Hi, I would greatly appreciate an answer to this problem. I have been stuck on it for the past few hours and attempt it, however am not sure how to do it. Please help!?
II. 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