1 / 19

Models of selection

Models of selection. How can we predict the speed of evolutionary change due to natural selection? - genotype and allele frequencies null models (no selection). Photo: C. Heiser Capsicum annuum. Foré genetics. A population living in Papua New Guinea

ace
Télécharger la présentation

Models of selection

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. Models of selection How can we predict the speed of evolutionary change due to natural selection? - genotype and allele frequencies null models (no selection) Photo: C. Heiser Capsicum annuum

  2. Foré genetics A population living in Papua New Guinea Young women and older women surveyed for genotypes at one locus (2 alleles: M or V). Genotype MM MV VV juvenile 31 72 37 adult 4 23 3

  3. Genotype frequencies • Consider a diploid population with two alleles at one locus, A and a • [NOTE: for this class, large letters DO NOT indicate dominance, just another allele!] • At time t, • C(t) = frequency of AA individuals • H(t) = frequency of Aa individuals • S(t) = frequency of aa individuals • C(t) + H(t) + S(t) = 1 • (sum of the frequencies of all possibilities will always equal 1)

  4. Genotype frequencies • The frequency of the A allele in this population, p(t), is given by the frequency of AA plus half of the frequency of Aa (since only half of the alleles in the heterozygote are A): • Similarly, the frequency of the a allele in this population, q(t) is given by the frequency of aa plus half of the frequency of Aa. • p(t) + q(t) = C(t) + H(t) + S(t) = 1 • The sum of the frequencies of the two possible alleles equals 1.

  5. Find the genotype and allele frequencies for this example. Number AA = Frequency = Number Aa = Frequency = Number aa = Frequency =  Total = 20 Total frequency = 1 Number A = Frequency = Number a = Frequency = Total = 40 Total frequency = 1

  6. What happens to these frequencies over one generation? We will assume none of the forces of evolution are operating:

  7. Random union Meiosis Model without selection Haploid stage (gametes) frequency: f(A) = p(t) = C(t) + 1/2 H(t) f(a) = q(t) = 1 - p(t) Meiosis: AA adults produce 100% A gametes Aa adults produce 50% A gametes aa adults produce 0% A gametes Diploid stage frequency: AA Aa aa C(t) H(t) S(t)

  8. Random union Meiosis Model without selection Haploid stage (gametes) frequency: f(A) = p(t) f(a) = q(t) If adults are equally fertile, then gamete frequencies equal the allele frequencies in the adults that produce them. Diploid stage frequency: A a p(t) q(t)

  9. Egg is drawn Sperm is drawn Offspring Random union Meiosis Allele A p Allele A p q Allele a Allele A p q Allele a q Allele a Random mating AA (prob p2) Aa (prob pq) aA (prob qp) aa (prob q2)

  10. Egg is drawn Sperm is drawn Offspring Random union Meiosis Allele A p Allele A p q Allele a Allele A p q Allele a q Allele a Random mating AA (prob p2) Aa (prob pq) aA (prob qp) At time t+1 C(t+1) = freq(AA) =p2 H(t+1) = freq(Aa) = 2pq S(t+1) = freq(aa) = q2 aa (prob q2)

  11. Effect of random mating At time t+1 C(t+1) = freq(AA) =p2 H(t+1) = freq(Aa) = 2pq S(t+1) = freq(aa) = q2

  12. Allele frequencies in next generation At time t+1 C(t+1) = freq(AA) =p2 H(t+1) = freq(Aa) = 2pq S(t+1) = freq(aa) = q2 so, p(t+1) = C(t+1) + 1/2 H(t+1) = p2 + 2pq / 2 = p(p + q) = p Likewise, q(t+1) = q

  13. Allele frequencies do not change in the absence of selection and with random mating. • What if one allele were dominant?

  14. Random union Meiosis Adults Selection Using Hardy-Weinberg to detect selection frequency: f(A) = p(t) f(a) = q(t) Selection affects aa genotype f(AA) = p2 f(Aa) = 2pq f(aa) < q2

  15. H-W in practice • In practice, we do not know p and q • We can calculate p and q for the adults. • Call these p' and q' (since they are after selection) • p' = f(AA) + 1/2 f(Aa) • q' = f(aa) + 1/2 f(Aa) • After selection p ≠ p‘ and q ≠ q'. • and f(AA) ≠ p'2 • f(Aa) ≠ 2p'q' • f(aa) ≠ q'2 We can use this to detect natural selection

  16. H-W example p = f(A) = 0.5, q = f(a) = 0.5 If we have 100 zygotes: p2 * 100 = 25 AA; f(AA) = 0.25 2pq * 100 = 50 Aa; f(Aa) = 0.5 q2 * 100 = 25 aa; f(aa) = 0.25 10 aa die. Adult population: 25 AA; f’(AA) = 25 / 90 = 0.28 50 Aa; f’(Aa) = 50 / 90 = 0.56 15 aa; f’(aa) = 15 / 90 = 0.17 p’ = f’(A) = 0.25 + (0.56 / 2) = 0.56 q’ = 0.17 + (0.56 / 2) = 0.45

  17. H-W example p = f(A) = 0.5, q = f(a) = 0.5 If we have 100 zygotes: p2 * 100 = 25 AA; f(AA) = 0.25 2pq * 100 = 50 Aa; f(Aa) = 0.5 q2 * 100 = 25 aa; f(aa) = 0.25 20 aa die. Adult population: 25 AA; f’(AA) = 25 / 80 = 0.313 50 Aa; f’(Aa) = 50 / 80 = 0.625 5 aa; f’(aa) = 15 / 80 = 0.063 p’ = f’(A) = 0.278 + (0.556 / 2) = 0.625 q’ = 0.167 + (0.556 / 2) = 0.375 Expected numbers of each genotype if population is in H-W? exp(AA) = p’2 * 90 = 31.3 exp(Aa) = 2p’q’ * 90 = 37.5 exp(aa) = 2q’2 * 90 = 11.3 Use 2 test to check: 2 =  (obs – exp)2 / exp = (25 – 31.3)2 / 27.8 + (50 – 37.5)2 / 44.5 + (15 – 11.3)2 / 18.2 = Critical value: 3.84

  18. H-W problem

  19. References, readings, and study questions See sections 6.1 to 6.3 in the text, (3rd edition: 5.1 to 5.3) and answer question 1.

More Related