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5: Chromosome Variation Playsan Important Role in Evolution

5: Chromosome Variation Playsan Important Role in Evolution

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188

Chapter 7

Concepts Summary
• Three basic types of chromosome mutations are:
(1) chromosome rearrangements, which are changes in the
structure of chromosomes; (2) aneuploidy, which is an increase
or decrease in chromosome number; and (3) polyploidy, which
is the presence of extra chromosome sets.
• Chromosome rearrangements include duplications, deletions,
inversions, and translocations.
• In individuals heterozygous for a duplication, the duplicated
region will form a loop when homologous chromosomes pair
in meiosis. Duplications often have pronounced effects on the
phenotype owing to unbalanced gene dosage.
• In individuals heterozygous for a deletion, one of the
chromosomes will loop out during pairing in meiosis. Deletions
may cause recessive alleles to be expressed.
• Pericentric inversions include the centromere; paracentric
inversions do not. In individuals heterozygous for an inversion,
the homologous chromosomes form inversion loops in
meiosis, with reduced recombination taking place within the
inverted region.
• In translocation heterozygotes, the chromosomes form
crosslike structures in meiosis.

• Fragile sites are constrictions or gaps that appear at particular
regions on the chromosomes of cells grown in culture and are
prone to breakage under certain conditions.
• Nullisomy is the loss of two homologous chromosomes;
monosomy is the loss of one homologous chromosome;
trisomy is the addition of one homologous chromosome;
tetrasomy is the addition of two homologous chromosomes.
• Aneuploidy usually causes drastic phenotypic effects because it
leads to unbalanced gene dosage.
• Primary Down syndrome is caused by the presence of three full
copies of chromosome 21, whereas familial Down syndrome is
caused by the presence of two normal copies of chromosome
21 and a third copy that is attached to another chromosome
through a translocation.
• All the chromosomes in an autopolyploid derive from one
species; chromosomes in an allopolyploid come from two or
more species.
• Chromosome variations have played an important role in the
evolution of many groups of organisms.

Important Terms
chromosome mutation (p. 168)
metacentric chromosome (p. 168)
submetacentric chromosome (p. 168)
acrocentric chromosome (p. 168)
telocentric chromosome (p. 168)
chromosome rearrangement (p. 170)
chromosome duplication (p. 170)
tandem duplication (p. 171)
displaced duplication (p. 171)
reverse duplication (p. 171)
chromosome deletion (p. 173)
pseudodominance (p. 174)
haploinsufficient gene (p. 174)
chromosome inversion (p. 174)
paracentric inversion (p. 174)

pericentric inversion (p. 174)
position effect (p. 174)
dicentric chromatid (p. 175)
acentric chromatid (p. 175)
dicentric bridge (p. 175)
translocation (p. 176)
nonreciprocal translocation (p. 176)
reciprocal translocation (p. 176)
Robertsonian translocation (p. 176)
fragile site (p. 178)
aneuploidy (p. 178)
polyploidy (p. 178)
nondisjunction (p. 178)
nullisomy (p. 178)
monosomy (p. 178)

trisomy (p. 178)
tetrasomy (p. 178)
Down syndrome (trisomy 21) (p. 180)
primary Down syndrome (p. 180)
familial Down syndrome (p. 180)
translocation carrier (p. 181)
Edward syndrome (trisomy 18) (p. 181)
Patau syndrome (trisomy 13) (p. 181)
trisomy 8 (p. 181)
autopolyploidy (p. 182)
allopolyploidy (p. 182)
unbalanced gametes (p. 183)
amphidiploid (p. 184)

Answers to Concept Checks
1. a
2. Pseudodominance is the expression of a recessive mutation.
It is produced when the wild-type allele in a heterozygous
individual is absent due to a deletion on one chromosome.
3. c
4. b
5. 37

6. Dosage compensation prevents the expression of additional
copies of X-linked genes in mammals, and there is little information
on the Y chromosome; so extra copies of the X and Y chromosomes
do not have major effects on development. In contrast, there is no
mechanism of dosage compensation for autosomes, and so extra
copies of autosomal genes are expressed, upsetting development
and causing the spontaneous abortion of aneuploid embryos.
7. c

Chromosome Variation

189

Worked Problems
1. A chromosome has the following segments, where • represents
the centromere.
ABCDE•FG
What types of chromosome mutations are required to change
this chromosome into each of the following chromosomes?
(In some cases, more than one chromosome mutation may
be required.)
a.
b.
c.
d.
e.

