# Genetics explained (genetic wizards help needed, to fill in gap)



## chewy86

Fill in gaps, not gap :-/

Hopefully there are more genetic wizards than useless at genetics people (like me) reading this. (Untill the threads loaded with knowledge atleast, then it needs stickying to help other numptys like me in the future. good referance if unsure on something in the future)

Long story short, Ive searched the forum and can only find info on the basic genetics, ie recessive genes x recessive genes and their hets, normals to and so on and get all of that. My problem starts when mixing in co-dom, dom, dbl co-dom and so on with other random multi gene morphs.
If you are one of these gifted genetic wizards and have time to fully explain it start to finish, displaying tables and workings out of all possible genetic out comes, please do below. Of course i dont mean all morphs just how to work out any genetic task that I/others could encounter in the future (snakes is my main interest although im sure they all tie in)

Massive thanks in advance and hope this thread becomes an absolute gold mine of genetic facts.

Gav.


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## JamesJ

Genetic Wizard - World of Ball Pythons try this genetic wizard


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## chewy86

James_and_Hana said:


> Genetic Wizard - World of Ball Pythons try this genetic wizard


These are good, and ive used one that has most snake species listed also but it means relying on a converter. Im after the knowledge to work any variation out myself even new morphs that may not be on the wizards. Plus as i say knowing how the wizard came to its answer.

ie what if I crossed a (retics) platinum sunfire het albino and genetic stripe to a tiger phantom ghost stripe het albino and pied (God i wish i owned those two snake lol) and i know those wouldnt be bred together, im just using extremes where i would like to be able to work out what it ended up with. 

Thanks for your input anyway.


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## JamesJ

Ah well then I can't be of more help, with codom / double codom royals I can work out all the possible combos but odds wise I have to use a generic calculator. It's like pinner squares I get simple ones but when you have lots of genetics and hets etc I get lost!

I'm sure someone more useful will pop up


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## chewy86

James_and_Hana said:


> Ah well then I can't be of more help, with codom / double codom royals I can work out all the possible combos but odds wise I have to use a generic calculator. It's like pinner squares I get simple ones but when you have lots of genetics and hets etc I get lost!
> 
> I'm sure someone more useful will pop up


Hopefully :2thumb:

Im like you i can work out the more common stuff but get lost with any of the big calculations and working out %.


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## chewy86

If a retic genetic wizard arrives could you also explain the outcome of the platy genes super froms being bred back to platys, normals and other morphs ie -

lucy x normal
ultra x normal
ivory x normal

All above x platy
All above cross albino/recessive morph
All above x other co-dom or super forms.


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## chewy86

Come on where are the genetic boffs?


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## Demonlude

Forget about punnet squares and genetic wizards etc.

Just breed 2 snakes, and see what comes out :lol:


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## chewy86

Demonlude said:


> Forget about punnet squares and genetic wizards etc.
> 
> Just breed 2 snakes, and see what comes out :lol:


lol Adam, I would like to have a better understanding for once.
Although im sure fun things can be made that way if you think well outside the box.


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## naja-naja

treat co-dominant morphs like recessives, i.e. pastel is a het, super pastel is homozygous.


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## chewy86

naja-naja said:


> treat co-dominant morphs like recessives, i.e. pastel is a het, super pastel is homozygous.


So how would you do a punnett square on my above example of retics.


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## Deano

The trouble is, you could write a book on this stuff, it would be impossible to explain from start to finish on a forum post! 

I'll try to cover the basics in a few posts, and we'll maybe end up with a multi gene punnet... let's see how this goes!

Each mutation or trait, be it a colour, or pattern change, and regardless of whether is a recessive, co-dom or dominant trait, is controlled by a pair of genes. 

The first step in building the punnet, is to donate a letter to describe each trait, example 'A' for Albino. You then use a capital letter to show that the Albino gene is present, and lower case where it is not (the gene at that location is 'Normal'

Remember, each individual animal has a pair of the gene that controls the mutation, so for the Recessive Albino Gene, the possible combinations are:

aa - Normal (2 normal genes).
Aa - Het Albino (1 Albino, 1 normal gene) This looks Normal if the gene is recessive. If we were working with a co-dom gene, such as Pastel, this would be a visual Pastel. If we were working with a dominant gene, such as Spider, is would also be visual.
aA - Het Albino, exactly the same as above
AA - Visual Albino (2 Albino genes). 2 mutant genes are required to visually display a recessive trait. For co-dom, you get the 'super' form, which looks different to the 'Het'. For dominant, the 'super' for looks identical to the Het.

I'll leave it there for now, and come back with some single gene punnets a little later


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## Deano

During breeding, each parent passes on one of the genes in the pair to it's offspring. The offspring get one from the female, and one from the male, to complete it's pair.

A Het Albino (Aa) can pass on the 'A', or the 'a'.
A visual albino (AA) can only pass on an 'A' (it's all it has), but it's important to remember that it can pass on the 1st 'A', or the 2nd.
A normal (aa) can pass on the 1st 'a', or the 2nd 'a'.

We can now build the single gene punnet.
You put the male on the left of the square, and list down the left axis the single gene possibilities he can pass on. 

You put the female across the top, and list across the top axis the single gene possibilities she can pass on. 

You create your square, and fill in each block the 2 gene combination that the offspring could inherit.

Below punnet is for a Visual Albino Male paired with a Het Abino Female:

__________ * Female - Het Albino (Aa)
_______________ A _____ a*

*Male _____A * _____ AA _____ Aa
*Albino
(AA) _____A* _____AA _____ Aa

For single gene punnets, each parent will have 2 possible gene 'types' to pass on (they may be the same), and the grid will contain 4 possible outcomes.

List the outcomes, and note what they are:

AA - Visual Albino
AA - Visual Albino
Aa - Het Albino
Aa - Het albino

So you have 2 in 4 chance in getting a Visual albino. To convert to a percentage, divide the number of Abinos in the square (2), by the number of possible outcomes (*ALWAYS* 4 for single gene punnet), and multiply by 100:
2 / 4 x 100 = 50%

They odds are PER EGG, meaning that each individual egg has a 50% chance of being an albino, just like when you flip a coin, you have a 50% chance of getting tails. 

If you have 4, it does not mean you will get 2 albinos, that's just the statistical probability. You could get 4, you could get none.


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## Deano

Still working with a single gene, let do het (Aa) to het (Aa) pairing. Remember, each parent can pass on an 'A', or an 'a':

__________ * Female - Het Albino (Aa)
_______________ A _____ a*

*Male _____A * _____ AA _____ Aa
*Het Albino
(Aa) _____a* _____aA _____ aa

'Aa' is the same as 'aA', so I have listed the 4 outcomes below with them both listed as 'Aa':

AA - Visual Albino
Aa - Het Albino 
Aa - Het albino
aa - Normal

We have 1 in 4 (25%) chance Albino, 2 in 4 (50%) chance het albino, 1 in 4 (25%) chance normal, per egg.

This is where POSSIBLE HET comes in for RECESSIVE genes. 
The Normals, and Het albinos all look normal, you cannot visibly tell them apart.

So out of 3 visibly normal snakes, statistically, 2 of the 3 are het albino. 2 / 3 x 100 = 66.6% possibility that each snake is het for albino. Again, it's per snake, if you had 3 then they could all be het, or they could all be genetically normal.

Each of these '66% possible het' snakes is actually either a Het albino, or a normal, but you have no way of knowing this until you breed them and 'Prove the het' by producing a visual albino. Once proven, the snake is then a 100% confirmed het.

If the 'possible het' is paired with a visual in the future, and no albinos are produced, it suggests that the snake is 'just' a normal. However, this is not definite, as there was only a 25% chance that each egg in the clutch was an albino (see the first punnet in above post). You may just have stinking luck!!
However, the more snakes you hatch from using the snake that do not show as visual albino, the more it suggests that the snake is not a Het.


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## Deano

Gonna stop there for now, but if people are still awake, I'll move onto multi gene punnets tomorrow :lol2:


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## paulh

chewy86 said:


> So how would you do a punnett square on my above example of retics.


I'm one of the genetics boffs. See my post #63 
http://www.reptileforums.co.uk/forums/genetics/2-learning-genetics-7.html

I wouldn't do a Punnett square on that example unless you put a pistol to my head. A branching system is my tool of choice. Sorry, I am not a retic genetics boff. I know royal pythons better.

1. figure out the genotype of each snake and convert them to symbols to save bandwidth. Write down the symbols and their definitions. +//a Pa//+ S//+ x +//a Pa//+ +//+
+ = normal gene
Pa = pastel (codominant to corresponding normal gene)
S = spider (either dominant or codominant to corresponding normal gene)

This is a pastel spider het albino royal python mated to a pastel het albino royal python

2. Do a one locus Punnett square or branching system for each locus. 
+//a x +//a --> 1/4 +//+, 2/4 +//a, 1/4 a//a
Pa//+ x Pa//+ --> 1/4 Pa//Pa, 2/4 Pa//+, 1/4 +//+
S//+ x +//+ --> 1/2 S//+, 1/2 +//+

3. Take the list of results for the first locus and the list of the results from the second locus. Match every entry from the second list with the first entry in the first list and multiple the fractions. Match every entry from the second list with the second entry in the first list and multiple the fractions. Repeat until you have a list made from every entry in the second list matched with each entry in the first list.

