升高的FBXL18促进RPS15A无处不在和SMAD3激活,以驱动HCC
Hong-Qiang Yu1, Feng Li2, HaoJun Xiong1
1Key Laboratory of Hepatobiliary and Pancreatic Surgery, Institute of Hepatobiliary Surgery, Southwest Hospital, Third Military Medical University (Army Medical University), Chongqing, P.R. China.
Hepatology communications
|June 28, 2023
概括
富含F盒和白的重复蛋白18 (FBXL18) 通过增加RPS15A稳定性和SMAD3水平来促进肝癌. 针对这种途径为肝细胞癌 (HCC) 提供了一个新的治疗策略.
科学领域:
- 分子生物学分子生物学
- 在瘤学瘤学.
- 生物化学 生物化学
背景情况:
- 富含F-box和白的重复蛋白18 (FBXL18) 与各种癌症有关.
- 它在肝癌发生中的作用尚不清楚.
研究的目的:
- 调查FBXL18和肝细胞癌 (HCC) 之间的相关性.
- 阐明FBXL18在HCC发育中的潜在分子机制.
主要方法:
- 在HCC组织中分析FBXL18表达和患者生存数据.
- FBXL18转基因小鼠模型用于HCC诱导.
- 调查FBXL18对RPS15A无化和SMAD3信号传递的影响.
- 评估RPS15A和SMAD3对HCC扩散的影响.
主要成果:
- FBXL18在HCC中高度表达,并与生存率差相关,作为一个独立的风险因素.
- 在转基因小鼠中,FBXL18促进HCC.
- FBXL18增强了RPS15A的K63连接的无处不在和稳定性,增加了SMAD3水平和核转位.
- 抑制RPS15A或SMAD3可以抑制FBXL18驱动的HCC扩散.
- 在临床HCC样本中,高FBXL18表达与RPS15A表达相关.
结论:
- FBXL18通过FBXL18/RPS15A/SMAD3通路促进肝细胞致癌.
- 准这种途径为HCC提供了一个新的治疗策略.
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