Genome 371 Lecture 18 - Genome 371 Lecture 18 1/26/11 4:18...

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1/26/11 4:18 PM Genome 371 Lecture 18 Page 1 of 18 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_18.html Lecture 18 Genetic analysis of a human disease 6 Dec 2010 Lecture 18 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 4:18 PM Genome 371 Lecture 18 Page 2 of 18 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_18.html Slide 3 Slide 4 To get around the issue of repeated sequences, one can use a "top-down" sequencing approach. In this approach, a genome is first cloned as a DNA library of large inserts, the genomic locations of the inserts is determined, and then the inserts are sequenced. Slide 5 Slide 6 A BAC is just a specialized vector for E. coli that it is capable of carrying large pieces of DNA as insert. (A typical plasmid of the sort we have seen previously can carry on the order of maybe 10-20 kb of insert DNA -- much smaller than what can be carried in a BAC.) A YAC is a vector for carrying even larger inserts, except that YACs are hosted by yeast cells, not bacteria. Also, YACs are linear, not circular.
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1/26/11 4:18 PM Genome 371 Lecture 18 Page 3 of 18 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_18.html Slide 7 STSs are unique locations in the genome that can be detected in a PCR assay. For example, suppose one does a PCR reaction with STS #24 (from the picture) using a BAC clone from the BAC library as template for the PCR reaction. If the PCR reaction results in a PCR product, the BAC must contain the sequence corresponding to STS #24. So now we know something about the insert that is in that BAC -- we know that it contains that portion of the genome that has STS 24. If one BAC tests positive for STSs 24 and 62, and another BAC is positive for 17 and 62, we can then deduce that the two BACs have inserts that overlap -- they have STS62 in common. Slide 8 Slide 9 Thus, by virtue of being unique in the genome, STSs serve as landmarks -- if a BAC tests positive for an STS we immediately know which portion of the source genome that BAC's insert came from.
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1/26/11 4:18 PM Genome 371 Lecture 18 Page 4 of 18 http://courses.washington.edu/au371mkr/resources/lecture_notes/lecture_18.html Slide 10 Ideally, the BAC library will contain lots of overlapping inserts. By doing PCR analysis (or restriction digest analysis), one can deduce which BAC overlaps with which. Based on that information, one can then figure out which minimum set of BACs spans a whole chromosome (and ultimately, the whole genome). Each BAC in this minimum "tiling path" is then broken up and sequenced. One key point to note is that since each BAC is tested for STSs, we always know which portion of the genome we are working with. Slide 11
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This note was uploaded on 03/31/2011 for the course GENOME 371 taught by Professor Unsure during the Spring '03 term at University of Washington.

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Genome 371 Lecture 18 - Genome 371 Lecture 18 1/26/11 4:18...

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