ABE•FG
AEDCB•FG
ABABCDE•FG
AF•EDCBG
ABCDEEDC•FG

• Solution
The types of chromosome mutations are identified by comparing
the mutated chromosome with the original, wild-type
chromosome.
a. The mutated chromosome (A B E • F G) is missing segment
C D; so this mutation is a deletion.
b. The mutated chromosome (A E D C B • F G) has one and
only one copy of all the gene segments, but segment B C D E
has been inverted 180 degrees. Because the centromere has not
changed location and is not in the inverted region, this
chromosome mutation is a paracentric inversion.
c. The mutated chromosome (A B A B C D E • F G) is longer
than normal, and we see that segment A B has been duplicated.
This mutation is a tandem duplication.
d. The mutated chromosome (A F • E D C B G) is normal
length, but the gene order and the location of the centromere
have changed; this mutation is therefore a pericentric inversion
of region (B C D E • F).
e. The mutated chromosome (A B C D E E D C • F G) contains
a duplication (C D E) that is also inverted; so this chromosome
has undergone a duplication and a paracentric inversion.
2. Species I is diploid (2n = 4) with chromosomes AABB; related
species II is diploid (2n = 6) with chromosomes MMNNOO. Give
the chromosomes that would be found in individuals with the
following chromosome mutations.

a. Autotriploidy in species I
b. Allotetraploidy including species I and II
c. Monosomy in species I
d. Trisomy in species II for chromosome M
e. Tetrasomy in species I for chromosome A
f. Allotriploidy including species I and II
g. Nullisomy in species II for chromosome N
• Solution
To work this problem, we should first determine the haploid
genome complement for each species. For species I, n = 2
with chromosomes AB and, for species II, n = 3 with
chromosomes MNO.
a. An autotriploid is 3n, with all the chromosomes coming from
a single species; so an autotriploid of species I would have
chromosomes AAABBB (3n = 6).
b. An allotetraploid is 4n, with the chromosomes coming from
more than one species. An allotetraploid could consist of 2n from
species I and 2n from species II, giving the allotetraploid (4n =
2 + 2 + 3 + 3 = 10) chromosomes AABBMMNNOO. An
allotetraploid could also possess 3n from species I and 1n from
species II (4n = 2+ 2 + 2 + 3 = 9; AAABBBMNO) or 1n from
species I and 3n from species II (4n = 2 + 3 + 3 + 3 = 11;
ABMMMNNNOOO).
c. A monosomic is missing a single chromosome; so a
monosomic for species I would be 2n – 1 = 4 – 1 = 3. The
monosomy might include either of the two chromosome pairs,
with chromosomes ABB or AAB.
d. Trisomy requires an extra chromosome; so a trisomic
of species II for chromosome M would be 2n + 1 = 6 + 1 =
7 (MMMNNOO).
e. A tetrasomic has two extra homologous chromosomes; so a
tetrasomic of species I for chromosome A would be 2n + 2 = 4 +
2 = 6 (AAAABB).
f. An allotriploid is 3n with the chromosomes coming from two
different species; so an allotriploid could be 3n = 2 + 2 + 3 =
7 (AABBMNO) or 3n = 2 + 3 + 3 = 8 (ABMMNNOO).
g. A nullisomic is missing both chromosomes of a homologous
pair; so a nullisomic of species II for chromosome N would be
2n – 2 = 6 – 2 = 4 (MMOO).

Comprehension Questions
Section 7.1
*1. List the different types of chromosome mutations and
define each one.

Section 7.2
*2. Why do extra copies of genes sometimes cause drastic
phenotypic effects?

190

Chapter 7

3. Draw a pair of chromosomes as they would appear during
synapsis in prophase I of meiosis in an individual
heterozygous for a chromosome duplication.
*4. What is the difference between a paracentric and a
pericentric inversion?
5. How do inversions cause phenotypic effects?
6. Explain why recombination is suppressed in individuals
heterozygous for paracentric inversions.
*7. How do translocations produce phenotypic effects?

