Originally Posted By: PaschaSuppose there is a breeding between two dogs. One carries the dominant black and tan and the recessive solid black genes. The other carries the blk/tan or blk/red double recessive dilution gene.
There must be two recessive solid black genes to produce a solid black or two recessive dilution genes to produce a blue.
Am I right in saying that there can only be blk/tan pups? OR if one of the dogs passes a solid black and the other passes a dilution, could there be some solid blues? Or could there be blue/tans?
So parent 1 is heterozygous Black & Tan/Black.
Parent 2 is homozygous Black & Tan. I'm unclear when you say "double dilution gene" if you mean it carries 2 copies of one dilution gene, either 2 blue or 2 liver, (in which case the dog would BE a dilute and every pup produced will be a carrier) or if it carries 1 copy of each of the 2 dilution genes, 1 blue and 1 liver.
Either way, ALL pups will be Black & Tan. There cannot be solid blacks without both parents carrying at least one gene for solid black. Statistically, half the pups would be homozygous Black & Tan and the other half would be heterozygous Black & Tan/Black.
As far as the dilutes, it depends on what you mean whe you say "double dilution gene".
If the parent carries 2 copies of one dilution gene, and therefore IS Blue or Liver & Tan, not Black & Tan, ALL pups will inherit one dilution gene. They will not be blue or liver, but all will be carriers of the gene.
If the parent carries 1 copy of each dilution gene (1 for blue, 1 for liver), most of the pups will probably inherit one or both of the dilution genes and be carriers, some will not, but none of them will be dilutes themselves.
Either way, none of the pups produced will be dilutes themselves, since only one parent carries dilute genes, but many will be carriers of dilute genes.