Population Genetics and Hardy-Weinberg Equilibrium

 

The Hardy-Weinberg Law can be used to determine the frequencies of alleles of a particular gene in a given population and predict allele frequencies in future generations.

 

The law requires the following assumptions:

the population mates randomly

no mutations occur

no selection occurs (no influence of “fitness”).

no migration (no gene flow)

large population size (no genetic drift)

 

For a given gene A with alleles frequencies p (dominant) and q (recessive) in a parental population the next generation will be in equilibrium (the gene frequencies p and q will not change over time).

            p = the frequency of the dominant allele

            q = the frequency of the recessive allele

AND

            p2 = the frequency of the homozygous dominant individuals

                2pq = the frequency of the heterozygous individuals

            q2 = the frequency of the homozygous recessive individuals

 

Hardy-Weinberg equilibrium is represented by the following equations:

 

p2 + 2pq + q2 = 1

 

p + q = 1

 

It is important to note the following:

 

The frequencies of alleles are represented by p and q.

The frequencies of individuals are represented by p2, 2pq, and q2

 

Under the appropriate conditions (random mating, no selection, no mutation, no migration, and large populations) there will be no change in gene frequencies.

 

The recessive trait does not die out, and it is possible that in fact the recessive trait may be very much more common than the dominant trait!

 

Remember that both the homozygous dominant and heterozygous individuals will express the dominant trait.  Given information about the population phenotypes, begin your calculations with q2, the frequency of the homozygous recessive individuals.

 

Greatly modified: Georgetown’s Hardy-Weinberg Page http://www.georgetown.edu/departments/biology/class/hardy/


 

How to solve Hardy-Weinberg problems:

How to find the frequency of alleles given phenotype frequency.

1.   Find the number of recessive homozygous individuals (those with the recessive trait)

2.   Find (count) the population size (the total number of individuals).

3.   Divide the number of individuals with the recessive trait by the total number of individuals.

      This is q2, the frequency of the homozygous individuals (expressed as a decimal).

4.   Take the square root of q2 to get q.

5.   Subtract q from 1 to get  p.

6.   Now that you know p and q, you can solve for 2pq (2*p*q) and p2 (p*p).

 

Hardy-Weinberg problems

 

In a class of 50 students, 18 students were non-tasters (the detected no flavor in the PTC paper) and 32 were tasters.  Tasting is dominant (T) and non-tasting is recessive (t).  Assume that the population is in Hardy-Weinberg equilibrium and solve for the following:

a.       What is t, the frequency of the recessive allele?

b.      What is T, the frequency of the dominant allele?

c.       What is p2, the frequency of the homozygous dominant individuals?

d.      What is 2pq, the frequency of the heterozygous individuals?

e.       What is q2, the frequency of the homozygous recessive individuals?

 

Solution to above example:

1.   Find the number of recessive homozygous individuals (those with the recessive trait).

      18

2.   Find (count) the population size (the total number of individuals).

      50

3.      Divide the number of individuals with the recessive trait by the total number of individuals.

18/50 = .36

      This is q2, the frequency of the homozygous individuals (expressed as a decimal).

4.   Take the square root of q2 to get q.

      .6

5.      Subtract q from 1 to get  p.

1-.6 = .4

6.   Now that you know p and q, you can solve for 2pq (2*p*q) and p2 (p*p).

    2pq = 2*0.4*0.6 = 0.48

      p2 = 0.4*0.4 = 0.16

 

CHECK YOUR WORK

p2 + 2pq + q2 = 1

0.16 + 0.48 + 0.36 = 1