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Recitation Week 7

Recitation Week 7 - Recitation 02 23 2011 Testing Linkage...

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Recitation 02 / 23 / 2011
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Testing Linkage Aa•Bb X aa•bb 93 Aabb 108 AaBb 104 aabb 95 aaBb note: nomenclature does not indicate if genes are linked from these test cross ratios can you tell if genes are unlinked? Q: you can’t tell just by looking … you have to use a χ 2 test A:
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The cross should give equal numbers of : (201-200) 2 200 (199-200) 2 200 + = 0.01 1. Test ratios at each locus (203-200) 2 200 (197-200) 2 200 + = 0.09 Aa and aa Bb and bb chance chance (93-100) 2 100 + (108-100) 2 100 + 2. Do the loci assort independently (104-100) 2 100 + (95-100) 2 100 The cross should give AaBb:Aabb:aaBb:aabb in 1:1:1:1 ratio = 1.54 chance ( tests linkage ) Conclusion: A and B are not linked ( rules out lethal alleles etc .)
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Three point testcross AABBCC X aabbcc AaBbCc X aabbcc AaBbCc 189 353 AaBbcc 189 349 AabbCc 189 21 Aabbcc 189 24 aaBbCc 189 21 aaBbcc 189 18 aabbCc 189 365 aabbcc 189 361 E O 1512 progeny
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Three point testcross AABBCC X aabbcc AaBbCc X aabbcc A a B b C c 189 353 A a B b c c 189 349 A a b b C c 189 21 A a b b c c 189 24 a a B b C c 189 21 a a B b c c 189 18 a a b b C c 189 365 a a b b c c 189 361 Look at each pair of loci and determine parental and nonparental Parental: AB AC BC ab ac bc Just look at alleles contributed by this parent Nonparental: Ab & aB : 21+24+18+21 = 84 84/1512 = 5.5% Ac & aC : 349+24+365+21 = 759 759/1512 = 50% Bc & bC : 349+18+365+21 = 753 753/1512
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Three point testcross AABBCC X aabbcc AaBbCc X aabbcc A a B b C c 189 353 A a B b c c 189 349 A a b b C c 189 21 A a b b c c 189 24 a a B b C c 189 21 a a B b c c 189 18 a a b b C c 189 365 a a b b c c 189 361 at each pair of loci and determine parental and nonparental Parental: AB AC BC ab ac bc Nonparental: Ab & aB : 21+24+18+21 = 84 84/1512 = 5.5% Ac & aC : 349+24+365+21 = 759 759/1512 = 50% Bc & bC : 349+18+365+21 = 753 753/1512 = 50% 2. Draw map A B C 5.5 cM
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Three point testcross cont. AAbbcc X aaBBCC AaBbCc X aabbcc A a B b C c 141 A a B b c c 7 A a b b C c 60 A a b b c c 888 a a B b C c 870 a a B b c c 68 a a b b C c 5 a a b b c c 133 note different parental configuration but same fully heterozygous F1 genotype
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Three point testcross cont. AAbbcc X aaBBCC AaBbCc X aabbcc A a B b C c 141 A a B b c c 7 A a b b C c 60 A a b b c c 888 a a B b C c 870 a a B b c c 68 a a b b C c 5 a a b b c c 133 Look at each pair of loci and determine parental and nonparental Parental: Ab Ac bc aB aC BC Nonparental:
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