1/4 +//+ -- 1/4 Pa//Pa = 1/16 +//+ Pa//Pa
1/4 +//+ -- 2/4 Pa//+ = 2/16 +//+ Pa//+
1/4 +//+ -- 1/4 +//+ = 1/16 +//+ +//+

2/4 +//a -- 1/4 Pa//Pa = 2/16 +//a Pa//Pa
2/4 +//a -- 2/4 Pa//+ = 4/16 +//a Pa//+
2/4 +//a -- 1/4 +//+ = 2/16 +//a +//+

1/4 a//a -- 1/4 Pa//Pa = 1/16 a//a Pa//Pa
1/4 a//a -- 2/4 Pa//+ = 2/16 a//a Pa//+
1/4 a//a -- 1/4 +//+ = 1/16 a//a +//+

4. Repeat step 3. The first list is the one made from the first two loci. The second list is the one made from the third locus. Continue until every locus has been used. Then assign phenotypes

1/16 +//+ Pa//Pa -- 1/2 S//+ = 1/32 +//+ Pa//Pa S//+ = 1/32 super pastel spider
1/16 +//+ Pa//Pa -- 1/2 +//+ = 1/32 +//+ Pa//Pa +//+ = 1/32 super pastel
2/16 +//+ Pa//+ -- 1/2 S//+ = 2/32 +//+ Pa//+ S//+ = 2/32 pastel spider
2/16 +//+ Pa//+ -- 1/2 +//+ = 2/32 +//+ Pa//+ +//+ = 2/32 pastel
1/16 +//+ +//+ -- 1/2 S//+ = 1/32 +//+ +//+ S//+ = 1/32 spider
1/16 +//+ +//+ -- 1/2 +//+ = 1/32 +//+ +//+ +//+ = 1/32 normal

2/16 +//a Pa//Pa -- 1/2 S//+ = 2/32 +//a Pa//Pa S//+ = 2/32 super pastel spider het albino
2/16 +//a Pa//Pa -- 1/2 +//+ = 2/32 +//a Pa//Pa +//+ = 2/32 super pastel het albino
4/16 +//a Pa//+ -- 1/2 S//+ = 4/32 +//a Pa//+ S//+ = 4/32 pastel spider het albino
4/16 +//a Pa//+ -- 1/2 +//+ = 4/32 +//a Pa//+ +//+ = 4/32 pastel het albino
2/16 +//a +//+ -- 1/2 S//+ = 2/32 +//a +//+ S//+ = 2/32 spider het albino
2/16 +//a +//+ -- 1/2 +//+ = 2/32 +//a +//+ +//+ = 2/32 normal looking het albino

1/16 a//a Pa//Pa -- 1/2 S//+ = 1/32 a//a Pa//Pa S//+ = 1/32 albino super pastel spider
1/16 a//a Pa//Pa -- 1/2 +//+ = 1/32 a//a Pa//Pa +//+ = 1/32 albino super pastel
2/16 a//a Pa//+ -- 1/2 S//+ = 2/32 a//a Pa//+ S//+ = 2/32 albino pastel spider
2/16 a//a Pa//+ -- 1/2 +//+ = 2/32 a//a Pa//+ +//+ = 2/32 albino pastel
1/16 a//a +//+ -- 1/2 S//+ = 1/32 a//a +//+ S//+ = 1/32 albino spider
1/16 a//a +//+ -- 1/2 +//+ = 1/32 a//a +//+ +//+ = 1/32 albino

Clear as mud? :lol2:


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## chewy86

Deano said:


> Gonna stop there for now, but if people are still awake, I'll move onto multi gene punnets tomorrow :lol2:


Im good to there and please continue tomo mate. Its the dbl co dom het recessives and 2 het 2 rcessives x to another and so on.

Example as an easy one. Platinum sunfire het albino and genetic stripe x the identical snake. How it is held in the snake on the punnett square? And then if its completely random like the above snake x tiger golden child het pied and het orange ghost stripe. just to make it hard although never going to happen.


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## chewy86

Reading that tomo Paul lol, too tired


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## bothrops

chewy86 said:


> So how would you do a punnett square on my above example of retics.


You wouldn't/don't!

Nobody does multi allele matings without a computer. I will do the odd three/four gene combo depending on the parents, but for more complicated combinations, I would use a wizard.





Deano said:


> During breeding, each parent passes on one of the genes in the pair to it's offspring. The offspring get one from the female, and one from the male, to complete it's pair.
> 
> A Het Albino (Aa) can pass on the 'A', or the 'a'.
> A visual albino (AA) can only pass on an 'A' (it's all it has), but it's important to remember that it can pass on the 1st 'A', or the 2nd.
> A normal (aa) can pass on the 1st 'a', or the 2nd 'a'.
> 
> .



You seem to have gone completely against convention with your symbols?

For recessive alleles, the convention is to use a capital for the dominant allele and a lower case for the recessive allele. In fact, for simple monohybrid and dihybrid crosses between simple recessive alleles, the lower case is used to DEFINE the mutation as recessive!

A = normal copy of the gene that codes for melanin
a = a mutation of the A allele that results in an amelanistic snake if two copies of it combine in the same animal. The fact that it is phenotypically hidden when paired with the A allele (i.e is heterozygous) defines it as recessive to A and therefore it is coded with a lower case 'a'

(I agree with everything else in your posts though!)


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## bothrops

You can do Punnett squares for many gene pairs, but it's really not practical for more than two.

I kinda do what Paul does, only I use percentages instead of fractions.


Basically, as people have said on here, it actually makes no difference to the actual calculations whether a gene is recessive, dominant or codominant, the way they are inherited is identical. The only difference is what each animal actually looks like.

Lets take an imaginary gene (called 'bob') and I going to code the normal version of the bob gene as '+' as this is the conventional, accepted term for the 'most commonly found allele in the normal, wild population (AKA 'wild type').

I'm going to call the mutated version of the bob gene 'B' (for Bob!). Note that my above statement about lower case and recessive genes is not applicable here as we have yet to define how the bob gene works!


Now, there are three possible combinations for any particular animal to have at the bob gene locus.

++ = normal wild type
+B = heterozygous bob
BB = homozygous bob


Now, regardless of what a 'het bob' or a '**** bob' looks like compared to the wild type animal, the gene will always be inherited in the same way:


____________________________________
wild type x wild type

++ x ++

100% ++ (wild type)

____________________________________
wild type x het bob

++ x +B

50% ++ (wild type)
50% +B (het bob)

_____________________________________
Het bob x het bob

+B x +B

25% ++ (wild type)
50% +B (het bob)
25% BB (**** bob)

_____________________________________
**** bob x het bob

BB x +B

50% +B (het bob)
50% BB (**** bob)

______________________________________
**** bob x **** bob

BB x BB

100% BB (**** bob)

_______________________________________



The above pairings hold true regardless of the 'status' of the bob gene.

It is what the different animals look like that will DEFINE the dominance of the bob allele.

if 

++ = normal
+B = normal
BB = different to normal

then 'bob' will be defined as recessive to normal

if 

++ = normal
+B = different to normal
BB = the same as +B

then 'bob' will be defined as dominant

if

++ = normal
+B = different from normal
BB = different from normal AND +B

then 'bob' will be defined as codominant.










When working out multiple gene combinations, you need to take each gene pair separately and then combine them.


Let's do your example......


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## chewy86

So how do we go about it with 2 visual traits in one animal ie -

Tiger albino x tiger albino?

25% normal albino
50% tiger albino
25% super tiger albino

Or have I gone off?

Genetic wizard says im right except they are het albino not albino! Why? :-(


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## bothrops

platinum sunfire het albino and genetic stripe x tiger phantom ghost stripe het albino and pied

Firstly I have to assume that all the following assumptions are correct. If any are not, or are actually found on the same locus, then my subsequent calculations will have errors.

platinum = co dominant (heterozygous ivory?)
sunfire = co dominant (heterozygous super sunfire)
albino = recessive
genetic stripe = recessive
tiger = codominant
phantom = co dom (AKA 'codominant ghost stripe' AKA heterozygous blue eyed lucy?)
ghost stripe = recessive
pied = recessive


With this sort of mating, I would never attempt a Punnet square.

With one gene a Punnet square has 4 boxes (monohybrid cross)
with two genes a Punnet square has 16 boxes (dihybrid cross)
with three genes a Punnet square has 64 boxes (tri hybrid cross)

You can see where this is going....

four genes = 256 boxes
five genes = 1024 boxes
six genes = 4096 boxes
seven genes = 16382
eight genes = *65536 *boxes!


