Hi Jamie,

Sorry it took me a while to get around to responding. First, a disclaimer and an apology -- HyPhy is always a work in progress and it is quite common to find standard analyses without documentation. BranchClassDNDS does the following fits a model that permits different sets of tree branches to have different selective regimes. The original impetus for the analysis was to permit the analysis of lineage-specific selection in the context of HIV transmission where you have two sets of clonal sequences (one from the source and one from the recipient) separated by the transmission branch. The tree would then be split into the patient clades and the transmission branch and fit a separate dN/dS to each branch class, i.e. all branches within each patient would have the same dN/dS and the transmission branch would have a separate dN/dS as in the attached picture (part of the archive). 

The inputs to the analysis are the alignment file and a HyPhy batch file (see attached) specifying the partitions. The analysis also includes several options for site-to-site rate variation.


HTH,
Sergei

Hi Jamie,

Good point about SlatkinMaddison.bf
If you are in Cambridge, then Simon is an excellent local resource on all things HyPhy.


Because a part of the SM test is a parsimony reconstruction of 'migration' events (the meaning of migration is very much context specific), you can automatically identify the lineages where such events took place, and the remainder of the lineages are placed in the corresponding 'population'.  However, SlatkinMaddison.bf requires that your sequences be named in a regular way, i.e. all of the sequences from all but one population must be identifiable by a unique regular expression. One of my postdocs has put together a script that converts color annotations from TreeFig into HyPhy partition files to make the process a bit more interactive. If you are interested in the latter, I'll see if I can track it down.

Sergei

Hi Dan,

I'd better finally start writing decent documentation
Take a look at Multimedia File Viewing and Clickable Links are available for Registered Members only!!  You need to Login

Let me know if this is helpful.

BranchSiteREL manuscript is currently in revision at MBE; I will post complete documentation shortly. In the meantime, look at Multimedia File Viewing and Clickable Links are available for Registered Members only!!  You need to Login

Sergei

Hi Dan,

Thanks for the reference -- a very nice paper! In terms of handling your analysis:

• Does the same tree apply to all genes?
• There are five compartments, but many (52) different ways to bin 5 rates (i.e. all the same, all different, paralog 1 separate all others the same etc). Do you want to test all such models?

There is no predefined analysis to perform this type of test, but SelectionLRT.bf can be modified,e.g. by prompting for two branches basal to Paralog 1 and Paralog 2 species, without too much hassle.

Sergei

Sergei

Hi Dan,

First, I must say that I always get confused by the paralog/ortholog terminology (not being a properly educated biologist:)

My general advice for answering your questions would be the following.

2). Is the easiest. Alternative model is "local dN/dS", Null Model dN/dS (MRCA Paralogs 1) = dN/dS (MRCA Paralogs 2). The test is a one D.F LRT.

For (1) and (3) the formulation is a bit more tricky. IF we are willing to assume that dN/dS is homogeneous within a given paralog clade (i.e. all branches in the paralog 1 ortholog subtree have the same dN/dS) then we could do the following.

1). Alternative model: deep branches have their own dN/dS, Paralog 1 clade, Paralog 2 clade and Unduplicated clade each have clade wide dN/dS, but those differ between clades (3 parameters, say omega1, omega2 and omegaUD). Several nulls could be tested (by a 1 or 2 D.F. LRT)

  (a) omega1 = omega2 = omegaUD
  (b) omega1 = omegaUD
  (c) omega1 = omega2
  (d) omega2 = omegaUD

3). Is analogous

The more challenging test would be to see if the distribution of dN/dS within clades are different. Roughly speaking, what happens if in paralog1 clade there are four branches all with dN/dS of 0.1, but in paralog 2 clade, 3 branches have dN/dS of 0 and 1 branch has dN/dS of 0.4. On average (which is what the above test would examine), there is probably not much difference between clades, but biologically there appears to be something quite different. There is not an easy way (that I know of) to examine this general distributional difference. If paralog clades have the same number of branches you could simply test the hypothesis that at least one pair of paralogs have different dN/dS. This can be done by using the local model as the alternative, and constraining a pair of branches (e.g. Paralog1_speciesX, Paralog2_speciesX) to have the same dN/dS.

Which is more appropriate in your case?

Sergei

Hi Dan,

The GUI approach will work for a few analyses (since you have to redefine the constrains for each data set); for batch processing I can write a simple script which will take an annotated list to determine which branches fall into which dN/dS class:

A,1
B,1
AB,2
C,3
D,3

for the tree ((A,B)AB,C,D)

Alternatively, something like ((A:1,B:1):2,C:3,D:3) could be used. Let me know if this will be useful; I can script it up fairly quickly and would only ask for a grant # acknowledgement if you ever publish based on it (the joys of NIH reporting).

To constrain ALL branches in the tree (or a clade) to have the same dN/dS in the GUI, use the ReplicateConstraint command as described around page 34 of Multimedia File Viewing and Clickable Links are available for Registered Members only!!  You need to Login

Serge