Lentiviral vector-mediated insertional mutagenesis screen identifies genes that influence androgen independent prostate cancer progression and predict clinical outcome

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• 作者：Arun K. Nalla ; Theodore F. Williams ; Casey P. Collins ; Dustin T. Rae and Grant D. Trobridge
• 刊名：Molecular Carcinogenesis
• 出版年：2016
• 出版时间：November 2016
• 年：2016
• 卷：55
• 期：11
• 页码：1761-1771
• 全文大小：798K
• ISSN：1098-2744

文摘

Prostate cancer (PC) is the second leading cause of cancer related deaths in US men. Androgen deprivation therapy (ADT) improves clinical outcome, but tumors often recur and progress to androgen independent prostate cancer (AIPC) which no longer responds to ADT. The progression to AIPC is due to genetic alterations that allow PC cancer cells to grow in the absence of androgen. Here we performed an insertional mutagenesis screen using a replication-incompetent lentiviral vector (LV) to identify the genes that promote AIPC in an orthotopic mouse model. Androgen sensitive PC cells, LNCaP, were mutagenized with LV and injected into the prostate of male mice. After tumor development, mice were castrated to select for cells that proliferate in the absence of androgen. Proviral integration sites and nearby dysregulated genes were identified in tumors developed in an androgen deficient environment. Using publically available datasets, the expression of these candidate androgen independence genes in human PC tissues were analyzed. A total of 11 promising candidate AIPC genes were identified: GLYATL1, FLNA, OBSCN, STRA13, WHSC1, ARFGAP3, KDM2A, FAM83H, CLDN7, CNOT6, and B3GNT9. Seven out the 11 candidate genes; GLYATL1, OBSCN, STRA13, KDM2A, FAM83H, CNOT6, and B3GNT6, have not been previously implicated in PC. An in vitro clonogenic assay showed that knockdown of KDM2A, FAM83H, and GLYATL1 genes significantly inhibited the colony forming ability of LNCaP cells. Additionally, we showed that a combination of four genes, OBSCN, FAM83H, CLDN7, and ARFGAP3 could significantly predicted the recurrence risk in PC patients after prostatectomy (P = 5.3 × 10⁻⁵).