17
Nucleic Acids,
Protein Synthesis,
and Heredity
C
hristine and Jeff Callahan have just had their first child, a healthy
baby boy whom they named Aiden. The day after Aiden is born, the nurse takes
a small blood sample from Aiden’s heel to test for an elevated concentration of the amino
acid phenylalanine in his blood. The nurse tells Christine that this test can detect phenylketonuria (PKU), a disease in which the body cannot convert phenylalanine into tyrosine.
Children with untreated PKU have high concentrations of phenylalanine in their blood,
causing severe, irreversible mental retardation in the first year of life. Therefore, it is critical
that PKU be diagnosed early so that babies with the disease can be fed a diet that contains
only the phenylalanine they need to build proteins.
Christine is puzzled about PKU: how does a baby catch this disease? The nurse tells her
that PKU is a genetic disorder, an error in the genetic code that determines a person’s
biological makeup. In PKU, the code for a specific enzyme that oxidizes phenylalanine is
incorrect, so the body produces an inactive version of the enzyme. Because this code occurs
on two different chromosomes, a baby will only have PKU if he or she inherits a defective
gene from both parents. A healthy adult may still have one defective copy of this gene, so it
is possible for a baby to have PKU even if neither parent has the disease.
The next day, the nurse tells Christine that Aiden’s phenylalanine level is in the normal
range, and a follow-up blood test two weeks later confirms that Aiden does not have PKU.
OUTLINE
17.1
17.2
17.3
17.4
17.5
17.6
17.7
Nucleotides
Structures of nucleic acids
DNA replication
Transcription and RNA processing
Translation and the genetic code
The mechanism of protein synthesis
LABORATORY REQUISITION—CHEMISTRY
c Sodium
c Potassium
c Creatinine
c BUN
c Glucose–fasting
c Glucose–random
c Hemoglobin A1C
NA
K
CREAT
BUN
GLUCF
GLUCR
HGBA1C
c Fructosamine
c PSA
c Chloride
c Calcium
c Phosphorus
c Phenylalanine
c Uric Acid
FRU/ALB
PSA
CL
CA
PHOS
PKU
URIC
Mutations and genetic disorders
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Diane Macdonald/Getty Images
c Total Bilirubin
c Neonate T. Bilirubin
c Serum Protein Elect.
c Ferritin
c Iron/TIBC
c Hgb Electrophoresis
c T4/FTI
93493_Ch17_722-765.indd 723
BILIT
BILITN
PEP
FERR
IRON/TIBC
HGB EP
T4S
c TSH
c Alk Phos
c SGOT (AST)
c Albumin
c SGPT (ALT)
c CPK (CK)
c CKMB
TSH
ALKP
AST
ALB
ALT
CK
CKMB
c Amylase
c Cholesterol
c HDL
c LDL–fasting
c Triglycerides–fasting
AMYL
CHOL
HDL
LDL
TRIG
All infants in the United States are
screened for PKU during their first week
of life.
10/15/10 11:41:13 AM
Online homework
for this chapter may
be assigned in OWL.
OBJECTIVES: Understand the
biological role of nucleotides, draw the
structure of a typical nucleotide, and
describe the differences between DNA and
RNA nucleotides.
Has anyone ever told you that you have your mother’s nose or your grandfather’s chin?
That you got your height from your father’s side of the family? That your singing ability
(or lack of it) runs in the family? We are all aware that people tend to resemble their
closest relations, and we know that we inherit many of our defining characteristics and
traits from our parents. However, this inheritance runs deeper than hair color or foot
size. We also inherit our basic anatomical design from our parents. No human parents
ever produced a chimpanzee, a lobster, or a chrysanthemum (contrary to what you may
have read in the supermarket checkout line).
Gregor Mendel, a Catholic monk and schoolteacher, uncovered the fundamental biological principles of heredity in the late 1800s. However, only in the last half-century
have researchers come to understand the chemical basis of heredity. The discovery of the
structure and function of DNA, the molecule that allows biological traits to be passed
from parent to child, has led to a true revolution in science, comparable in its impact to
Newton’s laws of motion or Darwin’s theory of evolution. From the medicine we take
to the food we eat, from solving crimes to diagnosing disease, the chemistry of heredity
has affected all of our lives.
In this chapter, we will explore the chemical basis of biological inheritance. Our starting point is a class of molecules called nucleic acids. We will look at the structures of
these compounds, and then we will explore how they control biological growth and
development.
17.1 Nucleotides
In 1869, a Swiss professor of physiology named Friedrich Miescher discovered an unusual substance in the nuclei of white blood cells. Over the next few years, he found that
this substance was a mixture of proteins and a previously unknown chemical compound,
which came to be called nucleic acid. Shortly afterward, a slightly different variety of
nucleic acid was discovered in yeast cells. Chemists found that one type of nucleic acid
contained the five-carbon sugar ribose, while the other contained a modified sugar called
deoxyribose. Because this seemed to be the main difference between the two nucleic
acids, these compounds were named ribonucleic acid (RNA) and deoxyribonucleic acid
(DNA).
Although most cells contain a good deal of nucleic acid, these compounds seemed to
have no particular function, so they were virtually ignored for many decades. However,
in 1944, Oswald Avery and his colleagues discovered that DNA is the carrier of genetic
information. This discovery triggered an explosion of research into the structure and
properties of nucleic acids, culminating with the announcement of the complete structure
of DNA by Francis Crick and James Watson in 1953. In this section we will examine the
building blocks of the nucleic acids, and in Section 17.2 we will look at how these building blocks are assembled into a molecule of DNA or RNA.
© WILDLIFE GmbH / Alamy
Nucleic Acids Are Built from Nucleotides
Children of all species closely resemble
their parents.
The building blocks of nucleic acids are called nucleotides. All nucleotides are built from
three compoments, a five-carbon sugar, an organic base, and one to three phosphate
groups. All organisms use two sugars (ribose and deoxyribose) and five bases (adenine,
cytosine, guanine, thymine, and uracil) to make their nucleic acids. Table 17.1 shows the
structures of these compounds.
The two types of nucleic acids use different sets of components. DNA nucleotides
contain the bases adenine, cytosine, guanine, and thymine, while RNA nucleotides contain uracil in place of thymine. The sugar in DNA nucleotides is deoxyribose, while that
in RNA is ribose.
The Phosphate and the Base Are Attached to the Sugar
in a Nucleotide
• Numbers without the prime symbol are
used to show the ring positions in the
bases.
Our starting point in looking at the structure of a nucleotide is the sugar. Recall from
Chapter 14 that the carbon atoms in a sugar are numbered starting from the right side
of the ring. In nucleic acids, we put a prime symbol beside each number, so the first
carbon is labeled 1⬘ (read “one prime”), the second is 2⬘ (“two prime”), and so forth.
The numbering system for ribose is shown in Figure 17.1.
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TABLE 17.1 The Building Blocks of Nucleotides
SUGARS IN NUCLEIC ACIDS
Structure
CH2OH
Name
Type of Nucleic Acid That
Contains This Sugar
Ribose
RNA
Deoxyribose
DNA
OH
O
H
H
OH
H
OH
H
CH2OH
OH
O
H
H
OH
H
H
H
BASES IN NUCLEIC ACIDS
Structure
NH2
N
Name and Abbreviation
Type of Nucleic Acid That
Contains This Base
Adenine (A)
RNA and DNA
Cytosine (C)
RNA and DNA
Guanine (G)
RNA and DNA
Thymine (T)
DNA
Uracil (U)
RNA
N
N
N
H
NH2
N
O
N
H
O
N
H
N
N
NH2
N
H
O
CH3
N
H
O
N
H
O
N
N
H
O
H
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5′ carbon
CH2OH
O
4′ carbon
3′ carbon
1′ carbon
H
H
OH
H
OH 2′ carbon
H
In a nucleotide, a base is attached to the 1⬘ carbon of the sugar, and the phosphate groups are attached to the 5⬘ carbon. The chemical bonds between these
groups are formed by condensation reactions involving the hydroxyl groups on
the sugar. The structure of a typical nucleotide, adenosine monophosphate, is
shown in Figure 17.2.
Chemists generally use abbreviations for the names of nucleotides. To designate a ribose nucleotide, we write the abbreviation of the base, followed by MP,
which stands for “monophosphate.” If the nucleotide contains deoxyribose, we
write a d before the abbreviation for the base. For example, the nucleotide that is
built from cytosine, deoxyribose, and one phosphate group is called dCMP.
OH
FIGURE 17.1 The numbering system for
five-carbon sugars.
Adenine
(a base)
NH2
N
N
N
P
O
OH
–
Phosphate
HO
CH2
O
OH
O
H
H
OH
H
OH
H
N
N
H
O
HO
NH2
N
HO
N
P
O
O
Ribose
–
CH2
N
+ 2 H2O
O
H
H
OH
H
OH
H
Adenosine monophosphate (AMP)
(a nucleotide)
FIGURE 17.2 Building a nucleotide from its components.
Sample Problem 17.1
Drawing the structure of a nucleotide
Draw the structure of dTMP.
SOLUTION
The abbreviation dTMP tells us that the nucleotide contains the base thymine (T),
deoxyribose, and a phosphate group. We make dTMP by connecting these three pieces
as shown here. Note that the phosphate group condenses with the OH group on the 5⬘
carbon atom of the sugar, and the base condenses with the OH group on the 1⬘ group of
the sugar.
Thymine
O
CH3
N
N
O
HO
CH3
OH
–
HO
CH2
O
H
H
N
H
O
O
HO
P
O
H
H
OH
Phosphate
OH
N
O
H
P
O
O
H
O
–
H
Deoxyribose
CH2
H
O
+ 2 H2O
H
H
H
OH
H
dTMP
TRY IT YOURSELF: Draw the structure of the nucleotide GMP.
For additional practice, try Core Problems 17.11 and 17.12.
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If we remove the phosphate group from a nucleotide, we produce a nucleoside. The
names of the nucleosides are derived from those of the bases and sugars they contain.
For example, the two possible nucleosides that contain adenine are called adenosine and
deoxyadenosine.
NH2
N
N
N
N
CH2OH
NH2
N
N
O
H
H
OH
H
OH
H
N
CH2OH
O
N
H
H
OH
H
H
H
Adenosine
(a nucleoside
containing ribose)
Deoxyadenosine
(a nucleoside
containing deoxyribose)
Nucleotides are not just found in nucleic acids. These molecules serve as components of
a variety of other important biomolecules, many of which you have already encountered.
For example, NAD⫹, FAD, and coenzyme A all contain AMP as part of their structure. If
we add two more phosphate groups to AMP, we get ATP, the universal energy-transfer
molecule. The nucleoside triphosphates UTP, GTP, and CTP also supply energy for certain
biochemical reactions. Nucleotides are ubiquitous in the living world.
NH2
Three phosphate groups
O
HO
O
P
O
O
–
O
P
O
N
O
P
O
–
O
CH2
N
N
Health Note: Although all
of the cells in our bodies require
nucleotides, neither ribose nor
any of the bases is an essential
nutrient. Our bodies can make all
of these molecules from glucose
and amino acids. However,
phosphate is an essential
nutrient for humans and all other
organisms.
Adenine
N
O
–
H
H
H
H
OH
Ribose
H
ATP
(adenosine triphosphate)
Table 17.2 summarizes the types of compounds that we have encountered in this
section.
TABLE 17.2 The Structures of Nucleosides and Nucleotides
Type of Compound
Molecular Components
Nucleoside
Sugar
⫹
base
Typical Example
NH2
N
CH2OH
N
O
H
H
OH
H
OH
H
N
N
adenosine
continued
17.1 | Nucleotides
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TABLE 17.2 The Structures of Nucleosides and Nucleotides—cont'd
Type of Compound
Molecular Components
Nucleoside
monophosphate*
Sugar
⫹
base
⫹
phosphate
Typical Example
NH2
N
O
HO
P
O
O
–
N
N
CH2
N
O
H
H
OH
H
OH
H
adenosine monophosphate (AMP)
Nucleoside diphosphate*
Sugar
⫹
base
⫹
two phosphates
NH2
N
O
HO
O
P
O
O
P
O
–
O
–
N
N
CH2
N
O
H
H
OH
H
OH
H
adenosine diphosphate (ADP)
Nucleoside triphosphate*
Sugar
⫹
base
⫹
three
phosphates
NH2
N
O
HO
O
P
O
O
–
O
P
O
O
P
O
–
O
CH2
–
N
N
N
O
H
H
OH
H
OH
H
adenosine triphosphate (ATP)
*One of the three types of nucleotides. Nucleoside monophosphates are the most common type of nucleotide in the nucleic acids.
All Core Problems are paired and the answers to the blue odd-numbered
problems appear in the back of the book.
17.1 What are the three components of a nucleotide?
17.2 Which component of a nucleotide is not present in a
17.8 How does the sugar in DNA differ from the sugar in
RNA?
17.9 Name the following nucleotide, using the standard
abbreviation:
O
nucleoside?
17.3 In a nucleotide, which carbon atom in the sugar is
N
attached to the base?
17.4 In a nucleotide, which carbon atom in the sugar is
O
attached to a phosphate group?
17.5 Which of the five bases occurs in DNA but not in RNA?
17.6 Which of the five bases occurs in RNA but not in DNA?
17.7 What is the sugar in RNA?
HO
N
P
O
O
–
CH2
N
H
NH2
O
H
H
OH
H
OH
H
N
continued
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17.10 Name the following nucleotide, using the standard
17.11 a) Draw the structure of dCMP.
abbreviation:
b) How does the structure of dCMP differ from the
structure of CMP?
c) How does the structure of dCMP differ from the
structure of dCTP?
