BIOL 2040 Lecture Notes - Lecture 18: Autosome, Centimorgan

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19 Apr 2016
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In order to phenotypically observe and detect the trait in progeny, always cross with other parent who is homozygous recessive (like a test cross**ensures the dominant allele stands out) 4 phenotypic outcomes: 150 a b a b (non recombinant); 120 a b a b (non recombinant) & 40 a b a b (recombinant); 35 a b a b (recombinant); therefore rf = 75/345 = 21. 7% 75 recombinant gametes therefore 150 chromosomes are actually involved! X chromosome = 150mbp long; therefore probability of crossing over = 1/150 million. **probability of cross over is determined by distance between genes; in theory genes can be so far apart that they behave as if they cross over at a max rf = 50% ** two phenotypic outcomes are indistiuguishible from each other when comparing whether they are on the same autosome, or just really far apart (linked vs. not linked) 1:1 f1 and 1:2:1 phenotypic f2 ratio is characteristic of complete linkage.

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