Molecular evolutionary analysis
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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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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.

Molecules as documents of evolutionary history


Palimpsest


“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


Assumption:

Life is monophyletic


ancestor

descendant 1

descendant 2

Any two organisms share a common ancestor in their past


(5 MYA)

ancestor

Some organisms have very recent ancestors.


(18 MYA)

ancestor

Some have less recent ancestors…


(120 MYA)

ancestor

…and less recent.


(1,500 MYA)

ancestor

But, any two organisms share a common ancestor in their past


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


Step 1:Sequence Alignment


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


Example:

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


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


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


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

Ki


r = rate of substitution = number of

substitutions per site per year

Ki


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