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[如何利用短串联重复序列作为遗传标记进行单亲亲子鉴定并得出结论]

[How to draw a conclusion in motherless parentage testing using short tandem repeats as genetic makers].

作者信息

Zhu Yun-Liang, Huang Yan-Mei, Wu Xin-Yao

机构信息

Department of Forensic Medicine, SUN Yat-sen University, Guangzhou 510080, China.

出版信息

Fa Yi Xue Za Zhi. 2006 Aug 15;22(4):281-4.

Abstract

OBJECTIVE

To calculate the exclusion power of STR loci in motherless parentage testing and to discuss how to draw a conclusion if there are inconsistent loci.

METHODS

Based on the law of inheritance and allele frequency, the powers of exclusion of STR loci in motherless parentage testing (PE(M)) were calculated. Based on the mean PE(M) and mutation rate of 13 CODIS loci. The probabilities of inconsistence under paternity and non-paternity were calculated respectively according to binomial theorem.

RESULTS

The PE(M) of locus having co-dominate alleles could be calculated as: PE(M) = (i = 1)sigma (n) p i 2(1-p (i))2+ (i < j)sigma (n) 2p (i)p (j)(1-p (i)-p (j))2. According to the formula, the average PE(M) of 13 CODIS was 0.411. Based on the mean PE(M) and mutation rate, the likelihood ratio of true father to random man (paternity index) was got using binomial theorem.

CONCLUSION

The conclusion in motherless parentage testing could be drawn based on the likelihood ratio (paternity index) derived from mean PE(M) and mutation ratio.

摘要

目的

计算STR基因座在无母亲亲子鉴定中的排除能力,并探讨当出现基因座不一致时如何得出结论。

方法

依据遗传规律和等位基因频率,计算STR基因座在无母亲亲子鉴定中的排除能力(PE(M))。基于13个CODIS基因座的平均PE(M)和突变率,根据二项式定理分别计算父权和非父权情况下不一致的概率。

结果

共显性等位基因基因座的PE(M)可计算为:PE(M) = (i = 1)∑(n) p i 2(1 - p (i))2 + (i < j)∑(n) 2p (i)p (j)(1 - p (i) - p (j))2。根据该公式,13个CODIS基因座的平均PE(M)为0.411。基于平均PE(M)和突变率,利用二项式定理得出亲生父亲与随机男性的似然比(父权指数)。

结论

可根据由平均PE(M)和突变率得出的似然比(父权指数)在无母亲亲子鉴定中得出结论。

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