Center for Radiobiology and Radiotherapy, Department of Nuclear Medicine and Radiobiology, Faculty of Medicine, Université de Sherbrooke, Sherbrooke, Québec, Canada J1H 5N4.
J Phys Chem B. 2011 Nov 24;115(46):13668-73. doi: 10.1021/jp205194g. Epub 2011 Nov 1.
The reaction of low-energy electrons (LEEs; 10 eV) with 5'-TpXpT-3' (TXT), where X is uracil (U), thymine (T), and 5-bromouracil (5BrU), was examined by HPLC-UV analysis. The presence of 5BrU increased total damage by >50%. The radiation products of T5BrUT included TUT (40%), free U, T, 5BrU (23%), and fragments (13%). These products may be explained by initial capture of LEEs by the nucleobase to form a transient anion, followed by transfer of the electron within the molecule and cleavage of susceptible bonds by dissociative electron attachment (C-Br, C-N, or C-O bonds). In addition, these products may arise from the uracilyl-5-yl (U-5-yl) radicals that undergo H-atom abstraction from the sugar moiety. Interestingly, several products contained two sites of cleavage (U, pUT, and TUp). The formation of these products was linear with dose, and thus, they arise from the single-electron reactions. To explain these products, we propose that the reaction of LEEs (10 eV) involves the coupling of two dissociative processes in the same molecule (for example, dissociative excitation and dissociative electron attachment). The latter reactions may contribute to the formation of clustered damage, which is the most deleterious damage induced by ionizing radiation.
低能电子(LEE;10 eV)与 5'-TpXpT-3'(TXT)的反应,其中 X 是尿嘧啶(U)、胸腺嘧啶(T)和 5-溴尿嘧啶(5BrU),通过 HPLC-UV 分析进行了检查。5BrU 的存在使总损伤增加了>50%。T5BrUT 的辐射产物包括 TUT(40%)、游离 U、T、5BrU(23%)和碎片(13%)。这些产物可以通过 LEE 最初被碱基捕获形成瞬态阴离子来解释,然后电子在分子内转移,并通过离解电子附着(C-Br、C-N 或 C-O 键)断裂敏感键。此外,这些产物可能来自尿嘧啶-5-基(U-5-基)自由基,它们从糖部分夺取 H 原子。有趣的是,一些产物包含两个断裂部位(U、pUT 和 TUp)。这些产物的形成与剂量呈线性关系,因此它们来自单电子反应。为了解释这些产物,我们提出 LEE(10 eV)的反应涉及同一分子中两个离解过程的耦合(例如,离解激发和离解电子附着)。后一种反应可能有助于形成聚集损伤,这是电离辐射诱导的最具危害性的损伤。