Hardy-Weinberg. I need help setting up a problem
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Hardy-Weinberg. I need help setting up a problem

[From: ] [author: ] [Date: 11-04-27] [Hit: ]
According to H-W equation #1, frequency of A = 1 - q or 1 - 0.4 = 0.6.Let this equal p.So,......
In a population of 100 rock pink plants, 84 individuals have red flowers while 16 have white flowers. Assume that white petals are inherited as a recessive trait (a) and red petals as a dominant trait (A). What are the frequencies p and q? In the next generation, what will be the equilibrium genotypic frequencies?

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Set it up like this:

The recessive phenotype has a frequency of 16/100 or 0.16. Therefore, 0.16 have the genotype aa, and the frequency of a is sqrt(0.16) = 0.4. Let this equal q. According to H-W equation #1, frequency of A = 1 - q or 1 - 0.4 = 0.6. Let this equal p.

So, p= 0.6 and q = 0.4.

The genotypic frequencies will be:
AA = p↑2 = 0.36
Aa = 2pq = 0.48
aa = q↑2 = 0.16

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16 white flowers = (aa) individuals......or flowers w/e =P
16/100 = 0.16 = Q^2(squared)
so q = 0.4
given p+q=1 ; p must equal 0.6

calculating geno freq. in next gen. p^2+2pq+q^2=1
AA = (.6)(.6) = .36
Aa= 2(.6)(.4) = .48
aa= (.4)(.4) = .16

p.s. under HW equil allele freq. do not change over time. (think about it)
gg
1
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