1 / 40

Given genotype frequencies, calculate allele frequencies in a gene pool !

Given genotype frequencies, calculate allele frequencies in a gene pool !. Alleles = A, a Genotypes = AA, Aa, aa Frequency of allele A: f (A) = f (AA) + 1/2 f (Aa) Frequency of allele a: f (a) = f (aa) + 1/2 f (Aa). f(A) + f(a) = 1.0 p + q = 1.0 (allele frequencies)

annis
Download Presentation

Given genotype frequencies, calculate allele frequencies in a gene pool !

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Given genotype frequencies, calculate allele frequencies in a gene pool ! Alleles = A, a Genotypes = AA, Aa, aa Frequency of allele A: f (A) = f (AA) + 1/2 f (Aa) Frequency of allele a: f (a) = f (aa) + 1/2 f (Aa) f(A) + f(a) = 1.0 p + q = 1.0 (allele frequencies) p = 1 - q or, q = 1 - p p2 + 2pq + q2 = 1 f [AA] + 2 [f(Aa)] + f [aa] = 1 A a A AA Aa Aa a aa 26a

  2. Hardy-Weinberg Equilibrium Parental generation: 2 alleles, r and R f (R) + f (r) = 1.0 p + q = 1.0 p = 0.1, q = 0.9 In the next generation (F1): p2 + 2pq + q2 = 1 predicts allele freqs. F1 genotypeGenotypeAllele freq. p2 (.01) RR p = 0.01+.18/2= .1 q2 (.81)rr q = 0.81+.18/2 = .9 2pq (.18) Rr 27a-1

  3. Hardy-Weinberg Equilibrium Parental allele frequencies p and q predict F1 generation genotype frequencies, by the formula p2 + 2pq + q2 = 1 Note: parental generation genotype frequencies do NOT predict F1 generation genotype frequencies!! 27a-2

  4. Hardy-Weinberg Equilibrium Conclusions: 1)Allele frequencies are conserved (i.e., the same) from one generation to the next. 2) genotype frequency reaches Hardy-Weinberg equilibrium in one generation 27a-2

  5. Hardy-Weinberg’s caveats: But • Hardy-Weinberg Law: allele frequencies in a population • remain constant from generation to generation ….. • IF random mating • IF all genotypes are equally viable • IF not disturbed by mutation, selection or whatever Only Bottom line: Only in an IDEAL population is genetic diversity conserved forever.

  6. Sex = Random sampling of a gene pool Population of 10 individuals (N = 10) Phenotypes RedWhite Genotypes RR,Rr, rr Allele frequencies R = 0.6 r = 0.4 Parental gene pool 10 genotypes 20 alleles Parental gametes Probability of F1 r = .4; R =.6 R r r R R R R r r R r r R R R r R r R R 25A -1

  7. F1 genotypes and phenotypes Genotype Frequency Phenotype Frequency Rr or rR .48 R_ .48+.36=.84 rr .16 rr .16 RR .36 ---------------  = 1 -------------  = 1 25A -2

  8. If allele frequencies are P and Q in the parental generation, how do we calculate what they will be in the F1 generation?

  9. If allele frequencies are P and Q in the parental generation, how do we calculate what they will be in the F1 generation? Genotype frequencies in F1 are calculated by: p2 + 2pq + q2 = 1 From which we can calculate p and q for F1

  10. If the fraction of the population with allele “A” at a given locus is .7, and the fraction of the population with “a” at the locus is .3, what will be the expected genotype frequencies in F1, , the next generation?

  11. If the fraction of the population with allele “A” at a given locus is .7, and the fraction of the population with “a” at the locus is .3, what will be the expected genotype frequencies in the F1? Allele frequencies in P (parental generation): A = .7 = p a = .3 = q Expected genotypes and their frequencies in F1: AA = p2 = .49 aa = q2 = .09 Aa = 2pq = .42

  12. What will be the expected phenotypes and their ratios in this example?

  13. What will be the expected phenotypes and their ratios in this example? Allele frequencies in P (parental generation): A = .7 = p a = .3 = q Expected genotypes in F1: AA = p2 = .49 aa = q2 = .09 Aa = 2pq = .42 Expected phenotypes in F1: A‑ = .49 + .42 = .91 aa = .09

  14. Heterozygosity defined H = % heterozygous genotypes for a particular locus = % heterozygous individuals for a particular locus = probability that a given individual randomly selected from the population will be heterozygous at a given locus 29f

  15. H = ? Aa aa aa AA aa aA aa AA aA AA aa aa AA aa aA H = 4/15

  16. Heterozygosity defined H (“H-bar”) = average heterozygosity for all loci in a population. H estimated = % heterozygous loci those examined H = 2pq 29f

  17. Calculating H (assuming simple dominance and Hardy-Weinberg Eq.) calculate H if q2 = 0.09 f (a) = 0.3 = q q2 = 0.09 f (A) = 0.7 = p p2 = 0.49 2 pq = 0.42 H = 2pq = 0.42 (for only 2 alleles) 29a - 1

  18. Calculating H (but if……) Codominance Genotype Phenotype N AA Red 50 Aa Pink 22 aa White 10 total = 82 H = 22/82 (don’t need Hardy - Weinberg) 29a - 1

  19. Calculating H for 3 alleles: p, q, r p q p = 0.5 q = 0.4 r = 0.1 H = 2pq + 2qr + 2pr = .40 + .08 + .10 = .58 r pp pq p pr qq q pq qr r pr qr rr 29a