NH2
N
17.12 a) Draw the structure of UMP.
O
N
P
HO
O
N
O
–
CH2
b) How does the structure of UMP differ from the
structure of dUMP?
c) How does the structure of UMP differ from the
structure of UDP?
N
O
H
H
OH
H
H
H
17.2 Structures of Nucleic Acids
OBJECTIVES: Know how nucleotides
are linked in a nucleic acid, and understand how complementary base-pairing
produces the double-helix structure of
DNA.
Both DNA and RNA are built from nucleotides, which are linked together to form an
unbranched chain. In this section, we will look at how nucleotides bond to one another,
and we will examine the structural differences between DNA and RNA.
When two nucleotides combine to form a single molecule, the phosphate group of one
nucleotide condenses with the 3⬘ hydroxyl group on the sugar of the other molecule. The
phosphate group forms a link between the 5⬘ carbon of one sugar and the 3⬘ carbon of
a second sugar. This connection is called a phosphodiester group, and the product of the
reaction is called a dinucleotide. Figure 17.3 shows the formation of a phosphodiester
link between two nucleotides.
The uppermost phosphate group in the dinucleotide of Figure 17.3 can link to the
sugar of a third nucleotide. We can repeat this condensation reaction over and over, adding additional nucleotides to our molecule. The resulting long chain is a nucleic acid.
Note that the condensation reactions that form the chain do not involve the bases.
Sugars and Phosphates Form the Backbone of a Nucleic Acid
Figure 17.4 shows the general structure of a nucleic acid chain. The repeating sugar–
phosphate chain is called the backbone of the nucleic acid. The end of the chain that has
a free phosphate group is the 5⬘ end, and the other end is the 3⬘ end. The bases project
from the backbone. The vast majority of nucleic acid molecules in living organisms contain an apparently random sequence of the four bases. However, as we will see, the exact
order of the bases in a nucleic acid is the key to its biological role.
OH
–
O
P
OH
O
CH2
O
base
O
H
H
OH
H
O
P
O
CH2
O
P
O
H
H
O
CH2
O
H
H
OH
–
base
O
H
O
H
H
3′ carbon
–
base
O
H
H
OH
H
–
O
P
O
O
H
H
CH2
base
O
H
H
OH
H
H
H
A phosphodiester
group
+ H2O
FIGURE 17.3 The formation of a phosphodiester group.
17.2 | Structures of Nucleic Acids
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The end with a
phosphate group
is the 5' end.
U
Bases
A
C
A
Sugar-phosphate
backbone
(shaded grey)
G
G
The end with a
sugar molecule
is the 3' end.
U
FIGURE 17.4 The general structure of a nucleic acid.
The Bases in Nucleic Acids Form Complementary Pairs
All of the bases in nucleic acids can form hydrogen bonds, so they attract each other. As
a result, whenever two nucleic acid chains come in contact with each other, each base in
one chain has a strong tendency to pair up with a base in the other chain. However,
because of the arrangement of hydrogen-bonding groups in each base, the sizes of the
bases, and the way in which the bases are attached to the backbone, some pairs are much
more stable than others. In general, A only pairs up with T or U, and C only pairs up
with G. These allowed combinations are called complementary base pairs. In these
pairs, each positively charged hydrogen faces a negatively charged oxygen or nitrogen.
The attraction of oppositely charged atoms produces a stable base pair. Figure 17.5
shows the two complementary base pairs in DNA, as well as an example of an unstable
combination.
The formation of complementary base pairs is particularly important to the structure
of DNA. All DNA molecules contain two nucleotide chains, lying side-by-side and attracted to each other by hydrogen bonds between their bases. The two backbones run in
opposite directions, with the 3⬘ end of one backbone facing the 5⬘ end of the other. Every
base in one strand forms a pair with a base in the other strand, and all of these base pairs
are complementary. As a result, if we know the sequence of bases in one DNA strand,
we can always tell the base sequence in the other, as shown in Figure 17.6.
When chemists write the base sequence in a single strand of a nucleic acid, they always start from the 5⬘ end of the strand. The base sequence in strand 1 in Figure 17.6 is
written ACCTGTA, and the sequence in strand 2 is TACAGGT.
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H
B
a
c
k
b
o
n
e
H
O
N
N
N
H
B
a
c
k
b
o
n
e
N
N
N
N
N
H
O
G
C
A
T
H
Guanine (G)
G
Cytosine (C)
C: a stable base pair
H
B
a
c
k
b
o
n
e
N
N
N
N
CH3 B
O
H
H
N
N
N
O
Adenine (A)
A
Thymine (T)
T: a stable base pair
H
B
a
c
k
b
o
n
e
a
c
k
b
o
n
e
H
N
N
N
H
H
N
N
N
N
N
O
Adenine (A)
A
B
a
c
k
b
o
n
e
A
C
Cytosine (C)
C: not a stable base pair
FIGURE 17.5 Complementary base-pairing in nucleic acids.
Strand 1
5′ end
3′ end
C
C
T
A
T
A
G
G
G
A
A
C
T
3′ end
T
5′ end
Strand 2
FIGURE 17.6 The arrangement of nucleotides in double-stranded DNA.
17.2 | Structures of Nucleic Acids
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Sample Problem 17.2
Writing a complementary base sequence
A strand of DNA has the base sequence GGCTTA. Write the base sequence of the other
strand.
SOLUTION
Matching each base in the original strand with its complement, we get the sequence
CCGAAT in the other strand. However, the two nucleotide chains in DNA run in
opposite directions. The original DNA strand is oriented in the 5⬘ to 3⬘ direction. Therefore, our complementary strand is oriented in the 3⬘ to 5⬘ direction.
Original strand
(5' end)
G
G
C
T
T
A
(3' end)
Complementary strand
(3' end)
C
C
G
A
A
T
(5' end)
Since we always write DNA sequences starting with the 5⬘ end, we must reverse the order
of the nucleotides. The correct sequence is TAAGCC.
TRY IT YOURSELF: A strand of DNA has the base sequence TTTGCAAA. Write the
base sequence of the other strand.
5' end of
strand 1
3' end of
strand 2
G
C
For additional practice, try Core Problems 17.19 and 17.20.
The Two Strands of DNA Form a Double Helix
C
G
A
T
T
A
G
A
C
C
T
G
G
G
T
C
A
T
A
A
T
G
C
T
C
C
C
G
A
G
3' end of
strand 1
5' end of
strand 2
FIGURE 17.7 The double-helix structure
of DNA.
The two strands of a DNA molecule coil around each other, forming a double helix, as shown in Figure 17.7. This structure allows the base pairs, which are somewhat hydrophobic, to avoid the surrounding water molecules. The sugar–
phosphate backbones lie on the outside of the double helix, in contact with water.
Although RNA molecules are normally single stranded, they can also form the
double-helix structure by looping back on themselves.
Every cell in our bodies contains DNA at the beginning of its life cycle, and most
cells retain this DNA throughout their lives. As we will see, these DNA molecules
are the genetic instructions that control our development from a single cell to an
adult human. Every cell in a person contains an identical set of DNA, but unless
someone is an identical twin, the base sequence in that person’s DNA is different
from that of anyone else. As a result, DNA sequences can be used to identify a
specific individual. In recent years, police have used the DNA sequences in such
materials as blood, hair, and semen to identify both the victims and the perpetrators
of crimes.
Most of our genetic information is contained in just 46 molecules of DNA. As
a result, DNA molecules are truly gigantic. The DNA molecules in our bodies contain between 50 million and 250 million base pairs, corresponding to formula
weights between 30 billion and 150 billion amu. RNA molecules are much smaller,
but they are still large; typical RNA molecules contain between 75 and 3000 bases
and have formula weights as high as 1 million amu.
In eukaryotes, most of the DNA is located in the nucleus of the cell, with a small
amount in the mitochondria (and in the chloroplasts of plants). Each DNA molecule in the nucleus is bound to positively charged proteins called histones, which
are attracted to the negatively charged phosphate groups in the DNA backbone.
The double-stranded DNA wraps around clusters of histones, which appear rather
like beads on a string, as shown in Figure 17.8. The DNA and histones then form
additional coils involving other proteins and some RNA, eventually producing a
compact mass of nucleic acid and protein called a chromosome. Chromosomes are
large enough to see with a microscope. Every cell in the human body contains 46
chromosomes, except for egg and sperm cells (which contain only 23 chromosomes) and red blood cells (which break down all of their DNA as they mature).
Each chromosome contains many protein molecules, but it has just one molecule
of DNA.
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DNA
Histone
cluster
Image copyright Mopic, 2010. Used under license from Shutterstock.com
FIGURE 17.8 The binding between DNA and histones.
Most of the RNA in a eukaryotic cell is in the cytoplasm. Roughly 80% of the RNA
is mixed with proteins in large clusters called ribosomes. The ribosomes are the catalysts
for protein synthesis, as we will see in Section 17.6. The rest of the RNA is dissolved in
the intracellular fluids. There are three main classes of RNA molecules, which differ from
one another in both structure and function. We will examine these three types of RNA
in Section 17.4.
A model showing the double helix
structure of DNA.
17.13 In the DNA backbone, does the phosphodiester
group connect two sugars to each other, two bases to
each other, or a sugar to a base?
17.14 In RNA, which two carbon atoms in successive
nucleotides are linked by the phosphodiester group?
Give their numbers.
17.15 Draw the structure of the dinucleotide that is formed
from two molecules of AMP.
17.16 Draw the structure of one of the two possible
dinucleotides that are formed from one molecule of
dTMP and one molecule of dGMP.
17.17 Which base forms a stable base pair with each of the
following bases in a molecule of DNA?
a) G
b) A
17.18 Which base forms a stable pair with each of the
following bases in a molecule of DNA?
a) T
b) C
17.19 A DNA strand has the base sequence AGTGGC.
a) What base is at the 5⬘ end of this strand?
b) What base is at the 5⬘ end of the complementary
strand?
c) Write the base sequence of the complementary
strand.
17.20 A DNA strand has the base sequence CTTAGA.
a) What base is at the 3⬘ end of this strand?
b) What base is at the 3⬘ end of the complementary
strand?
c) Write the base sequence of the complementary
strand.
17.21 a) What is a histone?
b) Are histones positively charged, negatively
charged, or electrically neutral?
17.22 a) What is a chromosome?
b) How many molecules of DNA are there in a chromosome?
17.23 Where is the DNA located in a eukaryotic cell?
17.24 Where is the RNA located in a eukaryotic cell?
17.2 | Structures of Nucleic Acids
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OBJECTIVES: Describe the mechanism
of DNA replication, and explain the role of
DNA replication in cell division.
17.3 DNA Replication
DNA is the master blueprint for an organism. In a multicelled organism (like you),
the DNA in each cell contains the instructions for building every cell in the organism.
This information is passed on whenever a cell divides, so the two daughter cells
contain identical sets of DNA. We are therefore faced with two fundamental
questions:
1. How is the information in DNA passed from one generation to the next?
2. How is this information used to build a complete organism?
© Medical-on-Line / Alamy
In this section, we will look at the answer to the first question. In Sections 17.4 through
17.6, we will explore the answer to the second question.
To begin, we need to know how DNA stores information. The fundamental principle is
that the sequence of bases in DNA is a set of instructions for building proteins. Every cell
can translate sequences of bases in its DNA into sequences of amino acids. This process is
analogous to translating written text from one language to another. For example, just as
gato in Spanish corresponds to cat in English, the sequence AAATTT in “DNA language”
corresponds to the sequence lysine–phenylalanine in “protein language.”
As we saw in Section 17.2, most cells in the human body contain 46 chromosomes. Of
these, 44 chromosomes (called autosomal chromosomes) occur in pairs, with each member
of the pair containing essentially the same collection of genes. The
final two chromosomes are the sex chromosomes. There are two
varieties of sex chromosomes, one of which (the X chromosome)
is rather large; the other (the Y chromosome) is much smaller. The
sex chromosomes determine your biological gender. If your cells
contain two X chromosomes, you are a woman, and if your cells
contain one X and one Y chromosome, you are a man. The entire
set of chromosomes, containing all of your genetic information, is
called your genome.
A cell contains only one set of chromosomes. When the cell
divides, however, each of the resulting cells (called daughter
cells) must contain a full set of chromosomes. Therefore,
when a cell divides, it makes a new copy of each DNA molecule, as shown in Figure 17.9. These copies are identical to the
The 46 chromosomes in a human being.
Daughter cells
(46 chromosomes in each)
Parent cell
Parent cell
(46 chromosomes)
(92 chromosomes)
Cytoplasm
Cell division
Chromosome
Nucleus
DNA replication
FIGURE 17.9 The role of DNA replication in cell division.
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5′ end
A
A
C
DNA
chain
C
C
C
T
T
3′ end
+
G
G
GTP
(an activated
molecule)
Removing two phosphate groups from
GTP supplies the energy needed to
make the new phosphodiester link.
FIGURE 17.10 Adding a nucleotide to a DNA chain.
original molecules, so both of the daughter cells can make all of the proteins they need
to survive.
