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free ends of a different broken chromosome, a translocation is produced. These exchanges of large
portions of chromosomes can have deleterious effects and are observed frequently in cancer cells.
C. Transposable Elements
Movable (or transposable) genetic elements, “jumping genes,” were first observed by Barbara
McClintock in the 1940s. Her work, initially greeted with skepticism, was ultimately accepted, and she
was awarded the Nobel Prize in 1983.
Transposons are segments of DNA that can move from their original position in the genome to a new
location (Fig. 13.19). They are found in all organisms. Transposons contain the gene for an enzyme called
a transposase, which is involved in cleaving the transposon from the genome and moving it from one
location to another.Retroposons are similar to transposons except that they involve an RNA molecule. Reverse
transcriptase (see below) makes a single-stranded DNA copy of the RNA which is converted to a doublestranded DNA. The double-stranded DNA is then inserted into the genome at multiple locations, forming a
repetitive element in the DNA. V. Reverse Transcriptase
Reverse transcriptase is an enzyme that uses a single-stranded RNA template and makes a DNA copy
(Fig. 13.20). The RNA template can be transcribed from DNA by RNA polymerase or obtained from
another source such as an RNA virus. The DNA copy of the RNA produced by reverse transcriptase is
known as complementary DNA (because it is complementary to the RNA template) or cDNA.
Retroviruses (RNA viruses) contain a reverse transcriptase, which copies the viral RNA genome. A
double-stranded cDNA is produced, which can become integrated into the human genome (see Fig.
12.19). After integration, the viral genes may be inactive, or they may be transcribed, sometimes causing
diseases such as AIDS or cancer (see Chapter 18). The integration event may also disrupt an adjacent
cellular gene, which also may lead to disease (see Chapter 18).CLINICAL COM M ENTS Isabel S. Isabel S. contracted HIV when she used needles contaminated with HIV to inject drugs
intravenously. Intravenous drug users account for about 10% of newly diagnosed HIV cases in the
United States. HIV mutates rapidly, and therefore, current treatment involves a combination of drugs that
affect different aspects of its life cycle (designated HAART, for highly active antiretroviral therapy). This
multidrug therapy lowers the viral titer (the number of viral particles found in a given volume of blood),
sometimes to undetectable levels. However, if treatment is not followed carefully (i.e., if the patient is not
“compliant”), the titer increases rapidly. Therefore, Isabel’s physician emphasized that she must carefully
follow her drug regimen and worked with her to ensure that she was able to take all her medications.
Dianne A. Dianne A.’s poorly controlled diabetes mellitus predisposed her to a urinary tract
infection because glucose in the urine serves as a “culture medium” for bacterial growth. The
kidney glomerulotubular unit reabsorbs filtered glucose so that, normally, the urine is glucose-free.
However, when serum blood glucose levels exceed 175 to 185 mg/dL(the tubular threshold for glucose),
the capacity for reabsorption is exceeded. In Ms. A.’s case, blood glucose levels frequently exceed this
threshold.
Calvin A. The average person has about 30 moles on the body surface, yet only about 20 people out
of every 100,000 develop a malignant melanoma. The incidence of malignant melanoma, however,
is rising rapidly. Because about 10% of patients with malignant melanoma die as a result of this cancer,
the physician’s decision to biopsy a pigmented mole with an irregular border and variation of color
probably saved Calvin’s life.
Michael T. Lung cancer currently accounts for about 15% of all cancers in men and women. The
overall 5-year survival rate is approximately 15%. Thankfully, cigarette smoking has declined in
the United States. Whereas 50% of men and 32% of women smoked in 1965, these figures have, in 2014,
fallen to 18.8% and 14.8%, respectively.BIOCHEM ICAL COM M ENTS
DNA Repair and Disease. DNA serves a unique role within a cell in that it produces the blueprint
for gene expression throughout the lifetime of the cell. However, DNA is present in limited copies
within cells (unlike RNA and proteins), and various agents often compromise its structural integrity. The
DNA, then, must be continually monitored for damage; when damage is found, repair mechanisms are
required to restore the DNA to its original structure. If the integrity of the DNA cannot be maintained,
deleterious mutations may accumulate in the genome, ultimately having a negative impact on the person as
a whole. Failures in DNA repair mechanisms will lead to disease, as indicated by the examples below.
