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Mendelian Probability and Allele Frequencies · Tutorial 5 of 989

Genotype Frequencies And Their Allele Marginals

Use genotype probabilities to calculate allele-copy proportions, verify that they form a probability distribution, and understand what information is lost when only allele frequencies are retained.

Beginner 10 min read

What You'll Learn

  • Define allele frequencies as marginal probabilities for a randomly selected allele copy
  • Calculate allele frequencies from biallelic genotype frequencies
  • Extend the calculation to loci with three or more alleles
  • Check genotype and allele frequencies using probability-sum identities
  • Explain why allele frequencies do not uniquely determine genotype frequencies

From Genotypes to Allele Copies

A genotype frequency describes the proportion of individuals with a particular genotype. An allele frequency describes the proportion of allele copies of a particular type. These are related quantities, but they do not have the same denominator: each diploid individual has two allele copies at a locus. To move from genotype frequencies to allele frequencies, count the copies contributed by each genotype.

This calculation is a probability marginal. Imagine selecting an individual at random and then selecting one of that individual's two allele copies at random. The probability that the selected copy is a particular allele is its allele frequency. A homozygote contributes two copies of the same allele; a heterozygote contributes one copy of each of its two alleles.

Definition: At a diploid locus, let \(f_{ij}\) denote the frequency of genotype \(A_iA_j\), with \(f_{ij}=f_{ji}\) for an unordered genotype. The allele frequency \(p_i\) is the proportion of all allele copies that are \(A_i\). Equivalently, it is the probability that a randomly selected allele copy from a randomly selected individual is \(A_i\). For a valid genotype-frequency distribution, all \(f_{ij}\) are nonnegative and their sum over distinct genotypes is 1.

For a locus with \(k\) possible alleles, a homozygote \(A_iA_i\) contributes two copies of \(A_i\), while each heterozygote \(A_iA_j\) contributes one. In frequency form, this gives

$$ p_i=f_{ii}+\frac{1}{2}\sum_{j\ne i} f_{ij}. $$

The factor \(1/2\) appears because each genotype frequency is a proportion of individuals, and each individual has two allele copies. A heterozygote containing \(A_i\) contributes one of its two copies to \(A_i\). The homozygote frequency \(f_{ii}\) enters without a factor of \(1/2\), because both copies are \(A_i\).

Biallelic Loci

For a locus with alleles \(A\) and \(a\), the possible genotypes are \(AA\), \(Aa\), and \(aa\). Write their frequencies as \(f_{AA}\), \(f_{Aa}\), and \(f_{aa}\). The two allele frequencies are

$$ p_A=f_{AA}+\frac{1}{2}f_{Aa}, \qquad p_a=f_{aa}+\frac{1}{2}f_{Aa}. $$

The genotype frequencies must satisfy \(f_{AA}+f_{Aa}+f_{aa}=1\). The allele frequencies must also sum to 1, since every allele copy is either \(A\) or \(a\). The relationship follows directly from the formulas:

$$ p_A+p_a =f_{AA}+f_{aa}+f_{Aa} =1. $$

Worked Example: Two Alleles From Genotype Frequencies

Suppose a hypothetical diploid population has genotype frequencies \(f_{AA}=0.36\), \(f_{Aa}=0.48\), and \(f_{aa}=0.16\). First check that the genotype frequencies sum to 1:

$$ 0.36+0.48+0.16=1.00. $$

The frequency of allele \(A\) is the homozygote frequency plus half the heterozygote frequency:

$$ p_A=f_{AA}+\frac{1}{2}f_{Aa} =0.36+\frac{1}{2}(0.48) =0.36+0.24 =0.60. $$

Likewise, the frequency of allele \(a\) is

$$ p_a=f_{aa}+\frac{1}{2}f_{Aa} =0.16+\frac{1}{2}(0.48) =0.16+0.24 =0.40. $$

The check is \(p_A+p_a=0.60+0.40=1.00\). Thus, 60% of the allele copies are \(A\) and 40% are \(a\). These are proportions of allele copies, not proportions of individuals with genotypes \(AA\) or \(aa\).

Allele Frequencies as Marginal Probabilities

The word marginal means that we focus on one component of a joint probability distribution while summing over the possible values of the other component. A diploid genotype describes the pair of allele copies in an individual. To obtain the marginal frequency of \(A_i\), count its share of the copies in every genotype, then add those contributions.

One way to formalize this is to define \(C_i\) as the number of \(A_i\) copies in an individual's genotype. For a homozygote \(A_iA_i\), \(C_i=2\); for a heterozygote containing \(A_i\), \(C_i=1\); and for other genotypes, \(C_i=0\). The average number of \(A_i\) copies per individual is therefore \(2f_{ii}+\sum_{j\ne i}f_{ij}\). Dividing by two copies per individual gives the allele frequency.

$$ p_i=\frac{E(C_i)}{2} =\frac{2f_{ii}+\sum_{j\ne i}f_{ij}}{2} =f_{ii}+\frac{1}{2}\sum_{j\ne i}f_{ij}. $$

This argument also establishes the normalization identity for any number of alleles. Each genotype has exactly two copies in total, so the sum of the allele-copy counts across all allele types is 2 for every individual. The expected total number of copies is therefore 2; dividing by 2 means the allele frequencies sum to 1.

$$ \sum_{i=1}^{k} p_i =\frac{1}{2}\sum_{i=1}^{k}E(C_i) =\frac{1}{2}E\left(\sum_{i=1}^{k}C_i\right) =\frac{1}{2}(2) =1. $$

The identity requires a complete genotype distribution for diploid individuals: every individual must have exactly two allele copies at the locus, and the listed genotypes must cover all possibilities. It does not require a particular mating pattern or a special relationship among genotype frequencies.