Therefore, to make it all simpler, I would just take each gene in turn.

platinum x 'not platinum' = 50% off all babies will be platinum, 50% will not be platinum

sunfire x not sunfire = 50% sunfire x 50% not sunfire

not tiger x tiger = 50% tiger x 50% not tiger

not phantom x phantom = 50% phantom x 50% not phantom

het albino x het albino = 75% poss het albino, 25% albino

not ghost stripe x ghost stripe = 100% het ghost stripe

het genetic stripe x not genetic stripe = 100% poss het stripe

not pied x het pied = 100% poss het pied



Now, 'all' we need to do is combine the above percentages.

start with the platinum

50% platinum
50% not platinum

then add sunfire

25% platinum sunfire
25% not platinum sunfire
25% platinum not sunfire
25% not platinum not sunfire

then add tiger

12.5% platinum sunfire tiger
12.5% platinum sunfire not tiger
12.5% not platinum sunfire tiger
12.5% not platinum sunfire not tiger
12.5% platinum not sunfire tiger
12.5% platinum not sunfire not tiger
12.5% not platinum not sunfire tiger
12.5% not platinum not sunfire not tiger


Then add phantom

6.25% platinum sunfire tiger phantom
6.25% platinum sunfire tiger not phantom
6.25% platinum sunfire not tiger phantom
6.25% platinum sunfire not tiger not phantom
6.25% not platinum sunfire tiger phantom
6.25% not platinum sunfire tiger not phantom
6.25% not platinum sunfire not tiger phantom
6.25% not platinum sunfire not tiger not phantom
6.25% platinum not sunfire tiger phantom
6.25% platinum not sunfire tiger not phantom
6.25% platinum not sunfire not tiger phantom
6.25% platinum not sunfire not tiger not phantom
6.25% not platinum not sunfire tiger phantom
6.25% not platinum not sunfire tiger not phantom
6.25% not platinum not sunfire not tiger phantom
6.25% not platinum not sunfire not tiger not phantom 

next the albino (I'm going to leave the '66% poss het' bit out to avoid confusion and just use 'poss het'!)

4.6875% platinum sunfire tiger phantom poss het albino
1.5625% platinum sunfire tiger phantom albino
4.6875% platinum sunfire tiger not phantom poss het albino
1.5625% platinum sunfire tiger not phantom albino
4.6875% platinum sunfire not tiger phantom poss het albino
1.5625% platinum sunfire not tiger phantom albino
4.6875% not platinum sunfire tiger phantom poss het albino
1.5625% not platinum sunfire tiger phantom albino
4.6875% not platinum sunfire tiger not phantom poss het albino
1.5625% not platinum sunfire tiger not phantom albino
4.6875% not platinum sunfire not tiger phantom poss het albino
1.5625% not platinum sunfire not tiger phantom albino
4.6875% not platinum sunfire not tiger not phantom poss het albino
1.5625% not platinum sunfire not tiger not phantom albino
4.6875% platinum not sunfire tiger phantom poss het albino
1.5625% platinum not sunfire tiger phantom albino
4.6875% platinum not sunfire tiger not phantom poss het albino
1.5625% platinum not sunfire tiger not phantom albino
4.6875% platinum not sunfire not tiger phantom poss het albino
1.5625% platinum not sunfire not tiger phantom albino
4.6875% platinum not sunfire not tiger not phantom poss het albino
1.5625% platinum not sunfire not tiger not phantom albino
4.6875% not platinum not sunfire tiger phantom poss het albino
1.5625% not platinum not sunfire tiger phantom albino
4.6875% not platinum not sunfire tiger not phantom poss het albino
1.5625% not platinum not sunfire tiger not phantom albino
4.6875% not platinum not sunfire not tiger phantom poss het albino
1.5625% not platinum not sunfire not tiger phantom albino
4.6875% not platinum not sunfire not tiger not phantom poss het albino
1.5625% not platinum not sunfire not tiger not phantom albino 



Then finally we can add the last three mutations to each without changing the percentages, as they all will have the same (again I'm removing the percentages of the possible hets)


4.6875% platinum sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% platinum sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% platinum sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% platinum sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% platinum sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% platinum sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum sunfire not tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum sunfire not tiger not phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% platinum not sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% platinum not sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% platinum not sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% platinum not sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% platinum not sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% platinum not sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% platinum not sunfire not tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% platinum not sunfire not tiger not phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum not sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum not sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum not sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum not sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum not sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum not sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
4.6875% not platinum not sunfire not tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
1.5625% not platinum not sunfire not tiger not phantom albino het ghost stripe poss het pied and genetic stripe




Last step (if you want to is to simplify the phenotypes - and if you want, re arrange from 'most normal' to 'most mutations' or 'highest percentage chance' to 'lowest percentage chance' or whatever you choose. (I'm not going to do this here! lol). You can also add back in the percentage for the possible hets, this I will do!

4.6875% platinum sunfire tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% platinum sunfire tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% platinum sunfire tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% platinum sunfire tiger albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% platinum sunfire phantom poss het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% platinum sunfire phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% sunfire tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% sunfire tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% sunfire tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% sunfire tiger albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% sunfire phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% sunfire phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% sunfire het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% sunfire albino het ghost stripe poss het pied and genetic stripe
4.6875% platinum tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% platinum tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% platinum tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% platinum tiger albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% platinum phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% platinum phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% platinum het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% platinum albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% tiger albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% phantom albino het ghost stripe 50% poss het pied and genetic stripe
4.6875% normal het ghost stripe 66% poss het albino 50% het pied and genetic stripe
1.5625% albino het ghost stripe 50% poss het pied and genetic stripe


----------



## bothrops

Obviously with the above I 'showed my working' so you can see how I would do it, in reality I wouldn't bother with the 'not phantom' 'not tiger' bit, I'd simply miss that out.

For your all your percentages, you just need the following data:


RECESSIVE:

normal to het = 100% normal (50% poss het)

het to het = 75% normal (66% poss het), 25% visual

normal x visual = 100% normal (100% het)

het to visual = 50% normal (100% het), 50% visual

visual to visual = 100% visual


DOMINANT:

normal = normal
visual heterozygote = 'one copy mutant' 
visual homozygote = 'two copy mutant'

so:

normal to visual het = 50% normal, 50% visual 

visual het x visual het = 25% normal, 75% visual (33% poss ****)

normal x visual **** = 100% visual het

visual het x visual **** = 100% visual (50% poss ****)

visual **** x visual **** = 100% visual ****

(note that this is exactly the same as the recessive gene, just 'backwards')




CODOMINANT:

normal = normal
het = visual
super = super form

normal x het = 50% normal, 50% het

het x het = 25% normal, 50% het, 25% super

super x normal = 100% het

het x super = 50% het, 50% super

super x super = 100% super








You should notice that actually, all three versions are identical from a genetics point of view - the only difference is which animals you can tell apart and which you can't!


----------



## bothrops

chewy86 said:


> So how do we go about it with 2 visual traits in one animal ie -
> 
> Tiger albino x tiger albino?
> 
> 25% normal albino
> 50% tiger albino
> 25% super tiger albino
> 
> Or have I gone off?
> 
> Genetic wizard says im right except they are het albino not albino! Why? :-(


You are right.


visual albino x visual albino will give 100% albino animals

tiger x tiger = 25% normal, 50% tiger and 25% super tiger.

Combine this and you get the result you got.



I don't know what wizard you used, but either it is wrong, or you forgot to put the albino into one of the parents.


----------



## chewy86

bothrops said:


> You are right.
> 
> 
> visual albino x visual albino will give 100% albino animals
> 
> tiger x tiger = 25% normal, 50% tiger and 25% super tiger.
> 
> Combine this and you get the result you got.
> 
> 
> 
> I don't know what wizard you used, but either it is wrong, or you forgot to put the albino into one of the parents.


I always use that traxxtec buts it down for some reason, so I searched and searched for ages till eventually I found another and it clearly said white tiger albino x lav tiger albino gave all hets. (As per conversation in other guys thread)This confused me as I had got the above result from my punnett, then felt I had missed something. Glad I havnt, I will never try to workout what you have above, if I could reach you I would give you one of these -










YOU LEGEND!!!

What genetic calculator would you use now I have lost mine and need a lets say more reliable one lol.


----------



## chewy86

bothrops said:


