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Number 207
How Science Works: Meselson and Stahl’s
Classic Experiment
n 1953 James Watson and Francis Crick built their model of the structure of DNA, which is still accepted today:
• DNA is an anti-parallel double stranded helix
• It consists of a sugar-phosphate backbone and a sequence of bases
• Complementary base pairing takes place
• A cytosine (C) base on one strand always pairs with a guanine (G) base on the other strand and an adenine (A) base on one strand
always pairs with a thymine (T) base on the other strand
This led them to propose a model for DNA replication in which each strand serves as a template for a newly synthesised strand; however
Watson and Crick had no evidence to back up this proposal.
In 1957, American geneticists and biologists, Matthew Meselson and Franklin Stahl provided evidence of the mechanism of DNA
replication in their classic experiment. Their research was important in showing how DNA replicates, recombines and is repaired in cells.
Fig. 1: Meselson (left) and Stahl (right)
Typical Exam Question
DNA replication can be described as being semi-conservative.
What does this mean?
When DNA replicates the two strands separate and each original
strand of the DNA molecule acts as a template and is copied;
therefore each new DNA molecule produced consists of one original
and one new DNA strand. This is known as semi-conservative
replication (Fig. 2).
Fig 2: Semi-conservative replication
original DNA
Newly synthersised
DNA
If DNA replication was conservative, both strands would still be copied, however the two original strands would join back together and
the two new strands would join together (Fig. 3). DNA does not replicate in this way, as Meselson and Stahl demonstrated.
Fig 3: Conservative replication
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207. How Science Works: Meselson and Stahl’s Classic Experiment
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How does DNA replicate semi-conservatively?
DNA replicates to preserve its information; this information can then be passed on to new cells. It occurs during interphase before mitosis
and meiosis in the nucleus. A cell can spend up to 90% of its life replicating DNA, i.e. it occurs almost all of the time. DNA replication begins
at a sequence of nucleotides called the origin of replication (Fig 4):
Fig 4: Semi-conservative DNA replication
5'
3'
5'
1
1. Helicase enzymes
unwind the DNA
double helix by
breaking hydrogen
bonds between the
bases.
3'
replication
bubble
3'
5'
3'
5'
3'
4. Exonuclease
removes the RNA
primers and ligase
inserts any
missing
phosphates to join
up the sugarphosphate
backbone.
4
new complementary DNA
nucleotides
5'
3'
5'
3'
5'
3. DNA polymerase then starts to
add new complementary DNA
nucleotides to both template
strands (from the 3’ end of the
primer only). This occurs at the
rate of approximately 80
nucleotides per second in
humans. Primase adds more
RNA primer in the gaps and DNA
polymerase then fills in the gaps
with new complementary
nucleotides.
3
3'
2. The strands separate forming one
of many replication ‘bubbles.’
These normally form in regions
rich in A-T base-pairs as there are
only two hydrogen bonds
between A and T, but three
between C and G. RNA primers
(single polynucleotide strands
approximately ten nucleotides
long) are attached by primase, an
RNA primer
RNA polymerase enzyme. They
signal where DNA polymerase is
to start adding new DNA
nucleotides.
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5'
5'
3'
5'
3'
5
5. Each of the new DNA molecules consists of one
new and one old strand.
3'
5'
3'
5'
2
207. How Science Works: Meselson and Stahl’s Classic Experiment
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How does the structure of DNA make it suited for
replication?
The Experiment
1. Meselson and Stahl grew the bacterium Escherichia coli for
fourteen generations in a medium containing 15NH Cl as the
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only source of nitrogen. This resulted in all of the DNA being
labelled with high density 15N. This was called generation 0.
(In exam questions always relate structure to function, even if it is
only for one mark).
• DNA is double stranded and each strand holds the same genetic
information; therefore both strands serve as templates for the
reproduction of the opposite strand. The template strand is
preserved in its entirety and the new strand is assembled from
new nucleotides.
• The hydrogen bonds holding the two strands together are weak
and split easily allowing replication to take place (they are
normally protected by being inside the double helix).
• Complementary base-pairing enables accurate copying, allowing
the information stored in the base sequence to be preserved.
