gsd colors - Page 2

LOL @ Beetree!!

My chart should be considered accurately effective, until evidence is provided that can disprove it.

What pod has proposed is that the a^s and a^t alleles are effectively the same, which is fine if you want to lump them together.  I don't.  ALL the agouti allele types are just mutational variants of the ONE original wild type (sable).  The bicolor phenotype is recessive to black/tan phenotype no matter how you want to dice it, and is most certainly controlled at the agouti locus.  Hence, the usefulness of differentiating between the two.  Call them a^t1 and a^t2 if it makes you feel more comfortable.

So, wherever you wish to draw a line between all manner of mutations of a single "original" allele that has evolved to distinguish it's various allele "types", so be it.  You can oversimplify, and end up with a lot of unexpected results outside and because of your willfully limited observational view, or you can even break these "types" down into even further subcategories, if you wish to focus on more precise detail of your phenotypic evaluations into genotypic predictions, such as my given example of sable with or w/o black ventral (belly) stipes.

Oversimplification at the cost of accuracy is the tradeoff you have to be willing to make.  But, understanding the agouti locus is going to contribute to furthering your understanding beyond simple color predictions.

Daryl, I do think that you should read the abstract.  

This is not my proposal, this is evidence supplied by the experts, in this published peer reviewd paper, who have been researching this for years.  It has been found that saddle and bicolour dogs are IDENTICAL at the A locus, they are all a^t.  There will of course be some variation due to co-dominance but that doesn't alter the fact that  a^t a^t genotype produces saddle, bicolour etc.  Don't you think that if there had been two different "allele types" for tanpoint producing different phenotypes, the reasearchers would have noticed and actually called them a^t and a^s?

Each of the Agouti alleles has variations within its own expression.  Sables are all different, wolf sables.  They don't all have the exact same pattern.  Some will have darker pigment, stronger mask expression, heavier shading and even recessive blacks.  These are not all solid black, there is variation in phaeomelanin contribution.  And if we go on to other patterns.... take merle.  No two merles are ever identical phenotypically, yet they are all Mm in genotype (MM produces a similar pattern with extensive white).  The merle mutation is also a SINE insertion.  I should add that there may be other factors at work with merle like developmental noise.

What this means is that these major genes lay down the foundation of a pattern and other genes and modifiers (and possibly environment) alter the expression by controlling the balance between pigments, and pigment shade.  This is why we don't see any clear distinction between saddle, B&T and bicolour.  It's what called a pattern of continuous variation and this requires the interaction of a number of genes to produce this and why just two A locus alleles doesn't fit.

And yes, of course saddle is dominant over bicolour.  These patterns definitely exist as distinct patterns, it's just that they are controlled elsewhere from the A locus.

This is the main problem I see, with NOT associating the bicolor to the A locus; say you have two sable dogs, each with a black recessive, (aw+a) and either one or both are carriers of the bicolor gene frome *some* other location.  Now, the offspring would have the potential of throwing three phenotypes; sable, bicolor, and black.  I cannot find a precedenting example of this, perhaps someone else can.  Only, the bicolor won't ever express itself until there is a homogenous carrier of the self-color alleles (a+a), and both parents must always be black or carriers of recessive black (a+a, at+a, aw+) at the agouti locus, in order for the bicolor to express itself.

In the study you cited, they only genotyped eight saddle backed dogs, not even all of the same breed.  The researchers did not sample any bicolor dogs.  Keep in mind, ALL the allele variants are derived from and still very similar to the wildtype original; the black allele, for example, is basically a sable allele with an early coding segment that negates the function of the entire allele, causing black or self-coloration. A bicolor allele at the agouti locus would not be drastically different by any measure than a black/tan allele after genotyping.

Very interesting banter. I'm learning a lot, thank you!

Daryl, 

"This is the main problem I see, with NOT associating the bicolor to the A locus; say you have two sable dogs, each with a black recessive, (aw+a) and either one or both are carriers of the bicolor gene frome *some* other location.  Now, the offspring would have the potential of throwing three phenotypes; sable, bicolor, and black."

No, bicolour is only possible if the a^t allele is present on the A locus.  The Agouti gene is responsible for laying the foundation of the pattern, the modifyers just adjust the pigment distribution by saying 'more black', 'more tan' etc.  It could be these exact same modifyers that control the pigment on sables and wolf sables.

The last sentence of that para, I don't get what you're saying.  Is this what you actually see happening?  Can you give an example.

Bicolour is not a term generally used outside of GSDs so any of the dogs described as black & tan in table 1 could have the pigment distribution referred to as bicolour in the GSD.  For example, any of the 9 GSDs, and other breeds such as Saluki, Finnish Lapphund, Corgis, Shetland Sheepdogs and most definitely the 4 Dachshunds and Dobermann.

Two of the alleles are almost identical, a^t and a.  They both have the SINE insertion and differ by just one point mutation.  That's just one nucleotide difference.

Daryl, here is an example of what you say. Check out the pedigree of Arko vom Goldwald, a bicolor littermate of sables. Both parents are sable. The bicolor comes over the father line (Capri).

No, bicolour is only possible if the a^t allele is present on the A locus.

Then the black & tan would be the phenotype, not bicolor, because it holds dominance.  If there is only sable/B&T/black at the agouti locus, and bicolor is elsewhere, the only genotype option at agouti is a+a, or you will never see the bicolor in phenotype.

Elkoor, you are saying that Arko's sable parents both carry a black recessive, AND one (or both) carry the bicolor gene?

Daryl, sorry I'm not following your line of thinking very well.  I'll just try to list the options.

a^wa^w,   a^wa^t,   a^wa - will produce sable with varying amounts of shading depending on modifying genes.

a^ta^t,   a^ta -  will produce saddle, black & tan, blanket B&T, bicolour.  The distribution of tan and black pigment is governed by modifyers.

aa - black

Ah, I think I get it now.  You're saying that a lower dominance allele has to be present at Agouti for higher dominance pattern modifiers to express.  No, think of these modifiers as separate in action.  They can't be dominated by the A locus, but they need the pattern to be laid down fist at A.  We need some lateral thinking here Daryl.

ETA: The allele for tanpoint( a^t) always has to be present at A locus for any of the B&T patterns (saddle, bicolour etc) to show.  The specific pigment distribution of bicolour is not goverened by a^t  but this doesn't mean that this allele, previously thought of as bicolour, is no longer there at A locus.

How is it that a solid black parent bred  to another solid black parent make bicolor puppies? I thought they could only have 2 color genes, but aren't both black and bicolor recessive? Is bicolor more so then black?

I'm sorry, I'm just trying to grasp this, myself....