Final answer:
In a Hardy-Weinberg equilibrium, the frequency of individuals homozygous for the recessive allele, given one allele frequency of 0.6, is 0.16 or 16%.
Step-by-step explanation:
If the frequency of one allele is 0.6 (representing p, the dominant allele), then the frequency of the other allele q (the recessive allele) is 0.4 (since p + q = 1). To find the frequency of individuals homozygous for the recessive allele (represented as q² in the Hardy-Weinberg equation), we must square the frequency of the recessive allele q. Therefore, q² = 0.4 × 0.4 = 0.16, so the frequency of individuals homozygous for the recessive allele is 16% if the gene is in Hardy-Weinberg equilibrium.