Genotype frequency

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title: "Genotype frequency" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["genetics-concepts", "population-genetics"] topic_path: "science/biology" source: "https://en.wikipedia.org/wiki/Genotype_frequency" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

::figure[src="https://upload.wikimedia.org/wikipedia/commons/5/5a/De_Finetti_diagram.svg" caption="Hardy–Weinberg equilibria]]."] ::

::figure[src="https://upload.wikimedia.org/wikipedia/commons/4/4d/Hardy–Weinberg_law_-_Punnett_square.svg" caption="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. Genotype frequency in a population is the number of individuals with a given genotype divided by the total number of individuals in the population. In population genetics, the genotype frequency is the frequency or proportion (i.e., 0

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 or even a birth defect. 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

: \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 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%.

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

:f(\mathbf{AA}) = p^2 (genotype frequency of homozygotes) :f(\mathbf{Aa}) = 2pq (genotype frequency of heterozygotes) :f(\mathbf{aa}) = q^2 (genotype frequency of homozygotes)

If p=0.7 and q=0.3, then :f(\mathbf{AA}) = p^2 = (0.7)2 = 0.49 :f(\mathbf{Aa}) = 2pq = 2×(0.7)×(0.3) = 0.42 :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%.

References

Notes

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. "Genomic profiling: the critical importance of genotype frequency". PHG Foundation.
  4. Shields. (1999). "Neural Tube Defects: an Evaluation of Genetic Risk". American Journal of Human Genetics.
  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

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