Section 7.3
8. List four major types of aneuploidy.
*9. What is the difference between primary Down syndrome
and familial Down syndrome? How does each type arise?

Section 7.4
*10. What is the difference between autopolyploidy and
allopolyploidy? How does each arise?
11. Explain why autopolyploids are usually sterile, whereas
allopolyploids are often fertile.

Application Questions and Problems
Section 7.1
*12. Which types of chromosome mutations
a. increase the amount of genetic material in a particular
chromosome?
b. increase the amount of genetic material in all
chromosomes?
c. decrease the amount of genetic material in a particular
chromosome?
d. change the position of DNA sequences in a single
chromosome without changing the amount of genetic
material?
e. move DNA from one chromosome to a nonhomologous
chromosome?

Section 7.2
*13. A chromosome has the following segments, where •
represents the centromere:
AB•CDEFG
What types of chromosome mutations are required to
change this chromosome into each of the following
chromosomes? (In some cases, more than one chromosome
mutation may be required.)
a. A B A B • C D E F G
b. A B • C D E A B F G
c. A B • C F E D G
d. A • C D E F G
e. A B • C D E
f. A B • E D C F G
g. C • B A D E F G
h. A B • C F E D F E D G
i. A B • C D E F C D F E G
14. The following diagram represents two nonhomologous
chromosomes:
AB•CDEFG
RS•TUVWX

What type of chromosome mutation would produce each of
the following chromosomes?
a. A B • C D
RS•TUVWXEFG
b. A U V B • C D E F G
RS•TWX
c. A B • T U V F G
RS•CDEWX
d. A B • C W G
RS•TUVDEFX
15. The green-nose fly normally has six chromosomes, two
metacentric and four acrocentric. A geneticist examines the
chromosomes of an odd-looking green-nose fly and
discovers that it has only five chromosomes; three of them
are metacentric and two are acrocentric. Explain how this
change in chromosome number might have taken place.
*16. A wild-type chromosome has the following segments:
ABC•DEFGHI
An individual is heterozygous for the following chromosome
mutations. For each mutation, sketch how the wild-type and
mutated chromosomes would pair in prophase I of meiosis,
showing all chromosome strands.
a. A B C • D E F D E F G H I
b. A B C • D H I
c. A B C • D G F E H I
d. A B E D • C F G H I
17. As discussed in this chapter, crossing over within a
DATA pericentric inversion produces chromosomes that have extra
copies of some genes and no copies of other genes. The
ANALYSIS
fertilization of gametes containing such duplication/
deficient chromosomes often results in children with
syndromes characterized by developmental delay, mental
retardation, abnormal development of organ systems, and
early death. Using a special two-color FISH (fluorescence in

Chromosome Variation

situ hybridization) analysis that revealed the presence of
crossing over within pericentric inversions, Maarit Jaarola
and colleagues examined individual sperm cells of a male
who was heterozygous for a pericentric inversion on
chromosome 8 and determined that crossing over took
place within the pericentric inversion in 26% of the meiotic
divisions (M. Jaarola, R. H. Martin, and T. Ashley. 1998.
American Journal of Human Genetics 63:218–224).
Assume that you are a genetic counselor and that a
couple seeks genetic counseling from you. Both the man
and the woman are phenotypically normal, but the woman
is heterozygous for a pericentric inversion on chromosome
8. The man is karyotypically normal. What is the probability
that this couple will produce a child with a debilitating
syndrome as the result of crossing over within the
pericentric inversion?

