Lecture 19 Flashcards
1
Q
Outgroup
A
- lacks all characters that are seen in other branches
- character state is 0
2
Q
Parsimony (2 concepts)
A
- methodological
- ontological
3
Q
Methodological Parsimony
A
- Ockham’s Razor
- best hypothesis is simplest i.e. makes fewest assumptions
4
Q
Ontological Parsimony
A
- nature prefers the simplest course of evolution (not necessarily true; what if nature doesn’t always prefer the fewest number of changes? homoplasies do exist-computers let you know how many exist in each tree)
- thus the best tree requires the fewest changes i.e. the fewest homoplasies
5
Q
Determining proper tree
A
- add more characters
- the longer the think is the less likely it is to be right
- want fewest number of informative steps
- want to find tree with shortest length, and reduce number of homoplasies
6
Q
Automorphy
A
- only present in one clade
- not informative
- not used in analysis at end
- homoplasy on these trees is a characteristic that arises twice
7
Q
Molecular Sequences
A
- loses relationship with extinct things but gains ability to see relationship between very different organisms like ears of corn or mushrooms as opposed to doing it morphologically
- genes don’t care about morphology
- don’t have to have any morphological characters in common to get relatedness
8
Q
Parsimony in Cladistics
A
- use only synaptomorphies
- character present in only one lineage not used in cladistics
- lose some data when you do it cladistically
- will use different data set depending on what method of analysis you use
9
Q
Why is there so much homoplasy in molecular data than in morphological data?
A
- mutation doesn’t always cause a morphological change
- more fundamental reason: only four possible bases for each position so you need a lot of sequence when you’re using this method because the homoplasy rate is so high
10
Q
How many bp in human genome?
A
-3 billion
11
Q
How do you estimate divergence times?
A
-fossil record
12
Q
Unrooted Tree
A
- don’t know topology of relationships seen in the data
- used to give you a rather large scale map
13
Q
Ordering States of Features for Analysis
A
- select features
- decide what are primitive and derived states for each feature
- generate data set
- application to cladogram inference
14
Q
Polarity Assignment
A
- Has to be determined empirically, as in examples; fossil data can help
- states can be assigned numbers for easy analysis
- e.g. 0 for primitive state, 1 for derived. If a second derived state exists can be called 2.
- with that, a table of taxa and character states can be prepared
15
Q
When Building a Simple Cladogram
A
- 0 means trait is absent 1 means trait is present
- computer doesn’t build a tree from the data it matches the data to all possible trees