Genotype Frequency Calculator

Calculate observed genotype frequencies and compare with Hardy-Weinberg expectations.

Population Data

Observed Genotype Frequencies

AA Frequency0.2000 (20.00%)
Aa Frequency0.5000 (50.00%)
aa Frequency0.3000 (30.00%)
Total Population1000

Allele Frequencies

p (A allele)
0.4500
q (a allele)
0.5500

Expected (Hardy-Weinberg)

Expected AA0.2025 (203)
Expected Aa0.4950 (495)
Expected aa0.3025 (303)

Genotype Frequency Calculator: Overview

The genotype frequency calculator converts the raw counts of three genotypes at a single biallelic locus into observed genotype frequencies, derives the underlying allele frequencies (p and q), and then predicts the genotype proportions you would expect under Hardy-Weinberg equilibrium. You simply enter how many homozygous dominant (AA), heterozygous (Aa), and homozygous recessive (aa) individuals you counted, and the tool does the rest. It is one of the most useful starting points in population genetics, because the gap between observed and expected genotype frequencies is exactly what reveals whether a population is evolving.

A genotype frequency is nothing more than the proportion of individuals in a population that carry a particular pair of alleles. Because every diploid individual has exactly one genotype at the locus, the three genotype frequencies always sum to 1. The calculator reports each frequency as a decimal to four places and as a percentage, alongside the total population size, so you can audit the arithmetic and use the values directly in a lab report, a genetics homework set, or a conservation survey.

What makes this genotype frequency calculator more than a simple proportion tool is the second half of its output: it counts the alleles hidden inside those genotypes, computes p (the frequency of the dominant A allele) and q (the frequency of the recessive a allele), and then applies the Hardy-Weinberg equation to show the expected number of each genotype. Comparing observed and expected counts is the classic first test for selection, non-random mating, migration, mutation, or genetic drift acting on the gene pool.

How the Genotype Frequency Calculation Works

The calculation runs in three connected stages. First, the calculator sums the three counts to find the total population size, N = AA + Aa + aa. It then divides each genotype count by that total to give the observed frequencies: the AA frequency is AA / N, the Aa frequency is Aa / N, and the aa frequency is aa / N. These three numbers always add up to 1 because every individual is counted exactly once.

Second, the tool reaches inside the genotypes to count alleles. Each diploid individual carries two alleles, so a population of N individuals holds 2N alleles at the locus. Each AA individual contributes two A alleles, each Aa individual contributes one A and one a, and each aa individual contributes two a alleles. Summing these gives the dominant allele frequency p = (2 × AA + Aa) / 2N and the recessive allele frequency q = (2 × aa + Aa) / 2N. By construction p + q = 1.

Third, the calculator applies the Hardy-Weinberg principle. Under random mating with no evolutionary forces, the expected genotype frequencies are p² for AA, 2pq for Aa, and q² for aa, and these three terms expand the binomial (p + q)² = 1. The tool reports each expected frequency and multiplies it by N to give an expected count rounded to the nearest individual, which you can compare directly with your observed counts.

Genotype and Allele Frequency Formulas

freq(AA) = AA/N freq(Aa) = Aa/N freq(aa) = aa/N p = (2·AA + Aa)/2N q = (2·aa + Aa)/2N Expected: p², 2pq, q²

Where:

  • AA= Count of homozygous dominant individuals
  • Aa= Count of heterozygous individuals
  • aa= Count of homozygous recessive individuals
  • N= Total population size (AA + Aa + aa); 2N is the total allele count
  • p= Frequency of the dominant allele A, between 0 and 1
  • q= Frequency of the recessive allele a, between 0 and 1 (p + q = 1)

Interpreting Observed vs Expected Frequencies

The first panel of results shows the observed genotype frequencies drawn straight from your data. A reading such as AA = 0.2000 (20%), Aa = 0.5000 (50%), aa = 0.3000 (30%) tells you precisely how the population is partitioned across the three genotypes. These values are descriptive statistics: they are always correct for the counts you entered and make no assumptions about how the population mates or evolves.

The allele frequency panel reports p and q. Notice that the heterozygotes split their contribution evenly between the two alleles, which is why a population can have a high heterozygote frequency without either allele dominating the gene pool. Knowing p and q lets you summarise the genetic state of the whole locus in a single number, since the two always sum to one.

The expected (Hardy-Weinberg) panel is where interpretation gets interesting. It shows what the genotype frequencies should look like if the population were in equilibrium. When the observed heterozygote count is noticeably lower than the expected 2pq value, the population may show a heterozygote deficit, often caused by inbreeding, population subdivision (the Wahlund effect), or selection against heterozygotes. An excess of heterozygotes can signal overdominance or recent admixture.

