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%

answer : haversian

[explanation: -(compact bone is organized as parallel columns, known as haversian systems, which run lengthwise down the axis of long bones. these columns are composed of lamellae, concentric rings of bone, surrounding a central channel, or haversian canal, that contains the nerves, blood vessels, and lymphatic system of the bone)]

answer///multiple alleles;