Alarm pheromones at two concentrations: Hart, Frank, Lopes et al. (2023), dandiset 000467¶
Paper: Hart T, Frank DD, Lopes LE et al. "Sparse and stereotyped encoding implicates a core glomerulus for ant alarm behavior", Cell 186 (2023), doi:10.1016/j.cell.2023.05.025.
Dataset: DANDI:000467, version 0.261004.1655 (doi:10.48324/dandi.000467/0.261004.1655), the only
published version; the paper cites none (see alarm_pheromones.ipynb).
Claims file: claims/000467/10.1016-j.cell.2023.05.025.yaml (accepted manuscript, PMC10334690).
This notebook tests one claim, figs5a-concentration-more-glomeruli, which needs whole-volume maps
(M1) of every pheromone trial at both of each ant's concentrations: 312 trials of 33 planes. It is
kept apart from alarm_pheromones.ipynb so that each notebook stays under the 30-minute limit.
| Claim | Here |
|---|---|
| figs5a-concentration-more-glomeruli | tested |
| all other claims | the other notebooks, or blocked (see the attempt record) |
import time
import numpy as np
import pandas as pd
from scipy import stats
import al_imaging as al
pd.set_option("display.width", 200)
pd.set_option("display.max_columns", 40)
T0 = time.time()
print("dandiset", al.DANDISET_ID, "version", al.DANDISET_VERSION)
dandiset 000467 version 0.261004.1655
assets = al.list_assets()
trials = al.trials(assets)
alarm_ants = sorted(trials[trials.odor == "heptanone"].ant.unique())
sel = trials[trials.ant.isin(alarm_ants) & trials.odor.isin(al.PHEROMONES) & (trials.nvol == 40)].copy()
pairs = sel.groupby("ant").conc.agg(lambda c: tuple(sorted(set(c))))
sel["level"] = np.where(sel.conc == sel.ant.map(pairs.map(max)), "higher", "lower")
print("ants", len(alarm_ants), "; concentration pairs:", pairs.value_counts().to_dict())
print(sel.groupby(["ant", "level", "odor"]).size().unstack("odor").to_string())
print(len(sel), "trials,", sum(len(b) for b in sel.blobs), "plane files")
ants 13 ; concentration pairs: {(0.03, 0.48): 5, (0.12, 0.48): 3, (0.0075, 0.12): 3, (0.03, 0.12): 2}
odor 6MHO heptanol heptanone hexanol
ant level
ant1-m4-d21-y2022 higher 3 3 3 3
lower 3 3 3 3
ant1-m4-d26-y2022 higher 3 3 3 3
lower 3 3 3 3
ant1-m4-d27-y2022 higher 3 3 3 3
lower 3 3 3 3
ant2-m4-d20-y2022 higher 3 3 3 3
lower 3 3 3 3
ant2-m4-d21-y2022 higher 3 3 3 3
lower 3 3 3 3
ant2-m4-d26-y2022 higher 3 3 3 3
lower 3 3 3 3
ant2-m4-d27-y2022 higher 3 3 3 3
lower 3 3 3 3
ant3-m4-d21-y2022 higher 3 3 3 3
lower 3 3 3 3
ant3-m4-d26-y2022 higher 3 3 3 3
lower 3 3 3 3
ant3-m4-d27-y2022 higher 3 3 3 3
lower 3 3 3 3
ant4-m4-d20-y2022 higher 3 3 3 3
lower 3 3 3 3
ant4-m4-d26-y2022 higher 3 3 3 3
lower 3 3 3 3
ant5-m4-d26-y2022 higher 3 3 3 3
lower 3 3 3 3
312 trials, 10296 plane files
t = time.time()
maps = al.trial_maps(sel, progress=lambda a, b: print(f"{a}/{b} files, {time.time() - t:.0f} s", flush=True) if a % 2500 == 0 else None)
print(f"maps in {time.time() - t:.0f} s")
2500/10296 files, 292 s
5000/10296 files, 590 s
7500/10296 files, 883 s
10000/10296 files, 1174 s
maps in 1209 s
shift = pd.Series({i: maps[i]["shift"].max() for i in sel.index})
print("largest frame shift per trial: median", shift.median(), "max", shift.max(), "pixels")
print("baseline F0 range", min(maps[i]["f0"].min() for i in sel.index), "to", max(maps[i]["f0"].max() for i in sel.index))
largest frame shift per trial: median 10.0 max 42 pixels baseline F0 range 0.0 to 8191.0
figs5a-concentration-more-glomeruli¶
Higher concentrations give more robust responses and a few more glomeruli passing the threshold, while the overall spatial pattern stays similar across concentrations.
Criterion: within each ant and pheromone, M2 counts at the higher minus the lower concentration. Reproduced when, pooled over pheromones, positive differences outnumber negative ones by a two-sided sign test at p < 0.05, and in more than half of ant and pheromone pairs every ROI that responds at the lower concentration also responds at the higher one. Shown under every M2 reading. The ROIs are those drawn at each ant's higher concentration (Methods: ROIs were drawn on the max projections "at the highest odor concentration for each ant").
