Frequency-Weighted Allele Burden Scores

Why Weight Allele Counts?

A variant-set score summarizes genotypes across several variants in one number per individual. An unweighted allele-count score gives each included allele copy the same contribution. A frequency-weighted score instead multiplies each variant's dosage by a weight chosen using its allele frequency. When the weights are larger for rarer variants, a copy of a rare allele contributes more to the score than a copy of a more common allele.

This choice changes the summary of genetic variation; it does not, by itself, establish that the set is associated with a phenotype. In particular, a large weighted score is not a p-value, an estimated effect on the phenotype, or evidence of causation. The goal here is to calculate and compare scores, leaving the derivation of an association test to the appropriate testing framework.

Definition: For individual $i$ and variant $j$, let $G_{ij}$ be the dosage of the designated minor allele, coded from 0 to 2. Let $q_j$ be that allele's frequency, with $0 \leq q_j \leq 0.5$, and let $w_j$ be a specified nonnegative weight. The frequency-weighted allele-count score is $$ B_i=\sum_{j=1}^{m} w_jG_{ij}. $$ Here $m$ is the number of variants included. The score is an allele-count summary on a scale determined by the weights; it is not generally an ordinary count.

Dosage coding and allele orientation must be consistent across variants. If $G_{ij}=1$, the individual carries one copy of the designated allele at variant $j$; a dosage of 2 represents two copies. For imputed genotypes, $G_{ij}$ may be a dosage between 0 and 2 rather than an integer. The frequency $q_j$ should refer to the same allele that is counted by $G_{ij}$. A mismatched allele orientation can make the frequency and dosage describe different alleles.

Common Frequency-Based Weights

Several weight choices are used in practice. They have different scales and emphasize frequency differently, so the formula and any rescaling should be reported rather than summarized only as “rare-variant weighting.”

Weight choice	Formula	Frequency behavior and interpretation
Unweighted dosage	$w_j=1$	Every allele copy contributes equally, regardless of frequency.
Inverse-square-root MAF	$w_j=1/\sqrt{q_j}$	Weight increases as the minor-allele frequency decreases.
Inverse genotype standard deviation	$w_j=1/\sqrt{2q_j(1-q_j)}$	Under Hardy–Weinberg equilibrium, this is the reciprocal of the dosage standard deviation.
Beta-density weight	$w_j=f(q_j)$, where $f$ is a Beta density	Can assign high relative weights near low frequencies; the chosen Beta parameters determine the shape.

The inverse-frequency formulas require $0<q_j\leq 0.5$. They are undefined at $q_j=0$, so they must not be applied unchanged to a variant with zero estimated frequency. A variant with no minor alleles observed in the analysis sample is monomorphic there and contributes no genotype variation. A practical default is to exclude such a variant from a variable-genotype score and record that decision. A documented frequency floor or regularization can make an inverse formula finite, but it does not create genotype variation; the choice must be prespecified and reported. It is not appropriate to silently substitute zero or infinity for an undefined weight.

A Beta-density weight is another common option. For example, the density for a Beta distribution with shape parameters 1 and 25 is $f(q)=25(1-q)^{24}$ for $0 \leq q \leq 1$. This weight is finite at zero, unlike the inverse formulas. However, a frequency of zero still signals a monomorphic variant in the sample, and its dosage cannot distinguish individuals. Beta-density weights can also have a different numerical scale from inverse-frequency weights; multiplying every weight by a common constant rescales all scores but does not change their relative weighting.

Why Inverse Genotype-Standard-Deviation Weighting Has This Form

The inverse genotype-standard-deviation choice is motivated by a simple variance calculation. Under Hardy–Weinberg equilibrium, the minor-allele dosage at a variant has a binomial distribution with two allele copies and success probability $q_j$. This assumption gives the following result for genotype dosage; it is not needed simply to compute a score from observed $G_{ij}$.

G_{ij}\sim\operatorname{Binomial}(2,q_j) \quad\Longrightarrow\quad \operatorname{Var}(G_{ij})=2q_j(1-q_j), \qquad \operatorname{SD}(G_{ij})=\sqrt{2q_j(1-q_j)}.

The variance follows from adding the variances of two independent Bernoulli allele indicators under the model: each has variance $q_j(1-q_j)$, so their sum has variance $2q_j(1-q_j)$. Taking the reciprocal of the standard deviation gives the stated weight. Thus, under these assumptions, multiplying dosage by this weight puts a one-unit dosage difference on a scale inversely related to its expected variability. It does not make the entire person-level score have variance one: variants may be correlated, and the score is a sum of weighted dosages.

The inverse-MAF formula $1/\sqrt{q_j}$ is not identical to inverse genotype-standard-deviation weighting. Their ratio depends on $1-q_j$ and the factor of 2. They behave similarly for very small $q_j$, because $1-q_j$ is then close to 1, but they should not be described as the same formula. The variance calculation also depends on Hardy–Weinberg equilibrium; the arithmetic definition of the score does not.

