Reptile & Amphibian Forums

In looking over offers on the web of snakes for sale, I keep running across breeders claiming that their snake is 66% het for some recessive trait. These claims come from a simple misunderstanding of the laws of chance and their application to genetics.

I am pretty sure that they are making their claim using the following logic. If they cross the F1 generation from a wild type and a recessive, then they know that both snakes being breed are het for the recessive trait. If one does a pundent square for their offspring, you expect to see a ratio 1:2:1, with 1/4th of the offspring homo for the recessive trait. Since this is an obvious phenotype, the only offspring whose genetic makeup will be in doubt are the wild types. It is here that the error comes in. Since they will have no doubt about the homo recessive snakes, they think that by subtracting that that portion from the mix, what is left are mix where 1/3rd will be wild and 2/3rd’s will be het. The fact is that just because you can tell which snakes turned out homo recessive has no influence on the genetics of any of the unknowns. The original odds were that 50% would be hetro, 25% would be homo recessive, and 25% would be homo wild types. Knowing the outcome for some of the offspring does not tell you anything about the other offspring that will be showing the wild phenotype. Each event is independent.

This is the same basic mistake that most of us make with the classic coin toss problem. If one flips a “fair” coin 99 times, and each times it comes up heads, what are the odds that it will come up tails on the next flip. Most of us would say it was a sure thing to come up tails because what are the odds of it happening 100 times in a row. We would think that it had to be due to come up tails. But the fact is that the odds are still the same, 50/50 because each previous event has no influence on the next toss. Each toss is an independent event.

From what I learned in college statistics class, the 66% probability of being heterozygous is accurate. I'd be interested in your probability figure and the logic behind it, though.

Paul Hollander

By the way, the coin flip analogy is a poor one. A closer coin flip analogy would be what is the probability of getting heads in a coin flip if the times that tails came up were not counted.

Paul Hollander

Hi Paul,

The point of the coin toss example is that we as humans naturally think of events as having connections.

If we go back to the original problem, and we crossed the F1 generation that are Aa for any independent trait, the expected outcome would be a mix of 1:2:1 where, if there were let us say, 20 offspring, 5 would be expected to be recessives, and 15 would be phenotypic wild types. Of the wild types, 10 would be expected to be hetro and 5 would be expected to be homo. So, if I were to choose any of the wild type offspring and ask myself the question “What are the chances of that one snake being homo for the trait that we are looking at?”, the answer would be 1 in 4. The fact that we know what some of the snakes are does not change the odds from the original event for any of the unknowns.

I hope that is clear. How do you get 66% from a single trait F1 cross?

You are connecting the original group of all babies with two subgroups of babies. The albinos are in one subgroup, and the nonalbinos are in the second subgroup. Each subgroup must be considered separately from each other and from the original group of all babies. The odds of pulling a heterozygote out of the nonalbino group is 10/15 = 2/3 = approximately 66%. The odds of pulling an albino out of the albino group is 5/5 = 1/1 = 100%. If the odds remained the same in both subgroups as in the original group of all babies, then you'd have to claim that the odds of pulling an albino out of the albino group is 25%.

Paul Hollander

Hi Paul,

Don't take up cards. (LOL) You are making the classic mistake that I was referring to in my original post. The outcome of any one offspring has no bearing on the outcome of any another offspring. Think about the problem like this. Each of the snake eggs was issued 4 cards, 1 king, 2 queens, and a jack. (The 1:2:1 ratio that you get when you do the square for Aa x Aa cross.) Then someone comes along and draws one of these four cards for each snake egg. The odds of getting a king on any of the draws are 1 in 4. For each draw, the odds do not change regardless of any of the past results because each event (draw) is independent of the other. The result for any one or more snakes in the clutch has no effect on any of the other snakes. So the odds of getting a hetro snake from picking any of the wild type offspring is still 50/50.

