Final answer:
To determine the frequency of allele R in a Hardy-Weinberg equilibrium population with a known frequency of heterozygotes (0.18), you can solve the quadratic equation 2p(1-p) = 0.18. This will yield two possible solutions for the frequency of allele R, from which the correct one can be chosen based on allele frequencies being between 0 and 1.
Step-by-step explanation:
Assuming a population is in Hardy-Weinberg equilibrium and knowing that the frequency of heterozygotes (Rr) is 0.18, we can compute the frequency of the R allele. In Hardy-Weinberg, the genotype frequencies are represented as p² for homozygous dominant (RR), 2pq for heterozygous (Rr), and q² for homozygous recessive (rr), where p is the frequency of the dominant allele R and q is the frequency of the recessive allele r.
Given the frequency of heterozygotes 2pq is 0.18, and knowing that p + q = 1, there are two unknowns in two equations, which we can solve with algebra. For example, if we assume p is the frequency of allele R, then we would set up the equation 2p(1-p) = 0.18 to find the value of p. By solving the quadratic equation p = 1 ± √((1 - 0.18) / 2), you can find p, which is the frequency of the R allele.
Since this is a quadratic equation, there would be two potential solutions for p and hence two possibilities for the frequency of allele R. It is important to remember that both p and q should be between 0 and 1, as they are frequencies. After calculating, we would pick the solution in that range for both p and q.