Why LD Errors Matter
Linkage disequilibrium (LD) summarizes how alleles at two loci occur together on chromosomes. A calculation can be arithmetically correct and still support a misleading conclusion: a large D-prime need not mean one variant predicts the other well, unphased genotypes do not always identify haplotypes, and an LD estimate depends on how alleles are coded and which population was sampled. These distinctions matter when interpreting correlated variants or using one variant as a proxy for another.
The checks in this tutorial use two biallelic loci. At the first locus, denote the alleles \(A\) and \(a\); at the second, denote them \(B\) and \(b\). Haplotype frequencies refer to chromosomes, not individuals. For estimates from sampled chromosomes, each frequency is the observed haplotype count divided by the total number of chromosomes. When genotypes are unphased, the chromosome haplotypes may not be directly observed.
D-prime Is Not a Measure of Predictive Strength
D-prime scales \(D\) by the largest magnitude permitted by the allele frequencies. It indicates how close the observed disequilibrium is to that frequency-dependent bound. By contrast, \(r^2\) is the squared correlation between binary indicators for \(A\) and \(B\). It is directly relevant to how well one allele predicts the other in a population. The two measures answer related but different questions.
For binary indicators \(X_A\) and \(X_B\), which equal 1 when a chromosome carries \(A\) or \(B\), respectively, their means are \(p_A\) and \(p_B\). Their covariance is \(p_{AB}-p_Ap_B=D\), and their variances are \(p_A(1-p_A)\) and \(p_B(1-p_B)\). Dividing covariance by the product of standard deviations gives their correlation; squaring it gives the formula for \(r^2\). This derivation also shows why \(r^2\), unlike D-prime, standardizes disequilibrium by both allele-frequency variances.
Worked Example: High D-prime but Modest Prediction
Suppose a hypothetical sample of 100 phased chromosomes has frequencies \(p_{AB}=0.10\), \(p_{Ab}=0.40\), \(p_{aB}=0\), and \(p_{ab}=0.50\). The counts sum to 100%, so the haplotype frequencies form a valid distribution. Calculate \(D\), D-prime, and \(r^2\).
The allele frequencies are \(p_A=p_{AB}+p_{Ab}=0.50\) and \(p_B=p_{AB}+p_{aB}=0.10\). Therefore,
Since \(D\) is positive, \(D_{\max}=\min\{(0.50)(0.90),(0.50)(0.10)\}=0.05\), so \(D'=0.05/0.05=1\). But
The observed disequilibrium reaches the maximum allowed for these allele frequencies, but \(r^2\) is only about 0.111. In this sample, knowing whether a chromosome carries \(A\) does not provide strong prediction of whether it carries the relatively uncommon allele \(B\). It would be an error to describe D-prime of 1 as perfect predictive correlation.
This distinction is especially important when evaluating proxy variants. If the question is how much one allele predicts another, report and interpret \(r^2\), not D-prime alone. D-prime remains useful for describing normalized disequilibrium, but its value is constrained by allele frequencies and does not by itself establish good tagging or strong prediction.
Phase Ambiguity: Genotypes Do Not Always Reveal Haplotypes
At two loci, an individual heterozygous at both has genotypes \(Aa\) and \(Bb\). Those genotypes are compatible with either haplotype pair \(AB/ab\) (coupling phase) or \(Ab/aB\) (repulsion phase). Without phase information, the individual's two chromosomes cannot be assigned uniquely. Treating one possible arrangement as if it had been observed can distort haplotype counts and LD estimates.
Worked Example: What Double Heterozygotes Leave Unknown
Imagine a hypothetical data set of 10 individuals, each with genotype \(Aa\) at the first locus and \(Bb\) at the second. The observed genotype data establish 10 copies of each allele at each locus across the 20 chromosomes, but they do not establish which alleles share a chromosome.
If all 10 individuals were in coupling phase, the compatible haplotype counts would be 10 \(AB\) and 10 \(ab\). Then \(p_A=p_B=0.50\), \(p_{AB}=0.50\), and
If instead all 10 were in repulsion phase, the counts would be 10 \(Ab\) and 10 \(aB\). The allele frequencies would still each be 0.50, but \(p_{AB}=0\), giving \(D=-0.25\) and \(r^2=1\). If five individuals were in each phase, the compatible counts would be five of each of the four haplotypes, giving \(D=0\) and \(r^2=0\). These are alternative chromosome assignments consistent with the same observed genotypes, not three conclusions established by the genotype data.
With real unphased data, haplotype-frequency estimation must account for the uncertainty over compatible phases. A likelihood-based procedure, such as the EM algorithm, uses the genotype data and its model to estimate frequencies while allowing ambiguous genotypes to contribute probabilistically. It does not turn an unobserved phase into a known one. The data set above is deliberately simple: it exposes the ambiguity without claiming that all phase assignments are equally supported under a particular population model.
A useful audit question is: were haplotypes observed, inferred, or assumed? If phase was inferred, the estimated frequencies and LD measures reflect that estimation, and uncertainty in the phase may matter. Reporting inferred haplotypes as directly observed chromosome counts hides a key part of the analysis.
