and the x,y directions are orthogonal, and right-handed. We are bending the rod in the xz-plane. We also assume that the center of mass of the rod lies on the center “neutral” axis. It is assumed that the stresses from the outside of the rod are small (negligible) compared to the stress inside the rod. The boundary conditions for the stress are:

27. The equation means that there can be no resultant i-th component of the force acting on the surface (for all “i” components). There is no strain in the “z” direction, and therefore n z =0. This means that

28. and the same for i=y,z. The normal to the surface where it is bending is also in the xz-plane, meaning that n y = 0. This means also that Now comes an assumption: it is assumed that because the rod is thin on both sides of the rod (both sides of the rod, meaning in the plane in which the rod is bending – n y = 0), and (because it is so small on the surface) throughout the entire interior of the rod. So, the only stress component left is

29. So, when bending a thin rod, there is stress only in the “z” direction. So, the rod bends, and there is only compression for stress in the rod, and there is no stress in the center axis. In the inside of the bend, there is compression and on the outside of the bend there is dilation (in the “z” direction). Using “x” as the coordinate in the direction of the bend, we can use the “radius of curvature”, R, to write this compression (dilation) Note that this assumes that the radius of curvature is much larger than the radius of the rod (or “x”) and much larger that the dz’ and dz values. The strain (u zz ) is therefore

30. So, now we can write the stress-strain relationship: Note also that the following is also true, where µ is Poisson’s ratio. We can show using this information (but it makes sense) that the “z” values anywhere in the rod become upon bending, and the cross-sections of the rod remain plane, but just become tilted. Due to Poisson’s ratio, the surface element component in the y directions will compress (dilate) and the surface element component in the x direction will become parabolic. The free energy per unit volume will become

31. The free energy integrated over the whole cross-section, is: Now, this is where we get the surface moment of inertia (I y ; for bending in the xz-plane): For a circular rod with radius “r”, we have For a square rod, there are two different axes perpendicular to “z”, and we have It is probably difficult to find the right area to integrate for the surface moment of inertia, because it is irregular. Maybe the square rod would be better, because this shows an anisotropy of bending.

32. The free energy of bending per unit length of the rod is: For the total free energy, including the torsional energy, we have:

33. Now let’s think about how well this applies to the situation of only four base pairs of DNA bending.

34. 1) The DNA is not a long thin rod. It is not clear whether all the assumptions in the derivation will hold. It is not sure whether it is isotropic, and that the only strain will be in the helical axis direction. Also, note that the strain is rather large, and the radius of curvature is not small compared to the size of the object. 2) The formulas for the Young’s modulus and for the shear modulus that use the average of long DNA molecules with certain sequences, very probably cannot really be very realistic, because it most certainly depends on the sequence context more than assumed. Critique of the model

35. 3) The shape changes hint that there will be some major electrostatic and solvent effects. The groove surface areas change quite a bit, and the phosphates distances changes. 4) Probably the dilation and compression in the z direction is not really symmetrical.