LIFESCI 4 Lecture Notes - Lecture 6: Chromosome, Centimorgan
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Lifesci 4 - lecture 6 - linkage and 3 factor cross. You do a test cross and get the following progeny: How many m. u separate them: 15 m. u, 20 m. u, 30 m. u, 50 m. u, they are not linked. For f1, the genotype is aabb x aabb = The most prominent progeny is 33 (a_bb) and 37 (aab_), which means the parental alleles would have to be ab and ab, giving instead of. , so recombinant alleles are (a_b_) and (aabb) In this example genes a and b are not linked, they map to separate chromosomes, but the student doing the experiment does not know that. What is the map distance the student will calculate between the two genes: 10 m. u, 20 m. u, 25 m. u, 50 m. u, none of the above. The ratio of f1 phenotype is 1:1:1:1 for ab : ab : ab : ab. 10 m. u: 2. 5, 5, 7. 5, 10, 20%
