If you expose HeLa cells to 3H-thymidine just as they enter S phase, then wash this material off the cells and let them go through a second S phase before looking at the chromosomes, how would you expect the 3H to be distributed over a pair of homologous chromosomes? Would the radioactivity be in (a) one chromatid of one homolog, (b) both chromatids of one homolog, (c) one chromatid each of both homologs, (d) both chromatids of both homologs, or (e) some other pattern?

Choose the correct answer and explain your reasoning.

D. Both chromatids of both homologs.

Explanation:

S phase is also known as synthetic phase of cell cycle. It occurs between G1 and G2 phase of the interphase. In this phase replication of DNA occurs. So, 3H-thymidine will incorporate into new DNA strands.

Chromatid is each of two strands into which chromosomes divides during cell division. Now, chromatid is a replicated DNA molecule and each new DNA molecule contains one new DNA strand (semi-conservative replication).

So, the radioactivity would be in both chromatids of both homologs.

(c) one chromatid each of both homologs.

Explanation:

The S phase is characterized in that the chromosomes are duplicated, in this way a copy is created and this copy is united forming chromatids. In meiosis it is characterized in that the homologs are assembled. According to the exercise, the S phase was incubated using 3H, presumably the chromosomes will have a chromatid to support the 3H residues. Later when the homologs assemble, each of those homologs would have a chromatid to withstand such radioactivity.