Population Genetics

Hardy-Weinberg Problem

Given: Population size which is 1000. 
       q^2 which is 0.38

Find: p=0.38
      q=0.62
      p^2=0.14
      2pq=0.47

How:

1.  To find q, square root the q^2. (√0.38=0.62)
2.  To find p, subtract one from q. (1-q=p, or 1-0.62=0.38)
3.  To find p^2, square p. (0.38^2=0,14)
4.  To find 2pq, multiply 2 to the factors of p and q. (2(0.38)(0.62)=0.47).

With these frequencies you can find the individual alleles for the population. 

Homozygous dominant individual:(p^2) 0.14*1000= 140 

Homozygous recessive individual: (q^2) 0.38*1000= 380

Heterozygous individual: (2pq) 0.47*1000=470 

The way to find the individuals of each allele is to multiply each of the frequencies to the population. 

In a total population of 1000 people we know that 14% of them are homozygous dominant. 38% of the population are homozygous recessive. 47% are heterozygous individuals. 62% have the recessive gene, and 38% have the dominant gene.