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) 0.022 = 0.0004
B) 0.982 = 0.9604
C) 0.02
D) 2(0.02 × 0.98) = 0.0392
Correct Answer:
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