Genetics Question - Page 3

Ibrahim:  I'd like to see how he did it as well.  Theres no reason that it cant exist, so it would be interesting to see a case of it.

I think it extremely likely that this dog is a dominant black:

http://www.pedigreedatabase.com/german_shepherd_dog/dog.html?id=522436

Look at the dogs he is bred to and what he produces; look at the dogs his parents were bred to. Look at the line of solid blacks going back through his pedigree.

It all goes back to this female:

http://www.pedigreedatabase.com/german_shepherd_dog/dog.html?id=522398

"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. 

I'm not confusing them. I'm saying they can look alike.



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? 

See my previous post


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. 

Hmmm. This is the first I've heard of this. This is how I've heard the dominant black gene described--is it scientifically proven to be the same for the recessive black or is this just how it has been interpreted? If it is true, then these bleedthrough blacks *cannot* be homozygous blacks.

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. 

You make a very good point--and you could be 100% correct. I know someone had posted a few months back that she was going to DNA test her "bleedthrough" black--one who looked much more like a bicolor than a solid black---I haven't heard yet what the DNA tests show. One could breed a bleedthrough black to a  sable with a black recessive, though, and if this breeding gave you bicolors, you would know that the dog wasn't a true recessive black. I don't know of any cases testing this, however--although I am positive they have been done. And anecdotally, at least, it seems, though, that bleedthrough blacks reproduce more like blacks than bicolors, though. (As in the example given above.)

It is conceivable that there are 3 types of black dogs in the GSD breed--

1. extremely rare dominant blacks
2. true homozygous blacks (no tan/brown between toes or one the legs)
3. bleedthrough blacks--which are really extremely melanistic bicolors

Christine

Ibrahim
ohhh sorry wrong link

here it is
http://www.pedigreedatabase.com/german_shepherd_dog/dog.html?id=472619 

Here it is Myret

Ibrahim

thanks
that are two almost black dogs with just a litle bit of tan wonder if they are called bicolor or black

Myret,

So far I don't think anyone can give a definate judgement, they look the black we know and are familiar with but having (bleeding through). As Christine said above if they produce solid blacks then they themselves are true solid blacks, if they do produce bi-colors too with the standard  pattern we know then they are not blacks. Personally I think they are both solid blacks and not bi-colors.
It is possible that the bleeding through may increase with age, remain the same or even disappear.

Ibrahim

There is a way to make sure whether they're blacks or not by looking at their vent, if it's all black then they are solid black.

Ibrahim

Ibrahim

I would say black to have not looked what they have produced but as far a I know thay both have all black progency beutiful dogs they are both of them