So __ 9% of the population or 9 out of 100 people have DRY earwax (ww). That means that 91 people out of 100 or 91% have wet earwax right? What 2 pieces of the H-W equation will add up to the 91% (select 2) 2pq p q q^2 p^2

To determine the two pieces of the Hardy-Weinberg equation that add up to 91%, we need to understand the Hardy-Weinberg equilibrium and the allele frequencies.<br /><br />The Hardy-Weinberg equilibrium states that the frequencies of alleles and genotypes in a population remain constant from generation to generation, assuming no mutations, migration, or other evolutionary forces.<br /><br />In this case, we are given that 9% of the population has DRY earwax (ww). This means that the frequency of the recessive allele (q) is 0.09.<br /><br />To find the frequency of the dominant allele (p), we can use the fact that the sum of the frequencies of the dominant and recessive alleles must equal 1:<br /><br />p + q = 1<br /><br />Substituting the value of q, we get:<br /><br />p + 0.09 = 1<br /><br />Solving for p, we find:<br /><br />p = 1 - 0.09 = 0.91<br /><br />Now, we can calculate the frequencies of the heterozygous genotype (2pq) and the homozygous dominant genotype ($p^{2}$):<br /><br />2pq = 2 * 0.91 * 0.09 = 0.1638<br />$p^{2}$ = (0.91)^2 = 0.8281<br /><br />Therefore, the two pieces of the Hardy-Weinberg equation that add up to 91% are:<br /><br />2pq = 0.1638<br />$p^{2}$ = 0.8281