?

Genotype frequency is the proportion of a particular genotype in a population, while allele frequency is the proportion of a particular allele in a population. Genotypes are the combination of alleles that an individual carries, so allele frequency is a factor that determines the genotype frequency.

Genotype frequency

AN Finetti’s Diagram visualizing genotype frequencies as distances to the edges of the triangle x (AA), y (Aa) and z (aa) in a ternary plot. The curved line is the Hardy-Weinberg Equilibrium.

AN Punnett square visualizing the genotypic frequencies of a Hardy-Weinberg Equilibrium as areas of a square. P (A) and what (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. At the population geneticsThe genotype frequency is the frequency or proportion (that is, 0 f < 1) of genotypes in a population.

Although allele and genotype frequencies are related, it is important to distinguish them clearly.

genotype frequency it can also be used in the future (for “genomic profiling”) to predict whether someone has a disease or even a birth defect. It can also be used to determine ethnic diversity.

Genotype frequencies can be represented by a Finetti’s Diagram.

numerical example

As an example, consider a population of 100 four-hour plants (jalapa mirabilis) with the following genotypes:

  • 49 red flowering plants with the genotype AA
  • 42 rose flowering plants with genotype aa
  • 9 white flowering plants with genotype aa

When calculating an allele frequency for a diploid species, remember that homozygous individuals have two copies of an allele, while heterozygous there’s only one. In our example, each of the 42 rose-flowered heterozygotes has a copy of the an allele, and each of the 9 white-flowered homozygotes has two copies. Therefore, the allele frequency for an (the white color allele) is equal to








f
(

an

)



=



(
AN
an
)
+
two
×
(
an
an
)


two
×
(
AN
AN
)
+
two
×
(
AN
an
)
+
two
×
(
an
an
)



=



42
+
two
×
9


two
×
49
+
two
×
42
+
two
×
9



=


60
200


=
0.3






\displaystyle \beginalignedf(a)&=(Aa)+2\times (aa) \more than 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\\\endaligned

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

Compare Genotype Frequency: Let’s now calculate the genotype frequency of aa homozygous (white flowering plants).








f
(

an
an

)



=


9

49
+
42
+
9



=


9
100


=
0.09
=
(
9
%
)






\displaystyle \beginalignedf(aa)&=9 \over 49+42+9=9 \over 100=0.09=(9\%)\\\endaligned

The allele and genotype frequencies always add up to one (100%).

Balance

The Hardy-Weinberg Law describes the relationship between allele frequencies and genotypes when a population is not evolving. Let’s examine the Hardy-Weinberg equation using the four-hour plant population we considered above:

if the allele AN frequency is indicated by the symbol P and the allele an frequency denoted by whatthen p+q=1. For example, if P=0.7, then what it should be 0.3. In other words, if the allele frequency of AN equal to 70%, the remaining 30% of the alleles must be anbecause together they add up to 100%.

For gene exists in two alleles, the Hardy-Weinberg equation states that (Ptwo) + (2because) + (whattwo) = 1. If we apply this equation to our flower color gene, then




f
(

AN
AN

)
=

P

two




\displaystyle f(\mathbf AA )=p^2

(genotypic frequency of homozygotes)




f
(

AN
an

)
=
two
P
what


\displaystyle f(\mathbf Aa )=2pq

(genotypic frequency of heterozygotes)




f
(

an
an

)
=

what

two




\displaystyle f(\mathbf aa )=q^2

(genotypic frequency of homozygotes)

if P=0.7 and what=0.3, then




f
(

AN
AN

)
=

P

two




\displaystyle f(\mathbf AA )=p^2

= (0.7)two = 0.49




f
(

AN
an

)
=
two
P
what


\displaystyle f(\mathbf Aa )=2pq

= 2×(0.7)×(0.3) = 0.42




f
(

an
an

)
=

what

two




\displaystyle f(\mathbf aa )=q^2

= (0.3)two = 0.09

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

References

Grades

  • Brooker R, Widmaier E, Graham L, Stiling P (2011). Biology (2nd ed.). New York: McGraw-Hill. ISBN 978-0-07-353221-9.
key concepts Selection
selection effects
in genomic variation
genetic drift Founders related topics


Source: Genotype frequency
Wikipedia

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Genotypic and allele frequencies

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