促进体结合的METTL3通过m6A-依赖的翻译控制来维持髓性白血病
Isaia Barbieri1, Konstantinos Tzelepis2, Luca Pandolfini1
1The Gurdon Institute and Department of Pathology, University of Cambridge, Tennis Court Road, Cambridge CB2 1QN, UK.
Nature
|November 30, 2017
概括
METTL3对于急性髓性白血病 (AML) 细胞生长至关重要. 这种酶通过RNA修饰来调节基因表达,为AML提供潜在的治疗点.
科学领域:
- 分子生物学
- 表观遗传学
- 癌症研究
背景情况:
- N6-甲基氨酸 (m6A) 是一种普遍存在的RNA修饰.
- m6A甲基转移酶复合体 (METTL3- METTL14) 在癌症中的作用尚不清楚.
研究的目的:
- 阐明METTL3在急性髓性白血病 (AML) 的特定功能.
- 确定METTL3作为AML的潜在治疗点.
主要方法:
- 鉴定AML细胞生长所需的基因.
- 染色体免疫沉和RNA测序以分析基因调节.
- 在体内评估细胞周期,分化和瘤形成.
主要成果:
- METTL3对于AML细胞的增殖和白血病状态的维持至关重要.
- 独立于METTL14,METTL3局部化到活跃的基因促进体,并被CEBPZ招募.
- METTL3通过缓解核糖体停滞,调节对AML至关重要的基因来增强mRNA翻译.
结论:
- 在AML中,METTL3调节一种基于色素的途径.
- METTL3是急性髓性白血病的一个有前途的治疗点.
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