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?

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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?

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Answer 1 · 2017-03-17T05:04:17

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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:

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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).

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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?

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