PS7a-11 - Bio D137 Human and Eukaryotic Genetics Prob Set 7...

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Bio D137 Human and Eukaryotic Genetics J.L.Marsh F'11 Prob Set 7 due Tues, Nov 15 Name/student ID:_______________ 1 1 1. You have recently isolated five new phage that contain DNA from a human genomic DNA library. You isolated these phage using a probe containing DNA from a region of chromosome #2, where your favorite gene resides. Genetic data tells you that this probe detects a region of DNA that is extremely close to your gene. Therefore, you isolated phage that hybridized to this probe. The following gel is an agarose gel of EcoR1 digested phage DNAs after electrophoresis. Lanes containing different phage DNAs are labeled A – E. A B C D E Ladder The rightmost lane is a 1 kb marker. Each band is 1 kb larger than the band below it, starting with 1 kb at the bottom, and ending with 10 kb at the top of the gel. NOTE: The two vector arms, which should appear in each lane as 10 and 9 kb fragments, are not shown on this gel. Darker bands may have multiple fragments. a) From these data, construct a map of the order of EcoRI fragments in the genome showing the overlap of the various phage. If any fragments cannot be definitively ordered, enclose them in parentheses. Answer: 4 points (2 points for diagram showing overlap and 2 points for correct order.) Key: 1kb band in A is darker, suggesting 2 fragments of same size. Treat this as a deletion problem, only here you’re dealing with numbers (sizes) instead of letters. Make a list of what fragments each phase has: A(1,4,1); B (6,3,2); C (1,4,5); D (6,3,1); E (4,3,1). I would start with 2 phage that share a lot of fragments such as B/D, which have 6 and 3 in common and therefore overlap at these sites. Since 6/3 are in common, 2 & 1 are to the opposite sides = 2, (6,3),1. If you add E to those two, E has frags 3 & 1 and a new one #4 so you can order the above to get: E BD: 2,6,3,1,4. Now add phage C to get 2,6,3,1,4,5. Phage A has 2 fragments of the same size so this is a bit more tricky since the order of A could possibly be 1,1,4 or 1,4,1. So we have to propose that ACE = 3,1,4,1,5 so the final order is 2,6,3,1,4,1,5.
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Bio D137 Human and Eukaryotic Genetics J.L.Marsh F'11 Prob Set 7 due Tues, Nov 15 Name/student ID:_______________ 2 2 2] 75% of Burkitt lymphomas are assoiciated with translocations
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This note was uploaded on 12/13/2011 for the course BIOSCI 137 taught by Professor Staff during the Fall '11 term at UC Irvine.

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PS7a-11 - Bio D137 Human and Eukaryotic Genetics Prob Set 7...

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