Genome 371 Lecture 15 - Genome 371 Lecture 15 1/26/11 4:17...

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1/26/11 4:17 PM Genome 371 Lecture 15 Page 1 of 14 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_15.html Lecture 15 Mapping genes in humans-2 22 Nov 2010 Lecture 15 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 4:17 PM Genome 371 Lecture 15 Page 2 of 14 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_15.html Slide 3 Slide 4 Status check: Can you interpret the restriction map and why we get the pattern shown on the gel? If not--ask for help! Slide 5 Status check: Can you figure out how we know the indicated parental types for individual II-1? If not ask yourself: What alleles did II-1 get from her mom? From her dad? Since the father (II-2) is homozygous A cp A cp all his gametes (the sperm) will have the genotype A cp . So, in the kids' genotypes listed below the pedigree, we can ignore the midle two rows of alleles (they represent the cp and the A from the dad) and look at the top and bottom rows of alleles to see what came from the mom (i.e., the egg genotype).
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1/26/11 4:17 PM Genome 371 Lecture 15 Page 3 of 14 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_15.html Slide 6 If the two loci (the polymorphic locus and the curvy pinky locus) are assorting independently, we expect all four possible egg genotypes to be equally frequent (i.e., 1/4 each). The observed proportions above look skewed from the 1/4-each predicted for independent assortment… but this sample size is really too small to make a reasonable judgment -- chi- square analysis (based on a null hypothesis of independent assortment) gives a P value just above 0.05. If there were more offspring we'd have more data and might be able to make a firm conclusion. .. but of course that's not an option with human pedigrees, we have to make do with the information we have. Slides 7-9 One strategy (which was developed as theory in the 1950s but didn't become practical until computers became cheap and powerful) to deal with the limited data in human pedigrees is to play the odds and ask: what would be more likely to give the result that I see, linkage between the two loci, or non-linkage? To make that comparison, we would have to predict the probability that this particular set of parental genotypes (II-1 and II-2 in this pedigree) would give rise to this particular set of offspring genotypes * (III-1 through III-12) for two kinds of situations -- one, if the two loci are unlinked, and two, if the loci are linked. It's easy to make that prediction if the loci are unlinked -- we know what the probability of each gamete genotype would be. But for linkage? How can we predict the outcome if we don't know a map distance? The solution is just to be systematic and make
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Genome 371 Lecture 15 - Genome 371 Lecture 15 1/26/11 4:17...

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