> platinum sunfire het albino and genetic stripe x tiger phantom ghost stripe het albino and pied
> 
> Firstly I have to assume that all the following assumptions are correct. If any are not, or are actually found on the same locus, then my subsequent calculations will have errors.
> 
> platinum = co dominant (heterozygous ivory?)
> sunfire = co dominant (heterozygous super sunfire)
> albino = recessive
> genetic stripe = recessive
> tiger = codominant
> phantom = co dom (AKA 'codominant ghost stripe' AKA heterozygous blue eyed lucy?)
> ghost stripe = recessive
> pied = recessive
> 
> 
> With this sort of mating, I would never attempt a Punnet square.
> 
> With one gene a Punnet square has 4 boxes (monohybrid cross)
> with two genes a Punnet square has 16 boxes (dihybrid cross)
> with three genes a Punnet square has 64 boxes (tri hybrid cross)
> 
> You can see where this is going....
> 
> four genes = 256 boxes
> five genes = 1024 boxes
> six genes = 4096 boxes
> seven genes = 16382
> eight genes = *65536 *boxes!
> 
> 
> Therefore, to make it all simpler, I would just take each gene in turn.
> 
> platinum x 'not platinum' = 50% off all babies will be platinum, 50% will not be platinum
> 
> sunfire x not sunfire = 50% sunfire x 50% not sunfire
> 
> not tiger x tiger = 50% tiger x 50% not tiger
> 
> not phantom x phantom = 50% phantom x 50% not phantom
> 
> het albino x het albino = 75% poss het albino, 25% albino
> 
> not ghost stripe x ghost stripe = 100% het ghost stripe
> 
> het genetic stripe x not genetic stripe = 100% poss het stripe
> 
> not pied x het pied = 100% poss het pied
> 
> 
> 
> Now, 'all' we need to do is combine the above percentages.
> 
> start with the platinum
> 
> 50% platinum
> 50% not platinum
> 
> then add sunfire
> 
> 25% platinum sunfire
> 25% not platinum sunfire
> 25% platinum not sunfire
> 25% not platinum not sunfire
> 
> then add tiger
> 
> 12.5% platinum sunfire tiger
> 12.5% platinum sunfire not tiger
> 12.5% not platinum sunfire tiger
> 12.5% not platinum sunfire not tiger
> 12.5% platinum not sunfire tiger
> 12.5% platinum not sunfire not tiger
> 12.5% not platinum not sunfire tiger
> 12.5% not platinum not sunfire not tiger
> 
> 
> Then add phantom
> 
> 6.25% platinum sunfire tiger phantom
> 6.25% platinum sunfire tiger not phantom
> 6.25% platinum sunfire not tiger phantom
> 6.25% platinum sunfire not tiger not phantom
> 6.25% not platinum sunfire tiger phantom
> 6.25% not platinum sunfire tiger not phantom
> 6.25% not platinum sunfire not tiger phantom
> 6.25% not platinum sunfire not tiger not phantom
> 6.25% platinum not sunfire tiger phantom
> 6.25% platinum not sunfire tiger not phantom
> 6.25% platinum not sunfire not tiger phantom
> 6.25% platinum not sunfire not tiger not phantom
> 6.25% not platinum not sunfire tiger phantom
> 6.25% not platinum not sunfire tiger not phantom
> 6.25% not platinum not sunfire not tiger phantom
> 6.25% not platinum not sunfire not tiger not phantom
> 
> next the albino (I'm going to leave the '66% poss het' bit out to avoid confusion and just use 'poss het'!)
> 
> 4.6875% platinum sunfire tiger phantom poss het albino
> 1.5625% platinum sunfire tiger phantom albino
> 4.6875% platinum sunfire tiger not phantom poss het albino
> 1.5625% platinum sunfire tiger not phantom albino
> 4.6875% platinum sunfire not tiger phantom poss het albino
> 1.5625% platinum sunfire not tiger phantom albino
> 4.6875% not platinum sunfire tiger phantom poss het albino
> 1.5625% not platinum sunfire tiger phantom albino
> 4.6875% not platinum sunfire tiger not phantom poss het albino
> 1.5625% not platinum sunfire tiger not phantom albino
> 4.6875% not platinum sunfire not tiger phantom poss het albino
> 1.5625% not platinum sunfire not tiger phantom albino
> 4.6875% not platinum sunfire not tiger not phantom poss het albino
> 1.5625% not platinum sunfire not tiger not phantom albino
> 4.6875% platinum not sunfire tiger phantom poss het albino
> 1.5625% platinum not sunfire tiger phantom albino
> 4.6875% platinum not sunfire tiger not phantom poss het albino
> 1.5625% platinum not sunfire tiger not phantom albino
> 4.6875% platinum not sunfire not tiger phantom poss het albino
> 1.5625% platinum not sunfire not tiger phantom albino
> 4.6875% platinum not sunfire not tiger not phantom poss het albino
> 1.5625% platinum not sunfire not tiger not phantom albino
> 4.6875% not platinum not sunfire tiger phantom poss het albino
> 1.5625% not platinum not sunfire tiger phantom albino
> 4.6875% not platinum not sunfire tiger not phantom poss het albino
> 1.5625% not platinum not sunfire tiger not phantom albino
> 4.6875% not platinum not sunfire not tiger phantom poss het albino
> 1.5625% not platinum not sunfire not tiger phantom albino
> 4.6875% not platinum not sunfire not tiger not phantom poss het albino
> 1.5625% not platinum not sunfire not tiger not phantom albino
> 
> 
> 
> Then finally we can add the last three mutations to each without changing the percentages, as they all will have the same (again I'm removing the percentages of the possible hets)
> 
> 
> 4.6875% platinum sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% platinum sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% platinum sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% platinum sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% platinum sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% platinum sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum sunfire not tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum sunfire not tiger not phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% platinum not sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% platinum not sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% platinum not sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% platinum not sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% platinum not sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% platinum not sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% platinum not sunfire not tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% platinum not sunfire not tiger not phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum not sunfire tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum not sunfire tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum not sunfire tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum not sunfire tiger not phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum not sunfire not tiger phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum not sunfire not tiger phantom albino het ghost stripe poss het pied and genetic stripe
> 4.6875% not platinum not sunfire not tiger not phantom poss het albino het ghost stripe poss het pied and genetic stripe
> 1.5625% not platinum not sunfire not tiger not phantom albino het ghost stripe poss het pied and genetic stripe
> 
> 
> 
> 
> Last step (if you want to is to simplify the phenotypes - and if you want, re arrange from 'most normal' to 'most mutations' or 'highest percentage chance' to 'lowest percentage chance' or whatever you choose. (I'm not going to do this here! lol). You can also add back in the percentage for the possible hets, this I will do!
> 
> 4.6875% platinum sunfire tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% platinum sunfire tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% platinum sunfire tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% platinum sunfire tiger albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% platinum sunfire phantom poss het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% platinum sunfire phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% sunfire tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% sunfire tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% sunfire tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% sunfire tiger albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% sunfire phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% sunfire phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% sunfire het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% sunfire albino het ghost stripe poss het pied and genetic stripe
> 4.6875% platinum tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% platinum tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% platinum tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% platinum tiger albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% platinum phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% platinum phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% platinum het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% platinum albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% tiger phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% tiger phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% tiger het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% tiger albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% phantom het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% phantom albino het ghost stripe 50% poss het pied and genetic stripe
> 4.6875% normal het ghost stripe 66% poss het albino 50% het pied and genetic stripe
> 1.5625% albino het ghost stripe 50% poss het pied and genetic stripe


All your genetic assumptions were correct, and although you didnt write tiger (het super tiger) I know/have seen you knew that but left it out of text)

If im honest iread it fast and got lost with the why you went platy x none platy and so on but im going to read again once had my morning cuppa and take it in much better.

Thanks (and just noticed your fairly local  

P.s have you studied genetics or just made it your mission to learn as I have now?


----------



## Demonlude

I think a little bit of my brain just melted.


----------



## eddygecko

Wow bothrops! That must have taken forever to type out haha. The only thing to note is that because the phantom ghost stripe and orange ghost stripe genes are on the same allele, the result is a homozygous (Not sure if thats the right word?). So a phantom stripe het orange ghost stripe makes a cow (Bel).


----------



## chewy86

eddygecko said:


> Wow bothrops! That must have taken forever to type out haha. The only thing to note is that because the phantom ghost stripe and orange ghost stripe genes are on the same allele, the result is a homozygous (Not sure if thats the right word?). So a phantom stripe het orange ghost stripe makes a cow (Bel).


well spotted Eddy (horrible creation that thing, you may disagree?)


----------



## chewy86

Demonlude said:


> I think a little bit of my brain just melted.


As did mine Adam, i had to go away drink tea and come back lol


----------



## eddygecko

chewy86 said:


> well spotted Eddy (horrible creation that thing, you may disagree?)


Yeah I'm not a fan either. Both the phantom and orange ghost stripes have incredible potential though. The cow could definitely be improved with a couple of genes added.


----------



## Deano

bothrops said:


> You seem to have gone completely against convention with your symbols?


Yeah sorry, I'm no scientist, I tend to adapt concepts slightly to make them simpler to understand. I use a capital letter to tell myself that the gene I am working with is present, regardless of whether it is recessive or dominant. Using convention, an 'A' has a different impact depending on whether the gene is recessive or dominant.

The import thing is, as long as you know what each symbol you use refers to, you'll be OK. 

This thread has moved on a bit from last night, some really interesting (and complex) information on here, nice one, I'll have a read through to enhance my own learning. 

I will, however, take it back a step with a 'simple' 2 gene punnet, as I feel this is where most genetic novices will find benefit, on the assumption that by the time they are working with 4, 5 + genes, their understanding is way past the basics : victory:


----------



## chewy86

Im very pleased with how the thread has turned out its exactly what i was aiming for.


----------



## Deano

Multiple genes.

The first point to understand, is that most gene pairs work* independently*. The pair of genes that control the Pastel (P) trait, are different to the 2 genes that control the Lesser (L) trait.

So start by listing the genes you are working with, and describe your adults.
If working with 2 genes, each animal is described with 4 letters or symbols.
Even if one of the adults only contains one of the genes, you use 2 letters to describe the 2 'normal' version of the gene you are working with.

So, for Pastel (P) and Lesser (L), animals could have:
PPLL - Super Pastel, Super Lesser
PPLl - Super Pastel, Lesser
PPll - Super Pastel (we are still describing the complete absence of the Lesser gene by including the 'll')
PpLL - Pastel, Super Lesser
PpLl - Pastel, Lesser
Ppll - Pastel
ppLL - Super Lesser (No Pastel at all, we still use 'pp')
ppLl - Lesser
ppll - Normal (no mutent genes present, but display the lower case for the 2 genes you are working with in your project

To build the 2 gene punnet, you again list the female at the top, male on the side, and build the axis with the possible genes each parent can pass on.

