Transcript Two-point Linkage Analysis: a brief outline of theory

Linkage Analyisis II
Phylogenic trees
Goals
 Given marker genes, we want to know the position of genes
linked to a given phenotype.
 We use genes with multiple alleles (for the marker). For example
a gene in which a SNP occurred and is present in some of the
population.
 We assume H.W distribution of the two alleles in the population,
and correlate the presence or absence of genes in pedigrees
with the known phenotype:
 Example:
In families where the father has the marker A and A’ (An
heterozygous father) the phenotype occurs in 35 % of the
children carrying A and 12 % of the children carrying A’.
What can we say about the distance between the gene
causing the disease and the marker ?
Linkage Analysis
 Assuming that there is gamete equilibrium at
the A and B loci, in parent 1 there is a
probability of 1/2 that alleles A1 and B1 will be
coupled, and a probability of 1/2 that they will
be in repulsion.
 In other words, we do not know in what
chromosome are the gene present.
Linkage Analysis
 (1) A1 and B1 are coupled,
The probability that parent (1) provides the
gametes A1B1 and A2B2 is (1-q)/2 and the
probability that this parent provides gametes
A1B2 and A2B1 is q/2. The probability that the
couple will have child of type (1) or (2) is (1q)/2, and that of their having a type (3) or
type (4) child is q/2.
 The probability of finding n1 children of type
(1), n2 of type (2), n3 of type (3) and n4 of
type (4) is therefore.
 [(1- q)/2]n1+n2 x (q/2)n3+n4.
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Linkage Analysis
 (2) A1 and B1 are in a state of repulsion.
The probability that parent (1) provides
the gametes A1B2 and A2B1 is (1-q)/2 and
the probability that this parent provides
gametes A1B1 and A2B2 is q/2.
 The probability of the previous
observation is therefore:
 (q/2)n1+n2 x[(1-q)/2]n3+n4.