>You are making the classic mistake that I was referring to in my original post. The outcome of any one offspring has no bearing on the outcome of any another offspring. Think about the problem like this. Each of the snake eggs was issued 4 cards, 1 king, 2 queens, and a jack. (The 1:2:1 ratio that you get when you do the square for Aa x Aa cross.) Then someone comes along and draws one of these four cards for each snake egg. The odds of getting a king on any of the draws are 1 in 4. For each draw, the odds do not change regardless of any of the past results because each event (draw) is independent of the other. The result for any one or more snakes in the clutch has no effect on any of the other snakes.

We agree 100% up to this point.

>So the odds of getting a hetro snake from picking any of the wild type offspring is still 50/50.

Disagree. Picking heterozygotes from the wild type-looking babies is a whole new card game. The new game is equivalent to issuing three cards, a king and two queens, to each normal-looking baby and then selecting one card. The odds in this game are 1/3 for a king and 2/3 for a queen. If you disagree, please list all possible outcomes and the odds for each, so that the total equals 1 (100%).

Paul Hollander

I think you are missing something there. It is twice as probable that the normal offspring will be hets rather than normal. As for your card analogy, lets try this:

king is homozygous recessive (albino)
queens are hets
jack is homozygous dominant (normal)

Draw one card for each egg. You cannot tell the difference between the queens and the jack. Each card is placed back after the draw so the first draw has no effect on the previous. Now, if you drew a queen or jack, you have a normal looking offspring. How likely is it that you drew a queen or a jack on that normal offspring? Well, there are two queens, and only one jack, so it is TWICE as likely that the normal is a het. 66% is twice of 33%, and approx. add up to 100%. If there was only one queen, then it would be 50/50. Half of your offspring are expected to be hets, and only 1/4 are expected to be normal. Much more likely that you have hets than normals.

Now, 50% hets come in when you breed het to normal. One albino gene and one normal gene from the het parent to disperse, so half pick up the albino gene.
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Exactly. I agree with Paul and metal. Why consider the "probability" of the albino gene in a group where 25% actually EXPRESS the gene. These are obviously proven albino. So, why include them in the craps shoot?
Now we have some normal looking animals, 1/3 of which will be all normal, and 2/3s of which will carry the gene, but not show it. If you were to scoop up one snake, and ask yourself "does this carry the gene for albino?" your answer should be maybe, and maybe not. This is where the 50/50 is coming from. The PROBABILITY that this specific snake in your hand will have the albino gene or wont have the gene is 66%, since 5 snakes are normal looking, and 10 snakes look normal, but carry the gene.
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Tom
TCJ Herps
"The more people I meet, the more I like my snakes"

>Now, 50% hets come in when you breed het to normal. One albino gene and one normal gene from the het parent to disperse, so half pick up the albino gene.

While this is true, it's a case of apples and oranges. The 50% heterozygotes in the early part of the thread referred to eggs. It's the probability of a given egg producing a heterozygote before the eggs hatched. In other words, before an egg from two heterozygotes hatches, it has a 25% chance of producing a normal, a 50% chance of producing a normal-looking heterozygote, and a 25% chance of producing an albino.

After the hatch, we talk about the babies, not the eggs. As you wrote, the albinos are deleted from the group in question, leaving only the normal-looking babies. Which means that a normal-looking baby from a heterozygous albino x heterozygous albino mating has an approximately 66% chance of being a heterozygous albino and an approximately 33% chance of being homozygous normal.

Paul Hollander

I think you misread the quote you used. I said 50% het is from a het to normal breeding, not het to het. 50% will be normal, 50% will be hets, and you won't be able to tell the difference. I understand the 66% het breeding from het to het.

Nick
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1.1 Cal Kings
1.1 Northern Pines
1.2 Honduran Boas
0.1 Tangerine Honduran Milks
1.2 Pueblan Milks
1.2 Blue Beauties
1.1 Irian Jaya Carpet Pythons

Pantherophisfan,

I reversed the AA and aa in my quickie diagram but it works like this.

Breed two hets together (het for ONE simple recessive trait so as not to further complicate the issue) and you get theoretically
25% simple recessive
75% normal in appearance but out of those two thirds will be het.

and it's Punnett, not Pundent.