Each parent will pass on 1 copy of each gene. That's *always* 4 possible pairs from each parent, *always* giving 16 squares in your punnet.

As 'Pastel Lesser' (PpLl) could pass on the following combinations:

PL
Pl
pL
PL

You always get these 4 possibilities correct, use this rule: Substitute the 4 letters for numbers, so 'Pp,Ll' becomes '12,34', and your 4 pairs are:
1,3 (PL)
1,4 (Pl)
2,3 (pL)
2,4 (pl)

I'll now do a punnet for a Pastel Lesser x Pastel pairing. I'm gonna do it on excel and post as a pic because the spacing on here is awkward!


----------



## Deano

Here we go:










I'll do a double het recessive, and explain the possible hets later : victory:


----------



## chewy86

Here goes -

Tiger phantom ghost stripe (two visual hets for super tiger and blue eyed lucy) x same -

TTPP - Super tiger blue eyed lucy (super phantom ghost stripe) 6.25%
TTPp - Super tiger phantom ghost stripe (het blue eyed lucy) 12.5%
TTpp - Super tiger (no hets) 6.25%
TtPP - Tiger (het super tiger) blue eyed lucy (super phantom ghost stripe) 12.5%
TtPp - Tiger (het super tiger) phantom ghost stripe (het blue eyed lucy) 25%
Ttpp - Tiger (het super tiger) (no phantom gene present) 12.50%
ttPP - Blue eyed lucy (super phantom ghost stripe) 6.25%
ttPp- Phantom ghost stripe (het blue eyed lucy) 12.50%
ttpp - Normal wild type (no hets) 6.25%

All % are chance per egg.


----------



## chewy86

Now how do you add a recessive gene, het recessive gene or multiple visual recessives (dbl, treble or quad) or multiple het recessive genes (dbl, treble or quad) or a visual of one recessive and het of another.

Thanks. (dont do it just explain how, i get it will be long winded)


----------



## chewy86

eddygecko said:


> Yeah I'm not a fan either. Both the phantom and orange ghost stripes have incredible potential though. The cow could definitely be improved with a couple of genes added.


How does mixing a recessive gene (ogs which is het nothing but itself being recessive) to a co dom gene (visual het for blue eyed lucy) make a cow? Its almost asif the ogs is like a dbl he. has cow x ogs or ogs x ogs been done? 

Help, you seem to understand this gene.


----------



## chewy86

Deano said:


> Here we go:
> 
> image
> 
> I'll do a double het recessive, and explain the possible hets later : victory:


Thanks, look forward to it. 

think ive got the punnett route but Pauls route im lost. Hooe he can dumb it down further lol.


----------



## Deano

chewy86 said:


> Now how do you add a recessive gene, het recessive gene or multiple visual recessives (dbl, treble or quad) or multiple het recessive genes (dbl, treble or quad) or a visual of one recessive and het of another.
> 
> Thanks. (dont do it just explain how, i get it will be long winded)


Genetically, and on the punnets, it does not matter whether a gene is recessive, Co-Dom, or dominant. They still work in pairs, the difference is whether the mutant genes present results in a visual difference in the animal.

Examples, co dominant Pastel (P) and Recessive Albino (A).

2 Pastel genes (PP) gives a visual Super Pastel
2 Albino genes (AA) give a visual albino

1 Pastel gene (Pp) gives a visual Pastel (As the gene is co dominant, this looks different to both normal, and the 'super' form)
1 Albino gene (Aa) gives a het Albino. But because it is recessive, it does not change the appearance of the animal.

As someone suggested earlier, for co-doms, try to think of 'Pastel' and 'Het Super Pastel'. 
The difference over a recessive trait is that you can 'see' the het.


----------



## Deano

This one is a double recessive: A Pied het Albino (PPAa) and a Normal double het Pied and Albino (PpAa):










As far as the punnet is concerned, no difference in concept so far - the statistical probabilities of the potential offspring are listed. 

However, as the genes are Recessive, we cannot visually identify the Hets, so we have to quote the probability that the Het is there:

The 2 'Pied Albino's are visually displaying both traits, so no doubt there.

The 2 visual 'Albino Het Pied's are 100% het Pied. They are the only visual Albinos on the potential list, and they MUST have inherited the pied Het from the adult male.

The 4 'Pied Het Albino's, and the 2 'Pied's, all visually look like Pieds, but 4 of the 6 carry the Albino Het, which you cannot see. So 4 / 6 x 100 = 66.6%. These snake are all described as 66% possible Het Albino, until proven otherwise.

The 4 'Het Pied, Het Albino's, and the 2 'Het Pied's, all visually look like Normals. All 6 snakes MUST have inerited the pied Het from the adult male, so they are 100% Het Pied. But 4 of the 6 carry the Albino Het, which you cannot see. These snake are all 66% possible het Albino, until proven otherwise. These normal looking snakes are therefore all described as Normal, 100% Het Pied, 66% possible Het Albino.


----------



## eddygecko

chewy86 said:


> How does mixing a recessive gene (ogs which is het nothing but itself being recessive) to a co dom gene (visual het for blue eyed lucy) make a cow? Its almost asif the ogs is like a dbl he. has cow x ogs or ogs x ogs been done?
> 
> Help, you seem to understand this gene.


The matter that one is codom and one is reccesive is irrelevant. Both have different visual forms in the heterozygous and homozygous form. You might be missing the fact the phantom is codom, and creates a super phantom, not a blue eyed lucy. A super phantom looks much like a the heterozygous form only with whiter sides much like a calico ball python. 

I'm not sure if those breedings have been done. I'm certain OGS X OGS would produce a whole clutch of OGS. 

Whereas cow x OGS could be interesting as there is still a lot to be learnt about this gene. I would expect it to produce 50% cows and 50% OGS.


----------



## bothrops

chewy86 said:


> P.s have you studied genetics or just made it your mission to learn as I have now?



I've studied it, and now I lecture in Animal Management and amongst other things I'm module leader for two genetics and breeding modules (one at level 3 (A level equivalent) and one at Level 5 (2nd year degree students))




eddygecko said:


> Wow bothrops! That must have taken forever to type out haha. The only thing to note is that because the phantom ghost stripe and orange ghost stripe genes are on the same allele, the result is a homozygous (Not sure if thats the right word?). So a phantom stripe het orange ghost stripe makes a cow (Bel).


Ah, didn't realise this.

Can you confirm the relationship between phantom ghost stripe and orange ghost stripe(if known)?


Assuming that the two genes share a locus, what do the following gene pairs look like?


phantom ghost stripe //normal = ?
phantom ghost stripe //phantom ghost stripe = ?
orange ghost stripe//normal = ?
orange ghost stripe//orange ghost stripe = ?
phantom ghost stripe//orange ghost stripe = ?


----------



## paulh

Deano said:


> This one is a double recessive: A Pied het Albino (PPAa) and a Normal double het Pied and Albino (PpAa):
> 
> image
> 
> (snip)


As done in a branching system.

PP x Pp --> 1/2 PP, 1/2 Pp (PP = pied, Pp = normal looking het pied, pp = normal)
Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa

............. 1/4 AA = 1/8 PP AA
1/2 PP <- 2/4 Aa = 2/8 PP Aa
............. 1/4 aa = 1/8 PP aa

............. 1/4 AA
1/2 Pp <- 2/4 Aa
............. 1/4 aa

Note that in the Punnett square, the top two lines are the same as the bottom two lines. The 1/2 PP stands for the left 2x2 squares, and the 1/2 Pp stands for the right 2x2 squares. 

Must log off.


----------



## eddygecko

> Ah, didn't realise this.
> 
> Can you confirm the relationship between phantom ghost stripe and orange ghost stripe(if known)?
> 
> 
> Assuming that the two genes share a locus, what do the following gene pairs look like?
> 
> 
> phantom ghost stripe //normal = ?
> phantom ghost stripe //phantom ghost stripe = ?
> orange ghost stripe//normal = ?
> orange ghost stripe//orange ghost stripe = ?
> phantom ghost stripe//orange ghost stripe = ?


Phantom ghost stripe (Heterozygous codom) with orange ghost stripe (Homozygous reccesive) 
Phantom is obviously the darker of the two. 










Super phantom ghost stripes (Homozygous codom) with cow/bels [Phantom ghost stripe het orange ghost stripe] (Heterozygous codom heterozygous recessive) 









Heterozygous orange ghost stripes do show a distinctly brighter colouration and commonly the net like pattern on the back is more of a line of drawn out circles/ovals.

There's still a lot to be learnt about this project but I think the animals it has already created are incredible. And its only the tip of the iceberg!


----------



## bothrops

eddygecko said:


> Phantom ghost stripe (Heterozygous codom) with orange ghost stripe (Homozygous reccesive)
> Phantom is obviously the darker of the two.
> 
> image
> 
> Super phantom ghost stripes (Homozygous codom) with cow/bels [Phantom ghost stripe het orange ghost stripe] (Heterozygous codom heterozygous recessive)
> image
> 
> Heterozygous orange ghost stripes do show a distinctly brighter colouration and commonly the net like pattern on the back is more of a line of drawn out circles/ovals.
> 
> There's still a lot to be learnt about this project but I think the animals it has already created are incredible. And its only the tip of the iceberg!



