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.^[1] Genotype frequency in a population is the number of individuals with a given genotype divided by the total number of individuals in the population.^[2] In population genetics, the genotype frequency is the frequency or proportion (i.e., 0 < f < 1) of genotypes in a population.

A De Finetti diagram visualizing genotype frequencies as distances to triangle edges x (AA), y (Aa) and z (aa) in a ternary plot. The curved line are the Hardy–Weinberg equilibria.

A Punnett square visualizing the genotype frequencies of a Hardy–Weinberg equilibrium as areas of a square. p (A) and q (a) are the allele frequencies.

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^[3] or even a birth defect.^[4] It can also be used to determine ethnic diversity.

Genotype frequencies may be represented by a De Finetti diagram.

Numerical example

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

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

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).

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

Allele and genotype frequencies always sum to one (100%).

Equilibrium

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%.^[5]

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

${\displaystyle f(\mathbf {AA} )=p^{2}}$  (genotype frequency of homozygotes)

${\displaystyle f(\mathbf {Aa} )=2pq}$  (genotype frequency of heterozygotes)

${\displaystyle f(\mathbf {aa} )=q^{2}}$  (genotype frequency of homozygotes)

If p=0.7 and q=0.3, then

${\displaystyle f(\mathbf {AA} )=p^{2}}$  = (0.7)² = 0.49

${\displaystyle f(\mathbf {Aa} )=2pq}$  = 2×(0.7)×(0.3) = 0.42

${\displaystyle f(\mathbf {aa} )=q^{2}}$  = (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%.^[6]

References

1. Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. 492
2. Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. G-14
3. Janssens; et al. "Genomic profiling: the critical importance of genotype frequency". PHG Foundation.
4. Shields; et al. (1999). "Neural Tube Defects: an Evaluation of Genetic Risk". American Journal of Human Genetics. 64 (4): 1045–1055. doi:10.1086/302310. PMC 1377828. PMID 10090889.
5. Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. 492
6. Brooker R, Widmaier E, Graham L, and Stiling P. Biology (2011): p. 493

Notes

• Brooker R, Widmaier E, Graham L, Stiling P (2011). Biology (2nd ed.). New York: McGraw-Hill. ISBN 978-0-07-353221-9.