Question

How did Meselson and Stahl demonstrate replication in E. coli takes place in a semi-conservative manner?

Answer 1

Meselson- Stahl Experiment:. Meselson and Stahl grow E. coli in NH₄Cl consisting of N¹⁵ (atomic mass = 15) isotope as the sole source of nitrogen for several generations to ensure that all the cells had N¹⁵ in their DNA (nitrogenous bases). They isolated DNA from these cells and run cesium chloride (CsCl) density gradient centrifugation. DNA settles at a level where its density becomes equal to that of CsCl along the vertically down length of the tube. Next, E. coli with N¹⁵ in their DNA were allowed to replicated once (a single cell division) in NH₄Cl consisting of N¹⁴ as sole source of nitrogen, being lighter than N¹⁵. When DNA isolated from these cells is subjected to density gradient centrifugation, it settles at a level of intermediate density (in between density of N¹⁵ and N¹⁴). The DNA isolated from N¹⁵ E. coli grown for two cell divisions in N¹⁴ medium settles at the density in between N¹⁴- N¹⁵ hybrid and N¹⁴ density.

They concluded that the intermediate density of N¹⁵DNA after one replication in N¹⁴ occurs because both the newly formed DNA copies has one strand from the parent DNA (N¹⁵) and the other strand is newly synthesized (N¹⁴) during replication. So, if N¹⁵ E. coli are grown in N¹⁴ medium for two generations, 25% of the total DNA would be parental and remaining 75% would have been newly synthesized. The experiment demonstrates the semiconservative mode of DNA replication over conservative and dispersive modes of DNA replication. The two new daughter DNA copies produced through semiconservative mode consist of one parental strand and one newly synthesized, thus each new copy has 50% DNA (one strand) from its parent DNA and 50% (the other strand) newly synthesized DNA.