rows = []
for ant in alarm_ants:
g = sel[sel.ant == ant]
mask = al.al_mask(maps, g.index)
for d in "ab":
rois = al.draw_rois(maps, g[g.level == "higher"], d, mask)
for v in ("i", "ii"):
for c in "xy":
for od in al.PHEROMONES:
rh, nh = al.responding(maps, rois, g[(g.odor == od) & (g.level == "higher")].index, v, c)
rl, nl = al.responding(maps, rois, g[(g.odor == od) & (g.level == "lower")].index, v, c)
rows.append({"ant": ant, "pair": pairs[ant], "reading": d + v + c, "odor": od, "higher": nh, "lower": nl,
"lower subset of higher": bool(np.all(rh[rl])) if len(rl) else True})
cc = pd.DataFrame(rows)
cc["diff"] = cc.higher - cc.lower
out = []
for k, g in cc.groupby("reading"):
pos, neg = int((g["diff"] > 0).sum()), int((g["diff"] < 0).sum())
p = stats.binomtest(pos, pos + neg, 0.5).pvalue if pos + neg else 1.0
out.append({"reading": k, "pairs": len(g), "higher > lower": pos, "higher < lower": neg, "ties": len(g) - pos - neg,
"sign test p": round(p, 4), "median diff": g["diff"].median(), "lower within higher": int(g["lower subset of higher"].sum()),
"sign test met": bool(pos > neg and p < 0.05), "subset met": bool(g["lower subset of higher"].sum() > len(g) / 2)})
out = pd.DataFrame(out)
out["met"] = out["sign test met"] & out["subset met"]
print(out.to_string(index=False))
print("\nPer ant and pheromone, reading bix (higher, lower):")
b = cc[cc.reading == "bix"]
print(b.pivot_table(index=["ant", "pair"], columns="odor", values=["higher", "lower"]).to_string())
print("\nBy concentration pair, reading bix: mean difference", b.groupby("pair")["diff"].mean().round(2).to_dict())
reading pairs higher > lower higher < lower ties sign test p median diff lower within higher sign test met subset met met
aiix 52 36 12 4 0.0007 3.000000 21 True False False
aiiy 52 39 11 2 0.0001 2.500000 23 True False False
aix 52 41 2 9 0.0000 2.000000 43 True True True
aiy 52 47 2 3 0.0000 2.500000 46 True True True
biix 52 14 12 26 0.8450 0.000000 37 False True False
biiy 52 31 13 8 0.0096 0.333333 36 True True True
bix 52 28 2 22 0.0000 1.000000 50 True True True
biy 52 43 2 7 0.0000 0.666667 50 True True True
Per ant and pheromone, reading bix (higher, lower):
higher lower
odor 6MHO heptanol heptanone hexanol 6MHO heptanol heptanone hexanol
ant pair
ant1-m4-d21-y2022 (0.12, 0.48) 1.0 1.0 2.0 3.0 0.0 0.0 0.0 0.0
ant1-m4-d26-y2022 (0.03, 0.48) 1.0 2.0 1.0 3.0 0.0 1.0 1.0 3.0
ant1-m4-d27-y2022 (0.0075, 0.12) 1.0 2.0 4.0 10.0 0.0 6.0 2.0 0.0
ant2-m4-d20-y2022 (0.03, 0.12) 0.0 0.0 1.0 3.0 0.0 0.0 0.0 1.0
ant2-m4-d21-y2022 (0.12, 0.48) 0.0 0.0 1.0 2.0 0.0 0.0 0.0 2.0
ant2-m4-d26-y2022 (0.03, 0.48) 0.0 2.0 3.0 5.0 0.0 0.0 1.0 3.0
ant2-m4-d27-y2022 (0.0075, 0.12) 1.0 0.0 0.0 2.0 0.0 0.0 0.0 0.0
ant3-m4-d21-y2022 (0.12, 0.48) 0.0 0.0 1.0 2.0 1.0 0.0 1.0 0.0
ant3-m4-d26-y2022 (0.03, 0.48) 0.0 0.0 2.0 3.0 0.0 0.0 0.0 2.0
ant3-m4-d27-y2022 (0.0075, 0.12) 0.0 0.0 0.0 5.0 0.0 0.0 0.0 0.0
ant4-m4-d20-y2022 (0.03, 0.12) 1.0 0.0 1.0 2.0 0.0 0.0 0.0 2.0
ant4-m4-d26-y2022 (0.03, 0.48) 2.0 2.0 1.0 8.0 0.0 2.0 1.0 3.0
ant5-m4-d26-y2022 (0.03, 0.48) 0.0 1.0 1.0 3.0 0.0 0.0 0.0 1.0
By concentration pair, reading bix: mean difference {(0.0075, 0.12): 1.42, (0.03, 0.12): 0.62, (0.03, 0.48): 1.1, (0.12, 0.48): 0.75}
Result: reproduced, resting on the adopted M2 reading. Under bix the count is higher at the higher concentration in 28 of 52 ant and pheromone pairs, lower in 2 and tied in 22 (two-sided sign test p < 0.0001), and every ROI responding at the lower concentration also responds at the higher in 50 of 52 pairs. aix, aiy, biy and biiy also meet the criterion; aiix and aiiy fail the subset condition (21 and 23 of 52) and biix the sign test (14 against 12, p = 0.845). The ROIs are drawn at the higher concentration, as in the Methods, which favours ROIs that respond there.
Summary¶
The outcomes of all claims, the differences from the paper's method and the findings about the dataset are summarised in alarm_pheromones.ipynb and the attempt record. Here: figs5a-concentration-more-glomeruli: reproduced, resting on the adopted M2 reading.
A failure to reproduce here may be a failure of this attempt, not of the paper: see the reasoning and the differences from the paper's method.
Runtime of this notebook:
print(f"notebook ran in {(time.time() - T0) / 60:.1f} min")
notebook ran in 20.3 min