Selection and Neutrality

1. Selection and Neutrality • Mutations (though rare) arise constantly in all organisms. • What happens to them over time? • Some changes are without any consequence - neutral. • Other changes are deleterious - negative selection. • Very rarely, some change is advantageous - positive selection.

2. Neutrality Any particular allele that confers no selective advantage or disadvantage (as homozygote or heterozygote) is called neutral. Examples? • most synonymous codon changes. • most changes in introns. • most changes in transposable element sequences.

3. Allele frequency • Consider one position in the genome (conventionally called a “site”). • Count up the nucleotide at that position in all members of a population (or the entire species). For example:

4. Two allele simplification • Most positions in the genome are all (or almost all) one nucleotide. • Called “monomorphic” (one form). • Most positions with differences are dominated by two nucleotides. • Called “dimorphic” - extensively modeled. For example: For example: (nearly) monomorphic (nearly) dimorphic

5. Two allele simplification By convention, allele frequencies are called p and q. Designate allele A (frequency p), allele a (frequency q). Genotype frequencies (with random mating) are: AA homozygote: p2 Aa heterozygote: 2pq aa homozygote: q2 Note - A and a don’t imply which one is “normal” or dominant, they are simply designators. (Hardy Weinberg equilibrium)

6. Tracking allele frequency over time (2-allele system) • At each generation, the gametes that make the next generation are drawn AT RANDOM from the current alleles. • As this random draw repeats, allele frequencies change. • Call the two alleles A and a. fraction of A allele (a allele is 1 - p) diploid population size = 1000 no selection, simply random gametes drawn at each generation.

7. A graphical representation of how the simulations work random draw of alleles for gametes that produce next generation What will the frequency of the red allele be? Population: ~100 diploids red allele 20% Possibly 19%, 20%, 21% etc. (in fact this value will have an approximately normal distribution with mean 20%)

8. diploid population size = 1000 At each generation, make a pool of 2,000 gametes drawn randomly according to the allele frequency. Repeat.

9. Tracking allele frequency over time (cont.) If you run multiple simulations, the result differs: diploid population size = 1000 Each line is from an identical independent simulation. This is called genetic drift.

10. Tracking allele frequency over time (cont.) The rate of drift depends only on population size: population 1,000 population 300 population 100

11. Allele extinction and fixation A allele fixed (a allele extinct) a allele fixed (A allele extinct)

12. Neutrality and Drift Now consider a brand new mutation (arose in one gamete of one parent): rarely the new allele drifts to fixation population size 50, single new allele Neutral mutations and drift to fixation underlie most genome change. most of the time the new allele quickly disappears

13. Most human SNPs are neutral! • Probably >90% of all known human SNPs (possibly 99%) are neutral. If a sequence has no function (evolves neutrally) it will usually have changed so much between mouse and human that you can no longer detect similarity. From UCSC browser: thin olive line - can’t be aligned to mouse positions of known SNPs Sum up over whole genome - is each SNP located in region conserved in mouse?

14. Neutrality and mapping SNPs You can now understand why this sort of pedigree makes sense - the Aand B alleles have no phenotype (they are neutral). The only way to detect them is by some molecular genotyping test. From lecture 12:

15. Negative (purifying) selection • Force that “keeps things the same”. • Eliminates deleterious sequence changes via phenotypic effect. • Accounts for sequence conservation over long times.

16. Negative (purifying) selection Eliminates deleterious sequence changes via phenotypic effect. population size 1,000 ¼ of AA homozygotes fail to reproduce

17. CFTR protein alignment: human, chimp, dog, mouse, rat, chicken, zebrafish (gap positions removed)

18. strong negative selection weaker negative selection Human: Chimp: Dog: Mouse: Rat: Chicken: Zebrafish: what about this site? probably neutral - it doesn’t matter what amino acid is here, so any mutation is acceptable but then why are the human-chimp and mouse-rat amino acids the same? very little divergence time - amino acid changing mutation just hasn’t happened

19. Negative selection gets very weak when allele frequency is very low. (for recessives) population size 10,000 ¼ of AA homozygotes fail to reproduce for a recessive, only these can be selected against

20. Disease alleles • Almost by definition, disease alleles are all under strong negative selection. Why do they still exist? • Primarily, balance between new mutation and effective removal by negative selection. • Rarely, disease alleles are advantageous in heterozygotes or under unusual conditions. Examples of heterozygote advantage include sickle cell trait.

