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7: Eukaryotic DNA Contains Several Classes of Sequence Variation

7: Eukaryotic DNA Contains Several Classes of Sequence Variation

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DNA: The Chemical Nature of the Gene

The other major class of repetitive DNA is highly repetitive DNA. These short sequences, often less than 10 bp in
length, are present in hundreds of thousands to millions of
copies that are repeated in tandem and clustered in certain
regions of the chromosome, especially at centromeres and
telomeres. Highly repetitive DNA is sometimes called satellite DNA, because its percentages of the four bases differ
from those of other DNA sequences and, therefore, it separates as a satellite fraction when centrifuged at high speeds.
Highly repetitive DNA is rarely transcribed into RNA.
Although these sequences may contribute to centromere and
telomere function, most highly repetitive DNA has no
known function.
Direct sequencing of eukaryotic genomes also tell us a
lot about how genetic information is organized within chromosomes. We now know that the density of genes varies
greatly among and within chromosomes. For example,
human chromosome 19 has a high density of genes, with
about 26 genes per million base pairs. Chromosome 13, on
the other hand, has only about 6.5 genes per million base
pairs. Gene density can also vary within different regions of
the same chromosome: some parts of the long arm of chro-

213

mosome 13 have only 3 genes per million base pairs, whereas
other parts have almost 30 genes per million base pairs. And
the short arm of chromosome 13 contains almost no genes,
consisting entirely of heterochromatin.

Concepts
Eukaryotic DNA comprises three major classes: unique-sequence
DNA, moderately repetitive DNA, and highly repetitive DNA.
Unique-sequence DNA consists of sequences that exist in one or
only a few copies; moderately repetitive DNA consists of
sequences that may be several hundred base pairs in length and
is present in thousands to hundreds of thousands of copies.
Highly repetitive DNA consists of very short sequences repeated
in tandem and present in hundreds of thousands to millions of
copies. The density of genes varies greatly among and even within
chromosomes.

✔ Concept Check 11
Most of the genes that encode proteins are found in
a. unique-sequence DNA.

c. highly repetitive DNA.

b. moderately repetitive DNA.

d. all of the above.

Concepts Summary
• Genetic material must contain complex information, be






replicated accurately, and have the capacity to be translated
into the phenotype.
Evidence that DNA is the source of genetic information came
from the finding by Avery, MacLeod, and McCarty that
transformation depended on DNA and from the
demonstration by Hershey and Chase that viral DNA is
passed on to progeny phages.
James Watson and Francis Crick proposed a new model for
the three-dimensional structure of DNA in 1953.
A DNA nucleotide consists of a deoxyribose sugar, a phosphate
group, and a nitrogenous base. RNA consists of a ribose sugar,
a phosphate group, and a nitrogenous base.
The bases of a DNA nucleotide are of two types: purines
(adenine and guanine) and pyrimidines (cytosine and thymine).
RNA contains the pyrimidine uracil instead of thymine.

• Nucleotides are joined together by phosphodiester linkages in



a polynucleotide strand. Each polynucleotide strand has a free
phosphate group at its 5Ј end and a free hydroxyl group at its
3Ј end.
DNA consists of two nucleotide strands that wind around
each other to form a double helix. The sugars and phosphates
lie on the outside of the helix, and the bases are stacked in the
interior. The two strands are joined together by hydrogen
bonding between bases in each strand. The two strands are
antiparallel and complementary.

• DNA molecules can form a number of different secondary





structures, depending on the conditions in which the DNA is
placed and on its base sequence.
The structure of DNA has several important genetic
implications. Genetic information resides in the base sequence
of DNA, which ultimately specifies the amino acid sequence of
proteins. Complementarity of the bases on DNA’s two strands
allows genetic information to be replicated.
The central dogma of molecular biology proposes that
information flows in a one-way direction, from DNA to RNA
to protein. Exceptions to the central dogma are now known.
Chromosomes contain very long DNA molecules that are
tightly packed. Supercoiling results from strain produced
when rotations are added to a relaxed DNA molecule or
removed from it.

