>>When you breed a snake with a single recessive (albino) to another single recessive (hypomelanistic), are all the babies double het?

>>-----

Assuming you mean breeding the two different homozygous morphs together (albino and hypo phenotypes, in your example) then yes, you're correct.



That's true, however, only so long as the two morphs occur on different gene pairs. So far, I'm not aware of any king or milk morphs that occur on the same gene pair, so that's a "problem" only in theory at this time, to the best of my knowledge.



(To elaborate, for those who care for any more detail than that: there are a number of gene pairs populated by the "normal"or wild type genes. A het for albino has one of those gene pairs occupied by one albino and one normal gene: if the two morphs occurred on that same gene pair, you could have double hets, but you couldn't have an albino het for hypo or hypo het for albino. As it turns out, of course, the two traits occur on different gene pairs. Before we produced triple-het hondurans the question hung in the air: what if two of those morphs (hypomelanistic; a melanistic; hypoerythristic) occurred on the same gene pair? That would preclude the existence of triple hets. The answer, of course, came when the first clutch of triple het x triple het breedings hatched with animals displaying each of the three phenotypes.)



Remember that on sex-linked morphs the mutation or genotype occurs on the same gene pair that determines gender: females are females, it turns out, because they get an extra chromosome, on that gene pair. That leaves only a single "slot" for a color-determining gene on that gene pair. Thus a male (having two slots available) that gets one gene for a sex-linked morph looks "normal" and is het, but a female getting one gene on her gene pair (which has only one open slot because the other is filled by the gender-determining gene) shows the phenotype.

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