Chapter 28, Problem 1P

Semiconservative or Conservative DNA Replication If ${}^{\text{15}}\text{N}$-Iabeled E. coli DNA has a density of 1.724 g/mL, ${}^{\text{14}}\text{N}$-labeled DNA has a density of 1.710 g/mL, and E. coli cells grown for many generations on ${}^{1\text{4}}{\text{NH}}_{\text{4}}{}^{+}$as a nitrogen source are transferred to media containing ${}^{1\text{5}}{\text{NH}}_{\text{4}}{}^{+}$as the sole N-source,

(a) What will be the density of the DNA after one generation, assuming replication is semiconservative?

(b) Suppose replication took place by a conservative mechanism in which the parental strands remained together and the two progeny strands were paired. Design an experiment that could distinguish between semiconservative and conservative modes of replication.

Interpretation Introduction

(a)

Interpretation:

The density of DNA of E.coli cells should be calculated after one generation that are grown for many generations on¹⁴NH₄+ and then transferred to a media containing¹⁵NH₄+ as only source of nitrogen assuming the replication is semiconservative given the density of${}^{15}N$-labeled E.coli DNA is 1.724 g/ml, density of¹⁴N-labeled E.coli DNA is 1.710g/ml.

Concept Introduction:

In model of semiconservative mode of replication parent DNA strands unwind and each one of these acts as a template for a new additional strand, thus synthesizes two DNA each with a pair of an old and a new strand.

Answer to Problem 1P

The average density after one generation of the transfer of media will be 1.717 g/ml.

Explanation of Solution

The density of DNA after one generation is calculated with arithmetic mean of densities of${}^{15}N$-labeled E.coli DNA (heavy strand) and the${}^{14}N$- labeled DNA (light strand).

Average density =$\rho =\frac{(1.724+1.710)}{2}=1.717$g/ml

After the transfer of E.coli cells from a light medium of¹⁴NH₄+ to a heavy medium of¹⁵NH₄⁺, the cells have¹⁵NH₄+ as only source of nitrogen. If a semi-conservative replication is assumed, replication produces two DNA each having an old strand of${}^{15}N$-labeled E.coli ($\rho$=1.724 g/ml ) and a new strand of${}^{14}N$- labeled E.coli.

Interpretation Introduction

(b)

Interpretation:

Assume a replication by a conservative mechanism in which parental strands stayed together and two daughter strands paired. A brief description of an experiment that can differentiate between semiconservative and conservative modes of replication should be given.

Concept Introduction:

For the DNA strands to differentiate there needs to be a system of tagging the parent DNA. This is done by changing the medium of the E.coli cells. Cells that are grown in¹⁵NH₄+ medium are transferred to a medium containing heavy nitrogen (¹⁴N)

Answer to Problem 1P

Grow E.coli cells for many generations in a medium containing¹⁵NH₄+ as a source of nitrogen. Extract DNA of the cells and calculate its density using “CsCSl” density gradient centrifugation. Feed E.coli for many generations on¹⁴NH₄+ medium and Similarly determine its density.

Now, transfer the cells grown on¹⁵NH₄+ medium to the media containing¹⁴N. The number of cells is carefully monitored by colony formation assay. The DNA is then extracted generation after generation. Extract the DNA after first generation and determine its density.

Explanation of Solution

The pair of progeny DNA produced has an equal density intermediate between the first group and the second group. This is not the case with conservative replication as density of progeny DNA is not equal.

In model of conservative replication, the pair of progeny DNA has one DNA containing two helical old strands and the other is newly fabricated. In contrast the semiconservative model has same density for both progeny DNA, intermediate between the¹⁴N and¹⁵N E.coli. The result fits with the latter model.