To calculate the frequencies related to sickle cell anemia, we can start by defining the given information:
- Total African-American population with the sickle cell trait = 2 million
- Frequency of the sickle cell trait in the population = 1 out of every 12
a. Frequency of the aa genotype:
- Since the trait is recessive, aa represents the sickle cell genotype.
- The aa genotype occurs when both alleles are recessive.
- The frequency of the aa genotype is calculated by squaring the fraction of individuals with the trait (1/12).
- Therefore, (1/12)^2 = 1/144 ≈ 0.00694 or 0.694%
b. Frequency of the Aa genotype:
- The Aa genotype is the carrier of the sickle cell trait.
- To find the frequency of carriers, we can use the complement of the aa genotype frequency.
- Frequency of Aa = 1 - Frequency of aa = 1 - 1/144 = 143/144 ≈ 0.99306 or 99.306%
c. Frequency of the AA genotype:
- The AA genotype does not have the sickle cell trait.
- Since the population only contains carriers (Aa) and individuals with the trait (aa), the frequency of AA = 0
d. Frequency of the dominant allele (A):
- The dominant allele (A) can be found using the frequencies of Aa and AA genotypes.
- Frequency of A = Frequency of Aa + 2 * Frequency of AA (to account for both alleles in the AA genotype).
- Frequency of A = 143/144 + 2 * 0 = 143/144 ≈ 0.99306 or 99.306%
e. Frequency of the recessive allele (a):
- Since a is the recessive allele, its frequency can be derived from the aa genotype frequency.
- Frequency of a = √(Frequency of aa) = √(1/144) = 1/12 = 0.0833 or 8.33%
f. Percentage of homozygous dominant (AA):
- As calculated earlier, the frequency of the AA genotype is 0, so the percentage of AA is 0%.
g. Percentage of heterozygous (Aa):
- The frequency of Aa was calculated as 143/144, which corresponds to 99.306% or 99.31% (rounded).
h. Percentage of homozygous recessive (aa):
- The frequency of the aa genotype was calculated as 1/144 or approximately 0.694%, representing the percentage of individuals with sickle cell anemia.
