评估乳病和自闭症谱系障碍之间的因果关系:一个双样本的门德尔随机化方法
Abiodun Fatoba1, Claire Simpson1
1Department of Genetics, Genomics and Informatics, University of Tennessee Health Science Centers, Memphis, Tennessee, USA.
概括
这项研究使用孟德尔随机化分析没有发现乳病和自闭症谱系障碍之间的因果关系. 来自欧洲人口的遗传数据证实,乳病不会导致自闭症谱系障碍.
科学领域:
- 遗传学和流行病学
- 神经发育障碍 神经发育障碍
- 胃肠病学 胃肠病学
背景情况:
- 关于乳病 (CD) 和自闭症谱系障碍 (ASD) 之间的潜在关联是有争议的,从观察性研究中得到了相互矛盾的证据.
- 澄清因果关系对于理解疾病机制和潜在干预措施至关重要.
研究的目的:
- 通过强大的双样本孟德尔随机化 (MR) 方法,研究乳病和自闭症谱系障碍之间的因果关系.
- 利用全基因组关联研究 (GWAS) 数据进行可靠的基因仪器变量选择.
主要方法:
- 两个样本的孟德尔随机化分析,使用来自欧洲祖先种群的汇总级GWAS数据.
- 工具变量 (IV) 根据全基因组显著性 (p < 5 × 10−8) 进行选择,并评估强度.
- 主要分析采用逆方差加权 (IVW) 方法,灵敏度分析包括加权中位数,MR-Egger,MR-PRESSO,异质性测试和留出一个分析以确保结果的有效性.
主要成果:
- 门德尔的随机化分析显示,乳病和自闭症谱系障碍之间没有统计学上显著的因果关系 (OR,0.994;95% CI,0.935-1.057;p = 0.859).
- 敏感性分析,包括MR-Egger回归,没有检测出显著的水平型 (p = 0.223) 或异质性 (p = 0.051).
- 遗漏分析,森林地块和分散地块证实了研究结果的稳定性,表明没有单一的SNP过度影响了总体结果.
结论:
- 这项研究提供了强有力的证据,反对乳病和自闭症谱系障碍之间的因果关系.
- 这些发现表明,在以前的观察性研究中观察到的关联可能是由于混因素而不是直接因果关系.
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