# het to het breeding



## chris2007 (Aug 18, 2010)

Hi, im picking up my het pied female at christmas to go with my het male, just wondering how many of you have had visual's from het to het breeding?? pics would be cool too!! 

regards chris


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## MrMike (Jun 28, 2008)

Crossing animals heterozygous for a particular trait will give you a 25% chance at homozygous offspring.


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## Brandan Smith (Nov 17, 2010)

per egg


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## pigglywiggly (Jul 19, 2008)

and the ones that arnt visual have a 2 in 3 chance of being het


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## paul k (Apr 15, 2007)

ive always wondered how that works? i mean a het to het breeding gives a 25% per hatchling of being visual and that makes sence (mathmaticly). but why are the siblings (not visual) given a 2 in 3 chance of being het ie66%?. surely they carry just as much chance of being completly normal??


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## Sziren (Mar 25, 2008)

I am still learning all this, but couldn't have gotten as far as I have without the help of the wonderful man that is alan1.... he has managed to explain all this to me, so that I actually understand it, and I am thick as sh**....lol
I am sure he will be able to explain this better than me, as I would just muddle it all up when trying to explain to someone else...lol


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## swad1000 (Nov 9, 2010)

paul k said:


> ive always wondered how that works? i mean a het to het breeding gives a 25% per hatchling of being visual and that makes sence (mathmaticly). but why are the siblings (not visual) given a 2 in 3 chance of being het ie66%?. surely they carry just as much chance of being completly normal??


Its a maths thing.

Say you get 4 eggs.

Outcome should be, statistically at least.

1 visual
2 het pied.
1 normal.

So the three visual normals, 2 are het pied, so 66% chance of any snake you pick is het pied.


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## Big Red One (Oct 17, 2007)

paul k said:


> ive always wondered how that works? i mean a het to het breeding gives a 25% per hatchling of being visual and that makes sence (mathmaticly). but why are the siblings (not visual) given a 2 in 3 chance of being het ie66%?. surely they carry just as much chance of being completly normal??


There's a 66% chance (1 in 3) as if it's not homozygous from a het to het breeding then it must be either heterozygous or have none of the particular gene passed down ( ie normal, genetically).
The chance of it being het are : 1 copy of the gene passed from
Mum, or 1 copy of the gene from Dad. That then leaves the '3rd option' of neither Mum or Dad passing on a copy of the gene.
So you have 3 options, two of which pass a single copy down and one that doesn't, hence 66% chance of being het.....33% of not.


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## tonydavo (Mar 10, 2008)

But what if................:whistling2:


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## paul k (Apr 15, 2007)

ok but surely a het snake carrys 1 normal gene and one mutant gene, so my point is i guess that any normal looking offspring has as much chance of being normal as it does of being a het!! ie 50%. 

im going for a cuppa, my head hurts!!!!


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## oakelm (Jan 14, 2009)

paul k said:


> ok but surely a het snake carrys 1 normal gene and one mutant gene, so my point is i guess that any normal looking offspring has as much chance of being normal as it does of being a het!! ie 50%.
> 
> im going for a cuppa, my head hurts!!!!


Well considering in the normal looking offspring that 2 out of 3 are hets then the odds are higher of picking a het for a morph than picking a normal, not by much but in simply maths terms a 2 in 3 chance is a 66% chance of picking a het. A 1 in 2 chance from for example a het let's say albino to a normal would result in 1 in a 1 in 2 chance of being a het albino. So a 1 in 2 chance when changed to percentages is 50%.

The actual animal is either het or it isn't, no in-between the odds are simply the chance of picking that het from the offspring produced. That's why people grasp on to het markers and many have looked and looked to try and find them, everyone has their own theories on what looks to be a marker but non have been proven conclusively so until then you simply buy a snake and take the chance. 50% and 66% hets will always be cheaper to acquire than 100% hets as they don't guarantee you have a het. If you like to take a chance go for one of these possible hets, but if you can afford it and what a more definate chance of producing the visual morph then stick with 100% hets or visuals, personally what I prefer to do. Saves wasting 3 years raising for it to be nothing but a normal.


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## Clarky_man (May 2, 2010)

This is the odds per egg.

Visual X Het = 50% Visual, 50% Hets

Visual X Normal = All Hets 100%

Het X Het = 25% Visual, 50% Hets, 25% Normals. So normal looking babies are classed as 66% Hets

Het X Normal = 50% Hets, 50% Normals. So all babies are classes as 50% Hets.


