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RecA kinetically selects homologous DNA by testing a five- or six-nucleotide matching sequence and deforming the second DNA
Quarterly Reviews of Biophysics ( IF 6.1 ) Pub Date : 2018-12-17 , DOI: 10.1017/s0033583518000094
Masayuki Takahashi 1
Affiliation  

RecA family proteins pair two DNAs with the same sequence to promote strand exchange during homologous recombination. To understand how RecA proteins search for and recognize homology, we sought to determine the length of homologous sequence that permits RecA to start its reaction. Specifically, we analyzed the effect of sequence heterogeneity on the association rate of homologous DNA with RecA/single-stranded DNA complex. We assumed that the reaction can start with equal likelihood at any point in the DNA, and that sequence heterogeneity abolishes some possible initiation sites. This analysis revealed that the effective recognition size is five or six nucleotides, larger than the three nucleotides recognized by a RecA monomer. Because the first DNA is elongated 1.5-fold by intercalation of amino acid residues of RecA every three bases, the second bound DNA must be elongated to pair with the first. Because this length is similar to estimates based on the strand-exchange reaction or DNA pair formation, the homology test is likely to occur primarily at the association step. The energetic difference due to the absence of hydrogen bonding is too small to discriminate single-nucleotide heterogeneity over a five- or six-nucleotide sequence. The selection is very likely to be made kinetically, and probably involves some structural factor other than Watson–Crick hydrogen bonding. It would be valuable to determine whether this is also the case for other biological reactions involving DNA base complementarity, such as replication, transcription, and translation.

中文翻译:

RecA 通过测试 5 个或 6 个核苷酸的匹配序列并对第二个 DNA 进行变形来动态选择同源 DNA

RecA 家族蛋白将两个具有相同序列的 DNA 配对,以促进同源重组过程中的链交换。为了了解 RecA 蛋白如何寻找和识别同源性,我们试图确定允许 RecA 开始其反应的同源序列的长度。具体来说,我们分析了序列异质性对同源DNA与RecA/单链DNA复合物结合率的影响。我们假设反应可以在 DNA 中的任何点以相同的可能性开始,并且序列异质性消除了一些可能的起始位点。该分析显示有效识别大小为五个或六个核苷酸,大于 RecA 单体识别的三个核苷酸。因为第一个 DNA 通过每三个碱基插入 RecA 的氨基酸残基而延长了 1.5 倍,第二个结合的 DNA 必须被拉长以与第一个配对。因为这个长度类似于基于链交换反应或 DNA 对形成的估计,同源性测试可能主要发生在关联步骤。由于不存在氢键而导致的能量差异太小,无法区分五或六核苷酸序列中的单核苷酸异质性。选择很可能是通过动力学进行的,并且可能涉及除 Watson-Crick 氢键之外的一些结构因素。确定其他涉及 DNA 碱基互补性的生物反应是否也是如此,例如复制、转录和翻译,这将是有价值的。因为这个长度类似于基于链交换反应或 DNA 对形成的估计,同源性测试可能主要发生在关联步骤。由于不存在氢键而导致的能量差异太小,无法区分五或六核苷酸序列中的单核苷酸异质性。选择很可能是通过动力学进行的,并且可能涉及除 Watson-Crick 氢键之外的一些结构因素。确定其他涉及 DNA 碱基互补性的生物反应是否也是如此,例如复制、转录和翻译,这将是有价值的。因为这个长度类似于基于链交换反应或 DNA 对形成的估计,同源性测试可能主要发生在关联步骤。由于不存在氢键而导致的能量差异太小,无法区分五或六核苷酸序列中的单核苷酸异质性。选择很可能是通过动力学进行的,并且可能涉及除 Watson-Crick 氢键之外的一些结构因素。确定其他涉及 DNA 碱基互补性的生物反应是否也是如此,例如复制、转录和翻译,这将是有价值的。由于不存在氢键而导致的能量差异太小,无法区分五或六核苷酸序列中的单核苷酸异质性。选择很可能是通过动力学进行的,并且可能涉及除 Watson-Crick 氢键之外的一些结构因素。确定其他涉及 DNA 碱基互补性的生物反应是否也是如此,例如复制、转录和翻译,这将是有价值的。由于不存在氢键而导致的能量差异太小,无法区分五或六核苷酸序列中的单核苷酸异质性。选择很可能是通过动力学进行的,并且可能涉及除 Watson-Crick 氢键之外的一些结构因素。确定其他涉及 DNA 碱基互补性的生物反应是否也是如此,例如复制、转录和翻译,这将是有价值的。
更新日期:2018-12-17
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