Patterns in Evolution

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Patterns in Evolution I. Phylogenetic II.
Morphological III. Historical (later) IV.
Biogeographical
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Patterns in Evolution I. Phylogenetic -
Determining the genealogical, familial patterns
among organisms, populations, species and higher
taxa - "family trees"
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Patterns in Evolution I. Phylogenetic A.
Systematics Taxonomy and Classification
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Patterns in Evolution I. Phylogenetic A.
Systematics Taxonomy and Classification 1.
Taxonomy - the naming of taxa (singular 'taxon")
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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• a. Rules for naming species
• Latin binomen (Drosophila melanogaster)
• italicized or underlined
• author recognized in some groups (insects)
• Genus - species agree in gender
• unambiguous within a kingdom
• if a species is named twice, priority counts
• based on a 'holotype' or 'type' specimen
• 'paratypes' show range of variation
• 'species' is both singular and plural genus
  (s.), genera (pl.)

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• b. Rules for renaming species
• if assigned to new genus, epithet stays
• new author name placed in parens

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• c. Rules for higher taxa
• Animal families end in "-idae" (Felidae)
• Animal sub-families end in "-inae" (Homininae)
• These are often derived from the same stem as the
  'type genus' - the first genus described for the
  family. (Felis)
• Plant families end in "-aceae" (Betulaceae)
• Higher taxa are capitalized, but not italicized
  (as above)
• adjectives are not capitalized ("hominids")

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• 2. Classification - determining the hierarchical
  position of each species within higher taxa.
• a. The Hierarchy....

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• 2. Classification - determining the hierarchical
  position of each species within higher taxa.
• a. The Hierarchy....
• b. Issues

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• 2. Classification - determining the hierarchical
  position of each species within higher taxa.
• a. The Hierarchy....
• b. Issues
• Cladogenesis you want the branching/"clade"
  pattern of taxa to reflect phylogenetic
  relationships Archosaurs for
• crocodilians and birds

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• 2. Classification - determining the hierarchical
  position of each species within higher taxa.
• a. The Hierarchy....
• b. Issues
• Cladogenesis you want the branching/"clade"
  pattern of taxa to reflect phylogenetic
  relationships Archosaurs for
• crocodilians and birds
• Anagenesis however, some evolutionary changes
  are so profound that we might honor the degree
  of difference ("Class Aves)

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• c. Terms
• Monophyletic taxon includes all (and only) the
  species descended from a common ancestor. Aves is
  good.

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c. Terms Monophyletic taxon includes all (and
only) the species descended from a common
ancestor. Aves is good. Paraphyletic taxon
includes all descendants of a common ancestor,
except for those placed in another taxon. So,
Reptilia is a paraphyletic group, as it
includes all diapsids and anapsids EXCEPT birds.
OR, it includes all amniotes EXCEPT mammals and
birds (this gets the synapsids).
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c. Terms Monophyletic taxon includes all (and
only) the species descended from a common
ancestor. Aves is good. Paraphyletic taxon
includes all descendants of a common ancestor,
except for those placed in another taxon. So,
Reptilia is a paraphyletic group, as it
includes all diapsids and anapsids EXCEPT birds.
OR, it includes all amniotes EXCEPT mammals and
birds (this gets the synapsids). Polyphyletic
taxon includes organisms that do not share a
common ancestor that is in the group. To be
avoided. Fliers (Birds, Pterosaurs)
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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• 1. Taxonomy - the naming of taxa (singular
  'taxon")
• 2. Classification - determining the hierarchical
  position of each species within higher taxa.
• a. The Hierarchy....
• b. Issues
• c. Terms
• d. Philosophy of Cladistics
• Term coined by Willi Hennig suggested that
  classification should only include monophyletic
  groups, and that phylogeny should be inferred
  from the analyses of shared derived traits.
• This gives strong preference to cladogenesis over
  anagenesis , such that birds really be
  classified as a derived group of
• dinosaurs or reptiles, not as separate from
  them.

