# super mojo x pied



## MrPhillips (Oct 18, 2009)

Help with Royal genetics needed please.
If i was to breed a super mojo to a pied would i get all mojave het pieds? or have i got that wrong?
If that is correct then what % of het would they be?
Also if that was the case then what would i get if i bred a mojave het pied to a pied?
I am trying to get the hang of genetics but the hets are confusing me :blush:
all help and explenations much appreciated.


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## Mehl (Jul 27, 2012)

If you are having problems with the royal morph genetics, check the genetics wizard at world of ballpythons. It will give you the percentages, just remember that these numbers are not a number of how the entire litter turns out rather that the percentages for each individual snake turning out normal, het og pied as in mentioned above. Or at least thats how I learned it was working ;-)

hope it helps


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## MrPhillips (Oct 18, 2009)

Thanks but i've already checked that site out but it doesn't give me the information i am after. I need somebody to explain it to me :blush:




Mehl said:


> If you are having problems with the royal morph genetics, check the genetics wizard at world of ballpythons. It will give you the percentages, just remember that these numbers are not a number of how the entire litter turns out rather that the percentages for each individual snake turning out normal, het og pied as in mentioned above. Or at least thats how I learned it was working ;-)
> 
> hope it helps


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## alan1 (Nov 11, 2008)

genes always come in pairs...

a 'regular' mojave - has 1x normal gene, 1x mojave gene 
paired to a normal, will yield a clutch of mojaves and normals (statistical odds = 50/50)

a super mojave - has 2x mojave genes
paired to a normal, will yield a whole clutch of mojaves


think of a VISUAL recessive as a 'super' form (acts the same) - in this case, is carrying 2x pied genes

soo...
super (mojave) x super (visual pied) = 1 copy of each gene from both parents will be passed to ALL offspring
therfore - ALL offspring from that pairing will be 'mojave 100% het pied'


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## MrPhillips (Oct 18, 2009)

Thank you alan1, that's easier for my poor brain to understand. What i would like to know now because this is the bit i can't work out - what would i get if i bred a mojave het pied to a pied? 
Would i be on my way to getting myself a mojave pied?


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## paulh (Sep 19, 2007)

MrPhillips said:


> Help with Royal genetics needed please.
> If i was to breed a super mojo to a pied would i get all mojave het pieds? or have i got that wrong?
> If that is correct then what % of het would they be?
> Also if that was the case then what would i get if i bred a mojave het pied to a pied?
> ...


See posts 63, 66, and 67 in 
http://www.reptileforums.co.uk/forums/genetics/2-learning-genetics-7.html

Super mojave x pied involves 2 pairs of genes:
super mojave snake's gene pair #1 = two mojave mutant genes
super mojave snake's gene pair #2 = two normal genes (normal version of the pied mutant gene)

pied snake's gene pair #1 = two normal genes (normal version of the mojave mutant gene)
pied snake's gene pair #2 = two pied mutant genes

Result = all babies have a mojave mutant gene paired with a copy of the corresponding normal gene and a pied mutant gene paired with a copy of the corresponding normal gene. This makes them all mojave het pieds.

100% het pied is a long way of writing het pied.

mojave het pied x pied involves 2 pairs of genes:
mojave het pied snake's gene pair #1 = one mojave mutant gene paired with one normal gene.
mojave het pied snake's gene pair #2 = one pied mutant gene paired with one normal gene.

pied snake's gene pair #1 = two normal genes (normal version of the mojave mutant gene)
pied snake's gene pair #2 = two pied mutant genes

Result =
1/4 mojave mutant gene paired with a normal gene and two pied genes. These snakes are mojave pieds.
1/4 mojave mutant gene paired with a normal gene and a pied mutant gene paired with a normal gene. These snakes are mojave het pieds.
1/4 two normal genes and two pied genes. The snake is a pied.
1/4 two normal genes and a pied mutant gene paired with a normal gene. These snakes are normal looking snakes that are het pieds.

The start and results of each mating are given here. Use the link to work through each problem and see if you can get the same results.


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## MrPhillips (Oct 18, 2009)

Paulh thanks very much :thumb: this should help me undertand more. I'll be able to see if i can work it out until i get the correct results. Thankyou.


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## ljb107 (Nov 16, 2008)

MrPhillips said:


> Thank you alan1, that's easier for my poor brain to understand. What i would like to know now because this is the bit i can't work out - what would i get if i bred a mojave het pied to a pied?
> Would i be on my way to getting myself a mojave pied?


My understanding of a mojave het pied x pied would give you 25% het pied, 25% pied, 25% mojave pied and 25% mojave het pied.

Lloyd :2thumb:


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## MrPhillips (Oct 18, 2009)

Thanks Lloyd, now that i am definitly understanding :2thumb:
I have another question now though, could anybody please tell me what % would the hets would be? I am thinking 100% because they will all carry a pied gene. Am i getting this right?
because from what i have read about hets they can be 50%, 66% or 100% etc how do you get the percentages?


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## bothrops (Jan 7, 2007)

MrPhillips said:


> Thanks Lloyd, now that i am definitly understanding :2thumb:
> I have another question now though, could anybody please tell me what % would the possible hets would be?
> because from what i have read about hets they can be 50%, 66% or 100% etc how do you get the percentages?



