breeding bay horses

Suppose we were to breed two bay horses together. We had a bay mare and stallion who were both of genotype E⁺e at the extension locus (E⁺ causes the production of the black eumelanin pigment) and of genotype A^AA^a at the agouti locus (which controls the distribution of black pigment).

The gametes may now be of four types, any of which are equally likely: E⁺A^A, E⁺A^a, eA^A or eA^a. The possible outcomes of the cross can be seen from a Punnett square:

Genetic contribution from mare:	Genetic contribution from stallion:
Genetic contribution from mare:	E⁺A^A	E⁺A^a	eA^A	eA^a
E⁺A^A	E⁺E⁺A^AA^A bay	E⁺E⁺A^AA^a bay	E⁺eA^AA^A bay	E⁺eA^AA^a bay
E⁺A^a	E⁺E⁺A^AA^a bay	E⁺E⁺A^aA^a black	E⁺eA^AA^a bay	E⁺eA^aA^a black
eA^A	E⁺eA^AA^A bay	E⁺eA^AA^a bay	eeA^AA^A chestnut	eeA^AA^a chestnut
eA^a	E⁺eA^AA^a bay	E⁺eA^aA^a black	eeA^AA^a chestnut	eeA^aA^a chestnut

There is a 9:3:4 of bay: black : chestnut. (The agouti allele in the chestnut horses is irrelevant to the phenotype since there is no black pigment to distribute, either uniformly or in the points.)

If either the mare or the stallion were of genotype E⁺E⁺ at the extension locus then only bay, black and brown foals would be possible according to the parents genotypes at the agouti locus.

Breeding bay horses with brown, black or chestnut horses could result in bay foals. Breeding together brown or/and black horses would only give brown or black foals.

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breeding bay horses