Tuesday, October 27, 2015

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. 

2 comments:

  1. Math is apeezy prefect. The only thing I will say is that you can find the individuals of the alleles(p and q).
    - Jessica Z

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  2. The math is nice and accurate good job on that. One thing that you could've included was giving "The way to find the individuals..." an example showing the work just like the others rather than it just being words but overall nicely done.

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