MIX - Mixed method parsimony

CONTENTS:

• DESCRIPTION
• INPUT FORMAT
• OUTPUT FORMAT
• TEST DATA SET
• TEST SET OUTPUT (with all numerical options on)

The Camin-Sokal parsimony method explains the data by assuming that changes 0 --> 1 are allowed but not changes 1 --> 0. Wagner parsimony allows both kinds of changes. (This under the assumption that 0 is the ancestral state, though the program allows reassignment of the ancestral state, in which case we must reverse the state numbers 0 and 1 throughout this discussion). The criterion is to find the tree which requires the minimum number of changes. The Camin-Sokal method is due to Camin and Sokal (1965) and the Wagner method to Eck and Dayhoff (1966) and to Kluge and Farris (1969).

Here are the assumptions of these two methods:

1. Ancestral states are known (Camin-Sokal) or unknown (Wagner).
2. Different characters evolve independently.
3. Different lineages evolve independently.
4. Changes 0 --> 1 are much more probable than changes 1 --> 0 (Camin- Sokal) or equally probable (Wagner).
5. Both of these kinds of changes are a priori improbable over the evolutionary time spans involved in the differentiation of the group in question.
6. Other kinds of evolutionary event such as retention of polymorphism are far less probable than 0 --> 1 changes.
7. Rates of evolution in different lineages are sufficiently low that two changes in a long segment of the tree are far less probable than one change in a short segment.

INPUT FORMAT

The input for MIX is the standard input for discrete characters programs, described above in the discrete characters programs documentation file. States "?", "P", and "B" are allowed.

The options are selected using a menu:

Mixed parsimony algorithm, version 3.5c

Settings for this run:
  U                 Search for best tree?  Yes
  X                     Use Mixed method?  No
  P                     Parsimony method?  Wagner
  J     Randomize input order of species?  No. Use input order
  O                        Outgroup root?  No, use as outgroup species  1
  T              Use Threshold parsimony?  No, use ordinary parsimony
  A   Use ancestral states in input file?  No
  M           Analyze multiple data sets?  No
  0   Terminal type (IBM PC, VT52, ANSI)?  ANSI
  1    Print out the data at start of run  No
  2  Print indications of progress of run  Yes
  3                        Print out tree  Yes
  4     Print out steps in each character  No
  5     Print states at all nodes of tree  No
  6       Write out trees onto tree file?  Yes

Are these settings correct? (type Y or the letter for one to change)

The option P toggles between the Camin-Sokal parsimony criterion and the default Wagner parsimony criterion. Option X invokes mixed-method parsimony. If the A option is invoked, the ancestor is not to be counted as one of the species.

In the input file the W (Weights) option is available, as usual. The A (Ancestral states) and X (Mixed methods) also require the options to be declared on the first line of the input file and other information to be present in the input file. The Ancestral States option also requires that you choose the A option in the input menu. Note that the X option requires that the option M (Mixture) be declared and the mixture information provided in the input file. This is admittedly confusing. The F (Factors) option is not available in this program, as it would have no effect on the result even if that information were provided in the input file.

OUTPUT FORMAT

Output is standard: a list of equally parsimonious trees, which will be printed as rooted or unrooted depending on which is appropriate, and, if the user chooses, a table of the number of changes of state required in each character. If the Wagner option is in force for a character, it may not be possible to unambiguously locate the places on the tree where the changes occur, as there may be multiple possibilities. If the user selects menu option 5, a table is printed out after each tree, showing for each branch whether there are known to be changes in the branch, and what the states are inferred to have been at the top end of the branch. If the inferred state is a "?" there will be multiple equally-parsimonious assignments of states; the user must work these out for themselves by hand.

If the Camin-Sokal parsimony method is invoked and the Ancestors option is also used, then the program will infer, for any character whose ancestral state is unknown ("?") whether the ancestral state 0 or 1 will give the fewest state changes. If these are tied, then it may not be possible for the program to infer the state in the internal nodes, and these will all be printed as ".". If this has happened and you want to know more about the states at the internal nodes, you will find helpful to use MOVE to display the tree and examine its interior states, as the algorithm in MOVE shows all that can be known in this case about the interior states, including where there is and is not amibiguity. The algorithm in MIX gives up more easily on displaying these states.

If the A option is not used, then the program will assume 0 as the ancestral state for those characters following the Camin-Sokal method, and will assume that the ancestral state is unknown for those characters following Wagner parsimony. If any characters have unknown ancestral states, and if the resulting tree is rooted (even by outgroup), a table will also be printed out showing the best guesses of which are the ancestral states in each character. You will find it useful to understand the difference between the Camin-Sokal parsimony criterion with unknown ancestral state and the Wagner parsimony criterion.

