On the delta Q-test of Templeton.

Using a modified form (delta Q [AQ]-test) of Pielou’s (1979) Q-test, Templeton (1985) reanalyzed Sibley and Ahlquist’s (1984) DNA-DNA hybridization data for hominoid species. He concluded that the phylogeny (fig. 1A) supported by Sibley and Ahlquist is not significantly better than the phylogeny (fig. 1B) favored by Templeton (1983). However, Templeton’s (1985) AQ-test has several statistical problems, and his conclusion does not seem to be justified. In brief, Pielou’s Q-test and Templeton’s AQ-test are as follows: Consider the distance matrix in table 1 A, where dij (i > j) denotes the evolutionary distance between species i and j. In the Q-test, dii is compared with dkl (k > I) with the restriction i dkl, and 0.5 when do = dkl, is considered. The Q-statistic is the sum of x for all possible comparisons of do and dkl. If we assume that phylogeny A in figure 1 is correct and that dij increases in proportion to evolutionary time without error, we will have the inequalitydz,<d3j<d4j<dsj(j= 1, l a* 4).InthiscaseQ=9+(2X7)+(3X4) = 35 for distance matrix A of table 1. This is the maximum value Q can take for the five-species case. In practice, dij may be subject to random errors, and, furthermore, the phylogeny A of figure 1 may not be true. Pielou’s ( 1979) original test is for examining whether or not there is a hierarchical structure of species (strata in her original problem), and the null hypothesis is that there is no hierarchical structure, i.e., that all species diverged at the same time. To test this null hypothesis, Pielou’s Q-test is justified. In evolutionary studies, however, there is almost always some hierarchical structure. For this reason, Templeton (1985) introduced the AQ-test. In this test the Q-statistic is computed for two different phylogenies (e.g., A and B in fig. l), and the difference in Q between the two phylogenies is called AQ. Templeton obtained the probability distribution of AQ and used this distribution for testing the statistical significance of AQ. However, this distribution is based on the assumption that there is no hierarchical structure. As mentioned earlier, the Q-value for matrix A of table 1 is 35. By contrast, the Q-value for matrix B, which corresponds to phylogeny B of figure 1 becomes 29-3 1, depending on the differences among ddl, d42, and d43 (see table 1 B). The distribution of AQ (Templeton’s table 2) indicates that AQ = 6 is required to achieve statistical significance at the 5% level, but other, smaller values are nonsignificant. If we apply the above test to Sibley and Ahlquist’s (1984) data, AQ becomes 4. Therefore, this test suggests that phylogeny A is not favored against phylogeny B. However, we note that AQ depends only on the ranks among d4j)s in the present case. With the inequality d2j < dsj < dbj < dsj (j = 1, l l l 4) maintained, AQ becomes 6 only when db3 is smaller than both ddl and dd2. Even when the strict rate constancy is assumed, where d4, = dd2 = d43, AQ = 5 (see table 1). Obviously, d4, , d42, and dJ3 may vary by chance effects even if species 4 is remotely related to species 1, 2, and 3, and AQ = 4 may be obtained with a high probability. It is therefore clear that the AQtest is inadequate for testing topological differences.