Haardy Weinberg: 1= q2+2qp+p2

Where q2= homozygote for the disease gene

2pq= heterozygote for the gene

In this inappropriately high prevalence of the gene you have to take into consideration p (p doesn't equal 1 here, instead we use 1=p+q and p will be 0.8) and q2 (severly affected and homozygotes for the autosomal dominant gene which are usually discarded due to their insignificantly small number)

So instead of calculating the prevalence of an autosomal domnant disease by 2q, we will use : q2+2pq= 0.2square + 2 x 0.2 x 0.8 = 36

So in a population of 100, 36 will be diseased and 64 will be healthy

-false +ve rate is 1/16, so false positives are 1/16 x 64 = 4

-false -ve rate is 0.0 (FN= 0.0) so true positive is 36

PPV= 36/(36+4) =0.9

First of all this disease is highly prevelant that it affects 36% of the population

It took me 15 minutes to figure it out so I would leave it in the exam...