标签：limit bsp stat between ima number phi tab reg
Tajima F (1989) Genetics 123:585-595
1 Average number of pairwise nucleotide differences
$$ \hat k = \frac{\sum\sum_{i<j} k_{ij}}{\binom{n}{2}} $$
\( k_{ij} \), the number of nucleotide differences between the $i$-th and $j$-th DNA sequences
2 D statistic
$$ D = \frac{\hat k - \frac{S}{a_1}}{\sqrt{e_1 S + e_2 S (S-1)}} $$
$$ a_1 = \sum \limits_{i=1}^{n-1} \frac{1}{i} $$
$$ a_2 = \sum \limits_{i=1}^{n-1} \frac{1}{i^2} $$
$$ b_1 = \frac{n+1}{3(n-1)} $$
$$ b_2 = \frac{2(n^2+n+3)}{9n(n-1)} $$
$$ c_1 = b_1 - \frac{1}{a_1} $$
$$ c_2 = b_2 - \frac{n+2}{a_1 n} + \frac{a_2}{a_1^2} $$
$$ e_1 = \frac{c_1}{a_1} $$
$$ e_2 = \frac{c_2}{a_1^2 + a_2} $$
\( S \), the number of segregating (or polymorphic) sites in the sample
3 计算
Tajima F (1989) Genetics 123:585-595
Carlson CS, et al. (2005) Genome Res 15:1553-1565
vcftools --vcf geno.cvf --TajimaD 100000
4 示例
一个 SNP 标记，5 个个体
A A C C C
两两个体间基因型差异
| i | j | d |
| 1 | 2 | 0 |
| 1 | 3 | 1 |
| 1 | 4 | 1 |
| 1 | 5 | 1 |
| 2 | 3 | 1 |
| 2 | 4 | 1 |
| 2 | 5 | 1 |
| 3 | 4 | 0 |
| 3 | 5 | 0 |
| 4 | 5 | 0 |
\( pi = n1*n2 / (n*(n-1)/2) = 2*n1*n2 / (n*(n-1)) = 2*2*3/(4*5) = 0.6 \)
标签：limit bsp stat between ima number phi tab reg
原文地址：https://www.cnblogs.com/hjbreg/p/11827788.html