Cells Use Their DNA to Determine the Base Sequence in New DNA
The process by which cells copy their DNA is called replication. To replicate DNA, a cell
needs a supply of the nucleoside triphosphates dATP, dCTP, dGTP, and dTTP, which
serve as the building blocks of the new DNA. The cell also needs a small amount of the
ribose-containing nucleoside triphosphates ATP, CTP, GTP, and UTP. Removing the two
extra phosphate groups from these high-energy molecules supplies the energy needed to
connect the nucleotide to the growing DNA chain. The cell cannot use the original nucleotides (which contain only one phosphate group) to make DNA, so the nucleoside
triphosphates are often called activated nucleotides. Figure 17.10 shows how an activated nucleotide is added to a growing DNA chain.
Figure 17.11 illustrates the specific steps that occur during DNA replication:
1. A protein called a helicase pulls the two DNA strands away from each other and
inserts itself between the two strands. In eukaryotic cells, this happens at many
locations along the DNA molecule.
2. An enzyme called primase matches a few bases on each strand of DNA with the
complementary RNA nucleotides (in their activated forms), and then it condenses
the RNA nucleotides to form short RNA strands called primers. These primers
remain attached to the original DNA strands.
3. The helicase unwinds more of the original DNA double helix, and an enzyme
called DNA polymerase starts making new strands by adding activated DNA
nucleotides to the primers. The added nucleotides must be the complements of the
original strand. DNA polymerase continues to add nucleotides until it has made a
complete complement to each strand of the original DNA.
4. Since eukaryotes start the replication process in many locations, they make the
complementary chains in multiple pieces. An enzyme called DNA ligase replaces
the RNA primers with the corresponding DNA nucleotides, and it seals the gaps
between the pieces.
DNA replication is both extraordinarily fast and extraordinarily accurate. It takes
only about one hour for a cell in your body to replicate all 46 chromosomes (around
6 billion base pairs). This feat requires the cooperative efforts of thousands of molecules of DNA polymerase working at locations scattered throughout the 46 chromosomes. However, speed would be useless if the daughter DNA did not have the correct
Health Note: Ciprofloxacin
and related antibiotics (called
quinolone antibiotics) interfere
with a protein that is involved in
DNA replication in bacteria.
Bacteria that cannot replicate
their DNA cannot reproduce and
eventually die. Some drugs that
are used in chemotherapy also
interfere with DNA replication,
so they have a disproportionate
effect on cells that are dividing
rapidly, including cancer cells.
Health Note: Scientists can
now replicate DNA rapidly in the
laboratory, using a process called
polymerase chain reaction (PCR).
PCR is used to diagnose genetic
disorders, to identify and
diagnose infectious diseases, and
to identify the functions of
specific genes.
17.3 | DNA Replication
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3'
5'
A
C
T
3'
G
G
T
T
T
A
G
C
T
G
C
C
A
A
A
3'
3'
T
C
C
G
T
C
C
C
G
T
C
U
G
T
C
A
A
T
T
G
G
T
A
C
T
C
T
3'
The parent DNA
molecule, before
replication.
5'
A
G
T
T
C
A
A
G
A
A
T
T
A
A
G
A
A
T
C
A
T
T
C
T
T
G
A
A
T
5'
3'
1 A helicase unwinds the
double helix and separates
the two DNA strands.
C
G
G
A
T
C
A
C
A
A
C
5'
T
A
T
T
G
C
A
G
5'
G
C
G
G
G
A
G
G
T
G
C
A
U
G
G
T
C
G
T
C
C
T
G
5'
A
5'
5'
A
G
G
A
C
A
C
C
A
3'
T
A
3'
2 Primase builds a short RNA strand (a
primer) on each DNA strand, using activated
nucleotides (ATP, CTP, GTP, and UTP).
FIGURE 17.11 The mechanism of DNA replication.
base sequence. DNA polymerase makes about one mistake per 100,000 bases, which
would produce around 60,000 mismatched base pairs during a single cycle of replication. However, DNA polymerase is part of a large cluster of proteins, one of which is
a proofreader. This protein recognizes mismatched base pairs and replaces the incorrect base in the new strand with the correct one. The proofreading catches about 99%
of the errors, leaving one mistake per 10 million bases (or 600 mismatched base pairs
per replication cycle). In addition, every cell contains enzymes that repair damaged
DNA, and these enzymes can replace mismatched bases with the correct complement.
The end result is that the new DNA typically contains less than one error per billion
base pairs. (This is comparable to typing the entire text of the Encyclopaedia Britannica three times, making only one mistake in the process.) Nonetheless, an occasional
error remains, and these errors become part of the permanent genetic blueprint of the
cell that inherits it. In Section 17.7, we will see that these changes in the DNA blueprint can have a profound impact on the human body, and that they have helped create
the diversity of life on Earth.
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Original strands
Direction
of chain
growth
3'
5'
A
C
G
T
C
T
T
T
A
G
C
U
C
5'
G
G
C
C
A
G
A
T
T
G
Direction
of chain
growth
A
T
T
C
A
T
3'
C
A
T
T
A
A
T
3'
5'
T
3'
The new strands are now complete, but they
contain gaps and some RNA nucleotides.
17.25 What is the function of DNA in an organism?
17.26 What is DNA replication, and why is it important?
17.27 a) How many chromosomes are there in a typical
cell in the human body?
b) How many of these are autosomal chromosomes?
17.28 a) How many sex chromosomes are there in a typical
cell in the human body?
b) What are the two types of sex chromosomes?
c) Which sex chromosomes are present in a man,
and which are present in a woman?
5'
A
T
A
G
T
T
C
A
A
G
A
T
C
A
T
C
A
A
G
A
C
C
G
T
T
G
G
+
A
T
A
C
C
A
C
G
A
T
A
C
5'
T
G
T
G
G
U
C
G
T
A
T
G
T
3 DNA polymerase adds nucleotides
to the 3′ end of each new strand, using
the activated nucleotides dATP, dCTP,
dGTP, and dTTP.
A
C
A
A
C
T
G
A
C
A
T
A
C
T
G
T
A
G
G
T
A
G
C
T
5'
T
A
+
C
A
A
A
C
A
C
C
G
C
A
G
T
C
T
T
G
G
G
A
C
A
G
G
T
A
3'
U
gap
C
C
C
A
C
5'
T
A
G
T
G
3'
U
G
T
G
A
G
G
C
T
T
C
C
A
C
C
5'
A
C
A
G
G
A
G
T
T
3'
G
G
A
C
C
C
C
A
G
G
5'
T
A
G
T
T
3'
U
G
A
C
C
C
C
A
G
G
5'
A
G
G
A
C
3'
T
C
A
5'
A
3'
U
G
New
strands
C
T
T
G
A
A
T
3'
T
A
5'
3'
4 Ligase replaces the RNA with DNA and
seals the gaps.
17.31 What is the function of each of the following proteins
in DNA replication?
a) primase
b) DNA ligase
17.32 What is the function of each of the following proteins
in DNA replication?
a) helicase
b) DNA polymerase
17.33 What is the function of the proofreader protein in
DNA replication?
17.34 Cells contain proteins that repair damaged DNA.
Why is this important to DNA replication?
17.29 What is an activated nucleotide?
17.30 Why do cells build DNA from the nucleoside
triphosphates (dATP, dCTP, dGTP, and dTTP)
rather than the normal nucleotides (dAMP, dCMP,
dGMP, and dTMP)?
17.3 | DNA Replication
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OBJECTIVES: Describe the mechanism
of transcription, and explain how cells
modify the three types of RNA after
transcription.
17.4 Transcription and RNA Processing
DNA is fundamentally an information molecule. The information in DNA is used to
make RNA, and the information in RNA is then used to make proteins. This flow of
information from DNA to RNA to protein is called the central dogma of molecular biology. Living organisms must also make a variety of other compounds, including carbohydrates, lipids, and coenzymes. These molecules are built by proteins, starting with the
materials that the organism finds in its environment. Figure 17.12 summarizes the flow
of information in a living organism.
RNA plays a key role in the information pathway of Figure 17.12. Cells contain three
main types of RNA, called messenger, transfer, and ribosomal RNA. Messenger RNA
(mRNA), like DNA, is a set of instructions: a molecule of mRNA contains the information needed to build one protein. Transfer RNA (tRNA) matches the coded information
in mRNA with the correct amino acids. Ribosomal RNA (rRNA) is a catalyst, allowing
the cell to link amino acids to form a protein chain. Let us now examine how a cell
makes the three types of RNA.
The Base Sequence in DNA Determines the Sequence in RNA
Making any type of RNA starts with a process called transcription. Transcription is
similar to DNA replication in that a cell uses the base sequence in a DNA molecule to
determine the sequence in the new strand of nucleic acid. However, in transcription, the
cell makes a single strand of RNA rather than a double-stranded DNA molecule. In addition, the cell copies its entire genome during replication, but it only transcribes short
regions of its DNA into RNA.
Figure 17.13 shows how a cell transcribes a section of a DNA molecule. First, the cell
unwinds part of the DNA double helix. Next, an enzyme called RNA polymerase builds
a molecule of RNA, using one of the two DNA strands as a template. The building
blocks for the new RNA are activated nucleotides (nucleoside triphosphates), just as they
were for DNA replication. Finally, the new RNA molecule is released, and the two DNA
strands reform the double helix.
When a cell makes a molecule of RNA, it only transcribes a short region of a DNA
molecule. However, the cell makes a great variety of RNA molecules, each of which is
transcribed from a different region of the cell’s DNA. Each section of DNA that is transcribed to make a single piece of RNA is called a gene. As shown in Figure 17.13, the
new RNA molecule is the complement of one of the DNA strands, called the template
strand. The other DNA strand is called the coding strand. The base sequence of the coding strand matches that of the new RNA, except that every T in the DNA is replaced
with a U.
Transcription
Translation
Information in
DNA is used to
build RNA.
Information in
RNA is used to
build proteins.
DNA
RNA
Proteins are
the catalysts
that build...
Proteins
All
biomolecules
FIGURE 17.12 Information flow in living organisms.
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Template
strand
3'
5'
A
C
G
A
C
C
A
C
T
T
G
C
C
G
C
T
T
T
C
G
A
T
T
A
A
T
5'
3'
The original DNA
double helix
C
C
T
A
C
A
C
3'
T
G
A
T
C
U
C
A
A
G
A
A
T
5'
3'
RNA polymerase builds a strand of
RNA, using the template strand of
the DNA to determine the RNA
base sequence.
+
T
G
T
C
A
A
G
C
A
G
T
T
C
C
G
C
T
G
A
C
A
U
G
A
G
A
T
T
G
T
C
A
G
T
A
3'
A
C
A
C
U
G
A
C
G
C
G
A
G
C
T
C
A
C
T
G
T
C
A
A
A
C
A
T
T
G
5'
T
C
G
A
5'
G
A
C
C
A
3'
C
U
G
G
5'
A
G
T
C
G
New
RNA
T
C
G
A
Coding
strand
G
A
A
G
T
U
G
G
A
Template
strand
5'
T
G
T
3'
Coding
strand
T
A
5'
3'
The DNA double helix re-forms, and
the RNA is released into solution.
(Note that the RNA sequence
matches that of the DNA coding
strand, except for the replacement
of T with U.)
FIGURE 17.13 The mechanism of transcription.
Sample Problem 17.3
Writing a base sequence of an RNA molecule after transcription
The coding strand of a DNA molecule has the base sequence ATCCGCT. If this piece of
DNA is transcribed, what will be the base sequence of the resulting RNA?
SOLUTION
The RNA has the same sequence as the coding strand of the DNA except that we must
replace T with U. Therefore, the RNA will have the sequence AUCCGCU. To show why
this is so, we need to first write the sequence of the DNA template strand. Then we can
write the RNA sequence, which will be the complement of the template strand.
Coding strand
(5' end)
A
T
C
C
G
C
T
(3' end)
Template strand
(3' end)
T
A
G
G
C
G
A
(5' end)
RNA
(5' end)
A
U
C
C
G
C
U
(3' end)
TRY IT YOURSELF: The coding strand of a DNA molecule has the base sequence ACAATGG.
If this piece of DNA is transcribed, what will be the base sequence of the resulting RNA?
For additional practice, try Core Problems 17.41 and 17.42.
17.4 | Transcription and RNA Processing
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All Types of RNA are Modified after Transcription
The RNA that is made during transcription is called the initial transcript. The initial
transcript must be modified before the cell can use it. Each type of RNA is modified in
a different way. Let us examine the ways in which cells process their RNA.
1. Ribosomal RNA. The initial transcript for rRNA always contains extra nucleotides, which the cell must remove. In addition, three of the four types of rRNA
molecules that our cells produce are contained within a single initial transcript.
Various enzymes cut apart the initial transcripts and remove the extra bases,
producing the final product rRNA. Figure 17.14 shows how cells process their
initial rRNA transcripts.
2. Transfer RNA. Like rRNA, tRNA is formed as part of a larger initial transcript.
The initial transcript normally contains several types of tRNA, and it can contain
one or more types of rRNA. Enzymes cut out the tRNA, just as they do the rRNA.
However, tRNA requires several additional modifications. A specific enzyme adds
the nucleotide sequence CCA to the 3⬘ end of the tRNA, and other enzymes modify
several of the bases in the tRNA. Figure 17.14 shows how the initial tRNA and
rRNA transcripts are processed.
3. Messenger RNA. The initial mRNA transcript undergoes a remarkable variety of
changes, as shown in Figure 17.15. Here is a summary of these changes:
• The cell adds an additional guanine nucleotide and a phosphate group to the
5⬘ end of the chain. This process is called capping, and it allows the cell to
recognize the molecule as mRNA.
• The cell removes several nucleotides from the 3⬘ end of the initial transcript, and it
replaces them with a long chain of adenine nucleotides called the poly(A) tail. The
length of the poly(A) tail determines the useful life span of the mRNA molecule.
