>I just put my snow with my salmon het albino together back at the end of november and been very curious what the offspring will be and the variations as well. Was hoping for some of the genetic gurus could tell me what to expect from this pairing, and has it been done before by anyone?
You have a mating here that involves three loci. Loci is the plural form of locus; it's Latin. A locus is the location in a chromosome where a gene resides. Salmon, albino, and anerythristic reside in different places in the chromosomes.
If we take each locus individually, we have the following:
Salmon locus: The salmon (AKA hypo) boa could have two salmon mutant genes (homozygous salmon, super salmon) or a salmon mutant gene paired with a normal gene (heterozygous salmon, salmon). Odds are you have the heterozygous salmon, but I'll do both. By the way, salmon is a dominant mutant gene because a snake with a salmon mutant paired with a normal gene shows the salmon appearance. If a normal looking snake is labelled a heterozygous salmon, it is labelled incorrectly.
The snow has a pair of normal genes at the salmon locus.
I like the FOIL technique. This takes each parent's genes and matches the first (F) gene from each pair, the outer (O) gene from each pair, the inner (I) gene from each pair, and the last (L) gene from each pair.
heterozygous salmon (genes: salmon//normal) x normal (genes: normal//normal) -->
F - 1/4 salmon//normal = heterozygous salmon
O - 1/4 salmon//normal = heterozygous salmon
I - 1/4 normal//normal = normal
L - 1/4 normal//normal = normal
combine the repeats, and the result is
heterozygous salmon x normal --> 1/2 heterozygous salmon, 1/2 normal
The alternative mating with a homozygous salmon x normal is
homozygous salmon x normal --> all heterozygous salmon
albino locus: the snow has a pair of albino mutant genes, and the salmon het albino has an albino mutant gene paired with a normal gene.
albino x heterozygous albino --> 1/2 albino, 1/2 normal looking, heterozygous albino
anerythristic locus: The snow has a pair of anerythristic mutant genes, and the salmon het albino has a pair of normal genes at this locus.
anerythristic x normal --> all normal looking heterozygous anerythristic
The easiest way to get all combinations of the results at the three loci is to draw a branching diagram. The first column is all the results from the first locus:
1/2 heterozygous salmon (which I will shorten to just salmon to save space)
1/2 normal
After each possibility, draw a fork with one to four branches. The number of branches is the number of possible outcomes for the mating at the second locus. In this case, there are two possible outcomes. One outcome is written in at the end of each branch of the fork:
...................1/2 albino
1/2 salmon <
...................1/2 het. albino
...................1/2 albino
1/2 normal <
...................1/2 het. albino
(The dots are to make the columns line up on this forum. They would not be used with pencil and paper.)
Repeat with the third locus. In this case, the fork has only one branch. As all the babies would be heterozygous albino, I use the fraction 1/1:
...................1/2 albino - 1/1 het. anerythristic
1/2 salmon <
...................1/2 het. albino - 1/1 het. anerythristic
...................1/2 albino - 1/1 het. anerythristic
1/2 normal <
...................1/2 het. albino - 1/1 het. anerythristic
The last step is to follow each possible branch to its conclusion and multiply the fractions:
1/4 salmon albino het. anerythrisitic (appearance = sunglow)
1/4 salmon het. albino het. anerythristic (appearance = salmon)
1/4 normal albino het. anerythrisitic (appearance = albino)
1/4 normal het. albino het. anerythristic (appearance = normal)
If the salmon het albino parent is actually homozygous salmon, then I'd use only the top three lines of the branching diagram, change the 1/2 salmon to 1/1 salmon, and redo the fractions.
Hope this helps.
Paul Hollander