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Natural Selection Can Create Linkage Disequilibrium

Natural Selection Can Create Linkage Disequilibrium. Natural Selection Can Create Linkage Disequilibrium. Detecting Postive Selection by Linkage Disequilibrium. Sabeti, et al . 2002. Admixture Can Create Linkage Disequilibrium. Recombination Reduces Linkage Disequilibrium. Chao’s Paper.

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Natural Selection Can Create Linkage Disequilibrium

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  1. Natural Selection Can CreateLinkage Disequilibrium

  2. Natural Selection Can CreateLinkage Disequilibrium

  3. Detecting Postive Selection byLinkage Disequilibrium Sabeti, et al. 2002

  4. Admixture Can CreateLinkage Disequilibrium

  5. Recombination ReducesLinkage Disequilibrium

  6. Chao’s Paper

  7. Bacteriophage 6 dsRNA

  8. dsRNA dsRNA dsRNA Müller’s Ratchet The fate of asexually reproducing organisms • Assume a single bacteriophage per bacterium = asexual reproduction

  9. dsRNA Experimental Müller’s Ratchet with 6 0.22 m filter Pseudomonas phaseolicola 40X

  10. dsRNA dsRNA dsRNA dsRNA Paired Growth and Fitness of 6 marked parent Müller parent wt parent marked parent Pseudomonas phaseolicola Pseudomonas alcaligenes Pseudomonas phaseolicola Pseudomonas phaseolicola Wt = Rt/R0 I.e., W37A = (150:200)/(200:200) = 0.75

  11. Fitness Results of 6

  12. dsRNA dsRNA Experimental Sex with 6 0.22 m filter Pseudomonas phaseolicola 40X

  13. Fitness of Improvement of selfed and hybrids

  14. Disequilibrium Disrupts HWE model

  15. Mendelian Genetics of Quantitative Traits

  16. Quantitative Results for Nicotiana longiflora

  17. Diopsids - Stalk-eyed Flies (Cyrtodiopsis dalmanni) • Wilkinson, Amitin & Johns • Integr. Comp. Biol., 2005

  18. QTL mapping: Mimulus phylogeny

  19. QTL mapping: Mimulus hybrids

  20. QTL mapping: predictions

  21. Selection gradients

  22. Skypilots

  23. Heritability of flower size 1/2h2=0.5

  24. Relative fitness

  25. Selection gradient = slope Selection gradient = 0.13 S (selection differential) = selection gradient*variance S = 0.13*5.66 mm = 0.74 mm S = 0.74 mm/14.2 mm mean flower size = 0.05

  26. Expected R = h2S = 1 x 0.05 = 0.05 Observed = 9% larger

  27. Heritability of abdominal bristles in D. melanogaster Clayton, et al. 1957

  28. Flies with >22 bristles bred to create generation 2 • Mean bristle number G1 = 19.3 • Mean bristle number of selected G1 = 22.7 • Mean bristle number G2 = 20.1

  29. G1 n = 115; mean bristle number = = 2221/115 = 19.3 G1 selected n = 22; mean bristles = 22.7 G2 n = 101; mean bristles = 20.1

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