推导IBS评分在无关个体对人群中概率分布的计算公式

Formula Derivation for the Probability Distribution of IBS Score in Unrelated Individual Pairs

目的推导通过STR等位基因频率计算无关个体对间状态一致性(identity by state,IBS)评分概率分布的计算公式。方法比较两名无关个体间某一STR基因座的基因型可以得到三种相互排斥的组合:(1)有2个相同的等位基因,此时令a_2=1(否则a_2=0);(2)有1个相同的等位基因,此时令a_1=1(否则a_1=0);(3)有0个相同的等位基因,此时令a_0=1(否则a_0=0);则该无关个体对在这1个STR基因座的IBS评分可采用ibs=2a_2+a_1计算。推导通过STR等位基因频率分别计算a_2=1、a_1=1和a_0=1的概率(p_2、p_1和p0)和的通用表达式,继而当该无关个体对得到n个相互独立的STR基因座分型结果时,可通过p_(2l)和P_(1l)计算该多重分型系统IBS评分的二项分布参数(l=1,2,…,n)。结果从p_2、p_1和p_0的基本概念出发,以f_i表示STR基因座第i个等位基因的频率(i=1,2,…,m),则p_2的通用计算公式为p_2=2(sum(f_i^2) from i=1 to m)~2-sum(f_i^4) from i=1 to m;p_1的通用计算公式为P_1=4sum(f_i^2) from i=1 to m-4sum(f_i^3) from i=1 to m+4sum(f_i^4) from i=1 to m;p_0的通用计算公式为p_0=1-4sum(f_i^2) from i=1 to m+2(sum(f_i^2) from i=1 to m)~2+4sum(f_i^3) from i=1 to m-3sum(f_i^4) from i=1 to m,p_2、p_1、p_0的和为1。IBS评分符合二项分布:IBS～B(2n,π)。其中总体率π的通用计算公式为π=1/n sum(p_(2l)) from l=1 to m+1/2n sum(p_(1l)) from l=1 to m。结论生物学全同胞鉴定中的原假设为两名被鉴定人系无关个体,对任意IBS评分所对应的原假设概率均可通过本文所推导的公式进行直接计算,计算结果是进行证据解释的基础。

Objective To derive the probability equation given by STR allele frequencies of identity by state(IBS)score shared by unrelated individual pairs.Methods By comparing the STR genotypes of two unrelated individuals,three mutually exclusive combinations could be obtained:(1)sharing 2 identical alleles,a2=1,otherwise a2=0;(2)sharing 1 identical allele,a1=1,otherwise a1=0;(3)sharing 0 identical allele,a0=1,otherwise a0=0.And the IBS score of the one STR locus in this unrelated individual pair could be given by the formula:ibs=2a2+a1.The probability of a2=1(p2),a1=1(p1)and a0=1(p0)were derived and expressed in powers of the allele frequencies.Subsequently,for a genotyping system including n independent STR loci,the characteristics of binomial distribution of IBS score shared by a pair of unrelated individuals could be given by p2l and p1l(l=1,2,…,n).Results All the general equations of p2,p1 and p0 were derived from the basic conceptions of a2,a1 and a0,respectively.Given fi(i=1,2,…,m)as the ith allele frequency of a STR locus,the general equations of p2,p1 and p0 could be respectively expressed in powers of fi:p2=2(■fi2)2-■fi4,p1=4■fi2-4(■fi2)2-4■fi3+4■fi4 and p0=1-4■fi2+2(■fi2)2+4■fi3-3■fi4.The sum of p2,p1 and p0 must be equal to 1.Then,the binomial distribution of IBS score shared by unrelated individual pairs genotyped with n independently STR loci could be written by:IBS^B(2n,π),and the general probability,π,could be given by the formula:π=■■p2l+■■p1l.Conclusion In the biological full sibling identification,the probability of null hypothesis corresponding to any specific IBS score can be directly calculated by the general equations presented in this study,which is the basement of the evidence explanation.