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RE: Triple het question (probably silly)

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Posted by: Paul Hollander at Mon Nov 3 18:31:38 2008   [ Email Message ] [ Show All Posts by Paul Hollander ]  
   

>Het hypo x het hypo =
>3/4 hypo (33% chance super)
>1/4 normal
>
>Het Albino x het Albino =
>1/4 albino
>3/4 normal (66% chance het)
>
>Het Anery x het Anery =
>1/4 Anery
>3/4 normal (66% chance het)


Here is an orderly way to work out all possible combinations. Take three strips of paper and draw a bar across the middle of each. Mark the three as I have here.

Strip #1:
3/4 hypo
-----------
1/4 normal

Strip #2:
1/4 albino
-----------
3/4 normal

Strip #3:
1/4 anerythristic
----------------------
3/4 normal

Tape the ends of the strips together to make 3 rings and put the rings on a finger in 1,2,3 order with 3/4 hypo, 1/4 albino, 1/4 anerythristic all showing together (see 1 below). Then rotate strip 3 halfway around the finger to make 2 below. Rotate strip 3 another half turn to its original position, and rotate strip 2 halfway around the finger to make 3 below. Rotate strip 3 another half turn to make 4 below. Rotate strip 3 another half turn to make a full rotation, which makes strip 2 rotate a half turn to make a full rotation, which makes strip 1 move to its second position, giving 5 below. And so on until strip 1 has made a full rotation and the three rings have returned to position 1 below.

1. 3/4 hypo x 1/4 albino x 1/4 anerythristic = 3/64 moonglow
2. 3/4 hypo x 1/4 albino x 3/4 normal = 9/64 sunglow
3. 3/4 hypo x 3/4 normal x 1/4 anerythristic = 9/64 ghost
4. 3/4 hypo x 3/4 normal x 3/4 normal = 27/64 hypo
5. 1/4 normal x 1/4 albino x 1/4 anerythristic = 1/64 snow
6. 1/4 normal x 1/4 albino x 3/4 normal = 3/64 albino
7. 1/4 normal x 3/4 normal x 1/4 anerythristic = 3/64 anerythristic
8. 1/4 normal x 3/4 normal x 3/4 normal = 9/64 normal

In other words, each time any ring makes a full rotation, the next ring to the left also cycles one place in the same direction.

This technique becomes second nature pretty quickly, and the actual rings can be dispensed with. It is much easier than a three locus Punnett square.

Paul Hollander


   

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