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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.

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## Chapter 6

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**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. • 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.**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**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**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)**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.**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**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+**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**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?**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.**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**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?**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**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**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**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**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**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**Assignment: Concept map, solved problems 1 and 2, all**other basic and challenging problems Continue with PubMed section of the Web tutorial.

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