Hardy-Weinberg Equation

Key Questions

p2 + 2pq + q2 = 1.

Explanation:

1. By the help of the Hardy-Weinberg equation, the geneticists examine and calculate the locus, which cotains two alleles. The two alleles are designated as A and a.
2. This equation is expressed as p2 + 2pq + q2 = 1. Where p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population.
3. The frequencies of genes remain constant in the population.
Thank

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.

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