OK, 

So:



phantom ghost stripe //normal = phantom ghost stripe

phantom ghost stripe //phantom ghost stripe = super phantom ghost stripe

orange ghost stripe//normal = normal (slightly brighter orange + slightly different pattern)

orange ghost stripe//orange ghost stripe = orange ghost stripe

phantom ghost stripe//orange ghost stripe = cow/bel


?


If this is the case then we need to slightly change our terminology.

The terms 'recessive/codom/dom' are not absolute. An allele isn't 'recessive' in it's own right, it is merely recessive in comparison to another allele.

For example, in corn snakes, motley is recessive to normal but dominant to stripe. Stripe is recessive to both motley and normal and normal is dominant to both.


This therefore means that the term 'heterozygous codom heterozygous recessive' is not technically accurate (though I see why you've used it!).


If the above is correct then:

phantom ghost stripe and orange ghost stripe share the same locus

phantom ghost stripe is codominant with normal and codominant with orange ghost stripe


orange ghost stripe is recessive to normal and codominant with phantom ghost stripe.



Actually, if a het ghost stripe can reliably be distinguished from a none heterozygous normal (due to the orange colour and the pattern) then technically, that too is codom to normal.



Certainly looks like an interesting gene combo!


----------



## eddygecko

bothrops said:


> OK,
> 
> So:
> 
> 
> 
> phantom ghost stripe //normal = phantom ghost stripe
> 
> phantom ghost stripe //phantom ghost stripe = super phantom ghost stripe
> 
> orange ghost stripe//normal = normal (slightly brighter orange + slightly different pattern)
> 
> orange ghost stripe//orange ghost stripe = orange ghost stripe
> 
> phantom ghost stripe//orange ghost stripe = cow/bel
> 
> 
> ?
> 
> 
> If this is the case then we need to slightly change our terminology.
> 
> The terms 'recessive/codom/dom' are not absolute. An allele isn't 'recessive' in it's own right, it is merely recessive in comparison to another allele.
> 
> For example, in corn snakes, motley is recessive to normal but dominant to stripe. Stripe is recessive to both motley and normal and normal is dominant to both.
> 
> 
> This therefore means that the term 'heterozygous codom heterozygous recessive' is not technically accurate (though I see why you've used it!).
> 
> 
> If the above is correct then:
> 
> phantom ghost stripe and orange ghost stripe share the same locus
> 
> phantom ghost stripe is codominant with normal and codominant with orange ghost stripe
> 
> 
> orange ghost stripe is recessive to normal and codominant with phantom ghost stripe.
> 
> 
> 
> Actually, if a het ghost stripe can reliably be distinguished from a none heterozygous normal (due to the orange colour and the pattern) then technically, that too is codom to normal.
> 
> 
> 
> Certainly looks like an interesting gene combo!


Yup all those are correct. I wasn't 100% sure on the terminology. I didn't do much genetics at school i've just learnt most of it as I go along. 

Although a het ghost stripe can normally be picked out from a group of normals. They are also quite similar to the heterozygous animals of a number of different mutations so i'm not sure where that leaves them. Genetic stripe, titanium and several newer mutations have obvious heterozygous forms and yet are still named as reccesive. 

Can't wait to see some more done with these genes anyway. Just look at the platinum phantom ghost stripe! And this is only after 2 sheds!


----------



## chewy86

eddygecko said:


> The matter that one is codom and one is reccesive is irrelevant. Both have different visual forms in the heterozygous and homozygous form. You might be missing the fact the phantom is codom, and creates a super phantom, not a blue eyed lucy. A super phantom looks much like a the heterozygous form only with whiter sides much like a calico ball python.
> 
> I'm not sure if those breedings have been done. I'm certain OGS X OGS would produce a whole clutch of OGS.
> 
> Whereas cow x OGS could be interesting as there is still a lot to be learnt about this gene. I would expect it to produce 50% cows and 50% OGS.


So in this case its the super phantom x super phantom breeding that is expected to make the blue eyed lucy?

And/or cow x super phantom or cow x cow?

If so this is where I have got lost.


----------



## chewy86

eddygecko said:


> Phantom ghost stripe (Heterozygous codom) with orange ghost stripe (Homozygous reccesive)
> Phantom is obviously the darker of the two.
> 
> image
> 
> Super phantom ghost stripes (Homozygous codom) with cow/bels [Phantom ghost stripe het orange ghost stripe] (Heterozygous codom heterozygous recessive)
> image
> 
> Heterozygous orange ghost stripes do show a distinctly brighter colouration and commonly the net like pattern on the back is more of a line of drawn out circles/ovals.
> 
> There's still a lot to be learnt about this project but I think the animals it has already created are incredible. And its only the tip of the iceberg!


cow x cow - possible bel
cow x super phantom - possible bel
super phantom x super phantom - possible bel

So there are super forms of super forms?


----------



## chewy86

paulh said:


> As done in a branching system.
> 
> PP x Pp --> 1/2 PP, 1/2 Pp (PP = pied, Pp = normal looking het pied, pp = normal)
> Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
> 
> ............. 1/4 AA = 1/8 PP AA
> 1/2 PP <- 2/4 Aa = 2/8 PP Aa
> ............. 1/4 aa = 1/8 PP aa
> 
> ............. 1/4 AA
> 1/2 Pp <- 2/4 Aa
> ............. 1/4 aa
> 
> Note that in the Punnett square, the top two lines are the same as the bottom two lines. The 1/2 PP stands for the left 2x2 squares, and the 1/2 Pp stands for the right 2x2 squares.
> 
> Must log off.


Right I get it upto the point where -

............. 1/4 AA
1/2 Pp <- 2/4 Aa
............. 1/4 aa

The above comes in if that part -

............. 1/4 AA = 1/8 PP AA
1/2 PP <- 2/4 Aa = 2/8 PP Aa
............. 1/4 aa = 1/8 PP aa

equals the above fractions then this part -

............. 1/4 AA
1/2 Pp <- 2/4 Aa
............. 1/4 aa

Has no fractions of a complete left. Or do you then half the fractions of -

............. 1/4 AA = 1/8 PP AA
1/2 PP <- 2/4 Aa = 2/8 PP Aa
............. 1/4 aa = 1/8 PP aa

To make the total when this part is added in? -

............. 1/4 AA
1/2 Pp <- 2/4 Aa
............. 1/4 aa

So finished fractions look like this -

............. 1/4 AA = 1/16 PP AA
1/2 PP <- 2/4 Aa = 2/16 PP Aa
............. 1/4 aa = 1/16 PP aa

............. 1/4 AA = 1/16 Pp AA
1/2 Pp <- 2/4 Aa = 2/16 Pp Aa
............. 1/4 aa = 1/16 Pp aa

Or am i missing it totally?


----------



## chewy86

Deano said:


> This one is a double recessive: A Pied het Albino (PPAa) and a Normal double het Pied and Albino (PpAa):
> 
> image
> 
> As far as the punnet is concerned, no difference in concept so far - the statistical probabilities of the potential offspring are listed.
> 
> However, as the genes are Recessive, we cannot visually identify the Hets, so we have to quote the probability that the Het is there:
> 
> The 2 'Pied Albino's are visually displaying both traits, so no doubt there.
> 
> The 2 visual 'Albino Het Pied's are 100% het Pied. They are the only visual Albinos on the potential list, and they MUST have inherited the pied Het from the adult male.
> 
> The 4 'Pied Het Albino's, and the 2 'Pied's, all visually look like Pieds, but 4 of the 6 carry the Albino Het, which you cannot see. So 4 / 6 x 100 = 66.6%. These snake are all described as 66% possible Het Albino, until proven otherwise.
> 
> The 4 'Het Pied, Het Albino's, and the 2 'Het Pied's, all visually look like Normals. All 6 snakes MUST have inerited the pied Het from the adult male, so they are 100% Het Pied. But 4 of the 6 carry the Albino Het, which you cannot see. These snake are all 66% possible het Albino, until proven otherwise. These normal looking snakes are therefore all described as Normal, 100% Het Pied, 66% possible Het Albino.


Up to this point i get it what im unsure of is how you would write ie -

pastel lesser het albino het pied and also how to add that to a chart.

is it PPLLAaPp? 

And would you workout its possibilities the same if bred to identical genetic snake so -

PPAaPp
PPAAPp
PPAaPP
PPAAPP
PPLlAaPp
PPLLAaPp
PPLLAAPp
PPLLAAPP
PpLlAAPP
PpLLAaPP

And so on? (above is just random example of what the above could make as there is too many to list)


----------



## eddygecko

chewy86 said:


> cow x cow - possible bel
> cow x super phantom - possible bel
> super phantom x super phantom - possible bel
> 
> So there are super forms of super forms?


You've lost it!

Cow = bel, they are the same thing! They start off as blue eyed lucies then develop spots and go all ugly! And start to moo! 

DOH!


----------



## chewy86

eddygecko said:


> You've lost it!
> 
> Cow = bel, they are the same thing! They start off as blue eyed lucies then develop spots and go all ugly! And start to moo!
> 
> DOH!


I know they are/do that ~(cows) but I thought there was hope for a pure white bel from one of those breedings? Is that s.p cx s.p?