To decide whether a difference is real or just sampling noise, geneticists follow up with a formal chi-square goodness-of-fit test that compares observed counts against the expected counts this tool provides. The genotype frequency calculator gives you both sets of numbers in one place, making it the natural launch pad for that statistical test.

A Step-by-Step Look at the Default Example

It helps to trace the calculator's logic through its own default values: AA = 200, Aa = 500, and aa = 300. The total population is N = 200 + 500 + 300 = 1000 individuals, which means the gene pool contains 2N = 2000 alleles.

The observed genotype frequencies are AA = 200/1000 = 0.2000, Aa = 500/1000 = 0.5000, and aa = 300/1000 = 0.3000. These sum to exactly 1.0000, a quick sanity check that the data was entered correctly.

To find allele frequencies, count the alleles. The A count is (2 × 200) + 500 = 900, so p = 900 / 2000 = 0.4500. The a count is (2 × 300) + 500 = 1100, so q = 1100 / 2000 = 0.5500. As expected, p + q = 1.0000.

The expected Hardy-Weinberg frequencies follow: expected AA = p² = 0.4500² = 0.2025, expected Aa = 2pq = 2 × 0.45 × 0.55 = 0.4950, and expected aa = q² = 0.5500² = 0.3025. Multiplying by N gives expected counts of 203 AA, 495 Aa, and 303 aa. Comparing these with the observed 200, 500, and 300 shows the population is very close to equilibrium, with only a slight excess of heterozygotes.

Genotype Observed count Observed freq Expected freq Expected count
AA 200 0.2000 0.2025 203
Aa 500 0.5000 0.4950 495
aa 300 0.3000 0.3025 303

Applications in Genetics and Conservation

Genotype frequencies sit at the heart of quantitative genetics, and this genotype frequency calculator turns up across teaching and research. In undergraduate population genetics courses it is the standard tool for working through Hardy-Weinberg problems, where students must move fluently between genotype counts, allele frequencies, and expected proportions.

  • Carrier and disease estimation: Public-health geneticists convert the observed frequency of a recessive disorder (the aa frequency) into carrier frequencies, which inform genetic counselling and screening programmes.
  • Conservation genetics: Researchers monitoring endangered species compare observed heterozygosity against Hardy-Weinberg expectations to detect inbreeding and loss of genetic diversity in small populations.
  • Detecting evolution: A persistent mismatch between observed and expected genotype frequencies flags selection, assortative mating, or migration acting on the locus.
  • Population structure: A heterozygote deficit across many loci often signals hidden subdivision (the Wahlund effect), a key concept when defining management units for wildlife.

By outputting observed frequencies, allele frequencies, and Hardy-Weinberg expectations side by side, the calculator produces exactly the inputs needed for downstream measures such as the inbreeding coefficient F, fixation indices like FST, and chi-square tests of equilibrium. It is a quick, reliable bridge from raw field or lab counts to interpretable genetic parameters.

Assumptions and Limitations

The observed genotype frequencies this tool reports are pure descriptive statistics and require no assumptions; they are simply counts divided by the total. The expected genotype frequencies, however, depend on the Hardy-Weinberg assumptions: a large, randomly mating population with no mutation, no migration, no selection, and negligible genetic drift. When any of these conditions fails, observed and expected counts diverge, which is precisely the signal the calculator is designed to highlight.

This calculator models a single biallelic, autosomal locus with two alleles, A and a, expressed through three genotypes. It does not handle loci with three or more alleles, such as the ABO blood-group system, nor sex-linked loci, where allele counts differ between males and females and the simple 2N denominator no longer applies. For such systems the same counting logic must be extended allele by allele or chromosome by chromosome.

Finally, frequencies estimated from small samples carry real sampling error. With only a handful of individuals, a single rare genotype can swing the result substantially, and the expected counts may round in ways that look misleading. Use this genotype frequency calculator for teaching, quick checks, and exploratory analysis, and pair larger or publication-grade datasets with a formal statistical package and a chi-square or exact test of Hardy-Weinberg equilibrium.

Worked Examples

Classic Equilibrium Population (Default Values)

Problem:

A population is scored at one locus: 200 AA, 500 Aa, and 300 aa individuals. Find the observed genotype frequencies, the allele frequencies, and the Hardy-Weinberg expected counts.

Solution Steps:

  1. 1Total population: N = 200 + 500 + 300 = 1000, so there are 2N = 2000 alleles.
  2. 2Observed frequencies: AA = 200/1000 = 0.2000, Aa = 500/1000 = 0.5000, aa = 300/1000 = 0.3000.
  3. 3Allele frequencies: p = (2×200 + 500)/2000 = 900/2000 = 0.4500; q = (2×300 + 500)/2000 = 1100/2000 = 0.5500.
  4. 4Expected frequencies: p² = 0.2025, 2pq = 0.4950, q² = 0.3025; expected counts = 203 AA, 495 Aa, 303 aa.

