Tuesday, October 27, 2015

Hardy-Weinbergin' The Easy Way

To truly understand how to Hardy-Weinberg, we must get q^2 first. This problem will have a population of 1,000.




  • Given that q^2 is .16 , we can proceed to find q. To do that I had to "unsquare" q^2 , which is basically taking the square root. 
        The square root of .16 is .4 , so q is equal to .4 
          √q^2 = q

          √.16 = .4


  • Since p+q = 1, by subtracting q from 1 we can find p, so p must be equal to .6 
        1 - .4 = .6 
  • Now to find p^2 we simply square the value of p, which would give us .36 
          .6^2 = .36 


  • Once this is done we can find 2pq by multiplying 2*p*q, which should give us .48 
          2 x .4 x .6 = .48

  • Or, we can use the values of p^2 and q^2 to solve for 2pq. Since p^2 + 2pq + q^2 = 1 , we can subtract q^2 and p^2 from 1 to solve for 2pq
         1 - .36 = .64 - .16 = .48

Now that we have all our variables we can determine what the question is asking for and what value you must use. 

p represents the frequency of the dominant allele, which is .6

p^2 represents the homozygous dominant individuals in the population,  for this instance, in our population of 1,000, 36% are homozygous dominant. To find out how many individuals there are in the population that are homozygous dominant we must multiply 

the population x p^2.
1,000 x .36 = 360 individuals

q represents the frequency of the recessive allele


q^2 represents the homozygous recessive individuals in the population. In our population of 1,000 , 16% of the population is homozygous recessive, so to find the number of individuals that are homozygous recessive, we multiply the population x q^2

1,000 x .16 = 160 individuals

2pq represents the heterozygous individuals in the population, in this case 48% of the population is heterozygous , so to find the number of individuals we multiple the population x 2pq

1,000 x .48 = 480 individuals

From all of our calculations we can say that the frequency of the dominant allele (p) is .6 and the frequency of the recessive allele (q) is .4 , in the population of 1,000 , there are 360 individuals that are homozygous dominant (p^2), 160 individuals that are homozygous recessive (q^2) and 480 individuals that are heterozygous (2pq). 

1 comment:

  1. I really liked your title! Way to make it unique! I thought your post was really easy to follow and very well organized. You have a real computer talent Mariel !

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