• A bacterial chromosome consists of a single, circular DNA


molecule that is bound to proteins and exists as a series of
large loops.
Each eukaryotic chromosome contains a single, long linear
DNA molecule that is bound to histone and nonhistone
chromosomal proteins. Euchromatin undergoes the normal
cycle of decondensation and condensation in the cell cycle.
Heterochromatin remains highly condensed throughout the
cell cycle.

• The nucleosome is a core of eight histone proteins and the
DNA that wraps around the core.

214

Chapter 8

• A nucleosome is folded into a 30-nm fiber that forms a series



of 300-nm-long loops; these loops are anchored at their bases
by proteins associated with the nuclear scaffold. The 300-nm
loops are condensed to form a fiber that is itself tightly coiled
to produce a chromatid.
Centromeres are chromosomal regions where spindle fibers
attach; chromosomes without centromeres are usually lost in
the course of cell division. Telomeres stabilize the ends of
chromosomes.

• Eukaryotic DNA exhibits three classes of sequences. Uniquesequence DNA exists in very few copies. Moderately repetitive
DNA consists of moderately long sequences that are repeated
from hundreds to thousands of times. Highly repetitive DNA
consists of very short sequences that are repeated in tandem
from many thousands to millions of times.

Important Terms
nucleotide (p. 195)
Chargaff ’s rules (p. 195)
transforming principle (p. 196)
isotope (p. 197)
X-ray diffraction (p. 199)
ribose (p. 200)
deoxyribose (p. 200)
nitrogenous base (p. 201)
purine (p. 201)
pyrimidine (p. 201)
adenine (A) (p. 201)
guanine (G) (p. 201)
cytosine (C) (p. 201)
thymine (T) (p. 201)
uracil (U) (p. 201)
nucleoside (p. 201)
phosphate group (p. 201)
deoxyribonucleotide (p. 201)
ribonucleotide (p. 201)
phosphodiester linkage (p. 202)

polynucleotide strand (p. 202)
5Ј end (p. 202)
3Ј end (p. 202)
antiparallel (p. 202)
complementary DNA strands (p. 202)
B-DNA (p. 204)
A-DNA (p. 204)
Z-DNA (p. 204)
transcription (p. 205)
translation (p. 205)
replication (p. 205)
central dogma (p. 205)
reverse transcription (p. 205)
RNA replication (p. 205)
supercoiling (p. 206)
relaxed state of DNA (p. 206)
positive supercoiling (p. 206)
negative supercoiling (p. 206)
topoisomerase (p. 206)
nucleoid (p. 207)

euchromatin (p. 208)
heterochromatin (p. 208)
nonhistone chromosomal protein (p. 208)
chromosomal scaffold protein (p. 208)
nucleosome (p. 209)
chromatosome (p. 209)
linker DNA (p. 209)
centromeric sequence (p. 211)
telomeric sequence (p. 211)
C value (p. 212)
unique-sequence DNA (p. 212)
gene family (p. 212)
repetitive DNA (p. 212)
moderately repetitive DNA (p. 212)
tandem repeat sequence (p. 212)
interspersed repeat sequence (p. 212)
short interspersed element (SINE) (p. 212)
long interspersed element (LINE) (p. 212)
highly repetitive DNA (p. 213)

Answers to Concept Checks
1. Without knowledge of the structure of DNA, an understanding of how genetic information was encoded or expressed was
impossible to achieve.
2. c
3. No, because carbon is found in both protein and nucleic acid.
4. b
5. d
6. Z-DNA has a left-handed helix; B-DNA has a right-handed
helix. The sugar–phosphate backbone of Z-DNA zigzags back

and forth, whereas the sugar–phosphate backbone of B-DNA
forms a smooth continuous ribbon.
7. b
8. Bacterial DNA is not complexed to histone proteins and is
circular.
9. b
10. d
11. a

DNA: The Chemical Nature of the Gene

215

Worked Problems
1. The percentage of cytosine in a double-stranded DNA
molecule is 40%. What is the percentage of thymine?

