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I've been trying to test for significant phylogenetic conservatism in some plant traits. So far I used R-package picante (Blomberg's K) and AOT in Phylocom. So far these two methods yield completely opposite results for the same traits, and I'm confused. Is it possible to get such very contrasting outputs from these two methods?

For example, one trait shows Blomberg's K = 0.2892 (P = 0.001), suggesting that the evolution of this trait overall has not been greatly constrained by phylogenetic relatedness. In contrast, for the same trait AOT yields variance of standardized contrasts across tree = 0.006, significance of contrast variance relative to null model (low rank) = 1, and significance of contrast variance relative to null model (high rank) = 1000 (i.e., P = 0.002), suggesting that treewise phylogenetic conservatism is significantly greater than the expectations of random tip swapping.

I've just started looking at these phylogenetic signals, so I may be overlooking something critical. I will greatly appreciate if you could kindly give me directions, thank you very much!

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I think I figured. Someone told me that the above results are the same. The absolute value of K = 0.29 indicates the magnitude of the signal is less than expected under a Brownian motion process, but it’s still more than random expectation (according to the P-value,P = 0.001). So both methods suggest that treewise phylogenetic conservatism is significantly greater than random expectations.

Actually the meanings of K statistic and P-value were clearly stated in the original paper (Blomberg et al. 2003). I should have looked at it first... but anyway, I think it is solved now.

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Glad you figured it out. In case you don't know, you are able to accept your own answer. –  kmm Apr 17 at 19:01
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