Question: Help with understanding PAML M0 output
0
gravatar for DNAngel
5 months ago by
DNAngel30
DNAngel30 wrote:

Hi all,

Could someone just clarify the following output for me?

Example output:

branch        t    N        S      dN/dS     dN      dS    N*dN   S*dS
38..1      0.431  2257.5   655.5  0.0546  0.0294  0.5379   66.3   352.6
38..39     0.068  2257.5   655.5  0.0546  0.0046  0.0843   10.4   55.2
39..40     0.026  2257.5   655.5  0.0546  0.0018  0.0323   4.0    21.1
tree length for dN:       0.1848
tree length for dS:       3.3855

I don't understand what t, N, and S are. I know dN/dS is the omega ratio, and dN and dS is the average nonsynonymous/synonymous substitutions from node..node. So what is N and what does NdN mean? same with S and SdS?

If I want to find the average dN and dS (individually) for particular branches, should I use these values from model M0 or M3? I essentially want to find out if some species are evolving at a faster rate compared to others.

Thank you all!

m0 paml • 280 views
ADD COMMENTlink modified 5 months ago by Dave Carlson120 • written 5 months ago by DNAngel30
0
gravatar for Dave Carlson
5 months ago by
Dave Carlson120
Stony Brook University, NY
Dave Carlson120 wrote:

N and S are the number of non-synonymous and synonymous sites in the alignment, respectively. I believe that t is the length of a particular branch (in substitutions per codon). If you're interested in looking for differences in dN and dS across branches (and not dN/dS, per se) then I believe you could use the M0 model. If you think you might be interested in the branch-specific dN/dS values, you could also run the M1 model (free-ratio) to get separate values of dN. dS, and omega for each branch. Ziheng Yang discourages use of this model because of the number of parameters, but in my experience people use it anyway.

ADD COMMENTlink written 5 months ago by Dave Carlson120

I am confused still, if there are 2000 N and only 650 S, how is the overall dN/dS less than 1?

ADD REPLYlink written 5 months ago by DNAngel30

N & S are the estimated number of sites in the alignment where a substitution would either be non-synonymous (N) or synonymous (S). This is calculated based on the genetic code and the codon usage model estimated from the data. It is not the same thing as the actual number of synonymous or non-synonymous changes. So in your alignment, there are about 2257 non-synonymous sites and 655 synonymous sites. This is roughly what you'd expect given the genetic code.

Despite there being more non-synonymous sites, the rate of non-synonymous changes is much lower than the rate of synonymous changes, leading to a dN/dS that is <<< 1. This suggests a significant degree of purifying selection acting on whatever gene you're looking at.

ADD REPLYlink written 5 months ago by Dave Carlson120

Also, why would model M1a be used to check for branch-specific dN and dS? Why not model M0 which doesn't constrain it since M1a does not allow positive selection anyways?

ADD REPLYlink written 5 months ago by DNAngel30

It's possible I got the model terminology wrong. If you set model = 1 in the control file, this will give you the free-ratio model, which allows each branch to have an independent dN/dS. Looking at the manual again, I see that this different from model M1a, which is a nearly neutral model that is used as a null model when looking for positive selection. Typically M1a is compared against M2a.

In your case, you can run M0 to get dN and dS rates across each branch, but the overall dN/dS rate will be constrained to be equal across the whole tree (this is rather non-intuitive, at least to me!). I was just suggesting you could also try the free-ratio model to get independent estimates of dN/dS at each branch as well.

ADD REPLYlink written 5 months ago by Dave Carlson120

Okay I can try the free ratio model as well, but I don't think it gives me individual dN and dS values with their individual branch lengths. I have to make a dN and dS tree where I can use their individual branch lengths and compare the differences visually in that manner. M0 gives individual dN and dS but the branch length (t) then is for the average dN/dS I believe.

ADD REPLYlink written 5 months ago by DNAngel30

The free ratio model actually will give you individual dN and dS values for each branch. The output file will have three different trees, one for dS, one for dN, and one of omega.

ADD REPLYlink written 5 months ago by Dave Carlson120
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