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just from wondering what would u get if a normal corn and a snow corn were to have babies????
1/2 normal + 1/2 snow..or
some a mixed colour?????
Thanx
1/2 normal + 1/2 snow..or
some a mixed colour?????
Thanx
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Normal+Snow=???????
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does anybody know what a snow and motley would make?
Snow (am + an) x Motley (mt) = All normals het am,an, and mt
(if I may represent this as AM/am AN/an MT/mt) in the F1.
In the F2, AM/am AN/an MT/mt x AM/am AN/an MT/mt should give you:
27/64 Normals (3/4 x 3/4 x 3/4)
9/64 Amels (1/4 x 3/4 x 3/4)
9/64 Anerys (3/4 x 1/4 x 3/4)
9/64 Motleys (3/4 x 3/4 x 1/4)
3/64 Snows (1/4 x 1/4 x 3/4)
3/64 Amel Motleys (1/4 x 3/4 x 1/4)
3/64 Anery Motleys (3/4 x 1/4 x 1/4)
1/64 Snow Motleys (1/4 x 1/4 x 1/4)
Snow (am + an) x Motley (mt) = All normals het am,an, and mt
(if I may represent this as AM/am AN/an MT/mt) in the F1.
In the F2, AM/am AN/an MT/mt x AM/am AN/an MT/mt should give you:
27/64 Normals (3/4 x 3/4 x 3/4)
9/64 Amels (1/4 x 3/4 x 3/4)
9/64 Anerys (3/4 x 1/4 x 3/4)
9/64 Motleys (3/4 x 3/4 x 1/4)
3/64 Snows (1/4 x 1/4 x 3/4)
3/64 Amel Motleys (1/4 x 3/4 x 1/4)
3/64 Anery Motleys (3/4 x 1/4 x 1/4)
1/64 Snow Motleys (1/4 x 1/4 x 1/4)
wow thats complicated
Nah, it only looks that way on the surface.
You just have to break it down a bit and it all makes sense.
Remember that (as far as we know) each of these traits works indepently and each is recessive to wild type/normal for that gene.
If you breed the progeny of the motley and snow, you are breeding triple het (amel, anery, mot) animals to each other.
Take each gene separately...let's start with amel.
Snake 1 can throw either <b>AM</b> (wild type/normal at amel locus) or<b> am</b> (amel mutation).
Snake 2 can throw either <b>AM</b> or <b>am </b>as well.
Possible outcomes:
<img src=http://pstr-m04.ygpweb.aol.com/data2/00B/00/55/FF/F8/pB5lMgZaFmLepoNuCwnQEfYM8+-ocTsg0174.jpg>
So with a simple het to het cross, we know that the possible phenotypic outcomes are: 3/4 Normals (2 of the 3 het for amel) and 1/4 Amels.
The same is going to hold true for the other 2 traits. The predicted outcomes are going to be 3/4 Normal to 1/4 Anery as well as 3/4 normal pattern to 1/4 motley.
Now it's just a matter of probability combinations.
If you are combining probabilities of something, you multiply the fraction (ratios) to get the result of both things happening.
For example, if the probability that the pair will throw an amel is 1/4 and the probability that the pair with throw an anery is 1/4, then the probability that the pair het for amel and anery will throw a snow (an animal that is both amel and anery) is:
1/4 X 1/4 = 1/16
The probability that they will throw an amel if both are het for amel and anery is:
1/4 X 3/4 = 3/16
because there is a 1/4 chance that amel will be thrown and a 3/4 chance that anery WILL NOT.
Same goes for anery...
3/4 X 1/4 = 1/16
because there is a 3/4 chance that amel will not show up and a 1/4 chance that anery will be.
OK, that does double hets...the same goes for triple hets.
Normal het am, an, and mot X Normal het am, an, and mot
give you:
Normals (3/4 x 3/4 x 3/4 = 27/64)
Amels (1/4 x 3/4 x 3/4 = 9/64)
Anerys (3/4 x 1/4 x 3/4 = 9/64)
Motleys (3/4 x 3/4 x 1/4 = 9/64)
Snows (1/4 x 1/4 x 3/4 = 3/64)
Amel Motleys (1/4 x 3/4 x 1/4 = 3/64)
Anery Motleys (3/4 x 1/4 x 1/4 = 3/64)
Snow Motleys (1/4 x 1/4 x 1/4 = 1/64)
(probability am x prob an x prob mot = prob total)
You can do really complicated crosses simply by breaking them down to the base genes involved. Or, I suppose you could write out a 4 square punnett square for 1 trait, a 16 square for 2 traits, 64 squares for 3 traits, 256 squares for 4 traits....
While punnetts give a great visualization of what's going on (once you crayon in all the squares in various pretty colors), they get quite tedious up above one or two traits.
