1. Molecular Evolutionary Analysis
The material basis of heredity
Dan Graur

2. 1959

4. Molecules as documents of evolutionary history
Q: We may ask the question where in the now living systems the greatest amount of information of their past history has survived and how it can be extracted?
A: Best fit are the macromolecules which carry the genetic information.

5. Palimpsest

6. “Searching for an objective reconstruction of the vanished past is surely the most challenging task in biology.”
“In one sense, everything in biology has already been ‘published’ in the form of DNA sequences of genomes.”
1991
Sydney Brenner

7. Assumption:
Life is monophyletic

8. Any two organisms share a common ancestor in their past
descendant 1
descendant 2

9. Some organisms have very recent ancestors.
(5 MYA)
ancestor
Some organisms have very recent ancestors.

10. Some have less recent ancestors…
(18 MYA)
ancestor
Some have less recent ancestors…

11. (120 MYA)
ancestor
…and less recent.

12. But, any two organisms share a common ancestor in their past
(1,500 MYA)
ancestor
But, any two organisms share a common ancestor in their past

13. The differences between 1 and 2 are the result of changes on the lineage leading to descendant 1 + those on the lineage leading to descendant 2.
ancestor
descendant 1
descendant 2

14. Step 1: Sequence Alignment

16. Step 2: Translating number of differences into number of changes (steps).

17. Example:
1. the number of nucleotide substitutions per site between two non-coding sequences (K) according to Kimura’s two parameter model

18. Example:
2. the number of synonymous nucleotide substitution per synonymous site between two coding sequences (KS) according to Jukes and Cantor’s one parameter model

19. Example: the number of nonsynonymous nucleotide substitution per nonsynonymous site between two coding sequences (KA) according to Jukes and Cantor’s one parameter model

20. Ki = number of substitutions (or replacements) of type i per number of sites of type i per 2T

21. Ki r = rate of substitution = number of
substitutions per site per year Ki