Respuesta :
Answer:
option E. [tex]\frac{4}{9}[/tex]
Explanation:
According to the question,
Allele frequency of the 'O' allele in the US population ,p = [tex]\frac{2}{3}[/tex]
Therefore, frequency for the rest of the patients, q = 1 - p = [tex]1 - \frac{2}{3} = \frac{1}{3}[/tex]
Hardy- Weinberg equilibrium condition states that in absence of the factors that can cause disturbance, the amount or extent of variation in genes in a population will not vary (constant) throughout the generations.
Therefore,
p + q = 1
In accordance to Hardy- Weinberg equilibrium eqn:
[tex]p^{2} + 2pq + q^{2} = 1[/tex] (1)
where,
[tex]p^{2}[/tex] = homozygous dominant
[tex]q^{2}[/tex] = homozygous recessive
2pq = heterozygous
Using eqn (1)
[tex](\frac{2}{3})^{2} + 2(\frac{2}{3})(\frac{1}{3}) + (\frac{1}{3})^{2} = 1[/tex]
[tex]\frac{4}{9} + \frac{4}{9} + \frac{1}{9} = 1[/tex]
Therefore, the frequency of people with blood type 'O' in the U.S. population is [tex]p^{2} = \frac{4}{9}[/tex]