http//palaeos.com/vertebrates/theropoda/dinosaurs
-birds.html
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Linnaean Classification of Apes
Pongidae
Hylobatidae
Hominidae
Apes primates (grasping hands, binocular
vision) with no tails
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Linnaean Classification of Apes
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Linnaean Classification of Apes
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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• morphological
• behavioral
• cellular (structural or chemical)
• genetic - nitrogenous base sequence amino acid
  sequence

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• morphological
• behavioral
• cellular (structural or chemical)
• genetic - nitrogenous base sequence amino acid
  sequence
• can be quantitative measurements, or qualitative
  "presence/absence"

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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• 2. Trees
• a. Unrooted trees show patterns among groups
  without specifying ancestral relationships

A B C D
Trait 1 0 0 1 1
Trait 2 0 0 1 1
Trait 3 0 1 1 0
Trait 4 1 1 0 0
Trait 5 1 1 1 1
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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• 2. Trees
• So, A and B share three traits that C and D
  don't have (1,2, 4) and are more similar to one
  another than they are to C and D.

A B C D
Trait 1 0 0 1 1
Trait 2 0 0 1 1
Trait 3 0 1 1 0
Trait 4 1 1 0 0
Trait 5 1 1 1 1
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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• 2. Trees
• Same for C and D.

A B C D
Trait 1 0 0 1 1
Trait 2 0 0 1 1
Trait 3 0 1 1 0
Trait 4 1 1 0 0
Trait 5 1 1 1 1
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• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• 2. Trees
• So, A and B share three traits that C and D
  don't have (1,2, 4) and are more similar to one
  another than they are to C and D.

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A B C D
Trait 1 0 0 1 1
Trait 2 0 0 1 1
Trait 3 0 1 1 0
Trait 4 1 1 0 0
Trait 5 1 1 1 1

• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• 2. Trees
• b. Rooted Trees Hypothetical patterns of
  descent that could be produced with this pattern.
  You might suppose it would have to be this

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A B C D
Trait 1 0 0 1 1
Trait 2 0 0 1 1
Trait 3 0 1 1 0
Trait 4 1 1 0 0
Trait 5 1 1 1 1

• Patterns in Evolution
• I. Phylogenetic
• A. Systematics Taxonomy and Classification
• B. Reconstructing Phylogenies
• 1. Characters
• 2. Trees
• b. Rooted Trees But it could easily be one of
  these, depending on whether the state 0 or 1
  for traits 1 and 2 were ancestral.

1 derived
0 derived
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Patterns in Evolution I. Phylogenetic A.
Systematics Taxonomy and Classification B.
Reconstructing Phylogenies 1. Characters 2.
Trees b. Rooted Trees SO, in order to access
ancestry, we need to compare the groups in
question to an "outgroup". An outgroup is a
sister taxon which should only share ancestral
traits with the group in question. So reptiles
would be the outgroup for comparisons among
diverse mammals, for example or a crocodile or
dinosaur would be the outgroup to a comparison
among diverse birds.
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• Now, we assume that spE expresses ANCESTRAL
  characters (plesiomorphies). Any different
  character state must have evolved FROM this
  ancestral state - and this evolved state is
  called DERIVED (apomorphy).

A B C D E
Trait 1 0 0 1 1 1
Trait 2 0 0 1 1 1
Trait 3 0 1 1 0 0
Trait 4 1 1 0 0 0
Trait 5 1 1 1 1 0
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• Now, all species in a clade might share
  plesiomorphies, because they are all ultimately
  derived from the same ancestor. So shared
  ancestral traits tell us nothing about patterns
  of relationship within the group. But DERIVED
  traits will only be shared by species that share
  a more recent common ancestor...

