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1) In peas, a gene controls flower color such that R = purple and r =...

1) In peas, a gene controls flower color such that R = purple and r = white. In an isolated pea patch, there are 36 purple-flowering plants and 64 white-flowering plants. Assuming Hardy-Weinberg equilibrium, what is the frequency of the white allele (q) for this population?

2) In an insect population at HWE where 16% of the population displays a recessive phenotype for green eyes conferred by gg, while the rest have black eyes (and are thus GG or Gg), what is the frequency of the g allele (q), and the G allele (p)?

3) What is the frequency of the GG and Gg individuals in the population (the fraction of gg is 0.16 or 16%)?

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Answer #1

We know that according to Hardy-Weinberg equilibrium law,

(p+q) = 1 [Where, p= frequency of dominant allele and q= frequency of recessive allele]

(p2+2pq+q2) = 1 [Where, p2 = frequency of the homozygous dominant, 2pq= frequency of heterozygous, and q2= frequency of homozygous recessive]

Question 1:  In peas, a gene controls flower color such that R = purple and r = white. In an isolated pea patch, there are 36 purple-flowering plants and 64 white-flowering plants.

So, the genotype for the plants that have white flowers must be homozygous recessive i.e. rr.

Therefore, the frequency of the genotype rr = q2 = {64/(64+36)} = 64/100

So, q = 8/10 = 0.8

The frequency of the white allele for this population is 0.8 [Answer]

Question 2: In the insect population,16% of the population displays a recessive phenotype for green eyes conferred by gg, while the rest have black eyes (and are thus GG or Gg).

Therefore, 16% of the population are homozygous recessive i.e. gg.

So, q2=16/100

So, q=0.4

The frequency of the g allele is 0.4 [Answer]

Now, p=(1-q)=(1-0.4)=0.6

The frequency of the G allele is 0.6 [Answer]

Question 3:

According to Hardy-Weinberg equilibrium law, (p2+2pq+q2) = 1

The frequency of the genotype gg=q2=0.16

So, q=0.4 and p=(1-q)=(1-0.4)=0.6

So, the frequency of GG=p2=(0.6)2=0.36

The frequency of Gg=2pq=(20.60.4)=0.48

So, (p2+2pq)=(0.36+0.48)= 0.84

Therefore, the total frequency of the GG and Gg individuals in the population is 0.84

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