Posted by:
Paul Hollander
at Mon Dec 15 17:07:13 2008 [ Email Message ] [ Show All Posts by Paul Hollander ]
>So Paul this dual mutant gene example would not be possible in a Homozygous Boa or Super form as in a true CoDom Boa like Motley or Jungle?
Such a dual mutant gene example is impossible in a homozygous boa. Because, by definition, when a gene pair is homozygous, the two genes are the same. I'm not sure what you mean by a "true CoDom Boa like motley". If you mean a boa with two copies of the motley mutant gene in the gene pair, then it is homozygous.
I picked the paradigm boa because it was the most likely to be familiar to readers of the boa constrictor forum. But there are plenty of other examples in a variety of species. And the two mutants do not need to be recessive to the normal gene. It's just that recessive mutants are more common than dominant and codominant mutants.
In the following gene pairs, the mutants are classed as dominant/codominant/recessive compared to the normal gene.
Heterozygous gene pairs that do not contain a normal gene:
Dominant mutant and dominant mutant -- example: A and B genes in the ABO blood type system in humans.
Dominant mutant and codominant mutant -- example: viable dominant yellow and dominant yellow in the lab mouse.
Dominant mutant and recessive mutant -- example: checker and barless in the domestic pigeon.
Codominant mutant and codominant mutant -- example: mojave and lesser platinum in the ball python.
Codominant mutant and recessive mutant -- example: dominant yellow and nonagouti in the lab mouse.
Recessive mutant and recessive mutant -- already covered with the paradigm boa.
Sorry I couldn't use more examples from reptiles, but there just aren't very many reptile genes with three or more forms (alleles). The lab mouse has over a hundred.
Paul Hollander
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- GENETICS - jscrick, Sun Dec 14 11:23:57 2008
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RE: Great example - Paul Hollander, Mon Dec 15 17:07:13 2008
