H - Weezy Problem

q ^2	(given) .41
q	.64
p	.36
p ^2	.13
2pq	(2)(.36)(.64) = .46

p^2 + 2pq + q^2	(.13) + (.46) + (.41) = 1
# of homo dominant	q^2 = .41 = 41% (q^2 x 1000) = (.41 x 1000) = 410
# of homo recessive	p^2 = .13 = 13% (p^2 x 1000 = (.13 x 1000) = 130
# of heterozygous	2pq = .46 = 46% (2pq x 1000) = (.46 x 1000) = 460

            We are given the value q^2 (homozygous dominant) of .41 and an initial population of 1000 individuals. 

Step 1: Find q by taking the square root of q^2 which equals .64

Step 2: Find p by knowing q + p = 1, therefore 1 - .64 = .36

Step 3: Find p^2 by taking p and squaring it, or multiplying it by itself. .36 x .36 = .13

Step 4: Find 2pq by plugging in the values p and q. (2)(.36)(.64) = .46

After finding all values, I then found the # of homo dom. individuals, # of homo rec. individuals, and # of heterozygotes in the population. 

In order to do so, I multiplied the value of q^2 (.41) ,which also equals 41% of the population, by the total population (1000) and got 410 individuals.

The number of individuals that contain the homozygous recessive alleles are 13% of the population which is 130 individuals, which is found by multiplying p^2 by the total population as well.

Lastly, the number of heterozygous individuals takes up 46% of the population which is 460 individuals. This is found by multiplying 2pq by 1000, the total population as well.