The Neighbor-Joining Method: A New Method for Reconstructing Phylogenetic Trees
Molecular Biology and Evolution, 1987 · paper · Biomedicine
Proposes the neighbor-joining method for reconstructing phylogenetic trees from evolutionary distance data. In computer simulations comparing five other methods, it and Sattath and Tversky's method were generally better at recovering the correct unrooted tree.
2 findings
Substance The worked example says "L_DF becomes L(1-2-3-4-5-6)Z − D(1-2-3-4)(5-6)/2 = 2 (fig. 3[f])". However, the D–F branch is labelled 1 in both Fig. 1 and Fig. 3(f), and Table 1 implies 1.
- p. 6 Worked example, last paragraph before 'Criterion for the Minimum-Evolution Tree'
D(1-2-3-4)(5-6)/2 = 2 (fig. 3[f])
- p. 2 Fig. 1, branch D–F
1
- p. 6 Fig. 3(f), interior branch to Z
1
- p. 2 Fig. 1, branch D–F
FIG. 1.—An unrooted tree of eight OTUs, 1–8.
Judge's note: I read the branch lengths off Fig. 1 on p. 407: 1A=5, 2A=2, AB=2, B3=1, BC=1, C4=3, CD=2, DE=2, E5=1, E6=4, DF=1, F7=2, F8=6. Fig. 3(f) on p. 411 also shows 1 on the branch to Z. Taking Table 1's D57=6 and D17=13 with the other branch lengths from Fig. 1 gives DF=1 both times. The text prints '= 2' (checked in the image) and then says all branch lengths are correctly reconstructed.
Second check: another model agreed
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Substance The varying-rate section says "The value of a and the expectation of D̄ over all branches were 0.01 and 0.016, respectively" and claims correspondence with Table 3, where the text gives "the D̄ for all OTUs is 0.16 for both model trees".
- p. 14 Varying Rate of Nucleotide Substitution, para 2
were 0.01 and 0.016, respectively
- p. 11 Text below Table 3
the D̄ for all OTUs is 0.16 for both model trees
- p. 14 Varying Rate section, paragraph 2
The value of a and the expectation of D over all branches were 0.01 and 0.016, respectively.
- p. 11 Text below Table 3
the D for all OTUs is 0.16 for both model trees
Only the authors can say what is right here.
Judge's note: The p. 419 image reads 0.016. On p. 416 the text gives D̄ = 0.16 for the Table 3 case (a=0.01, b=0.04, c=0.07), and p. 419 says the simulations correspond to that case. The branch lengths in Fig. 7(A) also rule out 0.016: the tip branches alone are 0.04–0.16, so pairwise distances are at least about 0.1, and the mean branch length is about 0.07. The value is a factor of 10 off from the case it is said to match; only the authors can confirm whether 0.16 was intended.
Second check: another model agreed
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3 minor (typos and the like; most checked by the judge only)
Wording The text says "If we conduct a similar test for the five remaining interior branches of the tree in figure 1", but Fig. 1 has five interior branches in all (AB, BC, CD, DE, DF), so four remain after CD.
- p. 8 Last paragraph before Fig. 4
test for the five remaining interior branches of the tree in figure 1
- p. 8 Criterion for the Minimum-Evolution Tree, last sentences
similar test for the five remaining interior branches of the tree in figure 1
- p. 8 last paragraph before Fig. 4
for the five remaining interior branches of the tree in figure 1
- p. 8 Criterion section, last paragraph
the five remaining interior branches of the tree in figure 1
Judge's note: On the upright image of p. 413, the text tests branch CD and then refers to the 'five remaining' branches. Fig. 1 on p. 407 has interior nodes A–F joined by five interior branches (N−3 = 5 for N = 8), so the count should be four.
Second check: another model agreed
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Typo The address footnote on p. 406 reads "Address for corespondence and reprints"; it should be "correspondence".
- p. 1 Footnote, correspondence address
Address for corespondence and reprints
Judge's note: The page image (crop C001-a) shows 'corespondence', so the misspelling is in the print.
Second check: none (judge only)
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Typo The text gives "(S_O = 39.28)", but the Table 1 distances sum to 275, so S_O = 275/7 = 39.2857, which rounds to 39.29.
- p. 6 Worked example, cycle 1
= 39.28) of branch lengths of the starting starlike topology
- p. 5 Table 1
Distance Matrix for the Tree in Figure 1
- p. 6 text on cycle 1
= 39.28
Judge's note: I added up the rows of Table 1 from the p. 410 image: 7, 13, 24, 38, 55, 53 and 85, a total of 275. The paper's S_O is the sum of the D_ij divided by N−1 = 7. The printed 39.28 is a truncation, while the other two-decimal values in the example are rounded (e.g. S12 = 36.67). This is a last-digit slip.
Second check: none (judge only)