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% is. Set q2 = .01
|
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 |
(.81) = P2 .18 = 2PQ and Q2 = .01 |
| The answer to the problem is 18% carry the recessive gene and 81% are free of the 'bad' gene. | |