Molecular variation. Joe Felsenstein. GENOME 453, Autumn Molecular variation p.1/32

Transcription

1 Molecular variation Joe Felsenstein GENOME 453, Autumn 2009 Molecular variation p.1/32

2 Views of genetic variation before 1966 The Classical view The Balancing Selection view Hermann Joseph Muller Most loci will be homozygous for the wild-type allele" but a few mutants will exist Theodosius Dobzhansky Most loci will be polymorphic due to balancing selection with strong selection Molecular variation p.2/32

3 Gel electrophoresis slots wick + tank of buffer solution gel of potato starch or polyacrilamide power supply Molecular variation p.3/32

4 After running the current wick slots + tank of buffer solution gel of potato starch or polyacrilamide power supply Molecular variation p.4/32

5 Making one locus visible by staining slots stained wick + tank of buffer solution gel of potato starch or polyacrilamide power supply AA AA Aa aa Aa AA Molecular variation p.5/32

6 A monomeric enzyme A A A a On gel: Molecular variation p.6/32

7 A dimeric enzyme A A A a On gel: Molecular variation p.7/32

8 Polymorphism on a gel Molecular variation p.8/32

9 Lewontin and Hubby s 1966 work Richard Lewontin, about 1980 Lewontin, R. C. and J. L. Hubby A molecular approach to the study of genic heterozygosity in natural populations. II. Amount of variation and degree of heterozygosity in natural populations of Drosophila pseudoobscura. Genetics 54: Molecular variation p.9/32

10 Measures of variability with multiple loci Lewontin and Hubby (Genetics, 1966) suggested two measures of variability: polymorphism and heterozygosity. Polymorphism is the fraction of all loci that have the most common allele less than 0.95 in frequency (i.e. all the rarer alleles together add up to less than Heterozygosity is the estimated fraction of all individuals who are heterozygous at a random locus. Molecular variation p.10/32

11 Computing the average heterozygosity If p i is the frequency in the sample of allele i at a locus, then the heterozygosity for that locus is estimated by taking the sum of squares of the gene frequencies (thus estimating the homozygosity) and subtracting from 1: H = 1 alleles i p 2 i Molecular variation p.11/32

12 An example: locus 1 1 locus 2 1 locus locus The heterozygosities are calculated as: locus = 0 locus = 0 locus 3 1 ( ) = 0.32 locus 4 1 ( ) = The average heterozygosity in this example is H = ( ) / 4 = Molecular variation p.12/32

13 Amounts of heterozygosity Molecular variation p.13/32

14 Kimura s neutral mutation theory Motoo Kimura and family, 1966 Tomoko Ohta, recently Kimura, M Evolutionary rate at the molecular level. Nature 217: Kimura, M., and T. Ohta Protein polymorphism as a phase of molecular evolution. Nature 229: Molecular variation p.14/32

15 Neutral mutation theory Kimura, 1968; Kimura and Ohta, 1971 assume: population size N, rate u of neutral mutations, all different Heterozygosity at any point is expected to be 4Nu 4Nu + 1 Molecular variation p.15/32

16 Crow and Kimura s theoretical calculation James F. Crow, about 1990 Kimura, M., and J. F. Crow The number of alleles that can be maintained in a finite population. Genetics 49: Molecular variation p.16/32

17 Expected heterozygosity with neutral mutation In a random mating population with neutral mutation, a fraction F of the pairs of copies will be homozygous. Suppose all mutations create completely new alleles, and the rate of these neutral mutations is u diploid population of size N F one generation ago now F F Molecular variation p.17/32

18 Expected heterozygosity with neutral mutation In a random mating population with neutral mutation, a fraction F of the pairs of copies will be homozygous. Suppose all mutations create completely new alleles, and the rate of these neutral mutations is diploid population of size N F u one generation ago 1/(2N) of the time 1 1/(2N) of the time now F F Molecular variation p.18/32

19 Expected heterozygosity with neutral mutation In a random mating population with neutral mutation, a fraction F of the pairs of copies will be homozygous. Suppose all mutations create completely new alleles, F = u and the rate of these neutral mutations is diploid population of size N F To be identical, both copies must not be new mutants, and the probability of this is (1 u) (1 u) 2 2 [(1/ (2N)) x 1 + (1 1/(2N)) F ] one generation ago 1/(2N) of the time 1 1/(2N) of the time now F F Molecular variation p.19/32

20 Expected heterozygosity with neutral mutation In a random mating population with neutral mutation, a fraction F of the pairs of copies will be homozygous. Suppose all mutations create completely new alleles, F = u and the rate of these neutral mutations is diploid population of size N one generation ago F To be identical, both copies must not be new mutants, and the probability of this is (1 u) 2 So if we have settled down to an equilibrium level of heterozygosity, F = F, so that F = (1 u) (1 u) 2 [(1/ (2N)) x 1 + (1 1/(2N)) F ] 2 [(1/ (2N)) x 1 + (1 1/(2N)) F ] 1/(2N) of the time 1 1/(2N) of the time now F F Molecular variation p.20/32

