>I was just trying to confirm that odds are more along the lines of getting a 1 in 8 sunglow arabesque as opposed to what I was told being a 1 in 16 snake.

You are correct. According to probability theory, the expectation is that 1/8 of the babies from the cross would be sunglow arabesque, not 1/16.

>But from what I'm being told, my work on paper came out right. It was actually really complicated. The only example I had to work off of in the book was a double recessive x double recessive, which through me off because neither of my snakes are double recessive. If you dont mind, I don't understand why you said you would put the arabesque as RrAaHH? I dont see where the HH comes in to play or what difference it would have made?

The answer was right, but the work was not done right. Most of the trouble is that you were trying to apply a two locus model to a three locus problem. In othe words, the albino mutant gene and its normal alternative are located at one place (locus) in the genome (the totality of the genetic information), arabesque and its normal alternative are located at a different place in the genome, and salmon (hypo) and its normal alternative are located at still a different place in the genome. That makes three different locations in the genome. Mushing arabesque and hypo into one locus is not kosher.

For simplicity, both snakes in the mating should have all the loci (plural of locus) listed except the ones where both parents are normal. As the gene pair at the anerythristic locus are normal genes in both snakes, they can be ignored. But one snake has a hypo and a normal gene. Even though the other snake has two normal genes at its hypo locus, those two normal genes have to be listed. And the same goes for the arabesque locus.

I'm not sure of all the abbreviations you used. So here's what I will use:
A = the normal gene at the albino locus
a = albino mutant gene

H = hypo mutant gene
h' = the normal gene at the salmon (hypo) locus

R = arabesque mutant gene
r' = the normal gene at the arabesque locus

Arabesque het albino - h'h' Rr' Aa (sex cells = h'RA, h'Ra, h'r'A, h'r'a)
Sunglow Phenotype - Hh' r'r' aa (sex cells = Hr'a, h'r'a)

I only listed the individual sperm and egg cell genotypes once. It does not matter which h', r', or a gene winds up in a particular sex cell when the source locus is homozygous. And this short cut reduces an 8x8 Punnett square (64 boxes!!!) to a manageable 4x2 Punnett square (8 boxes).

Mixing dominant and recessive genes is not all that difficult. In a sample mating, Aa Bb x Aa Bb, is A the mutant gene or is a the mutant gene? And is B the mutant gene or is b the mutant gene? The genotypes resulting from the mating are
1/16 AA BB
2/16 AA Bb
1/16 AA bb
2/16 Aa BB
4/16 Aa Bb
2/16 Aa bb
1/16 aa BB
2/16 aa Bb
1/16 aa bb

And the phenotypes are
9/16 A_ B_
3/16 A_ bb
3/16 aa B_
1/16 aa bb

(A_ means the phenotype for AA and Aa are the same.)

These results do not change whether A and B are the mutants, A and b are the mutants, a and B are the mutants, or a and b are the mutants.

Paul Hollander

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