Hardy weinburg 1 question?

Hardy weinburg equation is p^2+2pq+q^2=1, and also p+q=1. So zygous recessive phenotype aa is represented by q^2. Take the square root of .27 (approximately .519). This is the frequency of allele a. To get allele A, subtract this from one and you get roughly .481, or 48.1%. Double check with a real calculator to be sure