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Ernst Mayr. “It is altogether unlikely that two genes would have identical selective values under all the conditions in which they exit… cases of neutral polymorphism do not exist… it appears probable that random fixation is of negligible evolutionary importance’. Lewontin and Hubby, 1966.

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Ernst mayr
Ernst Mayr

“It is altogether unlikely that two genes would have identical selective values under all the conditions in which they exit… cases of neutral polymorphism do not exist… it appears probable that random fixation is of negligible evolutionary importance’


Lewontin and hubby 1966
Lewontin and Hubby, 1966

  • Calculated the proportion of polymorphic loci in Drosophila.

  • Argued that NS could not actively maintain so much genetic variation, and suggested that much of it might be selectively neutral.


Neutral theory of molecular evolution
Neutral Theory of Molecular Evolution

  • Kimura, 1968

    Holds that although a small minority of mutations in DNA or protein sequences are adv., and are fixed by NS, and although some are disadv. and are eliminated by ‘purifying’ NS, the great majority of mutations that are fixed are effectively neutral with respect to fitness, and are fixed by genetic drift.


THUS,

Most genetic variation at the molecular level is selectively neutral and lacks adaptive significance

Does not hold that the morphological, physiological and behavioral features of organisms evolve by RGD, such features evolve chiefly by NS


Selection does occur in nt
Selection does occur in NT

  • Most variation has little effect on fitness


Testing the neutral theory
Testing the Neutral theory

  • Synonymous vs Nonsynonymous substitutions

  • Microadaptation within protein coding genes

  • Types of selection “positive”


Evolutionary change in nucleotide sequence
Evolutionary change in Nucleotide sequence

  • Basic Process

    • Estimating rates of substitution

    • Reconstructing organism phylogeny



ancestor

Purines

Pyrimidines

A

G

C

T


Models of nucleotide sub
Models of Nucleotide Sub. ancestor

  • Jukes-Cantor

    • assumes that all nucleotides are present with equal frequencies

    • assumes equal probabilities for all possible nucleotide substitutions

  • Kimura 2-parameter

    • assumes that all nucleotides are present with equal frequencies

    • assumes Ti () and Tv (β) probabilities are different


3 Sub. Types ancestor

Tv, 2 Ti

Equal base frequencies

3 Sub. Types

2 Tv classes, Ti

2 Sub. Types

Tv vs. Ti

Equal base

frequencies

2 Sub. Types

Tv vs. Ti

Single sub. type

Equal base frequencies

Single sub. type

GTR

TrN

SYM

K3ST

HKY85

F84

F81

K2P

JC


Jukes and cantor 1969
Jukes and Cantor (1969) ancestor

  • If you have an A at site i it will change to G, T, C with equal probability

  • Thus the rate of substitution per unit time is 3.

  • The rate of sub. in each of the 3 possible directions of change is 


Jukes and cantor 1969 cont
Jukes and Cantor (1969) cont. ancestor

  • What is the prob. that this site is occupied by A at time t?

    PA(t)

  • The prob. that this site is occupied by A at time 0 isPA(0)=1and still having A time 1

    PA(1)= 1-3


Jukes and cantor 1969 cont1
Jukes and Cantor (1969) cont. ancestor

A

A

T=0

No sub.

sub.

Not A

A

T=1

No sub.

sub.

A

A

T=2

The prob. of A at time 2 is

PA(2) = (1-3) PA(1)+[1-PA(1)]



Kimura 2 parameter

Purines ancestor

Pyrimidines

Kimura 2 Parameter

A

G

β

β

C

T


Kimura scenario s
Kimura Scenario’s ancestor

A

A

A

A

T=0

No sub.

Ti.

Tv.

Tv.

T=1

G

A

C

T

No sub.

Ti.

Tv.

Tv.

T=2

A

A

A

A


3 Sub. Types ancestor

Tv, 2 Ti

Equal base frequencies

3 Sub. Types

2 Tv classes, Ti

2 Sub. Types

Tv vs. Ti

Equal base

frequencies

2 Sub. Types

Tv vs. Ti

Single sub. type

Equal base frequencies

Single sub. type

GTR

TrN

SYM

K3ST

HKY85

F84

F81

K2P

JC


Substitutions time 0
Substitutions Time 0 ancestor

Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG

Taxon 1) .....C......T..................

