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Dan Graur

Molecular Molecular Evolutionary Evolutionary AnalysisAnalysis

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1959

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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

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PalimpsestPalimpsest

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Sydney Brenner

“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

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Assumption: Assumption: Life is Life is

monophyleticmonophyletic

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Any two organisms share a common ancestor in their past

ancestor

descendant 1 descendant 2

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ancestor

Some organisms have very recent ancestors.

(5 MYA)

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ancestor(18 MYA)

Some have less recent ancestors…

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ancestor(120 MYA)

…and less recent.

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ancestor(1,500 MYA)

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

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The The differences differences between 1 between 1 and 2and 2 are the result of changes changes on the on the lineage lineage leading to leading to descendant descendant 11 ++ those those on the on the lineage lineage leading to leading to descendant descendant 2.2.

ancestor

descendant 1 descendant 2

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Step 1:Step 1:Sequence Alignment

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

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

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€

KS = −3

4ln 1−

4MS3NS

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

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

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€

KA = −3

4ln 1−

4MA3NA

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

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

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Ki = number of

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

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€

ri =Ki

2T

Ki

r = rate of substitution = number of substitutions per site per year