Hardy-Weinberg law problems

Question

1-One of the forms of glucosuria is inherited as an autosomal recessive trait and occurs with a frequency of 7: 1,000,000. Determine the frequency of occurrence of heterozygotes in the population.

2-The population of Europeans according to the rhesus blood group system contains 85% Rhesus positive individuals. Determine the saturation of a population with a recessive allele.

Answer 1

The two formulas you will need for both of these problems are the following:

allele frequencies --> p + q = 1 ; p is the dominant allele, q is the recessive allele

genotype frequencies --> p² + 2pq + q² = 1 ; p² are homozygous dominant, 2pq are heterozygous, and q² are homozygous recessive

1. The problem tells us that the recessive trait appears with a frequency of 7/1,000,000 (0.000007). This is the value of q² in the genotype frequency equation because in a recessive trait an individual needs 2 recessive alleles to display the trait. To solve for q, take the square root of 0.000007, which gets you 0.0026 rounded. This is the value of q, or the allele frequency for the recessive allele. To solve for the dominant allele p, 1 - 0.0026 = 0.9974 rounded. Now that we have the individual allele frequencies we can use the genotype frequency equation to solve for the heterozygote population (represented as 2pq). Using our previous values, 2*0.0026*0.9974 = 0.0052 for the frequency of heterozygotes.
2. The rhesus factor is an autosomal recessive trait. By "the saturation of individuals with a recessive allele", the question is really asking for the frequency of both homozygous recessive individuals (2 recessive alleles) and heterozygous individuals (1 recessive allele) combined. If 85% of the population is Rh+, then 15% of the population is Rh- (q² in the genotype equation). Following similar steps as before, square root 0.15 to get q = 0.39. The value of p is then 1 - 0.39 = 0.61. To find the frequency of heterozygotes, 2*0.61*0.39 = 0.48. We already know the value of q² from the start of the problem, so we can take 0.48 heterozygotes + 0.15 homozygous recessive = 0.63 individuals with a recessive allele. Depending on your teacher's preference this could also be written as a percentage (63%).