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Gene Substitution

Gene Substitution. A mutant allele replaces the predominant wild type allele Majority are lost after a few generations. Fixation Probability depends on:. Its frequency Its selective advantage or disadvantage (s) N e. New mutant arising as a single copy in a diploid population of size N.

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Gene Substitution

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  1. Gene Substitution • A mutant allele replaces the predominant wild type allele • Majority are lost after a few generations

  2. Fixation Probability depends on: • Its frequency • Its selective advantage or disadvantage (s) • Ne

  3. New mutant arising as a single copy in a diploid population of size N Neutral mutation i.e. s=0 Advantageous mutations i.e. s>0 Which is actually its frequency in the population

  4. Gene substitution-fixation prob. 1000 individuals (N=Ne) Neutral 0.05% chance of fixation (P=1/2N) 0.01 selection advantage = 2 % (P=2s) 0.001 selection disadvantage = 0.004% 0.01 sel. Advantage has a98% chance of being lost by chance

  5. Rate of gene substitution (K) • u = mutation rate • Number of mutants arising in a diploid population = 2Nu per generation • Prob. of fixation = 1/(2N) Rate of sub. of neutral alleles

  6. Rate of gene substitution (K) • Thus for neutral mutations, the rate of substitution is equal to the rate of mutation • Large pop. The number of new mutations arising every gen. is high, but fixation prob. is low. • Small pop. The number of new mutations arising every gen. is low, but fixation prob. is high • Thus the rate of substitution for neutral mutations is independent of population size

  7. Rate of gene substitution (K) • For advantageous mutations s > 0 • Depends on population size and selection advantage as well as mutation rate

  8. Fixation of a New Mutation • On avg. in a large pop. It takes 4Ne generations • If Ne is large >1000 it might take so long that other mutations occur in the interim

  9. 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’

  10. 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.

  11. 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.

  12. 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

  13. Selection does occur in NT • Most variation has little effect on fitness

  14. Testing the Neutral theory • Synonymous vs Nonsynonymous substitutions • Microadaptation within protein coding genes • Types of selection “positive”

  15. Evolutionary change in Nucleotide sequence • Basic Process • Estimating rates of substitution • Reconstructing organism phylogeny

  16. Compare two or more sequences descended from a common ancestor

  17. Purines     Pyrimidines  A G C T

  18. Models of Nucleotide Sub. • 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

  19. 3 Sub. Types 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

  20. Jukes and Cantor (1969) • 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 

  21. Jukes and Cantor (1969) cont. • 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

  22. Jukes and Cantor (1969) cont. 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)]

  23. Continuous time model

  24. Purines Pyrimidines Kimura 2 Parameter  A G   β β C T 

  25. Kimura Scenario’s 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

  26. Substitutions Time 0 Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG Taxon 1) .....C......T.................. Taxon 2) .....G......T........C......... Taxon 3) .....C...........A............. Taxon 4) .....G...........A........G....

  27. Substitutions Time 1 Outgroup) ATGTCAGGGACTCAGATCGAATGGGATCTAG Taxon 1) .....A......T.................. Taxon 2) .....G......G........C......... Taxon 3) .....G...........A............. Taxon 4) .....G...........A........G....

  28. Substitutions Time 2 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

  29. Hamming Distance or P=n/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%

  30. Substitutions Time 2 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%

  31. Nucleotide diff. between seq. Prob. at time t = PAA(t) For both seq. the prob. at time t = P2AA(t)

  32. I(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)

  33. Same as in the JC For 2 sequences Note that the prob. the 2 seq. are different at the site at time t is P = 1-I(t)

  34. JC model Problem, we do not know t

  35. 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

  36. JC model # of substitutions per site since the time of divergence

  37. K2 model

  38. Table 3.2 The one-parameter (jukes and Cantor 1969) and four-parameter (Blaisdell 1985) schemes of nucleotide substitution in matrix forma

  39. Table 3.1 General matrix of nucleotide substitutiona

  40. 3 Sub. Types 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

  41. So Which model? • Multiple assumptions (= nuc. freq. to start etc). • Sampling errors due to the use of logarithmic functions (zero).

  42. Comparison of Distance Measures

  43. Protein encoding genes • 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

  44. Multiple ways to calculate Ks & Ka • 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.

  45. Categorize degeneracy, • 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

  46. Why? • Study evolution • Positive selection • Negative selection

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