• The cell removes one or more sections from the interior of the original transcript and breaks these sections down into individual nucleotides. The remaining pieces are linked together to form the mature mRNA. The pieces that are
removed and destroyed are called introns, and the pieces that remain in the
mature mRNA are called exons.
Initial RNA transcript
+
Excess RNA
fragments
+
Ribosomal RNA
Nucleotides
Add CCA,
modify
bases
Transfer RNA
FIGURE 17.14 The processing of tRNA and rRNA.
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Exons
5'
3'
Initial
transcript
Introns
Capping
5' cap
3'
5'
Removal of 3' end
5'
3'
Addition of poly(A) tail
Poly(A) tail
5'
3'
Removal of introns
5'
3'
Mature
mRNA
FIGURE 17.15 mRNA processing.
Most known genes in the human genome contain introns, which means that a good
deal of our DNA does not code for any protein. For example, the gene for coagulation
factor VIII (a protein that plays a critical role in blood clotting) contains 140,359 DNA
base pairs, but the mature mRNA for this protein contains only 8787 nucleotides (not
including the poly(A) tail), of which 7056 are translated into the amino acid sequence of
factor VIII. This means that only 5% of the DNA in this gene actually codes for factor
VIII. The rest of the gene, including 23 separate introns, is noncoding DNA, which does
not correspond to any amino acid sequence.
Current estimates are that more than 98% of our DNA does not code for any protein.
More than a third of this DNA is transcribed to make tRNA, rRNA, the introns in
mRNA, and a variety of other small RNA molecules whose roles are still being worked
out. However, much of our DNA is apparently never transcribed. The function, if any,
of this noncoding DNA is an area of intense research and is one of the great mysteries
of biochemistry.
17.35 Which of the following is the best description of tran-
scription?
a) A cell makes a piece of RNA that is the complement of a piece of DNA.
b) A cell makes a piece of DNA that is the complement of a piece of RNA.
c) A cell makes a piece of RNA that is the complement of another piece of RNA.
Health Note: People with
the most common type of
hemophilia have a defective gene
for factor VIII, so their blood does
not clot when they are injured.
People with untreated
hemophilia suffer from
debilitating episodes of internal
bleeding after they are bruised,
along with a variety of potentially
life-threatening symptoms.
Treatment for hemophilia
involves giving regular injections
of factor VIII.
17.36 When a cell makes a single molecule of RNA, how
much of its DNA does it copy?
a) all of the DNA molecules in the cell
b) one complete DNA molecule
c) most of a DNA molecule
d) a short section of a DNA molecule
continued
17.4 | Transcription and RNA Processing
93493_Ch17_722-765.indd 741
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10/15/10 11:41:35 AM
17.37 Which of the following could a cell use to build
RNA?
a) adenosine
b) AMP
c) ADP
d) ATP
17.38 Which of the following could a cell use to build
RNA?
a) dTTP
17.42 A coding strand in a section of DNA has the base
sequence TAGGGCA. If this section of DNA is
transcribed, what will be the base sequence of the
resulting RNA?
17.43 Describe how the initial transcript of an rRNA
b) TTP
c) dUTP
d) UTP
17.39 Which DNA strand actually binds to the RNA
nucleotides during transcription, the coding strand or
the template strand?
17.40 RNA molecules have essentially the same base
sequence as one of the two DNA strands. Which
one?
17.41 The coding strand in a section of DNA has the base
sequence AACCTTG. If this section of DNA is
transcribed, what will be the base sequence of the
resulting RNA?
OBJECTIVE: Relate the base sequence
in DNA or mRNA to the primary structure
of a protein, using the genetic code.
molecule is processed.
17.44 Describe how the initial transcript of a tRNA
molecule is processed.
17.45 What is an intron, and what happens to introns
during mRNA processing?
17.46 What is an exon, and what happens to exons during
mRNA processing?
17.47 What is the 5⬘ cap in an mRNA molecule, and what
is its function?
17.48 What is the poly(A) tail in an mRNA molecule, and
what is its function?
17.5 Translation and the Genetic Code
In Section 17.3, you learned that the sequence of bases in DNA corresponds to the sequence of amino acids in all of the proteins that a cell can make. Cells use mRNA to
copy a gene (the code for a single protein) and make it available for decoding. Therefore,
mRNA plays a central role in translation, the process by which a cell converts the language of nucleic acids into the language of amino acids. In this section, we will examine
how a cell uses the information in its nucleic acids to build proteins.
The Genetic Code Relates RNA Sequences to Amino Acids
• In many proteins, the initial methionine is removed after translation.
The basic principle of translation is that a sequence of three bases in mRNA corresponds to one amino acid. These mRNA three-base sequences are called codons. Remarkably, virtually every organism on Earth translates its codons in the same way. For
example, the codon AAA corresponds to lysine in all known organisms. Biochemists
have worked out the relationship between the base sequence and the amino acid for
all possible mRNA codons. This relationship is called the genetic code and is shown
in Table 17.3.
There are 64 possible codons. Of these, 61 codons correspond to specific amino acids.
Since there are only 20 amino acids, most amino acids correspond to more than one
codon. For example, cysteine corresponds to the codons UGC and UGU, and leucine
corresponds to the codons CUA, CUC, CUG, CUU, UUA, and UUG. However, no codon
corresponds to more than one amino acid. Any given sequence of bases corresponds to
only one sequence of amino acids.
The genetic code contains four special codons. The codon AUG has two roles: it is the
codon for methionine, and it serves as a start codon. When a cell translates a piece of
mRNA, it always starts at the base sequence AUG, so all polypeptides contain methionine as their first amino acid. The three codons UAA, UAG, and UGA are called stop
codons or nonsense codons. These codons cannot be translated, so they serve as a signal
to stop making a protein. The cell translates the mRNA until it reaches a stop codon,
and then it ignores all of the mRNA beyond that point. Figure 17.16 shows how a
typical piece of mRNA would be translated.
When a cell builds a protein, it starts with the N-terminal amino acid. The last amino
acid to be added is the C-terminal amino acid, which corresponds to the three bases immediately before the stop codon. Sample Problem 17.4 shows how we can convert a
sequence of mRNA bases into the corresponding amino acid sequence.
742 CHAPTER 17 | Nucleic Acids, Protein Synthesis, and Heredity
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TABLE 17.3 The Genetic Code
FIRST
POSITION
(5⬘ END)
SECOND POSITION
A
C
G
U
A
AAA ⫽ Lys
AAC ⫽ Asn
AAG ⫽ Lys
AAU ⫽ Asn
ACA ⫽ Thr
ACC ⫽ Thr
ACG ⫽ Thr
ACU ⫽ Thr
AGA ⫽ Arg
AGC ⫽ Ser
AGG ⫽ Arg
AGU ⫽ Ser
AUA ⫽ Ile
AUC ⫽ Ile
AUG ⫽ Met/start
AUU ⫽ Ile
C
CAA ⫽ Gln
CAC ⫽ His
CAG ⫽ Gln
CAU ⫽ His
CCA ⫽ Pro
CCC ⫽ Pro
CCG ⫽ Pro
CCU ⫽ Pro
CGA ⫽ Arg
CGC ⫽ Arg
CGG ⫽ Arg
CGU ⫽ Arg
CUA ⫽ Leu
CUC ⫽ Leu
CUG ⫽ Leu
CUU ⫽ Leu
G
GAA ⫽ Glu
GAC ⫽ Asp
GAG ⫽ Glu
GAU ⫽ Asp
GCA ⫽ Ala
GCC ⫽ Ala
GCG ⫽ Ala
GCU ⫽ Ala
GGA ⫽ Gly
GGC ⫽ Gly
GGG ⫽ Gly
GGU ⫽ Gly
GUA ⫽ Val
GUC ⫽ Val
GUG ⫽ Val
GUU ⫽ Val
U
UAA ⫽ stop
UAC ⫽ Tyr
UAG ⫽ stop
UAU ⫽ Tyr
UCA ⫽ Ser
UCC ⫽ Ser
UCG ⫽ Ser
UCU ⫽ Ser
UGA ⫽ stop
UGC ⫽ Cys
UGG ⫽ Trp
UGU ⫽ Cys
UUA ⫽ Leu
UUC ⫽ Phe
UUG ⫽ Leu
UUU ⫽ Phe
The start codon is shown in green, and the stop codons are in red.
5' cap
Exons
5' G
Poly(A) tail
... AUG ...
... UAG ...
Start codon
Stop codon
AAAAAA ...
3'
Translated region
FIGURE 17.16 The structure and translation of mRNA.
Sample Problem 17.4
Translating an mRNA sequence
A piece of mRNA has the base sequence CGCAUUGUG. What is the amino acid
sequence in the corresponding polypeptide?
SOLUTION
To translate this mRNA into the amino acid sequence, we need to divide the base sequence into three-base codons:
CGC–AUU–GUG
Now we can use the genetic code to write the amino acid sequence. CGC codes for arginine (Arg), AUU codes for isoleucine (Ile), and GUG codes for valine (Val), so the amino
acid sequence is Arg–Ile–Val. Note that when the mRNA is written from 5⬘ to 3⬘, the
normal direction, the amino acid sequence will also be in the normal direction, with the
N-terminal amino acid on the left and the C-terminal amino acid on the right.
TRY IT YOURSELF: A piece of mRNA has the base sequence GGCCGGCCG. What is
the amino acid sequence in the corresponding polypeptide?
For additional practice, try Core Problems 17.57 and 17.58.
17.5 | Translation and the Genetic Code
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Coding strand 5'
A
C
T
G
A
A
T
C
T
3'
Template strand 3'
T
G
A
C
T
T
A
G
A
5'
A
C
U
G
A
A
U
C
U
3'
DNA
mRNA
Protein
5'
+
Threonine
NH3
Glutamic acid
Serine
CO2–
FIGURE 17.17 The relationship between a DNA sequence and an amino acid sequence.
Although the genetic code relates RNA base sequences to amino acids, we can also
use it to “translate” DNA sequences. However, we need to know which strand of the
DNA is the coding strand and which is the template strand. Chemists always write the
coding strand in the normal direction (from 5⬘ to 3⬘), so the corresponding mRNA also
reads from 5⬘ to 3⬘. Figure 17.17 shows how we can work out the amino acid sequence
that corresponds to a DNA sequence.
Sample Problem 17.5
Translating a DNA sequence
A section of a DNA coding strand has the base sequence ATCCATGACAGCTT. This
DNA contains the code for the beginning of a polypeptide. What are the first three amino
acids in this polypeptide?
SOLUTION
First, we must work out the base sequence in the corresponding mRNA. We do this in
two steps: we work out the sequence of the DNA template strand, and then we use the
template strand to find the sequence in the mRNA. When we write the mRNA sequence,
we must remember that RNA uses the base U instead of T.
DNA coding strand:
DNA template strand:
mRNA:
A T
C C A T G A C A G C
T A G G T A C
T G T
T
T
The template strand
is the complement
of the coding strand.
C G A A
A U C C A U G A C A G C U U
The mRNA is the
complement of the
template strand.
This mRNA contains the code for the beginning of a polypeptide, so it must contain a
start codon. To translate the mRNA, we need to divide the base sequence into three-base
codons, beginning with the start codon. The cell ignores the bases before the start codon.
A U C C A U G A C A G C U U
Ignored
Start
codon
Now we can use the genetic code to write the amino acid sequence. AUG codes for methionine (Met), ACA codes for threonine (Thr), and GCU codes for alanine (Ala), so the
first three amino acids are Met–Thr–Ala.
TRY IT YOURSELF: A section of a DNA coding strand has the base sequence TATCGGCATAAGC. This DNA contains the code for the end of a polypeptide. What are the
last two amino acids in the corresponding polypeptide?
For additional practice, try Core Problems 17.65 and 17.66.
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All forms of life on Earth share the genetic code. Researchers have found minor
variations in the code in certain microorganisms and in mitochondrial DNA, but these
are the exceptions rather than the rule. A human, a spider, a dandelion, and a Streptococcus bacterium will translate any mRNA sequence into the same protein. The consistency
of the genetic code across so many kinds of organisms is a striking reminder of the evolutionary kinship of all living things.
17.49 In translation, what molecule contains the coded
instructions?
17.50 What type of molecule is made during translation?
17.51 How many bases are there in one codon?
17.52 What molecule contains codons?
17.53 What is the function of a stop codon?
17.54 What are the two functions of the start codon?
17.55 A piece of mRNA has the base sequence AAUC-
GCUUA. What is the amino acid sequence of the
corresponding polypeptide?
17.56 A piece of mRNA has the base sequence UGCG-
GACCC. What is the amino acid sequence of the
corresponding polypeptide?
17.57 Using Table 17.3, answer the following questions:
a) How many different codons correspond to aspartic acid (Asp)?
b) Give one amino acid that has only one codon.
17.58 Using Table 17.3, answer the following questions:
a) How many different codons correspond to leucine (Leu)?
b) Give one amino acid that has exactly three codons.
17.59 A piece of an mRNA molecule has the following base
sequence:
Health Note: Because the
genetic code is virtually universal,
we can insert a gene for a
desirable protein into the DNA
of a microorganism and the
microorganism will decode it
correctly. For example, scientists
have been able to insert the
correct DNA sequence for human
insulin into the bacterial DNA, so
most insulin for treatment of
diabetes is now made by bacteria.
This mRNA contains the code for the beginning of a
polypeptide. What are the first four amino acids in
this polypeptide?
