# Royals Non visable Homozygous snakes



## bob63 (Jul 11, 2011)

Hi; I have a question for experienced breeders. 
Consider one has morph which does not have a visible super form such as a Spider. On genetics wizard at world of ball pythons it states that the outcome from breeding with a normal will produce 25% normal and 75% spiders. However if one does a punnett square it comes up with 25% normal 50% spider/normal or Heterozygous Spider and 25% Spider/Spider or Homozygous Spiders. Although all the spiders would be visibly the same the proof would be in the future breeding. The result of breeding a Homozygous Spider with a Normal would always be 100% visible Spiders ( 50% Heterozygous Spiders and 50% Homozygous ). The reason I ask is that if there are Homozygous Spiders ( or Homozygous versions of any other morphs with non visible supers ) it would make the punnett squares very different when breeding with other morph combos. I would be interested to know if this is something anyone has proven. Thanks in advance for any input.


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## bob63 (Jul 11, 2011)

Sorry - Think I made a mistake - a punnett square for a Homozygous Spider and a Normal would give 100% Heterozygous Spiders.


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## paulh (Sep 19, 2007)

bob63 said:


> Hi; I have a question for experienced breeders.
> Consider one has morph which does not have a visible super form such as a Spider. On genetics wizard at world of ball pythons it states that the outcome from breeding with a normal will produce 25% normal and 75% spiders. However if one does a punnett square it comes up with 25% normal 50% spider/normal or Heterozygous Spider and 25% Spider/Spider or Homozygous Spiders. (snip)


You are asking about dominant mutant genes. Dominant mutant genes are relatively rare in snakes. Salmon in boa constrictors, stripe in California king snakes, and probably pinstripe in royals are among the few I know of. But they are more common in species that have been domesticated for many generations. 

The result from a heterozygous x heterozygous mating would be expected to be 25% homozygous mutant, 50% heterozygous mutant, 25% homozygous normal. This is true whether the mutant gene is dominant, codominant, or recessive to the normal gene. The only difference among the three classes of mutant genes is in the appearance of the heterozygous mutant. 

This has been proven many times over. I've done it with the checker and spread mutant genes in pigeons.


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## bothrops (Jan 7, 2007)

I think the question here is about the spider gene specifically if I read it right.


To date, despite thousands of breedings world wide, a homozygous spider has NEVER been proven.

To me, this suggests that the homozygous spiders phenotype is 'does not develop at the zygote stage'.


This would mean that the actual outcome of a spider x spider mating would be 25% normal, 50% spider, 25% 'did not develop'.

This translates as each clutch of eggs will be 33% normal and 66% spider.


The trouble is that the nature of odds coupled with relatively small clutch sizes means their is no dicernable difference in any particular clutch between '33/66' and '25/75'. The only thing that could provide evidence would be heaps and heaps of data that doesn't exist (or at least, no-one has collated).


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## lovemysnakes (Apr 17, 2011)

wow, interesting stuff. I have soo much to learn. not that i plan to breed for morphs but still really interesting. Ill put it on my to do list for learning  
bothrops what you have said makes complete sense to me, even though i dont know much about genetics i know enough for that to make sense.


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## bob63 (Jul 11, 2011)

Thank you for your replies. The question was not Spider specific but more dominant morph specific ie. it would also include pinstripe or any morph which doesn't have a visibly different super form. If it is the case that these morphs do not develop a Homozygous form - it also would follow that they will not conform to standard punnett charts. If when bred together two Spiders - SN and SN only result in NN (Normal) or SN ( Het Spider) but not SS ( Homozygous Spider ) this throws off the standard punnett chart. Especially if it actually changes the odds from 25/75% to 33/66%. More importantly it must also throw off more complicated punnett charts like combining a Spider with a Queen Bee or a Pinstripe with a Pewter Blast. I know that, as the genetic dice are rolled for each egg, this is just an academic question but has anyone ever made a modified version of the punnett chart to show this ?


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## bothrops (Jan 7, 2007)

bob63 said:


> Thank you for your replies. The question was not Spider specific but more dominant morph specific ie. it would also include pinstripe or any morph which doesn't have a visibly different super form. If it is the case that these morphs do not develop a Homozygous form - it also would follow that they will not conform to standard punnett charts. If when bred together two Spiders - SN and SN only result in NN (Normal) or SN ( Het Spider) but not SS ( Homozygous Spider ) this throws off the standard punnett chart. Especially if it actually changes the odds from 25/75% to 33/66%. More importantly it must also throw off more complicated punnett charts like combining a Spider with a Queen Bee or a Pinstripe with a Pewter Blast. I know that, as the genetic dice are rolled for each egg, this is just an academic question but has anyone ever made a modified version of the punnett chart to show this ?


Ah, OK. In that case there is no need for a special punnett square as it is completely valid.

As far as I know the Spider mutation in royals is the only snake mutation that (possibly) has a lethal homozygote at the zygote stage. A few others produce weak homozygotes but they do develop - for example 'super motleys' in boas seem to die very young (a year or two) and super jaguars in carpets (leucistic) die stright out of the egg.

