Читать книгу Population Genetics - Matthew B. Hamilton - Страница 57

The first step is to hypothesize genotypes under the two models of inheritance, as shown in Tables 2.6 and 2.7 for blood groups. Then, these genotypes can be used to estimate allele frequencies (here, p indicates probability or frequency). For the hypothesis of two loci with two alleles each:

For the hypothesis of one locus with three alleles:

The expected numbers of each genotype as well as the differences between the observed and expected genotype frequencies are worked out in the tables. For the hypothesis of two loci with two alleles each, χ² = 0.266, whereas χ² = 19 688 for the hypothesis of one locus with three alleles. Both of these tests have one degree of freedom (4 genotypes −2 for estimated allele frequencies −1 for the test), giving a critical value of χ²_0.05,1 = 3.84 from Table 2.5. The deviations between observed and expected genotype frequencies could easily be due to chance under the hypothesis of two loci with two alleles each. However, the observed genotype frequencies are extremely unlikely under the hypothesis of three alleles at one locus since the deviations between observed and expected genotype frequencies are very large.

Phenotype	Genotype	Observed	Expected number of genotypes	Observed − expected	(Observed − expected)2/expected
Hypothesis 1: two loci with two alleles each
Purple/smooth	A_B_	2058	3816 (1–0.522)(1–0.512) = 2060.0	2.0	0.002
Purple/wrinkled	A_bb	728	3816 (1–0.522)(0.51)2 = 724.2	3.8	0.020
Yellow/smooth	aaB_	769	3816 (0.52)2(1–0.512) = 763.5	5.5	0.040
Yellow/wrinkled	Aabb	261	3816 (0.52)2(0.51)2 = 268.4	−7.4	0.204
Hypothesis 2: one locus with three alleles
Purple/smooth	AB	2058	3816 (2(0.48)(0.49)) = 1795	63.0	38.5
Purple/wrinkled	AA, AC	728	3816 ((0.48)2 + 2(0.48)(0.03)) = 989.1	−261.1	68.9
Yellow/smooth	BB, BC	769	3816 ((0.49)2 + 2(0.49)(0.03)) = 1028.4	− 259.4	63.7
Yellow/wrinkled	CC	261	3816 (0.03)2 = 3.4	257.6	19 517.0