Genotype and Allele Frequencies for a Population in Hardy-Weinberg Equilibrium

Given Information:

- 16% of the population has a recessive trait
- Conditions for Hardy-Weinberg equilibrium are met

Step 1: Calculate Allele Frequencies

- Let p = frequency of dominant allele
- Let q = frequency of recessive allele
- p + q = 1 (total frequency of alleles in population)

Since the recessive trait is caused by the homozygous recessive genotype (aa), and assuming that the population is in Hardy-Weinberg equilibrium, we can use the equation:

q^2 = frequency of homozygous recessive genotype (aa)

- q^2 = 0.16 (given)
- q = √0.16 = 0.4
- p = 1 - q = 0.6

Therefore, the allele frequencies for the population are:

- Frequency of dominant allele (p) = 0.6
- Frequency of recessive allele (q) = 0.4

Step 2: Calculate Genotype Frequencies

Using the Hardy-Weinberg equation:

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

where:

- p^2 = frequency of homozygous dominant genotype (AA)
- 2pq = frequency of heterozygous genotype (Aa)
- q^2 = frequency of homozygous recessive genotype (aa)

Substituting the allele frequencies we calculated:

- p^2 + 2(0.6)(0.4) + 0.16 = 1
- p^2 + 0.48 + 0.16 = 1
- p^2 = 0.36
- p = √0.36 = 0.6

Therefore, the genotype frequencies for the population are:

- Frequency of homozygous dominant genotype (AA) = 0.36
- Frequency of heterozygous genotype (Aa) = 0.48
- Frequency of homozygous recessive genotype (aa) = 0.16

Conclusion:

If the population is in Hardy-Weinberg equilibrium, the genotype frequencies calculated above should remain constant from generation to generation, as long as there is no mutation, migration, selection, or random genetic drift. These genotype and allele frequencies can be used as a baseline for comparison with other populations or for predicting the likelihood of certain traits or disorders in offspring.