**Question:**

*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 (that is, are heterozygous for) the recessive cf allele, which in homozygotes causes cystic fibrosis. Assuming that the frequency of this allele does not change as the population grows, what will the incidence of cystic fibrosis on your island be?*

**Answer:**

To answer the question, you first need to convert the population's genotypes into alleles. (Call 'C' the dominant, normal allele, and 'c' the recessive, potentialy cystic-fibrosis-causing allele.)

In other words, there are 20 people, so there will be 40 'C' or 'c' alleles. Two friends are heterozygous for the cystic firbosis allele, so they must both be Cc, giving us a total of 2 'c' alleles in the population. The remainder (38 alleles) must all be C.

So the frequency of the recessive allele will be 2/40. i.e. q = 0.05.

Lastly, we are asked for the incidence of cystic fibrosis. Cystic fibrosis is recessive - it only occurs if the individual happens to have TWO copies of the 'c' allele. In other words, the only homozygous recessive individuals will be affected. Recall that the homozygous recessive genotype is represented by the Hardy-Weinberg equation as q

^{2}.

We know that q = 0.05, so q

^{2}= 0.0025

And that's the incidence. You could also write it more accessibly by saying that the incidence is 2.5 in 1000.

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