Allele Frequency Calculator

An Allele Frequency Calculator is a powerful tool used in population genetics to determine the relative abundance of specific gene variants within a given population.

For example, consider a population of flowers where petal color is determined by two alleles: red (R) and white (r). If we observe 100 flowers, with 60 having red petals and 40 having white petals, the allele frequency calculator would help us determine the prevalence of each allele in the population.

Allele Frequency Calculator

Population	Trait	Dominant Homozygous	Heterozygous	Recessive Homozygous	Dominant Allele Frequency	Recessive Allele Frequency
Flowers	Petal Color	30 (RR)	50 (Rr)	20 (rr)	0.55	0.45
Fruit Flies	Wing Length	40 (LL)	45 (Ll)	15 (ll)	0.625	0.375
Cats	Fur Texture	60 (SS)	35 (Ss)	5 (ss)	0.775	0.225
Peas	Seed Shape	70 (RR)	25 (Rr)	5 (rr)	0.825	0.175
Dogs	Coat Color	50 (BB)	30 (Bb)	20 (bb)	0.65	0.35
Snakes	Scale Pattern	20 (PP)	40 (Pp)	40 (pp)	0.3	0.7
Mice	Ear Shape	45 (EE)	35 (Ee)	20 (ee)	0.625	0.375
Corn	Kernel Color	80 (YY)	15 (Yy)	5 (yy)	0.925	0.075
Frogs	Skin Texture	55 (TT)	30 (Tt)	15 (tt)	0.725	0.275
Cattle	Milk Production	60 (MM)	25 (Mm)	15 (mm)	0.775	0.225
Wheat	Grain Size	70 (GG)	20 (Gg)	10 (gg)	0.85	0.15
Fish	Fin Shape	25 (FF)	50 (Ff)	25 (ff)	0.5	0.5

Allele Frequency Formula

The allele frequency formula is fundamental to population genetics. It’s expressed as:

f(A) = (2 * AA + Aa) / (2N)

Where:

• f(A) represents the frequency of allele A
• AA is the number of homozygous dominant individuals
• Aa is the number of heterozygous individuals
• N is the total number of individuals in the population

In a population of 200 pea plants, we observe:

• 90 plants with round seeds (RR)
• 80 plants with wrinkled seeds (rr)
• 30 plants with heterozygous genotype (Rr)

To calculate the frequency of the R allele:

f(R) = (2 * 90 + 30) / (2 * 200) = 210 / 400 = 0.525 or 52.5%

How is Allele Frequency Calculated?

• Identify the alleles in question
• Count the number of each genotype in the population
• Determine the total number of alleles
• Apply the allele frequency formula

Suppose we have a population of 1000 individuals with the following ABO blood type distribution:

• 360 type A
• 130 type B
• 40 type AB
• 470 type O

To calculate the frequency of the A allele:

Count genotypes:

• AA or AO (type A): 360
• AB: 40

Total alleles: 2 * 1000 = 2000

Apply formula: f(A) = (2 * 360 + 40) / 2000 = 760 / 2000 = 0.38 or 38%

How to calculate genotype frequencies using hardy weinberg?

The formula for Hardy-Weinberg equilibrium is:

p^2 + 2pq + q^2 = 1

Where:

• p is the frequency of the dominant allele
• q is the frequency of the recessive allele

For example, if the frequency of a dominant allele (A) is 0.7 and the recessive allele (a) is 0.3:

• Frequency of AA (p^2) = 0.7^2 = 0.49
• Frequency of Aa (2pq) = 2 * 0.7 * 0.3 = 0.42
• Frequency of aa (q^2) = 0.3^2 = 0.09

What is the allele frequency rule

This rule is expressed as:

p + q + r + ... = 1

Where p, q, r, etc., represent the frequencies of different alleles for a gene.

The allele frequency rule is a fundamental concept in population genetics, stating that the sum of all allele frequencies for a given gene in a population must equal 1 (or 100%).

For instance, in a bi-allelic system: If the frequency of allele A is 0.6, then the frequency of allele a must be 0.4, as 0.6 + 0.4 = 1.

How to calculate allele frequency in G5?

To calculate allele frequency in G5:

Determine initial allele frequencies (G0)

Apply selection pressures or breeding strategies for each generation

Calculate new allele frequencies for each subsequent generation (G1, G2, G3, G4)

Compute the final allele frequencies in G5

Imagine we’re breeding mice for coat color, starting with an initial population where the frequency of the brown allele (B) is 0.3 and the white allele (b) is 0.7.

Assuming no selection pressure:

G0: f(B) = 0.3, f(b) = 0.7 G1-G5: Frequencies remain constant

If we apply selection favoring the brown allele, increasing its frequency by 10% each generation:

G1: f(B) = 0.3 * 1.1 = 0.33 G2: f(B) = 0.33 * 1.1 = 0.363 G3: f(B) = 0.363 * 1.1 = 0.399 G4: f(B) = 0.399 * 1.1 = 0.439 G5: f(B) = 0.439 * 1.1 = 0.483

The final allele frequency in G5 for the brown allele would be 0.483 or 48.3%.

References

• National Human Genome Research Institute. “Allele Frequency” https://www.genome.gov/genetics-glossary/Allele-Frequency
• Nature Education. “Population Genetics” https://www.nature.com/scitable/topic/population-genetics-13/
• Khan Academy. “Hardy-Weinberg equation” https://www.khanacademy.org/science/ap-biology/heredity/hardy-weinberg-equilibrium/a/hardy-weinberg-equation

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