>I understand that if I breed 2 BCI normal's het for albino, het for anery, one in 16 offspring would be a snow.

Statistically, that is correct. What actually happens could differ more or less. The difference between one litter and many, many litters also explains why casinos can pay off big winners and still make a profit.

>But what if I breed and albino and a normal het for albino, het for anery?

You can use a Punnett square. The genes could be symbolized as follows:
A = normal at the albino locus
a = albino
An = normal at the anerythristic locus
an = anerythristic
// = a pair of chromosomes. So A//a would be a pair of chromosomes with A in one chromosome and a in the same spot in the other chromosome.

The albino would be a//a An//An. All of its sex cells (sperm or eggs) would have a gene for albino and a gene for normal at the anerythristic location in the genome (locus), which is symbolized a An.

The heterozygous albino, heterozygous anerythristic would be A//a An//an. Half of the sex cells would have an albino mutant gene and half would have a normal gene at the albino locus. Half of the sex cells with the albino mutant gene would be An, and the other half of the sex cells with the albino mutant gene would be an. Same with the half of the sex cells that have the normal gene at the albino locus. In other words, this snake produces four types of sex cells -- 1/4 a An, 1/4 a an, 1/4 A An, 1/4 A an.

Combine each of the heterozygous albino, heterozygous anerythristic snake's four types of sex cells with the albino's single type of sex cell in a Punnett square. As Punnett squares are pretty hard to do in the kingsnake.com forums, I'll just give the results, which are
1/4 a//a An//An (albino)
1/4 a//a An//an (albino, heterozygous anerythristic)
1/4 A//a An//An (heterozygous albino)
1/4 A//a An//an (heterozygous albino, heterozygous anerythristic)

As the heterozygous anerythristic babies cannot be distinguished from the babies that are normal at the anerythristic locus, you could call all the babies 50% probability heterozygous anerythristic --
1/2 albino, 50% probability heterozygous anerythristic
1/2 heterozygous albino, 50% probability heterozygous anerythristic

By the way, 100% probability heterozygous albino is just a long way of writing heterozygous albino. Stamp out and eliminate redundancy!

Paul Hollander

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