Meselson-Stahl experiment

Question 1

Meselson-Stahl experiment

Answer 1

Proof for semiconservative replication

Question 2

Explain the Meselson-Stahl experiment

Answer 2

1. They grew E.coli in a medium that contained 15N labeled ammonium chloride
   > 15N = heavy nitrogen is a non-radioactive stable isotope of Nitrogen that contains one extra neutron
2. After several generations in this medium, nearly all the molecules in the bacterial cells that would contain nitrogen, contained heavy nitrogen
   [Includes purine and pyrimidine bases]
3. For the experiment to work, NB to be able to distinguish between the DNA which contains 15N and 14N (normal N)
4. Meselson-Stahl used sedimentation equilibrium centrifugation
   > centrifugation is used to drive DNA down a density gradient of a heavy metal salt (usually cesium chloride)
5. When the density of the DNA is the same as the density of the salt at that density gradient, the DNA will reach equilibrium and form a band.
6. Since 14N containing DNA is lighter, it will reach equilibrium first –> closer to top of tube
   DNA containing heavier isotope will reach equilibrium closer to the bottom of the tube
7. Use of a DNA binding dye ( such as ethidium bromide–> Used in electrophoresis) helps us visualize the DNA bands
   > Helps us distinguish between the two bands

Example:
1. DNA from cells that were grown in N15 for many generations
will contain heavy nitrogen in the bases of both strands= very
heavy DNA
> Therefore will form a heavy band near the base of the
centrifuge tube

2. When these cells were added to a medium containing only
   14N - all new molecules will be made using 14N
   = New DNA molecules contain 14N
   > If DNA is isolated after a single generation (20 min at 37 degrees C) then we would expect with semi-conservative replication an intermediate band where 1. Strand (parental strand) would still be labeled with 15N bases and the newly synthesized 2. Strand would have 14N

Question 3

how does the Meselson-Stahl experiment rule out conservative and dispersive replication?

Answer 3

1. We have 15N labeled DNA ( both strands contain the heavy N)
   - If we isolate at this point- expect to see a band of cells at the bottom of the tube.
2. If we take these cells and allow them to grow for one generation (20 min) in a medium containing 14N
   > we would expect that during replication (in a semi-conservative fashion) the DNA strands would denature and each of the original parental strands would provide a template for a new N14 containing strand
3. As a result, at the end, we have two identical DNA strands each containing a 15N and 14N strand
   = having an intermediate mass

This is what they saw after experiment one.
The hybrid DNA band moved up slightly higher in the centrifuge tube than the 15N DNA

4. What if we allow another generation to take place?
   - we would still have original parental strands which would attract nucleotides to make a new strand of light (14N) DNA
   - We would retain two intermediate strands
   HOWEVER:
   The newly synthesised DNA from the first generation would attract nucleotides which are again light = dual light stranded DNA
   Results: 2 hybrid DNA (14N/15N) and 2 light DNA (14N/14N)
   - in tube would have:
   > the hybrid band in the middle of the tube
   > Lighter band at top of tube.

So how does this rule out conservative and dispersive replication?
1. If we follow conservative replication:
After generation one: (in 14N medium)
- Would expect to see 15N parent DNA and 14N daughter DNA strands
- Two nads, one at the top and one at bottom of the tube
>This doesn’t happen- observed a hybrid band forming in
the tube

2.) If we follow dispersive replication:
- Hybrid band is consistent with dispersive replication - each DNA strand is a mix of light and heavy DNA
- If we run it through centrifugation- would expect hybrid band
- Meselson-Stahl isolated the DNA after a single round and heat treated it to denature the DNA
- If dispersive replication was taking place each strand would remain a hybrid of heavy and light DNA and would expect to see the hybrid band remaining
BUT: instead, they saw a band of heavy and a band of light DNA
WHY?
Because when the daughter cells separated, the parent cells showed up as a heavy DNA band and the daughter cells as a light DNA band.
This could rule out dispersive replication.

2.2.) There’s a second way:
After a second round of replication in (desp. rep.), we would again expect to see a hybrid band and in future generations, we would expect to see this hybrid band moving further and further towards the top of the tube as more and more light DNA was included
- This was not observed, they saw an accumulating light band which is consistent with semiconservative replication and not dispersive replication.

Question 4

A year before Meselson-Stahl published their findings, Taylor et al. published evidence on semi-conservative replication in eukaryotic organisms

Answer 4

They used the root tips of the broad bean Vicia faba:

• The root tips have very high levels of cell division and the chromosomes can be easily visualized (large).
• sister chromatids labeled with 3H-thymidine (radio-active)
• They then used autoradiography to follow the replication cycle

In autoradiography:

• Photographic film is placed over the sample.
• When the radioactive emits energy, it strikes the silver salt emulsion on the film and converts it into silver metal- forming a black spot on the film
• The same way photons do when film is subjected to light
• the developed film will thus show the location of the radioactive DNA
• Taylor grew the root tips for +/- for one cell generation in the
  presence of the radioisotope.
• Then placed them in an unlabeled medium where they could continue cell division
• after each cell division they stopped cell division at metaphase using Colchicine ( prevents formation of spindle fibres)
• Then they looked at the chromosomes using autoradiography

> Radio active thymidine was only found in chromatids which contained newly synthesised DNA
So after first round of replication, both sister chromatids would show radio activity
This indicates that each chromatid contains one old unlabeled strand and one new radioactive strand

• After the second round of replication in an unlabeled medium, only one of the two sister chromatids could be labeled.
• Because half of the original parental strands would be unlabeled
• This provided evidence for semi-conservative replication