• Solution
In double-stranded DNA, A pairs with T, whereas G pairs with C;
so the percentage of A equals the percentage of T, and the
percentage of G equals the percentage of C. If C ϭ 40%, then G
also must be 40%. The total percentage of C ϩ G is therefore
40% ϩ 40% ϭ 80%. All the remaining bases must be either A or
T; so the total percentage of A ϩ T ϭ 100% – 80% ϭ 20%;
because the percentage of A equals the percentage of T, the
percentage of T is 20%/2 ϭ 10%.
2. Which of the following relations will be true for the percentage
of bases in double-stranded DNA?
C
T
=
a. C ϩ T ϭ A ϩ G
b.
A
G

3. A diploid plant cell contains 2 billion base pairs of DNA.
a. How many nucleosomes are present in the cell?
b. Give the numbers of molecules of each type of histone protein
associated with the genomic DNA.

• Solution
Each nucleosome encompasses about 200 bp of DNA: from 145
to 147 bp of DNA wrapped around the histone core, from 20 to
22 bp of DNA associated with the H1 protein, and another 30 to
40 bp of linker DNA.
a. To determine how many nucleosomes are present in the cell,
we simply divide the total number of base pairs of DNA
(2 ϫ 109 bp) by the number of base pairs per nucleosome:
2 * 109 nucleotides
= 1 * 107 nucleosomes
2 * 102 nucleotides per nucleosome

• Solution
An easy way to determine whether the relations are true is to
arbitrarily assign percentages to the bases, remembering that, in
double-stranded DNA, A ϭ T and G ϭ C. For example, if the
percentages of A and T are each 30%, then the percentages of G
and C are each 20%. We can substitute these values into the
equations to see if the relations are true.
a. 20 ϩ 30 ϭ 30 ϩ 20. This relation is true.
b. 20΋30 Z 30΋20. This relation is not true.

Thus, there are approximately 10 million nucleosomes in the cell.
b. Each nucleosome includes two molecules each of H2A, H2B,
H3, and H4 histones. Therefore, there are 2 ϫ 107 molecules each
of H2A, H2B, H3, and H4 histones. Each nucleosome has
associated with it one copy of the H1 histone; so there are
1 ϫ 107 molecules of H1.

Comprehension Questions
Section 8.1
*1. What three general characteristics must the genetic material
possess?

Section 8.2
2. What is transformation? How did Avery and his colleagues
demonstrate that the transforming principle is DNA?
*3. How did Hershey and Chase show that DNA is passed to
new phages in phage reproduction?

Section 8.3
*4. Draw and identify the three parts of a DNA nucleotide.
5. How does an RNA nucleotide differ from a DNA
nucleotide?
*6. Draw a short segment of a single DNA polynucleotide
strand, including at least three nucleotides. Indicate the
polarity of the strand by identifying the 5Ј end and the
3Ј end.
7. What are some of the important genetic implications of the
DNA structure?

*8. What are the major transfers of genetic information?

Section 8.4
*9. How does supercoiling arise? What is the difference between
positive and negative supercoiling?
10. What functions does supercoiling serve for the cell?

Section 8.6
*11. Describe the composition and structure of the nucleosome.
How do core particles differ from chromatosomes?
12. Describe in steps how the double helix of DNA, which is
2 nm in width, gives rise to a chromosome that is 700 nm
in width.
*13. Describe the function and molecular structure of a
telomere.
14. What is the difference between euchromatin and
heterochromatin?

Section 8.7
*15. Describe the different types of DNA sequences that exist in
eukaryotes.

216

Chapter 8

Application Questions and Problems
Section 8.2
16. A student mixes some heat-killed type IIS Streptococcus
pneumoniae bacteria with live type IIR bacteria and injects
the mixture into a mouse. The mouse develops pneumonia
and dies. The student recovers some type IIS bacteria from
the dead mouse. It is the only experiment conducted by the
student. Has the student demonstrated that transformation
has taken place? What other explanations might explain the
presence of the type IIS bacteria in the dead mouse?

