pN, which is equal to (1-m)(pI) + (m)(pC). m = 600/800 = 0.75, which is the fraction of alleles from the mainland. pC = 0.60 and pI = 1.0, which are the frequencies of the A1 allele in the two populations. Thus, pN = (1 - 0.75)(1.0) + (0.75)(0.60) = 0.70
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