Posted by:
rainbowsrus
at Tue Jun 26 16:31:37 2007 [ Email Message ] [ Show All Posts by rainbowsrus ]
Thanks Paul, I figured I had not invented the wheel, just reinvented it. I was hoping you'd chime in.
The other thing I really like about this method is it ALSO works on codominant morphs. For example:
Albino Motley x het albino Motley
albino x het albino
1/2 albino
1/2 het albino
Motley x Motley
1/4 homozygous motley (aka super motley)
1/2 motley
1/4 normal
Total phenotypes = 2 x 3 = 6
Albino super motley = 1/2 x 1/4 = 1/8
Albino Motley = 1/2 x 1/2 = 1/4
Albino = 1/2 x 1/4 = 1/8
Het Albino super motley = 1/2 x 1/4 = 1/8
Het Albino Motley = 1/2 x 1/2 = 1/4
Het Albino = 1/2 x 1/4 = 1/8 ----- Thanks,
Dave Colling

www.rainbows-r-us-reptiles.com
0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)
LOL, to many snakes to list, last count:
21.29 BRB
19.19 BCI
And those are only the breeders 
lots.lots.lots feeder mice and rats   
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