Worked Example: Calculating an Inverse-Square-Root MAF Score

Suppose a hypothetical analysis includes three variants with minor-allele frequencies $q_1=0.20$, $q_2=0.02$, and $q_3=0.005$. For one individual, the minor-allele dosages are $(G_{i1},G_{i2},G_{i3})=(1,0,1)$. All three frequencies are positive, so the inverse-square-root weights are defined.

The weights are $$ w_1=\frac{1}{\sqrt{0.20}}=2.236,\qquad w_2=\frac{1}{\sqrt{0.02}}=7.071,\qquad w_3=\frac{1}{\sqrt{0.005}}=14.14. $$

Substitute the weights and dosages into the score: $$ B_i=(2.236)(1)+(7.071)(0)+(14.14)(1)=16.38. $$

The second variant contributes zero for this individual because its dosage is zero, despite its positive weight. The third contributes more than the first because it is rarer and the individual carries one copy at each. The value 16.38 is a weighted dosage score, not 16.38 allele copies.

Worked Example: Calculating Beta-Density Weights

Use the same hypothetical frequencies $0.20$, $0.02$, and $0.005$, the same dosage vector $(1,0,1)$, and the Beta-density weight $f(q)=25(1-q)^{24}$. Because $f(q)$ is finite at zero, this formula itself does not have the inverse-weight singularity. Here all frequencies are positive in any case.

The three weights, rounded to four significant figures, are $$ w_1=25(0.80)^{24}=0.1181,\qquad w_2=25(0.98)^{24}=15.39,\qquad w_3=25(0.995)^{24}=22.17. $$

The score is $$ B_i=(0.1181)(1)+(15.39)(0)+(22.17)(1)=22.29. $$

This weight function gives much larger values to the two rarer variants than to the variant at frequency 0.20. Its score is numerically larger than the previous example's score, but that does not mean it is stronger evidence of association: the two scores use different weight scales and neither is a test statistic.

Worked Example: Estimating Frequencies and Handling a Zero

Suppose a hypothetical reference sample of 8 unrelated individuals has 16 allele copies at each of three variants. The minor-allele counts are 4, 1, and 0. The estimated frequencies are therefore $$ \hat q_1=\frac{4}{16}=0.25,\qquad \hat q_2=\frac{1}{16}=0.0625,\qquad \hat q_3=\frac{0}{16}=0. $$

Use inverse genotype-standard-deviation weights for the first two variants. The third is monomorphic in this reference sample, so do not apply the inverse formula to it. Under the assumed Hardy–Weinberg model for the first two variants, $$ w_1=\frac{1}{\sqrt{2(0.25)(0.75)}}=\frac{1}{\sqrt{0.375}}=1.633, $$ $$ w_2=\frac{1}{\sqrt{2(0.0625)(0.9375)}}=\frac{1}{\sqrt{0.1171875}}=2.921. $$

For a hypothetical individual with dosages $(1,2,0)$, the score over the two variable variants is $$ B_i=(1.633)(1)+(2.921)(2)=7.475. $$

The third variant is excluded from this calculation because the stated frequency estimate is zero; assigning it an infinite inverse weight would be invalid. If a separate, well-justified frequency source indicates that it is variable in the target population, the analysis would need to reconcile that source with the monomorphic reference data before defining the score.

Assumptions, Frequency Estimates, and Practical Checks

A weighted score can be calculated mechanically, but interpreting it requires attention to how its inputs were obtained. At minimum, the variant set, allele coding, frequency source, weight formula, and handling of missing or monomorphic genotypes should be specified. Frequencies may be estimated from the analysis sample or obtained from a reference population. They should be relevant to the population being analyzed: frequency differences across ancestry groups can change the weights even when the underlying variant set is unchanged.

If frequencies are estimated from the analysis sample, rare-variant estimates can be imprecise because they may depend on very few observed allele copies. A small change in an estimated frequency can cause a substantial change in an inverse-frequency weight. The estimates should not be selected using the phenotype in a way that favors a desired association result. If a reference sample is used, differences in ancestry, sequencing, or quality control can make its frequencies a poor match for the analysis data.

Before relying on the scores, inspect the inputs and contributions. Useful checks include:

Confirm that each dosage counts the same allele whose frequency defines $q_j$, and that harmonization has not reversed an allele at one variant.
Check that every inverse-weighted variant has $q_j>0$. Report excluded monomorphic variants, or clearly state any prespecified frequency floor or regularization.
Inspect the largest $w_jG_{ij}$ contributions. If one variant dominates most individuals' scores, the result is effectively driven by that variant rather than a balanced summary of the set.
Check missingness and imputation quality. Treating a missing dosage as zero falsely records the individual as having no copies; use a declared missing-data procedure instead.
Compare frequency estimates with the intended target population and examine whether a few uncertain rare-allele counts determine the weights.

