Why Genotype Evidence Is Orthogonal
Transcriptome-based doublet methods ask whether a barcode’s expression profile resembles a mixture of cells. Genotype evidence asks a different question: do the reads in that barcode contain alleles that are incompatible with a single donor’s genotype? If two donors have different alleles at expressed loci, reads from both donors in one barcode can provide direct evidence of donor mixing. This evidence is orthogonal to a transcriptome score because it relies on allelic identity rather than on the overall expression profile.
The evidence is available only when the relevant genotypes differ and informative RNA reads are observed. A lack of incompatible alleles therefore does not establish that a barcode is a singlet. The barcode may have too few reads at informative loci, the donors may share the same genotype at those loci, or the doublet may consist of two cells from the same donor.
For a clear calculation, consider loci where donor A is homozygous for one allele and donor B is homozygous for the other. Call the allele expected from B the B-specific allele. Let \(\epsilon\) be the probability that a read is called as the other allele because of sequencing error. Assume \(0<\epsilon<0.5\), the genotypes are known correctly, and reads are independent conditional on their donor of origin. A singlet from A produces a B-specific read with probability \(\epsilon\); a read from B produces that allele with probability \(1-\epsilon\).
In a two-donor mixture, let \(q\) be the fraction of reads at these loci contributed by donor B. Under the simplified model, the probability of a B-specific read is
The equal-contribution case has \(q=0.5\), so \(p_B=0.5\). This is a modeling assumption for the calculation, not a claim that the two cells contribute equal numbers of RNA molecules. Cell size, expression, and capture can make their RNA contributions unequal.
Likelihood Evidence at Informative Loci
A likelihood is the probability of the observed reads under a specified model. Comparing the likelihood under a donor-mixture model with that under a singlet model quantifies how much better the observed alleles fit one explanation than the other. For \(y_B\) B-specific reads and \(y_A\) A-specific reads across independent, homozygous-discordant loci, the simplified likelihood under a mixture is
Under an A singlet, the corresponding likelihood is \(\epsilon^{y_B}(1-\epsilon)^{y_A}\). The likelihood ratio is the mixture likelihood divided by the singlet likelihood. A large ratio favors the mixture relative to that singlet model, conditional on the assumptions. It is not itself a posterior probability that the barcode is a doublet; posterior odds would also require prior odds and a model for other explanations.
The calculation needs care. Read independence can fail when reads are PCR duplicates or when several reads reflect the same original molecule. Mapping bias, allele-specific expression, genotype errors, and RNA editing can also alter the expected allele probabilities. In practice, evidence is more convincing when incompatible alleles are supported by multiple independent molecules at multiple informative loci, with technical artifacts considered.
Worked Example: Comparing a Two-Donor Mixture With Each Singlet
Suppose known genotypes show two independent, homozygous-discordant loci for donors A and B. At the first locus, four reads support B’s allele; at the second, two reads support A’s allele. Assume a sequencing error rate of \(\epsilon=0.01\), independent reads, and a fixed equal-contribution mixture. Compare that mixture with an A singlet and a B singlet.
Under the equal mixture, each observed allele has probability \(0.50\). The likelihood of the six observed reads is
Under an A singlet, each of the four B-specific reads is an error, while each of the two A-specific reads is correct:
Thus the likelihood ratio against an A singlet is
Under these specified models, the observed allele pattern is about 1,594,000 times as likely under the equal-mixture model as under the A-singlet model. This is not the probability that the barcode is a doublet. Against a B singlet, the likelihood is \((0.99)^4(0.01)^2=0.00009606\), giving a mixture-to-B-singlet likelihood ratio of \(0.015625/0.00009606\approx162.7\). The pattern favors the mixture over either singlet, but its strength depends on the stated error model and read-independence assumption.
What the Evidence Can and Cannot Establish
Genotype evidence is strongest when the candidate donors have several expressed, high-quality loci with clearly different genotypes. It can support the conclusion that genetic material from more than one donor is present in a barcode. It does not, by itself, show that the barcode contains exactly two intact cells, identify every cell type in it, or prove how the mixed material entered the barcode. A doublet from two cells of the same donor will generally be invisible to donor-genotype evidence.
A practical analysis starts by specifying the candidate donors and the genotype information used. At each informative locus, allele calls should be assessed with read quality and mapping quality in mind. Where possible, duplicate reads should not be treated as independent evidence. Across loci, linkage can also matter: nearby variants inherited together do not necessarily supply independent evidence. A likelihood that assumes independent loci can overstate support if it counts correlated observations as separate.
The following derivation shows why repeated incompatible reads can produce strong likelihood evidence under the simplified model. Suppose \(k\) independent reads each support the B-specific allele at loci where A is homozygous for the other allele. Under an equal mixture, the probability of this pattern is \((0.5)^k\); under an A singlet it is \(\epsilon^k\). Therefore,
When \(\epsilon=0.01\), each additional independent B-specific read multiplies this likelihood ratio by \(0.5/0.01=50\). The calculation explains the rapid accumulation of evidence, but its force depends on independence and on the error rate being realistic. If the reads are duplicates or the actual error probability is larger, this multiplication exaggerates the evidence.
