分析:病在1型糖尿病中的全球患病率
Sahand Karimzadhagh1, Elahe Abbaspour1, Maryam Shahriarinamin2
1Celiac Disease and Gluten Related Disorders Research Center, Research Institute for Gastroenterology and Liver Diseases, Shahid Beheshti University of Medical Sciences, Tehran, Iran.
Alimentary pharmacology & therapeutics
|November 5, 2024
概括
在全球范围内,大约每16名1型糖尿病 (T1D) 患者中就有1人患有乳病 (CD). 由于全球流行率各不相同,建议对所有T1D患者进行CD查.
科学领域:
- 内分泌学和免疫学.
- 胃肠病学 胃肠病学
- 公共卫生 公共卫生
背景情况:
- 结肠病 (CD) 经常被诊断为1型糖尿病 (T1D) 患者.
- 在T1D患者中CD的全球患病率因查协议的差异而有显著差异.
- 在过去的二十年中,T1D群体中对CD的查指南已大大发展.
研究的目的:
- 确定T1D患者中CD的全球患病率.
- 基于使用抗组织转质胺酶 (anti-tTG) 抗体查的研究来分析CD患病率.
- 调查影响不同国家和不同收入水平T1D患者CD患病率的因素.
主要方法:
- 在主要数据库 (PubMed,科学网,Embase,Scopus) 进行了全面的文献搜索,截至2023年12月.
- 与腹腔疾病和糖尿病相关的关键词被用来确定相关的查研究.
- 使用随机效应模型计算聚合流行率估计,并使用Cochran的Q测试和I2统计评估异质性.
主要成果:
- 分析包括106项研究,包括来自40个国家的65,102名T1D患者.
- 结合CD的血清流行率为9% (95% CI,8%-10%),确认CD的流行率为6% (95% CI,5%-7%).
- 患病率在女性和儿童中更高,在各国观察到显著差异,与人类发展指数 (HDI) 有负相关性.
结论:
- 全球大约每16名T1D患者中有1名,亚洲和中东地区每12名患者中有1名患有CD.
- 更高的人类发展指数 (HDI) 与T1D患者中较低的CD患病率相关.
- 强烈建议在所有被诊断为T1D的患者中进行CD的普遍查.
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