Linkage analysis
 With no additional information about the A1 and B1
phase, and assuming that the alleles at the A and B
loci are in a state of coupling equilibrium, the
probability of finding n1, n2, n3 and n4 children in
categories (1), (2), (3), (4) is: p(n1,n2,n3,n4/q)=1/2{[(1
-q)/2]n1+n2 x (q/2)n3+n4 + (q/2) n1+n2 x [(1-q)/2] n3+n4}
 So the liklihood of q for an observation n1, n2, n3, n4
can be written:
L(q/n1,n2,n3,n4)=1/2 {[(1-q)/2]n1+n2 (q/2)n3+n4 + (q/2) n1+n2
[(1-q)/2] n3+n4}
Special case: number of children n= 1
 Regardless of the category to which this child
belongs
L(q) = 1/2 [(1-q)/2] + 1/2 [q/2] = 1/4
 The likelihood of this observation for the
family does not depend on q. We can say that
such a family is not informative for q.
Informative families
 An "informative family" is a family for which the liklihood is a
variable function of q.
 One essential condition for a family to be informative is,
therefore, that it has more than one child. Furthermore, at least
one of the parents must be heterozygotic.
 Definition: if one of the parents is doubly heterozygotic and the
other is
 A double homozygote, we have a backcross
 A single homozygote, we have a simple backcross
 A double heterozygote, we have a double intercross
Definition Of The "Lod Score" Of A
Family
 Take a family of which we know the genotypes at the A and B loci of
each of the members.
 Let L(q) be the likelihood of a recombination fraction 0 and q < 1/2
 L(1/2) is the likelihood of q = 1/2, that is of independent segregation into
A and B.
 The lod score of the family in q is:
Z(q) = log10 [L(q)/L(1/2)]
 Z can be taken to be a function of q defined over the range [0,1/2].
 The likelihood of a value of q for a sample of independent families is the
product of the likelihoods of each family, and so the lod score of the
whole sample will be the sum of the lod scores of each family.
Test For Linkage
 Several methods have been proposed to detect linkage: "U
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scores", were suggested by Bernstein in 1931, "the sib pair test"
by Penrose in 1935, "likelihood ratios" by Haldane and Smith in
1947, "the lod score method" proposed by Morton in 1955 (1).
Morton’s method is the one most commonly used at present.
The test procedure in the lod score method is sequential (Wald,
1947 (2)). Information, i.e. the number of families in the sample,
is accumulated until it is possible to decide between the
hypotheses H0 and H1 :
H0 : genetic independence q = 1/2
H1: linkage of q1 0 < q1 < 1/2
The lod score of the q1 sample
Z(q1) = log10 [L(q1)/L(l/2)]
indicates the relative probabilities of finding that the sample is
H1 or H0. Thus, a lod score of 3 means that the probability of
finding that the sample is H1 is 1000 times greater than of
finding that it is H0
("lod = logarithm of the odds").
Test For Linkage
 The decision thresholds of the test are usually set at -
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2 and +3, so that if:
 Z(q1) > 3 H0 is rejected, and linkage is accepted.
 Z(q1) < -2 linkage of q1 is rejected.
 -2 < Z(q1) < 3 it is impossible to decide between
H0 and H1. It is necessary to go on accumulating
information.
For the thresholds chosen, -2 and +3, we can show
that:
The first degree error, (False negative) a < 10-3
The second degree error (false positive), b < 10-2
The reliability, 1-r > 0.95 " q1
r is the probability that we conclude that H1 is true
given H0
Recombination Fraction For A Disease
Locus And A Marker Locus
 Let us assume we are dealing with a disease
carried by a single gene, determined by an
allele, g0, located at a locus G (g0: harmful
allele, G0: normal allele).
 We would like to be able to situate locus G
relative to a marker locus T, which is known to
occupy a given locus on the genome. To do
this, we can use families with one or several
individuals affected and in which the genotype
of each member of the family is known with
regard to the marker T.
 In order to be able to use the lod scores
method described above, what is needed is to
be able to extrapolate from the phenotype of
the individuals (affected, not affected) to their
genotype at locus G (or their genotypical
probability at locus G).
Disease and Marker Locus
 What we need to know is:
 the frequency, g0.
 the penetration vector f1, f2,f3.
 f1 = Pr (affected /g0g0).
 f2 = Pr (affected /g0G0).
 f3 = Pr (affected /G0G0).
 It will often happen that the information available for
the marker is not also genotypic, but phenotypic in
nature. Once again, all possible genotypes must be
envisaged.
Disease and Marker Locus
 As a general rule, the information available about a
family concerns the phenotype. To calculate the
likelihood of q, we must envisage all the possible
genotype configurations at each of the loci, for this
family, writing the likelihood of q for each
configuration, weighting it by the probability of this
configuration, and knowing the phenotypes of
individuals in A and B.
 Knowledge of the genetic parameters at each of the
loci (gene frequency, penetration values) is therefore
necessary before we can estimate q.
Linkage Analysis For Three Loci :
Interference
 Now let us consider three loci A, B and C. Let the
recombination fraction between A and B be q1, that
between B and C be q2 and that between A and C be
q3.
 Let us consider the double recombinant event, firstly
between A and B, and secondly between B and C.
Let R12 be the probability of this event. If the
crossings-over occur independently in segments AB
and BC, then:
 R12 = q1q2
Interference
 If this is not the case, an
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interference phenomenon is
occurring and
R12 = C q1 q2 where C1
If C < 1 the interference is said
to be positive; and crossingsover in segment AB inhibit those
in segment BC.
If C >1 the interference is said to
be negative; and crossings-over
in segment AB promote those in
segment BC.
Let us consider the case of a
triple heterozygotic individual.
Such an individual can provide 8
types of gametes.
A1
A2
B2
B1
C1 
  R12
C2 
A1
A2
B2
B1
A1
A2
B1
B2
A1
A2
B1
B2

