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exam3 - sue 2 - You have five T4:11 mutants that will not...

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Unformatted text preview: _ You have five T4 :11 mutants that will not grow on E. coli K. (7L). You mixed tw mutants together in various Combinations (indicated below) and added them to E. s coh‘ K. (1.). The abilityof the mixture to grow and makeplaques on E. cat: K (it) was scored (indicated as a + in the table). ' - (10 points) ‘ 1 ' ' 2 3 4 5 1 -_ . + + ' .‘ ' + 2 - — .+ .- 3 ' - + 4 - ' + '5 (a) What event (recombination or complementation) is being scored in this . . analysis? ' ' .H, “Jr-‘0" cum-4‘?“ (b) How many complementation groups are identified in this analysis? [91 _ . a1 2‘ f 5 ran’ '3 wt g6! - . (0) Which mutants belong to the same complementation groups? \A' R ‘kbhfii. 4" out. Ate”? 2’3”? but“; 'l’o w“ A plasmid carrying genes for kanarnycm—resistance (Kan') and tetracyclme- resistance (Tet‘) is treated with the restriction enzyme Bglfl, which cleaves within the Kan’ gene. The plasmid DNA is ligated with a BgIII digest of yeast DNA and the ligated mixture is used to transform E. coli. ' (10 points) (a) What antibiotic (kanamycin or tetracycline) would you use in the grth medium to ensure that each colony has the plasmid? +L+rk£3thfi¢ (b) How will you identify colonies in which the plasmid contains the yeast DNA msert? 'fw "maid w 40 SH. which coins-'0; h, :n. 1m awrrwm'l' ”If _ . an: KaMMJLm‘ 11: m fl'MSWM Let gnarl-rug a, Khaafic-M, +4.“: in» “M- EJIA- mun rnsb’i’. 1t M3 an. cut-5W! 41 . 4:..‘4 t 4w "fl‘f'én: 2 ...
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