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PHGY 210 (301)
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PHGY 210
John White

January 20, 2014 BIOL-202 Lecture 7 AB/ab vs. A/a;B/b For the A/a;B/b, the A locus and the B locus genes will assort independently For AB/ab, the two genes are same chromosome, and in general may show evidence of linkage (excess of parental gamete types) Need to breed in disease-resistant for that gene -breed back to parent > back cross and BC1 X P1 -only thing left is disease resistant trait -know disease resistant trait is on chromosome > marker assistant back crossing -tells us position of marker in relation to trait Huntington’s disease is caused by too many CAG repeats Co-inheritance of restriction fragment (RFLP) -all affected individuals carry an extra band (piece of DNA co-inherited with this disease) -map to determine if patient has disease or not Centromere mapping with linear tetrads -difficult to map centromeres in humans -octad of four spore pairs = double tetrad -M I segregation pattern: no cross-over (close to centromere) > genes will segregate with the centromere -centromere mapping > cross-over > two genes have not segregated from one another after M1 (alleles together) > M2: segregation of alleles and then mitosis = M II segregation pattern -depends on which sister chromatids are involved in crosss-over, if A/a is in the “top” position = four possibilities of tetrads > four possibilities of octads = each is equally probable (only if cross-over) Centromere mapping with linear tetrads: M1: first two columns M2: 42 patterns -map distanc
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