Calculating allele frequency, part II

In a study of 989 members of the population from example 1, it was found that 11 people had the disease. Calculate the frequency of the recessive allele t.

Calculating genotype frequency

Using the information from example 1, consider the following: Create this table.

Suppose we had a population of 500 people, in which 9% were albino

(albinism is a recessive characteristic)

How many individuals in this population are heterozygous?

Using the equations: p + q = 1 and p² + 2pq + q² = 1

9% tells you the recessive phenotype, which is the q² value, the rest is logic and math.

• If q² = 0.09 ⇒ q = 0.3 (√ 0.09)
• If q = 0.3 ⇒ p = 0.7 (p + 0.3 = 1)
• If p = 0.7 ⇒ p² = 0.49 (0.7²)
• 2pq = 0.42 ⇒ (0.49 + 2pq + 0.09 = 1)

Substituting these numbers for frequencies and applying them to the original population shows that:

• 49% of people are homozygous dominant (AA), which is 245 individuals (0.49 × 500)
• 9% of people are homozygous recessive (aa), which is 45 individuals (0.09 × 500)
• 42% of people are heterozygous (Aa), which is 210 individuals (0.42 × 500)