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Hello to all, My last Anery possible het albino from Matt Salyer, just breed today with a DH for Snow from Terry Dunham. I put in with her a Albino male I think, but he did not pay any mind all he did was buck her off of him. I am going to put in a known male to see what happens. Question what can I expect from the pairing above??????? Just wondering.

Thanks
Tom Sierra

It all depends on if the Anery is in fact het albino.

Anery/ps albino x Double het snow "Anery/Albino"

From the clutch is you get a Snow or Albino you will know for sure the anery ph albino is definate het albino. If so you will have anery ph 2/3 albino, albinos definate het anery, normals het anery ph 2/3 albino. As for ratios of each it depends on clutch size and luck. If you do not produce a snow or albino it does not mean your anery is not het albino but you will not know for sure. So your percentage goes down from 2/3 chance to 50% or less. So if you do not produce a snow or albino all your normals will be possible double het snow, anery ph albino. I think i got it right. Lets what Terry thinks when he reads my response. Good Luck

Thanks for the response. I still have a hard time understanding all of this. But I hope to get it someday. Will keep you posted.

Tom

it will go like in the table here:
anery het albino X DH snow
aaBb X AaBb
a = anery
b = albino
i will have 1/8 chance to produce a snow

Tom,

JLambert is correct when he said : "From the clutch if you get a Snow or Albino you will know for sure the anery ph albino is definite het albino. If so you will have anery ph 2/3 albino, albinos definate het anery, normals het anery ph 2/3 albino".

To actually breakdown the possible odds in percentages of what the offsprings "could be" from breeding : Homozygous anery Het. albino with a double het for anery/albino

12.5% Wild type looking but definite Het. anery
25% Wild type looking but definite double het. anery/albino
12.5% Albino looking but het. anery
12.5% Anery looking with no het for anything (pure anery)
25% Anery looking but het. albino
12.5% Snow

To understand how he came up with anery 2/3 possible het for albino, look at my table above and you have a total 37.50% chance of getting visually anery hondos. The breakdown of that 37.50% is as follows: 25% with het albino (5th line) and 12.50% without any het (4th line). So, take the 25% and divide that by the total 37.50% and you will get 66.67% ( or 2/3 or 2 out of 3 ). In other words, if your anery parent is in fact het for albino, each of the anery hatchlings then will have a 66.67% chance of being het for akbino. Also, any albino hatching will be 100% definite anery because the table shows no other albinos with a different genetic makeup. Again, this is assuming that your anery parent is in fact het for albino. And if your produce snow or albino, then that anery parent is in fact het for albino, as jlambert said.

Hope that clarifies and helps you.

Ray

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RAY

.

She's definitely a thing of beauty and she's really growing fast and eats voraciously too. She'll definitely be ready to breed next early spring with my ghost male from Mike Alvarez.

Thanks again !
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RAY

First of all, looks like John got it right in his post.

I just wanted to be a cheerleader for the theory that this is NOT a challenging question. Any time you're calculating the results of multiple-morph breedings you just have to figure them out ONE AT A TIME:

1) re: anery
you know the one animal's anery and the other's het. So what's anery x het anery yield? the same as any other homozygous of a recessive x a het of it (works the same for albino or hypo) -- it gives you HALF babies that are homozygous (anerys) and half that are hets (anery).

the next step is to figure out the results of the OTHER morph, and then combine the two (which is where most people get confused, when they do, but i'm gonna try to make it simple)...

So let's take the case where your anery poss het amel IS het amel.

2) if that's the case, then IN TERMS OF AMEL you're breeding het x het, right? and het x het yields babies 1/4 of which are homozygous for that morph (amel, in this case) and 2/4 are het for it and 1/4 are "normal" for amel (no amel gene)--those latter 3/4 look alike so we consider that "3/4 of the babies will be POSSIBLE het, with 2/3 chance for each of them being het"

So now go back to your original (anery) results, where HALF are anery and HALF are het anery.
CONSIDER FIRST ONLY THOSE THAT ARE ANERY:
a) 1/4 of THOSE will be amel (by analysis immediately above)...anery & amel = snow, 1/4 of 1/2 of the babies = 1/8 of the TOTAL babies will be snows.
b) 3/4 of THOSE will be possible het amels, 1/2 x 3/4 = 3/8 so 3/8 of the TOTAL babies will be anerys poss het amel.

NOW CONSIDER THE OTHER HALF, THE ONES THAT ARE HET/ANERY:
a) 1/4 of those will be amel, so 1/4 x 1/2 = 1/8 so 1/8 of the TOTAL babies will be amels het/anery.
b) 3/4 of the het/anerys will be poss het amels, so 3/4 x 1/2 = 3/8 so 3/8 of the TOTAL babies will be het/anery poss het amel.

I HOPE this step by step shows how simple it really is, so people can understand how to calculate these things routinely. Remember:
1) calculate the results for ONE of the morphs
2) calculate the results of the OTHER morph
3) divide the first type results into additional subsets resulting from the 2nd type...I am not satisfied with how i'm saying this, but i hope the example above makes clear what i'm saying--if HALF the babies are gonna be anerys, how is that group gonna be further subdivided as a result of the OTHER morph, then do the same with the other half.

peace
terry
ps: anybody who gets lost in this, pls tell me precisely WHERE you got lost and i'll try to re-word it, see if i can make it clear.

>>Hello to all, My last Anery possible het albino from Matt Salyer, just breed today with a DH for Snow from Terry Dunham. I put in with her a Albino male I think, but he did not pay any mind all he did was buck her off of him. I am going to put in a known male to see what happens. Question what can I expect from the pairing above??????? Just wondering.
>>
>>Thanks
>>Tom Sierra

Thanks for making it easy to understand.

Tom