Susan

We've now held back a trio (1.2 - 1 male, 2 females) of these babies, fed and cared for them for 3 years, and finally bred them. After waiting what seems like forever, you finally get pippies. When all is said and done, you have from one clutch 16 normals and 5 amels. The second clutch produced 11 normals, 2 amels, 4 anerys and 2 snows. With this information, and knowing what the original pair was, you now know the genetics of your trio. The male is normal het amel and anery. So is the one female. The other female, however, is only het amel (or Murphy is playing games with you, which can happen, however unlikely).

To prove my theory with the punnett square:
Normal het snow X Normal het snow (AaBb X AaBb)

.......AB.......Ab.......aB.......ab
AB / AABB...AABb...AaBB...AaBb
Ab / AABb...AAbb...AaBb...Aabb
aB / AaBB...AaBb...aaBB...aaBb
ab / AaBb...Aabb...aaBb...aabb

Each snake has 4 different possible gene combinations. Each of these 4 has an equal chance of being paired with any of the 4 combos from the other snake. 4 X 4 = 16 possible combination chances.

According to the square, your odds for each genetic combination is as follows:
1/16 - AABB - normal
2/16 - AABb - normal het anery
2/16 - AaBB - normal het amel
4/16 - AaBb - normal het amel and anery
1/16 - AAbb - anery
2/16 - Aabb - anery het amel
1/16 - aaBB - amel
2/16 - aaBb - amel het anery
1/16 - aabb - snow (amel anery)

Phenotypically, you get 9/16 normal, 3/16 anery, 3/16 amel and 1/16 snow. The offspring from the second clutch follows these odds fairly well. When labeling these offspring, you again cannot be certain of the exact hets each hatchling can be carrying. Therefore, since 6 out of 9 (2 out of 3 or 66%)normal hatchlings could be het for either amel, anery or both, they would be labeled as normal 66% het amel and/or anery. The same goes for the amel and anery offspring (2 out of 3)...amel 66% het anery and anery 66% het amel.

The first clutch writes out as this:
Normal het snow X normal het amel (AaBb X AaBB)

.......AB.......aB
AB / AABB...AaBB
Ab / AABb...AaBb
aB / AaBB...aaBB
ab / AaBb...aaBb

One snake has 4 different gene combinations while the other only has 2 different combinations. 4 X 2 = 8 possible combination chances.

Again, according to the square, your odds are:
1/8 - AABB - normal
1/8 - AABb - normal het anery
2/8 - AaBB - normal het amel
2/8 - AaBb - normal het amel and anery
1/8 - aaBB - amel
1/8 - aaBb - amel het anery

Phenotypically, you 6/8 (3/4) normals and 2/8 (1/4) amels. The first clutches' hatchlings match the odds. Labeling these offspring would be a bit different than the other clutch. The normals have a 4 out of 6 (2 out of 3 or 66%) chance of being het for amel, but only 3 out of 6 (1 out of 2 or 50%) chance of being het for anery. Therefore, they are normal 66% het amel 50% het anery. The amel offspring are all 50% het anery (1 out of 2).

OK...I definately need more coffee!