Sampling Uncertainty in Rare Allele Counts

Why a Rare Allele Count Is Uncertain

A rare allele may be observed only once or twice in a sample, even when many chromosomes have been examined. The observed count is therefore not the frequency itself: it is a random result of sampling chromosomes from a population. To interpret the count, we need a model for how it varies from sample to sample.

This tutorial treats the count of copies of one allele at one biallelic site. It focuses on sampling uncertainty: uncertainty caused by which chromosomes enter the sample. It does not treat sequencing reads as independent allele copies or model errors in genotype calls. Those are separate sources of uncertainty.

Definition: Suppose $n$ allele copies are sampled, and each copy has probability $p$ of carrying the allele of interest. Let $X$ be the number of sampled copies carrying it. If the copies are independent and have the same probability $p$, then $X\sim\operatorname{Binomial}(n,p)$, with $$ P(X=x)=\binom{n}{x}p^x(1-p)^{n-x},\qquad x=0,1,\ldots,n. $$ The sample allele-frequency estimate is $\hat{p}=X/n$.

For example, if $N$ unrelated diploid individuals contribute two chromosomes each, then $n=2N$, provided the sampled allele copies can reasonably be treated as independent draws with a common frequency. The model concerns copies of the allele, not the number of individuals carrying at least one copy. The two counts are different: an individual can carry zero, one, or two copies.

Mean, Variance, and Standard Error

Write $X$ as the sum of $n$ indicator variables, one for each sampled copy: an indicator is 1 if that copy carries the allele and 0 otherwise. Each indicator has mean $p$ and variance $p(1-p)$. Independence makes the variance of the sum equal to the sum of the variances. Dividing the count by $n$ gives the sampling properties of the frequency estimate.

E(X)=np,\qquad \operatorname{Var}(X)=np(1-p)

E(\hat{p})=p,\qquad \operatorname{Var}(\hat{p})=\frac{p(1-p)}{n},\qquad SE(\hat{p})=\sqrt{\frac{p(1-p)}{n}}

The standard error describes the typical sampling variation in $\hat{p}$ over repeated samples under this model. Since the true $p$ is unknown, a commonly reported estimated standard error replaces it with $\hat{p}$:

\widehat{SE}(\hat{p})=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

This plug-in estimate is useful as a rough summary, but it can be misleading at very low counts. In particular, when $X=0$, it equals zero, even though a zero count does not establish that $p=0$. For sparse counts, an interval that respects the binomial model is more informative than relying on a normal approximation centered at $\hat{p}$.

Wilson Intervals for a Binomial Proportion

A Wilson confidence interval is obtained by inverting the binomial score test. For a 95% interval, use $z=1.96$, the standard normal cutoff with 2.5% in each tail. The interval is not a claim that the fixed population frequency has a 95% probability of lying between the calculated endpoints. Rather, under the model, the procedure produces intervals containing the true frequency in about 95% of repeated samples.

Result: For $x$ observed copies among $n$ sampled copies, let $\hat{p}=x/n$. The Wilson interval with normal cutoff $z$ has center $C$ and half-width $H$: $$ C=\frac{\hat{p}+\frac{z^2}{2n}}{1+\frac{z^2}{n}}, \qquad H=\frac{z}{1+\frac{z^2}{n}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}+\frac{z^2}{4n^2}}. $$ Its endpoints are $C-H$ and $C+H$. For a 95% interval, $z=1.96$.

The formula follows by asking which candidate frequencies $p$ would not be rejected by the score test. The accepted values satisfy

\frac{|\hat{p}-p|}{\sqrt{p(1-p)/n}}\le z.

Squaring both sides and multiplying by $n p(1-p)$, which is nonnegative for $0\le p\le1$, gives

n(\hat{p}-p)^2\le z^2p(1-p).

Collecting terms in $p$ yields a quadratic inequality:

(n+z^2)p^2-(2n\hat{p}+z^2)p+n\hat{p}^2\le0.

The two roots of this quadratic are the interval endpoints. Applying the quadratic formula and dividing numerator and denominator by $2n$ gives $C-H$ and $C+H$ above. Unlike the simple Wald interval $\hat{p}\pm z\widehat{SE}(\hat{p})$, the Wilson interval accounts for the fact that the variance depends on the candidate value $p$. It also remains useful at $x=0$ or $x=n$.

Worked Example: One Copy Among 200 Chromosomes

Suppose a hypothetical sample contains one copy of an allele among $n=200$ sampled chromosomes. The model assumes independent chromosome copies with a common allele frequency. The point estimate is

Estimate. $\hat{p}=x/n=1/200=0.005$, or 0.5%.

Estimated standard error. $$ \widehat{SE}(\hat{p})= \sqrt{\frac{0.005(1-0.005)}{200}} =\sqrt{0.000024875} \approx0.00499. $$

The estimated standard error, about 0.499 percentage points, is nearly as large as the estimate itself. That is a sign that one observed copy does not pin down the population frequency precisely.

95% Wilson interval. With $z=1.96$, $z^2=3.8416$, and $\hat{p}=0.005$, the center is $$ C=\frac{0.005+3.8416/(2\cdot200)}{1+3.8416/200} =\frac{0.014604}{1.019208} \approx0.01433. $$ The half-width is $$ H=\frac{1.96}{1.019208} \sqrt{\frac{0.005(0.995)}{200}+\frac{3.8416}{4(200)^2}} \approx0.01345. $$ Thus the interval is $0.01433\pm0.01345$, or approximately $(0.00088,\,0.02777)$. In percentage terms, its endpoints are about 0.088% and 2.777%. Under the sampling assumptions, the interval procedure has 95% coverage in repeated samples; this particular interval illustrates how compatible the data remain with a range of frequencies.

