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胞嘧啶、腺嘌呤和鸟嘌呤离子化碱基的碎裂机制。

Fragmentation mechanisms of cytosine, adenine and guanine ionized bases.

机构信息

Université de Lyon, Université Claude Bernard Lyon1, Institut des Sciences Analytiques, CNRS UMR 5280, 5 Rue de la Doua, 69100 Villeurbanne, France.

出版信息

Phys Chem Chem Phys. 2015 May 7;17(17):11813-26. doi: 10.1039/c5cp00104h.

DOI:10.1039/c5cp00104h
PMID:25869111
Abstract

The different fragmentation channels of cytosine, adenine and guanine have been studied through DFT calculations. The electronic structure of bases, their cations, and the fragments obtained by breaking bonds provides a good understanding of the fragmentation process that can complete the experimental approach. The calculations allow assigning various fragments to the given peaks. The comparison between the energy required for the formation of fragments and the peak intensity in the mass spectrum is used. For cytosine and guanine the elimination of the HNCO molecule is a major route of dissociation, while for adenine multiple loss of HCN or HNC can be followed up to small fragments. For cytosine, this corresponds to the initial bond cleavage of N3-C4/N1-C2, which represents the main dissociation route. For guanine the release of HNCO is obtained through the N1-C2/C5-C6 bond cleavage (reverse order also possible) leading to the largest peak of the spectrum. The corresponding energies of 3.5 and 3.9 eV are typically in the range available in the experiments. The loss of NH3 or HCN is also possible but requires more energy. For adenine, fragmentation consists of multiple loss of the HCN molecule and the main route corresponding to HC8N9 loss is followed by the release of HC2N1.

摘要

通过 DFT 计算研究了胞嘧啶、腺嘌呤和鸟嘌呤的不同碎片化通道。碱基、它们的阳离子以及通过键断裂获得的碎片的电子结构提供了对可以完成实验方法的碎片化过程的很好理解。计算允许将各种碎片分配给给定的峰。将形成碎片所需的能量与质谱中的峰强度进行比较。对于胞嘧啶和鸟嘌呤,HNCO 分子的消除是离解的主要途径,而对于腺嘌呤,可以通过多种方式失去 HCN 或 HNC 直到得到小碎片。对于胞嘧啶,这对应于 N3-C4/N1-C2 的初始键断裂,这代表主要的离解途径。对于鸟嘌呤,通过 N1-C2/C5-C6 键断裂(也可以反向)获得 HNCO 的释放,导致光谱中最大的峰。相应的能量为 3.5 和 3.9 eV,通常在实验中可用的范围内。也可以失去 NH3 或 HCN,但需要更多的能量。对于腺嘌呤,碎片化包括 HCN 分子的多次损失,并且遵循对应于 HC8N9 损失的主要途径,然后释放 HC2N1。

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