Posted by:
rainbowsrus
at Mon Mar 23 13:18:48 2009 [ Email Message ] [ Show All Posts by rainbowsrus ]
I've posted this part before.....
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 Albino)
Het Anery x Het Anery = ¼ Anery, ¾ Normal (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
One more, what are the odds on normal looking wild type baby (remember, you have to select one from each group) so......
Normal(Jungle) x Normal(Hypo) x Normal(Albino) x Normal(Anery)
¼ x ½ x ½ x ¾ = 3/64
Hope this helps, I know many get confused at first but once you grasp the concept, you'll always know how to figure out even the most complicated outcomes.
Looking at your specifically....
>>
>>I have 1.0 TH moonglow, 0.1 Pastel DH Ghost, 0.1 DH snow
TH Moonglow x Pastel DH Ghost...
Hypo x Hypo = 1/4 Normal, 3/4 Hypo (66% super)
Het Albino x normal = all 50% het Albino
Het Anery x het Anery = 1/4 Anery, 3/4 66% Het anery
Possible visual outcomes are 2 x 1 x 2 = 4
Ghost pos het Albino = 3/4 x 1/4 = 3/16
Anery pos het Albino = 1/4 x 1/4 = 1/16
Hypo, pos het Anery, pos het Albino = 3/4 x 3/4 = 9/16
Normal pos het Anery, pos het Albino = 1/4 x 3/4 = 3/16
(left Pastel out as it's variable and each baby may have some degree of pastel)
TH Moonglow x DH Snow
Hypo x normal = 1/2 Hypo, 1/2 normal
Het Albino x het Albino = 1/4 Albino, 3/4 normal (66% het Albino)
Het Anery x het Anery = 1/4 Anery, 3/4 normal (66% het Anery)
Possible visual outcomes are 2 x 2 x 2 = 8
Moonglow = 1/2 x 1/4 x 1/4 = 1/32
Sunglow pos het Anery = 1/2 x 1/4 x 3/4 = 3/32
Ghost pos het Albino = 1/2 x 1/4 x 3/4 = 3/32
Hypo pos het Albino, pos het Anery = 1/2 x 3/4 x 3/4 = 9/32
Snow = 1/2 x 1/4 x 1/4 = 1/32
Albino pos het Anery = 1/2 x 1/4 x 3/4 = 3/32
Anery pos het Albino = 1/2 x 1/4 x 3/4 = 3/32
Normal pos het Albino, pos het Anery = 1/2 x 3/4 x 3/4 = 9/32
Clear as mud right?
-----
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:
26.49 BRB
20.21 BCI
And those are only the breeders 
lots.lots.lots feeder mice and rats  
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RE: Any help is greatly appreciated - rainbowsrus, Mon Mar 23 13:18:48 2009