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Chapter 25
Phylogeny and
Systematics
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 25.1 A dragonfly fossil from Brazil, more
than 100 million years old
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Figure 25.2 An unexpected family tree
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Figure 25.3 Formation of sedimentary strata
containing fossils
1 Rivers carry sediment to the
ocean. Sedimentary rock layers
containing fossils form on the
ocean floor.
2 Over time, new strata are
deposited, containing fossils
from each time period.
3 As sea levels change and the seafloor
is pushed upward, sedimentary rocks are
exposed. Erosion reveals strata and fossils.
Younger stratum
with more recent
fossils
Older stratum
with older fossils
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Grand Canyon
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Volcanic eruption
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Figure 25.4 A gallery of fossil types
(c) Leaf fossil, about 40 million years old
(b) Petrified tree in Arizona, about
190 million years old
(a) Dinosaur bones being excavated
from sandstone
(d) Casts of ammonites,
about 375 million
years old
(f) Insects
preserved
whole in
amber
(g) Tusks of a 23,000-year-old mammoth,
frozen whole in Siberian ice
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(e) Boy standing in a 150-million-year-old
dinosaur track in Colorado
Figure 25.5 Convergent evolution of analogous
burrowing characteristics
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Figure 25.6 Aligning segments of DNA
1 Ancestral homologous
DNA segments are
identical as species 1
and species 2 begin to
diverge from their
common ancestor.
1 C C A T C A G A G T C C
2 C C A T C A G A G T C C
Deletion
2 Deletion and insertion
mutations shift what
had been matching
sequences in the two
species.
1 C C A T C A G A G T C C
2 C C A T C A G A G T C C
G T A
3 Homologous regions
(yellow) do not all align
because of these mutations.
4 Homologous regions
realign after a computer
program adds gaps in
sequence 1.
Insertion
1
C C A T
C A
2
C C A T
G T A
1
C C A T
2
C C A T
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A G T C C
G T A
C A G
A G T C C
C A
A G T C C
C A G
A G T C C
Figure 25.7 A molecular homoplasy
A C G G A T A G T C C A C T A G G C A C T A
T C A C C G A C A G G T C T T T G A C T A G
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Figure 25.8 Hierarchical classification
Species
Panthera
Genus
Felidae
Family
Carnivora
Order
Class
Phylum
Kingdom
Domain
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Panthera
pardus
Mammalia
Chordata
Animalia
Eukarya
Species
Panthera
Order
Family
Mephitis
Panthera
Canis
Canis
Lutra lutra
mephitis
pardus
familiaris
lupus
(European
(leopard) (striped skunk)
otter) (domestic dog) (wolf)
Genus
Figure 25.9 The connection between classification
and phylogeny
Felidae
Mephitis
Lutra
Mustelidae
Carnivora
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Canis
Canidae
Unnumbered Figure p.497
Leopard
Domestic cat
Common ancestor
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Unnumbered Figure p.497
Wolf
Leopard
Common ancestor
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Domestic cat
Figure 25.10 Monophyletic, paraphyletic, and
polyphyletic groupings
Grouping 2
Grouping 1
D
E
G
C
H
J
F
K
I
B
D
E
G
C
H
J
F
K
I
B
A
(a) Monophyletic. In this tree, grouping 1,
consisting of the seven species B– H, is a
monophyletic group, or clade. A monophyletic group is made up of an
ancestral species (species B in this case)
and all of its descendant species. Only
monophyletic groups qualify as
legitimate taxa derived from cladistics.
D
E
G
C
H
J
F
K
I
B
A
(b) Paraphyletic. Grouping 2 does not
meet the cladistic criterion: It is
paraphyletic, which means that it
consists of an ancestor (A in this case)
and some, but not all, of that ancestor’s
descendants. (Grouping 2 includes the
descendants I, J, and K, but excludes
B–H, which also descended from A.)
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Grouping 3
A
(c) Polyphyletic. Grouping 3 also fails the
cladistic test. It is polyphyletic, which
means that it lacks the common ancestor
of (A) the species in the group. Furthermore, a valid taxon that includes the
extant species G, H, J, and K would
necessarily also contain D and E, which
are also descended from A.
Figure 25.11 Constructing a cladogram
Lancelet
(outgroup)
Lamprey
Tuna
Salamander
Turtle
Leopard
CHARACTERS
TAXA
Hair
0
0
0
0
0
1
Amniotic (shelled) egg
0
0
0
0
1
1
Four walking legs
0
0
0
1
1
1
Hinged jaws
0
0
1
1
1
1
Vertebral column (backbone)
0
1
1
1
1
1
Turtle
(a) Character table. A 0 indicates that a character is absent; a 1
indicates that a character is present.
Leopard
Hair
Salamander
Amniotic egg
Tuna
Four walking legs
Lamprey
Hinged jaws
Lancelet (outgroup)
Vertebral column
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(b) Cladogram. Analyzing the distribution of these
derived characters can provide insight into vertebrate
phylogeny.
