Homologous 5,000 base pair DNA sequences are compared for several species of sma
ID: 64651 • Letter: H
Question
Homologous 5,000 base pair DNA sequences are compared for several species of small mammals. The matrix below represents the number of nucleotide substitutions observed (each value is shown just once to avoid redundancy). Use the UPGMA method (unweighted pair group method with arithmetic mean) to construct a phylogenetic tree based on this region of DNA. Indicate the average number of nucleotide substitutions for each branch point in the tree, as described in class and the textbook (1/2 credit each for species order and nucleotide substitution values).
Rat Mouse Shrew Mole 242 277 285 Chipmunk 0 Squirrel 0 255 263 270 105 191 179 27186 183 630) Chipmunk Rat Mouse Shrew Squirrel Mole 174 266 6Explanation / Answer
The pair of OTUs with the shortest pairwise distance is selected (i.e., A and B in this example). A subtree is then drawn for A and B with the branch length between them equal to 2 (Fig. 27.14).
This subtree is refined by positioning A and B as the tips and placing the node between them at the midpoint (e.g., 0.5 × 2 = 1) (Fig. 27.15). Note that by placing the node at the midpoint between the two groups, the UPGMA method implicitly assumes that the rate of change has been the same in each lineage since they diverged from a common ancestor. This assumption is invalid in many, if not most, cases and is one of the biggest limitations of the UPGMA method (discussed further below).
These two OTUs are merged and treated as one (AB). A new distance matrix is generated where the distance between each OTU and AB (Dx,AB ) is calculated as the average of the original distance from that OTU to A (Dx,A) and its original distance to B (Dx,B):
Dx,AB = 0.5 × (Dx,A + Dx,B).
The new distance matrix in this case would be that in Table 27.8.
These and the subsequent steps are summarized in Table 27.9 below.
One of the great advantages of UPGMA is that it is very fast. However, with speed comes some inaccuracy. Perhaps most critical is the assumption that evolution has been clocklike (i.e., that the rates of evolutionary change are uniform in different evolutionary branches). Given this clocklike constancy, then all of the tips in a tree must be equidistant from the root of the tree. Such a tree is called an ultrameric tree. In mathematical terms, this means that for any three OTUs (e.g., A, B, and C), the distance between any two (e.g., AC) is always less than or equal to the maximum distance between the other two (AB and BC):
DAC max(DAB, DBC).
This is illustrated in the tree in Figure 27.16. Looking at DAC, for example, we see that it passes the test because DAC = 6 and max(DAB, DBC) = 6. Likewise, all of the other possible OTU pairs also meet this criterion.
Although it would be nice if all evolutionary distances met this ultrameric criterion, this is not the case. For example, if the rate of evolution is not the same in all branches, as in the tree in Figure 27.17, the ultrameric criterion is not met for at least one pair of OTUs—for BC, in this case. Because DBC = 6 and the max of (DAB, DAC) is 4.5, DBC > max(DAB, DAC). When the ultrameric criterion is not met, UPGMA will return an incorrect tree.
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