Analysis of meiotic recombination pathways in the yeast Saccharomyces cerevisiae.
Analysis of meiotic recombination pathways in the yeast Saccharomyces cerevisiae.
复制标题
酿酒酵母减数分裂重组途径的分析。
DOI：
10.1093/genetics/144.1.71
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发表时间：
1996
期刊：
影响因子：
3.3
通讯作者：
Malone,RE
中科院分区：
文献类型：
--
作者：
Mao-Draayer,Y;Galbraith,AM;Pittman,DL;Cool,M;Malone,RE
在酿酒酵母中，有几个基因似乎在减数分裂早期重组中起作用。HOPland REDl就是这样的早期基因。本文的数据表明，arad52 spol3菌株产生的失活孢子都不能被挽救，这种表型有助于将这两个基因与其他早期减数分裂重组基因如SPO11、REC104或A4EI4区分开来。相反，任何一种突变都可以挽救arad50S spol3菌株；这种表型与其他早期重组基因(如REC104)的突变所赋予的表型相似。这两个不同的结果可以解释，因为这里提供的数据表明，arad50S突变不会减少减数分裂染色体内重组，类似于redlorhop所提供的突变表型。当然，REDlandHOPldo在正常的减数分裂染色体间重组途径中起作用；它们将染色体间重组减少到正常水平的约10%。我们证明，启动交换所需的基因(REC104)的突变对RED1的突变是完全上位的。最后，HOPlor RED1中的突变减少了在HIS2减数分裂重组热点观察到的双链断裂数量。
In the yeast,Saccharomyces cerevisiae, several genes appear to act early in meiotic recombination.HOPlandREDlhave been classified as such early genes. The data in this paper demonstrate that neither aredlnor ahoplmutation can rescue the inviable spores produced by arad52 spol3strain; this phenotype helps to distinguish these two genes from other early meiotic recombination genes such asSPOll, REC104, orA4EI4. In contrast, either aredlor ahoplmutation can rescue arad50S spol3strain; this phenotype is similar to that conferred by mutations in the other early recombination genes(e.g., REC104). These two different results can be explained because the data presented here indicate that arad50Smutation does not diminish meiotic intrachromosomal recombination, similar to the mutant phenotypes conferred byredlorhopl. Of course,REDlandHOPldo act in the normal meiotic interchromosomal recombination pathway; they reduce interchromosomal recombination to ~10% of normal levels. We demonstrate that a mutation in a gene(REC104)required for initiation of exchange is completely epistatic to a mutation inREDl. Finally, mutations in eitherHOPlorRED1reduce the number of doublestrand breaks observed at theHIS2meiotic recombination hotspot.