If the U (User Tree) option is used and more than one tree is supplied, the program also performs a statistical test of each of these trees against the best tree. This test, which is a version of the test proposed by Alan Templeton (1983) and evaluated in a test case by me ( 1985a). It is closely parallel to a test using log likelihood differences invented by Kishino and Hasegawa (1989), and uses the mean and variance of step differences between trees, taken across characters. If the mean is more than 1.96 standard deviations different then the trees are declared significantly different. The program prints out a table of the steps for each tree, the differences of each from the highest one, the variance of that quantity as determined by the step differences at individual sites, and a conclusion as to whether that tree is or is not significantly worse than the best one. It is important to understand that the test assumes that all the binary characters are evolving independently, which is unlikely to be true for many suites of morphological characters.

At the beginning of the program are a series of constants, which can be changed to help adapt the program to different computer systems. The constant "nmlngth" is the length of the species names. "maxtrees" is the maximum number of trees which the program will store for output. "maxuser" is the maximum number of user trees which can be input to the program. You may want to change these constants to reduce memory requirements.

The program is descended from earlier programs SOKAL and WAGNER which have since been removed from the PHYLIP package, since MIX has all their capabilites and more.

TEST DATA SET

     5    6
Alpha     110110
Beta      110000
Gamma     100110
Delta     001001
Epsilon   001110

Mixed parsimony algorithm, version 3.5c

Wagner parsimony method

Name         Characters
----         ----------

Alpha        11011 0
Beta         11000 0
Gamma        10011 0
Delta        00100 1
Epsilon      00111 0

     4 trees in all found

           +--Epsilon
     +-----4
     !     +--Gamma
  +--2
  !  !     +--Delta
--1  +-----3
  !        +--Beta
  !
  +-----------Alpha

  remember: this is an unrooted tree!

requires a total of      9.000

 steps in each character:
         0   1   2   3   4   5   6   7   8   9
     *-----------------------------------------
    0!       2   2   2   1   1   1

From    To     Any Steps?    State at upper node
                             ( . means same as in the node below it on tree)

          1                1?011 0
  1       2         no     .?... .
  2       4        maybe   .0... .
  4    Epsilon      yes    0.1.. .
  4    Gamma        no     ..... .
  2       3         yes    .?.00 .
  3    Delta        yes    001.. 1
  3    Beta        maybe   .1... .
  1    Alpha       maybe   .1... .

     +--------Gamma
     !
  +--2     +--Epsilon
  !  !  +--4
  !  +--3  +--Delta
--1     !
  !     +-----Beta
  !
  +-----------Alpha

  remember: this is an unrooted tree!

requires a total of      9.000

 steps in each character:
         0   1   2   3   4   5   6   7   8   9
     *-----------------------------------------
    0!       1   2   1   2   2   1

From    To     Any Steps?    State at upper node
                             ( . means same as in the node below it on tree)

          1                1?011 0
  1       2         no     .?... .
  2    Gamma       maybe   .0... .
  2       3        maybe   .?.?? .
  3       4         yes    001?? .
  4    Epsilon     maybe   ...11 .
  4    Delta        yes    ...00 1
  3    Beta        maybe   .1.00 .
  1    Alpha       maybe   .1... .

     +--------Epsilon
  +--4
  !  !  +-----Gamma
  !  +--2
--1     !  +--Delta
  !     +--3
  !        +--Beta
  !
  +-----------Alpha

  remember: this is an unrooted tree!

requires a total of      9.000

 steps in each character:

         0   1   2   3   4   5   6   7   8   9
     *-----------------------------------------
    0!       2   2   2   1   1   1

From    To     Any Steps?    State at upper node
                             ( . means same as in the node below it on tree)

          1                1?011 0
  1       4        maybe   .0... .
  4    Epsilon      yes    0.1.. .
  4       2         no     ..... .
  2    Gamma        no     ..... .
  2       3         yes    ...00 .
  3    Delta        yes    0.1.. 1
  3    Beta         yes    .1... .
  1    Alpha       maybe   .1... .

     +--------Gamma
  +--2
  !  !  +-----Epsilon
  !  +--4
--1     !  +--Delta
  !     +--3
  !        +--Beta
  !
  +-----------Alpha

  remember: this is an unrooted tree!

requires a total of      9.000

 steps in each character:
         0   1   2   3   4   5   6   7   8   9
     *-----------------------------------------
    0!       2   2   2   1   1   1

From    To     Any Steps?    State at upper node
                             ( . means same as in the node below it on tree)

          1                1?011 0
  1       2        maybe   .0... .
  2    Gamma        no     ..... .
  2       4        maybe   ?.?.. .
  4    Epsilon     maybe   0.1.. .
  4       3         yes    ?.?00 .
  3    Delta        yes    0.1.. 1
  3    Beta         yes    110.. .
  1    Alpha       maybe   .1... .