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Belgian Color Genetics:

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There’s a little bit of voodoo…

Hypothetical question: You’ve bred your black Belgian girl "Voodoo" to a grey Belgian boy. What’s the litter going to look like?

Note: updated to reflect our changed understanding of grey and how it is inherited to the I series rather than the Chinchilla series.

Base Color

The Belgians live in the agouti color family, also known as the "A series". The gene for the typical black color in Groenendaels, for example, is represented as As which is black without any tan markings. If the As gene is present in a Belgian, it will be black.

The gene for the typical fawn to red color range in Laekenois, Malinois, and Tervuren is represented as Ay, and referred to by geneticists as the color "sable". This is important to note, because in Belgians we use the word sable to mean an entirely different thing…a grey, sand, or biscuit colored Belgian. The Ay is recessive to the black gene (As) so it needs to be inherited as a matched set for a dog to be fawn or red.

It’s also obvious that there are some other more unusual color genes in the Belgian. The most common of these is the recessive black gene (a – the same black as is present in GSDs) and the At gene (black with tan points).

Greys

In addition to base color, we also have the additional complication of the grey Belgian, which we now understand to be influenced by the "I series" of genes, which is also known as the intensity series. The I series controls the expression of the yellow pigment in the hair, thus fawn or red in our breed. Most Belgians carry a matched set of the dominant I gene, for full expression of color. Some Belgians carry the i gene. This gene suppresses the expression of the yellow (fawn to red) pigment of the base color of the dog.

Since the intensity gene is recessive, it is only expressed when a dog has a matched set of intensity genes, i.e. ii AND it is only expressed when there is yellow (fawn or red) pigment to act on…so a black dog can carry the “i” gene, or even ii, but not show it since there’s no brown pigment to be suppressed.

Back to our Story…

So you’ve bred Voodoo to a grey Belgian boy. What’s the litter going to look like? To complicate things, let’s say your girl was born of two black parents, but in her litter there were four other black puppies, one red puppy, and one grey puppy. Hmmm.

We know Dad’s genotype absolutely (AyAyii), given his phenotype of grey. We aren’t positive of your girl’s genotype, but a likely candidate would be AsAyIi…

First (Two) Scenarios:

50% "Groenendael" (black) dogs
…all of which would carry the gene for "Terv" coloring (Ay); half of which would carry one I gene and one i gene, while the other half would actually genetically be ii but could not express this because they are black.

50% "Tervuren" (red or grey) dogs
In the first example, this "Terv" half would be grey/silver/sable or some variation of "grey." In the second example, the "Terv" half would be red and carry the chinchillation gene.

A Second Scenario

Now, let’s assume that we know that Voodoo is pure for black and non-i, i.e, doesn’t carry "Terv coloring gene" or "grey modifying gene" (AsAsII).

This would statistically result in:

100% "Groenendael" (black) dogs
…none of which would be pure for black. All would carry the gene for "Terv" coloring, and all would carry the ii set of alleles and thus be capable of producing grey when bred to the right partner.

Tada! That’s not so hard now, is it?

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