Question

The correct sequence of events during bacteriophage infection of a bacterial cell is

• A. landing → attachment → tail contraction → penetration and unplugging → DNA ejection
• B. attachment → landing → penetration and unplugging → tail contraction → DNA ejection
• C. landing → tail contraction → attachment → DNA ejection → penetration and unplugging
• D. attachment → tail contraction → landing → penetration and unplugging → DNA ejection

Hint:

- Remember phage infection sequence: Landing → Attachment → Tail contraction → Penetration → DNA injection.
- "Tail contraction" always occurs after firm attachment.

Answer 1

The Correct Option is A

Step 1: A bacteriophage first lands on the bacterial cell surface. This is called \textit{landing}.
Step 2: The phage attaches firmly to receptor sites on the bacterial surface (attachment).
Step 3: Once attached, the phage tail contracts, driving the tail tube into the bacterial cell wall (tail contraction).
Step 4: This is followed by penetration and unplugging of the base plate, creating a channel for DNA passage.
Step 5: Finally, the viral DNA is ejected into the bacterial cytoplasm (DNA ejection).
Final Answer: The correct sequence is option (A).
\[\boxed{\text{Correct Answer: (A)}}\]