gsd colors - Page 3

Two solid blacks (recessive black) can have only black offspring (except of course for recessives of other genes  eg brown, white).

Both tanpoint (bicolour) and black are recessive to sable, black is recessive to tanpoint.

They have only two colour alleles ('genes') on the A locus but multiples of other genes affect coat colour and these are responsible all the other variations we see.

Yes, but tell me, with your suggestion that bicolor is located elsewhere, something else then has to be on the A locus. To avoid repeating myself, why don't you explain exactly how bicolor becomes expressed (if not on the A locus).

Seems we're just debating in circles now. You are now saying bicolor is where I always have claimed it to be ( A locus ) and order of dominance works in the same way it always has before. Only difference is, your adding one more phenotype to the plate of possible results, when claiming it's a modifier to an At allele. This does not reflect what actually occurs.

Wow now I know what my next search for books needs to be on. I would apologize in advance for the stupidity of my question, but I get the impression that most of the questions asked of the two of you tend to appear to you to be thumb up the nose questions.  I "think" I have a very basic understanding of dominant, recessive.  Dawulf asked about breeding two black gsds.  In humans if both parents have blue eyes which is recessive they can still have a baby with brown if the brown gene is in there somewhere.  Since you can get an all black when neither parent is all black I would assume that some all blacks do carry other color genes.  So why couldn't you get the ocassional other color in the litter?

Ok, I'll make up a new locus.  This is probably an oversimplification, but - TP for tanpoint.  All dogs, no matter what their colour, will always have two alleles on A and TP.

TP - is dominant and gives a saddle phenotype.

TP^t - is recessive to TP and gives a B&T phenotype (less tan than saddle but more than bicolour).

TP^b - is recessive to both above and gives a bicolour phenotype.

 a^ta^t   TP TP   - saddle

a^ta^t    TP^t TP^{t -} Black & tan

a^ta^t TP^b TP^{b -} bicolour


aa   TP TP - black

aa   TP^t TP^t - black 

aa  TP^b TP^{b -} black


a^wa^w TP TP - wolf sable

a^wa^w TP^t TP^t - wolf sable

a^wa^w TP^b TP^b - wolf sable

You see when a^t isn't present, the TP alleles have nothing to express on.

Daryl, just to add TP is the modifying locus.  It is completely independant of A locus... may not even be on the same chromosome.  Does that make sense now?

Donner, human eye colour is a bit more complicated.  It is true that brown can very occasionally occur from blue eyed parents.  There is influence from another gene to cause that.

All blacks can't carry alleles (sable or B&T) for any other colour on the A locus but they can carry the modifyers to make lighter or darker shades of B&T.  But also, all colours do definitely carry many other genes for colour, eg a black can carry recessives for brown (liver), white, white markings etc.

I'm going to add a bit more on the basics that should help with understanding.

Any allele can only be dominant or recessive to other alleles on its own locus, like  a^t is dominant over a.  But these alleles can't be dominant or recessive to alleles on another locus.  So.... if the modifier for bicolour - TP^b is present on one locus, it can't override the A locus if aa is present there.  The dog will still be recessive black.

OK, just to play with this idea and work through some of the permutations, let's try to figure out the possible combinations that we commonly see....

a^ta^t -- saddle phenotype
a^ta -- "blanket back" phenotype; sometimes saddle phenotype (rare, but does occur)

a^wa^w -- homozygous sable, has black toemarks
a^wa^t  -- sable with black-tan pattern gene; varies from "sable" to "pattern sable" to heavily pigmented pattern sable
a^wa -- sable, black recessive, has black toemarks and black down front of legs almost always; may or may not be a "black sable"

aa -- solid black

I propose:
m = melanization, or extent of black coverage
m^h = heavy melanization
m = normal

Currently we don't know the melanization options, so I am making some stipulations for the sake of seeing how it all plays out. Amost always, in GSDs, it seems that  less melanization is dominant over more melanization, so I'm going to start there.

a^ta^t mm = normal saddle back
a^ta^t mm^h = large saddle/extended saddle coverage, but less than a "blanket back"
a^ta^t m^hm^h = "bright" bicolor (shows the bicolor pattern but has eyebrows, chest and throat patches, tarheels, toemarks) -- should never produce solid black
a^ta mm = saddle back, but can produce black dogs
a^ta mm^h = blanket back, may produce saddles, blankets, black, or bicolors
a^ta m^hm^h = bicolor pattern, may produce blankets, black, or bicolor

aa mm = solid black, shows bleedthrough tan on legs?
aa mm^h = solid black, shows bleedthrough tan on feet?
aa m^hm^h = solid black, shows no bleedthrough?

a^wa^t mm = sable with black-tan pattern gene, perhaps what we would normally call a "light" sable (not much black coverage)
a^wa^t mm^h = pattern sable, clearly defined patterning
a^wa^t m^hm^h = pattern sable, clearly defined patterning, dark areas are almost solid black (e.g., Elkoor's dog)
a^wa mm^h = sable, black recessive, very dark sable
a^wa m^hm^h = "black" sable


How it plays out in life...

EXAMPLE 1
a^ta^t mm^h -- A female who is a blanket-back dog, no black recessive, never produced black

---bred to
a^wa mm^h  -- sable with black recessive, dark but not a "black sable"; one parent was a pattern sable, other parent solid black

produced phenotypically
bicolors
blanket backs
sables, light patterning, no toemarks
sables, toemarks

--- bred to
a^ta m^hm^h =  bicolor dog, produced blacks

produced phenotypically:
bicolors
"bright" bicolors (no black recessive)
blanket back bt

(to be cont'd)


Christine

My post got truncated, here's the rest of what I had said....

EXAMPLE 2:
a^wa^t mm^h OR  a^wa^t m^hm^h = pattern sable, female, never produced black

a^ta m^hm^h = bicolor male

produced phenotypically:
bicolors
bright bicolors
pattern sables
dark sables

And I am out of time to do the actual combinations/Punnett squares, but assuming that this female was homozygous for melanization, the phenotypical results do support the idea of bicolor and saddle being the same allele.

Does this look about right, Pod?

Christine