Section 8.3
17. DNA molecules of different size are often separated with the
use of a technique called electrophoresis (see Chapter 14).
With this technique, DNA molecules are placed in a gel, an
electrical current is applied to the gel, and the DNA
molecules migrate toward the positive (ϩ) pole of the
current. What aspect of its structure causes a DNA molecule
to migrate toward the positive pole?
18. What aspects of its structure contribute to the stability of
the DNA molecule? Why is RNA less stable than DNA?
*19. Edwin Chargaff collected data on the proportions of
DATA
nucleotide bases from the DNA of a variety of different
organisms and tissues (E. Chargaff, in The Nucleic Acids:
ANALYSIS
Chemistry and Biology, vol. 1, E. Chargaff and J. N. Davidson,
Eds. New York: Academic Press, 1955). Data from the DNA
of several organisms analyzed by Chargaff are shown below.
Percent
Organism and tissue
Sheep thymus
Pig liver
Human thymus
Rat bone marrow
Hen erythrocytes
Yeast
E. coli
Human sperm
Salmon sperm
Herring sperm

A
29.3
29.4
30.9
28.6
28.8
31.7
26.0
30.9
29.7
27.8

G
21.4
20.5
19.9
21.4
20.5
18.3
24.9
19.1
20.8
22.1

C
21.0
20.5
19.8
20.4
21.5
17.4
25.2
18.4
20.4
20.7

T
28.3
29.7
29.4
28.4
29.2
32.6
23.9
31.6
29.1
27.5

a. For each organism, compute the ratio of (A ϩ G)/(T ϩ
C) and the ratio of (A ϩ T)/(C ϩ G).
b. Are these ratios constant or do they vary among the
organisms? Explain why.
c. Is the (A ϩ G)/(T ϩ C) ratio different for the sperm
samples? Would you expect it to be? Why or why not?

20. Boris Magasanik collected data on the amounts of the bases
DATA
of RNA isolated from a number of sources (shown below),
expressed relative to a value of 10 for adenine (B. Magasanik,
ANALYSIS
in The Nucleic Acids: Chemistry and Biology, vol. 1,
E Chargaff and J. N. Davidson, Eds. New York: Academic
Press, 1955).
Percent
Organism and tissue
Rat liver nuclei
Rabbit liver nuclei
Cat brain
Carp muscle
Yeast

A
10
10
10
10
10

G
14.8
13.6
14.7
21.0
12.0

C
14.3
13.1
12.0
19.0
8.0

U
12.9
14.0
9.5
11.0
9.8

a. For each organism, compute the ratio of (A ϩ G)/
(U ϩ C).
b. How do these ratios compare with the (A ϩ G)/(T ϩ C)
ratio found in DNA (see Problem 19)? Explain.
*21. Which of the following relations will be found in the
percentages of bases of a double-stranded DNA molecule?
A + G
= 1.0
a. A ϩ T ϭ G ϩ C
e.
C + T
A
G
=
b. A ϩ T ϭ T ϩ C
f.
C
T
A
T
=
c. A ϩ C ϭ G ϩ T
g.
G
C
A + T
A
G
=
d.
h.
C + G
T
C
*22. If a double-stranded DNA molecule is 15% thymine, what
are the percentages of all the other bases?
23. Heinz Shuster collected the following data on the base
DATA
composition of ribgrass virus (H. Shuster, in The Nucleic
Acids: Chemistry and Biology, vol. 3, E. Chargaff and J. N.
ANALYSIS
Davidson, Eds. New York: Academic Press, 1955). On the
basis of this information, is the hereditary information of
the ribgrass virus RNA or DNA? Is it likely to be single
stranded or double stranded?
Percent
Ribgrass virus

