For example,albinism occurs in 1/20,000 individuals.Let's say that this condition is due to a recessive allele a of a single gene that is in H-W equilibrium. f/a)=5×10-5=g2 g=1V5×10-5=7×10-3 f(A/a)=2pg≈2g=1.4×10-2 We will now calculate the fraction of alleles for albinism that are in individuals that are homozygous for albinism. Number of alleles in homozygotes≈2×N(q2) N=population size Number of alleles in heterozygotes N(2q) The ratio is: 2×N(q2) N(2q) =q Thus,for albinism(since q=7 x 10-3)the fraction of alleles in homozygotes is 7 x 10-3.That is,>99%of the alleles are in heterozygotes.For example, albinism occurs in 1/20,000 individuals. Let's say that this condition is due to a recessive allele a of a single gene that is in H-W equilibrium. f(a/a) = 5 x 10-5 = q2 q = 5 x 10-5 = 7 x 10-3 f (A/a) = 2pq ≈ 2q = 1.4 x 10-2 We will now calculate the fraction of alleles for albinism that are in individuals that are homozygous for albinism. Number of alleles in homozygotes ≈ 2 x N (q2) N = population size Number of alleles in heterozygotes ≈ N (2q) ) The ratio is: 2 x N (q2 = q N (2q) Thus, for albinism (since q = 7 x 10-3) the fraction of alleles in homozygotes is 7 x 10-3. That is, > 99% of the alleles are in heterozygotes