C2 

q

R

1
12 
C1 

  q3
C2 

  q 2  R12 
C1 

C1 
  1  q1  q 2  R12
C2 
Interference
 We can write that :
q3 = q1  q2 2 R12
q3 = q1  q2 2 Cq1 q2
 If C = 1
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q3 = q1 + q2 - 2q1 q2
The recombination fraction is a non-additive measurement.
However, we can write
(1-2 q3) = (1-2 q2)(1-2 q2)
if x(q) = k Log (1-2q) then we have x(q3) = x(q2) + x(q1)
and for k = -1/2, x(q)~q for small values of q. x(q) = -1/2 Log (1-2
q) is an additive measurement.
It is known as the genetic distance, and is measured in
Morgans. It can be shown that x measures the mean number of
crossings-over.
Test for the presence of interference
 Let us consider a sample of families with the
genotypes A, B and C. Let Lc be the greatest
likelihood for q1, q2, q3 and L1 the greatest
likelihood when we impose the constraint C=1
(i.e. q3 = q1 + q2 - 2q1 q2 )
 Then -2 Log (L1/Lc ) follows a c2 pattern, with
one degree of freedom.
Genetic Heterogeneity Of Localization
 The analysis of genetic linkage can be complicated
by the fact that mutations of several genes, located at
different places on the genome, can give rise to the
same disorder. This is known as genetic
heterogeneity of localization.
 One of the following two tests is used to identify
heterogeneity of this type:
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The "Predivided sample test"
The "Admixture Test".
 The first test is usually only appropriate if there is a
good family stratification criterion or if each family
individually has high informativity.
The Predivided Sample Test
This test is intended to demonstrate linkage
heterogeneity in different sub-groups of a sample of
families. The aim is to test whether the genetic
linkage between a disease and its marker(s) is the
same in all sub-groups. These groups are formed
ad hoc on the basis of clinical or geographical
criteria etc....
 Let us assume that the total sample of families has
been divided into n sub-groups (it is possible to test
for the existence of as many sub-groups as
families). qi denotes the true value of the
recombination fraction of sub-group i.
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The Predivided Sample Test
 We want to test the null hypothesis H0: q1= q2 = q3
= …= qn against the alternative hypothesis H1: the
values of qi are not all equal.
 Therefore, the quantity


ˆ
Q = 2 log 10   Z i (q i )  Z i (q ) 
 i

 Follows a c2 distribution with (n-1) degrees of
freedom.
The Admixture Test
 The "admixture test" is not based on an ad hoc
subdivision of the families. It is assumed that among
all the families studied genetic linkage between the
disease and the marker is found only in a proportion
a of the families, with a recombination fraction q <
1/2. In the remaining (l-a) families, it is assumed that
there is no linkage with the marker (q=1/2).
 For each family i of the sample, the likelihood is
calculated
Li(a, q) = a Li(q) + (1-a) Li(1/2),
The Admixture Test
 where Li(q) is the likelihood of q
for family i. The
likelihood of the couple (a, q) is defined by the
product of the likelihoods associated with all the
families :
L(a, q)= p Li(a, q)
 We test to find out whether a is significantly different
from 1 by comparing Lmax(a = 1, q), the maximized
likelihood for q assuming homogeneity, and Lmax(a,
q), the maximized likelihood for the two parameters a
and q (nested models).
Then variable Q =2[Ln Lmax (a, q) —Ln Lmax (a = 1,
q)]
follows a c2 distribution with one degree of freedom.
Generalization Of The Admixture Test
 In some single-gene diseases, several genes have
been shown to exist at different locations. This is true,
for example of multiple exostosis disease, for which 3
genes have been identified successively on 3
different chromosomes.
 The "admixture test" is then extended to determine
the proportion of families in which each of the three
genes is implicated , and the possibility that there is a
fourth gene.
 The three locations on chromosomes 8, 19 and 11
were reported as El, E2 and E3, and the proportions
of families concerned as a1, a2 and a3 respectively.
a4 was used to represent the proportion of the
families in which another location was involved.
Generalization Of The Admixture Test
 For each family i of the sample, the likelihood was
calculated using the observed segregation within the
family of the markers available in each of the three
regions, according to the clinical status of each of its
members.
Li(El, E2, E3,a1, a2, a3 |Fi) = a1 (L(E1|Fi)/L(E1=1/2 |Fi)] +
a2 (L(E2|Fi)/L(E2=1/2 |Fi)] + a3 [L(E3|Fi)/L(E3=1/2 |
Fi)]+ a4
 For all the families
Li(El, E2, E3,a1, a2, a3 |Ft) = i Li(El, E2, E3,a1, a2,
a3 | Fi)
Each ai can be tested to see if it is equal to 0, and
then the corresponding non null ai and Ei values are
estimated.
Generalization of The Admixture Test
-Results
 It is also possible to calculate the probability that the gene
implicated is at El, E2 or E3 for each of the families in the
sample. The post hoc probability makes use of the estimated ai
proportions, but also the specific observations in this family.
 The sample investigated has been shown to consist of three
types of families: in 48% of families, the gene is located on
chromosome 8, in 24% of them on chromosome 19, and in 28%
of families the gene is located on chromosome 11. There was no
evidence of a fourth location in this sample.
 The post hoc probabilities of belonging to one of these 3 subgroups were then estimated: the probability that the gene
implicated would be on chromosome 8 was over 90% for 5
families, that it would be on chromosome 19 for 3 of them, and
that it would be on chromosome 11 for 4 families. For the other
families, the situation was less clear-cut: the post-hoc
probabilities are similar to the ad hoc probabilities because of
the paucity of information provided by the markers used.
Phylogenic trees.
Classifying Organisms
 Nomenclature is the science of naming
organisms
 Names allow us to talk about groups of
organisms.

