# Double codom punnet squares



## yellow_python (May 14, 2007)

Hey,

I'd just like some clarification. Having thought I had a good grasp of genetics Ive been puzzled by how a double gene animal reproduces itself when bred to a normal.

Ive looked through several websites and can only see representations of het. x het., co-com x co-dom etc but not a punnet square for double co-dom to normal or double co-dom x co-dom.

thanks


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## Nix (Jan 23, 2008)

You need to do two punnet squares for this. First to find out all the outcomes for the double co dom parent and then one for the cross of the two parents. Please see attachment! 

The only thing that complicates this is that if the codoms form a super in any of the combinations it may visually mask a 3rd gene so the gene will be present but you may not be able to see it.


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## Nix (Jan 23, 2008)

double co dom gene x normal is easier.

Normal parent can only pass normal so all the combos have a normal normal in.

Parent double gene can pass one gene from each pair. As it is a codom gene we know each pair is codom + normal. 

So pair one can be co dom or normal giving us the following combos passed for parent one:

normal normal
gene 1 normal
normal gene 2
gene 1 gene 2

Therefore combos for cross are

NN NN Visual normal 
1N NN Visual co dom 1
N2 NN Visual co dom 2
12 NN Visual combination of gene 1 and gene 2 (say pastel enchi if it was a royal)


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## funky1 (Jun 18, 2008)

Don`t know if this will help but:

Say Co-Dom is CD, Dom is D and Recessive is R then:

CD, CD X D,D =

CD,D - CD, D
CD,D - CD, D

(ie SS x double dominant Enigma = 100% Mack Snow Enigmas - you could also use D for `normal` instead of enigma as normal is a dominant gene)

--------------

CD, CD X CD, R =

CD, R - CD, CD

CD, CD - CD, R

(ie SS X mack snow het recessive = 50% SS and 50% snow all 50% het recessive)

---------------

CD, R X CD, R =

CD, CD - CD, R

R, CD - R, R 

(Mack het Albino X Mack het Albino = 25% SS, 50% Mack, 25% Albino all non visual albinos are 66% het albino)

Hope that helps and I`ve not just confused things even more!!!


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## yellow_python (May 14, 2007)

i see where you are coming from and I get that if it was say pewter to normal i'd get 1/4 each normal, pastel, cinni and pewter.

What im failing to see is how both cinni and pastel get into one animal if following punnet square rules.

Male pewter will go along horizontal (x?) axis with normal going down vertical (y?)

so if you go around the 4 boxes starting top left going clockwise you should get one gene from each parent so one should get:

Box 1: Pastel from dad, normal from mum
Box 2: Cinni from dad, normal from mum
Box 3: Cinni from dad, normal from mum
Box 4: Pastel from dad, normal from mum

Hopefully you see what i mean, so how does both cinni and pastel get into the same animal from this pairing?? NERD website doesnt cover this only other variations.


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## ba1l3y76 (May 8, 2011)

I think i see what you are after, i had this peoblem when i first started to look into genectics!!
Hopefully below will help?

pewter dad x normal mum

N N
N N N

P PN PN

C CN CN

CP CPN CPN

The Pewter gene will be counted as one gene in the punnet square, so when put to a normal it will produce pewters, but if put to a spider for instance you will produce Pewter bees if the odd gods are in your favour?

This probably didn't help much!! Not the best at explaining:bash:
Just hope i'm right!!


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## eeji (Feb 22, 2006)

cinni and pastel are at different loci so can both be passed down from the pewter parent giving you pewter babies


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## Spuddy (Aug 24, 2009)

yellow_python said:


> i see where you are coming from and I get that if it was say pewter to normal i'd get 1/4 each normal, pastel, cinni and pewter.
> 
> What im failing to see is how both cinni and pastel get into one animal if following punnet square rules.
> 
> ...


 
I understand what you mean, Ive attatched a 'masterpiece' I created in Microsoft Paint to show you how its done rather than trying to explain. But as you can see I have chosen a Normal, which runs horizontally along the Punnett Square and a Pewter running vertically. 












Also note, I left out the 'normal' gene from the Pewter morph on the Punnet Square to make it less 'cramped', to illustrate the point I was trying to make. The normal gene would actually be present on the pewter side of the punnet square, so that pairing could end up with the 'Normal x Normal' combo been made, and thus' just hatching a normal.


Basically you have to count all individual morphs within a combo, other possible combo's and also the 'whole' morph itself. For example, lets say I chose Emperor Pin rather than a Pewter. Emperor Pin is a Lesser + Pastel + Pinstripe. The vertical squares on the punnet square would read....

Normal
Lesser
Pinstripe
Pastel
Lesser Pinstripe (Kingpin)
Lesser Pastel
Pinstripe Pastel (Lemonblast)
Emperor Pin 


Hopefully that makes more sense, let me know and I'll see if I can be of any help further.


Regards,
Spuddy.