Result:

Observed 0.2000 / 0.5000 / 0.3000 versus expected 0.2025 / 0.4950 / 0.3025 (p = 0.45, q = 0.55). The population is essentially in Hardy-Weinberg equilibrium.

Heterozygote Deficit Suggesting Inbreeding

Problem:

A sampled population shows 90 AA, 20 Aa, and 90 aa individuals. Does the heterozygote count match Hardy-Weinberg expectations?

Solution Steps:

  1. 1Total: N = 90 + 20 + 90 = 200, so 2N = 400 alleles.
  2. 2Allele frequencies: p = (2×90 + 20)/400 = 200/400 = 0.5000; q = (2×90 + 20)/400 = 200/400 = 0.5000.
  3. 3Expected frequencies: p² = 0.2500, 2pq = 0.5000, q² = 0.2500; expected counts = 50 AA, 100 Aa, 50 aa.
  4. 4Observed heterozygotes (20) are far below the expected 100, a strong heterozygote deficit.

Result:

Observed Aa frequency 0.1000 versus expected 0.5000. The large heterozygote deficit points to inbreeding, population subdivision, or selection against heterozygotes.

Estimating Carrier Frequency from a Recessive Trait

Problem:

In a population of 400 people, 16 show a recessive condition (aa), 144 are heterozygous carriers (Aa), and 240 are AA. What are the allele frequencies and expected carrier proportion?

Solution Steps:

  1. 1Total: N = 240 + 144 + 16 = 400, so 2N = 800 alleles.
  2. 2Allele frequencies: p = (2×240 + 144)/800 = 624/800 = 0.7800; q = (2×16 + 144)/800 = 176/800 = 0.2200.
  3. 3Observed frequencies: AA = 240/400 = 0.6000, Aa = 144/400 = 0.3600, aa = 16/400 = 0.0400.
  4. 4Expected carrier frequency 2pq = 2 × 0.78 × 0.22 = 0.3432, giving about 137 expected carriers.

Result:

p = 0.78, q = 0.22; expected carrier frequency 2pq ≈ 0.3432 (about 34%), close to the observed 0.3600 carrier frequency.

Tips & Best Practices

  • Always confirm your three observed frequencies sum to 1.0000; if they do not, recheck your counts.
  • Use whole-number counts of individuals rather than pre-computed percentages for the most accurate p and q.
  • Compare observed Aa against expected 2pq first; a heterozygote deficit is the most common red flag.
  • Remember p + q = 1, so once you know one allele frequency you immediately know the other.
  • For carrier estimation from a recessive disease, take the square root of the aa frequency to approximate q.
  • Follow up any large observed-versus-expected gap with a formal chi-square test using the expected counts.
  • Larger samples give more reliable frequencies; small samples are highly sensitive to a single rare genotype.

Frequently Asked Questions

Genotype frequency is the proportion of individuals carrying a specific pair of alleles, such as AA, Aa, or aa. Allele frequency is the proportion of a single allele (A or a) within the entire gene pool. This calculator reports both: it divides each genotype count by the total individuals for genotype frequencies, then counts alleles within those genotypes to derive p and q.
Each individual carries two alleles, so a population of N individuals has 2N alleles. The dominant allele count is 2×AA + Aa and the recessive allele count is 2×aa + Aa. The calculator divides each by 2N, giving p = (2×AA + Aa)/2N and q = (2×aa + Aa)/2N. By construction these always sum to one.
A meaningful gap between the observed genotype frequencies and the Hardy-Weinberg expected values suggests the population is not in equilibrium. Common causes include natural selection, non-random or assortative mating, inbreeding, migration, mutation, or genetic drift. A heterozygote deficit in particular often indicates inbreeding or hidden population subdivision.
Every diploid individual has exactly one genotype at a given locus, so each person is counted once and only once among AA, Aa, and aa. When you divide each genotype count by the same total population, the three resulting proportions necessarily sum to one (or 100%). If they do not, a data-entry error has occurred.
Yes, indirectly. The calculator gives you both the observed counts and the expected counts under Hardy-Weinberg equilibrium, which are exactly the two columns needed for a chi-square goodness-of-fit test. You would then compute the chi-square statistic and compare it against a critical value (with one degree of freedom for a biallelic locus) to decide if the deviation is statistically significant.
No. This genotype frequency calculator models a single biallelic, autosomal locus with two alleles, A and a, and three genotypes. Systems such as ABO blood groups have three or more alleles and require an extended multi-allele approach. Sex-linked loci also need special handling because males and females carry different numbers of the relevant chromosome.

Sources & References

Last updated: 2026-06-05

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This page is maintained as an educational calculator reference.

Source

Formula Source: Standard Mathematical References

by Various

UpdatedLast reviewed: May 2026
CheckedFormula checks are based on standard references and internal QA review.

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