Genetics and color genes - Page 1

My thinking is that the sire passes a color gene to both male and female off-spring and that the dam is only passing a color gene to the male off-spring. Can any genetic savy person elaberate on this?

In poultry I know this to be true. But was wondering if it is also true in canines. Specifally G.S. dogs.

mahon

wrong.  Each GSD carries two color genes.  One color gene from each parent is given to each puppy (=2 color genes per puppy).  There are three basic colors - sable (always dominant over black/tan and black), black/tan (recessive to sable, dominant over black) and black (recessive to sable and black/tan).  Each dog carries two color genes - if it is sable/sable, the dog will be a sable, and will only produce sables no matter the color of the mate.  The puppies will inherit a sable gene from the sable/sable parent, and a gene from the other parent which will be that puppies' recessive gene.  Sable/black dogs maybe dark or black sable in color - they may give a sable or a black gene to each puppy.  Black/tan dogs can have a black/tan or black recessive to give to each puppy.  (if a dog has a sable gene it will be a sable dog as sable is always dominant).  Black dogs have two black color genes - so must get one black gene from each parent.  So, breeding a black dog to a sable or black/tan does not mean you will get black puppies - only if the non-black parent has a black recessive to give. 

Hope that wasn't too confusing.

molly

Each parent contributes 2 genes, one for color, and one for pattern. Good explanation by eichenluft. Von Leistung

E and PP, your explanations together are far better to the point summary explanations than I have ever seen in any text on Dog or GSD genetics.   

ooooh im a bit stupid

if i mate my dark sable  bitch(her grandfather was black)  to a black dog will i get black pups? sorry ..

was one of your dark sable's PARENTS black?  Grandfather may have passed his black recessive along to her, or not.  But if one of her PARENTS are black, then you  know she does carry black (as the black dog has no other color gene to give  it's offspring).  Or, has she produced black puppies before?  If not, and if one of her parents were not black, then you don't know if she carries black or not.  If she is bred to a black dog and she does NOT carry black, then there will be no black puppies, but every puppy will carry black recessive from their one black parent.  If she does carry black, then puppies with a black parent will be black or sable/black.

molly

Depends one whether she inherited a black recessive or not.

I have personally produced black puppies out of blk/red WGR show line females who had NO black within their first 7+ generations of their pedigrees when bred with my black stud. So I would say there's a good chance your girl will produce black, but no guarantee.

A friend of mine bought a beautiful solid black female (she has a black sire & black dam) who was bred to a sable male (he has black sire & sable dam). The puppies produced (9) were 7 sables and 2 black. We had kinda expected and hoped for vice versa(we hoped for 7 blacks and 2 sables) but it was ok. I got one of the 2 black females produced in the litter so I was content.

Black is indeed a homozygous recessive color that must be carried by both parents in order to produce it.  A puppy has to inherit that black recessive gene from each parent in order to turn out that particular color.

A dog who is a sable, blk/red or blk/tan can either be homozygous dominant (meaning, they have inherited a dominant gene for a patterned color and are thus NOT carriers for black) or a carrier.
A homozygous dominant puppy's gene code would thus be genetically BB. This puppy is blk/red, blk/tan or sable and will not produce black pups, even if bred with a black partner.

A carrier puppy (a puppy who is a sable, blk/red or blk/tan but carries the black gene) is genetically Bb. This dog is patterned itself, but if bred to another black carrier, they have a small percentage possibility to produce black. The chances of this dog producing black is greater if it is bred to a black partner. Two carrier parents (meaning, 2 sables or 2 blk/red or blk/tan parents who are bred and are carriers of the black gene) can produce black puppies but it won't be very common. Here's an example of that: Jack vom Blender See : http://www.pedigreedatabase.com/gsd/pedigree/126955.html. He's a black progeny from 2 sable parents.

A homozygous recessive puppy (one who indeed turns out black) is genetically bb. This one is indeed black-no reason to explain that further. It can only produce it's color with a partner who carries the black gene OR is also black.

The chance isn't that minimal, though.

bb x bb

If both parents are black, (bb) in your example, then 100% will be black (barring an odd dog that carries a gene for something recessive to black--such as white).

Bb x bb

If one parent is black and one parent carries the recessive, statistically, 50% will be black.

Bb x Bb

Neither parent is black, but both carry the recessive-- 25% chance

But those are statistics--how it would turn out over a large number of repetitions. Within an individual litter, your numbers will vary.

Nothing I said was wrong. I've had some experience to show me that in reality, it seems the ACTUAL percentage for homozygous recessive things seems to tend to be lower than the predicted amount via Punnett Square figuring.

Yes, 2 carriers(Bb) have the potential as high as 1 in 4 (25%) to produce a black. But that doesn't mean 1 outta 4 pups will be black. It just means it's POSSIBLE. Of course, naturally, if you breed 2 blacks together, you're gonna get black. That's a given. 

In what I said earlier, a black female (bb) was bred to a carrier sable(Bb) and their pups were NOT 50/50 (2 out of 4 black) as predicted, but were in actuality 78% sables versus 22% black. We were far below the predicted outcome. It's the truly unpredictable beauty of genetics.

The long-coat gene works the exact same was as black; it's also a homozygous recessive condition. I've got 2 great dogs who both have to be carriers for the long coat gene because in their 3 litters together, they've produced ONE long-coat pup. So, with them both being carriers (Cc), the percentage for them to produce the long coat is as high as 25%(1 outta 4). But out of a total of 19 puppies in 3 litters they have  in fact produced, the REALITY is we've only produced one (solid black male at that) coated pup. So, while the predicted Punnett outcome is 25%, the ACTUAL outcome has been 5.2%- again, far BELOW the predicted amount.

The Punnett Square of Mendelian genetics way of looking at things is just a very general prediction and not to be considered how every outcome will precisely be. Over time and with more progeny produced, the figures should be closer to the predicted amount, but generally, it's just a loose percentage possibility.