DNA can undergo various types of damage within the cells aqueous environment. DNA undergoes
spontaneous hydrolysis of the N-glycosidic bonds, generating apurinic or apyrimidinic sites within the
DNA. Failure to repair these sites (via base excision repair) will lead to changes in nucleotide sequence
of the DNA. DNA, when exposed to UV light rapidly forms thymine dimers, which will interfere with
DNA replication unless repaired. DNA exposure to X-rays or ionizing radiation will lead to single-strand
or double-strand breaks within the DNA. Failure to repair these breaks will lead to replicative errors.
The DNA is also exposed to environmental toxins, leading to chemically modified bases, which need to
be repaired before DNA replication occurs; otherwise, the risk of inappropriate base-pairing during
replication is greatly increased.
DNA repair disorders, caused by mutations in single genes, often give rise to a cell that actively
accumulates mutations and can eventually turn into a cancer cell (see Chapter 18). A common theme in
DNA repair enzyme mutations is the clinical display of cancer. It is important to note that the mutations
themselves do not directly lead to cancer; rather, successive cell generations, each with accumulated
mutations caused by DNA repair defects, will eventually acquire a mutation that leads to growth
advantages and cancer. Distinct examples of such diseases are described in the following paragraph.
The first example is the mutations that give rise to XP. XP is primarily a defect in NER (see Fig.
13.15). There are at least 13 genes responsible for XP and its variants, all of
which are involved in NER
and/or transcription-coupled repair. The disorder is seen clinically as sun hypersensitivity resulting in
skin abnormalities. It leads to a significantly greater risk of developing skin cancer, in particular at a
younger age than in the general population. This results from an inability to remove UV-induced thymine
dimers in the DNA, leading to mismatches being created during DNA replication through the thymine
dimer.
Mutations in proteins specifically responsible for transcription-coupled repair lead to Cockayne
syndrome, which presents clinically as premature aging. Cells with these mutations cannot transcribe
damaged genes. If the DNA cannot be repaired because of the defect in transcription-coupled repair,
premature cell death can result from the reduction of gene expression. There are also specific mutations
within the XP constellation of genes that give rise to a phenotype that reflects traits of both XP and
Cockayne syndrome, indicating that there are commonalities in symptoms when either NER or
transcription-coupled repair is defective.
Hereditary nonpolyposis colon cancer (HNPCC) is caused by mutations in enzymes present in
intestinal epithelial cells that are responsible for mismatch repair. The inability to repair mismatches will
eventually lead to a series of mutations within the cells, leading to colon cancer (most commonly rightsided). This disorder is inherited, and the affected individuals have an increased risk of several cancers,with colorectal cancer being the most common. In addition, they have an early age of onset of the cancers
as compared to the general population.
Hereditary breast cancer (BRCA1 and BRCA2) results from the inheritance of mutations in proteins
responsible for the DNA repair of single-strand and double-strand breaks. Inheritance of mutations in
these genes will predispose the patients to an earlier age of onset of the disorder. The roles of BRCA1 and
BRCA2 will be discussed further in Chapter 18. KEY CONCEPTS
Replication of the genome requires DNA synthesis.
During replication, each of the two parental strands of DNA serves as a template for the synthesis of
a complementary strand.
The site at which replication is occurring is called the replication fork. Helicases and topoisomerases are required to unwind the DNA helix of the parental strands.
DNA polymerase is the major enzyme involved in replication.
DNA polymerase copies each parental template strand in the 3′-to-5′ direction, producing new
strands in a 5′-to-3′ direction.
The precursors for replication are deoxyribonucleotide triphosphates.
As DNA synthesis proceeds in the 5′-to-3′ direction, one parental strand is synthesized continuously,
whereas the other exhibits discontinuous synthesis, creating small fragments named Okazaki
fragments which are subsequently joined. This is necessary because DNA polymerase can only
synthesize DNA in the 5′-to-3′ direction.
