From Chromosome Counts to Frequencies
At two biallelic loci, label the alleles \(A\) and \(a\) at the first locus, and \(B\) and \(b\) at the second. A chromosome carries one allele at each locus, so its haplotype is one of \(AB\), \(Ab\), \(aB\), or \(ab\). For example, \(Ab\) means that the chromosome carries \(A\) at the first locus and \(b\) at the second.
A diploid individual has two chromosomes, and therefore two haplotypes at these loci. If the chromosome pairs are phased, each chromosome’s two-locus combination is known. Counting the four haplotypes across those chromosomes gives frequencies that describe the chromosome pool. This is different from counting genotypes: a two-locus genotype records the allele pair at each locus without necessarily distinguishing which alleles lie together on the same chromosome.
For \(N\) diploid individuals with both chromosomes counted, \(M=2N\). The denominator is the number of chromosome copies, not the number of individuals. Each haplotype frequency is between zero and one, and the four frequencies must sum to one. At each locus, the two allele frequencies must also sum to one: \(p_A+p_a=1\) and \(p_B+p_b=1\). These are basic consistency checks on the counts and calculations.
Worked Example: Calculating the Four Haplotype Frequencies
Worked Example: Calculating the Four Haplotype Frequencies
Suppose, in a hypothetical sample of 100 diploid individuals, both phased chromosomes are counted for every individual. The resulting counts are \(n_{AB}=80\), \(n_{Ab}=40\), \(n_{aB}=30\), and \(n_{ab}=50\). Calculate the haplotype and marginal allele frequencies and check their sums.
There are \(M=2(100)=200\) chromosome copies. Thus
The haplotype frequencies add to \(0.40+0.20+0.15+0.25=1.00\), as required. The first-locus allele frequencies are
The second-locus allele frequencies are
Both marginal pairs sum to one: \(0.60+0.40=1.00\) and \(0.55+0.45=1.00\). In this sample, 60% of the counted chromosome copies carry \(A\), and 55% carry \(B\). These are chromosome-level proportions, not percentages of individuals with a particular genotype.
Expected Genotypes Under Random Pairing
Haplotype frequencies alone describe the chromosome pool. To calculate genotype frequencies under random pairing, add an explicit pairing assumption: the two chromosomes in a diploid individual are drawn independently from that pool, with the four haplotype frequencies held fixed. The resulting values are expected frequencies under this model. They need not equal the observed genotype frequencies in a particular sample.
An ordered draw of one \(AB\) chromosome followed by one \(ab\) chromosome has probability \(f_{AB}f_{ab}\). For two different haplotypes \(h\) and \(k\), the unordered diplotype \(h/k\) can arise in either order, so its probability is \(f_hf_k+f_kf_h=2f_hf_k\). A diplotype here means the pair of haplotypes carried by an individual. If the two haplotypes are the same, there is only one pairing type and its probability is \(f_h^2\).
The factor of two for different haplotypes is essential. For example, \(AB/ab\) and \(ab/AB\) are the two possible ordered draws, but both describe the same unordered diplotype. When haplotypes are translated into genotypes, \(AB/ab\) gives \(AaBb\). In contrast, \(Ab/ab\) gives \(Aabb\), because it carries \(A\) and \(a\) at the first locus but \(b\) on both chromosomes at the second.
Worked Example: Aggregating Diplotypes into Genotypes
Worked Example: Aggregating Diplotypes into Genotypes
Use the frequencies from the preceding hypothetical chromosome counts: \(f_{AB}=0.40\), \(f_{Ab}=0.20\), \(f_{aB}=0.15\), and \(f_{ab}=0.25\). Find the expected two-locus genotype frequencies under random pairing.
First calculate the ten unordered diplotype frequencies. The four same-haplotype probabilities are \(f_h^2\), and the six different-haplotype probabilities are \(2f_hf_k\):
Now translate each diplotype into the genotype at each locus. The genotype \(AaBb\) arises from \(AB/ab\) and \(Ab/aB\), so its expected frequency is \(0.20+0.06=0.26\). The other genotype frequencies are obtained in the same way:
| Two-locus genotype | Diplotype contribution(s) | Expected frequency |
|---|---|---|
| AABB | AB/AB: 0.16 | 0.16 |
| AABb | AB/Ab: 0.16 | 0.16 |
| AAbb | Ab/Ab: 0.04 | 0.04 |
| AaBB | AB/aB: 0.12 | 0.12 |
| AaBb | AB/ab: 0.20; Ab/aB: 0.06 | 0.26 |
| Aabb | Ab/ab: 0.10 | 0.10 |
| aaBB | aB/aB: 0.0225 | 0.0225 |
| aaBb | aB/ab: 0.075 | 0.075 |
| aabb | ab/ab: 0.0625 | 0.0625 |
The genotype frequencies sum to \(0.16+0.16+0.04+0.12+0.26+0.10+0.0225+0.075+0.0625=1.00\). This check confirms that the genotype categories account for all the probability under the pairing model.