A B C D E
Trait 1 0 0 1 1 1
Trait 2 0 0 1 1 1
Trait 3 0 1 1 0 0
Trait 4 1 1 0 0 0
Trait 5 1 1 1 1 0
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• So, to reconstruct phylogenies and build a rooted
  tree, we don't just count shared traits... we
  count SHARED, DERIVED traits (synapomorphies)

A B C D E
Trait 1 0 0 1 1 1
Trait 2 0 0 1 1 1
Trait 3 0 1 1 0 0
Trait 4 1 1 0 0 0
Trait 5 1 1 1 1 0
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• So, A and B share 3 synapomorphies 1, 2, 4, and
  5 (they share these traits, and their state is
  different from the outgroup). B and C share 1
  synapomorphy (3).

A B C D E
Trait 1 0 0 1 1 1
Trait 2 0 0 1 1 1
Trait 3 0 1 1 0 0
Trait 4 1 1 0 0 0
Trait 5 1 1 1 1 0
A B C
B 4 - -
C 1 2 -
D 1 1 1
Number of synapomorphies
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• Now, there are a couple rooted trees that fit
  these data equally well
• First, our assumed tree

A B C
B 4 - -
C 1 2 -
D 1 1 1
In this case, the shared trait between B and C
must be interpreted as an instance of
"convergent/parallel evolution (CE)", in which
the trait evolved independently in both species
(not inherited from ancestor).
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A B C D E
Trait 1 0 0 1 1 1
Trait 2 0 0 1 1 1
Trait 3 0 1 1 0 0
Trait 4 1 1 0 0 0
Trait 5 1 1 1 1 0
1, 2, and 4
5
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• Now, there are a couple rooted trees that fit
  these data equally well
• But there is another

A B C
B 4 - -
C 1 2 -
D 1 1 1
In this case, the discrepancy between A, B, and C
is explained as an evolutionary "reversal" in A,
which has re-expressed the ancestral trait.
3
A B C D E
Trait 1 0 0 1 1 1
Trait 2 0 0 1 1 1
Trait 3 0 1 1 0 0
Trait 4 1 1 0 0 0
Trait 5 1 1 1 1 0
1, 2, and 4
3
5
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• In both cases, species share traits for reasons
  OTHER than inheritance for an immediate common
  ancestor. These are called homoplasies, and they
  obviously can confound the reconstruction of
  phylogenies. Both trees require 6 evolutionary
  events, so they are equally "parsimonious"
  (simple). We could envision lots of other trees,
  but they would require more reversions and
  convergent events. We apply Occam's Razor - a
  philosophical dictum that we will accept (and
  subsequently test) the simplest trees that
  express "maximum parsimony". So these two trees
  are our phylogenetic hypotheses to be tested by
  more data that explicitly addresses their
  differences.

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• The only trait we did not define was an
  autapomorphy - this is a trait unique to a
  species. In our examples above, each trait has
  only two character states. But consider
  nucleotides, where each trait (position) has 4
  possibilities. we can envision that a species
  might have a T whereas all other species in the
  tree have A, C, or G. This would be an
  autapomorphy, and obviously doesn't help us out
  in phylogeny reconstruction because it doesn't
  share this trait with anything else.

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Patterns in Evolution I. Phylogenetic A.
Systematics Taxonomy and Classification B.
Reconstructing Phylogenies 1. Characters 2.
Trees 3. Molecular Evolution and Algorithms DNA,
RNA, and protein sequence data - thousands of
characters - multiple parsimonious trees
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• 3. Molecular Evolution and Algorithms
• a. Synapomorphies and parsimony

Are cetaceans artiodactyls, or a sister group to
the Artiodactyla?
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• 3. Molecular Evolution and Algorithms
• a. Synapomorphies and parsimony

Exon 7 from the gene that encodes ß-casein, a
protein in milk. Shared derived traits with
cetaceans at positions 162, 166, 177
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• 3. Molecular Evolution and Algorithms
• a. Synapomorphies and parsimony

6 changes required at these positions 41 over
entire 60 base sequence
9 changes required at these positions 47 over
entire 60 base sequence
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• 3. Molecular Evolution and Algorithms
• a. Synapomorphies and parsimony

