Precision neutron diffraction structure determination of protein and nucleic acid components. XII. A study of hydrogen bonding in the purine-pyrimidine base pair 9-methyladenine · 1-methylthymine

Abstract

A neutron diffraction study of the 1 : 1 complex between 9‐methyladenine and 1‐methylthymine, ${\mathrm{C}}_6{\mathrm{H}}_7{\mathrm{N}}_5·{\mathrm{C}}_6{\mathrm{H}}_8{\mathrm{N}}_2{\mathrm{O}}_2$ , has been carried out. The structure is monoclinic, space group P2₁/m, with two base pairs per unit cell; cell parameters a =8.304(2), b =6.552(2), c =12.837(3)Å, and β=106.83(5)°. The structure has been refined by full‐matrix least squares techniques starting from the x‐ray results of Hoogsteen [K. Hoogsteen, Acta Crystallogr. 12, 822 (1959); Acta Crystallogr. 16, 907 (1963)]. All hydrogen atoms have been located with a precision better than 0.01 Å, with the exception of methyl group hydrogens. The thymine molecules appear to be slightly disordered by means of a 180° rotation about $\mathrm{N}\text{3 ···}\mathrm{C}6$ , which has the effect of interchanging N1 and C5 while leaving the positions of all other atoms approximately unchanged. Between 10% and 13% of the thymine molecules in the structure are disordered in this way. Average refined neutron scattering lengths for nitrogen and carbon are ${\mathrm{b̄}}_{\mathrm{N}}\text{=0.910(3)}$ and ${\mathrm{b̄}}_{\mathrm{C}}{\text{=0.657(3) × 10}}^{\text{−12}}\hspace{0.17em}\mathrm{cm}$ ; oxygen and hydrogen scattering lengths were fixed at $b_{\mathrm{H}}\text{=−0.372}$ and $b_0{\text{=0.575 × 10}}^{\text{−12}}\hspace{0.17em}\mathrm{cm}$ . The purine and pyrimidine bases are joined with N3 of thymine hydrogen bonded to N7 of adenine. Because of the disorder present in the structure, 87%–90% of the adenine · thymine pairs have the Hoogsteen configuration and the remainder have the reversed Hoogsteen configuration. These geometries differ from that of a Watson‐Crick adenine · thymine pair in which N3 of thymine donates a proton to N1 of adenine. The hydrogen bonding scheme in the crystalline complex consists of one $\mathrm{N–H}\mathrm{···}\mathrm{N}$ and two $\mathrm{N–H}\text{··· 0}$ bonds and is quite normal; there is no evidence for proton transfer in any of the hydrogen bonds which are formed.