# How can the hardy-weinberg equation be calculated?

Jul 31, 2015

To calculate Hardy-Weinberg equation you need to have the proportion of the studied genotype in order to calculate their frequence in the population from which you will find theorical frequency and then check if it matches reality.

#### Explanation:

To start let's recall the Wardy Weinberg equation :
${p}^{2} + 2 p q + {q}^{2} = 1$
with p the frequency of an allele A1 and q the frequence of an allele A2.
Let's try an example. In a population there are two alleles M and N with three possible genotypes:

homozygote MM : 1787 individuals
homozygot NN : 1303 individuals
heterozygot NM : 3039 individuals

Total population : 6129

We can calculate the frequency of each genotype :

F11 = frequency of MM $= \frac{1787}{6129} = 0.2915$
F22= frequency of NN $= \frac{1303}{6129} = 0.2126$
F12= frequency of NM$= \frac{3039}{6129} = 0.4958$

We then calculate the frequency p of M and q of N in the population:

$p = F 11 + F \frac{12}{2} = 0.2915 + \frac{0.4958}{2} = 0.54$
$q = F 22 + F \frac{12}{2} = 0.2126 + \frac{0.4958}{2} = 0.46$

Now that we have the Hardy Weinberg frequency, we can calculate the theorical frequency of the genotype by multiplying the frequency by the total population:

MM$= {p}^{2} = {0.54}^{2} = 0.2916$
theoretical frequency of MM$= 0.2916 \cdot 6129 = 1787.2$
NN$= {q}^{2} = {0.46}^{2} = 0.2116$
theoretical frequency of NN$= 0.2116 \cdot 6129 = 1286.9$
NM$= 2 p q = 2 \cdot 0.54 \cdot 0.46 = 0.4968$
theoretical frequency of NM$= 0.4968 \cdot 0.6129 = 3044.9$

Now that we have those theoretical frequencies, we can compare them to the real frequencies and check if the population is or isn't at the Hardy Weinberg equilibrium with, for instance, a ${\chi}^{2}$ test.

Apr 18, 2016

${p}^{2} + 2 p q + {q}^{2} = 1$ and $p + q = 1$

#### Explanation:

When calculating for a ratio in a species that is in HW equilibrium, the two important equations are ${p}^{2} + 2 p q + {q}^{2} = 1$ and $p + q = 1$

The $p$ is the percent of the species that is homozygous dominant for a trait, expressed as a number between 0 and 1. For example, if the percent of the species that is homozygous dominant is 6% then p would equal 0.6, and ${p}^{2}$ would be 0.36

The $q$ in the equations is the percent of the population that is homozygous recessive, and is expressed as a number between 0 and 1. The $2 p q$ is the variable for the percent of the population that is heterozygous. This number also is expressed as a number between 0 and 1.

If you know the value for either p or q and need to know the other, you can use the equation $p + q = 1$ by subtracting the known variable from one. You will often be told in problems to use this equation in order to complete the other equation.

Let's look at a quick example:
In a rat population that is in Hardy-Weinberg equilibrium, 60% of the population is homozygous dominant for the trait of brown fur. Calculate the percentage of the population that is heterozygous for the trait of brown fur.

$p + q = 1$
p=60% and therefore $p = 0.6$ and ${p}^{2} = 0.36$
$0.6 + q = 1$ so $q = 1 - 0.6$ and $q = 0.4$ and ${q}^{2} = 0.16$

${p}^{2} + 2 p q + {q}^{2} = 1$
$0.36 + 2 p q + 0.16 = 1$
$2 p q = 1 - \left(0.36 + 0.16\right)$
$2 p q = 0.44$ which would be your answer

P2 +2pq+q2=1

#### Explanation:

The two alleles p and q represent the frquencies of all alleles. Therefore, p+q=1 or p=1-q. If the equation P2 +2pq+q2=1, p is substituted by 1-q or q is substituted by 1-p. Thanks.