In Hardy-Weinberg equilibrium, which equations relate allele and genotype frequencies?

In Hardy-Weinberg equilibrium, which equations relate allele and genotype frequencies?

• A. p + q = 1 and p^2 + 2pq + q^2 = 1
• B. p^2 + q^2 = 1
• C. p + q = 2 and p^2 + q^2 = 1
• D. p^2 + q^2 + 2pq = 3

Answer: (A)

Think about what Hardy-Weinberg gives you: there are two alleles, A and a, with frequencies p and q. The sum of allele frequencies must be 1, so p + q = 1. The genotypes in the population are AA, Aa, and aa, with frequencies p^2, 2pq, and q^2 respectively, which come from squaring the binomial (p + q). Since p + q = 1, the genotype frequencies add up to p^2 + 2pq + q^2 = (p + q)^2 = 1. So the two relationships that relate allele and genotype frequencies are p + q = 1 and p^2 + 2pq + q^2 = 1. If you check the other forms: p^2 + q^2 = 1 would ignore the heterozygote class, which isn’t generally true unless one allele is absent (2pq = 0). Saying p + q = 2 can't be correct because allele frequencies must sum to 1. And p^2 + q^2 + 2pq = 3 would misstate the total, since genotype frequencies sum to 1, not 3.

Think about what Hardy-Weinberg gives you: there are two alleles, A and a, with frequencies p and q. The sum of allele frequencies must be 1, so p + q = 1. The genotypes in the population are AA, Aa, and aa, with frequencies p^2, 2pq, and q^2 respectively, which come from squaring the binomial (p + q). Since p + q = 1, the genotype frequencies add up to p^2 + 2pq + q^2 = (p + q)^2 = 1. So the two relationships that relate allele and genotype frequencies are p + q = 1 and p^2 + 2pq + q^2 = 1.

If you check the other forms: p^2 + q^2 = 1 would ignore the heterozygote class, which isn’t generally true unless one allele is absent (2pq = 0). Saying p + q = 2 can't be correct because allele frequencies must sum to 1. And p^2 + q^2 + 2pq = 3 would misstate the total, since genotype frequencies sum to 1, not 3.