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Another rephrasing...

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Posted by Serpwidgets on May 09, 2003 at 00:36:52:

In Reply to: Hmmm, let me see if I can rephrase one... posted by Hurley on May 08, 2003 at 10:53:22:

Heh, everyone knows that necessity is the mother of invention, but I say laziness is the father... ;)

Being an extremely lazy person, I always find the way to do something with the least amount of effort. (Even though it often takes 5 times as much effort to find the easy way, hehe.)

Picking out a specific combination from a huge clutch is as simple as Hurley puts it, but I tend to "see" it (the same exact thing) from a different angle. ;-)

The thing is, if you (anyone) cannot do a simple Punnet square (one trait, 4 squares, that's it) then you need to first get to the point where you understand that, and know it backwards and forwards. Or at the very least commit the outcomes of any single-trait cross to memory, like you did your times tables. Anybody with a bit of practice can do those in their head in a flash.

What makes it easy is this:

1- There are only 6 possible crosses involving a single trait:
AA X AA = Normal X Normal
AA X Aa = Normal X Het
AA X aa = Normal X Mutant
Aa X Aa = Het X Het
Aa X aa = Het X Mutant

aa X aa = Mutant X Mutant

2- You can pretty much ignore four of these:
- (AA X AA and aa X aa) = "only one outcome, who cares?"
- (AA X Aa and AA X aa) "only hets or possible hets" (only one outcome anyway, so who cares?)

When it's all done with, only two crosses count:
   -het X mutant (Aa X aa)
   -het X het (Aa X Aa)

As you already know or can easily figure out in your head (if you don't/can't the rest is a waste of time) the outcomes are
Aa X aa = het X mutant = 1/2 mutants (1/2 normal)
Aa X Aa = het X het = 1/4th mutants (3/4 normal)

That's it. Everybody got that? Good. :)

(If you don't get that, do not continue until you do.)

The only other thing you absolutely must know is what components go into any combinations you're using special terms for. "Snow" and "Butter" are 100% useless terms here because you need to deal with a single trait at a time. So throw all the combo names out the window for the moment and break EVERYTHING down into individual traits...

For example, you are not breeding a snow to a butter! Learn not to think in those terms because you will lose traits in the process. You are breeding an amel anery to an amel caramel, or if you prefer "aaCCee X aaccEE = aaEeCc" (Amels het anery and caramel)

My personal notation goes "ME X MR = M/ER" Amels het Anery & Caramel. If anyone wants a detailed explanation of how & why I use that notation (and why I think it makes things even easier) feel free to ask. :)


Now, to determine the odds on a specific phenotype, with ANY number of traits...

1 - "The answer is 1." Start off with 1 in your head.

2- start with the first trait, look at the genotype of both parents. What portion of the clutch will be the kind you want? (This is the "times tables" you memorized above, and it will always be either 0, 1, 1/2, 1/4, or 3/4.)

If the answer is zero, do not pass go, do not collect $200, the pair cannot produce these, period. You should already know before starting though, because it means one of the parents is neither het or homo for a trait you wanted. :-)

Otherwise multiply the number by your current answer, and hold onto that number.

3- Repeat step 2 for the next trait, until you're done.

That's it.

---------- Practice 1 --------
First, list out (in whatever way you prefer) the genotypes for each trait so you can match them against each other. My notation is easier for me, but I'll use the common notation because you had better understand common notation if you are this far already ;-)

Aa Bb cc dd Ee ff X
aa Bb Cc DD Ee ff

Now, how many will you get which are ABCD? (Expressing mutant traits A, B, C, and D)

Answer: none. First thing to check for is "is one parent non-het wild-type for any traits?"

--------- Practice 2 --------

Ok, let's try another one:
How many will you get which are expressing everything except trait D? (Do this on paper...)

Start with "1" then multiply:

A = 1/2 --> (1/2)
B = 1/4 --> (1/8)
C = 1/2 --> (1/16)
E = 1/4 --> (1/64)
F = 1 --> (1/64)

So 1 in 64 will be expressing ABCEF.

------------ Practice 3 -----------

But let's say you are doing a bunch of traits together to make kewl gray-only snakes with a nifty pattern, and trait B is "amelanism." You DON'T want amelanism because that wipes out the grays. So...

How many will express A, C, E, and F, and NOT trait B?
A = 1/2 --> (1/2)
C = 1/2 --> (1/4)
E = 1/4 --> (1/16)
F = 1 --> (1/16)

and NOT trait B?

Not B = 3/4, so you end up with 3 out of (4 x 16 = 64) so you get:

3/64ths of the clutch will be A, C, E, F. :)

Makes sense?


And from there you can go about determining the genotype... Which traits (that is is NOT expressing) will it be het or poss het for?

Since you know all one-trait Punnets, it's a simple matter for each trait one at a time:

-If only one parent is het, it's 50% possible het.

-If both parents are het, it's 66% poss het.

-If one parent is a mutant, it's het, period.

So you can also know the genotype just as easily. :)

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