Run on the reciprocal KMT2A-C antibody instead of the fusion-side KMT2A-N, the same rule gives zero AFF1-exclusive recurrent gene bodies against a null of 0.52 - the gene-scale class is fusion-arm specific
also called q154 C3 · the KMT2A-C control
Two antibodies are used: one sees the part of KMT2A that is kept in the fusion, the other sees the part that is lost. Repeating the whole analysis with the wrong-side antibody gives no AFF1-only genes at all. Whatever the six genes are, they are a property of the fusion side of the protein, not of the KMT2A locus in general.
The same thing, for a scientist
The pre-registered fusion-specificity control. The KMT2A-C1/C2 antibodies read the reciprocal C-terminal fragment, which is **not** retained in any KMT2A fusion; the KMT2A-N1/N2 antibodies read the N-terminal fragment that every fusion keeps. Running the identical gene-body rule on the C arm over the same 26 samples: the recurrent AFF1-exclusive class (>=3 of 8 AFF1, 0 of 18 non-AFF1) is **0**, against a permutation null of **0.58** unstratified and **0.52** under the exact institution-stratified permutation - so the C arm produces **fewer** such genes than chance, where the N arm produces **6 against 0.86** [E1]. At the singleton level the C arm gives **20 against a null of 43.7, p_lower = 0.0116** - significantly **below** chance [E1]. The C arm is not simply empty: its genome-wide pass rate is **0.469% of gene bodies per sample**, three-quarters of the N arm's 0.608% [E1]. **So the control is a real measurement of comparable volume that returns nothing, which is what a fusion-arm-specific class requires.** The pre-registered SURVIVES condition 3 - that the class be smaller in KMT2A-C than in KMT2A-N - is met in the strongest available form.
The numbers, both tails
| arm | class | observed | null mean | 5th | 95th | p_upper | p_lower |
|---|---|---|---|---|---|---|---|
| KMT2A-N | k>=3 of 8, 0 of 18, S1 | 6 | 0.88 | 0 | 3 | 0.0026 | 0.9996 |
| KMT2A-N | k>=3, S3 institution EXACT | 6 | 0.86 | 0 | 3 | 0.0011 | 1.0000 |
| KMT2A-C | k>=3 of 8, 0 of 18, S1 | 0 | 0.58 | 0 | 2 | 1.0000 | 0.6056 |
| KMT2A-C | k>=3, S3 institution EXACT | 0 | 0.52 | 0 | 2 | 1.0000 | 0.6450 |
| KMT2A-N | k>=1, S1 | 63 | 73.91 | 36 | 120 | 0.6120 | 0.4038 |
| KMT2A-C | k>=1, S1 | 20 | 43.68 | 24 | 68 | 0.9921 | 0.0116 |
| KMT2A-C | k>=1, S3 EXACT | 20 | 27.34 | 19 | 36 | 0.9460 | 0.0823 |
The C arm's singleton class being significantly below its own null is itself informative: the C-terminal fragment's occupancy is more uniform across samples than a random relabelling predicts, which is what a locus-wide, partner-independent readout should look like.
Why this is the strongest control in the beat, and what it still cannot do
It is a within-sample, within-library control. The C arm comes from the same nuclei, the same plate, the same sequencing run and the same depth regime as the N arm, so a batch effect, a depth effect, an institution effect or a lineage effect that produced the N-arm class would have to be antibody-specific to spare the C arm. None of the confounds this beat has controlled for - institution, lineage, call rate, patient pairing - can be antibody-specific.
What it cannot do is separate an address from a level. If the AFF1 samples simply carry more fusion protein on chromatin, the N arm would show a class and the C arm would not, for exactly the reason seen here - and that would be a level effect, not an address. So C3 rules out everything except the one confound that the member-level audit says is probably operating.
| script | analyses/q154-gene-scale-occupancy-null/analyze.py |
|---|---|
| status | run |
| result summary | KMT2A-C arm: 0 recurrent AFF1-exclusive gene bodies vs nulls of 0.58 and 0.52; 20 singletons vs 43.7 (p_lower 0.0116); pass rate 0.469% vs the N arm's 0.608%. |
inputs
[ "Q154 gene-body occupancy matrix, 19,402 x 133" ]
measurements
[
{
"quantity": "count_recurrent_partner_exclusive_gene_bodies_reciprocal_antibody",
"subject": "gene bodies occupied in >=3 of 8 AFF1 and 0 of 18 non-AFF1 scored on the KMT2A-C1/C2 antibodies",
"value": 0,
"units": "gene bodies",
"n": 19402,
"universe": "the same 19,402 gene bodies x 26 samples, identical rule, reciprocal C-terminal antibody pair",
"null": 0.52,
"note": "exact institution-stratified null; the fusion-side N arm gives 6 against 0.86 on the same matrix"
},
{
"quantity": "occupancy_rule_pass_rate_reciprocal_antibody",
"subject": "fraction of gene bodies called occupied per sample on the KMT2A-C arm",
"value": 0.469,
"units": "% of gene bodies per sample (median over 27 samples)",
"n": 19402,
"universe": "the same matrix, KMT2A-C1 and C2",
"null": 0.608,
"note": "the null quoted is the N arm's pass rate; the C arm is a measurement of comparable volume, not an empty one"
}
]- A control that rules out every confound except the one that is probably operating is the most dangerous kind, because it reads as decisive. This one excludes batch, depth, institution, lineage and patient pairing and does not exclude an oncoprotein-level gradient.
