version 3.5c
DNAINVAR -- Program to compute Lake's and Cavender's phylogenetic
invariants from nucleotide sequences
(c) Copyright 1986-1993 by Joseph Felsenstein and by the University of
Washington. Written by Joseph Felsenstein. Permission is granted to copy this
document provided that no fee is charged for it and that this copyright notice
is not removed.
This program reads in nucleotide sequences for four species and computes
the phylogenetic invariants discovered by James Cavender (Cavender and
Felsenstein, 1987) and James Lake (1987). Lake's method is also called by him
"evolutionary parsimony". I prefer Cavender's more mathematically precise term
"invariants", as the method bears somewhat more relationship to likelihood
methods than to parsimony. The invariants are mathematical formulas (in the
present case linear or quadratic) in the EXPECTED frequencies of site patterns
which are zero for all trees of a given tree topology, irrespective of branch
lengths. We can consider at a given site that if there are no ambiguities, we
could have for four species the nucleotide patterns (considering the same site
across all four species) AAAA, AAAC, AAAG, ... through TTTT, 256 patterns in
all.
The invariants are formulas in the expected pattern frequencies, not the
observed pattern frequencies. When they are computed using the observed
pattern frequencies, we will usually find that they are not precisely zero even
when the model is correct and we have the correct tree topology. Only as the
number of nucleotides scored becomes infinite will the observed pattern
frequencies approach their expectations; otherwise, we must do a statistical
test of the invariants.
Some explanation of invariants will be found in the above papers, and also
in my recent review article on statistical aspects of inferring phylogenies
(Felsenstein, 1988b). Although invariants have some important advantages,
their validity also depends on symmetry assumptions that may not be satisfied.
In the discussion below suppose that the possible unrooted phylogenies are I:
((A,B),(C,D)), II: ((A,C),(B,D)), and III: ((A,D),(B,C)).
Lake's Invariants, Their Testing and Assumptions
Lake's invariants are fairly simple to describe: the patterns involved are
only those in which there are two purines and two pyrimidines at a site. Thus
a site with AACT would affect the invariants, but a site with AAGG would not.
Let us use (as Lake does) the symbols 1, 2, 3, and 4, with the proviso that 1
and 2 are either both of the purines or both of the pyrimidines; 3 and 4 are
the other two nucleotides. Thus 1 and 2 always differ by a transition; so do 3
and 4. Lake's invariants, expressed in terms of expected frequencies, are the
three quantities:
(1) P(1133) + P(1234) - P(1134) - P(1233),
(2) P(1313) + P(1324) - P(1314) - P(1323),
(3) P(1331) + P(1342) - P(1341) - P(1332),
He showed that invariants (2) and (3) are zero under Topology I, (1) and (3)
are zero under topology II, and (1) and (2) are zero under Topology III. If,
for example, we see a site with pattern ACGC, we can start by setting 1=A.
Then 2 must be G. We can then set 3=C (so that 4 is T). Thus its pattern
type, making those substitutions, is 1323. P(1323) is the expected probability
of the type of pattern which includes ACGC, TGAG, GTAT, etc.
Lake's invariants are easily tested with observed frequencies. For
example, the first of them is a test of whether there are as many sites of
types 1133 and 1234 as there are of types 1134 and 1233; this is easily tested
with a chi-square test or, as in this program, with an exact binomial test.
Note that with several invariants to test, we risk overestimating the
significance of results if we simply accept the nominal 95% levels of
significance (Li and Guoy, 1990).
Lake's invariants assume that each site is evolving independently, and
that starting from any base a transversion is equally likely to end up at each
of the two possible bases (thus, an A undergoing a transversion is equally
likely to end up as a C or a T, and similarly for the other four bases from
which one could start. Interestingly, Lake's results do not assume that rates
of evolution are the same at all sites. The result that the total of 1133 and
1234 is expected to be the same as the total of 1134 and 1233 is unaffected by
the fact that we may have aggregated the counts over classes of sites evolving
at different rates.
Cavender's Invariants, Their Testing and Assumptions
Cavender's invariants (Cavender and Felsenstein, 1987) are for the case of
a character with two states. In the nucleic acid case we can classify
nucleotides into two states, R and Y (Purine and Pyrimidine) and then use the
two-state results. Cavender starts, as before, with the pattern frequencies.