Worked Example: Two Copies Among 200 Chromosomes

Now suppose a different hypothetical sample of $n=200$ chromosomes contains $x=2$ copies. Assume the same independent, equal-frequency sampling model. The estimated frequency is

Estimate. $\hat{p}=2/200=0.01$, or 1%.

Estimated standard error. $$ \widehat{SE}(\hat{p})= \sqrt{\frac{0.01(0.99)}{200}} =\sqrt{0.0000495} \approx0.00704. $$

The standard error is about 0.704 percentage points, roughly 70% of the point estimate. A two-copy count provides more information than a one-copy count in the same sample size, but its relative uncertainty is still substantial.

95% Wilson interval. The center is $$ C=\frac{0.01+3.8416/400}{1+3.8416/200} =\frac{0.019604}{1.019208} \approx0.01924. $$ The half-width is $$ H=\frac{1.96}{1.019208} \sqrt{\frac{0.01(0.99)}{200}+\frac{3.8416}{4(200)^2}} \approx0.01649. $$ The interval is approximately $0.01924\pm0.01649=(0.00275,\,0.03573)$, or about 0.275% to 3.573%. The point estimate is 1%, but the data do not support reporting it as though the frequency were known precisely.

What a Zero Count Can and Cannot Say

A sample with no observed copies has $\hat{p}=0$, but the count can be zero by chance even when the population frequency is positive. Under the binomial model, the probability of observing zero copies is $(1-p)^n$. This makes it possible to quantify how compatible a zero count is with particular candidate frequencies.

Worked Example: No Copies Among 100 Chromosomes

Consider a hypothetical sample with $x=0$ copies among $n=100$ chromosomes. Assume independent sampling. The sample estimate is $0/100=0$, but it is not proof that the population frequency is zero.

For instance, if the true frequency were $p=0.01$, then the probability of seeing no copies would be $$ P(X=0)=(1-0.01)^{100}=0.99^{100}\approx0.366. $$ So even at a true frequency of 1%, a zero count would occur in about 36.6% of samples of this size.

A one-sided 95% upper confidence bound $p_U$ can be found by setting the probability of zero copies at that boundary equal to 0.05: $$ (1-p_U)^{100}=0.05, \qquad p_U=1-0.05^{1/100}\approx0.0295. $$ Thus the one-sided interval is $[0,\,0.0295]$, or 0% to about 2.95%. Under the model, this construction has 95% coverage as an upper bound. It does not mean there is a 95% probability that the frequency is below 2.95%; it means the procedure’s upper bounds cover the fixed frequency in 95% of repeated samples.

Assumptions and Practical Interpretation

The binomial calculations are only as appropriate as their sampling model. They require a clearly defined set of $n$ sampled allele copies, the same probability $p$ for each copy, and independence among copies. These assumptions may be reasonable when unrelated individuals are sampled from a well-defined population and each contributes two chromosome copies under a model of random pairing. They may fail or need adjustment when samples include close relatives, population structure, nonrandom mating, or other dependencies that make copies correlated or their probabilities unequal.

The model also assumes that each sampled copy is classified correctly as carrying or not carrying the allele. If the count is based on sequencing evidence rather than trusted genotype calls, read depth, calling uncertainty, and error can affect the observed count. Those issues should not be silently folded into the binomial sampling uncertainty described here.

At low counts, uncertainty is naturally asymmetric and may extend far above the point estimate. A normal approximation can be especially misleading: the Wald interval $\hat{p}\pm1.96\widehat{SE}(\hat{p})$ can extend below zero, and for zero observed copies it collapses to the uninformative point $[0,0]$. Wilson or exact binomial intervals are more suitable for communicating uncertainty near the boundary. Regardless of interval method, report the count and denominator along with the estimated frequency so readers can see how much information supports it.

Common Mistakes

Wrong reasoning: “We observed no copies, so the allele frequency is zero.” Correction: Zero is the sample estimate, not proof of absence. A positive frequency can still produce a zero count, and the probability depends on both $p$ and $n$.
Wrong reasoning: “One copy among 200 chromosomes means the true frequency is exactly 0.5%.” Correction: 0.5% is the estimate. The Wilson interval in the example ranges from about 0.088% to 2.777%, reflecting substantial sampling uncertainty.
Wrong reasoning: “A small standard error relative to the sample size guarantees a precise rare-frequency estimate.” Correction: Compare uncertainty with the estimate as well as with the denominator. With one or two copies, the estimated standard error can be a large fraction of $\hat{p}$.
Wrong reasoning: “Every chromosome copy in the data is necessarily an independent draw.” Correction: Independence is a model assumption. Relatedness, population structure, and other dependencies can invalidate the simple binomial variance.

Key takeaway: Under independent sampling of allele copies, $X\sim\operatorname{Binomial}(n,p)$ and $\hat{p}=X/n$ has standard error $\sqrt{p(1-p)/n}$. When only one or two copies are observed, report the count, denominator, and an appropriate interval: the point estimate alone can conceal substantial uncertainty.

Check Your Understanding

Use the binomial sampling model and distinguish an observed count from the underlying population frequency.

If 3 copies are observed among 300 sampled chromosomes, what is $\hat{p}$, and what does it estimate?
State the assumptions that allow an allele-copy count to be modeled as binomial.
Why does the plug-in standard error equal zero when no copies are observed, and why is that not a useful statement of uncertainty?
For a zero count among 100 chromosomes, explain why a true frequency of 1% is still compatible with the observation.
What is the repeated-sampling interpretation of a 95% confidence interval?