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Hmmm, let me see if I can rephrase one...


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Posted by Hurley on May 08, 2003 at 10:53:22:

In Reply to: Wow... posted by Amanda E on May 08, 2003 at 07:11:46:

...since that post rambled on and I was just having fun working problems. LOL What's causing the jam?

Quick fraction review, feel free to ignore. :D
Basic math tells us that to multiply fractions, you multiply the numerators across for the answer's numerator, and the denominators across for the answer's denominator. So, for example, multiplying 2/3 times 3/4 = 6/12...3/6...1/2. Adding them requires the denominators to be equal, so convert 2/3 and 3/4 to a common denominator, and the problem becomes 8/12 + 9/12 which then equals 17/12.

OK, so...I'll lose my shorthand and express everything as aa bb cc.

Let's say:
a = amel recessive mutant
A = wild type at amel locus-->normal

n = anery recessive mutant
N = wild type at anery locus-->normal

If I want to figure numbers for a clutch from Normal het snow x Snow parents, I'd express the parents as: AaNn (female normal het snow) x aabb (male snow)

So, from this, we know that the chance of getting each gene is:

*AaNn *aann
------- -------
a = 1/2 a = 1/1
A = 1/2

and

n = 1/2 n = 1/1
N = 1/2

So, from this we know that the chances of getting
aa = 1/2 (1/2 x 1/1 = 1/2)
Aa = 1/2 (1/2 x 1/1 = 1/1)
AA = Not possible (1/2 x 0/0 = does not exist)

nn = 1/2
Nn = 1/2
NN = Not possible.


Alright, what are the possible combos using these 2 mutations?

Normal (AANN, AaNN, AANn, and AaNn)
Amel (aaNN and aaNn)
Anery (AAnn, Aann)
Snow (aann)

So, let's take the simplest one...snow.

Chance of getting a snow:

aa = 1/2, nn = 1/2.

1/2 x 1/2 = 1/4 aann.
aa x nn

Chance of getting an anery:


Chance of getting AA or Aa? 0/0 + 1/2 = 1/2
Chance of getting an nn? 1/2

1/2 x 1/2 = 1/4 Axnn.

Chance of getting an amel:

aa = 1/2
NN or Nn = 1/2

aaNx = 1/4


Chance of getting a normal:

AA or Aa = 1/2
NN or Nn = 1/2

AxNx = 1/4

Note: I use x to mean that it doesn't matter if that letter is 'A' or 'a'.

Summary:

Snow (aann) = 1/4
Anery (Axnn) = 1/4
Amel (aaNx) = 1/4
Normal (AxNx) = 1/4

OK, so I only took into account phenotypic outcome in this one...

Want to do one with figuring genotypes this time?

Same genes, let's do Normal het snow x Anery het amel, for spice.

AaNn x Aann
-------------
1/2 a ___1/2 a
1/2 A __1/2 A

1/2 n ___1/1 n
1/2 N

Chances on the A's
---------------
(1/2 A x 1/2 A) = 1/4 AA

(1/2 a x 1/2 A) + (1/2 A x 1/2 a) = 1/4 + 1/4 = 2/4 = 1/2 Aa

(1/2 a x 1/2 a) = 1/4 aa

Checking myself to make sure I got that right...1/4 + 1/2 + 1/4 = 4/4. Yep. :)

Chances on the N's
------------------
(1/2 N x 0/0 N) = NN not possible.

(1/2 n x 0/0 N) + (1/2 N x 1/1 n) = 0/0 + 1/2 = 1/2 Nn

(1/2 n x 1/1 n) = 1/2 nn

1/2 + 1/2 = 2/2...all there. :)

Chance of snows (aann) = 1/4 aa x 1/2 nn = 1/8 snows.

Chance of amels (aaNn + aaNN) = (1/4 aa x 1/2 Nn) + (aaNN not possible) = 1/8 amels (100% het for anery).

Chance of anerys (Aann + AAnn)
(1/2 Aa x 1/2 nn) + (1/4 AA x 1/2 nn)
= 1/4 Aann + 1/8 AAnn
= 2/8 + 1/8
= 3/8 anerys (67% het for amel).

The 3/8 anerys being comprised of 2/8 Anery het amel and 1/8 Anery ...so you can express the Anerys as 67% (2 of 3) chance het amel.

Chance of normals (AaNn, AANn, AaNN, AANN)

Aa + AA = 3/4, NN + Nn = 1/2
= 3/4 x 2/4 = 6/18
= 3/8 Normals

Of the 3/4, 2 are Aa, 1 is AA, so 67% chance het amel
Of the 2/4, 2 are Nn, 0 are NN, so 100% chance het anery

= 3/8 Normals het anery (67% chance het amel)


...and a last check to make sure everyone's accounted for: 1/8 + 1/8 + 3/8 + 3/8 = 8/8.

Summary:


1/8 Snow
1/8 Amel het anery
3/8 Anery (67% het amel)
3/8 Normal het anery (67% het amel)

Here's to hoping a different look at the same material lifts the logjam. :D After a while, you can just look at the crosses and do them in your head...and that's a great party trick. ;)

Hurley




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