Work out the individual probabilities of each trait and multiply them to get your final results. You can do this for other crosses as well. For example:
snow x anery het amel
(am/am an/an X AM/am an/an)
Looking at the above cross, we can see that the chance that this pair will throw an amel mutant is 1/2 and the chance that this pair will not throw an amel mutant (phenotypically...not predicting hets here) is 1/2. The chance that this pair will throw and anery mutant is 1/1, since neither parent has a normal gene at the anery locus. So what are our results of this cross?
Normal - 1/2 (AM) x 0 (AN) = <b>0</b>
Anery - 1/2 (AM) x 1/1 (an) = <b>1/2 anery</b>
Amel - 1/2 (am) x 0 (AN) = <b>0</b>
Snow - 1/2 (am) x 1/1 (an) = <b>1/2 snow</b>
So Snow x Anery het amel = 1/2 anerys, 1/2 snows.
You can do this as well if you want to figure hets...
Normal het anery and amel x Normal het anery and amel
AM/am AN/an x AM/am AN/an
Normal - 3/4 x 3/4 = 9/16 AM/AM or am + AN/AN or an
Amel - 1/4 x 3/4 = 3/16 am/am + AN/AN or AN/an
Anery - 3/4 x 1/4 = 3/16 AM/AM or AM/am + an/an
Snow - 1/4 x 1/4 = 1/16 am/am an/an
Phenotypically from above, we know the chances that anery, amel, and snow will be expressed (phenotype) is:
Normal = 9 / 16
Amel = 3 / 16
Anery = 3 / 16
Snow = 1 / 16
Now let's break it down to hets...
Normal no hets = 1/4 (AM/AM) x 1/4 (AN/AN) = 1/16
Normal het amel = 2/4 (AM/am) x 1/4 (AN/AN) = 2/8
Normal het anery = 1/4 (AM/AM) x 2/4 (AN/an) = 2/16
Normal double het = 2/4 (AM/am) x 2/4 (AN/an) = 4/16
(added up = 9/16 Normals )
Amel no hets = 1/4 (am/am) x 1/4 (AN/AN) = 1/16
Amel het anery = 1/4 (am/am) x 2/4 (AN/an) = 2/16
(which = 3/16 Amels)
Anery no hets = 1/4 (AM/AM) x 1/4 (an/an) = 1/16
Anery het amel = 2/4 (AM/am) x 1/4 (an/an) = 2/16
(3/16 Anerys)
Snows = 1/4 (am/am) x 1/4 (an/an) = 1/16
If you made it this far, congratulations. I do tend to ramble occasionally.
OK, now look back at the normals in the last example.
These normals have a 2/3 (67%) chance of carrying amel (look at the punnett square way above) and a 2/3 (67%) chance of carrying anery.
You can therefore express genotypic results of the above cross as:
9/16 Normals (67% het amel, 67% het anery)
3/16 Amels (67% het anery)
3/16 Anerys (67% het amel)
1/16 Snows
I hope that was ok to follow.
Nah, it only looks that way on the surface.
You just have to break it down a bit and it all makes sense.
Remember that (as far as we know) each of these traits works indepently and each is recessive to wild type/normal for that gene.
If you breed the progeny of the motley and snow, you are breeding triple het (amel, anery, mot) animals to each other.
Take each gene separately...let's start with amel.
Snake 1 can throw either <b>AM</b> (wild type/normal at amel locus) or<b> am</b> (amel mutation).
Snake 2 can throw either <b>AM</b> or <b>am </b>as well.
Possible outcomes:
<img src=http://pstr-m04.ygpweb.aol.com/data2/00B/00/55/FF/F8/pB5lMgZaFmLepoNuCwnQEfYM8+-ocTsg0174.jpg>
So with a simple het to het cross, we know that the possible phenotypic outcomes are: 3/4 Normals (2 of the 3 het for amel) and 1/4 Amels.
The same is going to hold true for the other 2 traits. The predicted outcomes are going to be 3/4 Normal to 1/4 Anery as well as 3/4 normal pattern to 1/4 motley.
Now it's just a matter of probability combinations.
If you are combining probabilities of something, you multiply the fraction (ratios) to get the result of both things happening.
For example, if the probability that the pair will throw an amel is 1/4 and the probability that the pair with throw an anery is 1/4, then the probability that the pair het for amel and anery will throw a snow (an animal that is both amel and anery) is:
1/4 X 1/4 = 1/16
The probability that they will throw an amel if both are het for amel and anery is:
1/4 X 3/4 = 3/16
because there is a 1/4 chance that amel will be thrown and a 3/4 chance that anery WILL NOT.
Same goes for anery...
3/4 X 1/4 = 1/16
because there is a 3/4 chance that amel will not show up and a 1/4 chance that anery will be.
