Why Phase Matters
At two loci, a diploid individual carries two copies of each locus, one on each homologous chromosome. A genotype call records the two alleles at each locus, but may not reveal which alleles lie together on the same chromosome. That missing chromosome-level arrangement is called phase. It matters because haplotype frequencies, and therefore linkage disequilibrium, are defined from alleles that occur together on chromosomes.
This tutorial considers two biallelic loci in diploid individuals, with alleles \(A\) and \(a\) at the first locus and \(B\) and \(b\) at the second. Assume the genotype calls are correct and both loci are observed for each individual. The four possible haplotypes are \(AB\), \(Ab\), \(aB\), and \(ab\). A pair of haplotypes carried by one individual is a diplotype; if phase is unobserved, it is an unordered pair whose members may not be known.
A genotype that is homozygous at either locus has no such ambiguity between these two loci. For example, \(AA,Bb\) must be \(AB/Ab\): both chromosomes carry \(A\), and one carries \(B\) while the other carries \(b\). By contrast, when both loci are heterozygous, either pairing of the alleles across the two chromosomes produces the same observed genotypes.
Enumerating Compatible Haplotype Pairs
The full genotype \(Aa,Bb\) says that the individual has one \(A\) and one \(a\), and one \(B\) and one \(b\). To form two chromosomes, each allele at the first locus must be paired with one allele at the second. Pairing \(A\) with \(B\) forces the remaining pair to be \(a\) with \(b\). Alternatively, pairing \(A\) with \(b\) forces \(a\) with \(B\). These exhaust the possibilities; swapping the order of chromosomes does not create a third diplotype.
Worked Example: Listing the Phases of a Double Heterozygote
A hypothetical individual has genotype \(Aa,Bb\), with no additional phase information. List the compatible haplotype pairs and state what each implies.
Coupling: the pair is \(AB/ab\). One chromosome carries \(A\) and \(B\), and the other carries \(a\) and \(b\).
Repulsion: the pair is \(Ab/aB\). One chromosome carries \(A\) and \(b\), and the other carries \(a\) and \(B\).
Both pairs give genotype \(Aa\) at the first locus and \(Bb\) at the second. Thus, the observed genotype identifies the allele counts within the individual, but does not identify which of these two chromosome arrangements is present.
This distinction also explains what can and cannot be recovered from unphased data. The allele counts at each locus are determined by the genotypes: every double heterozygote contributes one \(A\), one \(a\), one \(B\), and one \(b\). But the haplotype counts depend on the phase assignment. If there are \(m\) double heterozygotes and \(k\) are in coupling phase, their combined contribution to the haplotype counts is:
Changing \(k\) does not change any of the four allele counts contributed by these individuals. It does change how often particular alleles appear together on the same chromosome. This is the key source of phase ambiguity for two-locus linkage disequilibrium.
How Phase Changes Linkage Disequilibrium
For phased chromosomes, the haplotype frequency \(p_{AB}\) is the proportion of observed chromosomes that carry \(AB\). Likewise, \(p_A\) and \(p_B\) are the frequencies of \(A\) and \(B\) among chromosomes. The two-locus linkage disequilibrium coefficient is \(D=p_{AB}-p_Ap_B\). It compares the frequency of the \(AB\) haplotype with the frequency expected from the product of the two allele frequencies under independence.
If phase is unresolved, the marginal allele frequencies are still computable from genotypes, but \(p_{AB}\) may not be. Consequently, the data can leave \(D\) uncertain as well. The next example shows this directly: the genotype table and both allele frequencies are fixed, while compatible phase assignments yield different haplotype counts and different values of \(D\).
Worked Example: The Same Genotype Counts, Different Values of D
Consider a hypothetical sample of eight diploid individuals. The unphased genotype counts are four \(Aa,Bb\), two \(AA,Bb\), and two \(aa,Bb\). Compare the two extreme assignments in which all four double heterozygotes are in coupling or all four are in repulsion. The \(AA,Bb\) individuals must each be \(AB/Ab\), and the \(aa,Bb\) individuals must each be \(aB/ab\).
First, calculate the allele frequencies. There are \(16\) chromosomes. The four double heterozygotes contribute four copies of \(A\); the two \(AA,Bb\) individuals contribute four more; and the two \(aa,Bb\) individuals contribute none. Thus \(p_A=8/16=0.500\). All eight individuals are heterozygous at the second locus, so \(p_B=8/16=0.500\).