Section 7.3
18. Red–green color blindness is a human X-linked recessive
disorder. A young man with a 47,XXY karyotype (Klinefelter
syndrome) is color blind. His 46,XY brother also is color
blind. Both parents have normal color vision. Where did the
nondisjunction occur that gave rise to the young man with
Klinefelter syndrome?
*19. Bill and Betty have had two children with Down syndrome.
Bill’s brother has Down syndrome and his sister has two
children with Down syndrome. On the basis of these
observations, which of the following statements is most
likely correct? Explain your reasoning.
a. Bill has 47 chromosomes.
b. Betty has 47 chromosomes.
c. Bill and Betty’s children each have 47 chromosomes.
d. Bill’s sister has 45 chromosomes.
e. Bill has 46 chromosomes.
f. Betty has 45 chromosomes.
g. Bill’s brother has 45 chromosomes.
*20. In mammals, sex-chromosome aneuploids are more
common than autosomal aneuploids but, in fish, sexchromosome aneuploids and autosomal aneuploids are
found with equal frequency. Offer an explanation for these
differences in mammals and fish.

Section 7.4
*21. Species I has 2n ϭ 16 chromosomes. How many
chromosomes will be found per cell in each of the
following mutants in this species?
a. Monosomic
e. Double monosomic
b. Autotriploid
f. Nullisomic
c. Autotetraploid
g. Autopentaploid
d. Trisomic
h. Tetrasomic
22. Species I is diploid (2n ϭ 8) with chromosomes
AABBCCDD; related species II is diploid (2n ϭ 8) with

191

chromosomes MMNNOOPP. What types of chromosome
mutations do individual organisms with the following sets
of chromosomes have?
a. AAABBCCDD
e. AAABBCCDDD
b. MMNNOOOOPP
f. AABBDD
c. AABBCDD
g. AABBCCDDMMNNOOPP
d. AAABBBCCCDDD
h. AABBCCDDMNOP
23. Species I has 2n ϭ 8 chromosomes and species II has
2n ϭ 14 chromosomes. What would be the expected
chromosome numbers in individual organisms with the
following chromosome mutations? Give all possible answers.
a. Allotriploidy including species I and II
b. Autotetraploidy in species II
c. Trisomy in species I
d. Monosomy in species II
e. Tetrasomy in species I
f. Allotetraploidy including species I and II
24. Nicotiana glutinosa (2n ϭ 24) and N. tabacum (2n ϭ 48)
DATA are two closely related plants that can be intercrossed, but
the F1 hybrid plants that result are usually sterile. In 1925,
ANALYSIS
Roy Clausen and Thomas Goodspeed crossed N. glutinosa
and N. tabacum, and obtained one fertile F1 plant (R. E.
Clausen and T. H. Goodspeed. 1925 Genetics 10:278–284).
They were able to self-pollinate the flowers of this plant to
produce an F2 generation. Surprisingly, the F2 plants were
fully fertile and produced viable seed. When Clausen and
Goodspeed examined the chromosomes of the F2 plants,
they observed 36 pairs of chromosomes in metaphase I and
36 individual chromosomes in metaphase II. Explain the
origin of the F2 plants obtained by Clausen and Goodspeed
and the numbers of chromosomes observed.
25. What would be the chromosome number of progeny
resulting from the following crosses in wheat (see Figure
7.25)? What type of polyploid (allotriploid, allotetraploid,
etc.) would result from each cross?
a. Einkorn wheat and Emmer wheat
b. Bread wheat and Emmer wheat
c. Einkorn wheat and bread wheat
26. Karl and Hally Sax crossed Aegilops cylindrical (2n ϭ 28), a
DATA
wild grass found in the Mediterranean region, with Triticum
vulgare (2n ϭ 42), a type of wheat (K. Sax and H. J. Sax. 1924.
ANALYSIS
Genetics 9:454–464). The resulting F1 plants from this cross
had 35 chromosomes. Examination of metaphase I in the F1
plants revealed the presence of 7 pairs of chromosomes
(bivalents) and 21 unpaired chromosomes (univalents).
a. If the unpaired chromosomes segregate randomly, what
possible chromosome numbers will appear in the
gametes of the F1 plants?
b. What does the appearance of the bivalents in the F1
hybrids suggest about the origin of Triticum vulgare
wheat?

