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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

- 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

- 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

- Most variation has little effect on fitness

Testing the Neutral theory

- Synonymous vs Nonsynonymous substitutions
- Microadaptation within protein coding genes
- Types of selection “positive”

Evolutionary change in Nucleotide sequence

- Basic Process
- Estimating rates of substitution
- Reconstructing organism phylogeny

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) 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. 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) 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)]

Continuous time model ancestor

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 ancestor

Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG

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

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

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

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

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 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= 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 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. ancestor

Prob. at time t = PAA(t)

For both seq. the prob.

at time t = P2AA(t)

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 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 = 2(3t) where (3t) 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 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 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? 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 divergence between the two sequences

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 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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