Determine the frequencies of each genotype by taking each number over the total population.

-test if it follows p + q = 1 and p2 + 2pq + q2 = 1.

# What is the expected number of individuals belonging to the three respective phenotypes/genotypes? Is the population in HW equilibrium?

Genetic testing allows for investigators to distinguish between the three genotypes related to the sickle cell condition. The results from an analysis of members of a small fishing village in South Africa are given below:

Genotype/Phenotype/Number:

Normal-HbAHbA (109)

Sickle cell trait-HbAHbS (12)

Sickle cell disease- HbSHbS (23)

What is the expected number of individuals belonging to the three respective phenotypes/genotypes? Is the population in HW equilibirum? Why or why not?

Genotype/Phenotype/Number:

Normal-HbAHbA (109)

Sickle cell trait-HbAHbS (12)

Sickle cell disease- HbSHbS (23)

What is the expected number of individuals belonging to the three respective phenotypes/genotypes? Is the population in HW equilibirum? Why or why not?

## 2 Answers

- 00
- Genotype/Phenotype/Number:

Normal-HbAHbA (109)

Sickle cell trait-HbAHbS (12)

Sickle cell disease- HbSHbS (23)

There are 144 individuals and so 288 alleles in this sample. Use these numbers to calculate the frequencies of the two alleles:

Frequency HbA = 2(109) + 12 / 288 = 0.799 = p

Frequency HbS = 12 + 2(23) / 288 = 0.201 = q

Now, use these observed allele frequencies to calculate the expected numbers of each genotype in the population:

Number of HbAHbA:

(0.799)^2 X 144 = 91.8

Number HbAHbS:

2(0.799)(0.201) X 144 = 46.2

Number HbS HbS:

0.201^2 X 144 = 5.8

These expected numbers with each genotype in the population are very different than the observed numbers of each genotype. Therefore, the population IS NOT in H-W equilibrium.00

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