Answers.Body: <p><strong>Reversible</strong> At its simplist the GTR rate, is the General Time Reversible model and most importantly infers the matrix is &quot;reversible&quot;, thus as many A mutations will go to C as C mutations go to A and the number of mutations is relative to the G-T mutation rate. The G-T mutation rate is one of the least common mutations (transversion) so all other values are above 1. If you plotted this as a matrix the top right-hand matrix is identical to the bottom left hand matrix because its reversible, i.e. A &lt;-&gt; C . This is cool because you don't need to tell the algorithm which sequence is the root.</p>
  <p><strong>Directional</strong> The &quot;instantaneous&quot; matrix is a directional matrix, thus A -&gt; C is not the same as C -&gt; A. In order to obtain an accurate directional matrix you need to assign a root, its like lets start from the deepest node in the tree (the one between the root/outgroup and all ingroup sequences) and now - if that is correct - we can remove the restrictions of reversibility and impose directionality on the mutation process. This is needed if you are performing molecular clock type calculations. However, the key is assigning the root/outgroup - if it assigns it e.g. by mid-point rooting that is not necessarily a good idea.</p>
  <p><strong>Advice.</strong> My advice is if you are building a basic tree and haven't assigned a root/outgroup, obviously using GTR which is a good move, then you don't to worry about the intananeous matrix because it might not be meaningful. If you are running a molecular clock/dating/rate estimate type model and a very clear what the root/outgroup is and that is specified in the input, then the instananeous matrix becomes important.</p>
  <hr />
  <p>The mutation rates of GTR and instananeous are not directly comparable because the GTR rate is with respect to G &lt;-&gt; T mutation, whilst instananeous is primarily concerned with the directionality of mutation and is a 12-parameter model, as opposed to GTR which 6-perameters, where one of those parameters is the reference.</p>
  <p>Thus G-T is reversible in GTR (the reference) and so e.g. A&lt;-&gt;C is reversible with respect to G&lt;-&gt;T.
  In instananeous matrix G -&gt; T and T -&gt; G are two different rates as are A-&gt; C and C -&gt; A, so one approximation based on an assumption with respect to a second approximation based on the same assumption can't directly equate to a mutation rate which removes that assumption.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-06-20T15:27:37.203Z
Answers.Text: I'm running a rate estimate model. I have supplied the guide tree for it to follow. 
  According to the Instantaneous matrix, A->C is 4.38 (and C->A is 0.153) but how is this related to the A<->C of 1.95 which is a GTR rate parameter?
Answers.UserDisplayName: Anthony Guterres
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-06-21T02:29:21.257Z
Answers.Text: They are not directly comparable answered above because A<->C (GTR) is with respect to G<->T mutation rate (which is assigned to a nominal value of 1).
Answers.UserDisplayName: M__
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-06-21T08:32:30.863Z
Answers.Text: Okay and how do you obtain A<->C (GTR) from A->C and C->A rates? Is it like a weighted average (weighted wrt base frequency)?
Answers.UserDisplayName: Anthony Guterres
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-06-21T09:10:40.290Z
Answers.Text: I think it is a separate question, I would do it via molecular clock trees.
Answers.UserDisplayName: M__
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-06-21T10:43:09.163Z
Answers.Text: Last two clarifications
  1. In the matrix, N->C is high for all N. Does this mean that the a mutation most commonly results in 'C' vs any other base?
  2. Also, a value at (x,y) in the instantaneous matrix represent the probability of base x mutating to base y right?
Answers.UserDisplayName: Anthony Guterres
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-06-21T18:47:19.853Z
Answers.Text: First question, yes, for the root you have set for the tree (it can change if you used a different root). Second question, yes and likely represent (row, column) - which is important in a directional matrix.
Answers.UserDisplayName: M__
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-06-20T13:04:39.353Z
Answers.LastActivityDate: 2020-06-21T02:27:04.953Z
Answers.LastEditDate: 2020-06-21T02:27:04.953Z
Answers.UserDisplayName: M__
Body: <p>PhyML is a tool used for maximum likelihood estimation analysis.
  I was running a PhyML analysis to check the rate of nucleotide change in DNA sequences and in the PhyML output, there are rate parameters and an instantaneous rate matrix. Could you please clarify the difference between the two?</p>
  <p><a href="https://i.sstatic.net/BIP3T.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BIP3T.png" alt="enter image description here" /></a></p>
  <p>What I've understood so far is that the rate parameters indicate, for ex: that the A -C transition rate is 1.95 and so on. If so what does the instantaneous rate matrix tell us?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-06-20T12:54:12.777Z
Comments.Text: Hi @AntonyGuterres did you assign a root/outgroup to this tree before building it?
Comments.UserDisplayName: M__
Tags: <phylogenetics><phylogeny><hidden-markov-models><nucleotide-models>
Title: What does instantaneous rate matrix mean?
ThreadText: # What does instantaneous rate matrix mean?
  
