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Relatedness and Kinship Estimation · Tutorial 169 of 989

Linkage Disequilibrium And Effective Marker Counts

Learn to quantify redundancy among correlated markers and to assess what LD pruning or marker weighting does to the precision and stability of relatedness estimates.

Intermediate 10 min read

What You'll Learn

  • Distinguish SNP linkage disequilibrium from correlation among marker-level relatedness contributions
  • Derive an effective marker count from the variance of a correlated-marker average
  • Calculate how LD reduces the precision implied by a raw SNP count
  • Compare pruning and weighting as ways to limit redundant marker information
  • Recognize when an LD-based effective count is only an approximation
  • Avoid treating an effective marker count as a literal number of independent SNPs

Why the Raw Marker Count Can Mislead

A relatedness estimate may combine information across thousands of SNPs. It is tempting to think that multiplying the number of markers by four will reduce its standard error by a factor of two. That conclusion relies on the added markers contributing independent information. When nearby SNPs are in linkage disequilibrium (LD), they tend to carry overlapping information, so the gain in precision can be much smaller.

The practical issue is not simply whether SNP genotypes are correlated. It is how dependence across markers affects the marker-level contributions to the relatedness estimator. LD pruning removes some correlated markers; marker weighting reduces or redistributes their influence. Both can affect precision and the stability of estimates, but neither automatically improves every estimate.

Definition: Linkage disequilibrium is the non-random association of alleles at different loci in a population. For two biallelic loci, a common measure is \(r^2\), the squared correlation between their allele-count variables. An effective marker count is the number of independent, equally informative markers that would give the same variance as the correlated markers under a specified estimator and dependence model. It is a variance-equivalent summary, not a literal count of independent SNPs.

For relatedness estimation, let \(Z_l\) denote the contribution from marker \(l\), with common expected value \(\theta\) and common variance \(\tau^2\). Here \(\theta\) is the relatedness quantity the estimator targets, and the \(Z_l\) are understood to be on the estimator's scale. The correlations among these contributions are written \(\rho_{lm}=\operatorname{Corr}(Z_l,Z_m)\). In general, \(\rho_{lm}\) is not automatically equal to genotype \(r^2\): its value depends on how marker genotypes enter the estimator and on the individuals being compared. LD measures can be useful proxies, but the estimator's contribution covariance is the quantity that determines variance.

Variance and Effective Marker Count

Suppose the estimate is an equally weighted average of \(M\) marker contributions, each with variance \(\tau^2\). Its variance is the sum of all individual variances and all pairwise covariances, divided by \(M^2\):

$$ \hat{\theta}=\frac{1}{M}\sum_{l=1}^{M} Z_l,\qquad \operatorname{Var}(\hat{\theta}) =\frac{\tau^2}{M^2}\left(M+2\sum_{l<m}\rho_{lm}\right). $$

If the markers were independent, every \(\rho_{lm}\) would be zero, and the variance would be \(\tau^2/M\). Define \(M_{\mathrm{eff}}\) by setting that independent-marker variance equal to the variance with dependence. This gives the following result.

Result: With equal marker variances and equal marker weights, the variance-equivalent effective marker count is
$$ M_{\mathrm{eff}} =\frac{M^2}{M+2\sum_{l<m}\rho_{lm}}, \qquad \operatorname{Var}(\hat{\theta})=\frac{\tau^2}{M_{\mathrm{eff}}}. $$

If all distinct pairs have the same contribution correlation \(\rho\), there are \(M(M-1)/2\) such pairs. Substituting this number yields \(M_{\mathrm{eff}}=M/[1+(M-1)\rho]\). Positive correlations therefore reduce the effective count. The formula also shows why simply dividing the raw marker count by an informal “LD factor” can be unreliable: the dependence pattern and the estimator matter.

Worked Example: Four Correlated Marker Contributions

Worked Example: Four Correlated Marker Contributions

Suppose a hypothetical relatedness estimator averages four marker contributions, with observed values \(0.12, 0.14, 0.10,\) and \(0.16\). For a simple precision calculation, assume each contribution has standard deviation \(\tau=0.020\), and every pair of contributions has correlation \(\rho=0.60\). These assumptions describe the hypothetical dependence model; the four observed values alone do not establish it.

