Posted by:
Paul Hollander
at Sat Sep 17 10:49:10 2005 [ Report Abuse ] [ Email Message ] [ Show All Posts by Paul Hollander ]
Your anerythristic, heterozygous albino male can produce snows when mated to any one of the four following females: 1. anerythristic albino (snow) 2. anerythristic, heterozygous albino 3. albino, heterozygous anerythristic 4. heterozygous anerythristic, heterozygous albino
Which female or females chosen will depend on the proportion of snows you are willing to settle for, what is available, and on the amount of time and money you can spend on the project.
1. Anerythristic, heterozygous albino (AKA anery het snow) x anerythristic albino (AKA snow) --> 1/2 anerythristic, heterozygous albino 1/2 anerythristic albino (snow)
2. Anerythristic, heterozygous albino (AKA anery het snow) x anerythristic, heterozygous albino --> 3/4 anerythristic, (and 66% probability heterozygous albino) 1/4 anerythristic albino (snow)
3. Anerythristic, heterozygous albino (AKA anery het snow) x albino, heterozygous anerythristic --> 1/4 normal looking (heterozygous anerythristic, heterozygous albino) 1/4 anerythristic, heterozygous albino 1/4 albino, heterozygous anerythristic 1/4 anerythristic albino (snow)
4. Anerythristic, heterozygous albino (AKA anery het snow) x heterozygous anerythristic, heterozygous albino --> 3/8 normal looking (heterozygous anerythristic, probability 66% of being heterozygous albino) 3/8 anerythristic (probability 66% of being heterozygous albino) 1/8 albino, heterozygous anerythristic 1/8 anerythristic albino (snow)
Mating 1 produces the most snow babies but uses the most expensive female. Matings 2 and 3 can be expected to produce equal numbers of snows. I don't know where the prices would be for the females. Mating 4 would produce the fewest snows but uses the least expensive female.
Good luck.
Paul Hollander
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