17.61 A piece of an mRNA molecule has the following base
sequence:
GGCAGAGAGACUGACUA
This mRNA contains the code for the end of a polypeptide. What are the last three amino acids in this
polypeptide?
17.62 A piece of an mRNA molecule has the following base
sequence:
GUGGAACACGGCACUAACG
This mRNA contains the code for the end of a polypeptide. What are the last three amino acids in this
polypeptide?
17.63 The coding strand of a DNA molecule contains the
following sequence of bases:
ACATCGTTC
a) What is the base sequence in the mRNA when
this DNA is transcribed?
b) What is the amino acid sequence in the polypeptide when the mRNA is translated?
17.64 The coding strand of a DNA molecule contains the
CAAUGUUGGCAUACG
This mRNA contains the code for the beginning of a
polypeptide. What are the first four amino acids in
this polypeptide?
17.60 A piece of an mRNA molecule has the following base
sequence:
following sequence of bases:
TTGGAAACT
a) What is the base sequence in the mRNA when
this DNA is transcribed?
b) What is the amino acid sequence in the polypeptide when the mRNA is translated?
GCAUAUGGCCCUCAGGUC
17.5 | Translation and the Genetic Code
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17.6 The Mechanism of Protein Synthesis
OBJECTIVE: Describe the mechanism
of protein synthesis.
Francke, C, et al. 1982 Electron microscope visualization of a
discrete class of giant translation units in salivary gland cell of
Chironomous tentans, The EMBO Journal 1:59–62.
You now know how to relate the language of nucleic acids to the language of proteins.
However, we have not looked at how a cell actually accomplishes this translation. There
is no obvious structural relationship between amino acids and nucleic acids, and there is
little apparent organization to the genetic code. In this section, we will examine the
mechanism by which a cell translates the code in a piece of mRNA into a polypeptide
chain.
Recall that in eukaryotes the DNA is located in the nucleus of the cell. As a result, all
of the RNA is made and processed in the nucleus. The mature RNA then leaves the
nucleus, and all translation occurs in the cytoplasm of the cell. The catalysts for translation are ribosomes, large clusters of proteins and rRNA. The ribosomes bind to mRNA
molecules and use the base sequence of the mRNA to make polypeptides. Each ribosome
binds to the start codon and then moves toward the 3⬘ end of the mRNA until it reaches
a stop codon, at which point it separates from the mRNA and releases the completed
protein. Many ribosomes can bind to a single mRNA molecule, so many polypeptide
chains are being made at any given instant. Translation continues until the cell breaks
down the mRNA molecule. Figure 17.18 illustrates how an mRNA molecule can be
translated by many ribosomes at the same time.
tRNA Matches Each Codon with the Correct Amino Acid
Translation: The thin strand is the
mRNA, and the heavy dots are
ribosomes synthesizing protein.
The actual matching of mRNA codons with amino acids is done by tRNA. This is the
smallest of the three types of RNA, and it folds into a characteristic L shape, with the
nucleotide chain looping back on itself to form several short sections of double helix.
One end of the L contains an exposed three-base sequence called the anticodon. The
anticodon binds to the complementary mRNA codon during translation. The other end
of the L contains the two ends of the tRNA molecule, including the CCA sequence that
was added to the 3⬘ end of the chain. This end of the tRNA, called the acceptor stem,
becomes bonded to an amino acid. The general structure of a tRNA molecule is shown
in Figure 17.19.
A cell contains many types of tRNA. Each tRNA contains a unique anticodon, which
is able to form base pairs with a particular mRNA codon. In principle, a cell would need
exactly 61 kinds of tRNA, since there are 61 meaningful codons, but in practice the third
base in the anticodon need not be the exact complement of the third base in the codon.
For instance, the anticodon AAU (reading from 3⬘ to 5⬘) is able to bind to the codons
UUA (its normal complement) and UUG. As a result, cells only need one type of tRNA
to decode UUA and UUG. This is reflected in the genetic code: the codons UUA and UUG
correspond to the same amino acid, leucine.
Before translation can occur, each type of tRNA must be bonded to the correct amino
acid. Cells have 20 enzymes to do this, one for each amino acid. These enzymes, called
aminoacyl–tRNA synthetases, match each type of tRNA to its corresponding amino
acid. These remarkable enzymes are able to select and bind to one particular amino acid
and to all possible tRNAs for that amino acid. The enzyme then condenses the 3⬘ hy-
Completed
polypeptide
mRNA
5'
3'
Start codon
Stop codon
Ribosomes
FIGURE 17.18 Translation of an mRNA molecule.
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5'
C
C
A
3'
Acceptor stem
(An amino acid bonds here)
Only the
sugar-phosphate
backbone is shown
in this figure.
C
U
G
Anticodon
(Binds to mRNA)
FIGURE 17.19 The structure of tRNA.
Overall reation: tRNA + Alanine +
2 ATP
+
NH3
OH
tRNA
+
NH3
Alanyl-tRNA + 2 ADP + 2 Pi
CH3 O
CH
C
CH3 O
CH
–
O
C
O
Alanyl-tRNA
(an aminoacyl-tRNA)
Alanine
FIGURE 17.20 Forming the bond between tRNA and an amino acid.
droxyl group of the tRNA with the carboxyl group of the amino acid to form an ester.
The product of the reaction is called an aminoacyl-tRNA, and it is an activated form of
the amino acid. As we saw in Chapter 15, making an activated molecule requires energy,
which is supplied by ATP. This reaction is shown in Figure 17.20.
The Ribosome Moves along the mRNA during Translation
Figure 17.21 shows how the cell translates the mRNA code into an amino acid sequence.
Translation starts when the two parts of a ribosome bind to a molecule of mRNA. The
ribosome surrounds two codons, and it has a large space above the codons. Two molecules of tRNA, each bonded to its correct amino acid, enter the ribosome and bind to
the mRNA codons. Only the tRNAs with the correct anticodons can bind to the codons
and remain within the ribosome. Once the two tRNA molecules are in place, the ribo-
Health Note: Many
antibiotics, including tetracycline
and erythromycin, bind to
bacterial ribosomes and block
protein synthesis. Ribosomes of
humans (and all other
eukaryotes) have a different
structure from bacterial
ribosomes, so the body’s ability
to make proteins is not affected
by these antibiotics.
17.6 | The Mechanism of Protein Synthesis
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Ribosome
The ribosome
binds to the RNA,
and the first
AA-tRNA binds to
codon #1.
AA
tRNA
mRNA
5'
3'
Codon
number
1
2
3
4
5
6
7
8
9
(Start)
Ribosome
A second
AA-tRNA enters
the ribosome
and binds to
codon #2.
AA
AA
tRNA tRNA
5'
3'
1
2
3
4
5
6
7
8
9
(Start)
Ribosome
AA
The peptide bond
forms, and the
bond between
the first AA and
its tRNA breaks.
AA
tRNA tRNA
5'
3'
1
2
3
4
5
6
7
8
9
(Start)
Ribosome
AA
The ribosome
moves one
codon to the
right, releasing
the first tRNA.
AA
tRNA
tRNA
5'
3'
1
2
3
4
5
6
7
8
9
(Start)
FIGURE 17.21 The mechanism of translation.
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Ribosome
AA
AA
A third AA-tRNA
enters the
ribosome and binds
to codon #3.
AA
tRNA tRNA
5'
3'
1
2
3
4
5
6
7
8
9
(Start)
AA
Ribosome AA
A second peptide
bond forms.
AA
tRNA tRNA
5'
3'
1
2
3
4
5
6
7
8
9
(Start)
AA
AA Ribosome
AA
tRNA
tRNA
The ribosome
moves one codon
to the right,
releasing tRNA #2.
5'
3'
1
2
3
4
5
6
7
8
9
(Start)
some detaches the amino acid from the first tRNA and links the two amino acids with
a peptide bond.
After the peptide bond forms, the entire ribosome shifts to the right along the mRNA.
The first tRNA, now without its amino acid, leaves the ribosome. The second tRNA
remains attached, and a third tRNA moves into place and binds to the third mRNA
codon. The ribosome detaches the dipeptide from the second tRNA and links it to the
amino acid on the third tRNA, forming a tripeptide. This cycle then repeats over and
over, with the ribosome moving toward the 3⬘ end of the mRNA, until the ribosome
reaches a stop codon. At this point, the completed polypeptide is released and the ribosome detaches from the mRNA.
How much energy does it take to build a protein? In Figure 17.20, we saw that activating an amino acid requires the breakdown of two molecules of ATP. The cell must
also break down two molecules of GTP (equivalent to two molecules of ATP) when it
positions each tRNA in the ribosome. The result is that adding one amino acid to a
protein costs the equivalent of four molecules of ATP. Building an enzyme that contains
17.6 | The Mechanism of Protein Synthesis
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300 amino acids requires the breakdown of almost 1200 molecules of ATP. In addition,
a cell must make mRNA before it can build a protein. The initial transcript mRNA for
a typical enzyme might contain 20,000 nucleotides. Since adding a nucleotide to RNA
costs the equivalent of two molecules of ATP, the mRNA requires the breakdown of
40,000 molecules of ATP. A cell normally translates a single mRNA molecule many
times, but many mRNA molecules have lifetimes of 30 minutes or less, limiting the number of protein molecules the cell can make from one mRNA. Clearly, protein synthesis
17.65 Which type of RNA (mRNA, rRNA, or tRNA) fits
17.71 How many codons fit into a ribosome?
each of the following descriptions?
a) This type of RNA binds directly to amino acids.
b) This type of RNA contains codons.
c) This type of RNA helps catalyze protein
synthesis.
17.72 How many tRNA molecules can fit into a ribosome?
17.73 How many molecules of ATP must be broken down
to make the bond between tRNA and an amino acid?
17.74 What is the total number of molecules of ATP (or its
equivalent) that must be broken down to add one
amino acid to a polypeptide during translation?
17.66 Which type of RNA (mRNA, rRNA, or tRNA) fits
each of the following descriptions?
a) This type of RNA contains genetic information
transcribed from DNA.
b) This type of RNA contains an anticodon.
c) This type of RNA is found in the ribosomes.
17.75 Four codons correspond to the amino acid threonine
(ACA, ACC, ACG, and ACU), but only one tRNA
can bond to threonine. Why aren’t there four tRNAs
that bond to threonine?
17.67 What is the function of the ribosomes in a cell?
17.76 Can the tRNA that binds to the codon UGG also
bind to the codon UGA? Explain your answer. (Hint:
These codons are in Table 17.3.)
17.68 What is the function of aminoacyl–tRNA synthetases?
17.69 What is an anticodon, and what type of RNA
contains an anticodon?
17.70 What is an acceptor stem, and what type of RNA
contains an acceptor stem?
OBJECTIVES: Relate each of the four
types of genetic mutations to its effect on
protein sequence, describe some of the
causes of mutations, and explain the
relationship between a mutation and a
genetic disorder.
is an expensive undertaking for a cell. It is estimated that building proteins (and the
RNA needed to do so) consumes more than half of the ATP produced by our bodies.
17.7 Mutations and Genetic Disorders
Image copyright Jane McIlroy, 2010. Used under license from
Shutterstock.com
As we have seen, DNA replication is extremely accurate. However, it is not perfect.
When a cell divides, the daughter DNA usually differs slightly from the parent DNA. In
addition, DNA can be damaged in a variety of ways, some of which can change the base
sequence. Cells can repair most of the damage, but some changes inevitably become part
of the permanent genetic material of the cell. Permanent changes to DNA that can be
passed on from parent to offspring are called mutations. The
most common types of mutations are substitutions, additions,
deletions, and recombinations. Let us examine each of these
mutations and the effect that it has on a cell.
A Substitution Mutation Replaces One Base
with Another
This kangaroo has a genetic mutation that makes it unable to make
the normal pigment in its fur.
In the simplest type of mutation, a substitution mutation, one
base pair in the DNA is replaced by a different base pair. Substitution mutations can have a range of effects on a cell, depending
on where the mutation occurs. If the substitution occurs in a
region of DNA that is not part of a gene, it may have no effect.
However, if the substitution is close to the gene, it can dramatically affect the cell’s ability to make a particular protein, because
the cell uses the DNA sequences on either side of the gene to
determine where to begin and end transcription. In the worst
case, the cell may not be able to recognize the gene, so the cell
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cannot make the corresponding protein. If the cell requires the protein to survive, the mutation will kill the cell.
If the substitution occurs within an exon, it may change the amino acid sequence of the
corresponding protein. We can classify substitutions within an exon as silent mutations,
missense mutations, or nonsense mutations based on their impact on the amino acid sequence. Let us look at each of these kinds of mutations, using a DNA molecule that contains TTA in its coding strand to illustrate the effect of each type. When this DNA is
transcribed, the mRNA will contain the codon UUA (remember that the mRNA sequence
matches the coding strand of DNA, except that T is exchanged for U). The corresponding
protein will contain the amino acid leucine.
1. Silent mutation. In a silent mutation, the mutated codon and the original codon
correspond to the same amino acid. An example would be a mutation that changes
the third base of our original DNA from A to G, giving us the DNA sequence TTG.
The corresponding mRNA codon is UUG, which codes for leucine. Since the
original and mutated codons give the same amino acid, the primary structure of
the protein is not affected and the mutation has no effect on the cell.
2. Missense mutation. In a missense mutation, the mutated codon and the original
codon correspond to two different amino acids. For example, suppose a mutation
converts our original TTA to ATA. The corresponding mRNA codon is AUA,
which codes for isoleucine. The protein will have a different amino acid sequence
than the original version.
