The proportion of the population that should be heterozygous HbAIHbS is 0.40.
In a population that is in H-W equilibrium, the allelic frequencies in a locus are represented as p and q. Assuming a diallelic gene,
The allelic frequencies in a population are
• p is the dominant allele frequency,
• q is the recessive allele frequency.
The genotypic frequencies after one generation are
• p² (H0m0zyg0us dominant genotypic frequency),
• 2pq (Heter0zyg0us genotypic frequency),
• q² (H0m0zyg0us recessive genotypic frequency).
Population in H-W equilibrium
If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.
The addition of the allelic frequencies equals 1
p + q = 1.
The sum of genotypic frequencies equals 1
p² + 2pq + q² = 1
Thus, According to this framework, and following the problem statement, the frequency of individuals with the heterozygous HbAIHbS disease is 0.40.
To learn more about Hardy-Weinberg equilibrium click here:
https://brainly.com/question/16823644
#SPJ1
