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Posted by Hurley on May 07, 2003 at 19:23:51:
In Reply to: Results... posted by Amanda E on May 07, 2003 at 16:50:48:
Crossing the F2 normal offspring het for anery, amel, and motley to each other:
Possible phenotypes produced Normal, Anery, Amel, Motley, Anery Motley, Amel Motley, Snow, Snow Motley
Normal (3/4 AM x 3/4 AN x 3/4 MT = 27/64)
---> 27/64 Normals (67% chance het am, 67% het an, 67% het mt)
Anery (3/4 AM x 1/4 an x 3/4 MT = 9/64)
---> 9/64 Anerys (67% am, 67% mt)
Amel (1/4 am x 3/4 AN x 3/4 MT) = 9/64)
---> 9/64 Amels (67% an, 67% mt)
Motley (3/4 AM x 3/4 AN x 1/4 mt = 9/64)
---> 9/64 Motleys (67%am,67%an)
Snow (1/4 am x 1/4 an x 3/4 MT)
---> 3/64 Snows (67%mt)
Anery Motley (3/4 AM x 1/4 an x 1/4 mt)
---> 3/64 Anerys Mots (67%am)
Amel Motley (1/4 am x 3/4 AN x 1/4 mt)
---> 3/64 Amel Mots (67%an)
Snow Motley (1/4 am x 1/4 an x 1/4 mt = 1/64)
---> 1/64 Snow Mots
Lessee, that's 27 + 9 + 9 + 9 + 3 + 3 + 3 + 1 = 64 of 64...guess that's got em. :)
Sum up: F1 Normal, het amel, anery, motley x F1 Normal, het amel, anery, motley
F2 ratios expected:
27/64 Normals (67%am,an,mt)
9/64 Anerys (67% am,mt)
9/64 Amels (67% an,mt)
9/64 Motleys (67% am,an)
3/64 Snows (67% mt)
3/64 Amel Mots (67% an)
3/64 Anery Mots (67% am)
1/64 Snow Mots
Now, if you bred an F1 Normal het am,an,mt back to the anery mot...
(NO/am,an,mt x AN,MT/ )
Normals (1/1 AM x 1/2 AN x 1/2 MT = 1/4)
---> 1/4 Normals het anery and motley, 50% chance het amel
Anerys (1/1 AM x 1/2 an x 1/2 MT = 1/4)
---> 1/4 Anerys het motley, 50% chance het amel
Motleys (1/1 AM x 1/2 AN x 1/2 mt = 1/4)
---> 1/4 Motleys het anery, 50% chance het amel
Anery Motleys (1/1 AM x 1/2 an x 1/2 mt = 1/4)
---> 1/4 Anery Motleys, 50% chance het amel
...better odds of getting non-normal offspring than and F1 x F1 cross...depends on what you are after. The F1 x F1 has the potential for producing the most variety of morphs, but also has a higher potential of giving you normals. Shrug.
Of course, should you luck out and get a snow motley out of the deal, breeding it back to an F1 triple het gives a nice variety of morphs with improved ratios for getting an even mix.
F1 Normal/am,an,mt x Snow Motley
(NO/am,an,mt x AM,AN,MT/ )
All odds are 1/2 x 1/2 x 1/2 = 1/8 of each variety with known hets. I like this pairing the best of all for producing maximum variety with minimum number of breeders. :)
1/8 Normals (het an,am,mt)
1/8 Amels (het an,mt)
1/8 Anerys (het am,mt)
1/8 Motleys (het am,an)
1/8 Snows (het mt)
1/8 Amel Mots (het an)
1/8 Anery Mots (het am)
1/8 Snow Mots
Orrrr, you could use a predictor or calculator or punnetts or the like. Whichever works best for you. :) A buddy of mine got me converted from the punnetts to working the odds, and I have to say, it's a whole lot easier than drawing all of those squares and it helps me understand why certain types of crosses always give the same ratios so I don't have to work the numbers every time.
Oh, for clarification of the above if anyone has trouble deciphering:
In noting a cross like Normal het amel x Amel,
my shorthand for this cross would be NO/am x AM/ .
Capitals before the slash = Expressed phenotypes...NO = normal, AM = amel, and so on.
Lower case after the slash = Heterozygous genes not expressed, but known.
If there are 'possible hets', I put these in parentheses and add the percentage chance of being het if known.
So, for example:
AN/mt (67%am) reads Anery het for motley with a 67% chance of being het for amel.
Crossing NO/am x AM/ (normal het for amel x amel),
we know that the normal will throw "AM" (wild type) 1/2 the time,
and "am" (amel recessive) 1/2 the time.
We also know that the amel will throw "am" 2/2 = 1/1 = 100% of the time.
To figure the chances that a hatchling from this pair will be amelanistic, we multiply the chance that the normal will pass on the amel gene (1/2) times the chance the amel will throw the amel gene (1/1).
1 x 1 = 1
--- ----- ----
2 x 1 = 2
So, the chance that this pair will throw an amel is 1/2. And the chance that they'll throw a normal het amel is 1/2.
This is a little too much work to go through for just one trait...most of us commit that to memory, but if you are working with 5 traits, for example...it helps. How many squares is that in a punnett square? 1,024 squares? Uh, nothankyou. Hehe, but if you know the genotypes of the parents, you can figure your chance of getting a certain genotype hatchling in seconds.
If I bred a snow mot het for caramel to an amber mot het amel and anery, for example (as if I had these on hand...doesn't everyone? lol), that's 5 traits to deal with.
AM,AN,MT/ca x CA,HY,MT/am,an
Obviously, all the progeny should be motleys of some type, but what if I were to ask myself, "Self? What are my chances of getting a Butter Motley out of this mix?", and I was at work, away from a computer, limited to some scratch paper and a broken green crayon, and just dying to know...I could figure it relatively easily.
Butter Motley = AM + CA + MT
So, chances of getting AM, an, CA, hy, MT
(upper case = animal expresses the trait,
i.e. is homozygous for the mutation,
lower case = doesn't express, but may be het)
= 1/2 AM x 1/2 an x 1/2 CA x 1/1 hy x 1/1 MT
= 1/8 Butter Motleys
OK, if you actually read through all of that, you deserve a medal. :D Just blame it on my need to procrastinate, rather than packing up to move and cleaning the house. Hehehe. Plus, I admit it, I just like this stuff---don't mind me. ;) Oh, and only scanned through this once, so if there are typos, correct away. :p