Table 2 . Conditional genotype probabilities of a cow.

	Genotype sire
Genotype maternal grandsire	AA	AB	BB
AA	$P\left(A{A}_{\text{cow}}\right)=0.5+0.5p$	$P\left(A{A}_{\text{cow}}\right)=0.25+0.25p$	$P\left(A{A}_{\text{cow}}\right)=0$
	$P\left(A{B}_{\text{cow}}\right)=0.5-0.5p$	$P\left(A{B}_{\text{cow}}\right)=0.5$	$P\left(A{B}_{\text{cow}}\right)=0.5+0.5p$
	$P\left(B{B}_{\text{cow}}\right)=0$	$P\left(B{B}_{\text{cow}}\right)=0.25-0.25p$	$P\left(B{B}_{\text{cow}}\right)=0.5-0.5p$
AB	$P\left(A{A}_{\text{cow}}\right)=0.25+0.5p$	$P\left(A{A}_{\text{cow}}\right)=0.125+0.25p$	$P\left(A{A}_{\text{cow}}\right)=0$
	$P\left(A{B}_{\text{cow}}\right)=0.75-0.5p$	$P\left(A{B}_{\text{cow}}\right)=0.5$	$P\left(A{B}_{\text{cow}}\right)=0.25+0.5p$
	$P\left(B{B}_{\text{cow}}\right)=0$	$P\left(B{B}_{\text{cow}}\right)=0.375-0.25p$	$P\left(B{B}_{\text{cow}}\right)=0.75-0.5p$
BB	$P\left(A{A}_{\text{cow}}\right)=0.5p$	$P\left(A{A}_{\text{cow}}\right)=0.25p$	$P\left(A{A}_{\text{cow}}\right)=0$
	$P\left(A{B}_{\text{cow}}\right)=1-0.5p$	$P\left(A{B}_{\text{cow}}\right)=0.5$	$P\left(A{B}_{\text{cow}}\right)=0.5p$
	$P\left(B{B}_{\text{cow}}\right)=0$	$P\left(B{B}_{\text{cow}}\right)=0.5-0.25p$	$P\left(B{B}_{\text{cow}}\right)=1-0.5p$

The nine possible scenarios for the genotype probabilities of a cow conditional on the genotypes of the animal’s sire and maternal grandsire as well as the frequency of the rare allele A, $f(A)=p$ are shown. The probabilities were calculated by multiplying the respective transmission probabilities of the sire and dam of the cow under consideration.