A
29.3

G
25.8

C
18.0

T
0.0

U
27.0

24. For entertainment on a Friday night, a genetics professor
proposed that his children diagram a polynucleotide strand
of DNA. Having learned about DNA in preschool, his

DNA: The Chemical Nature of the Gene

5-year-old daughter was able to draw a polynucleotide
strand, but she made a few mistakes. The daughter’s diagram
(represented here) contained at least 10 mistakes.
a. Make a list of all the mistakes in the structure of this
DNA polynucleotide strand.
b. Draw the correct structure for the polynucleotide
strand.
O

H

base

H

H

H

O

*25. Chapter 1 considered the theory of the inheritance of
acquired characteristics and noted that this theory is no
longer accepted. Is the central dogma consistent with the
theory of the inheritance of acquired characteristics? Why
or why not?

Section 8.6
*26. Compare and contrast prokaryotic and eukaryotic
chromosomes. How are they alike and how do they differ?
*27. A diploid human cell contains approximately 6.4 billion
base pairs of DNA.
a. How many nucleosomes are present in such a cell?
(Assume that the linker DNA encompasses 40 bp.)
b. How many histone proteins are complexed to this DNA?

ϪO᎐᎐P᎐᎐OϪ

OH᎐᎐CH C

217

OH

ϪO᎐᎐P᎐᎐OϪ

OH᎐᎐CH C
H

base

H

H

H

OH

OH

Challenge Questions
Section 8.1

Section 8.3

*28. Suppose that an automated, unmanned probe is sent into
deep space to search for extraterrestrial life. After wandering
for many light-years among the far reaches of the universe,
this probe arrives on a distant planet and detects life. The
chemical composition of life on this planet is completely
different from that of life on Earth, and its genetic material
is not composed of nucleic acids. What predictions can you
make about the chemical properties of the genetic material
on this planet?

*30. Researchers have proposed that early life on Earth used RNA
as its source of genetic information and that DNA eventually
replaced RNA as the source of genetic information. What
aspects of DNA structure might make it better suited than
RNA to be the genetic material?

Section 8.2
29. How might 32P and 35S be used to demonstrate that the
transforming principle is DNA? Briefly outline an
experiment that would show that DNA rather than protein
is the transforming principle.

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9

DNA Replication
and Recombination
Preventing Train Wrecks
in Replication

D

avid had a tough childhood. It was bad enough that his fair skin
was densely covered with frecklelike spots, but what he really
hated were the long-sleeved shirts, long pants, and large straw hat that
his mother forced him to wear, even in the middle of summer, when the
other kids were in shorts and tee shirts. But, as David grew older, he
came to understand that his mother had not been unreasonable,
because David suffers from a rare genetic disease called xeroderma pigmentosum, characterized by acute sensitivity to sunlight and a predisposition to skin cancer triggered by exposure to the sun. Xeroderma
pigmentosum is an autosomal recessive disease that arises from a defect
in one of several genes that encode DNA synthesis and repair enzymes.
DNA polymerases—the enzymes that synthesize DNA—are beautiful and efficient molecular machines. Some of them operate at incredibly high speed, synthesizing DNA at a rate of more than 1000 nucleotides
per second, with less than one error per billion nucleotides. To achieve
this speed and accuracy, these DNA polymerases, like a high-speed train,
require a very smooth track. If the DNA template is damaged or
blocked—by, for example, distortions of structure induced by ultraviolet
(UV) light—the replication machinery comes to a grinding halt, resulting in gaps in the DNA, with disastrous consequences for the cell.
To overcome this problem, cells have evolved specialized, slower
DNA polymerases that are able to bypass distortions that normally
block the high-speed, high-fidelity polymerases that are the usual
workhorses of replication. But the use of these special “translesion”
polymerases comes at a price: they often make mistakes in those sections of DNA that they synthesize. However, most of the errors are corrected by DNA repair mechanisms, and the errors produced by the
Molecular model of DNA polymerase ␩, a translesion polylow-fidelity polymerases are not likely to be as detrimental as the gaps
merase that is able to bypass distortions in DNA structure but
in DNA left by failure to bypass the lesion.
often makes errors in DNA synthesis, resulting in mutations.
The importance of low-fidelity polymerases is revealed by people
[J. Trincao et al., Molecular Cell 8:417, 2001. Research Collaboratory for
with xeroderma pigmentosum. About 20% of those having the disease
Structural Bioinformatics. H. M. Berman et al., The Protein Data Bank
Nucleic Acids Research, 28:235–242, 2000. http://www.pdb.org/.]
have a defect in the POLH gene, which encodes DNA polymerase ␩, one
of the specialized (low-fidelity) DNA polymerases with the ability to
bypass distortions in the DNA template. One such distortion is the presence of bonds between
adjacent thymine bases on the same DNA strand; two thymine bases bonded together are
called a thymine dimer, which is produced by UV radiation. Because UV radiation is present
in sunlight, exposure to the sun causes thymine dimers to form. In most people, thymine
dimers are bypassed by specialized polymerases such as DNA polymerase ␩. Most of the
errors that are caused by DNA polymerase ␩ as it bypasses the lesion are later repaired by
219