6 changes required at these positions 41 over
entire 60 base sequence
9 changes required at these positions 47 over
entire 60 base sequence
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PROBLEMS WITH BASE DATA

• Scoring characters-its easy if its categorical
  (A, C, T, G), but very difficult if it is
  continuous. Need independent characters, so they
  are weighted evenly.
• Homoplasies are common - both as convergence or
  reversal.
• Ancient changes are obscured by more recent
  ones... A to G, then G to C, looks like it could
  be one change A to C.
• Rapid radiations mean that branches/subgroups may
  not have had time to evolve their own unique
  synapomorphies... and we have lots of species
  with autapomorphies (and are thus distinct) but
  it is difficult to group them.
• Trees of single genes may not "map" onto the
  phylogenetic tree among species. The loss of
  particular alleles may not parallel patterns of
  relationships.
• Hybridization and gene transfer - this can make
  populations look more similar at these loci than
  they really are across the whole genome.
• Rates of evolution of different characters and
  states differ...Some are "highly conserved' and
  don't change much... others change dramatically.
  This is called mosaic evolution. This affects the
  "branch lengths" that are used to represent the
  degree of departure (or the quantified number of
  genetic changes in that unique lineage.

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Patterns in Evolution I. Phylogenetic A.
Systematics Taxonomy and Classification B.
Reconstructing Phylogenies 1. Characters 2.
Trees 3. Molecular Evolution and Algorithms DNA,
RNA, and protein sequence data - thousands of
characters - multiple parsimonious trees a.
Synapomorphies and parsimony b. UPGMA
(unweighted pair group method with arithmetic
mean)
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• 3. Molecular Evolution and Algorithms
• b. UPGMA

- UPGMA assume constant mutation rates, and so
is the simplest likelihood model.
Unweighted Pair Group Method with Arithmetic Mean
These are the number of differences in AA
sequences between species-pairs.
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• 3. Molecular Evolution and Algorithms
• b. UPGMA

• The most similar sequences are those of humans
  and monkey (1 difference).
• This difference accumulated over TWO lineages
  since their divergence (constant mutation)
• So, the branch length of each is 1 difference / 2
  branches 0.5

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1. So, we join taxa B (human) and F (monkey). 2.
Then, we AVERAGE the differences between these
taxa and each other taxon and reduce the
matrix.... so, B differs from A by 19 AA's, and F
differs from A by 18 AA's. So the average
difference between A and new taxon 'BF' 18.5
(fusion of two orange boxes into one orange box
in the new and reduced matrix). (That's why this
is called UPGMA - unweighted pair-group method
using arithmetic averages)

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1. So, we join taxa B (human) and F (monkey). 2.
Then, we AVERAGE the differences between these
taxa and each other taxon and reduce the
matrix.... so, B differs from A by 19 AA's, and F
differs from A by 18 AA's. So the average
difference between A and new taxon 'BF' 18.5
(fusion of two orange boxes into one orange box
in the new and reduced matrix). 3. Now, in the
reduced matrix, we look for the most similar pair
(which is A and D 8 diffs). We halve the
difference to calculate each unique branch length
(4.0)

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1. So, we join taxa B (human) and F (monkey). 2.
Then, we AVERAGE the differences between these
taxa and each other taxon and reduce the
matrix.... so, B differs from A by 19 AA's, and F
differs from A by 18 AA's. So the average
difference between A and new taxon 'BF' 18.5
(fusion of two orange boxes into one orange box
in the new and reduced matrix). 3. Now, in the
reduced matrix, we look for the most similar pair
(which is A and D 8 diffs). We halve the
difference to calculate each unique branch length
(4.0) 4. Now repeat the averaging process with
other taxa to reduce the matrix.