The estimate is

$$ \hat{\theta}=\frac{0.12+0.14+0.10+0.16}{4} =\frac{0.52}{4}=0.130. $$

Under the equal-correlation model, the effective marker count is

$$ M_{\mathrm{eff}} =\frac{4}{1+(4-1)(0.60)} =\frac{4}{2.80}=1.429. $$

The corresponding standard error is

$$ SE(\hat{\theta})=\frac{0.020}{\sqrt{1.429}} =0.01673. $$

If the four contributions were incorrectly treated as independent, the calculated standard error would be \(0.020/\sqrt{4}=0.0100\). Thus, under the stated model, treating the raw count as four independent markers understates the standard error. The value \(M_{\mathrm{eff}}=1.429\) means that the variance matches that of about 1.429 independent contributions with the same variance; it does not mean that the data contain a fractional SNP.

Weighted Contributions and LD Pruning

The same reasoning applies when markers have unequal weights. For fixed weights \(w_l\), define the estimate as \(\hat{\theta}_w=\sum_l w_lZ_l/\sum_lw_l\). If the contributions have common variance \(\tau^2\), its variance-equivalent effective count is

$$ M_{\mathrm{eff},w} =\frac{\left(\sum_l w_l\right)^2} {\sum_l\sum_m w_lw_m\rho_{lm}}. $$

The double sum includes the diagonal terms, for which \(\rho_{ll}=1\). With independent markers, this reduces to \((\sum_lw_l)^2/\sum_lw_l^2\), which is no greater than the number of nonzero weights. Correlation can reduce it further. This calculation describes variance under the assumed contribution covariance; it does not establish that any particular weighting rule is optimal.

LD pruning usually removes one or more markers from groups that exceed a chosen LD threshold. It reduces redundancy and computation, and can make an analysis less dominated by dense clusters of correlated SNPs. However, pruning also discards data. A threshold such as a maximum pairwise \(r^2\) is a practical selection rule, not a direct calculation of \(M_{\mathrm{eff}}\) or a guarantee of a particular standard error. The result depends on the threshold, the region used, and the sample in which LD is assessed.

Weighting can instead limit the combined influence of dense regions. For example, assigning each region a fixed total weight means that a region with many typed SNPs does not automatically dominate one with fewer SNPs. This changes how the estimator combines its marker contributions. The weights should be stated and justified; a weighted estimate need not equal the unweighted estimate, and a different weighting scheme can have a different variance.

Worked Example: Pruning Two Correlated Blocks

Consider six hypothetical marker contributions in two blocks of three. The contributions are \(0.18, 0.16, 0.17\) in block A and \(0.10, 0.12, 0.11\) in block B. The mean of all six is

$$ \hat{\theta}=\frac{0.18+0.16+0.17+0.10+0.12+0.11}{6} =\frac{0.84}{6}=0.140. $$

Suppose each contribution has standard deviation \(0.020\), contributions within each block have correlation \(0.80\), and contributions across blocks are uncorrelated. There are three within-block pairs in each block, or six correlated pairs altogether. Therefore,

$$ \operatorname{Var}(\hat{\theta}) =\frac{0.020^2}{6^2}\left(6+2(6)(0.80)\right) =\frac{0.0004}{36}(15.6) =0.0001733. $$

This gives \(M_{\mathrm{eff}}=36/15.6=2.308\) and \(SE(\hat{\theta})=\sqrt{0.0001733}=0.01316\). Suppose pruning retains the first contribution from each block, \(0.18\) and \(0.10\). The pruned estimate is \((0.18+0.10)/2=0.140\), and, under the stated model of independent blocks, its standard error is \(0.020/\sqrt{2}=0.01414\). In this particular example the point estimate is unchanged, while pruning slightly increases the standard error because it retains only two contributions. Pruning can still be useful for reducing redundancy or stabilizing analysis choices; it is not a promise of a smaller standard error.

Worked Example: Giving Each LD Block Equal Weight

Suppose a hypothetical data set has four contributions in block A, \(0.18, 0.20, 0.22,\) and \(0.20\), and one contribution in block B, \(0.00\). The unweighted marker average is

$$ \hat{\theta}_{\mathrm{marker}} =\frac{0.18+0.20+0.22+0.20+0.00}{5} =\frac{0.80}{5}=0.160. $$

