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I'm wondering what the color outcome possibilities of a black & tan female with a sable male would be?

Female parents are black & tan with 3 black & tan grandparents and 1 solid black.

Male parents are father solid black, mother sable. 2 sable grandparents 2 black & tan.

Would at least 1 black puppy be likely?
 

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"I like Daffy" Moderator
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Possible.
 

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One of my males, the mother was b/t the father a very dark sable who carried the black factor.

I got a very very dark melanistic bi color. The litter was a split/ 2 b/t's/2 sables/2 bi's
 

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One of my males, the mother was b/t the father a very dark sable who carried the black factor.

I got a very very dark melanistic bi color. The litter was a split/ 2 b/t's/2 sables/2 bi's
One of the parents has to carry the bicolor....it doesn't appear with breeding black to b&t or sable.
 

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The father of my pup's litter was sable. The mother was a blanket b&t. The litter consisted of 4 blacks (2 male, 2 female), 3 sable (2 male, 1 female) and 1 male black and tan. One of the grandparents was black.
 

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How does sable work? A sable female (both her parents are also sable) was bred to my black and red male (both his parents are black and red). There were 8 dark sable puppies (4 males, 4 females) and 2 black and tan/red (both female) puppies. Does the male have to carry sable or does that make sense? I've never really thought much about color genetics since it didn't matter to us. We wanted rich color/pigment but did not care whether it was sable or black and tan/red.
 

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I'm curious about how the sable color or more coloring patterns work in the GSD as a whole I read the genetic chart but it didn't make much sense to me. I know how to breed the hounds towards a specific color scheme and not lose the brindle in their coats but if you don't breed the lighter colored dogs in every so often they will begin to come up a solid color which is a fault conformation wise. Looking at my girls parents and the gene code chart she should not have turned out the coloring she is unless I missed something
 

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Sable is dominant -- at least one parent has to be sable for there to be sable puppies. Black and tan is recessive to sable. Recessive genes such as black or black and tan can be hidden by the dominant gene for generations.

With Liesje's litter, although the dam is sable with two sable parents, you would need to look further back to find the black and tan recessive.
 

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From most dominant to most recessive:
Sable
Black/Tan
Bi-color
Black

All dogs have 2 genes for color, one from each parent. The dog will express his most dominant color gene, but can carry another color gene as his recessive. So a dog with a sable gene will BE sable, because it is most dominant. For his second color gene it could be a second gene for sable, in which case he can only produce sable offspring as the only color genes he can contribute to the offspring are sable. Or it could be a gene for another color, in which despite being sable himself he can produce the other color.

In Nikon's litter, given his lines he is most likely homozygous black/tan. The sable female is obviously not homozygous sable or all the pups would be sable. She could carry any of the other 3 colors as her recessive in order to produce black/tan pups. If her recessive is black/tan, then the pups are homozygous black/tan. If her recessive is bi or black, the pups would still be black/tan because that is the color gene they got from their sire and since it is more dominant it will trump her color genes and they will express black/tan.

Black is most recessive, so the only way for a black puppy to be born is if both parents carry the gene for black. They do not need to BE black themselves. If they carry black and any other color they will be the other color, but can still produce black puppies. So 2 black/tans can produce bi-color or black, if they carry those as their recessives. Same for a black/tan bred to a bi-color, black could be produced if both carry it as their recessives.
 

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From most dominant to most recessive:
Sable
Black/Tan
Bi-color
Black

All dogs have 2 genes for color, one from each parent. The dog will express his most dominant color gene, but can carry another color gene as his recessive. So a dog with a sable gene will BE sable, because it is most dominant. For his second color gene it could be a second gene for sable, in which case he can only produce sable offspring as the only color genes he can contribute to the offspring are sable. Or it could be a gene for another color, in which despite being sable himself he can produce the other color.

In Nikon's litter, given his lines he is most likely homozygous black/tan. The sable female is obviously not homozygous sable or all the pups would be sable. She could carry any of the other 3 colors as her recessive in order to produce black/tan pups. If her recessive is black/tan, then the pups are homozygous black/tan. If her recessive is bi or black, the pups would still be black/tan because that is the color gene they got from their sire and since it is more dominant it will trump her color genes and they will express black/tan.

Black is most recessive, so the only way for a black puppy to be born is if both parents carry the gene for black. They do not need to BE black themselves. If they carry black and any other color they will be the other color, but can still produce black puppies. So 2 black/tans can produce bi-color or black, if they carry those as their recessives. Same for a black/tan bred to a bi-color, black could be produced if both carry it as their recessives.
Excellent info! Thanks!

So where does white come into this? I know it is a 'masking' gene, but whites popping up after 3-4 generations of straight black-and-tan pairings means... what?

Thanks for the education here, I'm learning a lot!
 

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Thanks Daphne and Chris! Looking at the bitch's pedigree I see a black and tan dog as her grandmother (and great-grand father) so that must be where the black and tan females came from? All the other dogs in the 3-gen are sable but if you go farther back you start getting more black and tans.
 

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Very interesting. My pup is black, but his parents are b&t (her parents were black and b&t) and sable (his parents sable and black). Since Varik's parents are b&t and sable, does that mean he would only have b&t or sable pups? (Not that I'm going to breed him..this is theoretical as I try to understand).
 

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Falk von den Wolfen? He's a blanket B&T.
 

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Discussion Starter #20
That means the parents were black&tan+black recessive, and sable+black recessive. Because of the two black recessives that is how you got a black puppy. If bred with a sable+sable recessive you would get sable puppies, sable+b&t recessive, you would get sable or black&tan puppies, sable+bi-color recessive you would get sable or bi-color puppies. With sable+black recessive, you could get sable or black puppies.

Same for if bred with a black and tan, replace everything sable in the previous statement with black&tan.
 
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