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?

Answer 1

those who have it, or homozygous.
Heterozygous for the allele then is 2pq, or 2*.20*.80= 32 percent of the population.

Answer 2

https://www.houstonisd.org/cms/lib2/TX01001591/Centricity/Domain/5363/Hardy%20Weinberg%20Problem%20Set%20KEY.pdf

Answer 3

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%

Answer 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