Troy can you do a breakdown of a recessive gene?



Sure, it's the exact same method for what I did with the Superconda/Anaconda Codominant traits except that we're defining the genes differently.



Here's the FOIL method for Recessive traits:

(We're using Amelanism as an example but any Recessive trait will work)

Normal = AA

Amelanistic = aa

Normal 100% het for Amelanism = Aa



#1. AA x aa = Aa, Aa, Aa, Aa = 4 Normal 100% hets.

#2. AA x Aa = AA, Aa, AA, Aa = 2 Normal, 2 Normal 100% hets.

#3. Aa x Aa = AA, Aa, aA (Same as Aa), aa = 1 Normal, 2 Normal 100% hets, 1 Amelanistic.

#4. Aa x aa = Aa, Aa, aa, aa = 2 Normal 100% hets, 2 Amelanistic



This time I numbered the combinations because this is where it can get a little confusing as this is where we also learn about possible hets and understanding where 50% and 66% hets come from.



In examples #1 and #4, you are producing all Normals that are 100% het for Amelanism. There's no surprises here. Examples #2 and #3 are where the fun begins.



In example #2 you are producing all Normals but half are het for Amelanism and half are not. The problem is, you can't usually tell which ones might be carrying the gene or not. So we use ratios to determine the possibility of each one to carry the gene. If the method determines that we produce 2 out of 4 hets and 2 out of 4 non-hets, that gives us 50% odds whether each individual is het or not. The truth is, these are just statistical ratios though. In real life, anything is possible. You could actually end up with all hets or none at all.



In example #3 we actually produce an Amel out of the 4, but the problem is still determining the Normal non-het from the Normal hets. Being that there are 3 Normals, 1 non-het and 2 hets, we simply use ratios again. Since 1 out of 3 is 33%, we simply add up how many percentage points for each het which gives us 66%. So in this clutch, we'll produce one actual visual Amelanistic and 3 Normals which are 66% het for Amelanism.
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Troy Rexroth

Rextiles

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