Biology 130 – Molecular Biology and Genetics
Meselson and Stahl Discussion Questions
1. How did Meselson and Stahl distinguish between new and old DNA? Why did they
use N15 to label the DNA? How is this similar to or different from the use of P32 by
Hershey and Chase?
By growing the cells in media containing almost pure 15N up to time 0 (and then flooding
the cultures with excess 14N), M+S labeled the old DNA by making it denser (higher
molecular weight, same volume) then DNA made after time 0. 15N was a good choice
because it imparts to the DNA a property that allowed them to physically separate old
from new DNA based on density.
This is similar to the use of 32P by H & C in that both used a specific isotope to
label DNA and then track what happened to the DNA over the course of the experiment.
They differ in the details of how 15N and 32P are useful. 32P is radioactive, which means it
is possible to detect even tiny amount of 32P-labeled DNA. But it would be harder to
separate 32P-labeled DNA from non-labeled DNA. 15N does not make DNA any easier to
detect. But because there are multiple N atoms per nucleotide (as opposed to just one P),
the density difference is sufficient to separate on density gradient.
2. Why was it important for M & S to measure the rate of growth of their bacterial
cultures?
For several reasons:
1) They needed to know that their bacteria were dividing at a constant rate of the course
of their experiment. If for example, the growth rate slowed during the experiment, it
would make it much harder to interpret their data.
2) Similarly, they needed to know how long it took their cells to divide. This is the only
they could know what generation each sample corresponded to. (In Figure 4, for
example.) For example, only by knowing that the generation time was ~48 minutes
could they which samples corresponed to DNA taken 1 generation (or 2 generations, etc.)
after time 0.
3. How would Figure 4 and Figure 9 have looked different if the DNA were replicated
by the conservative model of replication?
Fig 4. By one generation, we would have seen 2 bands of DNA- one with the density of
pure 15N DNA, the other with the density of pure 14N DNA. Over generations, the 14N
would become more abundant while the 15N band would become fainter. By 4
generations, the 14N DNA would outnumber the 15N DNA 15:1.
Fig 9 would look exactly the same. Heating the DNA denatures it, separating the strands.
The conservative and semi-conservative differ only in how strands are paired, not in the
nature of the strands themselves. Therefore Figure 9b would still have the same 2 peaks.
4. How would Figure 4 and Figure 9 have looked different if the DNA were replicated by
the dispersive model of replication?
Fig 4. In the dispersive model, the 15N would be equally disitrubuted among subunits
every generation. So after 1 generation, all the DNA would be 50% 15N, and there would
be one band of intermediate density. (Exactly as happens with the semi-conservative
model). But over subsequent generations, rather than the sudden appearance of a
distinct14N band by 2 generations, there would a gradual migration of 1 band towards
lighter densities. By 4 generations, it would be close to (but not quite at) pure 14N
density.
Fig 9. Part B would have one band of intermediate density. This is because the
denatured strands would still each by a mix of 14N and 15N. Therefore there would by
one band, intermediate in density compared to pure 14N and pure 15N strands.