Hardy weinberg question?

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Hardy weinberg question?

in a poplulation at genetic equilibrium, the frequency of allele A is 0.5.

a) what is the frequency of the homozygous dominnant of genotype (AA)?

b) what is frequency of the heterozygous genotype (Aa)?

2007-02-12 09:54:44 · 2 answers · asked by atrankid 2 in Science & Mathematics ➔ Biology

2 answers

AA = p^2 = 0.5*0.5 = 0.25

Aa = 2pq = 2 * 0.5 * 0.5 = 0.50 (because p + q = 1)

2007-02-12 09:59:23 · answer #1 · answered by ecolink 7 · 0⤊ 0⤋

a) .25
b).5

definitely...

2007-02-12 09:58:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