首页> 中文期刊> 《中国骨质疏松杂志》 >骨质疏松性骨折与 OPG-RANK-RANKL 系统基因多态性的关联分析

骨质疏松性骨折与 OPG-RANK-RANKL 系统基因多态性的关联分析

         

摘要

目的从基因水平分析OPG-RANK-RANKL系统基因多态性与髋部骨质疏松性骨折的复杂关联关系。方法采用关联分析,以OPG-RANK-RANKL基因为候选基因,以700个无关汉族个体为研究对象,对候选基因的75个SNP进行了分型实验。根据分型结果,利用Haploview 进行了单体型识别,并对单体型识别的结果与髋部骨折表型进行关联分析。结果Haploview 分析识别出了OPG-RANK-RANKL基因每个单体域的单体型,并挑选出标签SNP。结论关联分析的结果表明,位于OPG基因的SNP位点rs2460985、位于RANK基因的rs1805034、位于RANKL基因的rs9525625与髋部骨折之间存在相关性(rs2460985,P=0.0299;rs1805034,P=0.0234;rs9525625,P=0.0144),这证明OPG-RANK-RANKL系统在调节骨代谢方面起了重要的作用。利用haploview进行的单体型关联分析结果发现,OPG单体域1内CACC,RANK单体域1内AAAAA,RANKL单体域1内CAAACC和单体域3内AAACACAA都与骨折存在关联,在病例组和对照组中的差异具有统计学意义( P<0.05)。%Objective To analyze the complex relationship between the OPG-RANK-RANKL gene polymorphisms and the hip osteoporotic fracture at gene level.Methods In this study, the method of correlation analysis was applied.OPG-RANK-RANKL gene was selected as candidate gene.The classification analysis of 75 single nucleotide polymorphisms ( SNPs ) , which were selected from OPG-RANK-RANKL genes, was performed in 700 unrelated Han individuals. According to the results, the haplotypes was identified using Haploview.And the correlation analysis of the relationship between the identified results and the phenotype of the hip fracture was performed.Results Each haplotype of OPG-RANK-RANKL gene was identified using Haploview.And the tagSNPs were selected.Conclusion The results of correlation analysis showed that 3 SNPs, rs2460985 locating in OPG gene, rs1805034 locating in RANK gene, and rs9525625 locating in RANKL gene, were correlated with the hip fracture (rs2460985, P=0.0299;rs1805034, P=0.0234; rs9525625, P=0.0144), indicating that OPG-RANK-RANKL system played an important role in the regulation of bone metabolism.The results also showed that haplotype block 1 ( CACC) in OPG gene, haplotype block 1 ( AAAAA) in RANK gene, and haplotype block 1 ( CAAACC) and haplotype block 3 ( AAACACA) in RANKL gene were associated with the fracture.The difference between the study group and the control group was significant ( P<0.05).

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