In each of the above cases, the mutation is actually 'co-dominant' as the homozygous form of the mutation is different in phenotype to the heterozygote.


With dominant genes the homozygous mutant is phenotypically exactly the same as the heterozygote. They develop exactly the same and are a valid part of the punnett square.


In all cases the punnett square is completely valid and doesn't need redoing. Remember that the punnett square simply describes the genotype. You still need to work out what the phenotype is from the genpotype. Therefore the punnett square is the same whether you are using lethal genes, co-dominant genes, recessive genes or whatever.


Pinstripe x pinstripe (using your coding system 'N' is not pinstripe, P is pinstripe)










As you can see, this gives us 25% normal, 50% pinstripe and 25% homozygous pinstripe.

So, the punnett square is fine, works fine and is valid. The homozygous pinstripe is theoretically completely valid. The only issue is that a homozygous pinstripe is yet to be proven. That doesn't mean they don't exist. For now, all you can do is sell all the visual pinstripe offspring as pinstripes or technically '33% poss homozygous pinstripe'.

Same goes for a trihybrid cross. The punnett square is valid, but the phenotype of each genotype is up for discussion...

pinstripe = P (not pinstripe = N)

pastel = T (not pastel = t)

Black pastel = B (not pastel = b)

so pinstripe x pewter blast = PNttbb x PNTtBb

punnett square looks like:











broken down

PPTtBb = pewterblast (homozygote pinstripe)
PPTtbb = lemon blast (homozygote pinstripe)
PPttBb = black pastel pin (homozygote pinstripe)
PPttbb = homozygous pinstripe
PNTtBb = pewter blast
PNTtbb = lemon blast
PNttBb = black pastel pin
PNttbb = pinstripe
PNTtBb = pewter blast
PNTtbb = lemon blast
PNttBb = black pastel pin
PNttbb = pinstripe
NNTtBb = pastel black pastel
NNTtbb = pastel
NNttBb = black pastel
NNttbb = normal


final cross....

3/16 pewter blast (poss **** pin)
3/16 lemonblast (poss **** pin)
3/16 black pastel pin (poss **** pin)
3/16 pinstripe (poss **** pin)
1/16 pastel black pastel
1/16 pastel
1/16 black pastel
1/16 normal


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## bob63 (Jul 11, 2011)

Thanks bothrops for taking the time to help clarify this in my bewildered head.


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## bothrops (Jan 7, 2007)

bob63 said:


> Thanks bothrops for taking the time to help clarify this in my bewildered head.


 
You're more than welcome. (And thank you for the 'thank you' - doesn't often happen on here!)




The main complication is the lack of a 'proven' homozygote in spiders and pinstripes.

In other animals where we have known dominant genes they are often simply refered to as 'one copy' (heterozygous) and 'two copy' (homozygous dominant) individuals.


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## paulh (Sep 19, 2007)

bob63 said:


> Thank you for your replies. The question was not Spider specific but more dominant morph specific ie. it would also include pinstripe or any morph which doesn't have a visibly different super form. If it is the case that these morphs do not develop a Homozygous form - it also would follow that they will not conform to standard punnett charts. If when bred together two Spiders - SN and SN only result in NN (Normal) or SN ( Het Spider) but not SS ( Homozygous Spider ) this throws off the standard punnett chart. Especially if it actually changes the odds from 25/75% to 33/66%. More importantly it must also throw off more complicated punnett charts like combining a Spider with a Queen Bee or a Pinstripe with a Pewter Blast. I know that, as the genetic dice are rolled for each egg, this is just an academic question but has anyone ever made a modified version of the punnett chart to show this ?


About all I can add to what Bothrops wrote is that the Punnett square is valid for spiders even if the homozygous spider embryo dies very early. 

SN x SN -->
1/4 SS 
2/4 SN
1/4 NN

SS spiders exist even if they die very early. It is only when calculating the fractions for living snakes that they must be modified.

By the way, only masochists do three gene pair Punnett squares. The branching system is much faster and less error prone. It is also much easier to modify for lethal mutants.


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## vetdebbie (Jan 4, 2008)

paulh said:


> By the way, only masochists do three gene pair Punnett squares. The branching system is much faster and less error prone. It is also much easier to modify for lethal mutants.


 
:lol2:

As far as I am concerned only masochists do punnett squares anyway - I am a firm advocate of the branching system


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## bothrops (Jan 7, 2007)

paulh said:


> By the way, only masochists do three gene pair Punnett squares. The branching system is much faster and less error prone. It is also much easier to modify for lethal mutants.





vetdebbie said:


> :lol2:
> 
> As far as I am concerned only masochists do punnett squares anyway - I am a firm advocate of the branching system


 

Couldn't agree more - but as the question specifically asked about punnett squares AND specifically mentioned 'pinstripe x pewter blast' I thought it only fair to answer in the manner I did :2thumb:...


(for the record I'd not use 'N' and 'P' to code for 'not pinstripe' and 'pinstripe' either!): victory:


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## cree (Jan 21, 2010)

vetdebbie said:


> :lol2:
> 
> As far as I am concerned only masochists do punnett squares anyway - I am a firm advocate of the branching system


 
Branching system????


