The recessive allele of a gene causes cystic fibrosis. For this gene among Caucasians, p = 0.98. If a Caucasian population is in Hardy-Weinberg equilibrium with respect to this gene, what proportion of babies is born homozygous recessive and therefore suffers cystic fibrosis?
A) (0.02)2 = 0.0004
B) 0.02
C) (0.98)2 = 0.9604
D) 2(0.02 × 0.98)= 0.0392
Answer: A
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