Allele-Balance Diagnostics for Heterozygous Calls

Why Allele Balance Matters

A diploid heterozygous genotype contains one copy of each of two alleles. At a biallelic sequencing site, reads from those alleles would be expected in roughly equal numbers if both copies are present and reads sample them without bias. The observed counts will not usually be exactly equal: sequencing depth is finite, and random sampling produces differences. The size and pattern of those differences can nevertheless help identify calls that deserve review.

The alternate-read fraction, also called the alternate allele fraction in this context, is a per-call summary of the proportion of usable reads supporting the alternate allele. It is not the same as an individual’s genotype dosage or a population allele frequency. It is a read-based diagnostic: a value far from one-half at a heterozygous call may reflect allele dropout, mapping bias, an unreliable call, or another technical feature.

Definition: For a biallelic call, let $R$ be the number of usable reads supporting the reference allele and $A$ the number supporting the alternate allele. The allele depth considered here is $n=R+A$. The alternate-read fraction is $$ \hat{f}_{\mathrm{alt}}=\frac{A}{R+A}=\frac{A}{n}, $$ provided $n>0$. A balanced heterozygous call has $\hat{f}_{\mathrm{alt}}$ near $0.5$, not necessarily exactly $0.5$.

Use counts for reads that can be assigned to the reference or alternate allele at the site. In a variant record, the total depth field and the sum of the allele-specific depths need not match, because filters or reads supporting other alleles may be handled differently. For this calculation, use the relevant reference and alternate counts and state which counts were used. The simple fraction above is intended for a biallelic heterozygous call. At a multiallelic site, each allele’s read fraction must be interpreted in relation to the called genotype; one-half is not automatically the expected fraction for every allele.

What Balance Should a Heterozygous Call Show?

A simple sampling model makes the one-half expectation precise. Suppose the genotype is truly heterozygous, each of the $n$ usable reads independently comes from either allele, and each allele has probability $0.5$ of being sampled. Then the alternate-read count $A$, conditional on $n$, follows a binomial distribution:

A\mid n\sim \operatorname{Binomial}(n,0.5).

This is a reference model, not a guarantee about real sequencing data. It assumes the genotype is correct, the two alleles are sampled equally, and reads behave like independent observations. Mapping preferences, amplification, duplicate reads, local sequence properties, and other technical features can violate those assumptions. The model explains why moderate imbalances can occur by chance and why a severe imbalance can be a warning; by itself, it does not identify the cause.

Result: Under the equal-sampling model, $E(A\mid n)=n/2$, and the alternate-read fraction has expected value $0.5$ and standard deviation $1/(2\sqrt{n})$. Thus the amount of ordinary sampling variation depends on depth: at greater depth, fractions tend to vary less around one-half under the model.

This result follows from the binomial mean and variance. If $A$ is binomial with $n$ trials and success probability $p$, then $E(A)=np$ and $\operatorname{Var}(A)=np(1-p)$. For $\hat{f}_{\mathrm{alt}}=A/n$, dividing the mean by $n$ gives $E(\hat{f}_{\mathrm{alt}})=p$, and dividing the variance by $n^2$ gives $\operatorname{Var}(\hat{f}_{\mathrm{alt}})=p(1-p)/n$. Setting $p=0.5$ yields the stated standard deviation.

E(\hat{f}_{\mathrm{alt}}\mid n)=0.5,\qquad \operatorname{SD}(\hat{f}_{\mathrm{alt}}\mid n) =\sqrt{\frac{0.5(1-0.5)}{n}} =\frac{1}{2\sqrt{n}}.

For example, at depth $n=40$, the model’s standard deviation for the fraction is $1/(2\sqrt{40})$, or about $0.0791$. This describes random variation under the stated model; it is not a universal acceptance range. A practical review should consider depth, call quality, the distribution of fractions across calls, and the assay’s technical characteristics.

Worked Example: Calculating One Call’s Alternate-Read Fraction

Suppose a hypothetical diploid sample has a heterozygous call with 18 usable reference-supporting reads and 22 usable alternate-supporting reads. Calculate the alternate-read fraction and interpret it.

The relevant allele depth is $n=18+22=40$ reads. Therefore,

\hat{f}_{\mathrm{alt}}=\frac{A}{R+A} =\frac{22}{18+22} =\frac{22}{40} =0.550.

The alternate allele accounts for 55.0% of these reads. The observed counts are 4 reads apart, rather than equal, but the fraction is reasonably close to one-half and is compatible with ordinary sampling variation under the simple model. This single value does not establish that the call is correct, nor does it prove that there is no technical bias; it is one diagnostic to consider alongside the depth and genotype-quality information.

Reading Allele-Balance Patterns

A single fraction can be noisy, especially at low depth. Looking across heterozygous calls helps distinguish an isolated fluctuation from a repeated pattern. Summaries such as the median, range, and a plot of fractions by variant are useful, but they should not replace inspection of the underlying counts. Fractions of $0.2$ based on 5 reads and 200 reads have the same numerical value but different precision under the simple sampling model.

A distribution concentrated near one-half is broadly consistent with balanced support, although it does not prove the absence of error. A repeated shift toward values below one-half means fewer alternate-supporting reads than expected under equal sampling. A shift above one-half means more alternate-supporting reads. Strong imbalance at a particular call can be consistent with allele dropout or a mapping preference; an incorrect genotype or other technical problem can also produce it. Balance alone does not establish which explanation is responsible.

