1 / 38

Chapter 6

Chapter 6. Genetic Recombination in Eukaryotes. 31 January, 2 February, 2005. Linkage and genetic diversity. Overview. In meiosis, recombinant products with new combinations of parental alleles are generated by: independent assortment (segregation) of alleles on nonhomologous chromosomes.

eros
Télécharger la présentation

Chapter 6

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. Chapter 6 Genetic Recombination in Eukaryotes 31 January, 2 February, 2005 Linkage and genetic diversity

  2. Overview • In meiosis, recombinant products with new combinations of parental alleles are generated by: • independent assortment (segregation) of alleles on nonhomologous chromosomes. • crossing-over in premeiotic S between nonsister homologs. • In dihybrid meiosis, 50% recombinants indicates either that genes are on different chromosomes or that they are far apart on the same chromosome. • Recombination frequencies can be used to map gene loci to relative positions; such maps are linear. • Crossing-over involves formation of DNA heteroduplex.

  3. Overview

  4. Recombination Detailed

  5. Independent Assortment

  6. A/A ; B/B  a/a ; b/b A/A ; b/b  a/a ; B/B A/a ; B/b A/a ; B/b ¼ A ; B P ¼ A ; b R ¼ a ; B R ¼ a ; b P ¼ A ; B R ¼ A ; b P ¼ a ; B P ¼ a ; b R Independent assortment (2) • For genes on different (nonhomologous) pairs of chromosomes, recombinant frequency is always 50% 50% recombinants

  7. Testcross of a Dihybrid

  8. Self Cross of a Dihybrid

  9. Independent assortment: multiple loci • Calculations can be made for any gene combination using predicted outcomes at single loci and the product rule P1A/a ; B/b ; C/c ; D/d P2a/a ; B/b ; C/c ; D/D

  10. Crossing-over • Breakage and rejoining of homologous DNA double helices • Occurs only between nonsister chromatids at the same precise place • Visible in diplotene as chiasmata • Occurs between linked loci on same chromosome • cis: recessive alleles on same homolog (AB/ab) • trans: recessive alleles on different homologs (Ab/aB)

  11. Linkage maps # observed 140 50 60 150 • RF is (60+50)/400=27.5%, clearly less than 50% • Map is given by: A B 27.5 m.u.

  12. Trihybrid testcross • Sometimes called three-point testcross • Determines gene order as well as relative gene distances • 8 categories of offspring • for linked genes, significant departure from 1:1:1:1:1:1:1:1 • Works best with large numbers of offspring, as in fungi, Drosophila

  13. Trihybrid testcross example v (vestigial) v+ (long) wings b (black) b+ (gray) body p (purple) p+ (red) eyes v+ v+ . b b . pp x vv. b+ b+ . p+ p+ v/ v+. b/ b+. p/ p+

  14. v/ v+. b/ b+. p/ p+ x v/v.b/b.p/p Progeny: v b+ p+ 580 Parental v+ b p 592 Parental v b p+ 45 v+ b+ p 40 v b p 89 v+ b+ p+ 94 v b+ p 3 v+ b p+ 5

  15. v.b and v+.b+ 45 + 40 + 89 + 94 = 268 / 1448 = .185 v.p and v+.p+ 89 + 94 + 3 + 5 = 191 / 1448 = .132 b.p+ and b+.p 45 + 40 + 3 + 5 = 93 / 1448 = .064 v 13.2 p 6.4 b Why do 13.2 + 6.4 = 19.6 instead of 18.5?

  16. The failure to count double crossover gametes as v.b recombinants results in an underestimation of the v.b distance. A better v.b number would be 268 + 3 + 3 + 5 + 5 = 284 / 1448 = 19.6. Aaah! In general, to minimize the effect of double crossovers, it is necessary to measure a number of small RF distances and sum to larger distances.

  17. Interference • Crossing-over in one region of chromosome sometimes influences crossing-over in an adjacent region • Interference = 1 – (coefficient of coincidence) • Usually, I varies from 0 to 1, but sometimes it is negative, meaning double crossing-over is enhanced

  18. In the previous example, expect (.132 x .064 = .0084)(1448) = 12 double crossovers, but saw 8, so I = 1-(8/12) = 4/12 = 1/3. In regions where double crossovers are forbidden, observed = 0, so I = 1. What do you think negative interference means?

  19. Genetic maps • Useful in understanding and experimenting with the genome of organisms • Available for many organisms in the literature and at Web sites • Maps based on RF are supplemented with maps based on molecular markers, segments of chromosomes with different nucleotide sequences

  20. Chi-squaretest • Statistical analysis of goodness of fit between observed data and expected outcome (null hypothesis) • Calculates the probability of chance deviations from expectation if hypothesis is true • 5% cutoff for rejecting hypothesis • may therefore reject true hypothesis • statistical tests never provide certainty, merely probability

  21. Chi-square application to linkage • Null hypothesis for linkage analysis • based on independent assortment, i.e., no linkage • no precise prediction for linked genes in absence of map • for all classes • Calculated from actual observed (O) and expected (E) numbers, not percentages

  22. What is E? For the AbBb testcross example from the text, E is not just 125 for each of the genotypes, as there may be allele effects on viability. Instead, get expected values from the data: A a B 142 112 254 b 113 133 246 255 245 So, E (AB) would be (255/500) x (254/500) x 500 = 129.54 Other E are found similarly, and Chi-square is 4.97

  23. Mechanism of meiotic crossing-over • Exact mechanism with no gain or loss of genetic material • Current model: heteroduplex DNA • hybrid DNA molecule of single strand from each of two nonsister chromatids • heteroduplex resolved by DNA repair mechanisms • May result in aberrant ratios in systems that allow their detection

  24. Recombination within a gene • Recombination between alleles at a single locus • In diploid heterozygous for mutant alleles of the same gene, recombination can generate wild-type and double mutant alleles • Rare event, 10-3 to 10-6, but in systems with large number of offspring, recombination can be used to map mutations within a gene a1/a2 a+ and a1,2

  25. Assignment: Concept map, solved problems 1 and 2, all other basic and challenging problems Continue with PubMed section of the Web tutorial.

More Related