21 Expected heterozygosity with neutral mutation In a random mating population with neutral mutation, a fraction F of the pairs of copies will be homozygous. Suppose all mutations create completely new alleles, F = u and the rate of these neutral mutations is diploid population of size N F To be identical, both copies must not be new mutants, and the probability of this is (1 u) 2 So if we have settled down to an equilibrium level of heterozygosity, F = F, so that F = (1 u) (1 u) 2 [(1/ (2N)) x 1 + (1 1/(2N)) F ] 2 [(1/ (2N)) x 1 + (1 1/(2N)) F ] one generation ago 1/(2N) of the time 1 1/(2N) of the time which is easily solved to give (1 u) F = 1 (1 u) 2 (1/ (2N)) 2 (1 1/ (2N)) now F F Molecular variation p.21/32

22 Expected heterozygosity with neutral mutation In a random mating population with neutral mutation, a fraction F of the pairs of copies will be homozygous. Suppose all mutations create completely new alleles, F = u and the rate of these neutral mutations is diploid population of size N F To be identical, both copies must not be new mutants, and the probability of this is (1 u) 2 So if we have settled down to an equilibrium level of heterozygosity, F = F, so that F = (1 u) (1 u) 2 [(1/ (2N)) x 1 + (1 1/(2N)) F ] 2 [(1/ (2N)) x 1 + (1 1/(2N)) F ] one generation ago 1/(2N) of the time 1 1/(2N) of the time which is easily solved to give (1 u) F = 1 (1 u) 2 (1/ (2N)) 2 (1 1/ (2N)) now F F or to good approximation: 1 F = N u heterozygosity is: 4N u 1 F = 1 + 4N u Molecular variation p.22/32

23 Heterozygosity in marine invertebrates (Valentine, 1975) species which is a samples/locus no. loci Het. Asterias vulgaris Northern sea star Cancer magister Dungeness crab Asterias forbesi Common sea star Lyothyrella notorcadensis brachiopod Homarus americanus lobster Crangon negricata shrimp Limulus polyphemus horseshoe crab Euphausia superba Antarctic krill Upogebia pugettensis blue mud shrimp Callianassa californiensis ghost shrimp Phoronopsis viridis horseshoe worm Crassostrea virginica Eastern oyster Euphausia mucronata small krill Asteriodea (4 spp.) deep sea stars Frielea halli brachiopod Ophiomusium lymani large brittlestar Euphausia distinguenda tropical krill Tridacna maxima giant clam Molecular variation p.23/32

24 An interesting case: Limulus polyphemus Carboniferous (300 mya) Jurassic (155 mya) today Selander, R.K., S.Y. Yang, R.C. Lewontin, W.E. Johnson Genetic variation in the horseshoe crab (Limulus polyphemus), a phylogenetic relic." Evolution 24: Molecular variation p.24/32

25 An interesting case Northern elephant seal Mirounga angustirostris Southern elephant seal Mirounga leonina Northern elephant seal: Population in 1890 s: 2-10? Population today: 150,000 or so ( 911? help! there s a monster dying on my beach") Bonnell, M.L., and R.K. Selander Elephant seals: genetic variation and near extinction. Science 184: Molecular variation p.25/32

26 A population cage" for Drosophila Molecular variation p.26/32

27 Yamazaki s population cage experiment Yamazaki, T Measurement of fitness at the esterase-5 locus in Drosophila melanogaster. Genetics 67: Molecular variation p.27/32

28 Explaining Electrophoretic Polymorphisms Can do it either way: By neutral mutation: If H = 0.15 then if N e = 1, 000, 000 we need 4N e µ = to predict this, so that implies µ = So we can explain the level of variation by a neutral mechanism. By selection: To be effective in a population with N e = 1, 000, 000 selection would need to be big enough that 4N e s > 1 so s > 1/4, 000, 000 which is quite small, and impossible to detect in laboratory settings. Molecular variation p.28/32

29 DNA sequencing reveals a similar picture Marty Kreitman Kreitman, M Nucleotide polymorphism at the alcohol-dehydrogenase Locus of Drosophila melanogaster. Nature 304: Molecular variation p.29/32

30 Kreitman s sample of 11 ADH gene sequences, front end Molecular variation p.30/32

31 Kreitman s sample of 11 ADH gene sequences, tail end Molecular variation p.31/32

32 SeattleSNPs data (Nickerson lab) Matrix Metalloproteinase 3 SNP data Molecular variation p.32/32