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Molecular Evolutionary Analysis

Molecular Evolutionary Analysis. Dan Graur. 1959. 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.

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Molecular Evolutionary Analysis

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  1. Molecular Evolutionary Analysis Dan Graur

  2. 1959

  3. 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. Molecules as documents of evolutionary history

  4. Palimpsest

  5. “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

  6. Assumption: Life is monophyletic

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

  8. (5 MYA) ancestor Some organisms have very recent ancestors.

  9. (18 MYA) ancestor Some have less recent ancestors…

  10. (120 MYA) ancestor …and less recent.

  11. (1,500 MYA) ancestor But, any two organisms share a common ancestor in their past

  12. 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. descendant 1 descendant 2 ancestor

  13. Step 1:Sequence Alignment

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

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

  16. 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

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

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

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

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