21. From lecture 1: Heterozygote advantage, homozygote lethal population size 500 AA homozygote reproduces 80% as often as Aa heterozygote, aa homozygote never reproduces. a allele similar to sickle-cell trait - heterozygote is relatively resistant to malaria, but homozygote is very sick. Reaches a stable allele frequency proportional the degree of heterozygote advantage.

22. Positive (Darwinian) selection • When an allele confers a selective advantage. • Very rare among newly arisen mutations. • Corresponds to the type of selection Darwin had in mind. Ultimately responsible for all “adaptive” change.

23. Positive selection is powerful Simulation started with a SINGLEA allele (mimics new mutation) population size 1,000 AA reproduces 100% of the time Aa reproduces 90% of the time aa reproduces 80% of the time notice relatively small advantage conferred by A allele

24. Lactase persistence - example of positive selection • The enzyme lactase permits digestion of lactose (mostly found in milk). • Encoded by a single gene - LCT. • Ancient humans expressed lactase only in infancy. • Many modern human populations express lactase throughout life. weaning off ancestral state LCT gene common in many current human populations weaning LCT gene

25. Frequency of lactase persistence little or no dairy farming The selection for lactase persistence probably started only about 5,000 to 7,000 years ago.

26. Course review session

27. Final exam: -8 questions - 200 points total -Questions focus almost exclusively on material covered in the second half of class (lecture 10 onwards)… - if you’ve forgotten the major concepts from the first half of the course (Mendelian segregation, complementation, epistasis, etc.) you will have trouble -if you didn’t work the problem sets, you may also be sorry -Todays review session will cover molecular markers, the use of molecular markers in linkage studies, LOD score analysis & constructing contigs (STS analysis, chr. walk)

28. What are polymorphic loci? A polymorphic site or locus… A location in the genome where at least two versions of the sequence exist in the population, each at a frequency of at least 1% e.g., UW student population— about 40,000  80,000 copies of (e.g.) chromosome 2 70,000 copies have A-T base pair 10,000 copies have C-G base pair each is at > 1% of total population, so this is a polymorphic site

29. * * * DNA from indiv. #1 DNA from indiv. #2 DNA from indiv. #3 Etc. Digest DNA with restriction enzyme How to find polymorphic sites? -Randomly select and test specific DNA sequences Use as a probe in southern blot

30. radioactive probe: restriction endonuclease digestion nylon filter with immobilzed DNA agarose gel electrophoresis large - Denature DNA with NaOH then transfer to nylon filter * * * * * * + small Southern blotting procedure = restriction site Hybridize radioactive single-stranded DNA probe Expose to film

31. ..AAAAGCTTAG.. ..TTTTCGAATC.. ..AATAGCTTAG.. ..TTATCGAATC.. Practice question What would homozygous genotypes look like on the Southern blot? Hind III restriction endonuclease cleavage site

32. Practice question What would homozygous genotypes look like on the Southern blot?

33. Practice question What would homozygous genotypes look like on the Southern blot?

34. * * * DNA from indiv. #1 DNA from indiv. #2 DNA from indiv. #3 Etc. Digest DNA with restriction enzyme How to find polymorphic sites? -Randomly select and test specific DNA sequences Use as a probe in southern blot Not polymorphic

35. * * * DNA from indiv. #1 DNA from indiv. #2 DNA from indiv. #3 Digest DNA with restriction enzyme How to find polymorphic sites? -Randomly select and test specific DNA sequences Etc. Repeat with a different probe OR the same probe and a different restriction enzyme digest Use as a probe in southern blot Not polymorphic Polymorphic !