Allele Coding Must Stay Consistent
The labels \(A\) and \(a\), or \(B\) and \(b\), are coding choices. Switching which allele is called the counted allele changes the sign of \(D\) when one locus is recoded, but leaves \(r^2\) unchanged because \(D\) is squared. Under the magnitude convention defined above, D-prime is also unchanged by a single-locus recoding: both the sign of \(D\) and the corresponding signed bound change. What changes is which haplotypes are described as coupling versus repulsion. As a result, signed \(D\) values should not be compared without checking allele coding.
Worked Example: Recode One Allele Before Comparing Results
Consider 100 hypothetical phased chromosomes with counts \(AB=40\), \(Ab=10\), \(aB=10\), and \(ab=40\). Initially, \(p_A=p_B=0.50\) and \(p_{AB}=0.40\). Thus,
Here \(D_{\max}=0.25\), so \(D'=0.15/0.25=0.60\). Now recode the second locus by calling the former \(b\) allele \(B\) and the former \(B\) allele \(b\). The new \(AB\) count is the old \(Ab\) count, or 10, so the new \(p_{AB}=0.10\); both allele frequencies remain 0.50. Therefore, the recoded \(D\) is \(0.10-0.25=-0.15\), and \(r^2\) remains \(0.36\). The new \(D_{\min}=-0.25\), so D-prime remains \( (-0.15)/(-0.25)=0.60\).
A report that one analysis found \(D=0.15\) and another found \(D=-0.15\) might look contradictory. These values agree if the second analysis reversed the allele coding at one locus. The sign can only be compared after harmonizing allele labels; \(r^2\) and the stated magnitude D-prime are unaffected in this example.
An LD Estimate Describes a Sampled Population
Population haplotype frequencies are not universal constants attached to a pair of locus names. An estimate calculated from sampled chromosomes describes the sampled population, subject to sampling uncertainty and the study's estimation method. It need not transfer unchanged to a population sampled elsewhere or at another time. Even when two samples have the same allele frequencies, their haplotype frequencies—and therefore their LD—can differ.
This is a scope-of-inference point, not a claim that a difference between samples has a particular cause. The estimate may differ because the populations genuinely differ, because of sampling variability, or because data processing and estimation choices differ. A careful interpretation names the population and sample to which the estimate refers rather than calling a locus pair universally “in LD.”
Worked Example: The Same Allele Frequencies, Different Sample LD
Suppose two hypothetical samples each contain 100 phased chromosomes. In sample 1, the counts are \(AB=45\), \(Ab=5\), \(aB=5\), and \(ab=45\). In sample 2, each haplotype occurs 25 times. Both samples have \(p_A=p_B=0.50\), but their \(AB\) frequencies differ.
For sample 1, \(p_{AB}=0.45\), so \(D=0.45-(0.50)(0.50)=0.20\). The denominator for \(r^2\) is \(0.50^4=0.0625\). Hence \(r^2=(0.20)^2/0.0625=0.64\), and \(D'=0.20/0.25=0.80\).
For sample 2, \(p_{AB}=0.25\), so \(D=0.25-0.25=0\), and \(r^2=0\) and \(D'=0\). The calculations are consistent with the stated counts. They show that identical marginal allele frequencies do not determine the joint haplotype distribution. The conclusions apply to these sample estimates; they do not establish that either value is a universal property of the loci.
A Practical Audit for LD Results
Before interpreting an LD estimate, identify what was estimated and what the number answers. This short sequence helps prevent the most consequential errors:
State which allele is counted at each locus, whether frequencies come from phased chromosomes or inferred haplotypes, and which sample is being described.
Verify that haplotype frequencies sum to 1 and that marginal allele frequencies agree with the haplotype counts. Then calculate \(D=p_{AB}-p_Ap_B\).
Use \(r^2\) to describe squared allelic correlation and predictive strength; use D-prime to describe disequilibrium relative to the frequency-dependent bound.
Interpret the result for the specified sample and population, and acknowledge phase inference or sampling uncertainty when relevant.
Common Mistakes
- “D-prime is 1, so one allele perfectly predicts the other.” This confuses a normalized disequilibrium bound with squared correlation. Calculate \(r^2\) to assess predictive strength.
- “Every double heterozygote has the same phase.” The unphased genotype \(Aa,Bb\) is compatible with two phase arrangements. Use an appropriate haplotype-estimation method rather than assigning phase without evidence.
- “The sign of \(D\) changed, so the biological result must have changed.” A single-locus allele recoding reverses the sign of \(D\). Harmonize allele coding before comparing signed values.
- “This locus pair has this LD value everywhere.” An estimate is tied to the sampled population and estimation procedure. State its scope and avoid treating it as a universal locus property.
Check Your Understanding
For each question, distinguish what the data establish from what would require an additional assumption or analysis.
- Why can D-prime equal 1 while \(r^2\) is far below 1?
- What two phase arrangements are compatible with an \(Aa,Bb\) genotype, and why can that genotype alone not identify its haplotypes?
- If one locus is recoded and the sign of \(D\) reverses, which coding information should be checked before comparing the two results?
- Two samples have the same allele frequencies but different haplotype frequencies. Which LD quantities can differ, and what should be stated when interpreting them?
- Which measure is more directly relevant when asking how well one allele predicts another: D-prime or \(r^2\)? Explain the distinction.