Figure 25.12 Phylogram
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65.5
Millions of
years ago
Proterozoic
542
Paleozoic
251
Mesozoic
Cenozoic
Figure 25.13 Ultrametric tree
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Figure 25.14 Trees with different likelihoods
Human
Human
Mushroom
0
Mushroom
Tulip
30%
40%
0
40%
Tulip
0
(a) Percentage differences between sequences
25%
15%
15%
15%
20%
10%
5%
5%
Tree 1: More likely
(b) Comparison of possible trees
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Tree 2: Less likely
Figure 25.15 Applying Parsimony to a Problem in
Molecular Systematics
APPLICATION In considering possible phylogenies for a group of species, systematists compare molecular data for the species. The most efficient
way to study the various phylogenetic hypotheses is to begin by first considering the most parsimonious—that is, which hypothesis requires the
fewest total evolutionary events (molecular changes) to have occurred.
TECHNIQUE Follow the numbered steps as we apply the principle of parsimony to a hypothetical phylogenetic problem involving four closely
related bird species.
1 First, draw the possible phylogenies for the species
(only 3 of the 15 possible trees relating these four
species are shown here).
Species
I
I
II
III
Species
II
IV
I
Species
III
III
II
Species
IV
I
IV
IV
II
Three possible phylogenetic hypothese
2 Tabulate the molecular data for the species (in this simplified
example, the data represent a DNA sequence consisting of
just seven nucleotide bases).
1
Sites in DNA sequence
4 5
3
7
2
6
I
A
G
G
G
G
G
T
II
G
G
G
A
G
G
G
III
G
A
G
G
A
A
T
IV
G
G
A
G
A
A
G
I
A
II
G
Species
3 Now focus on site 1 in the DNA sequence. A single basechange event, marked by the crossbar in the branch leading
to species I, is sufficient to account for the site 1 data.
III
G
G
G
Base-change
event
G
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IV
G
Bases at
site 1 for
each species
III
4 Continuing the comparison of bases at sites 2, 3, and 4
reveals that each of these possible trees requires a total
of four base-change events (marked again by crossbars).
Thus, the first four sites in this DNA sequence do not help
us identify the most parsimonious tree.
5 After analyzing sites 5 and 6, we find that the first tree requires
fewer evolutionary events than the other two trees (two base
changes versus four). Note that in these diagrams, we assume
that the common ancestor had GG at sites 5 and 6. But even if
we started with an AA ancestor, the first tree still would require
only two changes, while four changes would be required to
make the other hypotheses work. Keep in mind that parsimony
only considers the total number of events, not the particular
nature of the events (how likely the particular base changes
are to occur).
6 At site 7, the three trees also differ in the number of
evolutionary events required to explain the DNA data.
I
II
III
IV
I
III
II
IV
I
IV
II
III
I
GG
II
GG
III
AA
IV
AA
I
GG
III
AA
II
GG
IV
AA
I
GG
IV
AA
II
GG
III
AA
AA
GG
Two base
changes
GG
I
T
II
G
GG
GG
III
T
IV
G
T
T
GG
I
T
III
T
I
II
II
G
IV
G
G
I
T
IV
G
III
IV
I
III
II
G
III
T
T
T
T
8 events
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GG
T
T
RESULTS
To identify the most parsimonious tree, we
total all the base-change events noted in steps 3–6 (don’t
forget to include the changes for site 1, on the facing page).
We conclude that the first tree is the most parsimonious of
these three possible phylogenies. (But now we must complete
our search by investigating the 12 other possible trees.)
GG
GG
T
II
9 events
IV
I
IV
II
10 events
III
Figure 25.16 Parsimony and the analogy-versushomology pitfall
Bird
Lizard
Mammal
Four-chambered
heart
(a) Mammal-bird clade
Bird
Lizard
Mammal
Four-chambered
heart
Four-chambered
heart
(b) Lizard-bird clade
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Figure 25.17 Two types of homologous genes
Ancestral gene
Speciation
(a)
Orthologous genes
Ancestral gene
Gene duplication
Paralogous genes
(b)
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Fig. 26-18a
Ancestral gene
Ancestral species
Speciation with
divergence of gene
Species A
Orthologous genes
(a) Orthologous genes
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Species B
Fig. 26-18b
Species A
Gene duplication and divergence
Paralogous genes
Species A after many generations
(b) Paralogous genes
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Figure 25.18 The universal tree of life
Bacteria
Eukarya Archaea
0
4 Symbiosis of
chloroplast ancestor
with ancestor of
green plants
Billion years ago
1
3 Symbiosis of
mitochondrial
ancestor with
ancestor of
eukaryotes
4
2
3
2 Possible fusion of
bacterium and
archaean, yielding
ancestor of
eukaryotic cells
2
3
1
Origin of life
4
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1 Last common
ancestor of all
living things