Genetics Question - Page 2

Always if I am wrong please correct me. I'm always willing and trying to learn more to better myself.

Myret

They both are of the same dog 

Here we go again!!!!!

lol

"Black dogs in the GSD are usually black from the recessive black gene, which is different from teh gene that makes black dogs in most dogs--it seems to act somewhat as a masking gene--you can often tell the presence of the black recessive gene because it has a modifying effect (makes sables darker, makes black and tans into blanket backs usually).

Because of the way this black gene works, we do sometimes see genetically black GSDs who have a lot of bleed through on the legs."



This is partially correct and partially incorrect.  It does not act like a masking gene.  You can tell the presence of it usually but thats because, statistically, the black recessive trait carries a higher degree of penetrance with it.  The degree of this penetrance is a roll of the dice, typically.  What you are seeing here is a genetic trait called "penetrance".  Its a modification to the classical mendelian model.

This still doesnt negate the fact that a dog that is genetically black (homozygous recessive agouti) cannot have tan or red hairs.  The genotype is not programmed to call for those hair colors, so if your dog even has 1 red or tan hair on him, then he is not genetically black. 

"Bleedthrough" in "black" dogs is seldom anything more than a breeder trying to take an uneducated customer for what they can get.  There is no science behind it. 

yes tiff father has black recessive and mother has black and tan recessive.

This is partially correct and partially incorrect.  It does not act like a masking gene.  You can tell the presence of it usually but thats because, statistically, the black recessive trait carries a higher degree of penetrance with it.  The degree of this penetrance is a roll of the dice, typically.  What you are seeing here is a genetic trait called "penetrance".  Its a modification to the classical mendelian model.

What you just described is "acting like a masking gene" -- by which I meant it increases melanization. Yes, it acts like this because of the degree of penetrance.

This still doesnt negate the fact that a dog that is genetically black (homozygous recessive agouti) cannot have tan or red hairs.  The genotype is not programmed to call for those hair colors, so if your dog even has 1 red or tan hair on him, then he is not genetically black.

No, that's how DOMINANT black works--all hairs on the body are black. How do you know that the recessive black works that way? In fact our experience shows that it does NOT work that way. It is common for black GSDs to have a few tan/brown hairs between the toes. Less commonly you see the bleedthrough all the way up the legs.

If you breed two black dogs with bleedthrough together, and you get only blacks, then you know that those dog are what we in the GSD breed call "black". If they produce other colors, then they are not black, right? If the dog produces like a black dog, then it is what we consider black.

I believe that the dominant black gene still exists in GSDs--but that it is very uncommon and mostly comes in only through a few bloodlines.

well something that would screw up your hole charts..... the male of these puppies is a black sable who has sables and blacks and i think 1 bi-color in his pedigree. the mother is a creme sable who has black n tan white and sable.

Once you throw a white in the mix, you don't know what the color base is. The White gene conceals what patterns/colors are on the agouti locus. All the same, the colors you show don't contradict--

You have pictured:

sable/bicolor
sable/black-tan
black-tan/black
bicolor/black

"Does the "black" dog have tan bleed through on chest and face? If so, it's a bicolor."

No, however that doesn't make since. If he is truely a black dog WHY would he have any of this "bleed through"? What genetically causes it.

That was my point--that if it was bleedthrough on the face, it wasn't a black dog.

What is "bleed through"? What is the genetic reason for it? Again from what I understand the only way to get a black is through a homozygous recessive trait. Thus No tan should EVER pop up. If it does then the dominate gene by default would be bi-color with a black recessive please correct me if I am wrong. Again, trying to wrap my head around it. It just doesn't add up.

Why? The recessive black does not act like the dominant black gene. It's not on the same locus. It's one of the agouti variations, so it makes sense that it would act differently,

For the color combinations you mentioned. If a bi-color is bred to a sable yes you will end up with Sables due to that being the dominating trait. However just as the Sable can help determine what you get so can the bi-color depending on wether or not it has the Black Recessive trait. Please correct me if I'm wrong.

Yes--this does not contradict anything I said.

This is my observed experience.

HOWEVER, I have been told on this forum that current research indicates that the bicolor pattern is the same as the black-tan pattern and the difference is just because of modifying/masking genes. This doesn't match my experience, so we'll see where the thinking/theory ends up in a few years.

Also, please remember that there are things going on colorwise that we don't always know about--there are still some genes that mask or change or hide what we see phenotypically as well as potential mutations, so color/pattern doesn't always follow our rules.

Christine

BlackthornGSD,

I agree with you though I wouldn't be able to explain it as you did.
There is a breeder in Russia who produces dominant black but I'm not sure how he developed this dominance.

Ibrahim

"What you just described is "acting like a masking gene" -- by which I meant it increases melanization. Yes, it acts like this because of the degree of penetrance."


But its not a masking gene.  You are confusing two very distinct biological mechanisms here.  In a sense, you are comparing apples and oranges. 
While there is no biological reason the dominant black gene cant exist in the GSDs, can you please show the forum a case of a dominant black GSD?  We know the recessive black works that way because the nucleotide sequence in the agouti series doesnt contain the necessary information to produce tan hairs.  If tan hairs exist on the dog that cannot be contributed to environmental factors, then the dog is not genetically homozygous recessive.  The genetic information simply isnt there to support it.  Again, you are encouraged to show us an example that deductively shows a genetically black recessive GSD that is not fully black.  Check out the oxford journals Journal of Heredity. 

It seems that the simplest explanation is breeder miscategorization of the dogs coat.  The dog with a little tan on its toes is genetically a bi-color with a high likelihood of being a black gene carrier.  When bred to a black dog, the pups would have a 50% chance of being black and 50% chance of being bi-color.  The penetrance level in those bi-colors that receive the black gene is likely to be comparable to that of the bi-color parent who looks black except for the toes.  This misinterpretation leads breeders to idea of "bleed though" to explain away why this is happening as opposed to simply realizing that they were wrong about the genetics of the dog.

"If you breed two black dogs with bleedthrough together, and you get only blacks, then you know that those dog are what we in the GSD breed call "black". If they produce other colors, then they are not black, right? If the dog produces like a black dog, then it is what we consider black."

The fallacy here is that you are assuming that the dogs are genetically black, i.e. homozygous recessive.  Lets look at two examples here.  A black and a black can only produce black.  The homozygous traits wont allow for anything else.  Bi-colors have two possible genetic combinations; homozygous (bi-color/bi-color) or heterozygous (bi-color/black).  Now, if any of these 3 combinations are bred together, you will only end up with bi-colors and blacks, because no other trait is present in the pool.  Heres what happens if you falsely categorize a bi-color as a black with "bleed-through".... You still only get blacks and bi-colors, which you categorize as blacks with "bleed-through", therefore, to you, you would be getting "only blacks, just some with bleed-through".  Thats why this idea has been allowed to propagate.  The false assumption, if carried through, does not automatically contradict itself unless the genotype is tested.  Most breeders dont even know what the word "genotype" actually means, much less tests for it. 

A bi-color trait is the hardest phenotype of conclusively categorize because, in some cases, the line between B&T and bi-color is blurry, also because ideas like this, which has been proven to NOT be possible barring some sort of mutation, are allowed to spread. 

Back in the day, the idea that the world was flat sounded like a good idea and people believed it because it was consistant with what they observed.