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same as albino? such as 25% would be piebald 50 % het for piebald and 25% normal? thanks for responses scott

I really have no clue what :

"such as 25% would be piebald 50 % het for piebald and 25% normal?"

is...but if you understand the Albinism gentetics...they are the same.

I am assuming that you are saying you have a 100%het and you're breeding it to another 100% het?

therefore 25% will be Homo, 50% Het, and 25% Wild Type?
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yes 100% het to 100% het

Each baby would have a 25% chance of being homozygous from any heterozygous X heterozygous cross. Same for everything from piebald to albino and even pastel as long as you remember that the regular pastels are hets and the supers are homozygous.

Since each baby’s odds are independent there is no guarantee of the distribution of the clutch. The 25/50/25% odds are often misapplied to the distribution of the clutch when it’s really the odds for each egg. Your odds of getting at least one piebald from any four eggs from piebald X piebald is 68%. If you got 8 eggs your odds of getting at least one pied would go up to 90% (but still a 10% chance of none, however you would most likely get several).

thanks thats hat im lookin for!

I am confused now. Wouldn't the probabiltiy (odds are favorable to unfavorable) be 25% for a clutch of 4 from a het to het breeding? seeing as how theoretically 1 in 4 will be homo. And in a clutch of 8 the probability would be 12.5% for one of the hatchlings be homo.
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The odds for each egg from a het X het breeding being homozygous is 0.25 or 25%. The idea of applying these odds to the distribution of a clutch and hence that theoretically 1 in 4 will be homozygous is wrong. The 25% odds just apply to each baby, not to the clutch. So many people have applied the Punnetts square to clutches that it's hard for people to visualize that it's really just the distribution of odds for each egg.

When you want to figure the odds for a group you have to multiple the independent odds of each together. It really doesn't mater if you pick 4 eggs at random from 4 different clutches from het X het or if they are all eggs that happened to get laid in the same clutch.

It's easiest to figure the odds of getting at least one (i.e. the odds of not getting completely shut out). This calculation includes the odds of getting two, three, or all four.

Start with the odds of each baby not being homozygous. That is 1 - 0.25 = 0.75 (the 100% chance of being either homozygous or not minus the 25% chance of being homozygous lets you know that each baby has a 75% chance of not being a pied (could be either a het or not but not homozygous pied).

So, if you want to know the chance of all four eggs in a group of eggs from het X het not being pied you just multiply the chance of each together:

0.75 X 0.75 X 0.75 X 0.75 = (0.75)^4 = 0.3164

So the chance of getting at least one pied is:

1 - 0.3164 = 0.68 or 68%

If you want to figure a bigger clutch, say 8, it works like this for the odds of getting at least one:

1 - (0.75)^8 = 90%