LD Differences and Cross-Population Tagging

Why the Same Tag SNP Can Behave Differently Across Populations

A genome-wide association study may use a tag SNP as a proxy for a causal variant that was not directly measured. The proxy is informative when its alleles tend to occur with the causal allele on the same chromosomes. That tendency is linkage disequilibrium (LD). If the tendency differs across populations, the tag’s association with a trait can differ too, even when the causal variant has the same effect in both populations.

The key distinction is between the effect of a causal variant and the marginal association estimate for a tag SNP. A marginal estimate comes from relating the tag to the outcome without conditioning on the causal variant. It therefore reflects both the causal effect and how well the tag tracks the causal genotype in the population being analyzed.

Definition: For two biallelic SNPs, let $H_T$ and $H_C$ be indicators that a randomly sampled chromosome carries the designated allele at the tag SNP and causal SNP, respectively. Write their allele frequencies as $p_T=P(H_T=1)$ and $p_C=P(H_C=1)$. The LD coefficient is $D=P(H_T=1,H_C=1)-p_Tp_C$. The correlation between the chromosome-level allele indicators is $$ r_{TC}=\frac{D}{\sqrt{p_T(1-p_T)p_C(1-p_C)}}. $$ This correlation is defined when both SNPs are polymorphic. Its magnitude describes the strength of correlation, while its sign depends on the chosen allele coding.

For diploid additive genotypes, let $G_T$ and $G_C$ count the designated alleles, taking values 0, 1, or 2. Under random pairing of chromosomes, $\operatorname{Var}(G_T)=2p_T(1-p_T)$, $\operatorname{Var}(G_C)=2p_C(1-p_C)$, and $\operatorname{Cov}(G_T,G_C)=2D$. Thus the correlation between the two dosages is also $r_{TC}$. These variance and covariance formulas use the random-mating model; with real data, empirical genotype covariances can instead be estimated directly.

How LD Enters a Marginal Association

Consider one population at a time and the additive outcome model $Y=\alpha+\beta_C G_C+\varepsilon$, where $Y$ is a quantitative trait, $\beta_C$ is the causal-variant effect in trait units per causal allele, and the residual has mean zero conditional on the genotypes. Assume the model is correctly specified, the analyzed population is sufficiently homogeneous for the association model, and there is no remaining confounding of genotype and outcome. The simple marginal regression of $Y$ on $G_T$ has slope $\beta_T^{\mathrm{marg}}=\operatorname{Cov}(G_T,Y)/\operatorname{Var}(G_T)$.

By the outcome model and the conditional-mean assumption, $\operatorname{Cov}(G_T,\varepsilon)=0$. The covariance in the numerator is therefore $\beta_C\operatorname{Cov}(G_T,G_C)$. Substituting the diploid genotype covariance gives the relationship below. This is a population-level slope; in a finite sample, the estimated slope also varies because of sampling error.

\beta_T^{\mathrm{marg}} =\beta_C\frac{\operatorname{Cov}(G_T,G_C)}{\operatorname{Var}(G_T)} =\beta_C\frac{D}{p_T(1-p_T)} =\beta_C r_{TC}\sqrt{\frac{p_C(1-p_C)}{p_T(1-p_T)}}.

This identity clarifies what a cross-population comparison can and cannot establish. If the causal effect is unchanged but LD differs, the tag’s marginal estimate can change in magnitude or sign. Even when the LD correlation is the same, differences in allele frequencies can change the unstandardized slope through the variance ratio. A change in the tag estimate alone is not evidence that the causal effect changed.

Worked Example: Calculating Tagging in Two Populations

Suppose two hypothetical populations have the same tag-allele frequency $p_T=0.30$ and causal-allele frequency $p_C=0.20$. In Population A, the frequency of chromosomes carrying both designated alleles is $P(H_T=1,H_C=1)=0.14$. In Population B, it is 0.08. Assume the causal effect is $\beta_C=1.5$ trait units per causal allele in both populations, and the stated additive model assumptions hold.

