In humans, Rh-positive individuals have the Rh antigen on their red blood cells, while Rh-negative individuals do not. If the Rh-positive phenotype is produced by a dominant gene (A), and the Rh-negative phenotype is due to its recessive allele (a), what is the frequency of the Rh-positive allele of 84% of a population is Rh-positive

60%

Explanation:

In population genetics, the Hardy-Weinberg Principle states that the genetic composition of a population remains in equilibrium as long as no natural selection or other factors are active and no mutations occur.

The frequencies of the genotypes of an individual locus will be set to a particular equilibrium value. It also specifies that these equilibrium frequencies can be represented as a simple function of the allelic frequencies at that locus. In the simplest case, with a locus with two alleles A and a, with allele frequencies of p and q respectively, the principle predicts that the genotypic frequency for the dominant homozygous AA is p^2, that of the heterozygous Aa is 2pq and that of the recessive homozygous aa, is q^2. The allele frequency of a is represented as p and the frequency of recessive allele is q.

The three genotypes AA : Aa : aa appear in a ratio p² : 2pq : q². If we add them up, we get the unit:

p^2 + 2pq + q^2 = (p + q)^2 = 1  

And p + q = 1

Rh-positive genotypes are represented as AA (homozygous dominant) or Aa (heterozygous) since the presence of a single dominant allele is sufficient to express the phenotype. While Rh-negative genotypes are represented as aa.

If 84% of the population is Rh-positive, that means that this percentage includes all those who are AA and Aa. Then 16% are aa.

F(aa)=q^2=0.16

then F(a)=q=0.4

And since p + q = 1, p + 0.4 = 1, then p is 0.6

We can also calculate the rest, then F(AA)=p^2= 0.36

So F(Aa)= 2pq = 2 x 0.6 x 0.4 = 0.48

Notice that 0.36 + 0.16 + 0.48 = 1

that's all i luck!