This anerythristic will produce snows only if he is heterozygous albino (aka het for snow). If he is not, then he can't sire snows.

Assuming that he is heterozygous albino, mating him to the following females will produce snows. I've listed the four females in order of desirability, with the most desirable mating first.

1. snow -->
1/2 snow
1/2 anerythristic, heterozygous albino

2. albino, heterozygous anerythristic -->
1/4 snow
1/4 albino, heterozygous anerythristic
1/4 anerythristic, heterozygous albino
1/4 normal looking, heterozygous albino, heterozygous anerythristic

3. anerythristic, heterozygous albino -->
1/4 snow
2/4 anerythristic, heterozygous albino
1/4 anerythristic
(The anerythristics could be called 66% probability heterozygous albino.)

4. heterozygous albino, heterozygous anerythristic -->
1/8 snow
3/8 anerythristic, 66% probability heterozygous albino
1/8 albino, heterozygous anerythristic
3/8 normal looking, heterozygous anerythristic, 66% probability heterozygous albino

Matings 2 and 3 could be expected to produce the same fraction of snows. However, the anerythristic male is only a possible het for snow, if I'm reading this right. Seven normal babies without any snows or albinos from mating 2 means that the odds are 99% that he is not heterozygous albino. Mating 3 must produce 17 normal babies and no snows to get 99% odds.

Good luck.

Paul Hollander

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