p_{N}, which is equal to (1-m)(p_{I}) + (m)(p_{C}). m = 600/800 = 0.75, which is the fraction of alleles from the mainland. p_{C }= 0.60 and p_{I }= 1.0, which_{ }are the frequencies of the A_{1 }allele in the two populations. Thus, p_{N }= (1 - 0.75)(1.0) + (0.75)(0.60) = 0.70