192

Chapter 7

Challenge Questions
Section 7.3
27. Red–green color blindness is a human X-linked recessive
disorder. Jill has normal color vision, but her father is color
blind. Jill marries Tom, who also has normal color vision.
Jill and Tom have a daughter who has Turner syndrome
and is color blind.
a. How did the daughter inherit color blindness?
b. Did the daughter inherit her X chromosome from Jill or
from Tom?
28. Mules result from a cross between a horse (2n = 64) and a
DATA
donkey (2n = 62), have 63 chromosomes and are almost
ANALYSIS always sterile. However, in the summer of 1985, a female
mule named Krause who was pastured with a male donkey
gave birth to a newborn foal (O. A. Ryder et al. 1985. Journal
of Heredity 76:379–381). Blood tests established that the
male foal, appropriately named Blue Moon, was the offspring of Krause and that Krause was indeed a mule. Both
Blue Moon and Krause were fathered by the same donkey
(see the illustration). The foal, like his mother, had 63
chromosomes—half of them horse chromosomes and the
other half donkey chromosomes. Analyses of genetic
markers showed that, remarkably, Blue Moon seemed to
have inherited a complete set of horse chromosomes from
his mother, instead of a random mixture of horse and
donkey chromosomes that would be expected with normal
meiosis. Thus, Blue Moon and Krause were not only mother
and son, but also brother and sister.
I

II

III

Donkey
2n = 62

Horse
2n = 64
Mule “Krause”
2n = 63, XX
Mule “Blue Moon”
2n = 63, XY

a. With the use of a diagram, show how, if Blue Moon
inherited only horse chromosomes from his mother,
Blue Moon and Krause are both mother and son as well
as brother and sister.
b. Although rare, additional cases of fertile mules giving
births to offspring have been reported. In these cases,
when a female mule mates with a male horse, the offspring is horselike in appearance but, when a female
mule mates with a male donkey, the offspring is mulelike in appearance. Is this observation consistent with
the idea that the offspring of fertile female mules inherit
only a set of horse chromosomes from their mule mothers? Explain your reasoning.
c. Can you suggest a possible mechanism for how the offspring of fertile female mules might pass on a complete
set of horse chromosomes to their offspring?

Section 7.5
29. Humans and many other complex organisms are diploid,
possessing two sets of genes, one inherited from the mother
and one from the father. However, a number of eukaryotic
organisms spend most of their life cycles in a haploid state.
Many of these eukaryotes, such as Neurospora and yeast, still
undergo meiosis and sexual reproduction, but most of the
cells that make up the organism are haploid.
Considering that haploid organisms are fully capable of
sexual reproduction and generating genetic variation, why
are most complex eukaryotes diploid? In other words, what
might be the evolutionary advantage of existing in a diploid
state instead of a haploid state? And why might a few
organisms, such as Neurospora and yeast, exist as haploids?

8

DNA : The Chemical
Nature of the Gene
Neanderthal’s DNA

D

The remarkable stability of DNA facilitates the extraction and
analysis of DNA from ancient remains, including Neanderthal bones
that are more than 30,000 years old. [John Reader/Photo Researchers.]

NA, with its double-stranded spiral, is among the most
elegant of all biological molecules. But the double helix
is not just a beautiful structure; it also gives DNA incredible
stability and permanence, providing geneticists with a unique
window to the past.
In 1856, a group of men working a limestone quarry in the
Neander Valley of Germany discovered a small cave containing
a number of bones. The workers assumed that the bones were
those of a cave bear, but a local schoolteacher recognized them
as human, although they were clearly unlike any human bones
the teacher had ever seen. The bones appeared to be those of a
large person with great muscular strength, a low forehead, a
large nose with broad nostrils, and massive protruding brows.
Experts confirmed that the bones belonged to an extinct
human, who became known as Neanderthal. In the next 100
years, similar fossils were discovered in Spain, Belgium, France,
Croatia, and the Middle East.
Research has now revealed that Neanderthals roamed
Europe and western Asia for at least 200,000 years, disappearing abruptly 30,000 to 40,000 years ago. During the last years
of this period, Neanderthals coexisted with the direct ancestors
of modern humans, the Cro-Magnons. The fate of the
Neanderthals—why they disappeared—has captured the
imagination of scientists and laypersons alike. Did CroMagnons, migrating out of Africa with a superior technology,
cause the demise of the Neanderthal people, either through
competition or perhaps through deliberate extermination? Or
did the Neanderthals interbreed with Cro-Magnons, their
genes becoming assimilated into the larger gene pool of modern humans? Support for the latter hypothesis came from the
discovery of fossils that appeared to be transitional between
Neanderthals and Cro-Magnons. Unfortunately, the meager
fossil record of Neanderthals and Cro-Magnons did not allow
a definitive resolution of these questions.
Whether Neanderthals interbred with the ancestors of
modern humans became a testable question in 1997, when scientists extracted DNA from a bone of the original Neanderthal
specimen collected from the Neander Valley more than 140
years earlier. This Neanderthal bone was at least 30,000 years