- The C arm's singleton class is significantly BELOW its null (20 against 43.7, p_lower 0.0116). Is that a property of wild-type KMT2A occupancy being more uniform across patients than fusion occupancy, and is that measurable as a between-sample correlation rather than as a class count?
- Would an N-over-C ratio gate - the authors' criterion - produce a larger or a smaller AFF1-exclusive class than the N-only gate used here, on the same gene bodies?
- q142 found the C arm's replicate reproducibility is 0.62-0.73 against the N arm's 0.85-0.91. How much of the C arm's empty class is explained by requiring two replicates to agree, and what does a one-library C-arm rule give?
- E1strong confidence 4 of 5: Replicated in peer-reviewed work by more than one group.dataset
Running the identical two-antibody gene-body rule on the KMT2A-C1/C2 arm over the same 26 GSE252378 samples gives 0 AFF1-exclusive recurrent gene bodies against permutation nulls of 0.58 (unstratified) and 0.52 (exact institution-stratified), and 20 singleton AFF1-only gene bodies against a null of 43.68 with p_lower 0.0116; the C arm's genome-wide pass rate is a median 0.469% per sample against the N arm's 0.608%.
Q154 genome-wide gene-body occupancy matrix and label permutations · Iqbal Lab oncofusion knowledge graph, beat Q154 · this project 2026What this evidence cannot show
A zero observed against a null mean of 0.52 cannot be significant in the upper tail by construction, so the p_upper of 1.0 is uninformative and only the contrast with the N arm's 6 carries the argument. The C-terminal fragment is a real protein present in these cells - the reciprocal der(4) product and wild-type KMT2A both contribute - so this is a comparison between two occupancies, not between signal and blank.
Evidence that argues against this (1)
- q142 measured that the KMT2A-C arm's own replicate reproducibility is lower than the N arm's (C1~C2 rho 0.62-0.73 against N1~N2 0.85-0.91), so a two-replicate rule is harder to satisfy in the C arm for reasons unrelated to biology. Part of the C arm's empty class is reduced reproducibility, not reduced signal. Iqbal Lab oncofusion knowledge graph et al. 2026
- E2established confidence 5 of 5: Multiple independent lines of evidence; treated as settled in the field.paper
The KMT2A-N antibodies read the N-terminal fragment retained in every KMT2A fusion and the KMT2A-C antibodies read the reciprocal C-terminal fragment that no fusion retains, which is why the deposit's authors use the N-over-C ratio as their own definition of an oncoprotein target site.
MLL oncoprotein levels influence leukemia lineage identities · Janssens DH, Duran M, Otto DJ, Wu W, Xu Y, Kirkey D, Mullighan CG, Yi JS, Meshinchi S, Sarthy JF, Ahmad K, Henikoff S · Nature Communications 15:9341 2024What this evidence cannot show
The authors' criterion is a ratio of N to C within an interval with a Bayesian model and an FDR; this beat's C-arm control scores the C arm on its own against the same absolute gate, which is a different and simpler use of the same two antibodies.
Evidence that argues against this (1)
- Taspase1 cleaves wild-type KMT2A into N320 and C180 fragments that remain associated, so a KMT2A-C signal at a site does not establish that no fusion is there, and a KMT2A-N signal does not establish that the wild-type allele is absent. The arms are enriched for, not exclusive to, the two species. Janssens DH et al. 2024
corrects the kings cycle-144 misreading: the C arm control supports fusion-specificity
a second negative control on the same class, from the antibody side
uses the same reciprocal arm per gene rather than per class
same arm, per locus rather than per class; HOXA9 and MEIS1 validate the readout
the same rule on the reciprocal KMT2A-C antibody gives 0 against a null of 0.52, so the class is fusion-arm specific
- related to →KMT2A / MLL1 — the gene that breaks
the C-terminal fragment's gene-body occupancy is more uniform across patients than a random relabelling predicts