What is an example hardy-weinberg equation practice problem?

1 Answer
Jan 22, 2016

A question could look like:

If #mathbf98# out of #mathbf200# individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?

An explanation, walked through:

The Hardy-Weinberg equilibrium is a mathematical relationship of the alleles and genotypes in a population that meets certain characteristics. The relationships are as follow:

Alleles: #p+q=1#

#p="frequency of the dominant allele"#
#q="frequency of the recessive allele"#

Genotypes: #p^2+2pq+p^2=1#

#p^2="frequency of homozygous dominant genotype"#
#2pq="frequency of heterozygous genotype"#
#q^2="frequency of homozygous recessive genotype"#

From the question, we know that #98# of #200# individuals express the recessive phenotype. This means that these #98# also have the homozygous recessive genotype, the frequency of which is equal to #q^2#.

To determine what the actual frequency is, simply divide #98/200=0.49#. We now know that #q^2=0.49#.

However, we wish to find the frequency of the population that is heterozygous, which is equal to #2pq#. So, we must find both #p# and #q#.

Finding #mathbfq#:

#q^2=0.49#

Take the square root of both sides.

#q=0.7#

(This means that #70%# of the alleles in the system are recessive alleles.)

Now that we've found the value of #q#, we can find the value of #p# using the allele equation.

Finding #mathbfp#:

Through the equation #p+q=1#, substitute in #q=0.7#.

#p+0.7=1#

Subtract #0.7# from both sides to see that

#q=0.3#

Finding the frequency of heterozygotes:

#"frequency of heterozygous genotypes"=2pq#

Substitute the known values for #p# and #q#:

#"frequency of heterozygous genotypes"=2(0.3)(0.7)=0.42#

Converting this into a percent, we see that #42%# of the population is heterozygous.