Hi, Ok, here goes. If you breed a b/t to a b/t and each of those dogs has a ruby parent, you have a 1 in 4 (25%) chance of getting a dominant black and tan, a 2 in 4 (50%) chance of getting a black and tan who can produce rubies and a 1 in 4 (25%) chance of getting a ruby. If you then take one of the black and tans and breed it to a Blenheim here is what may happen. If the b/t is dominant your chances are high that you will get black and tan puppies but you may also get a tricolor. The black and tans may have white on them but more white dissapears than you will believe possible so don't get rid of what might be a good one just because it appears mismarked - unless there is a lot of white around the chin. Even then, consider using the dog or bitch for the next generation of breeding if it's got everything else that you like. If the black and tan carries for red and is not dominant, you could certainly get all four colors but will probably get more solids than particolors. The same applies about mismarked puppies. In general solids are dominant to particolors. It ultimately depends on how many particolor ancestors are behind the solid. I did a breeding recently between two English toys, a ruby and a black and tan. I ended up with a black and tan and a Blenheim. When I looked at the pedigrees I realized that there were lots of particolors behind the black and tan and none behind the ruby for 5 generations. On the other hand I bred a ruby dog some of you may remember named Kenjockety Red Anubis (Cavalier) who is 3/4 particolor bred. He had some white on his feet at birth which is very hard to find now. He went BOW at Trenton a few years ago in 97 degree heat. I wouldn't breed him to a particolor with any expectation that the solid coloring would stay. In other words: His mother was a mismarked b/t out of a tri bitch and b/t sire. His father was a tricolor with a Blenheim dam and tri sire. I hope that helps
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