PMC

Thus, all the 23 tree topologies have the same total dendritic surface area as well the same total dendritic length. A, For all the 23 tree topologies, the degree of bursting is shown as a function of the total length of the dendritic tree, plotted with an offset for clarity, and using somatic stimulation. The topologies are ordered as in , with the top line representing topology 1, and the bottom line topology 23. The total dendritic length of a given tree topology was varied by multiplying the lengths of all its segments by the same factor. B, Bursting as a function of topology at a total dendritic length of 1250 µm.

Natural selection inference of MYB genes with expected tree and reconstructed tree. (A) Inference of natural selection of RAD-like genes with the expected topology. (B) Inference of natural selection of RAD-like genes with the reconstructed topology. (C) Inference of natural selection of MYB59 genes with the expected topology. (D) Inference of natural selection of MYB59 genes with the reconstructed topology. The number on the branch is the ω value, and the branch with ω > 1 is marked with a thick line.

Colored lines represent gene lineages that trace back to a common ancestor along the branches of a species tree with topology (((AB)C)D). The figure illustrates how a gene tree can have a higher probability of having a symmetric topology, in this case ((AD)(BC)), than of having the topology that matches the species tree. If the internal branches of the species tree—x and y—are short so that coalescences occur deep in the tree, the two sequences of coalescences that produce a given symmetric gene tree topology together have higher probability than the single sequence that produces the topology that matches the species tree.
(a) and (b) Two coalescence sequences leading to gene tree topology ((AD)(BC)). In (a), the lineages from B and C coalesce more recently than those from A and D, and in (b), the reverse is true.
(c) The single sequence of coalescences leading to gene tree topology (((AB)C)D).

Supportive illustrations for the proof of Lemma 4.T^a corresponds to the topology in Figure (e) (replace z with a in tree T) in all the cases. (a)T^b corresponds to the topology in Figure (f) (replace z with b in tree T). (b)T^b corresponds to the topology in Figure (b) (replace z with b in tree T). (c)T^b corresponds to the topology in Figure (c) (replace z with b in tree T). (d)T^b corresponds to the topology in Figure (d) (replace z with b in tree T).