hardy-weinberg problem?

Question

in a population of 1000 indivuduals, 360 show the recessive phenotype. How many individuals would u expect to be the homozgous dominant and heterozygous fro the trait. (p^2 + 2pq +q^2 = 1)

how would you do this kind of problem?

Answer 1

360/1000 = q^2 => q = sqrt(0,36) = 0,6
p = 1-q = 0,4
2pq = 0,48
p^2=0,16

for N = 1000:
p^2*1000 = 160
2pq*1000 = 480

I Hope this will help you in the future.

Answer 2

p^2 + 2pq +q^2 = 1

The recessive phenotype is represented in the equation by q^2.
Homozygous dominant is the p^2.
Heterozygous is the 2pq.
Also, p + q = 1

q^2 = 360/1000 = 0.36
q = 0.6
p = 1 - 0.6 = 0.4 (Since p + q = 1)

Homozygous dominant = p^2 = 0.4 ^ 2 = 0.16
0.16 of 1000 = 160 individuals.

Heterozygous = 2pq = 2 * 0.4 * 0.6 = 0.48
0.48 of 1000 = 480 individuals

Total population = 360 + 160 + 480 = 1000