Quote:
|
Originally Posted by jsrocket
If I understand what I've read here, and assuming the salmon trait is dominant, here are the possibilities:
Super Salmon phenotype = S
Salmon phenotype = s
Wild type = w
S X S = 100% Super Salmon
S x s = 50% Super; 50% Salmon
S x w = 100% Salmon
s x w = 50% Salmon; 50% Wild type
|
IMO the best way to work these is to use the actual genotypes, since the phenotypes are not what are being crossed. It's like trying to learn mutiplication by simply memorizing times tables and never learning counting or addition. (Then if the number is not on the memorized table, you're screwed.
)
That is, use both genes present at the locus instead of one phenotype. The results are the same, but would make more sense to someone who is trying to understand what is going on underneath the phenotypes.
So instead of starting out by memorizing the results of "S X S" (and every other cross under the sun for every locus) you can use S for the dominant salmon mutant allele and + for the wild-type allele. There are then three possible genotypes at the salmon locus:
SS = "Super" Salmon
S+ = Salmon
++ = Wild-type
When you cross any two of these genotypes to each other, each baby receives one of the dad's genes, one of mom's genes. Using FOIL is an easy way to get the four possible outcomes. For example, if you are crossing 12 X 34 at a given locus you get:
First = 13
Outside = 14
Inside = 23
Last = 24
This is the same thing as doing a one-locus Punnett square:
<table border=1><tr><td></td><td>1</td><td>2</td></tr><tr><td>3</td><td>13</td><td>23</td></tr><tr><td>4</td><td>14</td><td>24</td></tr></table>
You can just substitute the 1 and 2 for the father's two genes, and 3 and 4 for the mother's two genes, and this will cover every cross at every (paired) locus, no matter what the parents' genotypes are.
-----
So, given that, the prediction for the original cross (Albino X hypo het albino) is:
At the Salmon Locus you are crossing ++ X S+
Results are: +S, ++, +S, ++ (half salmon, half normal)
At the Albino locus you are crossing aa x a+
Results are: aa, a+, aa, a+ (half albino, half normal)
You can then combine those two sets of results in a grid similar to a Punnett square:
<table border=1><tr><td></td><td><B>Salmon</B></td><td><B>Non-salmon</B></td></tr><tr><td><B>Albino</B></td><td>Albino, Salmon</td><td>Albino, Non-salmon</td></tr><tr><td><B>Non-albino</B></td><td>Non-albino, Salmon</td><td>Non-albino, Non-salmon</td></tr></table>