Meselson and Stahl designed an experiment to distinguish between three different
ID: 19794 • Letter: M
Question
Meselson and Stahl designed an experiment to distinguish between three different mechanisms for DNA replication. The adjacent image is a modified one from the scientific paper describing their results. The image is modified in the sense that I have expanded the horizontal scale so the bands are easier to distinguish. I have also rearranged the panels and 'moved' some bands for the purposes of this question. After these modifications, the relative intensities of the bands has no significance.
In the experiment E. coli was grown in a medium containing only heavy nitrogen (15N) then transferred to a medium containing light nitrogen (14N). After allowing time for only one DNA replication, the sample was centrifuged; the band(s) indicates the presence of DNA. The lightest DNA, with only 14N, will be on the left side of the panels, while the heavier forms of DNA will be on the right side of the image. A control panel showing DNA with only 15N and with only 14N is shown on the top.
a)Which panel shows the band location(s) that would support a hypothesis of semi-conservative replication, if only one round of replication occurred?
b)Which panel shows the band location(s) that would support a hypothesis of semi-conservative replication, if two rounds of replication occurred?
c)Which panel shows the band location(s) that would support a hypothesis of conservative replication, if two rounds of replication occurred?
Explanation / Answer
a) In first round of replication, two DNA molecules of N14-N15 DNA strands are formed since it is grown in N15 medium. so answer is 'B' b.) We get four molecules of DNA in which two molecules are having N14-N14 strands and two molecules are having N14-N15 strands. so answer is 'D' c.)For two conservative methods, we get four molecules of DNA in which there are three N14-N14 DNA molecules and one N15-N15 DNA molecule. so answer is 'E'