Approximately one out of every 2,500 caucasians in the united states is born with the recessive disease cystic fibrosis. According to the hardy-weinberg equilibrium equation, approximately how many people are carriers?.

Respuesta :

The disease cystic fibrosis is autosomal recessive. The prevalence of an autosomal recessive disease in a population is determined by Hardy Weinberg Equilibrium.

How is the prevalence of cystic fibrosis determined?

The disease cystic fibrosis is autosomal recessive. Hardy Weinberg Equilibrium states that q2, which in this case equals 1/2500 and q = 1/50, is the frequency of an autosomal recessive disease in a population. P = 49/50 1 because the frequency of the two alleles (p & q) must be equal to 1.

How is Hardy Weinberg equilibrium calculated?

P2 + 2pq + q2 = 1 is the Hardy-Weinberg equation used to calculate genotype frequencies. Where "p2" denotes the frequency of the homozygous dominant genotype (AA), "2pq" denotes the frequency of the heterozygous genotype (Aa), and "q2" denotes the frequency of the heterozygous dominant genotype.

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