In a population of 2500, how many babies would you expect to have cystic fibrosis, a homozygous recessive condition, if the frequency of the dominant allele is 0.9 and the population is at Hardy–Weinberg equilibrium?
a. 0.9 × 2500 = 2250
b. 2 × 0.9 × 0.1 × 2500 = 450
c. 0.9 × 0.1 × 2500 = 225
d. 0.1 x 0.1 x 2500 = 25