----------



## bothrops

chewy86 said:


> Up to this point i get it what im unsure of is how you would write ie -
> 
> pastel lesser het albino het pied and also how to add that to a chart.
> 
> is it PPLLAaPp?
> 
> And would you workout its possibilities the same if bred to identical genetic snake so -
> 
> PPAaPp
> PPAAPp
> PPAaPP
> PPAAPP
> PPLlAaPp
> PPLLAaPp
> PPLLAAPp
> PPLLAAPP
> PpLlAAPP
> PpLLAaPP
> 
> And so on? (above is just random example of what the above could make as there is too many to list)


 
Firstly, you wouldn't use P = pastel AND P = pied in the same problem as this leads to confusion, you'd need a different code for each.

Secondly, you've pretty much hit the nail on the head. The only real need for the codes is when you are doing Punnet squares, and nobody does Punnett squares for 4, 5, 6 gene problems as there are hundreds, sometimes thousands of possible gamete combinations.

Also, just to put a spanner in the works, you should only use single letters for alleles when you have two alleles, one locus and they are recessive/dominant in relationship. This is so you can distinguish the recessive and dominant alleles by capital and lower case letters.

When you are coding multiple alleles at a single locus or codominant relationships, you _should_ use superscript.

This means that instead of using Pp for pastel and PP for superpastel (which implies that pastel (P) is dominant to normal (p), or using Pp = pastel and pp = super pastel (which implies that pastel (p) is recessive to normal (P), you should instead code the locus itself (say C for 'colour') and then give a superscript letter to define the allele. 

Therefore Pastel becomes C^P and the normal version of the gene at that locus becomes C^+.

Because these forums don't allow subscript and superscript I have to use the '^' symbol to infer superscript, so it looks messy, but in proper writing it looks much better.

It also means you can more easily define multiple alleles at the same locus.

For example lets code the blue eyed lucy locus in royals as B for bel.

Therefore we have

B^+ = wild type
B^M = mojave 
B^L = lesser
B^B = butter
B^d = daddy
B^V = vin russo

and therefore

B^+//B^M = mojave
B^+//B^L = lesser
B^+//B^d = het daddy
B^L//B^M = lesser mojave Bel
B^M//B^M = super mojave
B^L//B^L = Bel
B^B//B^L = Bel
B^V//B^L = Bel
B^L//B^d = platty daddy


----------



## chewy86

So the Ivory retic would be -

I^+ - normal
I^P - platy

I^+//I^P - Platy (50% platy/normals?)
I^P//I^+ - platy (100%)
I^I//I^I - all ivory (if not polymorphic) so (ultra 25% ivory 50% lucy 25%)
I^P//I^P - Ivory (100%)

And if you added more genes to this you would just put ie for albino super tiger retic -

S^SAA and S^SAa super tiger het albino

or S^TAA albino tiger and S^TAa tiger het albino


----------



## paulh

chewy86 said:


> So the Ivory retic would be -
> 
> I^+ - normal
> I^P - platy
> 
> I^+//I^P - Platy (50% platy/normals?)
> I^P//I^+ - platy (100%)
> I^I//I^I - all ivory (if not polymorphic) so (ultra 25% ivory 50% lucy 25%)
> I^P//I^P - Ivory (100%)
> 
> And if you added more genes to this you would just put ie for albino super tiger retic -
> 
> S^SAA and S^SAa super tiger het albino
> 
> or S^TAA albino tiger and S^TAa tiger het albino


I have problems figuring out that post. First, I'm not sure whether "I^+ - normal" and "I^P - platy" mean different snakes or different genes. And what does "I^+//I^P - Platy (50% platy/normals?)" mean?

Definitions:
Genotype = the genes in a gene pair
phenotype = the appearance of a creature with a given gene pair.

Assuming 
I^+ - normal GENE
I^P - platy GENE

Assuming the following are gene pairs:
I^+//I^P genotype produces Platy phenotype (50% platy/normals? -- I do not understand this)
I^P//I^+ genotype produces platy phenotype (100% -- I do not understand this)
I^I//I^I - all ivory (if not polymorphic) so (ultra 25% ivory 50% lucy 25%) I do not understand this; there is no I^I gene.
I^P//I^P genotype produces ivory phenotype

And if you added more genes to this you would just put ie for albino super tiger retic -

S^SAA and S^SAa super tiger het albino 

or S^TAA albino tiger and S^TAa tiger het albino

In the two lines above, there is only one gene for the tiger gene pair. 
S^T = tiger gene (codominant to corresponding normal gene)
S^+ = normal gene

// = a pair of chromosomes, so S^T//S^+ means a pair of chromosomes with a tiger mutant gene in one chromosome and a normal gene in the same location in the other chromosome. In other words, S^T//S^+ is a gene pair. I put // between the genes of a gene pair and a space between gene pairs. I believe that this makes a genotype easier to read than a genotype that is all run together.

S^T//S^T genotype produces super tiger phenotype
S^T//S^+ genotype produces tiger phenotype

A^+ = normal gene
a = albino gene (recessive to corresponding normal gene)

A^+//a genotype (AKA het albino genotype) produces normal phenotype.
a//a genotype produces the albino phenotype.

S^T//S^+ a//a genotype produces the albino tiger phenotype
S^T//S^+ A^+//a genotype produces the tiger phenotype. The het albino genotype does not make the any further change in the phenotype.

A plea for uniformity: Over 100 years ago, Gregor Mendel established the convention that a recessive gene was symbolized in lower case and the dominant gene was symbolized with an upper case letter. (Mendel used only single letter symbols.) Various bodies have elaborated the rules for gene symbols, but that convention is, in essence, unchanged. Conforming to that convention will minimize confusion among the readers of these threads.


----------



## paulh

chewy86 said:


> Right I get it upto the point where -
> 
> ............. 1/4 AA
> 1/2 Pp <- 2/4 Aa
> ............. 1/4 aa
> 
> The above comes in if that part -
> 
> ............. 1/4 AA = 1/8 PP AA
> 1/2 PP <- 2/4 Aa = 2/8 PP Aa
> ............. 1/4 aa = 1/8 PP aa
> 
> equals the above fractions then this part -
> 
> ............. 1/4 AA
> 1/2 Pp <- 2/4 Aa
> ............. 1/4 aa
> 
> Has no fractions of a complete left. Or do you then half the fractions of -
> 
> ............. 1/4 AA = 1/8 PP AA
> 1/2 PP <- 2/4 Aa = 2/8 PP Aa
> ............. 1/4 aa = 1/8 PP aa
> 
> To make the total when this part is added in? -
> 
> ............. 1/4 AA
> 1/2 Pp <- 2/4 Aa
> ............. 1/4 aa
> 
> So finished fractions look like this -
> 
> ............. 1/4 AA = 1/16 PP AA
> 1/2 PP <- 2/4 Aa = 2/16 PP Aa
> ............. 1/4 aa = 1/16 PP aa
> 
> ............. 1/4 AA = 1/16 Pp AA
> 1/2 Pp <- 2/4 Aa = 2/16 Pp Aa
> ............. 1/4 aa = 1/16 Pp aa
> 
> Or am i missing it totally?


I was dragged away from the computer and could not finish that post the way I would have liked. 

The fractions in the finished branching system are wrong in the quote. They must add up to 1. The fractions are multiplied as you travel out each unique branch. 1/2 PP - 2/4 Aa = 1/2 x 2/4 PP AA = 2/8 PP Aa

Here is the finished branching system:
.............. 1/4 AA = 1/2 x 1/4 PP AA = 1/8 PP AA
1/2 PP <- 2/4 Aa = 1/2 x 2/4 PP Aa = 2/8 PP Aa
.............. 1/4 aa = 1/2 x 1/4 PP aa = 1/8 PP aa

.............. 1/4 AA = 1/2 x 1/4 Pp AA = 1/8 Pp AA
1/2 Pp <- 2/4 Aa = 1/2 x 1/4 Pp Aa = 2/8 Pp Aa
.............. 1/4 aa = 1/2 x 1/4 Pp aa = 1/8 Pp aa

Disregard the dots. They were put in to make the columns line up.

The Punnett square in the image is a 4x4 grid, with 16 boxes. So the fractions in the image have 16 as the denominator. Note that those fractions add up to 1. However, the top two rows are identical to the bottom two rows. So we can delete the bottom two rows leaving a 2x4 grid with 8 boxes. And if you look at the fractions, all of them can be reduced to fractions with 8 as the denominator. 2/16 = 1/8 and 4/16 = 2/8 as in my branching system.


----------



## eddygecko

chewy86 said:


> I know they are/do that ~(cows) but I thought there was hope for a pure white bel from one of those breedings? Is that s.p cx s.p?


Nope unfortunately not. Super phantom x super phantom makes a whole clutch of super phantoms if it follows the trend of the rest of the project. There is only one bel retic in existence at the minute and its nothing to do with this project or nerd. Another animal waiting to be proven out.


----------



## chewy86

eddygecko said:


> Nope unfortunately not. Super phantom x super phantom makes a whole clutch of super phantoms if it follows the trend of the rest of the project. There is only one bel retic in existence at the minute and its nothing to do with this project or nerd. Another animal waiting to be proven out.


A wild caught bel or a random snake that appeared in a clutch? Im sure I saw PP going on about one that came randomly out of a clutch, is it this/them or someone else?


----------



## chewy86

If S^S it super tiger How come I^I isnt Ivory?


----------



## eddygecko

chewy86 said:


> A wild caught bel or a random snake that appeared in a clutch? Im sure I saw PP going on about one that came randomly out of a clutch, is it this/them or someone else?