OK, that does double hets...the same goes for triple hets.
Normal het am, an, and mot X Normal het am, an, and mot
give you:
Normals (3/4 x 3/4 x 3/4 = 27/64)
Amels (1/4 x 3/4 x 3/4 = 9/64)
Anerys (3/4 x 1/4 x 3/4 = 9/64)
Motleys (3/4 x 3/4 x 1/4 = 9/64)
Snows (1/4 x 1/4 x 3/4 = 3/64)
Amel Motleys (1/4 x 3/4 x 1/4 = 3/64)
Anery Motleys (3/4 x 1/4 x 1/4 = 3/64)
Snow Motleys (1/4 x 1/4 x 1/4 = 1/64)
(probability am x prob an x prob mot = prob total)
You can do really complicated crosses simply by breaking them down to the base genes involved. Or, I suppose you could write out a 4 square punnett square for 1 trait, a 16 square for 2 traits, 64 squares for 3 traits, 256 squares for 4 traits....
While punnetts give a great visualization of what's going on (once you crayon in all the squares in various pretty colors), they get quite tedious up above one or two traits.
Work out the individual probabilities of each trait and multiply them to get your final results. You can do this for other crosses as well. For example:
snow x anery het amel
(am/am an/an X AM/am an/an)
Looking at the above cross, we can see that the chance that this pair will throw an amel mutant is 1/2 and the chance that this pair will not throw an amel mutant (phenotypically...not predicting hets here) is 1/2. The chance that this pair will throw and anery mutant is 1/1, since neither parent has a normal gene at the anery locus. So what are our results of this cross?
Normal - 1/2 (AM) x 0 (AN) = <b>0</b>
Anery - 1/2 (AM) x 1/1 (an) = <b>1/2 anery</b>
Amel - 1/2 (am) x 0 (AN) = <b>0</b>
Snow - 1/2 (am) x 1/1 (an) = <b>1/2 snow</b>
So Snow x Anery het amel = 1/2 anerys, 1/2 snows.
You can do this as well if you want to figure hets...
Normal het anery and amel x Normal het anery and amel
AM/am AN/an x AM/am AN/an
Normal - 3/4 x 3/4 = 9/16 AM/AM or am + AN/AN or an
Amel - 1/4 x 3/4 = 3/16 am/am + AN/AN or AN/an
Anery - 3/4 x 1/4 = 3/16 AM/AM or AM/am + an/an
Snow - 1/4 x 1/4 = 1/16 am/am an/an
Phenotypically from above, we know the chances that anery, amel, and snow will be expressed (phenotype) is:
Normal = 9 / 16
Amel = 3 / 16
Anery = 3 / 16
Snow = 1 / 16
Now let's break it down to hets...
Normal no hets = 1/4 (AM/AM) x 1/4 (AN/AN) = 1/16
Normal het amel = 2/4 (AM/am) x 1/4 (AN/AN) = 2/8
Normal het anery = 1/4 (AM/AM) x 2/4 (AN/an) = 2/16
Normal double het = 2/4 (AM/am) x 2/4 (AN/an) = 4/16
(added up = 9/16 Normals
)Amel no hets = 1/4 (am/am) x 1/4 (AN/AN) = 1/16
Amel het anery = 1/4 (am/am) x 2/4 (AN/an) = 2/16
(which = 3/16 Amels)
Anery no hets = 1/4 (AM/AM) x 1/4 (an/an) = 1/16
Anery het amel = 2/4 (AM/am) x 1/4 (an/an) = 2/16
(3/16 Anerys)
Snows = 1/4 (am/am) x 1/4 (an/an) = 1/16
If you made it this far, congratulations. I do tend to ramble occasionally.
OK, now look back at the normals in the last example.
These normals have a 2/3 (67%) chance of carrying amel (look at the punnett square way above) and a 2/3 (67%) chance of carrying anery.
You can therefore express genotypic results of the above cross as:
9/16 Normals (67% het amel, 67% het anery)
3/16 Amels (67% het anery)
3/16 Anerys (67% het amel)
1/16 Snows
I hope that was ok to follow.
Last edited:
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all i can say is ...
:
:
coralchris
New member
is that out of a book?
Susan
Go Ahead, Make My Day!
Once you understand the general principles behind simple recessive genes, figuring out the genetics of corn snakes is fairly easy. It's when you get into the not-so-simple genes that you can get confused (EX: zigzag, the motley-stripe linkage, etc.). Then you also have to deal with all the new genes coming to the surface...hypo A, B, C, D .................................................................................................................
So Pirate55
New member
Hypo is a whole different ballgame. I know there is different letters that stand for the diff types but which are which. Any help? which letters corolate to which common name? Std, Ultra, Lava, etc....?