Genotype Frequency - Numerical Example

Numerical Example

As an example, let's consider a population of 100 four-o-'clock plants (Mirabilis jalapa) with the following genotypes:

49 red-flowered plants with the genotype AA

42 pink-flowered plants with genotype Aa

9 white-flowered plants with genotype aa

When calculating an allele frequency for a diploid species, remember that homozygous individuals have two copies of an allele, whereas heterozygotes have only one. In our example, each of the 42 pink-flowered heterozygotes has one copy of the a allele, and each of the 9 white-flowered homozygotes has two copies. Therefore, the allele freguency for a (the white color allele) equals


\begin{align}
f({a}) & = { (Aa) + 2 \times (aa) \over 2 \times (AA) + 2 \times (Aa) + 2 \times (aa)} = { 42 + 2 \times 9 \over 2 \times 49 + 2 \times 42 + 2 \times 9 } = { 60 \over 200 } = 0.3 \\
\end{align}

This result tells us that the allele frequency of a is 0.3. In other words, 30% of the alleles for this gene in the population are the a allele.

Compare genotype frequency: let's now calculate the genotype frequency of aa homozygotes (white-flowered plants).


\begin{align}
f({aa}) & = { 9 \over 49 + 42 + 9 } = { 9 \over 100 } = 0.09 = (9%) \\
\end{align}

Allele and genotype frequencies always sum to less than or equal to one (in other words, less than or equal to 100%).

The Hardy-Weinberg law describes the relationship between allele and genotype frequencies when a population is not evolving. Let's examine the Hardy-Weinberg equation using the population of four-o'clock plants that we considered above:
if the allele A frequency is denoted by the symbol p and the allele a frequency denoted by q, then p+q=1. For example, if p=0.7, then q must be 0.3. In other words, if the allele frequency of A equals 70%, the remaining 30% of the alleles must be a, because together they equal 100%.


For a gene that exists in two alleles, the Hardy-Weinberg equation states that (p2) + (2pq) + (q2) = 1
If we apply this equation to our flower color gene, then

(genotype frequency of homozygotes)
(genotype frequency of heterozygotes)
(genotype frequency of homozygotes)

If p=0.7 and q=0.3, then

= (0.7)2 = 0.49
= 2×(0.7)×(0.3) = 0.42
= (0.3)2 = 0.09

This result tells us that, if the allele frequency of A is 70% and the allele frequency of a is 30%, the expected genotype frequency of AA is 49%, Aa is 42%, and aa is 9%.

Genotype frequencies may be represented by a De Finetti diagram.

Read more about this topic:  Genotype Frequency

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