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A hypothetical breeding question
You breed a 66% het for albino to a visual albino, and you get all het albinos. This is unlikely if he's actually het for albino, but it's possible. Would you still consider him 66% het for albino?
Say you would. You breed him again to a visual albino and, again, you get all het albinos. Is he still 66% het for albino? If so, how many times does he have to fail to produce an albino before you decide that he's not a het for albino?
Why do I ask? I've been reading a little about genetics and this is a question at occurred to me.
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Re: A hypothetical breeding question
Yes, this is an interesting problem.
66% is the raw statistical probability of a Het assuming basic genetic assortment and no additional data.
As soon as you produce a clutch: that 66% value *technically* should be adjusted based on statistical results.
If you produce no Homozygous (HO) offspring in a 66% Het vs. HO mating: then that Het probability percentage goes *downward* proportionally to the number of non-HO offspring produced--Likewise a single HO offspring adjusts the value to a proven 100%.
In addition, you may have a 66% Het that seemingly produces no proven offspring from one cross, but, there are several potential factors that could have prevented the Homozygous form from being viable (with no noticeable evidence). If you cross it to a different specimen then you may have very different results.
Anyhow, my
Last edited by Lord Sorril; 01-17-2025 at 08:52 PM.
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Re: A hypothetical breeding question
I guess that's possible, but why do that when you can breed to visual to an offspring so you're 100% sure that your het is actually a het. At least that's what I would do. I would keep breeding to the original het because it might not be an actual het. You know what the definition of insanity is? 
Also, I feel like this 66% het is just a marketing scam for other people buying hets.
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0.1 Reg. BP Het. Albino (Faye),
1.0 Albino BP (Henry),
0.1 Pastave BP Het. Pied (Kira)
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1.0 Bumble Bee BP (Izzy)
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Re: A hypothetical breeding question
 Originally Posted by Mr. Misha
I guess that's possible, but why do that when you can breed to visual to an offspring so you're 100% sure that your het is actually a het. At least that's what I would do. I would keep breeding to the original het because it might not be an actual het. You know what the definition of insanity is?
Also, I feel like this 66% het is just a marketing scam for other people buying hets.
66% Hets are critical if you are breeding multiple recessive genes with long shot odds: They can maximize your chances.
I would only buy 66% Hets if I could see some sort of pattern or color interaction which was blatantly indicative of the morph being present.
My original Het Piebald female was a 66%, she cost me $30 whereas a 100% would have cost $180 at the time.
Not sure if I told that story at one point, but, the breeder called me an 'idiot' for buying a 66% vs. 100% Het.
Last edited by Lord Sorril; 01-18-2025 at 12:16 AM.
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Re: A hypothetical breeding question
 Originally Posted by Lord Sorril
As soon as you produce a clutch: that 66% value *technically* should be adjusted based on statistical results.
If you produce no Homozygous (HO) offspring in a 66% Het vs. HO mating: then that Het probability percentage goes *downward* proportionally to the number of non-HO offspring produced--
How would that worK? Say the 66% het's first HO mating produced 4 het offspring. By how much would you reduce the 66%?
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Re: A hypothetical breeding question
 Originally Posted by Homebody
How would that worK? Say the 66% het's first HO mating produced 4 het offspring. By how much would you reduce the 66%?
Even though I ranked top of my class in statistical probability-I am lazy...
ChatGPT figure it out for me (note: it is using a sample size of *4* offspring so its calculations are a bit rounded).
TLDR: Reduce the 66% Het to ~50% Het
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Setup:
- First parent: 66% Het (66% chance of being heterozygous Hh and 34% chance of being not Het, meaning they are HH—homozygous dominant).
- Second parent: Homozygous recessive (hh), meaning they always pass on the h allele.
So, the first parent has a 66% chance of being Hh and a 34% chance of being HH.
Possible Offspring:
- If the first parent is Hh (66% chance):
- They can pass on either H (50% chance) or h (50% chance).
- The second parent is hh, so they always pass on h.
This gives the following possibilities:
- Hh (heterozygous) — if the first parent passes on H (50% of the time).
- hh (homozygous recessive) — if the first parent passes on h (50% of the time).
So, when the first parent is Hh:
- 50% chance of Hh offspring.
- 50% chance of hh offspring.
- If the first parent is HH (34% chance):
- The first parent can only pass on H.
- The second parent is hh, so they will always pass on h.
This results in:
- Hh (heterozygous) — 100% of the time.
So, when the first parent is HH:
- 100% chance of Hh offspring.
Expected Outcome:
- If the first parent is Hh (66% chance):
- 50% of the offspring will be Hh.
