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BackgroundObservational studies and some experimental investigations have indicated that gut microbiota are closely associated with the incidence and progression of chronic renal failure. However, the causal relationship between gut microbiota and chronic renal failure remains unclear. The present study employs a two-sample Mendelian randomization approach to infer the causal relationship between gut microbiota and chronic renal failure at the genetic level. This research aims to determine whether there is a causal effect of gut microbiota on the risk of chronic renal failure, aiming to provide new evidence to support targeted gut therapy for the treatment of chronic renal failure.MethodsEmploying genome-wide association study (GWAS) data from the public MiBioGen and IEU OpenGWAS platform, a two-sample Mendelian randomization analysis was conducted. The causal relationship between gut microbiota and chronic renal failure was inferred using five different methods: Inverse Variance Weighted, MR-Egger, Weighted Median, Simple Mode, and Weighted Mode. The study incorporated sensitivity analyses that encompassed evaluations for pleiotropy and heterogeneity. Subsequently, the results of the Mendelian randomization analysis underwent a stringent correction for multiple testing, employing the False Discovery Rate method to enhance the validity of our findings.ResultsAccording to the results from the Inverse Variance Weighted method, seven bacterial genera show a significant association with the outcome variable chronic renal failure. Of these, Ruminococcus (gauvreauii group) (OR = 0.82, 95% CI = 0.71–0.94, p = 0.004) may act as a protective factor against chronic renal failure, while the genera Escherichia-Shigella (OR = 1.22, 95% CI = 1.08–1.38, p = 0.001), Lactococcus (OR = 1.1, 95% CI = 1.02–1.19, p = 0.013), Odoribacter (OR = 1.23, 95% CI = 1.03–1.49, p = 0.026), Enterorhabdus (OR = 1.14, 95% CI = 1.00–1.29, p = 0.047), Eubacterium (eligens group) (OR = 1.18, 95% CI = 1.02–1.37, p = 0.024), and Howardella (OR = 1.18, 95% CI = 1.09–1.28, p |
