Integrative Biology 200B University of California, Berkeley Spring 2009 "Ecology and Evolution" NM HallinanAnswers to Lab 6: PIC Assignment 1. All tests are two tailed, because you have no reason to expect a positive or a negative correlation. You do not even know what these characters are. (1 point)2. Neither the data, nor the log transformed data fit the assumptions of multivariate normality, so I used a non-parametric test for correlation: Kendall's and/or Spearman's. (1 point)3.***The first number is for cor.table and the second is for cor.test. They're different, because cor.table has an error that I will give you a correction for. For Kendall's the 2-tailed ps are: (¾ point)V2-V3 4.164559e-06 6.069e-10V2-V4 6.320802e-15 2.2e-16V3-V4 1.552561e-05 4.308e-09Even better multiply by 3 again to correct for multiple tests: (+¼ point)V2-V3 1.249367e-05 1.8207e-09V2-V4 1.896240e-14 6.6e-16V3-V4 4.657684e-05 1.2924e-08For Spearman's, the same numbers are: (¾ point)V2-V3 5.858564e-13 3.625e-13V2-V4 2.094989e-29 2.2e-16V3-V4 9.687686e-12 5.818e-12+ Bonferroni: (+¼ point)V2-V3 1.757570e-12 1.0875e-12V2-V4 6.28497e-29 6.6e-16V3-V4 2.906306e-11 1.7454e-114. All non phylogenetic correlations are significant. V2 vs V4 is positive. V2 v V3 and V2 v V4 are negative. (1point)5. The untransformed data did not fit the assumptions of BM. In particular the contrasts vs Node value and contrasts vs node age were both significant. The log transformed data did fit the assumptions, so independent contrasts were derived from the log transformed data. (1point)6. The PICs did not meet the expectations of multivariate normality either, so I used non-parametric tests: (1point)7.***Only results from cor.table are valid, because they have correction for PICs. However, in reality all these numbers are bad, because cor.table has an error.***You get p-values greater than one, because the Bonferroni correction only works for small values. But it doesn't matter, because those results are insignificant anyways.***Don't forget to use cor.type= “c” Kendall's results: (¾ point)V2-V3 0.3248054 V2-V4 6.570840e-06V3-V4 0.3967225Bonferroni correction:(+¼ point)V2-V3 0.9744162 V2-V4 1.971252e-05V3-V4 1.1901675Spearman's results: (¾ point)V2-V3 0.1736300 V2-V4 1.517496e-09V3-V4 0.3955777Bonferroni correction:(+¼ point)V2-V3 0.52089 V2-V4 4.552488e-09V3-V4 1.186733Sign test results: (½ point)1-tailedV2-V3 0.02973169V2-V4 7.610913e-06V3-V4 0.23543952-tailed:(+¼ point)V2-V3 0.05946338 V2-V4 1.522183e-05V3-V4 0.470879Bonferroni correction:(+¼ point)V2-V3 0.1783901 V2-V4 4.566548e-05V3-V4 1.4126378. V2 vs V4 correlation is significant and positive. None of the other relationships are significant at all. (1 point)9. There are two ways to tell why you are getting this affect and they complement each other. First you can plot the characters against each other on a tree. This is a plot of log character 1 versus log character 2 that I made in Mesquite. As you can see character 1 has very low values in the clade at the bottom, while character 2 has very high values. This implies that both characters underwent a lot of change on the branch leading to this clade, and that produced a specious correlation in the data when the phylogeny was not taken into account.You can also plot the independent contrasts against each other. When you plot character 2 against either of the other characters, you will find most of the points along the axes. However, there are two contrasts that are negative for character 1 and positive for character 2.These are the 20th and the 1st contrast. The first contrast is of course the contrast for the root. It is easy to show, using identify or by looking at the $edge matrix that the 20th contrast is the contrast between the clade we identified in the previous plot and its sister clade. We would expect this contrast to show a relationship. The root node also shows a relationship, because it is the next node down from our clade, and thus still shows some affect from the large changes along that one branch.You can also see this pattern by plotting the logs of the data at the tips against each other, and coloring the tips from our crazy clade red. As you can see, that clade has excessively high values for character 1 and low for character 2.Thus, we can conclude that we see a large correlation in the untransformed data, because all three characters undergo a great deal of change along this one branch, which leads to correlation among the terminal taxa. However, this correlation is not found throughout the rest of the tree. This branch clearly underwent a great deal of evolution in all characters, and something interesting may have happened there.-0.004 -0.002 0.000 0.002 0.004-4e-04 -2e-04 0e+00 2e-04 4e-04PICs[, 1]PICs[, 2]120-1 0 1 2 3 4-3 -2 -1 0 1log(data[, 1])log(data[,
View Full Document
Unlocking...