193

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Chapter 8

old. Using a technique called the polymerase chain reaction (see Chapter 14), the scientists
amplified 378 nucleotides of the Neanderthal’s mitochondrial DNA a millionfold. They
then determined the base sequence of this amplified DNA, and compared it with mitochondrial sequences from living humans.
The mitochondrial DNA from the Neanderthal bone differed markedly from mitochondrial DNAs that were isolated from numerous humans throughout the world. DNA
from many different Neanderthal and Cro-Magnon specimens has provided clear evidence
that Neanderthals were genetically distinct from both present-day humans and ancient
Cro-Magnons. There is no Neanderthal mitochondrial DNA in the present-day human
gene pool or in DNA extracted from fossils of early Cro-Magnons, suggesting that interbreeding between the two forms did not take place. Recently, nuclear DNA has been
extracted from Neanderthal bones, and scientists are currently attempting to sequence the
entire Neanderthal genome, which will provide a more definitive test of possible interbreeding between early humans and Neanderthals.

T

hat Neanderthal’s DNA persists and reveals its genetic
affinities, even 40,000 years later, is testimony to the
remarkable stability of the double helix. This chapter focuses
on how DNA was identified as the source of genetic information and how this elegant molecule encodes the genetic
instructions. We begin by considering the basic requirements
of the genetic material and the history of our understanding
of DNA—how its relation to genes was uncovered and its
structure determined. The history of DNA illustrates several
important points about the nature of scientific research. As
with so many important scientific advances, the structure of
DNA and its role as the genetic material were not discovered
by any single person but were gradually revealed over a
period of almost 100 years, thanks to the work of many
investigators. Our understanding of the relation between
DNA and genes was enormously enhanced in 1953, when
James Watson and Francis Crick proposed a three-dimensional structure for DNA that brilliantly illuminated its role
in genetics. As illustrated by Watson and Crick’s discovery,
major scientific advances are often achieved, not through the
collection of new data, but through the interpretation of old
data in new ways.
After reviewing the history of DNA, we will examine
DNA structure. The structure of DNA is important in its
own right, but the key genetic concept is the relation between
the structure and the function of DNA—how its structure
allows it to serve as the genetic material.

8.1 Genetic Material Possesses
Several Key Characteristics
Life is characterized by tremendous diversity, but the coding
instructions of all living organisms are written in the same
genetic language—that of nucleic acids. Surprisingly, the
idea that genes are made of nucleic acids was not widely

accepted until after 1950. This skepticism was due in part to
a lack of knowledge about the structure of deoxyribonucleic
acid (DNA). Until the structure of DNA was understood,
how DNA could store and transmit genetic information was
unclear. Even before nucleic acids were identified as the
genetic material, biologists recognized that, whatever the
nature of the genetic material, it must possess three important characteristics.
1. Genetic material must contain complex information.
First and foremost, the genetic material must be capable
of storing large amounts of information—instructions
for all the traits and functions of an organism. This
information must have the capacity to vary, because
different species and even individual members of a
species differ in their genetic makeup. At the same time,
the genetic material must be stable, because most
alterations to the genetic instructions (mutations) are
likely to be detrimental.
2. Genetic material must replicate faithfully. A second
necessary feature is that genetic material must have the
capacity to be copied accurately. Every organism begins
life as a single cell, which must undergo billions of cell
divisions to produce a complex, multicellular creature like
yourself. At each cell division, the genetic instructions
must be transmitted to descendant cells with great
accuracy. When organisms reproduce and pass genes on
to their progeny, the coding instructions must be copied
with fidelity.
3. Genetic material must encode the phenotype. The
genetic material (the genotype) must have the capacity
to “code for” (determine) traits (the phenotype). The
product of a gene is often a protein; so there must be a
mechanism for genetic instructions to be translated into
the amino acid sequence of a protein.