36. Disease gene X normal allele of disease gene a polymorphic site in the genome could be 20,000,000bp (or more) ; the polymorphic marker may segregate with (be linked to) the disease gene, but is not the disease gene. The POINT: The polymorphic marker and the disease gene are two different entities

37. SNP resulting in RFLP marker Genomic DNA obtained from corresponding individuals in the pedigree. RFLP 2 RFLP genotype: 1,1 1,2 1,2 1,1 1,1 1,2 1,2 1,1 Using polymorphic markers to map a human gene Example: Using a Restriction Fragment Length Polymorphism (RFLP) marker to map a dominant (D) disease. RFLP allele 1: Restriction endonuclease cleavage site RFLP allele 2: Probe: 1 DNA is digested with restriction enzyme and subjected to southern blot analysis using probe shown above 1,2 1,1

38. 2 1 d d 2 1 D D Using polymorphic markers to map a human gene If linked, one of the RFLP alleles should segregate with the disease; the other RFLP allele should segregate with WT phenotype. RFLP 1 2 RFLP genotype: 1,2 1,1 1,1 1,2 1,2 1,1 1,1 1,2 1,2 1,1 Let’s say that we know the phase of I-1 and I-2: Which individuals in the pedigree appear to be recombinants?

39. 1 d 2 D Using polymorphic markers to map a human gene If linked, one of the RFLP alleles should segregate with the disease; the other RFLP allele should segregate with WT phenotype. RFLP 1 2 RFLP genotype: 1,2 1,1 1,1 1,2 1,2 1,1 1,1 1,2 1,2 1,1 Let’s say that we know the phase of I-1 and I-2: Map distance between disease gene and RFLP marker: 1/8 (100) = 12.5cM

40. 1 2 1,1 2,2 1,2 1,2 1,2 1,2 1,2 1,2 1,2 1,2 Using polymorphic markers to map a human gene If linked, one of the RFLP alleles should segregate with the disease; the other RFLP allele should segregate with WT phenotype. RFLP RFLP genotype: Is the marker linked to the disease? Can’t tell

41. R f r F X r f r f Practice question • In a certain plant species… • flower fragrance (F) is dominant over unscented (f) • blue flower color (B) is dominant over white (b) • rounded leaves (R) is dominant over pointy (r); and • thorny stems (T) is dominant over smooth stems (t). • From the following crosses, can you determine whether the fragrance gene is linked to any of the other genes; if so, at what map distance? The parental and recombinant types are the same! Need to be heterozygous at both loci Bb Ff x bb ff Rr ff x rr Ff Tt Ff x tt ff 270 blue, fragrant 281 blue, non-fragrant 268 white, fragrant 275 white, non-fragrant 219 rounded, fragrant 222 rounded, non-fragrant 209 pointy, fragrant 216 pointy, non-fragrant 333 thorny, fragrant 36 thorny, non-fragrant 39 smooth, fragrant 342 smooth, non-fragrant Can’t tell!

42. probability of observed genotypes if the loci are linked at X cM LOD=X% = log10 probability of observed genotypes if the loci are unlinked LOD score analysis

43. example Recombinants? 1/8*100 = 12.5% Recombination frequency?

44. probability of observed genotypes if the loci are linked at X cM LOD=X% = log10 probability of observed genotypes if the loci are unlinked example

45. probability of observed genotypes if the loci are linked at 12.5 cM LOD=12.5% = log10 probability of observed genotypes if the loci are unlinked example

46. example 0.125/2 = 0.0625 probablity of recombinant (12.5cM) = probablity of non-recombinant = (1-0.125)/2 = 0.4375

47. probability of observed genotypes if the loci are linked at 12.5 cM LOD=12.5% = log10 probability of observed genotypes if the loci are unlinked example 0.125/2 = 0.0625 probablity of recombinant (12.5cM) = probablity of non-recombinant = (1-0.125)/2 = 0.4375

48. example (0.4375)7*(0.0625) LOD=12.5% = log10 probability of observed genotypes if the loci are unlinked i.e. independent assortment 0.125/2 = 0.0625 probablity of recombinant (12.5cM) = probablity of non-recombinant = (1-0.125)/2 = 0.4375

49. example (0.4375)7*(0.0625) LOD=12.5% = = 1.099 log10 (0.25)8 0.125/2 = 0.0625 probablity of recombinant (12.5cM) = probablity of non-recombinant = (1-0.125)/2 = 0.4375

50. +3 and up: significant evidence in favor of linkage Does a LOD score value of 1.099 provide significant evidence in favor of linkage? evidence not significant Below -2: evidence against linkage