- Scientific names were originally
descriptive phrases; not practical
 Binomial nomenclature
 Developed by Linnaeus, a Swedish
naturalist
 Names are in Latin, formerly the
language of science
 binomials - names consisting of two parts
 The generic name is a noun.
 The epithet is a descriptive adjective.
 Thus a species' name is two words
e.g. Homo sapiens
Carolus Linnaeus (1707-1778)
Classifying Organisms
 Taxonomy is the science of the classification of organisms
 Taxonomy deals with the naming and ordering of taxa.

 The Linnaean hierarchy:

1. Kingdom
Taxonomic Classific ation

2. Division
of Man

3. Class
Homo sapiens

4. Order
Superkingdom: Eukaryota

5. Family
Kingdom: Metazoa

6. Genus
Phylum: Chordata
Class: Mammalia

7. Species
Order: Primata
Family: Hominidae
Genus: Homo
Species: sapiens
Subspecies: sapiens
Phylogenetics

Evolutionary theory states that
groups of similar organisms are
descended
from a common ancestor.
Phylogenetic systematics is a
method of taxonomic classification
based on their evolutionary
history.
 It was developed by Hennig, a German
entomologist, in 1950.

Willi Hennig (1913-1976)
Phylogenetics
Who uses phylogenetics? Some examples:
Evolutionary biologists (e.G. Reconstructing tree of life)
Systematists (e.G. Classification of groups)
Anthropologists (e.G. Origin of human populations)
Forensics (e.G. Transmission of HIV virus to a rape victim)
Parasitologists (e.G. Phylogeny of parasites, co-evolution)
Epidemiologists (e.G. Reconstruction of disease
transmission)
Genomics/Proteomics (e.G. Homology comparison of new
proteins)
Phylogenetic Trees
Node: a branchpoint in a tree (a presumed ancestral OTU)
Branch: defines the relationship between the taxa in terms of
descent and ancestry
Topology: the branching patterns of the tree
Branch length (scaled trees only): represents the number of
changes that have occurred in the branch
Root: the common ancestor of all taxa
Clade: a group of two or more taxa or DNA sequences that
includes both their common ancestor and all their descendents
Operational Taxonomic Unit (OTU): taxonomic level of
sampling selected by the user to be used in a study, such as
individuals, populations, species, genera, or bacterial strains
Phylogenetic Trees
Branch
Clade
Node
Root
Species A
Species B
Species C
Species D
Species E
Phylogenetic Trees
There are many ways of
drawing a tree
A
B
C
D
E
E
=
C
D
B
A
Phylogenetic Trees
A
B
C
D
E
E
=
C
D
B
A
=
A
B
C
D
A
E
=
B
C
D
E
Phylogenetic Trees
A
B
C
D
A
E
B
C
D
E
=
/
Bifurcation
Trifurcation
Bifurcation versus Multifurcation (e.g. Trifurcation)
Multifurcation (also called polytomy): a node in a tree that
connects more than three branches. If the tree is rooted, then
one of the branches represents an ancestral lineage and the
remaining branches represent descendent lineages. A
multifurcation may represent a lack of resolution because of too
few data available for inferring the phylogeny (in which case it is
said to be a soft multifurcation) or it may represent the
hypothesized simultaneous splitting of several lineages (in which
case it is said to be a hard multifurcation).