For Population A, $D=0.14-(0.30)(0.20)=0.0800$. The allele-indicator correlation is $$ r_{TC,A}=\frac{0.0800}{\sqrt{(0.30)(0.70)(0.20)(0.80)}}=\frac{0.0800}{0.1833}=0.4364. $$ The tag dosage variance is $2(0.30)(0.70)=0.4200$, and the tag-causal dosage covariance is $2D=0.1600$. Consequently, the marginal tag slope is $$ \beta_{T,A}^{\mathrm{marg}}=1.5\frac{0.1600}{0.4200}=0.5714 $$ trait units per tag allele.

For Population B, $D=0.08-(0.30)(0.20)=0.0200$, so $$ r_{TC,B}=\frac{0.0200}{0.1833}=0.1091. $$ The dosage variance of the tag remains 0.4200, while the dosage covariance is now $2D=0.0400$. The marginal estimate is $$ \beta_{T,B}^{\mathrm{marg}}=1.5\frac{0.0400}{0.4200}=0.1429 $$ trait units per tag allele. The causal effect was set equal in the two populations; the weaker tag-causal correlation in Population B accounts for its smaller marginal tag slope under these assumptions.

Correlation Is Not the Whole Unstandardized Estimate

The LD correlation is a useful measure of tagging strength, but the marginal slope in trait units per tag allele also depends on genotype variances. The following hypothetical comparison isolates that distinction. It is not a contradiction: a correlation is unitless, while a regression slope depends on the scale of its predictor.

Worked Example: Equal LD Correlations, Different Marginal Slopes

Suppose two hypothetical populations both have $r_{TC}=0.3000$, and the causal effect is $\beta_C=1.0$ trait unit per causal allele in each. In Population A, $p_T=0.10$ and $p_C=0.40$; in Population B, $p_T=0.40$ and $p_C=0.10$. For both populations, the LD coefficient required for this correlation is $$ D=(0.3000)\sqrt{(0.10)(0.90)(0.40)(0.60)}=0.04409. $$

In Population A, the tag dosage variance is $2(0.10)(0.90)=0.1800$, and the dosage covariance is $2D=0.08818$. Thus $$ \beta_{T,A}^{\mathrm{marg}}=1.0\frac{0.08818}{0.1800}=0.4899 $$ trait units per tag allele. In Population B, the tag dosage variance is $2(0.40)(0.60)=0.4800$, with the same dosage covariance. Therefore $$ \beta_{T,B}^{\mathrm{marg}}=1.0\frac{0.08818}{0.4800}=0.1837 $$ trait units per tag allele. Equal correlations do not imply equal unstandardized marginal slopes when allele frequencies, and hence genotype scales, differ.

LD Can Also Change the Direction of a Tag Association

The sign of $D$, and therefore of $r_{TC}$, depends on which allele is designated at each SNP. If allele coding is held consistent across populations, a change in the sign of LD can produce opposite-signed marginal tag effects even when the causal effect has the same sign. Such a result describes a change in proxy association, not necessarily a reversal of the biological effect at the causal variant.

Worked Example: Opposite-Signed Tag Estimates

Consider two hypothetical populations with $p_T=0.40$, $p_C=0.20$, and the same causal effect, $\beta_C=2.0$ trait units per causal allele. In Population A, let the joint haplotype frequency be 0.12; in Population B, let it be 0.04. The allele-frequency product is $(0.40)(0.20)=0.0800$, so $D_A=0.12-0.08=0.0400$ and $D_B=0.04-0.08=-0.0400$. The common denominator for the correlation is $\sqrt{(0.40)(0.60)(0.20)(0.80)}=0.19596$, giving $r_{TC,A}=0.2041$ and $r_{TC,B}=-0.2041$.

