Genotype frequency
Genetic variation in populations can be analyzed and quantified by the frequency of alleles. Two fundamental calculations are central to population genetics: allele frequencies and genotype frequencies.<ref>Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. 492</ref> Genotype frequency in a population is the number of individuals with a given genotype divided by the total number of individuals in the population.<ref>Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. G-14</ref> In population genetics, the genotype frequency is the frequency or proportion (i.e., 0 < f < 1) of genotypes in a population.
Although allele and genotype frequencies are related, it is important to clearly distinguish them.
Genotype frequency may also be used in the future (for "genomic profiling") to predict someone's having a disease<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> or even a birth defect.<ref>Template:Cite journal</ref> It can also be used to determine ethnic diversity.
Genotype frequencies may be represented by a De Finetti diagram.
Numerical exampleEdit
As an example, 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 frequency for a (the white color allele) equals
- <math>
\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} </math>
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).
- <math>
\begin{align} f({aa}) & = { 9 \over 49 + 42 + 9 } = { 9 \over 100 } = 0.09 = (9\%) \\ \end{align} </math>
Allele and genotype frequencies always sum to one (100%).
EquilibriumEdit
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%.<ref>Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. 492</ref>
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
- <math>f(\mathbf{AA}) = p^2</math> (genotype frequency of homozygotes)
- <math>f(\mathbf{Aa}) = 2pq</math> (genotype frequency of heterozygotes)
- <math>f(\mathbf{aa}) = q^2</math> (genotype frequency of homozygotes)
If p=0.7 and q=0.3, then
- <math>f(\mathbf{AA}) = p^2</math> = (0.7)2 = 0.49
- <math>f(\mathbf{Aa}) = 2pq</math> = 2×(0.7)×(0.3) = 0.42
- <math>f(\mathbf{aa}) = q^2</math> = (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%.<ref>Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. 493</ref>
ReferencesEdit
NotesEdit
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