Hom05-answers - HW5 BSCI410 Fall 2010 1. 2-72.8 cn is at...

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HW5 BSCI410 Fall 2010 1. 2-72.8 cn is at 2-57.5 and there are 85 recombinants between cn and hjm . c is at 2-75.5 and there are 15 recombinants between hjm and c . 57.5 + (85/100)(75.5-57.5) = 2-72.8 It's the cn c chromosome that is wild-type and the + + chromosome that is mutant ( cn c / hjm ), so the first line of the table lists 85 cn hjm + and the second lists 15 + hjm c recombinants. There are more recombinants between cn and hjm , so hjm lies closer to c . 2. 51, probably closest to 51C (answers from 50A to 51F were accepted) The following are relevant points of correlation: gene name genetic map cytology physical map mam 2-70.3 50C23-50D3 2R:9,878,686. .9,947,861 L 2-72.0 51A4 2R:10,368,708. .10,384,817 Asx 2-70 51A5-51A6 2R:10,391,514. .10,399,951 tra2 2-70 51B6 2R:10,489,509. .10,491,857 kn 2-72.3 51C2-51C3 2R:10,660,152. .10,694,566 Hex-C 51E7 2R:11,106,342. .11,107,917 scb 2-73 51E10-51E11 2R:11,136,290. .11,146,003 c 2-75.5 52D1-52D7 3. mam (50C-50C, 2-70.3) looks promising since it has previously been reported to be a negative regulator of neuroblast proliferation. Its lethal phenotype is that described for hkm . At 70.3, we would have expected 100 x (75.5-70.3)/(75.5-57.5) = 18 or 19 recombinants. The observed number of 15 is acceptably close.
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Hom05-answers - HW5 BSCI410 Fall 2010 1. 2-72.8 cn is at...

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