After graduation, you and 19 friends build a raft, sail to a deserted island, and start a new population, totally isolated from the world. Two of your friends carry(heterozygotes) the recessive allele c, which causes cystic fibrosis in homozygotes.
Assuming that the frequency of this allele does not change as the poulation grows, what will be the instance of cystic fibrosis on the island?
I'm thinking 1/4...but not sure..any ideas?
You neglect to give the gender of the 20 people. If they are all males or all females, there is no incidence, because they cannot reproduce.
In the first generation, there will also be no chance for being homozygous, if both of the carriers are of the same gender.
The male/female ratio is very important in determining the number of pairings available. What if there are only 2 males and 18 females, with the males being the carriers? What if two of the females in this ratio were the carriers?
I'm thinking that we need more information.
This is probably making the situation more confusing, but I hope it helps. Thanks for asking.
To determine the incidence of cystic fibrosis on the island, we need to consider the principles of population genetics. In this case, the recessive allele c causes cystic fibrosis only in homozygous individuals.
Given the initial information, where two friends carry the recessive allele c, we can assume that these individuals are heterozygous carriers. Let's denote the dominant allele as C and the recessive allele as c. The heterozygous carriers would have genotypes Cc.
To calculate the incidence of cystic fibrosis, we need to consider the chance of two carriers producing homozygous offspring (cc genotype). This depends on the probability of two heterozygous carriers (Cc) having a child with the recessive genotype (cc).
If we assume random mating, the probability of a heterozygous carrier (Cc) passing on the recessive allele c to their offspring is 1/2. So, the probability of two heterozygous individuals (Cc) having a child with the recessive genotype (cc) is (1/2) * (1/2) = 1/4.
Therefore, the incidence of cystic fibrosis on the island would be 1/4 if the two carriers have a child together. However, this assumes random mating without any factors affecting reproductive choices or genetic drift.
Keep in mind that this analysis only considers the first generation. To determine the long-term incidence of cystic fibrosis in subsequent generations, additional information about the population's reproductive choices, population growth, and genetic drift would be needed.