DNA: The Chemical Nature of the Gene

Concepts

Table 8.1

Base composition of DNA from
different sources and ratios of
bases

The genetic material must be capable of carrying large amounts of
information, replicating faithfully, and translating its coding
instructions into phenotypes.

Ratio

✔ Concept Check 1
Why was the discovery of the structure of DNA so important for
understanding genetics?

8.2 All Genetic Information
Is Encoded in the Structure
of DNA
Although our understanding of how DNA encodes genetic
information is relatively recent, the study of DNA structure
stretches back more than 100 years.

Early Studies of DNA
In 1868, Johann Friedrich Miescher graduated from medical
school in Switzerland. Influenced by an uncle who believed
that the key to understanding disease lay in the chemistry of
tissues, Miescher traveled to Tübingen, Germany, to study
under Ernst Felix Hoppe-Seyler, an early leader in the emerging field of biochemistry. Under Hoppe-Seyler’s direction,
Miescher turned his attention to the chemistry of pus, a substance of clear medical importance. Pus contains white
blood cells with large nuclei. Miescher determined that the
nucleus contained a novel substance that was slightly acidic
and high in phosphorus. This material consisted of DNA and
protein.
By 1887, researchers had concluded that the physical
basis of heredity lies in the nucleus. Chromatin was shown
to consist of nucleic acid and proteins, but which of these
substances is actually the genetic information was not clear.
In the late 1800s, further work on the chemistry of DNA was
carried out by Albrecht Kossel, who determined that DNA
contains four nitrogenous bases: adenine, cytosine, guanine,
and thymine (abbreviated A, C, G, and T).
Phoebus Aaron Levene later showed that DNA consists
of a large number of linked, repeating units called
nucleotides; each nucleotide contains a sugar, a phosphate,
and a base.
Base
Phosphate
Sugar
Nucleotide
Levene incorrectly proposed that DNA consists of a series of
four-nucleotide units, each unit containing all four bases—

Source
of DNA

A

T

G

C

A/T

G/C

(A ϩ G)/
(T ϩ C)

E. coli

26.0 23.9 24.9 25.2

1.09 0.99

1.04

Yeast

31.3 32.9 18.7 17.1

0.95 1.09

1.00

Sea urchin

32.8 32.1 17.7 18.4

1.02 0.96

1.00

Rat

28.6 28.4 21.4 21.5

1.01 1.00

1.00

Human

30.3 30.3 19.5 19.9

1.00 0.98

0.99

adenine, guanine, cytosine, and thymine—in a fixed
sequence. This concept, known as the tetranucleotide theory,
implied that the structure of DNA is not variable enough to
be the genetic material. The tetranucleotide theory contributed to the idea that protein is the genetic material
because, with its 20 different amino acids, protein structure
could be highly variable.
As additional studies of the chemistry of DNA were
completed in the 1940s and 1950s, this notion of DNA as a
simple, invariant molecule began to change. Erwin Chargaff
and his colleagues carefully measured the amounts of the
four bases in DNA from a variety of organisms and found
that DNA from different organisms varies greatly in base
composition. This finding disproved the tetranucleotide theory. They discovered that, within each species, there is some
regularity in the ratios of the bases: the amount of adenine is
always equal to the amount of thymine (A ϭ T), and the
amount of guanine is always equal to the amount of cytosine
(G ϭ C; Table 8.1). These findings became known as
Chargaff ’s rules.

Concepts
Details of the structure of DNA were worked out by a number of
scientists. At first, DNA was interpreted as being too regular in
structure to carry genetic information but, by the 1940s, DNA from
different organisms was shown to vary in its base composition.

DNA As the Source of Genetic
Information
While chemists were working out the structure of DNA, biologists were attempting to identify the source of genetic information. Mendel identified the basic rules of heredity in 1866,
but he had no idea about the physical nature of hereditary
information. By the early 1900s, biologists had concluded
that genes reside on chromosomes, which were known to

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