Answer:
cystic fibrosis births on the island would be approximately 16.95 times greater than on the original mainland.
Step-by-step explanation:
To calculate the incidence of cystic fibrosis on the island, we need to consider the Hardy-Weinberg principle. According to the principle, in a population where the allele frequencies do not change, the genotype frequencies can be predicted using the following equations:
p^2 + 2pq + q^2 = 1
where:
p^2 represents the frequency of homozygous dominant individuals (AA)
2pq represents the frequency of heterozygous individuals (Aa)
q^2 represents the frequency of homozygous recessive individuals (aa)
In this scenario, two out of the 20 friends (or 2/20 = 0.1) carry the recessive cf allele. This corresponds to the frequency of the recessive allele (q) in the population. Therefore, q = 0.1.
To find the frequency of the dominant allele (p), we subtract the recessive allele frequency from 1: p = 1 - q = 1 - 0.1 = 0.9.
Now we can calculate the incidence of cystic fibrosis (q^2) on the island:
q^2 = (0.1)^2 = 0.01
Therefore, the incidence of cystic fibrosis on the island would be 0.01 or 1%.
To determine the comparison with the original mainland, we need to calculate the frequency of cystic fibrosis births on the mainland. Given that the frequency of births with cystic fibrosis on the mainland is 0.059%, we can compare this with the incidence on the island:
Cystic fibrosis births on the island / Cystic fibrosis births on the mainland = 0.01 / 0.00059
This calculation shows that cystic fibrosis births on the island would be approximately 16.95 times greater than on the original mainland.