220

Chapter 9

other mechanisms. However, DNA polymerase ␩ is defective in some people who have xeroderma pigmentosum, and pyrimidine dimers are not bypassed in the normal manner, leading to numerous mutations that eventually produce skin cancer.

T

he synthesis of DNA is a complex process, fundamental
to cell function and health, in which dozens of proteins,
enzymes, and DNA structures take part in the copying of
DNA. A single defective component, such as DNA polymerase ␩, can disrupt the whole process and result in severe
disease symptoms.
This chapter focuses on DNA replication, the process by
which a cell doubles its DNA before division. We begin with
the basic mechanism of replication that emerged from the
Watson and Crick structure of DNA. We then examine several
different modes of replication, the requirements of replication, and the universal direction of DNA synthesis. We examine the enzymes and proteins that participate in this process
and conclude the chapter by considering the molecular details
of recombination, which is closely related to replication and is
essential for the segregation of homologous chromosomes, for
the production of genetic variation, and for DNA repair.

9.1 Genetic Information Must
Be Accurately Copied Every
Time a Cell Divides
In a schoolyard game, a verbal message, such as “John’s
brown dog ran away from home,” is whispered to a child,
who runs to a second child and repeats the message. The
message is relayed from child to child around the schoolyard
until it returns to the original sender. Inevitably, the last child
returns with an amazingly transformed message, such as “Joe
Brown has a pig living under his porch.” The more children
playing the game, the more garbled the message becomes.
This game illustrates an important principle: errors arise
whenever information is copied; the more times it is copied,
the greater the number of errors.
A complex, multicellular organism faces a problem
analogous to that of the children in the schoolyard game:
how to faithfully transmit genetic instructions each time its
cells divide. The solution to this problem is central to replication. A huge amount of genetic information and an enormous number of cell divisions are required to produce a
multicellular adult organism; even a low rate of error during
copying would be catastrophic. A single-celled human zygote
contains 6.4 billion base pairs of DNA. If a copying error was
made only once per million base pairs, 6400 mistakes would
be made every time a cell divided—errors that would be
compounded at each of the millions of cell divisions that
take place in human development.
Not only must the copying of DNA be astoundingly
accurate, it must also take place at breakneck speed. The

single, circular chromosome of E. coli contains about 4.6
million base pairs. At a rate of more than 1000 nucleotides
per minute, replication of the entire chromosome would
require almost 3 days. Yet, as already stated, these bacteria are
capable of dividing every 20 minutes. E. coli actually replicates its DNA at a rate of 1000 nucleotides per second, with
less than one error in a billion nucleotides. How is this extraordinarily accurate and rapid process accomplished?