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## bothrops (Jan 7, 2007)

I'm an FE/HE lecturer and teach a number of genetics modules to degree level and some of the stuff written above has got MY head hurting.

So many methods and, though many 'work', most lead to nothing but confusion and issues.




OP - Your terminology usage shows a few misunderstanding. Firstly the terms 'dominant', 'co-dominant' and 'recessive' only explain how the to alleles in a gene pair interact and had NO BEARING AT ALL on how the Punnett square should be written or how the genes are inherited. It is ALWAYS IDENTICAL. By using single letters to represent co-dom genes as some of the examples above have done, it leads to the confusion you have.

The term 'het' refers only to whether the two alleles in a gene pair are the same or not. It has no other meaning. A 'heterozygous' animal is one that has two DIFFERENT versions of a gene in a particular gene pair. 

Saying things like '....only find representations of het. x het., co-com x co-dom' makes it appear that you think the two things are comparable.


A mojave is 'het'
A lesser is 'het'




So, firstly:

Looking at a single gene breeding:
N = normal version
A = Mutant.


Three options for any particular snake:

NN = Normal animal
NA = Heterozygous animal
AA = homozygous mutant

All animals have two copies of each gene and therefore they can only have the above options. The 'heterozygous' version is the heterozygous version regardless of way the 'N' gene and the 'A' gene interact. How they interact defines the dominance of the gene.

i.e.

IF

NN = looks normal
NA = looks normal
AA = looks different

Then we describe the mutation as 'recessive'. In recessive mutations the heterozygous form is indistinguishable from the homozygous normal and only the homozygous mutation looks different.

however, IF

NN = looks normal
NA = Looks different from normal
AA = looks the same as NA

then we describe the mutation as 'Dominant'.

You will notice that actually, the above two examples are exactly opposite to each other and in fact the terms 'recessive' and 'dominant' are not mutually exclusive. So in the first example, the mutation is recessive but equally the normal gene is dominant. In the second example, normal is recessive and the mutation is dominant.


Finally, IF

NN = looks normal
NA = Looks different from normal
AA = looks different from normal AND different from NA

Then we describe the mutation as 'co-dominant'. In this case, as the heterozygous form is different from normal and often gets named as a morph, when a homozygous version of the mutation is bred, it gets given the term 'super'.


Now we have have the basics in place, we need to answer your question:
next instalment coming soon...


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## funky1 (Jun 18, 2008)

bothrops said:


> I'm an FE/HE lecturer and teach a number of genetics modules to degree level and some of the stuff written above has got MY head hurting.
> 
> So many methods and, though many 'work', most lead to nothing but confusion and issues.



You could at least let us know who passed Sir?! :lol2:

:grouphug:


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## bothrops (Jan 7, 2007)

Any questions?

:2thumb:


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## yellow_python (May 14, 2007)

Wow many thanks for that, greatly appreiciated.

How i originally explained it does sound confusing.

I think where I was going wrong is that the normal female only has one option for her egg whereas the male had 4 combination options and then when one gets to co-dom x co-dom both have 4 combinations that can be represented.

A great post though thanks again.

Just to add i originally put teh het. x het. and co-dom x co-dom bit to start with as i thought if i could see a punnet square showing co-dom to normal i could see for myself where i was going wrong in my head. I do realise that a pastel for example is a het its just the resulting babies would appear different depending if the gene is ressessive, co-dom or dom, but anyway thanks


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## gigasnake (Oct 2, 2011)

Bothrops - I wish you were my genetics lecturer.


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## bothrops (Jan 7, 2007)

gigasnake said:


> Bothrops - I wish you were my genetics lecturer.


 
I'm more than happy to take the role of 'Lecturer by proxy'....

..feel free to ask any questions you may have!


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## gigasnake (Oct 2, 2011)

bothrops said:


> I'm more than happy to take the role of 'Lecturer by proxy'....
> 
> ..feel free to ask any questions you may have!


Haha, you might regret that


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## alan1 (Nov 11, 2008)

>


simply replace *+* with *N *

Andy has seen the light (waits in anticipation on a 2 page reply) :Na_Na_Na_Na:


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## bothrops (Jan 7, 2007)

alan1 said:


> simply replace *+* with *N *
> 
> Andy has seen the light (waits in anticipation on a 2 page reply) :Na_Na_Na_Na:



Horses. Courses. : victory:


(p.s. this is how I've always done it. Your previous versions of the same discussions have involved using a single letter to represent the phenotype within punnett squares rather than the conventional (and _correct_ :Na_Na_Na_Na* pair* of letters or symbols per mutation, such as I've used here!....surely it's you whose seen the light? :whistling2


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## alan1 (Nov 11, 2008)

bothrops said:


> surely it's you whose seen the light? :whistling2:


nah - my way makes the Biro last longer


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## Nix (Jan 23, 2008)

I'm going for teachers pet/classroom assistant cause I was right


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