Phylogenetic Trees
Trees can be rooted or unrooted
A
B
C
A
D
C
D
E
B
E
Phylogenetic Trees
Trees can be scaled or unscaled (with or
without branch lengths)
A
A
B
B
C
unit
C
D
D
E
E
C
C
A
D
A
D
unit
B
B
E
E
Phylogenetic trees
Possible evolutionary trees
Taxa (n)
rooted
(2n-3)!/(2n-2(n-2)!)
unrooted
(2n-5)!/(2n-3(n-3)!)
2
1
1
3
3
1
4
15
3
5
105
15
6
954
105
7
10,395
954
8
135,135
10,395
9
2,027,025
135,135
10
34,459,425
2,027,025
Phylogenetic Trees
• Genes vs. Species.
• Relationships calculated from sequence data represent
the relationships between genes, this is not necessarily
the same as relationships between species.
• Your sequence data may not have the same
phylogenetic history as the species from which they
were isolated.
• Different genes evolve at different speeds, and there is
always the possibility of horizontal gene transfer
(hybridization, vector mediated DNA movement, or
direct uptake of DNA).
Phylogenetic Inference
helix suited to solve different problems:
Different genes will be abest
b sheet
- The RNA genomes of HIV viruses change
so quickly that
every person infected carries a different strain.
- Certain enzymes may evolve relatively fast to allow for
phylogeographic studies of species distribution postglaciation.
- mitochondrial DNA has a relatively fast substitution rate
(evolves quickly) – can be used to establish relatively
recent divergence.
- For establishing ‘deep phylogeny’ we need genes that
change very slowly (highly conserved ones).
Phylogenetic Inference
a helix
- Different sequences accumulate
changes at different rates b sheet
chose level of variation that is appropriate
to the group of
organisms being studied.
- Proteins (or protein coding DNAs) are constrained by natural
selection
- some sequences are highly variable (rRNA spacer regions,
immunoglobulin genes), while others are highly conserved
(actin, rRNA coding regions).
- Different regions within a single gene can evolve at different
rates (conserved vs. Variable domains).
Genetic Distance
Rat
0.0000
0.0646
0.1434
0.1456
0.3213
0.3213
0.7018
Mouse
0.0646
0.0000
0.1716
0.1743
0.3253
0.3743
0.7673
Rabbit
0.1434
0.1716
0.0000
0.0649
0.3582
0.3385
0.7522
Human
0.1456
0.1743
0.0649
0.0000
0.3299
0.2915
0.7116
Opossum
0.3213
0.3253
0.3582
0.3299
0.0000
0.3279
0.6653
Chicken
0.3213
0.3743
0.3385
0.2915
0.3279
0.0000
0.5721
Frog
0.7018
0.7673
0.7522
0.7116
0.6653
0.5721
0.0000
Tree building methods
Genetic Distance
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Unweighted Pair Group (UPGMA)
Neighbor-Joining
Fitch & Margoliash
–
Character-State
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Maximum Parsimony
Maximum Likelihood
Tree Building (Distance Based)
UPGMA
- The simplest of the distance methods is the UPGMA
(Unweighted pair group method using arithmetic
averages)
-
Many multiple alignment programs such as PILEUP use
a variant of UPGMA to create a dendrogram of DNA
sequences which is then used to guide the multiple
alignment algorithm
UPGMA Step 1
combine B and C
A
d
a
e
b
c
A
B
C
D
E
B
0
C
10
0
D
12
4
0
E
10
4
6
0
7
13
15
13
0
