G a and b 1654 sum all the offspring with recombinant

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Unformatted text preview: binant 131 phenotype of a+ b or a b+ (ignore c) 252 =131 + 252 + 241 + 118 + 9 + 9 + 13 +13 = 742 241 = 786 Text 1779 -divide by total number of offspring and 118 multiply by 100 13 = 742 x 100 = 17.7 m.u. a b 4197 = 786 x 100 4197 = 18.7 m.u. group that is missing the double recombinants, are the most distant genes 16 another example – X chromosome of Drosophila 3 possibilties: sc ec cv ec sc cv ec cv sc 17 another example – X chromosome of Drosophila 1. sc and ec Recombinants = sc ec+ and sc+ ec 2. sc anc cv recombinants = sc cv+ and sc+ cv = 163 + 130 + 192+ 148 = 633/3248*100 =19.5 m.u 3. ec and cv recombinants = 18 figure out genotype from F1 female and parentals doing mapping distances and correction. For practice parentals: p+j+r+ and pjr f1: p+j+r+/pjr 1. looking at p and j 52+46+4+2 = 104 /500 x 100 = 20.8 2. looking at p and r 52+46+22+22 = 142/500 = 28.4 3. looking at j and r 22+22+4+2=50/500 = 10 p and r...
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This document was uploaded on 02/25/2014.

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