Tajima's D 中性进化检验

Posted 2020-11-30 hjbreg

Tajima F (1989) Genetics 123:585-595

1  Average number of pairwise nucleotide differences

$$ hat k = frac{sumsum_{i<j} k_{ij}}{inom{n}{2}} $$

( 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 )