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Population Structure, Ancestry and Relatedness · Tutorial 109 of 795

Admixture Gradients and Intermediate PCA Positions

Learn why ancestry mixtures can yield intermediate PCA scores, how to verify the arithmetic, and why an intermediate position is not by itself proof of a particular ancestry history.

Intermediate 10 min read

What You'll Learn

  • Derive expected allele dosage from source-population allele frequencies and ancestry proportion.
  • Relate ancestry-weighted genotype means to intermediate PCA scores.
  • Distinguish affine centering from linear scaling and projection.
  • Interpret gradients while accounting for local ancestry and genotype sampling variation.
  • Recognize when intermediate PCA positions do not establish recent admixture.

Why an Admixed Individual Can Appear Between Clusters

A PCA plot often shows groups of individuals in separated regions, with some individuals between them or distributed along a gradient. One possible explanation is admixture: an individual's genome contains contributions from populations with different allele frequencies. The key mathematical point is that expected genotypes under a two-source mixture are weighted averages of the source expectations. A fixed PCA transformation carries that weighted-average relationship to the score coordinates.

This is an explanation of a geometric pattern, not a rule that every admixed individual must fall neatly between two visible clusters. Actual genotypes vary around their expectations, ancestry can vary across genomic regions, and the displayed principal components depend on the samples and variants used to construct them. An intermediate position alone does not identify ancestry proportions or prove a particular history.

A Two-Source Allele-Frequency Model

Consider a biallelic variant in diploid individuals. Let \(p_A\) and \(p_B\) be the frequencies of the counted allele in source populations A and B. Let \(G\) denote the individual's genotype dosage, with possible values 0, 1, or 2. Let \(q\) be the expected fraction of the individual's allele copies at this variant that come from source A; the remaining fraction \(1-q\) comes from source B. If an allele copy drawn from source A has probability \(p_A\) of carrying the counted allele, and one drawn from source B has probability \(p_B\), then the expected dosage is twice the mixture allele frequency.

Definition: Under the two-source mixture model, the allele frequency at a variant for ancestry fraction \(q\) is \(p(q)=q p_A+(1-q)p_B\), and the expected genotype dosage is \(E[G\mid q]=2p(q)\). Here \(0\leq q\leq 1\). The expectation is not necessarily an observed genotype: a realized dosage remains 0, 1, or 2.

The model is stated for one variant. Across many variants, the vector of expected dosages is a weighted average of the two source-population dosage-mean vectors, provided the same \(q\) applies at each variant. For recent admixture, local ancestry proportions can differ among genomic regions, so the genome-wide average \(q\) need not describe every variant equally well. The dosage expectation at a particular variant is most directly expressed using its local ancestry proportion.

The simple model assumes the source allele frequencies are appropriate for the populations contributing ancestry, the variant is measured consistently, and the ancestry mixture is meaningfully represented by the stated proportion. It does not require the observed genotype to equal its expectation. Random allele sampling, recombination, and finite-sample variation all create deviations.

Worked Example: Expected Dosages at Three Variants

Suppose three hypothetical variants have counted-allele frequencies \(p_A=(0.80,0.20,0.60)\) and \(p_B=(0.20,0.70,0.30)\). Consider an individual with ancestry fraction \(q=0.40\) from A and \(0.60\) from B at these variants.

At variant 1, the mixture frequency is \(0.40(0.80)+0.60(0.20)=0.32+0.12=0.44\), so the expected dosage is \(2(0.44)=0.88\). At variant 2 it is \(0.40(0.20)+0.60(0.70)=0.08+0.42=0.50\), giving expected dosage \(1.00\). At variant 3 it is \(0.40(0.60)+0.60(0.30)=0.24+0.18=0.42\), giving expected dosage \(0.84\).

The source mean-dosage vectors are \(\boldsymbol{\mu}_A=(1.60,0.40,1.20)\) and \(\boldsymbol{\mu}_B=(0.40,1.40,0.60)\). Their weighted average is \(0.40\boldsymbol{\mu}_A+0.60\boldsymbol{\mu}_B=(0.64+0.24,\;0.16+0.84,\;0.48+0.36)=(0.88,1.00,0.84)\), which agrees with the locus-by-locus calculation. These are expected values, not a claim that a single person's observed genotype vector has fractional entries.

