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Super x Super = ?
I know when you breed dom x dom you get a 25% chance of having the super form but if you breed Super x Super do you just get all do you get all supers? Or do you still get 25% chance and the rest dom form?
In the end, we will all die alone
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Re: Super x Super = ?
Theoretically, all supers.
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Re: Super x Super = ?
Bunches of little supers... 
Typically, they are called F2s, I believe, meaning they are the
2nd generation of supers.
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Re: Super x Super = ?
Dom x dom gets 75% doms and 25% normals. Co-doms can give you supers, but doms don't have a super form. Super x super of the same co-dom morph gets more supers. Super x super of 2 different morphs give you all a combo of the 2 co-dom mophs.
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Re: Super x Super = ?
Super supers!!!
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Re: Super x Super = ?
I think the non genotype terminology is very confusing.
What does “super” mean? I'm thinking “visibly different looking homozygous of a co-dominant mutation” but I'm not sure. If it's more general and is the same as “homozygous” then a dominant mutation could have a super too, it would just look the same as the hets for that dominant mutation. I prefer to use “homozygous” which has a more established meaning - having matching versions of the gene in question.
And what about “dom X dom”? Dominant refers to the mutation type, not a specific animal. I think the example was meant to indicate het X het for a mutation that happened to be the dominant type. But if we just call it het X het then it's easy to remember the same genotype inheritance rule that applies regardless of mutation type and was probably learned back with albino hets:
het X het = 25% chance homozygous mutant, 50% chance heterozygous, 25% chance homozygous normal.
It's just that once you apply the dominant mutation type to these results you see that the 25% homozygous mutant look like the 50% heterozygous so comprise 75% of the clutch looking the same. In this case you could refer to them as a group of 33% chance possible homozygous or even 66% chance possible hets I suppose. It will take some getting used to if we do eventually prove a dominant mutation.
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Re: Super x Super = ?
so once again how do you distinguish the 75% over the 25%
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Re: Super x Super = ?
 Originally Posted by RandyRemington
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It's just that once you apply the dominant mutation type to these results you see that the 25% homozygous mutant look like the 50% heterozygous so comprise 75% of the clutch looking the same. In this case you could refer to them as a group of 33% chance possible homozygous or even 66% chance possible hets I suppose. It will take some getting used to if we do eventually prove a dominant mutation.
Each egg from het X het with a dominant mutation type would have a 75% chance of being a visable morph and a 25% chance of looking normal. So you would distinguish the two types by looking at them.
Now within the 75% that look like morphs you would need breeding results to distinguish which of those 33% chance homozygous hit that chance and are actual homozygous and not just hets.
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Re: Super x Super = ?
so if the snakes are not old enough to breed how can one claim to have a 50% Het
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Re: Super x Super = ?
That's 50% _chance_ het. It either is or isn't a het but when you don't know (because it's a morph type where the hets don't show and it's too young to have breeding results) you assign a percentage to the chance that it is a het. You would know what the chance is based on the parentage. If one of the parents was a het that means it had one normal and one mutant copy of the gene in question so there was a 50% chance it passed on the mutant copy. There are even possible het pastel and spiders, it’s just that once the eggs hatch you know if they hit their chance or not because of the visible hets.
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