# Double Recessive Breeding, Is This Right?



## Joddy (Jan 27, 2014)

Hello all! I am a new member here and I thought that this would be a great place to get some advice on snake genetics! (This is also my first post on this forum!).

I am thinking of getting myself into snake breeding but first I want to make sure I understand snake genetics in order to not make any mistakes in the breeding process and, subsequently, lose quite a bit of money - which wouldn't help in the husbandry of the snakes.

I have learnt quite a bit and I am trying to work out probabilities of certain morphs but I'm not sure whether I am going about it the right way. I'll show you a worked example of the probability of achieving and Axanthic Piebald Python starting with a wildtype Het. Piebald and a wildtype Het. Axanthic - I understand that this is almost an impossible breed, but it was just to test to see if my methods are correct. If you can tell me where I go wrong - if at all - then I would be most grateful!!:2thumb:

First of all, breed the two wildtype hets together (a = recessive Axanthic allele while A = the dominant wildtype, and p = recessive Piebald allele while P = the dominant wildtype)

the Het. Axanthic would be: Aa
the Het. Piebald would be: Pp

Therefore:

--A - a
P PA Pa
p pA pa

which would result in 25% wildtype, 25% Het. Axanthic, 25% Het. Piebald and 25% Axanthic and Piebald Het.

I then would breed the Het. Axcanthic and Piebald with the Het. Piebald.

The Het. Axanthic and Piebald would be: PpAa
the Het. Piebald would be: PpAA

Therefore:

---PA----Pa---pA---pa
AA AAPA AAPa AApA AApa
AA AAPA AAPa AApA AApa
pA pAPA pApa pApA pApa
AA AAPA AAPa AApA AApa

The snake wanted from here would be the Piebald Het. Axanthic which would be the ppAa genes. For these there is 12.5% chance of the clutch being them.

Then you would want to breed the Piebald Het. Axanthic with the Het. Piebald and Axanthic from earlier.

Piebald Het. Axanthic would be: ppAa
Wildtype Het. Piebald and Axanthic would be: PpAa

Therefore:

---pA----pa---pA---pa
PA PApA PApa PapA PApa
Pa papA Papa PapA Papa
pA pApA pApa pApA pApa 
pa papA papa papA papa

Here the wanted snakes would be the Axanthic Piebalds which are represented by the genes papa (ppaa) and the chances of getting them from this clutch would be 12.5%.

There you have it! Axanthic Piebalds, however, the probability of this actually happening (using this line) is 0.4%

Working for the probability:

I used a probability tree for this,
0.25x0.125x0.125 = 0.0039

which in percent (I rounded the figure),
0.004x100 = 0.4%

In conclusion, if you started off with two snakes (Het. Piebald and Het. Axanthic) and you line bred them over the years to try to produce a double recessive Piebald Axanthic and you had some sort of magical powers to know which of the offspring had the right genes and alleles, then you would have 0.4% chance of gaining that result.

If anyone has spotted the slightest mistake in this working then please correct me, all criticism is appreciated.

Thanks for taking the time to read it!:2thumb:
Jordan!


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## rainbow.ben (Dec 11, 2007)

The theory looks about right but why not buy the actual visual animals and go from there. Therefore hightening the chances of producing the gene you want to achieve rather than it being more diluted/ potluck on hitting it.


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## bothrops (Jan 7, 2007)

Hi Jordan, welcome to the forum.

I've moved this to the genetics section as you'll get more focussed answers here than in the general snake section.



A couple of general points.

Firstly, if you are using two genes, your punnett squares should always contain both gene pairs. This makes it cleaner and clearer.

Therefore the first Punnett was not correct despite it producing the right answer.

Het axanthic = PPAa

Het Pied = PpAA


All animals from this mating would look normal, but the chances of any particular egg being normal, het axanthic, het pied or double het axanthic pied would indeed be 25% each respectively.


In the next part, you've made an error.

You put:


> ---PA----Pa---pA---pa
> AA AAPA AAPa AApA AApa
> AA AAPA AAPa AApA AApa
> pA pAPA pApa pApA pApa
> AA AAPA AAPa AApA AApa



In this Punnett square, some of the sperm/eggs (first column) have two 'Axanthic' genes and no 'pied' genes. As each sperm or egg can only have one copy of each gene and the must all have one of each gene the correct square for the mating would be:

dbl het pied axanthic = PpAa (possible gametes = PA, pA, Pa, pa)
het pied = PpAA (Possible gametes = PA, pA)

so Punnett should look this:

-----PA------pA-----Pa----pa 
PA- PPAA - PpAA - PPAa - PpAa
pA- PpAA - ppAA - PpAa - ppAa


Note that all offspring have a pair of each gene. Note that the order of the gene pairs is preserved throughout and note that the capital/dominant allele of each gene pair is always written first in the heterozygous form.

This gives the outcome of:

12.5% wildtype
12.5% het axanthic
25% het pied
25% dbl het axanthic pied
12.5% pied
12.5% pied het axanthic

Despite the errors in the Punnett, you arrived at the correct probability and genotype (ppAa) for a pied het axanthic



Your second Punnett is better, but still mixes the gene pairs. This is not the end of the world, but it is good practise to keep the gene pairs together. It is also good practise to always put the dominant allele of each gene pair first in the heterozygote.

Therefore



> ---pA----pa---pA---pa
> PA PApA PApa PapA PApa
> Pa papA Papa PapA Papa
> pA pApA pApa pApA pApa
> pa papA papa papA papa


Becomes:

---pA----pa---pA---pa
PA PpAA PpAa PpAA PpAa
Pa PpAa Ppaa PpAa Ppaa
pA ppAA ppAa ppAA ppAa 
pa ppAa ppaa ppAa ppaa


Also note that the second two columns are actually not necessary as they are repeats of the first two and as we are working percentages, it doesn't make a difference if we remove them:

---pA----pa--
PA PpAA PpAa
Pa PpAa Ppaa
pA ppAA ppAa 
pa ppAa ppaa

Leaving 

12.5% het pied
25% dbl het axanthic pied
12.5% axanthic het pied
12.5% pied
25% pied het axanthic
12.5% axanthic pied




Your final calculation was correct but ultimately meaningless as there is no magic wand to tell which of your animals would be carrying the recessive alleles and as such the odds of producing an axanthic pied from the original pair is far far less than your 0.4% (though assuming your magic wand did exist, your calculation would be correct.

:2thumb:


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## Joddy (Jan 27, 2014)

Thanks a lot for taking your time to reply! It's really helpful and much appreciated :2thumb: and thanks for moving it to the genetics section, I did try and look for one but I must of missed it!

Thanks,
Jordan


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## bothrops (Jan 7, 2007)

Any further questions - just ask! :2thumb:


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## eeji (Feb 22, 2006)

For a little help with the punnett squares have a little play on here: Punnett Square Calculator - Ians Vivarium Reptile Forum there is a link at the top of the page to the instructions and the various genotype formats it accepts.

At the bottom there is a copy/paste code that won't work properly on here, but hopefully the admins here will read this and set up some table bbcode so they can be used (hint hint Mr. Bothrops! :whistling2


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