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So scaleless, a true Co-Dom morph?
So most of us know, that what we call co-dom morphs technically fall under the definition of incomplete dominance. Incomplete Dominance is a blending of the two phenotypes (how the snake looks). When you have 2 normal genes or 2 lesser genes, the snake has a distinct phenotype (normal or BEL). have one normal and one lesser gene and you get blending of the phenotypes. not showing 1 or the other completely. Normal books give you the example of you mix a red a white flower together and you get pink.
Co-Dominance, will show both phenotypes in their entirety. Normal books give an example of you mix a red and white flower and you get a white flower with red spots. Normal and Scaleless and then the Scaleless Head, Which is Showing both Scaleless and Normal phenotypes.
So I think we have a true co-dom morph.
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Isn't the full scaleless a recessive?
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Re: So scaleless, a true Co-Dom morph?
 Originally Posted by Kodieh
Isn't the full scaleless a recessive?
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no. the heterozygous form has a few scales missing on the top of its head.
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Re: So scaleless, a true Co-Dom morph?
 Originally Posted by TheSnakeGeek
no. the heterozygous form has a few scales missing on the top of its head.
I guess I took the thread title as describing the full scales, not het scaleless.
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Last edited by Kodieh; 10-05-2013 at 02:00 PM.
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Re: So scaleless, a true Co-Dom morph?
 Originally Posted by Kodieh
I guess I took the thread title as describing the full scales, not het scaleless.
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one is heterozygous (scaleless head or het scaleless) one is homozygous (scaleless), same gene which is what I am referring to.
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Very interesting point!!! I believe you hit the nail on the head.
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why is the het scaleless not an incomplete dominant?
Just because it has some scales missing on top of the head means it's not a recessive trait?
Jerry Robertson

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i think this is true.
we have one gene, lets call it scaleless.
and in the heterozygous form it has a few scales missing on the head, lets call it "scaleless head" or "het scaleless". very visual, no guesswork.
and in the homozygous form we have the completely scaleless BP. we could call them "fully scaleless" or "super scaleless" or "OMFG what dark wizardry is that" 
seems to be a textbook example of incomplete dominant, or as we say codom.
especially if Brians assessment that scaleless head x scaleless head gives you 25% normals, 50% scaleless head, and 25% full scaleless holds true.
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Ok so I accept the premise that scaleless is codom, but indulge my curiosity: what would an incomplete dominant look like in this case? A snake completely covered in half-scales? Very weak/miniscule scales?
Last edited by MootWorm; 10-05-2013 at 03:00 PM.
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Registered User
Re: So scaleless, a true Co-Dom morph?
 Originally Posted by snakesRkewl
why is the het scaleless not an incomplete dominant?
Just because it has some scales missing on top of the head means it's not a recessive trait?
Recessive traits don't show up in the phenotype of the animal unless the gene is homozygous for that trait. Take brown eyes and blue eyes in people. Brown eyes are dominate.... if you have just one copy of the gene, you'll have brown eyes. Blues eyes are recessive because if you only have one copy of the gene, you will not have blue eyes. So the scaless head is incomplete dominant as the op suggested. Its more like the genes that control melanin production in people. Most of them are incomplete dominant genes.
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