H-Weezy Wiz

The problem: In a population of 1000 with 25% being recessive individuals, we can gather all of the variables in the Hardy-Weinberg equation.

The Variables include:

q^2: Homozygous recessive individuals.

q: recessive allele frequency

p: dominant allele frequency

p^2: Homozygous dominant individual

2pq: Heterozygous individual

Which make the equations:

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

1) q^2 is always the priority.
Referring to the problem, the homozygous recessive individuals or in other words the q^2=.25

2) Now, after knowing the q^2 we can figure out the next variable: q by square rooting .25
Equaling: .5

3) Considering the equation in which q belongs, both variables q and p need to equal one. So, that means we subtract the value of q from 1, which gives us p:  .5

4) In order to find p^2, we just square the value of p, and get .25

5) For the variable of 2pq, we simple plug in the values of p and q and multiply it times 2, and get .5

How do we know the number of individuals?

It's important to note that the variables representing the individuals in the population include q^2, p^2 , and 2pq

1) Because the population is 1000, we multiply each of the values of the variables representing individuals with 1000

2) Then we get,

p^2: 250 homozygous individuals

q^2: 250 homozygous recessive individuals

2pq: 500 heterozygous individuals