Constructing Dominance Relationship Matrices

Why Construct a Dominance Relationship Matrix?

A dominance relationship matrix summarizes similarity between individuals in their dominance-coded genotypes across loci. It is built from genotype contrasts that distinguish heterozygotes from the two homozygote classes, rather than from the linear allele-count contrast used for additive effects. The matrix is a genomic similarity measure under a specified coding and scaling; its entries depend on both.

The key construction decision is how to center and scale each locus. Centering makes the contrast have mean zero under a stated genotype-frequency model. Scaling determines how much each locus contributes: equal weight per locus and weight proportional to dominance variance are different choices, especially when allele frequencies vary. These decisions should be stated whenever the matrix is reported.

Dominance Contrasts at One Locus

At a biallelic locus, let \(G\) be the number of copies of a designated allele, so \(G\) is 0, 1, or 2. Let \(p\) be that allele's frequency and \(q=1-p\). For the basic derivation, assume Hardy–Weinberg equilibrium (HWE), so the genotype probabilities for \(G=0,1,2\) are \(q^2,2pq,p^2\), respectively. The coding below is a centered dominance contrast under this model.

Definition: The unstandardized dominance contrast \(h(G)\) takes the values \(-2p^2\) for \(G=0\), \(2pq\) for \(G=1\), and \(-2q^2\) for \(G=2\). Its HWE mean is zero and its HWE variance is \(4p^2q^2\). For \(0<p<1\), the standardized contrast is \(z(G)=h(G)/(2pq)\), which has HWE mean zero and variance one.

This coding is not simply the heterozygote indicator minus its mean. It is a particular contrast among all three genotype classes. Its scaling also matters: \(h\) is measured in the unstandardized contrast units, whereas \(z\) has unit variance under HWE.

Worked Example: Centering and Scaling One Locus

Suppose a hypothetical locus has counted-allele frequency \(p=0.30\), hence \(q=0.70\). Under HWE, genotype probabilities for \(G=0,1,2\) are \(0.49,0.42,0.09\). Find the dominance contrasts and verify their mean and variance.

The unstandardized values are \(h(0)=-2(0.30)^2=-0.18\), \(h(1)=2(0.30)(0.70)=0.42\), and \(h(2)=-2(0.70)^2=-0.98\). The probability-weighted mean is \(0.49(-0.18)+0.42(0.42)+0.09(-0.98)=-0.0882+0.1764-0.0882=0\).

The variance, since the mean is zero, is \(0.49(-0.18)^2+0.42(0.42)^2+0.09(-0.98)^2 =0.015876+0.074088+0.086436=0.1764\). The formula gives \(4p^2q^2=4(0.30)^2(0.70)^2=0.1764\), as required. Dividing each \(h\) value by \(2pq=0.42\) gives standardized values \(-0.4286,1,-2.333\). Their HWE mean is zero and their HWE variance is one.

Why the Contrast Is Orthogonal to the Additive Contrast

Under HWE, the additive genotype contrast is \(G-2p\), the deviation of allele count from its expected value. The dominance contrast above is orthogonal to it: their expected product is zero. This is a property of the coding under HWE, not a claim that every sample will have zero empirical covariance.

Derivation: Using genotype probabilities \(q^2,2pq,p^2\), the expected product of the additive and dominance contrasts is

\begin{aligned} E[(G-2p)h(G)] &=q^2(-2p)(-2p^2)+2pq(1-2p)(2pq)+p^2(2q)(-2q^2)\\ &=4p^2q^2\{p+(1-2p)-q\}\\ &=0, \end{aligned}

because \(q=1-p\). The mean and variance follow directly from the genotype probabilities:

\begin{aligned} E[h(G)]&=-2p^2q^2+4p^2q^2-2p^2q^2=0,\\ E[h(G)^2]&=4p^4q^2+8p^3q^3+4p^2q^4 =4p^2q^2(p+q)^2=4p^2q^2. \end{aligned}

Orthogonality provides a useful contrast to the additive coding at one locus. It does not by itself ensure that estimated additive and dominance components will be cleanly separated in a finite or structured sample. That depends on the data and model as well as on the coding.

