If p is 0.6 in a Hardy-Weinberg population, the frequency of heterozygotes is:

1. The equilibrium equation gives each genotype its own term. 2. The frequency of pq individuals is 2pq. 3. If p is 0.6 then q is 0.4, since the two allele frequencies sum to one. 4. So 2pq is 2 multiplied by 0.6 multiplied by 0.4. 5. That gives 0.48 of the population as heterozygotes. _Source: OpenStax Biology 2e (CC BY 4.0), Ch 19 'The Evolution of Populations', sections 19.1-19.3_