Read Counts in Non-UMI Plate-Based Assays

What a Non-UMI Count Measures

In a non-UMI plate-based assay, a gene-level count is formed by assigning sequenced reads or fragments to genes and tallying the assigned observations. The key point is that the tally is based on sequenced read or fragment observations, not on a label that identifies each original RNA molecule. If an original cDNA molecule is amplified many times, reads sampled from its amplified copies can contribute repeatedly to the gene’s count.

This distinction matters when interpreting counts. A larger count can reflect more RNA captured from a cell, but it can also reflect differences in reverse transcription, amplification, library representation, sequencing depth, and assignment. A non-UMI count therefore records observations produced by the assay and its analysis rules; it is not automatically an estimate of the number of original RNA molecules.

Definition: For a specified gene and well, a non-UMI read count is the number of sequenced read observations assigned to that gene under a stated assignment rule. A non-UMI fragment count is the number of sequenced library fragments assigned to that gene. In paired-end sequencing, a fragment is typically represented by two mate reads, so a pipeline that counts fragments counts the pair once, not once per mate. The exact unit must be identified from the pipeline’s counting convention.

A read is an observed sequence from one end of a library fragment. A fragment is the sequenced library molecule represented by one read in single-end sequencing or usually by a pair of mate reads in paired-end sequencing. Neither term means “one original RNA molecule.” Library preparation and amplification intervene before sequencing, and a single captured RNA molecule can give rise to multiple cDNA copies and multiple sequenced fragments.

From Alignment to a Gene Count

Once reads or fragments have been aligned, a counting procedure applies the chosen gene annotation and assignment rules. For a simple hard-assignment count, each eligible observation contributes one if it is assigned to the gene and zero otherwise. Let \(C_g\) be the count for gene \(g\), and let \(I_k(g)\) equal 1 when observation \(k\) is assigned to \(g\), and 0 when it is not. Then:

C_g=\sum_{k=1}^{N} I_k(g)

Here \(N\) is the number of observations considered by the procedure: reads if the pipeline counts reads, or fragments if it counts fragments. Under hard assignment, an observation that cannot be assigned unambiguously may be excluded, while an observation assigned to one gene contributes exactly one to that gene’s count. Some methods instead distribute ambiguous observations fractionally; the resulting counts need not be integers. Thus a reported “read count” is meaningful only with its unit and assignment policy specified.

In full-length plate-based protocols, reads can originate from different positions along a transcript. A read may overlap an exon, span an exon boundary, or fail to overlap the annotated region used by the counting procedure. The annotation and rules for exonic, intronic, or ambiguous alignments determine which observations enter a gene’s tally. These rules define the count; they do not recover a molecular identity that was never recorded with a UMI.

Worked Example: Tallying Assigned Fragments

Suppose a hypothetical well produces eight paired-end fragments considered by a counting pipeline. Five fragments are assigned to Gene A, two to Gene B, and one overlaps features of both genes and is excluded under the pipeline’s unique-assignment rule. Find the fragment counts for the two genes.

For Gene A, five of the eight fragments have \(I_k(A)=1\), and the remaining three have \(I_k(A)=0\). Therefore:

C_A=\sum_{k=1}^{8} I_k(A)=5

Likewise, two fragments are assigned to Gene B, so \(C_B=2\). The ambiguous fragment contributes to neither count under this rule. These are fragment counts: if each of the five Gene A fragments has two sequenced mates, the count is still 5 fragments, not 10 reads. A read-counting pipeline could report a different number because its counting unit is different.

Why Repeated Reads Are Not Deduplicated as Molecules

In UMI-based counting, reads can be grouped by a cell barcode, gene assignment, and UMI; after appropriate handling of UMI errors, each distinct UMI group is used as evidence for a captured molecule. A non-UMI plate-based library does not have that molecular tag. If two reads map to the same gene and have the same sequence, their similarity alone does not reveal whether they came from PCR copies of one starting molecule or from distinct molecules that happen to produce the same sequence.

This is the central difference in counting units. A read or fragment count tallies sampled library observations. A UMI count aims to tally distinct tagged molecules, subject to limitations such as UMI collisions, tag errors, and molecule loss. Neither count is a perfect census of all RNA molecules in the original cell, but only the UMI-based method has a tag designed to distinguish molecular origins.

