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. 2019 Aug 30;22(2):301–308. doi: 10.1038/s41436-019-0636-5

Table 2.

Probability of joint genotypes

	AA	AB	BB
AA	$p (p + q F_{1}) (p + q F_{2})$	$p (p + q F_{1}) q (1 - F_{2})$	0
AB	$p q (1 - F_{1}) (p + q F_{2})$	$p q (1 - F_{1}) (1 - F_{2})$	$p q (1 - F_{1}) (q + p F_{2})$
BB	0	$q (q + p F_{1}) p (1 - F_{2})$	$q (q + p F_{1}) (q + p F_{2})$

Each row is the genotype of the mother and each column is the genotype of the fetus.