In a population of mendel's garden peas, the frequency of dominant yellow-flowered plants is 50%. the population is in hardy-weinberg equilibrium. what is the frequency of the homozygous recessive genotype in the population? 0.71 0.25 0.5 the frequency cannot be determined from the data provided.

The correct answer is:

0.25

Explanation:

Since, the frequency of dominant yellow-flowered plants is 50% (i.e. 50/100 = 0.5) therefore, the frequency of dominant allele is 0.5 and recessive allele is 0.5 .

Lets say dominant allele is P and recessive allele is q.

According to the Hardy-Weinberg equilibrium, the frequency of dominant yellow-flowered plants is determined by the following formula:

2Pq

Substituting the frequency of dominant and recessive allele in the above formula: 2(0.5)(0.5) = 0.5

Similarly, according to the Hardy-Weinberg equilibrium, the frequency of  homozygous recessive genotype in the population is determined by the following formula:

q2

Substituting the frequency of dominant and recessive allele in the above formula: (0.5)2 = (0.5)(0.5) = 0.25

In this way, the frequency of homozygous recessive allele is calculated with the help of Hardy-Weinberg equilibrium.

i believe the answer is a hope it and is correct