Read Counts and Genotype Evidence at Low Frequency

From Sequencing Reads to Genotype Evidence

A rare variant may be observed in only one or a few people in a study, but the reads at that person's variant site still contain useful evidence about genotype. If the alternate allele appears in roughly half of the reads, that pattern is often consistent with a heterozygote. If it appears in nearly all reads, an alternate homozygote is more plausible; if it is absent or nearly absent, a reference genotype is more plausible. The strength of these patterns depends in part on how many reads cover the site.

A population allele frequency describes how common an allele is across a population; it does not determine the genotype of a particular person. A rare allele can occur in a heterozygous individual, and the evidence for that genotype comes from the reads for that individual. This tutorial considers a diploid, biallelic site with two possible alleles: reference and alternate.

Definition: The sequencing depth \(n\) at a sample and site is the number of usable reads covering that site. The alternate-read count \(k\) is the number of those reads that report the alternate allele. The allele balance is the observed fraction of reads reporting the alternate allele, \(k/n\), when \(n>0\).

Allele balance is a read fraction, not the fraction of a person's chromosomes that carry an allele. At a diploid site, a heterozygote has one alternate copy among two chromosome copies, so unbiased sampling is expected to put the alternate allele in about half the reads. The two homozygous genotypes instead have zero or two alternate copies. These are expectations: a finite collection of reads can depart from the expected fraction just by sampling variation.

A Simple Read-Sampling Model

Let \(X\) be the random number of alternate reads among \(n\) reads. Under a simple model, each read independently reports the alternate allele with probability \(q\). Then \(X\) has a binomial distribution. For an idealized diploid site with balanced sampling and no read misclassification, the expected alternate-read probabilities are 0 for a reference homozygote, \(1/2\) for a heterozygote, and 1 for an alternate homozygote.

X\sim\operatorname{Binomial}(n,q),\qquad P(X=k)=\binom{n}{k}q^k(1-q)^{n-k}.

For this model, the expected alternate-read count is \(nq\), and its standard deviation is \(\sqrt{nq(1-q)}\). Consequently, the expected allele balance is \(q\), and the standard deviation of the observed balance \(X/n\) is \(\sqrt{q(1-q)/n}\). At a heterozygote, this gives an expected count of \(n/2\) alternate reads and an approximate balance standard deviation of \(1/(2\sqrt{n})\). More depth generally makes random fluctuations in the fraction smaller.

E(X)=nq,\qquad \operatorname{SD}(X)=\sqrt{nq(1-q)},\qquad \operatorname{SD}(X/n)=\sqrt{\frac{q(1-q)}{n}}.

The ideal model is a starting point for interpreting patterns, not a complete description of every sequencing experiment. In practice, a calibrated model may allow a small probability \(\delta\) that a read reports the opposite allele. In that simplified model, the expected alternate-read probabilities are \(\delta\), \(1/2\), and \(1-\delta\) for reference homozygous, heterozygous, and alternate homozygous genotypes, respectively. The value of \(\delta\) here is an assumed calibration for illustration; it is not estimated from the read count being assessed.

Interpretation: Compare the observed count or allele balance with the pattern expected under each genotype, allowing for the amount of sampling variation at the observed depth. A count is evidence about genotype, not an automatic genotype call. The simple binomial assessment assumes independent reads with a common alternate-read probability; departures from those assumptions can make its uncertainty too small.

Worked Examples

Worked Example: A Heterozygous Pattern at Moderate Depth

Suppose a hypothetical diploid sample has 40 usable reads at a site, of which 19 report the alternate allele. Under the balanced heterozygote model, \(n=40\), \(q=0.5\), and \(X\sim\operatorname{Binomial}(40,0.5)\). The observed allele balance is

\frac{k}{n}=\frac{19}{40}=0.475.

The expected alternate-read count under that model is \(40(0.5)=20\). Its standard deviation is

\sqrt{40(0.5)(0.5)}=\sqrt{10}\approx 3.162\text{ reads}.

The observed count is 1 read below the expectation, or \(1/3.162\approx0.316\) standard deviations below it. A difference this small is readily consistent with ordinary binomial sampling variation. The 0.475 balance therefore supports a heterozygous pattern under the model, though it does not alone establish the genotype.

Worked Example: A Few Alternate Reads at a Reference-Like Site

In a hypothetical sample, 3 of 100 reads report the alternate allele. Suppose a calibrated simple model uses \(\delta=0.01\), so that a reference homozygote has alternate-read probability \(q=0.01\). Under that model, \(X\sim\operatorname{Binomial}(100,0.01)\). To assess how unusual 3 or more alternate reads would be under this reference-genotype model, calculate the upper-tail probability:

P(X\geq 3)=1-\{P(X=0)+P(X=1)+P(X=2)\}.

Substituting the binomial probabilities gives

\begin{aligned} P(X\geq 3) &=1-\left\{(0.99)^{100} +100(0.01)(0.99)^{99} +\binom{100}{2}(0.01)^2(0.99)^{98}\right\}\\ &\approx 1-(0.3660+0.3697+0.1849)\\ &\approx 0.0794. \end{aligned}

Thus, under the stated reference model, about 7.94% of samples would have at least 3 alternate reads. That count is not especially surprising under this model. By comparison, a balanced heterozygote model with \(q=0.5\) gives

P(X\leq3)=\frac{\binom{100}{0}+\binom{100}{1}+\binom{100}{2}+\binom{100}{3}}{2^{100}} =\frac{166{,}751}{2^{100}} \approx 1.32\times10^{-25}.