From Expected Genotypes to PCA Scores

PCA begins with a genotype matrix and represents each individual by coordinates along principal-component axes. In the usual centered and possibly scaled analysis, each variant is transformed using fixed values estimated from the analyzed data. For example, a genotype dosage \(G_j\) may be centered by subtracting \(c_j\), and divided by a positive scale \(s_j\). A component score is then a weighted sum of these transformed values. Once the centering constants, scales, and component loadings are fixed, this map from an individual's genotype vector to a score is affine.

An affine transformation has the form \(T(x)=Lx+b\), where \(L\) is a linear map and \(b\) is a fixed vector. Unlike a linear map, an affine map need not send zero to zero. Nevertheless, it preserves weighted averages when the weights sum to one. This detail matters because centering is a translation: it is affine, not generally linear. Scaling and projection are linear operations, but the whole centered-and-scaled score calculation includes that translation.

$$ T\{q\boldsymbol{\mu}_A+(1-q)\boldsymbol{\mu}_B\} =qT(\boldsymbol{\mu}_A)+(1-q)T(\boldsymbol{\mu}_B), \qquad q+(1-q)=1. $$

To verify the identity, write \(T(x)=Lx+b\). The left side is \(L\{q\boldsymbol{\mu}_A+(1-q)\boldsymbol{\mu}_B\}+b\). The right side is \(q(L\boldsymbol{\mu}_A+b)+(1-q)(L\boldsymbol{\mu}_B+b)\), which equals the same expression because \(qb+(1-q)b=b\). A PCA score is a coordinate of this transformed vector, so its expected value follows the same weighted-average relation when the PCA transformation is held fixed.

This does not guarantee that a point lies exactly on a line between two plotted cluster centers. The result concerns the transformed source mean vectors under a shared, fixed transformation. Cluster centers estimated from finite samples may not equal those source means, other sources of variation may contribute, and a plot may show only two components of a higher-dimensional representation.

Worked Example: Centering, Scaling, and Projecting

Take two hypothetical two-variant mean dosage vectors, \(\boldsymbol{\mu}_A=(1.60,0.40)\) and \(\boldsymbol{\mu}_B=(0.40,1.20)\), with \(q=0.25\). Center at \(\boldsymbol{c}=(1.00,0.80)\) and scale the coordinates by \(s_1=0.50\), \(s_2=0.40\). The mixture mean is \(0.25\boldsymbol{\mu}_A+0.75\boldsymbol{\mu}_B=(0.70,1.00)\).

The transformed A vector is \(((1.60-1.00)/0.50,\;(0.40-0.80)/0.40)=(1.20,-1.00)\). The transformed B vector is \(((0.40-1.00)/0.50,\;(1.20-0.80)/0.40)=(-1.20,1.00)\). Transforming the mixture directly gives \(((0.70-1.00)/0.50,\;(1.00-0.80)/0.40)=(-0.60,0.50)\).

The weighted average of the transformed endpoints is \(0.25(1.20,-1.00)+0.75(-1.20,1.00)=(-0.60,0.50)\), exactly the same. Now project onto the fixed unit direction \(\boldsymbol{v}=(0.60,0.80)\). The endpoint scores are \(1.20(0.60)-1.00(0.80)=-0.0800\) for A and \(-1.20(0.60)+1.00(0.80)=0.0800\) for B. The mixture score is \(-0.60(0.60)+0.50(0.80)=0.0400\), also equal to \(0.25(-0.0800)+0.75(0.0800)=0.0400\). The chosen direction illustrates the arithmetic; it is not asserted to be the PC estimated from these data.

Why Gradients Can Appear Rather Than Separate Groups

With two sources and a single ancestry fraction \(q\) shared across variants, expected genotype vectors trace a line segment as \(q\) ranges from zero to one. Their transformed scores likewise trace a line segment in the full score space under a fixed PCA map. A plot of two components shows a projection of that geometry. If the source contrast is represented mainly by the first component, changing \(q\) can produce a visible progression from one cluster toward the other.

A continuous-looking gradient can also arise when ancestry proportions vary across individuals, when there are more than two contributing populations, or when within-population structure is present. Conversely, recent admixture can produce scatter rather than a smooth line because local ancestry varies across the genome and each person's realized genotypes are discrete. The gradient is a consequence that is compatible with a mixture model, not a unique signature of it.

