Reptile & Amphibian Forums

I have a simple question about breeding a dbl het snow to a dbl het snow. I know in theory I should get out of 16 babies 3 amel 3 anery 1 snow and 9 poss hets. I know out of the possible hets 4 out of 9 should be dbl hets, that works out to 44%. In the corn snake world are these sold as 66% poss. het snows?? I know they are 66% poss het for anery or amel, but not snow. Am I wrong?? I ask this because I see half of the boa breeders out there marketing these babies as 66% poss het snow. I brought this to the cornsnake forum because I know you guys know more about gentics then us boa freaks in general. Thanks alot for the input. Craig Kade

If you breed a dbl het snow, to dbl het snow, you would have Amels poss het anery, Anerys poss het amel, Snows, and Normals possible het snow. I don't think that people would use a percentage for this. It gets too complicated beyond simple het situations.
-----
~Sasheena

I think the reason boa breeders market their snakes like this is b/c of the potential price for future offspring, which is higher than seen in corn morphs.
Now, according to Marcel's cornsnake program, a double hetXdouble het cross displays the following:
Offspring is predicted :
6.25% Amel
12.50% Amel, het. Anery A
12.50% Snow
25.00% het. Snow
6.25% het. Amel
12.50% normal
6.25% Anery A, het. Amel
12.50% het. Anery A
6.25% Anery A

genotype...
1/16 Amel
1/8 Amel, het. Anery A
1/8 Snow
1/4 het. Snow
1/16 het. Amel
1/8 normal
1/16 Anery A, het. Amel
1/8 het. Anery A
1/16 Anery A

Phenotypes..
1/5 Amel (is this 1/5 right?)
1/8 Snow
1/1 normal
1/8 Anery A

I also like to refer to Ralph Davis' website for genetics since it explains in simple wording for quick reference. I know how to use punnett squares for mendelian genetics but why if you don't need to...its like writing out math probles vs. using a calculator.

Another thing...I hate the use of 66% het...in my opinion its STILL only a 50% probability of producing a recessive trait when breeding two single het parents b/c theoretically, out of 4 babies: 1 will display the morph/recessive trait, 1 will be normal (not het for anything) and 2 will be '66%' hets. Now one problem I see is that reality doesn't always follow the statistical odds. Furthermore, you simply cannot truly know? Its still a 50/50 'flip of a coin' guess as to which ones will be true hets. I think the way to go is either sell 100% hets OR possible hets (instead of 50% or 66% OR whatever.lol). BUT as long as people are willing to pay more for 66% hets vs. 50% hets...it will still go on...mainly w/BP's and some boa morphs.

-----
Sincerely,
Jason
-----
0.2 Green Tree Pythons
2.2 Bismark Ringed Pythons
2.3 Ball Pythons
1.1 Brazilian Rainbow Boas
1.1 Argentine Boas
0.1 Solomon Island Ground Boa
1.2 Hogg Island Boas
1.1 Sonoran Desert Boas
2.2 Nicaraguan Boas
7.8 Cornsnake morphs
1.1 Northern Pine Snakes
2.2 Bairds Rat Snakes
1.2 White-Sided Black Rat Snakes
My Website & Pics (click here)

Just to clarify for anyone else reading this and had ?s about the term '66%' hets, b/c I probably made it more confusing than needed in my above reply...

If you bred: normal het albino male X normal het albino female
statistically, babies will be:
25% albino (1 in 4 or 1/4)
25% normal (1 in 4 or 1/4)
50% normal het albino 2 in 4 or 2/4)

So...if you produced only 4 babies, 1 should be albino leaving 3 babies that are 'normal' in appearance. Out of those 3 babies, 2 will supposedly carry the albino gene. Therefore you have a 2/3 (or 66%) chance of having a normal baby that is 100% heterozygous for albino.
-----
Sincerely,
Jason
-----
0.2 Green Tree Pythons
2.2 Bismark Ringed Pythons
2.3 Ball Pythons
1.1 Brazilian Rainbow Boas
1.1 Argentine Boas
0.1 Solomon Island Ground Boa
1.2 Hogg Island Boas
1.1 Sonoran Desert Boas
2.2 Nicaraguan Boas
7.8 Cornsnake morphs
1.1 Northern Pine Snakes
2.2 Bairds Rat Snakes
1.2 White-Sided Black Rat Snakes
My Website & Pics (click here)

