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Old 07-24-2013, 08:10 AM   #1 
Ilikebutterflies
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Black coppers

My understanding is that black coppers are black lace based. I'm wondering if using super black would be better. I have read that by breeding super black to copper I will get super black with varying degrees of irid. Would using offspring with the heaviest irid crossed back to the copper parent make for a darker black with more iridescence-hopefully beginning to be replaced by copper?

On a side note, will breeding a marbled DS to a regular DS (the same color as some of the marbling on the first fish) keep the marble pattern tighter or is the full body coverage more dominant? I have been hearing to breed a marble to a solid to keep the nice big splashes of color instead breeding marble to marble and the patches getting smaller and smaller like on a grizzle. I will be testing that theory. I have a steel blue that I will breed to a steel blue/green/opaque marble. I have already bred her to another marble.
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Old 07-24-2013, 11:48 PM   #2 
indjo
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Black copper, to my knowledge is a result of copper x melano. There are several opinions on super blacks which I need to experiment on. Some say it was a result of the above too. One thing for sure; SB x SB does not produce 100% clean SB. Many will carry irid, mainly on the fins. So copper x SB should give you various shades of black orchids and black copper - also depends on background.

Marbles are rather dominant and will be carried for generations. How dominant it is to the main color is unknown. Marbles are unpredictable - it may produce (eventually) the original color, but it may also cause splashes. . . . If you want definite splashes, it would be best to use colors that already show splashes. Working on to create new mutations can be frustrating.
I'm guessing DS is dragon scale : People are trying to create white body with blue fins (blue dragon). This pattern is thus far only possible by crossing regular patterned blue dragon to marbles. But that "perfect" pattern has yet to be crated let alone breed true.
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