Het pied confusion

So I have been seeing a lot of community post and for sale balls that list hatchling and juvenile as 100% het bied. None show any pied markings. All are to young to breed. In some cases, it was listed as both parents as pied or het pied. And some showed siblings being visual pieds.

  1. From my understanding, unless hets are breed and hatchling come out as pieds, this is the only way to truly list the parents as 100% het pied, not the hatchling.
  2. Are all 100% het listing, that have not proven out, listed incorrectly and should be labeled as 66%?
  3. Can something to young to prove out be listed as het pied if a sibling is pied?

I am not sure I fully understand your question but let me see if I can help clear it up

If one parent is a visual Pied that means it carries two copies of the pied gene. As such, all the offspring from that parent will be 100% het Pied because they all receive one copy of the gene from that parent

So any animals that are “too young to breed” that are listed as being 100% het Pied is because they came from a Pied parent.

Having a sibling that is Pied is not a guarantee of an animal being het Pied because you can get visual animals from a het x het pairing. Any non-Pied animals from such a pairing would be 66% possible hets because that is the statistical likelihood as determined by a Punnett square

Does that clear it up for you a little better?

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That kind of answers it.
So, in the case of Two het pieds, they produce a visual pied then that confirms that both hets are carrying the pied gene and makes both hets 100%. Even though they are not visual.
In this same case, Since both parents are now confirmed to be pied, are all non visual pied offspring guaranteed to be 100% het pied?

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Pied requires two copies of the gene to display the visual bold white patches. The same way you need two copies of the albino gene to get a visual white and yellow snake
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Nope. Both parents are not confirmed to be Pied, they are confirmed to be het Pied. So when you breed them together, any babies that are non-visual are 66% possible het

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I know my wording is making this very confusing, sorry about that. I get there needs to be 2 genes to be visible.

So then any listing of a young het pied as being 100% het would be inaccurate? It should be listed as 66%?

So the only 100% way to get visuals would be from visuals (not saying all would be visual, but chances are better). Would this statement be correct?

Asking this because I am looking into two young hets from different sellers, both listed as 100% het, one had visual siblings. And I am hoping to get visual pieds, since both are listed 100% het.

No.

This would probably be easier with some graphics. Give me a few to knock some together

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I guess what I am getting confused on is, how a non visual, non proven het be listed as 100%

If you breed a visual Pied to anything, all babies would be het for Pied because as was explained, a visual Pied carries two copies of the gene, so one must be thrown to every egg. If the partner is not het for Pied, then there would be no visual Pieds, yet all of the babies are still 100% het for Pied, guaranteed.

When you breed a het Pied to a het Pied, each Parent carries one copy of the gene. So they may or may not throw the gene. If both throw the gene, you get a visual Pied. If one throws the gene and the other does not, it is not visual, but it is still 100% het for Pied. If neither parent throws the gene. Then the baby is not het at all. The trouble with Het to Het pairings, is there is no way to know if the gene was thrown in non visual babies, until you prove it out. Once proven out, it is then a 100% het Pied animal. If it does not prove out over several attempts, then it likely does not carry the gene and is not het for Pied.

The reason it shows 66% het Pied is there are three possibilities, two of which could be het for Pied, and one where it is not. The father could throw the gene, the mother could throw the gene, or neither could throw the gene. Therefore, you have a 66% probability, of a non visual animal, from a het to het pairing, carrying the gene. Only proving it out will confirm or deny if the gene is present. Hope that helps.

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Alright… Let me see if I can brush my old lecture hat off and help a little better now

We will start with the basics of a Punnett square

This is a way to “visualize” the potential outcomes for any single offspring when you pair to animals together. Each parent, male and female, carry two copies of each gene. For the purposes of this conversation only, I am going to use a little non-conventional language and we will consider the wildtype gene to be the CAPITALIZED notation and the lowercase to be the mutant notation. So here is our basic Punnett for any given gene G/g:


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Since we are dealing with piebaldism, I will use P/p to designate the gene:

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Starting with the easiest combination first - take two Pieds and breed them together:

When you look at the genes alone, everything is lowercase because all the animals are carrying two copies of the mutant gene
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Applying a “visualization” of the gene, you get this:


Each parent can only donate a pied (p) to the equation, so all the babies are then also carrying only pied (pp) making them visual Pied:

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Now, we come to the first part of your confusion - breeding a visual Pied to a wildtype:


We now have both versions of the gene in play, the wildtype (P) and the mutant (p)
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Applying a “visualization” of the genes, gives you this:

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There are two important things you need to take note of here - ‘phenotype’ which is what the animal looks like, and ‘genotype’, which is what genes the animal is carrying. All the babies from this pairing have one copy of the normal gene (P) and one copy of the mutant gene (p) so their genotype then is (Pp). However, the only way you get a visual Piebald animal is if you have two copies of the mutant gene. Because none of these babies have two copies, their phenotype is that of the wildtype:


The term ‘het’ is hobby speak for ‘heterozygous’ which means they have two different forms of the gene which, in this case, are the wildtype form of piebald (P) and the mutant form (p). All of these babies are 100% het Pied because genetically they are (Pp)

So, any baby that had only one Pied animal as a parent will make babies that are guaranteed to carry the mutant gene (p) and so they would be listed as being 100% het
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Now, the second place you seem to be having a little confusion - Pairing two 100% hets together:


Again, we have both versions of the gene in play, the wildtype (P) and the mutant (p), but the dynamic here is a little more complicated
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And applying a “visualization” of the genes, gives you this:

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Remembering what I said above about ‘phenotype’ being what the animal looks like, versus ‘genotype’ being what genes the animal is carrying, you now have the complete range of genetics for babies from this pairing. Some will have two copies of the wildtype gene (P) giving them a genotype of (PP), some will have one copy of the wildtype gene (P) and one copy of the mutant gene (p) giving them a genotype that is (Pp), and some will have two copies of the mutant gene (p) giving them a genotype (pp). While we have three possible genotypes (PP, P/p, and pp), because you only get a visual Pied when there are two copies of the mutant gene in the same animal (pp), we only have two phenotypes - wildtype and visual:


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Obviously, we know the genotype of a visual Pied animal from this pairing. But because both the (PP) and the (Pp) genotype animals look the same, we cannot know what the genotype is for the wildtype-looking animal. This is where the term ‘possible heterozygous’ or “poss het” enters into the hobby
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So we can toss out the visual animal from this pairing (because, again, we already know the genes there for certain) and then focus only on the three animals I have circled:

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What we are left with is a statistical probability where there are two possibilities a wildtype-looking animal has the heterozygous genotype (Pp) and one possibility a wildtype-looking animal has the homozygous wild-type genotype (PP).

That 2:1 ratio is the same as saying 2/3 of a chance of a wildtype-looking baby being genetically heterozygous

2/3 = 66% so the wildtype-looking babies are then 66% possible het Pied


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Does that clear it up a little better for you?
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If you go back up to this slide:


You can see how it is possible for a clutch to produce a visual and still not get 100% hets.

If your concern is whether the animals you are buying are 100% het then you should ask the seller what the parents were. So long as one parent was a visual Pied then you are safe (assuming the seller is trustworthy, and that is why you check their reviews :+1:t4: :+1:t4:)

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After thinking about it and reading this, I believe I have it figured out. I was just stuck on "if not proven or visual, then it is 66. But I get it when it comes to this. My mind set was stuck in one place and I have to get something to snap me out of that way of thinking. Again, it was just me over thinking things again. Thank you @t_h_wyman & @graysnake

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@t_h_wyman and I love the charts you did. It is a “copy and save” item. Thanks again

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