>Now, according to Marcel's cornsnake program, a double hetXdouble het cross displays the following:

I don't have this program, but I thought that something was screwy with the results when I saw that 12.5% snow. And I was absolutely positive something was screwy when I saw that 1/1 normal in the phenotypes list. Here are the correct numbers.

>Offspring is predicted :
>6.25% Amel
>12.50% Amel, het. Anery A
>12.50% Snow
>25.00% het. Snow
>6.25% het. Amel
>12.50% normal
>6.25% Anery A, het. Amel
>12.50% het. Anery A
>6.25% Anery A
>
>genotype...
>1/16 Amel
>1/8 Amel, het. Anery A
>1/8 Snow
>1/4 het. Snow
>1/16 het. Amel
>1/8 normal
>1/16 Anery A, het. Amel
>1/8 het. Anery A
>1/16 Anery A

Genotypes:
heterozygous amelanistic x heterozygous amelanistic -->
1/4 normal
2/4 heterozygous amelanistic
1/4 amelanistic

heterozygous anerythristic x heterozygous anerythristic -->
1/4 normal
2/4 heterozygous anerythristic
1/4 anerythristic

Multiply the fractions to get the fraction for each combination ....

1/4 * 1/4 = 1/16 (6.25%) Amel
1/4 * 2/4 = 2/16 = 1/8 (12.5%) Amel, het. Anery A
1/4 * 1/4 = 1/16 (6.25%) Snow (= amelanistic, anerythristic)
2/4 * 2/4 = 4/16 = 1/4 (25%) het. Snow (= heterozygous amelanistic, heterozygous anerythristic)
2/4 * 1/4 = 2/16 = 1/8 (12.5%) het. Amel
1/4 * 1/4 = 1/16 (6.25%) normal
1/4 * 2/4 = 2/16 = 1/8 (12.5%) Anery A, het. Amel
1/4 * 2/4 = 2/16 = 1/8 (12.5%) het. Anery A
1/4 * 1/4 = 1/16 (6.25%) Anery A

>Phenotypes..
>1/5 Amel (is this 1/5 right?)
>1/8 Snow
>1/1 normal
>1/8 Anery A

Phenotypes:
heterozygous amelanistic x heterozygous amelanistic -->
3/4 normal
1/4 amelanistic

heterozygous anerythristic x heterozygous anerythristic -->
3/4 normal
1/4 anerythristic

Multiply the fractions to get the fraction for each combination ....

1/4 * 3/4 = 3/16 Amel
1/4 * 1/4 = 1/16 Snow
3/4 * 3/4 = 9/16 normal
3/4 * 1/4 = 3/16 Anery A

Paul Hollander

Thank you Paul...it was VERY late last night when I replied and I thought I was seeing things with the results but I'm glad you confirmed it was in fact incorrect. Also, thanks for posting the correct results.
-----
Sincerely,
Jason
-----
0.2 Green Tree Pythons
2.2 Bismark Ringed Pythons
2.3 Ball Pythons
1.1 Brazilian Rainbow Boas
1.1 Argentine Boas
0.1 Solomon Island Ground Boa
1.2 Hogg Island Boas
1.1 Sonoran Desert Boas
2.2 Nicaraguan Boas
7.8 Cornsnake morphs
1.1 Northern Pine Snakes
2.2 Bairds Rat Snakes
1.2 White-Sided Black Rat Snakes
My Website & Pics (click here)

The normals from a double het to double het breeding should not be sold as 66% possible het snow, because, as you say - the probability is that only 4 of the 9 normal appearing snakes in the clutch are het for both genes (44% chance that any given normal is het for both amel and anery). I am suprised that people can get away with selling them as such??