2. This DNA was centrifuged in caesium chloride. As expected,
they observed a band of DNA at the bottom of the tube (these
are the original strands):
Generation 0: All heavy DNA
How did Meselson and Stahl provide evidence for the
semi-conservative mechanism of DNA replication?
In their classic experiment Meselson and Stahl devised a strategy
that would enable original and new strands of DNA to be
distinguished.
• Nitrogen is a major constituent of DNA, it is found in all four of
the possible bases.
• The normal isotope of nitrogen, otherwise known as ‘light’
nitrogen, 14N, is by far the most abundant isotope of nitrogen.
This is the isotope of nitrogen normally found in DNA.
• DNA can also be made with a ‘heavy’ nitrogen isotope, 15N (see
Fig. 5 for an example of the DNA base thymine labelled with 15N)
3. E. coli with only 15N DNA were transferred to a 14N medium
(from this point on the only new DNA available for new strand
synthesis consisted of the normal lighter 14N nucleotides).
Again, the bacteria were allowed to divide once only. This was
called generation 1.
4. When centrifuged in caesium chloride the band in the tube was
in between that of 15N DNA and 14N DNA, i.e. of intermediate
density. What Meselson and Stahl learned from this is that
each of the original 15N DNA strands must have acted as a
template to which new 14N DNA was added. Hence replication
must have been semi-conservative:
Fig 5: Thymine labelled with 15N
H
H
15
N
O
C
C
C
15
N
H
CH3
Generation 0: All heavy DNA
C
O
• The 15N isotope is not radioactive, only denser than 14N.
• When centrifuged in caesium chloride, DNA settles out as bands
in the tube, the more dense the DNA, the further down the tube
it will be.
• Dense 15N DNA settles out as a band at the bottom of the tube
and normal less dense 14N DNA settles out as a band near the
top of the tube (see Fig. 6)
Generation 1:
All Inermediate DNA
Fig 6. 15N and 14N DNA settled out as bands when
centrifuged with caesium chloride
15
N
14
N
If DNA replication was conservative replication, Meselson and
Stahl would have observed equal amounts of DNA of the higher
and lower densities, i.e. a band at the top and bottom with no
intermediate band. They did not observe this and so rejected
the mechanism of conservative DNA replication.
5. To conclusively prove their findings Meselson and Stahl grew
E. coli for several more generations in the 14N medium only,
each time DNA was centrifuged in caesium chloride.
Interpretation of the DNA in the bands
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207. How Science Works: Meselson and Stahl’s Classic Experiment
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6. In generation 2 they observed two bands in the centrifuge tube, the same intermediate band in the middle of the tube and a new low
density band at the top of the tube with the size ratio of the low density band to the intermediate band being of 1:1. This meant that half
of the generation 2 bacteria would have one strand of the original 15N DNA along with one new 14N DNA strand (accounting for the
intermediate band), while the DNA in the other half of the bacteria would consist entirely of new 14N DNA; one strand synthesised in
generation 1, and the other in generation 2 (accounting for the low density band at the top):
Generation 1:
All Inermediate DNA
Generation 2:
Light DNA and intermediate
DNA in the ratio 1:1
7. In the generations 3 and 4 the same bands were seen as generation 2; however the band size ratio of low density to intermediate density
was different. In generation 3 it was 3:1 and in generation 4 it was 7:1. Each generation the bacteria reproduced, requiring new DNA.
The only DNA available for new strand synthesis was 14N DNA, therefore the proportion of 14N DNA in each successive generation
increased, i.e. the low density band size in the tube increased.
Generation 2:
Light DNA and intermediate
DNA in the ratio 1:1
Generation 3:
Light DNA and intermediate
DNA in the ratio 3:1
Generation 4:
Light DNA and intermediate
DNA in the ratio 7:1
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207. How Science Works: Meselson and Stahl’s Classic Experiment
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8. The Important thing to note is that the semi-conservative strands were still being produced in every generation. Meselson and Stahl
could therefore conclude that DNA replication was semi-conservative (Exam questions often show the centrifuge tubes blank and ask
for the appropriate bands to be put in each generation, it the test tubes on the question are graduated, remember to put in the
appropriate size bands).