The tag dosage variance is $2(0.40)(0.60)=0.4800$ in each population. The dosage covariances are $2D_A=0.0800$ and $2D_B=-0.0800$. Hence $$ \beta_{T,A}^{\mathrm{marg}}=2.0\frac{0.0800}{0.4800}=0.3333, \qquad \beta_{T,B}^{\mathrm{marg}}=2.0\frac{-0.0800}{0.4800}=-0.3333. $$ These are marginal estimates in trait units per tag allele. The sign difference follows from the opposite LD phases, given the fixed allele coding and model assumptions.

Diagnostics, Alternatives, and Limits

Before interpreting an ancestry-specific difference, check that LD was estimated in the populations relevant to the association analysis. An external reference panel may not match the analyzed sample’s ancestry composition, local ancestry, or allele frequencies. For an admixed cohort, a single pooled LD estimate can also obscure differences among ancestry backgrounds. Where appropriate, estimate or examine LD within ancestry strata or local ancestry backgrounds, while recognizing that small strata can yield imprecise estimates.

Check allele harmonization as well. Complementing the designated allele at one SNP changes the sign of $D$ and $r_{TC}$, and can change the reported sign of a slope, without changing the underlying biological relationship. Confirm strand and allele labels before comparing signs. Also inspect genotype quality, missingness, and sample size: noisy genotype calls or unstable frequency estimates can distort LD calculations, especially for uncommon alleles. A frequency table should be compatible with the reported joint frequency; for example, $P(H_T=1,H_C=1)$ cannot exceed either $p_T$ or $p_C$.

A single tag is not the only analysis option. Directly measuring the causal variant, when it is known and reliably genotyped, avoids relying on that tag’s LD as a proxy. When the causal variant is unknown, analyzing multiple variants or haplotypes can capture information that a single tag misses, but requires additional modeling and careful interpretation. Conditional analysis can ask whether a tag retains an association after accounting for a candidate causal variant; it does not make an imperfectly measured or incorrectly specified causal model certain.

These comparisons have limits. The simple relationship above assumes an additive, linear outcome model and no residual genotype-outcome confounding. Covariate adjustment, interactions, non-additive effects, or population structure can make an unadjusted covariance calculation insufficient for interpreting a fitted association. LD estimates can also be uncertain, and different LD does not by itself identify the causal variant. In particular, a weaker tag association is not proof of a smaller causal effect, while a strong tag association does not prove that the tag itself is causal.

Key takeaway: A tag SNP’s marginal association depends on its population-specific relationship with the causal genotype. Compare LD, allele frequencies, allele coding, and model assumptions before attributing cross-population differences to a change in the causal effect.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

If the causal effect is positive and $D=0$, what is the population marginal slope of the tag SNP under the stated additive model?
A tag and causal SNP have $D=0.0300$, $p_T=0.20$, and $p_C=0.30$. Calculate $r_{TC}$ to three significant figures.
Two populations have the same positive LD correlation and causal effect, but different tag allele frequencies. Must their unstandardized marginal tag slopes be equal? Explain.
With consistent allele coding, what does a negative tag slope imply about the causal effect if $D<0$ and the model assumptions hold?

Solutions. (1) The slope is zero because $\operatorname{Cov}(G_T,G_C)=2D=0$, so $\beta_T^{\mathrm{marg}}=\beta_C(0)/\operatorname{Var}(G_T)=0$. (2) $r_{TC}=0.0300/\sqrt{(0.20)(0.80)(0.30)(0.70)}=0.0300/0.1833=0.164$. (3) No. The slope is $\beta_C r_{TC}\sqrt{p_C(1-p_C)/[p_T(1-p_T)]}$, so a different tag frequency can change the genotype-scale factor. (4) Since $\beta_T^{\mathrm{marg}}=\beta_C D/[p_T(1-p_T)]$ and the denominator is positive, a negative tag slope with $D<0$ is consistent with a positive causal effect; the tag’s association direction is reversed by the LD phase.