DNA Structure Puzzle Number 4 - X-Ray Diffraction from a Fibre of DNA-polyarginine
Clive Delmonte
clivedelmonte at c-i-delmonte.freeserve.co.uk
Fri Oct 22 13:31:59 EST 1999
PUZZLE NUMBER 4 X-Ray Diffraction from a Fibre of DNA-polyarginine
Suwalski & Traub (8) studied a fibre of the DNA-polyarginine complex by
X-ray diffraction. Their diffraction pattern shows "pronounced streaks" in a
hexagonal lattice of side 4.15 nm and they remarked that the equatorial
reflections followed h x h + h x k + k x k = 3n.
This pattern of equatorial presences is equivalent to that of Fuller et al.
(Puzzle #3, Ref.6) where such presences follow h-k=3n in a cell whose side
can be reduced by root 3. (Squaring (h-k) and subtracting from the
expression in the previous paragraph, there is a difference of 3hk, a
quantity always divisible by 3.) Therefore Suwalski & Traub's value of 'a'
can reduced by root 3.
Suwalski and Traub decided that their pattern of systematic absence could be
explained by postulating axial translations of the molecules equal to c/3.
Now, the very same diffraction pattern of Suwalski and Traub with
DNA-polyarginine was re-indexed on an orthogonal net by Fita et al. (9) in a
unit cell of approximate section 2.6 nm x 3.6 nm (10) where equatorial
reflections followed h+k = 2n.
Langridge et al. (Puzzle #3, Ref. 7) have shown that this pattern of
presences would allow the halving of the long side of the cell section.
Fita et al. considered that the pattern of systematic absences in their
orthogonal net indicated that the axial translation of the molecules was
c/4.
So Suwalski & Traub considered that the fibre had axial displacements of the
molecules of c/3 in a hexagonal net, and Fita et al. found the axial
displacement to be c/4 in an orthogonal net.
But there was only one single fibre and only one diffraction pattern.
Therefore, whether a hexagonal or orthogonal array is chosen, the axial
dispacement can only have one value.
Marvin et al. (Puzzle #3, Ref. 5) considered that a random axial translation
of their molecules of c/2 would account for the streaks on their layer lines
in a hexagonal lattice.
The value common to both analyses is therefore c/2, but this is deducible
from the layer line streaks and not from the patterns of systematic absence.
Therefore the patterns of systematic absence in both the hexagonal and
orthogonal nets should be taken to show that, for both indexing schemes, a
smaller unit cell is the right choice, with a random axial displacement of
c/2.
However, if the unit cell sizes are reduced accordingly, the double helix
can no longer be fitted into the reduced dimensions of the original cells.
As we have seen in the earlier Puzzles, a true side-by-side duplex, where
each helix has a diameter of about 1.3 nm, and, in this case, has its
grooves at least partially filled with poly-arginine, will fit the new,
reduced unit cells (e-mail author for diagram). The helical diameter of
about 1.3 nm is a direct geometric consequence of the Watson-Crick base pair
width of 1.1 nm.
------------------------------------------
8 A Comparative X-ray Study of a Nucleoprotamine & DNA Complexes with
Polylysine & Polyarginine; M. Suwalski & W. Traub; Biopol. Vol 11 (1972)
2223 - 2231
9 X-ray Diffraction Study of DNA Complexes with Arginine Peptides and Their
Relation to Nucleoprotamine Structure; I. Fita, J.L. Campos, L.C. Puigjaner
& J.A. Subirana; J Mol Biol Vol 167 (1983) 157 - 177
10 X-ray Diffraction Studies of Nucleoprotamine Structure; P. Suau & J.A.
Subirana; J Mol Biol Vol 117 (1977) 909 - 926
Clive Delmonte
For a view of all the DNA Structure Puzzles
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