Missense mutations may change the ability of a protein to carry out its function,
depending on the location and properties of the changed amino acid. In this example, leucine and isoleucine are similar in size and both have nonpolar side chains.
Therefore, this mutation will probably not change the overall structure of the protein significantly.
An example of a substitution that has a dramatic effect on a protein is the sicklecell mutation, which affects one of the polypeptide chains in hemoglobin. In this
mutation, the DNA sequence GAG has been changed to GTG. The effect on the
protein is to replace glutamic acid with valine. These two amino acids have very
different properties, because glutamic acid is hydrophilic and is an anion at pH 7,
while valine is hydrophobic. The resulting hemoglobin is much less soluble in water,
and it forms insoluble crystals in the red cells when the oxygen concentration is low.
These crystals distort the red cells into a distinctive sickle shape. The affected cells
cannot transport oxygen efficiently, and they tend to clog the capillaries, producing
debilitating pain as the affected tissues are starved for oxygen.
Mutations that convert a stop codon into a meaningful codon are also considered missense mutations. However, since they eliminate the stop signal, these mutations produce a protein that contains extra amino acids. Such a protein will usually
be inactive.
3. Nonsense mutation. A nonsense mutation converts the original codon into a stop
codon. For instance, suppose that our mutation converts TTA into TAA. The
resulting mRNA codon is UAA, which is a stop codon. As a result, the cell will
TABLE 17.4 The Three Types of Substitution Mutations
Original DNA
(Coding Strand)
TTA
(codes for
leucine)
Type of
Mutation
Mutated DNA
(Coding Strand)*
mRNA
Amino Acid
Result of Mutation
Change base 3
from A to G
Silent
TTG
UUG
Leucine
The amino acid
sequence is unchanged.
Change base 1
from T to A
Missense
ATA
AUA
Isoleucine
Isoleucine is
substituted for leucine.
Change base 2
from T to A
Nonsense
TAA
UAA
None
(stop codon)
The protein lacks some
amino acids.
Mutation
*The incorrect base is shown in red.
17.7 | Mutations and Genetic Disorders
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Sample Problem 17.6
The effect of a substitution mutation on a protein
A mutation converts the DNA sequence in the coding strand from AAA to AAT. What
type of mutation is this? How will this mutation affect the primary structure of the
corresponding protein?
SOLUTION
First, we must work out the mRNA codons that are produced when the DNA is transcribed. The RNA sequence is the same as the DNA sequence, except that T is exchanged
for U. Therefore, the original mRNA contains the codon AAA, and the mutated codon
has the codon AAU.
Next, we need to translate each of these codons. AAA corresponds to Lys (lysine), and
AAU corresponds to Asn (asparagine). Since the mutation substituted one amino acid for
another in the protein, this is a missense mutation. The mutated protein will contain an
asparagine in place of a lysine.
TRY IT YOURSELF: A mutation converts the DNA sequence in the coding strand from
AAA to TAA. What type of mutation is this? How will this mutation affect the primary
structure of the corresponding protein?
For additional practice, try Core Problems 17.81 and 17.82.
make a protein that is missing all of the amino acids beyond the new stop codon.
Nonsense mutations normally produce proteins that are completely inactive, and
they can be lethal to the cell.
Table 17.4 summarizes the three examples of substitution mutations.
TABLE 17.5 The Categories of Substitution Mutations
Type of
Substitution
Description
Effect on a
Protein
Changes one
amino acid into
another
Missense
Changes a codon for
one amino acid into
a codon for a
different amino acid
Changes a stop
codon into a codon
for an amino acid
Produces a longer
protein
Silent
Changes a codon
into a different
codon for the same
amino acid, or
changes one stop
codon into another.
None
Nonsense
Changes a codon for
an amino acid into a
stop codon
Produces a
shorter protein
Number of
Possible
Mutations
of This Type
Percentage
of All
Possible
Mutations
392
68%
23
4%
138
24%
23
4%
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The most likely type of substitution is a missense mutation. However, nearly a quarter
of all possible substitutions are silent mutations, because many codons correspond to the
same amino acid. Table 17.5 shows the number of possible mutations for each type of
substitution.
Addition and Deletion Mutations Normally Produce Frameshifts
We can consider the next two types of mutations together, because they tend to have
similar effects on a protein. In an addition mutation, one or more base pairs are inserted
into a DNA molecule. A deletion mutation is the removal of one or more base pairs
from DNA. In either case, the result is usually a frameshift, in which all of the following
codons are interpreted incorrectly. For instance, suppose that we have a section of a
DNA coding strand that has the following sequence:
ACGATCATTACG
When the cell transcribes this region, it will produce the following mRNA:
ACGAUCAUUACG
The cell will then translate this mRNA as follows:
A C G A U C A U U A C G
Thr
Ile
Ile
Thr
Now, suppose that a mutation adds a T after the initial A in the original DNA. The
mutated DNA coding strand has the following sequence:
A T
C G A T
C A T
Health Note: People who
suffer from a genetic disorder
called cystic fibrosis produce
excess thick mucus in their lungs,
leading to lung infections and
damage to lung tissue. About
two-thirds of all cases of cystic
fibrosis are caused by a deletion
mutation that removes three
base pairs from the gene for the
chloride transport protein.
However, more than 1000
mutations causing cystic fibrosis
have been identified, including
substitutions, insertions, and
other deletions.
T A C G
The resulting mRNA and amino acid sequence will be:
Original codon separation points
A U C G A U C A U U A C G
Ile
Asp
His
Tyr
Sample Problem 17.7
The effect of a deletion mutation on a protein
A piece of a DNA coding strand has the base sequence ACGATCATTACG. A mutation
removes the third base from this DNA. What type of mutation is this? Will this mutation
produce a frameshift? Write the amino acid sequences of the original and the mutated
protein.
SOLUTION
This is a deletion mutation. The mutation will change the way all subsequent bases are
divided into codons, so it will produce a frameshift.
The original DNA corresponds to the mRNA sequence ACGAUCAUUACG. To translate this sequence, we must divide it into codons:
ACG–AUC–AUU–ACG
Translating these codons gives us the amino acid sequence Thr–Ile–Ile–Thr.
When we remove the third base from the DNA, we get the base sequence ACATCATTACG, which corresponds to the mRNA sequence ACAUCAUUACG. We can divide this
into codons:
ACA–UCA–UUA–CG
We can translate the three complete codons in this sequence. The fourth codon is incomplete (since we weren’t given the next base in the original DNA), but since all codons
continued
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that have CG as their first two bases correspond to arginine (Arg), we can write out four
amino acids in our mutated protein. The amino acid sequence is Thr–Ser–Leu–Arg.
TRY IT YOURSELF: A mutation removes the second, third, and fourth bases from the
original DNA coding strand in this sample problem. Write the amino acid sequence of
the mutated protein. Does this mutation produce a frameshift? Why or why not?
For additional practice, try Core Problems 17.87 (parts a and c) and 17.88 (parts a and c).
This amino acid sequence is completely different from the original one. Adding the
extra base changes the way the cell reads all subsequent codons. Additions and deletions
produce a frameshift unless the number of bases added or removed is a multiple of three.
Since frameshifts change many amino acids in a protein, they, like nonsense mutations,
normally produce an inactive protein.
Health Note: The bacteria
that produce cholera and
diphtheria are normally harmless
to humans. However, sometimes
viruses that insert some of their
DNA into the bacterial genome
infect these bacteria. The viral
DNA codes for highly toxic
proteins. Infected bacteria then
make these toxins, producing the
potentially fatal symptoms of
cholera and diphtheria.
Recombinations Change the Locations of Genes within the Genome
Recombination mutations are the most complex of the four types of mutations. In a
recombination, a large section of DNA moves from one DNA molecule to another, or it
moves from one position to another within the same DNA molecule. Figure 17.22 shows
two examples of recombinations.
Many recombinations simply move an entire gene from one place to another without
disturbing any of the coding regions of the DNA. Other recombinations disrupt a gene,
with potentially devastating effects on the cell. A recombination can remove a large section of a gene, or add a large section of extraneous DNA within a gene, or even reverse
the direction of a region of DNA within a gene. The majority of recombinations happen
within a single cell, but many bacteria can exchange sections of DNA with other members of their own species or even with members of different species.
Example 1: Recombination between two chromosomes
DNA #1
Gene exchange
DNA #2
Recombined
DNA #1
Recombined
DNA #2
Example 2: Recombination within a chromosome
Original
DNA
A gene moves
to a different
position
Recombined
DNA
FIGURE 17.22 Two types of recombinations.
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Recombinations are surprisingly common, and they serve a range of purposes. The
ability to move a gene from one chromosome to another increases the genetic diversity
of a species by creating new combinations of genetic traits. Cells can use recombinations
to replace damaged sections of DNA, a process that is particularly common in bacteria.
Many viruses are capable of inserting all or part of their DNA into the DNA of an infected cell, allowing the virus to hide from the host’s defense mechanisms. Our immune
system uses recombinations to produce a vast number of different antibodies from a
much smaller number of genes, allowing us to defend ourselves from a range of microorganisms. Bacteria that can absorb genes from other bacteria can gain useful new traits,
such as the ability to break down an antibiotic. In addition, scientists have discovered
recombinations that have no known function.
Chemicals and Radiation Can Produce Mutations
–
O
N
O
protein
Nitrite ion,
NO2–
N
N
Image copyright Joy Brown, 2010. Used
under license from Shutterstock.com
What causes mutations? Substitution mutations, additions, and deletions have two common causes: errors in replication and environmental damage. As we have seen, replication is extremely accurate, but it is not perfect. Any replication error that is not corrected
will produce a permanent mutation to the daughter cell’s DNA. In addition, a variety of
environmental factors can damage DNA, including certain chemical compounds (called
mutagens) and high-energy radiation (ultraviolet light and X-rays). For example, the
nitrites that are used to preserve meat react with proteins at high temperatures to form
compounds called nitrosamines.
O
A nitrosamine
Nitrosamines are mutagens because they can convert cytosine into uracil, changing a
C–G base pair into an unstable U–G base pair. The DNA repair enzymes then replace
the guanine with adenine (the complement of uracil) and the uracil with thymine. The
Frying bacon produces low
concentrations of nitrosamines, which
can cause mutations in DNA.
TABLE 17.6 Chemicals and Radiation That Cause Mutations
Mutagen
Source or Use
Effect
Nitrosamines
Formed when meats containing nitrites are
cooked
Modify the structures of C and G, producing
substitution mutations
Acridine orange
Used as a dye to detect bacteria
Inserts between bases in the double helix,
producing insertion mutations
Temozolomide
Used as a medication in the treatment of brain
tumors
Adds methyl groups to DNA bases, producing
substitution mutations
2-Amino-1-methyl6-phenylimidazo[4,5-b]
pyridine (PhIP)
Formed during cooking of foods containing
protein
Combines with G, producing substitution
mutations
Benzo[a]pyrene
Formed as a component of smoke from
cigarettes, fireplaces, barbecues, etc.
Combines with G, producing substitution
mutations
Aflatoxin B1
Formed by molds that grow on certain food
crops
Combines with DNA bases to produce
substitution mutations and appears to increase
recombination rates
Ultraviolet radiation
A component of sunlight
Forms covalent bonds between adjacent Ts,
producing a variety of errors in replication
X-rays
Used for diagnostic imaging in medicine
(including CT scans)
Produce a range of chemical changes in DNA
Gamma rays
Formed by radioactive elements and used in
some cancer treatments
Produce a range of chemical changes in DNA
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end result is to convert C–G into T–A. Table 17.6 lists some environmental factors that
cause mutations.
In contrast with the other types of mutations, recombinations are a normal part of
cell function. All cells contain enzymes that recombine DNA, moving genes around
within chromosomes and from one chromosome to another.
Mutations Can Affect an Entire Organism
Mutations clearly can have an enormous impact on a single cell. However, our bodies
are a collection of a vast number of cells. How does a change in just one of those cells
affect us? In general, a mutation in one cell within the human body has no discernible
impact on our health. Since the smallest organ contains millions of similar cells, a mutation in one of these cells does not affect the ability of the organ to do its job. Even if the
mutation kills the cell, we have many other cells of the same type to pick up the slack.
However, there are three occasions when a mutation can have a devastating impact
on a human. Each of the following depends on mutations becoming part of the DNA
blueprint for a cell and being passed on to all of the descendants of that cell:
1. The mutation damages the DNA that controls cell division. In our bodies, cells
only divide under specific conditions, which depend on the type of cell. Cell
division must be tightly controlled to ensure that each part of the body is the
correct size and shape. The instructions that tell cells when to divide are encoded in
each cell’s DNA. If this information is damaged by mutations, the cell may start to
divide uncontrollably, producing a mass of useless tissue that crowds out surrounding tissues. The result is cancer, the second-leading cause of death in the United
States.
Cancer does not result from a single mutation. Cells contain several genes that
work together to regulate cell division. If one of these genes is damaged, the others
can still control cell growth and division. However, if enough of these genes are
damaged, the cell will become cancerous.
2. The mutation occurs in the early stages of embryonic development. Each cell in an
embryo is the precursor of a vast number of cells in an adult. If one cell in the
embryo undergoes a lethal mutation, the embryo normally will not survive. Less
severe mutations can result in a variety of birth defects. However, not all mutations
in embryonic cells are harmful. Some people’s bodies contain a mixture of two
types of cells, one of which contains a harmless mutation. These people experienced a benign mutation in a single cell early in their embryonic development. All
of the cells in their bodies that are descendants of the mutated cell carry the
mutation, while the rest of their cells do not. Such people are called genetic
mosaics.