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• 3. Molecular Evolution and Algorithms
• b. UPGMA

Here, branch lengths are equal (and additive)
because averaging and constant mutation are
assumed. In other models, branch lengths vary
reflecting more complex models which accept
different substitution rates.
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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• In the UPGMA example, the
• Branch length is mean number
• of AA substitutions in cytochrome C.
• This protein has 104 AA in animals.
• 2) Typically, these raw data are
• Converted to nucleotide substitutions
• per site by dividing /length. Or, by
• Multiplying this by 100, as change.
• 18 differences.
• 18/104 AA 0.173 nucleotide substitutions per
  site
• 0.17 x 100 17.3 difference

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• In the UPGMA example, the
• Branch length is mean number
• of AA substitutions in cytochrome C.
• This protein has 104 AA in animals.
• 2) Typically, these raw data are
• Converted to nucleotide substitutions
• per site by dividing /length. Or, by
• Multiplying this by 100, as change.
• 18 differences.
• 18/104 AA 0.173 nucleotide substitutions per
  site
• 0.17 x 100 17.3 difference

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• 4) Evolutionary Modeling
• The relationship between difference
• and evolutionary divergence (substitution rate)
• may not be linear.
• - not all differences are indicative of change
• Even 2 random sequences will only differ by 75
• (just by chance there will be the same base at
  25 of sites).
• - some changes are more likely than others.
  Transition mutations (A to G, C to T) are more
  likely than transversions (A to C or T). So,
  models incorporate a transition/transversion
  ratio (2.0, above right).

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• 4) Evolutionary Modeling
• The relationship between difference
• and evolutionary divergence (substitution rate)
• May not be linear.
• - Our ability to detect change depends on
  existing degree of similarity. We are more likely
  to detect changes in sequences that are
  identical, than in sequences that are only 50
  similar, because many changes in that case will
  make the sequences MORE SIMILAR. So a change in
  similarity from 10-12 probably represents fewer
  mutations, and less genetic distance, than
  observed changes from 60-62. If sequences are
  60 different, a lot of mutations in one sequence
  will make it more similar to the otherthus the
  same NET change of 2 represents MORE
  evolutionary change (Distance).

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths

A
a
c
C
b
A B C
A 22 39
B 41
C
D
E
B

• a b 22
• a c 39
• b c 41
• 2 3 a b -2
• 5) 1 4 2a 20, so a 10.
• 6) The distance from A to B 22, so b 12, and
  C 29.

Hypothetical sequence differences
OR a ((AC BC) AB) / 2
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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths

A B C D E
A 22 39 39 41
B 41 41 43
C 18 20
D 10
E
Hypothetical sequence differences among 5 taxa
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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths

A B C D E
A 22 39 39 41
B 41 41 43
C 18 20
D 10
E

• Hypothetical sequence differences among 5 taxa
• D and E are most similar
• Calculate average distance from D and E to A, B,
  and C (reduce this to a 3-point problem)

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths

A B C D E
A 22 39 39 41
B 41 41 43
C 18 20
D 10
E

• Hypothetical sequence differences among 5 taxa
• D and E are most similar
• Calculate average distance from D and E to A, B,
  and C (D 32.6, E 34.6)
• So, E is 2 units farther away from node, and the
  distance between them is 10, so