Fig 7: Summary of Meselson and Stahl’s results
Generation
Centrifuge tube
Interpretation of DNA in the bands
Explanation
0
All heavy DNA
1
All intermediate DNA
2
Light DNA and intermediate DNA
in the ratio 1:1
3
Light DNA and intermediate DNA
in the ratio 3:1
4
Light DNA and intermediate DNA
in the ratio 7:1
Questions
1. The following drawing shows part of a DNA molecule:
G
phosphate
Samples of cells were removed from the radioactive solution
after one cell division. Chromosomes from the cells were tested
for radioactivity. The results are shown in the table below:
C
A
A
Chromosomes from cells
dividing in unlabelled
thymine
No radioactivity
B
Chromosomes from cells that
divided once in radioactive
thymine
All chromosomes
radioactive
T
C
G
Deoxyribose
G
C
Use your knowledge of semi-conservative DNA replication to
explain why all the chromosomes in B were radioactive (4 marks).
(a) Describe and explain three structural features shown in the
diagram that account for the ability of DNA to replicate (3 marks).
(b) Rapidly dividing cells were grown in a medium containing
unlabelled thymine. Some cells were transferred to a second
medium in which all of the thymine molecules were radioactively
labelled.
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207. How Science Works: Meselson and Stahl’s Classic Experiment
Answers
2. There are two forms of nitrogen. 15N is a heavier isotope than
the normal isotope 14N. In an investigation a culture of
Escherichia Coli was obtained in which only contained 15N
DNA. The bacteria were transferred to a medium containing
only 14N DNA and allowed to divide once. A sample of this first
generation was removed. The DNA was extracted and
centrifuged at high speed. This was repeated with samples of
the second and third generation. The diagram below shows the
results of this:
Generation 0
E.coli with 15N
labelled DNA
Generation 1
Generation 2
1. a. Two strands/double stranded, each strand acts as a template/
is copied;
Complementary base pairs present, allows accurate copying/
replication;
Weak hydrogen bonds, easily broken for replication;
b. DNA strands separate/hydrogen bonds are broken (a labelled
diagram could show this);
Each strand forms a template/is copied/one new strand & one
old (a labelled diagram could show this);
Complementary base pairing;
Radioactivity incorporated into (all) new strands;
Generation 3
cells replicate a
third time on 14N
medium
cells replicate a
second time on
14
N medium
cells replicate
once on 14N
medium
Bio Factsheet
2. (a) Base/any named base;
(b) Use of CsCl solution;
Concentration gradient in tube (set up by centrifugation);
DNA molecules move in the concentration gradient (until
they reach a place where their density equals that of the
caesium chloride)/DNA molecules separate according to
density/weight/mass;
DNA with 15N will be further down the tube (accept converse)
(a) Which component of DNA would contain nitrogen
(1 mark)?
(b) Explain how centrifugation allows DNA containing different
isotopes of nitrogen to be separated and distinguished (3
marks).
(c) Explain why the DNA from generation 1 is found in the
position shown in the diagram (2 marks).
(d) Complete the diagram to show the positions of the DNA
from generations 2 and 3 (2 marks).
(e) Select the letter or letters from the diagram below representing
the bacterial DNA in (3 marks):
i. Generation 1;
ii. Generation 2;
iii. Generation 3.
(c) Semi-conservative replication;
Contains one heavy and one light strand/half 15N and half 14N;
(d) Generation 2 - band at top of tube and band at middle of
tube;
Generation 3 - band at top of tube and band in middle
(e) i. A
ii. C
iii. C and E
(f) 28;
32, 28, 22.
represents
DNA with 15N
A
B
C
represents
DNA with 14N
D
E
F
f. In a further investigation, the DNA of the bacterium was isolated and separated into single strands. The percentage of each
nitrogenous base in each strand was found.
The table shows some of the results.
DNA Sample
Percentage of base present
Adenine Cytosine Guanine Thymine
Strand 1
22
Strand 2
18
32
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Acknowledgements:
This Factsheet was researched and written by John Greenhalgh.
Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU.
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Use your knowledge of base pairing to complete the table.
(2 marks)
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