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I'll get it started, but this is a complicated question. Assuming this is an autosomal trait, and assuming it only has one gene, and assuming it causes death before reproductive age: In order for someone to develop the phenotype, both parents have to be carriers. So, for the 1 out of 100,000 child born with the disease, you had to have 2 carriers mate. But there were more than 2 carriers in the 100,000. How many carriers do you have to have for 2 of them to mate? For that, you'd have to know how many people out of a population of 100,000 actually mate in the first place. You'd also have to know the average number of children resulting from a union. Let's say 75% of people mate, and the average number of children is 2. So of 100 people, 75 mate. But each union involves 2 people, so there are actually 37.5 pairings, producing 37.5 x 2 = 75 children. Now, assuming a frequency of carriers of F, what is the likelihood of 2 carriers mating? Let's say that everybody mates. So out of 100 people, there are 50 pairings. So the probability of mating with a carrier is approximately F.

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Q: A very severe genetic sickness occurs in about 1 in every 100000 individuals Calculate the frequency of carriers for this recessive disease?
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Related questions

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I don't know and don't care


Why are heterozygous individuals called carriers for non-sex-linked and X-linked recessive patterns of inheritance?

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