Worked Example: A One-Variant Gradient Across Ancestry Proportions

At a hypothetical variant, let \(p_A=0.75\) and \(p_B=0.25\). For individuals whose local ancestry proportions are \(q=0.20\), \(0.50\), and \(0.80\), the expected dosages are respectively:

For \(q=0.20\), \(2\{0.20(0.75)+0.80(0.25)\}=2(0.35)=0.70\). For \(q=0.50\), \(2\{0.50(0.75)+0.50(0.25)\}=2(0.50)=1.00\). For \(q=0.80\), \(2\{0.80(0.75)+0.20(0.25)\}=2(0.65)=1.30\).

The expected dosage increases by \(0.60\) as \(q\) increases by \(0.60\), a slope of \(1.00\) dosage unit per unit increase in \(q\). In general here, \(E[G\mid q]=2\{0.25+0.50q\}=0.50+1.00q\). A realized genotype at this one variant cannot equal 0.70 or 1.30. Across many informative variants, however, the average pattern can support a continuous score gradient, while individual scores retain sampling variation.

Reading the Pattern Carefully

The PCA coordinates are relative to the chosen samples, variants, preprocessing, and fitted axes. Thus “between” means between in the displayed coordinate system, not necessarily an intermediate value on every biological or historical dimension. A useful interpretation checks whether the proposed source populations are separated by allele-frequency differences at the analyzed variants and whether the observed direction of movement is consistent across informative components.

Several diagnostics can reveal when a simple two-source explanation is inadequate:

  • Inconsistent directions across components. If individuals that appear intermediate on one component are not intermediate on another, a single ancestry axis may not summarize the pattern.
  • Curvature or multiple branches. These may reflect more than two sources, continuous geographic structure, or other structure in the sampled populations.
  • Large individual departures from a gradient. Check genotype quality, missingness patterns, and whether a few unusual variants dominate the scores. The gradient should not be attributed to ancestry before technical artifacts are considered.
  • Changing results with the analyzed sample. PCA axes are estimated from the data. Adding or removing groups can change the directions and visual spacing, even though the fixed-map mixture identity remains true for any one fitted map.

The earlier material on genotype centering and scaling, PCA, and linkage-disequilibrium pruning is relevant when assessing these patterns: variant scaling affects each variant's contribution, and correlated variants can give one genomic region disproportionate influence. These choices affect the fitted representation, not the weighted-average algebra itself.

Common Mistake and Limits

Common mistake: “The point is between two clusters, so the person must have exactly the ancestry proportion suggested by its position.” This reasoning is too strong. Intermediate scores can be consistent with admixture, but score distances depend on the fitted PCA axes, variant scaling, the reference samples, and other sources of genetic variation. PCA position is not, by itself, a calibrated ancestry estimate or proof of recent admixture.

Another subtle mistake is calling centering a linear operation. Subtracting a nonzero centering value does not send zero to zero, so centering is affine. The intermediate-score relationship still holds because affine transformations preserve weighted averages whose weights sum to one; scaling and projection are linear, and the ancestry weights \(q\) and \(1-q\) sum to one.

The two-source model is deliberately simple. It does not distinguish recent from older admixture solely through a PCA score, model local ancestry segments, or account for every demographic process that can create allele-frequency gradients. The next analytical step depends on the scientific question; a PCA picture should be treated as a summary of genetic variation rather than a complete ancestry history.

Check Your Understanding

Answer each question, then compare with the brief solutions.

  1. If \(p_A=0.60\), \(p_B=0.20\), and \(q=0.25\), what is the expected genotype dosage?
  2. Why does a centered, scaled PCA score preserve the weighted-average relation even though centering is not linear?
  3. Does an expected dosage of 0.90 mean that the observed genotype is 0.90? Explain.
  4. Name two reasons an individual with admixture might not lie exactly on the line between two plotted cluster centers.

Solutions. For question 1, \(p(q)=0.25(0.60)+0.75(0.20)=0.15+0.15=0.30\), so \(E[G\mid q]=2(0.30)=0.60\). For question 2, the complete transformation is affine, \(T(x)=Lx+b\); applying it to a weighted average preserves that average when the weights sum to one, since \(qb+(1-q)b=b\). For question 3, no. An observed dosage is 0, 1, or 2; 0.90 is an expectation across the stated model, not a possible single diploid genotype dosage. For question 4, answers include local ancestry variation across the genome, random genotype variation, finite-sample differences between estimated cluster centers and source means, additional ancestral sources, or projection onto only two components.

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