Taxon 2) .....G......T........C.........

Taxon 3) .....C...........A.............

Taxon 4) .....G...........A........G....


Substitutions time 1
Substitutions Time 1 ancestor

Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG

Taxon 1) .....A......T..................

Taxon 2) .....G......G........C.........

Taxon 3) .....G...........A.............

Taxon 4) .....G...........A........G....


Substitutions time 2
Substitutions Time 2 ancestor

Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG

Taxon 1) .....G......T..................

Taxon 2) .....G......T........C.........

Taxon 3) .....G...........A.............

Taxon 4) .....G...........A........G....

Multiple Substitutions at the same site


Hamming distance or p n n 100
Hamming Distance or P= ancestorn/N*100

Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG

Taxon 1) .....C......T..................

Taxon 2) .....G......T........C.........

Taxon 3) .....C...........A.............

Taxon 4) .....G...........A........G....

N=31

P=2/31*100=6.45%


Substitutions time 21
Substitutions Time 2 ancestor

Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG

Taxon 1) .....G......T..................

Taxon 2) .....G......T........C.........

Taxon 3) .....G...........A.............

Taxon 4) .....G...........A........G....

A→C→G

P=2/31*100=6.45%


Nucleotide diff between seq
Nucleotide diff. between seq. ancestor

Prob. at time t = PAA(t)

For both seq. the prob.

at time t = P2AA(t)


I t prob that the nucleotide at a given site at time t is the same in both sequences
I ancestor(t) = Prob. That the nucleotide at a given site at time t is the same in both sequences

I(t) = P2AA(t) + P2 AT(t) P2AC(t) + P2AG(t)


Same as in the jc
Same as in the JC ancestor

For 2 sequences

Note that the prob. the 2 seq. are

different at the site at time t is

P = 1-I(t)


JC model ancestor

Problem, we do not know t


K the of substitutions per site since the time of divergence between the two sequences
K = the # of substitutions per site since the time of divergence between the two sequences

K = 2(3t) where (3t) is the number of sub. between a single lineage


JC model divergence between the two sequences

# of substitutions per site since

the time of divergence


K2 model
K2 model divergence between the two sequences


Table 3.2 divergence between the two sequences The one-parameter (jukes and Cantor 1969) and four-parameter (Blaisdell 1985) schemes of nucleotide substitution in matrix forma


Table 3 1 general matrix of nucleotide substitution a
Table 3.1 divergence between the two sequences General matrix of nucleotide substitutiona


3 Sub. Types divergence between the two sequences

Tv, 2 Ti

Equal base frequencies

3 Sub. Types

2 Tv classes, Ti

2 Sub. Types

Tv vs. Ti

Equal base

frequencies

2 Sub. Types

Tv vs. Ti

Single sub. type

Equal base frequencies

Single sub. type

GTR

TrN

SYM

K3ST

HKY85

F84

F81

K2P

JC


So which model
So Which model? divergence between the two sequences

  • Multiple assumptions (= nuc. freq. to start etc).

  • Sampling errors due to the use of logarithmic functions (zero).


Comparison of distance measures
Comparison of Distance Measures divergence between the two sequences


Protein encoding genes
Protein encoding genes divergence between the two sequences

  • Synonymous and Nonsynonymous

  • Very difficult as a site changes over time

    • CGG (arg) 3rd postion is syn. But if 1st pos mutates to T then the 3rd position of the resulting codon becoming Nonsynonymous

  • Many sites are not completely synonymous or nonsynonymous

    • Depending the type of mutation, a TI at the 3rd position of CGG (arg) is syn, whereas a TV is nonsynonymous


Multiple ways to calculate ks ka
Multiple ways to calculate Ks & Ka divergence between the two sequences

  • Li et al., 1985

    • Classify the nucleotides into:

    • nondegenerate: all changes at the site are nonsyn.

    • twofold degenerate: 1 of 3 is synonymous

    • fourfold degenerate: all 3 are syn.


  • Categorize degeneracy, divergence between the two sequences

  • Further separate on mutation types (transitional, or transversional) for each type of degeneracy.

  • Ks: the number of synonymous substitutions per synonymous site

  • Ka: the number of nonsynonymous substitutions per nonsynonymous site


Why? divergence between the two sequences

  • Study evolution

  • Positive selection

  • Negative selection


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