9.2 All DNA Replication Takes
Place in a Semiconservative
Manner
From the three-dimensional structure of DNA proposed by
Watson and Crick in 1953 (see Figure 8.12), several important
genetic implications were immediately apparent. The complementary nature of the two nucleotide strands in a DNA
molecule suggested that, during replication, each strand can
serve as a template for the synthesis of a new strand. The
specificity of base pairing (adenine with thymine; guanine
with cytosine) implied that only one sequence of bases can
be specified by each template, and so two DNA molecules
built on the pair of templates will be identical with the original. This process is called semiconservative replication,
because each of the original nucleotide strands remains
intact (conserved), despite no longer being combined in the
same molecule; the original DNA molecule is half (semi)
conserved during replication.
Initially, three alternative models were proposed for DNA
replication. In conservative replication (Figure 9.1a), the
entire double-stranded DNA molecule serves as a template for
a whole new molecule of DNA, and the original DNA molecule is fully conserved during replication. In dispersive replication (Figure 9.1b), both nucleotide strands break down
(disperse) into fragments, which serve as templates for the
synthesis of new DNA fragments, and then somehow reassemble into two complete DNA molecules. In this model, each
resulting DNA molecule is interspersed with fragments of old
and new DNA; none of the original molecule is conserved.
Semiconservative replication (Figure 9.1c) is intermediate
between these two models; the two nucleotide strands unwind
and each serves as a template for a new DNA molecule.
These three models allow different predictions to be
made about the distribution of original DNA and newly synthesized DNA after replication. With conservative replication,
after one round of replication, 50% of the molecules would
consist entirely of the original DNA and 50% would consist
entirely of new DNA. After a second round of replication,

DNA Replication and Recombination

(a) Conservative replication

(b) Dispersive replication

(c) Semiconservative replication

Original DNA

First replication

Second replication

9.1 Three proposed models of replication are conservative replication, dispersive
replication, and semiconservative replication.

25% of the molecules would consist entirely of the original
DNA and 75% would consist entirely of new DNA. With each
additional round of replication, the proportion of molecules
with new DNA would increase, although the number of molecules with the original DNA would remain constant.
Dispersive replication would always produce hybrid molecules, containing some original and some new DNA, but the
proportion of new DNA within the molecules would increase
with each replication event. In contrast, with semiconservative replication, one round of replication would produce two
hybrid molecules, each consisting of half original DNA and
half new DNA. After a second round of replication, half the
molecules would be hybrid, and the other half would consist
of new DNA only. Additional rounds of replication would
produce more and more molecules consisting entirely of new
DNA, and a few hybrid molecules would persist.

Meselson and Stahl distinguished between the heavy
N-laden DNA and the light 14N-containing DNA with
the use of equilibrium density gradient centrifugation
(Figure 9.2). In this technique, a centrifuge tube is filled with
15

A centrifuge tube is filled
with a heavy salt solution
and DNA fragments.

Meselson and Stahl’s Experiment
To determine which of the three models of replication
applied to E. coli cells, Matthew Meselson and Franklin Stahl
needed a way to distinguish old and new DNA. They did so
by using two isotopes of nitrogen, 14N (the common form)
and 15N (a rare, heavy form). Meselson and Stahl grew a culture of E. coli in a medium that contained 15N as the sole
nitrogen source; after many generations, all the E. coli cells
had 15N incorporated into the purine and pyrimidine bases
of DNA (see Figure 8.8). Meselson and Stahl took a sample
of these bacteria, switched the rest of the bacteria to a
medium that contained only 14N, and then took additional
samples of bacteria over the next few cellular generations. In
each sample, the bacterial DNA that was synthesized before
the change in medium contained 15N and was relatively
heavy, whereas any DNA synthesized after the switch contained 14N and was relatively light.

It is then spun in a centrifuge
at high speeds for several days.

DNA with

14N

DNA with

15N

A density gradient develops
within the tube. Heavy DNA
(with 15N) will move toward the
bottom; light DNA (with 14N)
will remain closer to the top.

9.2 Meselson and Stahl used equilibrium density gradient

centrifugation to distinguish between heavy, 15N-laden DNA
and lighter, 14N-laden DNA.

221