Question

Alu element at the PV92 locus on Chromosome 16. In this population, 436 people have a heterozygous genotype and 102 people have a homozygous negative genotype (meaning that neither copy of the PV92 has an Alu element). What is the value of q in this population (what is the proportion of alleles in this population that are negative for the Alu insertion)? What is the value of p in this population? Which proportion of this population is predicted to have a homozygous positive genotype if this gene is at a Hardy-Weinberg equilibrium?

Answer 1

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