The hereditary form of diabetes is due to the recessive autosomal gene. A survey of the population of a large isolated island showed that

The hereditary form of diabetes is due to the recessive autosomal gene. A survey of the population of a large isolated island showed that 1% of people living on this island suffer from congenital diabetes. What proportion (%) among the healthy population of the island are heterozygous carriers of this disease?

We introduce the notation of genes:
A – normal carbohydrate metabolism;
a – congenital diabetes mellitus.
– Let p be the frequency of the dominant allele in a given population, and q be the frequency of the recessive allele.
Then p + q = 1 (or 100%).
The trait under investigation is autosomal, i.e. male and female gametes carrying the allele A are formed in people inhabiting the island with the same frequency (probability) – p. Sperm and egg cells carrying the allele a are formed with a frequency (probability) of q. A random fusion of gametes leads to the appearance of descendants of three genotypic classes: AA, Aa and aa.
Since gametes A are formed with a frequency p, gametes a with a frequency q, the proportion of people with the AA genotype is p2, the Aa genotype is 2pq, and the aa genotype is q2.
– The condition says that the proportion of people suffering from a congenital form of diabetes is 0.01 (or 1%), i.e. q2 = 0.01. Therefore, q = √ ¯ 0.01 = 0.1 (or 10%). Then p = 1 – 0.1 = 0.9 (or 90%).
The healthy population of the island is 0.99 (or 99%).
Moreover, the carriers of the disease (Aa) are: 2pq = 2 × 0.9 × 0.1 = 0.18 (or 18%).
This means that the proportion of carriers among the healthy population is 0.18: 0.99 × 100% ≈ 18.2%.
Answer: the proportion of carriers of the congenital form of diabetes among the healthy population of the island is 18.2%.