Question

Radioactive thymine is added to media containing one actively dividing E. coli bacterium. Which of the following outcomes would be seen after a single cell division?

All four bases of the DNA would be radioactive.
DNA in both daughter cells would be radioactive.
Neither of the two daughter cells would be radioactive.
One of the daughter cells, but not the other, would have radioactive DNA.

Answer 1

To address this question, we need to consider how DNA replication works in bacteria like E. coli.

During DNA replication, each strand of the original DNA molecule serves as a template for the synthesis of a new complementary strand. When radioactive thymine (^3H-T or any radioactive form of thymine) is incorporated into the DNA, it will only be incorporated into the DNA strands that are synthesized in the presence of the radioactive thymine.

Since the E. coli cell divides via binary fission, after one cell division:

1. The original DNA, which contained the radioactive thymine, would have been replicated. When the DNA is replicated, one of the two resulting strands will have the radioactive thymine incorporated into it, while the complementary strand will contain only the normal (non-radioactive) thymine.

2. After this replication, each daughter cell will inherit one of the two original strands (with the radioactive thymine) and one newly synthesized strand (with non-radioactive thymine).

Thus, at the end of one cell division, each daughter cell will receive DNA that contains a mix of strands: one will have radioactive thymine (from the parent strand) and the other will have non-radioactive thymine (from the newly synthesized strand).

Consequently, the outcome will be that **DNA in both daughter cells would be radioactive** due to the presence of the strand that incorporated the radioactive thymine.

So the correct answer is:
**DNA in both daughter cells would be radioactive.**