DNA polymerase requires a free 3′-hydroxyl group of a nucleotide primer in order to replicate
DNA. The primer is synthesized by the enzyme primase, which provides an RNA primer. The enzyme telomerase synthesizes the replication of the ends of linear chromosomes (telomeres).
Errors during replication can lead to mutations, so error checking and repair
systems function to
maintain the integrity of the genome.
Table 13.4 summarizes the diseases discussed in this chapter.REVIEW QUESTIONS—CHAPTER 13
1.A variety of drugs can alter DNA replication in eukaryotic cells. Which one of the following steps
could be a target for such a drug? Choose the one best answer.
A. The enzyme family of DNA polymerases, which unwinds parental strands
B. The enzyme family of topoisomerases, which copy each parental strand in the 3′- to-5′ direction
C. The enzyme family of helicases, which copy each parental strand in the 5′-to-3′ direction
D. The finding that both strands of newly synthesized DNA always grow continuously E. The enzyme DNA ligase, which joins Okazaki fragments
2.There are a variety of DNA polymerases in both bacterial and eukaryotic cells. Targeting which one
of the following properties of DNA polymerases would result in inhibiting DNA synthesis?
A. The initiation de novo of the synthesis of new DNA strands
B. The formation of phosphodiester bonds through hydrogen bonding
C. The cleavage of released pyrophosphate that provides the energy for the polymerization reaction
D. The dissociation and reassociation of the enzyme with DNA as each nucleotide is added to an
existing DNA chain
E. The process of copying a template strand in its 3′-to-5′ direction, producing a new strand in the
5′-to-3′ direction
3.An antibiotic that inhibits bacterial DNA polymerases can damage human mitochondria owing to
which one of the following eukaryotic DNA polymerases being most similar to a prokaryotic DNApolymerase?
A. α B. β C. γ D. δ E. ε
4.A drug that inhibits DNA replication, but is inactivated by chromosomes containing telomeres,
would prove to be a very useful antibiotic. Telomeres can be best described by which one of the
following?
A. Telomeres are only present in circular chromosomes.
B. Before telomerase action, and after DNA replication, there is a 3′ overhang of the newly
synthesized strand.
C. Before telomerase action, and after DNA replication, there is a 5′ overhang of the strand being
replicated.
D. In human DNA, telomeres consist of repeating sequences of TTAGGT. E. Somatic eukaryotic cells do not contain telomeres.
5.Diseases caused by defects in DNA repair systems put the patient at risk for developing cancers.
DNA repair mechanisms can be best described by which one of the following? A. Proofreading works as the bases are paired and eliminates all base-pairing errors.
B. After replication, no further repairs are possible. C. Genes that produce mRNA have a unique repair system.
D. Genes that produce transfer RNA (tRNA) have a unique repair system.
E. DNA glycosylases recognize distortion of the DNA helix owing to bulky adducts being present on
a base within the DNA.
6.Translocations cause some of the most recognized genetic syndromes in human offspring. A
translocation can be best described by which one of the following?
A.They always produce cancer.
B.They always produce mental retardation.
C.They have to involve the exchange of an entire chromosome.
D.They can only occur in the presence of reverse transcriptase.
E.They can occur in somatic or stem cells.
7.The retroviruses, including HIV, use an RNA genome. In order to generate DNA from the genomic
RNA, the enzyme reverse transcriptase is required. Reverse transcriptase differs specifically from
DNA Pol δ by which one of the following?
A. Synthesizing DNA in the 5′-to-3′ direction B. Expressing 3′-to-5′ exonuclease activity
C. Using Watson–Crick base-pair rules during DNA synthesis D. Synthesizing DNA in the 3′-to-5′ direction
E. Inserting inosine into a growing DNA chain
8.The large DNA molecules in human chromosomes take more time to replicate than the smaller,
circular bacterial chromosomes. If a 1,000-kilobase (kb) fragment of DNA has 10 evenly spaced andsymmetric replication origins, and DNA polymerase moves at 1 kb per second, how many seconds
will it take to produce two daughter molecules? (Ignore potential problems at the ends of this linear
piece of DNA.) Assume that the 10 origins are evenly spaced from each other but not from the ends
of the chromosome. A. 20
B. 30 C. 40 D. 50 E. 100
9.DNA replication is a different process than DNA repair. Mutations in DNA repair enzymes can lead
to disease—especially certain forms of cancer. Primase is not required during DNA repair
processes because of which one of the following?
