Hello,

I am looking for a way to calculate the ancestral sequence for many independent sets of alignements in phylip format, by a reconstruction with NJ for each set and using a GY_JC codon model, in a simultaneous way.
What can I do for this?
A lot of thanks!!

Miguel.

Hi,

Miguel means batch ancestral reconstruction on multiple alignments (e.g., independent replicates from a simulation study). GY-JC should be GY94xJC.

Cheers,

D.

That is awesome Sergei, as usual! many thanks!

D.

Dear David,

Yes; HyPhy will only do a joint ML reconstruction by default. You have to twist its arm quite a bit to obtain node-by-node reconstruction.

Cheers,
Sergei

Dear Miguel,

Yes, MRCA is the node with the lowest number (in almost all cases).
The rule for naming the nodes is somewhat obscure - it's the order of the node in the pre-order traversal of the tree. If you want access to a specific node, or a different naming convention for internal nodes, that is reasonably easy to do from within HyPhy. For the first case, you just give the interior node a name in the input tree, e.g. ((a,b)myNode, c...); for the second, you need to do some HBL processing of trees, but I can send you the code if you have something specific in mind.

Cheers,
Sergei

Dear Miguel,

A more robust way to determine the MRCA (root) is to pick the first sequence from the reconstructed ancestral alignment. HyPhy always builds these sequences from the leaves up, and the root is invariable processed last, but the sequences are reported in the reversed order (root at the top, deep nodes further down).

HTH,
Sergei

Dear Sergei,

All my trees have an outgroup and they are rooted. I am interesting in the sequence of the penultimate most ancester node (the first descendant of the root or the MRCA of the ingroup). I attach a figure, the node that I am talking about, in this example, is the number 6.

What can I do to recognize that node in the output file?. Perhaps is the second sequence from the reconstructed ancestral alignment.

Cheers,

Miguel   

Dear Segei,

That is Right.
Let me ask you for a last problem that I have with the Ancestral Sequences Reconstruction (ASR).
I have rooted trees for the ASR, like the tree that I send you in the previous message. Then, for do the ASR I must write:
Tree givenTree = treeRoot;
(where treeRoot is my rooted tree, like the tree of the previous message, and givenTree is the tree for input in the ASR).

Well, I see that givenTree is different than treeRoot, it is for the MRCA node (givenTree has the MRCA node with 3 descedants to outgroup and ingroup, it seems to be an unrooted tree). I also see that it is necessary to have a tree in the format of the "givenTree" for doing the ASR (I tried to do the ASR directly with treeRoot (I just wrote: givenTree = treeRoot) and then ASR wrote me: LF: Not a valid tree ID: givenTree).

So, my questions are:
- Is it possible to do the ASR with my rooted tree (treeRoot)?
- I am interesting in the MRCA of the ingroup (node 6 in the tree of the previous message), so, is it true to say that the sequence reconstructed of the root of "givenTree" is the same than reconstructed sequence of the ingroup MRCA "treeRoot" (node 6 is the example)?

Cheers,
Miguel