Worked Example: Three Independent Incompatible Reads
At three independent, homozygous-discordant loci, suppose one B-specific read is observed at each locus. Assume \(\epsilon=0.01\) and compare a fixed equal mixture with an A singlet. The mixture likelihood is \((0.50)^3=0.125\), and the A-singlet likelihood is \((0.01)^3=0.000001\). Therefore,
Under these assumptions, the three-read pattern is 125,000 times as likely under the equal-mixture model as under the A-singlet model. The result is conditional on observing independent, correctly assigned reads at genuinely informative loci. It does not account for prior doublet prevalence or alternative sources of mixed alleles.
Worked Example: Estimating an Unequal RNA Contribution
Use the six-read pattern from the first example: four B-specific reads and two A-specific reads. Suppose the mixture contribution \(q\) from donor B is not fixed at 0.50, while \(\epsilon=0.01\) remains known. The likelihood is \(p_B^4(1-p_B)^2\). Its maximum over \(p_B\) occurs at the observed B-specific fraction, \(p_B=4/6=2/3\). This follows by differentiating the log likelihood:
Since \(p_B=\epsilon+q(1-2\epsilon)\), the estimated RNA contribution is
At this fitted value, the mixture likelihood is \((2/3)^4(1/3)^2=16/729\approx0.02195\). Dividing by the A-singlet likelihood \(0.000000009801\) gives a maximized likelihood ratio of approximately \(2.239\times10^6\). The fitted \(q\) describes the modeled fraction of reads contributed by B at these loci; it is not necessarily the fraction of cells or a direct measure of cell size. Estimating \(q\) also gives the mixture model more flexibility than the fixed equal-mixture model, so model complexity and the evidence being used should be considered when comparing models.
Diagnostics, Alternatives, and Limitations
Genotype evidence complements, rather than replaces, transcriptome-based doublet scores. A score-based method can flag a barcode whose expression resembles a combination of cell types even when donor genotypes are unavailable or uninformative. Genotype evidence can support donor mixing even when the expression profile is not an obvious blend. Agreement between the two types of evidence is useful, but disagreement is not automatically a failure of either method: they detect different features and have different blind spots. Cluster-level enrichment can provide another diagnostic context, but it does not turn a genotype likelihood ratio into a cell-level posterior probability.
Check for patterns that suggest the model is going wrong. A single low-quality incompatible read is weak evidence, especially if its locus is prone to mapping errors. Support confined to one locus may reflect a locus-specific artifact. Unexpected alleles at many barcodes can indicate a problem with genotype assignment, alignment, or sample labeling rather than many genuine donor mixtures. Conversely, no incompatible reads is inconclusive if coverage at informative loci is sparse.
More complete approaches can model locus-specific error rates, allele-specific expression, uncertain genotypes, read dependence, or more than two donors. Those models require data and assumptions beyond the simple likelihood calculation here. Regardless of method, genotype-based validation is limited by the number of informative loci expressed in the captured RNA and by the availability and accuracy of donor genotype information.
- Do not treat a likelihood ratio as a posterior probability. It compares likelihoods under specified models; prior odds and alternative explanations are not included.
- Do not interpret absence of incompatible alleles as proof of a singlet. There may be too little informative coverage, or the two cells may share a donor genotype.
- Do not count correlated or duplicated reads as independent evidence. Doing so can inflate the apparent support for a mixture.
- Do not equate donor mixing with a particular transcriptome profile. Allelic evidence and expression-based scores answer related but distinct questions.
Check Your Understanding
Answer each question using the assumptions stated in it.
- At two independent loci, one B-specific read is observed at each. If \(\epsilon=0.02\), what is the likelihood ratio of a fixed equal mixture to an A singlet?
- Why can genotype evidence fail to detect a doublet formed by two cells from the same donor?
- Does a likelihood ratio of 125,000 mean there is a 125,000-to-1 posterior probability that the barcode is a doublet? Explain.
- Name two diagnostics that can help distinguish genuine multi-donor evidence from technical artifact.
Solutions. (1) The mixture likelihood is \((0.50)^2=0.25\), while the A-singlet likelihood is \((0.02)^2=0.0004\). The ratio is \(0.25/0.0004=625\). (2) The two cells have the same donor genotype, so their reads need not contain alleles incompatible with a singlet from that donor; genotype evidence may therefore be uninformative. (3) No. The ratio compares the likelihoods of the observed data under two specified models. Posterior odds also depend on prior odds and on other possible explanations. (4) Examples include checking read and mapping quality, requiring support across multiple informative loci, checking for duplicate reads, and examining whether unexpected alleles recur across many barcodes in a way suggestive of a systematic error.