Worked Example: Comparing Balance Across Calls

At one hypothetical biallelic variant, six samples have been called heterozygous. Each has 60 usable reads. Their reference and alternate counts are shown below.

Sample	Reference reads $R$	Alternate reads $A$	Alternate-read fraction
1	28	32	0.533
2	39	21	0.350
3	47	13	0.217
4	30	30	0.500
5	52	8	0.133
6	33	27	0.450

For sample 3, for instance, the fraction is $13/(47+13)=13/60=0.217$. Calculating the same way for each row gives fractions ranging from $0.133$ to $0.533$. In sorted order they are $0.133,\ 0.217,\ 0.350,\ 0.450,\ 0.500,\ 0.533$, so the median is the average of the third and fourth values: $(0.350+0.450)/2=0.400$.

The combined alternate fraction is $(32+21+13+30+8+27)/360=131/360=0.364$. Both the median and the combined fraction are below one-half, and several samples have substantial imbalance. This pattern is a reason to investigate whether the variant has a technical bias or whether some calls are unreliable. It does not identify the cause. The combined fraction also conceals the spread: it should not be reported as if every sample had a fraction of $0.364$.

Severe Imbalance and the Limits of a Simple Model

At high enough depth, a fraction very close to zero or one is difficult to explain by independent, equal-probability sampling alone. The calculation below illustrates that point. It is a model-based check, not a general-purpose p-value for allele-balance quality control. Real reads may not be independent, and the assumed probability of $0.5$ may not hold for a particular assay or locus.

Worked Example: Reviewing an Extremely Imbalanced Call

A hypothetical call is labelled heterozygous, with $R=29$ reference reads and $A=1$ alternate read. The total allele depth is 30 reads, and the alternate-read fraction is

\hat{f}_{\mathrm{alt}}=\frac{1}{29+1}=\frac{1}{30}=0.0333.

Under the idealized equal-sampling model, $A\sim\operatorname{Binomial}(30,0.5)$. The probability of observing one or fewer alternate reads is

P(A\leq 1\mid n=30,p=0.5) =\frac{\binom{30}{0}+\binom{30}{1}}{2^{30}} =\frac{1+30}{1{,}073{,}741{,}824} \approx 2.887\times 10^{-8}.

Such a count is extremely unlikely under this particular model. It is therefore a strong diagnostic warning, not proof of a specific mechanism. Possible explanations include allele dropout, mapping bias against the alternate allele, or an unreliable genotype call. Review the read evidence and relevant quality information before deciding how to handle the call.

The model’s assumptions matter especially when a numerical tail probability is calculated. If reads are correlated, for example because many reads are duplicates or derive from the same underlying molecule, treating all 30 as independent can exaggerate the evidence against equal sampling. A locus-specific mapping preference can also make the true sampling probability differ from one-half. In practice, the simple binomial calculation clarifies the reference expectation; it should not be presented as a calibrated error probability unless the assumptions and calibration are appropriate.

A Practical Diagnostic Sequence

A consistent sequence of checks makes allele-balance summaries easier to interpret. The aim is to find calls or patterns that need review, not to diagnose a technical mechanism from a fraction alone.

Confirm the call and the counts.
Restrict the initial comparison to calls labelled heterozygous, and identify the reference and alternate read counts used. For a biallelic call, calculate $n=R+A$.

Calculate the fraction.
For each call with $n>0$, calculate $\hat{f}_{\mathrm{alt}}=A/n$. Keep the counts with the fraction so the role of depth remains visible.

Examine individual and group patterns.
Look for severe individual imbalances and repeated shifts across calls. Summarize the distribution with suitable statistics or a plot, without letting an aggregate hide outliers.

Interpret cautiously and investigate.
Consider sampling variation, depth, call reliability, and possible allele-specific technical effects. Treat imbalance as a flag for review rather than a definitive explanation.

Common Mistake

A common mistake is to reason, “A heterozygote must have exactly half alternate reads, so any fraction other than $0.5$ proves the call is wrong.” This confuses an expected proportion with a requirement for every finite read sample. Under the equal-sampling model, counts vary randomly; the size of that variation depends on depth. The correction is to calculate the fraction, consider its depth, and look for unusually strong or repeated patterns. Even an extreme imbalance is a diagnostic signal, not by itself a conclusion about the error mechanism or a definitive verdict on the genotype.

Key takeaway: For a biallelic heterozygous call, calculate alternate-read fraction as alternate reads divided by reference-plus-alternate reads. Equal sampling predicts a fraction near one-half, with more variation at lower depth. Persistent or severe imbalance warrants investigation, but allele balance alone cannot establish why it occurred.

Check Your Understanding

Use the definitions and assumptions above to answer these questions.

A heterozygous call has 24 reference reads and 16 alternate reads. What is its alternate-read fraction, and what is its total allele depth?
Under the equal-sampling model, why can a heterozygous call have a fraction other than exactly one-half?
Two calls both have alternate-read fraction $0.20$, but one is based on 5 reads and the other on 100 reads. Why should their diagnostic interpretation differ?
Across many heterozygous calls at one variant, fractions are repeatedly below one-half. Name two possible technical explanations and explain why the fractions do not distinguish them.
Why should a binomial tail probability based on independent reads be interpreted cautiously in sequencing data?