Coding purines as R and pyrimidines as Y, the patterns types are RRRR, RRRY,
and so on until YYYY, a total of 16 types. Cavender found quadratic functions
of the expected frequencies of these 16 types that were expected to be zero
under a given phylogeny, irrespective of branch lengths. Two invariants
(called K and L) were found for each tree topology. The L invariants are
particularly easy to understand. If we have the tree topology ((A,B),(C,D)),
then in the case of two symmetric states, the event that A and B have the same
state should be independent of whether C and D have the same state, as the
events determining these happen in different parts of the tree. We can set up
a contingency table:
C = D C =/= D
------------------------------
|
A = B | YYYY, YYRR, YYYR, YYRY,
| RRRR, RRYY RRYR, RRRY
|
A =/= B | YRYY, YRRR, YRYR, YRRY,
| RYYY, RYRR RYYR, RYRY
and we expect that the events C = D and A = B will be independent. Cavender's
L invariant for this tree topology is simply the negative of the crossproduct
difference,
P(A=/=B and C=D) P(A=B and C=/=D) - P(A=B and C=D) P(A=/=B and C=/=D).
One of these L invariants is defined for each of the three tree
topologies. They can obviously be tested simply by doing a chi-square test on
the contingency table. The one corresponding to the correct topology should be
statistically indistinguishable from zero. Again, there is a possible multiple
tests problem if all three are tested at a nominal value of 95%.
The K invariants are differences between the L invariants. When one of
the tables is expected to have crossproduct difference zero, the other two are
expected to be nonzero, and also to be equal. So the difference of their
crossproduct differences can be taken; this is the K invariant. It is not so
easily tested.
The assumptions of Cavender's invariants are different from those of
Lake's. One obviously need not assume anything about the frequencies of, or
transitions among, the two different purines or the two different pyrimidines.
However one does need to assume independent events at each site, and one needs
to assume that the Y and R states are symmetric, that the probability per unit
time that a Y changes into an R is the same as the probability that an R
changes into a Y, so that we expect equal frequencies of the two states. There
is also an assumption that all sites are changing between these two states at
the same expected rate. This assumption is not needed for Lake's invariants,
since expectations of sums are equal to sums of expectations, but for
Cavender's it is, since products of expectations are not equal to expectations
of products.
It is helpful to have both sorts of invariants available; with further
work we may appreciate what other invaraints there are for various models of
nucleic acid change.
INPUT FORMAT
The input data for DNAINVAR is standard. The first line of the input file
contains the number of species (which must always be 4 for this version of
DNAINVAR) and the number of sites. If the Weights option is being used, there
must also be a W in this first line to signal its presence. There is only one
option requiring information to be present in the input file, W (Weights). All
options other than W are invoked using the menu.
Next come the species data. Each sequence starts on a new line, has a
ten-character species name that must be blank-filled to be of that length,
followed immediately by the species data in the one-letter code. The sequences
must either be in the "interleaved" or "sequential" formats described in the
Molecular Sequence Programs document. The I option selects between them. The
sequences can have internal blanks in the sequence but there must be no extra
blanks at the end of the terminated line. Note that a blank is not a valid
symbol for a deletion.
The options are selected using an interactive menu. The menu looks like
this:
Nucleic acid sequence Invariants method, version 3.5c
Settings for this run:
M Analyze multiple data sets? No
I Input sequences interleaved? Yes
0 Terminal type (IBM PC, VT52, ANSI)? ANSI
1 Print out the data at start of run No
2 Print out the counts of patterns Yes
3 Print out the invariants Yes
Are these settings correct? (type Y or the letter for one to change)
The user either types "Y" (followed, of course, by a carriage-return) if the
settings shown are to be accepted, or the letter or digit corresponding to an
option that is to be changed.
The options M and 0 are the usual ones. It is described in the main
documentation file of this package. Option I is the same as in other molecular
sequence programs and is described in the documentation file for the sequence
programs.
OUTPUT FORMAT
The output consists first (if option 1 is selected) of a reprinting of the
input data, then (if option 2 is on) tables of observed pattern frequencies and
pattern type frequencies. A table will be printed out, in alphabetic order
AAAA through TTTT of all the patterns that appear among the sites and the
number of times each appears. This table will be invaluable for computation of
any other invariants. There follows another table, of pattern types, using the
1234 notation, in numerical order 1111 through 1234, of the number of times
each type of pattern appears. In this computation all sites at which there are
any ambiguities or deletions are omitted. Cavender's invariants could actually
be computed from sites that have only Y or R ambiguities; this will be done in
the next release of this program.