A. All of the primase is associated with replication origins. B. RNA would be highly mutagenic at a repair site.
C. Repair DNA polymerases do not require a primer.
D. Replicative DNA polymerases do not require a primer.
E. DNA polymerases (both repair and replicative) can use any 3′-OH for elongation.
10.The key mechanistic failure in patients with XP involves which one of the following?
A. Mutation in the primase gene
B. Inability to excise a section of the UV-damaged DNA C. Mutation of one of the mismatch repair components
D. Inability to synthesize DNA across the damaged region E. Loss of proofreading capacity
ANSWERS TO REVIEW QUESTIONS
1.The answer is E. Helicases and topoisomerases unwind the parental strands. DNA polymerases
copy each parental template in the 3′-to-5′ direction, producing new strands in a 5′- to-3′ direction.
One strand of newly synthesized DNA grows continuously, but the other strand is synthesized
discontinuously in short segments knows as Okazaki fragments. These fragments are subsequently
joined by DNA ligase. Targeting a single enzyme with a drug is more likely to succeed than
inhibiting an entire family of enzymes.
2.The answer is E. DNA polymerases catalyze the synthesis of DNA but cannot initiate the
synthesis of new strands de novo, because a short primer first must be synthesized by DNA
primase (a DNA-dependent RNA polymerase). Phosphodiester bonds, which link the backbone,
are covalent bonds and are not formed from hydrogen bonds. During the course of adding a
nucleotide to an existing DNA chain, pyrophosphate is released, and its subsequent hydrolysis
(pyrophosphate contains a high-energy bond) by the enzyme pyrophosphatase (not DNA polymerase) provides the energy that drives the polymerization reactions. DNA polymerases
exhibit processivity, in which the enzyme remains bound to the parental template strand as the
enzyme creates new phosphodiester bonds while reading the template. The enzyme does not
dissociate and reassociate after each nucleotide is added to the existing DNA strand. DNApolymerases copy a template in the 3′-to-5′ direction, producing new strands in a 5′-to-3′
direction.
3.The answer is C. Mitochondria in human cells are very similar to bacteria and are theorized to
have arisen from bacteria that developed a symbiotic relationship with the host cell. Polymerase γ
is located in mitochondria and replicates the DNA of this organelle. Polymerases δ and ε are the
major replicative enzymes in the eukaryotic nucleus. Polymerase α is involved in DNA repair.
Polymerase β participates in base excision repair.
4.The answer is B. Eukaryotic chromosomes are linear, and the ends of the chromosomes are called
telomeres. Bacteria have circular DNA and therefore have no telomeres. Telomeres in humans
consist of a repeating sequence of TTAGGG. The newly synthesized DNA strand, before telomerase action, is shorter at the 5′-end so that the strand being replicated has an overhang at the
3′-end. Somatic cells have telomeres, but as they age, their expression of telomerase decreases, so
the cells only survive for a fixed number of population doublings.
5.The answer is C. Proofreading, an inherent property of most DNA polymerase owing to its 3′
exonuclease activity, eliminates base-pairing errors as they occur during replication, but
proofreading does not eliminate all errors made by DNA polymerase. Postreplication error repair
systems replace mismatched bases that are missed by proofreading. Genes that produce mRNA
contain a unique transcription-coupled repair system (repair occurs as the genes are transcribed).
Nucleotide excision repair involves local distortion of the DNA helix, such as in bulky adducts,
whereas DNA glycosylases recognize damage to a single base.
6.The answer is E. Translocations can occur in both somatic and stem cells. Translocations occur
frequently and can be either beneficial or devastating. Some translocations can lead to
developmental delay, and some can lead to a higher risk of developing cancer, but they also can be
beneficial or have no discernable effect. Translocation consists of a portion of one chromosome
being exchanged for a portion of another chromosome. Reverse transcriptase is found in RNA
viruses and has no role in human translocation.
