Answer:
4% of the white population in the US are carriers of cystic fibrosis.
Step-by-step explanation:
According to the Hardy-Weinberg equilibrium equation, if q2 equals 1/2,500, then q equals .02, p equals .98, and 2pq equals approximately .04, or 4%.
Remember that the Hardy-Weinberg equilibrium equation is used when the population maintains its genetic variation constant.