Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A recessive autosomal condition affects in every births in a large population. The population mates at random and is in Hardy-Weinberg equilibrium for this gene, with alleles (normal) and (condition). A town in this population has people.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. Use and the genotype frequencies , and .
(a) , so and .
(b) Carriers: .
(c) carriers.
(d) Among healthy people (a fraction of the population), the carrier share is . This is a little more than because the affected people have been left out.
(e) Both parents must be carriers, with probability , and then the child is affected with probability . So the risk is , which is in births. A population with no knowledge of the parents would give in , so two healthy parents carry a lower risk than the population as a whole.
Check. Listing every pair of healthy genotypes ( or for each parent) with its frequency and adding the chance of an child gives the same in when exact fractions are used.
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