Department of Microbiology and Immunology, School of Basic Medical Sciences, Zhengzhou University, Zhengzhou, China.
Infection, Inflammation and Immunity Center, the Academy of Medical Sciences of Zhengzhou University, Zhengzhou, China.
Gut Microbes. 2023 Dec;15(2):2251646. doi: 10.1080/19490976.2023.2251646.
Inflammatory bowel disease (IBD) represents a prominent chronic immune-mediated inflammatory disorder, yet its etiology remains poorly comprehended, encompassing intricate interactions between genetics, immunity, and the gut microbiome. This study uncovers a novel colitis-associated risk gene, namely Ring1a, which regulates the mucosal immune response and intestinal microbiota. Ring1a deficiency exacerbates colitis by impairing the immune system. Concomitantly, Ring1a deficiency led to a genus-dominated pathogenic microenvironment, which can be horizontally transmitted to co-housed wild type (WT) mice, consequently intensifying dextran sodium sulfate (DSS)-induced colitis. Furthermore, we identified a potential mechanism linking the altered microbiota in Ring1aKO mice to decreased levels of IgA, and we demonstrated that metronidazole administration could ameliorate colitis progression in Ring1aKO mice, likely by reducing the abundance of the genus. We also elucidated the immune landscape of DSS colitis and revealed the disruption of intestinal immune homeostasis associated with Ring1a deficiency. Collectively, these findings highlight Ring1a as a prospective candidate risk gene for colitis and suggest metronidazole as a potential therapeutic option for clinically managing genus-dominated colitis.
炎症性肠病(IBD)是一种突出的慢性免疫介导的炎症性疾病,但其病因仍未被很好地理解,包括遗传、免疫和肠道微生物组之间的复杂相互作用。本研究揭示了一种新的结肠炎相关风险基因,即 Ring1a,它调节粘膜免疫反应和肠道微生物群。Ring1a 缺失通过损害免疫系统而加剧结肠炎。同时,Ring1a 缺失导致以属为主导的致病性微环境,可以水平传播给共同饲养的野生型(WT)小鼠,从而加剧葡聚糖硫酸钠(DSS)诱导的结肠炎。此外,我们确定了一种潜在的机制,将 Ring1aKO 小鼠中改变的微生物群与 IgA 水平降低联系起来,并且我们表明甲硝唑给药可以改善 Ring1aKO 小鼠的结肠炎进展,可能是通过减少属的丰度。我们还阐明了 DSS 结肠炎的免疫景观,并揭示了与 Ring1a 缺失相关的肠道免疫稳态的破坏。总之,这些发现强调了 Ring1a 作为结肠炎的有前途的候选风险基因,并表明甲硝唑可能是临床上治疗以属为主导的结肠炎的一种潜在治疗选择。