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X染色体失活可能解释了XO型人类和小鼠在生存能力上的差异。

X-chromosome inactivation may explain the difference in viability of XO humans and mice.

作者信息

Ashworth A, Rastan S, Lovell-Badge R, Kay G

机构信息

Chester Beatty Laboratories, Institute of Cancer Research, London, UK.

出版信息

Nature. 1991 May 30;351(6325):406-8. doi: 10.1038/351406a0.

Abstract

Only about 1% of human XO conceptuses survive to birth and these usually have the characteristics of Turner's syndrome, with a complex and variable phenotype including short stature, gonadal dysgenesis and anatomical defects. Both the embryonic lethality and Turner's syndrome are thought to be due to monosomy for a gene or genes common to the X and Y chromosomes. These genes would be expected to be expressed in females from both active and inactive X chromosomes to ensure correct dosage of gene product. Two genes with these properties are ZFX and RPS4X, both of which have been proposed to play a role in Turner's syndrome. In contrast to humans, mice that are XO are viable with no prenatal lethality (P. Burgoyne, personal communication) and are anatomically normal and fertile. We have devised a system to analyse whether specific genes on the mouse X chromosome are inactivated, and demonstrate that both Zfx and Rps4X undergo normal X-inactivation in mice. Thus the relative viability of XO mice compared to XO humans may be explained by differences between the two species in the way that dosage compensation of specific genes is achieved.

摘要

只有约1%的人类XO胚胎能存活至出生,这些胚胎通常具有特纳综合征的特征,其表型复杂且多变,包括身材矮小、性腺发育不全和解剖学缺陷。胚胎致死率和特纳综合征都被认为是由于X和Y染色体共有的一个或多个基因的单体性所致。预计这些基因会在雌性个体中从活跃和失活的X染色体上表达,以确保基因产物的正确剂量。具有这些特性的两个基因是ZFX和RPS4X,它们都被认为在特纳综合征中起作用。与人类不同,XO小鼠是可存活的,没有产前致死率(P. 伯戈因,个人交流),并且在解剖学上正常且可育。我们设计了一个系统来分析小鼠X染色体上的特定基因是否失活,并证明Zfx和Rps4X在小鼠中都经历正常的X染色体失活。因此,与XO人类相比,XO小鼠相对的生存能力可能是由于这两个物种在实现特定基因剂量补偿的方式上存在差异。

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