7.The answer is B. A DNA polymerase’s 3′-to-5′ exonuclease activity is required for proofreading
(check the base just inserted, and if it is incorrect, remove it), and reverse transcriptase does not
have this activity, whereas Pol δ does. Both reverse transcriptase and Pol δ synthesize DNA in the
5′-to-3′ directions (all DNA polymerases do this), and both follow standard
Watson–Crick basepairing rules (A with T or U, G with C). Neither polymerase can synthesize DNA in the wrong
direction (3′-to-5)′ or insert inosine into a growing DNA chain. Thus, the only difference between
the two polymerases is answer B. Pol δ is used primarily for lagging-strand synthesis during DNA
replication, although it also has repair functions.
8.The answer is D. In 50 seconds, each replication origin will have synthesized 100 kb of DNA (50
in each direction). Because there are 10 origins, 10 × 100 will yield the 1,000 kb needed to
replicate the DNA. The first origin will be 50 kb from one end, and the remaining 9 origins will
each be 100 kb apart.
9.The answer is E. The role of the primer is to provide a free 3′-OH group for DNA polymerase to
add the next nucleotide and form a phosphodiester bond. When DNA repair occurs, one of the
remaining bases in the DNA will have a free 3′-OH, which repair DNA polymerases (such asDNA pol I in bacteria) will use to begin extension of the DNA.
10.The answer is B. XP is a set of diseases all related to an inability to repair thymine dimers,
leading to an inability to excise UV-damaged DNA. It does not affect bypass polymerases, which
can synthesize across the damaged region, sometimes making mutations in its path. The primase
gene, or mismatch repair, is not involved in excising thymine dimers. Proofreading ability of DNA
polymerases is likewise not involved in this process.Transcription: Synthesis of RNA
For additional ancillary materials related to this chapter, please visit thePoint. Synthesis of RNA from a DNA template is called transcription. Genes are transcribed by enzymes
called RNA polymerases that generate a single-stranded RNA identical in sequence (with the exception
of U in place of T) to one of the strands of the double-stranded DNA. The DNA strand that directs the
sequence of nucleotides in the RNA by complementary base pairing is the template strand. The RNA
strand that is initially generated is the primary transcript. The DNA template is copied in the 3′-to-5′
direction, and the RNA transcript is synthesized in the 5′-to-3′ direction. RNA polymerases differ from
DNA polymerases in that they can initiate the synthesis of new strands in the absence of a primer.
In addition to catalyzing the polymerization of ribonucleotides, RNA polymerases must be able to
recognize the appropriate gene to transcribe, the appropriate strand of the double-stranded DNA to copy,
and the start point of transcription (Fig. 14.1). Specific sequences on DNA, called promoters, determine
where the RNA polymerase binds and how frequently it initiates transcription. Other regulatory
sequences, such as promoter-proximal elements and enhancers, also affect the frequency of
transcription.In bacteria, a single RNA polymerase produces the primary transcript precursors for all three major
classes of RNA: messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). Because
bacteria do not contain nuclei, ribosomes bind to mRNA as it is being transcribed, and protein synthesis
occurs simultaneously with transcription.
Eukaryotic genes are transcribed in the nucleus by three different RNA polymerases, each
principally responsible for one of the major classes of RNA. The primary transcripts
are modified and
trimmed to produce the mature RNAs. The precursors of mRNA (called pre-mRNA) have a guanosine
“cap” added at the 5′-end and a poly(A) “tail” at the 3′-end. Exons, which contain the coding sequences
for the proteins, are separated in pre-mRNA by introns, regions that have no coding function. During
splicing reactions, introns are removed and the exons connected to form the mature mRNA. In eukaryotes,
tRNA and rRNA precursors are also modified and trimmed, although not as extensively as pre-mRNA.
