Reptile & Amphibian Forums

Is this correct...

A piebald and an albino bred together would produce double hets for piebald and albino. Then if you breed the double hets you get (on average) 1 normal, 1 albino piebald, 3 albino, 3 piebald and some double hets?

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1.1 Normal Ball Pythons (Snakerton and Mathilda)
1.0 Het Albino Ball Python (Walter)
0.1 Albino Ball Python (Perrie)

genetics

Just sub axanthic with pied
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RicK @ BbI

Ball Boutique,Inc.
Proud sponsor of this forum

"Is this correct...
A piebald and an albino bred together would produce double hets for piebald and albino. Then if you breed the double hets you get (on average) 1 normal, 1 albino piebald, 3 albino, 3 piebald and some double hets?"

Assuming that we are talking about breeding a homozygous piebald X homozygous albino, then yes, the offspring would all be double hets; all would be heterozygous for piebald and for albino - but would all appear normal.

From a Double Het X Double Het cross, you can STATISTICALLY expect:

1/16 Normal (not heterozygous for anything)
2/16 Het Piebald (not het for albino)
1/16 Piebald (not het albino)
2/16 Het Albino (not het piebald)
4/16 Double Hets (just like the parents)
2/16 Piebald Het Albino
1/16 Albino (not het piebald)
2/16 Albino Het Piebald
1/16 Albino Piebald

I have intentionally not simplified those fractions to give a better picture of the proportions.

I suggest you draw out a Punnett Square of your own to get a feel for this. Punnet Squares tell you what you can statistically expect from a given breeding involving SIMPLY inherited traits (the gene seems to involve one allele). Nothing about breeding one pair of animals has to be what is statistically expected; statistics give you something with which you can better predict what would happen in say ten or more such breedings, where the results can be pooled together.

For the issue of selling the offspring from a Double Het X Double Het breeding, 9/16 or ~56.25% of the the offspring can be expected to appear normal. So with these 9/16, where you know 8/16 of the clutch can be expected to be heterozygous for one or another of these traits without showing a morph, we can say that we expect that 8/9 of the babies that appear normal are het for something. You can't guarantee that proportion to your buyer however, so DON'T call them "89% possible" hets! Of these 9/16, you can expect 6/9 to be Het for Albino, and 6/9 to be Het for Piebald, and could therefore call them "66% possible" hets for both Piebald and for Albino. The safest bet would probably be to draw up that Punnett Square and explain to your buyer where their chances lie.

You should have some offspring showing morphs! You can expect somewhere around 7/16 or ~43.75% of the offspring to show a morph of some kind. But what are those morphs really? I would hope that you could pick out your 1/16 Piebald Albino, so that leaves 6/16 more of the clutch that shows a morph, and could carry another. How do we market those? The popular practice seems to be again to say that of the 3/16 offspring that are Piebald, where we can expect 2/16 to be Piebald and Het Albino, to call all of the Piebald offspring "66% possible" Het Albino; similarly, all of the Albino offspring would be considered "66% possible" Het Piebald.

Either way, this is just speaking statistically, which gives you the chances of producing these animals. It is entirely possible for a Double Het X Double Het breeding to produce nothing but Normal offspring that are not heterozygous for either of these traits - possible, but not very likey; just as likely that the same cross would produce nothing but the targeted double morph, the Piebald Albinos. What is most likely is some kind of a distribution of traits that somewhat reflects what the Punnett Square will predict for you.

"Possible" Hets are still all or nothing; there is no such thing as a partial heterozygote! With the traits that we deal with in the Ball Python trade, the pattern of inheritance seems to indicate that all of our morphs of interest are inherited simply.

The moral of the story is of course, that if you cross Double Hets, you should be prepared for more than half of the offspring to appear Normal!

Good Luck, and I hope this helps!!
~Rebecca
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0.1 Dumeril's Boa '04 (Courtney)
1.2 Ball Pythons
[1.0 '05 Orange Hypo (Specter)]
[0.1 '05 Het Orange Hypo (Sylvia)]
[0.1 '03 Normal (Sue)]
0.2 American Pit Bull Terriers (40lb darling lap dogs:Brandy&Mara)

Rebecca
Did you see my link?
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RicK @ BbI

Ball Boutique,Inc.
Proud sponsor of this forum

Thank you so much! That is exactly what I was looking for. I have one follow up question if you have time. If you would breed 2 albino piebalds would they produce 100% albino piebald? And does anyone produce albino piebalds or double hets?

Thanks again!
Brian

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1.1 Normal Ball Pythons (Snakerton and Mathilda)
1.0 Het Albino Ball Python (Walter)
0.1 Albino Ball Python (Perrie)

Since the albino pied is the homozygous recessive form of both traits, then yes, breeding two albino pieds together should result in all albino pied babies, just like two albinos produce all albinos, etc. As for who produces such morphs, they're fairly new to the scene still and the prices would be...well, phenomenal I'd wager. Try contacting the big breeders or those who have posted pictures here to find out more.
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Chance Duncan
www.rivervalleyexotics.com

GORGEOUS albino by the way!!

I have to agree with other responders: I think double hets for albino and pied would be astronomically priced right now.

That being said, if you could get your hands on a pied female, and breed her to your albino, and you would be one of the few producing double hets!

Good luck!!
~Rebecca
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0.1 Dumeril's Boa '04 (Courtney)
1.2 Ball Pythons
[1.0 '05 Orange Hypo (Specter)]
[0.1 '05 Het Orange Hypo (Sylvia)]
[0.1 '03 Normal (Sue)]
0.2 American Pit Bull Terriers (40lb darling lap dogs:Brandy&Mara)