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## bothrops (Jan 7, 2007)

cree said:


> Branching system????


Easier if you are aware how genes are inherited but not as easy if you don't!

As I understand it, the 'branching system' takes each mutation in term and apply it 'in series'....

e.g.

Normal het albino x hypo motley (boas)



take albino first. 50% of the offspring will be normal, 50% will be het albino.

50% normal
50% het albino

Then take hypo. Half of the offspring will be hypo and half will be normal. Now however, we have to 'branch' that result from the previous one...


-------------- 50% hypo
50% normal <
-------------- 50% normal

-----------------50% hypo
50% het albino <
---------------- 50% normal


(obviously this is the long hand way to show you, but with practice this is usually done 'in your head') 

then multiply across

50% normal x 50% hypo = 25% hypo

50% normal x 50% normal = 25% normal

50% het albino x 50% hypo = 25% hypo het albino

50% het albino x 50% normal = 25% normal het albino

or


25% hypo
25% normal
25% hypo het albino
25% normal het albino



Finally we add the third mutation (motley)...half the offspring will be motley half normal.

(to save my having to work out how to type that....) its just a case of splitting each of the four above down again (each one above, split 50/50 with motley/not motley)

12.5% hypo motley
12.5% hypo

12.5% motley
12.5% normal

12.5% hypo motley het albino
12.5% hypo het albino

12.5% motley het albino
12.5% normal het albino



Of course in this case, we can 'recombine' all the 'het albinos' with their 'not het albino' equivalents and simply call them all '50% poss het albino

25% normal 50% poss het albino
25% hypo 50% poss het albino
25% motley 50% poss het albino
25% hypo motley 50% poss het albino




(at least thats what I know as 'the branching system' : victory


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## cree (Jan 21, 2010)

Thankyou Andy :notworthy:
I think I will stick to punnett squares for the time being although i do see the logic in it.


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## cree (Jan 21, 2010)

I guess this makes me a masochist then :gasp:


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## bothrops (Jan 7, 2007)

cree said:


> Thankyou Andy :notworthy:
> I think I will stick to punnett squares for the time being although i do see the logic in it.


 
You're welcome.

It's one of those systems that is very logical, but looks very long winded when explained. In practise, it actually ends up being much quicker than the punnett square. The difficultly is that you need to have a good strong grasp of whats happening to 'know its right'....

kinda like you need to know basic maths to use a calculator as firstly you need to know the numbers themselves then what to do with them and finally you need to have a bit of an idea what the answer might be so that you know if you've pressed the wrong buttons!


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## paulh (Sep 19, 2007)

A double het x double het mating Punnett square has 16 boxes but only 9 unique genotypes. So to get the percentages you have to identify and add the like boxes together. A triple het x triple het mating has 64 boxes and 27 unique genotypes. Identifying and adding the like genotypes together easily doubles the time required for the same problem done as a branching system.

And, IMO, the easiest way to determine the possible genotypes of the sperm and eggs in a two or more gene pair Punnett square is to use a branching system.

Then there is the arithmetic system:

Mating: Aa Bb x Aa Bb -->

Aa x Aa --> 1/4 AA + 2/4 Aa + 1/4 aa
Bb x Bb --> 1/4 BB + 2/4 Bb + 1/4 bb

1/4 AA + 2/4 Aa + 1/4 aa
1/4 BB + 2/4 Bb + 1/4 bb
-----------------------------
1/16 AA BB + 2/16 Aa BB + 1/16 aa BB +
2/16 AA Bb + 4/16 Aa Bb + 2/16 aa Bb +
1/16 AA bb + 2/16 Aa bb + 1/16 aa bb

Multiply the top row by 1/4 BB, then multiply the top row by 2/4 Bb, then multiply the top row by 1/4 bb.


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## cree (Jan 21, 2010)

Thanks guys :2thumb:
It took we a while to crack punnett squares which was really annoying as I understood how and why they worked and knew that I was doing something really silly wrong (please dont ask me what, its too shameful lol). After a couple of days I worked out what I was missing, Id become much like 'a dog with a bone' and refused to be beaten by what is essentially a basic graph! lol.
I will definitely give the branch system a try after your concise explanations! I can see how it would cut out alot of the leg work so to speak.
Thanks again : victory:


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## I am a scarecrow (Nov 18, 2009)

bothrops said:


> Ah, OK. In that case there is no need for a special punnett square as it is completely valid.
> 
> As far as I know the Spider mutation in royals is the only snake mutation that (possibly) has a lethal homozygote at the zygote stage. A few others produce weak homozygotes but they do develop - for example 'super motleys' in boas seem to die very young (a year or two) and super jaguars in carpets (leucistic) die stright out of the egg.
> 
> ...


*Looks at the screen, with a spazzy look on his face*
:lol2:


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## cree (Jan 21, 2010)

I am a scarecrow said:


> *Looks at the screen, with a spazzy look on his face*
> :lol2:


I know that look!!!!! :roll2:


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