  (Asked by: username_0 on 2020-06-20)
  
  PhyML is a tool used for maximum likelihood estimation analysis.
  I was running a PhyML analysis to check the rate of nucleotide change in DNA sequences and in the PhyML output, there are rate parameters and an instantaneous rate matrix. Could you please clarify the difference between the two?
  
  [![enter image description here](https://i.sstatic.net/BIP3T.png)](https://i.sstatic.net/BIP3T.png)
  
  What I've understood so far is that the rate parameters indicate, for ex: that the A -C transition rate is 1.95 and so on. If so what does the instantaneous rate matrix tell us?
  
  **Tags:** phylogenetics, phylogeny, hidden-markov-models, nucleotide-models
  
  > **Comments:**
  > **username_1**: Hi @AntonyGuterres did you assign a root/outgroup to this tree before building it?  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2020-06-20. Score: 1 | Accepted answer
  
  **Reversible** At its simplist the GTR rate, is the General Time Reversible model and most importantly infers the matrix is "reversible", thus as many A mutations will go to C as C mutations go to A and the number of mutations is relative to the G-T mutation rate. The G-T mutation rate is one of the least common mutations (transversion) so all other values are above 1. If you plotted this as a matrix the top right-hand matrix is identical to the bottom left hand matrix because its reversible, i.e. A <-> C . This is cool because you don't need to tell the algorithm which sequence is the root.
  
  **Directional** The "instantaneous" matrix is a directional matrix, thus A -> C is not the same as C -> A. In order to obtain an accurate directional matrix you need to assign a root, its like lets start from the deepest node in the tree (the one between the root/outgroup and all ingroup sequences) and now - if that is correct - we can remove the restrictions of reversibility and impose directionality on the mutation process. This is needed if you are performing molecular clock type calculations. However, the key is assigning the root/outgroup - if it assigns it e.g. by mid-point rooting that is not necessarily a good idea.
  
  **Advice.** My advice is if you are building a basic tree and haven't assigned a root/outgroup, obviously using GTR which is a good move, then you don't to worry about the intananeous matrix because it might not be meaningful. If you are running a molecular clock/dating/rate estimate type model and a very clear what the root/outgroup is and that is specified in the input, then the instananeous matrix becomes important.
  
  ---
  
  The mutation rates of GTR and instananeous are not directly comparable because the GTR rate is with respect to G <-> T mutation, whilst instananeous is primarily concerned with the directionality of mutation and is a 12-parameter model, as opposed to GTR which 6-perameters, where one of those parameters is the reference.
  
  Thus G-T is reversible in GTR (the reference) and so e.g. A<->C is reversible with respect to G<->T.
  In instananeous matrix G -> T and T -> G are two different rates as are A-> C and C -> A, so one approximation based on an assumption with respect to a second approximation based on the same assumption can't directly equate to a mutation rate which removes that assumption.
  
  > **Comments:**
  > **username_0**: I'm running a rate estimate model. I have supplied the guide tree for it to follow.  According to the Instantaneous matrix, A->C is 4.38 (and C->A is 0.153) but how is this related to the A<->C of 1.95 which is a GTR rate parameter?  
  > **username_1**: They are not directly comparable answered above because A<->C (GTR) is with respect to G<->T mutation rate (which is assigned to a nominal value of 1).  
  > **username_0**: Okay and how do you obtain A<->C (GTR) from A->C and C->A rates? Is it like a weighted average (weighted wrt base frequency)?  
  > **username_1**: I think it is a separate question, I would do it via molecular clock trees.  
  > **username_0**: Last two clarifications 1. In the matrix, N->C is high for all N. Does this mean that the a mutation most commonly results in 'C' vs any other base? 2. Also, a value at (x,y) in the instantaneous matrix represent the probability of base x mutating to base y right?  
  > **username_1**: First question, yes, for the root you have set for the tree (it can change if you used a different root). Second question, yes and likely represent (row, column) - which is important in a directional matrix.  
  
  ---
  
  