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Gene mapping

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  • 1. Genetics - Ch4 Gene Linkage & Genetic Mapping
  • 2. Gene Mapping determines order of genes & relative distances between them 1 map unit = 1 cM (centimorgan) A B A B A & B are Cis arrangement (coupled) Trans arrangement (repulsed) 1 locus; 2 loci a b a b
  • 3. Gene Distance
    • recombination frequencies between alleles:
    • determine relative distance between them
    • proportional to their distance apart
    1% recombination = 1 map unit = 1 cM A B A F a b a f 10% recombination 45% recombination
  • 4. % Recombination <50% recombination --> genes linked on same chromosome > 50% recombination --> genes are far apart on chromosome acts like genes are unlinked (indep. assort.)
  • 5. Genetic Distance vs Physical Distance
    • recombination “hot spots” overestimate physical length
    • low rates in heterochromatin and centromeres
    • underestimate actual physical length
    A a f F
  • 6. Genes on Separate Chromosomes + + a b a a b b +a X aa +a, +a, aa, aa 1/2 : 1/2 +b X bb +b, +b, bb, bb 1/2 : 1/2 +a+b : +abb : aa+b : aabb dihybrid pure recessive normal bent albino albino/bent expected 1/4 : 1/4 : 1/4 : 1/4 + = wt (normal for any trait) (A, B) a = albino (recessive mutant) b = bent (recessive mutant) + a + b X a a b b
  • 7. Chi-Square (X 2 )Analysis: “Goodness of Fit” +a+b : +abb : aa+b : aabb normal bent albino albino/bent 1/4 : 1/4 : 1/4 : 1/4 expected proportion observed # 21 26 27 26 = 100 total expected # 1/4x100 1/4x100 1/4x100 1/4x100 25 25 25 25 X 2 = (21-25) 2 + (26-25) 2 + (27-25) 2 + (26-25) 2 25 25 25 25 0.89 = 16 + 1 + 4 + 1 = 22 25 25 25 25 25
  • 8. Chi-Square Interpretation if X 2 < p.05, then observed and expected are not statistically different Therefore, 2 traits are not linked on same chromosomes and independent assortment occurs d.f. (degree of freedom) = # phenotypes - 1 d.f. = 4 - 1 = 3 probability .05 (95% confidence level) of being different from expected w/ d.f. 3 = 7.82 (read off a chart) X 2 = 0.89 0.89 < 7.82, therefore not linked
  • 9. If 2 Genes are linked 250 250 250 250 #1 make a dihybrid, then #2 cross with pure recessive
  • 10. Diploid Mapping wildtype albino/bent bent albino 430 437 65 68 = 1000 total + + a b a a b b + a + b X a a b b To determine distance between 2 genes: total # of recombinants/total # offspring = freq. of recombination freq. x 100 = % recomb. = M.U. = # cM 65 + 68 / 1000 = 0.133 x 100 = 13.3 % or 13.3 cM X