From Genotype Contrasts to a Matrix

For \(n\) individuals and \(m\) loci, let \(G_{il}\) denote the counted-allele dosage for individual \(i\) at locus \(l\), and let \(p_l\) and \(q_l=1-p_l\) be the allele frequencies used for that locus. Define \(z_{il}=h_l(G_{il})/(2p_lq_l)\). Collect these values in an \(n\)-by-\(m\) matrix \(Z_D\). One common convention gives every standardized locus equal weight:

Definition: The equal-locus standardized dominance relationship matrix is \(K_D=Z_DZ_D^{\mathsf T}/m\). Its \(i,j\) entry is the average, across the \(m\) loci, of the product of individuals \(i\)'s and \(j\)'s standardized dominance contrasts. This definition uses the same set of loci for every entry.

(K_D)_{ij}=\frac{1}{m}\sum_{l=1}^{m}z_{il}z_{jl}

A positive entry means that the two individuals tend to have dominance contrasts with the same sign and magnitude under this coding; a negative entry means their products tend to be negative. Neither sign alone establishes a biological interaction or a causal relationship. The diagonal is an individual's average squared standardized contrast and need not equal one in a finite data set.

Because \(K_D\) is a Gram matrix scaled by a positive constant, it is positive semidefinite: for any vector \(a\) of length \(n\), \(a^{\mathsf T}K_Da=(Z_D^{\mathsf T}a)^{\mathsf T}(Z_D^{\mathsf T}a)/m\geq0\). Under HWE, an unrelated individual's expected squared standardized contrast at each locus is one, so the expected diagonal is one. For two independent individuals, the expected cross-product is zero. Related individuals need not have zero expected cross-product.

Worked Example: Computing a One-Locus Relationship Matrix

For illustration, suppose three hypothetical individuals have dosages \(G=0,1,2\) at a locus with \(p=0.30\). Use the standardized coding from the first example and the equal-locus formula. With one locus, \(m=1\), so the matrix is the outer product of the three standardized values.

The standardized vector is \(z=(-0.4286,1,-2.333)^{\mathsf T}\), or exactly \((-3/7,1,-7/3)^{\mathsf T}\). Therefore

K_D=zz^{\mathsf T}= \begin{pmatrix} 9/49 & -3/7 & 1\\ -3/7 & 1 & -7/3\\ 1 & -7/3 & 49/9 \end{pmatrix} \approx \begin{pmatrix} 0.1837 & -0.4286 & 1.000\\ -0.4286 & 1.000 & -2.333\\ 1.000 & -2.333 & 5.444 \end{pmatrix}.

For example, the entry for individuals 1 and 3 is \((-3/7)(-7/3)=1\); their dominance contrasts have the same sign. The entry for individuals 1 and 2 is \((-3/7)(1)=-0.4286\). The large diagonal value for individual 3 is not an error: this individual has the rare, high-magnitude genotype contrast at this allele frequency. A one-locus matrix is an illustration, not a stable genome-wide estimate.

Scaling Choices and Allele Frequency

Equal-locus standardization gives each locus unit HWE variance before averaging. Consequently, a locus with a rare allele receives the same nominal weight as a common-allele locus. Another valid construction preserves the relative HWE dominance variances across loci:

K_D^{(v)}= \frac{HH^{\mathsf T}}{\sum_{l=1}^{m}4p_l^2q_l^2}, \qquad H_{il}=h_l(G_{il}).

Here locus \(l\)'s contribution is scaled according to its HWE variance \(4p_l^2q_l^2\). Equivalently, this matrix is a weighted average of standardized-locus products, with weights proportional to those variances. Under the stated HWE model, its expected diagonal is one when the denominator is positive. It is also positive semidefinite. The two definitions answer different weighting choices; neither is a universal normalization. In either case, the exact frequency estimates, locus set, and scaling should be reported.

