Ask the Star Pups

((please, if you know more about dog genetics, do weigh in!

To clarify on some points, the A locus controls the distribution of eumelanin (the melanin that controls black and brown pigmentation) and phaeomelanin (the melanin that controls red), not necessarily whether a dog is “black or red”. All dogs follow a distinct pattern depending on their alleles, from clear red to tan points. Sable (Ay) is the top dominant gene, and comes in three patterns: clear sable, tipped sable and shaded sable. Agouti (aw) consists of banded hairs, sometimes called “wolf grey.” Tan points (at) can be modified by a separate gene to recede into a “saddled” look as the dog ages. Recessive black (a) is a rare gene that keeps dogs from producing any phaeomelanin pigment. 

It’s the K locus that is actually primarily in charge of whether a dog displays its eumelanin, and how it will do so. The locus comes in three variations. (K), known as dominant black, is– as you might have guessed –the most dominant of the three alleles. A dog only needs one copy to be fully black, meaning they will show absolutely no phaeomelanin patterns. (kbr) is the genotype for a brindle dog. Brindle striping will appear on any phaeomelanin areas of the dog. It is recessive to dominant black, but dominant to non-black. (k) is the allele for non-black, and when homozgyous, will display whatever is on its A locus.

Weed is (kbrk), which I will expand on in a moment.

First, I want to say that the previous poster hit the nail on the head as far as dilution’s function, but I have to disagree with its usage. Weed is, for all intents and purposes, non-dilute (or black). Like stripes on a tabby cat, a brindle dog’s eumelanin can be figured out by looking at their stripes (or more easily the nose, but Ginga doesn’t really take that into account anyway. I can also excuse the confusion because GDW never drew Weed with any freakin’ stripes besides calling him “tora-ge” left and right.) 

Eumelanin and phaeomelanin, and both of their extensive variations, are the only “self” colours of dogs (”white” is caused by separate genes and is actually an absense of colour more than a colour itself, but I digress). Different genes will just control their shades and how they are expressed, if it all, but the genetic coding for them will never disappear. I bring this up because it’s “impossible” (in quotations because that’s a dangerous word to apply to genetics) for a “blue” dog to have “black” stripes. Black and blue are versions of one another, controlled by all the same genes that control eumelanin, to have both a blue coat and black stripes… they would need to be dilute, but also non-dilute, and dominant black but also brindle… we’d call that a highly, highly unlikely genetic abnormality at the very least.

This is a silver brindle Akita Inu (with white spotting, as is typical for most silver Akitas), which both Gin and Weed were based off of. The “silver” I’m referring to has actually nothing to do with black or the dilution gene, rather it is actually a shade of red (phaeomelanin).

The eumelanin (again, black) on this dog is black. The phaeomelanin (red) on this dog is silver. This dog is brindle, so it appears to have black stripes on a silver base. 

It is the I locus that is believed to control the intensity and richness of any phaeomelanin produced from the body; it is the gene that gives us the rich, dark red of Irish Setters and the ivory white of Samoyeds. That being said, this gene has absolutely no affect on any eumelanin areas of the dog in any way. So for example, if you have a black dog with tan-points, only the tan-points will be modified by this gene.

That’s where the Kishu Inu comes in. 

One can infer that the Kishu Inu is actually a very pale red dog, because red Kishu Inus are known to pop up every now and again even in strictly white lineages.

There are no traces of black on these dogs, so we can assume they are (kk) (non-black). However, this would mean that they would express what’s on their A locus, so what could it be?

It’s possible that they could be clear sables like the previous poster suggested. However, with the presence of the “sesame” pattern in Kishus, it is likely safer to assume that Kishus carry agouti or “wolf-grey” (aw) on their A locus instead. 

So what the hell’s going on?

There is another locus, called the E or Extension locus, that is responsible for the A locus’ patterning, controlling which areas of a dog produce eumelanin and which produce phaeomelanin.

(E) is the gene for normal extension, meaning that the locus makes no modification. Recessive to it is the (e) gene called recessive red, which prevents the body from producing any eumelanin on the fur at all. 

With different lineages of Kishus, and how dog genetics are still in good deal quite a mystery, it’s near impossible to tell at the moment whether it’s clear sable or recessive red that is causing white Kishus to not express any black. 

Regardless, earlier I mentioned that Weed is (kbrk), which I stand by. He inherited a brindle (kbr) gene from Gin and a non-black (k) gene from Sakura. Koyuki is neither black nor brindle, so I’d say she is (kk).

That’s three non-black genes and one brindle gene for Weed and Koyuki’s pups to fight over.

Sirius is silver with black brindle, so he, like Weed, is (kbrk). Orion is red, which is uncommon but not impossible (like the red Kishu example), so he is (kk) like his mother. Bellatrix is silver with black brindle, so she is (kbrk).

So how can one interpret Rigel’s design on a realistic level?

Rigel is (kbrk).

This is an example of “reverse brindle” on a boxer. All it is, is a dog with the brindle coat pattern that expresses more eumelanin distribution (or striping) than what is “average”.

And I have always head-canon’d personally that Gin’s family expresses white spotting, like the boxer above, because brindle DOES affect urajiro, unlike Ginga would have you believe.

Brindle without white-spotting (on the face):

Brindle with white-spotting:

A realistic interpretation of Rigel: