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Genome 371 Lecture 10 - Genome 371 Lecture 10 4:16 PM Home...

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1/26/11 4:16 PM Genome 371 Lecture 10 Page 1 of 13 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_10.html Lecture 10 DNA cloning 1 Nov 2010 Lecture 10 handout (pdf) Subscribe to podcast: Slide 1 Slide 2 Home Course mechanics Help hours Calendar Syllabus Lectures, Podcasts Quiz Sections Practice problems Exams GoPost Send email to class Useful links The Gradiator
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1/26/11 4:16 PM Genome 371 Lecture 10 Page 2 of 13 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_10.html Slide 3 Slide 4 Slide 5 When a mutation is behaving in a dominant fashion, the complementation test cannot be applied and becomes meaningless. So, we cannot conclude anything about how mutant 7 fits into the other complementation groups -- it could belong to one of the groups we've already discovered, or it could be in a new group altogether, we just can't tell. Slide 6 The double mutant ( aabb ) doesn't produce functional protein from either gene, and the phenotype of that strain is yellow flowers. So, if the pathway never gets off the ground , the flowers are yellow; so yellow must be the starting point. If only functional B is present, the phenotype is red -- at least part of the pathway must be functional because yellow has been changed to red. In contrast, when only A is present ( AAbb ), the phenotype is still yellow; the pathway still doesn't get off the ground. Those results tell us that B must be required before A in the pathway, and that red color is an intermediate: B A Yellow ---> Red ---> Purple
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1/26/11 4:16 PM Genome 371 Lecture 10 Page 3 of 13 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_10.html Continuing with this problem... if two AaBb plants were crossed to each other, what progeny phenotypes and proportions would you expect, assuming that the genes assort independently? Slide 7 To be sure you understand what is going on here, write out the full genotypes of each individual as best as you can. Look back through the various genetic pathways we have talked about until now and try to identify other cases of epistasis. Slide 8 Slide 9 Here, if the MC1-R gene is defective and MC1-R receptor is not made, it doesn't matter whether Agouti protein is made or not (if MC1-R is not made, there is nothing there for Agouti to act upon and so Agouti becomes irrelevant). Therefore, here MC1-R is epistatic to A or a . Notice that you cannot just blindly decide that the "upstream" gene will always be epistatic over the "downstream" gene. When both genes are acting positively to bring about an outcome (e.g., B and A in the flower example above) then the "upstream" gene is the epistatic one because if it is missing then the downstream gene is irrelevant. However, when there is a negative regulatory interaction -- one gene does something and another gene inhibits the first gene's effect -- then it's the "target" gene that is epistatic. The MC1-R gene promotes eumelanosome maturation, but its function can be inhibited by Agouti. So the it's the MC1-R gene -- that target of Agouti action -- that is epistatic, because without the target, the inhibitor (agouti) is irrelevant.
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