Practice question

When nine percent of population exhibits recessive phenotype, q squared equals 0.09, so q equals square root of 0.09 equals 0.3 and p equals 0.7 by complement. Carrier frequency follows Hardy-Weinberg heterozygote formula two pq equals two times 0.7 times 0.3 equals 0.42, about forty-two percent of population. This illustrates classic application: recessive phenotype incidence reveals q directly, then heterozygote frequency calculated. Because carriers harbor recessive allele in heterozygous state without disease, they are more common than affected homozygotes when q is moderate, guiding screening programs.