In a population of mendel's garden peas, the frequency of the dominant a (yellow flower) allele is 80%. let p represent the frequency of the a allele and q represent the frequency of the a allele. assuming that the population is in hardy-weinberg equilibrium, what are the genotype frequencies?
a) 16% aa, 40% aa, 44% aa
b) 75% aa, 15% aa, 10% aa
c) 64% aa, 32% aa, 4% aa
d) 80% aa, 10% aa, 10% aa
e) 50% aa, 25% aa, 25% aa

C) 64% AA, 32% Aa, 4% aa

Explanation:

The frequency of dominant allele A is = 80% = 0.8

Since the population is in Hardy-Weinberg equilibrium, p+ q =  1

Here, p= frequency of dominant allele

q = frequency of recessive allele

Given, p= 0.8

So, q = 1-p = 1-0.8= 0.2

Frequency of genotype AA = p2 = 0.8 x 0.8 = 0.64 = 64%

Frequency of genotype aa = q2 = 0.2 x 0.2 = 0.04 = 4%

Frequency of genotype Aa = 2pq = 2 x 0.8 x 0.2 = 0.32 = 32%

C) 64% AA, 32% Aa, 4% aa

Explanation:

got it right on test and bc

The frequency of dominant allele A is = 80% = 0.8

Since the population is in Hardy-Weinberg equilibrium, p+ q =  1

Here, p= frequency of dominant allele

q = frequency of recessive allele

Given, p= 0.8

So, q = 1-p = 1-0.8= 0.2

Frequency of genotype AA = p2 = 0.8 x 0.8 = 0.64 = 64%

Frequency of genotype aa = q2 = 0.2 x 0.2 = 0.04 = 4%

Frequency of genotype Aa = 2pq = 2 x 0.8 x 0.2 = 0.32 = 32%

true, i don't really know what you're asking but from the info you provided im guessing the answer is true.

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