Final answer:
The incorrect option regarding the mechanism that generates 3'-protruding ends for strand invasion is 'RecBCD binds 3' to the Chi site at a double-strand break and unwinds the duplex.' RecBCD unwinds DNA until it recognizes the Chi site and then preferentially cleaves one strand, but option d incorrectly suggests something else in the process.
Step-by-step explanation:
The series of events that do NOT generate 3'-protruding ends for strand invasion include option d, which states 'RecBCD binds 3' to the Chi site at a double-strand break and unwinds the duplex.' This statement is incorrect because RecBCD actually binds at a double-stranded break and unwinds the DNA duplex, cleaving both strands until it recognizes the Chi site. At the Chi site, RecBCD's nuclease activity is modulated, such that it cleaves preferentially the 5' strand, leaving a 3' overhang to which RecA can bind.
After the Chi site is recognized, the nuclease arm of RecB acts as an endonuclease (option a). Chi site recognition by RecC causes loss of helicase activity of the RecD subunit (option c), and the RecB nuclease arm degrades both strands (option f). On the other hand, RecBC does not cleave both strands 4-6 bp on the 5'-side of the Chi site (option b) and RecC's role in recognizing the Chi sequence (option f) does not directly generate 3'-protruding ends either.