BI111 Lecture Notes - Lecture 3: Coevolution, Frequency-Dependent Selection, Anemia
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F(b) = q = # b alleles/total # alleles = 147+147+93/788 = 387/788 = 0. 49. F(b) = q = # b alleles/total # alleles = 154+154+93/788 = 401/788 = 0. 51 bb 147. Bb 154 total 394x2 = 788 alleles total q + p = 1. 0, so 1. 0 q = p and vice versa. When random mating occurs, the frequency of p and q is not changed, since no cards are being created or destroyed. When random mating occurs, the genotypes change there were more bb and less bb and bb. F(bb) = f(b) x f(b) = p x p = (0. 49)(0. 49) = 0. 24. F(bb) = 2 x (f(b) x f(b)) = 2 x p x q = 2(0. 49)(0. 51) = 0. 50. F(bb) = f(b) x f(b) = q x q = (0. 51)(0. 51) = 0. 26. When added, all three genotype possibilities must equal 1.
