-- card: 51682 from stack: in
-- bmap block id: 126625
-- flags: 0000
-- background id: 2665
-- name: 


-- part 1 (field)
-- low flags: 00
-- high flags: 0002
-- rect: left=276 top=101 right=122 bottom=327
-- title width / last selected line: 0
-- icon id / first selected line: 0 / 0
-- text alignment: 0
-- font id: 3
-- text size: 12
-- style flags: 0
-- line height: 16
-- part name: p


-- part 2 (button)
-- low flags: 00
-- high flags: 8003
-- rect: left=342 top=123 right=145 bottom=442
-- title width / last selected line: 0
-- icon id / first selected line: 0 / 0
-- text alignment: 1
-- font id: 0
-- text size: 12
-- style flags: 0
-- line height: 16
-- part name: ACTIVATE
----- HyperTalk script -----
on mouseUp
  global pee
  set numberFormat to "0.0000"
  put the value of (card field p) into pee
  put the value of (1 - card field p) into card field q
  put the value of (pee^2) into card field pp
  put the value of (pee * (1-pee)) into card field pqa
  put the value of (pee * (1-pee)) into card field pqb
  put the value of ((1-pee)^2) into card field qq
end mouseUp



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-- high flags: 0000
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-- line height: 16
-- part name: pp


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-- low flags: 00
-- high flags: 0000
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-- icon id / first selected line: 0 / 0
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-- part name: pqb


-- part 5 (field)
-- low flags: 00
-- high flags: 0000
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-- title width / last selected line: 0
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-- part name: pqa


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-- low flags: 00
-- high flags: 0000
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-- title width / last selected line: 0
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-- part name: qq


-- part 7 (button)
-- low flags: 00
-- high flags: 8003
-- rect: left=341 top=71 right=92 bottom=424
-- title width / last selected line: 0
-- icon id / first selected line: 0 / 0
-- text alignment: 1
-- font id: 0
-- text size: 12
-- style flags: 0
-- line height: 16
-- part name: RESET
----- HyperTalk script -----
on mouseUp
  delete line 1 of card field p
  delete line 1 of card field q
  delete line 1 of card field pp
  delete line 1 of card field pqa
  delete line 1 of card field pqb
  delete line 1 of card field qq
  get the location of card field p
  click at it
end mouseUp



-- part 8 (button)
-- low flags: 00
-- high flags: 8003
-- rect: left=205 top=315 right=337 bottom=256
-- title width / last selected line: 0
-- icon id / first selected line: 0 / 0
-- text alignment: 1
-- font id: 0
-- text size: 12
-- style flags: 0
-- line height: 16
-- part name: NEXT
----- HyperTalk script -----
on mouseUp
  go to next card
end mouseUp



-- part 9 (button)
-- low flags: 00
-- high flags: 8003
-- rect: left=149 top=314 right=337 bottom=199
-- title width / last selected line: 0
-- icon id / first selected line: 0 / 0
-- text alignment: 1
-- font id: 0
-- text size: 12
-- style flags: 0
-- line height: 16
-- part name: PREV.
----- HyperTalk script -----
on mouseUp
  go back
end mouseUp



-- part 10 (field)
-- low flags: 00
-- high flags: 0002
-- rect: left=276 top=158 right=179 bottom=327
-- title width / last selected line: 0
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-- part name: q


-- part contents for background part 1
----- text -----
POPULATION GENETICS

-- part contents for background part 2
----- text -----
The emphasis in population genetics is how gene flow through a population from generation to generation is affected by both external and internal selection factors. A key analytical tool is the Hardy-Weinberg Law which in its simplest form can be seen at the right in diagram A.  It represents the relationship between gene and phenotypic frequencies in a population for a given set of genes. You will note that the expression describes the gene frequency in a Punnett Square. One can test this relationship and learn a great deal about the frequency of genotypes as a function of gene frequencies by running the little exercise on the right several times. Place in the "p" gene frequency box a number between 0 and 1. Then hit the activate button. Run the exercise several times with different p values and then answer the following questions. 
   
  1. p + q always equal what?
  Answer:


  2. Homozygous dominant genotypes in a population bear what relationship to the dominant gene frequency?
  Answer:



  3. Heterozygous genotypes bear what relationship to the dominant and recessive gene frequencies?
  Answer:



BELOW THIS POINT ARE THE ANSWERS. SCROLL DOWN THE WINDOW.

1. p + q =1
2. Homozygous dominant genotypes are the square of the dominant gene frequency. Homozygous recessive genotypes are the square of the recessive gene frequency. 
3. Heterozygous genotypes are twice the product of the dominant times recessive gene frequencies.

Although these calculations are quite straight forward, they can become difficult and confusing if continued for several generations when, for example, the recessive gene was being selected against by natural selection factors. Go to the next card for help in determining these changes.


-- part contents for background part 8
----- text -----
58

-- part contents for card part 10
----- text -----
0.8000

-- part contents for card part 1
----- text -----
.2

-- part contents for card part 3
----- text -----
0.0400

-- part contents for card part 5
----- text -----
0.1600

-- part contents for card part 4
----- text -----
0.1600

-- part contents for card part 6
----- text -----
0.6400