3. The mutation occurs in an egg or sperm cell before fertilization. Human egg and
sperm cells are unique in containing only 23 chromosomes, one of each of the
autosomal chromosomes plus one sex chromosome. When the egg and sperm fuse
during fertilization, the combined cell has a full set of chromosomes. Any mutation
in an egg or sperm becomes part of the DNA of the fertilized egg. This mutation is
then passed on to every cell in the body of the person that develops from the
fertilized egg.
Health Note: Medical
procedures that change a
person’s gender involve a variety
of lifestyle changes, hormone
treatments, and surgical
procedures, but they do not
change the person’s DNA.
If a mutation occurs in an autosomal chromosome, the mutation may not have any
impact on the health of the person who initially inherits it, because human cells contain
two of each of the autosomal chromosomes. As long as one of the two chromosomes
contains a functioning version of the gene, the body is able to make the correct version
of the protein that is affected by the mutation. However, any person who has this mutation is able to pass the mutated chromosome to his or her children. If two people whose
genomes contain this mutation have a child, there is a 25% chance that the child will
inherit two defective chromosomes, one from each parent. In such a case, the child cannot make the correct protein and suffers from an autosomal genetic disorder. These
diseases, which are equally likely in men and women, include cystic fibrosis, galactosemia, sickle-cell anemia, and phenylketonuria (PKU). Figure 17.23 shows how genetic
disorders are inherited.
If the mutation occurs in an X chromosome, the situation becomes more complex.
Women have two X chromosomes, so they suffer the effects of the mutation only when
both X chromosomes are defective. Since women inherit one X chromosome from each
parent, both parents must have the mutation, which is very unlikely. However, men in-
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© INTERFOTO / Alamy
herit only one X chromosome, so they have only one copy of
the genes on that chromosome. A man who inherits a defective
X from his mother (his Y chromosome always comes from his
father) has no backup copy of the defective gene, so he will suffer from the genetic disorder regardless of whether his father
carries the defective gene. As a result, genetic disorders involving the X chromosome affect men far more often than women.
Such diseases, called sex-linked genetic disorders, include hemophilia and the most severe form of muscular dystrophy, as
well as most forms of color blindness.
Mutations are closely linked to the process of evolution.
Biological evolution is generally believed to involve two processes. The first of these is the spontaneous arising of new traits
within individual members of any given species. The second
process is natural selection, the tendency of those individuals
best suited to their environment to reproduce more efficiently
than their competitors. Genetic mutations are the underlying
molecular cause of the new traits that drive evolution. To be
Nicholas II, last czar of Russia, with his wife Alexandra and their
children. Their son, Alexei, suffered from hemophilia, the result of
inheriting a defective X chromosome from his mother.
Autosomal disorder:
The mutation is on one of the
autosomal chromosomes.
1/4 of their children will
have two normal genes.
+
Half of their children will
have a copy of the
defective gene, but will
be healthy.
If each parent carries a copy
of the defective gene . . .
1/4 of their children will
have two defective genes
and will suffer from a
genetic disorder.
Sex-linked disorder:
The mutation is on one of the sex
chromosomes (usually the X).
Half of their male
children will have the
defective X and will
suffer from a genetic
disorder.
+
Half of their female
children will carry a
defective X, but none
will have the genetic
disorder.
If the woman carries one
defective X chromosome . . .
Key:
Normal autosomal
chromosome
Normal
X chromosome
Defective autosomal
chromosome
Defective
X chromosome
Y chromosome
FIGURE 17.23 The inheritance of genetic disorders.
continued
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17.77 What happens in each of the following types of
mutation?
a) substitution
17.87 a) Write the amino acid sequence that corresponds
to the following DNA coding strand:
b) deletion
17.78 What happens in each of the following types of
mutation?
a) recombination
b) addition
17.79 Explain why the addition of one base pair to a DNA
molecule produces a frameshift.
17.80 Explain why the substitution of one base pair with a
different base pair in a DNA molecule does not
produce a frameshift.
17.81 Classify each of the following substitutions as a
silent, missense, or nonsense mutation. (The base
sequences of the DNA coding strand are given.)
a) GAA → GAC
b) GAA → GAG
c) GAA → TAA
17.82 Classify each of the following substitutions as a
silent, missense, or nonsense mutation. (The base
sequences of the DNA coding strand are given.)
a) TAC → TAG
b) TAC → TAT
c) TAC → GAC
17.83 Which of the following is the most likely to be
harmful to a cell, and why?
a) replacement of one base pair by another in the
DNA
b) addition of one base pair in the DNA
17.84 Which of the following is the most likely to be
harmful to a cell, and why?
a) deletion of one base pair in the DNA
b) deletion of three consecutive base pairs in the
DNA
17.85 Which of the following substitution mutations is
most likely to be harmful to a cell, and which is least
likely to be harmful, based on their effects on the
corresponding polypeptides? Refer to Tables 17.4
and 13.1 as you answer this problem, and explain
your answers. (The base sequences of the DNA
coding strand are given.)
a) ACA → TCA
b) ACA → ACG
c) ACA → CCA
17.86 Which of the following substitution mutations is
most likely to be harmful to a cell, and which is least
likely to be harmful, based on their effects on the
corresponding polypeptides? Refer to Tables 17.4
and 13.1 as you answer this problem, and explain
your answers. (The base sequences of the DNA
coding strand are given.)
a) AGA → AAA
b) AGA → GGA
c) AGA → CGA
AAACATAAGTTG
b) A mutation replaces the fifth base in this DNA
with G. What will be the amino acid sequence
when this mutated DNA is transcribed and the
resulting mRNA is translated?
c) A mutation adds T between the fifth and sixth
bases in the original DNA (so the sequence is
now AAACATTAAGTTG). What will be the
amino acid sequence when this mutated DNA is
transcribed and the resulting mRNA is translated?
17.88 a) Write the amino acid sequence that corresponds
to the following DNA coding strand:
AGCGACTCTTAG
b) A mutation replaces the seventh base in this
DNA with C. What will be the amino acid sequence when this mutated DNA is transcribed
and the resulting mRNA is translated?
c) A mutation adds A between the second and third
bases in the original DNA (so the sequence is
now AGACGACTCTTAG). What will be the
amino acid sequence when this mutated DNA is
transcribed and the resulting mRNA is translated?
17.89 What is a mutagen?
17.90 List two types of radiation that can cause mutations.
17.91 A mutation in one cell in a developing embryo is
more harmful than a mutation in one cell in an adult.
Explain.
17.92 A mutation in a bacterium often kills the bacterium,
but a mutation in a cell in the human body usually
has no effect on the health of the person. Why is
this?
17.93 Many mutagens cause cancer, including all of the
chemicals and radiation types in Table 17.6. Explain.
17.94 Virtually all mutagens can cause birth defects
(genetic disorders in newborn children). Why is this?
17.95 What is an autosomal genetic disorder, and under
what circumstances will a person suffer from this
type of disorder?
17.96 What is a sex-linked genetic disorder, and under
what circumstances will a person suffer from this
type of disorder?
17.97 If a woman inherits a defective X chromosome from
one of her parents, she will normally suffer no ill
effects. Why is this?
17.98 If a man inherits a defective X chromosome from his
mother, he will normally suffer from a genetic
disorder. Why is this?
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The Human Genome Project (HGP), started in 1990, is a massive
effort to determine the base sequence of all of the DNA in a human,
some 3.3 billion base pairs spread over 23 pairs of chromosomes,
and to determine the locations and functions of all active genes.
The project has been funded primarily by the U.S. government and
carried out by researchers at laboratories around the world. By
2000, the HGP completed a rough map of the human genome, and
the project was declared effectively completed in 2003. However,
substantial portions of human DNA have not been sequenced to
this day. In particular, regions where chromosomes attach to one
another (the centromeres) and the ends of each chromosome (the
telomeres) cannot be sequenced using current technology. Around
92% of the genome has been sequenced to date, and some 24,000
genes have been identified.
Every human has a unique genome, with the exception (at least
initially) of identical twins. The DNA used by the HGP is a mixture
of the DNA from several anonymous individuals, selected at random from a larger number of volunteers. Thus, the HGP has not
determined the DNA sequence of any one person. However, part of
the work of the HGP has been to determine locations where DNA
sequences of different people differ from one another and discover
the relationship of these locations to observable variations in appearance and function.
The HGP has already proven to be beneficial to medicine. Many
human diseases have been traced to genetic defects whose position
within the human genome has been identified. Among these are
genes for cystic fibrosis, breast cancer (BRCA1 and BRCA2), muscular
dystrophy, Huntington’s disease, amyotrophic lateral sclerosis (ALS,
or Lou Gehrig’s disease), and fragile X-linked mental retardation. In
addition, scientists have isolated genes that are linked to increased
occurrence of diabetes, obesity, and affective disorders such as
schizophrenia and bipolar disorder. People with a familial history of
genetic disease who are considering having children can choose to
be tested to determine whether they carry the defective gene, and
fetuses can be tested to determine whether they have inherited a
genetic disorder such as hemophilia or cystic fibrosis. While these
diseases cannot be cured, they can be treated, and early diagnosis
allows affected children to begin treatment early.
Genetic testing raises a variety of ethical concerns. The first of
these is the effect of the test results on the individual being tested.
© Peter Arnold, Inc. / Alamy
The Human Genome Project and Genetic Screening
This couple is undergoing genetic counseling before starting a family.
Knowing that one is at high risk for cancer or dementia can be
stressful, particularly if there is no way to prevent the condition. A
second concern is the possibility that the information could be
used to discriminate against someone. For example, a person who
is at higher risk for alcoholism might be denied a job, or a person
who is unusually likely to develop heart disease could be denied
coverage by health insurers. A third concern is the possibility that
parents will screen embryos to select their children’s traits, choosing, for instance, to have only boys or to have children who are
likely to be unusually tall.
As technology has improved, we have seen the birth of a new
industry—personal genomics. It is possible for individuals to have
their DNA completely sequenced, although only the most affluent
can afford the $350,000 price tag. However, several companies offer a partial screen to scan for up to 1 million known DNA markers.
This type of genetic test is much less expensive, between $1000
and $2500, and it requires only a saliva swab. However, the U.S.
Federal Trade Commission and the Centers for Disease Control and
Prevention recommend against relying on such procedures on the
grounds that consumers are likely to misinterpret the meaning of
the results.
Key Terms
acceptor stem – 17.6
addition mutation – 17.7
anticodon – 17.6
autosomal chromosome – 17.3
autosomal genetic disorder – 17.7
backbone – 17.2
chromosome – 17.2
coding strand – 17.4
codon – 17.5
complementary base pair – 17.2
deletion mutation – 17.7
deoxyribonucleic acid (DNA) – 17.1
double helix – 17.2
exon – 17.4
frameshift – 17.7
gene – 17.4
genetic code – 17.5
initial transcript – 17.4
intron – 17.4
messenger RNA (mRNA) – 17.4
missense mutation – 17.7
mutagen – 17.7
mutation – 17.7
nonsense mutation – 17.7
nucleoside – 17.1
nucleotide – 17.1
recombination mutation – 17.7
replication – 17.3
ribonucleic acid (RNA) – 17.1
ribosomal RNA (rRNA) – 17.4
ribosome – 17.6
sex chromosome – 17.3
sex-linked genetic disorder – 17.7
silent mutation – 17.7
start codon – 17.5
stop codon (nonsense codon) – 17.5
substitution mutation – 17.7
template strand – 17.4
transcription – 17.4
transfer RNA (tRNA) – 17.4
translation – 17.5
| Connections
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SUMMARY OF OBJECTIVES
Now that you have read the chapter, test yourself on your knowledge of the objectives, using
this summary as a guide.
Section 17.1: Understand the biological role of nucleotides, draw the structure of a typical
nucleotide, and describe the differences between DNA and RNA nucleotides.
• Nucleic acids are long chains of nucleotides.
• Nucleotides contain one to three phosphate groups, a five-carbon sugar, and an organic base.
• DNA contains the sugar deoxyribose and the bases A, C, G, and T, while RNA contains ribose
and the bases A, C, G, and U.
Section 17.2: Know how nucleotides are linked in a nucleic acid, and understand how complementary base-pairing produces the double-helix structure of DNA.
• In nucleic acids, the phosphate group in each nucleotide condenses with the 3⬘ hydroxyl group
of the next nucleotide to form a phosphodiester group.
• The alternating sugars and phosphates form the backbone of a nucleic acid.
• Bases form complementary base pairs, held together by hydrogen bonds. Because of their
structures, A can only pair with T or U, and C can only pair with G.
• DNA molecules contain two separate chains of nucleotides, which are wound around each
other to form a double helix and held together by hydrogen bonding between complementary
bases.
• Each DNA molecule binds to histones and forms a tightly coiled structure called a chromosome.
Section 17.3: Describe the mechanism of DNA replication, and explain the role of DNA
replication in cell division.
• The base sequence of DNA corresponds to the amino acid sequence of proteins.
• Human cells contain 44 autosomal chromosomes (in 22 pairs) and two sex chromosomes.
• Cells replicate their DNA by separating the two strands and using each strand to determine
the base sequence in a new strand.
• During replication, cells first make RNA primers. They then add DNA nucleotides to the
primers, and finally they replace the primers with the corresponding DNA nucleotides.
• Cells have an elaborate mechanism to ensure the accuracy of DNA replication.
Section 17.4: Describe the mechanism of transcription, and explain how cells modify the three
types of RNA after transcription.