Three alternate reads out of 100 are therefore strongly inconsistent with the balanced heterozygote model as specified. This comparison does not prove the reference genotype: it assesses the count under particular assumptions and a particular calibration. Other explanations or model departures are not ruled out by this calculation.

Worked Example: An Alternate-Homozygous Pattern With Two Reference Reads

Suppose a hypothetical sample has 28 alternate reads and 2 reference reads at a site, for a total depth of 30. Its alternate allele balance is \(28/30\approx0.9333\). Under the illustrative calibrated model with \(\delta=0.01\), an alternate homozygote has alternate-read probability \(q=0.99\). Equivalently, the number \(Y\) of reference reads has distribution \(\operatorname{Binomial}(30,0.01)\). The probability of observing at least 2 reference reads under this model is

\begin{aligned} P(Y\geq2) &=1-\{P(Y=0)+P(Y=1)\}\\ &=1-\left\{(0.99)^{30}+30(0.01)(0.99)^{29}\right\}\\ &\approx 1-(0.7397+0.2242)\\ &\approx 0.0361. \end{aligned}

The observed two reference reads are somewhat unusual under this particular model, though not impossible. The high alternate balance is consistent with an alternate-homozygous pattern, but this calculation also shows why it is important to examine the count relative to depth and to state the model used. It would be incorrect to interpret the two reference reads as proof of a different genotype.

Why Depth Changes the Interpretation

The same allele balance can provide different amounts of evidence at different depths. A balance near 0.5 from 4 reads is much less informative about a heterozygous pattern than the same balance from 100 reads, because the smaller sample has greater sampling variability. Under \(q=0.5\), the standard deviation of allele balance is \(1/(2\sqrt{n})\): it is 0.25 at depth 4 and 0.05 at depth 100. Conversely, a small count at a low depth may be difficult to distinguish from a chance departure from the expected pattern.

A useful first check is to report both the count and the fraction, such as “19 of 40 reads, balance 0.475,” rather than reporting the fraction alone. The count makes depth visible, and the fraction makes the allelic pattern easy to compare with the expected values. For a diploid biallelic site, those rough reference points are 0, 0.5, and 1, with any calibrated allowance for read-level departures stated separately.

The binomial model also has assumptions that matter. It treats reads as independent draws with the same alternate-read probability. If reads are not independent, or if the probability differs systematically across reads, the binomial standard deviation may understate the actual variation. The model also presumes the relevant sample is diploid and the site is being treated as biallelic. A genotype pattern should not be interpreted mechanically when those conditions are in doubt.

A Short Derivation of the Balance Uncertainty

Write \(X\) as the sum of \(n\) indicator variables, one for each read, where an indicator equals 1 if that read reports the alternate allele and 0 otherwise. Under independent sampling, each indicator has mean \(q\) and variance \(q(1-q)\). The mean of the sum is the sum of the means, \(nq\); independence makes the variance of the sum the sum of the variances, \(nq(1-q)\). Dividing the count by \(n\) divides its variance by \(n^2\), so the balance has variance \(q(1-q)/n\), and hence standard deviation \(\sqrt{q(1-q)/n}\). This explains why the uncertainty in read fraction shrinks as depth increases.

\operatorname{Var}(X/n) =\frac{\operatorname{Var}(X)}{n^2} =\frac{nq(1-q)}{n^2} =\frac{q(1-q)}{n}.

Common Mistakes and Practical Use

Equating rarity with a heterozygous genotype. Population frequency does not determine an individual's genotype. Use the reads from that sample to assess its genotype pattern.
Treating 0.5 as an exact requirement. A heterozygote is expected to have a balance near 0.5 under balanced sampling, but finite read counts fluctuate. Compare the deviation with the depth-dependent variation.
Ignoring depth. A balance is not enough to describe the evidence. State \(k\) and \(n\), since the same fraction at different depths has different sampling uncertainty.
Calling a genotype from a threshold alone. A threshold can summarize a decision rule, but it does not show how compatible the data are with the relevant genotype patterns or assumptions.
Reading a model-based probability as certainty. A tail probability is calculated under the specified model; it is not the probability that a genotype is true.

Key takeaway: At a diploid biallelic site, alternate-read fractions near 0, 0.5, and 1 correspond to the expected read patterns for reference homozygous, heterozygous, and alternate homozygous genotypes under balanced sampling. Use the read count, depth, sampling assumptions, and any stated calibration together; allele balance is evidence, not a definitive call.

Check Your Understanding

Use the model assumptions stated in each question, and distinguish observed allele balance from genotype.

A sample has 12 alternate reads among 24 total reads. What is its allele balance, and which diploid genotype pattern is it most directly consistent with under balanced sampling?
Under a heterozygote model with \(q=0.5\), calculate the expected alternate-read count and the standard deviation of the count at depth 64.
Why does an allele balance of 0.5 based on 6 reads generally provide less evidence for a heterozygous pattern than the same balance based on 60 reads?
Under an illustrative reference model with \(q=0.01\), identify the distribution of the alternate-read count for depth 50. What does a calculated upper-tail probability describe?
Name one assumption of the simple binomial model that could make its uncertainty assessment unreliable if violated.