The supplied atoms form the three Watson-Crick hydrogen bonds of a real G-C base pair (guanine, chain A residue 2; cytosine, chain B residue 23) in a B-DNA dodecamer. Calculate all three donor-acceptor distances: N1(G)-N3(C), N2(G)-O2(C), and O6(G)-N4(C). Report each distance in angstrom to three decimal places. A Watson-Crick A-T pair instead forms only two hydrogen bonds (N1-N3 and N6-O4). Which base pair type would you expect to be more thermally stable, and why?

Coordinates are in angstrom (1 nm = 10 angstrom). XYZ atom rows are one-based after the two header lines. Use the supplied coordinate snapshot without optimization or periodic wrapping; this is a single deposited crystal model. Use S0009 only: the 6 hydrogen-bond atoms of the G(chain A residue 2)-C(chain B residue 23) base pair, in order row 1=N1(G), row 2=N2(G), row 3=O6(G), row 4=N3(C), row 5=O2(C), row 6=N4(C). Report each of the three distances in angstrom to three decimals, and answer the qualitative question in words.

6
PDB 1BNA B-DNA dodecamer, G(A2)-C(B23) Watson-Crick pair atoms in order N1(G),N2(G),O6(G),N3(C),O2(C),N4(C); angstrom
N 17.197 24.279 23.037
N 18.508 22.456 23.668
N 15.739 25.916 22.518
N 15.067 22.877 24.257
N 16.257 20.984 24.504
N 13.771 24.813 24.018


Attachments:
S0009: item-0001-asset-1.xyz