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It would be great if anyone can help because I am so confused.


komalshaheen

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The researchers who investigated bioluminescence and quorum sensing found that E. coli transformed with a plasmid containing a 9 kb fragment of V. fischeri DNA could glow when the cell population was dense. They mutagenized these E. coli cells and isolated many mutantions that map to the 9 kb fragment and prevented the cells from glowing. They then performed complementation testing with these mutants by transforming E. coli cells simultaneously with two plasmids, each containing the 9 kb fragment with one of these mutations. To ensure the E coli cells were transformed with both plasmids, one of the two plasmids had a gene conferring resistance to ampicillin, while the other plasmid had a gene conferring resistance to tetracycline, and cells were sellected on perti plates that had both antibiotics. Construct a 9 x 9 complementation table for the nine mutations described below, using "+" to indicate that cells would glow and "-" to indicate that cells would remain dark. (you only need to fill in half the table) Mutation 1: Encodes LuxA protein that cannot bind a substrate for the luciferase enzyme. Mutation 2: Encodes a LuxA protein that cannot associate with LuxB protein Mutation 3: Encodes a LuxB protein that cannot associate with LuxA protein Mutation 4: A null mutation in the LuxI gene Mutation 5: Encodes a LuxR promoter that prevents transcription Mutation 6: Encodes a LuxR protein that cannot bind DNA Mutation 7: Encodes a LuxR protein that cannot bind to the autoinducer Mutation 8: A mutation in the LuxICDABE promoter that prevents transcription Mutation 9: A mutation in the LuxICDABE promoter region that blocks binding of the LuxR protein

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