- 50% of the offspring will be hh.
- If the first parent is HH (34% chance):
- 100% of the offspring will be Hh.
Observed Outcome:
You observed 4 offspring, and they were all Hh (heterozygous).
This result suggests that the first parent must have been closer to HH than Hh, since you didn't observe any hh (homozygous recessive) offspring at all.
Adjusting the Het Percentage:
Given the 100% Hh offspring observed, it suggests that the first parent was much more likely to be HH (homozygous dominant) than initially thought. If the first parent were 66% Het (Hh), you would expect a 50/50 mix of Hh and hh offspring, but you saw only Hh offspring.
This means the 66% chance of being Hh should be lowered to explain the outcome of all Hh offspring. Based on the observation of all Hh offspring, the first parent likely had more of a 50/50 chance between being HH or Hh, so the 66% Het should be adjusted to something closer to 50% Het.
Conclusion:
The 66% Het should be reduced to around 50% Het to explain the observation of 100% Hh offspring, as this would make it more likely for the first parent to be HH
Last edited by Lord Sorril; 01-18-2025 at 02:04 PM.
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Re: A hypothetical breeding question
Thank you for this thoughtful response. I'm gonna have to let this bounce around in the old noodle a bit.
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Re: A hypothetical breeding question
 Originally Posted by Mr. Misha
Also, I feel like this 66% het is just a marketing scam for other people buying hets.
Noting the fact that an animal is 66% het can be a helpful flag to a potential buyer who wants to avoid the gene as much as possible. I took a long time sorting through ads for my last couple future breeder hognoses to avoid all the possible hets floating around from the last couple generations of moving fast and breaking things with that species. Yeah, not every hognose needs to be axanthic, folks.
To answer the original question, it is reasonable in an ad to list 66% het and then also note the number of offspring that did not confirm the het ("66% het, but all 24 offspring with a visual yielded no visuals" or something to that effect). My own interpretation is that the '66% het' designation is based on the parentage, and not on failures to confirm the het (because that's what '66% het' means -- it is a claim about the parentage). If someone sold me a snake that they called '50% het' and I found out that that snake was produced by breeding 2 hets together, I would feel mislead.
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Re: A hypothetical breeding question
I guess at the end of the day, if I were breeding hets, I'd just spend the money and either purchase 100% hets or purchase visuals. Breeding out 66% and guessing doesn't make much sense to me and may take a very long time.
Maybe to someone who is trying to breed out a certain rare gene this would make sense, but to me it just doesn't. Saving even a few hundred dollars in the initial price of the animal is not worth that time.
Now it does make sense avoiding the genes that you may not want.
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0.1 Reg. BP Het. Albino (Faye),
1.0 Albino BP (Henry),
0.1 Pastave BP Het. Pied (Kira)
1.0 Pied BP (Sam)
1.0 Bumble Bee BP (Izzy)
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When I worked in the university genetics lab, we had to do this sort of problem pretty commonly.
Problem: A snake is marketed as a 50% or 66% probability het for a recessive mutant gene. How can you be sure that it is het for that gene?
Answer: Use a breeding test. Mate the possible het to either a known het or to a visual. If any of the babies is a visual, then you are sure the questionable snake actually is a het.
Problem: A snake is marketed as a 50% or 66% probability het for a recessive mutant gene. In a breeding test, it has n babies, none of which is a visual. When do you stop the test?
Answer: In a het x visual mating, half the babies are expected to be visuals, and half the babies are expected to be normal-looking hets. So the probability of any baby being a het is 1/2 (.5). And the probability (P) of there being at least one visual in n eggs is P=1-.5^n. (^n means to the power n.) So the chance of getting 3 normal-looking babies in a clutch of 3 eggs is .5^3 = (.5x.5x.5) = .125, and the chance of getting at least one visual in a clutch of 3 eggs is 1-.125 = .875 or 87.5%. On the other hand, in a normal x visual mating, none of the babies will be visuals. Seven normals and no visuals gives probability of 99% that the possible het is not a het. That 99% point is the place to stop. You will never get a 0% probability that the possible het is a het.
In a possible het x het mating, a quarter (.25) of the babies are expected to be visuals, and the rest (.75) of the babies are expected to be normal-looking. The probability (P) of there being at least one visual in n eggs is P=1-.75^n. Seventeen normals and no visuals gives probability of 99% that the possible het is not a het. That 99% point is the place to stop. You will never get a 0% probability that the possible het is a het.
Breeding a possible het x possible het is a losing proposition unless absolutely necessary. Because the majority of the matings are either het x normal or normal x normal.
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