# Hardy-Weinberg Equation

## Key Questions

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.

• A question could look like:

If $m a t h b f 98$ out of $m a t h b f 200$ 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: $p + q = 1$

$p = \text{frequency of the dominant allele}$
$q = \text{frequency of the recessive allele}$

Genotypes: ${p}^{2} + 2 p q + {p}^{2} = 1$

${p}^{2} = \text{frequency of homozygous dominant genotype}$
$2 p q = \text{frequency of heterozygous genotype}$
${q}^{2} = \text{frequency of homozygous recessive genotype}$

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 $\frac{98}{200} = 0.49$. 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 $2 p q$. So, we must find both $p$ and $q$.

Finding $m a t h b f q$:

${q}^{2} = 0.49$

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 $m a t h b f p$:

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:

$\text{frequency of heterozygous genotypes} = 2 p q$

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

$\text{frequency of heterozygous genotypes} = 2 \left(0.3\right) \left(0.7\right) = 0.42$

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

This is an equation used to determine if a population is evolving or not.

#### Explanation:

The Hardy-Weinberg principle is not generally found in nature because it requires certain conditions in an environment. These conditions are the absence of the things that can cause evolution . In other words, if no mechanisms of evolution are acting on a population, evolution will not occur--the gene pool frequencies will remain unchanged. However, since it is highly unlikely that any of these seven conditions, let alone all of them, will happen in the real world, evolution is the inevitable result. The conditions are:

a. No mutation
b. No immigration or emigration
c. Infinitely large population size
d. No natural selection
e. All members of the population breed
f. All mating is totally random
g. Everyone produces the same number of offspring.
Hardy-Weinberg principle can be illustrated mathematically with the equation: p2+2pq+q2 = 1, where ‘p’ and ‘q’ represent the frequencies of alleles. P added to q always equals one (100%). The principle behind it is that, in a population where certain conditions are met (see below), the frequency of the alleles in the gene pool will be constant.