Hi,
HyPhy is a great tool for quickly calculating (pair-wise) relative rates in protein alignments (Standard Analysis -> PairwiseRelativeRate.bf). This method however requires to define an outgroup. Is there any way to calculate these rates without outgroups, but by using mid-point rooting instead?
Thanks

Hi lichgen,

I am not sure you can easily extend the test to use a mid-point root. Interpretation issues aside (under time reversible models, mid-point rooting is effectively uninformative), the standard relative rate test requires three sequences, one of which is designated as the outgroup. How does one define this sequence based on mid-point rooting?

It is possible to inquire whether path-lengths in the tree from the mid-point to any two taxa are of different lengths; this test is not the same as the relative rate test.

What is the biological question you are trying to answer?

Sergei

Hi Sergei,
I am trying to identify phylogenetically old genes in a symbiotic fungal species, by looking at reduced evolutionary rates, compared to homologues in a defined group of related fungal taxa. One of these reference taxa is more distant and used as outgroup. I considered a gene of my fungal species with a reduced rate, if it constantly showed the lowest rate in all triplet comparisons within the HyPhy rate test (I processed several thousand alignments, with exactly one sequence per species).  That worked well and let me identify a small number of genes.

The biological idea behind this approach is that symbiosis was acquired early in evolution, and some (not all) key genes haven't much evolved since then, in order to maintain the symbiotic lifestyle.

In a second approach I am now working with MCL clustered gene families, instead of well conserved homologue groups. These gene families are naturally more diverse, with usually more than one, non monophyletic outgroup sequence, and can therefore not be rooted. My idea was now to identify different path lengths (as you described it above) for each two species in an unrooted tree instead.

Since I have several thousand alignments it would be very helpful if a test existed, which could perform this mid-point path length calculation, without the need of building a tree first.

Olaf