While there is not enough market for snow corns for anyone to try to pull this, most buyers of the more expensive double or triple recessive morphs (hypo lavs or striped butters) seem to have enough understanding of genetics to not fall for skewed statistics.

In short, you are right - and in my experience, this deception would not occur often in the cornsnake world. Of course it is all chance anyway when you are buying possible hets - no recourse for the buyer if the odds don't turn out in your favor.

mary v.

What confuses many is that there is no more probability of a particular snake being het for a specific trait simply because there are more traits from which to draw. An animal is either het for amel, or it is not. That is the same probability of a coin being flipped and coming up heads -- 50/50.

However, because of the dispersal of genes within a particular clutch, the genetic appearance of "heads" will occur at random, with predictable numbers. So, we can say that we should (statistically speaking) expect one snow, three anery, three amel, and nine normal hatchlings from a dbl het to dbl het breeding. In reality we might get anything from all normals to all snows!

Each of the normals has a 50% probability of being het for either amel OR anery, but since the punnett square tells us that 4 of those nine projected normals SHOULD be het for both anery and amel, people say the equivalent of "OK, you have nine normals from which to choose, four of them should be het for snow, so you have a 44% chance of picking one out that is actually double het." In reality, it is a false premise, in my opinion, because there is no guarantee that ANY of them are het for anything at all, and it is the assumption that FOUR of them are actually double hets, upon which that 44% number was based.

However, there is no way that a normal from a dbl het x dbl het breeding can accurately be represented as being 66% possible het for snow, no matter what one's view of possible het percentages may be. The math just doesn't work out that way.

As for my personal oipinion, normals should either be sold as guaranteed hets (and the seller should be willing to stand behind that guarantee, by the way!) or as simple normals. Anything else confuses those who are most likely to be taken by a dishonest person. If someone knows enough to ASK me about the possible hets, I'll tell them, but I don't advertise possible hets anymore. I used to, but it just seems less than professional to me now. Of course, that's ONLY my opinion, and others who are VERY professional in their actions may see things otherwise. I've never taken a polll on the issue ... LOL
-----
Darin Chappell
Hillbilly Herps
PO Box 254
Rogersville, MO 65742

If I bred double het snow to double het snow, I might sell the normal offspring as possible het snow. But I would probably not do that. Since the only snake I have where I even KNOW what she is het for is my normal het motley, it's not that big of a deal. If I had some high-end morph where a 66% possible het is worth thousands more than a 50% possible het... well jeepers, I think it might make a difference. But in general, I don't sell anything where the possible hets make that big of a difference.

Soon to be for sale:

Normals het stripe
Striped-Motley Normals
Striped Normals het Anery
Normals het stripe and amel.

The striped normal papa might be het amel. If he is, I won't change the description on the offspring, because it wouldn't change anything important.

Anyway, just me babbling.
-----
~Sasheena

You stated exactly what I was thinking.lol.

What I don't get is how in the world an albino form of any snake can randomly appear in the wild. I can't imagine that both parents are simply heterozygous OR is it simply a random defect in that one particular offspring's genetic code. I understand how its easy to isolate a single recessive gene once it appears but whats to keep someone from producing an albino snake simply by mere luck. Albinoism had to start somewhere, right? I guess its the chicken vs. the egg argument and while odds are probably 1 in 10s of thousands of any single recessive trait popping up, it could still happen in captive populations (probably more so than in the wild...especially in certain areas/cities where there is limited sources for a species and risk of inbreeding increases). Well, sorry for the side thoughts..lol.
-----
Sincerely,
Jason
-----
0.2 Green Tree Pythons
2.2 Bismark Ringed Pythons
2.3 Ball Pythons
1.1 Brazilian Rainbow Boas
1.1 Argentine Boas
0.1 Solomon Island Ground Boa
1.2 Hogg Island Boas
1.1 Sonoran Desert Boas
2.2 Nicaraguan Boas
7.8 Cornsnake morphs
1.1 Northern Pine Snakes
2.2 Bairds Rat Snakes
1.2 White-Sided Black Rat Snakes
My Website & Pics (click here)