03-08-2007, 10:17 AM
D. This question uses Hardy-Weinberg population genetics. If a population is in Hardy-Weinberg equilibrium (which usually means that the affected gene does not alter the ability to produce children either directly or indirectly and that the affected gene is spread evenly through the population), two equations govern the prediction of what happens:
p2 + 2pq + q2 = 1
and, p + q = 1
where, p is the frequency of having the dominant gene, and q is the frequency of having the recessive gene at the same locus.
The frequency of people in the population having the characteristic associated with the recessive gene is q2, while the frequency of being a heterozygote (carrier) for the characteristic is 2pq. In our case, q2 is 1/100, so q is 1/10. Since p+q=1, p must be 9/10. The carrier rate, 2pq, is 2 x 9/10 x 1/10 = 18/100 or 18 out 100 people.
p2 + 2pq + q2 = 1
and, p + q = 1
where, p is the frequency of having the dominant gene, and q is the frequency of having the recessive gene at the same locus.
The frequency of people in the population having the characteristic associated with the recessive gene is q2, while the frequency of being a heterozygote (carrier) for the characteristic is 2pq. In our case, q2 is 1/100, so q is 1/10. Since p+q=1, p must be 9/10. The carrier rate, 2pq, is 2 x 9/10 x 1/10 = 18/100 or 18 out 100 people.