Quick PSA on allele statistics in reptile breeding.

[Tenevanica]

I’ve started breeding reptiles recently. As I’ve been shopping around for nice breeders, I’ve started to notice that people repeatedly make the same mistake over and over again when calculating the probability of an individual carrying a gene. I keep seeing reptiles being sold as “66% chance of being het for morph X,” and I keep wondering “how the hell are they getting 66%?” I’ve figured out that what they’re talking about is when an individual is the offspring of two parents who are heterozygous for morph X, and they know that individual isn’t the morph, so they figure that leaves three possibilities, and two of the three are Heterozygous, and 2/3 is 66%, so there must be a 66% chance of that individual being het. However, this is not actually how the statistics of this play out. I’ll explain, so let’s start with a punnet square I just drew up of the situation:


The outcomes here is that each offspring as a 1/4 chance of being homozygous dominant, a 1/4 chance of being homozygous recessive, and a 1/2 chance of being heterozygous. This means that every offspring has (had?) a 50% chance of being heterozygous. It seems that a lot of people think that if an individual is not homozygous recessive you can just eliminate that fourth box and base your statistical assumptions off of the remaining three options, but you can’t. Every individual has (had?) a 50% chance of being heterozygous, regardless of what they actually turned out to be. A homozygous recessive individual still has (had?) a 50% chance of being heterozygous; knowing that it is not heterozygous does not actually change its overall probability. Every offspring is a 50% for being het regardless of whether we know the actual genotype or not. If you have an individual who’s genotype is unknown, it is most accurate in this scenario to say that individual has a 50% chance of being heterozygous. Just because the genotype of a quarter of it’s siblings are known doesn’t change its probability of being heterozygous.

The statistics here work on the same principle as the famous “Monty Hall Problem,” in which you should always switch doors because knowing there is nothing behind door #X does not change the fact that there is a 2/3 probability of there being a prize behind the doors you didn’t pick. This breeding scenario is basically a four-door Monty Hall. Knowing what’s behind one door doesn’t change the probability of the others. Here’s a video that explains this statistical concept in context of the Monty Hall Problem, if you’re interested:

So, breeders: if you have an individual bred from two heterozygous parents, and that individual does not have the recessive trait, it is most accurate to say it has a 50% chance of being heterozygous, not a 66% chance.
Buyers: if someone tries to sell you an animal with a 66% chance of being heterozygous, you can pretty safely mentally correct that number to 50%.

 

[Patherophis]

I think that colleagues from genetics department wouldnt survive reading this

You cannot use Monty Hall problem here.

 

[Patherophis]

There are two differences. There are three options instead of two. And more importantly, Monty Hall problem is based on one not knowing what is behind any of doors in the moment of choosing, but here, continuing your M.H.P. analogy, one door is already open at the start, so it cannot be chosen.

 

[Tenevanica]

@Patherophis my eyes have been opened! Actually! Knowing the real answer to this is pretty important to me, given I’m getting into breeding. I actually ran a little experiment to try to wrap my head around this, which I’ll explain in a minute, and I realized what you were getting at as I was doing this. It appears I didn’t explain myself super well, so let me give you a graphic of my previous thinking so you see what I was getting at, and what I didn’t realize: (apologies for quality, I literally did this in Doodlebuddy on my phone)

I figured that each individual has a 1/2 shot of being either heterozygous or homozygous. And knowing that an individual is heterozygous recessive, (or, “seeing behind door 4”) does not change that probability (which it doesn’t, but that’s not why it’s wrong), and so a given individual still has a 50% chance of being Het regardless of whether you know a sibling has the recessive trait. (This has been my line of reasoning since I couldn’t figure out how anyone got 2/3 out of a four box punet square.)

But, for whatever stupid reason, I didn’t see one thing: you aren’t “seeing behind door #4,” door #4 doesn’t exist at all! In this situation, you’ve essentially selected against heterozygous recessive individuals, and you’re only counting the ones with at least one dominant copy of the gene. So, in this case there’s three “doors,” and none of them are open. That’s where the 2/3 is from. You only get 1/2 if you add the probabilities of the recessive individuals to the equation.

In this little experiment I did I took 24 little pieces of paper and marked 18 of them with a black mark and 6 of them with a green mark. On the backs of the papers I wrote their “genotypes” with the green papers labeled tt, 6 of the black papers labeled TT, and the other 12 black papers labeled Tt. I put them in a bag, shook it up, took out 10 random pieces without looking, discarded the green papers, and counted the genotypes. Repeat X10. I realized while doing this that I may as well have no green papers at all, and this seemingly obvious realization is what made this “click.” Oh, and 80 of the 100 papers I pulled were black. Of those, 51 were heterozygous and 29 were homozygous. That’s 63.75% heterozygous. Data is awesome!

Do me a favor and, so I know you’ve read this, reply to this post or like it or something. I’m going to try to get the thread deleted after that, no point in this being here.

 

[Patherophis]

I am glad we understand each other now.
Btw I really like your experiment with pieces of paper, it reminds me one we used to do for genetic drift.