If 20% of the individuals in a population have sickle-cell anemia, according to the Hardy Weinberg equation, what proportion of the population should be sickle-cell carriers?
https://www.houstonisd.org/cms/lib2/TX01001591/Centricity/Domain/5363/Hardy%20Weinberg%20Problem%20Set%20KEY.pdf
Let p be the recessive allele and q be the dominant allele
p^2=0.20
p=√(0.20)=0.447
p+q=1
1-0.447=0.553
Carrier=2pq=2(0.533)(0.447)=0.494
Or 49.4%
p+q=1
(p+q)(p+q)=p^2 +2pq+ q^2
The equation gives proportions of percentages.
p^2=homozygous recessive
q^2= homozygous dominant=(0.533)^2=0.306
2pq=heterozygous for both alleles
After plugging in values, the equation and numbers should equal 1.
p^2 +2pq+ q^2=0.20+0.494+0.306=1.00
To determine the proportion of the population that should be sickle-cell carriers according to the Hardy-Weinberg equation, you need to know the frequency of the recessive allele (q) in the population, where q^2 represents the frequency of individuals with sickle-cell anemia. In this case, let's assume that the frequency of the recessive allele (q) is 0.2 (as stated in the question).
According to the Hardy-Weinberg equation, the proportion of the population that are carriers (heterozygotes) is given by 2pq, where p is the frequency of the dominant allele and q is the frequency of the recessive allele.
To calculate the proportion of the population that are carriers, substitute the known values into the equation: 2 * (0.8) * (0.2).
Thus, the proportion of the population that should be sickle-cell carriers is 0.32, or 32%.
those who have it, or homozygous.
Heterozygous for the allele then is 2pq, or 2*.20*.80= 32 percent of the population.