  20. 1% of golden lion tamarins have diaphragmatic hernias, a condition expressed only in the homozygous recessive genotype. Calculate the number of heterozygous individuals in the wild population (N = 508). Assume Hardy-Weinberg equilibrium and simple dominance. Genotypes: AA Aa aa F1 generation p2 = ? = f (AA) 2pq = ? = f (Aa) = H q2 = .01 = f (aa) 29ez - 1

  21. q2 = .01 q = = 0.1 p + q = 1 so p = 1 - q p = 1 - .1 = .9 H = 2pq = 2 x .9 x .1 = 0.18 Nheterozygous = .18 x 508 = 91 29ez - 2

  22. H and P H = heterozygosity = the percent of heterozygous genotypes in the population for that locus H = 2pq (for 2 allele case) H = 2pq + 2qr + 2 pr (for 3 allele case) P = allelic diversity = percent polymorphism = percent of loci for which alternative alleles exist in the population 29A

  23. Gene pools Population 1 N = 8 (7 homozygous) 2N = 16 alleles f (blue) = 1/16 f (red) = 15/16 Population 2 N = 8 2N = 16 alleles f (blue) = 16/16 Polymorphic locus Monomorphic locus typical population P = approximately 0.25 individual H = approximately 0.07 5f

  24. The relationship of P to H Possible alleles = a, b, c aa bb ab ab cc ac ac ac H = 1/4 H = 4/4 N alleles = 3 N alleles = 3 Conclusion: H is not sensitive to the number of different alleles for the locus. 30f

  25. Population A, locus X alleles frequency a .5 b .5 H = 2pq = .50 P is low Population B, locus X alleles frequency a .7 b .05 c .05 d .05 e .05 f .05 g .05 H = .495, P is high The relationship of H to P 30A

  26. Uses of molecular genetics* in conservation 1) Parentage and kinship 2) Within-population genetic variability 3) Population structure and intraspecific phylogeny 4) Species boundaries, hybridization phenomena, and forensics 5) Species’ phylogenies and macroevolution *e.g., electrophoresis protein sequencing DNA fingerprinting immunological techniques 15 -2

  27. From Encarta

  28. Western Pyrénées National Park, France From Encarta

  29. Effective population size N = population size = total number of individuals Ne = effective population size = ideal population size that would have a rate of decrease in H equal to that of the actual population (N) number of individuals contributing gametes to the next generation 32A-1

  30. Effective population size Predictable loss of heterozygosity (H) in each generation for non-ideal populations GLT Ne = .32 N; N = 100, Ne = 32 Loss of H(N) = loss of H (Ne) If Ne/N 1, then rate of loss of H is minimum. The larger the Ne, the lower the rate of loss of H. 1 Rate of loss of H defined: 2Ne per generation 32b

  31. Examples of effective population size Taxon Ne Drosophila .48 to .71 N Humans .69 to .95 N a snail species .75 N plants lower golden lion tamarins .32 N (94 of 290) 32A-2

  32. Assumptions of an ideal population • Infinitely large population • random mating • no mutation • no selection • no migration 31a -1

  33. 5 causes of microevolution 1) genetic drift - stochastic variation in inheritance Expected F2: 9 - 3 - 3 - 1 Observed F2: 9 - 3 - 2.8 - 1.2 2) Assortative (nonrandom) mating 3) Mutation 4) Natural selection 5) Migration (gene flow) Random deviation 31a-2

  34. Sampling Error F1 allele frequencies = Parental allele frequencies Caused by, for example: Behavioral traits producing assortative mating Genetic stochasticity Results in Genetic Drift = random deviation from expected allele frequencies 34A-2

  35. Fixation of alleles Fn fixed Parental generation for many populations A = 1.0 a = 0 lost A = .5 a = .5 Genetic drift p = 1.0 lost A = 0 a = 1.0 p = q = .5 fragmentation fixed q = 1.0 time 34A-1

  36. What is a formula for calculating the effect of unequal numbers of males and females (non-random breeding)on Ne? Ne = 4 MF M = # of breeding males M + F F = # of breeding females Population A Population B M = 50 M = 10 F = 50 F = 90 N = 100 Ne = 4 x 50 x 50 = 4 x 10 x 90 50 + 50 10 + 90 = 100 = 36 10f

  37. The effect of non-random mating on H Given 2 cases, with N = 150 and Ne = 100 (population A) Ne = 36 (population B) Ht=1 = 1 - 1 Ht = the proportion of heterozygosity 2 Ne remaining in the next (t=1) generation Population A: Ht = 1 - 1 = 1 - .005 = .995 2 x 100 Population B: Ht = 1 - 1 = 1 - .014 = .986 2 x 36 * % H remaining after t=1 generations 36A-1

  38. Generalized equation: Ht = H0 1 - 1 t t = # of generations later 2Ne H0 = original heterozygosity What is H after 5 more generations? Population A: H5 = H0 (.995) 5 = .995 (.995)5 = .970 Population B: H5 = H0 (.986) 5 = .995 (.986)5 = .919 * 36A-2

  39. Formulae for calculating H and Ne 1 = proportion of H0 lost at each generation 2Ne 1 - 1 = proportion of H0 remaining after the first generation 2 Ne Ht = H0 1 - 1 t = the absolute amount of H0 remaining after 2Ne t generations Ne = 4 MF 1) unequal sex ratios or M + F 2) nonrandom breeding decrease Ne 37A

  40. Mutation Nondisjunctive point mutations over short term: not important in changing allele frequencies f (A1) = 0.5 mutation rate A1 --> A2 = 1 105 over 2000 generations, f (A1) = 0.49 If f (A2) increases rapidly, selection must be involved Long-term, over evolutionary time mutation is critical - providing raw material for natural selection Mutation rate is independent of H, P, Ne but mutation can increase H and increase P 36A1

More Related