Genotype frequency

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.

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.

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.