All four double heterozygotes in coupling: Their contribution is four \(AB\) and four \(ab\) chromosomes. Adding the unambiguous contributions gives \(AB=2+4=6\), \(Ab=2\), \(aB=2\), and \(ab=2+4=6\). The total is \(16\) chromosomes. Therefore:
All four double heterozygotes in repulsion: Their contribution is four \(Ab\) and four \(aB\) chromosomes. The counts are then \(AB=2\), \(Ab=2+4=6\), \(aB=2+4=6\), and \(ab=2\). Hence:
The unphased genotype counts are identical in both cases, and the marginal allele frequencies are identical. Nevertheless, the estimated \(D\) is positive in the first phase assignment and negative in the second. The genotype counts alone do not determine which value describes the sample.
A Short Derivation of the Ambiguity
The relationship between the number of coupling-phase individuals and \(D\) can be made explicit for the sample just considered. Let \(k\) of the four double heterozygotes be in coupling phase, so \(0\leq k\leq4\). The remaining \(4-k\) are in repulsion. The two \(AA,Bb\) individuals contribute two \(AB\) chromosomes, and the double heterozygotes add \(k\) more. Therefore, the total \(AB\) count is \(2+k\), out of \(16\) chromosomes. The allele frequencies remain \(p_A=p_B=0.500\) for every allowed value of \(k\). Substitution gives:
This derivation shows precisely what genotype data leave unresolved: as \(k\) varies from \(0\) to \(4\), \(D\) varies from \(-0.125\) to \(0.125\) in increments of \(1/16=0.0625\). Each value corresponds to a compatible assignment of phases. This is a set of possible sample values, not a confidence interval or a probability distribution over \(D\); no probabilities for the competing phases have been specified.
Worked Example: Finding the Values Allowed by Partial Phase Information
Suppose a hypothetical sample consists of six individuals, all with genotype \(Aa,Bb\). An independent observation establishes that four individuals are in coupling phase, but the phase of the other two remains unknown. Find the possible values of \(D\) consistent with this information.
Let \(k\) be the total number in coupling. The four resolved individuals are coupling, so \(k\) can be \(4\), \(5\), or \(6\), depending on the two unresolved phases. There are \(12\) chromosomes. Every individual is heterozygous at each locus, so \(p_A=p_B=6/12=0.500\). For any value of \(k\), the number of \(AB\) chromosomes is \(k\), giving \(p_{AB}=k/12\).
Substituting each allowed value:
Thus, the observed genotype data and the four resolved phases permit three values of \(D\). Resolving the remaining phases would select one of them; without that information, reporting a single value would require an additional assumption or method.
What Unphased Genotypes Do and Do Not Tell Us
The examples illustrate an important separation. Genotype counts can identify allele frequencies at each locus because each person's allele counts are observed, regardless of which alleles share a chromosome. In contrast, haplotype frequencies require chromosome pairings. Double heterozygotes create the ambiguity; genotypes homozygous at one locus have a determined pairing across the two loci.
This does not mean that every unphased data set leaves \(D\) equally uncertain. If there are no double heterozygotes, the two-locus haplotype pairs can be reconstructed from the genotypes under the stated diploid, two-locus setup. If there are double heterozygotes, the extent of ambiguity depends on their number and on the other genotype counts. As the second example showed, the compatible values can even have opposite signs.
Additional observations can sometimes resolve phase for particular individuals, such as chromosome-level sequence information or inheritance information from relatives. The central point here is not how to obtain or estimate phase, but that a two-locus unphased genotype by itself does not distinguish coupling from repulsion. Any reported haplotype frequencies or \(D\) based on such data must therefore account for the unresolved assignments rather than silently treating one phase as observed.
Common Mistakes
- Treating \(Aa,Bb\) as one definite haplotype pair. The genotype is compatible with both \(AB/ab\) and \(Ab/aB\). Unless phase information is available, neither should be assumed to be the observed arrangement.
- Assuming allele frequencies settle the phase. Both alternatives contain one copy of each allele per double heterozygote. Identical allele frequencies do not imply identical haplotype frequencies.
- Calculating \(D\) from genotype counts as if they were phased chromosomes. \(D\) uses the frequency of a haplotype such as \(AB\), not just the separate frequencies of \(A\) and \(B\). If that haplotype count is unresolved, \(D\) may also be unresolved.
- Calling the range of compatible values a confidence interval. The range found by enumerating possible phases reflects ambiguity about chromosome pairing. A confidence interval addresses sampling uncertainty under a specified estimation procedure; it is a different concept.
Check Your Understanding
Use the definitions and counting logic above to answer the following questions.
- List the two diplotypes compatible with genotype \(Aa,Bb\), and name each phase.
- Why does changing a double heterozygote from coupling to repulsion leave its allele counts unchanged?
- A sample contains only individuals with genotype \(Aa,Bb\). If all are in coupling, which two haplotypes occur? If all are in repulsion, which two occur?
- In the eight-person example, what values of \(D\) result if exactly two of the four double heterozygotes are in coupling? Show the calculation.
- Explain why a range of values compatible with unresolved phase is not, by itself, a confidence interval for \(D\).