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## alan1 (Nov 11, 2008)

a het pied has 1 copy of the normal gene (N), and 1 copy of the pied gene (p)
soo, a punnett square (my way) of a het x het breeding, would look like this

.... N .. .p
N.. NN .Np
p.. Np...pp

so as already stated, the chances per egg are as follows

1:4 normal (N)
2:4 het pied (Np)
1:4 visual pied (pp)

statistically speaking, 2 out of the 3 'normals' should be het for pied, with only one being totally 'normal'
hence the 66% probability

make sense?


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## alan1 (Nov 11, 2008)

just to add to ^ ^ ^

those percentages only apply to 'het to same gene het'

het x different gene het (ie; het pied x het albino) = whole clutch of 50% poss hets


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## paulh (Sep 19, 2007)

The above discussion has only been about recessive mutant genes. In such cases, a heterozygous creature looks normal and is more valuable as a breeder than a normal is. Concentrating on recessive mutants is not surprising, because most mutants are recessive to their respective normal counterpart genes.

But we must remember that "heterozygous" just means that two genes in a gene pair are not the same. A het may have a recessive mutant gene paired with a normal gene, a codominant mutant gene paired with a normal gene, a dominant mutant gene paired with a normal gene, or two different mutant genes in the pair. And there are cases of multiple alleles to muddy the waters even more.

I must log off and will continue this later on.


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## paulh (Sep 19, 2007)

paulh said:


> The above discussion has only been about recessive mutant genes. In such cases, a heterozygous creature looks normal and is more valuable as a breeder than a normal is. Concentrating on recessive mutants is not surprising, because most mutants are recessive to their respective normal counterpart genes.
> 
> But we must remember that "heterozygous" just means that two genes in a gene pair are not the same. A het may have a recessive mutant gene paired with a normal gene, a codominant mutant gene paired with a normal gene, a dominant mutant gene paired with a normal gene, or two different mutant genes in the pair. And there are cases of multiple alleles to muddy the waters even more.
> 
> I must log off and will continue this later on.





alan1 said:


> a het pied has 1 copy of the normal gene (N), and 1 copy of the pied gene (p)
> soo, a punnett square (my way) of a het x het breeding, would look like this
> 
> .... N .. .p
> ...


A heterozygous creature (a het) has a gene pair in which the two genes are not the same. In a het x het mating, both parents are hets. And the assumption is that the same gene pair in both parents is heterozygous.

There are a lot of ways to be heterozygous, and I've mentioned some of them. But it is a waste of time to go into all the possible ways to have a het to het mating. Because every one-gene-pair mating is analysed the same way. No exceptions. Here is how it's done:

1. Identify the two genes in the male's gene pair. One of those genes is in half the sperm, and the other is in the rest of the sperm.

2. Identify the two genes in the female's gene pair. One of those genes is in half the eggs, and the other is in the rest of the eggs.

3. Draw a Punnett square, as in the quote. Plug the sperm types into the top to define the columns. Plug the egg types into the left side to define the rows. (If you prefer to put the eggs on the top and the sperm on the left side, go ahead. It doesn't matter.)

4. Each cell of the square gets its column gene and its row gene to make a pair of genes. This is the genotype.

5. Gene order in the cells does not matter. Add the like pairs (if any) together.

6. Write in the appearance (phenotype) produced by that gene pair.

7. Add like phenotypes together.

8. (Optional) If at least two genotypes have the same phenotype, determine what fraction of the phenotype group can be assigned to each genotype.

Alan1 followed this procedure in the quote.

Here's a more complex example: 
1. Male royal python, blue-eyed leucistic. By pedigree, he has a lesser platinum (AKA lesser) mutant gene paired with a mojave mutant gene. Half the sperm have a lesser platinum mutant gene, and the other half of the sperm have a mojave mutant gene.

2. Female royal python, mojave. She has a mojave mutant gene paired with a normal gene. Half the eggs have a mojave mutant gene. The rest of the eggs have a normal gene.

3-4. L = lesser platinum gene, N = normal gene, M = mojave gene, 
.... L .. .M
M..LM .MM
N.. LN...MN

5. Genotypes:
1/4 lesser//mojave (LM)
1/4 lesser//normal (LN)
1/4 mojave//mojave (MM)
1/4 mojave//normal (MN)
(Fractions are odds per egg, not per clutch.)

6. Phenotypes:
1/4 lesser//mojave = blue-eyed leucistic
1/4 lesser//normal = lesser platinum
1/4 mojave//mojave = super mojave (mostly white with some darkening of top of head)
1/4 mojave//normal = mojave

7. Phenotypes cannot be added together.

8. No two genotypes have the same phenotype.


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