THE WAITING ROOM
Lisa N. is a 4-year-old girl of Mediterranean ancestry whose height and body weight are below the
20th percentile for girls of her age. She tires easily and complains of loss of appetite and shortness
of breath on exertion. A dull pain has been present in her right upper quadrant for the last 3 months and
she appears pale. Initial laboratory studies indicate a severe anemia (decreased red blood cell count)
with a hemoglobin of 7.0 g/dL(reference range = 12 to 16 g/dL). A battery of additional hematologic tests
reveals that Lisa has β+-thalassemia, intermediate type.
Isabel S., a patient with HIV (see Chapters 12 and 13), has developed a cough that produces a gray,
slightly blood-tinged sputum. A chest X-ray indicates a cavitary infiltrate in the right upper lung
field. A stain of sputum shows the presence of acid-fast bacilli, suggesting a diagnosis of pulmonary
tuberculosis caused by Mycobacterium tuberculosis.
Catherine T. picked mushrooms in a wooded area near her home. A few hours after eating one
small mushroom, she experienced mild nausea and diarrhea. She brought a mushroom with her to
the hospital emergency department. A poison expert identified it as Amanita phalloides (the “death cap”).
These mushrooms contain the toxin α-amanitin.Sarah L., a 28-year-old computer programmer, notes increasing fatigue, pleuritic chest pain, and a
nonproductive cough. In addition, she complains of joint pains, especially in her hands. A rash on
both cheeks and the bridge of her nose (“butterfly rash”) has been present for the last 6 months. Initial
laboratory studies reveal a subnormal white blood cell count and a mild reduction in hemoglobin. Tests
result in a diagnosis of systemic lupus erythematosus (SLE) (frequently called lupus).
The thalassemias are a heterogenous group of hereditary anemias that constitute the most
common gene disorder in the world, with a carrier rate of almost 7%. The disease was first
discovered in countries around the Mediterranean Sea and was named for the Greek word
thalassa, meaning “sea.” However, it is also present in areas extending into India and China that
are near the equator.
The thalassemia syndromes are caused by mutations that decrease or abolish the synthesis of
the α- or β-chains in the adult hemoglobin A tetramer. Individual syndromes are named according
to the chain whose synthesis is affected and the severity of the deficiency. Thus, in β0-thalassemia,
the superscript 0 denotes that none of the β-chain is present; in β+-thalassemia, the plus sign
denotes a partial reduction in the synthesis of the β-chain. More than 170 different mutations have
been identified that cause β-thalassemia; most of these interfere with the transcription of β-globin
mRNA or its processing or translation.
The measurement of hemoglobin levels in blood is important for the appropriate diagnosis
of many diseases such as anemia. Laboratories measure hemoglobin content by first exposing the sample (usually lysed blood cells to release the hemoglobin from the red blood
cells) to an oxidizing agent, which converts the ferrous iron in hemoglobin to its ferric state. The
level of ferric iron is then determined with a second reagent (either a cyanide or azide derivative),
which reacts with the ferric iron and generates a colored product whose concentration can be
determined spectrophotometrically. I. Action of RNA Polymerase
Transcription, the synthesis of RNA from a DNA template, is carried out by RNA polymerases (Fig.
14.2). Like DNA polymerases, RNA polymerases catalyze the formation of ester bonds between
nucleotides that base-pair with the complementary nucleotides on the DNA template. Unlike DNA
polymerases, RNA polymerases can initiate the synthesis of new chains in the absence of primers. They
also lack the 3′-to-5′ exonuclease activity found in DNA polymerases, although they do perform
rudimentary error-checking through a different mechanism. A strand of DNA serves as the template for
RNA synthesis and is copied in the 3′-to-5′ direction. Synthesis of the new RNA molecule occurs in the
5′-to-3′ direction. The ribonucleoside triphosphates adenosine triphosphate (ATP), guanosine triphosphate
(GTP), cytidine triphosphate (CTP), and uridine triphosphate (UTP) serve as the precursors. Each
nucleotide base sequentially pairs with the complementary deoxyribonucleotide base on the DNAtemplate (A, G, C, and U pair with T, C, G, and A, respectively). The polymerase forms an ester bond
between the α-phosphate on the ribose 5′-hydroxyl of the nucleotide precursor and the ribose 3′-hydroxyl
at the end of the growing RNA chain. The cleavage of a high-energy phosphate bond in the nucleotide
triphosphate and release of pyrophosphate (from the β- and γ-phosphates) provide the energy for this
polymerization reaction. Subsequent cleavage of the pyrophosphate by a pyrophosphatase also helps to
drive the polymerization reaction forward by removing a product. The overall error rate of RNA
polymerase is 1 in 100,000 bases.
