41. In a population, 99% of the people do not show a homozygous recessive disease. What percent of the population carry the bad gene?
If 99% of the population is not homozygous recessive, then 1% of the population is. To solve, set q2 = .01, q is = .1
p + q = 1
p = 1- (q)2 |
p2 + 2pQ + Q2 = 1 |
p = 1- (.01)2 |
(.9)2 + 2 (.9)(.1) + .01 =1 |
p = 1-.1 |
(.81) + 2(.09) + .01 =1 |
p = .9 |
p2 = .81, 2pq = .18 and q2 = .01 |
The answer to the problem is 18% carry the recessive gene and 81% are free of the 'bad' gene. |
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