Answer:
Explanation:
According to the exercise we can infer that for the Alu insert:
p: positive allelic frequency
q: negative allelic frequency
maintaining that the population is in equilibrium we can carry out the following formula
p + q = 1 and pp + 2pq + qq = 1
looking for the genotype frequency we clear and obtain the following data
genotype frequency of 2pq = 436/1000 = 0.436
The genotype frequency of qq = 102/1000 = 0.102
This is how we look now:
number of positive people for Alu = 1000- (436 + 102) = 1000- 538 = 46
In this way it is resolved that:
genotypic frequency of pp = 462/1000 = 0.462
p = 0.462 + (0.436 / 2) = 0.462 + 0.218 = 0.680
q = 0.102+ (0.436 / 2) = 0.102 + 0.218 = 0.320
According to the exercise carried out it is deduced:
The value of p in the population is = 0.68
We conclude that our prognosis showing a homozygous positive genotype is: 0.462