Worked Example: A Rare Allele and Large Standardized Values

Consider a hypothetical locus with counted-allele frequency \(p=0.02\) and \(q=0.98\). The standardized contrast is \(-p/q\) for \(G=0\), \(1\) for \(G=1\), and \(-q/p\) for \(G=2\). Thus the values are approximately \(-0.02041,1,-49.00\). Under HWE, the \(G=2\) genotype has probability \(p^2=0.0004\).

The expected squared standardized contrast is \(0.9604(0.02041)^2+0.0392(1)^2+0.0004(49)^2 =0.0004+0.0392+0.9604=1.000\), up to rounding. The rare homozygote is uncommon, but when observed it has a large standardized value and can contribute a large product to a matrix entry. For the unstandardized contrast, its value is \(-2q^2=-1.9208\), while the HWE variance of that contrast is only \(4p^2q^2=0.0015366\), approximately. This illustrates why equal-locus standardization can make rare-allele observations influential, whereas variance weighting reduces that locus's overall contribution.

Genotype Coding, Frequencies, and Practical Checks

The allele being counted must be specified, as must the frequencies used to center and scale. If the counted allele is switched, \(G\) becomes \(2-G\) and \(p\) becomes \(q\). Under the contrast defined here, each person's \(h\) and \(z\) value remains unchanged: the old \(G=2\) value becomes the new \(G=0\) value, the heterozygote value is unchanged, and the two homozygote formulas exchange consistently. Thus a pure allele-label swap does not change this matrix.

That invariance does not make arbitrary recoding harmless. Replacing the three-level contrast with an uncentered heterozygote indicator, or centering a genotype dosage but treating it as a dominance contrast, defines a different matrix. Likewise, using frequencies estimated in a sample with substantial ancestry structure or HWE departures may not produce empirical mean-zero contrasts. The HWE centering and variance statements are theoretical statements under HWE at the specified frequency; they do not guarantee that each observed sample has exactly those moments.

In applications, frequencies are often estimated from the analyzed genotypes or obtained from a reference set. State which choice is used and apply it consistently. Frequencies near zero or one make the standardized contrast unstable and undefined at the boundaries \(p=0\) or \(p=1\). Filtering rare variants or using variance-weighted scaling can limit their influence, but these choices alter the matrix and should be made deliberately. Genotype quality, consistent allele orientation, and a common locus set also matter for interpretable entries.

Key takeaway: A dominance relationship matrix is determined by the genotype contrast, allele frequencies, locus set, and normalization. Check the coding and scaling before interpreting its entries, and do not assume that every diagonal equals one or that rare alleles contribute little.

Common Mistakes

Using the heterozygote indicator as though it were this contrast. The coding here assigns a frequency-dependent value to each of the three genotype classes. Simply coding heterozygotes as one and homozygotes as zero does not reproduce the stated centered, additive-orthogonal contrast.
Assuming standardization makes all loci equally reliable. Unit HWE variance gives each locus equal nominal weight, but a rare genotype can still have a large standardized value. Examine frequency and genotype quality.
Reading a large diagonal as a universal scale failure. A diagonal is the individual's average squared coded contrast; it can exceed one, particularly in a small marker set or when a rare genotype occurs.
Changing scaling without changing interpretation. Equal-locus and variance-weighted matrices emphasize loci differently. Report which matrix was constructed instead of treating the normalizations as interchangeable.

Check Your Understanding

Use the definitions and assumptions above to answer the following questions.

At a locus with \(p=0.40\), what are the three unstandardized dominance contrast values for \(G=0,1,2\)?
Why does the standardized contrast have variance one under HWE, and what assumption is needed for that result?
What does a negative off-diagonal entry mean under the equal-locus matrix definition?
Why can a rare homozygote produce a large standardized matrix contribution even though it is uncommon?
How do equal-locus standardization and variance-weighted scaling differ in their treatment of loci with small \(4p_l^2q_l^2\)?