Now use the average contribution within each block, then give the two blocks equal weight. Block A has mean \((0.18+0.20+0.22+0.20)/4=0.200\), and block B has mean \(0.000\). Thus,

$$ \hat{\theta}_{\mathrm{block}} =\frac{0.200+0.000}{2}=0.100. $$

The estimates differ because the first weights each SNP equally, while the second weights each block equally. To see the precision implication under a simple model, assume the four contributions in block A are perfectly correlated, block B is independent of block A, and all contributions have variance \(\tau^2\). The block-weighted estimate averages one block-level contribution from each of two independent blocks, so its variance is \(\tau^2/2\), equivalent to two independent contributions. The marker-weighted estimate gives block A total weight \(4/5\) and block B weight \(1/5\). Its variance is

$$ \left[\left(\frac{4}{5}\right)^2+\left(\frac{1}{5}\right)^2\right]\tau^2 =\frac{17}{25}\tau^2, \qquad M_{\mathrm{eff}}=\frac{25}{17}=1.471. $$

In this model, equal block weighting increases the effective count from 1.471 to 2, but it also changes the estimate's weighting target. The calculation illustrates why the choice between marker weighting and block weighting should reflect the intended analysis, not just a desire to maximize an effective count.

A Short Derivation and Its Limits

The effective-count result follows directly from covariance addition. For the equally weighted estimate, each variance contributes \(\tau^2\). Each distinct pair contributes \(\operatorname{Cov}(Z_l,Z_m)=\rho_{lm}\tau^2\), and each pair appears twice when the variance of a sum is expanded. Hence

$$ \operatorname{Var}\left(\frac{1}{M}\sum_l Z_l\right) =\frac{1}{M^2}\left(\sum_l\tau^2 +2\sum_{l<m}\rho_{lm}\tau^2\right) =\frac{\tau^2}{M^2}\left(M+2\sum_{l<m}\rho_{lm}\right). $$

Equating this to \(\tau^2/M_{\mathrm{eff}}\) and solving for \(M_{\mathrm{eff}}\) gives the stated formula. Its assumptions matter: the contributions must have the common variance used in the calculation, and their covariance pattern must be represented adequately. If variances differ, replace the common-\(\tau^2\) covariance model with the actual variances and covariances. If marker weights are used, use the weighted formula rather than the equal-weight formula.

Genotype LD summaries such as \(r^2\) can help identify redundant markers, but they are not universally interchangeable with \(\rho_{lm}\) for relatedness contributions. In particular, dependence can differ across populations and estimator constructions. An effective count calculated from an LD summary is therefore best described as an approximation under a stated convention, not a universal property of a marker panel.

Why Precision and Stability Both Matter

Precision concerns the variability of an estimate under repeated sampling or comparable data. Redundant markers contribute less additional precision than independent ones, even if they make the raw marker count appear large. Stability concerns how sensitive results are to reasonable changes in marker selection, weighting, or sample composition. A dense LD block can have disproportionate influence on an unweighted estimate; pruning or block weighting can reveal whether that influence matters.

These procedures can also change results for reasons other than variance. Pruning selects a subset, and weighting changes the contribution of genomic regions. If marker contributions are not all centered on the same target, either choice can shift the estimate as well as its precision. Thus, compare estimates and uncertainty under clearly described choices; do not interpret agreement or disagreement as proof of a particular biological relationship by itself.

Key takeaway: The raw SNP count measures how many markers were included, not how many independent units of information they supplied. Use the covariance among marker-level contributions to quantify variance-equivalent information, and report pruning or weighting choices as part of the estimator specification.

Common Mistake

  • Wrong reasoning: “There are 10,000 SNPs, so the standard error is the single-marker standard deviation divided by \(\sqrt{10{,}000}\).”
  • Correction: That calculation assumes independent, equally variable contributions. LD can make the effective marker count much smaller than 10,000, so the standard error must reflect covariance among contributions or a clearly stated approximation to it.
  • Wrong reasoning: “Pruning always improves precision because it removes correlated SNPs.”
  • Correction: Pruning removes redundant information but also discards markers. It can improve balance or stability without reducing the variance of a particular estimate. Calculate or compare uncertainty under the actual procedure.

Check Your Understanding

Answer each question using the assumptions stated in it.

  1. Four equally variable marker contributions have pairwise correlation \(0.50\). What is their effective marker count?
  2. Why is genotype \(r^2\) not necessarily identical to the correlation between two markers’ contributions to a relatedness estimate?
  3. Under what assumptions does the effective-count formula \(M^2/[M+2\sum_{l<m}\rho_{lm}]\) apply?
  4. A pruning rule retains one SNP from each of two correlated blocks. Name one potential benefit and one cost of this choice.
  5. How can equal block weighting change both an estimate and its precision compared with equal SNP weighting?
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