Approximately 1 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?

Question

Victoria Angeli · Accepted Answer

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1 out of 2500 are homozygous individuals, expressing a recessive character (cystic fibrosis is inherited as autosomal recessive disease ).

According to Hardy Weinberg principle, recessive individuals are represented by #q^2# as in the following:

()

Hence, in this case
#q^2# = 1/2500, hence #q# = 1/50 = 0.02

So #p# = 1 - #q# = #1 - 0.02# = 0.98

#2pq# represents frequency of carriers in the population, which is
#2 xx 0.98 xx 0.02# = 0.0392 = 392/10000 = 3.92/100

Frequency of carriers is 3.92 in every hundred individuals. So, when about 98 individuals in a population of 2500 are carriers, there is possibility of birth of 1 affected individual (in 2500).

Julian Aguilar · Answer

undefined Approximately 1 in 25 Caucasians in the United States are carriers of the cystic fibrosis gene.