Calculating the frequency of dominant allele in a population

Given information:

- The population has 2 times heterozygotes as homozygous dominant.

Step 1: Determine the genotypes of the population

Let's assume the frequency of the dominant allele in the population is p, and the frequency of the recessive allele is q.

According to the Hardy-Weinberg equilibrium equation:

p^2 + 2pq + q^2 = 1

where:

- p^2 represents the frequency of the homozygous dominant genotype (AA)
- 2pq represents the frequency of the heterozygous genotype (Aa)
- q^2 represents the frequency of the homozygous recessive genotype (aa)
- 1 represents the total frequency of all genotypes in the population

Given that the population has 2 times heterozygotes as homozygous dominant, we can write:

2pq = 2p^2

Simplifying this equation, we get:

q = p/2

Substituting q in the Hardy-Weinberg equation, we get:

p^2 + 2p(p/2) + (p/2)^2 = 1

Simplifying this equation, we get:

2p^2 + p^2/4 = 1

Multiplying both sides by 4, we get:

8p^2 + p^2 = 4

Simplifying this equation, we get:

9p^2 = 4

Taking the square root of both sides, we get:

p = 2/3 or 0.67

Therefore, the frequency of the dominant allele in the population is 0.67.

Step 2: Check the answer

To check whether the answer is correct, we can calculate the frequency of the homozygous dominant genotype:

p^2 = (2/3)^2 = 4/9 or 0.44

And the frequency of the heterozygous genotype:

2pq = 2(2/3)(1/3) = 4/9 or 0.44

And the frequency of the homozygous recessive genotype:

q^2 = (1/3)^2 = 1/9 or 0.11

Adding up these frequencies, we get:

0.44 + 0.44 + 0.11 = 0.99

which is close to 1, confirming that the answer is reasonable.

Therefore, the correct answer is option D, 0.8.

I think 0.2 is the correct answer