How the Choices Differ—and What They Cannot Do

Unweighted dosage is a transparent baseline: each allele copy adds one. It avoids frequency-based amplification, but it does not give rare variants extra influence. Inverse-MAF and inverse-genotype-standard-deviation weights explicitly increase the contribution of low-frequency variants, though the latter has a variance-based interpretation under Hardy–Weinberg equilibrium. A Beta-density weight offers a tunable frequency profile and remains finite at zero, but its shape depends on the selected parameters. None is universally best; the choice should fit the scientific motivation and be set without looking for whichever formula produces a preferred result.

A carrier indicator summary, covered in the earlier material on carrier collapsing and indicator burden scores, records whether an individual carries at least one qualifying allele rather than summing dosages with frequency weights. It therefore answers a different summarization question, especially for individuals with multiple copies or multiple qualifying variants. A weighted dosage score retains dosage information and lets variants contribute unequally. The choice between them is not merely a change of notation.

Scores are also affected by linkage disequilibrium: correlated variants can contribute redundant information, and the sum does not adjust for that correlation. A single score compresses a potentially complex genotype pattern, so individuals with different patterns can have the same score. Finally, a frequency weight encodes a decision about which variants receive more influence; it does not establish that rarer variants have larger effects or that all effects point in the same direction. Those questions concern the association model and its assumptions, not the score's arithmetic.

Common Mistakes

Using an inverse weight at zero frequency. The formulas $1/\sqrt{q_j}$ and $1/\sqrt{2q_j(1-q_j)}$ are undefined when $q_j=0$. Exclude a monomorphic variant or use a documented, justified regularization; do not leave the case implicit.
Calling the weighted sum an allele count. With unequal weights, $B_i$ is on a weight-dependent scale. Describe it as a weighted allele-count score.
Assuming every “rare-variant weight” is equivalent. Inverse MAF, inverse genotype standard deviation, and Beta-density weights have different formulas and frequency behavior. State the one actually used.
Interpreting a large score as association evidence. The score is a genotype summary. Association requires an appropriate statistical model and test, with its assumptions and calibration.
Ignoring a dominant contribution. A high weight can make one observed allele account for much of a score. Inspect variant-level contributions rather than assuming the sum represents many variants equally.

Key takeaway: A frequency-weighted allele-count score is $B_i=\sum_jw_jG_{ij}$. Its meaning depends on dosage coding, frequency estimates, and the exact weight function. Inverse-frequency choices require positive frequencies; zero-frequency variants must be handled explicitly, not assigned an undefined weight.

Check Your Understanding

Answer each question, then compare with the solutions below.

For $q_1=0.10$, $q_2=0.01$, inverse-square-root MAF weights, and dosages $(2,1)$, calculate the weighted score to three significant figures.
Why is $1/\sqrt{2q_j(1-q_j)}$ undefined at $q_j=0$, and what is one defensible way to handle a variant with zero estimated frequency in the analysis sample?
Under Hardy–Weinberg equilibrium, derive the dosage variance for a minor allele with frequency $q$, and state which weight is its inverse standard deviation.
Can a Beta-density weight be finite at zero frequency while a zero-frequency variant still provides no genotype variation in the sample? Explain.
Give one reason that two different frequency-weighting formulas can produce scores with different numerical magnitudes without one being stronger evidence of association.

Solutions. (1) The weights are $1/\sqrt{0.10}=3.162$ and $1/\sqrt{0.01}=10$. Thus $B_i\approx16.3$. (2) The denominator contains $q_j$, so at zero it is zero and the reciprocal is undefined. Excluding the monomorphic variant is one defensible option; any frequency floor or regularization must be documented and justified. (3) Under Hardy–Weinberg equilibrium, dosage is the sum of two independent Bernoulli allele indicators, each with variance $q(1-q)$. Hence $\operatorname{Var}(G)=2q(1-q)$, and its inverse-standard-deviation weight is $1/\sqrt{2q(1-q)}$. (4) Yes. A Beta density such as $25(1-q)^{24}$ is finite at zero, but if every observed dosage is zero, the variant does not distinguish individuals in that sample. (5) The formulas assign different numerical scales and relative frequency profiles. A larger score value alone is not a test statistic or evidence of a stronger association.

Weight choice	Formula	Frequency behavior and interpretation
Unweighted dosage	\(w_j=1\)	Every allele copy contributes equally, regardless of frequency.
Inverse-square-root MAF	\(w_j=1/\sqrt{q_j}\)	Weight increases as the minor-allele frequency decreases.
Inverse genotype standard deviation	\(w_j=1/\sqrt{2q_j(1-q_j)}\)	Under Hardy–Weinberg equilibrium, this is the reciprocal of the dosage standard deviation.
Beta-density weight	\(w_j=f(q_j)\), where \(f\) is a Beta density	Can assign high relative weights near low frequencies; the chosen Beta parameters determine the shape.