More Than Two Alleles

The same copy-counting rule applies when there are three or more alleles. For each allele, include its homozygote frequency in full and each heterozygote frequency containing it at half weight. Genotype frequencies still describe individuals, while allele frequencies describe copies.

Worked Example: Three Alleles and a Normalization Check

Consider a hypothetical locus with alleles \(A_1\), \(A_2\), and \(A_3\), and genotype frequencies \(f_{11}=0.10\), \(f_{22}=0.20\), \(f_{33}=0.05\), \(f_{12}=0.15\), \(f_{13}=0.20\), and \(f_{23}=0.30\). First, confirm that the six possible genotype frequencies add to 1:

$$ 0.10+0.20+0.05+0.15+0.20+0.30=1.00. $$

For \(A_1\), the relevant genotypes are \(A_1A_1\), \(A_1A_2\), and \(A_1A_3\):

$$ p_1=f_{11}+\frac{1}{2}(f_{12}+f_{13}) =0.10+\frac{1}{2}(0.15+0.20) =0.275. $$

For \(A_2\), use its homozygote and the two heterozygotes that contain it:

$$ p_2=f_{22}+\frac{1}{2}(f_{12}+f_{23}) =0.20+\frac{1}{2}(0.15+0.30) =0.425. $$

For \(A_3\), the calculation is

$$ p_3=f_{33}+\frac{1}{2}(f_{13}+f_{23}) =0.05+\frac{1}{2}(0.20+0.30) =0.300. $$

Finally, \(p_1+p_2+p_3=0.275+0.425+0.300=1.000\). Each heterozygote appears in the calculation for both of its alleles, with half its frequency assigned to each. This makes the total contribution of each heterozygote equal to its full genotype frequency.

Checking Frequencies and Understanding the Reverse Calculation

There are two useful checks before interpreting an allele-frequency calculation. First, genotype frequencies must be nonnegative and sum to 1 if they describe a complete distribution. Second, the calculated allele frequencies must be nonnegative and sum to 1. These checks are necessary, though a sum of 1 alone does not guarantee that the frequencies were calculated correctly.

The operation is not generally reversible. Genotype frequencies determine allele frequencies, but allele frequencies alone do not usually determine genotype frequencies. Allele frequencies retain the overall proportions of copies while discarding information about how those copies are paired within individuals.

Worked Example: Different Genotype Frequencies, the Same Allele Frequencies

Compare two hypothetical biallelic genotype distributions. In distribution I, let \(f_{AA}=0.50\), \(f_{Aa}=0\), and \(f_{aa}=0.50\). In distribution II, let \(f_{AA}=0\), \(f_{Aa}=1.00\), and \(f_{aa}=0\). Both are valid because their genotype frequencies are nonnegative and sum to 1.

For distribution I, the allele frequencies are

$$ p_A=0.50+\frac{1}{2}(0)=0.50, \qquad p_a=0.50+\frac{1}{2}(0)=0.50. $$

For distribution II, they are

$$ p_A=0+\frac{1}{2}(1.00)=0.50, \qquad p_a=0+\frac{1}{2}(1.00)=0.50. $$

Thus, both populations have the same allele frequencies, even though one has only homozygotes and the other only heterozygotes. Knowing \(p_A=0.50\) and \(p_a=0.50\) does not tell us which genotype distribution applies. Additional assumptions or genotype data would be needed to specify genotype frequencies.

Common Mistakes and Why the Distinction Matters

  • Using genotype frequencies as allele frequencies. A genotype frequency has individuals as its denominator; an allele frequency has allele copies as its denominator. A heterozygote contributes one copy of each allele, not a whole copy of each genotype frequency.
  • Forgetting to halve heterozygote frequencies. A heterozygote's frequency is divided equally between its two allele types when calculating their copy proportions.
  • Halving a homozygote frequency. Every copy in a homozygote is the same allele, so its full frequency contributes to that allele's frequency.
  • Assuming allele frequencies uniquely give genotype frequencies. The reverse is not determined by copy proportions alone. As the example shows, distinct genotype distributions can have identical allele frequencies.
  • Skipping the sum check. If genotype frequencies do not sum to 1, they may be incomplete or incorrectly recorded. If the resulting allele frequencies do not sum to 1, check the genotype list and the copy weights.

Allele marginals matter because they provide a compact summary of the allele-copy composition at a locus. They are calculated directly from genotype frequencies, but they should not be mistaken for a full description of genotype composition. Keeping those two levels distinct makes probability calculations easier to interpret and helps identify what the available frequencies can, and cannot, establish.

Check Your Understanding

Use the copy-counting rule and the probability-sum checks to answer these questions.

  1. For \(f_{AA}=0.25\), \(f_{Aa}=0.50\), and \(f_{aa}=0.25\), calculate \(p_A\) and \(p_a\). Verify that both genotype and allele frequencies sum to 1.
  2. At a three-allele locus, explain why each heterozygote frequency is divided between two allele-frequency calculations.
  3. A genotype distribution has \(f_{11}=0.20\), \(f_{22}=0.30\), \(f_{12}=0.50\), and no other genotypes. Calculate \(p_1\) and \(p_2\).
  4. Can two populations with the same biallelic allele frequencies have different genotype frequencies? Explain using the information retained by allele marginals.
  5. State the assumptions about ploidy and the genotype list needed for the identity that allele frequencies sum to 1.
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