An animal either is or isn't het for a particular trait.


The percentages refer to the chance that a particular animal is or isn't heterzygous.

The percentages are used for recessive traits where both the normal and the het form look identical


For example:

In this case, the visual pied only has the pied version of the gene to give and so ALL it's offspring HAVE TO receive a copy of the pied gene from the pied parent. Therefore all offspring from a visual pied (or any other visual recessive) will be 100% het for pied (or whatever trait).


If you take one of these 100% hets and breed it to a normal then we start to get into the percentages....

Lets stick with pieds.

100% het pied x normal

The het pied parent has one copy of pied and one copy of normal. The normal parent has two copies of normal.

All offspring will get a normal version of the gene from the normal parent. Approximately Half of the offspring will get the pied version of the gene from the het pied parent and the other half will get the normal version from the het pied.

However, ALL the offspring will LOOK normal. As there is no way of telling which of the the offspring actually carry the pied gene and which ones don't, we simply label all the clutch as '50% poss het'. Meaning each animal has a 50% chance of being pied and a 50% chance of not being pied.



66% poss hets come in when you breed a het pied to a het pied.

The expected odds for this would be 

25% normal (normal gene from both parents)
25% het pied (pied gene from mom, normal from dad)
25% het pied (pied gene from dad, normal from mom)
25% pied (pied gene from both parents)

This simplifies to

25% normal
50% het pied
25% pied

Now, the pied individuals can easily be identified and pulled from the clutch.
The other 75% however will all LOOK normal. Looking at the odds, statistically, 2 out of 3 of the remaining normal looking animals will be het pied and the other third won't be. Therefore we describe all the normals in the clutch as '66% poss het' as that is the chance that they will be carrying the gene.


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## MrPhillips (Oct 18, 2009)

Thanks Bothrops :thumb: i've read it and re-read it and i think i'm starting to understand it all a little better now.
Thanks everyone who has helped me - very much appreciated :2thumb:


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## paulh (Sep 19, 2007)

IMO, it helps to think of it as a pie divided into four pieces, each a quarter of the pie.

50% heterozygous: Two of the four pieces are made with cherries lacking pits (= heterozygous, having a mutant gene paired with a normal gene). The other two pieces are made with cherries with pits (= homozygous normal, having a pair of normal genes). Which piece is heterozygous and which is homozygous normal? All pieces look the same. If you pick one piece, you have a 2 out of 4 (= 50%) chance of picking a piece without cherry pits (heterozygous) and a 2 out of 4 (= 50%) chance of picking the piece with cherry pits (homozygous normal).

66% heterozygous: One of the four pieces is made with blueberries (= homozygous mutant, having two mutant genes). Two of the four pieces are made with cherries lacking pits (= heterozygous, having a mutant gene paired with a normal gene). The fourth piece is made with cherries with pits (= homozygous normal, having a pair of normal genes). You take away the blueberry piece, leaving three pieces. Which piece is heterozygous and which is homozygous normal? All three pieces look the same. If you pick one piece, you have a 2 out of 3 (= 66%, rounded down) chance of picking a piece without cherry pits (heterozygous) and a 1 out of 3 (= 33%, rounded down) chance of picking the piece with cherry pits (homozygous normal).

100% heterozygous: All four pieces of the pie are made with cherries that lack pits.


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## MrPhillips (Oct 18, 2009)

paulh said:


> IMO, it helps to think of it as a pie divided into four pieces, each a quarter of the pie.
> 
> 50% heterozygous: Two of the four pieces are made with cherries lacking pits (= heterozygous, having a mutant gene paired with a normal gene). The other two pieces are made with cherries with pits (= homozygous normal, having a pair of normal genes). Which piece is heterozygous and which is homozygous normal? All pieces look the same. If you pick one piece, you have a 2 out of 4 (= 50%) chance of picking a piece without cherry pits (heterozygous) and a 2 out of 4 (= 50%) chance of picking the piece with cherry pits (homozygous normal).
> 
> ...


Thanks, the easier things are explained the better for me, i must be a bit thick lol because i have read over your posts on that link you put up for me and i am sorry to say but i had not got a clue what the majority of it was talking about, i totally lost the plot. To me it looked like some sort of science/maths degree paper :blush: Hopefully i'll get better with all this as time goes on, everyone has to learn from the beginning i suppose.


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## bothrops (Jan 7, 2007)

Have you read this?

http://www.reptileforums.co.uk/forums/genetics/814850-genetics-101-a.html

Could help with the basics.


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## MrPhillips (Oct 18, 2009)

bothrops said:


> Have you read this?
> 
> http://www.reptileforums.co.uk/forums/genetics/814850-genetics-101-a.html
> 
> Could help with the basics.


Thanks again bothrops, to be honest i had read through that a few times but did not really understand it but i have just read through it again and i am quite pleased with myself because i now understand it all.
It may take me a while to get there but i keep re-reading over all your posts until i understand it, i am trying to learn it all so that i don't have to constantly ask people a million questions. So a very big thanks to those that have helped me :2thumb:


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