UPGMA step 2
combine BC and D
A
d
a
2
e
b
2
c
A
BC
D
E
BC
0
D
11
0
E
10
5
0
7
14
13
0
UPGMA step 3
combine A and E
A
2.5
.5
d
a
2
e
b
2
c
A
BCD
E
BCD E
0 10.5
7
0 13.5
0
UPGMA step 4
combine AE and BCD
AE
2.5
3.5
3.5
.5
d
a
2
e
b
2
c
AE
BCD
BCD
0
12
0
UPGMA Result
2 .5
1.5
A
2.5
3.5
3.5
.5
d
a
2
e
b
2
c
A
B
C
D
E
B
0
C
10
0
D
12
4
0
E
10
4
6
0
7
13
15
13
0
Maximum Parsimony
1 2 3 4 5 6 7 8 9 10
Species 1 - A G G G T A A C T G
Species 2 - A C G A T T A T T A
Species 3 - A T A A T T G T C T
Species 4 - A A T G T T G T C G
How many possible unrooted trees?
Maximum Parsimony
How many possible unrooted trees?
1 2 3 4 5 6 7 8 9 10
Species 1 - A
Species 2 - A
Species 3 - A
Species 4 - A
G G G T A A C T G
C G A T T A T T A
T A A T T G T C T
A T G T T G T C G
1
3 1
2
1
3
2
4
3
4
4
2
Maximum Parsimony
How many substitutions?
1 change
tree
1
2
3
4
A
A
5 changes
G
A G
MP
G
A
A
G
G A
G
Maximum Parsimony
1 2 3 4 5 6 7 8 9 10
1-A G G G T A A C T G
2-A C G A T T A T T A
1
3
2
4
1
2
3
4
1
3
3-A T A A T T G T C T
0
0
0
4
2
4-A A T G T T G T C G
Maximum Parsimony
1 2 3 4 5 6 7 8 9 10
1-A G G G T A A C T G
2-A C G A T T A T T A
1
3
2
4
1
2
3
4
1
3
3-A T A A T T G T C T
0 3 4-A A T G T T G T C G
0 3
0 3
4
2
Maximum Parsimony
G1
C2
3T
C
4A
2C
C
G1
1-G
3
4A
3T
3
A4
4
2-C
G1
T3
3
C
2C
3-T
4-A
Maximum Parsimony
1 2 3 4 5 6 7 8 9 10
1-A G G G T A A C T G
2-A C G A T T A T T A
1
3
2
4
1
2
3
4
1
3
3-A T A A T T G T C T
0 3 2
0 3 2
0 3 2
4
2
4-A A T G T T G T C G
Maximum Parsimony
1 2 3 4 5 6 7 8 9 10
1
3
2
4
1
2
3
4
1
3
1-A G G G T A A C T G
0 3 2 2
3-A T A A T T G T C T
0 3 2 2
0 3 2 1
4
2
2-A C G A T T A T T A
4-A A T G T T G T C G
Maximum Parsimony
G1
A2
3A
A
G1
A3
4G
A1
3-A
2
4G
3G
1
A4
A
1-G
2-A
2A
A
2
4
2G
4-G
Maximum Parsimony
1 2 3 4 5 6 7 8 9 10
1-A G G G T A A C T G
1
3
2
4
1
2
3
4
1
3
2-A C G A T T A T T A
0 3 2 2
4-A A T G T T G T C G
0 3 2 2
0 3 2 1
4
2
3-A T A A T T G T C T
Maximum Parsimony
1
3
2
4
1
2
3
4
1
3
0 3 2 2 0 1 1 1 1 3 14
0 3 2 2 0 1 2 1 2 3 16
0 3 2 1 0 1 2 1 2 3 15
4
2
Maximum Parsimony
1 2 3 4 5 6 7 8 9 10
1-A G G G T A A C T G
2-A C G A T T A T T A
3-A T A A T T G T C T
1
2
3 4-A A T G T T G T C G
0 3 2 2 0 1 1 1 1 3 14
4
Assessing Phylogenetic Data
How much support is there for a particular clade?
Bootstrapping/Jack-knifing:
Lots of randomized data sets are produced by sampling the real data with replacement
(or in jackknifing, by removing some random proportion of the data);
Frequencies of occurrence of groups are a measure of support for those groups
16
59
26
21
71
16
Ochromonas
Symbiodinium
Prorocentrum
Loxodes
Spirostomumum
Tetrahymena
Euplotes
Tracheloraphis
Gruberia
59
71
Ochromonas
Symbiodinium
Prorocentrum
Loxodes
Spirostomumum
Tetrahymena
Euplotes
Tracheloraphis
Gruberia
Problems:
- Bootstrap proportions aren’t easily interpretable
- no indication for how good the data are but simply for how well the tree fits the data
Maximum Likelihood
C CC G
C CC G
= Prob
L
C CC G
+ ........+ Prob
A
T
A
 Find the most likely tree under a given model of
evolution by screening all possible trees.
 Sum log likelihood over all sites
 Model involves:




probability of finding different bases at root;
probability of different kinds of changes between
bases;
substitution rate(s)
length of time
T
Computing the Likelihood of a Tree
 Need a model of evolutionary change.
 To keep simple, assume :
 independence of evolutionary at different sites.
 expected # subs along any branch is function of
substitution rate and length of branch in evolutionary
time.
 subst rate same throughout tree (makes branch
lengths equivalent to evolutionary time)
 No consideration for deletion and insertion
The likelihood of the tree is the product of the
probabilities of change in each tree segment
L = ps0Ps0s6(v6) Ps6s1(v1)Ps6s2(v2)Ps0s8(v8)
Ps8s3(v3)Ps8s7(v7)Ps7s4(v4)Ps7s5(v5)
Where si is the state at point i on the trees, and v’s are the lengths of
the segments
Tree evaluation
Bootstrapping:
- Characters are resampled with replacement to create many bootstrap replicate
data sets
- Each bootstrap replicate data set is analysed (e.g. with parsimony, distance,
ML etc.)
- Agreement among the resulting trees is summarized with a majority-rule consensus tree
- Frequencies of occurrence of groups, bootstrap proportions (BPs), are a measure of
support for those groups
Ochromonas
Ochromonas
Symbiodinium
Symbiodinium
16
59
Prorocentrum
59
Prorocentrum
26
Loxodes
Loxodes
71
21
Spirostomumum
Spirostomumum
71
16
Tetrahymena
Tetrahymena
Euplotes
Euplotes
Tracheloraphis
Tracheloraphis
Gruberia
Gruberia
Jack-knifing
- Jack-knifing is very similar to bootstrapping and differs only in the character
resampling strategy
- Jack-knifing is not as widely available or widely used as bootstrapping
- tends to produce broadly similar results
Assessing Phylogenetic Data
Is there signal in the data?
Possible approach: Random Permutations
- Random permutation destroys any correlation among characters to that
expected by chance alone
- It preserves number of taxa, characters and character states in each
character (and the theoretical maximum and minimum tree lengths
taxa
characters
a
b
c
d
12345678
ATGTGTTG
AGGTGTTG
ACGACAAC
ACGACAAT
Original structured data with strong
correlations among characters
taxa
characters
a
b
c
d
12345678
ACGAGTAG
AGGTCATT
ATGACATC
ACGTGTAG
Randomly permuted data with any correlation
among characters due to chance
Assessing Phylogenetic Data
taxa
characters
a
b
c
d
12345678
ATGTGTTG
AGGTGTTG
ACGACAAC
ACGACAAT
Original structured data with strong
correlations among characters
taxa
characters
a
b
c
d
12345678
ACGAGTAG
AGGTCATT
ATGACATC
ACGTGTAG
Randomly permuted data with any correlation
among characters due to chance
Matrix Randomization Tests
Compare some measure of data quality/hierarchical structure for the real and
many randomly permuted data sets
This allows us to define a test statistic for the null hypothesis that the real data
are not better structured than randomly permuted and phylogenetically
uninformative data
PTP (permutation tail probability) test
Pass test
Fail test
Frequency
reject null
hypothesis
95%cutoff
Measure of data quality (e.g. tree length, ML ...)
good
bad
Null Hypothesis:
The length of the shortest tree is what you would see given random data
How it works:
Reject the null if the real data has shorter tree
(the real data is more internally consistent than random data)
Comments:
Even a little bit of signal can lead you to reject the null; does not mean phylogenetic signal
Frequency
g1 (skewness of tree length distribution)
random data showed that the distribution
of tree lengths tends to be normal
shortest tree
Frequency
Tree length
phylogenetically informative data is
expected to have a strongly skewed
distribution with few shortest trees
and few trees nearly as short
shortest tree
Tree length
Null Hypothesis:
The data has no signal and the shortest tree is just the shortest tree from a symmetric
distribution of tree lengths (distribution of tree lengths won’t be skewed if there is no signal)
How it works:
Significant left (negative) skew rejects the null of no signal
Comments:
Even a little bit of signal can lead you to reject the null; does not mean phylogenetic signal
g1 (skewness of tree length distribution)
random data
g1=-0.100478
real data
g1=-0.951947