Why the Probabilities Sum to One
The diplotype probabilities provide a useful general check. Let the sum run over the four possible haplotypes. The total probability over all unordered diplotypes is the sum of the same-haplotype probabilities and the different-haplotype probabilities:
The first equality follows by expanding the square: it contains each squared frequency once and each cross-product twice. The second follows because the haplotype frequencies sum to one. When diplotypes are grouped into genotypes, every diplotype belongs to exactly one two-locus genotype. Therefore aggregation preserves the total, so the expected genotype frequencies also sum to one.
A second check comes from looking at either locus alone. For the first locus, the probability of drawing an \(A\)-carrying chromosome is \(p_A=f_{AB}+f_{Ab}\). Under independent pairing, an individual is expected to have genotype \(AA\) with probability \(p_A^2\), genotype \(Aa\) with probability \(2p_Ap_a\), and genotype \(aa\) with probability \(p_a^2\). This follows because \(AA\) requires two \(A\)-carrying draws, while \(Aa\) can arise in either order. The same argument applies to \(B\) and \(b\). These marginal checks help detect errors in the two-locus genotype table.
Worked Example: Checking Marginal Genotype Frequencies
Worked Example: Checking Marginal Genotype Frequencies
In another hypothetical set of 30 diploid individuals, the phased chromosome counts are \(n_{AB}=18\), \(n_{Ab}=12\), \(n_{aB}=6\), and \(n_{ab}=24\). There are \(60\) chromosome copies, giving frequencies \(0.30\), \(0.20\), \(0.10\), and \(0.40\), respectively. Thus \(p_A=0.30+0.20=0.50\), \(p_a=0.50\), \(p_B=0.30+0.10=0.40\), and \(p_b=0.20+0.40=0.60\).
For a check, the expected marginal genotype frequencies at the first locus are
At the second locus they are
For example, the full two-locus random-pairing calculation gives \(P(AABB)=0.09\), \(P(AABb)=0.12\), and \(P(AAbb)=0.04\), which sum to \(P(AA)=0.25\). It gives \(P(AaBB)=0.06\), \(P(AaBb)=0.28\), and \(P(Aabb)=0.16\), which sum to \(P(Aa)=0.50\). The remaining \(aa\) genotypes have frequencies \(0.01\), \(0.08\), and \(0.16\), summing to \(P(aa)=0.25\). These marginal sums agree with the allele frequencies used in the calculation.
Assumptions and a Common Mistake
The arithmetic for haplotype frequencies requires chromosome counts classified into the four haplotypes. In diploid data, this means the two-locus phase must be known for the chromosomes being counted. If phase is unknown, a two-locus genotype may be compatible with more than one haplotype pair; assigning a single pair without justification can give incorrect counts. The frequencies calculated from a sample are sample proportions. Interpreting them as estimates of population frequencies additionally depends on how the chromosomes were sampled.
The genotype calculation requires more than the counts: it assumes that paired chromosomes are independent draws from the stated haplotype pool. If the pairing process differs from that model, the calculated genotype frequencies are model expectations, not a description guaranteed to match the observed individuals. State this assumption before interpreting the expected values.
- Wrong reasoning: “The frequencies of two diplotypes can be added only if they represent the same ordered chromosome pair.”
- Correction: Genotypes combine all diplotypes that produce the same allele pair at each locus. For instance, both \(AB/ab\) and \(Ab/aB\) produce \(AaBb\), so their probabilities are added.
- Another error to avoid: \(Ab/ab\) is not \(AaBb\). It produces \(Aabb\): the first-locus alleles are \(A,a\), and the second-locus alleles are \(b,b\).
- Denominator check: If \(N\) diploid individuals contribute both chromosomes, divide each haplotype count by \(2N\), not by \(N\).
Check Your Understanding
Use the definitions and random-pairing assumption above to answer these questions.
- In a sample of 40 diploid individuals with both phased chromosomes counted, what is the denominator for the four haplotype frequencies?
- Suppose \(f_{AB}=0.25\), \(f_{Ab}=0.15\), \(f_{aB}=0.35\), and \(f_{ab}=0.25\). Calculate \(p_A\), \(p_a\), \(p_B\), and \(p_b\), and check both marginal sums.
- Under random pairing, what is the expected frequency of the unordered diplotype \(AB/ab\) if \(f_{AB}=0.30\) and \(f_{ab}=0.20\)?
- Which two diplotypes produce genotype \(AaBb\), and how are their probabilities combined?
- Explain why haplotype frequencies calculated from phased chromosomes do not, by themselves, guarantee that observed genotype frequencies equal the random-pairing expectations.