I think P pets one died. Peter rice has a wild caught one.


----------



## chewy86

I thought he had a lemon lucy he had off peter in the recent sale of all his collection? Im refering to a decent blue eyed lucy like the platy gene black eyed but with blue eyes. (not a speck of colour on it)


----------



## eddygecko

A super lemon glow has never been produced, the lucy Peter owns has blue eyes. It might have a few yellow spots but that doesn't mean the babies will.


----------



## chewy86

eddygecko said:


> A super lemon glow has never been produced, the lucy Peter owns has blue eyes. It might have a few yellow spots but that doesn't mean the babies will.


Well the babies are all normal looking arnt they? Im sure thats what i saw in a thread on here from its first breeding.

The lemon glow x lemon glow is due to hatch next month i think but then again it may not produce a blue eyed and may be very similar to the platy gene.


----------



## bothrops

chewy86 said:


> If S^S it super tiger How come I^I isnt Ivory?


 
Because 'S^S' ISN'T super tiger!


If we code the gene that is associated with the tiger mutation as 'T', then the normal version of that gene will be T^+ and the tiger version of that gene will be T^G. (I've used G to avoid using two T's as one represents the locus and the other represents the allele at that locus)

As all animals need two copies of each gene then the following animals are possible:

T^+ // T^+ = normal retic

T^+ // T^G = tiger retic

T^G // T^G = super tiger retic.

Note the '//' *does not* mean 'mated with'. '//' represents two pairs of chromosomes at a particular locus. The symbols either side of the '//' tell you what the alleles are! Therefore the term 'T^+ // T^G' represents the *genotype* of a *single *animal.



So, if you are coding the 'platy' gene, then:

I^+ = the code given to normal version of the 'platy gene'
I^P = the code given to the mutated version of the gene which gives a 'platy' pattern

I^+ // I^+ = normal retic

I^+//I^P = a platy retic

I^P//I^+ = also a platy retic (the order of the alleles makes no difference to the phenotype)

I^I//I^I = This doesn't exist - you can't make up a new, undefined allele!

I^P//I^P = A homozygous platy AKA 'Ivory' 




To mate an ivory to a platinum you would write:

I^P//I^P *x* I^+//I^P

Which would give 50% I^+//I^P (platy) and 50% I^P//I^P (ivory)


----------



## chewy86

bothrops said:


> Because 'S^S' ISN'T super tiger!
> 
> 
> If we code the gene that is associated with the tiger mutation as 'T', then the normal version of that gene will be T^+ and the tiger version of that gene will be T^G. (I've used G to avoid using two T's as one represents the locus and the other represents the allele at that locus)
> 
> As all animals need two copies of each gene then the following animals are possible:
> 
> T^+ // T^+ = normal retic
> 
> T^+ // T^G = tiger retic
> 
> T^G // T^G = super tiger retic.
> 
> Note the '//' *does not* mean 'mated with'. '//' represents two pairs of chromosomes at a particular locus. The symbols either side of the '//' tell you what the alleles are! Therefore the term 'T^+ // T^G' represents the *genotype* of a *single *animal.
> 
> 
> 
> So, if you are coding the 'platy' gene, then:
> 
> I^+ = the code given to normal version of the 'platy gene'
> I^P = the code given to the mutated version of the gene which gives a 'platy' pattern
> 
> I^+ // I^+ = normal retic
> 
> I^+//I^P = a platy retic
> 
> I^P//I^+ = also a platy retic (the order of the alleles makes no difference to the phenotype)
> 
> I^I//I^I = This doesn't exist - you can't make up a new, undefined allele!
> 
> I^P//I^P = A homozygous platy AKA 'Ivory'
> 
> 
> 
> 
> To mate an ivory to a platinum you would write:
> 
> I^P//I^P *x* I^+//I^P
> 
> Which would give 50% I^+//I^P (platy) and 50% I^P//I^P (ivory)


Thats why I was going off course i did indeed read // as bred to. doh!

Also from ivory to ivory you get lucy,ivory and ultra ivory. How are they displayed?

L^P//L^P or is the I used for Ivory as thats what has made the lucy, although it still all started with a platy.


----------



## paulh

bothrops said:


> Because 'S^S' ISN'T super tiger!
> 
> 
> If we code the gene that is associated with the tiger mutation as 'T', then the normal version of that gene will be T^+ and the tiger version of that gene will be T^G. (I've used G to avoid using two T's as one represents the locus and the other represents the allele at that locus)
> 
> As all animals need two copies of each gene then the following animals are possible:
> 
> T^+ // T^+ = normal retic
> 
> T^+ // T^G = tiger retic
> 
> T^G // T^G = super tiger retic.
> 
> Note the '//' *does not* mean 'mated with'. '//' represents two pairs of chromosomes at a particular locus. The symbols either side of the '//' tell you what the alleles are! Therefore the term 'T^+ // T^G' represents the *genotype* of a *single *animal.
> 
> 
> 
> So, if you are coding the 'platy' gene, then:
> 
> I^+ = the code given to normal version of the 'platy gene'
> I^P = the code given to the mutated version of the gene which gives a 'platy' pattern
> 
> I^+ // I^+ = normal retic
> 
> I^+//I^P = a platy retic
> 
> I^P//I^+ = also a platy retic (the order of the alleles makes no difference to the phenotype)
> 
> I^I//I^I = This doesn't exist - you can't make up a new, undefined allele!
> 
> I^P//I^P = A homozygous platy AKA 'Ivory'
> 
> 
> 
> 
> To mate an ivory to a platinum you would write:
> 
> I^P//I^P *x* I^+//I^P
> 
> Which would give 50% I^+//I^P (platy) and 50% I^P//I^P (ivory)


Small correction:
Original = Note the '//' *does not* mean 'mated with'. '//' represents two pairs of chromosomes at a particular locus. The symbols either side of the '//' tell you what the alleles are! Therefore the term 'T^+ // T^G' represents the *genotype* of a *single *animal.

Revision = Note the '//' *does not* mean 'mated with'. '//' represents *one pair* of chromosomes at a particular locus. The symbols either side of the '//' tell you what the alleles are! Therefore the term 'T^+ // T^G' represents the *genotype* of a *single *animal.


There are two chromosomes in a pair of chromosomes. The T^+ gene is in one chromosome, and the T^G gene is in the same location in the other chromosome. T^+ // T^G represents one gene pair, the genotype at one locus of a single animal. For what it's worth, both the corn snake and boa constrictor have 18 pairs of chromosomes holding over 20000 gene pairs.


----------



## paulh

chewy86 said:


> Thats why I was going off course i did indeed read // as bred to. doh!
> 
> Also from ivory to ivory you get lucy,ivory and ultra ivory. How are they displayed?
> 
> L^P//L^P or is the I used for Ivory as thats what has made the lucy, although it still all started with a platy.


I do not think it is possible to get lucy, ivory and ultra ivory from an ivory x ivory cross. All the babies from that mating should be ivories. There are two possibilities, IMO.

There could be a second, unidentified gene pair in the ivory x ivory mating. Or the mating is ultra ivory x ultra ivory, and there are three alleles at the locus -- normal (L^+), platinum (L^P), and leucistic (L^L).

Three genes produce 6 possible gene pairs.
1. L^+//L^+ genotype produces normal phenotype
2. L^P//L^+ genotype produces platinum phenotype
3. L^L//L^+ genotype produces ________ (fill in the blank, if you can)
4. L^P//L^P genotype produces ivory phenotype
5. L^P//L^L genotype produces ultra ivory phenotype
6. L^L//L^L genotype produces leucistic phenotype

Comparing this to purple, lavender, and white albinos, ivory would match to purple albino, leucistic would match with white albino, and ultra ivory would match with lavender albino. This assumes that leucistic is whiter than ultra ivory and ultra ivory is whiter than ivory.


----------



## paulh

For making genetic symbols, section 3.1 of the rat/mouse guidelines is a good starting point.
MGI-Guidelines for Nomenclature of Genes, Genetic Markers, Alleles, & Mutations in Mouse & Rat

Breeding problems typically require only six steps to get a solution:
1. Identify the parents' genes and assign symbols.
2. Determine the genes in the parents' possible sperm and eggs.
3. Determine the genes in all possible fertilized eggs using Punnett square, branching system or some other method.
4. Combine identical genotypes.
5. Assign a phenotype to each genotype.
6. Combine identical phenotypes.

For a list of 216 matings involving 3 mutant genes, see posts 173-178 at
http://www.reptileforums.co.uk/forums/genetics/258989-boa-genetics-outcomes-albino-anery-18.html

See if you can start with the parent genotypes and work some of the matings independently.


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## bothrops

paulh said:


> Small correction:
> Original = Note the '//' *does not* mean 'mated with'. '//' represents two pairs of chromosomes at a particular locus. The symbols either side of the '//' tell you what the alleles are! Therefore the term 'T^+ // T^G' represents the *genotype* of a *single *animal.
> 
> Revision = Note the '//' *does not* mean 'mated with'. '//' represents *one pair* of chromosomes at a particular locus. The symbols either side of the '//' tell you what the alleles are! Therefore the term 'T^+ // T^G' represents the *genotype* of a *single *animal.


You're right of course! (Rushed it at lunchtime - you'll have to forgive the typo/brain hiccup!)


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