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GARD: sequences involved in recombination (Read 2833 times)
Maria
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GARD: sequences involved in recombination
Mar 1st, 2008 at 8:26am
 
Hi,
I am looking for evidence for hybridization among different species and each of them is represented by several sequences in my alignment. GARD detected break points in my alignment, but how do I know which sequence pairs are involved in recombination? I guess there is recombination within each species, and GARD may detect it, because I have several sequences for each species, but that is not what I am interested in. I would like to know whether sequences belonging to different species show evidence for recombination. Would it be reasonable to say that if a pair of sequences group together on any of the trees corresponding to different alignment fragments that means they have recombination? Do they need to be identical or just similar? Can it be that they group together just because this fragment is conserved and they share it because of common ancestry and not because of current recombination?

Thank you!
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Art Poon
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Re: GARD: sequences involved in recombination
Reply #1 - Mar 3rd, 2008 at 10:53pm
 
Hi Maria,
Wow, that's a lot of questions! The thing to keep in mind is that GARD is based on phylogenetic incogruence, I.e. that different parts of the alignment support different trees. In some cases this will be due to a recombination event such that part of a sequence is more closely related to a different sequence from the same species, or another species. This would be manifested by trees assigned to partitions of the alignment with incongruent topologies. The important thing to note here is that not all breakppints in GARD ---divisions of phylogenetically incongruent regions --- is not necessarily due to recombination. Trees can be incongruent due to branch lengths alone (rate variation over time). Thus, it is essential to inspect the trees in the detailed GARD output.
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Maria
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Some more questions :-)
Reply #2 - Mar 17th, 2008 at 4:44pm
 
Dear Artpoon,
thanks a lot for your reply. Yes, I know that GARD is based on phylogenetic incongruence. And I think that rate variation is quite probable in my data set. But how can I be sure that what GARD detected is due to recombination or due to rate variation? GARD has detected two break points. I have examined the trees and the first two trees seem to be due to rate variation. The first tree makes no sense: all sequences grouped together except few from different species placed on long branches. The second tree groups all taxa together and places the most divergent species on a long branch. The last tree looks plausible: the two taxa are grouped together (although the sequences are not identical) that could possibly hybridize, but I do not know it for sure, that’s what I wanted to check with GARD... So, your advice to examine the output trees closely is surely useful as it helps to sort out the trees which make no sense. But if there is a tree that does make sense, then how do I decide whether it is due to rate variation or to recombination?

Another question: I have tested my data set with GENECONV before, and it detected no significant recombination fragments. Can it be that GARD is more sensitive than GENECONV and can find recombination that actually happened but was not found by other methods? Or should I interpret these results so that there is in fact no recombination and phylogenetic incongruence detected by GARD is due to rate variation?
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Sergei
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Re: Some more questions :-)
Reply #3 - Mar 17th, 2008 at 10:48pm
 
Dear Maria,

I would run the KH test (which is implemented in HyPhy under Recomination/GARDProcessor.bf).
You need to feed the analysis the original alignment and the partitions file (available from the link
labeled [Raw] on the GARD results page). This tool will compute KH tests for all adjacent partitions; if KH-derived
p-values are small, then the topologies are quite different indeed. This should take care of
the rate variation case.

GENECONV may not detect recombinants if the parental strains are not in the sample; that may explain
the difference in results as well.

Cheers,
Sergei
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