Oncofusion

start / graph / exp-q120-an-mll1-c-terminal-chip-beside-the-n-terminal-one-in-one-human-line

Experimentmoderate confidence 3 of 5: One strong study, or several smaller consistent ones.Q120-h3k79me2-second-line-and-the-mll1-epitope

The cheap experiment nobody has run: an MLL1 C-TERMINAL ChIP beside the standard N-terminal one in a single human KMT2A-rearranged line under a menin inhibitor - the C-terminus is lost in the fusion, so the difference of the two tracks is the fusion, with no tag and no knock-in

also called the C-terminal subtraction · bound the masking without a tag

The fusion protein keeps the front half of KMT2A and loses the back half. So an antibody against the back half sees only the normal protein, and subtracting it from the usual antibody's signal leaves the fusion - one extra experiment in a run people already do.

The same thing, for a scientist

**The problem.** Every human occupancy statement about a KMT2A fusion under a menin inhibitor rests on Bethyl A300-086A, an N-terminal epitope retained in the fusion, which reads fusion plus wild-type MLL1 in unknown proportion. Q118 bounded the resulting masking at 3.54-4.00x using menin as a proxy; this beat measured 2.23x against a FLAG tag in mouse and then showed the mouse magnitude does not transfer, because the murine fusion is retrovirally over-expressed. **The design.** In RS4;11 or MOLM13, 330 nM VTP-50469 or DMSO for 3 days, three ChIPs from the same chromatin: **anti-MLL1 N-terminus** (A300-086A, the field's reagent), **anti-MLL1 C-terminus or anti-SET-domain** (an epitope 3' of the breakpoint, therefore absent from the fusion), and **anti-menin**, plus an input in BOTH conditions. Two biological replicates per condition, and a spike-in that is actually *used* - this beat's two source series ran or skipped spike-ins and neither applied one. **The readout.** At the N-terminal track's own top 0.1% of 2 kb bins: `fusion_signal = N_terminal - k * C_terminal`, with `k` set in DMSO by requiring the residual to be non-negative genome-wide, and the fusion's share ratio computed from the residual. The quantity the field wants - **what fraction of an MLL1-N ChIP signal at the fusion's sites is the fusion, in a human cell with an endogenous single-allele fusion** - falls straight out. **Why it has not been done.** Both datasets this project depends on used the same N-terminal reagent, and the control that would qualify it stopped being run once the field converged. It is one extra ChIP in an experiment people already perform.

Go deeper
Cost, and the honest weaknesses

Cost. One cell line, two conditions, four ChIPs plus two inputs, two replicates - about 12 libraries. No animals, no knock-in, no new reagent to raise. Weeks, not months, and cheaper by an order of magnitude than the tagged-allele experiment this graph already holds in exp-q118-a-tagged-kmt2a-aff1-chip-in-rs4-11-under-vtp-50469, which remains the definitive version.

Weakness 1. A C-terminal MLL1 antibody must be validated as not detecting the fusion, which needs a KMT2A-wild-type control cell and ideally a fusion-knockdown. If the C-terminal reagent cross-reacts the subtraction is worthless.

Weakness 2. The subtraction assumes the two epitopes are equally accessible in chromatin, which they are not guaranteed to be. k absorbs a global efficiency difference and not a locus-specific one, so the residual is a bound rather than a clean fusion track.

Weakness 3, and it is the one to state first. Wild-type MLL1 turns out to be strongly menin-dependent: this beat measured wild-type-dominated MLL1 sites in mouse falling to 0.1069, HARDER than the tagged fusion's 0.1793. So the C-terminal track will move under drug as well, and the subtraction must be done condition-by-condition rather than assuming a static wild-type component - which is exactly the assumption that made this beat's own two-component model insoluble.

Facts and figures
statusproposed
systemRS4;11 (KMT2A::AFF1) or MOLM13 (KMT2A::MLLT3), human, endogenous single-allele fusion
drugVTP-50469 330 nM, 3 days, with DMSO 0.33% control
n libraries estimate12
readoutfraction of the MLL1-N ChIP signal at the fusion's own sites attributable to the fusion, in a human endogenous-fusion cell
what it would settlewhether Q118's 3.54-4.00x human masking estimate or this beat's 2.23x murine measurement is nearer the truth for human cells, and therefore whether any published human 'the fusion is retained' statement stands
epigenetic readoutnone directly; this is an occupancy experiment
shown for fusionnobody has run it in any fusion; the murine FLAG substitute exists for MLL-AF9 only and is an over-expression model
absence classnever-measured
absence class noteNo archive holds an MLL1 C-terminal ChIP in a KMT2A-rearranged line under a menin inhibitor, and no re-analysis can create one. It needs bench work - though only weeks of it, which is why the class understates how cheap this particular never-measured quantity is.
absence class assigned2026-09-23 (Q120)
assays
[
  "anti-MLL1 N-terminus ChIP-seq (Bethyl A300-086A, the field's reagent)",
  "anti-MLL1 C-terminus or anti-SET ChIP-seq (epitope 3' of the breakpoint)",
  "anti-menin ChIP-seq (Bethyl A300-105A)",
  "input in BOTH conditions",
  "an applied spike-in normalisation"
]
How this could bite us back
  • If the subtraction shows the fusion is a small minority of the MLL1-N signal in human cells, then a decade of 'the fusion remains on chromatin' conclusions become unsupported at once, and the field will have to re-derive its pharmacodynamic model from menin occupancy instead.
  • If it shows the fusion is the majority, then the masking argument this project has built over three cycles - including this beat - is wrong, and the MLL1-N channel's flatness is the biology after all.
Still open
  1. Which commercial anti-MLL1 antibodies target an epitope 3' of the common breakpoint cluster region? A reagent survey was not run in this beat and is the first step; the answer may be that none is ChIP-validated, which would itself be worth recording.
  2. Could the same subtraction be done on existing public data? Only if some deposited series contains both an N-terminal and a C-terminal MLL1 ChIP in the same KMT2A-rearranged cells, and no such series is known to this project.
Every source behind this entry (1)
  1. E1strong confidence 4 of 5: Replicated in peer-reviewed work by more than one group.dataset

    The masking that this experiment would bound is measured: 2.23-fold at the FLAG-tagged fusion's own top 0.1% of murine bins (MLL1 0.3995 against FLAG 0.1793), 1.83-fold at Myb, and a naive fusion fraction of 0.7317 in mouse against 0.212-0.237 by the menin-proxy calculation in human RS4;11 and MOLM13 - a three-fold gap that cannot be resolved without a human-side measurement.

    What this evidence cannot show

    This is an experiment proposal: nothing in it has been run. The motivating measurement is real and comes from a murine over-expression model at a different dose (500 nM against 330 nM). Bethyl A300-086A, deposited as 'anti-MLL1n', an MLL1 N-TERMINUS epitope. The N-terminus is retained in every KMT2A fusion, so this track is the sum of the fusion and wild-type MLL1 in unknown proportion and CANNOT distinguish them. The immunogen residue range was not verified: three vendor routes were tried this session (fortislife.com two URL forms, bethyl.com/product/A300-086A) and all returned HTTP 404, so 'N-terminus' rests on the depositors' own target label and on the standard fact that MLL-FPs lack the C-terminal SET domain. NEB 14793S DYKDDDDK (D6W5B) reads the FLAG-tagged MLLAF9 construct only, so it is ground truth for the fusion and blind to wild-type MLL1. Neither it nor the MLL1 antibody can see a fusion molecule whose FLAG tag has been proteolysed, which would inflate the MLL1/FLAG ratio and mimic masking.

    Evidence that argues against this (1)
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