Key distinction: In a non-UMI assay, repeated reads assigned to a gene are retained as repeated read or fragment observations unless the pipeline applies a separate deduplication rule. Without molecular tags, sequence identity is generally insufficient to decide which observations are PCR copies of the same original molecule. In UMI counting, the counting unit is a distinct corrected UMI group, not every read supporting it.

A simple calculation shows why read counts can exceed the number of starting molecules. Suppose \(M_g\) captured molecules for gene \(g\) produce \(A_{gj}\) amplifiable library copies from molecule \(j\), and each copy has probability \(q_g\) of yielding an assigned sequenced observation. In this simplified model, conditional on the amplified library, the expected number of assigned observations is:

E(C_g\mid A_{g1},\ldots,A_{gM_g}) =q_g\sum_{j=1}^{M_g} A_{gj}

The expression follows by writing the count as a sum of indicators for whether each amplifiable copy yields an assigned observation. Each indicator has expectation \(q_g\), so the expectation of their sum is the sum of their expectations. The model is deliberately simplified: real libraries can have unequal amplification, fragment generation, sequencing probabilities, and assignment probabilities. Even so, it makes the distinction clear. The count depends on the number of amplified copies sampled and assigned, not just on \(M_g\), the number of captured starting molecules.

Worked Example: One Starting Molecule, Several Read Observations

Consider a hypothetical gene with two captured RNA molecules. After cDNA preparation and amplification, suppose the first molecule contributes 6 amplifiable library copies and the second contributes 3. If each copy independently has probability \(q_g=0.5\) of producing an assigned read observation, calculate the expected non-UMI count under the simplified model.

The total number of amplifiable copies is \(6+3=9\). Thus:

E(C_g\mid A)=0.5(6+3)=4.5

The expected count is 4.5 read observations, despite there being only two captured starting molecules. An observed count must be a whole number, but its expectation need not be. If the assay had used UMIs and both captured molecules were successfully tagged and distinguishable, they could contribute two molecule groups rather than a count based on how many amplified copies were sequenced. This comparison is illustrative, not a claim that every UMI count perfectly recovers every starting molecule.

Counting Conventions and Their Consequences

Counts from non-UMI assays are only comparable when their units and processing choices are understood. Important choices include whether the pipeline counts reads or fragments, which genomic features are eligible, how multi-mapping or overlapping alignments are treated, and whether ambiguous assignments are excluded or shared. A difference between two reported counts can arise from a different counting rule even when the underlying sequencing data are identical.

Counting choice	What contributes	Interpretive consequence
Read counting	Each eligible read observation	Paired-end mates may contribute separately if the procedure counts reads
Fragment counting	Each eligible library fragment	A paired-end read pair usually contributes one fragment
Hard assignment	Only observations assigned to one gene	Ambiguous observations may be omitted
Fractional assignment	A fraction of an observation assigned to each eligible gene	Counts may be non-integer
UMI grouping	Distinct corrected UMI groups supporting a gene	Targets molecules rather than counting every supporting read

Worked Example: Paired-End Reads and an Ambiguous Alignment

A hypothetical paired-end library has six read pairs considered for a gene. Four pairs are uniquely assigned to Gene C, one pair is assigned to Gene D, and one pair aligns ambiguously to Gene C and Gene D. The pipeline counts fragments and excludes ambiguous pairs. What count does it report for Gene C, and how would a separate read-counting convention change the interpretation?

Each read pair represents one fragment. Four fragments are uniquely assigned to Gene C, so the fragment count is \(C_C=4\). The ambiguous pair is excluded, so it adds zero under the stated rule. Counting the mates separately would make the four uniquely assigned pairs correspond to eight mate reads only if both mates are eligible and assigned under that read-counting rule. That would be a read tally, not evidence for eight original RNA molecules. The correct reported unit depends on the actual pipeline, and the example’s answer is four because the specified pipeline counts fragments.