• Transcription is the building of a single RNA strand, using one DNA strand as a template.
• A cell makes tRNA and rRNA as parts of larger RNA transcripts, and then it removes the extra
nucleotides. The cell also modifies several bases and adds CCA to the 3⬘ end of tRNA molecules.
• Cells process mRNA by adding a 5⬘ cap, removing introns, splicing the exons together, and
adding a poly(A) tail.
Section 17.5: Relate the base sequence in DNA or mRNA to the primary structure of a protein,
using the genetic code.
• The genetic code relates three-base codons in mRNA to amino acids.
• The genetic code contains three stop (nonsense) codons and one start codon.
• A cell translates only the region of mRNA between the start and the stop codons.
• The normal direction of information flow in a cell is from DNA to RNA to protein.
Section 17.6: Describe the mechanism of protein synthesis.
• Protein synthesis occurs in the ribosomes.
• Ribosomes move along the mRNA molecule during translation.
• tRNA matches each codon with its correct amino acid.
• Each type of tRNA contains an anticodon, which forms base pairs with the complementary
mRNA codon.
• Protein synthesis requires a large fraction of the energy consumed by the human body.
Section 17.7: Relate each of the four types of genetic mutations to its effect on protein
sequence, describe some of the causes of mutations, and explain the relationship between a
mutation and a genetic disorder.
• Mutations are changes in a DNA base sequence that can be passed on to daughter cells.
• Substitution mutations can be silent, missense, or nonsense.
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• Addition and deletion mutations normally produce a frameshift, which changes all subsequent
codons.
• Recombination moves a section of DNA to a different location, either within the same
chromosome or on a different chromosome.
• Mutagens are chemicals or radiation types that produce changes in a DNA sequence.
• Mutations in the genes that control cell division can cause cancer.
• Mutations in egg or sperm cells produce new genetic characteristics, which can be passed from
parent to offspring.
• Autosomal genetic disorders are caused by defects on one of the autosomal chromosomes.
Most autosomal disorders only occur when both copies of a particular chromosome are
defective.
• Sex-linked genetic disorders are a result of defects on a sex chromosome (normally the X), and
are far more common in males than in females.
Concept Questions
Online homework for this chapter may be assigned in OWL.
* indicates more challenging problems.
17.99
Describe the backbone of a nucleic acid molecule and
how the bases are attached to the backbone.
17.100 How do the structures of DNA and RNA differ from
each other?
17.101
a) Describe the double-helix structure of DNA.
b) What type of attractive force holds the two DNA
strands next to each other?
17.102 Why can each base in DNA only form a pair with one
of the four possible bases (e.g., why will A pair with T
but not with C, G, or another A)?
17.103 Describe the function of DNA in a living organism.
17.104 The terms gene, genome, and chromosome are all used
in describing DNA. How are these different from one
another?
17.105 Describe the steps that occur during DNA replication.
17.114
What types of mutations can produce frameshifts?
17.115
Which is more likely to be harmful to a cell, a mutation in an intron or a mutation in an exon? Explain
your answer.
17.116
A particular mutation makes one of the enzymes of the
citric acid cycle inactive. Explain why this mutation is
not likely to be harmful to an adult but may be lethal if
it occurs in a human embryo.
17.117
If the mutation in Problem 17.116 occurs in a bacterium, how damaging is it likely to be? Explain.
17.118
Under what circumstances is a mutation likely to have
a significant impact on human health?
17.119
How are mutations related to biological evolution?
Summary and Challenge Problems
17.120 Draw structures of the following:
a)
b)
c)
d)
UMP
dGMP
CDP
adenosine (the nucleoside that contains ribose and
adenine)
e) a dinucleotide that is built from dAMP and dTMP
17.106 Why does DNA replication require RNA nucleotides in
addition to DNA nucleotides?
17.107 How is the information in DNA used to build a living
organism?
17.108 How is transcription similar to replication, and how is
it different?
17.121
17.109 How is mRNA modified after it is made by a cell?
17.110
Describe the specific steps that occur when one amino
acid is added to a polypeptide during translation.
17.111
Humans make 49 types of tRNA, each with its own
anticodon. However, 61 different codons correspond to
amino acids. Why don’t people need 61 types of tRNA?
17.112
17.113
In the genetic code, an amino acid may correspond to
several codons. However, a codon never corresponds to
more than one amino acid. Why is this important?
a) Which is more likely to be harmful to a cell, an
error that occurs during replication or an error that
occurs during transcription? Explain your answer.
b) Which is more likely to be harmful to a cell, an error that occurs during transcription or an error that
occurs during translation? Explain your answer.
a) List the four bases that occur in DNA.
b) What base occurs in RNA but not in DNA?
c) What are the complementary base pairs in DNA?
17.122 Give the abbreviation for the following molecule:
O
CH3
O
HO
O
P
O
O
–
O
P
O
N
O
–
O
O
N
P
O
–
CH2
O
H
H
OH
H
H
H
| Summary and Challenge Problems
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17.123 a) Identify the bases and the sugar in the following nucleic acid molecule:
O
H
N
O
HO
P
O
O
N
O
CH2
O
H
–
O
N
N
P
O
N
H
OH
H
O
NH2
H
O
CH2
O
H
–
N
H
OH
O
N
N
P
O
O
H
H
O
N
O
CH2
–
H
NH2
N
O
H
H
OH
H
OH
H
b) Is this a molecule of DNA or a molecule of RNA? How can you tell?
c) Write the base sequence for this molecule. (Be sure to write the sequence in the correct direction.)
17.124 The structure of FADH2 is shown here. This molecule contains a nucleotide that is found in nucleic acids. Circle this
nucleotide and identify the base and the sugar in this nucleotide.
H
O
N
CH3
CH3
N
N
CH2
17.125
H
N
NH2
O
N
H
O
CH
CH
CH
OH
OH
OH
CH2
O
P
O
The following diagram shows two possible ways in
which DNA could be replicated. In this diagram, the
heavy blue lines represent the original DNA strands
and the thin green lines represent the new DNA
strands. Which of these is correct?
Option A
+
Option B
+
O
O
–
N
P
O
O
–
CH2
N
O
H
H
OH
H
OH
H
N
17.126 The following diagram shows two possible ways in
which DNA could be transcribed. In this diagram, the
blue lines represent DNA and the purple line represents
RNA. Which of these is correct?
Option A
+
Option B
+
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17.127
A piece of DNA has the following base sequence:
17.135
AATCAAGGC
a) If this DNA is the coding strand, what is the base
sequence in the template strand, and what is the
base sequence in the corresponding mRNA? Be sure
to write the base sequences in the correct direction.
b) If this DNA is the template strand, what is the base
sequence in the coding strand, and what is the base
sequence in the corresponding mRNA? Be sure to
write the base sequences in the correct direction.
17.128 What type of mature RNA has each of the following?
a)
b)
c)
d)
a 5⬘ cap
the sequence CCA at the 3⬘ end
a poly(A) tail
an anticodon
17.129 The following DNA sequence comes from a coding
strand, and it includes the code for the beginning of
cytochrome C, a protein that plays an important role in
the electron transport chain. Using this information,
determine the amino acid sequence for the beginning of
this polypeptide. Write as many amino acids as you can.
1
17.137
*Would you expect Chargaff’s rule (see Problem
16.136) to apply to RNA samples? Why or why not?
17.138
One type of tRNA in humans can bind to the codon
UGC. Which of the following codons should this tRNA
also be able to bind to, based on the genetic code?
17.130 The following DNA sequence comes from a coding
strand, and it includes the code for the end of cytochrome C:
TCAAAAAAGCTACTAATGAGTAATA
17.131
*Chemists can make mRNA with a random
sequence containing only A and C (e.g.,
AAACACCACCCCAC...). When this mRNA is
translated, the resulting polypeptide is built entirely
from 6 of the 20 amino acids. Which 6 are they?
17.132
*A protein contains the amino acid sequence Ile–Phe.
Write all mRNA sequences that could correspond to
this amino acid sequence.
17.133
*DNA samples that are extracted from cells usually
contain large amounts of Mg2⫹ ions bound to the
DNA. Explain why magnesium ions are so strongly
attracted to DNA.
17.134 *During research into the structure of DNA, a group
of chemists proposed a structure that had the backbones side by side, with the bases projecting outward,
as shown here. Explain why this is an unlikely
structure for DNA, based on the properties of the
backbone.
2
17.136 *In 1951, Erwin Chargaff discovered that the total
number of molecules of A and G in any sample of
DNA equals the total number of molecules of T and
C. This observation became known as Chargaff’s rule,
and it played a significant role in the discovery of the
structure of DNA. How is Chargaff’s rule related to
the double-helix structure of DNA?
ATTAAATATGGGTGATGTTGAGA
a) This DNA contains three sequences that correspond
to RNA stop codons. What are they, and where are
they located?
b) The third base sequence that you found in part a
(i.e., the closest one to the 3⬘ end of the DNA) is the
actual stop codon for this gene. Using this information, determine the amino acid sequence for the end
of cytochrome C. Write as many amino acids as you
can.
*Single-stranded RNA can form base-paired regions.
The following pictures show two ways in which this
can happen. Which of these is more likely, and why?
a) UGA
b) UGG
c) UGU
17.139 *A section of an mRNA molecule contains the
following base sequence:
AUUGUAUUUCCA
a) If this entire section of mRNA is translated, what is
the amino acid sequence of the resulting section of
polypeptide?
b) Is this section of the polypeptide hydrophilic, or is it
hydrophobic?
c) Would you expect this section of the polypeptide to
be on the inside or the outside of the folded protein?
17.140 *How many molecules of ATP (or its equivalent) must
be broken down to translate a piece of mRNA that
contains 150 coding bases (including the start codon
and the stop codon)?
17.141
A section of the coding strand of a DNA molecule
contains the following base sequence:
AAACGCAGTTTTAGAGCT
a) If this section is transcribed and the resulting
mRNA is translated, what will be the amino acid
sequence of the resulting polypeptide? (Assume that
the entire mRNA is translated.)
b) A mutation changes the fourth base in the DNA
from C to T. What will be the amino acid sequence
that corresponds to this mutated DNA?
c) A different mutation adds an extra T between the
third and fourth bases of the original DNA (so the
sequence is now AAATCGC...). What will be the
amino acid sequence that corresponds to this
mutated DNA?
d) Which of these two mutations will probably have
the larger impact on the activity of the protein, and
why?
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17.142 *Membrane lipids called gangliosides are broken down
by an enzyme called hexosaminidase A (HEXA), which
is encoded by a gene on chromosome 15. People who
inherit two mutated copies of this gene suffer from
Tay-Sachs disease, which produces progressive nerve
degeneration and is fatal before age 5. The following
list shows three substitution mutations that are known
to occur in the coding strand of the HEXA gene:
1) Codon 329: TGG → TGA
2) Codon 269: GGT → AGT
3) Codon 446: CCT → CCC
Match each of these mutations with one of the
following descriptions. (Hint: What effect does each
mutation have on the primary sequence of HEXA?)
a) This mutation has no effect on the person who
carries it.
b) This mutation produces a completely inactive
version of HEXA. People who have this mutation
on both copies of chromosome 15 have Tay-Sachs
disease.
c) This mutation produces a version of HEXA that has
low activity. People who have this mutation on both
copies of chromosome 15 have a less severe form of
the disease, called adult-onset Tay-Sachs.
17.143 *Blood cannot clot unless it contains a variety of
proteins called coagulation factors. One of these, called
coagulation factor VIII, contains 2409 amino acids.
The gene for coagulation factor VIII is located on the
X chromosome. Mutations in this gene can cause
hemophilia, a genetic disorder characterized by internal
bleeding and slow wound healing. The following two
mutations are known to occur in the gene for coagulation factor VIII:
1) Codon 80: GTT → GAT (produces severe
hemophilia)
2) Codon 85: GTC → GAC (produces mild
hemophilia)
a) What effect will each of these mutations have on
the primary structure of coagulation factor VIII?
b) One of these mutations affects an amino acid in the
active site of coagulation factor VIII. Which one is
it, and how can you tell?
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17.144 *a) List all DNA sequences that could be produced by
a single substitution mutation in the DNA
sequence CGA. (This sequence occurs in the
coding strand.)
b) How many of these will produce a silent
mutation?
c) How many of these will produce a missense
mutation?
d) How many of these will produce a nonsense
mutation?
17.145 *a) Is it possible to have an addition mutation within
the coding portion of a gene that does not produce
a frameshift? If so, how?
b) Is it possible to have a substitution mutation
within the coding portion of a gene that produces
a frameshift? If so, how?
17.146 Xeroderma pigmentosum (XP) is a genetic disorder
that is caused by a mutation in the gene for an enzyme
that repairs DNA that has been damaged by ultraviolet
radiation. People who have XP usually suffer from a
variety of skin cancers. Why is this?
17.147 XP (see Problem 17.146) is equally likely to occur in
males and in females. Based on this, is the gene for the
DNA repair enzyme located on a sex chromosome or
on an autosomal chromosome? How can you tell?
17.148 Genes A and B are both on chromosome 1 in Flora’s
body. However, Flora’s daughter Nancy inherited gene
A, but she did not inherit gene B. Explain how this
probably happened.
17.149 Gene C is on Charles’s X chromosome. (Charles is a
man, so he has only one X chromosome.)
a) Could Charles’s daughter Wilhelmina inherit gene
C? If so, how?
b) Could Charles’s son Aloysius inherit gene C? If so,
how?
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