Final answer:
In a population at Hardy-Weinberg equilibrium with 160 Rh+ out of 200 individuals, the frequency of the Rh+ allele is approximately 0.5528 and the Rh- allele is 0.4472.
Step-by-step explanation:
In a population that is in Hardy-Weinberg equilibrium, if 160 out of 200 individuals are Rh+, we can calculate the allele frequencies. First, we must determine the frequency of the Rh+ phenotype, which is the proportion of Rh+ individuals in the population. In this case, it is 160/200, which equals 0.8 or 80%. Let p represent the dominant allele (Rh) and q represent the recessive allele (Rh-). The frequency of the Rh+ phenotype (p2 + 2pq) is 0.8, and since we are looking for individuals who are homozygous dominant or heterozygous, we have to consider both p2 and 2pq.
The frequency of the homozygous recessive phenotype (q2) is then 1 - 0.8 = 0.2 or 20%, which is the proportion of Rh- individuals. To find q, we take the square root of 0.2, which gives us q = 0.4472. Subsequently, to find p, we use the equation p = 1 - q, which gives us p = 1 - 0.4472 = 0.5528. Therefore, the frequency of the Rh allele (p) is approximately 0.5528 and the frequency of the Rh- allele (q) is 0.4472.