If option 3 is on the invariants are then printed out, together with their
statistical tests. For Lake's invariants the two sums which are expected to be
equal are printed out, and then the result of an one-tailed exact binomial test
which tests whether the difference is expected to be this positive or more.
The P level is given (but remember the multiple-tests problem!).
For Cavender's L invariants the contingency tables are given. Each is
tested with a one-tailed chi-square test. It is possible that the expected
numbers in some categories could be too small for valid use of this test; the
program does not check for this. It is also possible that the chi-square could
be significant but in the wrong direction; this is not tested in the current
version of the program. To check it beware of a chi-square greater than 3.841
but with a positive invariant. The invariants themselves are computed, as the
difference of cross-products. Their absolute magnitudes are not important, but
which one is closest to zero may be indicative. Significantly nonzero
invariants should be negative if the model is valid. The K invariants, which
are simply differences among the L invariants, are also printed out without any
test on them being conducted. Note that it is possible to use the bootstrap
utility SEQBOOT to create multiple data sets, and from the output from sunning
all of these get the empirical variability of these quadratic invariants.
PROGRAM CONSTANTS
The constants that are defined at the beginning of the program include
"maxsp", which must always be 4 and should not be changed, and "nmlngth", the
name length.
The program is very fast, as it has rather little work to do; these
methods are just a little bit beyond the reach of hand tabulation. Execution
speed should never be a limiting factor.
FUTURE OF THE PROGRAM
In a future version I hope to allow for Y and R codes in the calculation
of the Cavender invariants, and to check for significantly negative cross-
product differences a future in them, which would indicate violation of the
model. By then there should be more known about invariants for larger number
of species, and any such advances will also be incorporated.
----------------------------TEST DATA SET--------------------------
4 13
Alpha AACGTGGCCAAAT
Beta AAGGTCGCCAAAC
Gamma CATTTCGTCACAA
Delta GGTATTTCGGCCT
--------------------------TEST SET OUTPUT--------------------------
Nucleic acid sequence Invariants method, version 3.5c
Name Sequences
---- ---------
Alpha AACGTGGCCA AAT
Beta ..G..C.... ..C
Gamma C.TT.C.T.. C.A
Delta GGTA.TT.GG CC.
Pattern Number of times
AAAC 1
AAAG 2
AACC 1
AACG 1
CCCG 1
CCTC 1
CGTT 1
GCCT 1
GGGT 1
GGTA 1
TCAT 1
TTTT 1
Symmetrized patterns (1, 2 = the two purines and 3, 4 = the two pyrimidines
or 1, 2 = the two pyrimidines and 3, 4 = the two purines)
1111 1
1112 2
1113 3
1121 1
1132 2
1133 1
1231 1
1322 1
1334 1
Tree topologies (unrooted):
I: ((Alpha,Beta),(Gamma,Delta))
II: ((Alpha,Gamma),(Beta,Delta))
III: ((Alpha,Delta),(Beta,Gamma))
Lake's linear invariants
(these are expected to be zero for the two incorrect tree topologies.
This is tested by testing the equality of the two parts
of each expression using a one-sided exact binomial test.
The null hypothesis is that the first part is no larger than the second.)
Tree Exact test P value Significant?
I 1 - 0 = 1 0.5000 no
II 0 - 0 = 0 1.0000 no
III 0 - 0 = 0 1.0000 no
Cavender's quadratic invariants (type L) using purines vs. pyrimidines
(these are expected to be zero, and thus have a nonsignificant
chi-square, for the correct tree topology)
They will be misled if there are substantially
different evolutionary rate between sites, or
different purine:pyrimidine ratios from 1:1.
Tree I:
Contingency Table
2 8
1 2
Quadratic invariant = 4.0
Chi-square = 0.23111 (not significant)
Tree II:
Contingency Table
1 5
1 6
Quadratic invariant = -1.0
Chi-square = 0.01407 (not significant)
Tree III:
Contingency Table
1 2
6 4
Quadratic invariant = 8.0
Chi-square = 0.66032 (not significant)
Cavender's quadratic invariants (type K) using purines vs. pyrimidines
(these are expected to be zero for the correct tree topology)
They will be misled if there are substantially
Different evolutionary rate between sites, or
different purine:pyrimidine ratios from 1:1.
No statistical test is done on them here.
Tree I: -9.0
Tree II: 4.0
Tree III: 5.0