Question

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

Answer 1

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

Step-by-step 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%

Answer 2

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

Step-by-step 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%