Assumptions, Diagnostics, and Limitations

The arithmetic count itself requires no statistical distributional assumption: given a set of observations and a fixed assignment rule, the tally is determined. Interpreting that tally as informative about RNA abundance requires more. At minimum, the library must be linked to the intended well and cell; alignment and gene assignment must be sufficiently reliable for the stated purpose; and comparisons must account for relevant differences in sequencing effort and technical processing. Even with these conditions, non-UMI counts can vary with amplification and capture efficiency, so they should not be equated directly with absolute transcript numbers.

Useful checks focus on whether the counting process matches its stated definition. Confirm whether the reported unit is reads or fragments, whether paired mates are combined, and whether the software excludes or fractionally assigns ambiguous alignments. Inspect a small number of alignments and count assignments manually when pipeline output is unexpected. Compare the number of input observations with the sum of assigned, unassigned, and excluded observations when those categories are available. Large unexplained discrepancies can indicate a unit mismatch, an annotation mismatch, or an assignment-rule error.

Counts can also be technically skewed. A small number of highly amplified starting molecules may contribute many reads, while other captured molecules contribute few or none. Increasing sequencing depth can raise observed read counts by sampling more of the amplified library, without increasing the number of original molecules in the well. Gene length and transcript structure can affect the chance that fragments overlap countable regions. These effects make read counts useful measurements of assay output, but they limit direct comparisons of raw counts across genes, cells, or protocols.

State the unit.
Determine whether the pipeline counts individual reads or library fragments; for paired-end data, do not assume that mates are counted separately.

State the assignment rule.
Record which alignments can be assigned, what happens to ambiguous observations, and which annotated regions are eligible.

Interpret the tally modestly.
Call it a read or fragment count unless a molecular tagging method supports a molecule-count interpretation.

Common Mistakes and the Main Alternative

A common mistake is to say, “These identical reads must be PCR duplicates, so I can collapse them into one molecule.” That conclusion does not follow from sequence identity alone. Distinct original molecules can produce indistinguishable reads, and PCR copies can differ in the observed fragments after fragmentation or sequencing. Without a molecular tag or a separately justified deduplication rule, collapsing identical reads can discard real observations without reliably recovering molecule counts.

The main alternative for molecule-oriented counting is UMI-based counting: reads sharing a gene assignment and a UMI are grouped, with the grouping procedure accounting for likely tag errors. This can reduce the impact of PCR amplification on the count, but it is not flawless. UMI collisions can merge distinct molecules, and UMI errors can split one molecule’s reads among apparent tags. Non-UMI read or fragment counts remain useful for measuring sequenced library representation and gene-associated evidence, but their unit must not be confused with a UMI-defined molecule count.

Wrong reasoning: “A gene has 100 reads, so the cell had exactly 100 molecules of that gene.”
Correction: The assay observed 100 assigned reads or fragments under its stated rules. Amplification and sampling can make that count differ substantially from the number of captured molecules.
Wrong reasoning: “Two identical reads must come from one molecule.”
Correction: Without molecular tags, sequence identity alone does not establish shared molecular origin.

Key takeaway: A non-UMI plate-based gene count is a tally of assigned sequencing observations—reads or fragments, depending on the pipeline. Amplification can make multiple observations arise from the same starting material, and the count should not be described as a direct count of original RNA molecules.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A paired-end pipeline reports fragment counts. Three read pairs are assigned to Gene A. What is the fragment count, and why is it not six?
A non-UMI gene has 20 assigned reads. Can this establish that 20 original RNA molecules were captured? Explain.
A pipeline excludes one alignment that overlaps two genes and uniquely assigns four other fragments to Gene B. What is Gene B’s hard-assignment fragment count?
Why can identical read sequences not generally be collapsed into reliable molecule counts in a non-UMI assay?

Solutions. (1) The count is 3 fragments; each paired-end read pair represents one fragment under the stated convention. (2) No. The 20 observations may include multiple reads derived from amplified copies of the same captured molecule, and the number also depends on sequencing and assignment. (3) The count is 4 fragments because the ambiguous alignment is excluded and the four unique fragments each contribute one. (4) Sequence identity does not identify molecular origin: distinct original molecules can yield identical reads, and repeated reads do not carry a tag proving they came from one original molecule.