A disease is due to a single autosomal recessive allele expr

A disease is due to a single autosomal recessive allele expressed according to classic Mendelian rules. In the following pedigree only individual 1 presents with the disease. Individual 2 has no family history of the disease as we can assume is homozygous dominant for the disease gene. 1/4 randomly selected individuals in your study population is a carrier for the disease. Given that individuals 5 and 6 are unaffected by the disease what is the probability that individual 4 is a carrier?

Solution

Let the diseased allele be represented by \'r\'; then genotype of individual 1 will be \'rr\'. Genotype of individual 2 will be \'RR\'. Individual 3 who is the offspring of individual 1 and 2 will definitely be a carrier withe the genotype \'Rr\'.

Individual 5 and 6 do not have the disease. This means there genotype is either \'RR\' or \'Rr\'.

1/4 population is a carrier for this disease.

If individual 4 is a carrier with Rr genotype, it will yield RR: Rr: rr in 1:2:1. ‘Rr’ becomes 50%.

On the other hand if individual 4 is homozygous dominant, then it will yield RR: Rr in 1:1. Here also “Rr” becomes 50%.

But generation 3 is shown to be 100% free of disease. This means that individual 4 is homozygous dominant and not heterozygous. The chances of individual 4 to be heterozygous are: 1/2 (for \'r\' from individual 3) * 1 (for R from individual 4) = 1/2. There are 50% chances of heterozygosity in individual 4.