RNA polymerases must be able to recognize the start point for transcription of each gene and the
appropriate strand of DNA to use as a template. They also must be sensitive to signals that reflect the
need for the gene product and control the frequency of transcription. A region of regulatory sequences
called the promoter (often composed of smaller sequences called boxes or elements), usually contiguous
with the transcribed region, controls the binding of RNA polymerase to DNA and identifies the start point
(see Fig. 14.1). The frequency of transcription is controlled by regulatory sequences within the promoter
and nearby the promoter (promoter-proximal elements) and by other regulatory sequences, such as
enhancers (also called distal-promoter elements), that may be located at considerable distances—
sometimes thousands of nucleotides—from the start point. Both the promoter-proximal
elements and the
enhancers interact with proteins that stabilize RNA polymerase binding to the promoter.
Patients with AIDS frequently develop tuberculosis. After Isabel S.’s sputum stain suggested that she had tuberculosis, a multidrug antituberculous regimen, which includes an
antibiotic of the rifamycin family (rifampin), was begun. A culture of her sputum was done to
confirm the diagnosis.
Rifampin inhibits bacterial RNA polymerase, selectively killing the bacteria that cause theinfection. The nuclear RNA polymerase from eukaryotic cells is not affected. Although rifampin
can inhibit the synthesis of mitochondrial RNA, the concentration required is considerably higher
than that used for treatment of tuberculosis. II. Types of RNA Polymerases
Bacterial cells have a single RNA polymerase that transcribes DNA to generate all of the different types
of RNA (mRNA, rRNAs, and tRNA. The RNA polymerase of Escherichia coli contains five subunits (2α,
β, β,′ and ω), which form the core enzyme. Another protein called a σ (sigma) factor binds the core
enzyme and directs binding of RNA polymerase to specific promoter regions of the DNA template. The σ
factor dissociates shortly after transcription begins. E. coli has several different σ factors that recognize
the promoter regions of different groups of genes. The major σ factor is σ70, a designation related to its
molecular weight of 70,000 Da.
In contrast to prokaryotes, eukaryotic cells have three RNA polymerases (Table 14.1). Polymerase I
produces most of the rRNAs, polymerase II produces mRNA and microRNAs (microRNAs regulate gene
expression and are discussed in more detail in Chapter 16), and polymerase III produces small RNAs,
such as tRNA and 5S rRNA. All of these RNA polymerases have the same mechanism of action.
However, they recognize different types of promoters.
The mushrooms picked by Catherine T. contained α-amanitin, an inhibitor of eukaryotic
RNA polymerases:
It is particularly effective at blocking the action of RNA polymerase II. This toxin initially causesgastrointestinal disturbances, then electrolyte imbalance and fever, followed by liver and kidney
dysfunction. Around 10% to 20% of individuals who ingest α-amanitin die within 10 days.
A. Sequences of Genes
Double-stranded DNA consists of a coding strand and a template strand (Fig. 14.3). The DNA template
strand is the strand that is used by RNA polymerase during the process of transcription. It is
complementary and antiparallel both to the coding (nontemplate) strand of the DNA and to the RNA
transcript produced from the template. Thus, the coding strand of the DNA is identical in base sequence
and direction to the RNA transcript except, of course, that wherever this DNA strand contains a T, the
RNA transcript contains a U. By convention, the nucleotide sequence of a gene is represented by the
letters of the nitrogenous bases of the coding strand of the DNA duplex. It is written from left to right in
the 5′-to-3′ direction.
During translation, mRNA is read 5′-to-3′ in sets of three bases, called codons, that determine the
amino acid sequence of the protein (see Fig. 14.3) Thus, the base sequence of the