Reptile & Amphibian Forums

I have been browsing the web and keep hitting dead ends. genetics wizard is down. and other ones lead to corn snakes

The Genetics Wizard, the only problem is that it's been down for
quite some period of time I believe.
Genetics Wizard

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Boas By Klevitz

I Support USark.org

yeah its down and has been down. :/

How to Manual: Multi morph and Punnet Squares....

Often I see where the same folks that can figure out a simple punnet square on one or two morphs are struggling to figure out the outcomes multiple morph genetics involved. It's actually fairly simple and straightforward.

There are two basic outcomes of interest.....

1) What visual phenotypes should/could pop out

2) What are the probabilities of any of those specific outcomes

Both are easily answered with just a little math....

For example Jungle Sunglow het Moonglow x Jungle het Snow (complicated on purpose)

Expand that out to show all the genetics involved......

Jungle, Hypo, Albino, het Anery x Jungle, het Albino, het anery

Now separate them into the individual traits, remember to show both animals for each trait......

Jungle x Jungle
Hypo x Normal
Albino x Het Albino
Het Anery x Het Anery

Then work out the visual phenotypes and odds for each trait.....

Jungle x Jungle = ¼ Super Jungle, ½ Jungle, ¼ Normal
Hypo x Normal = ½ Hypo, ½ Normal
Albino x Het Albino = ½ Albino, ½ Normal
Het Anery x Het Anery = ¼ Anery, ¾ Pos het Anery

The total number of visual phenotypes equals the product of the individual trait phenotypes (multiply them all together)....

#Jungle x #Hypo x #Albino x #Anery = 3 x 2 x 2 x 2 = 24

Which says if the litter were large enough and the odds gods were totally fair, there would be 24 different looks in all the babies.

To figure out the odds on any one of those outcomes (yeah, if you want all 24 you have to repeat the process 23 more times) Pick one phenotype from each of the visual traits. Then multiply the odds for each of those phenotypes.....

Examples....

Super Jungle, Hypo, Albino, Anery (yeah, might not able to visually ID all the traits with all that's going on, for the sake of this discussion, all are visible one way or another))
¼ x ½ x ½ x ¼

Or 1 x 1 x 1 x 1 over 4 x 2 x 2 x 4 = 1/64 or one in 64 babies from that pairing would be a Super Jungle Moonglow
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Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count (02/01/2010):
42.61 BRB
27.40 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

"Jungle x Jungle = ¼ Super Jungle, ½ Jungle, ¼ Normal
Hypo x Normal = ½ Hypo, ½ Normal
Albino x Het Albino = ½ Albino, ½ Normal
Het Anery x Het Anery = ¼ Anery, ¾ Pos het Anery

The total number of visual phenotypes equals the product of the individual trait phenotypes (multiply them all together)....

#Jungle x #Hypo x #Albino x #Anery = 3 x 2 x 2 x 2 = 24 '

This is where I get lost -- The Anery number should be 1 instead of 2 because its a het x het breeding. Albino x het Albino is 1/2 Albino, 1/2 het Albino --- I only say that because you mentioned pos hets in the Anery example.

"¼ x ½ x ½ x ¼

Or 1 x 1 x 1 x 1 over 4 x 2 x 2 x 4 = 1/64 or one in 64 babies from that pairing would be a Super Jungle Moonglow"

Where did you come up with the numbers 4x2x2x4 ? Wouldn't they be (in your exapmle 3x2x2x2 ? But in reality 3x2x2x1 -- 1 because of the het Anery x het Anery making 1 anery of 4 babies. I hope to understand this math .
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Snake room janitor

>>"Jungle x Jungle = ¼ Super Jungle, ½ Jungle, ¼ Normal
>>Hypo x Normal = ½ Hypo, ½ Normal
>>Albino x Het Albino = ½ Albino, ½ Normal
>>Het Anery x Het Anery = ¼ Anery, ¾ Pos het Anery
>>
>>The total number of visual phenotypes equals the product of the individual trait phenotypes (multiply them all together)....
>>
>>#Jungle x #Hypo x #Albino x #Anery = 3 x 2 x 2 x 2 = 24 '
>>
>>
>>This is where I get lost -- The Anery number should be 1 instead of 2 because its a het x het breeding. Albino x het Albino is
1/2 Albino, 1/2 het Albino --- I only say that because you mentioned pos hets in the Anery example.
That's the total possible number of VISUAL PHENOTYPES, not the odds of getting one, het anery x het Anery has two visual phenotypes,
Anery and Normal(pos hets)
>>
>>"¼ x ½ x ½ x ¼
>>
>>Or 1 x 1 x 1 x 1 over 4 x 2 x 2 x 4 = 1/64 or one in 64 babies from that pairing would be a Super Jungle Moonglow"
>>
>>Where did you come up with the numbers 4x2x2x4 ? Wouldn't they be (in your exapmle 3x2x2x2 ? But in reality 3x2x2x1 -- 1 because
of the het Anery x het Anery making 1 anery of 4 babies. I hope to understand this math .
>>-----
>>Snake room janitor

The example is Super Jungle Moonglow, Super Jungle = 1/4, Hypo = 1/2, Albino = 1/2 and Anery = 1/4

Note I highlighted the fractions above to show where they came from

Remember, there are two aspects to the math, how many different babies you can get and the odds of getting each specific combination.

Most are not as interested in how many different outcomes there can be as they are in the odds of getting the combination of traits they want.
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Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count (02/01/2010):
42.61 BRB
27.40 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

Thanks for the explanation -- very insightful, and helpful .
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Snake room janitor

I got the first part just fine but well im still lost on the second lol. ohwell I guess I will live without knowing the odds. I think you should work with your web guy and create a boa genetics calculator ^_^

ps. my brain numbers = NO BUENO!

I will DEFF copy and paste all this but we all know my brain doesnt work right dave lol. I should be able to figure it out hopefully lol.