Question

A bacterial cell exposed to UV radiation shows an increased rate of C-T transitions at specific sites. Later analysis reveals that the same replication fork also accumulated short DNA stretches synthesized by DNA Pol-I. Which DNA repair pathway was most likely activated?

• A. Photoreactivation
• B. Base excision repair
• C. Nucleotide excision repair
• D. Mismatch repair

Hint:

UV-induced bulky DNA lesions such as pyrimidine dimers are mainly repaired by nucleotide excision repair.

Answer 1

The Correct Option is C

Concept:
UV radiation commonly causes bulky DNA lesions such as pyrimidine dimers. These lesions distort the DNA helix.

Step 1: Effect of UV radiation.
UV radiation can produce thymine dimers and other bulky lesions in DNA.

Step 2: Repair mechanism needed.
Bulky DNA damage is usually repaired by nucleotide excision repair.

Step 3: Role of DNA Pol-I.
After the damaged segment is excised, DNA polymerase I fills the gap by synthesizing short DNA stretches.

Step 4: Final conclusion.
Since UV damage and DNA Pol-I gap filling are involved, the most likely pathway is nucleotide excision repair. \[ \therefore \text{Correct Answer is (C)} \]