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D
a
D and E are the closest sequences
c
A-C
b
A-C D E
A-C - 32.6 34.6
D - 10
E -
E
a 4 b 6
a ((AC BC) AB) / 2
Now lets recompute the complete distance matrix
A B C DE
A - 22 39 40
B - 41 42
C - 19
DE -
A B C D E
A 22 39 39 41
B 41 41 43
C 18 20
D 10
E
C and DE are the closet sequences
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D
a
D and E are the closest sequences
c
A-C
b
A-C D E
A-C - 32.6 34.6
D - 10
E -
E
a 4 b 6
a ((AC BC) AB) / 2
Now lets recompute the complete distance matrix
A B C DE
A - 22 39 40
B - 41 42
C - 19
DE -
A B C D E
A 22 39 39 41
B 41 41 43
C 18 20
D 10
E
Mean distance from C to AB 40, and mean
distance from DE to AB 41.
C and DE are the closet sequences
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C and DE are the closet sequences
C
a
b is not just for that segment, it represents the
complete distance from the connecting node to the
leaves
AB C DE
AB - 40 41
C - 19
DE -
c
A-B
b
a 9 b 10 (mean)
DE
So once again, there is one unit of branch length
difference to the node of C and DE, with a total
distance of 19.
a ((AC BC) AB) / 2
C
9
Now lets recompute the complate distance matrix
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A-B
A B C DE
A - 22 39 40
B - 41 42
C - 19
DE -
5
4
D
A B CDE
A - 22 39.5
B - 41.5
E -
6
E
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A
Now we are in thee trivial case of 3 sequences
a
b is not just for that segment, it represents the
complete distance from the connecting node to the
leaves
b
B
A B C-E
A - 22 39.5
B - 41.5
C-E -
c
CDE
a 10 b 12
a ((AC BC) AB) / 2
A
C
9
10
20
5
4
12
D
B
6
E
65
10
A
WHICH was the outgroup? Lets Say C
20
12
B
6
E
5
4
D
9
C
A
C
9
10
20
5
4
12
D
B
6
E
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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths
• e. Maximum Likelihood Models
• What evolutionary rates (in terms of transitions
  and tranversion, etc., are required to give us
  the pattern and rate (as measured in branch
  lengths) that we SEE?
• So, different models of evolution are tested.
  The models are probability matrices of
  substitution rates between bases.
• A tree is given. The branch lengths are given.
  The model of mutation changes, and the
  probabilities of generating the data (sequences)
  change with the model. The likelihood of a tree
  is the probability that it generates the data.

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths
• e. Maximum Likelihood Models
• Neighbor Joining
• Similar, but we dont prioritize which pair we
  group first. Rather, we repeat the tree
  formation using every possible pair-wise
  combination, and then pick the tree with the
  shortest total branch lengths (most conservative
  evolutionary tree). Repeat, using this pair as
  one node (like DE before).

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths
• e. Maximum Likelihood Models
• Neighbor Joining
• g. Bootstrapping
• Gain confidence in a node by subsampling the data
  and creating a tree. Is the node still there?
  How frequently is it present in 100 or 1000
  subsamples of the data set?

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Randomly sample characters (in this case, base
positions) WITH REPLACEMENT. Create the tree,
and report the frequency of a clade in the tree.
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Bootstrap using entire 1100 bases of casein gene,
N 1000. Whales are within the Artiodactyla in
99 of clades. Whales are in clade with deer,
hippo, cow (100)
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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths
• e. Maximum Likelihood Models
• Neighbor Joining
• g. Bootstrapping
• h. Bayesian inference

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• Must estimate the prior probability of trees
  based on external knowledge. Or, assume
  equality.
• Likelihood pp
• P(treedata) P(datatree) P(tree)
• P(data)
• Where P(data) SUM(tree likelihood x prior prob)
  across all trees considered.
• So, for the trees considered and given their
  prior probabilities, what is their fractional
  probability at which the data produces each tree?
  This is the posterior probability that we want.
  P(treedata).

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• Must estimate the prior probability of trees
  based on external knowledge. Or, assume
  equality.
• Likelihood pp
• P(treedata) P(datatree) P(tree)
• P(data)
• Where P(data) SUM(tree likelihood x prior prob)
  across all trees considered.
• So, for the trees considered and given their
  prior probabilities, what is their fractional
  probability at which the data produces each tree?
  This is the posterior probability.
• Clade credibility is the sum of the probabilities
  of the trees in which it occurs.

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• 3. Molecular Evolution and Algorithms
• Synapomorphies and parsimony
• b. UPGMA
• c. Branch Length Units
• d. Calculating Branch Lengths
• e. Maximum Likelihood Models
• Neighbor Joining
• g. Bootstrapping
• h. Baysian inference
• SINEs and LINEs
• - Short and Long interspersed sequences
  transposable elements.
• - Highly unlikely to end up in the same place in
  the genome by chance
• - Similarity is most likely a SHARED, DERIVED
  character.

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