Thanks Tom! :cheers:
No, that's not correct. The answers to your questions are complicated, but I'll try to make it as simple as I can.
Like all animals, baby snakes inherit one copy (allele) of a gene from each parent for every genetic trait. Every trait has a location (locus) on a chromosome where both contributed alleles reside. Most non-normal (mutant) corn pattern and color traits, like anerythrism (lacking red), are recessive. This means that the baby must inherit one copy of the same mutant gene from each parent for the mutant gene to be expressed (visible). When this happens, and the trait is expressed, that mutant gene is said to be in homozygous form (homo = same). When the snake receives a mutant gene for a trait from one parent, but the other parent contributes a non-mutant (normal) gene, the mutant gene is said to be in heterozygous form (hetero = different).
So in the case of this snake, a het anery, she received one anery gene from one parent, but a normal, non-mutant gene from the other. So the trait is not visible, but she is carrying it in het form. So at the anery locus, the two genes are not the same (one is normal, the other is anery). When she breeds, she can contribute a copy of either the normal gene or the mutant gene to each of her offspring. So there's a 50% chance that she will pass on the gene that is in het form (in this case anery) to each of her offspring.
So if she's paired with a snake that isn't even carrying anery in het form (and barring other mutant traits), the babies of this red snake WILL ALL be red.
If she's paired with a snake that's het for anery, each snake has a 50% chance of contributing a copy of the anery gene to each hatchling, and each hatchling will have a 25% chance of getting two anery genes and expressing anery. There's a 50% chance of each baby getting just one copy of the anery gene, and there's a 25% chance that each baby will not even get one copy. So statistically, 75% of the babies WILL be red.
If she's paired with a homozygous anery, the het anery female will still only have a 50% chance of contributing a copy of the anery gene. But the homozygous male can ONLY contribute an anery gene, since both copies of the gene at the anery locus are the mutant gene. In this case, there's a 50% chance for each baby to be het anery, and there's a 50% chance for each baby to be homozygous for anery. So even here, 50% of the clutch WILL be red.
I hope this helps. If you're interested in this stuff, you'd probably be better off finding a better reference than me.
